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4800	https://www.youtube.com/watch?v=cdXVx2ogWH0	Parallel Lines and Transversals (Geometry Made Easy) TabletClass Math 880000 subscribers 123 likes Description 10078 views Posted: 6 Jun 2018 Parallel lines and transversal form many angles in geometry. These angles include vertical angles, alternate interior angles, same side interior angles and corresponding angles. 9 comments Transcript: Intro okay in this video we're going to go ahead and talk about parallel lines and transversals this is an absolutely critical um Concept in Geometry so uh generally start studying this at the high school level high school geometry you probably May seen some of this at the Middle School uh level but it this is really really important stuff so let's go and get into it and let's first just Define a few things here so let's talk about parallel lines Parallel Lines okay so I have three lines here so in Geometry you can kind of name them with these lowercase uh um variables here L M and T there's other ways to uh name lines but I'm not going to get into that in this video but lines l and m okay here they appear they appear to be parallel now I'm using that word very carefully appear means they look to be what well they don't look like they're they're kind of running side by side that's what that word means parallel means that two lines kind of running side by side and they'll never intersect so just visually here they appear to uh to be in fact parallel however in mathematics we cannot assume based upon the way a figure looks that two lines are parallel you have to actually uh very specifically state that so the ways we can do that is we can say l is parallel to M line L is parallel to M that's one way of doing it another way you can do it is you can see a another arrow here on the line and you put another one down here okay so this indicates that these two lines are parallel without these uh this notation or this notation we cannot um assume that these two lines would be parallel now everything I'm going to talk to you about here I'm getting into transversals in a second depends upon these two lines in fact being parallel then there's a whole range of properties that um that follow from that but test like the sat and sat uh act they like to kind of quiz students sometimes they'll draw a figure that the lines look parallel but in fact they didn't state that then students kind of assume well those are parallel lines they'll work it out and they'll get the problem wrong and they're kind of set up that way so just be careful right so um you got to in fact uh look to see that the lines are in fact stated parallel now let's get let's get into this a little bit further so if you have two parallel lines two or more in fact but we're just going to stick with two and you have another line that kind of crosses through it like our line T here this line T would be referred to as a transversal okay so parallel line and transversal okay you can have one or more transversals we're just going to stick with one here now in this case if you have um parallel lines and a trans versial and this is the most basic setup here a lot of neat properties take care of uh or kind of come into play and these are actual um angle relationships so let's go and just talk about a whole bunch here it's something that you absolutely need to know and uh there's a lot of problems um in other words like on your final exam your chapter test or sat act GED whatever the case is that are that I'm sure going to in fact test you on your knowledge of these angle relationships they're not hard but there are many okay so I'm going to kind of go through these and I'll race and we'll just you know kind of go through them one by one so the first are vertical Vertical Angles angles so this angle and this angle here okay if you kind of look they kind of look the same well they are they're called vertical angles okay now there are many pairs of vertical angles here right this here and this here there's another pair vertical angles vertical angles and over here as well these two here are vertical angles vertical angles vertical angles uh vertical angles and vertical angles so vertical angles really don't have anything to do in fact with the parallel line condition in other words if I have a line and another line that kind of chopped through this way these are uh vertical angles are formed okay and vertical angles are congruent now the Congruent Angles word congruent just means that they're equal so all these angles here would be equal to one another so we can kind of specify that and actually what I'm going to do you're going to do it this way we'll give these angles some names one 2 3 four five 6 78 okay so let's kind of just kind of go down here we'll say vertical angles where are vertical angles angle one and four right so angle one is congruent to angle four why are these two angles equal to one another because they're vertical angles so in in Geometry mathematics you have to have a justification you just can't be like well they kind of look like they're equal no well in fact they're vertical angles that's why they're equal and so we can list the rest of them real quick angle two is congruent to angle three and we'll just do the rest here real quick five and 8 angle five is congruent to angle eight and angle 7 is congruent to angle six okay so those are all vertical angles now let's move on to some other I want to get into some things like alternate angles all that stuff maybe if you're watching this video probably already know a little bit about that we'll kind of hold on let's kind of take care of some of the more fundamentals so the vertical angles are fundamentals now straight lines are also considered angles so like angle three and four is what we call a straight line okay they're also supplementary which means they add up to 180 so from here you might say well this is an angle because there's like a bend in it right well what if I happens if I just can't and keep opening that bend up more and more and finally I open it up all the way there's a straight line you you might be saying to yourself well the angle disappeared well no the angle didn't disappear there's still an actual angle here right the angle of a St line is 180° okay so any two angles that make up if you add them or basically are adjacent that add up to 180° they form a straight line angle okay they're also supplementary so we can kind of build out some additional angle relationships here so let's kind of do this here so let's say straight line for example or supplementary so we can say angle three plus angle 4 is equal to what 180° right now there's a whole bunch of these straight line relationships here let me kind of erase this so you probably have already seen many more you probably say well seven and eight five and six one and two and you'd be correct but also this way as well one and three okay so one and three are supplementary so angle one plus angle 3 is 18 0° and so forth we can go on and on angle two plus angle 4 180° let's see here now we have three and four angle one and two angle one plus angle 2 is 180° uh 5 6 7 8 actually I'm not going to list out the rest of them but basically the same angles as we did in the bottom pair are going to apply to the top pair okay so all of these straight line angles this way this way this way this way they're all supplementary they all add up to 180° okay so now let's get into some Additional Angles additional angle properties now before we get into the next pair I want to kind of um uh just denote a few things here when we have parallel lines we consider these angles here these angles here okay inside the lines as interior angles okay interior angles so these would be exterior angles these guys here but 5 6 3 and four would be considered interior angles okay so I'm going to be using the notation here interior and I'm also going to be using well let me just do it this way um that angles on opposite side of the transversal would be alternating okay that's that word alternating okay so we have interior and alternating alternating are angles that are opposite side of the transversal and interior are those angles that are inside the lines okay so with that being said kind of scroll down here a little bit give ourselves some room let's talk about the properties between angles five and six so what would we how would we classify them well they're interior angles and are alterate also alternating angles so weal we we described these as alternate interior angles okay so alternate interior angles if you look here angles five and angle six they appear to be what congruent the same angle if you kind of look at them closely and in fact they are now this is only true if these two lines are parallel okay if they're not parallel then we can't make these um these claims okay but in this case we can so angle five is congruent to angle four and angle three is congruent to angle six okay why because they're alternate interior angles and I'm going to give you uh some example problems here uh in a second but let's just go and list all these out so we have alternate interior angles now we also have let's say these guys what do you think we' call these the these pairs together okay well we call those same side interior angles so same side interior angles they have kind of a unique relationship so let's say angle three plus angle five the deal with these guys is they are supplementary they add up to 180 de okay so angle four and angle six all right those are also same side interior angles angle 4 plus angle 6 and they also add up to 180° okay so let's just quickly review what we got so far we got vertical angles straight line angles alternate alternating uh interior angles and we have same side interior angles okay and now we have another um angle relationship and that is going to be corresponding angles corresponding angles so corresponding angles are pretty cool what they they look like is this let's take a look at angle one so can you find another angle that looks like angle one pretty much looks like the same angle measure well if you look kind of closely here angle five does okay so there it's kind of like in the same relative position if you kind of look at it this kind of corresponds to this well corresponding angles again all of this is predicated that the two lines were the that the angles being formed are from two lines that are parallel and are transversal so everything I'm stating here with the exception of vertical angles and straight line angles um depend upon the two lines being parallel okay so corresponding angles are congruent so that would be angle one there's a whole bunch of these guys angle one is congruent to angle five and let me just stop here again I know I'm kind of interc in here we use the word congruent uh actually I got to put the the word well no equal to 180 that's fine congruency in Geometry is meaning that that we don't like to use the word equal well you kind of use it but technically you're supposed to be congruent that means they have the same angle measure okay where the equal sign is usually when values like in algebra like x equals 3 or something like that then we like to use the equal sign but we're talking about angle measure or the the length of segments then you need to use the the congruent symbol okay so angle one is congruent to angle five corresponding angles and then we can see three and seven got a whole bunch let's go list them out here angle three is congruent to angle seven and then on this side we have angle two is congruent to angle six and let's see here angle four is congruent to angle 8 okay I think that pretty much covers most of uh of the angles being formed here okay with parallel lines and a transversal all right so knowing this Example Problem let's take a look at kind of like a typical kind of problem that you're going to see in your geometry course or on any other type of exam okay so let's suppose um I'm given uh let's say A B C D E let's call this uh 30° let's call this Y and let's call this um f okay so you might be asked to find all the angles here okay all the the value of the variables so you're saying well I only got one angle here but it's actually pretty simple okay so let's just start with angle a okay so how can we find the value of this guy right here well pretty simple actually um the way you can think about it is that these two guys angle a plus angle 30 is equal to what it's a straight line angle that's equal to 180° so a + 30 is equal to 180 you can solve this basic equation so a is equal to 150 okay 150° it's very simple right so this is 150 because 1 plus 30 he got to make up this entire straight line so now now knowing that you can be like oh okay these two are vertical angles right here these two right here are vertical angles so B must also be equal to 150 okay but why because the vertical angles and so is y y and 30 vertical angles so we can put a little 30° right here and now we can just keep going down the line so we got 50 right here well e angle e and 150 are corresponding angles so this has got to be 150 and c and 30 are also Co corresponding angles and there's different ways you can actually um approach these there's a ton of different uh you know options sometimes not all the times but properties that you can use to apply here and then e and D are vertical angles so this has got to be 150 C and F are vertical angles this has got to be 30 D and these are corresponding and then boom there you're all done now sometimes they'll have like like fancier algebraic expressions in there you might have to set up equations but this stuff is really not that difficult if you know how to do basic solve basic algebra equations and most importantly understand the angle relationships you can get these problems right but um couple things again to stress especially um for those of you that are are taking test I've seen this definitely like on the SAT or the ACT they'll try to kind of fool you that a you know the the figures look to be one way but in fact they draw the figure to kind of mislead you okay they're looking for your ability to to to actually see that a fact that you know a certain condition exist so this condition here that we need to be mindful of is parallel okay these two lines are sometimes the arrows look like this this is parallel okay so these two lines are parallel because we have this or this okay if it doesn't have that then you can't make that assumption all right so kind of a quick crash course on parallel lines and transversals but if you understand this much trust me you're going to you're going to do pretty well on these kind of angle relationship PRS and they do come up everywhere so anyways if you enjoyed the video please uh subscribe we do a ton of these kind of uh videos on my channel maybe give this a thumbs up and let me know how things are going with you in um in your mathematics journey and if you have a question what not I try to read as uh many comments as possible and and do some of this uh some of these free videos um just very quickly on my background is I'm a math teacher teaching for many many years middle school high school and Beyond and I really kind of enjoy putting a lot of uh free material out there but my aim is to really really give kind of like real uh quick reviews on some of the most important things that students really have to understand it's all important but if you understand at least this much you're going to go a a long way you don't have to understand everything to do pretty well in math but if you there are some things you absolutely have to understand at a at a you know and this is one of them if you absolutely understand the fundamentals here you'll do pretty well with this subject okay so best luck to you
4801	https://www.skyradar.com/blog/understanding-concept-of-range-and-bearing	written on Dec 19, 2019 by Peter Green # Understanding The Concepts of Range and Bearing ATC-Training Theory & Research Products Radars 8 GHz Pulse 24 GHz PSR 24 GHz FMCW Sonar Digital Signal Processing FreeScopes Basic I FreeScopes Basic II FreeScopes ATC I FreeScopes ATC II FreeScopes AI I System Monitoring & Control SkySMC - System Monitoring & Control (SMC) software SkyRack - Infrastructure for SMC Training Simulators SkySim Radar Simulator Radar & Tower Simulator Fibre Optics Laboratory Fire Fighting Simulator Security, Defense & Electronic Warfare SkySim Radar Simulator SkySim Deception Package I FreeScopes Disturbance Filtering I Solutions Civil Aviation Academies Military Academies Universities Technical Schools Pricing Price Packages Comparison of Features Resources & Blogs News Blog Videos Manuals Contact Contact About us Vision RADAR has become a key technology to help navigate aircraft. With a growing demand of efficient air space usage, the demand for RADAR systems is growing. RADAR is a technology that uses Radio waves to detect objects in the air or sea. Range and Bearing are one of two main characteristics that are measured by the RADAR. In order to understand these two parameters we need to understand how a RADAR works. RADAR Radar is an acronym for "radio detection and ranging." A radar system usually operates in the ultra-high-frequency (UHF) or microwave part of the radio-frequency (RF) spectrum, and is used to detect the position and/or movement of objects. Range Range in simpler words is the distance of an object from a particular point. In radar, range is derived through a formula. But why do you need a formula to get the range acquired through a RADAR? Because RADAR technology uses Radio waves that travel at the speed of light. As mentioned range is a distance of target from a particular object. To measure distance we use Newton's formula S=vt Let us measure distance of an object in this case an aircraft. The distance of an object from the RADAR is called slant range - it is along the line of sight. Slant range is the line of sight distance between the radar and the object illuminated. The ground range is the horizontal distance between the emitter and its target. Its calculation requires knowledge of the target's elevation. Since the radio waves are traveling at the speed of light let us call this speed Co The formula will be V=s/t Co =2R/t Whereas “2R” is the observation of the radar pulse traveling to the target and then coming back to the radar. “t” is the time taken. Now V= Co for the velocity of light which is C= 3·108 m/s Distance S=2R. T is the time taken. The formula derived for the range consequently is: R= Co t/2 The range formula is calculated by RADAR software module. Bearing In navigation “Bearing” describes the horizontal angle between the direction of an object and another object or between it and true north. It is measured in mils or degrees. Bearing is used in civil aviation, warfare as well as search and rescue. In RADAR technology bearing is the determination of the direction. The True Bearing (referenced to true north) of a radar target is the angle between the true north and a line pointed directly at the target. This angle is measured in the horizontal plane and in a clockwise direction from true north. The bearing angle to the radar target may also be measured in a clockwise direction from the centerline of your own ship or aircraft. It is referred to as the relative bearing. In aircraft navigation, an angle is normally measured from the aircraft's track or heading in a clockwise direction. Modern radar sets take on this task and with the help of the GPS satellites determine the north direction independently. RADAR has become a key technology to help navigate aircraft. With a growing demand of efficient air space usage, the demand for RADAR systems is growing. RADAR is a technology that uses Radio waves to detect objects in the air or sea. Range and Bearing are one of two main characteristics that are measured by the RADAR. In order to understand these two parameters we need to understand how a RADAR works. RADAR Radar is an acronym for "radio detection and ranging." A radar system usually operates in the ultra-high-frequency (UHF) or microwave part of the radio-frequency (RF) spectrum, and is used to detect the position and/or movement of objects. Range Range in simpler words is the distance of an object from a particular point. In radar, range is derived through a formula. But why do you need a formula to get the range acquired through a RADAR? Because RADAR technology uses Radio waves that travel at the speed of light. As mentioned range is a distance of target from a particular object. To measure distance we use Newton's formula S=vt Let us measure distance of an object in this case an aircraft. The distance of an object from the RADAR is called slant range - it is along the line of sight. Slant range is the line of sight distance between the radar and the object illuminated. The ground range is the horizontal distance between the emitter and its target. Its calculation requires knowledge of the target's elevation. Since the radio waves are traveling at the speed of light let us call this speed Co The formula will be V=s/t Co =2R/t Whereas “2R” is the observation of the radar pulse traveling to the target and then coming back to the radar. “t” is the time taken. Now V= Co for the velocity of light which is C= 3·108 m/s Distance S=2R. T is the time taken. The formula derived for the range consequently is: R= Co t/2 The range formula is calculated by RADAR software module. Bearing In navigation “Bearing” describes the horizontal angle between the direction of an object and another object or between it and true north. It is measured in mils or degrees. Bearing is used in civil aviation, warfare as well as search and rescue. In RADAR technology bearing is the determination of the direction. The True Bearing (referenced to true north) of a radar target is the angle between the true north and a line pointed directly at the target. This angle is measured in the horizontal plane and in a clockwise direction from true north. The bearing angle to the radar target may also be measured in a clockwise direction from the centerline of your own ship or aircraft. It is referred to as the relative bearing. In aircraft navigation, an angle is normally measured from the aircraft's track or heading in a clockwise direction. Modern radar sets take on this task and with the help of the GPS satellites determine the north direction independently. Peter Green subscribe Our Blog, delivered to your inbox. Subscribe Here! Subscribe related articles IFATSEA's 50th Annual Assembly in Prague - A Video-Message from SkyRadar's CEO (Video) Dennis Vasilev Video on SkyRack ATM - the ATSEP Training Infrastructure for System Monitoring & Control Ulrich Scholten, PhD Cybersecurity Challenges in Air Traffic Management Dawn Illing Other Related Articles: # ATC-Training # Theory & Research Links Home Contact Us Privacy Policy Term of Use About Us SkyRadar develops innovative radar training solutions and simulation systems, empowering education, research & professional training in aviation and defense sectors. social media Get in touch All rights reserved by SkyRadar 2008 - 2025 \| \| \| \| \| Understanding The Concepts of Range and Bearing RADAR has become a key technology to help navigate aircraft. With a growing demand of efficient air space usage, the demand for RADAR systems is growing. RADAR is a technology that uses Radio waves to detect objects in the air or sea. Range and Bearing are one of two main characteristics that are measured by the RADAR. In order to understand these two parameters we need to understand how a RADAR works. RADAR Radar is an acronym for "radio detection and ranging." A radar system usually operates in the ultra-high-frequency (UHF) or microwave part of the radio-frequency (RF) spectrum, and is used to detect the position and/or movement of objects. Range Range in simpler words is the distance of an object from a particular point. In radar, range is derived through a formula. But why do you need a formula to get the range acquired through a RADAR? Because RADAR technology uses Radio waves that travel at the speed of light. As mentioned range is a distance of target from a particular object. To measure distance we use Newton's formula S=vt Let us measure distance of an object in this case an aircraft. The distance of an object from the RADAR is called slant range - it is along the line of sight. Slant range is the line of sight distance between the radar and the object illuminated. The ground range is the horizontal distance between the emitter and its target. Its calculation requires knowledge of the target's elevation. Since the radio waves are traveling at the speed of light let us call this speed Co The formula will be V=s/t Co =2R/t Whereas “2R” is the observation of the radar pulse traveling to the target and then coming back to the radar. “t” is the time taken. Now V= Co for the velocity of light which is C= 3·108 m/s Distance S=2R. T is the time taken. The formula derived for the range consequently is: R= Co t/2 The range formula is calculated by RADAR software module. Bearing In navigation “Bearing” describes the horizontal angle between the direction of an object and another object or between it and true north. It is measured in mils or degrees. Bearing is used in civil aviation, warfare as well as search and rescue. In RADAR technology bearing is the determination of the direction. The True Bearing (referenced to true north) of a radar target is the angle between the true north and a line pointed directly at the target. This angle is measured in the horizontal plane and in a clockwise direction from true north. The bearing angle to the radar target may also be measured in a clockwise direction from the centerline of your own ship or aircraft. It is referred to as the relative bearing. In aircraft navigation, an angle is normally measured from the aircraft's track or heading in a clockwise direction. Modern radar sets take on this task and with the help of the GPS satellites determine the north direction independently. RADAR has become a key technology to help navigate aircraft. With a growing demand of efficient air space usage, the demand for RADAR systems is growing. RADAR is a technology that uses Radio waves to detect objects in the air or sea. Range and Bearing are one of two main characteristics that are measured by the RADAR. In order to understand these two parameters we need to understand how a RADAR works. RADAR Radar is an acronym for "radio detection and ranging." A radar system usually operates in the ultra-high-frequency (UHF) or microwave part of the radio-frequency (RF) spectrum, and is used to detect the position and/or movement of objects. Range Range in simpler words is the distance of an object from a particular point. In radar, range is derived through a formula. But why do you need a formula to get the range acquired through a RADAR? Because RADAR technology uses Radio waves that travel at the speed of light. As mentioned range is a distance of target from a particular object. To measure distance we use Newton's formula S=vt Let us measure distance of an object in this case an aircraft. The distance of an object from the RADAR is called slant range - it is along the line of sight. Slant range is the line of sight distance between the radar and the object illuminated. The ground range is the horizontal distance between the emitter and its target. Its calculation requires knowledge of the target's elevation. Since the radio waves are traveling at the speed of light let us call this speed Co The formula will be V=s/t Co =2R/t Whereas “2R” is the observation of the radar pulse traveling to the target and then coming back to the radar. “t” is the time taken. Now V= Co for the velocity of light which is C= 3·108 m/s Distance S=2R. T is the time taken. The formula derived for the range consequently is: R= Co t/2 The range formula is calculated by RADAR software module. Bearing In navigation “Bearing” describes the horizontal angle between the direction of an object and another object or between it and true north. It is measured in mils or degrees. Bearing is used in civil aviation, warfare as well as search and rescue. In RADAR technology bearing is the determination of the direction. The True Bearing (referenced to true north) of a radar target is the angle between the true north and a line pointed directly at the target. This angle is measured in the horizontal plane and in a clockwise direction from true north. The bearing angle to the radar target may also be measured in a clockwise direction from the centerline of your own ship or aircraft. It is referred to as the relative bearing. In aircraft navigation, an angle is normally measured from the aircraft's track or heading in a clockwise direction. Modern radar sets take on this task and with the help of the GPS satellites determine the north direction independently. \| \| \| Understanding The Concepts of Range and Bearing 29 / 08 / 2024 SkyRadar Consortium Email: sales@SkyRadar.com \| Website: www.SkyRadar.com \| © SkyRadar
4802	https://kconrad.math.uconn.edu/blurbs/galoistheory/algclosure.pdf	CONSTRUCTING ALGEBRAIC CLOSURES KEITH CONRAD Let K be a ﬁeld. We want to construct an algebraic closure of K, i.e., an algebraic extension of K which is algebraically closed. It will be built out of the quotient of a polynomial ring in a very large number of variables. Let P be the set of all nonconstant monic polynomials in K[X] and let A = K[tf]f∈P be the polynomial ring over K generated by a set of indeterminates indexed by P. This is a huge ring. For each f ∈K[X] and a ∈A, f(a) is an element of A. Let I be the ideal in A generated by the elements f(tf) as f runs over P. Lemma 1. The ideal I is proper: 1 ̸∈I. Proof. Every element of I has the form Pn i=1 aifi(tfi) for a ﬁnite set of f1, . . . , fn in P and a1, . . . , an in A. We want to show 1 can’t be expressed as such a sum. Construct a ﬁnite extension L/K in which f1, . . . , fn all have roots. There is a substitution homomorphism A = K[tf]f∈P →L sending each polynomial in A to its value when tfi is replaced by a root of fi in L for i = 1, . . . , n and tf is replaced by 0 for those f ∈P not equal to an fi. Under this substitution homomorphism, the sum Pn i=1 aifi(tfi) goes to 0 in L so this sum could not have been 1. □ Since I is a proper ideal, Zorn’s lemma guarantees that I is contained in some maximal ideal m in A. The quotient ring A/m is a ﬁeld and the natural composite homomorphism K →A →A/m of rings let us view the ﬁeld A/m as an extension of K (ring homomorphisms out of ﬁelds are always injective). Every nonconstant monic polynomial f ∈K[X] has a root in A/m: the coset tf = tf mod m is a root, since f(tf) = f(tf) = 0. Since each tf is algebraic over K and A/m is generated over K as a ring by the tf’s, A/m is an algebraic extension of K in which every monic polynomial in K[X] has a root. If K is not algebraically closed, the ﬁeld K′ := A/m is a larger ﬁeld than K because every polynomial in K[X] has a root in K′. If K′ is algebraically closed then we are done. If it is not then our construction can be iterated (producing a larger ﬁeld K′′ ⊃K′ whose relation to K′ is the same as that of K′ to K) over and over and a union of all iterations is taken. The union is an algebraic extension of the initial ﬁeld K since it is at the top of a tower of algebraic extensions. It can be proved [1, pp. 544-545] that this union contains an algebraic closure of K, and thus it is an algebraic closure of K since it’s algebraic over K. The interesting point is that there is no need to iterate the construction: K′ = A/m is already algebraically closed, and thus K′ is an algebraic closure of K. This requires some eﬀort to prove, but it is a nice illustration of various techniques (in particular, the use of perfect ﬁelds in characteristic p). The result follows from the next theorem and was inspired by . Theorem 2. Let L/K be an algebraic extension such that every nonconstant polynomial in K[X] has a root in L. Then every nonconstant polynomial in L[X] has a root in L, so L is an algebraic closure of K. 1 2 KEITH CONRAD Proof. It suﬃces to show every irreducible in L[X] has a root in L. First we will describe an incomplete attempt at a proof, just to make it clear where the diﬃculty in the proof lies. Pick an irreducible e π(X) in L[X]. We want to show it has a root in L, but all we know to begin with is that each irreducible in K[X] has a root in L. So let’s ﬁrst show e π(X) divides some irreducible of K[X] in L[X]. A root of e π(X) (in some extension of L) is algebraic over L, and thus is algebraic over K, so it has a minimal polynomial m(X) in K[X]. Then e π(X) \| m(X) in L[X] since e π(X) divides every polynomial in L[X] having a root in common with e π(X). Since m(X) ∈K[X], by hypothesis m(X) has a root in L. But this does not imply e π(X) has a root in L since we don’t know if the root of m(X) in L is a root of its factor e π(X) or is a root of some other irreducible factor of m(X) in L[X]. So we are stuck. It would have been much simpler if our hypothesis was that every irreducible polynomial in K[X] splits completely in L[X], since then m(X) would split completely in L[X] so its factor e π(X) would split completely in L[X] too: if a polynomial splits completely over a ﬁeld then so does every factor, but if a polynomial has a root in some ﬁeld then not every factor of it has to have a root in that ﬁeld. Thus, the diﬃculty with proving this theorem is working with the weaker hypothesis that polynomials in K[X] pick up a root in L rather than a full set of roots in L. It turns out that the stronger hypothesis we would rather work with is actually a conse-quence of the weaker hypothesis we are provided: if every irreducible polynomial in K[X] has a root in L then every irreducible polynomial in K[X] splits completely in L[X]. Once we prove this, the idea in the previous paragraph does show every irreducible in L[X] splits completely in L[X] and thus L is algebraically closed. First we will deal with the case when K has characteristic 0. We want to show that every irreducible polynomial in K[X] splits completely in L[X]. Let π(X) ∈K[X] be irreducible. Let Kπ denote a splitting ﬁeld of π over K. Since K has characteristic 0, it is perfect ﬁeld so by the primitive element theorem we can write Kπ = K(α) for some α. There is no reason to expect α is a root of π(X) (usually the splitting ﬁeld of π(X) over K is obtained by doing more than adjoining just one root of π(X) to K), but α does have some minimal polynomial over K. Denote it by m(X), so m(X) is an irreducible polynomial in K[X]. By hypothesis m(X) has a root in L, say β. Then the ﬁelds Kπ = K(α) and K(β) are both obtained by adjoining to K a root of the irreducible polynomial m(X) ∈K[X], so these ﬁelds are K-isomorphic. Since π(X) splits completely in Kπ[X] = K(α)[X] by the deﬁnition of a splitting ﬁeld, π(X) splits completely in K(β)[X] ⊂L[X]. Thus when K has characteristic 0, every irreducible in K[X] splits completely in L[X], which means the argument at the start of the proof shows L is algebraically closed. If K has characteristic p > 0, is the above argument still valid? The essential construction was a primitive element for the splitting ﬁeld Kπ/K for an irreducible π in K[X]. There is a primitive element for every ﬁnite extension of K provided K is perfect. In characteristic 0 this is no constraint at all. When K has characteristic p, it is perfect if and only if Kp = K. It may not be true for our K that Kp = K. We will ﬁnd a way to reduce ourselves to the case of a perfect base ﬁeld in characteristic p by replacing K with a larger base ﬁeld. Let F = {x ∈L : xpn ∈K for some n ≥1}. If xpn ∈K and ypn′ ∈K then let s = max(n, n′) and note (x ± y)ps = xps ± yps ∈K. So F is an additive subgroup of L and contains K. It is easy to see F is closed under multiplication and inversion of nonzero elements, so F is a ﬁeld between K and L. This ﬁeld is perfect: F p = F. To see this, choose x ∈F. For some n ≥1, xpn ∈K. Let a = xpn. The polynomial Xpn+1 −a is in K[X], so by the basic hypothesis of the theorem this polynomial has a root r in L. Since CONSTRUCTING ALGEBRAIC CLOSURES 3 rpn+1 = a is in K, r ∈F. Since xpn = a = (rp)pn, x = rp because the pth power map is injective for ﬁelds of characteristic p. Therefore every x ∈F is the pth power of an element of F, so F p = F. Since L/F is algebraic, each irreducible polynomial in L[X] divides some irreducible polynomial in F[X] and the latter polynomial is separable (F is perfect), so every irreducible polynomial in L[X] is separable. Thus L is perfect, so Lp = L. If we can show that every polynomial in F[X] has a root in L then our proof in char-acteristic 0 can be applied to the extension L/F, so we will be able to conclude that L is algebraically closed. Let g(X) ∈F[X], say g(X) = P ciXi. We want to show g(X) has a root in L. For some n, cpn i ∈K for all i. The polynomial P cpn i Xi is in K[X], so it has a root r ∈L by hypothesis. Since L = Lp, also L = Lpn, so r = zpn for some z ∈L. Then 0 = X i cpn i ri = X i (cizi)pn = X i cizi !pn = g(z)pn, so g(X) has a root z in L. □ For a generalization of this theorem, see . Remark 3. That every ﬁeld has an algebraic closure and that two algebraic closures of a ﬁeld are isomorphic were ﬁrst proved by Steinitz in 1910 in a long paper that created from scratch the general theory of ﬁelds as part of abstract algebra. The inﬂuence of this paper on the development of algebra was enormous; for an indication of this, see . Steinitz was hindered in this work by the primitive state of set theory at that time and he used the well-ordering principle rather than Zorn’s lemma (which only became widely known in the 1930s ). Steinitz’s proof of the existence of algebraic closures and their uniqueness up to isomorphism, together with his account of set theory, took up 20 pages [5, Sect. 19–21]. References D. Dummit, R. Foote, “Abstract Algebra,” 3rd ed., Wiley, New York, 2004. R. Gilmer, A Note on the Algebraic Closure of a Field, Amer. Math. Monthly 75, 1968, 1101-1102. I. M. Isaacs, Roots of Polynomials in Algebraic Extensions of Fields, Amer. Math. Monthly 87, 1980, 543-544. P. Roquette, In memoriam Ernst Steinitz (1871–1928), J. Reine Angew. Math. 648 (2010), 1–11. URL https:// www.mathi.uni-heidelberg.de/∼roquette/STEINITZ.pdf. E. Steinitz, Algebraische Theorie der K¨ orper, J. Reine Angew. Math. 137 (1910), 167–309. URL https:// eudml.org/doc/149323. M. Zorn, A Remark on a Method in Transﬁnite Algebra, Bull. Amer. Math. Society 41 (1935), 667–670.
4803	https://www.youtube.com/watch?v=W8Q5LN7-vn0	What is Successive Discount ? Exercise : Easy Trick : Math All About Math + Me 682 subscribers 12 likes Description 340 views Posted: 9 Apr 2025 What is Successive Discount ? Successive Discount More than one discount. Marked Price What is marked price ? What is listed price ? Selling Price Formulae with Easy Trick Selling Price Explain discount Statement sums of discount Statement sums of MP Statement sums of SP Listed price MP SP Word problems Converting percentage to decimal All about math Mathematics successivediscount #discount #markedprice #sp #mp #statement_questions #percentage #percentage_math #percentage_short_trick #percentageshorttrick #percentagemaths #wordproblems #statement_questions #units_and_measurements #calculate #conversion #measurement #measurementtools #class3maths #class4mathematics #class5maths #class6maths #class7maths #class8maths #class9maths #class10maths #maths #mathematics #grade4math #class11 #class12 #grade5maths #physics #chemisty #grade2 #class2 #grade3maths #grade6maths #grade7maths #grade2 #class2 #grade8math #grade9math #grade10maths #grade11 #grade12 #ctet #ptet #ncertmaths #cbse #cbsemaths #cambridgecurriculum #cambridge #allaboutmath #ncertsolutions #ncert #mathstricks #grade3maths #class3maths #lengths #weight #capacity #education #ibboard #imo #olympiad #matholympiad #profitandloss #profit #profitandlossbasics #profit_and_loss #profitpercent Transcript: [Music] What is successive discount? Successive discount means giving more than one discount one after another on a product. Each discount is applied on the price left after the previous discount. Let's understand this with the help of example. A shopkeeper gives 20% discount and then another 10% discount on a shirt whose price is 1,000 rupees. So what is successive discount? We have the two discounts here one after another. So this is the successive discount. Now how to calculate this? For that we'll not add 20 and 10 and we'll get 30% discount into our total. It's wrong. So this doesn't mean that the total discount is 30%. It means the second discount that is 10% is given on the reduced price after 20% discount. Let's solve this example and understand this definition better. So the question we have already done that is a shopkeeper gives 20% discount and then another 10% discount on a shirt whose mark price is given that is 1,000 rupees and we have to find the selling price. So we have 20% discount on the marked price that is 1,000 rupees and 10% discount is not on this marked price. So let's calculate the first discount first how much first discount we'll get. So for the first discount 20% discount is on 1,000. So 20% of 1,000. So 20 by 100 of instead of of we multiply 20 by 100 multly 1,000. Let's do the cutting zero cancels. We are left with 20 mult 10 rupees 200. So the first discount is 200 rupees. Now let's find the selling price after the first discount. So if there will be only one discount what we have to pay to get the shirt. So for that 1,000 was the mark price 20 after 20% discount we have 200 rupees. So 1,000US 200800 rupees. Now the second discount 10% will be applied on the selling price after discount. The reduced price which we get after the first discount 10% the second discount will not be marked on this 1,000 rupees. Remember this. So the second discount 10% of 800 rupees. So 10% 10 by 100 multiply 800 do the cutting and we are left with 80 mult 8 that is 80 rupees. Therefore let's find the final selling price after second discount. Now we have the second discount 800. Now finding the selling price this time the discount the selling price that is 800 minus 80 that will be rupees 720. So how we'll get it now? This time this 800 is our mark price. So for second discount we'll be using 800. So therefore the answer is the final selling price of shirt is 720 rupees. We have to pay just rupes 720 to get the shirt after two discounts. Uh let's do another example. A laptop has three successive discounts. This time we have three discounts 15%, 10% and 5%. If the match price is 50,000 rupees, what is the selling price? So we have three discounts. Worry not. We'll first find the first discount that is 15% of 50,000 because the marked price is 50,000. The first discount will be applied on the mark price. So let's calculate 15 by 100 multiply 50,000. Do the cutting. We are left with 50 multiply 500 and we'll get 7,500 rupees. Now selling price after the first discount do the subtraction part 50,000 minus the discount that is 7,500 and we have the answer 42,500 rupees. Now we are done with the first discount. For the second discount this 10% will be applied on this selling price that is the reduced price we get after the first discount. So the second discount 10% of rupees 42,500. Remember this step is really important. We'll not be finding the second discount on 50,000. Let's calculate this. 10 upon 100 42,500. Do the cutting. Then we are left with 10 425 and the answer is 4,215. It's not the answer. It's just the second discount. Now we'll find the selling price after the second discount. So this time we'll be getting this 42,500 minus 4,250 the discount and we'll get the selling price is 38,250. So now you can see every time we'll be using the reduced price not the marked price that means the original price. So this for this third discount we'll be using 38,250. So third discount 5% of 38,250 rupees. So do the calculations 5 by 100 multly 38,250. We'll not do cutting here because we want the calculations easy. Just multiply 5 38,250. We'll get uh 1 91250 upon 100. Now we can place the decimal before two digits and we have our answer that is 1,91250 rupees. So this is the third discount. So final selling price after third discount will be we'll be subtracting 38,250 minus 9 1,912.150 rupees. So after subtraction we get the selling price which is this. So the selling price we get is 36,33750 rupees. So it's very easy just remember the step. The first discount will be applied on the mark price. The second on the previous selling price we'll get after the discount and the third on the previous second discount that is a reduced price we'll get and we'll get our answer the final selling price. Do not do the addition in the successive discounts. We'll find step by step. So this exercise is for you to solve. Do some workout. The mark price of the jacket is 1,500 rupees. Two successive discounts are 10% and 5% are offered. What is the selling price? So find the first discount first that is 10% of 1,500. Calculator is then find the selling price 1,500 minus that discount which will come in rupees. Then 5% of the discount the selling price the reduced price you'll get after 10% and you'll get the sell final selling price after that. So, thank you for watching. Please subscribe and like and don't forget to hit the bell icon for the notifications of our upcoming videos. Bye-bye.
4804	https://english.stackexchange.com/questions/68844/date-format-in-uk-vs-us	Skip to main content Date format in UK vs US Ask Question Asked Modified 7 years, 1 month ago Viewed 249k times This question shows research effort; it is useful and clear 32 Save this question. Show activity on this post. Why is the most common date format in the US like mm/dd/yyyy, whereas in Europe (including the UK) it's more common to have dd/mm/yyyy? Looking around, I found that the US form is actually the more traditional Anglo-Saxon way, but the British adapted to using the European form in the early 20th Century. But I couldn't find a definitive discussion of the history of the different formats. Is it just conventional, or is there an official 'British date standard' (like with metric and imperial, for example). american-english british-english dates Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications edited May 24, 2012 at 12:04 Dan Blows asked May 24, 2012 at 11:49 Dan BlowsDan Blows 48122 gold badges55 silver badges1111 bronze badges 8 4 There are no "rules" -- but there is rigidly enforced convention in order that everyone knows what everyone else is talking about. Not quite sure whether this counts as ELU or not, though. Andrew Leach – Andrew Leach ♦ 05/24/2012 12:00:45 Commented May 24, 2012 at 12:00 2 Anglo-Saxon (i.e. English ca. 500-1100 AD) dates were written in full sentences (e.g. "fifteenth day before the calends of April" -- Anglo-Saxon Chronicle) rather than using any kind of notation. At least, that's been my experience. Is that what you really meant to say? user8850 – user8850 05/24/2012 16:20:59 Commented May 24, 2012 at 16:20 4 I think it depends on the context. In the UK, for example, for "information interchange" something similar to ISO8601 is ratified as BS ISO 8601:2004, BS EN 28601 standards (which are preferred). So the format yyyy-mm-dd became pretty common on official documents, manufacturing stuff or interfaces. oxygen – oxygen 04/11/2013 00:23:19 Commented Apr 11, 2013 at 0:23 4 I have no sources to quote as such, but personally prefer to use a logical format starting with the lowest unit (days) and ending with the highest (years), thus the 4th of July would be most logical (to me anyway) in European format as 04/07/13, not 07/04/13. user42857 – user42857 04/19/2013 23:36:28 Commented Apr 19, 2013 at 23:36 2 @user42857 actually the Japanese version is the most logical, considering that time is written in opposite order starting with the highest number and ending with the smallest. They have adapted this to the date as well, so its yyyy.mm.dd hh:mm:ss (not sure about the delimiter symbol). Otto Abnormalverbraucher – Otto Abnormalverbraucher 03/15/2018 09:25:02 Commented Mar 15, 2018 at 9:25 \| Show 3 more comments 4 Answers 4 Reset to default This answer is useful 18 Save this answer. Show activity on this post. Although there are people who will sometimes say: Today is Thursday, the 24th of May, 2012. There are also others who instead say the same thing this way: Today is Thursday, May 24th, 2012. Certainly in the United States, the second way of mentioning a date is more common than the first. The long form sounds more formal to us, as in “on the Fourth of July” being more formal and long-winded than simply saying “on July 4th”. It was pronouncing the month before the day out loud that gave to retaining that same original order when converted to digits: merely convert the month name to a natural number, and there you have your answer. What’s today’s date? It’s May 24th. Instead of writing May-24, we simply change the “May” to “5” and write 5-24 or ⁵⁄₂₄. That way it follows the natural language order and so requires no mental gymnastics to switch things around when speaking the date aloud. Similarly “September 11th” gets written ⁹⁄₁₁, etc. The full spoken form with the year, “May 24th, 2012”, then becomes the written shorthand “5/24/2012”, or often just “5/24/12”. “Christmas of 2001” can be, and somewhat annoying often is, written “12/25/1”, while “January 25th, 2012” becomes “1/25/12”. This isn’t usually any sort of problem because of universal consensus on how to interpret such things in the United States. If you write day/month/year in America, you will not be understood. Although I myself prefer the ISO notation, normal people do not use it in their daily affairs. Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Jul 31, 2018 at 14:59 answered May 24, 2012 at 12:47 tchrist♦tchrist 139k4949 gold badges379379 silver badges623623 bronze badges 21 31 I don't speak it that way. I say "Thursday the 24th of May" but then I'm British. Andrew Leach – Andrew Leach ♦ 05/24/2012 12:53:48 Commented May 24, 2012 at 12:53 6 Fortunately we all have our own culture-specific Date::Format routines :-) Andrew Leach – Andrew Leach ♦ 05/24/2012 13:02:29 Commented May 24, 2012 at 13:02 8 Re your edit, I didn't downvote (or upvote) but the problems may include the following. You say US order follows the 'natural language order' but I don't think that makes sense as those of us in other countries find the DMY order natural. You say that the US format is because people (in the US?) talk like that, but how do you know it's not that people in the US talk like that because that's the official date format? If the latter then you haven't answered the question. You say there are plenty of references on the web, perhaps you could provide links to the more persuasive ones? Gaston Ümlaut – Gaston Ümlaut 05/25/2012 01:36:37 Commented May 25, 2012 at 1:36 5 Because of the ambiguity that arises for the first 12 days of every month, I prefer either yyyy/mm/dd or dd/mm/yyyy so that one is either getting more or less specific as one goes along, not jumping from month to day to year. Being in the computer industry, I really prefer yyyy/mm/dd because it alphabetizes better. TecBrat – TecBrat 06/04/2012 04:02:37 Commented Jun 4, 2012 at 4:02 4 It makes no sense to justify the order of MM/DD/YYYY based on how the date would be spoken aloud in day-to-day conversation. If you're using numbers to represent a date, there should at least be some numeric logic (as opposed to aesthetic/heuristic logic) justifying the order of numbers. Personally I'd love nothing better than to see middle-endian date notation abolished and the US stick to a more sensible formula... if people find it more logical to put the month first, they can stick to writing "Oct 3 2013" or something similar instead... user45483 – user45483 06/05/2013 10:55:13 Commented Jun 5, 2013 at 10:55 \| Show 16 more comments This answer is useful 4 Save this answer. Show activity on this post. It's very possible that the US inherited this from an outdated English format - much like the length unit, after Henry III's foot and which the English have left behind in favour of the more logical metric system. One argument I've heard in favour of the American system of dating is that the numbers of months in a year is smaller than the number of days in a month which itself is smaller than the number of possible years. So you would have 12/31/2013, in ascending order. I don't really buy this argument, but OP might be interested in it anyway so here it is. Meanwhile, in Northern Europe they've moved on to an opposite, descending date standard: year/month/day. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Jul 5, 2013 at 17:56 CorinaCorina 85566 silver badges1010 bronze badges 4 2 That’s simply ISO 8601 format that uses YYYY-MM-DD. The rest of your diatribe against the traditional English measurement system is simply misplaced bigotry and has no place here. tchrist – tchrist ♦ 07/05/2013 18:27:58 Commented Jul 5, 2013 at 18:27 I argue that it's very relevant, if the US preserved various systems used in England before it achieved its independence. But ok? Corina – Corina 07/05/2013 18:32:22 Commented Jul 5, 2013 at 18:32 7 @tchrist: It didn't read as a diatribe or as bigoted to me. It was one sentence, with one concrete example (the foot), given to corroborate the plausibility of borrowing from the English. The closest thing to a "value judgment" in the whole thing is that the metric system is more logical than the old English (and current American) system. But the thing to that is, I don't even see Americans ever argue that feet/inches/miles/etc. are more logical than the metric system. Honestly, that judgment isn't in serious dispute, anywhere. John Y – John Y 07/05/2013 18:54:38 Commented Jul 5, 2013 at 18:54 Many English people still regularly use feet/inches, especially when referring to things such as height. All road signs are exclusively in miles, as well. Metric tends to be commonly recognised, but most prevalently used in scientific & technical contexts. Myles – Myles 02/07/2018 16:13:03 Commented Feb 7, 2018 at 16:13 Add a comment \| This answer is useful 3 Save this answer. Show activity on this post. The 29th of May Is Oak-Apple Day. Ring-a-ding-ding! Long live the King! That was an old rhyme celebrating the return of Charles II to Britain at the beginning of the Restoration in 1660. I like to abbreviate dates as day/month in Roman numerals/year, with periods/full stops rather than "/" between the numbers, such that today is 31.vii.2018. Sometimes I make the last "i" a "j" (an affectation, I know, and a throwback to ancient manuscripts), and drop the "20" in the year, so that I have "31.vij.18". A month-number in Roman numerals can't be mistaken for a day-number. My way is not perfect: I don't think O-W Kenobi would write "4.v. be with you", for example, but then, he wouldn't write "5/4" or "4/5" either. He would write out "May the 4th...". (I spent several years in the military when I was much younger, and got used to the dd/mm/yy format used in the US military. I surmise that the US armed forces adopted this format during World War I, to minimize confusion with allies, but I don't know for sure.) Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications answered Jul 31, 2018 at 16:56 tautophiletautophile 53711 gold badge44 silver badges66 bronze badges Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. Personally if I was to say a date I would do so in the format mentioned by Andrew Leach, or even 'the 24th of May'. I suppose it depends on what you're used to. The American date format often has me confused unless the month is spelled out. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered May 24, 2012 at 15:04 AlexAlex 11522 bronze badges 5 I think you should only use were in this context if it is contrary to the fact. For example 'if I were you' as opposed to 'if I was you'. Alex – Alex 05/24/2012 19:55:16 Commented May 24, 2012 at 19:55 1 I agree with you that using were is correct if the subjunctive is contrary to the fact, as furnished by the example 'if I were you', however the sentence '..if I was to say a date I would do so in the format...' does not contain anything to the contrary, hence was is acceptable. Alex – Alex 05/24/2012 20:45:06 Commented May 24, 2012 at 20:45 To re-iterate, 'were' should only be used if the subjunctive is contrary to the fact. So I used 'was' in this instance, which I feel is acceptable, you obviously don't. Well I aim to continue using 'was' where I feel it is appropriate. Alex – Alex 05/24/2012 21:00:42 Commented May 24, 2012 at 21:00 Wales and Welsh (but not fluently). That's how I distinguish between were and was. Perhaps we should agree to disagree as I am being warned to avoid extended discussions. Alex – Alex 05/24/2012 21:16:07 Commented May 24, 2012 at 21:16 1 @tchrist This is not relevant to the question and should be taken to chat. Anyway, what's your data to show that native English speakers don't use 'was' in that way? Gaston Ümlaut – Gaston Ümlaut 05/25/2012 01:40:39 Commented May 25, 2012 at 1:40 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions american-english british-english dates See similar questions with these tags. Linked 17 Is there any word in English for the date after interchanged day and month digit? 6 How do I formally speak dates? 2 How to write out dates correctly 0 What should I write under "signed" tag? 0 “25th December” vs “25 December”: Should I use ordinals or cardinals for the day of the month? 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Don't worry! We've compiled some resources that you can download and print for free. The arithmetic sequences in this set of pdfs have a finite number of terms. There are two ways to find the sum of the sequence. Direct students to use the relevant formula and calculate the sum of each sequence. With this bundle of sum of finite arithmetic series worksheets around, understanding the concept of finite arithmetic series is utter joy. These pdf worksheets are most recommended for students in grade 8 and high school. Finding Sums of Finite Arithmetic Series \| Worksheet #1 Get a grip on learning while practicing this pdf worksheet. Look at the grouping of numbers displayed in a specific order and with a clear starting point. Use the formula Sn=(n/2)((2a 1+d(n-1)) and evaluate each series by identifying the first term, common difference, and the number of terms. Finding Sums of Finite Arithmetic Series \| Worksheet #2 Deepen and strengthen the students' understanding of arithmetic series with this free printable pdf worksheet. Observe the progression of numbers to determine the common difference and plug in the values in the formula to calculate the sums of finite arithmetic series. Finding the Sum Using a 1, a n, and n \| Worksheet #1 Supplement the well-structured 8th grade school curriculum with this free find the sum of a finite arithmetic series worksheet pdf. Use the formula S n = (n/2) (a 1 +a n) and substitute the appropriate terms to find the sum of the given finite series. Finding the Sum Using a 1, a n, and n \| Worksheet #2 Provide a comprehensive review to high school students with this sum of a finite arithmetic series worksheet. Plug in the value of a 1, a n, and n in the formula and evaluate the series. Evaluating Arithmetic Series in Sigma Notation Recalibrate the way you find the summation of finite arithmetic series with these printable worksheets exclusively designed for students in grade 8 and high school. Each series is represented in a compact form called summation or sigma notation. Mixed Review \| Worksheet #1 How will you find the sum of a finite arithmetic series? Encourage learners to look for the common difference, first term, last term, or number of terms and plug in the values in the appropriate formula to find the sum. Mixed Review \| Worksheet #2 Here's yet another worksheet to review your skills on finding the sums of finite arithmetic series. Help your student to make the leap from the beginner to proficient by working out the wealth of sums in this printable worksheet. 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4806	https://graphicmaths.com/pure/hyperbolic-functions/arcosh/	arcosh function By Martin McBride, 2025-07-18 Tags: arcosh cosh logarithm exponential square root Categories: hyperbolic functions The arcosh function is a hyperbolic function. It is the inverse of the cosh, and is also known as the inverse hyperbolic cosine function. Equation and graph The arcosh function is defined as the inverse of cosh, ie if: then: There is a also a formula for finding arcosh directly: Here is a graph of the function: The function is valid for x >= 1. arcosh as inverse of cosh This animation illustrates the relationship between the cosh function and the arcosh function: The first, blue, curve is the cosh function. The grey dashed line is the line . The second, red, curve is the arcosh function. As with any inverse function, it is identical to the original function reflected in the line . Logarithm formula for arcosh arcosh can be calculated directly, using a logarithm function, like this: Here is a proof of the logarithm formula for arcosh. It is very similar to the proof for arsinh. We will use: The cosh of u will be x, because cosh is the inverse of arcosh: One form of the formula for cosh is: This gives us: Multiplying both sides by gives: This is a quadratic in (using the fact that ): We use the quadratic formula with , , : Simplifying and taking the positive solution (since must be positive) gives: Taking the logarithm of both sides gives: And since u is this gives: See also sinh function cosh function tanh function sech function cosech function coth function arsinh function artanh function Hyperbolic angle Join the GraphicMaths Newsletter Sign up using this form to receive an email when new content is added: Popular tags adder adjacency matrix alu and gate angle answers area argand diagram binary maths cartesian equation chain rule chord circle cofactor combinations complex modulus complex polygon complex power complex root cosh cosine cosine rule countable cpu cube decagon demorgans law derivative determinant diagonal directrix dodecagon e eigenvalue eigenvector ellipse equilateral triangle euclid euler eulers formula eulers identity exercises exponent exponential exterior angle first principles flip-flop focus gabriels horn galileo gradient graph hendecagon heptagon heron hexagon hilbert horizontal hyperbola hyperbolic function hyperbolic functions infinity integration integration by parts integration by substitution interior angle inverse hyperbolic function inverse matrix irrational irrational number irregular polygon isomorphic graph isosceles trapezium isosceles triangle kite koch curve l system lhopitals rule limit line integral locus logarithm maclaurin series major axis matrix matrix algebra mean minor axis n choose r nand gate net newton raphson method nonagon nor gate normal normal distribution not gate octagon or gate parabola parallelogram parametric equation pentagon perimeter permutation matrix permutations pi polar coordinates polynomial power probability probability distribution product rule proof pythagoras proof quadrilateral questions radians radius rectangle regular polygon rhombus root sech segment set set-reset flip-flop sine sine rule sinh sloping lines solving equations solving triangles square square root squeeze theorem standard curves standard deviation star polygon statistics straight line graphs surface of revolution symmetry tangent tanh transformation transformations translation trapezium triangle turtle graphics uncountable variance vertical volume volume of revolution xnor gate xor gate Pure mathematics Series Exponentials Number theory Complex numbers Numerical methods Coordinate systems Advanced trigonometry Maclaurin and Taylor series Polar coordinates Hyperbolic functions sinh cosh tanh sech cosech coth arsinh arcosh artanh Hyperbolic angle Differentiation Limits Integration Set theory Matrices Special functions Recent Tags Categories Site map Twitter Youtube Github Linkedin Leanpub Medium
4807	https://www.youtube.com/watch?v=__PwIB-DxVc	They all look the same!! How to solve this simple impossible puzzle - Triangle Vinco #toys #puzzle downtimefun 3170 subscribers 21 likes Description 604 views Posted: 26 Jul 2024 Triangle Vinco is an excellent puzzle that consists of 6 similar pieces, looks simple, yet very very challenging. The pieces look the same and very easy to be confused. Check it out to see how I solved it, and get a copy to challenge yourself! Vinco website: Subscribe for more - Get more puzzles here: Puzzle master - Puzzlocks - Crux Puzzles - Jpgames ltd - Follow my IG - Follow my Twitter - Like my FB: Your support will bring more puzzles and better production: For inquiries and suggestions email timothyyylo@gmail.com Subscribe to the channel - 8 comments Transcript: hello and welcome to downtime fun today I'm going to solve a puzzle from a Czech puzzle company vco called triangle vco the quality of the puzzle is very good for its price the listed price for this puzzle is 6os so do not expect similar quality as the cares from Japan however these are very challenging puzzles and I think they are even better value for money in terms of puzzling for this puzzle the designer rated it as a five circles the scale is slightly different as the maximum difficulty is five plus so if we were to convert it there should be around four out of five stars there are six pieces in this puzzle and the goal is to combine the pieces to make a roof like triangular prism let's quickly take a look at each of the pieces they look very similar but if we look closer each one is different consist of one big and one small triangular prism at first when I took out the pieces I thought to myself this should be pretty easy but very quickly I realized I was wrong there are so many possible combinations and given the pieces look so similar I find myself repeating a lot of combinations as I do the trial and error maybe I should have marked the pieces anyway like this that two pieces combined seems to be a legitimate placement then we can try to place a third piece like this one it looks good but when we flip the p to the other side we can see that there is a hole in the structure which means this is not the correct Arrangement the pieces look very similar and a bit difficult to memorize when solving it let's see if we do this let's see this one again this doesn't work nope how about this this doesn't seem to work either let's try another combination we can put this piece here no it doesn't work like this there are just so many possible combinations but a lot of it just won't work in the end I have tried for over 30 minutes and I have finally found an answer first we can place two pieces like so and then we can place the third piece with which is this one like no this one like this and we will have made the bottom layer of the prism roof and then we can place the fourth piece like this here very nice and then the pen an ultimate piece like this and finally place the six final piece in like so and it's solved and this is the triangle VIN code it looks not super difficult or special and it seems like it's just a six piece puzzle however I didn't realize it will take me around 30 minutes to find the real solution you can see that all the pieces are in different orientations which is even more difficult some are on sideways some linkwise I'll put the official website's Link in the description below so you can check it out and that's about it for this video thank you very much for watching if you enjoy this please like And subscribe and I will see you in the next video bye
4808	https://healy.econ.ohio-state.edu/kcb/Ma103/Notes/RandomWalk.pdf	Department of Mathematics Ma 3/103 KC Border Introduction to Probability and Statistics Winter 2021 Supplement 7: Simple Random Walk In 1950 William Feller published An Introduction to Probability Theory and Its Applications . According to Feller [11, p. vii], at the time “few mathematicians outside the Soviet Union recognized probability as a legitimate branch of mathemat-ics.” In 1957, he published a second edition, “which was in fact motivated principally by the unexpected discovery that [Chapter III’s] enticing material could be treated by elementary methods.” Here is an elementary treatment of some of these fun and possibly counterintuitive facts about random walks, the subject of Chapter III. It is based primarily on Feller [11, Chapter 3] and [12, Chapter 12], but I have tried to make it even easier to follow. S7.1 ⋆ What is the simple random walk? Let X1, . . . , Xt, . . . be a sequence of independent Rademacher random variables, Xt = ( 1 with probability 1/2 −1 with probability 1/2 so E Xt = 0, and Var Xt = 1 (t = 1, 2, . . . ) (Imagine a game played between Hetty and Taylor, in which a fair coin is tossed repeatedly. When Heads occurs, Taylor wins a dollar from Hetty, and when Tails occurs Hetty wins a dollar from Taylor. Then Xt is the change in Taylor’s net winnings on the tth coin toss.) The index t indicates a point in time. Feller uses the term epoch to denote a particular moment t, and reserves the use of the word “time” to refer to a duration or interval of time, rather than a point in time, and I shall adhere to his convention. The set of epochs is the set Z+ = {0, 1, 2, . . . } of nonnegative integers. The epoch 0 is the moment before any coin toss. For each t, define the running sums St = X1 + · · · + Xt. For convenience, we define S0 = 0. (The random variable St is Taylor’s total net winnings at epoch t, that is, after t coin tosses.) It follows that for each St, E St = 0 and Var St = t. The sequence of random variables S0, . . . , St, . . . , t ∈Z+ is a discrete-time stochastic process known as the simple random walk on the integers. It is both a martingale (E St+s St = St) and a stationary Markov chain (the distribution of St+s St = kt, . . . , S1 = k1 depends only on the value kt). S7.1.1 Remark The walk St = X1 + · · · + Xt can be “restarted” at any epoch n and it will have the same probabilistic properties. By this I mean that the process defined by ˆ St = Sn+t −Sn = Xn+1 + · · · + Xn+t, is also a simple random walk. KC Border v. 2020.10.21::10.29 Ma 3/103 Winter 2021 KC Border Random Walk S7–2 20 40 60 80 100 -100 -50 50 100 Figure S7.1. The areas bounded by ±t and by ± √ t. S7.2 ⋆ Asymptotics The Strong Law of Large Numbers tells us that St t − − − → t→∞0 a.s., and the Central Limit Theorem tells us that St √ t D − − − → t→∞N(0, 1). Recall that the probability that the absolute value of a mean-zero Normal random variable exceeds its standard deviation is 2 1 −Φ(1) = 0.317, where Φ is the standard normal cdf. The standard deviation of St is √ t, so there is about a two-thirds chance that St lies in the interval [− √ t, √ t]. See Figure S7.1. At each largish t, about two-thirds of the paths cross the vertical line at t in the red area. But the Strong Law of Large Numbers and the Central Limit Theorem are rather coarse and may mislead us about behavior of the random walk. They do not address interesting questions v. 2020.10.21::10.29 KC Border Ma 3/103 Winter 2021 KC Border Random Walk S7–3 such as, Which values can the walk assume?, What are the waiting times between milestones?, or What does a “typical” path look like? S7.3 ⋆ Paths as the sample space A natural way to think about the random walk is in term of paths. The outcome path s = (s1, s2, . . . ) can be identified with the sequence (t, st), t = 0, 1, . . . of ordered pairs, or better yet with the graph of the piecewise linear function that connects the points (t, st). See Figure S7.2. Figure S7.2. A sample path of a random walk. There are infinitely many paths, but it is also convenient to refer to an initial segment of a path as a path. (Technically, the initial segment defines a set, or equivalence class, of paths that agree through some epoch.) There are 2t paths the walk may take through epoch t, and each one has equal probability, namely 1/2t. Let us say that the path s visits k at epoch t if st = k. If there is a path s with st = k, we say that the path s reaches (t, k) and that (t, k) is reachable from the origin. More generally, if (t0, k0) and (t1, k1), where t1 > t0, are on the same path, then we say that (t1, k1) is reachable from (t0, k0). S7.3.1 Characterization of reachable points Which of the lattice points (t, k) ∈Z+ × Z can belong to a path? Or in other words, which points (t, k) are reachable? Not all lattice points are reachable. For instance, the points (1, 0) and (1, 2) are not reachable since S1 is either 1 or −1. S7.3.1 Proposition (Criterion for reachability) In order for (t, k) to be reachable, there must be nonnegative integers p and m, where p is the number of plus ones and m is the number of minus ones such that p + m = t and p −m = k, which implies p = t + k 2 and m = t −k 2 . (1) KC Border v. 2020.10.21::10.29 Ma 3/103 Winter 2021 KC Border Random Walk S7–4 Reachability implies that both t + k and t −k must be even, so that t and k must have the same parity. We must also have t ⩾\|k\|. But those are the only restrictions. ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 5 10 15 -15 -10 -5 5 10 15 Figure S7.3. Reachable points are the big dots. Many points can be reached by more than one path from the origin. v. 2020.10.21::10.29 KC Border Ma 3/103 Winter 2021 KC Border Random Walk S7–5 S7.3.2 Definition The number of initial segments of paths that reach the reachable point (t, k) is denoted Nt,k. If (t, k) is not reachable, then Nt,k = 0. S7.3.3 Proposition (Number of paths that reach (t, k)) If (t, k) is reachable, then Nt,k = t t+k 2 = t t−k 2 . (2) Proof: By Proposition S7.3.1, if (t, k) is reachable, there must be nonnegative integers p and m, where p is the number of plus ones and m is the number of minus ones such that (1) is satisfied. Since the p (1)’s and m (−1)’s can be arranged in any order, there are Nt,k = p + m p = p + m m = t t+k 2 = t t−k 2 paths with this property. Since there are 2t paths of length t, the probability is given by: Define pt,k = P(St = k). S7.3.4 Corollary (pt,k) If (t, k) is reachable, then pt,k = t t+k 2 2−t. (3) S7.3.5 Corollary If (t1, k1) is reachable from (t0, k0), then the number of sample paths con-necting them is Nt1−t0,k1−k0. (4) S7.3.2 The Reflection Principle Feller referred to “elementary methods” that simplified the analysis of the simple random walk. The procedure is this: Treat paths as piecewise linear curves in the plane. Use the simple geometric operations of cutting, joining, sliding, rotating, and reflecting to create new paths. Use this technique to demonstrate the one-to-one or two-to-one correspondence between events (sets of paths). If we can find a one-to-one correspondence between a set we care about and a set we can easily count, then we can compute its probability. The first example of this geometric manipulation approach is called the Reflection Principle. S7.3.6 The Reflection Principle Let (t1, k1) be reachable from (t0, k0) and on the same side of the time axis. Then there is a one-to-one correspondence between the set of paths from (t0, k0) to (t1, k1) that meet (touch or cross) the time axis and the set of all paths from (t0, −k0) to (t1, k1). KC Border v. 2020.10.21::10.29 Ma 3/103 Winter 2021 KC Border Random Walk S7–6 Proof: One picture is worth a thousand words, so Figure S7.4 should suﬀice for a proof. ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● (t0, k0) (t0, −k0) (t1, k1) t∗ Figure S7.4. The red path is the reflection of the blue path up until the first epoch t∗where the blue path touches the time axis. This establishes a one-to-one correspondence between paths from (t0, −k0) to (t1, k1) and paths from (t0, k0) to (t1, k1) that touch the time axis at some t0 < t < t1. q.e.d. A consequence is the following. S7.3.7 The Ballot Theorem If k > 0, then there are exactly k n Nn,k paths from the origin to (n, k) satisfying st > 0, t = 1, . . . , n. Proof: • If st > 0 for all t = 1, . . . , n, then s1 = 1. • By Corollary S7.3.5, the total number of paths from (1, 1) to (t1, k) is Nn−1,k−1. • Some of these paths though touch the time axis, and when they do, they do not satisfy st > 0. How many of these paths touch the time axis? By the Reflection Principle, it is as many as the paths from (1, −1) to (n, k), which by Corollary S7.3.5 is Nn−1,k+1. • Thus the number of paths from (1, 1) to (n, k) that do not touch the time axis is Nn−1,k−1 −Nn−1,k+1. • Let p and m be as defined by (1). Then p + m = n, p −m = k, n + k = 2p, so the following v. 2020.10.21::10.29 KC Border Ma 3/103 Winter 2021 KC Border Random Walk S7–7 “trite calculation, ” as Feller puts it, yields Nn−1,k−1 −Nn−1,k+1 = n −1 (n + k −2)/2 − n −1 (n + k)/2 = m + p −1 p −1 − m + p −1 p = (m + p −1)! m!(p −1)! −(m + p −1)! p!(m −1)! = p(m + p −1)! m!p! −m(m + p −1)! p!m! = (p −m)(m + p −1)! m!p! = p −m p + m (m + p)! m!p! = k n Nn,k. Why is this called the Ballot Theorem? S7.3.8 The Ballot Theorem, version 2 Suppose an election with n ballots cast has one candidate winning by k votes. Count the votes in random order. The probability the winning candidate always leads is k n. S7.3.9 The Ballot Theorem, version 3 Suppose an election has one candidate getting p votes and the other getting m votes with p > m. Count the votes in random order. The probability the winning candidate always leads is p −m p + m. S7.4 ⋆ Returns to zero S7.4.1 Definition We say that the walk equalizes or returns to zero at epoch t if St = 0. Epoch t must be even for equalization to occur, so let t = 2m. The number of paths from the origin to (2m, 0) is N2m,0, so the probability u2m defined by u2m = P(S2m = 0), satisfies u2m = N2m,0 22m = 2m m 1 22m (m ⩾0). (5) Recall Stirling’s approximation, Proposition 3.8.2, which states that for n ⩾1, n! = √ 2π nn+(1/2)e−n eεn where εn →0 as n →∞. KC Border v. 2020.10.21::10.29 Ma 3/103 Winter 2021 KC Border Random Walk S7–8 Stirling’s approximation applied to (5) implies that u2m ∼ 1 √πm, (6) where the notation am ∼bm means am/bm →1 as m →∞. We derived this in Example 3.8.1. S7.5 ⋆ The Main Lemma The next application of the geometric manipulation of paths approach is to prove the following mildly surprising result. (It is the distillation of Lemma 3.1 and the discussion following on pp. 76–77, and problem 3.10.7, p. 96 in Feller .) S7.5.1 Main Lemma The following are equal: P(S2m = 0) (= u2m) (7) P(S1 ̸= 0, . . . , S2m ̸= 0) (8) P(S1 ⩾0, . . . , S2m ⩾0) (9) P(S1 ⩽0, . . . , S2m ⩽0) (10) 2P(S1 > 0, . . . , S2m > 0) (11) 2P(S1 < 0, . . . , S2m < 0) (12) Proof: Start with the easy cases. • By symmetry, (9) = (10) and (11) = (12). • In order to have (S1 ̸= 0, . . . , S2m ̸= 0), either (S1 > 0, . . . , S2m > 0) or (S1 < 0, . . . , S2m < 0). Both are equally likely. So (8) = (11) = (12). Let Zt denote the set of paths satisfying st = 0, Pt denote the set of paths satisfying (s1 > 0, . . . , st > 0), Nt denote the set of paths satisfying (s1 ⩾0, . . . , st ⩾0). (The mnemonic is zero, positive, and nonnegative.) S7.5.2 Lemma There is a one-to-one correspondence between P2m and N2m−1: Proof: Every path s in P2m passes through (1, 1) and satisfies st ⩾1 for t = 1, . . . , 2m, so shifting the origin to (1, 1) creates a path s′ of length 2m −1 that satisfies s′ t ⩾0, t = 1, . . . , 2m −1. That is, s′ ∈N2m−1. See Figure S7.5. • Lemma S7.5.2 establishes a one-to-one correspondence between P2m and N2m−1, so # N2m−1 = # P2m. Thus P(N2m−1) = # N2m−1 22m−1 = 2# N2m−1 22m = 2# P2m 22m = 2P(P2m). v. 2020.10.21::10.29 KC Border Ma 3/103 Winter 2021 KC Border Random Walk S7–9 s s′ Figure S7.5. The paths s ∈P2m and s′ ∈N2m−1. Since 2m −1 is odd, and equalization occurs only in even epochs, we must have s′ 2m−1 > 0 for any s′ ∈N2m−1. There are two possible continuations of s′ and both of them will satisfy s′ 2m ⩾0. That is, # N2m = 2 # N2m−1, so P(N2m) = 2 # N2m−1/22m = # N2m−1/22m−1 = P(N2m−1). Thus P(N2m) = P(N2m−1) = 2P(P2m). • So far we have shown (8) = (9) = (10) = (11) = (12). • We complete the Main Lemma by showing (7) = (9), or P(S2m = 0) = P(S1 ⩾0, . . . , S2m ⩾0). We shall establish this with the following lemma. Feller attributes the construction used in this proof to Edward Nelson. S7.5.3 Nelson’s Lemma There is a one-to-one correspondence between Z2m and N2m. Moreover, a path in Z2m with minimum value −k corresponds to a path in N2m with terminal value 2k. Proof: Let s be a path in Z2m. It achieves a minimum value −k∗⩽0 at some t ⩽2m, perhaps more than once. Let t∗be the smallest t for which st = −k∗. If s also belongs to N2m, that is, if st ⩾0 for all t = 0, . . . , 2m, then k∗= 0 and t∗= 0, and we leave the path alone. If s does not belongs to N2m, that is, if st < 0 for some 0 < t < 2m, then k∗> 0 and 0 < t∗< 2m. See Figure S7.6. We create a new path s′ in N2m as follows: Take the path segment from (0, 0) to (t∗, −k∗), and reflect it about the vertical line t = t∗. Slide this reversed segment to the point (2m, 0). See Figure S7.7. KC Border v. 2020.10.21::10.29 Ma 3/103 Winter 2021 KC Border Random Walk S7–10 (t∗, −k∗)      k∗ Figure S7.6. The path s ∈Z2m. Epoch t∗is the first epoch at which the minimum −k∗occurs. k∗      (t∗, −k∗)      k∗ Figure S7.7. Reflect the initial segment around t = t∗, and slide it to the end. v. 2020.10.21::10.29 KC Border Ma 3/103 Winter 2021 KC Border Random Walk S7–11 Now slide the whole thing so that (t∗, −k∗) becomes the new origin. The path now ends at (2m, 2k∗), where k∗> 0. See Figure S7.8.                        2k∗ Figure S7.8. Now slide (t∗, −k∗) to the origin to get the path s′ ∈N2m. This process is invertible: Let s be a path in N2m. If S2m = 0, leave it alone. If s2m > 0, since s2m is even, write s2m = 2¯ k > 0. Let ¯ t be the last time st = ¯ k. See Figure S7.9.                        2¯ k ¯ t Figure S7.9. The path s ∈N2m satisfies s2m = 2¯ k > 0. Epoch ¯ t is the last epoch for which st = k. KC Border v. 2020.10.21::10.29 Ma 3/103 Winter 2021 KC Border Random Walk S7–12 Take the segment of the path from (¯ t, ¯ k) to (2m, 2¯ k), reflect it about the vertical line t = ¯ t, slide it to the origin, (so it juts out up and to the left). See Figure S7.10.      ¯ k      ¯ k Figure S7.10. Take the segment of the path from (¯ t, ¯ k) to (2m, 2¯ k), reflect it about the vertical line t = ¯ t, slide it to the origin. Now make the beginning the new origin. See Figure S7.11. This new path s′′ satisfies s2m = 0, and has a strictly negative minimum.      ¯ k Figure S7.11. The path s′′ ∈Z2m. In fact the procedure above inverts the first procedure. This establishes a one-to-one corre-spondence between Z2m and N2m. • The proof of the Main Lemma is now finished. S7.6 ⋆ First return to zero S7.6.1 Definition The first return to zero happens at epoch t = 2m if S1 ̸= 0, . . . , S2m−1 ̸= 0, and S2m = St = 0. Let ft = f2m denote the probability of this event, and define f0 = 0. The next result is equation (3.7), p. 78, . S7.6.2 Corollary The explicit formula for f2m is f2m = u2m−2 −u2m = 1 2m −1u2m = 1 2m −1 2m m 1 22m (m = 1, 2, . . . ). (13) v. 2020.10.21::10.29 KC Border Ma 3/103 Winter 2021 KC Border Random Walk S7–13 Proof: The event that the first return to zero occurs at epoch 2m is (S1 ̸= 0, . . . , S2m−2 ̸= 0, S2m = 0) = (S1 ̸= 0, . . . , S2m−2 ̸= 0) \ (S1 ̸= 0, . . . , S2m ̸= 0). Since (S1 ̸= 0, . . . , S2m ̸= 0) ⊂(S1 ̸= 0, . . . , S2m−2 ̸= 0) the probability satisfies P ((S1 ̸= 0, . . . , S2m−2 ̸= 0) \ (S1 ̸= 0, . . . , S2m ̸= 0)) = P(S1 ̸= 0, . . . , S2m−2 ̸= 0) −P(S1 ̸= 0, . . . , S2m ̸= 0), which by the Main Lemma is u2m−2 −u2m. Equation (5) implies u2m−2 = (2m −2)! (m −1)!(m −1)! 1 22m−2 = 4m2 2m(2m −1) (2m)! m!m! 1 22m so u2m−2 −u2m = 2m 2m−1 −1 u2m, which implies the second and third equalities of (13). S7.6.3 Corollary With probability 1, the random walk returns to zero. Consequently, with probability 1 it returns to zero infinitely often. Proof: The event that the walk returns to zero is the union over m of the disjoint events that the first return occurs at epoch 2m. Its probability is just the sum of the first return probabilities. According to (13), we have the telescoping series ∞ X m=1 f2m = ∞ X m=1 (u2(m−1) −u2m) = u0 = 1. (14) While the walk is certain to return to zero again and again, you wouldn’t want to hold your breath waiting for it. Cf. Theorem 1, p. 395, . S7.6.4 Proposition (Waiting time for first return to zero) Let W denote the epoch of the first return to zero. Then E W = ∞. Proof: From Corollary S7.6.2, E W = ∞ X m=1 2m f2m = ∞ X m=1 2m 2m −1u2m. (15) From (6), 2m 2m −1u2m ∼ 1 √πm. But ∞ X m=1 1 √πm ⩾ 1 √π ∞ X m=1 1 m →∞. So by the Limit Comparison Test [2, Theorem 10.9, p. 396], the series (15) also diverges to ∞. KC Border v. 2020.10.21::10.29 Ma 3/103 Winter 2021 KC Border Random Walk S7–14 S7.7 ⋆ Recurrence S7.7.1 Definition The value k is recurrent if P(St = k infinitely often) = 1. Corollary S7.6.3 proved that 0 was a recurrent value. An astonishingly simple consequence of this that every value is recurrent. S7.7.2 Corollary For every integer k, with probability 1 the random walk visits k. Con-sequently, with probability 1 it visits k infinitely often. Proof: The validity of this result depends on symmetry, and so does this proof. • For every k, the probability that the walk visits k is greater than zero. Indeed, for k ̸= 0, we have P(Sk = k) = 2−k > 0. • Once the walk visits k, the probability that it later visits zero must be one, otherwise the probability of visiting zero infinitely often could not be one. • But the probability of reaching 0 from k is the same as reaching k from 0. • Therefore the probability the walk visits k is one. • Once at k, the probability of visiting k again is the same as the probability of revisiting zero from the origin, which is one. Therefore k is recurrent. This fact is another illustration of the difference between impossibility and probability zero. The path st = t for all t never returns to zero or anything else, and it is certainly a possible path. But it has probability zero of occurring. S7.8 ⋆ The Arc Sine Law The next result appears as [11, Theorem 1, p. 79]. For each m, define the random variable L2m = the epoch of the last visit to zero, up to and including epoch 2m = max{t : 0 ⩽t ⩽2m & St = 0}. Note that L2m = 0 and L2m = 2m are allowed. For convenience, define α2k,2m = P(L2m = 2k). S7.8.1 The Arc Sine Law for Last Returns The probability mass function for L2m is given by P(L2m = 2k) = α2k,2m = u2ku2(m−k). (k = 0, . . . , m) (16) Proof: The event (L2m = 2k) can be written S2k = 0 \| {z } A , S2k+2 ̸= 0, . . . , S2m ̸= 0 \| {z } B . Recall that P(AB) = P(B A)P(A). Now P(A) is just u2k and P(B A) is just the probability that starting at 0, the next 2(m−k) values of St are nonzero, which is the same as the probability that St ̸= 0, t = 1, . . . , 2(m −k). By the Main Lemma S7.5.1 this is equal to u2(m−k). v. 2020.10.21::10.29 KC Border Ma 3/103 Winter 2021 KC Border Random Walk S7–15 Why is this called the Arc Sine Law?: From (6), u2k ∼ 1 √ πk, so for large enough k, m, α2k,2m = u2ku2(m−k) ≈ 1 π p k(m −k) = 1 m 1 π q k m 1 −k m . (17) See Figure S7.12. As you can see from the figure, the approximation is rather good for even modest values of k and m, and the highest probabilities are for k = 0 and k = 2m, with the minimum occurring around m. The function f(x) = 1 π p x(1 −x) . is a probability density on the unit interval, and the cumulative distribution function involves the arcsine function: For 0 ⩽ρ ⩽1, Z ρ 0 f(x) dx = 2 π arcsin √ρ. So for every 0 < ρ < 1, for m large enough, P(L2m ⩽ρ2m) = X k<ρm α2k,2m ≈ Z ρ 0 f(x) dx = 2 π arcsin √ρ. (18) The Arc Sine Law has the following remarkable consequence, [11, p. 78]. S7.8.2 Corollary For every m, P(the latest return to 0 through epoch 2m occurs no later than epoch m) = P(L2m ⩽m) = 1 2. In other words, the probability that no equalization has occurred in the last half of the history is 1/2 regardless of the length of the history. Proof: For t even, P(L2m = t) = utu2m−t, which is symmetric about m, so m is the median value of L2m. S7.9 ⋆ Time reversal and dual walks We would like to know about the probabilities of visiting points other than zero. To do that, we shall make use of the dual of a random walk. Recall that the walk S is given by St = X1 + · · · + Xt where the Xt’s are independent and identically distributed Rademacher random variables. S7.9.1 Definition Fix a length n, and create a new random walk S∗of length n by reversing the order of the Xt’s. That is, define X∗ 1 = Xn, . . . , X∗ n = X1, and define S∗ t = X∗ 1 + · · · + X∗ t , (t = 1, . . . , n). (19) In other words S∗ t = Xn + · · · + Xn−t+1 = Sn −Sn−t. This walk S∗ 0, S∗ 1, . . . , s∗ n is called the dual of S. KC Border v. 2020.10.21::10.29 Ma 3/103 Winter 2021 KC Border Random Walk S7–16 0.2 0.4 0.6 0.8 1.0 0.05 0.10 0.15 k m 0.2 0.4 0.6 0.8 1.0 0.02 0.04 0.06 0.08 0.10 0.12 k m Figure S7.12. Plots of the points (k/m, α2k,2m), k = 0, . . . , m and the function f(x) = 1 m 1 π p x (1 −x) , for x ∈[0, 1] for m = 10, 20. v. 2020.10.21::10.29 KC Border Ma 3/103 Winter 2021 KC Border Random Walk S7–17 Given a path s for S, the dual path s∗for S∗is gotten by rotating the path s one hundred eighty degrees around the origin (time reversal), so the left endpoint has a negative time coor-dinate, and then sliding the left endpoint to the origin to get s∗. See Figures S7.13 and S7.14. The most important think about the a walk and its is dual is this: Since C1, . . . , Cn are independent and identically distributed, the walk S0, . . . , Sn and the walk S∗ 0, . . . , S∗ n have the same distribution. That is, for any s1, . . . , sn, P(S1 = s1, . . . , Sn = sn) = P(S∗ 1 = s1, . . . , S∗ n = sn). (20) S7.10 ⋆ First visits This argument is taken from Feller [11, pp. 92–93]. Let k be greater than zero. Assume that (n, k) is reachable from the origin. That is, n−k ⩾0 and n−k is even. What is the probability that the first visit to k happens at epoch n? The event that the first visit to k happens at epoch n is just the event (Sn = k, S1 < Sn, S2 < Sn, . . . , Sn−1 < Sn). (21) It follows from (19) that the following events are the same: (Sn = k, Sn > S1, . . . , Sn > Sn−1) = S∗ n = k, S∗ 1 > 0, . . . , S∗ n−1 > 0 . (22) We already know the probability of the latter event for the dual walk. From (20) it is the same as the probability that S∗ n = k, S∗ 1 > 0, . . . , S∗ n−1 > 0 , which is given by the Ballot Theorem S7.3.7. Thus P(the first visit to k occurs at epoch n) = k n n n−k 2 1 2n , (23) provided n −k is a nonnegative even integer. (Otherwise, it is zero.) We know from Proposition S7.3.3 that the probability that St = k is P(St = k) = Nt,k 2t = t t−k 2 1 2t , provided (t, k) is reachable from the origin. (Otherwise, it is zero.) If n −k is a nonnegative even integer, write n = 2m + k. It follows from (23) and the fact that k is recurrent that for each k ⩾1, ∞ X m=0 k 2m + k 2m + k m 1 22m+k = 1, (24) but don’t ask me to prove it directly. Aside: You may have noticed a similarity between the values of f2m given in (13) and the terms in (24) KC Border v. 2020.10.21::10.29 Ma 3/103 Winter 2021 KC Border Random Walk S7–18 rotate s about the origin s slide the left endpoint to origin s∗ to get s∗ Figure S7.13. Transforming s to s∗. rotate s∗about the origin s∗ slide the left endpoint to origin s to get s Figure S7.14. Transforming s∗back to s by the same method, (s∗)∗= s. (This figure also demonstrates a one-to-one correspondence that proves (22).) v. 2020.10.21::10.29 KC Border Ma 3/103 Winter 2021 KC Border Random Walk S7–19 for k = 1. (When I first noticed it, it kept me awake until I could verify the following equalities.) 1 = ∞ X m=1 f2m equation (14) = ∞ X m=1 1 2m −1 2m m 1 22m equation (13) = ∞ X n=0 1 2n + 1 2n + 2 n + 1 1 22n+2 substitute n = m −1 = ∞ X n=0 1 2n + 1 2n + 1 n 1 22n+1 since 2n + 2 n + 1 = 2 2n + 1 n = 1 from (24) with k = 1. S7.11 ⋆ The number of visits to k before equalization The following highly counterintuitive result is Example (b) on p. 395 of Feller . Let k be nonzero, and let Mk = the count of epochs n for which Sn = k before the first return to zero. S7.11.1 Proposition For each k, E Mk = 1. Proof: Since k and −k are symmetric, it suﬀices to consider k > 0. Let V k n be the event that a visit to k occurs at epoch n before the first return to zero. That is, V k n = (Sn = k, S1 > 0, . . . , Sn−1 > 0). Then Mk = ∞ X n=1 1V k n , where 1V k n is the indicator function of the event V k n . By the Monotone Convergence Theorem 5.13.2, we have E Mk = ∞ X n=1 E 1V k n = ∞ X n=1 P(V k n ) In other words, the probability that the walk visits k at epoch n before returning to zero is the same as the probability that the first visit to k occurs at epoch n. (This is not obvious without dual walks.) Now P(V k n ) = P(Sn = k, S1 > 0, . . . , Sn−1 > 0) = P S∗ n = k, S∗ n > S∗ 1, . . . , S∗ n > S∗ n−1 (22) = P(Sn = k, Sn > S1, . . . , Sn > Sn−1) (20) = P(first visit to k occurs at epoch n) KC Border v. 2020.10.21::10.29 Ma 3/103 Winter 2021 KC Border Random Walk S7–20 Therefore E Mk = ∞ X n=1 P(V k n ) = ∞ X n=1 P(first visit to k occurs at epoch n) = P(walk visits k) = 1. The last equality holds because k is recurrent. S7.12 ⋆ Sign changes There is a sign change at epoch t if St−1 and St+1 have opposite signs. This requires that St = 0, and that t be even. S7.12.1 Theorem Let t = 2m + 1 be odd. Then P(there are exactly c sign changes before epoch t) = 2P(St = 2c + 1). (25) Proof: To save space, let Ct,c = (there are exactly c sign changes before epoch t). Now P(Ct,c) = P(Ct,c S1 = 1)P(S1 = 1) + P(Ct,c S1 = −1)P(S1 = −1) = P(Ct,c S1 = 1), where the last equality follows from symmetry. That is, the probability of Ct,c is independent of the value of S1, so we may assume that S1 = 1. Now P(Ct,c S1 = 1) is the number of paths starting at (1, 1) that have exactly c sign changes before epoch t = 2m+1 divided by the number of paths starting at epoch 1 and ending at epoch t = 2m + 1. Thus the theorem reduces to the following Lemma. S7.12.2 Lemma For every odd t = 2m + 1, there is a one-to-one correspondence between the sets of paths {s : s1 = 1 and s has exactly c sign changes before t} and {s : st = 2c + 1}. I don’t have time to write up the proof of the lemma, but you can find it in Feller [11, Section III.5, pp. 84–86]. S7.12.3 Corollary The probability of c sign changes decreases with c. Consequently the most likely number of sign changes is zero! S7.13 ⋆ More remarkable facts Blackwell, Deuel, and Freedman discovered the following theorem while validating some code for an IBM 7090 computation. S7.13.1 Theorem Let Vm,n be the event that there exists t satisfying m ⩽t < m + n and S2t = 0. That is, Vm,n is the event that there is an equalization between epochs 2m −1 and 2(m + n) −1. Then for all m, n ⩾1, P(Vm,n) + P(Vn,m) = 1. Note that the Corollary S7.8.2 of the Arcsine Law can be rewritten as the special case P(Vm,m) = 1/2. v. 2020.10.21::10.29 KC Border Ma 3/103 Winter 2021 KC Border Random Walk S7–21 S7.14 ⋆ Asymmetric random walks The remainder of this discussion relies heavily on the exposition by Frederick Mosteller [13, pp. 51–55]. The preceding results made heavy use of the symmetry that arose from the fact that upticks and downticks were exactly equally likely. What happens when that is not the case? Let X1, . . . , Xt, . . . be a sequence of independent Rademacher(p) random variables, Xt = ( 1 with probability p −1 with probability 1 −p so E Xt = 2p −1, and Var Xt = 4p(1 −p) (t = 1, 2, . . . ) It is convenient to consider starting walks at arbitrary integers, so let S0 = m, St = S0 + X1 + · · · + Xt (t ⩾1) denote the asymmetric random walk starting at m with uptick probability p. It is no longer a martingale, but it is a time-invariant Markov chain. S7.15 ⋆ Reaching zero For a symmetric random walk, every state is recurrent. This fails for the asymmetric random walk. Let’s calculate the probability zm of reaching zero starting at m > 0. Let’s start with S0 = m. In order to reach 0 from m, you must first reach m −1. Then from m −1, you must reach m −2, etc., all the way to reaching 0 from 1. Each of these steps looks exactly like the last. I also claim that the independence of the Xt’s means that the probability of all of these steps happening is the product of their probabilities. Thus zm = zm 1 , (m > 1). Now the trick is to calculate z1. Well, starting at 1, with probability 1 −p we reach 0 on the first step. With probability p we reach 2, and then with probability z2 = z2 1 we reach 0. Thus z1 = 1 −p + pz2 1. This is a quadratic that has two solutions, z1 = 1, z1 = 1 −p p . This makes sense, because z1 really depends on p. For p = 0 (never gain), clearly z1 = 1. And when p = 1 (never lose), z1 = 0. When p = 1/2 both roots agree and z1 = 1. Figure S7.15 shows a plot of 1 and (1 −p)/p against p. These are the candidates for z1(p). We know z1(p) at three points p = 0, 1/2, 1. So if z1(p) is a continuous function, we must have z1 = ( 1 p ⩽1/2 1−p p p ⩾1/2 See Figure S7.16. So if p > 1/2, the probability of reaching zero from m is zm(p) = 1 −p p m →0 as m →∞. But for p = 1/2, zm(p) is always 1. This is just one of the ways the simple random walk is special. KC Border v. 2020.10.21::10.29 Ma 3/103 Winter 2021 KC Border Random Walk S7–22 0.2 0.4 0.6 0.8 1.0 0.5 1.0 1.5 2.0 Figure S7.15. Plot of 1 and (1 −p)/p against p. 0.2 0.4 0.6 0.8 1.0 0.5 1.0 1.5 2.0 Figure S7.16. z1(p). v. 2020.10.21::10.29 KC Border Ma 3/103 Winter 2021 KC Border Random Walk S7–23 S7.16 ⋆ The Gambler’s Ruin problem The random walk is a model of the fortunes of a gambler who always make the same size bet. We saw that if the probability of winning is p > 1/2, then there is a positive probability that the gambler may never go bankrupt. 5 What happens when the gambler plays against a casino that has limited resources? What is the probability that the gambler “breaks the bank?” That is, the casino goes bankrupt before the gambler does? • One gambler, call him Bond, starts with fortune b. • The other, call him Goldfinger, start with fortune g. • They play until one is bankrupted. Aside: How do we know that with probability 1 one of the players will be bankrupted? This Markov chain has two absorbing states, 0 (Bond is broken) and m + n (Goldfinger is broken). These two absorbing states are recurrent and the other states are transient. • Let p be the probability Bond wins each bet. As Bond is the better gambler, assume that p ⩾1/2, and to simplify notation, let q = 1 −p. • What is the probability that Bond (the stronger player) breaks Goldfinger? • Let B denote the probability that Bond breaks Goldfinger. • Consider first the counterfactual that Bond is not playing against Goldfinger, but is playing against the Federal Reserve Bank, which can create money at will. We have seen that the probability the Fed breaks Bond is just (q/p)b. There two ways the Fed can break Bond: ◦ One is that Bond never attains b + g on his way to bankruptcy, ◦ the other is that he does attain b + g before bankruptcy, but is still broken by the Fed. • The event that Bond never attains b+g on his way to bankruptcy has the same probability that Goldfinger breaks Bond, namely 1 −B. • The event that Bond attains b+g before bankruptcy is the event that Bond breaks Goldfin-ger, which happens with probability B. • The probability that the Fed breaks Bond upon reaching b + g is just (q/p)b+g. • Thus 1 −p p b \| {z } prob Fed breaks Bond = (1 −B) \| {z } prob Goldfinger breaks Bond + B \|{z} prob Bond reaches b + g 1 −p p b+g \| {z } prob Fed breaks Bond from b + g • Solving for B gives Probability B that Bond breaks Goldfinger = 1 − 1−p p b 1 − 1−p p b+g . 5 The term bankrupt, meaning almost literally “broken bank,” is derived from the ancient Greek practice of punishing debtors who cannot repay their debts by breaking (rupturing) their workbench (bank). KC Border v. 2020.10.21::10.29 Ma 3/103 Winter 2021 KC Border Random Walk S7–24 • For the case p = 1/2, the formula gives 0/0, but evaluating it with l’Hôpital’s rule gives b b + g . • Note that when p = 1/2, the probability that Bond wins is homogeneous of degree zero in b and g, the initial wealths. That is, λb λb+λg = b b+g for any λ > 0. For p > 1/2, this is no longer true. Here is are chart showing the effect of p and b on the probability that Bond breaks Goldfinger, holding g fixed at 10 and at 100, holdidng the bet size constant at 1. 0.6 0.7 0.8 0.9 1.0 p 0.2 0.4 0.6 0.8 1.0 B g = 10 b=1 b=5 b=10 b=20 0.6 0.7 0.8 0.9 1.0 p 0.2 0.4 0.6 0.8 1.0 B g = 100 b=10 b=50 b=100 b=200 Note that Bond’s edge is greater when both gamblers have larger fortunes relative to the bet size. v. 2020.10.21::10.29 KC Border Ma 3/103 Winter 2021 KC Border Random Walk S7–25 A probability space for the random walk: The set Ωof such paths constitutes the sample space for a probability model. It is a large sample space, since there is a one-to-one correspondence between sample paths and infinite sequence of ±1s. There is a not quite one-to-one correspondence between the set {1, −1}N of sample paths and the unit interval I = [0, 1]: every real number in I has a binary expansion as a sequence of 0s and 1s, which can obviously be turned into a sequence of 1’s and −1’s. The reason this is not one-to-one is that some real numbers have more than one binary expansion. For instance 1/2 = 0.10 . . . = 0.01 . . .. It’s not important for this discussion, but there is a natural one-to-one correspondence between {1, −1}N and the Cantor ternary set, a subset of I. The point is that the sample space is uncountable, so we should not expect every subset to be an event. So what sort of subsets of Ωshould be events? S7.17 ⋆ Brownian motion Brownian motion is a continuous-time stochastic process on a time interval, say T = [0, 1] that is an idealized version of the simple random walk with “infinitely short” steps being taken “infinitely often” in any time interval. The summary is this. There is an uncountable probability space (Ω, Σ, P) and a real function B : T × Ω→R. For each t ∈T there is a random variable Bt : Ω→R defined by Bt(ω) = B(t, ω), and for each ω ∈Ω, there is a function Bω : T →R defined by Bω(t) = B(t, ω). The key properties are • For each t ∈T, the random variable Bt has a Normal(0, t) distribution. • Increments are stochastically independent. That is, whenever t0 < t1 < · · · < tk, the random variables Bt1 −Bt0, Bt2 −Bt1, . . . , Btk −Btk−1 are stochastically independent. • For every ω ∈Ω, the function Bω is continuous. These properties characterize Brownian motion. But how do we know such a stochastic process exists? One way to approach it is take as the sample space Ωthe set C[0, 1] of continuous function on [0, 1]. The random experiment consists of drawing a function ω ∈C[0, 1] at random according to some probability measure W on C[0, 1], where W has the desired properties. The measure W is known as Wiener measure after Norbert Wiener [15, pp. 214–234]. Now we just have to find W. So how can we draw a choose a continuous function at random? We have already done so. The graph of our simple random walk chooses a piecewise linear function at random. But this distribution does not gives us Brownian motion. But we can construct a sequence of distributions on the set of piecewise linear functions that does converge to the Wiener distribution. The nth distribution is gotten by speeding up the simple random walk by a factor of n and scaling down the step size by a factor 1/√n. That is, by epoch t = 1 we take n steps at random of length 1/√n. The reason for this rescaling is so that the value of the walk at time t has KC Border v. 2020.10.21::10.29 Ma 3/103 Winter 2021 KC Border Random Walk S7–26 variance t. So start with a simple random walk S0, S1, S2, . . . and for each n we will define a continuous-time stochastic process {B(n) t : t ∈[0, 1]} as follows. For time t = k/n set B(n) t = Sk √n, (t = k/n), and linearly interpolate these values: B(n) t = S[tn] + (tn −[tn])S[tn]+1, (0 ⩽t ⩽1), where [tn] is the greatest integer less than or equal to tn. Figure S7.17 show the graph of the function t 7→B(n) t for a fixed ω and n = 1, 4, 16. When tn = k is an integer, then the random ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 0.2 0.4 0.6 0.8 1.0 -1.0 -0.5 0.5 1.0 ●n=16 ■n=64 ◆n=256 Figure S7.17. Scaled initial paths of one sample of a random walk. variable B(n) t = Sk/√n (t = k/n) has mean 0 and variance k/n = t. For such t, the Central Limit Theorem implies B(n) t D − − − − → n→∞N(0, t). Also note that for each n, for t0 < t1 < · · · < tk, the random variables B(n) t1 −B(n) t0 , B(n) t2 −B(n) t1 , . . . , B(n) tk −B(n) tk−1 are stochastically independent (being the sums of disjoint sets of independent scaled Rademacher random variables). v. 2020.10.21::10.29 KC Border Ma 3/103 Winter 2021 KC Border Random Walk S7–27 At this point I am going to make use of a lot of results that are well beyond the scope of this course. For each n, B(n) defines a probability distribution on the set C[0, 1] of continuous functions on the interval as follows. For each subset A of C[0, 1], 6 define its probability by Pn(A) = P{ω ∈Ω: B(n)(ω) ∈A}. It is possible to extend the definition of convergence in distribution to spaces such as C[0, 1] (see for instance Aliprantis and Border [1, Chapter 15], Billingsley , or Parthasarathy ). Monroe Donsker proved that the probability distribution Pn on C[0, 1] associated with B(n) converges in distribution to Weiner measure W as n →∞. This result is known as Donsker’s Theorem. See Billingsley [4, § 10, p. 68]. Informally, this means that properties of Brownian motion can be deduced as limiting results from scaled random walks. As an example, Billingsley [4, pp. 80–83] uses the Arc Sine Law for random walks to prove the Arc Sine Law for Brownian motion on [0, 1]: If T is the last time Bt = 0, then P(T ⩽t) = 2 π arcsin √ t. Compare this with (18) on page S7–15 of these notes. S7.17.1 What is Brownian motion a model of? Brownian motion is named after Robert Brown, a 19th century botanist who observed the motion of pollen particles in water [5, p. 443]. We can generalize Brownian motion to several dimensions by considering random vector functions, where the components are independent Brownian motions. To cover such things as the two or three dimensional motion of particles colliding molecules of air or water. Brownian motion was the first good model of random diffusion. In 1900, Louis Bachelier [3, 8] developed a mathematical model for Brownian motion and argued that stock prices must follow a Brownian motion stochastic process. In 1973 Fischer Black and Myron Scholes independently rediscovered some of Bachelier’s results. Bibliography C. D. Aliprantis and K. C. Border. 2006. Infinite dimensional analysis: A hitchhiker’s guide, 3d. ed. Berlin: Springer–Verlag. T. M. Apostol. 1967. Calculus, Volume I: One-variable calculus with an introduction to linear algebra, 2d. ed. New York: John Wiley & Sons. L. J.-B. A. Bachelier. 1900. Théorie de la spéculation. Annales Scientifiques de l’École Normale Supérieure, 3 Série 17:21–86. P. Billingsley. 1968. Convergence of probability measures. Wiley Series in Probability and Mathematical Statistics. New York: Wiley. . 1979. Probability and measure. Wiley Series in Probability and Mathematical Statistics. New York: Wiley. 6 Actually we want to restrict A to be a “nice” subset of C[0, 1]. First turn C[0, 1] into a normed space with norm ∥f∥= maxx∈[0,1 f(x). This gives rise to the notion of open sets in C[0, 1]. The smallest σ-algebra that includes the open sets is called the Borel σ-algebra. We restrict the discussion to sets A that belong to the Borel σ-algebra. KC Border v. 2020.10.21::10.29 Ma 3/103 Winter 2021 KC Border Random Walk S7–28 F. Black and M. Scholes. 1973. The pricing of options and corporate liabilities. Journal of Political Economy 81(3):637–654. D. Blackwell, P. Deuel, and D. Freedman. 1964. The last return to equilibrium in a coin-tossing game. Annals of Mathematical Statistics 35:1344. M. Davis and A. Etheridge. 2011. Louis Bachelier’s Theory of Speculation: The origins of modern finance. Princeton, New Jersey: Princeton University Press. M. D. Donsker. 1951. An invariance principle for certain probability limit theorems. Mem-oirs of the American Mathematical Society 6. W. Feller. 1950. An introduction to probability theory and its applications, 1st. ed., vol-ume 1. New York: Wiley. . 1968. An introduction to probability theory and its applications, 3d. ed., volume 1. New York: Wiley. . 1971. An introduction to probability theory and its applications, 2d. ed., volume 2. New York: Wiley. F. Mosteller. 1987. Fifty challenging problems in probability with solutions. Mineola, New York: Dover. Reprint of the 1965 Addison-Wesley edition. K. R. Parthasarathy. 1967. Probability measures on metric spaces. Probability and Math-ematical Statistics. New York: Academic Press. N. Wiener. 1930. Generalized harmonic analysis. Acta Mathematica 55(1):117–258. DOI: 10.1007/BF02546511 v. 2020.10.21::10.29 KC Border
4809	https://www.elsevier.com/resources/anatomy/urogenital-system/urinary-system/urinary-bladder/17295	Urinary Bladder \| Complete Anatomy Skip to main content We use cookies that are necessary to make our site work. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. You can manage your cookie preferences using the “Cookie Settings” link. For more information, see our Cookie Policy. Accept Additional Cookies Cookie settings Unfortunately we don't fully support your browser. If you have the option to, please upgrade to a newer version or use Mozilla Firefox, Microsoft Edge, Google Chrome, or Safari 14 or newer. If you are unable to, and need support, please send us your feedback. Dismiss We'd appreciate your feedback.Tell us what you think! Academic & GovernmentAcademic & Government HealthHealth IndustryIndustry Elsevier ConnectInsights AboutAbout Customer supportSupport Publish with us Open Search Location Selector Show Menu Home Resources Complete Anatomy Articles Urogenital System Urinary Bladder Urogenital System Urinary Bladder Vesica urinaria Read more Quick Facts Structure/ Morphology Anatomical Relations Function Arterial Supply Venous Drainage Innervation Lymphatic Drainage List of Clinical Correlates References Slide Slide Quick Facts Location: Base of the pelvis. Arterial Supply: Superior and inferior vesical arteries. Venous Drainage: Superior (and inferior) vesical veins. Innervation: Via the vesical and hypogastric plexus: Sympathetic: sacral splanchnic nerves; Parasympathetic: pelvic splanchnic nerves; visceral afferent fibers. Lymphatic Drainage: Paravesical lymph nodes. Complete Anatomy The world's most advanced 3D anatomy platformTry it for Free See these models in 3D with Complete Anatomy App Related parts of the anatomy Urinary Bladder (Left Part) Urinary Bladder (Right Part) Trigone of Bladder Urinary Bladder (Posterior) Structure/Morphology The urinary bladder is essentially a hollow, muscular reservoir. The morphology changes as it becomes filled with urine, moving a portion of the bladder into the greater pelvis. Anatomical Relations The urinary bladder is directly posterior to the pubic symphysis and superior to the pelvic diaphragm. In females, the uterus, cervix, and vagina are posterior to the bladder. In males, the rectum is posterior to the bladder. Function The urinary bladder stores urine that has passed from the kidneys through the ureters. Urination occurs when the internal urethral sphincter (distal portion of the bladder) relaxes and the detrusor (smooth muscle) of the bladder contracts, expelling urine into the urethra. Arterial Supply The urinary bladder receives a dual blood supply that supplies the superior and inferior parts of the organ. Superior vesical arteries are branches off the umbilical artery. In males, the inferior vesical arteries branch directly from the internal iliac arteries. In females, the vaginal branch of the uterine artery serves the function of the inferior vesical and supplies the inferior bladder (Tubbs et al., 2016). Venous Drainage The bladder is drained by the vesical venous plexus, where blood is returned to the internal iliac vein via the superior (and in the male, the inferior) vesical veins. Innervation The nerves supplying the bladder arise from the pelvic plexuses. Sympathetic ﬁbers are derived from the sacral splanchnic nerve and synapse with the postganglionic neuronal cell bodies inside the celiac and mesenteric plexuses. From here the hypogastric plexuses descend into the pelvis and connect with the vesical plexus surrounding the bladder. Parasympathetic ﬁbers arise from the neuronal cell bodies in the pelvic splanchnic nerve. Vesical afferent nerves are concerned with pain and awareness of distension of the bladder (Burnstock, 1990). Lymphatic Drainage The paravesical lymph nodes are found in the subperitoneal tissue surrounding the urinary bladder. These nodes drain to the external (mostly) and internal (minimally) iliac lymph nodes, returning lymph to the lateral aortic and caval lymph nodes and the left and right lumbar lymph trunks (Földi et al., 2012). List of Clinical Correlates —Bladder cancer References Burnstock, G. (1990) 'Innervation of bladder and bowel', Ciba Found Symp, 151, pp. 2-18. Földi, M., Földi, E., Strößenreuther, R. and Kubik, S. (2012) Földi's Textbook of Lymphology: for Physicians and Lymphedema Therapists. Elsevier Health Sciences. Tubbs, R. S., Shoja, M. M. and Loukas, M. (2016) Bergman's Comprehensive Encyclopedia of Human Anatomic Variation. Wiley. Learn more about this topic from other Elsevier products Osmosis Ureter, bladder and urethra histology: Video, Causes, & Meaning Ureter, bladder and urethra histology: Symptoms, Causes, Videos & Quizzes \| Learn Fast for Better Retention! Explore on Osmosis Complete Anatomy The world's most advanced 3D anatomy platform Try it for Free Useful links Submit your paper Shop Books & Journals Open access View all products Elsevier Connect About About Elsevier Careers Global Press Office Advertising, reprints & supplements Modern slavery act statement Support Customer support Resource center Global \| English Copyright © 2025 Elsevier, its licensors, and contributors. All rights are reserved, including those for text and data mining, AI training, and similar technologies. Terms & Conditions Privacy policy Accessibility Cookie settings
4810	https://www.khanacademy.org/college-careers-more/bjc/x4c51c62a:2021-challenge/x4c51c62a:2021-finalists-and-popular-vote-top-scorers/v/the-tsiolkovsky-rocket-equation	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Top Voted
4811	https://n.ethz.ch/~ywigderson/math/static/BeyondEuclidNotes.pdf	Mathcamp 2021 Euclidean geometry beyond Euclid Yuval 1 Points and lines The most basic objects in Euclidean geometry are points and lines. The very ﬁrst of Euclid’s ﬁve axioms for plane geometry is about the fundamental relationship between points and lines: any two points deﬁne a (unique) line. In 1893—about 2200 years after Euclid—Sylvester asked the following question, which could easily have appeared in the ﬁrst book of Euclid’s Elements. Question. Is it possible to place ﬁnitely many non-collinear points in the plane so that whenever a line passes through two of them, it also passes through a third? Note that the non-collinearity is important, since n ≥3 points on a line certainly satisfy this property. Additionally, the requirement that we use ﬁnitely many points is necessary, as you’ll show on the homework. We can just try to place points in the plane so that every line through two of them goes through a third. Since we want them to not all be collinear, we can start with just a triangle of three points: Of course, this doesn’t yet satisfy our desired property, since all three lines only contain two points. We can ﬁx this by adding a point on each of these lines, which I’ll now draw as solid to indicate that they now have three points on them. We ﬁxed three lines, but created six new bad lines, which seems like negative progress. But we can actually ﬁx three of them by placing another point in the center of the triangle! Though we still have three bad lines. . . We could of course ﬁx them by placing three more points, but that would create some further bad lines. 1 Mathcamp 2021 Euclidean geometry beyond Euclid Yuval If you keep experimenting in this way, you might come to be convinced that the answer to Sylvester’s question is no, and that any ﬁnite set of non-collinear points in the plane has a line passing through exactly two of them. This is indeed true, and is known as the “Sylvester–Gallai” theorem, even though it was ﬁrst proved by Melchior in 1941. Theorem 1.1 (“The Sylvester–Gallai Theorem”). Any ﬁnite set of non-collinear points in the plane deﬁne a line containing exactly two of them. Proof (due to Kelly). Suppose for contradiction that this is false, and let P be the set of points, and let L be the set of lines they deﬁne. For a line ℓand a point p not on ℓ, we let d(p, ℓ) denote the distance from p to ℓ, i.e. the length of the segment passing through p and orthogonal to ℓ. We begin by picking a pair (p, ℓ) of a point and a line not through p whose distance is minimal. Formally, we consider the set of pairs {(p, ℓ) : p ∈P, ℓ∈L, p is not on ℓ}. By the assumption of non-collinearity, this set is non-empty. Moreover, since both P and L are ﬁnite, this set is ﬁnite. So we can pick p, ℓwith p not on ℓso that d(p, ℓ) is minimized. By assumption, ℓcontains at least three points of P, say a, b, c, and say they appear in this order. Drop a perpendicular from p to ℓ, and say it meets ℓat the point m. By the pigeonhole principle, at least two of the points a, b, c must be on the same side of the perpendicular from p. Say that these are b, c. Then the pair of points p, c ∈P deﬁne another line ℓ′ ∈L. ℓ m a b c p ℓ′ m′ We claim that d(b, ℓ′) < d(p, ℓ), which contradicts the fact that we chose (p, ℓ) to have minimal distance. So to ﬁnish the proof, it suﬃces to prove this claim, which is hopefully pretty intuitive from the picture above. To prove it rigorously, note that pmc and bm′c are both right triangles. Moreover, these two right triangles are similar, since they have the same angle at c. Finally, note that the length of the segment bc is at most the length of mc, and the length of mc is less than the length of pc since mc is a leg of the right triangle pmc with hypotenuse pc. So the length of bc is less than the length of pc, which shows that the triangle bm′c is smaller than the triangle pmc, where by smaller I mean that the constant of similarity is less than 1. But this implies that the length of bm′ is strictly less than the length of pm, as claimed. 2 Mathcamp 2021 Euclidean geometry beyond Euclid Yuval One important consequence of the Sylvester–Gallai theorem is the following theorem, due to de Bruijn and Erd˝ os from 1948. Theorem 1.2. Any n ≥3 non-collinear points in the plane deﬁne at least n lines. Proof. We proceed by induction on n. The base case is n = 3; in this case, three non-collinear points deﬁne a triangle, and thus three lines. For the inductive case, suppose we have a set P of n points in the plane. By the Sylvester– Gallai theorem, there are two points, say a, b ∈P, whose line contains no other point of P. We delete a from the conﬁguration to obtain a new set of n −1 points, which we call P ′. If P ′ is non-collinear, then by the inductive hypothesis it deﬁnes at least n −1 lines. Moreover, the line through a, b is not among these, since that line passes through b and no other point of P ′. Therefore, P deﬁnes at least one more line than P ′ does, and therefore P deﬁnes at least (n −1) + 1 = n lines. On the other hand, if P ′ is collinear, then there is some line ℓcontaining all the points of P ′. The point a must not lie on ℓ, by our assumption that P was non-collinear. For every point p in P ′, there is a distinct line passing through a and p, which yields \|P ′\| = n−1 lines. Together with the line ℓ, we get n lines in total. ℓ One extremely interesting topic, which you can explore on the homework, is the fact that there are other notions of geometry where statements like the Sylvester–Gallai and de Bruijn–Erd˝ os theorem make sense. In some of these more abstract “geometries”, the Sylvester–Gallai theorem is simply false. However, the de Bruijn–Erd˝ os theorem is always true; it captures a really fundamental property of geometry. In fact, one can prove the de Bruijn–Erd˝ os theorem simply from the axiom that every two points deﬁne a unique line, but the Sylvester–Gallai theorem cannot be proven in this way. Here is an alternative proof of the de Bruijn–Erd˝ os theorem, which does not use Sylvester– Gallai, or any other special properties of Euclidean geometry. This proof uses some linear algebra. On the homework, you can go through the de Bruijn and Erd˝ os’s original proof, which does not use linear algebra. Alternative proof of Theorem 1.2. Let p1, . . . , pn be a set of n points in the plane, and let the lines they deﬁne be ℓ1, . . . , ℓm. We wish to prove that m ≥n. For every 1 ≤i ≤n, we deﬁne a vector v(i) ∈Rm as follows. For every 1 ≤k ≤m, the kth coordinate of the vector v(i) is 1 if pi lies on the line ℓk, and is 0 otherwise. For example, consider the following conﬁguration of n = 4 and m = 6 lines. 3 Mathcamp 2021 Euclidean geometry beyond Euclid Yuval p1 p2 p3 p4 ℓ1 ℓ2 ℓ3 ℓ4 ℓ5 ℓ6 Then the four vectors v(1), v(2), v(3), v(4) are v(1) = 1 1 0 1 0 0 v(2) = 1 0 1 0 1 0 v(3) = 0 1 1 0 0 1 v(4) = 0 0 0 1 1 1 For instance, the ﬁrst coordinate in v(1) is 1 since p1 lies on ℓ1, but the ﬁrst coordinate of v(3) is 0 since p3 does not lie on ℓ1. To prove that m ≥n, we will show that the vectors v(1), . . . , v(n) are linearly independent. Since they are vectors in Rm, this immediately implies that n ≤m, since any collection of linearly independent vectors in Rm can have at most m elements. If you’ve never seen linear algebra, the rest of the proof (as well as the previous two sentences) will likely not make a lot of sense, which is OK! We ﬁrst claim that every vector v(i) has at least two 1s in it. Indeed, the number of 1s in v(i) is simply the number of lines containing the point pi. There is at least one 1 in v(i) since pi lies on at least one line. Moreover, if there is only one 1 in v(i), then pi lies on a unique line. But this means that there is a single line containing pi and pj for every j ̸= i, meaning that all the points are collinear, a contradiction. So each v(i) has at least two 1s. This, in turn, implies that v(i) · v(i) ≥2 for every i. Indeed, the dot product v(i) · v(i) is simply the sum of the squares of all the coordinates of v(i); since the coordinates of v(i) are just 0 and 1, and there are at least two 1s, this dot product is at least 2. On the other hand, for every i ̸= j, we claim that v(i) · v(j) = 1. Indeed, when computing v(i) · v(j), we add up the product of the kth coordinate of v(i) and the kth coordinate of v(j), for all 1 ≤k ≤m. The only way the product of the kth coordinates will be non-zero is if the kth coordinate of both v(i) and v(j) is 1. Therefore, v(i) · v(j) computes the number of lines ℓk containing both pi and pj. Since every pair of points lies on a unique line, this is exactly 1, so v(i) · v(j) = 1 as claimed. To prove that v(1), . . . , v(n) are linearly independent, suppose that there are real numbers c1, . . . , cn such that c1v(1) + · · · + cnv(n) = ⃗ 0, where ⃗ 0 is the zero vector in Rm, and assume for contradiction that not all the ci equal 0. Taking the dot product of this equation with v(1), we ﬁnd that 0 = c1(v(1) · v(1)) + c2(v(2) · v(1)) + · · · + cn(v(n) · v(1)) = c1(v(1) · v(1) −1) + (c1 + · · · + cn), 4 Mathcamp 2021 Euclidean geometry beyond Euclid Yuval using the fact that v(1) · v(j) = 1 for all j ≥2. Similarly, by taking the dot product with v(i), we ﬁnd that 0 = ci(v(i) · v(i) −1) + (c1 + · · · + cn) for all i. Recall that v(i) · v(i) ≥2, so v(i) · v(i) −1 is positive. If c1 + · · · + cn = 0, then this shows that ci = 0 for all i, which is a contradiction. If c1 + · · · + cn > 0, then we ﬁnd that ci < 0 for all i, which contradicts the fact that their sum is positive. Similarly, if c1 + · · · + cn < 0, then we ﬁnd that ci > 0 for all i, again a contradiction. 2 Higher dimensions While the most well-known geometry in Euclid’s Elements is planar, Euclid nonetheless devotes a few books to three-dimensional geometry. Are there analogues of the Sylvester– Gallai theorem and the de Bruijn–Erd˝ os theorem in higher dimensions? There are actually several ways of interpreting this question (depending on what you mean by “analogues”). For the Sylvester–Gallai theorem, the most natural high-dimensional analogue is the following statement, which is indeed true. Theorem 2.1 (High-dimensional Sylvester–Gallai). For every d ≥2 and n ≥3, there is no set of n non-collinear points in Rd such that every line passing through two of them also passes through a third. You’ll be asked to prove this on the homework. Once we have Theorem 2.1, we can prove the following high-dimensional de Bruijn–Erd˝ os theorem, using the exact same inductive argument as in our proof of Theorem 1.2. Theorem 2.2 (High-dimensional de Bruijn–Erd˝ os). For every d ≥2 and n ≥3, any set of n non-collinear points in Rd deﬁnes at least n lines. Although this can be proved from Theorem 2.1 by induction on n, there’s actually a totally diﬀerent proof that deduces the case for general d from the case d = 2. Proof of Theorem 2.2. Consider a set P of n non-collinear points in Rd, and let X denote the two-dimensional xy plane in Rd. Let π : Rd →R2 denote the projection onto X. Formally, recall that every point of Rd is a vector with d coordinates; then π is just the function that takes in a vector and throws away every coordinate except for the ﬁrst two. Let π(P) be the subset of R2 gotten by applying π to every point in P. 5 Mathcamp 2021 Euclidean geometry beyond Euclid Yuval X x y z p1 π(p1) p2 π(p2) Let’s imagine for a moment that we got extremely lucky. Speciﬁcally, let’s assume that no pair of points in P are mapped to the same point under π. Additionally, let’s assume that every triple of non-collinear points in P is also mapped to a triple of non-collinear points by π. If we really got this lucky, then π(P) is a set of n non-collinear points in R2. Additionally, there is a bijection between the lines deﬁned by P and the lines deﬁned by π(P). By the two-dimensional de Bruijn–Erd˝ os theorem, Theorem 1.2, π(P) deﬁnes at least n lines, which implies that P deﬁnes at least n lines as well. However, we obviously might not get so lucky. The trick to get around this is to pick X to be a random two-dimensional plane in Rd, rather than the speciﬁc xy plane. If we do this, then we will get lucky (in the sense above) with 100% probability. Proving this probabilistic statement is actually not so easy, and requires some background in measure theory, but I hope it’s intuitively reasonable. Indeed, there are only ﬁnitely many “problems” that can arise: there are n 2 pairs of points that might collide under π, and at most n 3 non-collinear triples that can be made collinear by π. But we have (uncountably) inﬁnitely many choices for the random plane X, so there’s a 0% probability that we’ll run into one of the ﬁnitely many problems. By making this argument rigorous, we can in particular ﬁnd some plane X where we get lucky (indeed, there will be inﬁnitely many such planes). If we project onto this plane X, then the argument above works: π(P) consists of n non-collinear points in R2, which deﬁne at least n lines by Theorem 1.2, which then implies that P deﬁnes at least n lines since there is a bijection between the lines made by P and those made by π(P). This proof shows that the high-dimensional statement Theorem 2.2 is true, but it actually tells us something more: this is a fundamentally two-dimensional statement. Even though we are dealing with points and lines in Rd, all the real mathematical structure comes from what happens when d = 2. Can we come up with some “genuinely d-dimensional” version of the de Bruijn–Erd˝ os the-orem? One natural thing to try is to use the new structure that exists in higher dimensions, namely planes, hyperplanes, and so on. 6 Mathcamp 2021 Euclidean geometry beyond Euclid Yuval Recall that in Rd, we have hyperplanes of every dimension 0 ≤i ≤d−1. Zero-dimensonal hyperplanes are just points, one-dimensional hyperplanes are just lines, two-dimensional hy-perplanes are our usual notion of planes. We don’t have special words for higher-dimensional hyperplanes, simply because our brains can’t really visualize what an 11-dimensional hy-perplane of R24 looks like. But these are perfectly natural and well-deﬁned mathematical notions. The main fact that we will need to know about hyperplanes is that for every 1 ≤i ≤d−1, every set of i + 1 points in Rd deﬁnes a unique i-dimensional hyperplane, unless this set of i+1 points lies in some (i−1)-dimensional hyperplane. This generalizes the well-known fact (i = 2) that any three points deﬁne a unique plane, unless they are collinear (i.e. contained in some 1-dimensional hyperplane). This also generalizes the fact that any two distinct points deﬁne a unique line, which is just the case i = 1. It is natural to expect that there is a higher-dimensional de Bruijn–Erd˝ os theorem, which tells us something about the number of i-dimensional hyperplanes deﬁned by a set of n points in Rd. Let’s try to come up with what such a statement might be. First, we must assume that our set of n points is “genuinely d-dimensional”. For in-stance, if we take n points in R100 that all lie on a two-dimensional plane, then they will not deﬁne any three-dimensional hyperplanes, nor any four-dimensional hyperplanes, nor any 99-dimensional hyperplanes. Of course, we already encountered essentially this same issue: in the de Bruijn–Erd˝ os theorem, we assumed that our n points were non-collinear, i.e. “genuinely 2-dimensional”. Deﬁnition 2.3. A set P of n points in Rd is called genuinely d-dimensional if P is not contained in any (d −1)-dimensional hyperplane. So from now on, let’s assume that we have a set P of n points in Rd, which is genuinely d-dimensional. This assumption automatically implies that P deﬁnes at least one line, at least one plane, at least one three-dimensional hyperplane, and so on. Again, this is the natural higher-dimensional analogue of our non-collinearity assumption in Theorem 1.2. Let Fi(P) denote the number of i-dimensional hyperplanes1 deﬁned by P. To get intuition for what type of results to expect, let’s do some examples. 1. Let P consist of the four vertices of a tetrahedron in R3. They deﬁne 6 lines and four planes (the edges and faces of the tetrahedron, respectively), so F0(P) = 4 F1(P) = 6 F2(P) = 4. 2. Let P consist of the eight vertices of a cube in R3, so F0(P) = 8. We again have that every pair of points deﬁnes a unique line, so F1(P) = 8 2 = 28. Counting the planes is a bit trickier: there are six faces of the cube, six additional planes containing four points (namely two opposite edges of the cube), and ﬁnally eight more planes containing only three points (there are two such planes orthogonal to any “long diagonal” of the cube). So in total, F0(P) = 8 F1(P) = 28 F2(P) = 20. 1The notation Fi is standard, since some people prefer the word “ﬂats” to “hyperplanes”. 7 Mathcamp 2021 Euclidean geometry beyond Euclid Yuval 3. Suppose we add the center of the cube to the previous set of points. Then we actually generate no new lines, since every line containing the center of the cube and one of its vertices passes through another vertex, and so was already counted. Similarly, one can check that we generate no new planes in this way. So for this example, F0(P) = 9 F1(P) = 28 F2(P) = 20. 4. The simplest genuinely d-dimensional set of points we can place in Rd is the vertices of the simplex, which is the d-dimensional analogue of a triangle, tetrahedron, etc. It consists of d + 1 points such that no triple is collinear, no quadruple is coplanar, and in general, no set of i + 1 points lies on an (i −1)-dimensional hyperplane. Concretely, we can take our points to be (0, 0, 0, . . . , 0), (1, 0, 0, . . . , 0), (0, 1, 0, . . . , 0), (0, 0, 1, . . . , 0), (0, 0, 0, . . . , 1). Then F0(P) = d + 1, since we have d + 1 points. Every pair of points deﬁnes a unique line, so F1(P) = d+1 2 = d2+d 2 . Similarly, every triple of points deﬁnes a unique plane, so F2(P) = d+1 3 . More generally, Fi(P) = d + 1 i + 1 for every 0 ≤i ≤d −1. Thus, the sequence Fi is simply the (d + 1)th row of Pascal’s triangle, without the 1 at the beginning and the end. So far, all the examples we’ve seen do have some common behavior. As guaranteed by the de Bruijn–Erd˝ os theorem, for instance, we always have F0(P) ≤F1(P). Additionally, in all the three-dimensional examples, we had F1(P) ≥F2(P). Additionally, we know that the binomial coeﬃcients d+1 i+1 are increasing until i + 1 = d+1 2 , and then are decreasing starting from i + 1 = d+1 2 . This is consistent with our three-dimensional examples, which are increasing from F0 to F1, and then decreasing from F1 to F2. It is natural to conjecture that this pattern continues, namely that for odd d, F0(P) ≤F1(P) ≤· · · ≤F d−1 2 (P) ≥F d+1 2 (P) ≥· · · ≥Fd−2(P) ≥Fd−1(P), ( ) and a similar thing for even d. However, the next example shows that this conjecture is too ambitious. 5. Let P be a set of n points in general position in Rd. General position means that no triple of points in P is collinear, no quadruple in P is coplanar, and so on: no set of i + 1 points in P lie on an (i −1)-dimensional hyperplane. It turns out that for every n ≥d + 1, there is a genuinely d-dimensional set in general position. We explicitly constructed such a set—the simplex—for n = d + 1, but it turns out that it’s not so easy to explicitly construct such sets for arbitrary n ≥d + 1. Nonetheless, we can again do it randomly! For example, we can let P consist of n randomly chosen points 8 Mathcamp 2021 Euclidean geometry beyond Euclid Yuval in the unit ball in Rd. Just as in our proof of Theorem 2.2, there are only ﬁnitely many “bad events” (e.g. there are only n 3 triples that might be collinear, n 4 quadruples that might be coplanar, etc.). Because of this, with 100% probability, we will get lucky and obtain a set in general position in this way. If we have a set P in general position, then Fi(P) = n i+1 for every 0 ≤i ≤d −1. Indeed, every set of i+1 points will deﬁne an i-dimensional hyperplane, and the general position assumption implies that this hyperplane will not contain any other point in P. Because of this, we will get n i+1 distinct hyperplanes. In particular, if n ≥2d, then the sequence n i+1 will simply be increasing for all i, i.e. we’ll have F0(P) < F1(P) < F2(P) < · · · < Fd−1(P). This demonstrates that the earlier conjecture can’t be true in general. Similarly, if d+1 < n < 2d, then the sequence will eventually start decreasing, but it won’t happen at the midpoint. Nonetheless, there is a reasonable conjecture we can salvage out of this data. Namely, it seems that our (false) conjecture ( ) can only fail “upwards”: if ( ) is false, then it’s because the sequence Fi(P) doesn’t start decreasing “when it’s supposed to”, at the halfway mark. This behavior is sometimes called top-heavy behavior, since there’s “more stuﬀat the top than at the bottom” of the sequence. Many people over the years observed that the naive conjecture is false, but that it seems to only be false in one way, which caused them to formulate a number of more reﬁned con-jectures. Maybe the most natural is the following, known as Rota’s unimodality conjecture. Conjecture 2.4 (Rota, 1971). Let P be a genuinely d-dimensional set of n points in Rd. Then there exists some 0 ≤m ≤d −1 such that F0(P) ≤F1(P) ≤· · · ≤Fm−1(P) ≤Fm(P) ≥Fm+1(P) ≥· · · ≥Fd−1(P). In other words, Rota’s conjecture says that the sequence Fi(P) goes up for a while, and then starts coming down. Note that if m = d −1, then it will actually never start coming down, which is the behavior we saw in our last example. Despite 50 years of eﬀort, and despite it being such a simple and basic problem, Rota’s conjecture remains open. Nonetheless, the past decade has seen a ﬂurry of activity on a number of related problems; essentially, results of the same form have been proved in a number of closely related contexts, though we still don’t know how to prove such a result for point sets in Rd. Note that Rota’s conjecture doesn’t say anything about the “upward failure” of ( ) that I alluded to above. That behavior was observed by Dowling and Wilson, who formulated the following top-heavy conjecture. 9 Mathcamp 2021 Euclidean geometry beyond Euclid Yuval Conjecture 2.5 (Dowling–Wilson, 1974). Let P be a genuinely d-dimensional set of n points in Rd. Then for every i < j with i + j ≤d −1, we have that Fi(P) ≤Fj(P). In particular, we have that F0(P) ≤F1(P) ≤· · · ≤F⌊d−1 2 ⌋(P) and Fi(P) ≤Fd−1−i(P) for every 0 ≤i ≤ d−1 2 . Note that for d = 2, the de Bruijn–Erd˝ os theorem proves both Rota’s conjecture and the Dowling–Wilson conjecture. Thus, both conjectures are natural generalizations of the de Bruijn–Erd˝ os theorem to higher dimensions. Unlike Rota’s conjecture, which is still open, the Dowling–Wilson conjecture was proved very recently, by Huh and Wang. Even more recently, Braden, Huh, Matherne, Proudfoot, and Wang gave a new proof of a more general theorem, which I won’t state; but roughly speaking, it proves the Dowling–Wilson conjecture in the most general possible setting (of so-called matroids), which form a vast generalization of point sets in Rd. Theorem 2.6 (Huh–Wang 2017, Braden–Huh–Matherne–Proudfoot–Wang 2020). Conjec-ture 2.5 is true (as are many generalizations of it). One astonishing thing about these proofs is their complexity. The conjecture is about the most basic objects in Euclidean geometry: points, lines, planes, etc. Moreover, the d = 2 case of this conjecture, namely the de Bruijn–Erd˝ os theorem, is pretty straightforward to prove. However, the original proof of Huh and Wang used some extraordinarily complicated mathematics, namely the theory of intersection cohomology for ℓ-adic perverse sheaves. These are algebraic structures that don’t obviously have much to do with point sets in Rd, so it’s already remarkable that Huh and Wang were able to ﬁnd a connection between these disparate areas of math. The more recent and more general proof, of Braden, Huh, Matherne, Proudfoot, and Wang, is in a certain sense simpler. Namely, they don’t need to use any of this intersection cohomology as a black box, and their paper is basically self-contained. However, it’s almost 100 pages long, and for good reason: they construct a “combinatorial Hodge theory”, which is some analogue to this intersection cohomology, and which is well-suited to deal with their problem directly. However, to use it, they need to prove that it satisﬁes certain complicated algebraic relations (the so-called K¨ ahler package), and these proofs are in some sense inspired by the corresponding proofs for ℓ-adic perverse sheaves. I don’t really understand either of the proofs of the Dowling–Wilson conjecture, and even if I did, it’d be impossible to cover a complicated 100-page paper in a three-day course. But at an immensely high level, one can view their proof(s) as a generalization of the linear-algebraic proof above of the de Bruijn–Erd˝ os theorem. Recall that there, we wished to prove that n ≤m, where n is the number of poitns in the plane and m is the number of lines they deﬁne. To do so, we deﬁned vectors v(1), . . . , v(n) ∈Rm, and showed that these vectors were linearly independent, which implied that n ≤m. The vectors were deﬁned in a very natural 10 Mathcamp 2021 Euclidean geometry beyond Euclid Yuval way, namely the kth coordinate of v(j) was 1 if the jth point was on the kth line, and 0 otherwise. Let’s say we wish to prove that Fi(P) ≤Fd−1−i(P) for some genuinely d-dimensional point set P in Rd. Let N = Fi(P) and M = Fd−1−i(P). Braden–Huh–Matherne–Proudfoot– Wang do eﬀectively deﬁne vectors v(1), . . . , v(N) ∈RM, and then prove that these vectors are linearly independent, implying that N ≤M as claimed. Moreover, the kth coordinate of the vector v(j) is non-zero if the jth i-dimensional hyperplane is a subset of the kth (d −1 −i)-dimensional hyperplane, and zero otherwise. However, it turns out that if we let the vectors v(j) have only 0 and 1 as their coordinates, then they will not be linearly independent in general. Instead, in order to prove the Dowling–Wilson conjecture, it is necessary to put some other non-zero numbers in the vectors v(1), . . . , v(N). And it really seems that the only way to come up with which numbers to put in the vectors is to build up some complicated algebraic machinery, as these authors do. I mentioned earlier that many results related to Rota’s conjecture have been proved, although Rota’s conjecture itself remains open. Most of these related results have come in the past decade, by several of the same authors I mentioned above, and also using techniques related to this sort of combinatorial Hodge theory. I personally wouldn’t be too surprised if Rota’s conjecture were proved in the next few years, using many of these same techniques. 3 Almost orthogonal lines Let us say that a collection of lines passing through a common point is pairwise orthogonal if every pair of them is orthogonal. It’s easy to show that three lines in the plane cannot all be pairwise orthogonal; moreover, this is basically an immediate consequence of the fourth of Euclid’s ﬁve axioms for plane geometry. In three dimensions, of course, we can have three pairwise orthogonal lines, but four lines cannot be pairwise orthogonal. The generalization of this, unsurprisingly, is the following. Theorem 3.1. The maximum number of pairwise orthogonal lines in Rn is n. Proof. One can prove this using linear algebra, but let’s prove it by induction on n. The base case n = 1 is simple: R1 consists of a single line, so the maximum number of distinct lines in R1 (let alone pairwise orthogonal) is 1. For the inductive step, suppose the statement is true for n −1. Say we are given any collection L of pairwise orthogonal lines in Rn. Fix any line ℓin the collection, and let H be the (n −1)-dimensional hyperplane orthogonal to ℓ. Let L′ = L \ {ℓ} be the collection of lines gotten by deleting ℓ. Since every line in L′ is orthogonal to ℓ, we see that L′ must lie in the hyperplane H. But since H is identical to Rn−1, we have that \|L′\| ≤n −1 by the inductive hypothesis. Therefore, \|L\| = \|L′\| + 1 ≤(n −1) + 1 = n. Although Euclid didn’t know about dimensions higher than 3, this proof was basically known to him, and is essentially given in Propositions XI.4 and XI.5 in the Elements. So far, we haven’t done anything too interesting. But we can get a new notion by slightly weakening our notion of pairwise orthogonality. 11 Mathcamp 2021 Euclidean geometry beyond Euclid Yuval Deﬁnition 3.2. Let us say that a collection of lines through a common point is almost orthogonal if for every pair, the angle between them is between 89◦and 91◦. Let f(n) denote the maximum number of almost orthogonal lines in Rn. Note that since a pairwise orthogonal collection is, in particular, almost orthogonal, we deﬁnitely have f(n) ≥n. In other words, a collection of n pairwise orthogonal lines in Rn is certainly almost orthogonal. Can f(n) be larger than n? We can try some small examples for intuition; for instance, I claim that f(2) = 2. Indeed, consider any line ℓ1 in R2. Any line which is almost orthogonal to ℓ1 must lie in the tiny shaded region in the following picture. ℓ1 We can put any line ℓ2 we want inside the shaded region, but once we pick such a line, we can’t ﬁnd another line ℓ3 that is nearly orthogonal to both ℓ1 and ℓ2. Similarly, one can prove that f(3) = 3, in much the same way. However, it’s pretty easy to believe that as the number of dimensions increases, the amount of wiggle room we have increases. My intuition suggests that once n is large enough, we can actually use this wiggle room and squeeze in one more line, so that f(n1) = n1 +1 for some suﬃciently large n1. It is then reasonable to imagine that for many more dimensions, n + 1 is the best we can do, but eventually we accumulate enough wiggle room so that f(n2) = n2 + 2 for some much larger n2. I would expect the pattern to continue in this fashion, so that as n tends to inﬁnity, we have that f(n) equals n plus some very slowly growing function of n. In particular, I’d expect that f(n) < 2n for all n. As it turns out, this intuition is wildly wrong, and high dimensions behave totally diﬀer-ently. For very large n, f(n) will be waaaaaay larger than n. 12 Mathcamp 2021 Euclidean geometry beyond Euclid Yuval Theorem 3.3. f(n) ≥⌊1.0007n⌋. In particular, in 100, 000-dimensional space, there are more than 1030 almost orthogonal lines. In million-dimensional space, there are more than a googol cubed almost orthogonal lines. To prove Theorem 3.3, we need to set up some background ﬁrst. Recall that for two vectors v, w ∈Rn, the dot product v ·w is deﬁned v ·w = Pn i=1 viwi, where v1, . . . , vn are the n coordinates of v, and similarly for w. The length of a vector v is deﬁned by ∥v∥= √v · v. The main fact we will need is essentially the same as the law of cosines. It says that for two vectors v, w, the angle θ between them satisﬁes cos θ = v · w ∥v∥∥w∥. One can check (e.g. on Wolfram Alpha) that an angle θ is between 89◦and 91◦if and only if \|cos θ\| ≤β, for some real number β ≈0.017452. So in order to prove Theorem 3.3, we will actually ﬁnd a set of M ≥⌊1.0007n⌋vectors v(1), . . . , v(M) in Rn such that \|v(i) · v(j)\| ≤β∥v(i)∥∥v(j)∥ for all distinct 1 ≤i, j ≤M. (1) It is actually not so easy to explicitly construct such a set of vectors. To get around this, we will use an immensely powerful technique called the probabilistic method. Basically, it turns out that if we pick these vectors at random, then property (1) holds with positive probability. In particular, since the probability that (1) holds is positive when the vectors are chosen randomly, there must exist some set of vectors for which it holds! Before we do the proof, we’ll collect a few probabilistic tools that we will need. 3.1 Probabilistic tools Let’s say we make some random choices, and let A be an event, i.e. a thing that can happen after the random choices. For instance, if our randomness is rolling a die, then the event A might be that die comes up 1, or that it comes up a prime number, or that it comes out even. If the randomness is that we deal a random card to everyone in this class, then the event A could be that we all receive red cards, or that I get the ace of spades, or that exactly two of us got queens. We denote by Pr(A) the probability of A, which is a number between 0 and 1. The ﬁrst basic result we’ll need about probability is the following, known as the union bound. Proposition 3.4 (Union bound). For any two events A, B, Pr(A happens or B happens (or both)) ≤Pr(A) + Pr(B). More generally, for any events A1, . . . , At, Pr(at least one of A1, . . . , At happens) ≤ t X i=1 Pr(Ai). 13 Mathcamp 2021 Euclidean geometry beyond Euclid Yuval Rigorously proving the union bound requires setting up the actual formal mathematical theory of probability, which we won’t do. But hopefully it is intuitively clear. For instance, if I pick a random card from a deck, the probability that it is either red or a king is at most the probability that it is red, plus the probability that it is a king. We say that two events A and B are independent if Pr(A and B) = Pr(A) Pr(B). Intu-itively, what independence “means” is that whether or not A happens has no inﬂuence on the probability that B happens. For example, if I ﬂip a fair coin twice, then the event that the ﬁrst ﬂip is heads is independent of the event that the second ﬂip is tails. If we have events A1, . . . , At, then we say that they are mutually independent if for every set I ⊆{1, . . . , t}, Pr(Ai happens for all i ∈I) = Y i∈I Pr(Ai). The technical condition is perhaps a bit tricky to wrap your head around, but again the intuition is that the events are mutually independent if they all have no inﬂuence on each other. A random variable is just any number whose value is determined by the randomness. For instance, if we roll a die, then the outcome of the die roll is a random variable. If we denote this random variable by X, then we have Pr(X = 1) = Pr(X = 2) = Pr(X = 3) = Pr(X = 4) = Pr(X = 5) = Pr(X = 6) = 1 6. This information is called the distribution of X: formally, the distribution is the list of probabilities Pr(X = x) for every value x that X can take. We say that two random variables X, Y are identically distributed if they have the same distribution: the probability that X takes on the value 5 equals the probability that Y takes on the value 5, and this holds for every possible choice of 5. Additionally, we say that random variables X, Y are independent if the event X = x is independent of the event Y = y for all x, y. We can analogously deﬁne what it means for a number of random variables X1, . . . , Xt to be mutually independent. Perhaps the most important result in all of probability theory is the central limit theorem, which was ﬁrst observed by Gauss and Laplace. The central limit theorem explains the shocking prevalence of the bell curve in a huge number of natural phenomena. The bell curve (or Gaussian distribution) is the curve in the plane given by the equation y = e−x2. x y I won’t state it rigorously or prove it, but the central limit theorem roughly says that if X1, X2, . . . are independent random variables that are identically distributed, then X1+· · ·+ Xt converges to a Gaussian distribution as t →∞. In real life, you could imagine that we are doing some scientiﬁc experiment, trying to ﬁnd the value of some experimentally determined 14 Mathcamp 2021 Euclidean geometry beyond Euclid Yuval number (e.g. the speed of light, the mass of an electron, etc.). Since our equipment is not perfect, we imagine that every experiment returns a random outcome, which is probably close to the truth, but with some random noise. We can also assume that in diﬀerent experiments, the random noise is independent. This implies that when we add up (or average) the results of the experiments, we are adding up independent, identically distributed random variables, which should then yield a bell curve by the central limit theorem. This is the reason why the bell curve shows up in many scientiﬁc experiments, as well as in many other contexts in real life. Instead of the central limit theorem, we will need the following result which can be seen as an “approximate” version of the central limit theorem. Results of this type are variously known as the Chernoﬀbound, Hoeﬀding’s inequality, and/or Azuma’s inequality. Rather than stating it for arbitrary random variables, we’ll only state it for variables that are +1 or −1 with probability 1/2. Proposition 3.5 (Chernoﬀbound). Let Z1, . . . , Zn be mutually independent random vari-ables, with Pr(Zi = 1) = Pr(Zi = −1) = 1 2 for all 1 ≤i ≤n. For any a > 0, we have Pr(Z1 + · · · + Zn > a) ≤e−a2/(2n) and Pr(Z1 + · · · + Zn < −a) ≤e−a2/(2n). Therefore, Pr(\|Z1 + · · · + Zn\| > a) ≤2e−a2/(2n). The reason that I say that this is an approximate form of the central limit theorem is that we see that quadratic behavior in the exponent, matching what is given by the bell curve y = e−x2. You’ll prove the Chernoﬀbound in the homework. 3.2 Back to almost orthogonal lines As we said earlier, in order to prove the existence of a set of vectors satisfying (1), we will pick the vectors v(i) randomly. Speciﬁcally, each coordinate of each vector will be +1 or −1 with probability 1 2, and all these choices will be made independently at random. The following two lemmas will allow us to prove (1). Lemma 3.6. Let X1, . . . , Xn be mutually independent, identically distributed random vari-ables with Pr(Xi = 1) = Pr(Xi = −1) = 1 2 for all i. Let v = (X1, . . . , Xn) be the random vector they deﬁne. Then ∥v∥= √n. Proof. By deﬁnition, ∥v∥2 = v · v = n X i=1 X2 i = n X i=1 1 = n. 15 Mathcamp 2021 Euclidean geometry beyond Euclid Yuval Lemma 3.7. Let X1, . . . , Xn, Y1, . . . , Yn be mutually independent, identically distributed ran-dom variables, each equal to +1 or −1 with probability 1 2. Let v = (X1, . . . , Xn) and w = (Y1, . . . , Yn). Then Pr(\|v · w\| > βn) ≤2e−β2n/2. Proof. Let Zi = XiYi for all 1 ≤i ≤n. Then by the deﬁnition of the dot product, v · w = Z1 + · · · + Zn. We ﬁrst observe that Z1, . . . , Zn are mutually independent, since all the Xi, Yi were mutually independent, and therefore the randomness in Zi has no eﬀect on the randomness in Zj for any j ̸= i. Additionally, we claim that Pr(Zi = 1) = Pr(Zi = −1) = 1 2. Indeed, Zi will be 1 if Xi and Yi are both either equal to 1 (which happens with probability 1 4) or both equal to −1 (which also happens with probability 1 4), so Pr(Zi = 1) = 1 2. By essentially the same argument, we see that Pr(Zi = −1) = 1 2. Therefore, we are in a position to apply Proposition 3.5, which tells us that for any a > 0, Pr(\|v · w\| > a) = Pr(\|Z1 + · · · + Zn\| > a) ≤2e−a2/(2n). Plugging in a = βn, we get the desired result. We are now ready to prove Theorem 3.3. Proof of Theorem 3.3. Let M = ⌊1.0007n⌋, and let v(1), . . . , v(M) be random vectors in Rn, each of which has all its coordinates equal to ±1 with probability 1 2, with all these choices made independently. By Lemma 3.6, we see that ∥v(i)∥= n for all 1 ≤i ≤M. For all 1 ≤i < j ≤M, let Aij be the event that \|v(i) · v(j)\| > βn. By Lemma 3.7, we know that Pr(Aij) ≤2e−β2n/2 for all i, j. Therefore, by the union bound, Pr(Aij happens for at least one pair i, j) ≤ X 1≤i<j≤M Pr(Aij) ≤ M 2 ·2e−β2n/2 < M 2e−β2n/2, where the ﬁnal inequality uses the fact that M 2 = M2−M 2 < M2 2 . To conclude, we note that by our choice of M, we have that M ≤1.0007n, and thus M 2 ≤1.0015n. Moreover, if we recall that β = cos(89◦) ≈0.017452, then we see that e−β2/2 < 0.99. Therefore, Pr(Aij happens for at least one pair i, j) < (1.0015 · 0.99)n < 1, since 1.0015 · 0.99 ≈0.9915 < 1. Because of this, we ﬁnd that with positive probability, none of the events Aij happen. Since this probability is positive, there must exist some vectors v(1), . . . , v(M) in Rn such that 16 Mathcamp 2021 Euclidean geometry beyond Euclid Yuval \|v(i) · v(j)\| ≤βn and ∥v(i)∥= n for all distinct i, j. Consider the set of M lines going through these vectors. By the law of cosines, the angle θij between any two of them satisﬁes cos θij = v(i) · v(j) ∥v(i)∥∥v(j)∥≤βn n = β. Thus, these M lines all have angles between 89◦and 91◦, and thus they are almost orthogonal, as claimed. The ﬁnal thing to remark is that the result about almost orthogonal lines in Rn is one instantiation of a very general phenomenon, often called the “curse of dimensionality”. Ba-sically, a bunch of weird things happen once one moves to very high dimensions, even though the basic rules of geometry are the same as in the dimensions we’re used to. Here are just a few other examples. • Volumes and distances get all weird in high dimensions. For instance, if you take the unit ball in Rn and inscribe it in a hypercube, then the volume of the ball is exponentially small in the volume of the cube. In other words, the vast majority of points with coordinates in [−1, 1] are really far from the origin. • It turns out that almost all the points on the unit sphere in Rn are extremely close to the equator: if you pick a random point on the unit sphere, then the probability that its latitude is greater than 1◦or less than −1◦is exponentially small. But this is extremely weird, since the sphere is symmetric, so this holds for all equators! So the sphere is somehow very squashed like a pancake, since almost all of it is very close to the equator. But it’s squashed in every direction, so it’s kind of like a spiky ball with spikes going out in every direction. It’s really hard to visualize! 17
4812	https://www.timeofcare.com/dopamine-effects-at-different-doses/	Dopamine effects at different doses \| Time of Care Home Pharmacy Mnemonics Resources OUTpatient Adults The Most Common Outpatient Conditions All Outpatient Adults Conditions OUTpatient Peds INpatient Adults The 25 Most Common Inpatient Conditions Less Common Inpatient Conditions All Inpatient Conditions Hospital Quick Reference INpatient Peds Differential Diagnosis Procedures Select Page OUTpatient Adults The Most Common Outpatient Conditions All Outpatient Adults Conditions OUTpatient Peds INpatient Adults The 25 Most Common Inpatient Conditions Less Common Inpatient Conditions All Inpatient Conditions Hospital Quick Reference INpatient Peds Differential Diagnosis Procedures Home Pharmacy Mnemonics Resources Dopamine effects at different doses Pharmacy and Therapeutics The effects of dopamine are based on the dosing rate EffectsDoseReceptorsComments Dopaminergic effects 0.5-2 mcg/kg/min Dopamine At low doses, like 1 to 2 mcg/kg/min,DA acts mainly on dopamine-1 receptors in the renal, mesenteric, cerebral, and coronary beds, resulting in selective vasodilation. Beta1 effects 2-10 mcg/kg/min Beta1, Dopamine-For does between2 and 5 mcg/kg/min, the effects on the hemodynamics vary for different patients -At doses between 5 to 10 mcg/kg/min, dopamine also stimulates Beta-1 adrenergic receptors and increases cardiac output (mainly by increasing stroke volume with variable effects on HR. Alpha effects>10 mcg/kg/min alpha, Beta, Dopamine At doses >10 mcg/kg/min, the main effect of dopamine is to stimulate alpha-adrenergic receptors and produce vasoconstriction with an increased SVR. However, the overall effect of DA on alpha-receptors is weaker than that of other vasopressors like NE and the beta-1 adrenergic stimulation of DA at doses > 2 mcg/kg/min can cause dose-limiting dysrhythmias. Reference print Related Search for: Well Child Check Well Adult Exam Low Income ACID-BASE DERMATOLOGY Forms & Charts History Templates PE Templates Prevention ECG Tips Guidelines Quick Physical Exam A good Assessment & Plan MV Tips EBM OB/GYN GERIATRICS PSYCHIATRY RADIOLOGY DIFFERENTIAL DIAGNOSIS Clinic Quick Reference Hospital Quick Reference InfoGraphics Book Chapter in Time of Care Notes Coding Guide Office Management Home Pharmacy Mnemonics Resources Facebook Twitter Tumblr Google Instagram Pinterest Dribbble Vimeo Linkedin Myspace Skype Youtube Flickr RSS Designed by Elegant Themes \| Powered by WordPress
4813	https://www.quora.com/What-is-the-formula-for-min-and-max-a-cos-x-b-sin-x	Something went wrong. Wait a moment and try again. Sine and Cosine Functions in Mathematics Mathematical Equations Formulas of Math Trigonometric Functions Trigonometry Maths Cosine (math function) 5 What is the formula for min and max acos(x)-bsin(x)? Jan van Delden MSc Math and still interested · Author has 4.8K answers and 6.5M answer views · 2y This expression can be seen as the inner product of the vectors [a,b]T and [cos(−x),sin(−x)]T, which achieves its extreme values if both vectors are parallel. That is cos(−x)=λa,sin(−x)=λb for some λ≠0 (assuming that not both a,b are 0 simultaneously). The inner product is thus equal to λ(a2+b2), with the condition that cos2(−x)+sin2(−x)=λ2(a2+b2)=1, thus λ=±1√a2+b2 The minimum and maximum value are given by ±√a2+b2 Related questions What is the formula for cos acos b? How does one prove sin ( 2 x ) sin ( x ) − cos ( 2 x ) cos ( x ) = sec ( x ) ? What is the formula for min and max sin(x) cos(x)? How can I prove ∫ π 0 cos ( sin ( sin x ) e cos x ) e e cos x cos ( sin x ) d x = π e ? What is the range of f(x): a cos(x) +b sin(x)? Francesco Amato Studied at Polytechnic University of Turin (Graduated 1977) · Author has 4.5K answers and 1M answer views · 2y f(x)=acosx−bsinx=√a2+b2cos(x−ϕ) where ϕ=tan−1(−ab) The stationary points are those for which the first derivative f′(x)=0 f′(x)=−√a2+b2sin(x−ϕ)=0 → xn=ϕ+nπ (n=0,±1,±2,⋯) To establish where the angles are maximum or minimum points the second derivative function must be calculated f′′(x)=−√a2+b2cos(x−ϕ) Maximum points −cos(xn−ϕ)<0 → −π/2+ϕ+2nπ<xn<π+ϕ+2nπ and y(xn)=√a2+b2 Minimum points −cos(xn−ϕ)>0 → π/2+ϕ+2nπ<xn<3π/2+ϕ+2nπ and y(xn)=−√a2+b2 Donald Hartig PhD in Mathematics, University of California, Santa Barbara (Graduated 1970) · Author has 7.4K answers and 2.8M answer views · 2y Originally Answered: How do I find the maximum and minimum values for acos(x)-bsin(x)? · acos(x)−bsin(x)=√a2+b2sin(ϕ−x) wherecos(ϕ)=b√a2+b2,sin(ϕ)=a√a2+b2. Consequently, the max/min are±√a2+b2. Mike Hirschhorn Honorary Associate Professor of Mathematics at UNSW · Author has 8.1K answers and 2.7M answer views · 2y Originally Answered: How do I find the maximum and minimum values for acos(x)-bsin(x)? · acosx−bsinx=√a2+b2cos(x+α). The max and min values are ±√a2+b2. Related questions Are there formulas for sin(bx) and cos(bx) for certain values of b? How do I prove that 1 + cos x + sin x 1 − cos x + sin x = cot ( x / 2 ) ? (x+a) (x-b) =xx-(b-a) x-ab. is this formula correct? What is the relationship between a and b independent of x if e sin x + e cos x = a , sin x − cos x = b ? If cos 2 ( x ) = 1 − sin 2 ( x ) , then does cos ( x ) = 1 − sin ( x ) ? Elaine Dawe BMath, in Mathematics & Computer Science, University of Waterloo (Graduated 1985) · Author has 5.4K answers and 6.6M answer views · 2y Originally Answered: How do I find the maximum and minimum values for acos(x)-bsin(x)? · f(x)=acosx−bsinx f′(x)=−asinx−bcosx Set derivative =0 −asinx−bcosx=0 −asinx=bcosx sinxcosx=b−a tanx=−ba (1)sinx=b√a2+b2,cosx=−a√a2+b2 f(x)=a(−a√a2+b2)−b(b√a2+b2)=−a2+b2√a2+b2=−√a2+b2 (2)sinx=−b√a2+b2,cosx=a√a2+b2 \qquad\quad f(x) = a \left(\dfrac{a}{\sqrt{a^2+b^2}}\right) - b\left(-\dfrac{b}{\sqrt{a f(x)=acosx−bsinx f′(x)=−asinx−bcosx Set derivative =0 −asinx−bcosx=0 −asinx=bcosx sinxcosx=b−a tanx=−ba (1)sinx=b√a2+b2,cosx=−a√a2+b2 f(x)=a(−a√a2+b2)−b(b√a2+b2)=−a2+b2√a2+b2=−√a2+b2 (2)sinx=−b√a2+b2,cosx=a√a2+b2 f(x)=a(a√a2+b2)−b(−b√a2+b2)=a2+b2√a2+b2=√a2+b2 Maximum value of acosx−bsinx=√a2+b2 Manimum value of acosx−bsinx=−√a2+b2 Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? 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M Shehzad Knows Urdu · Author has 106 answers and 18.2K answer views · 2y Originally Answered: How do I find the maximum and minimum values for acos(x)-bsin(x)? · Differentiate the expression and set it equal to 0. ddx(acos(x)−bsin(x))=−asin(x)−bcos(x)=0...(i) ⟹asin(x)=−bcos(x)⟹tan(x)=−ba there are two solutions to this: one in (π/2,π) and another in (3π/2,2π) In order to know which one is maximum and which one minimum, consider the 2nd derivative: d2dx2(acos(x)−bsin(x))=−acos(x)+bsin(x)=−acos(x)(1−batan(x)) =−acos(x)(1+b2a2) [tan(x)=−ba] the solution in (π/2,π) makes the expression above a positive number: a minimum. The solution in (3π/2,2π) makes the expression negati Differentiate the expression and set it equal to 0. ddx(acos(x)−bsin(x))=−asin(x)−bcos(x)=0...(i) ⟹asin(x)=−bcos(x)⟹tan(x)=−ba there are two solutions to this: one in (π/2,π) and another in (3π/2,2π) In order to know which one is maximum and which one minimum, consider the 2nd derivative: d2dx2(acos(x)−bsin(x))=−acos(x)+bsin(x)=−acos(x)(1−batan(x)) =−acos(x)(1+b2a2) [tan(x)=−ba] the solution in (π/2,π) makes the expression above a positive number: a minimum. The solution in (3π/2,2π) makes the expression negative: a maximum. Note: The expression is periodic, hence, minima and maxima also occur with the same period(2π) Subra Studied Mathematics at Indian Institute of Technology, Bombay (IITB) (Graduated 1979) · Author has 3.5K answers and 658K answer views · 2y f(x)=acos(x)-bsin(x)==> f’(x)=-asin(x)-bcos(x)=0==> asin(x)=bcos(x)==> tan(x)=b/a==> x=arctan(b/a)==> sin(x)=b/sqrt(a^2+b^2), cos(x)=a/sqrt(a^+b^2) f”(x)=-acos(x)+bsin(x) =(b^2-a^2)/sqrt(a^2+b^2)< 0 if ∣b∣<∣a∣, f’’(x) < 0 and f(x) attains f(x) its maximum if ∣b∣>∣a∣, f”(x) > 0 and f(x) attains its minimum, if ∣a∣=∣b∣, f”(x)= 0 and it is called a point of inflection Sponsored by CDW Corporation Want document workflows to be more productive? The new Acrobat Studio turns documents into dynamic workspaces. Adobe and CDW deliver AI for business. Jonathan Burros Studied at Stony Brook University · Author has 1.8K answers and 402.9K answer views · 2y Let cos ∅ = a/√(a^2 + b^2) sin ∅ = b/√(a^2 + b^2) Then f(x) = a cos(x) - b sin(x) = √(a^2 + b^2)[cos(x) cos ∅ - sin(x) sin ∅] = √(a^2 + b^2) cos(x + ∅) For all real x,∅ the maximum value of the cosine function is 1 the minimum of the cosine function is -1 So the maximum of f(x) is √(a^2 + b^2) the minimum of f(x) is -√(a^2 + b^2) Enrico Gregorio Associate professor in Algebra · Author has 18.4K answers and 16M answer views · 2y If a=b=0, there’s not much to say. Otherwise, look for r and φ such that a=rcosφ,b=rsinφ These exist, because we can see that a2+b2=r2cos2φ+r2sin2φ=r2 and we can take r=√a2+b2 and then there is a unique φ∈[0,2π) such that cosφ=ar,sinφ=br because (a/r,b/r) is a point on the unit circle. Thus f(x)=acosx−bsinx=r(cosxcosφ−sinxsinφ)=rcos(x+φ) and we see that the maximum and minimum are, respectively, r=√a2+b2,−r=−√a2+b2 Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Christopher Pellerito Neither pays for, nor charges for, Quora content · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 6.3K answers and 12M answer views · 5y Related What is the value of x where sin x − cos x is the maximum? f(x) = sin x - cos x f′(x) = cos x + sin x 0 = cos x + sin x For what value of x is cos x the opposite of sin x? 3π/4 and 7π/4 or 135° and 315°. Those should be the extremes. The sin of 135° is √2/2 and the cosine is −√2/2 for a difference of √2. f(x) = sin x - cos x f′(x) = cos x + sin x 0 = cos x + sin x For what value of x is cos x the opposite of sin x? 3π/4 and 7π/4 or 135° and 315°. Those should be the extremes. The sin of 135° is √2/2 and the cosine is −√2/2 for a difference of √2. Brian Bi uses math · Upvoted by Christian Brolin , Master of Science Mathematics & Computer Science, University of Gothenburg (1997) and Krzysztof Putyra , Ph.D. Mathematics, Columbia University (2012) · Author has 4.8K answers and 65.1M answer views · Updated 4y Related What is the minimum value of cos x-sin x? Two approaches have already been given: one using basic calculus, and one that uses trigonometric identities to rewrite the expression in terms of a single cosine. Here is a third approach. Instead of minimizing cosx−sinx we’ll maximize sinx−cosx. Notice that sinx−cosx=(cosx,sinx)⋅(−1,1). Here we have two vectors, where the first has magnitude 1 and the second has magnitude √2. According to the Cauchy–Schwarz inequality, this dot product attains its maximum value when the two vectors are parallel, and in that case it equals the product of the magnitudes. Si Two approaches have already been given: one using basic calculus, and one that uses trigonometric identities to rewrite the expression in terms of a single cosine. Here is a third approach. Instead of minimizing cosx−sinx we’ll maximize sinx−cosx. Notice that sinx−cosx=(cosx,sinx)⋅(−1,1). Here we have two vectors, where the first has magnitude 1 and the second has magnitude √2. According to the Cauchy–Schwarz inequality, this dot product attains its maximum value when the two vectors are parallel, and in that case it equals the product of the magnitudes. Since the first vector can point in any direction depending on x, there is some x that makes it parallel to the second. Therefore, the answer is √2. This means the original expression attains a minimum at −√2. This approach is ultimately not so different from the one that uses a trig identity. In a sense, it just sidesteps some algebra using geometric intuition to derive the same result a little bit more quickly. Brian Sittinger PhD in Mathematics, University of California, Santa Barbara (Graduated 2006) · Author has 8.5K answers and 21.1M answer views · 1y Related How do you find the max and min value of f(x) =sin^6 x +cos^6 x? We want to find the extreme values of the trigonometric function We first simplify the appearance of this function. By using the sum of cubes factorization formula combined with the Pythagorean identity, we obtain [math]\begin{align} f(x) &= (\sin^2{x} + \cos^2{x})(\sin^4{x} - \sin^2{x} \cos^2{x} + \cos^4{x})\ &= \sin^4{x} - \sin^2{x} \cos^2{x} + \cos^4{x}. \end{align} \tag{}[/math] Next, we complete the square on the middle term so we can apply the Pythagorean and double angle identities: [math]\begin{align} \displaystyle f(x) &= (\sin^4{x} + 2\sin^2{x} \cos^2{x} + \cos^4{x[/math] We want to find the extreme values of the trigonometric function [math]f(x) = \sin^6{x} + \cos^6{x}. \tag{}[/math] We first simplify the appearance of this function. By using the sum of cubes factorization formula combined with the Pythagorean identity, we obtain [math]\begin{align} f(x) &= (\sin^2{x} + \cos^2{x})(\sin^4{x} - \sin^2{x} \cos^2{x} + \cos^4{x})\ &= \sin^4{x} - \sin^2{x} \cos^2{x} + \cos^4{x}. \end{align} \tag{}[/math] Next, we complete the square on the middle term so we can apply the Pythagorean and double angle identities: [math]\begin{align} \displaystyle f(x) &= (\sin^4{x} + 2\sin^2{x} \cos^2{x} + \cos^4{x}) - 3\sin^2{x} \cos^2{x}\ &= (\sin^2{x} + \cos^2{x})^2 - \frac{3}{4} (2\sin{x} \cos{x})^2\ &= 1 - \frac{3}{4} \sin^2(2x). \end{align} \tag{}[/math] We can make one further simplification by using the half angle identity (in reduction form): [math]\begin{align} \displaystyle f(x) &= 1 - \frac{3}{4} \cdot \frac{1}{2}(1 - \cos(2 \cdot 2x))\ &= \frac{1}{8} (5 + 3\cos(4x)). \end{align} \tag{}[/math] Now, we can read off the extreme values of [math]f[/math] with minimal effort. Since [math]-1 \leq \cos(4x) \leq 1[/math] (where the bounds are achieved at [math]x = 0, \frac{\pi}{4}[/math] for instance), we conclude that the extreme values of [math]f[/math] are given by [math]\displaystyle \frac{1}{8} (5 + 3 \cdot (\pm 1)) = \boxed{\frac{1}{4}, \, 1}. \tag{}[/math] Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views · 8y Related What is the proof that cos (x + y) = cos (x) cos (y) − sin (x) sin (y)? This proof is so lovely and simple that I thought I must share it. We need to prove sin(A... Related questions What is the formula for cos acos b? How does one prove sin ( 2 x ) sin ( x ) − cos ( 2 x ) cos ( x ) = sec ( x ) ? What is the formula for min and max sin(x) cos(x)? How can I prove ∫ π 0 cos ( sin ( sin x ) e cos x ) e e cos x cos ( sin x ) d x = π e ? What is the range of f(x): a cos(x) +b sin(x)? Are there formulas for sin(bx) and cos(bx) for certain values of b? How do I prove that 1 + cos x + sin x 1 − cos x + sin x = cot ( x / 2 ) ? (x+a) (x-b) =xx-(b-a) x-ab. is this formula correct? What is the relationship between a and b independent of x if e sin x + e cos x = a , sin x − cos x = b ? If cos 2 ( x ) = 1 − sin 2 ( x ) , then does cos ( x ) = 1 − sin ( x ) ? If sin(x) and cos(x) are always between 1 and -1, what is the range for tan(x)? What is the formula of sin (A + B)- cos (A + B)? What is the formula for sin x? How do I evaluate ∫ 2 sin x + sin 2 x ( cos x − 1 ) √ cos x + cos 2 x + cos 3 x d x ? What is the formula of sinx(a-b)? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
4814	https://www.geeksforgeeks.org/chemistry/s-block-elements/	S Block Elements - GeeksforGeeks Skip to content Tutorials Python Java DSA ML & Data Science Interview Corner Programming Languages Web Development CS Subjects DevOps Software and Tools School Learning Practice Coding Problems Courses DSA / Placements ML & Data Science Development Cloud / DevOps Programming Languages All Courses Tracks Languages Python C C++ Java Advanced Java SQL JavaScript Interview Preparation GfG 160 GfG 360 System Design Core Subjects Interview Questions Interview Puzzles Aptitude and Reasoning Data Science Python Data Analytics Complete Data Science Dev Skills Full-Stack Web Dev DevOps Software Testing CyberSecurity Tools Computer Fundamentals AI Tools MS Excel & Google Sheets MS Word & Google Docs Maths Maths For Computer Science Engineering Mathematics Switch to Dark Mode Sign In Thermodynamics Analytical Chemistry Atomic Structure Classification of Organic Compounds Periodic Table of Elements Nuclear Force Importance of Chemistry Chemistry Notes Class 11 Chemistry Notes Class 8 Chemistry Notes Class 9 Chemistry Notes Class 10 Chemistry Notes Class 12 Sign In ▲ Open In App S Block Elements Last Updated : 23 Jul, 2025 Comments Improve Suggest changes Like Article Like Report S-block elements are those elements in which the last electron is present in the s-orbital. In the periodic table. They reside in the first 2 columns. S-block consists of 14 elements that include, Hydrogen (H), Lithium (Li), Helium (He), Sodium (Na), Beryllium (Be), Potassium (K), Magnesium (Mg), Rubidium (Rb), Calcium (Ca), Caesium (Cs), Strontium (Sr), Francium (Fr), Barium (Ba), and Radium (Ra). In this article, we will learn about s-block elements, their various properties, and others in detail. Table of Content What are s-block Elements? Properties of s-block Elements Atomic Mass of s-block Elements Diagonal Relationship of s-Block Elements Periodic Trends in s-Block Elements Compounds of s-block Elements What are s-block Elements? You all are well-familiar with the concept of the modern periodic table. In the modern periodic table, we learn about 4 kinds of blocks i.e. s-block, p-block, d-block & f-block. Here, we will learn the definition of s-block elements. It is the most intriguing block of the periodic table. The element which enters in the outermost s-orbital are known as s-block elements. S-block elements consists 2 groups i.e. Group 1 & Group 2. Group 1 consists elements which are called as alkali metals while Group 2 consists metals which are called as alkaline earth metals. Group 1- Alkali Metals Group 2- Alkaline Earth Metals What are Alkali Metals of Group 1? Have you ever wondered why sodium causes a fizzy explosion when it is dropped in water? We will uncover this further. Group 1 elements are called alkali metals because they make hydroxides, whenever reacts with water. You all know that water is strongly alkaline. Here you got your answer for the above question. Alkali metals are well-known for their extreme reactions & their easy bond with other elements. The elements of Group 1 have 1 valence electron in their outermost s-orbital. Learn more about,Alkali Metals What are Alkaline Earth Metals of Group 2? During any occasion, you buy fireworks right? And when you light them up, they starts glowing & gives a beautiful view. It's the magic of none other than our magnesium. Group 2 introduces us the alkaline earth metals which exists in the earth's crust. The interesting thing here is that their oxides & hydroxides are alkaline in nature. The elements of Group 2 have 2 valence electrons in their outermost s-orbital. Learn more about,Alkaline Earth Metals Table of S-block Elements & Their Symbol S-block comprises total 14 elements which are classifies into 2 groups: Group 1 & Group 2. The classification of these 14 elements are given below in tabular form: \| S-block elements \| \| Group 1 (Alkali Metals) \| Group 2 (Alkaline Earth Metals) \| \| Name of Elements \| Symbol \| Name of Elements \| Symbol \| \| Hydrogen \| H \| Beryllium \| Be \| \| Lithium \| Li \| Magnesium \| Mg \| \| Sodium \| Na \| Calcium \| Ca \| \| Potassium \| K \| Strontium \| Sr \| \| Rubidium \| Rb \| Barium \| Ba \| \| Cesium \| Cs \| Radium \| Ra \| \| Francium \| Fr \| \| \| \| Helium \| He \| \| \| Electronic Configuration of S-block elements Electronic Configuration of s-block elements is added in the image below, \| S-block Elements \| \| Group 1 (Alkali Metals) \| Group 2 (Alkaline Earth Metals) \| \| Name of elements \| Electronic Configuration \| Name of elements \| Electronic Configuration \| \| Hydrogen \| 1s 1 \| Beryllium \| [He]2s 2 \| \| Lithium \| [He]2s 1 \| Magnesium \| [Ne]3s 2 \| \| Sodium \| [Ne]3s 1 \| Calcium \| [Ar]4s 2 \| \| Potassium \| [Ar]4s 1 \| Strontium \| [Kr]5s 2 \| \| Rubidium \| [Kr]5s 1 \| Barium \| [Xe]6s 2 \| \| Cesium \| [Xe]6s 1 \| Radium \| [Rn]7s 2 \| \| Francium \| [Rn]7s 1 \| \| \| \| Helium \| 1s 2 \| \| \| Difference between Group 1A and Group 2A Elements Let's learn the basic differences between the elements of Group 1 & Group 2 on the basis of some characteristics: \| Characteristics \| Group 1 (Alkali Metals) \| Group 2 (Alkaline Earth Metals) \| --- \| Reactivity \| These are highly reactive elements. \| These are less reactive elements. \| \| Oxidation State \| These have oxidation state as +1 \| These have oxidation state as +2 \| \| Electrons in outermost shell \| These have only 1 electron. \| These have 2 electrons. \| \| Melting & Boiling Point \| These have low melting & boiling point. \| These have high melting & boiling point. \| Properties of s-block Elements The S-block elements possesses distinct properties. In this section, you'll learn about the properties of s-block elements. Their properties mainly categorized into 2 parts: Physical Properties Chemical Properties Physical Properties of s-block Elements Let's dive deep into the properties of s-block elements Melting Point: S-block elements consists different melting points. The Group 1 elements i.e. Alkali metals have low melting point. For example, Sodium can be melted around 98° celsius. When we talk about Group 2 elements i.e. Alkaline Earth Metals, they have high melting point. For example, Calcium needs 842° celsius temperature to melt which is 9 times more than sodium needs. Conductivity: Due to the presence of electrons that are loosely bound, these electrons can easily move through the metallic structure. This encourages the flow of electricity. That's why s-block elements have higher electrical conductivity. Nature of Bonds:In S-block, the Group 1 elements i.e. Alkali metals always form ionic bonds as these elements have low ionization enthalpies that decreases downwards to the group. In simple words, as we move down the group, the ionic characteristic increases. Density:The densities of s-block elements is relatively low due to their large size. Photoelectric effect:The s-block elements shows photoelectric effect. They emit electrons from their surface, when they come into the contact of light. Lithium is an exception here because of its higher ionization enthalpy. Chemical Properties of s-block Elements Now, it's time to learn about the chemical properties of S-block elements: Reactivity:S-block elements are highly reactive in nature. Group 1 elements especially Na and K are well-known for their high reactivity with water that results in producing hydrogen gas & hydroxides. Group 2 metals are less reactive as compared to Group 1 elements but still they show a notable reactivity when exposed to air or water. Formation of Compounds: S-block elements have high tendency to make compounds. They donate their electrons to form ionic compounds. Group 1 elements donate 1 valence electron while Group 2 elements donate 2 valence electrons to form a compound. Reducing Agents:Due to the strong tendency of loosing electrons, s-block elements are known as strong reducing agents. Atomic Mass of s-block Elements Atomic mass is the total mass of the protons & neutrons in an atom of the element. It is denoted by atomic mass per unit i.e. a.m.u or g/mol. Atomic mass of S-block elements refers to the average mass of the atoms of a particular element, taking into account the various isotopes and their relative abundances in nature. Atomic masses of the s-block elements changes as we go down the groups. The number of protons & neutrons increases and that's why the atomic mass also increases. s-block Elements & Their Atomic Masses The table added below shows the s-block elements and their atomic mass \| S-Block Elements \| \| Group 1 (Alkali Metals) \| Group 2 (Alkaline Earth Metals) \| \| Name of Element \| Atomic Mass \| Name of Element \| Atomic Mass \| \| Hydrogen \| 1.008 amu \| Beryllium \| 9.012 amu \| \| Lithium \| 22.99 amu \| Magnesium \| 24.305 amu \| \| Sodium \| 22.98 amu \| Calcium \| 40.08 amu \| \| Potassium \| 39.10 amu \| Strontium \| 87.62 amu \| \| Rubidium \| 85.47 amu \| Barium \| 137.33 amu \| \| Cesium \| 132.91 amu \| Radium \| 226 amu \| \| Francium \| 223 amu \| \| \| \| Helium \| 4.0026 amu \| \| \| Diagonal Relationship of s-Block Elements Diagonal relationship is the relationship which shows a similar behavior between certain elements. These elements are located diagonally across each other within the same group but different periods of a periodic table. For example. Lithium (Li) and Magnesium (Mg) have a diagonal relationship. Due to their similar ionic sizes and electronic configurations, they show similarities. Even after being in different periods, they have a very similar size of ions which forms a diagonal relationship between them. Some of the similarities observed in these diagonal relationships include: Similar Ionic radii- Certain elements have remarkably similar ionic sizes. For example, Lithium (Li) and Magnesium (Mg} & Beryllium (Be) and Aluminum (Al). Common chemical behavior- Diagonal elements show similarity in their chemical reactions. For example, Lithium (Li) and Magnesium (Mg} & Beryllium (Be) and Aluminum (Al). Complex formation- Due to their anonymous charge densities and electronic configurations, diagonal elements form similar kind of complexes. For example, Lithium (Li) and Magnesium (Mg}. Oxidation state- Diagonal elements tend to show similar oxidation state in some cases. For example,Beryllium (Be) and Aluminum (Al). Periodic Trends in s-Block Elements S-block elements show several kind of periodic trends. As you move down to the group: Atomic size of the elements increases due to the addition of electron shells. Metallic character of the s-block elements follows a certain pattern from left to right. The elements on left like Alkali metals easily lose electrons and shows strong metallic properties. While the elements on right like halogens and noble gases have more non-metallic traits. This happens due to the decreasing atomic sizes. As you move down the group, the reactivity of s-block elements increases due to the increase in their atomic sizes. Due to the increasing atomic size & shielding effect, the s-block elements need less energy to remove electrons which results in the decline of ionization energy. Compounds of s-block Elements Here are preparations & properties of some important compounds of S-block elements: Sodium Hydroxide (NaOH) Preparation: Sodium hydroxide is generally prepared through the electrolysis of aqueous sodium chloride or via the reaction of sodium carbonate with calcium hydroxide. Properties: It is a white solid with high solubility in water and form a strongly alkaline solution. It is a strong base which is used in various industries such paper production, detergents, and in the chemical industry. Calcium Carbonate (CaCO3) Preparation:Calcium carbonate is naturally occured and can be obtained from limestone, chalk, or marble. It's also prepared through precipitation reaction by mixing calcium chloride & sodium carbonate. Properties: It is a white, odorless powder with a variety of applications such as cement production, diet supplement, plastic manufacturing and as a filler in paints and coatings. Read More, Important Compounds of Sodium Important Compounds of Calcium Sample Questions on s-Block Elements Q1: Determine the atomic mass of following elements: Na Ra Solution: Atomic mass of the given elements is, Na= 6.94 a.m.u. Ra= 226 a.m.u. Q2: Which of the following is alkaline earth metal? Sodium Potassium Radium Francium Solution: Option is (3) Radium is Correct. Radium is an alkaline earth metal. Q3: Write the electronic configuration of following s-block elements: Calcium Francium Helium Solution: Electronic configuration of following s-block elements is: Calcium= [Ar]4s 2 Francium= [Rn]7s 1 Helium= 1s 2 Comment More info H hridyanshimishra Follow Improve Article Tags : Chemistry Chemistry-Class-11 Inorganic-Chemistry PCM JEE-Chemistry +1 More Explore Basic Concepts Importance of Chemistry in Everyday Life 10 min readWhat is Matter ? 9 min readProperties of Matter 9 min readMeasurement Uncertainty 9 min readLaws of Chemical Combination 7 min readDalton's Atomic Theory 8 min readGram Atomic and Gram Molecular Mass 7 min readMole Concept 10 min readPercentage Composition - Definition, Formula, Examples 5 min readStoichiometry and Stoichiometric Calculations 7 min read Structure of Atom Composition of an Atom 8 min readAtomic Structure 15+ min readDevelopments Leading to Bohr's Model of Atom 6 min readBohr's Model of the Hydrogen Atom 9 min readQuantum Mechanical Atomic Model 8 min read Classification of Periodicity Classification of Elements 8 min readPeriodic Classification of Elements 10 min readModern Periodic Law 6 min read118 Elements and Their Symbols 9 min readElectronic Configuration in Periods and Groups 9 min readElectron Configuration 8 min readS Block Elements 9 min readPeriodic Table Trends 13 min read Bonding and Molecular Structure Chemical Bonding 12 min readIonic Bond 8 min readBond Parameters - Definition, Order, Angle, Length 7 min readVSEPR Theory 9 min readValence Bond Theory 7 min readHybridization 7 min readMolecular Orbital Theory 7 min readHydrogen Bonding 13 min read Thermodynamics Basics Concepts of Thermodynamics 12 min readApplications of First Law of Thermodynamics 8 min readInternal Energy as a State of System 8 min readEnthalpy Change of a Reaction 9 min readEnthalpies for Different Types of Reactions 10 min readWhat is Spontaneity? - Definition, Types, Gibbs Energy 7 min readGibbs Energy Change and Equilibrium 10 min read Equilibrium Equilibrium in Physical Processes 11 min readEquilibrium in Chemical Processes 7 min readLaw of Chemical Equilibrium and Equilibrium Constant 8 min readDifference between Homogeneous and Heterogeneous Equilibria 7 min readApplications of Equilibrium Constants 6 min readWhat is the Relation between Equilibrium Constant, Reaction Quotient and Gibbs Energy? 8 min readFactors Affecting Chemical Equilibrium 8 min readIonic Equilibrium 5 min readAcids, Bases and Salts 15+ min readIonization of Acids and Bases 6 min readBuffer Solution 10 min readSolubility Equilibria 5 min read Redox Reactions Redox Reactions 14 min readRedox Reactions in terms of Electron Transfer 4 min readOxidation Number \| Definition, How To Find, Examples 13 min readRedox Reactions and Electrode Processes 8 min read Basic Principles and Techniques Organic Chemistry - Some Basic Principles and Techniques 10 min readWhat is Catenation and Tetravalency? 6 min readStructural Representations of Organic Compounds 5 min readClassification of Organic Compounds 12 min readIUPAC Nomenclature of Organic Compounds 13 min readIsomerism 6 min readFundamental Concepts in Organic Reaction Mechanism 15+ min readPurification of Organic Compounds 5 min readQualitative Analysis of Organic Compounds 10 min readWhat is Quantitative Analysis? 9 min read Hydrocarbons What are Hydrocarbons? 11 min readClassification of Hydrocarbons 10 min readAlkanes - Definition, Nomenclature, Preparation, Properties 7 min readAlkenes - Definition, Nomenclature, Preparation, Properties 6 min readAlkynes - Definition, Structure, Preparation, Properties 8 min readAromatic Compounds 9 min read Like Corporate & Communications Address: A-143, 7th Floor, Sovereign Corporate Tower, Sector- 136, Noida, Uttar Pradesh (201305) Registered Address: K 061, Tower K, Gulshan Vivante Apartment, Sector 137, Noida, Gautam Buddh Nagar, Uttar Pradesh, 201305 Company About Us Legal Privacy Policy Contact Us Advertise with us GFG Corporate Solution Campus Training Program Explore POTD Job-A-Thon Community Blogs Nation Skill Up Tutorials Programming Languages DSA Web Technology AI, ML & Data Science DevOps CS Core Subjects Interview Preparation GATE Software and Tools Courses IBM Certification DSA and Placements Web Development Programming Languages DevOps & Cloud GATE Trending Technologies Videos DSA Python Java C++ Web Development Data Science CS Subjects Preparation Corner Aptitude Puzzles GfG 160 DSA 360 System Design @GeeksforGeeks, Sanchhaya Education Private Limited, All rights reserved Improvement Suggest changes Suggest Changes Help us improve. 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4815	https://www.perioiap.org/publications/263-january-2022/263-subgingival-irrigation-with-phytotherapics-adjunct-to-scaling-and-root-planing-on-the-treatment-of-experimental-periodontal-disease-in-rats?downloadarticle=download	Journal of the International Academy of Periodontology 2022 24/1 : 62-73 © International Academy of Periodontology Subgingival irrigation with phytotherapics adjunct to scaling and root planing on the treatment of experimental periodontal disease in rats Carolina dos Santos Santinoni 1 , Marcela Lucio Caldeira 1,Taciane Menezes da Silveira 2, Bibiana Dalsasso Velasques 2 ,Natália Marcumini Pola 2, Christine Men Martins 2 ,Douglas Roberto Monteiro 1, Luciana Prado Maia 1 ,Edilson Ervolino 3 , Thiago Marchi Martins 21 Dental School of Presidente Prudente, Graduate Program in Dentistry (GPD - Master’s Degree), University of Western São Paulo, Presidente Prudente, Brazil; 2 Graduate Program in Den-tistry, Federal University of Pelotas, Pelotas, Brazil; 3Dental School of Araçatuba, Department of Basic Sciences, University Estadual Paulista, Ara çatuba, Brazil. Abstract Aim: To evaluate subgingival irrigation with Matricaria recutita (MAT) and Plantago major (PLA) adjunct to scaling and root planing (SRP) on treatment of experimental periodontitis (EP). Design: EP was induced in 72 rats. After 7 days, animals were randomly distributed in groups: SRP – SRP and irrigation with saline; MAT - SRP and irrigation with MAT solu-tion; and PLA - SRP and irrigation with PLA solution. Euthanasia was performed after 7, 15 and 30 days (n=8). It was evaluated colony-forming units (CFU), bone loss (BL), percentage of mature and immature collagen fibers, TRAP, RANKL and OPG (p <0.05). Results: Groups MAT and PLA had significantly lower number of CFU than Group SRP (15 days). Group PLA had significantly lower BL than Group MAT (7 days). Group MAT had significantly higher percentage of immature collagen fibers than groups SRP and PLA (15 days). Group PLA presented significantly higher OPG than Group SRP (7 days) and significantly lower RANKL than groups MAT and SRP (15 days). Conclusions: Combine MAT or PLA with SRP to treat EP presented additional antimi-crobial and anti-inflammatory effects when compared to SRP. However, PLA presented significantly higher collagen maturation and protective effect against bone resorption than MAT. Keywords: Periodontitis; herbal medicine; Plantago; Chamomile; immunohistochemistry. Introduction Periodontal disease is a chronic inflammatory disease considered a public health problem because it affects most of the population and because it is diagnosed as the main reason for edentulism (Papapanou et al. ,2018; Pradeep et al. , 2013). Periodontal diseases have beginning and progress related to the interaction be-tween an immune response to the host and coloniza-tion by periodontopathogenic microorganisms, such as Actinobacillus actinomycetemcomitans , Porphyromonas gingivalis and Prevotella intermedia . Also, it can be aggravated by environmental and behavioral factors (Nagasri et al. , 2015). They provoke local inflammato-ry responses but corroborate to trigger several systemic conditions in the body, such as arteriosclerosis and sev-eral other diseases (Kolte et al. , 2019). Correspondence to: Carolina dos Santos Santinoni E-mail: carolsantinoni@msn.com 63 Santinoni et al.: Subgingival irrigation with phytotherapics adjunct to scaling and root planing on the treatment of experimental periodontal disease in rats With purpose of achieve disease control, more con-ventional periodontal therapy (scaling and root planing or SRP), is the most common treatment used to reduce periodontopathogens and inflammation (Nagarakanti et al. , 2015; Bhatia et al. , 2014; Anuradha et al. , 2015). In some cases, it is necessary to associate an antimi-crobial agent to completely eliminate microorganisms present in deep and/or strait pockets, and furcation regions (Anuradha et al. , 2015; Matesanz-Pérez et al. ,2013). To avoid the use and adverse effects of antibi-otics, researchers seek alternative therapeutic resourc-es (Nagasri et al. , 2015; Almeida et al. , 2019; Behal et al. , 2011; Shah et al. , 2016; Hugar et al. , 2016). Herbal medicines, used since ancient times, have antimicrobi-al, antioxidant, antiseptic, anti-inflammatory and an-ti-collagenase properties (Almeida et al. , 2019; Behal et al. , 2011; Shah et al. , 2016; Hugar et al. , 2016; Lins et al. , 2013; Moro et al. , 2018). Contribute to improving the population’s access to prevent and treat formal eco-nomically viable diseases, in addition to having syner-gistic effects of its phytochemicals, a set of compounds composed of several molecules that are the target of studies of integrated performance of the organism, lower costs and easy access (Pai et al ., 2019). Some studies have pointed out several pharmaco-logical effects of Matricaria recutita L . and Plantago major L . due to their phytochemical aspects (Nardini et al ., 2019; Kumar et al ., 2009). Matricaria recutita L. , family Asteraceae , popularly known as chamomile, is considered a medicinal plant by anti-inflammatory, antimicrobial, antioxidant, anxiolytic, antimutagenic, healing, antidiabetic, antiseptic, spasmolytic, antidiar-rheal, neuroprotective and antiallergic effects (Kumar et al ., 2009; Cárcamo et al ., 2011). In its composition, there are volatile compounds, sesquiterpene lactones and phenolic compounds, such as flavonoids and cou-marins (Lucena et al ., 2009). The chemical constitu-ents actively present chamomile extracts are isolated phenolic compounds that are not eligible or bioactive (Lucena et al ., 2009). The main component of essential oil extracted from chamomile is the terpenoid α-bis-abol (Anushree et al ., 2015). Bioactivity of Plantago major L ., from the Plantaginaceae family, popularly known as Tanchagem major or Tansagem major , is at-tributed to its chemical compounds such as flavonoids, alkaloids, phenolic compounds, caffeic acids, polysac-charides, terpenoids, lipids, iridoid glycosides, fatty acids and chemicals (Navarro et al ., 1998). It has been widely used due to its gastroprotective, hepatoprotec-tive, antiulcerative, anti-diabetic, anti-inflammatory, anti-inflammatory, anti-cancer, anti-nociceptive, anti-oxidant, anti-hypertensive, anti-microbial and anti-vi-ral properties (Kumar et al ., 2009). Some studies have evaluated anti-inflammato-ry and antimicrobial effects of Matricaria recutita and Plantago major in Dentistry. Matricaria recutita demonstrated through randomized clinical trials to be as effective as chlorexidine to reduce plaque and bleed-ing indexes (Lins et al ., 2013; Cárcamo et al. , 2011; Lucena et al , 2009). Plantago major have shown to be as effective as triclosan in toothpaste to reduce biofilms of oral microorganisms in vitro (Anushree et al ., 2015) and its effectiveness can have residual effects in mouth-rinse for long periods in a serial cases report (Navarro et al ., 1998). Further studies are needed to evaluate the effective-ness of herbal medicines in reducing microorganisms and inflammatory reactions as well as its indication to periodontal treatment. Therefore, the purpose of the present study was to evaluate the influence of subgin-gival irrigation with Matricaria recutita (MAT) and Plantago major (PLA) coadjutant to scaling and root planing (SRP) on the treatment of experimental peri-odontitis (EP) in rats. Materials and Methods Ethical assessment and experimental model The research was carried out respecting the ethical principles of animal experimentation established by the Brazilian College of Animal Experimentation, and the ARRIVE guide (Animal Research: Reporting of in vivo Experiments). The experimental protocol was approved by the Ethics Committee on Animal Experimentation of the University of Western São Paulo - Unoeste (Protocol 4496). The animals were kept in shared ven-tilated cages with 3-4 animals/cage under a controlled environment with 12-hour cycles of light per day and temperature between 22-24ºC. Food and water were offered ad libitum. It was used 72 male rats ( Rattus norvegicus, albinus ,Wistar), weighing 250 to 300 g. The animals were ran-domly assigned to 3 experimental groups: SRP – SRP and irrigation with saline; MAT - SRP and irrigation with MAT solution; and PLA - SRP and irrigation with PLA solution. Each experimental group was sub-divided into 3 subgroups (n = 8) for euthanasia at 7, 15 or 30 postoperative days. Figure 1 shows experimental design of the study. To perform all procedures, animals were anes-thetized by intramuscular injection with ketamine (Dopalen, Agribands Purina do Brasil Ltda., Paulinia, SP, Brazil) (70 mg/kg) and xylazine (Coopazine, Coopers, São Paulo, São Paulo, Brazil) (6 mg/kg). Acute experimental periodontitis (EP) induc-tion in the mandibular left first molar of each rat (de Molon et al ., 2018; Johnson, 1975) and scaling and root planing (SRP) protocol were performed as previously described (Prietto et al ., 2020). Briefly, a cotton thread (Corrente Algodão no. 24, Coats Corrente, São Paulo, São Paulo, Brazil) was tied 64 Journal of the International Academy of Periodontology (2022) 24/1 : 62-73 around the tooth and kept for 7 days. If ligature was not in position after 7 days, the animal was exclud-ed. After 7 days, ligature was removed and SRP was performed with a curette (1−2 Min. Five curette, Hu-Friedy, Chicago, IL). It was performed ten dis-tal–medial and cervical–occlusal traction move-ments over the buccal and lingual surfaces, inter-proximal and furcation area. Figure 1. Scheme illustrating the experimental design of the study. Subgingival irrigation After SRP, animals from Group SRP received subgin-gival irrigation with 1 ml of saline solution. Animals from groups MAT and PLA received, respectively, sub-gingival irrigation with 1 ml of Matricaria recutita solu-tion and Plantago major solution produced in pharma-cy (Apothicário Farmácia de Manipulação, Araçatuba, SP, Brazil) by evaporation of ethanol/water. Solutions were inserted slowly into the periodontal pocket, using a 1 ml syringe and insulin needle without bevel. Microbiological analysis Biofilm samples were collected before (T0) and 7 days after (T1) the EP induction, and 7 (T2), 15 (T3) and 30 days (T4) after treatments, before the euthana-sia. For this, sterile absorbent paper tips number 20 (Dentsply Maillefer, Ballaigues, Switzerland) were in-serted into the periodontal pockets and kept for 1 min-ute. The tips with biofilm samples were transferred to microtubes containing 1 mL of Brain Heart Infusion (BHI). Microtubes were then vortexed for 10 seconds and the biofilm suspensions, serially diluted in saline solution. Afterwards, each dilution was plated in tripli-cate on BHI agar. The agar plates were aerobically incu-bated at 37°C and the number of colony-forming units (CFUs), counted after 48 h. Microbiological results were represented as Log10 CFU/mL. Euthanasia and laboratorial processing At 7, 15 or 30 days after treatments, animals were eu-thanized with an anesthesia (Thiopental, Cristália Produtos Químicos Farmacêuticos LTDA, Itapira, SP, Brasil) overdose (150 mg/kg) and laboratorial pro-cessing was performed. Histological sections of each experimental group and period were submitted to he-matoxylin and eosin (H.E.) staining (histomorpho-metric analysis), to picrosirius red staining (collagen maturation analysis) or to indirect immunoperoxidase method to detect tartrate-resistant acid phosphatase (TRAP), ligand of the nuclear factor kappa B activator receptor (RANKL) and osteoprotegerin (OPG) (os-teoclastogenesis analysis). For Picrosirius red staining, the histological sec-tions were deparaffinized, hydrated and immersed in a Sirius F3BA solution in aqueous picric acid for 1 hour. The colored sections were washed in two baths of 0.5% acetic acid solution for 1 minute. After dehydration, the sections protected with mounting medium and glass cover slip. For immunoperoxidase method, the histological sections were submitted to the same reactions de-scribed by Santinoni et al . (2020). Histomorphometric analysis Images of the furcation region of histological sections stained with H.E. were captured with a digital camera connected to a microscope. The area of bone loss (BL) in the furcation region was determined by an exam-iner calibrated and blind to the treatments, using an image analysis program (Image J - National Institutes of Health, Washington, DC, USA (ImageJ 1.51p Histochemical analysis Histological sections stained with Picrosirius red were analyzed under polarized light microscopy. Images of the furcation region of the histological sections stained with picrosirius red were captured with a digital camera connected to a polarized light microscope at 40x mag-nification. Using a color limit function of a software (Leica ICC50 HD, Wetzlar, Germany), it was selected furcation region which was the interest area. After, it was used the function “RGB Measure” that provide in-formation about red (R), green (G) and blue (B) of the 65 Santinoni et al.: Subgingival irrigation with phytotherapics adjunct to scaling and root planing on the treatment of experimental periodontal disease in rats circulated area. Values of red (R) were used to calculate percentage of mature collagen fibers in the furcation region and values of green (G) were used to calculate percentage of immature collagen fibers in the furcation region (Santinoni et al ., 2020). Immunohistochemical analyses Number of TRAP-positive cells in the furcation region was quantified. Immunostaining for OPG and RANKL in the furcation region were semi-quantified through scores following the criteria of Santinoni et al. (2020). Briefly, absence of immunostaining (score 1), low pattern of immunostaining (score 2), moderate pattern of immunostaining (score 3) and high pattern of immunostaining (score 4). Statistical analysis All tests considered a significance level of 5%. Data were separately analyzed. Microbiological data were evaluated by one software (SigmaPlot, ver-sion 12.0; Systat Software Inc., San Jose, USA). Histomorphometric, histochemical and immunohis-tochemical data were analyzed in other software (IBM SPSS Statistics for Windows, Version 21.0. Armonk, NY, USA: IBM Corp.). All data were submitted to normality verifica-tion by Shapiro-Wilk test. Subsequently, to verify the differences among groups, Student-Newman-Keuls test was performed for microbiological analysis; ANOVA followed by Tukey post-test for histomorphometric and histochemical analyses, and TRAP-positive cells; and Kurskal-Wallis for OPG and RANKL. Results Microbiological analysis Biofilm samples at T0 presented significantly higher CFU than biofilm samples at T1. For T3, treatments with MAT and PLA resulted in CFU counts sig-nificantly lower than that noted for the Group SRP (Figure 2). However, significant differences among treatments were not observed for T2 and T4 (Figure 2). Histological analysis In Group SRP, it was observed an intense inflamma-tory infiltrate in the connective tissue in the furcation region at 7 days. At 15 and 30 days, it was observed a moderate inflammatory infiltrate. Bone loss occupied approximately half the furcation region. In groups MAT and PLA, it was observed a mod-erate inflammatory infiltrate in the connective tissue in the furcation region at 7 days. At 15 and 30 days, it was observed low inflammatory infiltrate. Bone loss oc-cupied approximately a quarter or less of the furcation region (lower than Group SRP). Figure 2. Mean values of the logarithm of colony-forming unit per mL (Log10 CFU/ml) obtained from the biofilm collected for each experimental group (7, 15 and 30 days) after treatments. Abbreviations and symbol: MAT, Matricaria recutita; PLA, Plantago major; SRP, scaling and root planing; CFU, colony-forming unit; , significantly higher than T0; , significantly lower than SRP group, within 15 days; , significantly lower than SRP group, within 15 days. 66 Journal of the International Academy of Periodontology (2022) 24/1 : 62-73 Figure 3. Photomicrographs showing BL area in the furcation region in groups SRP, MAT and PLA, respectively, 7 (A-C), 15 (D-F) and 30 (G-I) days after treatments. Blue arrows show intense inflammatory infiltrate in the connective tissue in groups SRP and MAT. It can be also noted extensive alveolar bone loss in group SRP greater than groups MAT and PLA. Group PLA did not present high concentration of inflammatory cells. Abbreviations: ab, alveolar bone; MAT, Matricaria recutita; PLA, Plantago major; SRP, scaling and root planing. Hematolixin and eosin staining; 40x. Table 1. Parameters assessed in inflammatory infiltrate analysis of the mandibular first molar in all experimental groups. The inter-radicular septum was irregular for all experimental groups at 7 days. Table 1 shows param-eters assessed in inflammatory infiltrate analysis of the mandibular first molar in all experimental groups. Parameters followed study by Zuza et al . (2018). Figure 3 show representative images of each experimen-tal group and period. Histometric analysis Table 2 shows means and standard deviations of the percentage of BL in the furcation region in each ex-perimental group and period, as well as the results of the intergroup comparisons. In the intragroup com-parisons, no statistically significant differences were observed. Parameters and scores Percentage of animals Parameters and scores SRP MAT PLA 7d 15d 30d 7d 15d 30d 7d 15d 30d Intensity of local inflammatory infiltrate (0) Absence of inflammation 28.57 0033.33 37.5 66.67 42.86 42.86 33.33 (1) Small number of inflammatory cells 42.86 57.14 62.5 33.33 50 16.67 57.14 28.57 66.67 (2) Moderate number of inflammatory cells 28.57 42.86 37.5 33.33 12.5 16.67 028.57 0(3) Large number of inflammatory cells 000000000 Extension of local inflammatory infiltrate (0) Absence of inflammation 28.57 0033.33 37.5 66.67 42.86 42.86 33.33 (1) Extending to part of the connective tissue of the furcation area 71.43 100 100 66.67 62.5 33.33 57.14 57.14 66.67 (2) Extending to the whole connective tissue of the furcation area 000000000(3) Extending to the whole connective tissue and to the bone tissue of the furcation area 00000000067 Santinoni et al.: Subgingival irrigation with phytotherapics adjunct to scaling and root planing on the treatment of experimental periodontal disease in rats Table 2. Means ± standard deviations (SD) of the percentage (%) of bone loss (BL) in the furcation region for each experimental group and period, and the result of the intergroup comparisons (p value). Intergroup comparisons: Significantly lower than Group MAT at 7 days ( p=0.023). SRP MAT PLA 7d 15d 30d 7d 15d 30d 7d 15d 30d Mean 8.78 9.84 10.06 14.08 11.16 9.21 6.62 11.31 6.36 SD 5.76 5.03 5.91 7.73 5.42 4.85 3.32 5.42 2.86 Histochemical analysis (collagen maturation) Figures 4A and 4B shows mean and standard devia-tion of percentage of immature and mature collagen fibers for each experimental group and period, as well as the result of intergroup comparisons. Figures 4C, 4D and 4E present photomicrographs of histological sections stained with Picrosirius red under polarized light in groups SRP, MAT and PLA at 15 days, respec-tively. In the intragroup comparisons, no statistically significant differences were observed. Immunohistochemical analyses It was not observed statistically significant differenc-es among experimental groups and period regarding number of TRAP-positive cells. Figure 5A shows mean and standard deviation of number of TRAP-positive cells for each experimental group and period. Figures 5B-D show representative images of TRAP-immunolabeling. Table 3 presents scores observed for immunos-taining with both OPG and RANKL for each ex-perimental group and period, as well as the results of intergroup comparisons. Figures 5E-G present, respectively, photomicrographs showing immunola-beling for RANKL in groups SRP, MAT and PLA at 15 days and Figures 5H-J present, respectively, pho-tomicrographs showing immunolabeling for OPG in groups SRP, MAT and PLA at 7 days. Discussion The present study aimed to evaluate the influence of subgingival irrigation with two natural extracts associated with SRP in the treatment of experimen-tal periodontitis and to compare the results with the conventional treatment of experimental periodontitis (SRP and irrigation with saline). This objective was based on the need for local application of products that have antimicrobial and / or anti-inflammatory action within the periodontal pockets as an adjunct to mechanical debridement, specifically in sites with periodontitis that did not regress after conventional treatment (Nagarakanti et al ., 2015; Matesanz-Pérez et al ., 2013; Tan et al ., 2020). Among the advantages for the local use of chemical agents, including herbal products, the following can be evidenced: maximizing its effect in specific sites and the prevention of sys-temic toxicity and problems related to the patient’s lack of commitment (Batista et al .,2014; Kartini et al ., 2017; Gomes et al ., 2018). In the present study, groups treated with MAT and PLA showed less inflammatory infiltrate when compared to the control group in all experimental periods. In addition, these groups had less biofilm for-mation at 15 days postoperative and Group PLA pre-sented significantly lower bone loss than Group MAT at 7 days. Thus, it can be inferred that MAT and PLA solutions have the potential to improve the results of conventional periodontal treatment. Despite differences in experimental models, it can be inferred that results observed in Group MAT cor-roborate with other studies, which demonstrated that this plant has potential to improve periodontal healing through antimicrobial action and anti-inflammatory ef-fect (Lins et al ., 2013; Cárcamo et al ., 2011; Lucena et al ., 2009; Goes et al ., 2016). These clinical studies com-pared the effectiveness of the mouthwash with chlorhex-idine or MAT in patients with gingivitis. Results showed great reduction in plaque indexes with MAT that was so efficient as chlorhexidine. It is possible that benefits of MAT on periodontal healing are due to antioxidant activity of polyphenol and flavonoid content that down regulate both free -radical scavenging activity and ex-pression of matrix metalloproteinases (Al-Dabbagh et al ., 2019). In this context, it is also important consider that antioxidant activity of MAT demonstrated act in a dose dependent way (Al-Dabbagh et al ., 2019). Here, we used MAT extract produced in pharmacy by evapora-tion of ethanol/water. Although the concentration was not evaluated here and it is a limitation of the present study, considering previous studies that performed same extraction method, it can be inferred we had an estimat-ed concentration of 13.51% (Al-Dabbagh et al ., 2019; Roby et al ., 2013). However, the method to obtain the extract as well as the part of the plant may influence the result of substance used. Here, it was used all the plant in Group PLA. In in Group MAT, it was used only the flower, the same part used in the studies by Al-Dabbagh et al . (2019) and Roby et al . (2013). 68 Journal of the International Academy of Periodontology (2022) 24/1 : 62-73 Figure 4. Graphs showing the percentage of immature (A) and mature (B) collagen fibers, and photomicrographs showing the maturation of collagen in groups SRP (C), MAT (D) and PLA (E) 15 days after treatments. Note that the Group MTA has greenish color compared with groups SRP and PLA that have a more yellow color. Abbreviations and symbol: ab, alveolar bone; MAT, Matricaria recutita; PLA, Plantago major; SRP, scaling and root planing; , significantly higher than groups SRP and PLA at 15 days. Picrosirius red staining under polarized light; 40x. 69 Santinoni et al.: Subgingival irrigation with phytotherapics adjunct to scaling and root planing on the treatment of experimental periodontal disease in rats Figure 5. Graph (A) showing number of TRAP-positive cells in the experimental groups and period, in different analyzed periods. Photomicrographs showing immunolabeling for TRAP in groups SRP, MAT and PLA, respectively, 7 (B), 15 (C) and 30 (D) days after treatments presenting same immunolabeling pattern among groups. Photomicrographs showing RANKL-immunolabeling in groups SRP (E), MAT (F) and PLA (G) 15 days after treatments and photomicrographs showing OPG-immunolabeling in groups SRP (H), MAT (I) and PLA (J) 7 days after treatments. Groups SRP and MAT present similar pattern of RANKL-immunolabeling while Group PLA did not present immunolabeling in this histological section. Regarding OPG, Group PLA presents immunolabeling pattern higher than groups SRP and MAT. Abbreviations: ab, alveolar bone; ct, connective tissue; MAT, Matricaria recutita; PLA, Plantago major; SRP, scaling and root planing. Counterstaining with hematolixin; 400x. 70 Journal of the International Academy of Periodontology (2022) 24/1 : 62-73 Table 3. Scores observed for immunostaining with OPG and RANKL for each experimental group and period, in the different periods. Marker Score SRP MAT PLA 7d 15d 30d 7d 15d 30d 7d 15d 30d OPG 10/7 0/7 0/7 0/7 0/7 0/6 0/7 0/7 1/7 22/7 3/7 2/7 3/7 2/7 3/6 0/7 3/7 3/7 30/7 2/7 4/7 0/7 1/7 3/6 0/7 1/7 2/7 45/7 2/7 1/7 4/7 4/7 0/6 7/7 3/7 1/7 Mean 2.40 1.86 1.86 2.14 2.29 1.50 3.00 2.00 1.43 RANKL 11/7 1/6 1/6 2/6 0/7 1/6 4/7 6/7 3/7 23/7 4/6 5/6 2/6 6/7 4/6 3/7 1/7 3/7 32/7 1/6 1/6 1/6 1/7 1/6 0/7 0/7 1/7 41/7 0/6 0/6 1/6 0/7 0/6 0/7 0/7 0/7 Mean 1.42 0.63 0.63 1.16 1.14 1.00 0.42 0.28 0.71 Intergroup comparisons: Significantly higher than Group SRP at 7 days ( p=0.040). Significantly lower than groups MAT ( p= 0.011) and SRP ( p=0.045) at 15 days. Results observed with PLA also corroborate previ-ous studies demonstrated potential of its antimicrobial and anti-inflammatory effects to be used in Dentistry and Periodontics. Anushree et al . (2015) carried out an in vitro study to compare the antimicrobial effect of toothpaste containing triclosan or PLA and concluded that this herbal agent can be as effective as conventional antimicrobial agents already used. Navarro et al . (1998) revealed the effectiveness in reducing the biofilm indices of patients treated with PLA mouthwash and a residu-al effect of this plant even after 42 days (Navarro et al ., 1998). This characteristic is important to prevent recolo-nization (Goes et al ., 2016). PLA may have remained for a longer time than MAT in periodontal tissues, leading to better healing results. This characteristic of remain-ing for a long period in periodontal tissues is one of the required characteristics of an antimicrobial agents used for the treatment and prevention of periodontal diseas-es, known as substantivity (Matesanz-Pérez et al ., 2013). In addition to substantivity, PLA showed other desirable effects of antimicrobial agents, like significant reduction of bacterial biofilm and inflammation (Matesanz-Pérez et al ., 2013). Also, it has been demonstrated presence of oligosaccharides in PLA that have beneficial effects on human health (Lukova et al ., 2017; Adom et al ., 2017; Parhizgar et al ., 2018). The present study is the first to evaluate histochem-ically and immunohistochemically the effects of sub-gingival irrigation with MAT and PLA associated with SRP to treat EP in rats. Considering that the group treated with MAT showed a percentage of immature collagen fibers significantly higher than the groups PLA and SRP, it can be suggested that the PLA showed a better result on periodontal healing than MAT. Both histomorphometric and immunohistochemical results of the present study corroborate and reinforce this hypothesis. Group PLA presented significantly lower BL than Group MAT at 7 days. Also, Group PLA pre-sented significantly higher immunoexpression of OPG than Group SRP at 7 days and significantly lower ex-pression of RANKL than groups MAT and SRP at 15 days. OPG has a protective effect on bone tissue and RANKL stimulates osteoclastogenesis and bone re-sorption (Boyce et al ., 2008; Souza et al ., 2013; Harada et al ., 2011; Takahashi et al ., 2011). Presence of a specific component in the PLA com-position may explain differences in the results obtained by each treatment in the present study. PLA present caf-feic acid as one of its components (Navarro et al ., 1998) that have been associated with oxidative stress reduction and inflammation dampen (Stähli et al ., 2019; Li et al ., 2017). In vitro studies with other plants that contains caffeic acid showed it can reduce P orphyromonas gin-givalis and Prevotella Intermedia lipopolysaccharide pro-inflammatory action, and catalase antioxidant en-zyme gene expression through reduction of intracellular reactive oxygen species levels and the expression of genes encoding-producing enzymes (Le Sage et al ., 2017; Choi et al ., 2015). With similar methodology used in the present study, Yiğit et al . (2017) evaluated the effect of caffeic acid on alveolar bone loss, serum cytokines (in-terleukin (IL)-1β, IL-6, tumor necrosis factor-α and IL-10) and gingival apoptosis, as well as the levels of antioxidants. They also evaluated low dose doxycycline combined or not with caffeic acid. Group treated with caffeic acid presented lowest alveolar bone loss, inflam-matory infiltration and expression of serum cytokines among the experimental groups. The authors concluded that caffeic acid has more anti-inflammatory, antioxidant and anti-apoptotic effects than antibiotic evaluated. 71 Santinoni et al.: Subgingival irrigation with phytotherapics adjunct to scaling and root planing on the treatment of experimental periodontal disease in rats In addition to MAT and PLA, other herbal prod-ucts have also been evaluated as an adjunct treatment to conventional treatment of EP. Almeida et al . (2019) used assessed the effect of green tea extract on periodon-tal healing. They also performed histological analysis and immunohistochemical reactions for the detection of inflammatory proteins and osteoclasts in the furca-tion region. Promising results showed that the groups treated with green tea showed less inflammation, fewer osteoclasts and less BL, compared with control groups where only SRP was performed. Few studies are found in the literature about the use of herbal products in periodontal treatment. Moro et al . (2018) carried out a recent systematic review to assess the effect of the application of adjuvant herbal agents to SRP on clinical parameters of patients with periodontitis compared with SRP alone. The results showed that the local combination of phytotherapics with SRP can promote additional benefits in reduc-ing the probing depth and clinical attachment level. However, more studies are needed to better evaluate its application. Conclusions Within the limits of this study, it can be concluded that combine MAT or PLA with SRP to treat experi-mental periodontitis presented additional antimicrobi-al and anti-inflammatory effects when compared to SRP alone. 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Topical sodi-um alendronate combined or not with photody-namic therapy as an adjunct to scaling and root planing: Histochemical and immunohistochem-ical study in rats. Journal of Periodontal Research 2020; 55 : 850-858. Shah SA, Vijayakar HN, Rodrigues SV, Mehta CJ, Mitra DK and Shah RA. To compare the effect of the local delivery of hyaluronan as an adjunct to scaling and root planing versus scaling and root planing alone in the treatment of chronic periodontitis. Journal of Indian Society of Periodontology 2016; 20 : 549-556. Souza PP and Lerner UH. The role of cytokines in in-flammatory bone loss. Immunological Investigations 2013; 42 : 555-622. Stähli A, Maheen CU, Strauss FJ, Eick S, Sculean A and Gruber R. Caffeic acid phenethyl ester protects against oxidative stress and dampens inflammation via heme oxygenase 1. International Journal of Oral Science 2019; 11 : 6. Takahashi N, Maeda K, Ishihara A, Uehara S and Kobayashi Y. Regulatory mechanism of osteoclas-togenesis by RANKL and Wnt signals. Frontiers in Bioscience 2011; 16 : 21-30. Tan OL, Safii SH and Razali M. Commercial local phar-macotherapeutics and adjunctive agents for nonsur-gical treatment of periodontitis: a contemporary re-view of clinical efficacies and challenges. Antibiotics 2020; 9: 1-26. Yiğit U, Kırzıoğlu FY, Uğuz AC, Nazıroğlu M and Özmen Ö. Is caffeic acid phenethyl ester more pro-tective than doxycycline in experimental periodon-titis? Archives of Oral Biology 2017; 81 : 61-68. Zuza EP, Garcia VG, Theodoro LH, Ervolino E, Favero LFV, Longo M, Ribeiro FS, Martins AT, Spolidorio LC, Zuanon JAS, de Toledo BEC, Pires JR. Influence of obesity on experimental periodontitis in rats: histopathological, histomet-ric and immunohistochemical study. Clinical Oral Investigation. 2018; 22 : 1197-1208.
4816	https://www.youtube.com/watch?v=ajxcUlbk09E	Solving a 2x2 system of linear equations by substitution haraldim1 8870 subscribers 1 likes Description 80 views Posted: 11 Jan 2025 Master the method of solving a 2x2 system of linear equations using substitution in this clear and concise tutorial! I’ll walk you through the step-by-step process to find the values of both variables and explain the key concepts behind this approach. Perfect for anyone looking to strengthen their algebra skills. Don’t forget to like, subscribe, and visit www.charalampidisdimitris.com for more math tutorials! Transcript: hi there my name is m lais and in this video I'll show you how to solve a linear system of two equations and two unknowns by applying the substitution method we'll do we'll see everything through a worked example so here it is solve the system that consists of the linear equations 2x + y = 3 and 3x - 2 y = 1 I'll begin let's mention here that the first step in Step One We Begin by naming the two equations so from now on the top one here will be referred to as equation one while the other one will be equation two then it's time to move in step two in step two in this step I solve or let's better I go in any of these two equations and solve for any variable I want this means that I can go to the first equation to equation number one and solve for x I can go let's write that here as well I can go in one and solve for x I can go in one and solve for y I can go in two and solve for x or I can go in two and solve for y respectively but as a TP choose the one that has no coefficient in either the x or y so as a tip here choose to solve for the variable variable that doesn't have any coefficient in our case that would be the variable y in the first equation so in step two we say that I go in one and solve for y this means that the first equation which was 2x + y = 3 will now when solved for y will now become y = 3 - 2x this new equation will be named as three then it's time to move on to the next step that is step three go in the equation you didn't use in step two and substitute three so in step two in my example I did not I did use the first one so this means that I did not use the second one which means that in step three I'll go in the second equation and substitute this is why the method is called substitution method and substitute the third equation what do I mean by that I mean that the second equation which is the equation 3x - 2 y = 1 I should not go in it and substitute the third but the third says states that y = 3 - 2x which means that here I should substitute y by exactly that by three - 2x by doing so this will become 3 x - 2 3 - 2x and all that equals 1 what did I achieve under this substitution now this new substitution is a one variable equation which means that it can be easily solved so I'm going to do operations 3x - 6 + 4x = 1 1 which means that 3x + 4x = 1 + 6 which means that 7 x = 7 which means that X is equal to 1 the third step with have been go in the equation you didn't use in step two and substitute three and solve that equation now in step four so far we managed to find the unique value of x but remember that when we work with a system of equations we are looking for a pair depending here we're working with a system of two equations and two unknowns hence we're looking for a pair of numbers which combined both will be referred to as a solution we found that X should be equal to one but we still need to find the value of y how can we find that generally speaking I can use that value of x x = 1 and I can substitute it in any of the above three equations I can take xal 1 and substitute it in equation one I can take xal 1 and substitute it in equation number two or it would be easier here if we go directly in the third equation and the reason for that is that in the third equation we have already solved for y so if we substitute x = 1 that's all it takes we will directly calculate the value of y so in step four we can say that take the solution you found for one of the variables substitute it in equation three and find the value of the second variable as well so this means that I should now go in in the third one which is y = 3 - 2x X and say that for xal = 1 I get that Y is 3 - 2 1 which means 3 - 2 = 1 and that is the value of y finally in step five we need to properly present your solution in a pair X comma y so finally in step five my solution is 1A 1 so here it is again I name them this will allow me to easily navigate among the solution among the process of solving that I mean in step two it's totally up to me I go in any of them either in the first or in the second and solve for any variable I want as a tip if possible go in the one go and solve for the variable that does not have any clear number in front that doesn't have any or even better if you want to express it better go to the variable that has only number one in as a coefficient once you you do that this will be called equation number three now in the next step we need to take this three this equation number three and substitute it in the equation we didn't use in step two this will produce a new equation of one variable which can be easily solved and this solution will give us the value of one of the variables but remember that we need a pair of such values so having that value we will again substitute it in the third equation this will produce the new the value of the second variable and then both of them as a pair will be presented as the solution let's see however that in order for all of this to work before we even start we have to do what Express both in standard form remember remember that standard form is the form where both x's and y's are brought on the same side and then on the other side we have just the constant so with that said we can see here that 3x + 2 y = one is already in a standard form but the second one is not so I'm going to do 4x - y = 11 - 6 which means that we'll keep the first one as it is and we'll do some quick operations in the second one 4x - y = 5 now I'm going to name these as one and two and next I will go in any of them and solve for any variable I want the easiest here would be to go in the second one and solve for y so I go in two and solve for y this means that 4x - y = 5 will now become 4x - 5 = Y and this will be referred to as number three remember that I want to solve for y not just for minus y then in the first in the second step I used number two so now I'll go in one and substitute the third one this means that although the one was 3x + 2 y = 1 now I'm going to substitute three but three suggests that Y is 4x - 5 so this means that instead of y i should substitute 4 x - 5 and this will lead to 3x + 2 4 x - 5 = 1 and now I have to do operations 8 x - 10 = 1 so 11 x = 10 + 1 which means that 11 x = 11 so X is 1 that's half or that's part of my solution I still need to find the value of y as well this means that I will take this for x = 1 and I'll substitute it in three 3 becomes 4 1 - 5 = Y which means that1 = y so finally my Solutions which should be presented in X comma y form are 1 And1 and that is I understand that it might seem tricky it might seem hard trust me it's not it needs however a lot of practice and once you do that you'll see that the steps are not as many as you now think and definitely not as hard as you believe read them carefully do them again always start by doing the ones that we have already worked together because for these ones you already know the solutions in case however you need more uh clarification on something in case you have any question feel free and please do so leave this question as a comment below this video as always thank you for your time we'll meet again in one of the following videos bye-bye
4817	https://artofproblemsolving.com/wiki/index.php/AM-GM_Inequality?srsltid=AfmBOopIsEzrc8kZ_MvAVfy5Vg-I4L-QQ_fe4cOCV5CE0DttANp27wKz	Art of Problem Solving AM-GM Inequality - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki AM-GM Inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search AM-GM Inequality In algebra, the AM-GM Inequality, also known formally as the Inequality of Arithmetic and Geometric Means or informally as AM-GM, is an inequality that states that any list of nonnegative reals' arithmetic mean is greater than or equal to its geometric mean. Furthermore, the two means are equal if and only if every number in the list is the same. In symbols, the inequality states that for any real numbers , with equality if and only if . The AM-GM Inequality is among the most famous inequalities in algebra and has cemented itself as ubiquitous across almost all competitions. Applications exist at introductory, intermediate, and olympiad level problems, with AM-GM being particularly crucial in proof-based contests. Contents [hide] 1 Proofs 2 Generalizations 2.1 Weighted AM-GM Inequality 2.2 Mean Inequality Chain 2.3 Power Mean Inequality 3 Problems 3.1 Introductory 3.2 Intermediate 3.3 Olympiad 4 See Also Proofs Main article: Proofs of AM-GM All known proofs of AM-GM use induction or other, more advanced inequalities. Furthermore, they are all more complex than their usage in introductory and most intermediate competitions. AM-GM's most elementary proof utilizes Cauchy Induction, a variant of induction where one proves a result for , uses induction to extend this to all powers of , and then shows that assuming the result for implies it holds for . Generalizations The AM-GM Inequality has been generalized into several other inequalities. In addition to those listed, the Minkowski Inequality and Muirhead's Inequality are also generalizations of AM-GM. Weighted AM-GM Inequality The Weighted AM-GM Inequality relates the weighted arithmetic and geometric means. It states that for any list of weights such that , with equality if and only if . When , the weighted form is reduced to the AM-GM Inequality. Several proofs of the Weighted AM-GM Inequality can be found in the proofs of AM-GM article. Mean Inequality Chain Main article: Mean Inequality Chain The Mean Inequality Chain, also called the RMS-AM-GM-HM Inequality, relates the root mean square, arithmetic mean, geometric mean, and harmonic mean of a list of nonnegative reals. In particular, it states that with equality if and only if . As with AM-GM, there also exists a weighted version of the Mean Inequality Chain. Power Mean Inequality Main article: Power Mean Inequality The Power Mean Inequality relates all the different power means of a list of nonnegative reals. The power mean is defined as follows: The Power Mean inequality then states that if , then , with equality holding if and only if Plugging into this inequality reduces it to AM-GM, and gives the Mean Inequality Chain. As with AM-GM, there also exists a weighted version of the Power Mean Inequality. Problems Introductory For nonnegative real numbers , demonstrate that if then . (Solution) Find the maximum of for all positive . (Solution) Intermediate Find the minimum value of for . (Source) Olympiad Let , , and be positive real numbers. Prove that (Source) See Also Proofs of AM-GM Mean Inequality Chain Power Mean Inequality Cauchy-Schwarz Inequality Inequality Retrieved from " Categories: Algebra Inequalities Definition Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
4818	https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Ideal_Systems/Entropy_of_Mixing	Skip to main content Entropy of Mixing Last updated : Apr 2, 2023 Save as PDF Ideal Systems Ideal Gas Processes Page ID : 1945 ( \newcommand{\kernel}{\mathrm{null}\,}) A gas will always flow into a newly available volume and does so because its molecules are rapidly bouncing off one another and hitting the walls of their container, readily moving into a new allowable space. It follows from the second law of thermodynamics that a process will occur in the direction towards a more probable state. In terms of entropy, this can be expressed as a system going from a state of lesser probability (less microstates) towards a state of higher probability (more microstates). This corresponds to increasing the W in the equation S=kBlnW. The Mixing of Ideal Gases For our example, we shall again consider a simple system of two ideal gases, A and B, with a number of moles nA and nB, at a certain constant temperature and pressure in volumes of VA and VB, as shown in Figure 1A. These two gases are separated by a partition so they are each sequestered in their respective volumes. If we now remove the partition (like opening a window in the example above), we expect the two gases to randomly diffuse and form a homogenous mixture as we see in Figure 1B. To calculate the entropy change, let us treat this mixing as two separate gas expansions, one for gas A and another for B. From the statistical definition of entropy, we know that ΔS=nRlnV2V1. Now, for each gas, the volume V1 is the initial volume of the gas, and V2 is the final volume, which is both the gases combined, VA+VB. So for the two separate gas expansions, ΔSA=nARlnVA+VBVA ΔSB=nBRlnVA+VBVB So to find the total entropy change for both these processes, because they are happening at the same time, we simply add the two changes in entropy together. ΔmixS=ΔSA+ΔSB=nARlnVA+VBVA+nBRlnVA+VBVB Recalling the ideal gas law, PV=nRT, we see that the volume is directly proportional to the number of moles (Avogadro's Law), and since we know the number of moles we can substitute this for the volume: ΔmixS=nARlnnA+nBnA+nBRlnnA+nBnB Now we recognize that the inverse of the term nA+nBnA is the mole fraction χA=nAnA+nB, and taking the inverse of these two terms in the above equation, we have: ΔmixS=−nARlnnAnA+nB−nBRlnnAnA+nBχB=−nARlnχA−nBRlnχB since lnx−1=−lnx from the rules for logarithms. If we now factor out R from each term: ΔmixS=−R(nAlnχA+nBlnχB) represents the equation for the entropy change of mixing. This equation is also commonly written with the total number of moles: ΔmixS=−nR(χAlnχA+χBlnχB)(1) where the total number of moles is n=nA+nB Notice that when the two gases will be mixed, their mole fraction will be less than one, making the term inside the parentheses negative, and thus the entropy of mixing will always be positive. This observation makes sense, because as you add a component to another for a two-component solution, the mole fraction of the other component will decrease, and the log of a number less than 1 is negative. Multiplied by the negative in the front of the equation gives a positive quantity. Equation 1 applies to both ideal solutions and ideal gases. References Chang, Raymond. Physical Chemistry for the Biosciences. Sausalito, California: University Science Books, 2005. Outside Links Sattar, Simeen. "Thermodynamics of Mixing Real Gases." J. Chem. Educ. 2000 77 1361. Ideal Systems Ideal Gas Processes
4819	https://ocw.mit.edu/courses/18-900-geometry-and-topology-in-the-plane-spring-2023/mit18_900s23_lec3.pdf	THE SHOELACE FORMULA AND THE WINDING NUMBER 23 3. The shoelace formula and the winding number This is the last of our three lectures on areas of polygons. We introduce a formula for the area of a polygon, in terms of the coordinates of its vertices. Then, we subject this formula to destructive testing: • we look at increasingly complicated examples, and finally try cases that are outside the domain of applicability of the formula, because they aren’t polygons (they have self-intersections); • that will lead to the notion of winding number: our first topological invariant. (3a) Some coordinate geometry. The length of a vector v = (x, y) ∈R2 is (3.1) ∥v∥= p x2 + y2. Given v1 = (x1, y1) and v2 = (x2, y2), their scalar product and cross product are the numbers v1 · v2 = x1x2 + y1y2, (3.2) v1 × v2 = x1y2 −y1x2 = det x1 x2 y1 y2 . (3.3) You may be familiar with the cross product in three-dimensional space (where the outcome is again a vector). The two-dimensional version (which produces a number) is not common notation, but we find it convenient here. Both products are linear in each entry (satisfy the distributive law): (v1 + v2) · v3 = v1 · v3 + v2 · v3, v1 · (v2 + v3) = v1 · v2 + v1 · v3, (3.4) (v1 + v2) × v3 = v1 × v3 + v2 × v3, v1 × (v2 + v3) = v1 × v2 + v1 × v3. (3.5) They are also (anti)symmetric: v1 · v2 = v2 · v1, (3.6) v1 × v2 = −(v2 × v1). (3.7) Geometrically, if ∢(v1, v2) is the angle formed by the vectors, v1 · v2 = ∥v1∥∥v2∥cos(∢(v1, v2)), (3.8) v1 × v2 = ∥v1∥∥v2∥sin(∢(v1, v2)). (3.9) If one of the vectors is zero, both products are zero, so we don’t have to think about what we mean by angle. For two nonzero vectors, the sign of ∢(v1, v2) is important for the second formula: turning from v1 to v2 in anticlockwise direction is measured by a positive angle, while turning clockwise is measured by a negative angle. Two vectors are linearly dependent exactly when v1 × v2 is zero. A basis (v1, v2), consisting of two linearly independent vectors, is called positively oriented if v1 × v2 > 0 (meaning that one goes from v1 to v2 by an anticlockwise turn with angle 24 I. POLYGONS between 0 and π), and negatively oriented if v1 × v2 < 0. (3.10) v1 v2 v1 v2 (v1, v2) positively oriented (v1, v2) negatively oriented (3b) The shoelace formula. Take a polygon P with n vertices, having coordinates (3.11) v0 = (x0, y0), v1 = (x1, y1), . . . , vn−1 = (xn−1, yn−1), vn = (xn, yn) = (x0, y0) = v0. We repeat one vertex (index 0 and index n are the same), since that is convenient for writing down formulae. The shoelace formula is (3.12) area(P) = 1 2 v0 × v1 + v1 × v2 + · · · + vn−1 × vn . This formula is easiest to understand if P is convex and the origin o = (0, 0) lies in its interior. If we assume that the ordering of the vertices is anticlockwise, then each 1 2(vk−1 × vk) is positive, and equals the area of the triangle with vertices (o, vk−1, vk). Adding up those numbers yields the area of P. If the ordering of the vertices is clockwise, the same holds with the opposite signs (in the end, taking the absolute value will cancel out that overall sign change). Example 3.1. The following polygon has area 14 (three triangles of area 3, and two of area 5/2): (3.13) (3, 0) (2, 2) (−1, 2) (−2, −1) (1, −2) area = 1 2(−1, 2) × (−2, −1) = 5/2 We’ll now start analyzing what the formula does in more general situations. First of all, it’s not necessary that the origin should lie in the interior of P, because the entire expression is unchanged under translation by any vector w: (3.14) (v0 + w) × (v1 + w) + (v1 + w) × (v2 + w) + · · · + (vn−1 + w) × (vn + w) = (v0 × v1 + v0 × w + w × v1) + (v1 × v2 + v1 × w + w × v2) + · · · = (v0 × v1 + v1 × v2 + · · · + vn−1 × vn) + (v0 + · · · + vn−1) × w + w × (v1 + · · · + vn) = v0 × v1 + v1 × v2 + · · · + vn−1 × vn. It is also not necessary that P should be convex: the shoelace formula still applies, because the terms partially cancel. Example 3.2. In the following case, one can see how two of the triangles have pieces lying outside P, but those contribute with opposite signs, which provides the required partial cancellation. There is a part of P (shaded more darkly) that lies in 3 triangles, but again, cancellation means that it 3. THE SHOELACE FORMULA AND THE WINDING NUMBER 25 is effectively only counted once. (3.15) (3, 0) (2, 2) (−2, 1) (1, −2) 1 2(2, 2) × (2, 1) = −1 1 2(2, 1) × (−2, 1) = 2 1 2(3, 0) × (2, 2) = 3 (2, 1) Next, let’s look at situations which are not polygons, just polygonal loops. By that, we mean that we are given points (3.11) where the coordinates are arbitrary: points can repeat, the edges may intersect or overlap, and so on. To make that clear, we change the notation, and write p for polygonal loops (as opposed to P for polygons). A polygonal loop doesn’t really have an “inside”, so while we can plug the coordinates into the shoelace formula, it’s not obvious what the output means! Example 3.3. Working through the example below triangle-by-triangle, we see that the shoelace formula (omitting the absolute value, for simplicity) yields: the area of the light gray shaded region, plus twice the area of the dark gray shaded region, minus the area of the black region. This is easiest to see for the black region, which is part of only one of the triangles, yielding a negative contribution. (3.16) 1 2(−5, 1) × (−5, 2) = −5/2 (−5, 2) (−5, 1) The outcome we’ve been looking for is this (without the absolute value, which is more of a hindrance than a help): Theorem 3.4. Take a polygonal loop p, with vertices (v0, v1, . . . , vn = v0). Then (3.17) 1 2 v0 × v1 + v1 × v2 + · · · + vn−1 × vn = X R area(R) wind(p, some point in R). Here, the sum is over all regions R into which p divides the plane, and wind(p, ·) are the winding numbers of p. (Formally, we include the outermost “unbounded region” R, but that won’t matter since its winding number is zero.) 26 I. POLYGONS (3c) Winding numbers. The formula (3.17) involves a new notion, that of winding number (3.18) wind(p, q) ∈Z, for p a polygonal loop, and q a point not lying on p. The name describes the intuition correctly: we stand at the point q, and turn our heads to watch a train moving once around p. The winding number is how many full turns we have done at the end, with counterclockwise turns counting as +1 and clockwise ones as −1. Example 3.3 had winding numbers 1 (light gray shaded part), 2 (dark gray), and −1 (black). Example 3.5. The star (3.19) 2 1 1 1 1 1 0 has winding number 2 on the innermost pentagon region, winding number 1 on the triangle regions, and of course 0 for the unbounded region. Example 3.6. The winding numbers of the following loop take values from −2 to 2. Note the existence of a bounded region with winding number 0: (3.20) 1 2 0 -1 -2 0 Let’s return to the initial situation of a polygon P. There, the absolute value of (3.17) computes the area of (the inside of) the polygon. That happens because (3.21) wind(P, q) = ( ±1 if q lies inside P, 0 if q lies outside P. Let’s take a step back. Since the beginning of this course, we have used the fact that a polygon divides the plane into two regions, the inside and outside. The formula (3.21) gives us a way to check which of those two regions a point belongs to. Turning this on its head, we can use that idea to give a rigorous proof of the fact that the inside and outside are distinct regions. A similar observation concerns the sign in (3.21). When we think of a polygon as given by a list of numbered vertices, we choose one way of going around it (clockwise or anticlockwise). Clockwise yields a sign of −1, and anticlockwise yields +1. Again, one can use this as a mathematical definition of “clockwise” and “anticlockwise”. MIT OpenCourseWare 18.900 Geometry and Topology in the Plane Spring 2023 For information about citing these materials or our Terms of Use, visit:
4820	https://www.softouch.on.ca/kb/data/Algebra%20and%20Trigonometry%204E.pdf	Algebra and Trigonometry [FOURTH EDITION] Cynthia Y. Young PROFESSOR OF MATHEMATICS University of Central Florida VICE PRESIDENT & DIRECTOR Laurie Rosatone ACQUISITIONS EDITOR Joanna Dingle ASSISTANT DEVELOPMENT EDITOR Ryann Dannelly EDITORIAL ASSISTANT Giana Milazzo MARKETING MANAGER John LaVacca III SENIOR PRODUCT DESIGNER David Dietz PRODUCT DESIGN MANAGER Thomas Kulesa FREELANCE PROJECT EDITOR Anne Scanlan-Rohrer PRODUCTION SERVICES Cenveo® Publisher Services SENIOR CONTENT MANAGER Valerie Zaborski SENIOR PRODUCTION EDITOR Ken Santor SENIOR DESIGNER Maureen Eide SENIOR PHOTO EDITOR Billy Ray COVER PHOTO Jupiter Images/Getty Images This book was set in 10/12 Times by Cenveo® Publisher Services, and printed and bound by Quad/Graphics, Inc. The cover was printed by Quad/Graphics, Inc. This book is printed on acid free paper. 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Upon completion of the review period, please return the evaluation copy to Wiley. Return instructions and a free of charge return mailing label are available at www.wiley.com/go/returnlabel. If you have chosen to adopt this textbook for use in your course, please accept this book as your complimentary desk copy. Outside of the United States, please contact your local sales representative. The inside back cover will contain printing identification and country of origin if omitted from this page. In addition, if the ISBN on the back cover differs from the ISBN on this page, the one on the back cover is correct. ISBN: 978-1-119-32086-9 (ePub) ISBN: 978-1-119-27345-5 (LLPC) Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 for Christopher and Caroline Cynthia Y. Young is the Pegasus Professor of Mathematics and the Vice Provost for Faculty Excellence and UCF Global at the University of Central Florida (UCF) and the author of College Algebra, Trigonometry, Algebra and Trigonometry, and Precalculus. She holds a BA in Secondary Mathematics Education from the University of North Carolina (Chapel Hill), an MS in Mathematical Sciences from UCF, and both an MS in Electrical Engineering and a PhD in Applied Mathematics from the University of Washington. She has taught high school in North Carolina and Florida, developmental mathematics at Shoreline Community College in Washington, and undergraduate and graduate students at UCF. Dr. Young joined the faculty at UCF in 1997 as an assistant professor of mathematics, and her primary research area was the mathematical modeling of the atmospheric effects on propagating laser beams. Her atmospheric propagation research was recognized by the Office of Naval Research Young Investigator Award, and in 2017 she was selected as a fellow of the International Society for Optical Engineering. Her secondary area of research centers on improvement of student learning in mathematics. She has authored or co-authored over 60 books and articles and has served as the principal investigator or co-principal investigator on projects with more than $2.5 million in federal funding. Dr. Young was on the team at UCF that developed the UCF EXCEL program, which was originally funded by the National Science Foundation to support the increase in the number of students graduating with a degree in science, technology, engineering, and mathematics (STEM). The EXCEL learning community approach centered around core mathematics courses has resulted in a significant increase in STEM graduation rates and has been institutionalized at UCF. Dr. Young has been the recipient of many of UCF’s awards (Excellence in Undergraduate Teaching, Excellence in Research, Teaching Incentive Program, Research Incentive Program, Scholarship of Teaching and Learning award, and UCF’s highest honor, UCF Pegasus Professor). She has shared her techniques and experiences with colleagues around the country through talks at colleges, universities, and conferences. About the Author VI University of Central Florida University of Central Florida As a mathematics professor, I would hear my students say, “I understand you in class, but when I get home I am lost.” When I would probe further, students would continue with “I can’t read the book.” As a mathematician, I always found mathematics textbooks quite easy to read—and then it dawned on me: Don’t look at this book through a mathematician’s eyes; look at it through the eyes of students who might not view mathematics the same way that I do. What I found was that the books were not at all like my class. Students understood me in class, but when they got home they couldn’t understand the book. It was then that the folks at Wiley lured me into writing. My goal was to write a book that is seamless with how we teach and is an ally (not an adversary) to student learning. I wanted to give students a book they could read without sacrificing the rigor needed for conceptual understanding. The following quote comes from a reviewer when asked about the rigor of the book: I would say that this text comes across as a little less rigorous than other texts, but I think that stems from how easy it is to read and how clear the author is. When one actually looks closely at the material, the level of rigor is high. DISTINGUISHING FEATURES Four key features distinguish this book from others, and they came directly from my classroom. PARALLEL WORDS AND MATH Have you ever looked at your students’ notes? I found that my students were only scribbling down the mathematics that I would write—never the words that I would say in class. I started passing out handouts that had two columns: one column for math and one column for words. Each example would have one or the other; either the words were there and students had to fill in the math or the math was there and students had to fill in the words. If you look at the examples in this book, you will see that the words (your voice) are on the left and the mathematics is on the right. In most math books, when the author illustrates an example, the mathematics is usually down the center of the page, and if the students don’t know what mathematical operation was performed, they will look to the right for some brief statement of help. That’s not how we teach; we don’t write out an example on the board and then say, “Class, guess what I just did!” Instead we lead our students, telling them what step is coming and then performing that mathematical step together—and reading naturally from left to right. Student reviewers have said that the examples in this book are easy to read; that’s because your voice is right there with them, working through problems together. Preface VII SKILLS AND CONCEPTS (LEARNING OBJECTIVES AND EXERCISES) In my experience as a mathematics teacher/instructor/professor, I find skills to be on the micro level and concepts on the macro level of understanding mathematics. I believe that too often skills are emphasized at the expense of conceptual understanding. I have purposely separated learning objectives at the beginning of every section into two categories: skills objectives—what students should be able to do—and conceptual objectives—what students should understand. At the beginning of every class, I discuss the learning objectives for the day—both skills and concepts. These are reinforced with both skills exercises and conceptual exercises. Each subsection has a corresponding skills objective and conceptual objective. CATCH THE MISTAKE Have you ever made a mistake (or had a student bring you his or her homework with a mistake) and you’ve gone over it and over it and couldn’t find the mistake? It’s often easier to simply take out a new sheet of paper and solve it from scratch than it is to actually find the mistake. Finding the mistake demonstrates a higher level of understanding. I include a few Catch the Mistake exercises in each section that demonstrate a common mistake. Using these in class (with individuals or groups) leads to student discussion and offers an opportunity for formative assessment in real time. LECTURE VIDEOS BY THE AUTHOR I authored the videos to ensure consistency in the students’ learning experience. Throughout the book, wherever a student sees the video icon, that indicates a video. These videos provide mini lectures. The chapter openers and chapter summaries act as class discussions. The “Your Turn” problems throughout the book challenge the students to attempt a problem similar to a nearby example. The “worked-out example” videos are intended to come to the rescue for students if they get lost as they read the text and work problems outside the classroom. VIII PREFACE NEW TO THE FOURTH EDITION In the fourth edition, the main upgrades are updated applications throughout the text; new Skills and Conceptual objectives mapped to each subsection; new Concept Check questions in each subsection; and the substantially improved version of WileyPLUS, including ORION adaptive practice and interactive animations. PREFACE IX CONCEPT CHECK QUESTIONS SKILLS AND CONCEPTUAL OBJECTIVES APPLICATIONS TO BUSINESS, ECONOMICS, HEATH SCIENCES, AND MEDICINE X PREFACE FEATURE BENEFIT TO STUDENT Chapter-Opening Vignette Piques the student’s interest with a real-world application of material presented in the chapter. Later in the chapter, the concept from the vignette is reinforced. Chapter Overview, Flowchart, and Learning Objectives Allows students to see the big picture of how topics relate, and overarching learning objectives are presented. Skills and Conceptual Objectives Skills objectives represent what students should be able to do. Conceptual objectives emphasize a higher-level, global perspective of concepts. Clear, Concise, and Inviting Writing Style, Tone, and Layout Enables students to understand what they are reading, which reduces math anxiety and promotes student success. Parallel Words and Math Increases students’ ability to read and understand examples with a seamless representation of their instructor’s class (instructor’s voice and what they would write on the board). Common Mistakes Addresses a different learning style: teaching by counterexample. Demonstrates common mistakes so that students understand why a step is incorrect and reinforces the correct mathematics. Color for Pedagogical Reasons Particularly helpful for visual learners when they see a function written in red and then its corresponding graph in red or a function written in blue and then its corresponding graph in blue. Study Tips Reinforces specific notes that you would want to emphasize in class. Author Videos Gives students a mini class of several examples worked by the author. Your Turn Engages students during class, builds student confidence, and assists instructor in real-time assessment. Concept Checks Reinforces concept learning objectives, much as Your Turn features reinforce skill learning objectives. Catch the Mistake Exercises Encourages students to assume the role of teacher—demonstrating a higher mastery level. Conceptual Exercises Teaches students to think more globally about a topic. Inquiry-Based Learning Project (online only) Lets students discover a mathematical identity, formula, and the like that is derived in the book. Modeling Our World (online only) Engages students in a modeling project of a timely subject: global climate change. Chapter Review Presents key ideas and formulas section by section in a chart. Improves study skills. Chapter Review Exercises Improves study skills. Chapter Practice Test Offers self-assessment and improves study skills. Cumulative Test Improves retention. PREFACE XI INSTRUCTOR SUPPLEMENTS INSTRUCTOR’S SOLUTIONS MANUAL (ISBN: 978-1-119-27343-1) • Contains worked-out solutions to all exercises in the text. INSTRUCTOR’S MANUAL Authored by Cynthia Young, the manual provides practical advice on teaching with the text, including: • sample lesson plans and homework assignments • suggestions for the effective utilization of additional resources and supplements • sample syllabi • Cynthia Young’s Top 10 Teaching Tips & Tricks • online component featuring the author presenting these Tips & Tricks ANNOTATED INSTRUCTOR’S EDITION (ISBN: 978-1-119-27346-2) • Displays answers to the vast majority of exercise questions in the back of the book. • Provides additional classroom examples within the standard difficulty range of the in-text exercises, as well as challenge problems to assess your students’ mastery of the material. POWERPOINT SLIDES • For each section of the book, a corresponding set of lecture notes and worked- out examples are presented as PowerPoint slides, available on the Book Companion Site (www.wiley.com/college/young) and WileyPLUS. TEST BANK • Contains approximately 900 questions and answers from every section of the text. COMPUTERIZED TEST BANK Electonically enhanced version of the Test Bank that • contains approximately 900 algorithmically generated questions. • allows instructors to freely edit, randomize, and create questions. • allows instructors to create and print different versions of a quiz or exam. • recognizes symbolic notation. BOOK COMPANION WEBSITE (WWW.WILEY.COM/COLLEGE/YOUNG) • Contains all instructor supplements listed plus a selection of personal response system questions. WILEYPLUS • WileyPLUS online homework features a full-service, digital learning environment, including additional resources for students, such as lecture videos by the author, self-practice exercises, tutorials, integrated links between the online text and supplements, and new interactive animations and ORION adaptive practice. • WileyPLUS has been substantially revised and improved since the third edition of Algebra and Trigonometry. It now includes ORION, an adaptive practice engine built directly into WileyPLUS that can connect directly into the WileyPLUS gradebook, or into your campus Learning Management System gradebook if you select that option. Wiley has been incorporating ORION into WileyPLUS courses for over five years, including the Young Precalculus program. ORION brings the power of adaptive learning, which will continue to help students and instructors “bridge the gap.” XII PREFACE STUDENT SUPPLEMENTS STUDENT SOLUTIONS MANUAL (ISBN: 978-1-119-27342-4) • Includes worked-out solutions for all odd problems in the text. BOOK COMPANION WEBSITE (WWW.WILEY.COM/COLLEGE/YOUNG) • Provides additional resources for students to enhance the learning experience. PREFACE XIII ACKNOWLEDGMENTS I want to express my sincerest gratitude to the entire Wiley team. I’ve said this before, and I will say it again: Wiley is the right partner for me. There is a reason that my dog is named Wiley—she’s smart, competitive, a team player, and most of all, a joy to be around. There are several people within Wiley to whom I feel the need to express my appreciation: first and foremost to Laurie Rosatone, who convinced Wiley Higher Ed to invest in a young assistant professor’s vision for a series and who has been unwavering in her commitment to student learning. To my editor Joanna Dingle, whose judgment I trust in both editorial and preschool decisions; thank you for surpassing my greatest expectations for an editor. To the rest of the math editorial team (Jennifer Lartz, Anne Scanlan-Rohrer, and Ryann Dannelly), you are all first class! This revision was planned and executed exceptionally well thanks to you. To the math marketing manager John LaVacca, thank you for helping reps tell my story: you are outstanding at your job. To product designer David Dietz, many thanks for your role in developing the online course and digital assets. To Mary Sanger, thank you for your attention to detail. To Maureen Eide, thank you for the new design! And finally, I’d like to thank all of the Wiley reps: thank you for your commitment to my series and your tremendous efforts to get professors to adopt this book for their students. I would also like to thank all of the contributors who helped us make this an even better book. I’d first like to thank Mark McKibben. He is known as the author of the solutions manuals that accompany this series, but he is much more than that. Mark, thank you for making this series a priority, for being so responsive, and most of all for being my “go-to” person to think through ideas. I’d also like to thank Jodi B.A. McKibben, who is a statistician and teamed with Mark to develop the regression material. I’d like to thank Steve Davis, who was the inspiration for the Inquiry-Based Learning Projects (IBLPs) and a huge thanks to Lyn Riverstone, who developed all of the IBLPs. Special thanks to Laura Watkins for finding applications that are real and timely and to Ricki Alexander for updating all of the Technology Tips. I’d also like to thank Becky Schantz for her environmental problems (I now use AusPens because of Becky). Many thanks to Marie Vanisko, Jennifer Blue, and Diane Cook for accuracy checking of the text, exercises, and solutions. I’d also like to thank the following reviewers, whose input helped make this book even better. Text Reviewers Sandy Carlson, Pennsylvania College of Technology Ze-Li Dou, Texas Christian University Deborah Lynn Doucette, Erie Community College – North Campus Scott Gentile, Hunter College, CUNY Mike Kirby, Tidewater Community College Bridgett Lee, Georgia Southern University Lebbie Lee Ligon, Piedmont Technical College Kelly Pearson, Murray State University Colleen Quinn, Erie Community College Rachel Rader, Ohio Northern University Sean Smith, California State Polytechnic University, Pomona Animations Reviewers Monika H. Champion, Ivy Tech Community College Kim Christensen, Metropolitan Community College Yamir DeJesus-Decena, Dutchess Community College Marissa Ford, Ivy Tech Community College Barbara Hess, California University of Pennsylvania Christina K. Houston, Community College of Allegheny County Phyllis Lefton, Manhattanville College Carrie McCammon, Ivy Tech Community College Holly J. Middleton, University of North Florida Becky Moening, Ivy Tech Community College, Warsaw Denise Race, Eastfield College Edward Schwartz, Manhattanville College Mike Shirazi, Germanna Community College Misty Vorder, Hillsborough Community College, Brandon Contents Prerequisites and Review 2 0.1 REAL NUMBERS 4 0.2 INTEGER EXPONENTS AND SCIENTIFIC NOTATION 17 0.3 POLYNOMIALS: BASIC OPERATIONS 27 0.4 FACTORING POLYNOMIALS 36 0.5 RATIONAL EXPRESSIONS 46 0.6 RATIONAL EXPONENTS AND RADICALS 60 0.7 COMPLEX NUMBERS 70 Review 77 \| Review Exercises 79 \| Practice Test 81 Equations and Inequalities 82 1.1 LINEAR EQUATIONS 84 1.2 APPLICATIONS INVOLVING LINEAR EQUATIONS 93 1.3 QUADRATIC EQUATIONS 107 1.4 OTHER TYPES OF EQUATIONS 121 1.5 LINEAR INEQUALITIES 129 1.6 POLYNOMIAL AND RATIONAL INEQUALITIES 139 1.7 ABSOLUTE VALUE EQUATIONS AND INEQUALITIES 149 Review 157 \| Review Exercises 158 \| Practice Test 162 \| Cumulative Test 163 Graphs 164 2.1 BASIC TOOLS: CARTESIAN PLANE, DISTANCE, AND MIDPOINT 166 2.2 GRAPHING EQUATIONS, POINT-PLOTTING, INTERCEPTS, AND SYMMETRY 173 2.3 LINES 186 2.4 CIRCLES 202 2.5 LINEAR REGRESSION: BEST FIT 209 Review 231 \| Review Exercises 232 \| Practice Test 234 \| Cumulative Test 235 Functions and Their Graphs 236 3.1 FUNCTIONS 238 3.2 GRAPHS OF FUNCTIONS; PIECEWISE-DEFINED FUNCTIONS; INCREASING AND DECREASING FUNCTIONS; AVERAGE RATE OF CHANGE 255 3.3 GRAPHING TECHNIQUES: TRANSFORMATIONS 274 3.4 OPERATIONS ON FUNCTIONS AND COMPOSITION OF FUNCTIONS 287 3.5 ONE-TO-ONE FUNCTIONS AND INVERSE FUNCTIONS 298 3.6 MODELING FUNCTIONS USING VARIATION 312 Review 322 \| Review Exercises 324 \| Practice Test 328 \| Cumulative Test 329 fotografxx/Getty Images Hero Images/Getty Images, Inc. Fuse/Getty Images SuperStock/Alamy Stock Photo XIV CONTENTS XV Polynomial and Rational Functions 330 4.1 QUADRATIC FUNCTIONS 332 4.2 POLYNOMIAL FUNCTIONS OF HIGHER DEGREE 349 4.3 DIVIDING POLYNOMIALS: LONG DIVISION AND SYNTHETIC DIVISION 363 4.4 THE REAL ZEROS OF A POLYNOMIAL FUNCTION 372 4.5 COMPLEX ZEROS: THE FUNDAMENTAL THEOREM OF ALGEBRA 388 4.6 RATIONAL FUNCTIONS 396 Review 416 \| Review Exercises 418 \| Practice Test 422 \| Cumulative Test 423 Exponential and Logarithmic Functions 424 5.1 EXPONENTIAL FUNCTIONS AND THEIR GRAPHS 426 5.2 LOGARITHMIC FUNCTIONS AND THEIR GRAPHS 440 5.3 PROPERTIES OF LOGARITHMS 456 5.4 EXPONENTIAL AND LOGARITHMIC EQUATIONS 464 5.5 EXPONENTIAL AND LOGARITHMIC MODELS 474 Review 485 \| Review Exercises 487 \| Practice Test 490 \| Cumulative Test 491 Trigonometric Functions 492 6.1 ANGLES, DEGREES, AND TRIANGLES 494 6.2 DEFINITION 1 OF TRIGONOMETRIC FUNCTIONS: RIGHT TRIANGLE RATIOS 511 6.3 APPLICATIONS OF RIGHT TRIANGLE TRIGONOMETRY: SOLVING RIGHT TRIANGLES 525 6.4 DEFINITION 2 OF TRIGONOMETRIC FUNCTIONS: CARTESIAN PLANE 538 6.5 TRIGONOMETRIC FUNCTIONS OF NONACUTE ANGLES 552 6.6 RADIAN MEASURE AND APPLICATIONS 567 6.7 DEFINITION 3 OF TRIGONOMETRIC FUNCTIONS: UNIT CIRCLE APPROACH 584 6.8 GRAPHS OF SINE AND COSINE FUNCTIONS 594 6.9 GRAPHS OF OTHER TRIGONOMETRIC FUNCTIONS 623 Review 641 \| Review Exercises 647 \| Practice Test 650 \| Cumulative Test 651 Analytic Trigonometry 652 7.1 BASIC TRIGONOMETRIC IDENTITIES 654 7.2 VERIFYING TRIGONOMETRIC IDENTITIES 665 7.3 SUM AND DIFFERENCE IDENTITIES 675 7.4 DOUBLE-ANGLE IDENTITIES 689 7.5 HALF-ANGLE IDENTITIES 697 7.6 PRODUCT-TO-SUM AND SUM-TO-PRODUCT IDENTITIES 708 7.7 INVERSE TRIGONOMETRIC FUNCTIONS 716 7.8 TRIGONOMETRIC EQUATIONS 737 Review 754 \| Review Exercises 757 \| Practice Test 762 \| Cumulative Test 763 Focus On Sport/Getty Images, Inc. Francois Gohier/Science Source Richard Cummins / Getty Images, Inc. laurentiu iordache / Alamy Stock Photo XVI CONTENTS Additional Topics in Trigonometry 764 8.1 OBLIQUE TRIANGLES AND THE LAW OF SINES 766 8.2 THE LAW OF COSINES 780 8.3 THE AREA OF A TRIANGLE 791 8.4 VECTORS 798 8.5 THE DOT PRODUCT 813 8.6 POLAR (TRIGONOMETRIC) FORM OF COMPLEX NUMBERS 821 8.7 PRODUCTS, QUOTIENTS, POWERS, AND ROOTS OF COMPLEX NUMBERS; DE MOIVRE’S THEOREM 829 8.8 POLAR EQUATIONS AND GRAPHS 841 Review 856 \| Review Exercises 859 \| Practice Test 862 \| Cumulative Test 863 Systems of Linear Equations and Inequalities 864 9.1 SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 866 9.2 SYSTEMS OF LINEAR EQUATIONS IN THREE VARIABLES 881 9.3 PARTIAL FRACTIONS 892 9.4 SYSTEMS OF LINEAR INEQUALITIES IN TWO VARIABLES 903 9.5 THE LINEAR PROGRAMMING MODEL 914 Review 921 \| Review Exercises 922 \| Practice Test 924 \| Cumulative Test 925 Matrices 926 10.1 MATRICES AND SYSTEMS OF LINEAR EQUATIONS 928 10.2 MATRIX ALGEBRA 950 10.3 MATRIX EQUATIONS; THE INVERSE OF A SQUARE MATRIX 964 10.4 THE DETERMINANT OF A SQUARE MATRIX AND CRAMER’S RULE 976 Review 990 \| Review Exercises 993 \| Practice Test 996 \| Cumulative Test 997 Bermuda Miami, Florida Atlantic Ocean San Juan, Puerto Rico PAUL J. RICHARDS/ Getty Images, Inc. decodes sends sends decodes visitor web server encrypts receives encrypts receives CONTENTS XVII Analytic Geometry and Systems of Nonlinear Equations and Inequalities 998 11.1 CONIC BASICS 1000 11.2 THE PARABOLA 1003 11.3 THE ELLIPSE 1015 11.4 THE HYPERBOLA 1028 11.5 SYSTEMS OF NONLINEAR EQUATIONS 1039 11.6 SYSTEMS OF NONLINEAR INEQUALITIES 1049 11.7 ROTATION OF AXES 1057 11.8 POLAR EQUATIONS OF CONICS 1067 11.9 PARAMETRIC EQUATIONS AND GRAPHS 1078 Review 1086 \| Review Exercises 1089 \| Practice Test 1093 \| Cumulative Test 1095 Sequences, Series, and Probability 1096 12.1 SEQUENCES AND SERIES 1098 12.2 ARITHMETIC SEQUENCES AND SERIES 1109 12.3 GEOMETRIC SEQUENCES AND SERIES 1117 12.4 MATHEMATICAL INDUCTION 1128 12.5 THE BINOMIAL THEOREM 1133 12.6 COUNTING, PERMUTATIONS, AND COMBINATIONS 1141 12.7 PROBABILITY 1151 Review 1160 \| Review Exercises 1162 \| Practice Test 1166 \| Cumulative Test 1167 Answers to Odd-Numbered Exercises 1169 Applications Index 1261 Subject Index 1265 Ribeiro antonio/Shutterstock 2 3 4 5 6 Dealer’s Up Card A BASIC STRATEGY FOR BLACKJACK When surrender is allowed, surrender 9, 7 or 10, 6 vs 9, 10, A; 9, 6 or 10, 5 vs 10 When doubling down after splitting is allowed, split: 2’s, 3’s, 7’s vs 2-7; 4’s vs 5 or 6; 6’s vs 2-6 7 8 9 10 A S S S S S S S S S S S S S S S H H H H H S S S S S H H H H H S S S S S H H H H H S S S S S H H H H H H H S S S H H H H H D D D D D D D D D H D D D D D D D D H H H D D D D H H H H H H H H H H H H H H H S S S S S S S S S S S D D D D S S H H H H D D D D H H H H H H H D D D H H H H H H H D D D H H H H H H H H D D H H H H H H H H D D H H H H H SP SP SP SP SP SP SP SP SP SP S S S S S S S S S S SP SP SP SP SP S SP SP S S SP SP SP SP SP SP H H H H H SP SP SP SP H H H H H D D D D D D D D H H H H H H H H H H H H H H SP SP SP SP H H H H H 17+ 16 15 14 13 12 11 10 9 5 - 8 A, 8 - 10 A, 7 A, 6 A, 5 A, 4 A, 3 A, 2 A, A; 8, 8 10, 10 9, 9 7, 7 I II III Your Hand IV 6, 6 5, 5 4, 4 3, 3 2, 2 H SP SP SP SP H H H HIT STAND DOUBLE DOWN SPLIT H I wrote this text with careful attention to ways in which to make your learning experience more successful. If you take full advantage of the unique features and elements of this textbook, I believe your experience will be fulfilling and enjoyable. Let’s walk through some of the special book features that will help you in your study of College Algebra. A Note from the Author TO THE STUDENT XVIII Skills and Conceptual Objectives For every section, objectives are further divided by skills and concepts so you can see the difference between solving problems and truly understanding concepts. Chapter Introduction, Flowchart, Section Headings, and Objectives An opening vignette, flowchart, list of chapter sections, and chapter learning objectives give you an overview of the chapter. Prerequisites and Review [Chapter 0] A comprehensive review of prerequisite knowledge (intermediate algebra topics) in Chapter 0 provides a brushup on knowledge and skills necessary for success in the course. Clear, Concise, and Inviting Writing Special attention has been paid to presenting an engaging, clear, precise narrative in a layout that is easy to use and designed to reduce any math anxiety you may have. A NOTE FROM THE AUTHOR TO THE STUDENT XIX Study Tips and Caution Notes These marginal reminders call out important hints or warnings to be aware of related to the topic or problem. Parallel Words and Math This text reverses the common textbook presentation of examples by placing the explanation in words on the left and the mathematics in parallel on the right. This makes it easier to read through examples as the material flows more naturally from left to right and as commonly presented in class. Common Mistake/Correct Versus Incorrect In addition to standard examples, some problems are worked out both correctly and incorrectly to highlight common errors. Counterexamples like these are often an effective learning approach. Examples Examples pose a specific problem using concepts already presented and then work through the solution. These serve to enhance your understanding of the subject matter. Your Turn Immediately following many examples, you are given a similar problem to reinforce and check your understanding. This helps build confidence as you progress in the chapter. These are ideal for in-class activity or for preparing for homework later. Answers are provided in the margin for a quick check of your work. XX A NOTE FROM THE AUTHOR TO THE STUDENT Six Types of Exercises Every text section ends with Skills, Applications, Catch the Mistake, Conceptual, Challenge, and Technology exercises. The exercises gradually increase in difficulty and vary in skill and conceptual emphasis. Catch the Mistake exercises increase the depth of understanding and reinforce what you have learned. Conceptual and Challenge exercises specifically focus on assessing conceptual understanding. Technology exercises enhance your understanding and ability using scientific and graphing calculators. Video Icons Video icons appear on all chapter introductions, chapter and section reviews, as well as selected examples throughout the chapter to indicate that the author has created a video segment for that element. These video clips help you work through the selected examples with the author as your “private tutor.” A NOTE FROM THE AUTHOR TO THE STUDENT 1 Chapter Review, Review Exercises, Practice Test, Cumulative Test At the end of every chapter, a summary review chart organizes the key learning concepts in an easy-to-use one- or two-page layout. This feature includes key ideas and formulas, as well as indicating relevant pages and review exercises so that you can quickly summarize a chapter and study smarter. Review Exercises, arranged by section heading, are provided for extra study and practice. A Practice Test, without section headings, offers even more self-practice before moving on. A new Cumulative Test feature offers study questions based on all previous chapters’ content, thus helping you build upon previously learned concepts. Concept Check Questions Similar to how Your Turn features reinforce skill learning objectives, Concept Checks reinforce concept learning objectives. C H A P T E R LEARNING OBJECTIVES [ [ ■ ■Understand that rational and irrational numbers together constitute the real numbers. ■ ■Apply properties of exponents. ■ ■Perform operations on polynomials. ■ ■Factor polynomials. ■ ■Simplify expressions that contain rational exponents. ■ ■Simplify radicals. ■ ■Write complex numbers in standard form. Would you be able to walk successfully along a tightrope? Most people probably would say no because the foundation is “shaky.” Would you be able to walk successfully along a beam (4 inches wide)? Most people would probably say yes—even though for some of us it is still challenging. Think of this chapter as the foundation for your walk. The more solid your foundation is now, the more successful your walk through College Algebra will be. The purpose of this chapter is to review concepts and skills that you already have learned in a previous course. Mathematics is a cumulative subject in that it requires a solid foundation to proceed to the next level. Use this chapter to reaffirm your current knowledge base before jumping into the course. Prerequisites and Review 0 kasto80/Getty Images, Inc. fotografxx/Getty Images 3 [I N T HI S CHAPTER] Real numbers, integer exponents, and scientific notation will be discussed, followed by rational exponents and radicals. Simplification of radicals and rationalization of denominators will be reviewed. Basic operations such as addition, subtraction, and multiplication of polynomials will be discussed followed by a review of how to factor polynomials. Rational expressions will be discussed, and a brief overview of solving simple algebraic equations will be given. After reviewing all of these aspects of real numbers, this chapter will conclude with a review of complex numbers. PREREQUISITES AND REVIEW 0.1 REAL NUMBERS 0.2 INTEGER EXPONENTS AND SCIENTIFIC NOTATION 0.3 POLYNOMIALS: BASIC OPERATIONS 0.4 FACTORING POLYNOMIALS 0.5 RATIONAL EXPRESSIONS 0.6 RATIONAL EXPONENTS AND RADICALS 0.7 COMPLEX NUMBERS • The Set of Real Numbers • Approxima tions: Round ing and Truncation • Order of Operations • Properties of Real Numbers • Integer Exponents • Scientific Notation • Adding and Subtracting Polynomials • Multiplying Polynomials • Special Products • Greatest Common Factor • Factoring Formulas: Special Polynomial Forms • Factoring a Trinomial as a Product of Two Binomials • Factoring by Grouping • A Strategy for Factoring Polynomials • Rational Expressions and Domain Restrictions • Simplifying Rational Expressions • Multiplying and Dividing Rational Expressions • Adding and Subtracting Rational Expressions • Complex Rational Expressions • Square Roots • Other (nth) Roots • Rational Exponents • The Imaginary Unit, i • Adding and Subtracting Complex Numbers • Multiplying Complex Numbers • Dividing Complex Numbers • Raising Complex Numbers to Integer Powers 4 CHAPTER 0 Prerequisites and Review 0.1.1 The Set of Real Numbers A set is a group or collection of objects that are called members or elements of the set. If every member of set B is also a member of set A, then we say B is a subset of A and denote it as B ( A. For example, the starting lineup on a baseball team is a subset of the entire team. The set of natural numbers, 51, 2, 3, 4, . . .6, is a subset of the set of whole numbers, 50, 1, 2, 3, 4, . . .6, which is a subset of the set of integers, 5. . . , 24, 23, 22, 21, 0, 1, 2, 3, . . .6, which is a subset of the set of rational numbers, which is a subset of the set of real numbers. The three dots, called an ellipsis, indicate that the pattern continues indefinitely. If a set has no elements, it is called the empty set, or null set, and is denoted by the symbol [. The set of real numbers consists of two main subsets: rational and irrational numbers. DEFINITION Rational Number A rational number is a number that can be expressed as a quotient (ratio) of two integers, a b ,where the integer a is called the numerator and the integer b is called the denominator and where b 2 0. Rational numbers include all integers or all fractions that are ratios of integers. Note that any integer can be written as a ratio whose denominator is equal to 1. In decimal form, the rational numbers are those that terminate or are nonterminating with a repeated decimal pattern, which is represented with an overbar. Those decimals that do not repeat and do not terminate are irrational numbers. The numbers 5, 217, 1 3, !2, p, 1.37, 0, 2 19 17, 3.666, 3.2179 . . . are examples of real numbers, where 5, 217, 1 3, 1.37, 0, 2 19 17 , and 3.666 are rational numbers, and !2, p, and 3.2179 . . . are irrational numbers. It is important to note that the ellipsis following the last decimal digit denotes continuing in an irregular fashion, whereas the absence of such dots to the right of the last decimal digit implies that the decimal expansion terminates. RATIONAL NUMBER (FRACTION) CALCULATOR DISPLAY DECIMAL REPRESENTATION DESCRIPTION 7 2 3.5 3.5 Terminates 15 12 1.25 1.25 Terminates 2 3 0.666666666 0.6 Repeats 1 11 0.09090909 0.09 Repeats 0.1.1 S KILL Classify real numbers as rational or irrational. 0.1.1 C ON CEPTUAL Understand that rational and irrational numbers are mutually exclusive and complementary subsets of real numbers. S K I L L S O B J E C T I V E S ■ ■Classify real numbers as rational or irrational. ■ ■Round or truncate real numbers. ■ ■Simplify expressions and evaluate algebraic expressions using the correct order of operations. ■ ■Apply properties of real numbers and basic rules of algebra in simplifying and evaluating expressions. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that rational and irrational numbers are mutually exclusive and complementary subsets of real numbers. ■ ■Understand the difference between rounding and truncating decimal values and that the resulting approximations may or may not be equal. ■ ■Learn the order of operations for real numbers. ■ ■Know and understand the basic properties of real numbers and the basic rules of algebra. 0.1 REAL NUMBERS 0.1 Real Numbers 5 Notice that the overbar covers the entire repeating pattern. The following figure and table illustrate the subset relationship and examples of different types of real numbers. Real Numbers Irrational Numbers Integers Negative Counting Numbers Zero Rational Numbers Fractions Ratio of two nonzero integers that does not reduce to an integer Whole Numbers Natural Numbers SYMBOL NAME DESCRIPTION EXAMPLES N Natural numbers Counting numbers 1, 2, 3, 4, 5, . . . W Whole numbers Natural numbers and zero 0, 1, 2, 3, 4, 5, . . . Z Integers Whole numbers and negative natural numbers . . . , 25, 24, 23, 22, 21, 0, 1, 2, 3, 4, 5, . . . Q Rational numbers Ratios of integers: a b 1b 2 0 2 • Decimal representation terminates, or • Decimal representation repeats 217, 219 7 , 0, 1 3, 1.37, 3.666 , 5 I Irrational numbers Numbers whose decimal representation does not terminate or repeat !2, 1.2179. . . , p R Real numbers Rational and irrational numbers p, 5, 22 3, 17.25, !7 Since the set of real numbers can be formed by combining the set of rational numbers and the set of irrational numbers, then every real number is either rational or irrational. The set of rational numbers and the set of irrational numbers are mutually exclusive (no shared elements) and complementary sets. The real number line is a graph used to represent the set of all real numbers. –3 –2 –1 1 0 2 3 19 17 – √2 EXAMPLE 1 Classifying Real Numbers Classify the following real numbers as rational or irrational: 23, 0, 1 4, !3, p, 7.51, 1 3, 28 5, 6.66666 Solution: Rational: 23, 0, 1 4, 7.51, 1 3, 28 5, 26.66666 Irrational: !3, p Y OUR T UR N Classify the following real numbers as rational or irrational: 27 3, 5.9999, 12, 0, 25.27, !5, 2.010010001 . . . [CONCEPT CHECK] TRUE OR FALSE All integers are rational numbers. ANSWER True ▼ ▼ A N S W E R Rational: 27 3, 5.9999, 12, 0, 25.27 Irrational: !5 , 2.010010001 . . . ▼ STUDY TIP Every real number is either a rational number or an irrational number. 6 CHAPTER 0 Prerequisites and Review 0.1.2 Approximations: Rounding and Truncation Every real number can be represented by a decimal. When a real number is in decimal form, it can be approximated by either rounding off or truncating to a given decimal place. Truncation is “cutting off” or eliminating everything to the right of a certain decimal place. Rounding means looking to the right of the specified decimal place and making a judgment. If the digit to the right is greater than or equal to 5, then the specified digit is rounded up, or increased by one unit. If the digit to the right is less than 5, then the specified digit stays the same. In both of these cases all decimal places to the right of the specified place are removed. EXAMPLE 2 Approximating Decimals to Two Places Approximate 17.368204 to two decimal places by a. truncation b. rounding Solution: a. To truncate, eliminate all digits to the right of the 6. 17.36 b. To round, look to the right of the 6. Because “8” is greater than 5, round up (add 1 to the 6). 17.37 Y OUR TU R N Approximate 23.02492 to two decimal places by a. truncation b. rounding ▼ EXAMPLE 3 Approximating Decimals to Four Places Approximate 7.293516 to four decimal places by a. truncation b. rounding Solution: The “5” is in the fourth decimal place. a. To truncate, eliminate all digits to the right of 5. 7.2935 b. To round, look to the right of the 5. Because “1” is less than 5, the 5 remains the same. 7.2935 Y OUR TU R N Approximate 22.381865 to four decimal places by a. truncation b. rounding ▼ 0.1.2 SKI LL Round or truncate real numbers. 0.1.2 CO NCE PTUAL Understand the difference between rounding and truncating decimal values and know that the resulting approximations may or may not be equal. ▼ A N S W E R a. Truncation: 22.3818 b. Rounding: 22.3819 ▼ A N S W E R a. Truncation: 23.02 b. Rounding: 23.02 It is important to note that rounding and truncation sometimes yield the same approx-imation (Example 3), but not always (Example 2). 0.1.3 Order of Operations Addition, subtraction, multiplication, and division are called arithmetic operations. The results of these operations are called the sum, difference, product, and quotient, respectively. These four operations are summarized in the following table. OPERATION NOTATION RESULT Addition a 1 b Sum Subtraction a 2 b Difference Multiplication a # b or ab or 1a2 1b2 Product Division a b or a /b 1b 2 02 Quotient (Ratio) [CONCEPT CHECK] TRUE OR FALSE Truncating and rounding always have the same result. ANSWER False ▼ STUDY TIP When rounding, look to the right of the specified decimal place and use that digit (do not round that digit first). 5.23491 rounded to two decimal places is 5.23 (do not round the 4 to a 5 first). Since algebra involves variables such as x, the traditional multiplication sign 3 is not used. Three alternatives are shown in the preceding table. Similarly, the arithmetic sign for division 4 is often represented by vertical or slanted fractions. The symbol 5 is called the equal sign and is pronounced “equals” or “is.” It implies that the expression on one side of the equal sign is equivalent to (has the same value as) the expression on the other side of the equal sign. WORDS MATH The sum of seven and eleven equals eighteen: 7 1 11 5 18 Three times five is fifteen: 3 # 5 5 15 Four times six equals twenty-four: 4 162 5 24 Eight divided by two is four: 8 2 54 Three subtracted from five is two: 5 2 3 5 2 When evaluating expressions involving real numbers, it is important to remember the correct order of operations. For example, how do we simplify the expression 3 1 2 # 5? Do we multiply first and then add, or do we add first and then multiply? In mathematics, conventional order implies multiplication first and then addition: 3 1 2 # 5 5 3 1 10 5 13. Parentheses imply grouping of terms, and the necessary operations should always be performed inside them first. If there are nested parentheses, always start with the innermost parentheses and work your way out. Within parentheses follow the conventional order of operations. Exponents are an important part of the order of operations and will be discussed in Section 0.2. ORDER OF OPERATIONS 1. Start with the innermost parentheses (grouping symbols) and work outward. 2. Perform all indicated multiplications and divisions, working from left to right. 3. Perform all additions and subtractions, working from left to right. 0.1.3 S K IL L Simplify expressions and evaluate algebraic expressions using the correct order of operations. 0.1.3 C ON C E P T U A L Learn the order of operations for real numbers. EXAMPLE 4 Simplifying Expressions Using the Correct Order of Operations Simplify the expressions. a. 4 1 3⋅2 2 7⋅5 1 6 b. 7 2 6 2⋅3 1 8 Solution (a): Perform multiplication first. 4 1 3⋅2 2 7⋅5 1 6 Then perform the indicated additions and subtractions. 5 4 1 6 2 35 1 6 5 Solution (b): The numerator and the denominator are similar to expressions in parentheses. Simplify these separately first, following the correct order of operations. Perform multiplication in the denominator first. 7 2 6 2⋅3 1 8 Then perform subtraction in the numerator and addition in the denominator. YOUR T UR N Simplify the expressions. a. 27 1 4⋅5 2 2⋅6 1 9 b. 9 2 6 2⋅5 1 6 f 6 35 f 219 6 f 5 7 2 6 6 1 8 5 1 14 ▼ A N S W E R a. 10 b. 3 16 ▼ [CONCEPT CHECK] TRUE OR FALSE In Example 4(a), we could have also started with adding 4 1 3. ANSWER False ▼ 0.1 Real Numbers 7 8 CHAPTER 0 Prerequisites and Review Parentheses ( ) and brackets [ ] are the typical notations for grouping and are often used interchangeably. When nesting (groups within groups), use parentheses on the innermost and then brackets on the outermost. EXAMPLE 5 Simplifying Expressions That Involve Grouping Signs Using the Correct Order of Operations Simplify the expression 3 35⋅14 2 22 2 2⋅74. Solution: Simplify the inner parentheses. 3 35⋅ 14 2 22 2 2⋅74 5 3 35⋅2 2 2⋅74 Inside the brackets, perform the multiplication 5⋅2 5 10 and 2⋅7 5 14. 5 3 310 2 144 Inside the brackets, perform the subtraction. 5 3 3244 Multiply. 5 Y OUR TU R N Simplify the expression 2 323⋅ 113 2 52 1 4⋅34. 212 ▼ A N S W E R 224 ▼ EXAMPLE 6 Evaluating Algebraic Expressions Evaluate the algebraic expression 7x 1 2 for x 5 3. Solution: Start with the algebraic expression. 7x 1 2 Substitute x 5 3. 7 132 1 2 Perform the multiplication. 5 21 1 2 Perform the addition. 5 23 Y OUR TU R N Evaluate the algebraic expression 6y 1 4 for y 5 2. ▼ Algebraic Expressions Everything discussed until now has involved real numbers (explicitly). In algebra, however, numbers are often represented by letters (such as x and y), which are called variables. A constant is a fixed (known) number such as 5. A coefficient is the constant that is multiplied by a variable. Quantities within the algebraic expression that are separated by addition or subtraction are referred to as terms. DEFINITION Algebraic Expression An algebraic expression is the combination of variables and constants using basic operations such as addition, subtraction, multiplication, and division. Each term is separated by addition or subtraction. Algebraic Expression Variable Term Constant Term Coefficient 5x 1 3 5x 3 5 When we know the value of the variables, we can evaluate an algebraic expression using the substitution principle: Algebraic expression: 5x 1 3 Value of the variable: x 5 2 Substitute x 5 2: 5 122 1 3 5 10 1 3 5 13 ▼ A N S W E R 16 In Example 6, the value for the variable was specified in order for us to evaluate the algebraic expression. What if the value of the variable is not specified; can we simplify an expression like 3 12x 2 5y2? In this case, we cannot subtract 5y from 2x. Instead, we rely on the basic properties of real numbers, or the basic rules of algebra. 0.1.4 Properties of Real Numbers You probably already know many properties of real numbers. For example, if you add up four numbers, it does not matter in which order you add them. If you multiply five numbers, it does not matter in what order you multiply them. If you add 0 to a real number or multiply a real number by 1, the result yields the original real number. Basic properties of real numbers are summarized in the following table. Because these properties are true for variables and algebraic expressions, these properties are often called the basic rules of algebra. 0.1.4 S K I L L Apply properties of real numbers and basic rules of algebra in simplifying and evaluating expressions. 0.1.4 C O N C E P T U A L Know and understand the basic properties of real numbers and the basic rules of algebra. The properties in the previous table govern addition and multiplication. Subtraction can be defined in terms of addition of the additive inverse, and division can be defined in terms of multiplication by the multiplicative inverse (reciprocal). 0.1 Real Numbers 9 PROPERTIES OF REAL NUMBERS (BASIC RULES OF ALGEBRA) NAME DESCRIPTION MATH (LET a, b, AND c EACH BE ANY REAL NUMBER) EXAMPLE Commutative property of addition Two real numbers can be added in any order. a 1 b 5 b 1 a 3x 1 5 5 5 1 3x Commutative property of multiplication Two real numbers can be multiplied in any order. ab 5 ba y⋅3 5 3y Associative property of addition When three real numbers are added, it does not matter which two numbers are added first. 1a 1 b2 1 c 5 a 1 1b 1 c2 1x 1 52 1 7 5 x 1 15 1 72 Associative property of multiplication When three real numbers are multiplied, it does not matter which two numbers are multiplied first. 1ab2 c 5 a 1bc2 123x2 y 5 23 1xy2 Distributive property Multiplication is distributed over all the terms of the sums or differences within the parentheses. a 1b 1 c2 5 ab 1 ac a 1b 2 c2 5 ab 2 ac 5 1x 1 22 5 5x 1 10 5 1x 2 22 5 5x 2 10 Additive identity property Adding zero to any real number yields the same real number. a 1 0 5 a 0 1 a 5 a 7y 1 0 5 7y Multiplicative identity property Multiplying any real number by 1 yields the same real number. a⋅1 5 a 1⋅a 5 a 18x2 112 5 8x Additive inverse property The sum of a real number and its additive inverse (opposite) is zero. a 1 12a2 5 0 4x 1 (24x2 5 0 Multiplicative inverse property The product of a nonzero real number and its multiplicative inverse (reciprocal) is 1. a⋅1 a 5 1 a 2 0 1x 1 22⋅a 1 x 1 2b 5 1 x 2 22 SUBTRACTION AND DIVISION Let a and b be real numbers. MATH TYPE OF INVERSE WORDS Subtraction a 2 b 5 a 1 12b2 2b is the additive inverse or opposite of b Subtracting a real number is equal to adding its opposite. Division a 4 b 5 a⋅1 b b 2 0 1 b is the multiplicative inverse or reciprocal of b Dividing by a real number is equal to multiplying by its reciprocal. 10 CHAPTER 0 Prerequisites and Review EXAMPLE 7 Using the Distributive Property Use the distributive property to eliminate the parentheses. a. 3 1x 1 52 b. 2 1y 2 62 Solution (a): Use the distributive property. 31x 1 52 5 3 1x2 1 3 152 Perform the multiplication. 5 3x 1 15 Solution (b): Use the distributive property. 2 1y 2 62 5 2 1y2 2 2 162 Perform the multiplication. 5 2y 2 12 Y OUR TU R N Use the distributive property to eliminate the parentheses. a. 2 1x 1 32 b. 5 1y 2 32 ➤ ➤ ➤ ➤ ▼ You also probably know the rules that apply when multiplying a negative real number. For example, “a negative times a negative is a positive.” ▼ A N S W E R a. 2x 1 6 b. 5y 2 15 EXAMPLE 8 Using Properties of Negatives Eliminate the parentheses and perform the operations. a. 25 1 7 2 1222 b. 2 1232 1242 1262 Solution: a. Distribute the negative. 25 1 7 2 1222 5 25 1 7 1 2 Combine the three quantities. 5 4 b. Group the terms. 32 12324 3 1242 12624 Perform the multiplication inside the 3 4. 5 334 3244 Multiply. 5 72 12 f PROPERTIES OF NEGATIVES DESCRIPTION MATH (LET a AND b BE POSITIVE REAL NUMBERS) EXAMPLE A negative quantity times a positive quantity is a negative quantity. 12a2 1b2 5 2ab 1282 132 5 224 A negative quantity divided by a positive quantity is a negative quantity. or A positive quantity divided by a negative quantity is a negative quantity. 2a b 5 2 a b or a 2b 5 2 a b 216 4 5 24 or 15 23 5 25 A negative quantity times a negative quantity is a positive quantity. 12a2 12b2 5 ab 122x2 1252 5 10x A negative quantity divided by a negative quantity is a positive quantity. 2a 2b 5 a b 212 23 5 4 The opposite of a negative quantity is a positive quantity (subtracting a negative quantity is equivalent to adding a positive quantity). 2 12a2 5 a 2 1292 5 9 A negative sign preceding an expression is distributed throughout the expression. 2 1a 1 b2 5 2a 2 b 2 1a 2 b2 5 2a 1 b 23 1x 1 52 5 23x 2 15 23 1x 2 52 5 23x 1 15 We use properties of negatives to define the absolute value of any real number. The absolute value of a real number a, denoted a, is its magnitude. On a number line this is the distance from the origin, 0, to the point. For example, algebraically, the absolute value of 5 is 5, that is, 5 5 5; and the absolute value of 25 is 5, or 25 5 5. Graphically, the distance on the real number line from 0 to either 25 or 5 is 5. –5 0 5 5 5 Notice that the absolute value does not change a positive real number, but it changes a negative real number to a positive number. A negative number becomes a positive number if it is multiplied by 21. IF a IS A… a EXAMPLE Positive real number a 5 a 5 5 5 Negative real number a 5 2a 25 5 2 1252 5 5 Zero a 5 a 0 5 0 Properties of the absolute value will be discussed in Section 1.7. What is the product of any real number and zero? The answer is zero. This property also leads to the zero product property, which is the basis for factoring (one of the methods used to solve quadratic equations, which will be discussed in Section 1.3). EXAMPLE 9 Finding the Absolute Value of a Real Number Evaluate the expressions. a. 23 1 7 b. 2 2 8 Solution: a. 23 1 7 5 4 b. 2 2 8 5 26 5 4 5 6 [CONCEPT CHECK] TRUE OR FALSE a 2 b 5 b 2 a ANSWER True ▼ 0.1 Real Numbers 11 EXAMPLE 10 Using Properties of Negatives and the Distributive Property common mistake A common mistake is applying a negative only to the first term. YOUR T UR N Eliminate the parentheses. a. 22 1x 1 5y2 b. 2 13 2 2b2 ▼ ✖I N C O R R EC T Error: 22x 2 3y The negative 122 was not distributed through the second term. ✓COR R E C T 2 12x 2 3y2 5 2 12x2 2 123y2 5 22x 1 3y Eliminate the parentheses 2 12x 2 3y2. ▼ A N S W E R a. 22x 2 10y b. 23 1 2b 12 CHAPTER 0 Prerequisites and Review Fractions always seem to intimidate students. In fact, many instructors teach students to eliminate fractions in algebraic equations. It is important to realize that you can never divide by zero. Therefore, in the following table of fractional properties it is assumed that no denominators are zero. The least common multiple of two or more integers is the smallest integer that is evenly divisible by each of the integers. For example, the least common multiple (LCM) of 3 and 4 is 12. The LCM of 8 and 6 is 24. The reason the LCM of 8 and 6 is not 48 is that 8 and 6 have a common factor of 2. When adding and subtracting fractions, a common denominator can be found by multiplying the denominators. When there are common factors in the denominators, the LCM is the least common denominator (LCD) of the original denominators. PROPERTIES OF ZERO DESCRIPTION MATH (LET a BE A REAL NUMBER) EXAMPLE A real number multiplied by zero is zero. a⋅0 5 0 0⋅x 5 0 Zero divided by a nonzero real number is zero. 0 a 5 0 a 2 0 0 3 2 x 5 0 x 2 3 A real number divided by zero is undefined. a 0 is undefined x 1 2 0 is undefined ZERO PRODUCT PROPERTY DESCRIPTION MATH EXAMPLE If the product of two real numbers is zero, then one of those numbers has to be zero. If ab 5 0, then a 5 0 or b 5 0 If x 1x 1 22 5 0, then x 5 0 or x 1 2 5 0 therefore x 5 0 or x 5 22 Note: If a and b are both equal to zero, then the product is still zero. FRACTIONAL PROPERTIES DESCRIPTION MATH ZERO CONDITION EXAMPLE Equivalent fractions a b 5 c d if and only if ad 5 bc b 2 0 and d 2 0 y 2 5 6y 12 since 12y 5 12y Multiplying two fractions a b⋅c d 5 ac bd b 2 0 and d 2 0 3 5⋅x 7 5 3x 35 Adding fractions that have the same denominator a b 1 c b 5 a 1 c b b 2 0 x 3 1 2 3 5 x 1 2 3 Subtracting fractions that have the same denominator a b 2 c b 5 a 2 c b b 2 0 7 3 2 5 3 5 7 2 5 3 5 2 3 Adding fractions with different denominators using a common denominator a b 1 c d 5 ad bd 1 cb bd 5 ad 1 bc bd b 2 0 and d 2 0 1 2 1 5 3 5 112132 1 152122 122132 5 13 6 Subtracting fractions with different denominators using a common denominator a b 2 c d 5 ad bd 2 cb bd 5 ad 2 bc bd b 2 0 and d 2 0 1 3 2 1 4 5 112142 2 112132 132142 5 1 12 Dividing by a fraction is equivalent to multiplying by its reciprocal a b 4 c d 5 a b⋅d c b 2 0, c 2 0, and d 2 0 x 3 4 2 7 5 x 3⋅7 2 5 7x 6 EXAMPLE 11 Performing Operations with Fractions Perform the indicated operations involving fractions and simplify. a. 2 3 2 1 4 b. 2 3 4 4 c. x 2 1 3 5 Solution (a): Determine the LCD. 3⋅4 5 12 Rewrite fractions applying the LCD. 2 3 2 1 4 5 2⋅4 3⋅4 2 1⋅3 4⋅3 5 2142 2 1132 3142 Eliminate the parentheses. 5 8 2 3 12 Combine terms in the numerator. 5 5 12 Solution (b): Rewrite 4 with an understood 1 in the denominator. 5 2 3 4 4 1 Dividing by a fraction is equivalent to multiplying by its reciprocal. 5 2 3⋅1 4 Multiply numerators and denominators, respectively. 5 2 12 Reduce the fraction to simplest form. 5 1 6 Solution (c): Determine the LCD. 2 # 5 5 10 Rewrite fractions in terms of the LCD. x 2 1 3 5 5 5x 1 3122 122152 Simplify the numerator. 5 5x 1 6 10 Y OUR T UR N Perform the indicated operations involving fractions. a. 3 5 1 1 2 b. 1 5 4 3 10 c. 2 3 2 x 5 ▼ A N S W E R a. 11 10 b. 2 3 c. 10 2 3x 15 ▼ 0.1 Real Numbers 13 14 CHAPTER 0 Prerequisites and Review Subtraction and division can be defined in terms of addition and multiplication. • Subtraction: a 2 b 5 a 1 12b2 (add the opposite) • Division: a 4 b 5 a⋅1 b, where b 2 0 (multiply by the reciprocal) Properties of negatives were reviewed. If a and b are positive real numbers, then: • 12a2 1b2 5 2ab • 12a2 12b2 5 ab • 2 12a2 5 a • 2 1a 1 b2 5 2a 2 b and 2 1a 2 b2 5 2a 1 b • 2a b 5 2 a b • 2a 2b 5 a b Absolute value of real numbers: a 5 a if a is nonnegative, and a 5 2a if a is negative. Properties of zero were reviewed. • a⋅0 5 0 and 0 a 5 0 a 2 0 • a 0 is undefined • Zero product property: If ab 5 0, then a 5 0 or b 5 0 Properties of fractions were also reviewed. • a b 6 c d 5 ad 6 bc bd b 2 0 and d 2 0 • a b 4 c d 5 a b⋅d c b 2 0, c 2 0, and d 2 0 In this section, real numbers were defined as the set of all rational and irrational numbers. Decimals are approximated by either truncating or rounding. • Truncating: Eliminate all values after a particular digit. • Rounding: Look to the right of a particular digit. If the number is 5 or greater, increase the digit by 1; otherwise, leave it as is and eliminate all digits to the right. The order in which we perform operations is 1. parentheses (grouping); work from inside outward. 2. multiplication/division; work from left to right. 3. addition/subtraction; work from left to right. The properties of real numbers are employed as the basic rules of algebra when dealing with algebraic expressions. • Commutative property of addition: a 1 b 5 b 1 a • Commutative property of multiplication: ab 5 ba • Associative property of addition: 1a 1 b2 1 c 5 a 1 1b 1 c2 • Associative property of multiplication: 1ab2 c 5 a 1bc2 • Distributive property: a 1b 1 c2 5 ab 1 ac or a 1b 2 c2 5 ab 2 ac • Additive identity: a 1 0 5 a • Multiplicative identity: a⋅1 5 a • Additive inverse (opposite): a 1 12a2 5 0 • Multiplicative inverse (reciprocal): a⋅1 a 5 1 a 2 0 [SEC TION 0.1] S U M M A RY [SEC TION 0.1] E X E R CI SES • S K I L L S In Exercises 1–8, classify the following real numbers as rational or irrational. 1. 11 3 2. 22 3 3. 2.07172737. . . 4. p 5. 2.7766776677 6. 5.222222 7. !5 8. !17 In Exercises 9–16, approximate the real number to three decimal places by (a) rounding and (b) truncation. 9. 7.3471 10. 9.2549 11. 2.9949 12. 6.9951 13. 0.234492 14. 1.327491 15. 5.238473 16. 2.118465 In Exercises 17–40, perform the indicated operations in the correct order. 17. 5 1 2⋅3 2 7 18. 2 1 5⋅4 1 3⋅6 19. 2⋅ 15 1 7⋅4 2 202 20. 23⋅ 12 1 72 1 8⋅ 17 2 2⋅12 21. 2 2 3 34 12⋅3 1 524 22. 4⋅6 15 2 92 23. 8 2 1222 1 7 24. 210 2 1292 25. 23 2 1262 26. 25 1 2 2 1232 27. x 2 12y2 2 z 28. 2a 1 b 2 12c2 29. 2 13x 1 y2 30. 2 14a 2 2b2 31. 23 1521212 32. 2 12 12321242 33. 24 26 3 15 2 82 1424 34. 214 5 2 1222 35. 2 16x 2 4y2 2 13x 1 5y2 36. 24x 6 2 1222 37. 2 13 2 4x2 2 14x 1 72 38. 2 2 3 3 14x 2 52 2 3x 2 74 39. 24152 2 5 25 40. 26 12x 1 3y2 2 33x 2 12 2 5y24 In Exercises 41–56, write as a single fraction and simplify. 41. 1 3 1 5 4 42. 1 2 2 1 5 43. 5 6 2 1 3 44. 7 3 2 1 6 45. 3 2 1 5 12 46. 1 3 1 5 9 47. 1 9 2 2 27 48. 3 7 2 1242 3 2 5 6 49. x 5 1 2x 15 50. y 3 2 y 6 51. x 3 2 2x 7 52. y 10 2 y 15 53. 4y 15 2 123y2 4 54. 6x 12 2 7x 20 55. 3 40 1 7 24 56. 23 10 2 a27 12 b In Exercises 57–68, perform the indicated operation and simplify, if possible. 57. 2 7 ⋅14 3 58. 2 3 ⋅9 10 59. 2 7 4 10 3 60. 4 5 4 7 10 61. 4b 9 4 a 27 a 2 0 62. 3a 7 4 b 21 b 2 0 63. 3x 10 4 6x 15 x 2 0 64. 4 1 5 4 7 1 20 65. 3x 4 4 9 16y y 2 0 66. 14m 2 ⋅4 7 67. 6x 7 4 3y 28 y 2 0 68. 21 3 ⋅75 6 In Exercises 69–72, evaluate the algebraic expression for the specified values. 69. 2c 2d for c 5 24, d 5 3 70. 2l 1 2w for l 5 5, w 5 10 71. m1 ⋅m2 r 2 for m1 5 3, m2 5 4, r 5 10 72. x 2 m s for x 5 100, m 5 70, s 5 15 • A P P L I C A T I O N S On November 15, 2015, the United States debt was estimated at $18,152,809,942,589, and at that time the estimated population was 322,162,446 citizens. 73. U.S. National Debt. Round the debt to the nearest million. 74. U.S. Population. Round the number of citizens to the nearest thousand. 75. U.S. Debt. If the debt is distributed evenly to all citizens, what is the national debt per citizen? Round your answer to the nearest dollar. 76. U.S. Debt. If the debt is distributed evenly to all citizens, what is the national debt per citizen? Round your answer to the nearest cent. 0.1 Real Numbers 15 16 CHAPTER 0 Prerequisites and Review 81. Student athletes are a subset of the students in the honors program. 82. The students who are members of fraternities or sororities are a subset of the entire student population. 83. Every integer is a rational number. 84. A real number can be both rational and irrational. 85. What restrictions are there on x for the following to be true? 3 x 4 5 x 5 3 5 86. What restrictions are there on x for the following to be true? x 2 4 x 6 5 3 • C O N C E P T U A L In Exercises 81–84, determine whether each of the following statements is true or false. • C H A L L E N G E In Exercises 87 and 88, simplify the expressions. 87. 22 33 1x 2 2y2 1 74 1 33 12 2 5x2 1 104 2 7 322 1x 2 32 1 54 88. 22525 1y 2 x2 2 2 33 12x 2 52 1 7 122 2 44 1 36 1 7 In Exercises 77–80, explain the mistake that is made. • C A T C H T H E M I S T A K E 77. Round 13.2749 to two decimal places. Solution: The 9, to the right of the 4, causes the 4 to round to 5. 13.275 The 5, to the right of the 7, causes the 7 to be rounded to 8. 13.28 This is incorrect. What mistake was made? 78. Simplify the expression 2 3 1 1 9. Solution: Add the numerators and denominators. 2 1 1 3 1 9 5 3 12 Reduce. 5 1 4 This is incorrect. What mistake was made? 79. Simplify the expression 3 1x 1 52 2 2 14 1 y2. Solution: Eliminate parentheses. 3x 1 15 2 8 1 y Simplify. 3x 1 7 1 y This is incorrect. What mistake was made? 80. Simplify the expression 23 1x 1 22 2 11 2 y2. Solution: Eliminate parentheses. 23x 2 6 2 1 2 y Simplify. 23x 2 7 2 y This is incorrect. What mistake was made? • T E C H N O L O G Y 89. Use your calculator to evaluate !1260. Does the answer appear to be a rational or an irrational number? Why? 90. Use your calculator to evaluate Å 144 25 . Does the answer appear to be a rational or an irrational number? Why? 91. Use your calculator to evaluate !4489. Does the answer appear to be a rational or an irrational number? Why? 92. Use your calculator to evaluate Å 882 49 . Does the answer appear to be a rational or an irrational number? Why? 0.2 Integer Exponents and Scientific Notation 17 S K I L L S O B J E C T I V E S ■ ■Evaluate expressions using integer exponents and properties of exponents. ■ ■Convert numbers to and from scientific notation. C O N C E P T U A L O B J E C T I V ES ■ ■Understand that positive integer exponents represent repeated multiplication and that negative exponents are reciprocals. ■ ■Understand that scientific notation is an effective way to represent very large or very small real numbers. 0.2 INTEGER EXPONENTS AND SCIENTIFIC NOTATION 0.2.1 Integer Exponents Exponents represent repeated multiplication. For example, 2⋅2⋅2⋅2⋅2 5 25. The 2 that is repeatedly multiplied is called the base, and the small number 5 above and to the right of the 2 is called the exponent. DEFINITION Natural-Number Exponent Let a be a real number and n be a natural number (positive integer); then an is defined as an 5 a ? a ? a c a 1a appears as a factor n times2 where n is the exponent, or power, and a is the base. n factors f 0.2.1 S K IL L Evaluate expressions using inte-ger exponents and properties of exponents. 0.2.1 C O NC E P T U A L Understand that positive integer exponents represent repeated multiplication and that negative exponents are reciprocals. EXAMPLE 1 Evaluating Expressions Involving Natural-Number Exponents Evaluate the expressions. a. 43 b. 81 c. 54 d. A1 2B5 Solution: a. 43 5 4⋅4⋅4 5 64 c. 54 5 5⋅5⋅5⋅5 5 625 b. 81 5 8 d. A1 2B 5 5 1 2⋅1 2⋅1 2⋅1 2⋅1 2 5 1 32 Y OUR T UR N Evaluate the expressions. a. 63 b. A1 3B 4 ▼ A N S W E R a. 216 b. 1 81 ▼ We now include exponents in our order of operations: 1. Parentheses 2. Exponents 3. Multiplication/Division 4. Addition/Subtraction EXAMPLE 2 Evaluating Expressions Involving Natural-Number Exponents Evaluate the expressions. a. 12324 b. 234 c. 12223⋅52 Solution: a. 12324 5 1232123212321232 5 81 b. 234 5 213⋅3⋅3⋅32 5 281 c. 12223⋅52 5 122212221222 ⋅5⋅5 5 2200 Y OUR T UR N Evaluate the expression 243⋅23. 81 f 28 25 f f ▼ A N S W E R 2512 ▼ STUDY TIP an: “a raised to the nth power” a2: “a squared” a3: “a cubed” 18 CHAPTER 0 Prerequisites and Review So far, we have discussed only exponents that are natural numbers (positive integers). When the exponent is a negative integer, we use the following property. In other words, a base raised to a negative-integer exponent is equivalent to the reciprocal of the base raised to the opposite (positive) integer exponent. NEGATIVE-INTEGER EXPONENT PROPERTY Let a be any nonzero real number and n be a natural number (positive integer); then a2n 5 1 an a 2 0 EXAMPLE 3 Evaluating Expressions Involving Negative-Integer Exponents Evaluate the expressions. a. 224 b. 1 323 c. 423⋅1 224 d. 223⋅ 1 126222 Solution: a. 224 5 1 24 5 1 16 b. 1 323 5 1 a 1 33b 5 1 4 a 1 33b 5 1⋅33 1 5 33 5 27 c. 423⋅1 224 5 1 43⋅24 5 16 64 5 1 4 d. 223⋅ 1 126222 5 223⋅12622 5 12821362 5 2288 Y OUR TU R N Evaluate the expressions. a. 2 1 522 b. 1 322⋅622 28 36 ▼ A N S W E R a. 225 b. 1 4 ▼ f f Now we can evaluate expressions involving positive and negative exponents. How do we evaluate an expression with a zero exponent? We define any nonzero real number raised to the zero power as 1. [CONCEPT CHECK] Simplify the expression: a1 ab 24 . ANSWER a4 ▼ STUDY TIP A negative exponent implies a reciprocal. ZERO-EXPONENT PROPERTY Let a be any nonzero real number; then a0 5 1 a Þ 0 We now can evaluate expressions involving integer (positive, negative, or zero) exponents. What about when expressions involving integer exponents are multiplied, divided, or raised to a power? WORDS MATH When expressions with the same base are multiplied, the exponents are added. 23⋅24 5 2⋅2⋅2⋅2⋅2⋅2⋅2 5 2314 5 27 When expressions with the same base are divided, the exponents are subtracted. or 2523 5 22 When an expression involving an exponent is raised to a power, the exponents are multiplied. or 12322 5 23⋅2 5 26 5 64 The following table summarizes the properties of integer exponents. 23 f 24 f 25 23 5 2⋅2⋅2⋅2⋅2 2⋅2⋅2 5 2⋅2 1 5 22 12322 5 1822 5 64 0.2 Integer Exponents and Scientific Notation 19 EXAMPLE 4 Evaluating Expressions Involving Zero Exponents Evaluate the expressions. a. 50 b. 1 20 c. 12320 d. 240 Solution: a. 50 5 1 b. 1 20 5 1 1 5 1 c. 12320 5 1 d. 240 5 21 f 1 PROPERTIES OF INTEGER EXPONENTS NAME DESCRIPTION MATH (LET a AND b BE NONZERO REAL NUMBERS AND m AND n BE INTEGERS) EXAMPLE Product property When multiplying exponentials with the same base, add exponents. am⋅an 5 am1n x2⋅x5 5 x215 5 x7 Quotient property When dividing exponentials with the same base, subtract the exponents (numerator 2 denominator). am an 5 am2n x5 x3 5 x523 5 x2 x 2 0 Power property When raising an exponential to a power, multiply exponents. 1am2n 5 amn 1x224 5 x2⋅4 5 x8 Product to a power property A product raised to a power is equal to the product of each factor raised to the power. 1ab2n 5 anbn 12x23 5 23⋅x3 5 8x3 Quotient to a power property A quotient raised to a power is equal to the quotient of the factors raised to the power. aa bb n 5 an bn ax yb 4 5 x4 y4 y 2 0 20 CHAPTER 0 Prerequisites and Review Common Errors Made Using Properties of Exponents INCORRECT CORRECT ERROR x4⋅x3 5 x12 x4⋅x3 5 x7 Exponents should be added (not multiplied). x18 x6 5 x3 x18 x6 5 x12; x 2 0 Exponents should be subtracted (not divided). 1x223 5 x8 1x223 5 x6 Exponents should be multiplied (not raised to a power). 12x23 5 2x3 12x23 5 8x3 Both factors (the 2 and the x) should be cubed. 23⋅24 5 47 23⋅24 5 27 The original common base should be retained. 23⋅35 5 68 23⋅35 The properties of integer exponents require the same base. We will now use properties of integer exponents to simplify exponential expressions. An exponential expression is simplified when: ■ ■All parentheses (groupings) have been eliminated. ■ ■A base appears only once. ■ ■No powers are raised to other powers. ■ ■All exponents are positive. STUDY TIP It is customary not to leave negative exponents. Instead we use the negative exponent property to write exponential expressions with only positive exponents. ▼ A N S W E R a. 212x 4y 5 b. 227x 3y 9z6 c. 24x3 y4 EXAMPLE 5 Simplifying Exponential Expressions Simplify the expressions (assume all variables are nonzero). a. 122x2 y3215x3 y2 b. 12x2 yz323 c. 25x3 y6 25x5 y4 Solution (a): Parentheses imply multiplication. Group the same bases together. 122x2 y3215x3 y2 5 1222152 x2 x3 y3 y Apply the product property. 5 1222152x2 x3 y3 y Multiply the constants. 5 210x5y4 Solution (b): Apply the product to a power property. 12x2 yz323 5 12231x2231y231z323 Apply the power property. 5 8x2⋅3y1⋅3z3⋅3 Simplify. 5 8x6y3z9 Solution (c): Group the same bases together. 25x3 y6 25x5 y4 5 a 25 25b ax3 x5b ay6 y4b Apply the quotient property. 5 1252x325y624 5 25x22y2 Apply the negative exponent property. 5 25y2 x2 Y OUR TU R N Simplify the expressions (assume all variables are nonzero). a. 123x3 y2214xy32 b. 123xy3 z223 c. 216x4 y3 4xy7 x213 y311 f f ▼ EXAMPLE 6 Simplifying Exponential Expressions Write each expression so that all exponents are positive (assume all variables are nonzero). a. 13x2 z24223 b. 1x2 y2322 1x21 y4223 c. 122xy223 216xz322 Solution (a): Apply the product to a power property. 13x2 z24223 5 132231x22231z24223 Apply the power property. 5 323x26z12 Apply the negative-integer exponent property. 5 z12 33 x6 Evaluate 33. 5 z12 27x6 Solution (b): Apply the product to a power property. 1x2 y2322 1x21 y4223 5 x4 y26 x3 y212 Apply the quotient property. 5 x423 y26212122 Simplify. 5 xy6 Solution (c): Apply the product to a power property on both the numerator and denominator. Apply the power property. 5 28x3 y6 236x2 z6 Group constant terms and x terms. 5 a 28 236b ax3 x2b ay6 z6b Apply the quotient property. 5 a 8 36b 1x3222 ay6 z6b Simplify. 5 2xy6 9z6 YOUR T UR N Simplify the exponential expression and express it in terms of positive exponents 1tv2223 12t4 v3221. 122xy223 216xz322 5 122231x231y223 216221x221z322 ▼ ▼ A N S W E R 2t v3 0.2.2 Scientific Notation You are already familiar with base 10 raised to positive-integer powers. However, it can be inconvenient to write all the zeros out, so we give certain powers of 10 particular names: thousand, million, billion, trillion, and so on. For example, we say there are 322 million U.S. citizens as opposed to writing out 322,000,000 citizens. Or we say that the national debt is $18 trillion as opposed to writing out $18,000,000,000,000. The following table contains scientific notation for positive exponents and examples of some common prefixes and abbreviations. One of the fundamental applications of scientific notation is measurement. 0.2.2 S K IL L Convert numbers to and from scientific notation. 0.2.2 C ON C E P T U A L Understand that scientific notation is an effective way to represent very large or very small real numbers. 0.2 Integer Exponents and Scientific Notation 21 22 CHAPTER 0 Prerequisites and Review EXPONENTIAL FORM REAL NUMBER NUMBER OF ZEROS FOLLOWING THE 1 PREFIX ABBREVIATION EXAMPLE 101 10 1 102 100 2 103 1000 (one thousand) 3 kilo-k The relay-for-life team ran a total of 80 km (kilometers). 104 10,000 4 105 100,000 5 106 1,000,000 (one million) 6 mega-M Modern high- powered diesel–electric railroad locomotives typically have a peak power output of 3 to 5 MW (megawatts). 107 10,000,000 7 108 100,000,000 8 109 1,000,000,000 (one billion) 9 giga-G A flash drive typically has 1 to 4 GB (gigabytes) of storage. 1010 10,000,000,000 10 1011 100,000,000,000 11 1012 1,000,000,000,000 (one trillion) 12 tera-T Laser systems offer higher frequencies on the order of THz (terahertz). EXPONENTIAL FORM REAL NUMBER NUMBER OF PLACES DECIMAL (1.0) MOVES TO THE LEFT PREFIX ABBREVIATION EXAMPLE 1021 0.1 1 1022 0.01 2 1023 0.001 (one thousandth) 3 milli-m Excedrin Extra Strength tablets each have 250 mg (milligrams) of acetaminophen. 1024 0.0001 4 1025 0.00001 5 1026 0.000001 (one millionth) 6 micro-m A typical laser has a wavelength of 1.55 mm (micrometers). 1027 0.0000001 7 1028 0.00000001 8 1029 0.000000001 (one billionth) 9 nano-n PSA levels less than 4 ng/ml (nanogram per milliliter of blood) represent low risk for prostate cancer. 10210 0.0000000001 10 10211 0.00000000001 11 10212 0.000000000001 (one trillionth) 12 pico-p A single yeast cell weighs 44 pg (picograms). Notice that 108 is a 1 followed by 8 zeros; alternatively, you can start with 1.0 and move the decimal point eight places to the right (insert zeros). The same type of table can be made for negative-integer powers with base 10. To find the real number associated with exponential form, start with 1.0 and move the decimal a certain number of places to the left (fill in missing decimal places with zeros). In optics a micrometer is called a micron. Note that c is a real number between 1 and 10. Therefore, 22.5 3 103 is not in scientific notation, but we can convert it to scientific notation: 2.25 3 104. For example, there are approximately 50 trillion cells in the human body. We write 50 trillion as 50 followed by 12 zeros 50,000,000,000,000. An efficient way of writing such a large number is using scientific notation. Notice that 50,000,000,000,000 is 5 followed by 13 zeros, or in scientific notation, 5 3 1013. Very small numbers can also be written using scientific notation. For example, in laser communications a pulse width is 2 femtoseconds, or 0.000000000000002 second. Notice that if we start with 2.0 and move the decimal point 15 places to the left (adding zeros in between), the result is 0.000000000000002, or in scientific notation, 2 3 10215. SCIENTIFIC NOTATIONIN A positive real number can be written in scientific notation with the form c 3 10n, where 1 # c , 10 and n is an integer. [CONCEPT CHECK] The U.S. national debt can be represented in scientific notation using an exponent that is positive or negative? ANSWER Positive ▼ EXAMPLE 7 Expressing a Positive Real Number in Scientific Notation Express the numbers in scientific notation. a. 3,856,000,000,000,000 b. 0.00000275 Solution: a. Rewrite the number with the implied decimal point. 3,856,000,000,000,000. Move the decimal point to the left 15 places. 5 3.856 3 1015 b. Move the decimal point to the right six places. 0.00000275 5 2.75 3 1026 Y OUR T UR N Express the numbers in scientific notation. a. 4,520,000,000 b. 0.00000043 ▼ A N S W E R a. 4.52 3 109 b. 4.3 3 1027 ▼ EXAMPLE 8 Converting from Scientific Notation to Decimals Write each number as a decimal. a. 2.869 3 105 b. 1.03 3 1023 Solution: a. Move the decimal point 5 places to the right (add zeros in between). 286,900. or 286,900 b. Move the decimal point 3 places to the left (add zeros in between). 0.00103 YOUR T UR N Write each number as a decimal. a. 8.1 3 104 b. 3.7 3 1028 ▼ A N S W E R a. 81,000 b. 0.000000037 ▼ STUDY TIP Scientific notation is a number between 1 and 10 that is multiplied by 10 to a power. STUDY TIP Real numbers greater than 1 correspond to positive exponents in scientific notation, whereas real numbers greater than 0 but less than 1 correspond to negative exponents in scientific notation. 0.2 Integer Exponents and Scientific Notation 23 24 CHAPTER 0 Prerequisites and Review [SEC TION 0. 2] E X E R CI SE S • S K I L L S In Exercises 1–20, evaluate each expression. 1. 44 2. 53 3. 12325 4. 12422 5. 252 6. 272 7. 222⋅4 8. 232⋅5 9. 90 10. 28x0 11. 1021 12. a21 13. 822 14. 324 15. 26⋅52 16. 22⋅42 17. 8⋅223⋅5 18. 5⋅224⋅32 19. 26⋅322⋅81 20. 6⋅42⋅424 In this section we discussed properties of exponents. INT E G E R E X P ONE NT S The following table summarizes integer exponents. Let a be any real number and n be a natural number. NAME DESCRIPTION MATH Natural-number exponent Multiply n factors of a. an 5 a⋅a⋅a Na Negative-integer exponent property A negative exponent implies a reciprocal. a2n 5 1 an a 2 0 Zero-exponent property Any nonzero real number raised to the zero power is equal to 1. a0 5 1 a 2 0 f n factors P ROP E RT IE S OF I NT E G E R E X P ONENTS The following table summarizes properties of integer exponents. Let a and b be nonzero real numbers and m and n be integers. NAME DESCRIPTION MATH Product property When multiplying exponentials with the same base, add exponents. am⋅an 5 am1n Quotient property When dividing exponentials with the same base, subtract the exponents (numerator 2 denominator). am an 5 am2n Power property When raising an exponential to a power, multiply exponents. 1am2n 5 amn Product to a power property A product raised to a power is equal to the product of each factor raised to the power. 1ab2n 5 anbn Quotient to a power property A quotient raised to a power is equal to the quotient of the factors raised to the power. aa bb n 5 an bn SCIE NT IF I C NOTAT I ON Scientific notation is a convenient way of using exponents to represent either very small or very large numbers. Real numbers greater than 1 correspond to positive exponents in scientific notation, whereas real numbers greater than 0 but less than 1 correspond to negative exponents in scientific notation. Scientific notation offers the convenience of multiplying and dividing real numbers by applying properties of exponents. REAL NUMBER (DECIMAL FORM) PROCESS SCIENTIFIC NOTATION 2,357,000,000 Move the implied decimal point to the left 9 places 2.357 3 109 0.00000465 Move the decimal point to the right 6 places 4.65 3 1026 [SEC TION 0. 2] S U M MA RY In Exercises 21–50, simplify and write the resulting expression with only positive exponents. 21. x2⋅x3 22. y3⋅y5 23. x2x23 24. y3⋅y27 25. 1x223 26. 1y322 27. 14a23 28. 14x223 29. 122t23 30. 123b24 31. 15xy222 13x3y2 32. 14x2y2 12xy322 33. x5 y3 x7 y 34. y5 x2 y22 x25 35. 12xy22 122xy23 36. 123x3 y2 241x2 y323 37. ab 2b 24 38. ac 3b 22 39. 19a22b3222 40. 129x23y2224 41. a22 b3 a4 b5 42. x23 y2 y24 x5 43. 1x3 y2122 1xy2222 44. 1x3 y2222 1x4 y3223 45. 31x2 y23 121x22 y24 46. 124x2222 y3 z 12x3222 1y21z2 4 47. 1x24 y5222 3221x322 y2445 48. 22x2 122x325 49. c a212xy423 x412a3 y222d 3 50. c b2312x3 y224 y212b2 x52 3 d 5 51. Write 28⋅163⋅ 1642 as a power of 2 : 2? 52. Write 39⋅815⋅ 192 as a power of 3 : 3? In Exercises 53–60, express the given number in scientific notation. 53. 27,600,000 54. 144,000,000,000 55. 93,000,000 56. 1,234,500,000 57. 0.0000000567 58. 0.00000828 59. 0.000000123 60. 0.000000005 In Exercises 61–66, write the number as a decimal. 61. 4.7 3 107 62. 3.9 3 105 63. 2.3 3 104 64. 7.8 3 1023 65. 4.1 3 1025 66. 9.2 3 1028 • A P P L I C A T I O N S In Exercises 67 and 68, refer to the following: It is estimated that there are currently 5.0 3 109 cell phones being used worldwide. Assume that the average cell phone measures 5 inches in length and there are 5280 feet in a mile. 67. Cell Phones Spanning the Earth. a. If all of the cell phones currently in use were lined up next to each other tip to tip, how many feet would the line of cell phones span? Write the answer in scientific notation. b. The circumference of the Earth (measured at the equator) is approximately 25,000 miles. If the cell phones in part (a) were to be wrapped around the Earth at the equator, would they circle the Earth completely? If so, approximately how many times? 68. Cell Phones Reaching the Moon. a. If all of the cell phones currently in use were to be lined up next to each other tip to tip, how many miles would the line of cell phones span? Write the answer in scientific notation. b. The Moon traces an elliptical path around the Earth, with the average distance between them being approximately 239,000 miles. Would the line of cell phones in part (a) reach the Moon? 69. Astronomy. The distance from Earth to Mars on a particular day can be 200 million miles. Express this distance in scientific notation. 70. Astronomy. The distance from Mars to the Sun on a particular day can be 142 million miles. Express this distance in scientific notation. 71. Lasers. The wavelength of a typical laser used for communication systems is 1.55 microns 1or 1.55 3 1026 meters2. Express the wavelength in decimal representation in terms of meters. 72. Lasers. A ruby-red laser has a wavelength of 694 nanometers 1or 6.93 3 1027 meters2. Express the wavelength in decimal representation in terms of meters. 0.2 Integer Exponents and Scientific Notation 25 26 CHAPTER 0 Prerequisites and Review In Exercises 73–76, explain the mistake that is made. • C A T C H T H E M I S T A K E 73. Simplify 122y32 13x2y22. Group like factors together. 1222 132 x 2 y 3y 2 Use the product property. 26x2y6 This is incorrect. What mistake was made? 75. Simplify 122xy32 2 15x2y22. Apply the product to a power property. 5 12222x2 1y32 2 1522 1x222y2 Apply the power rule. 5 4x2y925x4y2 Group like factors. 5 142 1252x2x4y9y2 Apply the product property. 5 100x6y11 This is incorrect. What mistake was made? 74. Simplify 12xy2 2 3. Eliminate the parentheses. 12xy2 2 3 5 2x3y6 This is incorrect. What mistake was made? 76. Simplify 24x16 y9 8x2 y3 . Group like factors. 5 a2 4 8 b ax16 x2 b ay9 y3b Use the quotient property. 5 21 2x8 y3 This is incorrect. What mistake was made? • C O N C E P T U A L In Exercises 77–80, determine whether each of the following statements is true or false. 77. 22n 5 1222n, if n is an integer. 78. Any nonzero real number raised to the zero power is one. 79. xn11 xn 5 x for x 5 any real number. 80. x21 1 x22 5 x23 81. Simplify A1am2nB k. 82. Simplify A1a2m22nB2k. In Exercises 83–86, evaluate the expression for the given value. 83. 2a2 1 2ab for a 5 22, b 5 3 84. 2a3 2 7a2 for a 5 4 85. 216t2 1 100t for t 5 3 86. a3 2 27 a 2 4 for a 5 22 • C H A L L E N G E 87. The Earth’s population is approximately 7.3 3 109 people, and there are approximately 1.5 3 108 square kilometers of land on the surface of the Earth. If one square kilometer is approximately 247 acres, how many acres per person are there on Earth? Round to the nearest tenth of an acre. 89. Evaluate: 14 3 10223213 3 10122 16 3 102102 . Express your answer in both scientific and decimal notation. 88. The population of the United States is approximately 3.3 3 108 people, and there are approximately 3.79 3 106 square miles of land in the United States. If one square mile is approximately 640 acres, how many acres per person are there in the United States? Round to the nearest tenth of an acre. 90. Evaluate: 12 3 10217215 3 10132 11 3 10262 . Express your answer in both scientific and decimal notation. • T E C H N O L O G Y Scientific calculators have an EXP button that is used for scientific notation. For example, 2.5 3 103 can be input into the calculator by pressing 2.5 EXP 3. 91. Repeat Exercise 87 and confirm your answer with a calculator. 92. Repeat Exercise 88 and confirm your answer with a calculator. In Exercises 93 and 94, use a graphing utility or scientific calculator to evaluate the expression. Express your answer in scientific notation. 93. 17.35 3 10226212.19 3 10192 13.15 3 102212 94. 11.6849 3 10322 18.12 3 1016213.32 3 10292 0.3 Polynomials: Basic Operations 27 0.3.1 Adding and Subtracting Polynomials Polynomials in Standard Form The expressions 3x2 2 7x 2 1 4y3 2 y 5z are all examples of polynomials in one variable. A monomial in one variable, axk, is the product of a constant and a variable raised to a nonnegative-integer power. The constant a is called the coefficient of the monomial, and k is called the degree of the monomial. A polynomial is the sum of monomials. The monomials that are part of a polynomial are called terms. DEFINITION Polynomial A polynomial in x is an algebraic expression of the form anxn 1 an21xn21 1 an22xn22 1 c1 a2x2 1 a1x 1 a0 where a0, a1, a2, . . . , an are real numbers, with an 2 0, and n is a nonnegative integer. The polynomial is of degree n, an is the leading coefficient, and a0 is the constant term. Polynomials with one, two, and three terms are called monomials, binomials, and trinomials, respectively. Polynomials are typically written in standard form in order of decreasing degrees, and the degree of the polynomial is determined by the highest degree (exponent) of any single term. 0.3.1 S K IL L Add and subtract polynomials. 0.3.1 C ON C E P T U A L Recognize like terms. POLYNOMIAL STANDARD FORM SPECIAL NAME DEGREE DESCRIPTION 4x3 2 5x7 1 2x 2 6 25x7 1 4x3 1 2x 2 6 Polynomial 7 A seventh-degree polynomial in x 5 1 2y3 2 4y 2y3 2 4y 1 5 Trinomial 3 A third-degree polynomial in y 7z2 1 2 7z2 1 2 Binomial 2 A second-degree polynomial in z 217x5 217x5 Monomial 5 A fifth-degree monomial in x S K I L L S O B J E C T I V E S ■ ■Add and subtract polynomials. ■ ■Multiply polynomials. ■ ■Recognize special products and use them to perform operations on binomials. C O N C E P T U A L O B J E C T I V ES ■ ■Recognize like terms. ■ ■Understand how the distributive property is applied. ■ ■Learn formulas for special products. 0.3 POLYNOMIALS: BASIC OPERATIONS ▼ A N S W E R 24x3 1 17x2 2 x 1 5 Degree: 3 Leading coefficient: 24 Constant term: 5 EXAMPLE 1 Writing Polynomials in Standard Form Write the polynomials in standard form and state their degree, leading coefficient, and constant term. a. 4x 2 9x5 1 2 b. 3 2 x2 c. 3x2 2 8 1 14x3 2 20x8 1 x d. 27x3 1 25x Solution: Standard Form Degree Leading Coefficient Constant Term a. 29x5 1 4x 1 2 5 29 2 b. 2x2 1 3 2 21 3 c. 220x8 1 14x3 1 3x2 1 x 2 8 8 220 28 d. 27x3 1 25x 3 27 0 YOUR T UR N Write the polynomial in standard form and state its degree, leading coefficient, and constant term. 17x2 2 4x3 1 5 2 x ▼ 28 CHAPTER 0 Prerequisites and Review Adding and Subtracting Polynomials Polynomials are added and subtracted by combining like terms. Like terms are terms having the same variables and exponents. Like terms can be combined by adding their coefficients. WORDS MATH Identify like terms. 3x2 1 2x 1 4x2 1 5 Add coefficients of like terms. 7x2 1 2x 1 5 Note: The 2x and 5 could not be combined because they are not like terms. EXAMPLE 2 Adding Polynomials Find the sum and simplify 15x2 2 2x 1 32 1 13x3 2 4x2 1 72. Solution: Eliminate parentheses. 5x2 2 2x 1 3 1 3x3 2 4x2 1 7 Identify like terms. 5x2 2 2x 1 3 1 3x3 2 4x2 1 7 Combine like terms. x2 2 2x 1 10 1 3x3 Write in standard form. 3x3 1 x2 2 2x 1 10 Y OUR TU R N Find the sum and simplify: 13x2 1 5x 2 2x52 1 16x3 2 x2 1 112 ▼ EXAMPLE 3 Subtracting Polynomials Find the difference and simplify 13x3 2 2x 1 12 2 1x2 1 5x 2 92. common mistake Distributing the negative to only the first term in the second polynomial. Y OUR TU R N Find the difference and simplify: 127x2 2 x 1 52 2 12 2 x3 1 3x2 ▼ A N S W E R x3 2 7x2 2 4x 1 3 ✓C O R R EC T Eliminate the parentheses. 3x3 2 2x 1 1 2 x2 2 5x 1 9 Identify like terms. 3x3 2 2x 1 1 2 x2 2 5x 1 9 Combine like terms. 3x3 2 x2 2 7x 1 10 ✖I N C O R R EC T Error: 3x3 2 2x 1 1 2 x2 1 5x 2 9 Don’t forget to distribute the negative through the entire second polynomial. ▼ 0.3.2 Multiplying Polynomials The product of two monomials is found by using the properties of exponents (Section 0.2). For example, 125x32 19x22 5 1252 192x312 5 245x5 To multiply a monomial and a polynomial, we use the distributive property (Section 0.1). [CONCEPT CHECK] Determine the two like terms in: ax3 2 dx 2 b and kx4 1 cx3 ANSWER ax3 and cx3 ▼ 0.3.2 S KILL Multiply polynomials. 0.3.2 C ON CEPTUAL Understand how the distributive property is applied. STUDY TIP When subtracting polynomials, it is important to distribute the negative through all of the terms in the second polynomial. ▼ A N S W E R 22x5 1 6x3 1 2x2 1 5x 1 11 EXAMPLE 4 Multiplying a Monomial and a Polynomial Find the product and simplify 5x213x5 2 x3 1 7x 2 42. Solution: Use the distributive property. 5x213x5 2 x3 1 7x 2 42 5 5x213x52 2 5x21x32 1 5x2 17x2 2 5x2 142 Multiply each set of monomials. 5 15x7 2 5x5 1 35x3 2 20x2 YOUR T UR N Find the product and simplify 3x2 14x2 2 2x 1 12. ➤ ➤ ➤ ▼ A N S W E R 12x4 2 6x3 1 3x2 ▼ ➤ How do we multiply two polynomials if neither one is a monomial? For example, how do we find the product of a binomial and a trinomial such as 12x 2 52 1x2 2 2x 1 32? Notice that the binomial is a combination of two monomials. Therefore, we treat each monomial, 2x and 25, separately and then combine our results. In other words, use the distributive property repeatedly. WORDS MATH Apply the distributive property. 12x 2 521x2 2 2x 1 32 5 2x1x2 2 2x 1 32 2 51x2 2 2x 1 32 Apply the distributive property. 5 12x21x22 1 12x2122x2 1 12x2132 2 51x22 2 5 122x2 2 5 132 Multiply the monomials. 5 2x3 2 4x2 1 6x 2 5x2 1 10x 2 15 Combine like terms. 5 2x3 2 9x2 1 16x 2 15 EXAMPLE 5 Multiplying Two Polynomials Multiply and simplify 12x2 2 3x 1 121x2 2 5x 1 72. Solution: Multiply each term of the first trinomial by the entire second trinomial. 5 2x21x2 2 5x 1 72 2 3x1x2 2 5x 1 72 1 11x2 2 5x 1 72 Identify like terms. 5 2x4 2 10x3 1 14x2 2 3x3 1 15x2 2 21x 1 x2 2 5x 1 7 Combine like terms. 5 2x4 2 13x3 1 30x2 2 26x 1 7 Y OUR T UR N Multiply and simplify 12x3 1 2x 2 4213x2 2 x 1 52. ▼ A N S W E R 23x5 1 x4 1 x3 2 14x2 1 14x 2 20 ▼ 0.3.3 Special Products The method outlined for multiplying polynomials works for all products of polynomi-als. For the special case when both polynomials are binomials, the FOIL method can also be used. [CONCEPT CHECK] Multiply and Simplify (ax 1 b)(cx 2 d ) ANSWER acx2 1 (bc 2 ad )x 2 bd ▼ 0.3 Polynomials: Basic Operations 29 30 CHAPTER 0 Prerequisites and Review EXAMPLE 6 Multiplying Binomials Using the FOIL Method Multiply 13x 1 12 12x 2 52 using the FOIL method. Solution: Multiply the first terms. 13x2 12x2 5 6x2 Multiply the outer terms. 13x2 1252 5 215x Multiply the inner terms. 112 12x2 5 2x Multiply the last terms. 112 1252 5 25 Add the first, outer, inner, and last terms, and identify the like terms. 13x 1 1212x 2 52 5 6x2 2 15x 1 2x 2 5 Combine like terms. 5 6x2 2 13x 2 5 Y OUR TU R N Multiply 12x 2 32 15x 2 22. ▼ A N S W E R 10x2 2 19x 1 6 ▼ WORDS MATH Apply the distributive property. 15x 2 12 12x 1 32 5 5x 12x 1 32 2 1 12x 1 32 Apply the distributive property. 5 5x 12x2 1 5x 132 2 1 12x2 2 1 132 Multiply each set of monomials. 5 10x2 1 15x 2 2x 2 3 Combine like terms. 5 10x2 1 13x 2 3 The FOIL method finds the products of the First terms, Outer terms, Inner terms, and Last terms. Some products of binomials occur frequently in algebra and are given special names. Example 7 illustrates the difference of two squares and perfect squares. Let a and b be any real number, variable, or algebraic expression in the following special products. EXAMPLE 7 Multiplying Binomials Resulting in Special Products Find the following: a. 1x 2 52 1x 1 52 b. 1x 1 522 c. 1x 2 522 Solution: a. 1x 2 521x 1 52 5 x2 1 5x 2 5x 2 52 5 x2 2 52 5 x2 2 25 b. 1x 1 522 5 1x 1 521x 1 52 5 x2 1 5x 1 5x 1 52 5 x2 1 215x2 1 52 5 x2 1 10x 1 25 c. 1x 2 522 5 1x 2 521x 2 52 5 x2 2 5x 2 5x 1 52 5 x2 2 215x2 1 52 5 x2 2 10x 1 25 f First f Outer f Inner f Difference of two squares f Last f First f Outer f Last f Inner f First f Last f Outer f Inner 0.3.3 SKI LL Recognize special products and use them to perform operations on binomials. 0.3.3 CO NCE PTUAL Learn formulas for special products. STUDY TIP When the binomials are of the form (ax 1 b)(cx 1 d ), the outer and inner terms will be like terms and can be combined. 15x 2 12 12x 1 32 5 10x2 1 15x 2 2x 2 3 Product of First Terms 15x2 12x2 Product of Outer Terms 15x2 132 Product of Last Terms 1212 132 Product of Inner Terms 1212 12x2 EXAMPLE 9 Using Special Product Formulas Find the following: a. 12x 2 122 b. 13 1 2y22 c. 14x 1 32 14x 2 32 Solution (a): Write the square of a binomial difference formula. 1a 2 b22 5 a2 2 2ab 1 b2 Let a 5 2x and b 5 1. 12x 2 122 5 12x22 2 2 12x2 112 1 12 Simplify. 5 4x2 2 4x 1 1 Solution (b): Write the square of a binomial sum formula. 1a 1 b22 5 a2 1 2ab 1 b2 Let a 5 3 and b 5 2y. 13 1 2y22 5 1322 1 2 132 12y2 1 12y22 Simplify. 5 9 1 12y 1 4y2 Write in standard form. 5 4y2 1 12y 1 9 Solution (c): Write the difference of two squares formula. 1a 1 b2 1a 2 b2 5 a2 2 b2 Let a 5 4x and b 5 3. 14x 1 32 14x 2 32 5 14x22 2 32 Simplify. 5 16x2 2 9 YOUR T UR N Find the following: a. 13x 1 122 b. 11 2 3y22 c. 13x 1 22 13x 2 22 ▼ A N S W E R a. 9x2 1 6x 1 1 b. 9y2 2 6y 1 1 c. 9x2 2 4 ▼ EXAMPLE 8 Finding the Square of a Binomial Sum Find 1x 1 322. common mistake Forgetting the middle term, which is twice the product of the two terms in the binomial. ✓CO R R E CT 1x 1 322 5 1x 1 32 1x 1 32 5 x2 1 3x 1 3x 1 9 5 x2 1 6x 1 9 ✖I NC O R R EC T Error: 1x 1 322 2 x2 1 9 Don’t forget the middle term, which is twice the product of the two terms in the binomial. STUDY TIP (a 1 b)(a 2 b) 5 a2 2 ab 1 ab 2 b2 5 a2 2 b2 DIFFERENCE OF TWO SQUARES 1a 1 b21a 2 b2 5 a2 2 b2 PERFECT SQUARES Square of a binomial sum: 1a 1 b22 5 1a 1 b2 1a 1 b2 5 a2 1 2ab 1 b2 Square of a binomial difference: 1a 2 b22 5 1a 2 b2 1a 2 b2 5 a2 2 2ab 1 b2 0.3 Polynomials: Basic Operations 31 32 CHAPTER 0 Prerequisites and Review [CONCEPT CHECK] Multiply and simplify: (ax 1 by)(ax 2 by) ANSWER a2x2 2 b2y2 ▼ EXAMPLE 11 Applying the Special Product Formulas Find the following: a. 12x 1 123 b. 12x 2 523 Solution (a): Write the cube of a binomial sum formula. 1a 1 b23 5 a3 1 3a2b 1 3ab2 1 b3 Let a 5 2x and b 5 1. 12x 1 123 5 12x23 1 3 12x22 112 1 3 12x2 1122 1 13 Simplify. 5 8x3 1 12x2 1 6x 1 1 Solution (b): Write the cube of a binomial difference formula. 1a 2 b23 5 a3 2 3a2b 1 3ab2 2 b3 Let a 5 2x and b 5 5. 12x 2 523 5 12x23 2 3 12x22 152 1 3 12x2 1522 2 53 Simplify. 5 8x3 2 60x2 1 150x 2 125 Y OUR TU R N Find 13x 2 423. ▼ A N S W E R 27x3 2 108x2 1 144x 2 64 ▼ ▼ C A U T I O N 1x 1 223 2 x3 1 8 1x 2 223 2 x3 2 8 EXAMPLE 10 Cubing a Binomial Find the following: a. 1x 1 223 b. 1x 2 223 Solution (a): Write the cube as a product of 1x 1 223 5 1x 1 221x 1 221x 1 22 three binomials. Apply the perfect square formula. 5 1x 1 221x2 1 4x 1 42 Apply the distributive property. 5 x1x2 1 4x 1 42 1 21x2 1 4x 1 42 Apply the distributive property. 5 x3 1 4x2 1 4x 1 2x2 1 8x 1 8 Combine like terms. 5 x3 1 6x2 1 12x 1 8 Solution (b): Write the cube as a product of 1x 2 223 5 1x 2 221x 2 221x 2 22 three binomials. Apply the perfect square formula. 5 1x 2 22 1x2 2 4x 1 42 Apply the distributive property. 5 x 1x2 2 4x 1 42 2 2 1x2 2 4x 1 42 Apply the distributive property. 5 x3 2 4x2 1 4x 2 2x2 1 8x 2 8 Combine like terms. 5 x3 2 6x2 1 12x 2 8 f 1x 1 222 f 1x 2 222 PERFECT CUBES Cube of a binomial sum: 1a 1 b23 5 a3 1 3a2b 1 3ab2 1 b3 Cube of a binomial difference: 1a 2 b23 5 a3 2 3a2b 1 3ab2 2 b3 EXAMPLE 12 Applying the Special Product Formulas for Binomials in Two Variables Find 12x 2 3y22. Solution: Write the square of a binomial difference formula. 1a 2 b22 5 1a 2 b2 1a 2 b2 5 a2 2 2ab 1 b2 Let a 5 2x and b 5 3y. 12x 2 3y22 5 12x22 2 2 12x2 13y2 1 13y22 Simplify. 5 4x2 2 12xy 1 9y2 YOUR T UR N Find 13x 2 2y22. ▼ A N S W E R 9x2 2 12xy 1 4y2 ▼ PER F EC T S Q U A R ES Square of a binomial sum. 1a 1 b22 5 1a 1 b2 1a 1 b2 5 a2 1 2ab 1 b2 Square of a binomial difference. 1a 2 b22 5 1a 2 b2 1a 2 b2 5 a2 2 2ab 1 b2 PER F EC T C U B ES Cube of a binomial sum. 1a 1 b23 5 a3 1 3a2b 1 3ab2 1 b3 Cube of a binomial difference. 1a 2 b23 5 a3 2 3a2b 1 3ab2 2 b3 In this section, polynomials were defined. Polynomials with one, two, and three terms are called monomials, binomials, and trinomials, respectively. Polynomials are added and subtracted by combining like terms. Polynomials are multiplied by distributing the monomials in the first polynomial throughout the second polynomial. In the special case of the product of two binomials, the FOIL method can also be used. The following are special products of binomials. DIF FE R E NC E OF T W O S Q UA R ES 1a 1 b2 1a 2 b2 5 a2 2 b2 [SEC TION 0.3] S U M M A RY [SEC TION 0.3] E X E RC I S E S • S K I L L S In Exercises 1–8, write the polynomial in standard form and state the degree of the polynomial. 1. 5x2 2 2x3 1 16 2 7x4 2. 7x3 2 9x2 1 5x 2 4 3. 4x 1 3 2 6x3 4. 5x5 2 7x3 1 8x4 2 x2 1 10 5. 15 6. 214 7. y 2 2 8. x 2 5 In Exercises 9–24, add or subtract the polynomials, gather like terms, and write the simplified expression in standard form. 9. 12x2 2 x 1 72 1 123x2 1 6x 2 22 10. 13x2 1 5x 1 22 1 12x2 2 4x 2 92 11. 127x2 2 5x 2 82 2 124x 2 9x2 1 102 12. 18x3 2 7x2 2 102 2 17x3 1 8x2 2 9x2 13. 12x4 2 7x2 1 82 2 13x2 2 2x4 1 92 14. 14x2 2 9x 2 22 2 15 2 3x 2 5x22 15. 17z2 2 22 2 15z2 2 2z 1 12 16. 125y3 2 7y2 1 9y2 2 114y2 2 7y 1 22 17. 13y3 2 7y2 1 8y 2 42 2 114y3 2 8y 1 9y22 18. 12x2 1 3xy2 2 1x2 1 8xy 2 7y22 19. 16x 2 2y2 2 2 15x 2 7y2 20. 3a 2 32a2 2 15a 2 4a2 1 324 21. 12x2 2 22 2 1x 1 12 2 1x2 2 52 22. 13x3 1 12 2 13x2 2 12 2 15x 2 32 23. 14t 2 t2 2 t32 2 13t2 2 2t 1 2t32 1 13t3 2 12 24. 12z3 2 2z22 1 1z2 2 7z 1 12 2 14z3 1 3z2 2 3z 1 22 0.3 Polynomials: Basic Operations 33 34 CHAPTER 0 Prerequisites and Review In Exercises 25–64, multiply the polynomials and write the expressions in standard form. 25. 5xy2 17xy2 26. 6z 14z32 27. 2x3 11 2 x 1 x22 28. 24z2 12 1 z 2 z22 29. 22x2 15 1 x 2 5x22 30. 21 2z 12z 1 4z2 2 102 31. 1x2 1 x 2 222x3 32. 1x2 2 x 1 223x3 33. 2ab2 1a2 1 2ab 2 3b22 34. bc3d2 1b2c 1 cd3 2 b2d42 35. 12x 1 12 13x 2 42 36. 13z 2 12 14z 1 72 37. 1x 1 22 1x 2 22 38. 1 y 2 52 1y 1 52 39. 12x 1 32 12x 2 32 40. 15y 1 12 15y 2 12 41. 12x 2 12 11 2 2x2 42. 14b 2 5y2 14b 1 5y2 43. 12x2 2 32 12x2 1 32 44. 14xy 2 92 14xy 1 92 45. 17y 2 2y22 1 y 2 y2 1 12 46. 14 2 t22 16t 1 1 2 t22 47. 1x 1 12 1x2 2 2x 1 32 48. 1x 1 32 1x2 2 3x 1 92 49. 1t 2 222 50. 1t 2 322 51. 1z 1 222 52. 1z 1 322 53. 3 1x 1 y2 2 342 54. 12x2 1 3y22 55. 15x 2 222 56. 1x 1 12 1x2 1 x 1 12 57. y 13y 1 42 12y 2 12 58. p2 1p 1 12 1p 2 22 59. 1x2 1 12 1x2 2 12 60. 1t 2 522 1t 1 522 61. 1b 2 3a2 1a 1 2b2 1b 1 3a2 62. 1x 2 2y2 1x2 1 2xy 1 4y22 63. 1x 1 y 2 z2 12x 2 3y 1 5z2 64. 15b2 2 2b 1 12 13b 2 b2 1 22 • A P P L I C A T I O N S In Exercises 65–68, profit is equal to revenue minus cost: P 5 R 2 C. 65. Profit. Donna decides to sell fabric cord covers on eBay for $20 apiece. The material for each cord cover costs $9, and it costs her $100 a month to advertise on eBay. Let x be the number of cord covers sold. Write a polynomial representing her monthly profit. 66. Profit. Calculators are sold for $25 each. Advertising costs are $75 per month. Let x be the number of calculators sold. Write a polynomial representing the monthly profit earned by selling x calculators. 67. Profit. If the revenue associated with selling x units of a product is R 5 2x2 1 100x, and the cost associated with producing x units of the product is C 5 2100x 1 7500, find the polynomial that represents the profit of making and selling x units. 68. Profit. A business sells a certain quantity x of items. The revenue generated by selling x items is given by the equation R 5 21 2x2 1 50x. The costs are given by C 5 8000 2 150x. Find a polynomial representing the net profit of this business when x items are sold. 69. Volume of a Box. A rectangular sheet of cardboard is to be used in the construction of a box by cutting out squares of side length x from each corner and turning up the sides. Suppose the dimensions of the original rectangle are 15 inches by 8 inches. Determine a polynomial in x that would give the volume of the box. 70. Volume of a Box. Suppose a box is to be constructed from a square piece of material of side length x by cutting out a 2-inch square from each corner and turning up the sides. Express the volume of the box as a polynomial in the variable x. 71. Geometry. Suppose a running track is constructed of a rectangular portion that measures 2x feet wide by 2x 1 5 feet long. Each end of the rectangular portion consists of a semicircle whose diameter is 2x. Write a polynomial that determines the a. perimeter of the track in terms of the variable x. b. area of the track in terms of x. 2x + 5 2x 2x 72. Geometry. A right circular cylinder whose radius is r and whose height is 2r is surmounted by a hemisphere of radius r. a. Find a polynomial in the variable r that represents the volume of the “silo” shape. b. Find a polynomial in r that represents the total surface area of the “silo.” r r 2r 73. Engineering. The force of an electrical field is given by the equation F 5 k q1q2 r 2 . Suppose q1 5 x, q2 5 3x, and r 5 10x. Find a polynomial representing the force of the electrical field in terms of the variable x. 74. Engineering. If a football (or other projectile) is thrown upward, its height above the ground is given by the equation s 5 16t2 1 v0t 1 s0, where v0 and s0 are the initial velocity and initial height of the football, respectively, and t is the time in seconds. Suppose the football is thrown from the top of a building that is 192 feet tall, with an initial speed of 96 feet per second. a. Write the polynomial that gives the height of the football in terms of the variable t (time). b. What is the height of the football after 2 sec onds have elapsed? Will the football hit the ground after 2 seconds? In Exercises 77–80, determine whether each of the following statements is true or false. 77. All binomials are polynomials. 78. The product of two monomials is a binomial. 79. 1x 1 y23 5 x 3 1 y3 80. 1x 2 y2 2 5 x 2 1 y2 In Exercises 81 and 82, let m and n be real numbers and m + n. 81. What degree is the product of a polynomial of degree n and a polynomial of degree m? 82. What degree is the sum of a polynomial of degree n and a polynomial of degree m? • C O N C E P T U A L 0.3 Polynomials: Basic Operations 35 In Exercises 75 and 76, explain the mistake that is made. • C A T C H T H E M I S T A K E 75. Subtract and simplify 12x2 2 52 2 13x 2 x2 1 12. Solution: Eliminate the parentheses. 2x2 2 5 2 3x 2 x2 1 1 Collect like terms. x2 2 3x 2 4 This is incorrect. What mistake was made? 76. Simplify 12 1 x22. Solution: Write the square of the binomial as the sum of the squares. 12 1 x22 5 22 1 x2 Simplify. 5 x2 1 4 This is incorrect. What mistake was made? In Exercises 83–86, perform the indicated operations and simplify. 83. 17x 2 4y2 2 2 17x 1 4y 2 2 2 84. 13x 2 5y222 13x 1 5y222 85. 1x 2 a2 1x2 1 ax 1 a22 86. 1x 1 a2 1x2 2 ax 1 a22 • C H A L L E N G E • T E C H N O L O G Y 87. Use a graphing utility to plot the graphs of the three expressions 12x 1 321x 2 42, 2x2 1 5x 2 12, and 2x2 2 5x 2 12. Which two graphs agree with each other? 88. Use a graphing utility to plot the graphs of the three expressions 1x 1 522, x2 1 25, and x2 1 10x 1 25. Which two graphs agree? 36 CHAPTER 0 Prerequisites and Review In Section 0.3, we discussed multiplying polynomials. In this section, we examine the reverse of that process, which is called factoring. Consider the following product: 1x 1 321x 1 12 5 x2 1 4x 1 3 To factor the resulting polynomial, you reverse the process to undo the multiplication: x2 1 4x 1 3 5 1x 1 321x 1 12 The polynomials 1x 1 32 and 1x 1 12 are called factors of the polynomial x2 1 4x 1 3. The process of writing a polynomial as a product is called factoring. In Chapter 1 we will solve quadratic equations by factoring. In this section, we will restrict our discussion to factoring polynomials with integer coefficients, which is called factoring over the integers. If a polynomial cannot be factored using integer coefficients, then it is prime or irreducible over the integers. When a polynomial is written as a product of prime polynomials, then the polynomial is said to be factored completely. 0.4.1 Greatest Common Factor The simplest type of factoring of polynomials occurs when there is a factor common to every term of the polynomial. This common factor is a monomial that can be “factored out” by applying the distributive property in reverse: ab 1 ac 5 a1b 1 c2 For example, 2x2 2 6x can be written as 2x 1x2 2 2x 132. Notice that 2x is a common factor to both terms, so the distributive property tells us we can factor this polynomial to yield 2x 1x 2 32. Although 2 is a common factor and x is a common factor, the monomial 2x is called the greatest common factor. POLYNOMIAL GCF WRITE EACH TERM AS A PRODUCT OF GCF AND REMAINING FACTOR FACTORED FORM 7x 1 21 7 7 1x2 1 7 132 7 1x 1 32 3x2 1 12x 3x 3x 1x2 1 3x 142 3x 1x 1 42 4x3 1 2x 1 6 2 2 12x32 1 2x 1 2 132 2 12x3 1 x 1 32 6x4 2 9x3 1 12x2 3x2 3x2 12x22 2 3x2 13x2 1 3x2 142 3x2 12x2 2 3x 1 42 25x4 1 25x3 2 20x2 25x2 25x2 1x22 2 5x2 125x2 2 5x2 142 25x2 1x2 2 5x 1 42 0.4.1 S KILL Factor out the greatest common factor. 0.4.1 CO NCE PTUAL Understand that factoring has its basis in the distributive property. GREATEST COMMON FACTOR The monomial axk is called the greatest common factor (GCF) of a polynomial in x with integer coefficients if both of the following are true: ■ ■ a is the greatest integer factor common to all of the polynomial coefficients. ■ ■ k is the smallest exponent on x found in all the terms of the polynomial. S K I L L S O B J E C T I V E S ■ ■Factor out the greatest common factor. ■ ■Factor polynomials using special forms. ■ ■Factor a trinomial as a product of binomials. ■ ■Factor polynomials by grouping. ■ ■Follow a strategy to factor polynomials. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that factoring has its basis in the distributive property. ■ ■Learn formulas for factoring special forms. ■ ■Know the general strategy for factoring a trinomial as a product of binomials. ■ ■Be able to identify polynomials that may be factored by grouping. ■ ■Develop a general strategy for factoring polynomials. 0.4 FACTORING POLYNOMIALS 0.4 Factoring Polynomials 37 0.4.2 Factoring Formulas: Special Polynomial Forms The first step in factoring polynomials is to look for a common factor. If there is no common factor, then we look for special polynomial forms that we learned were special products in Section 0.3 and reverse the process. Difference of two squares a2 2 b2 5 1a 1 b2 1a 2 b2 Perfect squares a2 1 2ab 1 b2 5 1a 1 b22 a2 2 2ab 1 b2 5 1a 2 b22 Sum of two cubes a3 1 b3 5 1a 1 b2 1a2 2 ab 1 b22 Difference of two cubes a3 2 b3 5 1a 2 b2 1a2 1 ab 1 b22 0.4.2 S K I L L Factor polynomials using special forms. 0.4.2 C ON C E P T U A L Learn formulas for factoring special forms. EXAMPLE 1 Factoring Polynomials by Extracting the Greatest Common Factor Factor: a. 6x5 2 18x4 b. 6x5 2 10x4 2 8x3 1 12x2 Solution (a): Identify the greatest common factor. 6x4 Write each term as a product with the GCF as a factor. 6x5 2 18x4 5 6x4 1x2 2 6x4 132 Factor out the GCF. 5 6x4 1x 2 32 Solution (b): Identify the greatest common factor. 2x2 Write each term as a product with the GCF as a factor. 6x5 2 10x4 2 8x3 1 12x2 5 2x2 13x32 2 2x2 15x22 2 2x2 14x2 1 2x2 162 Factor out the GCF. 5 2x2 13x3 2 5x2 2 4x 1 62 YOUR T UR N Factor: a. 12x3 2 4x b. 3x5 2 9x4 1 12x3 2 6x2 ▼ A N S W E R a. 4x 13x2 2 12 b. 3x2 1x3 2 3x2 1 4x 2 22 ▼ [CONCEPT CHECK] Find the greatest common factor of the polynomial a3x6 1 a4x4 ANSWER a3x4 ▼ [CONCEPT CHECK] Factor: a2x4 2 b2y2 ANSWER (ax2 1 by)(ax2 2 by) ▼ EXAMPLE 2 Factoring the Difference of Two Squares Factor: a. x2 2 9 b. 4x2 2 25 c. x4 2 16 Solution (a): Rewrite as the difference of two squares. x2 2 9 5 x2 2 32 Let a 5 x and b 5 3 in a2 2 b2 5 1a 1 b2 1a 2 b2. 5 1x 1 32 1x 2 32 Solution (b): Rewrite as the difference of two squares. 4x2 2 25 5 12x22 2 52 Let a 5 2x and b 5 5 in a2 2 b2 5 1a 1 b2 1a 2 b2. 5 12x 1 52 12x 2 52 Solution (c): Rewrite as the difference of two squares. x4 2 16 5 1x222 2 42 Let a 5 x2 and b 5 4 in a2 2 b2 5 1a 1 b2 1a 2 b2. 5 1x2 1 421x2 2 42 Note that x2 2 4 is also a difference of two squares (part a of the following Your Turn). 5 1x 1 22 1x 2 22 1x2 1 42 Y OUR T UR N Factor: a. x2 2 4 b. 9x2 2 16 c. x4 2 81 ▼ ▼ A N S W E R a. 1x 1 22 1x 2 22 b. 13x 1 42 13x 2 42 c. 1x2 1 92 1x 2 32 1x 1 32 38 CHAPTER 0 Prerequisites and Review A trinomial is a perfect square if it has the form a2 6 2ab 1 b2. Notice that: ■ ■The first term and third term are perfect squares. ■ ■The middle term is twice the product of the bases of these two perfect squares. ■ ■The sign of the middle term determines the sign of the factored form: a2 6 2ab 1 b2 5 1a 6 b22 EXAMPLE 3 Factoring Trinomials That Are Perfect Squares Factor: a. x2 1 6x 1 9 b. x2 2 10x 1 25 c. 9x2 2 12x 1 4 Solution (a): Rewrite the trinomial so that the first and third terms are perfect squares. x2 1 6x 1 9 5 x2 1 6x 1 32 Notice that if we let a 5 x and b 5 3 in a2 1 2ab 1 b2 5 1a 1 b22, then the middle term 6x is 2ab. x2 1 6x 1 9 5 x2 1 2 13x2 1 32 5 1x 1 322 Solution (b): Rewrite the trinomial so that the first and third terms are perfect squares. x2 2 10x 1 25 5 x2 2 10x 1 52 Notice that if we let a 5 x and b 5 5 in a2 2 2ab 1 b2 5 1a 2 b22, then the middle term 210x is 22ab. x2 2 10x 1 25 5 x2 2 2 15x2 1 52 5 1x 2 522 Solution (c): Rewrite the trinomial so that the first and third terms are perfect squares. 9x2 2 12x 1 4 5 13x22 2 12x 1 22 Notice that if we let a 5 3x and b 5 2 in a2 2 2ab 1 b2 5 1a 2 b22, then the middle term 212x is 22ab. 9x2 2 12x 1 4 5 13x22 2 2 13x2 122 1 22 5 13x 2 222 Y OUR TU R N Factor: a. x2 1 8x 1 16 b. x2 2 4x 1 4 c. 25x2 2 20x 1 4 ▼ A N S W E R a. 1x 1 422 b. 1x 2 222 c. 15x 2 222 ▼ EXAMPLE 4 Factoring the Sum of Two Cubes Factor x3 1 27. Solution: Rewrite as the sum of two cubes. x3 1 27 5 x3 1 33 Write the sum of two cubes formula. a3 1 b3 5 1a 1 b2 1a2 2 ab 1 b22 Let a 5 x and b 5 3. x3 1 27 5 x3 1 33 5 1x 1 32 1x2 2 3x 1 92 0.4.3 Factoring a Trinomial as a Product of Two Binomials The first step in factoring is to look for a common factor. If there is no common factor, look to see whether the polynomial is a special form for which we know the factoring formula. If it is not of such a special form and if it is a trinomial, then we proceed with a general factoring strategy. We know that 1x 1 32 1x 1 22 5 x2 1 5x 1 6, so we say the factors of x2 1 5x 1 6 are 1x 1 32 and 1x 1 22. In factored form we have x2 1 5x 1 6 5 1x 1 32 1x 1 22. Recall the FOIL method from Section 0.3. The product of the last terms 13 and 22 is 6, and the sum of the products of the inner terms 13x2 and the outer terms 12x2 is 5x. Let’s pretend for a minute that we didn’t know this factored form but had to work with the general form: x2 1 5x 1 6 5 1x 1 a21x 1 b2 The goal is to find a and b. We start by multiplying the two binomials on the right. x2 1 5x 1 6 5 1x 1 a21x 1 b2 5 x2 1 ax 1 bx 1 ab 5 x2 1 1a 1 b2x 1 ab Compare the expression we started with on the left with the expression on the far right x2 1 5x 1 6 5 x2 1 1a 1 b2 x 1 ab. We see that ab 5 6 and 1a 1 b2 5 5. Start with the possible combinations of a and b whose product is 6, and then look among those for the combination whose sum is 5. ab 5 6 a, b: 1, 6 21, 26 2, 3 22, 23 a 1 b 7 27 5 25 All of the possible a, b combinations in the first row have a product equal to 6, but only one of those has a sum equal to 5. Therefore, the factored form is x2 1 5x 1 6 5 1x 1 a21x 1 b2 5 1x 1 221x 1 32 0.4.3 S K I L L Factor a trinomial as a product of binomials. 0.4.3 C ON C E P T U A L Know the general strategy for factoring a trinomial as a product of binomials. EXAMPLE 5 Factoring the Difference of Two Cubes Factor x3 2 125. Solution: Rewrite as the difference of two cubes. x3 2 125 5 x3 2 53 Write the difference of two cubes formula. a3 2 b3 5 1a 2 b2 1a2 1 ab 1 b22 Let a 5 x and b 5 5. x3 2 125 5 x3 2 53 5 1x 2 52 1x2 1 5x 1 252 YOUR T UR N Factor: a. x3 1 8 b. x3 2 64 ▼ A N S W E R a. 1x 1 22 1x2 2 2x 1 42 b. 1x 2 42 1x2 1 4x 1 162 ▼ EXAMPLE 6 Factoring a Trinomial Factor x2 1 10x 1 9. Solution: Write the trinomial as a product of two binomials in general form. x2 1 10x 1 9 5 1x 1 u2 1x 1 u2 0.4 Factoring Polynomials 39 40 CHAPTER 0 Prerequisites and Review EXAMPLE 7 Factoring a Trinomial Factor x2 2 3x 2 28. Solution: Write the trinomial as a product of two binomials in general form. x2 2 3x 2 28 5 1x 1 u2 1x 2 u2 Write all of the integers whose product is 228. Integers whose product is 228 1, 228 21, 28 2, 214 22, 14 4, 27 24, 7 Determine the sum of the integers. Integers whose product is 228 1, 228 21, 28 2, 214 22, 14 4, 27 24, 7 Sum 227 27 212 12 23 3 Select 4, 27 because the product is 228 (last term of the trinomial) and the sum is 23 (middle term coefficient of the trinomial). x2 2 3x 2 28 5 1x 1 42 1x 2 72 Check: 1x 1 42 1x 2 72 5 x2 2 7x 1 4x 2 28 5 x2 2 3x 2 28 3 Y OUR TU R N Factor x2 1 3x 2 18. ▼ A N S W E R 1x 1 62 1x 2 32 ▼ Write all of the integers whose product is 9. Integers whose product is 9 1, 9 21, 29 3, 3 23, 23 Determine the sum of the integers. Integers whose product is 9 1, 9 21, 29 3, 3 23, 23 Sum 10 210 6 26 Select 1, 9 because the product is 9 (last term of the trinomial) and the sum is 10 (middle term coefficient of the trinomial). x2 1 10x 1 9 5 1x 1 92 1x 1 12 Check: 1x 1 92 1x 1 12 5 x2 1 1x 1 9x 1 9 5 x2 1 10x 1 9 3 Y OUR TU R N Factor x2 1 9x 1 20. ▼ A N S W E R a. 1x 1 42 1x 1 52 ▼ In Example 6, all terms in the trinomial are positive. When the constant term is negative, then (regardless of whether the middle term is positive or negative) the factors will be opposite in sign, as illustrated in Example 7. When the leading coefficient of the trinomial is not equal to 1, then we consider all possible factors using the following procedure, which is based on the FOIL method in reverse. EXAMPLE 8 Factoring a Trinomial Whose Leading Coefficient Is Not 1 Factor 5x2 1 9x 2 2. Solution: STEP 1 Start with the first term. Note that 5x ? x 5 5x2. 15x 6 h2 1x 6 h2 STEP 2 The product of the last terms should yield 22. 21, 2 or 1, 22 STEP 3 Consider all possible factors based on Steps 1 and 2. 15x 2 12 1x 1 22 15x 1 12 1x 2 22 15x 1 22 1x 2 12 15x 2 22 1x 1 12 Since the outer and inner products must sum to 9x, the factored form m ust be: 5x2 1 9x 2 2 5 15x 2 12 1x 1 22 Check: 15x 2 12 1x 1 22 5 5x2 1 10x 2 1x 2 2 5 5x2 1 9x 2 2 3 YOUR T UR N Factor 2t2 1 t 2 3. ▼ A N S W E R 12t 1 32 1t 2 12 ▼ EXAMPLE 9 Factoring a Trinomial Whose Leading Coefficient Is Not 1 Factor 15x2 2 x 2 6. Solution: STEP 1 Start with the first term. 15x 6 h2 13x 6 h2 or 115x 6 h2 1x 6 h2 STEP 2 The product of the last terms should yield 26. 21, 6 or 1, 26 or 2, 23 or 22, 3 FACTORING A TRINOMIAL WHOSE LEADING COEFFICIENT IS NOT 1 Factors of a ax2 1 bx 1 c 5 1hx 1 h2 1hx 1 h2 Factors of c Step 1: Find two First terms whose product is the first term of the trinomial. Step 2: Find two Last terms whose product is the last term of the trinomial. Step 3: Consider all possible combinations found in Steps 1 and 2 until the sum of the Outer and Inner products is equal to the middle term of the trinomial. 0.4 Factoring Polynomials 41 42 CHAPTER 0 Prerequisites and Review STEP 3 Consider all possible factors 15x 2 12 13x 1 62 115x 2 12 1x 1 62 based on Steps 1 and 2. 15x 1 62 13x 2 12 115x 1 62 1x 2 12 15x 1 12 13x 2 62 115x 1 12 1x 2 62 15x 2 62 13x 1 12 115x 2 62 1x 1 12 15x 1 22 13x 2 32 115x 1 22 1x 2 32 15x 2 32 13x 1 22 115x 2 32 1x 1 22 15x 2 22 13x 1 32 115x 2 22 1x 1 32 15x 1 32 13x 2 22 115x 1 32 1x 2 22 Since the outer and inner products must sum to 2x, the factored form must be: 15x2 2 x 2 6 5 15x 1 32 13x 2 22 Check: 15x 1 32 13x 2 22 5 15x2 2 10x 1 9x 2 6 5 15x2 2 x 2 6 3 Y OUR TU R N Factor 6x2 1 x 2 12. ▼ A N S W E R 13x 2 42 12x 1 32 ▼ 0.4.4 Factoring by Grouping Much of our attention in this section has been on factoring trinomials. For polynomials with more than three terms we first look for a common factor to all terms. If there is no common factor to all terms of the polynomial, we look for a group of terms that have a common factor. This strategy is called factoring by grouping. 0.4.4 S KILL Factor polynomials by grouping. 0.4.4 CO NCE PTUAL Be able to identify polynomials that may be factored by grouping. [CONCEPT CHECK] TRUE OR FALSE A trinomial that is degree 2 and has integer coefficients with the leading coefficient equal to 1 is always factorable. ANSWER False ▼ EXAMPLE 11 Factoring a Polynomial by Grouping Factor x3 2 x2 1 2x 2 2. Solution: Group the terms that have a common factor. 5 1x3 2 x22 1 12x 2 22 Factor out the common factor in each pair of parentheses. 5 x2 1x 2 12 1 2 1x 2 12 Use the distributive property. 5 1x2 1 22 1x 2 12 STUDY TIP In Example 9, Step 3, we can eliminate any factors that have a common factor since there is no common factor to the terms in the trinomial. EXAMPLE 10 Identifying Prime (Irreducible) Polynomials Factor x2 1 x 2 8. Solution: Write the trinomial as a product of two binomials in general form. x2 1 x 2 8 5 1x 1 h2 1x 2 h2 Integers whose product is 28 1, 28 21, 8 4, 22 24, 2 Integers whose product is 28 1, 28 21, 8 4, 22 24, 2 Sum 27 7 2 22 The middle term of the trinomial is 2x, so we look for the sum of the integers that equals 21. Since no sum exists for the given combinations, we say that this polynomial is prime (irreducible) over the integers. Write all of the integers whose product is 28. Determine the sum of the integers. 0.4.5 A Strategy for Factoring Polynomials The first step in factoring a polynomial is to look for the greatest common factor. When specifically factoring trinomials, look for special known forms: a perfect square or a difference of two squares. A general approach to factoring a trinomial uses the FOIL method in reverse. Finally, we look for factoring by grouping. The following strategy for factoring polynomials is based on the techniques discussed in this section. 0.4.5 S K I L L Follow a strategy to factor polynomials. 0.4.5 C O N C E P T U A L Develop a general strategy for factoring polynomials. EXAMPLE 12 Factoring a Polynomial by Grouping Factor 2x2 1 2x 2 x 2 1. Solution: Group the terms that have a common factor. 5 12x2 1 2x2 1 12x 2 12 Factor out the common factor in each pair of parentheses. 5 2x 1x 1 12 2 1 1x 1 12 Use the distributive property. 5 12x 2 12 1x 1 12 Y OUR T UR N Factor x3 1 x2 2 3x 2 3. ▼ A N S W E R 1x 1 12 1x2 2 32 ▼ [CONCEPT CHECK] Factor ax2 2 ay2 ANSWER a(x 1 y)(x 2 y) ▼ [CONCEPT CHECK] TRUE OR FALSE Trinomials can not be factored by grouping ANSWER True ▼ STUDY TIP When factoring, always start by factoring out the GCF. STRATEGY FOR FACTORING POLYNOMIALS 1. Factor out the greatest common factor (monomial). 2. Identify any special polynomial forms and apply factoring formulas. 3. Factor a trinomial into a product of two binomials: 1ax 1 b2 1cx 1 d2. 4. Factor by grouping. 0.4 Factoring Polynomials 43 EXAMPLE 13 Factoring Polynomials Factor: a. 3x2 2 6x 1 3 b. 24x3 1 2x2 1 6x c. 15x2 1 7x 2 2 d. x3 2 x 1 2x2 2 2 Solution (a): Factor out the greatest common factor. 3x2 2 6x 1 3 5 3 1x2 2 2x 1 12 The trinomial is a perfect square. 5 3 1x 2 122 Solution (b): Factor out the greatest common factor. 24x3 1 2x2 1 6x 5 22x 12x2 2 x 2 32 Use the FOIL method in reverse to factor the trinomial. 5 22x 12x 2 32 1x 1 12 Solution (c): There is no common factor. 15x2 1 7x 2 2 Use the FOIL method in reverse to factor the trinomial. 5 13x 1 22 15x 2 12 Solution (d): Factor by grouping. x3 2 x 1 2x2 2 2 5 1x3 2 x2 1 12x2 2 22 5 x 1x2 2 12 1 2 1x2 2 12 5 1x 1 22 1x2 2 12 Factor the difference of two squares. 5 1x 1 22 1x 2 12 1x 1 12 44 CHAPTER 0 Prerequisites and Review In Exercises 1–12, factor each expression. Start by finding the greatest common factor (GCF). 1. 5x 1 25 2. x2 1 2x 3. 4t2 2 2 4. 16z2 2 20z 5. 2x3 2 50x 6. 4x2y 2 8xy2 1 16x2y2 7. 3x3 2 9x2 1 12x 8. 14x4 2 7x2 1 21x 9. x3 2 3x2 2 40x 10. 29y2 1 45y 11. 4x2y3 1 6xy 12. 3z3 2 6z2 1 18z In Exercises 13–20, factor the difference of two squares. 13. x2 2 9 14. x2 2 25 15. 4x2 2 9 16. 1 2 x4 17. 2x2 2 98 18. 144 2 81y2 19. 225x2 2 169y2 20. 121y2 2 49x2 In Exercises 21–32, factor the perfect squares. 21. x2 1 8x 1 16 22. y2 2 10y 1 25 23. x4 2 4x2 1 4 24. 1 2 6y 1 9y2 25. 4x2 1 12xy 1 9y2 26. x2 2 6xy 1 9y2 27. 9 2 6x 1 x2 28. 25x2 2 20xy 1 4y2 29. x4 1 2x2 1 1 30. x6 2 6x3 1 9 31. p2 1 2pq 1 q2 32. p2 2 2pq 1 q2 In Exercises 33–42, factor the sum or difference of two cubes. 33. t3 1 27 34. z3 1 64 35. y3 2 64 36. x3 2 1 37. 8 2 x3 38. 27 2 y3 39. y3 1 125 40. 64x 2 x4 41. 27 1 x3 42. 216x3 2 y3 In Exercises 43–52, factor each trinomial into a product of two binomials. 43. x2 2 6x 1 5 44. t2 2 5t 2 6 45. y2 2 2y 2 3 46. y2 2 3y 2 10 47. 2y2 2 5y 2 3 48. 2z2 2 4z 2 6 49. 3t2 1 7t 1 2 50. 4x2 2 2x 2 12 51. 26t2 1 t 1 2 52. 26x2 2 17x 1 10 In Exercises 53–60, factor by grouping. 53. x3 2 3x2 1 2x 2 6 54. x5 1 5x3 2 3x2 2 15 55. a4 1 2a3 2 8a 2 16 56. x4 2 3x3 2 x 1 3 57. 3xy 2 5rx 2 10rs 1 6sy 58. 6x2 2 10x 1 3x 2 5 59. 20x2 1 8xy 2 5xy 2 2y2 60. 9x5 2 a2x3 2 9x2 1 a2 [SEC TION 0.4] E X E R CI SE S • S K I L L S FA C T O R I N G A T R I N O M I A L A S A P R O D UC T O F T W O B I N O M I A L S x2 1 bx 1 c 5 1x 1 ?2 1x 1 ?2 1. Find all possible combinations of factors whose product is c. 2. Of the combinations in Step 1, look for the sum of factors that equals b. ax2 1 bx 1 c 5 1?x 1 ?2 1?x 1 ?2 1. Find all possible combinations of the first terms whose product is ax2. 2. Find all possible combinations of the last terms whose product is c. 3. Consider all possible factors based on Steps 1 and 2. FA C T O R I N G B Y GR O U PI N G • Group terms that have a common factor. • Use the distributive property. In this section, we discussed factoring polynomials, which is the reverse process of multiplying polynomials. Four main techniques were discussed. G REAT E S T COM M ON FAC T OR : axk • a is the greatest common factor for all coefficients of the polynomial. • k is the smallest exponent found on all the terms in the polynomial. FACT ORI NG FOR M ULAS : S P E C I A L P OLY NOMI AL FOR M S Difference of two squares: a2 2 b2 5 1a 1 b2 1a 2 b2 Perfect squares: a2 1 2ab 1 b2 5 1a 1 b22 a2 2 2ab 1 b2 5 1a 2 b22 Sum of two cubes: a3 1 b3 5 1a 1 b2 1a2 2 ab 1 b22 Difference of two cubes: a3 2 b3 5 1a 2 b2 1a2 1 ab 1 b22 [SEC TION 0.4] S U M M A RY In Exercises 61–92, factor each of the polynomials completely, if possible. If the polynomial cannot be factored, state that it is prime. 61. x2 2 4y2 62. a2 1 5a 1 6 63. 3a2 1 a 2 14 64. ax 1 b 1 bx 1 a 65. x2 1 16 66. x2 1 49 67. 4z2 1 25 68. 1 16 2 b4 69. 6x2 1 10x 1 4 70. x2 1 7x 1 5 71. 6x2 1 13xy 2 5y2 72. 15x 1 15xy 73. 36s2 2 9t2 74. 3x3 2 108x 75. a2b2 2 25c2 76. 2x3 1 54 77. 4x2 2 3x 2 10 78. 10x 2 25 2 x2 79. 3x3 2 5x2 2 2x 80. 2y3 1 3y2 2 2y 81. x3 2 9x 82. w3 2 25w 83. xy 2 x 2 y 1 1 84. a 1 b 1 ab 1 b2 85. x4 1 5x2 1 6 86. x6 2 7x3 2 8 87. x2 2 2x 2 24 88. 25x2 1 30x 1 9 89. x4 1 125x 90. x4 2 1 91. x4 2 81 92. 10x2 2 31x 1 15 • A P P L I C A T I O N S 93. Geometry. A rectangle has a length of 2x 1 4 and a width of x. Express the perimeter of the rectangle as a factored polynomial in x. 94. Geometry. The volume of a box is given by the expression x3 1 7x2 1 12x. Express the volume as a factored polynomial in the variable x. 95. Business. The profit of a business is given by the expression P 5 2x2 2 15x 1 4x 2 30. Express the profit as a factored polynomial in the variable x. 96. Business. The break-even point for a company is given by solving the equation 3x2 1 9x 2 4x 2 12 5 0. Factor the polynomial on the left side of the equation. 97. Engineering. The height of a projectile is given by the equation s 5 216t2 2 78t 1 10. Factor the expression on the right side of the equal sign. 98. Engineering. The electrical field at a point P between two charges is given by k 5 10x 2 x2 100 . Factor the numerator of this expression. • C A T C H T H E M I S T A K E In Exercises 99 and 100, explain the mistake that is made. 99. Factor x3 2 x2 2 9x 1 9. Solution: Group terms with common factors. 1x3 2 x22 1 129x 1 92 Factor out common factors. x2 1x 2 12 2 9 1x 2 12 Distributive property. 1x 2 12 1x2 2 92 Factor x2 2 9. 1x 2 12 1x 2 322 This is incorrect. What mistake was made? 100. Factor 4x2 1 12x 2 40. Solution: Factor the trinomial into a product of binomials. 12x 2 42 12x 1 102 Factor out a 2. 5 2 1x 2 22 1x 1 52 This is incorrect. What mistake was made? • C H A L L E N G E 105. Factor a2n 2 b2n completely, assuming a, b, and n are positive integers. 106. Find all the values of c such that the trinomial x2 1 cx 2 14 can be factored. 0.4 Factoring Polynomials 45 • C O N C E P T U A L 101. All trinomials can be factored into a product of two binomials. 102. All polynomials can be factored into prime factors with respect to the integers. 103. x2 2 y2 5 1x 2 y2 1x 1 y2 104. x2 1 y2 5 1x 1 y22 In Exercises 101–104, determine whether each of the following statements is true or false. • T E C H N O L O G Y 107. Use a graphing utility to plot the graphs of the three expressions 8x3 1 1, 12x 1 12 14x2 2 2x 1 12, and 12x 2 12 14x2 1 2x 1 12. Which two graphs agree with each other? 108. Use a graphing utility to plot the graphs of the three expressions 27x3 2 1, 13x 2 123, and 13x 2 12 19x2 1 3x 1 12. Which two graphs agree with each other? 46 CHAPTER 0 Prerequisites and Review 0.5.1 Rational Expressions and Domain Restrictions Recall that a rational number is the ratio of two integers with the denominator not equal to zero. Similarly, the ratio, or quotient, of two polynomials is a rational expression. Rational Numbers: 3 7 5 9 9 11 Rational Expressions: 3x 1 2 x 2 5 5x2 x2 1 1 9 3x 2 2 As with rational numbers, the denominators of rational expressions are never equal to zero. In the first and third rational expressions, there are values of the variable that would correspond to a denominator equal to zero; these values are not permitted: 3x 1 2 x 2 5 x 2 5 9 3x 2 2 x 2 2 3 In the second rational expression, 5x2 x2 1 1 , there are no real numbers that will correspond to a zero denominator. The set of real numbers for which an algebraic expression is defined is called the domain. Since a rational expression is not defined if its denominator is zero, we must eliminate from the domain those values of the variable that would result in a zero denominator. To find the domain of an algebraic expression, we ask the question, “What can x (the variable) be?” For rational expressions, the answer in general is “any values except those that make the denominator equal to zero.” S K I L L S O B J E C T I V E S ■ ■Find the domain of an algebraic expression. ■ ■Reduce a rational expression to lowest terms. ■ ■Multiply and divide rational expressions. ■ ■Add and subtract rational expressions. ■ ■Simplify complex rational expressions. C O N C E P T U A L O B J E C T I V E S ■ ■ Understand why rational expressions have domain restrictions. ■ ■Understand that we can cancel a common nonzero factor shared by the numerator and denominator because that ratio of a nonzero factor divided by itself is equal to 1. ■ ■Understand why we use the same rules to multiply or divide rational expressions that we use to multiply or divide rational numbers. ■ ■Understand the least common denominator method for rational expressions. ■ ■Know the two methods for simplifying complex rational expressions. 0.5 RATIONAL EXPRESSIONS 0.5.1 S KILL Find the domain of an algebraic expression. 0.5.1 CO NCE PTUAL Understand why rational expres-sions have domain restrictions. EXAMPLE 1 Finding the Domain of an Algebraic Expression Find the domain of the expressions. a. 2x2 2 5x 1 3 b. 2x 11 x 24 c. x x 2 11 d. 3x 11 x 0.5 Rational Expressions 47 Solution: ALGEBRAIC EXPRESSION TYPE DOMAIN NOTE a. 2x2 2 5x 1 3 Polynomial All real numbers The domain of all polynomials is the set of all real numbers. b. 2x 1 1 x 2 4 Rational expression All real numbers except x 5 4 When x 5 4, the rational expression is undefined. c. x x2 1 1 Rational expression All real numbers There are no real numbers that will result in the denominator being equal to zero. d. 3x 1 1 x Rational expression All real numbers except x 5 0 When x 5 0, the rational expression is undefined. YOUR T UR N Find the domain of the expressions. a. 3x 2 1 x 1 1 b. 5x 2 1 x c. 3x2 1 2x 2 7 d. 2x 1 5 x2 1 4 In this text, it will be assumed that the domain is the set of all real numbers except the real numbers shown to be excluded. ▼ A N S W E R a. x ∙ 21 b. x ∙ 0 c. all real numbers d. all real numbers ▼ [CONCEPT CHECK] Explain why 1 a2 1 1 does not have domain restrictions, but 1 a2 2 1 does have domain restrictions. ANSWER There are no real values that correspond to the denominator a 2 1 1 being equal to zero, but there are real values (a 5 21 or a 5 1) that correspond to the denominator a 2 2 1 being equal to zero. ▼ In this section, we will simplify rational expressions and perform operations on rational expressions such as multiplication, division, addition, and subtraction. The resulting expressions may not have explicit domain restrictions, but it is important to note that there are implicit domain restrictions because the domain restrictions on the original rational expression still apply. EXAMPLE 2 Excluding Values from the Domain of Rational Expressions Determine what real numbers must be excluded from the domain of the following rational expressions. a. 7x 1 5 x2 2 4 b. 3x 1 2 x2 2 5x Solution (a): Factor the denominator. 7x 1 5 x2 2 4 5 7x 1 5 1x 1 221x 2 22 Determine the values of x that will and must be make the denominator equal to zero. excluded from the domain. Solution (b): Factor the denominator. 3x 1 2 x2 2 5x 5 3x 1 2 x1x 2 52 Determine the values of x that will and must be make the denominator equal to zero. excluded from the domain. x 5 22 x 5 2 x 5 0 x 5 5 48 CHAPTER 0 Prerequisites and Review 0.5.2 Simplifying Rational Expressions Recall that a fraction is reduced when it is written with no common factors. 16 12 5 4 # 4 4 # 3 5 a4 4b # a4 3b 5 112 # a4 3b 5 4 3 or 16 12 5 4 # 4 4 # 3 5 4 3 Similarly, rational expressions are reduced to lowest terms, or simplified, if the numerator and denominator have no common factors other than 61. As with real numbers, the ability to write fractions in reduced form is dependent on your ability to factor. 0.5.2 S KILL Reduce a rational expression to lowest terms. 0.5.2 CO NCE PTUAL Understand that we can cancel a common nonzero factor shared by the numerator and denominator because that ratio of a nonzero factor divided by itself is equal to 1. EXAMPLE 3 Reducing a Rational Expression to Lowest Terms Simplify x2 2 x 2 2 2x 1 2 and state any domain restrictions. Solution: Factor the numerator and denominator. x2 2 x 2 2 2x 1 2 5 1x 2 221x 1 12 21x 1 12 State any domain restrictions. x 2 2 1 Cancel (divide out) the common factor, x 1 1. 5 1x 2 221x 1 12 21x 1 12 The rational expression is now in lowest terms (simplified). 5 Y OUR TU R N Simplify x2 1 x 2 2 2x 2 2 and state any domain restrictions. x 2 2 2 x 2 21 ▼ A N S W E R x 1 2 2 x 2 1 ▼ The following table summarizes two common mistakes made with rational expressions. CORRECT INCORRECT COMMENT x 1 5 y 1 5 is already simplified. Error: x 1 5 y 1 5 5 x y Factors can be divided out (canceled). Terms or parts of terms cannot be divided out. Remember to factor the numerator and denominator first, and then divide out common factors. x x2 1 x 5 x x 1x 1 12 x 2 0, x 2 21 5 1 x 1 1 x 2 0, x 2 21 Error: x x2 1 x 5 1 x 1 1 x 2 21 Note: Missing x 2 0. Determine the domain restrictions before dividing out common factors. STUDY TIP Factors can be divided out (canceled). Factor the numerator and denominator first, and then divide out common factors. STUDY TIP Determine domain restrictions of a rational expression before dividing out (canceling) common factors. REDUCING A RATIONAL EXPRESSION TO LOWEST TERMS (SIMPLIFYING) 1. Factor the numerator and denominator completely. 2. State any domain restrictions. 3. Cancel (divide out) the common factors in the numerator and denominator. 0.5.3 Multiplying and Dividing Rational Expressions The same rules that apply to multiplying and dividing rational numbers also apply to rational expressions. PROPERTY RESTRICTION DESCRIPTION a b ⋅ c d 5 ac bd b ∙ 0, d ∙ 0 Multiply numerators and denominators, respectively. a b 4 c d 5 a b ⋅ d c b ∙ 0, d ∙ 0, c ∙ 0 Dividing is equivalent to multiplying by a reciprocal. [CONCEPT CHECK] If a simplified rational expression results in 1 1x 2 12 where x is not equal to 1 or 3, what was the common factor that was divided out? ANSWER (x 2 3) ▼ ▼ A N S W E R 21x 1 5 2 x ∙ 5 0.5.3 S K I L L Multiply and divide rational expressions. 0.5.3 C ON C E P T U A L Understand why we use the same rules to multiply or divide rational expressions that we use to multiply or divide rational numbers. EXAMPLE 4 Simplifying Rational Expressions Reduce x2 2 x 2 6 x2 1 x 2 2 to lowest terms and state any domain restrictions. Solution: Factor the numerator and denominator. x2 2 x 2 6 x2 1 x 2 2 5 1x 2 321x 1 22 1x 2 121x 1 22 State domain restrictions. x 2 22, x 2 1 Divide out the common factor, x 1 2. 5 1x 2 321x 1 22 1x 2 121x 1 22 Simplify. 5 YOUR TURN Reduce x2 1 x 2 6 x2 1 2x 2 3 to lowest terms and state any domain restrictions. ▼ A N S W E R x 2 2 x 2 1 x 2 23, x 2 1 ▼ x 2 3 x 2 1 x 2 22, x 2 1 0.5 Rational Expressions 49 EXAMPLE 5 Simplifying Rational Expressions Reduce x2 2 4 2 2 x to lowest terms and state any domain restrictions. Solution: Factor the numerator and denominator. x2 2 4 2 2 x 5 1x 2 221x 1 22 12 2 x2 State domain restrictions. x 2 2 Factor out a negative in the denominator. 5 1x 2 221x 1 22 21x 2 22 Cancel (divide out) the common factor, x 2 2. 5 1x 2 221x 1 22 21x 2 22 Simplify. 5 YOUR T UR N Reduce x2 2 25 5 2 x to lowest terms and state any domain restrictions. 21x 1 22 x 2 2 ▼ 50 CHAPTER 0 Prerequisites and Review MULTIPLYING RATIONAL EXPRESSIONS 1. Factor all numerators and denominators completely. 2. State any domain restrictions. 3. Divide the numerators and denominators by any common factors. 4. Multiply the remaining numerators and denominators, respectively. DIVIDING RATIONAL EXPRESSIONS 1. Factor all numerators and denominators completely. 2. State any domain restrictions. 3. Rewrite division as multiplication by a reciprocal. 4. State any additional domain restrictions. 5. Divide the numerators and denominators by any common factors. 6. Multiply the remaining numerators and denominators, respectively. EXAMPLE 7 Dividing Rational Expressions Divide and simplify x2 2 4 x 4 3x3 2 12x 5x3 . Solution: Factor numerators and denominators. 1x 2 221x 1 22 x 4 3x1x 2 221x 1 22 5x3 State any domain restrictions. x 2 0 Write the quotient as a product. 5 1x 2 221x 1 22 x ⋅ 5x3 3x1x 2 221x 1 22 State any additional domain restrictions. x 2 22, x 2 2 EXAMPLE 6 Multiplying Rational Expressions Multiply and simplify 3x 1 1 4x2 1 4x ⋅x3 1 3x2 1 2x 9x 1 3 . Solution: Factor the numerators and 5 13x 1 12 4x1x 1 12 ⋅ x1x 1 121x 1 22 313x 1 12 denominators. State any domain restrictions. x 2 0, x 2 21, x 2 21 3 Divide the numerators and denominators by common factors. 5 13x 1 12 4x1x 1 12 ⋅ x1x 1 12 1x 1 22 313x 1 12 Simplify. 5 Y OUR TU R N Multiply and simplify 2x 1 1 3x2 2 3x ⋅ x3 1 2x2 2 3x 8x 1 4 . x 1 2 12 x 2 0, x 2 21, x 2 21 3 ▼ A N S W E R x 1 3 12 x 2 0, x 2 1, x 2 21 2 ▼ 0.5.4 Adding and Subtracting Rational Expressions The same rules that apply to adding and subtracting rational numbers also apply to rational expressions. PROPERTY RESTRICTION DESCRIPTION a b 6 c b 5 a 6 c b b 2 0 Adding or subtracting rational expressions when the denominators are the same a b 6 c d 5 ad 6 bc bd b 2 0 and d 2 0 Adding or subtracting rational expressions when the denominators are different [CONCEPT CHECK] Divide and simplify 1 x2 2 a2 4 1 x 2 a ANSWER 1 x 1 1 x 2 6 a ▼ 0.5.4 S K I L L Add and subtract rational expressions. 0.5.4 C ON C E P T U A L Understand the least common denominator method for rational expressions. EXAMPLE 8 Adding and Subtracting Rational Expressions: Equal Denominators Perform the indicated operation and simplify. a. x 1 7 1x 1 222 1 3x 1 1 1x 1 222 b. 6x 1 7 2x 2 1 2 2x 1 9 2x 2 1 Solution (a): Write as a single expression. 5 x 1 7 1 3x 1 1 1x 1 222 State any domain restrictions. x 2 22 Combine like terms in the numerator. 5 4x 1 8 1x 1 222 Factor out the common factor in the numerator. 5 41x 1 22 1x 1 222 Cancel (divide out) the common factor, x 1 2. 5 Solution (b): Write as a single expression. Use parentheses around the second numerator to ensure that the negative will be distributed throughout all terms. 5 6x 1 7 2 12x 1 92 2x 2 1 State any domain restrictions. x 2 1 2 4 x 1 2 x 2 22 [CONCEPT CHECK] Add 1 a 1 1 1 1 b 1 1 ANSWER a 1 b 1 2 1a 1 121b 1 12 ▼ STUDY TIP When subtracting a rational expression, distribute the negative of the quantity to be subtracted over all terms in the numerator. 0.5 Rational Expressions 51 5x 3 x 2 22, x 2 0, x 2 2 Divide out the common factors. 5 1x 2 221x 1 22 x ⋅ 5x3 3x 1x 2 221x 1 22 Simplify. 5 Y OUR T UR N Divide and simplify x2 2 9 x 4 2x3 2 18x 7x4 . ▼ A N S W E R 7x2 2 x 2 23, x 2 0, x 2 3 ▼ 52 CHAPTER 0 Prerequisites and Review EXAMPLE 9 Adding and Subtracting Rational Expressions: No Common Factors in Denominators Perform the indicated operation and simplify. a. 3 2 x 2x 1 1 1 x x 2 1 b. 1 x2 2 2 x 1 1 Solution (a): The common denominator is the product of the denominators. 3 2 x 2x 1 1 1 x x 2 1 5 13 2 x21x 2 12 12x 1 121x 2 12 1 x12x 1 12 1x 2 1212x 1 12 5 13 2 x21x 2 12 1 x12x 1 12 12x 1 121x 2 12 Eliminate parentheses in the numerator. 5 3x 2 3 2 x2 1 x 1 2x2 1 x 12x 1 121x 2 12 Combine like terms in the numerator. 5 Solution (b): The common denominator is the product of the denominators. 1 x2 2 2 x 1 1 5 1121x 1 12 2 21x22 x21x 1 12 Eliminate parentheses in the numerator. 5 x 1 1 2 2x2 x21x 1 12 5 22x2 1 x 1 1 x21x 1 12 5 212x2 2 x 2 12 x21x 1 12 Write the numerator in factored form to ensure 5 no further simplification is possible. Y OUR TU R N Perform the indicated operation and simplify. a. 2x 2 1 x 1 3 1 x 2x 1 1 b. 2 x2 1 1 2 1 x x2 1 5x 2 3 12x 1 121x 2 12 x 2 21 2, x 2 1 212x 1 121x 2 12 x21x 1 12 x 2 21, x 2 0 ▼ A N S W E R a. 5x2 1 3x 2 1 1x 1 3212x 1 12 x 2 23, x 2 2 1 2 b. 21x 2 122 x1x2 1 12 x 2 0 ▼ Eliminate parentheses. Distribute the negative. 5 6x 1 7 2 2x 2 9 2x 2 1 Combine like terms in the numerator. 5 4x 2 2 2x 2 1 Factor out the common factor in the numerator. 5 212x 2 12 2x 2 1 Divide out (cancel) the common factor, 2x 2 1. 5 2 x 2 1 2 STUDY TIP When adding or subtracting rational expressions whose denominators have no common factors, the least common denominator is the product of the two denominators. When combining two or more fractions through addition or subtraction, recall that the least common multiple, or least common denominator (LCD), is the smallest real number that all of the denominators divide into evenly (that is, the smallest of which all are factors). For example, 2 3 1 1 6 2 4 9 To find the LCD of these three fractions, factor the denominators into prime factors: 3 5 3 6 5 3⋅2 9 5 3 ⋅3 3⋅2⋅3 5 18 Rational expressions follow this same procedure, only now variables are also considered: 1 2x 1 1 x3 5 1 2x 1 1 x⋅x⋅x 2 x 1 1 2 x 2x 1 1 1 3 2 x 1x 1 122 5 2 1x 1 12 2 x 12x 1 12 1 3 2 x 1x 1 121x 1 12 The following box summarizes the LCD method for adding and subtracting rational expressions whose denominators have common factors. 2 3 1 1 6 2 4 9 5 12 1 3 2 8 18 5 7 18 LCD 5 2x3 LCD 5 1x 1 122 12x 1 12 THE LCD METHOD FOR ADDING AND SUBTRACTING RATIONAL EXPRESSIONS 1. Factor each of the denominators completely. 2. The LCD is the product of each of these distinct factors raised to the highest power to which that factor appears in any of the denominators. 3. Write each rational expression using the LCD for each denominator. 4. Add or subtract the resulting numerators. 5. Factor the resulting numerator to check for common factors. 0.5 Rational Expressions 53 EXAMPLE 10 Subtracting Rational Expressions: Common Factors in Denominators (LCD) Perform the indicated operation and write in simplified form. 5x 2x 2 6 2 7x 2 2 x2 2 x 2 6 Solution: Factor the denominators. 5 5x 21x 2 32 2 7x 2 2 1x 2 321x 1 22 Identify the LCD. LCD 5 2 1x 2 3 2 1x 1 2 2 Write each expression using the LCD as the denominator. Combine into one expression. Distribute the negative through the entire second numerator. 5 5x2 1 10x 2 14x 1 4 21x 2 321x 1 22 Simplify. 5 Y OUR T UR N Perform the indicated operation and write in simplified form. 2x 3x 2 6 2 5x 1 1 x2 1 2x 2 8 5x2 2 4x 1 4 21x 2 321x 1 22 x 2 22, x 2 3 ▼ A N S W E R 2x2 2 7x 2 3 31x 2 221x 1 42 x 2 24, x 2 2 ▼ 5 5x1x 1 22 21x 2 321x 1 22 2 217x 2 22 21x 2 321x 1 22 54 CHAPTER 0 Prerequisites and Review 0.5.5 Complex Rational Expressions A rational expression that contains another rational expression in either its numerator or denominator is called a complex rational expression. The following are examples of complex rational expressions. 1 x 2 5 2 1 x 2 2 x 4 1 3 x 2 1 3 x 2 7 6 2x 2 5 2 1 TWO METHODS FOR SIMPLIFYING COMPLEX RATIONAL EXPRESSIONS Procedure 1: Write a sum or difference of rational expressions that appear in either the numerator or denominator as a single rational expression. Once the complex rational expression contains a single rational expression in the numerator and one in the denominator, then rewrite the division as multiplication by the reciprocal. OR Procedure 2: Find the LCD of all rational expressions contained in both the numerator and denominator. Multiply the numerator and denominator by this LCD and simplify. 0.5.5 S KILL Simplify complex rational expressions. 0.5.5 CO NCE PTUAL Know the two methods for simplifying complex rational expressions. EXAMPLE 11 Simplifying a Complex Rational Expression Write the rational expression in simplified form. 2 x 1 1 1 1 1 x 1 1 Solution: State the domain restrictions. 2 x 1 1 x 2 0 , x 2 21 , and x 2 22 1 1 1 x 1 1 Procedure 1: Add the expressions in both the numerator and denominator. 5 2 x 1 x x x 1 1 x 1 1 1 1 x 1 1 5 2 1 x x 1x 1 12 1 1 x 1 1 Simplify. 5 2 1 x x x 1 2 x 1 1 Express the quotient as a product. 5 2 1 x x ⋅ x 1 1 x 1 2 Divide out the common factors. 5 2 1 x x ⋅ x 1 1 x 1 2 Write in simplified form. 5 x 1 1 x x 2 22, x 2 21, x 2 0 EXAMPLE 12 Simplifying a Complex Rational Expression Write the rational expression in simplified form. 1 x2 2 9 1 3 1 2 x 2x 1 6 x 2 26, 23, 3 Solution: Using Procedure 1 Factor the respective denominators. 1 1x 2 321x 1 32 1 3 1 2 x 21x 1 32 0.5 Rational Expressions 55 Procedure 2: Find the LCD of the numerator 2 x 1 1 1 1 1 x 1 1 and denominator. Identify the LCDs. Numerator LCD: x Denominator LCD: x 1 1 Combined LCD: x 1x 1 1 2 Multiply both numerator and denominator by their combined LCD. 5 2 x 1 1 1 1 1 x 1 1 ⋅ x1x 1 12 x1x 1 12 Multiply the numerators and denominators, respectively, applying the distributive property. 5 2 x ⋅x1x 1 12 1 1x1x 1 12 1⋅x1x 1 12 1 1 x 1 1 ⋅x1x 1 12 Divide out common factors. 5 2 x ⋅x1x 1 12 1 1x1x 1 12 x1x 1 12 1 1 x 1 1 ⋅x1x 1 12 Simplify. 5 21x 1 12 1 x1x 1 12 x1x 1 12 1 x Apply the distributive property. 5 2x 1 2 1 x2 1 x x2 1 x 1 x Combine like terms. 5 x2 1 3x 1 2 x2 1 2x Factor the numerator and denominator. 5 1x 1 221x 1 12 x1x 1 22 Divide out the common factor. 5 1x 1 221x 1 12 x1x 1 22 Write in simplified form. 5 x 1 1 x x 2 22, x 2 21, x 2 0 56 CHAPTER 0 Prerequisites and Review [CONCEPT CHECK] Simplify: 1 1 1 a 1 1 1 b ANSWER b1a 1 12 a1b 1 12 a 2 0, b 2 21, 0 ▼ Identify the LCDs. Numerator LCD: 1x 2 32 1x 1 32 Denominator LCD: 2 1x 1 32 Combined LCD: 21x 2 32 1x 1 32 Multiply both the numerator and the denominator by the combined LCD. 5 1 1x 2 321x 1 32 1 3 1 2 x 21x 1 32 ⋅ 21x 1 321x 2 32 21x 1 321x 2 32 Multiply the numerators and denominators, respectively, applying the distributive 5 21x 1 321x 2 32 1x 2 321x 1 32 1 3⋅21x 1 321x 2 32 21x 1 321x 2 32 2 x⋅21x 1 321x 2 32 21x 1 32 property. Simplify. 5 2 1 61x 1 321x 2 32 21x 1 321x 2 32 2 x1x 2 32 Eliminate the parentheses. 5 2 1 6x2 2 54 2x2 2 18 2 x2 1 3x Combine like terms. 5 6x2 2 52 x2 1 3x 2 18 Factor the numerator and denominator to make sure there are no 5 213x2 2 262 1x 1 621x 2 32 x 2 26, x 2 23, x 2 3 common factors. In this section, rational expressions were defined as quotients of polynomials. The domain of any polynomial is the set of all real numbers. Since rational expressions are ratios of polynomials, the domain of rational expressions is the set of all real numbers except those values that make the denominator equal to zero. In this section, rational expressions were simplified (written with no common factors), multiplied, divided, added, and subtracted. OPERATION EXAMPLE NOTE Multiplying rational expressions 2 x 1 1⋅ 3x x 2 1 5 6x x2 2 1 x 2 61 State domain restrictions. Dividing rational expressions 2 x 1 1 4 3x x 2 1 x 2 61 5 2 x 1 1⋅x 2 1 3x x 2 0 5 21x 2 12 3x1x 1 12 x 2 0, x 2 61 When dividing rational expressions, remember to check for additional domain restrictions once the division is rewritten as multiplication by a reciprocal. Adding/subtracting rational expressions with no common factors 2 x 1 1 1 3x x 2 1 x 2 61 LCD 5 1x 1 121x 2 12 5 21x 2 12 1 3x1x 1 12 1x 1 121x 2 12 5 2x 2 2 1 3x2 1 3x 1x 1 121x 2 12 5 3x2 1 5x 2 2 1x 1 121x 2 12 5 13x 2 121x 1 22 1x 1 121x 2 12 x 2 61 The LCD is the product of the two denominators. [SEC TION 0.5] S U M M A RY OPERATION EXAMPLE NOTE Adding/subtracting rational expressions with common factors 3 x1x 1 12 2 2 x1x 1 22 x 2 22, 21, 0 LCD 5 x1x 1 121x 1 22 5 31x 1 22 2 21x 1 12 x1x 1 121x 1 22 5 3x 1 6 2 2x 2 2 x1x 1 121x 1 22 5 x 1 4 x1x 1 121x 1 22 x 2 22, 21, 0 The LCD is the product of each of these distinct factors raised to the highest power that appears in any of the denominators. Complex rational expressions are simplified in one of two ways: 1. Combine the sum or difference of rational expressions in a numerator or denominator as a single rational expression. The result is a rational expression in the numerator and a rational expression in the denominator. Then write the division as multiplication by the reciprocal. 2. Multiply the numerator and denominator by the overall LCD (LCD for all rational expressions that appear). The result is a single rational expression. Then simplify, if possible. [SEC TION 0.5] E X ER C I S E S • S K I L L S In Exercises 1–10, state any real numbers that must be excluded from the domain of each rational expression. 1. 3 x 2. 5 x 3. 3 x 2 1 4. 6 y 2 1 5. 5x 2 1 x 1 1 6. 2x 3 2 x 7. 2p2 p2 2 1 8. 3t t2 2 9 9. 3p 2 1 p2 1 1 10. 2t 2 2 t2 1 4 In Exercises 11–30, reduce the rational expression to lowest terms and state any real numbers that must be excluded from the domain. 11. 1x 1 321x 2 92 21x 1 321x 1 92 12. 4y1y 2 821y 1 72 8y1y 1 721y 1 82 13. 1x 2 321x 1 12 21x 1 12 14. 12x 1 121x 2 32 31x 2 32 15. 213y 1 1212y 2 12 312y 2 1213y2 16. 712y 1 1213y 2 12 513y 2 1212y2 17. 15y 2 121y 1 12 25y 2 5 18. 12t 2 121t 1 22 4t 1 8 19. 13x 1 721x 2 42 4x 2 16 20. 1t 2 7212t 1 52 3t 2 21 21. x2 2 4 x 2 2 22. t3 2 t t 2 1 23. x 1 7 x 1 7 24. 2y 1 9 2y 1 9 25. x2 1 9 2x 1 9 26. x2 1 4 2x 1 4 27. x2 1 5x 1 6 x2 2 3x 2 10 28. x2 1 19x 1 60 x2 1 8x 1 16 29. 6x2 2 x 2 1 2x2 1 9x 2 5 30. 15x2 2 x 2 2 5x2 1 13x 2 6 In Exercises 31–48, multiply the rational expressions and simplify. State any real numbers that must be excluded from the domain. 31. x 2 2 x 1 1 ⋅ 3x 1 5 x 2 2 32. 4x 1 5 x 2 2 ⋅ 3x 1 4 4x 1 5 33. 5x 1 6 x ⋅ 2x 5x 2 6 34. 41x 2 221x 1 52 8x ⋅ 16x 1x 2 521x 1 52 35. 2x 2 2 3x ⋅ x2 1 x x2 2 1 36. 5x 2 5 10x ⋅ x2 1 x x2 2 1 37. 3x2 2 12 x ⋅ x2 1 5x x2 1 3x 2 10 38. 4x2 2 32x x ⋅ x2 1 3x x2 2 5x 2 24 39. t 1 2 3t 2 9 ⋅ t2 2 6t 1 9 t2 1 4t 1 4 40. y 1 3 3y 1 9 ⋅ y2 2 10y 1 25 y2 1 3y 2 40 41. t2 1 4 t 2 3 ⋅ 3t t 1 2 42. 7a2 1 21a 141a2 2 92 ⋅ a 1 3 7 0.5 Rational Expressions 57 58 CHAPTER 0 Prerequisites and Review 43. y2 2 4 y 2 3 ⋅ 3y y 1 2 44. t2 1 t 2 6 t2 2 4 ⋅ 8t 2t2 45. 3x2 2 15x 2x3 2 50x ⋅ 2x2 2 7x 2 15 3x2 1 15x 46. 5t 2 1 4t ⋅ 4t2 1 3t 16t2 2 9 47. 6x2 2 11x 2 35 8x2 2 22x 2 21 ⋅ 4x2 2 49 9x2 2 25 48. 3x2 2 2x 12x3 2 8x2 ⋅ x2 2 7x 2 18 2x2 2 162 In Exercises 49–66, divide the rational expressions and simplify. State any real numbers that must be excluded from the domain. 49. 3 x 4 12 x2 50. 5 x2 4 10 x3 51. 6 x 2 2 4 12 1x 2 221x 1 22 52. 51x 1 62 101x 2 62 4 201x 1 62 8 53. 1 x 2 1 4 5 x2 2 1 54. 5 3x 2 4 4 10 9x2 2 16 55. 2 2 p p2 2 1 4 2p 2 4 p 1 1 56. 4 2 x x2 2 16 4 12 2 3x x 2 4 57. 36 2 n2 n2 2 9 4 n 1 6 n 1 3 58. 49 2 y2 y2 2 25 4 7 1 y 2y 1 10 59. 3t3 2 6t2 2 9t 5t 2 10 4 6 1 6t 4t 2 8 60. x3 1 8x2 1 12x 5x2 2 10x 4 4x 1 8 x2 2 4 61. w2 2 w w 4 w3 2 w 5w3 62. y2 2 3y 2y 4 y3 2 3y2 8y 63. x2 1 4x 2 21 x2 1 3x 2 10 4 x2 2 2x 2 63 x2 1 x 2 20 64. 2y2 2 5y 2 3 2y2 2 9y 2 5 4 3y 2 9 y2 2 5y 65. 20x2 2 3x 2 2 25x2 2 4 4 12x2 1 23x 1 5 3x2 1 5x 66. x2 2 6x 2 27 2x2 1 13x 2 7 4 2x2 2 15x 2 27 2x2 1 9x 2 5 In Exercises 67–82, add or subtract the rational expression and simplify. State any real numbers that must be excluded from the domain. 67. 3 x 2 2 5x 68. 5 7x 2 3 x 69. 3 p 2 2 1 5p p 1 1 70. 4 9 1 x 2 5x x 2 2 71. 2x 1 1 5x 2 1 2 3 2 2x 1 2 5x 72. 7 2x 2 1 2 5 1 2 2x 73. 3y2 y 1 1 1 1 2 2y y 2 1 74. 3 1 2 x 1 4 x 2 1 75. 3x x2 2 4 1 3 1 x x 1 2 76. x 2 1 4 2 x2 2 x 1 1 2 1 x 77. x 2 1 x 2 2 1 x 2 6 x2 2 4 78. 2 y 2 3 1 7 y 1 2 79. 5a a2 2 b2 2 7 b 2 a 80. 1 y 1 4 y2 2 4 2 2 y2 2 2y 81. 7 1 1 x 2 3 82. 3 5y 1 6 2 4 y 2 2 1 y2 2 y 5y2 2 4y 2 12 In Exercises 83–90, simplify the complex rational expressions. State any real numbers that must be excluded from the domain. 83. 1 x 2 1 1 2 2 x 84. 3 y 2 5 4 2 2 y 85. 3 1 1 x 9 2 1 x2 86. 1 x 1 2 x2 9 x 2 5 x2 87. 1 x 2 1 1 1 1 2 1 x 1 1 88. 7 y 1 7 1 y 1 7 2 1 y 89. 1 x 2 1 1 1 1 x 1 1 1 1 90. 3 x 1 1 2 3 x 2 1 5 x2 2 1 91. Finance. The amount of payment made on a loan is given by the formula A 5 pi 1 2 1/11 1 i 2n, where p is the principal (amount borrowed), and i 5 r n, where r is the interest rate expressed as a decimal and n is the number of payments per year. Suppose n 5 5. Simplify the formula as much as possible. 92. Finance. Use the formula A 5 pi 1 2 1 11 1 i 2nt to calculate the amount your monthly payment will be on a loan of $150,000 at an interest rate of 6.5% for 30 years 1nt 5 360 2. 93. Circuits. If two resistors are connected in parallel, the combined resistance is given by the formula R 5 1 1/R1 1 1/R2 , where R1 and R2 are the individual resistances. Simplify the formula. 94. Optics. The focal length of a lens can be calculated by applying the formula ƒ 5 1 1/p 1 1/q, where p is the distance that the object is from the lens and q is the distance that the image is from the lens. Simplify the formula. • A P P L I C A T I O N S In Exercises 95 and 96, explain the mistake that is made. 95. Simplify x2 1 2x 1 1 x 1 1 . Solution: Factor the numerator. x2 1 2x 1 1 x 1 1 5 1x 1 121x 1 12 1x 1 12 Cancel the common factor, x 1 1. 5 1x 1 121x 1 12 1x 1 12 Write in simplified form. 5 x 1 1 This is incorrect. What mistake was made? 96. Simplify x 1 1 x2 1 2x 1 1. Solution: Cancel the common 1s. x 1 1 x2 1 2x 1 1 Factor the denominator. 5 x x1x 1 22 Cancel the common x. 5 x x1x 1 22 Write in simplified form. 5 1 x 1 2 x 2 22, x 2 0 This is incorrect. What mistake was made? 99. When adding or subtracting rational expressions, the LCD is always the product of all the denominators. 100. x 2 c c 2 x 5 21 for all values of x. In Exercises 97–100, determine whether each of the statements is true or false. • C A T C H T H E M I S T A K E • C O N C E P T U A L • C H A L L E N G E 101. Perform the operation and simplify (remember to state domain restrictions). x 1 a x 1 b 4 x 1 c x 1 d 102. Write the numerator as the product of two binomials. Divide out any common factors of the numerator and denominator. a2n 2 b2n an 2 bn • T E C H N O L O G Y 103. Utilizing a graphing technology, plot the expression y 5 x 1 7 x 1 7. Zoom in near x 5 27. Does this agree with what you found in Exercise 23? 104. Utilizing a graphing technology, plot the expression y 5 x2 2 4 x 2 2 . Zoom in near x 5 2. Does this agree with what you found in Exercise 21? In Exercises 105 and 106, for each given expression: (a) simplify the expression, (b) use a graphing utility to plot the expression and the answer in (a) in the same viewing window, and (c) determine the domain restriction(s) where the graphs will agree with each other. 105. 1 1 1 x 2 2 1 2 1 x 1 2 106. 1 2 2 x 1 3 1 1 1 x 1 4 97. x2 2 81 x 2 9 5 x 1 9 98. x 2 9 x2 2 81 5 1 x 1 9 x 2 29, 9 0.5 Rational Expressions 59 60 CHAPTER 0 Prerequisites and Review In Section 0.2, we discussed integer exponents and their properties. For example, 42 5 16 and x2 ? x3 5 x5. In this section, we expand our discussion of exponents to include any rational numbers. For example, 161/2 5 ? and 1x1/223/4 5 ?. We will first start with a more familiar notation (roots) and discuss operations on radicals, and then rational exponents will be discussed. 0.6.1 Square Roots It is important to note that the principal square root b is nonnegative. “The principal square root of 16 is 4” implies 42 5 16. Although it is also true that 124 2 2 5 16, the principal square root is defined to be nonnegative. It is also important to note that negative real numbers do not have real square roots. For example, !29 is not a real number because there are no real numbers that when squared yield 29. Since principal square roots are defined to be nonnegative, this means they must be zero or positive. The square root of zero is equal to zero: !0 5 0. All other nonnegative principal square roots are positive. S K I L L S O B J E C T I V E S ■ ■Evaluate and simplify square roots. ■ ■Simplify and combine radical expressions involving roots other than squares. ■ ■Simplify and factor expressions involving rational exponents. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that a radical implies one number (the principal root), not two (6 the principal root). ■ ■Know the properties of radicals. ■ ■Understand that radicals are equivalent to rational exponents. 0.6 RATIONAL EXPONENTS AND RADICALS DEFINITION Principal Square Root Let a be any nonnegative real number; then the nonnegative real number b is called the principal square root of a, denoted b 5 !a, if b2 5 a. The symbol ! is called a radical sign, and a is called the radicand. 0.6.1 S KILL Evaluate and simplify square roots. 0.6.1 CO NCE PTUAL Understand that a radical implies one number (the principal root), not (6 the principal root). EXAMPLE 1 Evaluating Square Roots Evaluate the square roots, if possible. a. !169 b. Å 4 9 c. !236 Solution: a. What positive real number squared results in 169? !169 5 13 Check: 132 5 169 b. What positive real number squared results in 4 9? Å 4 9 5 2 3 Check: A2 3B 2 5 4 9 c. What positive real number squared results in 236? No real number STUDY TIP Incorrect: !25 5 65 Correct: !25 5 5 The principal square root, ! , is defined as a nonnegative real number. SQUARE ROOTS OF PERFECT SQUARES Let a be any real number; then: !a2 5 ∙a∙ 0.6 Rational Exponents and Radicals 61 Simplifying Square Roots So far, only square roots of perfect squares have been discussed. Now we consider how to simplify square roots such as !12. We rely on the following properties. PROPERTIES OF SQUARE ROOTS Let a and b be nonnegative real numbers, then: Property !a⋅b 5 !a⋅!b Å a b 5 !a "b b 2 0 Description The square root of a product is the product of the square roots. The square root of a quotient is the quotient of the square roots. Example !20 5 !4⋅!5 5 2!5 Å 40 49 5 !40 !49 5 !4⋅!10 7 5 2!10 7 EXAMPLE 3 Simplifying Square Roots Simplify: a. "48x2 b. "28x3 c. !12x ⋅!6x d. "45x3 "5x Solution: a. "48x2 5 !48⋅"x2 5 !16⋅3⋅"x2 5 !16⋅!3⋅"x2 5 4 0 x0 !3 4 0 x 0 b. "28x3 5 !28⋅"x3 5 !4⋅7⋅"x2⋅x 5 !4⋅!7⋅"x2⋅!x 2 0 x 0 5 20 x0 !7 !x 5 20 x0 !7x 5 2x!7x since x $ 0 c. !12x⋅!6x 5 "72x2 5 "36⋅2⋅x2 5 !36⋅!2⋅"x2 5 6 0 x0 !2 6 0 x 0 5 6x!2 since x $ 0 d. "45x3 !5x 5 Å 45x3 5x 5 "9x2 5 !9⋅"x2 5 30 x0 5 3x since x . 0 3 0 x 0 Note: x 2 0, x . 0 Y OUR T UR N Simplify: a. "60x3 b. "125x5 "25x3 U U U U U U U U ▼ EXAMPLE 2 Finding Square Roots of Perfect Squares Evaluate the following: a. "62 b. "12722 c. "x2 Solution: a. "62 5 !36 5 6 b. "12722 5 !49 5 7 c. "x2 5 0 x 0 [CONCEPT CHECK] TRUE OR FALSE "a2 5 ∙a∙ and " 3 a3 5 a ANSWER True ▼ STUDY TIP If you have a single odd power under the square root, like "x3, the variable is forced to be nonnegative because a negative cubed results in a negative inside the radical; therefore, the absolute value is not necessary. ▼ A N S W E R a. 2x!15x b. x!5 62 CHAPTER 0 Prerequisites and Review 0.6.2 Other (nth) Roots We now expand our discussion from square roots to other nth roots. 0.6.2 S KILL Simplify and combine radical expressions involving roots other than squares. 0.6.2 CO NCE PTUAL Know the properties of radicals. DEFINITION Principal nth Root Let a be a real number and n be a positive integer. Then the real number b is called the principal nth root of a, denoted b 5 ! n a, if bn 5 a. If n is even, then a and b are nonnegative real numbers. The positive integer n is called the index. The square root corresponds to n 5 2, and the cube root corresponds to n 5 3. n a b EXAMPLE Even Positive Positive ! 4 16 5 2 because 24 5 16 Even Negative Not a real number ! 4 216 is not a real number Odd Positive Positive ! 3 27 5 3 because 33 5 27 Odd Negative Negative ! 3 2125 5 25 because 12523 5 2125 A radical sign, " , combined with a radicand is called a radical. EXAMPLE 4 Simplifying Radicals Simplify: a. " 3 224x5 b. " 4 32x5 Solution: a. " 3 224x5 5 " 3 1282132x3x2 5 ! 3 28 ⋅! 3 3 ⋅" 3 x3 ⋅" 3 x2 5 22x" 3 3x2 22 x b. " 4 32x5 5 " 4 16⋅2⋅x4⋅x 5 ! 4 16⋅! 4 2⋅" 4 x4⋅! 4 x 5 20 x 0 ! 4 2x 5 2x! 4 2x since x $ 0 2 0 x 0 u u u u PROPERTIES OF RADICALS Let a and b be real numbers, then PROPERTY DESCRIPTION EXAMPLE " n ab 5 " n a ⋅ " n b if " n a and " n b both exist The nth root of a product is the product of the nth roots. ! 3 16 5 ! 3 8 ⋅ ! 3 2 5 2! 3 2 Å n a b 5 " n a " n b b 2 0 if " n a and " n b both exist The nth root of a quotient is the quotient of the nth roots. Å 4 81 16 5 " 4 81 " 4 16 5 3 2 " n am 5 A ! n a B m The nth root of a power is the power of the nth root. " 3 82 5 A! 3 8B 2 5 1222 5 4 " n an 5 a n is odd When n is odd, the nth root of a raised to the nth power is a. " 3 x3 5 x " n an 5 ∙a∙ n is even When n is even, the nth root of a raised to the nth power is the absolute value of a. " 4 x4 5 ∙x∙ Combining Like Radicals We have already discussed properties for multiplying and dividing radicals. Now we focus on combining (adding or subtracting) radicals. Radicals with the same index and radicand are called like radicals. Only like radicals can be added or subtracted. EXAMPLE 5 Combining Like Radicals Combine the radicals if possible. a. 4!3 2 6!3 1 7!3 b. 2!5 2 3!7 1 6!3 c. 3!5 1 !20 2 2!45 d. ! 4 10 2 2! 3 10 1 3!10 Solution (a): Use the distributive property. 4!3 2 6!3 1 7!3 5 14 2 6 1 72 !3 Eliminate the parentheses. 5 5!3 Solution (b): None of these radicals are alike. The expression is in simplified form. 2!5 2 3!7 1 6!3 Solution (c): Write the radicands as products with a factor of 5. 3!5 1 !20 2 2!45 5 3!5 1 !4⋅5 2 2!9⋅5 The square root of a product is the product of square roots. 5 3!5 1 !4⋅!5 2 2!9⋅!5 Simplify the square roots of perfect squares. 5 3!5 1 2!5 2 2132 !5 6 All three radicals are now like radicals. 5 3!5 1 2!5 2 6!5 Use the distributive property. 5 13 1 2 2 62 !5 Simplify. 5 2!5 Solution (d): None of these radicals are alike because they have different indices. The expression is in simplified form. ! 4 10 2 2! 3 10 1 3!10 Y OUR T UR N Combine the radicals. a. 4! 3 7 2 6! 3 7 1 9! 3 7 b. 5!24 2 2!54 u ▼ A N S W E R a. 7! 3 7 b. 4!6 ▼ Rationalizing Denominators When radicals appear in a quotient, it is customary to write the quotient with no radicals in the denominator. This process is called rationalizing the denominator and involves multiplying by an expression that will eliminate the radical in the denominator. For example, the expression 1 !3 contains a single radical in the denominator. In a case like this, multiply the numerator and denominator by an appropriate radical expression, so that the resulting denominator will be radical free: 1 !3 ⋅ 1 !32 1 !32 5 !3 !3 ⋅!3 5 !3 3 1 If the denominator contains a sum of the form a 1 !b, multiply both the numerator and the denominator by the conjugate of the denominator, a 2 !b, which uses the difference of two squares to eliminate the radical term. Similarly, if the denominator U 0.6 Rational Exponents and Radicals 63 64 CHAPTER 0 Prerequisites and Review contains a difference of the form a 2 !b, multiply both the numerator and the denominator by the conjugate of the denominator, a 1 !b. For example, to rationalize 1 3 2 !5, take the conjugate of the denominator, which is 3 1 !5: 1 A3 2 !5B ⋅ A3 1 !5B A3 1 !5B 5 3 1 !5 32 1 3!5 2 3!5 2 A!5B 2 5 3 1 !5 9 2 5 5 3 1 !5 4 like terms In general, we apply the difference of two squares: A!a 1 !bBA!a 2 !bB 5 A!aB 2 2 A!bB 2 5 a 2 b Notice that the product does not contain a radical. Therefore, to simplify the expression 1 A!a 1 !bB multiply the numerator and denominator by A!a 2 !bB: 1 A!a 1 !bB ⋅ A!a 2 !bB A!a 2 !bB The denominator now contains no radicals: A!a 2 !bB 1a 2 b2 f EXAMPLE 6 Rationalizing Denominators Rationalize the denominators and simplify. a. 2 3!10 b. 5 3 2 !2 c. !5 !2 2 !7 Solution (a): Multiply the numerator and 5 2 3!10 ⋅ !10 !10 denominator by !10. Simplify. 5 2!10 3A!10B2 5 2!10 31102 5 2!10 30 Divide out the common 2 in the numerator and denominator. 5 !10 15 Solution (b): Multiply the numerator and denominator 5 5 A3 2 !2B ⋅ A3 1 !2B A3 1 !2B by the conjugate, 3 1 !2. 5 5A3 1 !2B A3 2 !2BA3 1 !2B The denominator now contains no radicals. 5 15 1 5!2 9 2 2 Simplify. 5 15 1 5!2 7 [CONCEPT CHECK] Simplify: 1 1 1 "a ANSWER 1 2 "a 1 2 a ▼ EXAMPLE 7 Expressing a Radical Expression in Simplified Form Express the radical expression in simplified form: Å 3 16x5 81y7 x $ 0, y . 0 Solution: Rewrite the expression so that the Ç 3 16x5 81y7 5 " 3 16x5 " 3 81y7 radical does not contain a fraction. Let 16 5 24 and 81 5 34. 5 " 3 24⋅x5 " 3 34⋅y7 Factors in both radicands are raised to powers greater than the index (3). Rewrite the expression so that 5 2x" 3 2x2 3y2" 3 3y each power in the radicand is less than the index. The denominator contains a radical. In order to 5 2x" 3 2x2 3y2" 3 3y ⋅ eliminate the radical in the denominator, we multiply the numerator and denominator by 3 " 9 y2. 5 2x" 3 18x2y2 3y2" 3 27y3 5 2x" 3 18x2y2 9y3 The radical expression now satisfies the 5 2x" 3 18x2y2 9y3 conditions for simplified form. 3 3 " 9 y2 " 9 y2 0.6 Rational Exponents and Radicals 65 Solution (c): Multiply the numerator and denominator 5 !5 A !2 2 !7B ⋅ A !2 1 !7B A !2 1 !7B by the conjugate, !2 1 !7. Multiply the numerators and 5 !5A !2 1 !7B A !2 2 !7BA !2 1 !7B denominators, respectively. The denominator now contains no radicals. 5 !10 1 !35 2 2 7 Simplify. 5 2 !10 1 !35 5 Y OUR T UR N Write the expression 7 1 2 !3 in simplified form. ▼ SIMPLIFIED FORM OF A RADICAL EXPRESSION A radical expression is in simplified form if ■ ■No factor in the radicand is raised to a power greater than or equal to the index. ■ ■The power of the radicand does not share a common factor with the index. ■ ■The denominator does not contain a radical. ■ ■The radical does not contain a fraction. ▼ A N S W E R 2 7A1 1 !3B 2 66 CHAPTER 0 Prerequisites and Review 0.6.3 S KILL Simplify and factor expressions involving rational exponents. 0.6.3 CO NCE PTUAL Understand that radicals are equivalent to rational exponents. 0.6.3 Rational Exponents We now use radicals to define rational exponents. EXAMPLE 8 Simplifying Expressions with Rational Exponents Simplify: a. 163/ 2 b. 12822/ 3 Solution: a. 163/2 5 1161/223 5 A !16B3 5 43 5 64 b. 12822/3 5 312821/342 5 12222 5 4 Y OUR TU R N Simplify 272/3. ▼ A N S W E R 9 ▼ The properties of exponents that hold for integers also hold for rational numbers: a21/n 5 1 a1/n and a2m/n 5 1 am/n a 2 0 EXAMPLE 9 Simplifying Expressions with Negative Rational Exponents Simplify 19x221/2 4x23/2 x . 0. Solution: Negative exponents correspond to positive exponents in the reciprocal. 19x221/2 4x23/2 5 x3/2 4⋅19x21/2 Eliminate the parentheses. 5 x3/2 4⋅91/2x1/2 Apply the quotient property on x. 5 x3/221/2 4⋅91/2 5 x1 4⋅3 Simplify. 5 x 12 Y OUR TU R N Simplify 9x3/2 14x221/2 x . 0. u 3 ▼ A N S W E R 18x2 ▼ RATIONAL EXPONENTS: 1 – n Let a be any real number and n be a positive integer, then a1/n 5 ! n a where 1 n is the rational exponent of a. ■ ■When n is even and a is negative, then a1/n and ! n a are not real numbers. ■ ■Furthermore, if m is a positive integer with m and n having no common factors, then am/n 5 1a1/n2 m 5 1am21/n 5 " n am Note: Any of the four notations can be used. EXAMPLE 10 Simplifying Algebraic Expressions with Rational Exponents Simplify 128x2 y21/3 19xy421/2 x . 0, y . 0. Solution: 128x2 y21/3 19xy421/2 5 12821/3 1x221/3 y1/3 91/2 x1/2 1y421/2 5 12821/3 x 2/3 y 1/3 91/2 x1/2 y2 5 a 22 3 bx2/321/2 y1/322 5 22 3x1/6 y25/3 Write in terms of positive exponents. 5 22x1/6 3y5/3 YOUR T UR N Simplify 116x3 y21/2 127x2 y321/3 and write your answer with only positive exponents. ▼ A N S W E R 4x5/6 3y1/2 ▼ EXAMPLE 11 Factoring Expressions with Rational Exponents Factor completely x8/3 2 5x5/3 2 6x2/3. Solution: Factor out the greatest common x8/3 2 5x5/3 2 6x2/3 5 x2/31x2 2 5x 2 62 factor x2/3. x2/3x6/3 x2/3x3/3 Factor the trinomial. 5 x2/31x 2 621x 1 12 Y OUR T UR N Factor completely x7/3 2 x4/3 2 2x1/3. ▼ A N S W E R x1/31x 2 221x 1 12 ▼ u u [CONCEPT CHECK] Write 1 " 3 a2 using rational exponents ANSWER a22/3 ▼ 0.6 Rational Exponents and Radicals 67 Radicals can be combined only if they are like radicals (same radicand and index). Quotients with radicals in the denominator are usually rewritten with no radicals in the denominator. Rational exponents were defined in terms of radicals: a1/n 5 ! n a. The properties for integer exponents we learned in Section 0.2 also hold true for rational exponents: am/n 5 1a1/n2m 5 A! n aB m and am/n 5 1am21/n 5 ! n am Negative rational exponents: a2m/n 5 1 am/n, for m and n positive integers with no common factors, a 2 0. In this section, we defined radicals as “b 5 ! n a means a 5 bn” for a and b positive real numbers when n is a positive even inte-ger, and a and b any real numbers when n is a positive odd integer. Properties of Radicals PROPERTY EXAMPLE ! n ab 5 ! n a⋅! n b ! 3 16 5 ! 3 8⋅! 3 2 5 2! 3 2 Å n a b 5 ! n a ! n b b 2 0 Å 4 81 16 5 ! 4 81 ! 4 16 5 3 2 ! n am 5 A! n aB m " 3 82 5 A! 3 8B 2 5 1222 5 4 ! n an 5 a n is odd " 3 x5 5 " 3 x3⋅" 3 x2 5 x" 3 x2 ! n an 5 ∙a∙ n is even " 4x6 5 " 4x4⋅" 4x2 5 ∙x∙" 4x2 E [SEC TION 0.6] S UM M A RY x 68 CHAPTER 0 Prerequisites and Review [SEC TION 0.6] E X E R C I SE S • S K I L L S In Exercises 1–24, evaluate each expression or state that it is not a real number. 1. !100 2. !121 3. 2!144 4. !2169 5. ! 3 2216 6. ! 3 2125 7. ! 3 343 8. 2! 3 227 9. ! 8 1 10. ! 7 21 11. ! 3 0 12. ! 5 0 13. !216 14. ! 5 21 15. 122721/3 16. 126421/3 17. 82/3 18. 126422/3 19. 123221/5 20. 1224321/3 21. 12121/3 22. 15/2 23. 93/2 24. 12722/3 In Exercises 25–40, simplify (if possible) the radical expressions. 25. !2 2 5!2 26. 3!5 2 7!5 27. 3!5 2 2!5 1 7!5 28. 6!7 1 7!7 2 10!7 29. !12⋅!2 30. 2!5⋅3!40 31. ! 3 12⋅! 3 4 32. ! 4 8⋅! 4 4 33. !3!7 34. !5!2 35. 8"25x2 36. 16"36y4 37. "4x2y 38. "16x3y 39. " 3 281x6y8 40. " 5 232x10y8 In Exercises 41–56, rationalize the denominators. 41. Ç 1 3 42. Ç 2 5 43. 2 3!11 44. 5 3!2 45. 3 1 2 !5 46. 2 1 1 !3 47. 1 1 !2 1 2 !2 48. 3 2 !5 3 1 !5 49. 3 !2 2 !3 50. 5 !2 1 !5 51. 4 3!2 1 2!3 52. 7 2!3 1 3!2 53. 4 1 !5 3 1 2!5 54. 6 3!2 1 4 55. !7 1 3 !2 2 !5 56. !y !x 2 !y In Exercises 57–64, simplify by applying the properties of rational exponents. Express your answers in terms of positive exponents. • A P P L I C A T I O N S 69. Gravity. If a penny is dropped off a building, the time it takes (seconds) to fall d feet is given by Ç d 16. If a penny is dropped off a 1280-foot-tall building, how long will it take until it hits the ground? Round to the nearest second. 70. Gravity. If a ball is dropped off a building, the time it takes (seconds) to fall d meters is approximately given by Ç d 5. If a ball is dropped off a 600-meter-tall building, how long will it take until it hits the ground? Round to the nearest second. 71. Kepler’s Law. The square of the period p (in years) of a planet’s orbit around the Sun is equal to the cube of the planet’s maximum distance from the Sun, d (in astronomical units, or AU). This relationship can be expressed mathematically as p2 5 d 3. If this formula is solved for d, the resulting equation is d 5 p2/3. If Saturn has an orbital period of 29.46 Earth years, calculate Saturn’s maximum distance from the Sun to the nearest hundredth of an AU. 72. Period of a Pendulum. The period (in seconds) of a pendulum of length L (in meters) is given by P 5 2⋅p⋅a L 9.8b 1/2 . If a certain pendulum has a length of 19.6 meters, determine the period P of this pendulum to the nearest tenth of a second. 57. 1x1/2 y2/32 6 58. 1y2/3y1/42 12 59. 1x1/3 y1/22 23 1x21/2 y1/422 60. 1x22/3y23/4222 1x1/3y1/424 61. x1/2 y1/5 x22/3 y29/5 62. 1y23/4x22/32 12 1y1/4x7/3224 63. 12x2/32 3 14x21/322 64. 12x22/323 14x24/322 In Exercises 65–68, factor each expression completely. 65. x7/3 2 x4/3 2 2x1/3 66. 8x1/4 1 4x5/4 67. 7x3/7 2 14x6/7 1 21x10/7 68. 7x21/3 1 70x • C A T C H T H E M I S T A K E In Exercises 73 and 74, explain the mistake that is made. 73. Simplify 14x1/2 y1/422 Solution: Use properties of exponents. 41x1/222 1y1/422 Simplify. 4xy1/2 This is incorrect. What mistake was made? 74. Simplify 2 5 2 !11. Solution: Multiply numerator and denominator by 5 2 !11. 2 5 2 !11 ⋅ A5 2 !11B A5 2 !11B Multiply numerators and denominators. 2A5 2 !11B 25 2 11 Simplify. 2A5 2 !11B 14 5 5 2 !11 7 This is incorrect. What mistake was made? In Exercises 75–78, determine whether each statement is true or false. 75. !121 5 611 76. "x2 5 x, where x is any real number. 77. "a2 1 b2 5 "a 1 "b 78. !24 5 22 In Exercises 79 and 80, a, m, n, and k are any positive real numbers. 79. Simplify 11am2n2k. 80. Simplify 1a2k221/k. In Exercises 81 and 82, evaluate each algebraic expression for the specified values. 81. "b2 2 4ac 2a for a 5 1, b 5 7, c 5 12 82. "b2 2 4ac for a 5 1, b 5 7, c 5 12 • C O N C E P T U A L 83. Rationalize the denominator and simplify: 1 A !a 1 !bB2. 84. Rationalize the denominator and simplify: !a 1 b 2 !a !a 1 b 1 !a. • C H A L L E N G E 88. Given 2 4!5 2 3!6 a. Rationalize the denominator. b. Use a graphing utility to evaluate the expression and the answer. c. Do they agree? 85. Use a calculator to approximate !11 to three decimal places. 86. Use a calculator to approximate ! 3 7 to three decimal places. 87. Given 4 5!2 1 4!3 a. Rationalize the denominator. b. Use a graphing utility to evaluate the expression and the answer. c. Do they agree? • T E C H N O L O G Y 0.6 Rational Exponents and Radicals 69 70 CHAPTER 0 Prerequisites and Review 0.7.1 The Imaginary Unit, i In Section 1.3, we will be studying equations whose solutions sometimes involve the square roots of negative numbers. In Section 0.6, when asked to evaluate the square root of a negative number, like !216, we said “it is not a real number” because there is no real number such that x2 5 216. To include such roots in the number system, mathematicians created a new expanded set of numbers, called the complex numbers. The foundation of this new set of numbers is the imaginary unit i. DEFINITION The Imaginary Unit i The imaginary unit is denoted by the letter i and is defined as i 5 !21 where i2 5 21. Recall that for positive real numbers a and b we defined the principal square root as b 5 !a which means b2 5 a Similarly, we define the principal square root of a negative number as !2a 5 i!a, since Ai!aB 2 5 i2 a 5 2a. We write i!a instead of !a i to avoid any confusion as to what is included in the radical. If 2a is a negative real number, then the principal square root of 2a is !2a 5 i!a where i is the imaginary unit and i2 5 21. S K I L L S O B J E C T I V E S ■ ■Write radicals with negative radicands as imaginary numbers. ■ ■Add and subtract complex numbers. ■ ■Multiply complex numbers. ■ ■Divide complex numbers. ■ ■Raise complex numbers to powers. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that real numbers and imaginary numbers are subsets of complex numbers. ■ ■Recognize the real and imaginary parts of a complex number. ■ ■Recognize that the square of i is 21. ■ ■Understand how to eliminate imaginary numbers in denominators. ■ ■Understand why i raised to a positive integer power can be reduced to 1, 21, i, or 2i . 0.7 COMPLEX NUMBERS 0.7.1 S KILL Write radicals with negative radicands as imaginary numbers. 0.7.1 CO NCE PTUAL Understand that real numbers and imaginary numbers are subsets of complex numbers. EXAMPLE 1 Using Imaginary Numbers to Simplify Radicals Simplify using imaginary numbers. a. !29 b. !28 Solution: a. !29 5 i!9 5 3i b. !28 5 i!8 5 i ⋅ 2!2 5 2i!2 Y OUR TU R N Simplify !2144. ▼ [CONCEPT CHECK] Write "2a2 as an imaginary number. ANSWER i0 a 0 ▼ ▼ A N S W E R 12i STUDY TIP !2a 5 !21⋅!a 5 i !a 0.7 Complex Numbers 71 DEFINITION Complex Number A complex number in standard form is defined as a 1 bi where a and b are real numbers and i is the imaginary unit. We denote a as the real part of the complex number and b as the imaginary part of the complex number. A complex number written as a 1 bi is said to be in standard form. If a 5 0 and b 2 0, then the resulting complex number bi is called a pure imaginary number. If b 5 0, then a 1 bi is a real number. The set of all real numbers and the set of all imaginary numbers are both subsets of the set of complex numbers. The following are examples of complex numbers. 17 2 2 3i 25 1 i 3 2 i!11 29i DEFINITION Equality of Complex Numbers The complex numbers a 1 bi and c 1 di are equal if and only if a 5 c and b 5 d. In other words, two complex numbers are equal if and only if both real parts are equal and both imaginary parts are equal. 0.7.2 Adding and Subtracting Complex Numbers Complex numbers in the standard form a 1 bi are treated in much the same way as binomials of the form a 1 bx. We can add, subtract, and multiply complex numbers the same way we performed these operations on binomials. When adding or subtracting complex numbers, combine real parts with real parts and combine imaginary parts with imaginary parts. Complex Numbers a 1bi Real Numbers a 1b 5 0 2 Imaginary Numbers bi 1a 5 0 2 0.7.2 S K I L L Add and subtract complex numbers. 0.7.2 C ON C E P T U A L Recognize the real and imaginary parts of a complex number. EXAMPLE 2 Adding and Subtracting Complex Numbers Perform the indicated operation and simplify. a. 13 2 2i 2 1 121 1 i 2 b. 12 2 i 2 2 13 2 4i 2 Solution (a): Eliminate the parentheses. 13 2 2i 2 1 121 1 i 2 5 3 2 2i 2 1 1 i Group real and imaginary numbers, respectively. 5 13 2 1 2 1 122i 1 i 2 Simplify. 5 Solution (b): Eliminate the parentheses (distribute the negative). 12 2 i 2 2 13 2 4i 2 5 2 2 i 2 3 1 4i Group real and imaginary numbers, respectively. 5 12 2 3 2 1 12i 1 4i 2 Simplify. 5 Y OUR T UR N Perform the indicated operation and simplify: 14 1 i 2 2 13 2 5i 2. 2 2 i 21 1 3i ▼ A N S W E R 1 1 6i ▼ [CONCEPT CHECK] Simplify: (a 2 bi ) 2 (a 1 bi ) ANSWER 22bi ▼ 72 CHAPTER 0 Prerequisites and Review 0.7.3 Multiplying Complex Numbers When multiplying complex numbers, you apply all the same methods as you did when multiplying binomials. It is important to remember that i2 5 21. WORDS MATH Multiply the complex numbers. 15 2 i2 13 2 4i2 Multiply using the distributive property. 5 5 132 1 5 124i2 2 i 132 2 1i2 124i2 Eliminate the parentheses. 5 15 2 20i 2 3i 1 4i2 Let i2 5 21. 5 15 2 20i 2 3i 1 4 1212 Simplify. 5 15 2 20i 2 3i 2 4 Combine real parts and imaginary parts, respectively. 5 11 2 23i 0.7.3 S KILL Multiply complex numbers. 0.7.3 CO NCE PTUAL Recognize that the square of i is 21. EXAMPLE 3 Multiplying Complex Numbers Multiply the complex numbers and express the result in standard form, a 6 bi. a. 13 2 i2 12 1 i2 b. i 123 1 i2 Solution (a): Use the distributive property. 13 2 i 2 12 1 i 2 5 3122 1 31i 2 2 i 122 2 i 1i 2 Eliminate the parentheses. 5 6 1 3i 2 2i 2 i2 Substitute i2 5 21. 5 6 1 3i 2 2i 2 1212 Group like terms. 5 16 1 12 1 13i 2 2i 2 Simplify. 5 Solution (b): Use the distributive property. i 123 1 i 2 5 23i 1 i2 Substitute i2 5 21. 5 23i 2 1 Write in standard form. 5 Y OUR TU R N Multiply the complex numbers and express the result in standard form, a 6 bi: 14 2 3i2 121 1 2i2. 7 1 i 21 2 3i ▼ A N S W E R 2 1 11i ▼ [CONCEPT CHECK] Multiply and simplify (a 1 bi )(a 2 bi ) ANSWER a2 1 b2 ▼ 0.7.4 Dividing Complex Numbers Recall the special product that produces a difference of two squares, 1a 1 b2 1a 2 b2 5 a2 2 b2. This special product has only first and last terms because the outer and inner terms cancel each other out. Similarly, if we multiply complex numbers in the same manner, the result is a real number because the imaginary terms cancel each other out. COMPLEX CONJUGATE The product of a complex number, z 5 a 1 bi, and its complex conjugate, z 5 a 2 bi, is a real number. zz 5 1a 1 bi21a 2 bi2 5 a2 2 b2 i2 5 a2 2 b21212 5 a2 1 b2 0.7.4 S KILL Divide complex numbers. 0.7.4 CO NCE PTUAL Understand how to eliminate imaginary numbers in denominators. STUDY TIP When multiplying complex numbers, remember that i 2 5 21. To write a quotient of complex numbers in standard form, a 1 bi, multiply the numerator and the denominator by the complex conjugate of the denominator. It is important to note that if i is present in the denominator, then the complex number is not in standard form. 0.7.5 Raising Complex Numbers to Integer Powers Note that i raised to the fourth power is 1. In simplifying imaginary numbers, we factor out i raised to the largest multiple of 4. i 5 !21 i2 5 21 i3 5 i2⋅ i 5 1212i 5 2i i4 5 i2⋅ i2 5 12121212 5 1 i5 5 i 4⋅ i 5 1121i2 5 i i 6 5 i 4⋅ i 2 5 1121212 5 21 i7 5 i 4⋅ i 3 5 11212i2 5 2i i8 5 1i 422 5 1 [CONCEPT CHECK] Write the quotient in standard form: 1 a 1 bi ANSWER a 1a2 1 b22 2 bi 1a2 1 b22 ▼ 0.7.5 S K I L L Raise complex numbers to powers. 0.7.5 C ON C E P T U A L Understand why i raised to a positive integer power can be reduced to 1, 21, i, or 2i. 0.7 Complex Numbers 73 EXAMPLE 4 Dividing Complex Numbers Write the quotient in standard form: 2 2 i 1 1 3i. Solution: Multiply numerator and denominator by the complex conjugate of the denominator, 1 2 3i. a 2 2 i 1 1 3ib a1 2 3i 1 2 3ib Multiply the numerators and denominators, respectively. 5 12 2 i211 2 3i2 11 1 3i211 2 3i2 Use the FOIL method (or distributive property). 5 2 2 6i 2 i 1 3i2 1 2 3i 1 3i 2 9i2 Combine imaginary parts. 5 2 2 7i 1 3i2 1 2 9i2 Substitute i2 5 21. 5 2 2 7i 2 3 1 2 91212 Simplify the numerator and denominator. 5 21 2 7i 10 Write in standard form. Recall that a 1 b c 5 a c 1 b c. 5 2 1 10 2 7 10i Y OUR T UR N Write the quotient in standard form: 3 1 2i 4 2 i . ▼ A N S W E R 10 17 1 11 17 i ▼ 74 CHAPTER 0 Prerequisites and Review EXAMPLE 5 Raising the Imaginary Unit to Integer Powers Simplify: a. i7 b. i13 c. i100 Solution: a. i7 5 i4⋅i3 5 112 12i 2 5 b. i13 5 i12⋅i 5 1i 4 2 3⋅i 5 13⋅i 5 c. i100 5 1i42 25 5 125 5 Y OUR TU R N Simplify i27. 2i i 1 ▼ EXAMPLE 6 Raising a Complex Number to an Integer Power Write 12 2 i 23 in standard form. Solution: Recall the formula for cubing a binomial. 1a 2 b23 5 a3 2 3a2b 1 3ab2 2 b3 Let a 5 2 and b 5 i. 12 2 i 23 5 23 2 3122 2 1i 2 1 3122 1i 22 2 i3 Let i2 5 21 and i3 5 2i. 5 23 2 3122 2 1i 2 1 132 122 1212 2 12i 2 Eliminate parentheses. 5 8 2 6 2 12i 1 i Combine the real parts and imaginary parts, respectively. 5 Y OUR TU R N Write 12 1 i 2 3 in standard form. 2 2 11i ▼ A N S W E R 2 1 11i ▼ [CONCEPT CHECK] Simplify i 4a where a 5 1, 2, 3, ... ANSWER 1 ▼ ▼ A N S W E R 2i Multiplying Complex Numbers • 1a 1 bi 2 1c 1 di 2 5 1ac 2 bd 2 1 1ad 1 bc2 i • Apply the same methods as for multiplying binomials. It is important to remember that i2 5 21. Dividing Complex Numbers • Complex conjugate of a 1 bi is a 2 bi. • In order to write a quotient of complex numbers in standard form, multiply the numerator and the denominator by the complex conjugate of the denominator: a 1 bi c 1 di ⋅ 1c 2 di2 1c 2 di2 The Imaginary Unit i • i 5 !21 • i2 5 21 Complex Numbers • Standard Form: a 1 bi, where a is the real part and b is the imaginary part. • The set of real numbers and the set of pure imaginary numbers are subsets of the set of complex numbers. Adding and Subtracting Complex Numbers • 1a 1 bi 2 1 1c 1 di 2 5 1a 1 c2 1 1b 1 d 2 i • 1a 1 bi 2 2 1c 1 di 2 5 1a 2 c2 1 1b 2 d 2 i • To add or subtract complex numbers, add or subtract the real parts and imaginary parts, respectively. [SEC TION 0.7 ] S U M MA RY In Exercises 1–12, write each expression as a complex number in standard form. Some expressions simplify to either a real number or a pure imaginary number. 1. !216 2. !2100 3. !220 4. !224 5. ! 3 264 6. ! 3 227 7. !264 8. !227 9. 3 2 !2100 10. 4 2 !2121 11. 210 2 !2144 12. 7 2 ! 3 2125 In Exercises 13–40, perform the indicated operation, simplify, and express in standard form. 13. 13 2 7i 2 1 121 2 2i 2 14. 11 1 i 2 1 19 2 3i 2 15. 13 2 4i 2 1 17 2 10i 2 16. 15 1 7i 2 1 1210 2 2i 2 17. 14 2 5i 2 2 12 2 3i 2 18. 122 1 i 2 2 11 2 i 2 19. 123 1 i 2 2 122 2 i 2 20. 14 1 7i 2 2 15 1 3i 2 21. 314 2 2i 2 22. 417 2 6i 2 23. 1218 2 5i 2 24. 23116 1 4i 2 25. 23116 2 9i 2 26. 5126i 1 32 27. 26 117 2 5i 2 28. 21218 1 3i 2 29. 11 2 i 2 13 1 2i 2 30. 123 1 2i 2 11 2 3i 2 31. 15 2 7i 2 123 1 4i 2 32. 116 2 5i 2 122 2 i 2 33. 17 2 5i 2 16 1 9i 2 34. 123 2 2i 2 17 2 4i 2 35. 112 2 18i 2 122 1 i 2 36. 124 1 3i 2 124 2 3i 2 37. A1 2 1 2iBA4 9 2 3iB 38. A23 4 1 9 16iBA2 3 1 4 9iB 39. 12i 1 17 2 12 1 3i 2 40. 123i 2 2 2 122 2 3i 2 In Exercises 41–48, for each complex number z, write the complex conjugate z and find zz. 41. z 5 4 1 7i 42. z 5 2 1 5i 43. z 5 2 2 3i 44. z 5 5 2 3i 45. z 5 6 1 4i 46. z 5 22 1 7i 47. z 5 22 2 6i 48. z 5 23 2 9i In Exercises 49–64, write each quotient in standard form. 49. 2 i 50. 3 i 51. 1 3 2 i 52. 2 7 2 i 53. 1 3 1 2i 54. 1 4 2 3i 55. 2 7 1 2i 56. 8 1 1 6i 57. 1 2 i 1 1 i 58. 3 2 i 3 1 i 59. 2 1 3i 3 2 5i 60. 2 1 i 3 2 i 61. 4 2 5i 7 1 2i 62. 7 1 4i 9 2 3i 63. 8 1 3i 9 2 2i 64. 10 2 i 12 1 5i In Exercises 65–76, simplify. 65. i15 66. i99 67. i40 68. i18 69. 15 2 2i 2 2 70. 13 2 5i 2 2 71. 12 1 3i 2 2 72. 14 2 9i 2 2 73. 13 1 i 2 3 74. 12 1 i 2 3 75. 11 2 i 2 3 76. 14 2 3i 2 3 [SEC TION 0.7] E X ERC I S E S • S K I L L S • A P P L I C A T I O N Electrical impedance is the ratio of voltage to current in ac circuits. Let Z represent the total impedance of an electrical circuit. If there are two resistors in a circuit, let Z1 5 3 2 6i ohms and Z2 5 5 1 4 i ohms. 77. Electrical Circuits in Series. When the resistors in the circuit are placed in series, the total impedance is the sum of the two impedances Z 5 Z1 1 Z2. Find the total impedance of the electrical circuit in series. 78. Electrical Circuits in Parallel. When the resistors in the circuit are placed in parallel, the total impedance is given by 1 Z 5 1 Z1 1 1 Z2 . Find the total impedance of the electrical circuit in parallel. 0.7 Complex Numbers 75 76 CHAPTER 0 Prerequisites and Review • C A T C H T H E M I S T A K E In Exercises 79 and 80, explain the mistake that is made. 79. Write the quotient in standard form: 2 4 2 i. Solution: Multiply the numerator 2 4 2 i ⋅ 4 2 i 4 2 i and the denominator by 4 2 i. Multiply the numerator using the distributive property and the denominator using the FOIL method. 8 2 2i 16 2 1 Simplify. 8 2 2i 15 Write in standard form. 8 15 2 2 15 i This is incorrect. What mistake was made? 80. Write the product in standard form: 12 2 3i 2 15 1 4i 2. Solution: Use the FOIL method to multiply the complex numbers. 10 2 7i 2 12i2 Simplify. 22 2 7i This is incorrect. What mistake was made? In Exercises 81–84, determine whether each statement is true or false. • C O N C E P T U A L 81. The product is a real number: 1a 1 bi 2 1a 2 bi 2. 82. Imaginary numbers are a subset of the complex numbers. 83. Real numbers are a subset of the complex numbers. 84. There is no complex number that equals its conjugate. 85. Factor completely over the complex numbers: x4 1 2x2 1 1. 86. Factor completely over the complex numbers: x4 1 18x2 1 81. • C H A L L E N G E In Exercises 87–90, use a graphing utility to simplify the expression. Write your answer in standard form. • T E C H N O L O G Y 87. 11 1 2i25 88. 13 2 i26 89. 1 12 2 i23 90. 1 14 1 3i22 CH A P TE R 0 R E VIE W [CH AP TER 0 REVIEW] SECTION CONCEPT KEY IDEAS/FORMULAS 0.1 Real numbers The set of real numbers Rational: a b, where a and b are integers or a decimal that terminates or repeats. Irrational: Nonrepeating/nonterminating decimal. Approximations: Rounding and truncation Rounding: Examine the digit to the right of the last desired digit. Digit , 5: Keep last desired digit as is. Digit $ 5: Round the last desired digit up 1. Truncating: Eliminate all digits to the right of the desired digit. Order of operations 1. Parentheses 2. Multiplication and Division 3. Addition and Subtraction Properties of real numbers ■ ■a 1b 1 c2 5 ab 1 ac ■ ■If xy 5 0, then x 5 0 or y 5 0 ■ ■a b 6 c d 5 ad 6 bc bd b 2 0 and d 2 0 ■ ■a b 4 c d 5 a b ⋅ d c b 2 0, c 2 0, and d 2 0 0.2 Integer exponents and scientific notation an 5 a⋅a⋅a . . . a n factors Integer exponents am⋅ an 5 am1n am an 5 am2n a0 5 1 1am2n 5 amn a2n 5 1 an 5 a1 ab n a 2 0 Scientific notation c 3 10n where c is a positive real number 1 # c , 10 and n is an integer. 0.3 Polynomials: Basic operations Adding and subtracting polynomials Combine like terms. Multiplying polynomials Distributive property Special products 1x 1 a2 1x 1 b2 5 x2 1 1a 1 b2 x 1 ab Perfect Squares 1a 1 b2 2 5 1a 1 b2 1a 1 b2 5 a2 1 2ab 1 b2 1a 2 b2 2 5 1a 2 b2 1a 2 b2 5 a2 2 2ab 1 b2 Difference of Two Squares 1a 1 b2 1a 2 b2 5 a2 2 b2 Perfect Cubes 1a 1 b2 3 5 a3 1 3a2b 1 3ab2 1 b3 1a 2 b2 3 5 a3 2 3a2b 1 3ab2 2 b3 0.4 Factoring polynomials Greatest common factor Factor out using distributive property: ax k Factoring formulas: Special polynomial forms Difference of Two Squares a2 2 b2 5 1a 1 b2 1a 2 b2 Perfect Squares a2 1 2ab 1 b2 5 1a 1 b2 2 a2 2 2ab 1 b2 5 1a 2 b2 2 Sum of Two Cubes a3 1 b3 5 1a 1 b2 1a2 2 ab 1 b22 Difference of Two Cubes a3 2 b3 5 1a 2 b2 1a2 1 ab 1 b22 Factoring a trinomial as a product of two binomials ■ ■x2 1 bx 1 c 5 1x 1 ?2 1x 1 ?2 ■ ■ax2 1 bx 1 c 5 1?x 1 ?2 1?x 1 ?2 µ Chapter Review 77 78 CHAPTER 0 Prerequisites and Review CH A P TE R 0 R E VI E W SECTION CONCEPT KEY IDEAS/FORMULAS Factoring by grouping Group terms with common factors. A strategy for factoring polynomials 1. Factor out any common factors. 2. Recognize any special products. 3. Use the FOIL method in reverse for trinomials. 4. Look for factoring by grouping. 0.5 Rational expressions Rational expressions and domain restrictions Note domain restrictions when denominator is equal to zero. Simplifying rational expressions Multiplying and dividing rational expressions ■ ■Use properties of rational numbers. ■ ■ State additional domain restrictions once division is rewritten as multiplication of a reciprocal. Adding and subtracting rational expressions Least common denominator (LCD) Complex rational expressions Two strategies: 1. Write sum/difference in numerator/denominator as a rational expression. 2. Multiply by the LCD of the numerator and denominator. 0.6 Rational exponents and radicals Square roots !25 5 5 Other (nth) roots b 5 ! n a means a 5 bn for a and b positive real numbers and n a positive even integer, or for a and b any real numbers and n a positive odd integer. ! n ab 5 ! n a⋅! n b Ä n a b 5 ! n a ! n b b 2 0 ! n am 5 A! n aB m ! n an 5 a n is odd ! n an 5 a n is even Rational exponents a1/n 5 ! n a am/n 5 1a1/n2m 5 A! n aB m a2m/n 5 1 am/n for m and n positive integers with no common factors, a 2 0. 0.7 Complex numbers The imaginary unit, i i 5 !21 Adding and subtracting complex numbers Complex Numbers: a 1 bi where a and b are real numbers. Combine real parts with real parts and imaginary parts with imaginary parts. Multiplying complex numbers Use the FOIL method and i2 5 21 to simplify. Dividing complex numbers If a 1 bi is in the denominator, then multiply the numerator and the denominator by a 2 bi. The result is a real number in the denominator. Raising complex numbers to integer powers i 5 !21 i 2 5 21 i 3 5 2i i 4 5 1 R E VI E W E XERCISES [CH AP TER 0 REVIEW EXE R C IS E S ] 0.1 Real Numbers Approximate to two decimal places by (a) rounding and (b) truncating. 1. 5.21597 2. 7.3623 Simplify. 3. 7 2 2⋅5 1 4⋅3 2 5 4. 22 15 1 32 1 713 2 2⋅52 5. 2 16 12221242 6. 231x 2 y2 1 4 13x 2 2y2 Perform the indicated operation and simplify. 7. x 4 2 x 3 8. y 3 1 y 5 2 y 6 9. 12 7 ⋅21 4 10. a2 b3 4 2a b2 0.2 Integer Exponents and Scientific Notation Simplify using properties of exponents. 11. 122z2 3 12. 124z223 13. 13x3 y222 21x2y24 14. 12x2 y322 14xy23 15. Express 0.00000215 in scientific notation. 16. Express 7.2 3 109 as a real number. 0.3 Polynomials: Basic Operations Perform the indicated operation and write the results in standard form. 17. 114z2 1 22 1 13z 2 42 18. 127y2 2 6y 1 22 2 1y2 1 3y 2 72 19. 136x2 2 4x 2 52 2 16x 2 9x2 1 102 20. 32x 2 14x2 2 7x24 2 33x 2 12x2 1 5x 2 424 21. 5xy2 13x 2 4y2 22. 22st2 12t 1 s 2 2st2 23. 1x 2 72 1x 1 92 24. 12x 1 12 13x 2 22 25. 12x 2 32 2 26. 15x 2 72 15x 1 72 27. 1x2 1 12 2 28. 11 2 x22 2 0.4 Factoring Polynomials Factor out the common factor. 29. 14x2y2 2 10xy3 30. 30x4 2 20x3 1 10x2 Factor the trinomial into a product of two binomials. 31. 2x2 1 9x 2 5 32. 6x2 2 19x 2 7 33. 16x2 2 25 34. 9x2 2 30x 1 25 Factor the sum or difference of two cubes. 35. x3 1 125 36. 1 2 8x3 Factor into a product of three polynomials. 37. 2x3 1 4x2 2 30x 38. 6x3 2 5x2 1 x Factor into a product of two binomials by grouping. 39. x3 1 x2 2 2x 2 2 40. 2x3 2 x2 1 6x 2 3 0.5 Rational Expressions State the domain restrictions on each of the rational expressions. 41. 4x2 2 3 x2 2 9 42. 1 x2 1 1 Simplify. 43. x2 2 4 x 2 2 44. x 2 5 x 2 5 45. t2 1 t 2 6 t2 2 t 2 2 46. z3 2 z z2 1 z Perform the indicated operation and simplify. 47. x2 1 3x 2 10 x2 1 2x 2 3 ⋅x2 1 x 2 2 x2 1 x 2 6 48. x2 2 x 2 2 x3 1 3x2 4 x 1 1 x2 1 2x 49. 1 x 1 1 2 1 x 1 3 50. 1 x 2 1 x 1 1 1 1 x 1 2 Simplify. 51. 2 1 1 x 2 3 1 5x 2 15 1 4 52. 1 x 1 2 x2 3 2 1 x2 Review Exercises 79 80 CHAPTER 0 Prerequisites and Review REV IEW E XE R CI SE S 0.6 Rational Exponents and Radicals Simplify. 53. !20 54. !80 55. " 3 2125x5 y4 56. " 4 32x4 y5 57. 3!20 1 5!80 58. 4!27x 2 8!12x 59. A2 1 !5BA1 2 !5B 60. A3 1 !xBA4 2 !xB 61. 1 2 2 !3 62. 1 3 2 !x 63. 13x2/322 14x1/322 64. 14x3/422 12x21/322 65. 51/2 51/3 66. 1x22/3y1/4212 0.7 Complex Numbers Simplify. 67. !2169 68. !232 69. i19 70. i9 Perform the indicated operation, simplify, and express in standard form. 71. 13 2 2i 2 1 15 2 4i 2 72. 124 1 7i 2 1 122 2 3i 2 73. 112 2 i 2 2 122 2 5i 2 74. 19 1 8i 2 2 14 2 2i 2 75. 12 1 2i 2 13 2 3i 2 76. 11 1 6i 2 11 1 5i 2 77. 14 1 7i 22 78. 17 2 i 22 Express the quotient in standard form. 79. 1 2 2 i 80. 1 3 1 i 81. 7 1 2i 4 1 5i 82. 6 2 5i 3 2 2i 83. 10 3i 84. 7 2i Technology Section 0.1 85. Use your calculator to evaluate "272.25. Does the answer appear to be a rational or an irrational number? Why? 86. Use your calculator to evaluate Å 1053 81 . Does the answer appear to be a rational or an irrational number? Why? Section 0.2 Use a graphing utility to evaluate the expression. Express your answer in scientific notation. 87. 18.2 3 1011211.167 3 102352 14.92 3 102182 88. 11.4805 3 10212 15.64 3 1026211.68 3 10292 Section 0.3 89. Use a graphing utility to plot the graphs of the three expressions 12x 1 323, 8x3 1 27, and 8x3 1 36x2 1 54x 1 27. Which two graphs agree? 90. Use a graphing utility to plot the graphs of the three expressions 1x 2 322, 8x2 1 9, and x2 2 6x 1 9. Which two graphs agree? Section 0.4 91. Use a graphing utility to plot the graphs of the three expressions x2 2 3x 1 18, 1x 1 621x 2 32, and 1x 2 621x 1 32. Which two graphs agree? 92. Use a graphing utility to plot the graphs of the three expressions x2 2 8x 1 16, 1x 1 422, and 1x 2 422. Which two graphs agree? Section 0.5 For each given expression: (a) simplify the expression, (b) use a graphing utility to plot the expression and the answer in (a) in the same viewing window, and (c) determine the domain restriction(s) where the graphs will agree with each other. 93. 1 2 14/x2 1 2 14/x22 94. 1 2 13/x2 1 1 19/x22 Section 0.6 95. Given 6 !5 2 !2 a. Rationalize the denominator. b. Use a graphing utility to evaluate the expression and the answer. c. Do they agree? 96. Given 11 2!6 1 !13 a. Rationalize the denominator. b. Use a graphing utility to evaluate the expression and the answer. c. Do they agree? Section 0.7 In Exercises 97 and 98, use a graphing utility to simplify the expression. Write your answer in standard form. 97. 13 1 5i 2 5 98. 1 11 1 3i24 99. Apply a graphing utility to simplify the expression and write your answer in standard form. 1 16 1 2i24 [CH AP TER 0 PRACTICE T E S T ] Simplify. 1. !16 2. " 3 54x6 3. 2312 1 522 1 213 2 72 2 132 2 12 4. " 5 232 5. "212x2 6. i17 7. 1x2 y23 z21222 1x21 y2 z321/2 8. 3!x 2 4!x 1 5!x 9. 3!18 2 4!32 10. A5!6 2 2!2BA !6 1 3!2B Perform the indicated operation and simplify. 11. 13y2 2 5y 1 72 2 1y2 1 7y 2 132 12. 12x 2 3 2 15x 1 7 2 Factor. 13. x2 2 16 14. 3x2 1 15x 1 18 15. 4x2 1 12xy 1 9y2 16. x4 2 2x2 1 1 17. 2x2 2 x 2 1 18. 6y2 2 y 2 1 19. 2t3 2 t2 2 3t 20. 2x3 2 5x2 2 3x 21. x2 2 3yx 1 4yx 2 12y2 22. x4 1 5x2 2 3x2 2 15 23. 81 1 3x3 24. 27x 2 x4 Perform the indicated operations and simplify. 25. 2 x 1 3 x 2 1 26. 5x x2 2 7x 1 10 2 4 x2 2 25 27. x 2 1 x2 2 1 ⋅ x2 1 x 1 1 x3 2 1 28. 4x2 2 9 x2 2 11x 2 60 ⋅ x2 2 16 2x 1 3 29. x 2 3 2x 2 5 4 x2 2 9 5 2 2x 30. 1 2 t 3t 1 1 4 t2 2 2t 1 1 7t 1 21t2 Write the resulting expression in standard form. 31. 11 2 3i 2 17 2 5i 2 32. 2 2 11i 4 1 i 33. Rationalize the denominator: 7 2 2!3 4 2 5!3. 34. Represent 0.0000155 in scientific notation. 35. Simplify 1 x 2 2 x 1 1 x 2 1 and state any domain restrictions. 36. For the given expression: 1 1 5 x 1 2 25 x2 a. Simplify the expression. b. Use a graphing utility to plot the expression and the answer in (a) in the same viewing window. c. Determine the domain restriction(s) where the graphs will agree with each other. 37. Apply a graphing utility to evaluate the expression. Round your answer to three decimal places. !5 !13 2 !7 Practice Test 81 PRACTICE TEST C H A P T E R [ [ LEARNING OBJECTIVES ■ ■Solve linear equations. ■ ■Solve application problems involving linear equations. ■ ■Solve quadratic equations. ■ ■Solve rational, polynomial, and radical equations. ■ ■Solve linear inequalities. ■ ■Solve polynomial and rational inequalities. ■ ■Solve absolute value equations and inequalities. Equations and Inequalities 1 Golf courses usually charge both greens fees (cost of playing the course) and cart fees (cost of renting a golf cart). Two friends who enjoy playing golf decide to investigate becoming members at a golf course. The course they enjoy playing the most charges $40 for greens fees and $15 for cart rental (per person), so it currently costs each of them $55 every time they play. The membership offered at that course costs $160 per month with no greens fees, but there is still the per person cart rental fee. How many times a month would they have to play golf in order for the membership option to be the better deal? This is just one example of how the real world can be modeled with equations and inequalities. See Section 1.5, Exercises 109 and 110. Hero Images/Getty Images, Inc. 83 [I N T HI S CHAPTER] You will solve linear and quadratic equations. You will then solve more complicated equations (polynomial, rational, radical, and absolute value) by first transforming them into linear or quadratic equations. Then you will solve linear, quadratic, polynomial, rational, and absolute value inequalities. Throughout this chapter you will solve applications of equations and inequalities. EQUATIONS AND INEQUALITIES 1.1 LINEAR EQUATIONS 1.2 APPLICATIONS INVOLVING LINEAR EQUATIONS 1.3 QUADRATIC EQUATIONS 1.4 OTHER TYPES OF EQUATIONS 1.5 LINEAR INEQUALITIES 1.6 POLYNOMIAL AND RATIONAL INEQUALITIES 1.7 ABSOLUTE VALUE EQUATIONS AND INEQUALITIES • Solving Linear Equations in One Variable • Solving Rational Equations That Are Reducible to Linear Equations • Solving Application Problems Using Mathematical Models • Geometry Problems • Interest Problems • Mixture Problems • Distance– Rate–Time Problems • Factoring • Square Root Method • Completing the Square • Quadratic Formula • Applications Involving Quadratic Equations • Radical Equations • Equations Quadratic in Form: u-Substitution • Factorable Equations • Graphing Inequalities and Interval Notation • Solving Linear Inequalities • Polynomial Inequalities • Rational Inequalities • Equations Involving Absolute Value • Inequalities Involving Absolute Value 84 CHAPTER 1 Equations and Inequalities 1.1.1 Solving Linear Equations in One Variable An algebraic expression (see Chapter 0) consists of one or more terms that are combined through basic operations such as addition, subtraction, multiplication, or division; for example: 3x 1 2 5 2 2y x 1 y An equation is a statement that says two expressions are equal. For example, the following are all equations in one variable, x: x 1 7 5 11 x2 5 9 7 2 3x 5 2 2 3x 4x 1 7 5 x 1 2 1 3x 1 5 To solve an equation means to find all the values of x that make the equation true. These values are called solutions, or roots, of the equation. The first of these state-ments shown above, x 1 7 5 11, is true when x 5 4 and false for any other values of x. We say that x 5 4 is the solution to the equation. Sometimes an equation can have more than one solution, as in x 2 5 9. In this case, there are actually two values of x that make this equation true, x 5 23 and x 5 3. We say the solution set of this equation is 523, 36. In the third equation, 7 2 3x 5 2 2 3x, no values of x make the statement true. Therefore, we say this equation has no solution. And the fourth equation, 4x 1 7 5 x 1 2 1 3x 1 5, is true for any values of x. An equation that is true for any value of the variable x is called an identity. In this case, we say the solution set is the set of all real numbers. Two equations that have the same solution set are called equivalent equations. For example, 3x 1 7 5 13 3x 5 6 x 5 2 are all equivalent equations because each of them has the solution set 526. Note that x 2 5 4 is not equivalent to these three equations because it has the solution set 522, 26. When solving equations, it helps to find a simpler equivalent equation in which the variable is isolated (alone). The following table summarizes the procedures for generating equivalent equations. 1.1.1 S KILL Solve linear equations in one variable. 1.1.1 CO NCE PTUAL Understand the definition of a linear equation in one variable. S K I L L S O B J E C T I V E S ■ ■Solve linear equations in one variable. ■ ■Solve rational equations that are reducible to linear equations. C O N C E P T U A L O B J E C T I V E S ■ ■Understand the definition of a linear equation in one variable. ■ ■Eliminate values that result in a denominator being equal to zero. 1.1 LINEAR EQUATIONS ORIGINAL EQUATION DESCRIPTION EQUIVALENT EQUATION 31x 2 6 2 5 6x 2 x n Eliminate parentheses. n Combine like terms on one or both sides of an equation. 3x 2 18 5 5x 7x 1 8 5 29 Add (or subtract) the same quantity to (from) both sides of an equation. 7x 1 8 2 8 5 29 2 8 7x 5 21 5x 5 15 Multiply (or divide) both sides of an equation by the same nonzero quantity: 5x 5 5 15 5 . x 5 3 27 5 x Interchange the two sides of the equation. x 5 27 Generating Equivalent Equations 1.1 Linear Equations 85 You probably already know how to solve simple linear equations. Solving a linear equation in one variable is done by finding an equivalent equation. In generating an equivalent equation, remember that whatever operation is performed on one side of an equation must also be performed on the other side of the equation. EXAMPLE 1 Solving a Linear Equation Solve the equation 3x 1 4 5 16. Solution: Subtract 4 from both sides of the equation. 3x 1 4 5 16 24 24 3x 5 12 Divide both sides by 3. 3x 3 5 12 3 The solution is x 5 4. x 5 4 The solution set is 546. YOUR T UR N Solve the equation 2x 1 3 5 9. ▼ Example 1 illustrates solving linear equations in one variable. What is a linear equation in one variable? ▼ A N S W E R The solution is x 5 3. The solution set is 536. DEFINITION Linear Equation A linear equation in one variable, x, can be written in the form ax 1 b 5 0 where a and b are real numbers and a 2 0. What makes this equation linear is that x is raised to the first power. We can also classify a linear equation as a first-degree equation. EQUATION DEGREE GENERAL NAME x 2 7 5 0 First Linear x2 2 6x 2 9 5 0 Second Quadratic x3 1 3x2 2 8 5 0 Third Cubic EXAMPLE 2 Solving a Linear Equation Solve the equation 5x 2 17x 2 42 2 2 5 5 2 13x 1 22. Solution: Eliminate the parentheses. 5x 2 17x 2 42 2 2 5 5 2 13x 1 22 Don’t forget to distribute the negative sign through both terms inside the parentheses. 5x 2 7x 1 4 2 2 5 5 2 3x 2 2 Combine x terms on the left, constants on the right. Add 3x to both sides. ▲ ▲ ▲ ▲ 22x 1 2 5 3 2 3x 13x 1 3x x 1 2 5 3 86 CHAPTER 1 Equations and Inequalities To solve a linear equation involving fractions, find the least common denominator (LCD) of all terms and multiply both sides of the equation by the LCD. We will first review how to find the LCD. To add the fractions 1 2 1 1 6 1 2 5, we must first find a common denominator. Some people are taught to find the lowest number that 2, 6, and 5 all divide evenly into. Others prefer a more systematic approach in terms of prime factors. ▼ A N S W E R The solution is x 5 2. The solution set is 526. ▼ A N S W E R The solution is m 5 218. The solution set is 52186. STUDY TIP Prime Factors 2 5 2 6 5 2 ? 3 5 5 ? 5 LCD 5 2 ? 3 ? 5 5 30 EXAMPLE 3 Solving a Linear Equation Involving Fractions Solve the equation 1 2p 2 5 5 3 4 p. Solution: Write the equation. 1 2 p 2 5 5 3 4 p Multiply each term in the equation 142 1 2 p 2 1425 5 142 3 4 p by the LCD, 4. The result is a linear equation with no fractions. 2p 2 20 5 3p Subtract 2p from both sides. 22p 22p 220 5 p p 5 220 Since p 5 220 satisfies the original equation, the solution set is 52206. Y OUR TU R N Solve the equation 1 4 m 5 1 12 m 2 3. ▼ STEP DESCRIPTION EXAMPLE 1 Simplify the algebraic expressions on both sides of the equation. 231x 2 22 1 5 5 71x 2 42 2 1 23x 1 6 1 5 5 7x 2 28 2 1 23x 1 11 5 7x 2 29 2 Gather all variable terms on one side of the equation and all constant terms on the other side. 23x 1 11 5 7x 2 29 13x 13x 11 5 10x 2 29 129 129 40 5 10x 3 Isolate the variable. 10x 5 40 x 5 4 Solving a Linear Equation in One Variable [CONCEPT CHECK] Which one of these is a linear equation in one variable? (A) 3x 1 2 5 11 (B) x 2 5 9 (C) y 5 x 1 1 ANSWER (A) 3x 1 2 5 11 (B) is nonlinear in one variable (C) is linear in two variables ▼ Subtract 2 from both sides. 2 2 2 2 x 5 1 Check to verify that x 5 1 is a 5 ? 1 2 17 ? 1 2 42 2 2 5 5 2 13 ? 1 1 22 solution to the original equation. 5 2 17 2 42 2 2 5 5 2 13 1 22 5 2 132 2 2 5 5 2 152 0 5 0 Since the solution x 5 1 makes the equation true, the solution set is 516. Y OUR TU R N Solve the equation 41x 2 12 2 2 5 x 2 31x 2 22. ▼ 1.1 Linear Equations 87 1.1.2 Solving Rational Equations That Are Reducible to Linear Equations A rational equation is an equation that contains one or more rational expressions (Chapter 0). Some rational equations can be transformed into linear equations that you can then solve, but as you will see momentarily, you must be certain that the solution to the linear equation also satisfies the original rational equation. 1.1.2 S K IL L Solve rational equations that are reducible to linear equations. 1.1.2 C ON C E P T U A L Eliminate values that result in a denominator being equal to zero. STUDY TIP Since dividing by 0 is not defined, we exclude values of the variable that correspond to a denominator equaling 0. EXAMPLE 4 Solving a Rational Equation That Can Be Reduced to a Linear Equation Solve the equation 2 3x 1 1 2 5 4 x 1 4 3. Solution: State the excluded values (those that make any denominator equal 0). 2 3x 1 1 2 5 4 x 1 4 3 x 2 0 Eliminate fractions by multiplying 6xa 2 3xb 1 6xa1 2b 5 6xa4 xb 1 6xa4 3b each term by the LCD, 6x. Simplify both sides. 4 1 3x 5 24 1 8x Subtract 4. 24 24 3x 5 20 1 8x Subtract 8x. 28x 28x 25x 5 20 Divide by 25. x 5 24 Since x 5 24 satisfies the original equation, the solution set is 5246. YOUR T UR N Solve the equation 3 y 1 2 5 7 2y. ▼ Extraneous solutions are solutions that satisfy a transformed equation but do not satisfy the original equation. It is important to first state any values of the variable that must be eliminated based on the original rational equation. Once the rational equation is transformed to a linear equation and solved, remove any excluded values of the variable. EXAMPLE 5 Solving Rational Equations That Can Be Reduced to Linear Equations Solve the equation 3x x 2 1 1 2 5 3 x 2 1. Solution: State the excluded values (those that make any denominator equal 0). 3x x 2 1 1 2 5 3 x 2 1 x 2 1 Eliminate the fractions by multiplying each term by 3x x 2 1 ⋅1x 2 12 1 2⋅1x 2 12 5 3 x 2 1 ⋅1x 2 12 the LCD, x 2 1. Simplify. 3x x 2 1 ⋅1x 2 12 1 2⋅1x 2 12 5 3 x 2 1 ⋅1x 2 12 3x 1 21x 2 12 5 3 ▼ A N S W E R The solution is y 5 1 4. The solution set is U1 4V. 88 CHAPTER 1 Equations and Inequalities We have reviewed finding the least common denominator (LCD) for real numbers. Now let us consider finding the LCD for rational equations that have different denominators. We multiply the denominators in order to get a common denominator. Rational expression: 1 x 1 2 x 2 1 LCD: x 1x 2 12 In order to find a least common denominator, it is useful to first factor the denominators to identify common multiples. Rational equation: 1 3x 2 3 1 1 2x 2 2 5 1 x2 2 x Factor the denominators: 1 31x 2 12 1 1 21x 2 12 5 1 x1x 2 12 LCD: 6x 1x 2 12 Distribute the 2. 3x 1 2x 2 2 5 3 Combine x terms on the left. 5x 2 2 5 3 Add 2 to both sides. 5x 5 5 Divide both sides by 5. x 5 1 It may seem that x 5 1 is the solution. However, the original equation had the restriction x 2 1. Therefore, x 5 1 is an extraneous solution and must be eliminated as a possible solution. Thus, the equation 3x x 2 1 1 2 5 3 x 2 1 has no solution . Y OUR TU R N Solve the equation 2x x 2 2 2 3 5 4 x 2 2. ▼ STUDY TIP When a variable is in the denom-inator of a fraction, the LCD will contain the variable. This sometimes results in an extraneous solution. ▼ A N S W E R no solution EXAMPLE 6 Solving Rational Equations Solve the equation 1 3x 1 18 2 1 2x 1 12 5 1 x2 1 6x. Solution: Factor the denominators. 1 31x 1 62 2 1 21x 1 62 5 1 x1x 1 62 State the excluded values. x 2 0, 26 Multiply the equation by the LCD, 6x 1x 1 62. 6x 1x 1 62 ⋅ 1 31x 1 62 2 6x1x 1 62 ⋅ 1 21x 1 62 5 6x1x 1 62 ⋅ 1 x1x 1 62 Divide out the common factors. 6x1x 1 62 ⋅ 1 31x 1 62 2 6x1x 1 62 ⋅ 1 21x 1 62 5 6x1x 1 62 ⋅ 1 x1x 1 62 Simplify. 2x 2 3x 5 6 Solve the linear equation. x 5 26 Since one of the excluded values is x 2 26, we say that x 5 26 is an extraneous solution. Therefore, this rational equation has no solution . Y OUR TU R N Solve the equation 2 x 1 1 x 1 1 5 2 1 x1x 1 12. ▼ ▼ A N S W E R no solution 1.1 Linear Equations 89 EXAMPLE 7 Solving Rational Equations Solve the equation 2 1x 2 32 5 23 12 2 x2. Solution: What values make either denominator equal to zero? The values x 5 2 and x 5 3 must be excluded from possible solutions to the equation. 2 1x 2 32 5 23 12 2 x2 x 2 2, x 2 3 Multiply the equation by 2 1x 2 32 1x 2 3212 2 x2 5 23 12 2 x2 1x 2 3212 2 x2 the LCD, 1x 2 32 12 2 x2. Divide out the common factors. 212 2 x2 5 231x 2 32 Eliminate the parentheses. 4 2 2x 5 23x 1 9 Collect x terms on the left, constants on the right. x 5 5 Since x 5 5 satisfies the original equation, the solution set is 556. YOUR T UR N Solve the equation 24 x 1 8 5 3 x 2 6. ▼ ▼ A N S W E R The solution is x 5 0. The solution set is 506. [CONCEPT CHECK] Which values of x must be eliminated as potential solutions of the rational equation 1 x 2 a 5 1 x 1 b? ANSWER x 5 a and x 5 2b ▼ EXAMPLE 9 Grades Dante currently has the following three test scores: 82, 79, and 90. If the score on the final exam is worth two test scores and his goal is to earn an 85 for his class average, what score on the final exam does Dante need to achieve his course goal? Solution: Let x equal final exam grade. Write the equation that determines 82 1 79 1 90 1 2x 5 5 85 the course grade. Simplify the numerator. 251 1 2x 5 5 85 Multiply the equation by 5 (or cross multiply). 251 1 2x 5 425 Solve the linear equation. x 5 87 Dante needs to score at least an 87 on the final exam. 5 5 5 e total of five test scores average scores 1, 2, and 3 final is worth two test scores EXAMPLE 8 Automotive Service A car dealership charges for parts and an hourly rate for labor. If parts cost $273, labor is $53 per hour, and the total bill is $458.50, how many hours did the dealer-ship spend working on your car? Solution: Let x equal the number of hours the dealership worked on your car. Write the cost equation. 53x 1 273 5 458.50 Subtract 273 from both sides of the equation. 53x 5 185.50 Divide both sides of the equation by 53. x 5 3.5 The dealership charged for 3.5 hours of labor. labor parts total cost e e e 90 CHAPTER 1 Equations and Inequalities In Exercises 1–36, solve for the indicated variable. 1. 5x 5 35 2. 4t 5 32 3. 23 1 n 5 12 4. 4 5 25 1 y 5. 24 5 23x 6. 250 5 25t 7. 1 5 n 5 3 8. 6 5 1 3 p 9. 3x 2 5 5 7 10. 4p 1 5 5 9 11. 9m 2 7 5 11 12. 2x 1 4 5 5 13. 5t 1 11 5 18 14. 7x 1 4 5 21 1 24x 15. 3x 2 5 5 25 1 6x 16. 5x 1 10 5 25 1 2x 17. 20n 2 30 5 20 2 5n 18. 14c 1 15 5 43 1 7c 19. 4 1x 2 32 5 2 1x 1 62 20. 512y 2 12 5 214y 2 32 21. 2314t 2 52 5 516 2 2t2 22. 2 13n 1 42 5 21n 1 22 23. 2 1x 2 12 1 3 5 x 2 31x 1 12 24. 4 1 y 1 62 2 8 5 2y 2 4 1 y 1 22 25. 5p 1 61 p 1 72 5 31 p 1 22 26. 31z 1 52 2 5 5 4z 1 71z 2 22 27. 7x 2 12x 1 32 5 x 2 2 28. 3x 2 14x 1 22 5 x 2 5 29. 2 2 14x 1 12 5 3 2 12x 2 12 30. 5 2 12x 2 32 5 7 2 13x 1 52 31. 2a 2 91a 1 62 5 61a 1 32 2 4a 32. 25 2 32 1 5y 2 31 y 1 224 5 2312y 2 52 2 351 y 2 12 2 3y 1 34 33. 32 2 34 1 6x 2 51x 1 424 5 4 13x 1 42 2 3613x 2 42 1 7 2 4x4 34. 12 2 33 1 4m 2 613m 2 224 5 2712m 2 82 2331m 2 22 1 3m 2 54 35. 20 2 4 3c 2 3 2 612c 1 324 5 513c 2 22 2 3217c 2 82 2 4c 1 74 36. 46 2 37 2 8y 1 916y 2 224 5 2714y 2 72 2 2 3612y 2 32 2 4 1 6y4 Exercises 37–48 involve fractions. Clear the fractions by first multiplying by the least common denominator, and then solve the resulting linear equation. 37. 1 5 m 5 1 60 m 1 1 38. 1 12 z 5 1 24 z 1 3 39. x 7 5 2x 63 1 4 40. a 11 5 a 22 1 9 41. 1 3 p 5 3 2 1 24 p 42. 3x 5 2 x 5 x 10 2 5 2 43. 5y 3 2 2y 5 2y 84 1 5 7 44. 2m 2 5m 8 5 3m 72 1 4 3 45. p 1 p 4 5 5 2 46. c 4 2 2c 5 5 4 2 c 2 47. x 2 3 3 2 x 2 4 2 5 1 2 x 2 6 6 48. 1 2 x 2 5 3 5 x 1 2 5 2 6x 2 1 15 In Exercises 49–70, specify any values that must be excluded from the solution set and then solve the equation. 49. 4 y 2 5 5 5 2y 50. 4 x 1 10 5 2 3x 51. 7 2 1 6x 5 10 3x 52. 7 6t 5 2 1 5 3t 53. 2 a 2 4 5 4 3a 54. 4 x 2 2 5 5 2x 55. x x 2 2 1 5 5 2 x 2 2 56. n n 2 5 1 2 5 n n 2 5 57. 2p p 2 1 5 3 1 2 p 2 1 58. 4t t 1 2 5 3 2 8 t 1 2 59. 3x x 1 2 2 4 5 2 x 1 2 60. 5y 2y 2 1 2 3 5 12 2y 2 1 [SEC TION 1.1] E X E R C I SES • S K I L L S Rational equations are solved by: 1. Determining any excluded values (denominator equals 0). 2. Multiplying the equation by the LCD. 3. Solving the resulting equation. 4. Eliminating any extraneous solutions. Linear equations, ax 1 b 5 0, are solved by: 1. Simplifying the algebraic expressions on both sides of the equation. 2. Gathering all variable terms on one side of the equation and all constant terms on the other side. 3. Isolating the variable. [SEC TION 1.1] S U M M A RY 1.1 Linear Equations 91 61. 1 n 1 1 n 1 1 5 21 n1n 1 12 62. 1 x 1 1 x 2 1 5 1 x 1x 2 12 63. 3 a 2 2 a 1 3 5 9 a1a 1 32 64. 1 c 2 2 1 1 c 5 2 c1c 2 22 65. n 2 5 6n 2 6 5 1 9 2 n 2 3 4n 2 4 66. 5 m 1 3 m 2 2 5 6 m1m 2 22 67. 2 5x 1 1 5 1 2x 2 1 68. 3 4n 2 1 5 2 2n 2 5 69. t 2 1 1 2 t 5 3 2 70. 2 2 x x 2 2 5 3 4 • A P P L I C A T I O N S 71. Temperature. To calculate temperature in degrees Fahrenheit, we use the formula F 5 9 5 C 1 32, where F is degrees Fahrenheit and C is degrees Celsius. Find the formula to convert from Fahrenheit to Celsius. 72. Geometry. The perimeter P of a rectangle is related to the length L and width W of the rectangle through the equation P 5 2L 1 2W. Determine the width in terms of the perimeter and length. 73. Costs: Cellular Phone. Your cell phone plan charges $15 a month plus 12 cents per minute. If your monthly bill is $25.08, how many minutes did you use? 74. Costs: Rental Car. Becky rented a car on her Ft. Lauderdale vacation. The car was $25 a day plus 10 cents per mile. She kept the car for 5 days and her bill was $185. How many miles did she drive the car? 75. Costs: Internet. When traveling in London, Charlotte decided to check her e-mail at an Internet café. There was a flat charge of $2 plus a charge of 10 cents a minute. How many minutes was she logged on if her bill was $3.70? 76. Sales: Income. For a summer job, Dwayne decides to sell magazine subscriptions. He will be paid $20 a day plus $1 for each subscription he sells. If he works for 25 days and makes $645, how many subscriptions did he sell? 77. Business. The operating costs for a local business per year are a fixed amount of $15,000 and $2,500 per day. a. Find C that represents operating costs for the company, which depend on the number of days open, x. b. If the business accrues $5,515,000 in annual operating costs, how many days did the business operate during the year? 78. Business. Negotiated contracts for a technical support pro-vider produce monthly revenue of $5000 and $0.75 per minute per phone call. a. Find R that represents the revenue for the technical support provider, which depends on the number of minutes of phone calls x. b. In one month the provider received $98,750 in revenue. How many minutes of technical support were provided? In Exercises 79 and 80 refer to the following: Medications are often packaged in liquid form (known as a suspension) so that a precise dose of a drug is delivered within a volume of inert liquid—for example, 250 milligrams amoxicillin in 5 milliliters of a liquid suspension. If a patient is prescribed a dose of a drug, medical personnel must compute the volume of the liquid with a known concentration to administer. The formula a 5 d c defines the relationship between the dose of the drug prescribed d, the concentration of the liquid suspension c, and the amount of the liquid administered a. 79. Medicine. A physician has ordered a 600-milligram dose of amoxicillin. The pharmacy has a suspension of amoxicillin with a concentration of 125 milligrams per 5 milliliters. How much liquid suspension must be administered to the patient? 80. Medicine. A physician has ordered a 600-milligram dose of carbamazepine. The pharmacy has a suspension of carbamazepine with a concentration of 100 milligrams per 5 milliliters. How much liquid suspension must be administered to the patient? 81. Speed of Light. The frequency ƒ of an optical signal in hertz (Hz) is related to the wavelength l in meters (m) of a laser through the equation ƒ 5 c l, where c is the speed of light in a vacuum and is typically taken to be c 5 3.0 3 108 meters per second (m/s). What values must be eliminated from the wavelengths? 82. Optics. For an object placed near a lens, an image forms on the other side of the lens at a distinct position determined by the distance from the lens to the object. The position of the image is found using the thin lens equation: 1 ƒ 5 1 do 1 1 di where do is the distance from the object to the lens, di is the distance from the lens to the image, and ƒ is the focal length of the lens. Solve for the object distance do in terms of the focal length and image distance. 2f f f f do di 2f Image Object 92 CHAPTER 1 Equations and Inequalities • C A T C H T H E M I S T A K E In Exercises 83–86, explain the mistake that is made. 83. Solve the equation 4x 1 3 5 6x 2 7. Solution: Subtract 4x and add 7 to the equation. 3 5 6x Divide by 3. x 5 2 This is incorrect. What mistake was made? 84. Solve the equation 3 1x 1 12 1 2 5 x 2 3 1x 2 12. Solution: 3x 1 3 1 2 5 x 2 3x 2 3 3x 1 5 5 22x 2 3 5x 5 28 x 5 28 5 This is incorrect. What mistake was made? 85. Solve the equation 4 p 2 3 5 2 5p. Solution: 1p 2 322 5 415p2 Cross multiply. 2p 2 6 5 20p 26 5 18p p 5 2 6 18 p 5 2 1 3 This is incorrect. What mistake was made? 86. Solve the equation 1 x 1 1 x 2 1 5 1 x 1x 2 12. Solution: Multiply by the LCD, x 1x 2 12. x 1x 2 12 x 1 x 1x 2 12 x 2 1 5 x 1x 2 12 x 1x 2 12 Simplify. 1x 2 12 1 x 5 1 x 2 1 1 x 5 1 2x 5 2 x 5 1 This is incorrect. What mistake was made? In Exercises 87–90, determine whether each of the statements is true or false. 87. The solution to the equation x 5 1 1/x is the set of all real numbers. 88. The solution to the equation 1 1x 2 121x 1 22 5 1 x2 1 x 2 2 is the set of all real numbers. 89. x 5 21 is a solution to the equation x2 2 1 x 2 1 5 x 1 1. 90. x 5 1 is a solution to the equation x2 2 1 x 2 1 5 x 1 1. 91. Solve for x, given that a, b, and c are real numbers and a 2 0: ax 1 b 5 c 92. Solve for x, given that a, b, and c are real numbers and c 2 0: a x 2 b x 5 c • C O N C E P T U A L 93. Solve the equation for x: b 1 c x 1 a 5 b 2 c x 2 a. Are there any restrictions given that a 2 0, x 2 0? 94. Solve the equation for y: 1 y 2 a 1 1 y 1 a 5 2 y 2 1. Does y have any restrictions? 95. Solve for x: 1 2 1/x 1 1 1/x 5 1. 96. Solve for t: t 1 1/t 1/t 2 1 5 1. 97. Solve the equation for x in terms of y: y 5 a 1 1 b/x 1 c 98. Find the number a for which y 5 2 is a solution of the equation y 2 a 5 y 1 5 2 3ay. • C H A L L E N G E In Exercises 99–106, graph the function represented by each side of the equation in the same viewing rectangle and solve for x. 99. 31x 1 22 2 5x 5 3x 2 4 100. 251x 2 12 2 7 5 10 2 9x 101. 2x 1 6 5 4x 2 2x 1 8 2 2 102. 10 2 20x 5 10x 2 30x 1 20 2 10 103. x1x 2 12 x2 5 1 104. 2x1x 1 32 x2 5 2 105. 0.035x 1 0.029 18706 2 x2 5 285.03 106. 1 0.75x 2 0.45 x 5 1 9 • T E C H N O L O G Y 1.2 Applications Involving Linear Equations 93 1.2.1 Solving Application Problems Using Mathematical Models In this section, we will use algebra to solve problems that occur in our day-to-day lives. You typically will read the problem in words, develop a mathematical model (equation) for the problem, solve the equation, and write the answer in words. S K I L L S O B J E C T I V E S ■ ■Solve application problems involving common formulas. ■ ■Solve geometry problems using linear equations. ■ ■Solve simple interest problems. ■ ■Solve mixture problems. ■ ■Solve distance–rate–time problems. C O N C E P T U A L O B J E C T I V ES ■ ■Understand the mathematical modeling process. ■ ■Estimate a practical solution (guess) prior to solving a problem and then check solution at the end. ■ ■Use intuition to confirm answers in multiple investment problems. ■ ■Use intuition to confirm answers to mixture problems. ■ ■Estimate distance–rate–time problem solutions prior to solving and then confirm with a check. 1.2 APPLICATIONS INVOLVING LINEAR EQUATIONS 1.2.1 S K I L L Solve application problems involving common formulas. 1.2.1 C ON C E P T U A L Understand the mathematical modeling process. Real-World Problem Mathematical Model Solve Mathematical Problem Solution to Real-World Problem Translate from Math to Words Translate Words to Math Solve Using Standard Methods You will have to come up with a unique formula to solve each kind of word problem, but there is a universal procedure for approaching all word problems. PROCEDURE FOR SOLVING WORD PROBLEMS Step 1: Identify the question. Read the problem one time and note what you are asked to find. Step 2: Make notes. Read until you can note something (an amount, a picture, anything). Continue reading and making notes until you have read the problem a second time. Step 3: Assign a variable to whatever is being asked for (if there are two choices, then let it be the smaller of the two). Step 4: Set up an equation. Step 5: Solve the equation. Step 6: Check the solution. Run the solution past the “common sense department” using estimation. Step 2 often requires multiple readings of the problem. 94 CHAPTER 1 Equations and Inequalities EXAMPLE 1 How Long Was the Trip? During a camping trip in North Bay, Ontario, a couple went one-third of the way by boat, 10 miles by foot, and one-sixth of the way by horse. How long was the trip? Solution: STEP 1 Identify the question. How many miles was the trip? STEP 2 Make notes. Read Write . . . one-third of the way by boat BOAT: 1 3 of the trip . . . 10 miles by foot FOOT: 10 miles . . . one-sixth of the way by horse HORSE: 1 6 of the trip STEP 3 Assign a variable. Distance of total trip in miles 5 x STEP 4 Set up an equation. The total distance of the trip is the sum of all the distances by boat, foot, and horse. Distance by boat 1 Distance by foot 1 Distance by horse 5 Total distance of trip Distance by boat 5 1 3x Distance by foot 5 10 miles Distance by horse 5 1 6x STEP 5 Solve the equation. 1 3x 1 10 1 1 6x 5 x Multiply by the LCD, 6. 2x 1 60 1 x 5 6x Collect x terms on the right. 60 5 3x Divide by 3. 20 5 x The trip was 20 miles. x 5 20 STEP 6 Check the solution. Estimate: The boating distance, 1 3 of 20 miles, is approximately 7 miles; the riding distance on horse, 1 6 of 20 miles, is approximately 3 miles. Adding these two distances to the 10 miles by foot gives a trip distance of 20 miles. Y OUR TU R N A family arrives at the Walt Disney World parking lot. To get from their car in the parking lot to the gate at the Magic Kingdom they walk 1 4 mile, take a tram for 1 3 of their total distance, and take a monorail for 1 2 of their total distance. How far is it from their car to the gate of Magic Kingdom? boat foot horse total 1 3x 1 10 1 1 6x 5 x 5 5 5 5 ▼ ▼ A N S W E R The distance from their car to the gate is 1.5 miles. [CONCEPT CHECK] TRUE OR FALSE In Example 1 we could have stopped at x 5 20. ANSWER False. A problem asked in words “How long was the trip?” needs an answer in words such as “The trip was 20 miles.” ▼ 1.2 Applications Involving Linear Equations 95 1.2.2 Geometry Problems Some problems require geometric formulas in order to be solved. The following geo-metric formulas may be useful. EXAMPLE 2 Find the Numbers Find three consecutive even integers so that the sum of the three numbers is 2 more than twice the third. Solution: STEP 1 Identify the question. What are the three consecutive even integers? STEP 2 Make notes. Examples of three consecutive even integers are 14, 16, 18 or 28, 26, 24 or 2, 4, 6. STEP 3 Assign a variable. Let n represent the first even integer. The next consecutive even integer is n 1 2 and the next consecutive even integer after that is n 1 4. n 5 1st integer n 1 2 5 2nd consecutive even integer n 1 4 5 3rd consecutive even integer STEP 4 Set up an equation. Read Write . . . sum of the three numbers n 1 1n 1 22 1 1n 1 42 . . . is 5 . . . two more than 12 . . . twice the third 21n 1 42 n 1 1n 1 22 1 1n 1 42 5 2 1 21n 1 42 STEP 5 Solve the equation. n 1 1n 1 22 1 1n 1 42 5 2 1 21n 1 42 Eliminate the parentheses. n 1 n 1 2 1 n 1 4 5 2 1 2n 1 8 Simplify both sides. 3n 1 6 5 2n 1 10 Collect n terms on the left and constants on the right. n 5 4 The three consecutive even integers are 4, 6, and 8. STEP 6 Check the solution. Substitute the solution into the problem to see whether it makes sense. The sum of the three integers 14 1 6 1 82 is 18. Twice the third is 16. Since 2 more than twice the third is 18, the solution checks. YOUR T UR N Find three consecutive odd integers so that the sum of the three integers is 5 less than 4 times the first. c c sum of the three numbers is 2 more twice the than third e 5 ▼ ▼ A N S W E R The three consecutive odd integers are 11, 13, and 15. 1.2.2 S K I L L Solve geometry problems using linear equations. 1.2.2 C ON C E P T U A L Estimate a practical solution (guess) prior to solving a problem and then check solution at the end. 96 CHAPTER 1 Equations and Inequalities Rectangle w l Perimeter Area P 5 2l 1 2w A 5 l⋅w Circle r Circumference Area C 5 2pr A 5 pr 2 Triangle a h height c b base               Perimeter Area P 5 a 1 b 1 c A 5 1 2 bh GEOMETRIC FORMULAS EXAMPLE 3 Geometry A rectangle 24 meters long has the same area as a square with 12-meter sides. What are the dimensions of the rectangle? Solution: STEP 1 Identify the question. What are the dimensions (length and width) of the rectangle? STEP 2 Make notes. Read Write/Draw A rectangle 24 meters long w l = 24 area of rectangle 5 l ? w 5 24w . . . a square with 12-meter sides 12 m 12 m area of square 5 12 ? 12 5 144 STEP 3 Assign a variable. Let w 5 width of the rectangle. STEP 4 Set up an equation. The area of the rectangle is equal to the area of the square. rectangle area 5 square area Substitute in known quantities. 24w 5 144 STEP 5 Solve the equation. Divide by 24. w 5 144 24 5 6 The rectangle is 24 meters long and 6 meters wide. [CONCEPT CHECK] If we were to guess an answer to Example 3, what would be our guess? (A) 24 m by 5 m (B) 24 m by 20 m (C) 24 m by 1 m ANSWER The answer is (A). Option (B) cannot be correct because that would have area much larger than a 12-m square, and (C) cannot be correct because that would have area much smaller than a 12-m square. ▼ 1.2 Applications Involving Linear Equations 97 1.2.3 Interest Problems In our personal or business financial planning, a particular concern we have is interest. Interest is money paid for the use of money; it is the cost of borrowing money. The total amount borrowed is called the principal. The principal can be the price of our new car; we pay the bank interest for loaning us money. The principal can also be the amount we keep in a CD or money market account; the bank uses this money and pays us interest. Typically interest rate, expressed as a percentage, is the amount charged for the use of the principal for a given time, usually in years. Simple interest is interest that is paid only on principal during a period of time. Later we will discuss compound interest, which is interest paid on both principal and the interest accrued over a period of time. STEP 6 Check the solution. A 24 meter by 6 meter rectangle has an area of 144 square meters. YOUR T UR N A rectangle 3 inches wide has the same area as a square with 9-inch sides. What are the dimensions of the rectangle? ▼ ▼ A N S W E R The rectangle is 27 inches long and 3 inches wide. DEFINITION Simple Interest If a principal of P dollars is borrowed for a period of t years at an annual interest rate r (expressed in decimal form), the interest I charged is I 5 Prt This is the formula for simple interest. 1.2.3 S K I L L Solve simple interest problems. 1.2.3 C ON C E P T U A L Use intuition to confirm answers in multiple investment problems. EXAMPLE 4 Simple Interest Through a summer job Morgan is able to save $2500. If she puts that money into a 6-month certificate of deposit (CD) that pays a simple interest rate of 3% a year, how much money will she have in her CD at the end of the 6 months? Solution: STEP 1 Identify the question. How much money does Morgan have after 6 months? STEP 2 Make notes. The principal is $2500. The annual interest rate is 3%, which in decimal form is 0.03. The time the money spends accruing interest is 6 months, or 1 2 of a year. STEP 3 Assign a variable. Label the known quantities. P 5 2500, r 5 0.03, and t 5 0.5 STEP 4 Set up an equation. Write the simple interest formula. I 5 Prt STEP 5 Solve the equation. I 5 Prt I 5 125002 10.032 10.52 5 37.5 98 CHAPTER 1 Equations and Inequalities The interest paid on the CD is $37.50. Adding this to the principal gives a total of $2500 1 $37.50 5 $2537.50 STEP 6 Check the solution. This answer agrees with our intuition. Had we made a mistake, say, of moving one decimal place to the right, then the interest would have been $375, which is much larger than we would expect on a principal of only $2500. EXAMPLE 5 Multiple Investments Theresa earns a full athletic scholarship for college, and her parents have given her the $20,000 they had saved to pay for her college tuition. She decides to invest that money with an overall goal of earning 11% interest. She wants to put some of the money in a low-risk investment that has been earning 8% a year and the rest of the money in a medium-risk investment that typically earns 12% a year. How much money should she put in each investment to reach her goal? Solution: STEP 1 Identify the question. How much money is invested in each (the 8% and the 12%) account? STEP 2 Make notes. Read Write/Draw Theresa has $20,000 to invest. If part is invested at 8% and the rest at 12%, how much should be invested at each rate to yield 11% on the total amount invested? STEP 3 Assign a variable. If we let x represent the amount Theresa puts into the 8% investment, how much of the $20,000 is left for her to put in the 12% investment? Amount in the 8% investment: x Amount in the 12% investment: 20,000 2 x STEP 4 Set up an equation. Simple interest formula: I 5 Prt INVESTMENT PRINCIPAL RATE TIME (YR) INTEREST 8% Account x 0.08 1 0.08x 12% Account 20,000 2 x 0.12 1 0.12 120,000 2 x2 Total 20,000 0.11 1 0.11120,0002 Adding the interest earned in the 8% investment to the interest earned in the 12% investment should earn an average of 11% on the total investment. 0.08x 1 0.12 120,000 2 x2 5 0.11120,0002 1.2 Applications Involving Linear Equations 99 1.2.4 Mixture Problems Mixtures are something we come across every day. Different candies that sell for different prices may make up a movie snack. New blends of coffees are developed by coffee connoisseurs. Chemists mix different concentrations of acids in their labs. Whenever two or more distinct ingredients are combined, the result is a mixture. Our choice at a gas station is typically 87, 89, and 93 octane. The octane number is the number that represents the percentage of iso-octane in fuel; 89 octane is significantly overpriced. Therefore, if your car requires 89 octane, it would be more cost effective to mix 87 and 93 octane. STEP 5 Solve the equation. Eliminate the parentheses. 0.08x 1 2400 2 0.12x 5 2200 Collect x terms on the left, constants on the right. 20.04x 5 2200 Divide by 20.04. x 5 5000 Calculate the amount at 12%. 20,000 2 5000 5 15,000 Theresa should invest $5000 at 8% and $15,000 at 12% to reach her goal. STEP 6 Check the solution. If money is invested at 8% and 12% with a goal of averaging 11%, our intuition tells us that more should be invested at 12% than 8%, which is what we found. The exact check is as follows: 0.08150002 1 0.12115,0002 5 0.11120,0002 400 1 1800 5 2200 2200 5 2200 Y OUR T UR N You win $24,000 and you decide to invest the money in two different investments: one paying 18% and the other paying 12%. A year later you have $27,480 total. How much did you originally invest in each account? ▼ [CONCEPT CHECK] If Teresa is going to invest in two accounts, one at 8% and one at 12% with a result of an average of 11% earnings, does your intuition think she should have (A) more at 8% than at 12% or (B) more at 12% than at 8%? ANSWER (B) because 11% is closer to 12% than 8% ▼ ▼ A N S W E R $10,000 is invested at 18% and $14,000 is invested at 12%. 1.2.4 S K I L L Solve mixture problems. 1.2.4 C ON C E P T U A L Use intuition to confirm answers to mixture problems. EXAMPLE 6 Mixture Problem The manual for your new car suggests using gasoline that is 89 octane. In order to save money, you decide to use some 87 octane and some 93 octane in combination with the 89 octane currently in your tank in order to have an approximate 89 octane mixture. Assuming you have 1 gallon of 89 octane remaining in your tank (your tank capacity is 16 gallons), how many gallons of 87 and 93 octane should be used to fill up your tank to achieve a mixture of 89 octane? Solution: STEP 1 Identify the question. How many gallons of 87 octane and how many gallons of 93 octane should be used? STEP 2 Make notes. Read Write/Draw Assuming you have one gallon of 89 octane remaining in your tank (your tank capacity is 16 gallons), how many gallons of 87 and 93 octane should you add? 89 octane + 87 octane + 93 octane = 89 octane [1 gallon] [? gallons] [? gallons] [16 gallons] 100 CHAPTER 1 Equations and Inequalities 1.2.5 Distance–Rate–Time Problems The next example deals with distance, rate, and time. On a road trip, you see a sign that says your destination is 90 miles away, and your speedometer reads 60 miles per hour. Dividing 90 miles by 60 miles per hour tells you that your arrival will be in 1.5 hours. Here is how you know. If the rate, or speed, is assumed to be constant, then the equation that relates distance (d), rate (r), and time (t) is given by d 5 r • t. In the above driving example, d 5 90 miles r 5 60 miles hour Substituting these into d 5 r⋅t, we arrive at 90 miles 5 c60 miles hour d ⋅t Solving for t, we get t 5 90 miles 60 miles hour 5 1.5 hours STEP 3 Assign a variable. x 5 gallons of 87 octane gasoline added at the pump 15 2 x 5 gallons of 93 octane gasoline added at the pump 1 5 gallons of 89 octane gasoline already in the tank STEP 4 Set up an equation. 0.89112 1 0.87x 1 0.93115 2 x2 5 0.891162 STEP 5 Solve the equation. 0.89112 1 0.87x 1 0.93115 2 x2 5 0.891162 Eliminate the parentheses. 0.89 1 0.87x 1 13.95 2 0.93x 5 14.24 Collect x terms on the left side. 20.06x 1 14.84 5 14.24 Subtract 14.84 from both sides of the equation. 20.06x 520.6 Divide both sides by 20.06. x 5 10 Calculate the amount of 93 octane. 15 2 10 5 5 Add 10 gallons of 87 octane and 5 gallons of 93 octane. STEP 6 Check the solution. Estimate: Our intuition tells us that if the desired mixture is 89 octane, then we should add approximately one part 93 octane and two parts 87 octane. The solution we found, 10 gallons of 87 octane and 5 gallons of 93 octane, agrees with this. Y OUR TU R N For a certain experiment, a student requires 100 milliliters of a solution that is 11% HCl (hydrochloric acid). The storeroom has only solutions that are 5% HCl and 15% HCl. How many milliliters of each available solution should be mixed to get 100 milliliters of 11% HCl? ▼ 1.2.5 S KILL Solve distance–rate–time problems. 1.2.5 CO NCE PTUAL Estimate distance–rate–time problem solutions prior to solving and then confirm with a check. ▼ A N S W E R 40 milliliters of 5% HCl and 60 milliliters of 15% HCl [CONCEPT CHECK] If a chemistry student has HCl concentrations of 5% and 15% and the desired solution is 11% HCl, do we expect (A) more 15% than 5% or (B) more 5% than 15%? ANSWER (A) because 11% is closer to 15% than 5%. ▼ 1.2 Applications Involving Linear Equations 101 EXAMPLE 7 Distance–Rate–Time It takes 8 hours to fly from Orlando to London and 9.5 hours to return. If an airplane averages 550 miles per hour in still air, what is the average rate of the wind blowing in the direction from Orlando to London? Assume the speed of the wind ( jet stream) is constant and the same for both legs of the trip. Round your answer to the nearest miles per hour. Solution: STEP 1 Identify the question. At what rate in mph is the wind blowing? STEP 2 Make notes. Read Write/Draw It takes 8 hours to fly from Orlando to London and 9.5 hours to return. If the airplane averages 550 miles per hour in still air . . . STEP 3 Assign a variable. w 5 wind speed STEP 4 Set up an equation. The formula relating distance, rate, and time is d 5 r ? t. The distance d of each flight is the same. On the Orlando to London flight the time is 8 hours due to an increased speed from a tailwind. On the London to Orlando flight the time is 9.5 hours, and the speed is decreased due to the headwind. Let w represent the wind speed. Orlando to London: d 5 1550 1 w2 8 London to Orlando: d 5 1550 2 w2 9.5 These distances are the same, so set them equal to each other: 1550 1 w28 5 1550 2 w29.5 STEP 5 Solve the equation. Eliminate the parentheses. 4400 1 8w 5 5225 2 9.5w Collect w terms on the left, constants on the right. 17.5w 5 825 Divide by 17.5. w 5 47.1429 < 47 The wind is blowing approximately 47 miles per hour in the direction from Orlando to London. STEP 6 Check the solution. Estimate: Going from Orlando to London, the tailwind is approximately 50 miles per hour, which added to the plane’s 550 miles per hour speed yields a ground speed of 600 miles per hour. The Orlando to London route took 8 hours. The distance of that flight is 1600 mph2 18 hr2, which is 4800 miles. The return trip experienced a headwind of approximately 50 miles per hour, so subtracting the 50 from 550 gives an average speed of 500 miles per hour. That route took 9.5 hours, so the distance of the London to Orlando flight was 1500 mph2 19.5 hr2, which is 4750 miles. Note that the estimates of 4800 and 4750 miles are close. YOUR T UR N A Cessna 150 averages 150 miles per hour in still air. With a tailwind it is able to make a trip in 2 1 3 hours. Because of the headwind, it is only able to make the return trip in 31 2 hours. What is the average wind speed? ▼ ▼ A N S W E R The wind is blowing 30 mph. 102 CHAPTER 1 Equations and Inequalities EXAMPLE 8 Work Connie can clean her house in 2 hours. If Alvaro helps her, they can clean the house in 1 hour and 15 minutes together. How long would it take Alvaro to clean the house by himself? Solution: STEP 1 Identify the question. How long would it take Alvaro to clean the house by himself? STEP 2 Make notes. Connie can clean her house in 2 hours, so Connie can clean 1 2 of the house per hour. Together Connie and Alvaro can clean the house in 1 hour and 15 minutes, or 5 4 of an hour. Therefore, together they can clean 1 5/4 5 4 5 of the house per hour. STEP 3 Assign a variable. Let x 5 number of hours it takes Alvaro to clean the house by himself. So Alvaro can clean 1 x of the house per hour. STEP 4 Set up an equation. Amount of house Connie can Amount of house Alvaro can Amount of house they can clean per hour clean per hour clean per hour if they work together 1 2 1 1 x 5 4 5 STEP 5 Solve the equation. Multiply 1 2 1 1 x 5 4 5 by the LCD, 10x. 5x 1 10 5 8x Solve for x. x 5 10 3 5 31 3 It takes Alvaro 3 hours and 20 minutes to clean the house by himself. STEP 6 Check the solution. Connie cleans the house in 2 hours. If Alvaro could clean it in 2 hours, then together it would take them 1 hour. Since together it takes them 1 hour and 15 minutes, we expect that it takes Alvaro more than 2 hours by himself. AMOUNT OF TIME AMOUNT OF JOB DONE TO DO ONE JOB PER UNIT OF TIME Connie 2 1 2 Alvaro x 1 x Together 5 4 4 5 ▲ 1 5 c c c [CONCEPT CHECK] If Connie cleans her house alone it takes 2 hours, and if Alvaro helps her it takes 1 hour and 15 minutes. Which of the following is true? (A) Alvaro cleans faster than Connie. (B) Alvaro and Connie clean at the same rate. (C) Alvaro cleans slower than Connie. ANSWER (C) is correct. (B) cannot be true or it would take them exactly one hour if they were working together. (A) cannot be true or the combined time would be less than one hour. ▼ 1.2 Applications Involving Linear Equations 103 [SEC TION 1. 2] E X E R C I S E S • A P P L I C A T I O N S 1. Discount Price. Donna redeems a 10% off coupon at her local nursery. After buying azaleas, bougainvillea, and bags of potting soil, her checkout price before tax is $217.95. How much would she have paid without the coupon? 2. Discount Price. The original price of a pair of binoculars is $74. The sale price is $51.80. How much was the markdown? 3. Cost: Fair Share. Jeff, Tom, and Chelsea order a large pizza. They decide to split the cost according to how much they will eat. Tom pays $5.16, Chelsea eats 1 8 of the pizza, and Jeff eats 1 2 of the pizza. How much did the pizza cost? 4. Event Planning. A couple decide to analyze their monthly spending habits. The monthly bills are 50% of their take-home pay, and they invest 20% of their take-home pay. They spend $560 on groceries, and 23% goes to miscellaneous. How much is their take-home pay per month? 5. Discounts. A builder of tract homes reduced the price of a model by 15%. If the new price is $125,000, what was its original price? How much can be saved by purchasing the model? 6. Markups. A college bookstore marks up the price it pays the publisher for a book by 25%. If the selling price of a book is $79, how much did the bookstore pay for the book? 7. Puzzle. Angela is on her way from home in Jersey City into New York City for dinner. She walks 1 mile to the train station, takes the train 3 4 of the way, and takes a taxi 1 6 of the way to the restaurant. How far does Angela live from the restaurant? 8. Puzzle. An employee at Kennedy Space Center (KSC) lives in Daytona Beach and works in the vehicle assembly building (VAB). She carpools to work with a colleague. She drives 7 miles from her house to the park-and-ride. Then she rides with her colleague from the park-and-ride in Daytona Beach to the KSC headquarters building, and then takes the KSC shuttle from the headquarters building to the VAB. The drive from the park-and-ride to the headquarters building is 5 6 of her total trip, and the shuttle ride is 1 20 of her total trip. How many miles does she travel from her house to the VAB on days when her colleague drives? 9. Puzzle. A typical college student spends 1 3 of her waking time in class, 1 5 of her waking time eating, 1 10 of her waking time working out, 3 hours studying, and 21 2 hours doing other things. How many hours of sleep does the typical college student get? 10. Diet. A particular 1550-calories-per-day diet suggests eating breakfast, lunch, dinner, and two snacks. Dinner is twice the calories of breakfast. Lunch is 100 calories more than breakfast. The two snacks are 100 and 150 calories. How many calories are each meal? 11. Budget. A company has a total of $20,000 allocated for mon t hly costs. Fixed costs are $15,000 per month and variable costs are $18.50 per unit. How many units can be manufactured in a month? 12. Budget. A woman decides to start a small business making monogrammed cocktail napkins. She can set aside $1870 for monthly costs. Fixed costs are $1329.50 per month and variable costs are $3.70 per set of napkins. How many sets of napkins can she afford to make per month? 13. Numbers. Find a number such that 10 less than 2 3 the number is 1 4 the number. 14. Numbers. Find a positive number such that 10 times the number is 16 more than twice the number. 15. Numbers. Find two consecutive even integers such that 4 times the smaller number is 2 more than 3 times the larger number. 16. Numbers. Find three consecutive integers such that the sum of the three is equal to 2 times the sum of the first two integers. 17. Geometry. Find the perimeter of a triangle if one side is 11 inches, another side is 1 5 the perimeter, and the third side is 1 4 the perimeter. 18. Geometry. Find the dimensions of a rectangle whose length is a foot longer than twice its width and whose perimeter is 20 feet. 19. Geometry. An NFL playing field is a rectangle. The length of the field (excluding the end zones) is 40 more yards than twice the width. The perimeter of the playing field is 260 yards. What are the dimensions of the field in yards? 20. Geometry. The length of a rectangle is 2 more than 3 times the width, and the perimeter is 28 inches. What are the dimensions of the rectangle? 21. Geometry. Consider two circles, a smaller one and a larger one. If the larger one has a radius that is 3 feet larger than that of the smaller circle and the ratio of the circumferences is 2:1, what are the radii of the two circles? 1. Identify the quantity you are to determine. 2. Make notes on any clues that will help you set up an equation. 3. Assign a variable. 4. Set up the equation. 5. Solve the equation. 6. Check the solution against your intuition. In the real world, many kinds of application problems can be solved through modeling with linear equations. The following six-step procedure will help you develop the model. Some prob-lems require development of a mathematical model, while others rely on common formulas. [SEC TION 1. 2] S U MM A RY 104 CHAPTER 1 Equations and Inequalities 22. Geometry. The perimeter of a semicircle is doubled when the radius is increased by 1. Find the radius of the semicircle. 23. Home Improvement. A man wants to remove a tall pine tree from his yard. Before he goes to Home Depot, he needs to know how tall an extension ladder he needs to purchase. He measures the shadow of the tree to be 225 feet long. At the same time he measures the shadow of a 4-foot stick to be 3 feet. Approximately how tall is the pine tree? 24. Home Improvement. The same man in Exercise 23 realizes he also wants to remove a dead oak tree. Later in the day he measures the shadow of the oak tree to be 880 feet long, and the 4-foot stick now has a shadow of 10 feet. Approximately how tall is the oak tree? 25. Biology: Alligators. It is common to see alligators in ponds, lakes, and rivers in Florida. The ratio of head size (back of the head to the end of the snout) to the full body length of an alligator is typically constant. If a 31 2-foot alligator has a head length of 6 inches, how long would you expect an alligator to be whose head length is 9 inches? 26. Biology: Snakes. In the African rainforest there is a snake called a Gaboon viper. The fang size of this snake is proportional to the length of the snake. A 3-foot snake typically has 2-inch fangs. If a herpetologist finds Gaboon viper fangs that are 2.6 inches long, how long a snake would she expect to find? 27. Investing. Ashley has $120,000 to invest and decides to put some in a CD that earns 4% interest per year and the rest in a low-risk stock that earns 7%. How much did she invest in each to earn $7800 interest in the first year? 28. Investing. You inherit $13,000 and you decide to invest the money in two different investments: one paying 10% and the other paying 14%. A year later your investments are worth $14,580. How much did you originally invest in each account? 29. Investing. Wendy was awarded a volleyball scholarship to the University of Michigan, so on graduation her parents gave her the $14,000 they had saved for her college tuition. She opted to invest some money in a privately held company that pays 10% per year and evenly split the remaining money between a money market account yielding 2% and a high-risk stock that yielded 40%. At the end of the first year she had $16,610 total. How much did she invest in each of the three? 30. Interest. A high school student was able to save $5000 by working a part-time job every summer. He invested half the money in a money market account and half the money in a stock that paid three times as much interest as the money market account. After a year he earned $150 in interest. What were the interest rates of the money market account and the stock? 31. Budget: Home Improvement. When landscaping their yard, a couple budgeted $4200. The irrigation system costs $2400 and the sod costs $1500. The rest they will spend on trees and shrubs. Trees each cost $32 and shrubs each cost $4. They plant a total of 33 trees and shrubs. How many of each did they plant in their yard? 32. Budget: Shopping. At the deli Jennifer bought spicy tur-key and provolone cheese. The turkey costs $6.32 per pound and the cheese costs $4.27 per pound. In total, she bought 3.2 pounds and the price was $17.56. How many pounds of each did she buy? 33. Chemistry. For a certain experiment, a student requires 100 milliliters of a solution that is 8% HCl (hydrochloric acid). The storeroom has only solutions that are 5% HCl and 15% HCl. How many milliliters of each available solution should be mixed to get 100 milliliters of 8% HCl? 34. Chemistry. How many gallons of pure alcohol must be mixed with 5 gallons of a solution that is 20% alcohol to make a solution that is 50% alcohol? 35. Automobiles. A mechanic has tested the amount of antifreeze in your radiator. He says it is only 40% antifreeze and the remainder is water. How many gallons must be drained from your 5 gallon radiator and replaced with pure antifreeze to make the mixture in your radiator 80% antifreeze? 36. Costs: Overhead. A professor is awarded two research grants, each having different overhead rates. The research project conducted on campus has a rate of 42.5% overhead, and the project conducted in the field, off campus, has a rate of 26% overhead. If she was awarded $1,170,000 total for the two projects with an average overhead rate of 39%, how much was the research project on campus and how much was the research project off campus? 37. Theater. On the way to the movies a family picks up a custom-made bag of candies. The parents like caramels 1$1.50/lb2 and the children like gummy bears 1$2.00/lb2. They bought a 1.25-pound bag of combined candies that cost $2.50. How much of each candy did they buy? 38. Coffee. Joy is an instructional assistant in one of the college labs. She is on a very tight budget. She loves Jamaican Blue Mountain coffee, but it costs $12 a pound. She decides to blend this with regular coffee beans that cost $4.20 a pound. If she spends $14.25 on 2 pounds of coffee, how many pounds of each did she purchase? 39. Communications. The speed of light is approximately 3.0 3 108 meters per second 1670,616,629 mph2. The distance from Earth to Mars varies because of the orbits of the planets around the Sun. On average, Mars is 100 million miles from Earth. If we use laser communication systems, what will be the delay between Houston and NASA astronauts on Mars? 40. Speed of Sound. The speed of sound is approximately 760 miles per hour in air. If a gun is fired 1 2 mile away, how long will it take the sound to reach you? 41. Business. Because of holiday travel during the month of November, the average price of gasoline rose 4.7% in the United States. If the average price of gasoline at the end of November was $3.21 per gallon, what was the price of gasoline at the beginning of November? 42. Business. During the Christmas shopping season, the average price of a flat screen television fell by 40%. A shopper purchased a 42-inch flat screen television for $299 in late November. How much would the shopper have paid, to the nearest dollar, for the same television if it was pur-chased in September? 43. Medicine. A patient requires an IV of 0.9% saline solution, also known as normal saline solution. How much distilled water, to the nearest milliliter, must be added to 1.2 Applications Involving Linear Equations 105 100 milliliters of a 3% saline solution to produce normal saline? 44. Medicine. A patient requires an IV of D5W, a 5% solution of dextrose (sugar) in water. To the nearest milliliter, how much D20W, a 20% solution of dextrose in water, must be added to 100 milliliters of distilled water to produce a D5W solution? 45. Boating. A motorboat can maintain a constant speed of 16 miles per hour relative to the water. The boat makes a trip upstream to a marina in 20 minutes. The return trip takes 15 minutes. What is the speed of the current? 46. Aviation. A Cessna 175 can average 130 miles per hour. If a trip takes 2 hours one way and the return takes 1 hour and 15 minutes, find the wind speed, assuming it is constant. 47. Exercise. A jogger and a walker cover the same distance. The jogger finishes in 40 minutes. The walker takes an hour. How fast is each exerciser moving if the jogger runs 2 miles per hour faster than the walker? 48. Travel. A high school student in Seattle, Washington, attended the University of Central Florida. On the way to UCF he took a southern route. After graduation he returned to Seattle via a northern trip. On both trips he had the same average speed. If the southern trek took 45 hours and the northern trek took 50 hours, and the northern trek was 300 miles longer, how long was each trip? 49. Distance–Rate–Time. College roommates leave for their first class in the same building. One walks at 2 miles per hour and the other rides his bike at a slow 6 miles per hour pace. How long will it take each to get to class if the walker takes 12 minutes longer to get to class and they travel on the same path? 50. Distance–Rate–Time. A long-distance delivery service sends out a truck with a package at 7 a.m. At 7:30 a.m., the manager realizes there was another package going to the same location. He sends out a car to catch the truck. If the truck travels at an average speed of 50 miles per hour and the car travels at 70 miles per hour, how long will it take the car to catch the truck? 51. Work. Christopher can paint the interior of his house in 15 hours. If he hires Cynthia to help him, they can do the same job together in 9 hours. If he lets Cynthia work alone, how long will it take her to paint the interior of his house? 52. Work. Jay and Morgan work in the summer for a landscaper. It takes Jay 3 hours to complete the company’s largest yard alone. If Morgan helps him, it takes only 1 hour. How much time would it take Morgan alone? 53. Work. Tracey and Robin deliver soft drinks to local convenience stores. Tracey can complete the deliveries in 4 hours alone. Robin can do it in 6 hours alone. If they decide to work together on a Saturday, how long will it take? 54. Work. Joshua can deliver his newspapers in 30 minutes. It takes Amber 20 minutes to do the same route. How long would it take them to deliver the newspapers if they worked together? 55. Music. A major chord in music is composed of notes whose frequencies are in the ratio 4:5:6. If the first note of a chord has a frequency of 264 hertz (middle C on the piano), find the frequencies of the other two notes. Hint: Set up two proportions using 4:5 and 4:6. 56. Music. A minor chord in music is composed of notes whose frequencies are in the ratio 10:12:15. If the first note of a minor chord is A, with a frequency of 220 hertz, what are the frequencies of the other two notes? 57. Grades. Danielle’s test scores are 86, 80, 84, and 90. The final exam will count as 2 3 of the final grade. What score does Danielle need on the final in order to earn a B, which requires an average score of 80? What score does she need to earn an A, which requires an average of 90? 58. Grades. Sam’s final exam will count as two tests. His test scores are 80, 83, 71, 61, and 95. What score does Sam need on the final in order to have an average score of 80? 59. Sports. In Super Bowl XXXVII, the Tampa Bay Buccaneers scored a total of 48 points. All of their points came from field goals and touchdowns. Field goals are worth 3 points and each touchdown was worth 7 points (Martin Gramatica was successful in every extra point attempt). They scored a total of 8 times. How many field goals and touchdowns were scored? 60. Sports. A tight end can run the 100-yard dash in 12 seconds. A defensive back can do it in 10 seconds. The tight end catches a pass at his own 20-yard line with the defensive back at the 15-yard line. If no other players are nearby, at what yard line will the defensive back catch up to the tight end? 61. Recreation. How do two children of different weights balance on a seesaw? The heavier child sits closer to the center and the lighter child sits farther away. When the product of the weight of the child and the distance from the center is equal on both sides, the seesaw should be horizontal to the ground. Suppose Max weighs 42 pounds and Maria weighs 60 pounds. If Max sits 5 feet from the center, how far should Maria sit from the center in order to balance the seesaw horizontal to the ground? 62. Recreation. Refer to Exercise 61. Suppose Martin, who weighs 33 pounds, sits on the side of the seesaw with Max. If their average distance to the center is 4 feet, how far should Maria sit from the center in order to balance the seesaw horizontal to the ground? 63. Recreation. If a seesaw has an adjustable bench, then the board can slide along the fulcrum. Maria and Max in Exercise 61 decide to sit on the very edge of the board on each side. Where should the fulcrum be placed along the board in order to balance the seesaw horizontally to the ground? Give the answer in terms of the distance from each child’s end. 64. Recreation. Add Martin (Exercise 62) to Max’s side of the seesaw and recalculate Exercise 63. In Exercises 65–68, refer to this lens law. (See Exercise 82 in Section 1.1.) The position of the image is found using the thin lens equation: 1 ƒ 5 1 do 1 1 di , 106 CHAPTER 1 Equations and Inequalities where do is the distance from the object to the lens, di is the distance from the lens to the image, and ƒ is the focal length of the lens. 2f f f f do di 2f Image Object 65. Optics. If the focal length of a lens is 3 centimeters and the image distance is 5 centimeters from the lens, what is the distance from the object to the lens? 66. Optics. If the focal length of the lens is 8 centimeters and the image distance is 2 centimeters from the lens, what is the distance from the object to the lens? 67. Optics. The focal length of a lens is 2 centimeters. If the image distance from the lens is half the distance from the object to the lens, find the object distance. 68. Optics. The focal length of a lens is 8 centimeters. If the image distance from the lens is half the distance from the object to the lens, find the object distance. In Exercises 69–76, solve each formula for the specified variable. 69. P 5 2l 1 2w for w 70. P 5 2l 1 2w for l 71. A 5 1 2 bh for h 72. C 5 2pr for r 73. A 5 lw for w 74. d 5 rt for t 75. V 5 lwh for h 76. V 5 pr2h for h • C O N C E P T U A L 77. Tricia and Janine are roommates and leave Houston on Interstate 10 at the same time to visit their families for a long weekend. Tricia travels west and Janine travels east. If Tricia’s average speed is 12 miles per hour faster than Janine’s, find the speed of each if they are 320 miles apart in 2 hours and 30 minutes. 78. Rick and Mike are roommates and leave Gainesville on Interstate 75 at the same time to visit their girlfriends for a long weekend. Rick travels north and Mike travels south. If Mike’s average speed is 8 miles per hour faster than Rick’s, find the speed of each if they are 210 miles apart in 1 hour and 30 minutes. • C H A L L E N G E 79. Suppose you bought a house for $132,500 and sold it 3 years later for $168,190. Plot these points using a graphing utility. Assuming a linear relationship, how much could you have sold the house for had you waited 2 additional years? 80. Suppose you bought a house for $132,500 and sold it 3 years later for $168,190. Plot these points using a graphing utility. Assuming a linear relationship, how much could you have sold the house for had you sold it 1 year after buying it? 81. A golf club membership has two options. Option A is a $300 monthly fee plus a $15 cart fee every time you play. Option B has a $150 monthly fee and a $42 fee every time you play. Write a mathematical model for monthly costs for each plan and graph both in the same viewing rectangle using a graphing utility. Explain when Option A is the better deal and when Option B is the better deal. 82. A phone provider offers two calling plans. Plan A has a $30 monthly charge and a $0.10 per minute charge on every call. Plan B has a $50 monthly charge and a $0.03 per minute charge on every call. Explain when Plan A is the better deal and when Plan B is the better deal. • T E C H N O L O G Y 1.3 Quadratic Equations 107 S K I L L S O B J E C T I V E S ■ ■Solve quadratic equations by factoring. ■ ■Use the square root method to solve quadratic equations. ■ ■Solve quadratic equations by completing the square. ■ ■Use the Quadratic Formula to solve quadratic equations. ■ ■Solve application problems using quadratic equations. C O N C E P T U A L O B J E C T I V ES ■ ■Understand the zero product property in factoring. ■ ■Understand that the square root method can only be used when there is no linear term in the quadratic equation. ■ ■Understand that completing the square transforms a standard quadratic equation into a perfect square. ■ ■Derive the Quadratic Formula. ■ ■Understand why eliminated nonphysical answers should be eliminated. 1.3 QUADRATIC EQUATIONS 1.3.1 Factoring In a linear equation, the variable is raised only to the first power in any term where it occurs. In a quadratic equation, the variable is raised to the second power in at least one term. Examples of quadratic equations, also called second-degree equations, are: x2 1 3 5 7 5x2 1 4x 2 7 5 0 x2 2 3 5 0 DEFINITION Quadratic Equation A quadratic equation in x is an equation that can be written in the standard form ax2 1 bx 1 c 5 0 where a, b, and c are real numbers and a 2 0. There are several methods for solving quadratic equations: factoring, the square root method, completing the square, and the Quadratic Formula. FACTORING METHOD The factoring method applies the zero product property: WORDS MATH If a product is zero, then at least If B ? C 5 0, then B 5 0 or C 5 0 one of its factors has to be zero. or both. Consider 1x 2 32 1x 1 22 5 0. The zero product property says that x 2 3 5 0 or x 1 2 5 0, which leads to x 5 22 or x 5 3. The solution set is 522, 36. When a quadratic equation is written in the standard form ax2 1 bx 1 c 5 0, it may be possible to factor the left side of the equation as a product of two first-degree polynomials. We use the zero product property and set each linear factor equal to zero. We solve the resulting two linear equations to obtain the solutions of the quadratic equation. 1.3.1 S K I L L Solve quadratic equations by factoring. 1.3.1 C ON C E P T U A L Understand the zero product property in factoring. 108 CHAPTER 1 Equations and Inequalities EXAMPLE 1 Solving a Quadratic Equation by Factoring Solve the equation x 2 2 6x 2 16 5 0. Solution: The quadratic equation is already in standard form. x 2 2 6x 2 16 5 0 Factor the left side into a product of two linear factors. 1x 2 82 1x 1 22 5 0 If a product equals zero, one of its factors has to be equal to zero. x 2 8 5 0 or x 1 2 5 0 Solve both linear equations. x 5 8 or x 5 22 The solution set is 522, 86 . Y OUR TU R N Solve the quadratic equation x 2 1 x 2 20 5 0 by factoring. ▼ EXAMPLE 2 Solving a Quadratic Equation by Factoring Solve the equation x 2 2 6x 1 5 5 24. common mistake A common mistake is to forget to put the equation in standard form first and then use the zero product property incorrectly. Y OUR TU R N Solve the quadratic equation 9p2 5 24p 2 16 by factoring. ▼ ✖I N C O R R EC T Factor the left side. 1x 2 521x 2 12 5 24 The error occurs here. x 2 5 5 24 or x 2 1 5 24 ✓C O R R EC T Write the original equation. x2 2 6x 1 5 5 24 Write the equation in standard form by adding 4 to both sides. x2 2 6x 1 9 5 0 Factor the left side. 1x 2 321x 2 32 5 0 Use the zero product property and set each factor equal to zero. x 2 3 5 0 or x 2 3 5 0 Solve each linear equation. x 5 3 cDon’t forget to put the quadratic equation in standard form first. Note: The equation has one solution, or root, which is 3. The solution set is 536. Since the linear factors were the same, or repeated, we say that 3 is a double root, or repeated root. ▼ C A U T I O N Don’t forget to put the quadratic equation in standard form first. ▼ A N S W E R The solution is x 5 25, 4. The solution set is 525, 46. ▼ A N S W E R The solution is p 5 4 3, which is a double root. The solution set is U4 3V. 1.3 Quadratic Equations 109 In Example 3, the root x 5 0 is lost when the original quadratic equation is divided by x. Remember to put the equation in standard form first and then factor. 1.3.2 Square Root Method The square root of 16, !16, is 4, not 64. In the review (Chapter 0) the principal square root was discussed. The solutions to x2 5 16, however, are x 5 24 and x 5 4. Let us now investigate quadratic equations that do not have a first-degree term. They have the form ax2 1 c 5 0 a 2 0 The method we use to solve such equations uses the square root property. ▼ C A U T I O N Do not divide by a variable (because the value of that variable may be zero). Bring all terms to one side first and then factor. [CONCEPT CHECK] If something times something is equal to zero, then _____ has to be zero: (A) One of those somethings; (B) Both of those somethings ANSWER (A) One of those somethings ▼ 1.3.2 S K I L L Use the square root method to solve quadratic equations. 1.3.2 C ON C E P T U A L Understand that the square root method can only be used when there is no linear term in the quadratic equation. SQUARE ROOT PROPERTY WORDS MATH If an expression squared is equal If x2 5 P, then x 5 6!P . to a constant, then that expression is equal to the positive or negative square root of the constant. Note: The variable squared must be isolated first (coefficient equal to 1). EXAMPLE 3 Solving a Quadratic Equation by Factoring Solve the equation 2x 2 5 3x. common mistake The common mistake here is dividing both sides by x, which is not allowed because x might be zero. ✖I N C O R R EC T Write the original equation. 2x2 5 3x The error occurs here when both sides are divided by x. 2x 5 3 ✓COR R E C T Write the equation in standard form by subtracting 3x. 2x2 2 3x 5 0 Factor the left side. x12x 2 32 5 0 Use the zero product property and set each factor equal to zero. x 5 0 or 2x 2 3 5 0 Solve each linear equation. x 5 0 or x 5 3 2 The solution set is U0, 3 2V . 110 CHAPTER 1 Equations and Inequalities If we alter Example 4 by changing subtraction to addition, we see in Example 5 that we get imaginary roots (as opposed to real roots), which we discussed in Chapter 0. EXAMPLE 4 Using the Square Root Property Solve the equation 3x 2 2 27 5 0. Solution: Add 27 to both sides. 3x 2 5 27 Divide both sides by 3. x 2 5 9 Apply the square root property. x 5 6!9 5 63 The solution set is 523, 36 . [CONCEPT CHECK] Which of the following can be solved using the square root method? (A) x 2 = x (B) x 2 = 9 ANSWER (B) x 2 = 9 can be solved by the square root method because it is in the form of x 2 = constant. (A) cannot be solved using the square root method. ▼ EXAMPLE 5 Using the Square Root Property Solve the equation 3x 2 1 27 5 0. Solution: Subtract 27 from both sides. 3x 2 5 227 Divide by 3. x 2 5 29 Apply the square root property. x 5 6!29 Simplify. x 5 6i!9 5 63i The solution set is 523i, 3i6 . Y OUR TU R N Solve the equations y2 2 147 5 0 and v2 1 64 5 0. ▼ ▼ A N S W E R The solution is y 5 67!3. The solution set is U27!3, 7!3V. The solution is v 5 68i. The solution set is 528i, 8i6. EXAMPLE 6 Using the Square Root Property Solve the equation 1x 2 22 2 5 16. Solution: Approach 1: If an expression squared is 16, then the expression equals 6!16. 1x 2 22 5 6!16 Separate into two equations. x 2 2 5 !16 or x 2 2 5 2!16 x 2 2 5 4 x 2 2 5 24 x 5 6 x 5 22 The solution set is 522, 66 . Approach 2: It is acceptable notation to keep the 1x 2 22 5 6!16 equations together. x 2 2 5 6 4 x 5 2 6 4 x 5 22, 6 1.3.3 Completing the Square Factoring and the square root method are two efficient, quick procedures for solving many quadratic equations. However, some equations, such as x2 2 10x 2 3 5 0, cannot be solved directly by these methods. A more general procedure to solve this kind of 1.3 Quadratic Equations 111 equation is called completing the square. The idea behind completing the square is to transform any standard quadratic equation ax 2 1 bx 1 c 5 0 into the form 1x 1 A2 2 5 B, where A and B are constants and the left side, 1x 1 A22, has the form of a perfect square. This last equation can then be solved by the square root method. How do we transform the first equation into the second equation? Note that the above-mentioned example, x 2 2 10x 2 3 5 0, cannot be factored into expressions in which all numbers are integers (or even rational numbers). We can, however, transform this quadratic equation into a form that contains a perfect square. WORDS MATH Write the original equation. x 2 2 10x 2 3 5 0 Add 3 to both sides. x 2 2 10x 5 3 Add 25 to both sides. x 2 2 10x 1 25 5 3 1 25 The left side can be written as a perfect square. 1x 2 52 2 5 28 Apply the square root method. x 2 5 5 6!28 Add 5 to both sides. x 5 5 6 2!7 Why did we add 25 to both sides? Recall that 1x 2 c2 2 5 x 2 2 2xc 1 c2. In this case c 5 5 in order for 22xc 5 210x. Therefore, the desired perfect square 1x 2 522 results in x2 2 10x 1 25. Applying this product, we see that 125 is needed. A systematic approach is to take the coefficient of the first-degree term x2 2 10x 2 3 5 0, which is 210. Take half of 12102, which is 1252, and then square it 12522 5 25. 1.3.3 S K I L L Solve quadratic equations by completing the square. 1.3.3 C ON C E P T U A L Understand that completing the square transforms a standard quadratic equation into a perfect square. [CONCEPT CHECK] The quadratic equation x 2 2 4x 1 3 5 0 can be transformed to which perfect square? (A) 1x 1 22 2 5 1 (B) 1x 2 22 2 5 1 ANSWER (B) ▼ EXAMPLE 7 Completing the Square Solve the quadratic equation x 2 1 8x 2 3 5 0 by completing the square. Solution: Add 3 to both sides. x 2 1 8x 5 3 Add Q1 2 ⋅8R 2 5 42 to both sides. x 2 1 8x 1 42 5 3 1 42 Write the left side as a perfect square and simplify the right side. 1x 1 42 2 5 19 Apply the square root method to solve. x 1 4 5 6!19 Subtract 4 from both sides. x 5 24 6!19 The solution set is U24 2 !19, 24 1 !19V . SOLVING A QUADRATIC EQUATION BY COMPLETING THE SQUARE WORDS MATH Express the quadratic equation in the following form. x 2 1 bx 5 c Divide b by 2 and square the result, then add the square to both sides. x2 1 bx 1 ab 2b 2 5 c 1 ab 2b 2 Write the left side of the equation as a perfect square. ax 1 b 2b 2 5 c 1 ab 2b 2 Solve using the square root method. 112 CHAPTER 1 Equations and Inequalities In Example 7, the leading coefficient (the coefficient of the x2 term) is 1. When the leading coefficient is not 1, start by first dividing the equation by that leading coefficient. EXAMPLE 8 Completing the Square When the Leading Coefficient Is Not Equal to 1 Solve the equation 3x2 2 12x 1 13 5 0 by completing the square. Solution: Divide by the leading coefficient, 3. x2 2 4x 1 13 3 5 0 Collect variables to one side of the x2 2 4x 5 213 3 equation and constants to the other side. Add A24 2B2 5 4 to both sides. x2 2 4x 1 4 5 213 3 1 4 Write the left side of the equation as a 1x 2 222 5 21 3 perfect square and simplify the right side. Solve using the square root method. x 2 2 5 6Å21 3 Simplify. x 5 2 6 iÅ 1 3 Rationalize the denominator (Chapter 0). x 5 2 6 i !3 ⋅!3 !3 Simplify. x 5 2 2 i!3 3 , x 5 2 1 i!3 3 The solution set is e 2 2 i!3 3 , 2 1 i!3 3 f . Y OUR TU R N Solve the equation 2x2 2 4x 1 3 5 0 by completing the square. ▼ ▼ A N S W E R The solution is x 5 1 6 i!2 2 . The solution set is e 1 2 i!2 2 , 1 1 i!2 2 f . 1.3.4 Quadratic Formula Let us now consider the most general quadratic equation: ax2 1 bx 1 c 5 0 a 2 0 We can solve this equation by completing the square. WORDS MATH Divide the equation by the x2 1 b ax 1 c a 5 0 leading coefficient a. Subtract c a from both sides. x2 1 b ax 5 2c a Square half of b a and add the result a b 2ab 2 to both sides. x2 1 b ax 1 a b 2ab 2 5 a b 2ab 2 2 c a 1.3.4 S KILL Use the Quadratic Formula to solve quadratic equations. 1.3.4 CO NCE PTUAL Derive the Quadratic Formula. STUDY TIP When the leading coefficient is not 1, start by first dividing the equation by that leading coefficient. 1.3 Quadratic Equations 113 Write the left side of the equation as a perfect square and the right side ax 1 b 2ab 2 5 b2 2 4ac 4a2 as a single fraction. Solve using the square root method. x 1 b 2a 5 6Å b2 2 4ac 4a2 Subtract b 2a from both sides x 5 2 b 2a 6 "b2 2 4ac 2a and simplify the radical. Write as a single fraction. x 5 2b6"b2 2 4ac 2a We have derived the Quadratic Formula. [CONCEPT CHECK] TRUE OR FALSE Completing the square is still necessary in certain cases when the Quadratic Formula cannot be used to solve a particular Quadratic Equation. ANSWER False. The Quadratic Formula was derived by solving the general Quadratic Equation by completing the square and can be used for solving any quadratic equation. ▼ We read this formula as negative b plus or minus the square root of the quantity b squared minus 4ac all over 2a. It is important to note that negative b could be positive (if b is negative). For this reason, an alternate form is “opposite b. . .” The Quadratic Formula should be memorized and used when simpler methods (factoring and the square root method) cannot be used. The Quadratic Formula works for any quadratic equation. STUDY TIP x 5 2b 6 "b2 2 4ac 2a Read as “negative b plus or minus the square root of the quantity b squared minus 4ac all over 2a.” STUDY TIP The Quadratic Formula works for any quadratic equation. STUDY TIP Using parentheses as placehold-ers helps avoid 6 errors. x 5 2b 6 "b2 2 4ac 2a x 5 21u2 6 "1u22 2 41u21u2 21u2 EXAMPLE 9 Using the Quadratic Formula and Finding Two Distinct Real Roots Use the Quadratic Formula to solve the quadratic equation x2 2 4x 2 1 5 0. Solution: For this problem a 5 1, b 5 24, and c 5 21. Write the Quadratic Formula. x 5 2b 6 "b2 2 4ac 2a Use parentheses to avoid losing a x 5 21u 2 6 "1u22 2 41u21u2 21u2 minus sign. Substitute values for a, b, and c x 5 21242 6 "12422 2 41121212 2112 into the parentheses. Simplify. x 5 4 6 "16 1 4 2 5 4 6 "20 2 5 4 6 2"5 2 5 4 2 6 2"5 2 5 2 6 "5 The solution set U2 2 !5, 2 1 !5V contains two distinct real numbers. YOUR T UR N Use the Quadratic Formula to solve the quadratic equation x2 1 6x 2 2 5 0. ▼ ▼ A N S W E R The solution is x 5 23 6!11. The solution set is U23 2 "11, 23 1 "11V. QUADRATIC FORMULA If ax2 1 bx 1 c 5 0, a 2 0, then the solution is x 5 2b 6 "b2 2 4ac 2a Note: The quadratic equation must be in standard form 1ax2 1 bx 1 c 5 0) in order to identify the parameters: a—coefficient of x2 b—coefficient of x c—constant 114 CHAPTER 1 Equations and Inequalities EXAMPLE 10 Using the Quadratic Formula and Finding Two Complex Roots Use the Quadratic Formula to solve the quadratic equation x2 1 8 5 4x. Solution: Write this equation in standard form x2 2 4x 1 8 5 0 in order to identify a 5 1, b 5 24, and c 5 8. Write the Quadratic Formula. x 5 2b 6 "b2 2 4ac 2a Use parentheses to avoid x 5 21u 2 6 "1u22 2 41u21u2 21u2 overlooking a minus sign. Substitute the values for a, b, x 5 21242 6 "12422 2 4112182 2112 and c into the parentheses. Simplify. x 5 4 6 !16 2 32 2 5 4 6 !216 2 5 4 6 4i 2 5 4 2 6 4i 2 5 2 6 2i The solution set 52 2 2i, 2 1 2i6 contains two complex numbers. Note that they are complex conjugates of each other. Y OUR TU R N Use the Quadratic Formula to solve the quadratic equation x2 1 2 5 2x. ▼ EXAMPLE 11 Using the Quadratic Formula and Finding One Repeated Real Root Use the Quadratic Formula to solve the quadratic equation 4x2 2 4x 1 1 5 0. Solution: Identify a, b, and c. a 5 4, b 5 24, c 5 1 Write the Quadratic Formula. x 5 2b 6 "b2 2 4ac 2a Use parentheses to avoid losing a x 5 21u2 6 "1u22 2 41u21u2 21u2 minus sign. Substitute values a 5 4, b 5 24, x 5 21242 6 "12422 2 4142112 2142 and c 5 1. Simplify. x 5 4 6 "16 2 16 8 5 4 6 0 8 5 1 2 The solution set is a repeated real root e 1 2 f . Note: This quadratic equation also could have been solved by factoring: 12x 2 122 5 0. Y OUR TU R N Use the Quadratic Formula to solve the quadratic equation 9x2 2 6x 1 1 5 0. ▼ ▼ A N S W E R The solution set is 51 2 i, 1 1 i6. ▼ A N S W E R U1 3V 1.3 Quadratic Equations 115 In Example 9, the discriminant is positive and the solution has two distinct real roots. In Example 10, the discriminant is negative and the solution has two complex (conjugate) roots. In Example 11, the discriminant is zero and the solution has one repeated real root. 1.3.5 Applications Involving Quadratic Equations In Section 1.2, we developed a procedure for solving word problems involving linear equations. The procedure is the same for applications involving quadratic equations. The only difference is that the mathematical equations will be quadratic, as opposed to linear. 1.3.5 S K I L L Solve application problems using quadratic equations. 1.3.5 C ON C E P T U A L Understand why nonphysical answers should be eliminated. EXAMPLE 12 Stock Value From March 1 to May 1 the price of Abercrombie & Fitch’s (ANF) stock was approximately given by P 5 23t2 1 6t 1 20, where P is the price of stock in dollars, t is in months, and t 5 0 corre-sponds to March 1. When was the value of the stock worth $22? Solution: STEP 1 Identify the question. When is the price of the stock equal to $22? STEP 2 Make notes. Stock price: P 5 23t2 1 6t 1 20 P 5 22 STEP 3 Set up an equation. 23t2 1 6t 1 20 5 22 STEP 4 Solve the equation. Subtract 22 from both sides. 23t2 1 6t 2 2 5 0 Solve for t using the t 5 2162 6 "62 2 412321222 21232 Quadratic Formula. Simplify. t 5 26 6 "62 2 412321222 21232 5 26 6 "12 26 < 1.57, 0.42 Feb Mar Apr May June July Aug Sep Nov Oct 15 20 25 30 35 TYPES OF SOLUTIONS The term inside the radical, b2 2 4ac, is called the discriminant. The discriminant gives important information about the corresponding solutions or roots of ax2 1 bx 1 c 5 0, where a, b, and c are real numbers. b2 2 4ac SOLUTIONS (ROOTS) Positive Two distinct real roots 0 One real root (a double or repeated root) Negative Two complex roots (complex conjugates) 116 CHAPTER 1 Equations and Inequalities Rounding these two numbers, we find that t < 1.5 and t < 0.5. Since t 5 1 corresponds to March 1, the value of t 5 0.5 corresponds to March 15, and the value t 5 1.5 corresponds to April 15. STEP 5 Check the solution. Look at the figure. The horizontal axis represents the month, and the vertical axis represents the stock price. Estimating when the stock price is approximately $22, we find March 15 and April 15. EXAMPLE 13 Pythagorean Theorem Hitachi makes a 60-inch HDTV that has a 60-inch diagonal. If the width of the screen is approximately 52 inches, what is the approximate height of the screen? Solution: STEP 1 Identify the question. What is the approximate height of the HDTV screen? STEP 2 Make notes. 60 inches 52 inches ? STEP 3 Set up an equation. Recall the Pythagorean theorem. a2 1 b2 5 c2 Substitute in the known values. h2 1 522 5 602 STEP 4 Solve the equation. Simplify the constants. h2 1 2704 5 3600 Subtract 2704 from both sides. h2 5 896 Solve using the square root method. h 5 6"896 < 630 Distance is positive, so the negative value is eliminated. The height is approximately 30 inches . STEP 5 Check the solution. 302 1 522 0 602 900 1 2704 0 3600 3604 < 3600 [CONCEPT CHECK] Which of the following would we NOT eliminate negative values for physical answers: (A) distance (B) height (C) width (D) temperature ANSWER (D) temperature. All others are defined as positive. ▼ STUDY TIP Dimensions such as length and width are distances, which are defined as positive quantities. Although the mathematics may yield both positive and negative values, the negative values are excluded. Formula and completing the square work for all quadratic equations can yield three types of solutions: two distinct real roots, one real root (repeated), or two complex roots (conjugates of each other). Quadratic Formula: x 5 2b 6 "b2 2 4ac 2a The four methods for solving quadratic equations ax2 1 bx 1 c 5 0 a 2 0 are factoring, the square root method, completing the square, and the Quadratic Formula. Factoring and the square root method are the quickest and easiest but cannot always be used. The Quadratic [SEC TION 1.3] S U M M A RY 1.3 Quadratic Equations 117 [SEC TION 1.3] E X E R C I S E S • S K I L L S In Exercises 1–22, solve by factoring. 1. x 2 2 5x 1 6 5 0 2. v 2 1 7v 1 6 5 0 3. p 2 2 8p 1 15 5 0 4. u2 2 2u 2 24 5 0 5. x 2 5 12 2 x 6. 11x 5 2x2 1 12 7. 16x 2 1 8x 5 21 8. 3x2 1 10x 2 8 5 0 9. 9y 2 1 1 5 6y 10. 4x 5 4x2 1 1 11. 8y 2 5 16y 12. 3A2 5 212A 13. 9p 2 5 12p 2 4 14. 4u 2 5 20u 2 25 15. x 2 2 9 5 0 16. 16v 2 2 25 5 0 17. x 1x 1 42 5 12 18. 3t 2 2 48 5 0 19. 2p 2 2 50 5 0 20. 5y 2 2 45 5 0 21. 3x 2 5 12 22. 7v 2 5 28 In Exercises 23–34, solve using the square root method. 23. p 2 2 8 5 0 24. y 2 2 72 5 0 25. x 2 1 9 5 0 26. v 2 1 16 5 0 27. 1x 2 32 2 5 36 28. 1x 2 12 2 5 25 29. 12x 1 32 2 5 24 30. 14x 2 12 2 5 216 31. 15x 2 22 2 5 27 32. 13x 1 82 2 5 12 33. 11 2 x2 2 5 9 34. 11 2 x2 2 5 29 In Exercises 35–44, what number should be added to complete the square of each expression? 35. x 2 1 6x 36. x 2 2 8x 37. x 2 2 12x 38. x 2 1 20x 39. x2 2 1 2x 40. x2 2 1 3x 41. x2 1 2 5x 42. x2 1 4 5x 43. x 2 2 2.4x 44. x 2 1 1.6x In Exercises 45–56, solve by completing the square. 45. x 2 1 2x 5 3 46. y 2 1 8y 2 2 5 0 47. t 2 2 6t 5 25 48. x2 1 10x 5 221 49. y 2 2 4y 1 3 5 0 50. x2 2 7x 1 12 5 0 51. 2p2 1 8p 5 23 52. 2x2 2 4x 1 3 5 0 53. 2x 2 2 7x 1 3 5 0 54. 3x 2 2 5x 2 10 5 0 55. x2 2 2 2x 5 1 4 56. t2 3 1 2t 3 1 5 6 5 0 In Exercises 57–68, solve using the Quadratic Formula. 57. t 2 1 3t 2 1 5 0 58. t 2 1 2t 5 1 59. s 2 1 s 1 1 5 0 60. 2s 2 1 5s 5 22 61. 3x 2 2 3x 2 4 5 0 62. 4x 2 2 2x 5 7 63. x 2 2 2x 1 17 5 0 64. 4m2 1 7m 1 8 5 0 65. 5x 2 1 7x 5 3 66. 3x 2 1 5x 5 211 67. 1 4x2 1 2 3x 2 1 2 5 0 68. 1 4x2 2 2 3x 2 1 3 5 0 In Exercises 69–74, determine whether the discriminant is positive, negative, or zero, and indicate the number and type of root to expect. Do not solve. 69. x 2 2 22x 1 121 5 0 70. x 2 2 28x 1 196 5 0 71. 2y 2 2 30y 1 68 5 0 72. 23y 2 1 27y 1 66 5 0 73. 9x 2 2 7x 1 8 5 0 74. 23x 2 1 5x 2 7 5 0 In Exercises 75–94, solve using any method. 75. v 2 2 8v 5 20 76. v 2 2 8v 5 220 77. t 2 1 5t 2 6 5 0 78. t 2 1 5t 1 6 5 0 79. 1x 1 32 2 5 16 80. 1x 1 32 2 5 216 81. 1 p 2 22 2 5 4p 82. 1u 1 52 2 5 16u 83. 8w 2 1 2w 1 21 5 0 84. 8w2 1 2w 2 21 5 0 85. 3p 2 2 9p 1 1 5 0 86. 3p2 2 9p 2 1 5 0 87. 2 3 t2 1 4 3 t 5 1 5 88. 1 2 x2 1 2 3 x 5 2 5 89. x 1 12 x 5 7 90. x 2 10 x 5 23 91. 41x 2 22 x 2 3 1 3 x 5 23 x1x 2 32 92. 5 y 1 4 5 4 1 3 y 2 2 93. x 2 2 0.1x 5 0.12 94. y 2 2 0.5y 5 20.06 118 CHAPTER 1 Equations and Inequalities • A P P L I C A T I O N S 95. Stock Value. From November 2014 to November 2015, Amazon.com stock was approximately worth P 5 6.25t2 2 35t 1 336, where P is the price of the stock in dollars, t is months, and t 5 0 corresponds to November 2014. During what months was the stock equal to $310? Nov Source: www.nasdaq.com/symbol/amzn/stock-chart Dec 2015 Feb Mar Apr May Jun Nov Jul Aug Sep Oct 275.00 300.00 325.00 350.00 375.00 400.00 575.00 625.00 675.00 $725.00 525.00 475.00 450.00 425.00 96. Stock Value. From November 2014 to November 2015, Apple stock was approximately worth P 5 20.39t2 1 4.29t 1 120.1, where P is the price of the stock in dollars, t is months, and t 5 0 corresponds to Novem-ber 2014. During what months was the stock equal to $124? Nov Source: www.nasdaq.com/symbol/aapl/stock-chart Dec 2015 Feb Mar Apr May Jun Nov Jul Aug Sep Oct 90.00 95.00 100.00 105.00 110.00 115.00 $140.00 135.00 130.00 125.00 120.00 In Exercises 97 and 98 refer to the following: Research indicates that monthly profit for Widgets R Us is modeled by the function P 5 2100 1 10.2q 2 32q where P is profit measured in millions of dollars and q is the quantity of widgets produced measured in thousands. 97. Business. Find the break-even point for a month to the nearest unit. 98. Business. Find the production level that produces a monthly profit of $40 million. In Exercises 99 and 100 refer to the following: In response to economic conditions, a local business explores the effect of a price increase on weekly profit. The function P 5 251x 1 32 1x 2 242 models the effect that a price increase of x dollars on a bottle of wine will have on the profit P measured in dollars. 99. Business/Economics. What is the smallest price increase that will produce a weekly profit of $460? 100. Business/Economics. What is the smallest price increase that will produce a weekly profit of $630? In Exercises 101 and 102 refer to the following: An epidemiological study of the spread of the flu in a small city finds that the total number P of people who contracted the flu t days into an outbreak is modeled by the function P 5 2t2 1 13t 1 130 1 # t # 6 101. Health/Medicine. After approximately how many days will 160 people have contracted the flu? 102. Health/Medicine. After approximately how many days will 172 people have contracted the flu? 103. Environment: Reduce Your Margins, Save a Tree. Let’s define the usable area of an 8.5-inch by 11-inch piece of paper as the rectangular space between the mar-gins of that piece of paper. Assume the default margins in a word processor in a college’s computer lab are set up to be 1.25 inches wide (top and bottom) and 1 inch wide (left and right). Answer the following questions using this information. a. Determine the amount of usable space, in square inches, on one side of an 8.5-inch by 11-inch piece of paper with the default margins of 1.25-inch and 1-inch. b. The Green Falcons, a campus environmental club, has convinced their college’s computer lab to reduce the default margins in their word-processing software by x inches. Create and simplify the quadratic expression that represents the new usable area, in square inches, of one side of an 8.5-inch by 11-inch piece of paper if the default margins at the computer lab are each reduced by x inches. c. Subtract the usable space in part (a) from the expression in part (b). Explain what this difference represents. d. If 10 pages are printed using the new margins and as a result the computer lab saved one whole sheet of paper, then by how much did the computer lab reduce the margins? Round to the nearest tenth of an inch. 104. Environment: Reduce Your Margins, Save a Tree. Repeat Exercise 103 assuming the computer lab’s default margins are 1 inch all the way around (left, right, top, and bottom). If 15 pages are printed using the new margins and as a result the computer lab saved one whole sheet of paper, then by how much did the computer lab reduce the margins? Round to the nearest tenth of an inch. 105. Television. A standard 32-inch television has a 32-inch diagonal and a 25-inch width. What is the height of the 32-inch television? 106. Television. A 42-inch LCD television has a 42-inch diagonal and a 20-inch height. What is the width of the 42-inch LCD television? 107. Numbers. Find two consecutive numbers such that their sum is 35 and their product is 306. 108. Numbers. Find two consecutive odd integers such that their sum is 24 and their product is 143. 1.3 Quadratic Equations 119 109. Geometry. The area of a rectangle is 135 square feet. The width is 6 feet less than the length. Find the dimensions of the rectangle. 110. Geometry. A rectangle has an area of 31.5 square meters. If the length is 2 more than twice the width, find the dimensions of the rectangle. 111. Geometry. A triangle has a height that is 2 more than 3 times the base and an area of 60 square units. Find the base and height. 112. Geometry. A square’s side is increased by 3 yards, which corresponds to an increase in the area by 69 square yards. How many yards is the side of the initial square? 113. Falling Objects. If a person drops a water balloon off the rooftop of a 100-foot building, the height of the water balloon is given by the equation h 5 216t2 1 100, where t is in seconds. When will the water balloon hit the ground? 114. Falling Objects. If the person in Exercise 113 throws the water balloon downward with a speed of 5 feet per second, the height of the water balloon is given by the equation h 5 216t2 2 5t 1 100, where t is in seconds. When will the water balloon hit the ground? 115. Gardening. A square garden has an area of 900 square feet. If a sprinkler (with a circular pattern) is placed in the center of the garden, what is the minimum radius of spray the sprinkler would need in order to water all of the garden? 116. Sports. A baseball diamond is a square. The distance from base to base is 90 feet. What is the distance from home plate to second base? 117. Volume. A flat square piece of cardboard is used to construct an open box. Cutting a 1-foot by 1-foot square off of each corner and folding up the edges will yield an open box (assuming these edges are taped together). If the desired volume of the box is 9 cubic feet, what are the dimensions of the original square piece of cardboard? 118. Volume. A rectangular piece of cardboard whose length is twice its width is used to construct an open box. Cutting a 1-foot by 1-foot square off of each corner and folding up the edges will yield an open box. If the desired volume is 12 cubic feet, what are the dimensions of the original rectangular piece of cardboard? 119. Gardening. A landscaper has planted a rectangular garden that measures 8 feet by 5 feet. He has ordered 1 cubic yard (27 cubic feet) of stones for a border along the outside of the garden. If the border needs to be 4 inches deep and he wants to use all of the stones, how wide should the border be? 120. Gardening. A gardener has planted a semicircular rose garden with a radius of 6 feet, and 2 cubic yards of mulch (1 cubic yard 5 27 cubic feet) are being delivered. Assuming she uses all of the mulch, how deep will the layer of mulch be? 121. Work. Lindsay and Kimmie, working together, can balance the financials for the Kappa Kappa Gamma sorority in 6 days. Lindsay by herself can complete the job in 5 days less than Kimmie. How long will it take Lindsay to complete the job by herself? 122. Work. When Jack cleans the house, it takes him 4 hours. When Ryan cleans the house, it takes him 6 hours. How long would it take both of them if they worked together? • C A T C H T H E M I S T A K E In Exercises 123–126, explain the mistake that is made. 123. t2 2 5t 2 6 5 0 1t 2 32 1t 2 22 5 0 t 5 2, 3 124. 12y 2 322 5 25 2y 2 3 5 5 2y 5 8 y 5 5 4 y 5 4 125. 16a2 1 9 5 0 16a2 5 29 a2 5 2 9 16 a 5 6" 9 16 a 5 6 3 4 126. 2x2 2 4x 5 3 21x2 2 2x2 5 3 21x2 2 2x 1 12 5 3 1 1 21x 2 122 5 4 1x 2 122 5 2 x 2 1 5 6"2 x 5 1 6 !2 In Exercises 127–130, determine whether the following statements are true or false. 127. The equation 13x 1 12 2 5 16 has the same solution set as the equation 3x 1 1 5 4. 128. The quadratic equation ax2 1 bx 1 c 5 0 can be solved by the square root method only if b 5 0. 129. All quadratic equations can be solved exactly. 130. The Quadratic Formula can be used to solve any quadratic equation. 131. Write a quadratic equation in general form that has x 5 a as a repeated real root. 132. Write a quadratic equation in general form that has x 5 bi as a root. 133. Write a quadratic equation in general form that has the solution set 52, 56. 134. Write a quadratic equation in general form that has the solution set 523, 06. • C O N C E P T U A L 120 CHAPTER 1 Equations and Inequalities In Exercises 135–138, solve for the indicated variable in terms of other variables. 135. Solve s 5 1 2gt2 for t. 136. Solve A 5 P 11 1 r2 2 for r. 137. Solve a2 1 b2 5 c2 for c. 138. Solve P 5 EI 2 RI2 for I. 139. Solve the equation by factoring: x4 2 4x2 5 0. 140. Solve the equation by factoring: 3x 2 6x2 5 0. 141. Solve the equation using factoring by grouping: x3 1 x2 2 4x 2 4 5 0. 142. Solve the equation using factoring by grouping: x3 1 2x2 2 x 2 2 5 0. 143. Show that the sum of the roots of a quadratic equation is equal to 2b a. 144. Show that the product of the roots of a quadratic equation is equal to c a. 145. Write a quadratic equation in general form whose solution set is U3 1 !5, 3 2 !5 V. 146. Write a quadratic equation in general form whose solution set is 52 2 i, 2 1 i6. 147. Aviation. An airplane takes 1 hour longer to go a distance of 600 miles flying against a headwind than on the return trip with a tailwind. If the speed of the wind is 50 miles per hour, find the speed of the plane in still air. 148. Boating. A speedboat takes 1 hour longer to go 24 miles up a river than to return. If the boat cruises at 10 miles per hour in still water, what is the rate of the current? 149. Find a quadratic equation whose two distinct real roots are the negatives of the two distinct real roots of the equation ax2 1 bx 1 c 5 0. 150. Find a quadratic equation whose two distinct real roots are the reciprocals of the two distinct real roots of the equation ax2 1 bx 1 c 5 0. 151. A small jet and a 757 leave Atlanta at 1 p.m. The small jet is traveling due west. The 757 is traveling due south. The speed of the 757 is 100 miles per hour faster than the small jet. At 3 p.m. the planes are 1000 miles apart. Find the average speed of each plane. 152. Two boats leave Key West at noon. The smaller boat is trav-eling due west. The larger boat is traveling due south. The speed of the larger boat is 10 miles per hour faster than the speed of the smaller boat. At 3 p.m. the boats are 150 miles apart. Find the average speed of each boat. • C H A L L E N G E 153. Solve the equation x2 2 x 5 2 by first writing it in standard form and then factoring. Now plot both sides of the equation in the same viewing screen 1 y1 5 x2 2 x and y2 5 22. At what x-values do these two graphs intersect? Do those points agree with the solution set you found? 154. Solve the equation x2 2 2x 5 22 by first writing it in standard form and then using the quadratic formula. Now plot both sides of the equation in the same viewing screen 1 y1 5 x2 2 2x and y2 5 222. Do these graphs intersect? Does this agree with the solution set you found? 155. a. Solve the equation x2 2 2x 5 b, b 5 8 by first writing it in standard form. Now plot both sides of the equation in the same viewing screen 1 y1 5 x2 2 2x and y2 5 b2. At what x values do these two graphs intersect? Do those points agree with the solution set you found? b. Repeat part (a) for b 5 23, 21, 0, and 5. 156. a. Solve the equation x2 1 2x 5 b, b 5 8 by first writing it in standard form. Now plot both sides of the equation in the same viewing screen 1 y1 5 x2 1 2x and y2 5 b2. At what x values do these two graphs intersect? Do those points agree with the solution set you found? b. Repeat part (a) for b 5 23, 21, 0, and 5. • T E C H N O L O G Y 1.4 Other Types of Equations 121 1.4.1 Radical Equations Radical equations are equations in which the variable is inside a radical (that is, under a square root, cube root, or higher root). Examples of radical equations follow. !x 2 3 5 2 !2x 1 3 5 x !x 1 2 1 !7x 1 2 5 6 Until now your experience has been with linear and quadratic equations. Often you can transform a radical equation into a simple linear or quadratic equation. Sometimes the transformation process yields extraneous solutions, or apparent solutions that may solve the transformed problem but are not solutions of the original radical equation. Therefore, it is very important to check your answers. S K I L L S O B J E C T I V E S ■ ■Solve radical equations. ■ ■Solve equations that are quadratic in form using u-substitutions. ■ ■Solve equations that are factorable. C O N C E P T U A L O B J E C T I V ES ■ ■Check for extraneous solutions. ■ ■Recognize the u-substitution required to transform the equation into a simpler quadratic equation. ■ ■Recognize when a polynomial equation or an equation with rational exponents can be factored either by grouping or by first factoring out a greatest common factor. 1.4 OTHER TYPES OF EQUATIONS 1.4.1 S K I L L Solve radical equations. 1.4.1 C ON C E P T U A L Check for extraneous solutions. EXAMPLE 2 Solving an Equation Involving a Radical Solve the equation !2x 1 3 5 x. Solution: Square both sides of the equation. A!2x 1 3B2 5 x2 Simplify. 2x 1 3 5 x2 Write the quadratic equation in standard form. x2 2 2x 2 3 5 0 Factor. 1x 2 32 1x 1 12 5 0 Use the zero product property. x 5 3 or x 5 21 EXAMPLE 1 Solving an Equation Involving a Radical Solve the equation !x 2 3 5 2. Solution: Square both sides of the equation. 1 !x 2 322 5 22 Simplify. x 2 3 5 4 Solve the resulting linear equation. x 5 7 The solution set is 576. Check: !7 2 3 5 !4 5 2 Y OUR T UR N Solve the equation !3p 1 4 5 5. ▼ ▼ A N S W E R p 5 7 or 576 When both sides of an equation are squared, extraneous solutions can arise. For example, take the equation x 5 2 If we square both sides of this equation, then the resulting equation, x2 5 4, has two solutions: x 5 22 and x 5 2. Notice that the value x 5 22 is not in the solution set of the original equation x 5 2. Therefore, we say that x 5 22 is an extraneous solution. In solving a radical equation, we square both sides of the equation and then solve the resulting equation. The solutions to the resulting equation can sometimes be extraneous in that they do not satisfy the original radical equation. STUDY TIP Extraneous solutions are common when we deal with radical equations, so remember to check your answers. [CONCEPT CHECK] Check the extraneous solution x 5 21 and explain why it does not satisfy the original equation. ANSWER !21212 1 3 0 21 !1 0 21 x 5 21 is extraneous. ▼ 122 CHAPTER 1 Equations and Inequalities What happened in Example 2? When we transformed the radical equation into a quadratic equation, we created an extraneous solution, x 5 21, a solution that appears to solve the original equation but does not. When solving radical equations, answers must be checked to avoid including extraneous solutions in the solution set. EXAMPLE 3 Solving an Equation That Involves a Radical Solve the equation 4x 2 2!x 1 3 5 210. Solution: Subtract 4x from both sides. 22!x 1 3 5 210 2 4x Divide both sides by 22. !x 1 3 5 2x 1 5 Square both sides. x 1 3 5 12x 1 52 12x 1 52 12x 1 522 Eliminate the parentheses. x 1 3 5 4x2 1 20x 1 25 Rewrite the quadratic equation in standard form. 4x2 1 19x 1 22 5 0 Factor. 14x 1 112 1x 1 22 5 0 Solve. x 5 211 4 and x 5 22 The apparent solutions are 211 4 and 22. Note that 211 4 does not satisfy the original equation; therefore it is extraneous. The solution is x 522 . The solution set is 5226. Y OUR TU R N Solve the equation 2x 2 4!x 1 2 5 26. v ▼ A N S W E R x 5 21 or 5216 ▼ In Examples 1 through 3 each equation only contained one radical each. The next example contains two radicals. Our technique will be to isolate one radical on one side of the equation with the other radical on the other side of the equation. common mistake ✖I N C O R R EC T Square the expression. A3 1 !x 1 2B 2 The error occurs here when only individual terms are squared. 2 9 1 1x 1 22 ✓C O R R EC T Square the expression. A3 1 !x 1 2B 2 Write the square as a product of two factors. A3 1 !x 1 2BA3 1 !x 1 2B Use the FOIL method. 9 1 6!x 1 2 1 1x 1 22 Check these values to see whether they both make the original equation statement true. x 5 3: !2132 1 3 5 3 1 !6 1 3 5 3 1 !9 5 3 1 3 5 3 3 x 5 21: !21212 1 3 5 21 1 !22 1 3 5 21 1 !1 5 21 1 1 2 21 X The solution is x 5 3 . The solution set is 536. Y OUR TU R N Solve the equation !12 1 t 5 t. Y OUR TU R N Solve the equation !2x 1 6 5 x 1 3. ▼ A N S W E R t 5 4 or 546 ▼ A N S W E R x 5 21 and x 5 23 or 523, 216 ▼ 1.4 Other Types of Equations 123 1.4.2 Equations Quadratic in Form: u-Substitution Equations that are higher order or that have fractional powers often can be transformed into a quadratic equation by introducing a u-substitution. When this is the case, we say that equations are quadratic in form. In the following table, the two original equations are quadratic in form because they can be transformed into a quadratic equation given the correct substitution. ORIGINAL EQUATION SUBSTITUTION NEW EQUATION x4 2 3x2 2 4 5 0 u 5 x2 u2 2 3u 2 4 5 0 t2/3 1 2t1/3 1 1 5 0 u 5 t1/3 u2 1 2u 1 1 5 0 2 y 2 1 !y 1 1 5 0 u 5 y21/2 2u2 2 u 1 1 5 0 EXAMPLE 4 Solving an Equation with More Than One Radical Solve the equation !x 1 2 1 !7x 1 2 5 6. Solution: Subtract !x 1 2 from both sides. !7x 1 2 5 6 2 !x 1 2 Square both sides. A!7x 1 2B 2 5 A6 2 !x 1 2B 2 Simplify. 7x 1 2 5 A6 2 !x 1 2BA6 2 !x 1 2B Multiply the expressions on the right side of the equation. 7x 1 2 5 36 2 12!x 1 2 1 1x 1 22 Isolate the term with the radical on the left side. 12!x 1 2 5 36 1 x 1 2 2 7x 2 2 Combine like terms on the right side. 12!x 1 2 5 36 2 6x Divide by 6. 2!x 1 2 5 6 2 x Square both sides. 41x 1 22 5 16 2 x22 Simplify. 4x 1 8 5 36 2 12x 1 x2 Rewrite the quadratic equation in standard form. x2 2 16x 1 28 5 0 Factor. 1x 2 142 1x 2 22 5 0 Solve. x 5 14 and x 5 2 The apparent solutions are 2 and 14. Note that x 5 14 does not satisfy the original equation; therefore, it is extraneous. The solution is x 5 2 . The solution set is 526. YOUR T UR N Solve the equation !x 2 4 5 5 2 !x 1 1. ▼ STUDY TIP Remember to check both solutions. ▼ A N S W E R x 5 8 or 586 1.4.2 S K I L L Solve equations that are quadratic in form using u-substitutions. 1.4.2 C ON C E P T U A L Recognize the u-substitution required to transform the equation into a simpler quadratic equation. PROCEDURE FOR SOLVING RADICAL EQUATIONS Step 1: Isolate the term with a radical on one side. Step 2: Raise both (entire) sides of the equation to the power that will eliminate this radical, and simplify the equation. Step 3: If a radical remains, repeat Steps 1 and 2. Step 4: Solve the resulting linear or quadratic equation. Step 5: Check the solutions and eliminate any extraneous solutions. Note: If there is more than one radical in the equation, it does not matter which radical is isolated first. 124 CHAPTER 1 Equations and Inequalities For example, the equation x4 2 3x2 2 4 5 0 is a fourth-degree equation in x. How did we know that u 5 x2 would transform the original equation into a quadratic equation? If we rewrite the original equation as 1x2 2 2 2 3 1x2 2 2 4 5 0, the expression in paren-theses is the u-substitution. Let us introduce the substitution u 5 x2. Note that squaring both sides implies u2 5 x4. We then replace x2 in the original equation with u, and x4 in the original equation with u2, which leads to a quadratic equation in u: u2 2 3u 2 4 5 0. WORDS MATH Solve for x. x4 2 3x2 2 4 5 0 Introduce u-substitution. u 5 x2 3Note that u2 5 x4.4 Write the quadratic equation in u. u2 2 3u 2 4 5 0 Factor. 1u 2 42 1u 1 12 5 0 Solve for u. u 5 4 or u 5 21 Transform back to x, u 5 x2. x2 5 4 or x2 5 21 Solve for x. x 5 62 or x 5 6i The solution set is 562, 6i 6. It is important to correctly determine the appropriate substitution in order to arrive at an equation quadratic in form. For example, t 2/3 1 2t1/3 1 1 5 0 is an original equa-tion given in the above table. If we rewrite this equation as 1t1/3 2 2 1 2 1t1/3 2 1 1 5 0, then it becomes apparent that the correct substitution is u 5 t1/3, which transforms the equation in t into a quadratic equation in u: u2 1 2u 1 1 5 0. ▼ A N S W E R The solution is x 5 21 2 or x 5 1 3. The solution set is U21 2, 1 3V. EXAMPLE 5 Solving an Equation Quadratic in Form with Negative Exponents Find the solutions to the equation x22 2 x21 2 12 5 0. Solution: Rewrite the original equation. 1x2122 2 1x212 2 12 5 0 Determine the u-substitution. u 5 x21 3Note that u2 5 x22.4 The original equation in x corresponds to a quadratic equation in u. u2 2 u 2 12 5 0 Factor. 1u 2 42 1u 1 32 5 0 Solve for u. u 5 4 or u 5 23 The most common mistake is forgetting to transform back to x. Transform back to x. Let u 5 x21. x21 5 4 or x21 5 23 Write x21 as 1 x. 1 x 5 4 or 1 x 5 23 Solve for x. x 5 1 4 or x 5 21 3 The solution set is U21 3, 1 4V. Y OUR TU R N Find the solutions to the equation x22 2 x21 2 6 5 0. ▼ PROCEDURE FOR SOLVING EQUATIONS QUADRATIC IN FORM Step 1: Identify the substitution. Step 2: Transform the equation into a quadratic equation. Step 3: Solve the quadratic equation. Step 4: Apply the substitution to rewrite the solution in terms of the original variable. Step 5: Solve the resulting equation. Step 6: Check the solutions in the original equation. 1.4 Other Types of Equations 125 1.4.3 Factorable Equations Some equations (both polynomial and with rational exponents) that are factorable can be solved using the zero product property. EXAMPLE 6 Solving an Equation Quadratic in Form with Fractional Exponents Find the solutions to the equation x2/3 2 3x1/3 2 10 5 0. Solution: Rewrite the original equation. 1x1/322 2 3x1/3 2 10 5 0 Identify the substitution as u 5 x1/3. u2 2 3u 2 10 5 0 Factor. 1u 2 52 1u 1 22 5 0 Solve for u. u 5 5 or u 5 22 Let u 5 x1/3 again. x1/3 5 5 x1/3 5 22 Cube both sides of the equations. 1x1/323 5 1523 1x1/323 5 12223 Simplify. x 5 125 x 5 28 The solution set is 528, 1256 , which a check will confirm. Y OUR T UR N Find the solution to the equation 2t 2 5t1/2 2 3 5 0. ▼ STUDY TIP Remember to transform back to the original variable. ▼ A N S W E R t 5 9 or 596. [CONCEPT CHECK] What do we let u be equal to in order to transform the equation 2t 2 5t1/2 2 3 5 0 into a quadratic equation? ANSWER u 5 t1/2 ▼ EXAMPLE 7 Solving an Equation with Rational Exponents by Factoring Solve the equation x7/3 2 3x4/3 2 4x1/3 5 0. Solution: Factor the left side of the equation. x1/31x2 2 3x 2 42 5 0 Factor the quadratic expression. x1/31x 2 42 1x 1 12 5 0 Apply the zero product property. x1/3 5 0 or x 2 4 5 0 or x 1 1 5 0 Solve for x. x 5 0 or x 5 4 or x 5 21 The solution set is 521, 0, 46. 1.4.3 S K I L L Solve equations that are factorable. 1.4.3 C ON C E P T U A L Recognize when a polynomial equation or an equation with rational exponents can be factored either by grouping or by first factoring out a greatest common factor. EXAMPLE 8 Solving a Polynomial Equation Using Factoring by Grouping Solve the equation x3 1 2x2 2 x 2 2 5 0. Solution: Factor by grouping (Chapter 0). 1x3 2 x2 1 12x2 2 22 5 0 Identify the common factors. x1x2 2 12 1 21x2 2 12 5 0 Factor. 1x 1 221x2 2 12 5 0 Factor the quadratic expression. 1x 1 22 1x 2 12 1x 1 12 5 0 [CONCEPT CHECK] How do we initially group the cubic equation in x3 1 x2 2 4x 2 4 5 0? ANSWER 1x3 2 4x2 1 1x2 2 42 5 0 ▼ 126 CHAPTER 1 Equations and Inequalities Apply the zero product property. x 1 2 5 0 or x 2 1 5 0 or x 1 1 5 0 Solve for x. x 5 22 or x 5 1 or x 5 21 The solution set is 522, 21, 16. Y OUR TU R N Solve the equation x3 1 x2 2 4x 2 4 5 0. ▼ ▼ A N S W E R x 5 21 or x 5 62 or 522, 21, 26 [SEC TION 1.4] E X E R C I SE S • S K I L L S In Exercises 1–40, solve the radical equation for the given variable. 1. "t 2 5 5 2 2. "2t 2 7 5 3 3. 14p 2 721/2 5 5 4. 11 5 121 2 p21/2 5. "u 1 1 5 24 6. 2"3 2 2u 5 9 7. 3 "5x 1 2 5 3 8. 3 "1 2 x 5 22 9. 14y 1 121/3 5 21 10. 15x 2 121/3 5 4 11. "12 1 x 5 x 12. x 5 "56 2 x 13. y 5 5!y 14. !y 5 y 4 15. s 5 3"s 2 2 16. 22s 5 "3 2 s 17. "2x 1 6 5 x 1 3 18. "8 2 2x 5 2x 2 2 19. "1 2 3x 5 x 1 1 20. "2 2 x 5 x 2 2 21. 3x 2 6!x 2 1 5 3 22. 5x 2 10!x 1 2 5 210 23. 3x 2 6!x 1 2 5 3 24. 2x 2 4!x 1 1 5 4 25. 3!x 1 4 2 2x 5 9 26. 2!x 1 1 2 3x 5 25 27. "x2 2 4 5 x 2 1 28. "25 2 x2 5 x 1 1 29. "x2 2 2x 2 5 5 x 1 1 30. "2x2 2 8x 1 1 5 x 2 3 31. !3x 1 1 2 !6x 2 5 5 1 32. !2 2 x 1 !6 2 5x 5 6 33. !x 1 12 1 !8 2 x 5 6 34. !5 2 x 1 !3x 1 1 5 4 35. "2x 2 1 2 "x 2 1 5 1 36. "8 2 x 5 2 1 "2x 1 3 37. "3x 2 5 5 7 2 "x 1 2 38. "x 1 5 5 1 1 "x 2 2 39. #2 1 "x 5 "x 40. #2 2 "x 5 "x In Exercises 41–70, solve the equations by introducing a substitution that transforms these equations to quadratic form. 41. x2/3 1 2x1/3 5 0 42. x1/2 2 2x1/4 5 0 43. x4 2 3x2 1 2 5 0 44. x4 2 8x2 1 16 5 0 45. 2x4 1 7x2 1 6 5 0 46. x8 2 17x4 1 16 5 0 47. 12x 1 12 2 1 5 12x 1 12 1 4 5 0 48. 1x 2 322 1 6 1x 2 32 1 8 5 0 49. 4 1t 2 122 2 9 1t 2 12 5 22 50. 211 2 y22 1 5 11 2 y2 2 12 5 0 51. x28 2 17x24 1 16 5 0 52. 2u22 1 5u21 2 12 5 0 53. 3y22 1 y21 2 4 5 0 54. 5a22 1 11a21 1 2 5 0 55. z2/5 2 2z1/5 1 1 5 0 56. 2x1/2 1 x1/4 2 1 5 0 57. 1x 1 32 5/3 5 32 58. 1x 1 224/3 5 16 59. 1x 1 12 2/3 5 4 60. 1x 2 72 4/3 5 81 61. 6t22/3 2 t21/3 2 1 5 0 ■ Equations Quadratic in Form: Identify the u-substitution that transforms the equation into a quadratic equa tion. Solve the quadratic equation and then remember to transform back to the original variable. ■ Factorable Equations: Look for a factor common to all terms or factor by grouping. Radical equations, equations quadratic in form, and factorable equations can often be solved by transforming them into simpler linear or quadratic equations. ■ Radical Equations: Isolate the term containing a radical and raise it to the appropriate power that will eliminate the radi-cal. If there is more than one radical, it does not matter which radical is isolated first. Raising radical equations to powers may cause extraneous solutions, so check each solution. [SEC TION 1.4] S U M M A RY 1.4 Other Types of Equations 127 62. t22/3 2 t21/3 2 6 5 0 63. 3 5 1 1x 1 122 1 2 1x 1 12 64. 1 1x 1 122 1 4 1x 1 12 1 4 5 0 65. a 1 2x 2 1b 2 1 a 1 2x 2 1b 2 12 5 0 66. 5 12x 1 122 2 3 12x 1 12 5 2 67. u4/3 2 5u2/3 5 24 68. u4/3 1 5u2/3 5 24 69. t 5 4 "t2 1 6 70. u 5 4 "22u2 2 1 In Exercises 71–86, solve by factoring. 71. x3 2 x2 2 12x 5 0 72. 2y3 2 11y2 1 12y 5 0 73. 4p3 2 9p 5 0 74. 25x3 5 4x 75. u5 2 16u 5 0 76. t 5 2 81t 5 0 77. x3 2 5x2 2 9x 1 45 5 0 78. 2p3 2 3p2 2 8p 1 12 5 0 79. y 1 y 2 523 2 141 y 2 522 5 0 80. v 1v 1 323 2 40 1v 1 322 5 0 81. x9/4 2 2x5/4 2 3x1/4 5 0 82. u7/3 1 u4/3 2 20u1/3 5 0 83. t 5/3 2 25t21/3 5 0 84. 4x9/5 2 9x21/5 5 0 85. y3/2 2 5y1/2 1 6y21/2 5 0 86. 4p5/3 2 5p2/3 2 6p21/3 5 0 • A P P L I C A T I O N S In Exercises 87 and 88 refer to the following: An analysis of sales indicates that demand for a product during a calendar year is modeled by d 5 3!t 1 1 2 0.75t where d is demand in millions of units and t is the month of the year where t 5 0 represents January. 87. Economics. During which month(s) is demand 3 million units? 88. Economics. During which month(s) is demand 4 million units? In Exercises 89 and 90 refer to the following: Body Surface Area (BSA) is used in physiology and medicine for many clinical purposes. BSA can be modeled by the function BSA 5 Å wh 3600 where w is weight in kilograms and h is height in centimeters. 89. Health. The BSA of a 72-kilogram female is 1.8. Find the height of the female to the nearest centimeter. 90. Health. The BSA of a 177-centimeter-tall male is 2.1. Find the weight of the male to the nearest kilogram. 91. Insurance: Health. Cost for health insurance with a private policy is given by C 5 !10 1 a, where C is the cost per day and a is the insured’s age in years. Health insurance for a 6-year-old, a 5 6, is $4 a day (or $1460 per year). At what age would someone be paying $9 a day (or $3285 per year)? 92. Insurance: Life. Cost for life insurance is given by C 5 !5a 1 1, where C is the cost per day and a is the insured’s age in years. Life insurance for a newborn, a 5 0, is $1 a day (or $365 per year). At what age would someone be paying $20 a day (or $7300 per year)? 93. Stock Value. The stock price of a certain pharmaceutical company from August to November can be approximately modeled by the equation P 5 5"t2 1 1 1 50, where P is the price of the stock in dollars and t is the month with t 5 0 corresponding to August. Assuming this trend continues, when would the stock be worth $85? Stock Price Sept Oct 40 50 60 70 94. Grades. The average combined math and verbal SAT score of incoming freshmen at a university is given by the equation S 5 1000 110!2t, where t is in years and t 5 0 corresponds to 1990. What year will the incoming class have an average SAT score of 1230? 95. Speed of Sound. A man buys a house with an old well but does not know how deep the well is. To get an estimate he decides to drop a rock in the opening of the well and time how long it takes until he hears the splash. The total elapsed time T given by T 5 t1 1 t2, is the sum of the time it takes for the rock to reach the water, t1, and the time it takes for the sound of the splash to travel to the top of the well, t2. The time (seconds) that it takes for the rock to reach the water is given by t1 5 "d 4 , where d is the depth of the well in feet. Since the speed of sound is 1100 ft/s, the time (seconds) it takes for the sound to reach the top of the well is t2 5 d 1100. If the splash is heard after 3 seconds, how deep is the well? 96. Speed of Sound. If the owner of the house in Exercise 95 forgot to account for the speed of sound, what would he have calculated the depth of the well to be? 97. Physics: Pendulum. The period (T ) of a pendulum is related to the length (L) of the pendulum and acceleration due to gravity (g) by the formula T 5 2pÅ L g. If gravity is 9.8 m/s2 and the period is 1 second, find the approximate length of the pendulum. Round to the nearest centimeter. Note: 100 cm 5 1 m. 98. Physics: Pendulum. The period (T ) of a pendulum is related to the length (L) of the pendulum and acceleration due to gravity (g) by the formula T 5 2pÅ L g. If gravity is 32 ft/s2 and the period is 1 second, find the approximate length of the pendulum. Round to the nearest inch. Note: 12 in. 5 1 ft. In Exercises 99 and 100, refer to the following: Einstein’s special theory of relativity states that time is relative: Time speeds up or slows down, depending on how fast one object is moving with respect to another. For example, a space probe traveling at a velocity v near the speed of light c will have “clocked” a time 128 CHAPTER 1 Equations and Inequalities t hours, but for a stationary observer on Earth that corresponds to a time t0. The formula governing this relativity is given by t 5 t0Å1 2 v2 c2 99. Physics: Special Theory of Relativity. If the time elapsed on a space probe mission is 18 years but the time elapsed on Earth during that mission is 30 years, how fast is the space probe traveling? Give your answer relative to the speed of light. 100. Physics: Special Theory of Relativity. If the time elapsed on a space probe mission is 5 years but the time elapsed on Earth during that mission is 30 years, how fast is the space probe traveling? Give your answer relative to the speed of light. • C A T C H T H E M I S T A K E In Exercises 101–104, explain the mistake that is made. 101. Solve the equation !3t 1 1 5 24. Solution: 3t 1 1 5 16 3t 5 15 t 5 5 This is incorrect. What mistake was made? 102. Solve the equation x 5 !x 1 2. Solution: x2 5 x 1 2 x2 2 x 2 2 5 0 1x 2 221x 1 12 5 0 x 5 21, x 5 2 This is incorrect. What mistake was made? 103. Solve the equation x2/3 2 x1/3 2 20 5 0. Solution: u 5 x1/3 u2 2 u 2 20 5 0 1u 2 52 1u 1 42 5 0 x 5 5, x 5 24 This is incorrect. What mistake was made? 104. Solve the equation x4 2 2x2 5 3. Solution: x4 2 2x2 2 3 5 0 u 5 x2 u2 2 2u 2 3 5 0 1u 2 32 1u 1 12 5 0 u 5 21, u 5 3 u 5 x2 x2 5 21, x2 5 3 x 5 61, x 5 63 This is incorrect. What mistake was made? In Exercises 105–108, determine whether each statement is true or false. 105. The equation 12x 2 126 1 412x 2 123 1 3 5 0 is quadratic in form. 106. The equation t25 1 2t5 1 1 5 0 is quadratic in form. 107. If two solutions are found and one does not check, then the other does not check. 108. Squaring both sides of !x 1 2 1 !x 5 !x 1 5 leads to x 1 2 1 x 5 x 1 5. • C O N C E P T U A L 109. Solve "x2 5 x. 110. Solve "x2 5 2x. 111. Solve the equation 3x2 1 2x 5 "3x2 1 2x without squar-ing both sides. 112. Solve the equation 3x7/12 2 x5/6 2 2x1/3 5 0. 113. Solve the equation !x 1 6 1 !11 1 x 5 5!3 1 x. 114. Solve the equation # 4 2x" 3 x!x 5 2. • C H A L L E N G E 115. Solve the equation !x 2 3 5 4 2 !x 1 2. Plot both sides of the equation in the same viewing screen, y1 5 !x 2 3 and y2 5 4 2 !x 1 2, and zoom in on the x-coordinate of the point of intersection. Does the graph agree with your solution? 116. Solve the equation 2!x 1 1 5 1 1 !3 2 x. Plot both sides of the equation in the same viewing screen, y1 5 2!x 1 1 and y2 5 1 1 !3 2 x, and zoom in on the x-coordinate of the points of intersection. Does the graph agree with your solution? 117. Solve the equation 24 5 !x 1 3. Plot both sides of the equation in the same viewing screen, y1 5 24 and y2 5 !x 1 3. Does the graph agree or disagree with your solution? 118. Solve the equation x1/4 5 24x1/2 1 21. Plot both sides of the equation in the same viewing screen, y1 5 x1/4 and y2 5 24x1/2 1 21. Does the point(s) of intersection agree with your solution? 119. Solve the equation x 1/2 5 24x1/4 1 21. Plot both sides of the equation in the same viewing screen, y1 5 x1/2 and y2 5 24x1/4 1 21. Does the point(s) of intersection agree with your solution? 120. Solve the equation x21 5 3x22 2 10. Plot both sides of the equation in the same viewing screen, y1 5 x21 and y2 5 3x22 2 10. Does the point(s) of intersection agree with your solution? 121. Solve the equation x22 5 3x21 2 10. Plot both sides of the equation in the same viewing screen, y1 5 x22 and y2 5 3x21 2 10. Does the point(s) of intersection agree with your solution? • T E C H N O L O G Y 1.5.1 Graphing Inequalities and Interval Notation An example of a linear equation is 3x 2 2 5 7, whereas 3x 2 2 # 7 is an example of a linear inequality. One difference between a linear equation and a linear inequality is that the equation has at most only one solution, or value of x, that makes the statement true, whereas the inequality can have a range or continuum of numbers that make the statement true. For example, the inequality x # 4 denotes all real numbers x that are less than or equal to 4. Four inequality symbols are used. SYMBOL IN WORDS , Less than . Greater than # Less than or equal to $ Greater than or equal to We call , and . strict inequalities. For any two real numbers a and b, one of three things must be true: a , b or a 5 b or a . b This property is called the trichotomy property of real numbers. If x is less than 5 1x , 52 and x is greater than or equal to 22 1x $ 222, then we can represent this as a double (or combined) inequality, 22 # x , 5, which means that x is greater than or equal to 22 and less than 5. We will express solutions to inequalities in four ways: an inequality, a solution set, an interval, and a graph. The following are ways of expressing all real numbers greater than or equal to a and less than b. 1.5.1 S K I L L Use interval notation. 1.5.1 C O N C E P T U A L Apply intersection and union concepts. S K I L L S O B J E C T I V E S ■ ■Use interval notation. ■ ■Solve linear inequalities in one variable. C O N C E P T U A L O B J E C T I V ES ■ ■Apply intersection and union concepts. ■ ■Understand that a linear inequality in one variable has an interval solution. 1.5 LINEAR INEQUALITIES Inequality Notation a # x , b Solution Set 5x a # x , b6 Interval Notation 3a, b2 In this example, a is referred to as the left endpoint and b is referred to as the right endpoint. If an inequality is a strict inequality 1, or .2, then the graph and interval notation use parentheses. If it includes an endpoint 1$ or #2, then the graph and interval notation use brackets. Number lines are drawn with either closed/open circles or brackets/parentheses. In this text, the brackets/parentheses notation will be used. Intervals are classified as follows: Open 1 , 2 Closed 3 , 4 Half open 1 , 4 or 3 , 2 LET x BE A REAL NUMBER. x IS... INEQUALITY SET NOTATION INTERVAL GRAPH greater than a and less than b a , x , b 5x a , x , b6 1a, b2 greater than or equal to a and less than b a # x , b 5x a # x , b6 3a, b2 1.5 Linear Inequalities 129 130 CHAPTER 1 Equations and Inequalities 1. Infinity 1q2 is not a number. It is a symbol that means continuing indefinitely to the right on the number line. Similarly, negative infinity 12q2 means continuing indefinitely to the left on the number line. Since both are unbounded, we use a parenthesis, never a bracket. 2. In interval notation, the lower number is always written to the left. Write the inequality in interval notation: 21 # x , 3. ✓COR R E C T ✖I NC O R R EC T 321, 32 13, 214 1 3 LET x BE A REAL NUMBER. x IS... INEQUALITY SET NOTATION INTERVAL GRAPH greater than a and less than or equal to b a , x # b 5x a , x # b6 1a, b4 greater than or equal to a and less than or equal to b a # x # b 5x a # x # b6 3a, b4 less than a x , a 5x x , a6 12q, a2 less than or equal to a x # a 5x x # a6 12q, a4 greater than b x . b 5x x . b6 1b, q2 greater than or equal to b x $ b 5x x $ b6 3b, q2 all real numbers R R 12q, q2 Since the solutions to inequalities are sets of real numbers, it is useful to discuss two operations on sets called intersection and union. EXAMPLE 1 Expressing Inequalities Using Interval Notation and a Graph Express the following as an inequality, an interval, and a graph. a. x is greater than 23. b. x is less than or equal to 5. c. x is greater than or equal to 21 and less than 4. d. x is greater than or equal to 0 and less than or equal to 4. Solution: Inequality Interval Graph a. x . 23 123, q2 b. x # 5 12q, 54 c. 21 # x , 4 321, 42 d. 0 # x # 4 30, 44 1.5 Linear Inequalities 131 As an example of intersection and union, consider the following sets of people: A 5 5Austin, Brittany, Jonathan6 B 5 5Anthony, Brittany, Elise6 Intersection: A ∩ B 5 5Brittany6 Union: A ∪ B 5 5Anthony, Austin, Brittany, Elise, Jonathan6 DEFINITION Union and Intersection The union of sets A and B, denoted A ∪ B, is the set formed by combining all the elements in A with all the elements in B. A ∪ B 5 5x 0 x is in A or B or both6 The intersection of sets A and B, denoted A ∩ B, is the set formed by the elements that are in both A and B. A ∩ B 5 5x 0 x is in A and B6 The notation “x x is in” is read “all x such that x is in.” The vertical line represents “such that.” EXAMPLE 2 Determining Unions and Intersections: Intervals and Graphs If A 5 323, 24 and B 5 11, 72, determine A ∪ B and A ∩ B. Write these sets in inter-val notation, and graph. Solution: YOUR T UR N If C 5 323, 32 and D 5 10, 54, find C ∪ D and C ∩ D. Express the intersection and union in interval notation, and graph. ▼ A N S W E R ▼ Set Interval notation Graph A 323, 24 B 11, 72 A ∪ B 323, 72 A ∩ B 11, 24 1.5.2 Solving Linear Inequalities As mentioned at the beginning of this section, if we were to solve the equation 3x 2 2 5 7, we would add 2 to both sides, divide by 3, and find that x 5 3 is the solution, the only value that makes the equation true. If we were to solve the linear inequality 3x 2 2 # 7, we would follow the same procedure: add 2 to both sides, divide by 3, and find that x # 3, which is an interval or range of numbers that make the inequality true. In solving linear inequalities, we follow the same procedures that we used in solv-ing linear equations with one general exception: if you multiply or divide an inequality by a negative number, then you must change the direction of the inequality sign. For example, if 22 x , 210, then the solution set includes real numbers such as x 5 6 and x 5 7. Note that real numbers such as x 5 26 and x 5 27 are not included in the solution set. Therefore, when this inequality is divided by 22, the inequality sign must also be reversed: x . 5. If a , b, then ac , bc if c . 0 and ac . bc if c , 0. [CONCEPT CHECK] If A is the set of all of the students who are enrolled in a math class and B is the set of all students who are enrolled in a history class, then which set is larger? (A) the intersection of A and B or (B) the union of A and B? Assume that the two classes do not contain exactly the same students. ANSWER (B) the set of students who are enrolled in either a math class or a history class. (A) is the smaller set because it is all of the students who are enrolled in BOTH math and history. ▼ 1.5.2 S K IL L Solve linear inequalities in one variable. 1.5.2 C ON C E P T U A L Understand that a linear inequal-ity in one variable has an interval solution. 132 CHAPTER 1 Equations and Inequalities The most common mistake that occurs when solving an inequality is forgetting to change the direction of, or reverse, the inequality symbol when the inequality is multiplied or divided by a negative number. EXAMPLE 3 Solving a Linear Inequality Solve and graph the inequality 5 2 3x , 23. Solution: Write the original inequality. 5 2 3x , 23 Subtract 5 from both sides. 23x , 18 Divide both sides by 23 and reverse the inequality sign. Simplify. x . 26 Solution set: 5x \| x . 266 Interval notation: 126, q2 Graph: Y OUR TU R N Solve the inequality 5 # 3 2 2x. Express the solution in set and interval notation, and graph. 23x 23 . 18 23 ▼ A N S W E R Solution set: 5x \| x # 216 Interval notation: 12q, 214 ▼ A common mistake is using cross multiplication to solve inequalities. Cross multiplication should not be used because the expression by which you are multiplying might be negative for some values of x, and that would require the direction of the inequality sign to be reversed. EXAMPLE 4 Solving Linear Inequalities with Fractions Solve the inequality 5x 3 # 4 1 3x 2 . common mistake STUDY TIP If you multiply or divide an inequality by a negative number, remember to change the direction of the inequality sign. INEQUALITY PROPERTIES Procedures That Do Not Change the Inequality Sign 1. Simplifying by eliminating parentheses 31x 2 62 , 6x 2 x and collecting like terms. 3x 2 18 , 5x 2. Adding or subtracting the same 7x 1 8 $ 29 quantity on both sides. 7x $ 21 3. Multiplying or dividing by the 5x # 15 same positive real number. x # 3 Procedures That Change (Reverse) the Inequality Sign 1. Interchanging the two sides of the inequality. 2. Multiplying or dividing by the same negative real number. x # 4 is equivalent to 4 $ x. 25x # 15 is equivalent to x $ 23. 1.5 Linear Inequalities 133 ✖I N C O R R EC T Cross multiply. 314 1 3x2 # 215x2 The error is in cross multiplying. ✓COR R E C T Eliminate the fractions by multiplying by the LCD, 6. 6 a5x 3 b # 6 a4 1 3x 2 b Simplify. 10x # 314 1 3x2 Eliminate the parentheses. 10x # 12 1 9x Subtract 9x from both sides. x # 12 ▼ C A U T I O N Cross multiplication should not be used in solving inequalities. Although it is not possible to “check” inequalities since the solutions are often inter-vals, it is possible to confirm that some points that lie in your solution do satisfy the inequality. It is important to remember that cross multiplication cannot be used in solving inequalities. For the remainder of this section we will use the shortcut method for solving inequalities. EXAMPLE 5 Solving a Double Linear Inequality Solve the inequality 22 , 3x 1 4 # 16. Solution: This double inequality can be written as two inequalities. 22 , 3x 1 4 # 16 Both inequalities must be satisfied. 22 3x 1 4 and 3x 1 4 " 16 Subtract 4 from both sides of each inequality. 26 3x and 3x " 12 Divide each inequality by 3. 22 x and x " 4 Combining these two inequalities gives us 22 x " 4 in inequality notation; in interval notation we have 122, H2 x 12H, 44 or 122, 44. Notice that the steps we took in solving these inequalities individually were identical. This leads us to a shortcut method in which we solve them together: Write the combined inequality. 22 , 3x 1 4 # 16 Subtract 4 from each part. 26 , 3x # 12 Divide each part by 3. 22 , x # 4 Interval notation: 122, 44 f f 134 CHAPTER 1 Equations and Inequalities EXAMPLE 6 Solving a Double Linear Inequality Solve the inequality 1 # 22 2 3x 7 , 4. Express the solution set in interval notation, and graph. Solution: Write the original double inequality. 1 # 22 2 3x 7 , 4 Multiply each part by 7. 7 # 22 2 3x , 28 Add 2 to each part. 9 # 23x , 30 Divide each part by 23 and reverse the inequality signs. 23 $ x . 210 Write in standard form. 210 , x # 23 Interval notation: 1210, 234 Graph: EXAMPLE 7 Solving a Double Linear Inequality Solve the inequality x 2 1 # 4x 2 4 # x 1 8. Express the solution in interval notation. Solution: Subtract x from all three parts. 21 # 3x 2 4 # 8 Add 4 to all three parts. 3 # 3x # 12 Divide all three parts by 3. 1 # x # 4 Express the solution in interval notation. 31, 44 YOUR TURN Solve the inequality 2x 1 1 , 4x 1 2 , 2x 1 5. Express the solution in interval notation. ▼ A N S W E R A21 2, 3 2B ▼ Applications Involving Linear Inequalities EXAMPLE 8 Temperature Ranges New York City on average has a yearly temperature range of 23 degrees Fahrenheit to 86 degrees Fahrenheit. What is the range in degrees Celsius given that the conversion relation is F 5 32 1 9 5C? Solution: The temperature ranges from 238F to 868F. 23 # F # 86 Replace F using the Celsius conversion. 23 # 32 1 9 5C # 86 Subtract 32 from all three parts. 29 # 9 5C # 54 Multiply all three parts by 5 9. 25 # C # 30 New York City has an average yearly temperature range of 258C to 308C . [CONCEPT CHECK] True or False: When expressing the solution to a linear equation or a linear inequality in one variable on a number line, a linear equation in one variable has a solution that is a single point, whereas a linear inequality in one variable has a solution that corresponds to a range or interval. ANSWER True ▼ 1.5 Linear Inequalities 135 EXAMPLE 9 Comparative Shopping Two car rental companies have advertised weekly specials on full-size cars. Hertz is advertising an $80 rental fee plus an additional $0.10 per mile. Thrifty is advertising $60 and $0.20 per mile. How many miles must you drive for the rental car from Hertz to be the better deal? Solution: Let x 5 number of miles driven during the week. Write the cost for the Hertz rental. 80 1 0.1x Write the cost for the Thrifty rental. 60 1 0.2x Write the inequality if Hertz is less than Thrifty. 80 1 0.1x , 60 1 0.2x Subtract 0.1x from both sides. 80 , 60 1 0.1x Subtract 60 from both sides. 20 , 0.1x Divide both sides by 0.1. 200 , x You must drive more than 200 miles for Hertz to be the better deal. In Exercises 1–16, rewrite in interval notation and graph. 1. x $ 3 2. x , 22 3. x # 25 4. x . 27 5. 22 # x , 3 6. 24 # x # 21 7. 23 , x # 5 8. 0 , x , 6 9. 0 # x # 0 10. 27 # x # 27 11. x # 6 and x $ 4 12. x . 23 and x # 2 13. x # 26 and x $ 28 14. x , 8 and x , 2 15. x . 4 and x # 22 16. x $ 25 and x , 26 In Exercises 17–24, rewrite in set notation. 17. 30, 22 18. 10, 34 19. 127, 222 20. 323, 24 21. 12q, 64 22. 15, q2 23. 12q, q2 24. 34, 44 In Exercises 25–32, write in inequality and interval notation. 25. 26. 27. 28. 0 5 10 10 5 0 1 1 5 3 4 8 [SEC TION 1.5] E X ERC I S E S • S K I L L S Linear inequalities are solved using the same procedures as linear equations with one exception: When you multiply or divide by a negative number, you must reverse the inequality sign. Note: Cross multiplication cannot be used with inequalities. The solutions of linear inequalities are solution sets that can be expressed four ways: 1. Inequality notation a , x # b 2. Set notation 5x \| a , x # b6 3. Interval notation 1a, b4 4. Graph (number line) [SEC TION 1.5] S U M M A RY 136 CHAPTER 1 Equations and Inequalities 29. 30. 31. 32. In Exercises 33–50, graph the indicated set and write as a single interval, if possible. 33. 125, 24 ∪ 121, 32 34. 12, 72 ∪ 325, 32 35. 326, 42 ∪ 322, 52 36. 323, 12 ∪ 326, 02 37. 12q, 14 ∩ 321, q2 38. 12q, 252 ∩ 12q, 74 39. 12q, 42 ∩ 31, q2 40. 123, q2 ∩ 325, q2 41. 325, 22 ∩ 321, 34 42. 324, 52 ∩ 322, 72 43. 12q, 42 ∪ 14, q2 44. 12q, 234 ∪ 323, q2 45. 12q, 234 ∪ 33, q2 46. 122, 22 ∩ 323, 14 47. 12q, q2 ∩ 123, 24 48. 12q, q2 ∪ 124, 72 49. 126, 222 ∩ 31, 42 50. 12q, 222 ∩ 121, q2 In Exercises 51–58, write in interval notation. 51. 52. 53. 54. 55. 56. 57. 58. In Exercises 59–90, solve and express the solution in interval notation. 59. x 2 3 , 7 60. x 1 4 . 9 61. 3x 2 2 # 4 62. 3x 1 7 $ 28 63. 25p $ 10 64. 24u , 12 65. 3 2 2x # 7 66. 4 2 3x . 217 67. 21.8x 1 2.5 . 3.4 68. 2.7x 2 1.3 , 6.8 69. 3 1t 1 12 . 2t 70. 2 1 y 1 52 # 3 1 y 2 42 71. 7 2 211 2 x2 . 5 1 31x 2 22 72. 4 2 312 1 x2 , 5 73. x 1 2 3 2 2 $ x 2 74. y 2 3 5 2 2 # y 4 75. t 2 5 3 # 24 76. 2p 1 1 5 . 23 77. 2 3 y 2 1 2 15 2 y2 ,5y 3 2 12 1y2 78. s 2 2 1s 2 32 3 . s 4 2 1 12 79. 22 , x 1 3 , 5 80. 1 , x 1 6 , 12 81. 28 # 4 1 2x , 8 82. 0 , 2 1 x # 5 83. 23 , 1 2 x # 9 84. 3 # 22 2 5x # 13 85. 0 , 2 2 1 3y , 4 86. 3 , 1 2A 2 3 , 7 87. 1 2 # 1 1 y 3 # 3 4 88. 21 , 2 2 z 4 # 1 5 89. 20.7 # 0.4x 1 1.1 # 1.3 90. 7.1 . 4.7 2 1.2x . 1.1 5 5 0 10 5 0 5 10 8 2 5 3 2 5 0 2 4 2 5 12 5 2 4 2 3 7 2 0 2 6 3 0 4 5 1 • A P P L I C A T I O N S 91. Weight. A healthy weight range for a woman is given by the following formula: ■ 110 pounds for the first 5 feet (tall) ■ 2–6 pounds per inch for every inch above 5 feet Write an inequality representing a healthy weight, w, for a 5 foot 9 inch woman. 92. Weight. NASA has more stringent weight allowances for its astronauts. Write an inequality representing allowable weight for a female 5 foot 9 inch mission specialist given 105 pounds for the first 5 feet, and 1–5 pounds per inch for every addi-tional inch. 93. Profit. A seamstress decides to open a dress shop. Her fixed costs are $4000 per month, and it costs her $20 to make each dress. If the price of each dress is $100, how many dresses does she have to sell per month to make a profit? 94. Profit. Labrador retrievers that compete in field trials typi-cally cost $2000 at birth. Professional trainers charge $400 to $1000 per month to train the dogs. If the dog is a champion by age 2, it sells for $30,000. What is the range of profit for a champion at age 2? 1.5 Linear Inequalities 137 In Exercises 95 and 96 refer to the following: The annual revenue for a small company is modeled by R 5 5000 1 1.75x where x is hundreds of units sold and R is revenue in thousands of dollars. 95. Business. Find the number of units (to the nearest 100) that must be sold to generate at least $10 million in revenue. 96. Business. Find the number of units (to the nearest 100) that must be sold to generate at least $7.5 million in revenue. In Exercises 97 and 98 refer to the following: The Target or Training Heart Rate (THR) is a range of heart rate (measured in beats per minute) that enables a person’s heart and lungs to benefit the most from an aerobic workout. THR can be modeled by the formula THR 5 1HRmax 2 HRrest2 3 I 1 HRrest where HRmax is the maximum heart rate that is deemed safe for the individual, HRrest is the resting heart rate, and I is the intensity of the workout that is reported as a percentage. 97. Health. A female with a resting heart rate of 65 beats per minute has a maximum safe heart rate of 170 beats per min-ute. If her target heart rate is between 100 and 140 beats per minute, what percent intensities of workout can she consider? 98. Health. A male with a resting heart rate of 75 beats per min-ute has a maximum safe heart rate of 175 beats per minute. If his target heart rate is between 110 and 150 beats per minute, what percent intensities of workout can he consider? 99. Cost: Cell Phones. A cell phone company charges $50 for an 800-minute monthly plan, plus an additional $0.22 per minute for every minute over 800. If a customer’s bill ranged from a low of $67.16 to a high of $96.86 over a 6-month period, what were the most minutes used in a single month? What were the least? 100. Cost: Internet. An Internet provider charges $30 per month for 1000 minutes of DSL service plus $0.08 for each additional minute. In a one-year period the customer’s bill ranged from $36.40 to $47.20. What were the most and least minutes used? 101. Grades. In your general biology class, your first three test scores are 67, 77, and 84. What is the lowest score you can get on the fourth test to earn at least a B for the course? Assume that each test is of equal weight and the minimum score required to earn a B is an 80. 102. Grades. In your Economics I class there are four tests and a final exam, all of which count equally. Your four test grades are 96, 87, 79, and 89. What grade on your final exam is needed to earn between 80 and 90 for the course? 103. Markups. Typical markup on new cars is 15–30%. If the sticker price is $27,999, write an inequality that gives the range of the invoice price (what the dealer paid the manufacturer for the car). 104. Markups. Repeat Exercise 103 with a sticker price of $42,599. 105. Lasers. A circular laser beam with a radius rT is transmitted from one tower to another tower. If the received beam radius rR fluctuates 10% from the transmitted beam radius due to atmospheric turbulence, write an inequality representing the received beam radius. 106. Electronics: Communications. Communication systems are often evaluated based on their signal-to-noise ratio (SNR), which is the ratio of the average power of received signal, S, to average power of noise, N, in the system. If the SNR is required to be at least 2 at all times, write an inequality representing the received signal power if the noise can fluctuate 10%. 107. Real Estate. The Aguileras are listing their house with a real estate agent. They are trying to determine a listing price, L, for the house. Their realtor advises them that most buyers traditionally offer a buying price, B, that is 85–95% of the listing price. Write an inequality that relates the buying price to the listing price. 108. Humidity. The National Oceanic and Atmospheric Admin-istration (NOAA) has stations on buoys in the oceans to measure atmosphere and ocean characteristics such as temperature, humidity, and wind. The humidity sensors have an error of 5%. Write an inequality relating the measured humidity hm, and the true humidity ht. 109. Recreation: Golf. Two friends enjoy playing golf. Their favorite course charges $40 for greens fees (to play the course) and a $15 cart rental (per person), so it currently costs each of them $55 every time they play. The membership offered at that course is $160 per month. The membership allows them to play as much as they want (no greens fees), but does still charge a cart rental fee of $10 every time they play. What is the least number of times they should play a month in order for the membership to be the better deal? 110. Recreation: Golf. The same friends in Exercise 109 have a second favorite course. That golf course charges $30 for greens fees (to play the course) and a $10 cart rental (per person), so it currently costs each of them $40 every time they play. The membership offered at that course is $125 per month. The membership allows them to play as much as they want (no greens fees), but does still charge a cart rental fee of $10. What is the least number of times they should play a month in order for the membership to be the better deal? TAX BRACKET # IF TAXABLE INCOME IS THE TAX IS: I $0 to $9,225 10% of amount over $0 II $9,226 to $37,450 $922.50 is 15% of the amount over $9,225 III $37,451 to $90,750 $5,156.25 plus 25% of the amount over $37,450 IV $90,751 to $189,300 $18,481.25 plus 28% of the amount over $90,750 V $189,301 to $411,500 $46,075.25 plus 33% of the amount over $189,300 VI $411,501 to $413,200 $119, 401.25 plus 35% of the amount over $411,500 VII $413,201 or more $119,996.25 plus 39.6% of the amount over $413,200 The following table is the 2015 Federal Tax Rate Schedule for people filing as single: 111. Federal Income Tax. What is the range of federal income taxes a person in tax bracket III will pay the IRS? 112. Federal Income Tax. What is the range of federal income taxes a person in tax bracket IV will pay the IRS? 138 CHAPTER 1 Equations and Inequalities In Exercises 113–116, explain the mistake that is made. • C A T C H T H E M I S T A K E 113. Rewrite in interval notation. 2 0 2 4 6 This is incorrect. What mistake was made? 114. Graph the indicated set and write as a single interval if possible. 2 0 2 4 6 This is incorrect. What mistake was made? 115. Solve the inequality 2 2 3p # 24 and express the solution in interval notation. Solution: This is incorrect. What mistake was made? 116. Solve the inequality 3 2 2x # 7 and express the solution in interval notation. Solution: This is incorrect. What mistake was made? 2 2 3p # 24 23p # 26 p # 2 12q, 24 3 2 2x # 7 22x # 4 x $ 22 12q, 224 129. a. Solve the inequality 2.7x 1 3.1 , 9.4x 22.5. b. Graph each side of the inequality in the same viewing screen. Find the range of x-values when the graph of the left side lies below the graph of the right side. c. Do (a) and (b) agree? 130. a. Solve the inequality 20.5x 1 2.7 . 4.1x 2 3.6. b. Graph each side of the inequality in the same viewing screen. Find the range of x-values when the graph of the left side lies above the graph of the right side. c. Do (a) and (b) agree? 131. a. Solve the inequality x 2 3 , 2x 2 1 , x 1 4. b. Graph all three expressions of the inequality in the same viewing screen. Find the range of x-values when the graph of the middle expression lies above the graph of the left side and below the graph of the right side. c. Do (a) and (b) agree? 132. a. Solve the inequality x 2 2 , 3x 1 4 # 2x 1 6. b. Graph all three expressions of the inequality in the same viewing screen. Find the range of x-values when the graph of the middle expression lies above the graph of the left side and on top of and below the graph of the right side. c. Do (a) and (b) agree? 133. a. Solve the inequality x 1 3 , x 1 5. b. Graph each side of the inequality in the same viewing screen. Find the range of x-values when the graph of the left side lies below the graph of the right side. c. Do (a) and (b) agree? 134. a. Solve the inequality 1 2 x 2 3 . 22 3 x 1 1. b. Graph each side of the inequality in the same viewing screen. Find the range of x-values when the graph of the left side lies above the graph of the right side. c. Do (a) and (b) agree? • C O N C E P T U A L In Exercises 117 and 118, determine whether each statement is true or false. 117. If x , a, then a . x. 118. If 2x $ a, then x $ 2 a. In Exercises 119–122, select any of the statements a 2 d that could be true. a. m . 0 and n . 0 b. m , 0 and n , 0 c. m . 0 and n , 0 d. m , 0 and n . 0 119. mn . 0 120. mn , 0 121. m n . 0 122. m n , 0 In Exercises 123 and 124, select any of the statements a 2 c that could be true. a. n 5 0 b. n . 0 c. n , 0 123. m 1 n , m 2 n 124. m 1 n $ m 2 n • C H A L L E N G E 125. Solve the inequality x # 2x mentally (without doing any algebraic manipulation). 126. Solve the inequality x . 2x mentally (without doing any algebraic manipulation). 127. Solve the inequality ax 1 b , ax 2 c, where 0 , b , c. 128. Solve the inequality 2ax 1 b , 2ax 1 c, where 0 , b , c. • T E C H N O L O G Y 1.6 Polynomial and Rational Inequalities 139 1.6.1 Polynomial Inequalities In this section we will focus primarily on quadratic inequalities, but the procedures outlined are also valid for higher degree polynomial inequalities. An example of a quadratic inequality is x2 1 x 2 2 , 0. This statement is true when the value of the polynomial on the left side is negative. For any value of x, a polynomial has a positive, negative, or zero value. A polynomial must pass through zero before its value changes from positive to negative or from negative to positive. Zeros of a polynomial are the values of x that make the polynomial equal to zero. These zeros divide the real number line into test intervals where the value of the polynomial is either positive or negative. For example, if we set the above polynomial equal to zero and solve: x2 1 x 2 2 5 0 1x 1 221x 2 12 5 0 x 5 22 or x 5 1 we find that x 5 22 and x 5 1 are the zeros. These zeros divide the real number line into three test intervals: 12q, 222, 122, 12, and 11, q2. Since the polynomial is equal to zero at x 5 22 and x 5 1, the value of the polynomial in each of these three intervals is either positive or negative. We select one real number that lies in each of the three intervals and test to see whether the value of the polynomial at each point is either positive or negative. In this example, we select the real numbers: x 5 23, x 5 0, and x 5 2. At this point, there are two ways we can determine whether the value of the polynomial is positive or negative on the interval. One approach is to substitute each of the test points into the polynomial x2 1 x 2 2. x 5 23 12322 1 1232 2 2 5 9 2 3 2 2 5 4 Positive x 5 0 1022 1 102 2 2 5 0 2 0 2 2 5 22 Negative x 5 2 1222 1 122 2 2 5 4 1 2 2 2 5 4 Positive The second approach is to simply determine the sign of the result as opposed to actually calculating the exact number. This alternate approach is often used when the expressions or test points get more complicated to evaluate. The polynomial is written as the product 1x 1 22 1x 2 12; therefore, we simply look for the sign in each set of parentheses. 1x 1 221x 2 12 x 523: 123 1 22 123 2 12 5 1212 1242 S 122 122 5 112 x 5 0: 10 1 22 10 2 12 5 122 1212 S 112 1225 122 x 52: 12 1 22 12 2 12 5 142 112 S 112 112 5 112 1.6.1 S K I L L Solve polynomial inequalities. 1.6.1 C ON C E P T U A L Understand zeros and test intervals. S K I L L S O B J E C T I V E S ■ ■Solve polynomial inequalities. ■ ■Solve rational inequalities. C O N C E P T U A L O B J E C T I V ES ■ ■Understand zeros and test intervals. ■ ■Realize that a rational inequality has an implied domain restriction on the variable. 1.6 POLYNOMIAL AND RATIONAL INEQUALITIES In this second approach we find the same result: 12q, 222 and 11, q2 correspond to a positive value of the polynomial, and 122, 12 corresponds to a negative value of the polynomial. y x x = 1 x = –2 x2 + x – 2 < 0 x2 + x – 2 > 0 x2 + x – 2 > 0 3 0 2 140 CHAPTER 1 Equations and Inequalities In this example, the statement x2 1 x 2 2 , 0 is true when the value of the polyno-mial (in factored form), 1x 1 22 1x 2 12, is negative. In the interval 122, 12, the value of the polynomial is negative. Thus, the solution to the inequality x2 1 x 2 2 , 0 is 122, 12. To check the solution, select any number in the interval and substitute it into the original inequality to make sure it makes the statement true. The value x 5 21 lies in the interval 122, 12. Upon substituting into the original inequality, we find that x 5 21 satisfies the inequality 1212 2 1 1212 2 2 5 22 , 0. STUDY TIP If the original polynomial is ,0, then the interval(s) that yield(s) negative products should be selected. If the original polynomial is .0, then the interval(s) that yield(s) positive products should be selected. Note: Be careful in Step 5. If the original polynomial is , 0, then the interval(s) that corr espond(s) to the value of the polynomial being negative should be selected. If the original polynomial is . 0, then the interval(s) that correspond(s) to the value of the polynomial being positive should be selected. EXAMPLE 1 Solving a Quadratic Inequality Solve the inequality x2 2 x . 12. Solution: STEP 1 Write the inequality in standard form. x2 2 x 2 12 . 0 Factor the left side. 1x 1 32 1x 2 42 . 0 STEP 2 Identify the zeros. 1x 1 32 1x 2 42 5 0 x 5 23 or x 5 4 STEP 3 Draw the number line with the zeros labeled. STEP 4 Determine the sign of 1x 1 32 1x 2 42 in each interval. STEP 5 Intervals in which the value of the polynomial is positive make this inequality true. 12, 232 or 14,2 STEP 6 Write the solution in interval notation. 12q, 232∪14, q2 YOUR TURN Solve the inequality x2 2 5x # 6 and express the solution in interval notation. ▼ 4 0 5 The inequality in Example 1, x2 2 x . 12, is a strict inequality, so we use paren-theses when we express the solution in interval notation 12q, 232 ∪ 14, q2. It is important to note that if we change the inequality sign from . to $, then the zeros x 5 23 and x 5 4 also make the inequality true. Therefore, the solution to x2 2 x $ 12 is 12q, 234 ∪ 34, q2. PROCEDURE FOR SOLVING POLYNOMIAL INEQUALITIES Step 1: Write the inequality in standard form. Step 2: Identify zeros. Step 3: Draw the number line with zeros labeled. Step 4: Determine the sign of the polynomial in each interval. Step 5: Identify which interval(s) make the inequality true. Step 6: Write the solution in interval notation. ▼ A N S W E R 321, 64 1.6 Polynomial and Rational Inequalities 141 ▼ C A U T I O N The square root method cannot be used for quadratic inequalities. Not all inequalities have a solution. For example, x2 , 0 has no real solution. Any real number squared is always nonnegative, so there are no real values that when squared will yield a negative number. The zero is x 5 0, which divides the real number line into two intervals: 12q, 02 and 10, q2. Both of these intervals, however, correspond to the value of x2 being positive, so there are no intervals that satisfy the inequality. We say that this inequality has no real solution. ✖I N C O R R EC T Error: Take the square root of both sides. x # 62 ✓COR R E CT STEP 1: Write the inequality in standard form. x2 2 4 # 0 Factor. 1x 2 221x 1 22 # 0 STEP 2: Identify the zeros. 1x 2 221x 1 22 5 0 x 5 2 and x 5 22 STEP 3: Draw the number line with the zeros labeled. –2 2 STEP 4: Determine the sign of 1x 2 221x 1 22, in each interval. –2 2 –3 0 3 (–)(–) = (+) (–)(+) = (–) (+)(+) = (+) STEP 5: Intervals in which the value of the polynomial is negative make the inequality true. 122, 22 The endpoints, x 5 22 and x 5 2, satisfy the inequality, so they are included in the solution. STEP 6: Write the solution in interval notation. 322, 24 When solving quadratic inequalities, you must first write the inequality in standard form and then factor to identify zeros. EXAMPLE 2 Solving a Quadratic Inequality Solve the inequality x2 # 4. common mistake Do not take the square root of both sides. You must write the inequality in standard form and factor. 142 CHAPTER 1 Equations and Inequalities STUDY TIP When solving quadratic inequalities, you must first write the inequality in standard form and then factor to identify zeros. EXAMPLE 3 Solving a Quadratic Inequality Solve the inequality x2 1 2x $ 23. Solution: STEP 1 Write the inequality in standard form. x2 1 2x 1 3 $ 0 STEP 2 Identify the zeros. x2 1 2x 1 3 5 0 Apply the quadratic formula. x 5 22 6 "22 2 4112132 2112 Simplify. x 5 22 6 !28 2 5 22 6 2i!2 2 5 21 6 i!2 Since there are no real zeros, the quadratic expression x2 1 2x 1 3 never equals zero; hence its value is either always positive or always negative. If we select any value for x, say, x 5 0, we find that 102 2 1 2 102 1 3 $ 0. Therefore, the quadratic expression is always positive, and so the solution is the set of all real numbers, 12q, q2 . ▼ C A U T I O N Do not divide inequalities by a variable. EXAMPLE 4 Solving a Quadratic Inequality Solve the inequality x2 . 25x. common mistake A common mistake is to divide by x. Never divide by a variable because the value of the variable might be zero. Always start by writing the inequality in standard form and then factor to determine the zeros. ✖I N C O R R EC T Write the original inequality. x2 . 25x ERROR: Divide both sides by x. x . 25 Dividing by x is the mistake. If x is negative, the inequality sign must be reversed. What if x is zero? ✓C O R R EC T STEP 1: Write the inequality in standard form. x2 2 5x . 0 Factor. x1x 1 52 . 0 STEP 2: Identify the zeros. x 5 0, x 5 25 STEP 3: Draw the number line with the zeros labeled. –5 0 STEP 4: Determine the sign of x 1x 1 52 in each interval. –5 0 –6 –1 1 (–)(–) = (+) (–)(+) = (–) (+)(+) = (+) STEP 5: Intervals in which the value of the polynomial is positive satisfy the inequality. 12H, 252 and 10, 2H2 STEP 6: Express the solution in interval notation. 12q, 252 ∪ 10, q2 1.6 Polynomial and Rational Inequalities 143 ▼ A N S W E R A2q, 1 2 !2 D ∪ C1 1 !2, qB EXAMPLE 5 Solving a Quadratic Inequality Solve the inequality x2 1 2x , 1. Solution: Write the inequality in standard form. x2 1 2x 2 1 , 0 Identify the zeros. x2 1 2x 2 1 5 0 Apply the quadratic formula. x 5 22 6 "22 2 41121212 2112 Simplify. x 5 22 6 "8 2 5 22 6 2"2 2 5 21 6 "2 Draw the number line with the intervals labeled. Note: 21 2 !2 < 22.41 21 1 !2 < 0.41 Test each interval. 12q, 21 2 !22 x 5 23: 12322 1 21232 2 1 5 2 . 0 121 2 !2, 21 1 !22 x 5 0: 1022 1 2102 2 1 5 21 , 0 121 1 !2, q2 x 5 1: 1122 1 2112 2 1 5 2 . 0 Intervals in which the value of the polynomial is negative make this inequality true. A21 2 !2, 21 1 !2B Y OUR T UR N Solve the inequality x2 2 2x $ 1. ▼ 2 ▼ A N S W E R 12q , 222 ∪ 10, 32 [CONCEPT CHECK] Solve: (x 2 a) (x 1 b) < 0, where a > 0 and b > 0. ANSWER (2b, a) ▼ EXAMPLE 6 Solving a Polynomial Inequality Solve the inequality x3 2 3x2 $ 10x. Solution: Write the inequality in standard form. x3 2 3x2 2 10x $ 0 Factor. x 1x 2 52 1x 1 22 $ 0 Identify the zeros. x 5 0, x 5 5, x 5 22 Draw the number line with the zeros (intervals) labeled. Test each interval. Intervals in which the value of the polynomial is positive make this inequality true. 322, 04 ∪ 35, q2 Y OUR T UR N Solve the inequality x3 2 x2 2 6x , 0. ▼ 144 CHAPTER 1 Equations and Inequalities 1.6.2 Rational Inequalities Rational expressions have numerators and denominators. Recalling the properties of negative real numbers (Chapter 0), we see that the following possible combinations correspond to either positive or negative rational expressions. 112 112 5 112 122 112 5 122 122 122 5 112 112 122 5 122 A rational expression can change signs if either the numerator or denominator changes signs. In order to go from positive to negative or vice versa, you must pass through zero. Therefore, to solve rational inequalities such as x 2 3 x2 2 4 $ 0 we use a similar procedure to the one used for solving polynomial inequalities, with one exception. You must eliminate values for x that make the denominator equal to zero. In this example, we must eliminate x 5 22 and x 5 2 because these values make the denominator equal to zero. Rational inequalities have implied domains. In this example, x 2 62 is a domain restriction and these values 1x 5 22 and x 5 22 must be eliminated from a possible solution. We will proceed with a similar procedure involving zeros and test intervals that was outlined for polynomial inequalities. However, in rational inequalities once expressions are combined into a single fraction, any values that make either the numerator or the denominator equal to zero divide the number line into intervals. 1.6.2 S KILLS Solve rational inequalities. 1.6.2 CO NCE PTUAL Realize that a rational inequality has an implied domain restriction on the variable. STUDY TIP Values that make the denomi-nator equal to zero are always excluded. ▼ A N S W E R 322, 12 EXAMPLE 7 Solving a Rational Inequality Solve the inequality x 2 3 x2 2 4 $ 0. Solution: Factor the denominator. 1x 2 32 1x 2 221x 1 22 $ 0 State the domain restrictions on the variable. x 2 2, x 2 22 Identify the zeros of numerator and denominator. x 5 22, x 5 2, x 5 3 Draw the number line and divide into intervals. Test the intervals. 3 0 2.5 4 Intervals in which the value of the rational 122, 22 and 13, H2 expression is positive satisfy this inequality. Since this inequality is greater than or equal to, we include x 5 3 in our solution because it satisfies the inequality. However, x 5 22 and x 5 2 are not included in the solution because they make the denominator equal to zero. The solution is 122, 22 ∪ 33, q2 . Y OUR TU R N Solve the inequality x 1 2 x 2 1 # 0. ▼ [CONCEPT CHECK] Which of the following has an implied domain restriction on the variable: (A) 1 x2 1 9 or (B) 1 x2 2 9 ANSWER (B) x 2 63. (A) does not have a domain restriction because x 2 1 9 is never equal to zero when x is a real number. ▼ 1.6 Polynomial and Rational Inequalities 145 ▼ A N S W E R 324, q2 ▼ C A U T I O N Rational inequalities should not be solved using cross multiplication. EXAMPLE 8 Solving a Rational Inequality Solve the inequality x 2 5 x2 1 9 , 0. Solution: Identify the zero(s) of the numerator. x 5 5 Note that the denominator is never equal to zero when x is any real number. Draw the number line and divide into intervals. Test the intervals. The denominator is always positive. Intervals in which the value of the rational expression is negative satisfy the inequality. 12q, 52 Notice that x 5 5 is not included in the solution because of the strict inequality. Y OUR T UR N Solve the inequality x 1 4 x2 1 25 $ 0. ▼ 5 5 () () () () () () EXAMPLE 9 Solving a Rational Inequality Solve the inequality x x 1 2 # 3. common mistake Do not cross multiply. The LCD or expression by which you are multiplying might be negative for some values of x, and that would require the direction of the inequality sign to be reversed. ✖I N C O R R EC T Error: Do not cross multiply. x # 31x 1 22 ✓COR R E C T Subtract 3 from both sides. x x 1 2 2 3 # 0 Write as a single rational expression. x 2 31x 1 22 x 1 2 # 0 Eliminate the parentheses. x 2 3x 2 6 x 1 2 # 0 Simplify the numerator. 22x 2 6 x 1 2 # 0 Factor the numerator. 221x 1 32 x 1 2 # 0 146 CHAPTER 1 Equations and Inequalities Applications Identify the zeros of the numerator and the denominator. x 5 23 and x 5 22 Draw the number line and test the intervals. 221x 1 32 x 1 2 # 0 Intervals in which the value of the rational expression is negative satisfy the inequality, 12H, 234 and 122, H2. Note that x 5 22 is not included in the solution because it makes the denominator zero, and x 5 23 is included because it satisfies the inequality. The solution is: 12q, 234 ∪ 122, q2 EXAMPLE 10 Stock Prices From June 2015 to November 2015, the price of Abercrombie and Fitch’s (ANF) stock was approximately given by P 5 0.5t2 2 2t 1 17, where P is the price of the stock in dollars, t is months, t 5 0 corresponds to June 2015, and t 5 5 corresponds to November 2015. During which months was the value of the stock worth no more than $21? June July Aug Sep Nov Oct 15 20 25 30 35 Solution: Set the price less than or equal to 21: 0.5t2 2 3t + 25 # 21, 0 # t # 5 Write in standard form: 0.5t2 2 3t 1 4 # 0 Multiply by 2: t2 − 6t 1 8 # 0 Factor. 1t − 42 1t 2 22 # 0 Identify zeros: t 5 2 and t 5 4 2 4 1 3 5 (–)(–) = (+) (–)(+) = (–) (+)(+) = (+) Test the intervals: 1t 2 42 1t 2 22 # 0 Negative interval satisfies the inequality: 32, 44 The Abercrombie & Fitch price was no more than $21 during August 2015, September 2015, and October 2015. 1.6 Polynomial and Rational Inequalities 147 In Exercises 1–28, solve the polynomial inequality and express the solution set in interval notation. 1. x2 2 3x 2 10 $ 0 2. x2 1 2x 2 3 , 0 3. u2 2 5u 2 6 # 0 4. u2 2 6u 2 40 . 0 5. p2 1 4p , 23 6. p2 2 2p $ 15 7. 2t2 2 3 # t 8. 3t2 $ 25t 1 2 9. 5v 2 1 . 6v2 10. 12t2 , 37t 1 10 11. 2s2 2 5s $ 3 12. 8s 1 12 # 2s2 13. y 2 1 2y $ 4 14. y2 1 3y # 1 15. x2 2 4x , 6 16. x2 2 2x . 5 17. u2 $ 3u 18. u2 # 24u 19. 22x # 2x2 20. 23x # x2 21. x2 . 9 22. x2 $ 16 23. t2 , 81 24. t2 # 49 25. z2 . 216 26. z2 $ 22 27. y2 , 24 28. y2 #225 In Exercises 29–58, solve the rational inequality and graph the solution on the real number line. 29. 2 3 x # 0 30. 3 x # 0 31. y y 1 3 . 0 32. y 2 2 y # 0 33. t 1 3 t 2 4 $ 0 34. 2t 2 5 t 2 6 , 0 35. s 1 1 4 2 s2 $ 0 36. s 1 5 4 2 s2 # 0 37. x 2 3 x2 2 25 $ 0 38. 1 2 x x2 2 9 # 0 39. 2u2 1 u 3 , 1 40. u2 2 3u 3 $ 6 41. 3t2 t 1 2 $ 5t 42. 22t 2 t2 4 2 t $ t 43. 3p 2 2p2 4 2 p2 , 3 1 p 2 2 p 44. 2 7p p2 2 100 # p 1 2 p 1 10 45. x2 5 1 x2 , 0 46. x2 5 1 x2 # 0 47. x2 1 10 x2 1 16 . 0 48. 2 x2 1 2 x2 1 4 , 0 49. v2 2 9 v 2 3 $ 0 50. v2 2 1 v 1 1 # 0 51. 2 t 2 3 1 1 t 1 3 $ 0 52. 1 t 2 2 1 1 t 1 2 # 0 53. 3 x 1 4 2 1 x 2 2 # 0 54. 2 x 2 5 2 1 x 2 1 $ 0 55. 1 p 1 4 1 1 p 2 4 . p2 2 48 p2 2 16 56. 1 p 2 3 2 1 p 1 3 # 2 57. 1 p 2 2 2 1 p 1 2 $ 3 p2 2 4 58. 2 2p 2 3 2 1 p 1 1 # 1 2p2 2 p 2 3 [SEC TION 1.6] E X ERC I S E S • S K I L L S Rational Inequality – Write as a single fraction. – Determine values that make the numerator or denominator equal to zero. Always exclude values that make the denominator 5 0. 3. Draw the number line labeling the intervals. 4. Test the intervals to determine whether they are positive or negative. 5. Select the intervals according to the sign of the inequality. 6. Write the solution in interval notation. The following procedure can be used for solving polynomial and rational inequalities. 1. Write in standard form—zero on one side. 2. Determine the zeros; if it is a rational function, note the domain restrictions. Polynomial Inequality – Factor if possible. – Otherwise, use the quadratic formula. [SEC TION 1.6] S U M M A RY 148 CHAPTER 1 Equations and Inequalities • A P P L I C A T I O N S 59. Profit. A Web-based embroidery company makes mono-grammed napkins. The profit associated with producing x orders of napkins is governed by the equation P 5 2x2 1 130x 2 3000 Determine the range of orders the company should accept in order to make a profit. 60. Profit. Repeat Exercise 59 using P 5 x2 2 130x 1 3600. 61. Car Value. The term “upside down” on car payments refers to owing more than a car is worth. Assume you buy a new car and finance 100% over 5 years. The difference between the value of the car and what is owed on the car is governed by the expression t t 2 3 where t is age (in years) of the car. Determine the time period when the car is worth more than you owe a t t 2 3 . 0b. When do you owe more than it’s worth a t t 2 3 , 0b? 62. Car Value. Repeat Exercise 61 using the expression 2 2 2 t 4 2 t. 63. Bullet Speed. A .22-caliber gun fires a bullet at a speed of 1200 feet per second. If a .22-caliber gun is fired straight upward into the sky, the height of the bullet in feet is given by the equation h 5 216t2 1 1200t, where t is the time in sec-onds with t 5 0 corresponding to the instant the gun is fired. How long is the bullet in the air? 64. Bullet Speed. A .38-caliber gun fires a bullet at a speed of 600 feet per second. If a .38-caliber gun is fired straight upward into the sky, the height of the bullet in feet is given by the equation h 5 216t2 1 600t. How many seconds is the bullet in the air? 65. Geometry. A rectangular area is fenced in with 100 feet of fence. If the minimum area enclosed is to be 600 square feet, what is the range of feet allowed for the length of the rectangle? 66. Stock Value. From November 2014 to November 2015, Amazon.com stock was approximately worth P 5 6.25t2 2 25t 1 325, where P is the price of the stock in dollars, t is months, and t 5 0 corresponds to November 2014. During what months was the stock worth no more than $525? Dec 2015 Feb Mar Apr May Jun Nov Jul Aug Sep Oct 350.00 325.00 300.00 275.00 375.00 400.00 425.00 450.00 475.00 $725.00 675.00 625.00 575.00 525.00 www.nasdaq.com/symbol/amzn/stock-chart In Exercises 67 and 68 refer to the following: In response to economic conditions, a local business explores the effect of a price increase on weekly profit. The function P 5 251x 1 321x 2 242 models the effect that a price increase of x dollars on a bottle of wine will have on the profit P measured in dollars. 67. Economics. What price increase will lead to a weekly profit of less than $460? 68. Economics. What price increases will lead to a weekly profit of more than $550? 69. Real Estate. A woman is selling a piece of land that she advertises as 400 acres 167 acres2 for $1.36 million. If you pay that price, what is the range of dollars per acre you have paid? Round to the nearest dollar. 70. Real Estate. A woman is selling a piece of land that she advertises as 1000 acres 1610 acres2 for $1 million. If you pay that price, what is the range of dollars per acre you have paid? Round to the nearest dollar. • C A T C H T H E M I S T A K E In Exercises 71–74, explain the mistake that is made. 71. Solve the inequality 3x , x2. Solution: Divide by x. 3 , x Write the solution in interval notation. 13, q2 This is incorrect. What mistake was made? 72. Solve the inequality u2 , 25. Solution: Take the square root of both sides. u , 25 Write the solution in interval notation. 12q, 252 This is incorrect. What mistake was made? 73. Solve the inequality x2 2 4 x 1 2 . 0. Solution: Factor the numerator and 1x 2 221x 1 22 1x 1 22 , 0 denominator. Cancel the 1x 1 22 common factor. x 2 2 . 0 Solve. x . 2 This is incorrect. What mistake was made? 74. Solve the inequality x 1 4 x , 21 3. Solution: Cross multiply. 3 1x 1 42 , 211x2 Eliminate the parentheses. 3x 1 12 , 2x Combine like terms. 4x , 212 Divide both sides by 4. x , 23 This is incorrect. What mistake was made? 1.7 Absolute Value Equations and Inequalities 149 1.7.1 Equations Involving Absolute Value The absolute value of a real number can be interpreted algebraically and graphically. Algebraically, the absolute value of 5 is 5, or in mathematical notation, 050 5 5; and the absolute value of 25 is 5 or 0250 5 5. Graphically, the absolute value of a real number is the distance on the real number line between the real number and the origin; thus, the distance from 0 to either 25 or 5 is 5. 5 5 5 5 • C O N C E P T U A L In Exercises 75 and 76, determine whether each statement is true or false. Assume that a is a positive real number. 75. If x , a2, then the solution is 12q, a2. 76. If x $ a2, then the solution is 3a, q2. 77. Assume the quadratic inequality ax2 1 bx 1 c , 0 is true. If b2 2 4ac , 0, then describe the solution. 78. Assume the quadratic inequality ax2 1 bx 1 c . 0 is true. If b2 2 4ac , 0, then describe the solution. In Exercises 79–82, solve for x given that a and b are both positive real numbers. 79. 2x2 # a2 80. x2 2 b2 x 1 b , 0 81. x2 1 a2 x2 1 b2 $ 0 82. a x2 , 2b • C H A L L E N G E 88. x2 1 3x 2 5 $ 2x2 1 2x 1 10 89. 2p 5 2 p . 1 90. 3p 4 2 p , 1 In Exercises 83–90, plot the left side and the right side of each inequality in the same screen and use the zoom feature to determine the range of values for which the inequality is true. 83. 1.4x2 2 7.2x 1 5.3 . 28.6x 1 3.7 84. 17x2 1 50x 2 19 , 9x2 1 2 85. 11x2 , 8x 1 16 86. 0.1x 1 7.3 . 0.3x2 2 4.1 87. x , x2 2 3x , 6 2 2x • T E C H N O L O G Y 1.7.1 S K I L L Solve absolute value equations. 1.7.1 C ON C E P T U A L Understand absolute value in terms of distance on the number line. S K I L L S O B J E C T I V E S ■ ■Solve absolute value equations. ■ ■Solve absolute value inequalities. C O N C E P T U A L O B J E C T I V ES ■ ■Understand absolute value in terms of distance on the number line. ■ ■Apply intersection and union concepts to expressing solutions to linear inequalities in one variable. 1.7 ABSOLUTE VALUE EQUATIONS AND INEQUALITIES DEFINITION Absolute Value The absolute value of a real number a, denoted by the symbol 0a0, is defined by 0a0 5 ba, if a $ 0 2a, if a , 0 150 CHAPTER 1 Equations and Inequalities The absolute value of a real number is never negative. When a 5 25, this definition says 0250 5 21252 5 5. DEFINITION Absolute Value Equation If 0x0 5 a, then x 5 2a or x 5 a, where a $ 0. Absolute value can be used to define the distance between two points on the real number line. EXAMPLE 1 Finding the Distance Between Two Points on the Number Line Find the distance between 24 and 3 on the real number line. Solution: The distance between 24 and 3 is given by the absolute value of the difference. 024 2 30 5 0270 5 7 Note that if we reverse the numbers the result is the same. 03 2 1242 0 5 070 5 7 We check this by counting the units between 24 and 3 on the number line. 4 3 7 EXAMPLE 2 Solving an Absolute Value Equation Solve the equation 0x 2 30 5 8 algebraically and graphically. Solution: Using the absolute value equation definition, we see that if the absolute value of an expression is 8, then that expression is either 28 or 8. Rewrite as two equations: x 2 3 5 28 or x 2 3 5 8 x 5 25 x 5 11 The solution set is 525, 116 . When absolute value is involved in algebraic equations, we interpret the definition of absolute value as follows. In words, “if the absolute value of a number is a, then that number equals 2a or a.” For example, the equation 0x0 5 7 is true if x 5 27 or x 5 7. We say the equation 0x0 5 7 has the solution set 527, 76. Note: 0x0 5 23 does not have a solution because there is no value of x such that its absolute value is 23. PROPERTIES OF ABSOLUTE VALUE For all real numbers a and b, 1. 0a0 $ 0 2. 02a0 5 0a0 3. 0ab0 5 0a0 0b0 4. a b 5 0 a 0 0 b 0 b 2 0 DISTANCE BETWEEN TWO POINTS ON THE REAL NUMBER LINE If a and b are real numbers, the distance between a and b is the absolute value of their difference given by 0a 2 b0 or 0b 2 a0. 1.7 Absolute Value Equations and Inequalities 151 Graph: The absolute value equation 0x 2 30 5 8 is interpreted as “what numbers are eight units away from 3 on the number line?” We find that eight units to the right of 3 is 11 and eight units to the left of 3 is 25. 5 3 11 8 8 YOUR T UR N Solve the equation 0x 1 50 5 7. ▼ STUDY TIP Rewrite an absolute value equation as two equations. ▼ A N S W E R x 5 23 or x 5 2. The solution set is 523, 26. ▼ A N S W E R x 5 24 or x 5 12. The solution set is 524, 126. EXAMPLE 3 Solving an Absolute Value Equation Solve the equation 01 2 3x0 5 7. Solution: If the absolute value of an expression is 7, 1 2 3x 5 27 or 1 2 3x 5 7 then that expression is 27 or 7. 23x 5 28 23x 5 6 x 5 8 3 x 5 22 The solution set is U22, 8 3V . YOUR T UR N Solve the equation 01 1 2x0 5 5. ▼ EXAMPLE 4 Solving an Absolute Value Equation Solve the equation 2 2 30x 2 10 5 240x 2 10 1 7. Solution: Isolate the absolute value expressions to one side. Add 40x 2 10 to both sides. 2 1 0x 2 10 5 7 Subtract 2 from both sides. 0x 2 10 5 5 If the absolute value of an expression x 2 1 5 25 or x 2 1 5 5 is equal to 5, then the expression is x 5 24 x 5 6 equal to either 25 or 5. The solution set is 524, 66 . YOUR T UR N Solve the equation 3 2 20x 2 40 5 230x 2 40 1 11. ▼ EXAMPLE 5 Finding That an Absolute Value Equation Has No Solution Solve the equation 01 2 3x0 5 27. Solution: The absolute value of an expression is never negative. Therefore, no values of x make this equation true. No solution ▼ A N S W E R x 5 2 or x 5 212. The solution set is 5212, 26. 152 CHAPTER 1 Equations and Inequalities 1.7.2 Inequalities Involving Absolute Value To solve the inequality 0x0 , 3, look for all real numbers that make this statement true. Some numbers that make it true are 22, 23 2, 21, 0, 1 5, 1, and 2. Some numbers that make it false are 27, 25, 23.5, 23, 3, and 4. If we interpret this inequality as distance, we ask what numbers are less than three units from the origin? We can represent the solution in the following ways. Inequality notation: 23 , x , 3 Interval notation: 123, 32 Graph: Similarly, to solve the inequality 0x 0 $ 3, look for all real numbers that make the statement true. If we interpret this inequality as a distance, we ask what numbers are at least three units from the origin? We can represent the solution in the following three ways. Inequality notation: x # 23 or x $ 3 Interval notation: 12q, 234 ∪ 33, q2 Graph: This discussion leads us to the following equivalence relations. 3 3 3 3 [CONCEPT CHECK] \|x 2 a\| 5 b is interpreted on the number line as (A) b units from a or (B) a units from b? ANSWER (A ) b units from a ▼ ▼ A N S W E R x 5 6"5 or x 5 63. The solution set is U6"5, 63V. 1.7.2 SKILLS Solve absolute value inequalities. 1.7.2 CO NCE PTUAL Apply intersection and union concepts to expressing solutions to linear inequalities in one variable. EXAMPLE 6 Solving a Quadratic Absolute Value Equation Solve the equation 05 2 x20 5 1. Solution: If the absolute value of an expression 5 2 x2 5 21 or 5 2 x2 5 1 is 1, that expression is either 21 or 1, which leads to two equations. 2x2 5 26 2x2 5 24 x2 5 6 x2 5 4 x 5 6"6 x 5 6"4 5 62 The solution set is U62, 6"6 V . Y OUR TU R N Solve the equation 07 2 x20 5 2. ▼ It is important to realize that in the above four properties the variable x can be any algebraic expression. PROPERTIES OF ABSOLUTE VALUE INEQUALITIES 1. 0x0 , a is equivalent to 2a , x , a 2. 0x 0 # a is equivalent to 2a # x # a 3. 0x 0 . a is equivalent to x , 2a or x . a 4. 0x 0 $ a is equivalent to x # 2a or x $ a Note: a . 0. 1.7 Absolute Value Equations and Inequalities 153 Notice that if we change the problem in Example 8 to 01 2 2x0 . 25, the answer is all real numbers because the absolute value of any expression is greater than or equal to zero. Similarly, 01 2 2x0 , 25 would have no solution because the absolute value of an expression can never be negative. ▼ A N S W E R Inequality notation: 26 , x , 5. Interval notation: 126, 52. ▼ A N S W E R Inequality notation: x # 2 or x $ 3. Interval notation: 12q, 24 ∪ 33, q2. STUDY TIP Less than inequalities can be written as a single statement. Greater than inequalities must be written as two statements. EXAMPLE 7 Solving an Inequality Involving an Absolute Value Solve the inequality 03x 2 20 # 7. Solution: We apply property (2) and squeeze the absolute value expression between 27 and 7. 27 # 3x 2 2 # 7 Add 2 to all three parts. 25 # 3x # 9 Divide all three parts by 3. 25 3 # x # 3 The solution in interval notation is C25 3, 3D . Graph: 3 0 3 5 Y OUR T UR N Solve the inequality 02x 1 10 , 11. ▼ EXAMPLE 8 Solving an Inequality Involving an Absolute Value Solve the inequality 01 2 2x0 . 5. Solution: Apply property (3). 1 2 2x , 25 or 12 2x . 5 Subtract 1 from all expressions. 22x , 26 22x . 4 Divide by 22 and reverse the inequality sign. x . 3 x , 22 Express the solution in interval notation. 12q, 222 ∪ 13, q2 Graph: Y OUR T UR N Solve the inequality 05 2 2x0 $ 1. 2 3 ▼ It is often helpful to note that for absolute value inequalities, ■ ■less than inequalities can be written as a single statement (see Example 7). ■ ■greater than inequalities must be written as two statements (see Example 8). [CONCEPT CHECK] If the solution to a linear inequality in one variable is all the values less than a OR greater than b (where a and b are both positive), then the appropriate notation is (A) intersection or (B) union. ANSWER (B) Union (2q, a) ∪ (b, q) ▼ 154 CHAPTER 1 Equations and Inequalities EXAMPLE 9 Solving an Inequality Involving an Absolute Value Solve the inequality 2 2 03x0 , 1. Solution: Subtract 2 from both sides. 203x0 , 21 Multiply by 1212 and reverse the inequality sign. 03x0 . 1 Apply property (3). 3x , 21 or 3x . 1 Divide both inequalities by 3. x , 21 3 x . 1 3 Express in interval notation. a2q, 21 3b ∪ a1 3, qb Graph. In Exercises 1–38, solve the equation. 1. 0x0 5 3 2. 0x0 5 2 3. 0x0 5 24 4. 0x0 5 22 5. 0t 1 30 5 2 6. 0t 2 30 5 2 7. 0p 2 70 5 3 8. 0p 1 70 5 3 9. 04 2 y0 5 1 10. 02 2 y0 5 11 11. 03x0 5 9 12. 05x0 5 50 13. 02x 1 70 5 9 14. 02x 2 50 5 7 15. 03t 2 90 5 3 16. 04t 1 20 5 2 17. 07 2 2x0 5 9 18. 06 2 3y0 5 12 19. 01 2 3y0 5 1 20. 05 2 x0 5 2 21. 04.7 2 2.1x0 5 3.3 22. 05.2x 1 3.70 5 2.4 23. 2 3x 2 4 7 5 5 3 24. 1 2x 1 3 4 5 1 16 25. 0x 2 50 1 4 5 12 26. 0x 1 30 2 9 5 2 27. 30x 2 20 1 1 5 19 28. 201 2 x0 2 4 5 2 29. 5 5 7 2 02 2 x0 30. 21 5 3 2 0x 2 30 31. 2 0p 1 30 2 15 5 5 32. 8 2 3 0p 2 40 5 2 33. 50y 2 20 2 10 5 40y 2 20 2 3 34. 3 2 0y 1 90 5 11 2 3 0y 1 90 35. 04 2 x20 5 1 36. 07 2 x20 5 3 37. 0x2 1 10 5 5 38. 0x2 2 10 5 5 [SEC TION 1.7] E X E R C I SE S • S K I L L S I N EQ U A L I T I ES 0 x 0 , A is equivalent to 2A , x , A 0 x 0 . A is equivalent to x , 2A or x . A Absolute value equations and absolute value inequalities are solved by writing the equations or inequalities in terms of two equations or inequalities. Note: A . 0. E Q UATI ONS 0 x 0 5 A is equivalent to x 5 2A or x 5 A [SEC TION 1.7] S U M M A RY 1.7 Absolute Value Equations and Inequalities 155 In Exercises 39–70, solve the inequality and express the solution in interval notation. 39. 0x0 , 7 40. 0y0 , 9 41. 0y0 $ 5 42. 0x0 $ 2 43. 0x 1 30 , 7 44. 0x 1 20 # 4 45. 0x 2 40 . 2 46. 0x 2 10 , 3 47. 04 2 x0 # 1 48. 01 2 y0 , 3 49. 02x0 . 23 50. 02x0 , 23 51. 02t 1 30 , 5 52. 03t 2 50 . 1 53. 07 2 2y0 $ 3 54. 06 2 5y0 # 1 55. 04 2 3x0 $ 0 56. 04 2 3x0 $ 1 57. 204x0 2 9 $ 3 58. 50x 210 1 2 # 7 59. 20x 1 10 2 3 # 7 60. 30x 2 10 2 5 . 4 61. 3 2 20x 1 40 , 5 62. 7 2 30x 1 20 $ 214 63. 9 2 02x0 , 3 64. 4 2 0x 1 10 . 1 65. 0 1 2 2x 0 , 1 2 66. 2 2 3x 5 $ 2 5 67. 02.6x 1 5.40 , 1.8 68. 03.7 2 5.5x0 . 4.3 69. 0x2 2 10 # 8 70. 0x2 1 40 $ 29 In Exercises 71–76, write an inequality that fits the description. 71. Any real numbers less than seven units from 2. 72. Any real numbers more than three units from 22. 73. Any real numbers at least 1 2 unit from 3 2. 74. Any real number no more than 5 3 units from 11 3 . 75. Any real numbers no more than two units from a. 76. Any real number at least a units from 23. • A P P L I C A T I O N S 77. Temperature. If the average temperature in Hawaii is 83°F 1615°2, write an absolute value inequality representing the temperature in Hawaii. 78. Temperature. If the average temperature of a human is 97.8°F 161.22, write an absolute value inequality describing normal human body temperature. 79. Sports. Two women tee off the green of a par-3 hole on a golf course. They are playing “closest to the pin.” If the first woman tees off and lands exactly 4 feet from the hole, write an inequality that describes where the second woman must land in order to win the hole. What equation would suggest a tie? Let d 5 the distance from where the second woman lands to the tee. 80. Electronics. A band-pass filter in electronics allows certain frequencies within a range (or band) to pass through to the receiver and eliminates all other frequencies. Write an abso-lute value inequality that allows any frequency ƒ within 15 Hertz of the carrier frequency ƒc to pass. In Exercises 81 and 82 refer to the following: A company is reviewing revenue for the prior sales year. The model for projected revenue and the model for actual revenue are Rprojected 5 200 1 5x Ractual 5 210 1 4.8x where x represents the number of units sold and R represents the revenue in thousands of dollars. Since the two revenue models are not identical, an error in projected revenue occurred. This error is represented by E 5 Rprojected 2 Ractual 81. Business. For what number of units sold was the error in projected revenue less than $5000? 82. Business. For what number of units sold was the error in projected revenue less than $3000? • C A T C H T H E M I S T A K E In Exercises 83–86, explain the mistake that is made. 83. Solve the absolute value equation 0 x 2 30 5 7. Solution: Eliminate the absolute value symbols. x 2 3 5 7 Add 3 to both sides. x 5 10 Check. 010 2 30 5 7 This is incorrect. What mistake was made? 84. Solve the inequality 0x 2 30 , 7. Solution: Eliminate the absolute x 2 3 , 27 or x 2 3 . 7 value symbols. Add 3 to both sides. x , 24 x . 10 The solution is 12q, 242 ∪ 110, q2. This is incorrect. What mistake was made? 156 CHAPTER 1 Equations and Inequalities In Exercises 87–90, determine whether each statement is true or false. 87. 20m0 # m # 0m0 88. 0n20 5 n2 89. 0m 1 n0 5 0m0 1 0n0 is true only when m and n are both nonnegative. 90. For what values of x does the absolute value equation 0x 2 70 5 x 2 7 hold? In Exercises 91–96, assuming a and b are real positive numbers, solve the equation or inequality and express the solution in interval notation. 91. 0x 2 a 0 , b 92. 0a 2 x 0 . b 93. 0x 0 $ 2a 94. 0x 0 # 2b 95. 0x 2 a 0 5 b 96. 0x 2 a 0 5 2b • C O N C E P T U A L 97. For what values of x does the absolute value equation 0x 1 10 5 4 1 0x 2 20 hold? 98. Solve the inequality 03x2 2 7x 1 20 . 8. • C H A L L E N G E 102. Solve the inequality 02.7x2 2 7.9x 1 50 # 05.3x2 2 9.20 by graphing both sides of the inequality and identify which x-values make this statement true. 103. Solve the inequality x x1 1 , 1 by graphing both sides of the inequality, and identify which x-values make this statement true. 104. Solve the inequality x x1 1 , 2 by graphing both sides of the inequality, and identify which x-values make this statement true. 99. Graph y1 5 0x 2 70 and y2 5 x 2 7 in the same screen. Do the x-values where these two graphs coincide agree with your result in Exercise 90? 100. Graph y1 5 0x 1 10 and y2 5 0x 2 20 1 4 in the same screen. Do the x-values where these two graphs coincide agree with your result in Exercise 97? 101. Graph y1 5 03x2 2 7x 1 20 and y2 5 8 in the same screen. Do the x-values where y1 lies above y2 agree with your result in Exercise 98? • T E C H N O L O G Y 85. Solve the inequality 05 2 2x0 # 1. Solution: Eliminate the absolute value symbols. 21 # 5 2 2x # 1 Subtract 5. 26 # 22x # 24 Divide by 22. 3 # x # 2 Write the solution in interval notation. 12q, 24 ∪ 33, q2 This is incorrect. What mistake was made? 86. Solve the equation 05 2 2x0 5 21. Solution: 5 2 2x 5 21 or 5 2 2x 5 1 22x 5 26 22x 5 24 x 5 3 x 5 2 The solution is 52, 36. This is incorrect. What mistake was made? CH A P TE R 1 R E VIE W [CH AP TER 1 REVIEW] Chapter Review 157 SECTION CONCEPT KEY IDEAS/FORMULAS 1.1 Linear equations ax 1 b 5 0 Solving linear equations in one variable Isolate variables on one side and constants on the other side. Solving rational equations that are reducible to linear equations Any values that make the denominator equal to 0 must be eliminated as possible solutions. 1.2 Applications involving linear equations Solving application problems using mathematical models Five-step procedure: Step 1: Identify the question. Step 2: Make notes. Step 3: Set up an equation. Step 4: Solve the equation. Step 5: Check the solution. Geometry problems Formulas for rectangles, triangles, and circles Interest problems Simple interest: I 5 Prt Mixture problems Whenever two distinct quantities are mixed, the result is a mixture. Distance–rate–time problems d 5 r ⋅ t 1.3 Quadratic equations ax2 1 bx 1 c 5 0 a 2 0 Factoring If 1x 2 h2 1x 2 k2 5 0, then x 5 h or x 5 k. Square root method If x2 5 P, then x 5 6"P. Completing the square Find half of b; square that quantity; add the result to both sides. Quadratic Formula x 5 2b 6 "b2 2 4ac 2a 1.4 Other types of equations Radical equations Check solutions to avoid extraneous solutions. Equations quadratic in form: u-substitution Use a u-substitution to write the equation in quadratic form. Factorable equations Extract common factor or factor by grouping. 1.5 Linear inequalities Solutions are a range of real numbers. Graphing inequalities and interval notation n a , x , b is equivalent to 1a, b2. n x # a is equivalent to 12q, a4. n x . a is equivalent to 1a, q2. Solving linear inequalities If an inequality is multiplied or divided by a negative number, the inequality sign must be reversed. 1.6 Polynomial and rational inequalities Polynomial inequalities Zeros are values that make the polynomial equal to 0. Rational inequalities The number line is divided into intervals. The endpoints of these intervals are values that make either the numerator or denominator equal to 0. Always exclude values that make the denominator 5 0. 1.7 Absolute value equations and inequalities 0 b 2 a 0 is the distance between points a and b on the number line. Equations involving absolute value If 0 x 0 5 a, then x 5 2a or x 5 a. Inequalities involving absolute value n 0 x 0 # a is equivalent to 2a # x # a. n 0 x 0 . a is equivalent to x , 2a or x . a. REV IEW E XE R CI SE S 158 CHAPTER 1 Equations and Inequalities [ C H AP T E R 1 REVIEW EXERC IS E S ] 1.1 Linear Equations Solve for the variable. 1. 7x 2 4 5 12 2. 13d 1 12 5 7d 1 6 3. 20p 1 14 5 6 2 5p 4. 4 1x 2 72 2 4 5 4 5. 31x 1 72 2 2 5 4 1x 2 22 6. 7c 1 31c 2 52 5 2 1c 1 32 2 14 7. 14 2 3231 y 2 42 1 94 5 3412y 1 32 2 64 1 4 8. 36 2 4x 1 21x 2 724 2 52 5 312x 2 42 1 63312x 2 32 1 64 9. 12 b 2 3 5 6 b 1 4 10. g 3 1 g 5 7 9 11. 13x 7 2 x 5 x 4 2 3 14 12. 5b 1 b 6 5 b 3 2 29 6 Specify any values that must be excluded from the solution set and then solve. 13. 1 x 2 4 5 3 x 2 5 14. 4 x 1 1 2 8 x 2 1 5 3 15. 2 t 1 4 2 7 t 5 6 t1t 1 42 16. 3 2x 2 7 5 22 3x 1 1 17. 3 2x 2 6 x 5 9 18. 3 2 15/m2 2 1 15/m2 5 1 19. 7x 2 12 2 4x2 5 3 326 1 14 2 2x 1 724 1 12 20. x 5 2 x 2 3 15 5 26 Solve for the specified variable. 21. Solve for x in terms of y: 3x 2 2 31 y 1 423 2 7 4 5 y 2 2x 1 6 1x 2 32 22. If y 5 x 1 3 1 1 2x, find y 1 2 1 2 2y in terms of x. 1.2 Applications Involving Linear Equations 23. Transportation. Maria is on her way from her home near Orlando to the Sun Dome in Tampa for a rock concert. She drives 16 miles to the Orlando park-n-ride, takes a bus 3 4 of the way to a bus station in Tampa, and then takes a cab 1 12 of the way to the Sun Dome. How far does Maria live from the Sun Dome? 24. Diet. A particular 2000 calorie per day diet suggests eating breakfast, lunch, dinner, and four snacks. Each snack is 1 4 the calories of lunch. Lunch has 100 calories less than dinner. Dinner has 1.5 times as many calories as breakfast. How many calories are in each meal and snack? 25. Numbers. Find a number such that 12 more than 1 4 the number is 1 3 the number. 26. Numbers. Find four consecutive odd integers such that the sum of the four numbers is equal to three more than three times the fourth integer. 27. Geometry. The length of a rectangle is one more than two times the width, and the perimeter is 20 inches. What are the dimensions of the rectangle? 28. Geometry. Find the perimeter of a triangle if one side is 10 inches, another side is 1 3 of the perimeter, and the third side is 1 6 of the perimeter. 29. Investments. You win $25,000 and you decide to invest the money in two different investments: one paying 20% and the other paying 8%. A year later you have $27,600 total. How much did you originally invest in each account? 30. Investments. A college student on summer vacation was able to make $5000 by working a full-time job every summer. He invested half the money in a mutual fund and half the money in a stock that yielded four times as much interest as the mutual fund. After a year he earned $250 in interest. What were the interest rates of the mutual fund and the stock? 31. Chemistry. For an experiment, a student requires 150 milliliters of a solution that is 8% NaCl (sodium chloride). The storeroom has only solutions that are 10% NaCl and 5% NaCl. How many milliliters of each available solution should be mixed to get 150 milliliters of 8% NaCl? 32. Chemistry. A mixture containing 8% salt is to be mixed with 4 ounces of a mixture that is 20% salt, in order to obtain a solution that is 12% salt. How much of the first solution must be used? 33. Grades. Going into the College Algebra final exam, which will count as two tests, Danny has test scores of 95, 82, 90, and 77. If his final exam is higher than his lowest test score, then it will count for the final exam and replace the lowest test score. What score does Danny need on the final in order to have an average score of at least 90? 34. Car Value. A car salesperson reduced the price of a model car by 20%. If the new price is $25,000, what was its original price? How much can be saved by purchasing the model? 1.3 Quadratic Equations Solve by factoring. 35. b2 5 4b 1 21 36. x 1x 2 32 5 54 37. x2 5 8x 38. 6y2 2 7y 2 5 5 0 Solve by the square root method. 39. q2 2 169 5 0 40. c2 1 36 5 0 41. 12x 2 42 2 5 264 42. 1d 1 72 2 2 4 5 0 Solve by completing the square. 43. x2 2 4x 2 12 5 0 44. 2x2 2 5x 2 7 5 0 45. x2 2 5 4 1 x 2 46. 8m 5 m2 1 15 R E VI E W E XERCISES Review Exercises 159 Solve by the Quadratic Formula. 47. 3t2 2 4t 5 7 48. 4x2 1 5x 1 7 5 0 49. 8ƒ 2 2 1 3ƒ 5 7 6 50. x2 5 26x 1 6 Solve by any method. 51. 5q2 2 3q 2 3 5 0 52. 1x 2 72 2 5 212 53. 2x2 2 3x 2 5 5 0 54. 1 g 2 22 1 g 1 52 5 27 55. 7x2 5 219x 1 6 56. 7 5 2b2 1 1 Solve for the indicated variable. 57. S 5 p2h for r 58. V 5 pr 3h 3 for r 59. h 5 vt 2 16t 2 for v 60. A 5 2pr 2 1 2prh for h 61. Geometry. Find the base and height of a triangle with an area of 2 square feet if its base is 3 feet longer than its height. 62. Falling Objects. A man is standing on top of a building 500 feet tall. If he drops a penny off the roof, the height of the penny is given by h 5 216t2 1 500, where t is in seconds. Determine how many seconds it takes until the penny hits the ground. 1.4 Other Types of Equations Solve the radical equation for the given variable. 63. ! 3 2x 2 4 5 2 64. !x 2 2 5 24 65. 12x 2 721/5 5 3 66. x 5 !7x 2 10 67. x 2 4 5 "x2 1 5x 1 6 68. !2x 2 7 5 !x 1 3 69. !x 1 3 5 2 2 !3x 1 2 70. 4 1 !x 2 3 5 !x 2 5 71. x 2 2 5 "49 2 x2 72. !2x 2 5 2 !x 1 2 5 3 73. 2x 5 !3 2 x 74. #15 1 2!x 2 4 1 !x 5 5 Solve the equation by introducing a substitution that transforms the equation to quadratic form. 75. 228 5 13x 2 222 2 1113x 2 22 76. x 4 2 6x2 1 9 5 0 77. a x 1 2 xb 2 5 15 2 2 a x 1 2 xb 78. 3 1x 2 42 4 2 111x 2 42 2 2 20 5 0 79. y22 2 5y21 1 4 5 0 80. p22 1 4p21 5 12 81. 3x1/3 1 2x2/3 5 5 82. 2x2/3 2 3x1/3 2 5 5 0 83. x22/3 1 3x21/3 1 2 5 0 84. y21/2 2 2y21/4 1 1 5 0 85. x4 1 5x2 5 36 86. 3 2 4x21/2 1 x21 5 0 Solve the equation by factoring. 87. x3 1 4x2 2 32x 5 0 88. 9t3 2 25t 5 0 89. p3 2 3p2 2 4p 1 12 5 0 90. 4x3 2 9x2 1 4x 2 9 5 0 91. p12p 2 522 2 312p 2 52 5 0 92. 2 1t2 2 923 2 20 1t2 2 922 5 0 93. y 2 81y21 5 0 94. 9x3/2 2 37x1/2 1 4x21/2 5 0 1.5 Linear Inequalities Rewrite using interval notation. 95. x # 24 96. 21 , x # 7 97. 2 # x # 6 98. x . 21 Rewrite using inequality notation. 99. 126, q2 100. 12q, 04 101. 323, 74 102. 125, 24 Express each interval using inequality and interval notation. 103. 104. Graph the indicated set and write as a single interval, if possible. 105. 14, 64 ∪ 35, q2 106. 12q, 232 ∪ 327, 24 107. 13, 124 ∩ 38, q2 108. 12q, 222 ∩ 322, 92 Solve and graph. 109. 2x , 5 2 x 110. 6x 1 4 # 2 111. 4 1x 2 12 . 2x 2 7 112. x 1 3 3 $ 6 113. 6 , 2 1 x # 11 114. 26 # 1 2 4 1x 1 22 # 16 115. 2 3 # 1 1 x 6 # 3 4 116. x 3 1 x 1 4 9 . x 6 2 1 3 REV IEW E XE R CI SE S 160 CHAPTER 1 Equations and Inequalities Applications 117. Grades. In your algebra class your first four exam grades are 72, 65, 69, and 70. What is the lowest score you can get on the fifth exam to earn a C for the course? Assume that each exam is equal in weight and a C is any score greater than or equal to 70. 118. Profit. A tailor decided to open a men’s custom suit business. His fixed costs are $8500 per month, and it costs him $50 for the materials to make each suit. If the price he charges per suit is $300, how many suits does he have to tailor per month to make a profit? 1.6 Polynomial and Rational Inequalities Solve the polynomial inequality and express the solution set using interval notation. 119. x2 # 36 120. 6x2 2 7x , 20 121. 4x # x2 122. 2x2 $ 9x 1 14 123. 2x2 , 27x 124. x2 , 24 125. 4x2 2 12 . 13x 126. 3x # x2 1 2 Solve the rational inequality and express the solution set using interval notation. 127. x x 2 3 , 0 128. x 2 1 x 2 4 . 0 129. x2 2 3x 3 $ 18 130. x2 2 49 x 2 7 $ 0 131. 3 x 2 2 2 1 x 2 4 # 0 132. 4 x 2 1 # 2 x 1 3 133. x2 1 9 x 2 3 $ 0 134. x , 5x 1 6 x 1.7 Absolute Value Equations and Inequalities Solve the equation. 135. 0 x 2 3 0 5 24 136. 0 2 1 x 0 5 5 137. 0 3x 2 4 0 5 1.1 138. 0 x2 2 6 0 5 3 Solve the inequality and express the solution using interval notation. 139. 0 x 0 , 4 140. 0 x 2 3 0 , 6 141. 0 x 1 4 0 . 7 142. 0 27 1 y 0 # 4 143. 0 2x 0 . 6 144. 4 1 2x 3 $ 1 7 145. 0 2 1 5x 0 $ 0 146. 0 1 2 2x 0 # 4 Applications 147. Temperature. If the average temperature in Phoenix is 858F 1610°2, write an inequality representing the average temperature T in Phoenix. 148. Blood Alcohol Level. If a person registers a 0.08 blood alcohol level, he will be issued a DUI ticket in the state of Florida. If the test is accurate within 0.007, write a linear inequality representing an actual blood alcohol level that will not be issued a ticket. Technology Exercises Section 1.1 Graph the function represented by each side of the question in the same viewing rectangle, and solve for x. 149. 0.031x 1 0.01714000 2 x2 5 103.14 150. 1 0.16x 2 0.2 x 5 1 4 Section 1.3 151. a. Solve the equation x2 1 4x 5 b, b 5 5 by first writing in standard form and then factoring. Now plot both sides of the equation in the same viewing screen 1 y1 5 x2 1 4x and y2 5 b2. At what x-values do these two graphs intersect? Do those points agree with the solution set you found? b. Repeat part (a) for b 5 25, 0, 7, and 12. 152. a. Solve the equation x2 2 4x 5 b, b 5 5 by first writing in standard form and then factoring. Now plot both sides of the equation in the same viewing screen 1 y1 5 x2 2 4x and y2 5 b2. At what x-values do these two graphs intersect? Do those points agree with the solution set you found? b. Repeat part (a) for b 5 25, 0, 7, and 12. Section 1.4 153. Solve the equation 2x1/4 5 2x1/2 1 6. Round your answer to two decimal places. Plot both sides of the equation in the same viewing screen, y1 5 2x1/4 and y2 5 2x1/2 1 6. Does the point(s) of intersection agree with your solution? 154. Solve the equation 2x21/2 5 x21/4 1 6. Plot both sides of the equation in the same viewing screen, y1 5 2x21/2 and y2 5 x21/4 1 6. Does the point(s) of intersection agree with your solution? Section 1.5 155. a. Solve the inequality 20.61x 1 7.62 . 0.24x 2 5.47. Express the solution set using interval notation. b. Graph each side of the inequality in the same viewing screen. Find the range of x-values when the graph of the left side lies above the graph of the right side. c. Do parts (a) and (b) agree? 156. a. Solve the inequality 21 2 x 1 7 , 3 4x 2 5. Express the solution set using interval notation. b. Graph each side of the inequality in the same viewing screen. Find the range of x-values when the graph of the left side lies below the graph of the right side. c. Do parts (a) and (b) agree? R E VI E W E XERCISES Review Exercises 161 Section 1.6 Plot the left side and the right side of each inequality in the same screen, and use the zoom feature to determine the range of values for which the inequality is true. 157. 0.2x2 2 2 . 0.05x 1 3.25 158. 12x2 2 7x 2 10 , 2x2 1 2x 2 1 159. 3p 7 2 2p . 1 160. 7p 15 2 2p , 1 Section 1.7 161. Solve the inequality 01.6x2 2 4.50 , 3.2 by graphing both sides of the inequality, and identify which x-values make this statement true. Express the solution using interval notation and round to two decimal places. 162. Solve the inequality 00.8x2 2 5.4x0 . 4.5 by graphing both sides of the inequality, and identify which x-values make this statement true. Express the solution using interval notation and round to two decimal places. PR ACTICE TEST 162 CHAPTER 1 Equations and Inequalities [ C H AP T E R 1 PRACTICE TEST ] Solve the equation. 1. 4p 2 7 5 6p 2 1 2. 22 1z 2 12 1 3 5 23z 1 3 1z 2 12 3. 3t 5 t2 2 28 4. 8x2 2 13x 5 6 5. 6x2 2 13x 5 8 6. 3 x 2 1 5 5 x 1 2 7. 5 y 2 3 1 1 5 30 y2 2 9 8. x4 2 5x2 2 36 5 0 9. !2x 1 1 1 x 5 7 10. 2x2/3 1 3x1/3 2 2 5 0 11. !3y 2 2 5 3 2 !3y 1 1 12. x 13x 2 523 2 2 13x 2 522 5 0 13. x7/3 2 8x4/3 1 12x1/3 5 0 Solve for the specified variable. 14. F 5 9 5C 1 32 for C 15. P 5 2L 1 2W for L Solve the inequality and express the solution in interval notation. 16. 7 2 5x . 218 17. 3x 1 19 $ 5 1x 2 32 18. 21 # 3x 1 5 , 26 19. 2 5 , x 1 8 4 # 1 2 20. 3x $ 2x2 21. 3p2 $ p 1 4 22. 0 5 2 2x 0 . 1 23. x 2 3 2x 1 1 # 0 24. x 1 4 x2 2 9 $ 0 25. Puzzle. A piling supporting a bridge sits so that 1 4 of the piling is in the sand, 150 feet is in the water, and 3 5 of the piling is in the air. What is the total height of the piling? 26. Real Estate. As a real estate agent you earn 7% of the sale price. The owners of a house you have listed at $150,000 will entertain offers within 10% of the list price. Write an inequality that models the commission you could make on this sale. 27. Costs: Cell Phones. A cell phone company charges $49 for a 600-minute monthly plan, plus an additional $0.17 per minute for every minute over 600. If a customer’s bill ranged from a low of $53.59 to a high of $69.74 over a 6-month period, write an inequality expressing the number of monthly minutes used over the 6-month period. 28. Television. Television and film formats are classified as ratios of width to height. Traditional televisions have a 4:3 ratio (1.33:1), and movies are typically made in widescreen format with a 21:9 ratio (2.35:1). If you own a traditional 25-inch television (20 inch 3 15 inch screen) and you play a widescreen DVD on it, there will be black bars above and below the image. What are the dimensions of the movie and of the black bars? 29. Solve the equation 1 0.75x 2 0.45 x 5 1 9. Graph the function represented by each side in the same viewing rectangle and solve for x. 30. Solve the inequality 0.3 1 02.4x2 2 1.50 # 6.3 by graphing both sides of the inequality, and identify which x-values make this statement true. Express the solution using interval notation. Andy Washnik 25 15 20 CU MU LA TIV E TEST Cumulative Test 163 [CH AP TER 1 CUMULATIVE T E S T ] Simplify. 1. 5 ⋅ 17 2 3 ⋅ 4 1 22 Simplify and express in terms of positive exponents. 2. 14x23b4223 3. 1x2y2223 1x2 y223 Perform the operations and simplify. 4. 12x 4 1 2x32 1 1x3 2 5x 2 62 2 15x4 1 4x3 2 6x 1 82 5. x2 1x 2 52 1x 2 32 Factor completely. 6. 3x3 2 3x2 2 60x 7. 2a3 1 2000 Perform the operations and simplify. 8. 3 2 x x2 2 1 4 5x 2 15 x 1 1 9. 6x x 2 2 2 5x x 1 2 Solve for x. 10. x3 2 x2 2 30x 5 0 11. 2 7x 5 1 8x 1 9 12. Perform the operation and express in standard form: 45 6 2 3i. Solve for x. 13. 6x 5 2 8x 3 5 4 2 7x 15 14. x 2 6 6 2 x 5 3 2 15. Tim can paint the interior of a condo in 9 hours. If Chelsea is hired to help him, they can do a similar condo in 5 hours. Working alone, how long will it take Chelsea to paint a similar condo? 16. Solve using the square root method: y 2 1 36 5 0. 17. Solve by completing the square: x2 1 12x 1 40 5 0. 18. Solve using the Quadratic Formula: x2 1 x 1 9 5 0. 19. Solve and check: "4 2 x 5 x 2 4. 20. Solve using substitution: 3x22 1 8x21 1 4 5 0. Solve and express the solution in interval notation. 21. 0 , 4 2 x # 7 22. 4x2 , 9x 2 11 23. x 1 2 9 2 x2 $ 0 24. 4 2 5x 7 $ 3 14 25. Solve for x: 1 5x 1 2 3 5 7 15. 26. Solve the equation x6 1 37 8 x3 5 27. Plot both sides of the equation in the same viewing screen, y1 5 x6 1 37 8 x3 and y2 5 27. Does the point(s) of intersection agree with your solution? 27. Solve the inequality 3x x 2 2 , 1 by graphing both sides of the inequality, and identify which x-values make this statement true. C H A P T E R LEARNING OBJECTIVES [ [ ■ ■Calculate the distance between two points and the midpoint of a line segment joining two points. ■ ■Sketch the graph of an equation using intercepts and symmetry as graphing aids. ■ ■Find the equation of a line. ■ ■Graph circles. ■ ■Find the line of best fit for a given set of data. Graphs are used in many ways. There is only one temperature that yields the same number in degrees Celsius and degrees Fahrenheit. Do you know what it is? The penguins are a clue. 2 See Section 2.3, Exercise 107. Optional Technology Required Section Fuse/Getty Images Emperor penguins walking in a line, Weddell Sea, Antarctica 100 200 100 50 ºF ºC The conversion between degrees Fahrenheit and degrees Celsius is a linear relationship. Notice that 08C corresponds to 328F. Graphs COKE 5–Minute 10 11 12 1 2 3 54.50 54.75 55.00 55.25 4:53 PM Stock prices fluctuating throughout the day Year 1994 2006 2010 2014 2002 1998 Antenatal clinic prevalence (%) 10 5 15 20 25 South Africa Thailand HIV infection rates (Data from countries/1W?page=4&display=default) 165 Optional Technology Required Section 165 [I N T HI S CHAPTER] You will review the Cartesian plane. You will calculate the distance between two points and find the midpoint of a line segment joining two points. You will then apply point-plotting techniques to sketch graphs of equations. Special attention is given to two types of equations: lines and circles. GRAPHS 2.1 BASIC TOOLS: CARTESIAN PLANE, DISTANCE, AND MIDPOINT 2.2 GRAPHING EQUATIONS: POINT-PLOTTING, INTERCEPTS, AND SYMMETRY 2.3 LINES 2.4 CIRCLES 2.5 LINEAR REGRESSION: BEST FIT • Cartesian Plane • Distance Between Two Points • Midpoint of a Line Segment Joining Two Points • Point-Plotting • Intercepts • Symmetry • Using Intercepts and Symmetry as Graphing Aids • Graphing a Line • Equations of Lines • Parallel and Perpendicular Lines • Standard Equation of a Circle • Transforming Equations of Circles to the Standard Form by Completing the Square • Scatterplots • Identifying Patterns • Linear Regression 166 CHAPTER 2 Graphs 2.1.1 Cartesian Plane HIV infection rates, stock prices, and temperature conversions are all examples of relationships between two quantities that can be expressed in a two-dimensional graph. Because it is two dimensional, such a graph lies in a plane. Two perpendicular real number lines, known as the axes in the plane, intersect at a point we call the origin. Typically, the horizontal axis is called the x-axis, and the vertical axis is denoted as the y-axis. The axes divide the plane into four quadrants, numbered by Roman numerals and ordered counterclockwise. Points in the plane are represented by ordered pairs, denoted 1x, y2. The first number of the ordered pair indicates the position in the horizontal direction and is often called the x-coordinate or abscissa. The second number indicates the position in the vertical direction and is often called the y-coordinate or ordinate. The origin is denoted 10, 02. Examples of other coordinates are given on the graph to the right. The point 12, 42 lies in quadrant I. To plot this point, start at the origin 10, 02 and move to the right two units and up four units. All points in quadrant I have positive coordinates, and all points in quadrant III have negative coordinates. Quadrant II has negative x-coordinates and positive y-coordinates; quadrant IV has positive x-coordinates and negative y-coordinates. This representation is called the rectangular coordinate system or Cartesian coordinate system, named after the French mathematician René Descartes (1596–1650). S K I L L S O B J E C T I V E S ■ ■Plot points on the Cartesian plane. ■ ■Calculate the distance between two points in the Cartesian plane. ■ ■Find the midpoint of a line segment joining two points in the Cartesian plane. C O N C E P T U A L O B J E C T I V E S ■ ■Expand the concept of a one-dimensional number line to a two-dimensional plane. ■ ■Derive the distance formula using the Pythagorean theorem. ■ ■Conceptualize the midpoint as the average of the x- and y-coordinates. 2.1 BASIC TOOLS: CARTESIAN PLANE, DISTANCE, AND MIDPOINT [CONCEPT CHECK] If the point (a, b) lies in quadrant I, then the point (2a, 2b) lies in which quadrant? ANSWER Quadrant III. ▼ x-axis Origin I II IV III y-axis x I x > 0, y > 0 II x < 0, y > 0 IV x > 0, y < 0 III x < 0, y < 0 y x x-coordinate y-coordinate (–4, 1) (–3, –2) (0, 0) (0, –2) (3, –4) (5, 0) (2, 4) y EXAMPLE 1 Plotting Points in a Cartesian Plane a. Plot and label the points 121, 242, 12, 22, 122, 32, 12, 232, 10, 52, and 123, 02 in the Cartesian plane. b. List the points and corresponding quadrant or axis in a table. Solution: a. x (2, 2) y (2, –3) (–1, –4) (–3, 0) (–2, 3) (0, 5) b. POINT QUADRANT 12, 22 I 122, 32 II 121, 242 III 12, 232 IV 10, 52 y-axis 123, 02 x-axis 2.1.1 SKI LL Plot points on the Cartesian plane. 2.1.1 CO NCE PTUAL Expand the concept of a one-dimensional number line to a two-dimensional plane. 2.1 Basic Tools: Cartesian Plane, Distance, and Midpoint 167 2.1.2 Distance Between Two Points Suppose you want to find the distance between any two points in the plane. In the previous graph, to find the distance between the points 12, 232 and 12, 22, count the units between the two points. The distance is 5. What if the two points do not lie along a horizontal or vertical line? Example 2 uses the Pythagorean theorem to help find the distance between any two points. 2.1.2 S K I L L Calculate the distance between two points in the Cartesian plane. 2.1.2 C O N C E P T U A L Derive the distance formula using the Pythagorean theorem. WORDS MATH For any two points, 1x1, y12 and 1x2, y2 2: The distance along the horizontal segment is the absolute value of the difference between the x-values. 0 x2 2 x10 x y (x1, y1) (x2, y2) \|x2 – x1\| \|y2 – y1\| d EXAMPLE 2 Finding the Distance Between Two Points Find the distance between the points 122, 212 and 11, 32. Solution: STEP 1 Plot and label the two points in the Cartesian plane and draw a line segment indicating the distance d between the two points. STEP 2 Form a right triangle by connecting the points to a third point, 11, 212. STEP 3 Calculate the length of the horizontal segment. 3 5 \|1 2 1222 \| Calculate the length of the vertical segment. 4 5 \|3 2 1212 \| STEP 4 Use the Pythagorean theorem to d2 5 32 1 42 calculate the distance d. d2 5 25 d 5 5 x (1, 3) d y (–2, –1) x (1, 3) y (–2, –1) (1, –1) d 4 3 168 CHAPTER 2 Graphs The distance along the vertical segment is the absolute value of the difference between the y-values. 0 y2 2 y10 Use the Pythagorean theorem to calculate the distance d. d2 5 0 x2 2 x10 2 1 0 y2 2 y10 2 0 a0 2 5 a2 for all real numbers a. d2 5 1x2 2 x122 1 1 y2 2 y122 Use the square root property. d 56 "1x2 2 x122 1 1y2 2 y122 Distance can be only positive. d 5 "1x2 2 x122 1 1y2 2 y122 DEFINITION Distance Formula The distance d between two points P1 5 1x1, y12 and P2 5 1x2, y22 is given by d 5 "1x2 2 x122 1 1y2 2 y122 The distance between two points is the square root of the sum of the square of the distance between the x-coordinates and the square of the distance between the y-coordinates. You will prove in the exercises that it does not matter which point you take to be the first point when applying the distance formula. EXAMPLE 3 Using the Distance Formula to Find the Distance Between Two Points Find the distance between 123, 72 and 15, 222. Solution: Write the distance formula. d 5 "3x2 2 x142 1 3y2 2 y142 Substitute 1x1, y12 5 123, 72 and 1x2, y22 5 15, 222. d 5 "35 2 123242 1 322 2 742 Simplify. d 5 "35 1 342 1 322 2 742 d 5 "82 1 12922 5 "64 1 81 5 "145 Solve for d. d 5 "145 Y OUR TU R N Find the distance between 14, 252 and 123, 222. ▼ A N S W E R d 5 "58 ▼ [CONCEPT CHECK] Which point is P1 and which point is P2? ANSWER It does not matter which point is labeled P1 and which point is labeled P2. ▼ 2.1.3 Midpoint of a Line Segment Joining Two Points The midpoint, 1xm, ym 2 , of a line segment joining two points 1x1, y1 2 and 1x2, y2 2 is defined as the point that lies on the segment that has the same distance d from both points. In other words, the midpoint of a segment lies halfway between the given endpoints. The coordinates of the midpoint are found by averaging the x-coordinates and averaging the y-coordinates. x P1 = (x1, y1) (xm, ym) P2 = (x2, y2) y d d STUDY TIP It does not matter which point is taken to be the first point or the second point. 2.1 Basic Tools: Cartesian Plane, Distance, and Midpoint 169 DEFINITION Midpoint Formula The midpoint, 1xm, ym 2 , of the line segment with endpoints 1x1, y12 and 1x2, y22 is given by 1xm, ym2 5 ax1 1 x2 2 , y1 1 y2 2 b The midpoint can be found by averaging the x-coordinates and averaging the y-coordinates. 2.1.3 S K I L L Find the midpoint of a line segment joining two points in the Cartesian plane. 2.1.3 C ON C E P T U A L Conceptualize the midpoint as the average of the x- and y-coordinates. EXAMPLE 4 Finding the Midpoint of a Line Segment Find the midpoint of the line segment joining the points 12, 62 and 124, 222 . Solution: Write the midpoint formula. 1xm, ym2 5 ax1 1 x2 2 , y1 1 y2 2 b Substitute 1x1, y12 5 12, 62 1xm, ym2 5 a2 1 1242 2 , 6 1 1222 2 b and 1x2, y22 5 124, 222. Simplify. 1xm, ym2 5 121, 22 One way to verify your answer is to plot the given points and the midpoint to make sure your answer looks reasonable. YOUR T UR N Find the midpoint of the line segment joining the points 13, 242 and 15, 82. x (2, 6) y (–4, –2) (–1, 2) ▼ A N S W E R Midpoint 5 14, 22 ▼ [CONCEPT CHECK] Graph the two points (3, –4) and (5, 8) and draw the line segment connecting these points. Take a “guess” at the midpoint. ANSWER Note the midpoint is (4, 2). ▼ y-axis II (+, +) (–, +) (+, –) (–, –) III IV I x-axis Origin CART E S IAN P LANE n Plotting coordinates: 1x, y2 n Quadrants: I, II, III, and IV n Origin: 10, 02 D I S TA N C E B ET W EEN T W O PO IN T S d 5 "1x2 2 x122 1 1y2 2 y122 M I D PO I N T O F L I N E S EGM E N T JO INING T W O PO I N T S Midpoint 5 1xm, ym2 5 ax1 1 x2 2 , y1 1 y2 2 b [SEC TION 2 .1] S U M M A RY 170 CHAPTER 2 Graphs In Exercises 1–6, give the coordinates for each point labeled. 1. Point A 2. Point B 3. Point C 4. Point D 5. Point E 6. Point F In Exercises 7 and 8, plot each point in the Cartesian plane and indicate in which quadrant or on which axis the point lies. 7. A: 122, 32 B: 11, 42 C: 123, 232 D: 15, 212 E: 10, 222 F: 14, 02 8. A: 121, 22 B: 11, 32 C: 124, 212 D: 13, 222 E: 10, 52 F: 123, 02 9. Plot the points 123, 12, 123, 42, 123, 222, 123, 02, 123, 242. Describe the line containing points of the form 123, y2. 10. Plot the points 121, 22, 123, 22, 10, 22, 13, 22, 15, 22. Describe the line containing points of the form 1x, 22. In Exercises 11–32, calculate the distance between the given points, and find the midpoint of the segment joining them. 11. 11, 32 and 15, 32 12. 122, 42 and 122, 242 13. 121, 42 and 13, 02 14. 123, 212 and 11, 32 15. 1210, 82 and 127, 212 16. 122, 122 and 17, 152 17. 123, 212 and 127, 22 18. 124, 52 and 129, 272 19. 126, 242 and 122, 282 20. 10, 272 and 124, 252 21. A21 2, 1 3B and A7 2, 10 3 B 22. A1 5, 7 3B and A9 5,22 3B 23. A22 3,2 1 5 B and A1 4, 1 3B 24. A7 5, 1 9B and A1 2, 27 3B 25. 121.5, 3.22 and 12.1, 4.72 26. 121.2, 22.52 and 13.7, 4.62 27. 1214.2, 15.12 and 116.3, 217.52 28. 11.1, 2.22 and 13.3, 4.42 29. 1!3, 5!22 and 1!3, !22 30. 13!5, 23!32 and 12!5, 2!32 31. 11, !32 and 12!2,222 32. 12!5, 42 and 11, 2!32 In Exercises 33 and 34, calculate (to two decimal places) the perimeter of the triangle with the following vertices: 33. Points A, B, and C 34. Points C, D, and E In Exercises 35–38, determine whether the triangle with the given vertices is a right triangle, an isosceles triangle, neither, or both. (Recall that a right triangle satisfies the Pythagorean theorem and an isosceles triangle has at least two sides of equal length.) 35. 10, 232, 13, 232, and 13, 52 36. 10, 22, 122, 222, and 12, 222 37. 11, 12, 13, 212, and 122, 242 38. 123, 32, 13, 32, and 123, 232 x F y E D C B A x C y D A B E [SEC TION 2 .1] E X E R C I SES • S K I L L S 2.1 Basic Tools: Cartesian Plane, Distance, and Midpoint 171 39. Cell Phones. A cellular phone company currently has three towers: one in Tampa, one in Orlando, and one in Gainesville to serve the central Florida region. If Orlando is 80 miles east of Tampa and Gainesville is 100 miles north of Tampa, what is the distance from Orlando to Gainesville? 40. Cell Phones. The same cellular phone company in Exercise 39 has decided to add towers at each “halfway” between cities. How many miles from Tampa is each “halfway” tower? 41. Travel. A retired couple who live in Columbia, South Carolina, decide to take their motor home and visit two children who live in Atlanta and in Savannah, Georgia. Savannah is 160 miles south of Columbia, and Atlanta is 215 miles west of Columbia. How far apart do the children live from each other? Albany Marietta Savannah Atlanta Columbia Macon Columbus Greenville Charleston 42. Sports. In the 1984 Orange Bowl, Doug Flutie, the 5 foot 9 inch quarterback for Boston College, shocked the world as he threw a “hail Mary” pass that was caught in the end zone with no time left on the clock, defeating the Miami Hurricanes 47–45. Although the record books have it listed as a 48-yard pass, what was the actual distance the ball was thrown? The following illustration depicts the path of the ball. x y (5, 50) (–10, 2) Midfeld End Zone 43. NASCAR Revenue. Action Performance Inc., the leading seller of NASCAR merchandise, recorded $260 million in revenue in 2002 and $400 million in revenue in 2004. Calculate the midpoint to estimate the revenue Action Performance Inc. recorded in 2003. Assume the horizontal axis represents the year and the vertical axis represents the revenue in millions. Revenue in Millions (2004, 400) (2002, 260) Year 2001 2004 2007 2010 200 100 300 400 500 44. Ticket Price. In 1993 the average Miami Dolphins ticket price was $28, and in 2001 the average price was $56. Find the midpoint of the segment joining these two points to estimate the ticket price in 1997. Average Cost of Miami Dolphins Ticket (2001, 56) (1993, 28) Year 1995 2000 40 20 60 In Exercises 45 and 46, refer to the following: It is often useful to display data in visual form by plotting the data as a set of points. This provides a graphical display between the two variables. The following table contains data on the average monthly price of gasoline. U.S. All Grades Conventional Retail Gasoline Prices, 2000–2015 (Dollars per Gallon) YEAR JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC 2000 1.319 1.409 1.538 1.476 1.496 1.645 1.568 1.480 1.562 1.546 1.533 1.458 2001 1.467 1.471 1.423 1.557 1.689 1.586 1.381 1.422 1.539 1.312 1.177 1.111 2002 1.134 1.129 1.259 1.402 1.394 1.380 1.402 1.398 1.403 1.466 1.424 1.389 2003 1.464 1.622 1.675 1.557 1.477 1.489 1.519 1.625 1.654 1.551 1.512 1.488 2004 1.595 1.654 1.728 1.794 1.981 1.950 1.902 1.880 1.880 1.993 1.973 1.843 2005 1.852 1.927 2.102 2.251 2.155 2.162 2.287 2.489 2.907 2.736 2.265 2.216 • A P P L I C A T I O N S 172 CHAPTER 2 Graphs The following graph displays the data for the year 2000. 47. Calculate the distance between 12, 72 and 19, 102. Solution: Write the distance formula. d 5 "1x2 2 x122 1 1y2 2 y122 Substitute 12, 72 and 19, 102. d 5 "17 2 222 1 110 2 922 Simplify. d 5 "1522 1 1122 5 "26 This is incorrect. What mistake was made? 48. Calculate the distance between 122, 12 and 13, 272. Solution: Write the distance formula. d 5 "1x2 2 x122 1 1y2 2 y122 Substitute 122, 12 and 13, 272. d 5 "13 2 222 1 127 2 122 Simplify. d 5 "1122 1 12822 5 "65 This is incorrect. What mistake was made? 49. Compute the midpoint of the segment with endpoints 123, 42 and 17, 92. Solution: Write the midpoint 1xm, ym2 5 ax1 1 x2 2 , y1 1 y2 2 b formula. Substitute 123, 42 1xm, ym2 5 a23 1 4 2 , 7 1 9 2 b and 17, 92. Simplify. 1xm, ym2 5 a1 2, 16 2 b 5 a1 2, 4b This is incorrect. What mistake was made? 50. Compute the midpoint of the segment with endpoints 121, 222 and 123, 242. Solution: Write the midpoint 1xm, ym2 5 ax1 2 x2 2 , y1 2 y2 2 b formula. Substitute 121, 222 and 123, 242. 1xm, ym2 5 a21 2 1232 2 , 22 2 1242 2 b Simplify. 1xm, ym2 5 11, 12 This is incorrect. What mistake was made? • C A T C H T H E M I S T A K E YEAR JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC 2006 2.343 2.293 2.454 2.762 2.873 2.849 2.964 2.952 2.548 2.258 2.254 2.328 2007 2.237 2.276 2.546 2.831 3.157 3.067 2.989 2.821 2.858 2.838 3.110 3.032 2008 3.068 3.064 3.263 3.468 3.783 4.038 4.051 3.789 3.760 3.065 2.153 1.721 2009 1.821 1.942 1.987 2.071 2.289 2.645 2.530 2.613 2.530 2.549 2.665 2.620 2010 2.730 2.657 2.793 2.867 2.847 2.733 2.728 2.733 2.727 2.816 2.866 3.004 2011 3.109 3.219 3.561 3.796 3.900 3.678 3.665 3.664 3.624 3.454 3.385 3.277 2012 3.388 3.576 3.827 3.893 3.698 3.515 3.433 3.724 3.859 3.714 3.444 3.322 2013 3.324 3.668 3.713 3.566 3.621 3.634 3.582 3.583 3.542 3.358 3.263 3.288 2014 3.331 3.382 3.546 3.665 3.678 3.699 3.615 3.501 3.431 3.202 2.957 2.575 2015 2.136 2.235 2.433 2.454 2.661 2.783 2.750 2.608 2.369 2.325 2.188 2.052 Source: U.S. Energy Information Administration (Feb. 2016). 2 4 6 8 12 10 $2.00 1.80 1.60 1.40 1.20 1.00 0.80 0.60 0.40 0.20 Month Dollars per Gallon (2000) 45. Economics. Create a graph displaying the price of gasoline for the year 2012. 46. Economics. Create a graph displaying the price of gasoline for the year 2014. In Exercises 47–50, explain the mistake that is made. 2.2 Graphing Equations, Point-Plotting, Intercepts, and Symmetry 173 • C O N C E P T U A L In Exercises 51–54, determine whether each statement is true or false. x y (0, 0) (b, 0) (a, c) (a + b, c) S K I L L S O B J E C T I V E S ■ ■Sketch graphs of equations by plotting points. ■ ■Find intercepts for graphs of equations. ■ ■Determine if the graph of an equation is symmetric about the x-axis, y-axis, or origin. ■ ■Use intercepts and symmetry as graphing aids. C O N C E P T U A L O B J E C T I V ES ■ ■Understand that if a point 1a, b2 satisfies the equation, then that point lies on its graph. ■ ■Understand that intercepts are points that lie on the graph and either the x-axis or y-axis. ■ ■Relate symmetry graphically and algebraically. ■ ■Understand that intercepts are good starting points, but not the only points, and that symmetry eliminates the need to find points in other quadrants. 2.2 GRAPHING EQUATIONS, POINT-PLOTTING, INTERCEPTS, AND SYMMETRY In this section, you will learn how to graph equations by plotting points. However, when we discuss graphing principles in Chapter 3, you will see that other techniques can be more efficient. 2.2.1 Point-Plotting Most equations in two variables, such as y 5 x2, have an infinite number of ordered pairs as solutions. For example, 10, 02 is a solution to y 5 x2 because when x 5 0 and y 5 0, the equation is true. Two other solutions are 121, 12 and 11, 12. 2.2.1 S K I L L Sketch graphs of equations by plotting points. 2.2.1 C ON C E P T U A L Understand that if a point 1a, b2 satisfies the equation, then that point lies on its graph. 51. The distance from the origin to the point 1a, b2 is d 5 "a2 1 b2. 52. The midpoint of the line segment joining the origin and the point 1a, a2 is aa 2, a 2b. 53. The midpoint of any segment joining two points in quadrant I also lies in quadrant I. 54. The midpoint of any segment joining a point in quadrant I to a point in quadrant III also lies in either quadrant I or III. 55. Calculate the length and the midpoint of the line segment joining the points 1a, b2 and 1b, a2. 56. Calculate the length and the midpoint of the line segment joining the points 1a, b2 and 12a, 2b2. 57. Assume that two points 1x1, y12 and 1x2, y22 are connected by a segment. Prove that the distance from the midpoint of the segment to either of the two points is the same. 58. Prove that the diagonals of a parallelogram in the figure intersect at their midpoints. 59. Assume that two points 1a, b2 and 1c, d 2 are the endpoints of a line segment. Calculate the distance between the two points. Prove that it does not matter which point is labeled as the “first” point in the distance formula. 60. Show that the points 121, 212, 10, 02, and 12, 22 are collinear (lie on the same line) by showing that the sum of the distance from 121, 212 to 10, 02 and the distance from 10, 02 to 12, 22 is equal to the distance from 121, 212 to 12, 22. • C H A L L E N G E • T E C H N O L O G Y In Exercises 61–64, calculate the distance between the two points. Use a graphing utility to graph the segment joining the two points and find the midpoint of the segment. 61. 122.3, 4.12 and 13.7, 6.22 62. 124.9, 23.22 and 15.2, 3.42 63. 11.1, 2.22 and 13.3, 4.42 64. 121.3, 7.22 and 12.3, 24.52 174 CHAPTER 2 Graphs The graph of an equation in two variables, x and y, consists of all the points in the xy-plane whose coordinates 1x, y2 satisfy the equation. A procedure for plotting the graphs of equations is outlined below and is illustrated with the example y 5 x2. WORDS MATH Step 1: In a table, list several pairs of coordinates that make the equation true. Step 2: Plot these points on a graph and connect the points with a smooth curve. Use arrows to indicate that the graph continues. In graphing an equation, first select arbitrary values for x and then use the equation to find the corresponding value of y, or vice versa. x y 5 x 2 (x, y ) 0 0 10, 02 21 1 121, 12 1 1 11, 12 22 4 122, 42 2 4 12, 42 x y (–2, 4) (–1, 1) (0, 0) (2, 4) (1, 1) EXAMPLE 1 Graphing an Equation of a Line by Plotting Points Graph the equation y 5 2x 2 1. Solution: STEP 1 In a table, list several pairs of coordinates that make the equation true. STEP 2 Plot these points on a graph and connect the points, resulting in a line. Y OUR TU R N The graph of the equation y 5 2x 1 1 is a line. Graph the line. x y 5 2x 2 1 (x, y ) 0 21 10, 212 21 23 121, 232 1 1 11, 12 22 25 122, 252 2 3 12, 32 x y (–1, –3) (0, –1) (–2, –5) (2, 3) (1, 1) ▼ ▼ A N S W E R x y (–1, 2) (1, 0) EXAMPLE 2 Graphing an Equation by Plotting Points Graph the equation y 5 x2 2 5. Solution: STEP 1 In a table, list several pairs of coordinates that make the equation true. STEP 2 Plot these points on a graph and connect the points with a smooth curve, indicating with arrows that the curve continues. This graph is called a parabola and will be discussed in further detail in Chapter 8. YOUR T UR N Graph the equation y 5 x2 2 1. x y (–3, 4) (3, 4) (–2, –1) (–1, –4) (0, –5) (2, –1) (1, –4) ▼ A N S W E R x y (–1, 0) (0, –1) (–2, 3) (2, 3) (1, 0) ▼ x y 5 x 2 2 5 (x, y ) 0 25 10, 252 21 24 121, 242 1 24 11, 242 22 21 122, 212 2 21 12, 212 23 4 123, 42 3 4 13, 42 EXAMPLE 3 Graph ing an Equation by Plotting Points Graph the equation y 5 x3. Solution: STEP 1 In a table, list several pairs of coordinates that satisfy the equation. STEP 2 Plot these points on a graph and connect the points with a smooth curve, indicating with arrows that the curve continues in both the positive and negative directions. x y 5 x 3 (x, y ) 0 0 10, 02 21 21 121, 212 1 1 11, 12 22 28 122, 282 2 8 12, 82 x y (2, 8) (–2, –8) (–1, –1) (0, 0) (1, 1) 4 8 –8 –4 [CONCEPT CHECK] Show the points (1, 0) and (2, 3) satisfy the equation y 5 x 2 2 1. ANSWER (1, 0): 0 5 12 – 1 and (2, 3): 3 5 2 2 2 1 ▼ 2.2 Graphing Equations, Point-Plotting, Intercepts, and Symmetry 175 176 CHAPTER 2 Graphs 2.2.2 Intercepts When point-plotting graphs of equations, which points should be selected? Points where a graph crosses (or touches) either the x-axis or y-axis are called intercepts, and identifying these points helps define the graph unmistakably. An x-intercept of a graph is a point where the graph intersects the x-axis. Specifically, an x-intercept is the x-coordinate of such a point. For example, if a graph intersects the x-axis at the point 13, 02, then we say that 3 is the x-intercept. Since the value for y along the x-axis is zero, all points corresponding to x-intercepts have the form 1a, 02. A y-intercept of a graph is a point where the graph intersects the y-axis. Specifically, a y-intercept is the y-coordinate of such a point. For example, if a graph intersects the y-axis at the point 10, 22, then we say that 2 is the y-intercept. Since the value for x along the y-axis is zero, all points corresponding to y-intercepts have the form 10, b2. It is important to note that graphs of equations do not have to have intercepts, and if they do have intercepts, they can have one or more of each type. x y One x-intercept Two y-intercepts x y No x-intercepts One y-intercept x y Three x-intercepts One y-intercept x y No x-intercepts No y-intercepts x y (–1, 0) (0, –1) (3, –2) (3, 2) (0, 1) Note: The origin 10, 02 corresponds to both an x-intercept and a y-intercept. The graph given to the left has two y-intercepts and one x-intercept. ■ ■The x-intercept is 21, which corresponds to the point 121, 02. ■ ■The y-intercepts are 21 and 1, which correspond to the points 10, 212 and 10, 12, respectively. Algebraically, how do we find intercepts from an equation? The graph in the margin corresponds to the equation x 5 y2 2 1. The x-intercepts are located on the x-axis, which corresponds to y 5 0. If we let y 5 0 in the equation x 5 y2 2 1 and solve for x, the result is x 5 21. This corresponds to the x-intercept we identified above. Similarly, the y-intercepts are located on the y-axis, which corresponds to x 5 0. If we let x 5 0 in the equation x 5 y2 2 1 and solve for y, the result is y 5 61. These correspond to the y-intercepts we identified above. 2.2.2 S KILL Find intercepts for graphs of equations. 2.2.2 CO NCE PTUAL Understand that intercepts are points that lie on the graph and either the x-axis or y-axis. EXAMPLE 4 Finding Intercepts from an Equation Given the equation y 5 x2 1 1, find the indicated intercepts of its graph, if any. a. x-intercept(s) b. y-intercept(s) STUDY TIP Identifying the intercepts helps define the graph unmistakably. ▼ A N S W E R a. x-intercepts: 22 and 2 b. y-intercept: 24 [CONCEPT CHECK] If the graph passes through the origin, then the point (0, 0) is (A) an x-intercept, (B) a y-intercept, or (C) both an x-intercept and a y-intercept. ANSWER (C) both ▼ Solution (a): Let y 5 0. 0 5 x2 1 1 Solve for x. x2 5 21 no real solution There are no x-intercepts. Solution (b): Let x 5 0. y 5 02 1 1 Solve for y. y 5 1 The y-intercept is located at the point 10, 12. Y OUR T UR N For the equation y 5 x2 2 4, a. find the x-intercept(s), if any. b. find the y-intercept(s), if any. ▼ 2.2.3 Symmetry The word symmetry conveys balance. Suppose you have two pictures to hang on a wall. If you space them equally apart on the wall, then you prefer a symmetric décor. This is an example of symmetry about a line. The word (water) written below is identical if you rotate the word 180 degrees (or turn the page upside down). This is an example of symmetry about a point. Symmetric graphs have the characteristic that their mirror image can be obtained about a reference, typically a line or a point. In Example 2, the points 122, 212 and 12, 212 both lie on the graph of y 5 x2 2 5, as do the points 121, 242 and 11, 242. Notice that the graph on the right side of the y-axis is a mirror image of the part of the graph to the left of the y-axis. This graph illustrates symmetry with respect to the y-axis 1the line x 5 02. In the graph below of the equation x 5 y2 2 1, the points 10, 12 and 10, 212 both lie on the graph, as do the points 13, 22 and 13, 222. Notice that the part of the graph above the x-axis is a mirror image of the part of the graph below the x-axis. This graph illustrates symmetry with respect to the x-axis 1the line y 5 02. In Example 3, the points 121, 212 and 11, 12 both lie on the graph. Notice that rotating this graph 180 degrees (or turning your page upside down) results in an identical graph. This is an example of symmetry with respect to the origin 10, 02. Symmetry aids in graphing by giving information “for free.” For example, if a graph is symmetric about the y-axis, then once the graph to the right of the y-axis is 2.2.3 S K I L L Determine if the graph of an equation is symmetric about the x-axis, y-axis, or origin. 2.2.3 C ON C E P T U A L Relate symmetry graphically and algebraically. x y (–3, 4) (3, 4) (–2, –1) (–1, –4) (0, –5) (2, –1) y = x2–5 (1, –4) x y (–1, 0) (0, –1) (3, –2) (3, 2) (0, 1) x = y2–1 x y (2, 8) (–2, –8) (–1, –1) (0, 0) (1, 1) 4 8 –8 –4 y = x3 2.2 Graphing Equations, Point-Plotting, Intercepts, and Symmetry 177 178 CHAPTER 2 Graphs found, the left side of the graph is the mirror image of that. If a graph is symmetric about the origin, then once the graph is known in quadrant I, the graph in quadrant III is found by rotating the known graph 180 degrees. It would be beneficial to know whether a graph of an equation is symmetric about a line or point before the graph of the equation is sketched. Although a graph can be symmetric about any line or point, we will discuss only symmetry about the x-axis, y-axis, and origin. These types of symmetry and the algebraic procedures for testing for symmetry are outlined below. Types and Tests for Symmetry TYPE OF SYMMETRY GRAPH IF THE POINT (a, b ) IS ON THE GRAPH, THEN THE POINT . . . ALGEBRAIC TEST FOR SYMMETRY Symmetric with respect to the x-axis 1a, 2b2 is on the graph. Replacing y with 2y leaves the equation unchanged. Symmetric with respect to the y-axis 12a, b2 is on the graph. Replacing x with 2x leaves the equation unchanged. Symmetric with respect to the origin 12a, 2b2 is on the graph. Replacing x with 2x and y with 2y leaves the equation unchanged. a –b b x y (a, b) (a, –b) x –a a b y (a, b) (–a, b) –a a –b b x y (0, 0) (a, b) (–a, –b) EXAMPLE 5 Testing for Symmetry with Respect to the Axes Test the equation y2 5 x3 for symmetry with respect to the axes. Solution: Test for symmetry with respect to the x-axis. Replace y with 2y. 12y22 5 x3 Simplify. y2 5 x3 The resulting equation is the same as the original equation, y2 5 x3. Therefore y2 5 x3 is symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis. Replace x with 2x. y2 5 12x23 Simplify. y2 5 2x3 The resulting equation, y2 5 2x3, is not the same as the original equation, y2 5 x3. Therefore, y2 5 x2 is not symmetric with respect to the y-axis. STUDY TIP Symmetry gives us information about the graph “for free.” When testing for symmetry about the x-axis, y-axis, and origin, there are five possibilities: ■ ■No symmetry ■ ■Symmetry with respect to the x-axis ■ ■Symmetry with respect to the y-axis ■ ■Symmetry with respect to the origin ■ ■Symmetry with respect to the x-axis, y-axis, and origin [CONCEPT CHECK] If the point (a, b) and the, point (a, 2b) are on the graph then the graph is symmetric about (A) the y-axis, (B) the x-axis, or (C) the origin. ANSWER (B) the x-axis ▼ EXAMPLE 6 Testing for Symmetry Determine what type of symmetry (if any) the graphs of the equations exhibit. a. y 5 x2 1 1 b. y 5 x3 1 1 Solution (a): Replace x with 2x. y 5 12x2 2 1 1 Simplify. y 5 x2 1 1 The resulting equation is equivalent to the original equation, so the graph of the equation y 5 x2 1 1 is symmetric with respect to the y-axis. Replace y with 2y. 12y2 5 x2 1 1 Simplify. y 5 2x2 2 1 The resulting equation y 5 2x2 2 1 is not equivalent to the original equation y 5 x2 1 1, so the graph of the equation y 5 x2 1 1 is not symmetric with respect to the x-axis. Replace x with 2x and y with 2y. 12y2 5 12x2 2 1 1 Simplify. 2y 5 x2 1 1 y 5 2x2 2 1 The resulting equation y 5 2x2 2 1 is not equivalent to the original equation y 5 x2 1 1, so the graph of the equation y 5 x2 1 1 is not symmetric with respect to the origin. The graph of the equation y 5 x2 1 1 is symmetric with respect to the y-axis. Solution (b): Replace x with 2x. y 5 12x2 3 1 1 Simplify. y 5 2x3 1 1 The resulting equation y 5 2x3 1 1 is not equivalent to the original equation y 5 x3 1 1. Therefore, the graph of the equation y 5 x3 1 1 is not symmetric with respect to the y-axis. Replace y with 2y. 12y2 5 x3 1 1 Simplify. y 5 2x3 2 1 The resulting equation y 5 2x3 2 1 is not equivalent to the original equation y 5 x3 1 1. Therefore, the graph of the equation y 5 x3 1 1 is not symmetric with respect to the x-axis. Replace x with 2x and y with 2y. 12y2 5 12x2 3 1 1 Simplify. 2y 5 2x3 1 1 y 5 x3 2 1 The resulting equation y 5 x3 2 1 is not equivalent to the original equation y 5 x3 1 1. Therefore, the graph of the equation y 5 x3 1 1 is not symmetric with respect to the origin. The graph of the equation y 5 x3 1 1 exhibits no symmetry. Y OUR T UR N Determine the symmetry (if any) for x 5 y2 2 1. ▼ ▼ A N S W E R The graph of the equation is symmetric with respect to the x-axis. 2.2 Graphing Equations, Point-Plotting, Intercepts, and Symmetry 179 180 CHAPTER 2 Graphs 2.2.4 Using Intercepts and Symmetry as Graphing Aids How can we use intercepts and symmetry to assist us in graphing? Intercepts are a good starting point—though not the only one. For symmetry, look back at Example 2, y 5 x2 2 5. We selected seven x-coordinates and solved the equation to find the corresponding y- coordinates. If we had known that this graph was symmetric with respect to the y-axis, then we would have had to find the solutions to only the positive x-coordinates, since we get the negative x-coordinates for free. For example, we found the point 11, 242 to be a solution to the equation. The rules of symmetry tell us that 121, 242 is also on the graph. 2.2.4 S KILL Use intercepts and symmetry as graphing aids. 2.2.4 CO NCE PTUAL Understand that intercepts are good starting points, but not the only points, and that symmetry eliminates the need to find points in other quadrants. EXAMPLE 7 Using Intercepts and Symmetry as Graphing Aids For the equation x2 1 y2 5 25, use intercepts and symmetry to help you graph the equation using the point-plotting technique. Solution: STEP 1 Find the intercepts. For the x-intercepts, let y 5 0. x2 1 02 5 25 Solve for x. x 5 65 The two x-intercepts correspond to the points 125, 02 and 15, 02. For the y-intercepts, let x 5 0. 02 1 y2 5 25 Solve for y. y 5 65 The two y-intercepts correspond to the points 10, 252 and 10, 52. STEP 2 Identify the points on the graph corresponding to the intercepts. STEP 3 Test for symmetry with respect to the y-axis, x-axis, and origin. Test for symmetry with respect to the y-axis. Replace x with 2x. 12x22 1 y2 5 25 Simplify. x2 1 y2 5 25 The resulting equation is equivalent to the original, so the graph of x2 1 y2 5 25 is symmetric with respect to the y-axis. Test for symmetry with respect to the x-axis. Replace y with 2y. x2 1 12y22 5 25 Simplify. x2 1 y2 5 25 The resulting equation is equivalent to the original, so the graph of x2 1 y2 5 25 is symmetric with respect to the x-axis. y x 5 –5 –5 5 [CONCEPT CHECK] If the graph is symmetric about the x-axis, y-axis, and the origin, then once you find a point (a, b) in quadrant I that lies on the graph what points do you get for free from the symmetry? (A) (2a, b), (B) (a, 2b), (C) (2a, 2b), or (D) all of the above ANSWER (D) all of the above. ▼ Test for symmetry with respect to the origin. Replace x with 2x and y with 2y. 12x22 1 12y22 5 25 Simplify. x2 1 y2 5 25 The resulting equation is equivalent to the original, so the graph of x2 1 y2 5 25 is symmetric with respect to the origin. Since the graph is symmetric with respect to the y-axis, x-axis, and origin, we need to determine solutions to the equation on only the positive x- and y-axes and in quadrant I because of the following symmetries: ■ Symmetry with respect to the y-axis gives the solutions in quadrant II. ■ Symmetry with respect to the origin gives the solutions in quadrant III. ■ Symmetry with respect to the x-axis yields solutions in quadrant IV. Solutions to x2 1 y2 5 25. Quadrant I: 13, 42, 14, 32 Additional points due to symmetry: Quadrant II: 123, 42 , 124, 32 Quadrant III: 123, 242 , 124, 232 Quadrant IV: 13, 242 , 14, 232 Connecting the points with a smooth curve yields a circle. We will discuss circles in more detail in Section 2.4. x (–3, 4) (3, 4) (–4, 3) (4, 3) (5, 0) (–5, 0) (0, –5) (0, 5) (4, –3) (–4, –3) (3, –4) (–3, –4) y 2.2 Graphing Equations, Point-Plotting, Intercepts, and Symmetry 181 Symmetry about the x-axis, y-axis, and origin is defined both algebraically and graphically. Intercepts and symmetry provide much of the information useful for sketching graphs of equations. Sketching graphs of equations can be accomplished using a point-plotting technique. Intercepts are defined as points where a graph intersects an axis or the origin. [SEC TION 2 . 2] S U M M A RY INT E R CE P T S INTERCEPTS POINT WHERE THE GRAPH INTERSECTS THE . . . HOW TO FIND INTERCEPTS POINT ON GRAPH x-intercept x-axis Let y 5 0 and solve for x. 1a, 02 y-intercept y-axis Let x 5 0 and solve for y. 10, b2 TY P E S AND T E S T S FOR S Y MM ETRY TYPE OF SYMMETRY GRAPH IF THE POINT (a, b ) IS ON THE GRAPH, THEN THE POINT . . . ALGEBRAIC TEST FOR SYMMETRY Symmetric with respect to the x-axis a –b b x y (a, b) (a, –b) 1a, 2b2 is on the graph. Replacing y with 2y leaves the equation unchanged. 182 CHAPTER 2 Graphs EXAMPLE7 [SEC TION 2 . 2] E X E R C I SE S • S K I L L S In Exercises 1–8, determine whether each point lies on the graph of the equation. Symmetric with respect to the y-axis x –a a b y (a, b) (–a, b) 12a, b2 is on the graph. Replacing x with 2x leaves the equation unchanged. Symmetric with respect to the origin –a a –b b x y (0, 0) (a, b) (–a, –b) 12a, 2b2 is on the graph. Replacing x with 2x and y with 2y leaves the equation unchanged. 1. y 5 3x 2 5 a. 11, 22 b. 122, 2112 3. y 5 2 5x 2 4 a. 15, 222 b. 125, 62 5. y 5 x2 2 2x 1 1 a. 121, 42 b. 10, 212 7. y 5 "x 1 2 a. 17, 32 b. 126, 42 2. y 5 22x 1 7 a. 121, 92 b. 12, 242 4. y 5 23 4x 1 1 a. 18, 52 b. 124, 42 6. y 5 x3 2 1 a. 121, 02 b. 122, 292 8. y 5 2 1 03 2 x0 a. 19, 242 b. 122, 72 In Exercises 9–14, complete the table and use the table to sketch a graph of the equation. 9. x y 5 2 1 x (x, y ) 22 0 1 10. x y 5 3x 2 1 (x, y ) 21 0 2 11. x y 5 x 2 2 x (x, y ) 21 0 1 2 1 2 12. x y 5 1 2 2x 2 x 2 (x, y ) 23 22 21 0 1 13. x y 5 "x 2 1 (x, y ) 1 2 5 10 14. x y 5 2 "x 1 2 (x, y ) 22 21 2 7 In Exercises 15–22, graph the equation by plotting points. 15. y 5 23x 1 2 16. y 5 4 2 x 17. y 5 x2 2 x 2 2 18. y 5 x2 2 2x 1 1 19. x 5 y2 2 1 20. x 5 0y 1 10 1 2 21. y 5 1 2x 2 3 2 22. y 5 0.5 0x 210 In Exercises 23–32, find the x-intercept(s) and y-intercepts(s) (if any) of the graphs of the given equations. 23. 2x 2 y 5 6 24. 4x 1 2y 5 10 25. y 5 x2 2 9 26. y 5 4x2 2 1 27. y 5 "x 2 4 28. y 5 " 3 x 2 8 29. y 5 1 x2 1 4 30. y 5 x2 2 x 2 12 x 31. 4x2 1 y2 5 16 32. x2 2 y2 5 9 In Exercises 33–38, match the graph with the corresponding symmetry. a. No symmetry b. Symmetry with respect to the x-axis c. Symmetry with respect to the y-axis d. Symmetry with respect to the origin e. Symmetry with respect to the x-axis, y-axis, and origin 33. 34. 35. 36. 37. 38. In Exercises 39–44, a point that lies on a graph is given along with that graph’s symmetry. State the other known points that must also lie on the graph. Point on a Graph The Graph Is Symmetric about the Point on a Graph The Graph Is Symmetric about the 39. 121, 32 x-axis 40. 122, 42 y-axis 41. 17, 2102 origin 42. 121, 212 origin 43. 13, 222 x-axis, y-axis, and origin 44. 121, 72 x-axis, y-axis, and origin In Exercises 45–58, test algebraically to determine whether the equation’s graph is symmetric with respect to the x-axis, y-axis, or origin. 45. x 5 y2 1 4 46. x 5 2y2 1 3 47. y 5 x3 1 x 48. y 5 x5 1 1 49. x 5 0y0 50. x 5 0y0 2 2 51. x2 2 y2 5 100 52. x2 1 2y2 5 30 53. y 5 x2/3 54. x 5 y2/3 55. x2 1 y3 5 1 56. y 5 "1 1 x2 57. y 5 2 x 58. x y 5 1 x (1, 1) (–1, –1) y x y x y x y x y x y (–2, –2) (–3, –3) (3, 3) (2, 2) (0, 0) 2.2 Graphing Equations, Point-Plotting, Intercepts, and Symmetry 183 184 CHAPTER 2 Graphs 73. Sprinkler. A sprinkler will water a grassy area in the shape of x2 1 y2 5 9. Apply symmetry to draw the watered area, assuming the sprinkler is located at the origin. 74. Sprinkler. A sprinkler will water a grassy area in the shape of x2 1 y2 9 5 1. Apply symmetry to draw the watered area, assuming the sprinkler is located at the origin. 75. Electronic Signals: Radio Waves. The received power of an electromagnetic signal is a fraction of the power transmitted. The relationship is given by P received 5 P transmitted . 1 R2 where R is the distance that the signal has traveled in meters. Plot the percentage of transmitted power that is received for R 5 100 m, 1 km, and 10,000 km. 76. Electronic Signals: Laser Beams. The wavelength l and the frequency ƒ of a signal are related by the equation f 5 c l where c is the speed of light in a vacuum, c 5 3.0 3 108 meters per second. For the values, l 5 0.001, l 5 1, and l 5 100 mm, plot the points corresponding to frequency, ƒ. What do you notice about the relationship between frequency and wavelength? Note that the frequency will have units Hz 5 1/seconds. 77. Profit. The profit associated with making a particular product is given by the equation y 5 2x2 1 6x 2 8 where y represents the profit in millions of dollars and x represents the number of thousands of units sold. 1x 5 1 corresponds to 1000 units and y 5 1 corresponds to $1M.2 Graph this equation and determine how many units must be sold to break even 1 profit 5 02. Determine the range of units sold that correspond to making a profit. 78. Profit. The profit associated with making a particular product is given by the equation y 5 2x2 1 4x 2 3 where y represents the profit in millions of dollars and x represents the number of thousands of units sold. 1x 5 1 corresponds to 1000 units and y 5 1 corresponds to $1M.2 Graph this equation and determine how many units must be sold to break even 1 profit 5 02. Determine the range of units sold that correspond to making a profit. 79. Economics. The demand for an electronic device is modeled by p 5 2.95 2 "0.01x 2 0.01 where x is thousands of units demanded per day and p is the price (in dollars) per unit. a. Find the domain of the demand equation. Interpret your result. b. Plot the demand equation. 80. Economics. The demand for a new electronic game is modeled by p 5 39.95 2 "0.01x 2 0.4 where x is thousands of units demanded per day and p is the price (in dollars) per unit. a. Find the domain of the demand equation. Interpret your result. b. Plot the demand equation. In Exercises 59–72, plot the graph of the given equation. 59. y 5 x 60. y 5 2 1 2x 1 3 61. y 5 x2 2 1 62. y 5 9 2 4x2 63. y 5 x3 2 64. x 5 y2 1 1 65. y 5 1 x 66. x y 5 21 67. y 5 0x0 68. 0x0 5 0y0 69. x2 1 y2 5 16 70. x2 4 1 y2 9 5 1 71. x2 2 y2 5 16 72. x2 2 y2 25 5 1 • A P P L I C A T I O N S 81. Graph the equation y 5 x2 1 1. Solution: x y (1, 2) (0, 1) This is incorrect. What mistake was made? x y 5 x 2 1 1 (x, y ) 0 1 10, 12 1 2 11, 22 • C A T C H T H E M I S T A K E In Exercises 81–84, explain the mistake that is made. • C O N C E P T U A L 85. If the point 1a, b2 lies on a graph that is symmetric about the x-axis, then the point 12a, b2 also must lie on the graph. 86. If the point 1a, b2 lies on a graph that is symmetric about the y-axis, then the point 12a, b2 also must lie on the graph. 87. If the point 1a, 2b2 lies on a graph that is symmetric about the x-axis, y-axis, and origin, then the points 1a, b2, 12a, 2b2, and 12a, b2 must also lie on the graph. 88. Two points are all that is needed to plot the graph of an equation. In Exercises 85–88, determine whether each statement is true or false. • C H A L L E N G E 89. Determine whether the graph of y 5 ax2 1 b cx3 has any sym-metry, where a, b, and c are real numbers. 82. Test y 5 2x2 for symmetry with respect to the y-axis. Solution: Replace x with 2x. y 5 212x22 Simplify. y 5 x2 The resulting equation is not equivalent to the original equation; y 5 2x2 is not symmetric with respect to the y-axis. This is incorrect. What mistake was made? 83. Test x 5 0y0 for symmetry with respect to the y-axis. Solution: Replace y with 2y. x 5 02y0 Simplify. x 5 0y0 The resulting equation is equivalent to the original equation; x 5 0y0 is symmetric with respect to the y-axis. This is incorrect. What mistake was made? 84. Use symmetry to help you graph x2 5 y 2 1. Solution: Replace x with 2x. 12x22 5 y 2 1 Simplify. x2 5 y 2 1 x2 5 y 2 1 is symmetric with respect to the x-axis. Determine points that lie on the graph in quadrant I. y x 2 5 y 2 1 (x, y ) 1 0 10, 12 2 1 11, 22 5 2 12, 52 Symmetry with respect to the x-axis implies that 10, 212, 11, 222, and 12, 252 are also points that lie on the graph. x y –5 5 –5 5 This is incorrect. What mistake was made? • T E C H N O L O G Y In Exercises 91–96, graph the equation using a graphing utility and state whether there is any symmetry. 91. y 5 16.7x4 2 3.3x2 1 7.1 94. 3.2x2 2 5.1y2 5 1.3 92. y 5 0.4x5 1 8.2x3 2 1.3x 95. 1.2x2 1 4.7y2 5 19.4 93. 2.3x2 5 5.5 0y0 96. 2.1y2 5 0.8 0x 1 10 2.2 Graphing Equations, Point-Plotting, Intercepts, and Symmetry 185 90. Find the intercepts of y 5 1x 2 a22 2 b2, where a and b are real numbers. 186 CHAPTER 2 Graphs 2.3.1 Graphing a Line What is the shortest path between two points? The answer is a straight line. In this section, we will discuss characteristics of lines such as slope and intercepts. We will also discuss types of lines such as horizontal, vertical, falling, and rising, and recognize relationships between lines such as perpendicular and parallel. At the end of this section, you should be able to find the equation of a line when given two specific pieces of information about the line. First-degree equations, such as y 5 22x 1 4, 3x 1 y 5 6, y 5 2, and x 5 23 have graphs that are straight lines. The first two equations given represent inclined or “slant” lines, whereas y 5 2 represents a horizontal line and x 5 23 represents a vertical line. One way of writing an equation of a straight line is called general form. 2.3.1 S KILL Calculate the slope of a line; graph a line; classify a line as rising, falling, vertical, or horizontal. 2.3.1 CO NCE PTUAL Understand slope as rate of change. S K I L L S O B J E C T I V E S ■ ■Calculate the slope of a line; graph a line; classify a line as rising, falling, vertical, or horizontal. ■ ■Find the equation of a line given either slope and a point that lies on the line or two points that lie on the line. ■ ■Find the equation of a line that is parallel or perpendicular to a given line. C O N C E P T U A L O B J E C T I V E S ■ ■Understand slope as rate of change. ■ ■Understand that the equation of a line given by y 5 mx 1 b corresponds to a line with slope m and y-intercept 10, b2. ■ ■Understand that parallel lines have the same slope, whereas perpendicular lines have slopes that are negative reciprocals of one another. 2.3 LINES Some books refer to this as standard form. x y (1, 4) (–1, 0) (0, 2) (–2, –2) x–intercept y–intercept x y (1, 4) (–1, 0) (0, 2) (–2, –2) x–intercept y–intercept EQUATION OF A STRAIGHT LINE: GENERAL FORM If A, B, and C are constants and x and y are variables, then the equation Ax 1 By 5 C is in general form, and its graph is a straight line. Note: A or B (but not both) can be zero. STUDY TIP x-intercept: Set y 5 0. y-intercept: Set x 5 0. The equation 2x 2 y 5 22 is a first-degree equation, so its graph is a straight line. To graph this line, list two solutions in a table, plot those points, and use a straight edge to draw the line. x y (1, 4) 2x – y = –2 (–2, –2) x y (x, y ) 22 22 122, 222 1 4 11, 42 Intercepts The point where a line crosses, or intersects, the x-axis is called the x-intercept. The point where a line crosses, or intersects, the y-axis is called the y-intercept. By inspecting the graph of the previous line, we see that the x-intercept is 121, 02 and the y-intercept is 10, 22 . 2.3 Lines 187 Intercepts were discussed in Section 2.2. There we found that the graphs of some equations could have no intercepts or one or multiple intercepts. Slant lines, however, have exactly one x-intercept and exactly one y-intercept. Horizontal lines and vertical lines, however, each have only one intercept. TYPE OF LINE EQUATION x-INTERCEPT y-INTERCEPT GRAPH Horizontal y 5 b None b Vertical x 5 a a None Note: The special cases of x 5 0 (y-axis) and y 5 0 (x-axis) have infinitely many y-intercepts and x-intercepts, respectively. x y (0, b) (a, 0) x y EXAMPLE 1 Determining x - and y -Intercepts Determine the x- and y-intercepts (if they exist) for the lines given by the following equations. a. 2x 1 4y 5 10 b. x 5 22 Solution (a): To find the x-intercept, set y 5 0. 2x 1 4102 5 10 Solve for x. 2x 5 10 x 5 5 The x-intercept corresponds to the point 15, 02 . To find the y-intercept, set x 5 0. 2102 1 4y 5 10 Solve for y. 4y 5 10 y 5 5 2 The y-intercept corresponds to the point A0, 5 2B . 2x 1 4y 5 10 DETERMINING INTERCEPTS Coordinates Axis Crossed Algebraic Method x-intercept: 1a, 02 x-axis Set y 5 0 and solve for x. y-intercept: 10, b2 y-axis Set x 5 0 and solve for y. x y (a, 0) (0, b) 188 CHAPTER 2 Graphs Solution (b): This vertical line consists of all points 122, y2. The graph shows that the x-intercept is 22. We also find that the line never crosses the y-axis, so the y-intercept does not exist. Y OUR TU R N Determine the x- and y-intercepts (if they exist) for the lines given by the following equations. a. 3x 2 y 5 2 b. y 5 5 x 5 22 x y x = –2 (–2, 0) ▼ A N S W E R a. x-intercept: A2 3, 0B; y-intercept: 10, 222 b. x-intercept does not exist; y-intercept: 10, 52 ▼ Slope If the graph of 2x – y 5 22 represented an incline that you were about to walk on, would you classify that incline as steep? In the language of mathematics, we use the word slope as a measure of steepness. Slope is the ratio of the change in y over the change in x. An easy way to remember this is rise over run. Let’s find the slope of our graph 2x 2 y 5 22. We’ll let 1x1, y12 5 122, 222 and 1x2, y22 5 11, 42 in the slope formula: m 5 y2 2 y1 x2 2 x1 5 34 2 12224 31 2 12224 5 6 3 5 2 Notice that if we had chosen the two intercepts 1x1, y12 5 10, 22 and 1x2, y22 5 121, 02 instead, we still would have found the slope to be m 5 2. ▼ C A U T I O N Interchanging the coordinates can result in a sign error in the slope. x y (1, 4) 2x – y = –2 (–2, –2) common mistake The most common mistake in calculating slope is writing the coordinates in the wrong order, which results in the slope being opposite in sign. SLOPE OF A LINE A nonvertical line passing through two points 1x1, y12 and 1x2, y22 has slope m, given by the formula m 5 y2 2 y1 x2 2 x1 , where x1 2 x2 or m 5 rise run 5 vertical change horizontal change Note: Always start with the same point for both the x-coordinates and the y-coordinates. x rise run y (x2, y2) (x1, y1) STUDY TIP To get the correct sign (6) for the slope, remember to start with the same point for both x and y. ✖I N C O R R EC T The ERROR is interchanging the coordinates of the first and second points. m 5 4 2 2 1 2 3 The calculated slope is INCORRECT by a negative sign. m 5 2 22 5 21 ✓COR R E C T Label the points. 1x1, y12 5 11, 22 1x2, y22 5 13, 42 Write the slope formula. m 5 y2 2 y1 x2 2 x1 Substitute the coordinates. m 5 4 2 2 3 2 1 Simplify. m 5 2 2 5 1 Find the slope of the line containing the two points 11, 22 and 13, 42. When interpreting slope, always read the graph from left to right. Since we have determined the slope to be 2, or 2 1, we can interpret this as rising two units and running (to the right) one unit. If we start at the point 122, 222 and move two units up and one unit to the right, we end up at the x-intercept, 121, 02. Again, moving two units up and one unit to the right puts us at the y-intercept, 10, 22. Another rise of two and run of one takes us to the point 11, 42. See the figure on the right. Lines fall into one of four categories: rising, falling, horizontal, or vertical. Line Slope Rising Positive 1m + 02 Falling Negative 1m 02 Horizontal Zero 1m 5 02 , hence y 5 b Vertical Undefined, hence x 5 a The slope of a horizontal line is 0 because the y-coordinates of any two points are the same. The change in y in the slope formula’s numerator is 0; hence m 5 0. The slope of a vertical line is undefined because the x-coordinates of any two points are the same. The change in x in the slope formula’s denominator is zero; hence m is undefined. x y a b x y (1, 4) (0, 2) (–2, –2) (–1, 0) Rise = 2 Run = 1 EXAMPLE 2 Graph, Classify the Line, and Determine the Slope Sketch a line through each pair of points, classify the line as rising, falling, vertical, or horizontal, and determine its slope. a. 121, 232 and 11, 12 b. 123, 32 and 13, 12 c. 121, 222 and 13, 222 d. 11, 242 and 11, 32 Solution (a): This line is rising, so its slope is positive. m 5 1 2 1232 1 2 1212 5 4 2 5 2 1 5 2 x y (1, 1) (–1, –3) 121, 232 and 11, 12 2.3 Lines 189 190 CHAPTER 2 Graphs Solution (b): This line is falling, so its slope is negative. m 5 3 2 1 23 2 3 5 22 6 5 2 1 3 Solution (c): This is a horizontal line, so its slope is zero. m 5 22 2 1222 3 2 1212 5 0 4 5 0 Solution (d): This is a vertical line, so its slope is undefined. m 5 3 2 1242 1 2 1 5 7 0, which is undefined. YOUR TURN For each pair of points classify the line that passes through them as rising, falling, vertical, or horizontal, and determine its slope. Do not graph. a. 12, 02 and 11, 52 b. 122, 232 and 12, 52 c. 123, 212 and 123, 42 d. 121, 22 and 13, 22 123, 32 and 13, 12 x y (3, 1) (–3, 3) 121, 222 and 13, 222 x y (3, –2) (–1, –2) 11, 242 and 11, 32 x y (1, –4) (1, 3) ▼ A N S W E R a. m 5 25, falling b. m 5 2, rising c. slope is undefined, vertical d. m 5 0, horizontal ▼ [CONCEPT CHECK] If two lines are rising from left to right, then the larger slope corresponds to the (A) steeper incline or (B) less steep incline. ANSWER (A) steeper incline ▼ 2.3.2 Equations of Lines Slope–Intercept Form As mentioned earlier, the general form for an equation of a line is Ax 1 By 5 C. A more standard way to write an equation of a line is in slope–intercept form because it identifies the slope and the y-intercept. 2.3.2 S KILL Find the equation of a line given either slope and a point that lies on the line or two points that lie on the line. 2.3.2 CO NCE PTUAL Understand that the equation of a line given by y 5 mx 1 b corresponds to a line with slope m and y-intercept 10, b2. EQUATION OF A STRAIGHT LINE: SLOPE–INTERCEPT FORM The slope–intercept form for the equation of a nonvertical line is y 5 mx 1 b Its graph has slope m and y-intercept b. For example, 2x 2 y 5 23 is in general form. To write this equation in slope–intercept form, we isolate the y variable: y 5 2x 1 3 The slope of this line is 2 and the y-intercept is 3. ▼ A N S W E R y 5 3 2 x 2 6 x y (0, –6) (2, –3) EXAMPLE 3 Using Slope–Intercept Form to Graph an Equation of a Line Write 2x 2 3y 5 15 in slope–intercept form and graph it. Solution: STEP 1 Write in slope–intercept form. Subtract 2x from both sides. 23y 5 22x 1 15 Divide both sides by 23. y 5 2 3x 2 5 STEP 2 Graph. Identify the slope and y-intercept. Slope: m 5 2 3 y-intercept: b 5 25 Plot the point corresponding to the y-intercept 10, 252. From the point 10, 252, rise two units and run (to the right) three units, which corresponds to the point 13, 232. Draw the line passing through the two points. YOUR T UR N Write 3x 2 2y 5 12 in slope–intercept form and graph it. x y (0, –5) (3, –3) ▼ Instead of starting with equations of lines and characterizing them, let us now start with particular features of a line and derive its governing equation. Suppose that you are given the y-intercept and the slope of a line. Using the slope–intercept form of an equation of a line, y 5 mx 1 b, you could find its equation. EXAMPLE 4 Using Slope–Intercept Form to Find the Equation of a Line Find the equation of a line that has slope 2 3 and y-intercept 10, 12. Solution: Write the slope–intercept form of an equation of a line. y 5 mx 1 b Label the slope. m 5 2 3 Label the y-intercept. b 5 1 The equation of the line in slope–intercept form is y 5 2 3x 1 1 . YOUR TURN Find the equation of the line that has slope 23 2 and y-intercept 10, 22. ▼ A N S W E R y 5 23 2x 1 2 ▼ 2.3 Lines 191 192 CHAPTER 2 Graphs Point–Slope Form Now, suppose that the two pieces of information you are given about an equation are its slope and one point that lies on its graph. You still have enough information to write an equation of the line. Recall the formula for slope: m 5 y2 2 y1 x2 2 x1 , where x2 2 x1 We are given the slope m, and we know a particular point that lies on the line 1x1, y12. We refer to all other points that lie on the line as 1x, y2. Substituting these values into the slope formula gives us m 5 y 2 y1 x 2 x1 Cross multiplying yields y 2 y1 5 m1x 2 x12 This is called the point–slope form of an equation of a line. EXAMPLE 5 Using Point–Slope Form to Find the Equation of a Line Find the equation of the line that has slope 21 2 and passes through the point 121, 22. Solution: Write the point–slope form of an equation of a line. y 2 y1 5 m1x 2 x12 Substitute the values m 5 21 2 and 1x1, y12 5 121, 22. y 2 2 5 21 2 1x 2 12122 Distribute. y 2 2 5 21 2x 2 1 2 Isolate y. y 5 21 2x 1 3 2 We can also express the equation in general form x 1 2y 5 3 . Y OUR TU R N Derive the equation of the line that has slope 1 4 and passes through the point A1, 21 2B. ▼ ▼ A N S W E R y 5 1 4x 2 3 4 or x 2 4y 5 3 Suppose the slope of a line is not given at all. Instead, two points that lie on the line are given. If we know two points that lie on the line, then we can calculate the slope. Then, using the slope and either of the two points, the equation of the line can be derived. EQUATION OF A STRAIGHT LINE: POINT–SLOPE FORM The point–slope form for the equation of a line is y 2 y1 5 m1x 2 x12 Its graph passes through the point 1x1, y12, and its slope is m. Note: This formula does not hold for vertical lines since their slope is undefined. EXAMPLE 6 Finding the Equation of a Line Given Two Points Find the equation of the line that passes through the points 122, 212 and 13, 22. Solution: Write the equation of a line. y 5 mx 1 b Calculate the slope. m 5 y2 2 y1 x2 2 x1 Substitute 1x1, y12 5 122, 212 and 1x2, y22 5 13, 22. m 5 2 2 1212 3 2 1222 5 3 5 Substitute 3 5 for the slope. y 5 3 5x 1 b Let 1x, y2 5 13, 22. 1Either point satisfies the equation.2 2 5 3 5 132 1 b Solve for b. b 5 1 5 Write the equation in slope–intercept form. y 5 3 5x 1 1 5 Write the equation in general form. 23x 1 5y 5 1 Y OUR T UR N Find the equation of the line that passes through the points 121, 32 and 12, 242. ▼ ▼ A N S W E R y 5 27 3x 1 2 3 or 7x 1 3y 5 2 EXAMPLE 7 Finding the Equation of a Horizontal or Vertical Line Find the equation of each of the following lines given their slope and a point that the line passes through: a. Slope: undefined; 123, 22 b. Slope: m 5 0; 124, 52 Solution (a): A vertical line has undefined slope. x 5 a The x-coordinate of the point the line passes through 123, 22 is 23. x 5 23 Graph of the line x 5 23 and the point 123, 22 indicated. Solution (b): A horizontal line has slope m 5 0. y 5 b The y-coordinate of the point the line passes through 124, 52 is 5. y 5 5 Graph of the line y 5 5 with the point 124, 52 indicated. Y OUR T UR N Find the equation of each of the following lines given their slope and a point that the line passes through: a. Slope: m 5 0; 13, 72 b. Slope: undefined; 15, 222 ▼ x y (–3, 2) x = 3 x y (–4, 5) y = 5 ▼ A N S W E R a. y 5 7 b. x 5 5 2.3 Lines 193 STUDY TIP When two points that lie on a line are given, first calculate the slope of the line, then use either point and the slope–intercept form (shown in Example 6) or the point–slope form: m 5 3 5, 13, 22 y 2 y1 5 m1x 2 x12 y 2 2 5 3 51x 2 32 5y 2 10 5 31x 2 32 5y 2 10 5 3x 2 9 23x 1 5y 5 1 194 CHAPTER 2 Graphs [CONCEPT CHECK] In what order are the slope and y-intercept found? (A) y-intercept first, then the slope, or (B) the slope and then the y-intercept? ANSWER (B) the slope and then the y-intercept ▼ 2.3.3 Parallel and Perpendicular Lines Two distinct nonintersecting lines in a plane are parallel. How can we tell whether the two lines in the graph on the left are parallel? Parallel lines must have the same steepness. In other words, parallel lines must have the same slope. The two lines shown on the right are parallel because they have the same slope, 2. DEFINITION Parallel Lines Two distinct lines in a plane are parallel if and only if their slopes are equal. In other words, if two lines in a plane are parallel, then their slopes are equal, and if the slopes of two lines in a plane are equal, then the lines are parallel. WORDS MATH Lines L1 and L2 are parallel. L10 0 L2 Two parallel lines have the same slope. m1 5 m2 2.3.3 S KILL Find the equation of a line that is parallel or perpendicular to a given line. 2.3.3 CO NCE PTUAL Understand that parallel lines have the same slope, whereas perpendicular lines have slopes that are negative reciprocals of one another. x y y = 2x – 1 y = 2x + 3 ▼ A N S W E R y 5 2x 1 5 EXAMPLE 9 Finding an Equation of a Parallel Line Find the equation of the line that passes through the point 11, 12 and is parallel to the line y 5 3x 1 1. Solution: Write the slope–intercept equation of a line. y 5 mx 1 b Parallel lines have equal slope. m 5 3 Substitute the slope into the equation of the line. y 5 3x 1 b Since the line passes through 11, 12, this point must satisfy the equation. 1 5 3112 1 b Solve for b. b 5 22 The equation of the line is y 5 3x 2 2 . Y OUR TU R N Find the equation of the line parallel to y 5 2x 2 1 that passes through the point 121, 32. ▼ EXAMPLE 8 Determining Whether Two Lines Are Parallel Determine whether the lines 2x 1 3y 5 23 and y 5 1 3x 2 6 are parallel. Solution: Write the first line in slope–intercept form. 2x 1 3y 5 23 Add x to both sides. 3y 5 x 2 3 Divide by 3. y 5 1 3x 2 1 Compare the two lines. y 5 1 3x 2 1 and y 5 1 3x 2 6 Both lines have the same slope, 1 3. These are distinct lines because they have different y-intercepts. Thus, the two lines are parallel . Two perpendicular lines form a right angle at their point of intersection. Notice the slopes of the two perpendicular lines in the figure to the right. They are 21 2 and 2, negative reciprocals of each other. It turns out that almost all perpendicular lines share this property. Horizontal 1m 5 02 and vertical 1m undefined2 lines do not share this property. DEFINITION Perpendicular Lines Except for the special case of a vertical and a horizontal line, two lines in a plane are perpendicular if and only if their slopes are negative reciprocals. In other words, if two lines in a plane are perpendicular, their slopes are negative reciprocals, provided their slopes are defined. Similarly, if the slopes of two lines in a plane are negative reciprocals, then the lines are perpendicular. WORDS MATH Lines L1 and L2 are perpendicular. L1 ' L2 Two perpendicular lines have negative reciprocal slopes. x y y = 2x y = – x 1 2 m1 5 2 1 m2 m1 2 0, m2 2 0 ▼ A N S W E R y 5 2x 2 7 EXAMPLE 10 Finding an Equation of a Line That Is Perpendicular to Another Line Find the equation of the line that passes through the point 13, 02 and is perpendicular to the line y 5 3x 1 1. Solution: Identify the slope of the given line y 5 3x 1 1. m1 5 3 The slope of a line perpendicular to the given line is the m2 5 2 1 m1 5 21 3 negative reciprocal of the slope of the given line. Write the equation of the line we are looking for in slope–intercept form. y 5 m2x 1 b Substitute m2 5 21 3 into y 5 m2x 1 b. y 5 2 1 3x 1 b Since the desired line passes through 13, 02, this point must satisfy the equation. 0 5 2 1 3 132 1 b 0 5 21 1 b Solve for b. b 5 1 The equation of the line is y 5 21 3x 1 1 . YOUR T UR N Find the equation of the line that passes through the point 11, 252 and is perpendicular to the line y 5 21 2 x 1 4. ▼ Applications Involving Linear Equations Slope is the ratio of the change in y over the change in x. In applications, slope can often be interpreted as the rate of change, as illustrated in the next example. 2.3 Lines 195 STUDY TIP If a line has slope equal to 3, then a line perpendicular to it has slope 21 3. 196 CHAPTER 2 Graphs In Example 11 we chose to use the points 11970, 202 and 12030, 352. We could have also used the point 11990, 252 with either of the other points. EXAMPLE 12 Service Charges Suppose that your two neighbors use the same electrician. One neighbor had a 2-hour job that cost her $100, and another neighbor had a 3-hour job that cost him $130. Assuming that a linear equation governs the service charge of this electrician, what will your cost be for a 5-hour job? Solution: STEP 1 Identify the question. Determine the linear equation for this electrician’s service charge and calculate the charge for a 5-hour job. STEP 2 Make notes. A 2-hour job costs $100 and a 3-hour job costs $130. STEP 3 Set up an equation. Let x equal the number of hours and y equal the service charge in dollars. Linear equation: y 5 mx 1 b Two points that must satisfy this equation are 12, 1002 and 13, 1302. STEP 4 Solve the equation. Calculate the rate of change (slope). m 5 130 2 100 3 2 2 5 30 1 5 30 Substitute the slope into the linear equation. y 5 30x 1 b EXAMPLE 11 Slope as a Rate of Change The average age that a person first marries has been increasing over the last several decades. In 1970 the average age was 20, in 1990 it was 25 years old, and in 2030 it is expected that the average age will be 35 at the time of a person’s first marriage. Find the slope of the line passing through these points. Describe what that slope represents. Solution: If we let x represent the year and y represent the age, then two points that lie on the line are 11970, 202 and 12030, 352. Write the slope formula. m 5 change in y change in x Substitute the points into the slope formula. The slope is 1 4 and can be interpreted as the rate of change of the average age when a person is first married. Every 4 years the average age at the first marriage is 1 year older. m 5 35 2 20 2030 2 1970 5 15 60 5 1 4 Year 1990 2010 2030 2050 1970 1950 1930 Age 15 10 5 20 25 30 35 40 45 [CONCEPT CHECK] Two parallel lines have the same slope and different y-intercepts, so the lines never intersect. Two perprendicular lines have negative reciprocal slopes (m and 21/m), and they intersect in (A) zero places, (B) one place, or (C) two places. ANSWER (B) one place ▼ ▼ A N S W E R $75 Either point must satisfy 100 5 30122 1 b the equation. Use 12, 1002. 100 5 60 1 b b 5 40 The service charge y is given by y 5 30x 1 40. Substitute x 5 5 into this equation for a 5-hour job. y 5 30152 1 40 5 190 The 5-hour job will cost $190 . STEP 5 Check the solution. The service charge y 5 30x 1 40 can be interpreted as a fixed cost of $40 for coming to your home and a $30 per hour fee for the job. A 5-hour job would cost $190. Additionally, a 5-hour job should cost less than the sum of a 2-hour job and a 3-hour job 1$100 1 $130 5 $2302, since the $40 fee is charged only once. YOUR TURN You decide to hire a tutor that some of your friends recommended. The tutor comes to your home, so she charges a flat fee per session and then an hourly rate. One friend prefers 2-hour sessions, and the charge is $60 per session. Another friend has 5-hour sessions that cost $105 per session. How much should you be charged for a 3-hour session? ▼ [SEC TION 2 .3] E X E RC I S E S • S K I L L S In Exercises 1–10, find the slope of the line that passes through the given points. 1. 11, 32 and 12, 62 2. 12, 12 and 14, 92 3. 122, 52 and 12, 232 4. 121, 242 and 14, 62 5. 127, 92 and 13, 2102 6. 111, 232 and 122, 62 7. 10.2, 21.72 and 13.1, 5.22 8. 122.4, 1.72 and 125.6, 22.32 9. A2 3, 21 4B and A5 6, 23 4B 10. A1 2, 3 5B and A 23 4, 7 5B 2.3 Lines 197 n Horizontal lines: m 5 0 n Vertical lines: m is undefined We found equations of lines, given either two points or the slope and a point. We found the point–slope form, y 2 y1 5 m 1x 2 x12, useful when the slope and a point are given. We also discussed both parallel (nonintersecting) and perpendicular (forming a right angle) lines. Parallel lines have the same slope. Perpendicular lines have negative reciprocal slopes, provided their slopes are defined. In this section, we discussed graphs and equations of lines. Lines are often expressed in two forms: n General Form: Ax 1 By 5 C n Slope–Intercept Form: y 5 mx 1 b All lines (except horizontal and vertical) have exactly one x-intercept and exactly one y-intercept. The slope of a line is a measure of steepness. n Slope of a line passing through 1x1, y12 and 1x2, y22: m 5 y2 2 y1 x2 2 x1 5 rise run [SEC TION 2 .3] S U M M A RY 198 CHAPTER 2 Graphs For each graph in Exercises 11–16, identify (by inspection) the x- and y-intercepts and slope if they exist, and classify the line as rising, falling, horizontal, or vertical. 11. 12. 13. 14. 15. 16. In Exercises 17–30, find the x- and y-intercepts if they exist and graph the corresponding line. 17. y 5 2x 2 3 18. y 5 23x 1 2 19. y 5 21 2 x 1 2 20. y 5 1 3 x 2 1 21. 2x 2 3y 5 4 22. 2x 1 y 5 21 23. 1 2x 1 1 2 y 5 21 24. 1 3 x 2 1 4 y 5 1 12 25. x 5 21 26. y 5 23 27. y 5 1.5 28. x 5 27.5 29. x 5 27 2 30. y 5 5 3 In Exercises 31–42, write the equation in slope–intercept form. Identify the slope and the y-intercept. 31. 2x 2 5y 5 10 32. 3x 2 4y 5 12 33. x 1 3y 5 6 34. x 1 2y 5 8 35. 4x 2 y 5 3 36. x 2 y 5 5 37. 12 5 6x 1 3y 38. 4 5 2x 2 8y 39. 0.2x 2 0.3y 5 0.6 40. 0.4x 1 0.1y 5 0.3 41. 1 2 x 1 2 3 y 5 4 42. 1 4 x 1 2 5 y 5 2 In Exercises 43–50, write the equation of the line, given the slope and intercept. 43. Slope: m 5 2 y-intercept: 10, 32 44. Slope: m 5 22 y-intercept: 10, 12 45. Slope: m 5 21 3 y-intercept: 10, 02 46. Slope: m 5 1 2 y-intercept: 10, 232 47. Slope: m 5 0 y-intercept: 10, 22 48. Slope: m 5 0 y-intercept: 10, 21.52 49. Slope: undefined x-intercept: A3 2, 0B 50. Slope: undefined x-intercept: 123.5, 02 In Exercises 51–60, write an equation of the line in slope–intercept form, if possible, given the slope and a point that lies on the line. 51. Slope: m 5 5 121, 232 52. Slope: m 5 2 11, 212 53. Slope: m 5 23 122, 22 54. Slope: m 5 21 13, 242 55. Slope: m 5 3 4 11, 212 56. Slope: m 5 21 7 125, 32 57. Slope: m 5 0 122, 42 58. Slope: m 5 0 13, 232 59. Slope: undefined 121, 42 60. Slope: undefined 14, 212 x y (–1, –3) (2, 3) x y (–4, –3) (0, 3) x y (2, –1) (–2, 3) x y (2, –2) (–3, 3) x y (3, 1) (–1, 1) x y (–4, –3) (–4, 4) In Exercises 61–80, write the equation of the line that passes through the given points. Express the equation in slope–intercept form or in the form x 5 a or y 5 b. 61. 122, 212 and 13, 22 62. 124, 232 and 15, 12 63. 123, 212 and 122, 262 64. 125, 282 and 17, 222 65. 120, 2372 and 1210, 2422 66. 128, 122 and 1220, 2122 67. 121, 42 and 12, 252 68. 122, 32 and 12, 232 69. A1 2, 3 4B and A3 2, 9 4B 70. A22 3, 21 2B and A7 3, 1 2B 71. 13, 52 and 13, 272 72. 125, 222 and 125, 42 73. 13, 72 and 19, 72 74. 122, 212 and 13, 212 75. 10, 62 and 125, 02 76. 10, 232 and 10, 22 77. 126, 82 and 126, 222 78. 129, 02 and 129, 22 79. A2 5, 23 4B and A2 5, 1 2B 80. A1 3, 2 5B and A1 3, 1 2B In Exercises 81–86, write the equation corresponding to each line. Express the equation in slope–intercept form. 81. 82. 83. 84. 85. 86. In Exercises 87–100, find the equation of the line that passes through the given point and also satisfies the additional piece of information. Express your answer in slope–intercept form, if possible. 87. 123, 12; parallel to the line y 5 2x 2 1 88. 11, 32; parallel to the line y 5 2x 1 2 89. 10, 02; perpendicular to the line 2x 1 3y 5 12 90. 10, 62; perpendicular to the line x 2 y 5 7 91. 13, 52; parallel to the x-axis 92. 13, 52; parallel to the y-axis 93. 121, 22; perpendicular to the y-axis 94. 121, 22; perpendicular to the x-axis 95. 122, 272; parallel to the line 1 2 x 2 1 3 y 5 5 96. 11, 42; perpendicular to the line 22 3x 1 3 2y 5 22 97. A22 3, 2 3B; perpendicular to the line 8x 1 10y 5245 98. A6 5, 3B; perpendicular to the line 6x 1 14y 5 7 99. A7 2, 4B; parallel to the line 215x 1 35y 5 7 100. A21 4,213 9 B; parallel to the line 10x 1 45y 5 29 x y x y x y x y x y x y 2.3 Lines 199 • A P P L I C A T I O N S 101. Budget: Home Improvement. The cost of having your bathroom remodeled is the combination of material costs and labor costs. The materials (tile, grout, toilet, fixtures, etc.) cost is $1200, and the labor cost is $25 per hour. Write an equation that models the total cost C of having your bath-room remodeled as a function of hours h. How much will the job cost if the worker estimates 32 hours? 102. Budget: Rental Car. The cost of a one-day car rental is the sum of the rental fee, $50, plus $0.39 per mile. Write an equation that models the total cost associated with the car rental. 103. Budget: Monthly Driving Costs. The monthly costs associ-ated with driving a new Honda Accord are the monthly loan payment plus $25 every time you fill up with gasoline. If you fill up 5 times in a month, your total monthly cost is $500. How much is your loan payment? 104. Budget: Monthly Driving Costs. The monthly costs associ-ated with driving a Ford Explorer are the monthly loan pay-ment plus the cost of filling up your tank with gasoline. If you fill up 3 times in a month, your total monthly cost is $520. If you fill up 5 times in a month, your total monthly cost is $600. How much is your monthly loan, and how much does it cost every time you fill up with gasoline? 200 CHAPTER 2 Graphs 105. Business. The operating costs for a local business are a fixed amount of $1300 plus $3.50 per unit sold, while revenue is $7.25 per unit sold. How many units does the business have to sell in order to break even? 106. Business. The operating costs for a local business are a fixed amount of $12,000 plus $13.50 per unit sold, while revenue is $27.25 per unit sold. How many units does the business have to sell in order to break even? 107. Weather: Temperature. The National Oceanic and Atmo-spheric Administration (NOAA) has an online conversion chart that relates degrees Fahrenheit, 8F, to degrees Celsius, 8C. 778F is equivalent to 258C, and 688F is equivalent to 208C. Assuming the relationship is linear, write the equation relating degrees Celsius C to degrees Fahrenheit F. What temperature is the same in both degrees Celsius and degrees Fahrenheit? 108. Weather: Temperature. According to NOAA, a “standard day” is 158C at sea level, and every 500 feet elevation above sea level corresponds to a 18C temperature drop. Assuming the relationship between temperature and elevation is linear, write an equation that models this relationship. What is the expected temperature at 2500 feet on a “standard day”? 109. Life Sciences: Height. The average height of a man has increased over the last century. What is the rate of change in inches per year of the average height of men? Year 2000 1900 1950 Height (inches) 68 67 66 65 69 70 71 72 110. Life Sciences: Height. The average height of a woman has increased over the last century. What is the rate of change in inches per year of the average height of women? Year 2000 1900 1950 Height (inches) 63 62 61 60 64 65 66 67 111. Life Sciences: Weight. The average weight of a baby born in 1900 was 6 pounds 4 ounces. In 2000, the average weight of a newborn was 6 pounds 10 ounces. What is the rate of change of birth weight in ounces per year? What do we expect babies to weigh at birth in 2040? 112. Sports. The fastest a man could run a mile in 1906 was 4 minutes and 30 seconds. In 1957, Don Bowden became the first American to break the 4-minute mile. Calculate the rate of change in mile speed per year. 113. Monthly Phone Costs. Mike’s home phone plan charges a flat monthly fee plus a charge of $0.05 per minute for long-distance calls. The total monthly charge is represented by y 5 0.05x 1 35, x $ 0, where y is the total monthly charge and x is the number of long-distance minutes used. Interpret the meaning of the y-intercept. 114. Cost: Automobile. The value of a Daewoo car is given by y 5 11,100 2 1850x, x $ 0, where y is the value of the car and x is the age of the car in years. Find the x-intercept and y-intercept and interpret the meaning of each. 115. Weather: Rainfall. The average rainfall in Norfolk, Virginia, for July was 5.2 inches in 2003. The average July rainfall for Norfolk was 3.8 inches in 2007. What is the rate of change of rainfall in inches per year? If this trend continues, what is the expected average rainfall in 2015? 116. Weather: Temperature. The average temperature for Boston in January 2005 was 438F. In 2007 the average January temperature was 44.58F. What is the rate of change of the temperature per year? If this trend continues, what is the expected average temperature in January 2016? 117. Environment. In 2000, Americans used approximately 380 billion plastic bags. In 2005, approximately 392 billion were used. What is the rate of change of plastic bags used per year? How many plastic bags will be expected to be used in 2016? 118. Finance: Debt. According to financial reports, the average household credit card debt in 2010 was $7768 and in 2012 was $7117. What is the rate of change of the credit card debt per year? If this trend were to continue, how much would you expect the average household credit card debt to be in 2016? 119. Business. A Web site that supplies Asian specialty foods to restaurants advertises a 64-ounce bottle of hoisin sauce for $16.00. Shipping cost for one bottle is $15.93. The shipping cost for two bottles is $19.18. The cost for five bottles, including shipping, is $111.83. Answer the following questions based on this scenario. Round to the nearest cent, when necessary. a. Write the three ordered pairs where x represents the number of bottles purchased and y represents the total cost (including shipping) for one, two, or five bottles purchased. b. Calculate the slope between the origin and the ordered pair that represents the purchase of one bottle of hoisin sauce. Explain what this amount means in terms of the sauce purchase. c. Calculate the slope between the origin and the ordered pair that represents the purchase of two bottles of hoisin (including shipping). Explain what this amount means in terms of the sauce purchase. d. Calculate the slope between the origin and the ordered pair that represents the purchase of five bottles of hoisin (including shipping). Explain what this amount means in terms of the sauce purchase. 120. Business. A Web site that supplies Asian specialty foods to restaurants advertises an 8-ounce bottle of plum sauce for $4.00, but shipping for one bottle is $14.27. The shipping cost for two bottles is $14.77. The cost for five bottles, including shipping, is $35.93. Answer the following questions based on this scenario. Round to the nearest cent, when necessary. a. Write the three ordered pairs where x represents the number of bottles purchased and y represents the total cost, including shipping for one, two, or five bottles purchased. b. Calculate the slope between the origin and the ordered pair that represents the purchase of one bottle of plum sauce. Explain what this amount means in terms of the sauce purchase. c. Calculate the slope between the origin and the ordered pair that represents the purchase of two bottles of plum sauce (including shipping). Explain what this amount means in terms of the sauce purchase. d. Calculate the slope between the origin and the ordered pair that represents the purchase of five bottles of plum sauce (including shipping). Explain what this amount means in terms of the sauce purchase. • C A T C H T H E M I S T A K E In Exercises 121–124, explain the mistake that is made. 121. Find the x- and y-intercepts of the line with equation 2x 2 3y 5 6. Solution: x-intercept: set x 5 0 and solve for y. 23y 5 6 y 5 22 The x-intercept is 10, 222. y-intercept: set y 5 0 and solve for x. 2x 5 6 x 5 3 The y-intercept is 13, 02. This is incorrect. What mistake was made? 122. Find the slope of the line that passes through the points 122, 32 and 14, 12. Solution: Write the slope formula. m 5 y2 2 y1 x2 2 x1 Substitute 122, 32 and 14, 12. m 5 1 2 3 22 2 4 5 22 26 5 1 3 This is incorrect. What mistake was made? 123. Find the slope of the line that passes through the points 123, 42 and 123, 72. Solution: Write the slope formula. m 5 y2 2 y1 x2 2 x1 Substitute 123, 42 and 123, 72. m 5 23 2 1232 4 2 7 5 0 This is incorrect. What mistake was made? 124. Given the slope, classify the line as rising, falling, horizontal, or vertical. a. m 5 0 b. m undefined c. m 5 2 d. m 5 21 Solution: a. vertical line b. horizontal line c. rising d. falling These are incorrect. What mistakes were made? • T E C H N O L O G Y For Exercises 137–142, determine whether the lines are parallel, perpendicular, or neither, and then graph both lines in the same viewing screen using a graphing utility to confirm your answer. 137. y1 5 17x 1 22 y2 5 2 1 17x 2 13 138. y1 5 0.35x 1 2.7 y2 5 0.35x 2 1.2 139. y1 5 0.25x 1 3.3 y2 5 24x 1 2 140. y1 5 1 2 x 1 5 y2 5 2x 2 3 141. y1 5 0.16x 1 2.7 y2 5 6.25x 2 1.4 142. y1 5 23.75x 1 8.2 y2 5 4 15 x 1 5 6 • C H A L L E N G E 131. Find an equation of a line that passes through the point 12B, A 1 12 and is parallel to the line Ax 1 By 5 C. Assume that B is not equal to zero. 132. Find an equation of a line that passes through the point 1B, A 212 and is parallel to the line Ax 1 By 5 C. Assume that B is not equal to zero. 133. Find an equation of a line that passes through the point 12A, B 212 and is perpendicular to the line Ax 1 By 5 C. Assume that A and B are both nonzero. 134. Find an equation of a line that passes through the point 1A, B 1 12 and is perpendicular to the line Ax 1 By 5 C. Assume that A and B are both nonzero. 135. Show that two lines with equal slopes and different y-intercepts have no point in common. Hint: Let y1 5 mx 1 b1 and y2 5 mx 1 b2 with b1 2 b2. What equation must be true for there to be a point of intersection? Show that this leads to a contradiction. 136. Let y1 5 m1x 1 b1 and y2 5 m2 x 1 b2 be two nonparallel lines 1 m1 2 m2 2. What is the x-coordinate of the point where they intersect? 2.3 Lines 201 • C O N C E P T U A L In Exercises 125–128, determine whether each statement is true or false. 125. A line can have at most one x-intercept. 126. A line must have at least one y-intercept. 127. If the slopes of two lines are 21 5 and 5, then the lines are parallel. 128. If the slopes of two lines are 21 and 1, then the lines are perpendicular. 129. If a line has slope equal to zero, describe a line that is perpendicular to it. 130. If a line has no slope, describe a line that is parallel to it. 202 CHAPTER 2 Graphs 2.4.1 Standard Equation of a Circle Most people understand the shape of a circle. The goal in this section is to develop the equation of a circle. DEFINITION Circle A circle is the set of all points in a plane that are a fixed distance from a point, the center. The center, C, is typically denoted by 1h, k2, and the fixed distance, or radius, is denoted by r. What is the equation of a circle? We’ll use the distance formula from Section 2.1. Distance formula: d 5 "1x2 2 x12 2 1 1y2 2 y12 2 The distance between the center 1h, k2 and any point 1x, y2 on the circle is the radius r. Substitute these values d 5 r, 1x1, y12 5 1h, k2 , and 1x2, y22 5 1x, y2 into the distance formula r 5 "1x 2 h2 2 1 1y 2 k2 2 and square both sides: 1x 2 h2 2 1 1y 2 k2 2 5 r 2. All circles can be written in standard form, which makes it easy to identify the center and radius. EQUATION OF A CIRCLE The standard form of the equation of a circle with radius r and center 1h, k2 is 1x 2 h22 1 1y 2 k22 5 r2 For the special case of a circle with center at the origin 10, 02, the equation simplifies to x2 1 y2 5 r2. UNIT CIRCLE A circle with radius 1 and center 10, 02 is called the unit circle: x2 1 y2 5 1 The unit circle plays an important role in the study of trigonometry. Note that if x2 1 y2 5 0, the radius is 0, so the “circle” is just a point. 2.4.1 S KILL Graph a circle. 2.4.1 CO NCE PTUAL Understand algebraic and graphical representations of circles. y x (h, k) (x, y) r S K I L L S O B J E C T I V E S ■ ■Graph a circle. ■ ■Transform general equations of circles to the standard form by completing the square. C O N C E P T U A L O B J E C T I V E S ■ ■Understand algebraic and graphical representations of circles. ■ ■Expand the concept of completing the square with quadratic equations to circles. 2.4 CIRCLES 2.4 Circles 203 EXAMPLE 1 Finding the Center and Radius of a Circle Identify the center and radius of the given circle and graph. 1x 2 222 1 1y 1 122 5 4 Solution: Rewrite this equation in standard form. 3x 2 24 2 1 3 y 21212 4 2 5 22 Identify h, k, and r by comparing this equation with the standard form of a circle: 1x 2 h22 1 1y 2 k22 5 r 2. h 5 2, k 5 21, and r 5 2 To draw the circle, label the center 12, 212. Label four additional points two units (the radius) away from the center: 14, 212, 10, 212, 12, 12, and 12, 232. Note that the easiest four points to get are those obtained by going out from the center both horizontally and vertically. Connect those four points with a smooth curve. YOUR T UR N Identify the center and radius of the given circle and graph. 1x 1 122 1 1y 1 222 5 9 Center 12, 212 and r 5 2 y x (2, –3) (0, –1) (4, –1) (2, –1) (2, 1) x y (–1, –5) (–4, –2) (2, –2) (–1, –2) (–1, 1) ▼ A N S W E R Center: 121, 222 Radius: 3 ▼ EXAMPLE 2 Graphing a Circle: Fractions and Radicals Identify the center and radius of the given circle and sketch its graph. ax 2 1 2b 2 1 ay 1 1 3b 2 5 20 Solution: If r 2 5 20, then r 5 "20 5 2"5. ax 2 1 2b 2 1 cy 2 a21 3b d 2 5 A2!5B2 Write the equation in standard form. Identify the center and radius. To graph the circle, we’ll use decimal approximations of the fractions and radicals: 10.5, 20.32 for the center and 4.5 for the radius. Four points on the circle that are 4.5 units from the center are 124, 20.32, 15, 20.32, 10.5, 4.22, and 10.5, 24.82. Connect them with a smooth curve. x y (0.5, –4.8) (–4.0, –0.3) (5.0, –0.3) (0.5, 4.2) (0.5, –0.3) Center a1 2, 2 1 3b and r 5 2"5 EXAMPLE 3 Determining the Equation of a Circle Given the Center and Radius Find the equation of a circle with radius 5 and center 122, 32. Graph the circle. Solution: Substitute 1h, k2 5 122, 32 and r 5 5 into the standard equation of a circle. 3x 2 1222 4 2 1 1 y 2 322 5 52 Simplify. 1x 1 222 1 1 y 2 322 5 25 204 CHAPTER 2 Graphs Let’s change the look of the equation given in Example 1. In Example 1 the equation of the circle was given as: 1x 2 222 1 1y 1 122 5 4 Eliminate the parentheses. x2 2 4x 1 4 1 y2 1 2y 1 1 5 4 Group like terms and subtract 4 from both sides. x2 1 y2 2 4x 1 2y 1 1 5 0 We have written the general form of the equation of the circle in Example 1. Suppose you are given a point that lies on a circle and the center of the circle. Can you find the equation of the circle? The general form of the equation of a circle is x2 1 y2 1 ax 1 by 1 c 5 0 To graph the circle, plot the center 122, 32 and four points 5 units away from the center: 127, 32, 13, 32, 122, 222, and 122, 82. Connect them with a smooth curve. YOUR TURN Find the equation of a circle with radius 3 and center 10, 12 and graph. x y (–2, –2) 5 (–7, 3) (3, 3) (–2, 3) (–2, 8) x y (0, –2) 3 (–3, 1) (3, 1) (0, 1) (0, 4) ▼ A N S W E R x2 1 1 y 2 12 2 5 9 ▼ EXAMPLE 4 Finding the Equation of a Circle Given Its Center and One Point The point 110, 242 lies on a circle centered at 17, 282. Find the equation of the circle in general form. Solution: This circle is centered at 17, 282, so its standard equation is 1x 2 72 2 1 1y 1 82 2 5 r 2. All that remains is to find the radius. Approach 1: Since the point 110, 242 lies on the circle, it must satisfy the equation of the circle. Substitute 1x, y2 5 110, 242. 110 2 722 1 124 1 822 5 r 2 Simplify. 32 1 42 5 r 2 The distance from 110, 242 to 17, 282 is five units. r 5 5 Approach 2: Find the distance between 110, 242 and 17, 282. r 5 d 5 "110 2 722 1 124 2 128222 5 "32 1 42 5 "25 5 5 Substitute r 5 5 into the standard equation. 1x 2 722 1 1y 1 822 5 52 Eliminate the parentheses and simplify. x2 2 14x 1 49 1 y2 1 16y 1 64 5 25 Write in general form. Y OUR TU R N The point 11, 112 lies on a circle centered at 125, 32. Find the equation of the circle in general form. x 2 2 4 10 12 14 4 6 8 10 12 14 y (7, –13) (2, –8) (10, –4) (12, –8) (7, –8) (7, –3) ▼ A N S W E R x2 1 y2 1 10x 2 6y 2 66 5 0 x2 1 y2 2 14x 1 16y 1 88 5 0 ▼ [CONCEPT CHECK] A circle described algebraically by (x 2 h)2 + (y 2 k )2 5 r 2 can be graphed by first going to the center (h, k ) and then going to the points: (A) (2h 6 r, 2k 6 r ) or (B) (h 6 r, k 6 r ) ANSWER (B) (h 6 r, k 6 r ) ▼ 2.4 Circles 205 2.4.2 Transforming Equations of Circles to the Standard Form by Completing the Square If the equation of a circle is given in general form, it must be rewritten in standard form in order to identify its center and radius. To transform equations of circles from general to standard form, complete the square (Section 1.3) on both the x- and y-variables. 2.4.2 S K I L L Transform general equations of circles to the standard form by completing the square. 2.4.2 C ON C E P T U A L Expand the concept of completing the square with quadratic equations to circles. EXAMPLE 5 Finding the Center and Radius of a Circle by Completing the Square Find the center and radius of the circle with the equation: x2 2 8x 1 y2 1 20y 1 107 5 0 Solution: Our goal is to transform this equation into standard form 1x 2 h22 1 1y 2 k22 5 r 2 Group x and y terms, respectively, on the left side of the equation; move constants to the right side. 1x2 2 8x2 1 1y2 1 20y2 5 2107 Complete the square on both the x and y expressions. 1x2 2 8x 1 u2 1 1 y2 1 20y 1 u2 5 2107 Add A28 2B 2 5 16 and A20 2 B2 5 100 to both sides. 1x2 2 8x 1 42 2 1 1 y2 1 20y 1 102 2 52107 1 16 1 100 16 100 Factor the perfect squares on the left side and simplify the right side. 1x 2 422 1 1 y 1 1022 5 9 Write in standard form. 1x 2 422 1 3 y 2 121024 2 5 32 Y OUR T UR N Find the center and radius of the circle with the equation: x2 1 y2 1 4x 2 6y 2 12 5 0 E E The center is 14, 2102 and the radius is 3. ▼ A N S W E R Center: 122, 32 Radius: 5 [CONCEPT CHECK] Show that (x 1 8)2 1 ( y 1 4)2 5 62 is equal to x 2 1 y 2 1 16x 1 8y 1 44 5 0 ANSWER (x 1 8)2 1 (y 1 4)2 5 62 x 2 1 16x 1 64 1 y 2 1 8y 1 16 5 36 x 2 1 y 2 1 16x 1 8y 1 80 5 36 x 2 1 y 2 1 16x 1 8y 1 44 5 0 ▼ ▼ ▼ C A U T I O N Don’t forget to add both constants to each side of the equation when completing the square for x and y. common mistake A common mistake is forgetting to add both constants to the right side of the equation. Identify the center and radius of the circle with the equation: x2 1 y2 1 16x 1 8y 1 44 5 0 ✖I N C O R R EC T Ax2 1 16x 1 64B 1 Ay2 1 8y 1 16B 5 244 ERROR: Don’t forget to add 16 and 64 to the right. ✓COR R E C T x2 1 y2 1 16x 1 8y 1 44 5 0 Ax2 1 16xB 1 Ay2 1 8yB 5 244 Ax2 1 16x 1 uB 1 Ay2 1 8y 1 uB 5 244 Ax2 1 16x 1 64B 1 Ay2 1 8y 1 16B 5 244 1 64 1 16 1x 1 822 1 1y 1 422 5 36 Center: 128, 242 Radius: 6 206 CHAPTER 2 Graphs [SEC TION 2 .4] E X E R C I SE S • S K I L L S In Exercises 1–20, write the equation of the circle in standard form. 1. Center 11, 22 r 5 3 2. Center 13, 42 r 5 5 3. Center 123, 242 r 5 10 4. Center 121, 222 r 5 4 5. Center 15, 72 r 5 9 6. Center 12, 82 r 5 6 7. Center 1211, 122 r 5 13 8. Center 16, 272 r 5 8 9. Center 10, 02 r 5 2 10. Center 10, 02 r 5 3 11. Center 10, 22 r 5 3 12. Center 13, 02 r 5 2 13. Center 10, 02 r 5 "2 14. Center 121, 22 r 5 "7 15. Center 15, 232 r 5 2"3 16. Center 124, 212 r 5 3"5 17. Center A2 3, 23 5B r 5 1 4 18. Center A21 3, 22 7B r 5 2 5 19. Center 11.3, 2.72 r 5 3.2 20. Center 123.1, 4.22 r 5 5.5 In Exercises 21–32, find the center and radius of the circle with the given equations. 21. 1x 2 122 1 1 y 2 322 5 25 22. 1x 1 122 1 1 y 1 322 5 11 23. 1x 2 222 1 1 y 1 522 5 49 24. 1x 1 322 1 1 y 2 722 5 81 25. 1x 2 422 1 1 y 2 922 5 20 26. 1x 1 122 1 1 y 1 222 5 8 27. Ax 2 2 5B 2 1 Ay 2 1 7B 2 5 4 9 28. Ax 2 1 2B 2 1 Ay 2 1 3B 2 5 9 25 29. 1x 2 1.522 1 1 y 1 2.722 5 1.69 30. 1x 1 3.122 1 1 y 2 7.422 5 56.25 31. x2 1 y2 2 50 5 0 32. x2 1 y2 2 8 5 0 In Exercises 33–50, state the center and radius of each circle. 33. x2 1 y2 1 4x 1 6y 2 3 5 0 34. x2 1 y2 1 2x 1 10y 1 17 5 0 35. x2 1 y2 1 6x 1 8y 2 75 5 0 36. x2 1 y2 1 2x 1 4y 2 9 5 0 37. x2 1 y2 2 10x 2 14y 2 7 5 0 38. x2 1 y2 2 4x 2 16y 1 32 5 0 39. x2 1 y2 2 2y 2 15 5 0 40. x2 1 y2 1 2x 2 8 5 0 41. x2 1 y2 2 2x 2 6y 1 1 5 0 42. x2 1 y2 2 8x 2 6y 1 21 5 0 43. x2 1 y2 2 10x 1 6y 1 22 5 0 44. x2 1 y2 1 8x 1 2y 2 28 5 0 45. x2 1 y2 2 6x 2 4y 1 1 5 0 46. x2 1 y2 2 2x 2 10y 1 2 5 0 47. x2 1 y2 2 x 1 y 1 1 4 5 0 48. x2 1 y2 2 x 2 2 3y 2 1 3 8 5 0 49. x2 1 y2 2 2.6x 2 5.4y 2 1.26 5 0 50. x2 1 y2 2 6.2x 2 8.4y 2 3 5 0 In Exercises 51–56, find the equation of each circle. 51. Centered at 121, 222 and passing through the point 11, 02. 52. Centered at 14, 92 and passing through the point 12, 52. 53. Centered at 122, 32 and passing through the point 13, 72. 54. Centered at 11, 12 and passing through the point 128, 252. 55. Centered at 122, 252 and passing through the point 11, 292. 56. Centered at 123, 242 and passing through the point 121, 282. n General form: x 2 1 y 2 1 ax 1 by 1 c 5 0. n Complete the square to transform the equation to standard form. The equation of a circle is given by n Standard form: 1x 2 h22 1 1 y 2 k22 5 r 2. n Center: 1h, k2 n Radius: r [SEC TION 2 .4] S U M M A RY 2.4 Circles 207 • A P P L I C A T I O N S 57. Cell Phones. If a cellular phone tower has a reception radius of 100 miles and you live 95 miles north and 33 miles east of the tower, can you use your cell phone while at home? 58. Cell Phones. Repeat Exercise 57, assuming you live 45 miles south and 87 miles west of the tower. 59. Construction/Home Improvement. A couple and their dog moved into a new house that does not have a fenced-in backyard. The backyard is square with dimensions 100 feet by 100 feet. If they put a stake in the center of the backyard with a long leash, write the equation of the circle that will map out the dog’s outer perimeter. 100 ft 100 ft 60. Construction/Home Improvement. Repeat Exercise 59 except that the couple put in a pool and a garden and want to restrict the dog to quadrant I. What coordinates represent the center of the circle? What is the radius? y r (h, k) x 50 –50 –50 50 61. Design. A university designs its campus with a master plan of two concentric circles. All of the academic buildings are within the inner circle (so that students can get between classes in less than 10 minutes), and the outer circle contains all the dormitories, the Greek park, cafeterias, the gymna-sium, and intramural fields. Assuming the center of campus is the origin, write an equation for the inner circle if the diam-eter is 3000 feet. x y 62. Design. Repeat Exercise 61 for the outer circle with a diameter of 6000 feet. 63. Cell Phones. A cellular phone tower has a reception radius of 200 miles. Assuming the tower is located at the origin, write the equation of the circle that represents the reception area. 64. Environment. In a state park, a fire has spread in the form of a circle. If the radius is 2 miles, write an equation for the circle. For Exercises 65 and 66, refer to the following: A cell phone provider is expanding its coverage and needs to place four cell phone towers to provide complete coverage of a 100-square-mile area formed by a 10 mile by 10 mile square. This area can be represented by a region on the Cartesian coordinate system; see figure. x y 10 10 Cell Phone Coverage Area The placement of the four towers is very important in that the cell phone provider needs to provide coverage of the entire 100-square-mile area. The cell phone towers being installed can process signals from cell phones within a 3.5-mile radius. 65. Engineering. One plan under consideration is to place the four towers in locations that correspond to the points 12.5, 2.52, 12.5, 7.52, 17.5, 2.52, and 17.5, 7.52 on the graph. a. Write an equation that describes the perimeter of the cell phone coverage for each of the four towers. b. Draw the coverage provided by each of these towers. Will this placement of towers provide the needed coverage? 66. Engineering. One plan under consideration is to place the four towers in locations that correspond to the points 13, 32, 13, 72, 17, 32, and 17, 72 on the graph. a. Write an equation that describes the perimeter of the cell phone coverage for each of the four towers. b. Draw the coverage provided by each of these towers. Will this placement of towers provide the needed coverage? 208 CHAPTER 2 Graphs • C A T C H T H E M I S T A K E In Exercises 67–70, explain the mistake that is made. 67. Identify the center and radius of the circle with equation 1x 2 422 1 1 y 1 322 5 25. Solution: The center is 14, 32 and the radius is 5. This is incorrect. What mistake was made? 68. Identify the center and radius of the circle with equation 1x 2 222 1 1 y 1 322 5 2. Solution: The center is 12, 232 and the radius is 2. This is incorrect. What mistake was made? 69. Graph the solution to the equation 1x 2 122 1 1 y 1 222 5 216. Solution: The center is 11, 222 and the radius is 4. x y (1, –6) 4 (–3, –2) (5, –2) (1, –2) (1, 2) This is incorrect. What mistake was made? 70. Find the center and radius of the circle with the equation x2 1 y2 2 6x 1 4y 2 3 5 0. Solution: Group like terms. 1x2 2 6x2 1 1 y2 1 4y2 5 3 Complete the 1x2 2 6x 1 92 1 1 y2 1 4y 1 42 5 12 square. 1x 2 322 1 1 y 1 222 5 12!322 The center is 13, 222 and the radius is 2!3. This is incorrect. What mistake was made? • C O N C E P T U A L In Exercises 71–74, determine whether each statement is true or false. 71. The equation whose graph is depicted has infinitely many solutions. x y (0, –5) 5 (–5, 0) (5, 0) (0, 5) 72. The equation 1x 2 722 1 1 y 1 1522 5 264 has no solution. 73. The equation 1x 2 22 2 1 1 y 1 522 5 220 has no solution. 74. The equation 1x 2 12 2 1 1 y 1 322 5 0 has only one solution. 75. Describe the graph (if it exists) of: x2 1 y2 1 10x 2 6y 1 34 5 0 76. Describe the graph (if it exists) of: x2 1 y2 2 4x 1 6y 1 49 5 0 77. Find the equation of a circle that has a diameter with endpoints 15, 22 and 11, 262. 78. Find the equation of a circle that has a diameter with endpoints 13, 02 and 121, 242. • C H A L L E N G E 79. For the equation x2 1 y2 1 ax 1 by 1 c 5 0, specify conditions on a, b, and c so that the graph is a single point. 80. For the equation x2 1 y2 1 ax 1 by 1 c 5 0, specify condi-tions on a, b, and c so that there is no corresponding graph. 81. Determine the center and radius of the circle given by the equation x2 1 y2 2 2ax 5 100 2 a2. 82. Determine the center and radius of the circle given by the equation x2 1 y2 1 2by 5 49 2 b2. 2.5 Linear Regression: Best Fit 209 In Exercises 83–86, use a graphing utility to graph each equation. Does this agree with the answer you gave in the Conceptual section? 83. 1x 2 222 1 1 y 1 522 5 220 (See Exercise 73 for comparison.) 84. 1x 2 122 1 1 y 1 322 5 0 (See Exercise 74 for comparison.) 85. x2 1 y2 1 10x 2 6y 1 34 5 0 (See Exercise 75 for comparison.) 86. x2 1 y2 2 4x 1 6y 1 49 5 0 (See Exercise 76 for comparison.) In Exercises 87–88, (a) with the equation of the circle in standard form, state the center and radius, and graph; (b) use the quadratic formula to solve for y; and (c) use a graphing utility to graph each equation found in (b). Does the graph in (a) agree with the graphs in (c)? 87. x2 1 y2 2 11x 1 3y 2 7.19 5 0 88. x2 1 y2 1 1.2x 2 3.2y 1 2.11 5 0 • T E C H N O L O G Y 2.5.1 Scatterplots An important aspect of applied research across disciplines is to discover and understand relationships between variables, and often how to use such a relationship to predict values of one variable in terms of another. You have likely encountered such issues while watching TV, reading a magazine or newspaper, or simply talking with friends. Some questions include ■ ■Is age predictive of texting speed? ■ ■Is the level of pollution in a country related to the prevalence of asthma in that country? ■ ■Do the ratings of car reliability necessarily increase with the price of the car? In this section, we focus on situations involving relationships between two variables x and y, so that the experimental data gathered consists of ordered pairs 1x1, y12, . . . , 1xn, yn2. A first step in understanding a data set of the form 5 1x1, y12, . . . , 1xn, yn2 6 is to create a pictorial representation of it. Identifying the first coordinates of these ordered pairs as values of an independent variable (or predictor variable) x and the second coordinates as the values of a dependent variable (or response variable) y, we simply plot them all on a single xy-plane. The resulting picture is called a scatterplot. S K I L L S O B J E C T I V E S ■ ■Draw a scatterplot. ■ ■Use linear regression to determine the line of best fit associated with some data. ■ ■Use the line of best fit to predict values of one variable from the values of another. C O N C E P T U A L O B J E C T I V ES ■ ■Recognize positive or negative association. ■ ■Recognize linear or nonlinear association. ■ ■Understand what “best fit” means. 2.5 LINEAR REGRESSION: BEST FIT Optional Technology Required Section. 2.5.1 S K IL L Draw a scatterplot. 2.5.1 C ON C E P T U A L Recognize positive or negative association. EXAMPLE 1 Drawing a Scatterplot of Olympic Decathlon Data The 2012 Men’s Olympic Decathlon consisted of the following 10 events: 100 meter, long jump, shot put, high jump, 400 meter, 110 meter hurdles, discus, pole vault, javelin throw, and 1500 meter. Actual scores are converted to a point system where points are assigned to each of these events based on performance. Events are equally weighted when converting to points. These points are then summed to obtain total scores and, in turn, medals are assigned based on these total scores. It would be interesting to know if certain events are more predictive of the total scores than are others. If someone does exceedingly well in the javelin throw, for example, is that person more likely to do well across all events and therefore obtain a large total points score? 210 CHAPTER 2 Graphs OLYMPIANS X100M LONG JUMP SHOT PUT HIGH JUMP X400M X110M HURDLE DISCUS POLE VAULT JAVELIN RANK TOTAL SCORE Eaton 10.35 8.03 14.66 2.05 46.90 13.56 42.53 5.20 61.96 1 8869 Hardee 10.42 7.53 15.28 1.99 48.11 13.54 48.26 4.80 66.65 2 8671 Suárez 11.27 7.52 14.50 2.11 49.04 14.45 45.7 4.70 76.94 3 8523 Alphen 11.05 7.64 15.48 2.05 49.18 14.89 48.28 4.80 61.69 4 8447 Warner 10.48 7.54 13.73 2.05 48.20 14.38 45.90 4.70 62.77 5 8442 Freimuth 10.65 7.21 14.87 1.90 48.06 13.89 49.11 4.90 57.37 6 8320 Kasyanov 10.56 7.55 14.45 1.99 48.44 14.09 46.72 4.60 54.87 7 8283 Sviridov 10.78 7.45 14.42 1.99 48.91 15.42 47.43 4.60 68.42 8 8219 Coertzen 11.09 7.17 13.79 2.05 48.56 14.15 43.58 4.50 64.79 9 8173 Behrenbruch 11.06 7.15 15.67 1.96 50.04 14.33 44.71 4.70 64.80 10 8126 Sintnicolaas 10.85 7.37 14.18 1.93 48.85 14.43 32.26 5.30 58.82 11 8034 Newdick 11.10 7.36 15.09 1.96 50.22 15.02 46.15 4.70 59.82 12 7988 Barroilhet 11.18 6.80 14.49 2.05 51.07 14.12 41.27 5.40 57.25 13 7972 García 10.80 6.75 14.48 1.99 48.76 14.24 42.27 4.60 59.85 14 7956 Mayer 11.32 7.17 14.05 2.05 48.76 15.59 41.20 4.70 62.41 15 7952 Shkurenyov 11.01 7.25 12.89 2.02 49.81 14.39 43.51 5.10 53.81 16 7948 Mikhan 10.74 6.94 14.75 1.93 48.42 14.15 44.42 4.40 55.69 17 7928 Karpov 10.91 7.21 16.47 1.99 49.83 14.40 44.93 5.10 49.93 18 7926 Alberto de Araújo 10.70 7.16 13.52 1.93 48.25 14.79 44.76 4.60 51.59 19 7849 Ushiro 11.32 6.86 13.59 1.99 50.78 15.47 46.66 4.90 66.38 20 7842 Vos 10.98 7.27 13.77 1.96 49.62 14.61 42.26 4.50 61.60 21 7805 Eri. nŠ 10.99 6.98 13.45 1.93 50.62 15.22 45.10 4.50 57.35 22 7649 Addy 10.89 6.90 14.97 1.93 48.64 14.23 45.61 4.20 50.36 23 7586 Szabó 11.15 6.96 13.93 1.90 50.83 14.92 45.14 4.60 58.84 24 7581 Draudvila 10.95 7.12 15.17 1.96 50.13 14.87 46.43 4.20 50.16 25 7557 Artikov 11.37 6.41 14.11 1.93 51.91 14.74 43.53 4.40 56.62 26 7203 Creating a scatterplot by hand can be tedious, especially for large data sets. You can also very easily lose precision and detail. Using technology to create a scatterplot is very appropriate and quite easy. Below are the procedures for how you would create the scatterplot shown in Example 1 using the TI-831 (or TI-84) and Excel 2010. 400 m Men’s 2012 Olympic Decathalon— 400 m vs. Total Points 47 49 51 53 55 Total Points 9000 8750 8500 8250 8000 7750 7500 7250 7000 x y Data from the Men’s 2012 Olympic Decathlon are presented below. Let’s consider the paired data set 5 1x, y2 6 where x 5 score on the 400 m and y 5 total score. One scatterplot using the 400 m and total points information from this data set is shown in the following graph. Several natural questions arise: How are different pairs of these data related? Are there any discernible patterns present, and if so, how strong are they? Is there a single curve that can be used to describe the general trend present in the data? We shall answer these questions one by one in this section. 2.5 Linear Regression: Best Fit 211 Creating a Scatterplot Using the TI-831 (or TI-84) INSTRUCTION SCREENSHOT Entering the Data Step 1 Press STAT, followed by 1:Edit. . . . Clear any data already present in columns L1 and L2 so that the screen looks like the one to the right. Step 2 Input the values of the x-variable (first entries in the ordered pairs) in column L1, pressing ENTER after each entry. Then, right arrow over to column L2 and input the values of the y-variable.The screen (starting from the beginning of the data set) should look like the one to the right when you are done. Plotting the Data Step 3 Press Y5 and then select Plot1 in the top row of the screen. If either Plot2 or Plot3 is darkened, move the cursor onto it and press ENTER to undarken it. The screen should look like the one to the right when you are done. Step 4 Press 2nd, followed by Y5 (for StatPlot). Select 1: and modify the entries to make the screen look like the one to the right. Step 5 Press 2nd, followed by Y5 and make certain that both Plot2 and Plot3 are OFF. The screen should look like the one to the right. Step 6 Make certain the ranges for the x and y values are appropriate for the given data set. Here, we use the window shown to the right. Step 7 Press GRAPH and you should get the scatterplot shown to the right. 212 CHAPTER 2 Graphs Creating a Scatterplot Using Excel 2010 INSTRUCTION SCREENSHOT Entering the Data Step 1 Open a new Excel spreadsheet. Input the values of the x-variable (first entries in the ordered pairs) in column A, starting with cell 1A. Then, input the values of the y-variable in column B, starting with cell 1B. The screen should look like the one to the right when you are done. Plotting the Data Step 2 Highlight the data. The screen should look like the one to the right when you are done. Step 3 Go to the Insert tab and select the icon labeled Scatter. A window list of five possible choices pops up. Step 4 Select the leftmost choice in the top row. Press ENTER. The scatterplot shown to the right should appear on the screen. 2.5 Linear Regression: Best Fit 213 INSTRUCTION SCREENSHOT Step 5 You can alter the format of the scatterplot with various bells and whistles by right-clicking anywhere near the data points and then selecting Format Plot Area at the bottom of the pop-up window. Y OUR T UR N Using the data in Example 1, identify x 5 score on the pole vault and z 5 total score. Use technology to create a scatterplot for the data set consisting of the ordered pairs 1x, z2. 2.5.2 Identifying Patterns While scatterplots show only clusters of ordered pairs, patterns of various types can emerge that can provide insight into how the variables x and y are related. Direction of Association This characteristic is analogous to the concept of slope of a line. If the cluster of points tends to rise from left to right, we say that x and y are positively associated, whereas if the cluster of points falls from left to right, we say that x and y are negatively associated. Certainly, the more closely packed together the points are to an identifiable curve, the easier it is to make such a determination. Some examples of scatterplots of varying degrees of positive and negative association are shown in the following table. ▼ A N S W E R Men's 2012 Olympic Decathlon– Pole Vault and Total Points 4.1 4.5 4.9 5.3 7000 7250 7750 7500 8000 8250 8500 8750 9000 Pole Vault Total Points 2.5.2 S K I L L Use linear regression to determine the line of best fit associated with some data. 2.5.2 C ON C E P T U A L Recognize linear or nonlinear association. SCATTERPLOT DIRECTION OF ASSOCIATION VERBAL DESCRIPTION y x x y Positive Association A shape that increases from left to right is very discernible in each case. y x The association is a bit loose, but you can still tell the data points tend to rise from left to right. ▼ 214 CHAPTER 2 Graphs SCATTERPLOT DIRECTION OF ASSOCIATION VERBAL DESCRIPTION y x No Association Haphazard scattering of points suggests neither positive nor negative association. y x Negative Association The association is a bit loose, but you can still tell the data points tend to fall from left to right. y x y x Negative Association A shape that decreases from left to right is very discernible in each case. Linearity Depending on the phenomena being studied and the actual sample being used, the data points comprising a scatterplot can conform very closely to an actual curve. If the curve is a line, we say that the relationship between x and y is linear; otherwise, we say the relationship is nonlinear. Some illustrative examples follow. SCATTERPLOT LINEARITY VERBAL DESCRIPTION y x Linear Perfect linear relationship; positive association y x Linear Perfect linear relationship; negative association [CONCEPT CHECK] If the points (x, y) of a scatterplot are tightly packed together and fall from left to right, we say that the relationship between x and y is a (A) strong negative linear relationship, (B) strong positive linear relationship, (C) strong negative nonlinear relationship, or (D) strong positive nonlinear relationship ANSWER (A) strong negative linear relationship ▼ 2.5 Linear Regression: Best Fit 215 SCATTERPLOT LINEARITY VERBAL DESCRIPTION y x Linear Fairly tight linear relationship; positive association y x Linear Fairly tight linear relationship; negative association y x Linear Rather loose, but still somewhat discernible, linear relationship; positive association y x Linear Rather loose, but still somewhat discernible, linear relationship; negative association y x Nonlinear Perfect nonlinear relationship y x Nonlinear Fairly tight nonlinear relationship y x No specifically identifiable relationship No discernible linear relationship or degree of association 216 CHAPTER 2 Graphs Strength of Linear Relationship The variability in the data can render it difficult to determine if there is a linear relationship between two variables. As such, it is useful to have a way of measuring how tightly a paired data set conforms to a linear shape. This measure is called the correlation coefficient, r, and is defined by the following formula: DEFINITION For a paired data set 5 1x1, y12, . . . , 1xn, yn2 6, the correlation coefficient, r, is defined by r 5 naxy 2 Q axRQ ayR Ånax2 2 Q axR 2 ~ Ånay2 2 Q ayR 2 The symbol az is a shorthand way of writing z1 1 c1 zn. So, for instance, ax2 5 x2 1 1 c1 x2 n. This is tedious to calculate by hand but is easily computed using technology. Below are the procedures to compute the correlation coefficient for the data set introduced in Example 1 using the TI-83+ (or TI-84) and Excel 2010. Computing a Correlation Coefficient Using the TI-831 (or TI-84) INSTRUCTION SCREENSHOT Step 1 Enter the data following the procedure outlined earlier in this section. The screen should look like the one to the right. EXAMPLE 2 Describing Patterns in a Data Set Describe the patterns present in the paired data set 1x, y2 considered in Example 1, where x 5 points on the 400 m and y 5 total score. Is this intuitive? Solution: We can surmise that the variable score on the 400 m has a relatively strong linear, negative association with the variable total score. This means that as the 400 m score decreases, the total score tends to increase. A negative relationship makes sense here in that 400 m scores reflect how much time it took to complete this race. So, lower scores (less time) reflect better performance and therefore more total points. Y OUR TU R N Using the data in Example 1, identify x 5 score on the pole vault and z 5 total score. Comment on the degree of association and linearity of the scatterplot consisting of the ordered pairs 1x, z2. Is this intuitive? ▼ A N S W E R The variable score the pole vault has a rather weak linear, positive association with the variable total score. This means that as the pole vault score increases, the total score tends to increase. In this case, a positive relationship makes sense in that pole vault scores reflect the height achieved. So, higher scores (greater height) reflect better performance and therefore more total points. ▼ 2.5 Linear Regression: Best Fit 217 INSTRUCTION SCREENSHOT Step 2 Set up what will display! In order for the desired output to display once we execute the commands to follow, we must tell the calculator to do so. As such, do the following: i. Press 2nd, followed by 0 to get CATALOG. ii. Scroll down until you get to DiagnosticOn. Press ENTER. Then, this command will appear on the home screen. Press ENTER again. The resulting screen should look like the one to the right. Step 3 Press STAT, followed by CALC, and then by 4:LinReg(ax1b). The resulting screen should look like the one to the right. Press ENTER. Step 4 Press ENTER again. After a brief moment, your screen should look like the one to the right. The value we want is in the bottom row of the screen, about r 5 20.7045. Note: Other information given will be pertinent once we define the best fit line in the next section. Computing a Correlation Coefficient Using Excel 2010 INSTRUCTION SCREENSHOT Step 1 Enter the data following the procedure outlined earlier in this section. The screen should look like the one to the right. Step 2 Select the Formulas tab at the top of the screen and then choose the More Functions within the Function Library grouping (on the left). The pull-down menu should be as shown to the right. Step 1 Step 2 218 CHAPTER 2 Graphs INSTRUCTION SCREENSHOT Step 3 From here, select Statistical, and then from this list, scroll down and choose CORREL. A pop-up window should appear, as shown to the right. Step 4 Enter A1:A26 in Array 1 and B1:B26 in Array 2, as shown to the right. Press OK. You will notice that the correlation coefficient appears directly beneath Array 2. In this case, r is about 20.7147. The square of the correlation coefficient is interpreted as a signed percentage of the variability among the y-values that is actually explained by the linear relationship, where the sign corresponds to the direction of the association. For instance, an r-value of 11 means that 100% of the variability among the y-values is explained by a line with positive slope; in such a case, all of the points in the data set actually lie on a single line. An r-value of 21 means the same thing, but the line has a negative slope. As the r-values get closer to zero, the more dispersed the points become from a line describing the pattern, so that an r-value very close to zero suggests no linear relationship is discernible. The following sample of scatterplots with the associated correlation coefficients should give you a feel for the strength of linearity suggested by various values of r. SCATTERPLOT CORRELATION COEFFICIENT r STRENGTH OF LINEARITY y x r 5 1.0 Perfect positive linear relationship 2.5 Linear Regression: Best Fit 219 SCATTERPLOT CORRELATION COEFFICIENT r STRENGTH OF LINEARITY y x r 5 21.0 Perfect negative linear relationship y x r 5 0.80 Reasonably strong, though not perfect, positive linear relationship y x r 5 20.45 Pretty weak, barely discernible, negative linear relationship r 5 0.10 No discernible linear relationship [CONCEPT CHECK] If the best-fit line for a set of data points (x, y) is horizontal, then we say (A) the relationship between x and y is positive linear, (B) the relationship between x and y is negative linear, or (C) there is no discernible relationship. ANSWER (C) There is no discernible relationship. ▼ EXAMPLE 3 Calculating the Correlation Coefficient Associated with a Data Set Use technology to calculate the correlation coefficient r for the paired data set 1x, y2 considered in Example 1, where x 5 points on the 400 m and y 5 total score. Interpret the strength of the linear relationship. Solution: We see that using either form of technology yields r 5 20.7147. This suggests that the data follow a relatively strong negative (i.e., negative slope) linear pattern. Y OUR T UR N Using the data in Example 1, identify x 5 score on the pole vault and z 5 total score, and calculate the correlation coefficient for the data set consisting of the ordered pairs 1x, z2 using technology. Interpret the strength of the linear relationship. ▼ A N S W E R The correlation coefficient is approximately r 5 0.45. This suggests that while the data follow a positive (i.e., positive slope) pattern, the degree to which an actual line describes the trend in the data is rather weak. ▼ 2.5.3 Linear Regression Determining the “Best Fit” Line Assuming that a data set follows a reasonably strong linear pattern, it is natural to ask which single straight line best describes this pattern. Having such a line would enable us not only to describe the relationship between the two variables x and y precisely, but also to predict values of y from values of x not present among the points of the data set. 2.5.3 S K I L L Use the line of best fit to predict values of one variable from the values of another. 2.5.3 C ON C E P T U A L Understand what “best fit” means. y x 220 CHAPTER 2 Graphs Consider the paired data set 1x, y2 from Example 1, where x 5 points on the 400 m and y 5 total score. You learned in Section 2.3 that between any two points there is a unique line whose equation can be determined. Three such lines passing through various pairs of points in the data set are illustrated below. line 1 line 2 line 3 400 m Men’s 2012 Olympic Decathalon— 400 m vs. Total Points 47 49 51 53 55 Total Points 9000 8750 8500 8250 8000 7750 7500 7250 7000 x y The unavoidable shortcoming of all of these lines, however, is that not all of the data points lie on a single one of them. Each has a negative slope, which is characteristic of the data set, and each of the lines is close to some of the data points but not close to others. In fact, we could draw infinitely many such lines and make a similar assessment. But which one best fits the data? The answer to this question depends on how you define “best.” Reasonably, for the line that best fits the data, the error incurred in using it to describe all of the points in the data set should be as small as possible. The conventional approach is to define this error by summing the n distances di between the y-coordinates of the data points and the corresponding y-value on the line y 5 Mx 1 B (that is, the y-value of the point on the line corresponding to the same x-value). These distances are, in effect, the error in making the approximation. This is illustrated below: x y x1 (x1, y1) (x3, y3) (x6, y6) (x8, y8) (x4, y4) (x7, y7) (x5, y5) (x2, y2) (x1, Mx1+B) (x2, Mx2+B) (x3, Mx3+B) y = Mx + B d1 d3 d4 d5 d6 d8 d7 x2 x3 x4 x5 x6 x7 x8 Using the distance formula, we find that di 5 "1xi 2 xi22 1 1 yi 2 1Mxi 1 B222 5 0 yi 2 1Mxi 1 B2 0. Note that di 5 0 precisely when the data point 1xi, yi2 lies directly on the line y 5 Mx 1 B, and that the closer di is to 0, the closer the point 1xi, yi2 is to the line y 5 Mx 1 B. As such, the goal is to determine the values of the slope M and y-intercept B for which the sum d1 1 c 1 dn is as small as possible. Then, the resulting straight line y 5 Mx 1 B best fits the data set 5 1x1, y12, c, 1xn, yn2 6. This is fine, in theory, but it turns out to be inconvenient to work with a sum of absolute value expressions. It is actually much more convenient to work with the 2.5 Linear Regression: Best Fit 221 squared distances d2 i . The values of M and B that minimize d1 1 c 1 dn are precisely the same as those that minimize d2 1 1 c 1 d2 n. Using calculus, it can be shown that the formulas for M and B are as follows: M 5 naxy 2 A axBA ayB nax2 2 A axB2 , B 5 ay n 2 M ax n The resulting line y 5 Mx 1 B is called the best fit least-squares regression line for the data set 5 1x1, y12, c, 1xn, yn2 6. Again, it is tedious to compute these by hand, but their values are produced easily using technology. EXAMPLE 4 Finding the Line of Best Fit by Linear Regression Find the line of best fit (best fit least-squares regression line) for the paired data set 1x, y2 from Example 1, where x 5 points on the 400 m and y 5 total score, using (a) the TI-831 (or TI-84) and (b) Excel 2010. Solution (a): Determining the Best Fit Least-Squares Regression Line Using the TI-831 (or TI-84). INSTRUCTION SCREENSHOT STEP 1 Enter the data following the procedure outlined earlier in this section. The screen should look like the one to the right. STEP 2 Press STAT, followed by CALC, and then by 4:LinReg(ax1b). The resulting screen should look like the one to the right. STEP 3 For this example, the data is stored in lists L1 and L2. And, since we will want to graph our best fit line on the scatterplot, it will need to be stored as a function of x, say as Y1. In order to do this, proceed as follows: Directly next to LinReg(ax1b) on the home screen, we need to type the following: L1, L2, Y1. Use the following key strokes: 2nd, 1, , 2nd, 2, , VARS, Y-VARS, 1:FUNCTION, Y1 The resulting screen should look like the one to the right. STEP 4 Press ENTER. The equation of the best fit least-squares line with the slope (labeled a) and the y-intercept (labeled b) appears on the screen as shown to the right. So, the equation of the best fit least-squares regression line is approximately y 5 2206.9x 1 18317.03. 222 CHAPTER 2 Graphs STEP 5 In order to obtain a graph of the scatterplot with the best fit line from Step 4 superimposed on it, press ZOOM, then 9:ZoomStat. The resulting screen should look like the one to the right. Solution (b): Determining the Best Fit Least-Squares Regression Line Using Excel 2010. INSTRUCTION SCREENSHOT STEP 1 Enter the data following the procedure outlined earlier in this section. The screen should look like the one to the right. STEP 2 Select the Formulas tab at the top of the screen and then choose the More Functions within the Function Library grouping (on the left). The pull-down menu should be as shown to the right. 2.5 Linear Regression: Best Fit 223 INSTRUCTION SCREENSHOT STEP 3 From here, select Statistical, and then from this list, scroll down and choose LINEST. A pop-up window should appear, as shown to the right. STEP 4 Enter B1:B26 in Known_y’s and A1:A26 in Known_x’s, as shown to the right. You will notice that a set of two values occurs directly beneath the entry boxes—the output is about 52232.03, 19,473.076. The first value is the slope M, and the second value is the y-intercept B of the best fit line. So, the equation of the best fit least-squares regression line is approximately y 5 2232.03x 1 19,437.07. STEP 5 In order to obtain a graph of the scatterplot with the best fit line from Step 4 superimposed on it, construct the scatterplot as before, right-click on the scatterplot near the data points, choose Add Trendline, and press ENTER. 224 CHAPTER 2 Graphs It is important to realize that the correlation coefficient is NOT equal, or even related, to the actual slope of the best fit line. In fact, two distinct perfectly linear, positively associated scatterplots will both have r 5 1, even though the actual lines that fit the data might have slope M 5 15 and M 5 0.04. Using the “Best Fit” Line for Prediction It is important to realize that for any scatterplot, no matter how haphazardly dispersed the data, a best fit least-squares regression line can be created. This is true even when the relationship between x and y is nonlinear. However, the utility of such a line in these instances is very limited. In fact, a best fit line should only be created when the linear relationship is reasonably strong, which means the correlation coefficient is “reasonably far away from 0.” This criterion can be made more precise using statistical methods, but for our present purposes, we shall make the blanket assumption that it makes sense to form the best fit line in all of the scenarios we present. Once we have the best fit line in hand, we can use it to predict y-values for values of x that do not correspond to any of the points in the data set. For instance, consider the following example. EXAMPLE 5 Making Predictions Using the Line of Best Fit Consider the data set from Example 1. a. Use the best fit line to predict the total score given that an Olympian scored 50.05 points in the 400 m. Is it reasonable to use the best fit line to make such a prediction? b. Use the best fit line to predict the total score given that an Olympian scored 40.05 points in the 400 m. Is it reasonable to use the best fit line to make such a prediction? Solution: a. Using the line y 5 2232.03x 1 19,473.07, we see that y is approximately 7961 when x 5 50.05. This means that if an Olympian were to score 50.05 on the INSTRUCTION SCREENSHOT STEP 6 The best fit line will appear on the scatterplot, along with a pop-up window allowing you to change the format of the line/ curve that is displayed. YOUR TURN Using the data in Example 1, identify x 5 score on the pole vault and z 5 total score, and determine the best fit least-squares regression line for the data set consisting of the ordered pairs 1x, z2. Superimpose the graph of this line on the scatterplot. ▼ A N S W E R Men’s 2012 Olympic Decathlon: Pole Vault and Total Points with Best Fit Line y = 545.58x + 5457.9 4 4.4 4.8 5.2 5.6 7000 7400 7800 8200 8600 9000 Pole Vault Total Points Here, the best fit line displays a positive relationship, and its equation is z 5 545.6x 1 5457.9. ▼ 2.5 Linear Regression: Best Fit 225 400 m, then his predicted total score would be approximately 7860. Using the best fit line to predict the total score in this case is reasonable because the value 50.05 is well within the range of x-values already present in the data set. b. Using the line y 5 2232.03x 1 19473.07, we see that y is approximately 10,030 when x 5 40.05. This means that if an Olympian were to score 40.05 on the 400 m, then his predicted total score would be approximately 10,180. This prediction is questionable because the x-value at which you are using the best fit line to predict y is far away from the rest of the data points that were used to construct the line. As such, there is no reason to believe that the line is valid for such x-values. Y OUR T UR N Using the data in Example 1, identify x 5 score on the pole vault and z 5 total score, and use the best fit line to predict the total score given that an Olympian scored 4.65 points on the pole vault and then, given that an Olympian scored 5.9 points on the pole vault. Comment on the validity of these predictions. ▼ A N S W E R Using the line z 5 545.6x 1 5457.9, we see that z is approximately 7995 when x 5 4.65. This means that if an Olympian were to score 4.65 on the pole vault, then his predicted total score would be approximately 7995. Using the best fit line to predict the total score in this case is reasonable because the value 4.65 is well within the range of x-values already present in the data set. Next, using the same line, we see that z is approximately 8,677 when x 5 5.9. This prediction is less reliable than the former one because 5.9 is outside of the range of x-values corresponding to the rest of the data points used to construct the line. But it isn’t too far outside this range, so there is a degree of validity to the prediction. ▼ [SEC TION 2 .5] E XE R C I S E S • S K I L L S In Exercises 1–4, for each of the following scatterplots, identify the pattern as a. having a positive association, negative association, or no identifiable association. b. being linear or nonlinear. 1. y x 2. y x 3. y x 4. y x In Exercises 5–8, match the following scatterplots with the following correlation coefficients. a. r 5 20.90 b. r 5 0.80 c. r 5 20.68 d. r 5 0.20 5. y x 6. y x 7. y x 8. y x measured using the correlation coefficient r. If r is sufficiently far from 0, a best fit least-squares regression line can be formed to precisely describe the linear relationship and used for reasonable prediction purposes. Two variables x and y can be related in different ways. A paired data set 51x1, y12, . . . , 1xn, yn26 obtained experimentally can be illus-trated using a scatterplot. Patterns concerning the direction of association and linearity can be used to describe the relationship between x and y, and the strength of the linear relationship can be [SEC TION 2 .5] S U M M A RY 226 CHAPTER 2 Graphs For each of the following data sets, a. create a scatterplot. b. guess the value of the correlation coefficient r. c. use technology to determine the equation of the best fit line and to calculate r. d. give a verbal description of the relationship between x and y. 9. x y 23 14 21 8 0 5 1 2 3 24 5 210 10. x y 28 216 26 212 24 28 22 24 1 2 3 6 11. x y 210 1 26 0 0 22 8 210 14 211 20 216 12. x y 21 217 21/2 211 21/4 25 21/10 0 0 1 1/10 1 1/5 8 1 12 13. x y 23 26 22 3 21 1 0 1 1 5 2 21 4 1 14. x y 26 21 23 4 21 3 1 0 2 26 5 24 8 1 In Exercises 15–18, for each of the data sets, a. use technology to create a scatterplot, to determine the best fit line, and to compute r. b. indicate whether or not the best fit line can be used for predictive purposes for the following x-values. For those for which it can be used, give the predicted value of y: i. x 5 0 ii. x 5 26 iii. x 5 12 iv. x 5 215 c. Using the best fit line, at what x-value would you expect y to be equal to 2? 15. x y 25 28 23 0 22 0 2 1.5 5 4 7 2 10 8 16. x y 5 0 5 3 6 3 7 6 7 9 8 9 8 15 9 9 9 15 10 15 10 18 17. x y 220 15 218 10 214 3 214 8 213 3 28 0 28 23 25 26 1 211 1 215 18. x y 215 4 215 12 215 16 213 3 210 4 210 8 210 12 25 4 22 3 22 22 2 3 2 6 4 21 4 0 4 4 7 22 7 3 10 24 10 22 10 3 For Exercises 19–22, a. use technology to create a scatterplot, to determine the best fit line, and to compute r for the entire data set. b. repeat (a), but with the data set obtained by removing the starred () data points. c. compare the r-values from (a) and (b), as well as the slopes of the best fit lines. Comment on any differences, whether they are substantive, and why this seems reasonable. 19. x y 23 14 21 8 0 5 1 2 3 24 5 210 20. x y 210 1 26 0 0 22 8 210 14 211 20 216 21. x y 23 14 21 8 0 5 1 2 3 24 5 215 6 216 22. x y 0 0 1 2 2 4 3 6 4 8 6 12 7 25 2.5 Linear Regression: Best Fit 227 • C O N C E P T U A L 23. Consider the data set from Exercise 17. a. Reverse the roles of x and y so that now y is the explanatory variable and x is the response variable. Create a scatterplot for the ordered pairs of the form 1 y, x2 using this data set. b. Compute r. How does it compare to the r-value from Exercise 17? Why does this make sense? c. The best fit line for the scatterplot in (a) will be of the form x 5 my 1 b. Determine this line. d. Using the line from (c), find the predicted x-value for the following y-values, if appropriate. If it is not appropriate, tell why. i. y 5 23 ii. y 5 2 iii. y 5 216 24. Consider the data set from Exercise 16. Redo the parts in Exercise 23. 25. Consider the following data set. x y 3 0 3 1 3 21 3 22 3 4 3 15 3 26 3 8 3 10 Guess the values of r and the best fit line. Then, check your answers using technology. What happens? Can you reason why this is the case? 26. Consider the following data set. x y 25 22 24 22 21 22 0 22 1 22 3 22 8 22 17 22 Guess the values of r and the best fit line. Then, check your answers using technology. What happens? Can you reason why this is the case? 27. The following screenshot was taken when using the TI-831 to determine the equation of the best fit line for paired data 1x, y2: Using the regression line, we observe that there is a strong positive linear association between x and y, and that for every unit increase in x, the y-value increases by about 1.257 units. 28. The following scatterplot was produced using the TI-831 for paired data 1x, y2. The equation of the best fit line was reported to be y 5 23.207x 1 0.971 with r2 5 0.9827. Thus, the correlation coefficient is given by r 5 0.9913, which indicates a strong linear association between x and y. • C A T C H T H E M I S T A K E 228 CHAPTER 2 Graphs For Exercises 29 and 30, refer to the data set in Example 1. 29. a. Examine the relationship between each of the decathlon events and the total points by computing the correlation coefficient in each case. b. Using the information from part (a), which event has the strongest relationship to the total points? c. What is the equation of the best fit line that describes the relationship between the event from part (b) and the total points? d. Using the best fit line, if you had a score of 40 for this event, what would the predicted total points score be? 30. a. Using the information from part (a), which event has the second strongest relationship to the total points? b. What is the equation of the best fit line that describes the relationship between the event in part (b) and the total points? c. Is it reasonable to expect the best fit line from part (c) to produce accurate predictions of total points using this event? d. Using the best fit line, if you had a score of 40 for this event, what would the total points score be? • A P P L I C AT I O N S 31. What is the relationship between the variables % residents immunized and % residents with influenza? a. Create a scatterplot to illustrate the relationship between % residents immunized and % residents with influenza. b. What is the correlation coefficient between % residents immunized and % residents with influenza? c. Describe the strength of the relationship between % residents immunized and % residents with influenza. d. What is the equation of the best fit line that describes the relationship between % residents immunized and % residents with influenza? e. Could you use the best fit line to produce accurate predictions of % residents with influenza using % residents immunized? 32. What is the impact of the outlier(s) on this data set? a. Identify the outlier in this data set. What is the nursing home number for this outlier? b. Remove the outlier and re-create the scatterplot to show the relationship between % residents immunized and % residents with influenza. c. What is the revised correlation coefficient between % residents immunized and % residents with influenza? d. By removing the outlier, is the strength of the relationship between % residents immunized and % residents with influenza increased or decreased? e. What is the revised equation of the best fit line that describes the relationship between % residents immunized and % residents with influenza? For Exercises 33–36, refer to the following data set: Amusement Park Rides. According to the International Association of Amusement Parks and Attractions (IAAPA), “There are more than 400 amusement parks and traditional attractions in the United States alone. . . . Amusement parks in the United States entertained 300 million visitors who safely enjoyed more than 1.7 billion rides.” Despite the popularity of amusement parks, the wait times, especially for the most popular rides, are not so highly regarded. There are different approaches and tactics that people take to get the most rides in their visit to the park. Now, there are even apps for the iPhone and Android to track waiting times at various amusement parks. One might ask, “Are the wait times worth it? Are the rides with the longest wait times, the most enjoyable?” Consider the following fictional data. For Exercises 31 and 32, refer to the following data set: Herd Immunity. According to the U.S. Department of Health and Human Services, herd immunity is defined as “a concept of protecting a community against certain diseases by having a high percentage of the community’s population immunized. Even if a few members of the community are unable to be immunized, the entire community will be indirectly protected because the disease has little opportunity for an outbreak. However, with a low percentage of population immunity, the disease would have great opportunity for an outbreak.” Suppose a study is conducted that looks at the outbreak of Haemophilus influenzae type b in the winter of the previous year across 22 nursing homes. We might look at the percentage of residents in each of the nursing homes that were immunized and the percentage of residents who were infected with this type of influenza. The fictional data set is as follows. NURSING HOME % RESIDENTS IMMUNIZED % RESIDENTS WITH INFLUENZA 1 70 11 2 68 9 3 80 8 4 10 34 5 12 30 6 18 31 7 27 22 8 64 18 9 73 6 10 9 31 11 35 19 12 56 16 13 57 22 14 83 10 15 74 13 16 64 15 17 16 28 18 23 25 19 29 24 20 33 20 21 82 28 22 67 9 2.5 Linear Regression: Best Fit 229 RIDE ID RIDE NAME AVG WAIT TIME AVG ENJOYMENT RATING PARK PARK LOCATION 1 Xoom 45 58 1 Florida 2 Accentuator 35 40 1 Florida 3 Wobbler 15 15 1 Florida 4 Arctic_Attack 75 75 1 Florida 5 Gusher 60 70 1 Florida 6 Alley_Cats 5 60 1 Florida 7 Moon_Swing 10 15 1 Florida 8 Speedster 70 50 1 Florida 9 Hailstorm 80 90 1 Florida 10 DragonFire 70 88 1 Florida 1 Xoom 50 10 2 California 2 Accentuator 35 40 2 California 3 Wobbler 20 75 2 California 4 Arctic_Attack 70 60 2 California 5 Gusher 70 80 2 California 6 Alley_Cats 10 18 2 California 7 Moon_Swing 15 80 2 California 8 Speedster 80 35 2 California 9 Hailstorm 95 40 2 California 10 DragonFire 55 60 2 California The data shows 10 popular rides in two sister parks located in Florida and California. For each ride in each park, average wait times (in minutes) in the summer of 2010 and the average rating of ride enjoyment (on a scale of 1–100) are provided. 33. What is the relationship between the variables average wait times and average rating of enjoyment? a. Create a scatterplot to show the relationship between average wait times and average rating of enjoyment. b. What is the correlation coefficient between average wait times and average rating of enjoyment? c. Describe the strength of the relationship between average wait times and average rating of enjoyment. d. What is the equation of the best fit line that describes the relationship between average wait times and average rating of enjoyment? e. Could you use the best fit line to produce accurate predictions of average wait times using average rating of enjoyment? 34. Examine the relationship between average wait times and average rating of enjoyment for Park 1 in Florida by repeating Exercise 33 for only Park 1. 35. Examine the relationship between average wait times and average rating of enjoyment for Park 2 in California by repeating Exercise 33 for only Park 2. 36. Compare the relationship between average wait times and average rating of enjoyment for Park 1 in Florida versus Park 2 in California. • C H A L L E N G E For Exercises 37–40, refer to the following: Exploring other types of best fit curves To describe the patterns that emerge in paired data sets, there are many more possibilities than best fit lines. Indeed, once you have drawn a scatterplot and are ready to identify the curve that best fits the data, there is a substantive collection of other curves that might more accurately describe the data. The following are listed among those in STATS/CALC on the TI-831, along with some comments: NAME OF REGRESSION CURVE FORM OF THE CURVE COMMENTS 5: QuadReg y 5 ax2 1 bx 1 c The data set must have at least 3 points to be able to select this option. 6: CubicReg y 5 ax3 1 bx2 1 cx 1 d The data set must have at least 4 points to be able to select this option. 7: QuartReg y 5 ax4 1 bx3 1 cx2 1 dx 1 e The data set must have at least 5 points to be able to select this option. 9: LnReg y 5 a 1 b ln x The data set must have at least 2 points to be able to select this option, and x cannot take on negative values. 0: ExpReg y 5 a bx The data set must have at least 2 points to be able to select this option, and y cannot take on the value of 0. A: PwrReg y 5 a x b The data set must have at least 2 points to be able to select this option. 230 CHAPTER 2 Graphs For each of the following data sets, a. Create a scatterplot. b. Use LinReg(ax1b) to determine the best fit line and r. Does the line seem to accurately describe the pattern in the data? c. For each of the choices listed in the preceding chart, find the equation of the best fit curve and its associated r 2 value. Of all of the curves, which seems to provide the best fit? Note: The r2-value reported in each case is NOT the linear correlation coefficient reported when running LinReg(ax1B). Rather, the value will typically change depending on the curve. The reason is that each time, the r 2-value is measuring how accurate the fit is between the data and that type of curve. A value of r 2 close to 1 still corresponds to a good fit with whichever curve you are fitting to the data. 37. x y 1 16.2 2 21 3 23.7 4 24.8 5 23.9 6 20.7 7 15.8 8 9.1 9 0.3 38. x y 0.5 1.20 1.0 0.760 1.5 0.412 2.1 0.196 2.9 0.131 3.3 0.071 39. x y 1 0.2 1.5 0.93 2 1.46 3 2.25 10 4.51 15 5.50 40. x y 1 32.3 2 8.12 3 216.89 5 245.2 6 0.89 8 62.1 CH A P TE R 2 R E VIE W [CH AP TER 2 REVIEW] Chapter Review 231 SECTION CONCEPT KEY IDEAS/FORMULAS 2.1 Basic tools: Cartesian plane, distance, and midpoint Two points in the xy-plane: 1x1, y12 and 1x2, y22 Cartesian plane x-axis, y-axis, origin, and quadrants Distance between two points d 5 "1x2 2 x122 1 1 y2 2 y122 Midpoint of a line segment joining two points 1xm, ym2 5 a x1 1 x2 2 , y1 1 y2 2 b 2.2 Graphing equations: Point-plotting, intercepts, and symmetry Point-plotting List a table with several coordinates that are solutions to the equation; plot and connect. Intercepts x-intercept: let y 5 0 y-intercept: let x 5 0 Symmetry The graph of an equation can be symmetric about the x-axis, y-axis, or origin. Using intercepts and symmetry as graphing aids If 1a, b2 is on the graph of the equation, then 12a, b2 is on the graph if symmetric about the y-axis, 1a, 2b2 is on the graph if symmetric about the x-axis, and 12a, 2b2 is on the graph if symmetric about the origin. 2.3 Lines General form: Ax 1 By 5 C Graphing a line Vertical: x 5 a Slant: Ax 1 By 5 C Horizontal: y 5 b where A 2 0 and B 2 0 Intercepts x-intercept 1a, 02 y-intercept 10, b2 Slope m 5 y2 2 y1 x2 2 x1 , where x1 2 x2 "rise" "run" Equations of lines Slope–intercept form: y 5 mx 1 b m is the slope and b is the y-intercept. Point–slope form: y 2 y1 5 m 1x 2 x12 Parallel and perpendicular lines L1 L2 if and only if m1 5 m2 (slopes are equal). L1 ' L2 if and only if m1 5 2 1 m2 e m1 2 0 m2 2 0 (slopes are negative reciprocals). 2.4 Circles Standard equation of a circle 1x 2 h22 1 1y 2 k22 5 r 2 C: 1h, k2 Transforming equations of circles to the standard form by completing the square General form: x2 1 y2 1 ax 1 by 1 c 5 0 2.5 Linear regression: Best fit Fitting data with a line Scatterplots Creating a scatterplot: n Using Microsoft Excel n Using a graphing calculator Identifying patterns Association Linearity n Positive n Negative n Linear n Nonlinear Correlation coefficient, r Linear regression n Determine the “best fit” line n Making predictions REV IEW E XE R CI SE S 232 CHAPTER 2 Graphs [ C H AP T E R 2 REVIEW EXERC IS E S ] 2.1 Basic Tools: Cartesian Plane, Distance, and Midpoint Plot each point and indicate which quadrant the point lies in. 1. 124, 22 2. 14, 72 3. 121, 262 4. 12, 212 Calculate the distance between the two points. 5. 122, 02 and 14, 32 6. 11, 42 and 14, 42 7. 124, 262 and 12, 72 8. A1 4, 1 12B and A1 3, 27 3B Calculate the midpoint of the segment joining the two points. 9. 12, 42 and 13, 82 10. 122, 62 and 15, 72 11. 12.3, 3.42 and 15.4, 7.22 12. 12a, 22 and 1a, 42 Applications 13. Sports. A quarterback drops back to pass. At the point 125, 2202 he throws the ball to his wide receiver located at 110, 302. Find the distance the ball has traveled. Assume the width of the football field is 3215, 15 4 and the length is 3250, 50 4 . Units of measure are yards. 14. Sports. Suppose that in the above exercise a defender was midway between the quarterback and the receiver. At what point was the defender located when the ball was thrown over his head? 2.2 Graphing Equations: Point-Plotting, Intercepts, and Symmetry Find the x-intercept(s) and y-intercept(s) if any. 15. x2 1 4y2 5 4 16. y 5 x2 2 x 1 2 17. y 5 "x2 2 9 18. y 5 x2 2 x 2 12 x 2 12 Use algebraic tests to determine symmetry with respect to the x-axis, y-axis, or origin. 19. x2 1 y3 5 4 20. y 5 x2 2 2 21. xy 5 4 22. y2 5 5 1 x Use symmetry as a graphing aid and point-plot the given equations. 23. y 5 x2 2 3 24. y 5 0 x 0 2 4 25. y 5 3 "x 26. x 5 y2 2 2 27. y 5 x"9 2 x2 28. x2 1 y2 5 36 Applications 29. Sports. A track around a high school football field is in the shape of the graph 8x2 1 y2 5 8. Graph by using symmetry and plotting points. 30. Transportation. A “bypass” around a town follows the graph y 5 x3 1 2, where the origin is the center of town. Graph the equation. 2.3 Lines Express the equation for each line in slope–intercept form. Identify the slope and y-intercept of each line. 31. 6x 1 2y 5 12 32. 3x 1 4y 5 9 33. 21 2x 2 1 3y 5 1 6 34. 22 3x 2 1 4y 5 1 8 Find the x- and y-intercepts and the slope of each line, if they exist, and graph. 35. y 5 4x 2 5 36. y 5 23 4x 2 3 37. x 1 y 5 4 38. x 5 24 39. y 5 2 40. 21 2x 2 1 2y 5 3 Write the equation of the line, given the slope and the intercepts. 41. Slope: m 5 4 y-intercept: 10, 232 42. Slope: m 5 0 y-intercept: 10, 42 43. Slope: m is undefined x-intercept: 123, 02 44. Slope: m 5 22 3 y-intercept: A0, 3 4B R E VI E W E XERCISES Review Exercises 233 Write an equation of the line, given the slope and a point that lies on the line. 45. m 5 22 123, 42 46. m 5 3 4 12, 162 47. m 5 0 124, 62 48. m is undefined 12, 252 Write the equation of the line that passes through the given points. Express the equation in slope–intercept form or in the form of x 5 a or y 5 b. 49. 124, 222 and 12, 32 50. 121, 42 and 122, 52 51. A23 4, 1 2B and A27 4, 5 2B 52. 13, 222 and 129, 22 Find the equation of the line that passes through the given point and also satisfies the additional piece of information. 53. 122, 212 parallel to the line 2x 2 3y 5 6 54. 15, 62 perpendicular to the line 5x 2 3y 5 0 55. A 23 4, 5 2B perpendicular to the line 2 3x 2 1 2y 5 12 56. 1a 1 2, b 2 12 parallel to the line Ax 1 By 5 C Applications 57. Grades. For a GRE prep class, a student must take a pretest and then a posttest after the completion of the course. Two students’ results are shown below. Give a linear equation to represent the given data. PRETEST POSTTEST 1020 1324 950 1240 58. Budget: Car Repair. The cost of having the air conditioner in your car repaired is the combination of material costs and labor costs. The materials (tubing, coolant, etc.) are $250, and the labor costs $38 per hour. Write an equation that models the total cost C of having your air conditioner repaired as a function of hours t. Graph this equation with t as the horizontal axis and C representing the vertical axis. How much will the job cost if the mechanic works 1.5 hours? 2.4 Circles Write the equation of the circle in standard form. 59. center 122, 32 r 5 6 60. center 126, 282 r 5 3"6 61. center A3 4, 5 2B r 5 2 5 62. center 11.2, 22.42 r 5 3.6 Find the center and the radius of the circle given by the equation. 63. 1x 1 222 1 1y 1 322 5 81 64. 1x 2 422 1 1y 1 222 5 32 65. Ax 1 3 4B 2 1 Ay2 1 2B 2 5 16 36 66. x2 1 y2 1 4x 2 2y 5 0 67. x2 1 y2 1 2y 2 4x 1 11 5 0 68. 3x2 1 3y2 2 6x 2 7 5 0 69. 9x2 1 9y2 2 6x 1 12y 2 76 5 0 70. x2 1 y2 1 3.2x 2 6.6y 2 2.4 5 0 71. Find the equation of a circle centered at 12, 72 and passing through 13, 62. 72. Find the equation of a circle that has the diameter with endpoints 122, 212 and 15, 52. Technology Exercises Section 2.1 Determine whether the triangle with the given vertices is a right triangle, isosceles triangle, neither, or both. 73. 1210, 252, 120, 2452, 110, 102 74. 14.2, 8.42, 124.2, 2.12, 16.3, 210.52 Section 2.2 Graph the equation using a graphing utility and state whether there is any symmetry. 75. y2 5 0 x2 2 4 0 76. 0.8x2 2 1.5y2 5 4.8 Section 2.3 Determine whether the lines are parallel, perpendicular, or neither, then graph both lines in the same viewing screen using a graphing utility to confirm your answer. 77. y1 5 0.875x 1 1.5 y2 5 28 7x 2 9 14 78. y1 5 20.45x 2 2.1 y2 5 5 6 2 9 20x Section 2.4 79. Use the Quadratic Formula to solve for y, and use a graphing utility to graph each equation. Do the graphs agree with the graph in Exercise 69? 9x2 1 9y2 2 6x 1 12y 2 76 5 0 80. Use the Quadratic Formula to solve for y, and use a graphing utility to graph each equation. Do the graphs agree with the graph in Exercise 70? x2 1 y2 1 3.2x 2 6.6y 2 2.4 5 0 PR ACTICE TEST 234 CHAPTER 2 Graphs [ C H AP T E R 2 PRACTICE TEST ] 1. Find the distance between the points 127, 232 and 12, 222. 2. Find the midpoint between 123, 52 and 15, 212. 3. Determine the length and the midpoint of a segment that joins the points 122, 42 and 13, 62. 4. Research Triangle. The Research Triangle in North Carolina was established as a collaborative research center among Duke University (Durham), North Carolina State University (Raleigh), and the University of North Carolina (Chapel Hill). Raleigh Durham Chapel Hill Gorman Cary New Hope Creedmoor Hillsborough Durham is 10 miles north and 8 miles east of Chapel Hill, and Raleigh is 28 miles east and 15 miles south of Chapel Hill. What is the perimeter of the research triangle? Round your answer to the nearest mile. 5. Determine the two values for y so that the point 13, y2 is five units away from the point 16, 52. 6. If the point 13, 242 is on a graph that is symmetric with respect to the y-axis, what point must also be on the graph? 7. Determine whether the graph of the equation x 2 y2 5 5 has any symmetry (x-axis, y-axis, and origin). 8. Find the x-intercept(s) and the y-intercept(s), if any: 4x2 2 9y2 5 36. Graph the following equations. 9. 2x2 1 y2 5 8 10. y 5 4 x2 1 1 11. Find the x-intercept and the y-intercept of the line x 2 3y 5 6. 12. Express the line in slope–intercept form: 4x 2 6y 5 12. 13. Express the line in slope–intercept form: 2 3x 2 1 4y 5 2. Find the equation of the line that is characterized by the given information. Graph the line. 14. Slope 5 4; y-intercept 10, 32 15. Passes through the points 123, 22 and 14, 92 16. Parallel to the line y 5 4x 1 3 and passes through the point 11, 72 17. Perpendicular to the line 2x 2 4y 5 5 and passes through the point 11, 12 18. x-intercept 13, 02; y-intercept 10, 62 For Exercises 19 and 20, write the equation of the line that corresponds to the graph. 19. 20. x y (1, 4) (–2, –2) x y (2, –1) (–2, 3) 21. Write the equation of a circle that has center 16, 272 and radius r 5 8. 22. Determine the center and radius of the circle x2 1 y2 2 10x 1 6y 1 22 5 0. 23. Find the equation of the circle that is centered at 14, 92 and passes through the point 12, 52. 24. Solar System. Earth is approximately 93 million miles from the Sun. Approximating Earth’s orbit around the Sun as circular, write an equation governing Earth’s path around the Sun. Locate the Sun at the origin. 25. Determine whether the triangle with the given vertices is a right triangle, isosceles triangle, neither, or both. 128.4, 16.82, 10, 37.82, 112.6, 8.42 26. Graph the given equation using a graphing utility and state whether there is any symmetry. 0.25y2 1 0.04x2 5 1 CU MU LA TIV E TEST [CH AP TERS 1–2 CUM UL AT IVE T E S T ] 1. Simplify 7 2 2 7 1 3. 2. Simplify and express in terms of positive exponents: A 5x3 / 4B4 25x21 / 4. 3. Perform the operation and simplify: 1x 2 422 1x 1 422. 4. Factor completely 8x3 2 27y3. 5. Perform the operations and simplify: 1/x 2 1/5 1/x 1 1/5. 6. Solve for x: x3 2 5x2 2 4x 1 20 5 0. 7. Perform the operations and write in standard form: "23615 2 2i2. Solve for x. 8. 15 2 35 1 3x 2 4 12x 2 62 4 5 4 16x 2 72 2 3 5 13x 2 72 2 6x 1 10 4 9. 5 4x 1 1 5 3 4x 2 1 10. Ashley inherited $17,000. She invested some money in a CD that earns 5% and the rest in a stock that earns 8%. How much was invested in each account, if the interest for the first year is $1075? 11. Solve by factoring: 5x2 5 45. 12. Solve by completing the square: 3x2 1 6x 5 7. 13. Use the discriminant to determine the number and type of roots: 5x2 1 2x 1 7 5 0. 14. Solve for r: p2 1 q2 5 r2. 15. Solve and check: "x2 1 3x 2 10 5 x 2 2. 16. Solve using substitution: 1 1x 1 222 2 5 x 1 2 1 4 5 0. Solve and express the solution in interval notation: 17. 6 , 1 4x 1 6 , 9 18. x2 2 x $ 20 19. 0 2 2 x 0 , 4 20. Solve for x: 0 5 2 4x 0 5 23. 21. Use algebraic tests to determine whether the graph of the equation y 5 4x is symmetric with respect to the x-axis, y-axis, or origin. 22. Write an equation of a line in slope–intercept form with slope m 5 4 5 that passes through the point 15, 12. 23. Write an equation of a line that is perpendicular to the x-axis and passes through the point 15, 32 . 24. Write an equation of a line in slope–intercept form that passes through the two points A1 7, 5 3B and A26 7, 22 3B. 25. Find the center and radius of the circle: 1x 1 522 1 1y 1 322 5 30. 26. Calculate the distance between the two points A 2!11, 5B and A2, !7B, and find the midpoint of the segment joining the two points. Round your answers to one decimal place. 27. Determine whether the lines y1 5 0.32x 1 1.5 and y2 5 2 5 16x 1 1 4 are parallel, perpendicular, or neither, then graph both lines in the same viewing screen using a graphing utility to confirm your answer. Cumulative Test 235 Getty Images C H A P T E R [ [ LEARNING OBJECTIVES ■ ■Find the domain and range of a function. ■ ■Sketch the graphs of common functions. ■ ■Sketch graphs of general functions employing translations of common functions. ■ ■Perform composition of functions. ■ ■Find the inverse of a function. ■ ■Model applications with functions using variation. On a sales rack of clothes at a department store, you see a shirt you like. The original price of the shirt was $100, but it has been discounted 30%. As a preferred shopper, you get an automatic additional 20% off the sale price at the register. How much will you pay for the shirt? Naïve shoppers might be lured into thinking this shirt will cost $50 because they add the 20% and 30% to get 50% off, but they will end up paying more than that. Experienced shoppers know that they first take 30% off of $100, which results in a price of $70, and then they take an additional 20% off of the sale price, $70, which results in a final discounted price of $56. Experienced shoppers have already learned the concept of composition of functions. A composition of functions can be thought of as a function of a function. One function takes an input (original price, $100) and maps it to an output (sale price, $70), and then another function takes that output as its input (sale price, $70) and maps that to an output (checkout price, $56). Functions and Their Graphs 3 SuperStock/Alamy Stock Photo 237 FUNCTIONS AND THEIR GRAPHS 3.1 FUNCTIONS 3.2 GRAPHS OF FUNCTIONS; PIECEWISE- DEFINED FUNCTIONS; INCREASING AND DECREASING FUNCTIONS; AVERAGE RATE OF CHANGE 3.3 GRAPHING TECHNIQUES: TRANSFORMATIONS 3.4 OPERATIONS ON FUNCTIONS AND COMPOSITION OF FUNCTIONS 3.5 ONE-TO-ONE FUNCTIONS AND INVERSE FUNCTIONS 3.6 MODELING FUNCTIONS USING VARIATION • Relations and Functions • Functions Defined by Equations • Function Notation • Domain of a Function • Recognizing and Classifying Functions • Increasing and Decreasing Functions • Average Rate of Change • Piecewise- Defined Functions • Horizontal and Vertical Shifts • Reflection about the Axes • Stretching and Compressing • Adding, Subtracting, Multiplying, and Dividing Functions • Composition of Functions • Determine Whether a Function Is One-to-One • Inverse Functions • Graphical Inter-pretation of Inverse Functions • Finding the Inverse Function • Direct Variation • Inverse Variation • Joint Variation and Combined Variation [I N T HI S CHAPTER] You will find that functions are part of our everyday thinking: converting from degrees Celsius to degrees Fahrenheit, DNA testing in forensic science, determining stock values, and the sale price of a shirt. We will develop a more complete, thorough understanding of functions. First, we will establish what a relation is, and then we will determine whether a relation is a function. We will discuss common functions, domain and range of functions, and graphs of functions. We will determine whether a function is increasing or decreasing on an interval and calculate the average rate of change of a function. We will perform operations on functions and composition of functions. We will discuss one-to-one functions and inverse functions. Finally, we will model applications with functions using variation. 238 CHAPTER 3 Functions and Their Graphs 3.1.1 Relations and Functions What do the following pairs have in common? ■ ■Every person has a blood type. ■ ■Temperature is some specific value at a particular time of day. ■ ■Every working household phone in the United States has a 10-digit phone number. ■ ■First-class postage rates correspond to the weight of a letter. ■ ■Certain times of the day are start times of sporting events at a university. They all describe a particular correspondence between two groups. A relation is a correspondence between two sets. The first set is called the domain, and the corresponding second set is called the range. Members of these sets are called elements. 3.1.1 S KILL Determine whether a relation is a function. 3.1.1 CO NCE PTUAL Understand that all functions are relations but not all relations are functions. S K I L L S O B J E C T I V E S ■ ■Determine whether a relation is a function. ■ ■Determine whether an equation represents a function. ■ ■Use function notation to evaluate functions for particular arguments. ■ ■Determine the domain and range of a function. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that all functions are relations but not all relations are functions. ■ ■Understand why the vertical line test determines if a relation is a function. ■ ■Think of function notation as a placeholder or mapping. ■ ■Understand the difference between implicit domain and explicit domain. 3.1 FUNCTIONS DEFINITION Relation A relation is a correspondence between two sets where each element in the first set, called the domain, corresponds to at least one element in the second set, called the range. WORDS MATH The domain is the set of all the 5Michael, Tania, Dylan, Trevor, Megan6 first components. The range is the set of all the 5A, AB, O6 second components. A relation in which each element in the domain corresponds to exactly one element in the range is a function. A relation is a set of ordered pairs. The domain is the set of all the first components of the ordered pairs, and the range is the set of all the second components of the ordered pairs. PERSON BLOOD TYPE ORDERED PAIR Michael A (Michael, A) Tania A (Tania, A) Dylan AB (Dylan, AB) Trevor O (Trevor, O) Megan O (Megan, O) 3.1 Functions 239 Note that the definition of a function is more restrictive than the definition of a relation. For a relation, each input corresponds to at least one output, whereas, for a function, each input corresponds to exactly one output. The blood-type example given is both a relation and a function. Also note that the range (set of values to which the elements of the domain cor-respond) is a subset of the set of all blood types. However, although all functions are relations, not all relations are functions. For example, at a university, four pri-mary sports typ ically overlap in the late fall: football, volleyball, soccer, and basketball. On a given Saturday, the accompanying table indicates the start times for the competitions. WORDS MATH The 1:00 start time corresponds (1:00 p.m., Football) to exactly one event, Football. The 2:00 start time corresponds (2:00 p.m., Volleyball) to exactly one event, Volleyball. The 7:00 start time corresponds (7:00 p.m., Soccer) to two events, Soccer and Basketball. (7:00 p.m., Basketball) Because an element in the domain, 7:00 p.m., corresponds to more than one element in the range, Soccer and Basketball, this is not a function. It is, however, a relation. DEFINITION Function A function is a correspondence between two sets where each element in the first set, called the domain, corresponds to exactly one element in the second set, called the range. Domain PEOPLE Michael Megan Dylan Trevor Tania Range BLOOD TYPE Function A O AB B Domain Range Not a Function Football Volleyball Soccer Basketball START TIME ATHLETIC EVENT 7:00 P.M. 1:00 P.M. 2:00 P.M. TIME OF DAY COMPETITION 1:00 p.m. Football 2:00 p.m. Volleyball 7:00 p.m. Soccer 7:00 p.m. Basketball [CONCEPT CHECK] If the domain consists of all physical (home) addresses in a particular county and the range is the persons living in that county, does this describe a relation? And if so, is that relation a function? ANSWER This is a relation but not a function. ▼ [ [ STUDY TIP All functions are relations but not all relations are functions. EXAMPLE 1 Determining Whether a Relation Is a Function Determine whether the following relations are functions. a. 5 123, 42, 12, 42, 13, 52, 16, 426 b. 5 123, 42, 12, 42, 13, 52, 12, 226 c. Domain 5 Set of all items for sale in a grocery store; Range 5 Price Solution: a. No x-value is repeated. Therefore, each x-value corresponds to exactly one y-value. This relation is a function. b. The value x 5 2 corresponds to both y 5 2 and y 5 4. This relation is not a function. c. Each item in the grocery store corresponds to exactly one price. This relation is a function. YOUR T UR N Determine whether the following relations are functions. a. 5 11, 22, 13, 22, 15, 62, 17, 626 b. 5 11, 22, 11, 32, 15, 62, 17, 826 c. 5 111:00 a.m., 838F2, 12:00 p.m., 898F2, 16:00 p.m., 858F26 ▼ A N S W E R a. function b. not a function c. function ▼ 240 CHAPTER 3 Functions and Their Graphs All of the examples we have discussed thus far are discrete sets in that they represent a countable set of distinct pairs of 1x, y2. A function can also be defined algebraically by an equation. 3.1.2 Functions Defined by Equations Let’s start with the equation y 5 x2 2 3x, where x can be any real number. This equation assigns to each x-value exactly one corresponding y-value. x y 5 x2 2 3x y 1 y 5 1122 2 3 112 22 5 y 5 1522 2 3 152 10 2 2 3 y 5A 22 3B 2 2 3A 22 3B 22 9 1.2 y 5 11.222 2 3 11.22 22.16 Since the variable y depends on what value of x is selected, we denote y as the dependent variable. The variable x can be any number in the domain; therefore, we denote x as the independent variable. Although functions are defined by equations, it is important to recognize that not all equations are functions. The requirement for an equation to define a function is that each element in the domain corresponds to exactly one element in the range. Throughout the ensuing discussion, we assume x to be the independent variable and y to be the dependent variable. Equations that represent functions of x: y 5 x2 y 5 0 x 0 y 5 x3 Equations that do not represent x 5 y2 x2 1 y2 5 1 x 5 0 y 0 functions of x: In the “equations that represent functions of x,” every x-value corresponds to exactly one y-value. Some ordered pairs that correspond to these functions are y 5 x2: 121, 12 10, 02 11, 12 y 5 0 x 0 : 121, 12 10, 02 11, 12 y 5 x3: 121, 212 10, 02 11, 12 The fact that x 5 21 and x 5 1 both correspond to y 5 1 in the first two examples does not violate the definition of a function. In the “equations that do not represent functions of x,” some x-values correspond to more than one y-value. Some ordered pairs that correspond to these equations are 3.1.2 S KILL Determine whether an equation represents a function. 3.1.2 CO NCE PTUAL Understand why the vertical line test determines if a relation is a function. ▼ C A U T I O N Not all equations are functions. RELATION SOLVE RELATION FOR y POINTS THAT LIE ON THE GRAPH x 5 y2 y 5 6!x 11, 212 10, 02 11, 12 x 5 1 maps to both y 5 21 and y 5 1 x2 1 y2 5 1 y 5 6"1 2 x2 10, 212 10, 12 121, 02 11, 02 x 5 0 maps to both y 5 21 and y 5 1 x 5 0 y0 y 5 6x 11, 212 10, 02 11, 12 x 5 1 maps to both y 5 21 and y 5 1 STUDY TIP We say that for x 5 y 2, y is not a function of x. However, if we reverse the independent and dependent variables, then for x 5 y 2, x is a function of y. 3.1 Functions 241 Let’s take any value for x, say x 5 a. The graph of x 5 a corresponds to a vertical line. A function of x maps each x-value to exactly one y-value; therefore, there should be at most one point of intersection with any vertical line. We see in the three graphs of the functions above that if a vertical line is drawn at any value of x on any of the three graphs, the vertical line only intersects the graph in one place. Look at the graphs of the three equations that do not represent functions of x. Let’s look at the graphs of the three functions of x: y = x2 x y y = \|x\| x y x y y = x3 x y x = y2 x y x2 + y2 = 1 x y x = \|y\| A vertical line can be drawn on any of the three graphs such that the vertical line will intersect each of these graphs at two points. Thus, there are two y-values that correspond to some x-value in the domain, which is why these equations do not define y as a function of x. DEFINITION Vertical Line Test Given the graph of an equation, if any vertical line that can be drawn intersects the graph at no more than one point, the equation defines y as a function of x. This test is called the vertical line test. [CONCEPT CHECK] Draw two vertical lines (x 5 0 and x 5 1) on the graph of x 5 y 2. What three points do these lines intersect on the graph of the equation? Explain why the equation x 5 y 2 cannot be a function. ANSWER (0, 0) (1, 21) and (1, 1): the vertical line x 5 1 intersects the graph at two places (fails the vertical line test). ▼ EXAMPLE 2 Using the Vertical Line Test Use the vertical line test to determine whether the graphs of equations define functions of x. a. x y b. x y 242 CHAPTER 3 Functions and Their Graphs EXPRESSING A FUNCTION VERBALLY NUMERICALLY ALGEBRAICALLY GRAPHICALLY Every real number has a corresponding absolute value. 5 123, 32, 121, 12, 10, 02, 11, 12, 15, 526 y 5 0x0 x y 3.1.3 Function Notation We know that the equation y 5 2x 1 5 defines y as a function of x because its graph is a nonvertical line and thus passes the vertical line test. We can select x-values (input) and determine unique corresponding y-values (output). The output is found by taking 2 times the input and then adding 5. If we give the function a name, say, “ƒ,” then we can use function notation: ƒ1x2 5 2x 1 5 The symbol ƒ1x2 is read “ƒ evaluated at x” or “ƒ of x” and represents the y-value that corresponds to a particular x-value. In other words, y 5 ƒ1x2. 3.1.3 S KILL Use function notation to evaluate functions for particular arguments. 3.1.3 CO NCE PTUAL Think of function notation as a placeholder or mapping. INPUT FUNCTION OUTPUT EQUATION x ƒ ƒ1x2 ƒ1x2 5 2x 1 5 Independent variable Mapping Dependent variable Mathematical rule It is important to note that ƒ is the function name, whereas ƒ1x2 is the value of the function. In other words, the function ƒ maps some value x in the domain to some value ƒ1x2 in the range. ▼ A N S W E R The graph of the equation is a circle, which does not pass the vertical line test. Therefore, the equation does not define a function. To recap, a function can be expressed one of four ways: verbally, numerically, algebra-ically, and graphically. This is sometimes called the Rule of 4. [ [ STUDY TIP If any x-value corresponds to more than one y-value, then y is not a function of x. Solution: Apply the vertical line test. a. x y b. x y a. Because the vertical line intersects the graph of the equation at two points, this equation does not represent a function. b. Because any vertical line will intersect the graph of this equation at no more than one point, this equation represents a function. Y OUR TU R N Determine whether the equation 1x 2 322 1 1y 1 222 5 16 is a function of x. ▼ 3.1 Functions 243 The independent variable is also referred to as the argument of a function. To evaluate functions, it is often useful to think of the independent variable, or argument, as a placeholder. For example, ƒ1x2 5 x2 2 3x can be thought of as ƒ1 2 5 1 22 2 31 2 In other words, “ƒ of the argument is equal to the argument squared minus 3 times the argument.” Any expression can be substituted for the argument: ƒ112 5 1122 2 3112 ƒ1x 1 12 5 1x 1 122 2 31x 1 12 ƒ12x2 5 12x22 2 312x2 It is important to note: ■ ■ƒ1x2 does not mean ƒ times x. ■ ■The most common function names are ƒ and F since the word function begins with an “f.” Other common function names are g and G, but any letter can be used. ■ ■The letter most commonly used for the independent variable is x. The letter t is also common because in real-world applications it represents time, but any letter can be used. ■ ■Although we can think of y and ƒ1x2 as interchangeable, the function notation is useful when we want to consider two or more functions of the same independent variable. x f (x) 5 2x 1 5 f (x) 0 ƒ102 5 2 102 1 5 ƒ102 5 5 1 ƒ112 5 2 112 1 5 ƒ112 5 7 2 ƒ122 5 2 122 1 5 ƒ122 5 9 Domain Range Function x 2x 5 f(0) 5 f(1) 7 f(2) 9 x 1 x 0 x 2 f f f [ [ STUDY TIP It is important to note that ƒ(x) does not mean ƒ times x. [CONCEPT CHECK] For the function in Example 3, find f A B. ANSWER f A B 5 2 3 2 3 2 1 6 ▼ EXAMPLE 4 Finding Function Values from the Graph of a Function The graph of ƒ is given on the right. a. Find ƒ102. b. Find ƒ112. c. Find ƒ122. d. Find 4ƒ132. e. Find x such that ƒ1x2 5 10. f. Find x such that ƒ1x2 5 2. x (1, 2) (2, 1) (3, 2) (4, 5) (0, 5) (5, 10) y 5 10 EXAMPLE 3 Evaluating Functions by Substitution Given the function ƒ1x2 5 2x3 2 3x2 1 6, find ƒ1212. Solution: Consider the independent variable x to be a placeholder. ƒ1 2 5 2 1 23 2 3 1 22 1 6 To find ƒ1212, substitute x 5 21 into the function. ƒ1212 5 2 12123 2 3 12122 1 6 Evaluate the right side. ƒ1212 5 22 2 3 1 6 Simplify. ƒ1212 5 1 244 CHAPTER 3 Functions and Their Graphs ▼ A N S W E R a. ƒ1212 5 2 b. ƒ102 5 1 c. 3ƒ122 5 221 d. x 5 1 ▼ C A U T I O N ƒ1x 1 12 2 ƒ1x2 1 ƒ112 ▼ A N S W E R g1x 2 12 5 x2 2 4x 1 6 Solution: Solution (a): The value x 5 0 corresponds to the value y 5 5. ƒ102 5 5 Solution (b): The value x 5 1 corresponds to the value y 5 2. ƒ112 5 2 Solution (c): The value x 5 2 corresponds to the value y 5 1. ƒ122 5 1 Solution (d): The value x 5 3 corresponds to the value y 5 2. 4ƒ132 5 4⋅2 5 8 Solution (e): The value y 5 10 corresponds to the value x 5 5 . Solution (f): The value y 5 2 corresponds to the values x 5 1 and x 5 3 . Y OUR TU R N For the following graph of a function, find: a. ƒ1212 b. ƒ102 c. 3ƒ122 d. the value of x that corresponds to ƒ1x2 5 0 x (0, 1) (–1, 2) (1, 0) (2, –7) (–2, 9) y –5 5 –10 10 ▼ EXAMPLE 5 Evaluating Functions with Variable Arguments (Inputs) For the given function ƒ1x2 5 x2 2 3x, evaluate ƒ1x 1 12 and simplify if possible. common mistake A common misunderstanding is to interpret the notation ƒ1x 1 12 as a sum: ƒ1x 1 12 2 ƒ1x2 1 ƒ112. Y OUR TU R N For the given function g 1x2 5 x2 2 2x 1 3, evaluate g 1x 2 12. ▼ ✖I N C O R R EC T The ERROR is in interpreting the notation as a sum. ƒ1x 1 12 2 ƒ1x2 1 ƒ112 2 x2 2 3x 2 2 ✓C O R R EC T Write the original function. ƒ1x2 5 x2 2 3x Replace the argument x with a placeholder. ƒ1 2 5 1 22 2 31 2 Substitute x 1 1 for the argument. ƒ1x 1 12 5 1x 1 122 2 31x 1 12 Eliminate the parentheses. ƒ1x 1 12 5 x2 1 2x 1 1 2 3x 2 3 Combine like terms. ƒ1x 1 12 5 x2 2 x 2 2 3.1 Functions 245 EXAMPLE 6 Evaluating Functions: Sums For the given function H 1x2 5 x2 1 2x, evaluate: a. H 1x 1 12 b. H 1x2 1 H 112 Solution (a): Write the function H in placeholder notation. H 1 2 5 1 22 1 2 1 2 Substitute x 1 1 for the argument of H. H 1x 1 12 5 1x 1 122 1 2 1x 1 12 Eliminate the parentheses on the right side. H 1x 1 12 5 x2 1 2x 1 1 + 2x 1 2 Combine like terms on the right side. H 1x 1 12 5 x2 1 4x 1 3 Solution (b): Write H 1x2. H 1x2 5 x2 1 2x Evaluate H at x 5 1. H 112 5 1122 1 2 112 5 3 Evaluate the sum H 1x2 1 H 112. H 1x2 1 H 112 5 x2 1 2x 1 3 H 1x2 1 H 112 5 x2 1 2x 1 3 Note: Comparing the results of part (a) and part (b), we see that H 1x 1 12 u H 1x2 1 H 112. EXAMPLE 7 Evaluating Functions: Negatives For the given function G 1t2 5 t2 2 t, evaluate: a. G 12t2 b. 2G 1t2 Solution (a): Write the function G in placeholder notation. G 1 2 5 1 22 2 1 2 Substitute 2t for the argument of G. G 12t2 5 12t22 2 12t2 Eliminate the parentheses on the right side. G 12t2 5 t2 1 t Solution (b): Write G 1t2. G 1t2 5 t2 2 t Multiply by 21. 2G 1t2 5 2 1t2 2 t2 Eliminate the parentheses on the right side. 2G1t2 5 2t2 1 t Note: Comparing the results of part (a) and part (b), we see that G12t2 u 2G1t2. EXAMPLE 8 Evaluating Functions: Quotients For the given function F 1x2 5 3x 1 5, evaluate: a. F a1 2b b. F 112 F 122 Solution (a): Write F in placeholder notation. F 1 2 5 3 1 2 1 5 Replace the argument with 1 2. F a1 2b 5 3a1 2b 1 5 Simplify the right side. F a1 2b 5 13 2 246 CHAPTER 3 Functions and Their Graphs Examples 6, 7, and 8 illustrate the following: ƒ1a 1 b2 2 ƒ1a2 1 ƒ1b2 ƒ12t2 2 2ƒ1t2 ƒaa bb 2 ƒ1a2 ƒ1b2 Now that we have shown that ƒ1x 1 h2 2 ƒ1x2 1 ƒ1h2, we turn our attention to one of the fundamental expressions in calculus: the difference quotient. ƒ1x 1 h2 2 ƒ1x2 h h 2 0 Example 9 illustrates the difference quotient, which will be discussed in detail in Section 3.2. For now, we will concentrate on the algebra involved when finding the difference quotient. In Section 3.2, the application of the difference quotient will be the emphasis. Solution (b): Evaluate F 112. F 112 5 3 112 1 5 5 8 Evaluate F 122. F 122 5 3 122 1 5 5 11 Divide F 112 by F 122. F 112 F 122 5 8 11 Note: Comparing the results of part (a) and part (b), we see that F a1 2b u F112 F122 . Y OUR TU R N Given the function G 1t2 5 3t 2 4, evaluate: a. G 1t 2 22 b. G 1t2 2 G 122 c. G112 G132 d. Ga1 3b ▼ ▼ A N S W E R 2x 1 h ▼ C A U T I O N ƒaa bb 2 ƒ1a2 ƒ1b2 ▼ A N S W E R a. G 1t 2 22 5 3t 210 b. G 1t2 2 G 122 5 3t 2 6 c. G112 G132 5 2 1 5 d. G a1 3b 5 23 EXAMPLE 9 Evaluating the Difference Quotient For the function ƒ1x2 5 x2 2 x, find ƒ1x 1 h2 2 ƒ1x2 h , h 2 0. Solution: Use placeholder notation for the function ƒ1x2 5 x2 2 x. ƒ1 2 5 1 22 2 1 2 Calculate ƒ1x 1 h2. ƒ1x 1 h2 5 1x 1 h22 2 1x 1 h2 Write the difference quotient. ƒ1x 1 h2 2 ƒ1x2 h Let ƒ1x 1 h2 5 1x 1 h22 2 1x 1 h2 and ƒ1x2 5 x2 2 x. ƒ1x 1 h2 ƒ1x2 ƒ1x 1 h2 2 ƒ1x2 h 5 C1x 1 h22 2 1x 1 h2D 2 Cx2 2 xD h h 2 0 Eliminate the parentheses inside the 5 3x2 1 2xh 1 h2 2 x 2 h4 2 3x2 2 x4 h first set of brackets. Eliminate the brackets in the numerator. 5 x2 1 2xh 1 h2 2 x 2 h 2 x2 1 x h Combine like terms. 5 2xh 1 h2 2 h h Factor the numerator. 5 h12x 1 h 2 12 h Divide out the common factor, h. 5 2x 1 h 2 1 h 2 0 Y OUR TU R N Evaluate the difference quotient for ƒ1x2 5 x2 2 1. ▼ e f 3.1 Functions 247 3.1.4 Domain of a Function Sometimes the domain of a function is stated explicitly. For example, ƒ1x2 5 0 x 0 x , 0 Here the explicit domain is the set of all negative real numbers 12q, 02. Every negative real number in the domain is mapped to a positive real number in the range through the absolute value function. If the expression that defines the function is given but the domain is not stated explicitly, then the domain is implied. The implicit domain is the largest set of real numbers for which the function is defined and the output value ƒ1x2 is a real number. For example, ƒ1x2 5 !x does not have the domain explicitly stated. There is, however, an implicit domain. Note that if the argument is negative, that is, if x , 0, then the result is an imaginary number. In order for the output of the function, ƒ1x2, to be a real number, we must restrict the domain to nonnegative numbers, that is, if x $ 0. 3.1.4 S K I L L Determine the domain and range of a function. 3.1.4 C ON C E P T U A L Understand the difference between implicit domain and explicit domain. domain Domain ( , 0) f(x) \|x\| Range (0, ) 1 7 4 1 7 4 FUNCTION IMPLICIT DOMAIN ƒ1x2 5 !x 30, q2 In general, we ask the question, “what can x be?” The implicit domain of a function excludes values that cause a function to be undefined or have outputs that are not real numbers. EXPRESSION THAT DEFINES THE FUNCTION EXCLUDED x-VALUES EXAMPLE IMPLICIT DOMAIN Polynomial None ƒ1x2 5 x3 2 4x2 All real numbers Rational x-values that make the denominator equal to 0 g1x2 5 2 x2 2 9 x 263 or 12q, 232 ∪ 123, 32 ∪ 13, q2 Radical x-values that result in a square (even) root of a negative number h1x2 5 !x 2 5 x $ 5 or 35, q2 EXAMPLE 10 Determining the Domain of a Function State the domain of the given functions. a. F 1x2 5 3 x2 2 25 b. H 1x2 5 ! 4 9 2 2x c. G 1x2 5 ! 3 x 2 1 Solution (a): Write the original equation. F 1x2 5 3 x2 2 25 Determine any restrictions on the values of x. x2 2 25 2 0 Solve the restriction equation. x2 2 25 or x 2 6!25 5 65 State the domain restrictions. x 2 65 Write the domain in interval notation. 12q, 252 ∪ 125, 52 ∪ 15, q2 [CONCEPT CHECK] Find the implicit domain for f 1x2 5 1 !x 2 a ANSWER (a, q ) ▼ e 248 CHAPTER 3 Functions and Their Graphs Applications Functions that are used in applications often have restrictions on the domains due to physical constraints. For example, the volume of a cube is given by the function V1x2 5 x3, where x is the length of a side. The function ƒ1x2 5 x3 has no restrictions on x, and therefore the domain is the set of all real numbers. However, the volume of any cube has the restriction that the length of a side can never be negative or zero. Solution (b): Write the original equation. H 1x2 5 ! 4 9 2 2x Determine any restrictions on the values of x. 9 2 2x $ 0 Solve the restriction inequality. 9 $ 2x State the domain restrictions. x # 9 2 Write the domain in interval notation. a 2q, 9 2 d Solution (c): Write the original equation. G1x2 5 ! 3 x 2 1 Determine any restrictions on the values of x. no restrictions State the domain. ℝ Write the domain in interval notation. 12q, q2 Y OUR TU R N State the domain of the given functions. a. ƒ1x2 5 !x 2 3 b. g 1x2 5 1 x2 2 4 ▼ EXAMPLE 11 Price of Gasoline Following the capture of Saddam Hussein in Iraq in 2003, gas prices in the United States escalated and then finally returned to their precapture prices. Over a 6-month period, the average price of a gallon of 87 octane gasoline was given by the function C 1x2 5 20.05x2 1 0.3x 1 1.7, where C is the cost function and x represents the number of months after the capture. a. Determine the domain of the cost function. b. What was the average price of gas per gallon 3 months after the capture? Solution (a): Since the cost function C 1x2 5 20.05x2 1 0.3x 1 1.7 modeled the price of gas only for 6 months after the capture, the domain is 0 # x # 6 or 30, 64 . Solution (b): Write the cost function. C 1x2 5 20.05x2 1 0.3x 1 1.7 0 # x # 6 Find the value of the function when x 5 3. C 132 5 20.05 1322 1 0.3 132 1 1.7 Simplify. C 132 5 2.15 The average price per gallon 3 months after the capture was $2.15 . ▼ A N S W E R a. x $ 3 or 33, q2 b. x 2 62 or 12q, 222 ∪ 122, 22 ∪ 12, q2 3.1 Functions 249 EXAMPLE 12 The Dimensions of a Pool Express the volume of a 30 ft 3 10 ft rectangular swimming pool as a function of its depth. Solution: The volume of any rectangular box is V 5 lwh, where V is the volume, l is the length, w is the width, and h is the height. In this example, the length is 30 ft, the width is 10 ft, and the height represents the depth d of the pool. Write the volume as a function of depth d. V 1d2 5 13021102d Simplify. V 1d2 5 300d Determine any restrictions on the domain. d . 0 [SEC TION 3.1] S U M MA RY Relations and Functions (Let x represent the independent variable and y the dependent variable.) All functions are relations, but not all relations are functions. Functions can be represented by equations. In the following table, each column illustrates an alternative notation. The domain is the set of all inputs (x-values), and the range is the set of all corresponding outputs (y-values). Placeholder notation is useful when evaluating functions. ƒ1x2 5 3x2 1 2x ƒ1 2 5 3 1 22 1 2 1 2 Explicit domain is stated, whereas implicit domain is found by excluding x-values that • make the function undefined (denominator 5 0). • result in a nonreal output (even roots of negative real numbers). TYPE MAPPING/CORRESPONDENCE EQUATION GRAPH Relation Every x-value in the domain maps to at least one y-value in the range. x 5 y2 x y Function Every x-value in the domain maps to exactly one y-value in the range. y 5 x2 x y Passes vertical line test INPUT CORRESPONDENCE OUTPUT EQUATION x Function y y 5 2x 1 5 Independent Variable Mapping Dependent Variable Mathematical Rule Argument ƒ ƒ1x2 ƒ1x2 5 2x 1 5 250 CHAPTER 3 Functions and Their Graphs In Exercises 1–24, determine whether each relation is a function. Assume that the coordinate pair (x, y) represents the independent variable x and the dependent variable y. 1. Domain Range 78°F 68°F October January April MONTH AVERAGE TEMPERATURE 2. Domain Range Mary Jason Chester PERSON (202) 555–1212 (307) 123–4567 (878) 799–6504 10-DIGIT PHONE # 3. Domain Range START TIME 1:00 P.M. 4:00 P.M. 7:00 P.M. • Bucs/Panthers • Bears/Lions • Falcons/Saints • Rams/Seahawks • Packers/Vikings NFL GAME 4. Domain Range Jordan Pat Chris Alex Morgan DATE THIS WEEKEND PERSON 5. Domain Range COURSE GRADE A B Carrie Michael Jennifer Sean PERSON 6. Domain Range MATH SAT SCORE 500 650 Carrie Michael Jennifer Sean PERSON 7. 5 10, 232, 10, 32, 123, 02, 13, 026 8. 5 12, 222, 12, 22, 15, 252, 15, 526 9. 5 10, 02, 19, 232, 14, 222, 14, 22, 19, 326 10. 5 10, 02, 121, 212, 122, 282, 11, 12, 12, 826 11. 5 10, 12, 11, 02, 12, 12, 122, 12, 15, 42, 123, 426 12. 5 10, 12, 11, 12, 12, 12, 13, 126 13. x2 1 y2 5 9 14. x 5 0y 0 15. x 5 y2 16. y 5 x3 17. y 5 0x 2 1 0 18. y 5 3 19. x y (0, –5) (0, 5) (–5, 0) (5, 0) 20. x y 21. x y 22. x y 23. x y 24. x y [SEC TION 3.1] E X E R C I SES • S K I L L S 3.1 Functions 251 In Exercises 25–32, use the given graphs to evaluate the functions. 25. y 5 ƒ1x2 x (0, 1) (1, 3) (2, 5) (–1, –1) (–2, –3) y –5 5 –5 5 a. ƒ122 b. ƒ102 c. ƒ1222 26. y 5 g1x2 x (0, –5) (3, –2) (5, 0) (–3, 1) y –5 5 –5 5 a. g 1232 b. g 102 c. g 152 27. y 5 p1x2 x (1, 5) (0, 2) (–1, 3) (–3, 5) (–5, 7) y –7 3 –3 7 a. p 1212 b. p 102 c. p 112 28. y 5 r 1x2 x (3, –5) (7, –3) (–4, 0) (–1, 4) (–6, –6) y –10 10 –10 10 a. r 1242 b. r 1212 c. r 132 29. y 5 C1x2 x (4, –5) (–4, –5) y –5 5 –10 a. C122 b. C102 c. C1222 30. y 5 q1x2 x (2, –4) (6, 0) y –4 6 –6 4 a. q1242 b. q102 c. q122 31. y 5 S1x2 x (6, 1) (2, –5) y –4 6 –8 2 a. S1232 b. S102 c. S122 32. y 5 T1x2 x (4, 3) (5, 1) (–3, 4) y –5 5 –5 5 a. T 1252 b. T 1222 c. T 142 33. Find x if ƒ1x2 5 3 in Exercise 25. 34. Find x if g 1x2 5 22 in Exercise 26. 35. Find x if p1x2 5 5 in Exercise 27. 36. Find x if C 1x2 5 27 in Exercise 29. 37. Find x if C1x2 5 25 in Exercise 29. 38. Find x if q 1x2 5 22 in Exercise 30. 39. Find x if S1x2 5 1 in Exercise 31. 40. Find x if T 1x2 5 4 in Exercise 32. In Exercises 41–56, evaluate the given quantities applying the following four functions. ƒ1x2 5 2x 2 3 F 1t2 5 4 2 t2 g1t2 5 5 1 t G1x2 5 x2 1 2x 2 7 41. ƒ1222 42. G 1232 43. g 112 44. F 1212 45. ƒ1222 1 g 112 46. G 1232 2 F 1212 47. 3ƒ1222 2 2g 112 48. 2F 1212 2 2G 1232 49. ƒ1222 g112 50. G1232 F1212 51. ƒ102 2 ƒ1222 g112 52. G102 2 G1232 F1212 53. ƒ1x 1 12 2 ƒ1x 2 12 54. F 1t 1 12 2 F 1t 2 12 55. g 1x 1 a2 2 ƒ1x 1 a2 56. G 1x 1 b2 1 F 1b2 In Exercises 57–64, evaluate the difference quotients using the same ƒ, F, G, and g given for Exercises 41–56. 57. ƒ1x 1 h2 2 ƒ1x2 h 58. F1t 1 h2 2 F1t2 h 59. g1t 1 h2 2 g1t2 h 60. G1x 1 h2 2 G1x2 h 61. ƒ122 1 h2 2 ƒ1222 h 62. F121 1 h2 2 F1212 h 63. g11 1 h2 2 g112 h 64. G123 1 h2 2 G1232 h In Exercises 65–96, find the domain of the given function. Express the domain in interval notation. 65. ƒ1x2 5 2x 2 5 66. ƒ1x2 5 22x 2 5 67. g(t) 5 t2 1 3t 68. h(x) 5 3x4 2 1 69. P1x2 5 x 1 5 x 2 5 70. Q1t2 5 2 2 t2 t 1 3 71. T1x2 5 2 x2 2 4 72. R1x2 5 1 x2 2 1 73. F1x2 5 1 x2 1 1 74. G1t2 5 2 t2 1 4 75. q1x2 5 !7 2 x 76. k1t2 5 !t 2 7 252 CHAPTER 3 Functions and Their Graphs 77. ƒ1x2 5 "2x 1 5 78. g1x2 5 "5 2 2x 79. G1t2 5 "t2 2 4 80. F1x2 5 "x2 2 25 81. F1x2 5 1 !x 2 3 82. G1x2 5 2 !5 2 x 83. ƒ1x2 5 ! 3 1 2 2x 84. g1x2 5 ! 5 7 2 5x 85. P1x2 5 1 " 5 x 1 4 86. Q1x2 5 x " 3 x2 2 9 87. R1x2 5 x 1 1 " 4 3 2 2x 88. p1x2 5 x2 "25 2 x2 89. H1t2 5 t "t2 2 t 2 6 90. ƒ1t2 5 t 2 3 " 4 t2 1 9 91. ƒ1x2 5 1x2 2 1621/2 92. g 1x2 5 12x 2 521/3 93. r 1x2 5 x2 13 2 2x221/2 94. p 1x2 5 1x 2 122 1x2 2 9223/5 95. ƒ1x2 5 2 5x 2 2 4 96. g1x2 5 2 3x2 2 1 6x 2 3 4 97. Let g 1x2 5 x2 2 2x 2 5 and find the values of x that correspond to g 1x2 5 3. 98. Let g1x2 5 5 6 x 2 3 4 and find the value of x that corresponds to g1x2 5 2 3. 99. Let ƒ1x2 5 2x 1x 2 523 2 12 1x 2 522 and find the values of x that correspond to ƒ1x2 5 0. 100. Let ƒ1x2 5 3x 1x 1 322 2 6 1x 1 323 and find the values of x that correspond to ƒ1x2 5 0. 101. Budget: Event Planning. The cost associated with a catered wedding reception is $45 per person for a reception for more than 75 people. Write the cost of the reception in terms of the number of guests and state any domain restrictions. 102. Budget: Long-Distance Calling. The cost of a local home phone plan is $35 for basic service and $0.10 per minute for any domestic long-distance calls. Write the cost of monthly phone service in terms of the number of monthly long-distance minutes and state any domain restrictions. 103. Temperature. The average temperature in Tampa, Florida, in the springtime is given by the function T 1x2 5 20.7x2 1 16.8x 2 10.8, where T is the temperature in degrees Fahrenheit and x is the time of day in military time and is restricted to 6 # x # 18 (sunrise to sunset). What is the temperature at 6 a.m.? What is the temperature at noon? 104. Falling Objects: Firecrackers. A firecracker is launched straight up, and its height is a function of time, h 1t2 5 216t2 1 128t, where h is the height in feet and t is the time in seconds with t 5 0 corresponding to the instant it launches. What is the height 4 seconds after launch? What is the domain of this function? 105. Collectibles. The price of a signed Alex Rodriguez base-ball card is a function of how many are for sale. When Rodriguez was traded from the Texas Rangers to the New York Yankees in 2004, the going rate for a signed baseball card on eBay was P1x2 5 10 1 !400,000 2 100x, where x represents the number of signed cards for sale. What was the value of the card when there were 10 signed cards for sale? What was the value of the card when there were 100 signed cards for sale? 106. Collectibles. In Exercise 105, what was the lowest price on eBay, and how many cards were available then? What was the highest price on eBay, and how many cards were available then? 107. Volume. An open box is constructed from a square 10-inch piece of cardboard by cutting squares of length x inches out of each corner and folding the sides up. Express the volume of the box as a function of x, and state the domain. 108. Volume. A cylindrical water basin will be built to harvest rainwater. The basin is limited in that the largest radius it can have is 10 feet. Write a function representing the volume of water V as a function of height h. How many additional gallons of water will be collected if you increase the height by 2 feet? Hint: 1 cubic foot 5 7.48 gallons. For Exercises 109–110, refer to the following: The weekly exchange rate of the U.S. dollar to the Japanese yen is shown in the graph as varying over an 8-week period. Assume the exchange rate E 1t2 is a function of time (week); let E 112 be the exchange rate during Week 1. Week 2 4 6 8 10 1 3 5 7 9 Japanese Yen to One U.S. Dollar 90 89 88 87 86 85 84 83 82 t E 109. Economics. Approximate the exchange rates of the U.S. dollar to the nearest yen during Weeks 4, 7, and 8. 110. Economics. Find the increase or decrease in the number of Japanese yen to the U.S. dollar exchange rate, to the nearest yen, from (a) Week 2 to Week 3 and (b) Week 6 to Week 7. For Exercises 111–112, refer to the following: An epidemiological study of the spread of malaria in a rural area finds that the total number P of people who contracted malaria t days into an outbreak is modeled by the function P 1t2 5 21 4 t 2 1 7t 1 180 1 # t # 14 • A P P L I C A T I O N S 3.1 Functions 253 111. Medicine/Health. How many people have contracted malaria 14 days into the outbreak? 112. Medicine/Health. How many people have contracted malaria 6 days into the outbreak? 113. Environment: Tossing the Envelopes. The average American adult receives 24 pieces of mail per week, usually of some combination of ads and envelopes with windows. Suppose each of these adults throws away a dozen envelopes per week. a. The width of the window of an envelope is 3.375 inches less than its length x. Create the function A(x) that represents the area of the window in square inches. Simplify, if possible. b. Evaluate A(4.5) and explain what this value represents. c. Assume the dimensions of the envelope are 8 inches by 4 inches. Evaluate A(8.5). Is this possible for this particular envelope? Explain. 114. Environment: Tossing the Envelopes. Each month, Jack receives his bank statement in a 9.5 inch by 6 inch envelope. Each month, he throws away the envelope after removing the statement. a. The width of the window of the envelope is 2.875 inches less than its length x. Create the function A(x) that represents the area of the window in square inches. Simplify, if possible. b. Evaluate A(5.25) and explain what this value represents. c. Evaluate A(10). Is this possible for this particular envelope? Explain. Refer to the table below for Exercises 115 and 116. It illus-trates the average federal funds rate for the month of January (2009 to 2017). YEAR FED. RATE 2009 5.45 2010 5.98 2011 1.73 2012 1.24 2013 1.00 2014 2.25 2015 4.50 2016 5.25 2017 3.50 115. Finance. Is the relation whose domain is the year and whose range is the average federal funds rate for the month of January a function? Explain. 116. Finance. Write five ordered pairs whose domain is the set of even years from 2009–2017 and whose range is the set of corresponding average federal funds rate for the month of January. For Exercises 117 and 118, use the following figure: Year Source: Kaiser Family Foundation Health Research and Education Trust. 2005 2000 2010 2015 Employer-Provided Health Insurance Premiums for Single Plans 1999-2015 4,000 2,000 6,000 8,000 10,000 $12,000 Employee contribution Employer contribution 117. Health Care Costs. Fill in the following table. Round dollars to the nearest $1000. YEAR TOTAL HEALTH CARE COST FOR FAMILY PLANS 1999 2003 2007 2011 2015 Write the five ordered pairs resulting from the table. 118. Health Care Costs. Using the table found in Exercise 115, let the years correspond to the domain and the total costs correspond to the range. Is this relation a function? Explain. For Exercises 119 and 120, use the following information: Year 1800 1850 1900 1950 2000 Metric Tons of Carbon/Year (in millions) 2,000 1,000 3,000 4,000 7,000 6,000 5,000 Total Global Fossil Carbon Emissions Petroleum Coal Natural Gas Cement Production Source: http:/www.naftc.wvu.edu Let the functions ƒ, F, g, G, and H represent the number of tons of carbon emitted per year as a function of year corresponding to cement production, natural gas, coal, petroleum, and the total amount, respectively. Let t represent the year, with t 5 0 corresponding to 1900. 119. Environment: Global Climate Change. Estimate (to the nearest thousand) the value of a. F 1502 b. g 1502 c. H 1502 120. Environment: Global Climate Change. Explain what the sum F 11002 1 g11002 1 G 11002 represents. 254 CHAPTER 3 Functions and Their Graphs • C A T C H T H E M I S T A K E In Exercises 121–126, explain the mistake that is made. 121. Determine whether the relationship is a function. Solution: Apply the horizontal line test. Because the horizontal line intersects the graph in two places, this is not a function. This is incorrect. What mistake was made? 122. Given the function H 1x2 5 3x 2 2, evaluate the quantity H 132 2 H 1212. Solution: H 132 2 H 1212 5 H 132 1 H 112 5 7 1 1 5 8 This is incorrect. What mistake was made? 123. Given the function ƒ1x2 5 x2 2 x, evaluate the quantity ƒ1x 1 12. This is incorrect. What mistake was made? Solution: ƒ1x 1 12 5 ƒ1x2 1 ƒ112 5 x2 2 x 1 0 ƒ1x 1 12 5 x2 2 x This is incorrect. What mistake was made? 124. Determine the domain of the function g1t2 5 !3 2 t and express it in interval notation. Solution: What can t be? Any nonnegative real number. 3 2 t . 0 3 . t or t , 3 Domain: 12q, 32 This is incorrect. What mistake was made? 125. Given the function G 1x2 5 x2, evaluate G121 1 h2 2 G1212 h . Solution: G121 1 h2 2G1212 h 5 G1212 1 G1h2 2 G1212 h 5 G1h2 h 5 h2 h 5 h This is incorrect. What mistake was made? 126. Given the functions ƒ1x2 5 0 x 2 A0 2 1 and ƒ112 5 21, find A. Solution: Since ƒ112 5 21, the point 121, 12 must satisfy the function. 21 5 0 21 2 A0 21 Add 1 to both sides of the equation. 0 21 2 A0 5 0 The absolute value of zero is zero, so there is no need for the absolute value signs: 21 2 A 5 0 1 A 5 21. This is incorrect. What mistake was made? In Exercises 127–130, determine whether each statement is true or false. 127. If a vertical line does not intersect the graph of an equation, then that equation does not represent a function. 128. If a horizontal line intersects a graph of an equation more than once, the equation does not represent a function. 129. If ƒ12a2 5 ƒ1a2, then ƒ does not represent a function. 130. If ƒ12a2 5 ƒ1a2, then ƒ may or may not represent a function. 131. If ƒ1x2 5 Ax2 2 3x and ƒ112 5 21, find A. 132. If g1x2 5 1 b 2 x and g 132 is undefined, find b. • C O N C E P T U A L 133. If F1x2 5 C 2 x D 2 x, F1222 is undefined, and F 1212 5 4, find C and D. 134. Construct a function that is undefined at x 5 5 and whose graph passes through the point 11, 212. In Exercises 135 and 136, find the domain of each function, where a is any positive real number. 135. ƒ1x2 5 2100 x2 2 a2 136. ƒ1x2 5 25"x2 2 a2 • C H A L L E N G E 137. Using a graphing utility, graph the temperature function in Exercise 103. What time of day is it the warmest? What is the temperature? Looking at this function, explain why this model for Tampa, Florida, is valid only from sunrise to sunset (6 to 18). 138. Using a graphing utility, graph the height of the firecracker in Exercise 104. How long after liftoff is the firecracker airborne? What is the maximum height that the firecracker attains? Explain why this height model is valid only for the first 8 seconds. 139. Using a graphing utility, graph the price function in Exer-cise 105. What are the lowest and highest prices of the cards? Does this agree with what you found in Exercise 106? 140. The makers of malted milk balls are considering increasing the size of the spherical treats. The thin chocolate coating on a malted milk ball can be approximated by the surface area, S 1r2 5 4pr2. If the radius is increased 3 mm, what is the resulting increase in required chocolate for the thin outer coating? 141. Let ƒ1x2 5 x2 1 1. Graph y1 5 ƒ1x2 and y2 5 ƒ1x 2 22 in the same viewing window. Describe how the graph of y2 can be obtained from the graph of y1. 142. Let ƒ1x2 5 4 2 x2. Graph y1 5 ƒ1x2 and y2 5 ƒ1x 1 22 in the same viewing window. Describe how the graph of y2 can be obtained from the graph of y1. • T E C H N O L O G Y x y x y 3.2 Graphs of Functions 255 S K I L L S O B J E C T I V E S ■ ■Classify functions as even, odd, or neither. ■ ■Determine whether functions are increasing, decreasing, or constant. ■ ■Calculate the average rate of change of a function. ■ ■Graph piecewise-defined functions. C O N C E P T U A L O B J E C T I V ES ■ ■Identify common functions and understand that even functions have graphs that are symmetric about the y-axis and odd functions have graphs that are symmetric about the origin. ■ ■Understand that intervals of increasing, decreasing, and constant correspond to the x-coordinates. ■ ■Understand that the difference quotient is just another form of the average rate of change. ■ ■Understand points of discontinutity and domain and range of piecewise-defined functions. 3.2 GRAPHS OF FUNCTIONS; PIECEWISE-DEFINED FUNCTIONS; INCREASING AND DECREASING FUNCTIONS; AVERAGE RATE OF CHANGE 3.2.1 Recognizing and Classifying Functions Common Functions Point-plotting techniques were introduced in Section 2.2, and we noted there that we would explore some more efficient ways of graphing functions in Chapter 3. The nine main functions you will read about in this section will constitute a “library” of functions that you should commit to memory. We will draw on this library of functions in the next section when graphing transformations are discussed. Several of these functions have been shown previously in this chapter, but now we will classify them specifically by name and identify properties that each function exhibits. In Section 2.3, we discussed equations and graphs of lines. All lines (with the exception of vertical lines) pass the vertical line test and hence are classified as functions. Instead of the traditional notation of a line, y 5 mx 1 b, we use function notation and classify a function whose graph is a line as a linear function. The domain of a linear function ƒ1x2 5 mx 1 b is the set of all real numbers R. The graph of this function has slope m and y-intercept b. LINEAR FUNCTION: f (x ) 5 mx 1 b SLOPE: m y-INTERCEPT: b ƒ1x2 5 2x 2 7 m 5 2 b 5 27 ƒ1x2 5 2x 1 3 m 5 21 b 5 3 ƒ1x2 5 x m 5 1 b 5 0 ƒ1x2 5 5 m 5 0 b 5 5 One special case of the linear function is the constant function 1m 5 02. The graph of a constant function ƒ1x2 5 b is a horizontal line. The y-intercept corresponds to the point 10, b2. The domain of a constant function is the set of all 3.2.1 S K I L L Classify functions as even, odd, or neither. 3.2.1 C ON C E P T U A L Identify common functions and understand that even functions have graphs that are symmetric about the y-axis and odd functions have graphs that are symmetric about the origin. LINEAR FUNCTION ƒ1x2 5 mx 1 b m and b are real numbers. CONSTANT FUNCTION ƒ1x2 5 b b is any real number. 256 CHAPTER 3 Functions and Their Graphs Another specific example of a linear function is the function having a slope of one 1m 5 12 and a y-intercept of zero 1b 5 02. This special case is called the identity function. real numbers R. The range, however, is a single value b. In other words, all x-values correspond to a single y-value. Points that lie on the graph of a constant function ƒ1x2 5 b are 125, b2 121, b2 10, b2 12, b2 14, b2 . . . 1x, b2 x y (0, b) (4, b) (–5, b) Domain: 12q, q2 Range: 3b, b4 or 5b6 x y (0, 0) (2, 2) (3, 3) (–2, –2) (–3, –3) Identity Function Domain: (–∞, ∞) Range: (–∞, ∞) x y (1, 1) (–1, 1) (2, 4) (–2, 4) Square Function Domain: (–∞, ∞) Range: [0, ∞) The graph of the identity function has the following properties: It passes through the origin, and every point that lies on the line has equal x- and y-coordinates. Both the domain and the range of the identity function are the set of all real numbers R. A function that squares the input is called the square function. The graph of the square function is called a parabola and will be discussed in further detail in Chapters 4 and 8. The domain of the square function is the set of all real numbers R. Because squaring a real number always yields a positive number or zero, the range of the square function is the set of all nonnegative numbers. Note that the intercept is the origin and the square function is symmetric about the y-axis. This graph is contained in quadrants I and II. A function that cubes the input is called the cube function. The domain of the cube function is the set of all real numbers R. Because cubing a negative number yields a negative number, cubing a positive number yields a positive number, and cubing 0 yields 0, the range of the cube function is also the set of all real numbers R. Note that the only intercept is the origin and the cube function is symmetric about the origin. This graph extends only into quadrants I and III. The next two functions are counterparts of the previous two functions: square root and cube root. When a function takes the square root of the input or the cube root of the input, the function is called the square root function or the cube root function, respectively. x y (2, 8) (–2, –8) Cube Function Domain: (–∞, ∞) Range: (–∞, ∞) 10 –10 –5 5 IDENTITY FUNCTION ƒ1x2 5 x SQUARE FUNCTION ƒ1x2 5 x2 CUBE FUNCTION ƒ1x2 5 x3 3.2 Graphs of Functions 257 In Section 3.1, we found the domain to be 30, q2. The output of the function will be all real numbers greater than or equal to zero. Therefore, the range of the square root function is 30, q2. The graph of this function will be contained in quadrant I. Some points that are on the graph of the absolute value function are 121, 12, 10, 02, and 11, 12. The domain of the absolute value function is the set of all real numbers R, yet the range is the set of nonnegative real numbers. The graph of this function is symmetric with respect to the y-axis and is contained in quadrants I and II. A function whose output is the reciprocal of its input is called the reciprocal function. In Section 3.1, we stated the domain of the cube root function to be 12q, q2. We see by the graph that the range is also 12q, q2. This graph is contained in quadrants I and III and passes through the origin. This function is symmetric about the origin. In Section 1.7, you read about absolute value equations and inequalities. Now we shift our focus to the graph of the absolute value function. The only restriction on the domain of the reciprocal function is that x 2 0. Therefore, we say the domain is the set of all real numbers excluding zero. The graph of the reciprocal function illustrates that its range is also the set of all real numbers except zero. Note that the reciprocal function is symmetric with respect to the origin and is contained in quadrants I and III. Even and Odd Functions Of the nine functions discussed above, several have similar properties of symmetry. The constant function, square function, and absolute value function are all symmetric with respect to the y-axis. The identity function, cube function, cube root function, and reciprocal function are all symmetric with respect to the origin. The term even is used to describe functions that are symmetric with respect to the y-axis, or vertical axis, and the term odd is used to describe functions that are symmetric with respect to the origin. Recall from Section 2.2 that symmetry can be determined both graphically and algebraically. The following box summarizes the graphic and algebraic characteristics of even and odd functions. x y 10 5 (4, 2) (9, 3) Square Root Function Domain: [0, ∞) Range: [0, ∞) x y (–8, –2) (8, 2) 10 5 Cube Root Function Domain: (–∞, ∞) Range: (–∞, ∞) x y (2, 2) (–2, 2) Absolute Value Function Domain: (–∞, ∞) Range: [0, ∞) x (1, 1) (–1, –1) y f(x) = 1 x Reciprocal Function Domain: (–∞, 0) (0, ∞) Range: (–∞, 0) (0, ∞) SQUARE ROOT FUNCTION ƒ1x2 5 !x or ƒ1x2 5 x1/2 CUBE ROOT FUNCTION ƒ1x2 5 ! 3 x or ƒ1x2 5 x1/3 ABSOLUTE VALUE FUNCTION ƒ1x2 5 0 x 0 RECIPROCAL FUNCTION ƒ1x2 5 1 x x 2 0 EVEN AND ODD FUNCTIONS Function Symmetric with Respect to On Replacing x with 2x Even y-axis, or vertical axis ƒ12x2 5 ƒ1x2 Odd origin ƒ12x2 5 2ƒ1x2 258 CHAPTER 3 Functions and Their Graphs The algebraic method for determining symmetry with respect to the y-axis, or vertical axis, is to substitute 2x for x. If the result is an equivalent equation, the function is symmetric with respect to the y-axis. Some examples of even functions are ƒ1x2 5 b, ƒ1x2 5 x2, ƒ1x2 5 x4, and ƒ1x2 5 ) x ). In any of these equations, if 2x is substituted for x, the result is the same; that is, ƒ12x2 5 ƒ1x2. Also note that, with the exception of the absolute value function, these examples are all even-degree polynomial equations. All constant functions are degree zero and are even functions. The algebraic method for determining symmetry with respect to the origin is to substitute 2x for x. If the result is the negative of the original function, that is, if ƒ12x2 5 2ƒ1x2, then the function is symmetric with respect to the origin and, hence, classified as an odd function. Examples of odd functions are ƒ1x2 5 x, ƒ1x2 5 x3, ƒ1x2 5 x5, and ƒ1x2 5 x1/3. In any of these functions, if 2x is substituted for x, the result is the negative of the original function. Note that with the exception of the cube root function, these equations are odd-degree polynomials. Be careful, though, because functions that are combinations of even- and odd-degree polynomials can turn out to be neither even nor odd, as we will see in Example 1. EXAMPLE 1 Determining Whether a Function Is Even, Odd, or Neither Determine whether the functions are even, odd, or neither. a. ƒ1x2 5 x2 2 3 b. g1x2 5 x5 1 x3 c. h1x2 5 x2 2 x Solution (a): Original function. ƒ1x2 5 x2 2 3 Replace x with 2x. ƒ12x2 5 12x22 2 3 Simplify. ƒ12x2 5 x2 2 3 5 ƒ1x2 Because ƒ12x2 5 ƒ1x2, we say that ƒ1x2 is an even function . Solution (b): Original function. g1x2 5 x5 1 x3 Replace x with 2x. g 12x2 5 12x25 1 12x23 Simplify. g12x2 5 2x5 2 x3 5 21x5 1 x32 5 2g1x2 Because g12x2 5 2g1x2, we say that g 1x2 is an odd function . Solution (c): Original function. h1x2 5 x2 2 x Replace x with 2x. h 12x2 5 12x22 2 12x2 Simplify. h12x2 5 x2 1 x h12x2 is neither 2h1x2 nor h1x2; therefore the function h 1x2 is neither even nor odd . In parts (a), (b), and (c), we classified these functions as even, odd, or neither, using the algebraic test. Look back at them now and reflect on whether these classifications agree with your intuition. In part (a), we combined two functions: the square function and the constant function. Both of these functions are even, and adding even functions yields another even function. In part (b), we combined two odd functions: the fifth-power function and the cube function. Both of these functions are odd, and adding two odd functions yields another odd function. In part (c), we combined two functions: the square function and the identity function. The square function is even, and the identity function is [CONCEPT CHECK] Classify the functions f (x) 5 x2n and g (x) 5 x (2n 1 1) where n is a positive integer (1, 2, 3, …) as Even, Odd, or Neither. ANSWER f (x) is even; g (x) is odd. ▼ ▼ A N S W E R a. even b. neither 3.2.2 S K IL L Determine whether functions are increasing, decreasing, or constant. 3.2.2 C O N C E P T U A L Understand that intervals of increasing, decreasing, and constant correspond to the x-coordinates. x y (6, 4) (0, 1) (–1, 1) (–2, –2) (2, –2) 3.2.2 Increasing and Decreasing Functions Look at the figure in the margin. Graphs are read from left to right. If we start at the left side of the graph and trace the red curve with our pen, we see that the function values (values in the vertical direction) are decreasing until arriving at the point 122, 222. Then, the function values increase until arriving at the point 121, 12. The values then remain constant 1y 5 12 between the points 121, 12 and 10, 12. Proceeding beyond the point 10, 12, the function values decrease again until the point 12, 222. Beyond the point 12, 222, the function values increase again until the point 16, 42. Finally, the function values decrease and continue to do so. When specifying a function as increasing, decreasing, or constant, the intervals are classified according to the x-coordinate. For instance, in this graph, we say the function is increasing when x is between x 5 22 and x 5 21 and again when x is between x 5 2 and x 5 6. The graph is classified as decreasing when x is less than 22 and again when x is between 0 and 2 and again when x is greater than 6. The graph is classified as constant when x is between 21 and 0. In interval notation, this is summarized as Decreasing Increasing Constant 12q, 222 ∪ 10, 22 ∪ 16, q2 122, 212 ∪ 12, 62 121, 02 An algebraic test for determining whether a function is increasing, decreasing, or constant is to compare the value ƒ1x2 of the function for particular points in the intervals. In addition to classifying a function as increasing, decreasing, or constant, we can determine the domain and range of a function by inspecting its graph from left to right: ■ ■The domain is the set of all x-values (from left to right) where the function is defined. ■ ■The range is the set of all y-values (from bottom to top) that the graph of the function corresponds to. ■ ■A solid dot on the left or right end of a graph indicates that the graph terminates there and the point is included in the graph. ■ ■An open dot indicates that the graph terminates there and the point is not included in the graph. ■ ■Unless a dot is present, it is assumed that a graph continues indefinitely in the same direction. (An arrow is used in some books to indicate direction.) odd. In this part, combining an even function with an odd function yields a function that is neither even nor odd and, hence, has no symmetry with respect to the vertical axis or the origin. Y OUR T UR N Classify the functions as even, odd, or neither. a. ƒ1x2 5 0 x 0 1 4 b. ƒ1x2 5 x3 2 1 ▼ 3.2 Graphs of Functions 259 INCREASING, DECREASING, AND CONSTANT FUNCTIONS 1. A function ƒ is increasing on an open interval I if for any x1 and x2 in I, where x1 , x2, then ƒ1x12 , ƒ1x22. 2. A function ƒ is decreasing on an open interval I if for any x1 and x2 in I, where x1 , x2, then ƒ1x12 . ƒ1x22. 3. A function ƒ is constant on an open interval I if for any x1 and x2 in I, then ƒ1x12 5 ƒ1x22. STUDY TIP • Graphs are read from left to right. • Intervals correspond to the x-coordinates. 260 CHAPTER 3 Functions and Their Graphs 3.2.3 Average Rate of Change How do we know how much a function is increasing or decreasing? For example, is the price of a stock slightly increasing or is it doubling every week? One way we determine how much a function is increasing or decreasing is by calculating its average rate of change. Let 1x1, y12 and 1x2, y22 be two points that lie on the graph of a function ƒ. Draw the line that passes through these two points 1x1, y12 and 1x2, y22. This line is called a secant line. Note that the slope of the secant line is given by m 5 y2 2 y1 x2 2 x1 , and recall that the slope of a line is the rate of change of that line. The slope of the secant line is used to represent the average rate of change of the function. x x1 x2 y secant (x1, y1) (x2, y2) f 3.2.3 S KILL Calculate the average rate of change of a function. 3.2.3 CO NCE PTUAL Understand that the difference quotient is just another form of the average rate of change. [CONCEPT CHECK] TRUE OR FALSE An even function has both increasing and decreasing intervals, but an odd function only has one or the other. ANSWER True ▼ EXAMPLE 2 Finding Intervals When a Function Is Increasing or Decreasing Given the graph of a function: a. State the domain and range of the function. b. Find the intervals when the function is increasing, decreasing, or constant. Solution (a): Domain: 325, q2 Range: 30, q2 Solution (b): Reading the graph from left to right, we see that the graph ■ ■decreases from the point 125, 72 to the point 122, 42. ■ ■is constant from the point 122, 42 to the point 10, 42. ■ ■decreases from the point 10, 42 to the point 12, 02. ■ ■increases from the point 12, 02 on. The intervals of increasing and decreasing correspond to the x-coordinates. We say that this function is ■ ■increasing on the interval A2, qB. ■ ■decreasing on the interval 125, 222 ∪ 10, 22. ■ ■constant on the interval 122, 02. Note: The intervals of increasing or decreasing are defined on open intervals. This should not be confused with the domain. For example, the point x 5 25 is included in the domain of the function but not in the interval where the function is classified as decreasing. x (2, 0) (0, 4) (–2, 4) (–5, 7) y –5 5 10 x (2, 0) (0, 4) (–2, 4) (–5, 7) y –5 5 10 Decreasing Decreasing Constant Increasing x (2, 0) (0, 4) (–2, 4) (–5, 7) y –5 5 10 Decreasing Decreasing Constant Increasing 3.2 Graphs of Functions 261 EXAMPLE 3 Average Rate of Change Find the average rate of change of ƒ1x2 5 x4 from: a. x 5 21 to x 5 0 b. x 5 0 to x 5 1 c. x 5 1 to x 5 2 Solution (a): Write the average rate of change formula. ƒ1x22 2 ƒ1x12 x2 2 x1 Let x1 5 21 and x2 5 0. 5 ƒ102 2 ƒ1212 0 2 1212 Substitute ƒ1212 5 12124 5 1 and ƒ102 5 04 5 0. 5 0 2 1 0 2 1212 Simplify. 5 21 Solution (b): Write the average rate of change formula. ƒ1x22 2 ƒ1x12 x2 2 x1 Let x1 5 0 and x2 5 1. 5 ƒ112 2 ƒ102 1 2 0 Substitute ƒ102 5 04 5 0 and ƒ112 5 1124 5 1. 5 1 2 0 1 2 0 Simplify. 5 1 Solution (c): Write the average rate of change formula. ƒ1x22 2 ƒ1x12 x2 2 x1 Let x1 5 1 and x2 5 2. 5 ƒ122 2 ƒ112 2 2 1 Substitute ƒ112 5 14 5 1 and ƒ122 5 1224 5 16. 5 16 2 1 2 2 1 Simplify. 5 15 AVERAGE RATE OF CHANGE Let 1x1, ƒ1x122 and 1x2, ƒ1x222 be two distinct points, 1x1 2 x22, on the graph of the function ƒ. The average rate of change of ƒ between x1 and x2 is given by Average rate of change 5 ƒ1x22 2 ƒ1x12 x2 2 x1 x x1 x2 x2 – x1 f (x2) – f (x1) y secant (x1, f (x1)) (x2, f (x2)) 262 CHAPTER 3 Functions and Their Graphs The average rate of change can also be written in terms of the difference quotient. WORDS MATH Let the difference between x1 and x2 be h. x2 2 x1 5 h Solve for x2. x2 5 x1 1 h Substitute x2 2 x1 5 h into the denominator and x2 5 x1 1 h into the numerator of the average rate of change. Let x1 5 x. 5 ƒ1x 1 h2 2 ƒ1x2 h Average rate of change 5 ƒ1x22 2 ƒ1x12 x2 2 x1 5 ƒ1x1 1 h2 2 ƒ1x12 h Graphical Interpretation: Slope of the Secant Line a. Average rate of change of ƒ from x 5 21 to x 5 0: Decreasing at a rate of 1 b. Average rate of change of ƒ from x 5 0 to x 5 1: Increasing at a rate of 1 c. Average rate of change of ƒ from x 5 1 to x 5 2: Increasing at a rate of 15 Y OUR TU R N Find the average rate of change of ƒ1x2 5 x2 from: a. x 5 22 to x 5 0 b. x 5 0 to x 5 2 ▼ A N S W E R a. 22 b. 2 x y 2 –2 3 –2 (0, 0) (–1, 1) x y 2 –2 3 –2 (0, 0) (1, 1) ▼ x y 2 –2 20 (0, 0) (1, 1) (2, 16) 3.2.4 Piecewise-Defined Functions Most of the functions that we have seen in this text are functions defined by polynomials. Sometimes the need arises to define functions in terms of pieces. For example, most plumbers charge a flat fee for a house call and then an additional hourly rate for the job. For instance, if a particular plumber charges $100 to drive out to your house and work for 1 hour and then an additional $25 an hour for every additional hour he or she works on your job, we would define this function in pieces. If we let h be the number of hours worked, then the charge is defined as Plumbing charge 5 b100 h # 1 100 1 25 1h 2 12 h . 1 If we were to graph this function, we would see that there is 1 hour that is constant and after that the function continually increases. Another piecewise-defined function is the absolute value function. The absolute value function can be thought of as two pieces: the line y 5 2x (when x is negative) and the line y 5 x (when x is nonnegative). We start by graphing these two lines on the same graph. When written in this form, the average rate of change is called the difference quotient. DEFINITION Difference Quotient The expression ƒ1x 1 h2 2 ƒ1x2 h , where h 2 0, is called the difference quotient. The difference quotient is more meaningful when h is small. In calculus the difference quotient is used to define a derivative. x x x + h y f (x) f (x + h) h 3.2.4 S K I L L Graph piecewise-defined functions. 3.2.4 C ON C E P T U A L Understand points of discontinutity and domain and range of piecewise-defined functions. [CONCEPT CHECK] Find the difference quotient for the line f (x) 5 mx 1 b. ANSWER m ▼ ▼ A N S W E R ƒ1x 1 h2 2 ƒ1x2 h 5 22x 2 h STUDY TIP Use brackets or parentheses around ƒ(x) to avoid forgetting to distribute the negative sign: f 1x 1 h2 2 3 f 1x24 h 3.2 Graphs of Functions 263 EXAMPLE 4 Calculating the Difference Quotient Calculate the difference quotient for the function ƒ1x2 5 2x2 1 1. Solution: Find ƒ1x 1 h2. ƒ1x 1 h2 5 21x 1 h22 1 1 5 21x2 1 2xh 1 h22 1 1 5 2x2 1 4xh 1 2h2 1 1 ƒ1x 1 h2 ƒ1x2 Find the difference ƒ1x 1 h2 2 ƒ1x2 h 5 2x2 1 4xh 1 2h2 1 1 2 12x2 1 12 h quotient. Simplify. ƒ1x 1 h2 2 ƒ1x2 h 5 2x2 1 4xh 1 2h2 1 1 2 2x2 2 1 h ƒ1x 1 h2 2 ƒ1x2 h 5 4xh 1 2h2 h Factor the numerator. ƒ1x 1 h2 2 ƒ1x2 h 5 h 14x 1 2h2 h Cancel (divide out) ƒ1x 1 h2 2 ƒ1x2 h 5 4x 1 2h h 2 0 the common h. YOUR T UR N Calculate the difference quotient for the function ƒ1x2 5 2x2 1 2. d g ▼ 264 CHAPTER 3 Functions and Their Graphs The absolute value function behaves like the line y 5 2x when x is negative (erase the blue graph in quadrant IV) and like the line y 5 x when x is positive (erase the red graph in quadrant III). Absolute value function ƒ1x2 5 0 x 0 5 b2x x 0 x x # 0 The next example is a piecewise-defined function given in terms of functions in our “library of functions.” Because the function is defined in terms of pieces of other functions, we draw the graph of each individual function, and then for each function, we darken the piece corresponding to its part of the domain. This is like the procedure above for the absolute value function. x f (x) y = x y = –x x f (x) f (x) = \|x\| EXAMPLE 5 Graphing Piecewise-Defined Functions Graph the piecewise-defined function, and state the domain, range, and intervals when the function is increasing, decreasing, or constant. G1x2 5 c x2 x , 21 1 21 # x # 1 x x . 1 Solution: Graph each of the functions on the same plane. Square function: ƒ1x2 5 x2 Constant function: ƒ1x2 5 1 Identity function: ƒ1x2 5 x The points to focus on in particular are the x-values where the pieces change over, that is, x 5 21 and x 5 1. Let’s now investigate each piece. When x , 21, this function is defined by the square function, ƒ 1x2 5 x2, so darken that particular function to the left of x 5 21. When 21 # x # 1, the function is defined by the constant function, ƒ 1x2 5 1, so darken that particular function between the x values of 21 and 1. When x . 1, the function is defined by the identity function, ƒ 1x2 5 x, so darken that function to the right of x 5 1. Erase everything that is not darkened, and the resulting graph of the piecewise-defined function is given on the right. This function is defined for all real values of x, so the domain of this function is the set of all real numbers. The values that this function yields in the vertical direction are all real numbers greater than or equal to 1. Hence, the range of this function is 31, q2. The intervals of increasing, decreasing, and constant are as follows: Decreasing: 12q, 212 Constant: 121, 12 Increasing: 11, q2 x y (–1, 1) (1, 1) f (x) = x2 f (x) = 1 f (x) = x –2 –1 1 2 1 2 x y (–1, 1) (1, 1) –5 5 –5 5 Plumber Charge Time (hours) 1 3 2 4 5 7 6 9 10 8 100 50 150 200 $250 The term continuous implies that there are no holes or jumps and that the graph can be drawn without picking up your pencil. A function that does have holes or jumps and cannot be drawn in one motion without picking up your pencil is classified as discontinuous, and the points where the holes or jumps occur are called points of discontinuity. The previous example illustrates a continuous piecewise-defined function. At the x 5 21 junction, the square function and constant function both pass through the point 121, 12. At the x 5 1 junction, the constant function and the identity function both pass through the point 11, 12. Since the graph of this piecewise-defined function has no holes or jumps, we classify it as a continuous function. The next example illustrates a discontinuous piecewise-defined function. 3.2 Graphs of Functions 265 EXAMPLE 6 Graphing a Discontinuous Piecewise-Defined Function Graph the piecewise-defined function, and state the intervals where the function is increasing, decreasing, or constant, along with the domain and range. ƒ1x2 5 c 1 2 x x , 0 x 0 # x , 2 21 x . 2 Solution: Graph these functions on the same plane. Linear function: ƒ 1x2 5 1 2 x Identity function: ƒ 1x2 5 x Constant function: ƒ 1x2 5 21 Darken the piecewise-defined function on the graph. For all values less than zero 1x 02 the function is defined by the linear function. Note the use of an open circle, indicating up to but not including x 5 0. For values 0 " x 2, the function is defined by the identity function. The circle is filled in at the left endpoint, x 5 0. An open circle is used at x 5 2. For all values greater than 2, x + 2, the function is defined by the constant function. Because this interval does not include the point x 5 2, an open circle is used. At what intervals is the function increasing, decreasing, or constant? Remember that the intervals correspond to the x-values. Decreasing: 12H, 02 Increasing: 10, 22 Constant: 12, H2 The function is defined for all values of x except x 5 2. Domain: 12q, 22 ∪ 12, q2 The output of this function (vertical direction) takes on the y-values y $ 0 and the additional single value y 5 21. Range: 321, 214 ∪ 30, q2 or 5216 ∪ 30, q2 We mentioned earlier that a discontinuous function has a graph that exhibits holes or jumps. In this example, the point x 5 0 corresponds to a jump because you would have to pick up your pencil to continue drawing the graph. The point x 5 2 corresponds to both a hole and a jump. The hole indicates that the function is not x y x y 266 CHAPTER 3 Functions and Their Graphs Piecewise-defined functions whose “pieces” are constants are called step functions. The reason for this name is that the graph of a step function looks like steps of a staircase. A common step function used in engineering is the Heaviside step function (also called the unit step function): H1t2 5 b0 t , 0 1 t $ 0 t H(t) 1 This function is used in signal processing to represent a signal that turns on at some time and stays on indefinitely. A common step function used in business applications is the greatest integer function. defined at that point, and there is still a jump because the identity function and the constant function do not meet at the same y-value at x 5 2. Y OUR TU R N Graph the piecewise-defined function, and state the intervals where the function is increasing, decreasing, or constant, along with the domain and range. ƒ1x2 5 c 2x x # 21 2 21 , x , 1 x x . 1 ▼ ▼ A N S W E R Increasing: 11, q2 Decreasing: 12q, 212 Constant: 121, 12 Domain: 12q, 12 ∪ 11, q2 Range: 31, q2 x y [CONCEPT CHECK] State the domain, range, and any points of discontinuity for the Heaviside function. ANSWER Domain: (2q, q) Range: 304 ∪ 314 Point of Discontinuity: x = 0 ▼ x 1.0 1.3 1.5 1.7 1.9 2.0 ƒ1x2 5 3 3 x4 4 1 1 1 1 1 2 x –5 5 –5 5 f(x) = x GREATEST INTEGER FUNCTION ƒ1x2 5 33x44 5 greatest integer less than or equal to x. [SEC TION 3. 2] S U M M A RY NAME FUNCTION DOMAIN RANGE GRAPH EVEN/ODD Linear ƒ1x2 5 mx 1 b, m 2 0 12q, q2 12q, q2 x y Neither 1unless y 5 x2 Constant ƒ1x2 5 c 12q, q2 3c, c4 or 5c6 x y Even Domain and Range of a Function • Implied domain: Exclude any values that lead to the function being undefined (dividing by zero) or imaginary outputs (square root of a negative real number). • Inspect the graph to determine the set of all inputs (domain) and the set of all outputs (range). Finding Intervals Where a Function Is Increasing, Decreasing, or Constant • Increasing: Graph of function rises from left to right. • Decreasing: Graph of function falls from left to right. • Constant: Graph of function does not change height from left to right. Average Rate of Change ƒ1x22 2 ƒ1x12 x2 2 x1 x1 2 x2 Difference Quotient ƒ1x 1 h2 2 ƒ1x2 h h 2 0 Piecewise-Defined Functions • Continuous: You can draw the graph of a function without picking up the pencil. • Discontinuous: Graph has holes and/or jumps. NAME FUNCTION DOMAIN RANGE GRAPH EVEN/ODD Identity ƒ1x2 5 x 12q, q2 12q, q2 x y Odd Square ƒ1x2 5 x2 12q, q2 30, q2 x y Even Cube ƒ1x2 5 x3 12q, q2 12q, q2 x y Odd Square Root ƒ1x2 5 !x 30, q2 30, q2 x y Neither Cube Root ƒ1x2 5 3 !x 12q, q2 12q, q2 x y Odd Absolute Value ƒ1x2 5 0 x 0 12q, q2 30, q2 x y Even Reciprocal ƒ1x2 5 1 x 12q, 02 ∪ 10, q2 12q, 02 ∪ 10, q2 x y Odd In Exercises 1–24, determine whether the function is even, odd, or neither. 1. G 1x2 5 x 1 4 2. h1x2 5 3 2 x 3. ƒ1x2 5 3x2 1 1 4. F1x2 5 x4 1 2x2 5. g1t2 5 5t3 2 3t 6. ƒ1x2 5 3x5 1 4x3 7. h1x2 5 x2 1 2x 8. G1x2 5 2x4 1 3x3 [SEC TION 3. 2] E X E R C I S E S • S K I L L S 3.2 Graphs of Functions 267 268 CHAPTER 3 Functions and Their Graphs 9. h 1x2 5 x1/3 2 x 10. g 1x2 5 x21 1 x 23. x y –5 5 10 24. x y –5 5 9 11. ƒ1x2 5 0 x0 1 5 12. ƒ1x2 5 0 x0 1 x2 13. ƒ1x2 5 0 x0 14. ƒ1x2 5 0 x30 15. G 1t2 5 0 t 2 30 16. g 1t2 5 0 t 1 20 17. G1t2 5 !t 2 3 18. ƒ1x2 5 !2 2 x 19. g1x2 5 "x2 1 x 20. ƒ1x2 5 "x2 1 2 21. h1x2 5 1 x 1 3 22. h1x2 5 1 x 2 2x In Exercises 25–36, state the (a) domain, (b) range, and (c) x-interval(s) where the function is increasing, decreasing, or constant. Find the values of (d) ƒ 102 , (e) ƒ 1222 , and (f ) ƒ 122 . 25. x f (x) y (–3, 3) (–2, –1) (–1, –1) –5 5 26. x f (x) y (–3, 3) (–2, 2) (0, –1) (1, –1) (2, 1) 27. x (0, 4) (–7, 1) (–4, –2) (2, –5) y –8 2 –5 5 28. x (3, –4) (–3, 4) (–6, 0) (6, 0) (0, 0) y –10 10 –5 5 29. x (4, 2) (–3, 2) y –5 5 –5 5 30. x y –5 5 –5 5 31. x (2, 0) (0, –4) (–2, 0) y –5 5 –5 5 32. x (3, 0) (–3, 0) y –5 5 –5 5 33. x (2, –3) (–2, 3) y –10 10 –10 10 34. x (4, 2) (–4, 0) (0, 4) y –5 5 –5 5 35. x (5, 0) (0, 5) (0, 7) (–5, 0) (–2, 3) y –5 5 –2 8 36. x (4, 3) (–5, 3) (–8, 0) (0, –4) y –10 10 –10 10 In Exercises 37–44, find the difference quotient ƒ1x 1 h2 2 ƒ1x2 h for each function. 37. ƒ1x2 5 x2 2 x 38. ƒ1x2 5 x2 1 2x 39. ƒ1x2 5 3x 1 x2 40. ƒ1x2 5 5x 2 x2 41. ƒ1x2 5 x2 2 3x 1 2 42. ƒ1x2 5 x2 2 2x 1 5 43. ƒ1x2 5 23x2 1 5x 2 4 44. ƒ1x2 5 24x2 1 2x 2 3 3.2 Graphs of Functions 269 In Exercises 45–52, find the average rate of change of the function from x 5 1 to x 5 3. 45. ƒ1x2 5 x3 46. ƒ1x2 5 1 x 47. ƒ1x2 5 0 x 0 48. ƒ1x2 5 2x 49. ƒ1x2 5 1 2 2x 50. ƒ1x2 5 9 2 x2 51. ƒ1x2 5 0 5 2 2x 0 52. ƒ1x2 5 "x2 2 1 In Exercises 53–78, graph the piecewise-defined functions. State the domain and range in interval notation. Determine the intervals where the function is increasing, decreasing, or constant. 53. ƒ1x2 5 bx x , 2 2 x $ 2 54. ƒ1x2 5 b2x x , 21 21 x $ 21 55. ƒ1x2 5 b1 x , 21 x2 x $ 21 56. ƒ1x2 5 bx2 x , 2 4 x $ 2 57. ƒ1x2 5 bx x , 0 x2 x $ 0 58. ƒ1x2 5 b2x x # 0 x2 x . 0 59. ƒ1x2 5 b2x 1 2 x , 1 x2 x $ 1 60. ƒ1x2 5 b2 1 x x # 21 x2 x . 21 61. ƒ1x2 5 b 5 2 2x x , 2 3x 2 2 x . 2 62. ƒ1x2 5 d 3 2 1 2 x x , 22 4 1 3 2 x x . 22 63. G1x2 5 c 21 x , 21 x 21 # x # 3 3 x . 3 64. G1x2 5 c 21 x , 21 x 21 , x , 3 3 x . 3 65. G1t2 5 c 1 t , 1 t2 1 # t # 2 4 t . 2 66. G1t2 5 c 1 t , 1 t2 1 , t , 2 4 t . 2 67. ƒ1x2 5 c 2x 2 1 x , 22 x 1 1 22 , x , 1 2x 1 1 x $ 1 68. ƒ1x2 5 c 2x 2 1 x # 22 x 1 1 22 , x , 1 2x 1 1 x . 1 69. G1x2 5 b0 x , 0 !x x $ 0 70. G1x2 5 b1 x , 1 3 !x x . 1 71. G1x2 5 c 0 x 5 0 1 x x 2 0 72. G1x2 5 c 0 x 5 0 2 1 x x 2 0 73. G1x2 5 c 2 3 !x x # 21 x 21 , x , 1 2!x x . 1 74. G1x2 5 c 2! 3 x x , 21 x 21 # x , 1 !x x . 1 75. ƒ1x2 5 c x 1 3 x # 22 0 x 0 22 , x , 2 x2 x $ 2 76. ƒ1x2 5 c 0 x 0 x , 21 1 21 , x , 1 0 x 0 x . 1 77. ƒ1x2 5 c x x # 21 x3 21 , x , 1 x2 x . 1 78. ƒ1x2 5 c x2 x # 21 x3 21 , x , 1 x x $ 1 270 CHAPTER 3 Functions and Their Graphs • A P P L I C A T I O N S For Exercises 79 and 80, refer to the following: A manufacturer determines that his profit and cost functions over one year are represented by the following graphs. 2 4 6 8 12 10 $20 18 16 14 12 10 8 6 4 2 t (time in months) P (proft in millions of dollars) 2 4 6 8 12 10 $5 4 3 2 1 t (time in months) C (cost in millions of dollars) 79. Business. Find the intervals on which profit is increasing, decreasing, and constant. 80. Business. Find the intervals on which cost is increasing, decreasing, and constant. 81. Budget: Costs. The Kappa Kappa Gamma sorority decides to order custom-made T-shirts for its Kappa Krush mixer with the Sigma Alpha Epsilon fraternity. If the sorority orders 50 or fewer T-shirts, the cost is $10 per shirt. If it orders more than 50 but less than or equal to 100, the cost is $9 per shirt. If it orders more than 100, the cost is $8 per shirt. Find the cost function C 1x2 as a function of the number of T-shirts x ordered. 82. Budget: Costs. The marching band at a university is ordering some additional uniforms to replace existing uniforms that are worn out. If the band orders 50 or fewer, the cost is $176.12 per uniform. If it orders more than 50 but less than or equal to 100, the cost is $159.73 per uniform. Find the cost function C 1x2 as a function of the number of new uniforms x ordered. 83. Budget: Costs. The Richmond rowing club is planning to enter the Head of the Charles race in Boston and is trying to figure out how much money to raise. The entry fee is $250 per boat for the first 10 boats and $175 for each additional boat. Find the cost function C 1x2 as a function of the number of boats x the club enters. 84. Phone Cost: Long-Distance Calling. A phone company charges $0.39 per minute for the first 10 minutes of an international long-distance phone call and $0.12 per minute every minute after that. Find the cost function C 1x2 as a function of the length of the phone call x in minutes. 85. Event Planning. A young couple are planning their wedding reception at a yacht club. The yacht club charges a flat rate of $1000 to reserve the dining room for a private party. The cost of food is $35 per person for the first 100 people and $25 per person for every additional person beyond the first 100. Write the cost function C 1x2 as a function of the number of people x attending the reception. 86. Home Improvement. An irrigation company gives you an estimate for an eight-zone sprinkler system. The parts are $1400, and the labor is $25 per hour. Write a function C 1x2 that determines the cost of a new sprinkler system if you choose this irrigation company. 87. Sales. A famous author negotiates with her publisher the monies she will receive for her next suspense novel. She will receive $50,000 up front and a 15% royalty rate on the first 100,000 books sold, and 20% on any books sold beyond that. If the book sells for $20 and royalties are based on the selling price, write a royalties function R 1x2 as a function of total number x of books sold. 88. Sales. Rework Exercise 87 if the author receives $35,000 up front, 15% for the first 100,000 books sold, and 25% on any books sold beyond that. 89. Profit. Some artists are trying to decide whether they will make a profit if they set up a Web-based business to market and sell stained glass that they make. The costs associated with this business are $100 per month for the Web site and $700 per month for the studio they rent. The materials cost $35 for each work in stained glass, and the artists charge $100 for each unit they sell. Write the monthly profit as a function of the number of stained-glass units they sell. 90. Profit. Philip decides to host a shrimp boil at his house as a fundraiser for his daughter’s AAU basketball team. He orders Gulf shrimp to be flown in from New Orleans. The shrimp costs $5 per pound. The shipping costs $30. If he charges $10 per person, write a function F 1x2 that represents either his loss or profit as a function of the number of people x that attend. Assume that each person will eat 1 pound of shrimp. 91. Postage Rates. The following table corresponds to first-class postage rates for the U.S. Postal Service. Write a piecewise-defined function in terms of the greatest integer function that models this cost of mailing flat envelopes first class. WEIGHT LESS THAN (OUNCES) FIRST-CLASS RATE (FLAT ENVELOPES) 1 $0.98 2 1.20 3 1.42 4 1.64 5 1.86 6 2.08 7 2.30 8 2.52 9 2.74 10 2.96 11 3.18 12 3.40 13 3.62 3.2 Graphs of Functions 271 92. Postage Rates. The following table corresponds to first-class postage rates for the U.S. Postal Service. Write a piecewise-defined function in terms of the greatest integer function that models this cost of mailing parcels first class. WEIGHT LESS THAN (OUNCES) PARCELS 1 $2.54 2 2.54 3 2.54 4 2.74 5 2.94 6 3.14 7 3.34 8 3.54 9 3.74 10 3.94 11 4.14 12 4.34 13 4.54 A square wave is a waveform used in electronic circuit testing and signal processing. A square wave alternates regularly and instantaneously between two levels. sciencephotos/Alamy 93. Electronics: Signals. Write a step function ƒ1t2 that represents the following square wave. t 5 –5 5 f (t) 94. Electronics: Signals. Write a step function ƒ1x2 that represents the following square wave, where x represents frequency in Hz. 1000 –1 1 For Exercises 95 and 96, refer to the following table: Global Carbon Emissions from Fossil Fuel Burning YEAR MILLIONS OF TONS OF CARBON 1990 500 1925 1000 1950 1500 1975 5000 2000 7000 95. Climate Change: Global Warming. What is the average rate of change in global carbon emissions from fossil fuel burning from a. 1900 to 1950? b. 1950 to 2000? 96. Climate Change: Global Warming. What is the average rate of change in global carbon emissions from fossil fuel burning from a. 1950 to 1975? b. 1975 to 2000? For Exercises 97 and 98, use the following information: The height (in feet) of a falling object with an initial velocity of 48 feet per second launched straight upward from the ground is given by h 1t2 5 216t2 1 48t, where t is time (in seconds). 97. Falling Objects. What is the average rate of change of the height as a function of time from t 5 1 to t 5 2? 98. Falling Objects. What is the average rate of change of the height as a function of time from t 5 1 to t 5 3? For Exercises 99 and 100, refer to the following: An analysis of sales indicates that demand for a product during a calendar year (no leap year) is modeled by d 1t2 5 3"t2 1 1 2 2.75t where d is demand in thousands of units and t is the day of the year and t 5 1 represents January 1. 99. Economics. Find the average rate of change of the demand of the product over the first quarter. 100. Economics. Find the average rate of change of the demand of the product over the fourth quarter. 272 CHAPTER 3 Functions and Their Graphs • C A T C H T H E M I S T A K E In Exercises 101–104, explain the mistake that is made. 101. Graph the piecewise-defined function. State the domain and range. ƒ1x2 5 b2x x , 0 x x . 0 Solution: Draw the graphs of ƒ1x2 5 2x and ƒ1x2 5 x. Darken the function ƒ1x2 5 2x when x , 0 and the function ƒ1x2 5 x when x . 0. This gives us the familiar absolute value graph. Domain: 12q, q2 or R Range: 30, q2 This is incorrect. What mistake was made? 102. Graph the piecewise-defined function. State the domain and range. ƒ1x2 5 b2x x # 1 x x . 1 Solution: Draw the graphs of ƒ1x2 5 2x and ƒ1x2 5 x. Darken the function ƒ1x2 5 2x when x , 1 and the function ƒ1x2 5 x when x . 1. The resulting graph is as shown. Domain: 12q, q2 or R Range: 121, q2 This is incorrect. What mistake was made? 103. The cost of airport Internet access is $15 for the first 30 minutes and $1 per minute for each additional minute. Write a function describing the cost of the service as a function of minutes used online. Solution: C 1x2 5 b15 x # 30 15 1 x x . 30 This is incorrect. What mistake was made? 104. Most money market accounts pay a higher interest with a higher principal. If the credit union is offering 2% on accounts with less than or equal to $10,000 and 4% on the additional money over $10,000, write the interest function I 1x2 that represents the interest earned on an account as a function of dollars in the account. Solution: I 1x2 5 b0.02x x # 10,000 0.02110,0002 1 0.04x x .10,000 This is incorrect. What mistake was made? x y x y x y x y 3.2 Graphs of Functions 273 In Exercises 105–108, determine whether each statement is true or false. 105. The identity function is a special case of the linear function. 106. The constant function is a special case of the linear function. 107. If an odd function has an interval where the function is increasing, then it also has to have an interval where the function is decreasing. 108. If an even function has an interval where the function is increasing, then it also has to have an interval where the function is decreasing. • C O N C E P T U A L In Exercises 109 and 110, for a and b real numbers, can the function given ever be a continuous function? If so, specify the value for a and b that would make it so. 109. ƒ1x2 5 bax x # 2 bx2 x . 2 110. ƒ1x2 5 d 21 x x , a 1 x x $ a • C H A L L E N G E 111. In trigonometry you will learn about the sine function, sin x. Plot the function ƒ1x2 5 sin x, using a graphing utility. It should look like the graph on the right. Is the sine function even, odd, or neither? 112. In trigonometry you will learn about the cosine function, cos x. Plot the function ƒ1x2 5 cos x, using a graphing utility. It should look like the graph on the right. Is the cosine function even, odd, or neither? 113. In trigonometry you will learn about the tangent function, tan x. Plot the function ƒ1x2 5 tan x, using a graphing utility. If you restrict the values of x so that 2p 2 , x , p 2, the graph should resemble the graph below. Is the tangent function even, odd, or neither? x y –1.5 1.5 –15 15 114. Plot the function ƒ1x2 5 sin x cos x. What function is this? 115. Graph the function ƒ1x2 5 333x44 using a graphing utility. State the domain and range. 116. Graph the function ƒ1x2 5 CC1 3xDD using a graphing utility. State the domain and range. • T E C H N O L O G Y x y –10 10 –1 1 x y 10 –10 –1 1 274 CHAPTER 3 Functions and Their Graphs 3.3.1 Horizontal and Vertical Shifts The focus of the previous section was to learn the graphs that correspond to particular functions such as identity, square, cube, square root, cube root, absolute value, and reciprocal. Therefore, at this point, you should be able to recognize and generate the graphs of y 5 x, y 5 x2, y 5 x3, y 5 !x, y 5 ! 3 x, y 5 0 x 0, and y 5 1 x. In this section, we will discuss how to sketch the graphs of functions that are very simple modifications of these functions. For instance, a common function may be shifted (horizontally or vertically), reflected, or stretched (or compressed). Collectively, these techniques are called transformations. Let’s take the absolute value function as an example. The graph of ƒ 1x2 5 0 x 0 was given in the last section. Now look at two examples that are much like this function: g1x2 5 0 x 0 1 2 and h1x2 5 0 x 2 10. Graphing these functions by point-plotting yields S K I L L S O B J E C T I V E S ■ ■Sketch the graph of a function using horizontal and vertical shifting of common functions. ■ ■Sketch the graph of a function by reflecting a common function about the x-axis or y-axis. ■ ■Sketch the graph of a function by stretching or compressing a common function. C O N C E P T U A L O B J E C T I V E S ■ ■Understand why a shift in the argument inside the function corresponds to a horizontal shift and a shift outside the function corresponds to a vertical shift. ■ ■Understand why a negative argument inside the function corresponds to a reflection about the y-axis and a negative outside the function corresponds to a reflection about the x-axis. ■ ■Understand the difference between rigid and nonrigid transformations. 3.3 GRAPHING TECHNIQUES: TRANSFORMATIONS 3.3.1 S KILL Sketch the graph of a function using horizontal and vertical shifting of common functions. 3.3.1 CO NCE PTUAL Understand why a shift in the argument inside the function corresponds to a horizontal shift and a shift outside the function corresponds to a vertical shift. x y f (x) g(x) h(x) x f (x ) 22 2 21 1 0 0 1 1 2 2 x g (x ) 22 4 21 3 0 2 1 3 2 4 x h (x ) 22 3 21 2 0 1 1 0 2 1 Instead of point-plotting the function g1x2 5 0 x 0 1 2, we could have started with the function ƒ 1x2 5 0x0 and shifted the entire graph up two units. Similarly, we could have generated the graph of the function h1x2 5 0 x 2 1 0 by shifting the function ƒ 1x2 5 0 x 0 to the right one unit. In both cases, the base or starting function is ƒ 1x2 5 0 x 0. Why did we go up for g 1x2 and to the right for h 1x2? Note that we could rewrite the functions g 1x2 and h 1x2 in terms of ƒ 1x2: g1x2 5 0x0 1 2 5 ƒ1x2 1 2 h1x2 5 0x 2 10 5 ƒ1x 2 12 3.3 Graphing Techniques: Transformations 275 In the case of g 1x2, the shift 1122 occurs “outside” the function—that is, outside the parentheses showing the argument. Therefore, the output for g1x2 is two more than the typical output for ƒ1x2. Because the output corresponds to the vertical axis, this results in a shift upward of two units. In general, shifts that occur outside the function correspond to a vertical shift corresponding to the sign of the shift. For instance, had the function been G 1x2 5 0 x 0 2 2, this graph would have started with the graph of the function ƒ1x2 and shifted down two units. In the case of h 1x2, the shift occurs “inside” the function—that is, inside the parentheses showing the argument. Note that the point 10, 02 that lies on the graph of ƒ1x2 was shifted to the point 11, 02 on the graph of the function h 1x2. The y-value remained the same, but the x-value shifted to the right one unit. Similarly, the points 121, 12 and 11, 12 were shifted to the points 10, 12 and 12, 12, respectively. In general, shifts that occur inside the function correspond to a horizontal shift opposite the sign. In this case, the graph of the function h 1x2 5 0 x 2 10 shifted the graph of the function ƒ1x2 to the right one unit. If, instead, we had the function H 1x2 5 0 x 1 10, this graph would have started with the graph of the function ƒ1x2 and shifted to the left one unit. x y f (x) g (x) f (x) = \|x\| g(x) = \|x\| + 2 x y f (x) h(x) f (x) = \|x\| h(x) = \|x – 1\| STUDY TIP Shifts outside the function are vertical shifts with the sign. Up (1) Down (2) STUDY TIP Shifts inside the function are horizontal shifts opposite the sign. Left (1) Right (2) VERTICAL SHIFTS Assuming that c is a positive constant, To Graph Shift the Graph of ƒ1x2 ƒ1x2 1 c c units upward ƒ1x2 2 c c units downward Adding or subtracting a constant outside the function corresponds to a vertical shift that goes with the sign. HORIZONTAL SHIFTS Assuming that c is a positive constant, To Graph Shift the Graph of ƒ1x2 ƒ1x 1 c2 c units to the left ƒ1x 2 c2 c units to the right Adding or subtracting a constant inside the function corresponds to a horizontal shift that goes opposite the sign. 276 CHAPTER 3 Functions and Their Graphs It is important to note that the domain and range of the resulting function can be thought of as also being shifted. Shifts in the domain correspond to horizontal shifts, and shifts in the range correspond to vertical shifts. ▼ A N S W E R a. x y –5 5 10 b. x y –5 5 10 [CONCEPT CHECK] For the functions F 1x2 5 !x 2 a and G 1x2 5 !x 2 a where a . 0 explain the shifts on y 5 !x. ANSWER F (x) is the graph of y 5 !x shifted a units to the right. G (x) is the graph of y 5 !x shifted a units down. ▼ EXAMPLE 2 Horizontal and Vertical Shifts and Changes in the Domain and Range Graph the functions using translations and state the domain and range of each function. a. g1x2 5 !x 1 1 b. G1x2 5 !x 2 2 Solution: In both cases, the function to start with is ƒ1x2 5 !x. Domain: 30, H2 Range: 30, H2 x y 10 5 (1, 1) (4, 2) (9, 3) f (x) = √x (0, 0) EXAMPLE 1 Horizontal and Vertical Shifts Sketch the graphs of the given functions using horizontal and vertical shifts: a. g 1x2 5 x2 2 1 b. H 1x2 5 1x 1 122 Solution: In both cases, the function to start with is ƒ 1x2 5 x2. a. g1x2 5 x2 2 1 can be rewritten as g 1x2 5 ƒ1x2 2 1. 1. The shift (one unit) occurs outside of the function. Therefore, we expect a vertical shift that goes with the sign. 2. Since the sign is negative, this corresponds to a downward shift. 3. Shifting the graph of the function ƒ 1x2 5 x2 down one unit yields the graph of g 1x2 5 x2 2 1. b. H1x2 5 1x 1 122 can be rewritten as H 1x2 5 ƒ1x 1 12. 1. The shift (one unit) occurs inside of the function. Therefore, we expect a horizontal shift that goes opposite the sign. 2. Since the sign is positive, this corresponds to a shift to the left. 3. Shifting the graph of the function ƒ 1x2 5 x2 to the left one unit yields the graph of H1x2 5 1x 1 122. Y OUR TU R N Sketch the graphs of the given functions using horizontal and vertical shifts. a. g 1x2 5 x2 1 1 b. H 1x2 5 1x 2 122 ▼ x y –2 2 4 f (x) (0, 0) (1, 1) (2, 4) x y –2 2 4 –1 f (x) (0, 0) (1, 1) g(x) (1, 0) (0, –1) x y –2 2 4 –1 f (x) H(x) 3.3 Graphing Techniques: Transformations 277 The previous examples have involved graphing functions by shifting a known function either in the horizontal or vertical direction. Let us now look at combinations of horizontal and vertical shifts. ▼ A N S W E R a. G1x2 5 !x 2 2 x y –2 8 –5 5 Domain: 32, q2 Range: 30, q2 b. h 1x2 5 0x0 1 1 x y –5 5 –5 5 Domain: 32q, q2 Range: 31, q2 EXAMPLE 3 Combining Horizontal and Vertical Shifts Sketch the graph of the function F 1x2 5 1x 1 122 2 2. State the domain and range of F. Solution: The base function is y 5 x2. 1. The shift (one unit) is inside the function, so it represents a horizontal shift opposite the sign. a. g1x2 5 !x 1 1 can be rewritten as g 1x2 5 ƒ1x 1 12. 1. The shift (one unit) is inside the function, which corresponds to a horizontal shift opposite the sign. 2. Shifting the graph of ƒ1x2 5 !x to the left one unit yields the graph of g1x2 5 !x 1 1. Notice that the point 10, 02, which lies on the graph of ƒ1x2, gets shifted to the point 121, 02 on the graph of g 1x2. Although the original function ƒ1x2 5 !x had an implicit restriction on the domain: 30, H2, the function g1x2 5 !x 1 1 has the implicit restriction that x $ 21. We see that the output or range of g 1x2 is the same as the output of the original function ƒ1x2. Domain: 321, H2 Range: 30, H2 b. G1x2 5 !x 2 2 can be rewritten as G1x2 5 ƒ1x2 2 2. 1. The shift (two units) is outside the function, which corresponds to a vertical shift with the sign. 2. The graph of G1x2 5 !x 2 2 is found by shifting ƒ1x2 5 !x down two units. Note that the point 10, 02, which lies on the graph of ƒ1x2, gets shifted to the point 10, 222 on the graph of G 1x2. The original function ƒ1x2 5 !x has an implicit restriction on the domain: 30, H2. The function G1x2 5 !x 2 2 also has the implicit restriction that x $ 0. The output or range of G 1x2 is always two units less than the output of the original function ƒ1x2. Domain: [0, H2 Range: [22, H2 YOUR T UR N Sketch the graph of the functions using shifts and state the domain and range. a. G1x2 5 !x 2 2 b. h1x2 5 0 x 0 1 1 ▼ x y 9 5 (1, 0) (0, 1) (3, 2) (8, 3) x y 10 3 –2 (0, 2) (4, 0) (0, 0) (4, 2) (9, 3) 278 CHAPTER 3 Functions and Their Graphs All of the previous transformation examples involve starting with a common function and shifting the function in either the horizontal or the vertical direction (or a combination of both). Now, let’s investigate reflections of functions about the x-axis or y-axis. 3.3.2 Reflection about the Axes To sketch the graphs of ƒ 1x2 5 x2 and g1x2 5 2x2, start by first listing points that are on each of the graphs and then connecting the points with smooth curves. 2. The 22 shift is outside the function, which represents a vertical shift with the sign. 3. Therefore, we shift the graph of y 5 x2 to the left one unit and down two units. For instance, the point 10, 02 on the graph of y 5 x2 shifts to the point 121, 222 on the graph of F1x2 5 1x 1 122 2 2. Domain: 12H, H2 Range: 322, H2 Y OUR TU R N Sketch the graph of the function ƒ1x2 5 0x 2 20 1 1. State the domain and range of ƒ. x y –5 5 8 (–1, –2) ▼ ▼ A N S W E R ƒ1x2 5 0 x 2 2 0 1 1 ƒ1x2 5 0 x 0 Domain: 12q, q2 Range: 31, q2 x y –5 5 10 3.3.2 S KILL Sketch the graph of a function by reflecting a common function about the x-axis or y-axis. 3.3.2 CO NCE PTUAL Understand why a negative argument inside the function corresponds to a reflection about the y-axis and a negative outside the function corresponds to a reflection about the x-axis. x y –5 5 5 –5 x f (x ) 22 4 21 1 0 0 1 1 2 4 x g (x ) 22 24 21 21 0 0 1 21 2 24 x y 5 –5 5 x f (x ) 0 0 1 1 4 2 9 3 x g (x ) 29 3 24 2 21 1 0 0 Note that if the graph of ƒ 1x2 5 x2 is reflected about the x-axis, the result is the graph of g1x2 5 2x2. Also note that the function g 1x2 can be written as the negative of the function ƒ1x2; that is, g 1x2 5 2ƒ1x2. In general, reflection about the x-axis is produced by multiplying a function by 21. Let’s now investigate reflection about the y-axis. To sketch the graphs of ƒ1x2 5 !x and g1x2 5 !2x start by listing points that are on each of the graphs and then connecting the points with smooth curves. 3.3 Graphing Techniques: Transformations 279 Note that if the graph of ƒ1x2 5 !x is reflected about the y-axis, the result is the graph of g1x2 5 !2x. Also note that the function g 1x2 can be written as g 1x2 5 ƒ12x2. In general, reflection about the y-axis is produced by replacing x with –x in the function. Notice that the domain of ƒ is 30, q2, whereas the domain of g is 12q, 04. [CONCEPT CHECK] For any even function f (x), describe the graph of f (2x). ANSWER The graph of f (2x) is identical to the graph of f (x) because for even functions f (2x) 5 f (x). ▼ EXAMPLE 5 Sketching the Graph of a Function Using Both Shifts and Reflections Sketch the graph of the function ƒ1x2 5 !2 2 x 1 1. Solution: Start with the square root function. g1x2 5 !x Shift the graph of g 1x2 to the left two units to arrive at the graph of g 1x 1 22. g1x 1 22 5 !x 1 2 Reflect the graph of g 1x 1 22 about the y-axis to arrive at the graph of g 12x 1 22. g12x 1 22 5 !2x 1 2 Shift the graph g 12x 1 22 up one unit to arrive at the graph of g 12x 1 22 1 1. g12x 1 22 1 1 5 !2 2 x 1 1 YOUR T UR N Use shifts and reflections to sketch the graph of the function ƒ1x2 5 2"x 2 1 1 2. State the domain and range of ƒ1x2. x y 5 –5 5 ▼ x y 9 5 –5 EXAMPLE 4 Sketching the Graph of a Function Using Both Shifts and Reflections Sketch the graph of the function G 1x2 5 2!x 1 1. Solution: Start with the square root function. ƒ1x2 5 !x Shift the graph of ƒ1x2 to the left one unit to arrive at the graph of ƒ1x 1 12. ƒ1x 1 12 5 !x 1 1 Reflect the graph of ƒ1x 1 12 about the x-axis to arrive at the graph of 2ƒ1x 1 12. 2ƒ1x 1 12 5 2!x 1 1 REFLECTION ABOUT THE AXES The graph of 2ƒ 1x2 is obtained by reflecting the graph of ƒ1x2 about the x-axis. The graph of ƒ 12x2 is obtained by reflecting the graph of ƒ1x2 about the y-axis. 280 CHAPTER 3 Functions and Their Graphs Look back at the order in which transformations were performed in Example 5: horizontal shift, reflection, and then vertical shift. Let us consider an alternate order of transformations. WORDS MATH Start with the square root function. g1x2 5 !x Shift the graph of g 1x2 up one unit to arrive at the graph of g 1x2 1 1. g1x2 1 1 5 !x 1 1 Reflect the graph of g 1x2 1 1 about the y-axis to arrive at the graph of g 12x2 1 1. g12x2 1 1 5 !2x 1 1 Replace x with x 2 2, which corresponds to a shift of the graph of g 12x2 1 1 to the right two units to arrive at the graph of g32 1x 2 224 1 1. g12x 1 22 1 1 5 "2 2 x 1 1 In the last step, we replaced x with x 2 2, which required us to think ahead knowing the desired result was 2 2 x inside the radical. To avoid any possible confusion, follow this order of transformations: 1. Horizontal shifts: ƒ1x 6 c2 2. Reflection: ƒ12x2 and/or 2ƒ1x2 3. Vertical shifts: ƒ1x2 6 c 3.3.3 Stretching and Compressing Horizontal shifts, vertical shifts, and reflections change only the position of the graph in the Cartesian plane, leaving the basic shape of the graph unchanged. These transformations (shifts and reflections) are called rigid transformations because they alter only the position. Nonrigid transformations, on the other hand, distort the shape of the original graph. We now consider stretching and compressing of graphs in both the vertical and the horizontal direction. A vertical stretch or compression of a graph occurs when the function is multiplied by a positive constant. For example, the graphs of the functions ƒ 1x2 5 x2, g 1x2 5 2 ƒ 1x2 5 2x2, and h1x2 5 1 2 ƒ1x2 5 1 2 x2 are illustrated below. Depending on if the constant is larger than 1 or smaller than 1 will determine whether it corresponds to a stretch (expansion) or compression (contraction) in the vertical direction. ▼ A N S W E R Domain: 31, q2 Range: 12q, 24 x y –5 5 10 x y 5 –5 5 3.3.3 SKI LL Sketch the graph of a function by stretching or compressing a common function. 3.3.3 C ON CEPTUAL Understand the difference between rigid and nonrigid transformations. x y –5 5 20 x f (x ) 22 4 21 1 0 0 1 1 2 4 x g (x ) 22 8 21 2 0 0 1 2 2 8 x h(x ) 22 2 21 1 2 0 0 1 1 2 2 2 3.3 Graphing Techniques: Transformations 281 Conversely, if the argument x of a function ƒ is multiplied by a positive real number c, then the result is a horizontal stretch of the graph of ƒ if 0 , c , 1. If c . 1, then the result is a horizontal compression of the graph of ƒ. Note that when the function ƒ 1x2 5 x2 is multiplied by 2, so that g 1x2 5 2 ƒ 1x2 5 2x2, the result is a graph stretched in the vertical direction. When the function ƒ 1x2 5 x2 is multiplied by 1 2, so that h1x2 5 1 2 ƒ1x2 5 1 2x2, the result is a graph that is compressed in the vertical direction. EXAMPLE 7 Vertically Stretching and Horizontally Compressing Graphs Given the graph of ƒ1x2, graph: a. 2ƒ1x2 b. ƒ12x2 x y 5 2 1 –2 –1 (, 0) 2 3 2 ( , –1) 2 ( , 1) [CONCEPT CHECK] Describe where the graphs of f (x) and a⋅f (x) intersect. ANSWER Only at the points when f (x) 5 0 (x-intercepts) ▼ x y 2 –2 –10 10 (–2, –2) (2, 2) EXAMPLE 6 Vertically Stretching and Compressing Graphs Graph the function h1x2 5 1 4x3. Solution: 1. Start with the cube function. ƒ 1x2 5 x3 2. Vertical compression is h1x2 5 1 4 x3 expected because 1 4 is less than 1. 3. Determine a few points that lie on the graph of h. 10, 02 12, 22 122, 222 VERTICAL STRETCHING AND VERTICAL COMPRESSING OF GRAPHS The graph of c ƒ1x2 is found by: ■ ■Vertically stretching the graph of ƒ1x2 if c . 1 ■ ■Vertically compressing the graph of ƒ1x2 if 0 , c , 1 Note: c is any positive real number. HORIZONTAL STRETCHING AND HORIZONTAL COMPRESSING OF GRAPHS The graph of ƒ1cx2 is found by: ■ ■Horizontally stretching the graph of ƒ1x2 if 0 , c , 1 ■ ■Horizontally compressing the graph of ƒ1x2 if c . 1 Note: c is any positive real number. 282 CHAPTER 3 Functions and Their Graphs In Example 8, we followed the same “inside out” approach with the functions to determine the order for the transformations: horizontal shift, vertical stretch, and reflection. Solution (a): Since the function is multiplied (on the outside) by 2, the result is that each y-value of ƒ1x2 is multiplied by 2, which corresponds to vertical stretching. Solution (b): Since the argument of the function is multiplied (on the inside) by 2, the result is that each x-value of ƒ1x2 is divided by 2, which corresponds to horizontal compression. Y OUR TU R N Graph the function g 1x2 5 4x3. x y 5 2 1 –2 –1 2 3 2 ( , –2) 2 ( , 2) x y 5 2 1 –2 –1 2 3 4 ( , –1) 4 ( , 1) 2 ( , 0) (, 0) ▼ ▼ A N S W E R Stretching of the graph ƒ 1x2 5 x3. x y g(x) f(x) –2 2 –40 40 x y –4 6 –5 5 EXAMPLE 8 Sketching the Graph of a Function Using Multiple Transformations Sketch the graph of the function H 1x2 5 22 1x 2 322. Solution: Start with the square function. ƒ 1x2 5 x2 Shift the graph of ƒ1x2 to the right three units to arrive at the graph of ƒ1x 2 32. ƒ 1x 2 32 5 1x 2 322 Vertically stretch the graph of ƒ1x 2 32 by a factor of 2 to arrive at the graph of 2ƒ1x 2 32. 2 ƒ 1x 2 32 5 21x 2 322 Reflect the graph 2 ƒ1x 2 32 about the x-axis to arrive at the graph of 22ƒ1x 2 32. 22 ƒ 1x 2 32 5 221x 2 322 3.3 Graphing Techniques: Transformations 283 In Exercises 1–12, match the function to the graph. 1. ƒ1x2 5 x2 1 1 2. ƒ1x2 5 1x 2 122 3. ƒ1x2 5 2 11 2 x22 4. ƒ1x2 5 2x2 2 1 5. ƒ1x2 5 2 1x 1 122 6. ƒ1x2 5 2 11 2 x22 1 1 7. ƒ1x2 5 !x 2 1 1 1 8. ƒ1x2 5 2!x 2 1 9. ƒ1x2 5 !1 2 x 2 1 10. ƒ1x2 5 !2x 1 1 11. ƒ1x2 5 2!2x 1 1 12. ƒ1x2 5 2!1 2 x 2 1 a. x y b. x y c. x y d. x y e. x y f. x y g. x y h. x y [SEC TION 3.3] E X E R C I S E S • S K I L L S [SEC TION 3.3] S U M M A RY TRANSFORMATION TO GRAPH THE FUNCTION… DRAW THE GRAPH OF f AND THEN… DESCRIPTION Horizontal shifts ƒ1x 1 c2 ƒ1x 2 c2 Shift the graph of ƒ to the left c units. Shift the graph of ƒ to the right c units. Replace x by x 1 c. Replace x by x 2 c. Vertical shifts ƒ1x2 1 c ƒ1x2 2 c Shift the graph of ƒ up c units. Shift the graph of ƒ down c units. Add c to ƒ1x2. Subtract c from ƒ1x2. Reflection about the x-axis 2 ƒ1x2 Reflect the graph of ƒ about the x-axis. Multiply ƒ1x2 by 21. Reflection about the y-axis ƒ12x2 Reflect the graph of ƒ about the y-axis. Replace x by 2x. Vertical stretch c ƒ1x2, where c . 1 Vertically stretch the graph of ƒ. Multiply ƒ1x2 by c. Vertical compression c ƒ1x2, where 0 , c , 1 Vertically compress the graph of ƒ. Multiply ƒ1x2 by c. Horizontal stretch ƒ1cx2, where 0 , c , 1 Horizontally stretch the graph of ƒ. Replace x by cx. Horizontal compression ƒ1cx2, where c . 1 Horizontally compress the graph of ƒ. Replace x by cx. 284 CHAPTER 3 Functions and Their Graphs i. x y j. x y k. x y l. x y In Exercises 13–18, write the function whose graph is the graph of y 5 0 x 0 but is transformed accordingly. 13. Shifted up three units 14. Shifted to the left four units 15. Reflected about the y-axis 16. Reflected about the x-axis 17. Vertically stretched by a factor of 3 18. Vertically compressed by a factor of 3 In Exercises 19–24, write the function whose graph is the graph of y 5 x3 but is transformed accordingly. 19. Shifted down four units 20. Shifted to the right three units 21. Shifted up three units and to the left one unit 22. Reflected about the x-axis 23. Reflected about the y-axis 24. Reflected about both the x-axis and the y-axis In Exercises 25–48, use the given graph to sketch the graph of the indicated functions. 25. x y a. y 5 ƒ1x 2 22 b. y 5 ƒ1x2 2 2 26. x y a. y 5 ƒ1x 1 22 b. y 5 ƒ1x2 1 2 27. x y a. y 5 ƒ1x2 2 3 b. y 5 ƒ1x 2 32 28. x y a. y 5 ƒ1x2 1 3 b. y 5 ƒ1x 1 32 29. x y a. y 5 2ƒ1x2 b. y 5 ƒ12x2 30. x y a. y 5 2ƒ1x2 b. y 5 ƒ12x2 31. x y a. y 5 2 ƒ1x2 b. y 5 ƒ12x2 32. x y a. y 5 2 ƒ1x2 b. y 5 ƒ12x2 3.3 Graphing Techniques: Transformations 285 33. y 5 ƒ1x 2 22 2 3 34. y 5 ƒ1x 1 12 2 2 35. y 5 2ƒ1x 2 12 1 2 36. y 5 22 ƒ1x2 1 1 x f (x) y 37. y 5 21 2 g1x2 38. y 5 1 4 g12x2 39. y 5 2g 12x2 40. y 5 gA1 2xB x g(x) y 41. y 5 1 2 F1x 2 12 1 2 42. y 5 1 2 F12x2 43. y 5 2F11 2 x2 44. y 5 2F1x 2 22 2 1 x F(x) y 45. y 5 2G1x 1 12 2 4 46. y 5 2G12x2 1 1 47. y 5 22G1x 2 12 1 3 48. y 5 2G1x 2 22 2 1 x G(x) y In Exercises 49–74, graph the function using transformations. 49. y 5 x2 2 2 50. y 5 x2 1 3 51. y 5 1x 1 122 52. y 5 1x 2 222 53. y 5 1x 2 322 1 2 54. y 5 1x 1 222 1 1 55. y 5 2 11 2 x22 56. y 5 2 1x 1 222 57. y 5 0 2x0 58. y 5 20 x0 59. y 520 x 1 20 2 1 60. y 5 0 1 2 x 0 1 2 61. y 5 2x2 1 1 62. y 5 20 x0 1 1 63. y 5 2!x 2 2 64. y 5 !2 2 x 65. y 5 2!2 1 x 2 1 66. y 5 !2 2 x 1 3 67. y 5 3 !x 2 1 1 2 68. y 5 3 !x 1 2 2 1 69. y 5 1 x 1 3 1 2 70. y 5 1 3 2 x 71. y 5 2 2 1 x 1 2 72. y 5 2 2 1 1 2 x 73. y 5 5!2x 74. y 5 21 5!x In Exercises 75–80, transform the function into the form ƒ1x2 5 c1x 2 h22 1 k, where c, k, and h are constants, by completing the square. Use graph-shifting techniques to graph the function. 75. y 5 x2 2 6x 1 11 76. ƒ1x2 5 x2 1 2x 2 2 77. ƒ1x2 5 2x2 2 2x 78. ƒ1x2 5 2x2 1 6x 2 7 79. ƒ1x2 5 2x2 2 8x 1 3 80. ƒ1x2 5 3x2 2 6x 1 5 • A P P L I C A T I O N S 81. Salary. A manager hires an employee at a rate of $10 per hour. Write the function that describes the current salary of the employee as a function of the number of hours worked per week, x. After a year, the manager decides to award the employee a raise equivalent to paying him for an additional 5 hours per week. Write a function that describes the salary of the employee after the raise. 82. Profit. The profit associated with St. Augustine sod in Florida is typically P 1x2 5 2x2 1 14,000x 2 48,700,000, where x is the number of pallets sold per year in a normal year. In rainy years Sod King gives away 10 free pallets per year. Write the function that describes the profit of x pallets of sod in rainy years. 83. Taxes. Every year in the United States each working American typically pays in taxes a percentage of his or her earnings (minus the standard deduction). Karen’s 2015 taxes were calculated based on the formula T 1x2 5 0.22 1x 2 63002. That year the standard deduction was $6300 and her tax bracket paid 22% in taxes. Write the function that will determine her 2016 taxes, assuming she receives the raise that places her in the 33% bracket. 84. Medication. The amount of medication that an infant requires is typically a function of the baby’s weight. The number of milliliters of an antiseizure medication A is given by A1x2 5 !x 1 2, where x is the weight of the infant in ounces. In emergencies there is often not enough time to weigh the infant, so nurses have to estimate the baby’s weight. What is the function that represents the actual amount of medication the infant is given if his weight is overestimated by 3 ounces? 286 CHAPTER 3 Functions and Their Graphs For Exercises 85 and 86, refer to the following: Body Surface Area (BSA) is used in physiology and medicine for many clinical purposes. BSA can be modeled by the function BSA 5 Å wh 3600 where w is weight in kilograms and h is height in centimeters. Since BSA depends on weight and height, it is often thought of as a function of both weight and height. However, for an individual adult height is generally considered constant; thus BSA can be thought of as a function of weight alone. 85. Health/Medicine. (a) If an adult female is 162 centimeters tall, find her BSA as a function of weight. (b) If she loses 3 kilograms, find a function that represents her new BSA. 86. Health/Medicine. (a) If an adult male is 180 centimeters tall, find his BSA as a function of weight. (b) If he gains 5 kilograms, find a function that represents his new BSA. • C A T C H T H E M I S T A K E In Exercises 87–90, explain the mistake that is made. 87. Describe a procedure for graphing the function ƒ1x2 5 !x 2 3 1 2. Solution: a. Start with the function ƒ1x2 5 !x. b. Shift the function to the left three units. c. Shift the function up two units. This is incorrect. What mistake was made? 88. Describe a procedure for graphing the function ƒ1x2 5 2!x 1 2 2 3. Solution: a. Start with the function ƒ1x2 5 !x. b. Shift the function to the left two units. c. Reflect the function about the y-axis. d. Shift the function down three units. This is incorrect. What mistake was made? 89. Describe a procedure for graphing the function ƒ1x2 5 0 3 2 x 0 1 1. Solution: a. Start with the function ƒ1x2 5 0x 0. b. Reflect the function about the y-axis. c. Shift the function to the left three units. d. Shift the function up one unit. This is incorrect. What mistake was made? 90. Describe a procedure for graphing the function ƒ1x2 5 22x2 1 1. Solution: a. Start with the function ƒ1x2 5 x2. b. Reflect the function about the y-axis. c. Shift the function up one unit. d. Expand in the vertical direction by a factor of 2. This is incorrect. What mistake was made? In Exercises 91–94, determine whether each statement is true or false. 91. The graph of y 5 02x 0 is the same as the graph of y 5 0x 0. 92. The graph of y 5 !2x is the same as the graph of y 5 !x. 93. If the graph of an odd function is reflected about the x-axis and then the y-axis, the result is the graph of the original odd function. 94. If the graph of y 5 1 x is reflected about the x-axis, it produces the same graph as if it had been reflected about the y-axis. • C O N C E P T U A L 95. The point 1a, b2 lies on the graph of the function y 5 ƒ1x2. What point is guaranteed to lie on the graph of ƒ1x 2 32 1 2? 96. The point 1a, b2 lies on the graph of the function y 5 ƒ1x2. What point is guaranteed to lie on the graph of 2ƒ12x2 1 1? • C H A L L E N G E 97. Use a graphing utility to graph: a. y 5 x2 2 2 and y 5 0x2 2 2 0 b. y 5 x3 2 1 and y 5 0x3 1 1 0 What is the relationship between ƒ1x2 and 0ƒ1x2 0? 98. Use a graphing utility to graph: a. y 5 x2 2 2 and y 5 0x 0 2 2 2 b. y 5 x3 1 1 and y 5 0x 0 3 1 1 What is the relationship between ƒ1x2 and ƒ1 0x 02? 99. Use a graphing utility to graph: a. y 5 !x and y 5 !0.1x b. y 5 !x and y 5 !10x What is the relationship between ƒ1x2 and ƒ1ax2, assuming that a is positive? 100. Use a graphing utility to graph: a. y 5 !x and y 5 0.1!x b. y 5 !x and y 5 10!x What is the relationship between ƒ1x2 and aƒ1x2, assuming that a is positive? 101. Use a graphing utility to graph y 5 ƒ1x2 5 330.5x44 11. Use transforms to describe the relationship between ƒ1x2 and y 5 33 x44. 102. Use a graphing utility to graph y 5 g1x2 5 0.533x44 1 1. Use transforms to describe the relationship between g 1x2 and y 5 33 x44. • T E C H N O L O G Y 3.4 Operations on Functions and Composition of Functions 287 S K I L L S O B J E C T I V E S ■ ■Add, subtract, multiply, and divide functions. ■ ■Evaluate composite functions and determine the corresponding domains. C O N C E P T U A L O B J E C T I V ES ■ ■Understand domain restrictions when dividing functions. ■ ■Realize that the domain of a composition of functions excludes values that are not in the domain of the inside function. 3.4 OPERATIONS ON FUNCTIONS AND COMPOSITION OF FUNCTIONS 3.4.1 S K I L L Add, subtract, multiply, and divide functions. 3.4.1 C ON C E P T U A L Understand domain restrictions when dividing functions. Two different functions can be combined using mathematical operations such as addition, subtraction, multiplication, and division. Also, there is an operation on functions called composition, which can be thought of as a function of a function. When we combine functions, we do so algebraically. Special attention must be paid to the domain and range of the combined functions. 3.4.1 Adding, Subtracting, Multiplying, and Dividing Functions Consider the two functions ƒ1x2 5 x2 1 2x 2 3 and g 1x2 5 x 1 1. The domain of both of these functions is the set of all real numbers. Therefore, we can add, subtract, or multiply these functions for any real number x. Addition: ƒ 1x2 1 g1x2 5 x2 1 2x 2 3 1 x 1 1 5 x2 1 3x 2 2 The result is in fact a new function, which we denote: 1ƒ 1 g2 1x2 5 x2 1 3x 2 2 This is the sum function. Subtraction: ƒ 1x2 2 g1x2 5 x2 1 2x 2 3 2 1x 1 12 5 x2 1 x 2 4 The result is in fact a new function, which we denote: 1ƒ 2 g2 1x2 5 x2 1 x 2 4 This is the difference function. Multiplication: ƒ 1x2 • g1x2 5 1x2 1 2x 2 32 1x 1 12 5 x3 1 3x2 2 x 2 3 The result is in fact a new function, which we denote: 1ƒ • g2 1x2 5 x3 1 3x2 2 x 2 3 This is the product function. Although both ƒ and g are defined for all real numbers x, we must restrict x so that x 2 21 to form the quotient ƒ g. Division: ƒ1x2 g1x2 5 x2 1 2x 2 3 x 1 1 , x 2 21 The result is in fact a new function, which we denote: aƒ gb 1x2 5 x2 1 2x 2 3 x 1 1 , x 2 21 This is called the quotient function. Two functions can be added, subtracted, and multiplied. The resulting function domain is therefore the intersection of the domains of the two functions. However, for division, any value of x (input) that makes the denominator equal to zero must be eliminated from the domain. The previous examples involved polynomials. The domain of any polynomial is the set of all real numbers. Adding, subtracting, and multiplying polynomials result in other polynomials, which have domains of all real numbers. Let’s now investigate operations applied to functions that have a restricted domain. The domain of the sum function, difference function, or product function is the intersection of the individual domains of the two functions. The quotient function has 288 CHAPTER 3 Functions and Their Graphs a similar domain in that it is the intersection of the two domains. However, any values that make the denominator zero must also be eliminated. FUNCTION NOTATION DOMAIN Sum 1ƒ 1 g2 1x2 5 ƒ1x2 1 g 1x2 5domain of ƒ6 " 5domain of g6 Difference 1ƒ 2 g2 1x2 5 ƒ1x2 2 g 1x2 5domain of ƒ6 " 5domain of g6 Product 1ƒ • g2 1x2 5 ƒ1x2 • g 1x2 5domain of ƒ6 " 5domain of g6 Quotient aƒ gb 1x2 5 ƒ1x2 g1x2 5domain of ƒ6 " 5domain of g6 " 5g1x2 2 06 We can think of this in the following way: Any number that is in the domain of both the functions is in the domain of the combined function. The exception to this is the quotient function, which also eliminates values that make the denominator equal to zero. EXAMPLE 1 Operations on Functions: Determining Domains of New Functions For the functions ƒ1x2 5 "x 2 1 and g1x2 5 "4 2 x, determine the sum func-tion, difference function, product function, and quotient function. State the domain of these four new functions. Solution: Sum function: ƒ1x2 1 g1x2 5 "x 2 1 1 "4 2 x Difference function: ƒ1x2 2 g1x2 5 "x 2 1 2 "4 2 x Product function: ƒ1x2 ⋅g1x2 5 !x 2 1⋅!4 2 x 5 "1x 2 1214 2 x2 5 "2x2 1 5x 2 4 Quotient function: ƒ1x2 g1x2 5 "x 2 1 "4 2 x 5 Å x 2 1 4 2 x The domain of the square root function is determined by setting the argument under the radical greater than or equal to zero. Domain of ƒ1x2: 31, q2 Domain of g 1x2: 12q, 44 The domain of the sum, difference, and product functions is 31, q2 ∩ 12q, 44 5 31, 44 The quotient function has the additional constraint that the denominator cannot be zero. This implies that x 2 4, so the domain of the quotient function is 31, 42. Y OUR TU R N Given the function ƒ1x2 5 !x 1 3 and g1x2 5 !1 2 x, find 1ƒ 1 g2 1x2 and state its domain. ▼ [CONCEPT CHECK] Let ƒ 1x2 5 !x 1 1 and g 1x2 5 1 x find the domain of g 1x2 ƒ1x2 . ANSWER 121, 02 ∪ 10, q2 ▼ ▼ A N S W E R 1ƒ 1 g2 1x2 5 "x 1 3 1"1 2 x Domain: 32 3, 14 3.4 Operations on Functions and Composition of Functions 289 3.4.2 Composition of Functions Recall that a function maps every element in the domain to exactly one corresponding element in the range as shown in the figure in the margin. Suppose there is a sales rack of clothes in a department store. Let x correspond to the original price of each item on the rack. These clothes have recently been marked down 20%. Therefore, the function g 1x2 5 0.80x represents the current sale price of each item. You have been invited to a special sale that lets you take 10% off the current sale price and an additional $5 off every item at checkout. The function ƒ1g 1x22 5 0.90g 1x2 2 5 determines the checkout price. Note that the output of the function g is the input of the function ƒ as shown in the figure below. x g(x) = 0.80 x g(x) Domain of g Domain of f Range of f Range of g f(x) = 0.90 g (x) − 5 f(g (x)) Sale price 20% off original price Additional 10% off sale price and $5 off at checkout Original price This is an example of a composition of functions, when the output of one function is the input of another function. It is commonly referred to as a function of a function. An algebraic example of this is the function y 5 "x2 2 2. Suppose we let g 1x2 5 x2 2 2 and ƒ1x2 5 "x. Recall that the independent variable in function notation is a placeholder. Since ƒ1u 2 5 "1u2, then ƒ1g1x22 5 "1g1x22. Substituting the expression for g 1x2, we find ƒ1g1x22 5 "x2 2 2. The function y 5 "x2 2 2 is said to be a composite function, y 5 ƒ1g 1x22. EXAMPLE 2 Quotient Function and Domain Restrictions Given the functions F1x2 5 !x and G 1x2 5 0 x 2 30 , find the quotient function, a F Gb 1x2, and state its domain. Solution: The quotient function is written as a F Gb 1x2 5 F1x2 G1x2 5 !x 0 x 2 3 0 Domain of F1x2: 30, q2 Domain of G1x2: 12q, q2 The real numbers that are in both the domain for F 1x2 and the domain for G 1x2 are represented by the intersection 30, q2 ∩ 12q, q2 5 30, q2. Also, the denominator of the quotient function is equal to zero when x 5 3, so we must eliminate this value from the domain. Domain of a F Gb 1x2: 30, 32 ∪ 13, q2 YOUR T UR N For the functions given in Example 2, determine the quotient function aG F b 1x2, and state its domain. ▼ A N S W E R aG F b 1x2 5 G1x2 F1x2 5 0 x 2 3 0 !x Domain: 10, q2 ▼ x f f(x) Domain Range 3.4.2 S K I L L Evaluate composite functions and determine the corresponding domains. 3.4.2 C ON C E P T U A L Realize that the domain of a composition of functions excludes values that are not in the domain of the inside. 290 CHAPTER 3 Functions and Their Graphs Note that the domain of g 1x2 is the set of all real numbers, and the domain of ƒ1x2 is the set of all nonnegative numbers. The domain of a composite function is the set of all x such that g 1x2 is in the domain of ƒ. For instance, in the composite function y 5 ƒ1g 1x22, we know that the allowable inputs into ƒ are all numbers greater than or equal to zero. Therefore, we restrict the outputs of g 1x2 $ 0 and find the corresponding x-values. Those x-values are the only allowable inputs and constitute the domain of the composite function y 5 ƒ1g 1x22. The symbol that represents composition of functions is a small open circle; thus 1ƒ + g21x2 5 ƒ1g1x22, which is read aloud as “ƒ of g.” It is important not to confuse this with the multiplication sign: 1ƒ • g2 1x2 5 ƒ1x2g 1x2. ▼ C A U T I O N ƒ + g 2 ƒ ⋅ g STUDY TIP Order is important: 1f + g2 1x2 5 f 1g1x22 1g + f 2 1x2 5 g1 f 1x22 STUDY TIP The domain of f + g is always a subset of the domain of g, and the range of f + g is always a subset of the range of f. ▼ A N S W E R g + ƒ 5 g 1ƒ1x22 5 x2 2 2 EXAMPLE 3 Finding a Composite Function Given the functions ƒ1x2 5 x2 1 1 and g 1x2 5 x 2 3, find 1ƒ + g21x2. Solution: Write ƒ1x2 using placeholder notation. ƒ1u2 5 1u22 1 1 Express the composite function ƒ + g. ƒ1g 1x22 5 1g 1x222 1 1 Substitute g 1x2 5 x 2 3 into ƒ. ƒ1g 1x22 5 1x 2 322 1 1 Eliminate the parentheses on the right side. ƒ1g 1x22 5 x2 2 6x 1 10 1ƒ + g21x2 5 ƒ1g1x22 5 x2 2 6x 1 10 Y OUR TU R N Given the functions in Example 3, find 1g + ƒ21x2. ▼ It is important to realize that there are two “filters” that allow certain values of x into the domain of 1ƒ + g21x2 5 ƒ1g1x22. The first filter is g 1x2. If x is not in the domain of g 1x2, it cannot be in the domain of 1ƒ + g21x2 5 ƒ1g1x22. Of those values for x that are in the domain of g 1x2, only some pass through because we restrict the output of g 1x2 to values that are allowable as input into ƒ. This adds an additional filter. The domain of ƒ + g is always a subset of the domain of g, and the range of ƒ + g is always a subset of the range of ƒ. x g(x) f(g(x)) ( f g)(x) = f(g(x)) COMPOSITION OF FUNCTIONS Given two functions ƒ and g, there are two composite functions that can be formed. NOTATION WORDS DEFINITION DOMAIN ƒ + g ƒ composed with g ƒ1g 1x22 The set of all real numbers x in the domain of g such that g 1x2 is also in the domain of ƒ. g + ƒ g composed with ƒ g 1ƒ 1x22 The set of all real numbers x in the domain of ƒ such that ƒ1x2 is also in the domain of g. 3.4 Operations on Functions and Composition of Functions 291 The domain of the composite function cannot always be determined by examining the final form of ƒ + g. EXAMPLE 4 Determining the Domain of a Composite Function Given the functions ƒ1x2 5 1 x 2 1 and g1x2 5 1 x, determine ƒ + g, and state its domain. Solution: Write ƒ1x2 using placeholder notation. ƒ1u2 5 1 1u2 2 1 Express the composite function ƒ + g. ƒ1g1x22 5 1 g1x2 2 1 Substitute g1x2 5 1 x into ƒ. ƒ1g1x22 5 1 1 x 2 1 Multiply the right side by x x. ƒ1g1x22 5 1 1 x 2 1 ⋅x x 5 x 1 2 x 1ƒ + g2 5 ƒ1g1x22 5 x 1 2 x What is the domain of 1ƒ + g2 1x2 5 ƒ1g 1x22? By inspecting the final result of ƒ1g 1x22, we see that the denominator is zero when x 5 1. Therefore, x 2 1. Are there any other values for x that are not allowed? The function g 1x2 has the domain x 2 0; therefore we must also exclude zero. The domain of 1ƒ + g2 1x2 5 ƒ1g 1x22 excludes x 5 0 and x 5 1 or, in interval notation, 12q, 02 ∪ 10, 12 ∪ 11, q2 YOUR T UR N For the functions ƒ and g given in Example 4, determine the composite function g + ƒ and state its domain. ▼ ▼ A N S W E R g1ƒ1x22 5 x 2 1. Domain of g + ƒ is x 2 1, or in interval notation, 12q, 12 ∪ 11, q2. ▼ C A U T I O N The domain of the composite function cannot always be determined by examining the final form of ƒ + g. [CONCEPT CHECK] For the functions f 1x2 5 x 2 a and g 1x2 5 1 x 1 a find g(f (x)) and state its domain. ANSWER g 1f 1x22 5 1 x and domain is all real numbers except x = 0. ▼ EXAMPLE 5 Determining the Domain of a Composite Function (Without Finding the Composite Function) Let ƒ1x2 5 1 x 2 2 and g1x2 5 !x 1 3. Find the domain of ƒ1g 1x22. Do not find the composite function. Solution: Find the domain of g. 323, q2 Find the range of g. 30, q2 In ƒ1g 1x22, the output of g becomes the input for ƒ. Since the domain of ƒ is the set of all real numbers except 2, we eliminate any values of x in the domain of g that correspond to g 1x2 5 2. Let g 1x2 5 2. !x 1 3 5 2 Square both sides. x 1 3 5 4 Solve for x. x 5 1 Eliminate x 5 1 from the domain of g, 323, q2. State the domain of ƒ1g 1x22. 323, 12 ∪ 11, q2 292 CHAPTER 3 Functions and Their Graphs Application Problems Recall the example at the beginning of this chapter regarding the clothes that are on sale. Often, real-world applications are modeled with composite functions. In the clothes example, x is the original price of each item. The first function maps its input (original price) to an output (sale price). The second function maps its input (sale price) to an output (checkout price). Example 7 is another real-world application of composite functions. Three temperature scales are commonly used: ■ ■The degree Celsius (8C) scale ■ ■This scale was devised by dividing the range between the freezing (08C) and boiling (1008C) points of pure water at sea level into 100 equal parts. This scale is used in science and is one of the standards of the “metric” (SI) system of measurements. EXAMPLE 6 Evaluating a Composite Function Given the functions ƒ1x2 5 x2 2 7 and g 1x2 5 5 2 x2, evaluate: a. ƒ1g 1122 b. ƒ1g 12222 c. g 1ƒ1322 d. g 1ƒ12422 Solution: One way of evaluating these composite functions is to calculate the two individual composites in terms of x: ƒ1g 1x22 and g 1ƒ1x22. Once those functions are known, the values can be substituted for x and evaluated. Another way of proceeding is as follows: a. Write the desired quantity. ƒ1g 1122 Find the value of the inner function g. g 112 5 5 2 12 5 4 Substitute g 112 5 4 into ƒ. ƒ1g 1122 5 ƒ142 Evaluate ƒ142. ƒ142 5 42 2 7 5 9 ƒ1g1122 5 9 b. Write the desired quantity. ƒ1g 12222 Find the value of the inner function g. g 1222 5 5 2 12222 5 1 Substitute g 1222 5 1 into ƒ. ƒ1g 12222 5 ƒ112 Evaluate ƒ112. ƒ112 5 12 2 7 5 26 ƒ1g12222 5 26 c. Write the desired quantity. g 1ƒ1322 Find the value of the inner function ƒ. ƒ132 5 32 2 7 5 2 Substitute ƒ132 5 2 into g. g 1ƒ1322 5 g 122 Evaluate g 122. g 122 5 5 2 22 5 1 g1ƒ 1322 5 1 d. Write the desired quantity. g 1ƒ12422 Find the value of the inner function ƒ. ƒ1242 5 12422 2 7 5 9 Substitute ƒ1242 5 9 into g. g 1ƒ12422 5 g 192 Evaluate g 192. g 192 5 5 2 92 5 276 g1ƒ 12422 5 276 Y OUR TU R N Given the functions ƒ1x2 5 x3 2 3 and g 1x2 5 1 1 x3, evaluate ƒ1g 1122 and g 1ƒ1122. ▼ ▼ A N S W E R ƒ1g 1122 5 5 and g 1ƒ1122 5 27 3.4 Operations on Functions and Composition of Functions 293 ■ ■The Kelvin (K) temperature scale ■ ■This scale shifts the Celsius scale down so that the zero point is equal to absolute zero (about 2273.158C), a hypothetical temperature at which there is a complete absence of heat energy. ■ ■Temperatures on this scale are called kelvins, not degrees kelvin, and kelvin is not capitalized. The symbol for the kelvin is K. ■ ■The degree Fahrenheit (8F) scale ■ ■This scale evolved over time and is still widely used mainly in the United States, although Celsius is the preferred “metric” scale. ■ ■With respect to pure water at sea level, the degrees Fahrenheit are gauged by the spread from 328F (freezing) to 2128F (boiling). The equations that relate these temperature scales are F 5 9 5 C 1 32 C 5 K 2 273.15 EXAMPLE 7 Applications Involving Composite Functions Determine degrees Fahrenheit as a function of kelvins. Solution: Degrees Fahrenheit is a function of degrees F 5 9 5 C 1 32 Celsius. Now substitute C 5 K 2 273.15 into the F 5 9 5 1K 2 273.152 1 32 equation for F. Simplify. F 5 9 5 K 2 491.67 1 32 F 5 9 5 K 2 459.67 Composition of Functions 1ƒ ° g21x2 5 ƒ1g1x22 The domain restrictions cannot always be determined simply by inspecting the final form of ƒ1g 1x22. Rather, the domain of the composite function is a subset of the domain of g 1x2. Values of x must be eliminated if their corresponding values of g 1x2 are not in the domain of ƒ. Operations on Functions Function Notation Sum 1ƒ 1 g2 1x2 5 ƒ1x2 1 g 1x2 Difference 1ƒ 2 g2 1x2 5 ƒ1x2 2 g 1x2 Product 1ƒ⋅g2 1x2 5 ƒ1x2⋅g 1x2 Quotient aƒ gb1x2 5 ƒ1x2 g1x2 g1x2 2 0 The domain of the sum, difference, and product functions is the intersection of the domains, or common domain shared by both ƒ and g. The domain of the quotient function is also the intersection of the domain shared by both ƒ and g with an additional restriction that g1x2 2 0. [SEC TION 3.4] S UM M A RY 294 CHAPTER 3 Functions and Their Graphs In Exercises 1–10, given the functions ƒ and g, find ƒ 1 g, ƒ 2 g, ƒ . g, and ƒ g, and state the domain of each. 1. ƒ1x2 5 2x 1 1 g 1x2 5 1 2 x 2. ƒ1x2 5 3x 1 2 g 1x2 5 2x 2 4 3. ƒ1x2 5 2x2 2 x g 1x2 5 x2 2 4 4. ƒ1x2 5 3x 1 2 g 1x2 5 x2 225 5. ƒ1x2 5 1 x g 1x2 5 x 6. ƒ1x2 5 2x 1 3 x 2 4 g1x2 5 x 2 4 3x 1 2 7. ƒ1x2 5 !x g1x2 5 2!x 8. ƒ1x2 5 !x 2 1 g 1x2 5 2x2 9. ƒ1x2 5 !4 2 x g1x2 5 !x 1 3 10. ƒ1x2 5 !1 2 2x g1x2 5 1 x In Exercises 11–20, for the given functions ƒ and g, find the composite functions ƒ + g and g + ƒ, and state their domains. 11. ƒ1x2 5 2x 1 1 g 1x2 5 x2 2 3 12. ƒ1x2 5 x2 2 1 g 1x2 5 2 2 x 13. ƒ1x2 5 1 x 2 1 g 1x2 5 x 1 2 14. ƒ1x2 5 2 x 2 3 g 1x2 5 2 1 x 15. ƒ1x2 5 0 x0 g1x2 5 1 x 2 1 16. ƒ1x2 5 0 x 2 10 g1x2 5 1 x 17. ƒ1x2 5 !x 2 1 g 1x2 5 x 1 5 18. ƒ1x2 5 !2 2 x g 1x2 5 x2 1 2 19. ƒ1x2 5 x3 1 4 g 1x2 5 1x 2 421/3 20. ƒ1x2 5 3 "x2 2 1 g 1x2 5 x2/3 1 1 In Exercises 21–38, evaluate the functions for the specified values, if possible. ƒ1x2 5 x2 1 10 g1x2 5 !x 2 1 21. 1ƒ 1 g2 122 22. 1ƒ 1 g2 1102 23. 1ƒ 2 g2 122 24. 1ƒ 2 g2 152 25. 1ƒ ? g2 142 26. 1ƒ ? g2 152 27. aƒ gb1102 28. aƒ gb122 29. ƒ1g 1222 30. ƒ1g 1122 31. g 1ƒ12322 32. g 1ƒ1422 33. ƒ1g 1022 34. g 1ƒ1022 35. ƒ1g 12322 36. gAƒA !7BB 37. 1ƒ + g2142 38. 1g + ƒ 21232 In Exercises 39–50, evaluate ƒ1g1122 and g1ƒ1222, if possible. 39. ƒ1x2 5 1 x, g1x2 5 2x 1 1 40. ƒ1x2 5 x2 1 1, g1x2 5 1 2 2 x 41. ƒ1x2 5 !1 2 x, g1x2 5 x2 1 2 42. ƒ1x2 5 !3 2 x, g1x2 5 x2 1 1 43. ƒ1x2 5 1 0 x 2 1 0 , g1x2 5 x 1 3 44. ƒ1x2 5 1 x, g1x2 5 0 2x 2 3 0 45. ƒ1x2 5 !x 2 1, g1x2 5 x2 1 5 46. ƒ1x2 5 3 !x 2 3, g1x2 5 1 x 2 3 47. ƒ1x2 5 1 x2 2 3, g1x2 5 !x 2 3 48. ƒ1x2 5 x 2 2 x, g1x2 5 4 2 x2 49. ƒ1x2 5 1x 2 121/3, g 1x2 5 x2 1 2x 1 1 50. ƒ1x2 5 11 2 x221/2, g 1x2 5 1x 2 321/3 In Exercises 51–60, show that ƒ1g1x22 5 x and g1ƒ1x22 5 x. 51. ƒ1x2 5 2x 1 1, g1x2 5 x 2 1 2 52. ƒ1x2 5 x 2 2 3 , g1x2 5 3x 1 2 53. ƒ1x2 5 !x 2 1, g1x2 5 x2 1 1 for x $ 1 54. ƒ1x2 5 2 2 x2, g1x2 5 !2 2 x for x # 2 55. ƒ1x2 5 1 x, g1x2 5 1 x for x 2 0 56. ƒ1x2 5 15 2 x21/3, g 1x2 5 5 2 x3 57. ƒ1x2 5 4x2 2 9, g1x2 5 !x 1 9 2 for x $ 0 58. ƒ1x2 5 3 !8x 2 1, g1x2 5 x3 1 1 8 59. ƒ1x2 5 1 x 2 1, g1x2 5 x 1 1 x for x 2 0, x 2 1 60. ƒ1x2 5 "25 2 x2, g1x2 5 "25 2 x2 for 0 # x # 5 [SEC TION 3.4] E X E R C I SE S • S K I L L S In Exercises 61–66, write the function as a composite of two functions ƒ and g. (There is more than one correct answer.) 61. ƒ1g 1x2 2 5 2 13x 2 122 1 5 13x 2 12 62. ƒ1g1x22 5 1 1 1 x2 63. ƒ1g1x22 5 2 0 x 2 3 0 64. ƒ1g1x22 5 "1 2 x2 65. ƒ1g1x22 5 3 !x 1 1 2 2 66. ƒ1g1x22 5 !x 3!x 1 2 • A P P L I C A T I O N S Exercises 67 and 68 depend on the relationship between degrees Fahrenheit, degrees Celsius, and kelvins: F 5 9 5 C 1 32 C 5 K 2 273.15 67. Temperature. Write a composite function that converts kelvins into degrees Fahrenheit. 68. Temperature. Convert the following degrees Fahrenheit to kelvins: 328F and 2128F. 69. Dog Run. Suppose that you want to build a square fenced-in area for your dog. Fencing is purchased in linear feet. a. Write a composite function that determines the area of your dog pen as a function of how many linear feet are purchased. b. If you purchase 100 linear feet, what is the area of your dog pen? c. If you purchase 200 linear feet, what is the area of your dog pen? 70. Dog Run. Suppose that you want to build a circular fenced-in area for your dog. Fencing is purchased in linear feet. a. Write a composite function that determines the area of your dog pen as a function of how many linear feet are purchased. b. If you purchase 100 linear feet, what is the area of your dog pen? c. If you purchase 200 linear feet, what is the area of your dog pen? 71. Market Price. Typical supply and demand relationships state that as the number of units for sale increases, the market price decreases. Assume that the market price p and the number of units for sale x are related by the demand equation: p 5 3000 2 1 2x Assume that the cost C 1x2 of producing x items is governed by the equation C1x2 5 2000 1 10x and the revenue R 1x2 generated by selling x units is governed by R1x2 5 100x a. Write the cost as a function of price p. b. Write the revenue as a function of price p. c. Write the profit as a function of price p. 72. Market Price. Typical supply and demand relationships state that as the number of units for sale increases, the market price decreases. Assume that the market price p and the number of units for sale x are related by the demand equation: p 5 10,000 2 1 4x Assume that the cost C 1x2 of producing x items is governed by the equation C1x2 5 30,000 1 5x and the revenue R 1x2 generated by selling x units is governed by R1x2 5 1000x a. Write the cost as a function of price p. b. Write the revenue as a function of price p. c. Write the profit as a function of price p. In Exercises 73 and 74, refer to the following: The cost of manufacturing a product is a function of the number of hours t the assembly line is running per day. The number of products manufactured n is a function of the number of hours t the assembly line is operating and is given by the function n 1t2. The cost of manufacturing the product C measured in thousands of dollars is a function of the quantity manufactured, that is, the function C 1n2. 73. Business. If the quantity of a product manufactured during a day is given by n1t2 5 50t 2 t2 and the cost of manufacturing the product is given by C1n2 5 10n 1 1375 a. Find a function that gives the cost of manufacturing the product in terms of the number of hours t the assembly line was functioning, C1n1t22. b. Find the cost of production on a day when the assembly line was running for 16 hours. Interpret your answer. 74. Business. If the quantity of a product manufactured during a day is given by n1t2 5 100t 2 4t2 and the cost of manufacturing the product is given by C1n2 5 8n 1 2375 3.4 Operations on Functions and Composition of Functions 295 296 CHAPTER 3 Functions and Their Graphs a. Find a function that gives the cost of manufacturing the product in terms of the number of hours t the assembly line was functioning, C1n1t22. b. Find the cost of production on a day when the assembly line was running for 24 hours. Interpret your answer. In Exercises 75 and 76, refer to the following: Surveys performed immediately following an accidental oil spill at sea indicate the oil moved outward from the source of the spill in a nearly circular pattern. The radius of the oil spill r measured in miles is a function of time t measured in days from the start of the spill, while the area of the oil spill is a function of radius, that is, the function A 1r2. 75. Environment: Oil Spill. If the radius of the oil spill is given by r1t2 5 10t 2 0.2t2 and the area of the oil spill is given by A1r2 5 pr2 a. Find a function that gives the area of the oil spill in terms of the number of days since the start of the spill, A1r1t22. b. Find the area of the oil spill to the nearest square mile 7 days after the start of the spill. 76. Environment: Oil Spill. If the radius of the oil spill is given by r1t2 5 8t 2 0.1t2 and the area of the oil spill is given by A1r2 5 pr2 a. Find a function that gives the area of the oil spill in terms of the number of days since the start of the spill, A1r1t22. b. Find the area of the oil spill to the nearest square mile 5 days after the start of the spill. 77. Environment: Oil Spill. An oil spill makes a circular pattern around a ship such that the radius in feet grows as a function of time in hours r1t2 5 150!t. Find the area of the spill as a function of time. 78. Pool Volume. A 20 foot by 10 foot rectangular pool has been built. If 50 cubic feet of water is pumped into the pool per hour, write the water-level height (feet) as a function of time (hours). 79. Fireworks. A family is watching a fireworks display. If the family is 2 miles from where the fireworks are being launched and the fireworks travel vertically, what is the distance between the family and the fireworks as a function of height above ground? 80. Real Estate. A couple are about to put their house up for sale. They bought the house for $172,000 a few years ago, and when they list it with a realtor they will pay a 6% commission. Write a function that represents the amount of money they will make on their home as a function of the asking price p. • C A T C H T H E M I S T A K E In Exercises 81–86, for the functions ƒ1x2 5 x 1 2 and g1x2 5 x2 2 4, find the indicated function and state its domain. Explain the mistake that is made in each problem. 81. g ƒ Solution: g1x2 ƒ1x2 5 x2 2 4 x 1 2 5 1x 2 221x 1 22 x 1 2 5 x 2 2 Domain: 12q, q2 This is incorrect. What mistake was made? 82. ƒ g Solution: ƒ1x2 g1x2 5 x 1 2 x2 2 4 5 x 1 2 1x 2 221x 1 22 5 1 x 2 2 5 1 x 2 2 Domain: 12q, 22 ∪ 12, q2 This is incorrect. What mistake was made? 83. ƒ + g Solution: ƒ + g 5 ƒ1x2g1x2 5 1x 1 221x2 2 42 5 x3 1 2x2 2 4x 2 8 Domain: 12q, q2 This is incorrect. What mistake was made? 84. Given the function ƒ1x2 5 x2 1 7 and g1x2 5 !x 2 3, find ƒ + g, and state the domain. Solution: ƒ + g 5 ƒ1g1x22 5 A !x 2 3B 2 1 7 5 ƒ1g1x22 5 x 2 3 1 7 5 x 2 4 Domain: 12q, q2 This is incorrect. What mistake was made? 85. 1ƒ 1 g2122 5 1x 1 2 1 x2 2 42122 5 1x2 1 x 2 22122 5 2x2 1 2x 2 4 Domain: 12q, q2 This is incorrect. What mistake was made? 86. ƒ1x2 2 g1x2 5 x 1 2 2 x2 2 4 5 2x2 1 x 2 2 Domain: 12q, q2 This is incorrect. What mistake was made? In Exercises 87–90, determine whether each statement is true or false. 87. When adding, subtracting, multiplying, or dividing two functions, the domain of the resulting function is the union of the domains of the individual functions. 88. For any functions ƒ and g, ƒ1g 1x22 5 g 1ƒ1x22 for all values of x that are in the domain of both ƒ and g. 89. For any functions ƒ and g, 1ƒ + g2 1x2 exists for all values of x that are in the domain of g 1x2, provided the range of g is a subset of the domain of ƒ. 90. The domain of a composite function can be found by inspection, without knowledge of the domain of the individual functions. • C O N C E P T U A L 91. For the functions ƒ1x2 5 x 1 a and g1x2 5 1 x 2 a, find g + ƒ and state its domain. 92. For the functions ƒ1x2 5 ax2 1 bx 1 c and g1x2 5 1 x 2 c, find g + ƒ and state its domain. 93. For the functions ƒ1x2 5 !x 1 a and g 1x2 5 x2 2 a, find g + ƒ and state its domain. 94. For the functions ƒ1x2 5 1 x a and g1x2 5 1 xb, find g + ƒ and state its domain. Assume a . 1 and b . 1. • C H A L L E N G E 95. Using a graphing utility, plot y15 !x 1 7 and y2 5 !9 2 x. Plot y3 5 y1 1 y2. What is the domain of y3? 96. Using a graphing utility, plot y1 5 3 !x 1 5, y2 5 1 !3 2 x, and y3 5 y1 y2 . What is the domain of y3? 97. Using a graphing utility, plot y1 5 "x2 2 3x 2 4, y2 5 1 x2 2 14, and y3 5 1 y2 1 2 14. If y1 represents a function ƒ and y2 represents a function g, then y3 represents the composite function g + ƒ. The graph of y3 is only defined for the domain of g + ƒ. State the domain of g + ƒ. 98. Using a graphing utility, plot y1 5 !1 2 x, y2 5 x2 1 2, and y3 5 y2 1 1 2. If y1 represents a function ƒ and y2 represents a function g, then y3 represents the composite function g + ƒ. The graph of y3 is only defined for the domain of g + ƒ. State the domain of g + ƒ. • T E C H N O L O G Y 3.4 Operations on Functions and Composition of Functions 297 298 CHAPTER 3 Functions and Their Graphs Every human being has a blood type, and every human being has a DNA sequence. These are examples of functions, where a person is the input and the output is blood type or DNA sequence. These relationships are classified as functions because each person can have one and only one blood type or DNA strand. The difference between these functions is that many people have the same blood type, but DNA is unique to each individual. Can we map backwards? For instance, if you know the blood type, do you know specifically which person it came from? No, but, if you know the DNA sequence, you know exactly to which person it corresponds. When a function has a one-to-one correspondence, like the DNA example, then mapping backwards is possible. The map back is called the inverse function. 3.5.1 Determine Whether a Function Is One-to-One In Section 3.1, we defined a function as a relationship that maps an input (contained in the domain) to exactly one output (found in the range). Algebraically, each value for x can correspond to only a single value for y. Recall the square, identity, absolute value, and reciprocal functions from our library of functions in Section 3.3. All of the graphs of these functions satisfy the vertical line test. Although the square function and the absolute value function map each value of x to exactly one value for y, these two functions map two values of x to the same value for y. For example, 121, 12 and 11, 12 lie on both graphs. The identity and reciprocal functions, on the other hand, map each x to a single value for y, and no two x-values map to the same y-value. These two functions are examples of one-to-one functions. 3.5.1 S KILL Determine whether a function is a one-to-one function. 3.5.1 CO NCE PTUAL Understand why a function that passes the horizontal line test is one-to-one. S K I L L S O B J E C T I V E S ■ ■Determine whether a function is a one-to-one function. ■ ■Verify that two functions are inverses of one another. ■ ■Graph the inverse function given the graph of the function. ■ ■Find the inverse of a function. C O N C E P T U A L O B J E C T I V E S ■ ■Understand why a function that passes the horizontal line test is one-to-one. ■ ■Visualize the relationships between the domain and range of a function and the domain and range of its inverse. ■ ■Understand why functions and their inverses are symmetric about y 5 x. ■ ■Understand why a function has to be one-to-one in order for its inverse to exist. 3.5 ONE-TO-ONE FUNCTIONS AND INVERSE FUNCTIONS DEFINITION One-to-One Function A function ƒ1x2 is one-to-one if no two elements in the domain correspond to the same element in the range; that is, if x1 2 x2, then ƒ1x12 2 ƒ1x22. In other words, it is one-to-one if no two inputs map to the same output. EXAMPLE 1 Determining Whether a Function Defined as a Set of Points Is a One-to-One Function For each of the three relations, determine whether the relation is a function. If it is a function, determine whether it is a one-to-one function. ƒ 5 510, 02, 11, 12, 11, 2126 g 5 5121, 12, 10, 02, 11, 126 h 5 5121, 212, 10, 02, 11, 126 3.5 One-to-One Functions and Inverse Functions 299 Just as there is a graphical test for functions, the vertical line test, there is a graphical test for one-to-one functions, the horizontal line test. Note that a horizontal line can be drawn on the square and absolute value functions so that it intersects the graph of each function at two points. The identity and reciprocal functions, however, will intersect a horizontal line in at most only one point. This leads us to the horizontal line test for one-to-one functions. DEFINITION Horizontal Line Test If every horizontal line intersects the graph of a function in at most one point, then the function is classified as a one-to-one function. [CONCEPT CHECK] Draw a horizontal line at y 5 4 on the graph of y 5 x2. What are the two points of intersection? Explain why the equation y 5 x2 cannot be a one-to-one function. ANSWER (22, 4) and (2, 4): Two different x values map to the same y value, so this graph fails the horizontal line test. ▼ EXAMPLE 2 Using the Horizontal Line Test to Determine Whether a Function Is One-to-One For each of the three relations, determine whether the relation is a function. If it is a function, determine whether it is a one-to-one function. Assume that x is the independent variable and y is the dependent variable. x 5 y2 y 5 x2 y 5 x3 Solution: x 5 y2 y 5 x2 y 5 x3 Y OUR T UR N Determine whether each of the functions is a one-to-one function. a. ƒ1x2 5 x 1 2 b. ƒ1x2 5 x2 1 1 ▼ x y 10 5 –5 x y –5 5 10 x y –5 5 –10 10 Not a function One-to-one function Function but not one-to-one (fails vertical line test) (passes vertical line test but fails horizontal line test) (passes both horizontal and vertical line tests) ▼ A N S W E R a. yes b. no Solution: h is a one-to-one function. f is not a function. g is a function but not one-to-one. 300 CHAPTER 3 Functions and Their Graphs Another way of writing the definition of a one-to-one function is: If ƒ1x12 5 ƒ1x22, then x1 5 x2. In the Your Turn following Example 2, we found (using the horizontal line test) that ƒ1x2 5 x 1 2 is a one-to-one function but that ƒ1x2 5 x2 1 1 is not a one-to-one function. We can also use this alternative definition to determine algebraically whether a function is one-to-one. WORDS MATH State the function. ƒ1x2 5 x 1 2 Let there be two real numbers, x1 and x2, such that ƒ1x12 5 ƒ1x22. x1 1 2 5 x2 1 2 Subtract 2 from both sides of the equation. x1 5 x2 ƒ1x2 5 x 1 2 is a one-to-one function. WORDS MATH State the function. ƒ1x2 5 x2 1 1 Let there be two real numbers, x1 and x2, such that ƒ1x12 5 ƒ1x22. x2 1 1 1 5 x2 2 1 1 Subtract 1 from both sides of the equation. x2 1 5 x2 2 Solve for x1. x1 5 ± x2 ƒ1x2 5 x2 1 2 is not a one-to-one function. EXAMPLE 3 Determining Algebraically Whether a Function Is One-to-One Determine algebraically whether the following functions are one-to-one: a. ƒ1x2 5 5x3 2 2 b. ƒ1x2 50 x 1 10 Solution (a): Find ƒ1x12 and ƒ1x22. ƒ1x12 5 5x3 1 2 2 and ƒ1x22 5 5x3 2 2 2 Let ƒ1x12 5 ƒ1x22. 5x3 1 2 2 5 5x3 2 2 2 Add 2 to both sides of the equation. 5x3 1 5 5x3 2 Divide both sides of the equation by 5. x3 1 5 x3 2 Take the cube root of both sides of the equation. Ax3 1B 1/3 5 Ax3 2B 1/3 Simplify. x1 5 x2 ƒ1x2 5 5x3 2 2 is a one-to-one function. Solution (b): Find ƒ1x12 and ƒ1x22. ƒ1x12 5 0 x1 1 10 and ƒ1x22 5 0 x2 1 10 Let ƒ1x12 5 ƒ1x22. 0 x1 1 10 5 0 x2 1 10 Solve the absolute value equation. 1x1 1 12 5 1x2 1 12 or 1x1 1 12 5 21x2 1 12 x1 5 x2 or x1 5 2x2 2 2 ƒ1x2 5 0 x 1 10 is not a one-to-one function. 3.5.2 Inverse Functions If a function is one-to-one, then the function maps each x to exactly one y, and no two x-values map to the same y-value. This implies that there is a one-to-one correspondence between the inputs (domain) and outputs (range) of a one-to-one function ƒ 1x2. In the special case of a one-to-one function, it would be possible to map from the output (range of ƒ ) back to the input (domain of ƒ ), and this mapping would also be a function. The function that maps the output back to the input of a function ƒ is called the inverse function and is denoted ƒ21 1x2. A one-to-one function ƒ maps every x in the domain to a unique and distinct corresponding y in the range. Therefore, the inverse function ƒ21 maps every y back to a unique and distinct x. The function notations ƒ 1x2 5 y and ƒ 21 1 y2 5 x indicate that if the point 1x, y2 sat-isfies the function, then the point 1 y, x2 satisfies the inverse function. For example, let the function h 1x2 5 5 121, 02, 11, 22, 13, 426. 3.5.2 S K I L L Verify that two functions are inverses of one another. 3.5.2 C ON C E P T U A L Visualize the relationships between the domain and range of a function and the domain and range of its inverse. x f y Range of f 1 Domain of f 1 Domain of f Range of f f 1 The inverse function undoes whatever the function does. For example, if ƒ1x2 5 5x, then the function ƒ maps any value x in the domain to a value 5x in the range. If we want to map backwards or undo the 5x, we develop a function called the inverse function that takes 5x as input and maps back to x as output. The inverse function is ƒ211x2 5 1 5x. Note that if we input 5x into the inverse function, the output is x: ƒ2115x2 5 1 515x2 5 x. DEFINITION Inverse Function If ƒ and g denote two one-to-one functions such that ƒ1g 1x22 5 x for every x in the domain of g and g 1 ƒ 1x22 5 x for every x in the domain of ƒ , then g is the inverse of the function ƒ. The function g is denoted by ƒ21 (read “f-inverse”). Domain of ƒ 5 range of ƒ21 and range of ƒ 5 domain of ƒ21 ƒ21 1ƒ 1x22 5 x and ƒ 1ƒ21 1x22 5 x Note: ƒ21 is used to denote the inverse of ƒ. The 21 is not used as an exponent and, therefore, does not represent the reciprocal of ƒ: 1 f . Two properties hold true relating one-to-one functions to their inverses: (1) the range of the function is the domain of the inverse, and the range of the inverse is the domain of the function, and (2) the composite function that results with a function and its inverse (and vice versa) is the identity function x. x f (x) = 5x 5x Domain of f Range of f ? x f(x) = 5x f 1(5x) = x 5x ▼ C A U T I O N f 21 2 1 f 3.5 One-to-One Functions and Inverse Functions 301 302 CHAPTER 3 Functions and Their Graphs [CONCEPT CHECK] If a one-to-one function f has domain [a, q) and range [b, q) then what are the domain and range of its inverse, f 21? ANSWER The inverse function, f 21, has domain [b, q) and range [a, q). ▼ EXAMPLE 4 Verifying Inverse Functions Verify that ƒ211x2 5 1 2x 2 2 is the inverse of ƒ 1x2 5 2x 1 4. Solution: Show that ƒ 21 1 ƒ 1x22 5 x and ƒ 1 ƒ 21 1x22 5 x. Write ƒ21 using placeholder notation. ƒ211u2 5 1 2 1u2 2 2 Substitute ƒ1x2 5 2x 1 4 into ƒ21. ƒ 21 1 ƒ 1x22 5 1 2 12x 1 42 2 2 Simplify. ƒ 21 1 ƒ 1x22 5 x 1 2 2 2 5 x ƒ 21 1 ƒ 1x22 5 x Write ƒ using placeholder notation. ƒ1u2 5 21u2 1 4 Substitute ƒ211x2 5 1 2x 2 2 into ƒ. ƒ 1 ƒ 21 1x22 5 2a 1 2 x 2 2 b 1 4 Simplify. ƒ 1 ƒ21 1x22 5 x 2 4 1 4 5 x ƒ 1 ƒ21 1x22 5 x Note the relationship between the domain and range of ƒ and ƒ21. DOMAIN RANGE ƒ 1x2 5 2x 1 4 12q, q2 12q, q2 f 211x2 5 1 2x 2 2 12q, q2 12q, q2 EXAMPLE 5 Verifying Inverse Functions with Domain Restrictions Verify that ƒ21 1x2 5 x2, for x $ 0, is the inverse of ƒ1x2 5 !x. Solution: Show that ƒ21 1 ƒ 1x22 5 x and ƒ 1 ƒ21 1x22 5 x. Write ƒ21 using placeholder notation. ƒ21 1u2 5 1u22 Substitute ƒ1x2 5 !x into ƒ21. ƒ 211ƒ1x22 5 1 !x22 5 x ƒ21 1 ƒ 1x22 5 x for x # 0 Write ƒ using placeholder notation. ƒ1u2 5"1u2 Substitute ƒ21 1x2 5 x2, x $ 0 into ƒ. ƒ 1 ƒ21 1x22 5 !x2 5 x, x # 0 ƒ 1 ƒ21 1x22 5 x for x # 0 DOMAIN RANGE ƒ1x2 5 !x 30, q2 30, q2 ƒ21 1x2 5 x2, x $ 0 30, q2 30, q2 3.5.3 Graphical Interpretation of Inverse Functions In Example 4, we showed that ƒ211x2 5 1 2x 2 2 is the inverse of ƒ 1x2 5 2x 1 4. Let’s now investigate the graphs that correspond to the function ƒ and its inverse ƒ 21. 3.5.3 S K I L L Graph the inverse function given the graph of the function. 3.5.3 C O N C E P T U A L Understand why functions and their inverses are symmetric about y 5 x. STUDY TIP If the point (a, b) is on the function, then the point (b, a) is on the inverse. Notice the interchanging of the x- and y-coordinates. x y (–2, 0) (–1, 2) (0, 4) (–3, –2) (–2, –3) (0, –2) (2, –1) (4, 0) f (x) x y 23 22 22 0 21 2 0 4 f 21(x) x y 22 23 0 22 2 21 4 0 Note that the point 123, 222 lies on the function and the point 122, 232 lies on the inverse. In fact, every point 1a, b2 that lies on the function corresponds to a point 1b, a2 that lies on the inverse. Draw the line y 5 x on the graph. In general, the point 1b, a2 on the inverse ƒ21 1x2 is the reflection 1about y 5 x2 of the point 1a, b2 on the function ƒ 1x2. In general, if the point 1a, b2 is on the graph of a function, then the point 1b, a2 is on the graph of its inverse. [CONCEPT CHECK] If the point (b, a ) lies on the graph of a one-to-one function f, then what point must lie on the graph of its inverse, f 21? ANSWER (a, b) ▼ ▼ A N S W E R x y EXAMPLE 6 Graphing the Inverse Function Given the graph of the function ƒ1x2, plot the graph of its inverse ƒ21 1x2. Solution: Because the points 123, 222, 122, 02, 10, 22, and 12, 42 lie on the graph of ƒ, then the points 122, 232, 10, 222, 12, 02, and 14, 22 lie on the graph of ƒ21. Y OUR T UR N Given the graph of a function ƒ, plot the inverse function. ▼ x f (x) y x y (–2, 0) (0, 2) (2, 4) (–3, –2) (–2, –3) (0, –2) (2, 0) (4, 2) x y 3.5 One-to-One Functions and Inverse Functions 303 304 CHAPTER 3 Functions and Their Graphs We have developed the definition of an inverse function and described properties of inverses. At this point, you should be able to determine whether two functions are inverses of one another. Let’s turn our attention to another problem: How do you find the inverse of a function? 3.5.4 Finding the Inverse Function If the point 1a, b2 lies on the graph of a function, then the point 1b, a2 lies on the graph of the inverse function. The symmetry about the line y 5 x tells us that the roles of x and y interchange. Therefore, if we start with every point 1x, y2 that lies on the graph of a function, then every point 1y, x2 lies on the graph of its inverse. Algebraically, this corresponds to interchanging x and y. Finding the inverse of a finite set of ordered pairs is easy: simply interchange the x- and y-coordinates. Earlier, we found that if h 1x2 5 5 121, 02, 11, 22, 13, 426, then h21 1x2 5 5 10, 212, 12, 12, 14, 326. But how do we find the inverse of a function defined by an equation? Recall the mapping relationship if ƒ is a one-to-one function. This relationship implies that ƒ 1x2 5 y and ƒ21 1 y2 5 x. Let’s use these two identities to find the inverse. Now consider the function defined by ƒ1x2 5 3x 2 1. To find ƒ21, we let ƒ 1x2 5 y, which yields y 5 3x 2 1. Solve for the variable x:x 5 1 3y 1 1 3. Recall that ƒ21 1 y2 5 x, so we have found the inverse to be ƒ211y2 5 1 3y 1 1 3. It is customary to write the independent variable as x, so we write the inverse as ƒ211 x2 5 1 3x 1 1 3. Now that we have found the inverse, let’s confirm that the properties ƒ21 1ƒ 1x22 5 x and ƒ 1ƒ 21 1x22 5 x hold. ƒAƒ211x2B 5 3a1 3 x 1 1 3b 2 1 5 x 1 1 2 1 5 x ƒ211ƒ1x22 5 1 313x 2 12 1 1 3 5 x 2 1 3 1 1 3 5 x 3.5.4 S KILL Find the inverse of a function. 3.5.4 CO NCE PTUAL Understand why a function has to be one-to-one in order for its inverse to exist. x f f 1 f (x) Domain of f Range of f f 1(y) y Range of f 1 Domain of f 1 FINDING THE INVERSE OF A FUNCTION Let ƒ be a one-to-one function. Then the following procedure can be used to find the inverse function ƒ 21 if the inverse exists. STEP PROCEDURE EXAMPLE 1 Let y 5 ƒ 1x2. ƒ1x2 5 23x 1 5 y 5 23x 1 5 2 Solve the resulting equation for x in terms of y (if possible). 3x 5 2y 1 5 x 5 21 3 y 1 5 3 3 Let x 5 ƒ21 1 y2. ƒ211 y2 5 21 3 y 1 5 3 4 Let y 5 x 1interchange x and y2. ƒ211x2 5 21 3 x 1 5 3 The same result is found if we first interchange x and y and then solve for y in terms of x. STEP PROCEDURE EXAMPLE 1 Let y 5 ƒ 1x2. ƒ1x2 5 23x 1 5 y 5 23x 1 5 2 Interchange x and y. x 5 23y 1 5 3 Solve for y in terms of x. 3y 5 2x 1 5 y 5 21 3 x 1 5 3 4 Let y 5 ƒ21 1x2 ƒ211x2 5 21 3x 1 5 3 EXAMPLE 7 The Inverse of a Square Root Function Find the inverse of the function ƒ1x2 5 !x 1 2. State the domain and range of both ƒ and ƒ21. Solution: ƒ 1x2 is a one-to-one function because it passes the horizontal line test. STEP 1 Let y 5 ƒ 1x2. y 5 !x 1 2 STEP 2 Interchange x and y. x 5 !y 1 2 STEP 3 Solve for y. Square both sides of the equation. x2 5 y 1 2 Subtract 2 from both sides. x2 2 2 5 y or y 5 x2 2 2 STEP 4 Let y 5 ƒ21 1x2. ƒ21 1x2 5 x2 2 2 Note any domain restrictions. (State the domain and range of both ƒ and ƒ21.) ƒ: Domain: 322, q2 Range: 30, q2 ƒ21: Domain: 30, q2 Range: 322, q2 The inverse of ƒ1x2 5 !x 1 2 is ƒ211x2 5 x2 2 2 for x $ 0 . Check: ƒ21 1ƒ 1x22 5 x for all x in the ƒ211ƒ1x22 5 A!x 1 2B 2 2 2 domain of ƒ. 5 x 1 2 2 2 for x $ 22 5 x ƒ 1ƒ21 1x22 5 x for all x in the ƒ1ƒ211x22 5 "Ax2 2 2B 1 2 domain of ƒ 21. 5 "x2 for x $ 0 5 x x y –5 5 –5 5 3.5 One-to-One Functions and Inverse Functions 305 Note the following: ■ ■Verify first that a function is one-to-one prior to finding an inverse (if it is not one-to-one, then the inverse does not exist). ■ ■State the domain restrictions on the inverse function. The domain of ƒ is the range of ƒ21 and vice versa. ■ ■To verify that you have found the inverse, show that ƒ 1 ƒ21 1x22 5 x for all x in the domain of ƒ21 and ƒ21 1 ƒ 1x22 5 x for all x in the domain of ƒ. STUDY TIP Had we ignored the domain and range in Example 7, we would have found the inverse function to be the square function f (x ) 5 x2 2 2, which is not a one-to-one function. It is only when we restrict the domain of the square function that we get a one-to-one function. 306 CHAPTER 3 Functions and Their Graphs Note that the function ƒ1x2 5 !x 1 2 and its inverse ƒ21 1x2 5 x2 2 2 for x $ 0 are symmetric about the line y 5 x. Y OUR TU R N Find the inverse of the given function. State the domain and range of the inverse function. a. ƒ 1x2 5 7x 2 3 b. g1x2 5 !x 2 1 ▼ x y –5 5 –5 5 f(x) f –1(x) ▼ A N S W E R a. ƒ211x2 5 x 1 3 7 Domain: 12q, q2, Range: 12q, q2 b. g21 1x2 5 x2 1 1 Domain: 30, q2, Range: 31, q2 STUDY TIP The range of a function is equal to the domain of its inverse function. EXAMPLE 9 Finding the Inverse Function The function ƒ1x2 5 2 x 1 3, x 2 23, is a one-to-one function. Find its inverse. Solution: STEP 1 Let y 5 ƒ1x2. y 5 2 x 1 3 STEP 2 Interchange x and y. x 5 2 y 1 3 STEP 3 Solve for y. Multiply the equation by 1y 1 32. xA y 1 3B 5 2 Eliminate the parentheses. xy 1 3x 5 2 Subtract 3x from both sides. xy 5 23x 1 2 Divide the equation by x. y 5 23x 1 2 x 5 23 1 2 x STEP 4 Let y 5 ƒ21AxB. ƒ211x2 5 23 1 2 x Note any domain restrictions on ƒ21 1 x2 . x 2 0 The inverse of the function ƒ1x2 5 2 x 1 3, x 2 23, is ƒ 211x2 5 23 1 2 x, x 2 0 . EXAMPLE 8 A Function That Does Not Have an Inverse Function Find the inverse of the function ƒ1x2 5 0 x0 if it exists. Solution: The function ƒ1x2 5 0 x0 fails the horizontal line test and therefore is not a one-to-one function. Because ƒ is not a one-to-one function, its inverse function does not exist. x y f (x) = \|x\| Check: ƒ211ƒ1x22 5 23 1 2 a 2 x 1 3b 5 23 1 1x 1 32 5 x, x 2 23 ƒAƒ211x2B 5 2 a 23 1 2 xb 1 3 5 2 a2 xb 5 x, x 2 0 YOUR T UR N The function ƒ1x2 5 4 x 2 1, x 2 1, is a one-to-one function. Find its inverse. ▼ A N S W E R ƒ211x2 5 1 1 4 x, x 2 0 ▼ Note in Example 9 that the domain of ƒ is 12 q , 232 ∪ 123, q 2 and the domain of ƒ 21 is 12q, 02 ∪ 10, q2. Therefore, we know that the range of ƒ is 12q, 02 ∪ 10, q2, and the range of ƒ 21 is 12 q , 232 ∪ 123, q 2. [CONCEPT CHECK] Explain why you cannot find the inverse of f (x) 5 x2 without restricting the domain? ANSWER f (x) = x2 is not a one-to-one function. ▼ 3.5 One-to-One Functions and Inverse Functions 307 x y –5 5 25 –25 f(x) = x2 f(x) = 3x (0, 0) EXAMPLE 10 Finding the Inverse of a Piecewise-Defined Function The function ƒ1x2 5 b3x x , 0 x2 x $ 0 is a one-to-one function. Find its inverse. Solution: From the graph of ƒ we can make a table with corresponding domain and range values. DOMAIN OF f RANGE OF f 12q, 02 12q, 02 30, q2 30, q2 From this information we can also list domain and range values for ƒ 21. DOMAIN OF f 21 RANGE OF f 21 12q, 02 12q, 02 30, q2 30, q2 ƒ1x2 5 3x on 12q, 02; find ƒ21 1x2 on 12q, 02. STEP 1 Let y 5 ƒ 1x2. y 5 3x STEP 2 Solve for x in terms of y. x 5 3y STEP 3 Solve for y. y 5 1 3x STEP 4 Let y 5 ƒ21 1x2. ƒ211x2 5 1 3x on 12q, 02 ƒ1x2 5 x2 on 30, q2; find ƒ21 1x2 on 30, q2. STEP 1 Let y 5 ƒ 1x2. y 5 x2 STEP 2 Solve for x in terms of y. x 5 y2 STEP 3 Solve for y. y 5 6!x STEP 4 Let y 5 ƒ21 1x2. ƒ211x2 5 6!x STEP 5 The range of ƒ21 is 30, q2 ƒ211x2 5 !x Combining the two pieces yields a piecewise-defined inverse function. ƒ 211x2 5 c 1 3 x x , 0 !x x $ 0 308 CHAPTER 3 Functions and Their Graphs In Exercises 1–16, determine whether the given relation is a function. If it is a function, determine whether it is a one-to-one function. 1. Domain Range 78°F 68°F AVERAGE TEMPERATURE MONTH October January April 2. Domain Range (202) 555-1212 (307) 123-4567 (878) 799-6504 10-DIGIT PHONE # PERSON Mary Jason Chester 3. Domain Range Chris Alex Morgan SPOUSE PERSON Jordan Pat Tim 4. Domain Range A B COURSE GRADE Carrie Michael Jennifer Sean PERSON 5. 5 10, 12, 11, 22, 12, 32, 13, 426 6. 5 10, 222, 12, 02, 15, 32, 125, 2726 7. 5 10, 02, 19, 232, 14, 222, 14, 22, 19, 326 8. 5 10, 12, 11, 12, 12, 12, 13, 126 9. 5 10, 12, 11, 02, 12, 12, 122, 12, 15, 42, 123, 426 10. 5 10, 02, 121, 212, 122, 282, 11, 12, 12, 826 11. x y (6, 4) (0, 1) (–1, 1) (–2, –2) (2, –2) 12. x y 13. x y [SEC TION 3.5] E X E R C I SE S • S K I L L S Properties of Inverse Functions 1. If ƒ is a one-to-one function, then ƒ21 exists. 2. Domain and range • Domain of ƒ 5 range of ƒ21 • Domain of ƒ21 5 range of ƒ 3. Composition of inverse functions • ƒ21 1ƒ 1x22 5 x for all x in the domain of ƒ. • ƒ 1ƒ21 1x22 5 x for all x in the domain of ƒ21. 4. The graphs of ƒ and ƒ21 are symmetric with respect to the line y 5 x. Procedure for Finding the Inverse of a Function 1. Let y 5 ƒ 1x2. 2. Interchange x and y. 3. Solve for y. 4. Let y 5 ƒ21 1x2. One-to-One Functions Each input in the domain corresponds to exactly one output in the range, and no two inputs map to the same output. There are three ways to test a function to determine whether it is a one-to-one function. 1. Discrete points: For the set of all points 1a, b2 verify that no y-values are repeated. 2. Algebraic equations: Let ƒ 1x12 5 ƒ 1x22; if it can be shown that x1 5 x2, then the function is one-to-one. 3. Graphs: Use the horizontal line test; if any horizontal line intersects the graph of the function in more than one point, then the function is not one-to-one. [SEC TION 3.5] S U M M A RY In Exercises 17–24, determine algebraically and graphically whether the function is one-to-one. 17. ƒ1x2 5 0 x 2 30 18. ƒ1x2 5 1x 2 222 1 1 19. ƒ1x2 5 1 x 2 1 20. ƒ1x2 5 3 !x 21. ƒ1x2 5 x2 2 4 22. ƒ1x2 5 !x 1 1 23. ƒ1x2 5 x3 2 1 24. ƒ1x2 5 1 x 1 2 In Exercises 25–34, verify that the function ƒ21 1x2 is the inverse of ƒ 1x2 by showing that ƒ 1 ƒ21 1x22 5 x and ƒ21 1 ƒ 1x22 5 x. Graph ƒ 1x2 and ƒ21 1x2 on the same axes to show the symmetry about the line y 5 x. 25. ƒ1x2 5 2x 1 1; ƒ 211x2 5 x 2 1 2 26. ƒ1x2 5 x 2 2 3 ; ƒ 211x2 5 3x 1 2 27. ƒ1x2 5 !x 2 1, x $ 1; ƒ 211x2 5 x2 1 1 , x $ 0 28. ƒ1x2 5 2 2 x2 , x $ 0 ; ƒ 211x2 5 !2 2 x, x # 2 29. ƒ1x2 5 1 x ; ƒ 211x2 5 1 x , x 2 0 30. ƒ 1x2 5 15 2 x21/3; ƒ21 1x2 5 5 2 x3 31. ƒ1x2 5 1 2x 1 6 , x 2 23 ; ƒ 211x2 5 1 2x 2 3, x 2 0 32. ƒ1x2 5 3 4 2 x , x 2 4; ƒ211x2 5 4 2 3 x , x 2 0 33. ƒ1x2 5 x 1 3 x 1 4 , x 2 24 ; ƒ 211x2 5 3 2 4x x 2 1 , x 2 1 34. ƒ1x2 5 x 2 5 3 2 x , x 2 3 ; ƒ211x2 5 3x 1 5 x 1 1 , x 2 21 In Exercises 35–42, graph the inverse of the one-to-one function that is given. 35. x y (1, 1) (–1, –3) 36. x y (3, 1) (–3, 3) 37. x y 8 –2 10 38. x y 10 5 –5 39. x y 2 –8 5 40. x y –5 5 –10 10 41. x y 10 5 42. x y –10 5 14. x y (2, –1) (–2, 3) 15. x y –5 5 10 16. x y 10 5 3.5 One-to-One Functions and Inverse Functions 309 310 CHAPTER 3 Functions and Their Graphs In Exercises 43–60, the function ƒ is one-to-one. Find its inverse, and check your answer. State the domain and range of both ƒ and ƒ21. 43. ƒ1x2 5 x 2 1 44. ƒ1x2 5 7x 45. ƒ1x2 5 23x 1 2 46. ƒ1x2 5 2x 1 3 47. ƒ1x2 5 x3 1 1 48. ƒ1x2 5 x3 2 1 49. ƒ1x2 5 !x 2 3 50. ƒ1x2 5 !3 2 x 51. ƒ1x2 5 x2 2 1, x $ 0 52. ƒ1x2 5 2x2 1 1, x $ 0 53. ƒ1x2 5 1x 1 222 2 3, x $ 22 54. ƒ1x2 5 1x 2 322 2 2, x $ 3 55. ƒ1x2 5 2 x 56. ƒ1x2 5 23 x 57. ƒ1x2 5 2 3 2 x 58. ƒ1x2 5 7 x 1 2 59. ƒ1x2 5 7x 1 1 5 2 x 60. ƒ1x2 5 2x 1 5 7 1 x In Exercises 61–64, graph the piecewise-defined function to determine whether it is a one-to-one function. If it is a one-to-one function, find its inverse. 61. G1x2 5 e 0 x , 0 !x x $ 0 62. G1x2 5 • 1 x x , 0 !x x $ 0 63. ƒ1x2 5 • x x # 21 x3 21 , x , 1 x x $ 1 64. ƒ1x2 5 • x 1 3 x # 22 0 x 0 22 , x , 2 x2 x $ 2 • A P P L I C A T I O N S 65. Temperature. The equation used to convert from degrees Celsius to degrees Fahrenheit is ƒ1x2 5 9 5 x 1 32. Determine the inverse function ƒ 21 1x2. What does the inverse function represent? 66. Temperature. The equation used to convert from degrees Fahrenheit to degrees Celsius is C1x2 5 5 9 1x 2 322. Determine the inverse function C 21 1x2. What does the inverse function represent? 67. Budget. The Richmond rowing club is planning to enter the Head of the Charles race in Boston and is trying to figure out how much money to raise. The entry fee is $250 per boat for the first 10 boats and $175 for each additional boat. Find the cost function C 1x2 as a function of the number of boats the club enters x. Find the inverse function that will yield how many boats the club can enter as a function of how much money it will raise. 68. Long-Distance Calling Plans. A phone company charges $0.39 per minute for the first 10 minutes of a long-distance phone call and $0.12 per minute every minute after that. Find the cost function C 1x2 as a function of the length of the phone call in minutes x. Suppose you buy a “prepaid” phone card that is planned for a single call. Find the inverse function that determines how many minutes you can talk as a function of how much you prepaid. 69. Salary. A student works at Target making $10 per hour, and the weekly number of hours worked per week x varies. If Target withholds 25% of his earnings for taxes and Social Security, write a function E 1x2 that expresses the student’s take-home pay each week. Find the inverse function E21 1x2. What does the inverse function tell you? 70. Salary. A grocery store pays you $8 per hour for the first 40 hours per week and time and a half for overtime. Write a piecewise-defined function that represents your weekly earn-ings E 1x2 as a function of the number of hours worked x. Find the inverse function E 21 1x2. What does the inverse function tell you? In Exercises 71–74, refer to the following: By analyzing available empirical data it was determined that during an illness a patient’s body temperature fluctuated during one 24-hour period according to the function T1t2 5 0.00031t 2 2423 1 101.70 where T represents that patient’s temperature in degrees Fahrenheit and t represents the time of day in hours measured from 12:00 a.m. (midnight). 71. Health/Medicine. Find the domain and range of the function T 1t2. 72. Health/Medicine. Find time as a function of temperature, that is, the inverse function t 1T2. 73. Health/Medicine. Find the domain and range of the function t 1T2 found in Exercise 72. 74. Health/Medicine. At what time, to the nearest hour, was the patient’s temperature 99.5°F? • C A T C H T H E M I S T A K E In Exercises 75–78, explain the mistake that is made. 75. Is x 5 y2 a one-to-one function? Solution: Yes, this graph represents a one-to-one function because it passes the horizontal line test. This is incorrect. What mistake was made? x y 3.5 One-to-One Functions and Inverse Functions 311 76. A linear one-to-one function is graphed below. Draw its inverse. Solution: Note that the points 13, 32 and 10, 242 lie on the graph of the function. By symmetry, the points 123, 232 and 10, 42 lie on the graph of the inverse. This is incorrect. What mistake was made? x y (3, 3) (0, –4) x y (3, 3) (0, –4) (0, 4) (–3, –3) 77. Given the function ƒ 1x2 5 x2, find the inverse function ƒ21 1x2. Solution: STEP 1: Let y 5 ƒ 1x2. y 5 x2 STEP 2: Solve for x. x 5 !y STEP 3: Interchange x and y. y 5 !x STEP 4: Let y 5 ƒ21 1x2. ƒ 211x2 5 !x Check: ƒAƒ 211x2B 5 A !x B 2 5 x and ƒ 211ƒ1x22 5 "x2 5 x. The inverse of ƒ 1x2 5 x2 is ƒ 211x2 5 !x. This is incorrect. What mistake was made? 78. Given the function ƒ1x2 5 !x 2 2, find the inverse function ƒ21 1x2, and state the domain restrictions on ƒ21 1x2. Solution: STEP 1: Let y 5 ƒ 1x2. y 5 !x 2 2 STEP 2: Interchange x and y. x 5 !y 2 2 STEP 3: Solve for y. y 5 x2 1 2 STEP 4: Let ƒ 21 1x2 5 y. ƒ21 1x2 5 x2 1 2 STEP 5: Domain restrictions: ƒ1x2 5 !x 2 2 has the domain restriction that x $ 2. The inverse of ƒ1x2 5 !x 2 2 is ƒ21 1x2 5 x2 1 2. The domain of ƒ21 1x2 is x $ 2. This is incorrect. What mistake was made? In Exercises 89–92, graph the following functions and determine whether they are one-to-one. 89. ƒ 1x2 5 0 4 2 x20 90. ƒ1x2 5 3 x3 1 2 91. ƒ 1x2 5 x1/3 2 x5 92. ƒ1x2 5 1 x1/2 In Exercises 93–96, graph the functions ƒ and g and the line y 5 x in the same screen. Do the two functions appear to be inverses of each other? 93. ƒ1x2 5 !3x 2 5; g1x2 5 x2 3 1 5 3 94. ƒ1x2 5 !4 2 3x; g1x2 5 4 3 2 x2 3 , x $ 0 95. ƒ 1x2 5 1x 2 721/3 1 2; g 1x2 5 x3 2 6x2 1 12x 2 1 96. ƒ1x2 5 ! 3 x 1 3 2 2; g1x2 5 x3 1 6x2 1 12x 1 6 • T E C H N O L O G Y 85. The unit circle is not a function. If we restrict ourselves to the semicircle that lies in quadrants I and II, the graph represents a function, but it is not a one-to-one function. If we further restrict ourselves to the quarter circle lying in quadrant I, the graph does represent a one-to-one function. Determine the equations of both the one-to-one function and its inverse. State the domain and range of both. 86. Find the inverse of ƒ1x2 5 c x , c 2 0. 87. Under what conditions is the linear function ƒ 1x2 5 mx 1 b a one-to-one function? 88. Assuming that the conditions found in Exercise 87 are met, determine the inverse of the linear function. • C H A L L E N G E In Exercises 79–82, determine whether each statement is true or false. 79. Every even function is a one-to-one function. 80. Every odd function is a one-to-one function. 81. It is not possible that ƒ 5 ƒ21. 82. A function ƒ has an inverse. If the function lies in quadrant II, then its inverse lies in quadrant IV. 83. If 10, b2 is the y-intercept of a one-to-one function ƒ, what is the x-intercept of the inverse ƒ21? 84. If 1a, 02 is the x-intercept of a one-to-one function ƒ, what is the y-intercept of the inverse ƒ21? • C O N C E P T U A L 312 CHAPTER 3 Functions and Their Graphs S K I L L S O B J E C T I V E S ■ ■Develop mathematical models using direct variation. ■ ■Develop mathematical models using inverse variation. ■ ■Develop mathematical models using joint variation and combined variation. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that direct variation implies that two things grow with one another (as one increases, the other increases). ■ ■Understand that inverse variation implies that two things grow opposite of one another (as one increases, the other decreases). ■ ■Understand the difference between combined variation and joint variation. 3.6 MODELING FUNCTIONS USING VARIATION In this section, we discuss mathematical models for different applications. Two quantities in the real world often vary with respect to one another. Sometimes they vary directly. For example, the more money we make, the more total dollars of federal income tax we expect to pay. Sometimes quantities vary inversely. For example, when interest rates on mortgages decrease, we expect the number of homes purchased to increase because a buyer can afford “more house” with the same mortgage payment when rates are lower. In this section, we discuss quantities varying directly, inversely, and jointly. 3.6.1 Direct Variation When one quantity is a constant multiple of another quantity, we say that the quantities are directly proportional to one another. In 2015, the national average cost of residential electricity was 13.06 ¢/kWh (cents per kilowatt-hour). For example, if a residence used 3400 kWh, then the bill would be $444.04, and if a residence used 2500 kWh, then the bill would be $326.50. 3.6.1 S KILL Develop mathematical models using direct variation. 3.6.1 CO NCE PTUAL Understand that direct variation implies that two things grow with one another (as one increases, the other increases). DIRECT VARIATION Let x and y represent two quantities. The following are equivalent statements: ■ ■y 5 kx, where k is a nonzero constant. ■ ■y varies directly with x. ■ ■y is directly proportional to x. The constant k is called the constant of variation or the constant of proportionality. Not all variation we see in nature is direct variation. Isometric growth, where the various parts of an organism grow in direct proportion to each other, is rare in living organisms. If organisms grew isometrically, young children would look just like adults, only smaller. In contrast, most organisms grow nonisometrically; the various parts of organisms do not increase in size in a one-to-one ratio. The relative proportions of a human body change dramatically as the human grows. Children have proportionately larger heads and shorter legs than adults. Allometric growth is the pattern of growth whereby different parts of the body grow at different rates with respect to each other. Some human body characteristics vary directly, and others can be mathematically modeled by direct variation with powers. EXAMPLE 1 Finding the Constant of Variation In the United States, the cost of electricity is directly proportional to the number of kilowatt-hours (kWh) used. If a household in Tennessee on average used 3098 kWh per month and had an average monthly electric bill of $320.02, find a mathematical model that gives the cost of electricity in Tennessee in terms of the number of kilowatt-hours used. Solution: Write the direct variation model. y 5 kx Label the variables and constant. x 5 number of kWh y 5 cost (dollars) k 5 cost per kWh Substitute the given data x 5 3098 kWh and y 5 $320.02 into y 5 kx. 320.02 5 3098k Solve for k. k 5 320.02 3098 5 0.1033 y 5 0.1033x In Tennessee the cost of electricity is 10.33 ¢/kWh . Y OUR T UR N Find a mathematical model that describes the cost of electricity in California if the cost is directly proportional to the number of kWh used and a residence that consumes 4000 kWh is billed $735.20. ▼ A N S W E R y 5 18.38x; the cost of electricity in California is 18.38 ¢/kWh. ▼ EXAMPLE 2 Direct Variation with Powers The following is a personal ad: Single professional male (6 ft/194 lb) seeks single professional female for long-term relationship. Must be athletic, smart, like the movies and dogs, and have height and weight similarly proportioned to mine. One example of direct variation with powers is height and weight of humans. Weight (in pounds) is directly proportional to the cube of height (feet). W 5 kH3 3.6 Modeling Functions Using Variation 313 [CONCEPT CHECK] Students’ final grades in College Algebra are directly proportional to: (A) their financial aid amount or (B) how many hours they spend doing homework and in their instructor’s office hours. ANSWER B ▼ DIRECT VARIATION WITH POWERS Let x and y represent two quantities. The following are equivalent statements: ■ ■y 5 kxn, where k is a nonzero constant. ■ ■y varies directly with the nth power of x. ■ ■y is directly proportional to the nth power of x. 314 CHAPTER 3 Functions and Their Graphs 3.6.2 Inverse Variation Two fundamental topics covered in economics are supply and demand. Supply is the quantity that producers are willing to sell at a given price. For example, an artist may be willing to paint and sell five portraits if each sells for $50, but that same artist may be willing to sell 100 portraits if each sells for $10,000. Demand is the quantity of a good that consumers are not only willing to purchase but also have the capacity to buy at a given price. For example, consumers may purchase 1 billion Big Macs from McDonald’s every year, but perhaps only 1 million filet mignons are sold at Outback. There may be 1 billion people who want to buy the filet mignon but don’t have the financial means to do so. Economists study the equilibrium between supply and demand. Demand can be modeled with an inverse variation of price: when the price increases, demand decreases, and vice versa. 3.6.2 S KILL Develop mathematical models using inverse variation. 3.6.2 CO NCE PTUAL Understand that inverse variation implies that two things grow opposite of one another (as one increases, the other decreases). Find a mathematical equation that describes the height and weight of the male who wrote the ad. How much would a 5'6" woman weigh who has the same proportionality as the male? Solution: Write the direct variation (cube) model for height versus weight. W 5 kH 3 Substitute the given data W 5 194 and H 5 6 into W 5 kH3. 194 5 k 162 3 Solve for k. k 5 194 216 5 0.898148 < 0.90 W 5 0.9H 3 Let H 5 5.5 ft. W 5 0.9 15.523 5 149.73 A woman 5960 tall with the same height and weight proportionality as the male would weigh 150 lb . Y OUR TU R N A brother and sister both have weight (pounds) that varies as the cube of height (feet) and they share the same proportionality constant. The sister is 6 feet tall and weighs 170 pounds. Her brother is 6 feet 4 inches. How much does he weigh? ▼ A N S W E R 200 pounds ▼ INVERSE VARIATION Let x and y represent two quantities. The following are equivalent statements: ■ ■y 5 k x, where k is a nonzero constant. ■ ■y varies inversely with x. ■ ■y is inversely proportional to x. The constant k is called the constant of variation or the constant of proportionality. Two quantities can vary inversely with the nth power of x. EXAMPLE 3 Inverse Variation The number of potential buyers of a house decreases as the price of the house increases (see graph on the right). If the number of potential buyers of a house in a particular city is inversely proportional to the price of the house, find a mathematical equation that describes the demand for houses as it relates to price. How many potential buyers will there be for a $2 million house? Solution: Write the inverse variation model. y 5 k x Label the variables and constant. x 5 price of house in thousands of dollars y 5 number of buyers Select any point that lies on the curve. 1200, 5002 Substitute the given data x 5 200 and y 5 500 into y 5 k x. 500 5 k 200 Solve for k. k 5 200 ? 500 5 100,000 y 5 100,000 x Let x 5 2000. y 5 100,000 2000 5 50 There are only 50 potential buyers for a $2 million house in this city. Y OUR T UR N In New York City, the number of potential buyers in the housing market is inversely proportional to the price of a house. If there are 12,500 potential buyers for a $2 million condominium, how many potential buyers are there for a $5 million condominium? ▼ A N S W E R 5000 ▼ Demand (number of potential buyers) Price of the house (in thousands of dollars) 800 600 400 200 200 400 600 800 1000 (100, 1000) (200, 500) (400, 250) (600, 167) [CONCEPT CHECK] Given a fixed distance, the time it takes you to drive that distance varies inversely with __? ANSWER rate (or speed) ▼ 3.6.3 S K I L L Develop mathematical models using joint variation and combined variation. 3.6.3 C ON C E P T U A L Understand the difference between combined variation and joint variation. If x and y are related by the equation y 5 k xn, then we say that y varies inversely with the nth power of x, or y is inversely proportional to the nth power of x. 3.6.3 Joint Variation and Combined Variation We now discuss combinations of variations. When one quantity is proportional to the product of two or more other quantities, the variation is called joint variation. When direct variation and inverse variation occur at the same time, the variation is called combined variation. 3.6 Modeling Functions Using Variation 315 316 CHAPTER 3 Functions and Their Graphs An example of a joint variation is simple interest (Section 1.2), which is defined as I 5 Prt where ■ ■I is the interest in dollars ■ ■P is the principal (initial) dollars ■ ■r is the interest rate (expressed in decimal form) ■ ■t is time in years The interest earned is proportional to the product of three quantities (principal, interest rate, and time). Note that if the interest rate increases, then the interest earned also increases. Similarly, if either the initial investment (principal) or the time the money is invested increases, then the interest earned also increases. An example of combined variation is the combined gas law in chemistry, P 5 k T V where ■ ■P is pressure ■ ■T is temperature (kelvins) ■ ■V is volume ■ ■k is a gas constant This relation states that the pressure of a gas is directly proportional to the temperature and inversely proportional to the volume containing the gas. For example, as the temperature increases, the pressure increases, but when the volume decreases, pressure increases. As an example, the gas in the headspace of a soda bottle has a fixed volume. Therefore, as temperature increases, the pressure increases. Compare the different pressures of opening a twist-off cap on a bottle of soda that is cold versus one that is hot. The hot one feels as though it “releases more pressure.” EXAMPLE 4 Combined Variation The gas in the headspace of a soda bottle has a volume of 9.0 ml, pressure of 2 atm (atmospheres), and a temperature of 298 K (standard room temperature of 778F). If the soda bottle is stored in a refrigerator, the temperature drops to approximately 279 K (428F). What is the pressure of the gas in the headspace once the bottle is chilled? Solution: Write the combined gas law. P 5 k T V Let P 5 2 atm, T 5 298 K, and V 5 9.0 ml. 2 5 k 298 9 Solve for k. k 5 18 298 Let k 5 18 298, T 5 279, and V 5 9.0 in P 5 k T V. P 5 18 298⋅279 9 < 1.87 Since we used the same physical units for both the chilled and room-temperature soda bottles, the pressure is in atmospheres. P 5 1.87 atm [CONCEPT CHECK] The area of a triangle, A 5 1 2 bh, is an example of what type of variation? (A) Combined; (B) Joint ANSWER B ▼ In Exercises 1–16, write an equation that describes each variation. Use k as the constant of variation. 1. y varies directly with x. 2. s varies directly with t. 3. V varies directly with x3. 4. A varies directly with x2. 5. z varies directly with m. 6. h varies directly with !t. 7. ƒ varies inversely with l. 8. P varies inversely with r2. 9. F varies directly with w and inversely with L. 10. V varies directly with T and inversely with P. 11. v varies directly with both g and t. 12. S varies directly with both t and d. 13. R varies inversely with both P and T. 14. y varies inversely with both x and z. 15. y is directly proportional to the square root of x. 16. y is inversely proportional to the cube of t. In Exercises 17–36, write an equation that describes each variation. 17. d is directly proportional to t. d 5 r when t 5 1. 18. F is directly proportional to m. F 5 a when m 5 1. 19. V is directly proportional to both l and w. V 5 6h when w 5 3 and l 5 2. 20. A is directly proportional to both b and h. A 5 10 when b 5 5 and h 5 4. 21. A varies directly with the square of r. A 5 9p when r 5 3. 22. V varies directly with the cube of r. V 5 36p when r 5 3. 23. V varies directly with both h and r2. V 5 1 when r 5 2 and h 5 4 p. 24. W is directly proportional to both R and the square of I. W 5 4 when R 5 100 and I 5 0.25. 25. V varies inversely with P. V 5 1000 when P 5 400. 26. I varies inversely with the square of d. I 5 42 when d 5 16. 27. F varies inversely with both l and L. F 5 20p/m2 when l 5 1 mm and L 5 100 kilometers. 28. y varies inversely with both x and z. y 5 32 when x 5 4 and z 5 0.05. 29. t varies inversely with s. t 5 2.4 when s 5 8. 30. W varies inversely with the square of d. W 5 180 when d 5 0.2. 31. R varies inversely with the square of I. R 5 0.4 when I 5 3.5. 32. y varies inversely with both x and the square root of z. y 5 12 when x 5 0.2 and z 5 4. 33. R varies directly with L and inversely with A. R 5 0.5 when L 5 20 and A 5 0.4. 34. F varies directly with m and inversely with d. F 5 32 when m 5 20 and d 5 8. 35. F varies directly with both m1 and m2 and inversely with the square of d. F 5 20 when m1 5 8, m2 5 16, and d 5 0.4. 36. w varies directly with the square root of g and inversely with the square of t. w 5 20 when g 5 16 and t 5 0.5. [SEC TION 3.6] E X ERC I S E S • S K I L L S 3.6 Modeling Functions Using Variation 317 Joint variation occurs when one quantity is directly proportional to two or more quantities. Combined variation occurs when one quantity is directly proportional to one or more quantities and inversely proportional to one or more other quantities. Direct, inverse, joint, and combined variation can be used to model the relationship between two quantities. For two quantities x and y, we say that • y is directly proportional to x if y 5 kx. • y is inversely proportional to x if y 5 k x. [SEC TION 3.6] S U M M A RY 318 CHAPTER 3 Functions and Their Graphs • A P P L I C A T I O N S 37. Wages. Jason and Valerie both work at Panera Bread and have the following paycheck information for a certain week. Find an equation that shows their wages W varying directly with the number of hours worked H. EMPLOYEE HOURS WORKED WAGES Jason 23 $172.50 Valerie 32 $240.00 38. Sales Tax. The sales tax in Orange and Seminole counties in Florida differs by only 0.5%. A new resident knows this but doesn’t know which of the counties has the higher tax. The resident lives near the border of the counties and is in the market for a new plasma television and wants to purchase it in the county with the lower tax. If the tax on a pair of $40 sneakers is $2.60 in Orange County and the tax on a $12 T-shirt is $0.84 in Seminole County, write two equations: one for each county that describes the tax T, which is directly proportional to the purchase price P. For Exercises 39 and 40, refer to the following: The ratio of the speed of an object to the speed of sound determines the Mach number. Aircraft traveling at a subsonic speed (less than the speed of sound) have a Mach number less than 1. In other words, the speed of an aircraft is directly proportional to its Mach number. Aircraft traveling at a supersonic speed (greater than the speed of sound) have a Mach number greater than 1. The speed of sound at sea level is approximately 760 miles per hour. 39. Military. The U.S. Navy Blue Angels fly F-18 Hornets that are capable of Mach 1.7. How fast can F-18 Hornets fly at sea level? 40. Military. The U.S. Air Force’s newest fighter aircraft is the F-35, which is capable of Mach 1.9. How fast can an F-35 fly at sea level? Exercises 41 and 42 are examples of the golden ratio, or phi, a proportionality constant that appears in nature. The numerical approximate value of phi is 1.618. From www.goldenratio.net. 41. Human Anatomy. The length of your forearm F (wrist to elbow) is directly proportional to the length of your hand H (length from wrist to tip of middle finger). Write the equation that describes this relationship if the length of your forearm is 11 inches and the length of your hand is 6.8 inches. 42. Human Anatomy. Each section of your index finger, from the tip to the base of the wrist, is larger than the preceding one by about the golden (Fibonacci) ratio. Find an equation that represents the ratio of each section of your finger related to the previous one if one section is eight units long and the next section is five units long. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 2 3 5 8 16 17 18 For Exercises 43 and 44, refer to the following: Hooke’s law in physics states that if a spring at rest (equilibrium position) has a weight attached to it, then the distance the spring stretches is directly proportional to the force (weight), according to the formula: F 5 kx where F is the force in Newtons (N), x is the distance stretched in meters (m), and k is the spring constant (N/m). Equilibrium position 2F 2x F x 43. Physics. A force of 30 N will stretch the spring 10 centimeters. How far will a force of 72 N stretch the spring? 44. Physics. A force of 30 N will stretch the spring 10 centimeters. How much force is required to stretch the spring 18 centimeters? 45. Business. A cell phone company develops a pay-as-you-go cell phone plan in which the monthly cost varies directly as the number of minutes used. If the company charges $17.70 in a month when 236 minutes are used, what should the company charge for a month in which 500 minutes are used? 46. Economics. Demand for a product varies inversely with the price per unit of the product. Demand for the product is 10,000 units when the price is $5.75 per unit. Find the demand for the product (to the nearest hundred units) when the price is $6.50. 47. Sales. Levi’s makes jeans in a variety of price ranges for juniors. The Flare 519 jeans sell for about $20, whereas the 646 Vintage Flare jeans sell for $300. The demand for Levi’s jeans is inversely proportional to the price. If 300,000 pairs of the 519 jeans were bought, approximately how many of the Vintage Flare jeans were bought? 48. Sales. Levi’s makes jeans in a variety of price ranges for men. The Silver Tab Baggy jeans sell for about $30, whereas the Offender jeans sell for about $160. The demand for Levi’s jeans is inversely proportional to the price. If 400,000 pairs of the Silver Tab Baggy jeans were bought, approximately how many of the Offender jeans were bought? Kim Steele/Getty Images, Inc. For Exercises 49 and 50, refer to the following: In physics, the inverse square law states that any physical quantity or strength is inversely proportional to the square of the distance from the source of that physical quantity. In particular, the intensity of light radiating from a point source is inversely proportional to the square of the distance from the source. Below is a table of average distances from the Sun: PLANET DISTANCE TO THE SUN Mercury 58,000 km Earth 150,000 km Mars 228,000 km 49. Solar Radiation. The solar radiation on the Earth is approximately 1400 watts per square meter 1w/m22. How much solar radiation is there on Mars? Round to the nearest hundred watts per square meter. 50. Solar Radiation. The solar radiation on the Earth is approximately 1400 watts per square meter. How much solar radiation is there on Mercury? Round to the nearest hundred watts per square meter. 51. Investments. Marilyn receives a $25,000 bonus from her company and decides to put the money toward a new car that she will need in two years. Simple interest is directly proportional to the principal and the time invested. She compares two different banks’ rates on money market accounts. If she goes with Bank of America, she will earn $750 in interest, but if she goes with the Navy Federal Credit Union, she will earn $1500. What is the interest rate on money market accounts at both banks? 52. Investments. Connie and Alvaro sell their house and buy a fixer-upper house. They made $130,000 on the sale of their previous home. They know it will take 6 months before the general contractor will start their renovation, and they want to take advantage of a 6-month CD that pays simple interest. What is the rate of the 6-month CD if they will make $3250 in interest? 53. Chemistry. A gas contained in a 4-milliliter container at a temperature of 300 K has a pressure of 1 atmosphere. If the temperature decreases to 275 K, what is the resulting pressure? 54. Chemistry. A gas contained in a 4-milliliter container at a temperature of 300 K has a pressure of 1 atmosphere. If the container changes to a volume of 3 millileters, what is the resulting pressure? • C A T C H T H E M I S T A K E In Exercises 55 and 56, explain the mistake that is made. 55. y varies directly with t and indirectly with x. When x 5 4 and t 5 2, then y 5 1. Find an equation that describes this variation. Solution: Write the variation equation. y 5 ktx Let x 5 4, t 5 2, and y 5 1. 1 5 k 122 142 Solve for k. k 5 1 8 Substitute k 5 1 8 into y 5 ktx. y 5 1 8tx This is incorrect. What mistake was made? 56. y varies directly with t and the square of x. When x 5 4 and t 5 1, then y 5 8. Find an equation that describes this variation. Solution: Write the variation equation. y 5 kt!x Let x 5 4, t 5 1, and y 5 8. 8 5 k112 !4 Solve for k. k 5 4 Substitute k 5 4 into y 5 kt!x. y 5 4t!x This is incorrect. What mistake was made? In Exercises 57 and 58, determine whether each statement is true or false. 57. The area of a triangle is directly proportional to both the base and the height of the triangle (joint variation). 58. Average speed is directly proportional to both distance and time (joint variation). In Exercises 59 and 60, match the variation with the graph. 59. Inverse variation 60. Direct variation a. 10 10 x y b. 10 10 x y • C O N C E P T U A L 3.6 Modeling Functions Using Variation 319 320 CHAPTER 3 Functions and Their Graphs Exercises 61 and 62 involve the theory governing laser propagation through the Earth’s atmosphere. The three parameters that help classify the strength of optical turbulence are: ■ ■C 2 n, index of refraction structure parameter ■ ■k, wave number of the laser, which is inversely proportional to the wavelength l of the laser: k 5 2p l ■ ■L, propagation distance The variance of the irradiance of a laser s2 is directly proportional to C2 n, k7/6, and L11/16. 61. When C 2 n 5 1.0 3 10213 m22/3, L 5 2 km, and l 5 1.55 mm, the variance of irradiance for a plane wave s2 pl is 7.1. Find the equation that describes this variation. 62. When C 2 n 5 1.0 3 10213 m22/3, L 5 2 km, and l 5 1.55 mm, the variance of irradiance for a spherical wave s2 sp is 2.3. Find the equation that describes this variation. • C H A L L E N G E For Exercises 63–66, refer to the following: Data from 1995 to 2006 for oil prices in dollars per barrel, the U.S. Dow Jones Utilities Stock Index, New Privately Owned Housing, and 5-year Treasury Constant Maturity Rate are given in the table. (Data are from Forecast Center’s Historical Economic and Market Home Page at www.neatideas.com/djutil.htm.) Use the calculator stat edit commands to enter the table with L1 as the oil price, L2 as the utilities stock index, L3 as number of housing units, and L4 as the 5-year maturity rate. • T E C H N O L O G Y JANUARY OF EACH YEAR OIL PRICE, $ PER BARREL U.S. DOW JONES UTILITIES STOCK INDEX NEW, PRIVATELY OWNED HOUSING UNITS 5-YEAR TREASURY CONSTANT MATURITY RATE 1995 17.99 193.12 1407 7.76 1996 18.88 230.85 1467 5.36 1997 25.17 232.53 1355 6.33 1998 16.71 263.29 1525 5.42 1999 12.47 302.80 1748 4.60 2000 27.18 315.14 1636 6.58 2001 29.58 372.32 1600 4.86 2002 19.67 285.71 1698 4.34 2003 32.94 207.75 1853 3.05 2004 32.27 271.94 1911 3.12 2005 46.84 343.46 2137 3.71 2006 65.51 413.84 2265 4.35 63. An increase in oil price in dollars per barrel will drive the U.S. Dow Jones Utilities Stock Index to soar. a. Use the calculator commands stat, linReg 1ax 1 b2, and statplot to model the data using the least-squares regression. Find the equation of the least-squares regression line using x as the oil price in dollars per barrel. b. If the U.S. Dow Jones Utilities Stock Index varies directly as the oil price in dollars per barrel, then use the calculator commands stat, PwrReg, and statplot to model the data using the power function. Find the variation constant and equation of variation using x as the oil price in dollars per barrel. c. Use the equations you found in (a) and (b) to predict the stock index when the oil price hits $72.70 per barrel in September 2006. Which answer is closer to the actual stock index of 417? Round all answers to the nearest whole number. 64. An increase in oil price in dollars per barrel will affect the interest rates across the board—in particular, the 5-year Treasury constant maturity rate. a. Use the calculator commands stat, linReg 1ax 1 b2, and statplot to model the data using the least-squares regression. Find the equation of the least-squares regression line using x as the oil price in dollars per barrel. b. If the 5-year Treasury constant maturity rate varies inversely as the oil price in dollars per barrel, then use the calculator commands stat, PwrReg , and statplot to model the data using the power function. Find the variation constant and equation of variation using x as the oil price in dollars per barrel. c. Use the equations you found in (a) and (b) to predict the maturity rate when the oil price hits $72.70 per barrel in September 2006. Which answer is closer to the actual maturity rate at 5.02%? Round all answers to two decimal places. 65. An increase in interest rates—in particular, the 5-year Treasury constant maturity rate—will affect the number of new, privately owned housing units. a. Use the calculator commands stat, linReg 1ax 1 b2, and statplot to model the data using the least-squares regression. Find the equation of the least-squares regression line using x as the 5-year rate. b. If the number of new, privately owned housing units varies inversely as the 5-year Treasury constant maturity rate, then use the calculator commands stat , PwrReg , and statplot to model the data using the power function. Find the variation constant and equation of variation using x as the 5-year rate. c. Use the equations you found in (a) and (b) to predict the number of housing units when the maturity rate is 5.02% in September 2006. Which answer is closer to the actual number of new, privately owned housing units of 1861? Round all answers to the nearest unit. 66. An increase in the number of new, privately owned housing units will affect the U.S. Dow Jones Utilities Stock Index. a. Use the calculator commands stat, linReg 1ax 1 b2, and statplot to model the data using the least-squares regression. Find the equation of the least-squares regression line using x as the number of housing units. b. If the U.S. Dow Jones Utilities Stock Index varies directly as the number of new, privately owned housing units, then use the calculator commands stat, PwrReg , and statplot to model the data using the power function. Find the variation constant and equation of variation using x as the number of housing units. c. Use the equations you found in (a) and (b) to pre-dict the utilities stock index if there are 1861 new, privately owned housing units in September 2006. Which answer is closer to the actual stock index of 417? Round all answers to the nearest whole number. For Exercises 67 and 68, refer to the following: Data for retail gasoline price in dollars per gallon for the period March 2006 to March 2015 are given in the following table. (Data are from Energy Information Administration, Official Energy Statistics from the U.S. government at gov/oog/info/gdu/gaspump.html.) Use the calculator stat edit command to enter the table below with L1 as the year 1x 5 1 for year 20062 and L2 as the gasoline price in dollars per gallon. MARCH OF EACH YEAR 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 RETAIL GASOLINE PRICE $ PER GALLON 2.353 2.599 3.216 1.984 2.768 3.617 3.954 3.888 3.601 2.710 67. a. Use the calculator commands stat LinReg to model the data using the least-squares regression. Find the equation of the least-squares regression line using x as the year 1x 5 1 for year 20062 and y as the gasoline price in dollars per gallon. Round all answers to three decimal places. b. Use the equation to predict the gasoline price in March 2018. Round all answers to three decimal places. Is the answer close to the actual price? c. Use the equation to predict the gasoline price in March 2020. Round all answers to three decimal places. 68. a. Use the calculator commands stat PwrReg to model the data using the power function. Find the variation constant and equation of variation using x as the year 1x 5 1 for year 20062 and y as the gasoline price in dollars per gallon. Round all answers to three decimal places. b. Use the equation to predict the gasoline price in March 2018. Round all answers to three decimal places. Is the answer close to the actual price? c. Use the equation to predict the gasoline price in March 2020. Round all answers to three decimal places. 3.6 Modeling Functions Using Variation 321 CH A P TE R 3 R E VI E W [ C H AP T E R 3 REVIEW] SECTION CONCEPT KEY IDEAS/FORMULAS 3.1 Functions Relations and functions All functions are relations, but not all relations are functions. Functions defined by equations A vertical line can intersect a function in at most one point. Function notation Placeholder notation: ƒ1x2 5 3x2 2 6x 1 2 ƒ1u2 5 31u22 2 61u2 1 2 Difference quotient: ƒ1x 1 h2 2 ƒ1x2 h ; h 2 0 Domain of a function Are there any restrictions on x? 3.2 Graphs of functions; piecewise-defined functions; increasing and decreasing functions; average rate of change Recognizing and classifying functions Common functions ƒ1x2 5 mx 1 b, ƒ1x2 5 x, ƒ1x2 5 x2, ƒ1x2 5 x3, ƒ1x2 5 !x, ƒ1x2 5 ! 3 x, ƒ1x2 5 0 x 0, ƒ1x2 5 1 x Even and odd functions Even: Symmetry about y-axis: ƒ12x2 5 ƒ1x2 Odd: Symmetry about origin: ƒ12x2 5 2ƒ1x2 Increasing and decreasing functions • Increasing: rises (left to right) • Decreasing: falls (left to right) Average rate of change ƒ1x22 2 ƒ1x12 x2 2 x1 x1 2 x2 Piecewise-defined functions Points of discontinuity 3.3 Graphing techniques: Transformations Shift the graph of ƒ1x2. Horizontal and vertical shifts ƒ1x 1 c2 c units to the left, c . 0 ƒ1x 2 c2 c units to the right, c . 0 ƒ1x2 1 c c units upward, c . 0 ƒ1x2 2 c c units downward, c . 0 Reflection about the axes 2ƒ1x2 Reflection about the x-axis ƒ12x2 Reflection about the y-axis Stretching and compressing cƒ1x2 if c . 1; stretch vertically cƒ1x2 if 0 , c , 1; compress vertically ƒ1cx2 if c . 1; compress horizontally ƒ1cx2 if 0 , c , 1; stretch horizontally 3.4 Operations on functions and composition of functions Adding, subtracting, multiplying, and dividing functions 1 ƒ 1 g2 1x2 5 ƒ1x2 1 g 1x2 1 ƒ 2 g2 1x2 5 ƒ1x2 2 g 1x2 1 ƒ ? g2 1x2 5 ƒ1x2 ? g 1x2 The domain of the resulting function is the intersection of the individual domains. a ƒ gb 1x2 5 ƒ1x2 g1x2, g1x2 2 0 The domain of the quotient is the intersection of the domains of ƒ and g, and any points when g 1x2 5 0 must be eliminated. Composition of functions 1ƒ + g21x2 5 ƒ1g1x22 The domain of the composite function is a subset of the domain of g 1x2. Values for x must be eliminated if their corresponding values g 1x2 are not in the domain of ƒ. 322 CHAPTER 3 Functions and Their Graphs SECTION CONCEPT KEY IDEAS/FORMULAS 3.5 One-to-one functions and inverse functions Determine whether a function is one-to-one • No two x-values map to the same y-value. If ƒ1x12 5 ƒ1x22, then x1 5 x2. • A horizontal line may intersect a one-to-one function in at most one point. Inverse functions • Only one-to-one functions have inverses. • ƒ21 1ƒ1x22 5 x and ƒ1ƒ21 1x22 5 x. • Domain of ƒ 5 range of ƒ21. Range of ƒ 5 domain of ƒ21. Graphical interpretation of inverse functions • The graph of a function and its inverse are symmetric about the line y 5 x. • If the point 1a, b2 lies on the graph of a function, then the point 1b, a2 lies on the graph of its inverse. Finding the inverse function 1. Let y 5 ƒ1x2. 2. Interchange x and y. 3. Solve for y. 4. Let y 5 ƒ21 1x2. 3.6 Modeling functions using variation Direct variation y 5 kx Inverse variation y 5 k x Joint variation and combined variation Joint: One quantity is directly proportional to the product of two or more other quantities. Combined: Direct variation and inverse variation occur at the same time. Chapter Review 323 CH A P TE R 3 R E VIE W REV IEW E XE R CI SE S [ C H AP T E R 3 REVIEW EXERC IS E S ] Evaluate the given quantities using the following three functions. ƒ1x2 5 4x 2 7 F 1t2 5 t2 1 4t 2 3 g 1x2 5 0 x2 1 2x 1 40 15. ƒ132 16. F 142 17. ƒ1272 . g 132 18. F102 g102 19. ƒ122 2 F122 g102 20. ƒ13 1 h2 21. ƒ13 1 h2 2 ƒ132 h 22. F1t 1 h2 2 F1t2 h Find the domain of the given function. Express the domain in interval notation. 23. ƒ1x2 5 23x 2 4 24. g 1x2 5 x2 2 2x 1 6 25. h1x2 5 1 x 1 4 26. F1x2 5 7 x2 1 3 27. G1x2 5 !x 2 4 28. H1x2 5 1 !2x 2 6 Challenge 29. If ƒ1x2 5 D x2 2 16, ƒ142 and ƒ1242 are undefined, and ƒ152 5 2, find D. 30. Construct a function that is undefined at x 5 23 and x 5 2 such that the point 10, 242 lies on the graph of the function. 3.2 Graphs of Functions Determine whether the function is even, odd, or neither. 31. ƒ1x2 5 2x 2 7 32. g 1x2 5 7x5 1 4x3 2 2x 33. h 1x2 5 x3 2 7x 34. ƒ1x2 5 x4 1 3x2 35. ƒ1x2 5 x1/4 1 x 36. ƒ1x2 5 !x 1 4 37. ƒ1x2 5 1 x3 1 3x 38. ƒ1x2 5 1 x2 1 3x4 1 0 x 0 Use the graph of the functions to find: a. Domain b. Range c. Intervals on which the function is increasing, decreasing, or constant. 39. x y –10 10 –10 10 40. x y –10 10 –10 10 41. Find the average rate of change of ƒ1x2 5 4 2 x2 from x 5 0 to x 5 2. 42. Find the average rate of change of ƒ1x2 5 0 2x 2 10 from x 5 1 to x 5 5. 3.1 Functions Determine whether each relation is a function. 1. Domain Range NAMES Allie Hannah Danny Ethan Vickie AGES 27 10 4 21 2. 5 11, 22, 13, 42, 12, 42, 13, 726 3. 5 122, 32, 11, 232, 10, 42, 12, 626 4. 5 14, 72, 12, 62, 13, 82, 11, 726 5. x2 1 y2 5 36 6. x 5 4 7. y 5 0 x 1 20 8. y 5 !x 9. x y 10. x y Use the graphs of the functions to find: 11. x y 12. x y 13. x y 14. x y a. ƒ1212 b. ƒ112 c. x, where ƒ1x2 5 0 a. ƒ1242 b. ƒ102 c. x, where ƒ1x2 5 0 a. ƒ1222 b. ƒ142 c. x, where ƒ1x2 5 0 a. ƒ1252 b. ƒ102 c. x, where ƒ1x2 5 0 324 CHAPTER 3 Functions and Their Graphs R E VI E W E XERCISES Graph the piecewise-defined function. State the domain and range in interval notation. 43. F1x2 5 bx2 x , 0 2 x $ 0 44. ƒ1x2 5 • 22x 2 3 x # 0 4 0 , x # 1 x2 1 4 x . 1 45. ƒ1x2 5 • x2 x # 0 2"x 0 , x # 1 0 x 1 2 0 x . 1 46. F1x2 5 • x2 x , 0 x3 0 , x , 1 20 x 0 2 1 x $ 1 Applications 47. Tutoring Costs. A tutoring company charges $25.00 for the first hour of tutoring and $10.50 for every 30-minute period after that. Find the cost function C 1x2 as a function of the length of the tutoring session. Let x 5 number of 30-minute periods. 48. Salary. An employee who makes $30.00 per hour also earns time and a half for overtime (any hours worked above the normal 40-hour work week). Write a function E 1x2 that describes her weekly earnings as a function of the number of hours worked x. 3.3 Graphing Techniques: Transformations Graph the following functions using graphing aids. 49. y 5 2 1x 2 222 1 4 50. y 5 0 2x 1 50 2 7 51. y 5 3 !x 2 3 1 2 52. y 5 1 x 2 2 2 4 53. y 5 21 2x3 54. y 5 2x2 1 3 Use the given graph to graph the following: 55. x y 56. x y y 5 ƒ1x 2 22 y 5 3ƒ1x2 57. x y 58. x y y 5 22ƒ1x2 y 5 ƒ1x2 1 3 Write the function whose graph is the graph of y 5 !x, but is transformed accordingly, and state the domain of the resulting function. 59. Shifted to the left three units 60. Shifted down four units 61. Shifted to the right two units and up three units 62. Reflected about the y-axis 63. Stretched by a factor of 5 and shifted down six units 64. Compressed by a factor of 2 and shifted up three units Transform the function into the form ƒ 1x2 5 c 1x 2 h22 1 k by completing the square and graph the resulting function using transformations. 65. y 5 x2 1 4x 2 8 66. y 5 2x2 1 6x 2 5 3.4 Operations on Functions and Composition of Functions Given the functions g and h, find g 1 h, g 2 h, g ? h, and g h, and state the domain. 67. g 1x2 5 23x 2 4 h 1x2 5 x 2 3 68. g 1x2 5 2x 1 3 h 1x2 5 x2 1 6 69. g1x2 5 1 x2 h1x2 5 !x 70. g1x2 5 x 1 3 2x 2 4 h1x2 5 3x 2 1 x 2 2 71. g1x2 5 !x 2 4 h1x2 5 !2x 1 1 72. g1x2 5 x2 2 4 h1x2 5 x 1 2 For the given functions ƒ and g, find the composite functions ƒ + g and g + ƒ, and state the domains. 73. ƒ1x2 5 3x 2 4 g 1x2 5 2x 1 1 74. ƒ1x2 5 x3 1 2x 2 1 g 1x2 5 x 1 3 75. ƒ1x2 5 2 x 1 3 g1x2 5 1 4 2 x 76. ƒ1x2 5 "2x2 2 5 g1x2 5 !x 1 6 77. ƒ1x2 5 !x 2 5 g1x2 5 x2 2 4 Review Exercises 325 REV IEW E XE R CI SE S 78. ƒ1x2 5 1 "x g1x2 5 1 x2 2 4 Evaluate ƒ 1 g 1322 and g 1ƒ 12122, if possible. 79. ƒ1x2 5 4x2 2 3x 1 2 g 1x2 5 6x 2 3 80. ƒ1x2 5 !4 2 x g 1x2 5 x2 1 5 81. ƒ1x2 5 x 0 2x 2 3 0 g1x2 5 0 5x 1 2 0 82. ƒ1x2 5 1 x 2 1 g1x2 5 x2 2 1 83. ƒ1x2 5 x2 2 x 1 10 g1x2 5 ! 3 x 2 4 84. ƒ1x2 5 4 x2 2 2 g 1x2 5 1 x2 2 9 Write the function as a composite ƒ 1 g 1x22 of two functions ƒ and g. 85. h 1x2 5 3 1x 2 222 1 4 1x 2 22 1 7 86. h1x2 5 3 "x 1 2 3 "x 87. h1x2 5 1 "x2 1 7 88. h1x2 5 "0 3x 1 4 0 Applications 89. Rain. A rain drop hitting a lake makes a circular ripple. If the radius, in inches, grows as a function of time, in minutes, r1t2 5 25"t 1 2, find the area of the ripple as a function of time. 90. Geometry. Let the area of a rectangle be given by 42 5 l . w, and let the perimeter be 36 5 2 . l 1 2 . w. Express the perimeter in terms of w. 3.5 One-to-One Functions and Inverse Functions Determine whether the given function is a one-to-one function. 91. Chris Harold Tom Danny Paula Vickie Renee Gabriel BROTHER SISTER Domain Range 92. Domain Range Function STUDENTS IN PRECALC GRADE IN PRECALCULUS COURSE A C D F Tonja Troy Maria Martin Bill Tracey 93. 5 12, 32, 121, 22, 13, 32, 123, 242, 122, 126 94. 5 123, 92, 15, 252, 12, 42, 13, 926 95. 5 122, 02, 14, 52, 13, 726 96. 5 128, 262, 124, 22, 10, 32, 12, 282, 17, 426 97. y 5 "x 98. y 5 x2 99. ƒ1x2 5 x3 100. ƒ1x2 5 1 x2 Verify that the function ƒ21 1x2 is the inverse of ƒ 1x2 by showing that ƒ 1 ƒ 21 1x22 5 x. Graph ƒ 1x2 and ƒ 21 1x2 on the same graph and show the symmetry about the line y 5 x. 101. ƒ1x2 5 3x 1 4; ƒ211x2 5 x 2 4 3 102. ƒ1x2 5 1 4x 2 7; ƒ211x2 5 1 1 7x 4x 103. ƒ1x2 5 "x 1 4; ƒ211x2 5 x2 2 4 x $ 0 104. ƒ1x2 5 x 1 2 x 2 7; ƒ211x2 5 7x 1 2 x 2 1 The function ƒ is one-to-one. Find its inverse and check your answer. State the domain and range of both ƒ and ƒ21. 105. ƒ1x2 5 2x 1 1 106. ƒ1x2 5 x5 1 2 107. ƒ1x2 5 "x 1 4 108. ƒ1x2 5 1x 1 422 1 3 x > 24 109. ƒ1x2 5 x 1 6 x 1 3 110. ƒ1x2 5 2 3 "x 2 5 2 8 Applications 111. Salary. A pharmaceutical salesperson makes $22,000 base salary a year plus 8% of the total products sold. Write a function S 1x2 that represents her yearly salary as a function of the total dollars worth of products sold x. Find S21 1x2. What does this inverse function tell you? 112. Volume. Express the volume V of a rectangular box that has a square base of length s and is 3 feet high as a func-tion of the square length. Find V21. If a certain volume is desired, what does the inverse tell you? 3.6 Modeling Functions Using Variation Write an equation that describes each variation. 113. C is directly proportional to r. C 5 2p when r 5 1. 114. V is directly proportional to both l and w. V 5 12h when w 5 6 and l 5 2. 115. A varies directly with the square of r. A 5 25p when r 5 5. 116. F varies inversely with both l and L. F 5 20p when l 5 10 mm and L 5 10 km. 326 CHAPTER 3 Functions and Their Graphs R E VI E W E XERCISES Applications 117. Wages. Cole and Dickson both work at the same museum and have the following paycheck information for a certain week. Find an equation that shows their wages (W) varying directly with the number of hours (H) worked. EMPLOYEE HOURS WORKED WAGE Cole 27 $229.50 Dickson 30 $255.00 118. Sales Tax. The sales tax in two neighboring counties differs by 1%. A new resident knows the difference but doesn’t know which county has the higher tax rate. The resident lives near the border of the two counties and wants to buy a new car. If the tax on a $50.00 jacket is $3.50 in County A and the tax on a $20.00 calculator is $1.60 in County B, write two equations (one for each county) that describe the tax (T), which is directly proportional to the purchase price (P). Technology Exercises Section 3.1 119. Use a graphing utility to graph the function and find the domain. Express the domain in interval notation. ƒ1x2 5 1 "x2 2 2x 2 3 120. Use a graphing utility to graph the function and find the domain. Express the domain in interval notation. ƒ1x2 5 x2 2 4x 2 5 x2 2 9 Section 3.2 121. Use a graphing utility to graph the function. State the (a) domain, (b) range, and (c) x intervals where the function is increasing, decreasing, and constant. ƒ1x2 5 • 1 2 x 33x44 x 1 1 x , 21 21 # x , 2 x . 2 122. Use a graphing utility to graph the function. State the (a) domain, (b) range, and (c) x intervals where the function is increasing, decreasing, and constant. ƒ1x2 5 • 0 x2 2 1 0 !x 2 2 1 4 22 , x , 2 x . 2 Section 3.3 123. Use a graphing utility to graph ƒ1x2 5 x2 2 x 2 6 and g1x2 5 x2 2 5x. Use transforms to describe the relationship between ƒ1x2 and g 1x2? 124. Use a graphing utility to graph ƒ1x2 5 2x2 2 3x 2 5 and g1x2 5 22x2 2 x 1 6. Use transforms to describe the relationship between ƒ1x2 and g 1x2? Section 3.4 125. Using a graphing utility, plot y1 5 !2x 1 3, y2 5 !4 2 x, and y3 5 y1 y2 . What is the domain of y3? 126. Using a graphing utility, plot y1 5 "x2 2 4, y2 5 x2 2 5, and y3 5 y1 2 2 5. If y1 represents a function ƒ and y2 represents a function g, then y3 represents the composite function g + ƒ. The graph of y3 is only defined for the domain of g + ƒ. State the domain of g + ƒ. Section 3.5 127. Use a graphing utility to graph the function and determine whether it is one-to-one. ƒ1x2 5 6 5 "x3 2 1 128. Use a graphing utility to graph the functions ƒ and g and the line y 5 x in the same screen. Are the two functions inverses of each other? ƒ1x2 5 ! 4 x 2 3 1 1, g1x2 5 x4 2 4x3 1 6x2 2 4x 1 3 Section 3.6 From December 2006 to December 2015, data for gold price in dollars per ounce are given in the table below (SOURCE: www .kitco.com/charts/historicalgold.html). Use the calculator stat edit commands to enter the table below with L1 as the year 1x 5 1 for year 20062 and L2 as the gold price in dollars per ounce. DECEMBER OF EACH YEAR GOLD PRICE PER OUNCE (AVERAGE) 2006 629.79 2007 803.20 2008 816.09 2009 1134.72 2010 1390.55 2011 1652.31 2012 1688.53 2013 1225.40 2014 1202.29 2015 1063.88 (Source: www.kitco.com/charts/historicalgold.html) 129. a. Use the calculator commands stat LinRege to model the data using the least-squares regression. Find the equation of the least-squares regression line using x as the year 1x 5 1 for year 20062 and y as the gold price in dollars per ounce. Round all answers to two decimal places. b. Use the equation to predict the gold price in December 2016. Round all answers to two decimal places. Is the answer close to the actual price? c. Use the equation to predict the gold price in December 2017. Round all answers to two decimal places. 130. a. Use the calculator commands stat PwrRegs to model the data using the power function. Find the varia-tion constant and equation of variation using x as the year 1x 5 1 for year 20062 and y as the gold price in dol-lars per ounce. Round all answers to two decimal places. b. Use the equation to predict the gold price in December 2016. Round all answers to two decimal places. Is the answer close to the actual price? c. Use the equation to predict the gold price in December 2017. Round all answers to two decimal places. Review Exercises 327 PR ACTICE TEST [ C H AP T E R 3 PRACTICE TEST ] Assuming that x represents the independent variable and y represents the dependent variable, classify the relationships as: a. not a function b. a function but not one-to-one c. a one-to-one function 1. ƒ1x2 5 02x 1 3 0 2. x 5 y2 1 2 3. y 5 3 !x 1 1 Use ƒ1x2 5 !x 2 2 and g 1x2 5 x2 1 11, and determine the desired quantity or expression. In the case of an expression, state the domain. 4. ƒ1112 22g 1212 5. aƒ g b1x2 6. ag ƒb1x2 7. g1ƒ1x22 8. 1ƒ 1 g2 162 9. ƒ1gA"7BB Determine whether the function is odd, even, or neither. 10. ƒ1x2 5 0x 0 2 x2 11. ƒ1x2 5 9x3 1 5x 2 3 12. ƒ1x2 5 2 x Graph the functions. State the domain and range of each function. 13. ƒ1x2 5 2!x 2 3 1 2 14. ƒ1x2 5 22 1x 2 122 15. ƒ1x2 5 • 2x x , 21 1 21 , x , 2 x2 x $ 2 Use the graphs of the function to find: 16. x y y f(x) 328 CHAPTER 3 Functions and Their Graphs Find ƒ 1x 1 h2 2 ƒ 1x2 h for: 19. ƒ1x2 5 3x2 2 4x 1 1 20. ƒ1x2 5 5 2 7x Find the average rate of change of the given functions. 21. ƒ1x2 5 64 2 16x2 for x 5 0 to x 5 2 22. ƒ1x2 5 !x 2 1 for x 5 2 to x 5 10 Given the function ƒ , find the inverse if it exists. State the domain and range of both ƒ and ƒ 21. 23. ƒ1x2 5 !x 2 5 24. ƒ 1x2 5 x2 1 5 25. ƒ1x2 5 2x 1 1 5 2 x 26. ƒ1x2 5 e 2x x # 0 2x2 x . 0 27. What domain restriction can be made so that ƒ1x2 5 x2 has an inverse? 28. If the point 122, 52 lies on the graph of a function, what point lies on the graph of its inverse function? 29. Discount. Suppose a suit has been marked down 40% off the original price. An advertisement in the newspaper has an “additional 30% off the sale price” coupon. Write a function that determines the “checkout” price of the suit. 30. Temperature. Degrees Fahrenheit (8F), degrees Celsius (8C), and kelvins (K) are related by the two equations: F 5 9 5C 1 32 and K 5 C 1 273.15. Write a function whose input is kelvins and output is degrees Fahrenheit. 31. Circles. If a quarter circle is drawn by tracing the unit circle in quadrant III, what does the inverse of that function look like? Where is it located? 32. Sprinkler. A sprinkler head malfunctions at midfield in a football field. The puddle of water forms a circular pattern around the sprinkler head with a radius in yards that grows as a function of time, in hours: r 1t2 5 10!t. When will the puddle reach the sidelines? (A football field is 30 yards from sideline to sideline.) 33. Internet. The cost of airport Internet access is $15 for the first 30 minutes and $1 per minute for each minute after that. Write a function describing the cost of the service as a function of minutes used. Use variation to find a model for the given problem. 34. y varies directly with the square of x. y 5 8 when x 5 5. 35. F varies directly with m and inversely with p. F 5 20 when m 5 2 and p 5 3. 36. Use a graphing utility to graph the function. State the (a) domain, (b) range, and (c) x intervals where the function is increasing, decreasing, and constant. ƒ1x2 5 e 5 24 # x , 22 0 3 1 2x 2 x2 0 22 , x # 4 37. Use a graphing utility to graph the function and determine whether it is one-to-one. y 5 x3 2 12x2 1 48x 2 65 a. ƒ132 b. ƒ102 c. ƒ1242 d. x, where ƒ1x2 5 3 e. x, where ƒ1x2 5 0 17. x y y g(x) a. g 132 b. g 102 c. g 1242 d. x, where g 1x2 5 0 18. x y y p(x) a. p 102 b. x, where p 1x2 5 0 c. p 112 d. p 132 CU MU LA TIV E TEST Cumulative Test 329 [CH AP TERS 1–3 CUM UL AT IVE T E S T ] 1. Simplify 2 3 2 "5 . 2. Factor completely: 10x2 2 29x 2 21. 3. Simplify and state the domain: x3 2 4x x 1 2 . 4. Solve for x: 1 6 x 5 21 5 x 1 11. 5. Perform the operation, simplify, and express in standard form: 18 2 9i2 18 1 9i2. 6. Solve for x, and give any excluded values: 5 x 2 10 5 10 3x. 7. The original price of a hiking stick is $59.50. The sale price is $35.70. Find the percent of the markdown. 8. Solve by factoring: x 16x 1 12 5 12. 9. Solve by completing the square: x2 2 2 x 5 1 5. 10. Solve and check: 3 "x 1 2 5 23. 11. Solve using substitution: x4 2 x2 2 12 5 0. Solve and express the solution in interval notation. 12. 27 , 3 2 2 x # 5 13. x x 2 5 , 0 14. 02.7 2 3.2x 0 # 1.3 15. Calculate the distance and midpoint between the segment joining the points 122.7, 21.42 and 15.2, 6.32. 16. Find the slope of the line passing through the points 10.3, 21.42 and 12.7, 4.32. 17. Write an equation of a line that passes through the points 11.2, 232 and 120.2, 232. 18. Transform the equation into standard form by completing the square, and state the center and radius of the circle: x2 1 y2 1 12x 2 18y 2 4 5 0. 19. Find the equation of a circle with center 122, 212 and passing through the point 124, 32. 20. If a cellular phone tower has a reception radius of 100 miles and you live 85 miles north and 23 miles east of the tower, can you use your cell phone at home? Explain. 21. Use interval notation to express the domain of the function g1x2 5 1 x 2 1. 22. Find the average rate of change for ƒ1x2 5 5x2, from x 5 2 to x 5 4. 23. Evaluate g 1ƒ12122 for ƒ1x2 5 06 2 x 0 and g 1x2 5 x2 2 3. 24. Find the inverse of the function ƒ1x2 5 x2 1 3 for x $ 0. 25. Write an equation that describes the variation: r is inversely proportional to t. r 5 45 when t 5 3. 26. Use a graphing utility to graph the function. State the (a) domain, (b) range, and (c) x intervals where the function is increasing, decreasing, and constant. ƒ1x2 5 e 1 2 0 x 0 21 # x , 1 1 2 0 x 2 2 0 1 , x # 3 27. Use a graphing utility to graph the function ƒ1x2 5 x2 2 3x and g1x2 5 x2 1 x 2 2 in the same screen. Find the function h such that g + h 5 ƒ. C H A P T E R LEARNING OBJECTIVES [ [ ■ ■Find the vertex (maximum or minimum) of the graph of a quadratic function. ■ ■Graph polynomial functions. ■ ■Divide polynomials using long division and synthetic division. ■ ■Develop strategies for searching for zeros of a polynomial function. ■ ■Understand that complex zeros come in conjugate pairs. ■ ■Graph rational functions. Indoor football stadiums are designed so that football punters will not hit the roof with the football. One of the greatest NFL punters of all time was Ray Guy, who played 14 seasons from 1973 to 1986. “One of his punts hit the giant TV screen hanging from the rafters in the Louisiana Superdome. Not only did Guy punt high and far—‘hang time’ came into the NFL lexicon during his tenure—once he even had an opponent take a ball he punted and test for helium!” (rayguy.net/fact-sheet; Ray Guy Fact Sheet) The path that punts typically follow is called a parabola and is classified as a quadratic function. The distance of a punt is measured in the horizontal direction. The yard line where the punt is kicked from and the yard line where the punt either hits the field or is caught are the zeros of the quadratic function. Zeros are the points where the function value is equal to zero. Polynomial and Rational Functions 4 Focus On Sport/Getty Images, Inc. Ray Guy 331 [I N T HI S CHAPTER] We will start by discussing quadratic functions (polynomial functions of degree 2) whose graphs are parabolas. We will find the vertex, which is the maximum or minimum point on the graph. Then we will expand our discussion to higher-degree polynomial functions. We will discuss techniques to find zeros of polynomial functions and strategies for graphing polynomial functions. Last we will discuss rational functions, which are ratios of polynomial functions. POLYNOMIAL AND RATIONAL FUNCTIONS 4.1 QUADRATIC FUNCTIONS 4.2 POLYNOMIAL FUNCTIONS OF HIGHER DEGREE 4.3 DIVIDING POLYNOMIALS: LONG DIVISION AND SYNTHETIC DIVISION 4.4 THE REAL ZEROS OF A POLYNOMIAL FUNCTION 4.5 COMPLEX ZEROS: THE FUNDAMENTAL THEOREM OF ALGEBRA 4.6 RATIONAL FUNCTIONS • Graphs of Quadratic Functions: Parabolas • Finding the Equation of a Parabola • Identifying Polynomial Functions • Graphing Polynomial Functions Using Transformations of Power Functions • Real Zeros of a Polynomial Function • Graphing General Polynomial Functions • Long Division of Polynomials • Synthetic Division of Polynomials • The Remainder Theorem and the Factor Theorem • The Rational Zero Theorem and Descartes’ Rule of Signs • Factoring Polynomials • The Interme-diate Value Theorem • Graphing Polynomial Functions • Complex Zeros • Factoring Polynomials • Domain of Rational Functions • Vertical, Horizontal, and Slant Asymptotes • Graphing Rational Functions 332 CHAPTER 4 Polynomial and Rational Functions S K I L L S O B J E C T I V E S ■ ■Graph a quadratic function given in either standard or general form. ■ ■Find the equation of a parabola. C O N C E P T U A L O B J E C T I V E S ■ ■Recognize characteristics of graphs of quadratic functions (parabolas): whether the parabola opens up or down; whether the vertex is a maximum or minimum; the axis of symmetry ■ ■Understand that as long as you know the vertex and a point that lies along the graph of a parabola, you can determine the equation of the parabola. 4.1 QUADRATIC FUNCTIONS 4.1.1 Graphs of Quadratic Functions: Parabolas In Chapter 3, we studied functions in general. In this chapter, we will learn about a special group of functions called polynomial functions. Polynomial functions are simple functions; often, more complicated functions are approximated by polynomial functions. Polynomial functions model many real-world applications such as the stock market, football punts, business costs, revenues and profits, and the flight path of NASA’s “vomit comet.” Let’s start by defining a polynomial function DEFINITION Polynomial Function Let n be a nonnegative integer, and let an, an21, . . . , a2, a1, a0 be real numbers with an 2 0. The function ƒ1x2 5 anxn 1 an21xn21 1 c1 a2x2 1 a1x 1 a0 is called a polynomial function of x with degree n. The coefficient an is called the leading coefficient, and a0 is the constant. Polynomials of particular degrees have special names. In Chapter 3, the library of functions included the constant function ƒ1x2 5 b, which is a horizontal line; the linear function ƒ1x2 5 mx 1 b, which is a line with slope m and y-intercept 10, b2; the square function ƒ1x2 5 x2; and the cube function ƒ1x2 5 x3. These are all special cases of a polynomial function. Here are more examples of polynomial functions of particular degree together with their names: POLYNOMIAL DEGREE SPECIAL NAME ƒ1x2 5 3 0 Constant function ƒ1x2 5 22x 1 1 1 Linear function ƒ1x2 5 7x2 2 5x 1 19 2 Quadratic function ƒ1x2 5 4x3 1 2x 2 7 3 Cubic function The leading coefficients of these functions are 3, 22, 7, and 4, respectively. The constants of these functions are 3, 1, 19, and 27, respectively. In Section 2.3, we discussed graphs of linear functions, which are first-degree polynomial functions. In this section, we will discuss graphs of quadratic functions, which are second-degree polynomial functions. In Section 3.2, the library of functions that we compiled included the square function ƒ 1x2 5 x2, whose graph is a parabola. See the graph on the left. In Section 3.3, we graphed functions using transformation techniques such as F1x2 5 1x 1 12 2 2 2, which can be graphed by starting with the square function y 5 x2 and shifting one unit to the left and down two units. See the graph on the left. 4.1.1 SKI LL Graph a quadratic function given in either standard or general form. 4.1.1 CO NCE PTUAL Recognize characteristics of graphs of quadratic functions (parabolas): whether the parabola opens up or down; whether the vertex is a maximum or minimum; the axis of symmetry. x y –5 5 10 f (x) = x2 x F(x) = (x+1)2 – 2 y –5 5 8 f (x) = x2 4.1 Quadratic Functions 333 Note that if we eliminate the parentheses in F1x2 5 1x 1 122 2 2 to get F1x2 5 x2 1 2x 1 1 2 2 5 x2 1 2x 2 1 the result is a function defined by a second-degree polynomial (a polynomial with x 2 as the highest-degree term), which is also called a quadratic function. DEFINITION Quadratic Function Let a, b, and c be real numbers with a 2 0. The function ƒ1x2 5 ax2 1 bx 1 c is called a quadratic function. The graph of any quadratic function is a parabola. If the leading coefficient a is positive, then the parabola opens up. If the leading coefficient a is negative, then the parabola opens down. The vertex (or turning point) is the minimum point, or low point, on the graph if the parabola opens up, whereas it is the maximum point, or high point, on the graph if the parabola opens down. The vertical line that intersects the parabola at the vertex is called the axis of symmetry. The axis of symmetry is the line x 5 h, and the vertex is located at the point 1h, k2, as shown in the following two figures. x y (h, k) Vertex x = h Axis of Symmetry h k f (x) = ax2 + bx + c a > 0 x y (h, k) Vertex x = h Axis of Symmetry h k f (x) = ax2 + bx + c a < 0 At the start of this section, the function F1x2 5 x2 1 2x 2 1, which can also be written F1x2 5 1x 1 12 2 2 2, was shown through graph-shifting techniques to have a graph of a parabola. Looking at the graph on top of the margin, we see that the parabola opens up 1a 5 1 . 02, has a vertex at the point 121, 222 and has an axis of symmetry of x 5 21. Graphing Quadratic Functions Given in Standard Form In general, writing a quadratic function in the form ƒ1x2 5 a 1x 2 h22 1 k allows the vertex 1h, k2 and the axis of symmetry x 5 h to be determined by inspection. This form is a convenient way to express a quadratic function in order to quickly deter-mine its corresponding graph. Hence, this form is called standard form. 334 CHAPTER 4 Polynomial and Rational Functions EXAMPLE 1 Graphing a Quadratic Function Given in Standard Form Graph the quadratic function ƒ1x2 5 1x 2 322 2 1. Solution: STEP 1 The parabola opens up. a 5 1, so a . 0 STEP 2 Determine the vertex. 1h, k2 5 13, 212 STEP 3 Find the y-intercept. ƒ102 5 12322 2 1 5 8 10, 82 corresponds to the y-intercept [CONCEPT CHECK] For the parabola y 5 2(x 2 2)2 1 1, will the parabola open up or down? Will it have x-intercepts? ANSWER Down; Yes ▼ Recall that graphing linear functions requires finding two points on the line, or a point and the slope of the line. However, for quadratic functions simply knowing two points that lie on its graph is no longer sufficient. Below is a general step-by-step procedure for graphing quadratic functions given in standard form. QUADRATIC FUNCTION: STANDARD FORM The quadratic function ƒ1x2 5 a1x 2 h22 1 k is in standard form. The graph of ƒ is a parabola whose vertex is the point 1h, k2. The parabola is symmetric with respect to the line x 5 h. If a . 0, the parabola opens up. If a , 0, the parabola opens down. GRAPHING QUADRATIC FUNCTIONS To graph ƒ1x2 5 a 1x 2 h22 1 k Step 1: Determine whether the parabola opens up or down. a . 0 up a , 0 down Step 2: Determine the vertex 1h, k2. Step 3: Find the y-intercept 1by setting x 5 02. Step 4: Find any x-intercepts 1by setting ƒ1x2 5 0 and solving for x2. Step 5: Plot the vertex and intercepts and connect them with a smooth curve. x y x y x y Note that Step 4 says to “find any x-intercepts.” Parabolas opening up or down will always have a y-intercept. However, they can have one, two, or no x-intercepts. The figures above illustrate this for parabolas opening up, and the same can be said about parabolas opening down. ▼ A N S W E R x y (1, –4) (0, –3) (–1, 0) (3, 0) STUDY TIP A quadratic function given in standard form can be graphed using the transformation techniques shown in Section 3.3 for the square function. STEP 4 Find any x-intercepts. ƒ1x2 5 1x 2 322 2 1 5 0 1x 2 322 5 1 Use the square-root method. x 2 3 5 61 Solve. x 5 2 or x 5 4 12, 02 and 14, 02 correspond to the x-intercepts STEP 5 Plot the vertex and intercepts 13, 212, 10, 82, 12, 02, 14, 02. Connect the points with a smooth curve opening up. The graph in Example 1 could also have been found by shifting the square function to the right three units and down one unit. Y OUR T UR N Graph the quadratic function ƒ1x2 5 1x 2 122 2 4. ▼ x y (3, –1) (2, 0) (4, 0) (0, 8) x y (3, –1) (2, 0) (4, 0) (0, 8) 4.1 Quadratic Functions 335 EXAMPLE 2 Graphing a Quadratic Function Given in Standard Form with a Negative Leading Coefficient Graph the quadratic function ƒ1x2 5 221x 2 122 2 3. Solution: STEP 1 The parabola opens down. a 5 22, so a , 0 STEP 2 Determine the vertex. 1h, k2 5 11, 232 STEP 3 Find the y-intercept. ƒ102 5 2212122 2 3 5 22 2 3 5 25 10, 252 corresponds to the y-intercept STEP 4 Find any x-intercepts. ƒ1x2 5 221x 2 122 2 3 5 0 221x 2 122 5 3 1x 2 122 5 23 2 The square root of a negative number is not a real number. No real solutions. There are no x-intercepts. x 2 1 5 6Å23 2 x 5 1 6 i Å 3 2 336 CHAPTER 4 Polynomial and Rational Functions ▼ A N S W E R x y (–2, –5) (0, –5) (–1, –2) Graphing Quadratic Functions in General Form A quadratic function is often written in one of two forms: Standard form: ƒ1x2 5 a 1x 2 h22 1 k General form: ƒ1x2 5 ax2 1 bx 1 c When the quadratic function is expressed in standard form, the graph is easily obtained by identifying the vertex 1h, k2 and the intercepts and by drawing a smooth curve that opens either up or down, depending on the sign of a. Typically, quadratic functions are expressed in general form and a graph is the ultimate goal, so we must first express the quadratic function in standard form. One technique for transforming a quadratic function from general form to standard form was introduced in Section 1.3 and is called completing the square. When graphing quadratic functions (parabolas), have at least 3 points labeled on the graph. ■ ■When there are x-intercepts (Example 1), label the vertex, y-intercept, and x-intercepts. ■ ■When there are no x-intercepts (Example 2), label the vertex, y-intercept, and another point. STEP 5 Plot the vertex 11, 232 and y-intercept 10, 252. Connect the points with a smooth curve. Note that the axis of symmetry is x 5 1. Because the point 10, 252 lies on the parabola, then by symmetry with respect to x 5 1, the point 12, 252 also lies on the graph. Y OUR TU R N Graph the quadratic function ƒ1x2 5 231x 1 122 2 2. ▼ x y (0, –5) (2, –5) (1, –3) EXAMPLE 3 Graphing a Quadratic Function Given in General Form Graph the quadratic function ƒ1x2 5 x2 2 6x 1 4. Solution: Express the quadratic function in standard form by completing the square. Write the original function. ƒ1x2 5 x2 2 6x 1 4 Group the variable terms together. 5 1x2 2 6x2 1 4 Complete the square. Half of 26 is 23; 23 squared is 9. Add and subtract 9 within the parentheses. 5 1x2 2 6x 1 9 2 92 1 4 Write the 29 outside the parentheses. 5 1x2 2 6x 1 92 2 9 1 4 Write the expression inside the parentheses as a perfect square and simplify. 5 1x 2 322 2 5 STUDY TIP Although either form (standard or general) can be used to find the intercepts, it is often more convenient to use the general form when finding the y- intercept and the standard form when finding the x-intercept. Now that the quadratic function is written in standard form, ƒ1x2 5 1x 2 322 2 5, we follow our step-by-step procedure for graphing a quadratic function in standard form. STEP 1 The parabola opens up. a 5 1, so a . 0 STEP 2 Determine the vertex. 1h, k2 5 13, 252 STEP 3 Find the y-intercept. ƒ102 5 1022 2 6102 1 4 5 4 10, 42 corresponds to the y-intercept STEP 4 Find any x-intercepts. ƒ1x2 5 0 ƒ1x2 5 1x 2 322 2 5 5 0 1x 2 322 5 5 x 2 3 5 6!5 x 5 3 6 !5 13 1 !5, 02 and 13 2 !5, 02 correspond to the x-intercepts. STEP 5 Plot the vertex and intercepts 13, 252, 10, 42, 13 1 !5, 02, and 13 2 !5, 02. Connect the points with a smooth parabolic curve. Note: 3 1 !5 < 5.24 and 3 2 !5 < 0.76. Y OUR T UR N Graph the quadratic function ƒ1x2 5 x2 2 8x 114. ▼ x y (3, –5) (0, 4) (0.76, 0) (5.24, 0) When the leading coefficient of a quadratic function is not equal to 1, the leading coefficient must be factored out before completing the square. EXAMPLE 4 Graphing a Quadratic Function Given in General Form with a Negative Leading Coefficient Graph the quadratic function ƒ1x2 5 23x2 1 6x 1 2. Solution: Express the function in standard form by completing the square. Write the original function. ƒ1x2 5 23x2 1 6x 1 2 Group the variable terms together. 5 123x2 1 6x2 1 2 Factor out 23 in order to make the coefficient of x2 equal to 1 inside the parentheses. 5 231x2 2 2x2 1 2 Add and subtract 1 inside the parentheses to create a perfect square. 5 231x2 2 2x 1 1 2 12 1 2 Regroup the terms. 5 231x2 2 2x 1 12 2312 12 1 2 Write the expression inside the parentheses as a perfect square and simplify. 5 231x 2 122 1 5 ▼ A N S W E R x y (5.4, 0) (2.6, 0) (4, –2) 4.1 Quadratic Functions 337 338 CHAPTER 4 Polynomial and Rational Functions Now that the quadratic function is written in standard form, ƒ1x2 5 231x 2 122 1 5, we follow our step-by-step procedure for graphing a quadratic function in standard form. STEP 1 The parabola opens down. a 5 23, therefore a , 0 STEP 2 Determine the vertex. 1h, k2 5 11, 52 STEP 3 Find the y-intercept using ƒ102 5 231022 1 6102 1 2 5 2 the general form. 10, 22 corresponds to the y-intercept STEP 4 Find any x-intercepts using the standard form. ƒ1x2 5 231x 2 122 1 5 5 0 231x 2 122 5 25 1x 2 122 5 5 3 x 2 1 5 6 Å 5 3 x 5 1 6 Å 5 3 x 5 1 6 !15 3 The x-intercepts are a1 1 !15 3 , 0b and a1 2 !15 3 , 0b. STEP 5 Plot the vertex and intercepts 11, 52, 10, 22, a1 1 !15 3 , 0b, and a1 2 !15 3 , 0b. Connect the points with a smooth curve. Note: 1 1 !15 3 < 2.3 and 1 2 !15 3 < 20.3. Y OUR TU R N Graph the quadratic function ƒ1x2 5 22x2 2 4x 1 1. ▼ ▼ A N S W E R x y –5 5 5 (–1, 3) (–2.2, 0) (0, 1) (0.2, 0) –5 x y (1, 5) (0, 2) (–0.3, 0) (2.3, 0) In Examples 3 and 4, the quadratic functions were given in general form, and they were transformed into standard form by completing the square. It can be shown (by completing the square) that the vertex of a quadratic function in general form, ƒ1x2 5 ax2 1 bx 1 c, is located at x 5 2 b 2a. Another approach to sketching the graphs of quadratic functions is to first find the vertex and then find additional points through point-plotting. ▼ A N S W E R x y –2 –2 8 8 (1, 1) (2, 4) (0, 4) EXAMPLE 5 Graphing a Quadratic Function Given in General Form Sketch the graph of ƒ1x2 5 22x2 1 4x 1 5. Solution: STEP 1 Find the vertex. x 5 2 b 2a 5 2 4 21222 5 1 ƒ112 5 221122 1 4112 1 5 5 7 Vertex 5 11, 72 STEP 2 The parabola opens down. a 5 22 STEP 3 Find additional points near the vertex. x 21 0 1 2 3 ƒ1x2 ƒ1212 5 21 ƒ102 5 5 ƒ112 5 7 ƒ122 5 5 ƒ132 5 21 STEP 4 Label the vertex and additional points, then sketch the graph. Y OUR T UR N Sketch the graph of ƒ1x2 5 3x2 2 6x 1 4. ▼ Let a 5 22, b 5 4, and c 5 5. x y (0, 5) (2, 5) (1, 7) (3, –1) (–1, –1) VERTEX OF A PARABOLA The graph of a quadratic function ƒ1x2 5 ax2 1 bx 1 c is a parabola with the vertex located at the point a2 b 2a, ƒa2 b 2ab b 5 a 2 b 2a, 4ac 2 b2 4a b GRAPHING A QUADRATIC FUNCTION IN GENERAL FORM Step 1: Find the vertex. Step 2: Determine whether the parabola opens up or down. ■ If a . 0, the parabola opens up. ■ If a , 0, the parabola opens down. Step 3: Find additional points near the vertex. Step 4: Sketch the graph with a parabolic curve. 4.1 Quadratic Functions 339 340 CHAPTER 4 Polynomial and Rational Functions 4.1.2 Finding the Equation of a Parabola It is important to understand that the equation y 5 x2 is equivalent to the quadratic function ƒ1x2 5 x2. Both have the same parabolic graph. Thus far, we have been given the equation and then asked to find characteristics (vertex and intercepts) in order to graph. We now turn our attention to the problem of determining the equation, or function, given certain characteristics. EXAMPLE 6 Finding the Equation of a Parabola Given the Vertex and Another Point Find the equation of a parabola whose graph has a vertex at 13, 42 and that passes through the point 12, 32. Express the quadratic function in both standard and general forms. Solution: Write the standard form of the equation of a parabola. ƒ1x2 5 a 1x 2 h22 1 k Substitute the coordinates of the vertex 1h, k2 5 13, 42. ƒ1x2 5 a 1x 2 322 1 4 Use the point 12, 32 to find a. The point 12, 32 implies ƒ122 5 3. ƒ122 5 a 12 2 322 1 4 5 3 Solve for a. a 12 2 322 1 4 5 3 a 12122 1 4 5 3 a 1 4 5 3 a 5 21 Write both forms of the equation of this parabola. Standard form: ƒ1x2 5 21x 2 322 1 4 General form: ƒ1x2 5 2x2 1 6x 2 5 Y OUR TU R N Find the standard form of the equation of a parabola whose graph has a vertex at 123, 252 and that passes through the point 122, 242. ▼ 4.1.2 SKI LL Find the equation of a parabola. 4.1.2 CO NCE PTUAL Understand that as long as you know the vertex and a point that lies along the graph of a parabola, you can determine the equation of the parabola. ▼ A N S W E R Standard form: ƒ1x2 5 1x 1 322 2 5 As we have seen in Example 6, once the vertex is known, the leading coefficient a can be found from any point that lies on the parabola. Application Problems That Involve Quadratic Functions ■ ■What is the minimum distance a driver has to maintain in order to be at a safe distance between her car and the car in front as a function of speed? ■ ■How many units produced will yield a maximum profit for a company? ■ ■Given the number of linear feet of fence, what rectangular dimensions will yield a maximum fenced-in area? ■ ■If a particular stock price has been shown to follow a quadratic trend, when will the stock achieve a maximum value? ■ ■If a gun is fired into the air, where will the bullet land? These are all problems that can be solved using quadratic functions. Because the vertex of a parabola represents either the minimum or maximum value of the quadratic function, in application problems it often suffices simply to find the vertex. In Example 6, the function F 1x2 5 2x2 1 6x 2 5, which can also be written F 1x2 5 21x 2 322 1 4, was shown to “open down” and has a vertex at the point 13, 42. Suppose this function represents profit, where x is the number (millions) of units made. The vertex would then represent the maximum profit. Instead of rewriting the function in standard form through completing the square, we use the vertex formula x 5 2 b 2a. WORDS MATH Quadratic function. F 1x2 5 2x2 1 6x 2 5 Coefficients. a 5 21, b 5 6, c 5 25 Find the x-coordinate of the vertex. x 5 2 b 2a 5 2 6 21212 5 3 Find the value o f the function at x 5 3. ƒ132 5 21322 1 6132 2 5 5 4 Therefore, the vertex is located at the point 13, 42 . [CONCEPT CHECK] TRUE OR FALSE If you know the vertex and a point (other than the vertex) that lies on the graph, you can determine the y-intercept. ANSWER True ▼ EXAMPLE 7 Finding the Minimum Cost of Manufacturing a Motorcycle A company that produces motorcycles has a daily production cost of C1x2 5 2000 2 15x 1 0.05x2 where C is the cost in dollars to manufacture a motorcycle and x is the number of motorcycles produced. How many motorcycles can be produced each day in order to minimize the cost of each motorcycle? What is the corresponding minimum cost? Solution: The graph of the quadratic function is a parabola. Rewrite the quadratic function in general form. C 1x2 5 0.05x2 2 15x 1 2000 The parabola opens up, because a is positive. a 5 0.05 . 0 Because the parabola opens up, the vertex of the parabola is a minimum. Find the x-coordinate of the vertex. x 5 2 b 2a 5 2 12152 210.052 5 150 The company keeps costs to a minimum when 150 motorcycles are produced each day. The minimum cost is $875 per motorcycle. C 11502 5 875 YOUR T UR N The revenue associated with selling vitamins is R1x2 5 500x 2 0.001x2 where R is the revenue in dollars and x is the number of bottles of vitamins sold. Determine how many bottles of vitamins should be sold to maximize the revenue. ▼ ▼ A N S W E R 250,000 bottles EXAMPLE 8 Finding the Dimensions That Yield a Maximum Area You have just bought a puppy and want to fence in an area in the backyard for her. You buy 100 linear feet of fence from Home Depot and have decided to make a rectangular fenced-in area using the back of your house as one side. Determine the dimensions of the rectangu lar pen that will maximize the area in which your puppy may roam. What is the maximum area of the rectangular pen? 4.1 Quadratic Functions 341 342 CHAPTER 4 Polynomial and Rational Functions ▼ A N S W E R 50 feet by 50 feet Solution: STEP 1 Identify the question. Find the dimensions of the rectangular pen. STEP 2 Draw a picture. STEP 3 Set up a function. If we let x represent the length of one side of the rectangle, then the opposite side is also of length x. Because there are 100 feet of fence, the remaining fence left for the side opposite the house is 100 2 2x. The area of a rectangle is equal to length times width: A1x2 5 x1100 2 2x2 STEP 4 Find the maximum value of the function. A1x2 5 x1100 2 2x2 5 22x2 1 100x Find the maximum of the parabola that corresponds to the quadratic function for area A1x2 5 22x2 1 100x. a 5 22 and b 5 100; therefore, the maximum occurs when x 5 2 b 2a 5 2 100 21222 5 25 Replacing x with 25 in our original diagram: The dimensions of the rectangle are 25 feet by 50 feet . The maximum area A1252 5 1250 is 1250 square feet . STEP 5 Check the solution. Two sides are 25 feet and one side is 50 feet, and together they account for all 100 feet of fence. Y OUR TU R N Suppose you have 200 linear feet of fence to enclose a rectangular garden. Determine the dimensions of the rectangle that will yield the greatest area. ▼ Pen House Pen 100 – 2x x x House Pen 100 – 2x = 100 – 2(25) = 50 x = 25 x = 25 House EXAMPLE 9 Path of a Punted Football The path of a particular punt follows the quadratic function: h1x2 5 21 8 1x 2 522 1 50, where h 1x2 is the height of the ball in yards and x corresponds to the horizontal distance in yards. Assume x 5 0 corresponds to midfield (the 50-yard line). For example, x 5 220 corresponds to the punter’s own 30-yard line, whereas x 5 20 corresponds to the other team’s 30-yard line. x = 0 x = –20 x = 20 Punter a. Find the maximum height the ball achieves. b. Find the horizontal distance the ball covers. Assume the height is zero when the ball is kicked and when the ball is caught. Solution (a): Identify the vertex since it is given in standard form. 1h, k2 5 15, 502 The maximum height of the punt occurs at the other team’s 45-yard line, and the height the ball achieves is 50 yards 1150 feet2 . Solution (b): The height when the ball is kicked or caught is zero. Solve for x. 1 8 1x 2 522 5 50 1x 2 522 5 400 1x 2 52 5 6!400 x 5 5 6 20 x 5 215 and x 5 25 The horizontal distance is the distance between these two points: 025 2 1215205 40 yards . h1x2 5 2 1 8 1x 2 522 1 50 5 0 4.1 Quadratic Functions 343 n When the quadratic function is given in general form, completing the square can be used to rewrite the function in standard form. n At least three points are needed to graph a quadratic function. n vertex n y-intercept n x-intercept(s) or other point(s) All quadratic functions ƒ1x2 5 ax2 1 bx 1 c or ƒ1x2 5 a 1x 2 h22 1 k have graphs that are parabolas: n If a . 0, the parabola opens up. n If a , 0, the parabola opens down. n The vertex is at the point 1h, k2 5 a2 b 2a, ƒa2 b 2abb 5 a2 b 2a, 4ac 2 b2 4a b [SEC TION 4 .1] S U M MA RY 344 CHAPTER 4 Polynomial and Rational Functions In Exercises 1–4, match the quadratic function with its graph. 1. ƒ1x2 5 31x 1 222 2 5 2. ƒ1x2 5 21x 2 122 1 3 3. ƒ1x2 5 21 21x 1 322 1 2 4. ƒ1x2 5 21 31x 2 222 1 3 a. b. c. d. In Exercises 5–8, match the quadratic function with its graph. 5. ƒ1x2 5 3x2 1 5x 2 2 6. ƒ1x2 5 3x2 2 x 2 2 7. ƒ1x2 5 2x2 1 2x 2 1 8. ƒ1x2 5 22x2 2 x 1 3 a. b. c. d. In Exercises 9–22, graph the quadratic function, which is given in standard form. 9. ƒ1x2 5 1x 1 122 2 2 10. ƒ1x2 5 1x 1 222 2 1 11. ƒ1x2 5 1x 2 222 2 3 12. ƒ1x2 5 1x 2 422 1 2 13. ƒ1x2 5 21x 2 322 1 9 14. ƒ1x2 5 21x 2 522 2 4 15. ƒ1x2 5 21x 1 122 2 3 16. ƒ1x2 5 21x 2 222 1 6 17. ƒ1x2 5 21x 2 222 1 2 18. ƒ1x2 5 231x 1 222 2 15 19. ƒ1x2 5 Ax 2 1 3B2 1 1 9 20. ƒ1x2 5 Ax 1 1 4B2 2 1 2 21. ƒ1x2 5 20.51x 2 0.2522 1 0.75 22. ƒ1x2 5 20.21x 1 0.622 1 0.8 In Exercises 23–32, rewrite the quadratic function in standard form by completing the square. 23. ƒ1x2 5 x2 1 6x 2 3 24. ƒ1x2 5 x2 1 8x 1 2 25. ƒ1x2 5 2x 2 2 10x 1 3 26. ƒ1x2 5 2x2 2 12x 1 6 27. ƒ1x2 5 2x2 1 8x 2 2 28. ƒ1x2 5 3x2 2 9x 1 11 29. ƒ1x2 5 24x 2 1 16x 2 7 30. ƒ1x2 5 25x2 1 100x 2 36 31. ƒ1x2 5 1 2x2 2 4x 1 3 32. ƒ1x2 5 21 3 x2 1 6x 1 4 In Exercises 33–40, graph the quadratic function. 33. ƒ1x2 5 x2 1 6x 2 7 34. ƒ1x2 5 x2 2 3x 1 10 35. ƒ1x2 5 x2 1 2x 2 15 36. ƒ1x2 5 2x2 1 3x 1 4 37. ƒ1x2 5 22x2 2 12x 2 16 38. ƒ1x2 5 23x2 1 12x 2 12 39. ƒ1x2 5 1 2x2 2 1 2 40. ƒ1x2 5 21 3 x2 1 4 3 In Exercises 41–48, find the vertex of the parabola associated with each quadratic function. 41. ƒ1x2 5 33x2 2 2x 1 15 42. ƒ1x2 5 17x2 1 4x 2 3 43. ƒ1x2 5 1 2x2 2 7x 1 5 44. ƒ1x2 5 21 3 x2 1 2 5 x 1 4 45. ƒ1x2 5 20.002x2 2 0.3x 1 1.7 46. ƒ1x2 5 0.05x2 1 2.5x 2 1.5 47. ƒ1x2 5 22 5 x2 1 3 7x 1 2 48. ƒ1x2 5 21 7 x2 2 2 3 x 1 1 9 x y –8 2 4 –6 x y –5 2 5 –5 x y –4 6 4 –6 y x –3 3 10 [SEC TION 4 .1] E X E R C I SES • S K I L L S y 4 –6 x –5 5 x y –5 5 4 –6 x y –5 5 2 –8 x y –5 5 6 –4 In Exercises 49–58, find the quadratic function that has the given vertex and goes through the given point. 49. vertex: 121, 42 point: 10, 22 50. vertex: 12, 232 point: 10, 12 51. vertex: 12, 52 point: 13, 02 52. vertex: 11, 32 point: 122, 02 53. vertex: 121, 232 point: 124, 22 54. vertex: 10, 222 point: 13, 102 55. vertex: A1 2, 23 4B point: A3 4, 0B 56. vertex: A 25 6, 2 3B point: 10, 02 57. vertex: 12.5, 23.52 point: 14.5, 1.52 58. vertex: 11.8, 2.72 point: 122.2, 22.12 4.1 Quadratic Functions 345 • A P P L I C A T I O N S 59. Business. The annual profit for a company that manufactures cell phone accessories can be modeled by the function P 1x2 5 20.0001x2 1 70x 1 12,500 where x is the number of units sold and P is the total profit in dollars. a. What sales level maximizes the company’s annual profit? b. Find the maximum annual profit for the company. 60. Business. A manufacturer of office supplies has daily production costs of C1x2 5 0.5x2 2 20x 1 1600 where x is the number of units produced measured in thousands and C is cost in hundreds of dollars. a. What production level will minimize the manufacturer’s daily production costs? b. Find the minimum daily production costs for the manufacturer. For Exercises 61 and 62, refer to the following: An adult male’s weight, in kilograms, can be modeled by the function W 1t2 5 2 2 3 t2 1 13 5 t 1 433 5 ; 1 # t # 18 where t measures months 1t 5 1 is January 2010, t 5 2 is February 2010, etc.2 and W is the male’s weight. 61. Health/Medicine. During which months was the male losing weight and gaining weight? 62. Health/Medicine. Find the maximum weight to the nearest kilogram of the adult male during the 18 months. Exercises 63 and 64 concern the path of a punted football. Refer to the diagram in Example 9. 63. Sports. The path of a particular punt follows the quadratic function h1x2 5 2 8 125 1x 1 522 1 40 where h 1x2 is the height of the ball in yards and x corresponds to the horizontal distance in yards. Assume x 5 0 corresponds to midfield (the 50-yard line). For example, x 5 220 corresponds to the punter’s own 30-yard line, whereas x 5 20 corresponds to the other team’s 30-yard line. a. Find the maximum height the ball achieves. b. Find the horizontal distance the ball covers. Assume the height is zero when the ball is kicked and when the ball is caught. 64. Sports. The path of a particular punt follows the quadratic function h1x2 5 2 5 40 1x 2 3022 1 50 where h 1x2 is the height of the ball in yards and x corresponds to the horizontal distance in yards. Assume x 5 0 corresponds to midfield (the 50-yard line). For example, x 5 220 corresponds to the punter’s own 30-yard line, whereas x 5 20 corresponds to the other team’s 30-yard line. a. Find the maximum height the ball achieves. b. Find the horizontal distance the ball covers. Assume the height is zero when the ball is kicked and when the ball is caught. 65. Ranching. A rancher has 10,000 linear feet of fencing and wants to enclose a rectangular field and then divide it into two equal pastures, with an internal fence parallel to one of the rectangular sides. What is the maximum area of each pasture? Round to the nearest square foot. 66. Ranching. A rancher has 30,000 linear feet of fencing and wants to enclose a rectangular field and then divide it into four equal pastures, with three internal fences parallel to one of the rectangular sides. What is the maximum area of each pasture? 346 CHAPTER 4 Polynomial and Rational Functions 67. Gravity. A person standing near the edge of a cliff 100 feet above a lake throws a rock upward with an initial speed of 32 feet per second. The height of the rock above the lake at the bottom of the cliff is a function of time and is described by h1t2 5 216t 2 1 32t 1 100 a. How many seconds will it take until the rock reaches its maximum height? What is that height? b. At what time will the rock hit the water? 100 feet 68. Gravity. A person holds a pistol straight upward and fires. The initial velocity of most bullets is around 1200 feet per second. The height of the bullet is a function of time and is described by h1t2 5 216t 2 1 1200t How long, after the gun is fired, does the person have to get out of the way of the bullet falling from the sky? 69. Zero Gravity. As part of their training, astronauts rode the “vomit comet,” NASA’s reduced gravity KC 135A aircraft that performed parabolic flights to simulate weightlessness. The plane started at an altitude of 20,000 feet and made a steep climb at 528 with the horizon for 20–25 seconds and then dove at that same angle back down, repeatedly. The equation governing the altitude of the flight is A1x2 5 20.0003x2 1 9.3x 2 46,075 where A1x2 is altitude and x is horizontal distance in feet. a. What is the maximum altitude the plane attains? b. Over what horizontal distance is the entire maneuver performed? (Assume the starting and ending altitude is 20,000 feet.) Courtesy NASA 70. Sports. A soccer ball is kicked from the ground at a 458 angle with an initial velocity of 40 feet per second. The height of the soccer ball above the ground is given by H 1x2 5 20.0128x2 1 x, where x is the horizontal distance the ball travels. a. What is the maximum height the ball reaches? b. What is the horizontal distance the ball travels? 71. Profit. A small company in Virginia Beach manufac tures handcrafted surfboards. The profit on selling x boards is given by P1x2 5 20,000 1 80x 2 0.4x2 a. How many boards should be made to maximize the profit? b. What is the maximum profit? 72. Environment: Fuel Economy. Gas mileage (miles per gallon, mpg) can be approximated by a quadratic function of speed. For a particular automobile, assume the vertex occurs when the speed is 50 miles per hour (mph); the mpg will be 30. a. Write a quadratic function that models this relationship, assuming 70 mph corresponds to 25 mpg. b. What gas mileage would you expect for this car driving 90 mph? Speed (mph) 90 10 70 50 30 MPG 5 15 25 35 45 (50, 30) (70, 25) For Exercises 73 and 74, use the following information: One function of particular interest in economics is the profit function. We denote this function by P 1x2. It is defined to be the difference between revenue R 1x2 and cost C1x2 so that P1x2 5 R1x2 2 C1x2 The total revenue received from the sale of x goods at price p is given by R1x2 5 px The total cost function relates the cost of production to the level of output x. This includes both fixed costs Cf and variable costs Cv (costs per unit produced). The total cost in producing x goods is given by C1x2 5 Cf 1 Cvx Thus, the profit function is P 1x2 5 px 2 Cf 2 Cvx Assume fixed costs are $1000, variable costs per unit are $20, and the demand function is p 5 100 2 x NASA’s “Vomit Comet” 73. Profit. How many units should the company produce to break even? 74. Profit. What is the maximum profit? 75. Cell Phones. The number of cell phones in the United States can be approximated by a quadratic function. In 1996 there were approximately 16 million cell phones, and in 2015 there were approximately 190.5 million. Let t be the number of years since 1996. The number of cell phones in 1996 is represented by 10, 162, and the number in 2015 is 119, 190.52. Let 10, 162 be the vertex. a. Find a quadratic function that represents the number of cell phones. b. Based on this model, how many cell phones were used in 2016? 76. Underage Smoking. The number of underage cigarette smokers (ages 10–17) has declined in the United States. The peak percent was in 1998 at 49%. In 2013 this had dropped to 15.7%. Let t be time in years after 1998 1t 5 0 corresponds to 19982. a. Find a quadratic function that models the percent of underage smokers as a function of time. Let 10, 492 be the vertex. b. Based on this model, what was the percent of underage smokers in 2016? 77. Drug Concentration. The concentration of a drug in the bloodstream, measured in parts per million, can be modeled with a quadratic function. In 50 minutes the concentration is 93.75 parts per million. The maximum concentration of the drug in the bloodstream occurs in 225 minutes and is 400 parts per million. a. Find a quadratic function that models the concentration of the drug as a function of time in minutes. b. In how many minutes will the drug be eliminated from the bloodstream? 78. Revenue. Jeff operates a mobile car washing business. When he charged $20 a car, he washed 70 cars a month. He raised the price to $25 a car, and his business dropped to 50 cars a month. a. Find a linear function that represents the demand equation (the price per car as a function of the number of cars washed). b. Find the revenue function R 1x2 5 xp. c. How many cars should he wash to maximize the revenue? d. What price should he charge to maximize revenue? 4.1 Quadratic Functions 347 In Exercises 79–82, explain the mistake that is made. There may be a single mistake, or there may be more than one mistake. • C A T C H T H E M I S T A K E 79. Plot the quadratic function ƒ1x2 5 1x 1 322 2 1. Solution: STEP 1: The parabola opens up because a 5 1 . 0. STEP 2: The vertex is 13, 212. STEP 3: The y-intercept is 10, 82. STEP 4: The x-intercepts are 12, 02 and 14, 02. STEP 5: Plot the vertex and intercepts, and connect the points with a smooth curve. x y (3, –1) (2, 0) (4, 0) This is incorrect. What mistake(s) was made? 80. Determine the vertex of the quadratic function ƒ1x2 5 2x2 2 6x 2 18. Solution: STEP 1: The vertex is given by 1h, k2 5 a2 b 2a, ƒa2 b 2abb. In this case, a 5 2 and b 5 6. STEP 2: The x-coordinate of the vertex is x 5 2 6 2122 5 2 6 4 5 2 3 2 STEP 3: The y-coordinate of the vertex is ƒa23 2b 5 22a23 2b 2 1 6a23 2b 2 18 5 22 a9 4b 2 18 2 2 18 5 2 9 2 2 9 2 18 5 2 63 2 This is incorrect. What mistake(s) was made? 81. Rewrite the following quadratic function in standard form: ƒ1x2 5 2x2 1 2x 1 3 Solution: STEP 1: Group the variables together. 12x2 1 2x2 1 3 STEP 2: Factor out a negative. 21x2 1 2x2 1 3 STEP 3: Add and subtract 1 inside the parentheses. 21x2 1 2x 1 1 2 12 1 3 STEP 4: Factor out the 21. 21x2 1 2x 1 12 1 1 1 3 STEP 5: Simplify. 21x 1 122 1 4 This is incorrect. What mistake(s) was made? 348 CHAPTER 4 Polynomial and Rational Functions 82. Find the quadratic function whose vertex is 12, 232 and whose graph passes through the point 19, 02. Solution: STEP 1: Write the quadratic function in standard form. ƒ1x2 5 a 1x 2 h22 1 k STEP 2: Substitute 1h, k2 5 12, 232. ƒ1x2 5 a 1x 2 222 2 3 STEP 3: Substitute the point 19, 02 and solve for a. ƒ102 5 a 10 2 222 2 3 5 9 4a 2 3 5 9 4a 5 12 a 5 3 The quadratic function sought is ƒ1x2 5 31x 2 222 2 3. This is incorrect. What mistake(s) was made? • C O N C E P T U A L In Exercises 83–86, determine whether each statement is true or false. 83. A quadratic function must have a y-intercept. 84. A quadratic function must have an x-intercept. 85. A quadratic function may have more than one y-intercept. 86. A quadratic function may have more than one x-intercept. 87. For the general quadratic equation, ƒ1x2 5 ax2 1 bx 1 c, show that the vertex is 1h, k2 5 a2 b 2a, ƒa2 b 2abb. 88. Given the quadratic function ƒ1x2 5 a 1x 2 h22 1 k, determine the x- and y-intercepts in terms of a, h, and k. 89. A rancher has 1000 feet of fence to enclose a pasture. a. Determine the maximum area if a rectangular fence is used. b. Determine the maximum area if a circular fence is used. 90. A 600-room hotel in Orlando is filled to capacity every night when the rate is $90 per night. For every $5 increase in the rate, 10 fewer rooms are filled. How much should the hotel charge to produce the maximum income? What is the maximum income? • C H A L L E N G E • T E C H N O L O G Y 91. On a graphing calculator, plot the quadratic function ƒ1x2 5 20.002x2 1 5.7x 2 23. a. Identify the vertex of this parabola. b. Identify the y-intercept. c. Identify the x-intercepts (if any). d. What is the axis of symmetry? 92. Determine the quadratic function whose vertex is 120.5, 1.72 and whose graph passes through the point 10, 42. a. Write the quadratic function in general form. b. Plot this quadratic function with a graphing calculator. c. Zoom in on the vertex and y-intercept. Do they agree with the given values? In Exercises 93 and 94: (a) use the calculator commands STAT QuadReg to model the data using a quadratic function; (b) write the quadratic function in standard form and identify the vertex; (c) plot this quadratic function with a graphing calculator and use the TRACE key to highlight the given points. Do they agree with the given values? 93. 94. For Exercises 95 and 96, refer to the following discussion of quadratic regression: The “least-squares” criterion used to create a quadratic regression curve y 5 ax2 1 bx 1 c that fits a set of n data points 1x1, y12, 1x2, y22, . . . , 1xn, yn2 is that the sum of the squares of the vertical distances from the points to the curve be minimum. This means that we need to determine values of a, b, and c for which a n i51 1yi 2 1axi 2 1 bxi 1 c222 is as small as possible. Calculus can be used to determine formulas for a, b, and c that do the job, but computing them by hand is tedious and unnecessary because the TI-831 has a built-in program called QuadReg that does this. In fact, this was introduced in Section 2.5, Problems 37–40. The following are application problems that involve experimental data for which the best fit curve is a parabola. Projectile Motion It is well known that the trajectory of an object thrown with initial velocity v0 from an initial height s0 is described by the quadratic function s1t2 5 216t2 1 v0t 1 s0. If we have data points obtained in such a context, we can apply the procedure outlined in Section 2.5 with QuadReg in place of LinReg 1ax1b2 to find a best fit parabola of the form y 5 ax2 1 bx 1 c. Each year during Halloween season, it is tradition to hold the Pumpkin Launching Contest where students literally hurl their pump-kins in the hope of throwing them the farthest horizontal distance. x 29 22 4 y 22.72 216.18 6.62 x 22 2 5 y 229.28 21.92 18.32 4.2 Polynomial Functions of Higher Degree 349 Ben claims that it is better to use a steeper trajectory since it will have more air time, while Rick believes in throwing the pumpkin with all of his might but with less inclination. The following data points that describe the pumpkin’s horizontal x and vertical y distances (measured in feet) were collected during the flights of their pumpkins: BEN’S DATA RICK’S DATA x y x y 0 4 0 4 1 14.2 1 8.5 2 28.4 2 10.6 3 30.1 3 13.3 4 35.9 4 16.2 5 37.8 5 17.3 6 41.1 6 19.3 7 38.2 7 19.5 95. a. Form a scatterplot for Ben’s data. b. Determine the equation of the best fit parabola and report the value of the associated correlation coefficient. c. Use the best fit curve from (b) to answer the following: i. What is the initial height of the pumpkin’s trajectory, and with what initial velocity was it thrown? ii. What is the maximum height of Ben’s pumpkin’s trajectory? iii. How much horizontal distance has the pumpkin traveled by the time it lands? 96. Repeat Exercise 95 for Rick’s data. S K I L L S O B J E C T I V E S ■ ■Identify a polynomial function and determine its degree. ■ ■Graph polynomial functions using transformations. ■ ■Identify real zeros of a polynomial function and their multiplicities. ■ ■Graph polynomial functions using x-intercepts, multiplicity of each zero, and end behavior. C O N C E P T U A L O B J E C T I V ES ■ ■Understand that polynomial functions are both continuous and smooth. ■ ■Understand that power functions of even powers have similar shapes and power functions of odd powers have similar shapes. ■ ■Understand that the real zeros of a polynomial function correspond to x-intercepts on its graph. ■ ■Understand that odd multiplicity of a zero corresponds to the graph crossing the x-axis and even multiplicity of a zero corresponds to the graph touching the x-axis, and that end behavior is a result of the leading term dominating. 4.2 POLYNOMIAL FUNCTIONS OF HIGHER DEGREE 4.2.1 Identifying Polynomial Functions Polynomial functions model many real-world applications. Often, the input is time, and the output of the function can be many things. For example, the number of active-duty military personnel in the United States, the number of new cases in the spread of an epidemic, and the stock price as a function of time t can all be modeled with polynomial functions. DEFINITION Polynomial Function Let n be a nonnegative integer and let an, an21, . . . , a2, a1, a0 be real numbers with an 2 0 . The function ƒ1x2 5 anxn 1 an21xn21 1 c1 a2x2 1 a1x 1 a0 is called a polynomial function of x with degree n. The coefficient an is called the leading coefficient. Note: If n is a nonnegative integer, then n 2 1, n 2 2, . . . , 2, 1, 0 are also nonnegative integers. 4.2.1 S K I L L Identify a polynomial function and determine its degree. 4.2.1 C O N C E P T U A L Understand that polynomial functions are both continuous and smooth. 350 CHAPTER 4 Polynomial and Rational Functions Whenever we have discussed a particular polynomial, we have graphed it too. The graph of a constant function (degree 0) is a horizontal line. The graph of a general linear function (degree 1) is a slant line. The graph of a quadratic function (degree 2) is a parabola. These functions are summarized in the table below. EXAMPLE 1 Identifying Polynomials and Their Degree For each of the functions given, determine whether the function is a polynomial function. If it is a polynomial function, then state the degree of the polynomial. If it is not a polynomial function, justify your answer. a. ƒ1x2 5 3 2 2x5 b. F1x2 5 !x 1 1 c. g 1x2 5 2 d. h 1x2 5 3x2 2 2x 1 5 e. H 1x2 5 4x512x 2 322 f. G 1x2 5 2x4 2 5x3 2 4x22 Solution: a. ƒ1x2 is a polynomial function of degree 5. b. F 1x2 is not a polynomial function. The variable x is raised to the power of 1 2, which is not an integer. c. g 1x2 is a polynomial function of degree zero, also known as a constant function. Note that g 1x2 5 2 can also be written as g 1x2 5 2 x0 1assuming x 2 02. d. h 1x2 is a polynomial function of degree 2. A polynomial function of degree 2 is called a quadratic function. e. H 1x2 is a polynomial function of degree 7. Note: 4x514x2 2 12x 1 92 5 16x7 2 48x6 1 36x5. f. G 1x2 is not a polynomial function. 24x22 has an exponent that is negative. Y OUR TU R N For each of the functions given, determine whether the function is a polynomial function. If it is a polynomial function, then state the degree of the polynomial. If it is not a polynomial function, justify your answer. a. ƒ1x2 5 1 x 1 2 b. g1x2 5 3x81x 2 2221x 1 123 ▼ ▼ A N S W E R a. ƒ1x2 is not a polynomial because x is raised to the power of 21, which is a negative integer. b. g 1x2 is a polynomial of degree 13. POLYNOMIAL DEGREE SPECIAL NAME GRAPH ƒ1x2 5 c 0 Constant function Horizontal line ƒ1x2 5 mx 1 b 1 Linear function Line • Slope 5 m • y-intercept: 10, b2 ƒ1x2 5 ax2 1 bx 1 c 2 Quadratic function Parabola • Opens up if a . 0. • Opens down if a , 0. How do we graph polynomial functions that are of degree 3 or higher, and why do we care? Polynomial functions model real-world applications, as mentioned earlier. One example is the percentage of fat in our bodies as we age. We can model the weight of a baby after it comes home from the hospital as a function of time. When a baby comes home from the hospital, it usually experiences weight loss. Then typically there is an increase in the percent of body fat when the baby is nursing. When infants start to walk, the increase in exercise is associated with a drop in the percentage of fat. Growth spurts in children are examples of the percent of body fat increasing and decreasing. Later in life, our metabolism slows down, and typically the percent of body fat increases. We will model this with a polynomial function. Other examples are stock prices, the federal funds rate, and yo-yo dieting as functions of time. Polynomial functions are considered simple functions. Graphs of all polynomial functions are both continuous and smooth. A continuous graph is one you can draw completely without picking up your pencil (the graph has no jumps or holes). A smooth graph has no sharp corners. The following graphs illustrate what it means to be smooth (no sharp corners or cusps) and continuous (no holes or jumps). The graph is not continuous. x y Jump The graph is not continuous. x y Hole The graph is continuous but not smooth. x y Corner The graph is continuous and smooth. x y All polynomial functions have graphs that are both continuous and smooth. 4.2.2 Graphing Polynomial Functions Using Transformations of Power Functions Recall from Chapter 3 that graphs of functions can be drawn by hand using graphing aids such as intercepts and symmetry. The graphs of polynomial functions can be graphed using these same aids. Let’s start with the simplest types of polynomial functions called power functions. Power functions are monomial functions of the form ƒ1x2 5 xn, where n is an integer greater than zero. DEFINITION Power Function Let n be a positive integer and the coefficient a ≠ 0 be a real number. The function ƒ1x2 5 axn is called a power function of degree n. Power functions with even powers look similar to the square function. Power functions with odd powers (other than n 5 1) look similar to the cube function. f (x) = x2 f (x) = x4 f (x) = x6 f (x) = x3 f (x) = x5 f (x) = x7 4.2.2 S K I L L Graph polynomial functions using transformations. 4.2.2 C O N C E P T U A L Understand that power functions of even powers have similar shapes and power functions of odd powers have similar shapes. [CONCEPT CHECK] TRUE OR FALSE It is possible for the graph of a polynomial function to have a hole. ANSWER False ▼ STUDY TIP All polynomial functions have graphs that are both continuous and smooth. [CONCEPT CHECK] TRUE OR FALSE An even power function has a domain and range that are both the set of all real numbers. ANSWER False ▼ 4.2 Polynomial Functions of Higher Degree 351 352 CHAPTER 4 Polynomial and Rational Functions All even power functions have similar characteristics to a quadratic function (parabola), and all odd 1n . 12 power functions have similar characteristics to a cubic function. For example, all even functions are symmetric with respect to the y-axis, whereas all odd functions are symmetric with respect to the origin. The following table summarizes their characteristics. We now have the tools to graph polynomial functions that are transformations of power functions. We will use the power functions summarized above, combined with our graphing techniques such as horizontal and vertical shifting and reflection (Section 3.3) EXAMPLE 2 Graphing Transformations of Power Functions Graph the function ƒ1x2 5 1x 2 123. Solution: STEP 1 Start with the graph of y 5 x3. STEP 2 Shift y 5 x3 to the right one unit to yield the graph of ƒ 1x2 5 1x 2 123. Y OUR TU R N Graph the function ƒ1x2 5 1 2 x4. x –5 5 y 10 –10 x –5 5 y 10 –10 ▼ ▼ A N S W E R ƒ1x2 5 1 2 x4 x y (0, 1) (–1, 0) (1, 0) –5 5 5 –5 CHARACTERISTICS OF POWER FUNCTIONS: f (x) 5 xn n EVEN n ODD Symmetry y-axis Origin Domain 12q, q2 12q, q2 Range 30, q2 12q, q2 Some key points that lie on the graph 121, 12, 10, 02, and 11, 12 121, 212, 10, 02, and 11, 12 Increasing 10, q2 12q, q2 Decreasing 12q, 02 N/A 4.2.3 Real Zeros of a Polynomial Function How do we graph general polynomial functions of degree greater than or equal to 3 if they cannot be written as transformations of power functions? We start by identifying the x-intercepts of the polynomial function. Recall that we determine the x-intercepts by setting the function equal to zero and solving for x. Therefore, an alternative name for an x-intercept of a function is a zero of the function. In our experience, to set a quadratic function equal to zero, the first step is to factor the quadratic expression into linear factors and then set each factor equal to zero. Therefore, there are four equivalent relationships, which are summarized in the following box. [CONCEPT CHECK] For the graph of a polynomial function f (x) 5 (x 2 a)(x 2 b), what do a and b correspond to? ANSWER x-intercepts ▼ 4.2.3 S K I L L Identify real zeros of a polynomial function and their multiplicities. 4.2.3 C O N C E P T U A L Understand that the real zeros of a polynomial function correspond to x-intercepts on its graph. STUDY TIP Real zeros correspond to x-intercepts. Let’s use a simple polynomial function to illustrate these four relationships. We’ll focus on the quadratic function ƒ1x2 5 x2 2 1. The graph of this function is a parabola that opens up and has as its vertex the point 10, 212. Illustration of Relationship 1 Set the function equal to zero, ƒ1x2 5 0. x2 2 1 5 0 Factor. 1x 2 12 1x 1 12 5 0 Solve. x 5 1 or x 5 21 x 5 21 and x 5 1 are solutions, or roots, of the equation x2 2 1 5 0. Illustration of Relationship 2 The x-intercepts are 121, 02 and 11, 02. Ill ustration of Relationship 3 The value of the function at x 5 21 and x 5 1 is 0. ƒ 1212 5 12122 21 5 0 ƒ 112 5 1122 21 5 0 Illustration of Relationship 4 ƒ1x2 5 x2 2 1 5 1x 2 121x 1 12 x 5 1 is a zero of the polynomial, so 1x 2 12 is a factor. x 5 21 is a zero of the polynomial, so 1x 1 12 is a factor. We have a good reason for wanting to know the x-intercepts, or zeros. When the value of a continuous function transitions from negative to positive and vice versa, it must pass through zero. x y (0, –1) (1, 0) (–1, 0) f (x) = x2 – 1 REAL ZEROS OF POLYNOMIAL FUNCTIONS If ƒ1x2 is a polynomial function and a is a real number, then the following statements are equivalent. Relationship 1: x 5 a is a solution, or root, of the equation ƒ1x2 5 0. Relationship 2: 1a, 02 is an x-intercept of the graph of ƒ1x2. Relationship 3: x 5 a is a zero of the function ƒ1x2. Relationship 4: 1x 2 a2 is a factor of ƒ1x2. 4.2 Polynomial Functions of Higher Degree 353 354 CHAPTER 4 Polynomial and Rational Functions DEFINITION Intermediate Value Theorem Let a and b be real numbers such that a , b and let ƒ be a polynomial function. If ƒ1a2 and ƒ1b2 have opposite signs, then there is at least one zero between a and b. The intermediate value theorem will be used later in this chapter to assist us in finding real zeros of a polynomial function. For now, it tells us that in order to change signs, the polynomial function must pass through the x-axis. In other words, once we know the zeros, then we know that between two consecutive zeros the polynomial is either entirely above the x-axis or entirely below the x-axis. This enables us to break the x-axis down into intervals that we can test, which will assist us in graphing polynomial functions. Keep in mind, though, that the existence of a zero does not imply that the function will change signs—as you will see in the subsection on graphing general polynomial functions. EXAMPLE 3 Identifying the Real Zeros of a Polynomial Function Find the zeros of the polynomial function ƒ1x2 5 x3 1 x2 2 2x. Solution: Set the function equal to zero. x3 1 x2 2 2x 5 0 Factor out an x common to all three terms. x 1x2 1 x 2 22 5 0 Factor the quadratic expression inside the parentheses. x 1x 1 22 1x 2 12 5 0 Apply the zero product property. x 5 0 or 1x 1 22 5 0 or 1x 2 12 5 0 Solve. x 5 2 2, x 5 0, and x 5 1 The zeros are 22, 0, and 1 . Y OUR TU R N Find the zeros of the polynomial function ƒ1x2 5 x3 2 7x2 1 12x. ▼ ▼ A N S W E R The zeros are 0, 3, and 4. Recall that when we were factoring a quadratic equation, if the factor was raised to a power greater than 1, the corresponding root, or zero, was repeated. For example, the quadratic equation x2 2 2x 1 1 5 0 when factored is written as 1x 2 122 5 0. The solution, or root, in this case is x 5 1, and we say that it is a repeated root. Similarly, when determining zeros of higher-order polynomial functions, if a factor is repeated, we say that the zero is a repeated, or multiple, zero of the function. The number of times that a zero repeats is called its multiplicity. DEFINITION Multiplicity of a Zero If 1x 2 a2n is a factor of a polynomial ƒ, then a is called a zero of multiplicity n of ƒ. EXAMPLE 4 Finding the Multiplicities of Zeros of a Polynomial Function Find the zeros, and state their multiplicities, of the polynomial function g1x2 5 1x 2 122Ax 1 3 5B 71x 1 52. Solution: 1 is a zero of multiplicity 2. 23 5 is a zero of multiplicity 7. 25 is a zero of multiplicity 1. ▼ A N S W E R 0 is a zero of multiplicity 2. 2 is a zero of multiplicity 3. 21 2 is a zero of multiplicity 5. 4.2.4 S K I L L Graph polynomial functions using x-intercepts; multiplicity of each zero, and end behavior. 4.2.4 C O N C E P T U A L Understand that odd multiplicity of a zero corresponds to the graph crossing the x-axis and even multiplicity of a zero corresponds to the graph touching the x-axis, and that end behavior is a result of the leading term dominating. Note: Adding the multiplicities yields the degree of the polynomial. The polynomial g 1x2 is of degree 10, since 2 1 7 1 1 5 10. YOUR T UR N For the polynomial h 1x2, determine the zeros and state their multiplicities. h1x2 5 x21x 2 223 ax 1 1 2b 5 ▼ EXAMPLE 5 Finding a Polynomial from Its Zeros Find a polynomial of degree 7 whose zeros are: 22 1multiplicity 22 0 1multiplicity 42 1 1multiplicity 12 Solution: If x 5 a is a zero, then 1x 2 a2 is a factor. ƒ1x2 5 1x 1 2221x 2 0241x 2 121 Simplify. 5 x41x 1 2221x 2 12 Square the binomial. 5 x41x2 1 4x 1 42 1x 2 12 Multiply the two polynomials. 5 x41x3 1 3x2 2 42 Distribute x4. 5 x7 1 3x6 2 4x4 4.2.4 Graphing General Polynomial Functions Let’s develop a strategy for sketching an approximate graph of any polynomial function. First, we determine the x- and y-intercepts. Then we use the x-intercepts, or zeros, to divide the domain into intervals where the polynomial is positive or negative so that we can find points in those intervals to assist in sketching a smooth and continuous graph. Note: It is not always possible to find x-intercepts. Sometimes there are no x-intercepts. STUDY TIP It is not always possible to find x-intercepts. Sometimes there are no x-intercepts. EXAMPLE 6 Using a Strategy for Sketching the Graph of a Polynomial Function Sketch the graph of ƒ1x2 5 1x 1 22 1x 2 122. Solution: STEP 1 Find the y-intercept. ƒ102 5 122 12122 5 2 1Let x 5 0.2 10, 22 is the y-intercept STEP 2 Find any x-intercepts. ƒ1x2 5 1x 1 22 1x 2 122 5 0 1Set ƒ1x2 5 0.2 x 5 22 or x 5 1 122, 02 and 11, 02 are the x-intercepts STEP 3 Plot the intercepts. x y (1, 0) (–2, 0) (0, 2) 4.2 Polynomial Functions of Higher Degree 355 356 CHAPTER 4 Polynomial and Rational Functions STEP 4 Divide the x-axis into intervals: 12q, 222, 122, 12, and 11, q2 The x-intercepts 122, 02 and 11, 02 divide the x-axis into three intervals similar to those created by zeros when we studied inequalities in Section 1.5. STEP 5 Select a number in each interval and test each interval. The function ƒ1x2 either crosses the x-axis at an x-intercept or touches the x-axis at an x-intercept. Therefore, we need to check each of these intervals to determine whether the function is positive (above the x-axis) or negative (below the x-axis). We do so by selecting numbers in the intervals and determining the value of the function at the corresponding points. x y (1, 0) (–2, 0) x = 2 x = –3 x = –1 (1, ) (–2, 1) (–, –2) x y (2, 4) (–1, 4) (–3, –16) INTERVAL 12q, 222 122, 12 11, q2 NUMBER SELECTED IN INTERVAL 23 21 2 VALUE OF FUNCTION ƒ1232 5 216 ƒ1212 5 4 ƒ122 5 4 POINT ON GRAPH 123, 2162 121, 42 12, 42 INTERVAL RELATION TO x-AXIS Below x-axis Above x-axis Above x-axis From the table, we find three additional points on the graph: 123, 2162, 121, 42, and 12, 42. The point 122, 02 is an intercept where the function crosses the x-axis because it is below the x-axis to the left of 22 and above the x-axis to the right of 22. The point 11, 02 is an intercept where the function touches the x-axis because it is above the x-axis on both sides of x 5 1. Connecting these points with a smooth curve yields the graph. STEP 6 Sketch a plot of the function. Y OUR TU R N Sketch the graph of ƒ1x2 5 x21x 1 322. ▼ A N S W E R x y –5 5 10 x y (1, 0) (–1, 4) (2, 4) (0, 2) ▼ STUDY TIP We do not know for sure that the points (21, 4) and (1, 0) are turning points. A graphing utility can confirm this, and later in calculus you will learn how to find relative maximum points and relative minimum points. In Example 6, we found that the function crosses the x-axis at the point 122, 02. Note that 22 is a zero of multiplicity 1. We also found that the function touches the x-axis at the point 11, 02. Note that 1 is a zero of multiplicity 2. In general, zeros with even multiplicity correspond to intercepts where the function touches the x-axis, and zeros with odd multiplicity correspond to intercepts where the function crosses the x-axis. STUDY TIP In general, zeros with even multiplicity correspond to intercepts where the function touches the x-axis and zeros with odd multiplicity correspond to intercepts where the function crosses the x-axis. Also in Example 6, we know that somewhere in the interval 122, 12 the function must reach a maximum and then turn back toward the x-axis because both points 122, 02 and 11, 02 correspond to x-intercepts. When we sketch the graph, it “appears” that the point 121, 42 is a turning point. The point 11, 02 also corresponds to a turning point. In general, if ƒ is a polynomial of degree n, then the graph of ƒ has at most n 2 1 turning points. The point 121, 42, which we call a turning point, is also a “high point” on the graph in the vicinity of the point 121, 42. Also note that the point 11, 02, which we call a turning point, is a “low point” on the graph in the vicinity of the point 11, 02. We call a “high point” on a graph a local (relative) maximum and a “low point” on a graph a local (relative) minimum. For quadratic functions we can find the maximum or minimum point by finding the vertex. However, for higher-degree polynomial functions, we rely on graphing utilities to locate such points. Later in calculus, techniques will be developed for finding such points exactly. For now, we use the zoom and trace features to locate such points on a graph, and we can use the table feature of a graphing utility to approximate relative minima or maxima. Let us take the polynomial ƒ1x2 5 x3 2 2x2 2 5x 1 6. Using methods discussed thus far, we can find that the x-intercepts of its graph are 122, 02, 11, 02, and 13, 02 and the y-intercept is the point 10, 62. We can also find additional points that lie on the graph such as 121, 82 and 12, 242. Plotting these points, we might “think” that the points 121, 82 and 12, 242 might be turning points, but a graphing utility reveals an approximate relative maximum at the point 120.7863, 8.20882072 and an approximate relative minimum at the point 12.1196331, 24.0606732. Intercepts and turning points assist us in sketching graphs of polynomial functions. Another piece of information that will assist us in graphing polynomial functions is knowledge of the end behavior. All polynomials eventually rise or fall without bound as x gets large in both the positive 1x S q2 and negative 1x S 2q2 directions. The highest-degree monomial within the polynomial dominates the end behavior. In other words, the highest power term is eventually going to overwhelm the other terms as x grows without bound. x y Turning points STUDY TIP If f is a polynomial of degree n, then the graph of f has at most n 2 1 turning points. Power functions behave much like a quadratic function (parabola) for even-degree polynomial functions and much like a cubic function for odd-degree polynomial functions. There are four possibilities because the leading coefficient can be positive or negative with either an odd or even power. Let y 5 anxn; then MULTIPLICITY OF A ZERO AND RELATION TO THE GRAPH OF A POLYNOMIAL If a is a zero of ƒ1x2, then: MULTIPLICITY OF a f (x) ON EITHER SIDE OF x 5 a GRAPH OF FUNCTION AT THE INTERCEPT Even Does not change sign Touches the x-axis (turns around) at point 1a, 02 Odd Changes sign Crosses the x-axis at point 1a, 02 END BEHAVIOR As x gets large in the positive 1x S q2 and negative 1x S 2q2 directions, the graph of the polynomial ƒ1x2 5 anxn 1 an21xn21 1 c1 a2x2 1 a1x 1 a0 has the same behavior as the power function y 5 anxn 4.2 Polynomial Functions of Higher Degree 357 358 CHAPTER 4 Polynomial and Rational Functions n Even Even Odd Odd an Positive Negative Negative Positive x u 2H (Left) The graph of the function rises The graph of the function falls The graph of the function rises The graph of the function falls x u H (Right) The graph of the function rises The graph of the function falls The graph of the function falls The graph of the function rises Graph x y an > 0 x y an < 0 x y an < 0 x y an > 0 EXAMPLE 7 Graphing a Polynomial Function Sketch a graph of the polynomial function ƒ1x2 5 2x4 2 8x2. Solution: STEP 1 Determine the y-intercept: 1x 5 02. ƒ102 5 0 The y-intercept corresponds to the point 10, 02. STEP 2 Find the zeros of the polynomial. ƒ1x2 5 2x4 2 8x2 Factor out the common 2x2. 5 2x21x2 2 42 Factor the quadratic binomial. 5 2x21x 2 22 1x 1 22 Set ƒ1x2 5 0. 5 2x21x 2 22 1x 1 22 5 0 0 is a zero of multiplicity 2. The graph will touch the x-axis. 2 is a zero of multiplicity 1. The graph will cross the x-axis. 22 is a zero of multiplicity 1. The graph will cross the x-axis. STEP 3 Determine the end behavior. ƒ1x2 5 2x4 2 8x2 behaves like y 5 2x4. y 5 2x4 is of even degree, and the leading coefficient is positive, so the graph rises without bound as x gets large in both the positive and negative directions. STEP 4 Sketch the intercepts and end behavior. STEP 5 Find additional points. x y (2, 0) (0, 0) (–2, 0) x 21 21 2 1 2 1 f (x) 26 215 8 215 8 26 [CONCEPT CHECK] The graph of the function f (x) 5 (x 2 a)3(x 2 b)2 crosses or touches at (a, 0) and (b, 0)? ANSWER Crosses at (a, 0) and touches at (b, 0). ▼ ▼ A N S W E R x y –5 5 10 –10 STEP 6 Sketch the graph. ■ ■Estimate additional points. ■ ■Connect with a smooth curve. Note the symmetry about the y-axis. This function is an even function: ƒ12x2 5 ƒ1x2. It is important to note that the local minimums occur at x 5 6!2 < 6 1.14 but at this time can only be illustrated using a graphing utility. YOUR T UR N Sketch a graph of the polynomial function ƒ1x2 5 x5 2 4x3. x y (2, 0) (–2, 0) 10 –10 ▼ 3. x-intercepts (real zeros) divide the x-axis into intervals. Test points in the intervals to determine whether the graph is above or below the x-axis. 4. Determine the end behavior by investigating the end behavior of the highest-degree monomial. 5. Sketch the graph with a smooth curve. In general, polynomials can be graphed in one of two ways: n Use graph-shifting techniques with power functions. n General polynomial function. 1. Identify intercepts. 2. Determine each real zero and its multiplicity, and ascertain whether the graph crosses or touches the x-axis there. [SEC TION 4 . 2] S U M M A RY 4.2 Polynomial Functions of Higher Degree 359 [SEC TION 4 . 2] E X E R C I S E S • S K I L L S In Exercises 1–10, determine which functions are polynomials, and for those that are, state their degree. 1. ƒ1x2 5 23x2 1 15x 2 7 2. ƒ1x2 5 2x5 2 x2 1 13 3. g1x2 5 1x 1 223Ax 2 3 5B2 4. g 1x2 5 x41x 2 1221x 1 2.523 5. h1x2 5 !x 1 1 6. h 1x2 5 1x 2 121/2 1 5x 7. F 1x2 5 x1/3 1 7x2 2 2 8. F 1x2 5 3x2 1 7x 2 2 3x 9. G1x2 5 x 1 1 x2 10. H1x2 5 x2 1 1 2 In Exercises 11–18, match the polynomial function with its graph. 11. ƒ1x2 5 23x 1 1 12. ƒ1x2 5 23x2 2 x 13. ƒ1x2 5 x2 1 x 14. ƒ1x2 5 22x3 1 4x2 2 6x 15. ƒ1x2 5 x3 2 x2 16. ƒ1x2 5 2x4 2 18x2 17. ƒ1x2 5 2x4 1 5x3 18. ƒ1x2 5 x5 2 5x3 1 4x a. b. c. d. x y –5 5 5 –5 x y –3 2 4 –1 x y –10 10 90 x y –10 10 50 –50 360 CHAPTER 4 Polynomial and Rational Functions e. f. g. h. In Exercises 19–26, graph each function by transforming a power function y 5 xn. 19. ƒ1x2 5 2x5 20. ƒ1x2 5 2x4 21. ƒ1x2 5 1x 2 224 22. ƒ1x2 5 1x 1 225 23. ƒ1x2 5 x5 1 3 24. ƒ1x2 5 2x4 2 3 25. ƒ1x2 5 3 2 1x 1 124 26. ƒ1x2 5 1x 2 325 2 2 In Exercises 27–38, find all the real zeros (and state their multiplicities) of each polynomial function. 27. ƒ1x2 5 21x 2 32 1x 1 423 28. ƒ1x2 5 231x 1 2231x 2 122 29. ƒ1x2 5 4x21x 2 7221x 1 42 30. ƒ1x2 5 5x31x 1 1241x 2 62 31. ƒ1x2 5 4x21x 2 1221x2 1 42 32. ƒ1x2 5 4x21x2 2 12 1x2 1 92 33. ƒ1x2 5 8x3 1 6x2 2 27x 34. ƒ1x2 5 2x4 1 5x3 2 3x2 35. ƒ1x2 5 22.7x3 2 8.1x2 36. ƒ1x2 5 1.2x6 2 4.6x4 37. ƒ1x2 5 1 3 x6 1 2 5 x4 38. ƒ1x2 5 2 7 x5 2 3 4 x4 1 1 2 x3 In Exercises 39–52, find a polynomial (there are many) of minimum degree that has the given zeros. 39. 23, 0, 1, 2 40. 22, 0, 2 41. 25, 23, 0, 2, 6 42. 0, 1, 3, 5, 10 43. 21 2, 2 3, 3 4 44. 23 4, 21 3, 0, 1 2 45. 1 2 !2, 1 1 !2 46. 1 2 !3, 1 1 !3 47. 22 1multiplicity 32, 0 1multiplicity 22 48. 24 1multiplicity 22, 5 1multiplicity 32 49. 23 1multiplicity 22, 7 1multiplicity 52 50. 0 1multiplicity 12, 10 1multiplicity 32 51. 2!3 1multiplicity 22, 21 1multiplicity 12, 0 1multiplicity 22, !3 1multiplicity 22 52. 2!5 1multiplicity 22, 0 1multiplicity 12, 1 1multiplicity 22, !5 1multiplicity 22 In Exercises 53–72, for each polynomial function given: (a) list each real zero and its multiplicity; (b) determine whether the graph touches or crosses at each x-intercept; (c) find the y-intercept and a few points on the graph; (d) determine the end behavior; and (e) sketch the graph. 53. ƒ1x2 5 2x2 2 6x 2 9 54. ƒ1x2 5 x2 1 4x 1 4 55. ƒ1x2 5 1x 2 223 56. ƒ1x2 5 21x 1 323 57. ƒ1x2 5 x3 2 9x 58. ƒ1x2 5 2x3 1 4x2 59. ƒ1x2 5 2x3 1 x2 1 2x 60. ƒ1x2 5 x3 2 6x2 1 9x 61. ƒ1x2 5 2x4 2 3x3 62. ƒ1x2 5 x5 2 x3 63. ƒ1x2 5 12x6 2 36x5 2 48x4 64. ƒ1x2 5 7x5 2 14x4 2 21x3 65. ƒ1x2 5 2x5 2 6x4 2 8x3 66. ƒ1x2 5 25x4 1 10x3 2 5x2 67. ƒ1x2 5 x3 2 x2 2 4x 1 4 68. ƒ1x2 5 x3 2 x2 2 x 1 1 69. ƒ1x2 5 21x 1 2221x 2 122 70. ƒ1x2 5 1x 2 2231x 1 123 71. ƒ1x2 5 x21x 2 2231x 1 322 72. ƒ1x2 5 2x31x 2 4221x 1 222 In Exercises 73–76, for each graph given: (a) list each real zero and its smallest possible multiplicity; (b) determine whether the degree of the polynomial is even or odd; (c) determine whether the leading coefficient of the polynomial is positive or negative; (d) find the y-intercept; and (e) write an equation for the polynomial function (assume the least degree possible). 73. 74. 75. 76. x y –5 5 5 –5 x y –5 5 10 –10 x y –5 5 5 –5 x y –5 5 5 –5 –5 5 x y –4 16 –5 5 x y –10 10 –5 5 x y –10 10 –5 5 x y –32 8 • A P P L I C A T I O N S For Exercises 77 and 78, refer to the following: The relationship between a company’s total revenue R (in millions of dollars) is related to its advertising costs x (in thousands of dollars). The relationship between revenue R and advertising costs x is illustrated in the graph. 100 200 300 400 600 500 50 45 40 35 30 25 20 15 10 5 x R Advertising Costs (in thousands of dollars) Revenue (in millions of dollars) 77. Business. Analyze the graph of the revenue function. a. Determine the intervals on which revenue is increasing and decreasing. b. Identify the zeros of the function. Interpret the meaning of zeros for this function. 78. Business. Use the graph to identify the maximum revenue for the company and the corresponding advertising costs that produce maximum revenue. For Exercises 79 and 80, refer to the following: During a cough, the velocity v (in meters per second) of air in the trachea may be modeled by the function v1r2 5 2120r3 1 80r2 where r is the radius of the trachea (in centimeters) during the cough. 79. Health/Medicine. Graph the velocity function and estimate the intervals on which the velocity of air in the trachea is increasing and decreasing. 80. Health/Medicine. Estimate the radius of the trachea that produces the maximum velocity of air in the trachea. Use this radius to estimate the maximum velocity of air in the trachea. 81. Weight. Jennifer has joined a gym to lose weight and feel better. She still likes to cheat a little and will enjoy the occasional bad meal with an ice cream dream dessert and then miss the gym for a couple of days. Given in the table is Jennifer’s weight for a period of 8 months. Her weight can be modeled as a polynomial. Plot these data. How many turning points are there? Assuming these are the minimum number of turning points, what is the lowest-degree polynomial that can represent Jennifer’s weight? MONTH WEIGHT 1 169 2 158 3 150 4 161 5 154 6 159 7 148 8 153 82. Stock Value. A day trader checks the stock price of Coca-Cola during a 4-hour period (given below). The price of Coca-Cola stock during this 4-hour period can be modeled as a polynomial function. Plot these data. How many turning points are there? Assuming these are the minimum number of turning points, what is the lowest-degree polynomial that can represent the Coca-Cola stock price? PERIOD WATCHING STOCK MARKET PRICE 1 $53.00 2 $56.00 3 $52.70 4 $51.50 83. Stock Value. The price of Tommy Hilfiger stock during a 4-hour period is given in the following table. If a third-degree polynomial models this stock, do you expect the stock to go up or down in the fifth period? PERIOD WATCHING STOCK MARKET PRICE 1 $15.10 2 $14.76 3 $15.50 4 $14.85 84. Stock Value. The stock prices for Coca-Cola during a 4-hour period on another day yield the following results. If a third-degree polynomial models this stock, do you expect the stock to go up or down in the fifth period? PERIOD WATCHING STOCK MARKET PRICE 1 $52.80 2 $53.00 3 $56.00 4 $52.70 For Exercises 85 and 86, the following graph illustrates the average annual federal funds rate in the month of January (2006–2014). Year 2006 2008 2010 2012 2014 Annual Average Federal Funds Rate 10.00% 9.00 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 4.2 Polynomial Functions of Higher Degree 361 362 CHAPTER 4 Polynomial and Rational Functions 85. Finance. If a polynomial function is used to model the federal funds rate data shown in the graph, determine the degree of the lowest-degree polynomial that can be used to model those data. 86. Finance. Should the leading coefficient in the polynomial found in Exercise 85 be positive or negative? Explain. • C A T C H T H E M I S T A K E 87. Find a fourth-degree polynomial function with zeros 22, 21, 3, 4. Solution: ƒ1x2 5 1x 2 22 1x 2 12 1x 1 32 1x 1 42 This is incorrect. What mistake was made? 88. Determine the end behavior of the polynomial function ƒ1x2 5 x 1x 2 223. Solution: This polynomial has similar end behavior to the graph of y 5 x3. End behavior falls to the left and rises to the right. This is incorrect. What mistake was made? 89. Graph the polynomial function ƒ1x2 5 1x 2 1221x 1 223. Solution: The zeros are 22 and 1, and therefore, the x-intercepts are 122, 02 and 11, 02. The y-intercept is 10, 82. Plotting these points and connecting with a smooth curve yield the graph on the right. This graph is incorrect. What did we forget to do? 90. Graph the polynomial function ƒ1x2 5 1x 1 1221x 2 122. Solution: The zeros are 21 and 1, so the x-intercepts are 121, 02 and 11, 02. The y-intercept is 10, 12. Plotting these points and connecting with a smooth curve yield the graph on the right. This graph is incorrect. What did we forget to do? x y –5 5 10 –10 x y –5 5 5 –5 In Exercises 87–90, explain the mistake that is made. In Exercises 91–94, determine if each statement is true or false. • C O N C E P T U A L 91. The graph of a polynomial function might not have any y-intercepts. 92. The graph of a polynomial function might not have any x-intercepts. 93. The domain of all polynomial functions is 12q, q2. 94. The range of all polynomial functions is 12q, q2. 95. What is the maximum number of zeros that a polynomial of degree n can have? 96 . What is the maximum number of turning points a graph of an nth-degree polynomial can have? • C H A L L E N G E 97. Find a seventh-degree polynomial that has the following graph characteristics: The graph touches the x-axis at x 5 21, and the graph crosses the x-axis at x 5 3. Plot this polynomial function. 98. Find a fifth-degree polynomial that has the following graph characteristics: The graph touches the x-axis at x 5 0 and crosses the x-axis at x 5 4. Plot the polynomial function. 99. Determine the zeros of the polynomial ƒ1x2 5 x3 1 1b 2 a2x2 2 abx for the positive real numbers a and b. 100. Graph the function ƒ1x2 5 x21x 2 a221x 2 b22 for the positive real numbers a, b, where b . a. 4.3 Dividing Polynomials: Long Division and Synthetic Division 363 • T E C H N O L O G Y In Exercises 101 and 102, use a graphing calculator or computer to graph each polynomial. From that graph, estimate the x-intercepts (if any). Set the function equal to zero, and solve for the zeros of the polynomial. Compare the zeros with the x-intercepts. 101. ƒ1x2 5 x4 1 2x2 1 1 102. ƒ1x2 5 1.1x3 2 2.4x2 1 5.2x For each polynomial in Exercises 103 and 104, determine the power function that has similar end behavior. Plot this power function and the polynomial. Do they have similar end behavior? 103. ƒ1x2 5 22x5 2 5x4 2 3x3 104. ƒ1x2 5 x4 2 6x2 1 9 In Exercises 105 and 106, use a graphing calculator or a computer to graph each polynomial. From the graph, estimate the x-intercepts and state the zeros of the function and their multiplicities. 105. ƒ1x2 5 x4 2 15.9x3 1 1.31x2 1 292.905x 1 445.7025 106. ƒ1x2 5 2x5 1 2.2x4 1 18.49x3 2 29.878x2 2 76.5x 1 100.8 In Exercises 107 and 108, use a graphing calculator or a computer to graph each polynomial. From the graph, estimate the coordinates of the relative maximum and minimum points. Round your answers to two decimal places. 107. ƒ1x2 5 2x4 1 5x3 2 10x2 2 15x 1 8 108. ƒ1x2 5 2x5 2 4x4 2 12x3 1 18x2 1 16x 2 7 S K I L L S O B J E C T I V E S ■ ■Divide polynomials with long division. ■ ■Divide polynomials with synthetic division. C O N C E P T U A L O B J E C T I V ES ■ ■Extend long division of real numbers to polynomials. ■ ■Understand that synthetic division can only be used when dividing a polynomial by a linear factor. 4.3 DIVIDING POLYNOMIALS: LONG DIVISION AND SYNTHETIC DIVISION To divide polynomials, we rely on the technique we use for dividing real numbers. For example, if you were asked to divide 6542 by 21, the long division method used is illustrated in the margin. This solution can be written two ways: 311 R11 or 311 1 11 21. In this example, the dividend is 6542, the divisor is 21, the quotient is 311, and the remainder is 11. We employ a similar technique (dividing the leading terms) when dividing polynomials. 4.3.1 Long Division of Polynomials Let’s start with an example whose answer we already know. We know that a quadratic expression can be factored into the product of two linear factors: x 2 1 4x 2 5 5 1x 1 52 1x 2 12. Therefore, if we divide both sides of the equation by 1x 2 12, we get x2 1 4x 2 5 x 2 1 5 x 1 5 We can state this by saying x2 1 4x 2 5 divided by x 2 1 is equal to x 1 5. Confirm this statement by long division: x 2 1qx2 1 4x 2 5 Note that although this is standard division notation, the dividend and the divisor are both polynomials that consist of multiple terms. The leading terms of each algebraic expression will guide us. 4.3.1 S K I L L Divide polynomials with long division. 4.3.1 C O N C E P T U A L Extend long division of real numbers to polynomials. 21q6542 311 263 221 24 11 32 221 364 CHAPTER 4 Polynomial and Rational Functions WORDS MATH Q: x times what quantity gives x2? x 2 1qx2 1 4x 2 5 A: x x x 2 1qx2 1 4x 2 5 x Multiply x 1x 2 12 5 x2 2 x. x 2 1qx2 1 4x 2 5 x2 2 x Subtract 1x2 2 x2 from x2 1 4x 2 5. x Note: 21x2 2 x2 5 2x2 1 x. x 2 1qx2 1 4x 2 5 Bring down the 25. x 1 5 Q: x times what quantity is 5x? x 2 1qx2 1 4x 2 5 A: 5 Multiply 51x 2 12 5 5x 2 5. x 1 5 x 2 1qx2 1 4x 2 5 Subtract 15x 2 52. Note: 215x 2 52 5 25x 1 5. As expected, the remainder is 0. By long division we have shown that x2 1 4x 2 5 x 2 1 5 x 1 5 Check: Multiplying the equation by x 2 1 yields x2 1 4x 2 5 5 1x 1 52 1x 2 12, which we know to be true. 2x2 1 x 5x 2 5 5x 2 5 2x2 1 x 5x 2 5 2x2 1 x 5x 2 5 25x 1 5 0 . . . . EXAMPLE 1 Dividing Polynomials Using Long Division; Zero Remainder Divide 2x3 2 9x 2 1 7x 1 6 by 2x 1 1. Solution: Multiply: x 2 12x 1 12. Subtract: Bring down the 7x. Multiply: 25x 12x 1 12. Subtract: Bring down the 6. Multiply: 6 12x 1 12. Subtract. Quotient: x2 2 5x 1 6 2x 1 1q2x3 2 9x2 1 7x 1 6 x2 2 5x 1 6 212 x3 1 x22 21210x2 2 5x2 210x2 1 7x 12x 1 6 0 2112 x 1 62 ▼ A N S W E R x2 1 2x 2 3, remainder 0. Check: 12x 1 12 1x2 2 5x 1 62 5 2x3 2 9x2 1 7x 1 6. Note: Since the divisor cannot be equal to zero, 2x 1 1 Z 0, then we say x Z 21 2. Y OUR T UR N Divide 4x3 1 13x2 2 2x 2 15 by 4x 1 5. ▼ Why are we interested in dividing polynomials? Because it helps us find zeros of polynomials. In Example 1, using long division, we found that 2x3 2 9x2 1 7x 1 6 5 12x 1 12 1x2 2 5x 1 62 Factoring the quadratic expression enables us to write the cubic polynomial as a product of three linear factors: 2x3 2 9x2 1 7x 1 6 5 12x 1 12 1x2 2 5x 1 62 5 12x 1 12 1x 2 32 1x 2 22 Set the value of the polynomial equal to zero, 12x 1 12 1x 2 32 1x 2 22 5 0, and solve for x. The zeros of the polynomial are 21 2, 2, and 3. In Example 1 and in the Your Turn, the remainder was 0. Sometimes there is a nonzero remainder (Example 2). ▼ A N S W E R 2x2 1 3x 2 1 R: 24 or 2x2 1 3x 2 1 2 4 x 2 1 x 1 1q6x2 2 x 2 2 6x 2 7 216x2 1 6x2 2127x 2 72 27x 2 2 1 5 EXAMPLE 2 Dividing Polynomials Using Long Division; Nonzero Remainder Divide 6x2 2 x 2 2 by x 1 1. Solution: Multiply 6x 1x 1 12. Subtract and bring down 22. Multiply 271x 1 12. Subtract and identify the remainder. Dividend Quotient Remainder 6x2 2 x 2 2 x 1 1 5 6x 2 7 1 5 x 1 1 x 2 21 Divisor Divisor Check: Multiply the quotient and 16x 2 72 1x 1 12 1 5 1x 1 12 ⋅1x 1 12 remainder by x 1 1. 5 6x2 2 x 2 7 1 5 The result is the dividend. 5 6x2 2 x 2 2 ✓ YOUR T UR N Divide 2x3 1 x2 2 4x 2 3 by x 2 1. ▼ In general, when a polynomial is divided by another polynomial, we express the result in the following form: P1x2 d1x2 5 Q1x2 1 r1x2 d1x2 where P 1x2 is the dividend, d1x2 2 0 is the divisor, Q 1x2 is the quotient, and r 1x2 is the remainder. Multiplying this equation by the divisor, d 1x2, leads us to the division algorithm. 4.3 Dividing Polynomials: Long Division and Synthetic Division 365 366 CHAPTER 4 Polynomial and Rational Functions [CONCEPT CHECK] To use long division to divide two polynomials, the degree of the numerator must be (greater/less) than the degree of the denominator. ANSWER greater than ▼ ▼ A N S W E R x2 1 x 1 1 EXAMPLE 3 Long Division of Polynomials with “Missing” Terms Divide x3 2 8 by x 2 2. Solution: Insert 0x2 1 0x for placeholders. Multiply x21x 2 22 5 x3 2 2x2. Subtract and bring down 0x. Multiply 2x 1x 2 22 5 2x2 2 4x. Subtract and bring down 28. Multiply 41x 2 22 5 4x 2 8. Subtract and get remainder 0. Since the remainder is 0, x 2 2 is a factor of x3 2 8. x3 2 8 x 2 2 5 x2 1 2x 1 4, x 2 2 Check: x3 2 8 5 1x2 1 2x 1 42 1x 2 22 5 x3 1 2x2 1 4x 2 2x2 2 4x 2 8 5 x3 2 8 ✓ Y OUR TU R N Divide x3 2 1 by x 2 1. x 2 2qx3 1 0x2 1 0x 2 8 x2 1 2x 1 4 21x3 2 2x22 212x2 2 4x2 2x2 1 0x 4x 2 8 214x 2 82 0 ▼ ▼ A N S W E R 2x2 1 6 1 11x2 1 18x 1 36 x3 2 3x 2 4 THE DIVISION ALGORITHM If P 1x2 and d 1x2 are polynomials with d1x2 2 0, and if the degree of P 1x2 is greater than or equal to the degree of d 1x2, then unique polynomials Q1x2 and r 1x2 exist such that P1x2 5 d1x2 ⋅Q1x2 1 r1x2 If the remainder r 1x2 5 0, then we say that d 1x2 divides P 1x2 and that d 1x2 and Q1x2 are factors of P 1x2. EXAMPLE 4 Long Division of Polynomials Divide 3x4 1 2x3 1 x2 1 4 by x2 1 1. Solution: Insert 0x as a placeholder in both the divisor and the dividend. Multiply 3x21x2 1 0x 1 12. Subtract and bring down 0x. Multiply 2x 1x2 1 0x 1 12. Subtract and bring down 4. Multiply 221x2 2 2x 1 12. Subtract and get remainder 22x 1 6. 3x4 1 2x3 1 x2 1 4 x2 1 1 5 3x2 1 2x 2 2 1 22x 1 6 x2 1 1 Y OUR TU R N Divide 2x5 1 3x2 1 12 by x3 2 3x 2 4. ▼ x2 1 0x 1 1q3x4 1 2x3 1 x2 1 0x 1 4 3x2 1 2x 2 2 213x4 1 0x3 1 3x22 212x3 1 0x2 1 2x2 2x3 2 2x2 1 0x 22x2 2 2x 1 4 2122x2 1 0x 2 22 22x 1 6 In Examples 1 through 4 the dividends, divisors, and quotients were all polynomials with integer coefficients. In Example 5, however, the resulting quotient has rational (noninteger) coefficients EXAMPLE 5 Long Division of Polynomials Resulting in Quotients with Rational Coefficients Divide 8x4 2 5x3 1 7x 2 2 by 2x2 1 1. Solution: Insert 0x2 as a placeholder in the dividend and 0x as a placeholder in the divisor. Multiply 4x212x2 1 0x 1 12. Subtract and bring down remaining terms. Multiply 25 2x12x2 1 0x 1 12. Subtract and bring down remaining terms. Multiply 2212x2 1 0x 1 12. Subtract and bring down the remainder 19 2 x. 8x4 2 5x3 1 7x 2 2 2x2 1 1 5 4x2 2 5 2x 2 2 1 19 2 x 2x2 1 1 YOUR T UR N Divide 10x4 2 3x3 1 5x 2 4 by 2x2 2 1. 2x2 1 0x 1 1q8x4 2 5x3 1 0x2 1 7x 2 2 4x2 2 5 2x 2 2 218x4 1 0x3 1 4x22 2125x3 1 0x2 2 5 2x2 25x3 2 4x2 1 7x 24x2 1 19 2 x 2 2 2124x2 1 0x 2 22 19 2 x ▼ ▼ A N S W E R 5x2 2 3 2x 1 5 2 1 7 2x 2 3 2 2x2 2 1 4.3.2 Synthetic Division of Polynomials In the special case when the divisor is a linear factor of the form x 2 a or x 1 a, there is another, more efficient way to divide polynomials. This method is called synthetic division. It is called synthetic because it is a contrived shorthand way of dividing a polynomial by a linear factor. A detailed step-by-step procedure is given below for synthetic division. Let’s divide x4 2 x3 2 2x 1 2 by x 1 1 using synthetic division. STEP 1 Write the division in synthetic form. ■ ■ List the coefficients of the dividend. Remember to use 0 for a placeholder. ■ ■ The divisor is x 1 1. Since x 1 1 5 0 x 5 21 is used. STEP 2 Bring down the first term 112 in the dividend. STEP 3 Multiply the 21 by this leading coefficient 112, and place the product up and to the right in the second column. Coefficients of Dividend 21 1 21 0 22 2 21 1 21 0 22 2 Bring down the 1 1 21 1 21 0 22 2 21 1 4.3.2 S K IL L Divide polynomials with synthetic division. 4.3.2 C ON C E P T U A L Understand that synthetic division can only be used when dividing a polynomial by a linear factor. 4.3 Dividing Polynomials: Long Division and Synthetic Division 367 368 CHAPTER 4 Polynomial and Rational Functions STUDY TIP If (x 2 a) is the divisor, then a is the number used in synthetic division. 21 1 21 0 22 2 21 ADD 1 22 21 1 21 0 22 2 21 2 22 4 1 22 2 24 6 21 1 21 0 22 2 21 2 22 4 1 22 2 24 6 Quotient Coefficient x3 2 2x2 1 2x 2 4 Remainder f STEP 4 Add the values in the second column. STEP 5 Repeat Steps 3 and 4 until all columns are filled. STEP 6 Identify the quotient by assigning powers of x in descending order, beginning with xn21 5 x421 5 x3. The last term is the remainder. We know that the degree of the first term of the quotient is 3 because a fourth-degree polynomial was divided by a first-degree polynomial. Let’s compare dividing x4 2 x3 2 2x 1 2 by x 1 1 using both long division and synthetic division. x 1 1qx4 2 x3 1 0x2 2 2x 1 2 x4 1 x3 22x3 1 0x2 2122x3 2 2x22 2x2 2 2x 212x2 1 2x2 24x 1 2 2124x 2 42 16 Both long division and synthetic division yield the same answer. x4 2 x3 2 2x 1 2 x 1 1 5 x3 2 2x2 1 2x 2 4 1 6 x 1 1 x3 2 2x2 1 2x 2 4 Long Division Synthetic Division 21 1 21 0 22 2 21 2 22 4 1 22 2 24 6 x3 2 2x2 1 2x 2 4 f EXAMPLE 6 Synthetic Division Use synthetic division to divide 3x5 2 2x3 1 x2 2 7 by x 1 2. Solution: STEP 1 Write the division in synthetic form. ■ ■ List the coefficients of the dividend. Remember to use 0 for a placeholder. ■ ■ The divisor of the original problem is x 1 2. If we set x 1 2 5 0, we find that x 5 22, so 22 is the divisor for synthetic division. 22 3 0 22 1 0 27 STUDY TIP Synthetic division can only be used when the divisor is of the form x 2 a or x 1 a. [CONCEPT CHECK] Can synthetic division be used to divide f (x) 5 x4 2 1 by g(x) 5 x3 1 1? ANSWER No. The degree of the denominator has to be one. ▼ STEP 2 Perform the synthetic division steps. STEP 3 Identify the quotient and remainder. 3x5 2 2x3 1 x2 2 7 x 1 2 5 3x4 2 6x3 1 10x2 2 19x 1 38 2 83 x 1 2 YOUR T UR N Use synthetic division to divide 2x3 2 x 1 3 by x 2 1. 22 3 0 22 1 0 27 26 12 220 38 276 3 26 10 219 38 283 22 3 0 22 1 0 27 26 12 220 38 276 3 26 10 219 38 283 f 3x4 2 6x3 1 10x2 2 19x 1 38 ▼ ▼ A N S W E R 2x2 1 2x 1 1 1 4 x 2 1 4.3 Dividing Polynomials: Long Division and Synthetic Division 369 EX PR ES S I N G R ES U LT S n Dividend Divisor 5 quotient 1 remainder divisor n Dividend 5 1quotient2 1divisor2 1 remainder W H EN R EM A I N D ER I S ZER O n Dividend 5 1quotient2 1divisor2 n Quotient and divisor are factors of the dividend. DIV IS ION OF P OLY NOM I ALS n Long division can always be used. n Synthetic division is restricted to when the divisor is of the form x 2 a or x 1 a. [SEC TION 4 .3] S U MM A RY In Exercises 1–30, divide the polynomials using long division. Use exact values. Express the answer in the form Q 1x2 5 ?, r 1x2 5 ?. 1. 12x2 1 5x 2 32 4 1x 1 32 2. 12x2 1 5x 2 32 4 1x 2 32 3. 1x2 2 5x 1 62 4 1x 2 22 4. 12x2 1 3x 1 12 4 1x 1 12 5. 13x2 2 9x 2 52 4 1x 2 22 6. 1x2 1 4x 2 32 4 1x 2 12 7. 13x2 2 13x 2 102 4 1x 1 52 8. 13x2 2 13x 2 102 4 1x 2 52 9. 1x2 2 42 4 1x 1 42 10. 1x2 2 92 4 1x 2 22 11. 19x2 2 252 4 13x 2 52 12. 15x2 2 32 4 1x 1 12 13. 14x2 2 92 4 12x 1 32 14. 18x3 1 272 4 12x 1 32 15. 111x 1 20x2 1 12x3 1 22 4 13x 1 22 16. 112x3 1 2 1 11x 1 20x22 4 12x 1 12 17. 14x3 2 2x 1 72 4 12x 1 12 18. 16x4 2 2x2 1 52 4 123x 1 22 19. 14x3 2 12x2 2 x 1 32 4 Ax 2 1 2B 20. 112x3 1 1 1 7x 1 16x22 4 Ax 1 1 3B [SEC TION 4 .3] E X E R C I S E S • S K I L L S 370 CHAPTER 4 Polynomial and Rational Functions 21. 122x5 1 3x4 2 2x22 4 1x3 2 3x2 1 12 22. 129x6 1 7x4 2 2x3 1 52 4 13x4 2 2x 1 12 23. x4 2 1 x2 2 1 24. x4 2 9 x2 1 3 25. 40 2 22x 1 7x3 1 6x4 6x2 1 x 2 2 26. 213x2 1 4x4 1 9 4x2 2 9 27. 23x4 1 7x3 2 2x 1 1 x 2 0.6 28. 2x5 2 4x3 1 3x2 1 5 x 2 0.9 29. 1x4 1 0.8x3 2 0.26x2 2 0.168x 1 0.04412 4 1x2 1 1.4x 1 0.492 30. 1x5 1 2.8x4 1 1.34x3 2 0.688x2 2 0.2919x 1 0.08822 4 1x2 2 0.6x 1 0.092 In Exercises 31–50, divide the polynomial by the linear factor with synthetic division. Indicate the quotient Q 1x2 and the remainder r 1x2 . 31. 13x2 1 7x 1 22 4 1x 1 22 32. 12x2 1 7x 2 152 4 1x 1 52 33. 17x2 2 3x 1 52 4 1x 1 12 34. 14x2 1 x 1 12 4 1x 2 22 35. 13x2 1 4x 2 x4 2 2x3 2 42 4 1x 1 22 36. 13x2 2 4 1 x32 4 1x 2 12 37. 1x4 1 12 4 1x 1 12 38. 1x4 1 92 4 1x 1 32 39. 1x4 2 162 4 1x 1 22 40. 1x4 2 812 4 1x 2 32 41. 12x3 2 5x2 2 x 1 12 4 Ax 1 1 2B 42. 13x3 2 8x2 1 12 4 Ax 1 1 3B 43. 12x4 2 3x3 1 7x2 2 42 4 Ax 2 2 3B 44. 13x4 1 x3 1 2x 2 32 4 Ax 2 3 4B 45. 12x4 1 9x3 2 9x2 2 81x 2 812 4 1x 1 1.52 46. 15x3 2 x2 1 6x 1 82 4 1x 1 0.82 47. x7 2 8x4 1 3x2 1 1 x 2 1 48. x6 1 4x5 2 2x3 1 7 x 1 1 49. 1x6 2 49x4 2 25x2 1 12252 4 Ax 2 !5B 50. 1x6 2 4x4 2 9x2 1 362 4 Ax 2 !3B In Exercises 51–60, divide the polynomials by either long division or synthetic division. 51. 16x2 2 23x 1 72 4 13x 2 12 52. 16x2 1 x 2 22 4 12x 2 12 53. 1x3 2 x2 2 9x 1 92 4 1x 2 12 54. 1x3 1 2x2 2 6x 2 122 4 1x 1 22 55. 1x5 1 4x3 1 2x2 2 12 4 1x 2 22 56. 1x4 2 x2 1 3x 2 102 4 1x 1 52 57. 1x4 2 252 4 1x2 2 12 58. 1x3 2 82 4 1x2 2 22 59. 1x7 2 12 4 1x 2 12 60. 1x6 2 272 4 1x 2 32 • A P P L I C A T I O N S 61. Geometry. The area of a rectangle is 6x4 1 4x3 2 x2 2 2x 2 1 square feet. If the length of the rectangle is 2x2 2 1 feet, what is the width of the rectangle? 62. Geometry. If the rectangle in Exercise 61 is the base of a rectangular box with volume 18x5 1 18x4 1 x3 2 7x2 2 5x 2 1 cubic feet, what is the height of the box? 63. Travel. If a car travels a distance of x3 1 60x2 1 x 1 60 miles at an average speed of x 1 60 miles per hour, how long does the trip take? 64. Sports. If a quarterback throws a ball 2x2 2 5x 1 50 yards in 5 2 x seconds, how fast is the football traveling? • C A T C H T H E M I S T A K E In Exercises 65–68, explain the mistake that is made. 65. Divide x3 2 4x2 1 x 1 6 by x2 1 x 1 1. Solution: x 2 3 x2 1 x 1 1qx3 2 4x2 1 x 1 6 x3 1 x2 1 x 23x2 1 2x 1 6 23x2 2 3x 2 3 2x 1 3 This is incorrect. What mistake was made? 66. Divide x4 2 3x2 1 5x 1 2 by x 2 2. Solution: 22 1 23 5 2 22 10 230 1 25 15 228 f x2 2 5x 1 15 This is incorrect. What mistake was made? 67. Divide x3 1 4x 2 12 by x 2 3. Solution: 3 1 4 212 3 21 1 7 9 This is incorrect. What mistake was made? 68. Divide x3 1 3x2 2 2x 1 1 by x2 1 1. Solution: 21 1 3 22 1 21 22 4 1 2 24 5 f x2 2 2x 2 4 This is incorrect. What mistake was made? f x 1 7 4.3 Dividing Polynomials: Long Division and Synthetic Division 371 • C O N C E P T U A L In Exercises 69–72, determine whether each statement is true or false. 69. A fifth-degree polynomial divided by a third-degree polynomial will yield a quadratic quotient. 70. A third-degree polynomial divided by a linear polynomial will yield a linear quotient. 71. Synthetic division can be used whenever the degree of the dividend is exactly one more than the degree of the divisor. 72. When the remainder is zero, the divisor is a factor of the dividend. 73. Is x 1 b a factor of x3 1 12b 2 a2x2 1 1b2 2 2ab2x 2 ab2? 74. Is x 1 b a factor of x4 1 1b2 2 a22x2 2 a2b2? 75. Divide x3n 1 x2n 2 xn 2 1 by xn 2 1. 76. Divide x3n 1 5x2n 1 8xn 1 4 by xn 1 1. • C H A L L E N G E • T E C H N O L O G Y 77. Plot 2x3 2 x2 1 10x 2 5 x2 1 5 . What type of function is it? Perform this division using long division, and confirm that the graph corresponds to the quotient. 78. Plot x3 2 3x2 1 4x 2 12 x 2 3 . What type of function is it? Perform this division using synthetic division, and confirm that the graph corresponds to the quotient. 79. Plot x4 1 2x3 2 x 2 2 x 1 2 . What type of function is it? Perform this division using synthetic division, and confirm that the graph corresponds to the quotient. 80. Plot x5 2 9x4 1 18x3 1 2x2 2 5x 2 3 x4 2 6x3 1 2x 1 1 . What type of function is it? Perform this division using long division, and confirm that the graph corresponds to the quotient. 81. Plot 26x3 1 7x2 1 14x 2 15 2x 1 3 . What type of function is it? Perform this division using long division, and confirm that the graph corresponds to the quotient. 82. Plot 23x5 2 4x4 1 29x3 1 36x2 2 18x 3x2 1 4x 2 2 . What type of function is it? Perform this division using long division, and confirm that the graph corresponds to the quotient. 372 CHAPTER 4 Polynomial and Rational Functions 4.4.1 The Remainder Theorem and the Factor Theorem The zeros of a polynomial function assist us in finding the x-intercepts of the graph of a polynomial function. How do we find the zeros of a polynomial function? For polynomial functions of degree 2, we have the quadratic formula, which allows us to find the two zeros. For polynomial functions whose degree is greater than 2, much more work is required.1 In this section, we focus our attention on finding the real zeros of a polynomial function. Later, in Section 4.5, we expand our discussion to complex zeros of polynomial functions. In this section, we start by listing possible rational zeros. As you will see, there are sometimes many possibilities. We can then narrow the search using Descartes’ rule of signs, which tells us possible combinations of positive and negative real zeros. We can narrow the search even further with the upper and lower bound rules. Once we have tested possible values and determined a zero, we will employ synthetic division to divide the polynomial by the linear factor associated with the zero. We will continue the process until we have factored the polynomial function into a product of either linear factors or irreducible quadratic factors. Last, we will discuss how to find irrational real zeros using the intermediate value theorem. If we divide the polynomial function ƒ1x2 5 x3 2 2x2 1 x 2 3 by x 2 2 using synthetic division, we find the remainder is 21. 2 1 22 1 23 2 0 2 1 0 1 21 Notice that if we evaluate the function at x 5 2, the result is 21. ƒ122 5 21 This leads us to the remainder theorem. There are complicated formulas for finding the zeros of polynomial functions of degree 3 and 4, but there are no such formulas for degree 5 and higher polynomials (according to the Abel–Ruffini theorem) S K I L L S O B J E C T I V E S ■ ■Apply the remainder theorem to evaluate a polynomial function and use the factor theorem to factor polynomials. ■ ■Use the rational zero (root) theorem to list possible rational zeros and Descartes’ rule of signs to determine the possible combination of positive and negative real zeros. ■ ■Express polynomials as a product of linear or irreducible quadratic factors. ■ ■Employ the intermediate value theorem to approximate an irrational zero. ■ ■Graph any polynomial function. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that a polynomial degree of n has at most n real zeros. ■ ■Understand that a real zero can be either rational or irrational and that irrational zeros will not be listed as possible zeros through the rational zero test. ■ ■Irreducible factors are quadratic expressions that are not factorable over the real numbers. ■ ■Realize that rational zeros can be found exactly, whereas irrational zeros must be approximated. ■ ■Understand that the rational root test and the intermediate value theorem make it possible for us to graph any polynomial function and that Descartes’ rule of signs and upper and lower bound rules help us determine the graphs more efficiently. 4.4 THE REAL ZEROS OF A POLYNOMIAL FUNCTION 4.4.1 SKI LL Apply the remainder theorem to evaluate a polynomial function and use the factor theorem to factor polynomials. 4.4.1 CO NCE PTUAL Understand that a polynomial degree of n has at most n real zeros. 4.4 The Real Zeros of a Polynomial Function 373 WORDS MATH Recall the division algorithm. P1x2 5 d1x2 ⋅Q1x2 1 r1x2 Let d1x2 5 x 2 a for any real number a. P1x2 5 1x 2 a2 ⋅Q1x2 1 r1x2 The degree of the remainder is always less than the degree of the divisor: therefore, the remainder must be a constant 1Call it r, r1x2 5 r2. P1x2 5 1x 2 a2 ⋅Q1x2 1 r Let x 5 a. P1a2 5 1a 2 a2 ⋅Q1x2 1 r Simplify. P1a2 5 r f 0 The remainder theorem tells you that polynomial division can be used to evaluate a polynomial function at a particular point. ▼ A N S W E R P 12 22 5 28 EXAMPLE 1 Two Methods for Evaluating Polynomials Let P 1x2 5 4x5 2 3x4 1 2x3 2 7x2 1 9x 2 5 and evaluate P 122 by a. Evaluating P 122 directly b. The remainder theorem and synthetic division Solution: a. P 122 5 41225 2 31224 1 21223 2 71222 1 9122 2 5 5 41322 2 31162 1 2182 2 7142 1 9122 2 5 5 128 2 48 1 16 2 28 1 18 2 5 5 81 b. YOUR T UR N Let P 1x2 5 2x3 1 2x2 2 5x 1 2 and evaluate P 1222 using the remainder theorem and synthetic division. 2 4 23 2 27 9 25 8 10 24 34 86 4 5 12 17 43 81 ▼ Recall that when a polynomial is divided by x 2 a, if the remainder is zero, we say that x 2 a is a factor of the polynomial. Through the remainder theorem, we now know that the remainder is related to evaluation of the polynomial at the point x 5 a. We are then led to the factor theorem. REMAINDER THEOREM If a polynomial P 1x2 is divided by x 2 a, then the remainder is r 5 P 1a2. FACTOR THEOREM If P 1a2 5 0, then x 2 a is a factor of P 1x2. Conversely, if x 2 a is a factor of P 1x2, then P 1a2 5 0. 374 CHAPTER 4 Polynomial and Rational Functions EXAMPLE 2 Using the Factor Theorem to Factor a Polynomial Determine whether x 1 2 is a factor of P 1x2 5 x3 2 2x2 2 5x 1 6. If so, factor P 1x2 completely. Solution: By the factor theorem, x 1 2 is a factor of P 1x2 5 x3 2 2x2 2 5x 1 6 if P 1222 5 0. By the remainder theorem, if we divide P 1x2 5 x3 2 2x2 2 5x 1 6 by x 1 2, then the remainder is equal to P 1222. STEP 1 Divide P 1x2 5 x3 2 2x2 2 5x 1 6 by x 1 2 using synthetic division. Since the remainder is zero, P 1222 5 0, x 1 2 is a factor of P 1x2 5 x3 2 2x2 2 5x 1 6. STEP 2 Write P 1x2 as a product. P 1x2 5 1x 1 22 1x2 2 4x 1 32 STEP 3 Factor the quadratic polynomial. P 1x2 5 1x 1 22 1x 2 32 1x 2 12 YOUR TURN Determine whether x 2 1 is a factor of P 1x2 5 x3 2 4x2 2 7x 1 10. If so, factor P 1x2 completely. ▼ 22 1 22 25 6 22 8 26 1 24 3 0 x2 2 4x 1 3 f ▼ A N S W E R 1x 2 12 is a factor; P 1x2 5 1x 2 52 1x 2 12 1x 1 22 [CONCEPT CHECK] If a fourth-degree polynomial can be factored into four distinct linear factors P(x) 5 (x 2 a)(x 2 b)(x 2 c)(x 2 d ), what are the x-intercepts of the graph of this polynomial? ANSWER (a, 0) (b, 0) (c, 0) (d, 0) ▼ EXAMPLE 3 Using the Factor Theorem to Factor a Polynomial Determine whether x 2 3 and x 1 2 are factors of P 1x2 5 x4 2 13x2 1 36. If so, factor P 1x2 completely. Solution: STEP 1 With synthetic division divide P 1x2 5 x4 2 13x2 1 36 by x 2 3. Because the remainder is 0, x 2 3 is a factor , and we can write the polynomial as P1x2 5 1x 2 321x3 1 3x2 2 4x 2 122 STEP 2 With synthetic division divide the remaining cubic polynomial 1x3 1 3x2 2 4x 2 122 by x 1 2. 3 1 0 213 0 36 3 9 212 236 1 3 24 212 0 f x3 1 3x2 2 4x 2 12 22 1 3 24 212 22 22 12 1 1 26 0 x2 1 x 2 6 f Because the remainder is 0, x 1 2 is a factor , and we can now write the polynomial as P1x2 5 1x 2 321x 1 221x2 1 x 2 62 STEP 3 Factor the quadratic polynomial: x2 1 x 2 6 5 1x 1 32 1x 2 22. STEP 4 Write P 1x2 as a product of linear factors: YOUR T UR N Determine whether x 2 3 and x 1 2 are factors of P 1x2 5 x4 2 x3 2 7x2 1 x 1 6. If so, factor P 1x2 completely. P1x2 5 1x 2 321x 2 221x 1 221x 1 32 ▼ ▼ A N S W E R 1x 2 32 and 1x 1 22 are factors; P 1x2 5 1x 2 32 1x 1 22 1x 2 12 1x 1 12 The Search for Real Zeros In all of the examples thus far, the polynomial function and one or more real zeros (or linear factors) were given. Now, we will not be given any real zeros to start with. Instead, we will develop methods to search for them. Each real zero corresponds to a linear factor, and each linear factor is of degree 1. Therefore, the largest number of real zeros a polynomial function can have is equal to the degree of the polynomial. The following functions illustrate that a polynomial function of degree n can have at most n real zeros. POLYNOMIAL FUNCTION DEGREE REAL ZEROS COMMENTS ƒ1x2 5 x2 2 9 2 x 5 63 Two real zeros ƒ1x2 5 x2 1 4 2 None No real zeros ƒ1x2 5 x3 2 1 3 x 5 1 One real zero ƒ1x2 5 x3 2 x2 2 6x 3 x 5 22, 0, 3 Three real zeros Now that we know the maximum number of real zeros a polynomial function can have, let us discuss how to find these zeros. 4.4.2 The Rational Zero Theorem and Descartes’ Rule of Signs When the coefficients of a polynomial are integers, then the rational zero theorem (rational root test) gives us a list of possible rational zeros. We can then test these possible values to determine whether they really do correspond to actual zeros. Descartes’ rule of signs tells us the possible combinations of positive real zeros and negative real zeros. Using Descartes’ rule of signs will help us narrow down the large list of possible zeros generated through the rational zero theorem to a (hopefully) shorter list of possible zeros. First, let’s look at the rational zero theorem; then we’ll turn to Descartes’ rule of signs. 4.4.2 S K I L L Use the rational zero (root) theorem to list possible rational zeros and Descartes’ rule of signs to determine the possible combination of positive and negative real zeros. 4.4.2 C O N C E P T U A L Understand that a real zero can be either rational or irrational and that irrational zeros will not be listed as possible zeros through the rational zero test. STUDY TIP The largest number of zeros a polynomial can have is equal to the degree of the polynomial. 4.4 The Real Zeros of a Polynomial Function 375 THE NUMBER OF REAL ZEROS A polynomial function cannot have more real zeros than its degree. 376 CHAPTER 4 Polynomial and Rational Functions To use this theorem, simply list all combinations of integer factors of both the constant term a0 and the leading coefficient term an and take all appropriate combinations of ratios. This procedure is illustrated in Example 4. Notice that when the leading coefficient is 1, then the possible rational zeros will simply be the possible integer factors of the constant term. STUDY TIP The remainder can be found by evaluating the function or synthetic division. For simple values like x 5 61, it is easier to evaluate the polynomial function. For other values, it is often easier to use synthetic division. EXAMPLE 4 Using the Rational Zero Theorem Determine possible rational zeros for the polynomial P 1x2 5 x4 2 x3 2 5x2 2 x 2 6 by the rational zero theorem. Test each one to find all rational zeros Solution: STEP 1 List factors of the constant a0 5 26 61, 62, 63, 66 and leading coefficient terms an 5 1 61 STEP 2 List possible rational zeros a0 an. There are three ways to test whether any of these are zeros: Substitute these values into the polynomial to see which ones yield zero; use either polynomial division or synthetic division to divide the polynomial by these possible zeros; and look for a zero remainder. STEP 3 Test possible zeros by looking for zero remainders. 1 is not a zero: P 112 5 1124 2 1123 2 51122 2 112 2 6 5 212 21 is not a zero: P 1212 5 12124 2 12123 2 512122 2 1212 2 6 5 28 We could continue testing with direct substitution, but let us now use synthetic division as an alternative. 2 is not a zero: 22 is a zero: 61 61, 62 61, 63 61, 66 61 5 61, 62, 63, 66 2 1 21 25 21 26 2 2 26 214 1 1 23 27 220 22 1 21 25 21 26 22 6 22 6 1 23 1 23 0 THE RATIONAL ZERO THEOREM (RATIONAL ROOT TEST) If the polynomial function P1x2 5 anxn 1 an21xn21 1 c1 a2x2 1 a1x 1 a0 has integer coefficients, then every rational zero of P 1x2 has the form: Rational zero 5 integer factors of a0 integer factors of an 5 integer factors of constant term integer factors of leading coefficient 5 6 positive integer factors of constant term positive integer factors of leading coefficient Since 22 is a zero, then x 1 2 is a factor of P 1x2, and the remaining quotient is x3 2 3x2 1 x 2 3. Therefore, if there are any other real roots remaining, we can now use the simpler x3 2 3x2 1 x 2 3 for the dividend. Also note that the rational zero theorem can be applied to the new dividend and possibly shorten the list of possible rational zeros. In this case, the possible rational zeros of F1x2 5 x3 23x2 1 x 23 are 61 and 63. 3 is a zero: We now know that 22 and 3 are confirmed zeros. If we continue testing, we will find that the other possible zeros fail. This is a fourth-degree polynomial, and we have found two rational real zeros. We see in the graph on the right that these two real zeros correspond to the x-intercepts. YOUR TURN List the possible rational zeros of the polynomial P 1x2 5 x4 1 2x3 2 2x2 1 2x 2 3, and determine rational real zeros. 3 1 23 1 23 3 0 3 1 0 1 0 ▼ A N S W E R Possible rational zeros: 61 and 63. Rational real zeros: 1 and 23. ▼ x y 20 5 –5 –20 Notice in Example 4 that the polynomial function P 1x2 5 x4 2 x3 2 5x2 2 x 2 6 had two rational real zeros, x 5 22 and x 5 3. This implies that x 1 2 and x 2 3 are factors of P 1x2. Also note in the last step that when we divided by the zero x 5 3, the quotient was x2 1 1. Therefore, we can write the polynomial in factored form as Notice that the first two factors are of degree 1, so we call them linear factors. The third expression, x2 1 1, is of degree 2 and cannot be factored in terms of real numbers. We will discuss complex zeros in the next section. For now, we say that a quadratic expres-sion, ax2 1 bx 1 c, is called irreducible if it cannot be factored over the real numbers. P 1x2 5 1x 1 22 1x 2 32 1x2 1 12 () () () linear factor linear factor irreducible quadratic factor EXAMPLE 5 Factoring a Polynomial Function Write the following polynomial function as a product of linear and/or irreducible quadratic factors: P 1x2 5 x4 2 4x3 1 4x2 2 36x 2 45. Solution: Use the rational zero theorem to list possible rational roots. x 5 61, 63, 65, 69, 615, 645 Test possible zeros by evaluating the function or by utilizing synthetic division. x 5 1 is not a zero. P 112 5 280 x 5 21 is a zero. P 1212 5 0 4.4 The Real Zeros of a Polynomial Function 377 STUDY TIP Notice in Step 3 that the polynomial F 1x2 5 x3 2 3x2 1 x 2 3 can be facto red by grouping: F 1x2 5 1x 2321x2 112. 378 CHAPTER 4 Polynomial and Rational Functions Divide P 1x2 by x 1 1. x 5 5 is a zero. The factor x2 1 9 is irreducible. Write the polynomial as a product of linear and/or irreducible quadratic factors. P 1x2 5 1x 2 52 1x 1 12 1x2 1 92 Notice that the graph of this polynomial function has x-intercepts at x 5 21 and x 5 5. Y OUR TU R N Write the following polynomial function as a product of linear and/or irreducible quadratic factors. P 1x2 5 x4 2 2x3 2 x2 2 4x 2 6 21 1 24 4 236 245 21 5 29 45 1 25 9 245 0 5 1 25 9 245 5 0 45 1 0 9 0 x2 1 9 (++)'+ ▼ A N S W E R P 1x2 5 1x 1 12 1x 2 32 1x2 1 22 ▼ The rational zero theorem lists possible zeros. It would be helpful if we could narrow that list. Descartes’ rule of signs determines the possible combinations of positive real zeros and negative real zeros through variations of sign. A variation in sign is a sign difference seen between consecutive coefficients. Sign Change 2 to 1 P1x2 5 2x6 2 5x5 2 3x4 1 2x3 2 x2 2 x 2 1 This polynomial experiences three sign changes or variations in sign. Sign Change 1 to 2 Sign Change 1 to 2 DESCARTES’ RULE OF SIGNS If the polynomial function P1x2 5 anx n 1 an21x n21 1 c1 a2x2 1 a1x 1 a0 has real coefficients and a0 2 0, then: ■ The number of positive real zeros of the polynomial is either equal to the number of variations of sign of P 1x2 or less than that number by an even integer. ■ The number of negative real zeros of the polynomial is either equal to the number of variations of sign of P 12x2 or less than that number by an even integer. Descartes’ rule of signs narrows our search for real zeros because we don’t have to test all of the possible rational zeros. For example, if we know there is one positive real zero, then if we find a positive rational zero we no longer need to continue to test possible positive zeros EXAMPLE 7 Using Descartes’ Rule of Signs to Find Possible Combinations of Real Zeros Determine the possible combinations of real zeros for P 1x2 5 x4 2 2x3 1 x2 1 2x 2 2. Solution: P 1x2 has 3 variations in sign. Apply Descartes’ rule of signs. P 1x2 has either 3 or 1 positive real zero. Find P 12x2. P 12x2 5 12x24 2 212x23 1 12x22 1 212x2 2 2 5 x4 1 2x3 1 x2 2 2x 2 2 P 12x2 has 1 variation in sign. P 12x2 5 x4 1 2x3 1 x2 2 2x 2 2 Apply Descartes’ rule of signs. P 1x2 has 1 negative real zero. Since P 1x2 5 x4 2 2x3 1 x2 1 2x 2 2 is a fourth-degree polynomial, there are at most 4 real zeros. P 1x2 has 1 negative real zero and could have 3 or 1 positive real zeros. Y OUR T UR N Determine the possible combinations of zeros for P 1x2 5 x4 1 2x3 1 x2 1 8x 2 12 Sign Change Sign Change P 1x2 5 x4 2 2x3 1 x2 1 2x 2 2 Sign Change Sign Change ▼ ▼ A N S W E R Positive real zeros: 1 Negative real zeros: 3 or 1 [CONCEPT CHECK] TRUE OR FALSE The rational zeros are a subset of the real zeros. ANSWER True ▼ EXAMPLE 6 Using Descartes’ Rule of Signs Determine the possible combinations of zeros for P 1x2 5 x3 2 2x2 2 5x 1 6. Solution: Determine the number of variations of sign in P 1x2. P 1x2 5 x3 2 2x2 2 5x 1 6 P 1x2 has 2 variations in sign. Apply Descartes’ rule of signs. P 1x2 has either 2 or 0 positive real zeros. Determine the number of 12x23 2 212x22 2 512x2 1 6 variations of sign in P 12x2. 2x3 2 2x2 1 5x 1 6 P 12x2 has 1 variation in sign. P 1x2 5 2x3 2 2x2 1 5x 1 6 Apply Descartes’ rule of signs. P 1x2 must have 1 negative real zero. Since P 1x2 5 x3 2 2x2 2 5x 1 6 is a third-degree polynomial, there are at most 3 real zeros. One zero is a negative real number, and there can be either 2 positive real zeros or 0 positive real zeros. Now look back at Example 2 and see that in fact there were 1 negative real zero and 2 positive real zeros. Sign Change Sign Change Sign Change 4.4 The Real Zeros of a Polynomial Function 379 380 CHAPTER 4 Polynomial and Rational Functions 4.4.3 SKI LL Express polynomials as a product of linear or irreducible quadratic factors. 4.4.3 CO NCE PTUAL Irreducible factors are quadratic expressions that are not factorable over the real numbers. 4.4.3 Factoring Polynomials Now let’s draw on the tests discussed in this section thus far to help us in finding all real zeros of a polynomial function. Doing so will enable us to factor polynomials. [CONCEPT CHECK] TRUE OR FALSE An irreducible quadratic factor corresponds to two additional x-intercepts. ANSWER False ▼ ▼ A N S W E R P 1x2 5 1x 2 22 1x 1 12 1x 2 12 1x2 1 22 EXAMPLE 8 Factoring a Polynomial Write the polynomial P 1x2 5 x5 1 2x4 2 x 2 2 as a product of linear and/or irreducible quadratic factors. Solution: STEP 1 Determine variations in sign. P 1x2 has 1 sign change. P 1x2 5 x5 1 2x4 2 x 2 2 P 12x2 has 2 sign changes. P 12x2 5 2x5 1 2x4 1 x 2 2 STEP 2 Apply Descartes’ rule of signs. STEP 3 Use the rational zero theorem to determine the possible rational zeros. 61, 62 We know (Step 2) that there is one positive real zero, so test the possible positive rational zeros first. STEP 4 Test possible rational zeros. 1 is a zero: Now that we have found the positive zero, we can test the other two possible negative zeros—because either they both are zeros or neither is a zero. 21 is a zero: At this point, from Descartes’ rule of signs we know that 22 must also be a zero, since there are either 2 or 0 negative zeros. Let’s confirm this. 22 is a zero: STEP 5 Three of the five zeros have been found to be zeros: 21, 22, and 1. STEP 6 Write the fifth-degree polynomial as a product of 3 linear factors and an irreducible quadratic factor. P1x2 5 1x 2 121x 1 121x 1 22Ax2 1 1B Y OUR TU R N Write the polynomial P 1x2 5 x5 2 2x4 1 x3 2 2x2 2 2x 1 4 as a product of linear and/or irreducible quadratic factors. Positive Real Zeros: 1 Negative Real Zeros: 2 or 0 1 1 2 0 0 21 22 1 3 3 3 2 1 3 3 3 2 0 21 1 3 3 3 2 21 22 21 22 1 2 1 2 0 22 1 2 1 2 22 0 22 1 0 1 0 x2 1 1 f ▼ The rational zero theorem gives us possible rational zeros of a polynomial, and Descartes’ rule of signs gives us possible combinations of positive and negative real zeros. Additional aids that help eliminate possible zeros are the upper and lower bound rules. These rules can give you an upper and lower bound on the real zeros of a polynomial function. If ƒ1x2 has a common monomial factor, you should factor it out first, and then follow the upper and lower bound rules. STUDY TIP If f (x) has a common monomial factor, it should be factored out before applying the bound rules. UPPER AND LOWER BOUND RULES Let ƒ1x2 be a polynomial with real coefficients and a positive leading coefficient. Suppose ƒ1x2 is divided by x 2 c using synthetic division. 1. If c . 0 and each number in the bottom row is either positive or zero, c is an upper bound for the real zeros of ƒ. 2. If c , 0 and the numbers in the bottom row are alternately positive and negative (zero entries count as either positive or negative), c is a lower bound for the real zeros of ƒ. 4.4 The Real Zeros of a Polynomial Function 381 EXAMPLE 9 Using Upper and Lower Bounds to Eliminate Possible Zeros Find the real zeros of ƒ1x2 5 4x3 2 x2 1 36x 2 9. Solution: STEP 1 The rational zero theorem gives possible rational zeros. Factors of 9 Factors of 4 5 61, 63, 69 61, 62, 64 5 61, 61 2, 61 4, 63 4, 63 2, 69 4, 63, 69 2 , 69 STEP 2 Apply Descartes’ rule of signs: ƒ1x2 has 3 sign variations. 3 or 1 positive real zeros ƒ12x2 has no sign variations. no negative real zeros STEP 3 Try x 5 1. x 5 1 is not a zero, but because the last row contains all positive entries, x 5 1 is an upper bound. Since we know there are no negative real zeros, we restrict our search to between 0 and 1. STEP 4 Try x 5 1 4. 1 4 is a zero and the quotient 4x2 1 36 has all positive coefficients; therefore, 1 4 is the only real zero . Note: If ƒ1x2 has a common monomial factor, it should be factored out first before applying the bound rules. 1 4 21 36 29 4 3 39 4 3 39 30 1 4 4 21 36 29 1 0 9 4 0 36 0 382 CHAPTER 4 Polynomial and Rational Functions 4.4.4 The Intermediate Value Theorem In our search for zeros, we sometimes encounter irrational zeros, as in, for example, the polynomial ƒ1x2 5 x5 2 x4 2 1 Descartes’ rule of signs tells us there is exactly one real positive zero. However, the rational zero test yields only x 5 61, neither of which is a zero. So if we know there is a real positive zero and we know it’s not rational, it must be irrational. Notice that ƒ112 5 21 and ƒ122 5 15. Since polynomial functions are continuous and the function goes from negative to positive between x 5 1 and x 5 2, we expect a zero somewhere in that interval. Generating a graph with a graphing utility, we find that there is a zero around x 5 1.3. The intermediate value theorem is based on the fact that polynomial functions are continuous. INTERMEDIATE VALUE THEOREM Let a and b be real numbers such that a , b and ƒ1x2 be a polynomial function. If ƒ1a2 and ƒ1b2 have opposite signs, then there is at least one real zero between a and b. x zero f (x) y a b f (b) f (a) If the intermediate value theorem tells us that there is a real zero in the interval 1a, b2, how do we approximate that zero? The bisection method is a root-finding algorithm that approximates the solution to the equation ƒ1x2 5 0. In the bisection method, the interval is divided in half, and then the subinterval that contains the zero is selected. This is repeated until the bisection method converges to an approximate root of ƒ. EXAMPLE 10 Approximating Real Zeros of a Polynomial Function Approximate the real zero of ƒ1x2 5 x5 2 x4 2 1. Note: Descartes’ rule of signs tells us that there are no real negative zeros and there is exactly one real positive zero. Solution: Find two consecutive integer values for x that have corresponding function values opposite in sign. Note that a graphing utility would have shown an x-intercept between x 5 1 and x 5 2. x f (x) 1 21 2 15 In calculus you will learn Newton’s method, which is a more efficient approximation technique for finding zeros. 4.4.4 SKI LL Employ the intermediate value theorem to approximate an irrational zero. 4.4.4 CO NCE PTUAL Realize that rational zeros can be found exactly, whereas irrational zeros must be approximated. [CONCEPT CHECK] TRUE OR FALSE If we know there is one positive zero and all of the potential rational roots have been found not to be zeros, then the real zero must be irrational. ANSWER True ▼ Apply the bisection method, with a 5 1 and b 5 2. c 5 a 1 b 2 5 1 1 2 2 5 3 2 Evaluate the function at x 5 c. ƒ11.52 < 1.53 Compare the values of ƒ at the endpoints and midpoint. ƒ112 5 21, ƒ11.52 < 1.53, ƒ122 5 15 Select the subinterval corresponding to the opposite signs of ƒ. 11, 1.52 Apply the bisection method again (repeat the algorithm). 1 1 1.5 2 5 1.25 Evaluate the function at x 5 1.25. ƒ11.252 < 20.38965 Compare the values of ƒ at the endpoints and midpoint. ƒ112 5 21, ƒ11.252 < 20.38965, ƒ11.52 < 1.53 Select the subinterval corresponding to the opposite signs of ƒ. 11.25, 1.52 Apply the bisection method again (repeat the algorithm). 1.25 1 1.5 2 5 1.375 Evaluate the function at x 5 1.375. ƒ11.3752 < 0.3404 Compare the values of ƒ at the endpoints and midpoint. ƒ11.252 < 20.38965, ƒ11.3752 < 0.3404, ƒ11.52 < 1.53 Select the subinterval corresponding to the opposite signs of ƒ. 11.25, 1.3752 We can continue this procedure (applying the bisection method ) to find that the zero is somewhere between ƒ11.322 < 20.285 and ƒ11.332 < 0.0326. We find that to three significant digits is an approximation to the real zero. 1.32 4.4.5 S K I L L Graph any polynomial function. 4.4.5 C O N C E P T U A L Understand that the rational root test and the intermediate value theorem make it possible for us to graph any polynomial function and Descartes’ rule of signs and that upper and lower bound rules help us determine the graphs more efficiently. 4.4.5 Graphing Polynomial Functions In Section 4.2, we graphed simple polynomial functions that were easily factored. Now that we have procedures for finding real zeros of polynomial functions (rational zero theorem, Descartes’ rule of signs, and upper and lower bound rules for rational zeros, and the intermediate value theorem and the bisection method for irrational zeros), let us return to the topic of graphing polynomial functions. Since a real zero of a polynomial function corresponds to an x-intercept of its graph, we now have methods for finding (or estimating) any x-intercepts of the graph of any polynomial function. EXAMPLE 11 Graphing a Polynomial Function Graph the function ƒ1x2 5 2x4 2 2x3 1 5x2 1 17x 2 22. Solution: STEP 1 Find the y-intercept. ƒ102 5 222 STEP 2 Find any x-intercepts (real zeros). Apply Descartes’ rule of signs. 3 sign changes correspond to 3 or 1 positive real zeros. ƒ1x2 5 2x4 2 2x3 1 5x2 1 17x 2 22 4.4 The Real Zeros of a Polynomial Function 383 384 CHAPTER 4 Polynomial and Rational Functions 1 sign change corresponds to 1 negative real zero. ƒ12x2 5 2x4 1 2x3 1 5x2 2 17x 2 22 Apply the rational zero theorem. Let a0 5 222 and an 5 2. Factors of a0 Factors of an 5 61 2, 61, 62, 611 2 , 611, 622 Test the possible zeros. x 5 1 is a zero. ƒ112 5 0 There are no other rational zeros. Apply the upper bound rule. Since x 5 1 is positive and all of the numbers in the bottom row are positive (or zero), x 5 1 is an upper bound for the real zeros. We know there is exactly one negative real zero, but none of the possible zeros from the rational zero theorem is a zero. Therefore, the negative real zero is irrational. Apply the intermediate value theorem and the bisection method. ƒ is positive at x 5 22. ƒ1222 5 12 ƒ is negative at x 5 21. ƒ1212 5 230 Use the bisection method to find the negative real zero between 22 and 21. x < 21.85 STEP 3 Determine the end behavior. y 5 2x4 STEP 4 Find additional points. x 22 21.85 21 0 1 2 f (x) 12 0 230 222 0 48 Point 122, 122 121.85, 02 121, 2302 10, 2222 11, 02 12, 482 STEP 5 Sketch the graph. 1 2 22 5 17 222 2 0 5 22 2 0 5 22 0 x y (1, 0) (2, 48) (–1, –30) (0, –22) (–1.85, 0) (–2, 12) 48 2 –2 –32 x y [CONCEPT CHECK] TRUE OR FALSE We now have the tools to find all x-intercepts (exactly or approximated). ANSWER True ▼ 4.4 The Real Zeros of a Polynomial Function 385 [SEC TION 4 .4] E X ER C I S E S • S K I L L S In Exercises 1–6, find the following values by using synthetic division. Check by substituting the value into the function ƒ1x2 5 3x4 2 2x3 1 7x2 2 8 g 1x2 5 2x3 1 x2 1 1 1. ƒ112 2. ƒ1212 3. g 112 4. g 1212 5. ƒ1222 6. g 122 In Exercises 7–10, determine whether the number given is a zero of the polynomial. 7. 27, P 1x2 5 x3 1 2x2 2 29x 1 42 8. 2, P 1x2 5 x3 1 2x2 2 29x 1 42 9. 23, P 1x2 5 x3 2 x2 2 8x 1 12 10. 1, P 1x2 5 x3 2 x2 2 8x 1 12 In Exercises 11–20, given a real zero of the polynomial, determine all other real zeros, and write the polynomial in terms of a product of linear and/or irreducible quadratic factors. Polynomial Zero Polynomial Zero 11. P 1x2 5 x3 2 13x 1 12 1 12. P 1x2 5 x3 1 3x2 2 10x 2 24 3 13. P 1x2 5 2x3 1 x2 2 13x 1 6 1 2 14. P 1x2 5 3x3 2 14x2 1 7x 1 4 21 3 15. P 1x2 5 x4 2 2x3 2 11x2 2 8x 2 60 23, 5 16. P 1x2 5 x4 2 x3 1 7x2 2 9x 2 18 21, 2 17. P 1x2 5 x4 2 5x2 1 10x 2 6 1, 23 18. P 1x2 5 x4 2 4x3 1 x2 1 6x 2 40 4, 22 19. P 1x2 5 x4 1 6x3 1 13x2 1 12x 1 4 22 1multiplicity 22 20. P 1x2 5 x4 1 4x3 2 2x2 2 12x 1 9 1 1multiplicity 22 In Exercises 21–28, use the rational zero theorem to list the possible rational zeros. 21. P 1x2 5 x4 1 3x2 2 8x 1 4 22. P 1x2 5 2x4 1 2x3 2 5x 1 4 23. P 1x2 5 x5 2 14x3 1 x2 2 15x 1 12 24. P 1x2 5 x5 2 x3 2 x2 1 4x 1 9 25. P 1x2 5 2x6 2 7x4 1 x3 2 2x 1 8 26. P 1x2 5 3x5 1 2x4 2 5x3 1 x 2 10 27. P 1x2 5 5x5 1 3x4 1 x3 2 x 2 20 28. P 1x2 5 4x6 2 7x4 1 4x3 1 x 2 21 PR O C ED U R E F O R FA C T O R I N G A PO LY N O M I A L F U N C T I O N n List possible rational zeros (rational zero theorem). n List possible combinations of positive and negative real zeros (Descartes’ rule of signs). n Test possible values until a zero is found. n Once a real zero is found, repeat testing on the quotient until linear and/or irreducible quadratic factors remain. n If there is a real zero but all possible rational roots have failed, then approximate the zero using the intermediate value theorem and the bisection method. Depending on the form of the quotient, upper and lower bounds may eliminate possible zeros. In this section, we discussed how to find the real zeros of a polynomial function. Once real zeros are known, it is possible to write the polynomial function as a product of linear and/or irreducible quadratic factors. THE NUMB E R OF Z E R OS n A polynomial of degree n has at most n real zeros. n Descartes’ rule of signs determines the possible combina-tions of positive and negative real zeros. n Upper and lower bounds help narrow the search for zeros. HOW T O FIND Z E R OS n Rational zero theorem: List possible rational zeros: Factors of constant, a0 Factors of leading coefficient, an n Irrational zeros: Approximate zeros by determining when the polynomial function changes sign (intermediate value theorem). [SEC TION 4 .4] S U M M A RY 386 CHAPTER 4 Polynomial and Rational Functions In Exercises 29–32, list the possible rational zeros, and test to determine all rational zeros. 29. P 1x2 5 x4 1 2x3 2 9x2 2 2x 1 8 30. P 1x2 5 x4 1 2x3 2 4x2 2 2x 1 3 31. P 1x2 5 2x3 2 9x2 1 10x 2 3 32. P 1x2 5 3x3 2 5x2 2 26x 2 8 In Exercises 33–44, use Descartes’ rule of signs to determine the possible number of positive real zeros and negative real zeros. 33. P 1x2 5 x4 2 32 34. P 1x2 5 x4 1 32 35. P 1x2 5 x5 2 1 36. P 1x2 5 x5 1 1 37. P 1x2 5 x5 2 3x3 2 x 1 2 38. P 1x2 5 x4 1 2x2 2 9 39. P 1x2 5 9x7 1 2x5 2 x3 2 x 40. P 1x2 5 16x7 2 3x4 1 2x 2 1 41. P 1x2 5 x6 2 16x4 1 2x2 1 7 42. P 1x2 5 27x6 2 5x4 2 x2 1 2x 1 1 43. P 1x2 5 23x4 1 2x3 2 4x2 1 x 2 11 44. P 1x2 5 2x4 2 3x3 1 7x2 1 3x 1 2 For each polynomial in Exercises 45–62: (a) use Descartes’ rule of signs to determine the possible combinations of positive real zeros and negative real zeros; (b) use the rational zero test to determine possible rational zeros; (c) test for rational zeros; and (d) factor as a product of linear and/or irreducible quadratic factors. 45. P(x) 5 x3 1 6x2 1 11x 1 6 46. P 1x2 5 x3 2 6x2 1 11x 2 6 47. P 1x2 5 x3 2 7x2 2 x 1 7 48. P 1x2 5 x3 2 5x2 2 4x 1 20 49. P 1x2 5 x4 1 6x3 1 3x2 2 10x 50. P 1x2 5 x4 2 x3 2 14x2 1 24x 51. P 1x2 5 x4 2 7x3 1 27x2 2 47x 1 26 52. P 1x2 5 x4 2 5x3 1 5x2 1 25x 2 26 53. P 1x2 5 10x3 2 7x2 2 4x 1 1 54. P 1x2 5 12x3 2 13x2 1 2x 2 1 55. P 1x2 5 6x3 1 17x2 1 x 2 10 56. P 1x2 5 6x3 1 x2 2 5x 2 2 57. P 1x2 5 x4 2 2x3 1 5x2 2 8x 1 4 58. P 1x2 5 x4 1 2x3 1 10x2 1 18x 1 9 59. P 1x2 5 x6 1 12x4 1 23x2 2 36 60. P 1x2 5 x4 2 x2 2 16x2 1 16 61. P 1x2 5 4x4 2 20x3 1 37x2 2 24x 1 5 62. P 1x2 5 4x4 2 8x3 1 7x2 1 30x 1 50 In Exercises 63–66, use the information found in Exercises 47, 51, 55, and 61 to assist in sketching a graph of each polynomial function. 63. See Exercise 47. 64. See Exercise 51. 65. See Exercise 55. 66. See Exercise 61. In Exercises 67–72, use the intermediate value theorem and the bisection method to approximate the real zero in the indicated interval. Approximate to two decimal places. 67. ƒ1x2 5 x4 2 3x3 1 4 31, 24 68. ƒ1x2 5 x5 2 3x3 1 1 30, 14 69. ƒ1x2 5 7x5 2 2x2 1 5x 2 1 30, 14 70. ƒ1x2 5 22x3 1 3x2 1 6x 2 7 322, 214 71. ƒ1x2 5 x3 2 2x2 2 8x 2 3 321, 04 72. ƒ1x2 5 x4 1 4x2 2 7x 2 13 322, 214 • A P P L I C A T I O N S 73. Profit. A mathematics honorary society wants to sell magazine subscriptions to Math Weekly. If there are x hundred subscribers, its monthly revenue and cost are given by: R1x2 5 46 2 3x2 and C1x2 5 20 1 2x a. Determine the profit function. Hint: P 5 R 2 C. b. Determine the number of subscribers needed in order to break even. 74. Profit. Using the profit equation P 1x2 5 x3 2 5x2 1 3x 1 6, when will the company break even if x represents the units sold? For Exercises 75 and 76, refer to the following: The demand function for a product is p1x2 5 28 2 0.0002x where p is the unit price (in dollars) of the product and x is the number of units produced and sold. The cost function for the product is C1x2 5 20x 1 1500 where C is the total cost (in dollars) and x is the number of units produced. The total profit obtained by producing and selling x units is P1x2 5 xp1x2 2 C1x2 75. Business. Find the total profit function when x units are produced and sold. Use Descartes’ rule of signs to determine possible combinations of positive zeros for the profit function 76. Business. Find the break-even point(s) for the product to the nearest unit. Discuss the significance of the break-even point(s) for the product. 77. Health/Medicine. During the course of treatment of an illness, the concentration of a dose of a drug (in mcg/mL) in the bloodstream fluctuates according to the mode C1t2 5 15.4 2 0.05t2 where t 5 0 is when the drug was administered. Assuming a single dose of the drug is administered, in how many hours (to the nearest hour) after being administered will the drug be eliminated from the bloodstream? 78. Health/Medicine. During the course of treatment of an illness, the concentration of a dose of a drug (in mcg/mL) in the bloodstream fluctuates according to the mode C1t2 5 60 2 0.75t2 where t 5 0 is when the drug was administered. Assuming a single dose of the drug is administered, in how many hours (to the nearest hour) after being administered will the drug be eliminated from the bloodstream? 4.4 The Real Zeros of a Polynomial Function 387 79. Use Descartes’ rule of signs to determine the possible combinations of zeros of P1x2 5 2x5 1 7x4 1 9x3 1 9x2 1 7x 1 2 Solution: No sign changes, so no positive real zeros. P1x2 5 2x5 1 7x4 1 9x3 1 9x2 1 7x 1 2 Five sign changes, so five negative real zeros P12x2 5 22x5 1 7x4 2 9x3 1 9x2 2 7x 1 2 This is incorrect. What mistake was made? 80. Determine whether x 2 2 is a factor of P 1x2 5 x3 2 2x2 2 5x 1 6. Solution: 22 1 22 25 6 22 8 26 1 24 3 0 Yes, x 2 2 is a factor of P 1x2. This is incorrect. What mistake was made? In Exercises 79 and 80, explain the mistake that is made. • C A T C H T H E M I S T A K E 81. All real zeros of a polynomial correspond to x-intercepts. 82. A polynomial of degree n, n . 0, must have at least one zero. 83. A polynomial of degree n, n . 0, can be written as a product of n linear factors over real numbers. 84. The number of sign changes in a polynomial is equal to the number of positive real zeros of that polynomial. • C O N C E P T U A L In Exercises 81–84 determine whether each statement is true or false. • C H A L L E N G E 85. Given that x 5 a is a zero of P 1x2 5 x3 2 1a 1 b 1 c2x2 1 1ab 1 ac 1 bc2x 2 abc, find the other two zeros, given that a, b, c are real numbers and a . b . c. 86. Given that x 5 a is a zero of p 1x2 5 x3 1 12a 1 b 2 c2x2 2 1ab 1 bc 2 ac2x 1 abc, find the other two real zeros, given that a, b, c are real positive numbers. In Exercises 87 and 88, determine all possible rational zeros of the polynomial. There are many possibilities. Instead of trying them all, use a graphing calculator or software to graph P 1x2 to help find a zero to test. 87. P 1x2 5 x3 2 2x2 1 16x 2 32 88. P 1x2 5 x3 2 3x2 1 16x 2 48 In Exercises 89 and 90: (a) determine all possible rational zeros of the polynomial, use a graphing calculator or software to graph P 1x2 to help find the zeros; and (b) factor as a product of linear and/or irreducible quadratic factors. 89. P1x2 5 12x4 1 25x3 1 56x2 2 7x 2 30 90. P1x2 5 – 3x3 2 x2 2 7x 2 49 • T E C H N O L O G Y 388 CHAPTER 4 Polynomial and Rational Functions S K I L L S O B J E C T I V E S ■ ■Factor a polynomial function given certain zeros. ■ ■Factor a polynomial function of degree n into n linear factors. C O N C E P T U A L O B J E C T I V E S ■ ■Understand why complex zeros occur in conjugate pairs. ■ ■Understand why an odd-degree polynomial must have at least one real zero. 4.5 COMPLEX ZEROS: THE FUNDAMENTAL THEOREM OF ALGEBRA 4.5.1 Complex Zeros In Section 4.4, we found the real zeros of a polynomial function. In this section we find the complex zeros of a polynomial function. In this chapter, we assume the coefficients of polynomial functions are real numbers. The domain of polynomial functions thus far has been the set of all real numbers. Now, we consider a more general case. In this section, the coefficients of a polynomial function and the domain of a polynomial function are complex numbers. Note that the set of real numbers is a subset of the complex numbers. (Choose the imaginary part to be zero.) It is important to note, however, that when we are discussing graphs of polynomial functions, we restrict the domain to the set of real numbers. A zero of a polynomial P 1x2 is the solution or root of the equation P 1x2 5 0. The zeros of a polynomial can be complex numbers. However, since the xy-plane represents real numbers, we interpret zeros as x-intercepts only when the zeros are real numbers. We can illustrate the relationship between real and complex zeros of polynomial functions and their graphs with two similar examples. Let’s take the two quadratic functions ƒ 1x2 5 x2 2 4 and g 1x2 5 x2 1 4. The graphs of these two functions are parabolas that open upward with ƒ 1x2 shifted down four units and g 1x2 shifted up four units as shown on the left. Setting each function equal to zero and solving for x, we find that the zeros for ƒ 1x2 are 22 and 2 and the zeros for g 1x2 are 22i and 2i. Notice that the x-intercepts for ƒ 1x2 are 122, 02 and 12, 02 and g 1x2 has no x-intercepts. The Fundamental Theorem of Algebra In Section 4.4, we were able to write a polynomial function as a product of linear and/or irreducible quadratic factors. Now, we consider factors over complex numbers. Therefore, what were irreducible quadratic factors over real numbers will now be a product of two linear factors over the complex numbers. What are the minimum and maximum number of zeros a polynomial can have? Every polynomial has at least one zero (provided the degree is greater than zero). The largest number of zeros a polynomial can have is equal to the degree of the polynomial. The fundamental theorem of algebra and the factor theorem are used to prove the following n zeros theorem. These two theorems are illustrated with five polynomials below. a. The first-degree polynomial ƒ1x2 5 x 1 3 has exactly one zero: x 5 23. b. The second-degree polynomial ƒ1x2 5 x2 1 10x 1 25 5 1x 1 52 1x 1 52 has exactly two zeros: x 5 25 and x 5 25. It is customary to write this as a single zero of multiplicity 2 or refer to it as a repeated root. x y (0, –4) (2, 0) (–2, 0) (0, 4) 8 8 THE FUNDAMENTAL THEOREM OF ALGEBRA Every polynomial P 1x2 of degree n . 0 has at least one zero in the complex number system. n ZEROS THEOREM Every polynomial P 1x2 of degree n . 0 can be expressed as the product of n linear factors. Hence, P 1x2 has exactly n zeros, not necessarily distinct. 4.5.1 S KILL Factor a polynomial function given certain zeros. 4.5.1 C ON CEPTUAL Understand why complex zeros occur in conjugate pairs. [ [ STUDY TIP The zeros of a polynomial can be complex numbers. Only when the zeros are real numbers do we interpret zeros as x-intercepts. STUDY TIP The largest number of zeros a polynomial can have is equal to the degree of the polynomial. [CONCEPT CHECK] TRUE OR FALSE Complex zeros correspond to x-intercepts. ANSWER False ▼ 4.5 Complex Zeros: The Fundamental Theorem of Algebra 389 c. The third-degree polynomial ƒ1x2 5 x3 1 16x 5 x 1x2 1 162 5 x 1x 1 4i2 1x 2 4i2 has exactly three zeros: x 5 0, x 5 24i, and x 5 4i. d. The fourth-degree polynomial ƒ1x2 5 x4 2 1 5 1x2 2 12 1x2 1 12 5 1x 2 12 1x 1 12 1x 2 i2 1x 1 i2 has exactly four zeros: x 5 1, x 5 21, x 5 i, and x 5 2i. e. The fifth-degree polynomial ƒ1x2 5 x5 5 x⋅x⋅x⋅x⋅x has exactly five zeros: x 5 0, which has multiplicity 5. The fundamental theorem of algebra and the n zeros theorem only tell you that the zeros exist—not how to find them. We must rely on techniques discussed in Section 4.4 and additional strategies discussed in this section to determine the zeros. Complex Conjugate Pairs Often, at a grocery store or a drugstore, we see signs for special offers—“buy one, get one free.” A similar phenomenon occurs for complex zeros of a polynomial function with real coefficients. If we restrict the coefficients of a polynomial to real numbers, complex zeros always come in conjugate pairs. In other words, if a zero of a polynomial function is a complex number, then another zero will always be its complex conjugate. Look at the third-degree polynomial in the above illustration, part (c), where two of the zeros were 24i and 4i, and in part (d), where two of the zeros were i and 2i. In general, if we restrict the coefficients of a polynomial to real numbers, complex zeros always come in conjugate pairs. EXAMPLE 1 Zeros That Appear as Complex Conjugates Find the zeros of the polynomial P 1x2 5 x2 2 4x 1 13. Solution: Set the polynomial equal to zero. P 1x2 5 x2 2 4x 1 13 5 0 Use the quadratic formula to solve for x. x 5 21242 6 "12422 2 41121132 2112 Simplify. The zeros are the complex conjugates 2 2 3i and 2 1 3i. Check: This is a second-degree polynomial, so we expect two zeros. x 5 2 6 3i EXAMPLE 2 Finding a Polynomial Given Its Zeros Find a polynomial of minimum degree that has the zeros: 22, 1 2 i, 1 1 i. Solution: Write the factors corresponding to each zero: 22: 1x 1 22 1 1 i: 3x 2 11 1 i24 1 2 i: 3x 2 11 2 i24 Express the polynomial as the product of the three factors. P1x2 5 1x 1 223x 2 11 1 i243x 2 11 2 i24 Regroup inner parentheses. P1x2 5 1x 1 2231x 2 12 2 i431x 2 12 1 i4 Use the difference of two squares formula 1a 2 b2 1a 1 b2 5 a2 2 b2 for the P1x2 5 1x 1 2231x 2 12 2 i431x 2 12 1 i4 product of the latter two factors. 1x 2 1222i2 g STUDY TIP If we restrict the coefficients of a polynomial to real numbers, complex zeros always come in conjugate pairs. COMPLEX CONJUGATE ZEROS THEOREM If a polynomial P 1x2 has real coefficients, and if a 1 bi is a zero of P 1x2, then its complex conjugate a 2 bi is also a zero of P 1x2. 390 CHAPTER 4 Polynomial and Rational Functions Simplify. P1x2 5 1x 1 2231x 2 122 2 i24 x2 2 2x + 1 11 P1x2 5 1x 1 221x2 2 2x 1 22 P1x2 5 x3 2 2x2 1 2x 1 2x2 2 4x 1 4 P1x2 5 x3 2 2x 1 4 Y OUR TU R N Find a polynomial of minimum degree that has the zeros: 1, 2 2 i, 2 1 i. d d ▼ A N S W E R P1x2 5 x3 2 5x2 1 9x 2 5 ▼ ▼ A N S W E R P 1x2 5 1x 2 2i2 1x 1 2i2 1x 2 12 1x 2 22 Note: The zeros of P 1x2 are 1, 2, 2i, and 22i. EXAMPLE 3 Factoring a Polynomial with Complex Zeros Factor the polynomial P 1x2 5 x4 2 x3 2 5x2 2 x 2 6 given that i is a zero of P 1x2. Since P 1x2 is a fourth-degree polynomial, we expect four zeros. The goal in this problem is to write P 1x2 as a product of four linear factors: P 1x2 5 1x 2 a2 1x 2 b2 1x 2 c2 1x 2 d2, where a, b, c, and d are complex numbers and represent the zeros of the polynomial. Solution: Write known zeros and linear factors. Since i is a zero, we know that 2i is a zero. x 5 i and x 5 2i We now know two linear factors of P 1x2. 1x 2 i2 and 1x 1 i2 Write P 1x2 as a product of four factors. P 1x2 5 1x 2 i2 1x 1 i2 1x 2 c2 1x 2 d2 Multiply the two known factors. 1x 1 i2 1x 2 i2 5 x2 2 i2 5 x2 2 1212 5 x2 1 1 Rewrite the polynomial. P 1x2 5 1x2 1 12 1x 2 c2 1x 2 d2 Divide both sides of the equation by x2 1 1. P1x2 x2 1 1 5 1x 2 c21x 2 d2 Divide P 1x2 by x2 1 1 using long division. x2 2 x 2 6 x2 1 0x11 qx4 2 x3 2 5x2 2 x 2 6 21x4 1 0x3 1 x22 2x3 2 6x2 2 x 212x3 1 0x2 2 x2 26x2 1 0x 2 6 2126x2 1 0x 2 62 0 Since the remainder is 0, x2 2 x 2 6 is a factor. P 1x2 5 1x2 1 12 1x2 2 x 2 62 Factor the quotient x2 2 x 2 6. x2 2 x 2 6 5 1x 2 32 1x 1 22 Write P 1x2 as a product of four linear factors. Check: P 1x2 is a fourth-degree polynomial and we found four zeros, two of which are complex conjugates. Y OUR TU R N Factor the polynomial P 1x2 5 x4 2 3x3 1 6x2 2 12x 1 8 given that x 2 2i is a factor. P 1x2 5 1x 2 i2 1x 1 i2 1x 2 32 1x 1 22 ▼ 4.5 Complex Zeros: The Fundamental Theorem of Algebra 391 Because an n-degree polynomial function has exactly n zeros and since complex zeros always come in conjugate pairs, if the degree of the polynomial is odd, there is guaranteed to be at least one zero that is a real number. If the degree of the polynomial is even, there is no guarantee that a zero will be real—all the zeros could be complex. EXAMPLE 4 Factoring a Polynomial with Complex Zeros Factor the polynomial P 1x2 5 x4 2 2x3 1 x2 1 2x 2 2 given that 1 1 i is a zero of P 1x2. Since P 1x2 is a fourth-degree polynomial, we expect four zeros. The goal in this problem is to write P 1x2 as a product of four linear factors: P 1x2 5 1x 2 a2 1x 2 b2 1x 2 c2 1x 2 d2, where a, b, c, and d are complex numbers and represent the zeros of the polynomial. Solution: STEP 1 Write known zeros and linear factors. Since 1 1 i is a zero, we know that 1 2 i is a zero. x 5 1 1 i and x 5 1 2 i We now know two linear factors of P 1x2. 3x 2 11 1 i24 and 3x 2 11 2 i24 STEP 2 Write P 1x2 as a product of four factors. P 1x2 5 3x 2 11 1 i24 3x 2 11 2 i24 1x 2 c2 1x 2 d 2 STEP 3 Multiply the first two terms. 3x 2 11 1 i24 3x 2 11 2 i24 First, group the real parts together in each bracket. 31x 2 12 2 i4 31x 2 12 1 i4 Use the special product 1x 2 122 2 i2 1a 2 b2 1a 1 b2 5 a2 2 b2, 1x2 2 2x 1 12 2 1212 where a is 1x 2 12 and b is i. x2 2 2x 1 2 STEP 4 Rewrite the polynomial. P 1x2 5 1x2 2 2x 1 22 1x 2 c2 1x 2 d 2 STEP 5 Divide both sides of the equation by x2 2 2x 1 2, and substitute in the original polynomial P 1x2 5 x4 2 2x3 1 x2 1 2x 2 2. STEP 6 Divide the left side of the equation using long division. STEP 7 Factor x2 2 1. 1x 2 12 1x 1 12 STEP 8 Write P 1x2 as a product of four linear factors. or YOUR TURN Factor the polynomial P 1x2 5 x4 2 2x2 1 16x 2 15 given that 1 1 2i is a zero. x4 2 2x3 1 x2 1 2x 2 2 x2 2 2x 1 2 5 1x 2 c2 1x 2 d 2 x4 2 2x3 1 x2 1 2x 2 2 x2 2 2x 1 2 5 x2 2 1 P 1x2 5 3x 2 11 1 i24 3x 2 11 2 i24 3x 2 14 3x 1 14 P 1x2 5 1x 2 1 2 i2 1x 2 1 1 i2 1x 2 12 1x 1 12 ▼ A N S W E R P 1x2 5 3x 2 11 1 2i24 # 3x 2 11 2 2i24 1x 2 12 1x 1 32 Note: The zeros of P 1x2 are 1, 23, 1 1 2i, and 1 2 2i. ▼ [ [ STUDY TIP Odd-degree polynomials have at least one real zero. 392 CHAPTER 4 Polynomial and Rational Functions 4.5.2 Factoring Polynomials Now let’s draw on the tests discussed in this chapter to help us find all the zeros of a polynomial. Doing so will enable us to write polynomials as a product of linear factors. Before reading Example 6, reread Section 4.4, Example 8. 4.5.2 S KILL Factor a polynomial function of degree n into n linear factors. 4.5.2 C ON CEPTUAL Understand why an odd-degree polynomial must have at least one real zero. EXAMPLE 5 Finding Possible Combinations of Real and Complex Zeros List the possible combinations of real and complex zeros for the given polynomials. a. 17x5 1 2x4 2 3x3 1 x2 2 5 b. 5x4 1 2x3 2 x 1 2 Solution: a. Since this is a fifth-degree polynomial, there are five zeros. Because complex zeros come in conjugate pairs, the table describes the possible five zeros. Applying Descartes’ rule of signs, we find that there are 3 or 1 positive real zeros and 2 or 0 negative real zeros. b. Because this is a fourth-degree polynomial, there are four zeros. Since complex zeros come in conjugate pairs, the table describes the possible four zeros. Applying Descartes’ rule of signs, we find that there are 2 or 0 positive real zeros and 2 or 0 negative real zeros. Y OUR TU R N List the possible combinations of real and complex zeros for P 1x2 5 x6 2 7x5 1 8x3 2 2x 1 1 REAL ZEROS COMPLEX ZEROS 1 4 3 2 5 0 POSITIVE REAL ZEROS NEGATIVE REAL ZEROS COMPLEX ZEROS 1 0 4 3 0 2 1 2 2 3 2 0 REAL ZEROS COMPLEX ZEROS 0 4 2 2 4 0 ▼ A N S W E R REAL ZEROS COMPLEX ZEROS 0 6 2 4 4 2 6 0 POSITIVE REAL ZEROS NEGATIVE REAL ZEROS COMPLEX ZEROS 0 0 4 2 0 2 0 2 2 2 2 0 ▼ 4.5 Complex Zeros: The Fundamental Theorem of Algebra 393 [CONCEPT CHECK] A polynomial of the form P(x) 5 (x 2 1 a2)(x 2 2 b2) where a and b are real numbers has how many real and how many imaginary zeros? ANSWER Two real ( 6 b) and two imaginary ( 6 ai) ▼ STUDY TIP From Step 2 we know there is one positive real zero, so test the positive possible rational zeros first in Step 4. EXAMPLE 6 Factoring a Polynomial Factor the polynomial P 1x2 5 x5 1 2x4 2 x 2 2. Solution: STEP 1 Determine variations in sign. P 1x2 has 1 sign change. P 1x2 5 x5 1 2x4 2 x 2 2 P 12x2 has 2 sign changes. P 12x2 5 2x5 1 2x4 1 x 2 2 STEP 2 Apply Descartes’ rule of signs and summarize the results in a table. POSITIVE REAL ZEROS NEGATIVE REAL ZEROS COMPLEX ZEROS 1 2 2 1 0 4 STEP 3 the rational zero theorem to determine the possible rational zeros. 61, 62 STEP 4 Test possible rational zeros. 1 is a zero: 21 is a zero: 22 is a zero: STEP 5 Write P 1x2 as a product of linear factors. 1 1 2 0 0 21 22 1 3 3 3 2 1 3 3 3 2 0 21 1 3 3 3 2 21 22 21 22 1 2 1 2 0 d 22 1 2 1 2 22 0 22 1 0 1 0 x2 1 1 5 1x 2 i2 1x 1 i2 P 1x2 5 1x 2 12 1x 1 12 1x 1 22 1x 2 i2 1x 1 i2 [ [ STUDY TIP P(x) is a fifth-degree polynomial, so we expect five zeros. n P 1x2 has at least one zero and no more than n zeros. n If a 1 bi is a zero, then a 2 bi is also a zero. n The polynomial can be written as a product of linear factors, not necessarily distinct. In this section, we discussed complex zeros of polynomial functions. A polynomial function, P 1x2, of degree n with real coefficients has the following properties: [SEC TION 4 .5] S U M M A RY 394 CHAPTER 4 Polynomial and Rational Functions In Exercises 1–8, find all zeros (real and complex). Factor the polynomial as a product of linear factors. 1. P 1x2 5 x2 1 4 2. P 1x2 5 x2 1 9 3. P 1x2 5 x2 2 2x 1 2 4. P 1x2 5 x2 2 4x 1 5 5. P 1x2 5 x4 2 16 6. P 1x2 5 x4 2 81 7. P 1x2 5 x4 2 25 8. P 1x2 5 x4 2 9 In Exercises 9–16, a polynomial function with real coefficients is described. Find all remaining zeros. 9. Degree: 3 Zeros: 21, i 10. Degree: 3 Zeros: 1, 2i 11. Degree: 4 Zeros: 2i, 3 2 i 12. Degree: 4 Zeros: 3i, 2 1 i 13. Degree: 6 Zeros: 2 1multiplicity 22, 1 2 3i, 2 1 5i 14. Degree: 6 Zeros: 22 1multiplicity 22, 1 2 5i, 2 1 3i 15. Degree: 6 Zeros: 2i, 1 2 i 1multiplicity 22 16. Degree: 6 Zeros: 2i, 1 1 i 1multiplicity 22 In Exercises 17–22, find a polynomial of minimum degree that has the given zeros. 17. 0, 1 2 2i, 1 1 2i 18. 0, 2 2 i, 2 1 i 19. 1, 1 2 5i, 1 1 5i 20. 2, 4 2 i, 4 1 i 21. 1 2 i, 1 1 i, 23i, 3i 22. 2i, i, 1 2 2i, 1 1 2i In Exercises 23–34, given a zero of the polynomial, determine all other zeros (real and complex), and write the polynomial in terms of a product of linear factors. Polynomial Zero Polynomial Zero 23. P 1x2 5 x4 2 2x3 2 11x2 2 8x 2 60 22i 24. P 1x2 5 x4 2 x3 1 7x2 2 9x 2 18 3i 25. P 1x2 5 x4 2 4x3 1 4x2 2 4x 1 3 i 26. P 1x2 5 x4 2 x3 1 2x2 2 4x 2 8 22i 27. P 1x2 5 x4 2 2x3 1 10x2 2 18x 1 9 23i 28. P 1x2 5 x4 2 3x3 1 21x2 2 75x 2 100 5i 29. P 1x2 5 x4 2 9x2 1 18x 2 14 1 1 i 30. P 1x2 5 x4 2 4x3 1 x2 1 6x 2 40 1 2 2i 31. P 1x2 5 x4 2 6x3 1 6x2 1 24x 2 40 3 2 i 32. P 1x2 5 x4 2 4x3 1 4x2 1 4x 2 5 2 1 i 33. P 1x2 5 x4 2 9x3 1 29x2 2 41x 1 20 2 2 i 34. P 1x2 5 x4 2 7x3 1 14x2 1 2x 2 20 3 1 i In Exercises 35–58, factor each polynomial as a product of linear factors. 35. P 1x2 5 x3 2 x2 1 9x 2 9 36. P 1x2 5 x3 2 2x2 1 4x 2 8 37. P 1x2 5 x3 2 5x2 1 x 2 5 38. P 1x2 5 x3 2 7x2 1 x 2 7 39. P 1x2 5 x3 1 x2 1 4x 1 4 40. P 1x2 5 x3 1 x2 2 2 41. P 1x2 5 x3 2 x2 2 18 42. P 1x2 5 x4 2 2x3 2 2x2 2 2x 2 3 43. P 1x2 5 x4 2 2x3 2 11x2 2 8x 2 60 44. P 1x2 5 x4 2 x3 1 7x2 2 9x 2 18 45. P 1x2 5 x4 2 4x3 2 x2 2 16x 2 20 46. P 1x2 5 x4 2 3x3 1 11x2 2 27x 1 18 47. P 1x2 5 x4 2 7x3 1 27x2 2 47x 1 26 47. P 1x2 5 x4 2 5x3 1 5x2 1 25x 2 26 49. P 1x2 5 2x4 2 3x3 1 x2 1 13x 1 10 50. P 1x2 5 2x4 2 x3 1 12x2 1 26x 1 24 51. P 1x2 5 x4 2 2x3 1 5x2 2 8x 1 4 52. P 1x2 5 x4 1 2x3 1 10x2 1 18x 1 9 53. P 1x2 5 x6 1 12x4 1 23x2 2 36 54. P 1x2 5 x6 2 2x5 1 9x4 2 16x3 1 24x2 2 32x 1 16 55. P 1x2 5 4x4 2 20x3 1 37x2 2 24x 1 5 56. P 1x2 5 4x4 2 44x3 1 145x2 2 114x 1 26 57. P 1x2 5 3x5 2 2x4 1 9x3 2 6x2 2 12x 1 8 58. P 1x2 5 2x5 2 5x4 1 4x3 2 26x2 1 50x 2 25 [SEC TION 4 .5] E X E R C I SE S • S K I L L S In Exercises 59–62, assume the profit model is given by a polynomial function P 1x2 where x is the number of units sold by the company per year. 59. Profit. If the profit function of a given company has all imaginary zeros and the leading coefficient is positive, would you invest in this company? Explain. 60. Profit. If the profit function of a given company has all imaginary zeros and the leading coefficient is negative, would you invest in this company? Explain. 61. Profit. If the profit function of a company is modeled by a third-degree polynomial with a negative leading coefficient and this polynomial has two complex conjugates as zeros and one positive real zero, would you invest in this company? Explain. • A P P L I C A T I O N S 4.5 Complex Zeros: The Fundamental Theorem of Algebra 395 • C A T C H T H E M I S T A K E In Exercises 67 and 68, explain the mistake that is made. 67. Given that 1 is a zero of P 1x2 5 x3 2 2x2 1 7x 2 6, find all other zeros. Solution: Step 1: P 1x2 is a third-degree polynomial, so we expect three zeros. Step 2: Because 1 is a zero, 21 is a zero, so two linear factors are 1x 2 12 and 1x 1 12. Step 3: Write the polynomial as a product of three linear factors. P 1x2 5 1x 2 12 1x 1 12 1x 2 c2 P 1x2 5 1x2 2 12 1x 2 c2 Step 4: To find the remaining linear factor, we divide P 1x2 by x2 2 1. x3 2 2x2 1 7x 2 6 x2 2 1 5 x 2 2 1 6x 2 8 x2 2 1 Which has a nonzero remainder? What went wrong? 68. Factor the polynomial P 1x2 5 2x3 1 x2 1 2x 1 1. Solution: Step 1: Since P 1x2 is an odd-degree polynomial, we are guaranteed one real zero (since complex zeros come in conjugate pairs). Step 2: Apply the rational zero test to develop a list of potential rational zeros. Possible zeros: 61 Step 3: Test possible zeros. 1 is not a zero: P 1x2 5 21123 1 1122 1 2112 1 1 5 6 21 is not a zero: P 1x2 5 212123 1 12122 1 21212 1 1 5 22 Note: 21 2 is the real zero. Why did we not find it? 62. Profit. If the profit function of a company is modeled by a third-degree polynomial with a positive leading coefficient and this polynomial has two complex conjugates as zeros and one positive real zero, would you invest in this company? Explain. For Exercises 63 and 64, refer to the following: The following graph models the profit P of a company where t is months and t $ 0. 1 2 3 4 6 5 15 12 9 6 3 t (months) P (proft in millions of dollars) 63. Business. If the profit function pictured is a third-degree polynomial, how many real and how many complex zeros does the function have? Discuss the implications of these zeros. 64. Business. If the profit function pictured is a fourth-degree polynomial with a negative leading coefficient, how many real and how many complex zeros does the function have? Discuss the implications of these zeros. For Exercises 65 and 66, refer to the following: The following graph models the concentration, C 1in mg/mL2 of a drug in the bloodstream; t is time in hours after the drug is admin-istered where t $ 0. 1 2 3 4 8 7 6 5 50 40 30 20 10 45 35 25 15 5 t (hours) C (mcg/mL) 65. Health/Medicine. If the concentration function pictured is a third-degree polynomial, how many real and how many complex zeros does the function have? Discuss the implications of these zeros. 66. Health/Medicine. If the concentration function pictured is a fourth-degree polynomial with a negative leading coefficient, how many real and how many complex zeros does the function have? Discuss the implications of these zeros. 396 CHAPTER 4 Polynomial and Rational Functions In Exercises 69–72, determine whether each statement is true or false. 69. If x 5 1 is a zero of a polynomial function, then x 5 21 is also a zero of the polynomial function. 70. All zeros of a polynomial function correspond to x-intercepts. 71. A polynomial function of degree n, n . 0 must have at least one zero. 72. A polynomial function of degree n, n . 0 can be written as a product of n linear factors. 73. Is it possible for an odd-degree polynomial to have all imaginary complex zeros? Explain. 74. Is it possible for an even-degree polynomial to have all imaginary zeros? Explain. • C O N C E P T U A L 4.6.1 Domain of Rational Functions So far in this chapter we have discussed polynomial functions. We now turn our attention to rational functions, which are ratios of polynomial functions. Ratios of integers are called rational numbers. Similarly, ratios of polynomial functions are called rational functions. DEFINITION Rational Function A function ƒ1x2 is a rational function if ƒ1x2 5 n1x2 d1x2 d1x2 2 0 where the numerator, n 1x2, and the denominator, d 1x2, are polynomial functions. The domain of ƒ1x2 is the set of all real numbers x such that d 1x2 Þ 0. Note: If d 1x2 is a constant, then ƒ1x2 is a polynomial function. S K I L L S O B J E C T I V E S ■ ■Find the domain of a rational function. ■ ■Determine vertical, horizontal, and slant asymptotes of rational functions. ■ ■Graph rational functions. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that the domain of a rational function is the set of all real numbers except those that correspond to the denominator being equal to zero. ■ ■Understand that the graph of a rational function can have either a horizontal asymptote or a slant asymptote but not both. ■ ■Any real number excluded from the domain of a rational function corresponds to either a vertical asymptote or a hole in the graph. 4.6 RATIONAL FUNCTIONS 4.6.1 S KILL Find the domain of a rational function. 4.6.1 C ON CEPTUAL Understand that the domain of a rational function is the set of all real numbers except those that correspond to the denominator being equal to zero. • T E C H N O L O G Y For Exercises 77 and 78, determine possible combinations of real and complex zeros. Plot P 1x2 and identify any real zeros with a graphing calculator or software. Does this agree with your list? 77. P 1x2 5 x4 1 13x2 1 36 78. P 1x2 5 x6 1 2x4 1 7x2 2 130x 2 288 For Exercises 79 and 80, find all zeros (real and complex). Factor the polynomial as a product of linear factors. 79. P 1x2 5 25x5 1 3x4 2 25x3 1 15x2 2 20x 1 12 80. P 1x2 5 x5 1 2.1x4 2 5x3 2 5.592x2 1 9.792x 2 3.456 • C H A L L E N G E In Exercises 75 and 76, assume a and b are nonzero real numbers. 75. Find a polynomial function that has degree 6 and for which bi is a zero of multiplicity 3. 76. Find a polynomial function that has degree 4 and for which a 1 bi is a zero of multiplicity 2. 4.6 Rational Functions 397 The domain of any polynomial function is the set of all real numbers. When we divide two polynomial functions, the result is a rational function, and we must exclude any values of x that make the denominator equal to zero. EXAMPLE 1 Finding the Domain of a Rational Function Find the domain of the rational function ƒ1x2 5 x 1 1 x2 2 x 2 6. Express the domain in interval notation. Solution: Set the denominator equal to zero. x2 2 x 2 6 5 0 Factor. 1x 1 22 1x 2 32 5 0 Solve for x. x 5 22 or x 5 3 Eliminate these values from the domain. x 2 22 or x 2 3 State the domain in interval notation. YOUR T UR N Find the domain of the rational function ƒ1x2 5 x 2 2 x2 2 3x 2 4. Express the domain in interval notation. 12q, 222 ∪ 122, 32 ∪ 13, q2 ▼ A N S W E R The domain is the set of all real numbers such that x 2 21 or x 2 4. Interval notation: 12q, 212 ∪ 121, 42 ∪ 14, q2 ▼ It is important to note that there are not always restrictions on the domain. For example, if the denominator is never equal to zero, the domain is the set of all real numbers. EXAMPLE 2 When the Domain of a Rational Function Is the Set of All Real Numbers Find the domain of the rational function g1x2 5 3x x2 1 9. Express the domain in interval notation. Solution: Set the denominator equal to zero. x2 1 9 5 0 Subtract 9 from both sides. x2 5 29 Solve for x. x 5 23i or x 5 3i There are no real solutions; therefore, the domain has no restrictions. R, the set of all real numbers State the domain in interval notation. YOUR T UR N Find the domain of the rational function g1x2 5 5x x2 1 4. Express the domain in interval notation. 12q, q2 ▼ A N S W E R The domain is the set of all real numbers. Interval notation: 12q, q2 ▼ It is important to note that ƒ1x2 5 x2 2 4 x 1 2 , where x 2 22 and g1x2 5 x 2 2 are not the same function. Although ƒ1x2 can be written in the factored form ƒ1x2 5 1x 2 221x 1 22 x 1 2 5 x 2 2, its domain is different. The domain of g 1x2 is the set of all real numbers, whereas the domain of ƒ1x2 is the set of all real numbers such that x 2 22. If we were to plot ƒ1x2 and g 1x2, they would both look like the line y 5 x 2 2. However, ƒ1x2 would have a hole, or discontinuity, at the point x 5 22. x y 5 5 –5 –5 hole 398 CHAPTER 4 Polynomial and Rational Functions 4.6.2 Vertical, Horizontal, and Slant Asymptotes If a function is not defined at a point, then it is still useful to know how the function behaves near that point. Let’s start with a simple rational function, the reciprocal function ƒ1x2 5 1 x. This function is defined everywhere except at x 5 0. We cannot let x 5 0 because that point is not in the domain of the function. We should, however, ask the question, “how does ƒ1x2 behave as x approaches zero?” Let us take values that get closer and closer to x 5 0, such as 1 10, 1 100, 1 1000, . . . (See the table above.) We use an arrow to represent the word approach, a positive superscript to represent from the right, and a negative superscript to represent from the left. A plot of this function can be generated using point-plotting techniques. The following are observations of the graph ƒ1x2 5 1 x. WORDS MATH As x approaches zero from the right, x S 01 the function ƒ 1x2 increases without bound. 1 x S As x approaches zero from the left, the x S 02 function ƒ 1x2 decreases without bound. 1 x S 2 As x approaches infinity (increases x S without bound), the function ƒ 1x2 approaches zero from above. 1 x S 01 As x approaches negative infinity x S 2 (decreases without bound), the function ƒ 1x2 approaches zero from below. 1 x S 02 The symbol q does not represent an actual real number. This symbol represents growing without bound. 1. Notice that the function is not defined at x 5 0. The y-axis, or the vertical line x 5 0, represents the vertical asymptote. 2. Notice that the value of the function is never equal to zero. The x-axis is never touched by the function. The x-axis, or y 5 0, is a horizontal asymptote. Asymptotes are lines that the graph of a function approaches. Suppose a football team’s defense is its own 8-yard line and the team gets an “offsides” penalty that results in loss of “half the distance to the goal.” Then the offense would get the ball on the 4-yard line. Suppose the defense gets another penalty on the next play that results in “half the distance to the goal.” The offense would then get the ball on the 2-yard line. If the defense received 10 more penalties, all resulting in “half the distance to the goal,” would the referees give the offense a touchdown? No, because although the offense may appear x 2 1 10 2 1 100 2 1 1000 0 1 1000 1 100 1 10 210 2100 21000 undefined 1000 100 10 f (x) 5 1 – x x 210 2 1 10 21 21 1 1 10 1 10 g g x approaching 0 from the left x approaching 0 from the right 4.6.2 S KILL Determine vertical, horizontal, and slant asymptotes of rational functions. 4.6.2 C ON CEPTUAL Understand that the graph of a rational function can have either a horizontal asymptote or a slant asymptote but not both. x y (–1, –1) (1, 1) x 1 f (x) = x y (–1, –1) (1, 1) x 1 f(x) = [CONCEPT CHECK] TRUE OR FALSE The domain of f 1x2 5 1 x2 1 a2 and the domain of g1x2 5 1 x2 1 b2 are equal. ANSWER True ▼ f (x) 5 1 – x 4.6 Rational Functions 399 to be snapping the ball from the goal line, technically it has not actually reached the goal line. Asymptotes utilize the same concept. We will start with vertical asymptotes. Although the function ƒ1x2 5 1 x had one vertical asymptote, in general, rational functions can have none, one, or several vertical asymptotes. We will first formally define a vertical asymptote and then discuss how to find it. Vertical asymptotes assist us in graphing rational functions since they essentially “steer” the function in the vertical direction. How do we locate the vertical asymptotes of a rational function? Set the denominator equal to zero. If the numerator and denominator have no common factors, then any numbers that are excluded from the domain of a rational function locate vertical asymptotes. A rational function ƒ1x2 5 n1x2 d1x2 is said to be in lowest terms if the numerator n 1x2 and denominator d 1x2 have no common factors. Let ƒ1x2 5 n1x2 d1x2 be a rational function in lowest terms; then any zeros of the numerator n 1x2 correspond to x-intercepts of the graph of ƒ, and any zeros of the denominator d 1x2 correspond to vertical asymptotes of the graph of ƒ. If a rational function does have a common factor (is not in lowest terms), then the common factor(s) should be canceled, resulting in an equivalent rational function R 1x2 in lowest terms. If 1x 2 a2p is a factor of the numerator and 1x 2 a2q is a factor of the denominator, then there is a hole in the graph at x 5 a provided p $ q and x 5 a is a vertical asymptote if p , q. DEFINITION Vertical Asymptotes The line x 5 a is a vertical asymptote for the graph of a function if ƒ1x2 either increases or decreases without bound as x approaches a from either the left or the right. x y x y x y x y x = a f (x) x = a f (x) x = a f (x) x = a f (x) LOCATING VERTICAL ASYMPTOTES Let ƒ1x2 5 n1x2 d1x2 be a rational function in lowest terms (that is, assume n 1x2 and d 1x2 are polynomials with no common factors); then the graph of ƒ has a vertical asymptote at any real zero of the denominator d 1x2. That is, if d 1a2 5 0, then x 5 a corresponds to a vertical asymptote on the graph of ƒ. Note: If ƒ is a rational function that is not in lowest terms, then divide out the common factors, resulting in a rational function R that is in lowest terms. Any common factor x 2 a of the function ƒ corresponds to a hole in the graph of ƒ at x 5 a provided the multiplicity of a in the numerator is greater than or equal to the multiplicity of a in the denominator. STUDY TIP The vertical asymptotes of a rational function in lowest terms occur at x-values that make the denominator equal to zero. 400 CHAPTER 4 Polynomial and Rational Functions We now turn our attention to horizontal asymptotes. As we have seen, rational functions can have several vertical asymptotes. However, rational functions can have at most one horizontal asymptote. Horizontal asymptotes imply that a function approaches a constant value as x becomes large in the positive or negative direction. Another difference between vertical and horizontal asymptotes is that the graph of a function never touches a vertical asymptote but, as you will see in the next box, the graph of a function may cross a horizontal asymptote, just not at the “ends” 1x S 6q2. EXAMPLE 3 Determining Vertical Asymptotes Locate any vertical asymptotes of the rational function ƒ1x2 5 5x 1 2 6x2 2 x 2 2. Solution: Factor the denominator. ƒ1x2 5 5x 1 2 12x 1 1213x 2 22 The numerator and denominator have no common factors. Set the denominator equal to zero. 2x 1 1 5 0 and 3x 2 2 5 0 Solve for x. x 5 21 2 and x 5 2 3 The vertical asymptotes are and . Y OUR TU R N Locate any vertical asymptotes of the following rational function: ƒ1x2 5 3x 2 1 2x2 2 x 2 15 x 5 21 2 x 5 2 3 ▼ A N S W E R x 5 2 5 2 and x 5 3 ▼ EXAMPLE 4 Determining Vertical Asymptotes When the Rational Function Is Not in Lowest Terms Locate any vertical asymptotes of the rational function ƒ1x2 5 x 1 2 x3 2 3x2 2 10x. Solution: Factor the denominator. x3 2 3x2 2 10x 5 x 1x2 2 3x 2 102 5 x 1x 2 52 1x 1 22 Write the rational function in factored form. ƒ1x2 5 1x 1 22 x1x 2 521x 1 22 Cancel (divide out) the common factor 1x 1 22. R1x2 5 1 x1x 2 52 x 2 22 Find the values when the denominator of R is equal to zero. x 5 0 and x 5 5 The vertical asymptotes are and . Note: x 5 22 is not in the domain of ƒ1x2, even though there is no vertical asymptote there. There is a “hole” in the graph at x 5 22. Graphing calculators do not always show such “holes.” Y OUR TU R N Locate any vertical asymptotes of the following rational function: ƒ1x2 5 x2 2 4x x2 2 7x 1 12 x 5 0 x 5 5 ▼ ▼ A N S W E R x 5 3 4.6 Rational Functions 401 How do we determine whether a horizontal asymptote exists? And, if it does, how do we locate it? We investigate the value of the rational function as x S q or as x S 2q. One of two things will happen: either the rational function will increase or decrease without bound or the rational function will approach a constant value. We say that a rational function is proper if the degree of the numerator is less than the degree of the denominator. Proper rational functions, like ƒ1x2 5 1 x, approach zero as x gets large. Therefore, all proper rational functions have the specific horizontal asymptote, y 5 0 (see Example 5a). We say that a rational function is improper if the degree of the numerator is greater than or equal to the degree of the denominator. In this case, we can divide the numerator by the denominator and determine how the quotient behaves as x increases without bound. ■ ■If the quotient is a constant (resulting when the degrees of the numerator and denominator are equal), then as x S q or as x S 2q, the rational function approaches the constant quotient (see Example 5b). ■ ■If the quotient is a polynomial function of degree 1 or higher, then the quotient depends on x and does not approach a constant value as x increases (see Example 5c). In this case, we say that there is no horizontal asymptote. We find horizontal asymptotes by comparing the degree of the numerator and the degree of the denominator. There are three cases to consider: 1. The degree of the numerator is less than the degree of the denominator. 2. The degree of the numerator is equal to the degree of the denominator. 3. The degree of the numerator is greater than the degree of the denominator. DEFINITION Horizontal Asymptote The line y 5 b is a horizontal asymptote of the graph of a function if ƒ1x2 approaches b as x increases or decreases without bound. The following are three examples: As x S q, ƒ1x2 S b x y y = b f (x) x y y = b f (x) x y y = b f (x) Note: A horizontal asymptote steers a function as x gets large. Therefore, when x is not large, the function may cross the asymptote. LOCATING HORIZONTAL ASYMPTOTES Let ƒ be a rational function given by ƒ1x2 5 n1x2 d1x2 5 anx n 1 an21x n21 1 c1 a1x 1 a0 bmx m 1 bm21x m21 1 c1 b1x 1 b0 where n 1x2 and d 1x2 are polynomials. 1. When n , m, the x-axis 1y 5 02 is the horizontal asymptote. 2. When n 5 m, the line y 5 an bm (ratio of leading coefficients) is the horizontal asymptote. 3. When n . m, there is no horizontal asymptote. 402 CHAPTER 4 Polynomial and Rational Functions In other words, 1. When the degree of the numerator is less than the degree of the denominator, then y 5 0 is the horizontal asymptote. 2. When the degree of the numerator is the same as the degree of the denominator, then the horizontal asymptote is the ratio of the leading coefficients. 3. If the degree of the numerator is greater than the degree of the denominator, then there is no horizontal asymptote. Thus far we have discussed linear asymptotes: vertical and horizontal. There are three types of lines: horizontal (slope is zero), vertical (slope is undefined), and slant (nonzero slope). Similarly, there are three types of linear asymptotes: horizontal, vertical, and slant. STUDY TIP There are three types of linear asymptotes: horizontal, vertical, and slant. EXAMPLE 5 Finding Horizontal Asymptotes Determine whether a horizontal asymptote exists for the graph of each of the given rational functions. If it does, locate the horizontal asymptote. a. ƒ1x2 5 8x 1 3 4x2 1 1 b. g1x2 5 8x2 1 3 4x2 1 1 c. h1x2 5 8x3 1 3 4x2 1 1 Solution (a): The degree of the numerator 8x 1 3 is 1. n 5 1 The degree of the denominator 4x2 1 1 is 2. m 5 2 The degree of the numerator is less than the degree of the denominator. n , m The x-axis is the horizontal asymptote for the graph of ƒ1x2. y 5 0 The line is the horizontal asymptote for the graph of ƒ1x2. Solution (b): The degree of the numerator 8x2 1 3 is 2. n 5 2 The degree of the denominator 4x2 1 1 is 2. m 5 2 The degree of the numerator is equal to the degree of the denominator. n 5 m The ratio of the leading coefficients is the horizontal asymptote for the graph of g 1x2. y 5 8 4 5 2 The line is the horizontal asymptote for the graph of g 1x2. If we divide the numerator by the denominator, the resulting quotient is the constant 2. g1x2 5 8x2 1 3 4x2 1 1 5 2 1 1 4x2 1 1 Solution (c): The degree of the numerator 8x3 1 3 is 3. n 5 3 The degree of the denominator 4x2 1 1 is 2. m 5 2 The degree of the numerator is greater than the degree of the denominator. n . m The graph of the rational function h 1x2 has . If we divide the numerator by the denominator, the resulting quotient is a linear function. h1x2 5 8x3 1 3 4x2 1 1 5 2x 1 22x 1 3 4x2 1 1 Y OUR TU R N Find the horizontal asymptote (if one exists) for the graph of the rational function ƒ1x2 5 7x3 1 x 2 2 24x3 1 1 . y 5 0 y 5 2 no horizontal asymptote ▼ ▼ A N S W E R y 5 27 4 is the horizontal asymptote. 4.6 Rational Functions 403 Recall that when dividing polynomials the degree of the quotient is always the difference between the degree of the numerator and the degree of the denominator. For example, a cubic (third-degree) polynomial divided by a quadratic (second-degree) polynomial results in a linear (first-degree) polynomial. A fifth-degree polynomial divided by a fourth-degree polynomial results in a first-degree (linear) polynomial. When the degree of the numerator is exactly one more than the degree of the denomi-nator, the quotient is linear and represents a slant asymptote. d EXAMPLE 6 Finding Slant Asymptotes Determine the slant asymptote of the rational function ƒ1x2 5 4x3 1 x2 1 3 x2 2 x 1 1 . Solution: Divide the numerator by the denominator with long division. 4x 1 5 x2 2 x 1 1q4x3 1 x2 1 0x 1 3 214x3 2 4x2 1 4x2 5x2 2 4x 1 3 215x2 2 5x 1 52 x 2 2 Note that as x S 6q, the rational ƒ1x2 5 4x 1 5 1 x 2 2 x2 2 x 1 1 expression approaches 0. S 0 as x S 6q The quotient is the slant asymptote. YOUR T UR N Find the slant asymptote of the rational function ƒ1x2 5 x2 1 3x 1 2 x 2 2 . y 5 4x 1 5 ▼ [CONCEPT CHECK] TRUE OR FALSE You can have both a horizontal and a slant asymptote. ANSWER False ▼ SLANT ASYMPTOTES Let ƒ be a rational function given by ƒ1x2 5 n1x2 d1x2, where n 1x2 and d 1x2 are polynomials and the degree of n 1x2 is one more than the degree of d 1x2. On dividing n 1x2 by d 1x2, the rational function can be expressed as ƒ1x2 5 mx 1 b 1 r1x2 d1x2 where the degree of the remainder r 1x2 is less than the degree of d 1x2 and the line y 5 mx 1 b is a slant asymptote for the graph of ƒ. Note that as x S 2q or x S q, ƒ1x2 S mx 1 b. ▼ A N S W E R y 5 x 1 5 404 CHAPTER 4 Polynomial and Rational Functions 4.6.3 Graphing Rational Functions We can now graph rational functions using asymptotes as graphing aids. The following box summarizes the five-step procedure for graphing rational functions. Graphing Rational Functions Let ƒ be a rational function given by ƒ1x2 5 n1x2 d1x2. Step 1: Find the domain of the rational function ƒ. Step 2: Find the intercept(s). n y-intercept: evaluate ƒ102. n x-intercept: solve the equation n 1x2 5 0 for x in the domain of ƒ. Step 3: Find any holes. n Factor the numerator and denominator. n Divide out common factors. n A common factor x 2 a corresponds to a hole in the graph of ƒ at x 5 a if the multiplicity of a in the numerator is greater than or equal to the multiplicity of a in the denominator. n The result is an equivalent rational function R1x2 5 p1x2 q1x2 in lowest terms. Step 4: Find any asymptotes. n Vertical asymptotes: solve q 1x2 5 0. n Compare the degree of the numerator and the degree of the denominator to determine whether either a horizontal or slant asymptote exists. If one exists, find it. Step 5: Find additional points on the graph of ƒ—particularly near asymptotes. Step 6: Sketch the graph; draw the asymptotes, label the intercept(s) and additional points, and complete the graph with a smooth curve between and beyond the vertical asymptotes. STUDY TIP Common factors need to be divided out first; then the remaining x-values corresponding to a denominator value of 0 are vertical asymptotes. 4.6.3 S KILL Graph rational functions. 4.6.3 C ON CEPTUAL Any real number excluded from the domain of a rational function corresponds to either a vertical asymptote or a hole in the graph. It is important to note that any real number eliminated from the domain of a rational function corresponds to either a vertical asymptote or a hole on its graph. STUDY TIP Any real number excluded from the domain of a rational function corresponds to either a vertical asymptote or a hole on its graph. EXAMPLE 7 Graphing a Rational Function Graph the rational function ƒ1x2 5 x x2 2 4. Solution: STEP 1 Find the domain. Set the denominator equal to zero. x2 2 4 5 0 Solve for x. x 5 62 State the domain. 12q, 222 ∪ 122, 22 ∪ 12, q2 STEP 2 Find the intercepts. y-intercept: ƒ102 5 0 24 5 0 y 5 0 x-intercepts: ƒ1x2 5 x x2 2 4 5 0 x 5 0 The only intercept is at the point . 10, 02 4.6 Rational Functions 405 STEP 3 Find any holes. ƒ1x2 5 x 1x 1 221x 2 22 There are no common factors, so ƒ is in lowest terms. Since there are no common factors, there are no holes on the graph of ƒ. STEP 4 Find any asymptotes. Vertical asymptotes: d 1x2 5 1x 1 22 1x 2 22 5 0 and Horizontal asymptote: Degree of numerator 5 1 Degree of denominator 5 2 Degree of numerator , Degree of denominator STEP 5 Find additional points on the graph. STEP 6 Sketch the graph; label the intercepts, asymptotes, and additional points and complete with a smooth curve approaching the asymptotes. Y OUR T UR N Graph the rational function ƒ1x2 5 x x2 2 1. x 5 22 x 5 2 y 5 0 x 23 21 1 3 f (x) 23 5 1 3 21 3 23 5 x y 5 5 –5 –5 ▼ A N S W E R x y –5 5 5 –5 ▼ EXAMPLE 8 Graphing a Rational Function with No Horizontal or Slant Asymptotes State the asymptotes (if there are any) and graph the rational function ƒ1x2 5 x4 2 x3 2 6x2 x2 2 1 . Solution: STEP 1 Find the domain. Set the denominator equal to zero. x2 2 1 5 0 Solve for x. x 5 61 State the domain. 12q, 212 ∪ 121, 12 ∪ 11, q2 STEP 2 Find the intercepts. y-intercept: ƒ102 5 0 21 5 0 x-intercepts: n 1x2 5 x4 2 x3 2 6x2 5 0 Factor. x 2 1x 2 32 1x 1 22 5 0 Solve. x 5 0, x 5 3, and x 5 22 The intercepts are the points 10, 02 , 13, 02 , and 122, 02 . 406 CHAPTER 4 Polynomial and Rational Functions STEP 3 Find any holes. ƒ1x2 5 x21x 2 321x 1 22 1x 2 121x 1 12 There are no common factors, so ƒ is in lowest terms. Since there are no common factors, there are no holes on the graph of ƒ. STEP 4 Find the asymptotes. Vertical asymptote: d 1x2 5 x2 2 1 5 0 Factor. 1x 1 12 1 x 2 12 5 0 Solve. x 5 21 and x 5 1 No horizontal asymptote: degree of n 1x2 . degree of d 1x2 34 . 24 No slant asymptote: degree of n 1x2 2 degree of d 1x2 . 1 34 2 2 5 2 . 14 The asymptotes are x 5 21 and x 5 1. STEP 5 Find additional points on the graph. STEP 6 Sketch the graph; label the intercepts and asymptotes, and complete with a smooth curve between and beyond the vertical asymptote. Y OUR TU R N State the asymptotes (if there are any) and graph the rational function ƒ1x2 5 x3 2 2x2 2 3x x 1 2 . x 23 20.5 0.5 2 4 f (x) 6.75 1.75 2.08 25.33 6.4 –5 5 –10 10 x y (3, 0) (–2, 0) x = 1 x = –1 ▼ A N S W E R Vertical asymptotes: x 5 22. No horizontal or slant asymptotes. –5 10 –25 50 x y (3, 0) (–1, 0) x = –2 ▼ EXAMPLE 9 Graphing a Rational Function with a Horizontal Asymptote State the asymptotes (if there are any) and graph the rational function ƒ1x2 5 4x3 1 10x2 2 6x 8 2 x3 Solution: STEP 1 Find the domain. Set the denominator equal to zero. 8 2 x3 5 0 Solve for x. x 5 2 State the domain. 12q, 22 ∪ 12, q2 4.6 Rational Functions 407 STEP 2 Find the intercepts. y-intercept: ƒ102 5 0 8 5 0 x-intercepts: n 1x2 5 4x3 1 10x2 2 6x 5 0 Factor. 2x 12x 2 12 1x 1 32 5 0 Solve. x 5 0, x 5 1 2, and x 5 23 The intercepts are the points 10, 02, A1 2, 0B, and 123, 02. STEP 3 Find the holes. ƒ1x2 5 2x12x 2 121x 1 32 12 2 x21x2 1 2x 1 42 There are no common factors, so ƒ is in lowest terms (no holes). STEP 4 Find the asymptotes. Vertical asymptote: d 1x2 5 8 2 x3 5 0 Solve. x 5 2 Horizontal asymptote: degree of n 1x2 5 degree of d 1x2 Use leading coefficients. y 5 4 21 5 24 The asymptotes are x 5 2 and y 5 24. STEP 5 Find additional points on the graph. STEP 6 Sketch the graph; label the intercepts and asymptotes and complete with a smooth curve. Y OUR T UR N Graph the rational function ƒ1x2 5 2x2 2 7x 1 6 x2 2 3x 2 4 . Give equations of the vertical and horizontal asymptotes and state the intercepts. x 24 21 1 4 1 3 f (x) 21 1.33 20.10 1.14 29.47 ▼ A N S W E R Vertical asymptotes: x 5 4, x 5 21 Horizontal asymptote: y 5 2 Intercepts: A0, 2 3 2B, A3 2, 0B, 12, 02 –4 6 x y –5 5 y = 2 x = –1 x = 4 ▼ –10 10 –10 10 x y (–3, 0) x = 2 y = –4 ( , 0) 1 2 408 CHAPTER 4 Polynomial and Rational Functions EXAMPLE 10 Graphing a Rational Function with a Slant Asymptote Graph the rational function ƒ1x2 5 x2 2 3x 2 4 x 1 2 . Solution: STEP 1 Find the domain. Set the denominator equal to zero. x 1 2 5 0 Solve for x. x 5 22 State the domain. 12q,222 ∪ 122, q2 STEP 2 Find the intercepts. y-intercept: ƒ102 5 24 2 5 22 x-intercepts: n 1x2 5 x2 2 3x 2 4 5 0 Factor. 1x 1 12 1x 2 42 5 0 Solve. x 5 21 and x 5 4 The intercepts are the points 10, 222, 121, 02, and 14, 02. STEP 3 Find any holes. ƒ1x2 5 1x 2 421x 1 12 1x 1 22 There are no common factors, so ƒ is in lowest terms. Since there are no common factors, there are no holes on the graph of ƒ. STEP 4 Find the asymptotes. Vertical asymptote: d 1x2 5 x 1 2 5 0 Solve. x 5 22 Slant asymptote: degree of n 1x2 2 degree of d 1x2 5 1 Divide n 1x2 by d 1x2. ƒ1x2 5 x2 2 3x 2 4 x 1 2 5 x 2 5 1 6 x 1 2 Write the equation of the asymptote. y 5 x 2 5 The asymptotes are x 5 22 and y 5 x 2 5. STEP 5 Find additional points on the graph. STEP 6 Sketch the graph; label the intercepts and asymptotes, and complete with a smooth curve between and beyond the vertical asymptote. Y OUR TU R N For the function ƒ1x2 5 x2 1 x 2 2 x 2 3 , state the asymptotes (if any exist) and graph the function. x 26 25 23 5 6 f (x) 212.5 212 214 0.86 1.75 –20 20 –20 20 x y (4, 0) (0, –2) (–1, 0) x = –2 y = x – 5 ▼ A N S W E R Horizontal asymptote: x 5 3 Slant asymptote: y 5 x 1 4 –10 10 –10 20 x y x = 3 y = x + 4 (–2, 0) (1, 0) (0, ) 2 3 ▼ 4.6 Rational Functions 409 ▼ A N S W E R x –7 3 y 5 –5 EXAMPLE 11 Graphing a Rational Function with a Hole in the Graph Graph the rational function ƒ1x2 5 x2 1 x 2 6 x2 2 x 2 2. Solution: STEP 1 Find the domain. Set the denominator equal to zero. x2 2 x 2 2 5 0 Solve for x. 1x 2 22 1x 1 12 5 0 x 5 21 or x 5 2 State the domain. 12q, 212 ∪ 121, 22 ∪ 12, q2 STEP 2 Find the intercepts. y-intercept: ƒ102 5 26 22 5 3 y 5 3 x-intercepts: n 1x2 5 x2 1 x 2 6 5 0 1x 1 32 1x 2 22 5 0 x 5 23 or x 5 2 The intercepts correspond to the points and . The point 12, 02 appears to be an x-intercept; however, x 5 2 is not in the domain of the function. STEP 3 Find any holes. ƒ1x2 5 1x 2 221x 1 32 1x 2 221x 1 12 Since x 2 2 is a common factor, there is a hole in the graph of ƒ at x 5 2. Dividing out the common factor generates R1x2 5 1x 1 32 1x 1 12 an equivalent rational function in lowest terms. STEP 4 Find the asymptotes. Vertical asymptotes: x 1 1 5 0 Horizontal asymptote: Since the degree of the numerator equals the degree of the denominator, use the leading coefficients. STEP 5 Find additional points on the graph. x 24 22 21 2 1 3 f (x) or R(x) 1 3 21 5 2 3 2 STEP 6 Sketch the graph; label the intercepts, asymptotes, and additional points and complete with a smooth curve approaching asymptotes. Recall the hole at x 5 2. Note that R122 5 5 3, so the open “hole” is located at the point 12, 5/32. YOUR T UR N Graph the rational function ƒ1x2 5 x2 2 x 2 2 x2 1 x 2 6. 10, 32 123, 02 Degree of numerator of ƒ 5 Degree of denominator of ƒ 5 2 and Degree of numerator of R 5 Degree of denominator of R 5 1 x 5 21 y 5 1 1 5 1 [CONCEPT CHECK] TRUE OR FALSE The only time a hole will exist on the graph of a rational function is when the numerator and the denominator have a common factor. ANSWER True ▼ x y 5 –5 –5 5 ▼ 410 CHAPTER 4 Polynomial and Rational Functions [SEC TION 4 .6] E X E R C I SE S • S K I L L S In Exercises 1–10, find the domain of each rational function 1. ƒ1x2 5 1 x 1 3 2. ƒ1x2 5 3 4 2 x 3. ƒ1x2 5 2x 1 1 13x 1 1212x 2 12 4. ƒ1x2 5 5 2 3x 12 2 3x21x 2 72 5. ƒ1x2 5 x 1 4 x2 1 x 2 12 6. ƒ1x2 5 x 2 1 x2 1 2x 2 3 7. ƒ1x2 5 7x x2 1 16 8. ƒ1x2 5 2 2x x2 1 9 9. ƒ1x2 5 231x2 1 x 2 22 21x2 2 x 2 62 10. ƒ1x2 5 51x2 2 2x 2 32 1x2 2 x 2 62 In Exercises 11–20, find all vertical asymptotes and horizontal asymptotes (if there are any). 11. ƒ1x2 5 1 x 1 2 12. ƒ1x2 5 1 5 2 x 13. ƒ1x2 5 7x3 1 1 x 1 5 14. ƒ1x2 5 2 2 x3 2x 2 7 15. ƒ1x2 5 6x5 2 4x2 1 5 6x2 1 5x 2 4 16. ƒ1x2 5 6x2 1 3x 1 1 3x2 2 5x 2 2 17. ƒ1x2 5 1 3x2 1 1 3x 2 1 4 x2 1 1 9 18. ƒ1x2 5 1 10 Ax2 2 2x 1 3 10B 2x 2 1 19. ƒ1x2 5 10.2x 2 3.1211.2x 1 4.52 0.71x 2 0.5210.2x 1 0.32 20. ƒ1x2 5 0.8x4 2 1 x2 2 0.25 1. If degree of the numerator 2 degree of the denominator 5 1, then there is a slant asymptote. 2. Divide the numerator by the denominator. The quotient corresponds to the equation of the line (slant asymptote). PR O C ED U R E F O R GR A PH I N G R AT IO N AL F U N C T I O N S 1. Find the domain of the function. 2. Find the intercept(s). n y-intercept n x-intercepts (if any) 3. Find any holes. n If x 2 a is a common factor of the numerator and denominator, then x 5 a corresponds to a hole in the graph of the rational function if the multiplicity of a in the numerator is greater than or equal to the multiplicity of a in the denominator. The result after the common factor is canceled is an equivalent rational function in lowest terms (no common factor). 4. Find any asymptotes. n Vertical asymptotes n Horizontal/slant asymptotes 5. Find additional points on the graph. 6. Sketch the graph: Draw the asymptotes and label the intercepts and points and connect with a smooth curve. In this section, rational functions were discussed. ƒ1x2 5 n1x2 d1x2 n Domain: All real numbers except the x-values that make the denominator equal to zero, d 1x2 5 0. n Vertical Asymptotes: Vertical lines, x 5 a, where d 1a2 5 a, after all common factors have been divided out. Vertical asymptotes steer the graph are never touched. n Horizontal Asymptotes: Horizontal lines, y 5 b, that steer the graph as x S 6q. 1. If degree of the numerator , degree of the denominator, then y 5 0 is a horizontal asymptote. 2. If degree of the numerator 5 degree of the denominator, then y 5 c is a horizontal asymptote where c is the ratio of the leading coefficients of the numerator and denominator, respectively. 3. If degree of the numerator . degree of the denominator, then there is no horizontal asymptote. n Slant Asymptotes: Slant lines, y 5 mx 1 b, that steer the graph as x S 6q. [SEC TION 4 .6] S U M M A RY 4.6 Rational Functions 411 In Exercises 21–26, find the slant asymptote corresponding to the graph of each rational function. 21. ƒ1x2 5 x2 1 10x 1 25 x 1 4 22. ƒ1x2 5 x2 1 9x 1 20 x 2 3 23. ƒ1x2 5 2x2 1 14x 1 7 x 2 5 24. ƒ1x2 5 3x3 1 4x2 2 6x 1 1 x2 2 x 2 30 25. ƒ1x2 5 8x4 1 7x3 1 2x 2 5 2x3 2 x2 1 3x 2 1 26. ƒ1x2 5 2x6 1 1 x5 2 1 In Exercises 27–32, match the function to the graph. 27. ƒ1x2 5 3 x 2 4 28. ƒ1x2 5 3x x 2 4 29. ƒ1x2 5 3x2 x2 2 4 30. ƒ1x2 5 2 3x2 x2 1 4 31. ƒ1x2 5 3x2 4 2 x2 32. ƒ1x2 5 3x2 x 1 4 a. b. c. –10 10 –10 10 x y –10 10 –10 10 x y –10 10 –150 150 x y d. e. f. –10 10 –10 10 x y –5 5 –5 5 x y –5 5 –5 5 x y In Exercises 33–58, use the graphing strategy outlined in this section to graph the rational functions. 33. ƒ1x2 5 2 x 1 1 34. ƒ1x2 5 4 x 2 2 35. ƒ1x2 5 2x x 2 1 36. ƒ1x2 5 4x x 1 2 37. ƒ1x2 5 x 2 1 x 38. ƒ1x2 5 2 1 x x 2 1 39. ƒ1x2 5 21x2 2 2x 2 32 x2 1 2x 40. ƒ1x2 5 31x2 2 12 x2 2 3x 41. ƒ1x2 5 x2 x 1 1 42. ƒ1x2 5 x2 2 9 x 1 2 43. ƒ1x2 5 2x3 2 x2 2 x x2 2 4 44. ƒ1x2 5 3x3 1 5x2 2 2x x2 1 4 45. ƒ1x2 5 x2 1 1 x2 2 1 46. ƒ1x2 5 1 2 x2 x2 1 1 47. ƒ1x2 5 7x2 12x 1 122 48. ƒ1x2 5 12x4 13x 1 124 49. ƒ1x2 5 1 2 9x2 11 2 4x223 50. ƒ1x2 5 25x2 2 1 116x2 2 122 51. ƒ1x2 5 3x 1 4 x 52. ƒ1x2 5 x 2 4 x 53. ƒ1x2 5 1x 2 122 1x2 2 12 54. ƒ1x2 5 1x 1 122 1x2 2 12 55. ƒ1x2 5 1x 2 121x2 2 42 1x 2 221x2 1 12 56. ƒ1x2 5 1x 2 121x2 2 92 1x 2 321x2 1 12 57. ƒ1x2 5 3x1x 2 12 x1x2 2 42 58. ƒ1x2 5 22x1x 2 32 x1x2 1 12 412 CHAPTER 4 Polynomial and Rational Functions In Exercises 59–62, for each graph of the rational function given determine: (a) all intercepts, (b) all asymptotes, and (c) equation of the rational function. 59. 60. 61. 62. x y –5 5 –5 5 x y –10 10 –10 10 x y –10 10 –10 10 x y –10 10 –10 10 63. Medicine. The concentration C of a particular drug in a person’s bloodstream t minutes after injection is given by C1t2 5 2t t2 1 100 a What is the concentration in the bloodstream after 1 minute? b. What is the concentration in the bloodstream after 1 hour? c. What is the concentration in the bloodstream after 5 hours? d. Find the horizontal asymptote of C1t2. What do you expect the concentration to be after several days? 64. Medicine. The concentration C of aspirin in the bloodstream t hours after consumption is given by C1t2 5 t t2 1 40. a. What is the concentration in the bloodstream after 1 2 hour? b. What is the concentration in the bloodstream after 1 hour? c. What is the concentration in the bloodstream after 4 hours? d. Find the horizontal asymptote for C 1t2. What do you expect the concentration to be after several days? 65. Typing. An administrative assistant is hired after graduating from high school and learns to type on the job. The number of words he can type per minute is given by N1t2 5 130t 1 260 t 1 5 t $ 0 where t is the number of months he has been on the job. a. How many words per minute can he type the day he starts? b. How many words per minute can he type after 12 months? c. How many words per minute can he type after 3 years? d. How many words per minute would you expect him to type if he worked there until he retired? 66. Memorization. A professor teaching a large lecture course tries to learn students’ names. The number of names she can remember N 1t2 increases with each week in the semester t and is given by the rational function: N1t2 5 600t t 1 20 How many students’ names does she know by the third week in the semester? How many students’ names should she know by the end of the semester (16 weeks)? According to this function, what are the most names she can remember? 67. Food. The amount of food that cats typically eat increases as their weight increases. A rational function that describes this is F1x2 5 10x2 x2 1 4, where the amount of food F1x2 is given in ounces and the weight of the cat x is given in pounds. Calculate the horizontal asymptote. How many ounces of food will most adult cats eat? 10 10 x y • A P P L I C A T I O N S 4.6 Rational Functions 413 68. Memorization. The Guinness Book of World Records, 2004 states that Dominic O’Brien (England) memorized on a single sighting a random sequence of 54 separate packs of cards all shuffled together (2808 cards in total) at Simpson’s-in-the-Strand, London, England, on May 1, 2002. He memorized the cards in 11 hours 42 minutes and then recited them in exact sequence in a time of 3 hours 30 minutes. With only a 0.5% margin of error allowed (no more than 14 errors), he broke the record with just 8 errors. If we let x represent the time (hours) it takes to memorize the cards and y represent the number of cards memorized, then a rational function that models this event is given by y 5 2800x2 1 x x2 1 2 . According to this model, how many cards could be memorized in an hour? What is the greatest number of cards that can be memorized? 69. Gardening. A 500-square-foot rectangular garden will be enclosed with fencing. Write a rational function that describes how many linear feet of fence will be needed to enclose the garden as a function of the width of the garden w. 70. Geometry. A rectangular picture has an area of 414 square inches. A border (matting) is used when framing. If the top and bottom borders are each 4 inches and the side borders are 3.5 inches, write a function that represents the area A1l2 of the entire frame as a function of the length of the picture l. For Exercises 71 and 72, refer to the following: The monthly profit function for a product is given by P1x2 5 2x3 1 10x2 where x is the number of units sold measured in thousands and P is profit measured in thousands of dollars. The average profit, which represents the profit per thousand units sold, for this product is given by P1x2 5 2x3 1 10x2 x where x is units sold measured in thousands and P is profit measured in thousands of dollars. 71. Business. Find the number of units that must be sold to produce an average profit of $16,000 per thousand units. Convert the answer to average profit in dollars per unit. 72. Business. Find the number of units that must be sold to produce an average profit of $25,000 per thousand units. Convert the answer to average profit in dollars per unit. For Exercises 73 and 74, refer to the following: Some medications, such as Synthroid, are prescribed as a maintenance drug because they are taken regularly for an ongoing condition, such as hypothyroidism. Maintenance drugs function by maintaining a therapeutic drug level in the bloodstream over time. The concentration of a maintenance drug over a 24-hour period is modeled by the function C1t2 5 221t 2 12 t2 1 1 1 24 where t is time in hours after the dose was administered and C is the concentration of the drug in the bloodstream measured in mg/mL. This medication is designed to maintain a consistent concentration in the bloodstream of approximately 25 mg/mL. Note: This drug will become inert; that is, the concentration will drop to 0 mg/mL, during the 25th hour after taking the medication. 73. Health/Medicine. Find the concentration of the drug, to the nearest tenth of mg/mL, in the bloodstream 15 hours after the dose is administered. Is this the only time this concentration of the drug is found in the bloodstream? At what other times is this concentration reached? Round to the nearest hour. Discuss the significance of this answer. 74. Health/Medicine. Find the time, after the first hour and a half, at which the concentration of the drug in the bloodstream has dropped to 25 mg/mL. Find the concentration of the drug 24 hours after taking a dose to the nearest tenth of a mgmL. Discuss the importance of taking the medication every 24 hours rather than every day. • C A T C H T H E M I S T A K E In Exercises 75–78, explain the mistake that is made. 75. Determine the vertical asymptotes of the function ƒ1x2 5 x 2 1 x2 2 1. Solution: Set the denominator equal to zero. x2 2 1 5 0 Solve for x. x 5 61 The vertical asymptotes are x 5 21 and x 5 1. The following is a correct graph of the function. Note that only x 5 21 is an asymptote. What went wrong? –10 10 –10 10 x y 414 CHAPTER 4 Polynomial and Rational Functions 76. Determine the vertical asymptotes of ƒ1x2 5 2x x2 1 1. Solution: Set the denominator equal to zero. x2 1 1 5 0 Solve for x. x 5 61 The vertical asymptotes are x 5 21 and x 5 1. The following is a correct graph of the function. Note that there are no vertical asymptotes. What went wrong? –10 10 –2 2 x y 77. Determine whether a horizontal or a slant asymptote exists for the function ƒ1x2 5 9 2 x2 x2 2 1. If one does, find it. Solution: Step 1: The degree of the numerator equals the degree of the denominator, so there is a horizontal asymptote. Step 2: The horizontal asymptote is the ratio of the lead coef-ficients: y 5 9 1 5 9. The horizontal asymptote is y 5 9. The following is a correct graph of the function. Note that there is no horizontal asymptote at y 5 9. What went wrong? –10 10 –30 20 x y 78. Determine whether a horizontal or a slant asymptote exists for the function ƒ1x2 5 x2 1 2x 2 1 3x3 2 2x2 2 1. If one does, find it. Solution: Step 1: The degree of the denominator is exactly one more than the degree of the numerator, so there is a slant asymptote. Step 2: Divide. 3x 2 8 x2 1 2x 2 1q3x3 2 2x2 1 0x 2 1 3x3 1 6x2 2 3x 2 8x2 1 3x 2 1 8x2 2 16x 1 8 19x 2 9 The slant asymptote is y 5 3x 2 8. The following is the correct graph of the function. Note that y 5 3x 2 8 is not an asymptote. What went wrong? –5 5 –10 10 x y • C O N C E P T U A L For Exercises 79–82, determine whether each statement is true or false. 79. A rational function can have either a horizontal asymptote or a slant asymptote but not both. 80. A rational function can have at most one vertical asymptote. 81. A rational function can cross a vertical asymptote. 82. A rational function can cross a horizontal or a slant asymptote. 4.6 Rational Functions 415 83. Determine the asymptotes of the rational function ƒ1x2 5 1x 2 a21x 1 b2 1x 2 c21x 1 d2. 84. Determine the asymptotes of the rational function ƒ1x2 5 3x2 1 b2 x2 1 a2 . 85. Write a rational function that has vertical asymptotes at x 5 23 and x 5 1 and a horizontal asymptote at y 5 4. 86. Write a rational function that has no vertical asymptotes, approaches the x-axis as a horizontal asymptote, and has an x-intercept of 13, 02. • C H A L L E N G E • T E C H N O L O G Y 87. Determine the vertical asymptotes of ƒ1x2 5 x 2 4 x2 2 2x 2 8. Graph this function utilizing a graphing utility. Does the graph confirm the asymptotes? 88. Determine the vertical asymptotes of ƒ1x2 5 2x 1 1 6x2 1 x 2 1. Graph this function utilizing a graphing utility. Does the graph confirm the asymptotes? 89. Find the asymptotes and intercepts of the rational function ƒ1x2 5 1 3x 1 1 2 2 x. Note: Combine the two expressions into a single rational expression. Graph this function utilizing a graphing utility. Does the graph confirm what you found? 90. Find the asymptotes and intercepts of the rational function ƒ1x2 5 2 1 x2 1 1 1 1 x. Note: Combine the two expressions into a single rational expression. Graph this function utilizing a graphing utility. Does the graph confirm what you found? For Exercises 91 and 92: (a) Identify all asymptotes for each function. (b) Plot ƒ 1x2 and g 1x2 in the same window. How does the end behavior of the function ƒ differ from g? (c) Plot g 1x2 and h 1x2 in the same window. How does the end behavior of g differ from h? (d) Combine the two expressions into a single rational expression for the functions g and h. Does the strategy of finding horizontal and slant asymptotes agree with your findings in (b) and (c)? 91. ƒ1x2 5 1 x 2 3, g1x2 5 2 1 1 x 2 3, h1x2 5 23 1 1 x 2 3 92. ƒ1x2 5 2x x2 2 1, g1x2 5 x 1 2x x2 2 1, h1x2 5 x 2 3 1 2x x2 2 1 CH A P TE R 4 R E VI E W 416 CHAPTER 4 Polynomial and Rational Functions [ C H AP T E R 4 REVIEW] SECTION CONCEPT KEY IDEAS/FORMULAS 4.1 Quadratic functions Graphs of quadratic functions: Parabolas Parabolas Graphing quadratic functions in standard form ƒ1x2 5 a 1x 2 h22 1 k n Vertex: 1h, k2 n Opens Up: a . 0 n Opens Down: a , 0 Graphing quadratic functions in general form ƒ1x2 5 ax2 1 bx 1 c, vertex is 1h, k2 5 a 2 b 2a, f a 2 b 2abb Finding the equation of a parabola 4.2 Polynomial functions of higher degree Identifying polynomial functions P1x2 5 anxn 1 an21xn21 1 c1 a2x2 1 a1x 1 a0 is a polynomial of degree n. Graphing polynomial functions using transformations of power functions Shift power functions y 5 xn behave similar to: n y 5 x2, when n is even. n y 5 x3, when n is odd. Real zeros of a polynomial function P 1x2 5 1x 2 a2 1x 2 b2n 5 0 n a is a zero of multiplicity 1. n b is a zero of multiplicity n. Graphing general polynomial functions Intercepts; zeros and multiplicities; end behavior 4.3 Dividing polynomials: Long division and synthetic division Use zero placeholders for missing terms. Long division of polynomials Can be used for all polynomial division. Synthetic division of polynomials Can only be used when dividing by 1x 6 a2. 4.4 The real zeros of a polynomial function P1x2 5 anxn 1 an21xn21 1 c1 a2x2 1 a1x 1 a0 If P 1c2 5 0, then c is a zero of P 1x2. The remainder theorem and the factor theorem If P 1x2 is divided by x 2 a, then the remainder r is r 5 P 1a2. The rational zero theorem and Descartes’ rule of signs Possible zeros 5 Factors of a0 Factors of an Number of positive or negative real zeros is related to the number of sign variations in P 1x2 or P 12x2. Factoring polynomials 1. List possible rational zeros (rational zero theorem). 2. List possible combinations of positive and negative real zeros (Descartes’ rule of signs). 3. Test possible values until a zero is found. 4. Once a real zero is found, use synthetic division. Then repeat testing on quotient until linear and/or irreducible quadratic factors are reached. 5. If there is a real zero but all possible rational roots have failed, then approximate the real zero using the intermediate value theorem/bisection method. The intermediate value theorem The intermediate value theorem and the bisection method are used to approximate irrational zeros. Graphing polynomial functions 1. Find the intercepts. 2. Determine end behavior. 3. Find additional points. 4. Sketch a smooth curve. CH A P TE R 4 R E VIE W Chapter Review 417 SECTION CONCEPT KEY IDEAS/FORMULAS 4.5 Complex zeros: The fundamental theorem of algebra P1x2 5 anxn 1 an21xn21 1 c1 a2x2 1 a1x 1 a0 P1x2 5 1x 2 c121x 2 c22 c 1x 2 cn2 n factors where the c’s represent complex (not necessarily distinct) zeros. The fundamental theorem of algebra n P 1x2 of degree n has at least one zero and at most n zeros. Complex zeros Complex conjugate pairs: n If a 1 bi is a zero of P 1x2, then a 2 bi is also a zero. Factoring polynomials The polynomial can be written as a product of linear factors, not necessarily distinct. 4.6 Rational functions ƒ1x2 5 n1x2 d1x2 d1x2 2 0 Domain of rational functions Domain: All real numbers except x-values that make the denominator equal to zero; that is, d 1x2 5 0. A rational function ƒ1x2 5 n1x2 d1x2 is said to be in lowest terms if n 1x2 and d 1x2 have no common factors. Horizontal, vertical, and slant asymptotes A rational function that has a common factor x 2 a in both the numerator and denominator has a hole at x 5 a in its graph if the multiplicity of a in the numerator is greater than or equal to the multiplicity of a in the denominator. Vertical Asymptotes A rational function in lowest terms has a vertical asymptote corresponding to any x-values that make the denominator equal to zero. Horizontal Asymptotes n y 5 0 if degree of n 1x2 , degree of d 1x2. n No horizontal asymptote if degree of n 1x2 . degree of d 1x2. n y 5 Leading coefficient of n1x2 Leading coefficient of d1x2 if degree of n 1x2 5 degree of d 1x2. Slant Asymptotes If degree of n 1x2 2 degree of d 1x2 5 1. Divide n 1x2 by d 1x2 and the quotient determines the slant asymptote; that is, y 5 quotient. Graphing rational functions 1. Find the domain of the function. 2. Find the intercept(s). 3. Find any holes. 4. Find any asymptotes. 5. Find additional points on the graph. 6. Sketch the graph: Draw the asymptotes and label the intercepts and points and connect with a smooth curve. d REV IEW E XE R CI SE S 418 CHAPTER 4 Polynomial and Rational Functions [ C H AP T E R 4 REVIEW EXERC IS E S ] 4.1 Quadratic Functions Match the quadratic function with its graph. 1. ƒ1x2 5 221x 1 622 1 3 2. ƒ1x2 5 1 4 1x 2 422 1 2 3. ƒ1x2 5 x2 1 x 2 6 4. ƒ1x2 5 23x2 2 10x 1 8 a. b. –10 10 –10 10 x y –10 10 –10 10 x y c. d. –10 10 –10 10 x y –10 10 x y –4 16 Graph the quadratic function given in standard form. 5. ƒ1x2 5 21x 2 722 1 4 6. ƒ1x2 5 1x 1 322 2 5 7. ƒ1x2 5 21 2Ax 2 1 3B2 1 2 5 8. ƒ1x2 5 0.61x 2 0.7522 1 0.5 Rewrite the quadratic function in standard form by completing the square. 9. ƒ1x2 5 x2 2 3x 2 10 10. ƒ1x2 5 x2 2 2x 2 24 11. ƒ1x2 5 4x2 1 8x 2 7 12. ƒ1x2 5 21 4 x2 1 2x 2 4 Graph the quadratic function given in general form. 13. ƒ1x2 5 x2 2 3x 1 5 14. ƒ1x2 5 2x2 1 4x 1 2 15. ƒ1x2 5 24x2 1 2x 1 3 16. ƒ1x2 5 20.75x2 1 2.5 Find the vertex of the parabola associated with each quadratic function. 17. ƒ1x2 5 13x2 2 5x 1 12 18. ƒ1x2 5 2 5 x2 2 4x 1 3 19. ƒ1x2 5 20.45x2 2 0.12x 1 3.6 20. ƒ1x2 5 23 4 x2 1 2 5x 1 4 Find the quadratic function that has the given vertex and goes through the given point. 21. vertex: 122, 32 point: 11, 42 22. vertex: 14, 72 point: 123, 12 23. vertex: 12.7, 3.42 point: 13.2, 4.82 24. vertex: A 25 2, 7 4B point: A1 2, 3 5B Applications 25. Profit. The revenue and the cost of a local business are given below as functions of the number of units x in thousands produced and sold. Use the cost and the revenue to answer the questions that follow. C1x2 5 1 3x 1 2 and R1x2 5 22x2 1 12x 2 12 a. Determine the profit function. b. State the break-even points. c. Graph the profit function. d. What is the range of units to make and sell that will correspond to a profit? 26. Geometry. Given the length of a rectangle is 2x 2 4 and the width is x 1 7, find the area of the rectangle. What dimensions correspond to the largest area? 27. Geometry. A triangle has a base of x 1 2 units and a height of 4 2 x units. Determine the area of the triangle. What dimensions correspond to the largest area? 28. Geometry. A person standing at a ridge in the Grand Canyon throws a penny upward and toward the pit of the canyon. The height of the penny is given by the function: h1t2 5 212t2 1 80t a. What is the maximum height that the penny will reach? b. How many seconds will it take the penny to hit the ground below? 4.2 Polynomial Functions of Higher Degree Determine which functions are polynomials, and for those, state their degree. 29. ƒ1x2 5 x6 2 2x5 1 3x2 1 9x 2 42 30. ƒ1x2 5 13x 2 4231x 1 622 31. ƒ1x2 5 3x4 2 x31 x2 1 ! 4 x 1 5 32. ƒ1x2 5 5x3 2 2x2 1 4x 7 2 3 R E VI E W E XERCISES Review Exercises 419 Match the polynomial function with its graph. 33. ƒ1x2 5 2x 2 5 34. ƒ1x2 5 23x2 1 x 2 4 35. ƒ1x2 5 x4 2 2x3 1 x2 2 6 36. ƒ1x2 5 x7 2 x5 1 3x4 1 3x 1 7 a. b. –5 5 –10 10 x y –5 5 x y –10 c. d. –5 5 –10 40 x y –10 10 x y 10 –10 Graph each function by transforming a power function y 5 xn. 37. ƒ1x2 5 2x7 38. ƒ1x2 5 1x 2 323 39. ƒ1x2 5 x4 2 2 40. ƒ1x2 5 26 2 1x 1 725 Find all the real zeros of each polynomial function, and state their multiplicities. 41. ƒ1x2 5 31x 1 4221x 2 625 42. ƒ1x2 5 7x 12x 2 4231x 1 52 43. ƒ1x2 5 x5 2 13x3 1 36x 44. ƒ1x2 5 4.2x4 2 2.6x2 Find a polynomial of minimum degree that has the given zeros. 45. 23, 0, 4 46. 2, 4, 6, 28 47. 22 5, 3 4, 0 48. 2 2 !5, 2 1 !5 49. 22 1multiplicity of 22, 3 1multiplicity of 22 50. 3 1multiplicity of 22, 21 1multiplicity of 22, 0 1multiplicity of 32 For each polynomial function given: (a) list each real zero and its multiplicity; (b) determine whether the graph touches or crosses at each x-intercept; (c) find the y-intercept and a few points on the graph; (d) determine the end behavior; and (e) sketch the graph. 51. ƒ1x2 5 x2 2 5x 2 14 52. ƒ1x2 5 21x 2 525 53. ƒ1x2 5 6x7 1 3x5 2 x2 1 x 2 4 54. ƒ1x2 5 2x413x 1 6231x 2 723 Applications 55. Salary. Tiffany has started tutoring students x hours per week. The tutoring job corresponds to the following additional income: ƒ1x2 5 1x 2 121x 2 321x 2 72 a. Graph the polynomial function. b. Give any real zeros that occur. c. How many hours of tutoring are financially beneficial to Tiffany? 56. Profit. The following function is the profit for Walt Disney World, where P 1x2 represents profit in millions of dollars and x represents the month 1x 5 1 corresponds to January2: P1x2 5 31x 2 2221x 2 5221x 2 1022 1 # x # 12 Graph the polynomial. When are the peak seasons? 4.3 Dividing Polynomials: Long Division and Synthetic Division Divide the polynomials with long division. If you choose to use a calculator, do not round off. Keep the exact values instead. Express the answer in the form Q 1x2 5 ?, r 1x2 5 ?. 57. 1x2 1 2x 2 62 4 1x 2 22 58. 12x2 2 5x 2 12 4 12x 2 32 59. 14x4 2 16x3 1 x 2 9 1 12x22 4 12x 2 42 60. 16x2 1 2x3 2 4x4 1 2 2 x2 4 12x2 1 x 2 42 Use synthetic division to divide the polynomial by the linear factor. Indicate the quotient Q 1x2 and the remainder r 1x2. 61. 1x4 1 4x3 1 5x2 2 2x 2 82 4 1x 1 22 62. 1x3 2 10x 1 32 4 12 1 x2 63. 1x6 2 642 4 1x 1 82 64. 12x5 1 4x4 2 2x3 1 7x 1 52 4 Ax 2 3 4B Divide the polynomials with either long division or synthetic division. 65. 15x3 1 8x2 2 22x 1 12 4 15x2 2 7x 1 32 66. 1x4 1 2x3 2 5x2 1 4x 1 22 4 1x 2 32 67. 1x3 2 4x2 1 2x 2 82 4 1x 1 12 68. 1x3 2 5x2 1 4x 2 202 4 1x2 1 42 Applications 69. Geometry. The area of a rectangle is given by the polynomial 6x4 2 8x3 2 10x2 1 12x 2 16. If the width is 2x 2 4, what is the length of the rectangle? 70. Volume. A 10-inch by 15-inch rectangular piece of cardboard is used to make a box. Square pieces x inches on a side are cut out from the corners of the cardboard, and then the sides are folded up. Find the volume of the box. REV IEW E XE R CI SE S 420 CHAPTER 4 Polynomial and Rational Functions 4.4 The Real Zeros of a Polynomial Function Find the following values by applying synthetic division. Check by substituting the value into the function. ƒ1x2 5 6x5 1 x4 2 7x2 1 x 2 1 g1x2 5 x3 1 2x2 2 3 71. ƒ1222 72. ƒ112 73. g 112 74. g 1212 Determine whether the number given is a zero of the polynomial. 75. 23, P 1x2 5 x3 2 5x2 1 4x 1 2 76. 2 and 22, P 1x2 5 x4 2 16 77. 1, P 1x2 5 2x4 2 2x 78. 4, P 1x2 5 x4 2 2x3 2 8x Given a zero of the polynomial, determine all other real zeros, and write the polynomial in terms of a product of linear or irreducible factors. Polynomial Zero 79. P 1x2 5 x4 2 6x3 1 32x 22 80. P 1x2 5 x3 2 7x2 1 36 3 81. P 1x2 5 x5 2 x4 2 8x3 1 12x2 0 82. P 1x2 5 x4 2 32x2 2 144 6 Use Descartes’ rule of signs to determine the possible number of positive real zeros and negative real zeros. 83. P 1x2 5 x4 1 3x3 2 16 84. P 1x2 5 x5 1 6x3 2 4x 2 2 85. P 1x2 5 x9 2 2x7 1 x4 2 3x3 1 2x 2 1 86. P 1x2 5 2x5 2 4x3 1 2x2 2 7 Use the rational zero theorem to list the possible rational zeros. 87. P 1x2 5 x3 2 2x2 1 4x 1 6 88. P 1x2 5 x5 2 4x3 1 2x2 2 4x 2 8 89. P 1x2 5 2x4 1 2x3 2 36x2 2 32x 1 64 90. P 1x2 5 24x5 2 5x3 1 4x 1 2 List the possible rational zeros, and test to determine all rational zeros. 91. P 1x2 5 2x3 2 5x2 1 1 92. P 1x2 5 12x3 1 8x2 2 13x 1 3 93. P 1x2 5 x4 2 5x3 1 20x 2 16 94. P 1x2 5 24x4 2 4x3 2 10x2 1 3x 2 2 For each polynomial: (a) use Descartes’ rule of signs to determine the possible combinations of positive real zeros and negative real zeros; (b) use the rational zero test to determine possible rational zeros; (c) use the upper and lower bound rules to eliminate possible zeros; (d) test for rational zeros; (e) factor as a product of linear and/or irreducible quadratic factors; and (f) graph the polynomial function. 95. P 1x2 5 x3 1 3x 2 5 96. P 1x2 5 x3 1 3x2 2 6x 2 8 97. P 1x2 5 x3 2 9x2 1 20x 2 12 98. P 1x2 5 x4 2 x3 2 7x2 1 x 1 6 99. P 1x2 5 x4 2 5x3 2 10x2 1 20x 1 24 100. P 1x2 5 x5 2 3x3 2 6x2 1 8x 4.5 Complex Zeros: The Fundamental Theorem of Algebra Find all zeros. Factor the polynomial as a product of linear factors. 101. P 1x2 5 x2 1 25 102. P 1x2 5 x2 1 16 103. P 1x2 5 x2 2 2x 1 5 104. P 1x2 5 x2 1 4x 1 5 A polynomial function is described. Find all remaining zeros. 105. Degree: 4 Zeros: 22i, 3 1 i 106. Degree: 4 Zeros: 3i, 2 2 i 107. Degree: 6 Zeros: i, 2 2 i 1multiplicity 22 108. Degree: 6 Zeros: 2i, 1 2 i 1multiplicity 22 Given a zero of the polynomial, determine all other zeros (real and complex) and write the polynomial in terms of a product of linear factors. Polynomial Zero 109. P 1x2 5 x4 2 3x3 2 3x2 2 3x 2 4 i 110. P 1x2 5 x4 2 4x3 1 x2 1 16x 2 20 2 2 i 111. P 1x2 5 x4 2 2x3 1 11x2 2 18x 1 18 23i 112. P 1x2 5 x4 2 5x2 1 10x 2 6 1 1 i Factor each polynomial as a product of linear factors. 113. P 1x2 5 x4 2 81 114. P 1x2 5 x3 2 6x2 1 12x 115. P 1x2 5 x3 2 x2 1 4x 2 4 116. P 1x2 5 x4 2 5x3 1 12x2 2 2x 2 20 R E VI E W E XERCISES Review Exercises 421 4.6 Rational Functions Determine the vertical, horizontal, or slant asymptotes (if they exist) for the following rational functions. 117. ƒ1x2 5 7 2 x x 1 2 118. ƒ1x2 5 2 2 x2 1x 2 123 119. ƒ1x2 5 4x2 x 1 1 120. ƒ1x2 5 3x2 x2 1 9 121. ƒ1x2 5 2x2 2 3x 1 1 x2 1 4 122. ƒ1x2 5 22x2 1 3x 1 5 x 1 5 Graph the rational functions. 123. ƒ1x2 5 2 2 x 2 3 124. ƒ1x2 5 5 x 1 1 125. ƒ1x2 5 x2 x2 1 4 126. ƒ1x2 5 x2 2 36 x2 1 25 127. ƒ1x2 5 x2 2 49 x 1 7 128. ƒ1x2 5 2x2 2 3x 2 2 2x2 2 5x 2 3 Technology Section 4.1 129. On a graphing calculator, plot the quadratic function: ƒ1x2 5 0.005x2 2 4.8x 2 59 a. Identify the vertex of this parabola. b. Identify the y-intercept. c. Identify the x-intercepts (if any). d. What is the axis of symmetry? 130. Determine the quadratic function whose vertex is 12.4, 23.12 and passes through the point 10, 5.542. a. Write the quadratic function in general form. b. Plot this quadratic function with a graphing calculator. c. Zoom in on the vertex and y-intercept. Do they agree with the given values? Section 4.2 In Exercises 131 and 132, use a graphing calculator or a computer to graph each polynomial. From the graph, estimate the x-intercepts and state the zeros of the function and their multiplicities. 131. ƒ1x2 5 5x3 2 11x2 2 10.4x 1 5.6 132. ƒ1x2 5 2x3 2 0.9x2 1 2.16x 2 2.16 Section 4.3 133. Plot 15x3 2 47x2 1 38x 2 8 3x2 2 7x 1 2 . What type of function is it? Perform this division using long division, and confirm that the graph corresponds to the quotient. 134. Plot 24x3 1 14x2 2 x 2 15 x 2 3 . What type of function is it? Perform this division using synthetic division, and confirm that the graph corresponds to the quotient. Section 4.4 In Exercises 135 and 136: (a) From the list of all possible rational zeros of the polynomial, use a graphing calculator or software to graph P 1x2 to find the rational zeros. (b) Factor as a product of linear and/or irreducible quadratic factors. 135. P1x2 5 x4 2 3x3 2 12x2 1 20x 1 48 136. P1x2 5 25x5 2 18x4 2 32x3 2 24x2 1 x 1 6 Section 4.5 Find all zeros (real and complex). Factor the polynomial as a product of linear factors. 137. P1x2 5 2x3 1 x2 2 2x 2 91 138. P1x2 5 22x4 1 5x3 1 37x2 2 160x 1 150 Section 4.6 In Exercises 139 and 140: (a) graph the function ƒ 1x2 utilizing a graphing utility to determine if it is a one-to-one function; (b) if it is, find its inverse; and (c) graph both functions in the same viewing window. 139. ƒ1x2 5 2x 2 3 x 1 1 140. ƒ1x2 5 4x 1 7 x 2 2 PR ACTICE TEST [ C H AP T E R 4 PRACTICE TEST ] 1. Graph the parabola y 5 21x 2 422 1 1. 2. Write the parabola in standard form y 5 2x2 1 4x 2 1. 3. Find the vertex of the parabola ƒ1x2 5 21 2x2 1 3x 2 4. 4. Find a quadratic function whose vertex is 123, 212 and whose graph passes through the point 124, 12. 5. Find a sixth-degree polynomial function with the given zeros: 2 of multiplicity 3 1 of multiplicity 2 0 of multiplicity 1 6. For the polynomial function ƒ1x2 5 x4 1 6x3 2 7x: a. List each real zero and its multiplicity. b. Determine whether the graph touches or crosses at each x-intercept. c. Find the y-intercept and a few points on the graph. d. Determine the end behavior. e. Sketch the graph. 7. Divide 24x4 1 2x3 2 7x2 1 5x 2 2 by 2x2 2 3x 1 1. 8. Divide 17x5 2 4x3 1 2x 2 10 by x 1 2. 9. Is x 2 3 a factor of x4 1 x3 2 13x2 2 x 1 12? 10. Determine whether 21 is a zero of P 1x2 5 x21 2 2x 18 1 5x12 1 7x3 1 3x2 1 2. 11. Given that x 2 7 is a factor of P 1x2 5 x3 2 6x2 2 9x 1 14, factor the polynomial in terms of linear factors. 12. Given that 3i is a zero of P 1x2 5 x4 2 3x3 1 19x2 2 27x 1 90, find all other zeros. 13. Can a polynomial have zeros that are not x-intercepts? Explain. 14. Apply Descartes’ rule of signs to determine the possible combinations of positive real zeros, negative real zeros, and complex zeros of P 1x2 5 3x5 1 2x4 2 3x3 1 2x2 2 x 1 1. 15. From the rational zero test, list all possible rational zeros of P 1x2 5 3x4 2 7x2 1 3x 1 12. In Exercises 16–18, determine all zeros of the polynomial function and graph. 16. P 1x2 5 2x3 1 4x 17. P 1x2 5 2x3 2 3x2 1 8x 2 12 18. P 1x2 5 x4 2 6x3 1 10x2 2 6x 1 9 19. Sports. A football player shows up in August at 300 pounds. After 2 weeks of practice in the hot sun, he is down to 285 pounds. Ten weeks into the season he is up to 315 pounds because of weight training. In the spring he does not work out, and he is back to 300 pounds by the next August. Plot these points on a graph. What degree polynomial could this be? 20. Profit. The profit of a company is governed by the polynomial P 1x2 5 x3 2 13x2 1 47x 2 35, where x is the number of units sold in thousands. How many units does the company have to sell to break even? 21. Interest Rate. The interest rate for a 30-year fixed mortgage fluctuates with the economy. In 1970, the mortgage interest rate was 8%, and in 1988 it peaked at 13%. In 2002, it dipped down to 4%, and in 2005, it was up to 6%. In 2015 it dipped back down to 4.25%. What is the lowest-degree polynomial that can represent this function? In Exercises 22–25, determine (if any) the: a. x- and y-intercepts b. vertical asymptotes c. horizontal asymptotes d. slant asymptotes e. graph 22. ƒ1x2 5 2x 2 9 x 1 3 23. g1x2 5 x x2 2 4 24. h1x2 5 3x3 2 3 x2 2 4 25. F1x2 5 x 2 3 x2 2 2x 2 8 26. Food. On eating a sugary snack, the average body almost doubles its glucose level. The percentage increase in glucose level y can be approximated by the rational function y 5 25x x2 1 50 , where x represents the number of minutes after eating the snack. Graph the function. 27. a. Use the calculator commands STAT QuadReg to model the data using a quadratic function. b. Write the quadratic function in standard form and identify the vertex. c. Find the x-intercepts. d. Plot this quadratic function with a graphing calculator. Do they agree with the given values? x 23 2.2 7.5 y 10.01 29.75 25.76 28. Find the asymptotes and intercepts of the rational function ƒ1x2 5 x12x 2 32 x2 2 3x 1 1. Note: Combine the two expressions into a single rational expression. Graph this function utilizing a graphing utility. Does the graph confirm what you found? 422 CHAPTER 4 Polynomial and Rational Functions CU MU LA TIV E TEST [CH AP TERS 1–4 CUM UL AT IVE T E S T ] 1. Simplify 15x21y2223 1210x2y222 and express in terms of positive exponents. 2. Factor 2xy 2 2x 1 3y 2 3. 3. Multiply and simplify 4x2 2 36 x2 ⋅ x3 1 6x2 x2 1 9x 1 18. 4. Solve for x: 0 2x 2 5 0 1 3 . 10. 5. Austin can mow a lawn in 75 minutes. The next week Stacey mows the same lawn in 60 minutes. How long would it take them to mow the lawn working together? 6. Use the discriminant to determine the number and type of roots: 24x2 1 3x 1 15 5 0. 7. Solve and check "16 1 x2 5 x 1 2. 8. Apply algebraic tests to determine whether the equation’s graph is symmetric with respect to the x-axis, y-axis, or origin for y 5 0 x 0 2 3. 9. Write an equation of the line that is parallel to the line x 2 3y 5 8 and has the point 14, 12. 10. Find the x-intercept and y-intercept and sketch the graph for 2y 2 6 5 0. 11. Write the equation of a circle with center 10, 62 and that passes through the point 11, 52. 12. Express the domain of the function ƒ1x2 5 !6x 2 7 with interval notation. 13. Determine whether the function g1x2 5 !x 1 10 is even, odd, or neither. 14. For the function y 5 21x 1 122 1 2, identify all of the transformations of y 5 x2. 15. Sketch the graph of y 5 !x 2 1 1 3 and identify all transformations. 16. Find the composite function f + g and state the domain for ƒ1x2 5 x2 2 3 and g1x2 5 !x 1 2. 17. Evaluate g 1 ƒ 1212 2 for ƒ1x2 5 7 2 2x2 and g 1x2 5 2x 2 10. 18. Find the inverse of the function ƒ1x2 5 1x 2 422 1 2, where x $ 4. 19. Find a quadratic function that has the vertex 122, 32 and point 121, 42. 20. Find all of the real zeros and state their multiplicities of the function ƒ1x2 5 23.7x4 2 14.8x3. 21. Use long division to find the quotient Q 1x2 and the remainder r 1x2 of 1220x3 2 8x2 1 7x 2 52 4 125x 1 32. 22. Use synthetic division to find the quotient Q 1x2 and the remainder r 1x2 of 12x3 1 3x2 2 11x 1 62 4 1x 2 32. 23. List the possible rational zeros, and test to determine all rational zeros for P 1x2 5 12x3 1 29x2 1 7x 2 6. 24. Given the real zero x 5 5 of the polynomial P 1x2 5 2x3 2 3x2 2 32x 2 15, determine all the other zeros and write the polynomial in terms of a product of linear factors. 25. Find all vertical and horizontal asymptotes of ƒ1x2 5 3x 2 5 x2 2 4. 26. Graph the function ƒ1x2 5 2x3 2 x2 2 x x2 2 1 . 27. Find the asymptotes and intercepts of the rational function ƒ1x2 5 5 2x 2 3 2 1 x. Note: Combine the two expressions into a single rational expression. Graph this function utilizing a graphing utility. Does the graph confirm what you found? 28. Find the asymptotes and intercepts of the rational function ƒ1x2 5 6x 3x 1 1 2 6x 4x 2 1. Note: Combine the two expressions into a single rational expression. Graph this function utilizing a graphing utility. Does the graph confirm what you found? Cumulative Test 423 C H A P T E R [ [ ■ ■Graph exponential functions. ■ ■Graph logarithmic functions. ■ ■Apply properties of logarithms. ■ ■Solve exponential and logarithmic equations. ■ ■Use exponential and logarithmic models to represent a variety of real-world phenomena. How do archaeologists and anthropologists date fossils? One method is carbon testing. The ratio of carbon 12 (the more stable kind of carbon) to carbon 14 at the moment of death is the same as the ratio in all living things while they are still alive, but the carbon 14 decays and is not replaced. By looking at the ratio of carbon 12 to carbon 14 in the sample and comparing it with the ratio in a living organism, it is possible to determine the age of a formerly living thing fairly precisely. The amount of carbon in a fossil is a function of how many years the organism has been dead. The amount is modeled by an exponential function, and the inverse of the exponential function, a logarithmic function, is used to determine the age of the fossil. Exponential and Logarithmic Functions 5 Francois Gohier/Science Source LEARNING OBJECTIVES [I N T HI S CHAPTER] We will discuss exponential functions and their inverses, logarithmic functions. We will graph these functions and use their properties to solve exponential and logarithmic equations. Last, we will discuss particular exponential and logarithmic models that represent phenomena such as compound interest, world populations, conservation biology models, carbon dating, pH values in chemistry, and the bell curve that is fundamental in statistics for describing how quantities vary in the real world. 425 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 5.1 EXPONENTIAL FUNCTIONS AND THEIR GRAPHS 5.2 LOGARITHMIC FUNCTIONS AND THEIR GRAPHS 5.3 PROPERTIES OF LOGARITHMS 5.4 EXPONENTIAL AND LOGARITHMIC EQUATIONS 5.5 EXPONENTIAL AND LOGARITHMIC MODELS • Evaluating Exponential Functions • Graphs of Exponential Functions • The Natural Base e • Applications of Exponential Functions • Evaluating Logarithms • Common and Natural Logarithms • Graphs of Logarithmic Functions • Applications of Logarithms • Properties of Logarithmic Functions • Change-of-Base Formula • Exponential Equations • Solving Logarithmic Equations • Exponential Growth Models • Exponential Decay Models • Gaussian (Normal) Distribution Models • Logistic Growth Models • Logarithmic Models 426 CHAPTER 5 Exponential and Logarithmic Functions 5.1.1 Evaluating Exponential Functions Most of the functions (polynomial, rational, radical, etc.) we have studied thus far have been algebraic functions. Algebraic functions involve basic operations, powers, and roots. In this chapter, we discuss exponential functions and logarithmic functions, which are called transcendental functions because they transcend our ability to define them with a finite number of algebraic expressions. The following table illustrates the difference between algebraic functions and exponential functions. FUNCTION VARIABLE IS IN THE CONSTANT IS IN THE EXAMPLE EXAMPLE Algebraic Base Exponent ƒ1x2 5 x2 g 1x2 5 x1/3 Exponential Exponent Base F 1x2 5 2x G1x2 5 a1 3b x DEFINITION Exponential Function An exponential function with base b is denoted by ƒ1x2 5 bx where b and x are any real numbers such that b . 0 and b 2 1. Note: ■ ■We eliminate b 5 1 as a value for the base because it merely yields the constant function ƒ1x2 5 1x 5 1. ■ ■We eliminate negative values for b because they would give non–real-number values such as 12921/2 5 !29 5 3i. ■ ■We eliminate b 5 0 because 0x corresponds to an undefined value when x is negative. Sometimes the value of an exponential function for a specific argument can be found by inspection as an exact number. x 23 21 0 1 3 F1x2 5 2x 223 5 1 23 5 1 8 221 5 1 21 5 1 2 20 5 1 21 5 2 23 5 8 If an exponential function cannot be evaluated exactly, then we find the decimal approximation using a calculator. Most calculators have either a base to a power button xy or a caret button for working with exponents. x 22.7 24 5 5 7 2.7 F1x2 5 2x 222.7 < 0.154 224/5 < 0.574 25/7 < 1.641 22.7 < 6.498 S K I L L S O B J E C T I V E S ■ ■Evaluate exponential functions. ■ ■Graph exponential functions. ■ ■Evaluate exponential functions of base e. ■ ■Apply exponential functions to economics and the natural sciences. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that irrational exponents lead to approximations. ■ ■Understand characteristics of exponential functions (implying domain, range, asymptotes, intercepts, etc.). ■ ■Understand that e is irrational and why it is the “natural” base. ■ ■Understand why compounding continuously results in higher interest than compounding daily. 5.1 EXPONENTIAL FUNCTIONS AND THEIR GRAPHS 5.1.1 S KILL Evaluate exponential functions. 5.1.1 CO NCE PTUAL Understand that irrational exponents lead to approximations. The domain of exponential functions, ƒ1x2 5 bx, is the set of all real numbers. It is important to note that in Chapter 0 we discussed exponents of the form bx, where x is an integer or a rational number. What happens if x is irrational? We can approximate the irrational number with a decimal approximation such as bp < b3.14 or b!2 < b1.41. Consider 7!3, and realize that the irrational number !3 is a decimal that never terminates or repeats: !3 5 1.7320508 . . . . We can show in advanced mathematics that there is a number 7!3, and although we cannot write it exactly, we can approximate the number. In fact, the closer the exponent is to !3, the closer the approximation is to 7!3. It is important to note that the properties of exponents (Chapter 0) hold when the exponent is any real number (rational or irrational). 71.7 < 27.3317 71.73 < 28.9747 71.732 < 29.0877 . . . 7!3 < 29.0906 x y 5 2x (x, y ) 22 222 5 1 22 5 1 4 a 22, 1 4 b 21 221 5 1 21 5 1 2 a 21, 1 2b 0 20 5 1 10, 12 1 21 5 2 11, 22 2 22 5 4 12, 42 3 23 5 8 13, 82 x y 5 22x (x, y ) 23 221232 5 23 5 8 123, 82 22 221222 5 22 5 4 122, 42 21 221212 5 21 5 2 121, 22 0 20 5 1 10, 12 1 221 5 1 21 5 1 2 a1, 1 2b 2 222 5 1 22 5 1 4 a2, 1 4b [CONCEPT CHECK] Find the approximation to f (x) 5 px for f (22.3); round to four decimal places. ANSWER 0.0719 ▼ 5.1.2 Graphs of Exponential Functions Let’s graph two exponential functions, y 5 2x and y 5 22x 5 A1 2B x, by plotting points. Notice that both graphs’ y-intercepts are 10, 12 (as shown to the right) and neither graph has an x-intercept. The x-axis is a horizontal asymptote for both graphs. The following box summarizes general characteristics of exponential functions. 5.1 Exponential Functions and Their Graphs 427 EXAMPLE 1 Evaluating Exponential Functions Let ƒ1x2 5 3x, g1x2 5 A1 4B x, and h1x2 5 10x22. Find the following values. If an approx-imation is required, approximate to four decimal places. a. ƒ122 b. ƒ1p2 c. gA2 3 2 B d. h12.32 e. ƒ102 f. g102 Solution: a. ƒ122 5 32 5 b. ƒ1p2 5 3p < 31.5443 c. gA23 2B 5 A1 4B 23/2 5 43/2 5 A!4B3 5 23 5 d. h12.32 5 102.322 5 100.3 < 1.9953 e. ƒ102 5 30 5 1 f. g102 5 A1 4B 0 5 1 Notice that parts (a) and (c) were evaluated exactly, whereas parts (b) and (d) required approximation using a calculator. YOUR T UR N Let ƒ1x2 5 2x and g1x2 5 A1 9B x and h1x2 5 5x22. Find the following values. Evaluate exactly when possible, and round to four decimal places when a calculator is needed. a. ƒ142 b. ƒ1p2 c. gA2 3 2B d. h12.92 In part (b), the p button on the calculator is selected. If we instead approximate p by 3.14, we get a slightly different approximation for the function value: ƒ1p2 5 3p < 33.14 < 31.4891 9 ▼ A N S W E R a. 16 b. 8.8250 c. 27 d. 4.2567 ▼ 8 x y (3, 8) (–3, 8) (1, 2) (–1, 2) (2, 4) (–2, 4) y = 2–x y = 2x 428 CHAPTER 5 Exponential and Logarithmic Functions Since exponential functions, ƒ1x2 5 bx, all go through the point 10, 12 and have the x-axis as a horizontal asymptote, we can find the graph by finding two additional points, as outlined in the following procedure. CHARACTERISTICS OF GRAPHS OF EXPONENTIAL FUNCTIONS ƒ1x2 5 bx, b . 0, b 2 1 ■ ■Domain: 12q, q2 ■ ■Range: 10, q2 ■ ■x-intercepts: none ■ ■y-intercept: 10, 12 ■ ■Horizontal asymptote: x-axis ■ ■The graph passes through 11, b2 and a 21, 1 bb. ■ ■As x increases, ƒ1x2 increases if b + 1 and decreases if 0 b 1. ■ ■The function ƒ is one-to-one. x y b > 1 0 < b < 1 PROCEDURE FOR GRAPHING ƒ1x2 5 bx Step 1: Label the point 10, 12 corresponding to the y-intercept ƒ102. Step 2: Find and label two additional points corresponding to ƒ1212 and ƒ112. Step 3: Connect the three points with a smooth curve, with the x-axis as the horizontal asymptote. EXAMPLE 2 Graphing Exponential Functions for b + 1 Graph the function ƒ1x2 5 5x. Solution: STEP 1: Label the y-intercept 10, 12. ƒ102 5 50 5 1 STEP 2: Label the point 11, 52. ƒ112 5 51 5 5 Label the point 121, 0.22. ƒ1212 5 521 5 1 5 5 0.2 STEP 3: Sketch a smooth curve through the three points with the x-axis as a horizontal asymptote. Domain: 12, 2 Range: 10,2 Y OUR TU R N Graph the function ƒ1x2 5 52x. T T T T T T ▼ x y (–1, 0.2) (0, 1) (1, 5) ▼ A N S W E R x y (1, 0.2) (0, 1) (–1, 5) 5.1.2 S KILL Graph exponential functions. 5.1.2 CO NCE PTUAL Understand characteristics of exponential functions (including domain, range, asymptotes, intercepts, etc.). 5.1 Exponential Functions and Their Graphs 429 Exponential functions, like all functions, can be graphed by point-plotting. We can also use transformations (horizontal and vertical shifting and reflection; Section 3.3) to graph exponential functions. EXAMPLE 3 Graphing Exponential Functions for b 1 Graph the function ƒ1x2 5 A2 5B x. Solution: STEP 1: Label the y-intercept 10, 12. ƒ102 5 a2 5b 0 5 1 STEP 2: Label the point 121, 2.52. ƒ1212 5 a2 5b 21 5 5 2 5 2.5 Label the point 11, 0.42. ƒ112 5 a2 5b 1 5 2 5 5 0.4 STEP 3: Sketch a smooth curve through the three points with the x-axis as a horizontal asymptote. Domain: 12q, q2 Range: 10, q2 x y (1, 0.4) (–1, 2.5) (0, 1) 2 5 –2 EXAMPLE 4 Graphing Exponential Functions Using a Horizontal or Vertical Shift a. Graph the function F1x2 5 2x21. State the domain and range of F. b. Graph the function G1x2 5 2x 1 1. State the domain and range of G. Solution (a): Identify the base function. ƒ 1x2 5 2x Identify the base function y-intercept and horizontal asymptote. 10, 12 and y 5 0 The graph of the function F is found by shifting the graph of the function ƒ to the right one unit. F 1x2 5 ƒ1x 2 12 Shift the y-intercept to the right one unit. 10, 12 shifts to 11, 12 The horizontal asymptote is not altered by a horizontal shift. y 5 0 Find additional points on the graph. F102 5 2021 5 221 5 1 2 y 2 intercept: A0, 1 2B F 122 5 2221 5 21 5 2 Sketch the graph of F1x2 5 2x21 with a smooth curve. Domain: 12,2 Range: 10, 2 x y f (x) = 2x F(x) = 2x–1 5 5 –5 (0, 1) (0, ) (1, 1) 1 2 [CONCEPT CHECK] State the y-intercept, domain, and range of F(x)5 bx 1 2 where b . 0 and b Þ 1. ANSWER y-intercept: (0, 3); domain: (2, ); range (2,) ▼ 430 CHAPTER 5 Exponential and Logarithmic Functions Solution (b): Identify the base function. ƒ 1x2 5 2x Identify the base function y-intercept and horizontal asymptote. 10, 12 and y 5 0 The graph of the function G is found by shifting the graph of the function ƒ up one unit. G1x2 5 ƒ1x2 1 1 Shift the y-intercept up one unit. 10, 12 shifts to 10, 22 Shift the horizontal asymptote up one unit. y 5 0 shifts to y 5 1 Find additional points on the graph. G112 5 21 1 1 5 2 1 1 5 3 G1212 5 221 1 1 5 1 2 1 1 5 3 2 Sketch the graph of G1x2 5 2x 1 1 with a smooth curve. Domain: 12,2 Range: 11, 2 x y f (x) = 2x G(x) = 2x + 1 5 5 –5 (0, 1) (0, 2) EXAMPLE 5 Graphing Exponential Functions Using Both Horizontal and Vertical Shifts Graph the function F1x2 5 3x11 2 2. State the domain and range of F. Solution: Identify the base function. ƒ 1x2 5 3x Identify the base function y-intercept and horizontal asymptote. 10, 12 and y 5 0 The graph of the function F is found by shifting the graph of the function ƒ to the left one unit and down two units. F 1x2 5 ƒ1x 1 12 2 2 Shift the y-intercept to the left one unit and down two units. 10, 12 shifts to 121, 212 Shift the horizontal asymptote down two units. y 5 0 shifts to y 5 22 Find additional points on the graph. F 102 5 3011 2 2 5 3 2 2 5 1 F 112 5 3111 2 2 5 9 2 2 5 7 Sketch the graph of F 1x2 5 3x11 2 2 with a smooth curve. Domain: 12, 2 Range: 122,2 Y OUR TU R N Graph ƒ1x2 5 2x13 2 1. State the domain and range of ƒ. ▼ x y f (x) = 3x F(x) = 3x+1 – 2 5 5 –5 –5 (0, 1) (–1, –1) ▼ A N S W E R Domain: 12,2 Range: 121, 2 x y (0, 7) (–1, 3) (–3, 0) (–2, 1) f (x) = 2x+3 – 1 5.1 Exponential Functions and Their Graphs 431 5.1.3 S K I L L Evaluate exponential functions of base e. 5.1.3 C ON C E P T U A L Understand that e is irrational and why it is the “natural” base. 5.1.3 The Natural Base e Any positive real number can serve as the base for an exponential function. A particular irrational number, denoted by the letter e, appears as the base in many applications, as you will soon see when we discuss continuous compounded interest. Although you will see 2 and 10 as common bases, the base that appears most often is e because e, as you will come to see in your further studies of mathematics, is the natural base. The exponential function with base e, ƒ1x2 5 ex, is called the exponential function or the natural exponential function. Mathematicians did not pull this irrational number out of a hat. The number e has many remarkable properties, but most simply, it comes from evaluating the expression a1 1 1 mb m as m increases without bound. Calculators have an ex button for approximating the natural exponential function. e < 2.71828 m a1 1 1 mb m 1 2 10 2.59374 100 2.70481 1000 2.71692 10,000 2.71815 100,000 2.71827 1,000,000 2.71828 Like all exponential functions of the form ƒ1x2 5 bx, ƒ 1x2 5 ex and ƒ 1x2 5 e2x have 10, 12 as their y-intercept and the x-axis as a horizontal asymptote, as shown in the figure below. x y (2, e2) (–2, e2) (1, e1) (–1, e1) f (x) = e x f (x) = e –x EXAMPLE 6 Evaluating the Natural Exponential Function Evaluate ƒ1x2 5 ex for the given x-values. Round your answers to four decimal places. a. x 5 1 b. x 5 21 c. x 5 1.2 d. x 5 20.47 Solution: a. ƒ112 5 e1 < 2.718281828 < 2.7183 b. ƒ1212 5 e21 < 0.367879441 < 0.3679 c. ƒ11.22 5 e1.2 < 3.320116923 < 3.3201 d. ƒ120.472 5 e20.47 < 0.625002268 < 0.6250 [CONCEPT CHECK] Since 2 , e , 3, what would 1/e lie between? ANSWER 1/2 and 1/3 ▼ 432 CHAPTER 5 Exponential and Logarithmic Functions 5.1.4 Applications of Exponential Functions Exponential functions describe either growth or decay. Populations and investments are often modeled with exponential growth functions, while the declining value of a used car and the radioactive decay of isotopes are often modeled with exponential decay functions. In Section 5.5, various exponential models will be discussed. In this section we discuss doubling time, half-life, and compound interest. A successful investment program, growing at about 7.2% per year, will double in size every 10 years. Let’s assume that you will retire at the age of 65. There is a saying: It’s not the first time your money doubles, it’s the last time that makes such a difference. As you may already know or as you will soon find, it is important to start investing early. Suppose Maria invests $5000 at age 25 and David invests $5000 at age 35. Let’s cal-culate how much will accrue from the initial $5000 investment by the time they each retire, assuming their money doubles every 10 years. AGE MARIA DAVID 25 $5,000 35 $10,000 $5,000 45 $20,000 $10,000 55 $40,000 $20,000 65 $80,000 $40,000 They each made a one-time investment of $5000. By investing 10 years sooner, Maria made twice what David made. A measure of growth rate is the doubling time, the time it takes for something to double. Often doubling time is used to describe populations. 5.1.4 S KILL Apply exponential functions to economics and the natural sciences. 5.1.4 CO NCE PTUAL Understand why compounding continuously results in higher interest than compounding daily. EXAMPLE 7 Graphing Exponential Functions with Base e Graph the function ƒ1x2 5 3 1 e2x. Solution: Note: The y-intercept is 10, 42 and the line y 5 3 is the horizontal asymptote. Y OUR TU R N Graph the function ƒ1x2 5 ex11 2 2. x y (1, 3 + e2) (0, 4) ▼ x f (x ) 5 3 1 e2x (x, y) 22 3.02 122, 3.022 21 3.14 11, 3.142 0 4 10, 42 1 10.39 11, 10.392 2 57.60 12, 57.602 ▼ A N S W E R x y (1, e2 – 2) (0, e1 – 2) (–1, –1) DOUBLING TIME GROWTH MODEL The doubling time growth model is given by P 5 P02t/d where P 5 Population at time t P0 5 Population at time t 5 0 d 5 Doubling time Note that when t 5 d, P 5 2P0 (population is equal to twice the original). 5.1 Exponential Functions and Their Graphs 433 The units for P and P0 are the same and can be any quantity (people, dollars, etc.). The units for t and d must be the same (years, weeks, days, hours, seconds, etc.). In the investment scenario with Maria and David, P0 5 $5000 and d 5 10 years, so the model used to predict how much money the original $5000 investment yielded is P 5 5000122t/10. Maria retired 40 years after the original investment, t 5 40, and David retired 30 years after the original investment, t 5 30. Maria: P 5 500012240/10 5 50001224 5 50001162 5 80,000 David: P 5 500012230/10 5 50001223 5 5000182 5 40,000 EXAMPLE 8 Doubling Time of Populations In 2004, the population in Kazakhstan, a country in Asia, reached 15 million. It is estimated that the population will double in 30 years. If the population continues to grow at the same rate, what will the population be 20 years from now? Round to the nearest million. Solution: Write the doubling model. P 5 P02t/d Substitute P0 5 15 million, d 5 30 years, and t 5 20 years. P 5 1512220/30 Simplify. P 5 151222/3 < 23.8110 In 20 years, there will be approximately 24 million people in Kazakhstan. YOUR T UR N What will the approximate population in Kazakhstan be in 40 years? Round to the nearest million. ▼ A N S W E R 38 million ▼ We now turn our attention from exponential growth to exponential decay, or neg-ative growth. Suppose you buy a brand-new car from a dealership for $24,000. The value of a car decreases over time according to an exponential decay function. The half-life of this particular car, or the time it takes for the car to depreciate 50%, is approximately 3 years. The exponential decay is described by A 5 A0 a1 2b t/h where A0 is the amount the car is worth (in dollars) when new (that is, when t 5 0), A is the amount the car is worth (in dollars) after t years, and h is the half-life in years. In our car scenario, A0 5 24,000 and h 5 3: A 5 24,000 a1 2b t/3 How much is the car worth after 3 years? 6 years? 9 years? 24 years? t 5 3: A 5 24,000 a1 2b 3/3 5 24,000 a1 2b 5 12,000 t 5 6: A 5 24,000 a1 2b 6/3 5 24,000 a1 2b 2 5 6000 t 5 9: A 5 24,000 a1 2b 9/3 5 24,000 a1 2b 3 5 3000 t 5 24: A 5 24,000 a1 2b 24/3 5 24,000 a1 2b 8 5 93.75 < 100 The car that was worth $24,000 new is worth $12,000 in 3 years, $6000 in 6 years, $3000 in 9 years, and about $100 in the junkyard in 24 years. 434 CHAPTER 5 Exponential and Logarithmic Functions EXAMPLE 9 Radioactive Decay The radioactive isotope of potassium 42K, which is used in the diagnosis of brain tumors, has a half-life of 12.36 hours. If 500 milligrams of potassium 42 are taken, how many milligrams will remain after 24 hours? Round to the nearest milligram. Solution: Write the half-life formula. A 5 A0 a1 2b t/h Substitute A0 5 500 mg, h 5 12.36 hours, t 5 24 hours. A 5 500 a1 2b 24/12.36 Simplify. A < 50010.26032 < 130.15 After 24 hours, there are approximately 130 milligrams of potassium 42 left. Y OUR TU R N How many milligrams of potassium are expected to be left in the body after 1 week? ▼ In Section 1.2, simple interest was discussed where the interest I is calculated based on the principal P, the annual interest rate r, and the time t in years, using the formula I 5 Prt. If the interest earned in a period is then reinvested at the same rate, future interest is earned on both the principal and the reinvested interest during the next period. Interest paid on both the principal and interest is called compound interest. COMPOUND INTEREST If a principal P is invested at an annual rate r compounded n times a year, then the amount A in the account at the end of t years is given by A 5 P a1 1 r nb nt The annual interest rate r is expressed as a decimal. The following list shows the typical number of times interest is compounded. Annually n 5 1 Monthly n 5 12 Semiannually n 5 2 Weekly n 5 52 Quarterly n 5 4 Daily n 5 365 EXAMPLE 10 Compound Interest If $3000 is deposited in an account paying 3% compounded quarterly, how much will you have in the account in 7 years? Solution: Write the compound interest formula. A 5 P a 1 1 r nb nt Substitute P 5 3000, r 5 0.03, n 5 4, and t 5 7. A 5 3000 a 1 1 0.03 4 b 142172 Simplify. A 5 300011.0075228 < 3698.14 You will have $3698.14 in the account. Y OUR TU R N If $5000 is deposited in an account paying 6% compounded annually, how much will you have in the account in 4 years? ▼ ▼ A N S W E R 0.04 mg 1less than 1 mg2 [CONCEPT CHECK] If we were to repeat Example 10 using interest compounding continuously, would you expect more or less money than if compounded quarterly? ANSWER More. See Example 11. ▼ ▼ A N S W E R $6312.38 5.1 Exponential Functions and Their Graphs 435 CONTINUOUS COMPOUND INTEREST If a principal P is invested at an annual rate r compounded continuously, then the amount A in the account at the end of t years is given by A 5 Pert The annual interest rate r is expressed as a decimal. Notice in the compound interest formula that as n increases the amount A also increases. In other words, the more times the interest is compounded per year, the more money you make. Ideally, your bank will compound your interest infinitely many times. This is called compounding continuously. We will now show the development of the compounding continuously formula, A 5 Pert. WORDS MATH Write the compound interest formula. A 5 P a1 1 r nb nt Note that r n 5 1 n/r and nt 5 an r brt. A 5 P a1 1 1 n/rb 1n/r2rt Let m 5 n r. A 5 P a 1 1 1 mb mrt Use the exponential property: xmrt 5 1xm2rt. A 5 Pc a 1 1 1 mb m d rt Earlier in this section, we showed that as m increases, a1 1 1 mb m approaches e. Therefore, as the number of times the interest is compounded approaches infinity, or as n S q, the amount in an account A 5 P a1 1 r nb nt approaches A 5 Pert. It is important to note that for a given interest rate, the highest return you can earn is by compounding continuously. EXAMPLE 11 Continuously Compounded Interest If $3000 is deposited in a savings account paying 3% a year compounded continuously, how much will you have in the account in 7 years? Solution: Write the continuous compound interest formula. A 5 Pert Substitute P 5 3000, r 5 0.03, and t 5 7. A 5 3000e10.032 172 Simplify. A < 3701.034 There will be $3701.03 in the account in 7 years. Note: In Example 10, we worked this same problem compounding quarterly, and the result was $3698.14. If the number of times per year interest is compounded increases, then the total interest earned that year also increases. YOUR T UR N If $5000 is deposited in an account paying 6% compounded continuously, how much will be in the account in 4 years? ▼ STUDY TIP If the number of times per year interest is compounded increases, then the total interest earned that year also increases. ▼ A N S W E R $6356.25 436 CHAPTER 5 Exponential and Logarithmic Functions In Exercises 1–10, evaluate exactly (without using a calculator). 1. 24 2. 34 3. 522 4. 423 5. 82/3 6. 272/3 7. A1 9B 23/2 8. A 1 16B 23/2 9. 250 10. 260 In Exercises 11–18, approximate with a calculator. Round your answer to four decimal places. 11. 41.2 12. 421.2 13. 5!2 14. 6!3 15. e2 16. e1/2 17. e2p 18. e2!2 In Exercises 19–26, for the functions f 1x2 5 3x, g1x2 5 A 1 16B x, and h1x2 5 10x11, find the function value at the indicated points. 19. ƒ132 20. h112 21. g1212 22. ƒ1222 23. gA2 1 2 B 24. gA23 2B 25. ƒ1e2 26. g1p2 [SEC TION 5.1] E X E R CI SES • S K I L L S Procedure for Graphing: ƒ1x2 5 bx Step 1: Label the point 10, 12 corresponding to the y-intercept ƒ102. Step 2: Find and label two additional points corresponding to ƒ1212 and ƒ112. Step 3: Connect the three points with a smooth curve with the x-axis as the horizontal asymptote. The Natural Exponential Function: ƒ1x2 5 ex • The irrational number e is called the natural base. • e 5 a1 1 1 mb m as m S q • e < 2.71828 Applications of Exponential Functions (all variables expressed in consistent units) 1. Doubling time: P 5 P02t/d • d is doubling time. • P is population at time t. • P0 is population at time t 5 0. 2. Half-life: A 5 A0A1 2B t/h • h is the half-life. • A is amount at time t. • A0 is amount at time t 5 0. 3. Compound interest 1P 5 principal, A 5 amount after t years, r 5 interest rate2 • Compounded n times a year: A 5 P a1 1 r nb nt • Compounded continuously: A 5 Pert In this section, we discussed exponential functions (constant base, variable exponent). General Exponential Functions: ƒ1x2 5 bx, b ∙1, and b . 0 1. Evaluating exponential functions • Exact (by inspection): ƒ1x2 5 2x ƒ132 5 23 5 8. • Approximate (with the aid of a calculator): ƒ1x2 5 2x ƒA!3B 5 2!3 < 3.322 2. Graphs of exponential functions • Domain: 12q, q2 and range: 10, q2. • The point 10, 12 corresponds to the y-intercept. • The graph passes through the points 11, b2 and a 21,1 bb. • The x-axis is a horizontal asymptote. • The function ƒ is one-to-one. ƒ1x2 5 bx, b . 0, b 2 1 [SEC TION 5.1] S U M MA RY x y b > 1 0 < b < 1 5.1 Exponential Functions and Their Graphs 437 In Exercises 27–32, match the graph with the function. 27. y 5 5x21 28. y 5 512x 29. y 5 25x 30. y 5 252x 31. y 5 1 2 52x 32. y 5 5x 2 1 a. x y (1, 4) b. x y (1, 0.8) (–1, –4) c. x y (0, 5) (2, 0.2) (1, 1) d. x y (0, –1) (–1, –5) e. x y (0, –1) (1, –5) (–1, –0.2) f. x y (2, 5) (1, 1) In Exercises 33–48, graph the exponential function using transformations. State the y-intercept, two additional points, the domain, the range, and the horizontal asymptote. 33. ƒ1x2 5 6x 34. ƒ1x2 5 7x 35. ƒ1x2 5 102x 36. ƒ1x2 5 42x 37. ƒ1x2 5 e2x 38. ƒ1x2 5 2ex 39. ƒ1x2 5 2x 2 1 40. ƒ1x2 5 3x 2 1 41. ƒ1x2 5 2 2 ex 42. ƒ1x2 5 1 1 e2x 43. ƒ1x2 5 ex11 2 4 44. ƒ1x2 5 ex21 1 2 45. ƒ1x2 5 3ex/2 46. ƒ1x2 5 2e2x 47. ƒ1x2 5 1 1 A1 2B x22 48. ƒ1x2 5 2 2 A1 3B x11 • A P P L I C A T I O N S 49. Population Doubling Time. In 2002, there were 7.1 million people living in London, England. If the population is expected to double by 2090, what is the expected population in London in 2050? 50. Population Doubling Time. In 2004, the population in Morganton, Georgia, was 43,000. The population in Morganton is expected to double by 2010. If the growth rate remains the same, what is the expected population in Morganton in 2020? 51. Investments. Suppose an investor buys land in a rural area for $1500 an acre and sells some of it 5 years later at $3000 an acre and the rest of it 10 years later at $6000. Write a function that models the value of land in that area, assuming the growth rate stays the same. What would the expected cost per acre be 30 years after the initial investment of $1500? 52. Salaries. Twin brothers, Collin and Cameron, get jobs immediately after graduating from college at the age of 22. Collin opts for the higher starting salary, $55,000, and stays with the same company until he retires at 65. His salary doubles every 15 years. Cameron opts for a lower starting salary, $35,000, but moves to a new job every 5 years; he doubles his salary every 10 years until he retires at 65. What is the annual salary of each brother upon retirement? 53. Radioactive Decay. A radioactive isotope of selenium 75Se used in the creation of medical images of the pancreas, has a half-life of 119.77 days. If 200 milligrams are given to a patient, how many milligrams are left after 30 days? 54. Radioactive Decay. The radioactive isotope indium-111 1111In2, used as a diagnostic tool for locating tumors associated with prostate cancer, has a half-life of 2.807 days. If 300 milligrams are given to a patient, how many milligrams will be left after a week? 55. Depreciation of Furniture. A couple buys a new bedroom set for $8000 and 10 years later sells it for $4000. If the depreciation continues at the same rate, how much would the bedroom set be worth in 4 more years? 56. Depreciation of a Computer. A student buys a new laptop for $1500 when she arrives as a freshman. A year later, the computer is worth approximately $750. If the depreciation continues at the same rate, how much would she expect to sell her laptop for when she graduates 4 years after she bought it? 57. Compound Interest. If you put $3200 in a savings account that earns 2.5% interest per year compounded quarterly, how much would you expect to have in that account in 3 years? 438 CHAPTER 5 Exponential and Logarithmic Functions 58. Compound Interest. If you put $10,000 in a savings account that earns 3.5% interest per year compounded annually, how much would you expect to have in that account in 5 years? 59. Compound Interest. How much money should you put in a savings account now that earns 5% a year compounded daily if you want to have $32,000 in 18 years? 60. Compound Interest. How much money should you put in a savings account now that earns 3.0% a year compounded weekly if you want to have $80,000 in 15 years? 61. Compound Interest. If you put $3200 in a savings account that pays 2% a year compounded continuously, how much will you have in the account in 15 years? 62. Compound Interest. If you put $7000 in a money market account that pays 4.3% a year compounded continuously, how much will you have in the account in 10 years? 63. Compound Interest. How much money should you deposit into a money market account that pays 5% a year compounded continuously to have $38,000 in the account in 20 years? 64. Compound Interest. How much money should you deposit into a certificate of deposit that pays 6% a year compounded continuously to have $80,000 in the account in 18 years? For Exercises 65 and 66, refer to the following: Exponential functions can be used to model the concentration of a drug in a patient’s body. Suppose the concentration of Drug X in a patient’s bloodstream is modeled by C1t2 5 C0e2rt where C1t2 represents the concentration at time t (in hours), C0 is the concentration of the drug in the blood immediately after injection, and r . 0 is a constant indicating the removal of the drug by the body through metabolism and/or excretion. The rate constant r has units of 1/time (1/hr). It is important to note that this model assumes that the blood concentration of the drug C0 peaks immediately when the drug is injected. 65. Health/Medicine. After an injection of Drug Y, the concentration of the drug in the bloodstream drops at the rate of 0.020 1/hr. Find the concentration, to the nearest tenth, of the drug 20 hours after receiving an injection with initial concentration of 5.0 mg/L. 66. Health/Medicine. After an injection of Drug Y, the concentration of the drug in the bloodstream drops at the rate of 0.009 1/hr. Find the concentration, to the nearest tenth, of the drug 4 hours after receiving an injection with initial concentration of 4.0 mg/L. For Exercises 67 and 68, refer to the following: The demand for a product, in thousands of units, can be expressed by the exponential demand function D1p2 5 230010.852 p where p is the price per unit. 67. Economics. Find the demand for the product by completing the following table. p (PRICE PER UNIT) D (p )—DEMAND FOR PRODUCT IN UNITS 1.00 5.00 10.00 20.00 40.00 60.00 80.00 90.00 68. Economics. Evaluate D(91) and interpret what this means in terms of demand. • C A T C H T H E M I S T A K E In Exercises 69–72, explain the mistake that is made. 69. Evaluate the expression 421/2. Solution: 421/2 5 42 5 16 The correct value is 1 2. What mistake was made? 70. Evaluate the function for the given x: ƒ1x2 5 4x for x 5 3 2. Solution: ƒa3 2b 5 43/2 5 43 42 5 64 16 5 4 The correct value is 8. What mistake was made? 71. If $2000 is invested in a savings account that earns 2.5% interest compounding continuously, how much will be in the account in one year? Solution: Write the compound continuous interest formula. A 5 Pert Substitute P 5 2000, r 5 2.5, and t 5 1. A 5 2000e 12.52 112 Simplify. A 5 24,364.99 This is incorrect. What mistake was made? 72. If $5000 is invested in a savings account that earns 3% interest compounding continuously, how much will be in the account in 6 months? Solution: Write the compound continuous interest formula. A 5 Pert Substitute P 5 5000, r 5 0.03, and t 5 6. A 5 5000e 10.032 162 Simplify. A 5 5986.09 This is incorrect. What mistake was made? 5.1 Exponential Functions and Their Graphs 439 73. The function ƒ1x2 5 2e2x has the y-intercept 10,12. 74. The function ƒ1x2 5 2e2x has a horizontal asymptote along the x-axis. 75. The functions y 5 32x and y 5 A1 3B x have the same graphs. 76. e 5 2.718. 77. Plot ƒ1x2 5 3x and its inverse on the same graph. 78. Plot ƒ1x2 5 ex and its inverse on the same graph. In Exercises 73–76, determine whether each statement is true or false. • C O N C E P T U A L 79. Graph ƒ1x2 5 e0x0. 80. Graph ƒ1x2 5 e20x0. 81. Find the y-intercept and horizontal asymptote of ƒ1x2 5 be2x11 2 a. 82. Find the y-intercept and horizontal asymptote of ƒ1x2 5 a 1 bex11. 83. Graph ƒ1x2 5 b0x0, b . 1, and state the domain. 84. Graph the function ƒ1x2 5 b ax x , 0 a2x x $ 0 where a . 1. • C H A L L E N G E 85. Plot the function y 5 a1 1 1 xb x . What is the horizontal asymptote as x increases? 86. Plot the functions y 5 2x, y 5 ex, and y 5 3x in the same viewing screen. Explain why y 5 ex lies between the other two graphs. 87. Plot y1 5 ex and y2 5 1 1 x 1 x2 2 1 x3 6 1 x4 24 in the same viewing screen. What do you notice? 88. Plot y1 5 e2x and y2 5 1 2 x 1 x2 2 2 x3 6 1 x4 24 in the same viewing screen. What do you notice? 89. Plot the functions ƒ1x2 5 a1 1 1 xb x , g 1x2 5 a1 1 2 xb x , and h 1x2 5 a1 1 2 xb 2x in the same viewing screen. Compare their horizontal asymptotes as x increases. What can you say about the function values of ƒ, g, and h in terms of the powers of e as x increases? 90. Plot the functions ƒ1x2 5 a1 1 1 xb x , g 1x2 5 a1 2 1 xb x , and h 1x2 5 a1 2 2 xb x in the same viewing screen. Compare their horizontal asymptotes as x increases. What can you say about the function values of ƒ, g, and h in terms of the powers of e as x increases? For Exercises 91 and 92, refer to the following: Newton’s Law of Heating and Cooling: Have you ever heated soup in a microwave and, upon taking it out, have it seem to cool considerably in the matter of minutes? Or has your ice-cold soda become tepid in just moments while outside on a hot summer’s day? This phenomenon is based on the so-called Newton’s Law of Heating and Cooling. Eventually, the soup will cool so that its temperature is the same as the temperature of the room in which it is being kept, and the soda will warm until its temperature is the same as the outside temperature. 91. Consider the following data: a. Form a scatterplot for this data. b. Use ExpReg to find the best fit exponential function for this data set, and superimpose its graph on the scatterplot. How good is the fit? c. Use the best fit exponential curve from (b) to answer the following: i. What will the predicted temperature of the soup be at 6 minutes? ii. What was the temperature of the soup the moment it was taken out of the microwave? d. Assume the temperature of the house is 72° F. According to Newton’s Law of Heating and Cooling, the temperature of the soup should approach 72°. In light of this, comment on the shortcomings of the best ﬁt exponential curve. 92. Consider the following data: a. Form a scatterplot for this data. b. Use ExpReg to find the best fit exponential function for this data set, and superimpose its graph on the scatterplot. How good is the fit? c. Use the best fit exponential curve from (b) to answer the following: i. What will the predicted temperature of the soda be at 10 minutes? ii. What was the temperature of the soda the moment it was taken out of the refrigerator? d. Assume the temperature of the house is 90° F. According to Newton’s Law of Heating and Cooling, the temperature of the soda should approach 90°. In light of this, comment on the shortcomings of the best fit exponential curve. TIME (IN MINUTES) 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 TEMPERATURE OF SOUP (IN DEGREES FAHRENHEIT) 203 200 195 188 180 171 160 151 TIME (IN MINUTES) 1 2 3 4 5 6 7 8 TEMPERATURE OF SODA (IN DEGREES FAHRENHEIT) 45 48 49 53 57 61 68 75 • T E C H N O L O G Y 440 CHAPTER 5 Exponential and Logarithmic Functions S K I L L S O B J E C T I V E S ■ ■Evaluate logarithmic expressions. ■ ■Approximate common and natural logarithms using a calculator. ■ ■Graph logarithmic functions. ■ ■Apply logarithmic functions to problems in the natural sciences and engineering. C O N C E P T U A L O B J E C T I V E S ■ ■Interpret logarithmic functions as inverses of exponential functions. ■ ■Recognize when a logarithm or the value of a logarithmic function can be evaluated exactly or when it must be approximated. ■ ■Understand the inverse relationship between the characteristics of logarithmic functions and exponential functions. ■ ■Understand that logarithmic functions allow very large ranges of numbers in science and engineering applications to be represented on a smaller scale. 5.2 LOGARITHMIC FUNCTIONS AND THEIR GRAPHS 5.2.1 Evaluating Logarithms In Section 5.1, we found that the graph of an exponential function ƒ1x2 5 bx passes through the point 10, 12, with the x-axis as a horizontal asymptote. The graph passes both the vertical line test (for a function) and the horizontal line test (for a one-to-one function), and therefore an inverse exists. We will now apply the technique outlined in Section 3.5 to find the inverse of ƒ1x2 5 bx: WORDS MATH Let y 5 ƒ1x2. y 5 bx Interchange x and y. x 5 by Solve for y. y 5 ? We see that y is the exponent that b is raised to in order to obtain x. We call this exponent a logarithm (or “log” for short). WORDS MATH x 5 b y is equivalent to y 5 logb x. y 5 logb x Let y 5 ƒ211x2. ƒ211x2 5 logb x DEFINITION Logarithmic Function For x . 0, b . 0, and b ∙ 1, the logarithmic function with base b is denoted ƒ1x2 5 logb x, where y 5 logb x if and only if x 5 by We read logbx as “log base b of x.” This definition says that x 5 by (exponential form) and y 5 logb x (logarithmic form) are equivalent. One way to remember this relationship is by adding arrows to the logarithmic form: logbx 5 y ⇔ by 5 x ➤ ➤ 5.2.1 S KILL Evaluate logarithmic expressions. 5.2.1 CO NCE PTUAL Interpret logarithmic functions as inverses of exponential functions. STUDY TIP logb x 5 y is equivalent to by 5 x. 5.2 Logarithmic Functions and Their Graphs 441 Some logarithms can be found exactly, whereas others must be approximated. Example 3 illustrates how to find the exact value of a logarithm. Example 4 illustrates approximating values of logarithms with a calculator. EXAMPLE 1 Changing from Logarithmic Form to Exponential Form Write each equation in its equivalent exponential form. a. log2 8 5 3 b. log9 3 5 1 2 c. log5 A 1 25B 5 22 Solution: a. log2 8 5 3 is equivalent to 23 5 8 b. log9 3 5 1 2 is equivalent to 91/2 5 3 c. log5 A 1 25B 5 22 is equivalent to 522 5 1 25 Y OUR T UR N Write each equation in its equivalent exponential form. a. log3 9 5 2 b. log16 4 5 1 2 c. log2 A1 8B 5 23 ▼ EXAMPLE 2 Changing from Exponential Form to Logarithmic Form Write each equation in its equivalent logarithmic form. a. 16 5 24 b. 9 5 !81 c. 1 9 5 322 d. xa 5 z Solution: a. 16 5 24 is equivalent to log2 16 5 4 b. 9 5 !81 5 811/2 is equivalent to log81 9 5 1 2 c. 1 9 5 322 is equivalent to log3A1 9B 5 22 d. xa 5 z is equivalent to logxz 5 a YOUR T UR N Write each equation in its equivalent logarithmic form. a. 81 5 92 b. 12 5 !144 c. 1 49 5 722 d. yb 5 w ▼ ▼ A N S W E R a. 32 5 9 b. 161/2 5 4 c. 223 5 1 8 ▼ A N S W E R a. log9 81 5 2 b. log144 12 5 1 2 c. log7 A 1 49B 5 22 d. logy w 5 b EXAMPLE 3 Finding the Exact Value of a Logarithm Find the exact value of: a. log3 81 b. log169 13 c. log5 A1 5B Solution (a): The logarithm has some value. Let’s call it x. log3 81 5 x Change from logarithmic to exponential form. 3x 5 81 3 raised to what power is 81? 34 5 81 x 5 4 Change from exponential to logarithmic form. log3 81 5 4 Solution (b): The logarithm has some value. Let’s call it x. log169 13 5 x Change from logarithmic to exponential form. 169x 5 13 169 raised to what power is 13? 1691/2 5 !169 5 13 x 5 1 2 Change from exponential to logarithmic form. log169 13 5 1 2 [CONCEPT CHECK] Write loga (x) 5 5 in exponential form. ANSWER a5 5 x ▼ 442 CHAPTER 5 Exponential and Logarithmic Functions 5.2.2 Common and Natural Logarithms Two logarithmic bases that arise frequently are base 10 and base e. The logarithmic function of base 10 is called the common logarithmic function. Since it is common, ƒ1x2 5 log10 x is often expressed as ƒ1x2 5 log x. Thus, if no explicit base is indicated, base 10 is implied. The logarithmic function of base e is called the natural logarithmic function. The natural logarithmic function ƒ1x2 5 loge x is often expressed as ƒ1x2 5 ln x. Both the LOG and LN buttons appear on scientific and graphing calculators. For the logarithms (not the functions) we say “the log” (for base 10) and “the natural log” (for base e). Earlier in this section, we evaluated logarithms exactly by converting to exponen-tial form and identifying the exponent. For example, to evaluate log10 100, we ask the question, 10 raised to what power is 100? The answer is 2. Calculators enable us to approximate logarithms. For example, evaluate log10 233. We are unable to evaluate this exactly by asking the question, 10 raised to what power is 233? Since 102 , 10x , 103, we know the answer x must lie between 2 and 3. Instead, we use a calculator to find an approximate value 2.367. Solution (c): The logarithm has some value. Let’s call it x. log5 a1 5b 5 x Change from logarithmic to exponential form. 5x 5 1 5 5 raised to what power is 1 5? 521 5 1 5 x 5 21 Change from exponential to logarithmic form. log5 a1 5b 5 21 Y OUR TU R N Evaluate the given logarithms exactly. a. log2 1 2 b. log100 10 c. log10 1000 ▼ 5.2.2 S KILL Approximate common and natural logarithms using a calculator. 5.2.2 CO NCE PTUAL Recognize when a logarithm or the value of a logarithmic function can be evaluated exactly or when it must be approximated. [CONCEPT CHECK] Find log(10), log(50), and log(100). ANSWER 1; 1.699; 2 ▼ ▼ A N S W E R a. log2 1 2 5 21 b. log100 10 5 1 2 c. log10 1000 5 3 STUDY TIP • log10 x 5 log x. No explicit base implies base 10. • loge x 5 ln x STUDY TIP Logarithms can only be evaluated for positive arguments. EXAMPLE 4 Using a Calculator to Evaluate Common and Natural Logarithms Use a calculator to evaluate the common and natural logarithms. Round your answers to four decimal places. a. log 415 b. ln 415 c. log 1 d. ln 1 e. log1222 f. ln1222 Solution: a. log14152 < 2.618048097 < 2.6180 b. ln14152 < 6.02827852 < 6.0283 c. log112 5 0 d. ln112 5 0 e. log1222 undefined f. ln1222 undefined Parts (c) and (d) in Example 4 illustrate that all logarithmic functions pass through the point (1, 0). Parts (e) and (f) in Example 4 illustrate that the domains of logarithmic functions are positive real numbers. 5.2 Logarithmic Functions and Their Graphs 443 5.2.3 Graphs of Logarithmic Functions The general logarithmic function y 5 logb x is defined as the inverse of the exponential function y 5 bx. Therefore, when these two functions are plotted on the same graph, they are symmetric about the line y 5 x. Notice the symmetry about the line y 5 x when y 5 bx and y 5 logbx are plotted on the same graph. 5.2.3 S K IL L Graph logarithmic functions. 5.2.3 C O N C E P T U A L Understand the inverse relationship between the characteristics of logarithmic functions and exponential functions. x y (1, 0) (0, 1) b > 1 y = logb x y = bx x y (1, 0) (0, 1) 0 < b < 1 y = logb x y = bx EXAMPLE 5 Finding the Domain of a Shifted Logarithmic Function Find the domain of each of the given logarithmic functions. a. ƒ1x2 5 logb1x 2 42 b. g1x2 5 logb15 2 2x2 Solution (a): Set the argument greater than zero. x 2 4 . 0 Solve the inequality. x . 4 Write the domain in interval notation. 14,2 Comparison of Inverse Functions: f (x ) 5 logb x and f 21(x ) 5 bx EXPONENTIAL FUNCTION LOGARITHMIC FUNCTION y 5 bx y 5 logbx y-intercept 10, 12 x-intercept 11, 02 Domain 12q, q2 Domain 10, q2 Range 10, q2 Range 12q, q2 Horizontal asymptote: x-axis Vertical asymptote: y-axis x y (0, 1) (–1, 1/b) (1, b) b > 1 x y (1, 0) (1/b, –1) (b, 1) b > 1 Additionally, the domain of one function is the range of the other, and vice versa. When dealing with logarithmic functions, special attention must be paid to the domain of the function. The domain of y 5 logbx is 10, 2. In other words, you can only take the log of a positive real number, x . 0. 444 CHAPTER 5 Exponential and Logarithmic Functions Recall from Section 3.3 that a technique for graphing general functions is transformation of known functions. For example, to graph ƒ1x2 5 1x 2 322 1 1, we start with the known parabola y 5 x2, whose vertex is at 10, 02, and we shift that graph to the right three units and up one unit. We use the same technique for graphing logarithmic functions. To graph y 5 logb1x 1 22 2 1, we start with the graph of y 5 logb1x2 and shift the graph to the left two units and down one unit. Solution (b): Set the argument greater than zero. 5 2 2x . 0 Solve the inequality. 22x . 25 2x , 5 x , 5 2 Write the domain in interval notation. a 2q, 5 2 b Y OUR TU R N Find the domain of the given logarithmic functions. a. ƒ1x2 5 logb1x 1 22 b. g1x2 5 logb13 2 5x2 ▼ It is important to note that when finding the domain of a logarithmic function, we set the argument strictly greater than zero and solve. EXAMPLE 6 Finding the Domain of a Logarithmic Function with a Complicated Argument Find the domain of each of the given logarithmic functions. a. ln Ax2 2 9B b. logA0 x 1 1 0 B Solution (a): Set the argument greater than zero. x2 2 9 . 0 Solve the inequality. 12q, 232 ∪ 13, q2 Solution (b): Set the argument greater than zero. 0 x 1 1 0 . 0 Solve the inequality. x 2 21 Write the domain in interval notation. 12q, 212 ∪ 121, q2 Y OUR TU R N Find the domain of each of the given logarithmic functions. a. ln 1x2 2 42 b. logA0 x 2 3 0 B ▼ ▼ A N S W E R a. 12q , 222 ∪ 12, q 2 b. 12 q , 32 ∪ 13, q 2 STUDY TIP Review solving inequalities in Sections 1.5, 1.6, and 1.7. ▼ A N S W E R a. 122,2 b. A2q, 3 5B 5.2 Logarithmic Functions and Their Graphs 445 EXAMPLE 7 Graphing Logarithmic Functions Using Horizontal and Vertical Shifts Graph the functions, and state the domain and range of each. a. y 5 log2 1x 2 32 b. log2 1x2 2 3 Solution: Identify the base function. y 5 log2x Label key features of y 5 log2x. x-intercept: 11, 02 Vertical asymptote: x 5 0 Additional points: 12, 12, 14, 22 x 10 5 –5 (1, 0) y = log2 x a. Shift the base function to the right three units. x-intercept: 14, 02 Vertical asymptote: x 5 3 Additional points: 15, 12, 17, 22 Domain: 13, 2 Range: 12, 2 x 10 5 –5 (4, 0) y = log2(x – 3) b. Shift the base function down three units. x-intercept: 11, 232 Vertical asymptote: x 5 0 Additional points: 12, 222, 14, 212 Domain: 10,2 Range: 12,2 x 10 5 –5 (1, –3) y = log2x – 3 Y OUR T UR N Graph the functions and state the domain and range of each. a. y 5 log3 x b. y 5 log3 1x 1 32 c. log3 1x2 1 1 ▼ All of the transformation techniques (shifting, reflection, and compression) discussed in Chapter 3 also apply to logarithmic functions. For example, the graphs of 2log2 x and log2 12x2 are found by reflecting the graph of y 5 log2 x about the x-axis and y-axis, respectively. ▼ A N S W E R a. Domain: 10, 2 Range: 12, 2 b. Domain: 123,2 Range: 12,2 c. Domain: 10, 2 Range: 12, 2 x x = –3 y (1, 0) (–2, 0) (1, 1) x (2, 1) (2, –1) (–4, 2) (4, 2) (4, –2) (–2, 1) 446 CHAPTER 5 Exponential and Logarithmic Functions 5.2.4 Applications of Logarithms Logarithms are used to make a large range of numbers manageable. For example, to create a scale to measure a human’s ability to hear, we must have a way to measure the sound intensity of an explosion, even though that intensity can be more than a trillion 110122 times greater than that of a soft whisper. Decibels in engineering and physics, pH in chemistry, and the Richter scale for earthquakes are all applications of logarithmic functions. The decibel is a logarithmic unit used to measure the magnitude of a physical quantity relative to a specified reference level. The decibel (dB) is employed in many engineering and science applications. The most common application is the intensity of sound. [CONCEPT CHECK] Find the x-intercept, domain, and range of logb(x 2 1). ANSWER x-intercept: (2, 0); Domain: (1,); Range: (2,). ▼ 5.2.4 S KILL Apply logarithmic functions to problems in the natural sciences and engineering. 5.2.4 CO NCE PTUAL Understand that logarithmic functions allow very large ranges of numbers in science and engineering applications to be represented on a smaller scale. EXAMPLE 8 Graphing Logarithmic Functions Using Transformations Graph the function ƒ1x2 5 2log21x 2 32 and state its domain and range. Solution: Graph y 5 log2 x. x-intercept: 11, 02 Vertical asymptote: x 5 0 Additional points: 12, 12, 14, 22 x y 10 5 –5 (4, 2) (2, 1) (1, 0) Graph y 5 log2 1x 2 32 by shifting y 5 log2 x to the right three units. x-intercept: 14, 02 Vertical asymptote: x 5 3 Additional points: 15, 12, 17, 22 x y 10 5 –5 (7, 2) (5, 1) (4, 0) Graph y 5 2log2 1x 2 32 by reflecting y 5 log2 1x 2 32 about the x-axis. x-intercept: 14, 02 Vertical asymptote: x 5 3 Additional points: 15, 212, 17, 222 x y 10 5 –5 (4, 0) (5, –1) (7, –2) Domain: 13, 2 Range: 12, `2 5.2 Logarithmic Functions and Their Graphs 447 DEFINITION Decibel (Sound) The decibel is defined as D 5 10 log a I IT b where D is the decibel level (dB), I is the intensity of the sound measured in watts per square meter, and IT is the intensity threshold of the least audible sound a human can hear. The human average threshold is IT 5 1 3 10212 W/m2. Notice that when I 5 IT, then D 5 10 log1 5 0 dB. People who work professionally with sound, such as acoustics engineers and medical hearing specialists, refer to this threshold level IT as “0 dB.” The following table illustrates typical sounds we hear and their corresponding decibel levels. SOUND SOURCE SOUND INTENSITY (W/m2) DECIBELS (dB) Threshold of hearing 1.0 3 10212 0 Vacuum cleaner 1.0 3 1024 80 iPod 1.0 3 1022 100 Jet engine 1.0 3 103 150 For example, a whisper (approximately 0 dB) from someone standing next to a jet engine (150 dB) might go unheard because when these are added, we get approximately 150 dB (the jet engine). 0 [CONCEPT CHECK] The sound intensity of a jet engine is ___ times the sound intensity of an iPhone, but it’s only _____ more decibels. ANSWER 100,000; 50 ▼ ▼ A N S W E R 160 dB EXAMPLE 9 Calculating Decibels of Sounds Suppose you have seats to a concert given by your favorite musical artist. Calculate the approximate decibel level associated with the typical sound intensity, given I 5 1 3 1022 W/m2. Solution: Write the decibel-scale formula. D 5 10 log a I IT b Substitute I 5 1 3 1022 W/m2 and IT 5 1 3 10212 W/m2. D 5 10 log a 1 3 1022 1 3 10212b Simplify. D 5 10 log110102 Recall that the implied base for log is 10. D 5 10 log10110102 Evaluate the right side. Clog10110102 5 10D D 5 10 ? 10 D 5 100 The typical sound level on the front row of a rock concert is 100 dB . YOU R T UR N Calculate the approximate decibels associated with a sound so loud it will cause instant perforation of the eardrums, I 5 1 3 104 W/m2. ▼ 448 CHAPTER 5 Exponential and Logarithmic Functions A logarithmic scale expresses the logarithm of a physical quantity instead of the quantity itself. In music, the pitch is the perceived fundamental frequency of sound. The note A above middle C on a piano has the pitch associated with a pure tone of 440 hertz (Hz). An octave is the interval between one musical pitch and another with either double or half its frequency. For example, if a note has a frequency of 440 Hz, then the note an octave above it has a frequency of 880 Hz, and the note an octave below it has a frequency of 220 Hz. Therefore, the ratio of two notes an octave apart is 2:1. The following table lists the frequencies associated with A notes. NOTE A1 A2 A3 A4 A5 A6 A7 Frequency (Hz) 55 110 220 440 880 1760 3520 Octave with respect to A4 23 22 21 0 11 12 13 The Richter scale (earthquakes) is another application of logarithms. DEFINITION Richter Scale The magnitude M of an earthquake is measured using the Richter scale M 5 2 3 log a E E0 b where M is the magnitude E is the seismic energy released by the earthquake (in joules) E0 is the energy released by a reference earthquake E0 5 104.4 joules EXAMPLE 10 Calculating the Magnitude of an Earthquake On October 17, 1989, just moments before game 3 of the World Series between the Oakland A’s and the San Francisco Giants was about to start—with 60,000 fans in Candlestick Park—a devastating earthquake erupted. Parts of interstates and bridges collapsed, and President George H. W. Bush declared the area a disaster zone. The earthquake released approximately 1.12 3 1015 joules of energy. Calculate the magnitude of the earthquake using the Richter scale. Solution: Write the Richter scale formula. M 5 2 3 log a E E0 b Substitute E 5 1.12 3 1015 and E0 5 104.4. M 5 2 3 log a1.12 3 1015 104.4 b Simplify. M 5 2 3 log A1.12 3 1010.6B Approximate the logarithm using a calculator. M < 2 3 110.652 < 7.1 The 1989 earthquake in California measured 7.1 on the Richter scale. Y OUR TU R N On May 3, 1996, Seattle experienced a moderate earthquake. The energy that the earthquake released was approximately 1.12 3 1012 joules. Calculate the magnitude of the 1996 Seattle earthquake using the Richter scale. ▼ ▼ A N S W E R 5.1 5.2 Logarithmic Functions and Their Graphs 449 Middle C A0 27.500 A0# 29.135 C1# 34.648 D1# 38.891 F1# 46.249 G1# 51.913 A1# 58.270 C2# 69.296 D2# 77.782 F2# 185.00 G2# 103.83 A2# 116.54 C3# 138.59 D3# 155.56 F3# 185.00 G3# 207.65 A3# 233.08 C4# 277.18 D4# 311.13 F4# 369.99 G4# 415.30 A4# 466.16 C5# 554.37 D5# 622.25 F5# 739.99 G5# 830.61 A5# 932.33 C6# 1108.7 D6# 1244.5 F6# 1480.0 G6# 1661.2 A6# 1864.7 C7# 2217.5 D7# 2489.0 F7# 2960.0 G7# 3322.4 A7# 3729.3 B0 30.868 C1 32.703 D1 36.708 E1 41.203 F1 43.654 G1 48.999 A1 55.000 B1 61.735 C2 65.406 D2 73.416 E2 82.407 F2 87.307 G2 97.999 A2 110.00 B2 123.47 C3 130.81 D3 146.83 E3 164.81 F3 174.61 G3 196.00 A3 220.00 B3 246.94 C4 261.63 D4 293.66 E4 329.63 F4 349.23 G4 392.00 A4 440.00 B4 493.88 C5 523.25 D5 587.33 E5 659.25 F5 698.46 G5 783.99 A5 880.00 B5 987.77 C6 1046.5 D6 1174.7 E6 1318.5 F6 1396.9 G6 1568.0 A6 1760.0 B6 1979.5 C7 2093.0 D7 2349.3 E7 2637.0 F7 2793.8 G7 3136.0 A7 3520.0 B7 3951.1 C8 4186.0 We can graph Frequency of note 440 Hz on the horizontal axis and the octave (with respect to A4) on the vertical axis. Octave 10 9 8 7 6 5 4 3 2 1 5 –5 Frequency of note 440 Hz ( ) (2, 1) (4, 2) (8, 3) If we instead graph the logarithm of this quantity, log cFrequency of note 440 Hz d, we see that using a logarithmic scale expresses octaves linearly (up or down an octave). In other words, an “octave” is a purely logarithmic concept. When a logarithmic scale is used, we typically classify a graph one of two ways: • Log-log plot (both the horizontal and vertical axes use logarithmic scales) • Semilog plot (one of the axes uses a logarithmic scale) The second graph with octaves on the vertical axis and the log of the ratio of frequencies on the horizontal axis is called a semilog plot. Octave Semilog Plot 0.1 –0.1 5 –5 Frequency of note 440 Hz Log10( ) EXAMPLE 11 Graphing Using a Logarithmic Scale Frequency is inversely proportional to the wavelength: In a vacuum ƒ 5 c l, where ƒ is the frequency (in hertz2, c 5 3.0 3 108 m/s is the speed of light in a vacuum, and l is the wavelength in meters. Graph frequency versus wavelength using a log-log plot. 450 CHAPTER 5 Exponential and Logarithmic Functions Solution: Let wavelength range from microns 110262 to hundreds of meters 11022 by powers of 10 along the horizontal axis. l f 5 3.0 3 108 l 1026 3.0 3 1014 1025 3.0 3 1013 1024 3.0 3 1012 (TeraHertz: THz) 1023 3.0 3 1011 1022 3.0 3 1010 1021 3.0 3 109 (GigaHertz: GHz) 100 3.0 3 108 101 3.0 3 107 102 3.0 3 106 (MegaHertz: MHz) The logarithmic scales allow us to represent a large range of numbers. In this graph, the x-axis ranges from microns, 1026 meters, to hundreds of meters, and the y-axis ranges from megahertz (MHz), 106 hertz, to hundreds of terahertz (THz), 1012 hertz. Hz 10–6 10–4 10–2 100 106 107 108 109 1010 1011 1012 1013 1014 102 m Lasers Log-Log Plot Short radio waves Broadcast band Wavelength (meters) Frequency (hertz) Evaluating Logarithms • Exact: Convert to exponential form first, then evaluate. • Approximate: Natural and common logarithms with calculators. In this section, logarithmic functions were defined as inverses of exponential functions. y 5 logb x is equivalent to x 5 by NAME EXPLICIT BASE IMPLICIT BASE Common logarithm ƒ1x2 5 log10x ƒ1x2 5 logx Natural logarithm ƒ1x2 5 logex ƒ1x2 5 lnx [SEC TION 5. 2] S U M MA RY Graphs of Logarithmic Functions EXPONENTIAL FUNCTION LOGARITHMIC FUNCTION y 5 bx y 5 logb x y-intercept: 10, 12 x-intercept: 11, 02 Domain: 1 2,2 Domain: 10, 2 Range: 10,2 Range: 1 2,2 Horizontal asymptote: x-axis Vertical asymptote: y-axis x y (0, 1) (1, b) b > 1 1 b (–1, ) y = bx x y (b, 1) (1, 0) 1 b ( , –1) b > 1 y = logb x 5.2 Logarithmic Functions and Their Graphs 451 In Exercises 1–20, write each logarithmic equation in its equivalent exponential form. 1. log5 125 5 3 2. log3 27 5 3 3. log81 3 5 1 4 4. log121 11 5 1 2 5. log2 A 1 32B 5 25 6. log3 A 1 81B 5 24 7. log 0.01 5 22 8. log 0.0001 5 24 9. log 10,000 5 4 10. log 1000 5 3 11. log1/4 1642 5 23 12. log1/6 1362 5 22 13. 21 5 ln A1 eB 14. 1 5 ln e 15. ln 1 5 0 16. log 1 5 0 17. ln 5 5 x 18. ln 4 5 y 19. z 5 logx y 20. y 5 logx z In Exercises 21–34, write each exponential equation in its equivalent logarithmic form. 21. 83 5 512 22. 26 5 64 23. 0.00001 5 1025 24. 100,000 5 105 25. 15 5 !225 26. 7 5 " 3 343 27. 8 125 5 A2 5B3 28. 8 27 5 A2 3B3 29. 3 5 A 1 27B21/3 30. 4 5 A 1 1024B21/5 31. ex 5 6 32. e2x 5 4 33. x 5 yz 34. z 5 yx In Exercises 35–46, evaluate the logarithms exactly (if possible). 35. log2 1 36. log5 1 37. log5 3125 38. log3 729 39. log 107 40. log 1022 41. log1/4 4096 42. log1/7 2401 43. log 0 44. ln 0 45. log121002 46. ln1212 In Exercises 47–54, approximate (if possible) the common and natural logarithms using a calculator. Round to two decimal places. 47. log 29 48. ln 29 49. ln 380 80. log 380 51. log 0 52. ln 0 53. ln 0.0003 54. log 0.0003 In Exercises 55–64, state the domain of the logarithmic function in interval notation. 55. ƒ1x2 5 log21x 1 52 56. ƒ1x2 5 log214x 2 12 57. ƒ1x2 5 log315 2 2x2 58. ƒ1x2 5 log315 2 x2 59. ƒ1x2 5 ln17 2 2x2 60. ƒ1x2 5 ln13 2 x2 61. ƒ1x2 5 log 0 x 0 62. ƒ1x2 5 log 0 x 1 1 0 63. ƒ1x2 5 log1x2 1 12 64. ƒ1x2 5 log11 2 x22 In Exercises 65–70, match the graph with the function. 65. y 5 log5 x 66. y 5 log512x2 67. y 5 2log512x2 68. y 5 log51x 1 32 2 1 69. y 5 log511 2 x2 2 2 70. y 5 2log513 2 x2 1 2 a. b. c. x y –5 5 5 –5 x y –5 5 5 –5 x y –5 5 –5 5 [SEC TION 5. 2] E X ERC I S E S • S K I L L S 452 CHAPTER 5 Exponential and Logarithmic Functions In Exercises 71–82, graph the logarithmic function using transformation techniques. State the domain and range of f. 71. ƒ1x2 5 log1x 2 12 72. ƒ1x2 5 log1x 1 22 73. ln x 1 2 74. ln x 2 1 75. ƒ1x2 5 log31x 1 22 2 1 76. ƒ1x2 5 log31x 1 12 2 2 77. ƒ1x2 5 2log1x2 1 1 78. ƒ1x2 5 log12x2 1 2 79. ƒ1x2 5 ln1x 1 42 80. ƒ1x2 5 ln14 2 x2 81. ƒ1x2 5 log12x2 82. ƒ1x2 5 2ln12x2 • A P P L I C A T I O N S For Exercises 83–86, refer to the following: Decibel: D 5 10 log a I IT b IT 5 1 3 10212 W/m2 83. Sound. Calculate the decibels associated with normal conversation if the intensity is I 5 1 3 1026 W/m2. 84. Sound. Calculate the decibels associated with the onset of pain if the intensity is I 5 1 3 101 W/m2. 85. Sound. Calculate the decibels associated with attending a football game in a loud college stadium if the intensity is I 5 1 3 1020.3 W/m2. 86. Sound. Calculate the decibels associated with a doorbell if the intensity is I 5 1 3 1024.5 W/m2. For Exercises 87–90, refer to the following: Richter Scale: M 5 2 3 log a E E0 b E0 5 104.4 joules 87. Earthquakes. On Good Friday 1964, one of the most severe North American earthquakes ever recorded struck Alaska. The energy released measured 1.41 3 1017 joules. Calculate the magnitude of the 1964 Alaska earthquake using the Richter scale. 88. Earthquakes. On January 22, 2003, Colima, Mexico, experienced a major earthquake. The energy released measured 6.31 3 1015 joules. Calculate the magnitude of the 2003 Mexican earthquake using the Richter scale. 89. Earthquakes. On December 26, 2003, a major earthquake rocked southeastern Iran. In Bam, 30,000 people were killed, and 85% of buildings were damaged or destroyed. The energy released measured 2 3 1014 joules. Calculate the magnitude of the 2003 Iran earthquake with the Richter scale. 90. Earthquakes. On November 1, 1755, Lisbon was destroyed by an earthquake, which killed 90,000 people and destroyed 85% of the city. It was one of the most destructive earthquakes in history. The energy released measured 8 3 1017 joules. Calculate the magnitude of the 1755 Lisbon earthquake with the Richter scale. For Exercises 91–96, refer to the following: The pH of a solution is a measure of the molar concentration of hydrogen ions, H1 in moles per liter, in the solution, which means that it is a measure of the acidity or basicity of the solution. The letters pH stand for “power of hydrogen,” and the numerical value is defined as pH 5 2log10CH 1D Very acidic corresponds to pH values near 1, neutral corresponds to a pH near 7 (pure water), and very basic corresponds to values near 14. In the next six exercises, you will be asked to calculate the pH value of wine, Pepto-Bismol, normal rainwater, bleach, and two fruits. List these six liquids and use your intuition to classify them as neutral, acidic, very acidic, basic, or very basic before you calculate their actual pH values. 91. Chemistry. If wine has an approximate hydrogen ion concentration of 5.01 3 1024, calculate its pH value. 92. Chemistry. Pepto-Bismol has a hydrogen ion concentration of about 5.01 3 10211. Calculate its pH value. 93. Chemistry. Normal rainwater is slightly acidic and has an approximate hydrogen ion concentration of 1025.6. Calculate its pH value. Acid rain and tomato juice have similar approximate hydrogen ion concentrations of 1024. Calculate the pH value of acid rain and tomato juice. 94. Chemistry. Bleach has an approximate hydrogen ion concentration of 5.0 3 10213. Calculate its pH value. d. e. f. x y 5 –5 5 –5 x y –5 5 5 –5 x y –5 5 5 –5 5.2 Logarithmic Functions and Their Graphs 453 95. Chemistry. An apple has an approximate hydrogen ion concentration of 1023.6. Calculate its pH value. 96. Chemistry. An orange has an approximate hydrogen ion concentration of 1024.2. Calculate its pH value. 97. Archaeology. Carbon dating is a method used to determine the age of a fossil or other organic remains. The age t in years is related to the mass C (in milligrams) of carbon 14 through a logarithmic equation: t 5 2 ln a C 500b 0.0001216 How old is a fossil that contains 100 milligrams of carbon 14? 98. Archaeology. Repeat Exercise 97, only now the fossil contains 40 milligrams of carbon 14. 99. Broadcasting. Decibels are used to quantify losses associated with atmospheric interference in a communication system. The ratio of the power (watts) received to the power transmitted (watts) is often compared. Often, watts are transmitted, but losses due to the atmosphere typically correspond to milliwatts being received: dB 5 10 log a Power received Power transmittedb If 1 W of power is transmitted and 3 mW is received, calculate the power loss in dB. 100. Broadcasting. Repeat Exercise 99 assuming 3 W of power is transmitted and 0.2 mW is received. For Exercises 101 and 102, refer to the following: The range of all possible frequencies of electromagnetic radiation is called the electromagnetic spectrum. In a vacuum, the frequency of electromagnetic radiation is modeled by ƒ 5 c l where c is 3.0 3 108 m/s and l is wavelength in meters. 101. Physics/Electromagnetic Spectrum. The radio spectrum is the portion of the electromagnetic spectrum that corresponds to radio frequencies. The radio spectrum is used for various transmission technologies and is government regulated. Ranges of the radio spectrum are often allocated based on usage—for example, AM radio, cell phones, and television. (Source: a. Complete the following table for the various usages of the radio spectrum. USAGE WAVELENGTH FREQUENCY Super Low Frequency— Communication with Submarines 10,000,000 m 30 Hz Ultra Low Frequency— Communication within Mines 1,000,000 m Very Low Frequency— Avalanche Beacons 100,000 m Low Frequency— Navigation, AM Long-wave Broadcasting 10,000 m Medium Frequency— AM Broadcasts, Amateur Radio 1000 m High Frequency— Shortwave broadcasts, Citizens Band Radio 100 m Very High Frequency— FM Radio, Television 10 m Ultra High Frequency— Television, Mobile Phones 0.050 m b. Graph the frequency within the radio spectrum (in Hertz) as a function of wavelength (in meters). 102. Physics/Electromagnetic Spectrum. The visible spectrum is the portion of the electromagnetic spectrum that is visible to the human eye. Typically, the human eye can see wavelengths between 390 and 750 nm (nanometers or 1029 m). a. Complete the following table for the following colors of the visible spectrum. COLOR WAVELENGTH FREQUENCY Violet 400 nm 750 3 1012 Hz Cyan 470 nm Green 480 nm Yellow 580 nm Orange 610 nm Red 630 nm b. Graph the frequency (in Hertz) of the colors as a function of wavelength (in meters) on a log-log plot. 454 CHAPTER 5 Exponential and Logarithmic Functions • C A T C H T H E M I S T A K E In Exercises 103–106, explain the mistake that is made. 103. Evaluate the logarithm log2 4. Solution: Set the logarithm equal to x. log2 4 5 x Write the logarithm in exponential form. x 5 24 Simplify. x 5 16 Answer: log2 4 5 16 This is incorrect. The correct answer is log2 4 5 2. What went wrong? 104. Evaluate the logarithm log100 10. Solution: Set the logarithm equal to x. log100 10 5 x Express the equation in exponential form. 10x 5 100 Solve for x. x 5 2 Answer: log100 10 5 2 This is incorrect. The correct answer is log100 10 5 1 2. What went wrong? 105. State the domain of the logarithmic function ƒ1x2 5 log21x 1 52 in interval notation. Solution: The domain of all logarithmic functions is x . 0. Interval notation: 10, q2 This is incorrect. What went wrong? 106. State the domain of the logarithmic function ƒ1x2 5 ln 0 x 0 in interval notation. Solution: Since the absolute value eliminates all negative numbers, the domain is the set of all real numbers. Interval notation: 12q, q2 This is incorrect. What went wrong? In Exercises 107–110, determine whether each statement is true or false. 107. The domain of the standard logarithmic function, y 5 ln x, is the set of nonnegative real numbers. 108. The horizontal axis is the horizontal asymptote of the graph of y 5 ln x. 109. The graphs of y 5 log x and y 5 ln x have the same x-intercept (1, 02. 110. The graphs of y 5 log x and y 5 ln x have the same vertical asymptote, x 5 0. • C O N C E P T U A L 111. State the domain, range, and x-intercept of the function ƒ1x2 5 2ln1x 2 a2 1 b for a and b real positive numbers. 112. State the domain, range, and x-intercept of the function ƒ1x2 5 log1a 2 x2 2 b for a and b real positive numbers. 113. Graph the function ƒ1x2 5 bln12x2 x , 0 ln1x2 x . 0. 114. Graph the function ƒ1x2 5 b2ln12x2 x , 0 2ln1x2 x . 0. • C H A L L E N G E 5.2 Logarithmic Functions and Their Graphs 455 higher the affinity. The higher the value of Km, the lower the affinity. d. The actual equation that governs the relationship between v and [S] is v 5 Vmax 1 1 Km / 3S4 , which is NOT linear. This is the simple Michaelis–Menten kinetics equation. i. Use LnReg to get a best fit logarithmic curve for this data. Although the relationship between v and [S] is not logarithmic (rather, it is logistic), the best fit logarithmic curve does not grow very quickly and so it serves as a reasonably good fit. ii. At what [S] value, approximately, is the velocity 100 mmol/min? 122. The Michaelis–Menten equation can be arranged into various other forms that give a straight line (rather than a logistic curve) when one variable is plotted against another. One such rearrangement is the double-reciprocal Lineweaver–Burk equation. This equation plots the data values of the reciprocal of velocity (1/v) versus the reciprocal of the substrate level (1/[S]). The equation is as follows: 1 v 5 Km Vmax 1 3S4 1 1 Vmax Think of y as 1 v and x as 1 3S4. a. What is the slope of the “line”? How about its y-intercept? b. Using the data from Exercise 121, we create two new columns for 1/v and 1/[S] to obtain the following data set: [S] (MOL/L) 1/[S] v (mMOL/MIN) 1/v 4.00E204 2.50E103 130 0.00769231 2.00E204 5.00E103 110 0.00909091 1.00E204 1.00E104 89 0.01123596 5.00E205 2.00E104 62 0.01612903 4.00E205 2.50E104 53 0.01886792 2.50E205 4.00E104 38 0.02631579 2.00E205 5.00E104 32 0.03125 Create a scatterplot for the new data, treating x as 1/[S] and y as 1/v. c. Determine the best fit line and value of r. d. Use the equation of the best fit line in (c) to calculate Vmax. e. Use the above information to determine Km. 115. Use a graphing utility to graph y 5 ex and y 5 ln x in the same viewing screen. What line are these two graphs symmetric about? 116. Use a graphing utility to graph y 5 10x and y 5 log x in the same viewing screen. What line are these two graphs symmetric about? 117. Use a graphing utility to graph y 5 log x and y 5 ln x in the same viewing screen. What are the two common characteristics? 118. Using a graphing utility, graph y 5 ln0x 0. Is the function defined everywhere? 119. Use a graphing utility to graph ƒ1x2 5 ln13x2, g1x2 5 ln 3 1 ln x, and h1x2 5 1ln 321ln x2 in the same viewing screen. Determine the domain where two of the functions give the same graph. 120. Use a graphing utility to graph ƒ1x2 5 ln1x2 2 42, g1x2 5 ln1x 1 22 1 ln1x 2 22, and h1x2 5 ln1x 1 22ln1x 2 22 in the same viewing screen. Determine the domain where two of the functions give the same graph. For Exercises 121 and 122, refer to the following: Experimental data are collected all the time in biology and chemistry labs as scientists seek to understand natural phenomena. In biochemistry, the Michaelis–Menten kinetics law describes the rates of enzyme reactions using the relationship between the rate of the reaction and the concentration of the substrate involved. The following data have been collected, where the velocity v is measured in mmol/min of the enzyme reaction and the substrate level [S] is measured in mol/L. [S] (IN MOL/L) v (IN mMOL/MIN) 4.0 3 1024 130 2.0 3 1024 110 1.0 3 1024 89 5.0 3 1025 62 4.0 3 1025 52 2.5 3 1025 38 2.0 3 1025 32 121. a. Create a scatterplot of these data by identifying [S] with the x-axis and v with the y-axis. b. The graph seems to be leveling off. Give an estimate of the maximum value the velocity might achieve. Call this estimate Vmax. c. Another constant of importance in describing the relationship between v and [S] is Km. This is the value of [S] that results in the velocity being half its maximum value. Estimate this value. Note: Km measures the affinity level of a particular enzyme to a particular substrate. The lower the value of Km, the • T E C H N O L O G Y 456 CHAPTER 5 Exponential and Logarithmic Functions S K I L L S O B J E C T I V E S ■ ■Use properties of logarithms to simplify logarithmic expressions. ■ ■Use the change-of-base formula to evaluate a logarithm of a general base (other than base 10 or e). C O N C E P T U A L O B J E C T I V E S ■ ■Understand that the properties of logarithms are derived from the properties of exponents and the properties of inverse functions. ■ ■Understand that it does not matter which logarithm (natural or common) is used in the change-of-base formula. 5.3 PROPERTIES OF LOGARITHMS 5.3.1 Properties of Logarithmic Functions Since exponential functions and logarithmic functions are inverses of one another, properties of exponents are related to properties of logarithms. We will start by reviewing properties of exponents and then proceed to properties of logarithms. In Chapter 0, properties of exponents were discussed. From these properties of exponents we can develop similar properties for logarithms. We list seven basic properties. We will devote this section to proving and illustrating these seven properties. The first two properties follow directly from the definition of a logarithmic function and properties of exponentials. Property 11): logb 1 5 0 since b0 5 1 Property 12): logb b 5 1 since b1 5 b The third and fourth properties follow from the fact that exponential functions and logarithmic functions are inverses of one another. Recall that inverse functions satisfy PROPERTIES OF EXPONENTS Let a, b, m, and n be any real numbers and m . 0 and n . 0, then the following are true. 1. bm ? bn 5 bm1n 2. b2m 5 1 bm 5 a1 bb m 3. bm bn 5 bm2n 4. 1bm2n 5 bmn 5. 1ab2m 5 am ? bm 6. b0 5 1 7. b1 5 b PROPERTIES OF LOGARITHMS If b, M, and N are positive real numbers, where b ∙ 1 and p and x are real numbers, then the following are true: 1. logb1 5 0 2. logbb 5 1 3. logbbx 5 x 4. blogb x 5 x x . 0 5. logbMN 5 logbM 1 logbN Product rule: Log of a product is the sum of the logs. 6. logb aM N b 5 logb M 2 logb N Quotient rule: Log of a quotient is the difference of the logs. 7. logbMp 5 p logbM Power rule: Log of a number raised to an exponent is the exponent times the log of the number. 5.3.1 S KILL Use properties of logarithms to simplify logarithmic expressions. 5.3.1 CO NCE PTUAL Understand that the properties of logarithms are derived from the properties of exponents and the properties of inverse functions. 5.3 Properties of Logarithms 457 the relationship that ƒ211ƒ1x22 5 x for all x in the domain of ƒ1x2, and ƒ1ƒ211x22 5 x for all x in the domain of ƒ21. Let ƒ1x2 5 bx and ƒ211x2 5 logb x. Property (3): Write the inverse identity. ƒ211ƒ1x22 5 x Substitute ƒ211x2 5 logb x. logb1ƒ1x22 5 x Substitute ƒ1x2 5 bx. logb bx 5 x Property (4): Write the inverse identity. ƒ1ƒ21 1x22 5 x Substitute ƒ1x2 5 bx. bƒ 211x2 5 x Substitute ƒ211x2 5 logb x x . 0. blogbx 5 x The first four properties are summarized for common and natural logarithms. COMMON AND NATURAL LOGARITHM PROPERTIES Common Logarithm (base 10) Natural Logarithm (base e) 1. log 1 5 0 1. ln 1 5 0 2. log 10 5 1 2. ln e 5 1 3. log 10x 5 x 3. ln ex 5 x 4. 10log x 5 x x . 0 4. eln x 5 x x . 0 EXAMPLE 1 Using Logarithmic Properties Use properties 11)–14) to simplify the following expressions. a. log10 10 b. ln 1 c. 10log(x18) d. eln(2x15) e. log 10x2 f. ln e x13 Solution: a. Use property (2). log10 10 5 1 b. Use property (1). ln 1 5 0 c. Use property (4). 10log1x182 5 x 1 8 x . 28 d. Use property (4). eln12x152 5 2x 1 5 x . 25 2 e. Use property (3). log 10x2 5 x2 f. Use property (3). ln ex13 5 x 1 3 The fifth through seventh properties follow from the properties of exponents and the definition of logarithms. We will prove the product rule and leave the proofs of the quotient and power rules for the exercises. Property 152: logb MN 5 logb M 1 logb N WORDS MATH Assume two logs that have the same base. Let u 5 logbM and v 5 logbN M . 0, N . 0 Change to equivalent exponential forms. bu 5 M and bv 5 N Write the log of a product. logb MN Substitute M 5 bu and N 5 bv. 5 logb1bu bv2 Use properties of exponents. 5 logb1bu1v2 Apply property 3. 5 u 1 v Substitute u 5 logbM, v 5 logbN. 5 logbM 1 logbN logbMN 5 logbM 1 logbN [CONCEPT CHECK] When we multiply exponential functions with the same base, we add their arguments (exponents); but when we add two logarithms with the same bases, we _ their arguments. ANSWER multiply ▼ 458 CHAPTER 5 Exponential and Logarithmic Functions In other words, the log of a product is the sum of the logs. Let us illustrate this property with a simple example. 3 2 5 log2 8 1 log2 4 5 log2 32 Notice that log2 8 1 log2 4 2 log2 12. r r r EXAMPLE 3 Writing a Sum of Logarithms as a Single Logarithmic Expression: The Right Way and the Wrong Way Use properties of logarithms to write the expression 2 logb 3 1 4 logb u as a single logarithmic expression. common mistake A common mistake is to write the sum of the logs as a log of the sum. logb M 1 logb N ∙ logb1M 1 N 2 Y OUR TU R N Express 2 ln x 1 3 ln y as a single logarithm. ▼ C A U T I O N logbM 1 logbN 5 logb1MN2 logbM 1 logbN ∙ logb1M 1 N2 ▼ A N S W E R ln1x2y32 ▼ ✖I N C O R R EC T Þ logb19 1 u42 ERROR ✓C O R REC T Use the power property (7). 2 logb 3 1 4 logb u 5 logb 32 1 logb u4 Simplify. logb 9 1 logb u4 Use the product property (5). 5 logb19u42 EXAMPLE 2 Writing a Logarithmic Expression as a Sum of Logarithms Use the logarithmic properties to write the expression logbAu2!vB as a sum of simpler logarithms. Solution: Convert the radical to exponential form. logbAu2!vB 5 logb1u2v1/22 Use the product property (5). 5 logbu2 1 logbv1/2 Use the power property (7). 5 2logb u 1 1 2logbv Y OUR TU R N Use the logarithmic properties to write the expression logbAx4 ! 3 yB as a sum of simpler logarithms. ▼ A N S W E R logbAx4! 3 yB 5 4 logb x 1 1 3 logb y ▼ 5.3 Properties of Logarithms 459 Another common mistake is misinterpreting the quotient rule. EXAMPLE 4 Writing a Logarithmic Expression as a Difference of Logarithms Write the expression ln ax3 y2b as a difference of logarithms. Solution: Apply the quotient property (6). ln ax3 y2b 5 ln1x32 2 ln1y22 Apply the power property (7). 5 3 ln x 2 2 ln y YOUR T UR N Write the expression log aa4 b5b as a difference of logarithms. ▼ EXAMPLE 5 Writing the Difference of Logarithms as a Logarithm of a Quotient Write the expression 2 3 ln x 2 1 2 ln y as a logarithm of a quotient. common mistake logb M 2 logb N 2 logb M logb N YOUR T UR N Write the expression 1 2 log a 2 3 log b as a single logarithm. ▼ ✖I N C O R R EC T ln x2/3 ln y1/2 ERROR ✓COR R E C T Use the power property (7). 2 3 ln x 2 1 2 ln y 5 ln x2/3 2 ln y1/2 Use the quotient property (6). ln ax2/3 y1/2b ▼ C A U T I O N logb M 2 logb N 5 logb aM N b logb M 2 logb N ∙ logb M logb N EXAMPLE 6 Combining Logarithmic Expressions into a Single Logarithm Write the expression 3 logb x 1 logb12x 1 12 2 2 logb 4 as a single logarithm. Solution: Use the power property (7) on the first and third terms. 5 logb x3 1 logb12x 1 12 2 logb 42 Use the product property (5) on the first two terms. 5 logb3x312x 1 12 4 2 logb 16 Use the quotient property (6). 5 logbcx312x 1 12 16 d YOUR TURN Write the expression 2 ln x 2 ln13y2 1 3 ln z as a single logarithm. ▼ ▼ A N S W E R ln ax2z3 3y b ▼ A N S W E R 4 log a 2 5 log b ▼ A N S W E R log aa1/2 b3 b 460 CHAPTER 5 Exponential and Logarithmic Functions 5.3.2 Change-of-Base Formula Recall that in the last section we were able to evaluate logarithms two ways: (1) exactly by writing the logarithm in exponential form and identifying the exponent and (2) using a calculator if the logarithms were base 10 or e. How do we evaluate a logarithm of general base if we cannot identify the exponent? We use the change-of-base formula. EXAMPLE 7 Expanding a Logarithmic Expression into a Sum or Difference of Logarithms Write ln c x2 2 x 2 6 x2 1 7x 1 6 d as a sum or difference of logarithms. Solution: Factor the numerator and denominator. 5 ln c 1x 2 321x 1 22 1x 1 621x 1 12 d Use the quotient property (6). 5 ln 31x 2 32 1x 1 22 4 2 ln 31x 1 62 1x 1 12 4 Use the product property (5). 5 ln1x 2 32 1 ln1x 1 22 2 3ln1x 1 62 1 ln1x 1 12 4 Eliminate brackets. 5 ln1x 2 32 1 ln1x 1 22 2 ln1x 1 62 2 ln1x 1 12 5.3.2 S KILL Use the change-of-base formula to evaluate a logarithm of a general base (other than base 10 or e). 5.3.2 CO NCE PTUAL Understand that it does not matter which logarithm (natural or common) is used in the change-of-base formula. EXAMPLE 8 Using Properties of Logarithms to Change the Base to Evaluate a General Logarithm Evaluate log3 8. Round the answer to four decimal places. Solution: Let y 5 log3 8. y 5 log3 8 Write the logarithm in exponential form. 3y 5 8 Take the log of both sides. log 3y 5 log 8 Use the power property (7). y log 3 5 log 8 Divide both sides by log 3. y 5 log 8 log 3 Let y 5 log3 8. log3 8 5 log 8 log 3 Example 8 illustrates our ability to use properties of logarithms to change from base 3 to base 10, which our calculators can handle. This leads to the general change-of-base formula. 5.3 Properties of Logarithms 461 Proof of Change-of-Base Formula WORDS MATH Let y be the logarithm we want to evaluate. y 5 logbM Write y 5 logb M in exponential form. by 5 M Let a be any positive real number 1where a 2 12. Take the log of base a of both sides of the equation. loga by 5 logaM Use the power rule on the left side of the equation. y logab 5 logaM Divide both sides of the equation by loga b. y 5 loga M loga b EXAMPLE 9 Using the Change-of-Base Formula Use the change-of-base formula to evaluate log4 17. Round to four decimal places. Solution: We will illustrate this two ways (choosing common and natural logarithms) using a scientific calculator. Common Logarithms Use the change-of-base formula with base 10. log4 17 5 log 17 log4 Approximate with a calculator. < 2.043731421 < 2.0437 Natural Logarithms Use the change-of-base formula with base e. log4 17 5 ln 17 ln 4 Approximate with a calculator. < 2.043731421 < 2.0437 YOUR T UR N Use the change-of-base formula to approximate log7 34. Round to four decimal places. ▼ CHANGE-OF-BASE FORMULA For any logarithmic bases a and b and any positive number M, the change-of-base formula says that logb M 5 loga M loga b In the special case when a is either 10 or e, this relationship becomes Common Logarithms Natural Logarithms logb M 5 log M log b or logb M 5 ln M ln b It does not matter what base we select (10, e, or any other base), the ratio will be the same. [CONCEPT CHECK] TRUE OR FALSE if In1a2 In1b2 5 log1a2 log1b2 then ln(a) 5 log(a) and ln(b) 5 log(b). ANSWER False. ▼ ▼ A N S W E R log7 34 < 1.8122 462 CHAPTER 5 Exponential and Logarithmic Functions In Exercises 1–18, apply the properties of logarithms to simplify each expression. Do not use a calculator. 1. log9 1 2. log69 1 3. log1/2A1 2B 4. log3.3 3.3 5. log10 108 6. ln e3 7. log10 0.001 8. log3 37 9. log2 !8 10. log5 ! 3 5 11. 8log8 5 12. 2log2 5 13. eln1x152 14. 10log13x212x112 15. 53 log5 2 16. 72 log7 5 17. 722 log7 3 18. e22 ln 10 In Exercises 19–32, write each expression as a sum or difference of logarithms. Example: logAm2n5B 5 2 log m 1 5 log n 19. logb1x3y52 20. logb1x23y252 21. logb1x1/2y1/32 22. logbA!r ! 3 tB 23. logb ar1/3 s1/2b 24. logb ar4 s2b 25. logb a x yzb 26. logb axy z b 27. logAx2!x 1 5B 28. log31x 2 32 1x 1 22 4 29. lnc x31x 2 222 "x2 1 5 d 30. lnc !x 1 3 ! 3 x 2 4 1x 1 124 d 31. logc x2 2 2x 1 1 x2 2 9 d 32. logc x2 2 x 2 2 x2 1 3x 2 4d In Exercises 33–44, write each expression as a single logarithm. Example: 2 log m 1 5 log n 5 log1m2n52 33. 3 logbx 1 5 logby 34. 2 logbu 1 3 logbv 35. 5 logbu 2 2 logbv 36. 3 logb x 2 logby 37. 1 2 logb x 1 2 3 logb y 38. 1 2 logb x 2 2 3 logb y 39. 2 log u 2 3 log v 2 2 log z 40. 3 log u 2 log2v 2 log z 41. ln1x 1 12 1 ln1x 2 12 2 2 ln1x2 1 32 42. ln!x 2 1 1 ln!x 1 1 2 2 ln1x2 2 12 43. 1 2 ln1x 1 32 2 1 3 ln1x 1 22 2 ln1x2 44. 1 3 ln1x2 1 42 2 1 2 ln1x2 2 32 2 ln1x 2 12 [SEC TION 5.3] E X E R CI SE S • S K I L L S • Product Property: logb MN 5 logb M 1 logb N • Quotient Property: logb aM N b 5 logb M 2 logb N • Power Property: logb Mp 5 p logb M GENERAL LOGARITHM COMMON LOGARITHM NATURAL LOGARITHM logb 1 5 0 log 1 5 0 ln 1 5 0 logb b 5 1 log 10 5 1 ln e 5 1 logb bx 5 x log 10x 5 x ln ex 5 x blogb x 5 x x . 0 10log x 5 x x . 0 eln x 5 x x . 0 logb M 5 loga M loga b logb M 5 log M log b logb M 5 ln M ln b Properties of Logarithms If a, b, M, and N are positive real numbers, where a 2 1, b 2 1, and p and x are real numbers, then the following are true: [SEC TION 5.3] S U M M A RY 5.3 Properties of Logarithms 463 • A P P L I C A T I O N S 55. Sound. Sitting in the front row of a rock concert exposes us to a sound pressure (or sound level) of 1 3 1021 W/m2 (or 110 dB), and a normal conversation is typically around 1 3 1026 W/m2 (or 60 dB). How many decibels are you exposed to if a friend is talking in your ear at a rock concert? Note: 160 dB causes perforation of the eardrums. Hint: Add the sound pressures and convert to dB. 56. Sound. A whisper corresponds to 1 3 10210 W/m2 (or 20 dB), and a normal conversation is typically around 1 3 1026 W/m2 (or 60 dB). How many decibels are you exposed to if one friend is whispering in your ear while the other one is talking at a normal level? Hint: Add the sound pressures and convert to dB. For Exercises 57 and 58, refer to the following: There are two types of waves associated with an earthquake: compression and shear. The compression, or longitudinal, waves displace material behind the earthquake’s path. Longitudinal waves travel at great speeds and are often called “primary waves” or simply “P” waves. Shear, or transverse, waves displace material at right angles to its path. Transverse waves do not travel as rapidly through the Earth’s crust and mantle as do longitudinal waves, and they are called “secondary” or “S” waves. 57. Earthquakes. If a seismologist records the energy of P waves as 4.5 3 1012 joules and the energy of S waves as 7.8 3 108 joules, what is the total energy? What would the combined effect be on the Richter scale? 58. Earthquakes. Repeat Exercise 57 assuming the energy associated with the P waves is 5.2 3 1011 joules and the energy associated with the S waves is 4.1 3 109 joules. • C A T C H T H E M I S T A K E In Exercises 59–62, simplify if possible and explain the mistake that is made. 59. 3 log 5 2 log 25 Solution: Apply the quotient property (6). 3 log 5 log 25 Write 25 5 52. 3 log 5 log 52 Apply the power property (7). 3 log 5 2 log 5 Simplify. 3 2 This is incorrect. The correct answer is log 5. What mistake was made? 60. ln 3 1 2 ln 4 2 3 ln 2 Solution: Apply the power property (7). ln 3 1 ln 42 2 ln 23 Simplify. ln 3 1 ln 16 2 ln 8 Apply property (5). ln13 1 16 2 82 Simplify. ln 11 This is incorrect. The correct answer is ln 64. What mistake was made? 61. log2x 1 log3y 2 log4z Solution: Apply the product property (5). log6 xy 2 log4 z Apply the quotient property (6). log24 xyz This is incorrect. What mistake was made? 62. 21log3 2 log52 Solution: Apply the quotient property (6). 2 alog 3 5b Apply the power property (7). alog 3 5b 2 Apply a calculator to approximate. < 0.0492 This is incorrect. What mistake was made? In Exercises 45–54, evaluate the logarithms using the change-of-base formula. Round to four decimal places. 45. log5 7 46. log4 19 47. log1/2 5 48. log5 1 2 49. log2.7 5.2 50. log7.2 2.5 51. logp 10 52. logp 2.7 53. log"3 8 54. log"2 9 464 CHAPTER 5 Exponential and Logarithmic Functions In Exercises 63–66, determine whether each statement is true or false. 63. log e 5 1 ln 10 64. ln e 5 1 log 10 65. ln 1xy23 5 1ln x 1 ln y23 66. ln a ln b 5 log a log b • C O N C E P T U A L 67. Prove the quotient rule: logb aM N b 5 logb M 2 logb N. Hint: Let u 5 logb M and v 5 logb N. Write both in exponential form and find the quotient logb aM N b. 68. Prove the power rule: logb M p 5 p logb M. Hint: Let u 5 logb M. Write this log in exponential form, and find logb M p. 69. Write in terms of simpler logarithmic forms. logb aÄ x2 y3z25 b 6 70. Show that logb a1 xb 5 2logb x. • C H A L L E N G E 75. Use a graphing calculator to plot y 5 ln1 x22 and y 5 2 ln x. Are they the same graph? 76. Use a graphing calculator to plot y 5 1ln x22 and y 5 2 ln x. Are they the same graph? 77. Use a graphing calculator to plot y 5 ln x and y 5 log x log e. Are they the same graph? 78. Use a graphing calculator to plot y 5 log x and y 5 ln x ln 10 . Are they the same graph? 71. Use a graphing calculator to plot y 5 ln12x2 and y 5 ln 2 1 ln x. Are they the same graph? 72. Use a graphing calculator to plot y 5 ln12 1 x2 and y 5 ln 2 1 ln x. Are they the same graph? 73. Use a graphing calculator to plot y 5 log x log 2 and y 5 logx 2 log2. Are they the same graph? 74. Use a graphing calculator to plot y 5 log ax 2b and y 5 logx 2 log2. Are they the same graph? • T E C H N O L O G Y S K I L L S O B J E C T I V E S ■ ■Use properties of one-to-one functions and properties of inverses to solve exponential equations. ■ ■Use properties of one-to-one functions and properties of inverses to solve logarithmic equations. ■ ■Solve application problems that involve exponential and logarithmic equations. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that properties of one-to-one functions enable us to solve some simple exponential equations; for other exponential equations, the properties of inverses must be used. ■ ■Understand why extraneous solutions can arise in logarithmic equations and how to eliminate extraneous solutions. ■ ■Use the rule of 70 to guide intuition on investment questions. 5.4 EXPONENTIAL AND LOGARITHMIC EQUATIONS 5.4 Exponential and Logarithmic Equations 465 5.4.1 Exponential Equations In this book, you have solved algebraic equations such as x2 2 9 5 0, in which the goal is to solve for x by finding the values of x that make the statement true. Exponential and logarithmic equations have the x buried within an exponent or a logarithm, but the goal is the same. Solve for x. Exponential equation: e2x11 5 5 Logarithmic equation: log13x 2 12 5 7 There are two methods for solving exponential and logarithmic equations that are based on the properties of one-to-one functions and inverses. To solve simple exponential and logarithmic equations, we will use one-to-one properties. To solve more complicated exponential and logarithmic equations, we will use properties of inverses. The following box summarizes the one-to-one and inverse properties that hold true when b . 0 and b ∙ 1. ONE-TO-ONE PROPERTIES bx 5 by if and only if x 5 y logb x 5 logb y if and only if x 5 y INVERSE PROPERTIES blogb x 5 x x . 0 logb bx 5 x The following strategies are outlined for solving simple and complicated exponen-tial equations using the one-to-one and inverse properties. 5.4.1 S K I L L Use properties of one-to-one functions and properties of inverses to solve exponential equations. 5.4.1 C ON C E P T U A L Understand that properties of one-to-one functions enable us to solve some simple exponential equations; for other exponential equations, the properties of inverses must be used. STRATEGIES FOR SOLVING EXPONENTIAL EQUATIONS Strategy for Solving Exponential Equations TYPE OF EQUATION STRATEGY EXAMPLE Simple 1. Rewrite both sides of the equation in terms of the same base. 2. Use the one-to-one property to equate the exponents. 3. Solve for the variable. 2x23 5 32 2x23 5 25 x 2 3 5 5 x 5 8 Complicated 1. Isolate the exponential expression. 2. Take the same logarithm of both sides. 3. Simplify using the inverse properties. 4. Solve for the variable. 3e2x 2 2 5 7 3e2x 5 9 e2x 5 3 ln e2x 5 ln 3 2x 5 ln 3 x 5 1 2 ln 3 Take the logarithm with base that is equal to the base of the exponent and use the property logbbx 5 x or take the natural logarithm and use the property ln M p 5 p ln M. 466 CHAPTER 5 Exponential and Logarithmic Functions EXAMPLE 1 Solving a Simple Exponential Equation Solve the exponential equations using the one-to-one property. a. 3x 5 81 b. 572x 5 125 c. A1 2B4y 5 16 Solution (a): Substitute 81 5 34. 3x 5 34 Use the one-to-one property to identify x. x 5 4 Solution (b): Substitute 125 5 53. 572x 5 53 Use the one-to-one property. 7 2 x 5 3 Solve for x. x 5 4 Solution (c): Substitute a1 2b 4y 5 a 1 24yb 5 224y. 224y 5 16 Substitute 16 5 24. 224y 5 24 Use the one-to-one property to identify y. y 5 21 Y OUR TU R N Solve the following equations. a. 2x21 5 8 b. A1 3B y 5 27 ▼ In Example 1, we were able to rewrite the equation in a form with the same bases so that we could use the one-to-one property. In Example 2, we will not be able to write both sides in a form with the same bases. Instead, we will use properties of inverses. EXAMPLE 2 Solving a More Complicated Exponential Equation with a Base Other Than 10 or e Solve the exponential equations exactly and then approximate answers to four decimal places. a. 53x 5 16 b. 43x12 5 71 Solution (a): Take the natural logarithm of both sides of the equation. ln 53x 5 ln 16 Use the power property on the left side of the equation. 3x ln 5 5 ln 16 Divide both sides of the equation by 3 ln 5. x 5 ln 16 3 ln 5 Use a calculator to approximate x to four decimal places. x < 0.5742 Solution (b): Rewrite in logarithmic form. 3x 1 2 5 log4 71 Subtract 2 from both sides. 3x 5 log4 71 2 2 Divide both sides by 3. x 5 log4 71 2 2 3 Use the change-of-base formula, log4 71 5 ln 71 ln 4 . x 5 ln 71 ln 4 2 2 3 Use a calculator to approximate x to four decimal places. x < 3.07487356 2 2 3 < 0.3583 [CONCEPT CHECK] TRUE OR FALSE In Solution (a) in Example 2, we could have used the log of any base (not just the natural log as shown). ANSWER True ▼ ▼ A N S W E R a. x 5 4 b. y 5 23 5.4 Exponential and Logarithmic Equations 467 We could have proceeded in an alternative way by taking either the natural log or the common log of both sides and using the power property (instead of using the change-of-base formula) to evaluate the logarithm with base 4. Take the natural logarithm of both sides. lnA43x12B 5 ln 71 Use the power property (7). 13x 1 22ln 4 5 ln 71 Divide by ln 4. 3x 1 2 5 ln 71 ln 4 Subtract 2 and divide by 3. x 5 ln 71 ln 4 2 2 3 Use a calculator to approximate x. x < 3.07487356 2 2 3 < 0.3583 Y OUR T UR N Solve the equation 5y2 5 27 exactly and then approximate the answer to four decimal places. ▼ A N S W E R y 5 6!log527 < 61.4310 ▼ EXAMPLE 3 Solving a More Complicated Exponential Equation with Base 10 or e Solve the exponential equation 4ex2 5 64 exactly and then round the answer to four decimal places. Solution: Divide both sides by 4. ex2 5 16 Take the natural logarithm (ln) of both sides. lnAex2B 5 ln 16 Simplify the left side with the property of inverses. x2 5 ln 16 Solve for x using the square-root method. x 5 6"ln 16 Use a calculator to approximate x to four decimal places. x < 61.6651 Y OUR T UR N Solve the equatio n 102x23 5 7 exactly and then round the answer to four decimal places. ▼ A N S W E R x 5 log107 1 3 2 < 1.9225 ▼ EXAMPLE 4 Solving an Exponential Equation Quadratic in Form Solve the equation e2x 2 4ex 1 3 5 0 exactly and then round the answer to four decimal places. Solution: Let u 5 ex. 1Note: u2 5 ex ? ex 5 e2x.2 u2 2 4u 1 3 5 0 Factor. 1u 2 32 1u 2 12 5 0 Solve for u. u 5 3 or u 5 1 Substitute u 5 ex. ex 5 3 or ex 5 1 Take the natural logarithm (ln) of both sides. ln1ex2 5 ln 3 or ln1ex2 5 ln 1 Simplify with the properties of logarithms. x 5 ln 3 or x 5 ln 1 Approximate or evaluate exactly the right sides. x < 1.0986 or x 5 0 YOU R T UR N Solve the equation 100x 2 10x 2 2 5 0 exactly and then round the answer to four decimal places. ▼ ▼ A N S W E R x 5 log 2 < 0.3010 468 CHAPTER 5 Exponential and Logarithmic Functions 5.4.2 Solving Logarithmic Equations We can solve simple logarithmic equations using the property of one-to-one functions. For more complicated logarithmic equations, we can employ properties of logarithms and properties of inverses. Solutions must be checked to eliminate extraneous solutions. 5.4.2 S KILL Use properties of one-to-one functions and properties of inverses to solve logarithmic equations. 5.4.2 CO NCE PTUAL Understand why extraneous solutions can arise in logarithmic equations and how to eliminate extraneous solutions. EXAMPLE 5 Solving a Simple Logarithmic Equation Solve the equation log412x 2 32 5 log41x2 1 log41x 2 22. Solution: Apply the product property (5) on the right side. log412x 2 32 5 log4 3x 1x 2 22 4 Apply the property of one-to-one functions. 2x 2 3 5 x 1x 2 22 Distribute and simplify. x2 2 4x 1 3 5 0 Factor. 1x 2 32 1x 2 12 5 0 Solve for x. x 5 3 or x 5 1 [CONCEPT CHECK] TRUE OR FALSE ALL of the logarithms must be undefined for a particular value in order for that value to be eliminated from the domain of the solution. ANSWER False ▼ Strategy for Solving Logarithmic Equations TYPE OF EQUATION STRATEGY EXAMPLE Simple 1. Combine logarithms on each side of the equation using properties. 2. Use the one-to-one property to equate the arguments. 3. Solve for the variable. 4. Check the results and eliminate any extraneous solutions. log1x 2 32 1 log x 5 log 4 log x1x 2 32 5 log 4 x1x 2 32 5 4 x2 2 3x 2 4 5 0 1x 2 421x 1 12 5 0 x 5 21, 4 Eliminate x 5 21 because log1212 is undefined. x 5 4 Complicated 1. Combine and isolate the logarithmic expressions. 2. Rewrite the equation in exponential form. 3. Solve for the variable. 4. Check the results and eliminate any extraneous solutions. log51x 1 22 2 log5 x 5 2 log5 ax 1 2 x b 5 2 x 1 2 x 5 52 x 1 2 5 25x 24x 5 2 x 5 1 12 log5 a 1 12 1 2b 2 log5 a 1 12b 5 log5 a25 12b 2 log5 a 1 12b 5 log5 c25/12 1/12 d 5 log5 3254 5 2 ✓ 5.4 Exponential and Logarithmic Equations 469 The possible solution x 5 1 must be eliminated because it is not in the domain of two of the logarithmic functions. x 5 1: log41212 0 log4112 1 log41212 x 5 3 YOUR T UR N Solve the equation ln1x 1 82 5 ln1x2 1 ln1x 1 32. undefined undefined b b ▼ STUDY TIP Solutions should be checked in the original equation to eliminate extraneous solutions. ▼ A N S W E R x 5 2 EXAMPLE 6 Solving a More Complicated Logarithmic Equation Solve the equation log319x2 2 log31x 2 82 5 4. Solution: Employ the quotient property (6) on the left side. log3 a 9x x 2 8b 5 4 Write in exponential form. logb x 5 y 1 x 5 by 9x x 2 8 5 34 Simplify the right side. 9x x 2 8 5 81 Multiply the equation by the LCD, x 2 8. 9x 5 811x 2 82 Eliminate parentheses. 9x 5 81x 2 648 Solve for x. 272x 5 2648 x 5 9 Check: log3 39 ? 9 4 2 log3 39 2 8 4 5 log3 381 4 2 log3 1 5 4 2 0 5 4 Y OUR T UR N Solve the equation log214x2 2 log2122 5 2. ▼ ▼ A N S W E R x 5 2 EXAMPLE 7 Solving a Logarithmic Equation with No Solution Solve the equation ln13 2 x22 5 7. Solution: Write in exponential form. 3 2 x2 5 e7 Simplify. x2 5 3 2 e7 3 2 e7 is negative. x2 5 negative real number There are no real numbers that when squared yield a negative real number. Therefore, there is no real solution . b 5.4.3 Applications In the chapter opener, we saw that archaeologists determine the age of a fossil by how much carbon 14 is present at the time of discovery. The number of grams of carbon 14 based on the radioactive decay of the isotope is given by A 5 A0e20.000124t where A is the number of grams of carbon 14 at the present time, A0 is the number of grams of carbon 14 while alive, and t is the number of years since death. Using the inverse properties, we can isolate t. 5.4.3 S K IL L Solve application problems that involve exponential and logarith-mic equations. 5.4.3 C ON C E P T U A L Use the rule of 70 to guide intu-ition on investment questions. 470 CHAPTER 5 Exponential and Logarithmic Functions WORDS MATH Divide by A0. A A0 5 e20.000124t Take the natural logarithm of both sides. ln a A A0b 5 lnAe20.000124tB Simplify the right side utilizing properties of inverses. ln a A A0b 5 20.000124t Solve for t. t 5 2 1 0.000124 ln a A A0b Let’s assume that animals have approximately 1000 mg of carbon 14 in their bodies when they are alive. If a fossil has 200 mg of carbon 14, approximately how old is the fossil? Substituting A 5 200 and A0 5 1000 into our equation for t, we find t 5 2 1 0.000124 ln a1 5b < 12,979.338 The fossil is approximately 13,000 years old. EXAMPLE 8 Calculating How Many Years It Will Take for Money to Double You save $1000 from a summer job and put it in a CD earning 5% compounding continuously. How many years will it take for your money to double? Round to the nearest year. Solution: Recall the compound continuous interest formula. A 5 Pert Substitute P 5 1000, A 5 2000, and r 5 0.05. 2000 5 1000e0.05t Divide by 1000. 2 5 e0.05t Take the natural logarithm of both sides. ln 2 5 ln Ae0.05tB Simplify with the property ln ex 5 x. ln 2 5 0.05t Solve for t. t 5 ln 2 0.05 < 13.8629 It will take almost 14 years for your money to double. Y OUR TU R N How long will it take $1000 to triple (become $3000) in a savings account earning 10% a year compounding continuously? Round your answer to the nearest year. ▼ ▼ A N S W E R approximately 11 years When an investment is compounded continuously, how long will it take for that investment to double? WORDS MATH Write the interest formula for compounding continuously. A 5 Pert Let A 5 2P (investment doubles). 2P 5 Pert Divide both sides of the equation by P. 2 5 ert Take the natural log of both sides of the equation. ln 2 5 ln ert Simplify the right side by applying the property ln ex 5 x. ln 2 5 rt Divide both sides by r. t 5 ln 2 r Approximate ln 2 < 0.7. t < 0.7 r 5.4 Exponential and Logarithmic Equations 471 Recall that an interest rate such as 4.5% is written in decimal form as r 5 0.045. Multiply the numerator and denominator by 100. t < 70 100r This is the “rule of 70.” If we divide 70 by the interest rate (compounding continuously), we get the approximate time for an investment to double. In Example 8, the interest rate (compounding continuously) is 5%. Dividing 70 by 5 yields 14 years. [CONCEPT CHECK] TRUE OR FALSE We do not need to know the principal to determine how long it will take for an investment to double; knowing the interest rate and how interest is compounded is sufficient. ANSWER True ▼ [SEC TION 5.4] E X E R C I S E S • S K I L L S In Exercises 1–18, solve the exponential equations exactly for x. 1. 3x 5 81 2. 5x 5 125 3. 7x 5 1 49 4. 4x 5 1 16 5. 2x2 5 16 6. 169x 5 13 7. A2 3Bx11 5 27 8 8. A3 5Bx11 5 25 9 9. e2x13 5 1 10. 10x221 5 1 11. 72x25 5 73x24 12. 125x 5 52x23 13. 2x2112 5 27x 14. 5x223 5 52x 15. 9x 5 3x224x 16. 16x21 5 2x2 17. e5x21 5 e x213 18. 10x228 5 100x In Exercises 19–44, solve the exponential equations exactly and then approximate your answers to three decimal places. 19. 102x23 5 81 20. 23x11 5 21 21. 3x11 5 5 22. 52x21 5 35 23. 27 5 23x21 24. 15 5 7322x 25. 3ex 2 8 5 7 26. 5ex 1 12 5 27 27. 9 2 2e0.1x 5 1 28. 21 2 4e0.1x 5 5 29. 213x2 2 11 5 9 30. 312x2 1 8 5 35 31. e3x14 5 22 32. ex2 5 73 33. 3e2x 5 18 34. 41103x2 5 20 35. e2x 1 7ex 2 3 5 0 36. e2x 2 4ex 2 5 5 0 37. 13x 2 32x22 5 0 38. 13x 2 32x2 13x 1 32x2 5 0 39. 2 ex 2 5 5 1 40. 17 e x 1 4 5 2 41. 20 6 2 e2x 5 4 42. 4 3 2 e3x 5 8 43. 4 102x 2 7 5 2 44. 28 10 x 1 3 5 4 [SEC TION 5.4] S U M M A RY Strategy for Solving Exponential Equations Strategy for Solving Logarithmic Equations Take the logarithm with the base that is equal to the base of the exponent and use the property logb bx 5 x, or take the natural logarithm and use the property ln Mp 5 p ln M. TYPE OF EQUATION STRATEGY Simple 1. Rewrite both sides of the equation in terms of the same base. 2. Use the one-to-one property to equate the exponents. 3. Solve for the variable. Complicated 1. Isolate the exponential expression. 2. Take the same logarithm of both sides. 3. Simplify using the inverse properties. 4. Solve for the variable. TYPE OF EQUATION STRATEGY Simple 1. Combine logarithms on each side of the equation using properties. 2. Use the one-to-one property to equate the arguments. 3. Solve for the variable. 4. Eliminate any extraneous solutions. Complicated 1. Combine and isolate the logarithmic expressions. 2. Rewrite the equation in exponential form. 3. Solve for the variable. 4. Eliminate any extraneous solutions. 472 CHAPTER 5 Exponential and Logarithmic Functions In Exercises 45–62, solve the logarithmic equations exactly. 45. log12x2 5 2 46. log15x2 5 3 47. log312x 1 12 5 4 48. log213x 2 12 5 3 49. log214x 2 12 5 23 50. log415 2 2x2 5 22 51. ln x2 2 ln 9 5 0 52. logx2 1 log x 5 3 53. log51x 2 42 1 log5 x 5 1 54. log21x 2 12 1 log21x 2 32 5 3 55. log1x 2 32 1 log1x 1 22 5 log14x2 56. log21x 1 12 1 log214 2 x2 5 log216x2 57. log14 2 x2 1 log1x 1 22 5 log13 2 2x2 58. log13 2 x2 1 log1x 1 32 5 log11 2 2x2 59. log414x2 2 log4 ax 4b 5 3 60. log317 2 2x2 2 log31x 1 22 5 2 61. log12x 2 52 2 log1x 2 32 5 1 62. log3110 2 x2 2 log31x 1 22 5 1 In Exercises 63–78, solve the logarithmic equations exactly, if possible; then approximate your answers to three decimal places. 63. ln x2 5 5 64. log 3x 5 2 65. log12x 1 52 5 2 66. ln14x 2 72 5 3 67. ln1x2 1 12 5 4 68. log1x2 1 42 5 2 69. ln12x 1 32 5 22 70. log13x 2 52 5 21 71. log12 2 3x2 1 log13 2 2x2 5 1.5 72. log213 2 x2 1 log211 2 2x2 5 5 73. ln1x2 1 ln1x 2 22 5 4 74. ln14x2 1 ln12 1 x2 5 2 75. log711 2 x2 2 log71x 1 22 5 log7 x 76. log51x 1 12 2 log51x 2 12 5 log51x2 77. ln!x 1 4 2 ln!x 2 2 5 ln!x 1 1 78. logA!1 2 x B 2 logA!x 1 2B 5 log x • A P P L I C A T I O N S 79. Health. After strenuous exercise, Sandy’s heart rate R (beats per minute) can be modeled by R1t2 5 151e20.055t, 0 # t # 15 where t is the number of minutes that have elapsed after she stops exercising. a. Find Sandy’s heart rate at the end of exercising (when she stops at time t 5 0). b. Determine how many minutes it takes after Sandy stops exercising for her heart rate to drop to 100 beats per minute. Round to the nearest minute. c. Find Sandy’s heart rate 15 minutes after she had stopped exercising. 80. Business. A local business purchased a new company van for $45,000. After 2 years the book value of the van is $30,000. a. Find an exponential model for the value of the van using V1t2 5 V0 ekt where V is the value of the van in dollars and t is time in years. b. Approximately how many years will it take for the book value of the van to drop to $20,000? 81. Money. If money is invested in a savings account earning 3.5% interest compounded yearly, how many years will pass until the money triples? 82. Money. If money is invested in a savings account earning 3.5% interest compounded monthly, how many years will pass until the money triples? 83. Money. If $7500 is invested in a savings account earning 5% interest compounded quarterly, how many years will pass until there is $20,000? 84. Money. If $9000 is invested in a savings account earning 6% interest compounded continuously, how many years will pass until there is $15,000? 85. Earthquakes. On September 25, 2003, an earthquake that measured 7.4 on the Richter scale shook Hokkaido, Japan. How much energy (joules) did the earthquake emit? 86. Earthquakes. Again, on that same day (September 25, 2003), a second earthquake that measured 8.3 on the Richter scale shook Hokkaido, Japan. How much energy ( joules) did the earthquake emit? 87. Sound. Matt likes to drive around campus in his classic Mustang with the stereo blaring. If his boom stereo has a sound intensity of 120 dB, how many watts per square meter does the stereo emit? 88. Sound. The New York Philharmonic has a sound intensity of 100 dB. How many watts per square meter does the orchestra emit? 89. Anesthesia. When a person has a cavity filled, the dentist typically administers a local anesthetic. After leaving the dentist’s office, one’s mouth often remains numb for several more hours. If a shot of anesthesia is injected into the bloodstream at the time of the procedure 1t 5 02, and the amount of anesthesia still in the bloodstream t hours after the initial injection is given by A 5 A0e20.5t, in how many hours will only 10% of the original anesthetic still be in the bloodstream? 90. Investments. Money invested in an account that compounds interest continuously at a rate of 3% a year is modeled by A 5 A0e0.03t, where A is the amount in the investment after t years and A0 is the initial investment. How long will it take the initial investment to double? 91. Biology. The U.S. Fish and Wildlife Service is releasing a population of the endangered Mexican gray wolf in a protected area along the New Mexico and Arizona border. It estimates the population of the Mexican gray wolf to be approximated by P1t2 5 200 1 1 24e20.2t How many years will it take for the population to reach 100 wolves? 92. Introducing a New Car Model. If the number of new model Honda Accord hybrids purchased in North America is given by N 5 100,000 1 1 10e22t , where t is the number of weeks after 5.4 Exponential and Logarithmic Equations 473 Honda releases the new model, how many weeks will it take after the release until there are 50,000 Honda hybrids from that batch on the road? 93. Earthquakes. A P wave measures 6.2 on the Richter scale, and an S wave measures 3.3 on the Richter scale. What is their combined measure on the Richter scale? 94. Sound. You and a friend get front row seats to a rock concert. The music level is 100 dB, and your normal conversation is 60 dB. If your friend is telling you something during the concert, how many decibels are you subjecting yourself to? • C A T C H T H E M I S T A K E In Exercises 95–98, explain the mistake that is made. 95. Solve the equation: 4ex 5 9. Solution: Take the natural log of both sides. ln14ex2 5 ln 9 Apply the property of inverses. 4x 5 ln 9 Solve for x. x 5 ln 9 4 < 0.55 This is incorrect. What mistake was made? 96. Solve the equation: log1x2 1 log132 5 1. Solution: Apply the product property (5). log13x2 5 1 Exponentiate (base 10). 10log(3x) 5 1 Apply the property of inverses. 3x 5 1 Solve for x. x 5 1 3 This is incorrect. What mistake was made? 97. Solve the equation: log1x2 1 log1x 1 32 5 1 for x. Solution: Apply the product property (5). log1x2 1 3x2 5 1 Exponentiate both sides (base 10). 10log1x213x2 5 101 Apply the property of inverses. x2 1 3x 5 10 Factor. 1x 1 52 1x 2 22 5 0 Solve for x. x 5 25 and x 5 2 This is incorrect. What mistake was made? 98. Solve the equation: log x 1 log 2 5 log 5. Solution: Combine the logarithms on the left. log1x 1 22 5 log 5 Apply the property of one-to-one functions. x 1 2 5 5 Solve for x. x 5 3 This is incorrect. What mistake was made? In Exercises 99–102, determine whether each statement is true or false. 99. The sum of logarithms with the same base is equal to the logarithm of the product. 100. A logarithm squared is equal to two times the logarithm. 101. elog x 5 x 102. ex 5 22 has no solution. 103. Solve for x in terms of b: 1 3 logb 1x3 2 1 1 2 logb 1x2 2 2x 1 12 5 2 104. Solve exactly: 2 logb1x2 1 2 logb11 2 x2 5 4. • C O N C E P T U A L 105. Solve y 5 3000 1 1 2e20.2t for t in terms of y. 106. State the range of values of x such that the following identity holds: eln1x22a2 5 x2 2 a. 107. A function called the hyperbolic cosine is defined as the average of exponential growth and exponential decay by ƒ1x2 5 ex 1 e2x 2 . Find its inverse. 108. A function called the hyperbolic sine is defined by ƒ1x2 5 ex 2 e2x 2 . Find its inverse. • C H A L L E N G E 109. Solve the equation ln 3x 5 ln1x2 1 12. Using a graphing calculator, plot the graphs y 5 ln13x2 and y 5 ln1x2 1 12 in the same viewing rectangle. Zoom in on the point where the graphs intersect. Does this agree with your solution? 110. Solve the equation 10x2 5 0.001x. Using a graphing calculator, plot the graphs y 5 10x2 and y 5 0.001x in the same viewing rectangle. Does this confirm your solution? 111. Use a graphing utility to help solve 3x 5 5x 1 2. 112. Use a graphing utility to help solve log x2 5 ln1x 2 32 1 2. 113. Use a graphing utility to graph y 5 ex 1 e2x 2 . State the domain. Determine if there are any symmetries and asymptotes. 114. Use a graphing utility to graph y 5 ex 1 e2x ex 2 e2x. State the domain. Determine if there are any symmetries and asymptotes. • T E C H N O L O G Y 474 CHAPTER 5 Exponential and Logarithmic Functions S K I L L S O B J E C T I V E S ■ ■Use exponential growth functions to model populations, economic appreciation of assets, and initial spreading of diseases. ■ ■Use exponential decay functions to model scenarios in medicine, economic depreciation of assets, archaeology, and forensic science. ■ ■Represent distributions using a Gaussian (normal) model. ■ ■Represent restricted growth with logistic models. ■ ■Use logistic growth models to determine time to pay off debt. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that exponential growth models assume “uninhibited” growth. ■ ■Understand that the horizontal asymptote corresponds to a limiting value, often zero. ■ ■Understand that the bell curve is used to represent standardized tests (IQ/SAT) as well as height/weight charts. ■ ■Understand that the asymptote of a logistic model corresponds to the carrying capacity of the system. ■ ■Understand why doubling your monthly payment will reduce the life of the loan by more than half. 5.5 EXPONENTIAL AND LOGARITHMIC MODELS The following table summarizes the five primary models that involve exponential and logarithmic functions. NAME MODEL GRAPH APPLICATIONS Exponential growth ƒ1x2 5 cekx k . 0 x y World populations, bacteria growth, appreciation, global spread of the HIV virus Exponential decay ƒ1x2 5 ce2kx k . 0 x y Radioactive decay, carbon dating, depreciation Gaussian (normal) distribution ƒ1x2 5 ce2(x2a)2/k x y Bell curve (grade distribution), life expectancy, height/weight charts, intensity of a laser beam, IQ tests Logistic growth ƒ1x2 5 a 1 1 ce2kx x y Conservation biology, learning curve, spread of virus on an island, carrying capacity Logarithmic ƒ1x2 5 a 1 c log x ƒ1x2 5 a 1 c ln x x y Population of species, anesthesia wearing off, time to pay off credit cards 5.5 Exponential and Logarithmic Models 475 5.5.1 Exponential Growth Models Quite often one will hear that something “grows exponentially,” meaning that it grows very fast and at increasing speed. In mathematics, the precise meaning of exponential growth is a growth rate of a function that is proportional to its current size. Let’s assume you get a 5% raise every year in a government job. If your starting annual salary out of college is $40,000, then your first raise will be $2000. Fifteen years later your annual salary will be approximately $83,000 and your next 5% raise will be around $4150. The raise is always 5% of the current salary, so the larger the current salary, the larger the raise. In Section 5.1 we saw that interest that is compounded continuously is modeled by A 5 Pert. Here A stands for amount and P stands for principal. There are similar models for populations; these take the form N1t2 5 N0ert, where N0 represents the number of people at time t 5 0, r is the growth rate, t is time in years, and N represents the number of people at time t. In general, any model of the form ƒ1x2 5 cekx, k . 0, models exponential growth. 5.5.1 S K I L L Use exponential growth functions to model populations, economic appreciation of assets, and initial spreading of diseases. 5.5.1 C O N C E P T U A L Understand that exponential growth models assume “uninhibited” growth. EXAMPLE 1 World Population Projections The world population is the total number of humans on Earth at a given time. In 2000 the world population was 6.1 billion, and in 2005 the world population was 6.5 billion. Find the relative growth rate and determine what year the population will reach 9 billion. Solution: Assume an exponential growth model. N1t2 5 N0ert Let t 5 0 correspond to 2000. N102 5 N0 5 6.1 In 2005, t 5 5, the population was 6.5 billion. 6.5 5 6.1e5r Solve for r. 6.5 6.1 5 e5r ln a6.5 6.1b 5 lnAe5rB ln a6.5 6.1b 5 5r r < 0.012702681 The relative growth rate is approximately 1.3% per year. Assuming the growth rate stays the same, write a population model. N1t2 5 6.1e0.013t Let N1t2 5 9. 9 5 6.1e0.013t Solve for t. e0.013t 5 9 6.1 ln1e0.013t2 5 ln a 9 6.1b 0.013t 5 ln a 9 6.1b t < 29.91813894 In 2030 the world population will reach 9 billion if the same growth rate is maintained. Y OUR T UR N The population of North America was 300 million in 1995 and 332 million in 2005. Find the relative growth rate and determine what year the population will reach 1 billion. ▼ A N S W E R 1% per year; 2115 ▼ [CONCEPT CHECK] TRUE OR FALSE Investments that compound continuously are examples of exponential growth. ANSWER True ▼ 476 CHAPTER 5 Exponential and Logarithmic Functions 5.5.2 Exponential Decay Models We mentioned radioactive decay briefly in Section 5.1. Radioactive decay is the process in which radioactive elements (atoms) lose energy by emitting radiation in the form of particles. This results in loss of mass. This process is random, but given a large number of atoms, the decay rate is directly proportional to the mass of the radioactive substance. Since the mass is decreasing, we say this represents exponential decay, m 5 m0e2rt, where m0 represents the initial mass at time t 5 0, r is the decay rate, t is time, and m represents the mass at time t. In general, any model of the form ƒ1x2 5 ce2kx, k . 0, models exponential decay. Typically, the decay rate r is expressed in terms of the half-life h. Recall (Section 5.1) that half-life is the time it takes for a quantity to decrease by half. WORDS MATH Write the radioactive decay model. m 5 m0e2rt Divide both sides by m0. m m0 5 e2rt The remaining mass is half of the initial mass when t 5 h. 1 2 5 e2rh Solve for r. Take the natural logarithm of both sides. ln a1 2b 5 lnAe2rhB Simplify. ln 1 2 ln 2 5 2rh rh 5 ln 2 r 5 ln 2 h 3 0 5.5.2 S KILL Use exponential decay functions to model scenarios in medicine, economic depreciation of assets, archaeology, and forensic science. 5.5.2 CO NCE PTUAL Understand that the horizontal asymptote corresponds to a limiting value, often zero. EXAMPLE 2 Radioactive Decay The radioactive isotope of potassium 42K, which is vital in the diagnosis of brain tumors, has a half-life of 12.36 hours. a. Determine the exponential decay model that represents the mass of 42K. b. If 500 milligrams of potassium-42 are taken, how many milligrams will remain after 48 hours? c. How long will it take for the original 500-milligram sample to decay to a mass of 5 milligrams? Solution (a): Write the relationship between rate of decay and half-life. r 5 ln 2 h Let h 5 12.36. r < 0.056 Write the exponential decay model for the mass of 42K. m 5 m0e20.056t Solution (b): Let m0 5 500 and t 5 48. m 5 500e210.05621482 < 34.00841855 There are approximately 34 milligrams of 42K still in the body after 48 hours. Note: Had we used the full value of r 5 0.056079868, the resulting mass would have been m 5 33.8782897, which is approximately 34 milligrams. [CONCEPT CHECK] TRUE OR FALSE The value of a car in age (how old the car is) is an example of exponential decay. ANSWER True ▼ 5.5 Exponential and Logarithmic Models 477 5.5.3 Gaussian (Normal) Distribution Models If your instructor plots the grades from the last test, typically you will see a Gaussian (normal) distribution of scores, otherwise known as the bell-shaped curve. Other examples of phenomena that tend to follow a Gaussian distribution are SAT scores, height distributions of adults, and standardized tests like IQ assessments. The graph to the right represents a Gaussian distribution of IQ scores. The average score, which for IQ is 100, is the x-value at which the maximum occurs. The typical probability distribution is F1x2 5 1 s!2p e21x2m22/ 2s2 where m is the average or mean value and the variance is s2. Any model of the form ƒ1x2 5 ce21x2a22/k is classified as a Gaussian model. Solution (c): Write the exponential decay model for the mass of 42K. m 5 m0e20.056t Let m 5 5 and m0 5 500. 5 5 500e20.056t Solve for t. Divide by 500. e20.056t 5 5 500 5 1 100 Take the natural logarithm of both sides. lnAe20.056tB 5 ln a 1 100b Simplify. 20.056t 5 ln a 1 100b Divide by 20.056 and approximate with a calculator. t < 82.2352 It will take approximately 82 hours for the original 500-milligram substance to decay to a mass of 5 milligrams. YOUR T UR N The radioactive element radon-222 has a half-life of 3.8 days. a. Determine the exponential decay model that represents the mass of radon-222. b. How much of a 64-gram sample of radon-222 will remain after 7 days? Round to the nearest gram. c. How long will it take for the original 64-gram sample to decay to a mass of 4 grams? Round to the nearest day. ▼ ▼ A N S W E R a. m 5 m0e20.1824t b. 18 grams c. 15 days 5.5.3 S K I L L Represent distributions using a Gaussian (normal) model. 5.5.3 C ON C E P T U A L Understand that the bell curve is used to represent standardized tests (IQ/SAT) as well as height/ weight charts. EXAMPLE 3 Weight Distributions Suppose each member of a Little League football team is weighed and the weight distribution follows the Gaussian model ƒ1x2 5 10e21x210022/25. a. Graph the weight distribution. b. What is the average weight of the members of this team? c. Approximately how many boys weigh 95 pounds? 55 Number of scores 70 85 100 Intelligence quotient (Score on Wechsler Adult Intelligence Scale) 115 130 145 2% 0.1% 0.1% 2% 14% 34% 34% 14% 68% 96% [CONCEPT CHECK] To be at approximately the 98th percentile in IQ, what score thresh-old would you have to meet? ANSWER 130 ▼ 478 CHAPTER 5 Exponential and Logarithmic Functions 5.5.4 Logistic Growth Models Earlier in this section, we discussed exponential growth models for populations that experience uninhibited growth. Now we will turn our attention to logistic growth, which models population growth when there are factors that impact the ability to grow, such as food and space. For example, if 10 rabbits are dropped off on an uninhabited island, they will reproduce and the population of rabbits on that island will experience rapid growth. The population will continue to increase rapidly until the rabbits start running out of space or food on the island. In other words, under favorable conditions the growth is not restricted, while under less favorable conditions the growth becomes restricted. This type of growth is represented by logistic growth models, ƒ1x2 5 a 1 1 ce2kx. Ultimately, the population of rabbits reaches the island’s carrying capacity, a. Solution: a. x y 150 60 10 b. 100 pounds c. ƒ1952 5 10e2195210022/25 5 10e225/25 5 10e21 < 3.6788 Approximately 4 boys weigh 95 pounds. 5.5.4 S KILL Represent restricted growth with logistic models. 5.5.4 CO NCE PTUAL Understand that the asymptote of a logistic model corresponds to the carrying capacity of the system. EXAMPLE 4 Number of Students on a College Campus In 2011, the University of Central Florida (UCF) was the second largest university in the country. The number of students can be modeled by the function: ƒ1x2 5 60,000 1 1 5e20.12t , where t is time in years and t 5 0 corresponds to 1970. a. How many students attended UCF in 1990? Round to the nearest thousand. b. How many students attended UCF in 2010? c. What is the carrying capacity of the UCF main campus? Round all answers to the nearest thousand. Solution (a): Let t 5 20. ƒ1x2 5 60,000 1 1 5e20.121202 < 41,000 Solution (b): Let t 5 40. ƒ1x2 5 60,000 1 1 5e20.121402 < 58,000 Solution (c): As t increases, the UCF student population approaches 60,000 . 10 (1980) 20 30 (2000) 40 60 50 (2020) 5,000 10,000 15,000 20,000 25,000 30,000 35,000 40,000 45,000 50,000 55,000 60,000 t (time in years) f (t) (number of students) 1990 1970 2010 [CONCEPT CHECK] Why would the spread of a disease on an island be modeled with a logistic growth model as opposed to an exponential growth model? ANSWER Once everyone on the island is infected, there are no more people to infect. ▼ 5.5 Exponential and Logarithmic Models 479 5.5.5 Logarithmic Models Homeowners typically ask the question, “If I increase my payment, how long will it take to pay off my current mortgage?” In general, a loan over t years with an annual interest rate r with n periods per year corresponds to an interest rate per period of i 5 r n. Typically loans are paid in equal payments consisting of the principal P plus total interest divided by the total number of periods over the life of the loan nt. The periodic payment R is given by R 5 P i 1 2 11 1 i22nt We can find the time (in years) it will take to pay off the loan as a function of periodic payment by solving for t. WORDS MATH Multiply both sides by 1 2 11 1 i 22nt. R31 2 11 1 i22nt4 5 Pi Eliminate the brackets. R 2 R 11 1 i22nt 5 Pi Subtract R. 2R 11 1 i22nt 5 Pi 2 R Divide by 2R. 11 1 i22nt 5 1 2 Pi R Take the natural log of both sides. ln11 1 i22nt 5 ln a1 2 Pi R b Use the power property for logarithms. 2nt ln11 1 i2 5 ln a1 2 Pi R b Isolate t. t 5 2ln11 2 Pi/R2 n ln11 1 i2 Let i 5 r n. t 5 2ln11 2 Pr/1nR22 n ln11 1 r/n2 5.5.5 S K I L L Use logistic growth models to determine time to pay off debt. 5.5.5 C O N C E P T U A L Understand why doubling your monthly payment will reduce the life of the loan by more than half. EXAMPLE 5 Paying Off Credit Cards James owes $15,000 on his credit card. The annual interest rate is 13% compounded monthly. a. Find the time it will take to pay off his credit card if he makes payments of $200 per month. b. Find the time it will take to pay off his credit card if he makes payments of $400 per month. Let P 5 15,000, r 5 0.13, and n 5 12. t 5 2 ln a1 2 15,00010.132 12R b 12 ln a1 1 0.13 12 b Solution (a): Let R 5 200. t 5 2 ln a1 2 15,00010.132 1212002 b 12 ln a1 1 0.13 12 b < 13 $200 monthly payments will allow James to pay off his credit card in about 13 years . [CONCEPT CHECK] TRUE OR FALSE If you double your payments on a loan, you will pay off the loan in half the time. ANSWER False ▼ 480 CHAPTER 5 Exponential and Logarithmic Functions Solution (b): Let R 5 400. t 5 2 ln a1 2 15,00010.132 1214002 b 12 ln a1 1 0.13 12 b < 4 $400 monthly payments will allow James to pay off the balance in approximately 4 years . It is important to note that doubling the payment reduced the time to pay off the balance to less than a third. [SEC TION 5.5] S U M M A RY In this section, we discussed five main types of models that involve exponential and logarithmic functions. NAME MODEL APPLICATIONS Exponential growth ƒ1x2 5 cekx, k . 0 Uninhibited growth (populations/inflation) Exponential decay ƒ1x2 5 ce2kx, k . 0 Carbon dating, depreciation Gaussian (normal) distributions ƒ1x2 5 ce21x2a22/k Bell curves (standardized tests, height/weight charts, distribution of power flux of laser beams) Logistic growth ƒ1x2 5 a 1 1 ce2kx Conservation biology (growth limited by factors like food and space), learning curve Logarithmic ƒ1x2 5 a 1 c log x ƒ1x2 5 a 1 c ln x or quotients of logarithmic functions Time to pay off credit cards, annuity planning EXAMPLE7 [SEC TION 5.5] E X E R C I SE S • S K I L L S In Exercises 1–6, match the function with the graph (a to f) and the model name (i to v). 1. ƒ1t2 5 5e2t 2. N 1t2 5 28e2t/2 3. T1x2 5 4e21x28022/10 4. P1t2 5 200 1 1 5e20.4t 5. D 1x2 5 4 1 log 1x 2 12 6. h 1t2 5 2 1 ln 1t 1 32 Model Name i. Logarithmic ii. Logistic iii. Gaussian iv. Exponential growth v. Exponential decay Graphs a. x y 100 60 10 b. x y 14 –6 200 c. x y 5 –5 10 5.5 Exponential and Logarithmic Models 481 d. x y 20 20 e. x y 5 –5 6 –4 f. x y 5 –5 5 –5 • A P P L I C A T I O N S 7. Population Growth. The population of the Philippines in 2003 was 80 million. Their population increases 2.36% per year. What is the expected population of the Philippines in 2020? Apply the formula N 5 N0ert, where N represents the number of people. 8. Population Growth. China’s urban population is growing at 2.5% a year, compounding continuously. If there were 13.7 million people in Shanghai in 1996, approximately how many people will there be in 2020? Apply the formula N 5 N0ert, where N represents the number of people. 9. Population Growth. Port St. Lucie, Florida, had the United States’ fastest growth rate among cities with a population of 100,000 or more between 2003 and 2004. In 2003, the population was 103,800 and increasing at a rate of 12% per year. In what year should the population reach 260,000? (Let t 5 0 correspond to 2003.) Apply the formula N 5 N0ert, where N represents the number of people. 10. Population Growth. San Francisco’s population has been declining since the “dot com” bubble burst. In 2002, the population was 776,000. If the population is declining at a rate of 1.5% per year, in what year will the population be 600,000? (Let t 5 0 correspond to 2002.) Apply the formula N 5 N0e2rt, where N represents the number of people. 11. Cellular Phone Plans. The number of cell phones in China is exploding. In 2007, there were 487.4 million cell phone subscribers, and the number was increasing at a rate of 16.5% per year. How many cell phone subscribers were there in 2016 according to this model? Use the formula N 5 N0ert, where N represents the number of cell phone subscribers. Let t 5 0 correspond to 2007. 12. Bacteria Growth. A colony of bacteria is growing exponentially. Initially, 500 bacteria were in the colony. The growth rate is 20% per hour. (a) How many bacteria should be in the colony in 12 hours? (b) How many in one day? Use the formula N 5 N0ert, where N represents the number of bacteria. 13. Real Estate Appreciation. In 2004, the average house in Las Vegas cost $185,000, and real estate prices were increasing at an amazing rate of 30% per year. What was the expected cost of an average house in Las Vegas in 2016? Use the formula N 5 N0ert, where N represents the average cost of a home. Round to the nearest thousand. 14. Real Estate Appreciation. The average cost of a single-family home in California in 2004 was $230,000. In 2005, the average cost was $252,000. If this trend continued, what was the expected cost in 2016? Use the formula N 5 N0ert, where N represents the average cost of a home. Round to the nearest thousand. 15. Oceanography (Growth of Phytoplankton). Phytoplankton are microscopic plants that live in the ocean. Phytoplankton grow abundantly in oceans around the world and are the foundation of the marine food chain. One variety of phytoplankton growing in tropical waters is increasing at a rate of 20% per month. If it is estimated that there are 100 million in the water, how many will there be in 6 months? Utilize formula N 5 N0ert, where N represents the population of phytoplankton. 16. Oceanography (Growth of Phytoplankton). In Arctic waters there are an estimated 50 million phytoplankton. The growth rate is 12% per month. How many phytoplankton will there be in 3 months? Utilize formula N 5 N0ert, where N represents the population of phytoplankton. 17. HIV/AIDS. In 2003, an estimated 1 million people had been infected with HIV in the United States. If the infection rate increases at an annual rate of 2.5% a year compounding continuously, how many Americans will be infected with the HIV virus by 2020? 18. HIV/AIDS. In 2003, there were an estimated 25 million people who had been infected with HIV in sub-Saharan Africa. If the infection rate increases at an annual rate of 9% a year compounding continuously, how many Africans will be infected with the HIV virus by 2020? 19. Anesthesia. When a person has a cavity filled, the dentist typically gives a local anesthetic. After leaving the dentist’s office, one’s mouth often is numb for several more hours. If 100 ml of anesthesia is injected into the local tissue at the time of the procedure 1t 5 02, and the amount of anesthesia still in the local tissue t hours after the initial injection is given by A 5 100e20.5t, how much is in the local tissue 4 hours later? 20. Anesthesia. When a person has a cavity filled, the dentist typically gives a local anesthetic. After leaving the dentist’s office, one’s mouth often is numb for several more hours. If 100 ml of anesthesia is injected into the local tissue at the time of the procedure 1t 5 02, and the amount of anesthesia still in the local tissue t hours after the initial injection is given by A 5 100e20.5t, how much is in the local tissue 12 hours later? 21. Business. The sales S (in thousands of units) of a new mp3 player after it has been on the market for t years can be modeled by S1t2 5 75011 2 e2kt 2 482 CHAPTER 5 Exponential and Logarithmic Functions a. If 350,000 units of the mp3 player were sold in the first year, find k to four decimal places. b. Use the model found in part (a) to estimate the sales of the mp3 player after it has been on the market for 3 years. 22. Business. During an economic downturn, the annual profits of a company dropped from $850,000 in 2013 to $525,000 in 2015. Assume the exponential model P1t2 5 P 0 ekt for the annual profit where P is profit in thousands of dollars and t is time in years. a. Find the exponential model for the annual profit. b. Assuming the exponential model is applicable in the year 2018, estimate the profit (to the nearest thousand dollars) for the year 2018. 23. Radioactive Decay. Carbon-14 has a half-life of 5730 years. How long will it take 5 grams of carbon-14 to be reduced to 2 grams? 24. Radioactive Decay. Radium-226 has a half-life of 1600 years. How long will it take 5 grams of radium-226 to be reduced to 2 grams? 25. Radioactive Decay. The half-life of uranium-238 is 4.5 billion years. If 98% of uranium-238 remains in a fossil, how old is the fossil? 26. Radioactive Decay. A drug has a half-life of 12 hours. If the initial dosage is 5 milligrams, how many milligrams will be in the patient’s body in 16 hours? In Exercises 27–30, use the following formula for Newton’s Law of Cooling. If you take a hot dinner out of the oven and place it on the kitchen countertop, the dinner cools until it reaches the temperature of the kitchen. Likewise, a glass of ice set on a table in a room eventually melts into a glass of water at that room temperature. The rate at which the hot dinner cools or the ice in the glass melts at any given time is proportional to the difference between its temperature and the temperature of its surroundings (in this case, the room). This is called Newton’s Law of Cooling (or warming) and is modeled by T 5 TS 1 1T0 2 TS2e2kt where T is the temperature of an object at time t, Ts is the temperature of the surrounding medium, T0 is the temperature of the object at time t 5 0, t is the time, and k is a constant. 27. Newton’s Law of Cooling. An apple pie is taken out of the oven with an internal temperature of 3258F. It is placed on a rack in a room with a temperature of 728F. After 10 minutes, the temperature of the pie is 2008F. What will be the temperature of the pie 30 minutes after coming out of the oven? 28. Newton’s Law of Cooling. A cold drink is taken out of an ice chest with a temperature of 388F and placed on a picnic table with a surrounding temperature of 758F. After 5 minutes the temperature of the drink is 458F. What will the temperature of the drink be 20 minutes after it is taken out of the chest? 29. Forensic Science (Time of Death). A body is discovered in a hotel room. At 7:00 a.m. a police detective found the body’s temperature to be 858F. At 8:30 a.m. a medical examiner measures the body’s temperature to be 828F. Assuming the room in which the body was found had a constant temperature of 748F, how long has the victim been dead? (Normal body temperature is 98.68F.) 30. Forensic Science (Time of Death). At 4 a.m. a body is found in a park. The police measure the body’s temperature to be 908F. At 5 a.m. the medical examiner arrives and determines the temperature to be 868F. Assuming the temperature of the park was constant at 608F, how long has the victim been dead? 31. Depreciation of Automobile. A new Lexus IS250 has a book value of $38,000, and after one year it has a book value of $32,000. What is the car’s value in 4 years? Apply the formula N 5 N0e2rt, where N represents the value of the car. Round to the nearest hundred. 32. Depreciation of Automobile. A new Hyundai Tiburon has a book value of $22,000, and after 2 years a book value of $14,000. What is the car’s value in 4 years? Apply the formula N 5 N0e2rt, where N represents the value of the car. Round to the nearest hundred. 33. Automotive. A new model BMW convertible coupe is designed and produced in time to appear in North America in the fall. BMW Corporation has a limited number of new models available. The number of new model BMW convertible coupes purchased in North America is given by N 5 100,000 1 1 10e22t, where t is the number of weeks after the BMW is released. a. How many new-model BMW convertible coupes will have been purchased 2 weeks after the new model becomes available? b. How many after 30 weeks? c. What is the maximum number of new model BMW con vertible coupes that will be sold in North America? 34. iPhone. The number of iPhones purchased is given by N 5 2,000,000 1 1 2e24t , where t is the time in weeks after they are made available for purchase. a. How many iPhones are purchased within the first 2 weeks? b. How many iPhones are purchased within the first month? 35. Spread of a Virus. The number of MRSA (methicillin- resistant Staphylococcus aureus) cases has been rising sharply in England and Wales since 1997. In 1997, 2422 cases were reported. The number of cases reported in 2003 was 7684. How many cases will be expected in 2020? (Let t 5 0 correspond to 1997.) Use the formula N 5 N0ert, where N represents the number of cases reported. 36. Spread of a Virus. Dengue fever, an illness carried by mosquitoes, is occurring in one of the worst outbreaks in decades across Latin America and the Caribbean. In 2004, 300,000 cases were reported, and 630,000 cases in 2007. How many cases might be expected in 2020? (Let t 5 0 be 2004.) Use the formula N 5 N0ert, where N represents the number of cases. 37. Carrying Capacity. The Virginia Department of Fish and Game stocks a mountain lake with 500 trout. Officials believe the lake can support no more than 10,000 trout. The number of trout is given by N 5 10,000 1 1 19e21.56t, where t is time in years. How many years will it take for the trout population to reach 5000? 5.5 Exponential and Logarithmic Models 483 38. Carrying Capacity. The World Wildlife Fund has placed 1000 rare pygmy elephants in a conservation area in Borneo. They believe 1600 pygmy elephants can be supported in this environment. The number of elephants is given by N 5 1600 1 1 0.6e20.14t, where t is time in years. How many years will it take the herd to reach 1200 elephants? 39. Lasers. The intensity of a laser beam is given by the ratio of power to area. A particular laser beam has an intensity function given by I 5 e2r2 mW/cm2, where r is the radius off the center axis given in cm. Where is the beam brightest (largest intensity)? 40. Lasers. The intensity of a laser beam is given by the ratio of power to area. A particular laser beam has an intensity function given by I 5 e2r2 mW/cm2, where r is the radius off the center axis given in cm. What percentage of the on-axis intensity 1r 5 02 corresponds to r 5 2 cm? 41. Grade Distribution. Suppose the first test in this class has a normal, or bell-shaped, grade distribution of test scores, with an average score of 75. An approximate function that models your class’s grades on test 1 is N1x2 5 10e21x27522/252, where N represents the number of students who received the score x. a. Graph this function. b. What is the average grade? c. Approximately how many students scored a 50? d. Approximately how many students scored 100? 42. Grade Distribution. Suppose the final exam in this class has a normal, or bell-shaped, grade distribution of exam scores, with an average score of 80. An approximate function that models your class’s grades on the exam is N1x2 5 10e21x28022/162, where N represents the number of students who received the score x. a. Graph this function. b. What is the average grade? c. Approximately how many students scored a 60? d. Approximately how many students scored 100? 43. Time to Pay Off Debt. Diana just graduated from medical school owing $80,000 in student loans. The annual interest rate is 9%. a. Approximately how many years will it take to pay off her student loan if she makes a monthly payment of $750? b. Approximately how many years will it take to pay off her loan if she makes a monthly payment of $1000? 44. Time to Pay Off Debt. Victor owes $20,000 on his credit card. The annual interest rate is 17%. a. Approximately how many years will it take him to pay off this credit card if he makes a monthly payment of $300? b. Approximately how many years will it take him to pay off this credit card if he makes a monthly payment of $400? For Exercises 45 and 46, refer to the following: A local business borrows $200,000 to purchase property. The loan has an annual interest rate of 8% compounded monthly and a minimum monthly payment of $1467. 45. Time to Pay Off Debt/Business. a. Approximately how many years will it take the business to pay off the loans if only the minimum payment is made? b. How much interest will the business pay over the life of the loan if only the minimum payment is made? 46. Time to Pay Off Debt/Business. a. Approximately how many years will it take the business to pay off the loan if the minimum payment is doubled? b. How much interest will the business pay over the life of the loan if the minimum payment is doubled? c. How much in interest will the business save by doubling the minimum payment (see Exercise 45, part b)? • C A T C H T H E M I S T A K E In Exercises 47 and 48, explain the mistake that is made. 47. The city of Orlando, Florida, has a population that is growing at 7% a year, compounding continuously. If there were 1.1 million people in greater Orlando in 2006, approximately how many people will there be in 2018? Apply the formula N 5 N0ert, where N represents the number of people. Solution: Use the population growth model. N 5 N0ert Let N0 5 1.1, r 5 7, and t 5 12. N 5 1.1e(7)(12) Approximate with a calculator. 2.8 3 1030 This is incorrect. What mistake was made? 48. The city of San Antonio, Texas, has a population that is growing at 5% a year, compounding continuously. If there were 1.3 million people in the greater San Antonio area in 2006, approximately how many people will there be in 2018? Apply the formula N 5 N0ert, where N represents the number of people. Solution: Use the population growth model. N 5 N0ert Let N0 5 1.3, r 5 5, and t 5 12. N 5 1.3e(5)(12) Approximate with a calculator. 6.7 3 1021 This is incorrect. What mistake was made? In Exercises 49–52, determine whether each statement is true or false. 49. When a species gets placed on an endangered species list, the species begins to grow rapidly and then reaches a carrying capacity. This can be modeled by logistic growth. 50. A professor has 400 students one semester. The number of names (of her students) she can memorize can be modeled by a logarithmic function. 51. The spread of lice at an elementary school can be modeled by exponential growth. 52. If you purchase a laptop computer this year 1t 5 02, then the value of the computer can be modeled with exponential decay. • C O N C E P T U A L 484 CHAPTER 5 Exponential and Logarithmic Functions In Exercises 53 and 54, refer to the logistic model f 1x2 5 a 1 1 ce2kx, where a is the carrying capacity. 53. As c increases, does the model reach the carrying capacity in less time or more time? 54. As k increases, does the model reach the carrying capacity in less time or more time? • C H A L L E N G E 55. Wing Shan just graduated from dental school owing $80,000 in student loans. The annual interest is 6%. Her time t to pay off the loans is given by t 5 2ln11 2 8000010.062/1nR22 n ln11 1 0.06/n2 where n is the number of payment periods per year and R is the periodic payment. a. Use a graphing utility to graph t1 5 2 ln11 2 8000010.062/112x22 12 ln11 1 0.06/122 as Y1 and t2 5 2 ln11 2 8000010.062/126x22 26 ln11 1 0.06/262 as Y2. Explain the difference in the two graphs. b. Use the TRACE key to estimate the number of years it will take Wing Shan to pay off her student loans if she can afford a monthly payment of $800. c. If she can make a biweekly payment of $400, estimate the number of years it will take her to pay off the loans. d. If she adds $200 more to her monthly or $100 more to her biweekly payment, estimate the number of years it will take her to pay off the loans. 56. Hong has a credit card debt in the amount of $12,000. The annual interest is 18%. His time t to pay off the loan is given by t 5 2 ln11 2 1200010.182/1nR22 n ln11 1 0.18/n2 where n is the number of payment periods per year and R is the periodic payment. a. Use a graphing utility to graph t1 5 2 ln11 2 1200010.182/112x22 12 ln11 1 0.18/122 as Y1 and t2 5 2 ln11 2 1200010.182/126x22 26 ln11 1 0.18/262 as Y2. Explain the difference in the two graphs. b. Use the TRACE key to estimate the number of years it will take Hong to pay off his credit card if he can afford a monthly payment of $300. c. If he can make a biweekly payment of $150, estimate the number of years it will take him to pay off the credit card. d. If he adds $100 more to his monthly or $50 more to his biweekly payment, estimate the number of years it will take him to pay off the credit card. • T E C H N O L O G Y CH A P TE R 5 R E VIE W Chapter Review 485 [CH AP TER 5 REVIEW] SECTION CONCEPT KEY IDEAS/FORMULAS 5.1 Expone ntial functions and their graphs Evaluating exponential functions ƒ1x2 5 bx b . 0, b 2 1 Graphs of exponential functions y-intercept 10, 12 Horizontal asymptote: y 5 0; the points 11, b2 and 121, 1/b2 The natural base e ƒ1x2 5 ex Applications of exponential functions Doubling time: P 5 P02t/d Compound interest: A 5 P a1 1 r nb nt Compounded continuously: A 5 Pert 5.2 Logarithmic functions and their graphs y 5 logb x x . 0 b . 0, b 2 1 Evaluating logarithms y 5 logb x and x 5 by Common and natural logarithms y 5 log x Common 1base 102 y 5 ln x Natural 1base e2 Graphs of logarithmic functions x-intercept 11, 02 Vertical asymptote: x 5 0; the points 1b, 12 and 11/b, 212 Domain: 10, q2 Applications of logarithms Decibel scale: D 5 10 log a I IT b IT 5 1 3 10212 W/m2 Richter scale: M 5 2 3 log a E E0 b E0 5 104.4 joules 5.3 Properties of logarithms Properties of logarithmic functions 1. logb 1 5 0 2. logb b 5 1 3. logb bx 5 x 4. blogbx 5 x x . 0 Product property: 5. logb MN 5 logb M 1 logb N Quotient property: 6. logb aM N b 5 logb M 2 logb N Power property: 7. logb Mp 5 p logb M Change-of-base formula logb M 5 log M log b or logb M 5 ln M ln b 5.4 Exponential and logarithmic equations Solving exponential equations Simple exponential equations 1. Rewrite both sides of the equation in terms of the same base. 2. Use the one-to-one property to equate the exponents. 3. Solve for the variable. Complicated exponential equations 1. Isolate the exponential expression. 2. Take the same logarithm of both sides. 3. Simplify using the inverse properties. 4. Solve for the variable. Solving logarithmic equations Simple logarithmic equations 1. Combine logarithms on each side of the equation using properties. 2. Use the one-to-one property to equate the exponents. 3. Solve for the variable. 4. Eliminate any extraneous solutions. CH A P TE R 5 R E VI E W 486 CHAPTR 5 Exponential and Logarithmic Functions SECTION CONCEPT KEY IDEAS/FORMULAS Complicated logarithmic equations 1. Combine and isolate the logarithmic expressions. 2. Rewrite the equation in exponential form. 3. Solve for the variable. 4. Eliminate any extraneous solutions. 5.5 Exponential and logarithmic models Exponential growth models ƒ1x2 5 cekx k . 0 Exponential decay models ƒ1x2 5 ce2kx k . 0 Gaussian (normal) distribution models ƒ1x2 5 ce21x2a22/k Logistic growth models ƒ1x2 5 a 1 1 ce2kx Logarithmic models ƒ1x2 5 a 1 c log x ƒ1x2 5 a 1 c ln x R E VI E W E XERCISES Review Exercises 487 [CH AP TER 5 REVIEW EXE R C IS E S ] 5.1 Exponential Functions and Their Graphs Approximate each number using a calculator and round your answer to two decimal places. 1. 84.7 2. p2/5 3. 4 ? 50.2 4. 1.21.2 Approximate each number using a calculator and round your answer to two decimal places. 5. e3.2 6. ep 7. e!p 8. e22.5!3 Evaluate each exponential function for the given values. 9. ƒ1x2 5 242x ƒ122.22 10. ƒ1x2 5 22x14 ƒ11.32 11. ƒ1x2 5 A2 5B 126x ƒA1 2B 12. ƒ1x2 5 A4 7B 5x11 ƒA1 5B Match the graph with the function. 13. y 5 2x22 14. y 5 2222x 15. y 5 2 1 3x12 16. y 5 22 2 322x a. b. x y –10 10 –10 x y –10 10 10 c. d. x y –10 20 x y 10 –20 State the y-intercept and the horizontal asymptote and graph the exponential function. 17. y 5 262x 18. y 5 4 2 3x 19. y 5 1 1 1022x 20. y 5 4x 2 4 State the y-intercept and horizontal asymptote, and graph the exponential function. 21. y 5 e22x 22. y 5 ex21 23. y 5 3.2ex/3 24. y 5 2 2 e12x Applications 25. Compound Interest. If $4500 is deposited into an account paying 4.5% compounding semiannually, how much will you have in the account in 7 years? 26. Compound Interest. How much money should be put in a savings account now that earns 4.0% a year compounded quarterly if you want $25,000 in 8 years? 27. Compound Interest. If $13,450 is put in a money market account that pays 3.6% a year compounded continuously, how much will be in the account in 15 years? 28. Compound Interest. How much money should be invested today in a money market account that pays 2.5% a year compounded continuously if you desire $15,000 in 10 years? 5.2 Logarithmic Functions and Their Graphs Write each logarithmic equation in its equivalent exponential form. 29. log4 64 5 3 30. log4 2 5 1 2 31. logA 1 100B 5 22 32. log16 4 5 1 2 Write each exponential equation in its equivalent logarithmic form. 33. 63 5 216 34. 1024 5 0.0001 35. 4 169 5 A 2 13B 2 36. ! 3 512 5 8 Evaluate the logarithms exactly. 37. log7 1 38. log4 256 39. log1/6 1296 40. log 1012 Approximate the common and natural logarithms utilizing a calculator. Round to two decimal places. 41. log 32 42. ln 32 43. ln 0.125 44. log 0.125 State the domain of the logarithmic function in interval notation. 45. ƒ1x2 5 log31x 1 22 46. ƒ1x2 5 log212 2 x2 47. ƒ1x2 5 log1x2 1 32 48. ƒ1x2 5 log13 2 x22 Match the graph with the function. 49. y 5 log7 x 50. y 5 2log7 12x2 51. y 5 log7 1x 1 12 2 3 52. y 5 2log7 11 2 x2 1 3 REV IEW E XE R CI SE S 488 CHAPTER 5 Exponential and Logarithmic Functions a. b. x y –10 10 10 x y –10 10 –10 c. d. x y –10 10 10 x y –10 10 –10 Graph the logarithmic function with transformation techniques. 53. ƒ1x2 5 log4 1x 2 42 1 2 54. ƒ1x2 5 log4 1x 1 42 2 3 55. ƒ1x2 5 2log4 1x2 2 6 56. ƒ1x2 5 22 log4 12x2 1 4 Applications 57. Chemistry. Calculate the pH value of milk, assuming it has a concentration of hydrogen ions given by H1 5 3.16 3 1027. 58. Chemistry. Calculate the pH value of Coca-Cola, assuming it has a concentration of hydrogen ions given by H1 5 2.0 3 1023. 59. Sound. Calculate the decibels associated with a teacher speaking to a medium-sized class if the sound intensity is 1 3 1027 W/m2. 60. Sound. Calculate the decibels associated with an alarm clock if the sound intensity is 1 3 1024 W/m2. 5.3 Properties of Logarithms Use the properties of logarithms to simplify each expression. 61. log2.5 2.5 62. log2 !16 63. 2.5log2.5 6 64. e23 ln 6 Write each expression as a sum or difference of logarithms. 65. logc xayb 66. log3 x2y23 67. logj ars t3 b 68. log xc !x 1 5 69. log c a1/2 b3/2c2/5d 70. log7 c c3d1/3 e6 d 1/3 Evaluate the logarithms using the change-of-base formula. 71. log8 3 72. log5 1 2 73. logp 1.4 74. log!3 2.5 5.4 Exponential and Logarithmic Equations Solve the exponential equations exactly for x. 75. 4x 5 1 256 76. 3x2 5 81 77. e3x24 5 1 78. e"x 5 e4.8 79. A1 3B x12 5 81 80. 100x223 5 10 Solve the exponential equation. Round your answer to three decimal places. 81. e2x13 2 3 5 10 82. 22x21 1 3 5 17 83. e2x 1 6ex 1 5 5 0 84. 4e0.1x 5 64 85. 12x 2 22x2 1 2x 1 22x2 5 0 86. 512x2 5 25 Solve the logarithmic equations exactly. 87. log13x2 5 2 88. log31x 1 22 5 4 89. log4 x 1 log4 2x 5 8 90. log6 x 1 log612x 2 12 5 log6 3 Solve the logarithmic equations. Round your answers to three decimal places. 91. ln x2 5 2.2 92. ln13x 2 42 5 7 93. log312 2 x2 2 log31x 1 32 5 log3 x 94. 4 log1x 1 12 2 2 log1x 1 12 5 1 5.5 Exponential and Logarithmic Models 95. Compound Interest. If Tania needs $30,000 a year from now for a down payment on a new house, how much should she put in a 1-year CD earning 5% a year compounding continuously so that she will have exactly $30,000 a year from now? 96. Stock Prices. Jeremy is tracking the stock value of Best Buy (BBY on the NYSE). In 2003, he purchased 100 shares at $28 a share. The stock did not pay dividends because the company reinvested all earnings. In 2005, Jeremy cashed out and sold the stock for $4000. What was the annual rate of return on BBY? 97. Compound Interest. Money is invested in a savings account earning 4.2% interest compounded quarterly. How many years will pass until the money doubles? 98. Compound Interest. If $9000 is invested in an investment earning 8% interest compounded continuously, how many years will pass until there is $22,500? 99. Population. Nevada has the fastest growing population according to the U.S. Census Bureau. In 2004, the population of Nevada was 2.62 million and increasing at an annual rate of 3.5%. What is the expected population in 2020? (Let t 5 0 be 2004.) Apply the formula N 5 N0ert, where N is the population. 100. Population. The Hispanic population is the fastest growing ethnic group in the United States. In 1996, there were an estimated 28.3 million Hispanics in the United States, and in 2000 there were an estimated 32.5 million. What is the expected population of Hispanics in the United States in 2020? (Let t 5 0 be 1996.) Apply the formula N 5 N0ert, where N is the population. R E VI E W E XERCISES Review Exercises 489 101. Bacteria Growth. Bacteria are growing exponentially. Initially, there were 1000 bacteria; after 3 hours there were 2500. How many bacteria should be expected in 6 hours? Apply the formula N 5 N0ert, where N is the number of bacteria. 102. Population. In 2003, the population of Phoenix, Arizona, was 1,388,215. In 2004, the population was 1,418,041. What is the expected population in 2020? (Let t 5 0 be 2003.) Apply the formula N 5 N0ert, where N is the population. 103. Radioactive Decay. Strontium-90 has a half-life of 28 years. How long will it take for 20 grams of this to decay to 5 grams? Apply the formula N 5 N0e2rt, where N is the number of grams. 104. Radioactive Decay. Plutonium-239 has a half-life of 25,000 years. How long will it take for 100 grams to decay to 20 grams? Apply the formula N 5 N0e2rt, where N is the number of grams. 105. Wild Life Population. The Boston Globe reports that the fish population of the Essex River in Massachusetts is declining. In 2003, it was estimated there were 5600 fish in the river, and in 2004, there were only 2420 fish. How many should there have been in 2010 assuming the same trend? Apply the formula N 5 N0e2rt, where N is the number of fish. 106. Car Depreciation. A new Acura TSX costs $28,200. In 2 years the value will be $24,500. What is the expected value in 6 years? Apply the formula N 5 N0e2rt, where N is the value of the car. 107. Carrying Capacity. The carrying capacity of a species of beach mice in St. Croix is given by M 5 100011 2 e20.035t2 where M is the number of mice and t is time in years 1t 5 0 corresponds to 1998). How many mice were there in 2010 according to this model? 108. Population. The city of Brandon, Florida, had 50,000 residents in 1970, and since the crosstown expressway was built, its population has increased 2.3% per year. If the growth continues at the same rate, how many residents will Brandon have in 2030? Technology Section 5.1 109. Use a graphing utility to graph the function ƒ1x2 5 a1 1 !2 x b x . Determine the horizontal asymptote as x increases. 110. Use a graphing utility to graph the functions y 5 e2x12 and y 5 3x 1 1 in the same viewing screen. Estimate the coordinates of the point of intersection. Round your answers to three decimal places. Section 5.2 111. Use a graphing utility to graph the functions y 5 log2.413x 2 12 and y 5 log0.81x 2 12 1 3.5 in the same viewing screen. Estimate the coordinates of the point of intersection. Round your answers to three decimal places. 112. Use a graphing utility to graph the functions y 5 log2.51x 2 12 1 2 and y 5 3.5x22 in the same viewing screen. Estimate the coordinates of the point(s) of intersection. Round your answers to three decimal places. Section 5.3 113. Use a graphing utility to graph ƒ1x2 5 log2 a x3 x2 2 1b and g1x2 5 3 log2 x 2 log21x 1 12 2 log21x 2 12 in the same viewing screen. Determine the domain where the two functions give the same graph. 114. Use a graphing utility to graph ƒ1x2 5 ln a9 2 x2 x2 2 1b and g1x2 5 ln 13 2 x2 1 ln 13 1 x2 2 ln 1x 1 12 2 ln 1x 2 12 in the same viewing screen. Determine the domain where the two functions give the same graph. Section 5.4 115. Use a graphing utility to graph y 5 ex 2 e2x ex 1 e2x. State the domain. Determine if there are any symmetries and asymptotes. 116. Use a graphing utility to graph y 5 1 ex 2 e2x. State the domain. Determine if there are any symmetries and asymptotes. Section 5.5 117. A drug with initial dosage of 4 milligrams has a half-life of 18 hours. Let 10, 4) and 118, 2) be two points. a. Determine the equation of the dosage. b. Use STAT CALC ExpReg to model the equation of the dosage. c. Are the equations in (a) and (b) the same? 118. In Exercise 105, let t 5 0 be 2003 and (0, 5600) and (1, 2420) be the two points. a. Use STAT CALC ExpReg to model the equation for the fish population. b. Using the equation found in (a), determine how many fish should have been expected in 2010? c. Does the answer in (b) agree with the answer in Exercise 105? PR ACTICE TEST 490 CHAPTER 5 Exponential and Logarithmic Functions [ C H AP T E R 5 PRACTICE TEST ] 1. Simplify log 10x3. 2. Use a calculator to evaluate log5 326 (round to two decimal places). 3. Find the exact value of log1/3 81. 4. Rewrite the expression ln c e5x x1x4 1 12 d in a form with no logarithms of products, quotients, or powers. In Exercises 5–20, solve for x, exactly if possible. If an approximation is required, round your answer to three decimal places. 5. ex221 5 42 6. e2x 2 5ex 1 6 5 0 7. 27e0.2x11 5 300 8. 32x21 5 15 9. 3 ln1x 2 42 5 6 10. log16x 1 52 2 log 3 5 log 2 2 log x 11. ln1ln x2 5 1 12. log213x 2 12 2 log21x 2 12 5 log21x 1 12 13. log6 x 1 log61x 2 52 5 2 14. ln1x 1 22 2 ln1x 2 32 5 2 15. ln x 1 ln1x 1 32 5 1 16. log2a2x 1 3 x 2 1 b 5 3 17. 12 1 1 2ex 5 6 18. ln x 1 ln1x 2 32 5 2 19. State the domain of the function ƒ1x2 5 logc x x2 2 1d . 20. State the range of x values for which the following is true: 10log (4x2a) 5 4x 2 a. In Exercises 21–24, find all intercepts and asymptotes, and graph. 21. ƒ1x2 5 32x 1 1 22. ƒ1x2 5 A1 2B x 2 3 23. ƒ1x2 5 ln12x 2 32 1 1 24. ƒ1x2 5 log11 2 x2 1 2 25. Interest. If $5000 is invested at a rate of 6% a year, compounded quarterly, what is the amount in the account after 8 years? 26. Interest. If $10,000 is invested at a rate of 5%, compounded continuously, what is the amount in the account after 10 years? 27. Sound. A lawn mower’s sound intensity is approximately 1 3 1023 W/m2. Assuming your threshold of hearing is 1 3 10212 W/m2, calculate the decibels associated with the lawn mower. 28. Population. The population in Seattle, Washington, has been increasing at a rate of 5% a year. If the population continues to grow at that rate, and in 2004 there were 800,000 residents, how many residents will there be in 2020? Hint: N 5 N0ert. 29. Earthquake. An earthquake is considered moderate if it is between 5 and 6 on the Richter scale. What is the energy range in joules for a moderate earthquake? 30. Radioactive Decay. The mass m(t) remaining after t hours from a 50-gram sample of a radioactive substance is given by the equation m(t) 5 50e20.0578t. After how long will only 30 grams of the substance remain? Round your answer to the nearest hour. 31. Bacteria Growth. The number of bacteria in a culture is increasing exponentially. Initially, there were 200 in the culture. After 2 hours there are 500. How many should be expected in 8 hours? Round your answer to the nearest hundred. 32. Carbon Decay. Carbon-14 has a half-life of 5730 years. How long will it take for 100 grams to decay to 40 grams? 33. Spread of a Virus. The number of people infected by a virus is given by N 5 2000 1 1 3e20.4t, where t is time in days. In how many days will 1000 people be infected? 34. Oil Consumption. The world consumption of oil was 76 million barrels per day in 2002. In 2004, the consumption was 83 million barrels per day. How many barrels should be expected to be consumed in 2020? 35. Use a graphing utility to graph y 5 ex 2 e2x 2 . State the domain. Determine if there are any symmetries and asymptotes. 36. Use a graphing utility to help solve the equation 432x 5 2x 2 1. Round your answer to two decimal places. Cumulative Test 491 1. Simplify x1/3y2/5 x21/2y28/5 and express in terms of positive exponents. 2. Simplify 2 x 2 2 1 3 3 2 2 x 2 2 . 3. Solve using the quadratic formula: 5x2 2 4x 5 3. 4. Solve and check: !2x 1 13 5 2 1 !x 1 3. 5. Solve and express the solution in interval notation: x 1 4 5 # 23. 6. Solve for x: 0 x2 2 4 0 5 9. 7. Write an equation of the line that is perpendicular to the line 4x 1 3y 5 6 and that passes through the point (7, 6). 8. Using the function ƒ1x2 5 4x 2 x2, evaluate the difference quotient ƒ1x 1 h2 2 ƒ1x2 h , h 2 0. 9. Given the piecewise-defined function ƒ1x2 5 c 5 2 2 !x x 2 3 22 , x # 0 0 , x , 4 x $ 4 find a. ƒ142 b. ƒ102 c. ƒ112 d. ƒ1242 e. State the domain and range in interval notation. f. Determine the intervals where the function is increasing, decreasing, or constant. 10. Sketch the graph of the function y 5 !1 2 x and identify all transformations. 11. Determine whether the function ƒ1x2 5 !x 2 4 is one-to-one. 12. Write an equation that describes the variation: R is inversely proportional to the square of d and R 5 3.8 when d 5 0.02. 13. Find the vertex of the parabola associated with the quadratic function ƒ1x2 5 24x2 1 8x 2 5. 14. Find a polynomial of minimum degree (there are many) that has the zeros x 5 25 (multiplicity 2) and x 5 9 (multiplicity 4). 15. Use synthetic division to find the quotient Q1x2 and remainder r 1x2 of 13x2 2 4x3 2 x4 1 7x 2 202 4 1x 1 42. 16. Given the zero x 5 2 1 i of the polynomial P1x2 5 x4 2 7x3 1 13x2 1 x 2 20, determine all the other zeros and write the polynomial as the product of linear factors. 17. Find the vertical and slant asymptotes of ƒ1x2 5 x2 1 7 x 2 3 . 18. Graph the rational function ƒ1x2 5 3x x 1 1. Give all asymptotes. 19. Without employing a calculator, give the exact value of A 1 25B 23/2. 20. If $5400 is invested at 2.75% compounded monthly, how much is in the account after 4 years? 21. Give the exact value of log3 243. 22. Write the expression 1 2 ln1x 1 52 2 2 ln1x 1 12 2 ln13x2 as a single logarithm. 23. Solve the logarithmic equation exactly: 102 log(4x19) 5 121. 24. Give an exact solution to the exponential equation 5x2 5 625. 25. If $8500 is invested at 4% compounded continuously, how many years will pass until there is $12,000? 26. Use a graphing utility to help solve the equation e322x 5 2x21. Round your answer to two decimal places. 27. Strontium-90 with an initial amount of 6 grams has a half-life of 28 years. a. Use STAT CALC ExpReg to model the equation of the amount remaining. b. How many grams will remain after 32 years? Round your answer to two decimal places. [CH AP TERS 1–5 CUM UL AT IVE T E S T ] CU MU LA TIV E TEST C H A P T E R LEARNING OBJECTIVES [ [ To the ancient Greeks, trigonometry was the study of right triangles. Trigonometric functions (sine, cosine, tangent, cotangent, secant, and cosecant) can be defined as right triangle ratios (ratios of sides of a right triangle). Thousands of years later, we still find applications of right triangle trigonometry in sports, surveying, navigation, and engineering. In the eighteenth century, the unit circle approach was formulated. It relies on the correspondence between the values of the cosine and sine functions and the x- and y-coordinates along the unit circle (circle with center at the origin and radius 1). The sine and cosine functions are used to represent periodic phenomena. Orbits, tide levels, the biological clock in animals and plants, and radio signals are all periodic (repetitive). ■ ■Understand degree measure. ■ ■Define the six trigonometric functions as ratios of lengths of the sides of right triangles. ■ ■Solve right triangles. ■ ■Define the six trigonometric functions as ratios of x- and y-coordinates and distances in the Cartesian plane. ■ ■Evaluate trigonometric functions for nonacute angles. ■ ■Relate degree and radian measures. Trigonometric Functions 6 x y (cos θ, sin θ) 1 θ 1 –1 1 –1 0.0 00.00 04.00 08.00 12.00 Mon 18 Jan Height (m) 16.00 20.00 00.00 04.00 08.00 12.00 Tue 19 Jan 16.00 20.00 00.00 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 Tidal height predictions for Pawleys Island Pier, South Carolina. Plot produced by the United Kingdom Hydrographic Office Admiralty Easytide, aspx?PortID=0406&PredictionLength=4; ©Crown Copyright 2016. Richard Cummins / Getty Images, Inc. 493 [I N T HI S CHAPTER] We will define the trigonometric functions in three ways (all of which are consistent with each other).We will solve right triangles using right triangle trigonometry. We will introduce radian measure, which allows us to formulate trigonometric functions of real numbers. We will graph the trigonometric functions, and we will solve applications involving trigonometric functions. ■ ■Draw the unit circle and label the sine and cosine values for special angles (in both degrees and radians). ■ ■Graph sine and cosine functions (amplitude, period, and translations). ■ ■Graph tangent, cotangent, secant, and cosecant functions. TRIGONOMETRIC FUNCTIONS 6.1 ANGLES, DEGREES, AND TRIANGLES 6.2 DEFINITION 1 OF TRIGONO-METRIC FUNCTIONS: RIGHT TRIANGLE RATIOS 6.3 APPLICA-TIONS OF RIGHT TRIANGLE TRIGONO-METRY: SOLVING RIGHT TRIANGLES 6.4 DEFINITION 2 OF TRIGONO-METRIC FUNCTIONS: CARTESIAN PLANE 6.5 TRIGONO-METRIC FUNCTIONS OF NONACUTE ANGLES 6.6 RADIAN MEASURE AND APPLICATIONS 6.7 DEFINITION 3 OF TRIGONO-METRIC FUNCTIONS: UNIT CIRCLE APPROACH 6.8 GRAPHS OF SINE AND COSINE FUNCTIONS 6.9 GRAPHS OF OTHER TRIGONO-METRIC FUNCTIONS • Angles and Degree Measure • Triangles • Special Right Triangles • Similar Triangles • Trigono-metric Functions: Right Trian-gle Ratios • Reciprocal Identities • Cofunctions • Evaluating Trigono-metric Functions Exactly for Special Angle Mea-sures: 308, 458, and 608 • Using Calculators to Evaluate (Approxi-mate) Trig-onometric Functions • Accuracy and Sig-nificant Digits • Solving a Right Triangle Given the Measure of an Acute Angle and a Side Length • Solving a Right Triangle Given the Length of Two Sides • Angles in Standard Position • Coterminal Angles • Trigono -metric Functions: The Cartesian Plane • Algebraic Signs of Trigono-metric Functions • Ranges of the Trigo-nometric Functions • Reference Angles and Reference Right Triangles • Evaluating Trigono-metric Functions for Nonacute Angles • The Radian Measure of an Angle • Converting Between Degrees and Radians • Arc Length • Area of a Circular Sector • Linear Speed • Angular Speed • Relationship Between Linear and Angular Speeds • Trigno- metric Functions and the Unit Circle (Circular Functions) • Properties of Circular Functions • The Graphs of Sinusoidal Functions • Graphing a Shifted Sinusoidal Function • Harmonic Motion • Graphing Sums of Functions: Addition of Ordinates • Graph-ing the Tangent, Cotangent, Secant, and Cosecant Functions • Transla-tions of Circular Functions 494 CHAPTER 6 Trigonometric Functions S K I L L S O B J E C T I V E S ■ ■Find the complement and supplement of an angle. ■ ■Use the Pythagorean theorem to solve a right triangle. ■ ■Solve 308-608-908 and 458-458-908 triangles. ■ ■Use similarity to determine the length of a side of a triangle. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that degrees are a measure of an angle. ■ ■Understand that the Pythagorean theorem applies only to right triangles. ■ ■Understand that to solve a triangle means to find all of the angle measures and side lengths. ■ ■Understand the difference between similar and congruent triangles. 6.1 ANGLES, DEGREES, AND TRIANGLES 6.1.1 Angles and Degree Measure The study of trigonometry relies heavily on the concept of angles. Before we define angles, let us review some basic terminology. A line is the straight path connecting two points (A and B) and extending beyond the points in both directions. The portion of the line between the two points (including the points) is called a line segment. A ray is the portion of the line that starts at one point (A) and extends to infinity (beyond B). A is called the endpoint of the ray. B A Line AB B A Segment AB B Ray AB A In geometry, an angle is formed when two rays share the same endpoint. The common endpoint is called the vertex. Vertex Angle In trigonometry, we say that an angle is formed when a ray is rotated around its endpoint. The ray in its original position is called the initial ray or the initial side of an angle. In the Cartesian plane, we usually assume the initial side of an angle is the positive x-axis. The ray after it is rotated is called the terminal ray or the terminal side of an angle. Rotation in a counterclockwise direction corresponds to a positive angle, whereas rotation in a clockwise direction corresponds to a negative angle. Positive Angle Terminal Side Initial Side Negative Angle Terminal Side Initial Side Lengths, or distances, can be measured in different units: feet, miles, and meters are three common units. In order to compare angles of different sizes, we need a standard unit of measure. One way to measure the size of an angle is with degree measure. We will discuss degrees now, and in Section 6.6 we will discuss another angle measure called radians. STUDY TIP Positive angle: counterclockwise Negative angle: clockwise 6.1.1 S KILL Find the complement and supplement of an angle. 6.1.1 C ON CEPTUAL Understand that degrees are a measure of an angle. [CONCEPT CHECK] TRUE OR FALSE The measure of an acute angle is greater than the measure of an obtuse angle. ANSWER False ▼ DEFINITION Degree Measure of Angles An angle formed by one complete counterclockwise rotation has measure 360 degrees, denoted 360°. One complete revolution = 360º WORDS MATH 360° represents 1 complete rotation. 360° 360° 5 1 180° represents a 1 2 rotation. 180° 360° 5 1 2 90° represents a 1 4 rotation. 90° 360° 5 1 4 The Greek letter u (theta) is the most common name for an angle. Other common names of angles are a (alpha), b (beta), and g (gamma). WORDS MATH An angle measuring exactly 90° is called a right angle. A right angle is often represented by the adjacent sides of a rectangle, indicating that the two rays are perpendicular. An angle measuring exactly 180° is called a straight angle. An angle measuring greater than 08 but less than 90° is called an acute angle. An angle measuring greater than 90° but less than 180° is called an obtuse angle. θ = 90º Right Angle: quarter rotation θ = 180º Straight Angle: half rotation θ Acute Angle 0º < θ < 90º θ Obtuse Angle 90º < θ < 180º STUDY TIP Greek letters are often used to denote angles in trigonometry. 6.1 Angles, Degrees, and Triangles 495 496 CHAPTER 6 Trigonometric Functions EXAMPLE 1 Finding Measures of Complementary and Supplementary Angles Find the measure of each angle. a. Find the complement of 50°. b. Find the supplement of 110°. c. Represent the complement of a in terms of a. d. Find two supplementary angles such that the first angle is twice as large as the second angle. Solution: a. The sum of complementary angles is 90°. u 1 50° 5 90° Solve for u. u 5 40° b. The sum of supplementary angles is 180°. u 1 110° 5 180° Solve for u. u 5 70° c. Let b be the complement of a. The sum of complementary angles is 90°. a 1 b 5 90° Solve for b. b 5 90° 2 a d. The sum of supplementary angles is 180°. a 1 b 5 180° Let b 5 2a. a 1 2a 5 180° Solve for a. 3a 5 180° a 5 60° Substitute a 5 60° into b 5 2a. b 5 120° The angles have measures 60° and 120° . Y OUR TU R N Find two supplementary angles such that the first angle is three times as large as the second angle. ▼ If the sum of the measures of two positive angles is 90°, the angles are called complementary. We say that a is the complement of b (and vice versa). If the sum of the measures of two positive angles is 180°, the angles are called supplementary. We say that a is the supplement of b (and vice versa) α β Complementary Angles α + β = 90º α β Supplementary Angles α + β = 180º ▼ A N S W E R The angles have measures 45° and 135°. It is important not to confuse an angle with its measure. In Example 1(d), angle a is a rotation and the measure of that rotation is 608. 6.1 Angles, Degrees, and Triangles 497 6.1.2 S K IL L Use the Pythagorean theorem to solve a right triangle. 6.1.2 C ON C E P T U A L Understand that the Pythagorean theorem applies only to right triangles. [CONCEPT CHECK] TRUE OR FALSE The hypotenuse is always longer than each of the legs of a right triangle. ANSWER True ▼ ANGLE SUM OF A TRIANGLE The sum of the measures of the angles of any triangle is 1808. α β γ α + β + γ = 180º EXAMPLE 2 Finding an Angle of a Triangle If two angles of a triangle have measures 32° and 68°, what is the measure of the third angle? Solution: The sum of the measures of all three angles is 180°. 32° 1 68° 1 a 5 180° Solve for a. a 5 80° YOUR T UR N If two angles of a triangle have measures 16° and 96°, what is the measure of the third angle? 32º 68º α ▼ ▼ A N S W E R 688 6.1.2 Triangles Trigonometry originated as the study of triangles, with emphasis on calculations involving the lengths of sides and the measures of angles. What do you already know about triangles? You know that triangles are three-sided, closed-plane figures. An important property of triangles is that the sum of the measures of the three angles of any triangle is 180°. STUDY TIP In this book when we say “equal angles,” this implies “equal angle measures.” Similarly, when we say an angle is x8, this implies that the angle’s measure is x8. Some triangles that may be familiar to you from geometry are equilateral, isosceles, and right. An equi lateral triangle has three equal sides and three equal angles 1608-608-6082. An isosceles triangle has two equal sides (legs) and two equal angles opposite those legs. The most important triangle that we will discuss in this course is the right triangle. A right triangle is a triangle in which one of the angles is a right angle 190°2. Since one angle is 90°, the other two angles must be complementary 1sum to 90°2, so that the sum of all three angles is 180°. The longest side of a right triangle, called the hy potenuse, is opposite the right angle. The other two sides are called the legs of the right triangle. Right Triangle Hypotenuse Leg Leg 498 CHAPTER 6 Trigonometric Functions The Pythagorean theorem relates the sides of a right triangle. It says that the sum of the squares of the lengths of the two legs is equal to the square of the length of the hypotenuse. It is important to note that length (a synonym of distance) is always positive. ▼ A N S W E R 15 feet PYTHAGOREAN THEOREM In any right triangle, the square of the length of the longest side (hypotenuse) is equal to the sum of the squares of the lengths of the other two sides (legs). b a c a2 1 b2 5 c2 It is important to note that the Pythagorean theorem applies only to right triangles. EXAMPLE 3 Using the Pythagorean Theorem to Find the Side of a Right Triangle Suppose you have a 10-foot ladder and want to reach a height of 8 feet to clean out the gutters on your house. How far from the base of the house should the base of the ladder be? 10 ft ? 8 ft Solution: Label the unknown side x. Apply the Pythagorean theorem. x2 1 82 5 102 Simplify. x2 1 64 5 100 Solve for x. x2 5 36 x 5 66 Length must be positive. x 5 6 The ladder should be 6 feet from the base of the house along the ground. Y OUR TU R N A steep ramp is being built for skateboarders. The height, horizontal ground distance, and ramp length form a right triangle. If the height is 9 feet and the horizontal ground distance is 12 feet, what is the length of the ramp? 10 8 x ▼ EXAMPLE 4 Using the Pythagorean Theorem with Radicals Use the Pythagorean theorem to solve for the unknown side length in the given right triangle. Express your answer exactly in terms of simplified radicals. Solution: Apply the Pythagorean theorem. 32 1 x2 5 72 Simplify known squares. 9 1 x2 5 49 Solve for x. x2 5 40 x 5 6"40 The side length x is a distance that is positive. x 5 "40 Simplify the radical. x 5 "4⋅10 5 "4⋅"10 5 2"10 YOUR T UR N Use the Pythagorean theorem to solve for the unknown side length in the given right triangle. Express your answer exactly in terms of simplified radicals. 7 3 x g 2 8 4 x ▼ ▼ A N S W E R 4"3 6.1.3 Special Right Triangles Right triangles whose sides are in the ratios of 3-4-5, 5-12-13, and 8-15-17 are examples of right triangles that are special because their side lengths are equal to whole numbers that satisfy the Pythagorean theorem. A Pythagorean triple consists of three positive integers that satisfy the Pythagorean theorem: 32 1 42 5 52, 52 1 122 5 132, and 82 1 152 5 172. There are two other special right triangles that warrant attention: the 308-608-908 triangle and the 458-458-908 triangle. Although in trigonometry we focus more on the angles than on the side lengths, we are interested in special relationships between the lengths of the sides of these right triangles. We will start with the 458-458-908 triangle. WORDS MATH A 458-458-908 triangle is an isosceles (two legs are equal) right triangle. Apply the Pythagorean theorem. x2 1 x2 5 hypotenuse2 Simplify the right side of the equation. 2x2 5 hypotenuse2 Solve for the hypotenuse. x x 45º 45º hypotenuse 5 6"2x2 5 6"2 k x k 6.1.3 S K IL L Solve 308-608-908 and 458-458-908 triangles. 6.1.3 C ON C E P T U A L Understand that to solve a triangle means to find all of the angle measures and side lengths. [CONCEPT CHECK] TRUE OR FALSE In a right triangle, if you know one of the acute angles and one side length, then it is possible to solve the triangle. ANSWER True ▼ When solving a right triangle exactly, simplification of radicals is often necessary. For example, if a side length of a triangle resulted in "17, the radical could not be simplified any further. However, if a side length of a triangle resulted in !20, the radical would be simplified: "20 5 "4⋅5 5 "4⋅"5 5 "22⋅"5 5 2"5 6.1 Angles, Degrees, and Triangles 499 500 CHAPTER 6 Trigonometric Functions x and the hypotenuse are lengths and must be positive. hypotenuse 5 !2 x The hypotenuse of a 458-458-908 triangle is !2 times the length of either leg. x x 45º 45º √2x STUDY TIP “To solve a triangle” means to find all of the angle measures and side lengths. If we let x 5 1, then the triangle will have legs with length equal to 1 and the hypotenuse will have length !2. Notice that these lengths satisfy the Pythagorean theorem: 12 1 12 5 A!2B 2 or 2 5 2. Later, when we discuss the unit circle approach, we will let the hypotenuse have length l. The legs will then have lengths "2 2 and "2 2 : a"2 2 b 2 1 a"2 2 b 2 5 1 2 1 1 2 5 12. EXAMPLE 5 Solving a 458-458-908 Triangle A house has a roof with a 45° pitch (the angle the roof makes with the house). If the house is 60 feet wide, what are the lengths of the sides of the roof that form the attic? Round to the nearest foot. Solution: Draw the 458-458-908 triangle. Let x represent the length of the unknown legs. Recall that the hypotenuse of a 458-458-908 triangle is !2 times the length of either leg. Let the hypotenuse equal 60 feet. 60 5 !2x Solve for x. x 5 60 "2 < 42.42641 Round to the nearest foot. x < 42 ft YOUR TURN A house has a roof with a 45° pitch. If the sides of the roof are 60 feet, how wide is the house? Round to the nearest foot. x x 45º 45º 60 ft √2x x x 45º 45º ▼ x x 45º 45º 60 ft ▼ A N S W E R 60!2 < 85 feet Let us now determine the relationship of the sides of a 308-608-908 triangle. We start with an equilateral triangle (equal sides and equal angles, 608). WORDS MATH Draw an equilateral triangle with sides 2x. Draw a line segment from one vertex that is perpendicular to the opposite side; this line segment represents the height of the triangle, h, and bisects the base. There are now two identical 308-608-908 triangles. Notice that in each triangle the hypotenuse is twice the shortest leg, which is opposite the 30° angle. To find the length h, use the Pythagorean theorem. h2 1 x2 5 12x22 h2 1 x2 5 4x2 Solve for h. h2 5 3x2 h 5 6"3x2 5 6!3 0 x 0 h and x are lengths and must be positive. h 5 !3x The hypotenuse of a 308-608-908 is two times the length of the leg opposite the 30° angle. The leg opposite the 60° angle is !3 times the length of the leg opposite the 30° angle. 60º 60º 60º 2x 2x 2x 60º 60º 30º 30º h x x 2x 2x 60º 30º h x 2x 60º 30º x 2x √3x If we let x 5 1, then the triangle will have legs with lengths 1 and !3 and hypotenuse of length 2. These satisfy the Pythagorean theorem: 12 1 A!3B 2 5 22 or 4 5 4. 6.1 Angles, Degrees, and Triangles 501 502 CHAPTER 6 Trigonometric Functions EXAMPLE 6 Solving a 308-608-908 Triangle Before a hurricane, it is wise to stake down trees for additional support during the storm. If the branches allow for the rope to be tied 15 feet up the tree and a desired angle between the rope and the ground is 60°, how much total rope is needed? How far from the base of the tree should each of the two stakes be hammered? Solution: Recall the relationship between the sides of a 308-608-908 triangle. In this case, the leg opposite the 60° angle is 15 feet. Solve for x. !3x 5 15 x 5 15 !3 < 8.7 Find the length of the hypotenuse. hypotenuse 5 2x 5 2 a 15 !3b < 17 The ropes should be staked approximately 8.7 feet from the base of the tree, and approximately 21172 5 34 total feet of rope will be needed. Y OUR TU R N Rework Example 6 with a height (where the ropes are tied) of 20 feet. How far from the base of the tree should each of the ropes be staked, and how much total rope will be needed? 15 ft 60º 30º 60º 30º x 2x √3x = 15 ft 15 ft 17 ft 17 ft 8.7 ft 60º 30º ▼ ▼ A N S W E R Each rope should be staked approximately 11.5 feet from the base of the tree. Approximately 46 total feet of rope will be needed. 6.1.4 S KILL Use similarity to determine the length of a side of a triangle. 6.1.4 C ON CEPTUAL Understand the difference between similar and congruent triangles. 6.1.4 Similar Triangles The word similar in mathematics means identical in shape, although not necessarily the same size. It is important to note that two triangles can have the exact same shape (same angles) but different sizes. DEFINITION Similar Triangles Similar triangles are triangles with equal corresponding angle measures (equal angles). [CONCEPT CHECK] TRUE OR FALSE Two similar triangles are congruent triangles, but two congruent triangles need not be similar triangles. ANSWER False ▼ STUDY TIP Although an angle and its measure are fundamentally different, out of convenience when “equal angles” is stated, this implies “equal angle measures.” The word congruent means equal in all corresponding parts; therefore, congruent triangles have all corresponding angles equal and all corresponding sides equal. Two triangles are congruent if they have exactly the same shape and size. In other words, if one triangle can be picked up and situated on top of another triangle so that the two triangles coincide, they are said to be congruent. STUDY TIP Similar triangles: exact same shape Congruent triangles: exact same shape and size DEFINITION Congruent Triangles Congruent triangles are triangles with equal corresponding angle measures (equal angles) and corresponding equal sides. It is important to note that all congruent triangles are also similar triangles, but not all similar triangles are congruent. Trigonometry (as you will see in Section 6.2) relies on the properties of similar triangles. Since similar triangles have the same shape (equal corresponding angles), the sides opposite the corresponding angles must be proportional. Given any 308-608-908 triangle (see top figure in the margin), if we let x 5 1 correspond to a triangle A and x 5 5 correspond to a triangle B, then we would have the following two triangles. Notice that the sides opposite the corresponding angles are proportional. This means that we will find that all three ratios of each side of triangle B to its corresponding side of triangle A are equal. Sides opposite 308: Triangle B Triangle A 5 5 1 5 5 Sides opposite 608: Triangle B Triangle A 5 5!3 !3 5 5 Sides opposite 908: Triangle B Triangle A 5 10 2 5 5 This proportionality property holds for all similar triangles. 60º 30º x 2x √3x 60º 30º 1 2 √3 A 60º 30º 5 10 5√3 B 6.1 Angles, Degrees, and Triangles 503 504 CHAPTER 6 Trigonometric Functions Let us now use the definition of similar triangles to determine lengths of sides of similar triangles. CONDITIONS FOR SIMILAR TRIANGLES One of the following must be verified in order for two triangles to be similar: n Corresponding angles must have the same measure. n Corresponding sides must be proportional (ratios must be equal). a ar 5 b br 5 c cr EXAMPLE 7 Finding Lengths of Sides in Similar Triangles Given that the two triangles are similar, find the length of each of the unknown sides 1b and c2. b 8 c 5 2 6 Solution: Solve for b. The corresponding sides are proportional. 8 2 5 b 5 Multiply by 5. b 5 4152 Solve for b. b 5 20 Solve for c. The corresponding sides are proportional. 8 2 5 c 6 Multiply by 6. c 5 4162 Solve for c. c 5 24 Check that the ratios are equal: 8 2 5 20 5 5 24 6 5 4 Y OUR TU R N Given that the two triangles are similar, find the length of each of the unknown sides 1a and b2. 24 27 15 a b 9 ▼ b a c b' a' c' ▼ A N S W E R a 5 5, b 5 8 Applications Involving Similar Triangles As you have seen, the common ratios associated with similar triangles are very useful: you can measure the sides of the smaller triangle and then calculate the measures of the sides of the larger triangle by measuring just one side. For example, you can quickly estimate the heights of flagpoles, trees, and any other tall objects by measuring their shadows along the ground because similar triangles are formed. Surveyors rely on the properties of similar triangles to determine distances that are difficult to measure. EXAMPLE 8 Calculating the Height of a Tree Billy wants to rent a lift to trim his tall trees. However, he must decide which lift he needs: one that will lift him 25 feet or a more expensive lift that will lift him 50 feet. His wife, Jeanine, hammered a stake into the ground and by measuring found its shadow to be 13 4 feet long and the tree’s shadow to be 19 feet. If the stake was standing 3 feet above the ground, how tall is the tree? Which lift should Billy rent? (Assume the tree and stake both are perpendicular to the ground.) x ft 19 ft 1.75 ft 3 ft (NOT TO SCALE) Solution: Rays of sunlight are straight and parallel to each other. Therefore, the rays make the same angle with the tree that they do with the stake. Draw and label the two similar triangles. The ratios of corresponding sides of similar triangles are equal. Solve for x. x 5 31192 1.75 Simplify. x < 32.57 The tree is approximately 33 feet tall. Billy should rent the more expensive lift to be safe. YOUR T UR N Billy’s neighbor decides to do the same thing. He borrows Jeanine’s stake and measures the shadows. If the shadow of his tree is 15 feet and the shadow of the stake (3 feet above the ground) is 1.2 feet, how tall is Billy’s neighbor’s tree? Round to the nearest foot. ▼ x 19 1.75 3 (NOT TO SCALE) x 3 5 19 1.75 ▼ A N S W E R approximately 38 feet 6.1 Angles, Degrees, and Triangles 505 506 CHAPTER 6 Trigonometric Functions In Exercises 1–10, specify the measure of the angle in degrees for the given rotations, using the correct algebraic sign 11 or 22. 1. 1 2 rotation counterclockwise 2. 1 4 rotation counterclockwise 3. 1 3 rotation clockwise 4. 2 3 rotation clockwise 5. 5 6 rotation counterclockwise 6. 7 12 rotation counterclockwise 7. 4 5 rotation clockwise 8. 5 9 rotation clockwise 9. 7 36 rotation clockwise 10. 7 36 rotation counterclockwise In Exercises 11–16, find 1a2 the complement and 1b2 the supplement of the given angles. 11. 18° 12. 39° 13. 42° 14. 57° 15. 89° 16. 75° In Exercises 17–20, find the measure of each angle. 17. 18. 19. Supplementary angles with measures 8x degrees and 4x degrees 20. Complementary angles with measures 3x 115 degrees and 10x 110 degrees In Exercises 21–26, refer to the triangle in the drawing. 21. If a 5 117° and b 5 33°, find g. 22. If a 5 110° and b 5 45°, find g. 23. If g 5 b and a 5 4b, find all three angles. 24. If g 5 b and a 5 3b, find all three angles. 25. If g 5 1 2 b and b 5 6a, find all three angles. 26. If g 5 2 3 b and b 5 3a, find all three angles. In Exercises 27–34, refer to the right triangle in the drawing. Express lengths exactly. 27. If a 5 4 and b 5 3, find c. 28. If a 5 3 and b 5 3, find c. 29. If a 5 6 and c 5 10, find b. 30. If b 5 7 and c 5 12, find a. 31. If a 5 8 and b 5 5, find c. 32. If a 5 6 and b 5 5, find c. 33. If a 5 24 and b 5 7, find c. 34. If c 5 20 and a 5 10, find b. (4x)º (6x)º (3x)º (15x)º α β γ α + β + γ = 180º b a c [SEC TION 6.1] E X E R C I SES • S K I L L S In this section, you have practiced working with angles and triangles. One unit of measurement for angles is the degree. An angle measuring exactly 908 is called a right angle. The sum of the three angles of any triangle is always 1808. Triangles that contain a right angle are called right triangles. With the Pythagorean theorem you can solve for one side of a right triangle given the other two sides. The right triangles 308-608-908 and 458-458-908 are special because of the simple numerical relations of their side lengths. Similar triangles (the same shape) have corresponding angles with equal measure, and the important thing to note is the consistency of their proportions. [SEC TION 6.1] S U M MA RY 60º 30º x 2x √3x x x 45º 45º √2x In Exercises 35–38, refer to the 458-458-908 triangle. 35. If the two legs have length 10 inches, how long is the hypotenuse? 36. If the two legs have length 8 meters, how long is the hypotenuse? 37. If the hypotenuse has length 2!2 centimeters, how long are the legs? 38. If the hypotenuse has length !10 feet, how long are the legs? In Exercises 39–44, refer to the 308-608-908 triangle. 39. If the shortest leg has length 5 meters, what are the lengths of the other leg and the hypotenuse? 40. If the shortest leg has length 9 feet, what are the lengths of the other leg and the hypotenuse? 41. If the longer leg has length 12 yards, what are the lengths of the other leg and the hypotenuse? 42. If the longer leg has length n units, what are the lengths of the other leg and the hypotenuse? 43. If the hypotenuse has length 10 inches, what are the lengths of the two legs? 44. If the hypotenuse has length 8 centimeters, what are the lengths of the two legs? In Exercises 45–52, calculate the specified lengths given that the two triangles are similar. 45. a 5 4, c 5 6, d 5 2, ƒ 5 ? 46. a 5 12, b 5 9, e 5 3, d 5 ? 47. d 5 5, e 5 2.5, b 5 7.5, a 5 ? 48. e 5 1.4, ƒ 5 2.6, c 5 3.9, b 5 ? 49. d 5 2.5 m, ƒ 5 1.1 m, a 5 26.25 km, c 5 ? 50. e 5 10 cm, ƒ 5 14 cm, c 5 35 m, b 5 ? 51. If a 5 c 5 9 2 in. and d 5 1 3 in., ƒ 5 ? 52. If b 5 e 5 4.2 cm and ƒ 5 1.1 cm, c 5 ? x x 45º 45º √2x 60º 30º x 2x √3x a c b e d f • A P P L I C A T I O N S 53. Clock. What is the measure (in degrees) of the angle the minute hand traces in 20 minutes? 54. Clock. What is the measure (in degrees) of the angle the minute hand traces in 25 minutes? 55. London Eye. The London Eye (similar to a bicycle wheel) makes 1 rotation in approximately 30 minutes. What is the measure of the angle (in degrees) that a cart (spoke) will rotate in 12 minutes? 56. London Eye. The London Eye (similar to a bicycle wheel) makes 1 rotation in approximately 30 minutes. What is the measure of the angle (in degrees) that a cart (spoke) will rotate in 5 minutes? 57. Revolving Restaurant. If a revolving restaurant overlooking a waterfall can rotate 270° in 45 minutes, how long does it take for the restaurant to make a complete revolution? 58. Revolving Restaurant. If a revolving restaurant overlooking a waterfall can rotate 72° in 9 minutes, how long does it take for the restaurant to make a complete revolution? 59. Field Trial. In a Labrador Retriever field trial, dogs are judged by the straight line they take to a fallen bird. They are required to go through the water, not along the shore. If the judge wants to calculate how far a dog will travel along a straight path, she walks the two legs of a right triangle as shown in the drawing and uses the Pythagorean theorem. How far would this dog travel (run and swim) if it traveled along the hypotenuse? 80 ft Pond 30 ft Jason Friend/PhotolibraryGetty Images, Inc. 6.1 Angles, Degrees, and Triangles 507 508 CHAPTER 6 Trigonometric Functions 60. Field Trial. How far would the dog in Exercise 59 run and swim if it traveled along the hypotenuse? The judge walks 25 feet along the shore and then 100 feet out to the bird. Pond 100 ft 25 ft 61. Christmas Lights. A couple want to put up Christmas lights along the roofline of their house. If the front of the house is 100 feet wide and the roof has a 45° pitch, how many linear feet of Christmas lights should the couple buy? Round to the nearest foot. 100 ft 45º 45º 62. Christmas Lights. Repeat Exercise 61 if the house is 60 feet wide. Round to the nearest foot. 63. Tree Stake. A tree needs to be staked down prior to a storm. If the ropes can be tied on the tree trunk 17 feet above the ground and the staked rope should make a 60° angle with the ground, how far from the base of the tree should each of the ropes be staked? 64. Tree Stake. What is the total amount of rope (feet) that extends to the two stakes supporting the tree in Exercise 63? 65. Tree Stake. A tree needs to be staked down prior to a storm. If the ropes can be tied on the tree trunk 10 feet above the ground and the staked rope should make a 30° angle with the ground, how far from the base of the tree should the ropes be staked? Round to the nearest foot. 66. Tree Stake. What is the total amount of rope (feet) that extends to the four stakes supporting the tree in Exercise 65? Round to the nearest foot. 67. Party Tent. Steve and Peggy want to rent a 40 foot 3 20 foot tent for their backyard to host a barbecue. The base of the tent is supported 7 feet above the ground by poles, and then roped stakes are used for support. The ropes make a 60° angle with the ground. How large a footprint in their yard would they need for this tent (and staked ropes)? In other words, what are the dimensions of the rectangle formed by the stakes on the ground? Round to the nearest foot. 68. Party Tent. Ashley’s parents are throwing a graduation party and are renting a 40 ft 3 80 ft tent for their back yard. The base of the tent is supported 7 feet above the ground by poles, and then roped stakes are used for support. The ropes make a 45° angle with the ground. How large a footprint (see Exer-cise 67) in their yard will they need for this tent (and staked ropes)? Round to the nearest foot. 69. Height of a Tree. The shadow of a tree measures 141 4 feet. At the same time of day, the shadow of a 4-foot pole measures 1.5 feet. How tall is the tree? 70. Height of a Flagpole. The shadow of a flagpole measures 15 feet. At the same time of day, the shadow of a 2-foot stake measures 3 4 foot. How tall is the flagpole? 71. Height of a Lighthouse. The Cape Hatteras Lighthouse on the Outer Banks of North Carolina is the tallest lighthouse in North America. If a 5-foot woman casts a 1.2-foot shadow and the lighthouse casts a 54-foot shadow, approximately how tall is the Cape Hatteras Lighthouse? 72. Height of a Man. If a 6-foot man casts a 1-foot shadow, how long a shadow will his 4-foot son cast? For Exercises 73–76, refer to the following: An archery target consists of a bullseye (at the very center) and concentric circles that form annular rings. The concentric circles have radii 4 inches, 8 inches, and 12 inches, respectively, and a remaining rectangular region that measures 8 inches vertically and horizontally with respect to the bullseye beyond the outer-most circle. 4 in 8 in 8 in 12 in Assume an archer releases the bow 50 feet from the target and that the bullseye is a point. Once an archer releases the arrow (assum-ing it actually hits the target), the score is assigned as follows: ■ 50 points: Hits the bullseye (yellow) ■ 30 points: Lies in the red ring (misses the bullseye but lies within 4 inches of the bullseye) ■ 20 points: Lies in the blue ring (between 4 and 8 inches of the bullseye) ■ 10 points: Lies in the black ring (between 8 and 12 inches of the bullseye) ■ 5 points: Hits the brown target outside all of the rings 73. Archery. Assume that at the moment prior to releasing the bow, the arrowhead is directly in line with the bullseye. If the force with which the archer releases the bow is such that the arrow falls 3 inches vertically for every 20 feet the arrow travels horizontally, what is her score for this shot? 74. Archery. In Exercise 73, if instead the force with which she releases the bow is such that the arrow rises 3 inches vertically for every 10 feet the arrow travels horizontally, what is her score for this shot? 75. Archery. In Exercise 73, assuming that the arrowhead is located at the origin of a coordinate system when released, what is the distance between its original position and the point at which it hits the target? Express your answer to five significant digits. 76. Archery. In Exercise 73, assuming that the arrowhead is located at the origin of a coordinate system when released, what is the distance between its original position and the point at which it hits the target? Express your answer to five significant digits. For Exercises 77–80, refer to the following: A collimator is a device used in radiation treatment that narrows beams or waves, causing the waves to be more aligned in a specific direction. The use of a collimator facilitates the focusing of radiation to treat an affected region of tissue beneath the skin. In the figure, ds is the distance from the radiation source to the skin, dt is the distance from the outer layer of skin to the targeted tissue, 2ƒs is the field size on the skin (diameter of the circular treated skin), and 2ƒd is the targeted field size at depth dt (the diameter of the targeted tissue at the specified depth beneath the skin surface). ds fd ds d fd f fs dt Radiation source Collimator 77. Health/Medicine. Radiation treatment is applied to a field size of 8 centimeters at a depth 2.5 centimeters below the skin surface. If the treatment head is positioned 80 centimeters from the skin, find the targeted field size to the nearest millimeter. 78. Health/Medicine. Radiation treatment is applied to a field size of 4 centimeters lying at a depth of 3.5 centimeters below the skin surface. If the field size on the skin is required to be 3.8 centimeters, find the distance from the skin that the radiation source must be located to the nearest millimeter. 79. Health/Medicine. Radiation treatment is applied to a field size on the skin of 3.75 centimeters to reach an affected region of tissue with field size of 4 centimeters at some depth below the skin. If the treatment head is positioned 60 centimeters from the skin surface, find the desired depth below the skin to the target area to the nearest millimeter. 80. Health/Medicine. Radiation treatment is applied to a field size of 4.15 centimeters on the skin to reach an affected area lying 4.5 centimeters below the skin surface. If the treatment head is positioned 60 centimeters from the skin surface, find the field size of the targeted area to the nearest millimeter. • C A T C H T H E M I S T A K E In Exercises 81 and 82, explain the mistake that is made. 81. In a 308-608-908 triangle, find the length of the side opposite the 60° angle if the side opposite the 30° angle is 10 inches. Solution: The length opposite the 60° angle is twice the length opposite the 30° angle. 21102 5 20 The side opposite the 60° angle has length 20 inches. This is incorrect. What mistake was made? 82. In a 458-458-908 triangle, find the length of the hypotenuse if each leg has length 5 centimeters. Solution: Use the Pythagorean theorem. 52 1 52 5 hypotenuse2 Simplify. 50 5 hypotenuse2 Solve for the hypotenuse. hypotenuse 5 65"2 The hypotenuse has length 65!2 centimeters. This is incorrect. What mistake was made? 6.1 Angles, Degrees, and Triangles 509 • C O N C E P T U A L In Exercises 83–88, determine whether each statement is true or false. 83. The Pythagorean theorem can be applied to any equilateral triangle. 84. The Pythagorean theorem can be applied to all isosceles triangles. 85. The two angles opposite the legs of a right triangle are complementary. 86. In a 308-608-908 triangle, the length of the side opposite the 60° angle is twice the length of the side opposite the 30° angle. 510 CHAPTER 6 Trigonometric Functions 87. If a triangle contains an obtuse angle, the Pythagorean theo-rem cannot apply. 88. The Pythagorean theorem applies only if a triangle has one right angle and exactly two acute angles. • C H A L L E N G E 89. What is the measure (in degrees) of the smaller angle the hour and minute hands make when the time is 12:20? 90. What is the measure (in degrees) of the smaller angle the hour and minute hands make when the time is 9:10? In Exercises 91–94, use the following figure: D C B A 91. If AB 5 3, AD 5 5, and AC 5 !58, find DC. 92. If AB 5 4, AD 5 5, and AC 5 !41, find DC. 93. If AC 5 30, AB 5 24, and DC 5 11, find AD. 94. If AB 5 60, AD 5 61, and DC 5 36, find AC. 95. Given a square with side length x, draw the two diagonals. The results are four special triangles. Describe these trian-gles. What are the angle measures? 96. Solve for x in the following triangle. 3x – 2 x 2x + 2 97. Consider the following diagram and determine y. y y y 8√2 98. Consider the following diagram and find y. y y 2 2 C D E B A y 2 y 2 99. Using the diagram in Exercise 98, find BE. 100. Using the diagram in Exercise 98, find AD. 101. Find x. 3 3 6 4 x 102. Find x and y. 4 5 20 18 x y • T E C H N O L O G Y 103. If the shortest leg of a 308-608-908 triangle has length 16.68 feet, what are the lengths of the other leg and the hypotenuse? Round answers to two decimal places. 104. If the longer leg of a 308-608-908 triangle has length 134.75 centimeters, what are the lengths of the other leg and the hypotenuse? Round answers to two decimal places. S K I L L S O B J E C T I V E S ■ ■Calculate trigonometric ratios of general acute angles. ■ ■Calculate cosecant, secant, and cotangent functions as reciprocals of sine, cosine, and tangent functions, respectively. ■ ■Express trigonometric functions in terms of their cofunctions. ■ ■Evaluate trigonometric functions exactly for special angles. ■ ■Evaluate (approximate) trigonometric functions using a calculator. C O N C E P T U A L O B J E C T I V ES ■ ■Understand that right triangle ratios are based on the properties of similar triangles. ■ ■Understand that if you learn the three main trigonometric functions (sine, cosine, and tangent), the other three trigonometric functions can always be calculated as reciprocals of these main three for any acute angle. ■ ■Understand that the trigonometric function of an angle is the cofunction of its complementary angle. ■ ■Learn the trigonometric functions as ratios of sides of a right triangle. ■ ■Understand the difference between evaluating trigonomet-ric functions exactly and using a calculator. 6.2 DEFINITION 1 OF TRIGONOMETRIC FUNCTIONS: RIGHT TRIANGLE RATIOS 6.2.1 S K IL L Calculate trigonometric ratios of general acute angles. 6.2.1 C ON C E P T U A L Understand that right triangle ratios are based on the properties of similar triangles. [CONCEPT CHECK] How many possible ratios of the length of the sides of a right triangle are there for an acute angle in a right triangle? ANSWER 6 ▼ 6.2.1 Trigonometric Functions: Right Triangle Ratios The word trigonometry stems from the Greek words trigonon, which means triangle, and metrein, which means to measure. Trigonometry began as a branch of geometry and was utilized extensively by early Greek mathematicians to determine unknown distances. The major trigonometric functions, including sine, cosine, and tangent, were first defined as ratios of sides in a right triangle. This is the way we will define them in this section. Since the two angles, besides the right angle, in a right triangle have to be acute, a second kind of definition was needed to extend the domain of trigonometric functions to nonacute angles in the Cartesian plane (Section 6.4). Starting in the eighteenth century, broader definitions of the trigonometric functions came into use, under which the functions are associated with points along the unit circle (Section 6.7). In Section 6.1, we learned that triangles with equal corresponding angles also have proportional sides and are called similar triangles. The concept of similar triangles, one of the basic insights in trigonometry, allows us to determine the length of a side of one triangle if we know the length of certain sides of a similar triangle. b a c b' a' c' Since the two right triangles above have equal angles, they are similar triangles, and the following ratios hold true: a ar 5 b br 5 c cr Separate the common ratio into three equations: a a9 5 b b9 b b9 5 c c9 a a9 5 c c9 t t t t t 6.2 Definition 1 of Trigonometric Functions: Right Triangle Ratios 511 t 512 CHAPTER 6 Trigonometric Functions WORDS MATH Start with the first ratio. a a9 5 b b9 Cross multiply. ab95 a9b Divide both sides by bbr. ab9 bb9 5 a9b bb9 Simplify. a b 5 a9 b9 Similarly, it can be shown that b c 5 b9 c9 and a c 5 a9 c9. Notice that even though the sizes of the triangles are different, since the corresponding angles are equal, the ratio of the lengths of the two legs of the large triangle is equal to the ratio of the lengths of the legs of the small triangle, or a b 5 a9 b9. Similarly, the ratios of the lengths of a leg and the hypotenuse of the large triangle and the corresponding leg and hypotenuse of the small triangle are also equal; that is, b c 5 b9 c9 and a c 5 a9 c9. For any right triangle, six possible ratios of sides can be calculated for each acute angle u: b c a c b a c b c a a b b a θ c These ratios are referred to as trigonometric ratios or trigonometric functions, since they depend on u, and each is given a name: FUNCTION NAME ABBREVIATION Sine sin Cosine cos Tangent tan Cosecant csc Secant sec Cotangent cot WORDS MATH The sine of u sin u The cosine of u cos u The tangent of u tan u The cosecant of u csc u The secant of u sec u The cotangent of u cot u Sine, cosine, tangent, cotangent, secant, and cosecant are names given to specific ratios of lengths of sides of right triangles. DEFINITION (1) Trigonometric Functions Let u be an acute angle in a right triangle, then sin u 5 b c cos u 5 a c tan u 5 b a csc u 5 c b sec u 5 c a cot u 5 a b b a θ c STUDY TIP You need to learn only the three main trigonometric ratios: sin u, cos u, and tan u. The other three can always be calculated as reciprocals of these main three for an acute angle u. The following terminology will be used throughout this text: ■ Side c is the hypotenuse. ■ Side b is the side (leg) opposite angle u. ■ Side a is the side (leg) adjacent to angle u. Also notice that since sin u 5 b c and cos u 5 a c, then tan u 5 sin u cos u 5 b c a c 5 b a. Using this terminology, we have an alternative definition that is easier to remember. STUDY TIP SOHCAHTOA SOH: sin u 5 opposite hypotenuse CAH: cos u 5 adjacent hypotenuse TOA: tan u 5 opposite adjacent STUDY TIP Trigonometric functions are functions of a specified angle. Always specify the angle. “Sin” alone means nothing. Sin u spec-ifies the angle dependency. The same is true for the other five trigonometric functions. DEFINITION (1) Trigonometric Functions (Alternate Form) sin u 5 opposite hypotenuse cos u 5 adjacent hypotenuse tan u 5 opposite adjacent and their reciprocals: csc u 5 1 sin u 5 hypotenuse opposite sec u 5 1 cos u 5 hypotenuse adjacent cot u 5 1 tan u 5 adjacent opposite b a θ c Adjacent Opposite Hypotenuse 6.2.2 Reciprocal Identities The three main trigonometric functions should be learned in terms of the ratios. sin u 5 opposite hypotenuse cos u 5 adjacent hypotenuse tan u 5 opposite adjacent The remaining three trigonometric functions can be derived from sin u, cos u, and tan u using the reciprocal identities. Recall that the reciprocal of x is 1 x for x 2 0. RECIPROCAL IDENTITIES csc u 5 1 sin u sec u 5 1 cos u cot u 5 1 tan u 6.2.2 S K I L L Calculate cosecant, secant, and cotangent functions as reciprocals of sine, cosine, and tangent functions, respectively. 6.2.2 C O N C E P T U A L Understand that if you learn the three main trigonometric functions (sine, cosine, and tangent), the other three trigonometric functions can always be calculated as reciprocals of these main three for any acute angle. EXAMPLE 1 Finding Trigonometric Function Values of an Acute Angle u in a Right Triangle For the given triangle, calculate sin u, tan u, and csc u. θ 3 4 6.2 Definition 1 of Trigonometric Functions: Right Triangle Ratios 513 514 CHAPTER 6 Trigonometric Functions Solution: STEP 1 Solve for the hypotenuse. Apply the Pythagorean theorem. 42 1 32 5 x2 16 1 9 5 x2 x2 5 25 x 5 65 Lengths of sides can be only positive. x 5 5 STEP 2 Label the sides of the triangle ■ ■with numbers representing lengths. ■ ■as hypotenuse, or as opposite or adjacent with respect to u. STEP 3 Set up the trigonometric functions as ratios. Sine is opposite over hypotenuse. sin u 5 opposite hypotenuse 5 4 5 Tangent is opposite over adjacent. tan u 5 opposite adjacent 5 4 3 Cosecant is the reciprocal of sine. csc u 5 1 sin u 5 1 4/5 5 5 4 Y OUR TU R N For the triangle in Example 1, calculate the values of cos u, sec u, and cot u. θ 3 4 x θ 3 4 5 Adjacent Opposite Hypotenuse ▼ ▼ A N S W E R cos u 5 3 5, sec u 5 5 3, and cot u 5 3 4 EXAMPLE 2 Finding Trigonometric Function Values of an Acute Angle u in a Right Triangle For the given triangle, calculate cos u, tan u, and sec u. Solution: STEP 1 Solve for the unknown leg. θ 7 √65 7 √65 x θ Apply the Pythagorean theorem. x2 1 72 5 A!65B 2 x2 1 49 5 65 x2 5 16 x 5 64 Lengths of sides can be only positive. x 5 4 STEP 2 Label the sides of the triangle ■ ■with numbers representing lengths. ■ ■as hypotenuse, or as opposite or adjacent with respect to u. STEP 3 Set up the trigonometric functions as ratios. Cosine is adjacent over hypotenuse. cos u 5 adjacent hypotenuse 5 4 !65 Tangent is opposite over adjacent. tan u 5 opposite adjacent 5 7 4 Secant is the reciprocal of cosine. sec u 5 1 cos u 5 1 4/!65 5 !65 4 Expressions that contain a radical in the denominator like 4 !65 can be rationalized by multiplying both the numerator and the denominator by the radical, !65. 4 !65 ⋅a !65 !65b 5 4!65 65 In this example, the cosine function can now be written as cos u 5 4!65 65 . In this text, denominators with rational expressions will be rationalized. YOUR T UR N For the triangle in Example 2, calculate sin u, csc u, and cot u. √65 Hypotenuse Adjacent 4 Opposite 7 θ 1 g ▼ ▼ A N S W E R sin u 5 7!65 65 , csc u 5 !65 7 , and cot u 5 4 7 6.2.3 Cofunctions Notice the co in cosine, cosecant, and cotangent functions. These cofunctions are based on the relationship of complementary angles. Let us look at a right triangle with labeled sides and angles. sin b 5 opposite of b hypotenuse 5 b c cos a 5 adjacent to a hypotenuse 5 b c sin b 5 cos a g b a β α c 6.2.3 S K IL L Express trigonometric functions in terms of their cofunctions. 6.2.3 C ON C E P T U A L Understand that the trigonometric function of an angle is the cofunction of its complementary angle. 6.2 Definition 1 of Trigonometric Functions: Right Triangle Ratios 515 516 CHAPTER 6 Trigonometric Functions Recall that the sum of the measures of angles in a triangle is 180°. In a right triangle, one angle is 90°, so the two acute angles are complementary angles (the measures sum to 90°). Therefore, in the triangle above, b and a are complementary angles. In other words, the sine of an angle is the same as the cosine of the complement of that angle. This is true for all trigonometric cofunction pairs. [CONCEPT CHECK] TRUE OR FALSE sin (u) 5 cos (908 2 u) ANSWER True ▼ COFUNCTION THEOREM A trigonometric function of an angle is always equal to the cofunction of the compl ement of the angle. If a 1 b 5 90°, then sin b 5 cos a sec b 5 csc a tan b 5 cot a COFUNCTION IDENTITIES sin u 5 cos190° 2 u2 cos u 5 sin190° 2 u2 tan u 5 cot190° 2 u2 cot u 5 tan190° 2 u2 sec u 5 csc190° 2 u2 csc u 5 sec190° 2 u2 b a c 90º – θ θ EXAMPLE 3 Writing Trigonometric Function Values in Terms of Their Cofunctions Write each function value in terms of its cofunction. a. sin 30° b. tan x c. csc 40° Solution (a): Cosine is the cofunction of sine. sinu 5 cos190° 2 u2 Substitute u 5 30°. sin 30° 5 cos190° 2 30°2 Simplify. sin 30° 5 cos 60° Solution (b): Cotangent is the cofunction of tangent. tan u 5 cot190° 2 u2 Substitute u 5 x. tan x 5 cot190° 2 x2 Solution (c): Secant is the cofunction of cosecant. csc u 5 sec190° 2 u2 Substitute u 5 40°. csc 40° 5 sec190° 2 40°2 Simplify. csc 40° 5 sec 50° Y OUR TU R N Write each function value in terms of its cofunction. a. cos 45° b. csc y ▼ ▼ A N S W E R a. sin 458 b. sec1908 2 y2 6.2 Definition 1 of Trigonometric Functions: Right Triangle Ratios 517 6.2.4 Evaluating Trigonometric Functions Exactly for Special Angle Measures: 308, 458, and 608 Thus far we have defined the trigonometric functions as ratios of sides of right trian-gles. We have demonstrated how to calculate the value of a trigonometric function for a general angle, but we have not yet discussed trigonometric function values for spe-cific angle measures. We now turn our attention to evaluating trigonometric functions for known angles. We will distinguish between evaluating a trigonometric function exactly and approximating the value of a trigonometric function with a calculator. Throughout this text, instructions will specify which is desired. There are three special acute angles that are very important in trigonometry: 30°, 45°, and 60°. In Section 6.1, we discussed two important triangles: 308-608-908 and 458-458-908. Recall the relationships between the sides of these two right triangles. x x 45º 45º √2x 60º 30º x 2x √3x We can combine these relationships with the trigonometric ratios developed in this section to evaluate the trigonometric functions for the special angle measures of 30°, 45°, and 60°. 6.2.4 S K I L L Evaluate trigonometric functions exactly for special angles. 6.2.4 C O N C E P T U A L Learn the trigonometric functions as ratios of sides of a right triangle. [CONCEPT CHECK] TRUE OR FALSE If you learn the values of sine and cosine for the special angles 308, 458, and 608, then you can calculate the other values (secant, cosecant, tangent, and cotangent) using the recipro-cal and quotient identities. ANSWER True ▼ EXAMPLE 4 Evaluating the Trigonometric Functions Exactly for 308 Evaluate the six trigonometric functions for an angle that measures 30°. Solution: Label the sides of the 308-608-908 triangle with respect to the 30° angle. Use the right triangle ratio definitions of sine, cosine, and tangent. sin 30° 5 opposite hypotenuse 5 x 2x 5 1 2 cos 30° 5 adjacent hypotenuse 5 "3x 2x 5 "3 2 tan 30° 5 opposite adjacent 5 x "3x 5 1 "3 5 1 "3 ⋅ "3 "3 5 "3 3 Use the reciprocal identities to obtain the values of the cosecant, secant, and cotangent functions. csc 30° 5 1 sin 30° 5 1 1 2 5 2 sec 30° 5 1 cos30° 5 1 !3 2 5 2 !3 5 2 !3 ⋅!3 !3 5 2!3 3 cot 30° 5 1 tan 30° 5 1 "3 3 5 3 !3 5 3 !3 ⋅ !3 !3 5 !3 30º 60º x 2x √3x Adjacent Opposite Hypotenuse STUDY TIP cos 30° 5 "3 2 is exact, whereas if we evaluate with a calculator, we get an approximation: cos 30° < 0.8660. This decimal approximation from a calculator is a way to “check” your “exact” answer. 518 CHAPTER 6 Trigonometric Functions The six trigonometric functions evaluated for an angle measuring 30° are sin 30° 5 1 2 cos 30° 5 !3 2 tan 30° 5 !3 3 csc 30° 5 2 sec 30° 5 2!3 3 cot 30° 5 !3 Y OUR TU R N Evaluate the six trigonometric functions for an angle that measures 60°. ▼ ▼ A N S W E R sin60° 5 !3 2 cos60° 5 1 2 tan60° 5 !3 csc60° 5 2!3 3 sec60° 5 2 cot 60° 5 !3 3 In comparing our answers in Example 4 and Your Turn, we see that the following cofunction relationships are true: sin 30° 5 cos 60° sec 30° 5 csc 60° tan 30° 5 cot 60° sin 60° 5 cos 30° sec 60° 5 csc 30° tan 60° 5 cot 30° which is expected, since 30° and 60° are complementary angles. STUDY TIP sin 45° 5 "2 2 is exact, whereas if we evaluate with a calculator, we get an approximation: sin 45° < 0.7071. This decimal approximation is a way to “check” your “exact” answer. EXAMPLE 5 Evaluating the Trigonometric Functions Exactly for 458 Evaluate the six trigonometric functions for an angle that measures 45°. Solution: Label the sides of the 458-458-908 right triangle with respect to one of the 45° angles. Use the right triangle ratio definitions of sine, cosine, and tangent. sin45° 5 opposite hypotenuse 5 x !2x 5 1 !2 5 1 !2 ⋅ !2 !2 5 !2 2 cos45° 5 adjacent hypotenuse 5 x !2x 5 1 !2 5 1 !2 ⋅ !2 !2 5 !2 2 tan45° 5 opposite adjacent 5 x x 5 1 Use the reciprocal identities to obtain the values of the cosecant, secant, and cotangent functions. csc 45° 5 1 sin 45° 5 1 !2 2 5 2 !2 5 2 !2 ⋅ !2 !2 5 !2 sec 45° 5 1 cos 45° 5 1 !2 2 5 2 !2 5 2 !2 ⋅ !2 !2 5 !2 cot 45° 5 1 tan 45° 5 1 1 5 1 x x 45º √2x Adjacent Opposite Hypotenuse 6.2 Definition 1 of Trigonometric Functions: Right Triangle Ratios 519 The six trigonometric functions evaluated for an angle measuring 45° are sin 45° 5 !2 2 cos 45° 5 !2 2 tan 45° 5 1 csc 45° 5 !2 sec 45° 5 !2 cot 45° 5 1 We see that the following cofunction relationships are true: sin 45° 5 cos 45° sec 45° 5 csc 45° tan 45° 5 cot 45° which is expected, since 45° and 45° are complementary angles. The trigonometric function values for the three special angle measures 1308, 458, and 6082 are summarized in the following table. It is important to learn the special values in red for sine and cosine. All other values in the table can be found through reciprocals or quotients of these two functions. Remember that the tangent function is the ratio of the sine function to the cosine function. sinu 5 opposite hypotenuse cosu 5 adjacent hypotenuse tanu 5 sinu cosu 5 opposite hypotenuse adjacent hypotenuse 5 opposite adjacent Note in the above table that sin 308 5 1 2 and cos 308 5 !3 2 so tan30° 5 sin30° cos30° 5 1/2 !3/2 5 1 !3 5 !3 3 . 6.2.5 Using Calculators to Evaluate (Approximate) Trigonometric Functions We now turn our attention to using calculators to evaluate trigonometric functions, which often results in an approximation. Scientific and graphing calculators have buttons for sine (sin), cosine (cos), and tangent (tan) functions. Let us start with what we already know and confirm it with our calculators. STUDY TIP SOHCAHTOA: • Sine is Opposite over Hypotenuse. • Cosine is Adjacent over Hypotenuse. • Tangent is Opposite over Adjacent. STUDY TIP If you memorize the values for sine and cosine for the angles given in the table, then the other trigonometric function values in the table can be found using the quotient and reciprocal identities. STUDY TIP In calculating secant, cosecant, and cotangent function values with a calculator, it is important not to round the number until after using the reciprocal function key 1/x. 6.2.5 S K I L L Evaluate (approximate) trigonometric functions using a calculator. 6.2.5 C O N C E P T U A L Understand the difference between evaluating trigonometric functions exactly and using a calculator. u sin u cos u tan u cot u sec u csc u 30° 1 2 !3 2 !3 3 !3 2!3 3 2 45° !2 2 !2 2 1 1 !2 !2 60° !3 2 1 2 !3 !3 3 2 2!3 3 Trigonometric Values for Special Angles 520 CHAPTER 6 Trigonometric Functions EXAMPLE 6 Evaluating Trigonometric Functions with a Calculator Use a calculator to find the values of a. sin 75° b. tan 67° c. sec 52° d. cos 30° Round your answers to four decimal places. Solution: a. 0.965925826 < 0.9659 b. 2.355852366 < 2.3559 c. 1 cos52° < 1 0.615661475 < 1.6247 d. 0.866025403 < 0.8660 Note: We know cos 30° 5 "3 2 < 0.8660. Y OUR TU R N Use a calculator to find the values of a. cos 22° b. tan 81° c. csc 37° d. sin 45° Round your answers to four decimal places. ▼ When calcu lating secant, cosecant, and cotangent function values with a calculator, it is important not to round the number until after using the reciprocal function key 1/x or x21 in order to be as accurate as possible. ▼ A N S W E R a. 0.9272 b. 6.3138 c. 1.6616 d. 0.7071 [CONCEPT CHECK] TRUE OR FALSE sin 458 = 0.7071 ANSWER False—that is an approximation. The exact value is !2 2 . ▼ special angles 1308, 458, and 6082, and calculators can be used to approximate trigonometric function values of any angle. In this section, we defined trigonometric functions as ratios of sides of right triangles. This approach is called right triangle trigonometry. This is the first of three definitions of trigonometric functions (others will follow in Sections 6.4 and 6.7). We now can find trigonometric functions of an acute angle by taking ratios of the three sides of a right triangle: “adjacent,” “opposite,” and “hypotenuse.” It is important to remember that “adjacent” and “opposite” are defined with respect to one of the acute angles we are considering. We learned that trigonometric functions of an angle are equal to the cofunctions of the complement to the angle. Trigonometric functions can be evaluated exactly for [SEC TION 6. 2] S U M MA RY u sin u cos u tan u 308 1 2 !3 2 !3 3 458 "2 2 "2 2 1 608 !3 2 1 2 !3 [SEC TION 6. 2] E X E R C I SE S • S K I L L S In Exercises 1–6, refer to the triangle in the drawing to find the indicated trigonometric function values. 1. sin u 2. cos u 3. csc u 4. sec u 5. tan u 6. cot u 8 6 10 θ In Exercises 7–12, refer to the triangle in the drawing to find the indicated trigonometric function values. Rationalize any denominators containing radicals that you encounter in the answers. 7. cos u 8. sin u 9. sec u 10. csc u 11. tan u 12. cot u For Exercises 13–18, refer to the triangle in the drawing to find the indicated trigonometric function values. Rationalize any denominators containing radicals that you encounter in the answers. 13. sin u 14. cos u 15. sec u 16. csc u 17. cot u 18. tan u In Exercises 19–24, use the cofunction identities to fill in the blanks. 19. sin 60° 5 cos _ 20. sin 45° 5 cos _ 21. cos x 5 sin __ 22. cot A 5 tan _ 23. csc 30° 5 sec __ 24. sec B 5 csc __ In Exercises 25–32, write the trigonometric function values in terms of its cofunction. 25. sin1x 1 y2 26. sin160° 2 x2 27. cos120° 1 A2 28. cos1A 1 B2 29. cot145° 2 x2 30. sec130° 2 u2 31. csc160° 2 u2 32. tan140° 1 u2 In Exercises 33–38, match the trigonometric function values. a. 1 2 b. !3 2 c. !2 2 33. sin 30° 34. sin 60° 35. cos 30° 36. cos 60° 37. sin 45° 38. cos 45° In Exercises 39–41, use the results in Exercises 33–38 and the trigonometric quotient identity tan u 5 sin u cos u to calculate the following values. 39. tan 30° 40. tan 45° 41. tan 60° In Exercises 42–50, use the results in Exercises 33–41 and the reciprocal identities csc u 5 1 sin u, sec u 5 1 cos u, and cot u 5 1 tan u to calculate the following values. 42. csc 30° 43. sec 30° 44. cot 30° 45. csc 60° 46. sec 60° 47. cot 60° 48. csc 45° 49. sec 45° 50. cot 45° In Exercises 51–64, use a calculator to evaluate the trigonometric functions for the indicated values. Round your answers to four decimal places. 51. sin 37° 52. sin 17.8° 53. cos82° 54. cos 21.9° 55. tan 54° 56. tan43.2° 57. sec8° 58. sec 75° 59. csc89° 60. csc 51° 61. cot 55° 62. cot 29° 63. tanA211 9 B° 64. cot A405 13 B° 2 1 θ 7 3 θ 6.2 Definition 1 of Trigonometric Functions: Right Triangle Ratios 521 • A P P L I C A T I O N S For Exercises 65 and 66, refer to the following: When traveling through air, a spherical drop of blood with diameter d maintains its spherical shape until hitting a flat surface. The direction of travel of the drop of blood dictates the directionality of the blood splatter on the surface. For this reason, the diameter of the blood drop is equal to the width of the blood splatter on the surface. The angle at which a spherical drop of blood is deposited on a surface, called angle of impact, is related to the width w and the length l of the splatter by sin u 5 w l . Top view Direction of travel Blood drop Blood splatter Side view θ w l Drop 522 CHAPTER 6 Trigonometric Functions 65. Forensic Science. If a drop of blood found at a crime scene has a width of 6 millimeters and a length of 12 millimeters, find the angle u that represents the directionality. 66. Forensic Science. If a drop of blood found at a crime scene has a width of !2 millimeters and a length of 2 millimeters, find the angle u that represents the directionality. For Exercises 67 and 68, refer to the following: The monthly profits of PizzaRia are a function of sales, that is, p(s). A financial analysis has determined that the sales s in thousands of dollars of PizzaRia are also related to monthly profits p in thousands of dollars by the relationship tan u 5 p s for 0 # s # 55 and 0 # p # 45 Profts (p) p(s) Sales (s) p s θ Based on sales and profits, it can be determined that the domain for angle u is 0° # u # 40° The ratio tan u represents the slope of the hypotenuse of the right triangle formed by sales s and profit p (see figure above). The angle u can be interpreted as a measure of the relative size of s to p. The larger the angle u is, the greater profit is relative to sales and, conversely, the smaller the angle u is, the smaller profit is relative to sales. 67. Business. If PizzaRia’s monthly sales are $25,000 and monthly profits are $10,000, find a. tan u b. cot u 68. Business. If PizzaRia’s monthly sales are s and monthly profits are p: a. Determine the hypotenuse in terms of s and p. b. Determine a formula for cos u in terms of s and p. For Exercises 69 and 70, refer to the following: A man lives in a house that borders a pasture. He decides to go to the grocery store to get some milk. He is trying to decide whether to drive along the roads in his car or take his all-terrain vehicle (ATV) across the pasture. His car is faster than the ATV, but the distance the ATV would travel is less than the distance he would travel in his car. θ 69. Shortcut. If sinu 5 3 5 and cosu 5 4 5 and he drove his car along the streets, it would be 14 miles round trip. How far would he have to go on his ATV round trip? 70. Shortcut. If tan u 5 1 and he drove his car along the streets, it would be 200 yards round trip. How far would he have to go on his ATV round trip? Round your answer to the nearest yard. For Exercises 71–74, refer to the following: Have you ever noticed that if you put a stick in the water, it looks bent? We know the stick didn’t bend. Instead, the light rays bent which made the image appear to bend. Light rays propagating from one medium (like air) to another medium (like water) experience refraction, or “bending,” with respect to the surface. Light bends according to Snell’s law, which states ni sin1ui2 5 nr sin1u r2 where ■ ni is the refractive index of the medium the light is leaving. ■ ui is the incident angle between the light ray and the normal (perpendicular) to the interface between mediums. ■ nr is the refractive index of the medium the light is entering. ■ ur is the refractive angle between the light ray and the normal (perpendicular) to the interface between mediums. Light Ray Incident Angle Surface Refractive Angle θiº θrº nr ni Calculate the index of refraction nr of the indicated refractive medium given the following assumptions. Round answers to three decimals. ■ The incident medium is air. ■ Air has an index of refraction of ni 5 1.00. ■ The incidence angle is ui 5 30°. 71. Optics. Diamond, ur 5 12° 72. Optics. Emerald, ur 5 18.5° 73. Optics. Water, ur 5 22° 74. Optics. Plastic, ur 5 20° For Exercises 75–78, refer back to Exercises 73–76 in Section 6.1 for an explanation of the setting. 75. Archery. A beginning archer stands 50 feet from the target. If u represents the angle of depression of the bow (where the horizontal is the line segment connecting the original position of the arrowhead, just before release, and the bullseye on the target), and it is known that cos u 5 0.9992, what would her score be on this shot? 76. Archery. A beginning archer stands 50 feet from the target. If b represents the angle at which the arrow strikes the target, and it is known that sin b 5 0.9995, what would her score be on this shot? 77. Archery. A beginning archer stands 50 feet from the target. If b represents the angle at which the arrow strikes the target, and it is known that tan b 5 0, what would her score be on this shot? 78. Archery. A beginning archer stands 50 feet from the target. If b represents the angle at which the arrow strikes the target, and it is known that cot b 5 1, what would her score be on this shot? For Exercises 79 and 80, refer to the following: X-ray crystallography is a method of determining the arrangement of atoms within a crystal. This method has revealed the structure and functioning of many biological molecules including vitamins, drugs, proteins, and nucleic acids (including DNA). The structure of a crystal can be determined experimentally using Bragg’s law: nl 5 2d sin u where l is the wavelength of X-ray (measured in angstroms), d is the distance between atomic planes (measured in angstroms), u is the angle of reflection (in degrees), and n is the order of Bragg reflection (a positive integer). 79. Physics/Life Sciences. A diffractometer was used to make a diffraction pattern for a protein crystal from which it was determined experimentally that X-rays of wavelength 1.54 angstroms produced an angle of reflection of 458 corre-sponding to a Bragg reflection of order 1. Find the distance between atomic planes for the protein crystal to the nearest hundredth of an angstrom. 80. Physics/Life Sciences. A diffractometer was used to make a diffraction pattern for a salt crystal from which it was determined experimentally that X-rays of wavelength 1.67 angstroms produced an angle of reflection of 71.38 corresponding to a Bragg reflection of order 4. Find the distance between atomic planes for the salt crystal to the nearest hundredth of an angstrom. 6.2 Definition 1 of Trigonometric Functions: Right Triangle Ratios 523 • C A T C H T H E M I S T A K E For Exercises 81–84, explain the mistake that is made. For the triangle in the drawing, calculate the indicated trigonomet-ric function values. 3 4 5 x y 81. Calculate sin y. Solution: Formulate sine in terms of trigonometric ratios. The opposite side is 4, and the hypotenuse is 5. This is incorrect. What mistake was made? 82. Calculate tan x. Solution: Formulate tangent in terms of trigonometric ratios. The adjacent side is 3, and the opposite side is 4. This is incorrect. What mistake was made? 83. Calculate sec x. Solution: Formulate sine in terms of trigonometric ratios. The opposite side is 4, and the hypotenuse is 5. Write secant as the reciprocal of sine. Simplify. sec x 5 1 4 5 5 5 4 This is incorrect. What mistake was made? 84. Calculate csc y. Solution: Formulate cosine in terms of trigonometric ratios. The adjacent side is 4, and the hypotenuse is 5. Write cosecant as the reciprocal of cosine. Simplify. csc y 5 1 4 5 5 5 4 This is incorrect. What mistake was made? sin y 5 opposite hypotenuse sin y 5 4 5 tan x 5 adjacent opposite tan x 5 3 4 sin y 5 opposite hypotenuse sin x 5 4 5 sec x 5 1 sin x cos y 5 adjacent hypotenuse cos y 5 4 5 csc y 5 1 cos y 524 CHAPTER 6 Trigonometric Functions In Exercises 85–92, use the special triangles 1308-608-908 and 458-458-9082 shown below to help you answer the questions: x x 45º 45º √2x 60º 30º x 2x √3x In Exercises 85 and 86, determine whether each statement is true or false. 85. sin 45° 5 cos 45° 86. sin 60° 5 cos 30° 87. Calculate sin 30° and cos 30°. 88. Calculate sin 60° and cos 60°. 89. Calculate tan 30° and tan 60°. 90. Calculate sin 45°, cos 45°, and tan 45°. 91. Calculate sec 45° and csc 45°. 92. Calculate tan 60° and cot 30°. • C O N C E P T U A L • C H A L L E N G E In Exercises 93–96, use trignometric ratios and the assumption that a is much larger than b. Thus far in this text we have discussed trigonometric values only for acute angles or for 0° , u , 90°. How do we determine these values when u is approximately 0° or 90°? We will formally consider these cases in Section 6.5, but for now, draw and label a right triangle that has one angle very close to 0°, so that the opposite side is very small compared to the adjacent side. Then the hypotenuse and the adjacent side will be very close to the same length. a b θ 93. Approximate sin 08 without using a calculator. 94. Approximate cos 08 without using a calculator. 95. Approximate cos 908 without using a calculator. 96. Approximate sin 908 without using a calculator. In Exercises 97 –104, use the following diagram. Make certain to rationalize the denominators of all answers involving radicals that you encounter. θ β y 2x 2x x 1 D C B A 97. Calculate BD. 98. Determine y. 99. Determine the exact numerical value for sin b. 100. Calculate sec190° 2 b2. 101. Calculate tan2 b 1 1. 102. True or False: cos2 b 2 sin2 b 5 1. 103. Calculate sin2 u 1 cos2 b. 104. Calculate csc190° 2 u2 ⋅sec 190° 2 b2. • T E C H N O L O G Y 105. Calculate sec 70° the following two ways: a. Find cos 70° to three decimal places and then divide 1 by that number. Write that number to five decimal places. b. In a calculator, enter 70, cos, 1/x, and round the result to five decimal places. 106. Calculate csc 40° the following two ways: a. Find sin 40°, write that down (round to three decimal places), and then divide 1 by that number. Write this last result to five decimal places. b. In a calculator, enter 40, sin, 1/x, and round the result to five decimal places. 107. Calculate cot 54.9° the following two ways: a. Find tan 54.9° to three decimal places and then divide 1 by that number. Write that number to five decimal places. b. In a calculator, enter 54.9, tan, 1/x, and round the result to five decimal places. 108. Calculate sec 18.6° the following two ways: a. Find cos 18.6° to three decimal places and then divide 1 by that number. Write that number to five decimal places. b. In a calculator, enter 18.6, cos, 1/x, and round the result to five decimal places. To solve a triangle means to find the measure of the three angles and three sides of the triangle. In this section, we will discuss only right triangles (therefore, we know one angle has a measure of 908). We will know some information (the measures of two sides or the measures of a side and an acute angle), and we will determine the measures of the unknown sides and angles. However, before we start solving right triangles and determining measures, we must first discuss accuracy and significant digits. 6.3.1 Accuracy and Significant Digits If we are upgrading our flooring and quickly measure a room as 10ft 3 12ft and want to calculate the diagonal length of the room, we use the Pythagorean theorem. WORDS MATH Use the Pythagorean theorem. 102 1 122 5 d 2 Simplify. d 2 5 244 Use the square root property. d 5 6!244 The length of the diagonal must be positive. d 5 !244 Approximate the radical with a calculator. d < 15.62049935 Would you say that the 10 ft 3 12 ft room has a diagonal of 15.62049935 feet? No, because the known room measurements were given only with an accuracy of 1 foot, and the diagonal above is calculated to eight decimal places. Your results are no more accurate than the least accurate number in your calculation. In this example, we round to the nearest foot, and hence we say that the diagonal of the 10 ft 3 12 ft room is about 16 feet. Significant digits are used to determine the precision of a measurement. 10 ft 12 ft d = ? ft S K I L L S O B J E C T I V E S ■ ■Identify the number of significant digits to express the lengths of sides and measures of angles when solving right triangles. ■ ■Solve right triangles given the measure of an acute angle and the length of a side. ■ ■Solve right triangles given two side lengths. C O N C E P T U A L O B J E C T I V ES ■ ■Understand that the least accurate number in your calculation determines the accuracy of your result. ■ ■Understand that when an acute angle is given, then the third angle can be found exactly. The remaining unknown side lengths can be found using right triangle trigonometry and the Pythagorean theorem. ■ ■Understand that the trigonometric inverse keys on a calculator can be used to approximate the measure of an angle given its trigonometric function value. 6.3 APPLICATIONS OF RIGHT TRIANGLE TRIGONOMETRY: SOLVING RIGHT TRIANGLES 6.3.1 S K IL L Identify the number of signifi-cant digits to express the lengths of sides and measures of angles when solving right triangles. 6.2.1 C ON C E P T U A L Understand that the least accurate number in your calculation deter-mines the accuracy of your result. STUDY TIP The least accurate number in your calculation determines the accuracy of your result. [CONCEPT CHECK] In solving a triangle, the (least/ greatest) number of significant digits of the given information is used when solving for the unknown side lengths and angle measures. ANSWER Least ▼ DEFINITION Significant Digits The number of significant digits in a number is found by counting all of the digits from left to right starting with the first nonzero digit. 6.3 Applications of Right Triangle Trigonometry: Solving Right Triangles 525 The question mark next to 8000 means that we don’t know. If 8000 is a result from rounding to the nearest thousand, then it has one significant digit. If 8000 is the result of rounding to the nearest ten, then it has three significant digits, and if there are exactly 8000 people surveyed, then 8000 is an exact value and it has four significant digits. In this text, we will assume that integers have the greatest number of significant digits. Therefore, 8000 has four significant digits and can be expressed in scientific notation: 8.000 3 103. In solving right triangles, we first determine which of the given measurements has the least number of significant digits and round our answers to the same number of significant digits. NUMBER SIGNIFICANT DIGITS 0.04 1 0.276 3 0.2076 4 1.23 3 17 2 17.00 4 17.000 5 6.25 3 8000 ? EXAMPLE 1 Identifying the Least Number of Significant Digits Determine the number of significant digits corresponding to the given information in the following triangle: measure of an acute angle and a side length. In solving this right triangle, what number of significant digits should be used to express the remaining measure and side lengths? Solution: Determine the significant digits corresponding to b 5 458. Two significant digits Determine the significant digits corresponding to b 5 27.3 ft. Three significant digits In solving this triangle, the remaining side lengths of a and c and the measure of a should be expressed to two significant digits. b = 27.3 ft a β = 45º c α 6.3.2 Solving a Right Triangle Given the Measure of an Acute Angle and a Side Length When solving a right triangle, we already know that one angle has measure 908. Let us now consider the case when the measure of an acute angle and a side length are given. Since the measure of one of the acute angles is given, the remaining acute angle can be found using the fact that the sum of three angles in a triangle is 1808. Right triangle trigonometry is then used to find a second side length, and then the Pythagorean theo-rem is used to find the remaining (third) side length. 6.3.2 SKI LL Solve right triangles given the measure of an acute angle and the length of a side. 6.3.2 CO NCE PTUAL Understand that when an acute angle is given, then the third angle can be found exactly. The remaining unknown side lengths can be found using right triangle trigonometry and the Pythago-rean theorem. 526 CHAPTER 6 Trigonometric Functions EXAMPLE 2 Solving a Right Triangle Given an Angle and a Side Solve the right triangle—find a, b, and a. Solution: STEP 1 Determine accuracy. Since the given quantities, 15 feet and 568, are both expressed to two significant digits, we will round all calculated values to two significant digits. STEP 2 Solve for a. Two acute angles in a right triangle are complementary. a 1 56° 5 90° Solve for a. a 5 34° STEP 3 Solve for a. Cosine of an angle is equal to the adjacent side over the hypotenuse. Solve for a. a 5 15 cos 56° Evaluate the right side of the expression using a calculator. a < 8.38789 Round a to two significant digits. a < 8.4 ft STEP 4 Solve for b. Notice that there are two ways to solve for b: trigonometric functions or the Pythagorean theorem. Although it is tempting to use the Pythagorean theorem, it is better to use the given information with trigonometric functions than to use a value that has already been rounded, which could make the results less accurate. Sine of an angle is equal to the opposite side over the hypotenuse. Solve for b. b 5 15 sin 56° Evaluate the right side of the expression using a calculator. b < 12.43556 Round b to two significant digits. b < 12 ft STEP 5 Check. Angles and sides are rounded to two significant digits. Check the trigonometric values of the specific angles by calculating the trigonometric ratios. sin 34° 0 8.4 15 cos 34° 0 12 15 tan 34° 0 8.4 12 0.5592 < 0.56 0.8290 < 0.80 0.6745 < 0.70 YOUR T UR N Solve the right triangle; find a, b, and u. a 15 ft b α β = 56º cos 56° 5 a 15 sin 56° 5 b 15 8.4 ft 15 ft 12 ft 56º 34º ▼ ▼ A N S W E R u 5 53°, a 5 26 in., b 5 20 in. a 33 in. b 37º θ [CONCEPT CHECK] In Your Turn, angle u can be calculated exactly. Can the side lengths be calculated exactly? ANSWER No, they must be approximated. ▼ 6.3 Applications of Right Triangle Trigonometry: Solving Right Triangles 527 6.3.3 Solving a Right Triangle Given the Length of Two Sides When solving a right triangle, we already know that one angle has measure 908. Let us now consider the case when the lengths of two sides are given. In this case the third side can be found using the Pythagorean theorem. If we can determine the measure of one of the acute angles, then we can find the measure of the third acute angle using the fact that the sum of the three angle measures in a triangle is 1808. How do we find the measure of one of the acute angles? Since we know the measure of the side lengths, we can use right triangle ratios to determine the trigonometric function (sine, cosine, or tangent) values and then ask ourselves: What angle corresponds to that value? Sometimes we may know the answer exactly. For example, if we determine that sinu 5 1 2, then we know that the acute angle u is 308 because sin30° 5 1 2. Other times we may not know the corresponding angle, such as sin u 5 0.9511. Calculators have three keys 1sin21, cos21, and tan212 that help us determine the unknown angle. For example, a calculator can be used to help us find what angle u corresponds to sin u 5 0.9511. sin2110.95112 5 72.00806419° At first glance, these three keys might appear to yield the reciprocal; however, the 21 superscript corresponds to an inverse function. We will learn more about inverse trigonometric functions in Chapter 7, but for now we waaaill use these three calculator keys to help us solve right triangles. 6.3.3 SKI LL Solve right triangles given two side lengths. 6.3.3 CO NCE PTUAL Understand that the trigonometric inverse keys on a calculator can be used to approximate the measure of an angle given its trigonometric function value. [CONCEPT CHECK] When two side lengths are given but the two acute angles in a right triangle are unknown, how do we find an approximation of the acute angles? ANSWER Inverse trigonometric function ▼ ▼ A N S W E R 518 EXAMPLE 3 Using a Calculator to Determine an Acute Angle Measure Use a calculator to find u. Round answers to the nearest degree. a. cos u 5 0.8734 b. tan u 5 2.752 Solution (a): Use a calculator to evaluate the inverse cosine function. cos2110.87342 5 29.143821968 Round to the nearest degree. u < 29° Solution (b): Use a calculator to evaluate the inverse tangent function. tan2112.7522 5 70.030267848 Round to the nearest degree. u < 70° Y OUR TU R N Use a calculator to find u, given sin u 5 0.7739. Round answer to the nearest degree. ▼ EXAMPLE 4 Solving a Right Triangle Given Two Sides Solve the right triangle—find a, a, and b. Solution: STEP 1 Determine accuracy. The given sides have four significant digits; therefore, round all calculated values to four significant digits. a 37.21 cm 19.67 cm α β 528 CHAPTER 6 Trigonometric Functions STEP 2 Solve for a. Cosine of an angle is equal to the adjacent side over hypotenuse. Evaluate the right side using a calculator. cos a < 0.528621338 Write the angle a in terms of the inverse cosine function. a < cos2110.5286213382 Use a calculator to evaluate the inverse cosine function. a < 58.08764854° Round a to the nearest hundredth of a degree. a < 58.09° STEP 3 Solve for b. The two acute angles in a right triangle are complementary. a 1 b 5 90° Substitute a < 58.09°. 58.09° 1 b < 90° Solve for b. b < 31.91° The answer is already rounded to the nearest hundredth of a degree. STEP 4 Solve for a. Use the Pythagorean theorem. a2 1 b2 5 c2 Substitute the given values for b and c. a2 1 19.672 5 37.212 Solve for a. a < 31.5859969 Round a to four significant digits. a < 31.59 cm STEP 5 Check. Check the trigonometric values of the specific angles by calculating the trigonometric ratio. sin 31.91° 5 ? 19.67 37.21 sin 58.09° 5 ? 31.59 37.21 0.5286 < 0.5286 0.8490 < 0.8490 YOUR T UR N Solve the right triangle; find a, a, and b. cos a 5 19.67 cm 37.21 cm 37.21 cm 19.67 cm 31.59 cm 58.09º 31.91º 17.2 mi a β α 23.5 mi ▼ ▼ A N S W E R a 5 16.0 miles, a 5 43.0°, b 5 47.0° STUDY TIP To find a length in a right triangle, use the sine, cosine, and tangent functions. To find an angle in a right triangle, given the proper ratio, use the inverse sine, inverse cosine, and inverse tangent functions. Applications In many applications of solving right triangles, you are given a side and an acute angle and are asked to find one of the other sides. Two common examples involve an observer (or point of reference) located on the horizontal and an object that is either above or below the horizontal. If the object is above the horizontal, then the angle made is called the angle of elevation, and if the object is below the horizontal, then the angle made is called the angle of depression. 6.3 Applications of Right Triangle Trigonometry: Solving Right Triangles 529 For example, if a race car driver is looking straight ahead (in a horizontal line of sight), then looking up is elevation and looking down is depression. Angle of Depression Angle of Elevation Horizontal John Harrelson/Stringer/Getty Images If the angle is a physical one (like a skateboard ramp), then the appropriate name is angle of inclination. Angle of Inclination EXAMPLE 5 Angle of Depression (NASCAR) In this picture, car 19 is behind the leader car 2. If the angle of depression is 18° from the car 19 driver’s eyes to the bottom of the 3-feet-high back end of car 2 (side opposite the angle of depression), how far apart are their bumpers? Assume that the horizontal distance from the car 19 driver’s eyes to the front of his car is 5 feet. Solution: Draw an appropriate right triangle and label the known quantities. Identify the tangent ratio. tan 18° 5 3 x Solve for x. x 5 3 tan 18° Evaluate the right side. x < 9.233 Round to the nearest foot. x < 9 ft Subtract 5 feet (eyes to end of car). 9 2 5 5 4 ft Their bumpers are 4 feet apart. 3 x 18º 530 CHAPTER 6 Trigonometric Functions Suppose NASA wants to talk with the International Space Station (ISS), which is traveling at a speed of 17,700 miles per hour, 400 kilometers (250 miles) above the surface of the Earth. If the antennas at the ground station in Houston have a pointing error of even 1/100 of a degree, that is, 0.018, the ground station will miss the chance to talk with the astronauts. EXAMPLE 6 Pointing Error Assume that the ISS (which is 108 meters long and 73 meters wide) is in a 400- kilometer low earth orbit. If the communications antennas have a 0.018 pointing error, how many meters off will the communications link be? Solution: Draw a right triangle that depicts this scenario. 400 km x 0.01° ISS (NOT TO SCALE) Identify the tangent ratio. tan 0.01° 5 x 400 Solve for x. x 5 1400 km2 tan 0.01° Evaluate the expression on the right. x < 0.06981317 km 400 kilometers is accurate to three significant digits, so we express the answer to three significant digits. The pointing error causes the signal to be off by 69.8 meters . Since the ISS is only 108 meters long, it is expected that the signal will be missed by the astronaut crew. In navigation, the word bearing means the direction toward which a vessel is pointing. Heading is the direction in which the vessel is actually traveling. Heading and bearing are synonyms only when there is no wind on land. Direction is often given as a bearing, which is the mea sure of an acute angle with respect to the north–south vertical line. “The plane has a bearing N 208 E” means that the plane is pointed 20° to the east of due north. E W N S 20º 6.3 Applications of Right Triangle Trigonometry: Solving Right Triangles 531 EXAMPLE 7 Bearing (Navigation) A jet takes off bearing N 28° E and flies 5 miles and then makes a left 190°2 turn and flies 12 miles farther. If the control tower operator wants to locate the plane, what bearing should she use? Solution: Draw a picture that represents this scenario. Identify the tangent ratio. tanu 5 12 5 Use the inverse tangent function to solve for u. u 5 tan21a12 5 b < 67.4° Subtract 28° from u to find the bearing, b. b < 67.4° 2 28° < 39.4° Round to the nearest degree. b < N 39° W N θ 12 mi 5 mi β 28º right triangle (find all unknown side and angle measurements). The least accurate number used in your calculations determines the appropriate number of significant digits for your results. In this section, we used right triangle trigonometry to solve right triangles. When either a side length and an acute angle measure are given or two side lengths are given, it is possible to solve the [SEC TION 6.3] S U M MA RY [SEC TION 6.3] E X E R C I SE S • S K I L L S In Exercises 1–4, determine the number of significant digits corresponding to each of the given angle measures and side lengths. 1. a 5 37.58 2. b 5 49.768 3. a 5 0.37 km 4. b 5 0.2 mi In Exercises 5–10, use a calculator to find the measure of angle u. Round answer to the nearest degree. 5. sin u 5 0.7264 6. sin u 5 0.1798 7. cos u 5 0.5674 8. cos u 5 0.8866 9. tan u 5 8.235 10. tan u 5 3.563 532 CHAPTER 6 Trigonometric Functions In Exercises 11–30, refer to the right triangle diagram and the given information to find the indicated measure. Write your answers for angle measures in decimal degrees. 11. b 5 35°, c 5 17 in.; find a. 12. b 5 35°, c 5 17 in.; find b. 13. a 5 55°, c 5 22 ft; find a. 14. a 5 55°, c 5 22 ft; find b. 15. a 5 20.5°, b 5 14.7 mi; find a. 16. b 5 69.3°, a 5 0.752 mi; find b. 17. b 5 25°, a 5 11 km; find c. 18. b 5 75°, b 5 26 km; find c. 19. a 5 48.25°, a 5 15.37 cm; find c. 20. a 5 29.80°, b 5 16.79 cm; find c. 21. a 5 29 mm, c 5 38 mm; find a. 22. a 5 89 mm, c 5 99 mm; find b. 23. b 5 2.3 m, c 5 4.9 m; find a. 24. b 5 7.8 m, c 5 13 m; find b. 25. a 5 21.283° , b 5 210.8 yd; find a. 26. b 5 27.35°, a 5 117.0 yd; find b. 27. b 5 15.33° , a 5 10.2 km; find c. 28. b 5 65.5° , b 5 18.6 km; find c. 29. a 5 40.4694° , a 5 12,522 km; find c. 30. a 5 28.5472°, b 5 17,986 km; find c. In Exercises 31–46, refer to the right triangle diagram and the given information to solve the right triangle. Write your answers for angle measures in decimal degrees. 31. a 5 32° and c 5 12 ft 32. a 5 65° and c 5 37 ft 33. b 5 72° and c 5 9.7 mm 34. b 5 45° and c 5 7.8 mm 35. a 5 54.2° and a 5 111 mi 36. b 5 47.2° and a 5 9.75 mi 37. a 5 28.383° and b 5 1734 ft 38. a 5 72.983° and a 5 2175 ft 39. a 5 42.5 ft and b 5 28.7 ft 40. a 5 19.8 ft and c 5 48.7 ft 41. a 5 35,236 km and c 5 42,766 km 42. b 5 0.1245 mm and c 5 0.8763 mm 43. b 5 25.4° and b 5 11.6 in. 44. b 5 39.21° and b 5 6.3 m 45. b 5 16.10° and a 5 4.00 cm 46. a 5 50.1° and b 5 4 cm b a β α c b a β α c • A P P L I C A T I O N S 47. Golf. If the flagpole that a golfer aims at on a green measures 5 feet from the ground to the top of the flag and a golfer measures a 18 angle from top to bottom, how far (in horizontal distance) is the golfer from the flag? Round to the nearest foot. 48. Golf. If the flagpole that a golfer aims at on a green measures 5 feet from the ground to the top of the flag and a golfer measures a 38 angle from top to bottom, how far (in horizontal distance) is the golfer from the flag? Round to the nearest foot. Exercises 49 and 50 illustrate a midair refueling scenario that military aircraft often enact. Assume the elevation angle that the hose makes with the plane being fueled is u 5 36°. Hose b a θ = 36º 49. Midair Refueling. If the hose is 150 feet long, what should be the altitude difference a between the two planes? Round to the nearest foot. 50. Midair Refueling. If the smallest acceptable altitude difference a between the two planes is 100 feet, how long should the hose be? Round to the nearest foot. Exercises 51–54 are based on the idea of a glide slope (the angle the flight path makes with the ground). Precision Approach Path Indicator (PAPI) lights are used as a visual approach slope aid for pilots landing aircraft. A typical glide path for commercial jet airliners is 38. The space shuttle had an outer glide approach of 188–208. PAPI lights are typically configured as a row of four lights. All four lights are on, but in different combinations of red or white. If all four lights are white, then the angle of descent is too high; if all four lights are red, then the angle of descent is too low; and if there are two white and two red, then the approach is perfect. 3° Ground Runway PAPI Altitude 51. Glide Path of a Commercial Jet Airliner. If a commercial jetliner is 5000 feet (about 1 mile) ground distance from the runway, what should the altitude of the plane be to achieve two red and two white PAPI lights? (Assume this corresponds to a 3° glide path.) 5 ft 1º (NOT TO SCALE) 6.3 Applications of Right Triangle Trigonometry: Solving Right Triangles 533 52. Glide Path of a Commercial Jet Airliner. If a commercial jetliner is at an altitude of 450 feet when it is 5200 feet from the runway (approximately 1 mile ground distance), what is the glide slope angle? Will the pilot see white lights, red lights, or both? 53. Glide Path of the Space Shuttle Orbiter. If the pilot of the space shuttle Orbiter was at an altitude of 3000 feet when she was 15,500 feet (approximately 3 miles) from the shuttle land-ing facility (ground distance), what was her glide slope angle (round to the nearest degree)? Was she too high or too low? 54. Glide Path of the Space Shuttle Orbiter. If the same pilot in Exercise 53 raised the nose of the gliding shuttle so that she droped only 500 feet by the time she was 7800 feet from the shuttle landing strip (ground distance), what was her glide angle then (round to the nearest degree)? Was she within the specs to land the shuttle? In Exercises 55 and 56, refer to the illustration below which shows a search and rescue helicopter with a 308 field of view with a searchlight. 30º 55. Search and Rescue. If the search and rescue helicopter is flying at an altitude of 150 feet above sea level, what is the diameter of the circle illuminated on the surface of the water? 56. Search and Rescue. If the search and rescue helicopter is flying at an altitude of 500 feet above sea level, what is the diameter of the circle illuminated on the surface of the water? For Exercises 57–60, refer to the following: Geostationary orbits are useful because they cause a satellite to appear stationary with respect to a fixed point on the rotating Earth. As a result, an antenna (dish TV) can point in a fixed direction and maintain a link with the satellite. The satellite orbits in the direction of the Earth’s rotation, at an altitude of approximately 35,000 kilometers. 57. Dish TV. If your dish TV antenna has a pointing error of 0.0002788, how long would the satellite have to be in order to maintain a link? Round your answer to the nearest meter. 58. Dish TV. If your dish TV antenna has a pointing error of 0.0001398, how long would the satellite have to be in order to maintain a link? Round your answer to the nearest meter. 59. Dish TV. If the satellite in a geosta tionary orbit (at 35,000 kilometers) was only 10 meters long, about how accurate would the pointing of the dish have to be? Give the answer in degrees to two significant digits. 60. Dish TV. If the satellite in a geosta tionary orbit (at 35,000 kilometers) was only 30 meters long, about how accurate would the pointing of the dish have to be? Give the answer in degrees to two significant digits. 61. Angle of Elevation (Traffic). A person driving in a sedan is driving too close to the back of an 18 wheeler on an interstate highway. He decides to back off until he can see the entire truck (to the top). If the height of the trailer is 15 feet and the sedan driver’s angle of elevation (to the top of the trailer from the horizontal line with the bottom of the trailer) is roughly 30°, how far is he sitting from the end of the trailer? 15 ft x 30º 62. Angle of Depression (Opera). The balcony seats at the opera house have an angle of depression of 55° to center stage. If the horizontal (ground) distance to the center of the stage is 50 feet, how far are the patrons in the balcony from the singer at center stage? Foto World/Stone/Getty Images, Inc. 63. Angle of Inclination (Skiing). The angle of inclination of a mountain with triple black diamond ski trails is 65°. If a skier at the top of the mountain is at an elevation of 4000 feet, how long is the ski run from the top to the base of the mountain? 64. Bearing (Navigation). If a plane takes off bearing N 33° W and flies 6 miles and then makes a right 190°2 turn and flies 10 miles farther, what bearing will the traffic controller use to locate the plane? For Exercises 65 and 66, refer to the following: With the advent of new technology, tennis racquets can now be constructed to permit a player to serve at speeds in excess of 120 mph (as demonstrated by Novak Djokovic and Rafael Nadal, to name just two). One of the most effective serves in tennis is a power serve that is hit at top speed directly at the top left corner of the right service court (or top right corner of the left service court). When attempting this serve, a player will toss the ball rather high into the air, bring the racquet back, and then make contact with the ball at the precise moment when the position of the ball in the air coincides with the top of the netted part of the racquet when the player’s arm is fully stretched over his or her head. 65. Tennis. Assume that the player is serving into the right service court and stands just 2 inches to the right of the center line behind the baseline. If, at the moment the racquet strikes the ball, the ball is 9 feet from the ground and the serve 35,000 km 534 CHAPTER 6 Trigonometric Functions actually hits the top left corner of the right service court, determine the angle at which the ball meets the ground in the right service court. Round to the nearest degree. 66. Tennis. Assume that the player is serving into the right service court and stands just 2 inches to the right of the center line behind the baseline. If the ball hits the top left corner of the right service court at an angle of 8°, at what height above the ground must the ball be struck? For Exercises 67 and 68, refer to the following: The structure of molecules is critical to the study of materials science and organic chemistry, and has countless applications to a variety of interesting phenomena. Trigonometry plays a critical role in deter mining the bonding angles of molecules. For instance, the structure of the 1FeCl4Br2223 ion (dibro matetetrachlorideferrate III) is shown in the figure below. 2.354 Br Br Fe Cl Cl Cl Cl θ 2.219 67. Chemistry. Determine the angle u (i.e., the angle between the axis containing the apical bromide atom (Br) and the segment connecting Br to Cl). 68. Chemistry. Now, suppose one of the chlorides (Cl) is removed. The resulting structure is triagonal in nature, resulting in the following structure. Does the angle u change? If so, what is its new value? 2.354 2.097 Br Br Fe Cl Cl Cl θ 120º 120º For Exercises 69–74, refer back to Exercises 75–78 in Section 6.1 for an explanation of the setting. 69. Archery. If the arrow hits the target 13 inches directly below the bullseye, and it has traveled 92 feet from its original position at release to the point of impact with the target, how far was the archer from the target? Express your answer in four significant digits. 70. Archery. In Exercise 69, what was the tangent of the angle of depression from the line of sight connecting the arrowhead to the bullseye? Express your answer in four significant digits. 71. Archery. Assuming that the angle of depression from the line of sight connecting the arrowhead to the bullseye is 0.78, how far should the archer stand from the target to ensure she hits the target directly below the bullseye where the red and blue rings meet? Express your answer in two significant digits. 72. Archery. In Exercise 71, how far has the arrow traveled? Express your answer in four significant digits. 73. Archery. If the archer stands 90 feet from the target and the angle at which the arrow hits the target is 88.48, what is her score? 74. Archery. In Exercise 73, what is the angle of depression of the arrowhead? Center Service Line Left Service Court Right Service Court Right Service Court Left Service Court Fore Court Back Court Net Side Line Side Screen Post Alley Line Service Line Base Line Back Screen Center Mark 42 ft 78 ft 21 ft 4 ft 6 in. 18 ft 21 ft 21 ft 18 ft 21 ft 13 ft 6 in. 13 ft 6 in. 27 ft Singles 36 ft Doubles 6.3 Applications of Right Triangle Trigonometry: Solving Right Triangles 535 75. Construction. Two neighborhood kids are planning to build a treehouse in tree 1 and connect it to tree 2, which is 40 yards away. The base of the treehouse will be 20 feet above the ground, and a platform will be nailed into tree 2, 3 feet above the ground. The plan is to connect the base of the treehouse on tree 1 to an anchor 2 feet above the platform on tree 2. 2 ft 3 ft 20 ft 40 yd Tree 1 Tree 2 How much zipline (in feet) will they need? Round your answer to the nearest foot. 76. Construction. In Exercise 75, what is the angle of depression b that the zipline makes with tree1? Express your answer in two significant digits. (Hint: Find an expression for tan b, and use your calculator to compute tan21 of this expression.) 77. Construction. A pool that measures 5 feet above the ground is to be placed between the trees directly in the path of the zipline in Exercise 75. Assuming that a rider of the zipline dangles at most 3 feet below the wire anywhere in route, what is the closest the edge of the pool can be placed to tree 2 so that the pool will not impede a rider’s trip down the zipline? 78. Construction. If the treehouse is to be built so that its base is now 22 feet above the base of tree 1, where should the anchor on tree 2 (to which the zipline is connected) be placed in order to ensure the same angle of depression found in Exercise 76? For Exercises 79 and 80, refer to the following: A canal constructed by a water-users association can be approximated by an isosceles triangle (see the figure). When the canal was originally constructed, the depth of the canal was 5.0 feet and the angle defining the shape of the canal was 60°. Width Depth θ 79. Environmental Science. If the width of the water surface today is 4.0 feet, find the depth of the water running through the canal. 80. Environmental Science. One year later a survey is performed to measure the effects of erosion on the canal. It is determined that when the water depth is 4.0 feet, the width of the water surface is 5.0 feet. Find the angle u defining the shape of the canal to the nearest degree. Has erosion affected the shape of the canal? Explain. For Exercises 81 and 82, refer to the following: After breaking a femur, a patient is placed in traction. The end of a femur of length l is lifted to an elevation forming an angle u with the horizontal (angle of elevation). Length Elevation θ 81. Health/Medicine. A femur 18 inches long is placed into traction, forming an angle of 158 with the horizontal. Find the height of elevation at the end of the femur. 82. Health/Medicine. A femur 18 inches long is placed in traction with an elevation of 6.2 inches. What is the angle of elevation of the femur? • C A T C H T H E M I S T A K E In Exercises 83 and 84, refer to the right triangle diagram below and explain the mistake that is made: b a β α c 83. If b 5 800 ft and a 5 10 ft, find b. Solution: Represent tangent as the opposite side over the adjacent side. Substitute b 5 800 ft and a 5 10 ft. Use a calculator to evaluate b. b 5 tan 80° < 5.67° This is incorrect. What mistake was made? 84. If b 5 56° and c 5 15 ft, find b and then find a. Solution: Write sine as the opposite side over the hypotenuse. Solve for b. b 5 15 sin 56° Use a calculator to approximate b. b < 12.4356 Round the answer to two significant digits. b < 12 ft Use the Pythagorean theorem to find a. a2 1 b2 5 c2 Substitute b 5 12 ft and c 5 15 ft. a2 1 122 5 152 Solve for a. a 5 9 Round the answer to two significant digits. a 5 9.0 ft Compare this with the results from Example 2. Why did we get a different value for a here? tan b 5 b a tan b 5 800 10 sin 56° 5 b 15 536 CHAPTER 6 Trigonometric Functions In Exercises 85–89, determine whether each statement is true or false. • C O N C E P T U A L 85. If you are given the measures of two sides of a right triangle, you can solve the right triangle. 86. If you are given the measures of one side and one acute angle of a right triangle, you can solve the right triangle. 87. If you are given the two acute angles of a right triangle, you can solve the right triangle. 88. If you are given the hypotenuse of a right triangle and the angle opposite the hypotenuse, you can solve the right triangle. 89. If both sin u and cos u are known for a given angle u in a right triangle, you can solve the triangle. • C H A L L E N G E 90. Use the information in the illustration below to determine the height of the mountain. 51º 38º y x 1900 ft 91. Two friends who are engineers at Kennedy Space Center (KSC) watch the rocket launch of the SpaceX Dragon CRS-8. Carolyn is at the Vehicle Assembly Building (VAB) 3 miles from the launch pad, and Jackie is across the Banana River, which is 8 miles from the launch pad. They call each other at liftoff, and after 10 seconds they each estimate the elevation with respect to the ground. Carolyn thinks the elevation is approximately 40°, and Jackie thinks the elevation is approximately 15°. Approx-imately how high is the SpaceX Dragon after 10 seconds? (Average their estimates and round to the nearest mile.) Image Bank/Getty Images, Inc. Photonica/Getty Images 40º 15º Jackie Carolyn 3 miles 5 miles 92. A crew sets out in search of a buried treasure. A diver has a device that informs him where on the ocean floor the treasure is buried, but it has difficulty determining the exact depth that it lies below the surface. The known information is given in the follow-ing diagram. Determine y, the depth at which the treasure lies. B B y 300 ft 600 ft Ship’s anchor Treasure Ship Surface Ocean foor For Exercises 93 and 94, consider the following diagram: A x y 2A 3x 4 1 93. Determine x. 94. Determine y. 95. Consider the following diagram and compute tan 1908 2 A2. 2 7 A 96. Consider the following diagram and compute sec A⋅csc B. Express your answer in terms of x. A B x x 2x 1 x 2 97. Use a calculator to find sin211sin 40°2. 98. Use a calculator to find cos211cos 17°2. 99. Use a calculator to find cos 1cos21 0.82. 100. Use a calculator to find sin1sin21 0.32. 101. Based on the result from Exercise 97, what would sin211sin u2 be for an acute angle u? 102. Based on the result from Exercise 98, what would cos211cos u2 be for an acute angle u? 103. Compute tan21 1tan 3582. What would you expect tan21 1tan u2 to be for an acute angle? 104. Compute tan 1tan2122. What would you expect tan 1tan21x2 to be for any real number? 105. Use a calculator to find sin 1sin21 0.7852. 106. Use a calculator to find tan 1tan21 5.3212. • T E C H N O L O G Y 6.3 Applications of Right Triangle Trigonometry: Solving Right Triangles 537 538 CHAPTER 6 Trigonometric Functions 6.4.1 Angles in Standard Position In Section 6.1 we introduced angles. A common unit of measure for angles is degrees. We discussed triangles and the fact that the sum of the measures of the three interior angles is always 180°. We also discussed the Pythagorean theorem, which relates the lengths of the three sides of a right triangle. Based on what we learned in Section 6.1, the following two angles, which correspond to a right angle or quarter rotation, both have a measurement of 90°. θ = 90º θ = 90º We need a frame of reference. In this section, we use the Cartesian plane as our frame of re ference by superimposing angles onto the Cartesian coordinate system, or rectan gular coordinate system (Section 2.1). Let us now bridge two concepts with which you are already familiar. First, recall that an angle is generated when a ray (the angle’s initial side) is rotated around an endpoint (which becomes the angle’s vertex). The ray, in its new position after it is rotated, is called the terminal side of the angle. Also recall the Cartesian (rectangular) coordinate system with the x-axis, y-axis, and origin. Using the positive x-axis combined with the origin as a frame of reference, we can graph angles in the Cartesian plane. If the initial side of the angle is aligned along the positive x-axis and the vertex of the angle is positioned at the origin, then the angle is said to be in standard position. S K I L L S O B J E C T I V E S ■ ■Sketch angles in standard position. ■ ■Identify coterminal angles. ■ ■Calculate trigonometric function values for nonacute and quadrantal angles. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that rotation in the counterclockwise direction corresponds to a positive angle measure, whereas rotation in the clockwise direction corresponds to a negative angle measure. ■ ■Understand that the measures of coterminal angles must differ by an integer multiple of 3608. ■ ■Understand that right triangle definitions of trigonometric functions for acute angles are consistent with definitions of trigonometric functions for all angles in the Cartesian plane. 6.4 DEFINITION 2 OF TRIGONOMETRIC FUNCTIONS: CARTESIAN PLANE 6.4.1 SKILL Sketch angles in standard position. 6.4.1 C ON CEPTUAL Understand that rotation in the counterclockwise direction corresponds to a positive angle measure, whereas rotation in the clockwise direction corresponds to a negative measure. STUDY TIP For a review of the rectangular (Cartesian) coordinate system, see Section 2.1. [CONCEPT CHECK] Match the following: When the angle is formed using a clockwise/counterclockwise rotation that corresponds to an angle with positive/negative measure. ANSWER Clockwise—negative and Counterclockwise—positive ▼ An angle is said to be in standard position if its initial side is along the positive x-axis and its vertex is at the origin. Standard Position DEFINITION Terminal side Initial side Vertex (0, 0) x y STUDY TIP The initial side of an angle is a ray (initial ray), and the terminal side of an angle is also a ray (terminal ray). We say that an angle lies in the quadrant in which its terminal side lies. For example, an acute angle 10° u 90°2 lies in quadrant I, whereas an obtuse angle 190° u 180°2 lies in quadrant II. Similarly, angles with measure 180° u 270° lie in quadrant III, and angles with measure 270° u 360° lie in quadrant IV. Angles in standard position with terminal sides along the x-axis or y-axis 190°, 180°, 270°, 360°, etc.2 are called quadrantal angles. An abbreviated way to represent an angle u that lies in quadrant I is u [ QI. Similarly, if an angle lies in quadrant II, we say u [ QII, and so forth. x y 90º < θ < 180º QII 0º < θ < 90º QI 180º < θ < 270º QIII 270º < θ < 360º QIV 90º 180º 0º or 360º 270º Recall that rotation in a counterclockwise direction corresponds to a positive angle, whereas rotation in a clockwise direction corresponds to a negative angle. ACUTE ANGLE Quadrant I x y OBTUSE ANGLE Quadrant II x y EXAMPLE 1 Sketching Angles in Standard Position Sketch the following angles in standard position, and state the quadrant in (or axis on) which the terminal side lies. a. 290° b. 210° Solution (a): The initial side lies on the positive x-axis. A negative angle indicates clockwise rotation. 90° is a right angle. The terminal side lies on the negative y-axis. Solution (b): The initial side lies on the positive x-axis. A positive angle indicates counterclockwise rotation. 180° represents a straight angle, and an additional 30° yields a 210° angle. The terminal side lies in quadrant III. YOUR T UR N Sketch the following angles in standard position, and state the quadrant in which the terminal side lies. a. 2300° b. 135° x y –90º x y 210º ▼ A N S W E R a. –300º x y Terminal side lies in quadrant I. b. 135º x y Terminal side lies in quadrant II. ▼ 6.4 Definition 2 of Trigonometric Functions: Cartesian Plane 539 540 CHAPTER 6 Trigonometric Functions Common Angles in Standard Position The common angles for which we determined the exact trigonometric function values in Section 6.2 are 30°, 45°, and 60°. Recall the relationships between the sides of 30°-60°-90° and 45°-45°-90° triangles. 60º 30º x 2x √3x x x 45º 45º √2x Let us assume the hypotenuse is equal to 1. Then we have the following triangles: 60º 30º 1 2 √3 2 1 45º 45º 1 2 √2 2 √2 We can position these triangles on the Cartesian plane with one leg along the positive x-axis, so that we have three angles 130°, 45°, and 60°2 in standard position. Remember that we are always assuming that the hypotenuse is equal to 1. Notice that the x- and y-coordinates of the points shown correspond to the side lengths. 60º 30º 30º angle 1 2 √3 2 √3 2 1 2 1 x y ( , ) 45º 45º 45º angle 1 2 √2 2 √2 2 √2 2 √2 x y ( , ) 60º 60º angle 30º 1 2 √3 2 1 2 √3 2 1 x y ( , ) If we graph the three angles 1308, 458, and 6082 on the same Cartesian coordinate system, we get the following in the first quadrant: 30º r = 1 r = 1 r = 1 60º 45º 2 √3 2 √2 2 √2 2 1 2 √3 2 1 x y ( , ) ( , ) ( , ) Using symmetry (Section 2.2) and the angles and coordinates in quadrant 1 (QI), we get the following angles and coordinates in QII, QIII, and QIV. Notice that all of these coordinate pairs satisfy the equation of the unit circle (radius equal to 1 and centered at the origin): x2 1 y2 5 1. 30º 330º 300º 240º 210º 150º 120º 60º x y 2 √3 2 1 (– , ) 2 √3 2 1 ( , ) 2 √3 2 1 (– , ) 2 √3 2 1 ( , ) 2 √3 2 1 (– , ) 2 √3 2 1 (– , ) – 2 √3 2 1 ( , ) – – 2 √3 2 1 ( , ) – 0° (1, 0) (0, 1) (0, –1) (–1, 0) 90° 315º 270° 225º 180° 135º 45º 2 √2 2 √2 ( , – ) 2 √2 2 √2 ( , ) 2 √2 2 √2 (– , – ) 2 √2 2 √2 (– , ) x y In Section 6.2, we defined trigonometric functions as ratios of side lengths of right triangles. This definition holds only for acute 10° , u , 90°2 angles, since the two angles in a right triangle other than the right angle must be acute. We now define trigonometric functions as ratios of x- and y-coordinates and distances in the Cartesian plane, which is consistent with right triangle trigonometry for acute angles. However, this second approach enables us to formulate trigonometric functions for quadrantal angles (whose terminal side lies along an axis) and nonacute angles. 6.4.2 Coterminal Angles STUDY TIP In Section 6.7, we will define the values of the sine and cosine functions as the y- and x-coordinates, respectively, along the unit circle. Two angles in standard position with the same terminal side are called coterminal angles. Coterminal Angles DEFINITION For example, 240° and 320° are coterminal; their terminal rays are identical, even though they are formed by rotation in opposite directions. The angles 60° and 420° are also coterminal; angles larger than 360° or less than 2360° are generated by continuing the rotation beyond a full rotation. Thus, all coterminal angles have the same initial side (positive x-axis) and the same terminal side, just different rotations. x y 320º –40º x y 420º 60º 6.4.2 S K I L L Identify coterminal angles. 6.4.2 C ON C E P T U A L Understand that the measures of coterminal angles must differ by an integer multiple of 3608. [CONCEPT CHECK] TRUE OR FALSE If an angle a has measure 508, then the angle b that has measure 23108 is coterminal with angle a. ANSWER True ▼ 6.4 Definition 2 of Trigonometric Functions: Cartesian Plane 541 542 CHAPTER 6 Trigonometric Functions To find measures of the smallest positive coterminal angles, if the given angle is positive and greater than 3608, subtract 360° repeatedly until the result is a positive angle less than or equal to 360°. If the given angle is nonpositive, add 360° repeatedly until the result is a positive angle less than or equal to 360°. EXAMPLE 2 Recognizing Coterminal Angles Determine whether the following pairs of angles are coterminal. a. a 5 120°, b 5 2180° b. a 5 20°, b 5 740° Solution (a): The terminal side of a is in QII. The terminal side of b is along the negative x-axis. a and b are not coterminal angles. Solution (b): Since 360° represents one rotation, 720° represents two rotations. Therefore, after 720° of rotation, the angle is again along the positive x-axis. An additional 20° of rotation achieves a 740° angle. Since a 5 20° and b 5 740° have the same terminal side, they are coterminal angles. Y OUR TU R N Determine whether the following pairs of angles are coterminal. a. a 5 240°, b 5 2120° b. a 5 20°, b 52380° α β α β ▼ EXAMPLE 3 Finding Measures of Coterminal Angles Determine the angle of the smallest possible positive measure that is coterminal with each of the following angles. a. 830° b. 2520° Solution (a): Since 830° is positive, subtract 360°. 830° 2 360° 5 470° Subtract 360° again. 470° 2 360° 5 110° The angle with measure 110° is the angle with the smallest positive measure that is coterminal with the angle with measure 830°. Solution (b): Since 2520° is negative, add 360°. 2520° 1 360° 5 2160° Add 360° again. 2160° 1 360° 5 200° The angle with measure 200° is the angle with the smallest positive measure that is coterminal with the angle with measure 2520°. Y OUR TU R N Determine the angle of the smallest possible positive measure that is coterminal with each of the following angles. a. 900° b. 2430° ▼ ▼ A N S W E R a. yes b. no STUDY TIP The measures of coterminal angles must differ by an integer multiple of 360°. ▼ A N S W E R a. 1808 b. 2908 6.4.3 Trigonometric Functions: The Cartesian Plane To define the trigonometric functions in the Cartesian plane, let us start with an acute angle u in standard position. Choose any point 1x, y2 on the terminal side of the angle as long as it is not the vertex (the origin). A right triangle can be drawn so that the right angle is made when a perpendicular segment connects the point 1x, y2 to the x-axis. Notice that the side opposite u has length y and the other leg of the right triangle has length x. (x, y) θ x y r x y WORDS MATH The distance r from the origin 10, 02 r 5 "1x 2 022 1 1y 2 022 to the point 1x, y2 can be found using the distance formula: r 5 "x2 1 y 2 Since r is a distance, it is always positive. r . 0 Using our first definition of trigonometric functions in terms of right triangle ratios (Section 6.2), we say that sin u 5 opposite hypotenuse. From this picture we see that sine can also be defined by the relation sin u 5 y r. Similar reasoning holds for all six trigonometric functions and leads us to the second definition of the trigonometric functions, in terms of ratios of coordinates and distances in the Cartesian plane. Let 1x, y2 be a point other than the origin on the terminal side of an angle u in standard position. Let r be the distance from the point 1x, y2 to the origin. Then the six trigonometric functions are defined as sin u 5 y r cos u 5 x r tan u 5 y x 1x 2 02 csc u 5 r y 1y 2 02 sec u 5 r x 1x 2 02 cot u 5 x y 1y 2 02 where r 5 "x2 1 y2, or x2 1 y2 5 r 2. The distance r is positive: r . 0. Trigonometric Functions DEFINITION (2) (x, y) θ x y 6.4.3 S K I L L Calculate trigonometric function values for nonacute and quadrantal angles. 6.4.3 C ON C E P T U A L Understand that right triangle definitions of trigonometric functions for acute angles are consistent with definitions of trigonometric functions for all angles in the Cartesian plane. [CONCEPT CHECK] TRUE OR FALSE The trigonometric function values for nonacute angles can be found using right triangle trigonometry for acute angles and then changing the algebraic sign (1/2) depending on the quadrant in which the terminal side lies. ANSWER True ▼ 6.4 Definition 2 of Trigonometric Functions: Cartesian Plane 543 544 CHAPTER 6 Trigonometric Functions We can now use this second definition of trigonometric functions to find values for nonacute angles (angles with measure greater than or equal to 908) as well as negative angles. EXAMPLE 4 Calculating Trigonometric Function Values for Acute Angles The terminal side of an angle u in standard position passes through the point (2, 5) Calculate the values of the six trigonometric functions for angle u. Solution: STEP 1 Draw the angle and label the point (2, 5). STEP 2 Calculate the distance r. r 5 "22 1 52 5 !29 STEP 3 Formulate the trigonometric functions in terms of x, y, and r. Let x 5 2, y 5 5, r 5 !29. sin u 5 y r 5 5 !29 cos u 5 x r 5 2 !29 tan u 5 y x 5 5 2 csc u 5 r y 5 !29 5 sec u 5 r x 5 !29 2 cot u 5 x y 5 2 5 STEP 4 Rationalize any denominators containing a radical. sin u 5 5 !29 ⋅ !29 !29 5 5!29 29 cos u 5 2 !29 ⋅ !29 !29 5 2!29 29 STEP 5 Write the values of the six trigonometric functions for u. sin u 5 5!29 29 cos u 5 2!29 29 tan u 5 5 2 csc u 5 !29 5 sec u 5 !29 2 cot u 5 2 5 Note: In Example 4, we could have used the values of the sine, cosine, and tangent functions along with the reciprocal identities to calculate the cosecant, secant, and cotangent function values. YOUR TURN The terminal side of an angle u in standard position passes through the point (3, 7). Calculate the values of the six trigonometric functions for angle u. x y θ r (2, 5) y = 5 x = 2 ▼ STUDY TIP There is no need to memorize definitions for secant, cosecant, and cotangent functions, since their values can be derived from the reciprocals of the sine, cosine, and tangent function values. ▼ A N S W E R sin u 5 7!58 58 cos u 5 3!58 58 tan u 5 7 3 csc u 5 !58 7 sec u 5 !58 3 cot u 5 3 7 EXAMPLE 5 Calculating Trigonometric Function Values for Nonacute Angles The terminal side of an angle u in standard position passes through the point (24, 27). Calculate the values of the six trigonometric functions for angle u. Solution: STEP 1 Draw the angle and label the point (24, 27). STEP 2 Calculate the distance r. r 5 "12422 1 12722 5 !65 STEP 3 Formulate the trigonometric functions in terms of x, y, and r. Let x 5 24, y 5 27, and r 5 !65. sin u 5 y r 5 27 !65 cos u 5 x r 5 24 !65 tan u 5 y x 5 27 24 5 7 4 csc u 5 r y 5 !65 27 sec u 5 r x 5 !65 24 cot u 5 x y 5 24 27 5 4 7 STEP 4 Rationalize the radical denominators in the sine and cosine functions. sin u 5 y r 5 27 !65 ⋅ !65 !65 5 27!65 65 cos u 5 x r 5 24 !65 ⋅ !65 !65 5 24!65 65 STEP 5 Write the values of the six trigonometric functions for u. sin u 5 2 7!65 65 cos u 5 2 4!65 65 tan u 5 7 4 csc u 5 2 !65 7 sec u 5 2 !65 4 cot u 5 4 7 YOUR T UR N The terminal side of an angle u in standard position passes through the point 123, 252. Calculate the values of the six trigonometric functions for angle u. x y θ (–4, –7) –7 r –4 ▼ ▼ A N S W E R sin u 5 25!34 34 cos u 5 23!34 34 tan u 5 5 3 csc u 5 2!34 5 sec u 5 2!34 3 cot u 5 3 5 6.4 Definition 2 of Trigonometric Functions: Cartesian Plane 545 546 CHAPTER 6 Trigonometric Functions If we say that the terminal side of an angle lies on a line that passes through the origin, we must specify which part of the line represents the terminal ray in order to determine the angle. For example, in the diagram to the left, if we specify x + 0, then we know that the terminal side lies in quadrant I. Alternatively, if we specify x 0, then we know that the terminal side lies in quadrant III. Once we know which part of the line represents the terminal side of the angle, it does not matter which point on the line we use to formulate the trigonometric function values because corresponding sides of similar triangles are proportional. x y x < 0 x > 0 EXAMPLE 6 Calculating Trigonometric Function Values for Nonacute Angles Calculate the values for the six trigonometric functions of angle u, given in standard position, if the terminal side of u lies on the line y 5 3x, x # 0. Solution: STEP 1 Draw the line and label a point on the terminal side. STEP 2 Calculate the distance r. r 5 "12122 1 12322 5 !10 STEP 3 Formulate the trigonometric functions in terms of x, y, and r. Let x 5 21, y 5 23, and r 5 !10. sin u 5 y r 5 23 !10 cos u 5 x r 5 21 !10 tan u 5 y x 5 23 21 5 3 csc u 5 r y 5 !10 23 sec u 5 r x 5 !10 21 cot u 5 x y 5 21 23 5 1 3 STEP 4 Rationalize the radical denominators in the sine and cosine functions. sin u 5 y r 5 23 !10⋅!10 !10 5 23!10 10 cos u 5 x r 5 21 !10⋅!10 !10 5 2 !10 10 STEP 5 Write the values of the six trigonometric functions for u. sin u 5 23!10 10 cos u 5 2 !10 10 tan u 5 3 csc u 5 2 !10 3 sec u 5 2!10 cot u 5 1 3 Y OUR TU R N Calculate the values for the six trigonometric functions of angle u, given in standard position, if the terminal side of u is defined by y 5 2x, x # 0. θ x y y = 3x (–1, –3) x ≤ 0 ▼ ▼ A N S W E R sin u 5 22!5 5 cos u 5 2!5 5 tan u 5 2 csc u 5 2!5 2 sec u 5 2!5 cot u 5 1 2 Thus far in Chapter 6 we have avoided evaluating expressions such as sin 90° because 90° is not an acute angle. However, with our second definition of trigonometric functions in the Cartesian plane, we now are able to evaluate the trigonometric functions for quadrantal angles (angles in standard position whose terminal sides coincide with an axis) such as 90°, 180°, 270°, and 360°. Notice that 90° and 270° lie along the y-axis and therefore have an x-coordinate value equal to 0. Similarly, 180° and 360° lie along the x-axis and have a y-coordinate value equal to 0. Some of the trigonometric functions are defined with x- or y-coordinates in the denominator, and since dividing by 0 is undefined in mathematics, not all trigonometric functions are defined for some quadrantal angles. EXAMPLE 7 Calculating Trigonometric Function Values for Quadrantal Angles Calculate the values for the six trigonometric functions when u 5 90°. Solution: STEP 1 Draw the angle and label a point on the terminal side. Note: A convenient point on the terminal side is 10, 12. STEP 2 Calculate the distance r. r 5 "1022 1 1122 5 !1 5 1 STEP 3 Formulate the trigonometric functions in terms of x, y, and r. Let x 5 0, y 5 1, and r 5 1. sin u 5 y r 5 1 1 cos u 5 x r 5 0 1 tan u 5 y x 5 1 0 csc u 5 r y 5 1 1 sec u 5 r x 5 1 0 cot u 5 x y 5 0 1 STEP 4 Write the values of the six trigonometric functions for u. sin u 5 1 cos u 5 0 tan u is undefined csc u 5 1 sec u is undefined cot u 5 0 YOUR T UR N Calculate the values for the six trigonometric functions when u 5 270°. (0, 1) θ = 90º x y ▼ ▼ A N S W E R sin u 5 21 cos u 5 0 tan u is undefined csc u 5 21 sec u is undefined cot u 5 0 6.4 Definition 2 of Trigonometric Functions: Cartesian Plane 547 548 CHAPTER 6 Trigonometric Functions EXAMPLE 8 Calculating Trigonometric Function Values for Quadrantal Angles Calculate the values for the six trigonometric functions when u 5 180°. Solution: STEP 1 Draw the angle and label a point on the terminal side. Note: A convenient point on the terminal side is 121, 02. STEP 2 Calculate the distance r. r 5 "12122 1 1022 5 !1 5 1 STEP 3 Formulate the trigonometric functions in terms of x, y, and r. Let x 5 21, y 5 0, and r 5 1. sin u 5 y r 5 0 1 cos u 5 x r 5 21 1 tan u 5 y x 5 0 21 csc u 5 r y 5 1 0 sec u 5 r x 5 1 21 cot u 5 x y 5 21 0 STEP 4 Write the values of the six trigonometric functions for u. sin u 5 0 cos u 5 21 tan u 5 0 csc u is undefined sec u 5 21 cot u is undefined Y OUR TU R N Calculate the values for the six trigonometric functions when u 5 360°. (–1, 0) θ = 180º x y ▼ ▼ A N S W E R sin u 5 0 cos u 5 1 tan u 5 0 csc u is undefined sec u 5 1 cot u is undefined u sin u cos u tan u cot u sec u csc u 08 0 1 0 undefined 1 undefined 908 1 0 undefined 0 undefined 1 1808 0 21 0 undefined 21 undefined 2708 21 0 undefined 0 undefined 21 3608 0 1 0 undefined 1 undefined To confirm these values, evaluate each of these functions with a calculator for the specified angles. Make sure the calculator is set in degree (not radian) mode. STUDY TIP Note: In Examples 7 and 8 we could have used the definitions of sine, cosine, and tangent along with the reciprocal identities to calculate cosecant, secant, and cotangent. The accompanying table summarizes the trigonometric function values for the quadrantal angles 08, 908, 1808, 2708, and 3608. An angle in the Cartesian plane is in standard position if its initial side lies along the positive x-axis and its vertex is located at the origin. Angles in standard position have terminal sides that lie either in one of the four quadrants or along one of the two axes. The special triangles, 308-608-908 and 458-458-908, with hypotenuses having measure 1, were used to develop the coordinates in quadrant I for the special angles, 308, 458, and 608. We then used symmetry to locate similar pairs of coordinates in the other quadrants. All right triangles with one vertex (not the right angle) located at the origin and hypotenuse equal to 1 have the other nonright angle vertex located along the unit circle 1x2 1 y2 5 12. Coterminal angles are angles in standard position that have the same initial side and the same terminal side. Trigonometric functions are defined in the Cartesian plane as ratios of coordinates and distances. Right triangle trigonometric definitions learned in Section 6.2 are consistent with these definitions. We now have the ability to evaluate trigonometric functions for nonacute angles. Trigonometric functions are not always defined for quadrantal angles. [SEC TION 6.4] S U M M A RY [SEC TION 6.4] E X ER C I S E S • S K I L L S In Exercises 1–22, state in which quadrant or on which axes the angles with given measure in standard position would lie. 1. 89° 2. 91° 3. 145° 4. 175° 5. 310° 6. 355° 7. 270° 8. 180° 9. 2540° 10. 2450° 11. 210.5° 12. 270.5° 13. 12.34° 14. 100.001° 15. 595° 16. 620° 17. 525° 18. 1085° 19. 2905° 20. 2640° 21. 1400.0001° 22. 23600° In Exercises 23–36, sketch each of the following angles with given measure in standard position. 23. 135° 24. 225° 25. 2405° 26. 2450° 27. 2225° 28. 2330° 29. 330° 30. 2150° 31. 510° 32. 2720° 33. 840° 34. 2380° 35. 2540° 36. 540° In Exercises 37–42, match the angles (a2f) with the coterminal angles (37–42). a. 30° b. 295° c. 185° d. 2560° e. 780° f. 75° 37. 2535° 38. 2690° 39. 60° 40. 265° 41. 2645° 42. 160° In Exercises 43–52, determine the angle of the smallest possible positive measure that is coterminal with each of the given angles. 43. 412° 44. 379° 45. 292° 46. 2187° 47. 2390° 48. 945° 49. 510° 50. 1395° 51. 1400.0001° 52. 612.34° In Exercises 53–72, the terminal side of an angle u in standard position passes through the indicated point. Calculate the values of the six trigonometric functions for angle u. 53. 11, 22 54. 12, 32 55. 13, 62 56. 18, 42 57. A1 2, 2 5B 58. A4 7, 2 3B 59. 122, 42 60. 121, 32 61. 124, 272 62. 129, 252 63. A2!2, !3B 64. A2!3, !2B 65. A2!5, 2!3B 66. A2!6, 2!5B 67. A210 3 , 24 3B 68. A22 9, 21 3B 69. 18, 242 70. 12.1, 24.22 71. A22 3, 1 3B 72. A22, 1 4B In Exercises 73–80, calculate the values for the six trigonometric functions of the angle u given in standard position, if the terminal side of u lies on each of the indicated lines. 73. y 5 2x, x $ 0 74. y 5 3x, x $ 0 75. y 5 1 2 x, x $ 0 76. y 5 1 2 x, x # 0 77. y 5 21 3 x, x $ 0 78. y 5 21 3 x, x # 0 79. 2x 1 3y 5 0, x # 0 80. 2x 1 3y 5 0, x $ 0 In Exercises 81–94, calculate (if possible) the values for the six trigonometric functions of the angle u given in standard position. 81. u 5 450° 82. u 5 540° 83. u 5 630° 84. u 5 720° 85. u 5 2270° 86. u 5 2180° 87. u 5 290° 88. u 5 2360° 89. u 5 2450° 90. u 5 2540° 91. u 5 2630° 92. u 5 2720° 93. u 5 21200° 94. u 5 1140° 6.4 Definition 2 of Trigonometric Functions: Cartesian Plane 549 550 CHAPTER 6 Trigonometric Functions • A P P L I C A T I O N S 95. Clock. What is the measure of the angle swept out by the second hand if it starts on the 3 and continues for 3 minutes and 20 seconds? 96. Clock. What is the measure of the angle swept out by the hour hand if it starts at 3 p.m. on Wednesday and continues until 5 p.m. on Thursday? For Exercises 97 and 98, refer to the following: The two-player game of Zim-Zam in which the players swat at a tennis ball tethered to a 4-foot-long string connected atop a 6-foot pole was popular in the mid-1980s. One player would be assigned the clockwise direction and the other the counterclockwise direction. A player’s goal was to use a sequence of hits to have the ball travel a full 7 revolutions around in the assigned direction. This was complicated by the fact that the person’s opponent was attempting to do the same thing in the opposite direction. For each complete revolution the ball made in a given direction, the corresponding player’s scoring device would increase by 1; this also had the effect of decreasing the opponent’s score by 1. The game was finished once a player reached a score of 7 points. 97. Sports. Suppose the scoring device for the player assigned the clockwise direction goes from 0 to 2 to 1 to 4 to 3 to 7. How many total degrees in the clockwise direction did the ball travel? 98. Sports. If during a game of Zim-Zam the ball traveled 9908 counterclockwise in one maneuver, how many points did the player assigned the counterclockwise direction earn? 99. Track. Don and Ron both started running around a circular track, starting at the same point, but Don ran counterclockwise and Ron ran clockwise. The paths they ran swept through angles of 900° and 2900°, respectively. Did they end up in the same spot when they finished? 100. Track. Dan and Stan both started running around a circular track, starting at the same point. The paths they ran swept through angles of 3640° and 1890°, respectively. Did they end up in the same spot when they finished? For Exercises 101 and 102, refer to the following: A common school locker combination lock is shown. The lock has a dial with 40 calibration marks numbered 0 to 39. A combination consists of three of these numbers (e.g., 5-35-20). To open the lock, the following steps are taken: iStockphoto n Turn the dial clockwise two full turns. n Stop at the first number of the combination. n Turn the dial counterclockwise one full turn. n Continue turning counterclockwise until the second number is reached. n Turn the dial clockwise again until the third number is reached. n Pull the shank and the lock will open. 101. Combination Lock. Given that the initial position of the dial is at zero (shown in the illustration), how many degrees is the dial rotated in total (sum of clockwise and counterclockwise rotations) in opening the lock if the combination is 35-5-20? 102. Combination Lock. Given that the initial position of the dial is at zero (shown in the illustration), how many degrees is the dial rotated in total (sum of clockwise and counterclockwise rotations) in opening the lock if the combination is 20-15-5? 103. Geometry. A right triangle is drawn in QI with one leg on the x-axis and its hypotenuse on the terminal side of angle u drawn in standard position. If sin u 5 7 25, what is tan u? 104. Geometry. A right triangle is drawn in QI with one leg on the x-axis and its hypotenuse on the terminal side of angle u drawn in standard position. If tan u 5 84 13, what is cos u? For Exercises 105 and 106, refer to the following figure: 105. Forestry. Let u be the angle of elevation from a point on the ground to the top of a tree. If cos u 5 11 61 and the distance from the point on the ground to the base of the tree is 22 feet, how high is the tree? 106. Forestry. Let u be the angle of elevation from a point on the ground to the top of a tree. If sin u 5 40 41 and the tree is 20 feet high, how far from the base of the tree is the point on the ground? For Exercises 107 and 108, refer to the following: The monthly revenues (meas-ured in thousands of dollars) of PizzaRia are a function of monthly costs (measured in thousands of dollars), that is, R1c2. Recall that the angle u can be interpreted as a meas-ure of the sizes of cost c and revenue R relative to each other, that is, tan u 5 revenue costs . The larger the angle u is, the greater revenue is relative to cost and, conversely the smaller the angle u is, the smaller revenue is relative to cost. 107. Business. An analysis of a month’s revenue and costs indicates sin u 5 8 10. Determine whether the company experiences a loss or profit for that month. 108. Business. An analysis of a month’s revenue and costs indicates cos u 5 3 4. Determine whether the company experiences a loss or profit for that month. For Exercises 109 and 110, refer to the following: A bunion is a progressive medical disorder of the foot in which the big toe gradually begins to lean toward the second toe. The condition can lead to a misalignment of the metatarsal and phalanges of the big toe in relation to the metatarsal and phalanges of the second toe (see the following figure). The angle u formed between the metatarsal of the big toe and the metatarsal of the second toe is called the first intermetatarsal angle and is normally less than 98. In the presence of a bunion, the intermetatarsal angle exceeds 98, which may be caused by the length of the first θ Revenue (R) Cost (c) R c θ metatarsal bone. Exercises 109 and 110 both correspond to the presence of a bunion. Phalanges Proximal end Distal end Metatarsals θ (Source: 109. Health/Medicine. If cos u 5 4.8 cm 5.0 cm, find the distance between the distal ends of the two metatarsals. Round to the nearest millimeter. 110. Health/Medicine. If sec u 5 5.0 cm 4.9 cm, find the distance between the distal ends of the two metatarsals. Round to the nearest millimeter. In Exercises 111 and 112, explain the mistake that is made. 111. The terminal side of an angle u in standard position passes through the point (1, 2). Calculate sin u. Solution: Label the coordinates. x 5 1, y 5 2 Calculate r. r 5 12 1 22 5 5 Use the definition of sine. sin u 5 y r Substitute y 5 2, r 5 5. sin u 5 2 5 This is incorrect. What mistake was made? 112. Evaluate sec 810° exactly. Solution: Find a coterminal angle that 810° 2 360° 5 450° lies between 0° and 360°. 450° 2 360° 5 90° 810° has the same terminal side as 90°. Evaluate cosine for 90°. cos 90° 5 0 Evaluate secant for 90°. sec 90° 5 1 cos 90° 5 1 0 5 0 This is incorrect. What mistake was made? • C A T C H T H E M I S T A K E • C O N C E P T U A L 113. sin 30° 1 sin 60° 5 sin 90° 114. tan 0° 1 tan 90° 5 tan 90° 115. cos u 5 cos1u 1 360° n2, where n is an integer. 116. sin u 5 sin1u 1 360° n2, where n is an integer. 117. cos190° 2 u2 5 sin1u 2 720°n2, where n is an integer. 118. csc1u 1 90°n2 5 sec190° 2 u 2 90°n2, where n is an integer. 119. If tan u 5 1, then u 5 45° 1 360°n, for some integer n. 120. If cot u 5 2!3, then u 5 150° 1 180°n, for some integer n. In Exercises 113–120, determine whether each statement is true or false. • C H A L L E N G E 121. If the terminal side of angle u passes through the point 123a, 4a2, find cos u. 122. If the terminal side of angle u passes through the point 123a, 4a2, find sin u. 123. If the line y 5 mx makes an angle u with the x-axis, which trigonometric function value for u is equal to m? 124. Find x if 1x, 222 is on the terminal side of angle u and csc u 5 2 !29 2 . 125. Find y if 123, y2 is on the terminal side of angle u and sec u 5 2 !17 3 . 126. Find the equation of the line with positive slope that passes through the point 1a, 02 and makes an acute angle u with the x-axis. The equation of the line will be in terms of x, a, and a trigonometric function of u. 127. Find the equation of the line with negative slope that passes through the point 1a, 02 and makes an acute angle u with the x-axis. The equation of the line will be in terms of x, a, and a trigonometric function of u. • T E C H N O L O G Y 128. sin 270° 129. cos 270° 130. tan 270° 131. cot 270° 132. sin12270°2 133. cos12270°2 134. csc12270°2 135. sec12270°2 136. sec 70° 137. csc 270° In Exercises 128–137, use a calculator to evaluate the following expressions. If you get an error, explain why. 6.4 Definition 2 of Trigonometric Functions: Cartesian Plane 551 552 CHAPTER 6 Trigonometric Functions S K I L L S O B J E C T I V E S ■ ■Determine the reference angle of a nonacute angle. ■ ■Determine the ranges of the six trigonometric functions. ■ ■Determine the reference angle for any angle whose terminal side lies in one of the four quadrants. ■ ■Evaluate trigonometric functions for nonacute and quadrantal angles. C O N C E P T U A L O B J E C T I V E S ■ ■ Understand which trigonometric functions are positive or negative in each of the four quadrants. ■ ■ Understand why sine and cosine range between 21 and 1. ■ ■Understand that reference angles are acute and have positive measure. ■ ■Understand that the value of a trigonometric function at an angle is the same as the trigonometric value of its reference angle, except there may be a difference in its algebraic sign 11/22. 6.5 TRIGONOMETRIC FUNCTIONS OF NONACUTE ANGLES In Section 6.4, we defined trigonometric functions in the Cartesian plane as ratios of x, y, and r. We calculated the trigonometric function values given a point on the terminal side of an angle in standard position. We also discussed the special case of a quadrantal angle. In this section, we now evaluate trigonometric functions exactly for common angles. Additionally, we will approximate trigonometric functions for nonacute angles using an acute reference angle and knowledge of algebraic signs of trigonometric functions in particular quadrants. 6.5.1 Algebraic Signs of Trigonometric Functions In Section 6.4, we defined trigonometric functions as ratios of x, y, and r. Since r is the distance from the origin to the point 1x, y2 and distance is never negative, r is always taken as the positive solution to r2 5 x2 1 y2, so r 5 "x2 1 y2. The x-coordinate is positive in quadrants I and IV and negative in quadrants II and III. The y-coordinate is positive in quadrants I and II and negative in quadrants III and IV. Recall the definition of the six trigonometric functions in the Cartesian plane: sin u 5 y r cos u 5 x r tan u 5 y x 1x 2 02 csc u 5 r y 1y 2 02 sec u 5 r x 1x 2 02 cot u 5 x y 1y 2 02 Therefore, the algebraic sign, 1 or 2, of each trigonometric function will depend on which quadrant contains the terminal side of angle u. Let us look at the three main trigonometric functions: sine, cosine, and tangent. In quadrant I, all three functions are positive since x, y, and r are all positive. However, in quadrant II, only sine is positive since y and r are both positive. In quadrant III, only tangent is positive, and in quadrant IV, only cosine is positive. The phrase “All Students Take Calculus” helps us remember which of the three main trigonometric functions are positive in each quadrant. PHRASE QUADRANT POSITIVE TRIGONOMETRIC FUNCTION All I All three: sine, cosine, and tangent Students II Sine Take III Tangent Calculus IV Cosine 6.5.1 S KILL Determine the reference angle of a nonacute angle. 6.5.1 C ON CEPTUAL Understand which trigonometric functions are positive or negative in each of the four quadrants. STUDY TIP All Students Take Calculus is a phrase that helps us remember which of the three (sine, cosine, tangent) functions are positive in quadrants I, II, III, and IV. x y QII QI QIII QIV x < 0 y > 0 r > 0 x > 0 y > 0 r > 0 x < 0 y < 0 r > 0 x > 0 y < 0 r > 0 x y A C T S Positivity Chart The following table indicates the algebraic sign of all six trigonometric functions according to the quadrant in which the terminal side of an angle u lies. Notice that the reciprocal functions have the same sign. TERMINAL SIDE OF u IN QUADRANT sin u cos u tan u cot u sec u csc u I 1 1 1 1 1 1 II 1 2 2 2 2 1 III 2 2 1 1 2 2 IV 2 1 2 2 1 2 6.5 Trigonometric Functions of Nonacute Angles 553 EXAMPLE 1 Evaluating a Trigonometric Function When the Quadrant of the Terminal Side Is Known If cos u 52 3 5 and the terminal side of u lies in quadrant III, find sin u. Solution: STEP 1 Draw some angle u in QIII. STEP 2 Identify known quantities from the information given. Recall that cos u 5 x r cos u 5 2 3 5 5 23 5 5 x r and r . 0. Identify x and r. x 5 23 and r 5 5 STEP 3 Since x and r are known, find y. Substitute x 5 23 and r 5 5 into x2 1 y2 5 r2. 12322 1 y2 5 52 Solve for y. 9 1 y2 5 25 y2 5 16 y 5 64 STEP 4 Select the sign of y based on quadrant information. Since the terminal side of u lies in quadrant III, y , 0 y 5 24 STEP 5 Find sin u. sin u 5 y r 5 24 5 sin u 5 24 5 YOUR T UR N If sin u 523 4 and the terminal side of u lies in quadrant III, find cos u. θ –4 (–3, –4) 5 –3 x y ▼ A N S W E R cos u 52 !7 4 ▼ [CONCEPT CHECK] In what quadrant are sine and cosine functions both negative? ANSWER III ▼ 554 CHAPTER 6 Trigonometric Functions We can also make a table showing the values of the trigonometric functions when the terminal side of u lies along each of the axes (i.e., when u is any of the quadrantal angles). We calculated this table for specific angles in Section 6.4, and we will develop it again here using the algebraic signs of the trigonometric functions in the different quadrants. When the terminal side lies along the x-axis, then y 5 0. When y 5 0, notice that r 5 "x2 1 y2 5 "x2 5 0 x 0. When the terminal side lies along the positive x-axis, x . 0; and when the terminal side lies along the negative x-axis, x , 0. Therefore, when the terminal side of the angle lies on the positive x-axis, then y 5 0, x . 0, and r 5 x; and when the terminal side lies along the negative x-axis, then y 5 0, x , 0, and r 5 0 x 0. A similar argument can be made for the y-axis that results in r 5 0 y 0. TERMINAL SIDE OF u LIES ALONG THE sin u cos u tan u cot u sec u csc u Positive x-axis 108 or 36082 0 1 0 undefined 1 undefined Positive y-axis 19082 1 0 undefined 0 undefined 1 Negative x-axis 118082 0 21 0 undefined 21 undefined Negative y-axis 127082 21 0 undefined 0 undefined 21 EXAMPLE 2 Working with Values of the Trigonometric Functions for Quadrantal Angles Evaluate the following expressions if possible. a. cos 540° 1 sin 270° b. cot 90° 1 tan1290°2 Solution (a): The terminal side of an angle with measure 540° lies along the negative x-axis. 540° 2 360° 5 180° Evaluate cosine of an angle whose terminal side lies along the negative x-axis. cos 540° 5 21 Evaluate sine of an angle whose terminal side lies along the negative y-axis. sin 270° 5 21 Sum the cosine and sine terms. cos 540° 1 sin 270° 5 21 1 1212 cos 540° 1 sin 270° 5 22 Check: Evaluate this expression with a calculator. Solution (b): Evaluate cotangent of an angle whose terminal side lies along the positive y-axis. cot 90° 5 0 The terminal side of an angle with measure 290° lies along the negative y-axis. tan1290°2 5 tan 270° The tangent function is undefined for an angle whose terminal side lies along the negative y-axis. tan1290°2 is undefined Even though cot 90° is defined, since tan1290°2 is undefined, the sum of the two expressions is also undefined. Y OUR TU R N Evaluate the following expressions if possible. a. csc90° 1 sec180° b. csc12630°2 1 sec12630°2 ▼ A N S W E R a. 0 b. undefined ▼ x y r = a (a, 0) r = \| –b \| = b (0, –b) 6.5.2 Ranges of the Trigonometric Functions Thus far, we have discussed what the algebraic sign of a trigonometric function value is in a particular quadrant, but we haven’t discussed how to find actual values of the trigonometric functions for nonacute angles. We will need to define reference angles and reference right triangles. However, before we proceed, let’s get a feel for the ranges (set of values of the functions) we will expect. Let us start with an angle u in quadrant I and the sine function defined as the ratio sin u 5 y r. sinu 5 y r If we keep the value of r constant, then as the measure of u increases toward 90°, y increases. Notice that the value of y approaches the value of r until they are equal when u 5 90°, and y can never be larger than r. A similar analysis can be conducted in quadrant IV as u approaches 290° from 0° (note that y is negative in quadrant IV). A result that is valid in all four quadrants is 0 y 0 # r. WORDS MATH Write the absolute value inequality as a double inequality. 2r # y # r Divide both sides by r. 21 # y r # 1 Let sinu 5 y r. 21 # sinu # 1 Similarly, by allowing u to approach 0° and 180°, we can show that 0 x 0 # r, which leads to the range of the cosine function: 21 # cos u # 1. Sine and cosine values θ x y r x y x y x y r θ 6.5.2 S K IL L Determine the ranges of the six trigonometric functions. 6.5.2 C ON C E P T U A L Understand why sine and cosine range between 21 and 1. [CONCEPT CHECK] The value of r is always (greater than/less than) the values of both x and y. ANSWER Greater than ▼ x y x y r u x y x y r θ r x y x y θ 6.5 Trigonometric Functions of Nonacute Angles 555 556 CHAPTER 6 Trigonometric Functions 6.5.3 Reference Angles and Reference Right Triangles Now that we know the trigonometric function ranges and their algebraic signs in each of the four quadrants, we can evaluate the trigonometric functions of nonacute angles. Before we do that, however, we first must discuss reference angles and reference right triangles. Every nonquadrantal angle in standard position has a corresponding reference angle and reference right triangle. We have already calculated the trigonometric function values for quadrantal angles. RANGES OF THE TRIGONOMETRIC FUNCTIONS For any angle u for which the trigonometric functions are defined, the six trigonometric functions have the following ranges: ■ 21 # sin u # 1 ■ csc u # 2 1 or csc u $ 1 ■ ■21 # cos u # 1 ■ sec u # 2 1 or sec u $ 1 ■ ■tan u and cot u can equal any real number. EXAMPLE 3 Determining Whether a Value Is Within the Range of a Trigonometric Function Determine whether or not each statement is possible. a. cos u 5 1.001 b. cotu 5 0 c. secu 5 !3 2 Solution (a): Not possible because 1.001 . 1. Solution (b): Possible because cot90° 5 0. Solution (c): Not possible because !3 2 < 0.866 , 1. Y OUR TU R N Determine whether or not each statement is possible. a. sinu 5 21.1 b. tanu 5 2 c. csc u 5 !3 ▼ ▼ A N S W E R a. not possible b. possible c. possible range between 21 and 1 and since secant and cosecant are reciprocals of the cosine and sine functions, respectively, their ranges are stated as sec u # 21 or sec u $ 1 csc u # 21 or csc u $ 1 Since tan u 5 y x and cot u 5 x y and since x , y, x 5 y, and x . y are all possible, the values of the tangent and cotangent functions can be any real numbers (positive, negative, or zero). The following box summarizes the ranges of the trigonometric functions. 6.5.3 SKI LL Determine the reference angle for any angle whose terminal side lies in one of the four quadrants. 6.5.3 CO NCE PTUAL Understand that reference angles are acute and have positive measure. The reference angle is the positive, acute angle that the terminal side makes with the x-axis. DEFINITION Reference Angle For angle u, 0° , u , 360°, in standard position whose terminal side lies in one of the four quadrants, there exists a reference angle a that is the acute angle with positive measure formed by the terminal side of u and the x-axis. QII θ α = 180º – θ α x y QIII θ α = θ – 180º α y x QIV θ α = 360º – θ α y x θ θ = α α x QI y STUDY TIP The reference angle is the acute angle that the terminal side makes with the x-axis, not the y-axis. DEFINITION Reference Right Triangle To form a reference right triangle for angle u, where 0° , u , 360°, drop a perpendicular line from the terminal side of the angle to the x-axis. The right triangle now has reference angle a as one of its angles. θ α = 180º – θ x y α x y r θ α = θ – 180º y x α x y r θ α = 360º – θ y x α x y r θ θ = α α x y x y r 6.5 Trigonometric Functions of Nonacute Angles 557 558 CHAPTER 6 Trigonometric Functions In Section 6.2, we first defined the trigonometric functions of an acute angle as ratios of lengths of sides of a right triangle. For example, sin u 5 opposite hypotenuse . The lengths of the sides of triangles are always positive. In Section 6.4, we defined the sine function of any angle as sin u 5 y r. For a nonacute angle u, sin u 5 y r; and for the acute reference angle a, sin a 5 k y k r . The only difference between these two expressions is the algebraic sign 11/22, since r is always positive and y is positive or negative depending on the quadrant. EXAMPLE 4 Finding Reference Angles Find the reference angle for each angle given. a. 210° b. 135° c. 422° Solution (a): The terminal side of u lies in quadrant III. The reference angle is made by the terminal side and the x-axis. 210° 2 180° 5 30° Solution (b): The terminal side of u lies in quadrant II. The reference angle is made by the terminal side and the negative x-axis. 180° 2 135° 5 45° Solution (c): The terminal side of u lies in quadrant I. The reference angle is made by the terminal side and the positive x-axis. 422° 2 360° 5 62° Y OUR TU R N Find the reference angle for each angle given. a. 160° b. 285° c. 600° x y 210º 30º x y 135º 45º x y 422º 62º ▼ A N S W E R a. 20° b. 75° c. 60° ▼ [CONCEPT CHECK] Which of the following is true about the reference angle? (A) It is the angle made by the terminal side and the x-axis (or negative x-axis). (B) It is positive. (C) It is an acute angle. (D) All of the above ANSWER (D) ▼ STUDY TIP To find the trigonometric function values of nonacute angles, first find the trigonometric values of the reference angle and then use the quadrant information to determine the algebraic sign. Therefore, to calculate the trigonometric function values for a nonacute angle, simply find the trigonometric values for the reference angle and determine the correct algebraic sign according to the quadrant in which the terminal side lies. 6.5.4 Evaluating Trigonometric Functions for Nonacute Angles Let’s look at a specific example before we generalize a procedure for evaluating trigonometric function values for non acute angles. Suppose we have the angles in standard position with measure 60°, 120°, 240°, and 300°. Notice that the reference angle for all of these angles is 608. x y 120º 60º 240º 300º 60º 60º 60º 60º If we draw reference triangles and let the shortest leg have length 1, we find that the other leg has length !3 and the hypotenuse has length 2. Notice that the legs of the triangles have lengths (always positive) 1 and !3; however, the coordinates are 161, 6!3 2. Therefore, when we calculate the trigo-nometric functions for any of the angles 60°, 120°, 240°, and 300°, we can simply calculate the trigonometric functions for the reference angle, 60°, and determine the algebraic sign 11 or 22 for the particular trigonometric function and quadrant. x y 120º 1 1 2 2 2 2 60º 240º 300º √3 √3 √3 √3 (1, –√3 ) (–1, –√3 ) (1, √3 ) (–1, √3 ) To find the value of cos 120°, we first recognize that the terminal side of an angle with 120º measure lies in quadrant II. We also know that cosine is negative in quadrant II. We then calculate the cosine of the reference angle, 60°. cos 60° 5 adjacent hypotenuse 5 1 2 Since we know cos 120° is negative because it lies in quadrant II, we know that cos120° 5 21 2 Similarly, we know that cos 240° 5 21 2 and cos 300° 5 1 2. 6.5.4 S K I L L Evaluate trigonometric functions for nonacute and quadrantal angles. 6.5.4 C O N C E P T U A L Understand that the value of a trigonometric function at an angle is the same as the trigonometric value of its reference angle, except there may be a difference in its algebraic sign 11/22. STUDY TIP The value of a trigonometric function of an angle is the same as the trigonometric value of its reference angle, except there may be an algebraic sign (1 or 2) difference between the two values. 6.5 Trigonometric Functions of Nonacute Angles 559 560 CHAPTER 6 Trigonometric Functions For any angle whose terminal side lies along one of the axes, we consult the table in this section for the values of the trigonometric functions for quadrantal angles. If the terminal side lies in one of the four quadrants, then the angle is said to be nonquadrantal and the following procedure can be used. The same attention must be given to the difference between evaluating trigonometric functions exactly and approximating with a calculator for nonacute angles as was done with acute angles. We follow the above procedure for all angles except when we get to Step 4. In Step 4 we evaluate exactly, if possible 1308, 458, 6082; otherwise, we use a calculator to approximate. PROCEDURE FOR EVALUATING FUNCTION VALUES FOR ANY NONQUADRANTAL ANGLE u Step 1: n If 0° , u , 360°, proceed to Step 2. n If u , 0°, add 360° as many times as needed to get a coterminal angle with measure between 0° and 360°. n If u . 360°, subtract 360° as many times as needed to get a coterminal angle with measure between 0° and 360°. Step 2: Find the quadrant in which the terminal side of the angle in Step 1 lies. Step 3: Find the reference angle a of the coterminal angle found in Step 1. Step 4: Find the trigonometric function values for the reference angle a. Step 5: Determine the correct algebraic signs 11 or 22 for the trigonometric function values based on the quadrant identified in Step 2. Step 6: Combine the trigonometric values found in Step 4 with the algebraic signs in Step 5 to give the trigonometric function values of u. EXAMPLE 5 Evaluating the Cosine Function of a Special Angle Exactly Find the exact value of cos 210°. Solution: The terminal side of u 5 210° lies in quadrant III. Find the reference angle for u 5 210°. 210° 2 180° 5 30° Find the value of the cosine of the reference angle. cos 30° 5 !3 2 Determine the algebraic sign for the cosine in quadrant III. Negative 122 Combine the algebraic sign of the cosine in quadrant III with the value of the cosine of the reference angle. cos 210° 5 2 !3 2 Y OUR TU R N Find the exact value of sin 330°. ▼ A N S W E R 2 1 2 ▼ x y 210º EXAMPLE 6 Evaluating the Tangent Function of a Special Angle Exactly Find the exact value of tan 495°. Solution: Subtract 360° to get a coterminal angle between 0° and 360°. 495° 2 360° 5 135° The terminal side of the angle lies in quadrant II. Find the reference angle for 135°. 180° 2 135° 5 45° Find the value of the tangent of the reference angle. tan 45° 5 1 Determine the algebraic sign for the tangent in quadrant II. Negative 122 Combine the algebraic sign of the tangent in quadrant II with the value of the tangent of the reference angle. tan 495° 5 21 YOUR T UR N Find the exact value of tan 660°. x y 135º ▼ A N S W E R 2!3 ▼ x y 150º EXAMPLE 7 Evaluating the Cosecant Function of a Special Angle Exactly Find the exact value of csc1221082. Solution: Add 360° to get a coterminal angle between 0° and 360°. 2210° 1 360° 5 150° The terminal side of the angle lies in quadrant II. Find the reference angle for 150°. 180° 2 150° 5 30° Find the value of the cosecant of the reference angle. csc 30° 5 1 sin 30° 5 1 1 2 5 2 6.5 Trigonometric Functions of Nonacute Angles 561 562 CHAPTER 6 Trigonometric Functions We can evaluate trigonometric functions exactly for quadrantal angles and any angle whose reference angle is a special angle 1308, 458, or 6082. For all other angles, we can approximate the trigonometric function value with a calculator. In the following examples, approximations will be rounded to four decimal places. Determine the algebraic sign for the cosecant in quadrant II. Positive 112 Combine the algebraic sign of the cosecant in quadrant II with the value of the cosecant of the reference angle. csc12210°2 5 2 Y OUR TU R N Find the exact value of sec12330°2. ▼ A N S W E R 2!3 3 ▼ EXAMPLE 8 Finding Exact Angle Measures Given Trigonometric Function Values Find all values of u where 0° # u # 360°, when sin u 5 2 !3 2 . Solution: Determine in what quadrants sine is negative. QIII and QIV Since the absolute value of sin u is !3 2 , the reference angle is 608. sin 60° 5 !3 2 Determine the angles between 1808 and 3608 in QIII and QIV with reference angle 60°. Quadrant III: 180° 1 60° 5 240° Quadrant IV: 360° 2 60° 5 300° The two angles are 240° and 300° . Y OUR TU R N Find all values of u where 0° # u # 360°, when cosu 5 2 !3 2 . ▼ A N S W E R 150° and 210° ▼ EXAMPLE 9 Using a Calculator to Evaluate Trigonometric Values of Angles Use a calculator to evaluate the following: a. tan 466° b. csc 12313º2 Solution (a): tan 4668 < 23.4874 Solution (b): sin 1231382 < 0.7314; 1 sin12313°2 < 1.3673 Y OUR TU R N Use a calculator to evaluate the following: a. sec12118°2 b. cot 226° ▼ A N S W E R a. 22.1301 b. 0.9657 ▼ [CONCEPT CHECK] If the terminal side lies in a quadrant other than I, and if cosine is positive and sine is the negative of the reference angle, then the terminal side lies in which quadrant? ANSWER IV ▼ EXAMPLE 10 Finding Approximate Angle Measures Given Trigonometric Function Values Find the measure of u (rounded to the nearest degree) if sin u 5 20.6293 and the terminal side of u (in standard position) lies in quadrant III where 0° # u # 360°. Solution: The sine of the reference angle is 0.6293. sin a 5 0.6293 Find the reference angle. a 5 sin2 110.62932 5 38.998° Round the reference angle to the nearest degree. a 5 39° Find u, which lies in quadrant III. 180° 1 39° 5 219° u 5 219° Check with a calculator. sin 219° 5 20.6293 YOUR T UR N Find the measure of u, the smallest positive angle (rounded to the nearest degree), if cos u 5 20.5299 and the terminal side of u (in standard position) lies in quadrant II. ▼ A N S W E R 1228 ▼ [SEC TION 6.5] S U M M A RY the trigonometric functions for quadrantal angles using the terminal side approach. When reference angles are 30°, 45°, or 60°, nonacute angles can be evaluated exactly; otherwise, we use a calculator to approximate trigonometric values for any angle. In this section, we identified the algebraic signs of the trigono-metric functions in each quadrant. The trigonometric functions can be evaluated for nonquadrantal angles using reference angles and knowledge of algebraic signs in each quadrant. We evaluated [SEC TION 6.5] E X E RC I S E S • S K I L L S In Exercises 1–10, indicate the quadrant in which the terminal side of u must lie in order for the information to be true. 1. cos u is positive and sin u is negative. 2. cos u is negative and sin u is positive. 3. tan u is negative and sin u is positive. 4. tan u is positive and cos u is negative. 5. sec u and csc u are both positive. 6. sec u and csc u are both negative. 7. cot u and cos u are both positive. 8. cot u and sin u are both negative. 9. tan u is positive and sec u is negative. 10. cot u is negative and csc u is positive. In Exercises 11–24, find the indicated trigonometric function values. 11. If cos u 5 2 3 5 and the terminal side of u lies in quadrant III, find sin u. 12. If tan u 5 2 5 12 and the terminal side of u lies in quadrant II, find cos u. 13. If sin u 5 60 61 and the terminal side of u lies in quadrant II, find tan u. 14. If cos u 5 40 41 and the terminal side of u lies in quadrant IV, find tan u. 15. If tan u 5 84 13 and the terminal side of u lies in quadrant III, find sin u. 6.5 Trigonometric Functions of Nonacute Angles 563 564 CHAPTER 6 Trigonometric Functions 16. If sin u 5 2 7 25 and the terminal side of u lies in quadrant IV, find cos u. 17. If sec u 5 22 and the terminal side of u lies in quadrant III, find tan u. 18. If cot u 5 1 and the terminal side of u lies in quadrant I, find sin u. 19. If sin u 5 !11 6 and the terminal side of u lies in quadrant II, find tan u. 20. If tan u 5 21 2 and the terminal side of u lies in quadrant II, find cos u. 21. If csc u 5 2 !3 and the terminal side of u lies in quadrant II, find cot u. 22. If sec u 5 213 5 and the terminal side of u lies in quadrant II, find csc u. 23. If cotu 5 2!3 and the terminal side of u lies in quadrant IV, find sec u. 24. If cot u 5 213 84 and the terminal side of u lies in quadrant II, find csc u. In Exercises 25–36, evaluate each expression, if possible: 25. cos12270°2 1 sin 450° 26. sin12270°2 1 cos 450° 27. sin 630° 1 tan12540°2 28. cos12720°2 1 tan 720° 29. cos 540° 2 sec12540°2 30. sin12450°2 1 csc 270° 31. csc12630°2 2 cot 630° 32. sec12540°2 1 tan 540° 33. tan 720° 1 sec 720° 34. cot 450° 2 cos12450°2 35. csc13600°2 1 sec123600°2 36. cot1600°2 1 tan12600°2 In Exercises 37–46, determine whether or not each statement is possible. 37. sin u 5 20.999 38. cos u 5 1.0001 39. cos u 5 2!6 3 40. sin u 5 !2 10 41. tan u 5 4!5 42. cot u 5 2 !6 7 43. sec u 5 2 4 !7 44. csc u 5 p 2 45. cot u 5 500 46. sec u 5 0.9996 In Exercises 47–58, evaluate the following expressions exactly. 47. cos 240° 48. cos 120° 49. sin 300° 50. sin 315° 51. tan 210° 52. sec 135° 53. tan12315°2 54. sec12330°2 55. csc 330° 56. csc12240°2 57. cot12315°2 58. cot1150°2 In Exercises 59–66, find all possible values of u, where 0° " u " 360°. 59. cosu 5 !3 2 60. sinu 5 !3 2 61. sin u 5 21 2 62. cos u 5 21 2 63. cos u 5 0 64. sin u 5 0 65. sin u 5 21 66. cos u 5 21 In Exercises 67–76, evaluate the trigonometric expression with a calculator. Round your result to four decimal places. 67. sin 237° 68. cos 317° 69. tan12265°2 70. tan 622° 71. sec 421° 72. sec12222°2 73. csc12111°2 74. csc 211° 75. cot 159° 76. cot1282°2 In Exercises 77–90, find the smallest positive measure of u (rounded to the nearest degree) if the indicated information is true. 77. sin u 5 0.9397 and the terminal side of u lies in quadrant II. 78. cos u 5 0.7071 and the terminal side of u lies in quadrant IV. 79. cos u 5 20.7986 and the terminal side of u lies in quadrant II. 80. sin u 5 20.1746 and the terminal side of u lies in quadrant III. 81. tan u 5 20.7813 and the terminal side of u lies in quadrant IV. 82. cos u 5 20.3420 and the terminal side of u lies in quadrant III. 83. tan u 5 20.8391 and the terminal side of u lies in quadrant II. 84. tan u 5 11.4301 and the terminal side of u lies in quadrant III. 85. sin u 5 20.3420 and the terminal side of u lies in quadrant IV. 86. sin u 5 20.4226 and the terminal side of u lies in quadrant III. 87. sec u 5 1.0001 and the terminal side of u lies in quadrant I. 88. sec u 5 23.1421 and the terminal side of u lies in quadrant II. 89. csc u 5 22.3604 and the terminal side of u lies in quadrant IV. 90. csc u 5 21.0001 and the terminal side of u lies in quadrant III. • A P P L I C A T I O N S In Exercises 91–94, refer to the following: When light passes from one substance to another, such as from air to water, its path bends. This is called refraction and is what is seen in eyeglass lenses, camera lenses, and gems. The rule governing the change in the path is called Snell’s law, named after the Dutch astronomer Willebrord Snellius (1580–1626): n1 sin u1 5 n2 sin u2, where the n1 and n2 are the indices of refraction of the different substances and the u1 and u2 are the respective angles that light makes with a line perpendicular to the surface at the boundary between substances. The figure shows the path of light rays going from air to water. Assume that the index of refraction in air is 1. Water surface θ1 θ2 91. If light rays hit the water’s surface at an angle of 30° from the perpendicular and are refracted to an angle of 22° from the perpendicular, then what is the refraction index for water? 92. If light rays hit a glass surface at an angle of 30° from the perpendicular and are refracted to an angle of 18° from the perpendicular, then what is the refraction index for that glass? 93. If the refraction index for a diamond is 2.4, then to what angle is light refracted if it enters the diamond at an angle of 30°? 94. If the refraction index for a rhinestone is 1.9, then to what angle is light refracted if it enters the rhinestone at an angle of 30°? For Exercises 95 and 96, refer to the following: An orthotic knee brace can be used to treat knee injuries by locking the knee at an angle u chosen to facilitate healing. The angle u is measured from the metal bar on the side of the brace on the thigh to the metal bar on the side of the brace on the calf (see the figure). To make working with the brace more convenient, rotate the image such that the thigh aligns with the positive x-axis (see the figure on the right below). Thigh Thigh Calf Calf x y θ θ 5 c 5 15 cm 15 cm 15 cm 5 cm 15 cm 95. Health/Medicine. If u 5 1658, find the measure of the reference angle. What is the physical meaning of the reference angle? 96. Health/Medicine. If u 5 1608, find the measure of the reference angle. What would an angle greater than 1808 represent? For Exercises 97 and 98, refer to the following: Water covers two-thirds of the Earth’s surface, and every living thing is dependent on it. For example, the human body is made up of over 70% water. The water molecule is composed of one oxygen atom and two hydrogen atoms and exhibits a bent shape with the oxygen molecule at the center. The angle u between the O-H bonds is 104.58. = 104.5º θ H O H 97. Chemistry. Sketch the water molecule in the xy-coordinate system in a convenient manner for illustrating angles. Find the reference angle. Illustrate both the angle u and the reference angle on the sketch. 98. Chemistry. Find cos1104.582 using the reference angle found in Exercise 97. (Source: reviews/pH/phwater.htm.) 6.5 Trigonometric Functions of Nonacute Angles 565 566 CHAPTER 6 Trigonometric Functions • C A T C H T H E M I S T A K E In Exercises 99 and 100, explain the mistake that is made. 99. Evaluate the expression sec 120° exactly. Solution: 120° lies in quadrant II. The reference angle is 30°. Find the cosine of the reference angle. cos 30° 5 !3 2 Cosine is negative in quadrant II. cos 120° 5 2 !3 2 Secant is the reciprocal of cosine. sec 120° 5 2 2 !3 5 2 2!3 3 This is incorrect. What mistake was made? 100. Find the measure of the smallest positive angle u (rounded to the nearest degree) if cos u 5 20.2388 and the terminal side of u (in standard position) lies in quadrant III. Solution: Evaluate with a calculator. u 5 cos 21120.23882 5 103.8157° Approximate to the nearest degree. u < 104° This is incorrect. What mistake was made? x y 120º 30º 107. If tan u 5 a b, where a and b are positive, and if u lies in quadrant III, find sin u. 108. If tanu 5 2 a b , where a and b are positive, and if u lies in quadrant II, find cos u. 109. If csc u 52 a b, where a and b are positive, and if u lies in quadrant IV, find cot u. 110. If sec u 52 a b, where a and b are positive, and if u lies in quadrant III, find tan u. • C H A L L E N G E 111. Use a calculator to evaluate sin 80° and sin 100°. Now use the calculator to evaluate sin21 10.98482. When sine is positive, in which of the quadrants, I or II, does the calculator assume the terminal side of the angle lies? 112. Use a calculator to evaluate sin 260° and sin 280°. Now use the calculator to evaluate sin21 120.98482. When sine is negative, in which of the quadrants, III or IV, does the calculator assume the terminal side of the angle lies? 113. Use a calculator to evaluate cos 75° and cos1275°2. Now use the calculator to evaluate cos2110.25882. When cosine is positive, in which of the quadrants, I or IV, does the calculator assume the terminal side of the angle lies? 114. Use a calculator to evaluate cos 105° and cos 255°. Now use the calculator to evaluate cos21120.25882. W hen cosine is negative, in which of the quadrants, II or III, does the calculator assume the terminal side of the angle lies? 115. Use a calculator to evaluate tan 65° and tan 245°. Now use the calculator to evaluate tan2112.14452. When tangent is positive, in which of the quadrants, I or III, does the calculator assume the terminal side of the angle lies? 116. Use a calculator to evaluate tan 136° and tan 316°. Now use the calculator to evaluate tan21120.96572. When tangent is negative, in which of the quadrants, II or IV, does the calculator assume the terminal side of the angle lies? • T E C H N O L O G Y In Exercises 101–106, determine whether each statement is true or false. • C O N C E P T U A L 101. It is possible for all six trigonometric functions of the same angle to have positive values. 102. It is possible for all six trigonometric functions of the same angle to have negative values. 103. The trigonometric function value for any angle with negative measure must be negative. 104. The trigonometric function value for any angle with positive measure must be positive. 105. sec2 u 2 1 can be negative for some value of u. 106. 1sec u2 1csc u2 is negative only when the terminal side of u lies in quadrant II or IV. 6.6 Radian Measure and Applications 567 6.6.1 The Radian Measure of an Angle In geometry and most everyday applications, angles are measured in degrees. However, radian measure is another way to measure angles. Using radian measure allows us to write trigonometric functions as functions not only of angles but also of real numbers in general. Recall that in Section 6.1 we defined one full rotation as an angle having measure 360°. Now we think of the angle in the context of a circle. A central angle is an angle that has its vertex at the center of a circle. When the intercepted arc’s length is equal to the radius, the measure of the central angle is 1 radian. From geometry we know that the ratio of the measures of two angles is equal to the ratio of the lengths of the arcs subtended by those angles (along the same circle). u1 u2 5 s1 s2 s1 s2 θ1 θ2 r r r r If u2 5 1 radian, then the length of the subtended arc is equal to the radius, s2 5 r. This leads to u1 5 s1 r , which is the general definition of radian measure. r θ = 1 radian r r S K I L L S O B J E C T I V E S ■ ■Calculate the radian measure of an angle. ■ ■Convert between degrees and radians. ■ ■Calculate the length of an arc along a circle. ■ ■Calculate the area of a circular sector. ■ ■Calculate linear speed. ■ ■Calculate angular speed. ■ ■Solve application problems involving both angular and linear speeds. C O N C E P T U A L O B J E C T I V ES ■ ■Realize that radian measure allows us to write trigonometric functions as functions of real numbers. ■ ■Understand that degrees and radians are both units for measuring angles. ■ ■Understand why the angle in the arc length formula must be in radian measure. ■ ■Understand that the central angle should be given in radians to use the area of a circular sector formula. ■ ■Understand that linear speed has units of length/time. ■ ■Understand that angular speed has units of radians/time. ■ ■Understand that angular speed and linear speed are related through the radius. 6.6 RADIAN MEASURE AND APPLICATIONS 6.6.1 S K I L L Calculate the radian measure of an angle. 6.6.1 C O N C E P T U A L Realize that radian measure allows us to write trigonometric functions as functions of real numbers. DEFINITION Radian Measure If a central angle u in a circle with radius r intercepts an arc on the circle of length s, then the measure of u, in radians, is given by u 1in radians2 5 s r Note: The formula is valid only if s (arc length) and r (radius) are expressed in the same units. s θ r r ▼ C A U T I O N To correctly calculate radians from the formula u 5 s/r, the radius and arc length must be expressed in the same units. 568 CHAPTER 6 Trigonometric Functions Note that both s and r are measured in units of length. When both are given in the same units, the units cancel, giving the number of radians as a dimensionless (unitless) real number. One full rotation corresponds to an arc length equal to the circumference 2pr of the circle with radius r. We see then that one full rotation is equal to 2p radians. ufull rotation 5 2pr r 5 2p [CONCEPT CHECK] What is the measure (in radians) of a central angle u that intercepts an arc of length A centimeters on a circle with radius A millimeters? ANSWER 10 radians ▼ EXAMPLE 1 Finding the Radian Measure of an Angle What is the measure (in radians) of a central angle u that intercepts an arc of length 4 feet on a circle with radius 10 feet? Solution: Write the formula relating radian measure u 5 s r to arc length and radius. Let s 5 4 ft and r 5 10 ft. u 5 4 ft 10 ft 5 0.4 rad Y OUR TU R N What is the measure (in radians) of a central angle u that intercepts an arc of length 3 inches on a circle with radius 50 inches? ▼ A N S W E R 0.06 rad ▼ EXAMPLE 2 Finding the Radian Measure of an Angle What is the measure (in radians) of a central angle u that intercepts an arc of length 6 centimeters on a circle with radius 2 meters? common mistake A common mistake is to forget to first put the radius and arc length in the same units. Y OUR TU R N What is the measure (in radians) of a central angle u that intercepts an arc of length 12 millimeters on a circle with radius 4 centimeters? ▼ A N S W E R 0.3 rad ▼ ✖I N C O R R EC T Substitute s 5 6 cm and r 5 2 m into the radian expression. u 5 6 cm 2 m Simplify. u 5 3 rad ERROR: Did not convert numerator and denominator to the same units. ✓C O R R EC T Write the formula relating radian measure to arc length and radius. u 1in radians2 5 s r Substitute s 5 6 cm and r 5 2 m into the radian expression. u 5 6 cm 2 m Convert the radius (2) meters to centimeters: 2 m 5 200 cm. u 5 6 cm 200 cm The units, cm, cancel and the result is a unitless real number. u 5 0.03 rad ▼ C A U T I O N Units for arc length and radius must be the same to use u 5 s r. In the above example the units, cm, canceled, therefore correctly giving radians as a unitless real number. Because radians are unitless, the word radians (or rad) is often omitted. If an angle measure is given simply as a real number, then radians are implied. WORDS MATH The measure of u is 4 degrees. u 5 4° The measure of u is 4 radians. u 5 4 6.6.2 Converting Between Degrees and Radians In order to convert between degrees and radians, we must first look for a relationship between them. We start by considering one full rotation around the circle. An angle corresponding to one full rotation is said to have measure 360°. Radians are defined as the ratio of the arc length that the angle intercepts on the circle to the radius of the circle. One full rotation corresponds to an arc length equal to the circumference of the circle. WORDS MATH Write the angle measure (in degrees) that corresponds to one full rotation. u 5 360° Write the angle measure (in radians) that corresponds to one full rotation. Arc length is the circumference of the circle. s 5 2pr Substitute s 5 2pr into u 1in radians2 5 s r. u 5 2pr r 5 2p radians Equate the measures corresponding to one full rotation. 360° 5 2p radians Divide by 2. 180° 5 p radians Divide by 1808 or p. 1 5 p 180° or 1 5 180° p This leads us to formulas that convert between degrees and radians. Let ud represent an angle measure given in degrees and ur represent the corresponding angle measure given in radians. 6.6.2 S K I L L Convert between degrees and radians. 6.6.2 C O N C E P T U A L Understand that degrees and radians are both units for measuring angles. CONVERTING DEGREES TO RADIANS To convert degrees to radians, multiply the degree measure by p 180° . ur 5 ud a p 180°b CONVERTING RADIANS TO DEGREES To convert radians to degrees, multiply the radian measure by 180° p . ud 5 ur a180° p b [CONCEPT CHECK] Which angle would have greater measure: B radians or B 8? ANSWER B radians ▼ 6.6 Radian Measure and Applications 569 570 CHAPTER 6 Trigonometric Functions Before we begin converting between degrees and radians, let’s first get a feel for radians. How many degrees is 1 radian? Use the technique outlined in the box above. WORDS MATH Multiply 1 radian by 180° p . 1 a180° p b Approximate p by 3.14. 1 a180° 3.14 b Use a calculator to evaluate and round to the nearest degree. < 57° 1rad < 57° A radian is much larger than a degree (almost 57 times larger). Let’s com-pare two angles, one measuring 30 radians and the other measuring 30°. Note that 30 rad 2p rad/rev < 4.77 revolutions, whereas 30° 5 1 12 revolution. x y 30º x y 30 radians EXAMPLE 3 Converting Degrees to Radians Convert 45° to radians. Solution: Multiply 45° by p 180°. 145°2 a p 180°b 5 45°p 180° Simplify. 5 p 4 radians Note: p 4 is the exact value. A calculator can be used to approximate this expression. Scientific and graphing calculators have a p button (on most scientific calculators it requires using a shift or second command). The decimal approximation rounded to three decimal places is 0.785. Exact value: p 4 Approximate value: 0.785 Y OUR TU R N Convert 60° to radians. ▼ A N S W E R p 3 or approximately 1.047 ▼ Since p 5 180° we know the following special angles p 2 5 90° p 3 5 60° p 4 5 45° p 6 5 30° and we can now draw the unit circle with the special angles in both degrees and radians. 60º = 3 π 45º = 4 π 30º = 6 π 360º = 2π 330º = 6 11π 315º = 4 7π 300º = 3 5π 270º = 2 3π 240º = 3 4π 225º = 4 5π 210º = 6 7π 180º = π 150º = 6 5π 135º = 4 3π 90º = 2 π 120º = 3 2π Next, we look at applications of radian measure that involve calculating arc lengths, areas of circular sectors, and angular and linear speeds. All of these applications are related to the definition of radian measure. EXAMPLE 4 Converting Degrees to Radians Convert 472° to radians. Solution: Multiply 472° by p 180°. 472° a p 180°b Simplify (factor out the common 4). 5 118 45 p Approximate with a calculator. < 8.238 YOUR T UR N Convert 460° to radians. ▼ A N S W E R 23 9 p or 8.029 ▼ EXAMPLE 5 Converting Radians to Degrees Convert 2p 3 to degrees. Solution: Multiply 2p 3 by 180° p . 2p 3 ⋅ 180° p Simplify. 5 120° YOUR T UR N Convert 3p 2 to degrees. ▼ A N S W E R 2708 ▼ 6.6 Radian Measure and Applications 571 572 CHAPTER 6 Trigonometric Functions 6.6.3 Arc Length From geometry we know that the length of an arc of a circle is proportional to its central angle. In this section, we have learned that for the special case when the arc is equal to the circumference of the circle, the angle measure in radians corresponding to one full rotation is 2p. Let us now assume that we are given the central angle and we want to find the arc length. WORDS MATH Write the definition of radian measure. u 5 s r Multiply both sides of the equation by r. r ⋅ u 5 s r ⋅ r Simplify. ru 5 s The formula s 5 ru is true only when u is in radians. To develop a formula when u is in degrees, we multiply u by p 180°. 6.6.3 SKI LL Calculate the length of an arc along a circle. 6.6.3 CO NCE PTUAL Understand why the angle in the arc length formula must be in radian measure. STUDY TIP To use the relationship s 5 r u the angle u must be in radians. DEFINITION Arc Length If a central angle u in a circle with radius r intercepts an arc on the circle of length s, then the arc length s is given by s 5 rur ur is in radians or s 5 rud a p 180°b ud is in degrees [CONCEPT CHECK] What are the units of arc length s if s 5 r ? ud ? p 180° where r is given in feet and ud is given in degrees? ANSWER Feet ▼ EXAMPLE 6 Finding Arc Length When the Angle Has Radian Measure In a circle with radius 10 centimeters, an arc is intercepted by a central angle with measure 7p 4 . Find the arc length. Solution: Write the formula for arc length when the angle has radian measure. s 5 rur Substitute r 5 10 cm and ur 5 7p 4 . s 5 110 cm2 a 7p 4 b Simplify. s 5 35p 2 cm Y OUR TU R N In a circle with radius 15 inches, an arc is intercepted by a central angle with measure p 3 . Find the arc length. ▼ ▼ A N S W E R 5p in. 6.6.4 Area of a Circular Sector A restaurant lists a piece of French Silk Pie as having 400 calories. How does the chef arrive at that number? She calculates the calories of all the ingredients that went into making the entire pie and then divides by the number of slices the pie yields. For example, if an entire pie has 3200 calories and it is sliced into 8 equal pieces, then each piece has 400 calories. Although this example involves volume, the idea is the same with areas of sectors of circles. Circular sectors can be thought of as a “piece of a pie.” Recall that arc lengths of a circle are proportional to the central angle (in radians) and the radius. Similarly, a circular sector is a portion of the entire circle. Let A repre-sent the area of the sector of the circle and ur represent the central angle (in radians) that forms the sector. Then let us consider the entire circle whose area is pr2 and the angle that represents one full rotation has measure 2p (radians). EXAMPLE 7 Finding Arc Length When the Angle Has Degree Measure The International Space Station (ISS) is in an approximately circular orbit 400 kilometers above the surface of the Earth. If the ground station tracks the space station when it is within a 458 central angle of this circular orbit above the tracking antenna, how many kilometers does the ISS cover while it is being tracked by the ground station? Assume that the radius of the Earth is 6400 kilometers. Round to the nearest kilometer. Solution: Write the formula for arc length when the angle has degree measure. s 5 r ud a p 180°b Substitute r 5 6400 1 400 5 6800 km and ud 5 45°. s 5 16800 km21 45°2 a p 180°b Evaluate with a calculator. s < 5340.708 km Round to the nearest kilometer. s < 5341 km The ISS travels approximately 5341 kilometers during the ground station tracking. YOUR T UR N If the ground station in Example 7 could track the ISS within a 608 central angle, how far would the ISS travel during the tracking? ▼ 400 km 6400 km 45º ISS ▼ A N S W E R 7121 km 6.6 Radian Measure and Applications 573 6.6.4 S K IL L Calculate the area of a circular sector. 6.6.4 C ON C E P T U A L Understand that the central angle should be given in radians to use the area of a circular sector formula. 574 CHAPTER 6 Trigonometric Functions WORDS MATH Write the ratio of the area of the sector to the area of the entire circle. A pr 2 Write the ratio of the central angle ur to the measure of one full rotation. ur 2p The ratios must be equal (proportionality of sector to circle). A pr 2 5 ur 2p Multiply both sides of the equation by pr 2. pr 2⋅A pr 2 5 ur 2p ⋅pr 2 Simplify. A 5 1 2 r 2ur s θ r r DEFINITION Area of a Circular Sector The area of a sector of a circle with radius r and central angle u is given by A 5 1 2 r 2ur ur is in radians or A 5 1 2 r 2ud a p 180°b ud is in degrees EXAMPLE 8 Finding the Area of a Sector When the Angle Has Radian Measure Find the area of the sector associated with a single slice of pizza if the entire pizza has a 14-inch diameter and the pizza is cut into eight equal pieces. Solution: The radius is half the diameter. r 5 14 2 5 7 in. Find the angle of each slice if a pizza is cut into eight pieces. ur 5 2p 8 5 p 4 Write the formula for circular sector area in radians. A 5 1 2 r 2 ur Substitute r 5 7 inches and ur 5 p 4 into the area equation. A 5 1 2 17 in.22 ap 4 b Simplify. A 5 49p 8 in.2 Approximate the area with a calculator (round to the nearest square inch). A < 19 in.2 Y OUR TU R N Find the area of a slice of pizza if the entire pizza has a 16-inch diameter. Assume the pizza is cut into eight equal pieces. ▼ STUDY TIP To use the relationship A 5 1 2 r 2 u the angle u must be in radians. [CONCEPT CHECK] If the area of a circular sector A is given by A 5 1 2 r2ur and r is given in feet, then what are the units of A? ANSWER Square feet ▼ ▼ A N S W E R 8p in.2 < 25 in.2 6.6.5 Linear Speed Although velocity and speed are often used as synonyms, speed is how fast you are traveling, whereas velocity is the speed at which you are traveling and the direction in which you are traveling. In physics the difference between speed and velocity is that velocity has direction and is written as a vector (Chapter 8), and speed is the magnitude of the velocity vector which results in a real number. In this chapter, we will only deal with speed. Recall the relationship between distance, rate (assumed to be constant), and time: d 5 rt. Rate is speed, and in words this formula can be rewritten as distance 5 speed⋅time or speed 5 distance time It is important to note that we assume speed is constant. If we think of a car driving around a circular track, the distance it travels is the arc length s; and if we let v represent speed and t represent time, we have the formula for speed around a circle (linear speed): v 5 s t s ▼ A N S W E R 450p ft2 < 1414 ft2 DEFINITION Linear Speed If a point P moves along the circumference of a circle at a constant speed, then the linear speed v is given by v 5 s t where s is the arc length and t is the time. 6.6.5 S K IL L Calculate linear speed. 6.6.5 C ON C E P T U A L Understand that linear speed has units of length/time. [CONCEPT CHECK] Which of the following is not a measure of linear speed? (A) Miles per hour (B) Feet per second (C) Degrees per second (D) Meters per minute ANSWER (C) ▼ EXAMPLE 9 Finding the Area of a Sector When the Angle Has Degree Measure Sprinkler heads come in all different sizes depending on the angle of rotation desired. If a sprinkler head rotates 90° and has enough pressure to keep a constant 25-foot spray, what is the area of the sector of the lawn that gets watered? Round to the nearest square foot. Solution: Write the formula for circular sector area in degrees. A 5 1 2 r 2ud a p 180°b Substitute r 5 25 ft and ud 5 908 into the area equation. A 5 1 2 125 ft2 2190°2 a p 180°b Simplify. A 5 a625p 4 b ft2 < 490.87 ft2 Round to the nearest square foot. A < 491 ft2 YOUR T UR N If a sprinkler head rotates 180° and has enough pressure to keep a constant 30-foot spray, what is the area of the sector of the lawn it can water? Round to the nearest square foot. ▼ 6.6 Radian Measure and Applications 575 576 CHAPTER 6 Trigonometric Functions 6.6.6 Angular Speed To calculate linear speed, we find how fast a position along the circumference of a circle is changing. To calculate angular speed, we find how fast the central angle is changing. EXAMPLE 10 Linear Speed A car travels at a constant speed around a circular track with circumference equal to 2 miles. If the car records a time of 15 minutes for 9 laps, what is the linear speed of the car in miles per hour? Solution: Calculate the distance traveled around the circular track. s 5 19 laps2 a 2 mi lap b 5 18 mi Substitute t 5 15 minutes and s 5 18 miles into v 5 s t. v 5 18 mi 15 min Convert the linear speed from miles per minute to miles per hour. v 5 a 18 mi 15 minb a60 min 1 hr b Simplify. v 5 72 mi mph Y OUR TU R N A car travels at a constant speed around a circular track with circumference equal to 3 miles. If the car records a time of 12 minutes for 7 laps, what is the linear speed of the car in miles per hour? ▼ ▼ A N S W E R 105 mph 6.6.6 S KILL Calculate angular speed. 6.6.6 C ON CEPTUAL Understand that angular speed has units of radians/time. DEFINITION Angular Speed If a point P moves along the circumference of a circle at a constant speed, then the central angle u that is formed with the terminal side passing through point P also changes over some time t at a constant speed. The angular speed v (omega) is given by v 5 u t where u is given in radians EXAMPLE 11 Angular Speed A lighthouse in the middle of a channel rotates its light in a circular motion with constant speed. If the beacon of light completes 1 rotation every 10 seconds, what is the angular speed of the beacon in radians per minute? Solution: Calculate the angle measure in radians associated with 1 rotation. u 5 2p s 6.6.7 Relationship Between Linear and Angular Speeds Angular speed and linear speed are related through the radius. WORDS MATH Write the definition of radian measure. u 5 s r Write the definition of arc length (u in radians). s 5 ru Divide both sides by t. s t 5 ru t Rewrite the right side of the equation. s t 5 r u t Recall the definitions of linear and angular speeds. v 5 s t and v 5 u t Substitute v 5 s t and v 5 u t into s t 5 r u t . v 5 rv Notice that tires of two different radii with the same angular speed have different linear speeds. The larger tire has the faster linear speed. ▼ A N S W E R v 5 3p rad/min Substitute u 5 2p and t 5 10 seconds into v 5 u t . v 5 2p 1rad2 10 sec Convert the angular speed from radians v 5 2p 1rad2 10 sec ⋅ 60 sec 1 min per second to radians per minute. Simplify. v 5 12p rad /min YOUR T UR N If the lighthouse in Example 11 is adjusted so that the beacon rotates 1 time every 40 seconds, what is the angular speed of the beacon in radians per minute? ▼ [CONCEPT CHECK] If we know that it takes a beacon A minutes to complete one full rotation, then what is its angular speed? ANSWER 2⋅p A radians per minute ▼ 6.6.7 S K I L L Solve application problems involving both angular and linear speeds. 6.6.7 C O N C E P T U A L Understand that angular speed and linear speed are related through the radius. RELATING LINEAR AND ANGULAR SPEEDS If a point P moves at a constant speed along the circumference of a circle with radius r, then the linear speed v and the angular speed v are related by v 5 rv or v 5 v r Note: This relationship is true only when u is given in radians. x y r P s θ STUDY TIP This relationship between linear speed and angular speed assumes the angle is given in radians. [CONCEPT CHECK] For a fixed angular speed, if the radius is increased, then the linear speed (increases/decreases)? ANSWER Increases ▼ 6.6 Radian Measure and Applications 577 578 CHAPTER 6 Trigonometric Functions EXAMPLE 12 Relating Linear and Angular Speeds A Ford F-150 truck comes standard with tires that have a diameter of 25.7 inches (17-inch rims). If the owner decides to upgrade to tires with a diameter of 28.2 inches (19-inch rims) without having the onboard computer updated, how fast will the truck actually be traveling when the speedometer reads 75 miles per hour? Solution: The computer onboard the F-150 “thinks” the tires are 25.7 inches in diameter and knows the angular speed. Use the programmed tire diameter and speedometer reading to calculate the angular speed. Then use that angular speed and the upgraded tire diameter to get the actual speed (linear speed). STEP 1 Calculate the angular speed of the tires. Write the formula for the angular speed. v 5 v r Substitute v 5 75 miles per hour and r 5 25.7 2 5 12.85 inches into the formula. v 5 75 mi/hr 12.85 in. 1 mile 5 5280 feet 5 63,360 inches. v 5 75163,3602 in./hr 12.85 in. Simplify. v < 369,805 rad hr STEP 2 Calculate the actual linear speed of the truck. Write the linear speed formula. v 5 rv Substitute r 5 28.2 2 5 14.1 inches and v < 369,805 radians per hour. v 5 114.1 in.2 a369,805 rad hr b Simplify. v < 5,214,251 in. hr 1 mile 5 5280 feet 5 63,360 inches v < 5,214,251 in. hr 63,360 in. mi v < 82.296 mi hr Although the speedometer indicates a speed of 75 miles per hour, the actual speed is approximately 82 miles per hour . Y OUR TU R N Suppose the owner of the Ford F-150 truck in Example 12 decides to downsize the tires from their original 25.7-inch diameter to a 24.4-inch diameter. If the speedometer indicates a speed of 65 miles per hour, what is the actual speed of the truck? 12.85 in. 14.1 in. ▼ STUDY TIP We could have solved Example 12 the following way: 75 mph 25.7 5 x 28.2 x 5 a28.2 25.7b 1752 < 82.296 mph ▼ A N S W E R Approximately 62 mph [SEC TION 6.6] S U M M A RY In this section, a second measure of angles, radians, was introduced. A central angle of a circle has radian measure equal to the ratio of the arc length to the radius of the circle. Radians and degrees are related by the relation p 5 180°. To convert from radians to degrees, multiply the radian measure by 180° p . Similarly, to convert from degrees to radians, multiply the degree measure by p 180°. Applications of radian measure are arc length, area of a circular sector, and angular speed. Linear speed (speed along the circumference of a circle) v is given by v 5 s t and is related to angular speed (speed of angle rotation) v by v 5 rv or v 5 v r APPLICATION ur IS IN RADIANS ud IS IN DEGREES Arc length s 5 r ur s 5 r ud a p 180°b Area of a circular sector A 5 1 2 r 2 ur A 5 1 2 r 2 ud a p 180°b Angular speed v 5 ur t In Exercises 1–12, find the measure (in radians) of a central angle u that intercepts an arc of length s on a circle with radius r. 1. r 5 10 cm, s 5 2 cm 2. r 5 20 cm, s 5 2 cm 3. r 5 22 in., s 5 4 in. 4. r 5 6 in., s 5 1 in. 5. r 5 100 cm, s 5 20 mm 6. r 5 1 m, s 5 2 cm 7. r 5 1 4 in., s 5 1 32 in. 8. r 5 3 4 cm, s 5 3 14 cm 9. r 5 2.5 cm, s 5 5 mm 10. r 5 1.6 cm, s 5 0.2 mm 11. r 5 2!3 ft, s 5 !3 in. 12. r 5 4!5 m, s 5 400!5 cm In Exercises 13–28, convert from degrees to radians. Leave the answers in terms of p. 13. 30° 14. 60° 15. 45° 16. 90° 17. 315° 18. 270° 19. 75° 20. 100° 21. 170° 22. 340° 23. 780° 24. 540° 25. 2210° 26. 2320° 27. 23600° 28. 1800° In Exercises 29–42, convert from radians to degrees. 29. p 6 30. p 4 31. 3p 4 32. 7p 6 33. 3p 8 34. 11p 9 35. 5p 12 36. 7p 3 37. 9p 38. 26p 39. 19p 20 40. 13p 36 41. 2 7p 15 42. 2 8p 9 In Exercises 43–50, convert from radian measure to degrees. Round your answers to the nearest hundredth of a degree. 43. 4 44. 3 45. 0.85 46. 3.27 47. 22.7989 48. 25.9841 49. 2!3 50. 5!7 [SEC TION 6.6] E X E R C I S E S • S K I L L S In Exercises 51–56, convert from degrees to radians. Round your answers to three significant digits. 51. 47° 52. 65° 53. 112° 54. 172° 55. 56.5° 56. 298.7° 6.6 Radian Measure and Applications 579 580 CHAPTER 6 Trigonometric Functions • A P P L I C A T I O N S 97. Low Earth Orbit Satellites. A low earth orbit (LEO) satellite is in an approximately circular orbit 300 kilometers above the surface of the Earth. If the ground station tracks the satellite when it is within a 45° central angle above the tracking antenna (directly overhead), how many kilometers does the satellite cover during the tracking? Assume the radius of the Earth is 6400 kilometers. Round to the nearest kilometer. 98. Low Earth Orbit Satellites. A low earth orbit (LEO) satellite is in an approximately circular orbit 250 kilometers above the surface of the Earth. If the ground station tracks the satellite when it is within a 30° central angle above the tracking antenna (directly overhead), how many kilometers does the satellite cover during the tracking? Assume the radius of the Earth is 6400 kilometers. Round to the nearest kilometer. 99. Big Ben. The famous clock tower in London has a minute hand that is 14 feet long. How far does the tip of the minute hand of Big Ben travel in 25 minutes? Round to two decimal places. In Exercises 57–66, find the exact length of the arc made by the indicated central angle and radius of each circle. 57. u 5 3, r 5 4 mm 58. u 5 4, r 5 5 cm 59. u 5 p 12, r 5 8 ft 60. u 5 p 8, r 5 6 yd 61. u 5 1 2, r 5 5 in. 62. u 5 3 4, r 5 20 m 63. u 5 22°, r 5 18 mm 64. u 5 14°, r 5 15 mm 65. u 5 8°, r 5 1500 km 66. u 5 3°, r 5 1800 km In Exercises 67–76, find the area of the circular sector given the indicated radius and central angle. Round your answers to three significant digits. 67. u 5 p 6 , r 5 7 ft 68. u 5 p 5 , r 5 3 in. 69. u 5 3p 8 , r 5 2.2 km 70. u 5 5p 6 , r 5 13 mi 71. u 5 5 2, r 5 2 yd 72. u 5 !3, r 5 4 m 73. u 5 56°, r 5 4.2 cm 74. u 5 27°, r 5 2.5 mm 75. u 5 1.2°, r 5 1.5 ft 76. u 5 14°, r 5 3.0 ft In Exercises 77–80, find the linear speed of a point that moves with constant speed in a circular motion if the point travels along the circle of arc length s in time t. 77. s 5 2 m, t 5 5 sec 78. s 5 12 ft, t 5 3 min 79. s 5 68,000 km, t 5 250 hr 80. s 5 7524 mi, t 5 12 days In Exercises 81–84, find the distance traveled (arc length) of a point that moves with constant speed v along a circle in time t. 81. v 5 2.8 m/sec, t 5 3.5 sec 82. v 5 6.2 km/hr, t 5 4.5 hr 83. v 5 4.5 mi/hr, t 5 20 min 84. v 5 5.6 ft/sec, t 5 2 min In Exercises 85–88, find the angular speed (radians/second) associated with rotating a central angle u in time t. 85. u 5 25p, t 5 10 sec 86. u 5 3p 4 , t 5 1 6 sec 87. u 5 200°, t 5 5 sec 88. u 5 60°, t 5 0.2 sec In Exercises 89–92, find the linear speed of a point traveling at a constant speed along the circumference of a circle with radius r and angular speed v. 89. v 5 2p rad 3 sec , r 5 9 in. 90. v 5 3p rad 4 sec , r 5 8 cm 91. v 5 p rad 20 sec, r 5 5 mm 92. v 5 5p rad 16 sec, r 5 24 ft In Exercises 93–96, find the distances a point travels along a circle over a time t, given the angular speed v and radius r of the circle. Round your answers to three significant digits. 93. r 5 5 cm, v 5 p rad 6 sec, t 5 10 sec 94. r 5 2 mm, v 5 6p rad sec, t 5 11 sec 95. r 5 5.2 in., v 5 p rad 15 sec, t 5 10 min 96. r 5 3.2 ft, v 5 p rad 4 sec, t 5 3 min Peter Adams/Digital Vision/Getty Images, Inc. 100. Big Ben. The famous clock tower in London has a minute hand that is 14 feet long. How far does the tip of the minute hand of Big Ben travel in 35 minutes? Round to two decimal places. 101. London Eye. The London Eye has 32 capsules and a diameter of 400 feet. What is the distance you will have traveled once you reach the highest point for the first time? Jason Friend/PhotolibraryGetty Images, Inc. 102. London Eye. (See Exercise 101.) Assuming every capsule is loaded and unloaded, what is the distance you will have traveled from the time you first get in the capsule to the point when the eye stops for the sixth time (not including the stop made for you)? 103. Gears. In the diagram below the smaller gear has a radius of 5 centimeters and the larger gear has a radius of 12.1 centimeters. If the smaller gear rotates 120°, how many degrees has the larger gear rotated? Round your answer to the nearest degree. 12.1 cm 5 cm 104. Gears. In the diagram above, the smaller gear has a radius of 3 inches, and the larger gear has a radius of 15 inches. If the smaller gear rotates 420°, how many degrees has the larger gear rotated? Round your answer to the nearest degree. 105. Bicycle Low Gear. If a bicycle has 26-inch-diameter wheels (the front chain drive has a radius of 2.2 inches and the back drive has a radius of 3 inches), how far does the bicycle travel every one rotation of the cranks (pedals)? Round to the nearest inch. 106. Bicycle High Gear. If a bicycle has 26-inch-diameter wheels (the front chain drive has a radius of 4 inches and the back drive has a radius of 1 inch), how far does the bicycle travel every one rotation of the cranks (pedals)? Round to the nearest inch. Stockbyte/Getty Images 107. Odometer. A Ford SUV comes standard with 17-inch rims (which corresponds to a tire with 25.7 inch diameter). Suppose you decide to later upgrade these tires for 19-inch rims (corresponds to a tire with 28.2 inch diameter). If you do not get your onboard computer reset for the new tires, the odometer will not be accurate. After your tires have actually driven 1000 miles, how many miles will the odometer report the SUV has been driven? Round to the nearest mile. 108. Odometer. For the same SUV in Exercise 107, after you have driven 50,000 miles, how many miles will the odometer report the SUV has been driven if the computer is not reset to account for the new oversized tires? Round to the nearest mile. 109. Sprinkler Coverage. A sprinkler has a 20-foot spray, and it covers an angle of 45°. What is the area that the sprinkler waters? Round to the nearest square foot. 110. Sprinkler Coverage. A sprinkler has a 22-foot spray, and it covers an angle of 60°. What is the area that the sprinkler waters? Round to the nearest square foot. 111. Windshield Wiper. A windshield wiper that is 12 inches long (blade and arm) rotates 70°. If the blade is 8 inches long, what is the area cleared by the wiper? Round your answer to the nearest square inch. 112. Windshield Wiper. A windshield wiper that is 11 inches long (blade and arm) rotates 65°. If the blade is 7 inches long, what is the area cleared by the wiper? Round your answer to the nearest square inch. For Exercises 113 and 114, refer to the following: Sniffers outside a chemical munitions disposal site monitor the atmosphere surrounding the site to detect any toxic gases. In the event that there is an accidental release of toxic fumes, the data provided by the sniffers make it possible to determine both the distance d that the fumes reach and the angle of spread u that sweeps out a circular sector. 113. Environment. If the maximum angle of spread is 1058 and the maximum distance at which the toxic fumes were detected was 9 miles from the site, find the area of the circular sector affected by the accidental release. 114. Environment. To protect the public from the fumes, officials must secure the perimeter of this area. Find the perimeter of the circular sector in Exercise 113. 6.6 Radian Measure and Applications 581 582 CHAPTER 6 Trigonometric Functions For Exercises 115 and 116, refer to the following: The structure of human DNA is a linear double helix formed of nucleotide base pairs (two nucleotides) that are stacked with spacing of 3.4 angstroms 13.4 3 10212m2, and each base pair is rotated 368 with respect to an adjacent pair and has 10 base pairs per helical turn. The DNA of a virus or a bacterium, however, is a circular double helix (see the figure below) with the structure varying among species. Twists (Source: PDFs/Education/Vologodskii.pdf.) 115. Biology. If the circular DNA of a virus has 10 twists (or turns) per circle and an inner diameter of 4.5 nanometers, find the arc length between consecutive twists of the DNA. 116. Biology. If the circular DNA of a virus has 40 twists (or turns) per circle and an inner diameter of 2.0 nanometers, find the arc length between consecutive twists of the DNA. 117. Tires. A car owner decides to upgrade from tires with a diam-eter of 24.3 inches to tires with a diameter of 26.1 inches. If she doesn’t update the onboard computer, how fast will she actually be traveling when the speedometer reads 65 mph? Round to the nearest mph. 118. Tires. A car owner decides to upgrade from tires with a diameter of 24.8 inches to tires with a diameter of 27.0 inches. If she doesn’t update the onboard computer, how fast will she actually be traveling when the speedometer reads 70 mph? Round to the nearest mph. 119. Planets. The Earth rotates every 24 hours (actually 23 hours, 56 minutes, and 4 seconds) and has a diameter of 7926 miles. If you’re standing on the equator, how fast are you traveling (how fast is the Earth spinning) in linear speed? Compute this using 24 hours and then with 23 hours, 56 minutes, 4 seconds. 120. Planets. The planet Jupiter rotates every 9.9 hours and has a diameter of 88,846 miles. If you’re standing on its equator, how fast are you traveling (linear speed)? 121. Carousel. A boy wants to jump onto a moving carousel that is spinning at the rate of 5 revolutions per minute. If the carousel is 60 feet in diameter, how fast must the boy run, in feet per second, to match the speed of the carousel and jump on? Round to the nearest foot per second. 122. Carousel. A boy wants to jump onto a playground carousel that is spinning at the rate of 30 revolutions per minute. If the carousel is 6 feet in diameter, how fast must the boy run, in feet per second, to match the speed of the carousel and jump on? Round to the nearest foot per second. 123. Music. Some people still have their phonograph collections and play the records on turntables. A phonograph record is a vinyl disc that rotates on the turntable. If a 12-inch- diameter record rotates at 33 1 3 revolutions per minute, what is the angular speed in radians per minute? Niall McDiarmid/Alamy 124. Music. Some people still have their phonograph collections and play the records on turntables. A phonograph record is a vinyl disc that rotates on the turntable. If a 12-inch- diameter record rotates at 33 1 3 revolutions per minute, what is the linear speed of a point on the outer edge? 125. Bicycle. How fast is a bicyclist traveling in miles per hour if his tires are 27 inches in diameter and his angular speed is 5p radians per second? 126. Bicycle. How fast is a bicyclist traveling in miles per hour if his tires are 22 inches in diameter and his angular speed is 5p radians per second? 127. Electric Motor. If a 2-inch-diameter pulley that is being driven by an electric motor and running at 1600 revolutions per minute is connected by a belt to a 5-inch-diameter pulley to drive a saw, what is the speed of the saw in revolutions per minute? 128. Electric Motor. If a 2.5-inch-diameter pulley that is being driven by an electric motor and running at 1800 revolutions per minute is connected by a belt to a 4-inch-diameter pulley to drive a saw, what is the speed of the saw in revolutions per minute? For Exercises 129 and 130, refer to the following: NASA explores artificial gravity as a way to counter the physiologic effects of extended weightlessness for future space exploration. NASA’s centrifuge has a 58-foot-diameter arm. Courtesy NASA 129. NASA. If two humans are on opposite (red and blue) ends of the centrifuge and their linear speed is 200 miles per hour, how fast is the arm rotating? 130. NASA. If two humans are on opposite (red and blue) ends of the centrifuge and they rotate one full rotation every second, what is their linear speed in feet per second? For Exercises 131 and 132, refer to the following: To achieve similar weightlessness as NASA’s centrifuge, ride the Gravitron at a carnival or fair. The Gravitron has a diameter of 14 meters, and in the first 20 seconds it achieves zero gravity and the floor drops. Patrick Reddy/America24-7/Getty Images, Inc. 131. Gravitron. If the Gravitron rotates 24 times per minute, find the linear speed of the people riding it in meters per second. 132. Gravitron. If the Gravitron rotates 30 times per minute, find the linear speed of the people riding it in kilometers per hour. 133. Biology. A hamster runs around a wheel at a constant rate such that the wheel goes through 1 revolution every 3 seconds. Find the angular speed in radians per minute. 134. Cooking. A chef can turn the hand crank of a pasta maker at a rate of 10p radians every 3 seconds. At this rate, a new batch of spaghetti can be made every 15 seconds. If 40 batches were made, how many revolutions of the hand crank were made? 135. The Price Is Right. On the TV show The Price is Right, the 8-foot-diameter wheel is divided into 20 sectors (corre-sponding to increments of 5 cents). The three most import-ant amounts on the wheel are the three abutting sectors with amounts $0.15, $1.00, and $0.05; the other amounts are ran-domly dispersed among the remaining sectors. A spin always starts at $1.00. A zealous contestant spins the wheel so that it makes four complete revolutions, landing on $1.00. What is the length of the arc created by this spin? 136. The Price Is Right. If a contestant on The Price Is Right lands on the $1.00 space on his first spin, he is granted a second spin. If this spin lands on either the $0.05 or $0.15 space, he wins $5,000; if it lands on the $1.00 space again, he wins $10,000. This time, he tries to control the spin so that the wheel makes exactly one complete revolution, again landing on $1.00. If the arc length of the spin is 84 11p feet, does the contestant win any money? • C A T C H T H E M I S T A K E In Exercises 137 and 138, explain the mistake that is made. 137. If the radius of a set of tires on a car is 15 inches and the tires rotate 180° per second, how fast is the car traveling (linear speed) in miles per hour? Solution: Write the formula for linear speed. v 5 rv Let r 5 15 inches and v 5 115 in.2 a180° sec b v 5 180° per second Simplify. v 5 2700 in. sec Let 1 mile 5 5280 feet 5 63,360 inches and v 5 a2700 ⋅3600 63,360 b mph 1 hour 5 3600 seconds. Simplify. v < 153.4 mph This is incorrect. The correct answer is approximately 2.7 mph. What mistake was made? 138. If a bicycle has tires with radius 10 inches and the tires rotate 90° per 1 2 second, how fast is the bicycle traveling (linear speed) in miles per hour? Solution: Write the formula for linear speed. v 5 rv Let r 5 10 inches and v 5 110 in.2 a180° sec b v 5 180° per second. Simplify. v 5 1800 in. sec Let 1 mile 5 5280 feet 5 63,360 inches and v 5 a1800⋅3600 63,360 b mph 1 hour 5 3600 seconds. Simplify. v < 102.3 mph This is incorrect. The correct answer is approximately 1.8 mph. What mistake was made? • C O N C E P T U A L In Exercises 139–142, determine whether each statement is true or false. 139. If the radius of a circle doubles, then the arc length (associated with a fixed central angle) doubles. 140. If the radius of a circle doubles, then the area of the sector (associated with a fixed central angle) doubles. 141. If the angular speed doubles, then the number of revolutions doubles. 142. If the central angle of a sector doubles, then the area cor-responding to the sector is double the area of the original sector. 6.6 Radian Measure and Applications 583 584 CHAPTER 6 Trigonometric Functions • C H A L L E N G E For Exercises 143 and 144, refer to the figure: The large gear has a radius of 6 centimeters, the medium gear has a radius of 3 centimeters, and the small gear has a radius of 1 centimeter. 6 cm 3 cm 1 cm 143. If the small gear rotates 1 revolution per second, what is the linear speed of a point traveling along the circumference of the large gear? 144. If the small gear rotates 1.5 revolutions per second, what is the linear speed of a point traveling along the circumference of the large gear? 145. A sector has a radius r and central angle u. If the radius is tripled, but the central angle is reduced by a factor of 3, how does the area of the resulting sector compare to the original one? 146. Doing “ladders” as an exercise is familiar to anyone who has participated on an athletic team. A version of this on a circu-lar track of diameter 400 feet is as follows: 1. Start at point A and run clockwise 1808 to point B. 2. Then run from point B counterclockwise 908 to point C. 3. Then run from point C clockwise 458 to point D. 4. Then run from point D counterclockwise 108 to point E. Then stop. How far has the athlete traveled in one such round of ladders? 147. Find the area of the shaded region below: x y –5 –1 1 –3 2 3 4 5 –5 –4 –3 –2 1 3 2 4 5 148. Find the perimeter of the shaded region in Exercise 143. • T E C H N O L O G Y In Exercises 149 and 150, find the measure (in degrees, min utes, and nearest seconds) of a central angle u that inter cepts an arc on a circle with radius r and indicated arc length s. With the TI calculator commands ANGLE and DMS , change to degrees, minutes, and seconds. 149. r 5 78.6 cm, s 5 94.4 cm 150. r 5 14.2 in., s 5 23.8 in. Recall that the first definition of trigonometric functions we developed was in terms of ratios of sides of right triangles (Section 6.2). Then in Section 6.4 we superimposed right triangles on the Cartesian plane, which led to a second definition of trigonometric functions (for any angle) in terms of ratios of x- and y-coordinates of a point and the distance from the origin to that point. In this section, we inscribe the right triangles into the unit circle in the Cartesian plane, which will yield a third definition of trigonometric functions. It is important to note that all three definitions are consistent with one another. S K I L L S O B J E C T I V E S ■ ■Draw the unit circle illustrating the special angles and label cosine and sine values. ■ ■Classify circular functions as even or odd. C O N C E P T U A L O B J E C T I V E S ■ ■Relate x-coordinates and y-coordinates of points on the unit circle to the values of cosine and sine functions. ■ ■Visualize the periodic properties of circular functions. 6.7 DEFINITION 3 OF TRIGONOMETRIC FUNCTIONS: UNIT CIRCLE APPROACH 6.7 Definition 3 of Trigonometric Functions: Unit Circle Approach 585 6.7.1 Trigonometric Functions and the Unit Circle (Circular Functions) Recall that the equation for the unit circle centered at the origin is given by x2 1 y2 5 1. We will use the term circular function later in this section, but it is important to note that a circle is not a function (it does not pass the vertical line test). If we form a central angle u in the unit circle such that the terminal side lies in quadrant I, we can use the previous two definitions of the sine and cosine functions when r 5 1 (i.e., in the unit circle). TRIGONOMETRIC FUNCTION RIGHT TRIANGLE TRIGONOMETRY CARTESIAN PLANE sin u opposite hypotenuse 5 y 1 5 y y r 5 y 1 5 y cos u adjacent hypotenuse 5 x 1 5 x x r 5 x 1 5 x Notice that the point 1x, y2 on the unit circle can be written as 1cos u, sin u2. If we recall the unit circle coordinate values for special angles (Section 6.4), we can now summa-rize the exact values for sine and cosine in the following illustration. 1x, y2 5 1cosu, sinu2, where u is the central angle whose terminal side intersects the unit circle at 1x, y2. x y 45º 4 360º 2 0º 0 0 315º 4 7 270º 2 3 225º 4 5 180º 135º 4 3 90º 2 (0, 1) (0, –1) (1, 0) (–1, 0) 2 √2 2 √2 ( , – ) 2 √2 2 √2 ( , ) 2 √2 2 √2 (– , – ) 2 √2 2 √2 (– , ) x y 60º 3 30º 6 360º 2 0º 0 0 330º 6 11 300º 3 5 270º 2 3 240º 3 4 210º 6 7 180º 150º 6 5 90º 2 120º 3 2 (0, 1) (0, –1) (1, 0) (–1, 0) 2 √3 2 1 ( , – ) 2 √3 2 1 ( , ) 2 √3 2 1 ( , – ) 2 √3 2 1 (– , – ) 2 √3 2 1 (– , – ) (– , ) 2 √3 2 1 2 √3 2 1 (– , ) 2 √3 2 1 ( , ) The following observations are consistent with properties of trigonometric functions we’ve studied already: ■ ■sin u . 0 in quadrant I and quadrant II ■ ■cos u . 0 in quadrant I and quadrant IV ■ ■The unit circle equation x2 1 y2 5 1 leads to the Pythagorean identity cos 2 u 1 sin 2 u 5 1 Circular Functions Using the unit circle relationship, 1x, y2 5 1cos u, sin u2, where u is the central angle whose terminal side intersects the unit circle at the point 1x, y2, we can now define the remaining trigonometric functions using this unit circle approach and the quotient and reciprocal identities. Because the trigonometric functions are defined in terms of the unit circle, the trigonometric functions are often called circular functions. x y (1, 0) (0, 1) (x, y) r = 1 x y s θ (0, –1) (–1, 0) 6.7.1 S K I L L Draw the unit circle illustrating the special angles and label cosine and sine values. 6.7.1 C O N C E P T U A L Relate x-coordinates and y-coordinates of points on the unit circle to the values of cosine and sine functions. STUDY TIP (cos u, sin u) represents a point (x, y) on the unit circle. [CONCEPT CHECK] If the point (2A, 2B) lies in quadrant III, then what must be true so that cos u 5 2A and sin u 5 2B? ANSWER A2 1 B2 5 1 ▼ 586 CHAPTER 6 Trigonometric Functions DEFINITION (3) Trigonometric Functions Unit Circle Approach Let 1x, y2 be any point on the unit circle. If u is a real number that represents the distance from the point 11, 02 along the circumference to the point 1x, y2, then sin u 5 y cos u 5 x tan u 5 y x x 2 0 csc u 5 1 y y 2 0 sec u 5 1 x x 2 0 cot u 5 x y y 2 0 EXAMPLE 1 Finding Exact Trigonometric (Circular) Function Values Find the exact values for a. sin a7p 4 b b. cos a5p 6 b c. tan a3p 2 b Solution (a): The angle 7p 4 corresponds to the coordinates a !2 2 , 2 !2 2 b on the unit circle. The value of the sine function is the y-coordinate. sin a7p 4 b 5 2 !2 2 Solution (b): The angle 5p 6 corresponds to the coordinate a2 !3 2 , 1 2b on the unit circle. The value of the cosine function is the x-coordinate. cos a5p 6 b 5 2 !3 2 Solution (c): The angle 3p 2 corresponds to the coordinate 10, 212 on the unit circle. The value of the cosine function is the x-coordinate. cos a3p 2 b 5 0 The value of the sine function is the y-coordinate. sin a3p 2 b 5 21 Tangent is the ratio of sine to cosine. tan a3p 2 b 5 sin a3p 2 b cos a3p 2 b Let cos a3p 2 b 5 0 and sin a3p 2 b 5 21. tan a3p 2 b 5 21 0 tan a3p 2 b is undefined. Y OUR TU R N Find the exact values for a. sin a5p 6 b b. cos a7p 4 b c. tan a2p 3 b ▼ ▼ A N S W E R a. 1 2 b. "2 2 c. 2"3 6.7.2 Properties of Circular Functions WORDS MATH For a point 1x, y2 that lies on the unit circle, x2 1 y2 5 1. 21 # x # 1 and 21 # y # 1 Since 1x, y2 5 1cos u, sin u2, the following holds. 21 # cos u # 1 and 21 # sin u # 1 State the domain and range of the cosine and sine functions. Domain: 12q, q2 Range: [21, 1] Since cot u 5 cos u sin u and csc u 5 1 sin u, the values for u that make sin u 5 0 must be eliminated from the domain of the cotangent and cosecant functions. Domain: u 2 np, n an integer Since tan u 5 sin u cos u and sec u 5 1 cos u, the values for u that make cos u 5 0 must be eliminated from the domain of the tangent and secant functions. Domain: u 2 12n 1 12p 2 , n an integer The following box summarizes the domains and ranges of the circular functions. ▼ A N S W E R u 5 2p 3 , 4p 3 EXAMPLE 2 Solving Equations Involving Circular Functions Use the unit circle to find all values of u, 0 # u # 2p, for which sin u 5 21 2. Solution: The value of sine is the y-coordinate. Since the value of sine is negative, u must lie in QIII or QIV. There are two values for u that are greater than or equal to zero and less than or equal to 2p that correspond to sin u 5 21 2. u 5 7p 6 , 11p 6 YOUR T UR N Find all values of u, 0 # u # 2p, for which cos u 5 21 2. ▼ (– , – ) x y 60º 3 π 45º 4 π 30º 6 π 360º 2π 0º 0 0 330º 6 11π 315º 4 7π 300º 3 5π 270º 2 3π 240º 3 4π 225º 4 5π 210º 6 7π π 180º 150º 6 5π 135º 4 3π 90º 2 π 120º 3 2π (0, 1) (0, –1) (1, 0) (–1, 0) 2 √3 2 1 ( , – ) 2 √3 2 1 ( , ) 2 √2 2 √2 ( , – ) 2 √2 2 √2 ( , ) 2 √3 2 1 ( , – ) 2 √3 2 1 2 √2 2 √2 (– , – ) 2 √3 2 1 (– , – ) 2 √3 2 1 (– , ) 2 √2 2 √2 (– , ) 2 √3 2 1 (– , ) 2 √3 2 1 ( , ) 6.7.2 S K IL L Classify circular functions as even or odd. 6.7.1 C ON C E P T U A L Visualize the periodic properties of circular functions. [CONCEPT CHECK] Which of the following is NOT equivalent to the others for all values of angle A? (A) sin (2p 1 A) (B) sin (A 2 2p) (C) sin (p 1 A) (D) sin (A 1 8p) ANSWER (C) ▼ 6.7 Definition 3 of Trigonometric Functions: Unit Circle Approach 587 588 CHAPTER 6 Trigonometric Functions Recall from algebra that even and odd functions have both an algebraic and a graphical interpretation. Even functions are functions for which f 12x2 5 f 1x2 and their graphs are symmetric about the y-axis. Odd functions are functions for which f 12x2 5 2f 1x2 and their graphs are symmetric about the origin. x y 1 –1 1 –1 θ –θ (x, y) = (cos θ, sin θ ) (x, –y) = (cos(–θ ), sin(–θ )) = (cos θ, –sin θ ) The cosine function is an even function. The sine function is an odd function. cos u 5 cos 12u2 sin12u2 5 2sin u DOMAINS AND RANGES OF THE CIRCULAR FUNCTIONS For any real number u and integer n: FUNCTION DOMAIN RANGE sin u 12q, q2 321, 14 cos u 12q, q2 321,14 tan u all real numbers such that u 2 12n 1 12p 2 12q, q2 cot u all real numbers such that u 2 np 12q, q2 sec u all real numbers such that u 2 12n 1 12p 2 12q, 214 ∪ 31, q2 csc u all real numbers such that u 2 np 12q, 214 ∪ 31, q2 EXAMPLE 3 Using Properties of Trigonometric (Circular) Functions Evaluate cos a25p 6 b. Solution: The cosine function is an even function. cos a 25p 6 b 5 cos a5p 6 b Use the unit circle to evaluate cosine. cos a5p 6 b 5 2 !3 2 cos a 25p 6 b 5 2 !3 2 Y OUR TU R N Evaluate sin a 25p 6 b. ▼ ▼ A N S W E R 21 2 Although circular functions can be evaluated exactly for some special angles, a calculator can be used to approximate circular functions for any angle. It is important to set the calculator to radian mode first, since u is a real number. EXAMPLE 4 Evaluating Trigonometric (Circular) Functions with a Calculator Use a calculator to evaluate sin a7p 12 b. Round the answer to four decimal places. ✖I N C O R R EC T Evaluate with a calculator. 0.031979376 ERROR (Calculator in degree mode) ✓COR R E C T Evaluate with a calculator. 0.965925826 Round to four decimal places. sin a7p 12 b < 0.9659 Many calculators automatically reset to degree mode after every calculation, so make sure to always check what mode the calculator indicates. Y OUR T UR N Use a calculator to evaluate tan a9p 5 b. Round the answer to four decimal places. ▼ EXAMPLE 5 Even and Odd Circular Functions Show that the secant function is an even function. Solution: Show that sec12u2 5 sec1u2. Secant is the reciprocal of cosine. sec12u2 5 1 cos 12u2 Cosine is an even function, cos 12u2 5 cos 1u2. sec12u2 5 1 cos 1u2 Secant is the reciprocal of cosine, sec u 5 1 cos u. sec12u2 5 1 cos 1u2 5 sec u Since sec12u2 5 sec1u2, the secant function is an even function . [SEC TION 6.7] S UM M A RY In this section, we have defined trigonometric functions as circular functions. Any point 1x, y2 that lies on the unit circle satisfies the equation x2 1 y2 5 1. The Pythagorean identity cos2 u 1 sin2 u 5 1 can also be represented on the unit circle where 1x, y2 5 1cos u, sin u2, and where u is the central angle whose terminal side intersects the unit circle at the point 1x, y2. The cosine function is an even function, cos12u2 5 cos u; and the sine function is an odd function, sin12u2 5 2sin u. STUDY TIP Set the calculator in radian mode before evaluating circular functions in radians. Alternatively, convert the radian measure to degrees before evaluating the trigonometric function value. ▼ A N S W E R 20.7265 6.7 Definition 3 of Trigonometric Functions: Unit Circle Approach 589 590 CHAPTER 6 Trigonometric Functions In Exercises 1–14, find the exact values of the indicated trigonometric functions using the unit circle. 1. sin a5p 3 b 2. cos a5p 3 b 3. cos a7p 6 b 4. sin a7p 6 b 5. sin a3p 4 b 6. cos a3p 4 b 7. tan a7p 4 b 8. cot a7p 4 b 9. sec 225° 10. csc 300° 11. tan 240° 12. cot 330° 13. csc 5p 6 14. cot 120° In Exercises 15–34, use the unit circle and the fact that sine is an odd function and cosine is an even function to find the exact values of the indicated functions. 15. sin a 2 2p 3 b 16. sin a 2 5p 4 b 17. sin a 2 p 3 b 18. sin a 2 7p 6 b 19. cos a 2 3p 4 b 20. cos a 2 5p 3 b 21. cos a 2 5p 6 b 22. cos a 2 7p 4 b 23. sin12225°2 24. sin12180°2 25. sin12270°2 26. sin1260°2 27. cos1245°2 28. cos12135°2 29. cos1290°2 30. cos12210°2 31. csc12150°2 32. sec12315°2 33. tan a2 11p 6 b 34. cot12330°2 In Exercises 35–54, use the unit circle to find all of the exact values of u that make the equation true in the indicated interval. 35. cos u 5 !3 2 , 0 # u # 2p 36. cos u 5 2 !3 2 , 0 # u # 2p 37. sin u 5 2 !3 2 , 0 # u # 2p 38. sin u 5 !3 2 , 0 # u # 2p 39. sin u 5 0, 0 # u # 4p 40. sin u 5 21, 0 # u # 4p 41. cos u 5 21, 0 # u # 4p 42. cos u 5 0, 0 # u # 4p 43. tan u 5 21, 0 # u # 2p 44. cot u 5 1, 0 # u # 2p 45. sec u 5 2!2, 0 # u # 2p 46. csc u 5 !2, 0 # u # 2p 47. csc u is undefined, 0 # u # 2p 48. sec u is undefined, 0 # u # 2p 49. tan u is undefined, 0 # u # 2p 50. cot u is undefined, 0 # u # 2p 51. csc u 5 22, 0 # u # 2p 52. cot u 5 2"3, 0 # u # 2p 53. sec u 5 2!3 3 , 0 # u # 2p 54. tan u 5 !3 3 , 0 # u # 2p x y 60º 3 π 45º 4 π 30º 6 π 360º 2π 0º 0 0 330º 6 11π 315º 4 7π 300º 3 5π 270º 2 3π 240º 3 4π 225º 4 5π 210º 6 7π π 180º 150º 6 5π 135º 4 3π 90º 2 π 120º 3 2π (0, 1) (0, –1) (1, 0) (–1, 0) 2 √3 2 1 ( , – ) 2 √3 2 1 ( , ) 2 √2 2 √2 ( , – ) 2 √2 2 √2 ( , ) 2 √3 2 1 ( , – ) (– , – ) 2 √3 2 1 (– , – ) 2 √2 2 √2 2 √3 2 1 (– , – ) 2 √3 2 1 (– , ) 2 √2 2 √2 (– , ) 2 √3 2 1 (– , ) 2 √3 2 1 ( , ) [SEC TION 6.7] E X E R C I SE S • S K I L L S 55. Atmospheric Temperature. The average daily temperature in Peoria, Illinois, can be predicted by the formula T 550 2 28 cos c 2p1x 2312 365 d , where x is the number of the day in the year (January 1 5 1, February 1 5 32, etc.) and T is in degrees Fahrenheit. What is the expected temperature on February 15? 56. Atmospheric Temperature. The average daily temperature in Peoria, Illinois, can be predicted by the formula T 550 228 cos c2p1x 2312 365 d , where x is the number of the day in the year (January 1 5 1, February 1 5 32, etc.) and T is in degrees Fahrenheit. What is the expected temperature on August 15? (Assume it is not a leap year.) 57. Body Temperature. The human body temperature normally fluctuates during the day. If a person’s body temperature can be predicted by the formula T 5 99.1 20.5 sin a x 1 p 12b, where x is the number of hours since midnight and T is in degrees Fahrenheit, then what is the person’s temperature at 6:00 a.m.? 58. Body Temperature. The human body temperature normally fluctuates during the day. If a person’s body temperature can be predicted by the formula T 5 99.1 2 0.5 sinax 1 p 12b, where x is the number of hours since midnight and T is in degrees Fahrenheit, then what is the person’s temperature at 9:00 p.m.? 59. Tides. The height of the water in a harbor changes with the tides. If the height of the water at a particular hour during the day can be determined by the formula h1x2 55 14.8 sinc p 6 1x 142d , where x is the number of hours since midnight and h is the height of the tide in feet, then what is the height of the tide at 3:00 p.m.? 60. Tides. The height of the water in a harbor changes with the tides. If the height of the water at a particular hour during the day can be determined by the formula h1x2 5 5 1 4.8 sin c p 6 1x 1 42 d , where x is the number of hours since midnight and h is the height of the tide in feet, then what is the height of the tide at 5:00 a.m.? 61. Yo-Yo Dieting. A woman has been yo-yo dieting for years. Her weight changes throughout the year as she gains and loses weight. Her weight in a particular month can be determined by the formula w1x2 5 145 1 10 cos ap 6 xb, where x is the month and w is in pounds. If x 5 1 corresponds to January, how much does she weigh in June? 62. Yo-Yo Dieting. How much does the woman in Exercise 61 weigh in December? 63. Seasonal Sales. The average number of guests visiting the Magic Kingdom at Walt Disney World per day is given by n1x2 5 33,000 1 20,000 sinc p 2 1x 1 12 d , where n is the number of guests and x is the month. If January corresponds to x 5 1, how many people on average are visiting the Magic Kingdom per day in February? 64. Seasonal Sales. How many guests are visiting the Magic Kingdom in Exercise 63 in December? • A P P L I C A T I O N S 6.7 Definition 3 of Trigonometric Functions: Unit Circle Approach 591 592 CHAPTER 6 Trigonometric Functions For Exercises 65 and 66, refer to the following: During the course of treatment of an illness, the concentration of a drug in the bloodstream in micrograms per microliter fluctuates during the dosing period of 8 hours according to the model C1t2 5 15.4 2 4.7 sin ap 4 t 1 p 2 b , 0 # t # 8 Note: This model does not apply to the first dose of the medication. 65. Health/Medicine. Find the concentration of the drug in the bloodstream at the beginning of a dosing period. 66. Health/Medicine. Find the concentration of the drug in the bloodstream 6 hours after taking a dose of the drug. In Exercises 67 and 68, refer to the following: By analyzing available empirical data, it has been determined that the body temperature of a particular species fluctuates during a 24-hour day according to the model T1t2 5 36.3 2 1.4 cos c p 12 1t 2 22d , 0 # t # 24 where T represents temperature in degrees Celsius and t represents time in hours measured from 12:00 a.m. (midnight). 67. Biology. Find the approximate body temperature at midnight. Round your answer to the nearest degree. 68. Biology. Find the approximate body temperature at 2:45 p.m. Round your answer to the nearest degree. • C A T C H T H E M I S T A K E In Exercises 69 and 70, explain the mistake that is made. In Exercises 71–76, determine whether each statement is true or false. • C O N C E P T U A L 71. sin12np 1 u2 5 sin u, n an integer. 72. cos 12np 1 u2 5 cos u, n an integer. 73. sin u 5 1 when u 5 12n 1 12p 2 , n an integer. 74. cos u 5 1 when u 5 np, n an integer. 75. tan 1u 1 2np2 5 tan u, n an integer. 76. tan u 5 0 if and only if u 5 12n 1 12p 2 , n an integer. 77. Is cosecant an even or an odd function? Justify your answer. 78. Is tangent an even or an odd function? Justify your answer. 69. Use the unit circle to evaluate tan a5p 6 b exactly. Solution: Tangent is the ratio of sine to cosine. tan a5p 6 b 5 sin a5p 6 b cos a5p 6 b Use the unit circle to identify sine and cosine. sin a5p 6 b 5 2 !3 2 and cos a5p 6 b 5 1 2 Substitute values for sine and cosine. tan a5p 6 b 5 2 !3 2 1 2 Simplify. tan a5p 6 b 5 2!3 This is incorrect. What mistake was made? 70. Use the unit circle to evaluate sec a11p 6 b exactly. Solution: Secant is the reciprocal of cosine. sec a11p 6 b 5 1 cos a11p 6 b Use the unit circle to evaluate cosine. cos a11p 6 b 5 2 1 2 Substitute the value for cosine. sec a11p 6 b 5 1 2 1 2 Simplify. sec a11p 6 b 5 22 This is incorrect. What mistake was made? 79. Find all the values of u, 0 # u # 2p, for which the equation sin u 5 cos u is true. 80. Find all the values of u 1u is any real number) for which the equation sin u 5 cos u is true. 81. Show that cscap 2 1 u 1 2npb 5 sec u, n an integer. 82. Show that secap 2 2 u 2 2npb 5 csc u, n an integer. • C H A L L E N G E 87. Use a calculator to approximate sin1423°2. What do you expect sin12423°2 to be? Verify your answer with a calculator. 88. Use a calculator to approximate cos 1227°2. What do you expect cos12227°2 to be? Verify your answer with a calculator. 89. Use a calculator to approximate tan181°2. What do you expect tan1281°2 to be? Verify your answer with a calculator. 90. Use a calculator to approximate csc1211°2. What do you expect csc12211°2 to be? Verify your answer with a calculator. For Exercises 91–94, refer to the following: A graphing calculator can be used to graph the unit circle with parametric equations (these will be covered in more detail in Section 8.7). For now, set the calculator in parametric and radian modes and let X1 5 cos T Y 1 5 sin T Set the window so that 0 # T # 2p, step 5 p 15, 22 # X # 2, and 22 # Y# 2. To approximate the sine or cosine of a T value, use the TRACE key, type in the T value, and read the corresponding coordinates from the screen. 91. Approximate cos ap 3 b, take 5 steps of p 15 each, and read the x-coordinate. 92. Approximate sin ap 3 b, take 5 steps of p 15 each, and read the y-coordinate. 93. Approximate sin a2p 3 b to four decimal places. 94. Approximate cos a5p 4 b to four decimal places. • T E C H N O L O G Y 83. Simplify csca2p 2 p 2 2 ub ⋅sec au 2 p 2 b ⋅sin12u2. 84. Simplify tan u⋅cot12p 2 u2. 85. Does there exist an angle 0 # u , 2p such that tan u 5 cot u? 86. Does there exist an angle 0 # u , 2p such that sec u 5 csc12 u2? 6.7 Definition 3 of Trigonometric Functions: Unit Circle Approach 593 594 CHAPTER 6 Trigonometric Functions S K I L L S O B J E C T I V E S ■ ■Determine the amplitude and period of sinusoidal functions. ■ ■Determine the phase shift of a sinusoidal function. ■ ■Solve harmonic motion problems. ■ ■Graph sums of trigonometric and other algebraic functions. C O N C E P T U A L O B J E C T I V E S ■ ■Understand why the graphs of the sine and cosine functions are called sinusoidal graphs. ■ ■Rewriting the sinusoidal function in standard form makes identifying the phase shift easier. ■ ■Visualize harmonic motion as a sinusoidal function. ■ ■Understand that the y-coordinates of the combined function are found by adding the y-coordinates of the individual functions. 6.8 GRAPHS OF SINE AND COSINE FUNCTIONS 6.8.1 The Graphs of Sinusoidal Functions The following are examples of things that repeat in a predictable way (i.e., they are roughly periodic): ■ ■heartbeat ■ ■tide levels ■ ■time of sunrise ■ ■average outdoor temperature for the time of year The trigonometric functions are strictly periodic. In the unit circle, the value of any of the trigonometric functions is the same for any coterminal angle (same initial and terminal sides no matter how many full rotations the angle makes). For example, if we add (or subtract) multiples of 2p to the angle u, the values for sine and cosine are unchanged. sin1u 1 2np2 5 sin u or cos1u 1 2np2 5 cos u 1n is any integer2 DEFINITION Periodic Function A function ƒ is called a periodic function if there is a positive number p such that ƒ1x 1 p2 5 ƒ1x2 for all x in the domain of ƒ If p is the smallest such number for which this equation holds, then p is called the fundamental period. You will see in this section and the next that sine, cosine, secant, and cosecant functions have fundamental period 2p, but that tangent and cotangent functions have fundamental period p. The Graph of f (x) 5 sinx Let us start by point-plotting the sine function. We select values for the sine function corresponding to some of the “special angles” we covered in the previous sections of this chapter. 6.8.1 S KILL Determine the amplitude and period of sinusoidal functions. 6.8.1 CO NCE PTUAL Understand why the graphs of the sine and cosine functions are called sinusoidal graphs. By plotting the above coordinates 1x, y2, we can obtain the graph of one period, or cycle, of the graph of y 5 sin x. Note that !2 2 < 0.7. 6.8 Graphs of Sine and Cosine Functions 595 x f (x) 5 sin x (x, y) 0 sin 0 5 0 10, 02 p 4 sinap 4 b 5 !2 2 ap 4, !2 2 b p 2 sinap 2 b 5 1 ap 2, 1b 3p 4 sina3p 4 b 5 !2 2 a3p 4 , !2 2 b p sin p 5 0 1p, 02 5p 4 sina5p 4 b 5 2 !2 2 a5p 4 , 2 !2 2 b 3p 2 sina3p 2 b 5 21 a3p 2 , 21b 7p 4 sina7p 4 b 5 2 !2 2 a7p 4 , 2 !2 2 b 2p sin 2p 5 0 12p, 02 STUDY TIP Note that either notation, y 5 sin x or f (x) 5 sin x, can be used. [CONCEPT CHECK] Which of the following is NOT a sinusoidal function: (A) f(x) 5 tan(x) (B) g(x) 5 sin(x) or (C) h(x) 5 cos(x) ANSWER (A) ▼ π π 2π 3π 2 2 (0, 0) (π, 0) (2π, 0) 3π 2 ( , –1) – π 2 ( , 1) ( , ) π 4 2 √2 3π 4 2 ( , ) √2 ( , ) ( , ) 5π 4 2 √2 – 7π 4 2 √2 –1 1 x y We can extend the graph horizontally in both directions (left and right) since the domain of the sine function is the set of all real numbers. π 4π 2π –π –2π 3π –1 –0.5 0.5 1 x y In this section, we are no longer showing angles on the unit circle but are now showing angles as real numbers in radians on the x-axis of the Cartesian graph. There-fore, we no longer illustrate a “terminal side” to an angle—the physical arcs and angles no longer exist; only their measures exist, as values of the x-coordinate. STUDY TIP Looking at the graph of f (x) 5 sin x, we are reminded that sin x . 0 when 0 , x , p and sin x , 0 when p , x , 2p. 596 CHAPTER 6 Trigonometric Functions If we graph the function f 1x2 5 sin x, the x-intercepts correspond to values of x at which the sine function is equal to zero. x f (x) 5 sin x (x, y) 0 sin 0 5 0 10, 02 p sin p 5 0 1p, 02 2p sin 12p2 5 0 12p, 02 3p sin13p2 5 0 13p, 02 4p sin 14p2 5 0 14p, 02 . . . . . . . . . np sin1np2 5 0 1np, 02 Notice that the point 10, 02 is both a y-intercept and an x-intercept but all x-intercepts have the form 1np, 02. The maximum value of the sine function is 1, and the minimum value of the sine function is 21; these values occur at odd multiples of p 2 . x f (x) 5 sin x (x, y) p 2 sin ap 2 b 5 1 ap 2, 1b 3p 2 sin a3p 2 b 5 21 a3p 2 , 21b 5p 2 sin a5p 2 b 5 1 a5p 2 , 1b 7p 2 sin a7p 2 b 5 21 a7p 2 , 21b . . . . . . . . . 12n 1 12p 2 sin a 12n 1 12p 2 b 5 61 a 12n 1 12p 2 , 61b The following box summarizes the sine function: n is an integer. n is an integer. SINE FUNCTION f (x) 5 sin x ■ ■Domain: 12q, q2 or 2q , x , q ■ ■Range: 321, 14 or 21 # y # 1 ■ ■The sine function is an odd function: ■ ■ symmetric about the origin ■ ■ sin12x2 5 2sinx ■ ■The sine function is a periodic function with fundamental period 2p. ■ ■ The x-intercepts, 0, 6p, 62p, . . . , are of the form np, where n is an integer. ■ ■The maximum 112 and minimum 1212 values of the sine function correspond to x-values of the form 12n 1 12p 2 such as 6 p 2 , 6 3p 2 , 6 5p 2 , . . . . π 2π –2π –π –1 1 x y 6.8 Graphs of Sine and Cosine Functions 597 The Graph of f (x) 5 cosx Let us start by point-plotting the cosine function. x f (x) 5 cos x (x, y) 0 cos 0 5 1 10, 12 p 4 cosap 4 b 5 !2 2 ap 4, !2 2 b p 2 cosap 2 b 5 0 ap 2, 0b 3p 4 cosa3p 4 b 5 2 !2 2 a3p 4 , 2 !2 2 b p cos p 5 21 1p, 212 5p 4 cosa5p 4 b 5 2 !2 2 a5p 4 , 2 !2 2 b 3p 2 cosa3p 2 b 5 0 a3p 2 , 0b 7p 4 cosa7p 4 b 5 !2 2 a7p 4 , !2 2 b 2p cos 12p2 5 1 12p, 12 By plotting the above coordinates 1x, y2 we can obtain the graph of one period, or cycle, of the graph of y 5 cos x. Note that !2 2 < 0.7. STUDY TIP Note that either notation, y 5 cos x or f (x) 5 cos x, can be used. π π 2π 3π 2 2 (0, 1) (π, –1) (2π, 1) 3π 2 ( , 0) π 2 ( , 0) π 4 2 ( , ) √2 7π 4 2 ( , ) √2 ( , ) 3π 4 2 – √2 ( , ) 5π 4 2 – √2 –1 1 x y π 4π 2π –π –2π 3π –1 1 x y STUDY TIP Looking at the graph of f (x) 5 cos x, we are reminded that cos x . 0 when 0 , p , p 2 and 3p 2 , x , 2p and cos x , 0 when p 2 , x , 3p 2 . We can extend the graph horizontally in both directions (left and right) since the domain of the cosine function is all real numbers. If we graph the function ƒ1x2 5 cos x, the x-intercepts correspond to values of x at which the cosine function is equal to zero. 598 CHAPTER 6 Trigonometric Functions x f (x) 5 cos x (x, y) p 2 cos ap 2 b 5 0 ap 2, 0b 3p 2 cos a3p 2 b 5 0 a3p 2 , 0b 5p 2 cos a5p 2 b 5 0 a5p 2 , 0b 7p 2 cos a7p 2 b 5 0 a7p 2 , 0b . . . . . . . . . 12n 1 12p 2 cos a 12n 1 12p 2 b 5 0 a 12n 1 12p 2 , 0b The point 10, 12 is the y-intercept, and there are several x-intercepts of the form a 12n 1 12p 2 , 0b. The maximum value of the cosine function is 1, and the minimum value of the cosine function is 21; these values occur at integer multiples of p; np. x f (x) 5 cos x (x, y) 0 cos 0 5 1 10, 12 p cos p 5 21 1p, 212 2p cos 2p 5 1 12p, 12 3p cos3p 5 21 13p, 212 4p cos 4p 5 1 14p, 12 . . . . . . . . . np cos np 5 61 1np, 612 The following box summarizes the cosine function: n is an integer. n is an integer. COSINE FUNCTION f (x) 5 cos x ■ ■Domain: 12q, q2 or 2q , x , q ■ ■Range: 321, 14 or 21 # y # 1 ■ ■The cosine function is an even function: ■ ■ symmetric about the y-axis ■ ■ cos12x2 5 cosx ■ ■The cosine function is a periodic function with fundamental period 2p. ■ ■The x-intercepts, 6 p 2 , 6 3p 2 , 6 5p 2 , . . . , are odd integer multiples of p 2 that have the form 12n 1 12p 2 , where n is an integer. ■ ■The maximum 112 and minimum 1212 values of the cosine function corre-spond to x-values of the form np, such as 0, 6p, 62p, . . . . π 2π –2π –π –1 1 x y 6.8 Graphs of Sine and Cosine Functions 599 The Amplitude and Period of Sinusoidal Graphs In mathematics, the word sinusoidal means “resembling the sine function.” Let us start by graphing ƒ1x2 5 sin x and ƒ1x2 5 cos x on the same graph. Notice that they have similar characteristics (domain, range, period, and shape). π 2π –2π –π –1 1 x y In fact, if we were to shift the cosine graph to the right p 2 units, the two graphs would be identical. For that reason, we refer to any graphs of the form y 5 cos x or y 5 sin x as sinusoidal functions. We now turn our attention to graphs of the form y 5 A sin1Bx2 and y 5 A cos1Bx2, which are graphs like y 5 sin x and y 5 cos x that have been stretched or compressed vertically and horizontally. EXAMPLE 1 Vertical Stretching and Compressing Plot the functions y 5 2 sin x and y 5 1 2 sin x on the same graph with y 5 sin x on the interval 24p # x # 4p. Solution: STEP 1 Make a table with the coordinate values of the graphs: x o p 2 p 3p 2 2p sin x 0 1 0 21 0 2 sin x 0 2 0 22 0 1 2 sin x 0 1 2 0 21 2 0 STEP 2 Label the points on the graph and connect with a smooth curve over one period, 0 # x # 2p. π π (π, 0) (0, 0) (2π, 0) 2π 2 –2 –1 1 2 x y 3π 2 ( , – ) 3π 2 ( , –2) π 2 ( , 1) π 2 1 2 1 2 ( , ) π 2 ( , 2) 3π 2 ( , –1) 600 CHAPTER 6 Trigonometric Functions Notice in Example 1 and the corresponding Your Turn that: ■ ■y 5 2 sin x has the shape and period of y 5 sin x but is stretched vertically. ■ ■y 5 1 2 sin x has the shape and period of y 5 sin x but is compressed vertically. ■ ■y 5 3 cos x has the shape and period of y 5 cos x but is stretched vertically. ■ ■y 5 1 3 cos x has the shape and period of y 5 cos x but is compressed vertically. In general, functions of the form y 5 A sin x and y 5 A cos x are stretched vertically when k A k . 1 and compressed vertically when k A k , 1. The amplitude of a periodic function is half the difference between the maxi-mum value of the function and the minimum value of the function. For the functions y 5 sin x and y 5 cos x, the maximum value is 1 and the minimum value is 21. There-fore, the amplitude of each of these two functions is k A k 5 1 2 k 1 2 1212 k 5 1. STEP 3 Extend the graph in both directions (repeat every 2p). Y OUR TU R N Plot the functions y 5 3 cos x and y 5 1 3 cos x on the same graph with y 5 cos x on the interval 22p # x # 2p. π 2π –2π –4π 3π 4π –2 –1 1 2 x y ▼ ▼ A N S W E R 2π π –π –2π –3 –2 –1 1 2 3 x y y = cosx y = 3cosx 1 3 y = cosx AMPLITUDE OF SINUSOIDAL FUNCTIONS For sinusoidal functions of the form y 5 A sin 1Bx2 and y 5 A cos 1Bx2, the amplitude is k A k . When k A k , 1, the graph is compressed vertically, and when k A k . 1, the graph is stretched vertically. EXAMPLE 2 Finding the Amplitude of Sinusoidal Functions State the amplitude of a. ƒ1x2 5 24 cos x b. g1x2 5 1 5 sin x Solution (a): The amplitude is the magnitude of 24. A 5 k 24 k 5 4 Solution (b): The amplitude is the magnitude of 1 5. A 5 1 5 5 1 5 6.8 Graphs of Sine and Cosine Functions 601 EXAMPLE 3 Horizontal Stretching and Compressing Plot the functions y 5 cos 12x2 and y 5 cosA1 2xB on the same graph with y 5 cos x on the interval 22p # x # 2p. Solution: STEP 1 Make a table with the coordinate values of the graphs. It is necessary only to select the points that correspond to x-intercepts, 1y 5 02, and maximum and minimum points, 1y 5 612. Usually, the period is divided into four subintervals (which you will see in Examples 5 to 7). x o p 4 p 2 3p 4 p 5p 4 3p 2 7p 4 2p cos x 1 0 21 0 1 cos 12x) 1 0 21 0 1 0 21 0 1 cos A1 2 xB 1 0 21 STEP 2 Label the points on the graph and connect with a smooth curve. π 2π –1 1 x y π 2 3π 2 1 2 y = cos x y = cos (2x) y = cos ( x) STEP 3 Extend the graph to cover the entire interval: 22p # x # 2p. π –π –2π 2π –1 1 x y 1 2 y = cos x y = cos (2x) y = cos ( x) Y OUR T UR N Plot the functions y 5 sin12x2 and y 5 sinA1 2 xB on the same graph with y 5 sin x on the interval 22p # x # 2p. ▼ ▼ A N S W E R y = sin x y = sin (2x) 1 2 y = sin ( x) –1 1 x y 2π π –π –2π 602 CHAPTER 6 Trigonometric Functions Notice in Example 3 and the corresponding Your Turn that: ■ ■y 5 cos12x2 has the shape and amplitude of y 5 cos x but is compressed horizontally. ■ ■y 5 cosA 1 2xB has the shape and amplitude of y 5 cos x but is stretched horizontally. ■ ■y = sin12x2 has the shape and amplitude of y 5 sin x but is compressed horizontally. ■ ■y 5 sinA1 2xB has the shape and amplitude of y 5 sin x but is stretched horizontally. In general, functions of the form y 5 sin1Bx2 and y 5 cos1Bx2, with B . 0, are compressed horizontally when B . 1 and stretched horizontally when 0 , B , 1. We will discuss negative arguments 1B , 02 later in this section in the context of reflections. The period of the functions y 5 sin x and y 5 cos x is 2p. To find the period of a function of the form y 5 A sin1Bx2 or y 5 A cos1Bx2, set Bx equal to 2p and solve for x. Bx 5 2p x 5 2p B EXAMPLE 4 Finding the Period of a Sinusoidal Function State the period of a. y 5 cos14x2 b. y 5 sinA 1 3xB Solution (a): Compare cos14x2 with cos 1Bx2 to identify B. B 5 4 Calculate the period of cos14x2, using p 5 2p B . p 5 2p 4 5 p 2 The period of cos14x2 is p 5 p 2 . Solution (b): Compare sin a1 3xb with sin 1Bx2 to identify B. B 5 1 3 Calculate the period of sina1 3 xb, using p 5 2p B . p 5 2p 1 3 5 6p The period of sina1 3xb is p 5 6p . Y OUR TU R N State the period of a. y 5 sin13x2 b. y 5 cosA1 2xB ▼ PERIOD OF SINUSOIDAL FUNCTIONS For sinusoidal functions of the form y 5 A sin 1Bx2 and y 5 A cos 1Bx2, with B . 0, the period is 2p B . When 0 , B , 1, the graph is stretched horizontally, and when B . 1, the graph is compressed horizontally since the period is less than 2p. ▼ A N S W E R a. p 5 2p 3 b. p 5 4p 6.8 Graphs of Sine and Cosine Functions 603 Now that you know the basic graphs of y 5 sin x and y 5 cos x, you can sketch one cycle (period) of these graphs with the following x-values: 0, p 2 , p, 3p 2 , 2p. For a period of 2p, we used steps of p 2. Therefore, for functions of the form y 5 A sin1Bx2 or y 5 A cos1Bx2, when we start at the origin and as long as we include these five basic values (corresponding to four equal intervals) during one period, we are able to sketch the graphs. STRATEGY FOR SKETCHING GRAPHS OF SINUSOIDAL FUNCTIONS To graph y 5 A sin1Bx2 or y 5 A cos1Bx2 with B . 0: Step 1: Find the amplitude k A k and period 2p B . Step 2: Divide the period into four subintervals of equal lengths. Step 3: Make a table and evaluate the function for x-values from Step 2 starting at x 5 0. Step 4: Draw the xy-plane (label the y-axis from 2 k A k to k A k 2 and plot the points found in Step 3. Step 5: Connect the points with a sinusoidal curve (with amplitude k A k 2. Step 6: Extend the graph over one or two additional periods in both directions (left and right). STUDY TIP Divide the period by 4 to get the key values along the x-axis for graphing. EXAMPLE 5 Graphing Sinusoidal Functions of the Form y 5 A sin(Bx) Use the strategy for graphing a sinusoidal function to graph y 5 3 sin12x2. Solution: STEP 1 Find the amplitude and period for k A k 5 k 3 k 5 3 A 5 3 and B 5 2. p 5 2p B 5 2p 2 5 p STEP 2 Divide the period p into four equal steps. p 4 STEP 3 Make a table starting at x 5 0 to the period x 5 p in steps of p 4 . x y 5 3 sin 2x (x, y) 0 33sin 04 5 3304 5 0 10, 02 p 4 3csin ap 2 b d 5 3314 5 3 ap 4 , 3b p 2 33sin p4 5 3304 5 0 ap 2 , 0b 3p 4 3csin a3p 2 b d 5 33214 5 23 a3p 4 , 23b p 33sin12p24 5 3304 5 0 1p, 02 604 CHAPTER 6 Trigonometric Functions STEP 4 Draw the xy-plane and label the points in the table. STEP 5 Connect the points with a sinusoidal curve. STEP 6 Repeat over several periods (to the left and right). Y OUR TU R N Use the strategy for graphing sinusoidal functions to graph y 5 2 sin13x2. –3 –2 –1 1 2 3 x y π π 2 π 4 3π 4 (π, 0) (0, 0) π 2 ( , 0) π 4 ( , 3) 3π 4 ( , –3) –3 –2 –1 1 2 3 x y π π 2 π 4 3π 4 –3 –2 –1 1 2 3 x y π 2π –π –2π ▼ ▼ A N S W E R –2 1 –1 2 2 x y π π –π –π 2 EXAMPLE 6 Graphing Sinusoidal Functions of the Form y 5 A cos(Bx) Use the strategy for graphing a sinusoidal function to graph y 5 22 cosA1 3 xB. Solution: STEP 1 Find the amplitude and period k A k 5 k 22 k 5 2 for A 5 22 and B 5 1 3. p 5 2p B 5 2p 1 3 5 6p STEP 2 Divide the period 6p into 6p 4 5 3p 2 four equal steps. 6.8 Graphs of Sine and Cosine Functions 605 Notice in Example 6 and the corresponding Your Turn that when A is negative, the result is a reflection of the original function (sine or cosine) about the x-axis. STEP 3 Make a table starting at x 5 0 and completing one period of 6p in steps of 3p 2 . x y 5 22 cos A1 3xB (x, y) 0 223cos 04 5 22314 5 22 10, 222 3p 2 22ccos ap 2 b d 5 22304 5 0 a3p 2 , 0b 3p 223cos p4 5 223214 5 2 13p, 22 9p 2 22ccos a3p 2 b d 5 22304 5 0 a9p 2 , 0b 6p 223cos12p24 5 22314 5 22 16p, 222 STEP 4 Draw the xy-plane and label the points in the table. STEP 5 Connect the points with a sinusoidal curve. STEP 6 Repeat over several periods (to the left and right). YOUR T UR N Use the strategy for graphing a sinusoidal function to graph y 5 23cos A1 2 xB. 3π 6π –2 –1 1 2 x y (0, –2) (3π, 2) (6π, –2) 3π 2 ( , 0) 9π 2 ( , 0) 3π 6π –2 –1 1 2 x y 6π 3π –6π –12π 9π 12π –2 –1 1 2 x y ▼ ▼ A N S W E R x 6π 4π 2π –6π –2π –3 –2 –1 1 2 3 606 CHAPTER 6 Trigonometric Functions 6.8.2 Graphing a Shifted Sinusoidal Function: y 5 A sin(Bx 1 C) 1 D and y 5 A cos(Bx 1 C) 1 D Recall from Section 3.3 that we graph functions using horizontal and vertical transla-tions (shifts) in the following way 1c . 02: ■ ■To graph ƒ1x 1 c2, shift ƒ1x2 to the left c units. ■ ■To graph ƒ1x 2 c2, shift ƒ1x2 to the right c units. ■ ■To graph ƒ1x2 1 c, shift ƒ1x2 up c units. ■ ■To graph ƒ1x2 2 c, shift ƒ1x2 down c units. EXAMPLE 7 Finding an Equation for a Sinusoidal Graph Find an equation for the graph. Solution: This graph represents a cosine function. y 5 Acos1Bx2 The amplitude is 4 (half the maximum spread). k A k 5 4 The period 2p B is equal to 4p. 2p B 5 4p Solve for B. B 5 1 2 Substitute A 5 4 and B 5 1 2 into y 5 A cos1Bx2. y 5 4 cos a1 2 xb Y OUR TU R N Find an equation for the graph. 2π –2π –4π 4π –4 –3 –2 –1 2 4 1 3 x y Period x –6 –4 –2 2 4 6 y π 4 π π –π –π 2 2 3π 4 ▼ ▼ A N S W E R y 5 6 sin12x2 6.8.2 S KILL Determine the phase shift of a sinusoidal function. 6.8.2 CO NCE PTUAL Rewriting the sinusoidal function in standard form makes identifying the phase shift easier. 6.8 Graphs of Sine and Cosine Functions 607 To graph functions of the form y 5 A sin1Bx 1 C2 1 D and y 5 A cos 1Bx 1 C2 1 D, utilize the strategy below. STRATEGY FOR GRAPHING y 5 A sin(Bx 1 C) 1 D AND y 5 A cos(Bx 1 C) 1 D A strategy for graphing y 5 A sin1Bx 1 C2 1 D is outlined below. The same strategy can be used to graph y 5 A cos1Bx 1 C2 1 D. Step 1: Find the amplitude k A k . Step 2: Find the period 2p B and phase shift 2C B. Step 3: Graph y 5 A sin1Bx 1 C2 over one period afrom 2C B to 2C B 1 2p B b. Step 4: Extend the graph over several periods. Step 5: Shift the graph of y 5 A sin1Bx 1 C2 vertically D units. Note: If we rewrite the function in standard form, we get y 5 A sin cBax 1 C Bb d 1 D which makes it easier to identify the phase shift. If B , 0, we can use properties of even and odd functions sin12x2 5 2sin x cos12x2 5 cos x to rewrite the function with B . 0. EXAMPLE 8 Graphing Functions of the Form y 5 A cos(Bx 6 C ) Graph y 5 5 cos14x 1 p2 over one period. Solution: STEP 1 Find the amplitude. k A k 5 k 5 k 5 5 STEP 2 Calculate the period and phase shift. The interval for one period is from 0 to 2p. 4x 1 p 5 0 to 4x 1 p 5 2p Solve for x. x 5 2p 4 to x 5 2p 4 1 p 2 Identify the phase shift. 2C B 5 2p 4 Identify the period 2p B 2p B 5 2p 4 5 p 2 STEP 3 Graph. Draw a cosine function starting at x 5 2 p 4 with period p 2 and amplitude 5. YOUR T UR N Graph y 5 3 cos12x 2 p2 over one period. π 4 –5 –4 –3 –2 –1 1 2 3 4 5 x y π 8 –π 4 ▼ ▼ A N S W E R 4 π π 2 –3 –2 –1 1 2 3 x y 3π π 4 STUDY TIP Rewriting in standard form y 5 A sin cBax 6 C Bb d makes identifying the phase shift easier. STUDY TIP An alternative method for finding the period and phase shift is to first write the function in standard form. y 5 5 cosc4ax 1 p 4 b d B 5 4 Period 5 2p B 5 2p 4 5 p 2 Phase shift 5 p 4 units to the left. [CONCEPT CHECK] Rewrite y 5 A sin(Bx 6 C) in standard form. ANSWER y 5 A sin cBax 6 C Bb d ▼ 608 CHAPTER 6 Trigonometric Functions EXAMPLE 9 Graphing Sinusoidal Functions Graph y 5 23 1 2 cos12x 2 p2. Solution: STEP 1 Find the amplitude. k A k 5 k 2 k 5 2 STEP 2 Find the phase shift and period. Set 2x 2 p equal to 0 and 2p. 2x 2 p 5 0 to 2x 2 p 5 2p Solve for x. x 5 p 2 to x 5 p 2 1 p Identify phase shift and period. 2C B 5 p 2 2p B 5 p STEP 3 Graph y 5 2 cos12x 2 p2 starting at x 5 p 2 over one period, p. STEP 4 Extend the graph of y 5 2 cos12x 2 p2 over several periods. STEP 5 Shift the graph of y 5 2 cos12x 2 p2 down three units to arrive at the graph of y 5 23 1 2 cos12x 2 p2. Y OUR TU R N Graph y 5 2 2 3 sin12x 1 p2. π π 2π 3π 2 2 –3 –2 –1 3 2 1 x y π π 2π 3π 2 2 –π –π 2 –3 –2 3 2 1 y x π π 2π 3π 2 2 –π –π 2 x y –5 –3 –2 –4 1 2 3 4 5 ▼ ▼ A N S W E R 2 –π 2 3π π π 2 x –5 –4 –3 –2 –1 1 2 3 4 5 y 6.8 Graphs of Sine and Cosine Functions 609 6.8.3 Harmonic Motion One of the most important applications of sinusoidal functions is in describing harmonic motion, which we define as the symmetric periodic movement of an object or quantity about a center (equilibrium) position or value. The oscillation of a pendulum is a form of harmonic motion. Other examples are the recoil of a spring balance scale when a weight is placed on the tray and the variation of current or voltage within an AC circuit. There are three types of harmonic motion: simple harmonic motion, damped harmonic motion, and resonance. Simple Harmonic Motion Simple harmonic motion is the kind of unvarying periodic motion that would occur in an ideal situation in which no resistive forces, such as friction, cause the amplitude of oscillation to decrease over time: the amplitude stays in exactly the same range in each period as time—the variable on the horizontal axis—increases. It will also occur if energy is being supplied at the correct rate to overcome resistive forces. Simple harmonic motion occurs, for example, in an AC electric circuit when a power source is consistently supplying energy. When you are swinging on a swing and “pumping” energy into the swing to keep it in motion at a constant period and amplitude, you are sustaining simple harmonic motion. Damped Harmonic Motion In damped harmonic motion, the amplitude of the periodic motion decreases as time increases. If you are on a moving swing and stop “pumping” new energy into the swing, the swing will continue moving with a constant period, but the amplitude— the height to which the swing will rise—will diminish with each cycle as the swing is slowed down by friction with the air or between its own moving parts. Resonance Resonance occurs when the amplitude of periodic motion increases as time increases. It is caused when the energy applied to an oscillating object or system is more than what is needed to oppose friction or other forces and sustain simple harmonic motion. Instead, the applied energy increases the amplitude of harmonic motion with each cycle. With resonance, eventually, the amplitude becomes unbounded and the result is disastrous. Bridges have collapsed because of resonance. Military soldiers know that when they march across a bridge, they must break cadence to prevent resonance. The following pictures are of the Tacoma Narrows Bridge (near Seattle, Washington) that was opened to traffic on July 1, 1940 and collapsed on November 7, 1940. The collapsing of the Tacoma Narrows bridge is often presented as an example of resonance (the wind providing an external periodic frequency that coincided with the bridges natural frequency) but many believe the actual cause of failure was torsional motion that twisted cubes. 6.8.3 S K I L L Solve harmonic motion problems. 6.8.3 C ON C E P T U A L Visualize harmonic motion as a sinusoidal function. time y time y time y 610 CHAPTER 6 Trigonometric Functions Examples of Harmonic Motion If we hang a weight from a spring, then while the resulting “system” is at rest we say it is in the equilibrium position. If we then pull down on the weight and release it, the elasticity in the spring pulls the weight up and causes it to start oscillating up and down. If we neglect friction and air resistance, we can imagine that the combination of the weight and the spring will oscillate indefinitely; the height of the weight with respect to the equilibrium position can be modeled by a simple sinusoidal function. This is an example of simple harmonic motion. SIMPLE HARMONIC MOTION The position of a point oscillating around an equilibrium position at time t is modeled by the sinusoidal function y 5 A sin1vt2 or y 5 Acos1vt2 Here k A k is the amplitude and the period is 2p v , where v . 0. Note: The symbol v (Greek lowercase Omega) represents the angular frequency. Keystone/Stringer/Getty Images © TopFoto/The Image Works 6.8 Graphs of Sine and Cosine Functions 611 EXAMPLE 10 Simple Harmonic Motion Let the height of the seat of a swing be equal to zero when the swing is at rest. Assume that a child starts swinging until she reaches the highest she can swing and keeps her effort constant. The height h1t2 of the seat is given by h1t2 5 8 sin ap 2 tb h where t is time in seconds and h is the height in feet. Note that positive h indicates height reached swinging forward and negative h indicates height reached swinging backward. Assume that t 5 0 is when the child passes through the equilibrium position swinging forward. a. Graph the height function h1t2 for 0 # t # 4. b. What is the maximum height above the resting level reached by the seat of the swing? c. What is the period of the swinging child? Solution (a): Make a table with integer values of t. 0 # t # 4 t (SECONDS) y 5 h(t ) 5 8 sinAp 2tB (FEET) (t, y) 0 8 sin 0 5 0 10, 02 1 8 sin ap 2 b 5 8 11, 82 2 8 sin p 5 0 12, 02 3 8 sin a3p 2 b 5 28 13, 282 4 8 sin 12p2 5 0 14, 02 (0, 0) (2, 0) (4, 0) (1, 8) (3, –8) 1 2 3 4 –8 –6 –4 –2 2 4 6 8 t y [CONCEPT CHECK] The scenario of a tennis ball being dropped vertically onto the ground and it continuing to bounce a few times before eventually coming to rest is an example of: (A) simple harmonic motion, (B) damped harmonic motion, or (C) resonance. ANSWER (B) ▼ 612 CHAPTER 6 Trigonometric Functions Damped harmonic motion can be modeled by a sinusoidal function whose amplitude decreases as time increases. If we again hang a weight from a spring so that it is suspended at rest and then pull down on the weight and release, the weight will oscillate about the equilibrium point. This time we will not neglect friction and air resistance: the weight will oscillate closer and closer to the equilibrium point over time until the weight eventually comes to rest at the equilibrium point. This is an example of damped harmonic motion. The product of any decreasing function and the original periodic function will describe damped oscillatory motion. Here are two examples of functions that describe damped harmonic motion: y 5 1 t sin1vt2 y 5 e2t cos1vt2 where e2t is a decreasing exponential function (exponential decay). Labeling the time and height on the original diagram, we see that the maximum height is 8 feet and the period is 4 seconds. Solutions (b) and (c): h1t2 5 8 sin ap 2 tb 5 8 sin a2p 4 tb The maximum height above the equilibrium is the amplitude. A 5 8 ft The period is 2p v . p 5 2p 2p 4 5 4 sec 8 ft t = 1 sec t = 3 sec t = 0 sec = 2 sec = 4 sec { A { v EXAMPLE 11 Damped Harmonic Motion Assume that the child in Example 10 decides to stop pumping and allows the swing to continue until she eventually comes to rest. Assume that h1t2 5 8 t cosap 2 tb where t is time in seconds and h is the height in feet above the resting position. Note that positive h indicates height reached swinging forward and negative h indicates height reached swinging backward, assuming that t 5 1 is when the child passes through the equilibrium position swinging backward and stops “pumping.” a. Graph the height function h1t2 for 1 # t # 8. b. What is the height above the resting level at 4 seconds? At 8 seconds? After 1 minute? 6.8 Graphs of Sine and Cosine Functions 613 Solution (a): Make a table with integer values of t. 1 # t # 8 t (SECONDS) y 5 h(t ) 5 8 t cosAp 2tB (FEET) (t, y) 1 8 1 cos ap 2 b 5 0 11, 02 2 8 2 cos p 5 24 12, 242 3 8 3 cos a3p 2 b 5 0 13, 02 4 8 4 cos 12p2 5 2 14, 22 5 8 5 cos a5p 2 b 5 0 15, 02 6 8 6 cos 13p2 5 24 3 a6, 24 3b 7 8 7 cos a7p 2 b 5 0 17, 02 8 8 8 cos 114p2 5 1 18, 12 (5, 0) (7, 0) (8, 1) 6 (2, –4) (4, 2) (1, 0) (3, 0) 2 4 6 8 –4 –3 –2 –1 1 2 3 4 3 4 t y ( , – ) Solution (b): 8 ft t = 6 sec t = 8 sec t = 4 sec t = 2 sec t = 1 sec = 3 sec = 5 sec = 7 sec The height is 2 feet when t is 4 seconds. 8 4 cos12p2 5 2 The height is 1 foot when t is 8 seconds. 8 8 cos 14p2 5 1 The height is 0.13 feet when t is 1 minute (60 seconds). 8 60 cos 130p2 5 0.1333 614 CHAPTER 6 Trigonometric Functions Resonance can be represented by the product of any increasing function and the original sinusoidal function. Here are two examples of functions that result in resonance as time increases: y 5 t cos1vt2 y 5 et sin1vt2 6.8.4 Graphing Sums of Functions: Addition of Ordinates Since you have the ability to graph sinusoidal functions, let us now consider graphing sums of functions such as y 5 x 2 sin a px 2 b y 5 sin x 1 cos x y 5 3 sin x 1 cos 12x2 The method for graphing these sums is called the addition of ordinates because we add the corresponding y-values (ordinates). The following table illustrates the ordinates (y-values) of the two sinusoidal functions sin x and cos x; adding the corresponding ordinates leads to the y-values of y 5 sin x 1 cos x. x sin x cos x y 5 sin x 1 cos x 0 0 1 1 p 4 !2 2 !2 2 !2 p 2 1 0 1 3p 4 !2 2 2 !2 2 0 p 0 21 21 5p 4 2 !2 2 2 !2 2 2!2 3p 2 21 0 21 7p 4 2 !2 2 !2 2 0 2p 0 1 1 Using a graphing utility, we can graph Y1 5 sin X, Y2 5 cos X, and Y3 5 Y1 1 Y2. π π 2π 2 3π 2 –2 –1 1 2 x y y = sinx y = cosx y = sinx + cosx EXAMPLE 12 Graphing Sums of Functions Graph y 5 x 2 sin a px 2 b on the interval 0 # x # 4. Solution: Let y1 5 x and y2 5 2sin apx 2 b. 6.8.4 S KILL Graph sums of trigonometric and other algebraic functions. 6.8.4 CO NCE PTUAL Understand that the y-coordinates of the combined function are found by adding the y-coordinates of the individual functions. 6.8 Graphs of Sine and Cosine Functions 615 State the amplitude and period of the graph of y2. k A k 5 k 21 k 5 1, p 5 4 Make a table of x- and y-values of y1, y2, and y 5 y1 1 y2. Graph y1 5 x, y2 52sin apx 2 b, and y 5 x 2 sin apx 2 b. x y1 5 x y2 5 2sinapx 2 b y 5 x 1 c 2sinapx 2 b d 0 0 0 0 1 1 21 0 2 2 0 2 3 3 1 4 4 4 0 4 x y –1 1 2 3 4 –1 1 2 3 4 (4, 4) (3, 4) (2, 2) EXAMPLE 13 Graphing Sums of Sine and Cosine Functions Graph y 5 3 sin x 1 cos 12x2 on the interval 0 # x # 2p. Solution: Let y1 5 3 sin x, and state the amplitude and period of its graph. k A k 5 3, p 5 2p Let y2 5 cos 12x2, and state the amplitude and period of its graph. k A k 5 1, p 5 p Make a table of x- and y-values of y1, y2, and y 5 y1 1 y2. x y1 5 3 sin x y2 5 cos (2x) y 5 3 sin x 1 cos (2x) 0 0 1 1 p 4 3!2 2 0 3!2 2 p 2 3 21 2 3p 4 3!2 2 0 3!2 2 p 0 1 1 5p 4 23!2 2 0 23!2 2 3p 2 23 21 24 7p 4 23!2 2 0 23!2 2 2p 0 1 1 616 CHAPTER 6 Trigonometric Functions Graph y1 5 3 sin x, y2 5 cos12x2, and y 5 3 sin x 1 cos12x2. π 2π –4 –2 –3 –1 2 4 1 3 x y π 2 3π 2 EXAMPLE 14 Graphing Sums of Cosine Functions Graph y 5 cos ax 2b 2 cos x on the interval 0 # x # 4p. Solution: Let y1 5 cos a x 2 b and state the amplitude and period of its graph. k A k 5 1, p 5 4p Let y2 52cos x and state the amplitude and period of its graph. k A k 5 k 21 k 5 1, p 5 2p Make a table of x- and y-values of y1, y2, and y 5 y1 1 y2. Graph y1 5 cos ax 2b, y2 5 2cos x, and y 5 cos ax 2b2cos x. π 4π 3π –2 –1 1 2 x y 2π (π, 1) (3π, 1) (2π, –2) x y1 5 cos Ax 2B y2 5 2cos x y 5 cos Ax 2B 1 [2cos x] 0 1 21 0 p 2 !2 2 0 !2 2 p 0 1 1 3p 2 2 !2 2 0 2 !2 2 2p 21 21 22 5p 2 2 !2 2 0 2 !2 2 3p 0 1 1 7p 2 !2 2 0 !2 2 4p 1 21 0 [CONCEPT CHECK] For addition of ordinates, divide the smallest period of the periodic functions by to get the corre-sponding ordinates to add. ANSWER 4 ▼ 6.8 Graphs of Sine and Cosine Functions 617 [SEC TION 6.8] E X E RC I S E S • S K I L L S To find an equation of a sinusoidal function given its graph, start by first finding the amplitude (half the distance between the maximum and minimum values) so you can find A. Then determine the period so you can find B. Graphs of the form y 5 A sin1Bx 1 C2 1 D and y 5 A cos1Bx 1 C2 1 D can be graphed using graph-shifting techniques. Harmonic motion is one of the primary applications of sinusoidal functions. To graph combinations of trigonometric functions, add the corresponding y-values of the individual functions. The sine function is an odd function, and its graph is symmetric about the origin. The cosine function is an even function, and its graph is symmetric about the y-axis. Graphs of the form y 5 A sin1Bx2 and y 5 A cos1Bx2 have amplitude 0 A 0 and period 2p B . To graph sinusoidal functions, point-plotting can be used. A more efficient way is to first determine the amplitude and period. Divide the period into four equal parts and choose the values of the division points starting at 0 for x. Make a table of those four points and graph them (this is the graph of one period) by labeling the four coordinates and drawing a smooth sinusoidal curve. Extend the graph to the left and right. [SEC TION 6.8] S UM M A RY In Exercises 1–10, match the function with its graph (a–j). 1. y 5 2sin x 2. y 5 sin x 3. y 5 cos x 4. y 5 2cos x 5. y 5 2 sin x 6. y 5 2 cos x 7. y 5 sin A1 2 x B 8. y 5 cos A 1 2 xB 9. y 522 cos A1 2xB 10. y 522 sinA 1 2 xB a. 2π π 4π 3π –2 –1 1 2 x y b. 2π π 4π 3π –2 –1 1 2 x y c. 2π π 4π 3π –2 –1 1 2 x y d. 2π π 4π 3π –2 –1 1 2 x y e. 2π π 4π 3π –2 –1 1 2 x y f. 2π π 4π 3π –2 –1 1 2 x y g. 2π π 4π 3π –2 –1 1 2 x y h. 2π π 4π 3π –2 –1 1 2 x y i. 2π π 4π 3π –2 –1 1 2 x y j. 2π π 4π 3π –2 –1 1 2 x y 618 CHAPTER 6 Trigonometric Functions In Exercises 11–20, state the amplitude and period of each function. 11. y 5 3 2 cos1 3x2 12. y 5 2 3 sin1 4x2 13. y 5 2sin15x2 14. y 5 2cos17x2 15. y 5 2 3 cos a3 2 xb 16. y 5 3 2 sin a2 3xb 17. y 5 23 cos1px2 18. y 5 22 sin1px2 19. y 5 5 sin ap 3 xb 20. y 5 4 cos ap 4 xb In Exercises 21–32, graph the given function over one period. 21. y 5 8 cos x 22. y 5 7 sin x 23. y 5 sin14x2 24. y 5 cos13x2 25. y 5 23 cos a1 2 xb 26. y 5 22 sin a1 4 xb 27. y 5 23 sin1px2 28. y 5 22 cos1px2 29. y 5 5 cos12px2 30. y 5 4 sin12px2 31. y 5 23 sin ap 4 xb 32. y 5 24 sin ap 2 xb In Exercises 33–40, graph the given function over the interval [22p, 2p], where p is the period of the function. 33. y 5 24 cos a1 2 xb 34. y 5 25sin a1 2 xb 35. y 5 2sin 16x2 36. y 5 2cos 14x2 37. y 5 3 cos ap 4 xb 38. y 5 4 sin ap 4 xb 39. y 5 sin14px2 40. y 5 cos16px2 In Exercises 41–48, find the equation for each graph. 41. x π 4 –π 4 –3π 4 3π 4 –1 1 y 42. x π 4 –π 4 –3π 4 3π 4 –2 –1 2 1 y 43. –4 –2 1 2 3 4 –1 1 x y 44. –4 –2 1 2 3 4 –1 1 x y 45. –4 –2 1 2 3 4 –2 –1 2 1 x y 46. –6 –4 –2 1 2 3 4 5 6 –3 –2 –1 1 2 3 x y 47. –1 –0.5 0.5 1 –1 1 x y 48. –1 –0.5 0.5 1 –1 1 x y In Exercises 49–60, state the amplitude, period, and phase shift (including direction) of the given function. 49. y 5 2 sin1p x 2 12 50. y 5 4 cos1x 1 p2 51. y 5 25 cos13x 1 22 52. y 5 27 sin14x 2 32 53. y 5 6 sin32p1x 1 224 54. y 5 3 sinc2 p 2 ax 2 1b d 55. y 5 3 sin12x 1 p2 56. y 5 24 cos12x 2 p2 57. y 5 2 1 4 cos a1 4 x 2 p 2 b 58. y 5 1 2 sin a1 3 x 1 pb 59. y 5 2 cos ap 2 1x 2 42b 60. y 5 25 sin12p 1x 1 122 6.8 Graphs of Sine and Cosine Functions 619 In Exercises 61–66, sketch the graph of the function over the indicated interval. 61. y 5 1 2 1 3 2 cos32x 1 p4, c 2 3p 2 , 3p 2 d 62. y 5 1 3 1 2 3 sin32 x 2 p4, c 23p 2 , 3p 2 d 63. y 5 1 2 2 1 2 sin c 1 2 x 2 p 4 d , c 2 7p 2 , 9p 2 d 64. y 5 2 1 2 1 1 2 cos c 1 2 x 1 p 4 d , c 29p 2 , 7p 2 d 65. y 5 23 1 4 sin 3p1x 2 224, 30, 44 66. y 5 4 2 3 cos 3p 1x 1 124, 321, 34 In Exercises 67–94, add the ordinates of the individual functions to graph each summed function on the indicated interval. 67. y 5 2x 2 cos1px2, 0 # x # 4 68. y 5 3x 2 2 cos1px2, 0 # x # 4 69. y 5 1 3 x 1 2 cos12x2, 0 # x # 2p 70. y 5 1 4 x 1 3 cos a x 2 b, 0 # x # 4p 71. y 5 x 2 cos a3p 2 xb, 0 # x # 6 72. y 5 22x 1 2 sin ap 2 xb, 22 # x # 2 73. y 5 1 4 x 2 1 2 cos3p1x 2 124, 2 # x # 6 74. y 5 2 1 3 x 1 1 3 sinc p 6 1x 1 22 d , 22 # x # 10 75. y 5 sin x 2 cos x, 0 # x # 2p 76. y 5 cos x 2 sin x, 0 # x # 2p 77. y 5 3 cos x 1 sin x, 0 # x # 2p 78. y 5 3 sin x 2 cos x, 0 # x # 2p 79. y 5 4 cos x 2 sin12x2, 0 # x # 2p 80. y 5 1 2 sin x 1 2 cos14x2, 2p # x # p 81. y 5 2 sin3p1x 2 124 2 2 cos3p1x 1 124, 21 # x # 2 82. y 5 sinc p 4 1x 1 22 d 1 3 cos c 3p 3 1x 2 12 d , 1 # x # 5 83. y 5 cos a x 2b 1 cos1 2x2, 0 # x # 4p 84. y 5 sin12x2 1 sin13x2, 2p # x # p 85. y 5 sin a x 2 b 1 sin12x2, 0 # x # 4p 86. y 5 2sin ap 4 xb 2 3 sin a5p 4 xb, 0 # x # 4 87. y 5 2 1 3 sin ap 6 xb 1 2 3 sin a5p 6 xb, 0 # x # 3 88. y 5 8 cos x 2 6 cos a1 2 xb, 22p # x # 2p 89. y 5 2 1 4 cos ap 6 xb 2 1 2 cos ap 3 xb, 0 # x # 12 90. y 5 2 cos a3 2 xb 2 cos a1 2 xb, 22p # x # 2p 91. y 5 2 sin a x 2b 2 cos12x2, 0 # x # 4p 92. y 5 2 cos a x 2b 1 sin 1 2x2, 0 # x # 4p 93. y 5 2 sin1p 1x 2 122 1 3 sinc2p ax 1 1 2b d , 22 # x # 2 94. y 5 2 1 2 cos ax 1 p 3 b 2 2 cos ax 2 p 6 b, 2p # x # p • A P P L I C A T I O N S For Exercises 95 and 96, refer to the following: An analysis of demand d for widgets manufactured by WidgetsRUs (measured in thousands of units per week) indicates that demand can be modeled by the graph below, where t is time in months since January 2017 (note that t 5 0 corresponds to January 2017). 95. Business. Find the amplitude of the graph. 96. Business. Find the period of the graph. 620 CHAPTER 6 Trigonometric Functions For Exercises 97 and 98, refer to the following: Researchers have been monitoring oxygen levels (milligrams per liter) in the water of a lake and have found that the oxygen levels fluctuate with an eight-week period. The following tables illustrate data from eight weeks. 97. Environment. Find the amplitude of the oxygen-level fluctuations. WEEK: t 0 (initial measurement) 1 2 3 4 5 6 7 8 OXYGEN LEVELS: MG/L 7 7.7 8 7.7 7 6.3 6 6.3 7 98. Environment. Find the amplitude of the oxygen-level fluctuations. WEEK: t 0 (initial measurement) 1 2 3 4 5 6 7 8 OXYGEN LEVELS: MG/L 7 8.4 9 8.4 7 5.6 5 5.6 7 For Exercises 99–102, refer to the following: A weight hanging on a spring will oscillate up and down about its equilibrium position after it is pulled down and released. This is an example of simple harmonic motion. This motion would continue forever if there were not any friction or air resistance. Simple harmonic motion can be described with the function y 5 A cos atÅ k mb, where 0 A 0 is the amplitude, t is the time in seconds, m is the mass of the weight, and k is a constant particular to the spring. 99. Simple Harmonic Motion. If the height of the spring is mea-sured in centimeters and the mass in grams, then what are the amplitude and mass if y 5 4 cos at!k 2 b? 100. Simple Harmonic Motion. If a spring is measured in cen-timeters and the mass in grams, then what are the amplitude and mass if y 5 3 cosA3t!kB? 101. Frequency of Oscillations. The frequency of the oscillations in cycles per second is determined by ƒ 5 1 p, where p is the period. What is the frequency for the oscillation modeled by y 5 3 cos a t 2b? 102. Frequency of Oscillations. The frequency of the oscillations ƒ is given by ƒ 5 1 p, where p is the period. What is the frequency of oscillation modeled by y 5 3.5 cos13t2? 103. Sound Waves. A pure tone created by a vibrating tuning fork shows up as a sine wave on an oscilloscope’s screen. A tuning fork vibrating at 256 hertz gives the tone middle C and can have the equation y 5 0.005 sin 2p1256t2, where the amplitude is in centimeters and the time t in seconds. What are the amplitude and frequency of the wave where the frequency is 1 p in cycles per second? 104. Sound Waves. A pure tone created by a vibrating tuning fork shows up as a sine wave on an oscilloscope’s screen. A tun-ing fork vibrating at 288 hertz gives the tone D and can have the equation y 5 0.005 sin 2p1288t2, where the amplitude is in centimeters and the time t in seconds. What are the ampli-tude and frequency of the wave where the frequency is 1 p in cycles per second? 105. Sound Waves. If a sound wave is represented by y 5 0.008 sin 1750 pt2 cm, what are its amplitude and frequency? See Exercise 103. 106. Sound Waves. If a sound wave is represented by y 5 0.006 cos 11000 pt2 cm, what are its amplitude and frequency? See Exercise 103. For Exercises 107–110, refer to the following: When an airplane flies faster than the speed of sound, the sound waves that are formed take on a cone shape, and where the cone hits the ground, a sonic boom is heard. If u is the angle of the vertex of the cone, then sin u 2 5 330 m/sec V 5 1 M, where V is the speed of the plane and M is the Mach number. θ 107. Sonic Booms. What is the speed of the plane if the plane is flying at Mach 2? 108. Sonic Booms. What is the Mach number if the plane is flying at 990 m/sec? 109. Sonic Booms. What is the speed of the plane if the cone angle is 60°? 110. Sonic Booms. What is the speed of the plane if the cone angle is 30°? For Exercises 111 and 112, refer to the following: With the advent of summer come fireflies. They are intriguing because they emit a flashing luminescence that beckons their mate to them. It is known that the speed and intensity of the flashing are related to the temperature—the higher the temperature, the quicker and more intense the flashing becomes. If you ever watch a single firefly, you will see that the intensity of the flashing is periodic with time. The intensity of light emitted is measured in candelas per square meter (of firefly). To give an idea of this unit 6.8 Graphs of Sine and Cosine Functions 621 • C A T C H T H E M I S T A K E In Exercises 113 and 114, explain the mistake that is made. 113. Graph the function y 5 22 cos x. Solution: Find the amplitude. k A k 5 k 22 k 5 2 The graph of y 5 22 cos x is similar to the graph of y 5 cos x with amplitude 2. x π –π –2π 2π –2 –1 2 1 y This is incorrect. What mistake was made? 114. Graph the function y 5 2sin 12x2. Solution: Make a table with values. x y 5 2sin(2x) (x, y) 0 y 5 2sin102 5 0 10, 02 p 2 y 5 2sin1p2 5 0 10, 02 p y 5 2sin12p2 5 0 10, 02 3p 2 y 5 2sin13p2 5 0 10, 02 2p y 5 2sin14p2 5 0 10, 02 Graph the function by plotting these points and connecting them with a sinusoidal curve. x π –π –2π 2π –1 1 y This is incorrect. What mistake was made? In Exercises 115–118, determine whether each statement is true or false. (A and B are positive real numbers.) 115. The graph of y 5 2A cos1Bx2 is the graph of y 5 A cos Bx reflected about the x-axis. 116. The graph of y 5 A sin12Bx2 is the graph of y 5 A sin1Bx2 reflected about the x-axis. 117. The graph of y 5 2A cos12Bx2 is the graph of y 5 A cos1Bx2. 118. The graph of y 5 2A sin12Bx2 is the graph of y 5 A sin1Bx2. In Exercises 119–122, A and B are positive real numbers. 119. Find the y-intercept of the function y 5 A cos Bx. 120. Find the y-intercept of the function y 5 A sin Bx. 121. Find the x-intercepts of the function y 5 A sin Bx. 122. Find the x-intercepts of the function y 5 A cos Bx. • C O N C E P T U A L of measure, the intensity of a picture on a typical TV screen is about 450 candelas. The measurement for the intensity of the light emitted by a typical firefly at its brightest moment is about 50 candelas. Assume that a typical cycle of this flashing is 4 seconds and that the intensity is essentially zero candelas at the beginning and ending of a cycle. 111. Bioluminescence in Fireflies. Find an equation that describes this flashing. What is the intensity of the flashing at 4 minutes? 112. Bioluminescence in Fireflies. Graph the equation from Exercise 111 for a period of 30 seconds. 622 CHAPTER 6 Trigonometric Functions 123. Find the y-intercept of y 5 2A sin aBx 1 p 6 b. 124. Find the y-intercept of y 5 A cos1Bx 2 p2 1 C. 125. Find the x-intercept(s) of y 5 A sin1Bx2 1 A. 126. Find an expression involving C and A that describes the values of C for which the graph of y 5 A cos1Bx2 1 C does not cross the x-axis. (Assume that A . 0.) 127. What is the range of y 5 2A sin1Bx 1 C 2 2 A 2? 128. Can the y-coordinate of a point on the graph of y 5 A sin1Bx2 1 3 A cos aB 2 xb exceed 4A? Explain. (Assume that A . 0.) • C H A L L E N G E 129. Use a graphing calculator to graph Y1 5 5 sin x and Y2 5 sin15x2. Is the following statement true based on what you see? y 5 sin 1cx2 has the same graph as y 5 c sin x. 130. Use a graphing calculator to graph Y1 5 3 cos x and Y2 5 cos13x2. Is the following statement true based on what you see? y 5 cos1cx2 has the same graph as y 5 c cos x. 131. Use a graphing calculator to graph Y1 5 sin x and Y2 5 cos ax 2 p 2 b. What do you notice? 132. Use a graphing calculator to graph Y1 5 cos x and Y2 5 sin ax 1 p 2 b. What do you notice? 133. Use a graphing calculator to graph Y1 5 cos x and Y2 5 cos 1x 1 c2, where a. c 5 p 3 , and explain the relationship between Y2 and Y1. b. c 5 2 p 3 , and explain the relationship between Y2 and Y1. 134. Use a graphing calculator to graph Y1 5 sin x and Y2 5 sin1x 1 c2, where a. c 5 p 3 , and explain the relationship between Y2 and Y1. b. c 5 2 p 3 , and explain the relationship between Y2 and Y1. For Exercises 135 and 136, refer to the following: Damped oscillatory motion, or damped oscillation, occurs when things in oscillatory motion experience friction or resistance. The friction causes the amplitude to decrease as a function of time. Mathematically, we can use a negative exponential function to damp the oscillations in the form of ƒ1t2 5 e2t sin t 135. Damped Oscillation. Graph the functions Y1 5 e2t, Y2 5 sin t, and Y3 5 e2t sin t in the same viewing window (let t range from 0 to 2p2. What happens as t increases? 136. Damped Oscillation. Graph Y1 5 e2t sin t, Y2 5 e22t sin t, and Y3 5 e24t sin t in the same viewing window. What happens to Y 5 e2kt sin t as k increases? 137. Use a graphing calculator to graph Y1 5 sin x and Y2 5 sin x 1 c, where a. c 5 1, and explain the relationship between Y2 and Y1. b. c 5 21, and explain the relationship between Y2 and Y1. 138. Use a graphing calculator to graph Y1 5 cos x and Y2 5 cos x 1 c, where a. c 5 1 2, and explain the relationship between Y2 and Y1. b. c 5 21 2, and explain the relationship between Y2 and Y1. 139. What is the amplitude of the function y 5 3 cos x 1 4 sin x? Use a graphing calculator to graph Y1 5 3 cos x, Y2 5 4 sin x, and Y3 5 3 cos x 1 4 sin x in the same viewing window. 140. What is the amplitude of the function y 5 !3 cos x 2 sin x? Use a graphing calculator to graph Y1 5!3 cos x, Y2 5 sin x, and Y3 5 !3 cos x 2 sin x in the same viewing window. • T E C H N O L O G Y 6.9 Graphs of Other Trigonometric Functions 623 6.9.1 Graphing the Tangent, Cotangent, Secant, and Cosecant Functions Section 6.8 focused on graphing sinusoidal functions (sine and cosine). We now turn our attention to graphing the other circular functions: tangent, cotangent, secant, and cosecant. We know the graphs of the sine and cosine functions, and we can get the graphs of the other circular functions from the sinusoidal functions. Recall the reciprocal and quotient identities: tan x 5 sin x cos x cot x 5 cos x sin x sec x 5 1 cos x csc x 5 1 sin x Recall that in graphing rational functions, a vertical asymptote corresponds to a denominator equal to zero (as long as the numerator and denominator have no common factors). As you will see in this section, tangent and secant functions have graphs with vertical asymptotes at the x-values where the cosine function is equal to zero, and cotangent and cosecant functions have graphs with vertical asymptotes at the x-values where the sine function is equal to zero. One important difference between the sinusoidal functions, y 5 sin x and y 5 cos x, and the other four trigonometric functions 1y 5 tan x, y 5 sec x, y 5 csc x, and y 5 cot x2 is that the sinusoidal functions have defined amplitudes, whereas the other four trigonometric functions do not (since they are unbounded vertically). The Tangent Function Since the tangent function is a quotient that relies on sine and cosine, let us start with a table of values for the quadrantal angles. S K I L L S O B J E C T I V E S ■ ■Graph basic tangent, cotangent, secant, and cosecant functions. ■ ■Graph translated tangent, cotangent, secant, and cosecant functions. C O N C E P T U A L O B J E C T I V ES ■ ■Understand the relationships between the graphs of cosine and secant functions and the sine and cosecant functions. ■ ■Understand that graph-shifting techniques for tangent and cotangent are consistent with translations used for sinusoidal functions; but for secant and cosecant functions, we first graph the horizontally translated sinusoidal functions and then we shift up or down depending on the vertical translations. 6.9 GRAPHS OF OTHER TRIGONOMETRIC FUNCTIONS 6.9.1 S K I L L Graph basic tangent, cotangent, secant, and cosecant functions. 6.9.1 C ON C E P T U A L Understand the relationships between the graphs of cosine and secant functions and the sine and cosecant functions. [CONCEPT CHECK] If the range of the graph of a sinusoidal function is [2A, A], what is the range of the corresponding secant or cosecant functions? ANSWER (2q, 2A] ∪ [A, q ) ▼ x sin x cos x tan x 5 sin x cos x (x, y ) OR ASYMPTOTE 0 0 1 0 (0, 0) p 2 1 0 undefined vertical asymptote: x 5 p 2 p 0 21 0 (p, 0) 3p 2 21 0 undefined vertical asymptote: x 5 3p 2 2p 0 1 0 (2p, 0) Notice that the x-intercepts correspond to integer multiples of p and that vertical asymptotes correspond to odd integer multiples of p 2 . 624 CHAPTER 6 Trigonometric Functions We know that the graph of the tangent function is undefined at the odd integer multiples of p 2 , so its graph cannot cross the vertical asymptotes. The question is, what happens between the asymptotes? We know the x-intercepts, so let us now make a table for special values of x. x y 2 –3π 2 –π 2 3π π 2 2π 0 –π π –2 –1 2 1 x y 2 –3π 2 –π 2 3π π 2 2π 0 –π π –100 –50 –75 1000 50 x y 2 –3π 2 –π 2 3π π 2 2π 0 –π π x sin x cos x y 5 tan x 5 sin x cos x (x, y ) p 6 1 2 !3 2 1 !3 5 !3 3 < 0.577 ap 6 , 0.577b p 4 !2 2 !2 2 1 ap 4 , 1b p 3 !3 2 1 2 !3 < 1.732 ap 3 , 1.732b 2p 3 !3 2 21 2 2!3 < 21.732 a2p 3 , 21.732b 3p 4 !2 2 2 !2 2 21 a3p 4 , 21b 5p 6 1 2 2 !3 2 1 2!3 5 2 !3 3 < 20.577 a5p 6 , 20.577b What happens to tan x as x approaches p 2 ? We know tan x is undefined at x 5 p 2 , but we must consider x-values both larger and smaller than p 2 < 1.571. The arrows on the graph on the right indicate increasing without bound (in the positive and negative directions). GRAPH OF y 5 tan x 1. The x-intercepts occur at integer multiples of p. 1np, 02 2. Vertical asymptotes occur at odd integer multiples of p 2 . x 5 12n 1 12p 2 3. The domain is the set of all real numbers except odd integer multiples of p 2 . x 2 12n 1 12p 2 4. The range is the set of all real numbers. 12q, q2 6.9 Graphs of Other Trigonometric Functions 625 The Cotangent Function The cotangent function is similar to the tangent function in that it is a quotient involving the sine and cosine functions. The difference is that cotangent has cosine in the numerator and sine in the denominator: cot x 5 cos x sin x. The graph of y 5 tan x has x-intercepts corresponding to integer multiples of p and vertical asymptotes correspond-ing to odd integer multiples of p 2. The graph of the cotangent function is the reverse in that it has x-intercepts corresponding to odd integer multiples of p 2 and vertical asymptotes corresponding to integer multiples of p. This is because the x-intercepts occur when the numerator, cos x, is equal to 0 and the vertical asymptotes occur when the denominator, sin x, is equal to 0. 5. y 5 tan x has period p. 6. y 5 tan x is an odd function (symmetric about the origin). tan12x2 5 2tan x 7. The graph has no defined amplitude, since the function is unbounded. Note: n is an integer. –5 –4 –3 –2 –1 1 2 3 4 5 x y π 2π –2π –π STUDY TIP The graphs of y 5 tan x and y 5 cot x both have period p, and neither of them has defined amplitude. GRAPH OF y 5 cot x 1. The x-intercepts occur at odd integer multiples of p 2. a 12n 1 12p 2 , 0b 2. Vertical asymptotes occur at integer multiples of p. x 5 np 3. The domain is the set of all real numbers except integer multiples of p. x 2 np 4. The range is the set of all real numbers. 12q, q2 5. y 5 cot x has period p. 6. y 5 cot x is an odd function (symmetric about the origin). cot12x2 5 2cot x 7. The graph has no defined amplitude, since the function is unbounded. Note: n is an integer. –5 –4 –3 –2 –1 1 2 3 4 5 x y π 2π –2π –π 626 CHAPTER 6 Trigonometric Functions The Secant Function Since y 5 cos x has period 2p, the secant function, which is the reciprocal of the cosine function, sec x 5 1 cos x, also has period 2p. We now illustrate values of the secant function with a table. x cos x y 5 sec x 5 1 cos x (x, y ) OR ASYMPTOTE 0 1 0 (0, 1) p 2 0 undefined vertical asymptote: x 5 p 2 p 21 21 (p, 21) 3p 2 0 undefined vertical asymptote: x 5 3p 2 2p 1 1 (2p, 1) x y 2 –3π 2 –π 2 3π π 2 (2π, 1) (0, 1) (–π, –1) (π, –1) 2π π –π –2π –1 1 Again, we ask the same question: What happens as x approaches the vertical asymptotes? The secant function grows without bound in either the positive or negative direction. If we graph y 5 cos x (the “guide” function) and y 5 sec x on the same graph, we notice the following: ■ ■The x-intercepts of y 5 cos x correspond to the vertical asymptotes of y 5 sec x. ■ ■The range of cosine is 321, 14, and the range of secant is 12q, 214 ∪ 31, q2. ■ ■When cosine is positive, secant is positive, and when one is negative, the other is negative. The cosine function is used as the guide function to graph the secant function. –5 –4 –3 –2 –1 1 2 3 4 5 x y π 2π –2π –π (2π, 1) (0, 1) (–π, –1) (π, –1) y = cosx y = secx GRAPH OF y 5 sec x 1. There are no x-intercepts. 1 cosx 2 0 2. Vertical asymptotes occur at odd integer multiples of p 2 . x 5 12n 1 12p 2 3. The domain is the set of all real numbers except odd integer multiples of p 2 . x 2 12n 1 12p 2 4. The range is 12q, 214 ∪ 31, q2. 5. y 5 sec x has period 2p. 6.9 Graphs of Other Trigonometric Functions 627 The Cosecant Function Since y 5 sin x has period 2p, the cosecant function, which is the reciprocal of the sine function, csc x 5 1 sin x, also has period 2p. We now illustrate values of cosecant with a table. x sin x y 5 csc x 5 1 sin x (x, y ) OR ASYMPTOTE 0 0 undefined vertical asymptote: x 5 0 p 2 1 1 a p 2 , 1b p 0 undefined vertical asymptote: x 5 p 3p 2 21 21 a 3p 2 , 2 1b 2p 0 undefined vertical asymptote: x 5 2p 6. y 5 sec x is an even function (symmetric about the y-axis). sec 12x2 5 sec x 7. The graph has no defined amplitude, since the function is unbounded. Note: n is an integer. –5 –4 –3 –2 –1 1 2 3 4 5 x y π 2π –2π –π Again, we ask the same question: What happens as x approaches the vertical asymptotes? The cosecant function grows without bound in either the positive or negative direction. If we graph y 5 sin x (the “guide” function) and y 5 csc x on the same graph, we notice the following: ■ ■The x-intercepts of y 5 sin x correspond to the vertical asymptotes of y 5 csc x. ■ ■The range of sine is 321, 14, and the range of cosecant is 12q, 214 ∪ 31, q2. ■ ■When sine is positive, cosecant is positive, and when one is negative, the other is negative. The sine function is used as the guide function to graph the cosecant function. –5 –4 –3 –2 –1 1 2 3 4 5 x y π 2π –2π –π 3π 2 ( , –1) π 2 ( , 1) y = cscx y = sinx 628 CHAPTER 6 Trigonometric Functions Graphing More General Tangent, Cotangent, Secant, and Cosecant Functions We use these basic functions as the starting point for graphing general tangent, cotangent, secant, and cosecant functions. GRAPH OF y 5 csc x 1. There are no x-intercepts. 1 sinx 2 0 2. Vertical asymptotes occur at integer multiples of p. x 5 np 3. The domain is the set of all real numbers except integer multiples of p. x 2 np 4. The range is 12q, 214 ∪ 31, q2. 5. y 5 csc x has period 2p. 6. y 5 csc x is an odd function (symmetric about the origin). csc12x2 5 2csc x 7. The graph has no defined amplitude, since the function is unbounded. Note: n is an integer. –5 –4 –3 –2 –1 1 2 3 4 5 x y π 2π –2π –π FUNCTION y 5 sin x y 5 cos x y 5 tan x y 5 cot x y 5 sec x y 5 csc x Graph of One Period Domain 12q, q2 12q, q2 x 2 12n 1 12p 2 x 2 np x 2 12n 1 12p 2 x 2 np Range 321, 14 321, 14 12q, q2 12q, q2 12q, 214 ∪ 31, q2 12q, 214 ∪ 31, q2 Amplitude 1 1 none none none none Period 2p 2p p p 2p 2p x-intercepts 1np, 02 a 12n 1 12p 2 , 0b 1np, 02 a 12n 1 12p 2 , 0b none none Vertical Asymptotes none none x 5 12n 1 12p 2 x 5 np x 5 12n 1 12p 2 x 5 np Note: n is an integer. 6.9 Graphs of Other Trigonometric Functions 629 GRAPHING TANGENT AND COTANGENT FUNCTIONS Graphs of y 5 A tan 1Bx2 and y 5 A cot1Bx2 can be obtained using the following steps 1assume B . 02: Step 1: Calculate the period: p B. Step 2: Find two neighboring vertical asymptotes. For y 5 A tan1Bx2: Bx 5 2p 2 and Bx 5 p 2. For y 5 A cot1Bx2: Bx 5 0 and Bx 5 p. Step 3: Find the x-intercept between the two asymptotes. For y 5 A tan1Bx2: Bx 5 0. For y 5 A cot1Bx2: Bx 5 p 2. Step 4: Draw the vertical asymptotes and label the x-intercept. Step 5: Divide the interval between the asymptotes into four equal parts. Set up a table with coordinates corresponding to the points in the interval. Step 6: Connect the points with a smooth curve. Use arrows to indicate the behavior toward the asymptotes. ■ ■If A . 0 ■ ■y 5 A tan1Bx2 increases from left to right. ■ ■y 5 A cot1Bx2 decreases from left to right. ■ ■If A , 0 ■ ■y 5 A tan1Bx2 decreases from left to right. ■ ■y 5 A cot1Bx2 increases from left to right. EXAMPLE 1 Graphing y 5 A tan(Bx) Graph y 5 23 tan 1 2x2 on the interval 2 p 2 # x # p 2 . Solution: A 5 23, B 5 2 STEP 1 Calculate the period. p B 5 p 2 STEP 2 Find two vertical asymptotes. Bx 5 2p 2 and Bx 5 p 2 Substitute B 5 2 and solve for x. x 5 2 p 4 and x 5 p 4 STEP 3 Find the x-intercept between Bx 5 0 the asymptotes. x 5 0 STEP 4 Draw the vertical asymptotes x 5 2 p 4 and x 5 p 4 and label the x-intercept (0, 0). x x = x = y π –π 4 π 4 π 2 4 –π 4 –π 2 (0, 0) –5 –4 –3 –2 –1 1 2 3 4 5 630 CHAPTER 6 Trigonometric Functions STEP 5 Divide the period p 2 into four equal parts, in steps of p 8 . Set up a table with coordinates corresponding to values of y 5 23 tan 12x2. STEP 6 Graph the points from the table and connect with a smooth curve. Repeat to the right and left until you reach the interval endpoints. Notice that the distance between the vertical asymptotes is the period length p 2. Y OUR TU R N Graph y 5 1 3 tanA1 2 xB on the interval 2p # x # p. ▼ x y 5 23 tan (2x ) (x, y ) 2p 4 undefined vertical asymptote, x 5 2p 4 2p 8 3 a2p 8, 3b 0 0 10, 02 p 8 23 ap 8, 23b p 4 undefined vertical asymptote, x 5 p 4 x y π –π 4 π 2 4 –π 2 (0, 0) π – 8 ( , 3) π 8 ( , –3) –5 –4 –3 –2 –1 1 2 3 4 5 ▼ A N S W E R 2 π π –π –π 2 –5 –4 –3 –2 1 2 3 4 5 x y EXAMPLE 2 Graphing y 5 A cot(Bx) Graph y 5 4 cotA1 2 xB on the interval 22p # x # 2p. Solution: A 5 4, B 5 1 2 STEP 1 Calculate the period. p B 5 2p STEP 2 Find two vertical asymptotes. Bx 5 0 and Bx 5 p Substitute B 5 1 2 and solve for x. x 5 0 and x 5 2p STEP 3 Find the x-intercept between the Bx 5 p 2 asymptotes. x 5 p STEP 4 Draw the vertical asymptotes x 5 0 and x 5 2p and label the x-intercept (p, 0). 6.9 Graphs of Other Trigonometric Functions 631 STEP 5 Divide the period 2p into four equal parts, in steps of p 2 . Set up a table with coordinates corresponding to values of y 5 4 cotA1 2 x2. STEP 6 Graph the points from the table and connect with a smooth curve. Repeat to the right and left until you reach the interval endpoints. Y OUR T UR N Graph y 5 2 cot12x2 on the interval 2p # x # p. ▼ x y π –π 2π –2π (π, 0) 3π 2 ( , –4) π 2 ( , 4) –5 –4 –3 –2 –1 1 2 3 4 5 x y 5 4 cot A1 2x B (x, y ) 0 undefined vertical asymptote, x 5 0 p 2 4 ap 2, 4b p 0 1p, 02 3p 2 24 a3p 2 , 24b 2p undefined vertical asymptote, x 5 2p GRAPHING SECANT AND COSECANT FUNCTIONS Graphs of y 5 A sec1Bx2 and y 5 A csc1 Bx2 can be obtained using the following steps: Step 1: Graph the corresponding guide function with a dashed curve. For y 5 A sec1Bx2, use y 5 A cos1Bx2 as a guide. For y 5 A csc1Bx2, use y 5 A sin1Bx2 as a guide. Step 2: Draw the asymptotes, which correspond to the x-intercepts of the guide function. Step 3: Draw the U shape between the asymptotes. If the guide function has a positive value between the asymptotes, the U opens upward; and if the guide function has a negative value, the U opens downward. ▼ A N S W E R 2 π π –π –π 2 –5 –4 –3 –2 1 2 3 4 5 x y 632 CHAPTER 6 Trigonometric Functions ▼ A N S W E R –5 –4 –3 –2 –1 1 2 3 4 5 x y –1 –0.5 0.5 1 EXAMPLE 3 Graphing y 5 A sec(Bx) Graph y 5 2 sec 1px2 on the interval 22 # x # 2. Solution: STEP 1 Graph the corresponding guide function with a dashed curve. For y 5 2 sec(px), use y 5 2 cos(px) as a guide. STEP 2 Draw the asymptotes, which correspond to the x-intercepts of the guide function. STEP 3 Use the U shape between the asymptotes. If the guide function is positive, the U opens upward, and if the guide function is negative, the U opens downward. Y OUR TU R N Graph y 5 2sec12px2 on the interval 21 # x # 1. ▼ –2 –1 1 2 –5 –4 –3 –2 –1 1 2 3 4 5 x y –2 –1 1 2 –5 –4 –3 –2 –1 1 2 3 4 5 x y –2 –1 1 2 –5 –4 –3 –2 –1 1 2 3 4 5 x y EXAMPLE 4 Graphing y 5 A csc(Bx) Graph y 5 23 csc12px2 on the interval 21 # x # 1. Solution: STEP 1 Graph the corresponding guide function with a dashed curve. For y 5 23 csc(2px), use y 5 23 sin(2px) as a guide. –1 –0.5 0.5 1 –5 –4 –3 –2 –1 1 2 3 4 5 x y 6.9 Graphs of Other Trigonometric Functions 633 6.9.2 Translations of Circular Functions Vertical translations and horizontal translations (phase shifts) of the tangent, cotangent, secant, and cosecant functions are graphed the same way as vertical and horizontal translations of sinusoidal graphs are drawn. For tangent and cotangent functions, we follow the same procedure as we did with sinusoidal functions. For secant and cotangent functions, we graph the guide function first and then translate up or down depending on the sign of the vertical shift. STEP 2 Draw the asymptotes, which correspond to the x-intercepts of the guide function. STEP 3 Use the U shape between the asymptotes. If the guide function is positive, the U opens upward, and if the guide function is negative, the U opens downward. Y OUR T UR N Graph y 5 1 2 csc 1px2 on the interval 21 # x # 1. ▼ ▼ A N S W E R –5 –4 –3 –2 –1 1 2 3 4 5 x y –1 –0.5 0.5 1 –1 –0.5 0.5 1 –5 –4 –3 –2 –1 1 2 3 4 5 x y x = 0 x = 1 x = –1 x = –1 2 x = 1 2 –1 –0.5 0.5 1 –5 –4 –3 –2 –1 1 2 3 4 5 x y EXAMPLE 5 Graphing y 5 A tan(Bx 1 C ) 1 D Graph y 5 1 2 tan ax 2 p 2 b on 2p # x # p. State the domain and range on the interval. There are two ways to approach graphing this function. Both will be illustrated. Solution (1): Plot y 5 tan x, and then do the following: ■ ■Shift the curve to the right p 2 units. y 5 tan ax 2 p 2 b ■ ■Reflect the curve about the x-axis (because of the negative sign). y 5 2tan ax 2 p 2 b ■ ■Shift the entire graph up one unit. y 5 1 2 tan ax 2 p 2 b 6.9.2 S K I L L Graph translated tangent, cotangent, secant, and cosecant functions. 6.9.2 C ON C E P T U A L Understand that graph-shifting techniques for tangent and cotangent are consistent with translations used for sinusoidal functions; but for secant and cosecant functions, we first graph the horizontally translated sinusoidal functions and then we shift up or down depending on the vertical translations. 634 CHAPTER 6 Trigonometric Functions Solution (2): Graph y 5 2tan ax 2 p 2 b, and then shift the entire graph up one unit because D 5 1. STEP 1 Calculate the period. p B 5 p STEP 2 Find two vertical asymptotes. x 2 p 2 5 2 p 2 and x 2 p 2 5 p 2 Solve for x. x 5 0 and x 5 p STEP 3 Find the x-intercept between x 2 p 2 5 0 the asymptotes. x 5 p 2 STEP 4 Draw the vertical asymptotes x 5 0 and x 5 p and label the x-intercept a p 2 , 0b. STEP 5 Divide the period p into four equal parts, in steps of p 4 . Set up a table with coordinates corresponding to values of y 5 2tanax 2 p 2 b between the two asymptotes. x y π –π 2 π 2 –π π 2 ( , 0) –5 –4 –3 –2 –1 1 2 3 4 5 x y 5 2tanax 2 p 2 b (x, y ) x 5 0 undefined vertical asymptote: x 5 0 p 4 1 ap 4, 1b p 2 0 ap 2, 0b 1x-intercept2 3p 4 21 a3p 4 , 21b x 5 p undefined vertical asymptote: x 5 p 6.9 Graphs of Other Trigonometric Functions 635 STEP 6 Graph the points from the table and connect with a smooth curve. Repeat to the right and left until reaching the interval endpoints. STEP 7 Shift the entire graph up one unit to arrive at the graph of y 5 1 2 tan ax 2 p 2 b. STEP 8 State the domain and range on the interval. Domain: 12p, 02 ∪ 10, p2 Range: 12q, q2 YOU R T UR N Graph y 5 21 1 cot ax 1 p 2 b on 2p # x # p. State the domain and range on the interval. ▼ x y π –π 2 π 2 –π π 2 ( , 0) π 4 ( , 1) 3π 4 ( , –1) –5 –4 –3 –2 –1 1 2 3 4 5 x y π –π 2 π 2 –π π 2 ( , 1) π 4 ( , 2) 3π 4 ( , 0) –5 –4 –3 –2 –1 1 2 3 4 5 ▼ A N S W E R Domain: c2p, 2 p 2 b ∪ a2 p 2 , p 2b ∪ ap 2 , pd Range: 12q, q2 2 π π –π –π 2 –5 –4 –3 –2 1 2 3 4 5 x y EXAMPLE 6 Graphing y 5 A csc(Bx 1 C ) 1 D Graph y 5 1 2 csc12x 2 p2 on 2p # x # p. State the domain and range on the interval. Solution: Graph y 5 2csc12x 2 p2, and shift the entire graph up one unit to arrive at the graph of y 5 1 2 csc12x 2 p2. STEP 1 Draw the guide function, y 5 2sin12x 2 p2. x y π –π 2 π 2 –π –5 –4 –3 –2 –1 1 2 3 4 5 [CONCEPT CHECK] If k . 0 , what is the last step in graphing the function y 5 2k 1 sec(p ? x )? ANSWER Shift the graph of sec(p ? x) down k units. ▼ 636 CHAPTER 6 Trigonometric Functions STEP 2 Draw the vertical asymptotes of y 5 2csc12x 2 p2 that correspond to the x-intercepts of y 5 2sin12x 2 p2. STEP 3 Use the U shape between the asymptotes. If the guide function is positive, the U opens upward, and if the guide function is negative, the U opens downward. STEP 4 Shift the entire graph up one unit to arrive at the graph of y 5 1 2 csc12x 2 p2. STEP 5 State the domain and range on the interval. Y OUR TU R N Graph y 5 22 1 sec1px 2 p2 on 21 # x # 1. State the domain and range on the interval. ▼ x x = x = y π –π 2 π 2 π 2 –π 2 –π –5 –4 –3 –2 –1 1 2 3 4 5 x y π –π 2 π 2 –π –5 –4 –3 –2 –1 1 2 3 4 5 ▼ A N S W E R Domain: C21, 21 2B ∪ A21 2, 1 2B ∪ A1 2, 1D Range: 12q, 234 ∪ 321, q2 –5 –4 –3 –2 –1 1 2 3 4 5 x y –1 –0.5 0.5 1 x y π –π 2 π 2 –π –5 –4 –3 –2 –1 1 2 3 4 5 Domain: a2p, 2p 2 b ∪ a2p 2 , 0b ∪ a0, p 2 b ∪ ap 2, pb Range: 12q, 04 ∪ 32, q2 (cosine and sine, respectively), and then label vertical asymptotes that correspond to x-intercepts of the guide function. The graphs of the secant and cosecant functions resemble the letter U opening up or down. The secant and cosecant functions are positive when their guide function is positive and negative when their guide function is negative. The tangent and cotangent functions have period p, whereas the secant and cosecant functions have period 2p. To graph the tangent and cotangent functions, first identify the vertical asymptotes and x-intercepts, and then find values of the function within a period (i.e., between the asymptotes). To find graphs of secant and cosecant functions, first graph their guide functions [SEC TION 6.9] S U M MA RY 6.9 Graphs of Other Trigonometric Functions 637 [SEC TION 6.9] E X E R C I S E S • S K I L L S In Exercises 1–8, match the graphs to the functions (a–h). 1. y 5 2tan x 2. y 5 2csc x 3. y 5 sec12x2 4. y 5 csc12x2 5. y 5 cot1p x2 6. y 5 2cot1p x2 7. y 5 3 sec x 8. y 5 3 csc x a. b. c. d. e. f. g. h. In Exercises 9–28, graph the functions over the indicated intervals. 9. y 5 tanA1 2 xB, 22p # x # 2p 10. y 5 cotA1 2xB, 22p # x # 2p 11. y 5 2cot12p x2, 21 # x # 1 12. y 5 2tan12p x2, 21 # x # 1 13. y 5 2 tan13x2, 2p # x # p 14. y 5 2 tanA1 3 xB, 23p # x # 3p 15. y 5 2 1 4 cot ax 2b, 22p # x # 2p 16. y 5 2 1 2 tan ax 4b, 24p # x # 4p 17. y 5 2tan ax 2 p 2 b, 2p # x # p 18. y 5 tan ax 1 p 4 b, 2p # x # p 19. y 5 2 tan ax 1 p 6 b, 2p # x # p 20. y 5 21 2 tan 1x 1 p2, 2p # x # p 21. y 5 cot ax 2 p 4 b, 2p # x # p 22. y 5 2cot ax 1 p 2 b, 2p # x # p 23. y 52 1 2 cot ax 1p 3b, 2p # x # p 24. y 5 3 cot ax 2 p 6 b, 2p # x # p 25. y 5 tan 12x 2 p2, 22p # x # 2p 26. y 5 cot12x 2 p2, 22p # x # 2p 27. y 5 cot ax 2 1 p 4 b, 2p # x # p 28. y 5 tan ax 3 2 p 3 b, 2p # x # p –5 –4 –3 –2 –1 1 2 3 4 5 x y π π 3π 4 π 2 4 x y π –π 4 π 2 4 –π 2 –5 –4 –3 –2 –1 1 2 3 4 5 x y –5 –4 –3 –2 –1 1 2 3 4 5 0.5 1 –5 –4 –3 –2 –1 1 2 3 4 5 x y π 3π 2π 2 π 2 x y –5 –4 –3 –2 –1 1 2 3 4 5 1 2 x y π 3π 2π 2 π 2 –5 –4 –3 –2 –1 1 2 3 4 5 –5 –4 –3 –2 –1 1 2 3 4 5 x y π 3π 2π 2 π 2 –5 –4 –3 –2 –1 1 2 3 4 5 x y π π 3π 4 π 2 4 638 CHAPTER 6 Trigonometric Functions In Exercises 29–46, graph the functions over the indicated intervals. 29. y 5 sec A1 2 xB, 22p # x # 2p 30. y 5 csc A1 2 xB, 22p # x # 2p 31. y 5 2csc12p x2, 21 # x # 1 32. y 5 2sec12p x2, 21 # x # 1 33. y 5 1 3 sec ap 2 xb, 24 # x # 4 34. y 5 1 2 csc ap 3 xb, 26 # x # 6 35. y 5 23 csc ax 3b, 26p # x # 0 36. y 5 24 sec ax 2b, 24p # x # 4p 37. y 5 2 sec 13x2, 0 # x # 2p 38. y 5 2 csc A1 3 xB, 23p # x # 3p 39. y 5 23 csc ax 2 p 2 b, over at least one period 40. y 5 5 sec ax 1 p 4 b, over at least one period 41. y 5 1 2 sec 1x 2 p2, over at least one period 42. y 5 24 csc 1x 1 p2, over at least one period 43. y 5 2 sec12x 2 p2, 22p # x # 2p 44. y 5 2 csc 12x 1 p2, 22p # x # 2p 45. y 5 21 4 sec13x 1 p2 46. y 5 2 2 3 csc a 4x2 p 2b, 2p# x # p In Exercises 47–56, graph the functions over at least one period. 47. y 5 3 2 2 sec ax 2 p 2 b 48. y 5 23 1 2 csc ax 1 p 2 b 49. y 5 1 2 1 1 2 tan ax 2 p 2 b 50. y 5 3 4 2 1 4 cot ax 1 p 2 b 51. y 5 22 1 3 csc12x 2 p2 52. y 5 21 1 4 sec12x 1 p2 53. y 5 21 2 sec a1 2 x 2 p 4 b 54. y 5 22 1 csc a1 2 x 1 p 4 b 55. y 5 22 2 3 cot a2x 2 p 4b, 2p # x # p 56. y 5 2 1 4 1 1 2 sec apx 1 p 4 b, 22 # x # 2 In Exercises 57–66, state the domain and range of the functions. 57. y 5 tan apx 2 p 2 b 58. y 5 cot ax 2 p 2 b 59. y 5 2 sec 15x2 60. y 5 24 sec 13x2 61. y 5 2 2 csc A1 2 x 2 pB 62. y 5 1 2 2 sec A1 2 x 1 pB 63. y 5 2 3 tan ap 4x 2 pb 1 1 64. y 5 1 4 cot a 2px 1 p 3 b 2 3 65. y 5 22 1 1 2 sec a px 1 p 2 b 66. y 5 1 2 2 1 3 csc a3x 2 p 2 b 6.9 Graphs of Other Trigonometric Functions 639 • A P P L I C A T I O N S For Exercises 67 and 68, refer to the following: When a chalet on the top of a mountain is viewed by a mountain climber from point A on the plain below, the measure of the angle of elevation is a degrees. When the mountain climber moves closer to the mountain, point B, the measure of the angle of elevation is b degrees. If the distance between points A and B is d, then the height of the mountain can be found with h 5 d cot a 2 cot b. Chalet A B d α β 67. Height of a Mountain. What is the height of the mountain if a 5 20°, b 5 25°, and d 5 2 miles? Round to the nearest mile. 68. Height of a Mountain. What is the height of the mountain if a 5 15°, b 5 35°, and d 5 3 miles? For Exercises 69 and 70, refer to the following: The area of a regular polygon that is inscribed by a circle can be determined by A 5 nr2 tan 180° n , where n is the number of sides of the polygon, and r is the radius of the inscribed circle. 69. Area of a Regular Polygon. Find the area of a regular hexagon if the radius of the inscribed circle is 4 inches. 4 in. 70. Area of a Regular Polygon. Find the area of a regular octagon if the radius of the inscribed circle is 2 feet. 2 ft • C A T C H T H E M I S T A K E In Exercises 71 and 72, explain the mistake that is made. 71. Graph y 5 3 csc12x2. Solution: Graph the guide function, y 5 sin12x2. Draw vertical asymptotes at x-values that correspond to x-intercepts of the guide function. Draw the cosecant function. This is incorrect. What mistake was made? –5 –4 –3 –2 –1 1 2 3 4 5 x y π 3π 2π 2 π 2 –5 –4 –3 –2 –1 1 2 3 4 5 x y π 3π 2π 2 π 2 640 CHAPTER 6 Trigonometric Functions 72. Graph y 5 tan14x2. Solution: Step 1: Calculate the period. p B 5 p 4 Step 2: Find two vertical asymptotes. 4x 5 0 and 4x 5 p Solve for x. x 5 0 and x 5 p 4 Step 3: Find the x-intercept between the asymptotes. 4x 5 p 2 x 5 p 8 Step 4: Draw the vertical asymptotes x 5 0 and x 5 p 4 and label the x-intercept ap 8, 0b. Step 5: Graph. This is incorrect. What mistake was made? x y π –π 4 π 2 4 –π 2 –5 –4 –3 –2 –1 1 2 3 4 5 π 8 ( , 0) • C O N C E P T U A L In Exercises 73 and 74, determine whether each statement is true or false. 73. secax 2 p 2 b 5 csc x 74. cscax 2 p 2 b 5 sec x 75. For what values of n do y 5 tan x and y 5 tan1x 2 np2 have the same graph? 76. For what values of n do y 5 csc x and y 5 csc1x 2 np2 have the same graph? 77. Solve the equation tan12x 2 p2 5 0 for x in the interval 32p, p4 by graphing. 78. Solve the equation csc12x 1 p2 5 0 for x in the interval 32p, p4 by graphing. 79. Find the x-intercepts of y 5 A tan1Bx 1 C2. 80. For what x-values does the graph of y 5 2A sec ap 2 xb lie above the x-axis? (Assume A . 0.) 81. How many solutions are there to the equation tan x 5 x? Explain. 82. For what values of A do the graphs of y 5 A sin 1Bx 1 C2 and y 5 22 csc ap 6 x 2 pb never intersect? 83. What is the amplitude of the function y 5 cos x 1 sin x? Use a graphing calculator to graph Y1 5 cos x, Y2 5 sin x, and Y3 5 cos x 1 sin x in the same viewing window. 84. Graph Y1 5 cos x 1 sin x and Y2 5 sec x 1 csc x in the same viewing window. Based on what you see, is Y1 5 cos x 1 sin x the guide function for Y2 5 sec x 1 csc x? 85. What is the period of the function y 5 tan x 1 cot x? Use a graphing calculator to graph Y15 tan x 1 cot x and Y3 5 2 csc12x2 in the same viewing window. 86. What is the period of the function y 5 tan a2x 1 p 2 b? Use a graphing calculator to graph Y1 5 tan a2x 1 p 2 b, Y2 5 tan a2x 2 p 2 b, and Y3 5 tan a22x 1 p 2 b in the same viewing window. Describe the relationships of Y1 and Y2 and Y2 and Y3. • T E C H N O L O G Y CH A P TE R 6 R E VIE W [CH AP TER 6 REVIEW] SECTION CONCEPT KEY IDEAS/FORMULAS 6.1 Angles, degrees, and triangles Angles and degree measure One complete rotation = 360º θ = 90º Right Angle: quarter rotation θ Acute Angle 0º < θ < 90º θ Obtuse Angle 90º < θ < 180º α β Complementary Angles α + β = 90º α β Supplementary Angles α + β = 180º Triangles α β γ α + β + γ = 180º b a c Right Triangle Pythagorean Theorem a2 + b2 = c2 Special right triangles x x 45º 45º 45º – 45º – 90º √2x 30º – 60º – 90º 60º 30º x 2x √3x Similar triangles a ar 5 b br 5 c cr b' a' c' b a c Chapter Review 641 CH A P TE R 6 R E VI E W 642 CHAPTER 6 Trigonometric Functions SECTION CONCEPT KEY IDEAS/FORMULAS 6.2 Definition 1 of trigonometric functions: Right triangle ratios Definition 1 defines trigonometric functions of acute angles as ratios of sides in a right triangle. Trigonometric functions: Right triangle ratios SOH sin u 5 opposite hypotenuse CAH cos u 5 adjacent hypotenuse TOA tan u 5 opposite adjacent b a θ c Adjacent Opposite Hypotenuse Reciprocal identities cot u 5 1 tan u csc u 5 1 sin u sec u 5 1 cos u Cofunctions If a 1 b 5 90°, then sin a 5 cos b sec a 5 csc b tan a 5 cot b Evaluating trigonometric functions exactly for special angle measures: 308, 458, and 608 u sin u cos u 30° 1 2 !3 2 45° !2 2 !2 2 60° !3 2 1 2 The other trigonometric functions can be found for these values using tan u 5 sin u cos u and reciprocal identities. Using calculators to evaluate (approximate) trigonometric functions Make sure the calculator is in degree mode. The sin, cos, and tan buttons can be combined with the reciprocal button, 1/x, to get csc, sec, and cot. 6.3 Applications of right triangle trigonometry: Solving right triangles Solving a right triangle given the measure of an acute angle and a side length • The third angle measure can be found exactly using a 1 b 1 g 5 180°. • Right triangle trigonometry is used to find the length of the second side. Solving a right triangle given the lengths of two sides • The length of the third side can be found using the Pythagorean theorem. • The measure of one of the acute angles can be found using right triangle trigonometry. CH A P TE R 6 R E VIE W SECTION CONCEPT KEY IDEAS/FORMULAS 6.4 Definition 2 of trigonometric functions: Cartesian plane Angles in standard position An angle is said to be in standard position if its initial side is along the positive x-axis and its vertex is at the origin. 30º 330º 315º 300º 240º 225º 210º 150º 135º 120º 60º 45º 2 √3 2 √2 2 √2 2 1 2 √3 2 1 ( , – ) 2 √3 2 1 ( , ) ( , – ) 2 √2 2 √2 ( , ) ( , – ) 2 √3 2 √2 2 √2 2 1 2 √3 2 1 (– , – ) (– , – ) (– , – ) 2 √3 2 √2 2 √2 2 1 2 √3 2 1 (– , ) (– , ) (– , ) 2 √3 2 1 ( , ) x y Coterminal angles Two angles in standard position with the same terminal side. Trigonometric functions: The Cartesian plane sin u 5 y r cos u 5 x r tan u 5 y x csc u 5 r y sec u 5 r x cot u 5 x y where x2 1 y2 5 r2. The distance r is positive: r . 0. u 08 908 1808 2708 sin u 0 1 0 21 cos u 1 0 21 0 tan u 0 Undefined 0 Undefined cot u Undefined 0 Undefined 0 sec u 1 Undefined 21 Undefined cse u Undefined 1 Undefined 21 Trigonometric function values for quadrantal angles. 6.5 Trigonometric functions of nonacute angles Algebraic signs of trigonometric functions u QI QII QIII QIV sin u 1 1 2 2 cos u 1 2 2 1 tan u 1 2 1 2 Ranges of the trigonometric functions 21 # sin u # 1 and 21 # cos u # 1 Reference angles and reference right triangles The reference angle a for angle u (between 08 and 3608) is given by n QI: a 5 u n QII: a 5 180° 2 u n QIII: a 5 u 2 180° n QIV: a 5 360° 2 u Evaluating trigonometric functions for nonacute angles Chapter Review 643 CH A P TE R 6 R E VI E W 644 CHAPTER 6 Trigonometric Functions SECTION CONCEPT KEY IDEAS/FORMULAS 6.6 Radian measure and applications The radian measure of an angle u 1in radians2 5 s r s (arc length) and r (radius) must have the same units. s θ r r Converting between degrees and radians Degrees to radians: ur 5 ud a p 180°b Radians to degrees: ud 5 ur a180° p b Arc length s 5 r ur ur is in radians. s 5 r ud a p 180°b ud is in degrees. Area of a circular sector A 5 1 2 r 2ur ur is in radians. A 5 1 2 r 2ud a p 180°b ud is in degrees. Linear speed Linear speed v is given by v 5 s t where s is the arc length and t is time. Angular speed Angular speed v is given by v 5 u t where u is given in radians. Relationship between linear and angular speeds Related through the radius of the circle: v 5 rv or v 5 v r It is important to note that these formulas hold true only when angular speed is given in radians per unit of time. 6.7 Definition 3 of trigonometric functions: Unit circle approach Trigonometric functions and the unit circle (circular functions) x y (1, 0) (0, 1) (cos θ, sin θ) r = 1 s θ (0, –1) (–1, 0) CH A P TE R 6 R E VIE W SECTION CONCEPT KEY IDEAS/FORMULAS x y 60º 3 π 45º 4 π 30º 6 π 360º 2π 0º 0 0 330º 6 11π 315º 4 7π 300º 3 5π 270º 2 3π 240º 3 4π 225º 4 5π 210º 6 7π π 180º 150º 6 5π 135º 4 3π 90º 2 π 120º 3 2π (0, 1) (0, –1) (1, 0) (–1, 0) 2 √3 2 1 ( , – ) 2 √3 2 1 ( , ) 2 √2 2 √2 ( , – ) 2 √2 2 √2 ( , ) 2 √3 2 1 ( , – ) 2 √3 2 1 (– , – ) 2 √2 2 √2 (– , – ) 2 √3 2 1 (– , – ) 2 √3 2 1 (– , ) 2 √2 2 √2 (– , ) 2 √3 2 1 (– , ) 2 √3 2 1 ( , ) Properties of circular functions Cosine is an even function. cos12u2 5 cos u Sine is an odd function. sin 12u2 5 2sin u 6.8 Graphs of sine and cosine functions The graph of ƒ1x2 5 sin x –1 1 x y 2π π –π –2π Odd function: ƒ12x2 5 2ƒ1x2 The graph of ƒ1x2 5 cos x –1 1 x y 2π π –π –2π Even function: ƒ12x2 5 ƒ1x2 The amplitude and period of sinusoidal graphs y 5 A sin Bx or y 5 A cos1Bx2, B . 0 Amplitude 5 0 A0 Period 5 2p B n 0A0 . 1 stretch vertically. n B . 1 compress horizontally. n 0A0 , 1 compress vertically. n B , 1 stretch horizontally. Graphing a shifted sinusoidal function: y 5 A sin1Bx 1 C 2 1 D and y 5 A cos1Bx 1 C 2 1 D n y 5 A sin 1Bx 6 C2 5 AsincB ax 6 C Bbd has period 2p B and a phase shift of C B unit to the left 112 or the right 122. n y 5 A cos 1Bx 6 C2 5 A coscB ax 6 C Bbd has period 2p B and a phase shift of C B unit to the left 112 or the right 122. n To graph y 5 A sin1Bx 1 C2 1 D or y 5 A cos1Bx 1 C2 1 D, start with the graph of y 5 A sin 1Bx 1 C2 or y 5 A cos1Bx 1 C2 and shift up or down D units. Harmonic motion n Simple n Damped n Resonance Graphing sums of functions: Addition of ordinates Chapter Review 645 CH A P TE R 6 R E VI E W 646 CHAPTER 6 Trigonometric Functions SECTION CONCEPT KEY IDEAS/FORMULAS 6.9 Graphs of other trigonometric functions The tangent function 2π π –π –2π –5 –4 –3 –2 –1 1 2 3 4 5 x y x-intercepts: x 5 np n is an integer Asymptotes: x 5 12n 1 12 2 p Period: p Amplitude: none The cotangent function 2π π –π –2π –5 –4 –3 –2 –1 1 2 3 4 5 x y x-intercepts: x 5 np n is an integer Asymptotes: x 5 12n 1 12 2 p Period: p Amplitude: none The secant function 2π π –π –2π –5 –4 –3 –2 –1 1 2 3 4 5 x y x-intercepts: none n is an integer Asymptotes: x 5 12n 1 12 2 p Period: 2p Amplitude: none The cosecant function 2π π –π –2π –5 –4 –3 –2 –1 1 2 3 4 5 x y x-intercepts: none n is an integer Asymptotes: x 5 np Period: 2p Amplitude: none Graphing tangent, cotangent, secant, and cosecant functions Translations of circular functions y 5 A tan1Bx 1 C2 or y 5 A cot1Bx 1 C2 To find asymptotes, set Bx 1 C equal to: n 2p 2 and p 2 for tangent. n 0 and p for cotangent. To find x-intercepts, set Bx 1 C equal to: n 0 for tangent. n p 2 for cotangent. y 5 A sec1Bx 1 C2 or y 5 A csc1Bx 1 C2 To graph y 5 A sec1Bx 1 C2, use y 5 A cos1Bx 1 C2 as the guide. To graph y 5 A csc1Bx 1 C2, use y 5 A sin1Bx 1 C2 as the guide. Intercepts on the guide function correspond to vertical asymptotes of secant or cosecant functions. R E VI E W E XERCISES [CH AP TER 6 REVIEW EXE R C IS E S ] 26. Clock. What is the measure (in degrees) of the angle that the second hand sweeps in exactly 15 seconds? 27. Height of a Tree. The shadow of a tree measures 9.6 meters. At the same time of day the shadow of a 4-meter basketball backboard measures 1.2 meters. How tall is the tree? 28. Height of a Man. If an NBA center casts a 1-foot 9-inch shadow and his 4-foot son casts a 1-foot shadow, how tall is the NBA center? 6.2 Definition 1 of Trigonometric Functions: Right Triangle Ratios Use the following triangle to find the indicated trigonometric functions. Rationalize any denominators that you encounter in the answers. 29. cos u 30. sin u 31. sec u 32. csc u 33. tan u 34. cot u Use the cofunction identities to fill in the blanks. 35. sin 30° 5 cos 36. cos A 5 sin 37. tan 45° 5 cot 38. csc 60° 5 sec 39. sec 30° 5 csc 40. cot 60° 5 tan Label each trigonometric function value with the corresponding value (a–c). a. !3 2 b. 1 2 c. !2 2 41. sin 30° 42. cos 30° 43. cos 60° 44. sin 60° 45. sin 45° 46. cos 45° Use a calculator to approximate the following trigonometric function values. Round the answers to four decimal places. 47. sin 42° 48. cos 57° 49. cos 17.3° 50. tan 25.2° 51. cot 33° 52. sec 16.8° 53. csc 40.25° 54. cot 19.76° 6.3 Applications of Right Triangle Trigonometry: Solving Right Triangles The following exercises illustrate a midair refueling scenario that U.S. military aircraft often use. Assume the elevation angle that the hose makes with the plane being fueled is u 5 30°. 55. Midair Refueling. If the hose is 150 feet long, what should the altitude difference a be between the two planes? 56. Midair Refueling. If the smallest acceptable altitude difference a between the two planes is 100 feet, how long should the hose be? 3 2 θ Hose b a θ = 30 6.1 Angles, Degrees, and Triangles Find (a) the complement and (b) the supplement of the given angles. 1. 28° 2. 17° 3. 35° 4. 78° 5. 89.01° 6. 0.013° Refer to the accompanying triangle for the following exercises. 7. If a 5 120° and b 5 35°, find g. 8. If a 5 105° and b 5 25°, find g. 9. If g 5 b and a 5 7b, find the measure of all three angles. 10. If g 5 b and a 5 6b, find the measure of all three angles. Refer to the accompanying right triangle for the following exercises. 11. If a 5 4 and c 5 12, find b. 12. If b 5 9 and c 5 15, find a. 13. If a 5 7 and b 5 4, find c. 14. If a 5 10 and b 5 8, find c. Refer to the accompanying 458-458-908 triangle for the following exercises. 15. If the two legs have length 12 yards, how long is the hypotenuse? 16. If the hypotenuse has length !8 feet, how long are the legs? Refer to the accompanying 308-608-908 triangle for the following exercises. 17. If the shortest leg has length 3 feet, what are the lengths of the other leg and the hypotenuse? 18. If the hypotenuse has length 12 kilometers, what are the lengths of the two legs? Calculate the specified lengths given that the two triangles are similar. 19. A 5 10, C 5 8, D 5 5, F 5 ? 20. A 5 15, B 5 12, E 5 4, D 5 ? 21. D 5 4.5 m, F 5 8.2 m, A 5 81 km, C 5 ? 22. E 5 8 cm, F 5 14 cm, C 5 84 m, B 5 ? 23. B 5 2!3 in., C 5 !3 in., F 5 5!3 in., E 5 ? 24. C 5 4.2 ft, F 5 0.2 ft, D 5 2 ft, A 5 ? Applications 25. Clock. What is the measure (in degrees) of the angle that the minute hand sweeps in exactly 25 minutes? α β γ α + β + γ = 180º b a c x x 45º 45º √2x x 60º 30º 2x √3x A C B E D F Review Exercises 647 REV IEW E XE R CI SE S 648 CHAPTER 6 Trigonometric Functions 6.4 Definition 2 of Trigonometric Functions: Cartesian Plane In the following exercises, the terminal side of an angle u in standard position passes through the indicated point. Calculate the values of the six trigonometric functions for angle u. 57. 16, 282 58. 1224, 272 59. 126, 22 60. 1240, 92 61. A!3, 1B 62. 129, 292 63. 1 1 2, 21 4 2 64. 123 4, 5 6 2 65. 121.2, 22.42 66. 10.8, 22.42 6.5 Trigonometric Functions of Nonacute Angles Evaluate the following expressions exactly. 67. sin 330° 68. cos12300°2 69. tan 150° 70. cot 315° 71. sec12150°2 72. csc 210° Evaluate the trigonometric expressions with a calculator. Round to four decimal places. 73. sin1214°2 74. cos 275° 75. tan 46° 76. cot 500° 77. sec 111° 78. csc 215° 79. cot1257°2 80. csc1259°2 6.6 Radian Measure and Applications Convert from degrees to radians. Leave your answers in terms of p. 81. 135° 82. 240° 83. 330° 84. 180° 85. 216° 86. 108° 87. 1620° 88. 900° Convert from radians to degrees. 89. p 3 90. 11p 6 91. 5p 4 92. 2p 3 93. 5p 9 94. 17p 10 95. 10p 96. 31p 2 97. A ladybug is clinging to the outer edge of a child’s spinning disk. The disk is 4 inches in diameter and is spinning at 60 revolutions per minute. How fast is the ladybug traveling? 98. How fast is a motorcyclist traveling in miles per hour if his tires are 30 inches in diameter and the angular speed of the tire is 10p radians per second? 6.7 Definition 3 of Trigonometric Functions: Unit Circle Approach Find each trigonometric function value in exact form. 99. tan a5p 6 b 100. cos a5p 6 b 101. sin a11p 6 b 102. sec a11p 6 b 103. cot a5p 4 b 104. csc a5p 4 b 105. sin a3p 2 b 106. cos a3p 2 b 107. cos p 108. tan 315° 109. cos 60° 110. sin 330° 6.8 Graphs of Sine and Cosine Functions Refer to the graph of the sinusoidal function to answer the questions. 111. Determine the period of the function. 112. Determine the amplitude of the function. 113. Write an equation for the sinusoidal function. Refer to the graph of the sinusoidal function to answer the questions. 114. Determine the period of the function. 115. Determine the amplitude of the function. 116. Write an equation for the sinusoidal function. Determine the amplitude and period of each function. 117. y 5 22 cos 12px2 118. y 5 1 3 sin p 2 x 119. y 5 1 5 sin 3x 120. y 5 27 6 cos 6x Graph each function from 22p to 2p. 121. y 5 22 sin x 2 122. y 5 3 sin 3x 123. y 5 1 2 cos 2x 124. y 5 21 4 cos x 2 State the amplitude, period, phase shift, and vertical shift of each function. 125. y 5 2 1 3 sin ax 2 p 2 b 126. y 5 3 2 1 2 sin ax 1 p 4 b 127. y 5 22 2 4 cos 3 ax 1 p 4 b 128. y 5 21 1 2 cos 2 ax 2 p 3 b 129. y 5 21 2 1 1 3 cos apx 2 1 2b 130. y 5 3 4 2 1 6 sin ap 6 x 1 p 3 b Graph each function from 2p to p. 131. y 5 3x 2 cos 2x 132. y 5 21 2 cos 4x 1 1 2 cos 2x 133. y 5 2 sin 1 3 x 2 3 sin 3x 134. y 5 5 cos x 1 3 sin x 2 x y π –π 2π –2π –5 –4 –3 –2 –1 1 2 3 4 5 x y π –π 2π –2π –5 –4 –3 –2 –1 1 2 3 4 5 R E VI E W E XERCISES 6.9 Graphs of Other Trigonometric Functions State the domain and range of each function. 135. y 5 4 tan ax 1 p 2 b 136. y 5 cot 2 ax 2 p 2 b 137. y 5 3 sec 12x2 138. y 5 1 1 2 csc x 139. y 5 21 2 1 1 4 secapx 2 2p 3 b 140. y 5 3 2 1 2 csc12x 2 p2 Graph each function on the interval 322p, 2p4. 141. y 5 2tan ax 2 p 4 b 142. y 5 1 1 cot12x2 143. y 5 2 1 sec1x 2 p2 144. y 5 2cscax 1 p 4 b 145. y 5 1 2 1 2 csca2x 2 p 2 b 146. y 5 21 2 1 2 secapx 2 3p 4 b Technology Exercises Section 6.1 Assume a 308–608–908 triangle. Round your answers to two decimal places. 147. If the shorter leg has length 41.32 feet, what are the lengths of the other leg and hypotenuse? 148. If the longer leg has length 87.65 centimeters, what are the lengths of the other leg and the hypotenuse? Section 6.2 149. Calculate csc 78.4° in the following two ways: a. Find sin 78.4° to three decimal places and then divide 1 by that number. Write that number to five decimal places. b. In a calculator: 78.4, sin, 1/x, round to five decimal places. 150. Calculate cot 34.8° in the following two ways: a. Find tan 34.8° to three decimal places and then divide 1 by that number. Write that number to five decimal places. b. In a calculator: 34.8, tan, 1/x, round to five decimal places. 151. Use a calculator to find tan1tan212.6122. 152. Use a calculator to find cos1cos210.1252. Use a calculator to evaluate the following expressions. If you get an error, explain why. 153. sec 180° 154. csc 180° Section 6.5 155. Use a calculator to evaluate csc 2188 and csc 3228. Now use the calculator to evaluate csc 21121.62432. When cosecant is negative, in which of the quadrants, III or IV, does the calculator assume the terminal side of the angle lies? 156. Use a calculator to evaluate sec 288 and sec 3328. Now use the calculator to evaluate sec 2111.13262. When secant is positive, in which of the quadrants, I or IV, does the calculator assume the terminal side of the angle lies? Section 6.6 Find the measure (in degrees, minutes, and nearest seconds) of a central angle u that intercepts an arc on a circle with radius r with indicated arc length s. Use the TI calculator comands ANGLE and DMS to change to degrees, minutes, and seconds. 157. r 5 11.2 ft, s 5 19.7 ft 158. r 5 56.9 cm, s 5 139.2 cm Section 6.7 In Exercises 159 and 160, refer to the following: A graphing calculator can be used to graph the unit circle with parametric equations (these will be covered in more detail in Section 8.7). For now, set the calculator in parametric and radian modes and let X1 5 cos T Y 1 5 sin T Set the window so that 0 # T # 2p, step 5 p 15, 22 # X # 2, and 22 # Y # 2. To approximate the sine or cosine of a T value, use the TRACE key, enter the T value, and read the corresponding coordinates from the screen. 159. Use the above steps to approximate cos a13p 12 b to four decimal places. 160. Use the above steps to approximate sin a5p 6 b to four decimal places. Section 6.8 161. Use a graphing calculator to graph Y 1 5 cos x and Y 2 5 cos 1x 1 c2, where a. c 5 p 6, and explain the relationship between Y 2 and Y 1. b. c 5 2p 6, and explain the relationship between Y 2 and Y 1. 162. Use a graphing calculator to graph Y 1 5 sin x and Y 2 5 sin x 1 c, where a. c 5 1 2, and explain the relationship between Y 2 and Y 1. b. c 5 21 2, and explain the relationship between Y 2 and Y1. Section 6.9 163. What is the amplitude of the function y 5 4 cos x 2 3 sin x? Use a graphing calculator to graph Y 1 5 4 cos x, Y 253 sin x, and Y 3 5 4 cos x 2 3 sin x in the same viewing window. 164. What is the amplitude of the function y 5 !3 sin x 1 cos x? Use a graphing calculator to graph Y 1 5 !3 sin x, Y 2 5 cos x, and Y 3 5 !3 sin x 1 cos x in the same viewing window. Review Exercises 649 PR ACTICE TEST [ C H AP T E R 6 PRACTICE TEST ] 650 CHAPTER 6 Trigonometric Functions 1. A 5-foot girl is standing in the Grand Canyon, and she wants to estimate the depth of the canyon. The Sun casts her shadow 6 inches along the ground. To measure the shadow cast by the top of the canyon, she walks the length of the shadow. She takes 200 steps and estimates that each step is roughly 3 feet. Approximately how tall is the Grand Canyon? 2. Fill in the values in the table. u sin u cos u tan u cot u sec u csc u 308 458 608 3. What is the difference between cos u 5 2 3 and cos u < 0.66? 4. Fill in the table with exact values for the quadrantal angles and the algebraic signs for the quadrants. 08 QI 908 QII 1808 QIII 2708 QIV 3608 sin u cos u 5. If cot u , 0 and sec u . 0, in which quadrant does the terminal side of u lie? 6. Evaluate sin 210° exactly. 7. Convert 13p 4 to degree measure. 8. Convert 260° to radian measure. Leave the answer in terms of p. 9. What is the area of the sector swept by the second hand of a clock in 25 seconds? Assume the radius of the sector is 3 inches. 10. What is the measure in radians of the smaller angle between the hour and minute hands at 10:10? 11. State the amplitude and period of y 5 25 sin13x2. 12. Graph y 5 22 cos A 1 2xB over 24p # x # 4p. 13. Graph y 5 tan apx 2 p 2 b over two periods. 14. The vertical asymptotes of y 5 2 csc13x 2 p2 correspond to the _ of y 5 2 sin 13x 2 p2. 15. State the x-intercepts of y 5 tan12x2 for all x. 16. State the phase shift and vertical shift for y 5 2cot ap 3 x 2 pb. 17. State the range of y 5 23 seca2x 1 p 3 b 2 1. 18. State the domain of y 5 tan a2x 2 p 6 b 1 3. 19. Graph y 5 22 cscax 1 p 2 b over two periods. 20. Find the x-intercept(s) of y 5 6 !3 2 3 sec a6x 2 5p 6 b. 21. True or False: The equation 2 sin u 5 2.0001 has no solution. 22. On what x-intervals does the graph of y 5 cos12x2 lie below the x‑axis? 23. Write the equation of a sine function that has amplitude 4, vertical shift 1 2 down, phase shift 3 2 to the left, and period p. 24. Write the equation of a cotangent function that has period p. vertical shift 0.01 up, and no phase shift. 25. Graph y 5 cos 3x 2 1 2 sin 3x for 0 # x # p. 26. If the longer leg of a 308-608-908 triangle has length 13.95 inches, what are the lengths of the other leg and the hypotenuse? Round your answer to two decimal places. 27. Calculate sec141.982 in the following two ways: a. Find cos141.982 to three decimal places and then divide 1 by that number. Write that number to five decimal places. b. In a calculator: 41.9, cos, 1/x, round to five decimal places. CU MU LA TIV E TEST [CH AP TERS 1–6 CUM UL AT IVE T E S T ] 1. Simplify 6x2 2 11x 1 5 30x 2 25 . 2. Perform the operation and write in standard form a 6 bi: 15 2 7i2 2 111 2 9i2. 3. Solve for x and give any excluded values: 5 x 2 5 x 1 1 5 25 x2 1 x. 4. Solve for the radius r: A 5 pr2. 5. Solve and express the solution in interval notation: 2t2 t 2 1 $ 3t. 6. Calculate the distance and midpoint between the segment joining the points A23 4, 1 6B and A1 5, 22 3B. 7. Write an equation of a line in slope–intercept form that passes through the two points 15.6, 6.22 and 13.2, 5.02. 8. Use interval notation to express the domain of the function ƒ1x2 5 "x2 2 25. 9. Find the average rate of change for ƒ1x2 5 5 x, from x 5 2 to x 5 4. 10. Sketch the graph of the function y 5 1 x 2 2 1 1, and identify all transformations of the function ƒ1x2 5 1 x. 11. Evaluate g1ƒ12122 for ƒ1x2 5 ! 3 x 2 7 and g1x2 5 5 3 2 x. 12. Find the inverse of the function ƒ1x2 5 5x 1 2 x 2 3 . 13. Find the quadratic function that has the vertex 10, 72 and goes through the point 12, 212. 14. Find all of the real zeros and state the multiplicity of the function ƒ1x2 5 1 7 x5 1 2 9 x3. 15. List the possible rational zeros, and test to determine all rational zeros for P1x2 5 2x4 1 7x3 2 18x2 2 13x 1 10. 16. Factor the polynomial P1x2 5 4x4 2 4x3 1 13x2 1 18x 1 5 as a product of linear factors. 17. Graph the exponential function ƒ1x2 5 52x 2 1, and state the y-intercept, the domain, the range, and the horizontal asymptote. 18. How much money should be put in a savings account now that earns 5.2% a year compounded daily if you want to have $50,000 in 16 years? 19. Graph the function ƒ1x2 5 ln1x 1 12 2 3 using transforma-tion techniques. 20. Evaluate log4.7 8.9 using the change-of-base formula. Round the answer to three decimal places. 21. Solve the equation 51102x2 5 37 for x. Round the answer to three decimal places. 22. Height of a tree. The shadow of a tree measures 151 3 feet. At the same time of day the shadow of a 6-foot pole measures 2.3 feet. How tall is the tree? 23. The angle u in standard position has its terminal side defined by the line 3x 1 2y 5 0, x # 0. Calculate the values for the six trigonometric functions of u. 24. Find the area of a circular sector with radius 3.4 m and central angle u 5 35°. Round the answer to three significant digits. 25. State the amplitude, period, and phase shift of the function y 5 24 sin12x 1 p2. Cumulative Test 651 C H A P T E R LEARNING OBJECTIVES [ [ When you dial a phone number on your iPhone®, how does the smart phone know which key you have pressed? Dual Tone Multi-Frequency (DTMF), also known as touch-tone dialing, was developed by Bell Labs in the 1960s. The Touch-Tone® system also introduced a standardized keypad layout. After testing 18 different layouts, Bell Labs eventually chose the one familiar to us today, with 1 in the upper left and 0 at the bottom between the star and the pound keys. The keypad is laid out in a 4 3 3 matrix, with each row representing a low frequency and each column representing a high frequency. ■ ■Review basic identities. ■ ■Verify a trigonometric identity. ■ ■Apply the sum and difference identities. ■ ■Apply the double-angle identities. ■ ■Apply half-angle identities. ■ ■Apply the product-to-sum and sum-to-product identities. ■ ■Graph inverse trigonometric functions. ■ ■Solve trigonometric equations. Analytic Trigonometry 7 FREQUENCY 1209 HZ 1336 HZ 1477 HZ 697 Hz 1 2 3 770 Hz 4 5 6 852 Hz 7 8 9 941 Hz 0 # When you press the number 8, the phone sends a sinusoidal tone that combines a low- frequency tone of 852 hertz and a high-frequency tone of 1336 hertz. The result can be found using sum-to-product trigonometric identities. laurentiu iordache/Alamy Stock Photo 653 [I N T HI S CHAPTER] We will discuss trigonometric identities and use them to simplify trigonometric expressions. We will verify trigonometric identities. Specific identities that we will discuss are sum and difference, double-angle and half-angle, and product-to-sum and sum-to-product. Musical tones and touch-tone keypads are applications of trigonometric identities. We will also define inverse trigonometric functions. Trigonometric identities and inverse trigonometric functions will be used in solving trigonometric equations. ANALYTIC TRIGONOMETRY 7.1 BASIC TRIGONO-METRIC IDENTITIES 7.2 VERIFYING TRIGONO-METRIC IDENTITIES 7.3 SUM AND DIFFERENCE IDENTITIES 7.4 DOUBLE-ANGLE IDENTITIES 7.5 HALF-ANGLE IDENTITIES 7.6 PRODUCT-TO- SUM AND SUM-TO- PRODUCT IDENTITIES 7.7 INVERSE TRIGONO-METRIC FUNCTIONS 7.8 TRIGONO-METRIC EQUATIONS • Reciprocal Identities • Quotient Identities • Pythag-orean Identities • Trigonometric Identities • Sum and Difference Identities for the Cosine Function • Sum and Difference Identities for the Sine Function • Sum and Difference Identities for the Tangent Function • Applying Double- Angle Identities • Applying Half-Angle Identities • Product-to- Sum Identities • Sum-to- Product Identities • Inverse Sine Function • Inverse Cosine Function • Inverse Tangent Function • Remaining Inverse Trigonometric Functions • Finding Exact Values for Expressions Involving Inverse Trigonometric Functions • Solving Trigonometric Equations by Inspection • Solving Trigonometric Equations Using Algebraic Techniques • Solving Trigonometric Equations Which Require the Use of Inverse Functions • Using Trigonometric Identities to Solve Trigonometric Equations 654 CHAPTER 7 Analytic Trigonometry 7.1.1 Reciprocal Identities In this section, we revisit the reciprocal and quotient identities and develop the Pythagorean identities. These identities will allow us to simplify trigonometric expressions. In mathematics, an identity is an equation that is true for all values of the variable for which the equation is defined. For example, x2 5 x⋅x is an identity because the statement is true no matter what values are selected for x. Similarly, 2x 5x 5 2 5 is an identity, but this identity holds for all x except 0 1x 2 02, since the expression on the left is not defined when x 5 0. In our discussions on identities, it is important to note that the identities hold for all values of the argument, typically chosen to be u, for which the expressions are defined. Recall that the reciprocal of a nonzero number x is 1 x. For example, the reciprocal of 5 is 1 5 and the reciprocal of 9 17 is 17 9 . There is no reciprocal of 0. In mathematics, the expression 1 0 is undefined. Calculators have a reciprocal key typically labeled 1/x or x21. This can be used to find the reciprocal of any number (except 0). In Section 6.4, the trigonometric functions were defined by the following, remembering that r . 0: sin u 5 y r cos u 5 x r tan u 5 y x 1x 2 02 csc u 5 r y 1y 2 02 sec u 5 r x 1x 2 02 cot u 5 x y 1y 2 02 Derivation of the Reciprocal Identities WORDS MATH Write the definition of the cosecant function. csc u 5 r y 1y 2 02 The reciprocal identity holds for y 2 0. csc u 5 r y 5 1 y r 1y 2 0, r . 02 Substitute the definition of the sine function, Write the condition y 2 0 in terms of sin u. sin u 2 0 State the reciprocal identity. csc u 5 1 sin u 1sin u 2 02 sin u 5 y r. csc u 5 r y 5 1 y r 5 1 sin u S K I L L S O B J E C T I V E S ■ ■Apply the reciprocal identities. ■ ■Apply the quotient identities. ■ ■Apply the Pythagorean identities. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that trigonometric reciprocal identities are not always defined. ■ ■Understand that trigonometric quotient identities are not always defined. ■ ■Understand that the second and third Pythagorean identities are not always defined. 7.1 BASIC TRIGONOMETRIC IDENTITIES 7.1.1 S KILL Apply the reciprocal identities. 7.1.1 C ON CEPTUAL Understand that trigonometric reciprocal identities are not always defined. [CONCEPT CHECK] For what values of u is cosecant( u ) not defined? ANSWER u 5 6np for n 5 0, 1, 2, 3, … ▼ 7.1 Basic Trigonometric Identities 655 An equivalent reciprocal identity can be written as sin u 5 1 csc u. There is no need to note the restriction that csc u 2 0, since csc u 5 r y is always greater than or equal to 1. The following box summarizes the reciprocal identities that are true for all values of u that do not correspond to the trigonometric function in the denominator having a zero value. RECIPROCAL IDENTITIES RECIPROCAL IDENTITIES EQUIVALENT FORMS DOMAIN RESTRICTIONS csc u 5 1 sin u sin u 5 1 csc u u 2 np n 5 integer sec u 5 1 cos u cos u 5 1 sec u u 2 np 2 n 5 odd integer cot u 5 1 tan u tan u 5 1 cot u u 2 np 2 n 5 integer STUDY TIP sin u 5 0 when u 5 np n 5 1, 2, 3, . . . cos u 5 0 when u 5 np 2 n 5 1, 3, 5, . . . tan u 5 sin u cos u 5 0 when u 5 np n 5 1, 2, 3, . . . cot u 5 cos u sin u 5 0 when u 5 np 2 n 5 1, 3, 5, . . . STUDY TIP Reciprocals always have the same algebraic sign (1 or 2). Reciprocals always have the same algebraic sign. Therefore, if sine is positive, then cosecant is positive. If cosine is negative, then secant is negative. In this section, we assume that the domain restrictions apply even though they are not explicitly stated when using the reciprocal identities. EXAMPLE 1 Using Reciprocal Identities a. If cos u 5 21 2, find sec u. b. If sin u 5 !3 2 , find csc u. c. If tan u 5 c, find cot u. Solution (a): The secant function is the reciprocal of the cosine function. sec u 5 1 cos u Substitute cos u 5 2 1 2 into the secant expression. sec u 5 1 2 1 2 Simplify. sec u 5 22 656 CHAPTER 7 Analytic Trigonometry Solution (b): The cosecant function is the reciprocal of the sine function. csc u 5 1 sin u Substitute sin u 5 !3 2 into the cosecant expression. csc u 5 1 !3 2 Simplify. csc u 5 2 !3 Rationalize the radical in the denominator. csc u 5 2 !3⋅!3 !3 csc u 5 2!3 3 Solution (c): The cotangent function is the reciprocal of the tangent function. cot u 5 1 tan u Substitute tan u 5 c into the cotangent expression. cot u 5 1 c State any restrictions on c. cot u 5 1 c c 2 0 Y OUR TU R N a. If cos u 5 21, find sec u. b. If sin u 5 x, find csc u. ▼ ▼ A N S W E R a. sec u 5 21 b. csc u 5 1 x 1x 2 02 7.1.2 Quotient Identities Let us now use the trigonometric definitions to derive the quotient identities. WORDS MATH Write the definition of the tangent function. tan u 5 y x 1x 2 02 Multiply the two sides of the equation by 1 r 1 r . tan u 5 y r x r 1x 2 0, r . 02 Substitute sin u 5 y r and cos u 5 x r. tan u 5 y r x r 5 sin u cos u Write x 2 0 in terms of u. cos u 2 0 State the quotient identity. tanu 5 sinu cos u 1cosu 2 02 A similar quotient identity can be written as cot u 5 cos u sin u , where sin u 2 0. 7.1.2 SKI LL Apply the quotient identities. 7.1.2 CO NCE PTUAL Understand that trigonometric quotient identities are not always defined. The following box summarizes the quotient identities that are true for all values of u that do not correspond to a zero value in the denominator. QUOTIENT IDENTITIES QUOTIENT IDENTITIES DOMAIN RESTRICTIONS tan u 5 sin u cos u cos u 2 0 or u 2 np 2 n 5 odd integer cot u 5 cos u sin u sin u 2 0 or u 2 np n 5 integer In this section, we assume that the domain restrictions apply even though they are not explicitly stated when using the quotient identities. EXAMPLE 2 Using the Quotient Identities If sin u 5 3 5 and cos u 5 24 5, find tan u and cot u. Solution: Write the quotient identity involving the tangent and cotangent functions. tan u 5 sin u cos u and cot u 5 cos u sin u Substitute sin u 5 3 5 and cos u 5 24 5. tan u 5 a3 5b a24 5b and cot u 5 a 24 5b a3 5b Simplify. tan u 5 23 4 and cot u 5 24 3 Note: We could have found tan u and then used the reciprocal identity, cot u 5 1 tan u, to find the cotangent function value. YOUR T UR N If sin u 5 4 5 and cos u 5 3 5, find tan u and cot u. ▼ 7.1.3 Pythagorean Identities Before we develop the Pythagorean identities, we first need to look at notation. OPERATION SHORTHAND NOTATION WORDS EXAMPLE 1sin u22 sin2 u “Sine squared of theta” sin u 5 !3 2 , so sin2 u 5 a !3 2 b 2 5 3 4 1cos u23 cos3 u “Cosine cubed of theta” cos u 5 21 2, so cos3 u 5 a21 2b 3 5 21 8 [CONCEPT CHECK] For what values of u is tan u not defined? ANSWER u 5 6(2n 1 1)p 2 for n = 0, 1, 2, 3, … ▼ ▼ A N S W E R tan u 5 4 3 and cot u 5 3 4 7.1.3 S K IL L Apply the Pythagorean identities. 7.1.3 C ON C E P T U A L Understand that the second and third Pythagorean identities are not always defined. 7.1 Basic Trigonometric Identities 657 658 CHAPTER 7 Analytic Trigonometry Two important things to note at this time are: ■ ■The value of u must remain the same in all expressions for this identity to hold. ■ ■The sum of the squares of the sine and cosine functions of the same angle is always 1. Let’s illustrate the Pythagorean identity with both a special angle and a calculation: Let us now derive our first Pythagorean identity. Recall that in the Cartesian plane, the relationship between the coordinates x and y and the distance r from the origin to the point 1x, y2 is given by x2 1 y2 5 r 2. WORDS MATH Write the distance relationship in the Cartesian plane. x2 1 y2 5 r2 Divide the equation by r 2. x2 1 y2 r2 5 r2 r2 Simplify. x2 r2 1 y2 r2 5 1 Use the power rule of exponents, a2 b2 5 aa bb 2 . ax rb 2 1 ay rb 2 5 1 Substitute the sine and cosine definitions, sin u 5 y r and cos u 5 x r, into the left side of the equation. 1cos u22 1 1sin u22 5 1 Use shorthand notation. cos2 u 1 sin2 u 5 1 Note: cos2 u 1 sin2 u 5 1 and sin2u 1 cos2u 5 1 are equivalent identities. u sin u cos u sin2 u 1 cos2 u 5 1 p 6 1 2 !3 2 a1 2b 2 1 a !3 2 b 2 5 1 4 1 3 4 5 1 278 0.4540 0.8910 10.454022 1 10.891022 5 0.999997 < 1 We can derive the second Pythagorean identity from the first. WORDS MATH Write the first Pythagorean identity. sin2 u 1 cos2 u 5 1 Divide both sides by cos2 u. sin2 u cos2 u 1 cos2 u cos2 u 5 1 cos2 u Use longhand notation. 1sin u22 1cos u22 1 1cos u22 1cos u22 5 1 1cos u22 Use the property of exponents. a sin u cos ub 2 1 acos u cos ub 2 5 a 1 cos ub 2 Use the quotient and reciprocal identities tan u 5 sin u cos u and sec u 5 1 cos u to write the expressions in terms of the tangent and secant functions. 1tan u22 1 1 5 1sec u22 Use shorthand notation. tan2 u 1 1 5 sec2 u (x, y) x y r x y θ In this derivation, if we had divided by sin2 u instead of cos2 u, we would have arrived at the third Pythagorean identity, 1 1 cot2 u 5 csc2 u. PYTHAGOREAN IDENTITIES sin2 u 1 cos2 u 5 1 tan2 u 1 1 5 sec2 u 1 1 cot2 u 5 csc2 u EXAMPLE 3 Using the Identities to Find Trigonometric Function Values Find cos u and tan u if sin u 5 4 5 and the terminal side of u lies in quadrant II. Solution: Write the first Pythagorean identity. sin2 u 1 cos2 u 5 1 Substitute sin u 5 4 5 into the equation. a4 5b 2 1 cos2 u 5 1 Eliminate the parentheses. 16 25 1 cos2 u 5 1 Subtract 16 25 from both sides of the equation. cos2 u 5 9 25 Apply the square root property. cos u 5 6Å 9 25 Simplify. cos u 5 63 5 The cosine function is negative in quadrant II. cos u 5 23 5 Write the quotient identity for the tangent function. tan u 5 sin u cos u Substitute sin u 5 4 5 and cos u 5 2 3 5. tan u 5 4 5 2 3 5 Simplify. tan u 5 2 4 3 Y OUR T UR N Find sin u and tan u if cos u 524 5 and the terminal side of u lies in quadrant III. ▼ STUDY TIP The second and third identities hold only for values of u for which the functions are defined. [CONCEPT CHECK] TRUE OR FALSE sin2 (x) 1 cos2 (y) 5 1, for all values of x and y. ANSWER False ▼ ▼ A N S W E R sin u 5 23 5 and tan u 5 3 4 7.1 Basic Trigonometric Identities 659 660 CHAPTER 7 Analytic Trigonometry EXAMPLE 5 Using Identities to Simplify Expressions Multiply 11 2 cos u2 11 1 cos u2 and simplify. Solution: Multiply using the FOIL method. 11 2 cos u2 11 1 cos u2 5 1 2 cos u 1 cos u 2 cos2 u Combine like terms. 5 1 2 cos u 1 cos u 2 cos2 u Rewrite the Pythagorean identity, sin2 u 1 cos2 u 5 1 as sin2 u 5 1 2 cos2 u. 5 1 2 cos2 u 11 2 cos u211 1 cos u2 5 sin2 u Y OUR TU R N Multiply 11 2 sin u2 11 1 sin u2 and simplify. sin2 u g ▼ ▼ A N S W E R cos2 u EXAMPLE 4 Using the Identities to Find Trigonometric Function Values Find sin u and cos u if tan u 5 3 4 and the terminal side of u lies in quadrant III. Solution: Write the second Pythagorean identity. tan2 u 1 1 5 sec2 u Substitute tan u 5 3 4 into the equation. a3 4b 2 1 1 5 sec2 u Eliminate the parentheses. 9 16 1 1 5 sec2 u Simplify. sec2 u 5 25 16 Apply the square root property. sec u 5 6Å 25 16 5 65 4 Substitute the reciprocal identity, sec u 5 1 cos u. 1 cos u 5 65 4 Solve for cos u. cos u 5 64 5 The cosine function is negative in quadrant III. cos u 5 24 5 Write the first Pythagorean identity. sin2 u 1 cos2 u 5 1 Substitute cos u 5 24 5. sin2 u 1 a24 5b 2 5 1 Simplify. sin2 u 5 9 25 Apply the square root property. sin u 5 6Å 9 25 5 63 5 The sine function is negative in quadrant III. sin u 5 23 5 PY T H A GO R EA N I D EN T I T I ES sin2 u 1 cos2 u 5 1 tan2 u 1 1 5 sec2 u 1 1 cot2 u 5 csc2 u These basic trigonometric identities (reciprocal, quotient, and Pythagorean) are used to calculate trigonometric function values given other trigonometric function values and information regarding the quadrant where the terminal side of the angle lies. RE C I P R OC AL I D E NT I T IE S csc u 5 1 sin u sec u 5 1 cos u cot u 5 1 tan u QUO T IE NT I D E NT I T IE S tan u 5 sin u cos u cot u 5 cos u sin u [SEC TION 7.1] S U M M A RY [SEC TION 7.1] E X ERC I S E S • S K I L L S In Exercises 1–12, use a reciprocal identity to find the function value indicated. Rationalize the denominator, if necessary. 1. If cos u 5 7 8, find sec u. 2. If sin u 5 23 7, find csc u. 3. If sin u 5 20.6, find csc u. 4. If cos u 5 0.8, find sec u. 5. If tan u 5 25, find cot u. 6. If tan u 5 0.5, find cot u. 7. If csc u 5 !5 2 , find sin u. 8. If sec u 5 !11 2 , find cos u. 9. If cot u 5 2 !7 5 , find tan u. 10. If cot u 5 3.5, find tan u. 11. If sec u 5 24, find cos u. 12. If csc u 5 2!5 find sin u. In Exercises 13–20, use a quotient identity to find the function value indicated. Rationalize the denominator, if necessary. 13. If sin u 5 21 2 and cos u 5 !3 2 , find tan u. 14. If sin u 5 21 2 and cos u 5 !3 2 , find cot u. 15. If sin u 5 20.6 and cos u 5 20.8, find cot u. 16. If sin u 5 20.6 and cos u 5 20.8, find tan u. 17. If sin u 5 2 !11 6 and cos u 5 25 6, find tan u. 18. If sin u 5 2 !11 6 and cos u 5 25 6, find cot u. 19. If sin u 5 a and cos u 5 b, find cot u and state any restrictions on a or b. 20. If sin u 5 a and cos u 5 b, find tan u and state any restrictions on a or b. In Exercises 21–26, find the indicated expression. 21. If sin u 5 !5 8 , find sin2 u. 22. If cos u 5 0.1, find cos3 u. 23. If csc u 5 22, find csc3 u. 24. If tan u 5 25, find tan2 u. 25. If cot u 5 2! 3 5 , find cot3 u. 26. If sec u 5 !7 2 , find sec4 u. In Exercises 27–46, use a Pythagorean identity to find the function value indicated. Rationalize the denominator, if necessary. 27. If sin u 5 21 2 and the terminal side of u lies in quadrant III, find cos u. 28. If sin u 5 23 5 and the terminal side of u lies in quadrant III, find cos u. 7.1 Basic Trigonometric Identities 661 662 CHAPTER 7 Analytic Trigonometry 29. If cos u 5 2 5 and the terminal side of u lies in quadrant IV, find sin u. 30. If cos u 5 2 7 and the terminal side of u lies in quadrant IV, find sin u. 31. If tan u 5 4 and the terminal side of u lies in quadrant III, find sec u. 32. If tan u 5 25 and the terminal side of u lies in quadrant II, find sec u. 33. If cot u 5 2 and the terminal side of u lies in quadrant III, find csc u. 34. If cot u 5 23 and the terminal side of u lies in quadrant II, find csc u. 35. If sin u 5 8 15 and the terminal side of u lies in quadrant II, find tan u. 36. If sin u 5 2 7 15 and the terminal side of u lies in quadrant III, find tan u. 37. If cos u 5 2 7 15 and the terminal side of u lies in quadrant III, find csc u. 38. If cos u 5 2 8 15 and the terminal side of u lies in quadrant II, find csc u. 39. If sec u 5 !13 3 and the terminal side of u lies in quadrant IV, find sin u. 40. If sec u 5 2 !61 5 and the terminal side of u lies in quadrant II, find sin u. 41. If sin u 5 2 !3 3 and the terminal side of u lies in quadrant IV, find sec u. 42. If cos u 5 0.15 and the terminal side of u lies in quadrant IV, find csc u. 43. If tan u 5 !5 and the terminal side of u lies in quadrant I, find sec u. 44. If cot u 5 25 and the terminal side of u lies in quadrant IV, find csc u. 45. If csc u 5 24!2 and the terminal side of u lies in quadrant III, find cos u. 46. If csc u 5 !11 and the terminal side of u lies in quadrant I, find cos u. In Exercises 47–56, use identities to find the function values indicated. Rationalize the denominator, if necessary. 47. Find sin u and cos u if tan u 5 24 3 and the terminal side of u lies in quadrant II. 48. Find sin u and cos u if tan u 5 23 4 and the terminal side of u lies in quadrant II. 49. Find sin u and cos u if tan u 5 2 and the terminal side of u lies in quadrant III. 50. Find sin u and cos u if tan u 5 5 and the terminal side of u lies in quadrant III. 51. Find sin u and cos u if tan u 5 0.6 and the terminal side of u lies in quadrant III. 52. Find sin u and cos u if tan u 5 0.8 and the terminal side of u lies in quadrant III. 53. Find sin u and cos u if cot u 5 25 2 and the terminal side of u lies in quadrant IV. 54. Find sin u and cos u if cot u 5 0.1 and the terminal side of u lies in quadrant III. 55. Find sin u and cos u if cot u 5 1 4 and the terminal side of u lies in quadrant I. 56. Find sin u and cos u if cot u 5 22.2 and the terminal side of u lies in quadrant II. In Exercises 57–70, perform the indicated operations and simplify your answers, if possible. Leave all answers in terms of sin u and cos u. 57. sec u cot u 58. csc u tan u 59. tan2 u 2 sec2 u 60. cot2 u 2 csc2 u 61. csc u 2 sin u 62. sec u 2 cos u 63. sec u tan u 64. csc u cot u 65. 1sin u 1 cos u22 66. 1sin u 2 cos u22 67. 11 1 tan u22 68. 11 1 cot u22 69. cot u 1 tan u sec u csc u 70. sin u cos u tan u 1 cot u For Exercises 71 and 72, refer to the following: The bifolium is a curve that can be drawn using either an algebraic equation or an equation involving trigonometric functions. Even though the trigonometric equation uses polar coordinates, it’s much easier to solve the trigonometric equation for function values than the algebraic equation. The graph of r 5 8 sin u cos2 u is shown below. You’ll see more on polar coordinates and graphs of polar equations in Chapter 8. The following graph is in polar coordinates. x y 71. Bifolium. The equation for the bifolium shown above is r 5 8 sin u cos2 u. Use a Pythagorean identity to rewrite the equation using only the function sin u. Then find r if u 5 30°, 60°, and 90°. 72. Bifolium. The equation for the bifolium shown on the left is r 5 8 sin u cos2 u. Use a Pythagorean identity to rewrite the equation using only the function sin u. Then find r if u 5 230°, 260°, and 290°. For Exercises 73 and 74, refer to the following: The monthly revenues (measured in thousands of dollars) of PizzaRia are a function of cost (measured in thousands of dollars), that is, R(c). Recall that the angle u can be interpreted as a measure of the sizes of c (cost) and R (revenue) relative to each other. The larger the angle u is, the greater revenue is relative to cost, and conversely, the smaller the angle u is, the smaller revenue is relative to cost. The ratio of revenue to costs can be approximated by the tangent function, tan u < R c. 73. Business. An analysis of the revenue and the costs of one month determined that cos u 5 3 5 and sin u 5 4 5. Find tan u and interpret its meaning. 74. Business. An analysis of the revenue and the costs of one month determined that cos u 5 2"5 5 and sin u 5 "5 5 . Find tan u and interpret its meaning. • A P P L I C A T I O N S • C A T C H T H E M I S T A K E In Exercises 75 and 76, explain the mistake that is made. 75. If sin u 5 21 3 and the terminal side of u lies in quadrant III, find cos u. Solution: Write the first Pythagorean identity. sin2 u 1 cos2 u 5 1 Substitute sin u 5 21 3 into the identity. Eliminate parentheses. 1 9 1 cos2 u 5 1 Subtract 1 9 from both sides. cos2 u 5 8 9 Take the square root of both sides. cos u 5 Ä 8 9 Simplify. cos u 5 2!2 3 This is incorrect. What mistake was made? 76. Find sin u and cos u if tan u 5 21 4 and the terminal side of u lies in quadrant II. Solution: Write the quotient identity. tan u 5 sin u cos u Substitute tan u 5 21 4 into the identity. In quadrant II, sine is positive and cosine is negative. Identify the sine and cosine values. sin u 5 1 and cos u 5 24 This is incorrect. What mistake was made? a2 1 3b 2 1 cos2 u 5 1 21 4 5 sin u cos u 1 24 5 sin u cos u 7.1 Basic Trigonometric Identities 663 664 CHAPTER 7 Analytic Trigonometry In Exercises 77 and 78, determine whether each statement is true or false. • C O N C E P T U A L 77. It is possible for sin u . 0 and csc u , 0. 78. It is possible for cos u , 0 and sec u . 0. 79. Find the measure of an angle u, 0° , u # 180°, which satisfies the equation sin u 5 csc u. 80. Find the measure of an angle u, 0° , u # 180°, which satisfies the equation cos u 5 sec u. 81. Write cot u only in terms of sin u. 82. Write csc u only in terms of cos u. • C H A L L E N G E 83. Let x 5 8 sin u in the expression "64 2 x2 and simplify. 84. Let x 5 6 cos u in the expression "36 2 x2 and simplify. 85. Let x 5 5 tan u in the expression "25 1 x2 and simplify. 86. Simplify cos2 u cos2 u 21⋅ 1 2 csc2 u tan2 u 2 sec2 u. In Exercises 87 and 88, determine whether each statement is true or false. 87. sec2 u $ 1, for all u. 88. 1 2 csc2 u . 0, for all u. • T E C H N O L O G Y 89. Verify the quotient tan u 5 sin u cos u numerically for u 5 22°. Use a calculator to find a. sin 22° b. cos 22° c. sin22° cos 22° d. tan 22° Are the results in (c) and (d) the same? 90. Is tan10° 5 sin160° cos16° ? Use a calculator to find a. sin160° b. cos16° c. sin160° cos16° d. tan10° Are the results in (c) and (d) the same? 91. Does sin 35° 5 "1 2 cos2 35°? Use a calculator to find: a. sin 35° b. 1 2 cos 35° c. "1 2 cos2 35° Which results are the same? 92. Does sec 68° 5 "1 1 tan2 68°? Use a calculator to find: a. sec 68° b. 1 1 tan68° c. "1 1 tan2 68° Which results are the same? 7.2.1 Trigonometric Identities Basic Identities In Section 7.1, we discussed the basic (fundamental) trigonometric identities: reciprocal, quotient, and Pythagorean. In mathematics, an identity is an equation that is true for all values of the variable for which the expressions in the equation are defined. If an equation is true for only some values of the variable, it is a conditional equation. If an equation is true for no values of the variable, it is a contradiction. The following are identities (true for all x for which the expressions are defined): S K I L L S O B J E C T I V E ■ ■Simplify trigonometric expressions using identities. C O N C E P T U A L O B J E C T I V E ■ ■Understand that there is more than one way to verify an identity. 7.2 VERIFYING TRIGONOMETRIC IDENTITIES 7.2.1 S K I L L Simplify trigonometric expressions using identities. 7.2.1 C O N C E P T U A L Understand that there is more than one way to verify an identity. IDENTITY TRUE FOR THESE VALUES OF x x2 1 3x 1 2 5 1x 1 221x 1 12 All real numbers tan x 5 sin x cos x All real numbers except x 5 np 2 , where n is an odd integer sin2 x 1 cos2 x 5 1 All real numbers The following are conditional equations (true only for particular values of x): EQUATION TRUE FOR ONLY THESE VALUES OF x x2 1 3x 1 2 5 0 x 5 22 and x 5 21 tan x 5 0 x 5 np, where n is an integer sin2 x 2 cos2 x 5 1 x 5 np 2 , where n is an odd integer EQUATION TRUE FOR THESE VALUES OF x x 1 5 5 x 1 7 none sin2 x 1 cos2 x 5 5 none The following are contradictions (not true for any values of x): The following boxes summarize the identities that were discussed in Section 7.1: RECIPROCAL IDENTITIES RECIPROCAL IDENTITIES EQUIVALENT FORMS DOMAIN RESTRICTIONS csc x 5 1 sin x sin x 5 1 csc x x 2 np n 5 integer sec x 5 1 cos x cos x 5 1 sec x x 2 np 2 n 5 odd integer cot x 5 1 tan x tan x 5 1 cot x x 2 np 2 n 5 integer STUDY TIP Just because an equation is true for some values of x does not mean it is an identity. 7.2 Verifying Trigonometric Identities 665 666 CHAPTER 7 Analytic Trigonometry QUOTIENT IDENTITIES QUOTIENT IDENTITIES DOMAIN RESTRICTIONS tan x 5 sin x cos x cos x 2 0 x 2 np 2 n 5 odd integer cot x 5 cos x sin x sin x 2 0 x 2 np n 5 integer PYTHAGOREAN IDENTITIES PYTHAGOREAN IDENTITIES DOMAIN RESTRICTIONS sin2 x 1 cos2 x 5 1 tan2 x 1 1 5 sec2 x cos x 2 0 or x 2 np 2 n 5 odd integer 1 1 cot2 x 5 csc2 x sin x 2 0 or x 2 np n 5 integer In previous chapters, we discussed even and odd functions that have these respective properties: TYPE OF FUNCTION ALGEBRAIC IDENTITY GRAPH Even ƒ12x2 5 ƒ1x2 Symmetry about the y-axis Odd ƒ12x2 5 2ƒ1x2 Symmetry about the origin We already learned in Chapter 6 that the sine function is an odd function and the cosine function is an even function. Combining this knowledge with the reciprocal and quotient identities, we arrive at the even-odd identities, which we can add to our list of basic identities. EVEN-ODD IDENTITIES sin12x2 5 2sin x csc12x2 5 2csc x cos 12x2 5 cos x Odd tan12x2 5 2tan x sec12x2 5 sec x cot12x2 5 2cot x g g Even Simplifying Trigonometric Expressions Using Identities In Section 7.1, we used the basic trigonometric identities to find values for trigonometric functions, and we simplified trigonometric expressions using the identities. We now will use the basic identities and algebraic manipulation to simplify more complicated trigonometric expressions. In simplifying trigonometric expressions, one approach is to first convert all expressions into sines and cosines and then simplify. EXAMPLE 1 Simplifying Trigonometric Expressions Simplify tan x sin x 1 cos x. Solution: Write the tangent function in terms of the sine tan x⋅sin x 1 cos x and cosine functions: tan x 5 sin x cos x. 5 a sin x cos x bsin x 1 cos x Simplify. 5 sin2 x cos x 1 cos x Write as a fraction with a single quotient by finding a common denominator, cos x. Use the Pythagorean identity: sin2 x 1 cos2 x 5 1. Use the reciprocal identity sec x 5 1 cos x. 5 sec x YOUR T UR N Simplify cot x cos x 1 sin x. 5 sin2 x 1 cos2 x cos x 5 1 cos x ▼ ▼ A N S W E R csc x In Example 1, tan x and sec x are not defined for odd integer multiples of p 2 . In Your Turn, cot x and csc x are not defined for integer multiples of p. Both the original expression and the simplified form are governed by the same restrictions. There are times when the original expression is subject to more domain restrictions than the simplified form, and thus special attention must be given to domain restrictions. For example, the algebraic expression x2 2 1 x 1 1 is under the domain restriction x 2 21 because that value for x makes the value of the denominator equal to zero. If we forget to state the domain restrictions, we might simplify the algebraic expression as x2 2 1 x 1 1 5 1x 2 121x 1 12 1x 1 12 5 x 2 1 and assume this is true for all values of x. The correct simplification is x2 2 1 x 1 1 5 x 2 1 for x 2 21. In fact, if we were to graph both the original expression y 5 x2 2 1 x 1 1 and the line y 5 x 2 1, they would coincide, except that the graph of the original expression would have a “hole” or discontinuity at x 5 1. In this chapter, it is assumed that the domain of the simplified expression is the same as the domain of the original expression. –5 5 x y y = x – 1 –5 5 y = x + 1 x2 – 1 hole 7.2 Verifying Trigonometric Identities 667 668 CHAPTER 7 Analytic Trigonometry EXAMPLE 2 Simplifying Trigonometric Expressions Simplify 1 csc2 x 1 1 sec2 x . Solution: Rewrite the expression in terms of quotients squared. Use the reciprocal identities: sin x 5 1 csc x and cos x 5 1 sec x. 5 sin2 x 1 cos2 x Use the Pythagorean identity: sin2 x 1 cos2 x 5 1. 5 1 Y OUR TU R N Simplify 1 cos2 x 2 1. 1 csc2 x 1 1 sec2 x 5 a 1 csc xb 2 1 a 1 sec xb 2 sin x g g cos x ▼ ▼ A N S W E R tan2 x Verifying Identities We will now use the trigonometric identities to verify, or establish, other trigonometric identities. For example, verify that 1sin x 2 cos x22 2 1 5 22 sin x cos x The good news is that we will know we are done when we get there, since we know the desired identity. But how do we get there? How do we verify that the identity is true? Remember that it must be true for all x, not just some x. Therefore, it is not enough to simply select values for x and show it is true for those specific values. WORDS MATH Start with one side of the equation (the more complicated side). 1sin x 2 cos x22 2 1 Remember that 1a 2 b22 5 a2 2 2ab 1 b2 and expand 1sin x 2 cos x22. 5 sin2 x 2 2 sin x cos x 1 cos2 x 2 1 Group the sin2 x and cos2 x terms and use the Pythagorean identity. 5 22 sin x cos x 1 1sin2 x 1 cos2 x2 2 1 Simplify. 5 22 sin x cos x g 1 When we arrive at the right side of the equation, then we have succeeded in verifying the identity. In verifying trigonometric identities, there is no one procedure that works for all identities. You must manipulate one side of the equation until it looks like the other side. Here are two suggestions that are generally helpful: 1. Convert all trigonometric expressions to sines and cosines. 2. Write all sums or differences of fractions (quotients) as a single fraction (quotient). The following suggestions help guide the way to verifying trigonometric identities. It is important to note that trigonometric identities must be valid for all values of the independent variable (usually x or u) for which the expressions in the equation are defined (domain of the equation). GUIDELINES FOR VERIFYING TRIGONOMETRIC IDENTITIES ■ Start with the more complicated side of the equation. ■ Combine all sums and differences of fractions (quotients) into a single fraction (quotient). ■ Use basic trigonometric identities. ■ Use algebraic techniques to manipulate one side of the equation until the other side of the equation is achieved. ■ Sometimes it is helpful to convert all trigonometric functions into sines and cosines. EXAMPLE 3 Verifying Trigonometric Identities Verify the identity 11 1 sin x231 1 sin12x24 5 cos2 x. Solution: Start with the more complicated side (left side). 11 1 sin x231 1 sin12x24 The sine function is odd: sin12x2 5 2sin x. 5 11 1 sin x211 2 sinx2 Eliminate the parentheses. 5 1 2 sin2 x Apply the Pythagorean identity, sin2 x 1 cos2 x 5 1. 5 cos2 x EXAMPLE 4 Verifying Trigonometric Identities Verify the identity tan x 2 cot x tan x 1 cot x 5 sin2 x 2 cos2 x. Solution: Start with the more complicated side of the equation. Use the quotient identities to write the tangent and cotangent functions in terms of the sine and cosine functions. Multiply by sin x cos x sin x cos x. 5 ± sin x cos x 2 cos x sin x sin x cos x 1 cos x sin x ≤asin x cos x sin x cos x b Simplify. 5 sin2 x 2 cos2 x sin2 x 1 cos2 x Use the Pythagorean identity: sin2 x 1 cos2 x 5 1. 5 sin2 x 2 cos2 x tan x 2 cot x tan x 1 cot x 5 sin x cos x 2 cos x sin x sin x cos x 1 cos x sin x 7.2 Verifying Trigonometric Identities 669 670 CHAPTER 7 Analytic Trigonometry EXAMPLE 5 Determining Whether a Trigonometric Equation Is an Identity Determine whether 11 2 cos2 x211 1 cot2 x2 5 0 is an identity, a conditional equation, or a contradiction. Solution: Use the quotient identity to write the cotangent function in terms of the sine and cosine functions. 11 2 cos2 x211 1 cot2 x2 5 11 2 cos2 x2 a1 1 cos2 x sin2 x b Combine the expression in the second parentheses so that it is a single quotient. Use the Pythagorean identity. 5 11 2 cos2 x2 asin2 x 1 cos2 x sin2 x b Eliminate the parentheses. 5 sin2 x sin2 x Simplify. 5 1 Since 1 2 0, this is not an identity. Therefore, it is either a conditional equation (true for some real values of x) or a contradiction (true for no real values of x). Note that for 11 2 cos2 x211 1 cot2 x2 5 0, then either 11 2 cos2 x2 5 0 or 11 1 cot 2 x2 5 0. Since 1 1 cot2 x . 0, then 12 cos2 x 5 0. cos2 x 5 1 when x 5 np. Since cot x is not defined for x 5 np, we say that this equation is a contradiction. 5 11 2 cos2 x2 asin2 x 1 cos2 x sin2 x b 1 g g sin2 x [CONCEPT CHECK] TRUE OR FALSE There is only one approach that can be used to verify an identity. ANSWER False ▼ EXAMPLE 6 Verifying Trigonometric Identities Verify that sin12x2 cos12x2 tan12x2 5 1. Solution: Start with the left side of the equation. sin12x2 cos12x2 tan12x2 Use the even-odd identities. 5 2sin x 2cos x tan x Simplify. 5 sin x cos x tan x Use the quotient identity to write the tangent function in terms of the sine and cosine functions. Divide out the cosine term in the denominator. 5 sin x sin x Simplify. 5 1 5 sin x cos x a sin x cos xb EXAMPLE 7 Verifying a More Complicated Identity Verify that 2 cot2 x 1 2 csc x 5 sin x 1 1 sin x . Solution: Start with the left side of the equation. 2 cot2 x 1 2 csc x Use the Pythagorean identity, cot2 x 5 csc2 x 2 1. 5 21csc2 x 2 12 1 2 csc x Distribute the negative throughout the numerator. 5 1 2 csc2 x 1 2 csc x Factor the numerator (difference of two squares). 5 11 2 csc x211 1 csc x2 1 2 csc x Divide out the 1 2 csc x in both the numerator and denominator. 5 1 1 csc x Use the reciprocal identity. 5 1 1 1 sin x Combine the two expressions into a single fraction. 5 sin x 1 1 sin x identities are (1) writing all of the trigonometric functions in terms of the sine and cosine functions and (2) combining sums or differences of quotients into a single quotient. When verifying trigonometric identities, we work with the more complicated side (keeping the other side in mind as our goal). We combined the basic trigonometric identities—reciprocal, quotient, Pythagorean, and even-odd—with algebraic techniques to simplify trigonometric expressions and verify more complex trigonometric identities. Two steps that we often use in both simplifying trigonometric expressions and verifying trigonometric [SEC TION 7. 2] S U M M A RY [SEC TION 7. 2] E X ERC I S E S • S K I L L S In Exercises 1–24, simplify each of the trigonometric expressions. 1. sin x csc x 2. tan x cot x 3. sec 12x2 cot x 4. tan12x2 cos12x2 5. csc12x2 sin x 6. cot12x2 tan1x2 7. sec1x2 cos12x2 1 tan2 x 8. sec12x2 tan12x2 cos12x2 9. 1sin2 x21cot2 x 1 12 10. 1cos2 x21tan2 x 1 12 11. 1sin x 2 cos x21sin x 1 cos x2 12. 1sin x 1 cos x22 13. csc x cot x 14. sec x tan x 15. 1 2 cot12x2 1 1 cot1x2 16. sec2 1x2 2 tan2 12x2 17. 1 2 cos4 x 1 1 cos2 x 18. 1 2 sin4 x 1 1 sin2 x 19. 1 2 cot4 x 1 2 cot2 x 20. 1 2 tan412x2 1 2 tan2 x 21. 1 2 sin2 x 1 2 cos x 22. 1 2 cos2 x 1 1 sin x 23. tan x 2 cot x tan x 1 cot x 1 2 cos2 x 24. tan x 2 cot x tan x 1 cot x 1 cos2 x 7.2 Verifying Trigonometric Identities 671 672 CHAPTER 7 Analytic Trigonometry In Exercises 25–50, verify each of the trigonometric identities. 25. 1sin x 1 cos x22 1 1sin x 2 cos x22 5 2 26. 11 2 sin x211 1 sin x2 5 cos2 x 27. 1csc x 1 121csc x 2 12 5 cot2 x 28. 1sec x 1 121sec x 2 12 5 tan2 x 29. tan x 1 cot x 5 csc x sec x 30. csc x 2 sin x 5 cot x cos x 31. 2 2 sin2 x cos x 5 sec x 1 cos x 32. 2 2 cos2 x sin x 5 csc x 1 sin x 33. 3cos12x2 2 1431 1 cos x4 5 2sin2 x 34. tan12x2 cot1x2 5 21 35. sec12x2 cot1x2 csc12x2 5 21 36. csc12x2 2 1 5 cot2 x csc12x2 1 1 37. 1 csc2 x 1 1 sec2 x 5 1 38. 1 cot2 x 2 1 tan2 x 5 sec2 x 2 csc2 x 39. 1 1 2 sin x 1 1 1 1 sin x 5 2 sec2 x 40. 1 1 2 cos x 1 1 1 1 cos x 5 2 csc2 x 41. sin2 x 1 2 cos x 5 1 1 cos x 42. cos2 x 1 2 sin x 5 1 1 sin x 43. sec x 1 tan x 5 1 sec x 2 tan x 44. csc x 1 cot x 5 1 csc x 2 cot x 45. csc x 2 tan x sec x 1 cot x 5 cos x 2 sin2 x sin x 1 cos2 x 46. sec x 1 tan x csc x 1 1 5 tan x 47. cos2 x 1 1 1 sin x cos 2 x 1 3 5 1 1 sin x 2 1 sin x 48. sin x 1 1 2 cos2 x cos2 x 5 sin x 1 2 sin x 49. sec x 1tan x 1 cot x2 5 csc x cos2 x 50. tan x 1csc x 2 sin x2 5 cos x In Exercises 51–64, determine whether each equation is an identity, a conditional equation, or a contradiction. 51. cos2 x 1tan x 2 sec x21tan x 1 sec x2 5 1 52. cos2 x 1tan x 2 sec x21tan x 1 sec x2 5 sin2 x 2 1 53. csc x cot x sec x tan x 5 cot3 x 54. sin x cos x 5 0 55. sin x 1 cos x 5 !2 56. sin2 x 1 cos2 x 5 1 57. tan2 x 2 sec2 x 5 1 58. sec2 x 2 tan2 x 5 1 59. sin x 5 "1 2 cos2 x 60. csc x 5 "1 1 cot2 x 61. "sin2 x 1 cos2 x 5 1 62. "sin2 x 1 cos2 x 5 sin x 1 cos x 63. 1sin x 2 cos x22 5 sin2 x 2 cos2 x 64. 3sin12x2 2 143sin12x2 1 14 5 cos2 x • A P P L I C A T I O N S For Exercises 65 and 66, refer to the following: In calculus, when integrating expressions such as "a2 2 x2, "a2 1 x2, and "x2 2 a2, trigonometric functions are used as “dummy” functions to eliminate the radical. Once the integration is performed, the trigonometric function is “unsubstituted.” The trigonometric substitutions shown (and corresponding trigonometric identities) are used to simplify these types of expressions. When simplifying, it is important to remember that x 5 e x if x . 0 2x if x , 0 65. Start with the expression "a2 2 x2 and let x 5 a sin u, assuming 2p 2 # u # p 2, and simplify the original expression so that it contains no radicals. 66. Start with the expression "a2 1 x2 and let x 5 a tan u, assuming 2p 2 , u , p 2, and simplify the original expression so that it contains no radicals. In Exercises 67–70, explain the mistake that is made. EXPRESSION SUBSTITUTION TRIGONOMETRIC IDENTITY "a2 2 x2 x 5 a sin u 2p 2 # u # p 2 1 2 sin2 u 5 cos2 u "a2 1 x2 x 5 a tan u 2p 2 , u , p 2 1 1 tan2 u 5 sec2 u "x2 2 a2 x 5 a sec u 0 # u , p 2 or p # u , 3p 2 sec2 u 2 1 5 tan2 u • C A T C H T H E M I S T A K E • C O N C E P T U A L In Exercises 71 and 72, determine whether each statement is true or false. 71. If an equation is true for some values (but not all values), then it is still an identity. 72. If an equation has an infinite number of solutions, then it is an identity. 73. In which quadrants is the equation cos u 5 "1 2 sin2 u true? 74. In which quadrants is the equation 2cos u 5 "1 2 sin2 u true? 75. In which quadrants is the equation csc u 5 2"1 1 cot2 u true? 76. In which quadrants is the equation sec u 5 "1 1 tan2 u true? 7.2 Verifying Trigonometric Identities 673 67. Verify the identity cos x 1 2 tan x 1 sin x 1 2 cot x 5 sin x 1 cos x. Solution: Start with the left side of the equation. Write the tangent and cotangent functions in terms of sines and cosines. Cancel the common cosine in the first term and sine in the second term. This is incorrect. What mistake was made? 68. Verify the identity cos3 x sec x 1 2 sin x 5 1 1 sin x. Solution: Start with the equation on the left. Rewrite secant in terms of sine. 5 cos3 x 1 sin x 1 2 sin x Simplify. 5 cos3 x 1 2 sin2 x Use the Pythagorean identity. 5 cos3 x cos2 x Simplify. 5 cos x This is incorrect. What mistakes were made? 69. Determine whether the equation is a conditional equation or an identity: tan x cot x 5 1. Solution: Start with the left side. tan x cot x Rewrite the tangent and cotangent functions in terms of sines and cosines. Simplify. 5 sin2 x cos2 x 5 tan2 x Let x 5 p 4. Note: tanap 4 b 5 1. 5 1 Since tan x cot x 5 1, this equation is an identity. This is incorrect. What mistake was made? 70. Determine whether the equation is a conditional equation or an identity: k sin x k 2 cos x 5 1. Solution: Start with the left side of the equation. Let x 5 np 2 , where n is an odd integer. Simplify. 61 2 0 5 1 Since sin x 2 cos x 5 0 5 1, this is an identity. This is incorrect. What mistake was made? cos x 1 2 tan x 1 sin x 1 2 cot x 5 cos x 1 2 sin x cos x 1 sin x 1 2 cos x sin x 5 1 1 2 sin x 1 1 1 2 cos x cos3 x sec x 1 2 sin x 5 sin x cos x cos x sin x sin x 2 cos x sinanp 2 b 2 cosanp 2 b 674 CHAPTER 7 Analytic Trigonometry • C H A L L E N G E 77. Simplify 1a sin x 1 b cos x22 1 1b sin x 2 a cos x22. 78. Simplify 1 1 cot3 x 1 1 cot x 1 cot x. 79. Do you think that sin1A 1 B2 5 sin A 1 sin B? Why? 80. Do you think that cosA1 2 AB 5 1 2 cos A? Why? 81. Do you think tan12 A2 5 2 tan A? Why? 82. Do you think cot1A22 5 1cot A22? Why? • T E C H N O L O G Y In the next section, you will learn the sum and difference identities. In Exercises 83–90, we illustrate these identities with graphing calculators. 83. Determine the correct sign 11 or 22 for cos1A 1 B2 5 cos A cos B 6 sin A sin B by graphing Y1 5 cos1A 1 B2, Y2 5 cos A cos B 1 sin A sin B, and Y3 5 cos A cos B 2 sin A sin B in the same viewing rectangle for several values of A and B. 84. Determine the correct sign 11 or 22 for cos1A 2 B2 5 cos A cos B 6 sin A sin B by graphing Y1 5 cos1A 2 B2, Y2 5 cos A cos B 1 sin A sin B, and Y3 5 cos A cos B 2 sin A sin B in the same viewing rectangle for several values of A and B. 85. Determine the correct sign 11 or 22 for sin1A 1 B2 5 sin A cos B 6 cos A sin B by graphing Y1 5 sin1A 1 B2, Y2 5 sin A cos B 1 cos A sin B, and Y3 5 sin A cos B 2 cos A sin B in the same viewing rectangle for several values of A and B. 86. Determine the correct sign 11 or 22 for sin1A 2 B2 5 sin A cos B 6 cos A sin B by graphing Y1 5 sin1A 2 B2, Y2 5 sinA cos B 1 cos A sin B, and Y3 5 sin A cos B 2 cos A sin B in the same viewing rectangle for several values of A and B. 87. Determine the correct sign 11 or 22 for cos2 A 5 1 6 cos12 A2 2 by graphing Y1 5 cos2 A, Y2 5 1 1 cos12 A2 2 , and Y3 5 1 2 cos12 A2 2 in the same viewing rectangle for several values of A. 88. Determine the correct sign 11 or 22 for sin2A 5 1 6 cos12 A2 2 by graphing Y1 5 sin2 A, Y2 5 1 1 cos12 A2 2 , and Y3 5 1 2 cos12 A2 2 in the same viewing rectangle for several values of A. 89. Determine the correct sign 11 or 22 for tanax 1 p 6 b 5 tan x 6 tanap 6 b 1 6 tan x tanap 6 b by graphing Y1 5 tanax 1 p 6 b, Y2 5 tan x 2 tanap 6 b 1 1 tan x tanap 6 b , and Y3 5 tan x 1 tanap 6 b 1 2 tan x tanap 6 b in the same viewing rectangle. Does Y1 5 Y2 or Y3? 90. Determine the correct sign 11 or 22 for tanax 2 p 3 b 5 tan x 6 tanap 3 b 1 6 tan x tanap 3 b by graphing Y1 5 tanax 1 p 3 b, Y2 5 tan x 2 tanap 3 b 1 1 tan x tanap 3 b , and Y3 5 tan x 1 tanap 3 b 1 2 tan x tanap 3 b in the same viewing rectangle. Does Y1 5 Y2 or Y3? g ? g ? g ? g ? g ? g ? g ? g ? S K I L L S O B J E C T I V E S ■ ■Find exact values for the cosine function using sum and difference identities. ■ ■Find exact values for the sine function using sum and difference identities. ■ ■Find exact values for the tangent function using the sum or difference identities for the tangent function. C O N C E P T U A L O B J E C T I V ES ■ ■Understand that some sums or differences of trigonometric functions can be written as a single cosine expression. ■ ■Understand that some sums or differences of trigonometric functions can be written as a single sine expression. ■ ■Understand that the sum and difference identities for sine and cosine functions are used in the derivation of the sum and difference identities for the tangent function. 7.3 SUM AND DIFFERENCE IDENTITIES In this section, we will consider trigonometric functions with arguments that are sums and differences. In general, ƒ1A 1 B2 2 ƒ1A2 1 ƒ1B2. First, note that function notation is not distributive: cos1A 1 B2 2 cos A 1 cos B This principle is easy to prove. Let A 5 p and B 5 0; then cos1A 1 B2 5 cos1p 1 02 5 cos p 5 21 cos A 1 cos B 5 cos p 1 cos 0 5 21 1 1 5 0 Since 21 2 0, we know that cos1A 1 B2 2 cos A 1 cos B. In this section, we will derive some new and important identities (sum and difference identities for the cosine, sine, and tangent functions) and revisit cofunction identities. We begin with the familiar distance formula, from which we can derive the sum and difference identities for the cosine function. From there we can derive the sum and difference formulas for the sine and tangent functions. Distance Formula Sum and Difference Identities for Cosine Sum and Difference Identities for Tangent Sum and Difference Identities for Sine Cofunction Identities Before we start deriving and working with trigonometric sum and difference identities, let us first discuss why these are important. Sum and difference (and later product-to-sum and sum-to-product) identities are important because they allow calculation in functional (analytic) form and often lead to evaluating expressions exactly (as opposed to approximating them with calculators). The identities developed in this chapter are useful in such applications as musical sound, where they allow the determination of the “beat” frequency. In calculus, these identities will simplify the integration and differentiation processes. 7.3 Sum and Difference Identities 675 7.3.1 Sum and Difference Identities for the Cosine Function Derivation of the Sum and Difference Identities for the Cosine Function Recall from Section 6.7 that the unit circle approach gave the relationship between the coordinates along the unit circle and the sine and cosine functions. Specifically, the x‑coordinate corresponded to the value of the cosine function, and the y-coordinate corresponded to the value of the sine function. x y (1, 0) (0, 1) (cos θ, sin θ) r = 1 θ (0, –1) (–1, 0) Let us now draw the unit circle with two angles a and b, realizing that the two terminal sides of these angles form a third angle, a 2 b. x y (1, 0) (0, 1) P2 P1 β α α – β (0, –1) (–1, 0) If we label the points P 1 5 1cos a, sin a2 and P 2 5 1cos b, sin b2, we c an then draw a segment connecting points P 1 and P 2. x y (1, 0) (0, 1) P2 = (cos β, sin β) P2 β α α – β (0, –1) (–1, 0) P1 = (cos α, sin α) P1 If we rotate the angle clockwise so that the central angle a 2 b is in standard position, then the two points where the initial and terminal sides intersect the unit circle are P 4 5 11, 02 and P 3 5 1cos 1a 2 b2, sin 1a 2 b22, respectively. 7.3.1 S KILL Find exact values for the cosine function using sum and difference identities. 7.3.1 C ON CEPTUAL Understand that some sums or differences of trigonometric functions can be written as a single cosine expression. 676 CHAPTER 7 Analytic Trigonometry x y P4 = (1, 0) (0, 1) (0, –1) (–1, 0) The distance from P 1 to P 2 is equal to the length of the segment. Similarly, the distance from P 3 to P 4 is equal to the length of the segment. Since the lengths of the segments are equal, we say that the distances ar e equal: d1P 1, P 22 5 d1P 3, P 42. We can now derive the sum identity for the cosine function from the difference identity for the cosine function and the properties of even and odd functions. WORDS MATH Replace b with 2b in the difference identity. cos Ca 2 12b2D 5 cos a cos12b2 1 sin a sin12b2 Simplify the left side and use properties of even and odd functions on the right side. cos1a 1 b2 5 cos a1cos b2 1 sin a12sin b2 Write the sum identity for the cosine function. cos1a 1 b2 5 cos a cos b 2 sin a sin b STUDY TIP The distance from point P 1 5 1x1, y12 to P 2 5 1x2, y22 is given by the distance formula d1P 1, P 22 5 "1x2 2 x122 1 1y2 2 y122. 7.3 Sum and Difference Identities 677 WORDS MATH Set the distances (segment lengths) equal. d1P 1, P 22 5 d1P 3, P 42 Apply the distance formula: "1x2 2 x122 1 1y2 2 y122 5 "1x4 2 x322 1 1y4 2 y322 Substitute P 1 5 1x1, y12 5 1cos a, sin a2 and P 2 5 1x2, y22 5 1cos b, sin b2 into the left side of the equation and P 3 5 1x3, y32 5 1cos 1a 2 b2, sin 1a 2 b22 and P 4 5 1x4, y42 5 11, 02 into the right side of the equation. "3cos b 2 cos a42 1 3sin b 2 sin a42 5 "31 2 cos1a 2 b242 1 30 2 sin1a 2 b242 Square both sides of the equation. 3cos b 2 cos a42 1 3sin b 2 sin a42 5 31 2 cos1a 2 b242 1 30 2 sin1a 2 b242 Eliminate the cos2 a 2 2 cos a cos b 1 cos2 b 1 sin2 a 2 2 sin a sin b 1 sin2 b brackets. 5 1 2 2 cos1a 2 b2 1 cos21a 2 b2 1 sin21a 2 b2 Regroup terms on cos2 a 1 sin2 a 2 2 cos a cos b 2 2 sin a sin b 1 cos2 b 1 sin2 b each side and use the Pythagorean identity. 5 1 2 2 cos1a 2 b2 1 cos2 1a 2 b2 1 sin2 1a 2 b2 Simplify. 2 2 2 cos a cos b 2 2 sin a sin b 5 2 2 2 cos1a 2 b2 Subtract 2 from both sides. 22 cos a cos b 2 2 sin a sin b 5 22 cos1a 2 b2 Divide by 22. cos a cos b 1 sin a sin b 5 cos1a 2 b2 Write the difference identity for the cosine function. cos1a 2 b2 5 cos a cos b 1 sin a sin b g 1 g 1 g 1 SUM AND DIFFERENCE IDENTITIES FOR THE COSINE FUNCTION Sum cos1A 1 B2 5 cos A cos B 2 sin A sin B Difference cos1A 2 B2 5 cos A cos B 1 sin A sin B [CONCEPT CHECK] TRUE OR FALSE sin A sinB 2 cos A cosB 5 2cos(A 1 B) ANSWER True ▼ EXAMPLE 1 Finding Exact Values for the Cosine Function Evaluate each of the following cosine expressions exactly. a. cos 15° b. cosa7p 12 b Solution (a): Write 15° as a difference of known “special” angles. cos 15° 5 cos145° 2 30°2 Write the difference identity for the cosine function. cos1A 2 B2 5 cos A cos B 1 sin A sin B Substitute A 5 45° and B 5 30°. cos 15° 5 cos 45° cos 30° 1 sin 45° sin 30° Evaluate the expressions on the right exactly. Simplify. cos 15° 5 !6 1 !2 4 Solution (b): Write 7p 12 as a sum of known “special” angles. Simplify. cosa7p 12 b 5 cosap 3 1 p 4 b Write the sum identity for the cosine function. cos1A 1 B2 5 cos A cos B 2 sin A sin B Substitute Evaluate the expressions on the right exactly. Simplify. cosa7p 12 b 5 !2 2 !6 4 Note: 7p 12 can also be represented as 105°. Y OUR TU R N Use the sum or difference identities for the cosine function to evaluate each cosine expression exactly. a. cos 75° b. cos a5p 12 b cos 15° 5 !2 2 !3 2 1 !2 2 1 2 cosa7p 12 b 5 cosa4p 12 1 3p 12 b A 5 p 3 and B 5 p 4 . cosa7p 12 b 5 cosap 3 b cosap 4 b 2 sinap 3 b sinap 4 b cosa7p 12 b 5 1 2 !2 2 2 !3 2 !2 2 ▼ ▼ A N S W E R a. "6 2 "2 4 b. "6 2 "2 4 678 CHAPTER 7 Analytic Trigonometry Example 1 illustrates an important characteristic of the sum and difference identities: that we can now find the exact trigonometric function value of angles that are multiples of 158 aor, equivalently, p 12b, since each of these can be written as a sum or difference of angles for which we know the trigonometric function values exactly. ▼ A N S W E R cos13x2 EXAMPLE 2 Writing a Sum or Difference as a Single Cosine Expression Use the sum or the difference identity for the cosine function to write each of the following expressions as a single cosine expression. a. sin15x2 sin12x2 1 cos15x2 cos12x2 b. cos x cos13x2 2 sin x sin13x2 Solution (a): Because of the positive sign, this will be a cosine of a difference. Reverse the expression and write the formula. cos A cos B 1 sin A sin B 5 cos1A 2 B2 Identify A and B. A 5 5x and B 5 2x Substitute A 5 5x and B 5 2x into the difference identity. cos15x2 cos12x2 1 sin15x2 sin12x2 5 cos15x 2 2x2 Simplify. cos15x2 cos12x2 1 sin15x2 sin12x2 5 cos13x2 Notice that if we had selected A 5 2x and B 5 5x instead, the result would have been cos123x2, but since the cosine function is an even function, this would have simplified to cos13x2. Solution (b): Because of the negative sign, this will be a cosine of a sum. Reverse the expression and write the formula. cos A cos B 2 sin A sin B 5 cos1A 1 B2 Identify A and B. A 5 x and B 5 3x Substitute A 5 x and B 5 3x into the sum identity. cos x cos13x2 2 sinx sin13x2 5 cos1x 1 3x2 Simplify. cos x cos13x2 2 sin x sin13x2 5 cos14x2 Y OUR T UR N Write cos14x2 cos17x2 1 sin14x2 sin17x2 as a single cosine expression. ▼ 7.3 Sum and Difference Identities 679 Cofunction Identities In Section 6.2, we discussed cofunction relationships for acute angles. Recall that a trigonometric function value of an angle is equal to the corresponding cofunction value of its complementary angle. Now we use the sum and difference identities for the cosine function to develop the cofunction identities for any angle, u. WORDS MATH Write the difference identity for the cosine function. cos1A 2 B2 5 cos A cos B 1 sin A sinB Let A 5 p 2 and B 5 u. cosap 2 2 ub 5 cosap 2 b cos u 1 sinap 2 bsin u Evaluate the known values for the sine and cosine functions. Simplify. cosap 2 2 ub 5 sin u Similarly, to determine the other corresponding cofunction identity, write the difference identity for the cosine function. cos1A 2 B2 5 cos A cos B 1 sin A sin B Let A 5 p 2 and B 5 p 2 2 u. cos cp 2 2 ap 2 2 ub d 5 cosap 2 b cosap 2 2 ub 1 sinap 2 b sinap 2 2 ub Evaluate the known values for the sine and cosine functions. Simplify. cos u 5 sin ap 2 2 ub cosap 2 2 ub 5 0⋅cos u 1 1⋅sin u cos1u2 5 0⋅cosap 2 2 ub 1 1⋅sinap 2 2 ub COFUNCTION IDENTITIES FOR THE SINE AND COSINE FUNCTIONS cosap 2 2 ub 5 sin u sinap 2 2 ub 5 cos u 7.3.2 Sum and Difference Identities for the Sine Function We can now use the cofunction identities together with the sum and difference identities for the cosine function to develop the sum and difference identities for the sine function. 7.3.2 SKI LL Find exact values for the sine function using sum and difference identities. 7.3.2 CO NCE PTUAL Understand that some sums or differences of trigonometric functions can be written as a single sine expression. 680 CHAPTER 7 Analytic Trigonometry SUM AND DIFFERENCE IDENTITIES FOR THE SINE FUNCTION Sum sin1A 1 B2 5 sin A cos B 1 cos A sin B Difference sin1A 2 B2 5 sin A cos B 2 cos A sin B WORDS MATH Start with the cofunction identity. sin u 5 cosap 2 2 ub Let u 5 A 1 B. sin1A 1 B2 5 cos cp 2 2 1A 1 B2 d Regroup the terms in the cosine expression. Use the difference identity for the cosine function. sin1A 1 B2 5 cosap 2 2 Abcos B 1 sinap 2 2 Ab sin B Use the cofunction identities. sin1A 1 B2 5 cosap 2 2 Abcos B 1 sinap 2 2 Ab sin B Simplify. sin1A 1 B2 5 sin A cos B 1 cos A sin B sin1A 1 B2 5 cos c ap 2 2 Ab 2 Bd g sin A g cos A Now we can derive the difference identity for the sine function using the sum identity for the sine function and the properties of even and odd functions. WORDS MATH Replace B with 2B in the sum identity. sin1A 1 12B22 5 sin A cos12B2 1 cos Asin12B2 Simplify using even and odd identities. sin1A 2 B2 5 sin AcosB 2 cos AsinB EXAMPLE 3 Finding Exact Values for the Sine Function Use the sum or the difference identity for the sine function to evaluate each sine expression exactly. a. sin 75° b. sina5p 12 b Solution (a): Write 75° as a sum of known “special” angles. sin 75° 5 sin 145° 1 30°2 Write the sum identity for the sine function. sin1A 1 B2 5 sin A cos B 1 cos A sin B Substitute A 5 45° and B 5 30°. sin 75° 5 sin 45° cos 30° 1 cos 45° sin 30° Evaluate the expressions on the right exactly. Simplify. sin 75° 5 !6 1 !2 4 sin 75° 5 a !2 2 b a !3 2 b 1 a !2 2 b a1 2b [CONCEPT CHECK] TRUE OR FALSE sin(A 2 B) 5 sin(B 2 A) ANSWER False ▼ 7.3 Sum and Difference Identities 681 Solution (b): Write 5p 12 as a sum of known “special” angles. Simplify. sina5p 12 b 5 sinap 6 1 p 4 b Write the sum identity for the sine function. sin1A 1 B2 5 sin A cos B 1 cos A sin B Substitute A 5 p 6 and B 5 p 4 . sina5p 12 b 5 sinap 6 b cosap 4 b 1 cosap 6 b sinap 4 b Evaluate the expressions on the right exactly. Simplify. sina5p 12 b 5 !2 1 !6 4 Note: 5p 12 can also be represented as 758. Y OUR TU R N Use the sum or the difference identity for the sine function to evaluate each sine expression exactly. a. sin 15° b. sina7p 12 b sina5p 12 b 5 sina2p 12 1 3p 12 b sina5p 12 b 5 a1 2b a !2 2 b 1 a !3 2 b a !2 2 b ▼ ▼ A N S W E R a. "6 2 "2 4 b. "6 1 "2 4 We see in Example 3 that the sum and difference identities allow us to calculate exact values for trigonometric functions of angles that are multiples of 158 aor, equivalently, p 12b, as we saw with the cosine function. EXAMPLE 4 Writing a Sum or Difference as a Single Sine Expression Graph y 5 3 sin x cos13x2 1 3 cos x sin13x2. Solution: Use the sum identity for the sine function to write the expression as a single sine expression. Factor out the common 3. y 5 33sin x cos13x2 1 cos x sin13x24 Write the sum identity for the sine function. sin A cos B 1 cos A sin B 5 sin1A 1 B2 Identify A and B. A 5 x and B 5 3x 682 CHAPTER 7 Analytic Trigonometry Substitute A 5 x and B 5 3x into the sum identity. sin x cos13x2 1 cos x sin13x2 5 sin1x 1 3x2 5 sin14x2 Simplify. y 5 33sin x cos13x2 1 cos x sin13x24 Graph y 5 3 sin14x2. g sin 4x –3 –2 –1 1 2 3 x y π 4 π π –π –π 2 2 3π 4 7.3.3 Sum and Difference Identities for the Tangent Function We now develop the sum and difference identities for the tangent function. WORDS MATH Start with the quotient identity. tan x 5 sin x cos x Let x 5 A 1 B. tan1A 1 B2 5 sin1A 1 B2 cos1A 1 B2 Use the sum identities for the sine and cosine functions. Multiply the numerator and denominator by 1 cos Acos B. Simplify. tan1A 1 B2 5 a sin A cos Ab 1 a sinB cosBb 1 2 a sin A cos Ab a sinB cosBb Write the expressions inside the parentheses in terms of the tangent tan1A 1 B2 5 tan A 1 tanB 1 2 tan A tanB function. Replace B with 2B. tan3A 1 12B24 5 tan1A 2 B2 5 tan A 1 tan12B2 1 2 tan A tan12B2 Since the tangent function is an odd function, tan12B2 5 2tan B. tan1A 1 B2 5 sinA cos B 1 cos A sin B cos A cos B 2 sin A sin B tan1A 1 B2 5 sin A cosB 1 cos A sinB cos A cosB cos A cosB 2 sin A sinB cos A cosB 5 sin A cosB cos A cosB 1 cosA sin B cosA cos B cos A cosB cos A cosB 2 sin A sin B cos A cos B tan1A 2 B2 5 tan A 2 tanB 1 1 tan A tanB 7.3.3 S K I L L Find exact values for the tangent function using the sum or difference identities for the tangent function. 7.3.3 C O N C E P T U A L Understand that the sum and difference identities for sine and cosine functions are used in the derivation of the sum and difference identities for the tangent function. 7.3 Sum and Difference Identities 683 SUM AND DIFFERENCE IDENTITIES FOR THE TANGENT FUNCTION Sum tan1A 1 B2 5 tan A 1 tan B 1 2 tan A tan B Difference tan1A 2 B2 5 tan A 2 tan B 1 1 tan A tan B EXAMPLE 5 Finding Exact Values for the Tangent Function Find the exact value of tan 1a 1 b2 if sin a 5 21 3 and cos b 5 21 4 and the terminal side of a lies in quadrant III and the terminal side of b lies in quadrant II. Solution: STEP 1 Write the sum identity for the tangent function. STEP 2 Find tan a. The terminal side of a lies in quadrant III. sin a 5 y r 5 21 3 Solve for x. 1x2 1 y2 5 r2.2 x2 1 12 5 32 x 5 6!8 Take the negative sign since we are in quadrant III. x 5 22!2 Find tan a. tan a 5 y x 5 21 22!2 5 1 2!2⋅!2 !2 5 !2 4 STEP 3 Find tan b. The terminal side of b lies in quadrant II. cos b 5 21 4 5 x r Solve for y. 1x2 1 y2 5 r2.2 12122 1 y2 5 42 y 5 6!15 Take the positive sign since we are in quadrant II. y 5 !15 Find tan b. tan b 5 y x 5 !15 21 5 2!15 tan1a 1 b2 5 tan a 1 tan b 1 2 tan a tan b x y 1 3 x α (x, –1) x y 1 4 y β (–1, y) [CONCEPT CHECK] TRUE OR FALSE tan(A 2 B) 5 2tan(B 2 A) ANSWER True ▼ 684 CHAPTER 7 Analytic Trigonometry STEP 4 Substitute tan a 5 !2 4 and tan b 5 2!15 into the sum identity for the tangent function. Multiply the numerator and the denominator by 4. 5 !2 2 4!15 4 1 !30 The expression tan 1a 1 b 2 5 !2 2 4!15 4 1 !30 can be simplified further if we rationalize the denominator. tan 1a 1 b2 5 !2 4 2 !15 1 2 a !2 4 b A2!15B tan 1a 1 b2 5 4 4 a !2 4 2 !15b a1 1 !30 4 b Note in Example 5 that right triangles have been superimposed in the Cartesian plane. The coordinate pair 1x, y2 can have positive or negative values, but the radius r is always positive. When right triangles are superimposed, with one vertex at the point 1x, y2 and another vertex at the origin, it is important to understand that triangles have positive side lengths. cos1A 1 B2 5 cos A cos B 2 sin A sin B sin1A 1 B2 5 sin A cos B 1 cos A sin B tan1A 1 B2 5 tan A 1 tan B 1 2 tan A tan B cos1A 2 B2 5 cos A cos B 1 sin A sin B sin1A 2 B2 5 sin A cos B 2 cos A sin B tan1A 2 B2 5 tan A 2 tan B 1 1 tan A tan B In this section, we derived the sum and difference identities for the cosine function using the distance formula. We then used these identities to derive the cofunction identities. The cofunction identities and sum and difference identities for the cosine function were used to derive the sum and difference identities for the sine function. We combined the sine and cosine sum and difference identities to determine the tangent sum and difference identities. The sum and difference identities enabled us to evaluate a trigonometric expression exactly for any integer multiple of 158 1i.e., p/122. [SEC TION 7.3] S UM MA RY [SEC TION 7.3] E X ERC I S E S • S K I L L S 1. sina p 12b 2. cosa p 12b 3. cosa25p 12 b 4. sina25p 12 b 5. tana2 p 12b 6. tana13p 12 b 7. sin 105° 8. cos 195° 9. tan12105°2 10. tan 165° 11. cota p 12b 12. cota25p 12 b 13. seca211p 12 b 14. seca213p 12 b 15. csc12255°2 16. csc1215°2 In Exercises 1–16, find the exact value for each trigonometric expression. 7.3 Sum and Difference Identities 685 In Exercises 17–28, write each expression as a single trigonometric function. 17. sin12x2 sin13x2 1 cos12x2 cos13x2 18. sin x sin12x2 2 cos x cos12x2 19. sin x cos12x2 2 cos x sin12x2 20. sin12x2 cos13x2 1 cos12x2 sin13x2 21. cos1p 2 x2 sin x 1 sin1p 2 x2 cos x 22. sinap 3 x b cosa2p 2 x b 2 cosap 3 xbsina2p 2 x b 23. 1sin A 2 sin B22 1 1cos A 2 cos B22 2 2 24. 1sin A 1 sin B22 1 1cos A 1 cos B22 2 2 25. 2 2 1sin A 1 cos B22 2 1cos A 1 sin B22 26. 2 2 1sin A 2 cos B22 2 1cos A 1 sin B22 27. tan 49° 2 tan 23° 1 1 tan 49° tan 23° 28. tan 49° 1 tan 23° 1 2 tan 49° tan 23° In Exercises 29–34, find the exact value of the indicated expression using the given information and identities. 29. Find the exact value of cos1a 1 b2 if cos a 5 21 3 and cos b 5 21 4 and the terminal side of a lies in quadrant III and the terminal side of b lies in quadrant II. 30. Find the exact value of cos1a 2 b2 if cos a 5 1 3 and cos b 5 21 4 and the terminal side of a lies in quadrant IV and the terminal side of b lies in quadrant II. 31. Find the exact value of sin1a 2 b2 if sin a 5 23 5 and sin b 5 1 5 and the terminal side of a lies in quadrant III and the terminal side of b lies in quadrant I. 32. Find the exact value of sin1a 1 b2 if sin a 5 23 5 and sin b 5 1 5 and the terminal side of a lies in quadrant III and the terminal side of b lies in quadrant II. 33. Find the exact value of tan1a 1 b2 if sin a 5 23 5 and cos b 5 21 4 and the terminal side of a lies in quadrant III and the terminal side of b lies in quadrant II. 34. Find the exact value of tan1a 2 b2 if sin a 5 23 5 and cos b 5 21 4 and the terminal side of a lies in quadrant III and the terminal side of b lies in quadrant II. In Exercises 35–52, determine whether each equation is a conditional equation or an identity. 35. sin1A 1 B2 1 sin1A 2 B2 5 2 sin A cos B 36. cos1A 1 B2 1 cos1A 2 B2 5 2 cos A cos B 37. sinax 2 p 2 b 5 cosax 1 p 2 b 38. sinax 1 p 2 b 5 cosax 1 p 2 b 39. !2 2 1sin x 1 cos x2 5 sinax 1 p 4 b 40. !3 cos x 1 sin x 5 2 cosax 1 p 3 b 41. sin2 x 5 1 2 cos12x2 2 42. cos2 x 5 1 1 cos12x2 2 43. sin12x2 5 2 sinx cosx 44. cos12x2 5 cos2 x 2 sin2 x 45. sin1A 1 B2 5 sin A 1 sin B 46. cos1A 1 B2 5 cos A 1 cos B 47. tan1p 1 B2 5 tan B 48. tan1A 2 p2 5 tan A 49. cot13p 1 x2 5 1 tan x 50. csc12x2 5 2 sec x csc x 51. 1 1 tan x 1 2 tan x 5 tanax 2 p 4 b 52. cotax 1 p 4 b 5 1 2 tan x 1 1 tan x 686 CHAPTER 7 Analytic Trigonometry • A P P L I C A T I O N S For Exercises 63 and 64, refer to the following: The difference quotient, ƒ1x 1 h2 2 ƒ1x2 h , is used to approximate the rate of change of the function f and will be used frequently in calculus. 63. Difference Quotient. Show that the difference quotient for ƒ1x2 5 sin x is cos xasin h h b 2 sin xa1 2 cos h h b. 64. Difference Quotient. Show that the difference quotient for ƒ1x2 5 cos x is 2sin xasin h h b 2 cos x a1 2 cos h h b. For Exercises 65 and 66, use the following: A nonvertical line makes an angle with the x-axis. In the figure, we see that the line L1 makes an acute angle u1 with the x-axis. Similarly, the line L2 makes an acute angle u2 with the x-axis. tan u1 5 slope of L1 5 m1 tan u2 5 slope of L2 5 m2 65. Angle Between Two Lines. Show that tan1u2 2 u12 5 m2 2 m1 1 1 m1m2 66. Relating Tangent and Slope. Show that tan1u1 2 u22 5 m1 2 m2 1 1 m1m2 For Exercises 67 and 68, refer to the following: An electric field E of a wave with constant amplitude A, propagating a distance z, is given by E 5 A cos 1kz 2 ct2 where k is the propagation wave number, which is related to the wavelength l by k 5 2p l and where c 5 3.0 3 108 m/s is the speed of light in a vacuum, and t is time in seconds. 67. Electromagnetic Wave Propagation. Use the cosine difference identity to express the electric field in terms of both sine and cosine functions. When the quotients of the propagation distance z and the wavelength l are equal to an integer, what do you notice? 68. Electromagnetic Wave Propagation. Use the cosine difference identity to express the electric field in terms of both sine and cosine functions. When t 5 0, what do you notice? 69. Biology. By analyzing available empirical data, it has been determined that the body temperature of a species fluctuates according to the model T1t2 5 38 2 2.5 cosS p 6 1t 2 32T , 0 # t # 24 where T represents temperature in degrees Celsius and t represents time (in hours) measured from 12:00 p.m. (noon). Use an identity to express T1t2 in terms of the sine function. 70. Health/Medicine. During the course of treatment of an illness, the concentration of a drug (in micrograms per milliliter) in the bloodstream fluctuates during the dosing period of 8 hours according to the model C1t2 5 15.4 2 4.7 sin ap 4 t 1 p 2 b, 0 # t # 8 Use an identity to express the concentration C1t2 in terms of the cosine function. Note: This model does not apply to the first dose of the medication as there will be no medication in the bloodstream. x y θ1 θ2 L1 L2 θ2 – θ1 7.3 Sum and Difference Identities 687 In Exercises 53–62, graph each of the functions by first rewriting it as a sine, cosine, or tangent of a difference or sum. 53. y 5 cosap 3 b sin x 1 cos x sina p 3 b 54. y 5 cosap 3 b sin x 2 cos x sinap 3 b 55. y 5 sin x sina p 4 b 1 cos x cosa p 4 b 56. y 5 sin x sina p 4 b 2 cos x cosap 4 b 57. y 5 2sin x cos 13x2 2 cos x sin 13x2 58. y 5 sin x sin 13x2 1 cos x cos 13x2 59. y 5 1 1 tan x 1 2 tan x 60. y 5 !3 2 tan x 1 1 !3 tan x 61. y 5 1 1 !3tan x !3 2 tan x 62. y 5 1 2 tan x 1 1 tan x In Exercises 73–76, determine whether each statement is true or false. • C O N C E P T U A L 77. Verify that sin1A 1 B 1 C2 5 sin A cos B cos C 1 cos A sin B cos C 1 cos A cos B sin C 2 sin A sin B sin C. 78. Verify that cos1A 1 B 1 C2 5 cos A cos B cos C 2 sin A sin B cos C 2 sin A cos B sin C 2 cos A sin B sin C. 79. Although in general the statement sin1A 2 B2 5 sin A 2 sin B is not true, it is true for some values. Determine some values of A and B that make this statement true. 80. Although in general the statement sin1A 1 B2 5 sin A 1 sin B is not true, it is true for some values. Determine some values of A and B that make this statement true. 73. cos115°2 5 cos145°2 2 cos130°2 74. sinap 2 b 5 sinap 3 b 1 sinap 6 b 75. tanax 1 p 4 b 5 1 1 tan x 76. cotap 4 2 xb 5 1 1 tan x 1 2 tan x • C H A L L E N G E 688 CHAPTER 7 Analytic Trigonometry • C A T C H T H E M I S T A K E In Exercises 71 and 72, explain the mistake that is made. 71. Find the exact value of tana5p 12 b. Solution: Write 5p 12 as a sum. tanap 4 1 p 6 b Distribute. tanap 4 b 1 tanap 6 b Evaluate the tangent function for p 4 and p 6 . 1 1 !3 3 This is incorrect. What mistake was made? 72. Find the exact value of tana27p 6 b. Solution: The tangent function is an even function. tana7p 6 b Write 7p 6 as a sum. tanap 1 p 6 b Use the tangent sum identity, tan1A 1 B2 5 tan A 1 tan B 1 2 tan A tan B. tan p 1 tanap 6 b 1 2 tan p tanap 6 b Evaluate the tangent functions on the right. Simplify. !3 3 This is incorrect. What mistake was made? 0 1 1 !3 1 2 0 • T E C H N O L O G Y 81. In Exercise 63, you showed that the difference quotient for ƒ1x2 5 sin x is cos x asin h h b 2 sin x a1 2 cos h h b. Plot Y1 5 cos x asin h h b 2 sin x a1 2 cos h h b for a. h 5 1 b. h 5 0.1 c. h 5 0.01 What function does the difference quotient for ƒ1x2 5 sin x resemble when h approaches zero? 82. Show that the difference quotient for ƒ1x2 5 cos x is 2sin x asin h h b 2 cos x a1 2 cos h h b. Plot Y1 5 2sin x asin h h b 2 cos x a1 2 cos h h b for a. h 5 1 b. h 5 0.1 c. h 5 0.01 What function does the difference quotient for ƒ1x2 5 cos x resemble when h approaches zero? 83. Show that the difference quotient for ƒ1x2 5 sin12x2 is cos12x2 c sin12h2 h d 2 sin12x2 c 1 2 cos12h2 h d . Plot Y1 5 cos12x2 c sin12h2 h d 2 sin12x2 c 1 2 cos12h2 h d for a. h 5 1 b. h 5 0.1 c. h 5 0.01 What function does the difference quotient for ƒ1x2 5 sin 12x2 resemble when h approaches zero? 84. Show that the difference quotient for ƒ1x2 5 cos12x2 is 2sin12x2 c sin12h2 h d 2 cos12x2 c 1 2 cos12h2 h d . Plot Y1 5 2sin12x2 c sin12h2 h d 2 cos12x2 c 1 2 cos12h2 h d for a. h 5 1 b. h 5 0.1 c. h 5 0.01 What function does the difference quotient for ƒ1x2 5 cos12x2 resemble when h approaches zero? S K I L L S O B J E C T I V E ■ ■Use double-angle identities in simplifying some trigonometric expressions. C O N C E P T U A L O B J E C T I V E ■ ■Understand that the double-angle identities are derived from the sum identities. 7.4 DOUBLE-ANGLE IDENTITIES 7.4.1 Applying Double-Angle Identities Throughout this text, much attention has been given to distinguishing between evaluating trigonometric functions exactly (for special angles) and approximating values of trigonometric functions with a calculator. In previous chapters, we could only evaluate trigonometric functions exactly for reference angles of 30°, 45°, and 60° or p 6 , p 4 , and p 3 ; note that as of the previous section, we now can include multiples of p 12 among these “special” angles. Now, we can use double-angle identities to also evaluate the trigonometric function values for other angles that are even integer multiples of the special angles or to verify other trigonometric identities. One important distinction now is that we will be able to find exact values of many functions using the double-angle identities without needing to know the actual value of the angle. 7.4.1 S K IL L Use double-angle identities in simplifying some trigonometric expressions. 7.4.1 C ON C E P T U A L Understand that the double-angle identities are derived from the sum identities. 7.4 Double-Angle Identities 689 Derivation of Double-Angle Identities To derive the double-angle identities, we let A 5 B in the sum identities. WORDS MATH Write the identity for the sine of a sum. sin1A 1 B2 5 sin A cos B 1 cos A sin B Let B 5 A. sin1A 1 A2 5 sin A cos A 1 cos A sin A Simplify. sin12A2 5 2 sin A cos A Write the identity for the cosine of a sum. cos1A 1 B2 5 cos A cos B 2 sin A sin B Let B 5 A. cos1A 1 A2 5 cos A cos A 2 sin A sin A Simplify. cos12A2 5 cos2 A 2 sin2 A We can write the double-angle identity for the cosine function two other ways if we use the Pythagorean identity: 1. Write the identity for the cosine function of a double angle. cos12A2 5 cos2 A 2 sin2 A Use the Pythagorean identity for the cosine function. cos12A2 5 cos2 A 2 sin2 A Simplify. cos12A2 5 1 2 2 sin2 A 2. Write the identity for the cosine function of a double angle. cos12A2 5 cos2 A 2 sin2 A Use the Pythagorean identity for the sine function. cos12A2 5 cos2 A 2 sin2 A Simplify. cos12A2 5 2 cos2 A 2 1 The tangent function can always be written as a quotient, tan12A2 5 sin12A2 cos12A2, if sin12A2 and cos12A2 are known. Here we write the double-angle identity for the tangent function in terms of only the tangent function. Write the tangent of a sum identity. tan1A 1 B2 5 tan A 1 tan B 1 2 tan A tan B Let B 5 A. tan1A 1 A2 5 tan A 1 tan A 1 2 tan A tan A Simplify. tan12A2 5 2 tan A 1 2 tan2 A 1 2sin2 A g g 1 2cos2 A [CONCEPT CHECK] TRUE OR FALSE The double-angle identities are derived by using the sum identities with B 5 A. ANSWER True ▼ DOUBLE-ANGLE IDENTITIES SINE COSINE TANGENT sin12A2 5 2 sin A cos A cos12A2 5 cos2 A 2 sin2 A tan12A2 5 2 tan A 1 2 tan2 A cos12A2 5 1 2 2 sin2 A cos12A2 5 2 cos2 A 2 1 690 CHAPTER 7 Analytic Trigonometry Applying Double-Angle Identities EXAMPLE 1 Finding Exact Values of Trigonometric Functions Using Double-Angle Identities If cos x 5 2 3, find sin12x2 given sin x , 0. Solution: Find sin x. Use the Pythagorean identity. sin2 x 1 cos2 x 5 1 Substitute cos x 5 2 3. sin2 x 1 a2 3b 2 5 1 Solve for sin x, which is negative. sin x 5 2Ä1 2 4 9 Simplify. sin x 5 2Å 5 9 5 2 !5 3 Find sin 2x. Use the double-angle identity for the sine function. sin12x2 5 2 sin x cos x Substitute sin x 5 2 !5 3 and cos x 5 2 3. sin12x2 5 2 a 2 !5 3 b a2 3b Simplify. sin12x2 5 24!5 9 YOUR T UR N If cos x 5 21 3, find sin12x2 given sin x , 0. ▼ ▼ A N S W E R sin12x2 5 4!2 9 x y 3 (–3, –4) x 4 5 x y 3 2 x 5 7.4 Double-Angle Identities 691 EXAMPLE 2 Finding Exact Values Using Double-Angle Identities If sin x 5 24 5 and cos x , 0, find sin12x2, cos12x2, and tan12x2. Solution: Solve for cos x. Use the Pythagorean identity. sin2 x 1 cos2 x 5 1 Substitute sin x 5 24 5. a24 5b 2 1 cos2 x 5 1 Simplify. cos2 x 5 9 25 Solve for cos x, which is negative. cos x 5 2Ä 9 25 5 23 5 Find sin12x2. Use the double-angle identity for the sine function. sin12x2 5 2sin x cos x Substitute sin x 5 24 5 and cos x 5 23 5. sin12x2 5 2a24 5b a23 5b Simplify. sin12x2 5 24 25 EXAMPLE 3 Verifying Trigonometric Identities Using Double-Angle Identities Verify the identity 1sin x 2 cos x22 5 1 2 sin12x2. Solution: Start with the left side of the equation. 1sin x 2 cos x22 Expand by squaring. 5 sin2 x 2 2 sin x cos x 1 cos2 x Group the sin2x and cos2x terms. 5 sin2 x 1 cos2 x 2 2 sin x cos x Apply the Pythagorean identity. 5 sin2 x 1 cos2 x 2 2 sin x cos x Apply the sine double-angle identity. 5 1 2 2 sin x cos x Simplify. 5 1 2 sin12x2 1 g g sin12x2 692 CHAPTER 7 Analytic Trigonometry Find cos12x2. Use the double-angle identity for the cosine function. cos12x2 5 cos2 x 2 sin2 x Substitute sin x 5 24 5 and cos x 5 23 5. cos12x2 5 a23 5b 2 2 a24 5b 2 Simplify. cos12x2 5 2 7 25 Find tan12x2. Use the quotient identity. tan u 5 sin u cos u Let u 5 2x. tan12x2 5 sin12x2 cos12x2 Substitute sin12x2 5 24 25 and cos12x2 5 2 7 25. tan12x2 5 24 25 2 7 25 Simplify. tan12x2 5 224 7 Note: We could also have found tan12x2 first by finding tan x 5 sin x cos x and then using the value for tan x in the double-angle identity, tan12A2 5 2 tan A 1 2 tan2 A. Y OUR TU R N If cos x 5 3 5 and sin x , 0, find sin12x2, cos12x2, and tan12x2. ▼ ▼ A N S W E R sin12x2 5 224 25, cos12x2 5 2 7 25, tan12x2 5 24 7 EXAMPLE 4 Verifying Multiple-Angle Identities, Using Double-Angle Identities Verify the identity cos13x2 5 11 2 4 sin2x2cos x. Solution: Write the cosine of a sum identity. cos1A 1 B2 5 cos Acos B 2 sin Asin B Let A 5 2x and B 5 x. cos12x 1 x2 5 cos12x2 cos x 2 sin12x2 sin x Apply the double-angle cos13x2 5 cos12x2 cosx 2 sin12x2 sinx identities. Simplify. cos13x2 5 cos x 2 2 sin2 x cos x 2 2 sin2 x cos x cos13x2 5 cos x 2 4 sin2 x cos x Factor out the common cosine term. cos13x2 5 11 2 4 sin2x2 cos x g 122sin2x 2sinx cosx g 7.4 Double-Angle Identities 693 EXAMPLE 5 Simplifying Trigonometric Expressions Using Double-Angle Identities Graph y 5 cot x 2 tan x cot x 1 tan x. Solution: Simplify y 5 cot x 2 tan x cot x 1 tan x first. Write the cotangent and tangent functions in terms of the sine and cosine functions. Multiply the numerator and the denominator by sin x cos x. Simplify. y 5 cos2 x 2 sin2 x cos2 x 1 sin2 x Use the double-angle and Pythagorean identities. y 5 cos12x2 Graph y 5 cos12x2. y 5 cos x sin x 2 sin x cos x cos x sin x 1 sin x cos x y 5 ± cos x sin x 2 sin x cos x cos x sin x 1 sin x cos x ≤asin x cos x sin x cos xb y 5 cos2 x 2 sin2 x cos2 x 1 sin2 x 1 cos12x2 g g π π –π –π –1 1 x y 2 2 694 CHAPTER 7 Analytic Trigonometry sin12A2 5 2sin Acos A cos12A2 5 cos2 A 2 sin2 A 5 1 2 2sin2 A 5 2cos2 A 2 1 tan12A2 5 2tan A 1 2 tan2A In this section, we derived the double-angle identities from the sum identities. We then used the double-angle identities to find exact values of trigonometric functions, to verify other trigonometric identities, and to simplify trigonometric expressions. There is no need to memorize the second and third forms of the cosine double-angle identity since they can be derived from the first using the Pythagorean identity. [SEC TION 7.4] S U M M A RY [SEC TION 7.4] E X E R C I SE S • S K I L L S In Exercises 1–12, use the double-angle identities to answer the following questions. 1. If sin x 5 1 !5 and cos x , 0, find sin12x2. 2. If sin x 5 1 !5 and cosx , 0, find cos12x2. 3. If cos x 5 5 13 and sin x , 0, find tan12x2. 4. If cos x 5 2 5 13 and sin x , 0, find tan12x2. 5. If tan x 5 12 5 and p , x , 3p 2 , find sin12x2. 6. If tan x 5 12 5 and p , x , 3p 2 , find cos12x2. 7. If sec x 5 !5 and sin x . 0, find tan12x2. 8. If sec x 5 !3 and sin x , 0, find tan12x2. 9. If csc x 5 22!5 and cos x , 0, find sin12x2. 10. If csc x 5 2!13 and cos x . 0, find sin12x2. 11. If cos x 5 212 13 and csc x , 0, find cot12x2. 12. If sin x 5 12 13 and cot x , 0, find csc12x2. In Exercises 13–24, simplify each expression. Evaluate the resulting expression exactly, if possible. 13. 2 tan15° 1 2 tan215° 14. 2 tanap 8 b 1 2 tan2ap 8 b 15. sinap 8 b cosap 8 b 16. sin15°cos15° 17. cos212x2 2 sin212x2 18. cos21x 1 22 2 sin21x 1 22 19. 2 tana5p 12 b 1 2 tan2a5p 12 b 20. 2 tanax 2b 1 2 tan2ax 2b 21. 1 2 2 sin2a7p 12 b 22. 2 sin2a25p 8 b 2 1 23. 2 cos2a27p 12 b 2 1 24. 1 2 2 cos2a2p 8 b In Exercises 25–40, verify each identity. 25. csc12A2 5 1 2 csc A sec A 26. cot12A2 5 1 2 1cot A 2 tan A2 27. 1sin x 2 cos x21cos x 1 sin x2 5 2cos12x2 28. 1sin x 1 cos x22 5 1 1 sin12x2 29. cos2 x 5 1 1 cos12x2 2 30. sin2 x 5 1 2 cos12x2 2 31. cos4 x 2 sin4 x 5 cos12x2 32. cos4 x 1 sin4 x 5 1 2 1 2 sin2 12x2 33. 8 sin2 x cos2 x 5 1 2 cos14x2 34. 3cos12x2 2 sin12x243sin12x2 1 cos12x24 5 cos14x2 35. 21 2 sec2 x 5 22 sin2 x csc2 12x2 36. 4 csc14 x2 5 sec x csc x cos 12x2 37. sin13 x2 5 sin x 14 cos2 x 2 12 38. tan13x2 5 tan x 13 2 tan2 x2 11 2 3 tan2 x2 39. 1 2 sin14 x2 5 2 sin x cos x 2 4 sin3x cos x 40. cos14 x2 5 3cos12 x2 2 sin12x2431cos12x2 1 sin12x24 In Exercises 41–50, graph the functions. 41. y 5 sin12x2 1 2 cos12x2 42. y 5 2 tan x 2 2 sec2 x 43. y 5 cot x 1 tan x cot x 2 tan x 44. y 5 1 2 tan x cot x sec x csc x 45. y 5 sin12x2cos12x2 46. y 5 3 sin13x2 cos123x2 47. y 5 1 2 tan x cot x sec x csc x 48. y 5 3 2 2sec12x2 csc12x2 49. y 5 sin12x2 cos x 2 3 cos12x2 50. y 5 2 1 sin12x2 cos x 2 3 cos12x2 • A P P L I C A T I O N S 51. Business/Economics. Annual cash flow of a stock fund (measured as a percentage of total assets) has fluctuated in cycles. The highs were roughly 112% of total assets, and lows were roughly 28% of total assets. This cash flow can be modeled by the function C1t2 5 12 2 20 sin2 t Use a double-angle identity to express C1t2 in terms of the cosine function. 52. Business. Computer sales are generally subject to seasonal fluctuations. An analysis of the sales of a computer manufacturer during 2015–2017 is approximated by the function s1t2 5 0.098 cos2 t 1 0.387 1 # t # 12 where t represents time in quarters 1t 5 1 represents the end of the first quarter of 20152, and s(t) represents computer sales (quarterly revenue) in millions of dollars. Use a double-angle identity to express s(t) in terms of the cosine function. For Exercises 53 and 54, refer to the following: An ore-crusher wheel consists of a heavy disk spinning on its axle. The normal (crushing) force F, in pounds, between the wheel and the inclined track is determined by F 5 W sin u 1 1 2 c 2 c C R 11 2 cos 2u2 1 A l sin 2ud , where W is the weight of the wheel, u is the angle of the axis, C and A are moments of inertia, R is the radius of the wheel, l is the distance from the wheel to the pin where the axle is attached, and c is the speed in rpm that the wheel is spinning. The optimum crushing force occurs when the angle u is between 45° and 90°. l θ 7.4 Double-Angle Identities 695 696 CHAPTER 7 Analytic Trigonometry 53. Ore-Crusher Wheel. Find F if the angle is 60°, W is 500 pounds, c is 200 rpm, C R 5 750, and A l 5 3.75. 55. With the information given in the diagram below, compute y. y B B 3 ft 1 ft 54. Ore-Crusher Wheel. Find F if the angle is 75°, W is 500 pounds, c is 200 rpm, C R 5 750, and A l 5 3.75. 56. With the information given in the diagram below, compute sin12b2. 3 in. β 7 in. • C A T C H T H E M I S T A K E In Exercises 57 and 58, explain the mistake that is made. 57. If cos x 5 1 3, find sin12x2 given sin x , 0. Solution: Write the double-angle identity for the sine function. sin12x2 5 2 sin x cos x Solve for sin x using the Pythagorean identity. sin2 x 1 a1 3b 2 5 1 sin x 5 2!2 3 Substitute cos x 5 1 3 and sin x 5 2!2 3 . sin12x2 5 2 a2!2 3 b a1 3b Simplify. sin12x2 5 4!2 9 This is incorrect. What mistake was made? 58. If sin x 5 1 3, find tan12x2 given cos x , 0. Solution: Use the quotient identity. tan12x2 5 sin12x2 cos x Use the double-angle formula for the sine function. tan12x2 5 2sin x cos x cos x Cancel the common cosine factors. tan12x2 5 2 sin x Substitute sin x 5 1 3. tan12x2 5 2 3 This is incorrect. What mistake was made? • C O N C E P T U A L For Exercises 59–67, determine whether each statement is true or false. 59. sin12A2 1 sin12A2 5 sin14A2 60. cos14A2 2 cos12A2 5 cos12A2 61. If tan x . 0, then tan12x2 . 0. 62. If sin x . 0, then sin12x2 . 0. 63. If 0 , cos x , sin x , 1, then cos12x2 , 0. 64. sin12x2. 0 if and only if the terminal side of x lies is quadrant I. 65. cos2 13x2 2 sin2 13x2 5 cos16x2 66. 1 2 2 sin2 ax 2b 5 2cos x 67. y 5 sin12x2 is an even function. 7.5 Half-Angle Identities 697 68. Express tan14x2 in terms of functions of tan x. 69. Express tan124 x2 in terms of functions of tan x. 70. Is the identity 2csc12 x2 5 1 1 tan2 x tan x true for x 5 p 2? Explain. 71. Is the identity tan12x2 5 2 tan x 1 2 tan2 x true for x 5 p 4? Explain. • C H A L L E N G E • T E C H N O L O G Y One cannot prove that an equation is an identity using technology, but rather one uses it as a first step to see whether or not the equation seems to be an identity. 72. With a graphing calculator, plot Y1 5 12x2 2 12x23 3! 1 12x25 5! and Y2 5 sin12x2 for x range 321, 14. Is Y1 a good approximation to Y2? 73. With a graphing calculator, plot Y1 5 1 2 12x22 2! 1 12x24 4! and Y2 5 cos12x2 for x range 321, 14. Is Y1 a good approximation to Y2? 74. With a graphing calculator, determine whether tan14x2 2 tan13x2 tan x 5 csc12x2 1 2 sec12x2 by plotting each side of the equation and seeing whether the graphs coincide. 7.5.1 Applying Half-Angle Identities We now use the double-angle identities from Section 7.4 to develop the half-angle identities. Like the double-angle identities, the half-angle identities will allow us to find certain exact values of trigonometric functions and to verify other trigonometric identities. The half-angle identities come directly from the double-angle identities. We start by rewriting the second and third forms of the cosine double-angle identity to obtain identities for the square of the sine and cosine functions, sin2 and cos2, otherwise known as the power reduction formulas. S K I L L S O B J E C T I V E ■ ■Use half-angle identities in simplifying some trigonometric expressions. C O N C E P T U A L O B J E C T I V E ■ ■Understand that the half-angle identities are derived from the double-angle identities. 7.5 HALF-ANGLE IDENTITIES 7.5.1 S K IL L Use half-angle identities in simplifying some trigonometric expressions. 7.5.1 C ON C E P T U A L Understand that the half-angle identities are derived from the double-angle identities. 75. With a graphing calculator, determine whether csc12x2 sec12x2 31cos12x2 2 sin12x224 5 1 2 2 sin2 x 2 2 sin x cos x 2 sin x cos x1cos2 x 2 sin2 x2 by plotting each side of the equation and seeing whether the graphs coincide. 76. With a graphing calculator, plot Y1 5 cos12x2, Y2 5 2 cos x, and Y3 5 2 cos2 x 2 1 in the same viewing rectangle 322p, 2p4 by [22, 2]. Which graphs are the same? 77. With a graphing calculator, plot Y1 5 sin12x2, Y2 5 2 sin x, and Y3 5 2 sin x cos x in the same viewing rectangle 322p, 2p4 by [22, 2]. Which graphs are the same? WORDS MATH Power reduction formula for the sine function Write the second form of the cosine double-angle identity. cos12A2 5 1 2 2 sin2 A Isolate the 2 sin2 A term on one side of the equation. 2 sin2 A 5 1 2 cos12 A2 Divide both sides by 2. sin2 A 5 1 2 cos12 A2 2 Power reduction formula for the cosine function Write the third form of the cosine double-angle identity. cos12 A2 5 2 cos2 A 2 1 Isolate the 2 cos2 A term on one side of the equation. 2 cos2 A 5 1 1 cos2 A Divide both sides by 2. cos2 A 5 1 1 cos12 A2 2 Power reduction formula for the tangent function Taking the quotient of these leads us to another identity. tan2 A 5 sin2 A cos2 A 5 1 2 cos12A2 2 1 1 cos12A2 2 5 1 2 cos12 A2 1 1 cos12 A2 These three identities for the squared functions—really, alternative forms of the double-angle identities—are used in calculus as power reduction formulas (identities that allow us to reduce the power of the trigonometric function from 2 to 1): sin2 A 5 1 2 cos12 A2 2 cos2 A 5 1 1 cos12 A2 2 tan2 A 5 1 2 cos12 A2 1 1 cos12 A2 Derivation of the Half-Angle Identities We can now use these forms of the double-angle identities to derive the half-angle identities. WORDS MATH Sine half-angle identity For the sine half-angle identity, start with the double-angle formula involving both the sine and cosine functions, cos12x2 5 1 2 2 sin2 x, and solve for sin2 x. sin2 x 5 1 2 cos12 x2 2 Solve for sin x. sin x 5 6 Å 1 2 cos12 x2 2 Let x 5 A 2. sinaA 2 b 5 6 å 1 2 cos 2aA 2 b 2 Simplify. sinaA 2 b 5 6 Ä 1 2 cos A 2 698 CHAPTER 7 Analytic Trigonometry Cosine half-angle identity For the cosine half-angle identity, start with the double-angle formula involving only the cosine function, cos12x2 5 2 cos2 x 2 1, and solve for cos2 x. cos2 x 5 1 1 cos12 x2 2 Solve for cos x. cos x 5 6Å 1 1 cos12 x2 2 Let x 5 A 2. cosaA 2 b 5 6å 1 1 cos 2aA 2 b 2 Simplify. cosaA 2 b 5 6Ä 1 1 cos A 2 Tangent half-angle identity For the tangent half-angle identity, start with the quotient identity. Substitute half-angle identities for the sine and cosine functions. tanaA 2 b 5 6Å 1 2 cos A 2 6Å 1 1 cos A 2 Simplify. tanaA 2 b 5 6Ä 1 2 cos A 1 1 cos A tanaA 2 b 5 sinaA 2 b cosaA 2 b HALF-ANGLE IDENTITIES SINE COSINE TANGENT sinaA 2 b 5 6Ä 1 2 cos A 2 cosaA 2 b 5 6Ä 1 1 cos A 2 tanaA 2 b 5 6Ä 1 2 cos A 1 1 cos A tanaA 2 b 5 sin A 1 1 cos A tanaA 2 b 5 1 2 cos A sin A Note: We can also find tanaA 2 b by starting with the identity tan2 x 5 1 2 cos12x2 1 1 cos12x2, solving for tan x, and letting x 5 A 2. The tangent function also has two other similar forms for tan A 2 (see Exercises 69 and 70). STUDY TIP The sign 1 or 2 is determined by what quadrant contains A 2 and what the sign of the particular trigonometric function is in that quadrant. 7.5 Half-Angle Identities 699 700 CHAPTER 7 Analytic Trigonometry Applying the Half-Angle Identities These identities hold for any real number A or any angle with either degree measure or radian measure A as long as both sides of the equation are defined. The sign 11 or 22 is determined by the sign of the trigonometric function in the quadrant which contains A 2. EXAMPLE 1 Finding Exact Values Using Half-Angle Identities Use a half-angle identity to find cos 15°. Solution: Write cos 15° in terms of a half-angle. cos 15° 5 cosa30° 2 b Write the half-angle identity for the cosine function. cosaA 2 b 5 6Ä 1 1 cos A 2 Substitute A 5 30°. cosa30° 2 b 5 6Ä 1 1 cos 30° 2 Simplify. cos 15° 5 6å 1 1 !3 2 2 15° is in quadrant I, where the cosine function is positive. cos 15° 5 Ç 2 1 !3 4 5 "2 1 !3 2 Y OUR TU R N Use a half-angle identity to find sin122.5°2. ▼ ▼ A N S W E R sin122.5°2 5 "2 2 !2 2 Look back at Example 1 in Section 7.3 and compare the result we found when evaluating cos158 using the cosine of a difference formula, !6 1 !2 4 , with the result in Example 1 above, "2 1 !3 2 . A calculator evaluates both expressions as 0.965925826. EXAMPLE 2 Finding Exact Values Using Half-Angle Identities Use a half-angle identity to find tana11p 12 b. Solution: Write tana11p 12 b in terms of a half-angle. tana11p 12 b 5 tan ° 11p 6 2 ¢ Write the half-angle identity for the tangent function. tan aA 2 b 5 1 2 cos A sin A Substitute A 5 11p 6 . tan° 11p 6 2 ¢ 5 1 2 cos a11 p 6 b sin a11 p 6 b 7.5 Half-Angle Identities 701 Simplify. tana11p 12 b 5 1 2 !3 2 21 2 tana11p 12 b 5 !3 2 2 11p 12 is in quadrant II, where the tangent function is negative. Notice that if we approximate tan a11p 12 b with a calculator, we find that tan a11p 12 b < 20.2679 and !3 2 2 < 20.2679. This form of the tangent half-angle identity was selected because of mathematical simplicity. If we had selected either of the other forms, we would have obtained an expression that had a square root of a square root or a radical in the denominator (requiring rationalization). Y OUR T UR N Use a half-angle identity to find tan ap 8 b. ▼ ▼ A N S W E R 2 2 !2 !2 or !2 2 1 EXAMPLE 3 Finding Exact Values Using Half-Angle Identities If cos x 5 3 5 and 3p 2 , x , 2p, find sinax 2b, cosax 2b, and tanax 2b. Solution: Determine in which quadrant x 2 lies. Since 3p 2 , x , 2p, we divide by 2. 3p 4 , x 2 , p x 2 lies in quadrant II; therefore, the sine function is positive and both the cosine and tangent functions are negative. Write the half-angle identity for the sine function. sinax 2b 5 6Å 1 2 cos x 2 Substitute cos x 5 3 5. sinax 2b 5 6ä 1 2 3 5 2 Simplify. sinax 2b 5 6Å 1 5 5 6 !5 5 Since x 2 lies in quadrant II, choose the positive sinax 2b 5 !5 5 value for the sine function. [CONCEPT CHECK] If angle A corresponds to a terminal side that lies in quadrant III, then sin(A/2) is positive. ANSWER True ▼ 702 CHAPTER 7 Analytic Trigonometry Write the half-angle identity for the cosine function. cosax 2b 5 6Å 1 1 cos x 2 Substitute cos x 5 3 5. cosax 2b 5 6ä 1 1 3 5 2 Simplify. cosax 2b 5 6Å 4 5 5 62!5 5 Since x 2 lies in quadrant II, choose the negative value for the cosine function. cosax 2b 5 22!5 5 Use the quotient identity for tangent. tanax 2b 5 sinax 2b cosax 2b Substitute sinax 2b 5 !5 5 and cosax 2b 5 22!5 5 . tanax 2b 5 !5 5 22!5 5 Simplify. tanax 2b 5 21 2 YOUR TURN If cos x 5 23 5 and p , x , 3p 2 , find sinax 2b, cosax 2b, and tanax 2b. ▼ ▼ A N S W E R sinax 2b 5 2!5 5 , cosa x 2b52!5 5 , tana x 2b 5 22 EXAMPLE 4 Using Half-Angle Identities to Verify Other Identities Verify the identity cos2ax 2b 5 tan x 1 sin x 2 tan x . Solution: Write the cosine half-angle identity. cosax 2b 5 6 Å 1 1 cos x 2 Square both sides of the equation. cos2ax 2b 5 1 1 cos x 2 Multiply the numerator and denominator on the right side by tan x. cos2ax 2b 5 a1 1 cos x 2 b atan x tan xb Simplify. cos2ax 2b 5 tan x 1 cos x tan x 2 tan x Note that cos x tan x 5 sin x. cos2 ax 2b 5 tan x 1 sin x 2 tan x An alternative solution is to start with the right-hand side. 7.5 Half-Angle Identities 703 Solution (alternative): Start with the right-hand side. tan x 1 sin x 2 tan x Write this expression as the sum of two expressions. 5 tan x 2 tan x 1 sin x 2 tan x Simplify. 5 1 2 1 1 2 sin x tan x Write tan x 5 sin x cos x. 5 1 2 1 1 2 sin x sin x cos x Simplify. 5 1 2 11 1 cos x2 5 cos2 ax 2b EXAMPLE 5 Using Half-Angle Identities to Verify Other Trigonometric Identities Verify the identity tan x 5 csc12x2 2 cot12x2. Solution: Write the third half-angle formula for the tangent function. tanaA 2 b 5 1 2 cos A sin A Write the right side as a difference of two expressions having the same denominator. tanaA 2 b 5 1 sin A 2 cos A sin A Substitute the reciprocal and quotient identities, respectively, on the right. tanaA 2 b 5 csc A 2 cot A Let A 5 2x. tan x 5 csc12x2 2 cot12x2 Notice that in Example 5 we started with the third half-angle identity for the tangent function. In Example 6, we will start with the second half-angle identity for the tangent function. In general, you select the form that appears to lead to the desired expression. EXAMPLE 6 Using Half-Angle Identities to Simplify Trigonometric Expressions Graph y 5 sin12px2 1 1 cos12px2 by first simplifying the trigonometric expression to a more recognizable form. Solution: Simplify the trigonometric expression using a half-angle identity for the tangent function. 704 CHAPTER 7 Analytic Trigonometry Write the second half-angle identity for the tangent function. tanaA 2 b 5 sin A 1 1 cos A Let A 5 2px. tan1px2 5 sin12px2 1 1 cos12px2 Graph y 5 tan1px2. –2 –1 1 2 –5 –2.5 2.5 5 x y y = tan(πx) y = sin(2πx) 1 + cos(2πx) Y OUR TU R N Graph y 5 1 2 cos1px2 sin1px2 . ▼ ▼ A N S W E R –2 2 –4 –3 –2 –1 1 2 3 4 x y y = tan( ) x 2 π We determine the sign, 1 or 2, by first deciding which quadrant contains A 2 and then finding the sign of the indicated trigonometric function in that quadrant. Recall that there are three forms of the tangent half-angle identity. There is no need to memorize the other forms of the tangent half-angle identity, since they can be derived by first using the Pythagorean identity and algebraic manipulation. In this section, we used the double-angle identities to derive the half-angle identities. We then used the half-angle identities to find certain exact values of trigonometric functions, verify other trigonometric identities, and simplify trigonometric expressions. sinaA 2 b 5 6Å 1 2 cos A 2 cosaA 2 b 5 6Å 1 1 cos A 2 tanaA 2 b 5 6Å 1 2 cos A 1 1 cos A [SEC TION 7.5] S U M M A RY [SEC TION 7.5] E X E R CI SE S • S K I L L S In Exercises 1–16, use the half-angle identities to find the exact values of the trigonometric expressions. 1. sin 15° 2. cos 22.5° 3. cosa11p 12 b 4. sinap 8 b 5. cos 75° 6. sin 75° 7. tan 67.5° 8. tan 202.5° 9. seca29p 8 b 10. csca9p 8 b 11. cota13p 8 b 12. cota7p 8 b 13. seca5p 8 b 14. csca25p 8 b 15. cot12135°2 16. cot105° 7.5 Half-Angle Identities 705 In Exercises 17–32, use the half-angle identities to find the desired function values. 17. If cos x 5 5 13 and sin x , 0, find sinax 2b. 18. If cos x 5 2 5 13 and sin x , 0, find cosax 2b. 19. If tan x 5 12 5 and p , x , 3p 2 , find sinax 2b. 20. If tan x 5 12 5 and p , x , 3p 2 , find cosax 2b. 21. If sec x 5 !5 and sin x . 0, find tanax 2b. 22. If sec x 5 !3 and sin x , 0, find tanax 2b. 23. If csc x 5 3 and cos x , 0, find sinax 2b. 24. If csc x 5 23 and cos x . 0, find cosax 2b. 25. If cos x 5 21 4 and csc x , 0, find cotax 2b. 26. If cos x 5 1 4 and cot x , 0, find cscax 2b. 27. If cot x 5 224 5 and p 2 , x , p, find cosax 2b. 28. If cot x 5 224 5 and p 2 , x , p, find sinax 2b. 29. If sin x 5 20.3 and sec x . 0, find tanax 2b. 30. If sin x 5 20.3 and sec x , 0, find cotax 2b. 31. If sec x 5 2.5 and tan x . 0, find cotax 2b. 32. If sec x 5 23 and cot x , 0, find tanax 2b. In Exercises 33–38, simplify each expression using half-angle identities. Do not evaluate. 33. ã 1 1 cosa5p 6 b 2 34. ã 1 2 cosap 4 b 2 35. sin 150° 1 1 cos 150° 36. 1 2 cos 150° sin 150° 37. ï 1 2 cos a5p 4 b 1 1 cosa5p 4 b 38. Å 1 2 cos 1 5° 1 1 cos 15° In Exercises 39–50, verify the identities. 39. sin2ax 2b 1 cos2ax 2b 5 1 40. cos2ax 2b 2 sin2ax 2b 5 cos x 41. sin12x2 5 22 sinax 2b cosax 2b 42. 2 cos2ax 4b 5 1 1 cosax 2b 43. tan2ax 2b 5 1 2 cos x 1 1 cos x 44. tan2ax 2b 5 1csc x 2 cot x22 45. tanaA 2 b 1 cotaA 2 b 5 2 csc A 46. cotaA 2 b 2 tanaA 2 b 5 2 cot A 47. csc2aA 2 b 5 211 1 cos A2 sin2 A 48. sec2aA 2 b 5 211 2 cos A2 sin2 A 49. cscaA 2 b 5 6 0csc A0 !2 1 2 cos A 50. secaA 2 b 5 6 0csc A0 !2 2 2 cos A In Exercises 51–58, graph the functions. 51. y 5 4 cos2ax 2b 52. y 5 26 sin2ax 2b 53. y 5 1 2 tan2ax 2b 1 1 tan2ax 2b 54. y 5 1 2 csinax 2b 1 cosax 2b d 2 55. y 5 4 sin2ax 2b 2 1 56. y 5 2 1 6 cos2ax 2b 1 2 57. y 5 Å 1 2 cos12x2 1 1 cos12x2 0 # x , p 58. y 5 Å 1 1 cos13x2 2 1 3 0 # x # p 3 706 CHAPTER 7 Analytic Trigonometry For Exercises 59 and 60, refer to the following: • A P P L I C A T I O N S Monthly profits can be expressed as a function of sales, that is, p(s). A financial analysis of a company has determined that the sales s in thousands of dollars are also related to monthly profits p in thousands of dollars by the relationship: tan u 5 p s for 0 # s # 50, 0 # p , 40 Based on sales and profits, it can be determined that the domain for angle u is 0 # u # 38°. Profts (p) p(s) Sales (s) p s θ 59. Business. If monthly profits are $3000 and monthly sales are $4000, find tanau 2b. 60. Business. If monthly profits are p and monthly sales are s (where p , s), find tanau 2b. 61. Area of an Isosceles Triangle. Consider the triangle below, where the vertex angle measures u, the equal sides measure a, the height is h, and half the base is b. (In an isosceles triangle, the perpendicular dropped from the vertex angle divides the triangle into two congruent triangles.) The two triangles formed are right triangles. h b b a a θ 2 θ 2 In the right triangles, sinau 2b 5 b a and cosau 2b 5 h a. Multiply each side of each equation by a to get b 5 a sinau 2b, h 5 a cosau 2b. The area of the entire isosceles triangle is A 5 1 2 12b2 h 5 bh. Substitute the values for b and h into the area formula. Show that the area is equivalent to aa2 2 bsin u. 62. Area of an Isosceles Triangle. Use the results from Exercise 61 to find the area of an isosceles triangle whose equal sides measure 7 inches and whose base angles each measure 75°. • C A T C H T H E M I S T A K E In Exercises 63 and 64, explain the mistake that is made. 63. If cos x 5 21 3, find sin ax 2b given p , x , 3p 2 . Solution: Write the half-angle identity for the sine function. sinax 2b 5 6Å 1 2 cos x 2 Substitute cos x 5 21 3. sinax 2b 5 6ä 1 1 1 3 2 Simplify. sinax 2b 5 6Å 4 3 5 6 2 !3 The sine function is negative. sinax 2b 5 2 2 !3 This is incorrect. What mistake was made? 64. If cos x 5 1 3, find tan2ax 2b. Solution: Use the quotient identity. tan2ax 2b 5 sin2 ax 2 b cos2 x Use the half-angle identity for the sine function. tan2ax 2 b 5 1 2 cos x 2 cos2 x Simplify. tan2ax 2 b 5 1 2a 1 cos2 x 2 cos x cos2 xb tan2 ax 2 b 5 1 2a 1 cos2 x 2 1 cos xb Substitute cos x 5 1 3. tan2 ax 2 b 5 1 2 ° 1 1 9 2 1 1 3 ¢ tan2 ax 2 b 5 3 This is incorrect. What mistake was made? 7.5 Half-Angle Identities 707 65. sin aA 2 b 1 sin aA 2 b 5 sin A 67. If tan x . 0, then tan ax 2b . 0. 69. Given tanaA 2 b 5 6 Å 1 2 cos A 1 1 cos A, verify tanaA 2 b 5 sin A 1 1 cos A. Substitute A 5 p into the identity and explain your results. 66. cos aA 2 b 1 cos aA 2 b 5 cos A 68. If sin x . 0, then sinax 2b . 0. 70. Given tanaA 2 b 5 6Å 1 2 cos A 1 1 cos A, verify tanaA 2 b 5 1 2 cos A sin A . Substitute A 5 p into the identity and explain your results. 71. Prove that cotaA 4 b 5 6 ï 1 1 cosaA 2 b 1 2 cosaA 2 b . 73. Find the values of x in the interval 30, 2p4 for which tanax 2b . 0. 72. Prove that cota2A 2 b secaA 2 b csca2A 2 b tanaA 2 b 5 2 csc A. 74. Find the values of x in the interval 30, 2p4 for which cotax 2b # 0. • C O N C E P T U A L In Exercises 65–68, determine whether each statement is true or false. • C H A L L E N G E • T E C H N O L O G Y 75. With a graphing calculator, plot Y1 5 ax 2b 2 ax 2b 3 3! 1 ax 2b 5 5! and Y2 5 sinax 2b for x range 321, 14. Is Y1 a good approximation to Y2? 77. With a graphing calculator, determine whether csc2ax 2b 1 sec2ax 2b 5 4 csc2 x is an identity by plotting each side of the equation and seeing whether the graphs coincide. 79. With a graphing calculator, plot Y1 5 cosax 2b, Y2 5 1 2 cos x, and Y3 5 Å 1 1 cos x 2 in the same viewing rectangle 32p, p4 by 321, 14. Which graphs are the same? 76. Using a graphing calculator, plot Y1 5 1 2 ax 2b 2 2! 1 ax 2b 4 4! and Y2 5 cosax 2b for x range 321, 14. Is Y1 a good approximation to Y2? 78. With a graphing calculator, determine whether tan2ax 2b 1 cot2ax 2b 5 22 cot2 x sec x is an identity by plotting each side of the equation and seeing whether the graphs coincide. 80. With a graphing calculator, plot Y1 5 sinax 2b, Y2 5 1 2 sin x, and Y3 5 Å 1 1 cos x 2 in the same viewing rectangle 30, 2p4 by 321, 14. Which graphs are the same? One cannot prove that an equation is an identity using technology, but rather one uses it as a first step to see whether or not the equation seems to be an identity. 708 CHAPTER 7 Analytic Trigonometry In calculus, often it is helpful to write products of trigonometric functions as sums of other trigonometric functions, and vice versa. In this section, we discuss the product-to-sum identities, which convert products of trigonometric functions to sums of trigonometric functions, and sum-to-product identities, which convert sums of trigonometric functions to products of trigonometric functions. S K I L L S O B J E C T I V E S ■ ■Express products of trigonometric functions as sums of trigonometric functions. ■ ■Express sums of trigonometric functions as products of trigonometric functions. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that the sum and difference identities are used to derive product-to-sum identities. ■ ■Understand that the product-to-sum identities are used to derive the sum-to-product identities. 7.6 PRODUCT-TO-SUM AND SUM-TO-PRODUCT IDENTITIES 7.6.1 Product-to-Sum Identities The product-to-sum identities are derived from the sum and difference identities. WORDS MATH Write the identity for the cosine of a sum. cos A cos B 2 sin A sin B 5 cos1A 1 B2 Write the identity for the cosine of a difference. cos A cos B 1 sin A sin B 5 cos1A 2 B2 Add the two identities. 2 cos A cos B 5 cos1A 1 B2 1 cos1A 2 B2 Divide both sides by 2. cos A cos B 5 1 2 3cos1A 1 B2 1 cos1A 2 B24 cos A cos B 1 sin A sin B 5 cos1A 2 B2 Subtract the sum identity from the 2cos A cos B 1 sin A sin B 5 2cos1A 1 B2 difference identity. 2 sin A sin B 5 cos1A 2 B2 2 cos1A 1 B2 Divide both sides by 2. sin A sin B 5 1 2 3cos1A 2 B2 2 cos1A 1 B24 Write the identity for the sine of a sum. sin A cos B 1 cos A sin B 5 sin1A 1 B2 Write the identity for the sine of a difference. sin A cos B 2 cos A sin B 5 sin1A 2 B2 Add the two identities. 2 sin Acos B 5 sin1A 1 B2 1 sin1A 2 B2 Divide both sides by 2. sin A cos B 5 1 2 3sin1A 1 B2 1 sin1A 2 B24 PRODUCT-TO-SUM IDENTITIES 1. cos A cos B 5 1 2 3cos1A 1 B2 1 cos1A 2 B24 2. sin A sin B 5 1 2 3cos1A 2 B2 2 cos1A 1 B24 3. sin A cos B 5 1 2 3sin1A 1 B2 1 sin1A 2 B24 7.6 Product-to-Sum and Sum-to-Product Identities 709 EXAMPLE 1 Illustrating a Product-to-Sum Identity for Specific Values Show that product-to-sum identity 132 is true when A 5 30° and B 5 90°. Solution: Write product-to-sum identity 132. sin A cos B 5 1 2 3sin1A 1 B2 1 sin1A 2 B24 Let A 5 30° and B 5 90°. sin 30° cos 90° 5 1 2 3sin130° 1 90°2 1 sin130° 2 90°24 Simplify. sin 30° cos 90° 5 1 2 3sin 120° 1 sin1260°24 Evaluate the trigonometric functions. 1 2⋅0 5 1 2 c !3 2 2 !3 2 d Simplify. 0 5 0 EXAMPLE 2 Convert a Product to a Sum Convert the product cos14x2 cos13x2 to a sum. Solution: Write product-to-sum identity 112. cos A cos B 5 1 2 3cos 1A 1 B2 1 cos 1A 2 B24 Let A 5 4x and B 5 3x. cos14x2 cos13x2 5 1 2 3cos14x 1 3x2 1 cos14x 2 3x24 Simplify. cos14x2 cos13x2 5 1 2 3cos17x2 1 cos1x24 YOUR T UR N Convert the product cos12x2 cos15x2 to a sum. ▼ 7.6.1 S K I L L Express products of trigonometric functions as sums of trigonometric functions. 7.6.1 C ON C E P T U A L Understand that the sum and difference identities are used to derive product-to-sum identities. [CONCEPT CHECK] TRUE OR FALSE sin(A) cos(B) = cos(A) sin(B), for all angles A and B ANSWER False ▼ EXAMPLE 3 Converting Products to Sums Express sin12x2 sin13x2 in terms of cosines. Solution: Write product-to-sum identity 122. sin A sin B 5 1 2 3cos 1A 2 B2 2 cos1A 1 B24 Let A 5 2x and B 5 3x. sin12x2 sin13x2 5 1 2 3cos12x 2 3x2 2 cos12x 1 3x24 Simplify. sin12x2 sin13x2 5 1 2 3cos12x2 2 cos15x24 The cosine function is an even function. sin12x2 sin13x2 5 1 2 3cos x 2 cos15x24 YOUR T UR N Express sin1x2 sin12x2 in terms of cosines. ▼ ▼ A N S W E R 1 2 3cos 17x2 1 cos13x24 ▼ A N S W E R 1 2 3cos x 2 cos 13x24 710 CHAPTER 7 Analytic Trigonometry 7.6.2 Sum-to-Product Identities The sum-to-product identities can be obtained from the product-to-sum identities. WORDS MATH Write the identity for the product of the sine and cosine functions. 1 2 3sin1x 1 y2 1 sin1x 2 y24 5 sin x cos y Let x 1 y 5 A and x 2 y 5 B; then x 5 A 1 B 2 and y 5 A 2 B 2 . Substitute these values into the identity. 1 2 3sin A 1 sin B4 5 sinaA 1 B 2 b cosaA 2 B 2 b Multiply by 2. sin A 1 sin B 5 2 sinaA 1 B 2 b cosaA 2 B 2 b The other three sum-to-product identities can be found similarly. All are summarized in the box below. 7.6.2 SKILL Express sums of trigonometric functions as products of trigonometric functions. 7.6.2 C ON CEPTUAL Understand that the product-to-sum identities are used to derive the sum-to-product identities. [CONCEPT CHECK] TRUE OR FALSE If B 5 2A, then cos(A) 2 cos(B) 5 0 and sin(A) 2 sin(B) 5 2sin(A) ANSWER True ▼ SUM-TO-PRODUCT IDENTITIES 4. sin A 1 sin B 5 2 sinaA 1 B 2 b cosaA 2 B 2 b 5. sin A 2 sin B 5 2 sin aA 2 B 2 b cosaA 1 B 2 b 6. cos A 1 cos B 5 2 cos aA 1 B 2 b cosaA 2 B 2 b 7. cos A 2 cos B 5 22 sinaA 1 B 2 b sinaA 2 B 2 b EXAMPLE 4 Illustrating a Sum-to-Product Identity for Specific Values Show that sum-to-product identity (7) is true when A 5 30° and B 5 90°. Solution: Write the sum-to-product identity (7). cos A 2 cos B 5 22 sinaA 1 B 2 b sinaA 2 B 2 b Let A 5 30° and cos 30° 2 cos 90° 5 22 sina30° 1 90° 2 b sina30° 2 90° 2 b B 5 90° Simplify. cos 30° 2 cos 90° 5 22 sin 60° sin1230°2 The sine function is an odd function. cos 30° 2 cos 90° 5 2 sin 60° sin 30° Evaluate the trigonometric functions. !3 2 2 0 5 2 a !3 2 b a1 2b Simplify. !3 2 5 !3 2 EXAMPLE 5 Convert a Sum to a Product Convert 29 3sin12x2 2 sin110x24, a trigonometric expression containing a sum, to a product. Solution: The expression inside the brackets is in the form of identity (5). sin A 2 sin B 5 2 sinaA 2 B 2 b cosaA 1 B 2 b Let A 5 2x and B 5 10x. sin12 x2 2 sin110 x2 5 2 sin a2x 2 10x 2 b cosa2x 1 10x 2 b Simplify. sin12 x2 2 sin110 x2 5 2 sin124 x2 cos16 x2 The sine function is an odd function. sin 12 x2 2 sin110 x2 5 22 sin14 x2 cos 16 x2 Multiply both sides by 29. 29 3sin12 x2 2 sin110 x24 5 18 sin14 x2 cos16 x2 EXAMPLE 6 Simplifying a Trigonometric Expression Using Sum-to-Product Identities Simplify the expression sinax 1 y 2 b cosax 2 y 2 b 1 sinax 2 y 2 b cos ax 1 y 2 b. Solution: Use identities (4) and (5). sinax 1 y 2 b cos ax 2 y 2 b 1 sinax 2 y 2 b cosax 1 y 2 b 5 1 2 sin x 1 1 2 sin y 1 1 2 sin x 2 1 2 sin y Simplify. 5 sin x g g 1 23sin x 1 sin y4 1 23sin x 2 sin y4 Applications In music, a tone is a fixed pitch (frequency) that is given a name. If two notes are sounded simultaneously, then they interfere and produce another tone, often called a “beat.” The beat frequency is the difference of the two frequencies. The more rapid the beat, the further apart the two frequencies of the tones are. When musicians tune their instruments, they use a tuning fork to sound a tone and then tune the instrument until the beat is eliminated; then the tuning fork and instrument are in tune with each other. Mathematically, a tone is represented as A cos12pƒt2, where A is the amplitude (loudness), ƒ is the frequency in Hz, and t is time in seconds. The following table summarizes common tones and frequencies. C 262 Hz D 294 Hz E 330 Hz F 349 Hz G 392 Hz A 440 Hz B 494 Hz 7.6 Product-to-Sum and Sum-to-Product Identities 711 712 CHAPTER 7 Analytic Trigonometry EXAMPLE 7 Music Express the musical tone that is heard when a C and G are simultaneously struck (assume they have the same loudness). Find the beat frequency, ƒ2 2 ƒ1. Assume uniform loudness, A 5 1. Solution: Write the mathematical description of the C tone. cos12pƒ1t2, ƒ1 5 262 Hz Write the mathematical description of the G tone. cos12pƒ2t2, ƒ2 5 392 Hz Add the two notes. cos1524pt2 1 cos1784pt2 Use sum-to-product identities. cos1524pt2 1 cos1784pt2 5 2 cosa524pt 1 784pt 2 b cosa524pt 2 784pt 2 b Simplify. 5 2 cos1654pt2 cos12130pt2 5 2 cos12⋅327⋅p⋅t2 cos1130 pt2 Identify the beat frequency. ƒ2 2 ƒ1 5 392 2 262 5 130 Hz Therefore, the tone of average frequency, 327 Hz, has a beat of 130 Hz (beats/sec). Notice that an average frequency results, cos12⋅327⋅p⋅t2, and that frequency is modulated by a beat frequency, 130 Hz. b average frequency In this section, we used the sum and difference identities (Section 7.3) to derive the product-to-sum identities. The product- to-sum identities allowed us to express products as sums. cos A cos B 5 1 2 3cos1A 1 B2 1 cos1A 2 B24 sin A sin B 5 1 2 3cos1A 2 B2 2 cos1A 1 B24 sin A cos B 5 1 2 3sin1A 1 B2 1 sin1A 2 B24 We then used the product-to-sum identities to derive the sum-to-product identities. The sum-to-product identities allow us to express sums as products. sin A 1 sin B 5 2 sinaA 1 B 2 b cosaA 2 B 2 b sin A 2 sin B 5 2 sinaA 2 B 2 b cosaA 1 B 2 b cos A 1 cos B 5 2 cosaA 1 B 2 b cosaA 2 B 2 b cos A 2 cos B 5 22 sinaA 1 B 2 b sinaA 2 B 2 b [SEC TION 7.6] S U M MA RY y = cos(524πt) y = cos(784πt) y = 2 cos(654πt) cos(130πt) –2 –1 1 2 y –1 –0.5 1 0.5 t [SEC TION 7.6] E X E R C I S E S • S K I L L S In Exercises 1–14, write each product as a sum or difference of sines and/or cosines. 1. sin12x2 cos1x2 2. cos110x2 sin15x2 3. 5 sin14x2 sin16x2 4. 23 sin12x2 sin14x2 5. 4 cos12x2 cos12x2 6. 28 cos 13x2 cos15x2 7. sina3x 2 b sina5x 2 b 8. sinapx 2 bsina5px 2 b 9. cosa2x 3 b cosa4x 3 b 10. sina2p 4 xb cosa2p 2 xb 11. 23 cos10.4x2 cos11.5x2 12. 2 sin12.1x2 sin13.4x2 13. 4 sinA2!3xB cosA3!3x2 14. 25 cosa2 !2 3 xb sina5!2 3 xb In Exercises 15–28, write each expression as a product of sines and/or cosines. 15. cos15x2 1 cos13x2 16. cos12x2 2 cos14x2 17. sin13x2 2 sin x 18. sin110x2 1 sin15x2 19. sinax 2b 2 sina5x 2 b 20. cosax 2b 2 cosa5x 2 b 21. cosa2 3 xb 1 cosa7 3 xb 22. sina2 3 xb 1 sina7 3 xb 23. sin10.4x2 1 sin10.6x2 24. cos10.3x2 2 cos10.5x2 25. sinA !5xB 2 sinA3!5xB 26. cosA23!7xB 2 cosA2!7xB 27. cosa2 p 4xb 1 cosap 6xb 28. sina3p 4 xb 1 sina5p 4 xb In Exercises 29–34, simplify the trigonometric expressions. 29. cos13x2 2 cos x sin13x2 1 sin x 30. sin14x2 1 sin12x2 cos14x2 2 cos12x2 31. cos x 2 cos13x2 sin13x2 2 sin x 32. sin14x2 1 sin12x2 cos14x2 1 cos12x2 33. cos15x2 1 cos12x2 sin15x2 2 sin12x2 34. sin17x2 2 sin12x2 cos17x2 2 cos12x2 In Exercises 35–42, verify the identities. 35. sin A 1 sin B cos A 1 cos B 5 tanaA 1 B 2 b 36. sin A 2 sin B cos A 1 cos B 5 tanaA 2 B 2 b 37. cos A 2 cos B sin A 1 sin B 5 2tanaA 2 B 2 b 38. cos A 2 cos B sin A 2 sin B 5 2tanaA 1 B 2 b 39. sin A 1 sin B sin A 2 sin B 5 tanaA 1 B 2 b cotaA 2 B 2 b 40. cos A 2 cos B cos A 1 cos B 5 2tanaA 1 B 2 b tanaA 2 B 2 b 41. cos1A 1 B2 1 cos1A 2 B2 sin1A 1 B2 1 sin1A 2 B2 5 cot A 42. cos1A 2 B2 2 cos1A 1 B2 sin1A 1 B2 1 sin1A 2 B2 5 tan B 7.6 Product-to-Sum and Sum-to-Product Identities 713 714 CHAPTER 7 Analytic Trigonometry • A P P L I C A T I O N S 43. Business. An analysis of the monthly costs and monthly revenues of a toy store indicates that monthly costs fluctuate (increase and decrease) according to the function C1t2 5 sinap 6 t 1 pb and monthly revenues fluctuate (increase and decrease) according to the function R1t2 5 sinap 6 t 1 5p 3 b Find the function that describes how the monthly profits fluctuate: P1t2 5 R1t2 2 C1t2. Using identities in this section, express P1t2 in terms of a cosine function. 44. Business. An analysis of the monthly costs and monthly revenues of an electronics manufacturer indicates that monthly costs fluctuate (increase and decrease) according to the function C1t2 5 cos ap 3 t 1 p 3 b and monthly revenues fluctuate (increase and decrease) according to the function R1t2 5 cosap 3 tb Find the function that describes how the monthly profits fluctuate: P1t2 5 R1t2 2 C1t2. Using identities in this section, express P1t2 in terms of a sine function. 45. Music. Write a mathematical description of a tone that results from simultaneously playing a G and a B. What is the beat frequency? What is the average frequency? 46. Music. Write a mathematical description of a tone that results from simultaneously playing an F and an A. What is the beat frequency? What is the average frequency? 47. Optics. Two optical signals with uniform 1A 5 12 intensities and wavelengths of 1.55 mm and 0.63 mm are “beat” together. What is the resulting sum if their individual signals are given by sina 2ptc 1.55 mmb and sina 2ptc 0.63 mmb, where c 5 3.0 3 108 m/s? 1Note: 1 mm 5 1026 m.2 48. Optics. The two optical signals in Exercise 47 are beat together. What are the average frequency and the beat frequency? For Exercises 49 and 50, refer to the following: Touch-tone keypads have the following simultaneous low and high frequencies: FREQUENCY 1209 Hz 1336 Hz 1477 Hz 697 Hz 1 2 3 770 Hz 4 5 6 852 Hz 7 8 9 941 Hz 0 # The signal given when a key is pressed is sin12pƒ1t2 1 sin12pƒ2t2, where ƒ1 is the low frequency and ƒ2 is the high frequency. 49. Touch-Tone Dialing. What is the mathematical function that models the sound of dialing 4? 50. Touch-Tone Dialing. What is the mathematical function that models the sound of dialing 3? 51. Area of a Triangle. A formula for finding the area of a triangle when given the measures of the angles and one side is Area 5 a2 sin B sin C 2 sin A , where a is the side opposite angle A. If the measures of angles B and C are 52.5° and 7.5°, respectively, and if a 5 10 feet, use the appropriate product-to-sum identity to change the formula so that you can solve for the area of the triangle exactly. a A C B 52. Area of a Triangle. If the measures of angles B and C in Exercise 51 are 75° and 45°, respectively, and if a 5 12 inches, use the appropriate product-to-sum identity to change the formula so that you can solve for the area of the triangle exactly. a A C B In Exercises 53 and 54, explain the mistake that is made. 53. Simplify the expression 1cos A 2 cos B22 1 1sin A 2 sin B22. Solution: Expand by squaring. cos2 A2 2 cos A cos B1cos2 B1sin2 A22 sin A sin B1sin2 B Group terms. cos2 A1sin2 A22 cos A cos B 22 sin A sin B1cos2 B1sin2 B Simplify using the Pythagorean identity. cos2 A 1 sin2 A 2 2 cos A cos B2 2 sin A sin B 1 cos2 B 1 sin2 B Factor the comon 2. 211 2 cos A cos B 2 sin A sin B2 Simplify. 211 2 cos AB 2 sin AB2 This is incorrect. What mistakes were made? 1 g 1 g 54. Simplify the expression 1sin A 2 sin B21cos A 1 cos B2. Solution: Multiply the expressions using the distributive property. sin A cos A 1 sin A cos B 2 sin B cos A 2 sin B cos B Cancel the second and third terms. sin A cos A 2 sin B cos B Use the product-to-sum identity. sin A cos A 2 sin B cos B Simplify. 5 1 2 sin 12 A2 2 1 2 sin 12 B2 This is incorrect. What mistake was made? 1 23sin1A 1 A2 1 sin1A 2 A24 g 1 23sin1B 1 B2 1 sin1B 2 B24 g • C A T C H T H E M I S T A K E • C O N C E P T U A L 55. cos A cos B 5 cos AB 56. sin A sin B 5 sin AB 57. The product of two cosine functions is a sum of two other cosine functions. 58. The product of two sine functions is a difference of two cosine functions. 59. Write sin A sin B sin C as a sum or difference of sines and cosines. 60. Write cos A cos B cos C as a sum or difference of sines and cosines. In Exercises 55–58, determine whether each statement is true or false. • C H A L L E N G E 61. Prove the addition formula cos1A 1 B2 5 cos A cos B 2 sin A sin B using the identities of this section. 62. Prove the difference formula sin1A 2 B2 5 sin A cos B 2 sin B cos A using the identities of this section. 63. Graph y 5 1 2 3 sin1px2 sina2 p 6 xb. 64. Graph y 5 4 sin12x 2 12 cos12 2 x2. 65. Graph y 5 2 cosa2p 3 xb cosa5p 6 xb. 66. Graph y 5 x 2 cos12x2 sin13x2. • T E C H N O L O G Y 67. Suggest an identity 4 sin x cos x cos12x2 5 __ by graphing Y1 5 4 sin x cos x cos12 x2 and determining the function based on the graph. 68. Suggest an identity 1 1 tan x tan12 x2 5 _ by graphing Y1 5 1 1 tan x tan12x2 and determining the function based on the graph. 69. With a graphing calculator, plot Y1 5 sin14x2 sin12x2, Y2 5 sin16x2, and Y3 5 1 2 3cos12x2 2 cos16x24 in the same viewing rectangle [0, 2p] by [21, 1]. Which graphs are the same? 70. With a graphing calculator, plot Y1 5 cos14x2 cos12x2, Y2 5 cos16x2, and Y3 5 1 2 3cos16x2 1 cos12x24 in the same viewing rectangle [0, 2p] by [21, 1]. Which graphs are the same? 7.6 Product-to-Sum and Sum-to-Product Identities 715 716 CHAPTER 7 Analytic Trigonometry In Section 3.5, we discussed one-to-one functions and inverse functions. Here we present a summary of that section. A function is one-to-one if it passes the horizontal line test: no two x-values map to the same y-value. Notice that the sine function does not pass the horizontal line test. However, if we restrict the domain to 2 p 2 # x # p 2 , then the restricted function is one-to-one. Recall that if y 5 ƒ1x2, then x 5 ƒ 211y2. The following are the properties of inverse functions: 1. If ƒ is a one-to-one function, then the inverse function ƒ 21 exists. 2. The domain of ƒ 21 5 the range of ƒ. The range of ƒ 21 5 the domain of ƒ. 3. ƒ 211ƒ 1x22 5 x for all x in the domain of ƒ. ƒ 1ƒ 211x22 5 x for all x in the domain of ƒ 21. 4. The graph of ƒ 21 is the reflection of the graph of ƒ about the line y 5 x. If the point 1a, b2 lies on the graph of a function, then the point 1b, a2 lies on the graph of its inverse. 7.7.1 Inverse Sine Function Let us start with the sine function with the restricted domain c2p 2 , p 2 d. y 5 sin x Domain: c2p 2 , p 2 d Range: 321, 14 x y 2p 2 21 2p 4 2 !2 2 0 0 p 4 !2 2 p 2 1 S K I L L S O B J E C T I V E S ■ ■Find exact values of an inverse sine function. ■ ■Find exact values of an inverse cosine function. ■ ■Find exact values of an inverse tangent function. ■ ■Find exact values of the cotangent, cosecant, and secant inverse functions. ■ ■Use identities to find exact values of trigonometric expressions involving inverse trigonometric functions. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that the domain of the sine function is restricted to c2p 2, p 2 d in order for the inverse sine function to exist. ■ ■Understand that the domain of the cosine function is restricted to [0, p] in order for the inverse cosine function to exist. ■ ■Understand that the domain of the tangent function is restricted to a2p 2, p 2 b in order for the inverse tangent function to exist. ■ ■Understand that the cotangent, cosecant, and secant inverse functions are not found from the reciprocal of the tangent, sine, and cosine functions, respectively, but rather from the inverse secant, inverse cosecant, and inverse cotangent identities. ■ ■Visualize the quadrants to find exact values of trigonometric expressions involving inverse trigonometric functions. 7.7 INVERSE TRIGONOMETRIC FUNCTIONS 2π π –π –2π x y = sin x –1 1 y y = 1 2 2π π –π –2π x –1 1 y 7.7.1 SKILL Find exact values of an inverse sine function. 7.7.1 C ON CEPTUAL Understand that the domain of the sine function is restricted to 2p 2, p 2 in order for the inverse sine function to exist. By the properties of inverse functions, the inverse sine function will have a domain of 321, 14 and a range of c2 p 2 , p 2 d. To find the inverse sine function, we interchange the x- and y-values of y 5 sin x. y 5 sin21 x Domain: 321, 14 Range: c2p 2 , p 2 d x y 21 2p 2 2 !2 2 2p 4 0 0 !2 2 p 4 1 p 2 Notice that the inverse sine function, like the sine function, is an odd function (symmetric about the origin). If the sine of an angle is known, what is the measure of that angle? The inverse sine function determines that angle measure. Another notation for the inverse sine function is arcsin x. 1.0 –1.0 x y y = sin–1 x or y = arcsin x π 2 π 2 π 4 – π 2 – π 4 –π 2 (0, 0) (–1, ) –π 4 π 4 2 ( , ) (1, ) √2 2 ( , ) –√2 STUDY TIP The inverse sine function gives an angle on the right half of the unit circle (QI and QIV). x y (0, 1) (0, –1) (–1, 0) (1, 0) INVERSE SINE FUNCTION y 5 sin21 x or y 5 arcsin x means x 5 sin y where 21 # x # 1 and 2p 2 # y # p 2 g g “y is the inverse sine of x” “y is the angle measure whose sine equals x” STUDY TIP Trigonometric functions take angle measures and return real numbers. Inverse trigonometric functions take real numbers and return angle measures. STUDY TIP sin21 x 2 1 sin x It is important to note that the 21 superscript indicates an inverse function. Therefore, the inverse sine function should not be interpreted as a reciprocal: sin21 x 2 1 sin x EXAMPLE 1 Finding Exact Values of an Inverse Sine Function Find the exact value of each of the following expressions: a. sin21a !3 2 b b. arcsina2 1 2b Solution (a): Let u 5 sin21a !3 2 b. sin u 5 !3 2 when 2p 2 # u # p 2 Which value of u, in the range 2 p 2 # u # p 2 , corresponds to a sine value of !3 2 ? n The range 2 p 2 # u # p 2 corresponds to quadrants I and IV. n The sine function is positive in quadrant I. n We look for a value of u in quadrant I that has a sine value of !3 2 . u 5 p 3 sin ap 3 b 5 !3 2 and p 3 is in the interval c2 p 2 , p 2 d. sin21a !3 2 b 5 p 3 STUDY TIP In Example 1, note that the graphs help identify the desired angles. a. π 60º or – 3 x y 1 1 – 2 , b. x y π –30º or – – – 6 1 1 – 2 , 7.7 Inverse Trigonometric Functions 717 718 CHAPTER 7 Analytic Trigonometry In Example 1, note that in part (a), both 60° and 120° correspond to the sine function equal to !3 2 , which is why the domain restrictions are necessary for inverse functions except for quadrantal angles. There are always two angles (values) from 0 to 360°or 0 to 2p cexcept for 90°ap 2 b and 270°a3p 2 b d that correspond to the sine function equal to a particular value. It is important to note that the inverse sine function has a domain 321, 14. For ex ample, sin21 3 does not exist because 3 is not in the domain of the inverse sine function. Notice that calculator evaluation of sin21 3 says error. Calculators can be used to evaluate inverse sine functions when an exact evaluation is not feasible, just as they are for the basic trigono metric functions. For example, sin21 0.3 < 17.46°, or 0.305 radians. We now state the properties relating the sine function and the inverse sine function which follow directly from properties of inverse functions. Calculator Confirmation: Since p 3 5 60°, if our calculator is set in degree mode, we should find that sin21a !3 2 bis equal to 60°. Solution (b): Let u 5 arcsin A2 1 2B. sin u 5 2 1 2 when 2 p 2 # u # p 2 Which value of u, in the range 2 p 2 # u # p 2 , corresponds to a sine value of 2 1 2? n The range 2 p 2 # u # p 2 corresponds to quadrants I and IV. n The sine function is negative in quadrant IV. n We look for a value of u in quadrant IV that has a sine value of 2 1 2. u 5 2 p 6 sina2 p 6 b 5 21 2 and 2 p 6 is in the interval c2 p 2 , p 2 d. arcsina2 1 2b 5 2 p 6 Calculator Confirmation: Since 2 p 6 5 230°, if our calculator is set in degree mode, we should find that sin21A2 1 2B is equal to 230°. Y OUR TU R N Find the exact value of each of the following expressions: a. sin21a2 !3 2 b b. arcsina1 2b ▼ ▼ A N S W E R a. 2 p 3 b. p 6 SINE-INVERSE SINE IDENTITIES sin211sin x2 5 x for 2 p 2 # x # p 2 sin1sin21 x2 5 x for 21 # x # 1 For example, sin21csina p 12b d 5 p 12, since p 12 is in the interval c2 p 2 , p 2 d. However, you must be careful not to overlook the domain restriction for which these identities hold, as illustrated in the next example. EXAMPLE 2 Using Inverse Identities to Evaluate Expressions Involving Inverse Sine Functions Find the exact value of each of the following trigonometric expressions: a. sin csin21a !2 2 b d b. sin21csina3p 4 b d Solution (a): Write the appropriate identity. sin1sin21x2 5 x for 21 # x # 1 Let x 5 !2 2 , which is in the interval [21, 1]. Since the domain restriction is met, the identity can be used. sin csin21 a !2 2 b d 5 !2 2 Solution (b): ✖I N C O R R EC T sin211sin x2 5 x ERROR Let x 5 3p 4 . (Forgot the domain restriction.) sin21 csina3p 4 b d 5 3p 4 INCORRECT ✓COR R E C T Write the appropriate identity. sin211sin x2 5 x for 2 p 2 # x # p 2 Let x 5 3p 4 , which is not in the interval c2 p 2, p 2 d . Since the domain restriction is not met, the identity cannot be used. Instead, we look for a value in the domain that corresponds to the same value of sine. x y 4 3π 4 π 2 √2 2 √2 (– , ) 2 √2 2 √2 ( , ) –1 –1 1 1 Substitute sin 3p 4 5 sin p 4 into the expression. sin21 csina3p 4 b d 5 sin21 csinap 4 b d Since p 4 is in the interval c2p 2, p 2 d , we can use the identity. sin21 csin a3p 4 b d 5 sin21 csinap 4 b d 5 p 4 common mistake Ignoring the domain restrictions on inverse identities. YOUR TURN Find the exact value of each of the following trigonometric expressions: a. sin csin21 a2 1 2b d b. sin21 asin 5p 6 b ▼ ▼ A N S W E R a. 2 1 2 b. p 6 7.7 Inverse Trigonometric Functions 719 [CONCEPT CHECK] Evaluate (if possible) sin[sin21(A)] when (A) A . 1 (B) A 5 0 (C) A 5 21 ANSWER (A) Does not exist (DNE) (B) 0 (C) 21 ▼ 720 CHAPTER 7 Analytic Trigonometry 7.7.2 Inverse Cosine Function The cosine function is not a one-to-one function, so we must restrict the domain in order to develop the inverse cosine function. y 5 cos x Domain: [0, p] Range: [21, 1] x y 0 1 p 4 !2 2 p 2 0 3p 4 2 !2 2 p 21 By the properties of inverses, the inverse cosine function will have a domain of 321, 14 and a range of 30, p4. To find the inverse cosine function, we interchange the x- and y-values of y 5 cos x. y 5 cos21 x Domain: [21, 1] Range: [0, p] x y 21 p 2 !2 2 3p 4 0 p 2 !2 2 p 4 1 0 Notice that the inverse cosine function, unlike the cosine function, is not symmetric about the y-axis, or the origin. Although the inverse sine and inverse cosine functions have the same domain, they behave differently. The inverse sine function increases on its domain (from left to right), whereas the inverse cosine function decreases on its domain (from left to right). If the cosine of an angle is known, what is the measure of that angle? The inverse cosine function determines that angle measure. Another notation for the inverse cosine function is arccos x. ( , ) x π 3π 4 π 2 π 4 (π, –1) (0, 1) 3π 4 2 – √2 –1 1 y π 4 2 ( , ) π 2 ( , 0) √2 0.5 –1.0 –0.5 x y y = cos–1 x or y = arccos x π π 2 3π 4 π 2 π 4 (1, 0) (–1, π) 3π 4 2 – √2 π 4 2 √2 INVERSE COSINE FUNCTION y 5 cos21 x or y 5 arccos x means x 5 cos y where 21 # x # 1 and 0 # y # p g “y is the inverse cosine of x” g “y is the angle measure whose cosine equals x” 7.7.2 SKILL Find exact values of an inverse cosine function. 7.7.2 C ON CEPTUAL Understand that the domain of the cosine function is restricted to 30, p4 in order for the inverse cosine function to exist. STUDY TIP The inverse cosine function gives an angle on the top half of the unit circle (QI and QII). x y (0, 1) (0, –1) (–1, 0) (1, 0) STUDY TIP cos21 x 2 1 cos x We now state the properties relating the cosine function and the inverse cosine function which follow directly from the properties of inverses. EXAMPLE 3 Finding Exact Values of an Inverse Cosine Function Find the exact value of each of the following expressions: a. cos21 a2 !2 2 b b. arccos 0 Solution (a): Let u 5 cos21 a2 !2 2 b. cos u 5 2 !2 2 when 0 # u # p Which value of u, in the range 0 # u # p, corresponds to a cosine value of 2 !2 2 ? n The range 0 # u # p corresponds to quadrants I and II. n The cosine function is negative in quadrant II. n We look for a value of u in quadrant II that has a cosine value of 2 !2 2 . u 5 3p 4 cosa3p 4 b 5 2 !2 2 and 3p 4 is in the interval 30, p4. cos21 a2 !2 2 b 5 3p 4 Calculator Confirmation: Since 3p 4 5 135°, if our calculator is set in degree mode, we should find that cos21a2 !2 2 b is equal to 135°. Solution (b): Let u 5 arccos 0. cos u 5 0 when 0 # u # p Which value of u, in the range 0 # u # p, corresponds to a cosine value of 0? u 5 p 2 cosap 2 b 5 0 and p 2 is in the interval 30, p4. arccos 0 5 p 2 Calculator Confirmation: Since p 2 5 90°, if our calculator is set in degree mode, we should find that cos21 0 is equal to 90°. Y OUR T UR N Find the exact value of each of the following expressions: a. cos21 a !2 2 b b. arccos 1 ▼ COSINE-INVERSE COSINE IDENTITIES cos211cos x2 5 x for 0 # x # p cos1cos21 x2 5 x for 21 # x # 1 ▼ A N S W E R a. p 4 b. 0 7.7 Inverse Trigonometric Functions 721 722 CHAPTER 7 Analytic Trigonometry As was the case with inverse identities for the sine function, you must be careful not to overlook the domain restrictions governing when each of these identities hold. EXAMPLE 4 Using Inverse Identities to Evaluate Expressions Involving Inverse Cosine Functions Find the exact value of each of the following trigonometric expressions: a. cos ccos21a2 1 2b d b. cos21ccosa7p 4 b d Solution (a): Write the appropriate identity. cos1cos21 x2 5 x for 21 # x # 1 Let x 5 2 1 2, which is in the interval 321, 14. Since the domain restriction is met, the identity can be used. cos ccos21a2 1 2b d 5 21 2 Solution (b): Write the appropriate identity. cos211cos x2 5 x for 0 # x # p Let x 5 7p 4 , which is not in the interval 30, p4. Since the domain restriction is not met, the identity cannot be used. x y 4 7π 4 π 2 √2 2 √2 ( , – ) 2 √2 2 √2 ( , ) –1 –1 1 1 Instead, we find another angle in the interval that has the same cosine value. cosa7p 4 b 5 cosap 4 b Substitute cosa7p 4 b 5 cosap 4 b cos21ccos a7p 4 b d 5 cos21ccosap 4 b d into the expression. Since p 4 is in the interval 30, p4, 5 p 4 we can use the identity. Y OUR TU R N Find the exact value of each of the following trigonometric expressions: a. cos ccos21 a1 2b d b. cos21ccosa2p 6 b d ▼ [CONCEPT CHECK] Evaluate (if possible) cos21(cos(A)) when (A) A 5 p 2 (B) A 5 2np 2 p 4 , where n is an integer. ANSWER (A) p 2 (B) p 4 ▼ ▼ A N S W E R a. 1 2 b. p 6 7.7.3 Inverse Tangent Function The tangent function is not a one-to-one function (it fails the horizontal line test). Let us start with the tangent function with a restricted domain: y 5 tan x Domain: a2p 2 , p 2 b Range: 12,2 x y 2p 2 22p 4 21 0 0 p 4 1 p 2 By the properties of inverse functions, the inverse tangent function will have a domain of 12q, q2 and a range of a2p 2 , p 2 b. To find the inverse tangent function, interchange x and y values. y 5 tan21 x Domain: 12,2 Range: a2p 2 , p 2 b x y 22p 2 21 2p 4 0 0 1 p 4 p 2 Notice that the inverse tangent function, like the tangent function, is an odd function (it is symmetric about the origin). The inverse tangent function allows us to answer the question: If the tangent of an angle is known, what is the measure of that angle? Another notation for the inverse tangent function is arctan x. INVERSE TANGENT FUNCTION y 5 tan21 x or y 5 arctan x means x 5 tan y where 2 p 2 , y , p 2 g “y is the inverse tangent of x” g “y is the angle measure whose tangent equals x” y = tan x π 2 π 4 –π 2 –π 4 x (0, 0) π 4 (– , –1) π 4 ( , 1) y –5 –1 –4 1 2 3 4 5 3 2 –3 –2 –1 x y y = tan–1 x or y = arctan x π 2 π 4 π 4 (0, 0) (1, ) – π 2 – π 4 –π 4 (–1, ) STUDY TIP The inverse tangent function gives an angle on the right half of the unit circle (QI and QIV). x y (0, 1) (0, –1) (–1, 0) (1, 0) 7.7.3 S K IL L Find exact values of an inverse tangent function. 7.7.3 C O N C E P T U A L Understand that the domain of the tangent function is restricted to 2p 2, p 2 in order for the inverse tangent function to exist. 7.7 Inverse Trigonometric Functions 723 724 CHAPTER 7 Analytic Trigonometry We now state the properties relating the tangent function and the inverse tangent function, which follow directly from the properties of inverses. EXAMPLE 5 Finding Exact Values of an Inverse Tangent Function Find the exact value of each of the following expressions: a. tan21A !3B b. arctan 0 Solution (a): Let u 5 tan21A !3B. tan u 5 !3 when 2 p 2 , u , p 2 Which value of u, in the range 2p 2 , u , p 2 , corresponds to a tangent value of !3? u 5 p 3 tan ap 3 b 5 !3 and p 3 is in the interval a 2 p 2 , p 2 b. tan21A !3B 5 p 3 Calculator Confirmation: Since p 3 5 60°, if our calculator is set in degree mode, we should find that tan21A !3B is equal to 60°. Solution (b): Let u 5 arctan 0. tan u 5 0 when 2 p 2 , u , p 2 Which value of u, in the range 2 p 2 , u , p 2 , corresponds to a tangent value of 0? u 5 0 tan 0 5 0, and 0 is in the interval a2p 2 , p 2 b. arctan 0 5 0 Calculator Confirmation: tan21 0 is equal to 0. TANGENT-INVERSE TANGENT IDENTITIES tan211tan x2 5 x for 2p 2 , x , p 2 tan1tan21 x2 5 x for 2q , x , q EXAMPLE 6 Using Inverse Identities to Evaluate Expressions Involving Inverse Tangent Functions Find the exact value of each of the following trigonometric expressions: a. tan1tan21 172 b. tan21ctan a2 p 3 b d Solution (a): Write the appropriate identity. tan1tan21 x2 5 x for 2q , x , q Let x 5 17, which is in the interval 12q, q2. Since the domain restriction is met, the identity can be used. tan1tan21 172 5 17 7.7.4 Remaining Inverse Trigonometric Functions The remaining three inverse trigonometric functions are defined similarly to the previous ones. ■ ■Inverse cotangent function: cot21 x or arccot x ■ ■Inverse secant function: sec21 x or arcsec x ■ ■Inverse cosecant function: csc21 x or arccsc x A table summarizing all six of the inverse trigonometric functions is given below: Solution (b): Write the appropriate identity. tan211tan x2 5 x for 2 p 2 , x , p 2 Let x 5 2p 3 , which is not in the interval a2 p 2 , p 2 b. Since the domain restriction is not met, the identity cannot be used. Instead, we find another angle in the interval that has the same tangent value. tana2p 3 b 5 tana2p 3 b Substitute tana 2p 3 b 5 tana2 p 3 b tan21ctana2p 3 b d 5 tan21ctana2 p 3 b d into the expression. Since 2p 3 is in the interval a 2 p 2 , p 2 b, tan21ctana2p 3 b d 5 2p 3 tan21ctana2p 3 b d 5 2 p 3 Y OUR T UR N Find the exact value of tan21ctana 7p 6 b d. ▼ [CONCEPT CHECK] Evaluate arctanatana2np 1 5p 4 bb, where n is an integer. ANSWER p 4 ▼ ▼ A N S W E R p 6 INVERSE FUNCTION y 5 sin21 x y 5 cos21 x y 5 tan21 x y 5 cot21 x y 5 sec21 x y 5 csc21 x DOMAIN 321, 14 321, 14 12q, q2 12q, q2 12q, 214 ∪ 31, q2 12q, 214 ∪ 31, q2 RANGE c2p 2, p 2 d 30, p4 a2p 2, p 2 b 10, p2 c0, p 2 b ∪ ap 2, pd c2 p 2, 0b ∪ a0, p 2 d GRAPH –1 1 x y π 2 π – 2 –1 1 x y π –1 1 x y π 2 π – 2 –1 1 y π x –2 –1 1 2 x y π π 2 y π 2 π – 2 x –2 –1 1 2 7.7.4 S K IL L Find exact values of the cotangent, cosecant, and secant inverse functions. 7.7.4 C O N C E P T U A L Understand that the cotangent, cosecant, and secant inverse functions are not found from the reciprocal of the tangent, sine, and cosine functions, respectively, but rather from the inverse secant, inverse cosecant, and inverse cotangent identities. 7.7 Inverse Trigonometric Functions 725 726 CHAPTER 7 Analytic Trigonometry EXAMPLE 7 Finding the Exact Value of Inverse Trigonometric Functions Find the exact value of the following expressions: a. cot211 !32 b. csc21A !2B c. sec21A2!2B Solution (a): Let u 5 cot21A !3B. cot u 5 !3 when 0 , u , p Which value of u, in the range 0 , u , p, corresponds to a cotangent value of !3? u 5 p 6 cot ap 6 b 5 !3 and p 6 is in the interval a2p 2 , p 2 b. cot21A !3B 5 p 6 Solution (b): Let u 5 csc21A !2B. csc u 5 !2 Which value of u, in the range c2p 2 , 0b ∪ a0, p 2 d, corresponds to a cosecant value of !2? u 5 p 4 csc ap 4 b 5 !2 and p 4 is in the interval c2p 2 , 0b ∪ a0, p 2 d. csc21A !2B 5 p 4 Solution (c): Let u 5 sec21A2!2B. sec u 5 2!2 Which value of u, in the range c0, p 2 b ∪ ap 2 , pd, corresponds u 5 3p 4 to a secant value of 2!2? seca3p 4 b 5 2!2 and 3p 4 is in the interval c0, p 2 b ∪ ap 2 , pd. sec21A2!2B 5 3p 4 How do we approximate the inverse secant, inverse cosecant, and inverse cotan-gent functions with a calculator? Scientific calculators have keys 1sin21, cos21, and tan212 for three of the inverse trigonometric functions but not for the other three. Recall that we find the cosecant, secant, and cotangent function values by taking sine, cosine, or tangent, and finding the reciprocal. csc x 5 1 sin x sec x 5 1 cos x cot x 5 1 tan x However, the reciprocal approach cannot be used for inverse functions. The three inverse trigonometric functions csc21 x, sec21 x, and cot21 x cannot be found by finding the reciprocal of sin21 x, cos21 x, or tan21 x. csc21 x 2 1 sin21 x sec21 x 2 1 cos21 x cot21 x 2 1 tan21 x [CONCEPT CHECK] Find the exact value (if possible) of csc21(A) for (A) A 5 0 and (B) A 5 2. ANSWER (A) DNE (B) p 6 ▼ Start with the inverse secant function. y 5 sec21 x for x # 21 or x $ 1 Write the equivalent secant expression. sec y 5 x for 0 # y , p 2 or p 2 , y # p Apply the reciprocal identity: sec y 5 1 cos y. 1 cos y 5 x Simplify using algebraic techniques. cos y 5 1 x Write the result in terms of the inverse cosine function. y 5 cos21a1 xb Therefore, we have the relationship: sec21 x 5 cos21a1 xb for x # 21 or x $ 1 The other relationships will be found in the exercises and are summarized below: Instead, we seek the equivalent sin21 x, cos21 x, or tan21 x values by algebraic means, always remembering to look within the correct domain and range. INVERSE SECANT, INVERSE COSECANT, AND INVERSE COTANGENT IDENTITIES sec21 x 5 cos21 a1 xb for x # 21 or x $ 1 csc21 x 5 sin21 a1 xb for x # 21 or x $ 1 cot21 x 5 µ tan21 a1 xb for x . 0 p 1 tan21 a1 xb for x , 0 EXAMPLE 8 Using Inverse Identities a. Find the exact value of sec21 2. b. Use a calculator to find the value of cot21 7. Solution (a): Let u 5 sec21 2. sec u 5 2 on c0, p 2 b∪ap 2 , pd Substitute the reciprocal identity. 1 cos u 5 2 Solve for cos u. cos u 5 1 2 The restricted interval c0, p 2 b∪ap 2 , pd corresponds to quadrants I and II. The cosine function is positive in quadrant I. u 5 p 3 cosap 3 b 5 1 2 sec21 2 5 cos21a1 2b 5 p 3 STUDY TIP sec21 x 2 1 cos21 x csc21 x 2 1 sin21 x cot21 x 2 1 tan21 x 7.7 Inverse Trigonometric Functions 727 728 CHAPTER 7 Analytic Trigonometry 7.7.5 Finding Exact Values for Expressions Involving Inverse Trigonometric Functions We will now find exact values of trigonometric expressions that involve inverse trigonometric functions. Solution (b): Since we do not know an exact value that would correspond to the cotangent function equal to 7, we proceed using identities and a calculator. Select the correct identity, given that x 5 7 . 0. cot21 x 5 tan21 a1 xb Let x 5 7. cot21 7 5 tan21a1 7b Evaluate the right side with a calculator. cot21 7 < 8.13° EXAMPLE 9 Finding Exact Values of Trigonometric Expressions Involving Inverse Trigonometric Functions Find the exact value of cosCsin21 A2 3BD. Solution: STEP 1 Let u 5 sin21 A2 3B. sin u 5 2 3 when 2 p 2 # u # p 2 The range 2 p 2 # u # p 2 corresponds to quadrants I and IV. The sine function is positive in quadrant I. STEP 2 Draw angle u in quadrant I. θ 3 2 x a y (a, 2) Label the sides known from the sine value. sin u 5 2 3 5 opposite hypotenuse STEP 3 Find the unknown side length a. a2 1 22 5 32 Solve for a. a 5 6!5 Since u is in quadrant I, a is positive. a 5 !5 7.7.5 SKI LL Use identities to find exact values of trigonometric expressions involving inverse trigonometric functions. 7.7.5 C ON CEPTUAL Visualize the quadrants to find exact values of trigonometric expressions involving inverse trigonometric functions. STEP 4 Find cosCsin21A2 3BD. Substitute u 5 sin21A2 3B. cos csin21a2 3b d 5 cos u Find cos u. θ 3 2 x y a = √5 (√5, 2) cos u 5 adjacent hypotenuse 5 !5 3 cos csin21a2 3b d 5 !5 3 YOUR T UR N Find the exact value of sinCcos21A1 3BD. ▼ ▼ A N S W E R 2!2 3 EXAMPLE 10 Finding Exact Values of Trigonometric Expressions Involving Inverse Trigonometric Functions Find the exact value of tan Ccos21A2 7 12BD. Solution: STEP 1 Let u 5 cos21A2 7 12B. cos u 5 2 7 12 when 0 # u # p The range 0 # u # p corresponds to quadrants I and II. The cosine function is negative in quadrant II. STEP 2 Draw angle u in quadrant II. π–θ θ –7 12 x y b (–7, b) Label the sides known from the cos u 5 2 7 12 5 adjacent hypotenuse cosine value. STEP 3 Find the length of the unknown side b. b2 1 12722 5 122 Solve for b. b 5 6!95 Since u is in quadrant II, b is positive. b 5 !95 [CONCEPT CHECK] TRUE OR FALSE sin(arctan(A)) , 0 if A , 0. ANSWER True ▼ 7.7 Inverse Trigonometric Functions 729 730 CHAPTER 7 Analytic Trigonometry STEP 4 Find tanCcos21A2 7 12BD. Substitute u 5 cos21A2 7 12B. tanccos 21a2 7 12b d 5 tan u Find tan u. θ –7 12 x y b = √95 (–7, √95) tan u 5 opposite adjacent 5 !95 27 tanccos21 a2 7 12b d 5 2 !95 7 Y OUR TU R N Find the exact value of tanCsin21A2 3 7BD. ▼ ▼ A N S W E R 23!10 20 EXAMPLE 11 Using Identities to Find Exact Values of Trigonometric Expressions Involving Inverse Trigonometric Functions Find the exact value of cos 3sin21 A3 5B 1 tan2114 . Solution: Recall the cosine sum identity: cos1A 1 B2 5 cos A cos B 2 sin A sinB Let A 5 sin21A3 5B and B 5 tan21 1. coscsin21a 3 5 b 1 tan21 1d 5 coscsin21a 3 5 b d cos1tan21 12 2 sincsin21a 3 5 b d sin1tan21 12 From the figure, 5 3 4 A x y A 5 sin21A3 5B 1 sin A 5 3 5 we see that cos csin21a 3 5 b d 5 cos A 5 4 5 sin csin21a 3 5 b d 5 sin A 5 3 5 From the figure, 2 B x y 2 2 B 5 tan211 12 1 tan B 5 1 we see that cos1tan21 12 5 cos B 5 "2 2 sin1tan21 12 5 sin B 5 "2 2 Substitute these values into the cosine sum identity: coscsin21a3 5b 1 tan211d 5 a4 5b a !2 2 b 2 a3 5b a !2 2 b Simplify. 5 !2 10 YOUR T UR N Find the exact value of sin3cos21A3 5B 1 tan2114. ▼ ▼ A N S W E R 7!2 10 ▼ A N S W E R u"u2 1 1 u2 1 1 EXAMPLE 12 Writing Trigonometric Expressions Involving Inverse Trigonometric Functions in Terms of a Single Variable Write the expression cos1tan21u2 as an equivalent expression in terms of only the variable u. Solution: Let u 5 tan21u; therefore, tan u 5 u 5 u 1. Realize that u can be positive or negative. Since the range of the inverse tangent function is a2p 2 , p 2 b, sketch the angle u in both quadrants I and IV and draw the corresponding two right triangles. Recalling that the tangent ratio is opposite over adjacent, we label those corresponding sides with u and 1, respectively. Then solving for the hypotenuse using the Pythagorean theorem gives "u2 1 1. Substitute u 5 tan21u into cos1tan21u2. cos1tan21u2 5 cos u Use the right triangle ratio for cosine: adjacent over hypotenuse. 5 1 "u2 1 1 Rationalize the denominator. cos1tan21 u2 5 1 "u2 1 1 ⋅"u2 1 1 "u2 1 1 5 "u2 1 1 u2 1 1 YOUR T UR N Write the expression sin1tan21 u2 as an equivalent expression in terms of only the variable u. ▼ 7.7 Inverse Trigonometric Functions 731 732 CHAPTER 7 Analytic Trigonometry trigonometric functions can be found when the function values are those of the special angles. Inverse trigonometric functions also provide a means for evaluating one trigonometric function when we are given the value of another. It is important to note that the 21 as a superscript indicates an inverse function, not a reciprocal. If a trigonometric function value of an angle is known, what is the measure of that angle? Inverse trigonometric functions determine the angle measure. To define the inverse trigonometric relations as functions, we first restrict the trigonometric functions to domains in which they are one-to-one functions. Exact values for inverse [SEC TION 7.7] S U M M A RY INVERSE FUNCTION y 5 sin21 x y 5 cos21 x y 5 tan21 x y 5 cot21 x y 5 sec21 x y 5 csc21 x DOMAIN 321, 14 321, 14 12q, q2 12q, q2 12q, 214∪ 31, q2 12q, 214∪31, q2 RANGE c2p 2, p 2 d 30, p4 a2p 2, p 2 b 10, p2 c0, p 2 b∪ap 2, pd c2 p 2, 0b∪a0, p 2 d GRAPH –1 1 x y π 2 π – 2 –1 1 x y π –1 1 x y π 2 π – 2 –1 1 y π x –2 –1 1 2 x y π π 2 y π 2 π – 2 x –2 –1 1 2 [SEC TION 7.7] E X E R CI SE S • S K I L L S In Exercises 1–16, find the exact value of each expression. Give the answer in radians. 1. arccosa !2 2 b 2. arccosa2 !2 2 b 3. arcsina2 !3 2 b 4. arcsina1 2b 5. cot211212 6. tan21a !3 3 b 7. arcseca2!3 3 b 8. arccsc1212 9. csc212 10. sec211222 11. arctanA2!3B 12. arccotA !3B 13. sin21 0 14. tan21 1 15. sec211212 16. cot21 0 In Exercises 17–32, find the exact value of each expression. Give the answer in degrees. 17. cos21a1 2b 18. cos21a2 !3 2 b 19. sin21a !2 2 b 20. sin21 0 21. cot21a2 !3 3 b 22. tan21A2!3B 23. arctana !3 3 b 24. arccot 1 25. arccsc1222 26. csc21a2 2!3 3 b 27. arcsecA2!2B 28. arccscA2!2B 29. sin211212 30. arctan1212 31. arccot 0 32. arcsec1212 In Exercises 33–42, use a calculator to evaluate each expression. Give the answer in degrees and round it to two decimal places. 33. cos2110.54322 34. sin2110.78212 35. tan2111.8952 36. tan2113.26782 37. sec2111.49732 38. sec2112.78642 39. csc21123.78932 40. csc21126.13242 41. cot21124.23192 42. cot21120.89772 In Exercises 43–52, use a calculator to evaluate each expression. Give the answer in radians and round it to two decimal places. 43. sin21120.58782 44. sin2110.86602 45. cos2110.14232 46. tan21120.92792 47. tan2111.32422 48. cot2112.41422 49. cot21120.57742 50. sec21121.04222 51. csc2113.23612 52. csc21122.92382 In Exercises 53–76, evaluate each expression exactly, if possible. If not possible, state why. 53. sin21csina 5p 12 b d 54. sin21csina 25p 12 b d 55. sin3sin2111.0324 56. sin3sin2111.124 57. sin21csin a27p 6 b d 58. sin21csina7p 6 b d 59. cos21ccosa 4p 3 b d 60. cos21ccosa2 5p 3 b d 61. cotCcot21A !3BD 62. cot21ccota5p 4 b d 63. sec21csec a2 p 3 b d 64. sec csec21a 1 2b d 65. csc ccsc 21a1 2b d 66. csc21ccsca7p 6 b d 67. cot 1cot 21 02 68. cot21ccota 2 p 4 b d 69. tan21ctana2 p 4 b d 70. tan21ctanap 4 b d 71. sec 1sec 21 02 72. csc21 3csc1p24 73. cot21ccota8p 3 b d 74. tan21 3tan18p24 75. csc21ccsca15p 4 b d 76. sec21cseca17p 2 b d In Exercises 77–96, evaluate each expression exactly. 77. cos csin21a3 4b d 78. sinccos21a2 3b d 79. sinctan21a12 5 b d 80. cos ctan21a 7 24b d 81. tancsin21a3 5b d 82. tanccos21a2 5b d 83. sec csin21a !2 5 b d 84. secccos21a !7 4 b d 85. csc ccos21a1 4b d 86. csc csin21a1 4b d 87. cotcsin21a60 61b d 88. cotcsec21a41 9 b d 89. cos ctan21a3 4b 2 sin21a4 5b d 90. cos ctan21a12 5 b 1 sin21a3 5b d 91. sinccos21a 5 13b 1 tan21a4 3b d 92. sinccos21a3 5b 2 tan21a 5 12b d 93. sinc2 cos21a3 5b d 94. cos c2 sin21a3 5b d 95. tanc2 sin21a 5 13b d 96. tanc2 cos21a 5 13b d For each of the following expressions, write an equivalent expression in terms of only the variable u. 97. cos1sin21 u2 98. sin1cos21 u2 99. tan1cos21 u2 100. tan1sin21 u2 7.7 Inverse Trigonometric Functions 733 734 CHAPTER 7 Analytic Trigonometry • A P P L I C A T I O N S For Exercises 101 and 102, refer to the following: Annual sales of a product are generally subject to seasonal fluctuations and are approximated by the function s1t2 5 4.3 cos ap 6 tb 1 56.2 0 # t # 11 where t represents time in months 1t 5 0 represents January) and s1t2 represents monthly sales of the product in thousands of dollars. 101. Business. Find the month(s) in which monthly sales are $56,200. 102. Business. Find the month(s) in which monthly sales are $51,900. For Exercises 103 and 104, refer to the following: Allergy sufferers’ symptoms fluctuate with pollen levels. Pollen levels are often reported to the public on a scale of 0–12, which is meant to reflect the levels of pollen in the air. For example, a pollen level between 4.9 and 7.2 indicates that pollen levels will likely cause symptoms for many individuals allergic to the pre dominant pollen of the season (Source: The pollen levels at a single location were measured and averaged for each month. Over a period of 6 months, the levels fluctuated according to the model p1t2 5 5.5 1 1.5 sin ap 6tb 0 # t # 6 where t is measured in months and p1t2 is the pollen level. 103. Biology/Health. In which month(s) was the monthly average pollen level 7.0? 104. Biology/Health. In which month(s) was the monthly average pollen level 6.25? 105. Alternating Current. Alternating electrical current in amperes (A) is modeled by the equation i 5 I sin12pƒt2, where i is the current, I is the maximum current, t is time in seconds, and ƒ is the frequency in hertz (Hz is the number of cycles per second). If the frequency is 5 Hz and maximum current is 115 A, what time t corresponds to a current of 85 A? Find the smallest positive value of t. 106. Alternating Current. If the frequency is 100 Hz and maximum current is 240 A, what time t corresponds to a current of 100 A? Find the smallest positive value of t. 107. Hours of Daylight. The number of hours of daylight in San Diego, California, can be modeled with H1t2 5 12 1 2.4 sin 10.017t 2 1.3772, where t is the day of the year (January 1, t 5 1, etc.). For what value of t is the number of hours of daylight equal to 14.4? If May 31 is the 151st day of the year, what month and day correspond to that value of t? 108. Hours of Daylight. Repeat Exercise 107. For what value of t is the number of hours of daylight equal to 9.6? What month and day correspond to the value of t? (You may have to count backwards.) 109. Money. A young couple get married and immediately start saving money. They renovate a house and are left with less and less saved money. They have children after 10 years and are in debt until their children are in college. They then save until retirement. A formula that represents the percentage of their annual income that they either save (positive) or are in debt (negative) is given by P1t2 5 12.5 cos 10.157t2 1 2.5, where t 5 0 corresponds to the year they were married. How many years into their marriage do they first accrue debt? Percentage of Annual Income Saved Time (in years) 50 40 30 20 10 –10 –5 5 10 15 110. Money. For the couple in Exercise 109, how many years into their marriage are they back to saving 15% of their annual income? 111. Viewing Angle of Painting. A museum patron whose eye level is 5 feet above the floor is studying a painting that is 8 feet in height and mounted on the wall 4 feet above the floor. If the patron is x feet from the wall, use tan1a 1 b2 to express tan1u2, where u is the angle that the patron’s eye sweeps from the top to the bottom of the painting. 5 ft x 4 ft 8 ft 112. Viewing Angle of Painting. Using the equation for tan1u2 in Exercise 111, solve for u using the inverse tangent. Then find the measure of the angles u for x 5 10 and x 5 20 (to the nearest degree). In Exercises 117–120, explain the mistake that is made. • C A T C H T H E M I S T A K E 117. Evaluate the expression exactly: sin21 csin a3p 5 b d . Solution: Use the identity sin211sin x2 5 x on 0 # x # p. Since 3p 5 is in the interval 30, p4, the identity can be used. sin21 csina 3p 5 b d 5 3p 5 This is incorrect. What mistake was made? 118. Evaluate the expression exactly: cos21ccosa2 p 5 b d . Solution: Use the identity cos 211cos x2 5 x on 2p 2 # x # p 2. Since 2p 5 is in the interval c2p 2, p 2 d , the identity can be used. cos21 ccos a 2p 5 b d 5 2p 5 This is incorrect. What mistake was made? 119. Evaluate the expression exactly: cot 2112.52. Solution: Use the reciprocal identity. cot2112.52 5 1 tan 2112.52 Evaluate tan2112.52 5 1.19. cot2112.52 5 1 1.19 Simplify. cot 2112.52 5 0.8403 This is incorrect. What mistake was made? 120. Evaluate the expression exactly: csc21a1 4b. Solution: Use the reciprocal identity. csc21a1 4b 5 1 sin 21a1 4b Evaluate sin21A1 4B 5 14.478. csc21a1 4b 5 1 14.478 Simplify. csc21a1 4b 5 0.0691 This is incorrect. What mistake was made? 113. Earthquake Movement. The horizontal movement of a point that is k kilometers away from an earthquake’s fault line can be estimated with M 5 ƒ 2 C1 2 2 tan21 ak db p S where M is the movement of the point in meters, ƒ is the total horizontal displacement occurring along the fault line, k is the distance of the point from the fault line, and d is the depth in kilometers of the focal point of the earthquake. If an earthquake produces a displacement ƒ of 2 meters and the depth of the focal point is 4 kilometers, then what is the movement M of a point that is 2 kilometers from the fault line? of a point 10 kilometers from the fault line? Focal Point Point Fault Line d k 114. Earthquake Movement. Repeat Exercise 113. If an earthquake produces a displacement ƒ of 3 meters and the depth of the focal point is 2.5 kilometers, then what is the movement M of a point that is 5 kilometers from the fault line? of a point 10 kilometers from the fault line? 115. Laser Communication. A laser communication system depends on a narrow beam, and a direct line of sight is necessary for communication links. If a transmitter/receiver for a laser system is placed between two buildings (see the figure) and the other end of the system is located on a low-earth-orbit satellite, then the link is operational only when the satellite and the ground system have a line of sight (when the buildings are not in the way). Find the angle u that corresponds to the system being operational (i.e., find the maximum value of u that permits the system to be operational). Express u in terms of inverse tangent functions and the distance from the shorter building. BJI/Blue Jean Images Getty Images, Inc. 150 ft 300 ft 200 ft 116. Laser Communication. Repeat Exercise 115, assuming that the ground system is on top of a 20-foot tower. 7.7 Inverse Trigonometric Functions 735 736 CHAPTER 7 Analytic Trigonometry • T E C H N O L O G Y 135. Use a graphing calculator to plot Y1 5 sin1sin21 x2 and Y2 5 x for the domain 21 # x # 1. If you then increase the domain to 23 # x # 3, you get a different result. Explain the result. 136. Use a graphing calculator to plot Y1 5 cos1cos21 x2 and Y2 5 x for the domain 21 # x # 1. If you then increase the domain to 23 # x # 3, you get a different result. Explain the result. 137. Use a graphing calculator to plot Y1 5 csc211csc x2 and Y2 5 x. Determine the domain for which the following statement is true: csc211csc x2 5 x . Give the domain in terms of p. 138. Use a graphing calculator to plot Y1 5 sec211sec x2 and Y2 5 x. Determine the domain for which the following statement is true: sec211sec x2 5 x . Give the domain in terms of p. 139. Given tan x 5 40 9 and p , x , 3p 2 : a. Find sin 12x2 using the double-angle identity. b. Use the inverse of tangent to find x in QIII and use a calculator to find sin 12x2. Round to five decimal places. c. Are the results in 1a2 and 1b2 the same? 140. Given sin x 5 2 1 !10 and 3p 2 , x , 2p: a. Find tan 12x2 using the double-angle identity. b. Use the inverse of sine to find x in QIV and find tan 12x2. c. Are the results in 1a2 and 1b2 the same? • C O N C E P T U A L 121. The inverse secant function is an even function. 122. The inverse cosecant function is an odd function. 123. csc211csc u2 5 u, for all u in the domain of cosecant. 124. sin2112x2 ⋅csc2112x2 5 1, for all x for which both functions are defined. 125. Explain why sec21A1 2B does not exist. 126. Explain why csc21A1 2B does not exist. In Exercises 121–124, determine whether each statement is true or false. • C H A L L E N G E 127. Find the expression that corresponds to sinccos21a1 xb d . 128. Determine the x-values for which sin21 c2 sina3x 2 bcosa3x 2 b d 5 3x 129. Evaluate exactly: sin12 sin2112. 130. Let ƒ1x2 5 2 2 4 sinax 2 p 2 b. a. State an accepted domain of ƒ1x2 so that ƒ1x2 is a one-to-one function. b. Find ƒ211x2 and state its domain. 131. Let ƒ1x2 5 3 1 cosax 2 p 4 b. a. State an accepted domain of ƒ1x2 so that ƒ1x2 is a one-to-one function. b. Find ƒ211x2 and state its domain. 132. Let ƒ1x2 5 1 2 tanax 1 p 3 b. a. State an accepted domain of ƒ1x2 so that ƒ1x2 is a one-to-one function. b. Find ƒ211x2 and state its domain. 133. Let ƒ1x2 5 2 1 1 4 cot a2x 2 p 6 b. a. State an accepted domain of ƒ1x2 so that ƒ1x2 is a one-to-one function. b. Find ƒ211x2 and state its domain. 134. Let ƒ1x2 5 2cscap 4x 2 1b. a. State an accepted domain of ƒ1x2 so that ƒ1x2 is a one-to-one function. b. Find ƒ211x2 and state its domain. Recall that in solving algebraic equations, the goal is to find the value for the variable that makes the equation true. For example, the linear equation 2x 2 5 5 7 has only one value, x 5 6, which makes the statement true. A quadratic equation, however, can have two solutions. The equation x2 5 9 has two values, x 5 6 3, which make the statement true. With trigonometric equations, the goal is the same: find the value (or values) that makes (or make) the equation true. 7.8.1 Solving Trigonometric Equations by Inspection To solve the equation sin x 1 cos x 5 1 0 # x , 2p we can draw the unit circle and ask the question: “Which points on the unit circle have x- and y-coordinates that sum to 1?” Inspecting the figure to the left, we see that x 5 0 and x 5 p 2 correspond to x- and y-coordinates along the unit circle that sum to 1. The solution to sin x 1 cos x 5 1 on the x interval 30, 2p4 is x 5 0, p 2 . This method of solving a trigonometric equation is called inspection. S K I L L S O B J E C T I V E S ■ ■Solve trigonometric equations by inspection. ■ ■Solve trigonometric equations by using algebraic techniques. ■ ■Solve trigonometric equations using inverse functions. ■ ■Solve trigonometric equations (involving more than one trigonometric function) using trigonometric identities. C O N C E P T U A L O B J E C T I V ES ■ ■When solving trigonometric equations by inspection, remember to find all solutions, not just the ones in an angle period. ■ ■Utilize the concept of substitution in solving trigonometric equations when the results are simple linear or quadratic equations. ■ ■Utilize a calculator to approximate solutions when solving trigonometric equations using inverse functions. ■ ■Using trigonometric identities can often help convert a trigonometric equation involving more than one trigonometric function into a trigonometric equation involving a single trigonometric function. 7.8 TRIGONOMETRIC EQUATIONS x y 60º 3 45º 4 30º 6 360º 2 0º 0 0 330º 6 11 315º 4 7 300º 3 5 270º 2 3 240º 3 4 225º 4 5 210º 6 7 180º 150º 6 5 135º 4 3 90º 2 120º 3 2 (0, 1) (0, –1) (1, 0) (–1, 0) 2 √3 2 1 2 1 ( , – ) 2 √3 2 1 2 1 ( , ) 2 √2 2 √2 ( , – ) 2 √2 2 √2 ( , ) 2 √3 2 1 ( , – ) 2 √3 (– , – ) 2 √2 2 √2 (– , – ) 2 √3 2 1 (– , – ) 2 √3 (– , ) 2 √2 2 √2 (– , ) 2 √3 2 1 (– , ) 2 √3 2 1 ( , ) 7.8.1 S K IL L Solve trigonometric equations by inspection. 7.8.1 C O N C E P T U A L When solving trigonometric equations by inspection, remember to find all solutions, not just the ones in an angle period. 7.8 Trigonometric Equations 737 738 CHAPTER 7 Analytic Trigonometry EXAMPLE 1 Solving a Trigonometric Equation by Inspection Solve each of the following equations over 30, 2p2: a. sin x 5 1 2 b. cos12x2 5 1 2 Solution (a): Ask the question, “sine of what angles is 1 2?’’ x 5 p 6 or x 5 5p 6 Solution (b): Ask the question, “cosine of what angles is 1 2?’’ In this case, the angle is equal to 2x. 2x 5 p 3 or 2x 5 5p 3 Solve for x: x 5 p 6 or x 5 5p 6 . Note: x 5 7p 6 and x 5 11p 6 are in the interval 30, 2p2 and also satisfy the equation cos12x2 5 1 2. x 5 p 6 , 5p 6 , 7p 6 , and 11p 6 Y OUR TU R N Solve each of the following equations over 30, 2p2: a. cos x 5 1 2 b. sin12x2 5 1 2 x y 60º 3 π 45º 4 π 30º 6 π 360º 2π 0º 0 0 330º 6 11π 315º 4 7π 300º 3 5π 270º 2 3π 240º 3 4π 225º 4 5π 210º 6 7π π 180º 150º 6 5π 135º 4 3π 90º 2 π 120º 3 2π (0, 1) (0, –1) (1, 0) (–1, 0) 2 √3 2 1 2 1 ( , – ) 2 √3 2 1 2 1 ( , ) 2 √2 2 √2 ( , – ) 2 √2 2 √2 ( , ) 2 √3 2 1 ( , – ) 2 √3 (– , – ) 2 √2 2 √2 (– , – ) 2 √3 2 1 (– , – ) 2 √3 (– , ) 2 √2 2 √2 (– , ) 2 √3 2 1 (– , ) 2 √3 2 1 ( , ) x y 60º 3 π 45º 4 π 30º 6 π 360º 2π 0º 0 0 330º 6 11π 315º 4 7π 300º 3 5π 270º 2 3π 240º 3 4π 225º 4 5π 210º 6 7π π 180º 150º 6 5π 135º 4 3π 90º 2 π 120º 3 2π (0, 1) (0, –1) (1, 0) (–1, 0) 2 √3 2 1 2 1 ( , – ) 2 √3 2 1 2 1 ( , ) 2 √2 2 √2 ( , – ) 2 √2 2 √2 ( , ) 2 √3 2 1 ( , – ) 2 √3 (– , – ) 2 √2 2 √2 (– , – ) 2 √3 2 1 (– , – ) 2 √3 (– , ) 2 √2 2 √2 (– , ) 2 √3 2 1 (– , ) 2 √3 2 1 ( , ) ▼ ▼ A N S W E R a. x 5 p 3 , 5p 3 b. x 5 p 12, 5p 12 , 13p 12 , 17p 12 EXAMPLE 2 Solving a Trigonometric Equation by Inspection Solve the equation sin x 5 !2 2 . x y 4 3π 4 π 2 √2 2 √2 (– , ) 2 √2 2 √2 ( , ) –1 –1 1 1 Solution: STEP 1 Solve over one period, 30, 2p2. Ask the question, “sine of what angles is !2 2 ?” The sine function is positive in quadrants I and II. STEP 2 Solve over all real numbers. Since the sine function has a period of 360° or 2p, adding integer multiples of 360° or 2p will give the other infinitely many solutions. DEGREES x 5 45° 1 360° n or x 5 135° 1 360° n RADIANS x 5 p 4 1 2np or x 5 3p 4 1 2np, where n is any integer Y OUR T UR N Solve the equation cos x 5 1 2. DEGREES x 5 45° or x 5 135° RADIANS x 5 p 4 or x 5 3p 4 ▼ ▼ A N S W E R DEGREES x 5 60° 1 360° n or x 5 300° 1 360° n RADIANS x 5 p 3 1 2np or x 5 5p 3 1 2np, where n is any integer STUDY TIP Find all solutions unless the domain is restricted. EXAMPLE 3 Solving a Trigonometric Equation by Inspection Solve the equation tan12x2 5 2!3. Solution: STEP 1 Solve over one period, 30, p2. Ask the question, “tangent of what angles is 2!3?” Note that the angle in this case is 2x. The tangent function is negative in quadrants II and IV. Since 30, p2 includes quadrants I and II, we find only the angle in quadrant II. (The solution corresponding to quadrant IV will be found when we extend the solution over all real numbers.) DEGREES 2x 5 120° RADIANS 2x 5 2p 3 [CONCEPT CHECK] Solve sin1x2 5 0 ANSWER x 5 np, where n is an integer ▼ Notice that the equations in Example 2 and Your Turn have an infinite number of solutions. Unless the domain is restricted, you must find all solutions. 7.8 Trigonometric Equations 739 740 CHAPTER 7 Analytic Trigonometry STEP 2 Solve over all x. Since the tangent function has a period of 180°, or p, adding integer multiples of 180° or p will give all of the other solutions. Solve for x by dividing by 2. Note: n There are infinitely many solutions. If we graph y 5 tan12x2 and y 5 2!3, we see that there are infinitely many points of intersection. n Had we restricted the domain to 0 # x , 2p, the solutions (in radians) would be the values given to the right in the table. Notice that only n 5 0, 1, 2, 3 yield x-values in the domain 0 # x , 2p. DEGREES 2x 5 120° 1 180° n RADIANS 2x 5 2p 3 1 np, where n is any integer DEGREES x 5 60° 1 90° n RADIANS x 5 p 3 1 np 2 , where n is any integer π –π –2 –1 1 2 x y π 2 –π 2 y = tan(2x) y = –√3 n x 5 p 3 1 np 2 0 x 5 p 3 1 x 5 5p 6 2 x 5 4p 3 3 x 5 11p 6 Notice that in Step 2 of Example 2, 2np was added to get all of the solutions, whereas in Step 2 of Example 3, we added np to the argument of the tangent function. We added 2np in Example 2 and np in Example 3 because the sine function has period 2p, whereas the tangent function has period p. 7.8.2 Solving Trigonometric Equations Using Algebraic Techniques We now will use algebraic techniques to solve trigonometric equations. Let us first start with linear and quadratic equations. For linear equations, we solve for the variable by isolating it. For quadratic equations, we often employ factoring or the quadratic formula. If we can let x represent the trigonometric function and the resulting equation is either linear or quadratic, then we use techniques learned in solving algebraic equations. EXAMPLE 4 Solving a Linear Trigonometric Equation Solve 4 sin u 2 2 5 24 on 0 # u , 2p. Solution: STEP 1 Solve for sin u. 4 sin u 2 2 5 24 Add 2. 4 sin u 5 22 Divide by 4. sin u 5 21 2 STEP 2 Find the values of u on 0 # u , 2p which satisfy the equation sin u 5 21 2. The sine function is negative in quadrants III and IV: sina7p 6 b 5 21 2 and sina11p 6 b 5 21 2. u 5 7p 6 or u 5 11p 6 YOUR T UR N Solve 2 cos u 1 1 5 2 on 0 # u , 2p. ▼ ▼ A N S W E R u 5 p 3 or 5p 3 EXAMPLE 5 Solving a Quadratic Trigonometric Equation Solve 2 cos2 u 1 cos u 2 1 5 0 on 0 # u , 2p. Solution: STEP 1 Solve for cos u. 2 cos2 u 1 cos u 2 1 5 0 Let x 5 cos u. 2 x2 1 x 2 1 5 0 Factor the quadratic equation. 12 x 2 121x 1 12 5 0 Set each factor equal to 0. 2x 2 1 5 0 or x 1 1 5 0 [CONCEPT CHECK] Solve cos21x2 5 1 ANSWER x 5 np, where n is an integer ▼ TYPE EQUATION SUBSTITUTION ALGEBRAIC EQUATION Linear trigonometric equation 4 sin u 2 2 5 24 x 5 sin u 4x 2 2 5 24 Quadratic trigonometric equation 2 cos2 u 1 cos u 2 1 5 0 x 5 cos u 2x2 1 x 2 1 5 0 It is not necessary to make the substitution, though it is convenient. Frequently, one can see how to factor a quadratic trigonometric equation without first converting it to an algebraic equation. In Example 4, we will not use the substitution. However, in Example 5, we will illustrate the use of a substitution. 7.8 Trigonometric Equations 741 7.8.2 S K IL L Solve trigonometric equations by using algebraic techniques. 7.8.2 C O N C E P T U A L Utilize the concept of substitution in solving trigonometric equations when the results are simple linear or quadratic equations. 742 CHAPTER 7 Analytic Trigonometry 7.8.3 Solving Trigonometric Equations That Require the Use of Inverse Functions Thus far, we have been able to solve the trigonometric equations exactly. Now we turn our attention to situations that require using a calculator and inverse functions to approximate a solution to a trigonometric equation. 7.8.3 SKI LL Solve trigonometric equations using inverse functions. 7.8.3 C ON CEPTUAL Utilize a calculator to approximate solutions when solving trigonometric equations using inverse functions. EXAMPLE 6 Solving a Trigonometric Equation That Requires the Use of Inverse Functions Solve tan2 u 2 tan u 5 6 on 0° # u , 180°. Solution: STEP 1 Solve for tan u. Subtract 6. tan2 u 2 tan u 2 6 5 0 Factor the quadratic trigonometric expression on the left. 1tan u 2 32 1tan u 1 22 5 0 Set the factors equal to 0. tan u 2 3 5 0 or tan u 1 2 5 0 Solve for tan u. tan u 5 3 or tan u 5 22 STEP 2 Solve tan u 5 3 on 0° # u , 180°. The tangent function is positive on 0° # u , 180° only in quadrant I. Write the equivalent inverse notation to tan u 5 3. u 5 tan21 3 Use a calculator to evaluate (approximate) u. u < 71.6° Solve each equation for x. x 5 1 2 or x 5 21 Substitute back to u: x 5 cos u. cos u 5 1 2 or cos u 5 21 STEP 2 Find the values of u on 0 # u , 2p that satisfy the equation cos u 5 1 2. The cosine function is positive in quadrants I and IV. cosap 3 b 5 1 2 and cosa5p 3 b 5 1 2. u 5 p 3 or u 5 5p 3 STEP 3 Find the values of u on 0 # u , 2p that satisfy the equation cos u 5 21. cos p 5 21 u 5 p The solutions to 2 cos2 u 1 cos u 2 1 5 0 on 0 # u , 2p are u 5 p 3 , u 5 5p 3 , or u 5 p. Y OUR TU R N Solve 2 sin2 u 2 sin u 2 1 5 0 on 0 # u , 2p. STUDY TIP If Example 5 asked for the solution to the trigonometric equation over all real numbers, then the solutions would be u 5 p 3 1 2np, 5p 3 1 2np, p 1 2np. ▼ A N S W E R u 5 p 2 , 7p 6 , or 11p 6 ▼ EXAMPLE 7 Solving a Quadratic Trigonometric Equation That Requires the Use of the Quadratic Formula and Inverse Functions Solve 2 cos2 u 1 5 cos u 2 6 5 0 on 0° # u , 360°. Solution: STEP 1 Solve for cos u. 2 cos2 u 1 5 cos u 2 6 5 0 Let x 5 cos u. 2x2 1 5x 2 6 5 0 Use the quadratic formula, a 5 2, b 5 5, c 5 26. x 5 25 6 "52 2 41221262 2122 Simplify. x 5 25 6 !73 4 Use a calculator to approximate the solution. x < 23.3860 or x < 0.8860 Substitute back to u: x 5 cos u. cos u < 23.3860 or cos u < 0.8860 STEP 2 Solve cos u 5 23.3860 on 0° # u , 360°. Recall that the range of the cosine function is 321, 14; therefore, the cosine function can never equal a number outside that range 123.3860 , 212. There is no solution with this equation. STEP 3 Solve tan u 5 22 on 0° # u , 180°. The tangent function is negative on 0° # u , 180° only in quadrant II. A calculator gives values of the inverse tangent in quadrants I and IV. We will call the reference angle in quadrant IV “a.” Write the equivalent inverse notation to tan a 5 22. a 5 tan211222 Use a calculator to evaluate (approximate) a. a < 263.4° To find the value of u in quadrant II, add 180°. u 5 a 1 180° u < 116.6° The solutions to tan2 u 2 tan u 5 6 on 0° # u , 180° are u 5 71.6° or u 5 116.6°. YOUR T UR N Solve tan2 u 1 tan u 5 6 on 0° # u , 180°. Recall that in solving algebraic quadratic equations, one method (when factoring is not obvious or possible) is to use the quadratic formula. ax2 1 bx 1 c 5 0 has solutions x 5 2b 6 "b2 2 4ac 2a a 2 0 [CONCEPT CHECK] Explain why cos21x2 1 cos1x2 5 6 does not have a solution. ANSWER The range of cosine is [21,1]. If you rewrite the equation as cos21x2 1 cos1x2 2 6 5 0 and factor the left side as 1cos x 2 321cos x 1 22, then neither cos x 5 22 nor cos x 5 3 has a solution. So there are no x-values that satisfy the original equation. ▼ ▼ A N S W E R u < 63.4° or 108.4° ▼ 7.8 Trigonometric Equations 743 744 CHAPTER 7 Analytic Trigonometry EXAMPLE 8 Using Trigonometric Identities to Solve Trigonometric Equations Solve sin x 1 cos x 5 1 on 0 # x , 2p. Solution: Square both sides. sin2 x 1 2 sin x cos x 1 cos2 x 5 1 Label the Pythagorean identity. sin2 x 1 cos2 x 1 2 sin x cos x 5 1 Subtract 1 from both sides. 2 sin x cos x 5 0 Use the zero product property. sin x 5 0 or cos x 5 0 Solve for x on 0 # x , 2p. x 5 0 or x 5 p or x 5 p 2 or x 5 3p 2 Because we squared the equation, we have to check for extraneous solutions. Check x 5 0. sin 0 1 cos 0 5 0 1 1 5 1 ✓ Check x 5 p. sin p 1 cos p 5 0 2 1 5 21 ✕ g 1 g g STEP 3 Solve cos u 5 0.8860 on 0° # u , 360°. The cosine function is positive in quadrants I and IV. Since a calculator gives inverse cosine values only in quadrants I and II, we will have to use a reference angle to get the quadrant IV solution. Write the equivalent inverse notation for cos u 5 0.8860. u 5 cos21 0.8860 Use a calculator to evaluate (approximate) the solution. u < 27.6° To find the second solution (in quadrant IV), subtract the reference angle from 360°. u 5 360° 2 27.6° u < 332.4° The solutions to 2 cos2 u 1 5 cos u 2 6 5 0 on 0° # u , 360° are u < 27.6° or u < 332.4°. Y OUR TU R N Solve 2 sin2 u 2 5 sin u 2 6 5 0 on 0° # u , 360°. 7.8.4 Using Trigonometric Identities to Solve Trigonometric Equations We now consider trigonometric equations that involve more than one trigonometric function. Trigonometric identities are an important part of solving these types of equations. The trigonometric equation in Example 8 was solved by inspection in the beginning of this section. We solve it again now using trigonometric identities. ▼ A N S W E R u < 242.4° or 297.6° 7.8.4 SKI LL Solve trigonometric equations (involving more than one trigonometric function) using trigonometric identities. 7.8.4 C ON CEPTUAL Using trigonometric identities can often help convert a trigonometric equation involving more than one trigonometric function into a trigonometric equation involving a single trigonometric function. ▼ EXAMPLE 9 Using Trigonometric Identities to Solve Trigonometric Equations Solve sin12x2 5 sin x on 0 # x , 2p. ✖I N C O R R EC T 2 sin x cos x 5 sin x Divide by sin x. ERROR 2 cos x 5 1 Solve for cos x. cos x 5 1 2 Missing solutions from sin x 5 0. ✓COR R E C T Use the double-angle formula for sine. sin12x2 5 sin x 2 sin x cos x Subtract sin x. 2 sin x cos x 2 sin x 5 0 Factor the common sin x. 1sin x212 cos x 2 12 5 0 Set each factor equal to 0. sin x 5 0 or 2 cos x 2 1 5 0 sin x 5 0 or cos x 5 1 2 Solve sin x 5 0 for x on 0 # x , 2p. x 5 0 or x 5 p Solve cos x 5 1 2 for x on 0 # x , 2p. x 5 p 3 or x 5 5p 3 g common mistake Dividing by a trigonometric function (which could be equal to zero). The solutions to sin12 x2 5 sin x are x 5 0, p 3, p, and 5p 3 . YOUR T UR N Solve sin12x2 5 cos x on 0 # x , 2p. ▼ Check x 5 p 2. sinap 2 b 1 cosap 2 b 5 1 1 0 5 1 ✓ Check x 5 3p 2 . sina3p 2 b 1 cosa3p 2 b 5 21 1 0 5 21 ✕ The solutions to sin x 1 cos x 5 1 on 0 # x , 2p are x 5 0 or x 5 p 2 . YOUR T UR N Solve sin x 2 cos x 5 1 on 0 # x , 2p. ▼ [CONCEPT CHECK] TRUE OR FALSE sin(2A) = cos(A) has the same solution as sin(2A) = cos(2A). ANSWER True ▼ ▼ C A U T I O N Do not divide equations by trigonometric functions, as they can sometimes equal zero. ▼ A N S W E R x 5 p 2 , 3p 2 , p 6 , or 5p 6 ▼ A N S W E R x 5 p 2 or x 5 p 7.8 Trigonometric Equations 745 746 CHAPTER 7 Analytic Trigonometry EXAMPLE 10 Using Trigonometric Identities to Solve Trigonometric Equations Solve sin x 1 csc x 5 22. Solution: Use the reciprocal identity. sin x 1 csc x 5 22 Add 2. sin x 1 2 1 1 sin x 5 0 Multiply by sin x. (Note sin x 2 0.) sin2 x 1 2 sin x 1 1 5 0 Factor as a perfect square. Asinx 1 1B2 5 0 Solve for sin x. sin x 5 21 Solve for x on one period of the sine function, 30, 2p2. x 5 3p 2 Add integer multiples of 2p to obtain all solutions. x 5 3p 2 1 2np g 1 sin x EXAMPLE 11 Using Trigonometric Identities and Inverse Functions to Solve Trigonometric Equations Solve 3 cos2 u 1 sin u 5 3 on 0° # u , 360°. Solution: Use the Pythagorean identity. 3 cos2 u 1 sin u 5 3 Subtract 3. 311 2 sin2 u2 1 sin u 2 3 5 0 Eliminate the parentheses. 3 2 3 sin2 u 1 sin u 2 3 5 0 Simplify. 23 sin2 u 1 sin u 5 0 Factor the common sin u. sin u11 2 3 sin u2 5 0 Set each factor equal to 0. sin u 5 0 or 1 2 3 sin u 5 0 Solve for sin u. sin u 5 0 or sin u 5 1 3 Solve sin u 5 0 for x on 0° # u , 360°. u 5 0° or u 5 180° Solve sin u 5 1 3 for x on 0° # u , 360°. The sine function is positive in quadrants I and II. A calculator gives inverse values only in quadrant I. Write the equivalent inverse notation for sin u 5 1 3. u 5 sin21 a1 3b Use a calculator to approximate the quadrant I solution. u < 19.5° To find the quadrant II solution, subtract the reference angle from 180°. u < 180° 2 19.5° u < 160.5° g 1 2 sin2 u Applications EXAMPLE 12 Applications Involving Trigonometric Equations Light bends (refracts) according to Snell’s law, which states ni sin1ui2 5 nr sin1ur2 where n ni is the refractive index of the medium the light is leaving. n ui is the incident angle between the light ray and the normal (perpendicular) to the interface between mediums. n nr is the refractive index of the medium the light is entering. n ur is the refractive angle between the light ray and the normal (perpendicular) to the interface between mediums. Assume that light is going from air into a diamond. Calculate the refractive angle ur if the incidence angle is ui 5 32° and the index of refraction values for air and diamond are ni 5 1.00 and nr 5 2.417, respectively. Solution: Write Snell’s law. ni sin1ui2 5 nr sin1ur2 Substitute ui 5 32°, ni 5 1.00, and nr 5 2.417. sin 32° 5 2.417 sin ur Isolate sin ur and simplify. sin ur 5 sin 32° 2.417 < 0.21925 Solve for ur using the inverse sine function. ur < sin 2110.219252 < 12.665° Round to the nearest degree. ur < 13° Janis Christie/Getty Images, Inc. Air Diamond 32º θi ni = 1.00 θr nr = 2.417 Musical tones can be represented mathematically with sinusoidal functions. Recall that a tone of 48 Hz (cycles per second) can be represented by the function A sin32p1482t4, where A is the amplitude (loudness) and 48 Hz is the frequency. If we play two musical tones simultaneously, the combined tone can be found using the sum-to-product identity. For example, if a tone of 48 Hz is simultaneously played with a tone of 56 Hz, the result (assuming uniform amplitude, A 5 12 is sin196 pt2 1 sin1112pt2 5 2 sin1104 pt2 cos18pt2, whose graph is given on the right. The sin1104pt2 term represents a sound of average frequency 52 Hz. The 2 cos18pt2 represents a time-varying amplitude and a “beat” frequency that corresponds to 8 beats per second. Notice that in the graph there are 4 beats in 1 2 second. 0.1 0.2 0.3 0.4 0.5 –2 –1 1 2 x y sec 1 2 7.8 Trigonometric Equations 747 748 CHAPTER 7 Analytic Trigonometry EXAMPLE 13 Tuning Fork A tuning fork is used to help musicians “tune” their instruments. They simultaneously listen to the vibrating fork and play a tone and adjust their instrument until the beat frequency is eliminated. If an E (659 Hz) tuning fork is used and the musician hears 4 beats per second, find the frequency of the tone from the instrument if the loudest beat is the third of each group. Assume that each tone has an amplitude of 1. Solution: Express the tone E mathematically. sin32p16592t4 Express the instrument’s tone mathematically. sin12pƒt2 Add the two to get the combined sound. sin32p16592t4 1 sin12pft2 Use the sum-to-product identity. 2 sin c2 aƒ 1 659 2 bptd cos c2a ƒ 2 659 2 bptd The beat frequency is 4 Hz. 2 sin c2aƒ 1 659 2 bptdcos34pt4 Since the cosine function is an even function, so that cos12x2 5 cos x, we equate the absolute value of the cosine arguments. 2 a ƒ 2 659 2 bpt 5 0 4pt 0 Solve for ƒ. ƒ 5 663 or ƒ 5 655 The loudest sound, 0 A 0 5 2, is heard on the third beat of each four, t 5 3 4 sec. 2 sin c1ƒ 1 6592 3 4pdcos c4p3 4 d 5 2 Simplify. sin c1ƒ 1 6592 3 4pd 5 21 The inverse does not exist unless we restrict the domain of the sine function. Therefore, we inspect the two choices, 655 Hz or 663 Hz , and both satisfy the equation. The instrument can be playing either frequency based on the information given. abzee/E+/Getty Images In this section, we began by solving basic trigonometric equations that contained only one trigonometric function. Some such equations can be solved exactly by inspection, and others can be solved exactly using algebraic techniques similar to those of linear and quadratic equations. Calculators and inverse functions are needed when exact values are not known. It is important to note that calculators give the inverse function in only one of the two relevant quadrants. The other quadrant solutions must be found using reference angles. Trigonometric identities are useful for solving equations that involve more than one trigonometric function. With trigonometric identities we can transform such equations into equations involving only one trigonometric function, and then we can apply algebraic techniques. [SEC TION 7.8] S U M M A RY [SEC TION 7.8] E X ERC I S E S • S K I L L S In Exercises 1–20, solve the given trigonometric equation exactly over the indicated interval. 1. cos u 5 2 !2 2 , 0 # u , 2p 2. sin u 5 2 !2 2 , 0 # u , 2p 3. csc u 5 22, 0 # u , 4p 4. sec u 5 22, 0 # u , 4p 5. tan u 5 0, all real numbers 6. cot u 5 0, all real numbers 7. sin12u2 5 21 2, 0 # u , 2p 8. cos12u2 5 !3 2 , 0 # u , 2p 9. sinau 2b 5 21 2, all real numbers 10. cosau 2b 5 21, all real numbers 11. tan12u2 5 !3, 22p # u , 2p 12. tan12u2 5 2!3, all real numbers 13. sec u 5 22, 22p # u , 0 14. csc u 5 2!3 3 , 2p # u , p 15. cot14u2 5 2 !3 3 , all real numbers 16. tan15u2 5 1, all real numbers 17. sec13u2 5 21, 22p # u # 0 18. sec14u2 5 !2, 0 # u # p 19. csc13u2 5 1, 22p # u # 0 20. csc16u2 5 22!3 3 , 0 # u # p In Exercises 21–40, solve the given trigonometric equation exactly on 0 " u 2p. 21. 2 sin12u2 5 !3 22. 2 cosau 2b 5 2!2 23. 3 tan12u2 2 !3 5 0 24. 4 tan au 2b 2 4 5 0 25. 2 cos12u2 1 1 5 0 26. 4 csc12u2 1 8 5 0 27. !3 cotau 2b 2 3 5 0 28. !3 sec12u2 1 2 5 0 29. tan2 u 2 1 5 0 30. sin2 u 1 2 sin u 1 1 5 0 31. 2 cos2 u 2 cos u 5 0 32. tan2 u 2 !3 tan u 5 0 33. csc2 u 1 3 csc u 1 2 5 0 34. cot2 u 5 1 35. sin2 u 1 2 sin u 2 3 5 0 36. 2 sec2 u 1 sec u 2 1 5 0 37. sec2 u 2 1 5 0 38. csc2 u 2 1 5 0 39. sec2 12u2 2 4 3 5 0 40. csc212u2 2 4 5 0 In Exercises 41–60, solve the given trigonometric equation on 0° " u 360° and express the answer in degrees to two decimal places. 41. sin12u2 5 20.7843 42. cos12u2 5 0.5136 43. tanau 2b 5 20.2343 44. secau 2b 5 1.4275 45. 5 cot u 2 9 5 0 46. 5 sec u 1 6 5 0 47. 4 sin u 1 !2 5 0 48. 3 cos u 2 !5 5 0 49. 4 cos2 u 1 5 cos u 2 6 5 0 7.8 Trigonometric Equations 749 750 CHAPTER 7 Analytic Trigonometry 50. 6 sin2 u 2 13 sin u 2 5 5 0 51. 6 tan2 u 2 tan u 2 12 5 0 52. 6 sec2 u 2 7 sec u 2 20 5 0 53. 15 sin212u2 1 sin12u2 2 2 5 0 54. 12 cos2au 2b 2 13 cos au 2b 1 3 5 0 55. cos2 u 2 6 cos u 1 1 5 0 56. sin2 u 1 3 sin u 2 3 5 0 57. 2 tan2 u 2 tan u 2 7 5 0 58. 3 cot2 u 1 2 cot u 2 4 5 0 59. csc213u2 2 2 5 0 60. sec2au 2b 2 2 5 0 In Exercises 61–88, solve the trigonometric equations exactly on the indicated interval, 0 " x 2p. 61. sin x 5 cos x 62. sin x 5 2cos x 63. sec x 1 cos x 5 22 64. sin x 1 csc x 5 2 65. sec x 2 tan x 5 !3 3 66. sec x 1 tan x 5 1 67. csc x 1 cot x 5 !3 68. csc x 2 cot x 5 !3 3 69. 2 sin x 2 csc x 5 0 70. 2 sin x 1 csc x 5 3 71. sin12x2 5 4 cos x 72. sin12x2 5 !3 sin x 73. !2 sin x 5 tan x 74. cos12x2 5 sin x 75. tan12x2 5 cot x 76. 3 cot12x2 5 cot x 77. !3 sec x 5 4 sin x 78. !3 tan x 5 2 sin x 79. sin2 x 2 cos12x2 5 21 4 80. sin2 x 2 2 sin x 5 0 81. cos2 x 1 2 sin x 1 2 5 0 82. 2 cos2 x 5 sin x 1 1 83. 2 sin2 x 1 3 cos x 5 0 84. 4 cos2 x 2 4 sin x 5 5 85. cos12x2 1 cos x 5 0 86. 2 cot x 5 csc x 87. 1 4 sec12x2 5 sin12x2 88. 2 1 4 csca1 2 xb 5 cosa1 2 xb In Exercises 89–98, solve each trigonometric equation on 0° " u 360°. Express solutions in degrees and round to two decimal places. 89. cos12x2 1 1 2 sin x 5 0 90. sec2 x 5 tan x 1 1 91. 6 cos2 x 1 sin x 5 5 92. sec2 x 5 2 tan x 1 4 93. cot2 x 2 3 csc x 2 3 5 0 94. csc2 x 1 cot x 5 7 95. 2 sin2 x 1 2 cos x 2 1 5 0 96. sec2 x 1 tan x 2 2 5 0 97. 1 16 csc2ax 4b 2 cos2ax 4b 5 0 98. 2 1 4 sec2ax 8b 1 sin2ax 8b 5 0 • A P P L I C A T I O N S Exercises 99 and 100, refer to the following: Computer sales are generally subject to seasonal fluctuations. The sales of QualComp computers during 2017–2019 is approximated by the function s1t2 5 0.120 sin10.790 t 2 2.3802 1 0.387 1 # t # 12 where t represents time in quarters 1t 5 1 represents the end of the first quarter of 20172, and s1t2 represents computer sales (quarterly revenue) in millions of dollars. 99. Business. Find the quarter(s) in which the quarterly sales are $472,000. 100. Business. Find the quarter(s) in which the quarterly sales are $507,000. For Exercises 101 and 102, refer to the following: Allergy sufferers’ symptoms fluctuate with the concentration of pollen in the air. At one location the pollen concentration, measured in grains per cubic meter, of grasses fluctuates throughout the day according to the function: p1t2 5 35 2 26 cosa p 12 t 2 7p 6 b, 0 # t # 24 where t is measured in hours and t 5 0 is 12:00 a.m. 101. Biology/Health. Find the time(s) of day when the grass pollen level is 41 grains per cubic meter. Round to the nearest hour. 102. Biology/Health. Find the time(s) of day when the grass pollen level is 17 grains per cubic meter. Round to the nearest hour. 103. Sales. Monthly sales of soccer balls are approximated by S 5 400 sinap 6 xb 1 2000, where x is the number of the month (January is x 5 1, etc.). During which month do sales reach 2400? 104. Sales. Monthly sales of soccer balls are approximated by S 5 400 sin ap 6 xb 1 2000, where x is the number of the month (January is x 5 1, etc.). During which two months do sales reach 1800? 105. Home Improvement. A rain gutter is constructed from a single strip of sheet metal by bending as shown below, so that the base and sides are the same length. Express the area of the cross section of the rain gutter as a function of the angle u (note that the expression will also involve x). θ θ x x x θ θ x x x h θ 106. Home Improvement. A rain gutter is constructed from a single strip of sheet metal by bending as shown above, so that the base and sides are the same length. When the area of the cross section of the rain gutter is expressed as a function of the angle u, you can then use calculus to determine the value of u that produces the cross section with the greatest possible area. The angle is found by solving the equation cos2 u 2 sin2 u 1 cos u 5 0. Which angle gives the maximum area? 107. Deer Population. The number of deer on an island is given by D 5 200 1 100 sinap 2 xb, where x is the number of years since 2015. Which is the first year after 2015 that the number of deer reaches 300? 108. Deer Population. The number of deer on an island is given by D 5 200 1 100 sinap 6 xb, where x is the number of years since 2015. Which is the first year after 2015 that the number of deer reaches 150? 109. Optics. Assume that light is going from air into a diamond. Calculate the refractive angle ur if the incidence angle is ui 5 75° and the index of refraction values for air and diamond are ni 5 1.00 and nr 5 2.417, respectively. Round to the nearest degree. (See Example 12 for Snell’s law.) 110. Optics. Assume that light is going from a diamond into air. Calculate the refractive angle ur if the incidence angle is ui 5 15° and the index of refraction values for diamond and air are ni 5 2.417 and nr 5 1.00, respectively. Round to the nearest degree. (See Example 12 for Snell’s law.) 111. Air in Lungs. If a person breathes in and out every 3 seconds, the volume of air in the lungs can be modeled by A 5 2 sinap 3 xb cosap 3 xb 1 3, where A is in liters of air and x is in seconds. How many seconds into the cycle is the volume of air equal to 4 liters? 112. Air in Lungs. For the function given in Exercise 111, how many seconds into the cycle is the volume of air equal to 2 liters? For Exercises 113 and 114, refer to the following: The figure below shows the graph of y 5 2 cos x 2 cos12x2. The maximum and minimum values of the curve occur at the turning points and are found in the solutions of the equation 22 sin x 1 2 sin12x2 5 0. 2π π –π –2π –4 –3 –2 –1 2 3 4 x y 113. Finding Turning Points. Solve for the coordinates of the turning points of the curve on 30, 2p4. 114. Finding Turning Points. Solve for the coordinates of the turning points of the curve on 30, 2p4. 115. Business. An analysis of a company’s costs and revenue shows that annual costs of producing their product as well as annual revenues from the sale of a product are generally subject to seasonal fluctuations and are approximated by the functions C1t2 5 2.3 1 0.25 sinap 6 tb 0 # t # 11 R1t2 5 2.3 1 0.5 cosap 6 tb 0 # t # 11 where t represents time in months 1t 5 0 represents January2, C1t2 represents the monthly costs of producing the product in millions of dollars, and R1t2 represents monthly revenue from sales of the product in millions of dollars. Find the month(s) in which the company breaks even. Hint: A company breaks even when its profit is zero. 7.8 Trigonometric Equations 751 752 CHAPTER 7 Analytic Trigonometry 116. Business. An analysis of a company’s costs and revenue shows that the annual costs of producing its product as well as annual revenues from the sale of a product are generally subject to seasonal fluctuations and are approximated by the functions C1t2 5 25.7 1 0.2 sinap 6 tb 0 # t # 11 R1t2 5 25.7 1 9.6 cosap 6 tb 0 # t # 11 where t represents time in months 1t 5 0 represents January2, C1t2 represents the monthly costs of producing the product in millions of dollars, and R1t2 represents monthly revenue from sales of the product in millions of dollars. Find the month(s) in which the company breaks even. Hint: A company breaks even when its profit is zero. For Exercises 117 and 118, refer to the following: By analyzing available empirical data, it has been determined that the body temperature of a species fluctuates according to the model T1t2 5 37.10 1 1.40 sina p 24 tb cosa p 24 tb 0 # t # 24 where T represents temperature in degrees Celsius and t represents time (in hours) measured from 12:00 a.m. (midnight). 117. Biology/Health. Find the time(s) of day the body temperature is 37.28 degrees Celsius. Round to the nearest hour. 118. Biology/Health. Find the time(s) of day the body temperature is 36.75 degrees Celsius. Round to the nearest hour. In Exercises 119–122, explain the mistake that is made. • C A T C H T H E M I S T A K E 119. Solve !2 1 sin u 5 sin u on 0 # u # 2p. Solution: Square both sides. 2 1 sin u 5 sin2 u Gather all terms to one side. sin2 u 2 sin u 2 2 5 0 Factor. 1sin u 2 221sin u 1 12 5 0 Set each factor equal to zero. sin u 2 2 5 0 or sin u 1 1 5 0 Solve for sin u. sin u 5 2 or sin u 5 21 Solve sin u 5 2 for u. no solution Solve sin u 5 21 for u. u 5 3p 2 This is incorrect. What mistake was made? 120. Solve !3 sin u 2 2 5 2sin u on 0 # u # 2p. Solution: Square both sides. 3 sin u 2 2 5 sin2 u Gather all terms to one side. sin2 u 2 3 sin u 1 2 5 0 Factor. 1sin u 2 221sin u 2 12 5 0 Set each factor equal to zero. sin u 2 2 5 0 or sin u 2 1 5 0 Solve for sin u. sin u 5 2 or sin u 5 1 Solve sin u 5 2 for u. no solution Solve sin u 5 1 for u. u 5 p 2 This is incorrect. What mistake was made? 121. Solve 3 sin12x2 5 2 cos x on 0° # u # 180°. Solution: Use the double-angle identity for the sine function. 3 sin 2x 5 2 cos x Simplify. 6 sin x cos x 5 2 cos x Divide by 2 cos x. 3 sin x 5 1 Divide by 3. sin x 5 1 3 Write the equivalent inverse notation. x 5 sin21 a1 3b Use a calculator to approximate the solution. x < 19.47°, QI solution The QII solution is: x < 180° 2 19.47° < 160.53° This is incorrect. What mistake was made? 122. Solve !1 1 sin x 5 cos x on 0 # x # 2p. Solution: Square both sides. 1 1 sin x 5 cos2 x Use the Pythagorean identity. 1 1 sin x 5 cos2 x Simplify. sin2 x 1 sin x 5 0 Factor. sin x 1sin x 1 12 5 0 Set each factor equal to zero. sin x 5 0 or sin x 1 1 5 0 Solve for sin x. sin x 5 0 or sin x 5 21 Solve for x. x 5 0, p, 3p 2 , 2p This is incorrect. What mistake was made? g 2 sin x cos x g 1 2 sin2 x • C O N C E P T U A L 123. Linear trigonometric equations always have one solution on 30, 2p4. 124. Quadratic trigonometric equations always have two solutions on 30, 2p4. 125. If a trigonometric equation has all real numbers as its solution, then it is an identity. 126. If a trigonometric equation has an infinite number of solutions, then it is an identity. In Exercises 123–126, determine whether each statement is true or false. • C H A L L E N G E 127. Solve 16 sin4 u 2 8 sin2 u 5 21 over 0 # u # 2p. 128. Solve cosau 1 p 4 b 5 !3 2 over all real numbers. 129. Solve for the smallest positive x that makes this statement true: sinax 1 p 4 b 1 sinax 2 p 4 b 5 !2 2 130. Solve for the smallest positive x that makes this statement true: cos x cos 15° 1 sin x sin 15° 5 0.7 131. Find all real numbers x such that 1 2 cosax 3b 1 1 cosax 3b 1 1 5 0. 132. Find all real numbers u such that sec4a1 3 ub 2 1 5 0. 133. Find all real numbers u such that csc4ap 4 u 2 pb 2 4 5 0. 134. Find all real numbers x such that 2 tan13x2 5 !3 2 !3 tan2 13x2. • T E C H N O L O G Y Graphing calculators can be used to find approximate solutions to trigonometric equations. For the equation ƒ1x) 5 g1x2, let Y1 5 ƒ1x2 and Y2 5 g1x2. The x-values that correspond to points of intersections represent solutions. 135. With a graphing utility, solve the equation sin u 5 cos 2u on 0 # u # p. 136. With a graphing utility, solve the equation csc u 5 sec u on 0 # u # p 2 . 137. With a graphing utility, solve the equation sin u 5 sec u on 0 # u # p. 138. With a graphing utility, solve the equation cos u 5 csc u on 0 # u # p. 139. With a graphing utility, find all the solutions to the equation sin u 5 eu for u $ 0. 140. With a graphing utility, find all the solutions to the equation cos u 5 eu for u $ 0. Find the smallest positive values of x that make the statement true. Give the answer in radians and round to two decimal places. 141. sec13x2 1 csc12x2 5 5 142. cot15x2 1 tan12x2 5 23 143. e x 2 tan x 5 0 144. e x 1 2 sin x 5 1 145. ln x 2 sin x 5 0 146. ln x 2 cos x 5 0 7.8 Trigonometric Equations 753 CH A P TE R 7 R E VI E W [ C H AP T E R 7 REVIEW] 754 CHAPTER 7 Analytic Trigonometry SECTION CONCEPT KEY IDEAS/FORMULAS 7.1 Basic trigonometric identities Reciprocal, quotient, and Pythagorean identities Reciprocal identities csc u 5 1 sin u sec u 5 1 cos u cot u 5 1 tan u Quotient identities tan u 5 sin u cos u cot u 5 cos u sin u Pythagorean identities sin2 u 1 cos2 u 5 1 tan2 u 1 1 5 sec2 u 1 1 cot2 u 5 csc2 u 7.2 Verifying trigonometric identities Identities must hold for all values of x (not just some values of x) for which both sides of the equation are defined. Trigonometric identities Simplifying trigonometric expressions using identities. Use the reciprocal, quotient, or Pythagorean identities to simplify trigonometric expressions. Verifying identities n Convert all trigonometric expressions to sines and cosines. n Write all sums or differences of fractions as a single fraction. 7.3 Sum and difference identities ƒ1A 6 B2 2 ƒ1A2 6 ƒ1B2 For trigonometric functions, we have the sum and difference identities. Sum and difference identities for the cosine function cos1A 1 B2 5 cos A cos B 2 sin A sin B cos1A 2 B2 5 cos A cos B 1 sin A sin B Sum and difference identities for the sine function sin1A 1 B2 5 sin A cos B 1 cos A sin B sin1A 2 B2 5 sin A cos B 2 cos A sin B Sum and difference identities for the tangent function tan1A 1 B2 5 tan A 1 tan B 1 2 tan A tan B tan1A 2 B2 5 tan A 2 tan B 1 1 tan A tan B 7.4 Double-angle identities sin12A2 5 2 sin A cos A cos12 A2 5 cos2 A 2 sin2 A 5 1 2 2 sin2 A 5 2 cos2 A 2 1 Applying double-angle identities tan12 A2 5 2 tan A 1 2 tan2 A 7.5 Half-angle identities Applying half-angle identities sinaA 2 b 5 6Å 1 2 cos A 2 cosaA 2 b 5 6 Å 1 1 cos A 2 tanaA 2 b 5 6 Å 1 2 cos A 1 1 cos A 5 sin A 1 1 cos A 5 1 2 cos A sin A 7.6 Product-to-sum and sum-to-product identities Product-to-sum identities cos Acos B 5 1 23cos1A 1 B2 1 cos1A 2 B24 sin Asin B 5 1 23cos1A 2 B2 2 cos1A 1 B24 sin Acos B 5 1 23sin1A 1 B2 1 sin1A 2 B24 CH A P TE R 7 R E VIE W SECTION CONCEPT KEY IDEAS/FORMULAS Sum-to-product identities sin A 1 sin B 5 2 sinaA 1 B 2 b cosaA 2 B 2 b sin A 2 sin B 5 2 sinaA 2 B 2 b cosaA 1 B 2 b cos A 1 cos B 5 2 cosaA 1 B 2 b cosaA 2 B 2 b cos A 2 cos B 5 22 sinaA 1 B 2 b sinaA 2 B 2 b 7.7 Inverse trigonometric functions sin21 x or arcsin x cos21 x or arccos x tan21 x or arctan x cot21 x or arccot x sec21 x or arcsec x csc21 x or arccsc x Inverse sine function Definition y 5 sin21x means x 5 sin y 21 # x # 1 and 2p 2 # y # p 2 Identities sin211sin x2 5 x for 2p 2 # x # p 2 sin1sin21 x2 5 x for 21 # x # 1 –π –1 –0.5 0.5 1 x y π 4 π 2 4 –π 2 y = sin–1x –π 2 (–1, ) π 2 (1, ) Inverse cosine function Definition y 5 cos21 x means x 5 cos y 21 # x # 1 and 0 # y # p Identities cos211cos x2 5 x for 0 # x # p cos1cos21 x2 5 x for 21 # x # 1 –1 –0.5 0.5 1 x y π 4 3π 4 π (–1, π) (1, 0) y = cos–1x π 2 (0, ) Inverse tangent function Definition y 5 tan21 x means x 5 tan y 2q , x , q and 2p 2 , y , p 2 Identities tan211tan x2 5 x for 2p 2 , x , p 2 tan1tan21 x2 5 x for 2q , x , q –π x y π 4 π 2 4 –π 2 3 2 1 –3 –2 –1 y = tan–1x Remaining inverse trigonometric functions Inverse cotangent function Definition y 5 cot21 x means x 5 cot y 2q , x , q and 0 , y , p Identity cot21x 5 µ tan21a1 xb, x . 0 p 1 tan21a1 xb, x , 0 –2 –1 1 2 x y π 4 3π 4 π y = cot–1x π 2 (0, ) Chapter Review 755 CH A P TE R 7 R E VI E W 756 CHAPTER 7 Analytic Trigonometry SECTION CONCEPT KEY IDEAS/FORMULAS Inverse secant function Definition y 5 sec21 x means x 5 sec y x # 21 or x $ 1 and 0 # y , p 2 or p 2 , y # p Identity sec21 x 5 cos21 a1 xb for x # 21 or x $ 1 –2 –1 2 x y π π 2 π 4 3π 4 (–1, π) y = sec–1x (1, 0) Inverse cosecant function Definition y 5 csc21 x means x 5 csc y x # 21 or x $ 1 and 2p 2 # y , 0 or 0 , y # p 2 Identity csc21 x 5 sin21a1 xb for x # 21 or x $ 1 –π –2 –1 1 2 x y π 4 π 2 4 –π 2 –π 2 (–1, ) π 2 (1, ) y = csc–1x Finding exact values for expressions involving inverse trigonometric functions 7.8 Trigonometric Equations Goal: Find the value(s) of the variable that make the equation true. Solving trigonometric equations by inspection Solve: sin u 5 !2 2 on 0 # u # 2p. Answer: u 5 p 4 or u 5 3p 4 . Solve: sin u 5 !2 2 on all real numbers. Answer: u 5 µ p 4 1 2np 3p 4 1 2np where n is an integer. Solving trigonometric equations using algebraic techniques Transform trigonometric equations into linear or quadratic algebraic equations by making a substitution such as x 5 sin u. Then use algebraic methods for solving linear and quadratic equations. Note: If an expression is squared, always check for extraneous solutions. Solving trigonometric equations that require the use of inverse functions Follow the same procedures outlined by inspection or algebraic methods. Finding the solution requires the use of inverse functions and a calculator. Be careful: Calculators only give one solution (the one in the range of the inverse function). Using trigonometric identities to solve trigonomatric equations n Use trigonometric identities to transform an equation with multiple trigonometric functions into an equation with only one trigonometric function. n Then use the methods outlined above. R E VI E W E XERCISES [CH AP TER 7 REVIEW EXE R C IS E S ] 7.1 Basic Trigonometric Identities Use trigonometric identities to find the indicated value(s). 1. If sin u 5 2 7 11, find csc u. 2. If cot u 5 3!5 5 , find tan u. 3. If sin u 5 8 17 and cos u 5 215 17, find tan u. 4. If sin u 5 212 13 and cos u 5 2 5 13, find cot u. 5. If sin u 5 2!3 9 , find sin2 u. 6. If cos u 5 20.45, find cos2 u. 7. If tan u 5 !3, find tan3 u. 8. If sin u 5 11 61 and the terminal side of u lies in QII, find tan u. 9. If cos u 5 7 25 and the terminal side of u lies in QIV, find cot u. 10. If tan u 5 21 and the terminal side of u lies in QIV, find sec u. 11. If csc u 5 13 12 and the terminal side of u lies in QII, find cot u. 12. Find sin u and cos u if tan u 5 224 7 and the terminal side of u lies in QII. 13. Find sin u and cos u if cot u 5 !3 3 and the terminal side of u lies in QIII. 14. Find sin u and cos u if cot u 5 !3 and the terminal side of u lies in QIII. Perform the operations and simplify. Write the answers in terms of sin u and cos u. 15. sec u cot2 u 16. sin u cot u 17. tan u sec u 18. 1cos u 1 sec u22 19. cot u1sec u 1 sin u2 20. sin2 u csc u 7.2 Verifying Trigonometric Identities Simplify the following trigonometric expressions. 21. tan x1cot x 1 tan x2 22. 1sec x 1 12 1sec x 2 12 23. tan4 x 2 1 tan2 x 2 1 24. sec2 x1cot2 x 2 cos2 x2 25. cos x3cos12x2 2 tan12x24 2 sin x 26. tan2 x 1 1 2 sec2 x 27. csc312x2 1 8 csc x 2 2 28. csc2 x 2 1 cot x Verify the trigonometric identities. 29. 1tan x 1 cot x22 2 2 5 tan2 x 1 cot2 x 30. csc2 x 2 cot2 x 5 1 31. 1 sin2 x 2 1 tan2 x 5 1 32. 1 csc x 1 1 1 1 csc x 2 1 5 2 tan x cos x 33. tan2 x 2 1 sec2 x 1 3 tan x 1 1 5 tan x 2 1 tan x 1 2 34. cot x1sec x 2 cos x2 5 sin x Determine whether each of the following equations is a conditional equation or an identity. 35. 2 tan2 x 1 1 5 1 1 sin2 x cos2 x 36. sin x 2 cos x 5 0 37. cot2 x 2 1 5 tan2 x 38. cos2 x11 1 cot2 x2 5 cot2 x 39. acot x 2 1 tan xb 2 5 0 40. csc x 1 sec x 5 1 sin x 1 cos x 7.3 Sum and Difference Identities Find the exact value for each trigonometric expression. 41. cos a7p 12 b 42. sin a p 12b 43. tan1215°2 44. cot 105° Write each expression as a single trigonometric function. 45. sin14x2 cos13x2 2 cos14x2 sin13x2 46. sin12x2 sin122x2 1 cos12x2 cos122x2 47. tan15x2 2 tan14x2 1 1 tan15x2 tan14x2 48. tan ap 4 b 1 tan ap 3 b 1 2 tan ap 4 b tan ap 3 b Review Exercises 757 REV IEW E XE R CI SE S 758 CHAPTER 7 Analytic Trigonometry Find the exact value of the indicated expression using the given information and identities. 49. Find the exact value of tan1a 2 b2 if sin a 5 2 3 5, sin b 5 224 25, the terminal side of a lies in quadrant IV, and the terminal side of b lies in quadrant III. 50. Find the exact value of cos1a 1 b2 if cos a 5 2 5 13, sin b 5 7 25, the terminal side of a lies in quadrant II, and the terminal side of b also lies in quadrant II. 51. Find the exact value of cos1a 2 b2 if cos a 5 9 41, cos b 5 7 25, the terminal side of a lies in quadrant IV, and the terminal side of b lies in quadrant I. 52. Find the exact value of sin1a 2 b2 if sin a 5 2 5 13, cos b 5 24 5, the terminal side of a lies in quadrant III, and the terminal side of b lies in quadrant II. Determine whether each of the following equations is a conditional equation or an identity. 53. 2 cos A cos B 5 cos1A 1 B2 1 cos1A 2 B2 54. 2 sin A sin B 5 cos1A 2 B2 2 cos1A 1 B2 Graph the following functions. 55. y 5 cos ap 2 b cos x 2 sin ap 2 b sin x 56. y 5 sin a2p 3 b cos x 1 cos a2p 3 b sin x 57. y 5 2 tan ax 3b 1 2 tan2 ax 3b 58. y 5 tan1px2 2 tan x 1 1 tan1px2 tan x 7.4 Double-Angle Identities Use double-angle identities to answer the following questions. 59. If sin x 5 3 5 and p 2 , x , p, find cos 12x2. 60. If cos x 5 7 25 and 3p 2 , x , 2p, find sin 12x2. 61. If cot x 5 211 61 and 3p 2 , x , 2p, find tan 12x2. 62. If tan x 5 212 5 and p 2 , x , p, find cos 12x2. 63. If sec x 5 25 24 and 0 , x , p 2 , find sin 12x2. 64. If csc x 5 5 4 and p 2 , x , p, find tan 12x2. Simplify each of the following expressions. Evaluate exactly, if possible. 65. cos2 15° 2 sin2 15° 66. 2 tana2 p 12b 1 2 tan2a2 p 12b 67. 6 sina p 12b cosa p 12b 68. 1 2 2 sin2ap 8 b Verify the following identities. 69. sin3 A 2 cos3 A 5 1sin A 2 cos A2C1 1 1 2 sin12 A2D 70. 2 sin A cos3 A 2 2sin3 A cos A 5 cos12 A2 sin12 A2 71. tan A 5 sin12 A2 1 1 cos12 A2 72. tan A 5 1 2 cos12 A2 sin12 A2 73. Launching a Missile. When launching a missile for a given range, the minimum velocity needed is related to the angle u of the launch, and the velocity is determined by V 5 2 cos12u2 1 1 cos12u2. Show that V is equivalent to 1 2 tan2 u. 74. Launching a Missile. When launching a missile for a given range, the minimum velocity needed is related to the angle u of the launch, and the velocity is determined by V 5 2 cos12u2 1 1 cos12u2. Find the value of V when u 5 p 6. 7.5 Half-Angle Identities Use half-angle identities to find the exact value of each of the following trigonometric expressions. 75. sin1222.5°2 76. cos 67.5° 77. cota3p 8 b 78. csca27p 8 b 79. sec12165°2 80. tan1275°2 Use half-angle identities to find each of the following values. 81. If sin x 5 2 7 25 and p , x , 3p 2 , find sinax 2b. 82. If cos x 5 24 5 and p 2 , x , p, find cosax 2b. 83. If tan x 5 40 9 and p , x , 3p 2 , find tanax 2b. 84. If sec x 5 17 15 and 3p 2 , x , 2p, find sinax 2b. R E VI E W E XERCISES Simplify each expression using half-angle identities. Do not evaluate. 85. å 1 2 cosap 6 b 2 86. ñ 1 2 cosa11p 6 b 1 1 cosa11p 6 b Verify each of the following identities. 87. csinaA 2 b 1 cosaA 2 b d 2 5 1 1 sin A 88. sec2aA 2 b 1 tan2aA 2 b 5 3 2 cos A 1 1 cos A 89. csc2aA 2 b 1 cot2aA 2 b 5 3 1 cos A 1 2 cos A 90. tan2aA 2 b 1 1 5 sec2aA 2 b Graph each of the following functions. 91. y 5 å 1 2 cosa p 12 xb 2 92. y 5 cos2ax 2b 2 sin2ax 2b 93. y 5 2Å 1 2 cos x 1 1 cos x 94. y 5 Å 1 1 cos13x 2 12 2 7.6 Product-to-Sum and Sum-to-Product Identities Write each product as a sum or difference of sines and/or cosines. 95. 6 sin15x2 cos12x2 96. 3 sin14x2 sin12x2 Write each expression as a product of sines and/or cosines. 97. cos15x2 2 cos13x2 98. sina5x 2 b 1 sina3x 2 b 99. sina4x 3 b 2 sina2x 3 b 100. cos17x2 1 cos x Simplify each trigonometric expression. 101. cos18x2 1 cos12x2 sin18x2 2 sin12x2 102. sin15x2 1 sin13x2 cos15x2 1 cos13x2 Verify the identities. 103. sin A 1 sin B cos A 2 cos B 5 2cotaA 2 B 2 b 104. sin A 2 sin B cos A 2 cos B 5 2cotaA 1 B 2 b 105. cscaA 2 B 2 b 5 2 sinaA 1 B 2 b cos B 2 cos A 106. secaA 1 B 2 b 5 2 sinaA 2 B 2 b sin A 2 sin B 7.7 Inverse Trigonometric Functions Find the exact value of each expression. Give the answer in radians. 107. arctan 1 108. arccsc1222 109. cos21 0 110. sin211212 111. sec21a 2 !3b 112. cot21A2!3B Find the exact value of each expression. Give the answer in degrees. 113. csc211212 114. arctan1212 115. arccota !3 3 b 116. cos21a !2 2 b 117. sin21a2 !3 2 b 118. sec21 1 Use a calculator to evaluate each expression. Give the answer in degrees and round to two decimal places. 119. sin21120.60882 120. tan2111.19182 121. sec2111.08242 122. cot21123.73212 Use a calculator to evaluate each expression. Give the answer in radians and round to two decimal places. 123. cos21120.17362 124. tan2110.15842 125. csc211210.01672 126. sec21121.12232 Review Exercises 759 REV IEW E XE R CI SE S 760 CHAPTER 7 Analytic Trigonometry Evaluate each expression exactly, if possible. If not possible, state why. 127. sin21csin a2 p 4b d 128. cos ccos21a2 !2 2 b d 129. tan Ctan21A2!3BD 130. cot21ccot a11p 6 b d 131. csc21ccsc a2p 3 b d 132. sec csec21 a22!3 3 b d Evaluate each expression exactly. 133. sinccos21a11 61b d 134. cos ctan21a40 9 b d 135. tanccot21a6 7b d 136. cotcsec21a25 7 b d 137. sec csin21a1 6b d 138. csc ccot21a 5 12b d Applications 139. Average Temperature. If the average temperature in degrees in Chicago, Illinois, can be modeled with the formula T1m2 5 26 sin10.48 m 2 1.842 1 47, where m is the month of the year (January corresponds to m 5 1, etc.), then during which month is the average temperature 738F? 140. Average Temperature. If the average temperature in degrees in Chicago, Illinois, can be modeled with the formula T1m2 5 26 sin10.48 m 2 1.842 1 47, where m is the month of the year (January corresponds to m 5 1, etc.), then during which month is the average temperature 218F? 7.8 Trigonometric Equations Solve the given trigonometric equation over the indicated interval. 141. sin12u2 5 2 !3 2 , 0 # u # 2p 142. secau 2b 5 2, 22p # u # 2p 143. sinau 2b 5 2 !2 2 , 22p # u # 2p 144. csc12u2 5 2, 0 # u # 2p 145. tana1 3 ub 5 21, 0 # u # 6p 146. cot14u2 5 2!3, 2p # u # p Solve each trigonometric equation exactly on 0 " u " 2p. 147. 4 cos12u2 1 2 5 0 148. !3 tanau 2b 2 1 5 0 149. 2 tan12u2 1 2 5 0 150. 2 sin2 u 1 sin u 2 1 5 0 151. tan2 u 1 tan u 5 0 152. sec2 u 2 3 sec u 1 2 5 0 Solve the given trigonometric equations on 0° " u " 360° and express the answer in degrees to two decimal places. 153. tan12u2 5 20.3459 154. 6 sin u 2 5 5 0 155. 4 cos2 u 1 3 cos u 5 0 156. 12 cos2 u 2 7 cos u11 5 0 157. csc2 u 2 3 csc u 2 1 5 0 158. 2 cot2 u 1 5 cot u 2 4 5 0 Solve each trigonometric equation exactly on the interval 0 " u " 2p. 159. sec x 5 2 sin x 160. 3 tan x 1 cot x 5 2!3 161. !3 tan x 2 sec x 5 1 162. 2 sin12x2 5 cot x 163. !3 tan x 5 2 sin x 164. 2 sin x 5 3 cot x 165. cos2 x 1 sin x 1 1 5 0 166. 2 cos2 x 2 !3 cos x 5 0 167. cos12x2 1 4 cos x 1 3 5 0 168. sin12x2 1 sin x 5 0 169. tan2A1 2 xB 2 1 5 0 170. cot2A1 3 xB 2 1 5 0 Solve each trigonometric equation on 0° " u " 360°. Give the answers in degrees and round to two decimal places. 171. csc2 x 1 cot x 5 1 172. 8 cos2 x 1 6 sin x 5 9 173. sin 2 x 1 2 5 2 cos x 174. cos12x2 5 3 sin x 2 1 175. cos x 2 1 5 cos12x2 176. 12 cos2 x 1 4 sin x 5 11 Technology Exercises Section 7.1 177. Is cos 73° 5 "1 2 sin2 73°? Use a calculator to find each of the following: a. cos 73° b. 1 2 sin 73° c. "1 2 sin2 73° Which results are the same? 178. Is csc 28° 5 "1 1 cot2 28°? Use a calculator to find each of the following. a. csc 28° b. 1 1 cot 28° c. "1 1 cot2 28° Which results are the same? R E VI E W E XERCISES Section 7.2 179. Determine the correct sign 11/22 for sina2x 1 p 4 b 5 sin12x2 cosap 4 b 6 cos12x2 sinap 4 b by graphing Y1 5 sina2x 1 p 4 b, Y2 5 sin12x2 cosap 4 b 1 cos12x2 sinap 4 b, and Y3 5 sin12x2 cosap 4 b 2 cos12x2 sinap 4 b in the same viewing rectangle. Does Y1 5 Y2 or Y3? 180. Determine the correct sign 11/22 for cosa2x 2 p 3 b 5 cos12x2 cosap 3 b 6 sin12x2 sin ap 3 b by graphing Y1 5 cosa2x 2 p 3 b, Y2 5 cos12x2 cosap 3 b 2 sin12x2 sinap 3 b, and Y3 5 cos12x2 cosap 3 b 1 sin12x2 sinap 3 b in the same viewing rectangle. Does Y1 5 Y2 or Y3? Section 7.3 Recall that the difference quotient for a function f is given by ƒ1x 1 h2 1 ƒ1x2 h . 181. Show that the difference quotient for ƒ1x2 5 sin13x2 is 3cos13x24 c sin13h2 h d 2 3sin13x24 c 1 2 cos13h2 h d . Plot Y1 5 3cos 13x24 c sin13h2 h d 2 3sin13x24 c 1 2 cos13h2 h d for a. h 5 1 b. h 5 0.1 c. h 5 0.01 What function does the difference quotient for ƒ1x2 5 sin13x2 resemble when h approaches zero? 182. Show that the difference quotient for ƒ1x2 5 cos13x2 is 32sin13x24 c sin13h2 h d 2 3cos13x24 c 1 2 cos13h2 h d . Plot Y1 5 32sin13x24 c sin13h2 h d 2 3cos13x24 c 1 2 cos13h2 h d for a. h 5 1 b. h 5 0.1 c. h 5 0.01 What function does the difference quotient for ƒ1x2 5 cos13x2 resemble when h approaches zero? Section 7.4 183. With a graphing calculator, plot Y1 5 tan12x2, Y2 5 2 tan x, and Y3 5 2 tan x 1 2 tan2 x in the same viewing rectangle 322p, 2p4 by 3210, 104. Which graphs are the same? 184. With a graphing calculator, plot Y1 5 cos12x2, Y2 5 2 cos x, and Y3 5 1 2 2 sin2 x in the same viewing rectangle 322p, 2p4 by 322, 24. Which graphs are the same? Section 7.5 185. With a graphing calculator, plot Y1 5 cosax 2b, Y2 5 1 2 cos x, and Y3 5 2Å 1 1 cos x 2 in the same viewing rectangle 3p, 2p4 by 321, 14. Which graphs are the same? 186. With a graphing calculator, plot Y1 5 sinax 2b, Y2 5 1 2 sin x, and Y3 5 2Å 1 2 cos x 2 in the same viewing rectangle 32p, 4p4 by 321, 14. Which graphs are the same? Section 7.6 187. With a graphing calculator, plot Y1 5 sin15x2 cos13x2, Y2 5 sin14x2, and Y3 5 1 2 3sin18x2 1 sin12x24 in the same viewing rectangle 30, 2p4 by 321, 14. Which graphs are the same? 188. With a graphing calculator, plot Y1 5 sin13x2 cos15x2, Y2 5 cos14x2, and Y3 5 1 2 3sin18x2 2 sin12x24 in the same viewing rectangle 30, 2p4 by 321, 14. Which graphs are the same? Section 7.7 189. Given cos x 5 2 1 !5 and p 2 , x , p: a. Find cos 12x2 using the double-angle identity. b. Use the inverse of cosine to find x in QII and to find cos 12x2. c. Are the results in (a) and (b) the same? 190. Given cos x 5 5 12 and 3p 2 , x , 2p: a. Find cosA1 2 xB using the half-angle identity. b. Use the inverse of cosine to find x in QIV and to find cosA1 2 xB. Round to five decimal places. c. Are the results in (a) and (b) the same? Section 7.8 Find the smallest positive value of x that makes each statement true. Give the answer in radians and round to four decimal places. 191. ln x 1 sin x 5 0 192. ln x 1 cos x 5 0 g ? g ? Review Exercises 761 PR ACTICE TEST [ C H AP T E R 7 PRACTICE TEST ] 762 CHAPTER 7 Analytic Trigonometry 1. For what values of x does the quotient identity tan x 5 sin x cos x not hold? 2. Is the equation "sin2 x 1 cos2 x 5 sin x 1 cos x a conditional equation or an identity? 3. Evaluate sina2p 8 b exactly. 4. Evaluate tana7p 12 b exactly. 5. If cos x 5 2 5 and 3p 2 , x , 2p, find sinax 2b. 6. If sin x 5 2 1 5 and p , x , 3p 2 , find cos12x2. 7. Write cos17x2 cos13x2 2 sin13x2 sin17x2 as a cosine or sine of a sum or difference. 8. Write 2 2 tan x 1 2 tan2 x as a single tangent function. 9. Write Å 1 1 cos1a 1 b2 2 as a single cosine function if a 1 b is an angle in QII, that is, p 2 , a 1 b , p. 10. Write 2 sinax 1 3 2 b cosax 2 3 2 b as a sum of two sine functions. 11. Write 10 cos13 2 x2 1 10 cos1x 1 32 as a product of two cosine functions. 12. In the expression "9 2 u2, let u 5 3 sin x. What is the resulting expression? Solve the trigonometric equations exactly, if possible. Otherwise, use a calculator to approximate solution(s). 13. 2 sin u 5 2!3 on all real numbers. 14. 2 cos2 u 1 cos u 2 1 5 0 on 0 # u # 2p 15. sin 2u 5 1 2 cos u over 0 # u # 360° 16. !sin x 1 cos x 5 21 over 0 # u # 2p 17. Determine whether 11 1 cot x22 5 csc2 x is a conditional or an identity. 18. Evaluate csca2 p 12b exactly. 19. If sin x 5 2 5 13 and p , x , 3p 2 , find cosax 2b. 20. If cos x 5 20.26 and p 2 , x , p, find sin12x2. 21. Express y 5 ñ 1 1 !2 2 ccosap 3xb 1 sinap 3 xb d 1 2 !2 2 ccosap 3 xb 1 sinap 3 xb d as a cotangent function. 22. Calculate cscAcsc21 !2B. 23. Determine an interval on which ƒ1x2 5 a 1 b csc1px 1 c2 is one-to-one, and determine the inverse of ƒ1x2 on this interval. Assume that a, b, and c are all positive. 24. Find the range of y 5 2p 4 1 arctan12x 2 32. 25. Solve cosap 4 ub 5 21 2, for all real numbers. 26. Solve Å 1 2 cos12px2 1 1 cos12px2 5 2 1 !3, for all real numbers. 27. Solve !3 cscax 3b 5 cosax 3b, for all real numbers. 28. Show that the difference quotient for ƒ1x2 5 cos A1 2 xB is 2sina1 2 xb Csina1 2 hb h S 2 cosa1 2 xb C1 2 cosa1 2 hb h S. Plot Y152sina1 2 xb Csina1 2 hb h S2 cos12x2 £1 2 cos a1 2 hb h § for a. h 5 1 b. h 5 0.1 c. h 5 0.01 What function does the difference quotient for ƒ1x2 5 cosA1 2 xB resemble when h approaches zero? 29. Given tan x 5 3 4 and p , x , 3p 2 : a. Find sin A1 2 xB using the half-angle identity. b. Use the inverse of tangent to find x in QIII and to find sin A1 2 xB. Round to five decimal places. c. Are the results in (a) and (b) the same? 1. Solve by completing the square: x2 2 6x 5 11. 2. Write an equation of a line that passes through the point 122, 52 and is parallel to the y-axis. 3. Write the equation of a circle with center 123, 212 and passing through the point 11, 22. 4. Determine whether the relation x2 2 y2 5 25 is a function. 5. Determine whether the function g1x2 5 "2 2 x2 is odd or even. 6. For the function y 5 51x 2 422, identify all of the transformations of y 5 x2. 7. Find the composite function, ƒ + g, and state the domain for ƒ1x2 5 x3 2 1 and g1x2 5 1 x. 8. Find the inverse of the function ƒ1x2 5 ! 3 x 2 1. 9. Find the vertex of the parabola associated with the quadratic function ƒ1x2 5 1 4 x2 1 3 5 x 2 6 25. 10. Find a polynomial of minimum degree (there are many) that has the zeros x 5 2!7 (multiplicity 2), x 5 0 (multiplicity 3), x 5 !7 (multiplicity 2). 11. Use long division to find the quotient Q 1x2 and the remain-der r 1x2 of 15x3 2 4x2 1 32 4 1x2 1 12. 12. Given the zero x 5 4i of the polynomial P 1x2 5 x4 1 2x3 1 x2 1 32x 2 240, determine all the other zeros and write the polynomial in terms of a product of linear factors. 13. Find the vertical and horizontal asymptotes of the function ƒ1x2 5 0.7x2 2 5x 1 11 x2 2 x 2 6 . 14. If $5400 is invested at 2.25% compounded continuously, how much is in the account after 4 years? 15. Use interval notation to express the domain of the function ƒ1x2 5 log41x 1 32. 16. Use properties of logarithms to simplify the expression log p 1. 17. Give an exact solution to the logarithmic equation log5 1x 1 22 1 log516 2 x2 5 log513x2. 18. If money is invested in a savings account earning 4% compounded continuously, how many years will it take for the money to triple? 19. Use a calculator to evaluate cos 62°. Round the answer to four decimal places. 20. Angle of Inclination (Skiing). The angle of inclination of a mountain with triple black diamond ski trails is 63º. If a skier at the top of the mountain is at an elevation of 4200 feet, how long is the ski run from the top to the base of the mountain? 21. Convert 2105º to radians. Leave the answer in terms of p. 22. Find all of the exact values of u, when tan u 5 1 and 0 # u # 2p. 23. Determine whether the equation cos2 x 2 sin2 x 5 1 is a conditional equation or an identity. 24. Simplify 2 tana2p 8 b 1 2 tan2a2 p 8 b and evaluate exactly. 25. Evaluate exactly the expression tancsin21 a 5 13b d . 26. With a graphing calculator, plot Y1 5 sin x cos13x2, Y2 5 cos14x2, and Y3 5 1 2 3sin14x2 2 sin12x24 in the same viewing rectangle 30, 2p4 by 321, 14. Which graphs are the same? 27. Find the smallest positive value of x that makes the statement true. Give the answer in radians and round to four decimal places. In x 2 sin12x2 5 0 [CH AP TERS 1–7 CUM UL AT IVE T E S T ] CU MU LA TIV E TEST Cumulative Test 763 C H A P T E R LEARNING OBJECTIVES [ [ 8 ■ ■Solve oblique triangles using the Law of Sines. ■ ■Solve oblique triangles using the Law of Cosines. ■ ■Find areas of oblique triangles. ■ ■Find the direction and magnitude of a vector; add and subtract vectors. ■ ■Perform scalar multiplication and dot products. ■ ■Express complex numbers in polar form. ■ ■Use De Moivre’s theorem to find a complex number raised to a power. ■ ■Graph polar equations. Additional Topics in Trigonometry In recent decades, many people have come to believe that an imaginary area called the “Bermuda Triangle,” located off the southeastern Atlantic coast of the United States, has been the site of a high incidence of losses of ships, small boats, and aircraft over the centuries. The U.S. Board of Geographic Names does not recognize the “Bermuda Triangle” as an official name and does not maintain an official file on the area. Assume for the moment, without judging the merits of the hypothesis, that the “Bermuda Triangle” has vertices either in Miami (Florida), San Juan (Puerto Rico), and Bermuda or in Norfolk (Virginia), Bermuda, and Santiago (Cuba). In this chapter, you will develop a formula that determines the area of a triangle from its perimeter and side lengths. Which “Bermuda Triangle” has a larger area: Miami– Bermuda–Puerto-Rico or Norfolk–Bermuda–Cuba? You will calculate the answer in this chapter. Section 8.3, Exercises 33 and 34. Bermuda Miami, Florida San Juan, Puerto Rico Atlantic Ocean Miami (Florida), Bermuda, Puerto Rico Bermuda Atlantic Ocean Norfolk, Virginia Santiago de Cuba Norfolk (Virginia), Bermuda, Cuba 765 [I N T HI S CHAPTER] We discuss oblique (nonright) triangles. We use the Law of Sines and the Law of Cosines to solve oblique triangles. Then on the basis of the Law of Sines and the Law of Cosines and with our knowledge of trigonometric identities, we develop formulas for calculating the area of an oblique triangle. We define vectors and use the Law of Cosines and the Law of Sines to determine the resulting velocity and force vectors. We define dot products (the product of two vectors) and see how they are applicable to physical problems such as calculating work. We then use trigonometry to define polar coordinates, and we examine polar equations and their corresponding graphs. ADDITIONAL TOPICS IN TRIGONOMETRY 8.1 OBLIQUE TRIANGLES AND THE LAW OF SINES 8.2 THE LAW OF COSINES 8.3 THE AREA OF A TRIANGLE 8.4 VECTORS 8.5 THE DOT PRODUCT 8.6 POLAR (TRIGONO- METRIC) FORM OF COMPLEX NUMBERS 8.7 PRODUCTS, QUOTIENTS, POWERS, AND ROOTS OF COMPLEX NUMBERS; DE MOIVRE’S THEOREM 8.8 POLAR EQUATIONS AND GRAPHS • Solving Oblique Triangles • Solving Oblique Triangles • The Area of a Triangle (SAS Case) • The Area of a Triangle (SSS Case) • Vectors: Magnitude and Direction • Vector Operations • Horizontal and Vertical • Components of a Vector • Unit Vectors • Resultant Vectors • Multiplying Two Vectors: The Dot Product • Angle Between Two Vectors • Work • Complex Numbers in Rectangular Form • Complex Numbers in Polar Form • Products of Complex Numbers • Quotients of Complex Numbers • Powers of Complex Numbers • Roots of Complex Numbers • Polar Coordinates • Converting Between Polar and Rectangular Coordinates • Graphs of Polar Equations 766 CHAPTER 8 Additional Topics in Trigonometry 8.1.1 Solving Oblique Triangles Thus far we have discussed only right triangles. There are, however, two types of triangles, right and oblique. An oblique triangle is any triangle that does not have a right angle. An oblique triangle is either an acute triangle, having three acute (less than 908) angles, or an obtuse triangle, having one obtuse (between 908 and 1808) angle. It is customary to label oblique triangles the following way: ■ ■angle a (alpha) opposite side a. ■ ■angle b (beta) opposite side b. ■ ■angle g (gamma) opposite side c. Remember that the sum of the three angles of any triangle must equal 1808. Recall that in Section 6.3 we solved right triangles. In this chapter we solve oblique triangles, which means we find the lengths of all three sides and the measures of all three angles. α β γ α + β + γ = 180º a b c Four Cases To solve an oblique triangle, we need to know the length of one side and one of the following three: ■ ■two angles ■ ■one angle and another side ■ ■the other two sides This requirement leads to the following four possible cases to consider: Oblique Triangles Acute Triangles TRIANGLES Right Triangles Obtuse Triangles 8.1.1 SKI LL Solve the AAS, ASA, and SSA right triangles using the Law of Sines. 8.1.1 CO NCE PTUAL Understand why an AAA case cannot be solved. S K I L L S O B J E C T I V E ■ ■Solve the AAS, ASA, and SSA right triangles using the Law of Sines. C O N C E P T U A L O B J E C T I V E ■ ■Understand why an AAA case cannot be solved. 8.1 OBLIQUE TRIANGLES AND THE LAW OF SINES 8.1 Oblique Triangles and the Law of Sines 767 Notice that there is no AAA case because two similar triangles can have the same angle measures but different side lengths. That is why at least one side must be known. In this section, we will derive the Law of Sines, which will enable us to solve Case 1 and Case 2 problems. In the next section, we will derive the Law of Cosines, which will enable us to solve Case 3 and Case 4 problems. REQUIRED INFORMATION TO SOLVE OBLIQUE TRIANGLES CASE WHAT’S GIVEN EXAMPLES/NAMES Case 1 One side and two angles AAS: Angle-Angle-Side ASA: Angle-Side-Angle Case 2 Two sides and the angle opposite one of them SSA: Side-Side-Angle Case 3 Two sides and the angle between them SAS: Side-Angle-Side Case 4 Three sides SSS: Side-Side-Side α β a b c α β a b STUDY TIP To solve triangles, at least one side must be known. β a b γ a b a b c 768 CHAPTER 8 Additional Topics in Trigonometry The Law of Sines Let us start with two oblique triangles, an acute triangle and an obtuse triangle. α β γ Acute Triangle a b c α β γ Obtuse Triangle a b c The following discussion applies to both triangles. First, construct an altitude (perpendicular) h from the vertex at angle g to the side (or its extension) opposite g. α β γ Acute Triangle a b h c α β γ Obtuse Triangle a b h c 180º – α WORDS MATH Formulate sine ratios for an acute triangle. sin a 5 h b and sin b 5 h a Formulate sine ratios for an obtuse triangle. sin1180° 2 a2 5 h b and sin b 5 h a For the obtuse triangle, apply sin1180° 2 a2 5 sin 180° cos a 2 cos 180° sin a the sine difference identity. 5 0⋅cos a 2 1212 sin a 5 sin a Therefore, in both triangles we find the same equations. sin a 5 h b and sin b 5 h a Solve for h in both equations. h 5 b sin a and h 5 a sin b Since h is equal to itself, equate the expressions for h. b sin a 5 a sin b Divide both sides by ab. b sin a ab 5 a sin b ab Divide out common factors. sin a a 5 sin b b In a similar manner, we can extend an altitude (perpendicular) from angle a, and we will find that sin g c 5 sin b b . Equating these two expressions leads us to the third ratio of the Law of Sines: sin a a 5 sin g c . THE LAW OF SINES For a triangle with sides a, b, and c and opposite angles a, b, and g, the following is true: sin a a 5 sin b b 5 sin g c In other words, the ratio of the sine of an angle in a triangle to its opposite side is equal to the ratios of the sines of the other two angles to their opposite sides. Some things to note before we begin solving oblique triangles are: ■ ■The angles and sides share the same progression of magnitude: • The longest side of a triangle is opposite the largest angle. • The shortest side of a triangle is opposite the smallest angle. ■ ■Draw the triangle and label the angles and sides. ■ ■If two angles are known, start by determining the third angle. ■ ■Whenever possible, in successive steps always return to given values rather than refer to calculated (approximate) values. Keeping these pointers in mind will help you determine whether your answers are reasonable. Case 1: Two Angles and One Side (AAS or ASA) STUDY TIP The longest side is opposite the largest angle; the shortest side is opposite the smallest angle. STUDY TIP Always use given values rather than calculated (approximated) values for better accuracy. EXAMPLE 1 Using the Law of Sines to Solve a Triangle (AAS) Solve the triangle. γ = 33º α = 110º β a = 7.0 m b c Solution: This is an AAS (angle-angle-side) case because two angles and a side are given and the side is opposite one of the angles. STEP 1 Find b. The sum of the measures of the angles in a triangle is 180°. a 1 b 1 g 5 180° Let a 5 110° and g 5 33°. 110° 1 b 1 33° 5 180° Solve for b. b 5 37° 8.1 Oblique Triangles and the Law of Sines 769 770 CHAPTER 8 Additional Topics in Trigonometry STEP 2 Find b. Use the Law of Sines with the known side, a. sin a a 5 sin b b Isolate b. b 5 a sin b sin a Let a 5 110°, b 5 37°, and a 5 7 m. b 5 7 sin 37° sin 110° Use a calculator to approximate b. b 5 4.483067 m Round b to two significant digits. b < 4.5 m STEP 3 Find c. Use the Law of Sines with the known side, a. sin a a 5 sin g c Isolate c. c 5 a sin g sin a Let a 5 110°, g 5 33°, and a 5 7 m. c 5 7 sin 33° sin 110° Use a calculator to approximate c. c 5 4.057149 m Round c to two significant digits. c < 4.1 m STEP 4 Draw and label the triangle. Y OUR TU R N Solve the triangle (round sides to the nearest foot and angles to the nearest degree). γ = 33º α = 110º β = 37º a = 7.0 m b ≈ 4.5 m c ≈ 4.1 m γ α = 105º β = 43º a b = 30 ft c ▼ ▼ A N S W E R g 5 32°, a < 42 ft, c < 23 ft γ = 32º α = 105º β = 43º a ≈ 42 ft b ≈ 30 ft c ≈ 23 ft STUDY TIP Notice in Step 3 that we used a which is given, as opposed to b which has been calculated (approximated). ▼ A N S W E R a 5 35°, b < 21 in., c < 18 in. c 18 in. a = 12 in. b 21 in. α = 35º γ = 60º β = 85º [CONCEPT CHECK] TRUE OR FALSE If you know all three angles (AAA), then you can solve the triangle using the Law of Sines. ANSWER False ▼ EXAMPLE 2 Using the Law of Sines to Solve a Triangle (ASA) Solve the triangle. Solution: This is an ASA (angle-side-angle) case because two angles and a side are given and the side is not opposite one of the angles. STEP 1 Find g. The sum of the measures of the angles in a triangle is 1808. a 1 b 1 g 5 180° Let a 5 80° and b 5 32°. 80° 1 32° 1 g 5 180° Solve for g. g 5 68° STEP 2 Find b. Write the Law of Sines to include the known side c. sin b b 5 sin g c Isolate b. b 5 c sin b sin g Let b 5 32°, g 5 68°, and c 5 17 miles. b 5 17 sin 32° sin 68° Use a calculator to approximate b. b 5 9.7161177 Round b to two significant digits. b < 9.7 mi STEP 3 Find a. Write the Law of Sines again incorporating the known side c. sin a a 5 sin g c Isolate a. a 5 c sin a sin g Let a 5 80°, g 5 68°, and c 5 17 miles. a 5 17 sin 80° sin 68° Use a calculator to approximate a. a 5 18.056539 Round a to two significant digits. a < 18mi STEP 4 Draw and label the triangle. Y OUR T UR N Solve the triangle. c = 17 mi a b α = 80º γ β = 32º ▼ c = 17 mi a 18 mi b 9.7 mi α = 80º γ = 68º β = 32º c a = 12 in. b α γ = 60º β = 85º 8.1 Oblique Triangles and the Law of Sines 771 772 CHAPTER 8 Additional Topics in Trigonometry Case 2 (Ambiguous Case): Two Sides and One Angle (SSA) If we are given two sides and an angle opposite one of the sides, then we call that Case 2, SSA (side-side-angle). This case is called the ambiguous case because the given information by itself can represent one triangle, two triangles, or no triangle at all. If the angle given is acute, then the possibilities are zero, one, or two triangles. If the angle given is obtuse, then the possibilities are zero or one triangle. The possibilities come from the fact that sin a 5 k, where 0 , k , 1, has two solutions for a: one in quadrant I (acute angle) and one in quadrant II (obtuse angle). In the figure on the left, note that ■ ■h 5 b sin a by the definition of the sine ratio; since 0 , sin a , 1, then h , b. ■ ■a may turn out to be smaller than, equal to, or larger than h. Given Angle (a) is Acute CONDITION PICTURE NUMBER OF TRIANGLES 0 , a , h sin b . 1 No Triangle α β b c a h 0 a 5 h sin b 5 1 Right Triangle α β γ b c a = h 1 h , a , b 0 , sin b , 1 Acute Triangle α β γ b c a h Obtuse Triangle α γ β b c a h 2 a $ b 0 , sin b , 1 Acute Triangle α β γ b c a h 1 b is the angle opposite side b. Given Angle (a) is Obtuse CONDITION PICTURE NUMBER OF TRIANGLES a # b sin b $ 1 No Triangle α β γ b c h a 0 a . b 0 , sin b , 1 One Triangle α β γ a b c h 1 α β b a h γ STUDY TIP Notice that if the angle given is obtuse, the side opposite that angle must be longer than the other given side (longest side opposite the largest angle). EXAMPLE 3 Solving the Ambiguous Case (SSA)—One Triangle Solve the triangle a 5 23 ft, b 5 11 ft, and a 5 122°. Solution: This is the ambiguous case because two sides and an angle opposite one of those sides are given. Since the given angle a is obtuse and a . b, we expect one triangle. STEP 1 Find b. Use the Law of Sines. sin a a 5 sin b b Isolate sin b. sin b 5 b sin a a Let a 5 23 ft, b 5 11 ft, and a 5 122°. sin b 5 111 ft2 sin 122° 23 ft Use a calculator to evaluate sin b. sin b 5 0.40558822 Solve for b using the inverse sine function. b 5 sin2110.405588222 Round to the nearest degree. b < 24° Note: b1 < 248 and there is a corresponding b2 < 1568, but there is no corresponding second triangle because a 1 b2 . 1808. STEP 2 Find g. The measures of angles in a triangle sum to 180°. a 1 b 1 g 5 180° Substitute a 5 122° and b < 24°. 122° 1 24° 1 g 5 180° Solve for g. g < 34° STEP 3 Find c. Use the Law of Sines. sin a a 5 sin g c Isolate c. c 5 a sin g sin a Substitute a 5 23 ft, a 5 122°, c 5 123 ft2 sin 34° sin 122° and g < 34°. Use a calculator to evaluate c. c < 15 ft STEP 4 Draw and label the triangle. Y OUR T UR N Solve the triangle a 5 133°, a 5 48 mm, and c 5 17 mm. c ≈ 15 ft a = 23 ft b = 11 ft α = 122º γ ≈ 34º β ≈ 24º ▼ ▼ A N S W E R b 5 32°, g 5 15°, b 5 35 mm c = 17 mm a = 48 mm b = 35 mm α = 133º γ = 15º β = 32º 8.1 Oblique Triangles and the Law of Sines 773 774 CHAPTER 8 Additional Topics in Trigonometry EXAMPLE 4 Solving the Ambiguous Case (SSA)—Two Triangles Solve the triangle a 5 8.1 m, b 5 8.3 m, and a 5 72°. Solution: This is the ambiguous case because two sides and an angle opposite one of those sides are given. Since the given angle a is acute and a , b, we expect two triangles. STEP 1 Find b. Write the Law of Sines for the given information. sin a a 5 sin b b Isolate sin b. sin b 5 b sin a a Let a 5 8.1 m, b 5 8.3 m, and a 5 72°. sin b 5 18.3 m2 sin 72° 8.1m Use a calculator to evaluate sin b. sin b < 0.974539393 Solve for b using the inverse sine function. Note that b can be acute or obtuse. b < sin2110.9745393932 This is the quadrant I solution 1b is acute). b1 < 77° The quadrant II (b is obtuse) solution is b2 5 180 2 b1. b2 < 103° STEP 2 Find g. The measures of the angles in a triangle sum to 180°. a 1 b 1 g 5 180° Solve for Triangle 1. Substitute a 5 72° and b1 < 77°. 72° 1 77° 1 g1 5 180° Solve for g1. g1 < 31° Solve for Triangle 2. Substitute a 5 72° and b2 < 103°. 72° 1 103° 1 g2 5 180° Solve for g2. g2 < 5° STEP 3 Find c. Use the Law of Sines. sin a a 5 sin g c Isolate c. c 5 a sin g sin a Solve for Triangle 1. Substitute a 5 8.1 m, a 5 72°, and g1 < 31°. c1 < 18.1 m2 sin 31° sin 72° Use a calculator to evaluate c1. c1 < 4.4 m Solve for Triangle 2. Substitute a 5 8.1 m, a 5 72°, and g2 < 5°. c2 5 18.1 m2 sin 5° sin 72° Use a calculator to evaluate c2. c2 < 0.74 m Applications Solving oblique triangles is rich with applications in astronomy, surveying, aircraft design, piloting, and many other areas. STEP 4 Draw and label the two triangles. c ≈ 4.4 m a = 8.1 m b = 8.3 m α = 72º γ = 31º β = 77º c ≈ 0.74 m a = 8.1 m b = 8.3 m α = 72º γ = 5º β = 103º EXAMPLE 5 Solving the Ambiguous Case (SSA)—No Triangle Solve the triangle a 5 107°, a 5 6, and b 5 8. Solution: This is the ambiguous case because two sides and an angle opposite one of those sides are given. Since the given angle a is obtuse and a , b, we know that there is no triangle. Notice that if we did not realize that there is no triangle and we had proceeded with the Law of Sines, the result would have been the sine function exceeding 1—which is not possible. Write the Law of Sines. sin a a 5 sin b b Isolate sin b. sin b 5 b sin a a Let a 5 107°, a 5 6, and b 5 8. sin b 5 8 sin 107° 6 Use a calculator to evaluate sin b. sin b < 1.28 Since the range of the sine function is 321, 14, there is no angle b such that sin b < 1.28. Therefore, there is no triangle with the given measurements. EXAMPLE 6 How Far Over Is the Tower of Pisa Leaning? The Tower of Pisa was originally built 56 meters tall. Because of poor soil in the foundation, it started to lean. At a distance 44 meters from the base of the tower, the angle of elevation is 55°. How much is the Tower of Pisa leaning away from the vertical position? 44 m 56 m 55º “?” Nathalie Michel/Photodisc/Getty Images, Inc. 8.1 Oblique Triangles and the Law of Sines 775 776 CHAPTER 8 Additional Topics in Trigonometry Solution: b 5 55°, c 5 44 m, and b 5 56 m is the given information, so this is an SSA problem. There is only one triangle because b . c. STEP 1 Find g. Use the Law of Sines. sin b b 5 sin g c Isolate sin g. sin g 5 c sin b b Let b 5 55°, c 5 44 m, sin g 5 44 sin 55° 56 and b 5 56 m. Evaluate the right side using a calculator. sin g < 0.643619463 Solve for g using the inverse sine function. g < sin2110.6436194632 Round to two significant digits. g < 40° STEP 2 Find a. The measures of angles in a triangle sum to 180°. a 1 b 1 g 5 180° Let b 5 55° and g < 40°. a 1 55° 1 40° < 180° Solve for a a < 85° The Tower of Pisa makes an angle of 85° with the ground. It is leaning at an angle of 5° . a c = 44 m b = 56 m γ β = 55º α [SEC TION 8.1] S U M M A RY In this section, we solved oblique triangles. When given three pieces of informatio n about a triangle, we classify the triangle according to the data (sides and angles). Four cases arise: ■ ■one side and two angles (AAS or ASA) ■ ■two sides and one angle opposite one of the sides (SSA) ■ ■two sides and one angle between sides (SAS) ■ ■three sides (SSS) The Law of Sines sin a a 5 sin b b 5 sin g c can be used to solve the first two cases (AAS or ASA, and SSA). It is important to note that the SSA case is called the ambiguous case because any one of three results is possible: no triangle, one triangle, or two triangles. In Exercises 1–6, classify each triangle problem as AAS, ASA, SAS, SSA, or SSS on the basis of the given information. 1. c, a, and a 2. c, a, and g 3. a, b, and c 4. a, b, and g 5. a, b, and c 6. b, g, and a In Exercises 7–16, solve the given triangles. 7. a 5 45°, b 5 60°, a 5 10 m 8. b 5 75°, g 5 60°, b 5 25 in. 9. a 5 46°, g 5 72°, b 5 200 cm 10. g 5 100°, b 5 40°, a 5 16 ft 11. a 5 16.3°, g 5 47.6°, c 5 211 yd 12. b 5 104.2°, g 5 33.6°, a 5 26 in. 13. a 5 30°, b 5 30°, c 5 12 m 14. a 5 45°, g 5 75°, c 5 9 in. 15. b 5 26°, g 5 57°, c 5 100 yd 16. a 5 80°, g 5 30°, b 5 3 ft In Exercises 17–34, two sides and an angle are given. Determine whether a triangle (or two) exist, and if so, solve the triangle(s). 17. a 5 4, b 5 5, a 5 16° 18. b 5 30, c 5 20, b 5 70° 19. a 5 12, c 5 12, g 5 40° 20. b 5 111, a 5 80, a 5 25° 21. a 5 21, b 5 14, b 5 100° 22. a 5 13, b 5 26, a 5 120° 23. a 5 30°, b 5 18, a 5 9 24. a 5 45°, b 5 !2, a 5 1 25. a 5 34°, b 5 7, a 5 10 26. a 5 71°, b 5 5.2, a 5 5.2 27. a 5 21.3°, b 5 6.18, a 5 6.03 28. a 5 47.3°, b 5 7.3, a 5 5.32 29. a 5 116°, b 5 4!3, a 5 5!2 30. a 5 51°, b 5 4!3, a 5 4!5 31. b 5 500, c 5 330, g 5 40° 32. b 5 16, a 5 9, b 5 137° 33. a 5 !2, b 5 !7, b 5 106° 34. b 5 15.3, c 5 27.2, g 5 11.6° [SEC TION 8.1] E X E R C I S E S • S K I L L S α β γ α + β + γ = 180º a b c For Exercises 35 and 36, refer to the following: • A P P L I C A T I O N S 195 ft 10º 1º a b Wire NASA Kennedy Space Center On the launch pad at Kennedy Space Center, when the Space Shuttle was in operation, there was an escape basket that could hold four astronauts. The basket slides down a wire that is attached 195 feet high, above the base of the launch pad. The angle of inclination measured from where the basket would touch the ground to the base of the launch pad is 1°, and the angle of inclination from that same point to where the wire is attached is 10°. 35. NASA. How long is the wire a? 36. NASA. How far from the launch pad does the basket touch the ground b? 37. Hot-Air Balloon. A hot-air balloon is sighted at the same time by two friends who are 1 mile apart on the same side of the balloon. The angles of elevation from the two friends are 20.5° and 25.5°. How high is the balloon? 20.5º 25.5º 1 mi 8.1 Oblique Triangles and the Law of Sines 777 778 CHAPTER 8 Additional Topics in Trigonometry 38. Hot-Air Balloon. A hot-air balloon is sighted at the same time by two friends who are 2 miles apart on the same side of the balloon. The angles of elevation from the two friends are 10° and 15°. How high is the balloon? 39. Rocket Tracking. A tracking station has two telescopes that are 1 mile apart. The telescopes can lock onto a rocket after it is launched and record the angles of elevation to the rocket. If the angles of elevation from telescopes A and B are 30° and 80°, respectively, then how far is the rocket from telescope A? U S A 80º 30º 1 mi B A 40. Rocket Tracking. Given the data in Exercise 39, how far is the rocket from telescope B? 41. Distance Across River. An engineer wants to construct a bridge across a fast-moving river. Using a straight-line segment between two points that are 100 feet apart along his side of the river, he measures the angles formed when sighting the point on the other side where he wants to have the bridge end. If the angles formed at points A and B are 65° and 15°, respectively, how far is it from point A to the point on the other side of the river? Round to the nearest foot. 65º 15º B A 100 ft 42. Distance Across River. Given the data in Exercise 41, how far is it from point B to the point on the other side of the river? Round to the nearest foot. 43. Lifeguard Posts. Two lifeguard chairs, labeled P and Q, are located 400 feet apart. A troubled swimmer is spotted by both lifeguards. If the lifeguard at P reports the swimmer at angle 358 (with respect to the line segment connecting P and Q) and the lifeguard at Q reports the swimmer at angle 418, how far is the swimmer from P? 44. Rock Climbing. A rock climbing enthusiast is creating a climbing route rated as 5.8 level (i.e., medium difficulty) on the wall at the local rock gym. Given the difficulty of the route, he wants to avoid placing any two holds on the same vertical or horizontal line on the wall. If he places holds at P, Q, and R such that lQPR 5 408, QR 5 6 feet, and QP 5 4.5 feet, how far is the hold at P from the hold at R? 45. Tennis. Two friends playing tennis are both 70 inches tall. After a rather long rally, Player II lobs the ball into Player I’s court, enabling him to hit an overhead smash such that the angle between the racquet head (at the point of contact with the ball) and his body is 568. The ball travels 20.3 feet to the other side of the court where Player II, with lightning-quick reflexes, volleys the ball 3 inches off the ground at an angle a such that the ball travels directly toward Player I’s head. The ball travels 19.4 feet during this return. Find angle a. 46. Tennis. Shocked by the move Player I made in Exercise 45, Player II is forced to quickly deflect the ball straight back directly over the middle of the net (which is 33 inches high) with great speed. Player I reaches behind himself and is able to contact the ball at the same height above the ground with which Player I initially hit it. If the angle between Player I’s racquet position at the end of the previous shot (i.e., a from Exercise 45) and its current position at the point of contact of this shot is 1308, and the angle with which it contacts the ball is 258, how far has the ball traveled horizontally as a result of Player II’s hit? 47. Archery. (Recall Exercise 73 in Section 6.1 for context.) An expert archer decides to shoot two arrows simultaneously in such a way that the two arrows hit directly above and below the bullseye where the blue and black rings meet (8 inches from the bullseye). If the angle each arrow makes with the target is 848, how far did each arrow travel? 48. Archery. An archer fires two arrows simultaneously toward the target. The topmost arrow hits the bullseye dead center. If this arrow traveled 74 feet, and the second arrow hit the target at an angle somewhere between 88.38 and 88.78, what was the archer’s score? 49. Bowling. The 6-8 split is common in bowling. To make this split, a bowler stands dead center and throws the ball hard and straight directly toward the right of the 6 pin. The distance from the ball at the point of release and the 8 pin is 63.2 feet. See the diagram. How far does the ball travel if it hits the 6 pin correctly? 72º 16º Bowler 63.2 ft 8 pin 6 pin 50. Bowling. A bowler is said to get a strike on the “Brooklyn side” of the head pin if he hits the head pin on the opposite side of the pocket. (For a right-handed bowler, the pocket is to the right of the head pin.) There is a small range of angle at which the ball must contact the head pin in order to convert all of the pins. If the measurements are as shown, how far does the ball travel (assuming it is thrown straight with no hook) before it contacts the head pin? For Exercises 51 and 52, refer to the following: To quantify the torque (rotational force) of the elbow joint of a human arm (see the figure below), it is necessary to identify angles A, B, and C as well as lengths a, b, and c. Measurements performed on an arm determine that the measure of angle C is 958, the measure of angle A is 828, and the length of the muscle a is 23 centimeters. b A B C c a Joint Muscle C 51. Health/Medicine. Find the length of the forearm from the elbow joint to the muscle attachment b. 52. Health/Medicine. Find the length of the upper arm from the muscle attachment to the elbow joint c. • C A T C H T H E M I S T A K E In Exercises 53 and 54, explain the mistake that is made. 53. Solve the triangle a 5 120°, a 5 7, and b 5 9. Solution: Use the Law of Sines to find b. sin a a 5 sin b b Let a 5 120°, a 5 7, sin 120° 7 5 sin b 9 and b 5 9. Solve for sin b. sin b 5 1.113 Solve for b. b 5 42° Sum the angle measures to 180°. 120° 1 42° 1 g 5 180° Solve for g. g 5 18° Use the Law of Sines to sin a a 5 sin g c find c. Let a 5 120°, a 5 7, sin 120° 7 5 sin 18° c and g 5 18°. Solve for c. c 5 2.5 a 5 120°, b 5 42°, g 5 18°, a 5 7, b 5 9, and c 5 2.5. This is incorrect. The longest side is not opposite the longest angle. There is no triangle that makes the original measure-ments work. What mistake was made? 54. Solve the triangle a 5 40°, a 5 7, and b 5 9. Solution: Use the Law of Sines to find b. sin a a 5 sin b b Let a 5 40°, a 5 7, sin 40° 7 5 sin b 9 and b 5 9. Solve for sin b. sin b 5 0.826441212 Solve for b. b 5 56° Find g. 40° 1 56° 1 g 5 180° g 5 84° Use the Law of Sines to find c. sin a a 5 sin g c Let a 5 40°, a 5 7, and g 5 84°. sin 40° 7 5 sin 84° c Solve for c. c 5 11 a 5 40°, b 5 56°, g 5 84°, a 5 7, b 5 9, and c 5 11. This is incorrect. What mistake was made? In Exercises 55–60, determine whether each statement is true or false. • C O N C E P T U A L 55. The Law of Sines applies only to right triangles. 56. If you are given two sides and any angle, there is a unique solution for the triangle. 57. An acute triangle is an oblique triangle. 58. An obtuse triangle is an oblique triangle. 59. If you are given two sides that have the same length, then there can be at most one triangle. 60. If a is obtuse and b 5 a 2, then the situation is unambiguous. 8.1 Oblique Triangles and the Law of Sines 779 780 CHAPTER 8 Additional Topics in Trigonometry 61. Mollweide’s Identity. For any triangle, the following identity is true. It is often used to check the solution of a triangle since all six pieces of information (three sides and three angles) are involved. Derive the identity using the Law of Sines. 1a 1 b2 sin A1 2 gB 5 c cosC1 21a 2 b2D 63. Use the Law of Sines to prove that all angles in an equilateral triangle must have the same measure. 62. The Law of Tangents. Use the Law of Sines and trigonometric identities to show that for any triangle, the following is true: a 2 b a 1 b 5 tanC1 21a 2 b2D tanC1 2 1a 1 b2D 64. Suppose that you have a triangle with side lengths a, b, and c, and angles a, b, and g, respectively, directly across from them. If it is known that a 5 1 !2 b, c 5 2, a is an acute angle, and b 5 2a, solve the triangle. • C H A L L E N G E 8.2.1 S KILL Solve SAS and the SSS triangles using the Law of Cosines. 8.2.1 CO NCE PTUAL Understand that the Pythagorean theorem is a special case of the Law of Cosines. 8.2.1 Solving Oblique Triangles In Section 8.1, we learned that to solve oblique triangles means to find all three side lengths and angle measures. At least one side length must be known. We need two additional pieces of information to solve a triangle (combinations of side lengths and/ or angles). We found that there are four cases: ■ ■Case 1: AAS or ASA (two angles and a side are given) ■ ■Case 2: SSA (two sides and an angle opposite one of the sides are given) ■ ■Case 3: SAS (two sides and the angle between them are given) ■ ■Case 4: SSS (three sides are given) We used the Law of Sines to solve Case 1 and Case 2 triangles. Now, we use the Law of Cosines to solve Case 3 and Case 4 triangles. A B C X Z Y 65. A 5 10, Y 5 40°, and Z 5 72° 66. B 5 42.8, X 5 31.6°, and Y 5 82.2° 67. A 5 22, B 5 17, and X 5 105° 68. B 5 16.5, C 5 9.8, and Z 5 79.2° 69. A 5 25.7, C 5 12.2, and X 5 65° 70. A 5 54.6, B 5 12.9, and Y 5 23° For Exercises 65–70, let A, B, and C be the lengths of the three sides with X, Y, and Z as the corresponding angles. Write a program to solve the given triangle with a calculator. • T E C H N O L O G Y S K I L L S O B J E C T I V E ■ ■Solve SAS and the SSS triangles using the Law of Cosines. C O N C E P T U A L O B J E C T I V E ■ ■Understand that the Pythagorean theorem is a special case of the Law of Cosines. 8.2 THE LAW OF COSINES WORDS MATH Start with a triangle. Drop a perpendicular line from g with height h. The result is two triangles within the larger triangle. Write the Pythagorean theorem for both right triangles. Triangle 1: x2 1 h2 5 b2 Triangle 2: 1c 2 x22 1 h2 5 a2 Solve for h2. Triangle 1: h2 5 b2 2 x2 Triangle 2: h2 5 a2 2 1c 2 x22 Since the segment of length h is shared, set h2 5 h2, for the two triangles. b2 2 x2 5 a2 2 1c 2 x22 Multiply out the squared binomial on the right. b2 2 x2 5 a2 2 1c2 2 2cx 1 x22 Eliminate the parentheses. b2 2 x2 5 a2 2 c2 1 2cx 2 x2 Add x2 to both sides. b2 5 a2 2 c2 1 2cx Isolate a2. a2 5 b2 1 c2 2 2cx Notice that cos a 5 x b. Let x 5 b cosa. a2 5 b2 1 c2 2 2bc cos a Note: If we instead drop the perpendicular line with length h from the angle a or the angle b, we can derive the other two parts of the Law of Cosines: b2 5 a2 1 c2 2 2ac cos b and c2 5 a2 1 b2 2 2ab cos g α β γ a b c α β a b h c 1 2 c – x x [CONCEPT CHECK] TRUE OR FALSE The Pythagorean theorem only applies to right triangles, whereas the Law of Sines and Law of Cosines can be applied to acute and obtuse triangles. ANSWER True ▼ THE LAW OF COSINES For a triangle with sides a, b, and c and opposite angles a, b, and g, the following are true: a2 5 b2 1 c2 2 2bc cosa b2 5 a2 1 c2 2 2ac cosb c2 5 a2 1 b2 2 2 ab cosg 8.2 The Law of Cosines 781 782 CHAPTER 8 Additional Topics in Trigonometry It is important to note that the Law of Cosines can be used to find side lengths or angles. As long as three of the four variables in any of the equations are known, the fourth can be calculated. Notice that in the special case of a right triangle 1say, g 5 90°2, c2 5 a2 1 b2 2 2ab cos 90° 0 one of the components of the Law of Cosines reduces to the Pythagorean theorem: c2 5 a2 1 b2 hyp leg leg The Pythagorean theorem can thus be regarded as a special case of the Law of Cosines. Case 3: Solving Oblique Triangles (SAS) We now solve SAS triangle problems where the angle between two sides is given. We start by using the Law of Cosines to solve for the side opposite the given angle. We then can apply either the Law of Sines or the Law of Cosines to find the second angle. α β γ a b c e e e e EXAMPLE 1 Using the Law of Cosines to Solve a Triangle (SAS) Solve the triangle a 5 13, c 5 6.0, and b 5 20°. Solution: Two sides and the angle between them are given (SAS). Notice that the Law of Sines can’t be used because it requires knowledge of at least one angle and the side opposite that angle. STEP 1 Find b. Apply the Law of Cosines that involves b. b2 5 a2 1 c2 2 2ac cos b Let a 5 13, c 5 6.0, and b 5 20°. b2 5 132 1 62 2 21132162 cos 20° Evaluate the right side with a calculator. b2 < 58.40795 Solve for b. b < 67.6425 Round to two significant digits; b can only be positive. b < 7.6 STEP 2 Find g. Use the Law of Sines. sin g c 5 sin b b Isolate sin g. sin g 5 c sin b b Let b 5 7.6, c 5 6.0, and b 5 20°. sin g 5 6 sin 20° 7.6 Apply the inverse sine function. g 5 sin21a6 sin 20° 7.6 b Evaluate the right side with a calculator. g < 15.66521° Round to the nearest degree. g < 16° c = 6.0 a = 13 b α γ β = 20º STUDY TIP The Pythagorean theorem is a special case of the Law of Cosines. Notice the steps we took in solving an SAS triangle: 1. Find the side opposite the given angle using the Law of Cosines. 2. Solve for the smaller angle using the Law of Sines. 3. Solve for the larger angle using properties of triangles. You may be thinking, “Would it matter if we had solved for a before solving for g?” Yes, it does matter— in this problem you cannot solve for a by the Law of Sines before finding g. The Law of Sines can be used only on the smaller angle (opposite the shortest side). If we had tried to use the Law of Sines with the obtuse angle a, the inverse sine would have resulted in a 5 36°. Since the sine function is positive in both quadrants I and II, we would not know whether that angle was a 5 36° or its supplementary angle a 5 144°. Notice that c , a; therefore, the angles opposite those sides must have the same relationship, g , a. We choose the smaller angle first. Alternatively, if we want to solve for the obtuse angle first, we can use the Law of Cosines to solve for a. If you use the Law of Cosines to find the second angle, you can choose either angle. The Law of Cosines can be used for either acute or obtuse angles. ▼ A N S W E R a 5 2.9, g 5 21°, b 5 124° STEP 3 Find a. The angle measures must sum to 180°. a 1 20° 1 16° < 180° Solve for a. a < 144° Y OUR T UR N Solve the triangle b 5 4.2, c 5 1.8, and a 5 35°. ▼ STUDY TIP Although the Law of Sines is sometimes ambiguous, the Law of Cosines is never ambiguous. EXAMPLE 2 Using the Law of Cosines in an Application (SAS) In an AKC (American Kennel Club)-sanctioned field trial, a judge sets up a mark (bird) to which the dog is required to swim (the dogs are judged on how closely they adhere to the straight line to the bird, not the time it takes to retrieve the bird). The judge is trying to calculate how far a swim this mark is for the dogs, so she walks off the two legs across the land and measures the angle as shown in the figure. How far will the dog swim from the starting line to the bird? Solution: Label the triangle. Apply the Law of Cosines. c2 5 a2 1 b2 2 2ab cos g Let a 5 176 yd, b 5 152 yd, and g 5 117°. c2 5 1762 1 1522 2 21176211522 cos 117° Simplify. c2 < 78370.3077 Solve for c and round to the nearest yard. c < 280 yd 152 yd Pond 176 yd 117º b = 152 yd Pond a = 176 yd γ = 117º c 8.2 The Law of Cosines 783 784 CHAPTER 8 Additional Topics in Trigonometry Case 4: Solving Oblique Triangles (SSS) We now solve oblique triangles when all three side lengths are given (the SSS case). In this case, start by finding the largest angle (opposite the longest side) using the Law of Cosines. Then apply the Law of Sines to find either of the remaining two angles. Last, find the third angle using the sum of the angles in a triangle. EXAMPLE 3 Using the Law of Cosines to Solve a Triangle (SSS) Solve the triangle a 5 8, b 5 6, and c 5 7. Solution: STEP 1 Identify the largest angle, which is a. Write the component of the Law of Cosines that involves a. a2 5 b2 1 c2 2 2bc cos a Let a 5 8, b 5 6, and c 5 7. 82 5 62 1 72 2 2162172 cos a Simplify and isolate cos a. cos a 5 62 1 72 2 82 2162172 5 0.25 Write the inverse cosine function. a 5 cos2110.252 Approximate with a calculator and round to the nearest degree. a < 76° STEP 2 Find either of the remaining angles to solve for b. Write the Law of Sines. sin a a 5 sin b b Isolate sin b. sin b 5 b sin a a Let a 5 8, b 5 6, and a < 76°. sin b < 6 sin 76° 8 Write the inverse sine function. b < sin21 a6 sin 76° 8 b Approximate with a calculator. b < 47° STEP 3 Find the third angle, g. The sum of the angle measures is 180°. 76° 1 47° 1 g < 180° Solve for g. g < 57° Y OUR TU R N Solve the triangle a 5 5, b 5 7, and c 5 8. ▼ ▼ A N S W E R a < 38.2°, b < 60.0°, g < 81.8° In the next example, instead of immediately substituting in values for the Law of Cosines equation, we will solve for the angle in general and then substitute in values. EXAMPLE 4 Using the Law of Cosines in an Application (SSS) Many people believe that an imaginary area called the Bermuda Triangle exists on the southeastern Atlantic coast of the United States. Assume that the Bermuda Triangle has vertices in Norfolk (Virginia), Bermuda, and Santiago (Cuba). Find the angles of the Bermuda Triangle given the following distances: LOCATION LOCATION DISTANCE (NAUTICAL MILES) Norfolk Bermuda 850 Bermuda Santiago 810 Norfolk Santiago 894 Ignore the curvature of the Earth in your calculations. Solution: Draw and label the triangle. STEP 1 Find b (the largest angle). Apply the Law of Cosines. b2 5 a2 1 c2 2 2ac cos b Isolate cos b. cos b 5 a2 1 c2 2 b2 2ac Apply the inverse cosine function to solve for b. b 5 cos21 aa2 1 c2 2 b2 2ac b Let a 5 810, b 5 894, and c 5 850. b 5 cos21 c8102 1 8502 2 8942 21810218502 d Approximate with a calculator. b < 65° STEP 2 Find a. Apply the Law of Sines. sin a a 5 sin b b Isolate sin a. sin a 5 a sin b b Apply the inverse sine function to solve for a. a 5 sin21 aa sin b b b Let a 5 810, b 5 894, and b 5 65°. a < sin21 a810 sin 65° 894 b Approximate with a calculator, and round to the nearest degree. a < 55° STEP 3 Find g. The angle measures must sum to 180°. 55° 1 65° 1 g < 180° Solve for g. g < 60° BERMUDA VIRGINIA CUBA γ α β c = 850 nm a = 810 nm b = 894 nm 8.2 The Law of Cosines 785 786 CHAPTER 8 Additional Topics in Trigonometry and the angle sum identity, or we apply a combination of the Law of Cosines, a2 5 b2 1 c2 2 2bc cos a b2 5 a2 1 c2 2 2ac cos b c2 5 a2 1 b2 2 2ab cos g the Law of Sines, and the angle sum identity. The table below summarizes the strategies for solving oblique triangles: We can solve any triangle given three pieces of information, as long as one of the pieces is a side length. Depending on the infor-mation given, either we apply the Law of Sines sin a a 5 sin b b 5 sin g c [SEC TION 8. 2] S U M M A RY OBLIQUE TRIANGLE WHAT’S KNOWN PROCEDURE FOR SOLVING AAS or ASA Two angles and a side Step 1: Find the remaining angle with a 1 b 1 g 5 180°. Step 2: Find the remaining sides with the Law of Sines. SSA Two sides and an angle opposite one of the sides This is the ambiguous case, so there is either no triangle, one triangle, or two triangles. If the given angle is obtuse, then there is either one or no triangle. If the given angle is acute, then there is no triangle, one triangle, or two triangles. Step 1: Apply the Law of Sines to find one of the angles. Step 2: Find the remaining angle with a 1 b 1 g 5 180°. Step 3: Find the remaining side with the Law of Sines. If two triangles exist, then the angle found in Step 1 can be either acute or obtuse, and Step 2 and Step 3 must be performed for each triangle. SAS Two sides and an angle between the sides Step 1: Find the third side with the Law of Cosines. Step 2: Find the smaller angle with the Law of Sines. Step 3: Find the remaining angle with a 1 b 1 g 5 180°. SSS Three sides Step 1: Find the largest angle with the Law of Cosines. Step 2: Find either remaining angle with the Law of Sines. Step 3: Find the last remaining angle with a 1 b 1 g 5 180°. For each of the triangles in Exercises 1–8, the angle sum identity, a 1 b 1 g 5 1808, will be used in solving the triangle. Label the problem as S if only the Law of Sines is needed to solve the triangle. Label the problem as C if the Law of Cosines is needed to solve the triangle. 1. a, b, and c are given. 2. a, b, and g are given. 3. a, b, and c are given. 4. a, b, and a are given. 5. a, b, and g are given. 6. a, b, and a are given. 7. b, c, and a are given. 8. b, c, and b are given. In Exercises 9–42, solve each triangle. 9. a 5 4, c 5 3, b 5 100° 10. a 5 6, b 5 10, g 5 80° 11. b 5 7, c 5 2, a 5 16° 12. b 5 5, a 5 6, g 5 170° 13. b 5 5, c 5 5, a 5 20° 14. a 5 4.2, b 5 7.3, g 5 25° 15. a 5 9, c 5 12, b 5 23° 16. b 5 6, c 5 13, a 5 16° 17. a 5 4, c 5 8, b 5 60° 18. b 5 3, c 5 !18, a 5 45° 19. a 5 8, b 5 5, c 5 6 20. a 5 6, b 5 9, c 5 12 21. a 5 4, b 5 4, c 5 5 22. a 5 17, b 5 20, c 5 33 23. a 5 8.2, b 5 7.1, c 5 6.3 24. a 5 1492, b 5 2001, c 5 1776 [SEC TION 8. 2] E X E R C I SE S • S K I L L S 25. a 5 4, b 5 5, c 5 10 26. a 5 1.3, b 5 2.7, c 5 4.2 27. a 5 12, b 5 5, c 5 13 28. a 5 4, b 5 5, c 5 !41 29. a 5 40°, b 5 35°, a 5 6 30. b 5 11.2, a 5 19.0, g 5 13.3° 31. a 5 31°, b 5 5, a 5 12 32. a 5 11, c 5 12, g 5 60° 33. a 5 !7, b 5 !8, c 5 !3 34. b 5 106°, g 5 43°, a 5 1 35. b 5 11, c 5 2, b 5 10° 36. a 5 25°, a 5 6, c 5 9 37. b 5 51.62°, a 5 48 5 , c 5 71 5 38. g 5 61.03°, a 5 22 9 , c 5 22 9 39. a 5 18.21, b 5 21.30, c 5 24.12 40. a 5 32 7 , b 5 29 8 , c 5 21 11 41. a 5 71.213°, b 5 5!2 7 , c 5 3!2 5 42. b 5 38.21°, a 5 !31, c 5 !31 • A P P L I C A T I O N S 43. Aviation. A plane flew due north at 500 mph for 3 hours. A second plane, starting at the same point and at the same time, flew southeast at an angle 150° clockwise from due north at 435 mph for 3 hours. At the end of the 3 hours, how far apart were the two planes? Round to the nearest mile. 150º 1500 mi 1305 mi 45. Aviation. A plane flew 308 NW at 350 mph for 2.5 hours. A second plane, starting at the same point and at the same time, flew 35° at an angle clockwise from due north at 550 mph for 2.5 hours. At the end of 2.5 hours, how far apart were the two planes? Round to the nearest mile. 44. Aviation. A plane flew due north at 400 mph for 4 hours. A second plane, starting at the same point and at the same time, flew southeast at an angle 120° clockwise from due north at 300 mph for 4 hours. At the end of the 4 hours, how far apart were the two planes? Round to the nearest mile. 120º 1600 mi 1200 mi 46. Aviation. A plane flew 308 NW at 350 mph for 3 hours. A second plane starts at the same point and takes off at the same time. It is known that after 3 hours, the two planes are 2100 miles apart. Find the original bearing of the second plane, to the nearest hundredth of a degree. 47. Baseball. A baseball diamond is actually a square that is 90 feet on each side. The pitcher’s mound is located 60.5 feet from home plate. How far is it from third base to the pitcher’s mound? Second Base Pitcher’s Mound First Base ? ? 90 ft 90 ft 90 ft 90 ft 60.5 ft Third Base Home Plate 45º 48. Aircraft Wing. Given the acute angle 114.482 and two sides 118 feet and 25 feet2 of the stealth bomber, what is the unknown length? ? ft 18 ft 25 ft 14.4 Stone/Getty Images, Inc. 8.2 The Law of Cosines 787 788 CHAPTER 8 Additional Topics in Trigonometry 49. Sliding Board. A 40-foot slide leaning against the bottom of a building’s window makes a 55° angle with the building. The angle formed with the building with the line of sight from the top of the window to the point on the ground where the slide ends is 40°. How tall is the window? 40 ft 40º 55º 50. Airplane Slide. An airplane door is 6 feet high. If a slide attached to the bottom of the open door is at an angle of 40° with the ground, and the angle formed by the line of sight from where the slide touches the ground to the top of the door is 45°, how long is the slide? 40º 45º 6 ft For Exercises 51 and 52, refer to the following: To quantify the torque (rotational force) of the elbow joint of a human arm (see the figure to the right), it is necessary to identify angles A, B, and C as well as lengths a, b, and c. Measurements performed on an arm determine that the measure of angle C is 1058, the length of the muscle a is 25.5 centimeters, and the length of the forearm from the elbow joint to the muscle attachment b is 1.76 centimeters. 51. Health/Medicine. Find the length of the upper arm from the muscle attachment to the elbow joint c. 52. Health/Medicine. Find the measure of angle B. For Exercises 53 and 54, refer to the following: The term triangulation is often used to pinpoint a location. A person’s cell phone always uses the closest tower. As the cell phone switches towers, authorities can locate the person’s location based on known distances from each of the towers. Similarly, firefighters can determine the location of a fire from known distances. 53. Triangulation. Two lead firefighters (Beth and Tim) are 300 yards apart, and they estimate their distances to the fire as approximately 150 yards (Beth) and 200 yards (Tim). What angle with respect to their line of sight with each other should each lead their team in order to reach the fire? 54. Triangulation. Two cell phone towers are 100 meters apart. When the cell phone switches from tower A to tower B, it is estimated that the phone is 60 meters from tower A and 50 meters from tower B. Authorities looking to locate the owner of the cell phone will head 50 meters from tower B at what angle (with respect to the line of sight between towers A and B)? 55. Law Enforcement. Two members of a SWAT team and the thief they are to apprehend are positioned as follows: 50º α γ 90 ft 120 ft Building 1 Building 2 Thief Cop 1 Cop 2 When the signal is given, Cop 2 shoots a zipline across to the window where the thief is spotted (at a 50° angle to Building 1) and Cop 1 shines a very bright light directly at the thief. Find the angle g at which Cop 1 holds the light to shine it directly at the thief. Round to the nearest hundredth degree. 56. Law Enforcement. In reference to Exercise 55, what angle does the zipline make with respect to Building 2? 57. Rock Climbing. (See Exercise 44 in Section 8.1 for the context.) In order to construct a more advanced route, holds are placed at P and R very close to each other, and a third hold is placed at Q at quite a distance away. Moving from either P or R will require a “dynamo” move! If the person setting the route wants PQ 5 7 feet, PR 5 1 foot, and lRPQ 5 258, find the angle between other holds on the route. 58. Bowling. Consider the following diagram: 21 in. 35º (P) Player (C) Center (H) Head pin (T) 10 pin 60 ft 63 ft From the diagram, we know that the ball is released at an angle of 358 to the line connecting the player to the 10 pin. Find the angle at which the ball hits the head pin, assuming the ball is thrown straight with no hook. Round to the nearest hundredth degree. 59. Bowling. Next, a bowler must convert a 7-8 spare and must hit the 8 pin directly on its right side. If the distance between the bowler and 7 pin is 63 feet, the distance between the bowler and 8 pin is 62.8 feet, and the distance between the b A C c B a Muscle Joint 7 and 8 pins is 7 inches, find the angle at which the ball is released (i.e., angle between the segment connecting the bowler to the 7 pin and the segment connecting the bowler to the 8 pin). 60. Surveying. A glaciologist needs to determine the length across a certain crevice on Mendenhall glacier in order to circumvent it with her team. She has the following measurements: 52º 480 yd 500 yd T (Team) Crevice P1 P2 Find the approximate length across the crevice. 61. Surveying. A glaciologist needs to determine the length across a certain crevice on Mendenhall glacier in order to circumvent it with his team. He has the following measurements: α 480 yd 500 yd 650 yd T (Team) Crevice P1 P2 Find the angle a. 62. Home Décor. A table in the shape of a regular octagon is to be constructed to fit on a round oriental rug 16 feet in diameter. Each guest sitting around the perimeter of the table requires 2.5 feet of space. How many people, approximately, can be seated at this table? • C A T C H T H E M I S T A K E In Exercises 63 and 64, explain the mistake that is made. 63. Solve the triangle b 5 3, c 5 4, and a 5 30°. Solution: Step 1: Find a. Apply the Law of Cosines. a2 5 b2 1 c2 2 2 bc cos a Let b 5 3, c 5 4, and a 5 30°. a2 5 32 1 42 2 2132142 cos 30° Solve for a. a < 2.1 Step 2: Find g. Apply the Law sin a a 5 sin g c of Sines. Solve for sin g. sin g 5 c sin a a Solve for g. g 5 sin21 ac sin a a b Let a 5 2.1, c 5 4, and a 5 30°. g < 72° Step 3: Find b. a 1 b 1 g 5 180° 30° 1 b 1 72° 5 180° Solve for b. b < 78° a < 2.1, b 5 3, c 5 4, a 5 30°, b < 78°, and g < 72° This is incorrect. The longest side is not opposite the largest angle. What mistake was made? 64. Solve the triangle a 5 6, b 5 2, and c 5 5. Solution: Step 1: Find b. Apply the Law of Cosines. b2 5 a2 1 c2 2 2ac cos b Solve for b . b 5 cos21 aa2 1 c2 2 b2 2 ac b Let a 5 6, b 5 2, c 5 5. b < 18° Step 2: Find a. Apply the Law sin a a 5 sin b b of Sines. Solve for a. a 5 sin21 aa sin b b b Let a 5 6, b 5 2, and b 5 18°. a < 68° Step 3: Find g. a 1 b 1 g 5 180° 68° 1 18° 1 g 5 180° g < 94° a 5 6, b 5 2, c 5 5, a < 68°, b < 18°, and g < 94° This is incorrect. The longest side is not opposite the largest angle. What mistake was made? 8.2 The Law of Cosines 789 790 CHAPTER 8 Additional Topics in Trigonometry 71. Show that cos a a 1 cos b b 1 cos g c 5 a2 1 b2 1 c2 2abc . Hint: Apply the Law of Cosines. 72. Show that a 5 c cos b 1 b cos g. Hint: Apply the Law of Cosines. 73. Consider the following diagram. Find cosaX 2 b. a a 2 a X 74. Using the diagram in Exercise 73, Find tanaX 2 b. In Exercises 65–70, determine whether each statement is true or false. • C O N C E P T U A L • C H A L L E N G E For Exercises 75–80, let A, B, and C be the lengths of the three sides with X, Y, and Z as the corresponding angles. Write a program using a TI calculator to solve the given triangle. • T E C H N O L O G Y A B C X Z Y 75. B 5 45, C 5 57, and X 5 43° 76. B 5 24.5, C 5 31.6, and X 5 81.5° 77. A 5 29.8, B 5 37.6, and C 5 53.2 78. A 5 100, B 5 170, and C 5 250 79. A 5 !12, B 5 !21, and Z 5 62.8° 80. A 5 1235, B 5 987, and C 5 1456 65. Given three sides of a triangle, there is insufficient information to solve the triangle. 66. Given three angles of a triangle, there is insufficient information to solve the triangle. 67. The Pythagorean theorem is a special case of the Law of Cosines. 68. The Law of Cosines is a special case of the Pythagorean theorem. 69. If an obtuse triangle is isosceles, then knowing the obtuse angle and a side adjacent to it is sufficient to solve the triangle. 70. All acute triangles can be solved using the Law of Cosines. 8.3.1 S K IL L Find the area of a triangle in the SAS case. 8.3.1 C ON C E P T U A L Understand how to derive a formula for the area of a triangle (SAS case) using the Law of Sines. In Sections 8.1 and 8.2, we applied the Law of Sines and the Law of Cosines to solve oblique triangles, which means to find all of the side lengths and angle measures. Now we use these laws to derive formulas for the area of a triangle (SAS and SSS). Our starting point for both cases is the standard formula for the area of a triangle: A 5 1 2 bh WORDS MATH Start with a triangle with base b and height h. Construct a rectangle with base b and height h around the triangle. The area of the rectangle is base times height. Arectangle 5 bh Triangles 1 and 2 have the same area. Triangles 3 and 4 have the same area. Therefore, the area of the triangle is half the area of the rectangle. Area of the triangle with base b and height h. Atriangle 5 1 2 bh 8.3.1 The Area of a Triangle (SAS Case) We now can use the general formula for the area of a triangle and the Law of Sines to develop a formula for the area of a triangle when two sides and the angle between them are given. b h b h b 1 3 2 4 h 8.3 The Area of a Triangle 791 S K I L L S O B J E C T I V E S ■ ■Find the area of a triangle in the SAS case. ■ ■Find the area of a triangle in the SSS case. C O N C E P T U A L O B J E C T I V ES ■ ■Understand how to derive a formula for the area of a triangle (SAS case) using the Law of Sines. ■ ■Understand how to derive a formula for the area of a triangle (SSS case) using the Law of Cosines. 8.3 THE AREA OF A TRIANGLE 792 CHAPTER 8 Additional Topics in Trigonometry WORDS MATH Start with a triangle, given b, c, and a. Write the sine ratio for angle a. sin a 5 h c Solve for h. h 5 c sin a Write the formula for area of a triangle. Atriangle 5 1 2 bh For the SAS case, substitute h 5 c sin a. ASAS case 5 1 2 bc sin a Now we can calculate the area of this triangle with the given information (two sides and the angle between them: b, c, and a.) Similarly, it can be shown that the other formulas for SAS triangles are ASAS case 5 1 2 ab sin g and ASAS case 5 1 2 ac sin b b c α h AREA OF A TRIANGLE (SAS) For any triangle where two sides and the angle between them are known, the area for that triangle is given by one of the following formulas (depending on which angle and sides are given): ASAS case 5 1 2 bc sin a when b, c, and a are known ASAS case 5 1 2 ab sin g when a, b, and g are known ASAS case 5 1 2 ac sin b when a, c, and b are known In other words, the area of a triangle equals one-half the product of two of its sides and the sine of the angle between them. EXAMPLE 1 Finding the Area of a Triangle (SAS Case) Find the area of the triangle a 5 7.0 ft, b 5 9.3 ft, and g 5 86°. Solution: Apply the area formula where a, b, and g are given. A 5 1 2 ab sin g Substitute a 5 7.0 ft, b 5 9.3 ft, and g 5 86°. A 5 1 2 17.0 ft2 19.3 ft2 sin 86° Approximate with a calculator. A 5 32.47071 ft2 Round to two significant digits. A 5 32 ft2 Y OUR TU R N Find the area of the triangle a 5 3.2 m, c 5 5.1 m, and b 5 498. ▼ ▼ A N S W E R 6.2 m2 [CONCEPT CHECK] TRUE OR FALSE The area of a triangle in the SAS case is found by dividing an oblique triangle into two right triangles and applying the Law of Sines. ANSWER True ▼ 8.3.2 The Area of a Triangle (SSS Case) We used the Law of Sines to develop a formula for the area of an SAS triangle. That formula, several trigonometric identities, and the Law of Cosines can be used to develop a formula for the area of an SSS triangle, called Heron’s formula. WORDS MATH Start with any of the formulas for SAS triangles. A 5 1 2 ab sin g Square both sides. A2 5 1 4 a2b2 sin2 g Isolate sin2 g. 4A2 a2b2 5 sin2 g Apply the Pythagorean identity. 4A2 a2b2 5 1 2 cos2 g Factor the difference of the 4A 2 a2b2 5 11 2 cos g2 11 1 cos g2 two squares on the right. Solve the Law of Cosines, cosg 5 a2 1 b2 2 c2 2ab c2 5 a2 1 b2 2 2ab cos g, for cos g. Substitute cosg 5 a2 1 b2 2 c2 2ab 8.3.2 S K IL L Find the area of a triangle in the SSS case. 8.3.2 C ON C E P T U A L Understand how to derive a formula for the area of a triangle (SSS case) using the Law of Cosines. [CONCEPT CHECK] TRUE OR FALSE Heron‘s formula can be used to find the area of a triangle when either all three sides of a triangle or all three angle measures are known. ANSWER False ▼ into 4A 2 a2b2 5 11 2 cosg2 11 1 cosg2. 4A2 a2b2 5 c1 2 a2 1 b2 2 c2 2ab d c1 1 a2 1 b2 2 c2 2ab d Combine the expressions in brackets. 4A2 a2b2 5 c2ab 2 a2 2 b2 1 c2 2ab d c2ab 1 a2 1 b2 2 c2 2ab d Group the terms in the numerators. 4A2 a2b2 5 c21a2 2 2ab 1 b22 1 c2 2ab d c 1a2 1 2ab 1 b22 2 c2 2ab d Write the numerators as the 4A2 a2b2 5 cc2 2 1a 2 b22 2ab d c 1a 1 b22 2 c2 2ab d difference of two squares. Factor the numerators: 4A2 a2b2 5 c 1c 2 3a 2 b421c 1 3a 2 b42 2ab d c 13a 1 b4 2 c213a 1 b4 1 c2 2ab d x2 2 y2 5 1x 2 y21x 1 y2. Simplify. 4A2 a2b2 5 c 1c 2 a 1 b21c 1 a 2 b2 2ab d c 1a 1 b 2 c21a 1 b 1 c2 2ab d 4A2 a2b2 5 1c 2 a 1 b21c 1 a 2 b21a 1 b 2 c21a 1 b 1 c2 4a2b2 Multiply by a2b2 4 . A2 5 1 16 1c 2 a 1 b21c 1 a 2 b21a 1 b 2 c21a 1 b 1 c2 The semiperimeter s is half the perimeter of the triangle. 2s 5 a 1 b 1 c Manipulate each of c 2 a 1 b 5 a 1 b 1 c 2 2a 5 2s 2 2a 5 21s 2 a2 the four factors: c 1 a 2 b 5 a 1 b 1 c 2 2b 5 2s 2 2b 5 21s 2 b2 a 1 b 2 c 5 a 1 b 1 c 2 2c 5 2s 2 2c 5 21s 2 c2 a 1 b 1 c 5 2s Substitute in these values for the four factors. A2 5 1 16⋅21s 2 a2 ⋅21s 2 b2 ⋅21s 2 c2 ⋅2s Simplify. A2 5 s1s 2 a21s 2 b21s 2 c2 Solve for A (area is always positive). A 5 "s1s 2 a21s 2 b21s 2 c2 8.3 The Area of a Triangle 793 794 CHAPTER 8 Additional Topics in Trigonometry EXAMPLE 2 Finding the Area of a Triangle (SSS Case) Find the area of the triangle a 5 5, b 5 6, and c 5 9. Solution: Find the semiperimeter s. s 5 a 1 b 1 c 2 Substitute a 5 5, b 5 6, and c 5 9. s 5 5 1 6 1 9 2 Simplify. s 5 10 Write the formula for the area of a triangle in the SSS case. A 5 "s1s 2 a21s 2 b21s 2 c2 Substitute a 5 5, b 5 6, c 5 9, and s 5 10. A 5 "10110 2 52110 2 62110 2 92 Simplify the radicand. A 5 !10⋅5⋅4⋅1 Evaluate the radical. A 5 10!2 < 14 sq.units Y OUR TU R N Find the area of the triangle a 5 3, b 5 5, and c 5 6. ▼ AREA OF A TRIANGLE (SSS CASE—HERON’S FORMULA) For any triangle where the lengths of the three sides are known, the area for that triangle is given by the following formula: ASSS case 5 "s1s 2 a21s 2 b21s 2 c2 where a, b, and c are the lengths of the sides of the triangle and s is half the perimeter of the triangle, called the semiperimeter. s 5 a 1 b 1 c 2 ▼ A N S W E R 2!14 < 7.5 sq. units The Law of Cosines was instrumental in developing a formula for the area of a triangle (SSS case) when all three sides are given. ASSS 5 "s1s 2 a21s 2 b21s 2 c2 where s 5 a 1 b 1 c 2 In this section, we derived formulas for calculating the areas of triangles (SAS and SSS cases). The Law of Sines leads to three area formulas for the SAS case depending on which angles and sides are given. ASAS 5 1 2 bc sin a ASAS 5 1 2 ab sin g ASAS 5 1 2 ac sin b [SEC TION 8.3] S U M M A RY In Exercises 1–32, find the area of each triangle described. 1. a 5 8, c 5 16, b 5 60° 2. b 5 6, c 5 4!3, a 5 30° 3. a 5 1, b 5 !2, a 5 45° 4. b 5 2!2, c 5 4, b 5 45° 5. a 5 6, b 5 8, g 5 80° 6. b 5 9, c 5 10, a 5 100° 7. a 5 4, c 5 7, b 5 27° 8. a 5 6.3, b 5 4.8, g 5 17° 9. b 5 100, c 5 150, a 5 36° 10. c 5 0.3, a 5 0.7, b 5 145° 11. a 5 !5, b 5 5!5, g 5 50° 12. b 5 !11, c 5 !11, a 5 21° 13. a 5 15, b 5 15, c 5 15 14. a 5 1, b 5 1, c 5 1 15. a 5 7, b 5 !51, c 5 10 16. a 5 9, b 5 40, c 5 41 17. a 5 6, b 5 10, c 5 9 18. a 5 40, b 5 50, c 5 60 19. a 5 14.3, b 5 15.7, c 5 20.1 20. a 5 146.5, b 5 146.5, c 5 100 21. a 5 14,000, b 5 16,500, c 5 18,700 22. a 5 !2, b 5 !3, c 5 !5 23. a 5 80, b 5 75, c 5 160 24. a 5 19, b 5 23, c 5 3 25. a 5 2!3, b 5 5!3, g 5 61.23° 26. a 5 7!5 3 , c 5 5!5 2 , b 5 16.36° 27. b 5 86 3 , c 5 51 4 , a 5 73° 28. a 5 17.61, c 5 17.61, b 5 20.03° 29. a 5 5.61, b 5 6.61, c 5 4.31 30. a 5 7.28, b 5 7.28, c 5 7.28 31. a 5 63 8 , b 5 50 7 , c 5 50 9 32. a 5 1 2, b 5 1 3, c 5 1 3 [SEC TION 8.3] E X E R C I S E S • S K I L L S • A P P L I C A T I O N S 33. Bermuda Triangle. Calculate the area of the so-called Bermuda Tri an gle, described in Example 4 of Section 8.2 if, as some people de fine it, its vertices are located in Norfolk, Bermuda, and Santiago. LOCATION LOCATION DISTANCE (NAUTICAL MILES) Norfolk Bermuda 850 Bermuda Santiago 810 Norfolk Santiago 894 Ignore the curvature of the Earth in your calculations. 34. Bermuda Triangle. Calculate the area of the Bermuda Triangle if, as other people define it, its vertices are located in Miami, Bermuda, and San Juan. LOCATION LOCATION DISTANCE (NAUTICAL MILES) Miami Bermuda 898 Bermuda San Juan 831 Miami San Juan 890 35. Triangular Tarp. A large triangular tarp is needed to cover a playground when it rains. If the sides of the tarp measure 160 feet, 140 feet, and 175 feet, what is the area of the tarp (to the nearest square foot)? 36. Flower Seed. A triangular garden measures 41 feet 3 16 feet 3 28 feet. You are going to plant wildflower seed that costs $4 per bag. Each bag of flower seed covers an area of 20 square feet. How much will the seed cost? (Assume you have to buy a whole bag—you can’t split one.) 37. Mural. Some students are painting a mural on the side of a building. They have enough paint for a 1000-square-foot-area triangle. If two sides of the triangle measure 60 feet and 120 feet, what angle (to the nearest degree) should the two sides form in order to create a triangle that uses up all the paint? 38. Mural. Some students are painting a mural on the side of a building. They have enough paint for a 500-square-foot-area triangle. If two sides of the triangle measure 40 feet and 60 feet, what angle (to the nearest degree) should the two sides form in order to create a triangle that uses up all the paint? 39. Insect Infestation. Some very destructive beetles have made their way into a forest preserve. The rangers are trying to keep track of their spread and how well preventative measures are working. In a triangular area that is 22.5 miles on one side, 28.1 miles on another, and 38.6 miles on the third, what is the total area the rangers are covering? 40. Real Estate. A real estate agent needs to determine the area of a triangular lot. Two sides of the lot are 150 feet and 60 feet. The angle between the two measured sides is 43°. What is the area of the lot? 8.3 The Area of a Triangle 795 796 CHAPTER 8 Additional Topics in Trigonometry For Exercises 41 and 42, refer to the following: A company is considering purchasing a triangular piece of property (see the figure below) for the construction of a new facility. The purchase is going to be “sight unseen” and the company is using old surveys to approximate area and costs. a c C A B b 41. Business. If the survey indicates that one side b is 275 feet, a second side a is 310 feet, and the angle between the two sides is 798, a. find the area of the property. b. if the company wants to offer the seller $2.13 per square foot, what is the total cost of the property? 42. Business. If the survey indicates that one side b is 475 feet, a second side c is 310 feet, and the angle between the two sides is 1188, a. find the area of the property. b. if the company wants to offer the seller $1.97 per square foot, what is the total cost of the property? 43. Parking Lot. A parking lot is to have the shape of a parallelogram that has adjacent sides measuring 200 feet and 260 feet. The angle between the two sides is 65°. What is the area of the parking lot? 65º 65º 200 ft 200 ft 260 ft 260 ft 44. Parking Lot. A parking lot is to have the shape of a parallelogram that has adjacent sides measuring 250 feet and 300 feet. The angle between the two sides is 55°. What is the area of the parking lot? 45. Regular Hexagon. A regular hexagon has sides measuring 3 feet. What is its area? Recall that the measure of an angle of a regular n-gon is given by the formula angle 5 180 1n 2 22 n . 5 in. 3 ft 46. Regular Decagon. A regular decagon has sides measuring 5 inches. What is its area? 47. Geometry. Three spokes Q1, Q2, and Q3 of lengths 8, 6, and 11, respec tively, emanate from a common point P with the following angles: 8 6 8 9 11 30º 25º 70º Q1 Q2 Q3 P Find the area of the triangles PQ1Q2, Q1Q2Q3, and PQ2Q3. Then, find the area of triangle PQ1Q3. Compare the sum of the areas of the smaller three triangles to the area of PQ1Q3. Are they equal? 48. Construction. A rectangular great room, 15 ft 3 25 ft, has an open beam ceiling. The two parts of the ceiling make angles 50° and 33° with the horizontal, as shown. Find the area of the ceiling. 33º 15 ft x a b 50º 25 ft 49. Geometry. A quadrilateral ABCD has sides of lengths AB 5 2, BC 5 3, CD 5 4, and DA 5 5. The angle between AB and BC is 1358. Find the area of ABCD. 50. Geometry. A quadrilateral ABCD has sides of lengths AB 5 5, BC 5 6, CD 5 7, and DA 5 8. The angle between AB and BC is 1358. Find the area of ABCD. • C A T C H T H E M I S T A K E In Exercises 51 and 52, explain the mistake that is made. 51. Calculate the area of the triangle a 5 2, b 5 6, and c 5 7. Solution: Find the semiperimeter. s 5 a 1 b 1 c 5 2 1 6 1 7 5 15 Write the formula for the area of the triangle. A 5 "s1s 2 a21s 2 b21s 2 c2 Let a 5 2, b 5 6, c 5 7, and s 5 15. A 5 !15115 2 22115 2 62115 2 72 Simplify. A 5 !14, 040 < 118 This is incorrect. What mistake was made? 52. Calculate the area of the triangle a 5 2, b 5 6, and c 5 5. Solution: Find the semiperimeter. s 5 a 1 b 1 c 2 5 2 1 6 1 5 2 5 6.5 Write the formula for area of the triangle. A 5 s1s 2 a21s 2 b21s 2 c2 Let a 5 2, b 5 6, c 5 5, and s 5 6.5. A 5 6.514.5210.5211.52 Simplify. A < 22 This is incorrect. What mistake was made? 59. Show that the area for an SAA triangle is given by A 5 a2 sin b sin g 2 sin a Assume that a, b, and a are given. 60. Show that the area of an isosceles triangle with equal sides of length s is given by Aisosceles 5 1 2 s2 sin u where u is the angle between the two equal sides. 61. Find the area of the shaded region. 5 5 40º A B O 62. Find the area of the shaded region. 120º A B O 3√2 3√2 63. Prove that the area of a rhombus that has side length s with an angle of 30° is 1 2s2. 64. Prove that the area of a parallelogram with side lengths s and 3s and an angle of 1358 is 3!2 2 s2. • C H A L L E N G E In Exercises 53–58, determine whether each statement is true or false. • C O N C E P T U A L 53. Heron’s formula can be used to find the area of right triangles. 54. Heron’s formula can be used to find the area of isosceles triangles. 55. If two triangles have the same side lengths, then they have the same area. 56. If two rhombi (i.e., quadrilateral with four congruent sides) have the same side length, then they have the same area. 57. If two parallelograms have the same area, then the corresponding sides must be the same. 58. The semiperimeter can be less than the length of the largest side. For Exercises 65–70, let A, B, and C be the lengths of the three sides with X, Y, and Z as the corresponding angles. Write a program using a TI calculator to calculate the area of the given triangle. • T E C H N O L O G Y A B C X Z Y 65. A 5 35, B 5 47, and Z 5 68° 66. A 5 1241, B 5 1472, and Z 5 56° 67. A 5 85, B 5 92, and C 5 123 68. A 5 !167, B 5 !113, and C 5 !203 69. A 5 145, B 5 172, and C 5 110 70. A 5 21.8, B 5 39.6, and C 5 124.5° 8.3 The Area of a Triangle 797 798 CHAPTER 8 Additional Topics in Trigonometry 8.4.1 Vectors: Magnitude and Direction What is the difference between velocity and speed? Speed has only magnitude, whereas velocity has both magnitude and direction. We use scalars, which are real numbers, to denote magnitudes such as speed and weight. We use vectors, which have magnitude and direction, to denote quantities such as velocity (speed in a certain direction) and force (weight in a certain direction). A vector quantity is geometrically denoted by a directed line segment, which is a line segment with an arrow representing direction. There are many ways to denote a vector. For example, the vector shown in the margin can be denoted as u, u S, or AB h, where A is the initial point and B is the terminal point. It is customary in books to use the bold letter to represent a vector and when hand written (as in your class notes and homework) to use the arrow on top to denote a vector. In this section, we will limit our discussion to vectors in a plane (two- dimensional). It is important to note that geometric representation can be extended to three dimensions and algebraic representation can be extended to any higher dimension, as you will see in the exercises and in later sections. Geometric Interpretation of Vectors The magnitude of a vector can be denoted one of two ways: k u k or 7u7. We will use the former notation. Two vectors have the same direction if they are parallel and point in the same direction. Two vectors have opposite direction if they are parallel and point in opposite directions. B A u MAGNITUDE: k U k The magnitude of a vector u, denoted k u k , is the length of the directed line segment which is the distance between the initial and terminal points of the vector. 8.4.1 S KILL Find the magnitude and direction of a vector. 8.4.1 CO NCE PTUAL Understand the difference between scalars and vectors. STUDY TIP The magnitude of a vector is the distance between the initial and terminal points of the vector. EQUAL VECTORS: U 5 V Two vectors u and v are equal 1u 5 v2 if and only if they have the same magnitude A k u k 5 k v k B and the same direction. [CONCEPT CHECK] Match: (1) Speed (2) Velocity to (A) Vector (B) Scalar ANSWER 1. B 2. A ▼ S K I L L S O B J E C T I V E S ■ ■Find the magnitude and direction of a vector. ■ ■Add and subtract vectors and perform scalar multiplication of a vector. ■ ■Express a vector in terms of its horizontal and vertical components. ■ ■Find unit vectors. ■ ■Find resultant vectors in application problems. C O N C E P T U A L O B J E C T I V E S ■ ■Understand the difference between scalars and vectors. ■ ■Relate the geometric and algebraic representations of operations on vectors. ■ ■Relate the right triangle trigonometric definitions of sine and cosine to the vertical and horizontal components of a vector. ■ ■Understand that unit vectors have a magnitude (length) equal to one. ■ ■Use resultant vectors to find actual velocities or actual forces. 8.4 VECTORS Equal Vectors u 5 v Same Magnitude but Opposite Direction u 5 2v Same Magnitude k u k 5 k v k Different Magnitude It is important to note that vectors do not have to coincide to be equal. VECTOR ADDITION: U 1 V Two vectors, u and v, can be added together using either of the following approaches: • The tail-to-tip (or head-to-tail) method: Sketch the initial point of one vector at the terminal point of the other vector. The sum, u 1 v, is the resultant vector from the tail end of u to the tip end of v. [or] • The parallelogram method: Sketch the initial points of the vectors at the same point. The sum u 1 v is the diagonal of the parallelogram formed by u and v. The difference, u 2 v, is the • Resultant vector from the tip of v to the tip of u, when the tails of v and u coincide. [or] • The other diagonal formed by the parallelogram method. u + v v u u v u + v v u u – v u v u – v Algebraic Interpretation of Vectors Since vectors that have the same direction and magnitude are equal, any vector can be translated to an equal vector with its initial point located at the origin in the Cartesian plane. Therefore, we will now consider vectors in a rectangular coordinate system. A vector with its initial point at the origin is called a position vector, or a vector in standard position. A position vector u with its terminal point at the point 1a, b2 is denoted: u 5 8a, b9 where the real numbers a and b are called the components of vector u. u x y a b (a, b) 8.4 Vectors 799 800 CHAPTER 8 Additional Topics in Trigonometry Notice the subtle difference between coordinate notation and vector notation. The point is denoted with parentheses, 1a, b2, whereas the vector is denoted with angled brackets, 8a, b9. The notation 8a, b9 denotes a vector whose initial point is 10, 02 and terminal point is 1a, b2. The vector with initial point 13, 42 and terminal point 18, 92 is equal to the vector 85, 59, which has initial point 10, 02 and terminal point 15, 52. Recall that the geometric definition of the magnitude of a vector is the length of the vector. MAGNITUDE: k U k The magnitude (or norm) of a vector, u 5 8a, b9, is 0 u 0 5 "a2 1 b2 x y a b (a, b) \|u\| DIRECTION ANGLE OF A VECTOR The positive angle between the x-axis and a position vector is called the direction angle, denoted u tan u 5 b a, where a 2 0 or u 5 tan21ab ab x y a b (a, b) \|u\| θ EXAMPLE 1 Finding the Magnitude of a Vector Find the magnitude of the vector u 5 83, 249. Solution: Write the formula for magnitude of a vector. 0u 0 5 "a2 1 b2 Let a 5 3 and b 5 24. 0u 0 5 "32 1 12422 Simplify. 0u 0 5 "25 5 5 Note: If we graph the vector u 5 83, 249, we see that the distance from the origin to the point 13, 242 is 5 units. Y OUR TU R N Find the magnitude of the vector v 5 821, 59. \|u\| = 5 x y (3, –4) ▼ ▼ A N S W E R !26 STUDY TIP 8a, b9 denotes a vector; 1a, b2 denotes a point. Recall that two vectors are equal if they have the same magnitude and direction. Algebraically, this corresponds to their corresponding components being equal. \|v\| x y (–1, 5) θ –78.7º EXAMPLE 2 Finding the Direction Angle of a Vector Find the direction angle of the vector v 5 821, 59. Solution: Start with tan u 5 b a and let a 5 21 and b 5 5. tan u 5 5 21 With a calculator, find tan211252. tan 211252 5 278.7° The calculator gives a quadrant IV angle. The point 121, 52 lies in quadrant II. Add 180°. u 5 278.7° 1 180° 5 101.3° u 5 101.3° YOUR T UR N Find the direction angle of the vector u 5 83, 249. ▼ ▼ A N S W E R 306.98 EQUAL VECTORS: U 5 V The vectors u 5 8a, b9 and v 5 8c, d9 are equal; that is, u 5 v, if and only if a5 c and b 5 d. 8.4.2 Vector Operations Vector addition is done geometrically with the tail-to-tip rule. Algebraically, vector addition is performed component by component. VECTOR ADDITION: U 1 V If u 5 8a, b9 and v 5 8c, d 9, then u 1 v 5 8a 1 c, b 1 d 9. EXAMPLE 3 Adding Vectors Let u 5 82, 279 and v 5 823, 49. Find u 1 v. Solution: Let u 5 82, 279 and v 5 823, 49 in the addition formula. u 1 v 5 82 1 1232, 27 1 49 Simplify. u 1 v 5 821, 239 Y OUR T UR N Let u 5 81, 29 and v 5 825, 249. Find u 1 v. ▼ 8.4.2 S K I L L Add and subtract vectors and perform scalar multiplication of a vector. 8.4.2 C O N C E P T U A L Relate the geometric and algebraic representations of operations on vectors. ▼ A N S W E R u 1 v 5 824, 229 8.4 Vectors 801 802 CHAPTER 8 Additional Topics in Trigonometry We now summarize vector operations. Addition and subtraction of vectors are performed algebraically component by component. Multiplication, however, is not as straightforward. To perform scalar multiplication of a vector (to multiply a vector by a real number), we multiply component by component. In Section 8.5 we will study a form of multiplication for two vectors that is defined as long as the vectors have the same number of components; it gives a result known as the dot product, and it is useful in solving common problems in physics. Scalar multiplication corresponds to ■ ■Increasing the length of the vector: 0 k 0 . 1 ■ ■Decreasing the length of the vector: 0 k 0 , 1 ■ ■Changing the direction of the vector: k , 0 u 2u u –u 1 2 The following box is a summary of vector operations: The zero vector, 0 5 80, 09, is a vector in any direction with a magnitude equal to zero. We now can state the algebraic properties (associative, commutative, and distributive) of vectors. SCALAR MULTIPLICATION: kU If k is a scalar (real number) and u 5 8a, b9, then k u 5 k 8a, b9 5 8k a, k b9 VECTOR OPERATIONS If u 5 8a, b9, v 5 8c, d 9, and k is a scalar, then u 1 v 5 8a 1 c, b 1 d 9 u 2 v 5 8a 2 c, b 2 d 9 k u 5 k8a, b9 5 8k a, k b9 DEFINITION Algebraic Properties of Vectors u 1 v 5 v 1 u 1u 1 v2 1 w 5 u 1 1v 1 w2 1k1k22u 5 k11k2u2 k1u 1 v2 5 ku 1 kv 1k1 1 k22u 5 k1u 1 k2u 0u 5 0 1u 5 u 21u 5 2u u 1 12u2 5 0 [CONCEPT CHECK] If a . c and b , d, then if a, b, c, and d are all positive, then u 2 v is a vector pointing into which quadrant? ANSWER IV ▼ 8.4.3 Horizontal and Vertical Components of a Vector The horizontal component a and vertical component b of a vector u are related to the magnitude of the vector, k u k , through the sine and cosine of the direction angle. cos u 5 a k u k sin u 5 b k u k x y a b (a, b) \|u\| θ 8.4.3 S K I L L Express a vector in terms of its horizontal and vertical components. 8.4.3 C O N C E P T U A L Relate the right triangle trigonometric definitions of sine and cosine to the vertical and horizontal components of a vector. HORIZONTAL AND VERTICAL COMPONENTS OF A VECTOR The horizontal and vertical components of vector u, with magnitude 0 u 0 and direction angle u, are given by horizontal component: a 5 k u k cos u vertical component: b 5 k u k sin u The vector u can then be written as u 5 8a, b9 5 8 k u k cos u, k u k sin u9. EXAMPLE 4 Finding the Horizontal and Vertical Components of a Vector Find the vector that has a magnitude of 6 and a direction angle of 15°. Solution: Write the horizontal and vertical components of a vector u. a 5 k u k cos u and b 5 k u k sin u Let k u k 5 6 and u 5 15°. a 5 6 cos 15° and b 5 6 sin 15° Evaluate the sine and cosine functions of 15°. a 5 5.8 and b 5 1.6 Let u 5 8a, b9. u 5 85.8, 1.69 YOUR TURN Find the vector that has a magnitude of 3 and direction angle of 75°. ▼ 8.4.4 Unit Vectors A unit vector is any vector with magnitude equal to 1, or k u k 5 1. It is often useful to be able to find a unit vector in the same direction of some vector v. A unit vector can be formed from any nonzero vector as follows: FINDING A UNIT VECTOR If v is a nonzero vector, then u 5 v k v k 5 1 k v k ⋅ v is a unit vector in the same direction as v. In other words, multiplying any nonzero vector by the reciprocal of its magnitude results in a unit vector. 8.4.4 S K I L L Find unit vectors. 8.4.4 C O N C E P T U A L Understand that unit vectors have a magnitude (length) equal to one. STUDY TIP Multiplying a nonzero vector by the reciprocal of its magnitude results in a unit vector. ▼ A N S W E R u 5 80.78, 2.99 [CONCEPT CHECK] Match: 1. u 5 n p 2. u 5 np 1 p 2 (A) vertical vector (B) horizontal vector ANSWER 1. B 2. A ▼ 8.4 Vectors 803 804 CHAPTER 8 Additional Topics in Trigonometry It is important to notice that since the magnitude is always a scalar, then the reciprocal of the magnitude is always a scalar. A scalar times a vector is a vector. Two important unit vectors are the horizontal and vertical unit vectors i and j. The unit vector i has an initial point at the origin and a terminal point at 11, 02. The unit vector j has an initial point at the origin and a terminal point at 10, 12. We can use these unit vectors to represent vectors algebraically. For example, the vector 83, 249 5 3i 2 4j. 8.4.5 Resultant Vectors Vectors arise in many applications. Velocity vectors and force vectors are two that we will discuss. For example, suppose that you are at the beach and “think” that you are swimming straight out at a certain speed (magnitude and direction). This is your apparent velocity with respect to the water. After a few minutes you turn around to look at the shore, and you are farther out than you thought and appear to have drifted down the beach. This is because of the current of the water. When the current velocity and the apparent velocity are added together, the result is the actual or resultant velocity. In Chapter 6 (Example 7) the navigation terms bearing (the direction a vessel is pointed) and heading (the direction the vessel is actually traveling) were used. In the context of vectors, bearing and heading are synonymous with apparent velocity and resultant velocity, respectively. EXAMPLE 5 Finding a Unit Vector Find a unit vector in the same direction as v 5 823, 249. Solution: Find the magnitude of the vector v 5 823, 249. k v k 5 "12322 1 12422 Simplify. k v k 5 5 Multiply v by the reciprocal of its magnitude. 1 k v k v Let 0 v 0 5 5 and v 5 823, 249. 1 5 823, 249 Simplify. h23 5, 24 5i Check: The unit vector, h23 5, 24 5i, should have a magnitude of 1. Ç a23 5b 2 1 a24 5b 2 5 Å 25 25 5 1 Y OUR TU R N Find a unit vector in the same direction as v 5 85, 2129. ▼ ▼ A N S W E R h 5 13, 2 12 13i [CONCEPT CHECK] TRUE OR FALSE A unit circle has radius equal to 1, and a unit vector has a length equal to 1. ANSWER True ▼ 8.4.5 SKI LL Find resultant vectors in application problems. 8.4.5 C ON CEPTUAL Use resultant vectors to find actual velocities or actual forces. x y j i (0, 1) (1, 0) In Example 6, the three vectors formed a right triangle. EXAMPLE 6 Resultant Velocities A boat’s speedometer reads 25 mph (which is relative to the water) and sets a course due east (908 from due north). If the river is moving 10 mph due north, what is the resultant (actual) velocity of the boat? Solution: Draw a picture. Label the horizontal and vertical components of the resultant vector. 825, 109 Determine the magnitude of the resultant vector. "252 1 102 5 5!29 < 27 mph Determine the direction angle. tan u 5 10 25 Solve for u u 5 tan21 a2 5b < 22° The actual velocity of the boat has magnitude 27 mph , and the boat is headed 22° north of east or 68° east of north. N S E W Resultant velocity 25 mph 10 mph θ [CONCEPT CHECK] The speedometer on a boat will correspond to the magnitude of which velocity—apparent or actual? ANSWER Apparent ▼ EXAMPLE 7 Resultant Velocities A speedboat traveling 30 mph has a compass heading of 100° east of north. The current velocity has a magnitude of 15 mph, and its heading is 22° east of north. Find the resultant (actual) velocity of the boat. Solution: Draw a picture. 30 mph 15 mph Resultant velocity 100º 22º N S E W 8.4 Vectors 805 806 CHAPTER 8 Additional Topics in Trigonometry Label the supplementary angles to 100°. Draw and label the oblique triangle. The magnitude of the actual (resultant) velocity is b. The heading of the actual (resultant) velocity is 100° 2 a. Use the Law of Sines and the Law of Cosines to solve for a and b. Find b: Apply the Law of Cosines. b2 5 a2 1 c2 2 2ac cos b Let a 5 15, c 5 30, and b 5 102°. b2 5 152 1 302 2 211521302 cos 102° Solve for b. b < 36 mph Find a: Apply the Law of Sines. sin a a 5 sin b b Isolate sin a. sin a 5 a b sin b Let a 5 15, b 5 36, sin a 5 15 36 sin 102° and b 5 102°. Apply the inverse sine function to solve for a a 5 sin21 c15 36 sin 1102°2 d Approximate a with a calculator. a < 24° Actual heading: 100° 2 a 5 100° 2 24° 5 76° The actual velocity vector of the boat has magnitude 36 mph , and the boat is headed 76° east of north . 80º 80º 30 mph 15 mph Resultant velocity 100º 22º N S E W c = 30 mph a = 15 mph b 100º N S E W α β = 102º γ Two vectors combine to yield a resultant vector. The opposite vector to the resultant vector is called the equilibrant. EXAMPLE 8 Finding an Equilibrant A skier is being pulled up a slope by a handle lift. Let F1 represent the vertical force due to gravity and F2 represent the force of the skier pushing against the side of the mountain, at an angle of 35° to the horizontal. If the weight of the skier is 145 pounds, that is, k F1 k 5 145, find the magnitude of the equilibrant force F3 required to hold the skier in place (that is, to keep the skier from sliding down the mountain). Assume that the side of the mountain is a frictionless surface. 35º F1 F2 F3 Solution: The angle between vectors F1 and F2 is 35°. The magnitude of vector F3 is the force required to hold the skier in place. Relate the magnitudes (side lengths) to the given angle using the sine ratio. sin 35° 5 k F3 k k F1 k Solve for k F3 k . k F3 k 5 k F1 k sin 35° Let k F1 k 5 145. k F3 k 5 145 sin 35° k F3 k 5 83.16858 A force of approximately 83 pounds is required to keep the skier from sliding down the hill. 35º 35º F2 55º 55º F3 F1 EXAMPLE 9 Resultant Forces A barge runs aground outside the channel. A single tugboat cannot generate enough force to pull the barge off the sandbar. A second tugboat comes to assist. The following diagram illustrates the force vectors, F1 and F2, from the tugboats. What is the resultant force vector of the two tugboats? 80º 80º 20º F1 = 15,000 lb F2 = 18,000 lb 8.4 Vectors 807 808 CHAPTER 8 Additional Topics in Trigonometry Solution: Using the tail-to-tip rule, we can add these two vectors and form a triangle: 80º 80º α 80º 18,000 lb Resulting force Resultant force 15,000 lb N N α γ β = 160º a = 15,000 lb c = 18,000 lb b Find b: Apply the Law of Cosines. b2 5 a2 1 c2 2 2ac cos b Let a 5 15,000, c 5 18,000, b2 5 15,0002 1 18,0002 and b 5 160°. 22115,0002118,0002 cos 160° Solve for b. b 5 32,503 lb Find a: Apply the Law of Sines. sin a a 5 sin b b Isolate sin a. sin a 5 a b sin b Let a 5 15,000, b 5 32,503, sin a 5 15,000 32,503 sin 160° and b 5 160°. Apply the inverse sine function to solve for a. a 5 sin21 c15,000 32,503 sin1160°2 d Approximate a with a calculator. a < 9.08° The resulting force is 32,503 lb at an angle of 9° from the tug pulling with a force of 18,000 lb . Vector addition is performed algebraically component by component. 8a, b9 1 8c, d9 5 8a 1 c, b 1 d9. The trigonometric functions are used to express the horizontal and vertical components of a vector. Horizontal component: a 5 k u k cos u Vertical component: b 5 k u k sin u Velocity and force vectors illustrate examples of the Law of Sines and the Law of Cosines. In this section, we discussed scalars (real numbers) and vectors. Scalars have only magnitude, whereas vectors have both magnitude and direction. Vector: u 5 8a, b9 Magnitude: k u k 5 "a2 1 b2 Direction 1u2: tan u 5 b a We defined vectors both algebraically and geometrically and gave interpretations of magnitude and vector addition in both ways. [SEC TION 8.4] S U M MA RY In Exercises 1–6, find the magnitude of the vector AB. 1. A 5 12, 72 and B 5 15, 92 2. A 5 122, 32 and B 5 13, 242 3. A 5 14, 12 and B 5 123, 02 4. A 5 121, 212 and B 5 12, 252 5. A 5 10, 72 and B 5 1224, 02 6. A 5 122, 12 and B 5 14, 92 In Exercises 7–16, find the magnitude and direction angle of the given vector. 7. u 5 83, 89 8. u 5 84, 79 9. u 5 85, 219 10. u 5 826, 229 11. u 5 824, 19 12. u 5 826, 39 13. u 5 828, 09 14. u 5 80, 79 15. u 5 H !3, 3I 16. u 5 825, 259 In Exercises 17–24, perform the indicated vector operation, given u 5 824, 39 and v 5 82, 259. 17. u 1 v 18. u 2 v 19. 3u 20. 22u 21. 2u 1 4v 22. 51u 1 v2 23. 61u 2 v2 24. 2u 2 3v 1 4u In Exercises 25–34, find the vector, given its magnitude and direction angle. 25. k u k 5 7, u 5 25° 26. k u k 5 5, u 5 75° 27. k u k 5 16, u 5 100° 28. k u k 5 8, u 5 200° 29. k u k 5 4, u 5 310° 30. k u k 5 8, u 5 225° 31. k u k 5 9, u 5 335° 32. k u k 5 3, u 5 315° 33. k u k 5 2, u 5 120° 34. k u k 5 6, u 5 330° In Exercises 35–44, find a unit vector in the direction of the given vector. 35. v 5 825, 2129 36. v 5 83, 49 37. v 5 860, 119 38. v 5 827, 249 39. v 5 824, 279 40. v 5 8210, 249 41. v 5 829, 2129 42. v 5 840, 299 43. v 5 H !2, 3!2 I 44. v 5 H24!3, 22!3 I In Exercises 45–50, express the vector in terms of unit vectors i and j. 45. 87, 39 46. 822, 49 47. 85, 239 48. 826, 229 49. 821, 09 50. 80, 29 In Exercises 51–56, perform the indicated vector operation. 51. 15i 2 2j2 1 123i 1 2j2 52. 14i 2 2j2 1 13i 2 5j2 53. 123i 1 3j2 2 12i 2 2j2 54. 1i 2 3j2 2 122i 1 j2 55. 15i 1 3j2 1 12i 2 3j2 56. 122i 1 j2 1 12i 2 4j2 [SEC TION 8.4] E X E R C I S E S • S K I L L S • A P P L I C A T I O N S 57. Bullet Speed. A bullet is fired from ground level at a speed of 2200 feet per second at an angle of 30° from the horizontal. Find the magnitude of the horizontal and vertical components of the velocity vector. 58. Weightlifting. A 50-pound weight lies on an inclined bench that makes an angle of 40° with the horizontal. Find the component of the weight directed perpendicular to the bench and also the component of the weight parallel to the inclined bench. 59. Weight of a Boat. A force of 630 pounds is needed to pull a speedboat and its trailer up a ramp that has an incline of 13°. What is the combined weight of the boat and its trailer? 13º 60. Weight of a Boat. A force of 500 pounds is needed to pull a speedboat and its trailer up a ramp that has an incline of 16°. What is the weight of the boat and its trailer? 61. Speed and Direction of a Ship. A ship’s captain sets a course due north at 10 mph. The water is moving at 6 mph due west. What is the actual velocity of the ship, and in what direction is it traveling? 8.4 Vectors 809 810 CHAPTER 8 Additional Topics in Trigonometry 62. Speed and Direction of a Ship. A ship’s captain sets a course due west at 12 mph. The water is moving at 3 mph due north. What is the actual velocity of the ship, and in what direction is it traveling? 63. Heading and Airspeed. A plane has a compass heading of 60° east of due north and an airspeed of 300 mph. The wind is blowing at 40 mph with a heading of 30° west of due north. What are the plane’s actual heading and airspeed? N 30º 60º 300 mph 40 mph 64. Heading and Airspeed. A plane has a compass heading of 30° east of due north and an airspeed of 400 mph. The wind is blowing at 30 mph with a heading of 60° west of due north. What are the plane’s actual heading and airspeed? 65. Sliding Box. A box weighing 500 pounds is held in place on an inclined plane that has an angle of 30°. What force is required to hold it in place? 30º 500 lb 66. Sliding Box. A box weighing 500 pounds is held in place on an inclined plane that has an angle of 10°. What force is required to hold it in place? 67. Baseball. A baseball player throws a ball with an initial velocity of 80 feet per second at an angle of 40° with the horizontal. What are the vertical and horizontal components of the velocity? 68. Baseball. A baseball pitcher throws a ball with an initial velocity of 100 feet per second at an angle of 5° with the horizontal. What are the vertical and horizontal components of the velocity? For Exercises 69 and 70, refer to the following: In a post pattern in football, the receiver in motion runs past the quarterback parallel to the line of scrimmage 1A2, runs 12 yards perpendicular to the line of scrimmage 1B2, and then cuts toward the goal post 1C2. 5 0 4 0 4 0 3 0 2 0 1 0 3 0 2 0 1 0 5 0 4 0 4 0 3 0 2 0 1 0 3 0 2 0 1 0 B C A A + B + C 30º θ 69. Football. A receiver runs the post pattern. If the magnitudes of the vectors are k A k 5 4 yards, k B k 5 12 yards, and k C k 5 20 yards, find the magnitude of the resultant vector A 1 B 1 C. 70. Football. A receiver runs the post pattern. If the magnitudes of the vectors are k A k 5 4 yards, k B k 5 12 yards, and k C k 5 20 yards, find the direction angle u. 71. Resultant Force. A force with a magnitude of 100 pounds and another with a magnitude of 400 pounds are acting on an object. The two forces have an angle of 60° between them. What is the direction of the resultant force with respect to the force of 400 pounds? 72. Resultant Force. A force with a magnitude of 100 pounds and another with a magnitude of 400 pounds are acting on an object. The two forces have an angle of 60° between them. What is the magnitude of the resultant force? 73. Resultant Force. A force of 1000 pounds is acting on an object at an angle of 45° from the horizontal. Another force of 500 pounds is acting at an angle of 240° from the horizontal. What is the magnitude of the resultant force? 74. Resultant Force. A force of 1000 pounds is acting on an object at an angle of 45° from the horizontal. Another force of 500 pounds is acting at an angle of 240° from the horizontal. What is the angle of the resultant force? 75. Resultant Force. Forces with magnitudes of 200 N and 180 N act on a hook. The angle between these two forces is 458. Find the direction and magnitude of the resultant of these forces. 76. Resultant Force. Forces with magnitudes of 100 N and 50 N act on a hook. The angle between these two forces is 308. Find the direction and magnitude of the resultant of these forces. 77. Exercise Equipment. A tether ball weighing 5 pounds is pulled outward from a pole by a horizontal force u S until the rope makes a 458 angle with the pole. Determine the resulting tension on the rope and magnitude of u S. 78. Exercise Equipment. A tether ball weighing 8 pounds is pulled outward from a pole by a horizontal force u S until the rope makes a 60° angle with the pole. Determine the resulting tension on the rope and magnitude of u S. 79. Recreation. A freshman wishes to sign up for four different clubs during orientation. Each club is positioned at a different table in the gym, and the clubs of interest to him are positioned at A, B, C, and D, as pictured below. He starts at the entrance way O and walks directly toward A, then to B, then to C, then to D, and then back to O. a. Find the resultant vector of all his movement. b. How far did he walk during this sign-up adventure? y x 5 4 3 2 1 –1 –3 5 4 3 2 1 –5 –4 –3 –2 –5 A B C D O 80. Recreation. A freshman wishes to sign up for three different clubs during orientation. Each club is positioned at a different table in the gym, and the clubs of interest to him are positioned at A, B, and C, as pictured below. He starts at the entrance way O at the far end of the gym, walks directly toward A, then to B, then to C, and then exits the gym through the exit P at the opposite end. a. Find the resultant vector of all his movement. b. How far did he walk during this sign-up adventure? A B C O y –6 –2 –10 10 8 6 4 2 10 8 6 4 2 –4 –6 –8 –10 P 81. Torque. Torque is the tendency for an arm to rotate about a pivot point. If a force F S is applied at an angle u to turn an arm of length L, as pictured below, then the magnitude of the torque 5 L k F S k sin u. θ Arm L Force F Pivot point Assume that a force of 45 N is applied to a bar 0.2 meters wide on a sewer shut-off valve at an angle 85°. What is the magnitude of the torque? 82. Torque. You walk through a swinging mall door to enter a department store. You exert a force of 40 N applied perpendicular to the door. The door is 0.85 meters wide. Assuming that you pushed the door at its edge and the hinge is the pivot point, find the magnitude of the torque. 83. Torque. You walk through a swinging mall door to enter a department store. You exert a force of 40 N applied at an angle 1108 to the door. The door is 0.85 meters wide. Assuming that you pushed the door at its edge and the hinge is the pivot point, find the magnitude of the torque. 84. Torque. Suppose that within the context of Exercises 82 and 83, the magnitude of the torque turned out to be 0 Nm. When can this occur? 85. Resultant Force. A person is walking two dogs fastened to separate leashes that meet in a connective hub, leading to a single leash that she is holding. Dog 1 applies a force 608 NW with a magnitude of 8, and dog 2 applies a force of 458 clockwise of N with a magnitude of 6. Find the magnitude and direction of the force w S that the walker applies to the leash in order to counterbalance the total force exerted by the dogs. 86. Resultant Force. A person is walking three dogs fastened to separate leashes that meet in a connective hub, leading to a single leash that she is holding. Dog 1 applies a force 608 NW with a magnitude of 8, dog 2 applies a force of 458 clockwise of N with a magnitude of 6, and dog 3 moves directly N with a magnitude of 12. Find the magnitude and direction of the force w S that the walker applies to the leash in order to counterbalance the total force exerted by the dogs. For Exercises 87 and 88, refer to the following: Muscle A and muscle B are attached to a bone as indicated in the figure below. Muscle A exerts a force on the bone at angle a, while muscle B exerts a force on the bone at angle b. Muscle B Muscle A Bone 87. Health/Medicine. Assume muscle A exerts a force of 900 N on the bone at angle a 5 8°, while muscle B exerts a force of 750 N on the bone at angle b 5 33°. Find the resultant force and the angle of the force due to muscle A and muscle B on the bone. 88. Health/Medicine. Assume muscle A exerts a force of 1000 N on the bone at angle a 5 9°, while muscle B exerts a force of 820 N on the bone at angle b 5 38°. Find the resultant force and the angle of the force due to muscle A and muscle B on the bone. 8.4 Vectors 811 812 CHAPTER 8 Additional Topics in Trigonometry • C A T C H T H E M I S T A K E In Exercises 89 and 90, explain the mistake that is made. 89. Find the magnitude of the vector 822,289. Solution: Factor the 21. 282, 89 Find the magnitude k 82, 89 k 5 "22 1 82 of 82, 89. 5 !68 5 2!17 Write the magnitude of 822,289. k 822, 289 k 5 22!17 This is incorrect. What mistake was made? 90. Find the direction angle of the vector 822, 289. Solution: Write the formula for the direction tan u 5 b a angle of 8a, b9. Let a 5 22 and b 5 28. tan u 5 28 22 Apply the inverse tangent function. u 5 tan214 Evaluate with a calculator. u 5 76° This is incorrect. What mistake was made? 99. Show that if u S is a unit vector in the direction of v S, then v S 5 k u S k u S. 100. Show that if u S 5 a i S 1 b j S is a unit vector, then 1a, b2 lies on the unit circle. 101. A vector u S is a linear combination of p S and q S if there exist constants c1 and c2 such that u S 5 c1p S 1 c2q S. Show that 826, 49 is a linear combination of 828, 49 and 81, 219. 102. Show that H2 2 9 a, 8 9bI is a linear combination of 8a, 3b9 and 82a, 2b9, for any real constants a and b. 103. Prove that u S 1 312v S 2 u S2 5 6v S 2 2u S, showing carefully how all relevant properties and definitions enter the proof. 104. Let u S 5 82a, a9,v S 5 82a, 22a9. Compute 2u S k v S k 2 3 v S k u S k . • C H A L L E N G E For Exercises 105–110, refer to the following: Vectors can be represented as column matrices. For example, the vector u 5 83, 249 can be represented as a 2 3 1 column matrix c 3 24d . With a TI-83 calculator, vectors can be entered as matrices in two ways, directly or via matrix . Directly: Matrix: Use a calculator to perform the vector operation given u 5 88,259 and v 5 827, 119. 105. u 1 3v 106. 291u 2 2v2 Use a calculator to find a unit vector in the direction of the given vector. 107. u 5 810, 2249 108. u 5 829, 2409 Use the graphing calculator sum command to find the magnitude of the given vector. Also, find the direction angle to the nearest degree. 109. 8233, 180 9 110. H220, 230!5I • T E C H N O L O G Y In Exercises 91–94, determine whether each statment is true or false. • C O N C E P T U A L 91. The magnitude of the vector i is the imaginary number i. 92. The arrow components of equal vectors must coincide. 93. The magnitude of a vector is always greater than or equal to the magnitude of its horizontal component. 94. The magnitude of a vector is always greater than or equal to the magnitude of its vertical component. 95. Would a scalar or a vector represent the following? The car is driving 72 mph due east (908 with respect to north). 96. Would a scalar or vector represent the following? The granite has a mass of 131 kilograms. 97. Find the magnitude of the vector 82a, b9 if a . 0 and b . 0. 98. Find the direction angle of the vector 82a, b9 if a . 0 and b . 0. 8.5 The Dot Product 813 S K I L L S O B J E C T I V E S ■ ■Find the dot product of two vectors. ■ ■Use the dot product to find the angle between two vectors. ■ ■Use the dot product to calculate the amount of work associated with a physical problem. C O N C E P T U A L O B J E C T I V ES ■ ■Understand that the dot product of two vectors is a scalar. ■ ■Understand why the dot product of two orthogonal (perpendicular) vectors is equal to zero. ■ ■Understand that work is done when a force causes an object to move a certain distance. 8.5 THE DOT PRODUCT 8.5.1 Multiplying Two Vectors: The Dot Product With two-dimensional vectors there are two types of multiplication defined for vectors: scalar multiplication and the dot product. Scalar multiplication (which we already demonstrated in Section 8.4) is multiplication of a scalar by a vector; the result is a vector. Now we discuss the dot product of two vectors. In this case, there are two important things to note: (1) the dot product of two vectors is defined only if the vectors have the same number of components, and (2) if the dot product does exist, then the result is a scalar. 8.5.1 S K IL L Find the dot products of two vectors. 8.5.1 C ON C E P T U A L Understand that the dot product of two vectors is a scalar. DOT PRODUCT The dot product of two vectors u 5 8a, b9 and v 5 8c, d 9 is given by u⋅v 5 ac 1 bd u⋅v is pronounced “u dot v.” PROPERTIES OF THE DOT PRODUCT 1. u⋅v 5 v⋅u 2. u⋅u 5 0 u 0 2 3. 0⋅u 5 0 4. k1u⋅v2 5 1ku2 ⋅v 5 u⋅1kv2 5. 1u 1 v2 ⋅w 5 u⋅w 1 v⋅w 6. u⋅1v 1 w2 5 u⋅v 1 u⋅w EXAMPLE 1 Finding the Dot Product of Two Vectors Find the dot product 827, 39⋅82, 59. Solution: Sum the products of the first components and the products of the second components. 827, 39⋅82, 59 5 1272122 1 132152 Simplify. 5 214 1 15 5 1 YOUR T UR N Find the dot product 86, 19⋅822, 39. ▼ [CONCEPT CHECK] TRUE OR FALSE The dot product of two vectors is a scalar, and the product of two scalars is a vector. ANSWER False ▼ STUDY TIP The dot product of two vectors is a scalar. ▼ A N S W E R 29 The following box summarizes the properties of the dot product: These properties are verified in the exercises. 814 CHAPTER 8 Additional Topics in Trigonometry 8.5.2 Angle Between Two Vectors We can use these properties to develop an equation that relates the angle between two vectors and the dot product of the vectors. WORDS MATH Let u and v be two vectors u v θ with the same initial point, and let u be the angle between them. The vector u 2 v is u v u – v θ opposite angle u. A triangle is formed with \|u\| \|v\| \|u – v\| θ side lengths equal to the magnitudes of the three vectors. Apply the Law of Cosines. 0u 2 v 0 2 5 0u 0 2 1 0v 0 2 2 2 0u 0 0v 0 cos u Use properties of the dot product to rewrite the left side of equation. Property (2): 0u 2 v 0 2 5 1u 2 v2 ⋅1u 2 v2 Property (6): 5 u⋅1u 2 v2 2 v⋅1u 2 v2 Property (6): 5 u⋅u 2 u⋅v 2 v⋅u 1 v⋅v Property (2): 5 0u 0 2 2 u⋅v 2 v⋅u 1 0v 0 2 Property (1): 5 0 u 0 2 2 21u⋅v2 1 0 v 0 2 Substitute this last expression for the left side of the original Law of Cosines equation. 0 u 0 2 2 21u⋅v2 1 0 v 0 2 5 0 u 0 2 1 0 v 0 2 2 20 u 0 0 v 0 cos u Simplify. 221u⋅v2 5 220 u 0 0 v 0 cos u Isolate cosu. cosu 5 u⋅v 0 u 0 0 v 0 Notice that u and v have to be nonzero vectors, since we divided by them in the last step. 8.5.2 S KILL Use the dot product to find the angle between two vectors. 8.5.2 C ON CEPTUAL Understand why the dot product of two orthogonal (perpendicular) vectors is equal to zero. ANGLE BETWEEN TWO VECTORS If u is the angle between two nonzero vectors u and v, where 0° # u #180°, then cos u 5 u⋅v 0 u 0 0 v 0 STUDY TIP The angle between two vectors u 5 cos21a u⋅v 0 u 0 0 v 0 b is an angle between 0° and 180° (the range of the inverse cosine function). 8.5 The Dot Product 815 In the Cartesian plane, there are two angles between two vectors, u and 360° 2 u. We assume that u is the “smaller” angle. ▼ A N S W E R 38° EXAMPLE 2 Finding the Angle Between Two Vectors Find the angle between 82, 239 and 824, 39. Solution: Let u 5 82, 239 and v 5 824, 39. STEP 1 Find u⋅v. u⋅v 5 82, 239⋅82 4, 39 5 1221242 1 1232132 5 217 STEP 2 Find 0 u 0. 0 u 0 5 !u⋅u 5 "22 1 12322 5 !13 STEP 3 Find 0 v 0. 0 v 0 5 !v⋅v 5 "12422 1 32 5 !25 5 5 STEP 4 Find u. cos u 5 u⋅v 0 u 00 v 0 5 217 5!13 Approximate u with a calculator. u 5 cos21 a 2 17 5!13b < 160.559965° u < 161° STEP 5 Draw a picture to confirm the answer. Draw the vectors 82, 239 and 824, 39. 161° appears to be correct. Y OUR T UR N Find the angle between 81, 59 and 822, 49. v u x y (–4, 3) (2, –3) 161º ▼ When two vectors are parallel, the angle between them is 0° or 180°. u v θ = 0º u v θ = 180º When two vectors are perpendicular (orthogonal), the angle between them is 90°. u v θ = 90º Note: We did not include 270° because the angle u between two vectors is taken to be the smaller angle. When two vectors u and v are perpendicular, u 5 90°. cos190°2 5 u⋅v 0u 0 0v 0 Substitute cos190°2 5 0. 0 5 u⋅v 0u 0 0v 0 Therefore, the dot product of u and v must be zero. u⋅v 5 0 816 CHAPTER 8 Additional Topics in Trigonometry 8.5.3 Work If you had to carry barbells with weights or pillows for 1 mile, which would you choose? You would probably pick the pillows over the barbell and weights because the pillows are lighter. It requires less work to carry the pillows than it does to carry the weights. If asked to carry either of them 1 mile or 10 miles, you would probably pick 1 mile because it’s a shorter distance and requires less work. Work is done when a force causes an object to move a certain distance. The simplest case is when the force is in the same direction as the displacement— for example, a stagecoach (the horses pull with a force in the same direction). In this case, the work is defined as the magnitude of the force times the magnitude of the displacement, distance d. W 5 0 F 0 d Notice that the magnitude of the force is a scalar, the distance d is a scalar, and hence the product is a scalar. If the horses pull with a force of 1000 pounds and they move the stagecoach 100 feet, the work done by the force is W 5 11000 lb2 1100 ft2 5 100,000 ft-lb In many physical applications, however, the force is not in the same direction as the displacement, and hence vectors (not just their magnitudes) are required. ORTHOGONAL VECTORS Two vectors u and v are orthogonal (perpendicular) if and only if their dot product is zero. u⋅v 5 0 8.5.3 S KILL Use the dot product to calculate the amount of work associated with a physical problem. 8.5.3 CO NCE PTUAL Understand that work is done when a force causes an object to move a certain distance. [CONCEPT CHECK] TRUE OR FALSE Work increases only if either the magnitude of the force or the distance increases. ANSWER True ▼ EXAMPLE 3 Determining Whether Vectors Are Orthogonal Determine whether each pair of vectors is orthogonal. a. u 5 82, 239 and v 5 83, 29 b. u 5 827, 239 and v 5 87, 39 Solution (a): Find the dot product u⋅v. u⋅v 5 122132 1 1232122 Simplify. u⋅v 5 0 Vectors u and v are orthogonal, since u⋅v 5 0. Solution (b): Find the dot product u⋅v. u⋅v 5 1272172 1 1232132 Simplify. u⋅v 5 258 Vectors u and v are not orthogonal, since u⋅v 2 0. [CONCEPT CHECK] Determine the relationship between a and b that guarantees that the vectors u 5 and are orthogonal. ANSWER a 5 6b ▼ 8.5 The Dot Product 817 We often want to know how much of a force is applied in a certain direction. For example, when your car runs out of gasoline and you try to push it, some of the force vector F1 you generate from pushing translates into the horizontal component F2; hence, the car moves horizontally. If we let u be the angle between the vectors F1 and F2, then the horizontal component of F1 is F2 where 0 F20 5 0 F10 cos u. F1 F2 θ If the woman in the picture pushes at an angle of 25° with a force of 150 pounds, then the horizontal component of the force vector F1 is 1150 lb2 1cos 25°2 < 136lb WORDS MATH To develop a generalized formula when the force exerted and the displacement are not in the same direction, we start with the formula for the angle between two vectors. cos u 5 u⋅v 0 u0 0 v0 We then isolate the dot product u⋅v. u⋅v 5 0 u0 0 v0 cosu Let u 5 F and v 5 d. W 5 F⋅d 5 0 F0 0 d0 cos u 5 0 F0 cos u ⋅ 0 d0 magnitude of force distance in direction of displacement g g WORK If an object is moved from point A to point B by a constant force, then the work associated with this displacement is W 5 F⋅d where d is the displacement vector and F is the force vector. Work is typically expressed in one of two units: SYSTEM FORCE DISTANCE WORK U.S. customary pound foot ft-lb SI newton meter N-m 818 CHAPTER 8 Additional Topics in Trigonometry ▼ A N S W E R 25 N-m [SEC TION 8.5] E X E R CI SE S • S K I L L S Orthogonal (perpendicular) vectors have an angle of 90° between them, and consequently the dot product of two orthogonal vectors is equal to zero. Work is the result of a force displacing an object. When the force and displacement are in the same direction, the work is equal to the product of the magnitude of the force and the distance (magnitude of the displacement). When the force and displacement are not in the same direction, work is the dot product of the force vector and displacement vector, W 5 F⋅d. In this section, we defined the dot product as a form of multipli-cation of two vectors. A scalar times a vector results in a vector, whereas the dot product of two vectors is a scalar. 8a, b9⋅8c, d9 5 ac 1 bd We developed a formula that determines the angle u between two vectors u and v. cos u 5 u⋅v 0u 0 0v 0 [SEC TION 8.5] S U M MA RY In Exercises 1–12, find the indicated dot product. 1. 84, 229⋅83, 59 2. 87, 89⋅82, 219 3. 825, 69⋅83, 29 4. 86, 239⋅82, 19 5. 827, 249⋅822, 279 6. 85, 229⋅821, 219 7. H !3, 22I ⋅H3!3, 21I 8. H4!2, !7 I ⋅H2!2, 2!7 I 9. 85, a9⋅823a, 29 10. 84 x, 3y9⋅82y, 25x9 11. 80.8, 20.59⋅82, 69 12. 8218, 39⋅810, 23009 In Exercises 13–24, find the angle (round to the nearest degree) between each pair of vectors. 13. 824, 39 and 825, 299 14. 82, 249 and 84, 219 15. 822, 239 and 823, 49 16. 86, 59 and 83, 229 17. 824, 69 and 826, 89 18. 81, 59 and 823, 229 19. H22, 2!3I and H2!3, 1I 20. H23!3, 23I and H22!3, 2I 21. H25!3, 25I and H!2, 2!2I 22. H25, 25!3I and H2, 2!2I 23. 84, 69 and 826, 299 24. 82, 89 and 8212, 39 In Exercises 25–36, determine whether each pair of vectors is orthogonal. 25. 826, 89 and 828, 69 26. 85, 229 and 825, 29 27. 86, 249 and 826, 299 28. 88, 39 and 826, 169 29. 80.8, 49 and 83, 269 30. 827, 39 and H1 7, 21 3I 31. 85, 20.49 and 81.6, 209 32. 812, 99 and 83, 249 33. H !3, !6I and H2!2, 1I 34. H !7, 2!3I and 83, 79 35. H4 3, 8 15I and H2 1 12, 5 24I 36. H5 6, 6 7I and H36 25, 2 49 36I EXAMPLE 4 Calculating Work How much work is done when a force (in pounds) F 5 82, 49 moves an object from 10, 02 to 15, 92 (the distance is in feet)? Solution: Find the displacement vector d. d 5 85, 99 Apply the work formula, W 5 F⋅d. W 5 82, 49⋅85, 99 Calculate the dot product. W 5 122152 1 142192 Simplify. W 5 46 ft-lb Y OUR TU R N How much work is done when a force (in newtons) F 5 81, 39 moves an object from 10, 02 to 14, 72 (the distance is in meters)? ▼ 8.5 The Dot Product 819 • A P P L I C A T I O N S 37. Lifting Weights. How much work does it take to lift 100 pounds vertically 4 feet? 38. Lifting Weights. How much work does it take to lift 150 pounds vertically 3.5 feet? 39. Raising Wrecks. How much work is done by a crane to lift a 2-ton car to a level of 20 feet? 40. Raising Wrecks. How much work is done by a crane to lift a 2.5-ton car to a level of 25 feet? 41. Work. To slide a crate across the floor, a force of 50 pounds at a 30° angle is needed. How much work is done if the crate is dragged 30 feet? 30º Direction dragged 42. Work. To slide a crate across the floor, a force of 800 pounds at a 20° angle is needed. How much work is done if the crate is dragged 50 feet? 43. Close a Door. A sliding door is closed by pulling a cord with a constant force of 35 pounds at a constant angle of 45°. The door is moved 6 feet to close it. How much work is done? 44. Close a Door. A sliding door is closed by pulling a cord with a constant force of 45 pounds at a constant angle of 55°. The door is moved 6 feet to close it. How much work is done? 45. Braking Power. A car that weighs 2500 pounds is parked on a hill in San Francisco with a slant of 40° from the horizontal. How much force will keep it from rolling down the hill? 46. Towing Power. A car that weighs 2500 pounds is parked on a hill in San Francisco with a slant of 40° from the horizontal. A tow truck has to remove the car from its parking spot and move it 120 feet up the hill. How much work is required? 47. Towing Power. A semi-trailer truck that weighs 40,000 pounds is parked on a hill in San Francisco with a slant of 10° from the horizontal. A tow truck has to remove the truck from its parking spot and move it 100 feet up the hill. How much work is required? 48. Braking Power. A truck that weighs 40,000 pounds is parked on a hill in San Francisco with a slant of 10° from the horizontal. How much force will keep it from rolling down the hill? 49. Business. Suppose that u S 5 82000, 50009 represents the number of units of battery A and B, respectively, produced by a company and that v S 5 88.40, 6.509 represents the price (in dollars) of a 10-pack of battery A and B, respectively. Compute and interpret u S⋅v S. 50. Demographics. Suppose that u S 5 8120, 809 represents the number of males and females in a high school class, and v S 5 87.2, 5.39 represents the average number of minutes it takes a male and female, respectively, to register. Compute and interpret u S⋅v S. 51. Geometry. Use vector methods to show that the diagonals of a rhombus are perpendicular to each other. 52. Geometry. Let u S be a unit vector, and consider the following diagram: \|u\| \|v\| \|w\| Compute u S⋅v S and u S⋅w S. 53. Geometry. Consider the following diagram: \|v\| = 〈v1, v2〉 \|w\| = 〈w1, w2〉 x y α β 8 7 6 5 4 3 2 1 9 8 7 6 5 4 3 2 1 –1 –2 a. Compute cos b, sin b, cos a, and sin a. b. Use (a) to show that cos1a 2b2 5 v S ⋅ w S "v S ⋅ v S "w S ⋅ w S. 54. Geometry. Consider the diagram in Exercise 53. a. Compute cos b, sin b, cos a, and sin a. b. Use (a) to show that cos1a 1 b2 5 v S⋅8w1, 2w29 !v S⋅v S !w S ⋅w S. 55. Tennis. A player hits an overhead smash at full arm exten-sion at the top of his racquet, which is 7 feet from the ground. The ball travels 16.3 feet. Consult the following diagram: y –10 –6 –2 –14 –18 –8 –10 x v = 〈v1, –7〉 u = 〈0, –7〉 \|v\| = 16.3 θ β a. Determine v1. b. Find the angle u with which the player hits this smash. 56. Tennis. In Exercise 55, use the dot product to determine the angle b with which the ball hits the ground. 57. Optimization. Let u S 5 8a, b9 be a given vector, and suppose that the head of n S 5 8n1, n29 lies on the circle x2 1 y2 5 r2. Find the vector n S such that u S⋅n S is as big as possible. Find the actual value of u S⋅n S in this case. 58. Optimization. Let u S 5 8a, b9 be a given vector, and suppose that the head of n S 5 8n1, n29 lies on the circle x2 1 y2 5 r2. Find the vector n S such that u S⋅n S is as small as possible. Find the actual value of u S⋅n S in this case. 820 CHAPTER 8 Additional Topics in Trigonometry • C A T C H T H E M I S T A K E In Exercises 61 and 62, explain the mistake that is made. 61. Find the dot product 823, 29⋅ 82, 59. Solution: Multiply component by component. 823, 29⋅ 82, 59 5 81232122, 122152 9 Simplify. 823, 29⋅ 82, 59 5 826, 109 This is incorrect. What mistake was made? 62. Find the dot product 811, 129⋅ 822, 39. Solution: Multiply the outer and inner components. 811, 129⋅ 822, 39 5 1112132 1 1122 1222 Simplify. 811, 129⋅ 822, 39 5 9 This is incorrect. What mistake was made? In Exercises 63–66, determine whether each statement is true or false. 63. A dot product of two vectors is a vector. 64. A dot product of two vectors is a scalar. 65. Orthogonal vectors have a dot product equal to zero. 66. If the dot product of two nonzero vectors is equal to zero, then the vectors must be perpendicular. For Exercises 67 and 68, refer to the following to find the dot product: The dot product of vectors with n components is 8a1, a2, . . . , an 9⋅ 8b1, b2, . . . , bn 9 5 a1b1 1 a2b2 1 . . . 1 anbn. 67. 83, 7, 259⋅822, 4, 19 68. 81, 0, 22, 39⋅85, 2, 3, 19 In Exercises 69–72, given u 5 8a, b9 and v 5 8c, d 9, show that the following properties are true. 69. u⋅v 5 v⋅u 70. u⋅u 5 0 u 0 2 71. 0⋅u 5 0 72. k1u⋅v2 5 1ku2 ⋅v 5 u⋅1kv2, k is a scalar • C O N C E P T U A L 73. Show that u S⋅1v S 1 w S2 5 u S⋅v S 1 u S⋅w S. 74. Show that 0 u S 2 v S 0 2 5 0 u S 0 2 1 0 v S 0 2 2 2Au S⋅v SB. 75. The projection of v S onto u S is defined by proju S v S 5 au S⋅v S 0 u S 0 2b u S. This vector is depicted below. Heuristically, this is the “shadow” of v S on u S. a. Compute proj2u S 2u S. b. What is proj2u S cu S for any c . 0? 76. a. Compute proju S 2u S. b. What is proju S cu S for any c . 0? 77. Suppose that you are given a vector u S. For what vectors v S does proju S v S 5 0 S? 78. True or False: proju S Av S 1 w SB 5 proju S v S 1 proju S w S. 79. If u S and v S are unit vectors, determine the maximum and minimum value of A22u SB ⋅A3v SB. 80. Assume that the angle between u S and v S is u 5 p 3. Show that Au S⋅v SBu S 0 v S 0 2 Av S⋅u SBv S 0 u 0 5 0 u S 0u S 2 0 v S 0 v S 2 . \|v\| \|u\| y x θ 10 8 6 4 2 –2 –6 10 8 6 4 2 –10 –8 –10 Projection • C H A L L E N G E 59. Pursuit Theory. Assume that the head of u S is restricted so that its tail is at the origin, and its head is on the unit circle in quadrant II or quadrant III. A vector v S has its tail at the origin, and its head must lie on the line y 5 2 2 x in quadrant I. Find the smallest value of u S⋅v S. 60. Pursuit Theory. Assume that the head of u S is restricted so that its tail is at the origin and its head is on the unit circle in quadrant I or quadrant IV. A vector v S has its tail at the origin, and its head must lie on the line y 5 2 2 x in quadrant I. Find the biggest value of u S⋅v S. 8.6 Polar (Trigonometric) Form of Complex Numbers 821 For Exercises 81 and 82, find the indicated dot product with a calculator. 81. 8211, 349⋅815, 2279 82. 823, 23509⋅845, 2029 83. A rectangle has sides with lengths 18 units and 11 units. Find the angle to one decimal place between the diagonal and the side with length of 18 units. (Hint: Set up a rectangular coordinate system, and use vectors 818, 09 to represent the side of length 18 units and 818, 119 to represent the diagonal.) 84. The definition of a dot product and the formula to find the angle between two vectors can be extended and applied to vectors with more than two components. A rectangular box has sides with lengths 12 feet, 7 feet, and 9 feet. Find the angle, to the nearest degree, between the diagonal and the side with length 7 feet. Use the graphing calculator SUM command to find the angle (round to the nearest degree) between each pair of vectors. 85. 8225, 429, 810, 359 86. 8212, 99, 8221, 2139 • T E C H N O L O G Y S K I L L S O B J E C T I V E S ■ ■Calculate the modulus of a complex number. ■ ■Convert complex numbers from rectangular form to polar form and vice versa. C O N C E P T U A L O B J E C T I V ES ■ ■Understand that the modulus, or magnitude, of a complex number is the distance from the origin to the point in the complex plane. ■ ■Understand that imaginary numbers lie along the vertical axis and that real numbers lie along the horizontal axis of the complex plane. 8.6 POLAR (TRIGONOMETRIC) FORM OF COMPLEX NUMBERS 8.6.1 Complex Numbers in Rectangular Form We are already familiar with the rectangular coordinate system, where the horizontal axis is called the x-axis and the vertical axis is called the y-axis. In our study of complex numbers, we refer to the standard (rectangular) form as a 1 bi, where a represents the real part and b represents the imaginary part. If we let the horizontal axis be the real axis and the vertical axis be the imaginary axis, the result is the complex plane. The point a 1 bi is located in the complex plane by finding the coordinates 1a, b2. a a + bi b Real axis Imaginary axis When b 5 0, the result is a real number, and therefore any numbers along the horizontal axis are real numbers. When a 5 0, the result is an imaginary number, so any numbers along the vertical axis are imaginary numbers. The variable z is often used to represent a complex number: z 5 x 1 iy. Complex numbers are analogous to vectors. Suppose that a vector z 5 8x, y9, whose initial point is the origin and whose terminal point is 1x, y2; then the magnitude of that vector is 0 z 0 5 "x2 1 y2. Similarly, the magnitude, or modulus, of a complex number is defined like the magnitude of a position vector in the xy-plane, as the distance from the origin 10, 02 to the point 1x, y2 in the complex plane. 8.6.1 S K I L L Calculate the modulus of a complex number. 8.6.1 C O N C E P T U A L Understand that the modulus, or magnitude, of a complex number is the distance from the origin to the point in the complex plane. x z = x + iy z y Real axis Imaginary axis 822 CHAPTER 8 Additional Topics in Trigonometry Recall that a complex number z 5 x 1 iy has a complex conjugate z 5 x 2 iy. The bar above a complex number denotes its conjugate. Notice that zz 5 1x 1 iy21x 2 iy2 5 x2 2 i2y2 5 x2 1 y2 and therefore the modulus can also be written as 0 z 0 5 !zz DEFINITION Modulus of a Complex Number The modulus, or magnitude, of a complex number z 5 x 1 iy is the distance from the origin to the point 1x, y2 in the complex plane given by 0 z 0 5 "x2 1 y2 EXAMPLE 1 Finding the Modulus of a Complex Number Find the modulus of z 5 23 1 2i. common mistake Including the i in the imaginary part. Y OUR TU R N Find the modulus of z 5 2 2 5i. ▼ ✖I N C O R R EC T Let x 5 23 and y 5 2i in 0 z 0 5 "x2 1 y2. 0 23 1 2i 0 5 "12322 1 12i22 ERROR The i is not included in the formula. Only the imaginary part (the coefficient of i) is used. ✓C O R REC T Let x 5 23 and y 5 2 in 0 z 0 5 "x2 1 y2. 0 23 1 2i 0 5 "12322 1 22 Eliminate the parentheses. 0 23 1 2i 0 5 !9 1 4 Simplify. 0 z 0 5 0 23 1 2i 0 5 !13 [CONCEPT CHECK] TRUE OR FALSE The modulus of a complex number is greater than or equal to the sum of the real part and the imaginary part of the complex number. ANSWER False ▼ ▼ A N S W E R 0 z 0 5 0 2 2 5i 0 5 !29 8.6.2 Complex Numbers in Polar Form We say that a complex number z 5 x 1 iy is in rectangular form because it is located at the point 1x, y2, which is expressed in rectangular coordinates, in the complex plane. Another convenient way of expressing complex numbers is in polar form. Recall from our study of vectors (Section 8.4) that vectors have both magnitude and a direction angle. The same is true of numbers in the complex plane. Let r represent the magnitude, or distance from the origin to the point 1x, y2, and u represent the direction angle; then we have the following relationships: r 5 "x2 1 y2 sin u 5 y r cos u 5 x r and tan u 5 y x 1x 2 02 Isolating x and y in the sinusoidal functions, we find x 5 r cos u y 5 r sin u Using these expressions for x and y, we can write a complex number in polar form. z 5 x 1 yi 5 1r cos u2 1 1r sin u2i 5 r 1cos u 1 i sin u2 8.6.2 S KILL Convert complex numbers from rectangular form to polar form and vice versa. 8.6.2 CO NCE PTUAL Understand that imaginary numbers lie along the vertical axis and that real numbers lie along the horizontal axis of the complex plane. 8.6 Polar (Trigonometric) Form of Complex Numbers 823 [CONCEPT CHECK] TRUE OR FALSE In the complex plane, a real number or purely imaginary number lies along one of the axes and all other complex numbers lie in one of the four quadrants. ANSWER True ▼ POLAR (TRIGONOMETRIC) FORM OF COMPLEX NUMBERS The following expression is the polar form of a complex number: z 5 r1cos u 1 i sin u2 where r represents the modulus (magnitude) of the complex number and u represents the argument of z. The following is standard notation for modulus and argument: r 5 mod z 5 0 z 0 and u 5 Arg z, 0 # u , 2p or 0° # u , 360° Converting Complex Numbers Between Rectangular and Polar Forms We can convert back and forth between rectangular and polar (trigonometric) forms of complex numbers using the modulus and trigonometric ratios. r 5 "x2 1 y2 sin u 5 y r cos u 5 x r and tan u 5 y x 1x 2 02 CONVERTING COMPLEX NUMBERS FROM RECTANGULAR FORM TO POLAR FORM Step 1: Plot the point z 5 x 1 yi in the complex plane (note the quadrant). Step 2: Find r. Use r 5 "x2 1 y2. Step 3: Find u. Apply tan u 5 y x, x 2 0, where u is in the quadrant found in Step 1. Step 4: Write the complex number in polar form: z 5 r 1cos u 1 isin u2. Realize that imaginary numbers, z 5 0 1 bi, lie on the imaginary axis. Therefore u 5 90° if b . 0 and u 5 270° if b , 0. EXAMPLE 2 Converting from Rectangular to Polar Form Express the complex number z 5 !3 2 i in polar form. Solution: STEP 1 Plot the point. The point lies in quadrant IV. STEP 2 Find r. Let x 5 !3 and y 5 21 in r 5 "x2 1 y2. r 5 "A !3B2 1 12122 Eliminate the parentheses. r 5 !3 1 1 Simplify. r 5 2 θ r Real axis QIV Imaginary axis z = 3 – i √ x θ z = x + iy y r Real axis Imaginary axis 824 CHAPTER 8 Additional Topics in Trigonometry You must be very careful in converting from rectangular to polar form. Remember that the inverse tangent function is a one-to-one function and will yield values in quadrants I and IV. If the point lies in quadrant II or III, add 180° to the angle found through the inverse tangent function. STEP 3 Find u. Let x 5 !3 and y 5 21 in tan u 5 y x. tan u 5 2 1 !3 Solve for u u 5 tan21 a 2 1 !3b 5 2p 6 Find the reference angle. reference angle 5 p 6 The complex number lies in quadrant IV. u 5 11p 6 STEP 4 Write the complex number in polar form. z 5 2 ccos a11p 6 b 1 i sin a11p 6 bd z 5 r 1cos u 1 i sin u2 Note: An alternative form is in degrees: z 5 21cos 330° 1 i sin 330°2. Y OUR TU R N Express the complex number z 5 1 2 i!3 in polar form. ▼ ▼ A N S W E R z 5 2 ccos a5p 3 b 1 isina5p 3 bd or 2 1cos 300° 1 isin 300°2 ✖I N C O R R EC T ✓C O R R EC T Step 1: Plot the point. The point lies in quadrant II. θ r Real axis QII Imaginary axis z = –2 + i Step 2: Find r. Let x 5 22 and y 5 1 in r 5 "x2 1 y2. r 5 "12222 1 12 Simplify. r 5 !5 EXAMPLE 3 Converting from Rectangular to Polar Form common mistake Forgetting to confirm the quadrant, which results in using the reference angle instead of the actual angle. Express the complex number z 5 22 1 i in polar form. 8.6 Polar (Trigonometric) Form of Complex Numbers 825 ▼ A N S W E R z 5 !5 3cos1116.6°2 1 isin1116.6°24 ▼ A N S W E R z 5 2!3 2 i ▼ A N S W E R z < 25.5904 2 4.2127i Evaluate the inverse function with a calculator. u 5 tan21 a 2 1 2b 5 226.565° Write the complex number in polar form. z 5 r 1cos u 1 i sin u2 z 5 !5 3cos1226.6°2 1i sin1226.6°24 Note: u 5 226.565° lies in quadrant IV, whereas the original point lies in quadrant II. Therefore, we should have added 180° to u in order to arrive at a point in quadrant II. Step 3: Find u. Let x 5 22 and y 5 1 in tan u 5 y x. tan u 5 21 2 u 5 tan21 a 2 1 2b 5 226.565° The complex number lies in quadrant II. u 5 226.6° 1 180° 5 153.4° Step 4: Write the complex number in polar form z 5 r 1cos u 1 i sin u2. z 5 !5 3cos1153.4°2 1 isin1153.4°24 Y OUR T UR N Express the complex number z 5 21 1 2i in polar form. ▼ To convert from polar to rectangular form, simply evaluate the trigonometric functions. EXAMPLE 4 Converting from Polar to Rectangular Form Express z 5 41cos 120° 1 i sin 120°2 in rectangular form. Solution: Evaluate the trigonometric z 5 4 acos 120° 1 i sin 120°b functions exactly. Distribute the 4. z 5 4 a21 2b 1 4 a !3 2 b i Simplify. z 5 22 1 2!3i Y OUR T UR N Express z 5 21cos 210° 1 i sin 210°2 in rectangular form. 21 2 3 3 !3 2 ▼ EXAMPLE 5 Using a Calculator to Convert from Polar to Rectangular Form Express z 5 31cos 109° 1 i sin 109°2 in rectangular form. Round to four decimal places. Solution: Use a calculator to evaluate z 5 3acos1109°2 1 i sin1109°2b the trigonometric functions. Simplify. z < 20.9767 1 2.8366 i Y OUR T UR N Express z 5 71cos 217° 1 i sin 217°2 in rectangular form. Round to four decimal places. 20.325568 3 3 0.945519 ▼ 826 CHAPTER 8 Additional Topics in Trigonometry [SEC TION 8.6] E X E R CI SE S • S K I L L S and tan u 5 y x, x 2 0 and 0 # u , 2p. It is important to note in which quadrant the point lies. To convert from polar to rectangular form, simply evaluate the trigonometric functions. x 5 r cos u and y 5 r sin u In the complex plane, the horizontal axis is the real axis and the vertical axis is the imaginary axis. We can express complex numbers in either rectangular form, z 5 x 1 iy, or polar form, z 5 r1cosu 1 i sinu2. The modulus of a complex number, z 5 x 1 iy, is given by 0 z 0 5 "x2 1 y2. To convert from rectangular to polar form, we use the relationships r 5 "x2 1 y2 [SEC TION 8.6] S U M MA RY In Exercises 1–8, graph each complex number in the complex plane. 1. 7 1 8i 2. 3 1 5i 3. 22 2 4i 4. 23 2 2i 5. 2 6. 7 7. 23i 8. 25i In Exercises 9–22, express each complex number in polar form. 9. 1 2 i 10. 2 1 2i 11. 1 1 !3i 12. 23 2 !3i 13. 24 1 4i 14. !5 2 !5i 15. !3 2 3i 16. 2!3 1 i 17. 3 1 0i 18. 22 1 0i 19. 2 1 2 2 1 2 i 20. 1 6 2 1 6 i 21. 5.32 2 0i 22. 26.48 1 0i In Exercises 23–38, use a calculator to express each complex number in polar form. 23. 3 2 7i 24. 2 1 3i 25. 26 1 5i 26. 24 2 3i 27. 25 1 12i 28. 24 1 7i 29. 8 2 6i 30. 23 1 4i 31. 2 1 2 1 3 4i 32. 2 5 8 2 11 4 i 33. 5.1 1 2.3i 34. 1.8 2 0.9i 35. 22!3 2 !5i 36. 2 4!5 3 1 !5 2 i 37. 4.02 2 2.11i 38. 1.78 2 0.12i In Exercises 39–48, express each complex number in rectangular form. 39. 51cos 180° 1 i sin 180°2 40. 21cos 135° 1 i sin 135°2 41. 21cos 315° 1 i sin 315°2 42. 31cos 270° 1 i sin 270°2 43. 241cos 60° 1 i sin 60°2 44. 241cos 210° 1 i sin 210°2 45. !31cos 150° 1 i sin 150°2 46. !31cos 330° 1 i sin 330°2 47. !2ccos ap 4b 1 i sin ap 4bd 48. 2ccos a5p 6 b 1 i sin a5p 6 bd In Exercises 49–58, use a calculator to express each complex number in rectangular form. 49. 51cos 295° 1 i sin 295°2 50. 41cos 35° 1 i sin 35°2 51. 31cos 100° 1 i sin 100°2 52. 61cos 250° 1 i sin 250°2 53. 271cos 140° 1 i sin 140°2 54. 251cos 320° 1 i sin 320°2 55. 3 ccos a11p 12 b 1 i sin a11p 12 bd 56. 2 ccos a4p 7 b 1 i sin a4p 7 bd 57. 22ccos a3p 5 b 1 i sin a3p 5 bd 58. 24 ccos a15p 11b 1 i sin a15p 11bd 8.6 Polar (Trigonometric) Form of Complex Numbers 827 • A P P L I C A T I O N S 59. Resultant Force. Force A, at 100 pounds, and force B, at 120 pounds, make an angle of 30° with each other. Represent their respective vectors as complex numbers written in trigonometric form, and solve for the resultant force. 100 lb Force A 120 lb Force B 30º 60. Resultant Force. Force A, at 40 pounds, and force B, at 50 pounds, make an angle of 45° with each other. Represent their respective vectors as complex numbers written in trigonometric form, and solve for the resultant force. 61. Resultant Force. Force A, at 80 pounds, and force B, at 150 pounds, make an angle of 30° with each other. Represent their respective vectors as complex numbers written in trigonometric form, and solve for the resultant angle. 62. Resultant Force. Force A, at 20 pounds, and force B, at 60 pounds, make an angle of 60° with each other. Represent their respective vectors as complex numbers written in trigonometric form, and solve for the resultant angle. 63. Resultant Force. Refer to Section 8.4, Exercise 85. Express the vectors u S, v S corresponding to the dogs as complex numbers written in trigonometric form, and find the resultant force w S to keep the system in equilibrium. 64. Resultant Force. Refer to Section 8.4, Exercise 86. Express the vectors u S, v S, w S as complex numbers written in trigonometric form, and find the resultant force z S to keep the system in equilibrium. 65. Resultant Force. Refer to Section 8.4, Exercise 75. Express the resultant vector R S as a complex number in trigonometric form. 66. Resultant Force. Refer to Section 8.4, Exercise 76. Express the resultant vector R S as a complex number in trigonometric form. • C A T C H T H E M I S T A K E In Exercises 67 and 68, explain the mistake that is made. 67. Express z 5 23 2 8i in polar form. Solution: Find r. r 5 "x2 1 y2 5 !9 1 64 5 !73 Find u. tan u 5 8 3 u 5 tan 21 a8 3b 5 69.44° Write the complex number in polar form. z 5 !73 1cos 69.44° 1 i sin 69.44°2 This is incorrect. What mistake was made? 68. Express z 5 23 1 8i in polar form. Solution: Find r. r 5 "x2 1 y2 5 !9 1 64 5 !73 Find u. tan u 5 2 8 3 u 5 tan 21 a2 8 3b 5 269.44° Write the complex number in polar form. z 5 !73 3cos1269.44°2 1 i sin1269.44°24 This is incorrect. What mistake was made? In Exercises 69–72, determine whether each statement is true or false. 69. In the complex plane, any point that lies along the horizontal axis is a real number. 70. In the complex plane, any point that lies along the vertical axis is an imaginary number. 71. The modulus of z and the modulus of z are equal. 72. The argument of z and the argument of z are equal. 73. Find the argument of z 5 a, where a is a positive real number. 74. Find the argument of z 5 bi, where b is a positive real number. 75. Find the modulus of z 5 bi, where b is a negative real number. 76. Find the modulus of z 5 a, where a is a negative real number. In Exercises 77 and 78, use a calculator to express the complex number in polar form. 77. a 2 2ai, where a . 0 78. 23a 2 4ai, where a . 0 • C O N C E P T U A L 828 CHAPTER 8 Additional Topics in Trigonometry 79. Suppose that a complex number z lies on the circle x2 1 y2 5 p2. If cos au 2b 5 1 2 and sin u , 0, find the rectangular form of z. 80. Suppose that a complex number z lies on the circle x2 1 y2 5 8. If sin au 2b 5 2 !3 2 and cos u , 0, find the rectangular form of z. 81. Consider the following diagram: x y θ 8 9 7 6 5 4 3 2 1 9 8 7 6 5 4 3 2 1 –1 –1 36º 110º Z d 6.5 4.1 Find z in trigonometric form. (Hint: Use the Law of Cosines.) 82. Consider the following diagram: y θ 1 –5 –7 –9 1 –3 –4 –5 –6 –7 –9 –8 80º 25º d x 3√2 2√3 Find z in trigonometric form. (Hint: Use the Law of Cosines.) • C H A L L E N G E Graphing calculators are able to convert complex numbers from rectangular to polar form using the abs command to find the modulus and the angle command to find the angle. 83. Find abs11 1 i2. Find angle11 1 i2. Write 1 1 i in polar form. 84. Find abs 11 2 i2. Find angle 11 2 i2. Write 1 2 i in polar form. A second way of using a graphing calculator to convert between rectangular and polar coordinates is with the Pol and Rec commands. 85. Find Pol(2, 1). Write 2 1 i in polar form. 86. Find Rec1345°2. Write 31cos 45° 1 i sin 45°2 in rectangular form. Another way of using a graphing calculator to represent complex numbers in rectangular form is to enter the real and imaginary parts as a list of two numbers and use the SUM command to find the modulus. 87. Write 28 2 21i in polar form using the SUM command to find its modulus, and round the angle to the nearest degree. 88. Write 2!21 1 10i in polar form using the SUM command to find its modulus, and round the angle to the nearest degree. • T E C H N O L O G Y 8.7 Products, Quotients, Powers, and Roots of Complex Numbers; De Moivre’s Theorem 829 S K I L L S O B J E C T I V E S ■ ■Find the product of two complex numbers. ■ ■Find the quotient of two complex numbers. ■ ■Raise a complex number to an integer power. ■ ■Find the nth root of a complex number. C O N C E P T U A L O B J E C T I V ES ■ ■Understand that when two complex numbers are multiplied, the magnitudes are multiplied and the arguments are added. ■ ■Understand that when two complex numbers are divided, the magnitudes are divided and the arguments are subtracted. ■ ■Understand that when a complex number is raised to an integer power n, the magnitude is raised to the power n and the argument is multiplied by n. ■ ■We can often solve a polynomial equation by finding complex roots. 8.7 PRODUCTS, QUOTIENTS, POWERS, AND ROOTS OF COMPLEX NUMBERS; DE MOIVRE’S THEOREM In this section, we will multiply complex numbers, divide complex numbers, raise complex numbers to powers, and find roots of complex numbers. 8.7.1 Products of Complex Numbers First, we will derive a formula for the product of two complex numbers. 8.7.1 S K I L L Find the product of two complex numbers. 8.7.1 C O N C E P T U A L Understand that when two complex numbers are multiplied, the magnitudes are multiplied and the arguments are added. WORDS MATH Start with two complex z1 5 r11cos u1 1 i sin u12 numbers z1 and z2. z2 5 r2 1cos u2 1 i sin u22 Multiply z1 and z2. z1z2 5 r1r2 1cos u1 1 i sin u12 1cos u2 1 i sin u22 Use the FOIL method to multiply the expressions in parentheses. z1z2 5 r1r21cos u1 cos u2 1 i cos u1 sin u2 1 i sin u1 cos u2 1 i 2 sin u1 sin u22 Group the real parts and the imaginary parts. z1z2 5 r1r2 31cos u1 cos u2 2 sin u1 sin u22 1 i1cos u1 sin u2 1 sin u1 cos u224 Apply the cosine and sine z1z2 5 r1r2 c 1cos u1 cos u2 2 sin u1 sin u22 1 i1cos u1 sin u2 1 sin u1 cos u22 d sum identities. Simplify. z1z2 5 r1r23cos1u1 1 u22 1 i sin1u1 1 u224 21 " cos1u1 1 u22 8 sin1u1 1 u22 8 PRODUCT OF TWO COMPLEX NUMBERS Let z1 5 r11cos u1 1 i sin u12 and z2 5 r2 1cos u2 1 i sin u22 be two complex numbers. The complex product z1z2 is given by z1z2 5 r1r2 3cos1u1 1 u22 1 i sin1u1 1 u224 In other words, when multiplying two complex numbers, multiply the magnitudes and add the arguments. STUDY TIP When two complex numbers are multiplied, the magnitudes are multiplied and the arguments are added. 830 CHAPTER 8 Additional Topics in Trigonometry EXAMPLE 1 Multiplying Complex Numbers Find the product of z1 5 31cos 35° 1 i sin 35°2 and z2 5 21cos 10° 1 i sin 10°2. Solution: Set up the product. z1z2 5 3 1cos 35° 1 i sin 35°2 ⋅2 1cos 10° 1 i sin 10°2 Multiply the magnitudes and add the arguments. z1z2 5 3⋅2 3cos135° 1 10°2 1 i sin135° 1 10°24 Simplify. z1z2 5 6 1cos 45° 1 i sin 45°2 The product is in polar form. To express the product in rectangular form, evaluate z1z2 5 6 a !2 2 1 i !2 2 b 5 3!2 1 3i !2 the trigonometric functions. Product in polar form: z1z2 5 6 1cos 45° 1 i sin 45°2 5 6 ccosap 4b 1 i sinap 4 bd Product in rectangular form: z1z2 5 3!2 1 3 i !2 Y OUR TU R N Find the product of z1 5 2 1cos 55° 1 i sin 55°2 and z2 5 5 1cos 65° 1 i sin 65°2. Express the answer in both polar and rectangular form. ▼ [CONCEPT CHECK] TRUE OR FALSE When two complex numbers are multiplied, their magnitudes and arguments are multiplied, respectively. ANSWER False ▼ ▼ A N S W E R z1z2 5 101cos 120° 1 i sin 120°2 or z1z2 5 25 1 5i!3 8.7.2 Quotients of Complex Numbers We now derive a formula for the quotient of two complex numbers. WORDS MATH Start with two complex numbers z1 and z2. z1 5 r11cos u1 1 i sin u12 and z2 5 r21cos u2 1 i sin u22 Divide z1 by z2. z1 z2 5 r11cos u1 1 i sin u12 r21cos u2 1 i sin u22 5 ar1 r2b acos u1 1 i sin u1 cos u2 1 i sin u2 b Multiply the numerator and the denominator of the second expression in parentheses by the conjugate of the denominator, z1 z2 5 ar1 r2b acos u1 1 i sin u1 cos u2 1 i sin u2 b acos u2 2 i sin u2 cos u2 2 i sin u2 b cos u2 2 i sin u2. Use the FOIL method to multiply the expressions in parentheses in the last z1 z2 5 ar1 r2b acos u1 cos u2 2 i2 sin u1 sin u2 1 i sin u1 cos u2 2 i sin u2 cos u1 cos 2 u2 2 i2 sin 2 u2 b two expressions. Substitute i2 5 21 and group the real parts and the z1 z2 5 ar1 r2b £ 1cos u1 cos u2 1 sin u1 sin u22 1 i 1sin u1 cos u2 2 sin u2 cos u12 cos2 u2 1 sin2 u2 § imaginary parts. Simplify. z1 z2 5 ar1 r2b 31cos u1 cos u2 1 sin u1 sin u22 1 i 1sin u1 cos u2 2 sin u2 cos u124 Use the cosine and sine difference identities. z1 z2 5 ar1 r2b 1cos u1 cos u2 1 sin u1 sin u22 1 i 1sin u1 cos u2 2 sin u2 cos u12 Simplify. z1 z2 5 r1 r2 3cos1u1 2 u22 1 i sin1u1 2 u224 1 5 cos1u1 2 u22 8 sin1u1 2 u22 8 8.7 Products, Quotients, Powers, and Roots of Complex Numbers; De Moivre’s Theorem 831 [CONCEPT CHECK] TRUE OR FALSE When two complex numbers are divided, their magnitudes and arguments are divided, respectively. ANSWER False ▼ 8.7.2 S K I L L Find the quotient of two complex numbers. 8.7.2 C ON C E P T U A L Understand that when two complex numbers are divided, the magnitudes are divided and the arguments are subtracted. It is important to notice that the argument difference is the argument of the numerator minus the argument of the denominator. QUOTIENT OF TWO COMPLEX NUMBERS Let z1 5 r11cos u1 1 i sin u12 and z2 5 r21cos u2 1 i sin u22 be two complex numbers. The complex quotient z1 z2 is given by z1 z2 5 r1 r2 3cos1u1 2 u22 1 i sin1u1 2 u224 In other words, when dividing two complex numbers, divide the magnitudes and subtract the arguments. It is important to note that the argument difference is the argument of the complex number in the numerator minus the argument of the complex number in the denominator. EXAMPLE 2 Dividing Complex Numbers Let z1 5 6 1cos 125° 1 i sin 125°2 and z2 5 3 1cos 65° 1 i sin 65°2. Find z1 z2. Solution: Set up the quotient. z1 z2 5 6 1cos 125° 1 i sin 125°2 3 1cos 65° 1 i sin 65°2 Divide the magnitudes and subtract the arguments. z1 z2 5 6 3 3cos1125° 2 65°2 1 i sin1125° 2 65°24 Simplify. z1 z2 5 2 1cos 60° 1 i sin 60°2 The quotient is in polar form. To express the quotient in rectangular form, evaluate z1 z2 5 2 a1 2 1 i !3 2 b 5 1 1 i!3 the trigonometric functions. Polar form: z1 z2 5 2 1cos 60° 1 i sin 60°2 Rectangular form: z1 z2 5 1 1 i!3 YOUR T UR N Let z1 5 10 1cos 275° 1 i sin 275°2 and z2 5 5 1cos 65° 1 i sin 65°2, and find z1 z2. Express the answers in both polar and rectangular form. ▼ When multiplying or dividing complex numbers, we have considered only those values of u such that 0° # u # 360°. When the value of u is negative or greater than 360°, find the coterminal angle in the interval 30°, 360°4. ▼ A N S W E R z1 z2 5 2 1cos 210° 1 i sin 210°2 or z1 z2 5 2!3 2 i 832 CHAPTER 8 Additional Topics in Trigonometry 8.7.3 Powers of Complex Numbers Raising a number to a positive integer power is the same as multiplying that number by itself repeated times. x3 5 x⋅x⋅x 1a 1 b22 5 1a 1 b2 1a 1 b2 Therefore, raising a complex number to a power that is a positive integer is the same as multiplying the complex number by itself multiple times. Let us illustrate this with the complex number z 5 r1cos u 1 i sin u2, which we will raise to positive integer powers (n). WORDS MATH Take the case n 5 2. z2 5 3r 1cos u 1 i sin u24 3r 1cos u 1 i sin u24 Apply the product rule (multiply the magnitudes and add the arguments). z2 5 r23cos12u2 1 i sin12u24 Take the case n 5 3. z3 5 z2z 5 5r23cos12u2 1 i sin12u246 3r 1cos u 1 i sin u24 Apply the product rule (multiply the magnitudes and add the arguments). z3 5 r33cos13u2 1 i sin13u24 Take the case n 5 4. z4 5 z3z 5 5r33cos13u2 1 i sin13u246 3r 1cos u 1 i sin u24 Apply the product rule (multiply the magnitudes and add the arguments). z4 5 r43cos14 u2 1 i sin1 4u24 The pattern observed for any n is: zn 5 r n3cos1 nu2 1 i sin1 nu24 Although we will not prove this generalized representation of a complex number raised to a power, it was proved by Abraham De Moivre 11667–17542, and hence its name. 8.7.3 SKI LL Raise a complex number to an integer power. 8.7.3 C ON CEPTUAL Understand that when a complex number is raised to an integer power n, the magnitude is raised to the power n and the argument is multiplied by n. [CONCEPT CHECK] TRUE OR FALSE When a complex number is raised to a positive integer power n, then the magnitude is raised to the nth power and the argument is multiplied by n. ANSWER True ▼ DE MOIVRE’S THEOREM If z 5 r 1cos u 1 i sin u2 is a complex number, then zn 5 rn3cos1nu2 1 i sin1nu24 when n is a positive integer 1n $ 12. In other words, when raising a complex number to a power n, raise the magnitude to the same power n and multiply the argument by n. Although De Moivre’s theorem was proved for all real numbers n, we will only use it for positive integer values of n and their reciprocals (nth roots). This is a very powerful theorem. For example, if asked to find 1 !3 1 i210, you have two choices: (1) multiply out the expression algebraically, which we will call the long way, or (2) convert to polar coordinates and use De Moivre’s theorem, which we will call the short way. We will use De Moivre’s theorem. 8.7 Products, Quotients, Powers, and Roots of Complex Numbers; De Moivre’s Theorem 833 8.7.4 Roots of Complex Numbers De Moivre’s theorem is the basis for the nth root theorem. Before we proceed, let us motivate it with a problem: Solve x3 2 1 5 0. Recall that a polynomial of degree n has n solutions (roots in the complex number system). So the polynomial P1x2 5 x3 2 1 is of degree 3 and has three solutions (roots). We can solve it algebraically. ▼ A N S W E R 2512 2 512i!3 WORDS MATH List the potential rational roots of the polynomial P1x2 5 x3 2 1. x 5 61 Use synthetic division to test x 5 1. 1 1 0 0 21 1 1 1 1 1 1 0 Since x 5 1 is a zero, then the polynomial can be written as a product of the linear P1x2 5 1x 2 121x2 1 x 1 12 factor 1x 2 12 and a quadratic factor. Use the quadratic formula on x2 1 x 1 1 5 0 x 5 21 6 "1 2 4 2 5 21 6 "23 2 5 21 2 6 i"3 2 to solve for x. So the three solutions to the equation x3 2 1 5 0 are x 5 1, x 5 2 1 2 1 i"3 2 , and x 5 2 1 2 2 i"3 2 . x2 1 x 1 1 3 8.7.4 S K I L L Find the nth root of a complex number. 8.7.4 C O N C E P T U A L We can often solve a polynomial equation by finding complex roots. An alternative approach to solving x3 2 1 5 0 is to use the nth root theorem to find the additional complex cube roots of 1. EXAMPLE 3 Finding a Power of a Complex Number Find A !3 1 iB10 and express the answer in rectangular form. Solution: STEP 1 Convert to polar form. 1!3 1 i210 5 321cos 30° 1 i sin 30°2410 STEP 2 Apply De Moivre’s A !3 1 iB10 5 321cos 30° 1 i sin 30°2410 theorem with n 5 10. 5 210 3cos110⋅30°2 1 i sin110⋅30°24 STEP 3 Simplify. A !3 1 iB10 5 210 1cos 300° 1 i sin 300°2 Evaluate 210 and the sine and cosine functions. 5 1024 a1 2 2 i !3 2 b Simplify. 5 512 2 512i!3 YOU R T UR N Find A1 1 i !3B10 and express the answer in rectangular form. ▼ 834 CHAPTER 8 Additional Topics in Trigonometry Derivation of the nth Root Theorem WORDS MATH Let z and w be complex numbers such that w is the w 5 z1/n or w 5 ! n z, where n is a positive nth root of z. integer Raise both sides of the equation to the nth power. wn 5 z Let z 5 r1cos u 1 i sin u2 and w 5 s1cos a 1 i sin a2. 3s1cos a 1 i sin a24n 5 r1cos u 1 i sin u2 Apply De Moivre’s theorem to the left side of the equation. sn3cos1 na2 1 i sin1na24 5 r 1cos u 1 i sin u2 For these two expressions to be equal, their magnitudes must be equal and their angles must be coterminal. sn 5 r and na 5 u 1 2kp, where k is any integer Solve for s and a. s 5 r 1/n and a 5 u 1 2kp n Substitute s 5 r1/n and a 5 u 1 2kp n into w 5 z1/n. z1/n 5 r1/nccos au 1 2kp n b 1 i sin au 1 2kp n bd Notice that when k 5 n, the arguments u n 1 2p and u n are coterminal. Therefore, to get distinct roots, let k 5 0, 1, . . . , n 2 1. If we let z be a given complex number and w be any complex number that satisfies the relationship z1/n 5 w or z 5 wn, where n $ 2, then we say that w is a complex nth root of z. NTH ROOT THEOREM The nth roots of the complex number z 5 r 1cos u 1 i sin u2 are given by wk 5 r1/n ccos au n 1 2kp n b 1 i sinau n 1 2kp n b d u in radians or wk 5 r1/n ccosau n 1 k⋅360° n b 1 i sinau n 1 k⋅360° n bd u in degrees where k 5 0, 1, 2, c, n 2 1.˙ 8.7 Products, Quotients, Powers, and Roots of Complex Numbers; De Moivre’s Theorem 835 EXAMPLE 4 Finding Roots of Complex Numbers Find the three distinct cube roots of 24 2 4i!3, and plot the roots in the complex plane. Solution: STEP 1 Write 24 2 4i!3 in polar form. 8 1cos 240° 1 i sin 240°2 STEP 2 Find the three cube roots. wk 5 r1/n ccos au n 1 k⋅360° n b 1 i sinau n 1 k⋅360° n b d u 5 240°, r 5 8, n 5 3, k 5 0, 1, 2 k 5 0: w0 5 81/3 ccosa240° 3 1 0⋅360° 3 b 1 i sina240° 3 1 0⋅360° 3 b d Simplify. w0 5 2 1cos 80° 1 i sin 80°2 k 5 1: w1 5 81/3 ccosa240° 3 1 1⋅360° 3 b 1 i sina240° 3 1 1⋅360° 3 b d Simplify. w1 5 2 1cos 200° 1 i sin 200°2 k 5 2: w2 5 81/3 ccosa240° 3 1 2⋅360° 3 b 1 i sina240° 3 1 2⋅360° 3 b d Simplify. w2 5 21cos 320° 1 i sin 320°2 STEP 3 Plot the three cube roots in the complex plane. Notice the following: ■ ■ The roots all have a magnitude of 2. ■ ■ The roots lie on a circle of radius 2. ■ ■ The roots are equally spaced around the circle 1120° apart). Y OUR T UR N Find the three distinct cube roots of 4 2 4i!3, and plot the roots in the complex plane. 2 –2 2i –2i Real axis Imaginary axis w0 w1 w2 80º 200º 320º ▼ ▼ A N S W E R w0 5 2 1cos 100° 1 i sin 100°2 w1 5 2 1cos 220° 1 i sin 220°2 w2 5 2 1cos 340° 1 i sin 340°2 2 –2 2i –2i w0 100º w1 220º w2 340º Real axis Imaginary axis 836 CHAPTER 8 Additional Topics in Trigonometry Solving Equations Using Roots of Complex Numbers Let us return to solving the equation x3 2 1 5 0. As stated, x 5 1 is the real solution to this cubic equation. However, there are two additional (complex) solutions. Since we are finding the zeros of a third-degree polynomial, we expect three solutions. Furthermore, when complex solutions arise in finding the roots of polynomials, they come in conjugate pairs. EXAMPLE 5 Solving Equations Using Complex Roots Find all complex solutions to x3 2 1 5 0. Solution: x3 5 1 STEP 1 Write 1 in polar form. 1 5 1 1 0 i 5 cos0° 1 i sin0° STEP 2 Find the three cube roots of 1. wk 5 r1/n ccosau n 1 k ⋅360° n b 1 i sinau n 1 k ⋅360° n b d r 5 1, u 5 0°, n 5 3, k 5 0, 1, 2 k 5 0: w0 5 11/3ccos a0° 3 1 0⋅360° 3 b 1 i sin a0° 3 1 0⋅360° 3 b d Simplify. w0 5 cos 0° 1 i sin 0° k 5 1: w1 5 11/3ccosa0° 3 1 1⋅360° 3 b 1 i sina0° 3 1 1⋅360° 3 b d Simplify. w1 5 cos 120° 1 i sin 120° k 5 2: w2 5 11/3ccosa0° 3 1 2⋅360° 3 b 1 i sina0° 3 1 2⋅360° 3 b d Simplify. w2 5 cos 240° 1 i sin 240° STEP 3 Write the roots in rectangular form. w0: w0 5 cos 0° 1 i sin 0° 5 1 w1: w1 5 cos 120° 1 i sin 120° 5 1 2 1 i !3 2 w2: w2 5 cos 240° 1 i sin 240° 5 21 2 2 i !3 2 STEP 4 Write the solutions to the equation x3 2 1 5 0. x 5 1 x 5 2 1 2 1 i !3 2 x 5 2 1 2 2 i !3 2 Notice that there is one real solution and there are two (nonreal) complex solutions and that the two complex solutions are complex conjugates. 3 1 3 0 3 21 2 3 !3 2 3 21 2 3 2!3 2 8.7 Products, Quotients, Powers, and Roots of Complex Numbers; De Moivre’s Theorem 837 It is always a good idea to check that the solutions indeed satisfy the equation. The equation x3 2 1 5 0 can also be written as x3 5 1, so the check in this case is to cube the three solutions and confirm that the result is 1. x 5 1: 13 5 1 ✓ x 5 2 1 2 1 i!3 2 : a2 1 2 1 i !3 2 b 3 5 a2 1 2 1 i !3 2 b 2 a2 1 2 1 i !3 2 b 5 a2 1 2 2 i !3 2 b a2 1 2 1 i !3 2 b 5 1 4 1 3 4 5 1 ✓ x 5 2 1 2 2 i!3 2 : a2 1 2 2 i !3 2 b 3 5 a2 1 2 2 i !3 2 b 2 a2 1 2 2 i !3 2 b 5 a2 1 2 1 i !3 2 b a2 1 2 2 i !3 2 b 5 1 4 1 3 4 5 1 ✓ 1 –1 i –i Real axis Imaginary axis w0 w1 w2 240º 120º [CONCEPT CHECK] TRUE OR FALSE The n distinct roots are equally spaced around the circle: u n far apart. ANSWER True ▼ In this section, we multiplied and divided complex numbers and, using De Moivre’s theorem, raised complex numbers to integer powers and found the nth roots of complex numbers, as follows: Let z1 5 r11cos u1 1 i sin u12 and z2 5 r21cos u2 1 i sin u22 be two complex numbers. The product z1z2 is given by z1z2 5 r1r2 3cos1u1 1 u22 1 i sin1u1 1 u224 The quotient z1 z2 is given by z1 z2 5 r1 r2 3cos1u1 2 u22 1 i sin1u1 2 u224 Let z 5 r 1cos u 1 i sin u2 be a complex number. Then for a positive integer n, z raised to a power n is given by z n 5 r n 3cos1nu2 1 i sin1nu24 The nth roots of z are given by wk 5 r1/nccos au n 1 k⋅360° n b 1 i sin au n 1 k⋅360° n b d where u is in degrees or wk 5 r1/nccos au n 1 k⋅2p n b 1 i sin au n 1 k⋅2p n b d where u is in radians k 5 0, 1, 2, c, n 2 1. [SEC TION 8.7] S U M M A RY 838 CHAPTER 8 Additional Topics in Trigonometry [SEC TION 8.7] E X E R CI SE S • S K I L L S In Exercises 1–10, find the product z1z2 and express it in rectangular form. 1. z1 5 41cos 40° 1 isin 40°2 and z2 5 31cos 80° 1 isin 80°2 2. z1 5 21cos100° 1 isin100°2 and z2 5 51cos 50° 1 isin 50°2 3. z1 5 41cos 80° 1 isin 80°2 and z2 5 21cos 145° 1 isin145°2 4. z1 5 31cos 130° 1 isin130°2 and z2 5 41cos170° 1 isin 170°2 5. z1 5 21cos 10° 1 isin10°2 and z2 5 41cos 80° 1 isin 80°2 6. z1 5 31cos 190° 1 isin 190°2 and z2 5 51cos 80° 1 i sin 80°2 7. z1 5 !3ccos ap 12b 1 i sin a p 12b d and z2 5 !27ccos ap 6b 1 i sin ap 6b d 8. z1 5 !5ccos a p 15b 1 i sin a p 15b d and z2 5 !5ccos a4p 15b 1 i sin a4p 15b d 9. z1 5 4ccosa3p 8 b 1 i sin a3p 8 b d and z2 5 3ccos ap 8 b 1 i sin ap 8 b d 10. z1 5 6ccosa2p 9 b 1 i sin a2p 9 b d and z2 5 5ccos ap 9 b 1 i sin ap 9 b d In Exercises 11–20, find the quotient z1 z2 and express it in rectangular form. 11. z1 5 61cos 100° 1 i sin 100°2 and z2 5 21cos 40° 1 i sin 40°2 12. z1 5 81cos 80° 1 isin 80°2 and z2 5 21cos 35° 1 isin 35°2 13. z1 5 101cos 200° 1 isin 200°2 and z2 5 51cos 65° 1 isin 65°2 14. z1 5 41cos 280° 1 isin 280°2 and z2 5 41cos 55° 1 isin 55°2 15. z1 5 !12 1cos 350° 1 isin 350°2 and z2 5 !3 1cos 80° 1 isin 80°2 16. z1 5 !40 1cos 110° 1 i sin 110°2 and z2 5 !10 1cos 20° 1 i sin 20°2 17. z1 5 9ccos a 5p 12 b 1 isin a5p 12 b d and z2 5 3ccosa p 12b 1 isin a p 12b d 18. z1 5 8ccosa5p 8 b 1 i sin a5p 8 b d and z2 5 4ccos a3p 8 b 1 i sin a3p 8 b d 19. z1 5 45ccosa22p 15 b 1 i sin a22p 15 b d and z2 5 9ccos a2p 15 b 1 i sin a2p 15 b d 20. z1 5 22ccosa11p 18 b 1 i sin a11p 18 b d and z2 5 11ccos a5p 18 b 1 i sin a5p 18 b d In Exercises 21–30, find the result of each expression using De Moivre’s theorem. Write the answer in rectangular form. 21. 121 1 i25 22. 11 2 i24 23. 12!3 1 i2 6 24. 1 !3 2 i28 25. 11 2 !3i2 4 26. 121 1 !3i2 5 27. 14 2 4i28 28. 123 1 3i210 29. 14!3 1 4i27 30. 125 1 5!3i27 In Exercises 31–40, find all nth roots of z. Write the answers in polar form, and plot the roots in the complex plane. 31. 2 2 2i!3, n 5 2 32. 2 1 2!3i, n 5 2 33. !18 2 !18i, n 5 2 34. 2!2 1 !2i, n 5 2 35. 4 1 4!3i, n 5 3 36. 227 2 1 27!3 2 i, n 5 3 37. !3 2 i, n 5 3 38. 4!2 1 4!2i, n 5 3 39. 8!2 2 8!2i, n 5 4 40. 2!128 1 !128 i, n 5 4 8.7 Products, Quotients, Powers, and Roots of Complex Numbers; De Moivre’s Theorem 839 In Exercises 41–56, find all complex solutions to the given equations. 41. x4 2 16 5 0 42. x3 2 8 5 0 43. x3 1 8 5 0 44. x3 1 1 5 0 45. x4 1 16 5 0 46. x6 1 1 5 0 47. x6 2 1 5 0 48. 4x2 1 1 5 0 49. x2 1 i 5 0 50. x2 2 i 5 0 51. x4 2 2i 5 0 52. x4 1 2i 5 0 53. x5 1 32 5 0 54. x5 2 32 5 0 55. x7 2 p14i 5 0 56. x7 1 p14 5 0 • A P P L I C A T I O N S 57. Complex Pentagon. When you graph the five fifth roots of 2 !2 2 2 !2 2 i and connect the points, you form a pentagon. Find the roots and draw the pentagon. 58. Complex Square. When you graph the four fourth roots of 16i and connect the points, you form a square. Find the roots and draw the square. 59. Hexagon. Compute the six sixth roots of 1 2 2 !3 2 i, and form a hexagon by connecting successive roots. 60. Octagon. Compute the eight eighth roots of 2i, and form an octagon by connecting successive roots. • C A T C H T H E M I S T A K E In Exercises 61–64, explain the mistake that is made. 61. Let z1 5 61cos 65° 1 isin 65°2 and z2 5 31cos 125° 1 isin 125°2. Find z1 z2 . Solution: Use the quotient formula. z1 z2 5 r1 r2 3cos1u1 2 u22 1 i sin1u1 2 u224 Substitute values. z1 z2 5 6 3 3cos1125° 2 65°2 1 isin1125° 2 65°24 Simplify. z1 z2 5 21cos 60° 1 isin 60°2 Evaluate the trigonometric functions. z1 z2 5 2a1 2 1 i!3 2 b 5 1 1 i!3 This is incorrect. What mistake was made? 62. Let z1 5 61cos 65° 1 isin 65°2 and z2 5 31cos 125° 1 isin 125°2. Find z1z2. Solution: Write the product. z1z2 5 6 1cos 65° 1 i sin 65°2 ⋅3 1cos 125° 1 isin 125°2 Multiply the magnitudes. z1z2 5 181cos 65° 1 isin 65°2 1cos 125° 1 isin 125°2 Multiply the cosine terms and sine terms (add the arguments). z1z2 5 183cos165° 1 125°2 1 i2sin165° 1 125°24 Simplify 1i2 5 212. z1z2 5 181cos 190° 2 sin 190°2 This is incorrect. What mistake was made? 63. Find A !2 1 i!2B6. Solution: Raise each term to the sixth power. A !2B6 1 i 6A!2B6 Simplify. 8 1 8i 6 Let i 6 5 i 4⋅i2 5 21. 8 2 8 5 0 This is incorrect. What mistake was made? 64. Find all complex solutions to x5 2 1 5 0. Solution: Add 1 to both sides. x5 5 1 Raise both sides to the fifth power. x 5 11/5 Simplify. x 5 1 This is incorrect. What mistake was made? 840 CHAPTER 8 Additional Topics in Trigonometry In Exercises 65–70, determine whether the statement is true or false. 65. The product of two complex numbers is a complex number. 66. The quotient of two complex numbers is a complex number. 67. There are always n distinct real solutions of the equation x n 2 a 5 0, where a is not zero. 68. There are always n distinct complex solutions of the equation xn 2 a 5 0, where a is not zero. 69. There are n distinct complex roots of 1 a 1 bi, where a and b are positive real numbers. 70. There exists a complex number for which there is no complex square root. • C O N C E P T U A L In Exercises 71–74, use the following identity: In calculus you will see an identity called Euler’s formula or identity, eiu 5 cos u 1 i sin u. Notice that when u 5 p, the identity reduces to eip 1 1 5 0, which is a beautiful identity in that it relates the five fundamental numbers 1e, p, 1, i, and 02 and the fundamental operations (multipli-cation, addition, exponents, and equality) in mathematics. 71. Let z1 5 r11cos u1 1 i sin u12 5 r1eiu1 and z2 5 r21cos u2 1 i sin u22 5 r2 eiu2 be two complex numbers. Use the properties of exponentials to show that z1z2 5 r1r23cos1u1 1 u22 1 i sin1u1 1 u224. 72. Let z1 5 r11cos u1 1 i sin u12 5 r1eiu1 and z2 5 r21cos u2 1 i sin u22 5 r2 eiu2 be two complex numbers. Use the properties of exponentials to show that z1 z2 5 r1 r2 3cos1u1 2 u22 1 i sin1u1 2 u224. 73. Let z 5 r 1cos u 1 i sin u2 5 reiu. Use the properties of exponents to show that z n 5 r n1cos n u 1 i sin n u2. 74. Let z 5 r 1cos u 1 i sin u2 5 reiu. Use the properties of exponents to show that wk 5 r1/nccosau n 1 2kp n b 1 isin au n 1 2kp n b d . 75. Use De Moivre’s theorem to prove the identity cos2u 5 cos2u 2 sin2u. 76. Use De Moivre’s theorem to derive an expression for sin3u. 77. Use De Moivre’s theorem to derive an expression for cos3u. 78. Calculate a1 2 1 !3 2 ib 14 a1 2 2 !3 2 ib 20. 79. Calculate 11 2 i2n⋅11 1 i2m, where n and m are positive integers. 80. Calculate 11 1 i2n 11 2 i2m, where n and m are positive integers. • C H A L L E N G E For Exercises 81–86, refer to the following: According to the nth root theorem, the first of the nth roots of the complex number z 5 r 1cos u 1 i sin u2 is given by w1 5 r1/nccos au n 1 2p n b 1 i sin au n 1 2p n bd, with u in radians or w1 5 r1/nccosau n 1 360° n b 1 i sinau n 1 360° n bd, with u in degrees. Using the graphing calculator to plot the n roots of a complex number z, enter r1 5 r, u min 5 u n, u max 5 2p 1 u n or 360° 1 u n, ustep 5 2p n or 360° n , xmin 5 2r, xmax 5 r, ymin 5 2r, ymax 5 r, and MODE in radians or degrees. 81. Find the fifth roots of !3 2 2 1 2i, and plot the roots with a calculator. 82. Find the fourth roots of 2 !2 2 1 !2 2 i, and plot the roots with a calculator. 83. Find the sixth roots of 2 1 2 2 !3 2 i, and draw the complex hexagon with a calculator. 84. Find the fifth roots of 24 1 4 i, and draw the complex pentagon with a calculator. 85. Find the cube roots of 27!2 2 1 27!2 2 i, and draw the complex triangle with a calculator. 86. Find the fifth roots of 8!2 1 !3 2 12 1 8!2 1 !3 2 12i, and draw the complex triangle with a calculator. • T E C H N O L O G Y 8.8 Polar Equations and Graphs 841 S K I L L S O B J E C T I V E S ■ ■Plot points in the polar coordinate system. ■ ■Convert between rectangular and polar coordinates. ■ ■Graph polar equations. C O N C E P T U A L O B J E C T I V ES ■ ■Understand that the name of a point 1r, u2 in the polar coordinate system is not unique. ■ ■Relate the rectangular coordinate system to the polar coordinate system. ■ ■Classify common shapes that arise from plotting certain types of polar equations. 8.8 POLAR EQUATIONS AND GRAPHS We have discussed the rectangular and the trigonometric (polar) form of complex numbers in the complex plane. We now turn our attention back to the familiar Cartesian plane, where the horizontal axis represents the x-variable, the vertical axis represents the y-variable, and points in this plane represent pairs of real numbers. It is often convenient to instead represent real-number plots in the polar coordinate system. 8.8.1 Polar Coordinates The polar coordinate system is anchored by a point, called the pole (taken to be the origin), and a ray with a vertex at the pole, called the polar axis. The polar axis is normally shown where we expect to find the positive x-axis in Cartesian coordinates. If you align the pole with the origin on the rectangular graph and the polar axis with the positive x-axis, you can label a point either with rectangular coordinates 1x, y2 or with an ordered pair 1r, u2 in polar coordinates. Typically, polar graph paper that gives the angles and radii is used. The graph below gives the angles in radians (the angle also can be given in degrees) and shows the radii from 0 through 5. 3 π 4 π 6 π 12 π 0 6 11π 12 17π 12 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π 1 3 5 When plotting points in the polar coordinate system, k r k represents the distance from the origin to the point. The following procedure guides us in plotting points in the polar coordinate system. 8.8.1 S K IL L Plot points in the polar coordinate system. 8.8.1 C ON C E P T U A L Understand that the name of a point 1r, u2 in the polar coordinate system is not unique. Polar Axis Pole x y x y x θ y r (x, y) (r, θ) POINT-PLOTTING POLAR COORDINATES To plot a point 1r, u2: 1. Start on the polar axis and rotate the terminal side of an angle to the value u. 2. If r . 0, the point is r units from the origin in the same direction of the terminal side of u. 3. If r , 0, the point is k r k units from the origin in the opposite direction of the terminal side of u. 842 CHAPTER 8 Additional Topics in Trigonometry In polar form it is important to note that 1r, u2, the name of the point, is not unique, whereas in rectangular form 1x, y2 it is unique. For example, 12, 30°2 5 122, 210°2. EXAMPLE 1 Plotting Points in the Polar Coordinate System Plot the points in a polar coordinate system. a. a3, 3p 4 b b. 122, 60°2 Solution (a): Start by placing a pencil along the polar axis. Rotate the pencil to the angle 3p 4 . Go out (in the direction of the pencil) 3 units. Solution (b): Start by placing a pencil along the polar axis. Rotate the pencil to the angle 60°. Go out (opposite the direction of the pencil) 2 units. Y OUR TU R N Plot the points in the polar coordinate system. a. a24, 3p 2 b b. 13, 330°2 3 π 4 π 6 π 12 π 0 6 11π 12 17π 12 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π 1 3 5 45º 30º 15º 360º 330º 255º 165º 195º 285º 345º 315º 300º 75º 105º 270º 240º 225º 210º 180º 150º 135º 90º 120º 1 3 5 60º (–2, 60º) ▼ ▼ A N S W E R 3 π 4 π 6 π 12 π 0 1 3 5 12 17π 12 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π a b 330º 8.8.2 Converting Between Polar and Rectangular Coordinates The relationships between polar and rectangular coordinates are the familiar ones: sin u 5 y r cos u 5 x r r 2 5 x2 1 y2 tan u 5 y x 1x 2 02 x y x θ y r (x, y) (r, θ) CONVERTING BETWEEN POLAR AND RECTANGULAR COORDINATES FROM TO IDENTITIES Polar 1r, u2 Rectangular 1x, y2 x 5 r cos u y 5 r sin u Rectangular 1x, y2 Polar 1r, u2 r 5 "x2 1 y2 tan u 5 y x 1x 2 02 Make sure that u is in the correct quadrant. 8.8.2 S KILL Convert between rectangular and polar coordinates. 8.8.2 C ON CEPTUAL Relate the rectangular coordinate system to the polar coordinate system. [CONCEPT CHECK] Match the following: (1) (x, y) (2) (r, u) (A) not unique (B) unique ANSWER 1. B 2. A ▼ [CONCEPT CHECK] TRUE OR FALSE The point (2a, 2b) in rectangular coordinates corresponds to the point (r, u), where r and u are both positive. ANSWER True ▼ 8.8 Polar Equations and Graphs 843 8.8.3 Graphs of Polar Equations We are familiar with equations in rectangular form such as y 5 3x 1 5 y 5 x2 1 2 x2 1 y2 5 9 (line) (parabola) (circle) We now discuss equations in polar form (known as polar equations) such as r 5 5u r 5 2 cos u r 5 sin15 u2 which you will learn to recognize in this section as typical equations whose plots are some general shapes. Our first example deals with two of the simplest forms of polar equations: when r or u is constant. The results are a circle centered at the origin and a line that passes through the origin, respectively. EXAMPLE 2 Converting Between Polar and Rectangular Coordinates a. Convert A21, !3B to polar coordinates. b. Convert A6!2, 135°B to rectangular coordinates. Solution (a): A21, !3B lies in quadrant II. Identify x and y. x 5 21, y 5 !3 Find r. r 5 "x2 1 y2 5 "12122 1 A !3B2 5 !4 5 2 Find u. tan u 5 !3 21 1u lies in quadrant II2 Identify u from the unit circle. u 5 2p 3 Write the point in polar coordinates. a2, 2p 3 b Note: Other polar coordinates like a2, 24p 3 b and a 22, 5p 3 b also correspond to the point A21, !3B. Solution (b): A6!2, 135°B lies in quadrant II. Identify r and u. r 5 6!2 u 5 135° Find x. x 5 r cos u 5 6!2 cos 135° 5 6!2 a 2 !2 2 b 5 26 Find y. y 5 r sin u 5 6!2 sin135° 5 6!2 a !2 2 b 5 6 Write the point in rectangular coordinates. 126, 62 x y 3 2π 2 √3 2 1 (– , ) (0, 1) (1, 0) (–1, 0) (0, –1) 8.8.3 S K I L L Graph polar equations. 8.8.3 C O N C E P T U A L Classify common shapes that arise from plotting certain types of polar equations. 844 CHAPTER 8 Additional Topics in Trigonometry EXAMPLE 3 Graphing a Polar Equation of the Form r 5 Constant or u 5 Constant Graph the polar equations. a. r 5 3 b. u 5 p 4 Solution (a): Constant value of r Approach 1 (polar coordinates): r 5 3 1u can take on any value2 Plot points for arbitrary u and r 5 3. Connect the points; a circle with radius 3. Approach 2 (rectangular coordinates): r 5 3 Square both sides. r 2 5 9 Remember that in rectangular coordinates r 2 5 x 2 1 y2. x 2 1 y2 5 32 This is a circle, centered at the origin, with radius 3. Solution (b): Constant value of u Approach 1: u 5 p 4 1r can take on any value, positive or negative2 Plot points for u 5 p 4 at several arbitrary values of r. Connect the points. The result is a line passing through the origin with slope 5 1cm 5 tanap 4 b d. Approach 2: u 5 p 4 Take the tangent of both sides. tan u 5 tanap 4 b 1 Use the identity tan u 5 y x. y x 5 1 Multiply by x. y 5 x The result is a line passing through the origin with slope 5 1. 3 π 4 π 6 π 12 π 0 6 11π 12 17π 12 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π 1 3 5 3 3 π 4 π 6 π 12 π 0 6 11π 12 17π 12 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π 1 3 5 [CONCEPT CHECK] Match the polar equations with their graphs: (1) r 5 constant (2) u 5 constant (3) r 5 constant ⋅u (A) spiral (B) circle (C ) line ANSWER 1. B 2. C 3. A ▼ 8.8 Polar Equations and Graphs 845 Rectangular equations that depend on varying (not constant) values of x or y can be graphed by point-plotting (making a table and plotting the points). We will use this same procedure for graphing polar equations that depend on varying (not constant) values of r and u. u r 5 4 cos u (r, u) 0 4112 5 4 14, 02 p 4 4a !2 2 b < 2.8 a2.8, p 4 b p 2 4102 5 0 a0, p 2 b 3p 4 4a2 !2 2 b < 22.8 a 22.8, 3p 4 b p 41212 5 24 124, p2 5p 4 4 a 2 !2 2 b < 22.8 a 22.8, 5p 4 b 3p 2 4102 5 0 a0, 3p 2 b 7p 4 4 a !2 2 b < 2.8 a22.8, 7p 4 b 2p 4112 5 4 124, 2p2 EXAMPLE 4 Graphing a Polar Equation of the Form r 5 c ? cos u or r 5 c ? sin u Graph r 5 4 cos u. Solution: STEP 1 Make a table and find several key values. STEP 2 Plot the points in polar coordinates. STEP 3 Connect the points with a smooth curve. Notice that 14, 02 and 124, p2 correspond to the same point. There is no need to continue with angles beyond p because the result would be to go around the same circle again. Y OUR T UR N Graph r 5 4 sin u. 3 π 4 π 6 π 12 π 0 6 11π 12 17π 12 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π 1 3 5 3 π 4 π 6 π 12 π 0 6 11π 12 17π 12 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π 1 3 5 ▼ ▼ A N S W E R 3 π 4 π 6 π 12 π 0 1 3 5 12 17π 12 11π 6 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π 846 CHAPTER 8 Additional Topics in Trigonometry Compare the result of Example 4, the graph of r 5 4 cos u, with the result of Your Turn, the graph of r 5 4 sin u. Notice that they are 90° out of phase (we simply rotate one graph 90° about the pole to get the other graph). In general, graphs of polar equations of the form r 5 a sin u and r 5 a cos u are circles. WORDS MATH r 5 a sin u r 5 a cos u Apply trigonometric ratios: sin u 5 y r and cos u 5 x r. r 5 a y r r 5 a x r Multiply the equations by r. r 2 5 ay r 2 5 ax Let r 2 5 x2 1 y2. x 2 1 y2 5 ay x 2 1 y2 5 ax Group x terms together and y terms together. x 2 1 1y2 2 ay2 5 0 1x 2 2 ax2 1 y 2 5 0 Complete the square on the expressions in parentheses. x 2 1 cy 2 2 ay 1 aa 2b 2 d 5 aa 2b 2 cx 2 2 ax 1 aa 2b 2 d 1 y 2 5 aa 2b 2 x 2 1 ay 2 a 2b 2 5 aa 2b 2 ax 2 a 2b 2 1 y 2 5 aa 2b 2 Graph of a circle. Center: a0, a 2b Radius: a 2 Center: aa 2, 0b Radius: a 2 EXAMPLE 5 Graphing a Polar Equation of the Form r 5 c ? sin (2 u) or r 5 c ? cos (2 u) Graph r 5 5 sin12u2. Solution: STEP 1 Make a table and find key values. Since the argument of the sine function is doubled, the period is halved. Therefore, instead of steps of p/4, take steps of p/8. u r 5 5 sin (2u) (r, u) 0 5102 5 0 10, 02 p 8 5a !2 2 b < 3.5 a3.5, p 8 b p 4 5112 5 5 a5, p 4 b 3p 8 5a !2 2 b < 3.5 a3.5, 3p 8 b p 2 5102 5 0 a0, p 2 b STUDY TIP Graphs of r 5 a sin u and r 5 a cos u are circles. 8.8 Polar Equations and Graphs 847 Compare the result of Example 5, the graph of r 5 5 sin12u2, with the result of Your Turn, the graph of r 5 5 cos12u2. Notice that they are 45° out of phase (we rotate one graph 45° about the pole to get the other graph). In general, for r 5 a sin1nu2 or r 5 a cos1nu2, the graph has n leaves (petals) if n is odd and 2n leaves (petals) if n is even. As r increases, the leaves (petals) get longer. The next class of graphs are called limaçons, which have equations of the form r 5 a 6 b cos u or r 5 a 6 b sin u. When a 5 b, the result is a cardioid (heart shape). STEP 2 Label the polar coordinates. The values in the table represent what happens in quadrant I. The same pattern repeats in the other three quadrants. The result is a four-leaved rose. 3 π 4 π 6 π 12 π 0 6 11π 12 17π 12 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π 3 5 1 STEP 3 Connect the points with smooth curves. YOUR T UR N Graph r 5 5 cos12u2. ▼ ▼ A N S W E R 3 π 4 π 6 π 12 π 0 1 3 5 12 17π 12 11π 6 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π STUDY TIP Graphs of r 5 a sin1nu2 and r 5 a cos1nu2 are roses with n leaves if n is odd and 2n leaves if n is even. EXAMPLE 6 The Cardioid as a Polar Equation Graph r 5 2 1 2 cos u. Solution: STEP 1 Make a table and find key values. This behavior repeats in quadrant III and quadrant IV because the cosine function has corresponding values in quadrant I and quadrant IV and in quadrant II and quadrant III. u r 5 2 1 2 cos u (r, u) 0 2 1 2112 5 4 14, 02 p 4 2 1 2a !2 2 b < 3.4 a3.4, p 4 b p 2 2 1 2102 5 2 a2, p 2 b 3p 4 2 1 2a 2 !2 2 b < 0.6 a0.6, 3p 4 b p 2 1 21212 5 0 10, p2 848 CHAPTER 8 Additional Topics in Trigonometry STEP 2 Plot the points in polar coordinates. STEP 3 Connect the points with a smooth curve. The curve is a cardioid, a term formed from Greek roots meaning “heart-shaped.” 3 π 4 π 6 π 12 π 0 6 11π 12 17π 12 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π 3 1 5 EXAMPLE 7 Graphing a Polar Equation of the Form r 5 c ? u Graph r 5 0.5 u. Solution: STEP 1 Make a table and find key values. STEP 2 Plot the points in polar coordinates. STEP 3 Connect the points with a smooth curve. The curve is a spiral. u r 5 0.5 u (r, u) 0 0.5102 5 0 10, 02 p 2 0.5ap 2 b < 0.8 a0.8, p 2 b p 0.51p2 < 1.6 11.6, p2 3p 2 0.5a3p 2 b < 2.4 a2.4, 3p 2 b 2p 0.512p2 < 3.1 13.1, 2p2 3 π 4 π 6 π 12 π 0 6 11π 12 17π 12 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π 3 1 5 8.8 Polar Equations and Graphs 849 Converting Equations Between Polar and Rectangular Form It is not always advantageous to plot an equation in the form in which it is given. It is sometimes easier to first convert to rectangular form and then plot. For example, to plot r 5 2 cos u 1 sin u, we could make a table with values. However, as you will see in Example 9, it is much easier to convert this equation to rectangular coordinates. EXAMPLE 8 Graphing a Polar Equation of the Form r 2 5 c ? sin (2 u) or r 2 5 c ? cos (2 u) Graph r2 5 4 cos12 u2. Solution: STEP 1 Make a table and find key values. Solving for r yields r 5 62!cos12u2. All coordinates 12r, u2 can be expressed as 1r, u 1 p2. The following table does not have values for p 4 , u , 3p 4 because the corresponding values of cos12u2 are negative, and hence r is an imaginary number. The table also does not have values for u . p because 2u . 2p and the corresponding points are repeated. u cos (2u) r 5 62!cos12u2 (r, u) 0 1 r 5 62 12, 02 and 122, 02 5 12, p2 p 6 0.5 r < 61.4 a1.4, p 6 b and a21.4, p 6 b 5 a1.4, 7p 6 b p 4 0 r 5 0 a0, p 4 b 3p 4 0 r 5 0 a0, 3p 4 b 5p 6 0.5 r < 61.4 a1.4, 5p 6 b and a21.4, 5p 6 b 5 a1.4, 11p 6 b p 1 r 5 62 12, p2 and 122, p2 5 12, 2p2 STEP 2 Plot the points in polar coordinates. STEP 3 Connect the points with a smooth curve. The resulting curve is known as a lemniscate. 3 π 4 π 6 π 12 π 0 6 11π 12 17π 12 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π 1 2 850 CHAPTER 8 Additional Topics in Trigonometry EXAMPLE 9 Converting an Equation from Polar Form to Rectangular Form Graph r 5 2 cos u 1 sin u. Solution: Multiply the equation by cos u 1 sin u. r 1cos u 1 sin u2 5 2 Eliminate parentheses. r cos u 1 r sin u 5 2 Convert the result to rectangular form. rcosu 1 rsinu 5 2 x y Simplify. The result is a straight line. y 5 2x 1 2 Graph the line. Y OUR TU R N Graph r 5 2 cos u 2 sin u. 3 3 1 –1 2 3 4 5 x y –2 1 2 3 4 5 ▼ ▼ A N S W E R y 5 x 2 2 –5 –4 –3 –2 1 2 3 4 5 1 –3 –1 –5 2 3 4 5 x y Graph polar coordinates 1r, u2 in the polar coordinate system first by rotating a ray to get the terminal side of the angle. Then if r is positive, go out r units from the origin in the direction of the angle. If r is negative, go out 0 r 0 units in the opposite direction of the angle. Conversions between polar and rectangular forms are given by FROM TO IDENTITIES Polar 1r, u2 Rectangular 1x, y2 x 5 r cos u y 5 r sin u Rectangular 1x, y2 Polar 1r, u2 r 5 "x2 1 y2 tan u 5 y x, x 2 0 Be careful to note the proper quadrant for u. We can graph polar equations by point-plotting. Common shapes that arise are given in the following table. Sine and cosine curves have the same shapes (just rotated). If more than one equation is given, then the top equation corresponds to the actual graph. In this table, a and b are assumed to be positive. CLASSIFICATION SPECIAL NAME POLAR EQUATIONS GRAPH Line Radial line u 5 a Circle Circle centered at the origin r 5 a [SEC TION 8.8] S U M M A RY 8.8 Polar Equations and Graphs 851 CLASSIFICATION SPECIAL NAME POLAR EQUATIONS GRAPH Circle Circle that touches the pole and whose center is on the polar axis r 5 a cos u Circle Circle that touches the pole and whose center is on the line u 5 p 2 r 5 a sin u Limaçon Cardioid r 5 a 1 a cos u r 5 a 1 a sin u Limaçon Without inner loop a . b r 5 2a 2 b cos u r 5 a 1 b sin u Limaçon With inner loop a , b r 5 a 1 b sin u r 5 a 1 b cos u Lemniscate r 2 5 a2 cos12 u2 r 2 5 a2 sin12 u2 Rose Three rose petals r 5 a sin13 u2 r 5 a cos13 u2 Rose Four rose petals r 5 a sin12 u2 r 5 a cos12 u2 Spiral r 5 a u In the argument nu, if n is odd, there are n petals (leaves), and if n is even, there are 2n petals (leaves). 852 CHAPTER 8 Additional Topics in Trigonometry [SEC TION 8.8] E X E R C I SE S • S K I L L S In Exercises 1–10, plot each indicated point in a polar coordinate system. 1. a3, 5p 6 b 2. a2, 5p 4 b 3. a4, 11p 6 b 4. a1, 2p 3 b 5. a 22, p 6 b 6. a 24, 7p 4 b 7. 124, 270°2 8. 13, 135°2 9. 14, 225°2 10. 122, 60°2 In Exercises 11–20, convert each point to exact polar coordinates. Assume that 0 # u , 2p. 11. A2, 2!3B 12. 13, 232 13. A21, 2!3B 14. A6, 6!3B 15. 124, 42 16. A0, !2B 17. 13, 02 18. 127, 272 19. A2!3, 21B 20. A2!3, 22B In Exercises 21–30, convert each point to exact rectangular coordinates. 21. a4, 5p 3 b 22. a2, 3p 4 b 23. a 21, 5p 6 b 24. a 22, 7p 4 b 25. a0, 11p 6 b 26. 16, 02 27. 12, 240°2 28. 123, 150°2 29. 121, 135°2 30. 15, 315°2 In Exercises 31–34, match the polar graphs with their corresponding equations. 31. r 5 4 cos u 32. r 5 2 u 33. r 5 3 1 3 sin u 34. r 5 3 sin 12 u2 a. 3 π 4 π 6 π 12 π 0 6 11π 12 17π 12 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π 6 2 10 b. 3 π 4 π 6 π 12 π 0 6 11π 12 17π 12 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π 9 3 15 c. 3 π 4 π 6 π 12 π 0 6 11π 12 17π 12 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π 3 5 1 d. 3 π 4 π 6 π 12 π 0 6 11π 12 17π 12 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π 3 5 1 In Exercises 35–50, graph each equation. 35. r 5 5 36. u 5 2p 3 37. r 5 2 cos u 38. r 5 3 sin u 39. r 5 4 sin12u2 40. r 5 5 cos12u2 41. r 5 3 sin13u2 42. r 5 4 cos13u2 43. r2 5 9 cos12u2 44. r2 5 16 sin12u2 45. r 5 22 cos u 46. r 5 23 sin13u2 47. r 5 4u 48. r 5 22u 49. r 5 23 1 2 cos u 50. r 5 2 1 3 sin u In Exercises 51–54, convert the equation from polar to rectangular form. Identify the resulting equation as a line, parabola, or circle. 51. r 1sin u 1 2 cos u2 5 1 52. r 1sin u 2 3 cos u2 5 2 53. r2cos2u 2 2r cos u 1 r2sin2u 5 8 54. r2cos2u 2 r sin u 5 22 In Exercises 55–60, graph the polar equation. 55. r 5 21 3u 56. r 5 1 4u 57. r 5 4 sin 15u2 58. r 5 23 cos 14u2 59. r 5 22 2 3 cos u 60. r 5 4 2 3 sin u 8.8 Polar Equations and Graphs 853 • A P P L I C A T I O N S 61. Halley’s Comet. Halley’s comet travels an elliptical path that can be modeled with the polar equation r 5 0.58711 1 0.9672 1 2 0.967 cos u . Sketch the graph of the path of Halley’s comet. 62. Dwarf Planet Pluto. The dwarf planet Pluto travels in an elliptical orbit that can be modeled with the polar equation r 5 29.6211 1 0.2492 1 2 0.249 cos u . Sketch the graph of Pluto’s orbit. 63. Archimedes Spiral. Spirals are seen in nature, as in the swirl of a pine cone; they are also used in machinery to convert motions. An Archimedes spiral has the general equation r 5 a u. A more general form for the equation of a spiral is r 5 a u1/n, where n is a constant that determines how tightly the spiral is wrapped. Compare the Archimedes spiral r 5 u with the spiral r 5 u1/2 by graphing both on the same polar graph. 64. Archimedes Spiral. Spirals are seen in nature, as in the swirl of a pine cone; they are also used in machinery to convert motions. An Archimedes spiral has the general equation r 5 a u. A more general form for the equation of a spiral is r 5 a u1/n, where n is a constant that determines how tightly the spiral is wrapped. Compare the Archimedes spiral r 5 u with the spiral r 5 u4/3 by graphing both on the same polar graph. For Exercises 65 and 66, refer to the following: The lemniscate motion occurs naturally in the flapping of birds’ wings. The bird’s vertical lift and wing sweep create the distinctive figure-eight pattern. The patterns vary with the different wing profiles. 65. Flapping Wings of Birds. Compare the following two possible lemniscate patterns by graphing them on the same polar graph: r 2 5 4 cos12u2 and r 2 5 1 4 cos12u2. 66. Flapping Wings of Birds. Compare the following two possible lemniscate patterns by graphing them on the same polar graph: r 2 5 4 cos12u2 and r 2 5 4 cos12u 1 22. For Exercises 67 and 68, refer to the following: Many microphone manufacturers advertise that their microphones’ exceptional pickup capabilities isolate the sound source and minimize background noise. These microphones are described as cardioid microphones because of the pattern formed by the range of the pickup. 67. Cardioid Pickup Pattern. Graph the cardioid curve r 5 2 1 2 sin u to see what the range looks like. 68. Cardioid Pickup Pattern. Graph the cardioid curve r 5 24 2 4 sin u to see what the range looks like. For Exercises 69 and 70, refer to the following: The sword artistry of the Samurai is legendary in Japanese folklore and myth. The elegance with which a samurai could wield a sword rivals the grace exhibited by modern figure skaters. In more modern times, such legends have been rendered digitally in many different video games (e.g., Onimusha). In order to make the characters realistically move across the screen, and in particular, wield various sword motions true to the legends, trigonometric functions are extensively used in constructing the underlying graphics module. One famous movement is a figure eight, swept out with two hands on the sword. The actual path of the tip of the blade as the movement progresses in this figure-eight motion depends essentially on the length L of the sword and the speed with which it is swept out. Such a path is modeled using a polar equation of the form r 21u2 5 L cos1Au2 or r 21u2 5 L sin1Au2, u1 # u # u2 whose graphs are called lemniscates. 3 π 4 π 6 π 12 π 0 6 11π 12 17π 12 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π 3 5 1 69. Video Games. Graph the following equations: a. r 21u2 5 5 cos u, 0 # u # 2p b. r 21u2 5 5 cos12 u2, 0 # u # p c. r 21u2 5 5 cos14 u2, 0 # u # p 2 What do you notice about all of these graphs? Suppose that the movement of the tip of the sword in a game is governed by these graphs. Describe what happens if you change the domain in (b) and (c) to 0 # u # 2p. 70. Video Games. Write a polar equation that would describe the motion of a sword 12 units long that makes 8 complete motions in 30, 2p4. 71. Home Improvement. The owner of a garden maze has decided to replace the square central plot of the maze, which is 60 feet 3 60 feet, with a section in the shape of a spiral to make the participants literally feel as though they are going in circles as they reach the tall slide in the center of the maze (that leads to the exit). Assume that the center of the maze is at the origin. If the walkway must be 3 feet wide to accommodate a participant, and the bushes on either side of the walkway are 1.5 feet thick, how many times can the spiral wrap around before the center is reached? Graph the resulting spiral. 72. Home Improvement. Consider the garden maze in Exercise 71, but now suppose that the owner wants the spiral to wrap around exactly 3 times. How wide can the walkway then be throughout the spiral? Graph it. 73. Magnetic Pendulum. A magnetic bob is affixed to an arm of length L, which is fastened to a pivot point. Three magnets of equal strength are positioned on a plane 8 inches from the 854 CHAPTER 8 Additional Topics in Trigonometry center; one is placed on the x-axis, one at 120° with respect to the positive x-axis, and the other at 240° with respect to the positive x-axis. The path swept out is a three-petal rose, as shown below: 3 π 4 π 6 π 12 π 0 6 11π 12 17π 12 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π 10 6 a. Find the equation of this path. b. How many times does the path retrace itself on the interval [0, 100p]? 74. Magnetic Pendulum. In reference to the context of Exercise 73, now position 8 magnets, each 8 units from the origin and at the vertices of a regular octagon, one being on the x-axis. Assume that the path of the pendulum is an eight-petal rose. a. Find the equation of this path. b. Graph this equation. • C A T C H T H E M I S T A K E In Exercises 75 and 76, explain the mistake that is made. 75. Convert 122, 222 to polar coordinates. Solution: Label x and y. x 5 22, y 5 22 Find r. r 5 "x2 1 y2 5 !4 1 4 5 !8 5 2!2 Find u. tan u 5 22 22 5 1 u 5 tan 21112 5 p 4 Write the point in polar coordinates. a2!2, p 4 b This is incorrect. What mistake was made? 76. Convert A2!3, 1B to polar coordinates. Solution: Label x and y. x 5 2!3, y 5 1 Find r. r 5 "x2 1 y2 5 !3 1 1 5 !4 5 2 Find u. tan u 5 1 2!3 5 2 1 !3 u 5 tan 2 1a 2 1 !3b 5 2p 4 Write the point in polar coordinates. a2, 2 p 4 b This is incorrect. What mistake was made? In Exercises 77 and 78, determine whether each statement is true or false. 77. All cardioids are limaçons, but not all limaçons are cardioids. 78. All limaçons are cardioids, but not all cardioids are limaçons. 79. Find the polar equation that is equivalent to a vertical line, x 5 a. 80. Find the polar equation that is equivalent to a horizontal line, y 5 b. 81. Give another pair of polar coordinates for the point 1a, u2. 82. Convert 12a, b2 to polar coordinates. Assume that a . 0, b . 0. • C O N C E P T U A L 8.8 Polar Equations and Graphs 855 83. Determine the values of u at which r 5 4 cos u and r cos u 5 1 intersect. Graph both equations. 84. Find the Cartesian equation for r 5 a sin u 1 b cos u, where a and b are positive. Identify the type of graph. • C H A L L E N G E 85. Find the Cartesian equation for r 5 a sin12u2 cos3 u 2 sin3 u. 86. Identify an equation for the following graph: 3 π 4 π 6 π 12 π 0 6 11π 12 17π 12 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π 10 87. Consider the equation r 5 2a cos 1u 2 b2. Sketch the graph for various values of a and b, and then give a general description of the graph. 88. Consider the equation r 5 a sin 1bu2, where a, b . 0. Determine the smallest number M for which the graph starts to repeat. 89. Given r 5 cosau 2b, find the u-intervals for the inner loop above the x-axis. 90. Given r 5 2 cosa3u 2 b, find the u-intervals for the petal in the first quadrant. 91. Given r 5 1 1 3 cos u, find the u-intervals for the inner loop. 92. Given r 5 1 1 sin12u2 and r 5 1 2 cos12u2, find all points of intersection. 93. Given r 5 2 1 sin 14u2 and r 5 1, find the angles of all the points of intersection. 94. Given r 5 2 2 cos 13u2 and r 5 1.5, find the angles of all points of intersection. • T E C H N O L O G Y CH A P TE R 8 R E VI E W 856 CHAPTER 8 Additional Topics in Trigonometry [ C H AP T E R 8 REVIEW] SECTION CONCEPT KEY IDEAS/FORMULAS 8.1 Oblique triangles and the Law of Sines Solving oblique triangles Four cases ■ AAS/ASA ■ SSA ■ SAS ■ SSS Oblique (Nonright) Triangles α β γ Acute Triangle a b c α β γ Obtuse Triangle a b c The Law of Sines sin a a 5 sin b b 5 sin g c ■ AAS (or ASA) triangles ■ SSA triangles (ambiguous case) 8.2 The Law of Cosines a2 5 b2 1 c2 2 2bc cos a b2 5 a2 1 c2 2 2ac cos b c2 5 a2 1 b2 2 2ab cos g Solving oblique triangles ■ SAS triangles ■ SSS triangles 8.3 The area of a triangle The area of a triangle (SAS case) Use the Law of Sines for the SAS case: ASAS case 5 1 2 bc sin a when b, c, and a are known. ASAS case 5 1 2 ab sin g when a, b, and g are known. ASAS case 5 1 2 ac sin b when a, c, and b are known. The area of a triangle (SSS case) Use Heron’s formula for the SSS case: ASSS case 5 !s1s 2 a2 1s 2 b2 1s 2 c2 where a, b, and c are the lengths of the sides of the triangle and s is half the perimeter of the triangle, called the semiperimeter. s 5 a 1 b 1 c 2 8.4 Vectors Vector u or AB r B A u Vectors: Magnitude and Direction u 5 8a, b9 Geometric Algebraic Magnitude length of a vector 0 u 0 5"a2 1 b2 Direction u tan u 5 b a Addition tail-to-tip u v u + v u 5 8a, b9 and v 5 8c, d 9 u 1 v 5 8a 1 c, b 1 d 9 CH A P TE R 8 R E VIE W Chapter Review 857 SECTION CONCEPT KEY IDEAS/FORMULAS Vector operations Scalar multiplication: k 8a, b9 5 8k a, k b9 Horizontal and vertical components of a vector Horizontal component: a 5 0 u 0 cos u Vertical component: b 5 0 u 0 sin u Unit vectors u 5 v 0 v 0 Resultant vectors 8.5 The dot product ■ The product of a scalar and a vector is a vector. ■ The dot product of two vectors is a scalar. Multiplying two vectors: the dot product u 5 8a, b9 and v 5 8c, d 9 u⋅v 5 ac 1 bd Angle between two vectors If u is the angle between two nonzero vectors u and v, where 0° # u # 180°, then cos u 5 u⋅v 0 u 0 0 v 0 Orthogonal (perpendicular) vectors: u⋅v 5 0 Work When force and displacement are in the same direction: W 5 0 F 0 0 d 0. When force and displacement are not in the same direction: W 5 F⋅d. 8.6 Polar (trigonometric) form of complex numbers Complex numbers in rectangular form The modulus, or magnitude, of a complex number z 5 x 1 iy is the distance from the origin to the point 1x, y2 in the complex plane given by 0 z 0 5 "x2 1 y2 Complex numbers in polar form The polar form of a complex number is z 5 r1cos u 1 i sin u2 where r represents the modulus (magnitude) of the complex number and u represents the argument of z. Converting complex numbers between rectangular and polar forms: Step 1: Plot the point z 5 x 1 yi in the complex plane (note the quadrant). Step 2: Find r. Use r 5 "x2 1 y2. Step 3: Find u. Use tan u 5 y x, x 2 0, where u is in the quadrant found in Step 1. 8.7 Products, quotients, powers, and roots of complex numbers Products of complex numbers Let z1 5 r11cos u1 1 i sin u12 and z2 5 r21cos u2 1 i sin u22 be two complex numbers. The product z1z2 is given by z1z2 5 r1r23cos1u1 1 u22 1 i sin1u1 1 u224 Multiply the magnitudes and add the arguments. CH A P TE R 8 R E VI E W 858 CHAPTER 8 Additional Topics in Trigonometry SECTION CONCEPT KEY IDEAS/FORMULAS Quotients of complex numbers Let z1 5 r11cos u1 1 i sin u12 and z2 5 r21cos u2 1 i sin u22 be two complex numbers. The quotient z1 z2 is given by z1 z2 5 r1 r2 3cos1u1 2 u22 1 i sin1u1 2 u224 Divide the magnitudes and subtract the arguments. Powers of complex numbers De Moivre’s theorem If z 5 r 1cos u 1 i sin u2 is a complex number, then z n 5 r n3cos1n u2 1 i sin1 n u24, n $ 1, where n is an integer. Roots of complex numbers The nth roots of the complex number z 5 r 1cos u 1 i sin u2 are given by wk 5 r1/n ccos au n 1 k ⋅360° n b 1 i sin au n 1 k ⋅360° n bd u in degrees, where k 5 0, 1, 2, c, n 2 1. 8.8 Polar equations and graphs x y x θ y r (x, y) (r, θ) Polar coordinates To plot a point 1r, u2: ■ Start on the polar axis and rotate a ray to form the terminal side of an angle u. ■ If r . 0, the point is r units from the origin in the same direction as the terminal side of u. ■ If r , 0, the point is 0 r 0 units from the origin in the opposite direction of the terminal side of u. 3 π 4 π 6 π 12 π 0 1 3 5 12 17π 12 11π 6 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π Converting between polar and rectangular coordinates From polar 1r, u2 to rectangular 1x, y2: x 5 r cos u y 5 r sin u From rectangular 1x, y2 to polar 1r, u2: r 5 "x2 1 y2 tan u 5 y x x 2 0 Graphs of polar equations Radial line, circle, spiral, rose petals, lemniscate, and limaçon R E VI E W E XERCISES [CH AP TER 8 REVIEW EXE R C IS E S ] Review Exercises 859 8.1 Oblique Triangles and the Law of Sines Solve the given triangles. 1. a 5 10°, b 5 20°, a 5 4 2. b 5 40°, g 5 60°, b 5 10 3. a 5 5°, b 5 45°, c 5 10 4. b 5 60°, g 5 70°, a 5 20 5. g 5 11°, a 5 11°, c 5 11 6. b 5 20°, g 5 50°, b 5 8 7. a 5 45°, g 5 45°, b 5 2 8. a 5 60°, b 5 20°, c 5 17 9. a 5 12°, g 5 22°, a 5 99 10. b 5 102°, g 5 27°, a 5 24 Two sides and an angle are given. Determine whether a triangle (or two) exist and, if so, solve the triangle. 11. a 5 7, b 5 9, a 5 20° 12. b 5 24, c 5 30, b 5 16° 13. a 5 10, c 5 12, a 5 24° 14. b 5 100, c 5 116, b 5 12° 15. a 5 40, b 5 30, b 5 150° 16. b 5 2, c 5 3, g 5 165° 17. a 5 4, b 5 6, a 5 10° 18. c 5 25, a 5 37, g 5 4° 8.2 The Law of Cosines 19. How Far from Home? Gary walked 8 miles due north, made a 130° turn to the southeast, and walked for another 6 miles. How far was Gary from home? 20. How Far from Home? Mary walked 10 miles due north, made a 55° turn to the northeast, and walked for another 3 miles. How far was Mary from home? Solve each triangle. 21. a 5 40, b 5 60, g 5 50° 22. b 5 15, c 5 12, a 5 140° 23. a 5 24, b 5 25, c 5 30 24. a 5 6, b 5 6, c 5 8 25. a 5 !11, b 5 !14, c 5 5 26. a 5 22, b 5 120, c 5 122 27. b 5 7, c 5 10, a 5 14° 28. a 5 6, b 5 12, g 5 80° 29. b 5 10, c 5 4, a 5 90° 30. a 5 4, b 5 5, g 5 75° 31. a 5 10, b 5 11, c 5 12 32. a 5 22, b 5 24, c 5 25 33. b 5 16, c 5 18, a 5 100° 34. a 5 25, c 5 25, b 5 9° 35. b 5 12, c 5 40, a 5 10° 36. a 5 26, b 5 20, c 5 10 37. a 5 26, b 5 40, c 5 13 38. a 5 1, b 5 2, c 5 3 39. a 5 6.3, b 5 4.2, a 5 15° 40. b 5 5, c 5 6, b 5 35° 8.3 The Area of a Triangle Find the area of each triangle described. 41. b 5 16, c 5 18, a 5 100° 42. a 5 25, c 5 25, b5 9° 43. a 5 10, b 5 11, c 5 12 44. a 5 22, b 5 24, c 5 25 45. a 5 26, b 5 20, c 5 10 46. a 5 24, b 5 32, c 5 40 47. b 5 12, c 5 40, a 5 10° 48. a 5 21, c 5 75, b560° Applications 49. Area of Inscribed Triangle. The area of a triangle inscribed in a circle can be found if you know the lengths of the sides of the triangle and the radius of the circle: A 5 abc 4r . Find the radius of the circle that circumscribes the triangle if all the sides of the triangle measure 9 inches and the area of the triangle is 35 square inches. 50. Area of Inscribed Triangle. The area of a triangle inscribed in a circle can be found if you know the lengths of the sides of the triangle and the radius of the circle: A 5 abc 4r . Find the radius of the circle that circumscribes the triangle if the sides of the triangle measure 9, 12, and 15 inches and the area of the triangle is 54 square inches. 8.4 Vectors Find the magnitude of vector AB r. 51. A 5 14, 232 and B 5 128, 22 52. A 5 122, 112 and B 5 12, 82 53. A 5 10, 232 and B 5 15, 92 54. A 5 13, 2112 and B 5 19, 232 Find the magnitude and direction angle of the given vector. 55. u 5 8210, 249 56. u 5 825, 2129 57. u 5 816, 2129 58. u 5 80, 39 Perform the vector operation, given that u 5 87, 229 and v 5 824, 59. 59. 2u 1 3v 60. u 2 v 61. 6u 1 v 62. 231u 1 2v2 Find the vector, given its magnitude and direction angle. 63. 0 u 0 5 10, u 5 75° 64. 0 u 0 5 8, u 5 225° 65. 0 u 0 5 12, u 5105° 66. 0 u 0 5 20, u 5 15° Find a unit vector in the direction of the given vector. 67. v 5 8!6, 2!69 68. v 5 8211, 609 Perform the indicated vector operation. 69. 13i 2 4j2 1 12i 1 5j2 70. 126i 1 j2 2 19i 2 j2 8.5 The Dot Product Find the indicated dot product. 71. 86, 239⋅81, 49 72. 826, 59⋅824, 29 73. 83, 39⋅83, 269 74. 822, 289⋅821, 19 75. 80, 89⋅81, 29 76. 84, 239⋅821, 09 REV IEW E XE R CI SE S 860 CHAPTER 8 Additional Topics in Trigonometry Find the angle (round to the nearest degree) between each pair of vectors. 77. 83, 49 and 825, 129 78. 824, 59 and 85, 249 79. 81, !2 9 and 821, 3!29 80. 87, 2249 and 826, 89 81. 83, 59 and 824, 249 82. 821, 69 and 82, 229 Determine whether each pair of vectors is orthogonal. 83. 88, 39 and 823, 129 84. 826, 29 and 84, 129 85. 85, 269 and 8212, 2109 86. 81, 19 and 824, 49 87. 80, 49 and 80, 249 88. 827, 29 and H1 7, 21 2I 89. 86z, a 2 b9 and 8a 1 b, 26z9 90. 8a 2 b, 219 and 8a 1 b, a2 2 b2 9 8.6 Polar (Trigonometric) Form of Complex Numbers Graph each complex number in the complex plane. 91. 26 1 2i 92. 5i Express each complex number in polar form. 93. !2 2 !2i 94. !3 1 i 95. 28i 96. 28 2 8i With a calculator, express each complex number in polar form. 97. 260 1 11i 98. 9 2 40i 99. 15 1 8i 100. 210 2 24i Express each complex number in rectangular form. 101. 61cos 300° 1 i sin 300°2 102. 41cos 210° 1 i sin 210°2 103. !21cos 135°1i sin 135°2 104. 41cos 150° 1 i sin 150°2 With a calculator, express each complex number in rectangular form. 105. 41cos 200° 1 i sin 200°2 106. 31cos 350° 1 i sin 350°2 8.7 Products, Quotients, Powers, and Roots of Complex Numbers Find the product z1z2. 107. 31cos 200° 1 i sin 200°2 and 41cos 70° 1 i sin 70°2 108. 31cos 20° 1 i sin 20°2 and 41cos 220° 1 i sin 220°2 109. 71cos 100° 1 i sin 100°2 and 31cos 140° 1 i sin 140°2 110. 1cos 290° 1 i sin 290°2 and 4 1cos 40° 1 i sin 40°2 Find the quotient z1 z2. 111. !61cos 200° 1 i sin 200°2 and !61cos 50° 1 i sin 50°2 112. 181cos 190° 1 i sin 190°2 and 2 1cos 100° 1 i sin 100°2 113. 241cos 290° 1 i sin 290°2 and 41cos 110° 1 i sin 110°2 114. !2001cos 93° 1 i sin 93°2 and !21cos 48° 1 i sin 48°2 Find the result of each expression using De Moivre’s theorem. Write the answer in rectangular form. 115. 13 1 3i24 116. 13 1 !3i24 117. 11 1 !3i25 118. 122 2 2i27 Find all nth roots of z. Write the answers in polar form, and plot the roots in the complex plane. 119. 2 1 2!3i, n 5 2 120. 28 1 8!3i, n 5 4 121. 2256, n 5 4 122. 218i, n 5 2 Find all complex solutions to the given equations. 123. x3 1 216 5 0 124. x4 2 1 5 0 125. x4 1 1 5 0 126. x3 2 125 5 0 8.8 Polar Equations and Graphs Convert each point to exact polar coordinates (assuming that 0 " u 2p2, and then graph the point in the polar coordinate system. 127. 122, 22 128. 14, 24!32 129. 125!3, 252 130. 1 !3, !32 131. 10, 222 132. 111, 02 Convert each point to exact rectangular coordinates. 133. a23, 5p 3 b 134. a4, 5p 4 b 135. a2, p 3 b 136. a6, 7p 6 b 137. a1, 4p 3 b 138. a 23, 7p 4 b Graph each equation. 139. r 5 4 cos12u2 140. r 5 sin13u2 141. r 5 2u 142. r 5 4 2 3 sin u Technology Exercises Section 8.1 Let A, B, and C be the lengths of the three sides of a triangle with X, Y, and Z as the corresponding angles. Write a program using a TI calculator to solve the given triangle. 143. A 5 31.6, C 5 23.9, X 5 42° 144. A 5 137.2, B 5 125.1, Y 5 54° Section 8.2 Let A, B, and C be the lengths of the three sides of a triangle with X, Y, and Z as the corresponding angles. Write a program using a TI calculator to solve the given triangle. 145. A 5 !33, B 5 !29, Z 5 41.6° 146. A 5 3412, B 5 2178, C 5 1576 147. A 5 312, B 5 267, C 5 189 148. A 5 12.7, B 5 29.9, Z 5 104.8° Section 8.4 With the graphing calculator SUM command, find the magnitude of the given vector. Also, find the direction angle to the nearest degree. 149. 825, 2609 150. 8270, 10!159 R E VI E W E XERCISES Review Exercises 861 Section 8.5 With the graphing calculator SUM command, find the angle (round to the nearest degree) between each pair of vectors. 151. 814, 379, 89, 2269 152. 8223, 289, 818, 2329 Section 8.6 Another way of using a graphing calculator to represent complex numbers in rectangular form is to enter the real and imaginary parts as a list of two numbers and use the SUM command to find the modulus. 153. Write 2!23 2 11i in polar form using the SUM command to find its modulus, and round the angle to the nearest degree. 154. Write 11 1 !23i in polar form using the SUM command to find its modulus, and round the angle to the nearest degree. Section 8.7 155. Find the fourth roots of 28 1 8!3i, and draw the complex rectangle with the calculator. 156. Find the fourth roots of 8!3 1 8i, and draw the complex rectangle with the calculator. Section 8.8 157. Given r 5 1 2 2 sin13u2, find the angles of all points of intersection (where r 5 02. 158. Given r 5 1 1 2 cos13u2, find the angles of all points of intersection (where r 5 02. PR ACTICE TEST [ C H AP T E R 8 PRACTICE TEST ] 862 CHAPTER 8 Additional Topics in Trigonometry Solve the triangles, if possible. 1. a 5 30°, b 5 40°, b 5 10 2. a 5 47°, b 5 98°, g 5 35° 3. a 5 7, b 5 9, c 5 12 4. a 5 45°, a 5 8, b 5 10 5. a 5 1, b 5 1, c 5 2 6. a 5 23 7 , c 5 5 7, b 5 61.2° 7. a 5 110°, b 5 20°, a 5 5 8. b 5 !5 2 , c 5 3!5, a 5 45° In Exercises 9 and 10, find the areas of the given triangles. 9. g 5 72°, a 5 10, b 5 12 10. a 5 7, b 5 10, c 5 13 11. Find the magnitude and direction angle of the vector u 5 825, 129. 12. Find a unit vector pointing in the same direction as v 5 823, 249. 13. Perform the indicated operations: a. 2 821, 49 2 3 84, 19 b. 827, 219⋅82, 29 14. In a post pattern in football, the receiver in motion runs past the quarterback parallel to the line of scrimmage (A), runs 12 yards perpendicular to the line of scrimmage (B), and then cuts toward the goal post (C). 5 0 4 0 4 0 3 0 2 0 1 0 3 0 2 0 1 0 5 0 4 0 4 0 3 0 2 0 1 0 3 0 2 0 1 0 B C A A + B + C 30º θ A receiver runs the post pattern. If the magnitudes of the vectors are 0A 0 5 3 yards, 0 B 0 5 12 yards, and 0 C 0 5 18 yards, find the magnitude of the resultant vector A 1 B 1 C and the direction angle u. For Exercises 15–17, use the complex number z 5 161cos 120° 1 i sin 120°2. 15. Find z4. 16. Find the four distinct fourth roots of z. 17. Convert the point 13, 210°2 to rectangular coordinates. 18. Graph r 5 6 sin1 2u2. 19. Graph r 2 5 9 cos1 2u2. 20. Find x such that 8x, 19 is perpendicular to 3i S2 4 j S. 21. Prove that u S⋅1v S 2 w S2 5 u S⋅v S 2 u S⋅w S. 22. Construct a unit vector in the opposite direction of 83, 59. 23. Compute u S⋅v S if 0 u S 0 5 4, 0 v S 0 5 10, and u 5 2p 3 . 24. Determine whether u S and v S are parallel, perpendicular, or neither: u S 5 8sin u, cos u9, v S 5 82cos u, sin u9. 25. Find the magnitude of 2i S 2 j S. 26. Determine u, when a streetlight is formed as follows: θ 2 ft 4 ft 5 ft 27. Find the lengths of a, b, c, and d in the following diagram: 2 d a b c 50º 110º 95º 28. True or False: If u S 1 v S is perpendicular to u S 2 v S, then 0 u S 0 5 0 v S 0. 29. Solve z4 1 256i 5 0. 30. Convert to a Cartesian equation: r 2 5 tan u. 31. With the graphing calculator SUM command, find the angle (round to the nearest degree) between the pair of vectors: 828, 2119 and 8216, 269. 32. Find the fourth roots of 28!3 2 8i, and draw the complex rectangle with the calculator. CU MU LA TIV E TEST [CH AP TERS 1–8 CUM UL AT IVE T E S T ] Cumulative Test 863 1. Solve using the square root method: 12 2 x22 5 216. 2. Solve for x, and express the solution in interval notation: 25x 2 3 # 12. 3. Find the slope of the line that passes through the points A1 3, 3 4B and A21 6, 21 4B. 4. Transform the equation into standard form by completing the square. State the center and radius of the circle: x2 1 y2 2 3.2x 1 4.4y 2 0.44 5 0. 5. Using the function ƒ1x2 5 5 2 x2, evaluate the difference quotient ƒ1x 1 h2 2 ƒ1x2 h . 6. Given the piecewise-defined function ƒ1x2 5 • x2 x , 0 2x 2 1 0 # x , 5 5 2 x x $ 5 find: a. ƒ102 b. ƒ142 c. ƒ152 d. ƒ1242 e. State the domain and range in interval notation. f. Determine the intervals where the function is increasing, decreasing, or constant. 7. Evaluate ƒ1g12122, g1x2 5 ! 3 x , and ƒ1x2 5 9 x 1 1. 8. Find a quadratic function that has the vertex 11.5, 2.52 and point 120.5, 20.52. 9. For the function ƒ1x2 5 5x21x 1 2231x2 1 72, find all of the real zeros and state the multiplicity. 10. List the possible rational zeros, and test to determine all rational zeros for P 1x2 5 6x3 1 x2 2 5x 2 2. 11. Graph the rational function ƒ1x2 5 x2 1 3 x 2 2 . Give all asymptotes. 12. Graph the exponential function ƒ1x2 5 5x21, and state the y-intercept, the domain, the range, and the horizontal asymptote. 13. How much money should be put in a savings account now that earns 5.5% a year compounded continuously, if you want to have $85,000 in 15 years? 14. Graph the function ƒ1x2 5 2log1x2 1 2, using transformation techniques. 15. Use properties of logarithms to simplify the expression 1023 log 10. 16. Solve for x: ln !6 2 3x 2 1 2 ln1x 1 22 5 ln1x2. 17. In a 45°-45°-90° triangle, if the two legs have length 15 feet, how long is the hypotenuse? 18. Find the positive measure of u (rounded to the nearest degree) when tan u 5 1.4285 and the terminal side of u lies in quadrant III. 19. Graph y 5 3 cos12x2 over one period. 20. If sin u 5 8 11 and the terminal side of u lies in quadrant II, find the exact value of sec u. 21. Find the exact value of cos1a 2 b2 when cos a 5 1 5, cos b 5 23 5, the terminal side of a lies in quadrant IV, and the terminal side of b in quadrant II. 22. Solve the trigonometric equation cos12u2 5 2 1 2 exactly, over the interval 0 # u , 2p. 23. Solve the triangle with b 5 8, a 5 13, and g 5 21°. 24. Given u 5 85, 239 and v 5 822, 69, find 41v 2 u2. 25. Express the complex number 2!2 acos 3p 4 1 i sin 3p 4 b in rectangular form. 26. With the graphing calculator SUM command, find the magnitude of vector 840, 2969 and the direction angle to the nearest degree. 27. Given r 5 2 2 5 sin1 2u2, find the angles of all points of intersection 1when r 5 02. Round the angles to the nearest degree. C H A P T E R LEARNING OBJECTIVES [ [ On August 29, 2005, Hurricane Katrina unleashed its fury in Southeast Louisiana. The most significant number of deaths occurred in New Orleans, which flooded when its levee system failed. The Federal Emergency Management Agency (FEMA) coordinated the relief effort, including trucks that hauled generators, water, and tarps. FEMA was faced with an optimization problem in mathematics that involved maximizing the number of people aided by the relief subject to the constraints of the weight of the relief items and how much space they would take on the trucks. For example, how many generators, cases of water, and tarps could each truck haul, and how many of each would bring aid to the greatest number of people? 9 Systems of Linear Equations and Inequalities ■ ■Solve systems of linear equations in two variables with the substitution method and the elimina-tion method. ■ ■Solve systems of lin-ear equations in three variables employing a combination of the elimination and substitution methods. ■ ■Perform partial-fraction decomposition. ■ ■Solve a system of linear inequalities by finding the overlapping shaded regions. ■ ■Use the linear programming model to solve optimization problems subject to constraints. PAUL J. RICHARDS/ Getty Images, Inc. Mario Tama/Getty Images, Inc. Justin Sullivan/Getty Images, Inc. 865 [I N T HI S CHAPTER] We will solve systems of linear equations in two variables and in three variables. We will then perform partial-fraction decomposition, which will be valuable in calculus and is made possible by techniques of solving systems of linear equations. We will discuss linear inequalities in two variables and systems of linear inequalities. Systems of nonlinear equations and inequalities will be discussed in Chapter 11. Finally, we will discuss the linear programming model, which will enable you to address the FEMA relief problem as a result of Florida’s multiple hurricanes. 865 SYSTEMS OF LINEAR EQUATIONS AND INEQUALITIES 9.1 SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 9.2 SYSTEMS OF LINEAR EQUATIONS IN THREE VARIABLES 9.3 PARTIAL FRACTIONS 9.4 SYSTEMS OF LINEAR INEQUALITIES IN TWO VARIABLES 9.5 THE LINEAR PROGRAMMING MODEL • Solving Systems of Linear Equations • Three Methods and Three Types of Solutions • Solving Systems of Linear Equations in Three Variables • Types of Solutions • Performing Partial-Fraction Decomposition • Linear Inequalities in Two Variables • Systems of Linear Inequalities in Two Variables • Solving an Optimization Problem 866 CHAPTER 9 Systems of Linear Equations and Inequalities 9.1.1 Solving Systems of Linear Equations Overview We learned in Section 2.3 that a linear equation in two variables is given in standard form by Ax 1 By 5 C and that the graph of this linear equation is a line, provided that A and B are not both equal to zero. In this section we discuss systems of linear equations, which can be thought of as simultaneous equations. To solve a system of linear equations means to find the solution that satisfies both equations: x 1 2y 5 6 3x 2 y 5 11 We can interpret the solution to this system of equations both algebraically and graphically. S K I L L S O B J E C T I V E S ■ ■Solve systems of linear equations in two variables. ■ ■Solve applications involving systems of linear equations. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that a system of linear equations has either one solution (two lines intersecting at a single point), no solution (parallel lines), or infinitely many solutions (same line). ■ ■Understand how to model a real-world problem using systems of linear equations in two variables. 9.1 SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 9.1.1 SKI LL Solve systems of linear equations in two variables. 9.1.1 CO NCE PTUAL Understand that a system of linear equations has either one solution (two lines intersecting at a single point), no solution (parallel lines), or infinitely many solutions (same line). ALGEBRAIC GRAPHICAL Solution x 5 4 and y 5 1 14, 12 Check x y 5 6 –4 (4, 1) 3x – y = 11 x + 2y = 6 Interpretation x 5 4 and y 5 1 satisfy both equations. The point 14, 12 lies on both lines. Equation 1 x 1 2y 5 6 142 1 2112 5 6 3 Equation 2 3x 2 y 5 11 3142 2 1 5 11 3 This particular example had one solution. There are also systems of equations that have no solution or infinitely many solutions. We give these three systems special names: independent, inconsistent, and dependent, respectively. INDEPENDENT SYSTEM INCONSISTENT SYSTEM DEPENDENT SYSTEM One solution No solution Infinitely many solutions x y x y y x Lines have different slopes. Lines are parallel (same slope and different y-intercepts). Lines coincide (same slope and same y-intercept). In this section, we discuss three methods for solving systems of linear equations: substitution, elimination, and graphing. We use the algebraic methods—substitution and elimination—to find solutions exactly; we then look at a graphical interpretation of the solution (two lines that intersect at one point, parallel lines, or coinciding lines). We will illustrate each method with the example given earlier: x 1 2y 5 6 Equation (1) 3x 2 y 5 11 Equation (2) Substitution Method The following box summarizes the substitution method for solving systems of linear equations in two variables. EXAMPLE 1 Determining by Substitution That a System Has One Solution Use the substitution method to solve the following system of linear equations: x 1 y 5 8 E quation (1) 3x 2 y 5 4 Equation (2) Solution: STEP 1 Solve Equation (2) for y in terms of x. y 5 3x 2 4 STEP 2 Substitute y 5 3x 2 4 into Equation (1). x 1 13x 2 42 5 8 STEP 3 Solve for x. x 1 3x 2 4 5 8 4x 5 12 x 5 3 STEP 4 Back-substitute x 5 3 into the equation found in Step 1. y 5 3132 2 4 y 5 5 x y –2 8 8 –2 x + y = 8 3x – y = 4 (3, 5) SUBSTITUTION METHOD Step 1: Solve one of the equations for one Equation (2): y 5 3x 2 11 variable in terms of the other variable. Step 2: Substitute the expression found in Equation (1): x 1 213x 2 112 5 6 Step 1 into the other equation. The result is an equation in one variable. Step 3: Solve the equation obtained in Step 2. x 1 6x 2 22 5 6 7x 5 28 x 5 4 Step 4: Back-substitute the value found in y 5 3142 2 11 Step 3 into the expression found y 5 1 in Step 1. Step 5: Check that the solution satisfies both Equation (1): x 1 2y 5 6 equations. 142 1 2112 5 6 3 Substitute 14, 12 into both equations. Equation (22: 3x 2 y 5 11 3142 2 1 5 11 3 9.1 Systems of Linear Equations in Two Variables 867 868 CHAPTER 9 Systems of Linear Equations and Inequalities STEP 5 Check that 13, 52 satisfies both equations. Equation (1): x 1 y 5 8 3 1 5 5 8 Equation (2): 3x 2 y 5 4 3132 2 5 5 4 Note: The graphs of the two equations are two lines that intersect at the point 13, 52. x –1 9 –5 y 5 x – y = 2 2x – 2y = 10 EXAMPLE 2 Determining by Substitution That a System Has No Solution Use the substitution method to solve the following system of linear equations: x 2 y 5 2 Equation (1) 2x 2 2y 5 10 Equation (2) Solution: STEP 1 Solve Equation (1) for y in terms of x. y 5 x 2 2 STEP 2 Substitute y 5 x 2 2 into Equation (2). 2x 2 21x 2 22 5 10 STEP 3 Solve for x. 2x 2 2x 1 4 5 10 4 5 10 4 5 10 is never true, so this is called an inconsistent system. There is no solution to this system of linear equations. Note: The graphs of the two equations are parallel lines. EXAMPLE 3 Determining by Substitution That a System Has Infinitely Many Solutions Use the substitution method to solve the following system of linear equations: x 2 y 5 2 Equation (1) 2x 1 y 5 22 Equation (2) Solution: STEP 1 Solve Equation (1) for y in terms of x. y 5 x 2 2 STEP 2 Substitute y 5 x 2 2 into Equation (2). 2x 1 1x 2 22 5 22 STEP 3 Solve for x. 2x 1 x 2 2 5 22 22 5 22 22 5 22 is always true, so this is called a dependent system. Notice, for instance, that the points 12, 02, 14, 22, and 17, 52 all satisfy both equations. In fact, there are infinitely many solutions to this system of linear equations. All solutions are in the form 1x, y2, where y 5 x 2 2 (the graphs of these two equations are the same line). x –5 5 –5 x – y = 2 –x + y = –2 y 5 9.1 Systems of Linear Equations in Two Variables 869 Elimination Method We now turn our attention to another method, elimination, which is often preferred over substitution and will later be used in higher-order systems. In a system of two linear equations in two variables, the equations can be combined, resulting in a third equation in one variable, thus eliminating one of the variables. The following is an example of when elimination would be preferred because the y terms sum to zero when the two equations are added together: 2x 2 y 5 5 2x 1 y 5 22 x 5 3 When you cannot eliminate a variable simply by adding the two equations, multiply one equation by a constant that will cause the coefficients of some variable in the two equations to match and be opposite in sign. The following box summarizes the elimination method, also called the addition method, for solving systems of linear equations in two variables using the example given earlier: x 1 2y 5 6 Equation (1) 3x 2 y 5 11 Equation (2) ELIMINATION METHOD Step 1: Multiply the coefficients of one or both of the Multiply Equation (2) by 2: equations so that one of the variables will be 6x 2 2y 5 22 eliminated when the two equations are added. Step 2: Eliminate one of the variables by adding the x 1 2y 5 6 expression found in Step 1 to the other equation. 6x 2 2y 5 22 The result is an equation in one variable. 7x 5 28 Step 3: Solve the equation obtained in Step 2. 7x 5 28 x 5 4 Step 4: Back-substitute the value found in Step 3 142 1 2y 5 6 into either of the two original equations. 2y 5 2 y 5 1 Step 5: Check that the solution satisfies both equations. Equation (1): Substitute 14, 12 into both equations. x 1 2y 5 6 142 1 2112 5 6 3 Equation (2): 3x 2 y 5 11 3142 2 1 5 11 3 Step 1 is not necessary in cases where corresponding terms already sum to zero. If we let x 5 a, then y 5 a 2 2. In other words, all of the points 1a, a 2 22 where a is any real number are solutions to this system of linear equations. YOUR TURN Use the substitution method to solve each system of linear equations. a. 2x 1 y 5 3 b. x 2 y 5 2 c. x 1 2y 5 1 4x 1 2y 5 4 4x 2 3y 5 10 2x 1 4y 5 2 ▼ A N S W E R a. no solution b. 14, 22 c. infinitely many solutions: aa, 1 2 a 2 b ▼ 870 CHAPTER 9 Systems of Linear Equations and Inequalities In Example 4, we eliminated the variable y simply by adding the two equations. Sometimes it is necessary to multiply one (Example 5) or both (Example 6) equations by constants prior to adding. EXAMPLE 4 Determining by the Elimination Method That a System Has One Solution Use the elimination method to solve the following system of linear equations: 2x 2 y 5 25 Equation (1) 4x 1 y 5 11 Equation (2) Solution: STEP 1 Not necessary. STEP 2 Eliminate y by adding Equation (1) to Equation (2). 2x 2 y 5 25 4x 1 y 5 11 6x 5 6 STEP 3 Solve for x. x 5 1 STEP 4 Back-substitute x 5 1 into Equation (2). 4112 1 y 5 11 Solve for y. y 5 7 STEP 5 Check that 11, 72 satisfies both equations. Equation (1): 2x 2 y 5 25 2112 2 172 5 25 3 Equation (2): 4x 1 y 5 11 4112 1 172 5 11 3 Note: The graphs of the two given equations correspond to two lines that intersect at the point 11, 72. STUDY TIP You can eliminate one variable from the system by addition when (1) the coefficients are equal and (2) the signs are opposite. x –5 5 –2 y 8 2x – y = –5 4x + y = 11 (1, 7) EXAMPLE 5 Applying the Elimination Method When Multiplying One Equation by a Constant Is Necessary Use the elimination method to solve the following system of linear equations: 24x 1 3y 5 23 Equation (1) 12x 1 5y 5 1 Equation (2) Solution: STEP 1 Multiply Equation (1) by 3. 212x 1 9y 5 69 STEP 2 Eliminate x by adding the modified Equation (1) 212x 1 9y 5 69 to Equation (2). 12x 1 5y 5 1 14y 5 70 STEP 3 Solve for y. y 5 5 STUDY TIP Be sure to multiply the entire equation by the constant. In Example 5, we eliminated x simply by multiplying the first equation by a constant and adding the result to the second equation. In order to eliminate either of the variables in Example 6, we will have to multiply both equations by constants prior to adding. STEP 4 Back-substitute y 5 5 into Equation (2). 12x 1 5152 5 1 Solve for x. 12x 1 25 5 1 12x 5 224 x 5 22 STEP 5 Check that 122, 52 satisfies both equations. Equation (1): 241222 1 3152 5 23 8 1 15 5 23 3 Equation (2): 121222 1 5152 5 1 224 1 25 5 1 3 Note: The graphs of the two given equations correspond to two lines that intersect at the point 122, 52 x –5 5 –2 y 8 –4x + 3y = 23 12x + 5y = 1 (–2, 5) EXAMPLE 6 Applying the Elimination Method When Multiplying Both Equations by Constants Is Necessary Use the elimination method to solve the following system of linear equations: 3x 1 2y 5 1 Equation (1) 5x 1 7y 5 9 Equation (2) Solution: STEP 1 Multiply Equation (1) by 5 and Equation (2) by 23. 15x 1 10y 5 5 215x 2 21y 5 227 STEP 2 Eliminate x by adding the modified Equation (1) 15x 1 10y 5 5 to the modified Equation (2). 215x 2 21y 5 227 2 11y 5 222 STEP 3 Solve for y. y 5 2 STEP 4 Back-substitute y 5 2 into Equation (1). 3x 1 2122 5 1 Solve for x. 3x 5 23 x 5 21 STEP 5 Check that 121, 22 satisfies both equations. Equation (1): 3x 1 2y 5 1 31212 1 2122 5 1 3 Equation (2): 5x 1 7y 5 9 51212 1 7122 5 9 3 Note: The graphs of the two given equations correspond to two lines that intersect at the point 121, 22. x –5 5 –5 y 5 3x + 2y = 1 5x + 7y = 9 (–1, 2) 9.1 Systems of Linear Equations in Two Variables 871 872 CHAPTER 9 Systems of Linear Equations and Inequalities Notice in Example 6 that we could have also eliminated y by multiplying the first equation by 7 and the second equation by 22. Typically, the choice is dictated by which approach will keep the coefficients as simple as possible. In the event that the coefficients contain fractions or decimals, first rewrite the equations in standard form with integer coefficients and then make the decision. EXAMPLE 7 Determining by the Elimination Method That a System Has No Solution Use the elimination method to solve the following system of linear equations: 2x 1 y 5 7 Equation (1) 2x 2 2y 5 4 Equation (2) Solution: STEP 1 Multiply Equation (1) by 2. 22x 1 2y 5 14 STEP 2 Eliminate y by adding the modified 22x 1 2y 5 14 Equation (1) found in Step 1 to Equation (2). 2x 2 2y 5 4 0 5 18 This system is inconsistent since 0 5 18 is never true. Therefore, there are no values of x and y that satisfy both equations. We say that there is no solution to this system of linear equations. Note: The graphs of the two equations are two parallel lines. x –5 5 y 8 –x + y = 7 2x – 2y = 4 –2 EXAMPLE 8 Determining by the Elimination Method That a System Has Infinitely Many Solutions Use the elimination method to solve the following system of linear equations: 7x 1 y 5 2 Equation (1) 214x 2 2y 5 24 Equation (2) Solution: STEP 1 Multiply Equation (1) by 2. 14x 1 2y 5 4 STEP 2 Add the modified Equation (1) found in 14x 1 2y 5 4 Step 1 to Equation (2). 214x 2 2y 5 24 0 5 0 This system is dependent since 0 5 0 is always true. We say that there are infinitely many solutions to this system of linear equations, and these can be represented by the points 1a, 2 2 7a2. Note: The graphs of the two equations are the same line. Y OUR TU R N Apply the elimination method to solve each system of linear equations. a. 2x 1 3y 5 1 b. x 2 5y 5 2 c. x 2 y 5 14 4x 2 3y 5 27 210x 1 50y 5 220 2x 1 y 5 9 ▼ STUDY TIP Systems of linear equations in two variables have one solution, no solution, or infinitely many solutions. x –5 5 –5 y 7x + y = 2 –14x – 2y = –4 5 [CONCEPT CHECK] Solve the system of linear equations: Ax 2 By 5 C 2Ax 1 By 5 C Are these two lines parallel, or do they coincide? ANSWER No Solution; parallel ▼ ▼ A N S W E R a. 121, 12 b. infinitely many solutions: aa, a 2 2 5 b c. no solution Graphing Method A third way to solve a system of linear equations in two variables is to graph the two lines. If the two lines intersect, then the point of intersection is the solution. Graphing is the most labor-intensive method for solving systems of linear equations in two variables. The graphing method is typically not used to solve systems of linear equations when an exact solution is desired. Instead, it is used to interpret or confirm the solution(s) found by the other two methods (substitution and elimination). If you are using a graphing calculator, however, you will get as accurate an answer using the graphing method as you will when applying the other methods. The following box summarizes the graphing method for solving systems of linear equations in two variables using the example given earlier: x 1 2y 5 6 Equation (1) 3x 2 y 5 11 Equation (2) EXAMPLE 9 Determining by Graphing That a System Has One Solution Use graphing to solve the following system of linear equations: x 1 y 5 2 Equation (1) 3x 2 y 5 2 Equation (2) Solution: STEP 1 Write each equation in y 5 2x 1 2 Equation (1) slope–intercept form. y 5 3x 2 2 Equation (2) STEP 2 Plot both lines on the same graph. x y (1, 1) 5 5 –5 –5 9.1 Systems of Linear Equations in Two Variables 873 GRAPHING METHOD Step 1: Write the equations in slope–intercept form. Equation (1): Equation (2): y 5 21 2x 1 3 y 5 3x 2 11 Step 2: Graph the two lines. x y 5 6 –4 (4, 1) y = 3x –11 y = x + 3 1 2 Step 3: Identify the point of intersection. 14, 12 Step 4: Check that the solution satisfies both equations. Equation (1): Equation (2): x 1 2y 5 6 3x 2 y 5 11 142 1 2112 5 6 3 3142 2 1 5 11 3 Step 1 is not necessary when the lines are already in slope–intercept form. 874 CHAPTER 9 Systems of Linear Equations and Inequalities STEP 3 Identify the point of intersection. 11, 12 STEP 4 Check that the point 11, 12 x 1 y 5 2 satisfies both equations. 1 1 1 5 2 3 Equation (1) 3x 2 y 5 2 3112 2 112 5 2 3 Equation (2) Note: There is one solution because the two lines intersect at one point. EXAMPLE 10 Determining by Graphing That a System Has No Solution Use graphing to solve the following system of linear equations: 2x 2 3y 5 9 Equation (1) 24x 1 6y 5 12 Equation (2) Solution: STEP 1 Write each equation in y 5 2 3 x 2 3 Equation (1) slope–intercept form. y 5 2 3 x 1 2 Equation (2) STEP 2 Plot both lines on the same graph. x y 5 5 –5 –5 STEP 3 Identify the point of intersection. None The two lines are parallel because they have the same slope but different y-intercepts. For this reason there is no solution —two parallel lines do not intersect. EXAMPLE 11 Determining by Graphing That a System Has Infinitely Many Solutions Use graphing to solve the following system of linear equations: 3x 1 4y 5 12 Equation (1) 3 4x 1 y 5 3 Equation (2) Solution: STEP 1 Write each equation in y 5 23 4 x 1 3 y 5 23 4 x 1 3 slope–intercept form. 9.1.2 Three Methods and Three Types of Solutions Given any system of two linear equations in two variables, any of the three methods (substitution, elimination, or graphing) can be utilized. If you find that it is easy to eliminate a variable by adding multiples of the two equations, then elimination is the preferred choice. If you do not see an obvious elimination, then solve the system by substitution. For exact solutions, choose one of these two algebraic methods. You should typically use graphing to confirm the solution(s) you have found by applying the other two methods or when you are using a graphing utility. STEP 2 Plot both lines on the same graph. x y 5 5 –5 –5 STEP 3 Identify the point of intersection. Infinitely many points There are infinitely many solutions , since the two lines are identical and coincide. The points that lie along the line are aa, 2 3 4a 1 3b. YOUR T UR N Utilize graphing to solve each system of linear equations. a. x 2 2y 5 1 b. x 2 2y 5 1 c 2x 1 y 5 3 2x 2 4y 5 2 2x 1 y 5 7 2x 1 y 5 7 ▼ 9.1.2 S K I L L Solve applications involving systems of linear equations. 9.1.2 C O N C E P T U A L Understand how to model a real-world problem using systems of linear equations in two variables. EXAMPLE 12 Identifying Which Method to Use State which of the two algebraic methods (elimination or substitution) would be the preferred method to solve each system of linear equations. a. x 2 2y 5 1 b. x 5 2y 2 1 c. 7x 2 20y 5 1 2x 1 y 5 2 2x 2 y 5 4 5x 1 3y 5 18 Solution: a. Elimination: Because the x variable is eliminated when the two equations are added. b. Substitution: Because the first equation is easily substituted into the second equation (for x). c. Either: There is no preferred method, as both elimination and substitution require substantial work. Regardless of which method is used to solve systems of linear equations in two variables, in general, we can summarize the three types of solutions both algebraically and graphically. 9.1 Systems of Linear Equations in Two Variables 875 ▼ A N S W E R a. infinitely many solutions solutions: aa, a 2 1 2 b b. 13, 12 c. no solution 876 CHAPTER 9 Systems of Linear Equations and Inequalities Applications Suppose you have two job offers that require sales. One pays a higher base, while the other pays a higher commission. Which job do you take? THREE TYPES OF SOLUTIONS TO SYSTEMS OF LINEAR EQUATIONS NUMBER OF SOLUTIONS GRAPHICAL INTERPRETATION One solution The two lines intersect at one point. No solution The two lines are parallel. (Same slope/different y-intercepts.) Infinitely many solutions The two lines coincide. (Same slope/same y-intercept.) [CONCEPT CHECK] When determining which sales job will result in more money, the higher base/lower commision job is better when sales are low and the lower base/higher commision job is better when sales are high. The point of intersection of the two linear salary models corresponds to ___?. ANSWER The sales level that corresponds to the two salary models being equal. ▼ EXAMPLE 13 Deciding Which Job to Take Suppose that upon graduation you are offered a job selling biomolecular devices to laboratories studying DNA. The Beckman-Coulter Company offers you a job selling its DNA sequencer with an annual base salary of $20,000 plus 5% commis-sion on total sales. The MJ Research Corporation offers you a job selling its PCR Machine that makes copies of DNA with an annual base salary of $30,000 plus 3% commission on sales. Determine what the total sales would have to be to make the Beckman-Coulter job the better offer. Solution: STEP 1 Identify the question. When would these two jobs have equal compensations? STEP 2 Make notes. Beckman-Coulter 20,000 1 5% MJ Research 30,000 1 3% STEP 3 Set up the equations. Let x 5 total sales and y 5 compensation. Equation (1) Beckman-Coulter: y 5 20,000 1 0.05x Equation (2) MJ Research: y 5 30,000 1 0.03x STEP 4 Solve the system of equations. Substitution method Substitute Equation (1) into Equation (2). 20,000 1 0.05x 5 30,000 1 0.03x Solve for x. 0.02x 5 10,000 x 5 500,000 If you make $500,000 worth of sales per year, the jobs will yield equal compensations. If you sell less than $500,000, the MJ Research job is the better offer, and more than $500,000 , the Beckman-Coulter job is the better offer. The elimination method could also have been used. STEP 5 Check the solution. Equation (1) Beckman-Coulter: y 5 20,000 1 0.051500,0002 5 $45,000 Equation (2) MJ Research: y 5 30,000 1 0.031500,0002 5 $45,000 EXAMPLE 14 Deciding How Many Pounds of Each Meat to Buy at the Deli The Chi Omega sorority is hosting a party, and the membership committee would like to make sandwiches for the new members. They already have bread and condiments but ask Tara to run to the deli to buy sliced turkey. The membership chair has instructed Tara to buy 5 pounds of the best turkey she can find, and the treasurer has given her $37 and told her to spend it all. When she arrives at Publix supermarket, she has a choice of two types of turkey: Boar’s Head ($8 per pound) and the store brand ($7 per pound). How much of each kind should Tara buy to spend the entire $37 on the best quality of turkey? Solution: STEP 1 Identify the question. How much of each type of sliced turkey should Tara buy? STEP 2 Make notes. Store brand turkey costs $7/pound. Boar’s Head turkey costs $8/pound. Tara has $37 to spend on 5 pounds of turkey. STEP 3 Set up the equations. Let x 5 number of pounds of Boar’s Head turkey and y 5 number of pounds of store brand turkey. The system of equations therefore has to include one equation giving a total of 5 pounds and one giving a total of $37. Equation (1): x 1 y 5 5 Equation (2): 8x 1 7y 5 37 STEP 4 Solve the system of equations. Elimination method Multiply Equation (1) by 27. 27x 2 7y 5 235 Add this result to Equation (2). 27x 2 7y 5 235 8x 1 7y 5 37 x 5 2 Substitute x 5 2 into the original Equation (1). 2 1 y 5 5 Solve for y. y 5 3 Tara should buy 2 pounds of Boar’s Head and 3 pounds of store brand turkey. The substitution method could also have been used to solve this system of linear equations. STEP 5 Check the solution. Equation (1): 2 1 3 5 5 Equation (2): 8122 1 7132 5 37 9.1 Systems of Linear Equations in Two Variables 877 878 CHAPTER 9 Systems of Linear Equations and Inequalities EXAMPLE7 [SEC TION 9.1] E X E R C I SES • S K I L L S In Exercises 1–22, solve each system of linear equations by substitution. 1. x 2 y 5 1 x 1 y 5 1 2. x 2 y 5 2 x 1 y 5 22 3. x 1 y 5 7 x 2 y 5 9 4. x 2 y 5 210 x 1 y 5 4 5. 2x 2 y 5 3 x 2 3y 5 4 6. 4x 1 3y 5 3 2x 1 y 5 1 7. 3x 1 y 5 5 2x 2 5y 5 28 8. 6x 2 y 5 215 2x 2 4y 5 216 9. 2u 1 5v 5 7 3u 2 v 5 5 10. m 2 2n 5 4 3m 1 2n 5 1 11. 2x 1 y 5 7 22x 2 y 5 5 12. 3x 2 y 5 2 3x 2 y 5 4 13. 4r 2 s 5 1 8r 2 2s 5 2 14. 23p 1 q 5 24 6p 2 2q 5 8 15. 5r 2 3s 5 15 210r 1 6s 5 230 16. 25p 2 3q 5 21 10p 1 6q 5 2 17. 2x 2 3y 5 27 3x 1 7y 5 24 18. 4x 2 5y 5 27 3x 1 8y 5 30 19. 1 3x 2 1 4y 5 0 22 3x 1 3 4y 5 2 20. 1 5x 1 2 3y 5 10 21 2x 2 1 6y 5 27 21. 7.2x 2 4.1y 5 7.0 23.5x 1 16.5y 5 2.4 22. 6.3x 1 1.5y 5 10.5 20.4x 1 2.2y 5 28.7 In Exercises 23–42, solve each system of linear equations by elimination. 23. x 2 y 5 2 x 1 y 5 4 24. x 1 y 5 2 x 2 y 5 22 25. x 2 y 5 23 x 1 y 5 7 26. x 2 y 5 210 x 1 y 5 8 27. 5x 1 3y 5 23 3x 2 3y 5 221 28. 22x 1 3y 5 1 2x 2 y 5 7 29. 2x 2 7y 5 4 5x 1 7y 5 3 30. 3x 1 2y 5 6 23x 1 6y 5 18 31. 2x 1 5y 5 7 3x 2 10y 5 5 32. 6x 2 2y 5 3 23x 1 2y 5 22 33. 2x 1 5y 5 5 24x 2 10y 5 210 34. 11x 1 3y 5 3 22x 1 6y 5 6 35. 3x 2 2y 5 12 4x 1 3y 5 16 36. 5x 2 2y 5 7 3x 1 5y 5 29 37. 6x 2 3y 5 215 7x 1 2y 5 212 38. 7x 2 4y 5 21 3x 2 5y 5 16 39. 0.02x 1 0.05y 5 1.25 20.06x 2 0.15y 5 23.75 40. 20.5x 1 0.3y 5 0.8 21.5x 1 0.9y 5 2.4 41. 1 3 x 1 1 2 y 5 1 1 5 x 1 7 2 y 5 2 42. 1 2x 2 1 3y 5 0 3 2x 2 1 2y 5 3 4 The algebraic methods are preferred for exact solutions, and the graphing method is typically used to give a visual interpretation and confirmation of the solution. There are three types of solutions to systems of linear equations: one solution, no solution, or infinitely many solutions. In this section we discussed two algebraic techniques for solving systems of linear equations in two variables: • Substitution method • Elimination method [SEC TION 9.1] S U M MA RY INDEPENDENT SYSTEM INCONSISTENT SYSTEM DEPENDENT SYSTEM One solution No solution Infinitely many solutions x y x y x y Lines have different slopes. Lines are parallel (same slope and different y-intercepts). Lines coincide (same slope and same y-intercept). In Exercises 43–46, match the systems of equations with the graphs. 43. 3x 2 y 5 1 3x 1 y 5 5 44. 2x 1 2y 5 21 2x 1 y 5 7 45. 2x 1 y 5 3 2x 1 y 5 7 46. x 2 2y 5 1 2x 2 4y 5 2 a. x y –5 5 5 –5 b. x y –5 5 5 –5 c. x y –5 5 5 –5 d. x y –5 5 5 –5 In Exercises 47–54, solve each system of linear equations by graphing. 47. y 5 2x y 5 x 48. x 2 3y 5 0 x 1 3y 5 0 49. 2x 1 y 5 23 x 1 y 5 22 50. x 2 2y 5 21 2x 2 y 5 25 51. 1 2x 2 2 3y 5 4 1 4x 2 y 5 6 52. 1 5x 2 5 2y 5 10 1 15x 2 5 6y 5 10 3 53. 1.6x 2 y 5 4.8 20.8x 1 0.5y 5 1.5 54. 1.1x 2 2.2y 5 3.3 23.3x 1 6.6y 5 26.6 • A P P L I C A T I O N S 55. Environment. Approximately 2 million dry erase markers are disposed of by teachers each year. Traditional dry erase markers are toxic and nonbiodegradable. EcoSmart World sells AusPens, which are dry erase markers that are recyclable, refillable, and nontoxic. The markers are available in a kit containing x AusPens and x refill ink bottles, each a different color. One refill ink bottle can refill up to 40 AusPens, and one kit is equivalent to y dry erase markers. If one kit is equivalent to 246 traditional dry erase markers, find the number of AusPens that are in each kit. 56. Pharmacy. A pharmacy technician receives an order for 454 grams of a 3% zinc oxide cream. If the pharmacy has 1% and 10% zinc oxide creams in stock, how much of each should be mixed to fill the order? 57. Event Planning. You are in charge of event planning for a major corporation that is having a reception. Your boss has given you a budget of $1000 for gifts and wants you to order Montblanc pens for VIPs and Cross pens for other guests. A Montblanc pen costs $72 and a Cross pen costs $10. You must have 69 pens for guests. How many pens of each type should you order in order to stay within the budget? 58. Health Club Management. A fitness club has a budget of $915 to purchase two different types of dumbbell sets. One set costs $30 each, and the deluxe model dumbbell set costs $45 each. The club wants to purchase 24 new dumbbell sets to use in its gym. How many sets of each type can it purchase? 59. Mixture. In chemistry lab, Stephanie has to make a 37 milliliter solution that is 12% HCl. All that is in the lab is 8% and 15% HCl. How many milliliters of each solution should she use to obtain the desired mix? 60. Mixture. A mechanic has 340 gallons of gasoline and 10 gallons of oil to make gas/oil mixtures. He wants one mixture to be 4% oil and the other mixture to be 2.5% oil. If he wants to use all of the gas and oil, how many gallons of gas and oil are in each of the resulting mixtures? 61. Salary Comparison. Upon graduation with a degree in management of information systems (MIS), you decide to work for a company that buys data from states’ departments of motor vehicles and sells to banks and car dealerships customized reports detailing how many cars at each dealership are financed through particular banks. Autocount Corporation offers you a $15,000 base salary and 10% commission on your annual sales. Polk Corporation offers you a base salary of $30,000 plus a 5% commission on your annual sales. How many sales would you have to make per year to make more money at Autocount? 62. Salary Comparison. Two types of residential real estate agents are those who sell existing houses (resale) and those who sell new homes for developers. Resale of existing homes typically earns 6% commission on every sale, and representing developers in selling new homes typically earns a base salary of $15,000 per year plus a 1.5% commission because agents are required to work 5 days a week on site in a new development. How many dollars would an agent have to sell per year to make more money in resale than in new homes? 63. Gas Mileage. A Honda Accord gets approximately 26 mpg on the highway and 19 mpg in the city. You drove 349.5 miles on a full tank (16 gallons) of gasoline. Approximately how many miles did you drive in the city and how many on the highway? 64. Wireless Plans. AT&T is offering a 600-minute peak plan with free mobile-to-mobile and weekend minutes at $59 per month plus $0.13 per minute for every minute over 600. The next plan up is the 800-minute plan that costs $79 per month. You think you may go over 600 minutes but are not sure you need 800 minutes. How many minutes would you have to talk for the 800-minute plan to be the better deal? 65. Distance/Rate/Time. A direct flight on Delta Air Lines from Atlanta to Paris is 4000 miles and takes approximately 8 hours going east (Atlanta to Paris) and 10 hours going west (Paris to Atlanta). Although the plane averages the same air speed, there is a headwind while traveling west and a tailwind while traveling east, resulting in different air speeds. What is the average air speed of the plane, and what is the average wind speed? 9.1 Systems of Linear Equations in Two Variables 879 880 CHAPTER 9 Systems of Linear Equations and Inequalities 66. Distance/Rate/Time. A private pilot flies a Cessna 172 on a trip that is 500 miles each way. It takes her approximately 3 hours to get there and 4 hours to return. What is the approximate air speed of the Cessna, and what is the approximate wind speed? 67. Investment Portfolio. Leticia has been tracking two vol-atile stocks. Stock A over the last year has increased 10%, and stock B has increased 14% (using a simple interest model). She has $10,000 to invest and would like to split it between these two stocks. If the stocks continue to perform at the same rate, how much should she invest in each to result in a balance of $11,260? 68. Investment Portfolio. Toby split his savings into two different investments, one earning 5% and the other earning 7%. He put twice as much in the investment earning the higher rate. In 1 year, he earned $665 in interest. How much money did he invest in each account? 69. Break-Even Analysis. A company produces CD players for a unit cost of $15 per CD player. The company has fixed costs of $120. If each CD player can be sold for $30, how many CD players must be sold to break even? Determine the cost equation first. Next, determine the revenue equation. Use the two equa-tions you have found to determine the break-even point. 70. Managing a Lemonade Stand. An elementary-school-age child wants to have a lemonade stand. She wants to sell each glass of lemonade for $0.25. She has determined that each glass of lemonade costs about $0.10 to make (for lemons and sugar). It costs her $15.00 for materials to make the lemonade stand. How many glasses of lemonade must she sell to break even? • C A T C H T H E M I S T A K E In Exercises 71–74, explain the mistake that is made. 71. Solve the system of equations by elimination. 2x 1 y 5 23 3x 1 y 5 8 Solution: Multiply Equation (1) by 21. 2x 2 y 5 23 Add the result to Equation (2). 3x 1 y 5 8 5x 5 5 Solve for x. x 5 1 Substitute x 5 1 into Equation (2). 3112 1 y 5 8 y 5 5 The answer 11, 52 is incorrect. What mistake was made? 72. Solve the system of equations by elimination. 4x 2 y 5 12 4x 2 y 5 24 Solution: Multiply Equation (1) by 21. 24x 1 y 5 212 Add the result to Equation (2). 24x 1 y 5 212 4x 2 y 5 24 0 5 12 Answer: Infinitely many solutions. This is incorrect. What mistake was made? 73. Solve the system of equations by substitution. x 1 3y 5 24 2x 1 2y 5 26 Solution: Solve Equation (1) for x. x 5 23y 2 4 Substitute x 5 23y 2 4 into Equation (2). 2123y 2 42 1 2y 5 26 Solve for y. 3y 2 4 1 2y 5 26 5y 5 22 y 5 22 5 Substitute y 5 22 5 into Equation (1). x 1 3a 2 2 5 b 5 24 Solve for x. x 5 214 5 The answer A22 5, 214 5 B is incorrect. What mistake was made? 74. Solve the system of equations by graphing. 2x 1 3y 5 5 4x 1 6y 5 10 Solution: Write both equations in slope–intercept form. y 5 2 2 3 x 1 5 3 y 5 2 2 3 x 1 5 3 Since these lines have the same slope, they are parallel lines. Parallel lines do not intersect, so there is no solution. This is incorrect. What mistake was made? • C O N C E P T U A L In Exercises 75–78, determine whether each statement is true or false. 75. A system of equations represented by a graph of two lines with the same slope always has no solution. 76. A system of equations represented by a graph of two lines with slopes that are negative reciprocals always has one solution. 77. If two lines do not have exactly one point of intersection, then they must be parallel. 78. The system of equations Ax 2 By 5 1 and 2Ax 1 By 5 21 has no solution. • T E C H N O L O G Y 83. Use a graphing utility to graph the two equations y 5 21.25x 1 17.5 and y 5 2.3x 2 14.1. Approximate the solution to this system of linear equations. 84. Use a graphing utility to graph the two equations y 5 14.76x 1 19.43 and y 5 2.76x 1 5.22. Approximate the solution to this system of linear equations. 85. Use a graphing utility to graph the two equations and determine the solution set: 23x 1 15y 5 7 and 46x 1 30y 5 14. 86. Use a graphing utility to graph the two equations and determine the solution set: 23x 1 7y 5 2 and 6x 2 14y 5 3. 87. Use a graphing utility to graph the two equations 1 3x 2 5 12y 5 5 6 and 3 7x 1 1 14 y 5 29 28. Approximate the solution to this system of linear equations. 88. Use a graphing utility to graph the two equations 5 9x 1 11 13y 5 2 and 3 4x 1 5 7y 5 13 14. Approximate the solution to this system of linear equations. • C H A L L E N G E 79. The point 12, 232 is a solution to the system of equations Ax 1 By 5 229 Ax 2 By 5 13 Find A and B. 80. If you graph the lines x 2 50y 5 100 x 2 48y 5 298 they appear to be parallel lines. However, there is a unique solution. Explain how this might be possible. 81. Energy Drinks. A nutritionist wishes to market a new vitamin-enriched fruit drink and is preparing two versions of it to distribute at a local health club. She has 100 cups of pineapple juice and 4 cups of super vitamin-enriched pome-granate concentrate. One version of the drink is to contain 2% pomegranate and the other version 4% pomegranate. How much of each drink can she create? 82. Easter Eggs. A family is coloring Easter eggs and wants to make two shades of purple, “light purple” and “deep purple.” They have 30 tablespoons of deep red solution and 2 table-spoons of blue solution. If “light purple” consists of 2% blue solution and “deep purple” consists of 10% blue solution, how much of each version of purple solution can be created? S K I L L S O B J E C T I V E S ■ ■Solve systems of linear equations in three variables using a combination of the elimination method and the substitution method. ■ ■Identify three types of solutions: one solution (point), no solution, or infinitely many solutions (a single line in three-dimensional space). C O N C E P T U A L O B J E C T I V ES ■ ■Understand that a graph of a linear equation in three variables corresponds to a plane. ■ ■Understand parametric representation of a line in three dimensions. 9.2 SYSTEMS OF LINEAR EQUATIONS IN THREE VARIABLES 9.2.1 Solving Systems of Linear Equations in Three Variables In Section 9.1, we solved systems of linear equations in two variables. Graphs of linear equations in two variables correspond to lines. Now we turn our attention to linear equations in three variables. A linear equation in three variables x, y, and z, is given by Ax 1 By 1 Cz 5 D where A, B, C, and D are real numbers that are not all equal to zero. All three variables have degree equal to one, which is why this is called a linear equation in three variables. The graph of any equation in three variables requires a three-dimensional coordinate system. The x-axis, y-axis, and z-axis are each perpendicular to the other two. For the three-dimensional coordinate system on the next page, a point 1x, y, z2 5 12, 3, 12 is 9.2.1 S K I L L Solve systems of linear equations in three variables using a combination of the elimination method and the substitution method. 9.2.1 C ON C E P T U A L Understand that a graph of a linear equation in three variables corresponds to a plane. 9.2 Systems of Linear Equations in Three Variables 881 882 CHAPTER 9 Systems of Linear Equations and Inequalities found by starting at the origin, moving two units to the right, three units up, and one unit out toward you. In two variables, the graph of a linear equation is a line. In three variables, however, the graph of a linear equation is a plane. A plane can be thought of as an infinite sheet of paper. When solving systems of linear equations in three variables, we find one of three possibilities: one solution, no solution, and infinitely many solutions. x z y (2, 3, 1) There are many ways to solve systems of linear equations in more than two variables. One method is to combine the elimination and substitution methods, which will be discussed in this section. Other methods involve matrices, which will be discussed in the next chapter. We now outline a procedure for solving systems of linear equations in three variables, which can be extended to solve systems of more than three variables. Solutions are usually given as ordered triples of the form 1x, y, z2. SOLVING SYSTEMS OF LINEAR EQUATIONS IN THREE VARIABLES USING ELIMINATION AND SUBSTITUTION Step 1: Reduce the system of three equations in three variables to two equations in two (of the same) variables by applying elimination. Step 2: Solve the resulting system of two linear equations in two variables by applying elimination or substitution. Step 3: Substitute the solutions in Step 2 into any of the original equations and solve for the third variable. Step 4: Check that the solution satisfies all three original equations. [CONCEPT CHECK] TRUE OR FALSE In systems of linear equations in three variables, all three planes may intersect at a single point or along a line (infinitely many points). ANSWER True ▼ EXAMPLE 1 Solving a System of Linear Equations in Three Variables Solve the system: 2x 1 y 1 8z 5 21 Equation (1) x 2 y 1 z 5 22 Equation (2) 3x 2 2y 2 2z 5 2 Equation (3) Solution: Inspecting the three equations, we see that y is easily eliminated when Equations (1) and (2) are added because the coefficients of y, 11 and 21, are equal in magnitude and opposite in sign. We can also eliminate y from Equation (3) by adding Equation (3) to either 2 times Equation (1) or 22 times Equation (2). Therefore, our plan of attack is to eliminate y from the system of equations, so the result will be two equations in two variables x and z. One Solution Solution No Solution or Infinitely Many Solutions Solution (line of intersection) 9.2 Systems of Linear Equations in Three Variables 883 In Example 1 and the Your Turn, the variable y was eliminated by adding the first and second equations. In practice, any of the three variables can be eliminated, but typically we select the most convenient variable to eliminate. If a variable is missing from one of the equations (has a coefficient of 02, then we eliminate that variable from the other two equations. STEP 1 Eliminate y in Equation (1) and Equation (2). Equation (1): 2x 1 y 1 8z 5 21 Equation (2): x 2 y 1 z 5 22 Add. 3x 1 9z 5 23 Eliminate y in Equation (2) and Equation (3). Multiply Equation (2) by 22. 22x 1 2y 2 2z 5 4 Equation (3): 3x 2 2y 2 2z 5 2 Add. x 2 4z 5 6 STEP 2 Solve the system of two linear equations in two variables. 3x 1 9z 5 23 x 2 4z 5 6 Substitution method: x 5 4z 1 6 314z 1 62 1 9z 5 23 Distribute. 12z 1 18 1 9z 5 23 Combine like terms. 21z 5 221 Solve for z. z 5 21 Substitute z 5 21 into x 5 4z 1 6. x 5 41212 1 6 5 2 x 5 2 and z 5 21 are the solutions to the system of two equations. STEP 3 Substitute x 5 2 and z 5 21 into any of the three original equations and solve for y. Substitute x 5 2 and z 5 21 into Equation (2). 2 2 y 2 1 5 22 Solve for y. y 5 3 STEP 4 Check that x 5 2, y 5 3, and z 5 21 satisfy all three equations. Equation (1): 2122 1 3 1 81212 5 4 1 3 2 8 5 21 Equation (2): 2 2 3 2 1 5 22 Equation (3): 3122 2 2132 2 21212 5 6 2 6 1 2 5 2 The solution is x 5 2, y 5 3, z 5 21, or 12, 3, 212 . Elimination method could also be used. YOUR T UR N Solve the system: 2x 2 y 1 3z 5 21 x 1 y 2 z 5 0 3x 1 3y 2 2z 5 1 ▼ ▼ A N S W E R x 5 21, y 5 2, z 5 1 STUDY TIP Eliminate the same variable. EXAMPLE 2 Solving a System of Linear Equations in Three Variables When One Variable Is Missing Solve the system: x 1 z 5 1 Equation (1) 2x 1 y 2 z 5 23 Equation (2) x 1 2y 2 z 5 21 Equation (3) 884 CHAPTER 9 Systems of Linear Equations and Inequalities 9.2.2 Types of Solutions There are three types of solutions: independent, dependent, and inconsistent. There are three corresponding outcomes: one solution, infinitely many solutions, or no solution. Examples 1 and 2 each had one solution. Examples 3 and 4 illustrate systems with infinitely many solutions and no solution, respectively. Solution: Since y is missing from Equation (1), y is the variable to be eliminated in Equation (2) and Equation (3). STEP 1 Eliminate y. Multiply Equation (2) by 22. 24x 2 2y 1 2z 5 6 Equation (3): x 1 2y 2 z 5 21 Add. 23x 1 z 5 5 STEP 2 Solve the system of two equations: x 1 z 5 1 Equation (1) and the resulting equation in Step 1. 23x 1 z 5 5 Multiply the second equation by x 1 z 5 1 1212 and add to first equation. 3x 2 z 5 25 4x 5 24 Solve for x. x 5 21 Substitute x 5 21 into Equation (1). 21 1 z 5 1 Solve for z. z 5 2 STEP 3 Substitute x 5 21 and z 5 2 into one of the original equations (Equation 2 or Equation 3) and solve for y. Substitute x 5 21 and z 5 2 into x 1 2y 2 z 5 21. 1212 1 2y 2 2 5 21 Gather like terms. 2y 5 2 Solve for y. y 5 1 STEP 4 Check that x 5 21, y 5 1, and z 5 2 satisfy all three equations. Equation (1): 1212 1 2 5 1 Equation (2): 21212 1 112 2 122 5 23 Equation (3): 1212 1 2112 2 122 5 21 The solution is x 5 21, y 5 1, z 5 2 . Y OUR TU R N Solve the system: x 1 y 1 z 5 0 2x 1 z 5 21 x 2 y 2 z 5 2 ▼ A N S W E R x 5 1, y 5 2, z 5 23 ▼ EXAMPLE 3 A Dependent System of Linear Equations in Three Variables (Infinitely Many Solutions) Solve the system: 2x 1 y 2 z 5 4 Equation (1) x 1 y 5 2 Equation (2) 3x 1 2y 2 z 5 6 Equation (3) 9.2.2 S KILL Identify three types of solutions: one solution (point), no solution, or infinitely many solutions (a single line in three-dimensional space). 9.2.2 CO NCE PTUAL Understand parametric rep resentation of a line in three dimensions. [CONCEPT CHECK] For example 3, the infinitely many points that lie along a line are given by the parametric representation (a 1 2, 2a, a). Find three points that lie along the line (use a 5 0, 1, 2, 3). ANSWER (2, 0, 0) (3, 21, 1) (4 ,22, 2) (5, 23, 3) ▼ ▼ A N S W E R 1a 2 1, a 1 1, a2 Solution: Since z is missing from Equation (2), z is the variable to be eliminated from Equation (1) and Equation (3). STEP 1 Eliminate z. Multiply Equation (1) by 1212. 22x 2 y 1 z 5 24 Equation (3): 3x 1 2y 2 z 5 6 Add. x 1 y 5 2 STEP 2 Solve the system of two equations: Equation (2) and the resulting equation in Step 1. x 1 y 5 2 x 1 y 5 2 Multiply the first equation by 1212 2x 2 y 5 22 and add to second equation. x 1 y 5 2 0 5 0 This statement is always true; therefore, there are infinitely many solutions . The original system has been reduced to a system of two identical linear equations. Therefore, the equations are dependent (share infinitely many solutions). Typically, to define those infinitely many solutions, we let z 5 a, where a stands for any real number, and then find x and y in terms of a. The resulting ordered triple showing the three variables in terms of a is called a parametric representation of a line in three dimensions. STEP 3 Let z 5 a and find x and y in terms of a. Solve Equation (2) for y. y 5 2 2 x Let y 5 2 2 x and z 5 a in Equation (1). 2x 1 12 2 x2 2 a 5 4 Solve for x. 2x 1 2 2 x 2 a 5 4 x 2 a 5 2 x 5 a 1 2 Let x 5 a 1 2 in Equation (2). 1a 1 22 1 y 5 2 Solve for y. y 5 2a The infinitely many solutions are written as 1a 1 2, 2a, a2 . STEP 4 Check that x 5 a 1 2, y 5 2a, and z 5 a satisfy all three equations. Equation (1): 21a 1 22 1 12a2 2 a 5 2a 1 4 2 a 2 a 5 4 3 Equation (2): 1a 1 22 1 12a2 5 a 1 2 2 a 5 2 3 Equation (3): 31a 1 22 1 212a2 2 a 5 3a 1 6 2 2a 2 a 5 6 3 Y OUR T UR N Solve the system: x 1 y 2 2z 5 0 x 2 z 5 21 x 2 2y 1 z 5 23 ▼ EXAMPLE 4 An Inconsistent System of Linear Equations in Three Variables (No Solution) Solve the system: x 1 2y 2 z 5 3 Equation (1) 2x 1 y 1 2z 5 21 Equation (2) 22x 2 4y 1 2z 5 5 Equation (3) 9.2 Systems of Linear Equations in Three Variables 885 886 CHAPTER 9 Systems of Linear Equations and Inequalities So far in this section, we have discussed only systems of three linear equations in three variables. What happens if we have a system of two linear equations in three variables? The two linear equations in three variables will always correspond to two planes in three dimensions. The possibilities are no solution (the two planes are parallel) or infinitely many solutions (the two planes intersect in a line). Solution: STEP 1 Eliminate x. Multiply Equation (1) by 22. 22x 2 4y 1 2z 5 26 Equation (2): 2x 1 y 1 2z 5 21 Add. 23y 1 4z 5 27 Equation (2): 2x 1 y 1 2z 5 21 Equation (3): 22x 2 4y 1 2z 5 5 Add. 23y 1 4z 5 4 STEP 2 Solve the system of two equations: 23y 1 4z 5 27 23y 1 4z 5 4 Multiply the top equation by 1212 3y 2 4z 5 7 and add to the second equation. 23y 1 4z 5 4 0 5 11 This is a contradiction, or inconsistent statement, and therefore there is no solution . NO SOLUTION INFINITELY MANY SOLUTIONS (LINE) x z y x z y EXAMPLE 5 Solving a System of Two Linear Equations in Three Variables Solve the system of linear equations: x 2 y 1 z 5 7 Equation (1) x 1 y 1 2z 5 2 Equation (2) Solution: Eliminate y by adding the two equations. x 2 y 1 z 5 7 x 1 y 1 2z 5 2 2x 1 3z 5 9 Therefore, Equation (1) and Equation (2) are both true if 2x 1 3z 5 9. Since we know there is a solution, it must be a line. To define the line of intersection, we again turn to parametric representation. Let z 5 a , where a is any real number. 2x 1 3a 5 9 Solve for x. x 5 9 2 2 3 2 a Modeling with a System of Three Linear Equations Suppose you want to model a stock price as a function of time and based on the data you feel a quadratic model would be the best fit. Therefore, the model is given by P1t2 5 at2 1 bt 1 c where P1t2 is the price of the stock at time t. If we have data corresponding to three distinct points 1t, P1t22, the result is a system of three linear equations in three variables a, b, and c. We can solve the resulting system of linear equations, which determines the coefficients a, b, and c of the quadratic model for stock price. Substitute z 5 a and x 5 9 2 2 3 2 a into a9 2 2 3 2 ab 2 y 1 a 5 7 Equation (1). Solve for y. y 5 21 2 a 2 5 2 The solution is the line in three dimensions given by A9 2 2 3 2 a,21 2 a 2 5 2, aB , where a is any real number. Note: Every real number a corresponds to a point on the line of intersection. a 19 2 2 3 2a, 21 2a 2 5 2, a2 21 16, 22, 212 0 A9 2, 2 5 2, 0 B 1 13, 23, 12 EXAMPLE 6 Stock Value The Oracle Corporation’s stock (ORCL) over 3 days (Wednesday, October 13, to Friday, October 15) can be approximately modeled by a quadratic function: ƒ1t2 5at2 1 bt 1 c. If Wednesday corresponds to t 5 1, where t is in days, then the following data points approximately correspond to the stock value. t f 1t2 DAYS 1 $12.20 Wednesday 2 $12.00 Thursday 3 $12.20 Friday ORCL 15-Minute T W TH F 11.9 12.1 12.0 12.2 12.3 12.4 6:11 PM Determine the function that models this behavior. Solution: Substitute the points 11, 12.202, 12, 12.002, and 13, 12.202 into ƒ1t2 5 at2 1 bt 1 c. a1122 1 b112 1 c 5 12.20 a1222 1 b122 1 c 5 12.00 a1322 1 b132 1 c 5 12.20 Simplify to a system of three equations in three variables 1a, b, and c2. a 1 b 1 c 5 12.20 Equation (1) 4a 1 2b 1 c 5 12.00 Equation (2) 9a 1 3b 1 c 5 12.20 Equation (3) Solve for a, b, and c by applying the technique of this section. 9.2 Systems of Linear Equations in Three Variables 887 888 CHAPTER 9 Systems of Linear Equations and Inequalities STEP 1 Eliminate c. Multiply Equation 112 by 1212. 2a 2 b 2 c 5 212.20 Equation 122: 4a 1 2b 1 c 5 12.20 Add. 3a 1 b 5 20.20 Multiply Equation 112 by 21. 2a 2 b 2 c 5 212.20 Equation 132: 9a 1 3b 1 c 5 12.20 Add. 8a 1 2b 5 0 STEP 2 Solve the system of two equations. 3a 1 b 5 20.20 8a 1 2b 5 0 Multiply the first equation by 22 and 26a 2 2b 5 0.40 add to the second equation. 8a 1 2b 5 0 Add. 2a 5 0.4 Solve for a. a 5 0.2 Substitute a 5 0.2 into 8a 1 2b 5 0. 810.22 1 2b 5 0 Simplify. 2b 5 21.6 Solve for b. b 5 20.8 STEP 3 Substitute a 5 0.2 and b 5 20.8 into one of the original three equations. Substitute a 5 0.2 and b 5 20.8 into a 1 b 1 c 5 12.20. 0.2 2 0.8 1 c 5 12.20 Gather like terms. 20.6 1 c 5 12.20 Solve for c. c 5 12.80 STEP 4 Check that a 5 0.2, b 5 20.8, and c 5 12.80 satisfy all three equations. Equation (1): a 1 b 1 c 5 0.2 2 0.8 1 12.8 5 12.2 Equation (2): 4a 1 2b 1 c 5 410.22 1 2120.82 1 12.80 5 0.8 2 1.6 1 12.8 5 12.00 Equation (3): 9a 1 3b 1 c 5 910.22 1 3120.82 1 12.80 5 1.8 2 2.4 1 12.8 5 12.20 The model is given by ƒ1t2 5 0.2t2 2 0.8t 1 12.80 . When the solution to a system of three linear equations is a line in three dimensions, we use parametric representation to express the solution. Graphs of linear equations in two variables are lines, whereas graphs of linear equations in three variables are planes. Systems of linear equations in three variables have one of three outcomes: • One solution (intersection point of the three planes) • No solution (no intersection of all three planes) • Infinitely many solutions (planes intersect along a line) [SEC TION 9. 2] S U M MA RY [SEC TION 9. 2] E X E RC I S E S • S K I L L S In Exercises 1–30, solve each system of linear equations. 1. x 2 y 1 z 5 6 2x 1 y 1 z 5 3 2x 2 y 2 z 5 0 2. 2x 2 y 1 z 5 21 2x 1 y 2 z 5 3 x 2 y 2 z 5 5 3. x 1 y 2 z 5 2 2x 2 y 2 z 5 23 2x 1 y 2 z 5 6 4. x 1 y 1 z 5 21 2x 1 y 2 z 5 3 2x 2 y 1 z 5 8 5. 2x 1 y 2 z 5 21 x 2 y 2 z 5 3 x 1 y 2 z 5 9 6. x 2 y 2 z 5 2 2x 2 y 1 z 5 4 2x 1 y 2 z 5 6 7. 2x 2 3y 1 4z 5 23 2x 1 y 1 2z 5 1 5x 2 2y 2 3z 5 7 8. x 2 2y 1 z 5 0 22x 1 y 2 z 5 25 13x 1 7y 1 5z 5 6 9. 3y 2 4x 1 5z 5 2 2x 2 3y 2 2z 5 23 3z 1 4y 2 2x 5 1 10. 2y 1 z 2 x 5 5 2x 1 3z 2 2y 5 0 22z 1 y 2 4x 5 3 11. x 2 y 1 z 5 21 y 2 z 5 21 2x 1 y 1 z 5 1 12. 2y 1 z 5 1 x 2 y 1 z 5 21 x 2 y 2 z 5 21 13. 3x 2 2y 2 3z 5 21 x 2 y 1 z 5 24 2x 1 3y 1 5z 5 14 14. 3x 2 y 1 z 5 2 x 2 2y 1 3z 5 1 2x 1 y 2 3z 5 21 15. 23x 2 y 2 z 5 2 x 1 2y 2 3z 5 4 2x 2 y 1 4z 5 6 16. 2x 2 3y 1 z 5 1 x 1 4y 2 2z 5 2 3x 2 y 1 4z 5 23 17. 3x 1 2y 1 z 5 4 24x 2 3y 2 z 5 215 x 2 2y 1 3z 5 12 18. 3x 2 y 1 4z 5 13 24x 2 3y 2 z 5 215 x 2 2y 1 3z 5 12 19. 2x 1 2y 1 z 5 22 3x 2 2y 1 z 5 4 2x 2 4y 2 2z 5 4 20. 2x 2 y 5 1 2x 1 z 5 22 22x 1 y 5 21 21. x 2 z 2 y 5 10 2x 2 3y 1 z 5 211 y 2 x 1 z 5 210 22. 2x 1 z 1 y 5 23 2y 2 z 1 x 5 0 x 1 y 1 2z 5 5 23. 3x1 1 x2 2 x3 5 1 x1 2 x2 1 x3 5 23 2x1 1 x2 1 x3 5 0 24. 2x1 1 x2 1 x3 5 21 x1 1 x2 2 x3 5 5 3x1 2 x2 2 x3 5 1 25. 2x 1 5y 5 9 x 1 2y 2 z 5 3 23x 2 4y 1 7z 5 1 26. x 2 2y 1 3z 5 1 22x 1 7y 2 9z 5 4 x 1 z 5 9 27. 2x1 2 x2 1 x3 5 3 x1 2 x2 1 x3 5 2 22x1 1 2x2 2 2x3 5 24 28. x1 2 x2 2 2x3 5 0 22x1 1 5x2 1 10x3 5 23 3x1 1 x2 5 0 29. 2x 1 y 2 z 5 2 x 2 y 2 z 5 6 30. 3x 1 y 2 z 5 0 x 1 y 1 7z 5 4 • A P P L I C A T I O N S 31. Business. A small company has an assembly line that produces three types of widgets. The basic widget is sold for $10 per unit, the midprice widget for $12 per unit, and the top-of-the-line widget for $15 per unit. The assembly line has a daily capacity of producing 300 widgets that may be sold for a total of $3700. Find the quantity of each type of widget produced on a day when the number of basic widgets and top-of-the-line widgets is the same. 32. Business. A small company has an assembly line that produces three types of widgets. The basic widget is sold for $10 per unit, the midprice widget for $12 per unit, and the top-of-the-line widget for $15 per unit. The assembly line has a daily capacity of producing 325 widgets that may be sold for a total of $3825. Find the quantity of each type of widget produced on a day when twice as many basic widgets as top-of-the-line widgets are produced. 33. Football. On September 1, 2007, the Appalachian State Mountaineers defeated the University of Michigan Wolverines by a score of 34 to 32. The points came from a total of these types of plays: touchdowns (six points), extra points (one point), and field goals (three points). There were a total of 18 scoring plays. There were four more touchdowns than field goals. How many touchdowns, extra points, and field goals were scored in this football game? 34. Basketball. On April 4, 2004, the University of Connecticut Huskies finished the season the same way they started it—as the number one men’s basketball team in the NCAA. They defeated the Georgia Tech Yellow Jackets 82–73 in the Final Four championship game. The points came from three types of scoring plays: two-point shots, three-point shots, and one-point free throws. There were seven more two-point shots made than there were one-point free throws completed. The number of successful two-point shots was four more than four times the number of successful three-point shots. How many two-point, three-point, and one-point free throw shots were made in the finals of the 2004 Final Four NCAA tournament? Exercises 35 and 36, rely on a selection of sandwiches whose nutrition information is given in the table. Suppose you are going to eat only sandwiches for a week (7 days) for lunch and dinner (total of fourteen meals). SANDWICH CALORIES FAT (GRAMS) Sriracha Chicken 350 18 Tuna Salad 430 19 Roast Beef 290 5 35. Diet. Your goal is a total of 4840 calories and 190 grams of fat. How many of each sandwich would you eat that week to obtain this goal? 9.2 Systems of Linear Equations in Three Variables 889 890 CHAPTER 9 Systems of Linear Equations and Inequalities 36. Diet. Your goal is a total of 4380 calories and 123 grams of fat. How many of each sandwich would you eat that week to obtain this goal? Exercises 37 and 38 involve vertical motion and the effect of gravity on an object. Because of gravity, an object that is projected upward will eventually reach a maximum height and then fall to the ground. The equation that determines the height h of a projectile t seconds after it is shot upward is given by h 5 1 2 at2 1 v0t 1 h0 where a is the acceleration due to gravity, h0 is the initial height of the object at time t 5 0, and v0 is the initial velocity of the object at time t 5 0. Note that a projectile follows the path of a parabola opening down, so a , 0. 37. Vertical Motion. An object is thrown upward, and the following table depicts the height of the ball t seconds after the projectile is released. Find the initial height, initial velocity, and acceleration due to gravity. t (SECONDS) HEIGHT (FEET) 1 36 2 40 3 12 38. Vertical Motion. An object is thrown upward, and the following table depicts the height of the ball t seconds after the projectile is released. Find the initial height, initial velocity, and acceleration due to gravity. t (SECONDS) HEIGHT (FEET) 1 84 2 136 3 156 39. Data Curve-Fitting. The number of minutes that an average person of age x spends driving a car can be modeled by a quadratic function y 5 ax2 1 bx 1 c, where a , 0 and 18 # x # 65. The following table gives the average number of minutes per day that a person spends driving a car. Determine the quadratic function that models this quantity. AGE AVERAGE DAILY MINUTES DRIVING 20 30 40 60 60 40 40. Data Curve-Fitting. The average age when a woman gets married began increasing during the last century. In 1930 the average age was 18.6, in 1950 the average age was 20.2, and in 2002 the average age was 25.3. Find a quadratic function y 5 ax2 1 bx 1 c, where a . 0 and 18 , y , 35, that models the average age y when a woman gets married as a function of the year x 1x 5 0 corresponds to 19302. What will the average age be in 2020? 41. Money. Tara and Lamar decide to place $20,000 of their savings into investments. They put some in a money market account earning 3% interest, some in a mutual fund that has been averaging 7% a year, and some in a stock that rose 10% last year. If they put $6000 more in the money market than in the mutual fund and the mutual fund and stocks have the same growth in the next year as they did in the previous year, they will earn $1180 in a year. How much money did they put in each of the three investments? 42. Money. Tara talks Lamar into putting less money in the money market and more money in the stock. They place $20,000 of their savings into investments. They put some in a money market account earning 3% interest, some in a mutual fund that has been averaging 7% a year, and some in a stock that rose 10% last year. If they put $6000 more in the stock than in the mutual fund and the mutual fund and stock have the same growth in the next year as they did in the previous year, they will earn $1680 in a year. How much money did they put in each of the three investments? 43. Ski Production. A company produces three types of skis: regular model, trick ski, and slalom ski. It needs to fill a customer order of 110 pairs of skis. There are two major production divisions within the company: labor and finishing. Each regular model of skis requires 2 hours of labor and 1 hour of finishing. Each trick ski model requires 3 hours of labor and 2 hours of finishing. Finally, each slalom ski model requires 3 hours of labor and 5 hours of finishing. Suppose the company has only 297 labor hours and 202 finishing hours. How many of each type ski can be made under these restrictions? 44. Automobile Production. An automobile manufacturing company produces three types of automobiles: compact, intermediate, and luxury models. The company has the capability of producing 500 automobiles. Suppose that each compact-model car requires 200 units of steel and 30 units of rubber, each intermediate model requires 300 units of steel and 20 units of rubber, and each luxury model requires 250 units of steel and 45 units of rubber. The number of units of steel available is 128,750, and the number of units of rubber available is 15,625. How many of each type of automobile can be produced with these constraints? 45. Computer versus Man. The Seattle Times reported a story on November 18, 2006, about a game of Scrabble played between a human and a computer. The best Scrabble player in the United States was pitted against a computer program designed to play the game. Remarkably, the human beat the computer in the best of two out of three games competition. The total points scored by both computer and the man for all three games was 2591. The difference between the first game’s total and second game’s total was 62 points. The dif-ference between the first game’s total and the third game’s total was only 2 points. Determine the total number of points scored by both computer and the man for all three contests. 46. Brain versus Computer. Can the human brain perform more calculations per second than a supercomputer? The calculating speed of the three top supercomputers, IBM’s Blue Gene/L, IBM’s BGW, and IBM’s ASC Purple, has been determined. The speed of IBM’s Blue Gene/L is 245 teraflops more than IBM’s BGW. The computing speed of IBM’s BGW is 22 teraflops more than IBM’s ASC Purple. The combined speed of all three top supercomputers is 568 teraflops. Determine the com puting speed (in teraflops) of each supercomputer. A teraflop is the measure of a computer’s speed and can be expressed as one trillion floating-point operations per second. By comparison, it is estimated that the human brain can perform 10 quadrillion calculations per second. • C A T C H T H E M I S T A K E In Exercises 47 and 48, explain the mistake that is made. 47. Solve the system of equations. Equation (1): 2x 2 y 1 z 5 2 Equation (2): x 2 y 5 1 Equation (3): x 1 z 5 1 Solution: Equation (2): x 2 y 5 1 Equation (3): x 1 z 5 1 Add Equation (2) and Equation (3). 2y 1 z 5 2 Multiply Equation (1) by 1212. 22x 1 y 2 z 5 22 Add. 22x 5 0 Solve for x. x 5 0 Substitute x 5 0 into Equation (2). 0 2 y 5 1 Solve for y. y 5 21 Substitute x 5 0 into Equation (3). 0 1 z 5 1 Solve for z. z 5 1 The answer is x 5 0, y 5 21, and z 5 1. This is incorrect. Although x 5 0, y 5 21, and z 5 1 does satisfy the three original equations, it is only one of infinitely many solutions. What mistake was made? 48. Solve the system of equations. Equation (1): x 1 3y 1 2z 5 4 Equation (2): 3x 1 10y 1 9z 5 17 Equation (3): 2x 1 7y 1 7z 5 17 Solution: Multiply Equation (1) by 23. 23x 2 9y 2 6z 5 212 Equation (2): 3x 1 10y 1 9z 5 17 Add. y 1 3z 5 5 Multiply Equation (1) by 22. 22x 2 6y 2 4z 5 28 Equation (3): 2x 1 7y 1 7z 5 17 Add. y 1 3z 5 9 Solve the system of two equations. y 1 3z 5 5 y 1 3z 5 9 Infinitely many solutions. Let z 5 a, then y 5 5 2 3a. Substitute z 5 a and y 5 5 2 3a into x 1 3y 1 2z 5 4 Equation (1). x 1 315 2 3a2 1 2a 5 4 Eliminate parentheses. x 1 15 2 9a 1 2a 5 4 Solve for x. x 5 7a 2 11 The answer is x 5 7a 2 11, y 5 5 2 3a, and z 5 a. This is incorrect. There is no solution. What mistake was made? In Exercises 49 and 50, determine whether each statement is true or false. 49. A system of linear equations that has more variables than equations cannot have a unique solution. 50. A system of linear equations that has the same number of equations as variables always has a unique solution. 51. Geometry. The circle given by the equation x2 1 y2 1 ax 1 by 1 c 5 0 passes through the points 122, 42, 11, 12, and 122, 222. Find a, b, and c. 52. Geometry. The circle given by the equation x2 1 y2 1 ax 1 by 1 c 5 0 passes through the points 10, 72, 16, 12, and 15, 42. Find a, b, and c. • C O N C E P T U A L 9.2 Systems of Linear Equations in Three Variables 891 53. A fourth-degree polynomial, ƒ1x2 5 ax4 1 bx3 1 cx2 1 dx 1 e, with a , 0, can be used to represent the following data on the number of deaths per year due to lightning strikes. Assume 2012 corresponds to x 5 22 and 2016 corresponds to x 5 2. Use the data to determine a, b, c, d, and e. Year 2016 2012 2014 US deaths by lightning 45 43 47 49 51 54. A copy machine accepts nickels, dimes, and quarters. After 1 hour, there are 30 coins and their total value is $4.60. If there are four more quarters than nickels, how many nickels, quarters, and dimes are in the machine? In Exercises 55–58, solve the system of linear equations. 55. 2y 1 z 5 3 4x 2 z 5 23 7x 2 3y 2 3z 5 2 x 2 y 2 z 5 22 56. 22x 2 y 1 2z 5 3 3x 2 4z 5 2 2x 1 y 5 21 2x 1 y 2 z 5 28 57. 3x1 2 2x2 1 x3 1 2x4 5 22 2x1 1 3x2 1 4x3 1 3x4 5 4 x1 1 x2 1 x3 1 x4 5 0 5x1 1 3x2 1 x3 1 2x4 5 21 58. 5x1 1 3x2 1 8x3 1 x4 5 1 x1 1 2x2 1 5x3 1 2x4 5 3 4x1 1 x3 2 2x4 5 23 x2 1 x3 1 x4 5 0 • C H A L L E N G E 892 CHAPTER 9 Systems of Linear Equations and Inequalities In Exercises 59 and 60, employ a graphing calculator to solve the system of linear equations (most graphing calculators have the capability of solving linear systems with the user entering the coefficients). 59. x 2 z 2 y 5 10 2x 2 3y 1 z 5 211 y 2 x 1 z 5 210 60. 2x 1 z 1 y 5 23 2y 2 z 1 x 5 0 x 1 y 1 2z 5 5 61. Some graphing calculators and graphing utilities have the ability to graph in three dimensions (3D) as opposed to the traditional two dimensions (2D). The line must be given in the form z 5 ax 1 by 1 c. Rewrite the system of equations in Exercise 59 in this form and graph the three lines in 3D. What is the point of intersection? Compare that with your answer in Exercise 59. 62. Some graphing calculators and graphing utilities have the ability to graph in three dimensions (3D) as opposed to the traditional two dimensions (2D). The line must be given in the form z 5 ax 1 by 1 c. Rewrite the system of equations in Exercise 60 in this form and graph the three lines in 3D. What is the point of intersection? Compare that with your answer in Exercise 60. In Exercises 63 and 64, employ a graphing calculator to solve the system of equations. 63. 0.2x 2 0.7y 1 0.8z 5 11.2 21.2x 1 0.3y 2 1.5z 5 0 0.8x 2 0.1y 1 2.1z 5 6.4 64. 1.8x 2 0.5y 1 2.4z 5 1.6 0.3x 2 0.6z 5 0.2 • T E C H N O L O G Y S K I L L S O B J E C T I V E S ■ ■Decompose rational expressions into sums of partial fractions when the denominators contain distinct linear factors, repeated linear factors, distinct irreducible quadratic factors, or repeated irreducible quadratic factors. C O N C E P T U A L O B J E C T I V E S ■ ■Understand the connection between partial-fraction decomposition and systems of linear equations. 9.3 PARTIAL FRACTIONS 9.3.1 Performing Partial-Fraction Decomposition In Chapter 4 we studied polynomial functions, and in Section 4.5 we discussed ratios of polynomial functions, called rational functions. Rational expressions are of the form: n1x2 d1x2 d1x2 2 0 where the numerator n1x2 and the denominator d 1x2 are polynomials. Examples of rational expressions are 4x 2 1 2x 1 3 2x 1 5 x2 2 1 3x4 2 2x 1 5 x 2 1 2x 1 4 Suppose we are asked to add two rational expressions: 2 x 1 1 1 5 x 2 3. We already possess the skills to accomplish this. We first identify the least common denominator 1x 1 12 1x 2 32 and combine the fractions into a single expression. 2 x 1 1 1 5 x 2 3 5 21x 2 32 1 51x 1 12 1x 1 121x 2 32 5 2x 2 6 1 5x 1 5 1x 1 121x 2 32 5 7x 2 1 x2 2 2x 2 3 9.3.1 SKI LL Decompose rational expressions into sums of partial fractions when the denominators contain distinct linear factors, repeated linear factors, distinct irreducible quadratic factors, or repeated irreducible quadratic factors. 9.3.1 C ON CEPTUAL Understand the connection between partial-fraction decomposition and systems of linear equations. 9.3 Partial Fractions 893 How do we do this in reverse? For example, how do we start with 7x 2 1 x2 2 2x 2 3 and write this expression as a sum of two simpler expressions? 7x 2 1 x2 2 2x 2 3 5 2 x 1 1 1 5 x 2 3 Each of the two expressions on the right is called a partial fraction. The sum of these fractions is called the partial-fraction decomposition of 7x 2 1 x2 2 2x 2 3. Partial-fraction decomposition is an important tool in calculus. Calculus operations such as differentiation and integration are often made simpler if you apply partial fractions. Partial fractions were not discussed until now because partial-fraction decomposition requires the ability to solve systems of linear equations. Since partial-fraction decomposition is made possible by the techniques of solving systems of linear equations, we consider partial fractions an important application of systems of linear equations. As mentioned earlier, a rational expression is the ratio of two polynomial expressions n1x2/d1x2, and we assume that n1x2 and d 1x2 are polynomials with no common factors. If the degree of n1x2 is less than the degree of d 1x2, then the rational expression n1x2/d1x2 is said to be proper. If the degree of n1x2 is greater than or equal to the degree of d 1x2, the rational expression is said to be improper. If the rational expression is improper, it should first be divided using long division. n1x2 d1x2 5 Q1x2 1 r1x2 d1x2 The result is the sum of a quotient Q1x2 and a rational expression, which is the ratio of the remainder r1x2 and the divisor d 1x2. The rational expression r1x2/d1x2 is proper, and the techniques outlined in this section can be applied to its partial-fraction decomposition. Partial-fraction decomposition of proper rational expressions always begins with factoring the denominator d 1x2. The goal is to write d 1x2 as a product of distinct linear factors, but that may not always be possible. Sometimes d 1x2 can be factored into a product of linear factors, where one or more are repeated. And sometimes the factored form of d 1x2 contains irreducible quadratic factors, such as x2 1 1. There are times when the irreducible quadratic factors are repeated, such as 1x2 1 122. A procedure is now outlined for partial-fraction decomposition. f f Partial-Fraction Decomposition f Partial Fraction Partial Fraction PARTIAL-FRACTION DECOMPOSITION To write a rational expression n1x2 d1x2 as a sum of partial fractions: Step 1: Determine whether the rational expression is proper or improper. • Proper: degree of n1x2 , degree of d 1x2 • Improper: degree of n1x2 $ degree of d 1x2 Step 2: If proper, proceed to Step 3. If improper, divide n1x2 d1x2 using polynomial (long) division and write the result as n1x2 d1x2 5 Q1x2 1 r 1x2 d1x2. Then proceed to Step 3 with r1x2 d1x2. 894 CHAPTER 9 Systems of Linear Equations and Inequalities Step 4 depends on which cases, or types of factors, arise. It is important to note that these four cases are not exclusive and combinations of different types of factors will appear. Distinct Linear Factors Step 3: Factor d 1x2. One of four possible cases will arise: Case 1: Distinct (nonrepeated) linear factors: 1ax 1 b2 d1x2 5 13x 2 121x 1 22 Case 2: One or more repeated linear factors: 1ax 1 b2m m $ 2 d1x2 5 1x 1 52 21x 2 32 Case 3: One or more distinct irreducible 1ax2 1 bx 1 c 5 0 has no real roots2 quadratic factors: 1ax2 1 bx 1 c2 d1x2 5 1x2 1 421x 1 121x 2 22 Case 4: One or more repeated irreducible quadratic factors: 1ax2 1 bx 1 c2m m $ 2 d1x2 5 1x2 1 x 1 1221x 1 121x 2 22 Step 4: Decompose the rational expression into a sum of partial fractions according to the procedure outlined in each case in this section. CASE 1 d(x) Has Only Distinct (Nonrepeated) Linear Factors If d 1x2 is a polynomial of degree p, and it can be factored into p linear factors: d1x2 5 1ax 1 b21cx 1 d2 c where no two factors are the same, then the partial-fraction decomposition of n1x2 d1x2 can be written as n1x2 d1x2 5 A 1ax 1 b2 1 B 1cx 1 d2 1 c where the numerators, A, B, and so on are constants to be determined. f p linear factors The goal is to write a proper rational expression as the sum of proper rational expressions. Therefore, if the denominator is a linear factor (degree 1), then the numerator is a constant (degree 0). EXAMPLE 1 Partial-Fraction Decomposition with Distinct Linear Factors Find the partial-fraction decomposition of 5x 1 13 x 2 1 4x 2 5. Solution: Factor the denominator. 5x 1 13 1x 2 121x 1 52 In Example 1, we started with a rational expression that had a numerator of degree 1 and a denominator of degree 2. Partial-fraction decomposition enabled us to write that rational expression as a sum of two rational expressions with degree 0 numerators and degree 1 denominators. Repeated Linear Factors Express as a sum of two partial fractions. 5x 1 13 1x 2 121x 1 52 5 A 1x 2 12 1 B 1x 1 52 Multiply the equation by the LCD 1x 2 121x 1 52. 5x 1 13 5 A1x 1 52 1 B 1x 2 12 Eliminate the parentheses. 5x 1 13 5 Ax 1 5A 1 Bx 2 B Group the x’s and constants on the right. 5x 1 13 5 1A 1 B2x 1 15A 2 B2 Identify like terms. 5x 1 13 5 1A 1 B2x 1 15A 2 B2 Equate the coefficients of x. 5 5 A 1 B Equate the constant terms. 13 5 5A 2 B Solve the system of two linear equations using any method to solve for A and B. A 5 3, B 5 2 Substitute A 5 3, B 5 2 into partial-fraction decomposition. 5x 1 13 1x 2 121x 1 52 5 3 1x 2 12 1 2 1x 1 52 Check by adding the partial fractions. 3 1x 2 12 1 2 1x 1 52 5 31x 1 52 1 21x 2 12 1x 2 121x 1 52 5 5x 1 13 x2 1 4x 2 5 YOU R T UR N Find the partial-fraction decomposition of 4x 2 13 x2 2 3x 2 10. ▼ ▼ A N S W E R 4x 2 13 x2 2 3x 2 10 5 3 x 1 2 1 1 x 2 5 CASE 2 d(x) Has at Least One Repeated Linear Factor If d 1x2 can be factored into a product of linear factors, then the partial-fraction decomposition will proceed as in Case 1, with the exception of a repeated factor 1ax 1 b2m, m $ 2. Any linear factor repeated m times will result in the sum of m partial fractions: A 1ax 1 b2 1 B 1ax 1 b22 1 C 1ax 1 b23 1 c1 M 1ax 1 b2m where the numerators, A, B, C, . . . , M are constants to be determined. Note that if d 1x2 is of degree p, the general form of the decomposition will have p partial fractions. If some numerator constants turn out to be zero, then the final decomposition may have fewer than p partial fractions. EXAMPLE 2 Partial-Fraction Decomposition with a Repeated Linear Factor Find the partial-fraction decomposition of 23x2 1 13x 2 12 x3 2 4x2 1 4x . Solution: Factor the denominator. 23x2 1 13x 2 12 x1x 2 222 9.3 Partial Fractions 895 896 CHAPTER 9 Systems of Linear Equations and Inequalities Express as a sum of three partial fractions. 23x2 1 13x 2 12 x1x 2 222 5 A x 1 B 1x 2 22 1 C 1x 2 222 Multiply the equation by the LCD x 1x 2 222. 23x2 1 13x 2 12 5 A1x 2 222 1 Bx1x 2 22 1 Cx Eliminate the parentheses. 23x2 1 13x 2 12 5 Ax2 2 4Ax 1 4A 1 Bx2 2 2Bx 1 Cx Group like terms on the right. 23x2 1 13x 2 12 5 1A 1 B2x2 1 124A 2 2B 1 C2x 1 4A Identify like terms on both sides. 23x2 1 13x 2 12 5 1A 1 B2x2 1 124A 2 2B 1 C2x 1 4A Equate the coefficients of x2. 23 5 A 1 B (1) Equate the coefficients of x. 13 5 24A 2 2B 1 C (2) Equate the constant terms. 212 5 4A (3) Solve the system of three equations for A, B, and C. Solve (3) for A. A 5 23 Substitute A 5 23 into (1). B 5 0 Substitute A 5 23 and B 5 0 into (2). C 5 1 Substitute A 5 23, B 5 0, C 5 1 into the partial-fraction decomposition. 23x2 1 13x 2 12 x1x 2 222 5 23 x 1 0 1x 2 22 1 1 1x 2 222 23x2 1 13x 2 12 x3 2 4x2 1 4x 5 23 x 1 1 1x 2 222 Check by adding the partial fractions. 23 x 1 1 1x 2 222 5 231x 2 222 1 11x2 x1x 2 222 5 23x2 1 13x 2 12 x3 2 4x2 1 4x Y OUR TU R N Find the partial-fraction decomposition of x2 1 1 x3 1 2x2 1 x. ▼ ▼ A N S W E R x 2 1 1 x 3 1 2x 2 1 x 5 1 x 2 2 1x 1 122 EXAMPLE 3 Partial-Fraction Decomposition with Multiple Repeated Linear Factors Find the partial-fraction decomposition of 2x3 1 6x2 1 6x 1 9 x4 1 6x3 1 9x2 . Solution: Factor the denominator. 2x3 1 6x2 1 6x 1 9 x21x 1 322 Express as a sum of four 2x3 1 6x2 1 6x 1 9 x21x 1 322 5 A x 1 B x2 1 C 1x 1 32 1 D 1x 1 322 partial fractions. Multiply the equation by the LCD x2 1x 1 322. 2x3 1 6x2 1 6x 1 9 5 Ax1x 1 322 1 B1x 1 322 1 Cx21x 1 32 1 Dx2 Eliminate the parentheses. 2x3 1 6x2 1 6x 1 9 5 Ax3 1 6Ax2 1 9Ax 1 Bx2 1 6Bx 1 9B 1 Cx3 1 3Cx2 1 Dx2 Group like terms on the right. 2x3 1 6x2 1 6x 1 9 5 1A 1 C2x3 1 16A 1 B 1 3C 1 D2x2 1 19A 1 6B2x 1 9B ▼ A N S W E R 2x3 1 2x 1 1 x4 1 2x3 1 x2 5 1 x2 1 2 1x 1 12 2 3 1x 1 122 Distinct Irreducible Quadratic Factors There will be times when a polynomial cannot be factored into a product of linear factors with real coefficients. For example, x2 1 4, x2 1 x 1 1, and 9x2 1 3x 1 2 are all examples of irreducible quadratic expressions. The general form of an irreducible quadratic factor is given by: ax2 1 bx 1 c where ax2 1 bx 1 c 5 0 has no real roots CASE 3 d(x) Has a Distinct Irreducible Quadratic Factor If the factored form of d 1x2 contains an irreducible quadratic factor ax2 1 bx 1 c, then the partial-fraction decomposition will contain a term of the form: Ax 1 B ax2 1 bx 1 c where A and B are constants to be determined. 9.3 Partial Fractions 897 Identify like terms on both sides. 2x3 1 6x2 1 6x 1 9 5 1A 1 C2x3 1 16A 1 B 1 3C 1 D2x2 1 19A 1 6B2x 1 9B Equate the coefficients of x3. 2 5 A 1 C (1) Equate the coefficients of x2. 6 5 6A 1 B 1 3C 1 D (2) Equate the coefficients of x. 6 5 9A 1 6B (3) Equate the constant terms. 9 5 9B (4) Solve the system of four equations for A, B, C, and D. Solve Equation (4) for B. B 5 1 Substitute B 5 1 into Equation (3) and solve for A. A 5 0 Substitute A 5 0 into Equation (1) and solve for C. C 5 2 Substitute A 5 0, B 5 1, and C 5 2 into Equation (2) and solve for D. D 5 21 Substitute A 5 0, B 5 1, C 5 2, D 5 21 into the partial-fraction decomposition. 2x3 1 6x2 1 6x 1 9 x21x 1 322 5 0 x 1 1 x2 1 2 1x 1 32 1 21 1x 1 322 2x3 1 6x2 1 6x 1 9 x21x 1 322 5 1 x2 1 2 1x 1 32 2 1 1x 1 322 Check by adding the partial fractions. 1 x2 1 2 1x 1 32 2 1 1x 1 322 5 1x 1 322 1 2x21x 1 32 2 1Ax2B x21x 1 322 5 2x3 1 6x2 1 6x 1 9 x4 1 6x3 1 9x2 Y OUR T UR N Find the partial-fraction decomposition of 2x3 1 2x 1 1 x4 1 2x3 1 x2. ▼ 898 CHAPTER 9 Systems of Linear Equations and Inequalities Recall that for a proper rational expression, the numerator is less than the denominator. For irreducible quadratic (degree 2) denominators we assume a linear (degree 1) numerator. For example, 7x2 1 2 12x 1 121x2 1 12 5 A 12x 1 12 1 Bx 1 C 1x2 1 12 A constant is used in the numerator when the denominator consists of a linear expression and a linear expression is used in the numerator when the denominator consists of a quadratic expression. f f Constant numerator Linear factor 1 Linear numerator Quadratic factor STUDY TIP The degree of the numerator is always 1 less than the degree of the denominator. ▼ A N S W E R 22x2 1 x 1 6 1x 2 121x2 1 42 5 1 x 2 1 2 3x 1 2 x2 1 4 EXAMPLE 4 Partial-Fraction Decomposition with an Irreducible Quadratic Factor Find the partial-fraction decomposition of 7x2 1 2 12x 1 121x2 1 12. Solution: The denominator is already in factored form. 7x2 1 2 12x 1 121x2 1 12 Express as a sum of two partial fractions. 7x2 1 2 12x 1 121x2 1 12 5 A 12x 1 12 1 Bx 1 C 1x2 1 12 Multiply the equation by the LCD 12x 1 12 1x2 1 12. 7x2 1 2 5 A1x2 1 12 1 1Bx 1 C212x 1 12 Eliminate the parentheses. 7x2 1 2 5 Ax2 1 A 1 2Bx2 1 Bx 1 2Cx 1 C Group like terms on the right. 7x2 1 2 5 1A 1 2B2x2 1 1B 1 2C2x 1 1A 1 C2 Identify like terms on both sides. 7x2 1 0x 1 2 5 1A 1 2B2x2 1 1B 1 2C2x 1 1A 1 C2 Equate the coefficients of x2. 7 5 A 1 2B Equate the coefficients of x. 0 5 B 1 2C Equate the constant terms. 2 5 A 1 C Solve the system of three equations for A, B, and C. A 5 3, B 5 2, C 5 21 Substitute A 5 3, B 5 2, C 5 21 into the partial-fraction decomposition. 7x2 1 2 12x 1 121x2 1 12 5 3 12x 1 12 1 2x 2 1 1x2 1 12 Check by adding the partial fractions. 3 12x 1 12 1 2x 2 1 1x2 1 12 5 31x2 1 12 1 12x 2 1212x 1 12 12x 1 121x2 1 12 5 7x2 1 2 12x 1 121x2 1 12 Y OUR TU R N Find the partial-fraction decomposition of 22x2 1 x 1 6 1x 2 121x2 1 42. ▼ Repeated Irreducible Quadratic Factors CASE 4 d(x) Has a Repeated Irreducible Quadratic Factor If the factored form of d 1x2 contains an irreducible quadratic factor 1ax2 1 bx 1 c2 m, where b2 2 4ac , 0 and m $ 2, then the partial-fraction decomposition will contain a series of terms of the form: A1x 1 B1 ax2 1 bx 1 c 1 A2x 1 B2 1ax2 1 bx 1 c22 1 A3x 1 B3 1ax2 1 bx 1 c23 1c 1 Amx 1 Bm 1ax2 1 bx 1 c2 m where Ai and Bi, i 5 1, 2, . . . , m, are constants to be determined. ▼ A N S W E R 3x3 1 x2 1 4x 2 1 1x2 1 422 5 3x 1 1 x2 1 4 2 8x 1 5 1x 2 1 422 9.3 Partial Fractions 899 EXAMPLE 5 Partial-Fraction Decomposition with a Repeated Irreducible Quadratic Factor Find the partial-fraction decomposition of x3 2 x2 1 3x 1 2 1x2 1 122 . Solution: The denominator is x3 2 x2 1 3x 1 2 1x2 1 122 already in factored form. Express as a sum of two partial fractions. x3 2 x2 1 3x 1 2 1x2 1 122 5 Ax 1 B x2 1 1 1 Cx 1 D 1x2 1 122 Multiply the equation by the LCD 1x2 1 122. x3 2 x2 1 3x 1 2 5 1Ax 1 B2 1x2 1 12 1 Cx 1 D Eliminate the parentheses. x3 2 x2 1 3x 1 2 5 Ax3 1 Bx2 1 Ax 1 B 1 Cx 1 D Group like terms on the right. x3 2 x2 1 3x 1 2 5 Ax3 1 Bx2 1 1A 1 C2x 1 1B 1 D2 Identify like terms on both sides. x3 2 x2 1 3x 1 2 5 Ax3 1 Bx2 1 1A 1 C2x 1 1B 1 D2 Equate the coefficients of x3. 1 5 A (1) Equate the coefficients of x2. 21 5 B (2) Equate the coefficients of x. 3 5 A 1 C (3) Equate the constant terms. 2 5 B 1 D (4) Substitute A 5 1 into Equation (3) and solve for C. C 5 2 Substitute B 5 21 into Equation (4) and solve for D. D 5 3 Substitute A 5 1, B 5 21, C 5 2, D 5 3 into the partial-fraction decomposition. x3 2 x2 1 3x 1 2 1x2 1 122 5 x 2 1 x2 1 1 1 2x 1 3 1x2 1 122 Check by adding the partial fractions. x 2 1 x2 1 1 1 2x 1 3 1x2 1 122 5 1x 2 121x2 1 12 1 12x 1 32 1x2 1 122 5 x3 2 x2 1 3x 1 2 1x2 1 122 YOUR T UR N Find the partial-fraction decomposition of 3x3 1 x2 1 4x 2 1 1x2 1 422 . ▼ 900 CHAPTER 9 Systems of Linear Equations and Inequalities Combinations of All Four Cases As you probably can imagine, there are rational expressions that have combinations of all four cases, which can lead to a system of several equations when solving for the numerator constants. EXAMPLE 6 Partial-Fraction Decomposition Find the partial-fraction decomposition of x5 1 x4 1 4x3 2 3x2 1 4x 2 8 x21x2 1 222 . Solution: The denominator is already in factored form. x5 1 x4 1 4x3 2 3x2 1 4x 2 8 x21x2 1 222 Express as a sum of partial fractions. There are repeated linear and irreducible quadratic factors. x5 1 x4 1 4x3 2 3x2 1 4x 2 8 x21x2 1 222 5 A x 1 B x2 1 Cx 1 D 1x2 1 22 1 Ex 1 F 1x2 1 222 Multiply the equation by the LCD x21x2 1 222. x5 1 x4 1 4x3 2 3x2 1 4x 2 8 5 Ax1x2 1 222 1 B1x2 1 222 1 1Cx 1 D2x21x2 1 22 1 1Ex 1 F 2x2 Eliminate the parentheses. x5 1 x4 1 4x3 2 3x2 1 4x 2 8 5 Ax514Ax314Ax1Bx414Bx214B1Cx512Cx3 1Dx412Dx21Ex31Fx2 Group like terms on the right. x5 1 x4 1 4x3 2 3x2 1 4x 2 8 5 1A1C2x51 1B1D2x4 1 14A1 2C1 E2x3 1 14B 1 2D 1F2x2 1 4Ax 1 4B Equating the coefficients of like terms A 1 C 5 1 leads to six equations. B 1 D 5 1 4A 1 2C 1 E 5 4 4B 1 2D 1 F 5 23 4A 5 4 4B 5 28 Solve this system of equations. A 5 1, B 5 22, C 5 0, D 5 3, E 5 0, F 5 21 Substitute A 5 1, B 5 22, C 5 0, D 5 3, E 5 0, F 5 21 into the partial-fraction decomposition. x5 1 x4 1 4x3 2 3x2 1 4x 2 8 x21x2 1 222 5 1 x 1 22 x2 1 0x 1 3 1x2 1 22 1 0x 1 21 1x2 1 222 x5 1 x4 1 4x3 2 3x2 1 4x 2 8 x21x2 1 222 5 1 x 1 22 x2 1 3 1x2 1 22 2 1 1x2 1 222 Check by adding the partial fractions. [CONCEPT CHECK] TRUE OR FALSE: Partial-fraction decomposition relies on solving systems of linear equations in order to determine the unknown coefficients. ANSWER True ▼ In Exercises 1–6, match the rational expression (1–6) with the form of the partial-fraction decomposition (a–f). 1. 3x 1 2 x1x2 2 252 2. 3x 1 2 x1x2 1 252 3. 3x 1 2 x21x2 1 252 4. 3x 1 2 x21x2 2 252 5. 3x 1 2 x1x2 1 2522 6. 3x 1 2 x21x2 1 2522 a. A x 1 B x2 1 Cx 1 D x2 1 25 b. A x 1 Bx 1 C x2 1 25 1 Dx 1 E 1x2 1 2522 c. A x 1 Bx 1 C x2 1 25 d. A x 1 B x 1 5 1 C x 2 5 e. A x 1 B x2 1 Cx 1 D x2 1 25 1 Ex 1 F 1x2 1 2522 f. A x 1 B x2 1 C x 1 5 1 D x 2 5 In Exercises 7–14, write the form of the partial-fraction decomposition. Do not solve for the constants. 7. 9 x2 2 x 2 20 8. 8 x2 2 3x 2 10 9. 2x 1 5 x3 2 4x2 10. x2 1 2x 2 1 x4 2 9x2 11. 2x3 2 4x2 1 7x 1 3 1x2 1 x 1 52 12. 2x3 1 5x2 1 6 1x2 2 3x 1 72 13. 3x3 2 x 1 9 1x2 1 1022 14. 5x3 1 2x2 1 4 1x2 1 1322 In Exercises 15–40, find the partial-fraction decomposition for each rational function. 15. 1 x1x 1 12 16. 1 x1x 2 12 17. x x1x 2 12 18. x x1x 1 12 19. 9x 2 11 1x 2 321x 1 52 20. 8x 2 13 1x 2 221x 1 12 21. 3x 1 1 1x 2 122 22. 9y 2 2 1y 2 122 23. 4x 2 3 x2 1 6x 1 9 24. 3x 1 1 x2 1 4x 1 4 25. 4x2 2 32x 1 72 1x 1 121x 2 522 26. 4x2 2 7x 2 3 1x 1 221x 2 122 27. 5x2 1 28x 2 6 1x 1 421x2 1 32 28. x2 1 5x 1 4 1x 2 221x2 1 22 29. 22x2 2 17x 1 11 1x 2 7213x2 2 7x 1 52 30. 14x2 1 8x 1 40 1x 1 5212x2 2 3x 1 52 31. x3 1x2 1 922 32. x2 1x2 1 922 33. 2x3 2 3x2 1 7x 2 2 1x2 1 122 34. 2x3 1 2x2 2 3x 1 15 1x2 1 822 35. 3x 1 1 x4 2 1 36. 2 2 x x4 2 81 37. 5x2 1 9x 2 8 1x 2 121x2 1 2x 2 12 38. 10x2 2 5x 1 29 1x 2 321x2 1 4x 1 52 39. 3x x3 2 1 40. 5x 1 2 x3 2 8 [SEC TION 9.3] E X E RC I S E S • S K I L L S 9.3 Partial Fractions 901 2. Repeated linear factors 2x 1 5 1x 2 3221x 1 12 5 A x 2 3 1 B 1x 2 322 1 C x 1 1 3. Distinct irreducible quadratic factors 1 2 x 1x2 1 12 5 Ax 1 B x2 1 1 4. Repeated irreducible quadratic factors 4x2 2 3x 1 2 1x2 1 122 5 Ax 1 B x2 1 1 1 Cx 1 D 1x2 1 122 A rational expression n1x2 d1x2 is • Proper: If the degree of the numerator is less than the degree of the denominator. • Improper: If the degree of the numerator is equal to or greater than the degree of the denominator. Partial-Fraction Decomposition of Proper Rational Expressions 1. Distinct (nonrepeated) linear factors 3x 2 10 1x 2 521x 1 42 5 A x 2 5 1 B x 1 4 [SEC TION 9.3] S U M MA RY 902 CHAPTER 9 Systems of Linear Equations and Inequalities • A P P L I C A T I O N S 41. Optics. The relationship between the distance of an object to a lens do, the distance to the image di, and the focal length ƒ of the lens is given by ƒ1di 1 do2 dido 5 1 2f f f do di 2f Image Object Use partial-fraction decomposition to write the lens law in terms of sums of fractions. What does each term represent? 42. Sums. Find the partial-fraction decomposition of 1 n1n 1 12, and apply it to find the sum of 1 1⋅2 1 1 2⋅3 1 1 3⋅4 1 . . . 1 1 999⋅1000. • C A T C H T H E M I S T A K E In Exercises 43 and 44, explain the mistake that is made. 43. Find the partial-fraction decomposition of 3x2 1 3x 1 1 x1x2 1 12 . Solution: Write the partial-fraction decomposition form. 3x2 1 3x 1 1 x1x2 1 12 5 A x 1 B x2 1 1 Multiply by the LCD x1x2 1 12. 3x2 1 3x 1 1 5 A1x2 1 12 1 Bx Eliminate the parentheses. 3x2 1 3x 1 1 5 Ax2 1 Bx 1 A Matching like terms leads to three equations. A 5 3, B 5 3, and A 5 1 This is incorrect. What mistake was made? 44. Find the partial-fraction decomposition of 3x4 2 x 2 1 x1x 2 12 . Solution: Write the partial-fraction decomposition form. 3x4 2 x 2 1 x1x 2 12 5 A x 1 B x 2 1 Multiply by the LCD x 1x 2 12. 3x4 2 x 2 1 5 A1x 2 12 1 Bx Eliminate the parentheses and group like terms. 3x4 2 x 2 1 5 1A 1 B2x 2 A Compare like coefficients. A 5 1, B 5 22 This is incorrect. What mistake was made? In Exercises 45 and 46, determine whether each statement is true or false. 45. Partial-fraction decomposition can only be employed when the degree of the numerator is greater than the degree of the denominator. 46. The degree of the denominator of a reducible rational expression is equal to the number of partial fractions in its decomposition. • C O N C E P T U A L For Exercises 47–52, find the partial-fraction decomposition. 47. x2 1 4x 2 8 x3 2 x2 2 4x 1 4 48. ax 1 b x2 2 c2 a, b, c are real numbers. 49. 2x3 1 x2 2 x 2 1 x4 1 x3 50. 2x3 1 2x 2 2 x5 2 x4 51. x5 1 2 1x2 1 123 52. x2 2 4 1x2 1 123 • C H A L L E N G E 9.4 Systems of Linear Inequalities in Two Variables 903 57. Use a graphing utility to graph y1 5 2x3 2 8x 1 16 1x 2 2221x2 1 42 and y2 5 1 x 2 2 1 2 1x 2 222 1 x 1 4 x2 1 4 in the same viewing rectangle. Is y2 the partial-fraction decomposition of y1? 58. Use a graphing utility to graph y1 5 3x3 1 14x2 1 6x 1 51 1x2 1 3x 2 421x2 1 2x 1 52 and y2 5 2 x 2 1 2 1 x 1 4 1 2x 2 3 x2 1 2x 1 5 in the same viewing rectangle. Is y2 the partial-fraction decomposition of y1? 53. Use a graphing utility to graph y1 5 5x 1 4 x2 1 x 2 2 and y2 5 3 x 2 1 1 2 x 1 2 in the same viewing rectangle. Is y2 the partial-fraction decomposition of y1? 54. Use a graphing utility to graph y1 5 2x2 1 2x 2 5 x3 1 5x and y2 5 3x 1 2 x2 1 5 2 1 x in the same viewing rectangle. Is y2 the partial-fraction decomposition of y1? 55. Use a graphing utility to graph y1 5 x9 1 8x 2 1 x51x2 1 123 and y2 5 4 x 2 1 x5 1 2 x2 1 1 2 3x 1 2 1x2 1 122 in the same viewing rectangle. Is y2 the partial-fraction decomposition of y1? 56. Use a graphing utility to graph y1 5 x3 1 2x 1 6 1x 1 32Ax2 2 4B3 and y2 5 2 x 1 3 1 x 1 3 1x2 2 423 in the same viewing rectangle. Is y2 the partial-fraction decomposition of y1? • T E C H N O L O G Y S K I L L S O B J E C T I V E S ■ ■Graph a linear inequality in two variables. ■ ■Graph a system of linear inequalities in two variables. C O N C E P T U A L O B J E C T I V ES ■ ■Interpret the difference between solid and dashed lines. ■ ■Interpret an overlapped shaded region as a solution. 9.4 SYSTEMS OF LINEAR INEQUALITIES IN TWO VARIABLES 9.4.1 Linear Inequalities in Two Variables To graph linear inequalities in two variables, we will bridge together two concepts that we have already learned: linear inequalities (Section 1.5) and lines (Section 2.3). Recall that in Section 1.5 we discussed linear inequalities in one variable. For example, 3x 2 1 , 8 has a solution x , 3, which can be represented graphically on a number line where the red colored area to the left of 3 represents the solution. Recall in Section 2.3 that y 5 2x 1 1 is an equation in two variables whose graph is a line in the xy-plane. We now turn our attention to linear inequalities in two variables. For example, if we change the 5 in y 5 2x 1 1 to , we get y , 2x 1 1. The solution to this inequality in two variables is the set of all points 1x, y2 that make this inequality true. Some solutions to this inequality are (22, 25), (0, 0), (3, 4), (5, 21), . . . In fact, the entire region below the line y 5 2x 1 1 satisfies the inequality y 2x 1 1. If we reverse the sign of the inequality to get y + 2x 1 1, then the entire region above the line y 5 2x 1 1 represents the solution to the inequality. Any line divides the xy-plane into two half-planes. For example, the line y 5 2 x 1 1 divides the xy-plane into two half-planes represented as y + 2x 1 1 and y 2x 1 1. Recall that with inequalities in one variable we used the notation of parentheses and brackets to denote the type of inequality (strict or nonstrict). We use a similar notation with linear inequalities in two variables. If the inequality is a strict inequality, , or ., then the line is dashed, and, if the inequality includes the equal sign, # or $, then a solid line is used. The following box summarizes the procedure for graphing a linear inequality in two variables. 9.4.1 S K I L L Graph a linear inequality in two variables. 9.4.1 C ON C E P T U A L Interpret the difference between solid and dashed lines. ) 3 x y y = 2x + 1 (3, 4) (–2, –5) (5, –1) (0, 0) 904 CHAPTER 9 Systems of Linear Equations and Inequalities GRAPHING A LINEAR INEQUALITY IN TWO VARIABLES Step 1: Change the sign. Change the inequality sign, ,, #, $, or ., to an equal sign, 5. Step 2: Draw the line that corresponds to the resulting equation in Step 1. • If the inequality is strict, , or ., use a dashed line. • If the inequality is not strict, # or $, use a solid line. Step 3: Test a point. • Select a point in one half-plane and test to see whether it satisfies the inequality. If it does, then so do all the points in that region (half-plane). If not, then none of the points in that half-plane satisfy the inequality. • Repeat this step for the other half-plane. Step 4: Shade the half-plane that satisfies the inequality. x y y > 2x + 1 y < 2x + 1 STUDY TIP A dashed line means that the points that lie on the line are not included in the solution of the linear inequality. EXAMPLE 1 Graphing a Strict Linear Inequality in Two Variables Graph the inequality 3x 1 y , 2. Solution: STEP 1 Change the inequality sign to an equal sign. 3x 1 y 5 2 STEP 2 Draw the line. Convert from standard form to slope–intercept form. y 5 23x 1 2 Since the inequality , is a strict inequality, use a dashed line. x y y = –3x + 2 STEP 3 Test points in each half-plane. Substitute 13, 02 into 3x 1 y , 2. 3132 1 0 , 2 The point 13, 02 does not satisfy the inequality. 9 , 2 Substitute 122, 02 into 3x 1 y , 2. 31222 1 0 , 2 The point 122, 02 does satisfy the inequality. 26 , 2 STEP 4 Shade the region containing the point 122, 02. x y 3x + y < 2 or y < –3x + 2 (3, 0) (–2, 0) Y OUR TU R N Graph the inequality 2x 1 y . 21. ▼ ▼ A N S W E R x y –x + y > –1 or y > x – 1 [CONCEPT CHECK] TRUE OR FALSE: A dashed line indicates that the points along the line satisfy the inequality. ANSWER False ▼ 9.4.2 Systems of Linear Inequalities in Two Variables Systems of linear inequalities are similar to systems of linear equations. In systems of linear equations, we sought the points that satisfied all of the equations. The solution set of a system of inequalities contains the points that satisfy all of the inequalities. The graph of a system of inequalities can be obtained by simultaneously graphing each individual inequality and finding where the shaded regions intersect (or overlap), if at all. ▼ A N S W E R x y x – 2y ≤ 6 or y ≥ x – 3 1 2 EXAMPLE 2 Graphing a Nonstrict Linear Inequality in Two Variables Graph the inequality 2x 2 3y $ 6. Solution: STEP 1 Change the inequality sign to an equal sign. 2x 2 3y 5 6 STEP 2 Draw the line. Convert from standard form to slope–intercept form. y 5 2 3x 2 2 Since the inequality $ is not a strict x y y = x – 2 2 3 inequality, use a solid line. STEP 3 Test points in each half-plane. Substitute 15, 02 into 2x 2 3y $ 6. 2152 2 3102 $ 6 The point 15, 02 satisfies the inequality. 10 $ 6 Substitute 10, 02 into 2x 2 3y $ 6. 2102 2 3102 $ 6 0 $ 6 The point 10, 02 does not satisfy the inequality. STEP 4 Shade the region containing x y (5, 0) 2x –3y ≥ 6 or y ≤ x – 2 2 3 the point 15, 02. YOUR T UR N Graph the inequality x 2 2y # 6. ▼ 9.4 Systems of Linear Inequalities in Two Variables 905 9.4.2 S K IL L Graph a system of linear inequalities in two variables. 9.4.2 C O N C E P T U A L Interpret an overlapped shaded region as a solution. 906 CHAPTER 9 Systems of Linear Equations and Inequalities EXAMPLE 3 Graphing a System of Two Linear Inequalities Graph the system of inequalities: x 1 y $ 22 x 1 y # 2 Solution: STEP 1 Change the inequality signs to equal signs. x 1 y 5 22 x 1 y 5 2 STEP 2 Draw the two lines. Because the inequality signs are x y y = –x + 2 y = –x – 2 not strict, use solid lines. STEP 3 Test points for each inequality. x 1 y # 22 Substitute 124, 02 into x 1 y $ 22. 24 $ 22 The point 124, 02 does not satisfy the inequality. Substitute 10, 02 into x 1 y $ 22. 0 $ 22 The point 10, 02 does satisfy the inequality. x 1 y " 2 Substitute 10, 02 into x 1 y # 2. 0 # 2 The point 10, 02 does satisfy the inequality. Substitute 14, 02 into x 1 y # 2. 4 # 2 The point 14, 02 does not satisfy the inequality. STEP 4 For x 1 y # 22, shade the region For x 1 y " 2, shade the region above that includes 10, 02. below that includes 10, 02. x y y ≥ –x – 2 x y y ≤ –x + 2 STEP 5 The overlapping region x y y ≤–x + 2 y ≥–x – 2 is the solution. Notice that the points 10, 02, 121, 12, and 11, 212 all lie in the shaded region and all three satisfy both inequalities. ▼ A N S W E R a. No solution. b. x y y > x – 1 y < x + 1 EXAMPLE 4 Graphing a System of Two Linear Inequalities with No Solution Graph the system of inequalities: x 1 y # 22 x 1 y $ 2 Solution: STEP 1 Change the inequality signs to equal signs. x 1 y 5 22 x 1 y 5 2 STEP 2 Draw the two lines. Because the inequality signs are not x y y = –x + 2 y = –x – 2 strict, use solid lines. STEP 3 Test points for each inequality. x 1 y " 22 Substitute 124, 02 into x 1 y # 22. 24 # 22 The point 124, 02 does satisfy the inequality. Substitute 10, 02 into x 1 y # 22. 0 # 22 The point 10, 02 does not satisfy the inequality. x 1 y # 2 Substitute 10, 02 into x 1 y $ 2. 0 $ 2 The point 10, 02 does not satisfy the inequality. Substitute 14, 02 into x 1 y $ 2. 4 $ 2 The point 14, 02 does satisfy the inequality. STEP 4 For x 1 y " 22, shade the region x y y ≥–x + 2 y ≤–x – 2 (4, 0) (–4, 0) below that includes 124, 02. For x 1 y " 2, shade the region above that includes 14, 02. STEP 5 There is no overlapping region. Therefore, no points satisfy both inequalities. We say there is no solution . Y OUR T UR N Graph the solution to the system of inequalities. a. y . x 1 1 b. y , x 1 1 y , x 2 1 y . x 2 1 ▼ 9.4 Systems of Linear Inequalities in Two Variables 907 [CONCEPT CHECK] TRUE OR FALSE Only the points that lie in the shaded region satisfy both linear inequalities. ANSWER True ▼ 908 CHAPTER 9 Systems of Linear Equations and Inequalities Thus far we have addressed only systems of two linear inequalities. Systems with more than two inequalities are treated in a similar manner. The solution is the set of all points that satisfy all of the inequalities. When there are more than two linear inequali-ties, the solution may be a bounded region. We can algebraically determine where the lines intersect by setting the y-values equal to each other. EXAMPLE 5 Graphing a System of Multiple Linear Inequalities y # x Solve the system of inequalities: y $ 2x y , 3 Solution: STEP 1 Change the inequalities to equal signs. y 5 x y 5 2x y 5 3 STEP 2 Draw the three lines. To determine the points of intersection, set the y-values equal. Point where y 5 x and y 5 2x intersect: x 5 2x x 5 0 Substitute x 5 0 into y 5 x. 10, 02 Point where y 5 2x and 2x 5 3 x y y = x y = –x y = 3 (3, 3) (3, 3) y 5 3 intersect: x 5 23 123, 32 Point where y 5 3 and x 5 3 y 5 x intersect: 13, 32 STEP 3 Test points to determine the shaded x y y < 3 y ≤x y ≥–x half-planes corresponding to y " x, y # 2x, and y 3. STEP 4 Shade the overlapping region. x y y ≥–x y ≤x y < 3 Applications Systems of linear inequalities arise in many applications, for example, the target zone for heart rate during exercise, normal weight ranges for humans, capacity of a room for an event, and return on investments. EXAMPLE 6 Cost of a Wedding Reception A couple has invited 300 guests to their wedding. The fixed costs (such as formal wear, entertainment, flowers, and invitations) are $7000, and the variable costs (party favors, chair covers, food, and drinks) range between $25 and $50 per person, depending on the menu. Assuming at least 200 and at most 300 people attend, graph the cost of the wedding as a system of inequalities. Solution: Let x represent the number of people attending the wedding and y represent the cost of the wedding. There are four linear inequalities: At least 200 guests attend: x # 200 No more than 300 guests attend: x " 300 Minimum cost of the wedding: y # 7000 1 25x Maximum cost of the wedding: y " 7000 1 50x Graph the system of inequalities. Number of Guests 400 500 100 300 200 Wedding Cost 10500 7000 14000 17500 $21000 y = 7000 + 50x y = 7000 + 25x This solution implies that the wedding cost will be approximately between $12,000 and $22,000. In economics, the point where the supply and demand curves intersect is called the equilibrium point. Consumer surplus is a measure of the amount that consumers benefit by being able to purchase a product for a price less than the maximum they would be willing to pay. Producer surplus is a measure of the amount that producers benefit by selling at a mar-ket price that is higher than the least they would be willing to sell for. Quantity Market Price Consumer Surplus Producer Surplus Equilibrium Supply curve Demand curve 9.4 Systems of Linear Inequalities in Two Variables 909 910 CHAPTER 9 Systems of Linear Equations and Inequalities EXAMPLE 7 Consumer Surplus and Producer Surplus The Tesla Motors Roadster is the first electric car that is able to travel 245 miles on a single charge. The price of the first model of this car was approximately $90,000 (including tax and incentives). Stefan Falke/Laif/ Redux Pictures Suppose the supply and demand equations for this electric car are given by P 5 90,000 2 0.1x 1Demand2 P 5 10,000 1 0.3x 1Supply2 where P is the price in dollars and x is the number of cars produced. Calculate the consumer surplus and the producer surplus for these two equations. Solution: Find the equilibrium point. 90,000 2 0.1x 5 10,000 1 0.3x 0.4x 5 80,000 x 5 200,000 Let x 5 200,000 in either the P 5 90,000 2 0.11200,0002 5 70,000 supply or demand equation. P 5 10,000 1 0.31200,0002 5 70,000 According to these models, if the price of a Tesla Motors Roadster is $70,000, then 200,000 cars will be sold and there will be no surplus. Write the systems of linear inequalities that correspond to consumer surplus and producer surplus. CONSUMER SURPLUS PRODUCER SURPLUS P # 90,000 2 0.1x P $ 10,000 1 0.3x P $ 70,000 P # 70,000 x $ 0 x $ 0 Number of Cars 250,000 150,000 50,000 Price $100,000 90,000 80,000 70,000 60,000 50,000 40,000 30,000 20,000 10,000 Consumer Surplus Producer Surplus (200,000, 70,000) The consumer surplus is the area of the red triangle. A 5 1 2 bh 5 1 2 1200,0002120,0002 The consumer surplus is $2B. 5 2,000,000,000 The producer surplus is the area of the blue triangle. A 5 1 2 bh 5 1 2 1200,0002160,0002 The producer surplus is $6B. 5 6,000,000,000 The graphs of the systems of linear inequalities in Examples 6 and 7 are said to be bounded, whereas the graphs of the systems of linear inequalities in Examples 3, 4, and 5 are said to be unbounded. Any points that correspond to boundary lines intersecting are called corner points or vertices. In Example 7, the vertices corresponding to the consumer surplus are the points 10, 90,0002, 10, 70,0002, and 1200,000, 70,0002, and the vertices corresponding to the producer surplus are the points 10, 70,0002, 10, 10,0002, and 1200,000, 70,0002. 9.4 Systems of Linear Inequalities in Two Variables 911 S Y S T EM O F L I N EA R I N EQ U ALIT IE S • Draw the individual linear inequalities. • The overlapping shaded region is the solution. LINE AR I NE Q UALI T Y 1. Change the inequality sign to an equal sign. 2. Draw the line (dashed for strict inequalities and solid for nonstrict inequalities). 3. Test a point. (Select a point in one-half plane and test the inequality. Repeat this step for the other half‑plane.) 4. Shade the half-plane that satisfies the linear inequality. [SEC TION 9.4] S U M MA RY [SEC TION 9.4] E X ERC I S E S • S K I L L S In Exercises 1–4, match the linear inequality with the correct graph. 1. y . x 2. y $ x 3. y , x 4. y # x a. x y b. x y c. x y d. x y In Exercises 5–20, graph each linear inequality. 5. y . x 2 1 6. y $ 2x 1 1 7. y # 2x 8. y . 2x 9. y # 23x 1 2 10. y , 2x 1 3 11. y # 22x 1 1 12. y . 3x 2 2 13. 3x 1 4y , 2 14. 2x 1 3y . 26 15. 5x 1 3y , 15 16. 4x 2 5y # 20 17. 4x 2 2y $ 6 18. 6x 2 3y $ 9 19. 6x 1 4y # 12 20. 5x 2 2y $ 10 In Exercises 21–48, graph each system of inequalities or indicate that the system has no solution. 21. y $ x 2 1 y # x 1 1 22. y . x 1 1 y , x 2 1 23. y . 2x 1 1 y , 2x 2 1 24. y # 2x 2 1 y $ 2x 1 1 25. y $ 2x y # 2x 26. y . 2x y , 2x 27. x . 22 x , 4 28. y , 3 y . 0 29. x $ 2 y # x 30. y # 3 y $ x 31. y . x x , 0 y , 4 32. y # x x $ 0 y # 1 912 CHAPTER 9 Systems of Linear Equations and Inequalities 33. x 1 y . 2 y , 1 x . 0 34. x 1 y , 4 x . 0 y $ 1 35. 2x 1 y . 1 y , 3 x . 0 36. x 2 y . 2 y , 4 x $ 0 37. x 1 3y . 6 y , 1 x $ 1 38. x 1 2y . 4 y , 1 x $ 0 39. y $ x 2 1 y # 2x 1 3 y , x 1 2 40. y , 4 2 x y . x 2 4 y . 2x 2 4 41. x 1 y . 24 2x 1 y , 2 y $ 21 y # 1 42. y , x 1 2 y . x 2 2 y , 2x 1 2 y . 2x 2 2 43. y , x 1 3 x 1 y $ 1 y $ 1 y # 3 44. y # 2x 1 2 y 2 x $ 23 y $ 22 y # 1 45. y 1 x , 2 y 1 x $ 4 y $ 22 y # 1 46. y 2 x , 3 y 1 x . 3 y # 22 y $ 24 47. 2x 2 y , 2 2x 1 y . 2 y , 2 48. 3x 2 y . 3 3x 1 y , 3 y , 22 • A P P L I C A T I O N S 49. Area. Find the area enclosed y . 0 x 0 by the system of inequalities. y , 2 50. Area. Find the area enclosed y , 0 x 0 by the system of inequalities. x $ 0 y $ 0 x , 3 51. Area. Find the area enclosed 5x 1 y # 10 by the system of linear inequalities x $ 0 1assume y $ 02. x # 1 52. Area. Find the area enclosed 25x 1 y # 0 by the system of linear inequalities x $ 1 1assume y $ 02. x # 2 53. Hurricanes. After back-to-back-to-back-to-back hurricanes (Charley, Frances, Ivan, and Jeanne) in Florida in the summer of 2004, FEMA sent disaster relief trucks to Florida. Floridians mainly needed drinking water and generators. Each truck could carry no more than 6000 pounds of cargo or 2400 cubic feet of cargo. Each case of bottled water takes up 1 cubic foot of space and weighs 25 pounds. Each generator takes up 20 cubic feet and weighs 150 pounds. Let x represent the number of cases of water and y represent the number of generators, and write a system of linear inequalities that describes the number of generators and cases of water each truck can haul to Florida. 54. Hurricanes. Repeat Exercise 53 with a smaller truck and different supplies. Suppose the smaller trucks that can haul 2000 pounds and 1500 cubic feet of cargo are used to haul plywood and tarps. A case of plywood is 60 cubic feet and weighs 500 pounds. A case of tarps is 10 cubic feet and weighs 50 pounds. Letting x represent the number of cases of plywood and y represent the number of cases of tarps, write a system of linear inequalities that describes the number of cases of tarps and plywood each truck can haul to Florida. Graph the system of linear inequalities. 55. Health. A diet must be designed to provide at least 275 units of calcium, 125 units of iron, and 200 units of Vitamin B. Each ounce of food A contains 10 units of calcium, 15 units of iron, and 20 units of vitamin B. Each ounce of food B contains 20 units of calcium, 10 units of iron, and 15 units of vitamin B. a. Find a system of inequalities to describe the different quantities of food that may be used (let x 5 the number of ounces of food A and y 5 the number of ounces of food B). b. Graph the system of inequalities. c. Using the graph found in part (b), find two possible solutions (there are infinitely many). 56. Health. A diet must be designed to provide at least 350 units of calcium, 175 units of iron, and 225 units of Vitamin B. Each ounce of food A contains 15 units of calcium, 25 units of iron, and 20 units of vitamin B. Each ounce of food B contains 25 units of calcium, 10 units of iron, and 10 units of vitamin B. a. Find a system of inequalities to describe the different quantities of food that may be used (let x 5 the number of ounces of food A and y 5 the number of ounces of food B). b. Graph the system of inequalities. c. Using the graph found in part (b), find two possible solutions (there are infinitely many). 57. Business. A manufacturer produces two types of computer mouse: a USB wireless mouse and a Bluetooth mouse. Past sales indicate that it is necessary to produce at least twice as many USB wireless mice than Bluetooth mice. To meet demand, the manufacturer must produce at least 1000 computer mice per hour. a. Find a system of inequalities describing the production levels of computer mice. Let x be the production level for the USB wireless mouse and y be the production level for the Bluetooth mouse. b. Graph the system of inequalities describing the production levels of computer mice. c. Use your graph in part (b) to find two possible solutions. 58. Business. A manufacturer produces two types of mechanical pencil lead: 0.5 millimeter and 0.7 millimeter. Past sales indicate that it is necessary to produce at least 50% more 0.5 millimeter lead than 0.7 millimeter lead. To meet demand, the manufacturer must produce at least 10,000 pieces of pencil lead per hour. a. Find a system of inequalities describing the production levels of pencil lead. Let x be the production level for 0.5 millimeter pencil lead and y be the production level for 0.7 millimeter pencil lead. b. Graph the system of inequalities describing the production levels of pencil lead. c. Use your graph in part (b) to find two possible solutions. For Exercises 59–62, employ the following supply and demand equations: Demand: P 5 80 2 0.01x Supply: P 5 20 1 0.02x 59. Consumer Surplus. Write a system of linear inequalities corresponding to the consumer surplus. 60. Producer Surplus. Write a system of linear inequalities corresponding to the producer surplus. 61. Consumer Surplus. Calculate the consumer surplus given the supply and demand equations. 62. Producer Surplus. Calculate the producer surplus given the supply and demand equations. • C A T C H T H E M I S T A K E In Exercises 63 and 64, explain the mistake that is made. 63. Graph the inequality y $ 2x 1 1. Solution: Graph the line y 5 2x 1 1 with a solid line. Since the inequality is $, shade to the right. This is incorrect. What mistake was made? 64. Graph the inequality y , 2x 1 1. Solution: Graph the line y 5 2x 1 1 with a solid line. Since the inequality is ,, shade below. This is incorrect. What mistake was made? x y y < 2x + 1 x y y = 2x + 1 x y y ≥ 2x + 1 x y y = 2x + 1 In Exercises 65–68, determine whether each statement is true or false. 65. A linear inequality always has a solution that is a half-plane. 66. A dashed curve is used for strict inequalities. 67. A solid curve is used for strict inequalities. 68. A system of linear inequalities always has a solution. • C O N C E P T U A L For the system of linear inequalities (Exercises 69 and 70), assume a, b, c, and d are real numbers. x $ a x , b y . c y # d 69. Describe the solution when a , b and c , d. 70. What will the solution be if a . b and c . d, given the system of linear inequalities? Use the following system of linear inequalities for Exercises 71 and 72: y # ax 1 b y $ 2ax 1 b 71. If a and b are positive real numbers, graph the solution. 72. If a and b are negative real numbers, graph the solution. • C H A L L E N G E In Exercises 75 and 76, use a graphing utility to graph each system of inequalities or indicate that the system has no solution. 75. 20.05x 1 0.02y $ 0.12 76. y # 2x 1 3 0.01x 1 0.08y # 0.08 y . 20.5x 1 5 In Exercises 73 and 74, use a graphing utility to graph the following inequalities. 73. 4x 2 2y $ 6 1Check with your answer to Exercise 17.2 74. 6x 2 3y $ 9 1Check with your answer to Exercise 18.2 • T E C H N O L O G Y 9.4 Systems of Linear Inequalities in Two Variables 913 914 CHAPTER 9 Systems of Linear Equations and Inequalities 9.5.1 Solving an Optimization Problem Often we seek to maximize or minimize a function subject to constraints. This process is called optimization. For example, in the chapter opener about Hurricane Katrina, FEMA had to determine how many generators, cases of water, and tarps should be in each truck to maximize the number of Louisianians given help, yet at the same time factor in the weight and space constraints on the trucks. When the function we seek to minimize or maximize is linear and the constraints are given in terms of linear inequalities, a graphing approach to such problems is called linear programming. In linear programming, we start with a linear equation, called the objective function, that represents the quantity that is to be maximized or minimized, for example, the number of Louisianians aided by FEMA. The number of people aided, however, is subject to constraints represented as linear inequalities, such as how much weight each truck can haul and how much space each truck has for cargo. The goal is to minimize or maximize the objective function z 5 Ax 1 By subject to constraints. In other words, find the points 1x, y2 that make the value of z the largest (or smallest). The constraints are a system of linear inequalities, and the common shaded region represents the feasible (possible) solutions. If the constraints form a bounded region, the vertices represent the coordinates 1x, y2 that correspond to a maximum or minimum value of the objective function z 5 Ax 1 By. If the region is not bounded, then if an optimal solution exists, it will occur at a vertex. A procedure for solving linear programming problems is outlined below. S K I L L S O B J E C T I V E S ■ ■Solve the optimization problem, which combines minimizing or maximizing a function subject to constraints, using linear programming. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that linear programming is a graphical method for solving optimization problems in which vertices represent maxima or minima. 9.5 THE LINEAR PROGRAMMING MODEL SOLVING AN OPTIMIZATION PROBLEM USING LINEAR PROGRAMMING Step 1: Write the objective function. This expression represents the quantity that is to be minimized or maximized. Step 2: Write the constraints. This is a system of linear inequalities. Step 3: Graph the constraints. Graph the system of linear inequalities and shade the common region, which contains the feasible solutions. Step 4: Identify the vertices. The corner points of the shaded region represent possible maximum or minimum values of the objective function. Step 5: Evaluate the objective function for each vertex. For each corner point of the shaded region, substitute the coordinates into the objective function and list the value of the objective function. Step 6: Identify the optimal solution. The largest (maximum) or smallest (minimum) value of the objective function in Step 5 is the optimal solution if the feasible region is bounded. 9.5.1 S KILL Solve the optimization problem, which combines minimizing or maximizing a function subject to constraints, using linear programming. 9.5.1 CO NCE PTUAL Understand that linear programming is a graphical method for solving optimization problems in which vertices represent maxima or minima. x y 5 5 Feasible (possible) solutions [CONCEPT CHECK] In Example 1, the point (5,1) would correspond to a z value of 11 (which is greater than 9). Why is the point (5,1) not a maximum in Example 1? ANSWER Because (5,1) does not satisfy the constraints ▼ EXAMPLE 1 Maximizing an Objective Function Find the maximum value of z 5 2x 1 y, subject to the constraints: x $ 1 x # 4 x 1 y # 5 y $ 0 Solution: STEP 1 Write the objective function. z 5 2x 1 y 9.5 The Linear Programming Model 915 ▼ A N S W E R The maximum value of z is 7, which occurs when x 5 1 and y 5 2. STEP 2 Write the constraints. x # 1 x " 4 y " 2x 1 5 y # 0 STEP 3 Graph the constraints. x y x ≥1 x ≤4 y ≥0 y ≤–x + 5 Feasible Solutions (1, 0) (4, 0) (1, 4) (4, 1) STEP 4 Identify the vertices. 11, 42, 14, 12, 11, 02, 14, 02 STEP 5 Evaluate the objective function for each vertex. VERTEX x y OBJECTIVE FUNCTION: z 5 2x 1 y 11, 42 1 4 2112 1 4 5 6 14, 12 4 1 2142 1 1 5 9 11, 02 1 0 2112 1 0 5 2 14, 02 4 0 2142 1 0 5 8 STEP 6 The maximum value of z is 9, which occurs when x 5 4 and y 5 1. Y OUR T UR N Find the maximum value of z 5 x 1 3y subject to the constraints: x $ 1 x # 3 y # 2x 1 3 y $ 0 ▼ STUDY TIP The bounded region is the region that satisfies all of the constraints. Only vertices of the bounded region correspond to possible solutions. Even though y 5 2x 1 5 and y 5 0 intersect at x 5 5, that point of intersection is outside the shaded region and therefore is not one of the vertices. EXAMPLE 2 Minimizing an Objective Function Find the minimum value of z 5 4x 1 5y, subject to the constraints: x $ 0 2x 1 y # 6 x 1 y # 5 y $ 0 Solution: STEP 1 Write the objective function. z 5 4x 1 5y STEP 2 Write the constraints. x # 0 y " 22x 1 6 y " 2 x 1 5 y # 0 916 CHAPTER 9 Systems of Linear Equations and Inequalities ▼ A N S W E R The minimum value of z is 8, which occurs when x 5 4 and y 5 0. STEP 3 Graph the constraints. x y x ≥0 y ≥0 y ≤–x + 5 y ≤ –2x + 6 Feasible Solutions (0, 0) (3, 0) (0, 5) (1, 4) STEP 4 Identify the vertices. 10, 02, 10, 52, 11, 42, 13, 02 STEP 5 Evaluate the objective function for each vertex. VERTEX x y OBJECTIVE FUNCTION: z 5 4x 1 5y 10, 02 0 0 4102 1 5102 5 0 10, 52 0 5 4102 1 5152 5 25 11, 42 1 4 4112 1 5142 5 24 13, 02 3 0 4132 1 5102 5 12 STEP 6 The minimum value of z is 0, which occurs when x 5 0 and y 5 0. Y OUR TU R N Find the minimum value of z 5 2x 1 3y subject to the constraints: x $ 1 2x 1 y # 8 x 1 y $ 4 ▼ STUDY TIP Maxima or minima of objective functions occur at the vertices of the shaded region corresponding to the constraints. EXAMPLE 3 Solving an Optimization Problem Using Linear Programming: Unbounded Region Find the maximum value and minimum value of z 5 7x 1 3y, subject to the constraints: y $ 0 22x 1 y # 0 2x 1 y $ 24 Solution: STEP 1 Write the objective function. z 5 7x 1 3y STEP 2 Write the constraints. y # 0 y " 2x y # x 2 4 STEP 3 Graph the constraints. x y y ≥0 y ≥x – 4 Feasible Solutions (0, 0) (4, 0) y ≤2x STEP 4 Identify the vertices. 10, 02, 14, 02 When the feasible solutions are contained in a bounded region, then a maximum and a minimum exist and are each located at one of the vertices. If the feasible solutions are contained in an unbounded region, then if a maximum or minimum exists, it is located at one of the vertices. STEP 5 Evaluate the objective function for each vertex. VERTEX x y OBJECTIVE FUNCTION: z 5 7x 1 3y 10, 02 0 0 7102 1 3102 5 0 14, 02 4 0 7142 1 3102 5 28 STEP 6 The minimum value of z is 0, which occurs when x 5 0 and y 5 0. There is no maximum value because if we select a point in the shaded region, say 13, 32, the objective function at 13, 32 is equal to 30, which is greater than 28. 9.5 The Linear Programming Model 917 EXAMPLE 4 Maximizing the Number of People Aided by a Hurricane Relief Effort After Hurricane Katrina, FEMA sent disaster relief trucks to Louisiana. Each truck could carry no more than 6000 pounds of cargo. Each case of bottled water weighs 25 pounds, and each generator weighs 150 pounds. If each generator helps one household and five cases of water help one household, determine the maximum number of Louisiana households aided by each truck and how many generators and cases of water should be sent in each truck. Due to the number of trucks and the supply of water and generators nationwide, each truck must contain at least 10 times as many cases of water as generators. Solution: Let x represent the number of cases of water. Let y represent the number of generators. Let z represent the number of households aided per truck. STEP 1 Write the objective function. z 5 1 5x 1 y STEP 2 Write the constraints. Number of cases of water is nonnegative. x $ 0 Number of generators is nonnegative. y $ 0 At least 10 times as many cases of water as generators. x # 10y Weight capacity of truck: 25x 1 150y " 6000 STEP 3 Graph the constraints and determine feasible solutions. Number of Generators (150, 15) (240, 0) Cases of Water 250 200 150 100 50 20 10 30 40 50 y ≤ x 1 10 y ≤– x + 40 1 6 STEP 4 Identify the vertices. 10, 02, 1150, 152, 1240, 02 918 CHAPTER 9 Systems of Linear Equations and Inequalities STEP 5 Evaluate the objective function for each vertex. VERTEX x y OBJECTIVE FUNCTION: z 5 1 5x 1 y 10, 02 0 0 0 1150, 152 150 15 45 1240, 02 240 0 48 STEP 6 A maximum of 48 households are aided by each truck. This maximum occurs when there are 240 cases of water and 0 generators on each truck. 4. Identify the vertices. 5. Evaluate the objective function for each vertex. 6. Identify the optimal solution. In this section, the linear programming model was discussed. 1. Write the objective function. 2. Write the constraints. 3. Graph the constraints. [SEC TION 9.5] S U M MA RY [SEC TION 9.5] E X E R C I SE S • S K I L L S In Exercises 1–4, find the value of the objective function at each of the vertices. What is the maximum value of the objective function? What is the minimum value of the objective function? 1. Objective function: z 5 2x 1 3y 2. Objective function: z 5 3x 1 2y 3. Objective function: z 5 1.5x 1 4.5y 4. Objective function: z 5 2 3x 1 3 5y In Exercises 5–12, minimize or maximize each objective function subject to the constraints. 5. Minimize z 5 7x 1 4y subject to: 6. Maximize z 5 3x 1 5y subject to: 7. Maximize z 5 4x 1 3y subject to: x $ 0 y $ 0 2x 1 y # 4 x $ 0 y $ 0 2x 1 y $ 4 x $ 0 y # 2x 1 4 y $ 2x 8. Minimize z 5 4x 1 3y subject to: 9. Minimize z 5 2.5x 1 3.1y subject to: 10. Maximize z 5 2.5x 2 3.1y subject to: x $ 0 y $ 0 x $ 0 y $ 0 x # 4 x $ 1 y # 7 x # 3 x 1 y # 10 x 1 y $ 0 2x 1 y # 2 x 1 y # 6 2x 1 y $ 2 x 1 y $ 6 11. Maximize z 5 1 4x 1 2 5y subject to: 12. Minimize z 5 1 3x 2 2 5y subject to: x 1 y $ 5 x 1 y # 7 x 1 y $ 6 x 1 y # 8 2x 1 y # 5 2x 1 y $ 3 2x 1 y # 6 2x 1 y $ 4 x y (2, 4) (–1, 4) (1, –1) (–2, –1) • A P P L I C A T I O N S 13. Hurricanes. After the 2004 back-to-back-to-back-to-back hurricanes in Florida, a student at Valencia Community Col-lege decided to create two T-shirts to sell. One T-shirt says, “I survived Charley on Friday the Thirteenth,” and the second says, “I survived Charley, Frances, Ivan, and Jeanne.” The Charley T-shirt costs him $7 to make and he sells it for $13. The other T-shirt costs him $5 to make and he sells it for $10. He does not want to invest more than $1000. He estimates that the total demand will not exceed 180 T-shirts. Find the number of each type of T-shirt he should make to yield maximum profit. 14. Hurricanes. After Hurricane Charley devastated central Florida unexpectedly, Orlando residents prepared for Hurricane Frances by boarding up windows and filling up their cars with gas. It took 5 hours of standing in line to get plywood, and lines for gas were just as time consuming. A student at Seminole Community College decides to do a spoof of the “Got Milk” ads and creates two T-shirts: “Got Plywood,” with a line of people in a home improvement store, and “Got Gas,” with a street lined with cars waiting to pump gasoline. The “Got Plywood” shirts cost $8 to make, and she sells them for $13. The “Got Gas” shirts cost $6 to make, and she sells them for $10. She decides to limit her costs to $1400. She estimates that demand for these T-shirts will not exceed 200 T-shirts. Find the number of each type of T-shirt she should make to yield maximum profit. 15. Computer Business. A computer science major and a business major decide to start a small business that builds and sells a desktop computer and a laptop computer. They buy the parts, assemble them, load the operating system, and sell the computers to other students. The costs for parts, time to assemble the computer, and profit are summarized in the following table: DESKTOP LAPTOP Cost of Parts $700 $400 Time to Assemble (hours) 5 3 Profit $500 $300 They were able to get a small business loan in the amount of $10,000 to cover costs. They plan on making these computers over the summer and selling them the first day of class. They can dedicate at most only 90 hours to assembling these computers. They estimate that the demand for laptops will be at least three times as great as the demand for desktops. How many of each type of computer should they make to maximize profit? 16. Computer Business. Repeat Exercise 15 if the two students are able to get a loan for $30,000 to cover costs and they can dedicate at most 120 hours to assembling the computers. 17. Passenger Ratio. The Eurostar is a high-speed train that travels between London, Brussels, and Paris. There are 30 cars on each departure. Each train car is designated first class or second class. Based on demand for each type of fare, there should always be at least two but no more than four first-class train cars. The management wants to claim that the ratio of first-class to second-class cars never exceeds 1:8. If the profit on each first-class train car is twice as much as the profit of each second-class train car, find the number of each class of train car that will generate a maxi-mum profit. 18. Passenger Ratio. Repeat Exercise 17. This time, assume that there has to be at least one first-class train car and that the profit from each first-class train car is 1.2 times as much as the profit from each second-class train car. The ratio of first class to second class cannot exceed 1:10. 19. Production. A manufacturer of skis produces two models: a regular ski and a slalom ski. A set of regular skis produces a $25 profit, and a set of slalom skis produces a profit of $50. The manufacturer expects a customer demand of at least 200 pairs of regular skis and at least 80 pairs of slalom skis. The maximum number of pairs of skis that can be produced by this company is 400. How many of each model of skis should be produced to maximize profits? 20. Donut Inventory. A well-known donut store makes two popular types of donuts: crème-filled and jelly-filled. The manager knows from past statistics that the number of dozens of donuts sold is at least 10, but no more than 30. To prepare the donuts for frying, the baker needs (on the average) 3 minutes for a dozen crème-filled and 2 minutes for jelly-filled. The baker has at most 2 hours available per day to prepare the donuts. How many dozens of each type should be prepared to maximize the daily profit if there is a $1.20 profit for each dozen crème-filled and $1.80 profit for each dozen jelly-filled donuts? • C A T C H T H E M I S T A K E In Exercises 21 and 22, explain the mistake that is made. 9.5 The Linear Programming Model 919 21. Maximize the objective function z 5 2x 1 y subject to the following constraints: x $ 0 y $ 0 2x 1 y $ 0 x 1 y # 2 Solution: Graph the constraints. x y Identify the vertices. 10, 22, 10, 02, 11, 12 Comparing the y-coordinates of the vertices, the largest y-value is 2. The maximum value occurs at 10, 22. The objective function at that point is equal to 2. This is incorrect. The maximum value of the objective function is not 2. What mistake was made? 920 CHAPTER 9 Systems of Linear Equations and Inequalities 22. Maximize the objective function z 5 2x 1 y subject to the following constraints: x $ 0 y $ 0 2x 1 y # 0 x 1 y # 2 Solution: Graph the constraints. x y Identify the vertices. 10, 22, 10, 02, 11, 12 VERTEX x y OBJECTIVE FUNCTION: z 5 2x 1 y 10, 22 0 2 2102 1 2 5 2 10, 02 0 0 2102 1 0 5 0 11, 12 1 1 2112 1 1 5 3 The maximum, 3, is located at the point 11, 12. This is incorrect. What mistake was made? In Exercises 23 and 24, determine whether each statement is true or false. • C O N C E P T U A L 23. An objective function always has a maximum or minimum. 24. An objective function subject to constraints that correspond to a bounded region always has a maximum and a minimum. 25. Maximize the objective function z 5 2x 1 y subject to the conditions, where a . 2. ax 1 y $ 2a 2ax 1 y # a ax 1 y # a 2ax 1 y $ 2a 26. Maximize the objective function z 5 x 1 2y subject to the conditions, where a . b . 0. x 1 y $ a 2x 1 y # a x 1 y # a 1 b 2x 1 y $ a 2 b • C H A L L E N G E • T E C H N O L O G Y In Exercises 27–32, employ a graphing utility to determine the region of feasible solutions, and apply the intersect feature to find the approximate coordinates of the vertices to solve the linear programming problems. 27. Minimize z 5 2.5x 1 3.1y subject to: x $ 0 y $ 0 x # 4 2x 1 y # 2 x 1 y # 6 Compare with your answer to Exercise 9. 28. Maximize z 5 2.5x 1 3.1y subject to: x $ 0 y $ 0 x # 4 2x 1 y # 2 x 1 y # 6 Compare with your answer to Exercise 10. 29. Maximize z 5 17x 1 14y subject to: y $ 4.5 2x 1 y # 3.7 x 1 y # 11.2 30. Minimize z 5 1.2x 1 1.5y subject to: 2.3x 1 y # 14.7 22.3x 1 y # 14.7 25.2x 1 y # 3.7 22.3x 1 y $ 1.5 31. Maximize z 5 4.5x 1 1.8y subject to: 23.1x 1 y # 11.4 1.4x 1 y $ 1.5 5x 1 y # 15 32. Minimize z 5 5.4x 2 1.6y subject to: y $ 1.8x 2 3.6 y $ 22.2x 1 1.8 y # 1.8x 1 4.2 y # 22.2x 1 10.8 CH A P TE R 9 R E VIE W [CH AP TER 9 REVIEW] SECTION CONCEPT KEY IDEAS/FORMULAS 9.1 Systems of linear equations in two variables A1x 1 B1y 5 C1 A2x 1 B2y 5 C2 Solving systems of linear equations Substitution method Solve for one variable in terms of the other and substitute that expression into the other equation. Elimination method Eliminate a variable by adding multiples of the equations. Graphing method Graph the two lines. The solution is the point of intersection. Parallel lines have no solution, and identical lines have infinitely many solutions. Three methods and three types of solutions One solution, no solution, infinitely many solutions. 9.2 Systems of linear equations in three variables Planes in three-dimensional coordinate system. Solving systems of linear equations in three variables Step 1: Reduce the system to 2 equations and 2 unknowns. Step 2: Solve the resulting system from Step 1. Step 3: Substitute solutions found in Step 2 into any of the equations to find the third variable. Step 4: Check. Types of solutions One solution (point), no solution, infinitely many solutions (line). 9.3 Partial fractions Performing partial-fraction decomposition on a rational expression in proper form n1x2 d1x2 Factor d1x2 Distinct linear factors n1x2 d1x2 5 A 1ax 1 b2 1 B 1cx 1 d2 1 . . . Repeated linear factors n1x2 d1x2 5 A 1ax 1 b2 1 B 1ax 1 b22 1 . . . 1 M 1ax 1 b2m Distinct irreducible quadratic factors n1x2 d1x2 5 Ax 1 B ax2 1 bx 1 c 1 Cx 1 D dx2 1 ex 1 f 1 . . . Repeated irreducible quadratic factors n1x2 d1x2 5 A1x 1 B1 ax2 1 bx 1 c 1 A2x 1 B2 1ax2 1 bx 1 c22 1 A3x 1 B3 1ax2 1 bx 1 c23 1 c1 Amx 1 Bm 1ax2 1 bx 1 c2m 9.4 Systems of linear inequalities in two variables Linear inequalities in two variables n # or $ use solid lines n , or . use dashed lines Systems of linear inequalities in two variables Solutions are determined graphically by finding the common shaded regions. 9.5 The linear programming model Solving an optimization problem Finding optimal solutions. Minimizing or maximizing a function subject to constraints (linear inequalities). Chapter Review 921 REV IEW E XE R CI SE S 922 CHAPTER 9 Systems of Linear Equations and Inequalities [ C H AP T E R 9 REVIEW EXERC IS E S ] 9.1 Systems of Linear Equations in Two Variables Solve each system of linear equations. 1. r 2 s 5 3 2. 3x 1 4y 5 2 r 1 s 5 3 x 2 y 5 6 3. 24x 1 2y 5 3 4. 0.25x 2 0.5y 5 0.6 4x 2 y 5 5 0.5x 1 0.25y 5 0.8 5. x 1 y 5 3 6. 3x 1 y 5 4 x 2 y 5 1 2x 1 y 5 1 7. 4c 2 4d 5 3 8. 5r 1 2s 5 1 c 1 d 5 4 r 2 s 5 23 9. y 5 21 2 x 10. 2x 1 4y 5 22 y 5 1 2x 1 2 4x 2 2y 5 3 11. 1.3x 2 2.4y 5 1.6 12. 1 4x 2 3 4y 5 12 0.7x 2 1.2y 5 1.4 1 2y 1 1 4x 5 1 2 13. 5x 2 3y 5 21 14. 6x 2 2y 5 22 22x 1 7y 5 220 4x 1 3y 5 16 15. 10x 2 7y 5 224 16. 1 3x 2 2 9 y 5 2 9 7x 1 4y 5 1 4 5x 1 3 4y 5 23 4 Match each system of equations with its graph. 17. 2x 2 3y 5 4 18. 5x 2 y 5 2 x 1 4y 5 3 5x 2 y 5 22 19. x 1 2y 5 2 6 20. 5x 1 2y 5 3 2x 1 4y 5 212 4x 2 2y 5 6 a. x y –5 5 5 –5 b. x y –5 5 5 –5 c. x y –3 7 5 –5 d. x y –5 5 2 –8 Applications 21. Chemistry. In chemistry lab, Alexandra needs to make a 42 milliliter solution that is 15% NaCl. All that is in the lab is 6% and 18% NaCl. How many milliliters of each solution should she use to obtain the desired mix? 22. Gas Mileage. A Nissan Sentra gets approximately 32 mpg on the highway and 18 mpg in the city. Suppose 265 miles were driven on a full tank (12 gallons) of gasoline. Approximately how many miles were driven in the city and how many on the highway? 9.2 Systems of Linear Equations in Three Variables Solve each system of linear equations. 23. x 1 y 1 z 5 1 24. x 2 2y 1 z 5 3 x 2 y 2 z 5 23 2x 2 y 1 z 5 24 2x 1 y 1 z 5 3 3x 2 3y 2 5z 5 2 25. x 1 y 1 z 5 7 26. x 1 z 5 3 x 2 y 2 z 5 17 2x 1 y 2 z 5 21 y 1 z 5 5 x 1 y 1 z 5 5 Applications 27. Fitting a Curve to Data. The average number of flights on a commercial plane that a person takes per year can be modeled by a quadratic function y 5 ax2 1 bx 1 c, where a , 0, and x represents age: 16 # x # 65. The following table gives the average number of flights per year that a person takes on a commercial airline. Determine the quadratic function that models this quantity. AGE NUMBER OF FLIGHTS PER YEAR 16 2 40 6 65 4 28. Investment Portfolio. Danny and Paula decide to invest $20,000 of their savings. They put some in an IRA account earning 4.5% interest, some in a mutual fund that has been averaging 8% a year, and some in a stock that earned 12% last year. If they put $4000 more in the IRA than in the mutual fund and the mutual fund and stock have the same growth in the next year as they did in the previous year, they will earn $1525 in a year. How much money did they put in each of the three investments? 9.3 Partial Fractions Write the form of each partial-fraction decomposition. Do not solve for the constants. 29. 4 1x 2 1221x 1 321x 2 52 30. 7 1x 2 9213x 1 5221x 1 42 31. 12 x14x 1 5212x 1 122 32. 2 1x 1 121x 2 521x 2 922 33. 3 x2 1 x 2 12 34. x2 1 3x 2 2 x3 1 6x2 35. 3x3 1 4x2 1 56x 1 62 1x2 1 1722 36. x3 1 7x2 1 10 1x2 1 1322 Find the partial-fraction decomposition for each rational function. 37. 9x 1 23 1x 2 121x 1 72 38. 12x 1 1 13x 1 2212x 2 12 R E VI E W E XERCISES 39. 13x2 1 90x 2 25 2x3 2 50x 40. 5x2 1 x 1 24 x3 1 8x 41. 2 x2 1 x 42. x x1x 1 32 43. 5x 2 17 x2 1 4x 1 4 44. x3 1x2 1 6422 9.4 Systems of Linear Inequalities in Two Variables Graph each linear inequality. 45. y $ 22x 1 3 46. y , x 2 4 47. 2x 1 4y . 5 48. 5x 1 2y # 4 49. y $ 23x 1 2 50. y , x 2 2 51. 3x 1 8y # 16 52. 4x 2 9y # 18 Graph each system of inequalities or indicate that the system has no solution. 53. y $ x 1 2 54. y $ 3x y # x 2 2 y # 3x 55. x # 22 56. x 1 3y $ 6 y . x 2x 2 y # 8 57. 3x 2 4y # 16 58. x 1 y . 24 5x 1 3y . 9 x 2 y , 3 y $ 22 x # 8 9.5 The Linear Programming Model Minimize or maximize the objective function subject to the constraints. 59. Minimize z 5 2x 1 y subject to: x $ 0 y $ 0 x 1 y # 3 60. Maximize z 5 2x 1 3y subject to: x $ 0 y $ 0 2x 1 y # 0 x # 3 61. Maximize z 5 2.5x 1 3.2y subject to: x $ 0 y $ 0 x 1 y # 8 2x 1 y $ 0 62. Minimize z 5 5x 1 11y subject to: 2x 1 y # 10 x 1 y $ 2 x # 2 63. Minimize z 5 3x 2 5y subject to: 2x 1 y $ 6 2x 2 y # 6 x $ 0 64. Maximize z 5 22x 1 7y subject to: 3x 1 y # 7 x 2 2y $ 1 x $ 0 Applications For Exercises 65 and 66, refer to the following: An art student decides to hand-paint coasters and sell sets at a flea market. She decides to make two types of coaster sets: an ocean watercolor and black-and-white geometric shapes. The cost, profit, and time it takes her to paint each set are sum marized in the following table. OCEAN WATERCOLOR GEOMETRIC SHAPES Cost $4 $2 Profit $15 $8 Hours 3 2 65. Profit. If her costs cannot exceed $100 and she can spend only 90 hours painting the coasters, determine the number of each type she should make to maximize her profit. 66. Profit. If her costs cannot exceed $300 and she can spend only 90 hours painting, determine the number of each type she should make to maximize her profit. Technology Exercises Section 9.1 67. Use a graphing utility to graph the two equations 0.4x 1 0.3y 5 20.1 and 0.5x 2 0.2y 5 1.6. Find the solution to this system of linear equations. 68. Use a graphing utility to graph the two equations 1 2x 1 3 10y 5 1 5 and 25 3x 1 1 2y 5 4 3. Find the solution to this system of linear equations. Section 9.2 In Exercises 69 and 70, employ a graphing calculator to solve the system of equations. 69. 5x 2 3y 1 15z 5 21 22x 1 0.8y 2 4z 5 28 2.5x 2 y 1 7.5z 5 12 70. 2x 2 1.5y 1 3z 5 6.5 0.5x 2 0.375y 1 0.75z 5 1.5 Section 9.3 71. Apply a graphing utility to graph y1 5 x2 1 4 x4 2 x2 and y2 5 2 4 x2 1 5/2 x 2 1 2 5/2 x 1 1 in the same viewing rectangle. Is y2 the partial-fraction decomposition of y1? 72. Use a graphing utility to graph y1 5 x3 1 6x2 1 27x 1 38 1x2 1 8x 1 1721x2 1 6x 1 132 and y2 5 2x 1 1 x2 1 8x 1 17 2 x 2 3 x2 1 6x 1 13 in the same viewing rectangle. Is y2 the partial-fraction decomposition of y1? Section 9.4 In Exercises 73 and 74, use a graphing utility to graph each system of inequalities or indicate that the system has no solution. 73. 2x 1 5y $ 215 74. y # 0.5x y # 22 3x 2 1 y . 21.5x 1 6 Section 9.5 75. Maximize z 5 6.2x 1 1.5y subject to: 4x 2 3y # 5.4 2x 1 4.5y # 6.3 3x 2 y $ 210.7 76. Minimize z 5 1.6x 2 2.8y subject to: y $ 3.2x 2 4.8 x $ 22 y # 3.2x 1 4.8 x # 4 Review Exercises 923 PR ACTICE TEST 924 CHAPTER 9 Systems of Linear Equations and Inequalities [ C H AP T E R 9 PRACTICE TEST ] Solve each system of linear equations. 1. x 2 2y 5 1 2. 3x 1 5y 5 22 2x 1 3y 5 2 7x 1 11y 5 26 3. x 2 y 5 2 4. 3x 2 2y 5 5 22x 1 2y 5 24 6x 2 4y 5 0 5. x 1 y 1 z 5 21 6. 6x 1 9y 1 z 5 5 2x 1 y 1 z 5 0 2x 2 3y 1 z 5 3 2x 1 y 1 2z 5 0 10x 1 12y 1 2z 5 9 7. x 2 2y 1 3z 5 5 8. x 1 z 5 1 22x 1 y 1 4z 5 21 x 1 y 5 21 3x 2 5y 1 z 5 214 y 1 z 5 0 Write each rational expression as a sum of partial fractions. 9. 2x 1 5 x2 1 x 10. 3x 2 13 1x 2 522 11. 7x 1 5 1x 1 222 12. 1 2x2 1 5x 2 3 13. 5x 2 3 x1x2 2 92 14. 5x 2 3 x1x2 1 92 Graph the inequalities. 15. 22x 1 y , 6 16. 4x 2 y $ 8 In Exercises 17–20, graph the system of inequalities. 17. x 1 y # 4 18. x 1 3y # 6 2x 1 y $ 22 2x 2 y # 4 19. 5x 2 2y # 8 20. 2x 1 y . 6 3x 1 4y , 12 x 1 y # 4 21. Minimize the function z 5 5x 1 7y, subject to the constraints: x $ 0 y $ 0 x 1 y # 3 2x 1 y $ 1 22. Find the maximum value of the objective function z 5 3x 1 6y, given the constraints: x $ 0 y $ 0 x 1 y # 6 2x 1 2y # 4 23. Stock Investment. On starting a new job, Cameron gets a $30,000 signing bonus and decides to invest the money. She invests some in a money market account earning 3% and some in two different stocks. The aggressive stock rose 12% last year, and the conservative stock rose 6% last year. She invested $1000 more in the aggressive stock than in the conservative stock, and in 1 year she earned $1890 on the investment. How much did she put in the money market, and how much was invested in each stock? 24. Ranch. A rancher buys a rectangular parcel of property that is 245,000 square feet. She fences the entire border and then adds three internal fences so that there are four equal rectangular pastures. If the fence required 3150 linear feet, what are the dimensions of the entire property? 25. Employ a graphing calculator to solve the system of equations. 5x 2 3y 1 3z 5 1 2x 2 y 1 z 5 4 4x 2 2y 1 3z 5 9 26. Minimize z 5 x 2 1.2y subject to: 3.2x 1 y # 15.9 0.4x 2 y # 0.3 x $ 1.0 CU MU LA TIV E TEST [CH AP TERS 1–9 CUM UL AT IVE T E S T ] 1. Simplify 6x2 2 11x 1 5 30x 2 25 . 2. Perform the operation and write in standard form: 15 2 7i2 2 111 2 9i2. 3. Solve for x and give any excluded values: 5 x 2 5 x 1 1 52 5 x2 1 x. 4. Solve for the radius r: A 5 pr 2. 5. Solve and express the solution in interval notation: 2t 2 t 2 1 $ 3t. 6. Calculate the distance and midpoint between the line segment joining the points A23 4, 1 6B and A1 5, 22 3B. 7. Write an equation of a line in slope–intercept form that passes through the two points 15.6, 6.22 and 13.2, 5.02. 8. Use interval notation to express the domain of the function ƒ1x2 5 "x2 2 25. 9. Find the average rate of change for ƒ1x2 5 5 x, from x 5 2 to x 5 4. 10. Sketch the graph of the function y 5 1 x 2 2 1 1 and identify all transformations. 11. Evaluate g1ƒ12122 for ƒ1x2 5 ! 3 x 2 7 and g1x2 5 5 3 2 x. 12. Find the inverse of the function ƒ1x2 5 5x 1 2 x 2 3 . 13. Find the quadratic function that has the vertex 10, 72 and goes through the point 12, 212. 14. Find all of the real zeros and state the multiplicity of the function ƒ1x2 5 4x5 1 x3. 15. List the possible rational zeros and test to determine all rational zeros for P1x2 5 2x4 1 7x3 2 18x2 2 13x 1 10. 16. Factor the polynomial P1x2 5 4x4 2 4x3 1 13x2 1 18x 1 5 as a product of linear factors. 17. Graph the exponential function ƒ1x2 5 52x 2 1; state the y-intercept, the domain, the range, and the horizontal asymptote. 18. How much money should be put in a savings account now that earns 5.2% a year compounded daily if you want to have $50,000 in 16 years? 19. Graph the function ƒ1x2 5 ln1x 1 12 2 3 using transformation techniques. 20. Evaluate log4.7 8.9 using the change-of-base formula. Round the answer to three decimal places. 21. Solve the equation 51102x2 5 37 for x. Round the answers to three decimal places. 22. At a coffee shop, one order of 4 tall coffees and 6 donuts is $14.10. A second order of 5 tall coffees and 4 donuts is $13.11. Find the cost of a tall coffee and the cost of a donut. 23. Solve the system of linear equations. x 1 6y 2 z 5 23 2x 2 5y 1 z 5 9 x 1 4y 1 2z 5 12 24. Find the partial-fraction decomposition for the rational expression 21 x2 2 2x 1 1. 25. Graph the system of linear inequalities. y . 2x y $ 23 x # 3 26. Employ a graphing calculator to solve the system of equations. 2x 2 9y 1 8z 5 221.8 5x 1 6y 2 11z 5 41.7 3x 2 y 1 9z 5 211.7 27. Apply a graphing utility to graph y1 5 3x2 1 x 1 8 x3 1 4x and y2 5 2 x 1 x 1 1 x2 1 4 in the same viewing rectangle. Is y 2 the partial-fraction decomposition of y1? Cumulative Test 925 C H A P T E R LEARNING OBJECTIVES [ [ ■ ■Use Gauss–Jordan elimination to solve systems of linear equations. ■ ■Perform matrix operations: addition, subtraction, and multiplication. ■ ■Utilize matrix algebra and inverses of matrices to solve systems of linear equations. ■ ■Use Cramer’s rule (determinants) to solve systems of linear equations. Cryptography is the practice and study of encryption and decryption—encoding data so that it can be decoded only by specific individuals. In other words, it turns a message into gibberish so that only the person who has the deciphering tools can turn that gibberish back into the original message. ATM cards, online shopping sites, and secure military communications all depend on coding and decoding of information. Matrices are used extensively in cryptography. A matrix is used as the “key” to encode the data, and then its inverse matrix is used as the “key” to decode the data. Section 10.3, Exercises 49–54. Matrices 10 decodes sends sends decodes visitor web server encrypts receives encrypts receives 927 [I N T HI S CHAPTER] We will use matrices to express systems of linear equations. We will discuss three ways that matrices can be used to solve systems of linear equations: augmented matrices (Gaussian elimination and Gauss–Jordan elimination), inverse matrices (matrix algebra), and Cramer’s rule (determinants). 927 MATRICES 10.1 MATRICES AND SYSTEMS OF LINEAR EQUATIONS 10.2 MATRIX ALGEBRA 10.3 MATRIX EQUATIONS; THE INVERSE OF A SQUARE MATRIX 10.4 THE DETERMINANT OF A SQUARE MATRIX AND CRAMER’S RULE • Matrices • Augmented Matrices • Row Operations on a Matrix • Row–Echelon Form of a Matrix • Gaussian Elimination with Back-Substitution • Gauss–Jordan Elimination • Inconsistent and Dependent Systems • Equality of Matrices • Matrix Addition and Subtraction • Scalar and Matrix Multiplication • Matrix Equations • Finding the Inverse of a Matrix • Solving Systems of Linear Equations Using Matrix Algebra and Inverses of Square Matrices • Determinant of a 2 3 2 Matrix • Determinant of an n 3 n Matrix • Cramer’s Rule: Systems of Linear Equations in Two Variables • Cramer’s Rule: Systems of Linear Equations in Three Variables 928 CHAPTER 10 Matrices 10.1.1 Matrices Some information is best displayed in a table. For example, the number of calories burned per half hour of exercise depends upon the person’s weight, as illustrated in the following table. Note that the rows correspond to activities and the columns correspond to weight. ACTIVITY 127–137 LB 160–170 LB 180–200 LB Walking/4 mph 156 183 204 Volleyball 267 315 348 Jogging/5 mph 276 345 381 Another example is the driving distance in miles from cities in Arizona (columns) to cities outside the state (rows). CITY FLAGSTAFF PHOENIX TUCSON YUMA Albuquerque, NM 325 465 440 650 Las Vegas, NV 250 300 415 295 Los Angeles, CA 470 375 490 285 If we selected only the numbers in each of the preceding tables and placed brackets around them, the result would be a matrix. Calories: £ 156 183 204 267 315 348 276 345 381 § Miles: £ 325 465 440 650 250 300 415 295 470 375 490 285 § S K I L L S O B J E C T I V E S ■ ■Determine the order of a matrix. ■ ■Write a system of linear equations as an augmented matrix. ■ ■Perform row operations on an augmented matrix. ■ ■Write a matrix in reduced row–echelon form. ■ ■Solve systems of linear equations using Gaussian elimination with back-substitution. ■ ■Solve systems of linear equations using Gauss–Jordan elimination. ■ ■Recognize matrices that correspond to inconsistent and dependent systems. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that aij corresponds to the element in row i and column j of matrix A. ■ ■Visualize an augmented matrix as a system of linear equations. ■ ■Understand that row operations correspond to generating equivalent systems of linear equations. ■ ■Understand the difference between row–echelon and reduced row–echelon forms. ■ ■Understand that solving systems with augmented matrices is equivalent to solving by the method of elimination. ■ ■Understand that when you move down the augmented matrix you use rows above and on the way back up you use rows below. ■ ■Understand the difference between matrices that correspond to inconsistent or dependent systems. 10.1 MATRICES AND SYSTEMS OF LINEAR EQUATIONS 10.1.1 SKILL Determine the order of a matrix. 10.1.1 CO NC EPTUAL Understand that aij corresponds to the element in row i and column j of matrix A. 10.1 Matrices and Systems of Linear Equations 929 When the number of rows equals the number of columns (that is, when m 5 n), the matrix is a square matrix of order n. In a square matrix, the entries a11, a22, a33, . . . , ann are the main diagonal entries. The matrix A433 5 ≥ a32 ¥ has order (dimensions) 4 3 3, since there are four rows and three columns. The element a32 is in the third row and second column. [CONCEPT CHECK] How many calories correspond to the a23 element? ANSWER 348 ▼ A matrix is a rectangular array of numbers written within brackets. a11 a12 c a1j c a1n a21 a22 c a2j c a2n ( ( c ( c ( ai1 ai2 c aij c ain ( ( c ( c ( am1 am2 c amj c amn Each number aij in the matrix is called an element of the matrix. The first subscript i is the row index, and the second subscript j is the column index. This matrix contains m rows and n columns, and is said to be of order m 3 n. V F STUDY TIP The order of a matrix is always given as the number of rows by the number of columns. EXAMPLE 1 Finding the Order of a Matrix Determine the order of each matrix given. a. c2 1 3 0d b. c 1 22 5 21 3 4d c. £ 22 5 4 1 21 3 0 3 8 1 § d. C4 9 21 2 3D e. ≥ 3 5 0 7 22 1 22 3 6 ¥ Solution: a. This matrix has 2 rows and 2 columns, so the order of the matrix is 2 3 2 . b. This matrix has 2 rows and 3 columns, so the order of the matrix is 2 3 3 . c. This matrix has 3 rows and 3 columns, so the order of the matrix is 3 3 3 . d. This matrix has 1 row and 4 columns, so the order of the matrix is 1 3 4 . e. This matrix has 4 rows and 2 columns, so the order of the matrix is 4 3 2 . 930 CHAPTER 10 Matrices A matrix with only one column is called a column matrix, and a matrix that has only one row is called a row matrix. Notice that in Example 1 the matrices given in parts (a) and (c) are square matrices and the matrix given in part (d) is a row matrix. You can use matrices as a shorthand way of writing systems of linear equations. There are two ways we can represent systems of linear equations with matrices: as augmented matrices or with matrix equations. In this section we will discuss augmented matrices and solve systems of linear equations using two methods: Gaussian elimination with back-substitution and Gauss–Jordan elimination. 10.1.2 Augmented Matrices A coefficient matrix is a matrix whose elements are the coefficients of a system of linear equations. A particular type of matrix that is used in representing a system of linear equations is an augmented matrix. It resembles a coefficient matrix with an additional vertical line and column of numbers, hence the name augmented. The following table illustrates examples of augmented matrices that represent systems of linear equations. SYSTEM OF LINEAR EQUATIONS AUGMENTED MATRIX 3x 1 4y 5 1 x 2 2y 5 7 c3 4 1 22 1 7d x 2 y 1 z 5 2 2x 1 2y 2 3z 5 23 x 1 y 1 z 5 6 £ 1 21 1 2 2 23 1 1 1 † 2 23 6 § x 1 y 1 z 5 0 3x 1 2y 2 z 5 2 c1 1 1 3 2 21 0 2d Note the following: ■ ■Each row represents an equation. ■ ■The vertical line represents the equal sign. ■ ■The first column represents the coefficients of the variable x. ■ ■The second column represents the coefficients of the variable y. ■ ■The third column (in the second and third systems) represents the coefficients of the variable z. ■ ■The coefficients of the variables are on the left of the equal sign (vertical line), and the constants are on the right. ■ ■Any variable that does not appear in an equation has an implied coefficient of 0. 10.1.2 SKILL Write a system of linear equations as an augmented matrix. 10.1.2 CO NC EPTUAL Visualize an augmented matrix as a system of linear equations. EXAMPLE 2 Writing a System of Linear Equations as an Augmented Matrix Write each system of linear equations as an augmented matrix. a. 2x 2 y 5 5 b. 3x 2 2y 1 4z 5 5 c. x1 2 x2 1 2x3 2 3 5 0 2x 1 2y 5 3 y 2 3z 5 22 x1 1 x2 2 3x3 1 5 5 0 7x 2 z 5 1 x1 2 x2 1 x3 2 2 5 0 [CONCEPT CHECK] TRUE OR FALSE The entries to the right of the solid vertical line in an augmented matrix correspond to the constants in the system of linear equations. ANSWER True ▼ 10.1.3 Row Operations on a Matrix Row operations on a matrix are used to solve a system of linear equations when the system is written as an augmented matrix. Recall from the elimination method in Chapter 9 that we could interchange equations, multiply an entire equation by a nonzero constant, and add a multiple of one equation to another equation to produce equivalent systems. Because each row in a matrix represents an equation, the operations that produced equivalent systems of equations that were used in the elimination method will also produce equivalent augmented matrices. Solution: a. c 2 21 21 2 5 3d b. Note that all missing terms have a 0 coefficient. 3x 2 2y 1 4z 5 5 0x 1 y 2 3z 5 22 7x 1 0y 2 z 5 1 3 22 4 £ 0 1 23 7 0 21 † 5 22 1 § c. Write the constants on the right side of the x1 2 x2 1 2x3 5 3 vertical line in the matrix. x1 1 x2 2 3x3 5 25 x1 2 x2 1 x3 5 2 1 21 2 £ 1 1 23 1 21 1 † 3 25 2 § Y OUR T UR N Write each system of linear equations as an augmented matrix. a. 2x 1 y 2 3 5 0 b. y 2 x 1 z 5 7 x 2 y 5 5 x 2 y 2 z 5 2 z 2 y 5 21 ▼ ▼ A N S W E R a. c2 1 1 21 3 5d b. £ 21 1 1 1 21 21 0 21 1 † 7 2 21 § ROW OPERATIONS The following operations on an augmented matrix will yield an equivalent matrix: 1. Interchange any two rows. 2. Multiply a row by a nonzero constant. 3. Add a multiple of one row to another row. The following symbols describe these row operations: 1. Ri 4 Rj Interchange row i with row j. 2. cRi S Ri Multiply row i by the constant c. 3. cRi 1 Rj S Rj Multiply row i by the constant c and add to row j, writing the results in row j. 10.1.3 S K I L L Perform row operations on an augmented matrix. 10.1.3 C ON C E P T U A L Understand that row operations correspond to generating equiva-lent systems of linear equations. STUDY TIP Each missing term in the system of linear equations is represented with a zero in the augmented matrix. 10.1 Matrices and Systems of Linear Equations 931 932 CHAPTER 10 Matrices 10.1.4 Row–Echelon Form of a Matrix We can solve systems of linear equations using augmented matrices with two procedures: Gaussian elimination with back-substitution, which uses row operations to transform a matrix into row–echelon form, and Gauss–Jordan elimination, which uses row operations to transform a matrix into reduced row–echelon form. EXAMPLE 3 Applying a Row Operation to an Augmented Matrix For each matrix, perform the given operation. a. c2 21 0 2 3 1d R1 4 R2 b. £ 21 0 1 3 21 2 0 1 3 † 22 3 1 § 2R3 S R3 c. c1 2 0 2 0 1 2 3 2 5d R1 2 2R2 S R1 Solution: a. Interchange the first row with c2 21 0 2 3 1d R1 4 R2 c0 2 2 21 1 3d the second row. b. Multiply the third row by 2. £ 21 0 1 3 21 2 0 1 3 † 22 3 1 § 2R3 S R3 £ 21 0 1 3 21 2 0 2 6 † 22 3 2 § c. From row 1 subtract 2 times row 2, and write the answer in row 1. Note that finding row 1 minus 2 times row 2 is the same as adding row 1 to the product of 22 with row 2. R1 2 2R2 S R1 c1 2 2102 2 2 2112 0 2 2122 2 2 2132 0 1 2 3 2 2 2152 5 d c1 0 24 24 0 1 2 3 28 5 d Y OUR TU R N Perform the operation R1 1 2R3 S R1 on the matrix. 1 0 22 £ 0 1 2 0 0 1 † 23 3 2 § ▼ [CONCEPT CHECK] The notation R1 1 2R3 S R1 corresponds to what row operations? ANSWER Add row 1 to two times row 3 and put the answer in row 1. ▼ ▼ A N S W E R £ 1 0 0 0 1 2 0 0 1 † 1 3 2 § 10.1.4 SKILL Write a matrix in reduced row–echelon form. 10.1.4 CO NC EPTUAL Understand the difference between row–echelon and reduced row–echelon forms. [CONCEPT CHECK] TRUE OR FALSE A matrix that does not satisfy row–echelon form does not satisfy reduced row–echelon form. ANSWER True ▼ Row–Echelon Form A matrix is in row–echelon form if it has all three of the following properties: 1. Any rows consisting entirely of 0s are at the bottom of the matrix. 2. For each row that does not consist entirely of 0s, the first (leftmost) nonzero entry is 1 (called the leading 1). 3. For two successive nonzero rows, the leading 1 in the higher row is farther to the left than the leading 1 in the lower row. Reduced Row–Echelon Form If a matrix in row–echelon form has the following additional property, then the matrix is in reduced row–echelon form: 4. Every column containing a leading 1 has zeros in every position above and below the leading 1. 10.1.5 Gaussian Elimination with Back-Substitution Gaussian elimination with back-substitution is a method that uses row operations to transform an augmented matrix into row–echelon form and then uses back-substitution to find the solution to the system of linear equations. EXAMPLE 4 Determining Whether a Matrix Is in Row–Echelon Form Determine whether each matrix is in row–echelon form. If it is in row–echelon form, determine whether it is in reduced row–echelon form. a. 1 3 2 £ 0 1 4 0 0 1 † 3 2 21 § b. £ 1 3 2 0 1 1 0 0 0 † 3 3 0 § c. c1 0 3 0 1 21 2 5d d. c1 0 0 3 1 1d e. £ 1 0 0 0 1 0 0 0 1 † 3 5 7 § f. £ 1 3 2 0 0 1 0 1 0 † 3 2 23 § Solution: The matrices in (a), (b), (c), and (e) are in row–echelon form. The matrix in (d) is not in row–echelon form, by condition 2; the leading nonzero entry is not a 1 in each row. If the “3” were a “1,” the matrix would be in row–echelon form. The matrix in (f ) is not in row–echelon form because of condition 3; the leading 1 in row 2 is not to the left of the leading 1 in row 3. The matrices in (c) and (e) are in reduced row–echelon form because in the columns containing the leading 1s there are zeros in every position (above and below the leading 1). GAUSSIAN ELIMINATION WITH BACK-SUBSTITUTION Step 1: Write the system of linear equations as an augmented matrix. Step 2: Use row operations to rewrite the augmented matrix in row–echelon form. Step 3: Write the system of linear equations that corresponds to the matrix in row–echelon form found in Step 2. Step 4: Use the system of linear equations found in Step 3 together with back-substitution to find the solution of the system. The order in which we perform row operations is important. You should move from left to right. Here is an example of Step 2 in the procedure. 10.1.5 S K I L L Solve systems of linear equations using Gaussian elimination with back-substitution. 10.1.5 C ON C E P T U A L Understand that solving systems with augmented matrices is equivalent to solving by the method of elimination. STUDY TIP For row–echelon form, get 1s along the main diagonal and 0s below these 1s. £ 1 † § S £ 1 0 0 † § S £ 1 0 1 0 † § S £ 1 0 1 0 0 † § S £ 1 0 1 0 0 1 † § Matrices are not typically used for systems of linear equations in two variables because the methods from Chapter 9 (substitution and elimination) are more efficient. Example 5 illustrates this procedure with a simple system of linear equations in two variables. 10.1 Matrices and Systems of Linear Equations 933 934 CHAPTER 10 Matrices EXAMPLE 5 Using Gaussian Elimination with Back-Substitution to Solve a System of Linear Equations in Two Variables Apply Gaussian elimination with back-substitution to solve the system of linear equations. 2x 1 y 5 28 x 1 3y 5 6 Solution: STEP 1 Write the system of linear equations as an augmented matrix. c2 1 1 3 28 6d STEP 2 Use row operations to rewrite the matrix in row–echelon form. Get a 1 in the top left. Interchange rows 1 and 2. c2 1 1 3 28 6d R1 4 R2 c1 3 2 1 6 28d Get a 0 below the leading 1 in row 1. c1 3 2 1 6 28d R2 2 2R1 S R2 c1 3 0 25 6 220d Get a leading 1 in row 2. Make the “25” a “1” by dividing by 25. Dividing by 25 is the same as multiplying by its reciprocal 21 5. c1 3 0 25 6 220d 2 1 5R2 S R2 c1 3 0 1 6 4d The resulting matrix is in row–echelon form. STEP 3 Write the system of linear equations corresponding to the row–echelon form of the matrix resulting in Step 2. c1 3 0 1 6 4d S x 1 3y 5 6 y 5 4 STEP 4 Use back-substitution to find the solution to the system. Let y 5 4 in the first equation x 1 3y 5 6. x 1 3142 5 6 Solve for x. x 5 26 The solution to the system of linear equations is x 5 26, y 5 4 . [CONCEPT CHECK] Solve the system of linear equations in Example 5 using the elimination method. ANSWER x 5 26, y 5 4 ▼ EXAMPLE 6 Using Gaussian Elimination with Back-Substitution to Solve a System of Linear Equations in Three Variables Use Gaussian elimination with back-substitution to solve the system of linear equations. 2x 1 y 1 8z 5 21 x 2 y 1 z 5 22 3x 2 2y 2 2z 5 2 Solution: STEP 1 Write the system of linear equations £ 2 1 8 1 21 1 3 22 22 † 21 22 2 § as an augmented matrix. 10.1.6 Gauss–Jordan Elimination In Gaussian elimination with back-substitution, we used row operations to rewrite the matrix in an equivalent row–echelon form. If we continue using row operations until the matrix is in reduced row–echelon form, this eliminates the need for back-substitution, and we call this process Gauss–Jordan elimination. STEP 2 Use row operations to rewrite the matrix in row–echelon form. Get a 1 in the top left. Interchange rows 1 and 2. R1 4 R2 £ 1 21 1 2 1 8 3 22 22 † 22 21 2 § Get 0s below the leading 1 in row 1. R2 2 2R1 S R2 £ 1 21 1 0 3 6 3 22 22 † 22 3 2 § R3 2 3R1 S R3 £ 1 21 1 0 3 6 0 1 25 † 22 3 8 § Get a leading 1 in row 2. Make the “3” a “1” by dividing by 3. 1 3R2 S R2 £ 1 21 1 0 1 2 0 1 25 † 22 1 8 § Get a zero below the leading 1 in row 2. R3 2 R2 S R3 £ 1 21 1 0 1 2 0 0 27 † 22 1 7 § Get a leading 1 in row 3. Make the “27” a “1” by dividing by 27. 2 1 7R3 S R3 £ 1 21 1 0 1 2 0 0 1 † 22 1 21 § STEP 3 Write the system of linear equations corresponding to the row–echelon form of the matrix resulting in Step 2. x 2 y 1 z 5 22 y 1 2z 5 1 z 5 21 STEP 4 Use back-substitution to find the solution to the system. Let z 5 21 in the second equation y 1 2z 5 1. y 1 21212 5 1 Solve for y. y 5 3 Let y 5 3 and z 5 21 in the first equation x 2 y 1 z 5 22. x 2 132 1 1212 5 22 Solve for x. x 5 2 The solution to the system of linear equations is x 5 2, y 5 3, and z 5 21 . Y OUR T UR N Use Gaussian elimination with back-substitution to solve the system of linear equations. x 1 y 2 z 5 0 2x 1 y 1 z 5 1 2x 2 y 1 3z 5 21 ▼ ▼ A N S W E R x 5 21, y 5 2, and z 5 1 10.1.6 S KI L L Solve systems of linear equations using Gauss-Jordan elimination. 10.1.6 C O N C E P T U A L Understand that when you move down the augmented matrix you use rows above and on the way back up you use rows below. 10.1 Matrices and Systems of Linear Equations 935 936 CHAPTER 10 Matrices The order in which we perform row operations is important. You should move from left to right. Think of this process as climbing down a set of stairs first and then back up the stairs second. On the way down the stairs always use operations with rows above where you currently are, and on the way back up the stairs always use rows below where you currently are. GAUSS–JORDAN ELIMINATION Step 1: Write the system of linear equations as an augmented matrix. Step 2: Use row operations to rewrite the augmented matrix in reduced row–echelon form. Step 3: Write the system of linear equations that corresponds to the matrix in reduced row–echelon form found in Step 2. The result is the solution to the system. Down the stairs: £ 1 † § S £ 1 0 0 † § S £ 1 0 1 0 † § S £ 1 0 1 0 0 † § S £ 1 0 1 0 0 1 † § Up the stairs: £ 1 0 1 0 0 1 † § S £ 1 0 1 0 0 0 1 † § S £ 1 0 0 1 0 0 0 1 † § S £ 1 0 0 0 1 0 0 0 1 † § EXAMPLE 7 Using Gauss–Jordan Elimination to Solve a System of Linear Equations in Three Variables Apply Gauss–Jordan elimination to solve the system of linear equations. x 2 y 1 2z 5 21 3x 1 2y 2 6z 5 1 2x 1 3y 1 4z 5 8 Solution: STEP 1 Write the system as an augmented matrix. £ 1 21 2 3 2 26 2 3 4 † 21 1 8 § STEP 2 Utilize row operations to rewrite the matrix in reduced row–echelon form. There is already a 1 in the first row/first column. Get 0s below the leading 1 in row 1. R2 2 3R1 S R2 R3 2 2R1 S R3 £ 1 21 2 0 5 212 0 5 0 † 21 4 10 § [CONCEPT CHECK] TRUE OR FALSE Going down the stairs corresponds to transforming the augmented matrix to row–echelon form, and going up the stairs corresponds to transforming the augmented matrix to reduced row–echelon form. ANSWER True ▼ STUDY TIP For reduced row–echelon form, get 1s along the main diagonal and 0s above and below these 1s. Get a 1 in row 2/column 2. R2 4 R3 £ 1 21 2 0 5 0 0 5 212 † 21 10 4 § 1 5R2 S R2 £ 1 21 2 0 1 0 0 5 212 † 21 2 4 § Get a 0 in row 3/column 2. R3 2 5R2 S R3 £ 1 21 2 0 1 0 0 0 212 † 21 2 26 § Get a 1 in row 3/column 3. 2 1 12R3 S R3 £ 1 21 2 0 1 0 0 0 1 † 21 2 1 2 § Now, go back up the stairs. Get 0s above the 1 in row 3/column 3. R1 2 2R3 S R1 £ 1 21 0 0 1 0 0 0 1 † 22 2 1 2 § Get a 0 in row 1/column 2. R1 1 R2 S R1 £ 1 0 0 0 1 0 0 0 1 † 0 2 1 2 § STEP 3 Identify the solution. x 5 0, y 5 2, z 5 1 2 Y OUR T UR N Use an augmented matrix and Gaussian elimination to solve the system of equations. x 1 y 2 z 5 22 3x 1 y 2 z 5 24 2x 2 2y 1 3z 5 3 ▼ ▼ A N S W E R x 5 21, y 5 2, and z 5 3 EXAMPLE 8 Solving a System of Linear Equations in Four Variables Solve the system of equations with Gauss–Jordan elimination. x1 1 x2 2 x3 1 3x4 5 3 3x2 2 2x4 5 4 2x1 2 3x3 5 21 4x4 1 2x1 5 26 Solution: STEP 1 Write the system as an augmented matrix. D 1 1 21 3 0 3 0 22 2 0 23 0 2 0 0 4 4 3 4 21 26 T STEP 2 Use row operations to rewrite the matrix in reduced row–echelon form. There is already a 1 in the first row/first column. Get 0s below the 1 in row 1/column 1. R3 2 2R1 S R3 R4 2 2R1 S R4 D 1 1 21 3 0 3 0 22 0 22 21 26 0 22 2 22 4 3 4 27 212 T STUDY TIP Careful attention should be paid to order of terms, and zeros should be used for missing terms. 10.1 Matrices and Systems of Linear Equations 937 938 CHAPTER 10 Matrices Get a 1 in row 2/column 2. R2 4 R4 D 1 1 21 3 0 22 2 22 0 22 21 26 0 3 0 22 4 3 212 27 4 T 2 1 2R2 4 R2 D 1 1 21 3 0 1 21 1 0 22 21 26 0 3 0 22 4 3 6 27 4 T Get 0s below the 1 in row 2/column 2. R3 1 2R2 S R3 R4 2 3R2 S R4 D 1 1 21 3 0 1 21 1 0 0 23 24 0 0 3 25 4 3 6 5 214 T Get a 1 in row 3/column 3. 2 1 3R3 S R3 D 1 1 21 3 0 1 21 1 0 0 1 4 3 0 0 3 25 4 3 6 25 3 214 T Get a 0 in row 4/column 3. R4 2 3R3 S R4 D 1 1 21 3 0 1 21 1 0 0 1 4 3 0 0 0 29 4 3 6 25 3 29 T Get a 1 in row 4/column 4. 2 1 9R4 S R4 D 1 1 21 3 0 1 21 1 0 0 1 4 3 0 0 0 1 4 3 6 25 3 1 T Now go back up the stairs. Get 0s above the 1 in row 4/column 4. R3 2 4 3R4 S R3 R2 2 R4 S R2 R1 2 3R4 S R1 D 1 1 21 0 0 1 21 0 0 0 1 0 0 0 0 1 4 0 5 23 1 T Get 0s above the 1 in row 3/column 3. R2 1 R3 S R2 R1 1 R3 S R1 D 1 1 0 0 0 1 0 0 0 0 1 0 0 0 0 1 4 23 2 23 1 T Get a 0 in row 1/column 2. R1 2 R2 S R1 D 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 4 25 2 23 1 T STEP 3 Identify the solution. x1 5 25, x2 5 2, x3 5 23, x4 5 1 10.1.7 Inconsistent and Dependent Systems Recall that systems of linear equations can be independent, inconsistent, or dependent systems and therefore have one solution, no solution, or infinitely many solutions. All of the systems we have solved so far in this section have been independent systems (unique solution). When solving a system of linear equations using Gaussian elimination or Gauss–Jordan elimination, the following will indicate the three possible types of solutions. 10.1.7 S K I L L Recognize matrices that correspond to inconsistent and dependent systems. 10.1.7 C ON C E P T U A L Understand the difference between matrices that correspond to inconsistent or dependent systems. SYSTEM TYPE OF SOLUTION MATRIX DURING GAUSS–JORDAN ELIMINATION EXAMPLE Independent One (unique) solution Diagonal entries are all 1s and 0s elsewhere. £ 1 0 0 0 1 0 0 0 1 † 1 23 2 § or x 5 1 y 5 23 z 5 2 Inconsistent No solution One row will have only zero entries for coefficients and a nonzero entry for the constant. £ 1 0 0 0 1 0 0 0 0 † 1 23 2 § or x 5 1 y 5 23 0 5 2 Dependent Infinitely many solutions One row will be entirely 0s when the number of equations equals the number of variables. £ 1 0 22 0 1 1 0 0 0 † 1 23 0 § or x 2 2z 5 1 y 1 z 5 23 0 5 0 EXAMPLE 9 Solving an Inconsistent System: No Solution Solve the system of equations. x 1 2y 2 z 5 3 2x 1 y 1 2z 5 21 22x 2 4y 1 2z 5 5 Solution: STEP 1 Write the system of equations as an augmented matrix. £ 1 2 21 2 1 2 22 24 2 † 3 21 5 § STEP 2 Apply row operations to rewrite the matrix in row–echelon form. Get 0s below the 1 in column 1. R2 2 2R1 S R2 R3 1 2R1 S R3 £ 1 2 21 0 23 4 0 0 0 † 3 27 11 § There is no need to continue because 0x 1 0y 1 0z 5 11 or 0 5 11 row 3 is a contradiction. Since this is inconsistent, there is no solution to this system of equations. [CONCEPT CHECK] TRUE OR FALSE An inconsistent system of linear equations in three variables corresponds to three planes that do not have a common point or line of intersection. The dependent system of linear equations in three variables corresponds to three planes that intersect along a line. ANSWER True ▼ 10.1 Matrices and Systems of Linear Equations 939 940 CHAPTER 10 Matrices A common mistake that is made is to identify a unique solution as no solution when one of the variables is equal to zero. For example, what is the difference between the following two matrices? £ 1 0 2 0 1 3 0 0 3 † 1 2 0 § and £ 1 0 2 0 1 3 0 0 0 † 1 2 3 § The first matrix has a unique solution, whereas the second matrix has no solution. The third row of the first matrix corresponds to the equation 3z 5 0, which implies that z 5 0. The third row of the second matrix corresponds to the equation 0x 1 0y 1 0z 5 3 or 0 5 3, which is inconsistent, and therefore the system has no solution. EXAMPLE 10 Solving a Dependent System: Infinitely Many Solutions Solve the system of equations. x 1 z 5 3 2x 1 y 1 4z 5 8 3x 1 y 1 5z 5 11 Solution: STEP 1 Write the system of equations as an augmented matrix.£ 1 0 1 2 1 4 3 1 5 † 3 8 11 § STEP 2 Use row operations to rewrite the matrix in reduced row–echelon form. Get the 0s below the 1 in column 1. R2 2 2R1 S R2 R3 2 3R1 S R3 £ 1 0 1 0 1 2 0 1 2 † 3 2 2 § Get a 0 in row 3/column 2. R3 2 R2 S R3 £ 1 0 1 0 1 2 0 0 0 † 3 2 0 § This matrix is in reduced row–echelon form. This matrix corresponds to a dependent system of linear equations and has infinitely many solutions. STEP 3 Write the augmented matrix as a system of linear equations. x 1 z 5 3 y 1 2z 5 2 Let z 5 a, where a is any real number, and substitute this into the two equations. We find that x 5 3 2 a and y 5 2 2 2a. The general solution is x 5 3 2 a, y 5 2 2 2a, z 5 a for a any real number. Note that 12, 0, 12 and 13, 2, 02 are particular solutions when a 5 1 and a 5 0, respectively. STUDY TIP In a system with three variables, say x, y, and z, we typically let z 5 a (where a is called a parameter) and then solve for x and y in terms of a. Applications The following table gives nutritional information for sandwiches. SANDWICH CALORIES FAT (g) CARBOHYDRATES (g) PROTEIN (g) Veggie 350 18 17 36 Chicken Salad 430 19 46 20 Ham & Cheese 290 5 45 19 ▼ A N S W E R x 5 3a 1 2, y 5 24a 2 2, z 5 a, where a is any real number. EXAMPLE 11 Solving a Dependent System: Infinitely Many Solutions Solve the system of linear equations. 2x 1 y 1 z 5 8 x 1 y 2 z 5 23 Solution: STEP 1 Write the system of equations as an augmented matrix. c2 1 1 1 1 21 8 23d STEP 2 Use row operations to rewrite the matrix in reduced row–echelon form. Get a 1 in row 1/column 1. R1 4 R2 c1 1 21 2 1 1 23 8d Get a 0 in row 2/column 1. R2 2 2R1 S R2 c1 1 21 0 21 3 23 14d Get a 1 in row 2/column 1. 2R2 S R2 c1 1 21 0 1 23 23 214d Get a 0 in row 1/column 2. R1 2 R2 S R1 c1 0 2 0 1 23 11 214d This matrix is in reduced row–echelon form. STEP 3 Identify the solution. x 1 2z 5 11 y 2 3z 5 214 Let z 5 a, where a is any real number. Substituting z 5 a into these two equations gives the infinitely many solutions x 5 11 2 2a, y 5 3a 2 14, z 5 a . Y OUR T UR N Solve the system of equations using an augmented matrix. x 1 y 1 z 5 0 3x 1 2y 2 z 5 2 ▼ 10.1 Matrices and Systems of Linear Equations 941 942 CHAPTER 10 Matrices EXAMPLE 12 Meal Nutrition Suppose you are going to eat only sandwiches for a week (7 days) for both lunch and dinner (total of 14 meals). If your goal is to eat 388 grams of protein and 4900 calories in those 14 sandwiches, how many of each sandwich should you eat that week? Solution: STEP 1 Determine the system of linear equations. Let three variables represent the number of each type of sandwich you eat in a week. x 5 number of Veggie sandwiches y 5 number of Chicken Salad sandwiches z 5 number of Ham & Cheese sandwiches The total number of sandwiches eaten is 14. x 1 y 1 z 5 14 The total number of calories consumed is 4900. 350x 1 430y 1 290z 5 4900 The total number of grams of protein consumed is 388. 36x 1 20y 1 19z 5 388 Write an augmented matrix representing this system of linear equations. 1 1 1 £ 350 430 290 36 20 19 † 14 4900 388 § STEP 2 Utilize row operations to rewrite the matrix in reduced row–echelon form. R2 2 350R1 S R2 R3 2 36R1 S R3 1 1 1 £ 0 80 260 0 216 217 † 14 0 2116 § 1 80R2 S R2 1 1 1 £ 0 1 23 4 0 216 217 † 14 0 2116 § R3 1 16R2 S R3 1 1 1 £ 0 1 23 4 0 0 229 † 14 0 2116 § 2 1 29R3 S R3 1 1 1 £ 0 1 23 4 0 0 1 † 14 0 4 § R2 1 3 4R3 S R2 R1 2 R3 S R1 1 1 0 £ 0 1 0 0 0 1 † 10 3 4 § R1 2 R2 S R1 1 0 0 £ 0 1 0 0 0 1 † 7 3 4 § STEP 3 Identify the solution. x 5 7, y 5 3, z 5 4 You should eat 7 Veggie, 3 Chicken Salad, and 4 Ham & Cheese sandwiches . Many times in the real world we see a relationship that looks like a particular function such as a quadratic function and we know particular data points, but we do not know the function. We start with the general function, fit the curve to particular data points, and solve a system of linear equations to determine the specific function parameters. EXAMPLE 13 Fitting a Curve to Data The amount of money awarded in medical malpractice suits is rising. This can be modeled with a quadratic function y 5 at2 1 bt 1 c, where t . 0 and a . 0. Determine a quadratic function that passes through the three points shown on the graph. Solution: Let 2006 correspond to t 5 0 and y represent the number of dollars awarded for malpractice suits. The following data are reflected in the illustration above: YEAR t y (THOUSANDS OF DOLLARS) (t, y) 2007 1 253 11, 2532 2011 5 287 15, 2872 2016 10 431 110, 4312 Substitute the three points 11, 2532, 15, 2872, and 110, 4312 into the general quadratic equation: y 5 at2 1 bt 1 c. POINT y 5 at 2 1 bt 1 c SYSTEM OF EQUATIONS 11, 2532 253 5 a 1122 1 b 112 1 c a 1 b 1 c 5 253 15, 2872 287 5 a 1522 1 b 152 1 c 25a 1 5b 1 c 5 287 110, 4312 431 5 a 11022 1 b 1102 1 c 100a 1 10b 1 c 5 431 STEP 1 Write this system of linear equations as an augmented matrix. £ 1 1 1 25 5 1 100 10 1 † 253 287 431 § STEP 2 Apply row operations to rewrite the matrix in reduced row–echelon form. R2 2 25R1 S R2 £ 1 1 1 0 220 224 100 10 1 † 253 26,038 431 § R3 2 100R1 S R3 £ 1 1 1 0 220 224 0 290 299 † 253 26,038 224,869 § 2 1 20R2 S R2 £ 1 1 1 0 1 6 5 0 290 299 † 253 3,019 10 224,869 § Year 2016 2007 2010 2013 Median damages awarded in medical malpractice jury trials (adjusted to 2016 dollars) 200,000 100,000 300,000 400,000 $500,000 $253,000 $287,000 $431,000 10.1 Matrices and Systems of Linear Equations 943 944 CHAPTER 10 Matrices R3 1 90R2 S R3 £ 1 1 1 0 1 6 5 0 0 9 † 253 3,019 10 2,302 § 1 9R3 S R3 £ 1 1 1 0 1 6 5 0 0 1 † 253 3,019 10 2,302 9 § R2 2 6 5R3 S R2 £ 1 1 1 0 1 0 0 0 1 † 253 2 151 30 2,302 9 § R1 2 R3 S R1 £ 1 1 0 0 1 0 0 0 1 † 2 25 9 2 151 30 2,302 9 § R1 2 R2 S R1 £ 1 0 0 0 1 0 0 0 1 † 203 90 2 151 30 2,302 9 § STEP 3 Identify the solution. a 5 203 90 b 5 2151 30 c 5 2,302 9 Substituting a 5 203/90, b 5 2151/30, c 5 2302/9 into y 5 at2 1 bt 1 c, we find that the thousands of dollars spent on malpractice suits as a function of year is given by y 5 203 90 t2 2 151 30 t 1 2302 9 2006 is t 5 0 Notice that all three points lie on this curve. If a matrix in row–echelon form has the following additional property, then the matrix is in reduced row–echelon form: 4. Every column containing a leading 1 has zeros in every position above and below the leading 1. The two methods used for solving systems of linear equations represented as augmented matrices are Gaussian elimination with back-substitution and Gauss–Jordan elimination. In both cases, we represent the system of linear equations as an augmented matrix and then use row operations to rewrite in row–echelon form. With Gaussian elimination we then stop and perform back-substitution to solve the system, and with Gauss–Jordan elimination we continue with row operations until the matrix is in reduced row–echelon form and then identify the solution to the system. In this section, we used augmented matrices to represent a system of linear equations. a1x 1 b1y 1 c1z 5 d1 a2x 1 b2y 1 c2z 5 d2 a3x 1 b3y 1 c3z 5 d3 3 £ a1 b1 c1 a2 b2 c2 a3 b3 c3 † d1 d2 d3 § Any missing terms correspond to a 0 in the matrix. A matrix is in row–echelon form if it has all three of the following properties: 1. Any rows consisting entirely of 0s are at the bottom of the matrix. 2. For each row that does not consist entirely of 0s, the first (leftmost) nonzero entry is 1 (called the leading 1). 3. For two successive nonzero rows, the leading 1 in the higher row is farther to the left than the leading 1 in the lower row. [SEC TION 10.1] S U M M A RY In Exercises 1–6, determine the order of each matrix. 1. c21 3 4 2 7 9d 2. £ 0 1 3 9 7 8 § 3. C1 2 3 4D 4. D 3 7 21 10 T 5. 304 6. D 21 3 6 8 2 9 7 3 5 4 22 210 6 3 1 5 T In Exercises 7–14, write the augmented matrix for each system of linear equations. 7. 3x 2 2y 5 7 8. 2x 1 y 5 2 9. 2x 2 3y 1 4z 5 23 10. x 2 2y 1 z 5 0 24x 1 6y 5 23 x 2 y 5 24 2x 1 y 1 2z 5 1 22x 1 y 2 z 5 25 5x 2 2y 2 3z 5 7 13x 1 7y 1 5z 5 6 11. x 1 y 5 3 12. x 2 y 5 24 13. 3y 2 4x 1 5z 2 2 5 0 14. 2y 1 z 2 x 2 3 5 2 x 2 z 5 2 y 1 z 5 3 2x 2 3y 2 2z 5 23 2x 1 3z 2 2y 5 0 y 1 z 5 5 3z 1 4y 2 2x 2 1 5 0 22z 1 y 2 4x 2 3 5 0 In Exercises 15–20, write the system of linear equations represented by the augmented matrix. Utilize the variables x, y, and z. 15. c23 7 1 5 2 8d 16. £ 21 2 4 7 9 3 4 6 25 † 4 23 8 § 17. £ 21 0 0 7 9 3 4 6 25 † 4 23 8 § 18. c2 3 24 7 21 5 6 9d 19. c1 0 0 1 a bd 20. £ 3 0 5 0 24 7 2 21 0 † 1 23 8 § In Exercises 21–30, indicate whether each matrix is in row–echelon form. If it is, determine whether it is in reduced row–echelon form. 21. c1 0 1 1 3 2d 22. c0 1 1 0 3 2d 23. c1 0 21 0 1 3 23 14d 24. c1 0 0 0 1 3 23 14d 25. £ 1 0 1 0 0 0 0 1 2 † 3 0 2 § 26. £ 1 0 1 0 1 2 0 0 0 † 3 2 0 § 27. £ 1 0 0 0 1 0 0 0 1 † 3 2 5 § 28. £ 21 0 0 0 21 0 0 0 21 † 3 2 5 § 29. D 1 0 0 1 0 1 0 3 0 0 1 0 0 0 0 1 4 3 2 5 0 T 30. D 1 0 0 1 0 1 0 3 0 0 1 0 0 0 0 0 4 3 2 5 0 T In Exercises 31–40, perform the indicated row operations on each augmented matrix. 31. c1 22 2 3 23 21d R2 2 2R1 S R2 32. c2 23 1 2 24 5d R1 4 R2 33. £ 1 22 21 2 1 23 3 22 5 † 3 6 28 § R2 2 2R1 S R2 34. £ 1 22 1 0 1 22 23 0 21 † 3 6 25 § R3 1 3R1 S R3 35. D 1 22 5 21 0 3 0 21 0 22 1 22 0 0 1 21 4 2 22 5 26 T R3 1 R2 S R2 36. D 1 0 5 210 0 1 2 23 0 2 23 0 0 0 1 21 4 15 4 21 23 T R2 2 1 2R3 S R3 [SEC TION 10.1] E X E RC I S E S • S K I L L S 10.1 Matrices and Systems of Linear Equations 945 946 CHAPTER 10 Matrices 37. D 1 0 5 210 0 1 2 23 0 2 23 0 0 23 2 21 4 25 22 21 23 T R3 2 2R2 S R3 R4 1 3R2 S R4 38. D 1 0 4 0 0 1 2 0 0 0 1 0 0 0 0 1 4 1 22 0 23 T R2 2 2R3 S R2 R1 2 4R3 S R1 39. D 1 0 4 8 0 1 2 23 0 0 1 6 0 0 0 1 4 3 22 3 23 T R3 2 6R4 S R3 R2 1 3R4 S R2 R1 2 8R4 S R1 40. D 1 0 21 5 0 1 2 3 0 0 1 22 0 0 0 1 4 2 25 2 1 T R3 1 2R4 S R3 R2 2 3R4 S R2 R1 2 5R4 S R1 In Exercises 41–50, use row operations to transform each matrix to reduced row–echelon form. 41. c1 2 2 3 4 2d 42. c 1 21 23 2 3 2d 43. £ 1 21 1 0 1 21 21 1 1 † 21 21 1 § 44. £ 0 21 1 1 21 1 1 21 21 † 1 21 21 § 45. £ 3 22 23 1 21 1 2 3 5 † 21 24 14 § 46. £ 3 21 1 1 22 3 2 1 23 † 2 1 21 § 47. c2 1 26 1 22 2 4 23d 48. c23 21 2 21 22 1 21 23d 49. £ 21 2 1 3 22 1 2 24 22 † 22 4 4 § 50. £ 2 21 0 21 0 1 22 1 0 † 1 22 21 § In Exercises 51–70, solve the system of linear equations using Gaussian elimination with back-substitution. 51. 2x 1 3y 5 1 52. 3x 1 2y 5 11 x 1 y 5 22 x 2 y 5 12 53. 2x 1 2y 5 3 54. 3x 2 y 5 21 2x 2 4y 5 26 2y 1 6x 5 2 55. 2 3x 1 1 3y 5 8 9 56. 0.4x 2 0.5y 5 2.08 1 2x 1 1 4y 5 3 4 20.3x 1 0.7y 5 1.88 57. x 2 z 2 y 5 10 58. 2x 1 z 1 y 5 23 2x 2 3y 1 z 5 211 2y 2 z 1 x 5 0 y 2 x 1 z 5 210 x 1 y 1 2z 5 5 59. 3x1 1 x2 2 x3 5 1 60. 2x1 1 x2 1 x3 5 21 x1 2 x2 1 x3 5 23 x1 1 x2 2 x3 5 5 2x1 1 x2 1 x3 5 0 3x1 2 x2 2 x3 5 1 61. 2x 1 5y 5 9 62. x 2 2y 1 3z 5 1 x 1 2y 2 z 5 3 22x 1 7y 2 9z 5 4 23x 2 4y 1 7z 5 1 x 1 z 5 9 63. 2x1 2 x2 1 x3 5 3 64. x1 2 x2 2 2x3 5 0 x1 2 x2 1 x3 5 2 22x1 1 5x2 1 10x3 5 23 22x1 1 2x2 2 2x3 5 24 3x1 1 x2 5 0 65. 2x 1 y 2 z 5 2 66. 3x 1 y 2 z 5 0 x 2 y 2 z 5 6 x 1 y 1 7z 5 4 67. 2y 1 z 5 3 68. 22x 2 y 1 2z 5 3 4x 2 z 5 23 3x 2 4z 5 2 7x 2 3y 2 3z 5 2 2x 1 y 5 21 x 2 y 2 z 5 22 2x 1 y 2 z 5 28 69. 3x1 2 2x2 1 x3 1 2x4 5 22 70. 5x1 1 3x2 1 8x3 1 x4 5 1 2x1 1 3x2 1 4x3 1 3x4 5 4 x1 1 2x2 1 5x3 1 2x4 5 3 x1 1 x2 1 x3 1 x4 5 0 4x1 1 x3 2 2x4 5 23 5x1 1 3x2 1 x3 1 2x4 5 21 x2 1 x3 1 x4 5 0 In Exercises 71–86, solve the system of linear equations using Gauss–Jordan elimination. 71. x 1 3y 5 25 72. 5x 2 4y 5 31 73. x 1 y 5 4 74. 3x 2 4y 5 12 22x 2 y 5 0 3x 1 7y 5 219 23x 2 3y 5 10 26x 1 8y 5 224 75. x 2 2y 1 3z 5 5 76. x 1 2y 2 z 5 6 77. x 1 y 1 z 5 3 78. x 2 2y 1 4z 5 2 3x 1 6y 2 4z 5 212 2x 2 y 1 3z 5 213 x 2 z 5 1 2x 2 3y 2 2z 5 23 2x 2 4y 1 6z 5 16 3x 2 2y 1 3z 5 216 y 2 z 5 24 1 2 x 1 1 4 y 1 z 522 79. x 1 2y 1 z 5 3 80. x 1 2y 1 z 5 3 81. 3x 2 y 1 z 5 8 82. x 2 2y 1 3z 5 10 2x 2 y 1 3z 5 7 2x 2 y 1 3z 5 7 x 1 y 2 2z 5 4 23x 1 z 5 9 3x 1 y 1 4z 5 5 3x 1 y 1 4z 5 10 83. 4x 2 2y 1 5z 5 20 84. y 1 z 5 4 85. x 2 y 2 z 2 w 5 1 86. x 2 3y 1 3z 2 2w 5 4 x 1 3y 2 2z 5 6 x 1 y 5 8 2x 1 y 1 z 1 2w 5 3 x 1 2y 2 z 5 23 x 2 2y 2 2z 2 3w 5 0 x 1 3z 1 2w 5 3 3x 2 4y 1 z 1 5w 5 23 y 1 z 1 5w 5 6 • A P P L I C A T I O N S 87. Football. In Super Bowl XXXVIII, the New England Patriots defeated the Carolina Panthers 32–29. The points came from a total of four types of plays: touchdowns (6 points), extra points (1 point), two-point conversions (2 points), and field goals (3 points). There were a total of 16 scoring plays. There were four times as many touchdowns as field goals, and there were five times as many extra points as 2-point conversions. How many touchdowns, extra points, 2-point conversions, and field goals were scored in Super Bowl XXXVIII? 88. Basketball. In the 2004 Summer Olympics in Athens, Greece, the U.S. men’s basketball team, consisting of NBA superstars, was defeated by the Puerto Rican team 92–73. The points came from three types of scoring plays: 2-point shots, 3-point shots, and 1-point free throws. There were six more 2-point shots made than there were 1-point free throws. The number of successful 2-point shots was three less than four times the number of successful 3-point shots. How many 2-point and 3-point shots, and 1-point free throws were made in that Olympic competition? Exercises 89 and 90 rely on a selection of sandwiches whose nutrition information is given in the following table. Suppose you are going to eat only sandwiches for a week (7 days) for lunch and dinner (a total of 14 meals). SANDWICH CALORIES FAT (g) CARBOHYDRATES (g) PROTEIN (g) Egg Salad 350 18 17 36 Club 430 19 46 20 Turkey 290 5 45 19 Roast Beef 430 27 20 34 89. Diet. Your goal is a low-fat diet consisting of 526 grams of carbohydrates, 168 grams of fat, and 332 grams of protein. How many of each sandwich would you eat that week to obtain this goal? 90. Diet. Your goal is a low-carb diet consisting of 5180 calories, 335 grams of carbohydrates, and 263 grams of fat. How many of each sandwich would you eat that week to obtain this goal? Exercises 91 and 92 involve vertical motion and the effect of gravity on an object. Because of gravity, an object that is projected upward will eventually reach a maximum height and then fall to the ground. The equation that relates the height h of a projectile t seconds after it is shot upward is given by h 5 1 2 at2 1 v0t 1 h0 where a is the acceleration due to gravity, h0 is the initial height of the object at time t 5 0, and v0 is the initial velocity of the object at time t 5 0. Note that a projectile follows the path of a parabola opening down, so a , 0. 91. Vertical Motion. An object is thrown upward, and the following table depicts the height of the ball t seconds after the projectile is released. Find the initial height, initial velocity, and acceleration due to gravity. t (SECONDS) HEIGHT (FEET) 1 34 2 36 3 6 92. Vertical Motion. An object is thrown upward, and the following table depicts the height of the ball t seconds after the projectile is released. Find the initial height, initial velocity, and acceleration due to gravity. t (SECONDS) HEIGHT (FEET) 1 54 2 66 3 46 93. Data Curve-Fitting. The average number of minutes that a person spends driving a car can be modeled by a quadratic function y 5 ax2 1 bx 1 c, where a , 0 and 15 , x , 65. The following table gives the average number of minutes 10.1 Matrices and Systems of Linear Equations 947 948 CHAPTER 10 Matrices a day that a person spends driving a car. Determine the quadratic function that models this quantity. AGE AVERAGE DAILY MINUTES DRIVING 16 25 40 64 65 40 94. Data Curve-Fitting. The average age when a woman gets married has been increasing during the last century. In 1920 the average age was 18.4, in 1960 the average age was 20.3, and in 2002 the average age was 25.30. Find a quadratic function y 5 ax2 1 bx 1 c, where a . 0 and 18 , x , 35, that models the average age y when a woman gets married as a function of the year x 1x 5 0 corresponds to 19202. What will the average age be in 2020? 95. Chemistry/Pharmacy. A pharmacy receives an order for 100 milliliters of 5% hydrogen peroxide solution. The pharmacy has a 1.5% and a 30% solution on hand. A technician will mix the 1.5% and 30% solutions to make the 5% solution. How much of the 1.5% and 30% solutions, respectively, will be needed to fill this order? Round to the nearest ml. 96. Chemistry/Pharmacy. A pharmacy receives an order for 60 grams of a 0.7% hydrocortisone cream. The pharmacy has 1% and 0.5% hydrocortisone creams as well as a Eucerin cream for use as a base (0% hydrocortisone). The technician must use twice as much 0.5% hydrocortisone cream than the Eucerin base. How much of the 1% and 0.5% hydrocortisone creams and Eucerin cream are needed to fill this order? 97. Business. A small company has an assembly line that produces three types of widgets. The basic widget is sold for $12 per unit, the midprice widget for $15 per unit, and the top-of-the-line widget for $18 per unit. The assembly line has a daily capacity of producing 375 widgets that may be sold for a total of $5250. Find the quantity of each type of widget produced on a day when twice as many basic widgets as midprice widgets are produced. 98. Business. A small company has an assembly line that produces three types of widgets. The basic widget is sold for $10 per unit, the midprice widget for $12 per unit, and the top-of-the-line widget for $15 per unit. The assembly line has a daily capacity of producing 350 widgets that may be sold for a total of $4600. Find the quantity of each type of widget produced on a day when twice as many top-of-the-line widgets as basic widgets are produced. 99. Money. Gary and Ginger decide to place $10,000 of their savings into investments. They put some in a money market account earning 3% interest, some in a mutual fund that has been averaging 7% a year, and some in a stock that rose 10% last year. If they put $3000 more in the money market than in the mutual fund and the mutual fund and stocks have the same growth in the next year as they did in the previous year, they will earn $540 in a year. How much money did they put in each of the three investments? 100. Money. Ginger talks Gary into putting less money in the money market and more money in the stock. They place $10,000 of their savings into investments. They put some in a money market account earning 3% interest, some in a mutual fund that has been averaging 7% a year, and some in a stock that rose 10% last year. If they put $3000 more in the stock than in the mutual fund and the mutual fund and stock have the same growth in the next year as they did in the previous year, they will earn $840 in a year. How much money did they put in each of the three investments? 101. Manufacturing. A company produces three products x, y, and z. Each item of product x requires 20 units of steel, 2 units of plastic, and 1 unit of glass. Each item of product y requires 25 units of steel, 5 units of plastic, and no units of glass. Each item of product z requires 150 units of steel, 10 units of plastic, and 0.5 units of glass. The available amounts of steel, plastic, and glass are 2400, 310, and 28, respectively. How many items of each type can the company produce and utilize all the available raw materials? 102. Geometry. Find the values of a, b, and c such that the graph of the quadratic function y 5 ax2 1 bx 1 c passes through the points 11, 52, 122, 2102, and 10, 42. 103. Ticket Sales. One hundred students decide to buy tickets to a football game. There are three types of tickets: general admission, reserved, and end zone. Each general admission ticket costs $20, each reserved ticket costs $40, and each end zone ticket costs $15. The students spend a total of $2375 for all the tickets. There are 5 more reserved tickets than general admission tickets and 20 more end zone tickets than general admission tickets. How many of each type of ticket were purchased by the students? 104. Exercise and Nutrition. Ann would like to exercise one hour per day to burn calories and lose weight. She would like to engage in three activities: walking, step-up exercise, and weight training. She knows she can burn 85 calories walking at a certain pace in 15 minutes, 45 calories doing the step-up exercise in 10 minutes, and 137 calories by weight training for 20 minutes. (a) Determine the number of calories per minute she can burn doing each activity. (b) Suppose she has time to exercise for only one hour (60 minutes). She sets a goal of burning 358 calories in one hour and would like to weight train twice as long as walking. How many minutes must she engage in each exercise to burn the required number of calories in one hour? 105. Geometry. The circle given by the equation x2 1 y2 1 ax 1 by 1 c 5 0 passes through the point 14, 42, 123, 212, and 11, 232. Find a, b, and c. 106. Geometry. The circle given by the equation x2 1 y2 1 ax 1 by 1 c 5 0 passes through the point 10, 72, 16, 12, and 15, 42. Find a, b, and c. • C A T C H T H E M I S T A K E In Exercises 107–110, explain the mistake that is made. 107. Solve the system of equations using the augmented matrices. y 2 x 1 z 5 2 x 2 2z 1 y 5 23 x 1 y 1 z 5 6 Solution: Step 1: Write as an augmented matrix. £ 1 21 1 1 22 1 1 1 1 † 2 23 6 § Step 2: Reduce the matrix using Gaussian elimination. £ 1 21 1 0 1 0 0 0 0 † 2 5 26 § Step 3: Identify the solution. Row 3 is inconsistent, so there is no solution. This is incorrect. The correct answer is x 5 1, y 5 2, z 5 3. What mistake was made? 108. Perform the indicated row operations on the matrix. £ 1 21 1 2 23 1 3 1 2 † 2 4 26 § a. R2 2 2R1 S R2 b. R3 2 3R1 S R3 Solution: a. £ 1 21 1 0 23 1 3 1 2 † 2 4 26 § b. £ 1 21 1 2 23 1 0 1 2 † 2 4 26 § This is incorrect. What mistake was made? 109. Solve the system of equations using an augmented matrix. 3x 2 2y 1 z 5 21 x 1 y 2 z 5 3 2x 2 y 1 3z 5 0 Solution: Step 1: Write the system as an augmented matrix. £ 3 22 1 1 1 21 2 21 3 † 21 3 0 § Step 2: Reduce the matrix using Gaussian elimination. £ 1 0 0 0 1 0 0 0 1 † 1 2 0 § Step 3: Identify the answer. Row 3 is inconsistent 1 5 0, therefore there is no solution. This is incorrect. What mistake was made? 110. Solve the system of equations using an augmented matrix. x 1 3y 1 2z 5 4 3x 1 10y 1 9z 5 17 2x 1 7y 1 7z 5 17 Solution: Step 1: Write the system as an augmented matrix. £ 1 3 2 3 10 9 2 7 7 † 4 17 17 § Step 2: Reduce the matrix using Gaussian elimination. £ 1 0 27 0 1 3 0 0 0 † 211 5 4 § Step 3: Identify the answer: x 5 7t 2 11 Infinitely many solutions. y 5 23t 1 5 z 5 t This is incorrect. What mistake was made? 111. A nonsquare matrix cannot have a unique solution. 112. The procedure for Gaussian elimination can be used only for square matrices. 113. A square matrix that has a unique solution has a reduced matrix with 1s along the diagonal and 0s above and below the 1s. 114. A square matrix with an all-zero row has infinitely many solutions. Determine whether each of the following statements is true or false. • C O N C E P T U A L 10.1 Matrices and Systems of Linear Equations 949 950 CHAPTER 10 Matrices 10.2.1 Equality of Matrices In Section 10.1, we defined a matrix with m rows and n columns to have order m 3 n: A 5 D a11 a12 c a1n a21 a22 c a2n ( ( c ( am1 am2 c amn T 115. A fourth-degree polynomial ƒ1x2 5 ax4 1 bx3 1 cx2 1 dx 1 e, with a , 0, can be used to represent the following data on the number of deaths per year due to lightning strikes 1assume 2012 corresponds to x 5 02. Use the data to determine a, b, c, d, and e. Year 2016 2012 2014 U.S. deaths by lightning 46 44 48 50 52 116. A copy machine accepts nickels, dimes, and quarters. After 1 hour, there are 30 coins total, and their value is $4.60. How many nickels, quarters, and dimes are in the machine? • C H A L L E N G E S K I L L S O B J E C T I V E S ■ ■Use equality of matrices. ■ ■Add and subtract matrices. ■ ■Perform scalar and matrix multiplication. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that only matrices of the same order can be equal. ■ ■Understand that you add or subtract matrices of the same order element by element. ■ ■Understand why matrix multiplication is not commutative. 10.2 MATRIX ALGEBRA 10.2.1 SKILL Use equality of matrices. 10.2.1 CO NC EPTUAL Understand that only matrices of the same order can be equal. 117. In Exercise 57, you were asked to solve this system of equations using an augmented matrix. x 2 z 2 y 5 10 2x 2 3y 1 z 5 211 y 2 x 1 z 5 210 A graphing calculator or graphing utility can be used to solve systems of linear equations by entering the coefficients of the matrix. Solve this system and confirm your answer with the calculator’s answer. 118. In Exercise 58, you were asked to solve this system of equations using an augmented matrix. 2x 1 z 1 y 5 23 2y 2 z 1 x 5 0 x 1 y 1 2z 5 5 A graphing calculator or graphing utility can be used to solve systems of linear equations by entering the coefficients of the matrix. Solve this system and confirm your answer with the calculator’s answer. In Exercises 119 and 120, you are asked to model a set of three points with a quadratic function y 5 ax2 1 bx 1 c and determine the quadratic function. a. Set up a system of equations and use a graphing utility or graphing calculator to solve the system by entering the coefficients of the augmented matrix. b. Use the graphing calculator commands STAT QuadReg to model the data using a quadratic function. Round your answers to two decimal places. 119. 126, 282, 12, 72, 17, 12 120. 129, 202, 12, 2182, 111, 162 • T E C H N O L O G Y 10.2 Matrix Algebra 951 Capital letters are used to represent (or name) a matrix, and lowercase letters are used to represent the entries (elements) of the matrix. The subscripts are used to denote the location (row/column) of each entry. The order of a matrix is often written as a subscript of the matrix name: Am3n. Other words like “size” and “dimension” are used as synonyms of “order.” Matrices are a convenient way to represent data. Average Marriage Age on the Rise A survey conducted by the Census Bureau found that the average marriage age in the United States is rising. 1960 2015 Men 22.8 29 Women 20.3 27 These data can be represented by the 2 3 2 matrix c22.8 29 20.3 27d , with two rows (gender) and two columns (year). Opinion on Gun Control According to a Gallup poll conducted in 2015, adult Americans were asked the question, “Do you favor or oppose laws governing gun control that are more strict than current laws?” The results were along party lines: REPUBLICAN (%) INDEPENDENT (%) DEMOCRAT (%) Favor 27 56 77 Oppose 73 44 23 These data can be represented with the 2 3 3 matrix c27 56 77 73 44 23d, with two rows (opinions on gun control) and three columns (registered political party). There is an entire field of study called matrix algebra that treats matrices similarly to functions and variables in traditional algebra. This section serves as an introduction to matrix algebra. We will discuss how to ■ ■determine whether two matrices are equal, ■ ■add and subtract matrices, ■ ■multiply a matrix by a scalar, ■ ■multiply two matrices. It is important to pay special attention to the order of a matrix because it determines whether certain operations are defined. Two matrices are equal if and only if they have the same order, m 3 n, and all of their corresponding elements are equal. Comstock Images/Getty Images, Inc. 952 CHAPTER 10 Matrices 10.2.2 Matrix Addition and Subtraction Two matrices A and B can be added or subtracted only if they have the same order. Suppose A and B are both of order m 3 n; then the sum A 1 B is found by adding corresponding elements, or taking aij 1 bij. The difference A 2 B is found by subtracting the elements in B from the corresponding elements in A, or finding aij 2 bij. DEFINITION Equality of Matrices Two matrices A and B are equal, written as A 5 B if and only if both of the following are true: ■ ■A and B have the same order m 3 n, ■ ■Every pair of corresponding elements is equal: aij 5 bij for all i 5 1, 2, . . . , m and j 5 1, 2, . . . , n. EXAMPLE 1 Equality of Matrices Referring to the definition of equality of matrices, find the indicated elements. £ a11 a12 a13 a21 a22 a23 a31 a32 a33 § 5 £ 2 27 1 0 5 23 21 8 9 § Find the main diagonal elements: a11, a22, and a33. Solution: Since the matrices are equal, their corresponding elements are equal. a11 5 2 a22 5 5 a33 5 9 EXAMPLE 2 Equality of Matrices Given that c2x y 2 1 1 2 d 5 c6 7 1 2d, solve for x and y. Solution: Equate corresponding elements. 2x 5 6 y 2 1 5 7 Solve for x and y. x 5 3 y 5 8 [CONCEPT CHECK] TRUE OR FALSE A matrix with three rows and two columns can be equal to a matrix with two rows and three columns. ANSWER False ▼ DEFINITION Matrix Addition and Matrix Subtraction If A is an m 3 n matrix and B is an m 3 n matrix, then their sum A 1 B is an m 3 n matrix whose elements are given by aij 1 bij and their difference A 2 B is an m 3 n matrix whose elements are given by aij 2 bij 10.2.2 SKILL Add and subtract matrices. 10.2.2 CO NC EPTUAL Understand that you add or subtract matrices of the same order element by element. 10.2 Matrix Algebra 953 It is important to note that only matrices of the same order can be added or subtracted. For example, if A 5 c21 3 4 25 2 0d and B 5 c 5 23 12 1d, the sum and difference of these matrices is undefined because A233 and B232 do not have the same order. A matrix whose elements are all equal to 0 is called a zero matrix. The following are examples of zero matrices: 2 3 2 square zero matrix c0 0 0 0d 3 3 2 zero matrix £ 0 0 0 0 0 0 § 1 3 4 zero matrix 30 0 0 04 If A, an m 3 n matrix, is added to the m 3 n zero matrix, the result is A. A 1 0 5 A EXAMPLE 3 Adding and Subtracting Matrices Given that A 5 c21 3 4 25 2 0d and B 5 c2 1 23 0 25 4d, find the following: a. A 1 B b. A 2 B Solution: Since A233 and B233 have the same order, they can be added or subtracted. a. Write the sum. A 1 B 5 c21 3 4 25 2 0d 1 c2 1 23 0 25 4d Add the corresponding elements. 5 c21 1 2 3 1 1 4 1 1232 25 1 0 2 1 1252 0 1 4 d Simplify. 5 c 1 4 1 25 23 4d b. Write the difference. A 2 B 5 c21 3 4 25 2 0d 2 c2 1 23 0 25 4d Subtract the corresponding elements. 5 c21 2 2 3 2 1 4 2 1232 25 2 0 2 2 1252 0 2 4 d Simplify. 5 c23 2 7 25 7 24d Y OUR T UR N Perform the indicated matrix operations, if possible. A 5 c24 1 0 2d B 5 c 2 24 3 0d C 5 32 9 5 214 D 5 D 0 23 4 2 T a. B 2 A b. C 1 D c. A 1 B d. A 1 D ▼ ▼ A N S W E R a. c 6 3 25 22d b. not defined c. c22 3 23 2d d. not defined STUDY TIP Only matrices of the same order can be added or subtracted. 954 CHAPTER 10 Matrices For example, c1 23 2 5d 1 c0 0 0 0d 5 c1 23 2 5d Because of this result, an m 3 n zero matrix is called the additive identity for m 3 n matrices. Similarly, for any matrix A, there exists an additive inverse 2A such that every element of 2A is the negative of every element of A. For example, A 5 c1 23 2 5d and 2A 5 c21 3 22 25d, and adding these two matrices results in a zero matrix: A 1 12A2 5 0. The same properties that hold for adding real numbers also hold for adding matrices, provided that addition of matrices is defined. PROPERTIES OF MATRIX ADDITION If A, B, and C are all m 3 n matrices and 0 is the m 3 n zero matrix, then the following are true: Commutative property: A 1 B 5 B 1 A Associative property: 1A 1 B2 1 C 5 A 1 1B 1 C2 Additive identity property: A 1 0 5 A Additive inverse property: A 1 12A2 5 0 10.2.3 Scalar and Matrix Multiplication Two types of multiplication involve matrices: scalar multiplication and matrix multiplication. A scalar is any real number. Scalar multiplication is the multiplication of a matrix by a scalar, or real number, and is defined for all matrices. Matrix multiplication is the multiplication of two matrices and is defined only for certain pairs of matrices, depending on the order of each matrix. Scalar Multiplication To multiply a matrix A by a scalar k, multiply every element in A by k. 3c21 0 4 7 5 22d 5 c31212 3102 3142 3172 3152 31222 d 5 c23 0 12 21 15 26d Here, the scalar is k 5 3. [CONCEPT CHECK] TRUE OR FALSE When performing matrix subtraction A 2 B, we subtract each element in B, bij, from each corresponding element in A, aij. ANSWER True ▼ 10.2.3 SKILL Perform scalar and matrix multiplication. 10.2.3 CO NC EPTUAL Understand why matrix multiplication is not commutative. DEFINITION Scalar Multiplication If A is an m 3 n matrix and k is any real number, then their product kA is an m 3 n matrix whose elements are given by kaij In other words, every element aij that is in the ith row and jth column of A is multiplied by k. In general, uppercase letters are used to denote a matrix, and lowercase letters are used to denote scalars. Notice that the elements of each matrix are also represented with lowercase letters, since they are real numbers. 10.2 Matrix Algebra 955 Matrix Multiplication On the one hand, scalar multiplication is straightforward in that it is defined for all matrices and is performed by multiplying every element in the matrix by the scalar. Addition of matrices is also an element-by-element operation. Matrix multiplication, on the other hand, is not as straightforward in that we do not multiply the corresponding elements and it is not defined for all matrices. Matrices are multiplied using a row-by-column method. Before we even try to find the product AB of two matrices A and B, we first have to determine whether the product is defined. For the product AB to exist, the number of columns in the first matrix A must equal the number of rows in the second matrix B. In other words, if the matrix Am3n has m rows and n columns and the matrix Bn3p has n rows and p columns, then the product 1AB2 m3p is defined and has m rows and p columns. Matrix: Order: EXAMPLE 4 Multiplying a Matrix by a Scalar Given that A 5 c21 2 23 4d and B 5 c 0 1 22 3d, perform the following operations. a. 2A b. 23B c. 2A 2 3B Solution (a): Write the scalar multiplication. 2A 5 2c21 2 23 4d Multiply all elements of A by 2. 2A 5 c21212 2122 21232 2142 d Simplify. 2A 5 c22 4 26 8d Solution (b): Write the scalar multiplication. 23B 5 23 c 0 1 22 3d Multiply all elements of B by 23. 23B 5 c23102 23112 231222 23132 d Simplify. 23B 5 c0 23 6 29d Solution (c): Add the results of parts (a) and (b). 2A 2 3B 5 2A 1 123B2 2A 2 3B 5 c22 4 26 8d 1 c0 23 6 29d Add the corresponding elements. 2A 2 3B 5 c22 1 0 4 1 1232 26 1 6 8 1 1292 d Simplify. 2A 2 3B 5 c22 1 0 21d YOUR T UR N For the matrices A and B given in Example 4, find 25A 1 2B. ▼ ▼ A N S W E R 25A 1 2B 5 c 5 28 11 214d STUDY TIP When we multiply matrices, we do not multiply corresponding elements. STUDY TIP For the product AB of two matrices A and B to be defined, the number of columns in the first matrix must equal the number of rows in the second matrix. 956 CHAPTER 10 Matrices Now that we can determine whether a product of two matrices is defined and if so, what the order of the resulting product is, let us turn our attention to how to multiply two matrices. [CONCEPT CHECK] TRUE OR FALSE In matrix multiplication, the two matrices must be the same order. ANSWER False ▼ EXAMPLE 5 Determining Whether the Product of Two Matrices Is Defined Given the matrices A 5 c1 22 0 5 21 3d B 5 £ 2 3 0 7 4 9 § C 5 c6 21 5 2d D 5 323 224 state whether each of the following products exists. If the product exists, state the order of the product matrix. a. AB b. AC c. BC d. CD e. DC Solution: Label the order of each matrix: A233, B332, C232, and D132. a. AB is defined, because A has 3 columns and B has 3 rows. A233B332 AB is order 2 3 2 . 1AB2232 b. AC is not defined , because A has 3 columns and C has 2 rows. c. BC is defined, because B has 2 columns and C has 2 rows. B332C232 BC is order 3 3 2 . 1BC2332 d. CD is not defined , because C has 2 columns and D has 1 row. e. DC is defined, because D has 2 columns and C has 2 rows. D132C232 DC is order 1 3 2 . 1DC2132 Notice that in part (d) we found that CD is not defined, but in part (e) we found that DC is defined. Matrix multiplication is not commutative. Therefore, the order in which matrices are multiplied is important in determining whether the product is defined or undefined. For the product of two matrices to exist, the number of columns in the first matrix A must equal the number of rows in the second matrix B. Y OUR TU R N For the matrices given in Example 5, state whether the following products exist. If the product exists, state the order of the product matrix. a. DA b. CB c. BA ▼ ▼ A N S W E R a. DA exists and is order 1 3 3. b. CB does not exist. c. BA exists and is order 3 3 3 DEFINITION Matrix Multiplication If A is an m 3 n matrix and B is an n 3 p matrix, then their product AB is an m 3 p matrix whose elements are given by 1ab2ij 5 ai1b1j 1 ai2b2j 1 c1 ainbnj In other words, the element 1ab2ij, which is in the ith row and jth column of AB, is the sum of the products of the corresponding elements in the ith row of A and the jth column of B. 10.2 Matrix Algebra 957 Compare the products obtained in Example 6 and the preceding Your Turn. Note that AB 2 BA. Therefore, there is no commutative property for matrix multiplication. EXAMPLE 6 Multiplication of Two 2 3 2 Matrices Given A 5 c1 2 3 4d and B 5 c5 6 7 8d, find AB. common mistake Do not multiply element by element. YOUR T UR N For matrices A and B given in Example 6, find BA. ▼ ✖I N C O R R EC T Multiply the corresponding elements. ERROR AB 2 c 112152 122162 132172 142182d ✓COR R E C T Write the product of the two matrices A and B. AB 5 c1 2 3 4d c5 6 7 8d Perform the row-by-column multiplication. AB 5 c 112152 1 122172 112162 1 122182 132152 1 142172 132162 1 142182d Simplify. AB 5 c19 22 43 50d ▼ A N S W E R BA 5 c5 6 7 8d c1 2 3 4d 5 c23 34 31 46d EXAMPLE 7 Multiplying Matrices For A 5 c21 2 23 22 0 4d and B 5 £ 2 0 1 3 21 22 § , find AB. Solution: Since A is order 2 3 3 and B is order 3 3 2, the product AB is defined and has order 2 3 2. A233B332 5 1AB2232 Write the product of the two matrices. AB 5 c21 2 23 22 0 4d £ 2 0 1 3 21 22 § 958 CHAPTER 10 Matrices Although we have shown repeatedly that there is no commutative property for matrices, matrices do have an associative property of multiplication, as well as a distributive property of multiplication similar to real numbers. Perform the row-by-column multiplication. AB 5 c 1212122 1 122112 1 12321212 1212102 1 122132 1 12321222 1222122 1 102112 1 1421212 1222102 1 102132 1 1421222 d Simplify. AB 5 c 3 12 28 28d Y OUR TU R N For A 5 c 1 0 2 23 21 4d and B 5 £ 0 21 1 2 0 22 § , find AB. ▼ ▼ A N S W E R AB 5 c 0 25 21 27d ▼ A N S W E R a. AB 5 £ 4 5 8 10 12 15 § b. does not exist EXAMPLE 8 Multiplying Matrices For A 5 c 1 0 3 22 5 21d and B 5 £ 22 0 1 23 21 4 0 2 5 § , find AB. Solution: Since A is order 2 3 3 and B is order 3 3 3, the product AB is defined and has order 2 3 3. A233B333 5 1AB2233 Write the product of the two matrices. AB 5 c 1 0 3 22 5 21d £ 22 0 1 23 21 4 0 2 5 § Perform the row-by-column multiplication. AB 5 c 1121222 1 1021232 1 132102 112102 1 1021212 1 132122 12221222 1 1521232 1 1212102 1222102 1 1521212 1 1212122 112112 1 102142 1 132152 1222112 1 152142 1 1212152d Simplify. AB 5 c 22 6 16 211 27 13d Y OUR TU R N Given A 5 £ 1 2 3 § and B 5 34 54: a. Find AB, if it exists. b. Find BA, if it exists. ▼ PROPERTIES OF MATRIX MULTIPLICATION If A, B, and C are all matrices for which AB, AC, and BC are all defined, then the following properties are true: Associative property: A1BC2 5 1AB2C Distributive property: A1B 1 C2 5 AB 1 AC or 1A 1 B2C 5 AC 1 BC 10.2 Matrix Algebra 959 EXAMPLE 9 Application of Matrix Multiplication The following table gives fuel and electric requirements per mile associated with gasoline and electric automobiles. NUMBER OF GALLONS/MILE NUMBER OF KW-HR/MILE Gas Car 0.05 0 Hybrid Car 0.02 0.1 Electric Car 0 0.25 The following table gives an average cost for gasoline and electricity. Cost per gallon of gasoline $3.00 Cost per kW-hr of electricity $0.05 a. Let matrix A represent the gasoline and electricity consumption and matrix B represent the costs of gasoline and electricity. b. Find AB and describe what the elements of the product matrix represent. c. Assume you drive 12,000 miles per year. What are the yearly costs associated with driving the three types of cars? Solution (a): A has order 3 3 2. A 5 £ 0.05 0 0.02 0.1 0 0.25 § B has order 2 3 1. B 5 c$3.00 $0.05d Solution (b): Find the order of the product matrix AB. A332B231 5 1AB2331 AB 5 £ 0.05 0 0.02 0.1 0 0.25 § c$3.00 $0.05d Calculate AB. 5 £ 10.0521$3.002 1 1021$0.052 10.0221$3.002 1 10.121$0.052 1021$3.002 1 10.2521$0.052 § AB 5 £ $0.15 $0.065 $0.0125 § Interpret the product matrix. AB 5 £ Cost per mile to drive the gas car Cost per mile to drive the hybrid car Cost per mile to drive the electric car § Solution (c): Find 12,000 AB. 12,000£ $0.15 $0.065 $0.0125 § 5 £ $1800 $780 $150 § GAS/ELECTRIC COSTS PER YEAR ($) Gas Car 1800 Hybrid Car 780 Electric Car 150 960 CHAPTER 10 Matrices [SEC TION 10. 2] E X E R C I SE S • S K I L L S In Exercises 1–10, state the order of each matrix. 1. c21 2 4 7 23 9d 2. £ 3 5 2 6 21 24 § 3. c24 5 0 1d 4. 324 5 3 74 5. £ 23 4 1 10 8 0 22 5 7 § 6. D 1 2 3 4 T 7. 3224 8. c21 3 6 9 2 5 27 8d 9. D 23 6 0 5 4 29 2 7 1 8 3 6 5 0 24 11 T 10. 33 7 21 5 84 In Exercises 11–16, solve for the indicated variable. 11. c2 x y 3d 5 c2 25 1 3d 12. c23 17 x y d 5 c23 17 10 12d 13. cx 1 y 3 x 2 y 9d 5 c25 z 21 9d 14. cx 24 y 7d 5 c2 1 y 24 5 7 d 15. c3 4 0 12d 5 cx 2 y 4 0 2y 1 xd 16. c 9 2b 1 1 25 16 d 5 c a2 9 2a 1 1 b2d In Exercises 17–30, perform the indicated operations for each expression, if possible. A 5 c21 3 0 2 4 1d B 5 c0 2 1 3 22 4d C 5 £ 0 1 2 21 3 1 § D 5 £ 2 23 0 1 4 22 § 17. A 1 B 18. C 1 D 19. C 2 D 20. A 2 B 21. B 1 C 22. A 1 D 23. D 2 B 24. C 2 A 25. 2A 26. 4D 27. 25C 28. 22B 29. 2A 1 3B 30. 2B 2 3A MATRIX MULTIPLICATION IS NOT COMMUTATIVE: AB u BA OPERATION ORDER REQUIREMENT Equality Same: Am3n 5 Bm3n Addition Same: Am3n 1 Bm3n Subtraction Same: Am3n 2 Bm3n Scalar multiplication None: kAm3n Matrix multiplication Am3n Bn3p 5 1AB2m3p Matrices can be used to represent data. Operations such as equality, addition, subtraction, and scalar multiplication are performed element by element. Two matrices can be added or subtracted only if they have the same order. Matrix multiplication, however, requires that the number of columns in the first matrix is equal to the number of rows in the second matrix and is performed using a row-by-column procedure. [SEC TION 10. 2] S U M M A RY 10.2 Matrix Algebra 961 In Exercises 31–50, perform the indicated operations for each expression, if possible. A 5 £ 1 2 21 0 3 1 5 0 22 § B 5 32 0 234 C 5 c21 7 2 3 0 1d D 5 £ 3 0 1 21 2 5 § E 5 £ 21 0 1 2 1 4 23 1 5 § F 5 £ 1 0 21 § G 5 c1 2 3 4d 31. CD 32. BF 33. DC 34. 1A 1 E2D 35. DG 36. 2A 1 3E 37. GD 38. ED 1 C 39. 24BD 40. 23ED 41. B1A 1 E2 42. GC 1 5C 43. FB 1 5A 44. A2 45. G2 1 5G 46. C⋅12E2 47. 12E2⋅F 48. CA 1 5C 49. DF 50. AE • A P P L I C A T I O N S 51. Smoking. A survey of adult smokers in the United States asked, “Have you ever tried to quit smoking?” 70% said yes and 30% said no. Write a 2 3 1 matrix—call it A—that represents those smokers. When asked what consequences smoking would have on their lives, 89% believed it would increase their chance of getting lung cancer and 84% believed smoking would shorten their lives. Write a 2 3 1 matrix—call it B—that represents those smokers. If there are 46 million adult smokers in the United States: a. What does 46A tell us? b. What does 46B tell us? 52. Women in Science. According to the study of science and engineering indicators by the National Science Foundation (www.nsf.gov), the number of female graduate students in science and engineering disciplines has increased over the last 30 years. In 1991, 32% of mathematics graduate students and 21% of computer science graduate students were female. In 2001, 38% of mathematics graduate students and 30% of computer science graduate students were female. In 2011, 43% of mathematics graduate students were female and 18% of computer science graduate students were female. Write three 2 3 1 matrices representing the percentage of female graduate students. A 5 c% female2math2 1991 % female2C.S.2 1991d B 5 c% female2math22001 % female2C.S.22001d C 5 c% female2math22011 % female2C.S.22011d What does C 2 B tell us? What does B 2 A tell us? What can you conclude about the number of women pursuing mathematics and computer science graduate degrees? Note: C.S. 5 computer science. 53. Registered Voters. According to the U.S. Census Bureau (www.census.gov), in the 2012 national election, 69.3% of the men over the age of 18 were registered voters, but only 59.7% voted; and 72.9% of women over the age of 18 were registered voters, but only 63.7% actually voted. Write a 2 3 2 matrix with the following data: A 5 ≥ Percentage of registered Percentage of registered male voters female voters Percent of males Percent of females that voted that voted ¥ If we let B be a 2 3 1 matrix representing the total population of males and females over the age of 18 in the United States, or B 5 c100 M 110 Md, what does AB tell us? 54. Job Application. A company has two rubrics for scoring job applicants based on weighting education, experience, and the interview differently. Matrix A Education Experience Interview £ Rubric 1 Rubric 2 0.5 0.6 0.3 0.1 0.2 0.3 § Applicants receive a score from 1 to 10 in each category (education, experience, and interview). Two applicants are shown in matrix B: Matrix B Applicant 1 Applicant 2 c Education Experience Interview 8 7 5 6 8 8 d What is the order of BA? What does each entry in BA tell us? 55. Taxes. The IRS allows an individual to deduct business expenses in the following way: $0.575 per mile driven, 50% of entertainment costs, and 100% of actual expenses. Represent these deductions as a row matrix A. In 2017, Jamie had the following business expenses: $2700 in entertainment, $15,200 actual expenses, and he drove 7523 miles. Represent Jamie’s expenses as a column matrix B. Multiply these matrices to find the total amount of business expenses Jamie can claim on his 2017 federal tax form: AB. 56. Tips on Service. Marilyn decides to go to the Safety Harbor Spa for a day of pampering. She is treated to a Hot Stone 962 CHAPTER 10 Matrices Massage 1$852, a manicure and pedicure 1$752, and a haircut and style 1$1002. Represent the costs of the individual services as a row matrix A. She decides to tip her masseur 25%, her nail tech 20%, and her hair stylist 15%. Represent the tipping percentages as a column matrix B. Multiply these matrices to find the total amount in tips AB she needs to add to her final bill. Use the following nutritional chart for Exercises 57 and 58: SANDWICH CALORIES FAT (g) CARBOHYDRATES (g) PROTEIN (g) Veggie Wrap 230 3 44 9 Tuna Salad 430 19 46 20 Turkey 290 5 45 19 Roast Beef 330 5 47 24 57. Nutrition. Utilize the table to write a 4 3 4 matrix A. Find 2A. What would 2A represent? Find 0.5A. What would 0.5A represent? 58. Nutrition. Don decides to film a documentary similar to Supersize Me, but instead of eating only at McDonald’s, he decides to eat only at a local sandwich shop. He is hoping to lose weight and name the film Minimize Me. He is going to eat only three meals a day from the four options listed in the preceding table. Each week he will consume the following sandwiches: 7 veggie wraps, 5 tuna salad, 8 turkey, and 1 roast beef. Let A be the matrix (found in Exercise 57) that represents the nutrition values and let B be a row matrix that represents the number of each type of sandwich consumed in a week. Find BA. What does BA represent? Find 1 7 BA. What does 1 7 BA represent? Use the following tables for Exercises 59 and 60: The following table gives fuel and electric requirements per mile associated with gasoline and electric automobiles. NUMBER OF GALLONS/MILE NUMBER OF KW-HR/MILE SUV full size 0.06 0 Hybrid car 0.02 0.1 Electric car 0 0.3 The following table gives an average cost for gasoline and electricity. Cost per gallon of gasoline $3.80 Cost per kW-hr of electricity $0.05 59. Environment. Let matrix A represent the gasoline and electricity consumption and matrix B represent the costs of gasoline and electricity. Find AB and describe what the elements of the product matrix represent. 60. Environment. Assume you drive 12,000 miles per year. What are the yearly costs associated with driving the three types of cars in Exercise 59? For Exercises 61 and 62, refer to the following: The results of a nutritional analysis of one serving of three foods A, B, and C were Carbohydrates (g) Protein (g) Fat (g) X 5 £ 5 0 2 5 6 5 8 4 4 § A B C It is possible to find the nutritional content of a meal consisting of a combination of the foods A, B, and C by multiplying the matrix X by a second matrix N 5 £ r s t § , that is, XN, where r is the number of servings of food A, s is the number of servings of food B, and t is the number of servings of food C. 61. Health/Nutrition. Find the matrix N that represents a meal consisting of two servings of food A and one serving of food B. Find the nutritional content of that meal. 62. Health/Nutrition. Find the matrix N that represents a meal consisting of one serving of food A and two servings of food C. Find the nutritional content of that meal. For Exercises 63 and 64, refer to the following: Cell phone companies charge users based on the number of minutes talked, the number of text messages sent, and the number of megabytes of data used. The costs for three cell phone providers are given in the following table: MINUTES TEXT MESSAGES MEGABYTES OF DATA C1 $0.04 $0.05 $0.15 C2 $0.06 $0.05 $0.18 C3 $0.07 $0.07 $0.13 It is possible to find the cost to a cell phone user for each of the three providers by creating a matrix X whose rows are the rows of data in the table and multiplying the matrix X by a second matrix N 5 £ m t d § , that is, XN, where m is the number of minutes talked, t is the number of text messages sent, and d is the megabytes of data used. 63. Telecommunications/Business. A local business is looking at providing an employee a cell phone for business use. Find the matrix N that represents expected normal cell phone usage of 200 minutes, 25 text messages, and no data usage. Find and interpret XN. Which is the better cell phone provider for this employee? 64. Telecommunications/Business. A local business is looking at providing an employee a cell phone for business use. Find the matrix N that represents expected normal cell phone usage of 125 minutes, 125 text messages, and 320 megabytes of data usage. Find and interpret XN. Which is the better cell phone provider for this employee? 10.2 Matrix Algebra 963 • C A T C H T H E M I S T A K E In Exercises 65 and 66, explain the mistake that is made. 65. Multiply c3 2 1 4d c21 3 22 5d . Solution: Multiply corresponding elements. c3 2 1 4d c21 3 22 5d 5 c 1321212 122132 1121222 142152 d Simplify. c3 2 1 4d c21 3 22 5d 5 c23 6 22 20d This is incorrect. What mistake was made? 66. Multiply c3 2 1 4d c21 3 22 5d . Solution: Multiply using column-by-row method. c3 2 1 4d c21 3 22 5d 5 c 1321212 1 112132 1221212 1 142132 1321222 1 112152 1221222 1 142152 d Simplify. c3 2 1 4d c21 3 22 5d 5 c 0 10 21 16d This is incorrect. What mistake was made? In Exercises 67–70, determine whether the statements are true or false. 67. If A 5 ca11 a12 a21 a22 d and B 5 cb11 b12 b21 b22 d , then AB 5 ca11b11 a12b12 a21b21 a22b22 d . 68. If AB is defined, then AB 5 BA. 69. AB is defined only if the number of columns in A equals the number of rows in B. 70. A 1 B is defined only if A and B have the same order. 71. For A 5 ca11 a12 a21 a22 d , find A2. 72. In order for A2 m3n to be defined, what condition (with respect to m and n) must be met? • C O N C E P T U A L 73. For A 5 c1 1 1 1d find A, A2, A3, . . . . What is An? 74. For A 5 c1 0 0 1d find A, A2, A3, . . . . What is An? 75. If Am3nBn3p is defined, explain why 1Am3n Bn3p22 is not defined for m 2 p. 76. If Am3n 5 Bm3n and Cn3m, explain why AC 2 CB, if m 2 n. • C H A L L E N G E In Exercises 77–82, apply a graphing utility to perform the indicated matrix operations, if possible. A 5 D 1 7 9 2 23 26 15 11 0 3 2 5 9 8 24 1 T B 5 D 7 9 8 6 24 22 3 1 T 77. AB 78. BA 79. BB 80. AA A 5 £ 2 1 1 23 0 2 4 26 0 § 81. A2 82. A5 • T E C H N O L O G Y 964 CHAPTER 10 Matrices S K I L L S O B J E C T I V E S ■ ■Write a system of linear equations as a matrix equation. ■ ■Find the inverse of a square matrix. ■ ■Solve systems of linear equations using inverse matrices. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that only a system of linear equations can be represented with a matrix equation. ■ ■Understand that only a square matrix can have an inverse and that not all square matrices have inverses. ■ ■Matrix algebra can only be used to solve systems of linear equations that have a unique solution. 10.3 MATRIX EQUATIONS; THE INVERSE OF A SQUARE MATRIX 10.3.1 Matrix Equations Matrix equations are another way of writing systems of linear equations. WORDS MATH Start with a matrix equation. c2 23 1 5d cx yd 5 c27 9d Multiply the two matrices on the left. c2x 2 3y x 1 5yd 5 c27 9d Apply equality of two matrices. 2x 2 3y 5 27 x 1 5y 5 9 Let A be a matrix with m rows and n columns, which represents the coefficients in the system; let X represent the variables in the system; and let B represent the constants in the system. Then, a system of linear equations can be written as AX 5 B. SYSTEM OF LINEAR EQUATIONS A X B MATRIX EQUATION: AX 5 B 3x 1 4y 5 1 x 2 2y 5 7 c3 4 1 22d cx yd c1 7d c3 4 1 22d c x y d 5 c1 7d x 2 y 1 z 5 2 2x 1 2y 2 3z 5 23 x 1 y 1 z 5 6 £ 1 21 1 2 2 23 1 1 1 § £ x y z § £ 2 23 6 § £ 1 21 1 2 2 23 1 1 1 § £ x y z § 5 £ 2 23 6 § x 1 y 1 z 5 0 3x 1 2y 2 z 5 2 c1 1 1 3 2 21d £ x y z § c0 2d c1 1 1 3 2 21d £ x y z § 5 c0 2d 10.3.1 SKILL Write a system of linear equations as a matrix equation. 10.3.1 CON CEPTUAL Understand that only a system of linear equations can be represented with a matrix equation. EXAMPLE 1 Writing a System of Linear Equations as a Matrix Equation Write each system of linear equations as a matrix equation. a. 2x 2 y 5 5 b. 3x 2 2y 1 4z 5 5 c. x1 2 x2 1 2x3 2 3 5 0 2x 1 2y 5 3 y 2 3z 5 22 x1 1 x2 2 3x3 1 5 5 0 7x 2 z 5 1 x1 2 x2 1 x3 2 2 5 0 Solution: a. c 2 21 21 2d cx yd 5 c5 3d 10.3 Matrix Equations; the Inverse of a Square Matrix 965 10.3.2 Finding the Inverse of a Matrix Before we discuss solving systems of linear equations in the form AX 5 B, let us first recall how we solve ax 5 b, where a and b are real numbers (not matrices). WORDS MATH Write the linear equation in one variable. ax 5 b Multiply both sides by a21 (same as dividing by a), provided a 2 0. a21ax 5 a21b Simplify. a21ax 5 a21b 1 x 5 a21b Recall that a21, or 1 a, is the multiplicative inverse (Chapter 0) of a because a21a 5 1. And we call 1 the multiplicative identity because any number multiplied by 1 is itself. Before we solve matrix equations, we need to define the multiplicative identity matrix and the multiplicative inverse matrix. A square matrix of order n with 1s along the main diagonal 1aii2 and 0s for all other elements is called the multiplicative identity matrix In. I2 5 c1 0 0 1d I3 5 £ 1 0 0 0 1 0 0 0 1 § I4 5 D 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 T Since a real number multiplied by 1 is itself 1a ⋅ 1 5 a2, we expect that a matrix mul-tiplied by the appropriate identity matrix should result in itself: Am3nIn 5 Am3n and Im Am3n 5 Am3n f b. Note that all missing terms have 0 coefficients. 3x 2 2y 1 4z 5 5 0x 1 y 2 3z 5 22 7x 1 0y 2 z 5 1 c. Write the constants on the right side of the equal sign. x1 2 x2 1 2x3 5 3 x1 1 x2 2 3x3 5 25 x1 2 x2 1 x3 5 2 Y OUR T UR N Write each system of linear equations as a matrix equation. a. 2x 1 y 2 3 5 0 b. y 2 x 1 z 5 7 x 2 y 5 5 x 2 y 2 z 5 2 z 2 y 5 21 £ 3 22 4 0 1 23 7 0 21 § £ x y z § 5 £ 5 22 1 § £ 1 21 2 1 1 23 1 21 1 § £ x1 x2 x3 § 5 £ 3 25 2 § ▼ A N S W E R a. c2 1 1 21d cx yd 5 c3 5d b. £ 21 1 1 1 21 21 0 21 1 § £ x y z § 5 £ 7 2 21 § ▼ [CONCEPT CHECK] TRUE OR FALSE Matrices can be used to solve systems of linear and nonlinear equations. ANSWER False ▼ 10.3.2 S KI L L Find the inverse of a square matrix. 10.3.2 C ON C E P T U A L Understand that only a square matrix can have an inverse and that not all square matrices have inverses. 966 CHAPTER 10 Matrices The identity matrix In will assist us in developing the concept of an inverse of a square matrix. DEFINITION Inverse of a Square Matrix Let A be a square n 3 n matrix. If there exists a square n 3 n matrix A21 such that AA 21 5 In and A 21A 5 In then A21, stated as “A inverse,” is the inverse of A. It is important to note that only a square matrix can have an inverse. Even then, not all square matrices have inverses. EXAMPLE 2 Multiplying a Matrix by the Multiplicative Identity Matrix In For A 5 c22 4 1 3 7 21d, find I2 A. Solution: Write the two matrices. A 5 c22 4 1 3 7 21d I2 5 c1 0 0 1d Find the product I2 A. I2A 5 c1 0 0 1d c22 4 1 3 7 21d I2A 5 c 1121222 1 102132 112142 1 102172 112112 1 1021212 1021222 1 112132 102142 1 112172 102112 1 1121212 d I2A 5 c22 4 1 3 7 21d 5 A Y OUR TU R N For A in Example 2, find AI3. ▼ A N S W E R AI35 c22 4 1 3 7 21d 5 A ▼ STUDY TIP ■ ■Only a square matrix can have an inverse. ■ ■Not all square matrices have inverses. EXAMPLE 3 Multiplying a Matrix by Its Inverse Verify that the inverse of A 5 c1 3 2 5d is A 21 5 c25 3 2 21d. Solution: Show that AA21 5 I2 and A21A 5 I2. Find the product AA21. AA 21 5 c1 3 2 5d c25 3 2 21d 5 c 1121252 1 132122 112132 1 1321212 1221252 1 152122 122132 1 1521212 d 5 c1 0 0 1d 5 I2 Find the product A21A. A21A 5 c25 3 2 21d c1 3 2 5d 5 c 1252112 1 132122 1252132 1 132152 122112 1 1212122 122132 1 1212152 d 5 c1 0 0 1d 5 I2 Y OUR TU R N Verify that the inverse of A 5 c1 4 2 9d is A 21 5 c 9 24 22 1d. ▼ A N S W E R AA21 5 A21A 5 I2 ▼ Now that we can show that two matrices are inverses of one another, let us describe the process for finding an inverse, if it exists. If an inverse A21 exists, then the matrix A is said to be nonsingular. If the inverse does not exist, then the matrix A is said to be singular. Let A 5 c1 21 2 23d and the inverse be A 21 5 cw x y zd, where w, x, y, and z are variables to be determined. A matrix and its inverse must satisfy the identity AA21 5 I2. WORDS MATH The product of a matrix and its inverse is the identity matrix. c1 21 2 23d cw x y zd 5 c1 0 0 1d Multiply the two matrices on the left. c w 2 y x 2 z 2w 2 3y 2x 2 3zd 5 c1 0 0 1d Equate corresponding matrix elements. w 2 y 5 1 2w 2 3y 5 0 and x 2 z 5 0 2x 2 3z 5 1 Notice that there are two systems of equations, both of which can be solved by several methods (elimination, substitution, or augmented matrices). We will find that w 5 3, x 5 21, y 5 2, and z 5 21. Therefore, we know the inverse is A 21 5 c3 21 2 21d. But, instead, let us use augmented matrices in order to develop the general procedure. Write the two systems of equations as two augmented matrices: w y x z c1 21 2 23 1 0d c1 21 2 23 0 1d Since the left side is the same for each augmented matrix, we can combine these two matrices into one matrix, thereby simultaneously solving both systems of equations: c1 21 2 23 1 0 0 1d Notice that the right side of the vertical line is the identity matrix I2. Using Gauss–Jordan elimination, transform the matrix on the left to the identity matrix. c1 21 2 23 1 0 0 1d R2 2 2R1 S R2 c1 21 0 21 1 0 22 1d 2R2 S R2 c1 21 0 1 1 0 2 21d R1 1 R2 S R1 c1 0 0 1 3 21 2 21d The matrix on the right of the vertical line is the inverse A 21 5 c3 21 2 21d. FINDING THE INVERSE OF A SQUARE MATRIX To find the inverse of an n 3 n matrix A: Step 1: Form the augmented matrix 3A 0 In4. Step 2: Use row operations to transform this entire matrix to 3In 0 A214. This is done by applying Gauss–Jordan elimination to reduce A to the identity matrix In (which is in reduced row–echelon form). If this is not possible, then A is a singular matrix and no inverse exists. Step 3: Verify the result by showing that AA21 5 In and A21A 5 In. 10.3 Matrix Equations; the Inverse of a Square Matrix 967 968 CHAPTER 10 Matrices This procedure for finding an inverse of a square matrix is used for all square matrices of order n. For the special case of a 2 3 2 matrix, there is a formula (that will be derived in Exercises 65 and 66) for finding the inverse. Let A 5 ca b c dd represent any 2 3 2 matrix; then the inverse matrix is given by The denominator ad 2 bc is called the determinant and will be discussed in Section 10.4. We found the inverse of A 5 c1 2 3 5d in Example 4. Let us now find the inverse using this formula. WORDS MATH Write the formula for A21. A 21 5 1 ad 2 bc c d 2b 2c ad Substitute a 5 1, b 5 2, c 5 3, d 5 5 into the formula. A 21 5 1 112152 2 122132 c 5 22 23 1d Simplify. A 21 5 1212 c 5 22 23 1d A 21 5 c25 2 3 21d The result is the same as what we found in Example 4. A 21 5 1 ad 2 bc c d 2b 2c ad ad 2 bc 2 0 EXAMPLE 4 Finding the Inverse of a 2 3 2 Matrix Find the inverse of A 5 c1 2 3 5d. Solution: STEP 1 Form the matrix 3A 0 I24. c1 2 3 5 1 0 0 1d STEP 2 Use row operations to transform R2 2 3R1 S R2 c1 2 0 21 1 0 23 1d A into I2. 2R2 S R2 c1 2 0 1 1 0 3 21d R1 2 2R2 S R1 c1 0 0 1 25 2 3 21d Identify the inverse. STEP 3 Check. AA 21 5 c1 2 3 5d c25 2 3 21d 5 c1 0 0 1d 5 I2 A 21A 5 c25 2 3 21d c1 2 3 5d 5 c1 0 0 1d 5 I2 Y OUR TU R N Find the inverse of A 5 c2 3 5 8d. A21 5 c25 2 3 21d ▼ A N S W E R A 21 5 c 8 23 25 2d ▼ EXAMPLE 5 Finding That No Inverse Exists: Singular Matrix Find the inverse of A 5 c 1 25 21 5d. Solution: STEP 1 Form the matrix 3A 0 I24. c 1 25 21 5 1 0 0 1d STEP 2 Apply row operations to transform R2 1 R1 S R2 c1 25 0 0 1 0 1 1d A into I2. We cannot convert the left-hand side of the augmented matrix to I2 because of the all-zero row on the left-hand side. Therefore, A is not invertible . [CONCEPT CHECK] TRUE OR FALSE A matrix has to be square in order for its inverse to exist; some square matrices are singular and do not have an inverse. ANSWER True ▼ EXAMPLE 6 Finding the Inverse of a 3 3 3 Matrix Find the inverse of A 5 £ 1 2 21 0 1 21 21 0 22 § . Solution: STEP 1 Form the matrix 3A 0 I34. £ 1 2 21 0 1 21 21 0 22 † 1 0 0 0 1 0 0 0 1 § STEP 2 Apply row operations to transform A into I3. R3 1 R1 S R3 £ 1 2 21 0 1 21 0 2 23 † 1 0 0 0 1 0 1 0 1 § R3 2 2R2 S R3 £ 1 2 21 0 1 21 0 0 21 † 1 0 0 0 1 0 1 22 1 § 2R3 S R3 £ 1 2 21 0 1 21 0 0 1 † 1 0 0 0 1 0 21 2 21 § R2 1 R3 S R2 R1 1 R3 S R1 £ 1 2 0 0 1 0 0 0 1 † 0 2 21 21 3 21 21 2 21 § R1 2 2R2 S R1 £ 1 0 0 0 1 0 0 0 1 † 2 24 1 21 3 21 21 2 21 § Identify the inverse. A 21 5 £ 2 24 1 21 3 21 21 2 21 § 10.3 Matrix Equations; the Inverse of a Square Matrix 969 970 CHAPTER 10 Matrices 10.3.3 Solving Systems of Linear Equations Using Matrix Algebra and Inverses of Square Matrices We can solve systems of linear equations using matrix algebra. We will use a system of three equations and three variables to demonstrate the procedure. However, it can be extended to any square system. Linear System of Equations Matrix Form of the System a1x 1 b1y 1 c1z 5 d1 a2x 1 b2y 1 c2z 5 d2 a3x 1 b3y 1 c3z 5 d3 £ a1 b1 c1 a2 b2 c2 a3 b3 c3 § £ x y z § 5 £ d1 d2 d3 § A X B Recall that a system of linear equations has a unique solution, no solution, or infinitely many solutions. If a system of n equations in n variables has a unique solution, it can be found using the following procedure. WORDS MATH Write the system of linear equations as a matrix equation. An3n Xn31 5 Bn31 Multiply both sides of the equation by A21. A21AX 5 A21B A matrix times its inverse is the identity matrix. In X 5 A21B A matrix times the identity matrix is equal to itself. X 5 A21B Notice the order in which the right side is multiplied, Xn31 5 A21 n3nBn31, and remember that matrix multiplication is not commutative. f f f SOLVING A SYSTEM OF LINEAR EQUATIONS USING MATRIX ALGEBRA: UNIQUE SOLUTION If a system of linear equations is represented by AX 5 B, where A is a nonsingular square matrix, then the system has a unique solution given by X 5 A 21B STEP 3 Check. AA 21 5 £ 1 2 21 0 1 21 21 0 22 § £ 2 24 1 21 3 21 21 2 21 § 5 £ 1 0 0 0 1 0 0 0 1 § 5 I3 A 21A 5 £ 2 24 1 21 3 21 21 2 21 § £ 1 2 21 0 1 21 21 0 22 § 5 £ 1 0 0 0 1 0 0 0 1 § 5 I3 Y OUR TU R N Find the inverse of A 5 £ 1 1 0 21 0 1 2 0 21 § . ▼ A N S W E R A 21 5 £ 0 1 1 1 21 21 0 2 1 § ▼ 10.3.3 SKILL Solve systems of linear equations using inverse matrices. 10.3.3 CON CEPTUAL Matrix algebra can only be used to solve systems of linear equations that have a unique solution. EXAMPLE 7 Solving a System of Linear Equations Using Matrix Algebra Solve the system of equations using matrix algebra. x 1 y 1 z 5 2 x 1 z 5 1 x 2 y 2 z 5 24 Solution: Write the system in matrix form. AX 5 B A 5 £ 1 1 1 1 0 1 1 21 21 § X 5 £ x y z § B 5 £ 2 1 24 § Find the inverse of A. Form the matrix 3A 0 I34. £ 1 1 1 1 0 1 1 21 21 † 1 0 0 0 1 0 0 0 1 § R2 2 R1 S R2 R3 2 R1 S R3 £ 1 1 1 0 21 0 0 22 22 † 1 0 0 21 1 0 21 0 1 § 2R2 S R2 £ 1 1 1 0 1 0 0 22 22 † 1 0 0 1 21 0 21 0 1 § R3 1 2R2 S R3 £ 1 1 1 0 1 0 0 0 22 † 1 0 0 1 21 0 1 22 1 § 21 2R3 S R3 £ 1 1 1 0 1 0 0 0 1 † 1 0 0 1 21 0 21 2 1 21 2 § R1 2 R3 S R1 £ 1 1 0 0 1 0 0 0 1 † 3 2 21 1 2 1 21 0 21 2 1 21 2 § R1 2 R2 S R1 £ 1 0 0 0 1 0 0 0 1 † 1 2 0 1 2 1 21 0 21 2 1 21 2 § Identify the inverse. A 21 5 £ 1 2 0 1 2 1 21 0 21 2 1 21 2 § The solution to the system is X 5 A21B. X 5 A 21B 5 £ 1 2 0 1 2 1 21 0 21 2 1 21 2 § £ 2 1 24 § Simplify. X 5 £ x y z § 5 £ 21 1 2 § Y OUR T UR N Solve the system of equations using matrix algebra. x 1 y 2 z 5 3 y 1 z 5 1 2x 1 3y 1 z 5 5 x 5 21, y 5 1, z 5 2 ▼ A N S W E R x 5 0, y 5 2, z 5 21 ▼ [CONCEPT CHECK] TRUE OR FALSE Matrix algebra can be used to solve any system of linear equations. ANSWER False ▼ 10.3 Matrix Equations; the Inverse of a Square Matrix 971 972 CHAPTER 10 Matrices Cryptography Applications Cryptography is the practice of hiding information, or secret communication. Let’s assume you want to send your ATM PIN code over the Internet, but you don’t want hackers to be able to retrieve it. You can represent the PIN code in a matrix and then multiply that PIN matrix by a “key” matrix so that it is encrypted. If the person you send it to has the “inverse key” matrix, he can multiply the encrypted matrix he receives by the inverse key matrix and the result will be the original PIN matrix. Although PIN numbers are typically four digits, we will assume two digits to illustrate the process. WORDS MATH Suppose the two-digit ATM PIN is 13. P 5 31 34132 Apply any 2 3 2 nonsingular matrix as the “key” (encryption) matrix. K 5 c2 3 5 8d Multiply the PIN and encryption matrices. PK 5 31 34132c2 3 5 8d 232 5 31122 1 3152 1132 1 31824 5 317 274 The receiver of the encrypted matrix sees only 317 274132. The decoding “key” is the inverse matrix K21. K21 5 c 8 23 25 2d Any receiver who has the decoding key can multiply the received encrypted matrix by the decoding “key” matrix. The result is the original transmitted PIN number. 317 274132c 8 23 25 2d 5 317182 1 271252 171232 1 271224 5 31 34 STUDY TIP K 5 c2 3 5 8d K 21 5 1 122182 2 132152 c 8 23 25 2d 5 c 8 23 25 2d Systems of linear equations can be solved using matrix equations. SYSTEM OF LINEAR EQUATIONS A X B MATRIX EQUATION: AX 5 B x 2 y 1 z 5 2 2x 1 2y 2 3z 5 23 x 1 y 1 z 5 6 £ 1 21 1 2 2 23 1 1 1 § £ x y z § £ 2 23 6 § £ 1 21 1 2 2 23 1 1 1 § £ x y z § 5 £ 2 23 6 § [SEC TION 10.3] S U M M A RY If this system of linear equations has a unique solution, then it is represented by X 5 A 21B A 21 is the inverse of A, that is, AA 21 5 A 21A 5 I, and is found by CAn3n 0 InD S CIn 0 An3n 21 D [SEC TION 10.3] E XE R C I S E S • S K I L L S In Exercises 1–8, write the system of linear equations as a matrix equation. (Do not solve the system.) 1. 22x 1 5y 5 10 2. 4x 2 8y 5 10 3. x 2 2y 5 8 4. 7x 2 2y 5 28 7x 2 2y 5 24 3x 1 5y 5 15 23x 1 y 5 6 3x 1 7y 5 42 5. 3x 1 5y 2 z 5 2 6. x 2 y 1 z 5 12 7. 3x 1 z 5 10 8. x 1 y 2 2z 1 w 5 11 x 1 2z 5 17 2x 1 y 2 3z 5 6 y 2 2z 5 4 2x 2 y 1 3z 5 17 2x 1 y 2 z 5 4 23x 1 2y 1 z 5 18 x 1 2y 5 6 2x 1 2y 2 3z 1 4w 5 12 y 1 4z 1 6w 5 19 In Exercises 9–18, determine whether B is the multiplicative inverse of A using AA21 5 I. 9. A 5 c 8 211 25 7d B 5 c7 11 5 8d 10. A 5 c 7 29 23 4d B 5 c4 9 3 7d 11. A 5 c3 1 1 22d B 5 c 2 7 1 7 1 7 23 7 d 12. A 5 c2 3 1 21d B 5 c 1 5 3 5 1 5 2 2 5 d 13. A 5 c1 2 3 4d B 5 c 4 22 23 1d 14. A 5 c1 2 3 4d B 5 c1 1 2 1 3 1 4 d 15. A 5 £ 1 21 1 1 0 21 0 1 21 § B 5 £ 1 0 1 1 21 2 1 21 1 § 16. A 5 £ 21 0 21 21 1 22 21 1 21 § B 5 £ 21 1 21 21 0 1 0 21 1 § 17. A 5 £ 2 0 1 0 3 1 0 2 21 § B 5 £ 0 2 1 0 3 0 2 0 2 § 18. A 5 £ 1 0 0 0 2 0 0 0 3 § B 5 £ 1 0 0 0 1 2 0 0 0 1 3 § In Exercises 19–32, find A21, if possible. 19. A 5 c 2 1 21 0d 20. A 5 c3 1 2 1d 21. A 5 c 1 3 2 5 3 4 d 22. A 5 c 1 4 2 1 3 2 3 d 23. A 5 c1.3 2.4 5.3 1.7d 24. A 5 c22.3 1.1 4.6 23.2d 25. A 5 £ 1 1 1 1 21 21 21 1 21 § 26. A 5 £ 1 21 1 1 1 1 21 2 23 § 27. A 5 £ 1 0 1 0 1 1 1 21 0 § 28. A 5 £ 1 2 23 1 21 21 1 0 24 § 29. A 5 £ 2 4 1 1 1 21 1 1 0 § 30. A 5 £ 1 0 1 1 1 21 2 1 21 § 31. A 5 £ 1 1 21 1 21 1 2 21 21 § 32. A 5 £ 1 21 21 1 1 23 3 25 1 § In Exercises 33–46, apply matrix algebra to solve the system of linear equations. 33. 2x 2 y 5 5 x 1 y 5 1 34. 2x 2 3y 5 12 x 1 y 5 1 35. 4x 2 9y 5 21 7x 2 3y 5 5 2 36. 7x 2 3y 5 1 4x 2 5y 5 27 5 37. 3 4x 2 2 3y 5 5 21 2x 2 5 3y 5 3 38. 2 5x 1 3 7y 5 1 21 2x 2 1 3y 5 1 6 39. x 1 y 1 z 5 1 x 2 y 2 z 5 21 2x 1 y 2 z 5 21 40. x 2 y 1 z 5 0 x 1 y 1 z 5 2 2x 1 2y 2 3z 5 1 41. x 1 z 5 3 y 1 z 5 1 x 2 y 5 2 42. x 1 2y 2 3z 5 1 x 2 y 2 z 5 3 x 2 4z 5 0 43. 2x 1 4y 1 z 5 25 x 1 y 2 z 5 7 x 1 y 5 0 44. x 1 z 5 3 x 1 y 2 z 5 23 2x 1 y 2 z 5 25 45. x 1 y 2 z 5 4 x 2 y 1 z 5 2 2x 2 y 2 z 5 23 46. x 2 y 2 z 5 0 x 1 y 2 3z 5 2 3x 2 5y 1 z 5 4 10.3 Matrix Equations; the Inverse of a Square Matrix 973 974 CHAPTER 10 Matrices • A P P L I C A T I O N S 47. NCAA. University of Florida apparel sales associated with the Final Four Basketball Tournament and the BCS Championship Game in Football are represented with the following matrix: Sweatshirts T-shirts Final Four BCS c 20,000 100,000 100,000 50,000 d 5 A The revenue generated by the sales of these T-shirts and sweatshirts is given by the following matrix: Final Four BCS c$3,000,000 $6,000,000d 5 B a. Find A 21B. b. What does A 21B represent? 48. Basketball. A huge national women’s basketball weekend tournament is hosted annually and involves two big games, one on Saturday and one on Sunday. The numbers of pennants and sweatshirts sold at each of these games are represented by the following matrix: Pennants Sweatshirts Saturday Sunday c30,000 2,000 50,000 5,000d 5 A The revenue generated by the sales of these pennants and sweatshirts is given by the following matrix: Saturday Sunday c $750,000 $1,375,000d 5 B a. Find A 21B. b. What does A 21B represent? For Exercises 49–54, apply the following decoding scheme: The encoding matrix is £ 1 1 0 21 0 1 2 0 21 §. The encrypted matrices are given below. For each of the following, determine the three-letter word that is originally transmitted. Hint: All four words are parts of the body. 49. Cryptography. 355 10 2224 50. Cryptography. 331 8 274 51. Cryptography. 321 12 224 52. Cryptography. 39 1 54 53. Cryptography. 3210 5 204 54. Cryptography. 340 5 2174 For Exercises 55 and 56, refer to the following: The results of a nutritional analysis of one serving of three foods A, B, and C were: Carbohydrates (g) Protein (g) Fat (g) Y 5 £ 8 4 6 6 10 5 10 4 8 § A B C The nutritional content of a meal consisting of a combination of the foods A, B, and C is the product of the matrix Y and a second matrix N 5 £ r s t § , that is, YN, where r is the number of servings of food A, s is the number of servings of food B, and t is the number of servings of food C. 55. Health/Nutrition. Use the inverse matrix technique to find the number of servings of foods A, B, and C necessary to create a meal of 18 grams of carbohydrates, 21 grams of protein, and 22 grams of fat. 56. Health/Nutrition. Use the inverse matrix technique to find the number of servings of foods A, B, and C necessary to create a meal of 14 grams of carbohydrates, 25 grams of protein, and 16 grams of fat. For Exercises 57 and 58, refer to the following: Cell phone companies charge users based on the number of minutes talked, the number of text messages sent, and the number of megabytes of data used. The costs for three cell phone providers are given in the table: MINUTES TEXT MESSAGES MEGABYTES OF DATA C1 $0.03 $0.06 $0.15 C2 $0.04 $0.05 $0.18 C3 $0.05 $0.07 $0.13 The cost to a cell phone user for each of the three providers is the product of the matrix X whose rows are the rows of data in the table and the matrix N 5 £ m t d § where m is the number of minutes talked, t is the number of text messages sent, and d is the megabytes of data used. 57. Telecommunications/Business. A local business is looking at providing an employee a cell phone for business use. The business solicits estimates for its normal monthly usage from three cell phone providers. Company 1 estimates the cost to be $49.50, Company 2 estimates the cost to be $52.00, and Company 3 estimates the cost to be $58.50. Use the inverse matrix technique to find the normal monthly usage for the employee. 58. Telecommunications/Business. A local business is looking at providing an employee a cell phone for business use. The business solicits estimates for its normal monthly usage from three cell phone providers. Company 1 estimates the cost to be $82.50, Company 2 estimates the cost to be $85.00, and Company 3 estimates the cost to be $92.50. Use the inverse matrix technique to find the normal monthly usage for the employee. 1 A 10 J 19 S 2 B 11 K 20 T 3 C 12 L 21 U 4 D 13 M 22 V 5 E 14 N 23 W 6 F 15 O 24 X 7 G 16 P 25 Y 8 H 17 Q 26 Z 9 I 18 R, In Exercises 59 and 60, explain the mistake that is made. • C A T C H T H E M I S T A K E 59. Find the inverse of A 5 £ 1 0 1 21 0 21 1 2 0 § . Solution: Write the matrix 3A 0 I34.£ 1 0 1 21 0 21 1 2 0 3 1 0 0 0 1 0 0 0 1 § Use Gaussian elimination to reduce A. R2 1 R1 S R2 R3 2 R1 S R3 £ 1 0 1 0 0 0 0 2 21 † 1 0 0 1 1 0 21 0 1 § R2 4 R3 £ 1 0 1 0 2 21 0 0 0 † 1 0 0 21 0 1 1 1 0 § 1 2R2 S R2 £ 1 0 1 0 1 21 2 0 0 0 † 1 0 0 21 2 0 1 2 1 1 0 § A 21 5 £ 1 0 0 21 2 0 1 2 1 1 0 § is incorrect because AA 21 2 I3. What mistake was made? 60. Find the inverse of A given that A 5 c2 5 3 10d . Solution: A 21 5 1 A A 21 5 1 c2 5 3 10d Simplify. A 21 5 c 1 2 1 5 1 3 1 10 d This is incorrect. What mistake was made? • C O N C E P T U A L 61. If A 5 c a11 a12 a21 a22d , then A 21 5 ≥ 1 a11 1 a21 1 a12 1 a22 ¥ . 62. All square matrices have inverses. 63. For what values of x does the inverse of A not exist, given A 5 c x 6 3 2d ? 64. Let A 5 £ a 0 0 0 b 0 0 0 c § . Find A21. • C H A L L E N G E 65. Verify that A21 5 1 ad 2 bc c d 2b 2c ad is the inverse of A 5 ca b c dd , provided ad 2 bc 2 0. 66. Let A 5 ca b c dd and form the matrix 3A 0 I24. Apply row operations to transform into 3I2 0 A214, where A21 5 1 ad 2 bc c d 2b 2c ad such that ad 2 bc 2 0. 67. Why does the square matrix A 5 c2 3 4 6d not have an inverse? 68. Why does the square matrix A 5 £ 1 2 21 2 4 22 0 1 3 § not have an inverse? In Exercises 61 and 62, determine whether each statement is true or false. 10.3 Matrix Equations; the Inverse of a Square Matrix 975 976 CHAPTER 10 Matrices • T E C H N O L O G Y In Exercises 69 and 70, apply a graphing utility to perform the indicated matrix operations. A 5 D 1 7 9 2 23 26 15 11 0 3 2 5 9 8 24 1 T 69. Find A21. 70. Find AA21. In Exercises 71–74, apply a graphing utility and matrix algebra to solve the system of linear equations. 71. 2.7x 2 3.1y 5 9.82 1.5x 1 2.7y 5 21.62 72. 3.7x 2 2.5y 5 31.77 25.1x 1 1.3y 5 239.07 73. 5.1x 1 7.3y 1 1.2z 5 12.51 2.3x 2 1.5y 1 4.5z 5 53.96 28.1x 1 5.4y 2 9.4z 5 2130.35 74. 12.4x 2 5.8y 1 2.7z 5 260.92 23.9x 1 1.9y 2 0.6z 5 18.73 6.4x 2 4.3y 1 8.5z 5 262.79 S K I L L S O B J E C T I V E S ■ ■nFind the determinant of a 2 3 2 matrix. ■ ■nFind the determinant of a n 3 n matrix. ■ ■n Use Cramer’s rule to solve a square system of linear equations in two variables. ■ ■ Use Cramer’s rule to solve a square system of linear equations in three variables. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that only a square matrix has a determinant. ■ ■Understand that it does not matter which row or which column is chosen to expand the determinant. ■ ■Derive Cramer’s rule. ■ ■If the determinant of the coefficient matrix is equal to zero, then the system has no unique solution 10.4 THE DETERMINANT OF A SQUARE MATRIX AND CRAMER’S RULE In Section 10.1, we discussed Gauss–Jordan elimination as a way to solve systems of linear equations using augmented matrices. Then in Section 10.3, we employed matrix algebra and inverses to solve systems of linear equations that are square (same number of equations as variables). In this section we will describe another method, called Cramer’s rule, for solving systems of linear equations. Cramer’s rule is applicable only to square systems. Determinants of square matrices play a vital role in Cramer’s rule and indicate whether a matrix has an inverse. 10.4.1 Determinant of a 2 3 2 Matrix Every square matrix A has a number associated with it called its determinant, denoted det 1A2 or 0 A 0. Although the symbol for determinant, 0 0, looks like absolute value bars, the determinant can be any real number (positive, negative, or zero). The determinant of a 2 3 2 matrix is found by finding the product of the main diagonal entries (top left to bottom right) and subtracting the product of the entries along the other diagonal (bottom left to top right). a c b d 5 ad 2 bc 10.4.1 SKI LL Find the determinant of a 2 3 2 matrix. 10.4.1 CO NC EPTUAL Understand that only a square matrix has a determinant. DEFINITION Determinant of a 2 3 2 Matrix The determinant of the 2 3 2 matrix A 5 ca b c dd is given by det1A2 5 0 A 0 5 a b c d 5 ad 2 bc. " " " " STUDY TIP The determinant of a 2 3 2 matrix is found by finding the product of the main diagonal entries and subtracting the product of the other diagonal entries. 10.4 The Determinant of a Square Matrix and Cramer’s Rule 977 10.4.2 Determinant of an n 3 n Matrix In order to define the determinant of a 3 3 3 or a general n 3 n 1where n $ 32 matrix, we first define minors and cofactors of a square matrix. ELEMENT, aij MINOR, Mij COFACTOR, Cij a11 5 1 For M11, delete the first row and first column: C11 5 1212111M11 5 1121232 5 23 £ 1 4 5 23 21 22 2 0 3 § M11 5 21 0 22 3 5 23 2 0 5 23 a32 5 22 For M32, delete the third row and second column: £ 1 23 2 4 21 0 5 22 3 § M32 5 1 2 4 0 5 0 2 8 5 28 C32 5 1212312M32 5 12121282 5 8 DEFINITION Minor and Cofactor Let A be a square matrix of order n 3 n. Then ■ ■The minor Mij of the element aij is the determinant of the 1n 2 12 3 1n 2 12 matrix obtained when the ith row and jth column of A are deleted. ■ ■The cofactor Cij of the element aij is given by Cij 5 1212i1jMij. EXAMPLE 1 Finding the Determinant of a 2 3 2 Matrix Find the determinant of each matrix. a. c 2 25 21 3d b. c 0.5 0.2 23.0 24.2d c. c 2 3 1 2 3d Solution: a. 2 25 21 3 5 122132 2 12121252 5 6 2 5 5 1 b. 0.5 0.2 23 24.2 5 10.52124.22 2 123210.22 5 22.1 1 0.6 5 21.5 c. 2 3 1 2 3 5 a2 3b 132 2 122112 5 2 2 2 5 0 In Example 1, we see that determinants are real numbers that can be positive, negative, or zero. Although evaluating determinants of 2 3 2 matrices is a simple process, one common mistake is reversing the difference: a b c d 2 bc 2 ad. Y OUR T UR N Evaluate the determinant 22 1 23 2 . ▼ A N S W E R 21 ▼ [CONCEPT CHECK] TRUE OR FALSE A determinant of a 2 3 2 matrix can be positive, negative, or zero. ANSWER True ▼ A 5 £ 1 23 2 4 21 0 5 22 3 § 10.4.2 S KI L L Find the determinant of an n 3 n matrix. 10.4.2 C O N C E P T U A L Understand that it does not matter which row or which column is chosen to expand the determinant. 978 CHAPTER 10 Matrices Notice that the cofactor is simply the minor multiplied by either 1 or 21, depending on whether i 1 j is even or odd. Therefore, we can make the following sign pattern for 3 3 3 and 4 3 4 matrices and obtain the cofactor by multiplying the minor with the appropriate sign 111 or 212. £ 1 2 1 2 1 2 1 2 1 § D 1 2 1 2 2 1 2 1 1 2 1 2 2 1 2 1 T Combining the definitions of minors, cofactors, and determinants, we now give a general definition for the determinant of a 3 3 3 matrix. Row 1 expansion: † a1 a2 a3 b1 b2 b3 c1 c2 c3 † 5 a1 b2 b3 c2 c3 2 b1 a2 a3 c2 c3 1 c1 a2 a3 b2 b3 Column 1 expansion: † a1 a2 a3 b1 b2 b3 c1 c 2 c3 † 5 a1 b2 b3 c2 c 3 2 a2 b1 b3 c1 c3 1 a3 b1 b2 c1 c2 Whichever row or column is expanded, an alternating sign scheme is used (see sign arrays above). Notice that in either of the expansions above, the 2 3 2 determinant obtained is found by crossing out the row and column containing the element that is multiplying the determinant. DEFINITION Determinant of an n 3 n Matrix Let A be an n 3 n matrix. Then the determinant of A is found by summing the elements in any row of A (or column of A), multiplied by each element’s respective cofactor. If A is a 3 3 3 matrix, the determinant can be given by det 1A2 5 a11C11 1 a12C12 1 a13C13; this is called expanding the determinant by the first row. It is important to note that any row or column can be used. Typically, the row or column with the most zeros is selected because it makes the arithmetic simpler. EXAMPLE 2 Finding the Determinant of a 3 3 3 Matrix For the given matrix, expand the determinant by £ 2 1 3 21 5 22 23 7 4 § the first row. Solution: Expand the determinant by the first row. Remember the alternating sign. † 2 1 3 21 5 22 23 7 4 † 5 1 2 5 22 7 4 2 1 21 22 23 4 1 3 21 5 23 7 Evaluate the resulting 2 3 2 determinants. 5 2 3152 142 2 172 12224 2 131212 142 2 1232 12224 1 3 31212 172 2 1232 1524 5 2 320 1 144 2 324 2 64 1 3 327 1 154 Simplify. 5 2 1342 2 12102 1 3182 5 68 1 10 1 24 5 102 Y OUR TU R N For the given matrix, expand the determinant by the first row. £ 1 3 22 2 5 4 7 21 6 § ▼ A N S W E R 156 ▼ STUDY TIP The determinant by the third column is also 102. It does not matter on which row or column the expansion occurs. Determinants can be expanded by any row or column. Typically, the row or column with the most zeros is selected. EXAMPLE 3 Finding the Determinant of a 3 3 3 Matrix Find the determinant of the matrix 3 21 4 5 2 7 3 0 1 0 † . Solution: Since there are two 0s in the third column, expand the £ 1 2 1 2 1 2 1 2 1 § determinant by the third column. Recall the sign array. 3 21 2 0 4 7 1 5 3 0 † 5 1 0 4 7 5 3 2 1 21 2 5 3 1 0 21 2 4 7 There is no need to calculate the two determinants that are multiplied by 0s, since 0 times any real number is zero. 3 21 4 5 2 7 3 0 1 0 † 5 0 2 1 21 5 2 3 1 0 Simplify. 5 2112132 5 13 Y OUR T UR N Evaluate the determinant 3 1 22 1 21 0 3 24 0 2 † . f 23210 ▼ A N S W E R 20 ▼ [CONCEPT CHECK] TRUE OR FALSE When finding the determinant of an n 3 n matrix, only the first row or first column can be chosen for expanding. ANSWER False ▼ EXAMPLE 4 Finding the Determinant of a 4 3 4 Matrix Find the determinant of the matrix 4 1 22 3 4 24 0 21 0 23 9 6 5 25 7 2 1 4 . Solution: Since there are two 0s in the second row, expand the determinant by the second row. Recall the sign array for a 4 3 4 matrix. D 1 2 1 2 2 1 2 1 1 2 1 2 2 1 2 1 T D 1 22 3 4 24 0 21 0 23 9 6 5 25 7 2 1 T 5 21242 3 22 3 4 9 6 5 7 2 1 † 1 0 2 1212 3 1 22 4 23 9 5 25 7 1 † 1 0 10.4 The Determinant of a Square Matrix and Cramer’s Rule 979 980 CHAPTER 10 Matrices 10.4.3 Cramer’s Rule: Systems of Linear Equations in Two Variables Let’s now apply determinants of 2 3 2 matrices to solve systems of linear equations in two variables. We begin by solving the general system of two linear equations in two variables: 112 a1x 1 b1y 5 c1 122 a2x 1 b2 y 5 c2 Solve for x using elimination (eliminate y). Multiply (1) by b2. b2a1x 1 b2b1y 5 b2c1 Multiply (2) by 2b1. 2b1a2x 2 b1b2y 5 2b1c2 Add the two new equations to eliminate y. 1a1b2 2 a2b12 x 5 1b2c1 2 b1c22 Divide both sides by 1a1b2 2 a2b12. x 5 1b2c1 2 b1c22 1a1b2 2 a2b12 Write both the numerator and x 5 c1 b1 c2 b2 a1 b1 a2 b2 the denominator as determinants. 10.4.3 SKI LL Use Cramer’s rule to solve a square system of linear equations in two variables. 10.4.3 CO NC EPTUAL Derive Cramer’s rule. Evaluate the two 3 3 3 determinants. 3 22 3 4 9 6 5 7 2 1 † 5 22 6 5 2 1 2 3 9 5 7 1 1 4 9 6 7 2 5 22 16 2 102 2 319 2 352 1 4 118 2 422 5 22 1242 2 312262 1 4 12242 5 8 1 78 2 96 5 210 3 1 22 4 23 9 5 25 7 1 † 5 1 9 5 7 1 2 1222 23 5 25 1 1 4 23 9 25 7 5 119 2 352 1 2 123 1 252 1 4 1221 1 452 5 226 1 2 1222 1 4 1242 5 226 1 44 1 96 5 114 D 1 22 3 4 24 0 21 0 23 9 6 5 25 7 2 1 T 5 4 3 22 3 4 9 6 5 7 2 1 † 1 3 1 22 4 23 9 5 25 7 1 † 5 412102 1 114 5 74 f 114 f 210 Solve for y using elimination (eliminate x). Multiply (1) by 2a2. 2a2a1x 2 a2b1y 5 2a2c1 Multiply (2) by a1. a1a2x 1 a1b2y 5 a1c2 Add the two new equations to eliminate x. 1a1b2 2 a2b12y 5 1a1c2 2 a2c12 Divide both sides by 1a1b2 2 a2b12. y 5 1a1c2 2 a2c12 1a1b2 2 a2b12 Write both the numerator and y 5 a1 c1 a2 c2 a1 b1 a2 b2 the denominator as determinants. Notice that the solutions for x and y involve three determinants. If we let D 5 a1 b1 a2 b2 Dx 5 c1 b1 c2 b2 Dy 5 a1 c1 a2 c2 then x 5 Dx D and y 5 Dy D . Notice that the real number D is the determinant of the coefficient matrix of the system and cannot equal zero 1D 2 02 or there will be no unique solution. These formulas for solving a system of two linear equations in two variables are known as Cramer’s rule. Notice that the determinants Dx and Dy are similar to the determinant D. A three-step procedure is outlined for setting up the three determinants for a system of two linear equations in two variables. a1x 1 b1y 5 c1 a2x 1 b2y 5 c2 Step 1: Set up D. Apply the coefficients of x and y. D 5 a1 b1 a2 b2 CRAMER’S RULE FOR SOLVING SYSTEMS OF TWO LINEAR EQUATIONS IN TWO VARIABLES For the system of linear equations a1x 1 b1y 5 c1 a2x 1 b2y 5 c2 let D 5 a1 b1 a2 b2 Dx 5 c1 b1 c2 b2 Dy 5 a1 c1 a2 c2 If D 2 0, then the solution to the system of linear equations is x 5 Dx D , y 5 Dy D If D 5 0, then the system of linear equations has either no solution or infinitely many solutions. STUDY TIP Cramer’s rule is only applicable to square systems of linear equations. 10.4 The Determinant of a Square Matrix and Cramer’s Rule 981 982 CHAPTER 10 Matrices [CONCEPT CHECK] TRUE OR FALSE Cramer’s rule for systems of linear equations in two variables is derived by solving a general system of linear equations in two variables using the elimination method. ANSWER True ▼ Recall from Section 6.1 that systems of two equations and two variables led to one of three possible outcomes: a unique solution, no solution, and infinitely many solutions. When D 5 0, Cramer’s rule does not apply and the system is either inconsistent (has no solution) or contains dependent equations (has infinitely many solutions). 10.4.4 Cramer’s Rule: Systems of Linear Equations in Three Variables Cramer’s rule can also be used to solve higher order systems of linear equations. The following box summarizes Cramer’s rule for solving a system of three equations in three variables. 10.4.4 SKI LL Use Cramer’s rule to solve a square system of linear equations in three variables. 10.4.4 CO NC EPTUAL If the determinant of the coefficient matrix is equal to zero, then the system has no unique solution. Step 2: Set up Dx. Start with D and replace the coefficients of x (column 1) with the constants on the right side of the equal sign. Dx 5 c1 b1 c2 b2 Step 3: Set up Dy. Start with D and replace the coefficients of y (column 2) with the constants on the right side of the equal sign. Dy 5 a1 c1 a2 c2 EXAMPLE 5 Using Cramer’s Rule to Solve a System of Two Linear Equations Apply Cramer’s rule to solve the system. x 1 3y 5 1 2x 1 y 5 23 Solution: Set up the three determinants. D 5 1 3 2 1 Dx 5 1 3 23 1 Dy 5 1 1 2 23 Evaluate the determinants. D 5 1 2 6 5 25 Dx 5 1 2 1292 5 10 Dy 5 23 2 2 5 25 Solve for x and y. x 5 Dx D 5 10 25 5 22 y 5 Dy D 5 25 25 5 1 Y OUR TU R N Apply Cramer’s rule to solve the system. 5x 1 4y 5 1 23x 2 2y 5 23 x 5 22, y 5 1 ▼ A N S W E R x 5 5, y 5 26 ▼ CRAMER’S RULE: SOLUTION FOR SYSTEMS OF THREE EQUATIONS IN THREE VARIABLES The system of linear equations a1x 1 b1y 1 c1z 5 d1 a2x 1 b2y 1 c2z 5 d2 a3x 1 b3y 1 c3z 5 d3 has the solution x 5 Dx D y 5 Dy D z 5 Dz D D 2 0 where the determinants are given as follows: WORDS MATH Display the coefficients of x, y, and z. D 5 3 a1 b1 c1 a2 b2 c2 a3 b3 c3 † Replace the coefficients of x (column 1) in D with the constants on the right side of the equal sign. Dx 5 3 d1 b1 c1 d2 b2 c2 d3 b3 c3 † Replace the coefficients of y (column 2) in D with the constants on the right side of the equal sign. Dy 5 3 a1 d1 c1 a2 d2 c2 a3 d3 c3 † Replace the coefficients of z (column 3) in D with the constants on the right side of the equal sign. Dz 5 3 a1 b1 d1 a2 b2 d2 a3 b3 d3 † EXAMPLE 6 Using Cramer’s Rule to Solve a System of Three Linear Equations Use Cramer’s rule to solve the system. 3x 2 2y 1 3z 5 23 5x 1 3y 1 8z 5 22 x 1 y 1 3z 5 1 Solution: Set up the four determinants. D contains the coefficients of x, y, and z. D 5 3 3 22 3 5 3 8 1 1 3 † Replace a column with constants on the right side of the equation. Dx 5 3 23 22 3 22 3 8 1 1 3 † Dy 5 3 3 23 3 5 22 8 1 1 3 † Dz 5 3 3 22 23 5 3 22 1 1 1 † 10.4 The Determinant of a Square Matrix and Cramer’s Rule 983 984 CHAPTER 10 Matrices As was the case in two equations, when D 5 0, Cramer’s rule does not apply and the system of three equations is either inconsistent (no solution) or contains dependent equations (infinitely many solutions). All of the coefficients and constants in the systems we have solved have been nice (rational). One of the advantages of Cramer’s rule is its ability to help you easily solve systems of linear equations with irrational coefficients. Evaluate the determinants. D 5 319 2 82 2 1222 115 2 82 1 315 2 32 5 23 Dx 5 2319 2 82 2 1222 126 2 82 1 3122 2 32 5 246 Dy 5 3126 2 82 2 1232 115 2 82 1 315 1 22 5 0 Dz 5 313 1 22 2 1222 15 1 22 2 315 2 32 5 23 Solve for x, y, and z. x 5 Dx D 5 246 23 5 22 y 5 Dy D 5 0 23 5 0 z 5 Dz D 5 23 23 5 1 x 5 22, y 5 0, z 5 1 Y OUR TU R N Use Cramer’s rule to solve the system. 2x 1 3y 1 z 5 21 x 2 y 2 z 5 0 23x 2 2y 1 3z 5 10 ▼ A N S W E R x 5 1, y 5 22, z 5 3 ▼ EXAMPLE 7 Using Cramer’s Rule to Solve a System of Linear Equations with Irrational Coefficients and Constants Apply Cramer’s rule to solve the system. 5x 2 ey 1 z 5 2!2 !3x 1 2y 2 pz 5 1 4x 2 7y 1 !5z 5 0 Solution: Set up the four determinants. D contains the coefficients of x, y, and z. D 5 † 5 2e 1 !3 2 2p 4 27 !5 † Replace a column with constants on the right side of the equation. Dx 5 † 2!2 2e 1 1 2 2p 0 27 !5 † Dy 5 † 5 2!2 1 !3 1 2p 4 0 !5 † Dz 5 † 5 2e 2!2 !3 2 1 4 27 0 † [CONCEPT CHECK] TRUE OR FALSE Cramer’s rule can only be used to solve a system of linear equations if there is a unique solution. If D = 0, then you have either no solution or infinitely many solutions. ANSWER True ▼ Evaluate the determinants (along the third row because of the zero). Dx 5 0 2 1272 2!2 1 1 2p 1 !5 2!2 2e 1 2 5 7C A2!2B12p2 2 112112D 1 !5 CA 2!2B122 2 11212e2D 5 7Cp!2 2 1D 1 !5 C22!2 1 eD 5 7p!2 2 7 2 2!10 1 e!5 Dy 5 4 2!2 1 1 2p 2 0 1 !5 5 2!2 !3 1 5 4CA2!2B12p2 2 112112D 1 !5 C152112 2 A !3BA2!2BD 5 4Cp!2 2 1D 1 !5 C5 1 !6D 5 4p!2 2 4 1 5!5 1 !30 Dz 5 4 2e 2!2 2 1 2 1272 5 2!2 !3 1 1 0 5 4C12e2112 2 122A2!2BD 1 7C152112 2 A !3BA2!2BD 5 4C2e 1 2!2D 1 7C5 1 !6D 5 24e 1 8!2 1 35 1 7!6 D 5 4 2e 1 2 2p 2 1272 5 1 !3 2p 1 !5 5 2e !3 2 5 4312e212p2 2 1221124 1 7C15212p2 2 A !3B112D 1 !5 C152122 2 A !3B12e2D 5 43ep 2 24 1 7C25p 2 !3 D 1 !5 C10 1 e!3 D 5 4ep 2 8 2 35p 2 7!3 1 10!5 1 e!15 Solve for x, y, and z. x 5 Dx D 5 7p!2 2 7 2 2!10 1 e!5 4ep 2 8 2 35p 2 7!3 1 10!5 1 e!15 y 5 Dy D 5 4p!2 2 4 1 5!5 1 !30 4ep 2 8 2 35p 2 7!3 1 10!5 1 e!15 z 5 Dz D 5 24e 1 8!2 1 35 1 7!6 4ep 2 8 2 35p 2 7!3 1 10!5 1 e!15 In this section, determinants were discussed for square matrices. ORDER DETERMINANT ARRAY 2 3 2 det1A2 5 0 A 0 5 a b c d 5 ad 2 bc 3 3 3 † a1 b1 c1 a2 b2 c2 a3 b3 c3 † 5 a1 b2 c2 b3 c3 2 b1 a2 c2 a3 c3 1 c1 a2 b2 a3 b3 Expansion by first row (any row or column can be used) £ 1 2 1 2 1 2 1 2 1 § [SEC TION 10.4] S U M M A RY 10.4 The Determinant of a Square Matrix and Cramer’s Rule 985 986 CHAPTER 10 Matrices Cramer’s rule was developed for 2 3 2 and 3 3 3 matrices, but it can be extended to general n 3 n matrices. When the coefficient determinant is equal to zero 1D 5 02, then the system is either inconsistent (and has no solution) or represents dependent equations (and has infinitely many solutions) and Cramer’s rule does not apply. SYSTEM ORDER SOLUTION DETERMINANTS a1x 1 b1y 5 c1 a2x 1 b2 y 5 c2 2 3 2 x 5 Dx D y 5 Dy D D 5 a1 b1 a2 b2 2 0 Dx 5 c1 b1 c2 b2 Dy 5 a1 c1 a2 c2 a1x 1 b1y 1 c1z 5 d1 a2x 1 b2 y 1 c2z 5 d2 a3x 1 b3y 1 c3z 5 d3 3 3 3 x 5 Dx D y 5 Dy D z 5 Dz D D 5 3 a1 b1 c1 a2 b2 c2 a3 b3 c3 3 2 0 Dx 5 3 d1 b1 c1 d2 b2 c2 d3 b3 c3 3 Dy 5 3 a1 d1 c1 a2 d2 c2 a3 d3 c3 3 Dz 5 3 a1 b1 d1 a2 b2 d2 a3 b3 d3 3 11. x 1 y 5 21 x 2 y 5 11 12. x 1 y 5 21 x 2 y 5 29 13. 3x 1 2y 5 24 22x 1 y 5 5 14. 5x 1 3y 5 1 4x 2 7y 5 218 15. 3x 2 2y 5 21 5x 1 4y 5 231 16. x 2 4y 5 27 3x 1 8y 5 19 17. 7x 2 3y 5 229 5x 1 2y 5 0 18. 6x 2 2y 5 24 4x 1 7y 5 41 19. 3x 1 5y 5 16 y 2 x 5 0 20. 22x 2 3y 5 15 7y 1 4x 5 233 21. 3x 2 5y 5 7 26x 1 10y 5 221 22. 3x 2 5y 5 7 6x 2 10y 5 14 23. 2x 2 3y 5 4 210x 1 15y 5 220 24. 2x 2 3y 5 2 10x 2 15y 5 20 25. 3x 1 1 2y 5 1 4x 1 1 3y 5 5 3 26. 3 2x 1 9 4y 5 9 8 1 3x 1 1 4y 5 1 12 27. 0.3x 2 0.5y 5 20.6 0.2x 1 y 5 2.4 28. 0.5x 2 0.4y 5 23.6 10x 1 3.6y 5 214 29. y 5 17x 1 7 y 5 215x 1 7 30. 9x 5 245 2 2y 4x 5 23y 2 20 31. 2 x 2 3 y 5 2 5 x 2 6 y 5 7 32. 2 x 2 3 y 5 212 3 x 1 1 2y 5 7 In Exercises 1–10, evaluate each 2 3 2 determinant. In Exercises 11–32, use Cramer’s rule to solve each system of equations, if possible. [SEC TION 10.4] E X E RC I SE S • S K I L L S 1. 1 2 3 4 2. 1 22 23 24 3. 7 9 25 22 4. 23 211 7 15 5. 0 7 4 21 6. 0 0 1 0 7. 21.2 2.4 20.5 1.5 8. 21.0 1.4 1.5 22.8 9. 3 4 1 3 2 8 9 10. 21 2 1 4 2 3 28 9 33. 3 3 1 0 2 0 21 24 1 0 † 34. 3 1 1 0 0 2 21 0 23 5 † 35. 3 2 1 25 3 0 21 4 0 7 † 36. 3 2 1 25 3 27 0 4 26 0 † 37. 3 1 1 25 3 27 24 4 26 9 † 38. 3 23 2 25 1 8 2 4 26 9 † 39. 3 1 3 4 2 21 1 3 22 1 † 40. 3 27 2 5 7 8 3 4 21 4 6 † 41. 3 23 1 5 2 0 6 4 7 29 † 42. 3 1 21 5 3 23 6 4 9 0 † 43. 3 22 1 27 4 22 14 0 1 8 † 44. 3 5 22 21 4 29 23 2 8 26 † 45. 3 3 4 21 0 0 1 5 212 8 0 22 † 46. 3 0.2 0 3 5 21.4 2 21 0 23 † In Exercises 33–46, evaluate each 3 3 3 determinant. For Exercises 61–64, the area of a triangle with vertices 1x1, y1 2 , 1x2, y2 2 , and 1x3, y32 is given by Area 5 61 2 3 x1 y1 1 x2 y2 1 x3 y3 1 † where the sign is chosen so that the area is positive. 61. Geometry. Apply determinants to find the area of a triangle with vertices 13, 22, 15, 22, and 13, 242. Check your answer by plotting these vertices in a Cartesian plane and using the area of a right triangle. 62. Geometry. Apply determinants to find the area of a triangle with vertices 12, 32, 17, 32, and 17, 72. Check your answer by plotting these vertices in a Cartesian plane and using the area of a right triangle. 63. Geometry. Apply determinants to find the area of a triangle with vertices 11, 22, 13, 42, and 122, 52. 64. Geometry. Apply determinants to find the area of a triangle with vertices 121, 222, 13, 42, and 12, 12. 65. Geometry. An equation of a line that passes through two points 1x1, y12 and 1x2, y22 can be expressed as a determinant: 3 x y 1 x1 y1 1 x2 y2 1 † 5 0 Apply the determinant to write an equation of the line passing through the points 11, 22 and 12, 42. Expand the determinant and express the equation of the line in slope–intercept form. 66. Geometry. If three points 1x1, y12, 1x2, y22, and 1x3, y32 are collinear (lie on the same line), then the following determinant must be satisfied: 3 x1 y1 1 x2 y2 1 x3 y3 1 † 5 0 Determine whether 10, 52, 12, 02, and 11, 22 are collinear. • A P P L I C A T I O N S 47. x 1 y 2 z 5 0 x 2 y 1 z 5 4 x 1 y 1 z 5 10 48. 2x 1 y 1 z 5 24 x 1 y 2 z 5 0 x 1 y 1 z 5 2 49. 3x 1 8y 1 2z 5 28 22x 1 5y 1 3z 5 34 4x 1 9y 1 2z 5 29 50. 7x 1 2y 2 z 5 21 6x 1 5y 1 z 5 16 25x 2 4y 1 3z 5 25 51. 3x 1 5z 5 11 4y 1 3z 5 29 2x 2 y 5 7 52. 3x 2 2z 5 7 4x 1 z 5 24 6x 2 2y 5 10 53. x 1 y 2 z 5 5 x 2 y 1 z 5 21 22x 2 2y 1 2z 5 210 54. x 1 y 2 z 5 3 x 2 y 1 z 5 22 22x 2 2y 1 2z 5 26 55. x 1 y 1 z 5 9 x 2 y 1 z 5 3 2x 1 y 2 z 5 5 56. x 1 y 1 z 5 6 x 2 y 2 z 5 0 2x 1 y 1 z 5 7 57. x 1 2y 1 3z 5 11 22x 1 3y 1 5z 5 29 4x 2 y 1 8z 5 19 58. 8x 2 2y 1 5z 5 36 3x 1 y 2 z 5 17 2x 2 6y 1 4z 5 22 59. x 2 4y 1 7z 5 49 23x 1 2y 2 z 5 217 5x 1 8y 2 2z 5 224 60. 1 2x 2 2y 1 7z 5 25 x 1 1 4 y 2 4z 5 22 24x 1 5y 5 256 In Exercises 47–60, apply Cramer’s rule to solve each system of equations, if possible. 10.4 The Determinant of a Square Matrix and Cramer’s Rule 987 988 CHAPTER 10 Matrices 67. Electricity: Circuit Theory. The following equations come from circuit theory. Find the currents I1, I2, and I3. I1 5 I2 1 I3 16 5 4I1 1 2I3 24 5 4I1 1 4I2 68. Electricity: Circuit Theory. The following equations come from circuit theory. Find the currents I1, I2, and I3. I1 5 I2 1 I3 24 5 6I1 1 3I3 36 5 6I1 1 6I2 16 V 8.0 V 3.0 Ω 1.0 Ω 3.0 Ω 2.0 Ω 1.0 Ω I1 I2 I3 24 V 12 V 4.0 Ω 1.0 Ω 5.0 Ω 3.0 Ω 2.0 Ω I1 I2 I3 • C A T C H T H E M I S T A K E In Exercises 69–72, explain the mistake that is made. 69. Evaluate the determinant 3 2 1 3 23 0 2 1 4 21 † . Solution: Expand the 3 3 3 determinant in terms of the 2 3 2 determinants. † 2 1 3 23 0 2 1 4 21 † 5 2 0 2 4 21 1 1 23 2 1 21 1 3 23 0 1 4 Expand the 2 3 2 determinants. 5 210 2 82 1 113 2 22 1 31212 2 02 Simplify. 5 216 1 1 2 36 5 251 This is incorrect. What mistake was made? 70. Evaluate the determinant 3 2 1 3 23 0 2 1 4 21 † . Solution: Expand the 3 3 3 determinant in terms of the 2 3 2 determinants. 3 2 1 3 23 0 2 1 4 21 † 5 2 0 2 4 21 2 1 23 2 1 21 1 3 23 2 1 21 Expand the 2 3 2 determinants. 5 210 2 82 2 113 2 22 1 313 2 22 Simplify. 5 216 2 1 1 3 5 214 This is incorrect. What mistake was made? 71. Solve the system of linear equations. 2x 1 3y 5 6 2x 2 y 5 23. Solution: Set up the determinants. D 5 2 3 21 21 , Dx 5 2 6 21 23 , and Dy 5 6 3 23 21 Evaluate the determinants. D 5 1, Dx 5 0, and Dy 5 3 Solve for x x 5 Dx D 5 0 1 5 0 and y 5 Dy D 5 3 1 5 3 and y. x 5 0, y 5 3 is incorrect. What mistake was made? 72. Solve the system of linear equations. 4x 2 6y 5 0 4x 1 6y 5 4 Solution: Set up the determinants. D 5 4 26 4 6 , Dx 5 0 26 4 6 , and Dy 5 4 0 4 4 Evaluate the determinants. D 5 48, Dx 5 24, and Dy 5 16 Solve for x x 5 D Dx 5 48 24 5 2 and y 5 Dy D 5 48 16 5 3 and y. x 5 2, y 5 3 is incorrect. What mistake was made? In Exercises 73–76, determine whether each statement is true or false. 73. The value of a determinant changes sign if any two rows are interchanged. 74. If all the entries in any column are equal to zero, the value of the determinant is 0. 75. † 2 6 4 0 2 8 4 0 10 † 5 2 † 1 3 2 0 1 4 2 0 5 † 76. 3 3 1 2 0 2 8 3 1 2 † 5 0 77. Calculate the determinant 3 a 0 0 0 b 0 0 0 c † . 78. Calculate the determinant 3 a1 b1 c1 0 b2 c2 0 0 c3 † . • C O N C E P T U A L 79. Evaluate the determinant: 4 1 22 21 3 4 0 1 2 0 3 2 4 1 23 5 24 4 80. For the system of equations 3x 1 2y 5 5 ax 2 4y 5 1 find a that guarantees no unique solution. • C H A L L E N G E 81. Show that 3 a1 b1 c1 a2 b2 c2 a3 b3 c3 † 5 a1b2c3 1 b1c2a3 1 c1a2b3 2 a3b2c1 2 b3c2a1 2 b1a2c3 by expanding down the second column. 82. Show that 3 a1 b1 c1 a2 b2 c2 a3 b3 c3 † 5 a1b2c3 1 b1c2a3 1 c1a2b3 2 a3b2c1 2 b3c2a1 2 c3b1a2 by expanding across the third row. In Exercises 83–86, use a graphing utility to evaluate the determinants. 83. 3 1 1 25 3 27 24 4 26 9 † Compare with your answer to Exercise 37. 84. 3 23 2 25 1 8 2 4 26 9 † Compare with your answer to Exercise 38. 85. 4 23 2 21 3 4 1 5 2 17 2 2 8 13 24 10 211 4 86. 4 23 21 19 3 4 1 16 2 17 31 2 5 13 24 10 2 4 In Exercises 87 and 88, apply Cramer’s rule to solve each system of equations and a graphing utility to evaluate the determinants. 87. 3.1x 1 1.6y 2 4.8z 5 233.76 5.2x 2 3.4y 1 0.5z 5 236.68 0.5x 2 6.4y 1 11.4z 5 25.96 88. 29.2x 1 2.7y 1 5.1z 5 289.2 4.3x 2 6.9y 2 7.6z 5 38.89 2.8x 2 3.9y 2 3.5z 5 34.08 • T E C H N O L O G Y 10.4 The Determinant of a Square Matrix and Cramer’s Rule 989 990 CHAPTER 10 Matrices [ C H AP T E R 10 REVIEW] CH A P TE R 10 R E VI E W SECTION CONCEPT KEY IDEAS/FORMULAS 10.1 Matrices and systems of linear equations Matrices F Column 1 Column 2 . . . Column j . . . Column n a11 a12 . . . a1j . . . a1n a21 a22 . . . a2j . . . a2n ( ( . . . ( . . . ( ai1 ai2 . . . aij . . . ain ( ( . . . ( . . . ( am1 am2 . . . amj . . . amn V Order: m 3 n Row 1 Row 2 ( Row i ( Row m Augmented matrices a1x 1 b1y 1 c1z 5 d1 a2x 1 b2y 1 c2z 5 d2 a3x 1 b3y 1 c3z 5 d3 1 £ a1 b1 c1 a2 b2 c2 a3 b3 c3 † d1 d2 d3 § Row operations on a matrix 1. Ri 4 Rj Interchange row i with row j. 2. cRi S Ri Multiply row i by the constant c. 3. cRi 1 Rj S Rj Multiply row i by the constant c and add to row j, writing the results in row j. Row–echelon form of a matrix A matrix is in row–echelon form if it has all three of the following properties: 1. Any rows consisting entirely of 0s are at the bottom of the matrix. 2. For each row that does not consist entirely of 0s, the first (leftmost) nonzero entry is 1 (called the leading 1). 3. For two successive nonzero rows, the leading 1 in the higher row is farther to the left than the leading 1 in the lower row. If a matrix in row–echelon form has the following additional property, then the matrix is in reduced row–echelon form: 4. Every column containing a leading 1 has zeros in every position above and below the leading 1. Gaussian elimination with back-substitution Step 1: Write the system of equations as an augmented matrix. Step 2: Apply row operations to transform the matrix into row–echelon form. Step 3: Apply back-substitution to identify the solution. Gauss–Jordan elimination Step 1: Write the system of equations as an augmented matrix. Step 2: Apply row operations to transform the matrix into reduced row–echelon form. Step 3: Identify the solution. Inconsistent and dependent systems No solution or infinitely many solutions CH A P TE R 10 R E VIE W SECTION CONCEPT KEY IDEAS/FORMULAS 10.2 Matrix algebra Equality of matrices The orders must be the same: Am3n and Bm3n. Matrix addition and subtraction The orders must be the same: Am3n and Bm3n. Scalar and matrix multiplication Scalar Multiplication Perform operation element by element. Each element is multiplied by the scalar. Matrix Multiplication • The orders must satisfy the relationship: Am 3 n and Bn 3 p, resulting in 1AB2 m 3 p. • Perform multiplication row by column. • Matrix multiplication is not commutative: AB 2 BA. 10.3 Matrix equations; the inverse of a square matrix Matrix equations Linear system: AX 5 B. Finding the inverse of a matrix Only square matrices, n 3 n, can have inverses. A21A 5 In Step 1: Form the matrix 3A 0 In4. Step 2: Use row operations to transform this matrix to 3In 0 A214. Note: Not every square matrix has an inverse. Solving systems of linear equations using matrix algebra and inverses of square matrices AX 5 B Step 1: Find A21. Step 2: X 5 A21 B. 10.4 The determinant of a square matrix and Cramer’s rule Cramer’s rule can be used to solve only a system of linear equations with a unique solution. Determinant of a 2 3 2 matrix a b c d 5 ad 2 bc Determinant of an n 3 n matrix Let A be a square matrix of order n 3 n; then • The minor Mij of the element aij is the determinant of the 1n 2 12 3 1n 2 12 matrix obtained when the ith row and j th column of A are deleted. • The cofactor Cij of the element aij is given by Cij 5 1212 i1jMij. £ 1 23 2 4 21 0 5 22 3 § M11 5 221 0 22 32 5 23 2 0 5 23 C11 5 1212111M11 5 1121232 5 23 Sign Array of a 3 3 3 matrix: £ 1 2 1 2 1 2 1 2 1 § If A is a 3 3 3 matrix, the determinant can be given by det 1A2 5 a11C11 1 a12C12 1 a13C13. This is called expanding the determinant by the first row. (Note that any row or column can be used.) † a1 b1 c1 a2 b2 c2 a3 b3 c3 † 5 a1 b2 c2 b3 c3 2 b1 a2 c2 a3 c3 1 c1 a2 b2 a3 b3 Chapter Review 991 CH A P TE R 10 R E VI E W 992 CHAPTER 10 Matrices SECTION CONCEPT KEY IDEAS/FORMULAS Cramer’s rule: Systems of linear equations in two variables The system a1x 1 b1y 5 c1 a2x 1 b2y 5 c2 has the solution x 5 Dx D y 5 Dy D if D 2 0 where D 5 a1 b1 a2 b2 Dx 5 c1 b1 c2 b2 Dy 5 a1 c1 a2 c2 Cramer’s rule: Systems of linear equations in three variables The system a1x 1 b1y 1 c1z 5 d1 a2x 1 b2y 1 c2z 5 d2 a3x 1 b3y 1 c3z 5 d3 has the solution x 5 Dx D y 5 Dy D z 5 Dz D if D 2 0 where D 5 † a1 b1 c1 a2 b2 c2 a3 b3 c3 † Dx 5 † d1 b1 c1 d2 b2 c2 d3 b3 c3 † Dy 5 † a1 d1 c1 a2 d2 c2 a3 d3 c3 † Dz 5 † a1 b1 d1 a2 b2 d2 a3 b3 d3 † R E VI E W E XERCISES [CH AP TER 10 REVIEW EXE R C IS E S ] Review Exercises 993 10.1 Matrices and Systems of Linear Equations Write the augmented matrix for each system of linear equations. 1. 5x 1 7y 5 2 2. 2.3x 2 4.5y 5 6.8 3x 2 4y 5 22 20.4x 1 2.1y 5 29.1 3. 2x 2 z 5 3 4. 2y 2 x 1 3z 5 1 y 2 3z 5 22 4z 2 2y 1 3x 5 22 x 1 4z 5 23 x 2 y 2 4z 5 0 Indicate whether each matrix is in row–echelon form. If it is, state whether it is in reduced row–echelon form. 5. c1 1 0 1 0 2d 6. c1 2 0 0 0 1d 7. £ 2 0 1 0 22 0 0 0 2 † 1 2 3 § 8. D 1 0 1 0 0 0 1 1 0 1 0 0 0 0 0 1 4 2 23 2 1 T Perform the indicated row operations on each matrix. 9. c1 22 0 22 1 2d 21 2 R2 S R2 10. c1 4 2 22 1 3d R2 2 2R1 S R2 11. £ 1 22 0 0 22 3 0 1 24 † 1 22 8 § R2 1 R1 S R1 12. D 1 1 1 6 0 2 22 3 0 0 1 22 0 21 3 23 4 0 22 4 3 T 22R1 1 R2 S R1 R4 1 R3 S R4 Apply row operations to transform each matrix to reduced row–echelon form. 13. c1 3 3 4 0 1d 14. £ 1 2 21 0 1 21 22 0 1 † 0 21 22 § 15. £ 4 1 22 1 0 21 22 1 1 † 0 0 12 § 16. £ 2 3 2 0 21 1 1 1 21 † 1 22 6 § Solve the system of linear equations using augmented matrices. 17. 3x 2 2y 5 2 18. 2x 2 7y 5 22 22x 1 4y 5 1 x 1 5y 5 223 19. 5x 2 y 5 9 20. 8x 1 7y 5 10 x 1 4y 5 6 23x 1 5y 5 42 21. x 2 2y 1 z 5 3 22. 3x 2 y 1 4z 5 18 2x 2 y 1 z 5 24 5x 1 2y 2 z 5 220 3x 2 3y 2 5z 5 2 x 1 7y 2 6z 5 238 23. x 2 4y 1 10z 5 261 24. 4x 2 2y 1 5z 5 17 3x 2 5y 1 8z 5 252 x 1 6y 2 3z 5 217 2 25x 1 y 2 2z 5 8 22x 1 5y 1 z 5 2 25. 3x 1 y 1 z 5 24 26. 2x 2 y 1 3z 5 6 x 2 2y 1 z 5 26 3x 1 2y 2 z 5 12 Applications 27. Fitting a Curve to Data. The average number of flights on a commercial plane that a person takes a year can be modeled by a quadratic function y 5 ax2 1 bx 1 c, where a , 0 and x represents age: 16 , x , 65. The following table gives the average number of flights per year that a person takes on a commercial airline. Determine the quadratic function that models this quantity. AGE NUMBER OF FLIGHTS PER YEAR 16 2 40 6 65 0 28. Investment Portfolio. Danny and Paula decide to invest $20,000 of their savings. They put some in an IRA account earning 4.5% interest, some in a mutual fund that has been averaging 8% a year, and some in a stock that earned 12% last year. If they put $3000 more in the mutual fund than in the IRA and the mutual fund and stock have the same growth in the next year as they did in the previous year, they will earn $1877.50 in a year. How much money did they put in each of the three investments? 10.2 Matrix Algebra Calculate the given expression, if possible. A 5 c2 23 0 1d B 5 c1 5 21 3 7 2d C 5 £ 5 0 1 2 21 4 0 3 6 § D 5 c5 2 9 7d E 5 c2 0 3 4 1 21d 29. A 1 C 30. B 1 A 31. B 1 E 32. A 1 D 33. 2A 1 D 34. 3E 1 B 35. 2D 2 3A 36. 3B 2 4E 37. 5A 2 2D 38. 5B 2 4E 39. AB 40. BC 41. DA 42. AD 43. BC 1 E 44. DB 45. EC 46. CE REV IEW E XE R CI SE S 994 CHAPTER 10 Matrices 10.3 Matrix Equations; The Inverse of a Square Matrix Determine whether B is the multiplicative inverse of A using AA21 5 I. 47. A 5 c6 4 4 2d B 5 c20.5 1 1 21.5d 48. A 5 c1 22 2 24d B 5 c1 2 2 22d 49. A 5 £ 1 22 6 2 3 22 0 21 1 § B 5 £ 21 7 4 7 2 2 7 21 7 22 2 7 21 7 21 § 50. A 5 £ 0 7 6 1 0 24 22 1 0 § B 5 £ 1 1 1 22 22 22 2 0 6 § Find A21, if it exists. 51. A 5 c 1 2 23 4d 52. A 5 c22 7 24 6d 53. A 5 c 0 1 22 0d 54. A 5 c 3 21 22 2d 55. A 5 £ 1 3 22 2 1 21 0 1 23 § 56. A 5 £ 0 1 0 4 1 2 23 22 1 § 57. A 5 £ 21 1 0 22 1 2 1 2 4 § 58. A 5 £ 24 4 3 1 2 2 3 21 6 § Solve the system of linear equations using matrix algebra. 59. 3x 2 y 5 11 60. 6x 1 4y 5 15 5x 1 2y 5 33 23x 2 2y 5 21 61. 5 8x 2 2 3y 5 23 62. x 1 y 2 z 5 0 3 4x 1 5 6y 5 16 2x 2 y 1 3z 5 18 3x 2 2y 1 z 5 17 63. 3x 2 2y 1 4z 5 11 64. 2x 1 6y 2 4z 5 11 6x 1 3y 2 2z 5 6 2x 2 3y 1 2z 5 211 2 x 2 y 1 7z 5 20 4x 1 5y 1 6z 5 20 10.4 The Determinant of a Square Matrix and Cramer’s Rule Evaluate each 2 3 2 determinant. 65. 2 4 3 2 66. 22 24 23 2 67. 2.4 22.3 3.6 21.2 68. 21 4 4 3 4 24 Employ Cramer’s rule to solve each system of equations, if possible. 69. x 2 y 5 2 70. 3x 2 y 5 217 x 1 y 5 4 2x 1 5y 5 43 71. 2x 1 4y 5 12 72. 2x 1 y 5 4 x 2 2y 5 6 2x 2 6y 5 25 73. 23x 5 40 2 2y 74. 3x 5 20 1 4y 2x 5 25 1 y y 2 x 5 26 Evaluate each 3 3 3 determinant. 75. † 1 2 2 0 1 3 2 21 0 † 76. † 0 22 1 0 23 7 1 210 23 † 77. † a 0 2b 2a b c 0 0 2d † 78. † 22 24 6 2 0 3 21 2 3 4 † Employ Cramer’s rule to solve each system of equations, if possible. 79. x 1 y 2 2z 5 22 80. 2x 2 y 1 z 5 3 2x 2 y 1 z 5 3 x 1 2y 2 2z 5 8 x 1 y 1 z 5 4 2x 1 y 1 4z 5 24 81. 3x 1 4z 5 21 82. x 1 y 1 z 5 0 x 1 y 1 2z 5 23 2x 2 3y 1 5z 5 22 y 2 4z 5 29 2x 1 y 2 3z 5 24 Applications 83. Apply determinants to find the area of a triangle with vertices 12, 42, 14, 42, and 124, 32. 84. If three points 1x1, y12, 1x2, y22, and 1x3, y32 are collinear (lie on the same line), then the following determinant must be satisfied: † x1 y1 1 x2 y2 1 x3 y3 1 † 5 0 Determine whether 10, 232, 13, 02, and 11, 62 are collinear. Technology Exercises Section 10.1 In Exercises 85 and 86, refer to the following: You are asked to model a set of three points with a quadratic function y 5 ax2 1 bx 1 c and determine the quadratic function. a. Set up a system of equations; use a graphing utility or graphing calculator to solve the system by entering the coefficients of the augmented matrix. b. Use the graphing calculator commands STAT QuadReg to model the data using a quadratic function. Round your answers to two decimal places. R E VI E W E XERCISES 85. 1210, 12.52, 13, 22.82, 19, 8.52 86. 124, 102, 12.5, 29.52, 113.5, 12.62 Section 10.2 Apply a graphing utility to perform the indicated matrix operations, if possible. A 5 £ 26 0 4 1 3 5 2 21 0 § B 5 £ 5 1 0 2 28 4 § C 5 c4 23 0 1 2 5d 87. ABC 88. CAB Section 10.3 Apply a graphing utility and matrix algebra to solve the system of linear equations. 89. 6.1x 2 14.2y 5 75.495 22.3x 1 7.2y 5 236.495 90. 7.2x 1 3.2y 2 1.7z 5 5.53 21.3x 1 4.1y 1 2.8z 5 223.949 Section 10.4 Apply Cramer’s rule to solve each system of equations and use a graphing utility to evaluate the determinants. 91. 4.5x 2 8.7y 5 272.33 21.4x 1 5.3y 5 31.32 92. 1.4x 1 3.6y 1 7.5z 5 42.08 2.1x 2 5.7y 2 4.2z 5 5.37 1.8x 2 2.8y 2 6.2z 5 29.86 Review Exercises 995 PR ACTICE TEST 996 CHAPTER 10 Matrices [ C H AP T E R 1 0 PRACTIC E TES T ] Write each of the following systems of linear equations as an augmented matrix. 1. x 2 2y 5 1 2x 1 3y 5 2 2. 3x 1 5y 5 22 7x 1 11y 5 26 3. 6x 1 9y 1 z 5 5 2x 2 3y 1 z 5 3 10x 1 12y 1 2z 5 9 4. 3x 1 2y 2 10z 5 2 x 1 y 2 z 5 5 5. Perform the following row operations: £ 1 3 5 2 7 21 23 22 0 § R2 2 2R1 S R2 R3 1 3R1 S R3 6. Rewrite the following matrix in reduced row–echelon form: £ 2 21 1 1 1 21 3 2 22 † 3 0 1 § In Exercises 7 and 8, solve the systems of linear equations using augmented matrices. 7. 6x 1 9y 1 z 5 5 2x 2 3y 1 z 5 3 10x 1 12y 1 2z 5 9 8. 3x 1 2y 2 10z 5 2 x 1 y 2 z 5 5 9. Multiply the matrices, if possible. c1 22 5 0 21 3d £ 0 4 3 25 21 1 § 10. Add the matrices, if possible. c1 22 5 0 21 3d 1 £ 0 4 3 25 21 1 § 11. Find the inverse of c4 3 5 21d , if it exists. 12. Find the inverse of £ 1 23 2 4 2 0 21 2 5 § , if it exists. 13. Find the inverse of c3 1 0 5 2 21d , if it exists. 14. Solve the system of linear equations with matrix algebra (inverses). x 1 3y 5 21 22x 2 5y 5 4 15. Solve the system of linear equations with matrix algebra (inverses). 3x 2 y 1 4z 5 18 x 1 2y 1 3z 5 20 24x 1 6y 2 z 5 11 Calculate the determinant. 16. 7 25 2 21 17. 3 1 22 21 3 25 2 4 21 0 † In Exercises 18 and 19, solve the system of linear equations using Cramer’s rule. 18. x 2 2y 5 1 2x 1 3y 5 2 19. 3x 1 5y 2 2z 5 26 7x 1 11y 1 3z 5 2 x 2 y 1 z 5 4 20. A company has two rubrics for scoring job applicants based on weighting education, experience, and the interview differently. Rubric 1 Rubric 2 Matrix A: Education 0.4 0.6 Experience £ 0.5 0.1 Interview 0.1 0.3 § Applicants receive a score from 1 to 10 in each category (education, experience, and interview). Two applicants are shown in the matrix B: Education Experience Interview Matrix B: Applicant 1 Applicant 2 c 4 7 3 6 5 4 d What is the order of BA? What does each entry in BA tell us? 21. A college student inherits $15,000 from his favorite aunt. He decides to invest it with a diversified scheme. He divides the money into a money market account paying 2% annual simple interest, a conservative stock that rose 4% last year, and an aggressive stock that rose 22% last year. He places $1000 more in the conservative stock than in the money market and twice as much in the aggressive stock than in the money market. If the stocks perform the same way this year, he will make $1790 in one year, assuming a simple interest model. How much did he put in each of the three investments? 22. You are asked to model a set of three points with a quadratic function y 5 ax2 1 bx 1 c and determine the quadratic function. a. Set up a system of equations; use a graphing utility or graphing calculator to solve the system by entering the coefficients of the augmented matrix. b. Use the graphing calculator commands STAT QuadReg to model the data using a quadratic function. 123, 62, 11, 122, 15, 72 23. Apply a graphing utility and matrix algebra to solve the system of linear equations. 5.6x 2 2.7y 5 87.28 24.2x 1 8.4y 5 2106.26 24. Apply Cramer’s rule to solve the system of equations and a graphing utility to evaluate the determinants. 1.5x 1 2.6y 5 18.34 22.3x 1 1.5y 5 28.94 CU MU LA TIV E TEST [CH AP TERS 1–10 CUM ULAT IVE T E S T ] 1. Solve by completing the square: x2 2 6x 5 11. 2. Write an equation of a line that passes through the point 122, 52 and is parallel to the y-axis. 3. Write the equation of a circle with center 123, 212 and passing through the point 11, 22. 4. Determine whether the relation x2 2 y2 5 25 is a function. 5. Determine whether the function g1x2 5 "2 2 x2 is odd or even. 6. For the function y 5 5 1x 2 422, identify all of the transformations of y 5 x2. 7. Find the composite function ƒ + g, and state the domain, for ƒ1x2 5 x3 2 1 and g1x2 5 1 x. 8. Find the inverse of the function ƒ1x2 5 ! 3 x 2 1. 9. Find the vertex of the parabola associated with the quadratic function ƒ1x2 5 1 4x2 1 3 5 x 2 6 25. 10. Find a polynomial of minimum degree (there are many) that has the zeros x 5 2!7 1multiplicity 22, x 5 0 1multiplicity 32, x 5 !7 1multiplicity 22. 11. Use long division to find the quotient Q 1x2 and the remainder r 1x2 of 15x3 2 4x2 1 32 4 1x2 1 12. 12. Given the zero x 5 4i of the polynomial P 1x2 5 x4 1 2x3 1 x2 1 32x 2 240, determine all the other zeros and write the polynomial in terms of a product of linear factors. 13. Find the vertical and horizontal asymptotes of the function ƒ1x2 5 0.7x2 2 5x 1 11 x2 2 x 2 6 . 14. If $5400 is invested at 2.25% compounded continuously, how much is in the account after 4 years? 15. Use interval notation to express the domain of the function ƒ1x2 5 log4 1x 1 32. 16. Use the properties of logarithms to simplify the expression logp1. 17. Give an exact solution to the logarithmic equation log5 1x 1 22 1 log5 16 2 x2 5 log5 13x2. 18. If money is invested in a savings account earning 4% compounded continuously, how many years will pass until the money triples? 19. Solve the following system of linear equations: 2x 1 3y 5 6 x 5 21.5y 2 5.5 20. At the student union, one group of students bought 6 deluxe burgers, 5 orders of fries, and 5 sodas for $24.34. A second group of students ordered 8 deluxe burgers, 4 orders of fries, and 6 sodas for $28.42. A third group of students ordered 3 deluxe burgers, 2 orders of fries, and 4 sodas for $13.51. Determine the price of each food item. 21. Maximize the objective function z 5 5x 1 7y, subject to the constraints: 2x 1 5y # 10 x $ 0 5x 1 2y # 10 y $ 0 22. Solve the system with Gauss–Jordan elimination. x 2 2y 1 3z 5 11 4x 1 5y 2 z 5 28 3x 1 y 2 2z 5 1 23. Given A5 c3 4 27 0 1 5d B 5 c8 22 6 9 0 21d C 5 c9 0 1 2d find 2A 1 CB. 24. Write the matrix equation, find the inverse of the coefficient matrix, and solve the system using matrix algebra. 2x 1 5y 5 21 2x 1 4y 5 7 25. Apply Cramer’s rule to solve the system of equations. 7x 1 5y 5 1 2x 1 4y 5 21 26. Use a graphing calculator to graph the given polynomial. From the graph, estimate x-intercepts and the coordinates of the relative maximum and minimum points. Round your answers to two decimal places. f 1x2 5 x3 1 5x2 2 12x 2 13 27. Given A 5 c0 21 6 4 3 1d B 5 £ 27 24 1 0 6 2 § find AB and 1AB221. Cumulative Test 997 C H A P T E R LEARNING OBJECTIVES [ [ 11 ■ ■Determine if a general second-degree equation in two variables corresponds to a parabola, ellipse, or hyperbola. ■ ■Graph a parabola in rectangular coordinates. ■ ■Graph an ellipse in rectangular coordinates. ■ ■Graph a hyperbola in rectangular coordinates. ■ ■Solve systems of nonlinear equations and interpret some graphically in terms of conics. In this chapter, we will study the three types of conic sections (conics): the parabola, the ellipse, and the hyperbola. We have already studied the circle (Section 2.4), which is a special form of the ellipse. We see these shapes all around us: satellite dishes (parabolas), cooling towers (hyperbolas), and planetary orbits (ellipses). Analytic Geometry and Systems of Nonlinear Equations and Inequalities Fall Fall Spring Spring Winter Winter Summer Summer The Earth’s orbit around the Sun is an ellipse. Ribeiro antonio/Shutterstock A satellite dish is a paraboloid. Bloomberg/Getty Images, Inc. Some buildings have a hyperbolic shape. 999 [I N T HI S CHAPTER] We define the three conic sections: the parabola, the ellipse, and the hyperbola. Algebraic equations and the graphs of these conics are discussed. We solve systems of nonlinear equations and inequalities involving parabolas, ellipses, and hyperbolas. We then will rotate the axes and identify the graph of a general second-degree equation as one of the three conics. We will discuss the equations of the conics first in rectangular coordinates and then in polar coordinates. Last, we will look at parametric equations, which give orientation along a plane curve. ANALYTIC GEOMETRY AND SYSTEMS OF NONLINEAR EQUATIONS AND INEQUALITIES 11.1 CONIC BASICS 11.2 THE PARABOLA 11.3 THE ELLIPSE 11.4 THE HYPERBOLA 11.5 SYSTEMS OF NONLINEAR EQUATIONS 11.6 SYSTEMS OF NONLINEAR INEQUALITIES 11.7 ROTATION OF AXES 11.8 POLAR EQUATIONS OF CONICS 11.9 PARAMETRIC EQUATIONS AND GRAPHS • Three Types of Conics • Parabola with a Vertex at the Origin • Parabola with a Vertex at the Point (h, k) • Applications • Ellipse Centered at the Origin • Ellipse Centered at the Point (h, k) • Applications • Hyperbola Centered at the Origin • Hyperbola Centered at the Point (h, k) • Applications • Solving a System of Nonlinear Equations • Nonlinear Inequalities in Two Variables • Systems of Nonlinear Inequalities • Rotation of Axes Formulas • The Angle of Rotation Necessary to Trans- form a General Second- Degree Equation into an Equation of a Conic • Equations of Conics in Polar Coordinates • Parametric Equations of a Curve • Applications of Parametric Equations ■ ■Solve systems of nonlinear inequalities and interpret some graphically in terms of conics. ■ ■Graph general second-degree polynomial functions by rotation of axes. ■ ■Graph parabolas, ellipses, and hyperbolas in polar coordinates. ■ ■Use parametric equations to describe orientation along a plane curve. 1000 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities S K I L L S O B J E C T I V E ■ ■Classify a conic as an ellipse, a parabola, or a hyperbola. C O N C E P T U A L O B J E C T I V E ■ ■Understand that conics are graphs that correspond to a second-degree equation in two variables. 11.1 CONIC BASICS 11.1.1 Three Types of Conics Names of Conics The word conic is derived from the word cone. Let’s start with a (right-circular) double cone (see figure on the left). Conic sections are curves that result from the intersection of a plane and a double cone. The four conic sections are a circle, an ellipse, a parabola, and a hyperbola. “Conics” is an abbreviation for conic sections. Circle Ellipse Parabola Hyperbola In Section 2.4, circles were discussed, and it will be shown that a circle is a particular type of an ellipse. Now we will discuss parabolas, ellipses, and hyperbolas. There are two ways in which we usually describe conics: graphically and algebraically. An entire section will be devoted to each of the three conics, but in the present section we will summarize the definitions of a parabola, ellipse, and hyperbola, and show how to identify the equations of these conics. Definitions You already know that a circle consists of all points equidistant (at a distance equal to the radius) from a point (the center). Ellipses, parabolas, and hyperbolas have similar definitions in that they all have a constant distance (or a sum or difference of distances) to some reference point. A parabola is the set of all points that are equidistant from both a line and a point. An ellipse is the set of all points, the sum of whose distances to two fixed points is constant. A hyperbola is the set of all points, the difference of whose distances to two fixed points is a constant. d1 d2 d1 d2 Parabola d2 d3 d1 d1 + d2 = d3 + d4 d4 Ellipse d2 d4 d1 \|d2 – d1\| = \|d4 – d3\| d3 Hyperbola The general form of a second-degree equation in two variables x and y is given by Ax2 1 Bxy 1 Cy2 1 Dx 1 Ey 1 F 5 0 If we let A 5 1, B 5 0, C 5 1, D 5 0, E 5 0, and F 5 2r2, this general equation reduces to the equation of a circle centered at the origin: x2 1 y2 5 r2. In fact, all three 11.1.1 SKILL Classify a conic as an ellipse, a parabola, or a hyperbola. 11.1.1 CON CEPTUAL Understand that conics are graphs that correspond to a second-degree equation in two variables. STUDY TIP A circle is a special type of ellipse. All circles are ellipses, but not all ellipses are circles. 11.1 Conic Basics 1001 conics (parabolas, ellipses, and hyperbolas) are special cases of the general second- degree equation. Recall from Section 1.3 (Quadratic Equations) that the discriminant b2 2 4ac determines what types of solutions result from solving a second-degree equation in one variable. If the discriminant is positive, the solutions are two distinct real roots. If the discriminant is zero, the solution is a real repeated root. And if the discriminant is negative, the solutions are two complex conjugate roots. The concept of discriminant has been generalized to conic sections. The discriminant determines the shape of the conic section. CONIC DISCRIMINANT Ellipse B2 2 4AC , 0 Parabola B2 2 4AC 5 0 Hyperbola B2 2 4AC . 0 Using the discriminant to identify the shape of the conic will not work for degenerate cases (when the polynomial can be factored). For example, 2x2 2 xy 2 y2 5 0. At first glance one may think it is a hyperbola because B2 2 4AC . 0, but this is a degenerative case: 12x 1 y2 1x 2 y2 5 0 2x 1 y 5 0 or x 2 y 5 0 y 5 22x or y 5 x The graph is two intersecting lines. We now identify conics from the general form of a second-degree equation in two variables. EXAMPLE 1 Determining the Type of Conic Determine what type of conic corresponds to each of the following equations. a. x2 a2 1 y2 b2 5 1 b. y 5 x2 c. x2 a2 2 y2 b2 5 1 Solution: Write the general form of the second-degree equation: Ax2 1 Bxy 1 Cy2 1 Dx 1 Ey 1 F 5 0 a. Identify A, B, C, D, E, and F A 5 1 a2, B 5 0, C 5 1 b2, D 5 0, E 5 0, F 5 21 Calculate the discriminant. B2 2 4AC 5 2 4 a2b2 , 0 Since the discriminant is negative, the equation x2 a2 1 y2 b2 5 1 is that of an ellipse . Notice that if a 5 b 5 r, then this equation of an ellipse reduces to the general equation of a circle x2 1 y2 5 r2, centered at the origin with radius r. b. Identify A, B, C, D, E, and F. A 5 1, B 5 0, C 5 0, D 5 0, E 5 21, F 5 0 Calculate the discriminant. B2 2 4AC 5 0 Since the discriminant is zero, the equation y 5 x2 is a parabola . STUDY TIP All circles are ellipses since B2 2 4AC , 0. [CONCEPT CHECK] TRUE OR FALSE The graph of a third-degree equation in two variables is also a conic section. ANSWER False ▼ 1002 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities c. Identify A, B, C, D, E, and F . A 5 1 a2, B 5 0, C 5 2 1 b2, D 5 0, E 5 0, F 5 21 Calculate the discriminant. B2 2 4AC 5 4 a2b2 . 0 Since the discriminant is positive, the equation x2 a2 2 y2 b2 5 1 is a hyperbola . Y OUR TU R N Determine what type of conic corresponds to each of the following equations. a. 2x2 1 y2 5 4 b. 2x2 5 y2 1 4 c. 2y2 5 x ▼ A N S W E R a. ellipse b. hyperbola c. parabola ▼ In the next three sections, we will discuss the standard forms of equations and the graphs of parabolas, ellipses, and hyperbolas. [SEC TION 11.1] E X E R C I SE S • S K I L L S In Exercises 1–12, identify the conic section as a parabola, ellipse, circle, or hyperbola. 1. x2 1 xy 2 y2 1 2x 5 23 2. x2 1 xy 1 y2 1 2x 5 23 3. 2x2 1 2y2 5 10 4. x2 2 4x 1 y2 1 2y 5 4 5. 2x2 2 y2 5 4 6. 2y2 2 x2 5 16 7. 5x2 1 20y2 5 25 8. 4x2 1 8y2 5 30 9. x2 2 y 5 1 10. y2 2 x 5 2 11. x2 1 y2 5 10 12. x2 1 y2 5 100 In this section we defined the three conic sections and determined their general equations with respect to the general form of a second-degree equation in two variables: Ax2 1 Bxy 1 Cy2 1 Dx 1 Ey 1 F 5 0 The following table summarizes the three conics: ellipse, parabola, and hyperbola. CONIC GEOMETRIC DEFINITION: THE SET OF ALL POINTS DISCRIMINANT Ellipse The sum of whose distances to two fixed points is constant. Negative: B2 2 4AC , 0 Parabola Equidistant to both a line and a point. Zero: B2 2 4AC 5 0 Hyperbola The difference of whose distances to two fixed points is a constant. Positive: B2 2 4AC . 0 It is important to note that a circle is a special type of ellipse. [SEC TION 11.1] S U M M A RY 11.2 The Parabola 1003 11.2.1 Parabola with a Vertex at the Origin Recall from Section 4.1 that the graphs of quadratic functions such as ƒ1x2 5 a1x 2 h22 1 k or y 5 ax2 1 bx 1 c were parabolas that opened either upward or downward. We now expand our discussion of parabolas to parabolas that open to the right or left. We did not discuss these types of parabolas before because they are not functions (they fail the vertical line test). S K I L L S O B J E C T I V E S ■ ■Find the equation of a parabola whose vertex is at the origin. ■ ■Find the equation of a parabola whose vertex is at the point 1h, k2. ■ ■Solve applied problems that involve parabolas. C O N C E P T U A L O B J E C T I V ES ■ ■Understand that a parabola is the set of all points that are equidistant from a fixed line (directrix) and a fixed point not on that line (focus). ■ ■Employ completing the square to transform an equation into the standard form of a parabola. ■ ■Understand that it is the focus that is the key to applications. 11.2 THE PARABOLA Let’s consider a parabola with the vertex at the origin and the focus on the positive x-axis. Let the distance from the vertex to the focus be p. Therefore, the focus is located at the point 1 p, 02. Since the distance from the vertex to the focus is p, the distance from the vertex to the directrix must also be p. Since the axis of symmetry is the positive axis, the directrix must be perpendicular to the positive axis. Therefore, the directrix is given by x 5 2p. Any point 1x, y2 must have the same distance to the focus 1 p, 02 as it does to the directrix 12p, y2. x y x = y2 x = –y2 11.2.1 S KI L L Find the equation of a parabola whose vertex is at the origin. 11.2.1 C ON C E P T U A L Understand that a parabola is the set of all points that are equidistant from a fixed line (directrix) and a fixed point not on that line (focus). A parabola is the set of all points in a plane that are equidistant from a fixed line, the directrix, and a fixed point not on the line, the focus. The line through the focus and perpendicular to the directrix is the axis of symmetry. The vertex of the parabola is located at the midpoint between the directrix and the focus along the axis of symmetry. Axis of symmetry Directrix Vertex Focus Parabola p p Here p is the distance along the axis of symmetry from the directrix to the vertex and from the vertex to the focus. Parabola DEFINITION x = –p Directrix (p, 0) Focus x y 1004 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities Derivation of the Equation of a Parabola WORDS MATH Calculate the distance from 1x, y2 to 1 p, 02 with the distance formula. d 5 "1x 2 p22 1 y2 Calculate the distance from 1x, y2 to 12p, y2 with the distance formula. d 5 "1x 2 12p222 1 02 Set the two distances equal to one another. 1Definition of a parabola.2 "1x 2 p22 1 y2 5 "1x 1 p22 Recall that "x2 5 x. "1x 2 p22 1 y2 5 x 1 p Square both sides of the equation. 1x 2 p22 1 y2 5 1x 1 p22 Square the binomials inside the parentheses. x2 2 2px 1 p2 1 y2 5 x2 1 2px 1 p2 Simplify. y2 5 4px The equation y2 5 4px represents a parabola opening right with the vertex at the origin. The following box summarizes parabolas that have a vertex at the origin and a focus along either the x-axis or the y-axis. EQUATION OF A PARABOLA WITH VERTEX AT THE ORIGIN The standard (conic) form of the equation of a parabola with vertex at the origin is given by: Equation y2 5 4px x2 5 4py Vertex 10, 02 10, 02 Focus 1 p, 02 10, p2 Directrix x 5 2p y 5 2p Axis of symmetry x-axis y-axis p + 0 Opens to the right Opens upward p 0 Opens to the left Opens downward Graph 1p + 02 Directrix x = –p Focus (p, 0) y2 = 4px p p x y Directrix y = –p Focus (0, p) x2 = 4py p p x y [CONCEPT CHECK] TRUE OR FALSE The directrix is perpendicular to the axis along which the focus of a parabola lies. ANSWER True ▼ EXAMPLE 1 Finding the Focus and Directrix of a Parabola Whose Vertex Is Located at the Origin Find the focus and directrix of a parabola whose equation is y2 5 8x. Solution: Compare this parabola with the general equation of a parabola. y2 5 4px y2 5 8x Solve for p. 4p 5 8 p 5 2 The focus of a parabola of the form y2 5 4px is 1 p, 02. Focus: 12, 02 The directrix of a parabola of the form y2 5 4px is x 5 2p. Directrix: x = 22 Y OUR TU R N Find the focus and directrix of a parabola whose equation is y2 5 16x. ▼ ▼ A N S W E R The focus is 14, 02, and the directrix is x 5 24. 11.2 The Parabola 1005 Graphing a Parabola Whose Vertex Is at the Origin When a seamstress starts with a pattern for a custom-made suit, the pattern is used as a guide. The pattern is not sewn into the suit but rather removed once it is used to determine the exact shape and size of the fabric to be sewn together. The focus and directrix of a parabola are similar to the pattern used by a seamstress. Although the focus and directrix define a parabola, they do not appear on the graph of a parabola. An approximate sketch of a parabola whose vertex is at the origin can be drawn with three pieces of information. We know that the vertex is located at 10, 02. Additional information that we seek is the direction in which the parabola opens and approximately how wide or narrow to draw the parabolic curve. The direction toward which the parabola opens is found from the equation. An equation of the form y2 5 4px opens either left or right. It opens right if p . 0, and it opens left if p , 0. An equation of the form x2 5 4py opens either up or down. It opens up if p . 0, and it opens down if p , 0. How narrow or wide should the parabolic curve be drawn? If we select a few points that satisfy the equation, we can use those as graphing aids. In Example 1, we found that the focus of that parabola is located at 12, 02. If we select the x-coordinate of the focus x 5 2, and substitute that value into the equation of the parabola y2 5 8x, we find the corresponding y values to be y 5 24 and y 5 4. If we plot the three points 10, 02, 12, 242, and 12, 42 and then connect the points with a parabolic curve, we get the graph on the right. The line segment that passes through the focus 12, 02 is parallel to the directrix x 5 22, and the line segment whose endpoints are on the parabola is called the latus rectum. The latus rectum in this case has length 8. x y Latus rectum Focus (2, 0) 8 (2, 4) (2, –4) The latus rectum is a graphing aid that helps us find the width of a parabola. In general, the points on a parabola of the form y2 5 4px that lie above and below the focus 1 p, 02 satisfy the equation y2 5 4p2 and are located at 1 p, 22p2 and 1 p, 2p2. The latus rectum will have length 4p. Similarly, a parabola of the form x2 5 4py will have a horizontal latus rectum of length 4p. We will use the latus rectum as a graphing aid to determine the parabola’s width. STUDY TIP The focus and directrix define a parabola but do not appear on its graph. x y (2, 4) (0, 0) (2, –4) EXAMPLE 2 Graphing a Parabola Whose Vertex Is at the Origin Using the Focus, Directrix, and Latus Rectum as Graphing Aids Determine the focus, directrix, and length of the latus rectum of the parabola x2 5 212y. Employ these to assist in graphing the parabola. Solution: Compare this parabola with the general equation of a parabola. x2 5 4py x2 5 212y Solve for p. 4p 5 212 p 5 23 1006 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities Finding the Equation of a Parabola Whose Vertex Is at the Origin Thus far we have started with the equation of a parabola and then determined its focus and directrix. Let’s now reverse the process. For example, if we know the focus and directrix of a parabola, how do we find the equation of the parabola? If we are given the focus and directrix, then we can find the vertex, which is the midpoint between the focus and the directrix. If the vertex is at the origin, then we know the general equation of the parabola that corresponds with the focus. A parabola of the form x2 5 4py has focus 10, p2, directrix y 5 2p, and a latus rectum of length 4p. For this parabola, p 5 23; therefore, the focus is 10, 232 , the directrix is y 5 3 , and the length of the latus rectum is 12 . Y OUR TU R N Find the focus, directrix, and length of the latus rectum of the parabola y2 5 28x, and use these to graph the parabola. ▼ x y Focus (0, –3) (6, –3) (–6, –3) Directrix y = 3 Latus rectum: 12 10 10 –10 –10 ▼ A N S W E R The focus is 122, 02. The directrix is x 5 2. The length of the latus rectum is 8. x y Latus rectum: 8 Directrix x = 2 Focus (–2, 0) EXAMPLE 3 Finding the Equation of a Parabola Given the Focus and Directrix When the Vertex Is at the Origin Find the equation of a parabola whose focus is at the point A0, 1 2B and whose directrix is y 5 21 2. Graph the equation. Solution: The midpoint of the segment joining the focus and the directrix along the axis of symmetry is the vertex. Calculate the midpoint between A0, 1 2B and A0, 21 2B. Vertex 5 £0 1 0 2 , 1 2 2 1 2 2 ≥5 10, 02. A parabola with vertex at 10, 02, focus at 10, p2, and directrix y 5 2p corresponds to the equation x2 5 4py. Identify p given that the focus is 10, p2 5 A0, 1 2B. p 5 1 2 Substitute p 5 1 2 into the standard equation of a parabola with the vertex at the origin x2 5 4py. x2 5 2y Now that the equation is known, a few points can be selected, and the parabola can be point-plotted. Alternatively, the length of the latus rectum can be calculated to sketch the approximate width of the parabola. Axis of symmetry Directrix Vertex Focus 11.2 The Parabola 1007 Before we proceed to parabolas with general vertices, let’s first make a few observations: The larger the latus rectum, the wider the parabola. An alternative approach for graphing the parabola is to plot a few points that satisfy the equation of the parabola, which is the approach in most textbooks. 11.2.2 Parabola with a Vertex at the Point (h, k) Recall (Section 2.4) that the graph of x2 1 y2 5 r2 is a circle with radius r centered at the origin, whereas the graph of 1x 2 h22 1 1 y 2 k22 5 r2 is a circle with radius r centered at the point 1h, k2. In other words, the center is shifted from the origin to the point 1h, k2. This same translation (shift) can be used to describe parabolas whose vertex is at the point 1h, k2. To graph x2 5 2y, first calculate the latus rectum. 40 p 0 5 4 a1 2b 5 2 Y OUR T UR N Find the equation of a parabola whose focus is at the point 125, 02 and whose directrix is x 5 5. ▼ x y x2 = 2y 2 (–2, 2) (2, 2) ▼ A N S W E R y2 5 220x EQUATION OF A PARABOLA WITH VERTEX AT THE POINT (h, k) The standard (conic) form of the equation of a parabola with vertex at the point 1h, k2 is given by: Equation 1 y 2 k22 5 4p 1x 2 h2 1x 2 h22 5 4p 1 y 2 k2 Vertex 1h, k2 1h, k2 Focus 1 p 1 h, k2 1h, p 1 k2 Directrix x 5 2p 1 h y 5 2p 1 k Axis of symmetry y 5 k x 5 h p + 0 Opens to the right Opens upward p 0 Opens to the left Opens downward In order to find the vertex of a parabola given a general second-degree equation, first complete the square (Section 1.3) to identify 1h, k2. Then determine whether the parabola opens up, down, left, or right. Identify points that lie on the graph of the parabola. Intercepts are often the easiest points to find, since one of the variables is set equal to zero. STUDY TIP When 1h, k2 5 10, 02, the vertex of the parabola is located at the origin. 11.2.2 S K I L L Find the equation of a parabola whose vertex is at the point 1h, k2. 11.2.2 C O N C E P T U A L Employ completing the square to transform an equation into the standard form of a parabola. EXAMPLE 4 Graphing a Parabola with Vertex (h, k) Graph the parabola given by the equation y2 2 6y 2 2x 1 8 5 0. Solution: Transform this equation into the form 1y 2 k22 5 4p1x 2 h2, since this equation is of second degree in y and first degree in x. We know this parabola opens either to the left or right. [CONCEPT CHECK] For the equation y 2 1 4y 2 5x 2 5 5 0, on which variable do you complete the square? ANSWER y ▼ 1008 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities Complete the square on y: y2 2 6y 2 2x 1 8 5 0 Isolate the y terms. y2 2 6y 5 2x 2 8 Add 9 to both sides to complete the square. y2 2 6y 1 9 5 2x 2 8 1 9 Write the left side as a perfect square. 1y 2 322 5 2x 1 1 Factor a 2 out on the right side. 1y 2 322 5 2 ax 1 1 2b Compare with 1y 2 k22 5 4p1x 2 h2 and 1h, k2 5 a 21 2, 3b identify 1h, k2 and p. 4p 5 2 1 p 5 1 2 The vertex is at the point A21 2, 3B, and since p 5 1 2 is positive, the parabola opens to the right. Since the parabola’s vertex lies in quadrant II and it opens to the right, we know there are two y-intercepts and one x-intercept. Apply the general equation y2 2 6y 2 2x 1 8 5 0 to find the intercepts. Find the y-intercepts 1set x 5 02. y2 2 6y 1 8 5 0 Factor. 1y 2 221y 2 42 5 0 Solve for y. y 5 2 or y 5 4 Find the x-intercept 1set y 5 02. 22x 1 8 5 0 Solve for x. x 5 4 Label the following points and connect them with a smooth curve: Vertex: A21 2, 3B y-intercepts: 10, 22 and 10, 42 x-intercept: 14, 02 Y OUR TU R N For the equation y2 1 4y 2 5x 2 5 5 0 identify the vertex and the intercepts, and graph. ▼ x y (0, 4) (0, 2) (4, 0) Vertex (– , 3) 1 2 ▼ A N S W E R Vertex: A29 5, 22B x-intercept: 121, 02 y-intercepts: 10, 252 and 10, 12 x y (0, –5) (0, 1) (–1, 0) Vertex (– , –2) 9 5 STUDY TIP It is often easier to find the intercepts by converting to the general form of the equation. EXAMPLE 5 Graphing a Parabola with Vertex (h, k) Graph the parabola given by the equation x2 2 2x 2 8y 2 7 5 0. Solution: Transform this equation into the form 1x 2 h22 5 4p1y 2 k2, since this equation is second degree in x and first degree in y. We know this parabola opens either upward or downward. Complete the square on x: x2 2 2x 2 8y 2 7 5 0 Isolate the x terms. x2 2 2x 5 8y 1 7 Add 1 to both sides to complete the square. x2 2 2x 1 1 5 8y 1 7 1 1 Write the left side as a perfect square. 1x 2 122 5 8y 1 8 Factor out the 8 on the right side. 1x 2 122 5 81y 1 12 Compare with 1x 2 h22 5 4p1y 2 k2 and identify 1h, k2 and p. 1h, k2 5 11, 212 4p 5 8 1 p 5 2 11.2 The Parabola 1009 The vertex is at the point 11, 212, and since p 5 2 is positive, the parabola opens upward. Since the parabola’s vertex lies in quadrant IV and it opens upward, we know there are two x‑intercepts and one y-intercept. Use the general equation x2 2 2x 2 8y 2 7 5 0 to find the intercepts. Find the y-intercept 1set x 5 02. 28y 2 7 5 0 Solve for y. y 5 27 8 Find the x-intercepts 1set y 5 02. x2 2 2x 2 7 5 0 Solve for x. x 5 2 6 !4 1 28 2 5 2 6 !32 2 5 2 6 4!2 2 5 1 6 2!2 Label the following points and connect with a smooth curve: Vertex: 11, 212 y-intercept: A0, 27 8B x-intercepts: A1 2 2!2, 0B and A1 1 2!2, 0B YOUR TURN For the equation x2 1 2x 1 8y 2 7 5 0 identify the vertex and the intercepts, and graph. ▼ x y Vertex (1, –1) (0, – ) 7 8 (1 – 2√2, 0) (1 + 2√2, 0) ▼ A N S W E R Vertex: 121, 12 x-intercepts: A21 2 2!2, 0B and A21 1 2!2, 0B y-intercept: A0, 7 8B x y Vertex (–1, 1) (0, ) 7 8 (–1 + 2√2, 0) (–1 – 2√2, 0) EXAMPLE 6 Finding the Equation of a Parabola with Vertex (h, k) Find the equation of a parabola whose vertex is located at the point 12, 232 and whose focus is located at the point 15, 232. Solution: Draw a Cartesian plane and label the vertex and focus. The vertex and focus share the same axis of symmetry y 5 23 and indicate a parabola opening to the right. Write the standard (conic) equation of a parabola opening to the right. 1y 2 k22 5 4p1x 2 h2 p . 0 Substitute the vertex 1h, k2 5 12, 232 into the standard equation. 3y 2 123242 5 4p1x 2 22 Find p. The general form of the vertex is 1h, k2, and the focus is 1h 1 p, k2. For this parabola, the vertex is 12, 232, and the focus is 15, 232. Find p by taking the difference of the x coordinates. p 5 3 Substitute p 5 3 into 3y 2 123242 5 4p1x 2 22. 1y 1 322 5 41321x 2 22 Eliminate parentheses. y2 1 6y 1 9 5 12x 2 24 Simplify. y2 1 6y 2 12x 1 33 5 0 Y OUR T UR N Find the equation of the parabola whose vertex is located at 12, 232 and whose focus is located at 10, 232. ▼ x y Vertex (2, –3) Focus (5, –3) Axis of symmetry ▼ A N S W E R y2 1 6y 1 8x 2 7 5 0 1010 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities 11.2.3 Applications If we start with a parabola in the xy-plane and rotate it around its axis of symmetry, the result will be a three-dimensional paraboloid. Solar cookers illustrate the physical property that the rays of light coming into a parabola should be reflected to the focus. A flashlight reverses this process in that its light source at the focus illuminates a parabolic reflector to direct the beam outward. x z y x z y A satellite dish is in the shape of a paraboloid. Functioning as an antenna, the parabolic dish collects all incoming signals and reflects them to a single point, the focal point, which is where the receiver is located. In Examples 7 and 8, and in the appli cations exercises, it is not intended to find the three-dimensional equation of the paraboloid, but rather the equation of the plane parabola that is rotated to generate the paraboloid. Photodisc/Getty Images, Inc. Satellite dish 11.2.3 SKI LL Solve applied problems that involve parabolas. 11.2.3 CO NC EPTUAL Understand that it is the focus that is the key to applications. EXAMPLE 7 Finding the Location of the Receiver in a Satellite Dish A satellite dish is 24 feet in diameter at its opening and 4 feet deep in its center. Where should the receiver be placed? Solution: Draw a parabola with a vertex at the origin representing the center cross section of the satellite dish. Write the standard equation of a parabola opening upward with vertex at 10, 02. x2 5 4py The point 112, 42 lies on the parabola, so substitute 112, 42 into x2 5 4py. 11222 5 4p142 Simplify. 144 5 16p Solve for p. p 5 9 Substitute p 5 9 into the focus 10, p2. focus 5 10, 92 The receiver should be placed 9 feet from the vertex of the dish. x y (–12, 4) 24 ft 4 ft (12, 4) Parabolic antennas work for sound in addition to light. Have you ever wondered how the sound of the quarterback calling audible plays is heard by the sideline crew? The crew holds a parabolic system with a microphone at the focus. All of the sound in the direction of the parabolic system is reflected toward the focus where the microphone amplifies and records the sound. Francis Specker/Icon SMI/NewsCom [CONCEPT CHECK] The more concave the parabolic dish, the closer/further the focus is from the vertex. ANSWER Closer ▼ 11.2 The Parabola 1011 EXAMPLE 8 Finding the Equation of a Parabolic Sound Dish If the parabolic sound dish the sideline crew is holding has a 2-foot diameter at the opening and the microphone is located 6 inches from the vertex, find the equation that governs the center cross section of the parabolic sound dish. Solution: Write the standard equation of a parabola opening to the right with the vertex at the origin 10, 02. y2 5 4px The focus is located 6 inches A1 2 footB from the vertex. 1p, 02 5 a1 2, 0b Solve for p. p 5 1 2 Let p 5 1 2 in y2 5 4px. y2 5 4 a1 2bx Simplify. y2 5 2x In this section, we discussed parabolas whose vertex is at the origin. Equation y2 5 4px x2 5 4py Vertex 10, 02 10, 02 Focus 1 p, 02 10, p2 Directrix x 5 2p y 5 2p Axis of symmetry x-axis y-axis p + 0 Opens to the right Opens upward p 0 Opens to the left Opens downward Graph 1p + 02 Directrix x = –p Focus (p, 0) y2 = 4px p p x y Directrix y = –p Focus (0, p) x2 = 4py p p x y For parabolas whose vertex is at the point 1h, k2: Equation 1 y 2 k22 5 4p1x 2 h2 1x 2 h22 5 4p1y 2 k2 Vertex 1h, k2 1h, k2 Focus 1 p 1 h, k2 1h, p 1 k2 Directrix x 5 2p 1 h y 5 2p 1 k Axis of symmetry y 5 k x 5 h p + 0 Opens to the right Opens upward p 0 Opens to the left Opens downward [SEC TION 11. 2] S U M M A RY 1012 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities In Exercises 1–4, match the parabola to the equation. 1. y2 5 4x 2. y2 5 24x 3. x2 5 24y 4. x2 5 4y a. x y –10 5 –5 b. x y –5 5 10 c. x 10 y 5 –5 d. x y –5 5 –10 In Exercises 5–8, match the parabola to the equation. 5. 1 y 2 122 5 41x 2 12 6. 1 y 1 122 5 241x 2 12 7. 1x 1 122 5 241 y 1 12 8. 1x 2 122 5 41 y 2 12 a. x y –6 4 –10 (–1, –1) b. x y –4 6 10 (1, 1) c. x y 6 10 –4 (1, 1) d. x y –10 4 –6 (1, –1) In Exercises 9–20, find an equation for the parabola described. 9. vertex at 10, 02; focus at 10, 32 10. vertex at 10, 02; focus at 12, 02 11. vertex at 10, 02; focus at 125, 02 12. vertex at 10, 02; focus at 10, 242 13. vertex at 13, 52; focus at 13, 72 14. vertex at 13, 52; focus at 17, 52 15. vertex at 12, 42; focus at 10, 42 16. vertex at 12, 42; focus at 12, 212 17. focus at 12, 42; directrix at y 5 22 18. focus at 12, 222; directrix at y 5 4 19. focus at 13, 212; directrix at x 5 1 20. focus at 121, 52; directrix at x 5 5 In Exercises 21–24, write an equation for each parabola. 21. x y (1, –2) (1, 6) Vertex (–1, 2) 22. (–2, 1) (6, 1) Vertex (2, –1) x y 23. x y (–2, –3) (6, –3) Vertex (2, –1) 24. x y (–3, –2) (–3, 6) Vertex (–1, 2) In Exercises 25–32, find the focus, vertex, directrix, and length of the latus rectum and graph the parabola. 25. x2 5 8y 26. x2 5 212y 27. y2 5 22x 28. y2 5 6x 29. x2 5 16y 30. x2 5 28y 31. y2 5 4x 32. y2 5 216x [SEC TION 11. 2] E X E RC I SE S • S K I L L S 11.2 The Parabola 1013 In Exercises 33–44, find the vertex and graph the parabola. 33. 1 y 2 222 5 41x 1 32 34. 1 y 1 222 5 241x 2 12 35. 1x 2 322 5 281 y 1 12 36. 1x 1 322 5 281 y 2 22 37. 1x 1 522 5 22y 38. y2 5 2161x 1 12 39. y2 2 4y 2 2x 1 4 5 0 40. x2 2 6x 1 2y 1 9 5 0 41. y2 1 2y 2 8x 2 23 5 0 42. x2 2 6x 2 4y 1 10 5 0 43. x2 2 x 1 y 2 1 5 0 44. y2 1 y 2 x 1 1 5 0 • A P P L I C A T I O N S 45. Satellite Dish. A satellite dish measures 8 feet across its opening and 2 feet deep at its center. The receiver should be placed at the focus of the parabolic dish. Where is the focus? 46. Satellite Dish. A satellite dish measures 30 feet across its opening and 5 feet deep at its center. The receiver should be placed at the focus of the parabolic dish. Where is the focus? 47. Eyeglass Lens. Eyeglass lenses can be thought of as very wide parabolic curves. If the focus occurs 2 centimeters from the center of the lens and the lens at its opening is 5 centimeters, find an equation that governs the shape of the center cross section of the lens. 48. Optical Lens. A parabolic lens focuses light onto a focal point 3 centimeters from the vertex of the lens. How wide is the lens 0.5 centimeter from the vertex? Exercises 49 and 50 are examples of solar cookers. Parabolic shapes are often used to generate intense heat by collecting Sun rays and focusing all of them at a focal point. 49. Solar Cooker. The parabolic cooker MS-ST10 is delivered as a kit, handily packed in a single carton, with complete assembly instructions and even the necessary tools. Thanks to the reflector diameter of 1 meter, it develops an immense power. One liter of water boils in significantly less than half an hour. If the rays are focused 40 centimeters from the vertex, find the equation for the parabolic cooker. 50. Le Four Solaire at Font-Romeur “Mirrors of the Solar Furnace.” There is a reflector in the Pyrenees Mountains that is eight stories high. It cost $2M, and it took 10 years to build. It is made of 9000 mirrors arranged in a parabolic formation. It can reach 6000°F just from the Sun! If the diameter of the parabolic mirror is 100 meters and the sunlight is focused 25 meters from the vertex, find the equation for the parabolic dish. 51. Sailing under a Bridge. A bridge with a parabolic shape has an opening 80 feet wide at the base of the bridge (where the bridge meets the water), and the height in the center of the bridge is 20 feet. A sailboat whose mast reaches 17 feet above the water is traveling under the bridge 10 feet from the center of the bridge. Will it clear the bridge without scrap-ing its mast? Justify your answer. 52. Driving under a Bridge. A bridge with a parabolic shape reaches a height of 25 feet in the center of the road, and the width of the bridge opening at ground level is 20 feet combined (both lanes). If an RV is 10 feet tall and 8 feet wide, it won’t make it under the bridge if it hugs the center line. What if it straddles the center line? Will it make it? Justify your answer. 53. Parabolic Telescope. The Arecibo radio telescope in Puerto Rico has an enormous reflecting surface, or radio mirror. The huge “dish” is 1000 feet in diameter and 167 feet deep and covers an area of about 20 acres. Using these dimensions, determine the focal length of the telescope. Find the equation for the dish portion of the telescope. 54. Suspension Bridge. If one parabolic segment of a suspension bridge is 300 feet and if the cables at the vertex are suspended 10 feet above the bridge, whereas the height of the cables 150 feet from the vertex reaches 60 feet, find the equation of the parabolic path of the suspension cables. 60 ft 300 ft 10 ft 55. Business. The profit, in thousands of dollars, for a product is P1x2 52x2 1 60x 2 500 where x is the production level in hundreds of units. Find the production level that maximizes the profit. Find the maximum profit. 56. Business. The profit, in thousands of dollars, for a product is P1x2 52x2 1 80x 2 1200 where x is the production level in hundreds of units. Find the production level that maximizes the profit. Find the maximum profit. SAURABH DAS/ASSOCIATED PRESS Solar cooker, Ubuntu Village, Johannesburg, South Africa 20 feet 80 feet 25 feet 20 feet Mark Antman/The Image Works Solar furnace, Odellio, France 1014 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities In Exercises 57 and 58, explain the mistake that is made. 57. Find an equation for a parabola whose vertex is at the origin and whose focus is at the point 13, 02. Solution: Write the general equation for a parabola whose vertex is at the origin. x2 5 4py The focus of this parabola is 1 p, 02 5 13, 02. p 5 3 Substitute p 5 3 into x2 5 4py. x2 5 12y This is incorrect. What mistake was made? 58. Find an equation for a parabola whose vertex is at the point 13, 22 and whose focus is located at 15, 22. Solution: Write the equation associated with a parabola whose vertex is 13, 22. 1x 2 h22 5 4p1 y 2 k2 Substitute 13, 22 into 1x 2 h22 5 4p1 y 2 k2. 1x 2 322 5 4p1 y 2 22 The focus is located at 15, 22; therefore, p 5 5. Substitute p 5 5 into 1x 2 322 5 4p1 y 2 22. 1x 2 322 5 201 y 2 22 This is incorrect. What mistake(s) was made? • C A T C H T H E M I S T A K E • C H A L L E N G E 63. Derive the standard equation of a parabola with vertex at the origin, opening upward, x2 5 4py. 3Calculate the distance d1 from any point on the parabola 1x, y2 to the focus 10, p2. Calculate the distance d2 from any point on the parabola 1x, y2 to the directrix 1x, 2p2. Set d1 5 d2.4 64. Derive the standard equation of a parabola opening right, y2 5 4px. 3Calculate the distance d1 from any point on the parabola 1x, y2 to the focus 1 p, 02. Calculate the distance d2 from any point on the parabola 1x, y2 to the directrix 12p, y2. Set d1 5 d2.4 • T E C H N O L O G Y 65. Using a graphing utility, plot the parabola x2 2 x 1 y 2 1 5 0. Compare with the sketch you drew for Exercise 43. 66. Using a graphing utility, plot the parabola y21 y 2 x 1 1 5 0. Compare with the sketch you drew for Exercise 44. 67. In your mind, picture the parabola given by 1 y 1 3.522 5 101x 2 2.52. Where is the vertex? Which way does this parabola open? Now plot the parabola using a graphing utility. 68. In your mind, picture the parabola given by 1x 1 1.422 5 251 y 1 1.72. Where is the vertex? Which way does this parabola open? Now plot the parabola using a graphing utility. 69. In your mind, picture the parabola given by 1 y 2 1.522 5 281x 2 1.82. Where is the vertex? Which way does this parabola open? Now plot the parabola using a graphing utility. 70. In your mind, picture the parabola given by 1x 1 2.422 5 61 y 2 3.22. Where is the vertex? Which way does this parabola open? Now plot the parabola using a graphing utility. • C O N C E P T U A L 59. The vertex lies on the graph of a parabola. 60. The focus lies on the graph of a parabola. 61. The directrix lies on the graph of a parabola. 62. The endpoints of the latus rectum lie on the graph of a parabola. In Exercises 59–62, determine whether each statement is true or false. 11.3 The Ellipse 1015 S K I L L S O B J E C T I V E S ■ ■Find the equation of an ellipse centered at the origin. ■ ■Find the equation of an ellipse centered at the point 1h, k2. ■ ■Solve applied problems that involve ellipses. C O N C E P T U A L O B J E C T I V ES ■ ■Understand that an ellipse is the set of all points in a plane, the sum of whose distances from two fixed points (foci) is constant. ■ ■Employ completing the square to transform an equation into the standard form of an ellipse. ■ ■Understand that the endpoints and the foci are important for application problems involving ellipses. 11.3 THE ELLIPSE 11.3.1 Ellipse Centered at the Origin Definition of an Ellipse If we were to take a piece of string, tie loops at both ends, and tack the ends down so that the string had lots of slack, we would have the picture on the right. If we then took a pencil and pulled the string taut and traced our way around for one full rotation, the result would be an ellipse. See the second figure in the right margin. An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points is constant. These two fixed points are called foci (plural of focus). A line segment through the foci called the major axis inter-sects the ellipse at the vertices. The midpoint of the line segment joining the vertices is called the center. The line segment that intersects the center and joins two points on the ellipse and is perpendicular to the major axis is called the minor axis. Ellipse DEFINITION Vertices Foci Center Major axis Minor axis Let’s start with an ellipse whose center is located at the origin. Using graph-shifting techniques, we can later extend the characteristics of an ellipse centered at a point other than the origin. Ellipses can vary from the shape of circles to something quite elongated, either horizontally or vertically, that resembles the shape of a racetrack. We say that the ellipse has either greater or lesser eccentricity; as we will see, there is a simple mathematical definition of eccentricity. It can be shown that the standard equation of an ellipse with its center at the origin is given by one of two forms, depending on whether the orientation of the major axis of the ellipse is horizontal or vertical. For a . b . 0, if the major axis is horizontal, then the equation is given by x2 a2 1 y2 b2 5 1, and if the major axis is vertical, then the equation is given by x2 b2 1 y2 a2 5 1. Let’s consider an ellipse with center at the origin and the foci on the x-axis. Let the distance from the center to the focus be c. Therefore, the foci are located at the points 12c, 02 and 1c, 02. The line segment containing the foci is called the major axis, and it lies along the x-axis. The sum of the two distances from the foci to any point 1x, y2 must be constant. 11.3.1 S KI L L Find the equation of an ellipse centered at the origin. 11.3.1 C ON C E P T U A L Understand that an ellipse is the set of all points in a plane, the sum of whose distances from two fixed points (foci) is constant. x y (–c, 0) (c, 0) (x, y) 1016 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities Derivation of the Equation of an Ellipse WORDS MATH Calculate the distance from 1x, y2 to 12c, 02 applying the distance formula. "1x 2 12c222 1 y2 Calculate the distance from 1x, y2 to 1c, 02 applying the distance formula. "1x 2 c22 1 y2 The sum of these two distances is equal to a constant 12a for convenience2. "1x 2 12c222 1 y2 1 "1x 2 c22 1 y2 5 2a Isolate one radical. "1x 2 12c222 1 y2 5 2a 2 "1x 2 c22 1 y2 Square both sides of the equation. 1x 1 c22 1 y2 5 4a2 2 4a"1x 2 c22 1 y2 1 1x 2 c22 1 y2 Square the binomials inside the x2 1 2cx 1 c2 1 y2 5 4a2 2 4a"1x 2 c22 1 y2 parentheses. 1 x2 2 2cx 1 c2 1 y2 Simplify. 4cx 2 4a2 5 24a"1x 2 c22 1 y2 Divide both sides of the equation by –4. a2 2 cx 5 a"1x 2 c22 1 y2 Square both sides of the equation. 1a2 2 cx2 2 5 a231x 2 c22 1 y24 Square the binomials inside the parentheses. a4 2 2a2cx 1 c2x2 5 a23x2 2 2cx 1 c2 1 y24 Distribute the a2 term. a4 2 2a2cx 1 c2x2 5 a2x2 2 2a2cx 1 a2c2 1 a2y2 Group x and y terms together, respectively, on one side and constants on the other side. c2x2 2 a2x2 2 a2y2 5 a2c2 2 a4 Factor out common factors. 1c2 2 a22x2 2 a2y2 5 a21c2 2 a22 Multiply both sides of the equation by –1. 1a2 2 c22x2 1 a2y2 5 a21a2 2 c22 We can make the argument that a . c in order for a point to be on the ellipse (and not on the x-axis). Thus, since a and c represent distances, and therefore are positive, we know that a2 . c2, or a2 2 c2 . 0. Hence, we can divide both sides of the equation by a2 2 c2, since a2 2 c2 2 0. x2 1 a2y2 1a2 2 c22 5 a2 Let b2 5 a2 2 c2. x2 1 a2y2 b2 5 a2 Divide both sides of the equation by a2. x2 a2 1 y2 b2 5 1 11.3 The Ellipse 1017 The equation x2 a2 1 y2 b2 5 1 represents an ellipse with its center at the origin and with the foci along the x-axis, since a . b. The following box summarizes ellipses that have their center at the origin and foci along either the x-axis or y-axis. EQUATION OF AN ELLIPSE WITH CENTER AT THE ORIGIN The standard form of the equation of an ellipse with its center at the origin is given by: Orientation of Major Axis Horizontal along the x-axis Vertical along the y-axis Equation x2 a2 1 y2 b2 5 1 a . b . 0 x2 b2 1 y2 a2 5 1 a . b . 0 Graph x y (–a, 0) (–c, 0) (c, 0) (0, 0) (a, 0) (0, b) (0, –b) x y (0, –a) (–b, 0) (b, 0) (0, a) (0, c) (0, –c) (0, 0) Foci 12c, 02 1c, 02 10, 2c2 10, c2 Vertices 12a, 02 1a, 02 10, 2a2 10, a2 Other Intercepts 10, b2 10, 2b2 1b, 02 12b, 02 In both cases, the value of c, the distance along the major axis from the center to the focus, is given by c2 5 a2 2 b2. The length of the major axis is 2a, and the length of the minor axis is 2b. Notice that when a 5 b, the equation x2 a2 1 y2 b2 5 1 simplifies to x2 a2 1 y2 a2 5 1 or x2 1 y2 5 a2, which corresponds to a circle. The vertices correspond to intercepts when an ellipse is centered at the origin. One of the first things we notice about an ellipse is its eccentricity. The eccentricity, denoted e, is given by e 5 c a , where 0 , e , 1. The circle is a limiting form of an ellipse, c 5 0. In other words, if the eccentricity is close to 0, then the ellipse resembles a circle, whereas if the eccentricity is close to 1, then the ellipse is quite elongated, or eccentric. Graphing an Ellipse with Center at the Origin The equation of an ellipse in standard form can be used to graph an ellipse. Although an ellipse is defined in terms of the foci, the foci are not part of the graph. Notice that if the divisor of x2 is larger than the divisor of y2, then the ellipse is elongated horizontally. x y a a c b [CONCEPT CHECK] TRUE OR FALSE In an ellipse, the foci are always closer to the center than the vertices. ANSWER True ▼ 1018 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities EXAMPLE 1 Graphing an Ellipse with a Horizontal Major Axis Graph the ellipse given by x2 25 1 y2 9 5 1. Solution: Since 25 . 9, the major axis is horizontal. a2 5 25 and b2 5 9 Solve for a and b. a 5 5 and b 5 3 Identify the vertices: 12a, 02 and 1a, 02. 125, 02 and 15, 02 Identify the endpoints (y-intercepts) on the minor axis: 10, 2b2 and 10, b2. 10, 232 and 10, 32 Graph by labeling the points 125, 02, 15, 02, 10, 232, and 10, 32 and connecting them with a smooth curve. x y (0, 3) (–5, 0) (5, 0) (0, –3) EXAMPLE 2 Graphing an Ellipse with a Vertical Major Axis Graph the ellipse given by 16x2 1 y2 5 16. Solution: Write the equation in standard form by dividing by 16. x2 1 1 y2 16 5 1 Since 16 . 1, this ellipse is elongated vertically. a2 5 16 and b2 5 1 Solve for a and b. a 5 4 and b 5 1 Identify the vertices: 10, 2a2 and 10, a2. 10, 242 and 10, 42 Identify the x-intercepts on the minor axis: 12b, 02 and 1b, 02. 121, 02 and 11, 02 Graph by labeling the points 10, 242, 10, 42, 121, 02, and 11, 02 and connecting them with a smooth curve. Y OUR TU R N Graph the ellipses. a. x2 9 1 y2 4 5 1 b. x2 9 1 y2 36 5 1 x y (0, 4) (–1, 0) (1, 0) (0, –4) ▼ ▼ A N S W E R a. x y (0, 2) (–3, 0) (3, 0) (0, –2) x2 9 y2 4 + = 1 b. x y (0, 6) (–3, 0) (3, 0) (0, –6) x2 9 y2 36 + = 1 STUDY TIP If the divisor of x2 is larger than the divisor of y2, then the major axis is horizontal along the x-axis, as in Example 1. If the divisor of y2 is larger than the divisor of x2, then the major axis is vertical along the y-axis, as in Example 2. 11.3 The Ellipse 1019 Finding the Equation of an Ellipse Centered at the Origin What if we know the vertices and the foci of an ellipse and want to find the equation to which it corresponds? The axis on which the foci and vertices are located is the major axis. Therefore, we will have the standard equation of an ellipse, and a will be known (from the vertices). Since c is known from the foci, we can use the relation c2 5 a2 2 b2 to determine the unknown b. EXAMPLE 3 Finding the Equation of an Ellipse Centered at the Origin Find the standard form of the equation of an ellipse with foci at 123, 02 and 13, 02 and vertices 124, 02 and 14, 02. Solution: The major axis lies along the x-axis, since it contains the foci and vertices. Write the corresponding general equation of an ellipse. x2 a2 1 y2 b2 5 1 Identify a from the vertices: Match vertices 124, 02 5 12a, 02 and 14, 02 5 1a, 02. a 5 4 Identify c from the foci: Match foci, 123, 02 5 12c, 02 and 13, 02 5 1c, 02. c 5 3 Substitute a 5 4 and c 5 3 into b2 5 a2 2 c2. b2 5 42 2 32 Simplify. b2 5 7 Substitute a2 5 16 and b2 5 7 into x2 a2 1 y2 b2 5 1. x2 16 1 y2 7 5 1 YOUR TURN Find the standard form of the equation of an ellipse with vertices at 10, 262 and 10, 62 and foci 10, 252 and 10, 52. 11.3.2 Ellipse Centered at the Point (h, k) We can use graph-shifting techniques to graph ellipses that are centered at a point other than the origin. For example, to graph 1x 2 h22 a2 1 1y 2 k22 b2 5 1 (assuming h and k are positive constants), start with the graph of x2 a2 1 y2 b2 5 1 and shift to the right h units and up k units. The center, the vertices, the foci, and the major and minor axes all shift. In other words, the two ellipses are identical in shape and size, except that the ellipse 1x 2 h22 a2 1 1y 2 k22 b2 5 1 is centered at the point 1h, k2. The following table summarizes the characteristics of ellipses centered at a point other than the origin. 11.3.2 S K I L L Find the equation of an ellipse centered at the point 1h, k2. 11.3.2 C ON C E P T U A L Employ completing the square to transform an equation into the standard form of an ellipse. ▼ ▼ A N S W E R x2 11 1 y2 36 5 1 1020 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities EQUATION OF AN ELLIPSE WITH CENTER AT THE POINT (h, k) The standard form of the equation of an ellipse with its center at the point 1h, k2: Orientation of Major Axis Horizontal (Parallel to the x-axis) Vertical (Parallel to the y-axis) Equation 1x 2 h22 a2 1 1y 2 k22 b2 5 1 1x 2 h22 b2 1 1y 2 k22 a2 5 1 Graph x y (h – a, k) (h – c, k) (h + c, k) (h + a, k) (h, k) (h, k + b) (h, k – b) x y (h – b, k) (h + b, k) (h, k) (h, k + a) (h, k + c) (h, k – a) (h, k – c) Foci 1h 2 c, k2 1h 1 c, k2 1h, k 2 c2 1h, k 1 c2 Vertices 1h 2 a, k2 1h 1 a, k2 1h, k 2 a2 1h, k 1 a2 In both cases, a . b . 0, c2 5 a2 2 b2, the length of the major axis is 2a, and the length of the minor axis is 2b. EXAMPLE 4 Graphing an Ellipse with Center (h, k) Given the Equation in Standard Form Graph the ellipse given by 1x 2 222 9 1 1y 1 122 16 5 1. Solution: Write the equation in the form 1x 2 h22 b2 1 1y 2 k22 a2 5 1. 1x 2 222 32 1 1y 2 121222 42 5 1 Identify a, b, and the center 1h, k2. a 5 4, b 5 3, and 1h, k2 5 12, 212 Draw a graph and label the center: 12, 212. Since a 5 4, the vertices are up four and down four units from the center: 12, 252 and 12, 32. Since b 5 3, the endpoints of the minor axis are to the left and right three units: 121, 212 and 15, 212. Y OUR TU R N Graph the ellipse given by 1x 1 122 9 1 1y 2 322 1 5 1. x y (2, –1) (5, –1) (–1, –1) (2, –5) (2, 3) ▼ Recall that when we are given the equation of a circle in general form, we first complete the square in order to express the equation in standard form, which allows the center and radius to be identified. That same approach is used when the equation of an ellipse is given in a general form. ▼ A N S W E R x y (–1, 2) (–1, 3) (–1, 4) (2, 3) (–4, 3) 11.3 The Ellipse 1021 11.3.3 Applications There are many examples of ellipses all around us. On Earth we have racetracks, and in our solar system, the planets travel in elliptical orbits with the Sun as a focus. Satellites are in elliptical orbits around the Earth. Most communications satellites are in a geosynchronous (GEO) orbit—they orbit the Earth once each day. And in order to stay over the same spot on Earth, a geostationary satellite has to be directly above the equator; it circles the Earth in exactly the time it takes the Earth to turn once on its axis, and its orbit has to follow the path of the equator as the Earth rotates. Otherwise, from the Earth the satellite would appear to move in a north–south line every day. x y (–3, 1) (–8, 1) (2, 1) (–3, 3) (–3, –1) EXAMPLE 5 Graphing an Ellipse with Center (h, k) Given an Equation in General Form Graph the ellipse given by 4x2 1 24x 1 25y2 2 50y 2 39 5 0. Solution: Transform the general equation into standard form. Group x terms together and y terms together and add 39 to both sides. 14x2 1 24x2 1 125y2 2 50y2 5 39 Factor out the 4 common to the x terms and the 25 common to the y terms. 41x2 1 6x2 1 251y2 2 2y2 5 39 Complete the square on x and y. 41x2 1 6x 1 92 1 251y2 2 2y 1 12 5 39 1 4192 1 25112 Simplify. 41x 1 322 1 251y 2 122 5 100 Divide by 100. 1x 1 322 25 1 1y 2 122 4 5 1 Since 25 . 4, this is an ellipse with a horizontal major axis. Now that the equation of the ellipse is in standard form, compare it to 1x 2 h22 a2 1 1y 2 k22 b2 5 1 and identify a, b, h, k. a 5 5, b 5 2, center at 1h, k2 5 123, 12 Since a 5 5, the vertices are five units to the left and right of the center. 128, 12 and 12, 12 Since b 5 2, the endpoints of the minor axis are up and down two units from the center. 123, 212 and 123, 32 Graph. YOUR T UR N Write the equation 4x2 1 32x 1 y2 2 2y 1 61 5 0 in standard form. Identify the center, vertices, and endpoints of the minor axis, and graph. ▼ A N S W E R 1x 1 422 1 1 1y 2 122 4 5 1 Center: 124, 12 Vertices: 124, 212 and 124, 32 Endpoints of minor axis: 125, 12 and 123, 12 x y (–3, 1) (–4, 3) (–4, –1) (–5, 1) (–4, 1) [CONCEPT CHECK] An ellipse of the form Ax 2 1 Bx 1 Cy 2 2 Dy 2 E 5 0, where A, B, C, D and E are all positive has a center that lies in which quadrant? ANSWER Quadrant II ▼ 11.3.3 S KI L L Solve applied problems that involve ellipses. 11.3.3 C ON C E P T U A L Understand that the endpoints and the foci are important for application problems involving ellipses. ▼ 1022 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities If we start with an ellipse in the xy-plane and rotate it around its major axis, the result is a three-dimensional ellipsoid. A football and a blimp are two examples of ellipsoids. The ellipsoidal shape allows for a more aerodynamic path. cscredon/iStockphoto Joseph Sohm/Visions of America/Age Fotostock America, Inc. [CONCEPT CHECK] Would the foci of a football be closer to the center/vertices than the foci of a basketball? ANSWER Vertices ▼ EXAMPLE 6 An Official NFL Football A longitudinal section (that includes the two vertices and the center) of an official Wilson NFL football is an ellipse. The longitudinal section is approximately 11 inches long and 7 inches wide. Write an equation governing the elliptical longitudinal section. Solution: Locate the center of the ellipse at the origin, and orient the football horizontally. Write the general equation of a circle centered at the origin. x2 a2 1 y2 b2 5 1 The length of the major axis is 11 inches. 2a 5 11 Solve for a. a 5 5.5 The length of the minor axis is 7 inches. 2b 5 7 Solve for b. b 5 3.5 Substitute a 5 5.5 and b 5 3.5 into x2 a2 1 y2 b2 5 1. x2 5.52 1 y2 3.52 5 1 or x2 30.25 1 y2 12.25 5 1 11.3 The Ellipse 1023 In this section, we first analyzed ellipses that are centered at the origin. Orientation of Major Axis Horizontal along the x-axis Vertical along the y-axis Equation x2 a2 1 y2 b2 5 1 a . b . 0 x2 b2 1 y2 a2 5 1 a . b . 0 Graph x y (–a, 0) (–c, 0) (c, 0) (0, 0) (a, 0) (0, b) (0, –b) x y (0, –a) (–b, 0) (b, 0) (0, a) (0, c) (0, –c) (0, 0) Foci 12c, 02 1c, 02 10, 2c2 10, c2 Vertices 12a, 02 1a, 02 10, 2a2 10, a2 Other Intercepts 10, 2b2 10, b2 12b, 02 1b, 02 c2 5 a2 2 b2 For ellipses centered at the origin, we can graph an ellipse by finding all four intercepts. For ellipses centered at the point 1h, k2, the major and minor axes and endpoints of the ellipse all shift accordingly. Orientation of Major Axis Horizontal (Parallel to the x-axis) Vertical (Parallel to the y-axis) Equation 1x 2 h22 a2 1 1y 2 k22 b2 5 1 1x 2 h22 b2 1 1y 2 k22 a2 5 1 Graph x y (h – a, k) (h – c, k) (h + c, k) (h + a, k) (h, k) (h, k + b) (h, k – b) x y (h – b, k) (h + b, k) (h, k) (h, k + a) (h, k + c) (h, k – a) (h, k – c) Foci 1h 2 c, k2 1h 1 c, k2 1h, k 2 c2 1h, k 1 c2 Vertices 1h 2 a, k2 1h 1 a, k2 1h, k 2 a2 1h, k 1 a2 When a 5 b, the ellipse is a circle. [SEC TION 11.3] S U M M A RY 1024 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities [SEC TION 11.3] E X E R C I SE S • S K I L L S In Exercises 1–4, match the equation to the ellipse. 1. x2 36 1 y2 16 5 1 2. x2 16 1 y2 36 5 1 3. x2 8 1 y2 72 5 1 4. 4x2 1 y2 5 1 a. x y –10 10 10 –10 b. x y –10 10 10 –10 c. x y –1 1 1 –1 d. x y –10 10 10 –10 In Exercises 5–16, graph each ellipse. 5. x2 25 1 y2 16 5 1 6. x2 49 1 y2 9 5 1 7. x2 16 1 y2 64 5 1 8. x2 25 1 y2 144 5 1 9. x2 100 1 y2 5 1 10. 9x2 1 4y2 5 36 11. 4 9x2 1 81y2 5 1 12. 4 25x2 1 100 9 y2 5 1 13. 4x2 1 y2 5 16 14. x2 1 y2 5 81 15. 8x2 1 16y2 5 32 16. 10x2 1 25y2 5 50 In Exercises 17–24, find the standard form of the equation of an ellipse with the given characteristics. 17. foci: 124, 02 and 14, 02 vertices: 126, 02 and 16, 02 18. foci: 121, 02 and 11, 02 vertices: 123, 02 and 13, 02 19. foci: 10, 232 and 10, 32 vertices: 10, 242 and 10, 42 20. foci: 10, 212 and 10, 12 vertices: 10, 222 and 10, 22 21. Major axis vertical with length of 8, minor axis length of 4 and centered at 10, 02. 22. Major axis horizontal with length of 10, minor axis length of 2 and centered at 10, 02. 23. Vertices 10, 272 and 10, 72 and endpoints of minor axis 123, 02 and 13, 02. 24. Vertices 129, 02 and 19, 02 and endpoints of minor axis 10, 242 and 10, 42. In Exercises 25–28, match each equation with the ellipse. 25. 1x 2 322 4 1 1y 1 222 25 5 1 26. 1x 1 322 4 1 1y 2 222 25 5 1 27. 1x 2 322 25 1 1y 1 222 4 5 1 28. 1x 1 322 25 1 1y 2 222 4 5 1 a. x y –8 2 7 –3 b. x y –2 8 3 –7 c. x y –3 7 3 –7 d. x y –6 4 7 –3 11.3 The Ellipse 1025 In Exercises 29–38, graph each ellipse. 29. 1x 2 122 16 1 1 y 2 222 4 5 1 30. 1x 1 122 36 1 1 y 1 222 9 5 1 31. 101x 1 322 1 1 y 2 422 5 80 32. 31x 1 322 1 121y 2 422 5 36 33. x2 1 4y2 2 24y 1 32 5 0 34. 25x2 1 2y2 2 4y 2 48 5 0 35. x2 2 2x 1 2y2 2 4y 2 5 5 0 36. 9x2 2 18x 1 4y2 2 27 5 0 37. 5x2 1 20x 1 y2 1 6y 2 21 5 0 38. 9x2 1 36x 1 y2 1 2y 1 36 5 0 In Exercises 39–46, find the standard form of an equation of the ellipse with the given characteristics. 39. foci: 122, 52 and 16, 52 vertices: 123, 52 and 17, 52 40. foci: 12, 222 and 14, 222 vertices: 10, 222 and 16, 222 41. foci: 14, 272 and 14, 212 vertices: 14, 282 and 14, 02 42. foci: 12, 262 and 12, 242 vertices: 12, 272 and 12, 232 43. Major axis vertical with length of 8, minor axis length of 4 and centered at 13, 22. 44. Major axis horizontal with length of 10, minor axis length of 2 and centered at 124, 32. 45. Vertices 121, 292 and 121, 12 and endpoints of minor axis 124, 242 and 12, 242. 46. Vertices 122, 32 and 16, 32 and endpoints of minor axis 12, 12 and 12, 52. • A P P L I C A T I O N S 47. Carnival Ride. The Zipper, a favorite carnival ride, maintains an elliptical shape with a major axis of 150 feet and a minor axis of 30 feet. Assuming it is centered at the origin, find an equation for the ellipse. Ingram Publishing/Photolibrary Zipper 48. Carnival Ride. A Ferris wheel traces an elliptical path with both a major and minor axis of 180 feet. Assuming it is centered at the origin, find an equation for the ellipse (circle). Tina Buckman/Photolibrary/Getty Images, Inc. Ferris wheel, Barcelona, Spain For Exercises 49 and 50, refer to the following information. A high school wants to build a football field surrounded by an elliptical track. A regulation football field must be 120 yards long and 30 yards wide. 30 yards 120 yards 49. Sports Field. Suppose the elliptical track is centered at the origin and has a horizontal major axis of length 150 yards and a minor axis length of 40 yards. a. Write an equation for the ellipse. b. Find the width of the track at the end of the field. Will the track completely enclose the football field? 50. Sports Field. Suppose the elliptical track is centered at the origin and has a horizontal major axis of length 150 yards. How long should the minor axis be in order to enclose the field? For Exercises 51 and 52, refer to orbits in our solar system. The planets have elliptical orbits with the Sun as one of the foci. Pluto (orange), the planet farthest from the Sun, has a very elongated, or flattened, elliptical orbit, whereas Earth (royal blue) has almost a circular orbit. Because of Pluto’s flattened path, it is not always the planet farthest from the Sun. Pluto Earth 1026 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities 51. Planetary Orbits. The orbit of the dwarf planet Pluto has approximately the following characteristics (assume the Sun is the focus): • The length of the major axis 2a is approximately 11,827,000,000 kilometers. • The distance from the dwarf planet to the Sun is 4,447,000,000 kilometers. Determine the equation for Pluto’s elliptical orbit around the Sun. 52. Planetary Orbits. The Earth’s orbit has approximately the following characteristics (assume the Sun is the focus): • The length of the major axis 2a is approximately 299,700,000 kilometers. • The distance from the Earth to the Sun is 147,100,000 kilometers. Determine the equation for the Earth’s elliptical orbit around the Sun. For Exercises 53 and 54, refer to the following information. Asteroids orbit the Sun in elliptical patterns and often cross paths with the Earth’s orbit, making life a little tense now and again. A few asteroids have orbits that cross the Earth’s orbit—called “Apollo asteroids” or “Earth-crossing asteroids.” In recent years asteroids have passed within 100,000 kilometers of the Earth! 53. Asteroids. Asteroid 433, or Eros, is the second largest near-Earth asteroid. The semimajor axis is 150 million kilometers and the eccentricity is 0.223, where eccentricity is defined as e 5 Å1 2 b2 a2, where a is the semimajor axis, or 2a is the major axis, and b is the semiminor axis, or 2b is the minor axis. Find the equation of Eros’s orbit. Round to the nearest million kilometers. 54. Asteroids. The asteroid Toutatis is the largest near-Earth asteroid. The semimajor axis is 350 million kilometers and the eccentricity is 0.634, where eccentricity is defined as e 5 Å1 2 b2 a2, where a is the semiminor axis, or 2a is the major axis, and b is the semimajor axis, or 2b is the minor axis. On September 29, 2004, it missed Earth by 961,000 miles. Find the equation of Toutatis’s orbit. 55. Halley’s Comet. The eccentricity of Halley’s Comet is approximately 0.967. If a comet had e 5 1, what would its orbit appear to be from Earth? 56. Halley’s Comet. The length of the semimajor axis is 17.8 AU (astronomical unit) and the eccentricity is approxi-mately 0.967. Find the equation of Halley’s Comet. (Assume 1 AU 5 150 million km.) For Exercises 57 and 58, refer to the following: An elliptical trainer is an exercise machine that can be used to simulate stair climbing, walking, or running. The stride length is the length of a step on the trainer (forward foot to rear foot). The minimum step-up height is the height of a pedal at its lowest point, while the maximum step-up height is the height of a pedal at its highest point. 57. Health/Exercise. An elliptical trainer has a stride length of 16 inches. The maximum step-up height is 12.5 inches, while the minimum step-up height is 2.5 inches. a. Find the equation of the ellipse traced by the pedals assuming the origin lies at the pedal axle (the center of the ellipse is at the origin). b. Use the approximation to the perimeter of an ellipse p 5 p"21a2 1 b22 to find the distance, to the nearest inch, traveled in one complete step (revolution of a pedal). c. How many steps, to the nearest step, are necessary to travel a distance of one mile? 58. Health/Exercise. An elliptical trainer has a stride length of 18 inches. The maximum step-up height is 13.5 inches, while the minimum step-up height is 3.5 inches. Find the equation of the ellipse traced by the pedals assuming the origin lies at the pedal axle. a. Find the equation of the ellipse traced by the pedals assuming the origin lies at the pedal axle (the center of the ellipse is at the origin). b. Use the approximation to the perimeter of an ellipse p 5 p"21a2 1 b22 to find the distance, to the nearest inch, traveled in one complete step (revolution of a pedal). c. How many steps, to the nearest step, are necessary to travel a distance of one mile? 11.3 The Ellipse 1027 In Exercises 59 and 60, explain the mistake that is made. • C A T C H T H E M I S T A K E 59. Graph the ellipse given by x2 6 1 y2 4 5 1. Solution: Write the standard form of the equation of an ellipse. x2 a2 1 y2 b2 5 1 Identify a and b. a 5 6, b 5 4 Label the vertices and the endpoints of the minor axis, 126, 02, 16, 02, 10, 242, 10, 42, and connect with an elliptical curve. This is incorrect. What mistake was made? 60. Determine the foci of the ellipse x2 16 1 y2 9 5 1. Solution: Write the general equation of a horizontal ellipse. x2 a2 1 y2 b2 5 1 Identify a and b. a 5 4, b 5 3 Substitute a 5 4, b 5 3 into c2 5 a2 1 b2. c2 5 42 1 32 Solve for c. c 5 5 Foci are located at 125, 02 and 15, 02. The points 125, 02 and 15, 02 are located outside of the ellipse. What mistake was made? x y (0, 4) (–6, 0) (6, 0) (0, –4) • C O N C E P T U A L 61. If you know the vertices of an ellipse, you can determine the equation for the ellipse. 62. If you know the foci and the endpoints of the minor axis, you can determine the equation for the ellipse. 63. Ellipses centered at the origin have symmetry with respect to the x-axis, y-axis, and the origin. 64. All ellipses are circles, but not all circles are ellipses. In Exercises 61–64, determine whether each statement is true or false. • C H A L L E N G E 65. The eccentricity of an ellipse is defined as e 5 c a. Compare the eccentricity of the orbit of Pluto to that of Earth (refer to Exercises 51 and 52). 66. The eccentricity of an ellipse is defined as e 5 c a. Since a . c . 0, then 0 , e , 1. Describe the shape of an ellipse when a. e is close to zero b. e is close to one c. e 5 0.5 • T E C H N O L O G Y 67. Graph the following three ellipses: x2 1 y2 5 1, x2 1 5y2 5 1, and x2 1 10y2 5 1. What can be said to happen to the ellipse x2 1 cy2 5 1 as c increases? 68. Graph the following three ellipses: x2 1 y2 5 1, 5x2 1 y2 5 1, and 10x2 1 y2 5 1. What can be said to happen to the ellipse cx2 1 y2 5 1 as c increases? 69. Graph the following three ellipses: x2 1 y2 5 1, 5x2 1 5y2 5 1, and 10x2 1 10y2 5 1. What can be said to happen to the ellipse cx2 1 cy2 5 1 as c increases? 70. Graph the equation x2 9 2 y2 16 5 1. Notice what a difference the sign makes. Is this an ellipse? 71. Graph the following three ellipses: x2 1 y2 5 1, 0.5x2 1 y2 5 1, and 0.05x2 1 y2 5 1. What can be said to happen to ellipse cx2 1 y2 5 1 as c decreases? 72. Graph the following three ellipses: x2 1 y2 5 1, x2 1 0.5y2 5 1, and x2 1 0.05y2 5 1. What can be said to happen to ellipse x2 1 cy2 5 1 as c decreases? 1028 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities 11.4.1 Hyperbola Centered at the Origin The definition of a hyperbola is similar to the definition of an ellipse. An ellipse is the set of all points, the sum of whose distances from two points (the foci) is constant. A hyperbola is the set of all points whose difference of the distances from two points (the foci) is constant. What distinguishes their equations is a minus sign. Ellipse centered at the origin: x2 a2 1 y2 b2 5 1 Hyperbola centered at the origin: x2 a2 2 y2 b2 5 1 S K I L L S O B J E C T I V E S ■ ■Find the equation of a hyperbola centered at the origin. ■ ■Find the equation of a hyperbola centered at the point 1h, k2. ■ ■Solve applied problems that involve hyperbolas. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that a hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points (foci) is a positive constant. ■ ■Employ completing the square to transform an equation into the standard form of a hyperbola. ■ ■Understand that it is the difference in time of signals remaining constant that allows ships to navigate along a hyperbolic curve to shore. 11.4 THE HYPERBOLA 11.4.1 SKI LL Find the equation of a hyperbola centered at the origin. 11.4.1 CO NC EPTUAL Understand that a hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points (foci) is a positive constant. A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points is a positive constant. These two fixed points are called foci. The hyperbola has two separate curves called branches. The two points where the hyperbola intersects the line joining the foci are called vertices. The line segment joining the vertices is called the transverse axis of the hyperbola. The midpoint of the transverse axis is called the center. Branches Transverse axis Conjugate axis Vertices Foci Center Hyperbola DEFINITION Let’s consider a hyperbola with the center at the origin and the foci on the x-axis. Let the distance from the center to the focus be c. Therefore, the foci are located at the points 12c, 02 and 1c, 02. The difference of the two distances from the foci to any point 1x, y2 must be constant. We then can follow a similar analysis as done with an ellipse. (–c, 0) (c, 0) (x, y) 11.4 The Hyperbola 1029 Derivation of the Equation of a Hyperbola WORDS MATH The difference of these two distances is equal to a constant (2a for convenience). "1x 2 12c222 1 y2 2 "1x 2 c22 1 y2 5 62a Following the same procedure that we did with an ellipse leads to: Ac2 2 a2Bx2 2 a2y2 5 a2Ac2 2 a2B We can make the argument that c . a in order for a point to be on the hyperbola (and not on the x-axis). Therefore, since a and c represent distances and therefore are positive, we know that c2 . a2, or c2 2 a2 . 0. Hence, we can divide both sides of the equation by c2 2 a2, since c2 2 a2 2 0. x2 2 a2y2 Ac2 2 a2B 5 a2 Let b2 5 c2 2 a2. x2 2 a2y2 b2 5 a2 Divide both sides of the equation by a2. x2 a2 2 y2 b2 5 1 The equation x2 a2 2 y2 b2 5 1 represents a hyperbola with center at the origin and the foci along the x-axis. The following box summarizes hyperbolas that have their center at the origin and foci along either the x-axis or y-axis. EQUATION OF A HYPERBOLA WITH CENTER AT THE ORIGIN The standard form of the equation of a hyperbola with its center at the origin: Orientation of Transverse Axis Horizontal along the x-axis Vertical along the y-axis Equation x2 a2 2 y2 b2 5 1 y2 a2 2 x2 b2 5 1 Graph x y (–c, 0) (c, 0) (–a, 0) (a, 0) x y (0, –c) (0, –a) (0, c) (0, a) Foci 12c, 02 1c, 02 10, 2c2 10, c2 Asymptotes y 5 b ax y 5 2b ax y 5 a bx y 5 2a bx Vertices 12a, 02 1a, 02 10, 2a2 10, a2 Transverse Axis Horizontal length 2a Vertical length 2a where c2 5 a2 1 b2 [CONCEPT CHECK] TRUE OR FALSE In a hyperbola the foci are always closer to the center than the vertices. ANSWER False ▼ 1030 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities Note that for x2 a2 2 y2 b2 5 1, if x 5 0, then 2y2 b2 5 1, which yields an imaginary number for y. But when y 5 0, x2 a2 5 1, and therefore x 56a. The vertices for this hyperbola are 12a, 02 and 1a, 02. EXAMPLE 1 Finding the Foci and Vertices of a Hyperbola Given the Equation Find the foci and vertices of the hyperbola given by x2 9 2 y2 4 5 1. Solution: Compare to the standard equation of a hyperbola, x2 a2 2 y2 b2 5 1. a2 5 9, b2 5 4 Solve for a and b. a 5 3, b 5 2 Substitute a 5 3 into the vertices, 12a, 02 and 1a, 02. 123, 02 and 13, 02 Substitute a 5 3, b 5 2 into c2 5 a2 1 b2. c2 5 32 1 22 Solve for c. c2 5 13 c 5 !13 Substitute c 5 !13 into the foci, 12c, 02 and 1c, 02. 12!13, 02 and 1 !13, 02 The vertices are 123, 02 and 13, 02 , and the foci are 12!13, 02 and 1 !13, 02 . Y OUR TU R N Find the vertices and foci of the hyperbola y2 16 2 x2 20 5 1. ▼ ▼ A N S W E R Vertices: 10, 242 and 10, 42 Foci: 10, 262 and 10, 62 ▼ A N S W E R x2 4 2 y2 12 5 1 EXAMPLE 2 Finding the Equation of a Hyperbola Given Foci and Vertices Find the equation of a hyperbola whose vertices are located at 10, 242 and 10, 42 and whose foci are located at 10, 252 and 10, 52. Solution: The center is located at the midpoint of the segment joining the vertices. a0 1 0 2 , 24 1 4 2 b 5 10, 02 Since the foci and vertices are located on the y-axis, the standard equation is given by: y2 a2 2 x2 b2 5 1 The vertices 10, 6a2 and the foci 10, 6c2 can be used to identify a and c. a 5 4, c 5 5 Substitute a 5 4, c 5 5 into b2 5 c2 2 a2. b2 5 52 2 42 Solve for b. b2 5 25 2 16 5 9 b 5 3 Substitute a 5 4 and b 5 3 into y2 a2 2 x2 b2 5 1. y2 16 2 x2 9 5 1 Y OUR TU R N Find the equation of a hyperbola whose vertices are located at 122, 02 and 12, 02 and whose foci are located at 124, 02 and 14, 02. ▼ 11.4 The Hyperbola 1031 Graphing a Hyperbola Centered at the Origin To graph a hyperbola, we use the vertices and asymptotes. The asymptotes are found by the equations y 5 6 b ax or y 5 6 a bx, depending on whether the transverse axis is horizontal or vertical. An easy way to draw these graphing aids is to first draw the rectangular box that passes through the vertices and the points 10, 6b2 or 16b, 02. The conjugate axis is perpendicular to the transverse axis and has length 2b. The asymptotes pass through the center of the hyperbola and the corners of the rectangular box. x2 a2 2 y2 b2 5 1 y2 a2 2 x2 b2 5 1 x y (–c, 0) (0, –b) (0, b) (c, 0) (–a, 0) (a, 0) y = x b a y = – x b a x y (–b, 0) (0, –c) (0, –a) (0, c) (0, a) (b, 0) y = x a b y = – x a b EXAMPLE 3 Graphing a Hyperbola Centered at the Origin with a Horizontal Transverse Axis Graph the hyperbola given by x2 4 2 y2 9 5 1. Solution: Compare x2 22 2 y2 32 5 1 to the general equation x2 a2 2 y2 b2 5 1. Identify a and b. a 5 2 and b 5 3 The transverse axis of this hyperbola lies on the x-axis. Label the vertices 12a, 02 5 122, 02 and 1a, 02 5 12, 02 and the points 10, 2b2 5 10, 232 and 10, b2 5 10, 32. Draw the rectangular box that passes through those points. Draw the asymptotes that pass through the center and the corners of the rectangle. x y (–2, 0) (0, –3) (0, 3) (2, 0) Draw the two branches of the hyperbola, each passing through a vertex and guided by the asymptotes. x y (–2, 0) (2, 0) 1032 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities In Example 3, if we let y 5 0, then x2 4 5 1 or x 562. So the vertices are 122, 02 and 12, 02, and the transverse axis lies along the x-axis. Note that if x 5 0, y 563i. EXAMPLE 4 Graphing a Hyperbola Centered at the Origin with a Vertical Transverse Axis Graph the hyperbola given by y2 16 2 x2 4 5 1. Solution: Compare y2 42 2 x2 22 5 1 to the general equation y2 a2 2 x2 b2 5 1. Identify a and b. a 5 4 and b 5 2 The transverse axis of this hyperbola lies along the y-axis. Label the vertices 10, 2a2 5 10, 242 and 10, a2 5 10, 42 and the points 12b, 02 5 122, 02 and 1b, 02 5 12, 02. Draw the rectangular box that passes through those points. Draw the asymptotes that pass through the center and the corners of the rectangle. x y (0, –4) (0, 4) (–2, 0) (2, 0) Draw the two branches of the hyperbola, each passing through a vertex and guided by the asymptotes. x y (–2, 0) (0, –4) (0, 4) (2, 0) Y OUR TU R N Graph the hyperbolas. a. y2 1 2 x2 4 5 1 b. x2 4 2 y2 1 5 1 ▼ ▼ A N S W E R a. x y (–2, 0) (0, –1) (0, 1) (2, 0) b. x y (–2, 0) (0, –1) (0, 1) (2, 0) 11.4.2 Hyperbola Centered at the Point (h, k) We can use graph-shifting techniques to graph hyperbolas that are centered at a point other than the origin, say, 1h, k2. For example, to graph 1x 2 h22 a2 2 1y 2 k22 b2 5 1, start with the graph of x2 a2 2 y2 b2 5 1 and shift to the right h units and up k units. The center, the vertices, the foci, the transverse and conjugate axes, and the asymptotes all shift. The follow ing table summarizes the characteristics of hyperbolas centered at a point other than the origin. 11.4.2 SKI LL Find the equation of a hyperbola centered at the point 1h, k2. 11.4.2 CO NC EPTUAL Employ completing the square to transform an equation into the standard form of a hyperbola. 11.4 The Hyperbola 1033 EXAMPLE 5 Graphing a Hyperbola with Center Not at the Origin Graph the hyperbola 1y 2 222 16 2 1x 2 122 9 5 1. Solution: Compare 1y 2 222 42 2 1x 2 122 32 5 1 to the general equation 1y 2 k22 a2 2 1x 2 h22 b2 5 1. Identify a, b, and 1h, k2. a 5 4, b 5 3, and 1h, k2 5 11, 22 The transverse axis of this hyperbola lies along x 5 1, which is parallel to the y-axis. Label the vertices 1h, k 2 a2 5 11, 222 and 1h, k 1 a2 5 11, 62, and the points 1h 2 b, k2 5 122, 22 and 1h 1 b, k2 5 14, 22. Draw the rectangular box that passes through those points. Draw the asymptotes that pass through the center 1h, k2 5 11, 22 and the corners of the rectangle. Draw the two branches of the hyperbola, each passing through a vertex and guided by the asymptotes. x y (–2, 2) (1, –2) (1, 6) (4, 2) EQUATION OF A HYPERBOLA WITH CENTER AT THE POINT (h, k) The standard form of the equation of a hyperbola with its center at the point 1h, k2 is given by: Orientation of Transverse Axis Horizontal Parallel to the x-axis Vertical Parallel to the y-axis Equation 1x 2 h22 a2 2 1 y 2 k22 b2 5 1 1 y 2 k22 a2 2 1x 2 h22 b2 5 1 Graph Vertices 1h 2a, k2 1h 1 a, k2 1h, k 2a2 1h, k 1 a2 Foci 1h 2 c, k2 1h 1 c, k2 1h, k 2 c2 1h, k 1 c2 where c2 5 a2 1 b2 1034 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities 11.4.3 Applications Nautical navigation is assisted by hyperbolas. For example, suppose two radio stations on a coast are emitting simultaneous signals. If a boat is at sea, it will be slightly closer to one station than the other station, which results in a small time difference between the received signals from the two stations. Recall that a hyperbola is the set of all points whose differences of the distances from two points (the foci—or the radio stations) are constant. Therefore, if the boat follows the path associated with a constant time difference, that path will be hyperbolic. The synchronized signals would intersect one another in associated hyperbolas. Each time difference corresponds to a different path. The radio stations are the foci of the hyperbolas. This principle forms the basis of a hyperbolic radio navigation system known as loran (LOng-RAnge Navigation). There are navigational charts that correspond to different time differences. A ship selects the hyperbolic path that will take it to the desired port, and the loran chart lists the corresponding time difference. EXAMPLE 6 Transforming an Equation of a Hyperbola to Standard Form Graph the hyperbola 9x2 2 16y2 2 18x 1 32y 2 151 5 0. Solution: Complete the square on the x 91x2 2 2x2 2 161y2 2 2y2 5 151 terms and y terms, respectively. 91x2 2 2x 1 12 2 161y2 2 2y 1 12 5 151 1 9 2 16 91x 2 122 2 161y 2 122 5 144 1x 2 122 16 2 1y 2 122 9 5 1 Compare 1x 2 122 16 2 1y 2 122 9 5 1 to the general form 1x 2 h22 a2 2 1y 2 k22 b2 5 1. Identify a, b, and 1h, k2. a 5 4, b 5 3, and 1h, k2 5 11, 12 The transverse axis of this hyperbola lies along y 5 1. Label the vertices 1h 2 a, k2 5 123, 12 and 1h 1 a, k2 5 15, 12 and the points 1h, k 2 b2 5 11, 222 and 1h, k 1 b2 5 11, 42. Draw the rectangular box that passes through these points. Draw the asymptotes that pass through the center 11, 12 and the corners of the box. Draw the two branches of the hyperbola, each passing through the vertex and guided by the asymptotes. x y 6 –4 –4 6 (5, 1) (1, 4) (1, –2) (–3, 1) [CONCEPT CHECK] A hyperbola of the form Ax 2 2 Bx 2 Cy 2 1 Dy 2 E 5 0 where A, B, C, D, E are all positive has a center that lies in which quadrant? ANSWER Quadrant I ▼ Radio stations Water 11.4.3 SKI LL Solve applied problems that involve hyperbolas. 11.4.3 CO NC EPTUAL Understand that it is the differ-ence in time of signals remaining constant that allows ships to navigate along a hyperbolic curve to shore. 11.4 The Hyperbola 1035 In this section, hyperbolas centered at the origin were discussed: Equation x2 a2 2 y2 b2 5 1 y2 a2 2 x2 b2 5 1 Transverse axis Horizontal (x-axis), length 2a Vertical ( y-axis), length 2a Conjugate axis Vertical ( y-axis), length 2b Horizontal (x-axis), length 2b Vertices 12a, 02 and 1a, 02 10, 2a2 and 10, a2 Foci c2 5 a2 1 b2 12c, 02 and 1c, 02 10, 2c2 and 10, c2 Asymptote y 5 b ax and y 5 2b ax y 5 a bx and y 5 2a bx Graph x y (–c, 0) (0, –b) (0, b) (c, 0) (–a, 0) (a, 0) x y (–b, 0) (0, –c) (0, –a) (0, c) (0, a) (b, 0) For a hyperbola centered at 1h, k2, the vertices, foci, and asymptotes all shift accordingly. Orientation of Transverse Axis Horizontal Parallel to the x-axis Vertical Parallel to the y-axis Equation 1x 2 h22 a2 2 1 y 2 k22 b2 5 1 1 y 2 k22 a2 2 1x 2 h22 b2 5 1 Graph Vertices 1h 2a, k2 1h 1 a, k2 1h, k 2a2 1h, k 1 a2 Foci 1h 2 c, k2 1h 1 c, k2 1h, k 2 c2 1h, k 1 c2 where c2 5 a2 1 b2 [SEC TION 11.4] S UM M A RY 1036 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities [SEC TION 11.4] E X E R C I SE S • S K I L L S In Exercises 1–4, match each equation with the corresponding hyperbola. 1. x2 36 2 y2 16 5 1 2. y2 36 2 x2 16 5 1 3. x2 8 2 y2 72 5 1 4. 4y2 2 x2 5 1 a. x y –10 10 10 –10 b. x y –10 10 5 –5 c. x y –2 2 2 –2 d. x y –10 10 20 –20 In Exercises 5–16, graph each hyperbola. 5. x2 25 2 y2 16 5 1 6. x2 49 2 y2 9 5 1 7. y2 16 2 x2 64 5 1 8. y2 144 2 x2 25 5 1 9. x2 100 2 y2 5 1 10. 9y2 2 4x2 5 36 11. 4y2 9 2 81x2 5 1 12. 4 25x2 2 100 9 y2 5 1 13. 4x2 2 y2 5 16 14. y2 2 x2 5 81 15. 8y2 2 16x2 5 32 16. 10x2 2 25y2 5 50 In Exercises 17–24, find the standard form of an equation of the hyperbola with the given characteristics. 17. vertices: 124, 02 and 14, 02 foci: 126, 02 and 16, 02 18. vertices: 121, 02 and 11, 02 foci: 123, 02 and 13, 02 19. vertices: 10, 232 and 10, 32 foci: 10, 242 and 10, 42 20. vertices: 10, 212 and 10, 12 foci: 10, 222 and 10, 22 21. center: 10, 02; transverse: x-axis; asymptotes: y 5 x and y 5 2x 22. center: 10, 02; transverse: y-axis; asymptotes: y 5 x and y 5 2x 23. center: 10, 02; transverse axis: y-axis; asymptotes: y 5 2x and 24. center: 10, 02; transverse axis: x-axis; asymptotes: y 5 2x and y 5 22x y 5 22x In Exercises 25–28, match each equation with the hyperbola. 25. 1x 2 322 4 2 1y 1 222 25 5 1 26. 1x 1 322 4 2 1y 2 222 25 5 1 27. 1y 2 322 25 2 1x 1 222 4 5 1 28. 1y 1 322 25 2 1x 2 222 4 5 1 a. –10 10 10 –10 x y b. –10 10 10 –10 x y c. x y –10 10 10 –10 d. x y –10 10 10 –10 In Exercises 29–38, graph each hyperbola. 29. 1x 2 122 16 2 1y 2 222 4 5 1 30. 1y 1 122 36 2 1x 1 222 9 5 1 31. 101y 1 322 2 1x 2 422 5 80 32. 31x 1 322 2 121y 2 422 5 36 11.4 The Hyperbola 1037 33. x2 2 4x 2 4y2 5 0 34. 29x2 1 y2 1 2y 2 8 5 0 35. 29x2 2 18x 1 4y2 2 8y 2 41 5 0 36. 25x2 2 50x 2 4y2 2 8y 2 79 5 0 37. x2 2 6x 2 4y2 2 16y 2 8 5 0 38. 24x2 2 16x 1 y2 2 2y 2 19 5 0 In Exercises 39–42, find the standard form of an equation of the hyperbola with the given characteristics. 39. vertices: 122, 52 and 16, 52 foci: 123, 52 and 17, 52 40. vertices: 11, 222 and 13, 222 foci: 10, 222 and 14, 222 41. vertices: 14, 272 and 14, 212 foci: 14, 282 and 14, 02 42. vertices: 12, 262 and 12, 242 foci: 12, 272 and 12, 232 • A P P L I C A T I O N S 43. Ship Navigation. Two loran stations are located 150 miles apart along a coast. If a ship records a time difference of 0.0005 second and continues on the hyperbolic path corresponding to that difference, where will it reach shore? 44. Ship Navigation. Two loran stations are located 300 miles apart along a coast. If a ship records a time difference of 0.0007 second and continues on the hyperbolic path corresponding to that difference, where will it reach shore? Round to the nearest mile. 45. Ship Navigation. If the captain of the ship in Exercise 43 wants to reach shore between the stations and 30 miles from one of them, what time difference should he look for? 46. Ship Navigation. If the captain of the ship in Exercise 44 wants to reach shore between the stations and 50 miles from one of them, what time difference should he look for? 47. Light. If the light from a lamp casts a hyperbolic pattern on the wall due to its lampshade, calculate the equation of the hyperbola if the distance between the vertices is 2 feet and the foci are half a foot from the vertices. 48. Special Ops. A military special ops team is calibrating its recording devices used for passive ascertaining of enemy location. They place two recording stations, alpha and bravo, 3000 feet apart (alpha is due east of bravo). The team detonates small explosives 300 feet west of alpha and records the time it takes each station to register an explosion. The team also sets up a second set of explosives directly north of the alpha station. How many feet north of alpha should the team set off the explosives if it wants to record the same times as on the first explosion? 3000 ft Alpha Explosive Explosive Bravo 300 ft ? ft For Exercises 49 and 50, refer to the following: Nuclear cooling towers are typically built in the shape of a hyperboloid. The cross section of a cooling tower forms a hyperbola. The cooling tower pictured is 450 feet tall and modeled by the equation x2 8100 2 y2 16,900 5 1 x y –200 200 200 –300 49. Engineering/Design. Find the diameter of the top of the cooling tower to the nearest foot. 50. Engineering/Design. Find the diameter of the base of the tower to the nearest foot. 1038 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities 51. Graph the hyperbola y2 4 2 x2 9 5 1. Solution: Compare the equation to the standard form and solve for a and b. a 5 2, b 5 3 Label the vertices 12a, 02 and 1a, 02. 122, 02 and 12, 02 Label the points 10, 2b2 and 10, b2. 10, 232 and 10, 32 Draw the rectangle connecting these four points and align the asymptotes so that they pass through the center and the corner of the boxes. Then draw the hyperbola using the vertices and asymptotes. x y (–2, 0) (0, –3) (0, 3) (2, 0) This is incorrect. What mistake was made? 52. Graph the hyperbola x2 1 2 y2 4 5 1. Solution: Compare the equation to the general form and solve for a and b. a 5 2, b 5 1 Label the vertices 12a, 02 and 1a, 02. 122, 02 and 12, 02 Label the points 10, 2b2 and 10, b2. 10, 212 and 10, 12 Draw the rectangle connecting these four points and align the asymptotes so that they pass through the center and the corner of the boxes. Then draw the hyperbola using the vertices and asymptotes. x y (–2, 0) (0, –1) (0, 1) (2, 0) This is incorrect. What mistake was made? • C A T C H T H E M I S T A K E In Exercises 51 and 52, explain the mistake that is made. • C O N C E P T U A L 53. If you know the vertices of a hyperbola, you can determine the equation for the hyperbola. 54. If you know the foci and vertices, you can determine the equation for the hyperbola. 55. Hyperbolas centered at the origin have symmetry with respect to the x-axis, y-axis, and the origin. 56. The center and foci are part of the graph of a hyperbola. In Exercises 53–56, determine whether each statement is true or false. • C H A L L E N G E 57. Find the general equation of a hyperbola whose asymptotes are perpendicular. 58. Find the general equation of a hyperbola whose vertices are 13, 222 and 121, 222 and whose asymptotes are the lines y 5 2x 2 4 and y 5 22x. • T E C H N O L O G Y 59. Graph the following three hyperbolas: x2 2 y2 5 1, x2 2 5y2 5 1, and x2 2 10y2 5 1. What can be said to happen to the hyperbola x2 2 cy2 5 1 as c increases? 60. Graph the following three hyperbolas: x2 2 y2 5 1, 5x2 2 y2 5 1, and 10x2 2 y2 5 1. What can be said to happen to the hyperbola cx2 2 y2 5 1 as c increases? 61. Graph the following three hyperbolas: x2 2 y2 5 1, 0.5x2 2 y2 5 1, and 0.05x2 2 y2 5 1. What can be said to happen to hyperbola cx2 2 y2 5 1 as c decreases? 62. Graph the following three hyperbolas: x2 2 y2 5 1, x2 2 0.5y2 5 1, and x2 2 0.05y2 5 1. What can be said to happen to hyperbola x2 2 cy2 5 1 as c decreases? 11.5 Systems of Nonlinear Equations 1039 11.5.1 Solving a System of Nonlinear Equations In Chapters 9 and 10, we discussed solving systems of linear equations. In Chapter 9, we applied elimination and substitution to solve systems of linear equations in two variables, and in Chapter 10 we employed matrices to solve systems of linear equations. Recall that a system of linear equations in two variables has one of three types of solutions: One Solution Two lines that intersect at one point x y No Solution Two parallel lines (that never intersect) x y Infinitely Many Solutions Two lines that coincide (s ame line) x y Notice that systems of linear equations in two variables always corresponded to lines. Now, we turn our attention to systems of nonlinear equations in two variables. If any of the equations in a system of equations is nonlinear, then the system is a nonlinear system. The following are systems of nonlinear equations: by 5 x2 1 1 1Parabola2 y 5 2x 1 2 1Line2 r bx2 1 y2 5 25 1Circle2 y 5 x 1Line2 r e x2 9 1 y2 4 5 1 1Ellipse2 y2 16 2 x2 25 5 1 1Hyperbola2 u To find the solution to these systems, we ask the question “at what point(s)—if any— do the graphs of these equations intersect?” Since some nonlinear equations represent conics, this is a convenient time to discuss systems of nonlinear equations. 11.5.1 S KI L L Solve systems of nonlinear equations. 11.5.1 C ON C E P T U A L Understand the types of solutions to systems of nonlinear equations—distinct number of solutions, no solutions, and infinitely many solutions— and interpret the solutions graphically. S K I L L S O B J E C T I V E ■ ■Solve systems of nonlinear equations C O N C E P T U A L O B J E C T I V E ■ ■Understand the types of solutions to systems of nonlinear equations—distinct number of solutions, no solutions, and infinitely many solutions—and interpret the solutions graphically 11.5 SYSTEMS OF NONLINEAR EQUATIONS How many points of intersection do a line and a parabola have? The answer depends on which line and which parabola. As we see in the following graphs, the answer can be one, two, or none. How many points of intersection do a parabola and an ellipse have? One, two, three, four, or no points of intersection correspond to one solution, two solutions, three solutions, four solutions, or no solution, respectively. How many points of intersection do a parabola and a hyperbola have? The answer depends on which parabola and which hyperbola. As we see in the following graphs, the answer can be one, two, three, four, or none. Using Elimination to Solve Systems of Nonlinear Equations The first three examples in this section use elimination to solve systems of two nonlinear equations. In linear systems, we can eliminate either variable. In nonlinear systems, the variable to eliminate is the one that is raised to the same power in both equations. x y x y x y One solution Two solutions No solution x y x y x y x y x y One solution Four solutions No solution Two solutions Three solutions x y x y x y One solution Two solutions Three solutions x y x y Four solutions No solution 1040 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities EXAMPLE 2 Solving a System of Two Nonlinear Equations Using Elimination: More Than One Solution Solve the system of equations and graph the corresponding parabola and circle to verify the answer. Equation (1): 2x2 1 y 5 27 Equation (2): x2 1 y2 5 9 Solution: Equation (1): 2x2 1 y 5 27 Equation (2): x2 1 y2 5 9 Add. y2 1 y 5 2 Gather all terms to one side. y2 1 y 2 2 5 0 Factor. 1y 1 221y 2 12 5 0 Solve for y. y 5 22 or y 5 1 EXAMPLE 1 Solving a System of Two Nonlinear Equations Using Elimination: One Solution Solve the system of equations and graph the corresponding line and parabola to verify the answer. Equation (1): 2x 2 y 5 3 Equation (2): x2 2 y 5 2 Solution: Equation (1): 2x 2 y 5 3 Multiply Equation (2) by 21. 2x2 1 y 5 22 Add. 2x 2 x2 5 1 Gather all terms to one side. x2 2 2x 1 1 5 0 Factor. 1x 2 122 5 0 Solve for x. x 5 1 Substitute x 5 1 into original Equation (1). 2112 2 y 5 3 Solve for y. y 5 21 The solution is x 5 1, y 5 21, or 11, 212. Graph the line y 5 2x 2 3 and the parabola y 5 x2 2 2 and confirm that the point of intersection is 11, 212. x y (1, –1) 11.5 Systems of Nonlinear Equations 1041 Substitute y 5 22 into Equation (2). x2 1 12222 5 9 Solve for x. x 5 6!5 Substitute y 5 1 into Equation (2). x2 1 1122 5 9 Solve for x. x 5 6!8 5 62!2 There are four solutions: A2!5, 22B, A !5, 22B, A22!2, 1B, and A2!2, 1B . Graph the parabola y 5 x2 2 7 and the circle x2 1 y2 5 9 and confirm the four points of intersection. EXAMPLE 3 Solving a System of Two Nonlinear Equations Using Elimination: No Solution Solve the system of equations and graph the corresponding parabolas to verify the answer. Equation (1): x2 1 y 5 3 Equation (2): 2x2 1 y 5 5 Solution: Equation (1): x2 1 y 5 3 Equation (2): 2x2 1 y 5 5 Add. 2y 5 8 Solve for y. y 5 4 Substitute y 5 4 into Equation (1). x2 1 4 5 3 Simplify. x2 5 21 x2 5 21 has no real solution. There is no solution to this system of nonlinear equations. Graph the parabola x2 1 y 5 3 and the parabola y 5 x2 1 5 and confirm that there are no points of intersection. x y 1042 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities EXAMPLE 4 Solving a System of Nonlinear Equations with Elimination Solve the system of nonlinear equations with elimination. Equation (1): x2 4 1 y2 5 1 Equation (2): x2 2 y2 5 1 Solution: Add Equations (1) and (2) to eliminate y2. x2 4 1 y2 5 1 x2 2 y2 5 1 5 4x2 5 2 Solve for x. x2 5 8 5 5 6Å 8 5 Let x 5 6"8 5 in Equation (2). ¢6 Å 8 5 ≤ 2 2 y2 5 1 Solve for y. y2 5 8 5 2 1 5 3 5 y 5 6Å 3 5 There are four solutions: A2"8 5, 2 "3 5B, A2"8 5, "3 5B, A"8 5, 2 "3 5B, and A"8 5, "3 5B . We can approximate the radicals "8 5 < 1.26 and "3 5 < 0.77 to find the four points of intersection of the ellipse x2 4 1 y2 5 1 and the hyperbola x2 2 y2 5 1. Y OUR T UR N Solve the following systems of nonlinear equations. a. 2x 1 y 5 3 b. x2 1 y 5 2 c. x2 9 1 y2 4 5 1 x2 2 y 5 21 2x 1 y 5 3 y2 9 2 x2 16 5 1 ▼ A N S W E R a. 121, 22 and 12, 52 b. No solution c. No solution x (–1, 2) (2, 5) y x y x y ▼ 11.5 Systems of Nonlinear Equations 1043 Using Substitution to Solve Systems of Nonlinear Equations Elimination is based on the idea of eliminating one of the variables and solving the resulting equation in one variable. This is not always possible with nonlinear systems. For example, a system consisting of a circle and a line x2 1 y2 5 5 2x 1 y 5 1 cannot be solved with elimination because both variables are raised to different powers in each equation. We now turn to the substitution method. It is important to always check solutions because extraneous solutions are possible. STUDY TIP Extraneous solutions are possible when you have one equation with a linear (odd) power and one equation with a second-degree (even) power. EXAMPLE 5 Solving a System of Nonlinear Equations Using Substitution Solve the system of equations and graph the corresponding circle and line to verify the answer. Equation (1): x2 1 y2 5 5 Equation (2): 2x 1 y 5 1 Solution: Equation (1): x2 1 y2 5 5 Rewrite Equation (2) with y isolated. y 5 x 1 1 Substitute Equation (2) into Equation (1). x2 1 1x 1 122 5 5 Eliminate the parentheses. x2 1 x2 1 2x 1 1 5 5 Gather like terms. 2x2 1 2x 2 4 5 0 Divide by 2. x2 1 x 2 2 5 0 Factor. 1x 1 221x 2 12 5 0 Solve for x. x 5 22 or x 5 1 Substitute x 5 22 into Equation (1). 12222 1 y2 5 5 Solve for y. y 5 21 or y 5 1 Substitute x 5 1 into Equation (1). 1122 1 y2 5 5 Solve for y. y 5 22 or y 5 2 There appear to be four solutions: 122, 212, 122, 12, 11, 222, and 11, 22. But a line can intersect a circle in no more than two points. Therefore, at least two solutions are extraneous. All four points satisfy Equation (1), but only 122, 212 and 11, 22 also satisfy Equation (2). The answer is 122, 212 and 11, 22 . Graph the circle x2 1 y2 5 5 and the line y 5 x 1 1 and confirm the two points of intersection. Note: After solving for x, had we substituted back into the linear Equation (2) instead of Equation (1), extraneous solutions would not have appeared. In general, substitute back into the lowest-degree equation and always check solutions. Y OUR TU R N Solve the system of equations x2 1 y2 5 13 and x 1 y 5 5. x (–2, –1) (1, 2) y ▼ A N S W E R 12, 32 and 13, 22 ▼ STUDY TIP Substitute back into the lowest-degree equation and always check solutions. In Example 6, the equation xy 5 2 can also be shown to be a rotated hyperbola. Although we will not discuss rotated conics in this book, we can express this equation in terms of a reciprocal function y 5 2 x, which was discussed in Section 3.2. 1044 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities EXAMPLE 6 Solving a System of Nonlinear Equations with Substitution Solve the system of equations. Equation (1): x2 1 y2 5 5 Equation (2): xy 5 2 Solution: Since Equation (2) tells us that xy 5 2, we know neither x nor y can be zero. Solve Equation (2) for y. y 5 2 x Substitute y 5 2 x into Equation (1). x2 1 a2 xb 2 5 5 Eliminate parentheses. x2 1 4 x2 5 5 Multiply by x2. x4 1 4 5 5x2 Collect terms to one side. x4 2 5x2 1 4 5 0 Factor. 1x2 2 421x2 2 12 5 0 Solve for x. x 5 62 or x 5 61 Substitute x 5 22 into Equation (2) and solve for y. y 5 21 Substitute x 5 2 into Equation (2) and solve for y. y 5 1 Substitute x 5 21 into Equation (2) and solve for y. y 5 22 Substitute x 5 1 into Equation (2) and solve for y. y 5 2 Check to see that there are four solutions: 122, 212, 121, 222, 12, 12, and 11, 22 . Note: It is important to check the solutions either algebraically or graphically (see graph on the right). Y OUR T UR N Solve the system of equations. x2 1 y2 5 2 xy 5 1 ▼ [CONCEPT CHECK] TRUE OR FALSE A system of nonlinear equations that correspond to the instersection of a hyperbola and an ellipse can have five solutions. ANSWER False ▼ Applications EXAMPLE 7 Calculating How Much Fence to Buy A couple buys a rectangular piece of property advertised as 10 acres (approximately 400,000 square feet). They want two fences to divide the land into an internal grazing area and a surrounding riding path. If they want the riding path to be 20 feet wide, one fence will enclose the property and one internal fence will sit 20 feet inside the outer fence. If the internal grazing field is 237,600 square feet, how many linear feet of fencing should they buy? Solution: Use the five-step procedure for solving word problems from Section 1.2 and use two variables. x y (1, 2) (2, 1) (–2, –1) (–1, –2) ▼ A N S W E R 121, 212 and 11, 12 11.5 Systems of Nonlinear Equations 1045 STEP 1 Identify the question. How many linear feet of fence should they buy? Or what is the sum of the perimeters of the two fences? STEP 2 Make notes. STEP 3 Set up the equations. x 5 length of property x 2 40 5 length of internal field y 5 width of property y 2 40 5 width of internal field Equation (1): xy 5 400,000 Equation (2): 1x 2 4021y 2 402 5 237,600 STEP 4 Solve the system of equations. Substitution Method Since Equation (1) tells us that xy 5 400,000, we know that neither x nor y can be zero. Solve Equation (1) for y. y 5 400,000 x Substitute y 5 400,000 x into Equation (2). 1x 2 402 a400,000 x 2 40b 5 237,600 Eliminate parentheses. 400,000 2 40x 2 16,000,000 x 1 1600 5 237,600 Multiply by the LCD, x. 400,000x 2 40x2 2 16,000,000 1 1600x 5 237,600x Collect like terms on one side. 40x2 2 164,000x 1 16,000,000 5 0 Divide by 40. x2 2 4100x 1 400,000 5 0 Factor. 1x 2 400021x 2 1002 5 0 Solve for x. x 5 4000 or x 5 100 Substitute x 5 4000 into the original Equation (1). 4000y 5 400,000 Solve for y. y 5 100 Substitute x 5 100 into the original Equation (1). 100y 5 400,000 Solve for y. y 5 4000 The two solutions yield the same dimensions: 4000 3 100. The inner field has the dimensions 3960 3 60. Therefore, the perimeter of both fences is: 2140002 1 211002 1 2139602 1 21602 5 8000 1 200 1 7920 1 120 5 16,240 The couple should buy 16,240 linear feet of fencing. STEP 5 Check the solution. The point 14000, 1002 satisfies both Equation (1) and Equation (2). 1046 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities It is important to note that some nonlinear equations are not conic sections (exponential, logarithmic, and higher-degree polynomial equations). These systems of nonlinear equations are typically solved by the substitution method (see Exercises). [SEC TION 11.5] E X E RC I S E S • S K I L L S In Exercises 1–12, solve the system of equations by applying the elimination method. 1. x2 2 y 5 22 2. x2 1 y 5 2 3. x2 1 y 5 1 4. x2 2 y 5 2 2x 1 y 5 4 2x 1 y 5 21 2x 1 y 5 2 22x 1 y 5 23 5. x2 1 y 5 25 6. x2 2 y 5 27 7. x2 1 y2 5 1 8. x2 1 y2 5 1 2x 1 y 5 3 x 1 y 5 22 x2 2 y 5 21 x2 1 y 5 21 9. x2 1 y2 5 3 10. x2 1 y2 5 6 11. x2 1 y2 5 26 12. x2 1 y2 5 5 4x2 1 y 5 0 27x2 1 y 5 0 22x2 1 y 5 7 3x2 1 y 5 9 In Exercises 13–24, solve the system of equations by applying the substitution method. 13. x 1 y 5 2 x2 1 y2 5 2 14. x 2 y 5 22 x2 1 y2 5 2 15. xy 5 4 x2 1 y2 5 10 16. xy 5 23 x2 1 y2 5 12 17. y 5 x2 2 3 y 5 24x 1 9 18. y 5 2x2 1 5 y 5 3x 2 4 19. x2 1 xy 2 y2 5 5 x 2 y 5 21 20. x2 1 xy 1 y2 5 13 x 1 y 5 21 21. 2x 2 y 5 3 x2 1 y2 2 2x 1 6y 5 29 22. x2 1 y2 2 2x 2 4y 5 0 22x 1 y 5 23 23. 4x2 1 12xy 1 9y2 5 25 22x 1 y 5 1 24. 24xy 1 4y2 5 8 3x 1 y 5 2 In Exercises 25–34, solve the system of equations by applying any method. 25. x3 2 y3 5 63 x 2 y 5 3 26. x3 1 y3 5 226 x 1 y 5 22 27. 4x2 2 3xy 5 25 2x2 1 3xy 5 8 28. 2x2 1 5xy 5 2 x2 2 xy 5 1 29. logx12y2 5 3 logx1y2 5 2 30. logx1y2 5 1 logx12y2 5 1 2 31. 1 x3 1 1 y2 5 17 1 x3 2 1 y2 521 32. 2 x2 1 3 y2 5 5 6 4 x2 2 9 y2 5 0 33. 2x2 1 4y4 5 22 6x2 1 3y4 5 21 34. x2 1 y2 5 22 x2 1 y2 5 21 In Exercises 35–38, graph each equation and find the point(s) of intersection. 35. The parabola y 5 x2 2 6x 1 11 and the line y 5 2x 1 7. 36. The circle x2 1 y2 2 4x 2 2y 1 5 5 0 and the line 2x 1 3y 5 6. 37. The ellipse 9x2 2 18x 1 4y2 1 8y 2 23 5 0 and the line 23x 1 2y 5 1. 38. The parabola y 5 2x2 1 2x and the circle x2 1 6x 1 y2 2 4y 1 12 5 0. In this section systems of two equations were discussed when at least one of the equations is nonlinear (e.g., conics). The substitution method and elimination method from Section 9.1 can sometimes be applied to nonlinear systems. When graphing the two equations, the points of intersection are the solutions of the system. Systems of nonlinear equations can have more than one solution. Also, extraneous solutions can appear, so it is important to always check solutions. [SEC TION 11.5] SU M M A RY 11.5 Systems of Nonlinear Equations 1047 In Exercises 49 and 50, explain the mistake that is made. • C A T C H T H E M I S T A K E 49. Solve the system of equations: x2 1 y2 5 4 x 1 y 5 2 Solution: Multiply the second equation by 1212 and add to the first equation. x2 2 x 5 2 Subtract 2. x2 2 x 2 2 5 0 Factor. 1x 1 121x 2 22 5 0 Solve for x. x 5 21 and x 5 2 Substitute x 5 21 and x 5 2 into x 1 y 5 2. 21 1 y 5 2 and 2 1 y 5 2 Solve for y. y 5 3 and y 5 0 The answer is 121, 32 and 12, 02. This is incorrect. What mistake was made? 50. Solve the system of equations: x2 1 y2 5 5 2x 2 y 5 0 Solution: Solve the second equation for y. y 5 2x Substitute y 5 2x into the first equation. x2 1 12x22 5 5 Eliminate the parentheses. x2 1 4x2 5 5 Gather like terms. 5x2 5 5 Solve for x. x 5 21 and x 5 1 Substitute x 5 21 into the first equation. 12122 1 y2 5 5 Solve for y. y 5 22 and y 5 2 Substitute x 5 1 into the first equation. 1122 1 y2 5 5 Solve for y. y 5 22 and y 5 2 The answers are 121, 222, 121, 22, 11, 222, and 11, 22. This is incorrect. What mistake was made? • A P P L I C A T I O N S 39. Numbers. The sum of two numbers is 10, and the difference of their squares is 40. Find the numbers. 40. Numbers. The difference of two numbers is 3, and the difference of their squares is 51. Find the numbers. 41. Numbers. The product of two numbers is equal to the reciprocal of the difference of their reciprocals. The product of the two numbers is 72. Find the numbers. 42. Numbers. The ratio of the sum of two numbers to the difference of the two numbers is 9. The product of the two numbers is 80. Find the numbers. 43. Geometry. A rectangle has a perimeter of 36 centimeters and an area of 80 square centimeters. Find the dimensions of the rectangle. 44. Geometry. Two concentric circles have perimeters that add up to 16p and areas that add up to 34p. Find the radii of the two circles. 45. Horse Paddock. An equestrian buys a 5-acre rectangular parcel (approximately 200,000 square feet) and is going to fence in the entire property and then divide the parcel into two halves with a fence. If 2200 linear feet of fencing is required, what are the dimensions of the parcel? 46. Dog Run. A family moves into a new home and decides to fence in the yard to give its dog room to roam. If the area that will be fenced in is rectangular and has an area of 11,250 square feet, and the length is twice as much as the width, how many linear feet of fence should they buy? 47. Footrace. Your college algebra professor and Kirani James (2012 Olympic Gold Medalist in the men’s 400 meter) decided to race. The race was 400 meters. Kirani gave your professor a 1-minute head start and still crossed the finish line 1 minute 40 seconds before your professor. If Kirani ran five times faster than your professor, what was each person’s average speed? 48. Footrace. You decided to race Kirani James for 800 meters. At that distance, Kirani runs approximately twice as fast as you. He gave you a 1-minute head start and crossed the finish line 20 seconds before you. What were each of your average speeds? 1048 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities 55. A circle and a line have at most two points of intersection. A circle and a parabola have at most four points of intersection. What is the greatest number of points of intersection that a circle and an nth-degree polynomial can have? 56. A line and a parabola have at most two points of intersection. A line and a cubic function have at most three points of intersection. What is the greatest number of points of intersection that a line and an nth-degree polynomial can have? 57. Find a system of equations representing a line and a parabola that has only one real solution. 58. Find a system of equations representing a circle and a parabola that has only one real solution. • C H A L L E N G E Use a graphing utility to solve the following systems of equations. 59. y 5 ex 60. y 5 10x 61. 2x3 1 4y2 5 3 62. 3x4 2 2xy 1 5y2 5 19 y 5 ln x y 5 log x xy3 5 7 x4y 5 5 63. 5x3 1 2y2 5 40 64. 4x4 1 2xy 1 3y2 5 60 x3y 5 5 x4y 5 8 2 3x4 • T E C H N O L O G Y 53. The elimination method can always be used to solve systems of two nonlinear equations. 54. The substitution method always works for solving systems of nonlinear equations. In Exercises 51–54, determine whether each statement is either true or false. • C O N C E P T U A L 51. A system of equations representing a line and a parabola can intersect in at most three points. 52. A system of equations representing a line and a cubic function can intersect in at most three places. 11.6 Systems of Nonlinear Inequalities 1049 11.6.1 Nonlinear Inequalities in Two Variables Linear inequalities are expressed in the form Ax 1 By # C. Examples of nonlinear inequalities in two variables are 9x2 1 16y2 $ 1 x2 1 y2 . 1 y # 2x2 1 3 and x2 20 2 y2 81 , 1 We follow the same procedure as we did with linear inequalities. We change the inequality to an equal sign, graph the resulting nonlinear equation, test points from the two regions, and shade the region that makes the inequality true. For strict inequalities, , or ., dashed curves are used; and for nonstrict inequalities, # or $, solid curves are used. S K I L L S O B J E C T I V E S ■ ■Graph a nonlinear inequality in two variables. ■ ■Graph a system of nonlinear inequalities in two variables. C O N C E P T U A L O B J E C T I V ES ■ ■Understand that a nonlinear inequality in two variables may be represented by either a bounded or an unbounded region. ■ ■Interpret an overlapping shaded region as a solution. 11.6 SYSTEMS OF NONLINEAR INEQUALITIES 11.6.1 S KI L L Graph a nonlinear inequality in two variables. 11.6.1 C O N C E P T U A L Understand that a nonlinear inequality in two variables may be represented by either a bounded or an unbounded region. EXAMPLE 1 Graphing a Strict Nonlinear Inequality in Two Variables Graph the inequality x2 1 y2 . 1. Solution: STEP 1 Change the inequality sign to an equal sign. x2 1 y2 5 1 STEP 2 Draw the graph of the circle. The center is 10, 02 and the radius is 1. Since the inequality . is a strict inequality, draw the circle as a dashed curve. STEP 3 Test points in each region (outside the circle and inside the circle). Substitute 12, 02 into x2 1 y2 . 1. 4 . 1 The point 12, 02 satisfies the inequality. Substitute 10, 02 into x2 1 y2 . 1. 0 . 1 The point 10, 02 does not satisfy the inequality. STEP 4 Shade the region containing the point 12, 02. x y x2 + y2 = 1 2 2 –2 –2 x y x2 + y2 > 1 (2, 0) EXAMPLE 2 Graphing a Nonstrict Nonlinear Inequality in Two Variables Graph the inequality y # 2x2 1 3. Solution: STEP 1 Change the inequality sign to an equal sign. y 5 2x2 1 3 The equation is that of a parabola. STEP 2 Graph the parabola. Reflect the base function ƒ1x2 5 x2 about the x-axis and shift up three units. Since the inequality # is a nonstrict inequality, draw the parabola as a solid curve. x y y = –x2 + 3 [CONCEPT CHECK] TRUE OR FALSE A nonlinear inequality with an ellipse as a boundary can have a bounded or an unbounded region for a solution, depending on the inequality. ANSWER True ▼ 1050 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities 11.6.2 Systems of Nonlinear Inequalities To solve a system of inequalities, first graph the inequalities and shade the region containing the points that satisfy each inequality. The overlap of all the shaded regions is the solution. STEP 3 Test points in each region (inside the parabola and outside the parabola). Substitute 13, 02 into y # 2x2 1 3. 0 # 26 The point 13, 02 does not satisfy the inequality. Substitute 10, 02 into y # 2x2 1 3. 0 # 3 The point 10, 02 does satisfy the inequality. STEP 4 Shade the region containing the point 10, 02. YOUR T UR N Graph the following inequalities. a. x2 1 y2 # 9 b. y . 2x2 1 2 ▼ A N S W E R a. b. x y y > –x2 + 2 11.6.2 S KI L L Graph a system of nonlinear inequalities in two variables. 11.6.2 C ON C E P T U A L Interpret an overlapping shaded region as a solution. EXAMPLE 3 Graphing a System of Inequalities Graph the solution to the system of inequalities: y $ x2 2 1 y , x 1 1 Solution: STEP 1 Change the inequality signs to equal signs. y 5 x2 2 1 y 5 x 1 1 STEP 2 The resulting equations represent a parabola (to be drawn solid) and a line (to be drawn dashed). Graph the two equations. To determine the points of intersection, set the y values equal. x2 2 1 5 x 1 1 x2 2 x 2 2 5 0 Factor. 1x 2 221x 1 12 5 0 Solve for x. x 5 2 or x 5 21 Substitute x 5 2 into y 5 x 1 1. 12, 32 Substitute x 5 21 into y 5 x 1 1. 121, 02 x y y = x2 – 1 y = x + 1 (2, 3) (–1, 0) 11.6 Systems of Nonlinear Inequalities 1051 ▼ STEP 3 Test points and shade regions. STEP 4 Shade the common region. Y OUR TU R N Graph the solution to the system of inequalities: x2 1 y2 , 9 y . 0 ▼ ▼ A N S W E R x y x2 + y2 < 9 y > 0 EXAMPLE 4 Solving a System of Inequalities Solve the system of inequalities: x2 1 y2 , 2 y $ x2 Solution: STEP 1 Change the inequality signs to equal signs. x2 1 y2 5 2 y 5 x2 STEP 2 The resulting equations correspond to a circle (to be drawn dashed) and a parabola (to be drawn solid). To determine the points of intersection, solve the system of equations by substitution. x2 1 1x222 5 2 x4 1 x2 2 2 5 0 Factor. 1x2 1 221x2 2 12 5 0 Solve for x. x2 5 22 or x2 5 1 The points of intersection are 121, 12 and 11, 12. { y no solution x 5 61 c b x y y = x2 x2 + y2 = 2 1052 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities STEP 3 Test points and shade regions. STEP 4 Identify the common region as the solution. Y OUR T UR N Solve the system of inequalities: x2 1 y2 , 2 y , x2 x y ▼ ▼ A N S W E R x y y < x2 x2 + y2 < 2 It is important to note that any inequality based on an equation whose graph is not a line is considered a nonlinear inequality. EXAMPLE 5 Solving a System of Nonlinear Inequalities Solve the system of inequalities: 1x 2 122 1 y2 4 , 1 y $ !x Solution: STEP 1 Change the inequality signs to equal signs. 1x 2 122 1 y2 4 5 1 y 5 "x STEP 2 The resulting equations correspond to an ellipse (to be drawn dashed) and the square-root function (to be drawn solid). Graph the two inequalities. To determine the points of intersection, solve the system of equations by substitution. 1x 2 122 1 A !xB2 4 5 1 Multiply by 4. 41x 2 122 1 x 5 4 Expand the binomial squared. 41x2 2 2x 1 12 1 x 5 4 Distribute. 4x2 2 8x 1 4 1 x 5 4 Combine like terms and gather terms to one side. 4x2 2 7x 5 0 Factor. x14x 2 72 5 0 Solve for x. x 5 0 and x 5 7 4 } y [CONCEPT CHECK] A system of nonlinear inequalities corresponding to y . x 2 1 A and y # x 2 2 A where A . 0 has no solution. ANSWER True ▼ 11.6 Systems of Nonlinear Inequalities 1053 [SEC TION 11.6] E X E R C I SE S • S K I L L S In Exercises 1–12, match the nonlinear inequality with the correct graph. 1. x2 1 y2 , 25 2. x2 1 y2 # 9 3. x2 9 1 y2 16 $ 1 4. x2 4 1 y2 9 . 1 5. y $ x2 2 3 6. x2 $ 16y 7. x $ y2 2 4 8. x2 9 1 y2 25 $ 1 9. 9x2 1 9y2 , 36 10. 1x 2 222 1 1y 1 322 # 9 11. x2 4 2 y2 9 $ 1 12. y2 16 2 x2 9 , 1 a. b. c. d. x y 5 –5 –5 5 x y 10 –10 –10 10 x y 5 –5 –5 5 x y 5 –5 –5 5 1e.g., x2 1 y2 . 12. When solving systems of inequalities, we first graph each of the inequalities separately and then look for the intersection (overlap) of all shaded regions. In this section, we discussed nonlinear inequalities in two variables. Sometimes these result in bounded regions 1e.g., x2 1 y2 # 12, and sometimes these result in unbounded regions [SEC TION 11.6] S U MM A RY The points of intersection are 10, 02 and a7 4, !7 2 b. STEP 3 Shade the solution. √7 1054 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities e. f. g. h. x y 5 –5 –5 5 x 10 –10 –10 10 x y 5 –5 –5 5 x y 5 –5 –5 5 i. j. k. l. x y 5 –5 –5 5 x y 5 –5 –5 5 x y 5 –5 –5 5 x y 5 –5 –5 5 In Exercises 13–30, graph the nonlinear inequality. 13. y # x2 2 2 14. y $ 2x2 1 3 15. x2 1 y2 . 4 16. x2 1 y2 , 16 17. x2 1 y2 2 2x 1 4y 1 4 $ 0 18. x2 1 y2 1 2x 2 2y 2 2 # 0 19. 3x2 1 4y2 # 12 20. 1x 2 222 9 1 1y 1 122 25 . 1 21. 9x2 1 16y2 2 18x 1 96y 1 9 . 0 22. 1x 2 222 4 2 1y 1 322 1 $ 1 23. 9x2 2 4y2 $ 36 24. 1y 1 122 9 2 1x 1 222 16 , 1 25. 36x2 2 9y2 $ 324 26. 25x2 2 36y2 1 200x 1 144y 2 644 $ 0 27. y $ ex 28. y # ln x 29. y , 2x3 30. y . 2x4 In Exercises 31–50, graph each system of inequalities or indicate that the system has no solution. 31. y , x 1 1 y # x2 32. y , x2 1 4x y # 3 2 x 33. y $ 2 1 x y # 4 2 x2 34. y $ 1x 2 222 y # 4 2 x 35. y # 21x 1 222 y . 2 5 1 x 36. y $ 1x 2 122 1 2 y # 10 2 x 37. 2x2 1 y . 21 x2 1 y , 1 38. x , 2y2 1 1 x . y2 2 1 39. y $ x2 x $ y2 40. y , x2 x . y2 41. x2 1 y2 , 36 2x 1 y . 3 42. x2 1 y2 , 36 y . 6 43. x2 1 y2 , 25 y $ 6 1 x 44. 1x 2 122 1 1y 1 222 # 36 y $ x 2 3 45. x2 1 y2 # 9 y $ 1 1 x2 46. x2 1 y2 $ 16 x2 1 1y 2 322 # 9 47. x2 2 y2 , 4 y . 1 2 x2 48. x2 4 2 y2 9 # 1 y $ x 2 5 49. y , e x y . ln x x . 0 50. y , 10 x y . log x x . 0 • A P P L I C A T I O N S 51. Find the area enclosed by the system x2 1 y2 , 9 of inequalities: x . 0 52. Find the area enclosed by the system x2 1 y2 # 5 of inequalities: x # 0 y $ 0 11.6 Systems of Nonlinear Inequalities 1055 • C A T C H T H E M I S T A K E 53. Graph the system of inequalities: x2 1 y2 , 1 x2 1 y2 . 4 Solution: Draw the circles x2 1 y2 5 1 and x2 1 y2 5 4 Shade outside x2 1 y2 5 1 and inside x2 1 y2 5 4 This is incorrect. What mistake was made? 54. Graph the system of inequalities: x . 2y2 1 1 x , y2 2 1 Solution: Draw the parabolas x 5 2y2 1 1 and x 5 y2 2 1 and shade the region between the curves. x y 5 –5 2.5 –2.5 This is incorrect. What mistake was made? x y x2 + y2 = 1 x2 + y2 = 4 x y x2 + y2 < 1 x2 + y2 > 4 In Exercises 53 and 54, explain the mistake that is made. In Exercises 55 and 56, determine whether each statement is true or false. • C O N C E P T U A L 55. A nonlinear inequality always represents a bounded region. 56. A system of inequalities always has a solution. • C H A L L E N G E 57. For the system of nonlinear inequalities x2 1 y2 $ a2 x2 1 y2 # b2, what restriction must be placed on the values of a and b for this system to have a solution? Assume a and b are real numbers. 58. Can x2 1 y2 , 21 ever have a real solution? What types of numbers would x and/or y have to be to satisfy this inequality? • T E C H N O L O G Y Use a graphing utility to graph the following inequalities. 59. x2 1 y2 2 2x 1 4y 1 4 $ 0 60. x2 1 y2 1 2x 2 2y 2 2 # 0 61. y $ ex 62. y # ln x 63. y , ex y . ln x x . 0 64. y , 10x y . log x x . 0 65. x2 2 4y2 1 5x 2 6y 1 18 $ 0 66. x2 2 2xy 1 4y2 1 10x 2 25 # 0 1056 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities 11.7 Rotation of Axes 1057 11.7.1 Rotation of Axes Formulas In Sections 11.1 through 11.4 we learned the general equations of parabolas, ellipses, and hyperbolas that were centered at any point in the Cartesian plane and whose vertices and foci were aligned either along or parallel to either the x-axis or the y-axis. We learned, for example, that the general equation of an ellipse centered at the origin is given by x2 a2 1 y2 b2 5 1 where the major and minor axes are, respectively, either the x- or the y-axis, depending on whether a is greater than or less than b. Now let us look at an equation of a conic section whose graph is not aligned with the x- or y-axis: the equation 5x2 2 8xy 1 5y2 2 9 5 0. –5 –3 2 3 4 5 y Y –5 –3 –2 –4 2 3 4 5 x X 45º This graph can be thought of as an ellipse that started with the major axis along the x-axis and the minor axis along the y-axis and then was rotated counterclockwise 45°. A new XY-coordinate system can be introduced such that this system has the same origin, but the XY-coordinate system is rotated by a certain amount from the standard xy-coordinate system. In this example, the major axis of the ellipse lies along the new X-axis, and the minor axis lies along the new Y-axis. We will see that we can write the equation of this ellipse as X2 9 2 Y2 1 5 1 We will now develop the rotation of axes formulas, which allow us to transform the generalized second-degree equation in xy, that is, Ax2 1 Bxy 1 Cy2 1 Dx 1 Ey 1 F 5 0, into an equation in XY of a conic that is familiar to us. S K I L L S O B J E C T I V E S ■ ■Transform general second-degree equations into recognizable equations of conics by analyzing rotations of axes. ■ ■Determine the angle of rotation that will transform a general second-degree equation into a familiar equation of a conic section. C O N C E P T U A L O B J E C T I V ES ■ ■Understand how the equation of a conic section is altered by rotation of axes. ■ ■Understand how the angle of rotation formula is derived. 11.7 ROTATION OF AXES 11.7.1 S KI L L Transform general second-degree equations into recognizable equations of conics by analyzing rotations of axes. 11.7.1 C ON C E P T U A L Understand how the equation of a conic section is altered by rotation of axes. 1058 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities Let the new XY-coordinate system be displaced from the xy-coordinate system by rotation through an angle u. Let P represent some point a distance r from the origin. We can represent the point P as the point 1x, y2 or the point 1X, Y 2. We define the angle a as the angle r makes with the X-axis and a 1 u as the angle r makes with the x-axis. We can represent the point P in polar coordinates using the following relationships: x 5 r cos1a 1 u2 y 5 r sin1a 1 u2 X 5 r cos a Y 5 r sin a WORDS MATH Start with the x-term and write the cosine x 5 rcos1a 1 u2 identity for a sum. 5 r 1cos a cos u 2 sin a sin u2 Eliminate the parentheses and group r with the a-terms. x 5 1rcos a2 cos u 2 1rsin a2sin u Substitute according to the relationships X 5 r cos a and Y 5 r sin a. x 5 X cos u 2 Ysin u Start with the y-term and write the sine y 5 rsin1a 1 u2 identity for a sum. 5 r1sin a cos u 1 cos a sin u2 Eliminate the parentheses and group r with the a-terms. y 5 1rsin a2 cos u 1 1rcos a2sin u Substitute according to the relationships X 5 r cos a and Y 5 r sin a. y 5 Ycos u 1 Xsin u y r P Y x X θ α y r y x P Y Y x X X θ α By treating the highlighted equations for x and y as a system of linear equations in X and Y, we can then solve for X and Y in terms of x and y. The results are summarized in the following box: 11.7 Rotation of Axes 1059 ROTATION OF AXES FORMULAS Suppose that the x- and y-axes in the rectangular coordinate plane are rotated through an acute angle u to produce the X- and Y-axes. Then, the coordinates (x, y) and (X, Y) are related according to the following equations: x 5 X cos u 2 Y sin u X 5 x cos u 1 y sin u or y 5 X sin u 1 Y cos u Y 5 2x sin u 1 y cos u EXAMPLE 1 Rotating the Axes If the xy-coordinate axes are rotated 60°, find the XY-coordinates of the point 1x, y2 5 123, 42. Solution: Start with the rotation formulas. X 5 x cos u 1 y sinu Y 5 2x sinu 1 y cosu Let x 5 23, y 5 4, and u 5 60°. X 5 23 cos160°2 1 4 sin160°2 Y 5 21232 sin160°2 1 4 cos160°2 Simplify. X 5 23 cos160°2 1 4 sin160°2 1 2 !3 2 Y 5 3 sin160°2 1 4 cos160°2 !3 2 1 2 X 5 23 2 1 2!3 Y 5 3!3 2 1 2 The XY-coordinates are a23 2 1 2!3, 3!3 2 1 2b . Y OUR T UR N If the xy-coordinate axes are rotated 30°, find the XY-coordinates of the point 1x, y2 5 13, 242. • • • • ▼ EXAMPLE 2 Rotating an Ellipse Show that the graph of the equation 5x2 2 8xy 1 5y2 2 9 5 0 is an ellipse aligning with coordinate axes that are rotated by 45°. Solution: Start with the rotation formulas. x 5 X cos u 2 Y sinu y 5 X sinu 1 Y cosu Let u 5 45°. x 5 X cos145°2 2 Y sin145°2 !2 2 !2 2 y 5 X sin145°2 1 Y cos145°2 !2 2 !2 2 • • • • ▼ A N S W E R a3!3 2 2 2, 23 2 2 2!3b 1060 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities 11.7.2 The Angle of Rotation Necessary to Transform a General Second-Degree Equation into an Equation of a Conic In Section 11.1 we stated that the general second-degree equation Ax2 1 Bxy 1 Cy2 1 Dx 1 Ey 1 F 5 0 corresponds to a graph of a conic. Which type of conic it is depends on the value of the discriminant, B2 2 4AC. In Sections 11.2–11.4 we discussed graphs of parabolas, ellipses, and hyperbolas with vertices along either the axes or lines parallel (or perpendicular) to the axes. In all cases the value of B was taken to be zero. When the value of B is nonzero, the result is a conic with vertices along the new XY-axes (or, respectively, parallel and perpendicular to them), which are the original xy-axes rotated through an angle u. If given u, we can determine the rotation equations as illustrated in Example 2, but how do we find the angle u that represents the angle of rotation? To find the angle of rotation, let us start with a general second-degree polynomial equation: Ax2 1 Bxy 1 Cy2 1 Dx 1 Ey 1 F 5 0 Simplify. x 5 !2 2 1X 2 Y2 y 5 !2 2 1X 1 Y2 Substitute x 5 !2 2 1X 2 Y2 and y 5 !2 2 1X 1 Y2 into 5x2 2 8xy 1 5y2 2 9 5 0. 5c !2 2 1X 2 Y2 d 2 2 8c !2 2 1X 2 Y2 d c !2 2 1X 1 Y2 d 1 5c !2 2 1X 1 Y2 d 2 2 9 5 0 Simplify. 5 2 1X 2 2 2XY 1 Y 22 2 41X 2 2 Y 22 1 5 2 1X 2 1 2XY 1 Y 22 2 9 5 0 5 2 X 2 2 5XY 1 5 2 Y 2 2 4X 2 1 4Y 2 1 5 2 X 2 1 5XY 1 5 2 Y 2 5 9 Combine like terms. X 2 1 9Y 2 5 9 Divide by 9. X 2 9 1 Y 2 1 5 1 This (as discussed earlier) is an ellipse whose major axis is along the X-axis. The vertices are at the points 1X, Y2 5 1 6 3, 02. –5 –3 2 3 4 5 y Y –5 –3 –2 –4 2 3 4 5 x X 45º 11.7.2 SKI LL Determine the angle of rotation that will transform a general second-degree equation into a familiar equation of a conic section. 11.7.2 CO NC EPTUAL Understand how the angle of rotation formula is derived. 11.7 Rotation of Axes 1061 We want to transform this equation into an equation in X and Y that does not contain an XY-term. Suppose we rotate our coordinates by an angle u and then use the rotation equations x 5 X cos u 2 Y sin u y 5 X sin u 1 Y cos u in the general second-degree polynomial equation; then the result is A1X cos u 2 Y sin u22 1 B1X cos u 2 Y sin u21X sin u 1 Y cos u2 1 C1X cos u 1 Y sin u22 1 D1X cos u 2 Y sin u2 1 E1X sin u 1 Y cos u2 1 F 5 0 If we expand these expressions and collect like terms, the result is an equation of the form aX2 1 bXY 1 cY2 1 dX 1 eY 1 f 5 0 where a 5 A cos2 u 1 B sin u cos u 1 C sin2 u b 5 B1cos2 u 2 sin2 u2 1 21C 2 A2 sin u cos u c 5 A sin2 u 2 B sin u cos u 1 C cos2 u d 5 D cos u 1 E sin u e 5 2D sin u 1 E cos u ƒ 5 F WORDS MATH We do not want this new equation to have an XY-term, so we set b 5 0. B1cos2 u 2 sin2 u2 1 21C 2 A2 sin u cos u 5 0 We can use the double-angle B1cos2 u 2 sin2 u2 1 1C 2 A2 2 sin u cos u 5 0 formulas to simplify. Subtract the sine term. B cos 2u 5 1A 2 C2 sin 2u Divide by B sin 2u. B cos 2u B sin 2u 5 1A 2 C2 sin 2u B sin 2u Simplify. cot 2u 5 A 2 C B µ f cos 2u sin 2u ANGLE OF ROTATION FORMULA To transform the equation of a conic Ax2 1 Bxy 1 Cy2 1 Dx 1 Ey 1 F 5 0 into an equation in X and Y without an XY-term, rotate the xy-axes by an acute angle u that satisfies the equation cot 2u 5 A 2 C B 1062 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities Notice that the trigonometric equation cot 2u 5 A 2 C B can be solved exactly for some values of u (Example 3) and will have to be approximated with a calculator for other values of u (Example 4). EXAMPLE 3 Determining the Angle of Rotation I: The Value of the Cotangent Function Is That of a Known (Special) Angle Determine the angle of rotation necessary to transform the following equation into an equation in X and Y with no XY-term. 3x2 1 2!3xy 1 y2 1 2x 2 2!3y 5 0 Solution: Identify the A, B, and C 3x2 1 2!3 xy 1 1y2 1 2x 2 2!3y 5 0 parameters in the equation. A B C Write the rotation formula. cot 2u 5 A 2 C B Let A 5 3, B 5 2!3, and C 5 1. cot 2u 5 3 2 1 2!3 Simplify. cot 2u 5 1 !3 Apply the reciprocal identity. tan 2u 5 !3 From our knowledge of trigonometric exact values, we know that 2u 5 60° or u 5 30° . { • e EXAMPLE 4 Determining the Angle of Rotation II: The Argument of the Cotangent Function Needs to Be Approximated with a Calculator Determine the angle of rotation necessary to transform the following equation into an equation in X and Y with no XY-term. Round to the nearest tenth of a degree. 4x2 1 2xy 2 6y2 2 5x 1 y 2 2 5 0 Solution: Identify the A, B, and C 4x2 1 2 xy 2 6y2 2 5x 1 y 2 2 5 0 parameters in the equation. A B C Write the rotation formula. cot 2u 5 A 2 C B Let A 5 4, B 5 2, and C 5 26. cot 2u 5 4 2 1262 2 Simplify. cot 2u 5 5 Apply the reciprocal identity. tan 2u 5 1 5 5 0.2 Write the result as an inverse tangent function. 2u 5 tan 21 0.2 With a calculator evaluate the right side of the equation. 2u < 11.31° Solve for u and round to the nearest tenth of a degree. u 5 5.7° { { e 11.7 Rotation of Axes 1063 Special attention must be given when evaluating the inverse tangent function on a calculator, as the result is always in quadrant I or IV. If 2u turns out to be negative, then 180° must be added so that 2u is in quadrant II (as opposed to quadrant IV). Then u will be an acute angle lying in quadrant I. Recall that we stated (without proof) in Section 11.1 that we can identify a general equation of the form Ax2 1 Bxy 1 Cy2 1 Dx 1 Ey 1 F 5 0 as that of a particular conic depending on the discriminant. Parabola B2 2 4AC 5 0 Ellipse B2 2 4AC , 0 Hyperbola B2 2 4AC . 0 EXAMPLE 5 Graphing a Rotated Conic For the equation x2 1 2xy 1 y2 2 !2x 2 3!2y 1 6 5 0: a. Determine which conic the equation represents. b. Find the rotation angle required to eliminate the XY-term in the new coordinate system. c. Transform the equation in x and y into an equation in X and Y. d. Graph the resulting conic. Solution (a): Identify A, B, and C. 1x2 1 2 xy 1 1y2 2 !2 x 2 3!2 y 1 6 5 0 A B C A 5 1, B 5 2, C 5 1 Compute the discriminant. B2 2 4AC 5 22 2 4112112 5 0 Since the discriminant equals zero, the equation represents a parabola. Solution (b): Write the rotation formula. cot 2u 5 A 2 C B Let A 5 1, B 5 2, and C 5 1. cot 2u 5 1 2 1 2 Simplify. cot 2u 5 0 Write the cotangent function in terms of cos 2u sin 2u 5 0 the sine and cosine functions. The numerator must equal zero. cos 2u 5 0 From our knowledge of trigonometric exact values, we know that 2u 5 90° or u 5 45° . { { { 1064 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities Solution (c): Start with the equation x 5 X cos145°2 2 Y sin145°2 5 !2 2 1X 2 Y2 x2 1 2xy 1 y2 2 !2x 2 3!2y 1 6 5 0, and use the rotation formulas with u 5 45°. y 5 X sin145°2 1 Y cos145°2 5 !2 2 1X 1 Y2 Find x2, xy, and y2. x2 5 c !2 2 1X 2 Y 2 d 2 5 1 2 1X2 2 2XY 1 Y22 xy 5 c !2 2 1X 2 Y 2 d c !2 2 1X 1 Y 2 d 5 1 2 1X2 2 Y22 y2 5 c !2 2 1X 1 Y 2 d 2 5 1 2 1X2 1 2XY 1 Y22 Substitute the values x2 1 2xy 1 y2 2 !2x 2 3!2y 1 6 5 0 for x, y, x2, xy, and y2 into the 1 2 1X2 2 2XY 1 Y22 1 21 2 1X2 2 Y22 original equation. 1 1 2 1X2 1 2XY 1 Y22 2 !2 c !2 2 1X 2 Y 2 d 2 3!2c !2 2 1X 1 Y 2 d 1 6 5 0 Eliminate the parentheses and combine like terms. 2X2 2 4X 2 2Y 1 6 5 0 Divide by 2. X2 2 2X 2 Y 1 3 5 0 Add Y. Y 5 AX2 2 2XB 1 3 Complete the square on X. Y 5 1X 2 122 1 2 Solution (d): This is a parabola opening upward in the XY-coordinate system shifted to the right one unit and up two units. –5 –3 2 3 4 5 y Y –5 –3 –2 –4 2 4 5 x X 45º [SEC TION 11.7] S U M M A RY In this section, we found that the graph of the general second- degree equation Ax2 1 Bxy 1 Cy2 1 Dx 1 Ey 1 F 5 0 can represent conics in a system of rotated axes. The following are the rotation formulas relating the xy-coordinate system to a rotated coordinate system with axes X and Y x 5 X cosu 2 Y sinu y 5 X sinu 1 Y cosu where the rotation angle u is found from the equation cot 2u 5 A 2 C B 11.7 Rotation of Axes 1065 In Exercises 1–8, the coordinates of a point in the xy-coordinate system are given. Assuming that the XY-axes are found by rotating the xy-axes by an angle u, find the corresponding coordinates for the point in the XY-system. [SEC TION 11.7] E X E R C I S E S • S K I L L S 1. 12, 42, u 5 45° 2. 15, 12, u 5 60° 3. 123, 22, u 5 30° 4. 124, 62, u 5 45° 5. 121, 232, u 5 60° 6. 14, 242, u 5 45° 7. 10, 32, u 5 60° 8. 122, 02, u 5 30° In Exercises 9–24, (a) identify the type of conic from the discriminant, (b) transform the equation in x and y into an equation in X and Y (without an XY-term) by rotating the x- and y-axes by an angle u to arrive at the new X- and Y-axes, and (c) graph the resulting equation (showing both sets of axes). 9. xy 2 1 5 0, u 5 45° 10. xy 2 4 5 0, u 5 45° 11. x2 1 2xy 1 y2 1 !2x 2 !2y 2 1 5 0, u 5 45° 12. 2x2 2 4xy 1 2y2 2 !2x 1 1 5 0, u 5 45° 13. y2 2 !3xy 1 3 5 0, u 5 30° 14. x2 2 !3xy 2 3 5 0, u 5 60° 15. 7x2 2 2!3xy 1 5y2 2 8 5 0, u 5 60° 16. 4x2 1 !3xy 1 3y2 2 45 5 0, u 5 30° 17. 3x2 1 2!3xy 1 y2 1 2x 2 2!3y 2 2 5 0, u 5 30° 18. x2 1 2!3xy 1 3y2 2 2!3x 1 2y 2 4 5 0, u 5 60° 19. 7x2 1 4!3xy 1 3y2 2 9 5 0, u 5 p 6 20. 37x2 1 42!3xy 1 79y2 2 400 5 0, u 5 p 3 21. 7x2 2 10!3xy 2 3y2 1 24 5 0, u 5 p 3 22. 9x2 1 14!3xy 2 5y2 1 48 5 0, u 5 p 6 23. x2 2 2xy 1 y2 2 !2x 2 !2y 2 8 5 0, u 5 p 4 24. x2 1 2xy 1 y2 1 3!2x 1 !2y 5 0, u 5 p 4 In Exercises 25–38, determine the angle of rotation necessary to transform the equation in x and y into an equation in X and Y with no XY-term. 25. x2 1 4xy 1 y2 2 4 5 0 26. 3x2 1 5xy 1 3y2 2 2 5 0 27. 2x2 1 !3xy 1 3y2 2 1 5 0 28. 4x2 1 !3xy 1 3y2 2 1 5 0 29. 2x2 1 !3xy 1 y2 2 5 5 0 30. 2!3x2 1 xy 1 3!3y2 1 1 5 0 31. !2x2 1 xy 1 !2y2 2 1 5 0 32. x2 1 10xy 1 y2 1 2 5 0 33. 12!3x2 1 4xy 1 8!3y2 2 1 5 0 34. 4x2 1 2xy 1 2y2 2 7 5 0 35. 5x2 1 6xy 1 4y2 2 1 5 0 36. x2 1 2xy 1 12y2 1 3 5 0 37. 3x2 1 10xy 1 5y2 2 1 5 0 38. 10x2 1 3xy 1 2y2 1 3 5 0 In Exercises 39–48, graph the second-degree equation. (Hint: Transform the equation into an equation that contains no xy-term.) 39. 21x2 1 10!3xy 1 31y2 2 144 5 0 40. 5x2 1 6xy 1 5y2 2 8 5 0 41. 8x2 2 20xy 1 8y2 1 18 5 0 42. 3y2 2 26!3xy 2 23x2 2 144 5 0 43. 3x2 1 2!3xy 1 y2 1 2x 2 2!3y 2 12 5 0 44. 3x2 2 2!3xy 1 y2 2 2x 2 2!3y 2 4 5 0 45. 37x2 2 42!3xy 1 79y2 2 400 5 0 46. 71x2 2 58!3xy 1 13y2 1 400 5 0 47. x2 1 2xy 1 y2 1 5!2x 1 3!2y 5 0 48. 7x2 2 4!3xy 1 3y2 2 9 5 0 In Exercises 49–52, determine whether each statement is true or false. 49. The graph of the equation x2 1 kxy 1 9y2 5 5, where k is any positive constant less than 6, is an ellipse. 50. The graph of the equation x2 1 kxy 1 9y2 5 5, where k is any constant greater than 6, is a parabola. 51. The reciprocal function is a rotated hyperbola. 52. The equation !x 1 !y 5 3 can be transformed into the equation X2 1 Y2 5 9. • C O N C E P T U A L 1066 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities 53. Determine the equation in X and Y that corresponds to x2 a2 1 y2 b2 5 1 when the axes are rotated through a. 90° b. 180° 54. Determine the equation in X and Y that corresponds to x2 a2 2 y2 b2 5 1 when the axes are rotated through a. 90° b. 180° 55. Identify the conic section with equation y2 1 ax2 5 x for various values of a. 56. Identify the conic section with equation x2 2 ay2 5 y for various values of a. • C H A L L E N G E For Exercises 57–62, refer to the following: To use a TI-83 or TI-83 Plus (function-driven software or graphing utility) to graph a general second-degree equation, you need to solve for y. Let us consider a general second-degree equation Ax2 1 Bxy 1 Cy2 1 Dx 1 Ey 1 F 5 0. Group y2 terms together, y terms together, and the remaining terms together. Ax2 1 Bxy 1 Cy2 1 Dx 1 Ey 1 F 5 0 Cy2 1 1Bxy 1 Ey2 1 1Ax2 1 Dx 1 F2 5 0 Factor out the common y in the first set of parentheses. Cy2 1 y 1Bx 1 E2 1 1Ax2 1 Dx 1 F2 5 0 Now this is a quadratic equation in y: ay2 1 by 1 c 5 0. Use the quadratic formula to solve for y. Cy2 1 y 1Bx 1 E2 1 1Ax2 1 Dx 1 F2 5 0 a 5 C, b 5 Bx 1 E, c 5 Ax2 1 Dx 1 F y 5 2b 6 "b2 2 4ac 2a y 5 21Bx 1 E2 6 "1Bx 1 E22 2 41C21Ax2 1 Dx 1 F2 21C2 y 5 21Bx 1 E2 6 "B2x2 1 2BEx 1 E2 2 4ACx2 2 4CDx 2 4CF 2C y 5 21Bx 1 E2 6 "1B2 2 4AC2x2 1 12BE 2 4CD2x 1 1E2 2 4CF2 2C Case I: B2 2 4AC 5 0 S The second-degree equation Ax2 1 Bxy 1 Cy2 1 Dx 1 Ey 1 F 5 0 is a parabola. y 5 21Bx 1 E2 6 "12BE 2 4CD2x 1 1E2 2 4CF2 2C Case II: B2 2 4AC , 0 S The second-degree equation Ax2 1 Bxy 1 Cy2 1 Dx 1 Ey 1 F 5 0 is an ellipse. y 5 21Bx 1 E2 6 "1B2 2 4AC2x2 1 12BE 2 4CD2x 1 1E2 2 4CF2 2C Case III: B2 2 4AC . 0 S The second-degree equation Ax2 1 Bxy 1 Cy2 1 Dx 1 Ey 1 F 5 0 is a hyperbola. y 5 21Bx 1 E2 6 "1B2 2 4AC2x2 1 12BE 2 4CD2x 1 1E2 2 4CF2 2C • T E C H N O L O G Y 57. Use a graphing utility to explore the second-degree equation 3x2 1 2!3xy 1 y2 1 Dx 1 Ey 1 F 5 0 for the following values of D, E, and F: a. D 5 1, E 5 3, F 5 2 b. D 5 21, E 5 23, F 5 2 Show the angle of rotation to the nearest degree. Explain the differences. 58. Use a graphing utility to explore the second-degree equation x2 1 3xy 1 3y2 1 Dx 1 Ey 1 F 5 0 for the following values of D, E, and F: a. D 5 2, E 5 6, F 5 21 b. D 5 6, E 5 2, F 5 21 Show the angle of rotation to the nearest degree. Explain the differences. 11.8 Polar Equations of Conics 1067 11.8.1 Equations of Conics in Polar Coordinates In Section 11.1 we discussed parabolas, ellipses, and hyperbolas in terms of geometric definitions. Then in Sections 11.2–11.4 we examined the rectangular equations of these conics. The equations for the conics are simpler when their centers are at the origin than when they are not (when conics are shifted). In Section 8.8, we discussed polar coordinates and graphing of polar equations. In this section, we develop a more unified definition of the three conics in terms of a single focus and a directrix. You will see in this section that if the focus is located at the origin, then equations of conics are simpler when written in polar coordinates. Alternative Definition of Conics Recall that when we work with rectangular coordinates we define a parabola (Sections 11.1 and 11.2) in terms of a fixed point (focus) and a line (directrix), whereas we define an ellipse and hyperbola (Sections 11.1, 11.3, and 11.4) in terms of two fixed points (the foci). However, it is possible to define all three conics in terms of a single focus and a directrix. The following alternative representation of conics depends on a parameter called eccentricity. 59. Use a graphing utility to explore the second-degree equation 2x2 1 3xy 1 y2 1 Dx 1 Ey 1 F 5 0 for the following values of D, E, and F: a. D 5 2, E 5 1, F 5 22 b. D 5 2, E 5 1, F 5 2 Show the angle of rotation to the nearest degree. Explain the differences. 60. Use a graphing utility to explore the second-degree equation 2!3x2 1 xy 1 !3y2 1 Dx 1 Ey 1 F 5 0 for the following values of D, E, and F: a. D 5 2, E 5 1, F 5 21 b. D 5 2, E 5 6, F 5 21 Show the angle of rotation to the nearest degree. Explain the differences. 61. Use a graphing utility to explore the second-degree equation Ax2 1 Bxy 1 Cy2 1 2x 1 y 2 1 5 0 for the following values of A, B, and C: a. A 5 4, B 5 24, C 5 1 b. A 5 4, B 5 4, C 5 21 c. A 5 1, B 5 24, C 5 4 Show the angle of rotation to the nearest degree. Explain the differences. 62. Use a graphing utility to explore the second-degree equation Ax2 1 Bxy 1 Cy2 1 3x 1 5y 2 2 5 0 for the following values of A, B, and C: a. A 5 1, B 5 24,C 5 4 b. A 5 1, B 5 4, D 5 24 Show the angle of rotation to the nearest degree. Explain the differences. S K I L L S O B J E C T I V E ■ ■Express equations of conics in polar form and graph. C O N C E P T U A L O B J E C T I V E ■ ■Define all conics in terms of a focus and a directrix. 11.8 POLAR EQUATIONS OF CONICS ALTERNATIVE DESCRIPTION OF CONICS Let D be a fixed line (the directrix), F be a fixed point (a focus) not on D, and e be a fixed positive number (eccentricity). The set of all points P such that the ratio of the distance from P to F to the distance from P to D equals the constant e defines a conic section. d1P, F2 d1P, D2 5 e ■ ■If e 5 1, the conic is a parabola. ■ ■If e , 1, the conic is an ellipse. ■ ■If e . 1, the conic is a hyperbola. 11.8.1 S K I L L Express equations of conics in polar form and graph. 11.8.1 C O N C E P T U A L Define all conics in terms of a focus and a directrix. 1068 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities When e 5 1, the result is a parabola, described by the same definition we used previously in Section 11.1. When e 2 1, the result is either an ellipse or a hyperbola. The major axis of an ellipse passes through the focus and is perpendicular to the directrix. The transverse axis of a hyperbola also passes through the focus and is perpendicular to the directrix. If we let c represent the distance from the focus to the center and a represent the distance from the vertex to the center, then eccentricity is given by e 5 c a In polar coordinates, if we locate the focus of a conic at the pole and the directrix is either perpendicular or parallel to the polar axis, then we have four possible scenarios: ■ ■The directrix is perpendicular to the polar axis and p units to the right of the pole. ■ ■The directrix is perpendicular to the polar axis and p units to the left of the pole. ■ ■The directrix is parallel to the polar axis and p units above the pole. ■ ■The directrix is parallel to the polar axis and p units below the pole. Let us take the case in which the directrix is perpendicular to the polar axis and p units to the right of the pole. x Polar axis y D F P In polar coordinates 1r, u2, we see that the distance from the focus to a point P is equal to r, that is, d1P, F2 5 r, and the distance from P to the closest point on the directrix is d1P, D2 5 p 2 r cos u. x Polar axis y D F P p d(P, D) r cos θ r θ x = p WORDS MATH Substitute d1P, F2 5 r and d1P, D2 5 p 2 r cos u into the formula for eccentricity, d1P, F2 d1P, D2 5 e. r p 2 r cos u 5 e Multiply the result by p 2 r cos u. r 5 e1 p 2 r cos u2 Eliminate the parentheses. r 5 ep 2 er cos u Add er cos u. r 1 er cos u 5 ep Factor out the common r. r11 1 e cos u2 5 ep Divide by 1 1 e cos u. r 5 ep 1 1 e cos u 11.8 Polar Equations of Conics 1069 We need not derive the other three cases here, but note that if the directrix is perpendicular to the polar axis and p units to the left of the pole, the resulting polar equation is r 5 ep 1 2 e cos u If the directrix is parallel to the polar axis, the directrix is either above 1y 5 p2 or below 1y 5 2p2 the polar axis and we get the sine function instead of the cosine function, as summarized in the following box: POLAR EQUATIONS OF CONICS The following polar equations represent conics with one focus at the origin and with eccentricity e. It is assumed that the positive x-axis represents the polar axis. EQUATION DESCRIPTION r 5 ep 1 1 e cos u The directrix is vertical and p units to the right of the pole. r 5 ep 1 2 e cos u The directrix is vertical and p units to the left of the pole. r 5 ep 1 1 e sin u The directrix is horizontal and p units above the pole. r 5 ep 1 2 e sin u The directrix is horizontal and p units below the pole. ECCENTRICITY THE CONIC IS A _ THE _____ IS PERPENDICULAR TO THE DIRECTRIX e 5 1 Parabola Axis of symmetry e , 1 Ellipse Major axis e . 1 Hyperbola Transverse axis ▼ A N S W E R r 5 3 12 6 cos u EXAMPLE 1 Finding the Polar Equation of a Conic Find a polar equation for a parabola that has its focus at the origin and whose directrix is the line y 5 3. Solution: The directrix is horizontal and above the pole. r 5 ep 1 1 e sin u A parabola has eccentricity e 5 1 and p 5 3. r 5 3 1 1 sin u YOUR T UR N Find a polar equation for a parabola that has its focus at the origin and whose directrix is the line x 5 23. ▼ 1070 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities In Example 2 we found that the polar equation r 5 10 3 1 2 cos u is an ellipse with major axis along the x-axis. We will graph this ellipse in Example 3. EXAMPLE 2 Identifying a Conic from Its Equation Identify the type of conic represented by the equation 10 3 1 2 cos u. Solution: To identify the type of conic, we need to rewrite the r 5 ep 1 6 e cos u equation in the form: Divide the numerator and denominator by 3. 10 3 a1 1 2 3 cos ub Identify e in the denominator. 5 10 3 a1 1 2 3 cos ub The numerator is equal to ep. 5 5⋅2 3 a1 1 2 3 cos ub Since e 5 2 3 , 1, the conic is an ellipse . The directrix is x 5 5, so the major axis is along the x-axis (perpendicular to the directrix). Y OUR TU R N Identify the type of conic represented by the equation r 5 10 2 2 10 sin u { e { e { p { e ▼ ▼ A N S W E R hyperbola, e 5 5, with transverse axis along the y-axis 11.8 Polar Equations of Conics 1071 EXAMPLE 3 Graphing a Conic from Its Equation The graph of the polar equation r 5 10 3 1 2 cos u is an ellipse. a. Find the vertices. b. Find the center of the ellipse. c. Find the lengths of the major and minor axes. d. Graph the ellipse. Solution (a): From Example 2 we see that e 5 2 3, which corresponds to an ellipse, and x 5 5 is the directrix. The major axis is perpendicular to the directrix. Therefore, the major axis lies along the polar axis. To find the vertices (which lie along the major axis), let u 5 0 and u 5 p. u 5 0: r 5 10 3 1 2 cos u 5 10 5 5 2 u 5 p: r 5 10 3 1 2 cos p 5 10 1 5 10 The vertices are the points V1 5 12, 02 and V2 5 110, p2 . Solution (b): The vertices in rectangular coordinates are V1 5 12, 02 and V2 5 1210, 02. The midpoint (in rectangular coordinates) between the two vertices is the point 124, 02, which corresponds to the point 14, p2 in polar coordinates. Solution (c): The length of the major axis, 2a, is the distance between the vertices. 2a 5 12 The length a 5 6 corresponds to the distance from the center to a vertex. Apply the formula e 5 c a with a 5 6 c 5 ae 5 6 2 3 5 4 and e 5 2 3 to find c. Let a 5 6 and c 5 4 in b2 5 a2 2 c2. b2 5 62 2 42 5 20 Solve for b. b 5 2!5 The length of the minor axis is 2b 5 4!5 . Solution (d): Graph the ellipse. 2√5 V2 V1 –8 –10 –6 –4 –2 1 2 y –5 –3 –2 –1 –4 2 1 3 4 5 x Polar axis 1072 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities EXAMPLE 4 Identifying and Graphing a Conic from Its Equation Identify and graph the conic defined by the equation r 5 2 2 1 3 sin u. Solution: Rewrite the equation in the form r 5 ep 1 1 e sin u. r 5 2 2 1 3 sin u 5 a2 3b a3 2b 1 1 a3 2b sin u The conic is a hyperbola since e 5 3 2 . 1. The directrix is horizontal and 2 3 unit above the pole (origin). To find the vertices, let u 5 p 2 and u 5 3p 2 . u 5 p 2 : r 5 2 2 1 3 sinap 2 b 5 2 5 u 5 3p 2 : r 5 2 2 1 3 sina3p 2 b 5 2 21 5 22 The vertices in polar coordinates are a2 5, p 2 b and a22, 3p 2 b. The vertices in rectangular coordinates are V1 5 A0, 2 5B and V2 5 10, 22. The center is the midpoint between the vertices: A0, 6 5B. The distance from the center to a focus is c 5 6 5. Apply the formula e 5 c a with c 5 6 5 and e 5 3 2 to find a. a 5 c e 5 6 5 3 2 5 4 5 Let a 5 4 5 and c 5 6 5 in b2 5 c2 2 a2. b2 5 a6 5b 2 2 a4 5b 2 5 20 25 Solve for b. b 5 2!5 5 The asymptotes are given by y 5 6a b 1x 2 h2 1 k, where a 5 4 5, b 5 2!5 5 , and 1h, k2 5 a0, 6 5b. y 5 6 2 !5 x 1 6 5 e e e p e e –2 –1 1 2 y –1 1 2 3 4 x Polar axis b = 5 2√5 11.8 Polar Equations of Conics 1073 It is important to note that although we relate specific points (vertices, foci, etc.) to rectangular coordinates, another approach to finding a rough sketch is to simply point-plot the equation in polar coordinates. EXAMPLE 5 Graphing a Conic by Point-Plotting in Polar Coordinates Sketch a graph of the conic r 5 4 1 2 sin u. Solution: STEP 1 The conic is a parabola because the equation is in the form r 5 142112 1 2 112 sin u Make a table with key values for u and r. u r 5 4 1 2 sin u (r, u) 0 r 5 4 1 2 sin 0 5 4 1 5 4 14, 02 p 2 r 5 4 1 2 sin p 2 5 4 1 2 1 5 4 0 undefined p r 5 4 1 2 sin p 5 4 1 5 4 14, p2 3p 2 r 5 4 1 2 sin 3p 2 5 4 1 2 1212 5 4 2 5 2 a2, 3p 2 b 2p r 5 4 1 2 sin 2p 5 4 1 5 4 14, 2p2 STEP 2 Plot the points on a polar graph and connect them with a smooth parabolic curve. 3 π 4 π 6 π 12 π 0 6 11π 12 17π 12 11π 12 13π 12 19π 12 23π 4 7π 3 5π 12 5π 12 7π 2 3π 3 4π 4 5π 6 7π π 6 5π 4 3π 2 π 3 2π 3 1 5 1074 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities In this section, we found that we could graph polar equations of conics by identifying a single focus and the directrix. There are four possible equations in terms of eccentricity e: EQUATION DESCRIPTION r 5 ep 1 1 e cos u The directrix is vertical and p units to the right of the pole. r 5 ep 1 2 e cos u The directrix is vertical and p units to the left of the pole. r 5 ep 1 1 e sin u The directrix is horizontal and p units above the pole. r 5 ep 1 2 e sin u The directrix is horizontal and p units below the pole. [SEC TION 11.8] S U MM A RY [SEC TION 11.8] E X E R CI SE S • S K I L L S In Exercises 1–14, find the polar equation that represents the conic described (assume that a focus is at the origin). Conic Eccentricity Directrix Conic Eccentricity Directrix 1. Ellipse e 5 1 2 y 5 25 2. Ellipse e 5 1 3 y 5 3 3. Hyperbola e 5 2 y 5 4 4. Hyperbola e 5 3 y 5 22 5. Parabola e 5 1 x 5 1 6. Parabola e 5 1 x 5 21 7. Ellipse e 5 3 4 x 5 2 8. Ellipse e 5 2 3 x 5 24 9. Hyperbola e 5 4 3 x 5 23 10. Hyperbola e 5 3 2 x 5 5 11. Parabola e 5 1 y 5 23 12. Parabola e 5 1 y 5 4 13. Ellipse e 5 3 5 y 5 6 14. Hyperbola e 5 8 5 y 5 5 In Exercises 15–26, identify the conic (parabola, ellipse, or hyperbola) that each polar equation represents. 15. r 5 4 1 1 cos u 16. r 5 3 2 2 3 sin u 17. r 5 2 3 1 2 sin u 18. r 5 3 2 2 2 cos u 19. r 5 2 4 1 8 cos u 20. r 5 1 4 2 cos u 21. r 5 7 3 1 cos u 22. r 5 4 5 1 6 sin u 23. r 5 40 5 1 5 sin u 24. r 5 5 5 2 4 sin u 25. r 5 1 1 2 6 cos u 26. r 5 5 3 2 3 sin u 11.8 Polar Equations of Conics 1075 In Exercises 27–40, for the given polar equations: (a) identify the conic as a parabola, an ellipse, or a hyperbola; (b) find the eccentricity and vertex (or vertices); and (c) graph. 27. r 5 2 1 1 sin u 28. r 5 4 1 2 cos u 29. r 5 4 1 2 2 sin u 30. r 5 3 3 1 8cos u 31. r 5 2 2 1 sin u 32. r 5 1 3 2 sin u 33. r 5 1 2 2 2sin u 34. r 5 1 1 2 2sin u 35. r 5 4 3 1 cos u 36. r 5 2 5 1 4 sin u 37. r 5 6 2 1 3sin u 38. r 5 6 1 1 cos u 39. r 5 2 5 1 5 cos u 40. r 5 10 6 2 3cos u • A P P L I C A T I O N S For Exercises 41 and 42, refer to the following: Planets travel in elliptical orbits around a single focus, the Sun. Pluto (orange), the dwarf planet furthest from the Sun, has a pronounced elliptical orbit, whereas Earth (royal blue) has an almost circular orbit. The polar equation of a planet’s orbit can be expressed as r 5 a11 2 e22 11 2 e cosu2 where e is the eccentricity and 2a is the length of the major axis. It can also be shown that the perihelion distance (minimum distance from the Sun to a planet) and the aphelion distance (maximum distance from the Sun to the planet) can be represented by r 5 a11 2 e2 and r 5 a11 1 e2, respectively. 41. Planetary Orbits. Mercury’s orbit is summarized in the picture below. Find the eccentricity of Mercury’s orbit. Find the polar equation that governs Mercury’s orbit. Mercury’s Orbit The Sun and Mercury are not to scale in this drawing. Mercury has a very elliptical orbit, which is highly inclined with respect to the plane of the ecliptic. Perihelion 46,000,000 km Aphelion 69,820,000 km 42. Planetary Orbits. Earth’s orbit is summarized in the picture below. Find the eccentricity of Earth’s orbit. Find the polar equation that governs Earth’s orbit. Earth's Orbit The Sun and Earth are not to scale in this drawing. Perihelion 147,100,000 km Aphelion 152,600,000 km 1076 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities For Exercises 43 and 44, refer to the following: Asteroids, meteors, and comets all orbit the Sun in elliptical patterns and often cross paths with Earth’s orbit, making life a little tense now and again. Asteroids are large rocks (bodies under 1000 kilometers across), meteors range from sand particles to rocks, and comets are masses of debris. A few asteroids have orbits that cross the Earth’s orbits—called Apollos or Earth-crossing asteroids. In recent years, asteroids have passed within 100,000 kilometers of Earth! 43. Asteroids. The asteroid 433 or Eros is the second largest near-Earth asteroid. The semimajor axis of its orbit is 150 million kilometers, and the eccentricity is 0.223. Find the polar equation of Eros’s orbit. 44. Asteroids. The asteroid Toutatis is the largest near-Earth asteroid. The semimajor axis of its orbit is 350 million kilometers, and the eccentricity is 0.634. On September 29, 2004, it missed Earth by 961,000 miles. Find the polar equation of Toutatis’s orbit. 45. When 0 , e , 1, the conic is an ellipse. Does the conic become more elongated or elliptical as e approaches 1 or as e approaches 0? 46. Show that r 5 ep 1 2 e sin u is the polar equation of a conic with a horizontal directrix that is p units below the pole. 47. Convert from rectangular to polar coordinates to show that the equation of a hyperbola, x2 a2 2 y2 b2 5 1, in polar form is r2 5 2 b2 1 2 e2 cos2 u. 48. Convert from rectangular to polar coordinates to show that the equation of an ellipse, x2 a2 1 y2 b2 5 1, in polar form is r2 5 b2 1 2 e2 cos2 u. • C O N C E P T U A L 49. Find the major diameter of the ellipse with polar equation r 5 ep 1 1 e cos u in terms of e and p. 50. Find the minor diameter of the ellipse with polar equation r 5 ep 1 1 e cos u in terms of e and p. 51. Find the center of the ellipse with polar equation r 5 ep 1 1 e cos u in terms of e and p. 52. Find the length of the latus rectum of the parabola with polar equation r 5 p 1 1 cos u. Assume that the focus is at the origin. • C H A L L E N G E 53. Let us consider the polar equations r 5 ep 1 1 e cos u and r 5 ep 1 2 e cos u with eccentricity e 5 1. With a graphing utility, explore the equations with p 5 1, 2, and 6. Describe the behavior of the graphs as p S q and also the difference between the two equations. 54. Let us consider the polar equations r 5 ep 1 1 e sin u and r 5 ep 1 2 e sin u with eccentricity e 5 1. With a graphing utility, explore the equations with p 5 1, 2, and 6. Describe the behavior of the graphs as p S q and also the difference between the two equations. 55. Let us consider the polar equations r 5 ep 1 1 e cos u and r 5 ep 1 2 e cos u with p 5 1. With a graphing utility, explore the equations with e 5 1.5, 3, and 6. Describe the behavior of the graphs as e S q and also the difference between the two equations. 56. Let us consider the polar equations r 5 ep 1 1 e sin u and r 5 ep 1 2 e sin u with p 5 1. With a graphing utility, explore the equations with e 5 1.5, 3, and 6. Describe the behavior of the graphs as e S q and also the difference between the two equations. 57. Let us consider the polar equations r 5 ep 1 1 e cos u and r 5 ep 1 2 e cos u with p 5 1. With a graphing utility, explore the equations with e 5 0.001, 0.5, 0.9, and 0.99. Describe the behavior of the graphs as e S 1 and also the difference between the two equations. Be sure to set the window parameters properly. 58. Let us consider the polar equations r 5 ep 1 1 e sin u and r 5 ep 1 2 e sin u with p 5 1. With a graphing utility, explore the equations with e 5 0.001, 0.5, 0.9, and 0.99. Describe the behavior of the graphs as e S 1 and also the difference between the two equations. Be sure to set the window parameters properly. • T E C H N O L O G Y 11.8 Polar Equations of Conics 1077 59. Let us consider the polar equation r 5 5 5 1 2 sin u. Explain why the graphing utility gives the following graphs with the specified window parameters: a. 322, 24 by 322, 24 with u step 5 p 2 b. 322, 24 by 322, 24 with u step 5 p 3 60. Let us consider the polar equation r 5 2 1 1 cos u. Explain why a graphing utility gives the following graphs with the specified window parameters: a. 322, 24 by 324, 44 with u step 5 p 2 b. 322, 24 by 324, 44 with u step 5 p 3 61. Let us consider the polar equation r 5 6 1 1 3 sin u. Explain why a graphing utility gives the following graphs with the specified window parameters: a. 328, 84 by 322, 44 with u step 5 p 2 b. 324, 84 by 322, 64 with u step 5 0.4p 62. Let us consider the polar equation r 5 2 1 2 sin u. Explain why a graphing utility gives the following graphs with the specified window parameters: a. 324, 44 by 322, 44 with u step 5 p 3 b. 324, 44 by 322, 64 with u step 5 0.8p 1078 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities 11.9.1 Parametric Equations of a Curve Thus far we have talked about graphs in planes. For example, the equation x2 1 y2 5 1 when graphed in a plane is the unit circle. Similarly, the function ƒ1x2 5 sin x when graphed in a plane is a sinusoidal curve. Now, we consider the path along a curve. For example, if a car is being driven on a circular racetrack, we want to see the movement along the circle. We can determine where (position) along the circle the car is at some time t using parametric equations. Before we define parametric equations in general, let us start with a simple example. Let x 5 cos t and y 5 sin t and t $ 0. We then can make a table of some corresponding values. t SECONDS x 5 cos t y 5 sin t (x, y ) 0 x 5 cos 0 5 1 y 5 sin 0 5 0 11, 02 p 2 x 5 cos a p 2b 5 0 y 5 sin ap 2b 5 1 10, 12 p x 5 cos p 5 21 y 5 sin p 5 0 121, 02 3p 2 x 5 cos a3p 2 b 5 0 y 5 sin a3p 2 b 5 21 10, 212 2p x 5 cos 12p2 5 1 y 5 sin 12p2 5 0 11, 02 If we plot these points and note the correspondence to time ( by converting all numbers to decimals), we will be tracing a path counterclockwise along the unit circle. TIME (SECONDS) t 5 0 t 5 1.57 t 5 3.14 t 5 4.71 POSITION 11, 02 10, 12 121, 02 10, 212 Notice that at time t 5 6.28 seconds we are back to the point 11, 02. We can see that the path represents the unit circle, since x2 1 y2 5 cos2 t 1 sin2 t 5 1. Parametric equations are useful for showing movement along a curve. We insert arrows in the graph to show direction, or orientation, along the curve as t increases. S K I L L S O B J E C T I V E S ■ ■Graph parametric equations. ■ ■Graph cycloids and the curves representing projectile motion. C O N C E P T U A L O B J E C T I V E S ■ ■Understand that the result of increasing the value of the parameter reveals the orientation of a curve or the direction of motion along it. ■ ■Use time as a parameter in parametric equations such as cycloids and projectile motion. 11.9 PARAMETRIC EQUATIONS AND GRAPHS DEFINITION Parametric Equations Let x 5 ƒ1t2 and y 5 g1t2 be functions defined for t on some interval. The set of points 1x, y2 5 1ƒ1t2, g1t22 represents a plane curve. The equations x 5 ƒ1t2 and y 5 g1t2 are called parametric equations of the curve. The variable t is called the parameter. x y t = 0 t = 3.14 t = 4.71 t = 1.57 (1, 0) (–1, 0) (0, 1) (0, –1) t = 6.28 11.9.1 SKI LL Graph parametric equations. 11.9.1 CO NC EPTUAL Understand that the result of increasing the value of the parameter reveals the orientation of a curve or the direction of motion along it. 11.9 Parametric Equations and Graphs 1079 EXAMPLE 1 Graphing a Curve Defined by Parametric Equations Graph the curve defined by the parametric equations x 5 t2 y 5 1t 2 12 t in 322, 24 Indicate the orientation with arrows. Solution: STEP 1 Make a table and find values for t, x, and y. t x 5 t 2 y 5 (t 2 1) (x, y ) t 5 22 x 5 12222 5 4 y 5 122 2 12 5 23 14, 232 t 5 21 x 5 12122 5 1 y 5 121 2 12 5 22 11, 222 t 5 0 x 5 02 5 0 y 5 10 2 12 5 21 10, 212 t 5 1 x 5 12 5 1 y 5 11 2 12 5 0 11, 02 t 5 2 x 5 22 5 4 y 5 12 2 12 5 1 14, 12 STEP 2 Plot the points in the xy-plane. STEP 3 Connect the points with a smooth curve and use arrows to indicate direction. The shape of the graph appears to be a parabola. The parametric equations are x 5 t 2 and y 5 1t 2 12. If we solve the second equation for t, getting t 5 y 1 1, and substi tute this expression into x 5 t 2, the result is x 5 1y 1 122. The graph of x 5 1y 1 122 is a parabola with vertex at the point 10, 212 and opening to the right. YOUR T UR N Graph the curve defined by the parametric equations x 5 t 1 1 y 5 t2 t in 322, 24 Indicate the orientation with arrows. –1 1 2 3 4 x y –2 –3 –1 1 2 t = –2 t = –1 t = 0 t = 1 t = 2 (4, –3) (1, –2) (0, –1) (1, 0) (4, 1) –1 1 2 3 4 x y –2 –3 –1 1 2 t = –2 t = –1 t = 0 t = 1 t = 2 (4, –3) (1, –2) (0, –1) (1, 0) (4, 1) ▼ ▼ A N S W E R –1 1 2 3 4 –1 1 2 3 4 x y t = –2 t = –1 t = 0 t = 1 t = 2 (–1, 4) (0, 1) (1, 0) (2, 1) (3, 4) Sometimes it is easier to show the rectangular equivalent of the curve and eliminate the parameter. 1080 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities 11.9.2 Applications of Parametric Equations Parametric equations can be used to describe motion in many applications. Two that we will discuss are the cycloid and a projectile. Suppose that you paint a red X on a bicycle tire. As the bicycle moves in a straight line, if you watch the motion of the red X, you will see that it follows the path of a cycloid. X STUDY TIP For open curves the orientation can be determined from two values of t. However, for closed curves three points should be chosen to ensure clockwise or counterclockwise orientation. 11.9.2 SKI LL Graph cycloids and the curves representing projectile motion. 11.9.2 CO NC EPTUAL Use time as a parameter in parametric equations such as cycloids and projectile motion. EXAMPLE 2 Graphing a Curve Defined by Parametric Equations by First Finding an Equivalent Rectangular Equation Graph the curve defined by the parametric equations x 5 4 cos t y 5 3 sin t t is any real number Indicate the orientation with arrows. Solution: One approach is to point-plot as in Example 1. A second approach is to find the equivalent rectangular equation that represents the curve. We apply the Pythagorean identity. sin2 t 1 cos2 t 5 1 Find sin2 t from the parametric equation for y. y 5 3 sin t Square both sides. y2 5 9 sin2 t Divide by 9. sin2 t 5 y2 9 Similarly, find cos2 t. x 5 4 cos t Square both sides. x2 5 16 cos2 t Divide by 16. cos2 t 5 x2 16 Substitute sin2 t 5 y2 9 and cos2 t 5 x2 16 into sin2 t 1 cos2 t 5 1. y2 9 1 x2 16 5 1 The curve is an ellipse centered at the origin and elongated horizontally. The orientation is counterclockwise. For example, when t 5 0, the position is 14, 02, when t 5 p 2 , the position is 10, 32, and when t 5 p, the position is 12 4, 02. 1 –3 –1 –5 2 3 4 5 x y –5 –4 –3 –2 1 2 3 4 5 t = 0 t = 2π π 2 t = π t = 3π 2 t = (0, –3) (4, 0) (–4, 0) (0, 3) 11.9 Parametric Equations and Graphs 1081 STUDY TIP A cycloid is a curve that does not have a simple rectangular equation. The only convenient way to describe its path is with parametric equations. EXAMPLE 3 Graphing a Cycloid Graph the cycloid given by x 5 21t 2 sin t2 and y 5 211 2 cos t2 for t in 30, 4p4. Solution: STEP 1 Make a table and find key values for t, x, and y. t x 5 2(t 2 sin t) y 5 2(1 2 cos t) (x, y ) t 5 0 x 5 210 2 02 5 0 y 5 211 2 12 5 0 10, 02 t 5 p x 5 21p 2 02 5 2p y 5 231 2 12124 5 4 12p, 42 t 5 2p x 5 212p 2 02 5 4p y 5 211 2 12 5 0 14p, 02 t 5 3p x 5 213p 2 02 5 6p y 5 231 2 12124 5 4 16p, 42 t 5 4p x 5 214p 2 02 5 8p y 5 211 2 12 5 0 18p, 02 STEP 2 Plot points in a plane and connect them with a smooth curve. 8π 4π 2π 6π –4 1 –3 2 –2 3 –1 4 x y t = 0 t = 2π t = 4π t = π t = 3π The parametric equations that define a cycloid are x 5 a1t 2 sin t2 and y 5 a11 2 cos t2 where t is any real number. Another example of parametric equations describing real-world phenomena is projectile motion. The accompanying photo of a golfer hitting a golf ball illustrates an example of a projectile. Fitzer/Getty Images 1082 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities Let v0 be the initial velocity of an object, u be the initial angle of inclination with the horizontal, and h be the initial height above the ground. Then the parametric equations describing the projectile motion (which will be developed in calculus) are x 5 1v0 cos u2 t and y 5 21 2 gt2 1 1v0 sin u2 t 1 h where t is the time and g is the constant acceleration due to gravity (9.8 meters per square second or 32 feet per square second). EXAMPLE 4 Graphing Projectile Motion Suppose a golfer hits his golf ball with an initial velocity of 160 feet per second at an angle of 30° with the ground. How far is his drive, assuming the length of the drive is from the tee to where the ball first hits the ground? Graph the curve representing the path of the golf ball. Assume that he hits the ball straight off the tee and down the fairway. Solution: STEP 1 Find the parametric equations that describe the golf ball that the golfer drove. First, write the parametric equations for projectile motion. x 5 1v0 cos u2 t and y 5 21 2gt2 1 1v0 sin u2 t 1 h Let g 5 32 ft/sec2, v0 5 160 ft/sec, h 5 0, and u 5 30°. x 5 1160 # cos 30°2 t and y 5 216t2 1 1160 # sin 30°2 t Evaluate the sine and cosine functions and simplify. x 5 80!3 t and y 5 216t2 1 80t STEP 2 Graph the projectile motion. t x 5 80!3 t y 5 216t 2 1 80t 1x, y2 t 5 0 x 5 80!3 102 5 0 y 5 2161022 1 80102 5 0 10, 02 t 5 1 x 5 80!3 112 < 139 y 5 2161122 1 80112 5 64 1139, 642 t 5 2 x 5 80!3 122 < 277 y 5 2161222 1 80122 5 96 1277, 962 t 5 3 x 5 80!3 132 < 416 y 5 2161322 1 80132 5 96 1416, 962 t 5 4 x 5 80!3 142 < 554 y 5 2161422 1 80142 5 64 1554, 642 t 5 5 x 5 80!3 152 < 693 y 5 2161522 1 80152 5 0 1693, 02 100 100 300 300 500 500 700 700 = 2 t = 5 t = 3 t = 1 t = 0 t = 4 t y x We can see that we selected our time increments well (the last point, 1693, 02, corresponds to the ball hitting the ground 693 feet from the tee). 11.9 Parametric Equations and Graphs 1083 STEP 3 Identify the horizontal distance from the tee to where the ball first hits the ground. Algebraically, we can determine the distance of the tee shot by setting the height y equal to zero. y 5 216t2 1 80t 5 0 Factor (divide) the common, 216t. 216t1t 2 52 5 0 Solve for t. t 5 0 or t 5 5 The ball hits the ground after 5 seconds. Let t 5 5 in the horizontal distance, x 5 80!3 t. x 5 80!3 152 < 693 The ball hits the ground 693 feet from the tee. With parametric equations, we can also determine when the ball lands (5 seconds). orientation along the curve can be determined by finding points corresponding to different t-values. Two important applications are cycloids and projectiles, whose paths we can trace using parametric equations. Parametric equations are a way of describing the path an object takes along a curve in the xy-plane. Parametric equations have equivalent rectangular equations. Typically, the method of graphing a set of parametric equations is to eliminate t and graph the corresponding rectangular equation. Once the curve is found, [SEC TION 11.9] S U M M A RY In Exercises 1–30, graph the curve defined by the parametric equations. 1. x 5 t 1 1, y 5 !t, t $ 0 2. x 5 3t, y 5 t2 2 1, t in 30, 44 3. x 5 23t, y 5 t2 1 1, t in 30, 44 4. x 5 t2 2 1, y 5 t2 1 1, t in 323, 34 5. x 5 t2, y 5 t3, t in 322, 24 6. x 5 t3 1 1, y 5 t3 2 1, t in 322, 24 7. x 5 !t, y 5 t, t in 30, 104 8. x 5 t, y 5 "t2 1 1, t in 30, 104 9. x 5 1t 1 122, y 5 1t 1 223, t in 30, 14 10. x 5 1t 2 123, y 5 1t 2 222, t in 30, 44 11. x 5 et, y 5 e2t, 2ln 3 # t # ln 3 12. x 5 e22t, y 5 e2t 1 4, 2ln 2 # t # ln 3 13. x 5 2t4 2 1, y 5 t8 1 1, 0 # t # 4 14. x 5 3t6 2 1, y 5 2t3, 21 # t # 1 15. x 5 t1t 2 223, y 5 t1t 2 223, 0 # t # 4 16. x 5 2t! 3 t, y 5 2 5t8 2 2, 23 # t # 3 17. x 5 3 sin t, y 5 2 cos t, t in 30, 2p4 18. x 5 cos12t2, y 5 sin t, t in 30, 2 4 p 19. x 5 sin t 1 1, y 5 cos t 2 2, t in 30, 2 4 p 20. x 5 tan t, y 5 1, t in c2 p 4, p 4 d 21. x 5 1, y 5 sin t, t in 322p, 2p4 22. x 5 sin t, y 5 2, t in 30, 2p4 23. x 5 sin2 t, y 5 cos2 t, t in 30, 2p4 24. x 5 2 sin2 t, y 5 2 cos2 t, t in 30, 2p4 25. x 5 2 sin 13t2, y 5 3 cos 12t2, t in 30, 2p4 26. x 5 4 cos12t2, y 5 t, t in 30, 2p4 27. x 5 cos a t 2b 2 1, y 5 sin a t 2b 1 1, 22p # t # 2p 28. x 5 sin a t 3b 1 3, y 5 cos a t 3 b 2 1, 0 # t # 6p 29. x 5 2 sin at 1 p 4 b, y 5 22 cos at 1 p 4 b, 2 p 4 # t # 7p 4 30. x 5 23 cos213t2, y 5 2 cos13t2, 2 p 3 # t # p 3 [SEC TION 11.9] E X E R C I S E S • S K I L L S 1084 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities In Exercises 31–40, the given parametric equations define a plane curve. Find an equation in rectangular form that also corresponds to the plane curve. 31. x 5 1 t , y 5 t2 32. x 5 t2 2 1, y 5 t2 1 1 33. x 5 t3 1 1, y 5 t3 2 1 34. x 5 3t, y 5 t2 2 1 35. x 5 t, y 5 "t2 1 1 36. x 5 sin2 t, y 5 cos2 t 37. x 5 2 sin2 t, y 5 2 cos2 t 38. x 5 sec2 t, y 5 tan2 t 39. x 5 41t2 1 12, y 5 1 2 t2 40. x 5 !t 2 1, y 5 !t • A P P L I C A T I O N S For Exercises 41–50, recall that the flight of a projectile can be modeled with the parametric equations x 5 1v0 cos u2 t y 5 216t2 1 1v0 sin u2 t 1 h where t is in seconds, v0 is the initial velocity, u is the angle with the horizontal, and x and y are in feet. 41. Flight of a Projectile. A projectile is launched from the ground at a speed of 400 feet per second at an angle of 45° with the horizontal. After how many seconds does the projectile hit the ground? 42. Flight of a Projectile. A projectile is launched from the ground at a speed of 400 feet per second at an angle of 45° with the horizontal. How far does the projectile travel (what is the horizontal distance), and what is its maximum altitude? 43. Flight of a Baseball. A baseball is hit at an initial speed of 105 mph and an angle of 20° at a height of 3 feet above the ground. If home plate is 420 feet from the back fence, which is 15 feet tall, will the baseball clear the back fence for a home run? 44. Flight of a Baseball. A baseball is hit at an initial speed of 105 mph and an angle of 20° at a height of 3 feet above the ground. If there is no back fence or other obstruction, how far does the baseball travel (horizontal distance), and what is its maximum height? 45. Bullet Fired. A gun is fired from the ground at an angle of 60°, and the bullet has an initial speed of 700 feet per second. How high does the bullet go? What is the horizontal (ground) distance between the point where the gun is fired and the point where the bullet hits the ground? 46. Bullet Fired. A gun is fired from the ground at an angle of 60°, and the bullet has an initial speed of 2000 feet per second. How high does the bullet go? What is the horizontal (ground) distance between the point where the gun is fired and the point where the bullet hits the ground? 47. Missile Fired. A missile is fired from a ship at an angle of 30°, an initial height of 20 feet above the water’s surface, and a speed of 4000 feet per second. How long will it be before the missile hits the water? 48. Missile Fired. A missile is fired from a ship at an angle of 40°, an initial height of 20 feet above the water’s surface, and a speed of 5000 feet per second. Will the missile be able to hit a target that is 2 miles away? 49. Path of a Projectile. A projectile is launched at a speed of 100 feet per second at an angle of 35° with the horizontal. Plot the path of the projectile on a graph. Assume that h 5 0. 50. Path of a Projectile. A projectile is launched at a speed of 150 feet per second at an angle of 55° with the horizontal. Plot the path of the projectile on a graph. Assume that h 5 0. For Exercises 51 and 52, refer to the following: Modern amusement park rides are often designed to push the envelope in terms of speed, angle, and ultimately gs, and usually take the form of gargantuan roller coasters or skyscraping towers. However, even just a couple of decades ago, such creations were depicted only in fantasy-type drawings, with their creators never truly believing their construction would become a reality. Nevertheless, thrill rides capable of nauseating any would-be rider were still able to be constructed; one example is the Calypso. This ride is a not-too-distant cousin of the more well-known Scrambler. It consists of four rotating arms (instead of three like the Scrambler), and on each of these arms, four cars (equally spaced around the circumference of a circular frame) are attached. Once in motion, the main piston to which the four arms are connected rotates clockwise, while each of the four arms themselves rotates counterclockwise. The combined motion appears as a blur to any onlooker from the crowd, but the motion of a single rider is much less chaotic. In fact, a single rider’s path can be modeled by the following graph: 2 –6 –2 –10 4 6 8 10 x y –10 –8 –6 –4 2 4 6 8 10 The equation of this graph is defined parametrically by x1t2 5 A cos t 1 B cos123t2 y1t2 5 A sin t 1 B sin123t2 0 # t # 2p 51. Amusement Rides. What is the location of the rider at t 5 0, t 5 p 2, t 5 p, t 5 3p 2 , and t 5 2p? 52. Amusement Rides. Suppose that the ride conductor was rather sinister and speeded up the ride to twice the speed. How would you modify the parametric equations to model such a change? Now vary the values of A and B. What do you think these parameters are modeling in this problem? 11.9 Parametric Equations and Graphs 1085 • C A T C H T H E M I S T A K E In Exercises 53 and 54, explain the mistake that is made. 53. Find the rectangular equation that corresponds to the plane curve defined by the parametric equations x 5 t 1 1 and y 5 !t. Describe the plane curve. Solution: Square y 5 !t. y2 5 t Substitute t 5 y2 into x 5 t 1 1. x 5 y2 1 1 The graph of x 5 y2 1 1 is a parabola opening to the right with vertex at 11, 02. This is incorrect. What mistake was made? 54. Find the rectangular equations that correspond to the plane curve defined by the parametric equations x 5 !t and y 5 t 2 1. Describe the plane curve. Solution: Square x 5 !t. x2 5 t Substitute t 5 x2 into y 5 t 2 1. y 5 x2 2 1 The graph of y 5 x2 2 1 is a parabola opening up with vertex at 10, 212. This is incorrect. What mistake was made? In Exercises 55 and 56, determine whether each statement is true or false. • C O N C E P T U A L 55. Curves given by equations in rectangular form have orientation. 56. Curves given by parametric equations have orientation. 57. Determine what type of curve the parametric equations x 5 !t and y 5 !1 2 t define. 58. Determine what type of curve the parametric equations x 5 ln t and y 5 t define. • C H A L L E N G E 59. Prove that x 5 a tan t, y 5 b sec t, 0 # t # 2p, t 2 p 2, 3p 2 are parametric equations for a hyperbola. Assume that a and b are nonzero constants. 60. Prove that x 5 a csc a t 2b, y 5 b cot a t 2b, 0 # t # 4p, t 2 p, 3p are parametric equations for a hyperbola. Assume that a and b are nonzero constants. 61. Consider the parametric curve x 5 a sin2 t 2 b cos2 t, y 5 b cos2 t 1 a sin2 t, 0 # t # p 2. Assume that a and b are nonzero constants. Find the Cartesian equation for this curve. 62. Consider the parametric curve x 5 a sin t 1 a cos t, y 5 a cos t 2 a sin t, 0 # t # 2p. Assume that a is not zero. Find the Cartesian equation for this curve. 63. Consider the parametric curve x 5 eat, y 5 bet, t . 0. Assume that a is a positive integer and b is a positive real number. Determine the Cartesian equation. 64. Consider the parametric curve x 5 a ln t, y 5 ln1bt2, t . 0. Assume that b is a positive integer and a is a positive real number. Determine the Cartesian equation. • T E C H N O L O G Y 65. Consider the parametric equations: x 5 asin t 2 sin1at2 and y 5 a cos t 1 cos1at2. With a graphing utility, explore the graphs for a 5 2, 3, and 4. 66. Consider the parametric equations: x 5 a cos t 2 b cos1at2 and y 5 asin t 1 sin1at2. With a graphing utility, explore the graphs for a 5 3 and b 5 1, a 5 4 and b 5 2, and a 5 6 and b 5 2. Find the t-interval that gives one cycle of the curve. 67. Consider the parametric equations: x 5cos1at2 and y 5 sin(bt). With a graphing utility, explore the graphs for a 5 2 and b 5 4, a 5 4 and b 5 2, a 5 1 and b 5 3, and a 5 3 and b 5 1. Find the t-interval that gives one cycle of the curve. 68. Consider the parametric equations: x 5 a sin1at2 2 sin t and y 5 a cos1at2 2 cos t. With a graphing utility, explore the graphs for a 5 2 and 3. Describe the t-interval for each case. 69. Consider the parametric equations x 5 a cos1at2 2 sin t and y 5 a sin1at2 2 cos t. With a graphing utility, explore the graphs for a 5 2 and 3. Describe the t-interval for each case. 70. Consider the parametric equations x 5 a sin1at2 2 cos t and y 5 a cos1at2 2 sin t. With a graphing utility, explore the graphs for a 5 2 and 3. Describe the t-interval for each case. CH A P TE R 11 R E VI E W 1086 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities [ C H AP T E R 1 1 REVIEW] SECTION CONCEPT KEY IDEAS/FORMULAS 11.1 Conic basics Three types of conics Parabola: Distance to a reference point (focus) and a reference line (directrix) is constant. Ellipse: Sum of the distances between the point and two reference points (foci) is constant. Hyperbola: Difference of the distances between the point and two reference points (foci) is a constant. 11.2 The parabola Parabola with a vertex at the origin Parabola with a vertex at the point 1h, k2 Applications Satellite Dish 11.3 The ellipse Ellipse centered at the origin Equation 1 y 2 k22 5 4p1x 2 h2 1x 2 h22 5 4p1y 2 k2 Vertex 1h, k2 1h, k2 Focus 1 p 1 h, k2 1h, p 1 k2 Directrix x 5 2p 1 h y 5 2p 1 k Axis of symmetry y 5 k x 5 h p + 0 Opens to the right Opens upward p 0 Opens to the left Opens downward Directrix y = –p Focus (0, p) x2 = 4py p p x y Up: p > 0 Down: p < 0 Directrix x = –p Focus (p, 0) y2 = 4px p p x y Right: p > 0 Left: p < 0 x y (–a, 0) (–c, 0) (c, 0) (a, 0) (0, b) (0, –b) x2 a2 y2 b2 = 1 + c2 = a2 – b2 x y (0, –a) (–b, 0) (b, 0) (0, a) (0, c) (0, –c) x2 b2 y2 a2 = 1 + c2 = a2 – b2 CH A P TE R 11 R E VIE W Chapter Review 1087 SECTION CONCEPT KEY IDEAS/FORMULAS Ellipse centered at the point 1h, k2 Applications Blimps, football, and orbits 11.4 The hyperbola Hyperbola centered at the origin Hyperbola centered at the point 1h, k2 Applications Loran Orientation of Major Axis Horizontal (Parallel to the x-axis) Vertical (Parallel to the y-axis) Equation 1x 2 h22 a2 1 1y 2 k22 b2 5 1 1x 2 h22 b2 1 1y 2 k22 a2 5 1 Graph Foci 1h 2 c, k2 1h 1 c, k2 1h, k 2 c2 1h, k 1 c2 Vertices 1h 2 a, k2 1h 1 a, k2 1h, k 2 a2 1h, k 1 a2 x y (–c, 0) (0, –b) (0, b) (c, 0) (–a, 0) (a, 0) y = x b a y = – x b a x2 a2 y2 b2 = 1 – c2 = a2 + b2 x y (–b, 0) (0, –c) (0, –a) (0, c) (0, a) (b, 0) y = x a b y = – x a b x2 b2 y2 a2 = 1 – c2 = a2 + b2 Orientation of Transverse Axis Horizontal Parallel to the x-axis Vertical Parallel to the y-axis Equation 1x 2 h22 a2 2 1 y 2 k22 b2 5 1 1 y 2 k22 a2 2 1x 2 h22 b2 5 1 Graph Vertices 1h 2a, k2 1h 1 a, k2 1h, k 2a2 1h, k 1 a2 Foci 1h 2 c, k2 1h 1 c, k2 1h, k 2 c2 1h, k 1 c2 CH A P TE R 11 R E VI E W 1088 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities SECTION CONCEPT KEY IDEAS/FORMULAS 11.5 Systems of nonlinear equations There is no procedure guaranteed to solve nonlinear equations. Solving a system of nonlinear equations Elimination: Eliminate a variable by either adding one equation to or subtracting one equation from the other. Substitution: Solve for one variable in terms of the other and substitute into the second equation. 11.6 Systems of nonlinear inequalities Solutions are determined graphically by finding the common shaded regions. ■ # or $ use solid curves ■ , or . use dashed curves Nonlinear inequalities in two variables Step 1: Rewrite the inequality as an equation. Step 2: Graph the equation. Step 3: Test points. Step 4: Shade. Systems of nonlinear inequalities Graph the individual inequalities and the solution in the common (overlapping) shaded region. 11.7 Rotation of axes Rotation of axes formulas x 5 X cos u 2 Y sin u x 5 X sin u 1 Y cos u The angle of rotation necessary to transform a general second-degree equation into an equation of a conic cot 2u 5 A 2 C B 11.8 Polar equations of conics All three conics (parabolas, ellipses, and hyperbolas) are defined in terms of a single focus and a directrix. Equations of conics in polar coordinates The directrix is vertical and p units to the right of the pole. r 5 ep 1 1 e cos u The directrix is vertical and p units to the left of the pole. r 5 ep 1 2 e cos u The directrix is horizontal and p units above the pole. r 5 ep 1 1 e sin u The directrix is horizontal and p units below the pole. r 5 ep 1 2 e sin u 11.9 Parametric equations and graphs Parametric equations of a curve Parametric equations: x 5 ƒ1t2 and y 5 g1t2 Plane curve: 1x, y2 5 1ƒ1t2, g1t22 Applications of parametric equations Cycloid: x 5 a1t 2 sin t2 y 5 a11 2 cost2 Projectile motion: x 5 1n0cosu2t y 5 21 2 gt 2 1 1n0 sin u2t 1 h R E VI E W E XERCISES Review Exercises 1089 [CH AP TER 11 REVIEW EXE R C IS E S ] 11.1 Conic Basics Determine whether each statement is true or false. 1. The focus is a point on the graph of the parabola. 2. The graph of y2 5 8x is a parabola that opens upward. 3. x2 9 2 y2 1 5 1 is the graph of a hyperbola that has a horizontal transverse axis. 4. 1 x 1 122 9 1 1 y 2 322 16 5 1 is a graph of an ellipse whose center is 11, 32. 11.2 The Parabola Find an equation for the parabola described. 5. vertex at 10, 02; focus at 13, 02 6. vertex at 10, 02; focus at 10, 22 7. vertex at 10, 02; directrix at x 5 5 8. vertex at 10, 02; directrix at y 5 4 9. vertex at 12, 32; focus at 12, 52 10. vertex at 121, 222; focus at 11, 222 11. focus at 11, 52; directrix at y 5 7 12. focus at 12, 22; directrix at x 5 0 Find the focus, vertex, directrix, and length of the latus rectum, and graph the parabola. 13. x2 5 212y 14. x2 5 8y 15. y2 5 x 16. y2 5 26x 17. 1 y 1 222 5 41 x 2 22 18. 1 y 2 222 5 241 x 1 12 19. 1 x 1 322 5 281 y 2 12 20. 1 x 2 322 5 281 y 1 22 21. x2 1 5x 1 2y 1 25 5 0 22. y2 1 2y 2 16x 1 1 5 0 Applications 23. Satellite Dish. A satellite dish measures 10 feet across its opening and 2 feet deep at its center. The receiver should be placed at the focus of the parabolic dish. Where should the receiver be placed? 24. Clearance under a Bridge. A bridge with a parabolic shape reaches a height of 40 feet in the center of the road, and the width of the bridge opening at ground level is 30 feet combined (both lanes). If an RV is 14 feet tall and 8 feet wide, will it make it through the tunnel? 11.3 The Ellipse Graph each ellipse. 25. x2 9 1 y2 64 5 1 26. x2 81 1 y2 49 5 1 27. 25x2 1 y2 5 25 28. 4x2 1 8y2 5 64 Find the standard form of an equation of the ellipse with the given characteristics. 29. foci: 123, 02 and 13, 02 vertices: 125, 02 and 15, 02 30. foci: 10, 222 and 10, 22 vertices: 10, 232 and 10, 32 31. Major axis vertical with length of 16, minor axis length of 6 and centered at 10, 02. 32. Major axis horizontal with length of 30, minor axis length of 20 and centered at 10, 02. Graph each ellipse. 33. 1 x 2 722 100 1 1 y 1 522 36 5 1 34. 201 x 1 322 1 1 y 2 422 5 120 35. 4x2 2 16x 1 12y2 1 72y 1 123 5 0 36. 4x2 2 8x 1 9y2 2 72y 1 147 5 0 Find the standard form of an equation of the ellipse with the given characteristics. 37. foci: 121, 32 and 17, 32 vertices: 122, 32 and 18, 32 38. foci: 11, 232 and 11, 212 vertices: 11, 242 and 11, 02 Applications 39. Planetary Orbits. Jupiter’s orbit is summarized in the picture. Utilize the fact that the Sun is a focus to determine an equation for Jupiter’s elliptical orbit around the Sun. Round to the nearest hundred thousand kilometers. Jupiter's Orbit The Sun and Jupiter are not to scale in this drawing. Perihelion 740,900,000 km Aphelion 815,700,000 km 40. Planetary Orbits. Mars’s orbit is summarized in the picture below. Utilize the fact that the Sun is a focus to determine an equation for Mars’s elliptical orbit around the Sun. Round to the nearest million kilometers. Mars' Orbit The Sun and Mars are not to scale in this drawing. Perihelion 207,000,000 km Aphelion 249,000,000 km 11.4 The Hyperbola Graph each hyperbola. 41. x2 9 2 y2 64 5 1 42. x2 81 2 y2 49 5 1 43. x2 2 25y2 5 25 44. 8y2 2 4x2 5 64 REV IEW E XE R CI SE S 1090 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities Find the standard form of an equation of the hyperbola with the given characteristics. 45. vertices: 123, 02 and 13, 02 foci: 125, 02 and 15, 02 46. vertices: 10, 212 and 10, 12 foci: 10, 232 and 10, 32 47. center: 10, 02; transverse: y-axis; asymptotes: y 5 3x and y 5 23x 48. center: 10, 02; transverse axis: y-axis; asymptotes: y 5 1 2x and y 5 21 2x Graph each hyperbola. 49. 1 y 2 122 36 2 1 x 2 222 9 5 1 50. 31x 1 322 2 121y 2 422 5 72 51. 8x2 2 32x 2 10y2 2 60y 2 138 5 0 52. 2x2 1 12x 2 8y2 1 16y 1 6 5 0 Find the standard form of an equation of the hyperbola with the given characteristics. 53. vertices: 10, 32 and 18, 32 foci: 121, 32 and 19, 32 54. vertices: 14, 222 and 14, 02 foci: 14, 232 and 14, 12 Applications 55. Ship Navigation. Two loran stations are located 220 miles apart along a coast. If a ship records a time difference of 0.00048 second and continues on the hyperbolic path corresponding to that difference, where would it reach shore? Assume the speed of radio signals is 186,000 miles per second. 56. Ship Navigation. Two loran stations are located 400 miles apart along a coast. If a ship records a time difference of 0.0008 second and continues on the hyperbolic path corresponding to that difference, where would it reach shore? 11.5 Systems of Nonlinear Equations Solve the system of equations with the elimination method. 57. x2 1 y 5 23 58. x2 1 y2 5 4 x 2 y 5 5 x2 1 y 5 2 59. x2 1 y2 5 5 60. x2 1 y2 5 16 2x2 2 y 5 0 6x2 1 y2 5 16 Solve the system of equations with the substitution method. 61. x 1 y 5 3 62. xy 5 4 x2 1 y2 5 4 x2 1 y2 5 16 63. x2 1 xy 1 y2 5 212 64. 3x 1 y 5 3 x 2 y 5 2 x 2 y2 5 29 Solve the system of equations by applying any method. 65. x3 2 y3 5 219 66. 2x2 1 4xy 5 9 x 2 y 5 21 x2 2 2xy 5 0 67. 2 x2 1 1 y2 5 15 68. x2 1 y2 5 2 1 x2 2 1 y2 5 23 11.6 Systems of Nonlinear Inequalities Graph the nonlinear inequality. 69. y $ x2 1 3 70. x2 1 y2 . 16 71. y # ex 72. y , 2x3 1 2 73. y $ ln1x 2 12 74. 9x2 1 4y2 # 36 Solve each system of inequalities and shade the region on a graph or indicate that the system has no solution. 75. y $ x2 2 2 76. x2 1 y2 # 4 y # 2x2 1 2 y # x 77. y $ 1x 1 122 2 2 78. 3x2 1 3y2 # 27 y # 10 2 x y $ x 2 1 79. 4y2 2 9x2 # 36 80. 9x2 1 16y2 # 144 y $ x 1 1 y $ 1 2 x2 11.7 Rotation of Axes The coordinates of a point in the xy-coordinate system are given. Assuming the X- and Y-axes are found by rotating the x- and y-axes by an angle u, find the corresponding coordinates for the point in the XY-system. 81. 123, 22, u 5 60° 82. 14, 232, u 5 45° Transform the equation of the conic into an equation in X and Y (without an XY-term) by rotating the x- and y-axes through an angle u. Then graph the resulting equation. 83. 2x2 1 4!3xy 2 2y2 2 16 5 0, u 5 30° 84. 25x2 1 14xy 1 25y2 2 288 5 0, u 5 p 4 Determine the angle of rotation necessary to transform the equation in x and y into an equation in X and Y with no XY-term. 85. 4x2 1 2!3xy 1 6y2 2 9 5 0 86. 4x2 1 5xy 1 4y2 2 11 5 0 Graph the second-degree equation. 87. x2 1 2xy 1 y2 1 !2x 2 !2y 1 8 5 0 88. 76x2 1 48!3xy 1 28y2 2 100 5 0 11.8 Polar Equations of Conics Find the polar equation that represents the conic described. 89. An ellipse with eccentricity e 5 3 7 and directrix y 5 27 90. A parabola with directrix x 5 2 x2 1 y2 5 4 R E VI E W E XERCISES Review Exercises 1091 Identify the conic (parabola, ellipse, or hyperbola) that each polar equation represents. 91. r 5 6 4 2 5 cos u 92. r 5 2 5 1 3 sin u For the given polar equations, find the eccentricity and vertex (or vertices), and graph the curve. 93. r 5 4 2 1 cos u 94. r 5 6 1 2 sin u 11.9 Parametric Equations and Graphs Graph the curve defined by the parametric equations. 95. x 5 sin t, y 5 4 cos t for t in 32p, p4 96. x 5 5 sin2 t, y 5 2 cos2 t for t in 32p, p4 97. x 5 4 2 t2, y 5 t2 for t in 323, 34 98. x 5 t 1 3, y 5 4 for t in 324, 44 The given parametric equations define a plane curve. Find an equation in rectangular form that also corresponds to the plane curve. 99. x 5 4 2 t2, y 5 t 100. x 5 5 sin2 t, y 5 2 cos2 t 101. x 5 2 tan2 t, y 5 4 sec2 t 102. x 5 3t2 1 4, y 5 3t2 2 5 Technology Exercises Section 11.2 103. In your mind, picture the parabola given by 1x 2 0.622 5 241 y 1 1.22. Where is the vertex? Which way does this parabola open? Now plot the parabola with a graphing utility. 104. In your mind, picture the parabola given by 1y 2 0.222 5 31x 2 2.82. Where is the vertex? Which way does this parabola open? Now plot the parabola with a graphing utility. 105. Given is the parabola y2 1 2.8y 1 3x 2 6.85 5 0. a. Solve the equation for y, and use a graphing utility to plot the parabola. b. Transform the equation into the form 1y 2 k22 5 4p1x 2 h2. Find the vertex. Which way does the parabola open? c. Do (a) and (b) agree with each other? 106. Given is the parabola x2 2 10.2x 2 y 1 24.8 5 0. a. Solve the equation for y, and use a graphing utility to plot the parabola. b. Transform the equation into the form 1x 2 h22 5 4p1y 2 k2. Find the vertex. Which way does the parabola open? c. Do (a) and (b) agree with each other? Section 11.3 107. Graph the following three ellipses: 4x2 1 y2 5 1, 412x22 1 y2 5 1, and 413x22 1 y2 5 1. What can be said to happen to ellipse 41cx22 1 y2 5 1 as c increases? 108. Graph the following three ellipses: x2 1 4y2 5 1, x2 1 412y22 5 1, and x2 1 413y22 5 1. What can be said to happen to ellipse x2 1 41cy22 5 1 as c increases? Section 11.4 109. Graph the following three hyperbolas: 4x2 2 y2 5 1, 412x22 2 y2 5 1, and 413x22 2 y2 5 1. What can be said to happen to hyperbola 41cx22 2 y2 5 1 as c increases? 110. Graph the following three hyperbolas: x2 2 4y2 5 1, x2 2 412y22 5 1, and x2 2 413y22 5 1. What can be said to happen to hyperbola x2 2 41cy22 5 1 as c increases? Section 11.5 With a graphing utility, solve the following systems of equations. 111. 7.5x2 1 1.5y2 5 12.25 x2y 5 1 112. 4x2 1 2xy 1 3y2 5 12 x3y 5 3 2 3x3 Section 11.6 With a graphing utility, graph the following systems of nonlinear inequalities. 113. y $ 10x 2 1 114. x2 1 4y2 # 36 y # 1 2 x2 y $ ex Section 11.7 115. With a graphing utility, explore the second-degree equation Ax2 1 Bxy 1 Cy2 1 10x 2 8y 2 5 5 0 for the following values of A, B, and C: a. A 5 2, B 5 23, C 5 5 b. A 5 2, B 5 3, C 5 25 Show the angle of rotation to one decimal place. Explain the differences. 116. With a graphing utility, explore the second-degree equation Ax2 1 Bxy 1 Cy2 1 2x 2 y 5 0 for the following values of A, B, and C: a. A 5 1, B 5 22, C 5 21 b. A 5 1, B 5 2, D 5 1 Show the angle of rotation to the nearest degree. Explain the differences. Section 11.8 117. Let us consider the polar equation r 5 8 4 1 5 sin u. Explain why a graphing utility gives the following graph with the specified window parameters: 326, 64 by 323, 94 with u step 5 p 4 REV IEW E XE R CI SE S 1092 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities 118. Let us consider the polar equation r 5 9 3 2 2 sin u. Explain why a graphing utility gives the following graph with the specified window parameters: 326, 64 by 323, 94 with u step 5 p 2 Section 11.9 119. Consider the parametric equations x 5 a cos at 1 b sin bt and y 5 a sin at 1 b cos bt. Use a graphing utility to explore the graphs for 1a, b2 5 12, 32 and 1a, b2 5 13, 22. Describe the t-interval for each case. 120. Consider the parametric equations x 5 a sin at 2 b cos bt and y 5 a cos at 2 b sin bt. Use a graphing utility to explore the graphs for 1a, b2 5 11, 22 and 1a, b2 5 12, 12. Describe the t-interval for each case. PRACTICE TEST [CH AP TER 11 PRACTICE T E S T ] Match the equation to the graph. 1. x 5 16y2 2. y 5 16x2 3. x2 1 16y2 5 1 4. x2 2 16y2 5 1 5. 16x2 1 y2 5 1 6. 16y2 2 x2 5 1 a. b. c. d. e. f. Find the equation of the conic with the given characteristics. 7. Parabola vertex: 10, 02 focus: 124, 02 8. Parabola vertex: 10, 02 directrix: y 5 2 9. Parabola vertex: 121, 52 focus: 121, 22 10. Parabola vertex: 12, 232 directrix: x 5 0 11. Ellipse center: 10, 02 vertices: 10, 242, 10, 42 foci: 10, 232, 10, 32 12. Ellipse center: 10, 02 vertices: 123, 02, 13, 02 foci: 121, 02, 11, 02 13. Ellipse vertices: 12, 262, 12, 62 foci: 12, 242, 12, 42 14. Ellipse vertices: 127, 232, 124, 232 foci: 126, 232, 125, 232 15. Hyperbola vertices: 121, 02 and 11, 02 asymptotes: y 5 22x and y 5 2x 16. Hyperbola vertices: 10, 212 and 10, 12 asymptotes: y 5 21 3x and y 5 1 3x 17. Hyperbola foci: 12, 262, 12, 62 vertices: 12, 242, 12, 42 18. Hyperbola foci: 127, 232, 124, 232 vertices: 126, 232, 125, 232 Graph the following equations. 19. 9x2 1 18x 2 4y2 1 16y 2 43 5 0 20. 4x2 2 8x 1 y2 1 10y 1 28 5 0 21. y2 1 4y 2 16x 1 20 5 0 22. x2 2 4x 1 y 1 1 5 0 23. Eyeglass Lens. Eyeglass lenses can be thought of as very wide parabolic curves. If the focus occurs 1.5 centimeters from the center of the lens and the lens at its opening is 4 centimeters across, find an equation that governs the shape of the lens. 24. Planetary Orbits. The planet Uranus’s orbit is described in the following picture, with the Sun as a focus of the elliptical orbit. Write an equation for the orbit. Uranus' Orbit The Sun and Uranus are not to scale in this drawing. Perihelion 2,739,000,000 km Aphelion 3,003,000,000 km Graph the following nonlinear inequalities. 25. y , x3 1 1 26. y2 $ 16x Graph the following systems of nonlinear inequalities. 27. y # 4 2 x2 28. y # e2x 16x2 1 25y2 # 400 y $ x2 2 4 29. Identify the conic represented by the equation r 5 12 3 1 2 sin u. State the eccentricity. 30. Use rotation of axes to transform the equation in x and y into an equation in X and Y that has no XY-term: 6!3x2 1 6xy 1 4!3y2 5 21!3. State the rotation angle. 31. A golf ball is hit with an initial speed of 120 feet per second at an angle of 458 with the ground. How long will the ball stay in the air? How far will the ball travel (horizontal distance) before it hits the ground? –5 5 5 –5 x y x y –1 1 10 x y 10 1 –1 x y –1 1 1 –1 –5 5 5 –5 x y x y –1 1 1 –1 Practice Test 1093 PR ACTICE TEST 1094 CHAPTER 11 Analytic Geometry and Systems of Nonlinear Equations and Inequalities 32. Describe (classify) the plane curve defined by the parametric equations x 5 !1 2 t and y 5 !t for t in 30, 14. 33. Use a graphing utility to graph the following nonlinear inequality: x2 1 4xy 2 9y2 2 6x 1 8y 1 28 # 0 34. Use a graphing utility to solve the following systems of equations: 0.1225x2 1 0.0289y2 5 1 y3 5 11x Round your answers to three decimal places. 35. Given is the parabola x 2 1 4.2x 2 y 1 5.61 5 0. a. Solve the equation for y and use a graphing utility to plot the parabola. b. Transform the equation into the form 1x 2 h22 5 4p 1 y 2 k2. Find the vertex. Which way does the parabola open? c. Do (a) and (b) agree with each other? 36. With a graphing utility, explore the second-degree equation Ax 2 1 Bx y 1 Cy2 1 10x 2 8y 2 5 5 0 for the following values of A, B, and C: a. A 5 2, B 5 2!3, C 5 1 b. A 5 2, B 5 !3, C 5 21 Show the angle of rotation to one decimal place. Explain the differences. CU MU LA TIV E TEST Cumulative Test 1095 [CH AP TERS 1–11 CUM ULAT IVE T E S T ] 1. Solve for x: 1x 1 222 2 1x 1 22 2 20 5 0. 2. Find an equation of a circle centered at 15, 12 and passing through the point 16, 222. 3. Evaluate the difference quotient ƒ1x 1 h2 2 ƒ1x2 h for the function ƒ1x2 5 8 2 7x. 4. Write an equation that describes the following variation: I is directly proportional to both P and t, and I 5 90 when P 5 1500 and t 5 2. 5. Find the quadratic function that has vertex 17, 72 and goes through the point 110, 102. 6. Compound Interest. How much money should you put in a savings account now that earns 4.7% interest a year compounded weekly if you want to have $65,000 in 17 years? 7. Solve the logarithmic equation exactly: log x2 2 log 16 5 0. 8. In a 30°-60°-90° triangle, if the shortest leg has length 8 inches, what are the lengths of the other leg and the hypotenuse? 9. Use a calculator to evaluate cot1227°2. Round your answer to four decimal places. 10. Sound Waves. If a sound wave is represented by y 5 0.007 sin 850pt cm, what are its amplitude and frequency? 11. For the trigonometric expression tan u 1csc u 1 cos u2, perform the operations and simplify. Write the answer in terms of sin u and cos u. 12. Find the exact value of cosa2 11p 12 b. 13. Solve the trigonometric equation 4 cos2 x 1 4 cos 2x 1 1 5 0 exactly over the interval 0 # u # 2p. 14. Airplane Speed. A plane flew due north at 450 mph for 2 hours. A second plane, starting at the same point and at the same time, flew southeast at an angle of 1358 clockwise from due north at 375 mph for 2 hours. At the end of 2 hours, how far apart were the two planes? Round to the nearest mile. 15. Find the vector with magnitude u 5 15 and direction angle u 5 110°. 16. Given z1 5 5 3cos115°2 1 i sin115°24 and z2 5 2 3cos175°2 1 i sin175°24, find the product z1z2 and express it in rectangular form. 17. At a food court, 3 medium sodas and 2 soft pretzels cost $6.77. A second order of 5 medium sodas and 4 soft pretzels costs $12.25. Find the cost of a soda and the cost of a soft pretzel. 18. Find the partial fraction decomposition for the rational expression 3x 1 5 1x 2 321x2 1 52. 19. Graph the system of inequalities or indicate that the system has no solution. y $ 3x 2 2 y # 3x 1 2 20. Solve the system using Gauss–Jordan elimination. x 2 2y 1 z 5 7 23x 1 y 1 2z 5 211 21. Given A 5 c3 4 27 0 1 5d, B 5 c8 22 6 9 0 21d , and C 5 c9 0 1 2d , find 2B 2 3A. 22. Use Cramer’s rule to solve the system of equations. 25x 1 40y 5 212 75x 2 105y 5 69 23. Find the standard form of the equation of an ellipse with foci 16, 22 and 16, 262 and vertices 16, 32 and 16, 272. 24. Find the standard form of the equation of a hyperbola with vertices 15, 222 and 15, 02 and foci 15, 232 and 15, 12. 25. Solve the system of equations. x 1 y 5 6 x2 1 y2 5 20 26. Use a graphing utility to graph the following equation: x2 2 3xy 1 10y2 2 1 5 0 27. Use a graphing utility to graph the following system of nonlinear inequalities: y $ e20.3x 2 3.5 y # 4 2 x2 C H A P T E R LEARNING OBJECTIVES [ [ 12 ■ ■Find the general nth term of a sequence or series. ■ ■Evaluate a finite arithmetic series. ■ ■Determine if an infinite geometric series converges or diverges. ■ ■Prove a mathematical statement using induction. ■ ■Use the binomial theorem to expand a binomial raised to a positive integer power. ■ ■Understand the difference between permutations and combinations. ■ ■Calculate the probability of an event. Have you ever been to a casino and played blackjack? It is the only game in the casino that you can win based on the law of large numbers. In the early 1990s, a group of math and science majors from the Massachusetts Institute of Technology (MIT) devised a foolproof scheme to win at blackjack. A professor at MIT developed a basic strategy outlined in the figure on the right that is based on the probability of combinations of particular cards being dealt, given certain cards already showing. To play blackjack (also called 21), each person is dealt two cards with the option of taking additional cards. The goal is to get a combination of cards that is worth 21 points (or less) without going over (called a bust). You have to avoid going over 21 or staying too far below 21. All face cards (jacks, queens, and kings) are worth 10 points, and an ace in blackjack is worth either 1 or 11 points. The students used the professor’s strategy along with a card-counting technique to place higher bets when there were more high-value cards left in the deck. It is reported that in 1992 the team won $4 million from Las Vegas casinos. The casinos caught on, and the students were all banned within 2 years. The 2008 movie 21 was based on this event. Sequences, Series, and Probability 2 3 4 5 6 Dealer’s Up Card A BASIC STRATEGY FOR BLACKJACK When surrender is allowed, surrender 9, 7 or 10, 6 vs 9, 10, A; 9, 6 or 10, 5 vs 10 When doubling down after splitting is allowed, split: 2’s, 3’s, 7’s vs 2-7; 4’s vs 5 or 6; 6’s vs 2-6 7 8 9 10 A S S S S S S S S S S S S S S S H H H H H S S S S S H H H H H S S S S S H H H H H S S S S S H H H H H H H S S S H H H H H D D D D D D D D D H D D D D D D D D H H H D D D D H H H H H H H H H H H H H H H S S S S S S S S S S S D D D D S S H H H H D D D D H H H H H H H D D D H H H H H H H D D D H H H H H H H H D D H H H H H H H H D D H H H H H SP SP SP SP SP SP SP SP SP SP S S S S S S S S S S SP SP SP SP SP S SP SP S S SP SP SP SP SP SP H H H H H SP SP SP SP H H H H H D D D D D D D D H H H H H H H H H H H H H H SP SP SP SP H H H H H 17+ 16 15 14 13 12 11 10 9 5 - 8 A, 8 - 10 A, 7 A, 6 A, 5 A, 4 A, 3 A, 2 A, A; 8, 8 10, 10 9, 9 7, 7 I II III Your Hand IV 6, 6 5, 5 4, 4 3, 3 2, 2 H SP SP SP SP H H H HIT STAND DOUBLE DOWN SPLIT H 1097 [I N T HI S CHAPTER] We will discuss counting and probability in addition to three other topics: sequences and series, mathematical induction, and the binomial theorem. SEQUENCES, SERIES, AND PROBABILITY 12.1 SEQUENCES AND SERIES 12.2 ARITHMETIC SEQUENCES AND SERIES 12.3 GEOMETRIC SEQUENCES AND SERIES 12.4 MATHEMATICAL INDUCTION 12.5 THE BINOMIAL THEOREM 12.6 COUNTING, PERMUTATIONS, AND COMBINATIONS 12.7 PROBABILITY • Sequences • Factorial Notation • Recursion Formulas • Sums and Series • Arithmetic Sequences • The General (nth) Term of an Arithmetic Sequence • The Sum of an Arithmetic Sequence • Geometric Sequences • The General (nth) Term of a Geometric Sequence • Geometric Series • Mathematical Induction • Binomial Coefficients • Binomial Expansion • Pascal’s Triangle • Finding a Particular Term of a Binomial Expansion • The Fundamental Counting Principle • Permutations • Combinations • Permutations with Repetition • Sample Space • Probability of an Event • Probability of an Event Not Occurring • Mutually Exclusive Events • Independent Events 1098 CHAPTER 12 Sequences, Series, and Probability 12.1.1 Sequences The word sequence means an order in which one thing follows another in succession. In mathematics, it means the same thing. For example, if we write x, 2x2, 3x3, 4x4, 5x5, ?, what would the next term in the sequence be, the one where the question mark now stands? The answer is 6x6. S K I L L S O B J E C T I V E S ■ ■Find terms of a sequence given the general term. ■ ■Apply factorial notation. ■ ■Apply recursion formulas. ■ ■Use summation (sigma) notation to represent a series and evaluate finite series and infinite series (if possible). C O N C E P T U A L O B J E C T I V E S ■ ■Recognize patterns in a sequence in order to identify the general term. ■ ■Understand why (mn)! Is not equal to m!n!. ■ ■Understand why the Fibonacci sequence is a recursion formula. ■ ■Understand why all finite series sum and why infinite series may or may not sum. 12.1 SEQUENCES AND SERIES 12.1.1 SKILL Find terms of a sequence given the general term. 12.1.1 CO NC EPTUAL Recognize patterns in a sequence in order to identify the general term. DEFINITION A Sequence A sequence is a function whose domain is a set of positive integers. The function values, or terms, of the sequence are written as a1, a2, a3, c, an, c Rather than using function notation, sequences are usually written with subscript (or index) notation, asubscript. A finite sequence has the domain 51, 2, 3, c, n6 for some positive integer n. An infinite sequence has the domain of all positive integers 51, 2, 3, c6. There are times when it is convenient to start the indexing at 0 instead of 1: a0, a1, a2, a3, c, an, c Sometimes a pattern in the sequence can be obtained and the sequence can be written using a general term. In the previous example, x, 2x2, 3x3, 4x4, 5x5, 6x6, c, each term has the same exponent and coefficient. We can write this sequence as an 5 nx n, n 5 1, 2, 3, 4, 3, 6, c, where an is called the general term. EXAMPLE 1 Finding the Sequence, Given the General Term Find the first four 1n 5 1, 2, 3, 42 terms of the sequences, given the general term. a. an 5 2n 2 1 b. bn 5 1212n n 1 1 Solution (a): an 5 2n 2 1 Find the first term, n 5 1. a1 5 2112 2 1 5 1 Find the second term, n 5 2. a2 5 2122 2 1 5 3 Find the third term, n 5 3. a3 5 2132 2 1 5 5 Find the fourth term, n 5 4. a4 5 2142 2 1 5 7 The first four terms of the sequence are 1, 3, 5, 7 . 12.1 Sequences and Series 1099 Solution (b): bn 5 1212n n 1 1 Find the first term, n 5 1. b1 5 12121 1 1 1 5 21 2 Find the second term, n 5 2. b2 5 12122 2 1 1 5 1 3 Find the third term, n 5 3. b3 5 12123 3 1 1 5 21 4 Find the fourth term, n 5 4. b4 5 12124 4 1 1 5 1 5 The first four terms of the sequence are 21 2, 1 3, 21 4, 1 5 . Y OUR T UR N Find the first four terms of the sequence an 5 1212n n2 . ▼ EXAMPLE 2 Finding the General Term, Given Several Terms of the Sequence Find the general term of the sequence, given the first five terms. a. 1, 1 4, 1 9, 1 16, 1 25, c b. 21, 4, 29, 16, 225, c Solution (a): Write 1 as 1 1. 1 1, 1 4, 1 9, 1 16, 1 25, c Notice that each denominator is an integer squared. 1 12, 1 22, 1 32, 1 42, 1 52, c Identify the general term. an 5 1 n2 n 5 1, 2, 3, 4, 5, c Solution (b): Notice that each term includes an integer squared. 212, 22, 232, 42, 252, c Identify the general term. bn 5 1212nn2 n 5 1, 2, 3, 4, 5, c Y OUR T UR N Find the general term of the sequence, given the first five terms. a. 21 2, 1 4, 21 6, 1 8, 2 1 10, c b. 1 2, 1 4, 1 8, 1 16, 1 32, c ▼ ▼ A N S W E R 21, 1 4, 21 9, 1 16 ▼ A N S W E R a. an 5 1212n 2n b. an 5 1 2n [CONCEPT CHECK] What is the next term in the sequence: 1, 23, 9, 227, ? ANSWER 81 ▼ STUDY TIP (21)n or (21)n11 is a way to represent an alternating sequence. Parts (b) in both Example 1 and Example 2 are called alternating sequences because the terms alternate signs (positive and negative). If the odd terms, a1, a3, a5, c are negative and the even terms, a2, a4, a6, c, are positive, we include 1212n in the general term. If the opposite is true, and the odd terms are positive and the even terms are negative, we include 1212n11 in the general term. 1100 CHAPTER 12 Sequences, Series, and Probability The values of n! for the first six nonnegative integers are 0! 5 1 1! 5 1 2! 5 2⋅1 5 2 3! 5 3⋅2⋅1 5 6 4! 5 4⋅3⋅2⋅1 5 24 5! 5 5⋅4⋅3⋅2⋅1 5 120 Notice that 4! 5 4⋅3⋅2⋅1 5 4⋅3!. In general, we can apply the formula n! 5 n31n 2 12!4. Often the brackets are not used, and the notation n! 5 n1n 2 12! implies calculating the factorial 1n 2 12! and then multiplying that quantity by n. For example, to find 6!, we employ the relationship n! 5 n1n 2 12! and set n 5 6: 6! 5 6⋅5! 5 6⋅120 5 720 DEFINITION Factorial If n is a positive integer, then n! (stated as “n factorial”) is the product of all positive integers from n down to 1. n! 5 n1n 2 121n 2 22 c 3⋅2⋅1 n $ 2 and 0! 5 1 and 1! 5 1. EXAMPLE 3 Finding the Terms of a Sequence Involving Factorials Find the first four terms of the sequence, given the general term an 5 x n n!. Solution: Find the first term, n 5 1. a1 5 x1 1! 5 x Find the second term, n 5 2. a2 5 x2 2! 5 x2 2⋅1 5 x2 2 Find the third term, n 5 3. a3 5 x3 3! 5 x3 3⋅2⋅1 5 x3 6 Find the fourth term, n 5 4. a4 5 x4 4! 5 x4 4⋅3⋅2⋅1 5 x4 24 The first four terms of the sequence are x, x2 2 , x3 6 , x4 24 . 12.1.2 Factorial Notation Many important sequences that arise in mathematics involve terms that are defined with products of consecutive positive integers. The products are expressed in factorial notation. 12.1.2 SKILL Apply factorial notation. 12.1.2 CO NC EPTUAL Understand why (mn)! Is not equal to m!n!. 12.1.3 Recursion Formulas Another way to define a sequence is recursively, or using a recursion formula. The first few terms are listed, and the recursion formula determines the remaining terms based on previous terms. For example, the famous Fibonacci sequence is 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, c. Each term in the Fibonacci sequence is found by adding the previous two terms. 1 1 1 5 2 1 1 2 5 3 2 1 3 5 5 3 1 5 5 8 5 1 8 5 13 8 1 13 5 21 13 1 21 5 34 21 1 34 5 55 34 1 55 5 89 We can define the Fibonacci sequence using a general term: a1 5 1, a2 5 1, and an 5 an22 1 an21 n $ 3 EXAMPLE 4 Evaluating Expressions with Factorials Evaluate each factorial expression. a. 6! 2!⋅3! b. 1n 1 12! 1n 2 12! Solution (a): Expand each factorial in the numerator and denominator. 6! 2!⋅3! 5 6⋅5⋅4⋅3⋅2⋅1 2⋅1⋅3⋅2⋅1 Cancel the 3⋅2⋅1 in both the numerator and denominator. 5 6⋅5⋅4 2⋅1 Simplify. 5 6⋅5⋅2 1 5 60 6! 2!⋅3! 5 60 Solution (b): Expand each factorial in the 1n 1 12! 1n 2 12! 5 1n 1 121n21n 2 121n 2 22 c3⋅2⋅1 1n 2 121n 2 22 c3⋅2⋅1 numerator and denominator. Cancel the 1n 2 121n 2 22 c3⋅2⋅1 1n 1 12! 1n 2 12! 5 1n 1 121n2 in the numerator and denominator. Alternatively, 1n 1 12! 1n 2 12! 5 1n 1 121n21n 2 12! 1n 2 12! 1n 1 12! 1n 2 12! 5 n2 1 n common mistake In Example 4 we found 6! 2!⋅3! 5 60. It is important to note that 2!⋅3! 2 6! Y OUR T UR N Evaluate the factorial expressions. a. 3!⋅4! 2!⋅6!. b. 1n 1 22! n! ▼ [CONCEPT CHECK] TRUE OR FALSE (mn)! 5 m!n! ANSWER False ▼ STUDY TIP In general, m!n! 2 (mn)! ▼ A N S W E R a. 1 10 b. n2 1 3n 1 2 12.1.3 S K I L L Apply recursion formulas. 12.1.3 C ON C E P T U A L Understand why the Fibonacci sequence is a recursion formula. 12.1 Sequences and Series 1101 1102 CHAPTER 12 Sequences, Series, and Probability The Fibonacci sequence is found in places we least expect them (for example, pineapples, broccoli, and flowers). The number of petals in a flower is a Fibonacci number. For example, a wild rose has 5 petals, lilies and irises have 3 petals, and daisies have 34, 55, or even 89 petals. The number of spirals in an Italian broccoli is a Fibonacci number (13). EXAMPLE 5 Using a Recursion Formula to Find a Sequence Find the first four terms of the sequence: a1 5 2 and an 5 2an21 2 1, n $ 2. Solution: Write the first term, n 5 1. a1 5 2 Find the second term, n 5 2. a2 5 2a1 2 1 5 2122 2 1 5 3 Find the third term, n 5 3. a3 5 2a2 2 1 5 2132 2 1 5 5 Find the fourth term, n 5 4. a4 5 2a3 2 1 5 2152 2 1 5 9 The first four terms of the sequence are 2, 3, 5, 9 . Y OUR TU R N Find the first four terms of the sequence: a1 5 1 and an 5 an21 n! n $ 2 ▼ 12.1.4 Sums and Series When we add the terms in a sequence, the result is a series. DEFINITION Series Given the infinite sequence a1, a2, a3, . . . , an, . . . , the sum of all of the terms in the infinite sequence is called an infinite series and is denoted by a1 1 a2 1 a3 1 ? ? ? 1 an 1 ? ? ? and the sum of only the first n terms is called a finite series, or nth partial sum, and is denoted by Sn 5 a1 1 a2 1 a3 1 ? ? ? 1 an The capital Greek letter g (sigma) corresponds to the capital S in our alphabet. Therefore, we use g as a shorthand way to represent a sum (series). For example, the sum of the first five terms of the sequence 1, 4, 9, 16, 25, . . . , n2, . . . can be represented using sigma (or summation) notation: a 5 n51 n2 5 1122 1 1222 1 1322 1 1422 1 1522 5 1 1 4 1 9 1 16 1 25 This is read “the sum as n goes from 1 to 5 of n2.” The letter n is called the index of summation, and often other letters are used instead of n. It is important to note that the sum can start at other numbers besides 1. If we wanted the sum of all of the terms in the sequence, we would represent that infinite series using summation notation as a q n51 n2 5 1 1 4 1 9 1 16 1 25 1 c STUDY TIP If an 5 an21 1 an22, then a100 5 a99 1 a98. [CONCEPT CHECK] What is the next term in the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ? ANSWER 144 ▼ ▼ A N S W E R 1, 1 2, 1 12, 1 288 12.1.4 SKI LL Use summation (sigma) notation to represent a series and evaluate finite series and infinite series (if possible). 12.1.4 CON CEPTUAL Understand why all finite series sum and why infinite series may or may not sum. STUDY TIP S is a regular Greek letter, but when used to represent the mathematical sum operation, we oversize it. STUDY TIP When seen in running text, we often use the following notation: a 5 n51 5 a 5 n51 STUDY TIP A series can start at any integer (not just 1). Now that we are comfortable with sigma (summation) notation, let’s turn our attention to evaluating a series (calculating the sum). You can always evaluate a finite series. However, you cannot always evaluate an infinite series. EXAMPLE 6 Writing a Series Using Sigma Notation Write the following series using sigma notation. a. 1 1 1 1 1 2 1 1 6 1 1 24 1 1 120 b. 8 1 27 1 64 1 125 1 c Solution (a): Write 1 as 1 1. 1 1 1 1 1 1 1 2 1 1 6 1 1 24 1 1 120 Notice that we can write the denominators using factorials. 5 1 1 1 1 1 1 1 2! 1 1 3! 1 1 4! 1 1 5! Recall that 0! 5 1 and 1! 5 1. 5 1 0! 1 1 1! 1 1 2! 1 1 3! 1 1 4! 1 1 5! Identify the general term. an 5 1 n! n 5 0, 1, 2, 3, 4, 5 Write the finite series using sigma notation. a 5 n50 1 n! Solution (b): Write the infinite series as a sum of 8 1 27 1 64 1 125 1 c terms cubed. 5 23 1 33 1 43 1 53 1 c Identify the general term of the series. an 5 n3 n $ 2 Write the infinite series using sigma notation. a q n52 n3 Y OUR T UR N Write the following series using sigma notation. a. 1 2 1 2 1 1 6 2 1 24 1 1 120 2 c b. 4 1 8 1 16 1 32 1 64 1 c ▼ ▼ A N S W E R a. a q n51 1212n11 n! b. a q n52 2n ▼ A N S W E R 23 EXAMPLE 7 Evaluating a Finite Series Evaluate the series a 4 i50 12i 1 12. Solution: 1i 5 12 1i 5 32 T T Write out the partial sum. a 4 i50 12i 1 12 5 1 1 3 1 5 1 7 1 9 c c c 1i 5 02 1i 5 22 1i 5 42 5 25 Simplify. a 4 i50 12i 1 12 5 25 YOUR T UR N Evaluate the series a 5 n51 1212nn. ▼ 12.1 Sequences and Series 1103 1104 CHAPTER 12 Sequences, Series, and Probability Infinite series may or may not have a finite sum. For example, if we keep adding 1 1 1 1 1 1 1 1 c, then there is no single real number that the series sums to because the sum continues to grow without bound. However, if we add 0.9 1 0.09 1 0.009 1 0.0009 1 c, this sum is 0.9999 c 5 0.9, which is a rational number, and it can be proven that 0.9 5 1. EXAMPLE 8 Evaluating an Infinite Series, If Possible Evaluate the following infinite series, if possible. a. a q n51 3 10n b. a q n51 n2 Solution (a): Expand the series. a q n51 3 10n 5 3 10 1 3 100 1 3 1000 1 3 10,000 1 c Write in decimal form. a q n51 3 10n 5 0.3 1 0.03 1 0.003 1 0.0003 1 c Calculate the sum. a q n51 3 10n 5 0.3333333 5 1 3 a q n51 3 10n 5 1 3 Solution (b): Expand the series. a q n51 n2 5 1 1 4 1 9 1 16 1 25 1 36 1 c This sum is infinite since it continues to grow without any bound. In part (a) we say that the series converges to 1 3, and in part (b) we say that the series diverges. Y OUR TU R N Evaluate the following infinite series, if possible. a. a q n51 2n b. a q n51 6A 1 10B n ▼ STUDY TIP The sum of a finite series always exists. The sum of an infinite series may or may not exist. Applications The annual sales at an electronics retailer from 2015 to 2017 can be approximated by the model an 5 45.7 1 9.5n 2 1.6n2, where an is the yearly sales in billions of dollars and n 5 0, 1, 2. ▼ A N S W E R a. Series diverges. b. Series converges to 2 3. [CONCEPT CHECK] TRUE OR FALSE An infinite series may or may not converge. ANSWER True ▼ YEAR n an 5 45.7 1 9.5n 2 1.6n2 TOTAL SALES IN BILLIONS 2015 0 a0 5 45.7 1 9.5102 2 1.61022 $45.7 2016 1 a1 5 45.7 1 9.5112 2 1.61122 $53.6 2017 2 a2 5 45.7 1 9.5122 2 1.61222 $58.3 What does the finite series 1 3 a 2 n50 an tell us? It tells us the average yearly sales over 3 years. Year 2015 2016 2017 $45.7 $53.6 $58.3 Annual Sales (in billions) 30.0 10.0 50.0 70.0 $90.0 [SEC TION 12 .1] E X E RC I S E S • S K I L L S In Exercises 1–12, write the first four terms of the sequence. Assume n starts at 1. 1. an 5 n 2. an 5 n2 3. an 5 2n 2 1 4. an 5 xn 5. an 5 n 1n 1 12 6. an 5 1n 1 12 n 7. an 5 2n n! 8. an 5 n! 1n 1 12! 9. an 5 1212nxn11 10. an 5 1212n11n2 11. an 5 1212n 1n 1 121n 1 22 12. an 5 1n 2 122 1n 1 122 In Exercises 13–20, find the indicated term of the sequence. 13. an 5 a1 2b n a9 5 ? 14. an 5 n 1n 1 122 a15 5 ? 15. an 5 1212nn! 1n 1 22! a19 5 ? 16. an 5 1212n111n 2 121n 1 22 n a13 5 ? 17. an 5 a1 1 1 nb 2 a100 5 ? 18. an 5 1 2 1 n2 a10 5 ? 19. an 5 log 10n a23 5 ? 20. an 5 elnn a49 5 ? In Exercises 21–28, write an expression for the nth term of the given sequence. 21. 2, 4, 6, 8, 10, . . . 22. 3, 6, 9, 12, 15, . . . 23. 1 2⋅1, 1 3⋅2, 1 4⋅3, 1 5⋅4, 1 6⋅5, c 24. 1 2, 1 4, 1 8, 1 16, 1 32, c 25. 2 2 3, 4 9, 2 8 27, 16 81, c 26. 1 2, 3 4, 9 8, 27 16, 81 32, c 27. 1, 21, 1, 21, 1, c 28. 1 3, 2 2 4 , 3 5, 2 4 6, 5 7, c In Exercises 29–40, simplify the ratio of factorials. 29. 9! 7! 30. 4! 6! 31. 29! 27! 32. 32! 30! 33. 75! 77! 34. 100! 103! 35. 97! 93! 36. 101! 98! 37. 1n 2 12! 1n 1 12! 38. 1n 1 22! n! 39. 12n 1 32! 12n 1 12! 40. 12n 1 22! 12n 2 12! The sum of a finite series is always finite. The sum of an infinite series is either ■ ■convergent or ■ ■divergent. Sigma notation is used to express a series. ■ ■Finite series: a n i51 ai 5 a1 1 a2 1 c1 an ■ ■Infinite series: a q n51 an 5 a1 1 a2 1 a3 1 c In this section, we discussed finite and infinite sequences and series. When the terms of a sequence are added together, the result is a series. Finite sequence: a1, a2, a3, c, an Infinite sequence: a1, a2, a3, c, an, c Finite series: a1 1 a2 1 a3 1 c1 an Infinite series: a1 1 a2 1 a3 1 c1 an 1 c Factorial notation was also introduced: n! 5 n⋅1n 2 12 ⋅c⋅3⋅2⋅1 n $ 2 and 0! 5 1 and 1! 5 1. [SEC TION 12 .1] S U M M A RY 12.1 Sequences and Series 1105 1106 CHAPTER 12 Sequences, Series, and Probability In Exercises 41–50, write the first four terms of the sequence defined by the recursion formula. Assume the sequence begins at 1. 41. a1 5 7 an 5 an21 1 3 42. a1 5 2 an 5 an21 1 1 43. a1 5 1 an 5 n⋅an21 44. a1 5 2 an 5 1n 1 12 ⋅an21 45. a1 5 100 an 5 an21 n! 46. a1 5 20 an 5 an21 n2 47. a1 5 1, a2 5 2 an 5 an21⋅an22 48. a1 5 1, a2 5 2 an 5 an22 an21 49. a1 5 1, a2 5 21 an 5 1212nCan21 2 1 an22 2 D 50. a1 5 1, a2 5 21 an 5 1n 2 12 an21 1 1n 2 22an22 In Exercises 51–64, evaluate the finite series. 51. a 5 n51 2 52. a 5 n51 7 53. a 4 n50 n2 54. a 4 n51 1 n 55. a 6 n51 12n 2 12 56. a 6 n51 1n 1 12 57. a 4 n50 1n 58. a 4 n50 2n 59. a 3 n50 12x2n 60. a 3 n50 12x2n11 61. a 5 k50 2k k! 62. a 5 k50 1212k k! 63. a 4 k50 xk k! 64. a 4 k50 1212kxk k! In Exercises 65–68, evaluate the infinite series, if possible. 65. a q j50 2⋅10.12 j 66. a q j50 5⋅A 1 10B j 67. a q j50 n j n $ 1 68. a q j50 1j In Exercises 69–76, apply sigma notation to write the sum. 69. 1 2 1 2 1 1 4 2 1 8 1 c 1 1 64 70. 1 1 1 2 1 1 4 1 1 8 1 c 1 1 64 1 c 71. 1 2 2 1 3 2 4 1 5 2 6 1 c 72. 1 1 2 1 3 1 4 1 5 1 c 1 21 1 22 1 23 73. 2⋅1 1 1 3⋅2⋅1 1 1 4⋅3⋅2⋅1 2⋅1 1 5⋅4⋅3⋅2⋅1 3⋅2⋅1 1 6⋅5⋅4⋅3⋅2⋅1 4⋅3⋅2⋅1 74. 1 1 2 1 1 22 2⋅1 1 23 3⋅2⋅1 1 24 4⋅3⋅2⋅1 1 c 75. 1 2 x 1 x2 2 2 x3 6 1 x4 24 2 x5 120 1 c 76. x 1 x2 1 x3 2 1 x4 6 1 x5 24 1 x6 120 • A P P L I C A T I O N S 77. Money. Upon graduation Jessica receives a commission from the U.S. Navy to become an officer and a $20,000 signing bonus for selecting aviation. She puts the entire bonus in an account that earns 6% interest compounded monthly. The balance in the account after n months is An 5 20,000 a 1 1 0.06 12 b n n 5 1, 2, 3, c Her commitment to the Navy is 6 years. Calculate A72. What does A72 represent? 78. Money. Dylan sells his car in his freshman year and puts $7000 in an account that earns 5% interest compounded quarterly. The balance in the account after n quarters is An 5 7000 a 1 1 0.05 4 b n n 5 1, 2, 3, c Calculate A12. What does A12 represent? 79. Salary. An attorney is trying to calculate the costs associated with going into private practice. If she hires a paralegal to assist her, she will have to pay the paralegal $20 per hour. To be competitive with most firms, she will have to give her paralegal a $2 per hour raise per year. Find a general term of a sequence an, which represents the hourly salary of a paralegal with n years of experience. What will be the paralegal’s salary with 20 years of experience? 80. NFL Salaries. A player in the NFL typically has a career that lasts 3 years. The practice squad makes the league minimum of $275,000 (2004) in the first year, with a $75,000 raise per year. Write the general term of a sequence an that represents the salary of an NFL player making the league minimum during his entire career. Assuming n 5 1 corresponds to the first year, what does a 3 n51 an represent? 81. Salary. Upon graduation Sheldon decides to go to work for a local police department. His starting salary is $30,000 per year, and he expects to get a 3% raise per year. Write the recursion formula for a sequence that represents his salary n years on the job. Assume n 5 0 represents his first year making $30,000. 82. Escherichia coli. A single cell of bacteria reproduces through a process called binary fission. Escherichia coli cells divide into two every 20 minutes. Suppose the same rate of division is maintained for 12 hours after the original cell enters the body. How many E. coli bacteria cells would be in the body 12 hours later? Suppose there is an infinite nutrient source so that the E. coli bacteria maintain the same rate of division for 48 hours after the original cell enters the body. How many E. coli bacteria cells would be in the body 48 hours later? 83. AIDS/HIV. A typical person has 500 to 1500 T cells per drop of blood in the body. HIV destroys the T cell count at a rate of 50–100 cells per drop of blood per year, depending on how aggressive it is in the body. Generally, the onset of AIDS occurs once the body’s T cell count drops below 200. Write a sequence that represents the total number of T cells in a person infected with HIV. Assume that before infection the person has a 1000 T cell count 1a0 5 10002 and the rate at which the infection spreads corresponds to a loss of 75 T cells per drop of blood per year. How much time will elapse until this person has full-blown AIDS? 84. Company Sales. A national furniture store reported total sales from 2016 through 2017 in the billions. The sequence an 5 3.8 1 1.6n represents the total sales in billions of dollars. Assuming n 5 3 corresponds to 2016, what were the reported sales in 2016 and 2017? What does 1 2⋅a 4 n53 an represent? 85. Cost of Eating Out. A college student tries to save money by bringing a bag lunch instead of eating out. He will be able to save $100 per month. He puts the money into his savings account, which draws 1.2% interest and is compounded monthly. The balance in his account after n months of bagging his lunch is An 5 100,000311.0012n 2 14 n 5 1, 2, c Calculate the first four terms of this sequence. Calculate the amount after 3 years (36 months). 86. Cost of Acrylic Nails. A college student tries to save money by growing her own nails out and not spending $50 per month on acrylic fills. She will be able to save $50 per month. She puts the money into her savings account, which draws 1.2% interest and is compounded monthly. The balance in her account after n months of natural nails is An 5 50,000311.0012n 2 14 n 5 1, 2, c Calculate the first four terms of this sequence. Calculate the amount after 4 years (48 months). 87. Math and Engineering. The formula a q n50 xn n! can be used to approximate the function y 5 ex. Compute the first five terms of this formula to approximate ex. Apply the result to find e2 and compare this result with the calculator value of e2. 88. Home Prices. If the inflation rate is 3.5% per year and the average price of a home is $195,000, the average price of a home after n years is given by An 5 195,00011.0352n. Find the average price of the home after 6 years. 89. Approximating Functions. Polynomials can be used to approximate transcendental functions such as ln 1x2 and ex, which are found in advanced mathematics and engineering. For example, a q n501212n 1x 2 12n11 n 1 1 can be used to approximate ln 1x2, where x is close to 1. Use the first five terms of the series to approximate ln 1x2. Next, find ln 11.12 and compare with the value given by your calculator. 90. Future Value of an Annuity. The future value of an ordinary annuity is given by the formula FV 5 PMT 3111 1 i 2n 2 12/i 4, where PMT 5 amount paid into the account at the end of each period, i 5 interest rate per period, and n 5 number of compounding periods. If you invest $5000 at the end of each year for 5 years, you will have an accumulated value of FV as given in the above equation at the end of the nth year. Determine how much is in the account at the end of each year for the next 5 years if i 5 0.06. • C A T C H T H E M I S T A K E In Exercises 91–94, explain the mistake that is made. 91. Simplify the ratio of factorials: 13!215!2 6! . Solution: Express 6! in factored form. 13!215!2 13!212!2 Cancel the 3! in the numerator and 15!2 12!2 denominator. Write out the factorials. 5⋅4⋅3⋅2⋅1 2⋅1 Simplify. 5⋅4⋅3 5 60 13!215!2 13!212!2 2 60. What mistake was made? 92. Simplify the factorial expression: 2n12n 2 22! 12n 1 22! . Solution: Express factorials in factored form. 2n12n 2 2212n 2 4212n 2 62 c 12n 1 2212n212n 2 2212n 2 4212n 2 62 c Cancel common terms. 1 2n 1 2 This is incorrect. What mistake was made? 12.1 Sequences and Series 1107 1108 CHAPTER 12 Sequences, Series, and Probability 93. Find the first four terms of the sequence defined by an 5 1212n11n2. Solution: Find the n 5 1 term. a1 5 21 Find the n 5 2 term. a2 5 4 Find the n 5 3 term. a3 5 29 Find the n 5 4 term. a4 5 16 The sequence 21, 4, 29, 16, c, is incorrect. What mistake was made? 94. Evaluate the series a 3 k501212kk2. Solution: Write out the sum. a 3 k50 1212kk2 5 21 1 4 2 9 Simplify the sum. a 3 k56 1212kk2 5 26 This is incorrect. What mistake was made? 95. a n k50 cxk 5 c a n k50 xk 96. a n i51 ai 1 bi 5 a n i51 ai 1 a n i51 bi 97. a n k51 akbk 5 a n k51 ak ⋅a n k51 bk 98. 1a!21b!2 5 1ab2! In Exercises 95–98, determine whether each statement is true or false. • C O N C E P T U A L 99. Write the first four terms of the sequence defined by the recursion formula: a1 5 C an 5 an21 1 D D 2 0 100. Write the first four terms of the sequence defined by the recursion formula: a1 5 C an 5 Dan21 D 2 0 101. Fibonacci Sequence. An explicit formula for the nth term of the Fibonacci sequence is: Fn 5 A1 1 !5Bn 2 A1 2 !5 Bn 2n!5 Apply algebra (not your calculator) to find the first two terms of this sequence and verify that these are indeed the first two terms of the Fibonacci sequence. 102. Let an 5 !an21 for n $ 2 and a1 5 7. Find the first five terms of this sequence and make a generalization for the nth term. • C H A L L E N G E 103. The sequence defined by an 5 a1 1 1 nb n approaches the number e as n gets large. Use a graphing calculator to find a100, a1000, a10,000, and keep increasing n until the terms in the sequence approach 2.7183. 104. The Fibonacci sequence is defined by a1 5 1, a2 5 1, and an 5 an22 1 an21 for n $ 3. The ratio an11 an is an approximation of the golden ratio. The ratio approaches a constant f (phi) as n gets large. Find the golden ratio using a graphing utility. 105. Use a graphing calculator “SUM” to sum a 5 k50 2k k! . Compare it with your answer to Exercise 61. 106. Use a graphing calculator “SUM” to sum a 5 k50 1212k k! . Compare it with your answer to Exercise 62. • T E C H N O L O G Y 12.2 Arithmetic Sequences and Series 1109 12.2.1 Arithmetic Sequences The word arithmetic (with emphasis on the third syllable) often implies adding or subtracting of numbers. Arithmetic sequences are sequences whose terms are found by adding a constant to each previous term. The sequence 1, 3, 5, 7, 9, . . . is arithmetic because each successive term is found by adding 2 to the previous term. S K I L L S O B J E C T I V E S ■ ■Determine if a sequence is arithmetic. ■ ■Find the general nth term of an arithmetic sequence. ■ ■Evaluate a finite arithmetic series. C O N C E P T U A L O B J E C T I V ES ■ ■Understand that the common difference of an arithmetic sequence can be either positive or negative. ■ ■Derive the formula for the nth term of an arithmetic sequence in terms of n, the first term, and the common difference. ■ ■Derive the formula for the sum of a finite arithmetic sequence. 12.2 ARITHMETIC SEQUENCES AND SERIES 12.2.1 S K I L L Determine if a sequence is arithmetic. 12.2.1 C ON C E P T U A L Understand that the common difference of an arithmetic sequence can be either positive or negative. DEFINITION Arithmetic Sequences A sequence is arithmetic if each term in the sequence is found by adding a real number d to the previous term, so that an11 5 an 1 d. Because an11 2 an 5 d, the number d is called the common difference. EXAMPLE 1 Identifying the Common Difference in Arithmetic Sequences Determine whether each sequence is arithmetic. If so, find the common difference for each of the arithmetic sequences. a. 5, 9, 13, 17, c b. 18, 9, 0, 29, c c. 1 2, 5 4, 2, 11 4 , c Solution (a): Label the terms. a1 5 5, a2 5 9, a3 5 13, a4 5 17, c Find the difference d 5 an11 2 an. d 5 a2 2 a1 5 9 2 5 5 4 Check that the difference for the d 5 a3 2 a2 5 13 2 9 5 4 next two terms is also 4. d 5 a4 2 a3 5 17 2 13 5 4 There is a common difference of 4 . Therefore, this sequence is arithmetic, and each successive term is found by adding 4 to the previous term. Solution (b): Label the terms. a1 5 18, a2 5 9, a3 5 0, a4 5 29, c Find the difference d 5 an11 2 an. d 5 a2 2 a1 5 9 2 18 5 29 Check that the difference for d 5 a3 2 a2 5 0 2 9 5 29 the next two terms is also 29. d 5 a4 2 a3 5 29 2 0 5 29 There is a common difference of 29 . Therefore, this sequence is arithmetic, and each successive term is found by subtracting 9 from (that is, adding 29 to) the previous term. [CONCEPT CHECK] TRUE OR FALSE If the second term is less than the first term in an arithmetic sequence, then the common difference is negative. If the second term is greater than the first term in an arithmetic sequence, then the common difference is positive. ANSWER True ▼ 1110 CHAPTER 12 Sequences, Series, and Probability 12.2.2 The General (nth) Term of an Arithmetic Sequence To find a formula for the general, or nth, term of an arithmetic sequence, write out the first several terms and look for a pattern. First term, n 5 1. a1 Second term, n 5 2. a2 5 a1 1 d Third term, n 5 3. a3 5 a2 1 d 5 1a1 1 d2 1 d 5 a1 1 2d Fourth term, n 5 4. a4 5 a3 1 d 5 1a1 1 2d2 1 d 5 a1 1 3d In general, the nth term is given by an 5 a1 1 1n 2 12d. Solution (c): Label the terms. a1 5 1 2, a2 5 5 4, a3 5 2, a4 5 11 4 , c Find the difference d 5 an11 2 an. d 5 a2 2 a1 5 5 4 2 1 2 5 3 4 Check that the difference for the next two terms is also 3 4. d 5 a3 2 a2 5 2 2 5 4 5 3 4 d 5 a4 2 a3 5 11 4 2 2 5 3 4 There is a common difference of 3 4 . Therefore, this sequence is arithmetic, and each successive term is found by adding 3 4 to the previous term. Y OUR TU R N Find the common difference for each of the arithmetic sequences. a. 7, 2, 23, 28, c b. 1, 5 3, 7 3, 3, c ▼ STUDY TIP We check several terms to confirm that a sequence is arithmetic. ▼ A N S W E R a. 25 b. 2 3 12.2.2 SKILL Find the general nth term of an arithmetic sequence. 12.2.2 CO NC EPTUAL Derive the formula for the nth term of an arithmetic sequence in terms of n, the first term, and the common difference. THE nTH TERM OF AN ARITHMETIC SEQUENCE The nth term of an arithmetic sequence with common difference d is given by an 5 a1 1 1n 2 12d for n $ 1 EXAMPLE 2 Finding the nth Term of an Arithmetic Sequence Find the 13th term of the arithmetic sequence 2, 5, 8, 11, . . . . Solution: Identify the common difference. d 5 5 2 2 5 3 Identify the first 1n 5 12 term. a1 5 2 Substitute a1 5 2 and d 5 3 into an 5 a1 1 1n 2 12d. an 5 2 1 31n 2 12 Substitute n 5 13 into an 5 2 1 31n 2 12. a13 5 2 1 3113 2 12 5 38 Y OUR TU R N Find the 10th term of the arithmetic sequence 3, 10, 17, 24, . . . . ▼ ▼ A N S W E R 66 12.2.3 The Sum of an Arithmetic Sequence What is the sum of the first 100 counting numbers 1 1 2 1 3 1 4 1 c1 99 1 100 5 ? If we write this sum twice (one in ascending order and one in descending order) and add, we get 100 pairs of 101. 1 1 2 1 3 1 4 1 ? ? ? 1 99 1 100 100 1 99 1 98 1 97 1 ? ? ? 1 2 1 1 101 1 101 1 101 1 101 1 ? ? ? 1 101 1 101 5 100 11012 Since we added twice the sum, we divide by 2. 1 1 2 1 3 1 4 1 c1 99 1 100 5 1101211002 2 5 5050 Now, let us develop the sum of a general arithmetic series. The sum of the first n terms of an arithmetic sequence is called the nth partial sum, or finite arithmetic series, and is denoted by Sn. An arithmetic sequence can be found by starting at the first term and adding the common difference to each successive term, and so the nth partial sum, or finite series, can be found the same way, but terminating the sum at the nth term: Sn 5 a1 1 a2 1 a3 1 a4 1 c Sn 5 a1 1 1a1 1 d2 1 1a1 1 2d2 1 1a1 1 3d2 1 c1 1an2 Similarly, we can start with the nth term and find terms going backward by subtracting the common difference until we arrive at the first term: Sn 5 an 1 an21 1 an22 1 an23 1 c Sn 5 an 1 1an 2 d2 1 1an 2 2d2 1 1an 2 3d2 1 c 1 1a12 EXAMPLE 3 Finding the Arithmetic Sequence The 4th term of an arithmetic sequence is 16, and the 21st term is 67. Find a1 and d and construct the sequence. Solution: Write the 4th and 21st terms. a4 5 16 and a21 5 67 Adding d 17 times to a4 results in a21. a21 5 a4 1 17d Substitute a4 5 16 and a21 5 67. 67 5 16 1 17d Solve for d. d 5 3 Substitute d 5 3 into an 5 a1 1 1n 2 12d. an 5 a1 1 31n 2 12 Let a4 5 16. 16 5 a1 1 314 2 12 Solve for a1. a1 5 7 The arithmetic sequence that starts at 7 and has a common difference of 3 is 7, 10, 13, 16, . . . . YOU R T UR N Construct the arithmetic sequence whose 7th term is 26 and whose 13th term is 50. ▼ [CONCEPT CHECK] For the arithmetic sequence 3, 10, 17, 24, . . . what is the general term for the nth term? ANSWER an 5 3 1 7(n 2 1) ▼ ▼ A N S W E R 2, 6, 10, 14, . . . 12.2.3 S K I L L Evaluate a finite arithmetic series. 12.2.3 C ON C E P T U A L Derive the formula for the sum of a finite arithmetic sequence. 12.2 Arithmetic Sequences and Series 1111 1112 CHAPTER 12 Sequences, Series, and Probability Add these two representations of the nth partial sum. Notice that the d terms are eliminated: Sn 5 a1 1 1a1 1 d2 1 1a1 1 2d2 1 1a1 1 3d2 1 c1 1an2 Sn 5 an 1 1an 2 d2 1 1an 2 2d2 1 1an 2 3d2 1 c 1 1a12 2Sn 5 1a1 1 an2 1 1a1 1 an2 1 1a1 1 an2 1 c 1 1a1 1 an2 n1a1 1 an2 2Sn 5 n1a1 1 an2 or Sn 5 n 2 1a1 1 an2 Let an 5 a1 1 1n 2 12d. Sn 5 n 2 3a1 1 a1 1 1n 2 12d4 5 n 2 32a1 1 1n 2 12d 4 5 na1 1 n1n 2 12d 2 ('''''''''''')'''''''''''' DEFINITION Evaluating a Finite Arithmetic Series The sum of the first n terms of an arithmetic sequence, called a finite arithmetic series, is given by the formula Sn 5 n 2 32a1 1 1n 2 12d4 or Sn 5 n 2 1a1 1 an2 n $ 2 STUDY TIP Sn can also be written as Sn 5 n 2 32a1 1 1n 2 12d4. EXAMPLE 4 Evaluating a Finite Arithmetic Series Evaluate the finite arithmetic series a 100 k51 k. Solution: Expand the arithmetic series. a 100 k51 k 5 1 1 2 1 3 1 c1 99 1 100 This is the sum of an arithmetic sequence of numbers with a common difference of 1. Identify the parameters of the arithmetic sequence. a1 5 1, an 5 100, and n 5 100 Substitute these values into Sn 5 n 2 1a1 1 an2. S100 5 100 2 11 1 1002 Simplify. S100 5 5050 The sum of the first 100 natural numbers is 5050. Y OUR TU R N Evaluate the following finite arithmetic series. a. a 30 k51 k b. a 20 k51 12k 1 12 ▼ ▼ A N S W E R a. 465 b. 440 Applications EXAMPLE 5 Finding the nth Partial Sum of an Arithmetic Sequence Find the sum of the first 20 terms of the arithmetic sequence 3, 8, 13, 18, 23, . . . . Solution: Recall the partial-sum formula. Sn 5 n 2 1a1 1 an2 Find the 20th partial sum of this arithmetic sequence. S20 5 20 2 1a1 1 a202 Recall that the general nth term of an arithmetic sequence is given by: an 5 a1 1 1n 2 12d Note that the first term of the arithmetic sequence is 3. a1 5 3 This is an arithmetic sequence with a common difference of 5. d 5 5 Substitute a1 5 3 and d 5 5 into an 5 a1 1 1n 2 12d. an 5 3 1 1n 2 125 Substitute n 5 20 to find the 20th term. a20 5 3 1 120 2 125 5 98 Substitute a1 5 3 and a20 5 98 into the partial sum. S20 5 1013 1 982 5 1010 The sum of the first 20 terms of this arithmetic sequence is 1010. Y OUR T UR N Find the sum of the first 25 terms of the arithmetic sequence 2, 6, 10, 14, 18, . . . . ▼ [CONCEPT CHECK] Derive the formula for the sum of an arithmetic sequence given by 2, 6, 10, 14, 18, . . . ANSWER Sn 5 2n2 ▼ ▼ A N S W E R 1250 EXAMPLE 6 Marching Band Formation Suppose a band has 18 members in the first row, 22 members in the second row, and 26 members in the third row and continues with that pattern for a total of nine rows. How many marchers are there all together? David Young-Wolff/PhotoEdit UC Berkeley marching band 12.2 Arithmetic Sequences and Series 1113 1114 CHAPTER 12 Sequences, Series, and Probability Solution: The number of members in each row forms an arithmetic sequence with a common difference of 4, and the first row has 18 members. a1 5 18 d 5 4 Calculate the nth term of the sequence an 5 a1 1 1n 2 12d. an 5 18 1 1n 2 124 Find the 9th term n 5 9. a9 5 18 1 19 2 124 5 50 Calculate the sum Sn 5 n 2 1a1 1 an2 of the nine rows. S9 5 9 2 1a1 1 a92 5 9 2 118 1 502 5 9 2 1682 5 306 There are 306 members in the marching band. Y OUR TU R N Suppose a bed of tulips is arranged in a garden so that there are 20 tulips in the first row, 26 tulips in the second row, and 32 tulips in the third row and the rows continue with that pattern for a total of 8 rows. How many tulips are there all together? ▼ ▼ A N S W E R 328 an 5 a1 1 1n 2 12d n $ 1 Sn 5 n 2 1a1 1 an2 5 n 2 32a1 1 1n 2 12d 4 In this section, arithmetic sequences were defined as sequences of which each successive term is found by adding the same constant d to the previous term. Formulas were developed for the general, or nth, term of an arithmetic sequence, and for the nth partial sum of an arithmetic sequence, also called a finite arithmetic series. [SEC TION 12 . 2] S U M M A RY [SEC TION 12 . 2] E X E R C I SE S • S K I L L S In Exercises 1–10, determine whether the sequence is arithmetic. If it is, find the common difference. 1. 2, 5, 8, 11, 14, c 2. 9, 6, 3, 0, 23, 26, c 3. 12 1 22 1 32 1 c 4. 1! 1 2! 1 3! 1 c 5. 3.33, 3.30, 3.27, 3.24, c 6. 0.7, 1.2, 1.7, 2.2, c 7. 4, 14 3 , 16 3 , 6, c 8. 2, 7 3, 8 3, 3, c 9. 101, 102, 103, 104, c 10. 120, 60, 30, 15, c In Exercises 11–20, find the first four terms of the sequence. Determine whether the sequence is arithmetic, and if so, find the common difference. 11. an 5 22n 1 5 12. an 5 3n 2 10 13. an 5 n2 14. an 5 n2 n! 15. an 5 5n 2 3 16. an 5 24n 1 5 17. an 5 101n 2 12 18. an 5 8n 2 4 19. an 5 1212nn 20. an 5 1212n112n In Exercises 21–28, find the general, or nth, term of the arithmetic sequence given the first term and the common difference. 21. a1 5 11 d 5 5 22. a1 5 5 d 5 11 23. a1 5 24 d 5 2 24. a1 5 2 d 5 24 25. a1 5 0 d 5 2 3 26. a1 5 21 d 5 23 4 27. a1 5 0 d 5 e 28. a1 5 1.1 d 5 20.3 In Exercises 29–34, find the specified term for each arithmetic sequence. 29. The 10th term of the sequence 7, 20, 33, 46, c 30. The 19th term of the sequence 7, 1, 25, 211, c 31. The 100th term of the sequence 9, 2, 25, 212, c 32. The 90th term of the sequence 13, 19, 25, 31, c 33. The 21st term of the sequence 1 3, 7 12, 5 6, 13 12, c 34. The 33rd term of the sequence 1 5, 8 15, 13 15, 6 5, c In Exercises 35–40, for each arithmetic sequence described, find a1 and d and construct the sequence by stating the general, or nth, term. 35. The 5th term is 44 and the 17th term is 152. 36. The 9th term is 219 and the 21st term is 255. 37. The 7th term is 21 and the 17th term is 241. 38. The 8th term is 47 and the 21st term is 112. 39. The 4th term is 3 and the 22nd term is 15. 40. The 11th term is 23 and the 31st term is 213. In Exercises 41–52, find the sum. 41. a 23 k51 2k 42. a 20 k50 5k 43. a 30 n51 122n 1 52 44. a 17 n50 13n 2 102 45. a 14 j53 0.5j 46. a 33 j51 j 4 47. 2 1 7 1 12 1 17 1 c1 62 48. 1 2 3 2 7 2 c2 75 49. 4 1 7 1 10 1 c 1 151 50. 2 1 0 2 2 2 c2 56 51. 1 6 2 1 6 2 1 2 2 c 2 13 2 52. 11 12 1 7 6 1 17 12 1 c1 14 3 In Exercises 53–58, find the partial sum of the arithmetic series. 53. The first 18 terms of 1 1 5 1 9 1 13 1 c. 54. The first 21 terms of 2 1 5 1 8 1 11 1 c. 55. The first 43 terms of 1 1 1 2 1 0 2 1 2 2 c. 56. The first 37 terms of 3 1 3 2 1 0 2 3 2 2 c. 57. The first 18 terms of 29 1 1 1 11 1 21 1 31 1 c. 58. The first 21 terms of 22 1 8 1 18 1 28 1 c. • A P P L I C A T I O N S 59. Comparing Salaries. Colin and Camden are twin brothers graduating with B.S. degrees in biology. Colin takes a job at the San Diego Zoo making $28,000 for his first year with a $1500 raise per year every year after that. Camden accepts a job at Florida Fish and Wildlife making $25,000 with a guaranteed $2000 raise per year. How much will each of the brothers have made in a total of 10 years? 60. Comparing Salaries. On graduating with a Ph.D. in optical sciences, Jasmine and Megan choose different career paths. Jasmine accepts a faculty position at the University of Arizona making $80,000 with a guaranteed $2000 raise every year. Megan takes a job with the Boeing Corporation making $90,000 with a guaranteed $5000 raise each year. Calculate how many total dollars each woman will have made after 15 years. 61. Theater Seating. You walk into the premiere of Chris Pine’s new movie, and the theater is packed, with almost every seat filled. You want to estimate the number of people in the theater. You quickly count to find that there are 22 seats in the front row, and there are 25 rows in the theater. Each row appears to have 1 more seat than the row in front of it. How many seats are in that theater? 62. Field of Tulips. Every spring the Skagit County Tulip Festival plants more than 100,000 bulbs. In honor of the Tri-Delta sorority that has sent 120 sisters from the University of Washington to volunteer for the festival, Skagit County has planted tulips in the shape of DDD. In each of the triangles there are 20 rows of tulips, each row having one less than the row before. How many tulips are planted in each delta if there is 1 tulip in the first row? 63. World’s Largest Champagne Fountain. From December 28 to 30, 1999, Luuk Broos, director of Maison Luuk-Chalet Fontaine, constructed a 56-story champagne fountain at the Steigenberger Kurhaus Hotel, Scheveningen, Netherlands. The fountain consisted of 30,856 champagne glasses. Assuming there was one glass at the top and the number of glasses in each row forms an arithmetic sequence, how many were on the bottom row (story)? How many fewer glasses did each successive row (story) have? Assume each story is one row. 64. Stacking of Logs. If 25 logs are laid side by side on the ground, and 24 logs are placed on top of those, and 23 logs are placed on the 3rd row, and the pattern continues until there is a single log on the 25th row, how many logs are in the stack? 65. Falling Object. When a skydiver jumps out of an airplane, she falls approximately 16 feet in the 1st second, 48 feet during the 2nd second, 80 feet during the 3rd second, 112 feet during the 4th second, and 144 feet during the 5th second, and this pattern continues. If she deploys her parachute after 10 seconds have elapsed, how far will she have fallen during those 10 seconds? 12.2 Arithmetic Sequences and Series 1115 1116 CHAPTER 12 Sequences, Series, and Probability 66. Falling Object. If a penny is dropped out of a plane, it falls approximately 4.9 meters during the 1st second, 14.7 meters during the 2nd second, 24.5 meters during the 3rd second, and 34.3 meters during the 4th second. Assuming this pattern continues, how many meters will the penny have fallen after 10 seconds? 67. Grocery Store. A grocer has a triangular display of oranges in a window. There are 20 oranges in the bottom row, and the number of oranges decreases by one in each row above this row. How many oranges are in the display? 68. Salary. Suppose your salary is $45,000 and you receive a $1500 raise for each year you work for 35 years. a. How much will you earn during the 35th year? b. What is the total amount you earned over your 35-year career? 69. Theater Seating. At a theater, seats are arranged in a triangular pattern of rows with each succeeding row having one more seat than the previous row. You count the number of seats in the fourth row and determine that there are 26 seats. a. How many seats are in the first row? b. Now, suppose there are 30 rows of seats. How many total seats are there in the theater? 70. Mathematics. Find the exact sum of 1 e 1 3 e 1 5 e 1 c 1 23 e • C A T C H T H E M I S T A K E In Exercises 71–74, explain the mistake that is made. 71. Find the general, or nth, term of the arithmetic sequence 3, 4, 5, 6, 7, . . . . Solution: The common difference of this sequence is 1. d 5 1 The first term is 3. a1 5 3 The general term is an 5 a1 1 nd. an 5 3 1 n This is incorrect. What mistake was made? 72. Find the general, or nth, term of the arithmetic sequence 10, 8, 6, . . . . Solution: The common difference of this sequence is 2. d 5 2 The first term is 10. a1 5 10 The general term is an 5 a1 1 1n 2 12d. an 5 10 1 21n 2 12 This is incorrect. What mistake was made? 73. Find the sum a 10 k50 2n 1 1. Solution: The sum is given by Sn 5 n 2 1a1 1 an2, where n 5 10. Identify the 1st and 10th terms. a1 5 1 a10 5 21 Substitute a1 5 1, a10 5 21, and n 5 10 into Sn 5 n 2 1a1 1 an2. S10 5 10 2 11 1 212 5 110 This is incorrect. What mistake was made? 74. Find the sum 3 1 9 1 15 1 21 1 27 1 33 1 c1 87. Solution: This is an arithmetic sequence with common difference of 6. d 5 6 The general term is given by an 5 a1 1 1n 2 12d. an 5 3 1 1n 2 126 87 is the 15th term of the series. a15 5 3 1 115 2 126 5 87 The sum of the series is Sn 5 n 2 1an 2 a12. S15 5 15 2 187 2 32 5 630 This is incorrect. What mistake was made? 75. An arithmetic sequence and a finite arithmetic series are the same. 76. The sum of all infinite and finite arithmetic series can always be found. 77. An alternating sequence cannot be an arithmetic sequence. 78. The common difference of an arithmetic sequence is always positive. In Exercises 75–78, determine whether each statement is true or false. • C O N C E P T U A L 12.3 Geometric Sequences and Series 1117 12.3.1 Geometric Sequences In Section 12.2, we discussed arithmetic sequences, where successive terms had a common difference. In other words, each term was found by adding the same constant to the pre vious term. In this section we discuss geometric sequences, where successive terms have a common ratio. In other words, each term is found by multiplying the previous term by the same constant. The sequence 4, 12, 36, 108, c is geometric because each successive term is found by multiplying the previous term by 3. S K I L L S O B J E C T I V E S ■ ■Determine if a sequence is geometric. ■ ■Find the general, nth, term of a geometric sequence. ■ ■Evaluate finite geometric series and infinite geometric series (if possible). C O N C E P T U A L O B J E C T I V ES ■ ■Understand that the common ratio of a geometric sequence can be positive or negative. ■ ■Derive the formula for the nth term of a geometric sequence in terms of n, the first term of the sequence, and the common ratio. ■ ■Understand why it is not possible to evaluate all infinite geometric series. 12.3 GEOMETRIC SEQUENCES AND SERIES DEFINITION Geometric Sequences A sequence is geometric if each term in the sequence is found by multiplying the previous term by a number r, so that an11 5 r⋅an. Because an11 an 5 r, the number r is called the common ratio. 12.3.1 S KI L L Determine if a sequence is geometric. 12.3.1 C ON C E P T U A L Understand that the common ratio of a geometric sequence can be positive or negative. 79. Find the sum a 1 1a 1 b2 1 1a 1 2b2 1 c1 1a 1 nb2. 80. Find the sum a 30 k5 229 ln ek. 81. The wave number, v (reciprocal of wave length), of certain light waves in the spectrum of light emitted by hydrogen is given by v 5 R a 1 k 2 2 1 n2b, n . k, where R 5 109,678. A series of lines is given by holding k constant and varying the value of n. Suppose k 5 2 and n 5 3, 4, 5, c ⋅ Find the limit of the wave number of the series. 82. In a certain arithmetic sequence a1 5 24 and d 5 6. If Sn 5 570, find the value of n. • C H A L L E N G E 83. Use a graphing calculator “SUM” to sum the natural numbers from 1 to 100. 84. Use a graphing calculator to sum the even natural numbers from 1 to 100. 85. Use a graphing calculator to sum the odd natural numbers from 1 to 100. 86. Use a graphing calculator to sum a 30 n51 122n 1 52. Compare it with your answer to Exercise 43. 87. Use a graphing calculator to sum a 100 n51 3259 1 51n 2 124. 88. Use a graphing calculator to sum a 200 n51 C218 1 4 5 1n 2 12D. • T E C H N O L O G Y 1118 CHAPTER 12 Sequences, Series, and Probability 12.3.2 The General (nth) Term of a Geometric Sequence To find a formula for the general, or nth, term of a geometric sequence, write out the first several terms and look for a pattern. First term, n 5 1. a1 Second term, n 5 2. a2 5 a1⋅r Third term, n 5 3. a3 5 a2⋅r 5 1a1⋅r2 ⋅r 5 a1⋅r 2 Fourth term, n 5 4. a4 5 a3⋅r 5 1a1⋅r 22 ⋅r 5 a1⋅r 3 In general, the nth term is given by an 5 a1⋅r n21. EXAMPLE 1 Identifying the Common Ratio in Geometric Sequences Find the common ratio for each of the geometric sequences. a. 5, 20, 80, 320, c b. 1, 21 2, 1 4, 21 8, c c. $5000, $5500, $6050, $6655, c Solution (a): Label the terms. a1 5 5, a2 5 20, a3 5 80, a4 5 320, c Find the ratio r 5 an11 an . r 5 a2 a1 5 20 5 5 4 r 5 a3 a2 5 80 20 5 4 r 5 a4 a3 5 320 80 5 4 The common ratio is 4. Solution (b): Label the terms. a1 5 1, a2 5 21 2, a3 5 1 4, a4 5 21 8, c Find the ratio r 5 an11 an . r 5 a2 a1 5 21/2 1 5 21 2 r 5 a3 a2 5 1/4 21/2 5 21 2 r 5 a4 a3 5 21/8 1/4 5 21 2 The common ratio is 21 2. Solution (c): Label the terms. a1 5 $5000, a2 5 $5500, a3 5 $6050, a4 5 $6655, c Find the ratio r 5 an11 an . r 5 a2 a1 5 $5500 $5000 5 1.1 r 5 a3 a2 5 $6050 $5500 5 1.1 r 5 a4 a3 5 $6655 $6050 5 1.1 The common ratio is 1.1. Y OUR TU R N Find the common ratio of each geometric sequence. a. 1, 23, 9, 227, c b. 320, 80, 20, 5, c ▼ [CONCEPT CHECK] TRUE OR FALSE If the geometric sequence is alternating, then the common ratio is negative. ANSWER True ▼ ▼ A N S W E R a. 23 b. 1 4 or 0.25 12.3.2 SKI LL Find the general, nth, term of a geometric sequence. 12.3.2 CON CEPTUAL Derive the formula for the nth term of a geometric sequence in terms of n, the first term of the sequence, and the common ratio. EXAMPLE 2 Finding the nth Term of a Geometric Sequence Find the 7th term of the sequence 2, 10, 50, 250, . . . . Solution: Identify the common ratio. r 5 10 2 5 50 10 5 250 50 5 5 Identify the first 1n 5 12 term. a1 5 2 Substitute a1 5 2 and r 5 5 into an 5 a1⋅r n21. an 5 2⋅5n21 Substitute n 5 7 into an 5 2⋅5n21. a7 5 2⋅5721 5 2⋅56 5 31,250 The 7th term of the geometric sequence is 31,250. YOUR T UR N Find the 8th term of the sequence 3, 12, 48, 192, . . . . ▼ EXAMPLE 3 Finding the Geometric Sequence Find the geometric sequence whose 5th term is 0.01 and whose common ratio is 0.1. Solution: Label the common ratio and 5th term. a5 5 0.01 and r 5 0.1 Substitute a5 5 0.01, n 5 5, and r 5 0.1 into an 5 a1⋅r n21. 0.01 5 a1⋅10.12521 Solve for a1. a1 5 0.01 10.124 5 0.01 0.0001 5 100 The geometric sequence that starts at 100 and has a common ratio of 0.1 is 100, 10, 1, 0.1, 0.01, . . . . Y OUR T UR N Find the geometric sequence whose 4th term is 3 and whose common ratio is 1 3. ▼ ▼ A N S W E R 49,152 ▼ A N S W E R 81, 27, 9, 3, 1, . . . [CONCEPT CHECK] For the geometric sequence 3, 12, 48, 192, . . . find the general nth, term of the sequence. ANSWER an 5 3 3 4(n21) ▼ 12.3.3 Geometric Series The sum of the terms of a geometric sequence is called a geometric series. a1 1 a1⋅r 1 a1⋅r 2 1 a1⋅r 3 1 c If we only sum the first n terms of a geometric sequence, the result is a finite geometric series given by Sn 5 a1 1 a1⋅r 1 a1⋅r 2 1 a1⋅r 3 1 c 1 a1⋅r n21 12.3.3 S K I L L Evaluate finite geometric series and infinite geometric series (if possible). 12.3.3 C O N C E P T U A L Understand why it is not possible to evaluate all infinite geometric series. THE nTH TERM OF A GEOMETRIC SEQUENCE The nth term of a geometric sequence with common ratio r is given by an 5 a1⋅r n21 for n $ 1 12.3 Geometric Sequences and Series 1119 1120 CHAPTER 12 Sequences, Series, and Probability To develop a formula for the nth partial sum, we multiply the above equation by r: r⋅Sn 5 a1⋅r 1 a1⋅r 2 1 a1⋅r 3 1 c 1 a1⋅r n21 1 a1⋅r n Subtracting the second equation from the first equation, we find that all of the terms on the right side drop out except the first term in the first equation and the last term in the second equation: Sn 5 a1 1 a1⋅r 1 a1⋅r 2 1 c 1 a1r n21 2rSn 5 2 a1⋅r 2 a1⋅r 2 2 c 2 a1r n21 2 a1r n Sn 2 rSn 5 a1 2a1r n Factor the Sn out of the left side and the a1 out of the right side: Sn11 2 r2 5 a111 2 r n2 Divide both sides by 11 2 r2, assuming r 2 1. The result is a general formula for the sum of a finite geometric series: Sn 5 a1 11 2 r n2 11 2 r2 r 2 1 EVALUATING A FINITE GEOMETRIC SERIES The sum of the first n terms of a geometric sequence, called a finite geometric series, is given by the formula Sn 5 a1 11 2 r n2 11 2 r2 r 2 1 It is important to note that a finite geometric series can also be written in sigma (summation) notation: Sn 5 a n k51 a1⋅r k21 5 a1 1 a1⋅r 1 a1⋅r 2 1 a1⋅r 3 1 c 1 a1⋅r n21 STUDY TIP The underscript k 5 1 applies only when the summation starts at the a1 term. It is important to note which term is the starting term. EXAMPLE 4 Evaluating a Finite Geometric Series Evaluate the finite geometric series. a. a 13 k51 3⋅10.42k21 b. The first nine terms of the series 1 1 2 1 4 1 8 1 16 1 32 1 64 1 c Solution (a): Identify a1, n, and r. a1 5 3, n 5 13, and r 5 0.4 Substitute a1 5 3, n 5 13, and r 5 0.4 into sn 5 a1 11 2 r n2 11 2 r2 . S13 5 3 A1 2 0.413B 11 2 0.42 Simplify. S13 < 4.99997 Solution (b): Identify the first term and common ratio. a1 5 1 and r 5 2 Substitute a1 5 1 and r 5 2 into Sn 5 a1 11 2 r n2 11 2 r2 . Sn 5 11 2 2n2 11 2 22 To sum the first nine terms, let n 5 9. S9 5 A1 2 29B 11 2 22 Simplify. S9 5 511 The sum of an infinite geometric sequence is called an infinite geometric series. Some infinite geometric series converge (yield a finite sum), and some diverge (do not have a finite sum). For example, 1 2 1 1 4 1 1 8 1 1 16 1 1 32 1 c 1 1 2n 1 c 5 1 1converges2 2 1 4 1 8 1 16 1 32 1 c 1 2n 1 c 1diverges2 For infinite geometric series that converge, the partial sum Sn approaches a single number as n gets large. The formula used to evaluate a finite geometric series Sn 5 a1 11 2 r n2 11 2 r2 can be extended to an infinite geometric series for certain values of r. If 0 r 0 , 1, then when r is raised to a power, it continues to get smaller, approaching 0. For those values of r, the infinite geometric series converges to a finite sum. Let n S q; then a1 11 2 r n2 11 2 r2 S a1 11 2 02 11 2 r2 5 a1 1 1 2 r, if 0 r 0 , 1. EVALUATING AN INFINITE GEOMETRIC SERIES The sum of an infinite geometric series is given by the formula a q n50 a1⋅r n 5 a1 1 11 2 r2 0 r 0 , 1 STUDY TIP The formula used to evaluate an infinite geometric series is: 1First term2 1 2 1Ratio2 EXAMPLE 5 Determining Whether the Sum of an Infinite Series Exists Determine whether the sum exists for each of the geometric series. a. 3 1 15 1 75 1 375 1 c b. 8 1 4 1 2 1 1 1 1 2 1 1 4 1 1 8 1 c Solution (a): Identify the common ratio. r 5 5 Since 5 is greater than 1, the sum does not exist . r 5 5 . 1 Solution (b): Identify the common ratio. r 5 1 2 Since 1 2 is less than 1, the sum exists . r 5 1 2 , 1 Y OUR T UR N Determine whether the sum exists for the following geometric series. a. 81, 9, 1, 1 9, c b. 1, 5, 25, 125, c ▼ Do you expect 1 4 1 1 12 1 1 36 1 1 64 1 c and 1 4 2 1 12 1 1 36 2 1 64 1 c to sum to the same number? The answer is no, because the second series is an alternating series and terms are both added and subtracted. Hence, we would expect the second series to sum to a smaller number than the first series sums to. ▼ A N S W E R a. yes b. no 12.3 Geometric Sequences and Series 1121 1122 CHAPTER 12 Sequences, Series, and Probability It is important to note the restriction on the common ratio r. The absolute value of the common ratio has to be strictly less than 1 for an infinite geometric series to converge. Otherwise the infinite geometric series diverges. EXAMPLE 6 Evaluating an Infinite Geometric Series Evaluate each infinite geometric series. a. 1 1 1 3 1 1 9 1 1 27 1 c b. 1 2 1 3 1 1 9 2 1 27 1 c Solution (a): Identify the first term and the common ratio. a1 5 1 r 5 1 3 Since 0 r 0 5 1 3 , 1, the sum of the series exists. Substitute a1 5 1 and r 5 1 3 into a q n50 a1⋅r n 5 a1 1 11 2 r2. 1 1 2 1/3 Simplify. 5 1 2/3 5 3 2 1 1 1 3 1 1 9 1 1 27 1 c5 3 2 Solution (b): Identify the first term and the common ratio. a1 5 1 r 5 2 1 3 Since 0 r 0 5 0 21 3 0 , 1, the sum of the series exists. Substitute a1 5 1 and r 5 2 1 3 into a q n50 a1⋅r n 5 a1 1 11 2 r2. Simplify. 5 1 1 2 121/32 5 1 1 1 11/32 5 1 4/3 5 3 4 1 2 1 3 1 1 9 2 1 27 1 c5 3 4 Notice that the alternating series summed to 3 4, whereas the positive series summed to 3 2. Y OUR TU R N Find the sum of each infinite geometric series. a. 1 4 1 1 12 1 1 36 1 1 108 1 c b. 1 4 2 1 12 1 1 36 2 1 108 1 c ▼ ▼ A N S W E R a. 3 8 b. 3 16 [CONCEPT CHECK] TRUE OR FALSE All infinite series with a negative common ratio converge. ANSWER False ▼ STUDY TIP a q n50 a1⋅r n 5 First term 1 2 Ratio 5 1 12 1/3 STUDY TIP a q n50 a1⋅r n 5 First term 1 2 Ratio 5 1 12 121/32 EXAMPLE 7 Evaluating an Infinite Geometric Series Evaluate the infinite geometric series, if possible. a. a q n50 2a21 4b n b. a q n51 3⋅122n21 Solution (a): Identify a1 and r. a q n50 2 a21 4b n 5 2 2 1 2 1 1 8 2 1 32 1 1 128 2c r 5 2 1 4 r 5 2 1 4 r 5 2 1 4 e e e a1 { Applications Suppose you are given a job offer with a guaranteed percentage raise per year. What will your annual salary be 10 years from now? That answer can be obtained using a geometric sequence. Suppose you want to make voluntary contributions to a retirement account directly debited from your paycheck every month. Suppose the account earns a fixed percentage rate: How much will you have in 30 years if you deposit $50 a month? What is the difference in the total you will have in 30 years if you deposit $100 a month instead? These important questions about your personal finances can be answered using geometric sequences and series. Since 0 r 0 5 0 2 1 4 0 5 1 4 , 1, the infinite a q n50 a1⋅r n 5 a1 11 2 r2 geometric series converges. Let a1 5 2 and r 5 2 1 4. 5 2 31 2 121/424 Simplify. 5 2 1 1 1/4 5 2 5/4 5 8 5 This infinite geometric series converges. a q n50 2 a21 4b n 5 8 5 Solution (b): Identify a1 and r. a q n51 3⋅122n21 5 3 1 6 1 12 1 24 1 48 1 c Since r 5 2 . 1, this infinite geometric series diverges. r 5 2 r 5 2 e e a1 { EXAMPLE 8 Future Salary: Geometric Sequence Suppose you are offered a job as an event planner for the PGA Tour. The starting salary is $45,000, and employees are given a 5% raise per year. What will your annual salary be during the 10th year with the PGA Tour? Solution: Every year the salary is 5% more than the previous year. Label the year 1 salary. a1 5 45,000 Calculate the year 2 salary. a2 5 1.05⋅a1 Calculate the year 3 salary. a3 5 1.05⋅a2 5 1.0511.05⋅a12 5 11.0522a1 Calculate the year 4 salary. a4 5 1.05⋅a3 5 1.0511.0522a1 5 11.0523a1 Identify the year n salary. an 5 1.05n21a1 Substitute n 5 10 and a1 5 45,000. a10 5 11.0529⋅45,000 Simplify. a10 < 69,809.77 During your 10th year with the company your salary will be $69,809.77. YOUR T UR N Suppose you are offered a job with AT&T at $37,000 per year with a guaranteed raise of 4% after every year. What will your annual salary be after 15 years with the company? ▼ STUDY TIP a10 5 45,00011.0529 < 69,809.77 ▼ A N S W E R $64,072.03 12.3 Geometric Sequences and Series 1123 1124 CHAPTER 12 Sequences, Series, and Probability EXAMPLE 9 Savings Growth: Geometric Series Karen has maintained acrylic nails by paying for them with money earned from a part-time job. After hearing a lecture from her economics professor on the importance of investing early in life, she decides to remove the acrylic nails, which cost $50 per month, and do her own manicures. She has that $50 automatically debited from her checking account on the first of every month and put into a money market account that receives 3% interest compounded monthly. What will the balance be in the money market account exactly 2 years from the day of her initial $50 deposit? Solution: Recall the compound interest formula. A 5 P a1 1 r nb nt Substitute r 5 0.03 and n 5 12 into the compound interest formula. A 5 P a1 1 0.03 12 b 12t 5 P11.0025212t Let t 5 n 12, where n is the number of months of the investment: An 5 P11.00252n The first deposit of $50 will gain interest for 24 months. A24 5 5011.0025224 The second deposit of $50 will gain interest for 23 months. A23 5 5011.0025223 The third deposit of $50 will gain interest for 22 months. A22 5 5011.0025222 The last deposit of $50 will gain interest for 1 month. A1 5 5011.002521 Sum the amounts accrued from the 24 deposits. A1 1 A2 1 c1 A24 5 5011.00252 1 5011.002522 1 5011.002523 1 c1 5011.0025224 Identify the first term and common ratio. a1 5 5011.00252 and r 5 1.0025 Sum the first n terms of a geometric series. Sn 5 a1 11 2 r n2 11 2 r2 Substitute n 5 24, a1 5 5011.00252, and r 5 1.0025. S24 5 5011.00252 A1 2 1.002524B 11 2 1.00252 Simplify. S24 < 1238.23 Karen will have $1238.23 saved in her money market account in 2 years. Y OUR TU R N Repeat Example 9 with Karen putting $100 (instead of $50) in the same money market. Assume she does this for 4 years (instead of 2 years). ▼ ▼ A N S W E R $5105.85 than or equal to 1, the infinite geometric series diverges and the sum does not exist. Many real-world applications involve geometric sequences and series, such as growth of salaries and annuities through percentage increases. Finite Geometric Series: a n k50 a1r k 5 a1 11 2 r n2 11 2 r2 Infinite Geometric Series: a q k50 a1r k 5 a1 1 11 2 r2 0 r 0 , 1 In this section, we discussed geometric sequences, in which each successive term is found by multiplying the previous term by a constant, so that an11 5 r⋅an. That constant, r, is called the common ratio. The nth term of a geometric sequence is given by an 5 a1r n21, n $ 1 or an11 5 a1r n, n $ 0. The sum of the terms of a geometric sequence is called a geometric series. Finite geometric series converge to a number. Infinite geometric series converge to a number if the absolute value of the common ratio is less than 1. If the absolute value of the common ratio is greater [SEC TION 12 .3] S U MM A RY [SEC TION 12 .3] E X E R C I S E S • S K I L L S In Exercises 1–8, determine whether the sequence is geometric. If it is, find the common ratio. 1. 1, 3, 9, 27, c 2. 2, 4, 8, 16, c 3. 1, 4, 9, 16, 25, c 4. 1, 1 4, 1 9, 1 16, c 5. 8, 4, 2, 1, c 6. 8, 24, 2, 21, c 7. 800, 1360, 2312, 3930.4, c 8. 7, 15.4, 33.88, 74.536, c In Exercises 9–16, write the first five terms of the geometric series. 9. a1 5 6 r 5 3 10. a1 5 17 r 5 2 11. a1 5 1 r 5 24 12. a1 5 23 r 5 22 13. a1 5 10,000 r 5 1.06 14. a1 5 10,000 r 5 0.8 15. a1 5 2 3 r 5 1 2 16. a1 5 1 10 r 5 21 5 In Exercises 17–24, write the formula for the nth term of the geometric series. 17. a1 5 5 r 5 2 18. a1 5 12 r 5 3 19. a1 5 1 r 5 23 20. a1 5 24 r 5 22 21. a1 5 1000 r 5 1.07 22. a1 5 1000 r 5 0.5 23. a1 5 16 3 r 5 21 4 24. a1 5 1 200 r 5 5 In Exercises 25–30, find the indicated term of the geometric sequence. 25. 7th term of the sequence 22, 4, 28, 16, c 26. 10th term of the sequence 1, 25, 25, 2225, c 27. 13th term of the sequence 1 3, 2 3, 4 3, 8 3, c 28. 9th term of the sequence 100, 20, 4, 0.8, c 29. 15th term of the sequence 1000, 50, 2.5, 0.125, c 30. 8th term of the sequence 1000, 2800, 640, 2512, c In Exercises 31–40, find the sum of the finite geometric series. 31. 1 3 1 2 3 1 22 3 1 c1 212 3 32. 1 1 1 3 1 1 32 1 1 33 1 c1 1 310 33. 2 1 6 1 18 1 54 1 c1 2A39B 34. 1 1 4 1 16 1 64 1 c1 49 35. a 10 n50 210.12n 36. a 11 n50 310.22n 37. a 8 n51 2132n21 38. a 9 n51 2 3 152n21 39. a 13 k50 2k 40. a 13 k50 a1 2b k In Exercises 41–54, find the sum of the infinite geometric series, if possible. 41. a q n50 a1 2b n 42. a q n51 a1 3b n 43. a q n51 a 2 1 3 b n 44. a q n50 a2 1 2 b n 45. a q n50 1n 46. a q n50 1.01n 47. a q n50 29 a1 3b n 48. a q n50 28 a 2 1 2 b n 49. a q n50 10,00010.052n 50. a q n50 20010.042n 51. a q n51 0.4n 52. 0.3 1 0.03 1 0.003 1 0.0003 1 c 53. a q n50 0.99n 54. a q n50 a5 4b n • A P P L I C A T I O N S 55. Salary. Jeremy is offered a government job with the Department of Commerce. He is hired on the “GS” scale at a base rate of $34,000 with 2.5% increases in his salary per year. Calculate what his salary will be after he has been with the Department of Commerce for 12 years. 56. Salary. Alison is offered a job with a small start-up company that wants to promote loyalty to the company with incentives for employees to stay with the company. The company offers her a starting salary of $22,000 with a guaranteed 15% raise per year. What will her salary be after she has been with the company for 10 years? 57. Depreciation. Brittany, a graduating senior in high school, receives a laptop computer as a graduation gift from her Aunt Jeanine so that she can use it when she gets to the University of Alabama. If the laptop costs $2000 new and depreciates 50% per year, write a formula for the value of the laptop n years after it was purchased. How much will the laptop be worth when Brittany graduates from college (assuming she will graduate in 4 years)? How much will it be worth when she finishes graduate school? Assume graduate school is another 3 years. 58. Depreciation. Derek is deciding between a new Honda Accord and the BMW 325 series. The BMW costs $35,000 and the Honda costs $25,000. If the BMW depreciates at 20% per year and the Honda depreciates at 10% per year, find formulas for the value of each car n years after it is purchased. Which car is worth more in 10 years? 12.3 Geometric Sequences and Series 1125 1126 CHAPTER 12 Sequences, Series, and Probability 59. Bungee Jumping. A bungee jumper rebounds 70% of the height jumped. Assuming the bungee jump is made with a cord that stretches to 100 feet, how far will the bungee jumper travel upward on the fifth rebound? Laurence Fordyce/Eye Ubiquitous/Corbis Images 60. Bungee Jumping. A bungee jumper rebounds 65% of the height jumped. Assuming the bungee cord stretches 200 feet, how far will the bungee jumper travel upward on the eighth rebound? 61. Population Growth. One of the fastest-growing universities in the country is the University of Central Florida. The student populations each year starting in 2000 were 36,000, 37,800, 39,690, 41,675, c. If this rate continued, how many students were at UCF in 2010? 62. Website Hits. The website for a band has noticed that every week the number of hits to its website increases 5%. If there were 20,000 hits this week, how many will there be exactly 52 weeks from now? 63. Rich Man’s Promise. A rich man promises that he will give you $1000 on January 1, and every day after that, he will pay you 90% of what he paid you the day before. How many days will it take before you are making less than $1? How much will the rich man pay out for the entire month of January? Round answers to the nearest dollar. 64. Poor Man’s Clever Deal. A poor man promises to work for you for $0.01 the first day, $0.02 on the second day, $0.04 on the third day; his salary will continue to double each day. If he started on January 1, how much would he be paid to work on January 31? How much total would he make during the month? Round answers to the nearest dollar. 65. Investing Lunch. A newlywed couple decides to stop going out to lunch every day and instead brings their lunch. They estimate it will save them $100 per month. They invest that $100 on the first of every month into an account that is compounded monthly and pays 5% interest. How much will be in the account at the end of 3 years? 66. Pizza as an Investment. A college freshman decides to stop ordering late-night pizzas (for both health and cost reasons). He realizes that he has been spending $50 a week on pizzas. Instead, he deposits $50 into an account that compounds weekly and pays 4% interest. How much money will be in the account in 52 weeks? 67. Tax-Deferred Annuity. Dr. Schober contributes $500 from her paycheck (weekly) to a tax-deferred investment account. Assuming the investment earns 6% and is compounded weekly, how much will be in the account in 26 weeks? 52 weeks? 68. Saving for a House. If a new graduate decides she wants to save for a house and she is able to put $300 every month into an account that earns 5% compounded monthly, how much will she have in the account in 5 years? 69. House Values. In 2017, you buy a house for $240,000. The value of the house appreciates 3.1% per year, on the average. How much is the house worth after 15 years? 70. The Bouncing Ball Problem. A ball is dropped from a height of 9 feet. Assume that on each bounce, the ball rebounds to one-third of its previous height. Find the total distance that the ball travels. 71. Probability. A fair coin is tossed repeatedly. The probability that the first head occurs on the nth toss is given by the function p1n2 5 A1 2B n, where n $ 1. Show that a q n51 a1 2b n 5 1.0 72. Salary. Suppose you work for a supervisor who gives you two different options to choose from for your monthly pay. Option 1: The company pays you $0.01 for the first day of work, $0.02 for the second day, $0.04 for the third day, $0.08 for the fourth day, and so on for 30 days. Option 2: You can receive a check right now for $10 million. Which pay option is better? How much better is it? • C A T C H T H E M I S T A K E In Exercises 73–76, explain the mistake that is made. 73. Find the nth term of the geometric sequence: 21, 1 3, 21 9, 1 27, c. Solution: Identify the first term and common ratio. a1 5 21 and r 5 1 3 Substitute a1 5 21 and r 5 1 3 into an 5 a1⋅r n21. an 5 1212 ⋅a1 3b n21 Simplify. an 5 21 3n21 This is incorrect. What mistake was made? 74. Find the sum of the first n terms of the finite geometric series: 2, 4, 8, 16, c. Solution: Write the sum in sigma notation. a n k51 122k Identify the first term and common ratio. a1 5 1 and r 5 2 Substitute a1 5 1 and r 5 2 into Sn 5 a1 11 2 r n2 11 2 r2 . Sn 5 1 11 2 2n2 11 2 22 Simplify. Sn 5 2n 2 1 This is incorrect. What mistake was made? 75. Find the sum of the finite geometric series a 8 n51 4 1232n. Solution: Identify the first term and common ratio. a1 5 4 and r 5 23 Substitute a1 5 4 and r 5 23 into Sn 5 a1 11 2 r n2 11 2 r2 . Sn 5 4 31 2 1232n4 31 2 12324 5 4 31 2 1232n4 4 Simplify. Sn 5 31 2 1232n4 Substitute n 5 8. S8 5 31 2 123284 5 2 6,560 This is incorrect. What mistake was made? 76. Find the sum of the infinite geometric series a q n51 2⋅3n21. Solution: Identify the first term and common ratio. a1 5 2 and r 5 3 Substitute a1 5 2 and r 5 3 into Sq 5 a1 1 11 2 r2. Sq 5 2 1 11 2 32 Simplify. Sq 5 21 This is incorrect. The series does not sum to 21. What mistake was made? 77. An alternating sequence cannot be a geometric sequence. 78. All finite and infinite geometric series can always be evaluated. 79. The common ratio of a geometric sequence can be positive or negative. 80. An infinite geometric series can be evaluated if the common ratio is less than or equal to 1. In Exercises 77–80, determine whether each statement is true or false. • C O N C E P T U A L 81. State the conditions for the sum a 1 a⋅b 1 a⋅b2 1 c1 a⋅bn 1 c to exist. Assuming those conditions are met, find the sum. 82. Find the sum of a 20 k50 log 102k. 83. Represent the repeating decimal 0.474747. . . as a fraction (ratio of two integers). 84. Suppose the sum of an infinite geometric series is S 5 2 1 2 x, where x is a variable. a. Write out the first five terms of the series. b. For what values of x will the series converge? • C H A L L E N G E 85. Sum the series: a 50 k51 1222k21. Apply a graphing utility to confirm your answer. 86. Does the sum of the infinite series a q n50A1 3Bn exist? Use a graphing calculator to find it. 87. Apply a graphing utility to plot y1 5 1 1 x 1 x2 1 x3 1 x4 and y2 5 1 1 2 x, and let the range of x be 320.5, 0.54. Based on what you see, what do you expect the geometric series a q n50 xn to sum to? 88. Apply a graphing utility to plot y1 5 1 2 x 1 x2 2 x3 1 x4 and y2 5 1 1 1 x, and let x range from 320.5, 0.54. Based on what you see, what do you expect the geometric series a q n50 1212nxn to sum to? 89. Apply a graphing utility to plot y1 5 1 1 2x 1 4x2 1 8x3 1 16x4 and y2 5 1 1 2 2x, and let x range from 320.3, 0.34. Based on what you see, what do you expect the geometric series a q n50 12x2n to sum to? • T E C H N O L O G Y 12.3 Geometric Sequences and Series 1127 1128 CHAPTER 12 Sequences, Series, and Probability 12.4.1 Mathematical Induction Is the expression n2 2 n 1 41 always a prime number if n is a natural number? Your instinct may lead you to try a few values for n. It appears that the statement might be true for all natural numbers. However, what about when n 5 41? n2 2 n 1 41 5 14122 2 41 1 41 5 412 We find that when n 5 41, n2 2 n 1 41 is not prime. The moral of the story is that just because a pattern seems to exist for some values, the pattern is not necessarily true for all values. We must look for a way to show whether a statement is true for all values. In this section we talk about mathematical induction, which is a way to show a statement is true for all values. Mathematics is based on logic and proof (not assumptions or belief). One of the most famous mathematical statements was Fermat’s last theorem. Pierre de Fermat (1601–1665) conjectured that there are no positive integer values for x, y, and z such that xn 1 yn 5 zn, if n $ 3. Although mathematicians believed that this theorem was true, no one was able to prove it until 350 years after the assumption was made. Professor Andrew Wiles at Princeton University received a $50,000 prize for successfully proving Fermat’s last theorem in 1994. Mathematical induction is a technique used in college algebra and even in very advanced mathematics to prove many kinds of mathematical statements. In this section, you will use it to prove statements like “if x . 1, then xn . 1 for all natural numbers n.” The principle of mathematical induction can be illustrated by a row of standing dominos, as in the picture. We make two assumptions: 1. The first domino is knocked down. 2. If a domino is knocked down, then the domino immediately following it will also be knocked down. If both of these assumptions are true, then it is also true that all of the dominos will fall. S K I L L S O B J E C T I V E ■ ■Prove mathematical statements using mathematical induction. C O N C E P T U A L O B J E C T I V E ■ ■Understand that just because there appears to be a pattern in a mathematical expression, the pattern is not necessarily true for all values. 12.4 MATHEMATICAL INDUCTION 12.4.1 SKILL Prove mathematical statements using mathematical induction. 12.4.1 CO NC EPTUAL Understand that just because there appears to be a pattern in a mathematical expression, the pattern is not necessarily true for all values. [CONCEPT CHECK] TRUE OR FALSE If you can show that a statement is true for n 5 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10, then we can claim it is true for all n. ANSWER False ▼ n n2 2 n 1 41 PRIME? 1 41 Yes 2 43 Yes 3 47 Yes 4 53 Yes 5 61 Yes PRINCIPLE OF MATHEMATICAL INDUCTION Let Sn be a statement involving the positive integer n. To prove that Sn is true for all positive integers, the following steps are required. Step 1: Show that S1 is true. Step 2: Assume Sk is true and show that Sk11 is true 1k 5 positive integer2. Combining Steps 1 and 2 proves the statement is true for all positive integers. 12.4 Mathematical Induction 1129 EXAMPLE 1 Using Mathematical Induction Apply the principle of mathematical induction to prove this statement: If x . 1, then x n . 1 for all natural numbers n. Solution: STEP 1 Show the statement is true for n 5 1. x1 . 1 because x . 1 STEP 2 Assume the statement is true for n 5 k. x k . 1 Show the statement is true for k 1 1. Multiply both sides by x. x k⋅x . 1⋅x (Since x . 1, this step does not reverse the inequality sign.) Simplify. x k11 . x Recall that x . 1. x k11 . x . 1 Therefore, we have shown that x k11 . 1. This completes the induction proof. Thus, the following statement is true. “If x . 1, then x n . 1 for all natural numbers n.” EXAMPLE 2 Using Mathematical Induction Use mathematical induction to prove that n2 1 n is divisible by 2 for all natural numbers n. Solution: STEP 1 Show the statement we are testing is true for n 5 1. 12 1 1 5 2 2 is divisible by 2. 2 2 5 1 STEP 2 Assume the statement is true for n 5 k. k2 1 k 2 5 an integer Show it is true for k 1 1 where k $ 1. 1k 1 122 1 1k 1 12 2 0 an integer k2 1 2k 1 1 1 k 1 1 2 0 an integer Regroup terms. Ak2 1 kB 1 21k 1 12 2 0 an integer Ak2 1 kB 2 1 21k 1 12 2 0 an integer We assumed k2 1 k 2 5 an integer. an integer 1 1k 1 12 0 an integer Since k is a natural number, an integer 1 an integer 5 an integer This completes the induction proof. The following statement is true: “n2 1 n is divisible by 2 for all natural numbers n.” 1130 CHAPTER 12 Sequences, Series, and Probability Mathematical induction is often used to prove formulas for partial sums. is to (1) show the statement is true for n 5 1, then (2) assume the statement is true for n 5 k and show the statement must be true for n 5 k 1 1. The combination of Steps 1 and 2 proves the statement. Just because we believe something is true does not mean that it is. In mathematics we rely on proof. In this section, we discussed mathematical induction, a process of proving mathematical statements. The two-step procedure for mathematical induction [SEC TION 12 .4] S U MM A RY EXAMPLE 3 Proving a Partial-Sum Formula with Mathematical Induction Apply mathematical induction to prove the following partial-sum formula: 1 1 2 1 3 1 c1 n 5 n1n 1 12 2 for all positive integers n Solution: STEP 1 Show the formula is true for n 5 1. 1 5 111 1 12 2 5 2 2 5 1 STEP 2 Assume the formula is true for n 5 k. 1 1 2 1 3 1 c1 k 5 k1k 1 12 2 Show it is true for n 5 k 1 1. 1 1 2 1 3 1 c1 k 1 1k 1 12 0 1k 1 121k 1 22 2 1 1 2 1 3 1 c1 k 1 1k 1 12 0 1k 1 121k 1 22 2 k1k 1 12 2 1 1k 1 12 0 1k 1 121k 1 22 2 k1k 1 12 1 21k 1 12 2 0 1k 1 121k 1 22 2 k2 1 3k 1 2 2 0 1k 1 121k 1 22 2 1k 1 121k 1 22 2 5 1k 1 121k 1 22 2 This completes the induction proof. The following statement is true: “1 1 2 1 3 1 c1 n 5 n1n 1 12 2 for all positive integers n.” k1k 1 12 2 µ [SEC TION 12 .4] E X E RC I S E S • S K I L L S In Exercises 1–24, prove the statements using mathematical induction for all positive integers n. 1. n2 # n3 2. If 0 , x , 1, then 0 , xn , 1. 3. 2n # 2n 4. 5n , 5n11 5. n! . 2n n $ 4 (Show it is true for n 5 4, instead of n 5 1.2 6. 11 1 c2n $ nc c . 1 7. n1n 1 121n 2 12 is divisible by 3. 8. n3 2 n is divisible by 3. 9. n2 1 3n is divisible by 2. 10. n1n 1 121n 1 22 is divisible by 6. 11. 2 1 4 1 6 1 8 1 c1 2n 5 n1n 1 12 12. 1 1 3 1 5 1 7 1 c1 12n 2 12 5 n2 13. 1 1 3 1 32 1 33 1 c1 3n 5 3n11 2 1 2 14. 2 1 4 1 8 1 c1 2n 5 2n11 2 2 15. 12 1 22 1 32 1 c1 n2 5 n1n 1 1212n 1 12 6 16. 13 1 23 1 33 1 c1 n3 5 n21n 1 122 4 17. 1 1⋅2 1 1 2⋅3 1 1 3⋅4 1 c1 1 n1n 1 12 5 n n 1 1 18. 1 2⋅3 1 1 3⋅4 1 c1 1 1n 1 121n 1 22 5 n 21n 1 22 19. 11⋅22 1 12⋅32 1 13⋅42 1 c1 n1n 1 12 5 n1n 1 121n 1 22 3 20. 11⋅32 1 12⋅42 1 13⋅52 1 c1 n1n 1 22 5 n1n 1 1212n 1 72 6 21. 1 1 x 1 x2 1 x3 1 c1 xn21 5 1 2 x n 1 2 x x 2 1 22. 1 2 1 1 4 1 1 8 1 c1 1 2n 5 1 2 1 2n 23. The sum of an arithmetic sequence: a1 1 1a1 1 d2 1 1a1 1 2d 2 1 c1 3a1 1 1n 2 12d4 5 n 2 32a1 1 1n 2 12d4. 24. The sum of a geometric sequence: a1 1 a1r 1 a1r 2 1 c1 a1r n21 5 a1 a1 2 r n 1 2 r b. • A P P L I C A T I O N S The Tower of Hanoi. This is a game with three pegs and n disks (largest on the bottom and smallest on the top). The goal is to move this entire tower of disks to another peg (in the same order). The challenge is that you may move only one disk at a time, and at no time can a larger disk be resting on a smaller disk. You may want to first go online to www.mazeworks.com/hanoi/index/htm and play the game. Andy Washnik Tower of Hanoi 12.4 Mathematical Induction 1131 1132 CHAPTER 12 Sequences, Series, and Probability 25. What is the smallest number of moves needed if there are three disks? 26. What is the smallest number of moves needed if there are four disks? 27. What is the smallest number of moves needed if there are five disks? 28. What is the smallest number of moves needed if there are n disks? Prove it by mathematical induction. 29. Telephone Infrastructure. Suppose there are n cities that are to be connected with telephone wires. Apply mathematical induction to prove that the number of telephone wires required to connect the n cities is given by n1n 2 12 2 . Assume each city has to connect directly with any other city. 30. Geometry. Prove, with mathematical induction, that the sum of the interior angles of a regular polygon of n sides is given by the formula: 1n 2 2211802 for n $ 3. Hint: Divide a polygon into triangles. For example, a four-sided polygon can be divided into two triangles. A five-sided polygon can be divided into three triangles. A six-sided polygon can be divided into four triangles, and so on. 31. Assume Sk is true. If it can be shown that Sk11 is true, then Sn is true for all n, where n is any positive integer. 32. Assume S1 is true. If it can be shown that S2 and S3 are true, then Sn is true for all n, where n is any positive integer. In Exercises 31 and 32, determine whether each statement is true or false. • C O N C E P T U A L 33. Apply mathematical induction to prove: a n k51 k4 5 n1n 1 1212n 1 12A3n2 1 3n 2 1B 30 34. Apply mathematical induction to prove: a n k51 k5 5 n21n 1 122A2n2 1 2n 2 1B 12 35. Apply mathematical induction to prove: a 1 1 1 1b a1 1 1 2b a1 1 1 3b c a1 1 1 nb 5 n 1 1 36. Apply mathematical induction to prove that x 1 y is a factor of x 2n 2 y 2n. 37. Apply mathematical induction to prove: ln1c1⋅c2⋅c3 c cn2 5 lnc1 1 lnc2 1 c1 lncn • C H A L L E N G E 38. Use a graphing calculator to sum the series 11 • 22 1 12 • 32 1 13 • 42 1 ? ? ? 1 n 1n 1 12 on the left side, and evaluate the expression n1n 1 121n 1 22 3 on the right side for n 5 200. Do they agree with each other? Do your answers confirm the proof for Exercise 19? 39. Use a graphing calculator to sum the series 1 2 1 1 4 1 1 8 1 ? ? ? 1 1 2n on the left side, and evaluate the expression 1 2 1 2n on the right side for n 5 8. Do they agree with each other? Do your answers confirm the proof for Exercise 22? • T E C H N O L O G Y 12.5 The Binomial Theorem 1133 12.5.1 Binomial Coefficients A binomial is a polynomial that has two terms. The following are all examples of binomials: x2 1 2y a 1 3b 4x2 1 9 In this section, we will develop a formula for raising a binomial to a power n, where n is a positive integer. 1x2 1 2y26 1a 1 3b24 14x2 1 925 To begin, let’s start by writing out the expansions of 1a 1 b2n for several values of n. 1a 1 b20 5 1 1a 1 b21 5 a 1 b 1a 1 b22 5 a2 1 2ab 1 b2 1a 1 b23 5 a3 1 3a2b 1 3ab2 1 b3 1a 1 b24 5 a4 1 4a3b 1 6a2b2 1 4ab3 1 b4 1a 1 b25 5 a5 1 5a4b 1 10a3b2 1 10a2b3 1 5ab4 1 b5 All of the binomial expansions have several patterns. 1. The number of terms in each resulting polynomial is always one more than the power of the binomial n. Thus, there are n 1 1 terms in each expansion. n 5 3: 1a 1 b23 5 a3 1 3a2b 1 3ab2 1 b3 2. Each expansion has symmetry. For example, a and b can be interchanged, and you will arrive at the same expansion. Furthermore, the powers of a decrease by 1 in each successive term, and the powers of b increase by 1 in each successive term. 1a 1 b23 5 a3b0 1 3a2b1 1 3a1b2 1 a0b3 3. The sum of the powers of each term in the expansion is n. n 5 3: 1a 1 b23 5 a3b0 1 3a2b1 1 3a1b2 1 a0b3 4. The coefficients increase and decrease in a symmetric manner. 1a 1 b25 5 1a5 1 5a4b 1 10a3b2 1 10a2b3 1 5ab4 1 1b5 Using these patterns, we can develop a generalized formula for 1a 1 b2n. 1a 1 b2 n 5 an 1 an21b 1 an22b2 1 . . . 1 a2bn22 1 abn21 1 bn Where is a placeholder for a constant to be determined. four terms f 31053 21153 11253 01353 f f f f S K I L L S O B J E C T I V E S ■ ■Evaluate a binomial coefficient. ■ ■Use the binomial theorem to expand a binomial raised to a positive integer power. ■ ■Evaluate a binomial coefficient using Pascal’s triangle. ■ ■Find a particular term of a binomial expansion. C O N C E P T U A L O B J E C T I V ES ■ ■Understand why n has to be greater than or equal to k in order for “n choose k” to be defined. ■ ■Recognize patterns in binomial expansions that lead to the binomial theorem. ■ ■Understand why the nth row of Pascal’s triangle has n 1 1 coefficients. ■ ■Understand that it is not necessary to perform an entire binomial expansion if you only are looking for a single term. 12.5 THE BINOMIAL THEOREM 12.5.1 S KI L L Evaluate a binomial coefficient. 12.5.1 C ON C E P T U A L Understand why n has to be greater than or equal to k in order for “n choose k” to be defined. We know that there are n 1 1 terms in the expansion. We also know that the sum of the powers of each term must equal n. The powers increase and decrease by 1 in each successive term, and if we interchanged a and b, the result would be the same expansion. The question that remains is, what coefficients go in the blanks? We know that the coefficients must increase and then decrease in a symmetric order (similar to walking up and then down a hill). It turns out that the binomial coefficients are represented by a symbol that we will now define. DEFINITION Binomial Coefficients For nonnegative integers n and k, where n $ k, the symbol an kb is called the binomial coefficient and is defined by an kb 5 n! 1n 2 k2!k! an kb is read “n choose k.” You will see in the following sections that “n choose k” comes from combinations. [CONCEPT CHECK] TRUE OR FALSE ”n choose k” is always greater than or equal to 1 when n is greater than or equal to k. ANSWER True ▼ ▼ A N S W E R a. 84 b. 28 EXAMPLE 1 Evaluating a Binomial Coefficient Evaluate the following binomial coefficients. a. a6 4b b. a5 5b c. a4 0b d. a10 9 b Solution: Select the top number as n and the bottom number as k and substitute into the binomial coefficient formula an kb 5 n! 1n 2 k2!k!. a. a6 4b 5 6! 16 2 42!4! 5 6! 2!4! 5 6⋅5⋅4⋅3⋅2⋅1 12⋅1214⋅3⋅2⋅12 5 6⋅5 2 5 15 b. a5 5b 5 5! 15 2 52!5! 5 5! 0!5! 5 1 0! 5 1 1 5 1 c. a4 0b 5 4! 14 2 02!0! 5 4! 4!0! 5 1 0! 5 1 d. a10 9 b 5 10! 110 2 92!9! 5 10! 1!9! 5 10⋅9! 9! 5 10 Y OUR TU R N Evaluate the following binomial coefficients. a. a9 6b b. a8 6b ▼ Parts (b) and (c) of Example 1 lead to the general formulas: an nb 5 1 and an 0b 5 1 1134 CHAPTER 12 Sequences, Series, and Probability 12.5.2 Binomial Expansion Let’s return to the question of the binomial expansion and how to determine the coefficients: 1a 1 b2 n 5 an 1 an21b 1 an22b2 1 … 1 a2bn22 1 abn21 1 bn The symbol an kb is called a binomial coefficient because the coefficients in the blanks in the binomial expansion are equivalent to this symbol. 12.5.2 S KI L L Use the binomial theorem to expand a binomial raised to a positive integer power. 12.5.2 C ON C E P T U A L Recognize patterns in binomial expansions that lead to the binomial theorem. THE BINOMIAL THEOREM Let a and b be real numbers; then for any positive integer n, 1a 1 b2n 5 an 0ban1 an 1ban21b 1 an 2ban22b2 1 P 1 a n n2 2ba2bn22 1 a n n 2 1babn21 1 an nbbn or in sigma (summation) notation as 1a 1 b2n 5 a n k50 an kban2kbk EXAMPLE 2 Applying the Binomial Theorem Expand 1x 1 223 with the binomial theorem. Solution: Substitute a 5 x, b 5 2, n 5 3 into the equation of the binomial theorem. Expand the summation. 5 a3 0bx3 1 a3 1bx2 ⋅ 2 1 a3 2bx ⋅ 22 1 a3 3b23 Find the binomial coefficients. 5 x3 1 3x2⋅2 1 3x⋅22 1 23 Simplify. 5 x3 1 6x2 1 12x 1 8 YOUR T UR N Expand 1x 1 524 with the binomial theorem. 1x 1 223 5 a 3 k50 a3 kbx32k2k ▼ ▼ A N S W E R x4 1 20x3 1 150x2 1 500x 1 625 EXAMPLE 3 Applying the Binomial Theorem Expand 12x 2 324 with the binomial theorem. Solution: Substitute a 5 2x, b 5 23, n 5 4 into the equation of the binomial theorem. Expand the summation. 5 a4 0b 12x24 1 a4 1b 12x231232 1 a4 2b 12x2212322 1 a4 3b 12x2123231 a4 4b 12324 12x 2 324 5 a 4 k50 a4 kb 12x242k1232k [CONCEPT CHECK] In the binomial expansion of (3x 2 2)4, what is the first term and last term? ANSWER 81x4; 16 ▼ 12.5 The Binomial Theorem 1135 Find the binomial coefficients. 5 12x24 1 412x231232 1 612x2212322 1 412x212323 1 12324 Simplify. 5 16x4 2 96x3 1 216x2 2 216x 1 81 Y OUR TU R N Expand 13x 2 224 with the binomial theorem. ▼ 12.5.3 Pascal’s Triangle Instead of writing out the binomial theorem and calculating the binomial coefficients using factorials every time you want to do a binomial expansion, we now present an alternative, more convenient way of remembering the binomial coefficients, called Pascal’s triangle. Notice that the first and last number in every row is 1. Each of the other numbers is found by adding the two numbers directly above it. For example, 3 5 2 1 1 4 5 1 1 3 10 5 6 1 4 Let’s arrange values of an kb in a triangular pattern. a0 0b a1 0b a1 1b a2 0b a2 1b a2 2b a3 0b a3 1b a3 2b a3 3b a4 0b a4 1b a4 2b a4 3b a4 4b a5 0b a5 1b a5 2b a5 3b a5 4b a5 5b a6 0b a6 1b a6 2b a6 3b a6 4b a6 5b a6 6b It turns out that these numbers in Pascal’s triangle are exactly the coefficients in a binomial expansion. 1 1a 1 1b 1a2 1 2ab 1 1b2 1a3 1 3a2b 1 3ab2 1 1b3 1a4 1 4a3b 1 6a2b2 1 4ab3 1 1b4 1a5 1 5a4b 1 10a3b2 1 10a2b3 1 5ab4 1 1b5 ? 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 n 5 0 row n 5 1 row n 5 2 row n 5 3 row n 5 4 row n 5 5 row n 5 6 row n 5 0 row n 5 1 row n 5 2 row n 5 3 row n 5 4 row n 5 5 row n 5 6 row ▼ A N S W E R 81x42216x31216x2296x116 12.5.3 SKI LL Evaluate a binomial coefficient using Pascal’s triangle. 12.5.3 CO NC EPTUAL Understand why the nth row of Pascal’s triangle has n 1 1 coefficients. [CONCEPT CHECK] The top row of Pascal’s triangle corresponds to (a 1 b)0 (zero power or n 5 0 row). The next row of Pascal’s triangle corresponds to (a 1 b)1 (1st power or n 5 1 row). The next row of Pascal's triangle corresponds to (a 1 b)2 (2nd power or n 5 2 row) which is equal to a2 1 2ab 1 b2. So the n 5 0 row has one term, the n 5 1 row has two terms, and the n 5 2 row has three terms. How many terms does the nth row of Pascal's triangle have? ANSWER n 1 1 ▼ 1136 CHAPTER 12 Sequences, Series, and Probability The top row is called the zero row because it corresponds to the binomial raised to the zero power, n 5 0. Since each row in Pascal’s triangle starts and ends with a 1 and all other values are found by adding the two numbers directly above it, we can now easily calculate the sixth row. 1n 5 5 row2 1n 5 6 row2 EXAMPLE 4 Applying Pascal’s Triangle in a Binomial Expansion Use Pascal’s triangle to determine the binomial expansion of 1x 1 225. Solution: Write the binomial expansion with blanks for coefficients. 1x 1 22 5 5 x5 1 x4⋅2 1 x3⋅22 1 x2⋅23 1 x⋅24 1 25 Write the binomial coefficients in the fifth row of Pascal’s triangle. 1, 5, 10, 10, 5, 1 Substitute these coefficients into the blanks of the binomial expansion. 1x 1 225 5 1x5 1 5x4⋅2 1 10x3⋅22 1 10x2⋅23 1 5x⋅24 1 1⋅25 Simplify. 1x 1 225 5 x5 1 10x4 1 40x3 1 80x2 1 80x 1 32 YOUR T UR N Apply Pascal’s triangle to determine the binomial expansion of 1x 1 324. ▼ ▼ A N S W E R x4 1 12x3 1 54x2 1 108x 1 81 EXAMPLE 5 Applying Pascal’s Triangle in a Binomial Expansion Use Pascal’s triangle to determine the binomial expansion of 12x 1 524. Solution: Write the binomial expansion with blanks for coefficients. 12x 1 52 4 5 12x2 4 1 12x2 3⋅5 1 12x2 2⋅52 1 12x2⋅53 1 54 Write the binomial coefficients in the fourth row of Pascal’s triangle. 1, 4, 6, 4, 1 Substitute these coefficients into the blanks of the binomial expansion. 12x 1 524 5 112x24 1 412x23⋅5 1 612x22⋅52 1 412x2 ⋅53 1 1⋅54 Simplify. 12x 1 524 5 16x4 1 160x3 1 600x2 1 1000x 1 625 YOUR T UR N Use Pascal’s triangle to determine the binomial expansion of: a. 13x 1 223 b. 13x 2 225 ▼ STUDY TIP Since the top row of Pascal’s triangle is called the zero row, the fifth row is the row with 6 coefficients. The nth row is the row with n 1 1 coefficients. ▼ A N S W E R a. 27x3 1 54x2 1 36x 1 8 b. 243x5 2 810x4 1 1080x3 2 720x2 1 240x 2 32 12.5 The Binomial Theorem 1137 12.5.4 Finding a Particular Term of a Binomial Expansion What if we don’t want to find the entire expansion, but instead want just a single term? For example, what is the fourth term of 1a 1 b25? 12.5.4 SKI LL Find a particular term of a binomial expansion. 12.5.4 CO NC EPTUAL Understand that it is not necessary to perform an entire binomial expansion if you only are looking for a single term. WORDS MATH Recall the sigma notation. 1a 1 b2n 5 a n k50 an kban2kbk Let n 5 5. 1a 1 b25 5 a 5 k50 a5 kba52kbk Expand. 1a 1 b25 5 a5 0ba5 1 a5 1ba4b 1 a5 2ba3b2 1 a5 3ba2b3 1 a5 4bab4 1 a5 5bb5 Simplify the fourth term. 10a2b3 fourth term f [CONCEPT CHECK] TRUE OR FALSE To find a particular term of a binomial expansion, you must always first perform the entire binomial expansion. ANSWER False ▼ FINDING A PARTICULAR TERM OF A BINOMIAL EXPANSION The 1r 1 12 term of the expansion 1a 1 b2n is an rban2rbr. EXAMPLE 6 Finding a Particular Term of a Binomial Expansion Find the 5th term of the binomial expansion of 12x 2 726. Solution: Recall that the r 1 1 term of 1a 1 b2n is an rban2rbr. For the 5th term, let r 5 4. an 4ban24b4 For this expansion, let a 5 2x, b 5 27, n 5 6. a6 4b 12x262412724 Note that a6 4b 5 15. 1512x2212724 Simplify. 144,060x2 Y OUR TU R N What is the third term of the binomial expansion of 13x 2 225? ▼ ▼ A N S W E R 1080x3 1138 CHAPTER 12 Sequences, Series, and Probability In Exercises 1–10, evaluate the binomial coefficients. 1. a7 3b 2. a8 2b 3. a10 8 b 4. a23 21b 5. a17 0 b 6. a100 0 b 7. a99 99b 8. a52 52b 9. a48 45b 10. a29 26b In Exercises 11–32, expand the expression using the binomial theorem. 11. 1x 1 224 12. 1x 1 325 13. 1y 2 325 14. 1y 2 424 15. 1x 1 y25 16. 1x 2 y26 17. 1x 1 3y23 18. 12x 2 y23 19. 15x 2 223 20. 1a 2 7b23 21. a1 x 1 5yb 4 22. a2x 1 3 yb 4 23. 1x2 1 y224 24. 1r3 2 s323 25. 1ax 1 by25 26. 1ax 2 by25 27. A !x 1 2B6 28. A3 1 !yB4 29. 1a3/4 1 b1/424 30. 1x2/3 1 y1/323 31. Ax1/4 1 2!yB4 32. A !x 2 3y1/4B8 In Exercises 33–36, expand the expression using Pascal’s triangle. 33. 1r 2 s24 34. 1x2 1 y227 35. 1ax 1 by26 36. 1x 1 3y24 In Exercises 37–44, find the coefficient C of the term in the binomial expansion. Binomial Term Binomial Term Binomial Term Binomial Term 37. 1x 1 2210 Cx6 38. 13 1 y29 Cy5 39. 1y 2 328 Cy4 40. 1x 2 1212 Cx5 41. 12x 1 3y27 Cx3y4 42. 13x 2 5y29 Cx2y7 43. 1x2 1 y28 Cx8y4 44. 1r 2 s2210 Cr6s8 are that every row begins and ends with 1 and all other numbers are found by adding the two numbers above the entry. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 Last, a formula was given for finding a particular term of a binomial expansion; the 1r 1 12 term of 1a 1 b2n is an rb an2rbr. In this section, we developed a formula for raising a binomial expression to an integer power, n $ 0. The patterns that surfaced were that the expansion displays symmetry between the two terms; that is, every expansion has n 1 1 terms, the powers sum to n, and the coefficients, called binomial coefficients, are ratios of factorials: 1a 1 b2n 5 a n k50 an kban2kbk an kb 5 n! 1n 2 k2! k! Also, Pascal’s triangle, a shortcut method for evaluating the binomial coefficients, was discussed. The patterns in the triangle [SEC TION 12 .5] S U M M A RY [SEC TION 12 .5] E X E R C I S E S • S K I L L S In later sections, you will learn the “n choose k” notation for combinations. • A P P L I C A T I O N S 45. Lottery. In a state lottery in which six numbers are drawn from a possible 40 numbers, the number of possible six- number combinations is equal to a40 6 b. How many possible combinations are there? 46. Lottery. In a state lottery in which six numbers are drawn from a possible 60 numbers, the number of possible six- number combinations is equal to a60 6 b. How many possible combinations are there? 12.5 The Binomial Theorem 1139 • C A T C H T H E M I S T A K E In Exercises 49 and 50, explain the mistake that is made. 49. Evaluate the expression a7 5b. Solution: Write out the binomial coefficient in terms of factorials. Write out the factorials. Simplify. a7 5b 5 7! 5! 5 7⋅6 1 5 42 This is incorrect. What mistake was made? 50. Expand 1x 1 2y24. Solution: Write out with blanks. 1x 1 2y2 4 5 x4 1 x3y 1 x2y2 1 x y3 1 y4 Write out the terms from the fifth row of Pascal’s triangle. 1, 4, 6, 4, 1 Substitute these coefficients into the binomial expansion. 1x 1 2y24 5 x4 1 4x3y 1 6x2y2 1 4xy3 1 y4 This is incorrect. What mistake was made? a7 5b 5 7! 5! a7 5b 5 7! 5! 5 7⋅6⋅5⋅4⋅3⋅2⋅1 5⋅4⋅3⋅2⋅1 • C O N C E P T U A L In Exercises 51–54, determine whether each statement is true or false. 51. The binomial expansion of 1x 1 y210 has 10 terms. 52. The binomial expansion of 1x2 1 y2215 has 16 terms. 53. an nb 5 1 54. a n 2nb 5 21 • C H A L L E N G E 55. Show that an kb 5 a n n 2 kb, if 0 # k # n. 56. Show that if n is a positive integer, then: an 0b 1 an 1b 1 an 2b 1 c1 an nb 5 2n Hint: Let 2n 5 11 1 12n and use the binomial theorem to expand. • T E C H N O L O G Y 57. With a graphing utility, plot y1 5 1 2 3x 1 3x2 2 x3, y2 5 21 1 3x 2 3x2 1 x3, and y3 5 11 2 x23 in the same viewing screen. What is the binomial expansion of 11 2 x23? 58. With a graphing utility, plot y1 5 1x 1 324, y2 5 x4 1 4x3 1 6x2 1 4x 1 1, and y3 5 x4 1 12x3 1 54x2 1 108x 1 81. What is the binomial expansion of 1x 1 324? 59. With a graphing utility, plot y1 5 1 2 3x, y2 5 1 2 3x 1 3x2, y3 5 1 2 3x 1 3x2 2 x3, and y4 5 11 2 x23 for 21 , x , 1. What do you notice happening each time an additional term is added? Now, let 1 , x , 2. Does the same thing happen? 60. With a graphing utility, plot y1 5 1 2 3 x, y2 5 1 2 3 x 1 3 x2, y3 5 1 2 3 x 1 3 x2 2 1 x3, and y4 5 a1 2 1 xb 3 for 1 , x , 2. What do you notice happening each time an additional term is added? Now, let 0 , x , 1. Does the same thing happen? 61. With a graphing utility, plot y1 5 1 1 3 x, y2 5 1 1 3 x 1 3 x2, y3 5 1 1 3 x 1 3 x2 2 1 x3, and y4 5 a1 1 1 xb 3 for 1 , x , 2. What do you notice happening each time an additional term is added? Now, let 0 , x , 1. Does the same thing happen? 62. With a graphing utility, plot y1 5 1 1 x 1!, y2 5 1 1 x 1! 1 x2 2!, y3 5 1 1 x 1! 1 x2 2! 2 x3 3!, and y4 5 e x for 21 , x , 1. What do you notice happening each time an additional term is added? Now, let 1 , x , 2. Does the same thing happen? 47. Poker. With a deck of 52 cards, 5 cards are dealt in a game of poker. There are a total of a52 5 b different 5-card poker hands that can be dealt. How many possible hands are there? 48. Canasta. In the card game canasta, two decks of cards including the jokers are used and 11 cards are dealt to each person. There are a total of a108 11 b different 11-card canasta hands that can be dealt. How many possible hands are there? 1140 CHAPTER 12 Sequences, Series, and Probability S K I L L S O B J E C T I V E S ■ ■Apply the fundamental counting principle to solve counting problems. ■ ■Apply permutations to solve counting problems. ■ ■Apply combinations to solve counting problems. ■ ■Apply distinguishable permutations to solve counting problems. C O N C E P T U A L O B J E C T I V ES ■ ■Understand that the number of ways successive things can occur is found by multiplying the number of ways each thing can occur. ■ ■Understand that in a permutation order matters. ■ ■Understand that in combinations order does not matter. ■ ■Understand that permutation with repetition means that some objects are nondistinguishable from each other. 12.6 COUNTING, PERMUTATIONS, AND COMBINATIONS 12.6.1 The Fundamental Counting Principle You are traveling through Europe for the summer and decide the best packing option is to select separates that can be mixed and matched. You pack one pair of shorts and one pair of khaki pants. You pack a pair of Teva sport sandals and a pair of sneakers. You have three shirts (red, blue, and white). How many different outfits can be worn using only the clothes mentioned above? The answer is 12. There are two options for bottoms (pants or shorts), three options for shirts, and two options for shoes. The product of these is 2⋅3⋅2 5 12 The general formula for counting possibilities is given by the fundamental counting principle. FUNDAMENTAL COUNTING PRINCIPLE Let E1 and E2 be two independent events. The first event E1 can occur in m1 ways. The second event E2 can occur in m2 ways. The number of ways that the combination of the two events can occur is m1⋅m2. In other words, the number of ways in which successive things can occur is found by multiplying the number of ways each thing can occur. 12.6.1 S K I L L Apply the fundamental counting principle to solve counting problems. 12.6.1 C O N C E P T U A L Understand that the number of ways successive things can occur is found by multiplying the number of ways each thing can occur. [CONCEPT CHECK] If you packed four shirts, a pair of pants and a pair of shorts, and two types of shoes (sandals and sneakers), how many different outfits can be worn on your vacation? ANSWER 4⋅2⋅2 5 16 ▼ STUDY TIP The fundamental counting principle can be extended to more than two events. 12.6 Counting, Permutations, and Combinations 1141 EXAMPLE 1 Possible Meals Served at a Restaurant A restaurant is rented for a retirement party. The owner offers an appetizer, an entrée, and a dessert for a set price. The following are the choices that people attending the party may choose from. How many possible dinners could be served that night? Appetizers: calamari, stuffed mushrooms, or caesar salad Entrées: tortellini alfredo, shrimp scampi, eggplant parmesan, or chicken marsala Desserts: tiramisu or flan Solution: There are three possible appetizers, four possible entrées, and two possible desserts. Write the product of possible options. 3⋅4⋅2 5 24 There are 24 possible dinners for the retirement party. Y OUR TU R N In Example 1, the restaurant will lower the cost per person for the retirement party if the number of appetizers and entrées is reduced. Suppose the appetizers are reduced to either soup or salad and the entrées are reduced to either tortellini or eggplant parmesan. How many possible dinners could be served at the party? ▼ ▼ A N S W E R 8 ▼ A N S W E R 8 billion EXAMPLE 2 Telephone Numbers (When to Require 10-Digit Dialing) In many towns in the United States, residents can call one another using a 7-digit dialing system. In some large cities, 10-digit dialing is required because two or more area codes coexist. Determine how many telephone numbers can be allocated in a 7-digit dialing area. Solution: With 7-digit telephone numbers, the first number cannot be a 0 or a 1, but each of the following six numbers can be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. First number: 2, 3, 4, 5, 6, 7, 8, or 9. 8 possible digits Second number: 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. 10 possible digits Third number: 10 possible digits Fourth number: 10 possible digits Fifth number: 10 possible digits Sixth number: 10 possible digits Seventh number: 10 possible digits Counting principle: 8⋅10⋅10⋅10⋅10⋅10⋅10 Possible telephone numbers: 8,000,000 Eight million 7-digit telephone numbers can be allocated within one area code. Y OUR TU R N If the first digit of an area code cannot be 0 or 1, but the second and third numbers of an area code can be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9, how many 10-digit telephone numbers can be allocated in the United States? ▼ 1142 CHAPTER 12 Sequences, Series, and Probability The fundamental counting principle applies when an event can occur more than once. We now introduce two other concepts, permutations and combinations, which allow individual events to occur only once. For example, in Example 2, the allowable telephone numbers can include the same number in two or more digit places, as in 555-1212. However, in many state lottery games, once a number is selected, it cannot be used again. An important distinction between a permutation and a combination is that in a permutation order matters, but in a combination order does not matter. For example, the Florida winning lotto numbers one week could be 2–3–5–11–19–27. This would be a combination because the order in which they are drawn does not matter. However, if you were betting on a trifecta at the Kentucky Derby, to win you must not only select the first, second, and third place horses, you must select them in the order in which they finished. This would be a permutation. 12.6.2 Permutations DEFINITION Permutation A permutation is an ordered arrangement of distinct objects without repetition. 12.6.2 S KI L L Apply permutations to solve counting problems. 12.6.2 C ON C E P T U A L Understand that in a permutation order matters. EXAMPLE 3 Finding the Number of Permutations of n Objects How many permutations are possible for the letters A, B, C, and D? Solution: ABCD ABDC ACBD ACDB ADCB ADBC BACD BADC BCAD BCDA BDCA BDAC CABD CADB CBAD CBDA CDAB CDBA DABC DACB DBCA DBAC DCAB DCBA There are 24 (or 4!) possible permutations of the letters A, B, C, and D. Notice that in the first row of permutations in Example 3, A was selected for the first space. That left one of the remaining three letters to fill the second space. Once that was selected there remained two letters to choose from for the third space, and then the last space was filled with the unselected letter. In general, there are n! ways to order n objects. NUMBER OF PERMUTATIONS OF n OBJECTS The number of permutations of n objects is n! 5 n⋅1n 2 12 ⋅1n 2 22 c 2⋅1 STUDY TIP In a permutation of objects, order matters. That is, the same objects arranged in a different order are considered to be a distinct permutation. [CONCEPT CHECK] TRUE OR FALSE In a permutation, an event only occurs once and order matters. ANSWER True ▼ 12.6 Counting, Permutations, and Combinations 1143 EXAMPLE 4 Running Order of Dogs In a field trial sponsored by the American Kennel Club (AKC), the dogs compete in random order. If there are nine dogs competing in the trials, how many possible running orders are there? Courtesy Cynthia Young, University of Central Florida Solution: There are nine dogs that will run. n 5 9 The number of possible running orders is n! n! 5 9! 5 362,880 There are 362,880 different possible running orders of nine dogs. Y OUR TU R N Five contestants in the Miss America pageant reach the live television interview round. In how many possible orders can the contestants compete in the interview round? ▼ ▼ A N S W E R 120 In Examples 3 and 4, we were interested in all possible permutations. Sometimes, we are interested in only some permutations. For instance, 20 horses usually run in the Kentucky Derby. If you bet on a trifecta, you must pick the top three places in the correct order to win. Therefore, we do not consider all possible permutations of 20 horses finishing (places 1–20). Instead, we only consider the possible permuta-tions of first-, second-, and third-place finishes of the 20 horses. We would call this a permutation of 20 objects taken 3 at a time. In general, this ordering is called a permutation of n objects taken r at a time. If 20 horses are entered in the Kentucky Derby, there are 20 possible first-place finishers. We consider the permutations with one horse in first place. That leaves 19 possi ble horses for second place and then 18 possible horses for third place. Therefore, there are 20⋅19⋅18 5 6840 possible winning scenarios for the trifecta. This can also be represented as 20! 120 2 32! 5 20! 17!. 1144 CHAPTER 12 Sequences, Series, and Probability NUMBER OF PERMUTATIONS OF n OBJECTS TAKEN r AT A TIME The number of permutations of n objects taken r at a time is nP r 5 n! 1n 2 r2! 5 n1n 2 121n 2 22 c 1n 2 r 1 12 EXAMPLE 5 Starting Lineup for a Volleyball Team The starters for a six-woman volleyball team have to be listed in a particular order (1–6). If there are 13 women on the team, how many possible starting lineups are there? Solution: Identify the total number of players. n 5 13 Identify the total number of starters in the lineup. r 5 6 Substitute n 5 13 and r 5 6 into nP r 5 n! 1n 2 r2!. 13P 6 5 13! 113 2 62! 5 13! 7! 5 13⋅12⋅11⋅10⋅9⋅8 5 1,235,520 There are 1,235,520 possible combinations. YOUR T UR N A softball team has 12 players, 10 of whom will be in the starting lineup (batters 1–10). How many possible starting lineups are there for this team? ▼ ▼ A N S W E R 239,500,800 12.6.3 Combinations The difference between a permutation and a combination is that a permutation has an order associated with it, whereas a combination does not have an order associated with it. DEFINITION Combination A combination is an arrangement, without specific order, of distinct objects without repetition. The six winning Florida lotto numbers and the NCAA men’s Final Four basketball tournament are examples of combinations (six numbers and four teams) without regard to order. The number of combinations of n objects taken r at a time is equal to the binomial coefficient an rb. NUMBER OF COMBINATIONS OF n OBJECTS TAKEN r AT A TIME The number of combinations of n objects taken r at a time is nCr 5 an rb 5 n! 1n 2 r2!r! 12.6.3 S KI L L Apply combinations to solve counting problems. 12.6.3 C ON C E P T U A L Understand that in combinations order does not matter. 12.6 Counting, Permutations, and Combinations 1145 Compare the number of permutations nP r 5 n! 1n 2 r2! and the number of combinations nCr 5 n! 1n 2 r2!r!. It makes sense that the number of combinations is less than the number of permutations. The denominator is larger because there are no separate orders associated with a combination. [CONCEPT CHECK] In permutations and combinations, an event can only occur once and in permutations order of events matters, whereas in combinations order does not matter, so do we expect a greater number of permutations or combinations given the same n object taken r at a time? ANSWER Permutations ▼ EXAMPLE 6 Possible Combinations to the Lottery If there are a possible 59 numbers and the lottery officials draw 6 numbers, how many possible combinations are there? Solution: Identify how many numbers are in the drawing. n 5 59 Identify how many numbers are chosen. r 5 6 Substitute n 5 59 and r 5 6 into nCr 5 n! 1n 2 r2!r!. Simplify. 5 59⋅58⋅57⋅56⋅55⋅54⋅1532! 53!⋅6! 5 59⋅58⋅57⋅56⋅55⋅54 6⋅5⋅4⋅3⋅2 5 45,057,474 There are 45,057,474 possible combinations. Y OUR TU R N What are the possible combinations for a lottery with 49 possible numbers and 6 drawn numbers? 59C6 5 59! 159 2 62!6! ▼ ▼ A N S W E R 13,983,816 12.6.4 Permutations with Repetition Permutations and combinations are arrangements of distinct (nonrepeated) objects. A permutation in which some of the objects are repeated is called a permutation with repetition or a nondistinguishable permutation. For example, if a sack has three red marbles, two blue marbles, and one white marble, how many possible permutations would there be when drawing six marbles, one at a time? This is a different problem from writing the numbers 1 through 6 on pieces of paper, putting them in a hat, and drawing them out. The reason the problems are different is that the two blue balls are indistinguishable and the three red balls are also indistinguishable. The possible permutations for drawing numbers out of the hat are 6!, whereas the possible permutations for drawing balls out of the sack are given by 6! 3!⋅2!⋅1! 12.6.4 SKILL Apply distinguishable permutations to solve counting problems. 12.6.4 CON CEPTUAL Understand that permutation with repetition means that some objects are nondistinguishable from each other. NUMBER OF DISTINGUISHABLE PERMUTATIONS If a set of n objects has n1 of one kind of object, n2 of another kind of object, n3 of a third kind of object, and so on for k different types of objects so that n 5 n1 1 n2 1 c1 nk, then the number of distinguishable permutations of the n objects is n! n1!⋅n2!⋅n3! c nk! 1146 CHAPTER 12 Sequences, Series, and Probability In our sack of marbles, there were six marbles n 5 6. Specifically, there were three red marbles 1n1 5 32, two blue marbles 1n2 5 22, and one white marble 1n3 5 12. Notice that n 5 n1 1 n2 1 n3 and that the number of distinguishable permutations is equal to 6! 3!⋅2!⋅1! 5 60 [CONCEPT CHECK] Is the number of permutations of n objects that have some objects that are repeated or nondistinguishable greater than or less than the number of permutations of n distinguishable (nonrepeated) objects? ANSWER Less than ▼ EXAMPLE 7 Peg Game at Cracker Barrel The peg game on the tables at Cracker Barrel is a triangle with 15 holes drilled in it, in which pegs are placed. There are 5 red pegs, 5 white pegs, 3 blue pegs, and 2 yellow pegs. If all 15 pegs are in the holes, how many different ways can the pegs be aligned? Solution: There are four different colors of pegs (red, white, blue, and yellow). 5 red pegs: n1 5 5 5 white pegs: n2 5 5 3 blue pegs: n3 5 3 2 yellow pegs: n4 5 2 There are 15 pegs total: n 5 15 Substitute n 5 15, n1 5 5, n2 5 5, n3 5 3, and n4 5 2 into n! n1!⋅n2!⋅n3!c nk!. 15! 5!⋅5!⋅3!⋅2! Simplify. 5 7,567,560 There are a possible 7,567,560 ways to insert the 15 colored pegs at the Cracker Barrel. YOUR TURN Suppose a similar game to the peg game at Cracker Barrel is set up with only 10 holes in a triangle. With 5 red pegs, 2 white pegs, and 3 blue pegs, how many different permutations can fill that board? Andy Washnik ▼ ▼ A N S W E R 2520 12.6 Counting, Permutations, and Combinations 1147 In Exercises 1–8, use the formula for nP r to evaluate each expression. 1. 6P 4 2. 7P 3 3. 9P 5 4. 9P 4 5. 8P 8 6. 6P 6 7. 13P 3 8. 20P 3 In Exercises 9–18, use the formula for nCr to evaluate each expression. 9. 10C5 10. 9C4 11. 50C6 12. 50C10 13. 7C7 14. 8C8 15. 30C4 16. 13C5 17. 45C8 18. 30C4 • A P P L I C A T I O N S [SEC TION 12 .6] E X E R C I SE S • S K I L L S 19. Computers. At the www.dell.com website, a customer can “build” a system. If there are four monitors to choose from, three different computers, and two different keyboards, how many possible system configurations are there? 20. Houses. In a “new home” community, a person can select from one of four models, five paint colors, three tile selections, and two landscaping options. How many different houses (interior and exterior) are there to choose from? 21. Wedding Invitations. An engaged couple is ordering wedding invitations. The wedding invitations come in white or ivory. The writing can be printed, embossed, or engraved. The envelopes can come with liners or without. How many possible designs of wedding invitations are there to choose from? 22. Dinner. Siblings are planning their father’s 65th birthday din-ner and have to select one of four main courses (baked chicken, grilled mahi-mahi, beef Wellington, or lasagna), one of two starches (rosemary potatoes or rice), one of three vegetables (green beans, carrots, or zucchini), and one of five appetizers (soup, salad, pot stickers, artichoke dip, or calamari). How many possible dinner combinations are there? 23. PIN Number. Most banks require a four-digit ATM PIN code for each customer’s bank card. How many possible four-digit PIN codes are there to choose from? 24. Password. All e-mail accounts require passwords. If a four-character password is required that can contain letters (but no numbers), how many possible passwords can there be? (Assume letters are not case sensitive.) 25. Leadership. There are 15 professors in a department, and there are four leadership positions (chair, assistant chair, undergraduate coordinator, and graduate coordinator). How many possible leadership teams are there? C O M B I N AT I O N S ■ ■Objects cannot be repeated. ■ ■ Order does not matter. ■ ■ Number of combinations of n objects taken r at a time: nCr 5 n! 1n 2 r2!r! N O N D I S T I N GU I S H A B L E PER M U TAT IO NS ■ ■Some objects are repeated because they are not distinguishable. ■ ■For n objects with k different types of objects: n! n1!n2!n3! c nk! In this section, we discussed the fundamental counting principle, permutations, and combinations. THE F UNDAM E NTAL C OUNT I NG PR I NC I PL E IS AP P LIC AB LE W H E N ■ ■Objects can be repeated. ■ ■The objects can occur in any order. ■ ■The first event E1 can occur m1 ways, and the second event E2 can occur m2 ways: the number of ways successive events can occur is m1m2 ways. P E R MUTAT I ONS ■ ■Objects cannot be repeated. ■ ■Order matters. ■ ■Number of permutations of n objects: n!. ■ ■Number of permutations of n objects taken r at a time: nP r 5 n! 1n 2 r2! [SEC TION 12 .6] S U M M A RY 1148 CHAPTER 12 Sequences, Series, and Probability 26. Fraternity Elections. A fraternity is having elections. There are three men running for president, two men running for vice president, four men running for secretary, and one man running for treasurer. How many possible outcomes do the elections have? 27. Multiple-Choice Tests. There are 20 questions on a multiple -choice exam, and each question has four possible answers (A, B, C, and D). Assuming no answers are left blank, how many different ways can you answer the questions on the exam? 28. Multiple-Choice Tests. There are 25 questions on a multiple-choice exam, and each question has five possible answers (A, B, C, D, and E). Assuming no answers are left blank, how many different ways can you answer the questions on the exam? 29. Zip Codes. In the United States a five-digit zip code is used to route mail. How many five-digit zip codes are possible? (All numbers can be used.) If 0s were eliminated from the first and last digits, how many possible zip codes would there be? 30. License Plates. In a particular state, there are six characters in a license plate: three letters followed by three numbers. If 0s and 1s are eliminated from possible numbers and Os and Is are eliminated from possible letters, how many different license plates can be made? 31. Class Seating. If there are 30 students in a class and there are exactly 30 seats, how many possible seating charts can be made, assuming all 30 students are present? 32. Season Tickets. Four friends buy four season tickets to the Green Bay Packers. To be fair, they change the seating arrangement every game. How many different seating arrangements are there for the four friends? 33. Combination “Permutation” Lock. A combination lock on most lockers will open when the correct choice of three numbers (1 to 40) is selected and entered in the correct order. Therefore, a combination lock should really be called a permutation lock. How many possible permutations are there, assuming no numbers can be repeated? 34. Safe. A safe will open when the correct choice of three numbers (1 to 50) is selected and entered in the correct order. How many possible permutations are there, assuming no numbers can be repeated? 35. Raffle. A fundraiser raffle is held to benefit cystic fibrosis research, and 1000 raffle tickets are sold. Three prizes are raffled off. First prize is a round-trip ticket on Delta Air Lines, second prize is a round of golf for four people at a local golf course, and third prize is a $50 gift certificate to Chili’s. How many possible winning scenarios are there if all 1000 tickets are sold to different people? 36. Ironman Triathlon. If 100 people compete in an ironman triathlon, how many possible placings are there (first, second, and third place)? 37. Lotto. If a state lottery picks from 53 numbers and 6 numbers are selected, how many possible 6-number combinations are there? 38. Lotto. If a state lottery picks from 53 numbers and 5 numbers are selected, how many possible 5-number combinations are there? 39. Cards. In a deck of 52 cards, how many different 5-card hands can be dealt? 40. Cards. In a deck of 52 cards, how many different 7-card hands can be dealt? 41. Blackjack. In a single-deck blackjack game (52 cards), how many different 2-card combinations are there? 42. Blackjack. In a single deck, how many two-card combinations are there that equal 21: ace (worth 11) and a 10 or face card—jack, queen, or king? 43. March Madness. Every spring, the NCAA men’s basketball tournament starts with 64 teams. After two rounds, it is down to the Sweet Sixteen, and after two more rounds, it is reduced to the Final Four. Once 64 teams are selected (but not yet put in brackets), how many possible scenarios are there for the Sweet Sixteen? 44. March Madness. Every spring, the NCAA men’s basketball tournament starts with 64 teams. After two rounds, it is down to the Sweet Sixteen, and after two more rounds, it is reduced to the Final Four. Once the 64 teams are identified (but not yet put in brackets), how many possible scenarios are there for the Final Four? 45. NFL Playoffs. There are 32 teams in the National Football League (16 AFC and 16 NFC). How many possible combinations are there for the Superbowl? (Assume one team from the AFC plays one team from the NFC in the Superbowl.) 46. NFL Playoffs. After the regular season in the National Football League, 12 teams make the playoffs (6 from the AFC and 6 from the NFC). How many possible combinations are there for the Superbowl once the 6 teams in each conference are identified? 47. Survivor. On the television show Survivor, one person is voted off the island every week. When it is down to six contestants, how many possible voting combinations remain, if no one will vote him- or herself off the island? Assume that the order (who votes for whom) makes a difference. How many total possible voting outcomes are there? 48. American Idol. On the television show American Idol, a young rising star is eliminated from the competition every week. The first week, each of the 12 contestants sings one song. How many possible ways could the contestants be ordered 1–12? How many possible ways could 6 men and 6 women be ordered to alternate female and male contestants? 49. Dancing with the Stars. In the popular TV show Dancing with the Stars, 12 entertainers (6 men and 6 women) compete in a dancing contest. The first night, the show decides to select 3 men and 3 women. How many ways can this be done? 50. Dancing with the Stars. See Exercise 49. How many ways can six male celebrities line up for a picture alongside six female celebrities? 12.6 Counting, Permutations, and Combinations 1149 In Exercises 51 and 52, explain the mistake that is made. • C A T C H T H E M I S T A K E 51. In a lottery that picks from 30 numbers, how many five- number combinations are there? Solution: Let n 5 30 and r 5 5. Calculate nP r 5 n! 1n 2 r2!. 30P 5 5 30! 25! Simplify. 30P 5 5 17,100,720 This is incorrect. What mistake was made? 52. A homeowners association has 12 members on the board of directors. How many ways can the board elect a president, vice president, secretary, and treasurer? Solution: Let n 5 12 and r 5 4. Calculate nCr 5 n! 1n 2 r2!⋅r!. 12C4 5 12! 8!⋅4! Simplify. 12C4 5 495 This is incorrect. What mistake was made? • C O N C E P T U A L In Exercises 53–56, determine whether each statement is true or false. 53. The number of permutations of n objects is always greater than the number of combinations of n objects if the objects are distinct. 54. The number of permutations of n objects is always greater than the number of combinations of n objects even when the objects are indistinguishable. 55. The number of four-letter permutations of the letters A, B, C, and D is equal to the number of four-letter permutations of ABBA. 56. The number of possible answers to a true/false question is a permutation problem. • C H A L L E N G E 57. What is the relationship between nCr and nCr11? 58. What is the relationship between nP r and nP r21? 59. Simplify the expression nCr⋅r!. 60. What is the relationship between nCr and nCn2r? • T E C H N O L O G Y 61. Employ a graphing utility with a nP r feature and compare it with answers to Exercises 1–8. 62. Employ a graphing utility with a nCr feature and compare it with answers to Exercises 9–18. 63. Use a graphing calculator to evaluate: a. 10P 4 b. 4!110C42 c. Are answers in (a) and (b) the same? d. Why? 64. Use a graphing calculator to evaluate: a. 12P 5 b. 5!112C52 c. Are answers in (a) and (b) the same? d. Why? 1150 CHAPTER 12 Sequences, Series, and Probability 12.7.1 Sample Space You are sitting at a blackjack table at Caesar’s Palace, and the dealer is showing a 7. You have a 9 and a 7; should you hit? Will it rain today? What will the lotto numbers be this week? Will the coin toss at the Superbowl result in a head or a tail? Will Derek Jeter get a hit at his next trip to the plate? These are all questions where probability is used to guide us. Anything that happens for which the result is uncertain is called an experiment. Each trial of an experiment is called an outcome. All of the possible outcomes of an experiment constitute the sample space. The term event is used to describe the kind of possible outcomes. For example, a coin toss is an experiment. Every outcome is either heads or tails. The sample space of a single toss is {heads, tails}. The result of one experiment has no certain outcome. However, if the experiment is performed many times, the results will produce regular patterns. For example, if you toss a fair coin, you don’t know whether it will come up heads or tails. You can toss a coin 10 times and get 10 heads. However, if you made 1,000,000 tosses, you would get about 500,000 heads and 500,000 tails. Therefore, since we assume a head is equally likely as a tail and there are only two possible events (heads or tails), we assign a probability of a head equal to 1 2 and a probability of a tail equal to 1 2. S K I L L S O B J E C T I V E S ■ ■Determine the sample space of an outcome. ■ ■Find the probability of an event. ■ ■Find the probability that an event will not occur. ■ ■Find the probability of mutually exclusive events. ■ ■Find the probability of independent events. C O N C E P T U A L O B J E C T I V ES ■ ■Understand that all possible outcomes of an experiment constitute the sample space. ■ ■Understand that the probability of an event occurring is the ratio of the number of outcomes in an event divided by the number of outcomes in the sample space. ■ ■Understand that the sum of the probability of an event occur-ring and the probability of an event not occurring is one. ■ ■Understand the difference between the probability of: event 1 and event 2; event 1 or event 2. ■ ■Understand that the probability of two independent events both occurring is the product of the individual probabilities. 12.7 PROBABILITY 12.7.1 S K I L L Determine the sample space of an outcome. 12.7.1 C ON C E P T U A L Understand that all possible outcomes of an experiment constitute the sample space. ▼ A N S W E R S 5 5 GGG, GGB, GBG, GBB, BBB, BBG, BGB, BGG6 EXAMPLE 1 Finding the Sample Space Find the sample space for each of the following outcomes. a. Tossing a coin once b. Tossing a coin twice c. Tossing a coin three times Solution (a): Tossing a coin one time will result in one of two events: heads (H) or tails (T ). The sample space S is written as S 5 5H, T6 . Solution (b): Tossing a coin twice can result in one of four possible outcomes. The sample space consists of all possible outcomes. S 5 5HH, HT, TH, TT6 Note that TH and HT are two different outcomes. Solution (c): There are eight possible outcomes when a coin is tossed three times. S 5 5HHH, HHT, HTH, HTT, TTT, TTH, THT, THH6 Y OUR T UR N Find the sample space associated with having three children (B boys or G girls). ▼ [CONCEPT CHECK] TRUE OR FALSE The sample space includes all possible outcomes and order matters (for example, Heads then Tails is not the same as Tails then Heads, even though the result of two rolls is one head and one tail). ANSWER True ▼ 12.7 Probability 1151 STUDY TIP If the coin is tossed n times, there are 2n possible outcomes. 1152 CHAPTER 12 Sequences, Series, and Probability 12.7.2 Probability of an Event To calculate the probability of an event, start by counting the number of outcomes in the event and the number of outcomes in the sample space. The ratio is equal to the probability if all outcomes are equally likely. 12.7.2 SKILL Find the probability of an event. 12.7.2 CON CEPTUAL Understand that the probability of an event occurring is the ratio of the number of outcomes in an event divided by the number of outcomes in the sample space. DEFINITION Probability of an Event If an event E has n1E2 equally likely outcomes and its sample space S has n1S2 equally likely outcomes, then the probability of event E, denoted P1E2, is P1E2 5 n1E2 n1S2 5 Number of outcomes in event E Number of outcomes in sample space S EXAMPLE 2 Finding the Probability of Two Girls If two children are born, what is the probability that they are both girls? Solution: The event is both children being girls. E 5 5GG6 The sample space is all four possible outcomes. S 5 5BB, BG, GB, GG6 The number of outcomes in the event is 1. n1E2 5 1 The number of events in the sample space is 4. n1S2 5 4 Compute the probability using P1E2 5 n1E2 n1S2 . P1E2 5 1 4 The probability that both children are girls is 1 4, or 0.25. Since the number of outcomes in an event must be less than or equal to the number of outcomes in the sample space, the probability of an event must be a number between 0 and 1 or equal to 0 or 1; to be precise, 0 # P1E2 # 1. If P1E2 5 0, then the event can never happen, and if P1E2 5 1, the event is certain to happen. EXAMPLE 3 Finding the Probability of Drawing a Face Card Find the probability of drawing a face card (jack, queen, or king) out of a 52-card deck. Solution: There are 12 face cards in a deck. n1E2 5 12 There are 52 cards in a deck. n1S2 5 52 Compute the probability using P1E2 5 n1E2 n1S2 . P1E2 5 12 52 5 3 13 The probability of drawing a face card out of a 52-card deck is 3 13 or <0.23. YOUR TURN a. Find the probability that an ace is drawn from a deck of 52 cards. b. Find the probability that a jack, queen, king, or ace is drawn from a deck of 52 cards. ▼ [CONCEPT CHECK] Would the probability of having a boy and a girl be greater or less than the probability of having two boys? ANSWER Greater than ▼ ▼ A N S W E R a. 4 52 5 1 13 b. 16 52 5 4 13 12.7.3 Probability of an Event Not Occurring The sum of the probabilities of all possible outcomes is 1. For example, when a die is rolled, if the outcomes are equally likely, then the probabilities are all 1 6: P112 5 1 6, P122 5 1 6, P132 5 1 6, P142 5 1 6, P152 5 1 6, P162 5 1 6 The sum of these six probabilities is 1. Since the sum of the probabilities of all possible outcomes sums to 1, we can find the probability that an event won’t occur by subtracting the probability that the event will occur from 1. P1E2 1 P1not E2 5 1 or P1not E2 5 1 2 P1E2 The complement of an event E is the collection of all outcomes in the sample space that are not in E. The complement of E is denoted Er or E. PROBABILITY OF AN EVENT NOT OCCURRING The probability that an event E will not occur is equal to 1 minus the probability that E will occur. P1not E2 5 1 2 P1E2 EXAMPLE 4 Finding the Probability of Rolling a 7 or an 11 If you bet on the “pass line” at a craps table and the person’s first roll is a 7 or an 11, using a pair of dice, then you win. Find the probability of winning a pass line bet on the first roll. Solution: The fundamental counting principle tells us that there will be 6⋅6 5 36 possible rolls of the pair of dice. n1S2 5 36 Draw a table listing possible sums of the two dice. DICE VALUE 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12 Of the 36 rolls, there are 8 rolls that will produce a 7 or an 11. n1E2 5 8 Compute the probability using P1E2 5 n1E2 n1S2 . P1E2 5 8 36 5 2 9 The probability of winning a pass line bet is 2 9 or <0.22 (22%). Y OUR T UR N If a 2, 3, or 12 is rolled on the first roll, then the pass line bet loses. Find the probability of losing a pass line bet on the first roll. ▼ PASS LINE PASS LINE NO CALL BETS DON’T PASS BAR COME 6: 8 4 5 8 SIX NINE 10 2 3 4 9 10 11 12 Pass Line ▼ A N S W E R 4 36 5 1 12 < 8.3% 12.7.3 S K I L L Find the probability that an event will not occur. 12.7.3 C ON C E P T U A L Understand that the sum of the probability of an event occuring and the probability of an event not occurring is one. 12.7 Probability 1153 1154 CHAPTER 12 Sequences, Series, and Probability 12.7.4 Mutually Exclusive Events Recall the definition of union and intersection in Section 1.5. The probability of one event E1 or a second event E2 occurring is given by the probability of the union of the two events. P1E1 ∪ E22 If there is any overlap between the two events, we must be careful not to count those twice. For example, what is the probability of drawing either a face card or a spade out of a deck of 52 cards? We must be careful not to count twice any face cards that are spades. PROBABILITY OF THE UNION OF TWO EVENTS If E1 and E2 are two events in the same sample space, the probability of either E1 or E2 occurring is given by P1E1 ∪ E22 5 P1E12 1 P1E22 2 P1E1 ∩ E22. If E1 and E2 are disjoint, P1E1 ∩ E22 5 0, then E1 and E2 are mutually exclusive. In that case, the probability of either E1 or E2 occurring is given by P1E1 ∪ E22 5 P1E12 1 P1E22. STUDY TIP The following three notations are equivalent. • P1Er2 5 1 2 P1E2 • P1E2 5 1 2 P1E2 • P1not E2 5 1 2 P1E2 EXAMPLE 5 Finding the Probability of Not Winning the Lottery Find the probability of not winning the lottery if six numbers are selected from 1 to 49. Solution: Calculate the number of possible six-number combinations. 49C6 5 49! 43!⋅6! 5 13,983,816 Calculate the probability of winning. P1winning2 5 1 13,983,816 Calculate the probability of P1not winning2 5 1 2 P1winning2 not winning. P1not winning2 5 1 2 1 13,983,816 P1not winning2 5 13,983,816 13,983,816 2 1 13,983,816 < 0.999999928 The probability of not winning the lottery is very close to 1. Y OUR TU R N Find the probability of not winning the lottery if 6 numbers are selected from 1 to 39. ▼ ▼ A N S W E R 0.9999997 [CONCEPT CHECK] P(Event occurring) 1 P(Event not occurring) 5 ? ANSWER 1 ▼ 12.7.4 SKILL Find the probability of mutually exclusive events. 12.7.4 CO NC EPTUAL Understand the difference between the probability of: event 1 and event 2; event 1 or event 2. EXAMPLE 6 Finding the Probability of Drawing a Face Card or a Spade Find the probability of drawing either a face card or a spade out of a deck of 52 cards. Solution: The deck has 12 face cards. P1face card2 5 12 52 The deck has 13 spades. P1spade2 5 13 52 The deck has 3 face cards that are spades. P1face card and spade2 5 3 52 Apply the probability formula. P1E1 ∪ E22 5 P1E12 1 P1E22 2 P1E1 ∩ E22 P1face card or a spade2 5 12 52 1 13 52 2 3 52 Simplify. P1face card or a spade2 5 22 52 5 11 26 The probability of either a face card or a spade being drawn is 11 26 < 0.42. Y OUR T UR N Find the probability of drawing either a heart or an ace out of a deck of 52 cards. Spades Face Cards ▼ ▼ A N S W E R 4 13 [CONCEPT CHECK] Do we expect for a single draw from a deck of cards that P(ace and heart) will be greater than or less than the P(ace or heart)? ANSWER Less than ▼ 12.7 Probability 1155 1156 CHAPTER 12 Sequences, Series, and Probability 12.7.5 Independent Events Suppose you have two children. The sex of the second child is not affected by the sex of the first child. For example, if your first child is a boy, then it is no less likely that the second child will be a boy. We say that two events are independent if the occurrence of either of them has no effect on the probability of the other occurring. Scientists used to believe that one gene controlled human eye color. Each parent gives one chromosome (either blue, green, or brown), and the result is a child with an eye-color gene composed of combinations, with brown being dominant over blue and blue being dominant over green. The genetic basis for eye color is actually far more complex, but we use this simpler model in the next example. PROBABILITIES OF INDEPENDENT EVENTS If E1 and E2 are independent events, then the probability of both occurring is the product of the individual probabilities: P1E1 ∩ E22 5 P1E12 ⋅P1E22 ▼ A N S W E R 7 9 < 0.78 12.7.5 SKILL Find the probability of independent events. 12.7.5 CON CEPTUAL Understand that the probability of two independent events both occurring is the product of the individual probabilities. EXAMPLE 8 Probabilities of Blue-Eyed Children of Brown-Eyed Parents If two brown-eyed parents have a blue-eyed child, then the parents must each have one blue- and one brown-eye gene. In order to have blue eyes, the child must get the blue-eye genes from both parents. What is the probability that two brown-eyed parents can have three children, all with blue eyes? The deck has two jokers. P1joker2 5 2 54 Apply the probability formula. P1E1 ∪ E22 5 P1E12 1 P1E22 P1ace or a joker2 5 4 54 1 2 54 Simplify. P1ace or a joker2 5 6 54 5 1 9 < 0.111 The probability of drawing either an ace or a joker is 1 9 < 0.11. Y OUR TU R N If there are 10 women mathematicians, 8 men mathematicians, 6 women engineers, and 12 men engineers, what is the probability that a selected person is either a woman or an engineer? ▼ EXAMPLE 7 Finding the Probability of Mutually Exclusive Events Find the probability of drawing either an ace or a joker in a 54-card deck (a deck with two jokers). Solution: Drawing an ace and drawing a joker are two mutually exclusive events, since a card cannot be both an ace and a joker. The deck has four aces. P1ace2 5 4 54 [CONCEPT CHECK] If a litter of puppies from two black Labs produces both yellow and black Labs and a repeat litter happens, what is the probability that the entire litter is yellow if there are n puppies, assuming Lab colors are equally likely? ANSWER 1 12n2 ▼ Solution: The sample space for children’s eye-color genes from these parents is S 5 5Blue/Blue, Blue/Brown, Brown/Brown, Brown/Blue6 The only way for a child to have blue eyes is if that child inherits two blue-eyed genes. P1blue-eyed child2 5 1 4 The eye color of each child is independent of that of the other. The probability of having three blue-eyed children is the product of the three individual probabilities. P1all three children with blue eyes2 5 a1 4b ⋅a1 4b ⋅a1 4b 5 1 64 The probability of two brown-eyed parents having three blue-eyed children is 1 64 < 0.016. YOUR T UR N Find the probability of the brown-eyed parents in Example 8 having three brown-eyed children. ▼ ▼ A N S W E R 27 64 < 0.422 [SEC TION 12 .7] SUM M A RY The probability of one event or another event occurring is found by adding the individual probabilities of each event and subtracting the probability of both: “or” P1E1 ∪ E22 5 P1E12 1 P1E22 2 P1E1 ∩ E22 If two events are mutually exclusive, they have no outcomes in common: P1E1 ∪ E22 5 P1E12 1 P1E22 The probability of two events occurring is the product of the individual probabilities, provided the two events are independent or do not affect one another: “and” P1E1 ∩ E22 5 P1E12 ⋅P1E22 In this section, we discussed the probability, or likelihood, of an event. It is found by dividing the total number of possible equally likely outcomes in the event by all of the possible outcomes in the sample space. P1E2 5 n1E2 n1S2 5 Number of outcomes in event E Number of outcomes in sample space S Probability is a number between 0 and 1 or equal to 0 or 1: 0 # P1E2 # 1 The probability of an event not occurring is 1 minus the probability of the event occurring: P1not E2 5 1 2 P1E2 In Exercises 1–6, find the sample space for each experiment. 1. The sum of two dice rolled simultaneously. 2. A coin tossed three times in a row. 3. The sex (boy or girl) of four children born to the same parents. 4. Tossing a coin and rolling a die. 5. Two balls selected from a container that has 3 red balls, 2 blue 6. The grade (freshman, sophomore, or junior) of two high balls, and 1 white ball. school students who work at a local restaurant. Heads or Tails. In Exercises 7–10, find the probability for the experiment of tossing a coin three times. 7. Getting all heads. 8. Getting exactly one heads. 9. Getting at least one heads. 10. Getting more than one heads. [SEC TION 12 .7] E X E RC I S E S • S K I L L S 12.7 Probability 1157 1158 CHAPTER 12 Sequences, Series, and Probability Tossing a Die. In Exercises 11–16, find the probability for the experiment of tossing two dice. 11. The sum is 3. 12. The sum is odd. 13. The sum is even. 14. The sum is prime. 15. The sum is more than 7. 16. The sum is less than 7. Drawing a Card. In Exercises 17–20, find the probability for the experiment of drawing a single card from a deck of 52 cards. 17. Drawing a nonface card. 18. Drawing a black card. 19. Drawing a 2, 4, 6, or 8. 20. Drawing a 3, 5, 7, 9, or ace. In Exercises 21–26, let P1E12 5 1 4 and P1E22 5 1 2 and find the probability of the event. 21. Probability of E1 not occurring. 22. Probability of E2 not occurring. 23. Probability of either E1 or E2 occurring if E1 and E2 are 24. Probability of either E1 or E2 occurring if E1 and E2 are mutually exclusive. not mutually exclusive and P1E1 ∩ E22 5 1 8. 25. Probability of both E1 and E2 occurring if E1 and E2 are 26. Probability of both E1 and E2 occurring if E1 and E2 are mutually exclusive. independent. • A P P L I C A T I O N S 27. Cards. A deck of 52 cards is dealt. a. How many possible combinations of four-card hands are there? b. What is the probability of having all spades? c. What is the probability of having four of a kind? 28. Blackjack. A deck of 52 cards is dealt for blackjack. a. How many possible combinations of two-card hands are there? b. What is the probability of having 21 points (ace with a 10 or face card)? 29. Cards. With a 52-card deck, what is the probability of drawing a 7 or an 8? 30. Cards. With a 52-card deck, what is the probability of drawing a red 7 or a black 8? 31. Cards. By drawing twice, what is the probability of drawing a 7 and then an 8? 32. Cards. By drawing twice, what is the probability of drawing a red 7 and then a black 8? 33. Children. What is the probability of having five daughters in a row and no sons? 34. Children. What is the probability of having four sons in a row and no daughters? 35. Children. What is the probability that of five children at least one is a boy? Note: P1at least one boy2 5 1 2 P1no boys2. 36. Children. What is the probability that of six children at least one is a girl? Note: P1at least one girl2 5 1 2 P1no girls2. 37. Roulette. In roulette, there are 38 numbered slots (1–36, 0, and 00). Eighteen are red, 18 are black, and the 0 and 00 are green. What is the probability of having 4 reds in a row? 38. Roulette. What is the probability of having 2 greens in a row on a roulette table? 39. Item Defectiveness. For a particular brand of DVD players, 10% of the ones on the market are defective. If a company has ordered 8 DVD players, what is the probability that none of the 8 DVD players is defective? 40. Item Defectiveness. For a particular brand of generators, 20% of the ones on the market are defective. If a company buys 10 generators, what is the probability that none of the 10 generators is defective? 41. Number Generator. A random-number generator (computer program that selects numbers in no particular order) is used to select two numbers between 1 and 10. What is the probability that both numbers are even? 42. Number Generator. A random-number generator is used to select two numbers between 1 and 15. What is the probability that both numbers are odd? In Exercises 43 and 44, assume each deal is from a complete (shuffled) single deck of cards. 43. Blackjack. What is the probability of being dealt a blackjack (any ace and any face card) with a single deck? 44. Blackjack. What is the probability of being dealt two blackjacks in a row with a single deck? 45. Sports. With the salary cap in the NFL, it is said that on “any given Sunday” any team could beat any other team. If we assume every week a team has a 50% chance of winning, what is the probability that a team will go 16–0? 46. Sports. With the salary cap in the NFL, it is said that on “any given Sunday” any team could beat any other team. If we assume every week a team has a 50% chance of winning, what is the probability that a team will have at least 1 win? 47. Genetics. Suppose both parents have the brown/blue pair of eye-color genes, and each parent contributes one gene to the child. Suppose the brown eye-color gene is dominant so that if the child has at least one brown gene, the color will dominate and the eyes will be brown. a. List the possible outcomes (sample space). Assume each outcome is equally likely. b. What is the probability that the child will have the blue/ blue pair of genes? c. What is the probability that the child will have brown eyes? 48. Genetics. Refer to Exercise 47. In this exercise, the father has a brown/brown pair of eye-color genes, while the mother has a brown/blue pair of eye-color genes. a. List the possible outcomes (sample space). Assume each outcome is equally likely. b. What is the probability that the child will have a blue/blue pair of genes? c. What is the probability that the child will have brown eyes? 49. Playing Cards. How many 5-card hands can be drawn from a 52-card deck (no jokers)? 50. Playing Cards. a. How many ways can you select 2 aces and 3 other cards (non-aces) from a standard deck of 52 cards (no jokers)? b. What is the probability that you draw a 5-card hand with 2 aces and 3 non-aces? 51. Playing Cards. Find the probability of drawing 5 clubs from a standard deck of 52 cards. 52. Poker. Find the probability of getting 2 fives and 3 kings when drawing 5 cards from a standard deck of 52 cards. 63. Use a random-number generator on a graphing utility to select two numbers between 1 and 10. Run this generator 50 times. How many times (out of 50 trials) were both of the two numbers even? Compare with your answer from Exercise 41. 64. Use a random-number generator on a graphing utility to select two numbers between 1 and 15. Run this 50 times. How many times (out of 50 trials) were both of the two numbers odd? Compare with your answer from Exercise 42. In Exercises 65 and 66, when a die is rolled once, the probability of getting a 2 is 1 6 and the probability of not getting a 2 is 5 6. If a die is rolled n times, the probability of getting a 2 exactly k times can be found by using the binomial theorem: nCka1 6b k a5 6b n2k 65. If a die is rolled 10 times, find the probability of getting a 2 exactly two times. Round your answer to four decimal places. 66. If a die is rolled 8 times, find the probability of getting a 2 at most two times. Round your answer to four decimal places. • T E C H N O L O G Y 59. If two people are selected at random, what is the probability that they have the same birthday? Assume 365 days per year. 60. If 30 people are selected at random, what is the probability that at least two of them will have the same birthday? 61. If one die is weighted so that 3 and 4 are the only numbers that the die will roll, and the other die is fair, what is the probability of rolling two dice that sum to 2, 5, or 6? 62. If one die is weighted so that 3 and 4 are the only numbers that the dice will roll, and 3 comes up twice as often as 4, what is the probability of rolling a 3? • C H A L L E N G E In Exercises 55–58, determine whether each statement is true or false. • C O N C E P T U A L 55. If P1E12 5 0.5 and P1E22 5 0.4, then P1E32 must equal 0.1 if there are three possible events and they are all mutually exclusive. 56. If two events are mutually exclusive, then they cannot be independent. 57. If two events are independent, then they are not mutually exclusive. 58. The probability of having five sons and no daughters is 1 minus the probability of having five daughters and no sons. • C AT C H T H E M I S TA K E In Exercises 53 and 54, explain the mistake that is made. 53. Calculate the probability of drawing a 2 or a spade from a deck of 52 cards. Solution: The probability of drawing a 2 from a deck of 52 cards is 4 52. The probability of drawing a spade from a deck of 52 cards is 13 52. The probability of drawing a 2 or a spade is 4 52 1 13 52 5 17 52. This is incorrect. What mistake was made? 54. Calculate the probability of having two boys and one girl. Solution: The probability of having a boy is 1 2. The probability of having a girl is 1 2. These are independent, so the probability of having two boys and a girl is A1 2BA1 2BA1 2B 5 1 8. This is incorrect. What mistake was made? 12.7 Probability 1159 CH A P TE R 12 R E VI E W 1160 CHAPTER 12 Sequences, Series, and Probability [ C HA P TE R 1 2 REVIEW] SECTION CONCEPT KEY IDEAS/FORMULAS 12.1 Sequences and series Sequences a1, a2, a3, c, an, c an is the general term. Factorial notation n! 5 n1n 2 121n 2 22 c 3⋅2⋅1 n $ 2 0! 5 1 and 1! 5 1 Recursion formulas When an is defined by previous terms an21. Sums and series Sigma Notation: a q n51 an 5 a1 1 a2 1 a3 1 c1 an 1 c Finite series, or nth partial sum, Sn. Sn 5 a1 1 a2 1 a3 1 c1 an 12.2 Arithmetic sequences and series Arithmetic sequences an11 5 an 1 d or an11 2 an 5 d d is called the common difference. The general (nth) term of an arithmetic sequence an 5 a1 1 1n 2 12d for n $ 1 The sum of an arithmetic sequence Sn 5 n 2 1a1 1 an2 12.3 Geometric sequences and series Geometric sequences an11 5 r⋅an or an11 an 5 r r is called the common ratio. The general (nth) term of a geometric sequence an 5 a1⋅rn21 for n $ 1 Geometric series Finite series: Sn 5 a1 11 2 r n2 11 2 r2 r 2 1 Infinite series: a q n50 a1r n 5 a1 1 11 2 r2 0 r 0 , 1 12.4 Mathematical induction Mathematical induction Prove that Sn is true for all positive integers. Step 1: Show that S1 is true. Step 2: Assume Sn is true for Sk and show it is true for Sk11 1k 5 integer). 12.5 The Binomial theorem Binomial coefficients an kb 5 n! 1n 2 k2!k! an nb 5 1 and an 0b 5 1 Binomial expansion 1a 1 b2n 5 a n k50 an kban2kbk Pascal’s triangle Shortcut way of remembering binomial coefficients. Each term is found by adding the two numbers above it. Finding a particular term of a binomial expansion The 1r 1 12 term of the expansion 1a 1 b2n is an rb an2rbr. CH A P TE R 12 R E VIE W Chapter Review 1161 SECTION CONCEPT KEY IDEAS/FORMULAS 12.6 Counting, permutations, and combinations The fundamental counting principle The number of ways in which successive independent things can occur is found by multiplying the number of ways each thing can occur. Permutations The number of permutations of n objects is n! 5 n⋅1n 2 12⋅1n 2 22 c 2⋅1 The number of permutations of n objects taken r at a time is nP r 5 n! 1n 2 r2! 5 n1n 2 121n 2 22 c 1n 2 r 1 12 Combinations The number of combinations of n objects taken r at a time nCr 5 n! 1n 2 r2!r! Permutations with repetition The number of distinguishable permutations of n objects is n! n1!⋅n2!⋅n3! c nk! 12.7 Probability Sample space All of the possible outcomes of an experiment. Probability of an event The probability of event E, denoted P1E2, is P1E2 5 n1E2 n1S2 5 number of outcomes in event E number of outcomes in sample space S Probability of an event not occurring P1not E2 5 1 2 P1E2 Mutually exclusive events The probability of either E1 or E2 occurring is P1E1 ∪ E22 5 P1E12 1 P1E22 2 P1E1 ∩ E22 If E1 and E2 are mutually exclusive, P1E1 ∩ E22 5 0. Independent events If E1 and E2 are independent events, then the probability of both occurring is the product of the individual probabilities: P1E1 ∩ E22 5 P1E12⋅P1E22 REV IEW E XE R CI SE S 1162 CHAPTER 12 Sequences, Series, and Probability [ C H AP T E R 1 2 REVIEW EXER C IS E S ] 12.1 Sequences and Series Write the first four terms of the sequence. Assume n starts at 1. 1. an 5 n3 2. an 5 n! n 3. an 5 3n 1 2 4. an 5 1212n x n12 Find the indicated term of the sequence. 5. an 5 a2 3b n a5 5 ? 6. an 5 n2 3n a8 5 ? 7. an 5 1212n1n 2 12! n1n 1 12! a15 5 ? 8. an 5 1 1 1 n a10 5 ? Write an expression for the nth term of the given sequence. 9. 3, 26, 9, 212, c 10. 1, 1 2, 3, 1 4, 5, 1 6, 7, 1 8, c 11. 21, 1, 21, 1, c 12. 1, 10, 102, 103, c Simplify the ratio of factorials. 13. 8! 6! 14. 20! 23! 15. n1n 2 12! 1n 1 12! 16. 1n 2 22! n! Write the first four terms of the sequence defined by the recursion formula. 17. a1 5 5 an 5 an21 2 2 18. a1 5 1 an 5 n2⋅an21 19. a1 5 1, a2 5 2 an 5 1an2122⋅1an222 20. a1 5 1, a2 5 2 an 5 an22 1an2122 Evaluate the finite series. 21. a 5 n51 3 22. a 4 n51 1 n2 23. a 6 n51 13n 1 12 24. a 5 k50 2k11 k! Use sigma (summation) notation to write the sum. 25. 21 1 1 2 2 1 4 1 1 8 1 c2 1 64 26. 2 1 4 1 6 1 8 1 10 1 c1 20 27. 1 1 x 1 x2 2 1 x3 6 1 x4 24 1 c 28. x 2 x2 1 x3 2 2 x4 6 1 x5 24 2 x6 120 1 c Applications 29. Marines Investment. With the prospect of continued fighting in Iraq, in December 2004, the Marine Corps offered bonuses of as much as $30,000—in some cases, tax-free—to persuade enlisted personnel with combat experience and training to reenlist. Suppose a Marine put her entire $30,000 reenlistment bonus in an account that earns 4% interest compounded monthly. The balance in the account after n months is An 5 30,000 a1 1 0.04 12 b n n 5 1, 2, 3, c Her commitment with the Marines is 5 years. Calculate A60. What does A60 represent? 30. Sports. The NFL minimum salary for a rookie is $450,000. Suppose a rookie comes into the league making the minimum and gets a $30,000 raise every year he plays. Write the general term an of a sequence that represents the salary of an NFL player making the league minimum during his entire career. Assuming n 5 1 corresponds to the first year, what does a 4 n51an represent? 12.2 Arithmetic Sequences and Series Determine whether the sequence is arithmetic. If it is, find the common difference. 31. 7, 5, 3, 1, 21, c 32. 13 1 23 1 33 1 c 33. 1, 3 2, 2, 5 2, c 34. an 5 2n 1 3 35. an 5 1n 1 12! n! 36. an 5 51n 2 12 Find the general, or nth, term of the arithmetic sequence given the first term and the common difference. 37. a1 5 24 d 5 5 38. a1 5 5 d 5 6 39. a1 5 1 d 5 22 3 40. a1 5 0.001 d 5 0.01 For each arithmetic sequence described below, find a1 and d and construct the sequence by stating the general, or nth, term. 41. The 5th term is 13 and the 17th term is 37. 42. The 7th term is 214 and the 10th term is 223. 43. The 8th term is 52 and the 21st term is 130. 44. The 11th term is 230 and the 21st term is 280. Find the sum. 45. a 20 k51 3k 46. a 15 n51 n 1 5 47. 2 1 8 1 14 1 20 1 c1 68 48. 1 4 2 1 4 2 3 4 2 c2 31 4 R E VI E W E XERCISES Review Exercises 1163 Applications 49. Salary. Upon graduating with M.B.A.s, Bob and Tania opt for different career paths. Bob accepts a job with the U.S. Department of Transportation making $45,000 with a guaranteed $2000 raise every year. Tania takes a job with Templeton Corporation making $38,000 with a guaranteed $4000 raise every year. Calculate how many total dollars both Bob and Tania will have each made after 15 years. 50. Gravity. When a skydiver jumps out of an airplane, she falls approximately 16 feet in the 1st second, 48 feet during the 2nd second, 80 feet during the 3rd second, 112 feet during the 4th second, 144 feet during the 5th second, and this pattern continues. If she deploys her parachute after 5 seconds have elapsed, how far will she have fallen during those 5 seconds? 12.3 Geometric Sequences and Series Determine whether the sequence is geometric. If it is, find the common ratio. 51. 2, 24, 8, 216, c 52. 1, 1 22, 1 32, 1 42, c 53. 20, 10, 5, 5 2, c 54. 1 100, 1 10, 1, 10, c Write the first five terms of the geometric series. 55. a1 5 3 r 5 2 56. a1 5 10 r 5 1 4 57. a1 5 100 r 5 24 58. a1 5 260 r 5 21 2 Write the formula for the nth term of the geometric series. 59. a1 5 7 r 5 2 60. a1 5 12 r 5 1 3 61. a1 5 1 r 5 22 62. a1 5 32 5 r 5 21 4 Find the indicated term of the geometric sequence. 63. 25th term of the sequence 2, 4, 8, 16, c 64. 10th term of the sequence 1 2, 1, 2, 4, c 65. 12th term of the sequence 100, 220, 4, 20.8, c 66. 11th term of the sequence 1000, 2500, 250, 2125, c Evaluate the geometric series, if possible. 67. 1 2 1 3 2 1 32 2 1 c1 38 2 68. 1 1 1 2 1 1 22 1 1 23 1 c1 1 210 69. a 8 n51 5132n21 70. a 7 n51 2 3 152n 71. a q n50 a2 3b n 72. a q n51 a 2 1 5 b n11 Applications 73. Salary. Murad is fluent in four languages and is offered a job with the U.S. government as a translator. He is hired on the “GS” scale at a base rate of $48,000 with 2% increases in his salary per year. Calculate what his salary will be after he has been with the U.S. government for 12 years. 74. Boat Depreciation. Upon graduating from Auburn University, Philip and Steve get jobs at Disney Ride and Show Engineering and decide to buy a ski boat together. If the boat costs $15,000 new, and depreciates 20% per year, write a formula for the value of the boat n years after it was purchased. How much will the boat be worth when Philip and Steve have been working at Disney for 3 years? 12.4 Mathematical Induction Prove the statements using mathematical induction for all positive integers n. 75. 3n # 3n 76. 4n , 4n11 77. 2 1 7 1 12 1 17 1 c1 15n 2 32 5 n 2 15n 2 12 78. 2n2 . 1n 1 122 n $ 3 12.5 The Binomial Theorem Evaluate the binomial coefficients. 79. a11 8 b 80. a10 0 b 81. a22 22b 82. a47 45b Expand the expression using the binomial theorem. 83. 1x 2 524 84. 1x 1 y25 85. 12x 2 523 86. 1x2 1 y324 87. A!x 1 1B5 88. 1x2/3 1 y1/326 Expand the expression using Pascal’s triangle. 89. 1r 2 s25 90. 1ax 1 by24 Find the coefficient C of the term in the binomial expansion. Binomial Term 91. 1x 2 228 Cx6 92. 13 1 y27 Cy4 93. 12x 1 5y26 Cx2y4 94. 1r2 2 s28 Cr8s4 REV IEW E XE R CI SE S 1164 CHAPTER 12 Sequences, Series, and Probability Applications 95. Lottery. In a state lottery in which 6 numbers are drawn from a possible 53 numbers, the number of possible 6-number combinations is equal to a53 6 b. How many possible combinations are there? 96. Canasta. In the card game canasta, two decks of cards including the jokers are used, and 13 cards are dealt to each person. A total of a108 13 b different 13-card canasta hands can be dealt. How many possible hands are there? 12.6 Counting, Permutations, and Combinations Use the formula for nP r to evaluate each expression. 97. 7P 4 98. 9P 9 99. 12P 5 100. 10P 1 Use the formula for nCr to evaluate each expression. 101. 12C7 102. 40C5 103. 9C9 104. 53C6 Applications 105. Car Options. A new Honda Accord comes in three models (LX, VX, and EX). Each of those models comes with either a cloth or a leather interior, and the exterior comes in silver, white, black, red, or blue. How many different cars (models, interior seat upholstery, and exterior color) are there to choose from? 106. E-mail Passwords. All e-mail accounts require passwords. If a six-character password is required that can contain letters (but no numbers), how many possible passwords can there be if letters can be repeated? (Assume no letters are case-sensitive.) 107. Team Arrangements. There are 10 candidates for the board of directors, and there are four leadership positions (president, vice president, secretary, and treasurer). How many possible leadership teams are there? 108. License Plates. In a particular state, there are six charac-ters in a license plate consisting of letters and numbers. If 0s and 1s are eliminated from possible numbers and Os and Is are eliminated from possible letters, how many different license plates can be made? 109. Seating Arrangements. Five friends buy five season tickets to the Philadelphia Eagles. To be fair, they change the seating arrangement every game. How many different seating arrangements are there for the five friends? How many seasons would they have to buy tickets in order to sit in all of the combinations (each season has eight home games)? 110. Safe. A safe will open when the correct choice of three numbers (1 to 60) is selected in a specific order. How many possible permutations are there? 111. Raffle. A fundraiser raffle is held to benefit the Make a Wish Foundation, and 100 raffle tickets are sold. Four prizes are raffled off. First prize is a round-trip ticket on American Airlines, second prize is a round of golf for four people at a Links golf course, third prize is a $100 gift certificate to the Outback Steakhouse, and fourth prize is a half-hour massage. How many possible winning scenarios are there if all 100 tickets are sold to different people? 112. Sports. There are 117 Division 1-A football teams in the United States. At the end of the regular season is the Bowl Championship Series, and the top two teams play each other in the championship game. Assuming that any two Division 1-A teams can advance to the championship, how many possible matchups are there for the championship game? 113. Cards. In a deck of 52 cards, how many different 6-card hands can be dealt? 114. Blackjack. In a game of two-deck blackjack (104 cards), how many 2-card combinations are there that equal 21, that is, ace and a 10 or face card—jack, queen, or king? 12.7 Probability 115. Coin Tossing. For the experiment of tossing a coin four times, what is the probability of getting all heads? 116. Dice. For an experiment of tossing two dice, what is the probability that the sum of the dice is odd? 117. Dice. For an experiment of tossing two dice, what is the probability of not rolling a combined 7? 118. Cards. For a deck of 52 cards, what is the probability of drawing a diamond? For Exercises 119–122, let P1E12 5 1 3 and P1E22 5 1 2 and find the probability of the event. 119. Probability. Find the probability of an event E1 not occurring. 120. Probability. Find the probability of either E1 or E2 occurring if E1 and E2 are mutually exclusive. 121. Probability. Find the probability of either E1 or E2 occurring if E1 and E2 are not mutually exclusive and P1E1 ∩ E22 5 1 4. 122. Probability. Find the probability of both E1 and E2 occurring if E1 and E2 are independent. 123. Cards. With a 52-card deck, what is the probability of drawing an ace or a 2? 124. Cards. By drawing twice, what is the probability of drawing an ace and then a 2? (Assume that after the first card is drawn it is not put back into the deck.) 125. Children. What is the probability that in a family of five children at least one is a girl? 126. Sports. With the salary cap in the NFL, it is said that on “any given Sunday” any team could beat any other team. If we assume every week a team has a 50% chance of winning, what is the probability that a team will go 11–1? R E VI E W E XERCISES Review Exercises 1165 Technology Section 12.1 127. Use a graphing calculator “SUM” to find the sum of the series a 6 n51 1 n2. 128. Use a graphing calculator “SUM” to find the sum of the infinite series a q n51 1 n, if possible. Section 12.2 129. Use a graphing calculator to sum a 75 n51 C3 2 1 6 7 1n 2 12D. 130. Use a graphing calculator to sum a 264 n51 C219 1 1 3 1n 2 12D. Section 12.3 131. Apply a graphing utility to plot y1 5 1 2 2x 1 4x2 2 8x3 1 16x4 and y2 5 1 1 1 2x, and let x range from 320.3, 0.34. Based on what you see, what do you expect the geometric series a q n50 1212n12x2n to sum to in this range of x values? 132. Does the sum of the infinite series a q n50a e pb n exist? Use a graphing calculator to find it and round to four decimal places. Section 12.4 133. Use a graphing calculator to sum the series 2 1 7 1 12 1 17 1 c 1 15n 2 32 on the left side, and evaluate the expression n 2 15n 2 12 on the right side for n 5 200. Do they agree with each other? Do your answers confirm the proof for Exercise 77? 134. Use a graphing calculator to plot the graphs of y1 5 2x2 and y2 5 1x 1 122 in the 3100, 10004 by 310,000, 2,500,0004 viewing rectangle. Do your answers confirm the proof for Exercise 78? Section 12.5 135. With a graphing utility, plot y1 5 1 1 8x, y2 5 1 1 8x 1 24x2, y3 5 1 1 8x 1 24x2 1 32x3, y4 5 1 1 8x 1 24x2 1 32x3 1 16x4, and y5 5 11 1 2x24 for 20.1 , x , 0.1. What do you notice happening each time an additional term is added? Now, let 0.1 , x , 1. Does the same thing happen? 136. With a graphing utility, plot y1 5 1 2 8x, y2 5 1 2 8x 1 24x2, y3 5 1 2 8x 1 24x2 2 32x3, y4 5 1 2 8x 1 24x2 2 32x3 1 16x4, and y5 5 11 2 2x24 for 20.1 , x , 0.1 What do you notice happening each time an additional term is added? Now, let 0.1 , x , 1. Does the same thing happen? Section 12.6 137. Use a graphing calculator to evaluate: a. 1 6P7 7! b. 16C7 c. Are answers in (a) and (b) the same? d. Why? 138. Use a graphing calculator to evaluate: a. 52P 6 6! b. 52C6 c. Are answers in (a) and (b) the same? d. Why? Section 12.7 In Exercises 139 and 140, when two dice are rolled, the probability of getting a sum of 9 is 1 9 and the probability of not getting a sum of 9 is 8 9. If two dice are rolled n times, the probability of getting a sum of 9 exactly k times can be found by using the binomial theorem nCk a1 9b k a8 9b n2k . 139. If two dice are rolled 10 times, find the probability of getting a sum of 9 exactly three times. Round your answer to four decimal places. 140. If a die is rolled 8 times, find the probability of getting a sum of 9 at least two times. Round your answer to four decimal places. PR ACTICE TEST [ C H AP T E R 1 2 PRACTIC E TES T ] For Exercises 1 –5, use the sequence 1, x, x2, x3, P 1. Write the nth term of the sequence. 2. Classify this sequence as arithmetic, geometric, or neither. 3. Find the nth partial sum of the series Sn. 4. Assuming this sequence is infinite, write the series using sigma notation. 5. Assuming this sequence is infinite, what condition would have to be satisfied in order for the sum to exist? 6. Find the following sum: 1 3 1 1 9 1 1 27 1 1 81 1 c. 7. Find the following sum: a 10 n51 3⋅A1 4B n. 8. Find the following sum: a 50 k51 12k 1 12. 9. Write the series using sigma notation, then find its sum: 2 1 7 1 12 1 17 1 c1 497. 10. Use mathematical induction to prove that 2 1 4 1 6 1 c1 2n 5 n2 1 n. 11. Evaluate 7! 2!. 12. Find the third term of 12x 1 y25. In Exercises 13–16, evaluate the expressions. 13. a15 12b 14. ak kb 15. 14P 3 16. 200C3 17. Expand the expression ax2 1 1 xb 5 . 18. Use the binomial theorem to expand the binomial 13x 2 224. 19. Explain why there are always more permutations than combinations. 20. What is the probability of not winning a trifecta (selecting the first-, second-, and third-place finishers) in a horse race with 15 horses? For Exercises 21–23, refer to a roulette wheel with 18 red, 18 black, and 2 green slots. 00 1 13 36 24 3 15 34 22 5 17 32 20 7 11 30 26 9 28 0 2 14 35 23 4 16 33 21 6 18 31 19 8 12 29 25 10 27 21. Roulette. What is the probability of the ball landing in a red slot? 22. Roulette. What is the probability of the ball landing in a red slot 5 times in a row? 23. Roulette. If the four previous rolls landed on red, what is the probability that the next roll will land on red? 24. Marbles. If there are four red marbles, three blue marbles, two green marbles, and one black marble in a sack, find the probability of pulling out the following order: black, blue, red, red, green. 25. Cards. What is the probability of drawing an ace or a diamond from a deck of 52 cards? 26. Human Anatomy. Vasopressin is a relatively simple protein that is found in the human liver. It consists of eight amino acids that must be joined in one particular order for the effective functioning of the protein. a. How many different arrangements of the eight amino acids are possible? b. What is the probability of randomly selecting one of these arrangements and obtaining the correct arrangement to make vasopressin? 27. Find the constant term in the expression ax3 1 1 x3b 20 . 28. Use a graphing calculator to sum a 125 n51C211 4 1 5 6 1n 2 12D. 1166 CHAPTER 12 Sequences, Series, and Probability CU MU LA TIV E TEST Cumulative Test 1167 [CH AP TERS 1–12 CUM ULAT IVE T E S T ] 1. Simplify 10x 1 1 3x 2 2 1 5 1 4x 2 2 3x. 2. The length of a rectangle is 5 less than twice the width and the perimeter is 38 inches. What are the dimensions of the rectangle? 3. Solve using the quadratic formula: 2x2 2 5x 1 11 5 0. 4. Solve and express the solution in interval notation: 0 x 2 5 0 . 3. 5. Write an equation of the line with slope undefined and x-intercept 128, 02. 6. Find the x-intercept and y-intercept and slope of the line 3x 2 5y 5 15. 7. Using the function ƒ1x2 5 x2 2 3x 1 2, evaluate the difference quotient ƒ1x 1 h2 2 ƒ1x2 h . 8. Find the composite function f + g, and state the domain for ƒ1x2 5 4 2 1 x and g1x2 5 1 2x 1 1. 9. Find the vertex of the parabola ƒ1x2 5 20.04x2 1 1.2x 2 3. 10. Factor the polynomial P1x2 5 x4 1 8x2 2 9 as a product of linear factors. 11. Find the vertical and horizontal asymptotes of the function: ƒ1x2 5 5x 2 7 3 2 x 12. How much money should be put in a savings account now that earns 4.7% a year compounded weekly, if you want to have $65,000 in 17 years? 13. Evaluate log2 6 using the change-of-base formula. Round your answer to three decimal places. 14. Solve ln 15x 2 62 5 2. Round your answer to three decimal places. 15. Solve the system of linear equations. 8x 2 5y 5 15 y 5 8 5 x 1 10 16. Solve the system of linear equations. 2x 2 y 1 z 5 1 x 2 y 1 4z 5 3 17. Maximize the objective function z 5 4x 1 5y, subject to the constraints x 1 y # 5, x $ 1, y $ 2. 18. Solve the system using Gauss–Jordan elimination. x 1 5y 2 2z 5 3 3x 1 y 1 2z 5 23 2x 2 4y 1 4z 5 10 19. Given A 5 c3 0 4 1 27 5d B 5 c8 9 22 0 6 21d C 5 c9 0 1 2d find C 1A 1 B2. 20. Calculate the determinant. 3 2 1 22 5 4 1 21 0 3 3 21. Find the equation of a parabola with vertex (3, 5) and directrix x 5 7. 22. Graph x2 1 y2 , 4. 23. Find the sum of the finite series a 4 n51 2n21 n! . 24. Classify the sequence as arithmetic, geometric, or neither: 5, 15, 45, 135, . . . 25. There are 10 true/false questions on a quiz. Assuming no answers are left blank, how many different ways can you answer the questions on the quiz? 26. Use a graphing calculator to sum a 164 n51 S1 4 1 3 2 1n 2 12T. 27. Find the constant term in the expression ax3 2 1 x3b 14 . Answers to Odd-Numbered Exercises [ [ CH A P T E R 0 Section 0.1 1. rational 3. irrational 5. rational 7. irrational 9. a. 7.347 b. 7.347 11. a. 2.995 b. 2.994 13. a. 0.234 b. 0.234 15. a. 5.238 b. 5.238 17. 4 19. 26 21. 2130 23. 17 25. 3 27. x 1 y 2 z 29. 23x 2 y 31. 3 5 33. 68 35. 29x 2 y 37. 210 39. 5 41. 19 12 43. 1 2 45. 23 12 47. 1 27 49. x 3 51. x 21 53. 61y 60 55. 11 30 57. 4 3 59. 3 35 61. 12b a 63. 3 4 65. 4xy 3 67. 8x y 69. 2 3 71. 3 25 73. $18,152,810,000,000 75. $56,347 77. The mistake is rounding the number that is used in the rounding. Look to the right of the number; if it is less than 5, round down. 79. The 22 did not distribute to the second term. 81. false 83. true 85. x 2 0 87. 27x 1 12y 2 75 89. irrational 91. rational Section 0.2 1. 256 3. 2243 5. 225 7. 216 9. 1 11. 0.1 13. 1 64 15. 2150 17. 5 19. 254 21. x5 23. 1 x 25. x6 27. 64a3 29. 28t3 31. 75x5y5 33. y2 x2 35. 2 1 2xy 37. 16 b4 39. a4 81b6 41. 1 a6b2 43. x8y2 45. x14 4y 47. y10 232x22 49. 2 y24 a12x3 51. 226 53. 2.76 3 107 55. 9.3 3 107 57. 5.67 3 1028 59. 1.23 3 1027 61. 47,000,000 63. 23,000 65. 0.000041 67. a. 2.08 3 109 ft b. Yes—almost 16 times 69. 2.0 3 108 miles 71. 0.00000 155 meter 73. In the power property, exponents are added 1not multiplied). 75. In the first step, 1 y322 5 y6. Exponents should have been multiplied (not raised to power). 77. false 79. false 81. amnk 83. 216 85. 156 87. 3.4 acres/person 89. 2.0 3 1021 or 0.2 91. same 93. 5.11 3 1014 Section 0.3 1. 27x4 2 2x3 1 5x2 1 16 degree 4 3. 26x3 1 4x 1 3 degree 3 5. 15 degree 0 7. y 2 2 degree 1 9. 2x2 1 5x 1 5 11. 2x2 2 x 2 18 13. 4x4 2 10x2 2 1 15. 2z2 1 2z 2 3 17. 211y 3 2 16y2 1 16y 2 4 19. 24x 1 12y 21. x2 2 x 1 2 23. 24t 2 1 6t 2 1 25. 35x2y3 27. 2x5 2 2x4 1 2x3 29. 10x4 2 2x3 2 10x2 31. 2x5 1 2x4 2 4x3 33. 2a3b2 1 4a2b3 2 6ab4 35. 6x2 2 5x 2 4 37. x2 2 4 39. 4x2 2 9 41. 24x2 1 4x 2 1 43. 4x4 2 9 45. 2y4 2 9y3 1 5y2 1 7y 47. x3 2 x2 1 x 1 3 49. t 2 2 4t 1 4 51. z 2 1 4z 1 4 53. x 2 1 2xy 1 y2 2 6x 2 6y 1 9 55. 25x2 2 20x 1 4 57. 6y3 1 5y2 2 4y 59. x4 2 1 61. ab2 1 2b3 2 9a3 2 18a2b 63. 2x2 2 3y2 2 xy 1 3xz 1 8yz 2 5z2 65. P 5 11x 2 100 67. P 5 2x2 1 200x 2 7500 69. V 5 115 2 2x2 18 2 2x2x 5 4x3 2 46x2 1 120x 71. a. P 5 12px 1 4x 1 102 feet b. A 5 1px2 1 4x2 1 10x2 feet2 73. F 5 3k 100 75. The negative was not distributed through the second polynomial. 77. true 79. false 81. m 1 n 83. 2401x4 2 1568x2y4 1 256y8 85. x3 2 a3 87. 12x 1 32 1x 2 42 and 2x2 2 5x 2 12 Answers that require a proof, graph, or otherwise lengthy solution are not included. 1169 1170 Answers to Odd-Numbered Exercises Section 0.4 1. 51x 1 52 3. 212t2 2 12 5. 2x 1x 2 52 1x 1 52 7. 3x1x2 2 3x 1 42 9. x 1x 2 82 1x 1 52 11. 2xy12xy2 1 32 13. 1x 2 32 1x 1 32 15. 12x 2 32 12x 1 32 17. 21x 2 72 1x 1 72 19. 115x 2 13y2 115x 1 13y2 21. 1x 1 422 23. 1x2 2 222 25. 12x 1 3y22 27. 1x 2 322 29. 1x2 1 122 31. 1 p 1 q22 33. 1t 1 32 1t2 2 3t 1 92 35. 1 y 2 42 1y2 1 4y 1 162 37. 12 2 x2 14 1 2x 1 x22 39. 1 y 1 52 1y 2 2 5y 1 252 41. 13 1 x2 19 2 3x 1 x22 43. 1x 2 52 1x 2 12 45. 1 y 2 32 1 y 1 12 47. 12y 1 12 1 y 2 32 49. 13t 1 12 1t 1 22 51. 123t 1 22 12t 1 12 53. 1x2 1 22 1x 2 32 55. 1a3 2 82 1a 1 22 5 1a 1 22 1a 2 221a2 1 2a 1 42 57. 13y 2 5r2 1x 1 2s2 59. 14x 2 y2 15x 1 2y2 61. 1x 2 2y2 1x 1 2y2 63. 13a 1 72 1a 2 22 65. prime 67. prime 69. 213x 1 22 1x 1 12 71. 13x 2 y2 12x 1 5y2 73. 912s 2 t2 12s 1 t2 75. 1ab 2 5c2 1ab 1 5c2 77. 1x 2 22 14x 1 52 79. x 13x 1 12 1x 2 22 81. x 1x 2 32 1x 1 32 83. 1x 2 12 1 y 2 12 85. 1x2 1 321x2 1 22 87. 1x 2 62 1x 1 42 89. x 1x 1 52 1x2 2 5x 1 252 91. 1x 2 32 1x 1 32 1x2 1 92 93. 213x 1 42 95. 1x 1 22 12x 2 152 97. 2218t 2 12 1t 1 52 99. Last step. 1x 2 12 1x2 2 92 5 1x 2 12 1x 2 32 1x 1 32 101. false 103. true 105. 1an2 bn21an1 bn2 107. 8x3 1 1 and 12x 1 12 14x2 2 2x 1 12 Section 0.5 1. x 2 0 3. x 2 1 5. x 2 21 7. p 2 61 9. no restrictions 11. 1x 2 92 21x 1 92; x 2 29, 23 13. x 2 3 2 ; x 2 21 15. 213y 1 12 9y ; y 2 0, 1 2 17. y 1 1 5 ; y 2 1 5 19. 3x 1 7 4 ; x 2 4 21. x 1 2; x 2 2 23. 1; x 2 27 25. x2 1 9 2x 1 9; x 2 29 2 27. x 1 3 x 2 5; x 2 22, 5 29. 3x 1 1 x 1 5 ; x 2 25, 1 2 31. 3x 1 5 x 1 1 ; x 2 21, 2 33. 215x 1 62 5x 2 6 ; x 2 0, 6 5 35. 2 3; x 2 0, 61 37. 31x 1 22; x 2 25, 0, 2 39. t 2 3 31t 1 22; t 2 22, 3 41. 3t 1t2 1 42 1t 2 321t 1 22; t 2 22, 3 43. 3y1y 2 22 y 2 3 ; y 2 22, 3 45. 12x 1 321x 2 52 2x1x 1 522 ; x 2 0, 65 47. 12x 2 7212x 1 72 13x 2 5214x 1 32 ; x 2 2 3 4, 7 2, 65 3 49. x 4; x 2 0 51. x 1 2 2 ; x 2 62 53. x 1 1 5 ; x 2 61 55. 1 211 2 p2; p 2 2, 61 57. 6 2 n n 2 3; n 2 26, 63 59. 2t1t 2 32 5 ; t 2 21, 2 61. 5w2 w 1 1; w 2 0, 61 63. 1x 2 321x 2 42 1x 2 221x 2 92; x 2 27, 25, 2, 4, 9 65. x 5x 1 2; x 2 25 3, 21 4, 0, 62 5 67. 13 5x; x 2 0 69. 5p2 2 7p 1 3 1 p 2 221 p 1 12; p 2 21, 2 71. 4 5x 2 1; x 2 1 5 73. 3y3 2 5y2 2 y 1 1 y2 2 1 ; y 261 75. x2 1 4x 2 6 x2 2 4 ; x 262 77. x 1 4 x 1 2; x 2 62 79. 12 1 7b a2 2 b2 ; a 2 6b 81. 7x 2 20 x 2 3 ; x 2 3 83. 1 2 x x 2 2; x 2 0, 2 85. x 3x 2 1; x 2 0, 61 3 87. x 1 1 x 2 1; x 2 0, 61 89. x1x 1 12 1x 2 121x 1 22; x 2 22, 61 91. A 5 p11 1 i25 11 1 i25 2 1 93. R1 R2 R2 1 R1 95. x 1 1 where x 2 21. 97. false 99. false 101. 1x 1 a21x 1 d2 1x 1 b21x 1 c2; x 2 2b, 2c, 2d 103. yes Answers to Odd-Numbered Exercises 1171 105. a. 1x 2 121x 1 22 1x 2 221x 1 12; x 2 22, 21, 2 b. c. agree as long as x 2 21, 62 Section 0.6 1. 10 3. 212 5. 26 7. 7 9. 1 11. 0 13. not a real number 15. 23 17. 4 19. 22 21. 21 23. 27 25. 24!2 27. 8!5 29. 2!6 31. 2" 3 6 33. !21 35. 40 0x0 37. 20 x 0 !y 39. 23x2y2 " 3 3y2 41. !3 3 43. 2!11 33 45. 3 1 3!5 24 47. 23 2 2!2 49. 23A!2 1 !3B 51. 2A3!2 2 2!3B 3 53. 5!5 2 2 11 55. A !7 1 !3BA !2 1 !5B 23 57. x3y4 59. 1 y2 61. x7/6y2 63. x8/3 2 65. x1/31x 2 22 1x 1 12 67. 7x3/711 2 2x3/7 1 3x2 69. 9 sec 71. d < 9.54 astronomical units. 73. The 4 should also have been squared. 75. false 77. false 79. amnk 81. 1 2 83. a 2 2!ab 1 b 1a 2 b22 85. 3.317 87. a. 10!2 2 8!3 b. 0.2857291632 c. yes Section 0.7 1. 4i 3. 2i!5 5. 24 7. 8i 9. 3 2 10i 11. 210 2 12i 13. 2 2 9i 15. 10 2 14i 17. 2 2 2i 19. 21 1 2i 21. 12 2 6i 23. 96 2 60i 25. 248 1 27i 27. 2102 1 30i 29. 5 2 i 31. 13 1 41i 33. 87 1 33i 35. 26 1 48i 37. 56 9 2 11 18i 39. 37 1 49i 41. 4 2 7i; 65 43. 2 1 3i; 13 45. 6 2 4i; 52 47. 22 1 6i; 40 49. 22i 51. 3 10 1 1 10 i 53. 3 13 2 2 13 i 55. 14 53 2 4 53 i 57. 2i 59. 2 9 34 1 19 34 i 61. 18 53 2 43 53 i 63. 66 85 1 43 85 i 65. 2i 67. 1 69. 21 2 20i 71. 25 1 12i 73. 18 1 26i 75. 22 2 2i 77. 8 2 2i ohms 79. multiplied by the denominator 4 2 i instead of the conjugate 4 1 i 81. true 83. true 85. 1x 1 i221x 2 i22 87. 41 2 38i 89. 2 125 1 11 125i Review Exercises 1. a. 5.22 b. 5.21 3. 4 5. 22 7. 2 x 12 9. 9 11. 28z3 13. 9 2x2 15. 2.15 3 1026 17. 14z2 1 3z 2 2 19. 45x2 2 10x 2 15 21. 15x2y2 2 20xy3 23. x2 1 2x 2 63 25. 4x2 2 12x 1 9 27. x4 1 2x2 1 1 29. 2xy2 17x 2 5y2 31. 1x 1 52 12x 2 12 33. 14x 1 52 14x 2 52 35. 1x 1 52 1x2 2 5x 1 252 37. 2x 1x 1 52 1x 2 32 39. 1x2 2 22 1x 1 12 41. x 2 63 43. x 1 2; x 2 2 45. t 1 3 t 1 1; t 2 21, 2 47. 1x 1 521x 1 22 1x 1 322 ; x 2 23, 1, 2 49. 2 1x 1 121x 1 32; x 2 21, 23 51. 10x 2 25 20x 2 59; x 2 3, 59 20 53. 2!5 55. 25xy" 3 x2y 57. 26!5 59. 23 2 !5 61. 2 1 "3 63. 9x 2@3 16 65. 5 1 @6 67. 13i 69. 2i 71. 8 2 6i 73. 14 1 4i 75. 12 77. 233 1 56i 79. 2 5 1 1 5 i 81. 38 41 2 27 41 i 83. 210 3 i 85. rational 87. 1.945 3 1026 89. The solid curve represents the graph of y 5 12x 1 323 and y 5 8x3 1 36x2 1 54x 1 27. 91. None, because x2 2 3x 1 18 is prime. 93. a. x 1x 2 42 1x 1 221x 2 22 b. c. x 2 0, 62 95. a. 2!5 1 2!2 b. 7.30056308 c. yes 97. 2868 2 6100i 99. 7 40000 2 3 5000 i 1172 Answers to Odd-Numbered Exercises Practice Test 1. 4 3. 297 5. 2i0 x 0 !3 7. y5z1/2 x7/2 9. 27!2 11. 2y2 2 12y 1 20 13. 1x 2 42 1x 1 42 15. 12x 1 3y22 17. 12x 1 12 1x 2 12 19. t1t 1 12 12t 2 32 21. 1x 1 4y2 1x 2 3y2 23. 313 1 x2 19 2 3x 1 x 22 25. 5x 2 2 x1x 2 12 ; x 2 0, 1 27. 1 1x 1 121x 2 12; x 2 6 1 29. 2 1 x 1 3; x 2 5 2, 6 3 31. 28 2 26i 33. 2 2 27!3 59 35. 2 1 x1x 1 12 ; x 2 0, 61 37. 2.330 C HA P TE R 1 Section 1.1 1. x 5 7 3. n 5 15 5. x 5 28 7. n 5 15 9. x 5 4 11. m 5 2 13. t 5 7 5 15. x 5 210 17. n 5 2 19. x 5 12 21. t 5 215 2 23. x 5 21 25. p 5 29 2 27. x 5 1 4 29. x 5 23 2 31. a 5 28 33. x 5 215 35. c 5 235 13 37. m 5 60 11 39. x 5 36 41. p 5 8 43. y 5 22 45. p 5 2 47. no solution 49. y 2 0, y 5 3 10 51. x 2 0, x 5 1 2 53. a 2 0, a 5 1 6 55. x 2 2, no solution 57. p 2 1, no solution 59. x 2 22, x 5 210 61. n 2 21, 0, no solution 63. a 2 0, 23, no solution 65. n 2 1, n 5 53 11 67. x 2 21 5, 1 2, x 5 23 69. t 2 1, no solution 71. C 5 5 9 F 2 160 9 73. 84 min 75. 17 min 77. a. C 5 15,000 1 2500x b. 2200 days 79. 24 ml of the 125 mg/5 ml suspension of amoxicillin 81. l 2 0 83. The error is forgetting to do to both sides of any equation what is done to one side. The correct answer is x 5 5. 85. You cannot cross multiply with the 23 in the problem. You must find a common denominator first. The correct solution is p 5 6 5. 87. false 89. true 91. x 5 c 2 b a 93. x 5 ba c , x 2 6a 95. no solution 97. x 5 by a 2 y 2 cy, x 2 0, 2 b c 1 1 99. x 5 2 101. all real numbers 103. no solution 105. x 5 5426 Section 1.2 1. $242.17 3. $13.76 5. $147,058.82 1$22,058.82 saved2 7. 12 miles 9. 9 hours of sleep 11. 270 units 13. 24 15. 8, 10 17. 20 inches 19. width 5 30 yards, length 5 100 yards 21. r1 5 3 feet, r2 5 6 feet 23. 300 feet 25. The body is 63 inches or 5.25 feet 27. $20,000 at 4%, $100,000 at 7% 29. $3000 at 10%, $5500 at 2%, $5500 at 40% 31. 6 trees, 27 shrubs 33. 70 ml of 5% HCl, 30 ml of 15% HCl 35. 31 3 gallons 37. The entire bag was gummy bears. 39. 9 minutes 41. $3.07 per gallon 43. 233 ml 45. 2.3 mph 47. walker: 4 mph, jogger: 6 mph 49. bicyclist: 6 minutes, walker: 18 minutes 51. 22.5 hours 53. 2.4 hours 55. 330 hertz, 396 hertz 57. 77.5, 92.5 59. 2 field goals and 6 touchdowns 61. 3.5 feet from the center 63. Fulcrum is 0.4 units from Maria and 0.6 units from Max 65. 7.5 cm in front of lens 67. Image distance is 3 centimeters, object distance is 6 centimeters. 69. P 2 2l 2 5 w 71. 2A b 5 h 73. A l 5 w 75. V lw 5 h 77. Janine: 58 mph; Tricia: 70 mph 79. $191,983.35 81. Plan B is better for 5 or fewer plays/month. Plan A is better for 6 or more plays/month. 20 40 60 80 400 600 800 1000 A B x y Section 1.3 1. x 5 3 or x 5 2 3. p 5 5 or p 5 3 5. x 5 24 or x 5 3 7. x 5 21 4 9. y 5 1 3 11. y 5 0 or y 5 2 13. p 5 2 3 15. x 5 23 or x 5 3 17. x 5 26 or x 5 2 19. p 5 25 or p 5 5 21. x 5 22 or x 5 2 23. p 5 62!2 25. x 5 63i 27. x 5 23, 9 29. x 5 23 6 2i 2 31. x 5 2 6 3!3 5 Answers to Odd-Numbered Exercises 1173 33. x 5 22, 4 35. x2 1 6x 1 9 37. x2 2 12x 1 36 39. x2 2 1 2x 1 1 16 41. x2 1 2 5x 1 1 25 43. x2 2 2.4 1 1.44 45. x 5 23, 1 47. t 5 1, 5 49. y 5 1, 3 51. p 5 24 6 !10 2 53. x 5 1 2, 3 55. x 5 4 6 3!2 2 57. t 5 23 6 !13 2 59. s 5 21 6 !3i 2 61. x 5 3 6 !57 6 63. x 5 1 6 4i 65. x 5 27 6 !109 10 67. x 5 24 6 !34 3 69. 1 real solution 71. 2 real solutions 73. 2 complex solutions 75. v 5 22, 10 77. t 5 26, 1 79. x 5 27, 1 81. p 5 4 6 2!3 83. w 5 21 6 !167 i 8 85. p 5 9 6 !69 6 87. t 5 10 6 !130 10 89. x 5 3 or x 5 4 91. x 5 23 4 or x 5 2 93. x 5 20.3, 0.4 95. t 5 2 1January 20152 and t 5 4 1March 20152 97. 31,000 units 99. $1 per bottle 101. 3 days 103. a. 55.25 square inches b. 4x2 1 30x 1 55.25 c. 4x2 1 30x represents the increase in useable area of the paper. d. x < 0.2 inches 105. 20 inches 107. 17, 18 109. Length: 15 ft, width: 9 ft 111. Base is 6 units and height is 20 units. 113. 2.5 seconds after it is dropped. 115. 21.2 feet 117. 5 ft 3 5 ft 119. 2.3 feet 121. 10 days 123. The problem is factored incorrectly. The correction would be t 5 6 and t 5 21. 125. When taking the square root of both sides, the i is missing from the right side. The correction would be a 5 63 4 i. 127. false 129. true 131. x2 2 2ax 1 a2 5 0 133. x2 2 7x 1 10 5 0 135. t 5 6Å 2s g 137. c 5 6"a2 1 b2 139. x 5 0, 62 141. x 5 21, 62 143. 2b 2a 1 "b2 2 4ac 2a 2 b 2a 2 "b2 2 4ac 2a 5 22b 2a 5 2b a 145. x2 2 6x 1 4 5 0 147. 250 mph 149. ax2 2 bx 1 c 5 0 151. Small jet: 300 mph, 757: 400 mph 153. x 5 21, 2 155. a. x 5 22, 4 b. b 5 23: x 5 1 6 !2 i b 5 21, x 5 1 b 5 0, x 5 0, 2 b 5 5, x 5 1 6 !6 Section 1.4 1. t 5 9 3. p 5 8 5. no solution 7. x 5 5 9. y 5 21 2 11. x 5 4 13. y 5 0, 25 15. s 5 3, 6 17. x 5 23, 21 19. x 5 0 21. x 5 1 and x 5 5 23. x 5 7 25. x 5 23 and x 5 215 4 27. x 5 5 2 29. no solution 31. x 5 1 33. x 5 4 and x 5 28 35. x 5 1, 5 37. x 5 7 39. x 5 4 41. x 5 0, x 5 28 43. x 5 61, x 5 6!2 45. x 5 6!6 i 2 , x 5 6!2 i 47. x 5 25 2, x 5 21 49. t 5 5 4, t 5 3 51. x 5 61, 6i, x 5 61 2, 61 2 i 53. y 5 23 4, y 5 1 55. z 5 1 1174 Answers to Odd-Numbered Exercises 57. x 5 5 59. x 5 29 or x 5 7 61. t 5 227, t 5 8 63. x 5 24 3, x 5 0 65. x 5 3 8, x 5 2 3 67. u 5 68, u 5 61 69. t 5 !3 71. x 5 0, 23, 4 73. p 5 0, 63 2 75. u 5 0, 62, 62i 77. x 5 63, 5 79. y 5 22, 5, 7 81. x 5 0, 3 1x 5 21 does not check2 83. t 5 65 85. y 5 2, 3 87. January and September 89. 162 cm 91. a 5 71 years old 93. t 5 4!3 < 7 months 1Oct. 20042 95. 132 feet 97. 25 cm 99. 80% of the speed of light 101. no solution 103. The error is not converting u back to x using the substitution. 105. true 107. false 109. 30, q2 111. x 5 0, x 5 22 3, x 5 21, x 5 1 3 113. x 5 22 115. x 5 313 64 > 4.891 117. no solution 119. x 5 81 y1 5 x1/2, y2 5 2 4x1/4 1 21 121. no real solution y1 5 x22, y2 5 3x21 2 10 Section 1.5 1. 33, q2 … 0 1 2 3 4 5 6 … 3. 12q, 254 … 27 26 25 24 23 22 21 0 … 5. 322, 32 … 23 22 21 0 1 2 3 4 … 7. 123, 54 … 23 22 21 0 1 2 3 4 5 … 9. 30, 04 … 23 22 21 0 1 2 3 … 11. 34, 64 … 3 4 5 6 7 … 13. 328, 264 … 29 28 27 26 25 … 15. [ … 23 22 21 0 1 2 3 … 17. Ex 0 0 # x , 2F 19. Ex 0 27 , x , 22F 21. Ex 0 x # 6F 23. Ex 0 2q , x , qF 25. 23 , x # 7 123, 74 27. 3 # x , 5 33, 52 29. 22 # x 322, q2 31. 2q , x , 8 12q, 82 33. 125, 32 … 25 24 23 22 21 0 1 2 3 … 35. 326, 52 … 27 26 25 … 0 1 2 3 4 5 … 37. 321, 14 … 22 21 0 1 2 3 … 39. 31, 42 … 0 1 2 3 4 … 41. 321, 22 … 22 21 0 1 2 3 … 43. 12q, 42 ∪ 14, q2 … 2 3 4 5 6 7 … 45. 12q, 234 ∪ 33, q2 … 23 22 21 0 1 2 3 … 47. 123, 24 … 24 23 22 21 0 1 2 3 4 … 49. [ … 23 22 21 0 1 2 3 … 51. 12q, 22 ∪ 33, 52 53. 12q, 244 ∪ 12, 54 55. 324, 222 ∪ 13, 74 57. 126, 234 ∪ 30, 42 59. 12q, 102 61. 12q, 24 63. 12q, 224 65. 322, q2 67. 12q, 20.52 69. 123, q2 71. 12q, 62 73. 12q, 284 75. 12q, 274 77. 12q, 12 79. 125, 22 81. 326, 22 83. 328, 42 85. 126, 62 87. S1 2, 5 4T 89. 324.5, 0.54 91. 128 # w # 164 93. More than 50 dresses 95. 285,700 units 97. between 33% and 71% intensities 99. Least: 878 minutes, most: 1013 minutes 101. 92 103. $21,537.69 , invoice price , $24,346.96 Answers to Odd-Numbered Exercises 1175 105. 0.9 rT # rR # 1.1 rT 107. 0.85L # B # 0.95L 109. 4 times 111. $5,156.25 # T # $18,481 113. The correction should be 321, 42, and the graph is not correct. 115. You must reverse the sign. 117. true 119. a and b 121. a and b 123. c 125. 12q, 04 127. no solution 129. a. x . 0.83582 1rounded2 b. c. yes 131. a. 122, 52 b. c. yes 133. a. 12q, q2 b. c. yes Section 1.6 1. 12q, 224 ∪ 35, q2 3. 321, 64 5. 123, 212 7. C21, 3 2D 9. A1 3, 1 2B 11. A2q,21 2D ∪ 33, q2 13. A2q, 21 2 !5D ∪ C21 1 !5, qB 15. A2 2 !10, 2 1 !10B 17. 12q, 04 ∪ 33, q2 19. 30, 24 21. 12q, 232 ∪ 13, q2 23. 129, 92 25. 12q, q2 27. no real solution 29. 10, q2 31. 12q, 232 ∪ 10, q2 33. 12q, 234 ∪ 14, q2 35. 12q, 222 ∪ 321, 22 37. 125, 34 ∪ 15, q2 39. A23 2, 1B 41. 12q, 254 ∪ 122, 04 43. 122, 22 45. no real solution 47. 12q, q2 49. 323, 32 ∪ 13, q2 51. 123, 214 ∪ 13, q2 53. 12q, 242 ∪ 12, 54 55. 126, 242 ∪ 14, 82 57. 12q, 222 ∪ 12, q2 59. Between 30 and 100 orders 61. For years 3–5, the car is worth more than you owe. In the first 3 years you owe more than the car is worth. 63. 75 seconds 65. Between 20 and 30 feet 67. A price increase less than $1 per bottle or greater than $20 per bottle 69. $3342 to $3461 per acre 71. Cannot divide by x. 12q, 02 ∪ 13, q2 73. You must consider the x 5 22 as a critical point. 75. false 77. either no real solution or infinitely many solutions 79. 12q, q2 81. 12q, q2 83. 12q, q2 85. 120.8960, 1.62332 87. 122, 02 89. A5 3, 5B Section 1.7 1. x 5 23 or x 5 3 3. no solution 5. t 5 25 or t 5 21 7. p 5 10 or p 5 4 9. y 5 5 or y 5 3 11. x 5 23 or x 5 3 13. x 5 28 or x 5 1 15. t 5 4 or t 5 2 17. x 5 8 or x 5 21 19. y 5 0 or y 5 2 3 21. x 5 80 21 or x 5 2 3 23. x 5 223 14 or x 5 47 14 25. x 513 or x 5 23 27. x 5 24 or x 5 8 29. x 5 0 or x 5 4 31. p 5 7 or p 5 213 33. y 5 9 or y 5 25 35. x 5 6!5 or x 5 6!3 37. x 5 62 39. 127, 72 41. 12q, 254 ∪ 35, q2 43. 1210, 42 45. 12q, 22 ∪ 16, q2 47. 33, 54 49. 12q, q2 51. 124, 12 53. 12q, 24 ∪ 35, q2 55. 12q, q2 57. A2q, 23 2D ∪ C3 2, qB 59. 326, 44 61. 12q, q2 63. 12q, 232 ∪ 13, q2 65. A1 4, 3 4B 67. 122.769, 21.3852 69. 323, 34 71. 0x 2 20 , 7 73. 0 x 2 3 20 $ 1/2 75. 0x 2 a0 # 2 77. 0T 2 830 # 15 79. Win: d , 4, tie: d 5 4 81. When the number of units sold was between 25 and 75 units 83. The mistake was that x 2 3 5 27 was not considered. 85. Did not reverse the inequality sign when divided by a negative. 87. true 89. false 91. 1a 2 b, a 1 b2 93. 12q, q2 95. x 5 a 2 b, x 5 a 1 b 97. no solution 99. x $ 7, yes 1176 Answers to Odd-Numbered Exercises 101. A2q, 22 3B ∪ A3, qB, yes 103. A21 2, qB Review Exercises 1. x 5 16 7 3. p 5 2 8 25 5. x 5 27 7. y 5 217 5 9. b 5 6 7 11. x 5 2 6 17 13. x 5 2, x 2 0 15. t 5 234 5 17. x 5 21 2 19. x 5 29 17 21. x 5 8 2 7y 23. 96 miles 25. 144 27. Width: 3 inches, length: 7 inches 29. $5000 @ 20%, $20,000 @ 8% 31. 60 ml of 5%, 90 ml of 10% 33. At least 91 35. b 5 23, 7 37. x 5 0, 8 39. q 5 613 41. x 5 2 64i 43. x 5 22, 6 45. x 5 1 6 !33 2 47. t 5 21, 7 3 49. f 5 1 6 !337 48 51. q 5 3 6 !69 10 53. x 5 21, 5 2 55. x 5 23, 2 7 57. r 5 Å S ph 59. v 5 h 1 16t2 t 5 h t 1 16t 61. h 5 1 ft, b 5 4 ft 63. x 5 6 65. x 5 125 67. no solution 69. 36 6 "1664 8 , x > 20.6 71. 4 6 "376 4 , x > 5.85 73. 21 6 "13 2 , x > 22.303 75. x 5 2, x 5 3 77. x 5 5 4, x 5 3 4 79. y 5 1 4, y 5 1 81. x 5 2125 8 , x 5 1 83. x 5 21 8, x 5 21 85. x 5 63i, x 5 62 87. x 5 0, 28, 4 89. p 5 62, 3 91. p 5 21 2, 5 2, 3 93. y 5 69 95. 12q, 244 97. 32, 64 99. x . 26 101. 23 # x # 7 103. x $ 24, 324, q2 105. , 14, q2 … 0 1 2 3 4 5 6 … 107. , 38, 124 … 7 8 9 10 11 12 … 109. A2q, 5 3B, … 0 1 3 2 3 1 4 3 5 3 … 111. A 23 2, qB, … 22 23 2 21 21 2 0 … 113. 14, 94, … 3 4 5 6 7 8 9 10 … 115. C3, 7 2D, … 25 2 3 7 2 4 … 117. x $ 74 119. 326, 64 121. 12q, 04 ∪ 34, q2 123. 12q, 02 ∪ 17, q2 125. A2q, 23 4B ∪ 14, q2 127. 10, 32 129. 12q, 264 ∪ 39, q2 131. 12q, 22 ∪ 14, 54 133. 13, q2 135. no solution 137. x 5 1.7 and x 5 0.9667 139. 124, 42 141. 12q, 2112 ∪ 13, q2 143. 12q, 232 ∪ 13, q2 145. 12q, q2 147. 75 # T # 95 or 0T 2 850 # 10 149. x 5 2,510 151. a. x 5 25, 1 b. b 5 25: x 5 22 6 i b 5 0: x 5 0, 24 b 5 7: x 5 22 6 !11 b 5 12: x 5 26, 2 Answers to Odd-Numbered Exercises 1177 153. x 5 A21 1 !7B4 < 7.34 155. a. 12q, 15.42 b. y1 5 20.61x 1 7.62, y2 5 0.24x 2 5.47 c. agree 157. 12q, 252 ∪ 15.25, q2 159. A7 5, 7 2B 161. 122.19, 20.92 ∪ 10.9, 2.192 Practice Test 1. 23 5 p 3. t 5 24, 7 5. x 5 21 2, 8 3 7. y 5 28 9. x 5 4 11. y 5 1 13. x 5 0, 2, 6 15. L 5 P 2 2W 2 17. 12q, 174 19. Q232 5 , 26T 21. 12q, 214 ∪ S4 3, qR 23. Q21 2, 3T 25. 1000 feet 27. 627 # minutes # 722 29. x 5 7.95 Cumulative Test 1. 215 3. x12 y3 5. x4 1 2x3 2 15x2 7. 21a 1 102 1a2 2 10a 1 1002 9. x2 1 22x x2 2 4 11. x 5 56 13. x 5 24 15. 11.25 hours 17. x 5 26 6 2i 19. x 5 4 21. 323, 42 23. 12q, 232 ∪ 322, 32 25. x 5 21, x 5 217 3 27. Q21, 1 2R y1 5 3x x 2 2 , y2 5 1 C H A P T E R 2 Section 2.1 1. 14, 22 3. 123, 02 5. 10, 232 7. A: quadrant II B: quadrant I C: quadrant III D: quadrant IV E: on negative y-axis F: on positive x-axis y1 5 2x1/4 y2 5 2x1/2 1 6 1178 Answers to Odd-Numbered Exercises 9. 11. d 5 4, 13, 32 13. d 5 4!2, 11, 22 15. d 5 3!10, A217 2 , 7 2B 17. d 5 5, A 25, 1 2B 19. d 5 4!2, 124, 262 21. d 5 5, A3 2, 11 6 B 23. d 5 !4049 60 , A 2 5 24, 1 15B 25. d 5 3.9, 10.3, 3.952 27. d 5 44.64, 11.05, 21.22 29. d 5 4!2, A!3, 3!2B 31. d 5 #10 1 2!2 1 4!3, a1 2 !2 2 , 22 1 !3 2 b 33. The perimeter of the triangle rounded to two decimal places is 21.84. 35. right triangle 37. isosceles 39. 128.06 miles 41. 268 miles 43. 12003, 3302; $330 million 45. 47. x values were subtracted from y values. The correct distance would be d 5 !58. 49. The values were not used in the correct positions. The correct midpoint would be A2, 13 2 B. 51. true 53. true 55. d 5 !2 0 a 2 b 0, aa 1 b 2 , b 1 a 2 b 57. d 5 Ç ax1 2 x1 1 x2 2 b 2 1 ay1 2 y1 1 y2 2 b 2 5 Ç a 2x1 2 x1 2 x2 2 b 2 1 a 2y1 2 y1 2 y2 2 b 2 5 Ç a x1 2 x2 2 b 2 1 a y1 2 y2 2 b 2 5 1 2 "1x1 2 x222 1 1 y1 2 y222 Using 1x2, y22 with the midpoint yields the same result. 59. P1 5 1a, b2, P2 5 1c, d2, d1 5 distance from P1 to P2, and d2 5 distance from P2 to P1, d1 5 "1a 2 c22 1 1b 2 d 22 d2 5 "1c 2 a22 1 1d 2 b 22 since 1c 2 a22 5 121a 2 c222 5 1a 2 c22, 1d 2 b22 5 121b 2 d222 5 1b 2 d22 [ d1 5 d2, so does not matter what point is labeled “first.” 61. d > 6.357 10.7, 5.152 Section 2.2 1. a. no b. yes 3. a. yes b. no 5. a. yes b. no 7. a. yes b. no 9. 11. 1 2 3 4 5 6 7 8 9 10 11 12 0.5 1 1.5 2 2.5 3 3.5 4 Month Dollars per Gallon (2012) 63. d > 3.111 12.2, 3.32 x y 5 2 1 x 1x, y2 22 0 122, 02 0 2 10, 22 1 3 11, 32 x y 5 x2 2 x 1x, y2 21 2 121, 22 0 0 10, 02 1 2 21 4 A1 2, 21 4B 1 0 11, 02 2 2 12, 22 Answers to Odd-Numbered Exercises 1179 13. 15. 17. 19. 21. 23. x-intercept: 13, 02, y-intercept: 10, 262 25. x-intercept: 163, 02, y-intercept: 10, 292 27. x-intercept: 14, 02, no y-intercept 29. no x-intercept, y-intercept: A0, 1 4B 31. x-intercept: 162, 02, y-intercept: 10, 642 33. d 35. a 37. b 39. 121, 232 41. 127, 102 43. 13, 22, 123, 22, 123, 222 45. x-axis 47. origin 49. x-axis 51. x-axis, y-axis, origin 53. y-axis 55. y-axis 57. origin 59. 61. 63. 65. 67. 69. 71. 73. 75. 77. Break even units: 2000 2000 , x , 4000 79. x $ 1 or 31, q2; the demand model is defined when at least 1000 units per day are demanded. 81. The equation is not linear 2 you need more than two points to plot the graph. x y 5 !x 2 1 1x, y2 1 0 11, 02 2 1 12, 12 5 2 15, 22 10 3 110, 32 pr pt km 0.00002 0.00004 0.00006 0.00008 0.00010 0.00012 0.00014 2000 6000 10,000 2 4 6 8 1 2 3 x (thousands of units demanded) p (price per unit) 1180 Answers to Odd-Numbered Exercises 83. You are checking to see whether the graph is symmetric with respect to the y-axis. Thus the substitution shown is incorrect. The correct substitution would be plugging 2x for x into the equation, not 2y for y. 85. false 87. true 89. origin 91. y-axis 93. x-axis, y-axis, and origin 95. x-axis, y-axis, and origin Section 2.3 1. m 5 3 3. m 5 22 5. m 5 219 10 7. m < 2.379 9. m 5 23 11. x-intercept: 10.5, 02, y-intercept: 10, 212, m 5 2, rising 13. x-intercept: 11, 02, y-intercept: 10, 12, m 5 21, falling 15. x-intercept: none, y-intercept: 10, 12, m 5 0, horizontal 17. x-intercept: A3 2, 0B y-intercept: 10, 232 19. x-intercept: 14, 02 y-intercept: 10, 22 21. x-intercept: 12, 02 y-intercept: A0, 24 3 B 23. x-intercept: 122, 02 y-intercept: 10, 222 25. x-intercept: 121, 02 y-intercept: nonev 27. x-intercept: none y-intercept: 10, 1.52 29. x-intercept: A27 2, 0B y-intercept: none 31. y 5 2 5 x 2 2, m 5 2 5, y-intercept: 10, 222 33. y 5 21 3 x 1 2, m 5 21 3, y-intercept: 10, 22 35. y 5 4x 2 3, m 5 4, y-intercept: 10, 232 37. y 5 22x 1 4, m 5 22, y-intercept: 10, 42 39. y 5 2 3 x 2 2, m 5 2 3, y-intercept: 10, 222 41. y 5 23 4 x 1 6, m 5 23 4, y-intercept: 10, 62 43. y 5 2x 1 3 45. y 5 21 3x 47. y 5 2 49. x 5 3 2 51. y 5 5x 1 2 53. y 5 23x 2 4 55. y 5 3 4x 2 7 4 57. y 5 4 59. x 5 21 61. y 5 3 5 x 1 1 5 63. y 5 25x 2 16 65. y 5 1 6 x 2 121 3 67. y 5 23x 1 1 69. y 5 3 2x 71. x 5 3 73. y 5 7 75. y 5 6 5 x 1 6 77. x 5 26 79. x 5 2 5 81. y 5 x 2 1 83. y 5 22x 1 3 85. y 5 21 2 x 1 1 87. y 5 2x 1 7 89. y 5 3 2 x 91. y 5 5 93. y 5 2 95. y 5 3 2 x 2 4 97. y 5 5 4x 1 3 2 99. y 5 3 7x 1 5 2 101. 32-hour job will cost $2,000 103. $375 105. 347 units 107. F 5 9 5 C 1 32, 2408C 5 2408F 109. The rate of change in inches per year is 1 50. 111. 0.06 ounces per year. In 2040 we expect a baby to weigh 6 pounds 12.4 ounces. 113. y-intercept 0.35 is the flat monthly charge of $35. 115. 20.35 in./yr, 1 inch 117. 2.4 plastic bags per year 1in billions2, 418.4 billion 119. a. 11, 31.932 12, 51.182 15, 111.832 b. m 5 25.59. This means that when you buy one bottle of hoisin it costs $31.93 per bottle. c. m 5 31.93. This means that when you buy two bottles of hoisin it costs $25.59 per bottle. d. m 5 22.366. This means that when you buy five bottles of hoisin it costs $22.37 per bottle. 121. The correction that needs to be made is that for the x-intercept, y 5 0, and for the y-intercept, x 5 0. Answers to Odd-Numbered Exercises 1181 123. Values for the numerator and denominator reversed. 125. true 127. false 129. The line perpendicular is vertical and has undefined slope. 131. y 5 2A B x 1 1 133. y 5 B A x 1 12B 2 12 135. Let y1 5 mx 1 b1 and y2 5 mx 1 b2, assuming that b1 Þ b2. At a point of intersection of these two lines, y1 5 y2. This is equivalent to mx 1 b1 5 mx 1 b2, which implies b1 5 b2, which contradicts our assumption. Hence, there are no points of intersection. 137. perpendicular 139. perpendicular 141. neither Section 2.4 1. 1x 2 122 1 1 y 2 222 5 9 3. 1x 1 322 1 1 y 1 422 5 100 5. 1x 2 522 1 1 y 2 722 5 81 7. 1x 1 1122 1 1 y 2 1222 5 169 9. x2 1 y2 5 4 11. x2 1 1 y 2 222 5 9 13. x2 1 y2 5 2 15. 1x 2 522 1 1 y 1 322 5 12 17. Ax 2 2 3B 2 1 Ay 1 3 5B 2 5 1 16 19. 1x 2 1.322 1 1 y 2 2.722 5 10.24 21. C11, 32, r 5 5 23. C12, 252, r 5 7 25. C14, 92, r 5 2!5 27. C A2 5, 1 7B, r 5 2 3 29. C11.5, 22.72, r 5 1.3 31. C10, 02, r 5 5!2 33. C122, 232, r 5 4 35. C123, 242, r 5 10 37. C15, 72, r 5 9 39. C10, 12, r 5 4 41. C11, 32, r 5 3 43. C15, 232, r 5 2!3 45. C13, 22, r 5 2!3 47. CA1 2, 2 1 2B, r 5 1 2 49. C11.3, 2.72, r 5 3.2 51. 1x 1 122 1 1 y 1 222 5 8 53. 1x 1 222 1 1 y 2 322 5 41 55. 1x 1 222 1 1 y 1 522 5 25 57. no A"952 1 332 ≅ 100.568B 59. x2 1 y2 5 2500 61. x2 1 y2 5 2,250,000 63. x2 1 y2 5 40,000 65. a. Tower 1: 1x 2 2.522 1 1 y 2 2.522 5 3.52 Tower 2: 1x 2 2.522 1 1 y 2 7.522 5 3.52 Tower 3: 1x 2 7.522 1 1 y 2 2.522 5 3.52 Tower 4: 1x 2 7.522 1 1 y 2 7.522 5 3.52 b. This placement of cell phone towers will provide cell phone coverage for the entire 10 mile by 10 mile square. 67. The standard form of an equation of a circle is 1x 2 h22 1 1 y 2 k22 5 25, which would lead to a center of 14, 232, not 14, 32. 69. This equation is not a circle. The standard equation of a circle is 1x 2 h22 1 1 y 2 k22 5 r2. 71. true 73. true 75. single point 125, 32 77. 1x 2 322 1 1 y 1 222 5 20 79. 4c 5 a2 1 b2 81. C1a, 02, r 5 10 83. no graph (because no solution) 85. single point 125, 32 87. a. 1x 2 5.522 1 1 y 1 1.522 5 39.69, 15.5, 21.52, r 5 6.3 b. y 5 21.5 6 "39.69 2 1x 2 5.522 c. d. The graphs in (a) and (c) are the same. Section 2.5 1. Negative linear association because the data closely cluster around what is reasonably described as a linear with negative slope. 3. Although the data seem to comprise two linear segments, the over-all data set cannot be described as having a positive or negative direction of association. Moreover, the pattern of the data is not linear, per se; rather, it is nonlinear and conforms to an identifiable curve (an upside down V called the absolute value function). 5. B because the association is positive, thereby eliminating choices A and C. And, since the data are closely clustered around a linear curve, the bigger of the two correlation coefficients, 0.80 and 0.20, is more appropriate. 7. C because the association is negative, thereby eliminating choices B and D. And the data are more loosely clustered around a linear curve than are those pictured in number 6. So, the correlation coefficient is the negative choice closer to 0. 2 4 6 8 10 2 4 6 8 10 x y 1182 Answers to Odd-Numbered Exercises 9. a. b. The data seem to be nearly perfectly aligned to a line with negative slope. So, it is reasonable to guess that the correlation coefficient is very close to 21. c. The equation of the best fit line is y 5 23x 1 5 with a correlation coefficient of r 5 21. d. There is a perfect negative linear association between x and y. 11. a. b. The data tend to fall from left to right, so that the correlation coefficient should be negative. Also, the data do not seem to stray too far from a linear curve, so the r value should be reasonably close to 21, but not equal to it. A reasonable guess would be around 20.90. c. The equation of the best fit line is approximately y 5 20.5844x 2 3.801 with a correlation coefficient of about r 5 0.9833. d. There is a strong (but not perfect) negative linear relationship between x and y. 13. a. b. The data seem to rise from left to right, but it is difficult to be certain about this relationship since the data stray considerably away from an identifiable line. As such, it is reasonable to guess that r is a rather small value close to 0, say around 0.30. c. The equation of the best fit line is approximately y 5 0.5x 1 0.5 with a correlation coefficient of about 0.349. d. There is a very loose (bordering on unidentifiable) positive linear relationship between x and y. 15. a. The equation of the best fit line is about y 5 0.7553x 2 0.4392 with a correlation coefficient of about r 5 0.868. b. The values x 5 0 and x 5 26 are within the range of the data set, so that using the best fit line for predictive purposes is reasonable. This is not the case for the values x 5 12 and x 5 215. The predicted value of y when x 5 0 is approximately 20.4392, and the predicted value of y when x 5 26 is 24.971. c. Solve the equation 2 5 0.7553x 2 0.4392 for x to obtain: 2.4392 5 0.7553x so that x 5 3.229. So, using the best fit line, you would expect to get a y-value of 2 when x is approximately 3.229. 17. a. The equation of the best fit line is about y 5 21.2631x 2 11.979 with a correlation coefficient of about r 5 20.980. b. The values x 5 215, 26, and 0 are within the range of the data set, so that using the best fit line for predictive purposes is reasonable. This is not the case for the value x 5 12. The predicted value of y when x 5 215 is approximately 6.9675, the predicted value of y when x 5 26 is about 24.4004, and the predicted value of y when x 5 0 is 211.979. c. Solve the equation 2 5 21.2631x 2 11.979 for x to obtain: 13.979 5 21.2631x so that x 5 211.067. So, using the best fit line, you would expect to get a y-value of 2 when x is approximately 211.067. 19. a. The scatterplot for the entire data set is: -15 -10 -5 0 5 10 15 20 -4 -2 0 2 4 6 -18 -16 -14 -12 -10 -8 -6 -4 -2 0 2 -20 -10 0 10 20 30 -8 -6 -4 -2 0 2 4 6 -4 -2 0 2 4 6 -10 -8 -6 -4 -2 0 2 4 6 8 10 -10 -5 0 5 10 15 -20 -15 -10 -5 0 5 10 15 20 -25 -20 -15 -10 -5 0 5 -15 -10 -5 0 5 10 15 20 -4 -2 0 2 4 6 Answers to Odd-Numbered Exercises 1183 The equation of the best fit line is y 5 23x 1 5 with a correlation coefficient of r 5 21. b. The scatterplot for the data set obtained by removing the starred data point 15, 2102 is: The equation of the best fit line of this modified data set is y 5 23x 1 5 with a correlation coefficient of r 5 21. c. Removing the data point did not result in the slightest change in either the equation of the best fit line or the correlation coefficient. This is reasonable since the relationship between x and y in the original data set is perfectly linear, so that all of the points lie ON the same line. As such, removing one of them has no effect on the line itself. 21. a. The scatterplot for the entire data set is: The equation of the best fit line is y 5 23.4776x 1 4.6076 with a correlation coefficient of about r 5 20.993. b. The scatterplot for the data set obtained by removing the starred data points 13, 242 and 16, 2162 is: The equation of the best fit line of this modified data set is y 5 23.6534x 1 4.2614 with a correlation coefficient of r 5 20.995. c. Removing the data point did change both the best fit line and the correlation coefficient, but only very slightly. 23. a. b. The correlation coefficient is approximately r 5 20.980. This is identical to the r-value from Problem 17. This makes sense because simply interchanging the x- and y-values does not change how the points cluster together in the xy-plane. c. The equation of the best fit line for the paired data 1 y, x2 is x 5 20.7607y 2 9.4957. d. It is not reasonable to use the best fit line in (c) to find the predicted value of x when y 5 23 because this value falls outside the range of the given data. However, it is okay to use the best fit line to find the predicted values of x when y 5 2 or y 5 216. Indeed, the predicted value of x when y 5 2 is about 211.0171, and the predicted value of x when y 5 216 is about 2.6755. 25. First, note that the scatterplot is given by The paired data all lie identically on the vertical line x 5 3. As such, you might think that the square of the correlation coefficient would be 1 and the best fit line is, in fact, x 5 3. However, since there is absolutely no variation in the x-values for this data set, it turns out that in the formula for the correlation coefficient r 5 na xy 2 a a xb a a yb Åna x2 2 a a xb 2 ⋅Åna y2 2 a a yb 2 the quantity Åna x2 2 a a xb 2 turns out to be zero. (Check this on Excel for this data set!) As such, there is no meaningful r-value for this data set. Also, the best fit line is definitely the vertical line x 5 3, but the technology cannot provide it because its slope is undefined. 27. The y-intercept 1.257 is mistakenly interpreted as the slope. The correct interpretation is that for every unit increase in x, the y-value increases by about 5.175. -6 -4 -2 0 2 4 6 8 10 12 14 16 -4 -2 0 2 4 -20 -15 -10 -5 0 5 10 15 20 -4 -2 0 2 4 6 8 -20 -15 -10 -5 0 5 10 15 20 -4 -2 0 2 4 6 -25 -20 -15 -10 -5 0 5 -20 -10 0 10 20 -10 -5 0 5 10 15 20 0 1 2 3 4 1184 Answers to Odd-Numbered Exercises 29. a. Here is a table listing all of the correlation coefficients between each of the events and the total points: EVENT r 100 m 20.567 Long Jump 0.842 Shot Put 0.227 High Jump 0.587 400 m 20.715 110 m hurdle 20.488 Discus 0.173 Pole Vault 0.446 Javelin 0.542 Long jump has the strongest relationship to the total points. b. The correlation coefficient between long jump and total events is r 5 0.842. c. The equation of the best fit line between the two events in (b) is y 5 939.98x 1 1261.2. d. Evaluate the equation in (c) at x 5 40 to get the total points are about 38.860. 31. a. A scatterplot illustrating the relationship between % residents immunized and % residents with influenza is shown below. b. The correlation coefficient between % residents immunized and % residents with influenza is r 5 20.812. c. Based on the correlation coefficient 1r 5 20.8122, we would believe that there is a strong relationship between % residents immunized and % residents with influenza. d. The equation of the best fit line that describes the relationship between % residents immunized and % residents with influenza is y 5 20.27x 1 32.20. e. Since r 5 20.812 indicates a strong relationship between % residents immunized and % residents with influenza, we can make a reasonably accurate prediction. However it will not be completely accurate. 33. a. A scatterplot illustrating the relationship between average wait times and average rating of enjoyment is shown below. b. The correlation coefficient between average wait times and average rating of enjoyment is r 5 0.348. c. The correlation coefficient 1r 5 0.3482 indicates a somewhat weak relationship between average wait times and average rating of enjoyment. d. The equation of the best fit line that describes the relationship between average wait times and average rating of enjoyment is y 5 0.31x 1 37.83. e. No, given the somewhat weak correlation coefficient between average wait times and average rating of enjoyment, one could not use the best fit line to produce accurate predictions. 35. a. A scatterplot illustrating the relationship between average wait times and average rating of enjoyment for Park 2 is shown below. b. The correlation coefficient between average wait times and average rating of enjoyment is r 5 20.064. c. The correlation coefficient 1r 5 20.0642 indicates a weak relationship between average wait times and average rating of enjoyment for Park 2. d. The equation of the best fit line that describes the relationship between average wait times and average rating of enjoyment is y 5 20.05x 1 52.53. e. No, given the weak correlation coefficient between average wait times and average rating of enjoyment for Park 2, one could not use the best fit line to produce accurate predictions. Answers to Odd-Numbered Exercises 1185 37. a. The scatterplot for this data set is given by b. The equation of the best fit line is y 5 21.9867x 1 27.211 with a correlation coefficient of r 5 20.671. This line does not seem to accurately describe the data because some of the points rise as you move left to right, while others fall as you move left to right; a line cannot capture both types of behavior simultaneously. Also, r being negative has no meaning here. c. The best fit is provided by QuadReg. The associated equation of the best fit quadratic curve, the correlation coefficient, and associated scatterplot are: 39. a. The scatterplot for this data set is given by b. The equation of the best fit line is y 5 0.3537x 1 0.5593 with a correlation coefficient of r 5 0.971. This line seems to provide a very good fit for this data, although not perfect. c. The best fit is provided by LnReg. The associated equation of the best fit logarithmic curve, the correlation coefficient, and associated scatterplot are: Review Exercises 1. quadrant II 3. quadrant III 5. d 5 3!5 7. d 5 !205 9. A5 2, 6B 11. 13.85, 5.32 13. d < 52.20 units 15. x-intercepts: 162, 02, y-intercepts: 10, 612 17. x-intercepts: 163, 02, no y-intercepts 19. y-axis 21. origin 23. 25. 27. 29. 31. y 5 23x 1 6 m 5 23 y-intercept: 10, 62 33. y 5 23 2 x 21 2 m 5 23 2 y-intercept: A0, 21 2B 35. x-intercepts A5 4, 0B, y-intercepts: 10, 252, m 5 4 37. x-intercepts: 14, 02, y-intercepts: 10, 42, m 5 21 39. no x-intercepts, y-intercepts: 10, 22, m 5 0 41. y 5 4x 2 3 43. x 5 23 45. y 5 22x 2 2 47. y 5 6 49. y 5 5 6 x 1 4 3 51. y 5 22x 2 1 53. y 5 2 3x 1 1 3 55. y 5 23 4x 1 31 16 57. y 5 1.2x 1 100 (x is pretest score, y is posttest score) 59. 1x 1 222 1 1 y 2 322 5 36 61. Ax 2 3 4 B 2 1 Ay 2 5 2B 2 5 4 25 63. C122, 232, r 5 9 65. C A2 3 4, 1 2B, r 5 2 3 0 5 10 15 20 25 30 0 2 4 6 8 10 0 1 2 3 4 5 6 0 2 4 6 8 10 12 14 16 1186 Answers to Odd-Numbered Exercises 67. not a circle 69. C A1 3, 22 3B, r 5 3 71. 1x 2 222 1 1 y 2 722 5 2 73. right triangle 75. x-axis, y-axis, origin 77. perpendicular 79. y 5 22 3 6"9 2 Ax 2 1 3B2 Practice Test 1. d 5 !82 3. d 5 !29, A1 2, 5B 5. y 5 1, 9 7. x-axis 9. 11. x-intercepts: 16, 02; y-intercepts: 10, 222 13. y 5 8 3 x 2 8 15. y 5 x 1 5 17. y 5 22x 1 3 19. y 5 2x 1 2 21. 1x 2 622 1 1 y 1 722 5 64 23. 1x 2 422 1 1 y 2 922 5 20 25. both Cumulative Test 1. 1 2 3. x4 2 32x2 1 256 5. 5 2 x 5 1 x 7. 12 1 30i 9. x 5 1 11. x 5 63 13. discriminant is negative; two complex (conjugate) roots 15. x 5 2 17. 10, 122 19. 122, 62 21. origin 23. x 5 5 25. C125, 232, r 5 !30 27. neither C H A P T E R 3 Section 3.1 1. function 3. not a function 5. function 7. not a function 9. not a function 11. function 13. not a function 15. not a function 17. function 19. not a function 21. function 23. not a function 25. a. 5 b. 1 c. 23 27. a. 3 b. 2 c. 5 29. a. 25 b. 25 c. 25 31. a. 2 b. 28 c. 25 33. 1 35. 23 and 1 37. 324, 44 39. 6 41. 27 43. 6 45. 21 47. 233 49. 27 6 51. 2 3 53. 4 55. 8 2 x 2 a 57. 2 59. 1 61. 2 63. 1 65. 12q, q2 67. 12q, q2 69. 12q, 52 < 15, q2 71. 12q, 222 < 122, 22 < 12, q2 73. 12q, q2 75. 12q, 74 77. C25 2, qB 79. 12q, 224 < 32, q2 81. 13, q2 83. 12q, q2 85. 12q, 242 < 124, q2 87. A2q, 3 2B 89. 12q, 222 < 13, q2 91. 12q, 244 < 34, q2 93. A2q, 3 2B 95. 12q, q2 97. x 5 22, 4 99. x 5 21, 5, 6 101. y 5 45x, 175, q2 103. T 162 5 64.88F, T 1122 5 908F 105. P1102 > $641.66, P11002 > $634.50 107. V1x2 = x 110 2 2x22, 10, 52 109. E 142 > 84 Yen E 172 > 84 Yen E 182 > 83 Yen 111. 229 people 113. a. A1x2 5 x 1x − 3.3752 5 x2 − 3.375x b. A14.52 5 5.0625 window 1area of window < 5 sq in.2. c. A18.52 5 43.5625 (not possible because the window cannot be larger than the envelope). 115. Yes, because every input (year) corresponds to exactly one output (federal funds rate). 117. 11999, 25002, 12003, 37002, 12007, 47002, 12011, 59002, 12015, 66002 119. a. F1502 5 0 b. g1502 5 1000 c. H1502 5 2000 Answers to Odd-Numbered Exercises 1187 121. You must apply the vertical line test instead of the horizontal line test. Applying the vertical line test would show that the graph given is actually a function. 123. ƒ1x 112 2 ƒ1x2 1 ƒ112, in general. 125. G1211 h2 2 G1212 1 G1h2, in general. 127. false 129. false 131. A 5 2 133. C 5 25, D 5 22 135. 12q, 2a2 < 12a, a2 < 1a, q2 137. Warmest at noon: 908 F. Outside the interval 36, 184 the temperatures are too low. 139. lowest price $10, highest $642.38 agrees (if there is only 1 card for sale.) 141. Graph of y2 can be obtained by shifting the graph of y1 two units to the right. Section 3.2 1. neither 3. even 5. odd 7. neither 9. odd 11. even 13. even 15. neither 17. neither 19. neither 21. neither 23. neither 25. a. 12q, q2 b. 321, q2 c. increasing: 121, q2, decreasing: 123, 222, constant: 12q, 232 < 122, 212 d. 0 e. 21 f. 2 27. a. 327, 24 b. 325, 44 c. increasing: 124, 02, decreasing: 127, 242 < 10, 22, constant: nowhere d. 4 e. 1 f. 25 29. a. 12q, q2 b. 12q, q2 c. increasing: 12q, 232 < 14, q2, decreasing: nowhere, constant: 123, 42 d. 2 e. 2 f. 2 31. a. 12q, q2 b. 324, q2 c. increasing: 10, q2, decreasing: 12q, 02, constant: nowhere d. 24 e. 0 f. 0 33. a. 12q, 02 < 10, q2 b. 12q, 02 < 10, q2 c. increasing: 12q, 02 < 10, q2, decreasing: nowhere, constant: nowhere d. undefined e. 3 f. 23 35. a. 12q, 02 < 10, q2 b. 12q, 52 < 374 c. increasing: 12q, 02, decreasing: 15, q2, constant: 10, 52 d. undefined e. 3 f. 7 37. 2x 1 h 21 39. 2x 1 h 1 3 41. 2x 1 h 2 3 43. 26x 2 3h 1 5 45. 13 47. 1 49. 22 51. 21 53. domain: 12q, q2 range: 12q, 24 increasing: 12q, 22 decreasing: nowhere constant: 12, q2 55. domain: 12q, q2 range: 30, q2 increasing: 10, q2 decreasing: 121, 02 constant: 12q, 212 57. domain: 12q, q2 range: 12q, q2 increasing: 12q, q2 decreasing: nowhere constant: nowhere 59. domain: 12q, q2 range: 31, q2 increasing: 11, q2 decreasing: 12q, 12 constant: nowhere 61. -3 -2 -1 0 1 2 3 4 5 1 2 3 4 5 6 7 8 9 10 -3 -2 -1 0 1 2 3 4 5 1 2 3 4 5 6 7 8 9 10 5-2x 3x-2 domain: 12q, 22 < 12, q2 range: 11, q2 increasing: 12, q2 decreasing: 12q, 22 63. domain: 12q, q2 range: 321, 34 increasing: 121, 32 decreasing: nowhere constant: 12q, 212 < 13, q2 65. domain: 12q, q2 range: 31, 44 increasing: 11, 22 decreasing: nowhere constant: 12q, 12 < 12, q2 67. open holes 122,12, 122, 212, 11, 22 closed hole 11, 02 domain: 12q, 222 < 122, q2 range: 12q, q2 increasing: 122, 12 decreasing: 12q, 222 < 11, q2 constant: nowhere 1188 Answers to Odd-Numbered Exercises 69. domain: 12q, q2 range: 30, q2 increasing: 10, q2 decreasing: nowhere constant: 12q, 02 71. closed hole 10, 02 domain: 12q, q2 range: 12q, q2 increasing: nowhere decreasing: 12q, 02 < 10, q2 constant: nowhere 73. open holes 121, 212, 11, 12, 11, 212 graph of 2! 3 x on 12q, 212, closed hole 121, 12 domain: 12q, 12 < 11, q2 range: 12q, 212 < 121, q2 increasing: 121, 12 decreasing: 12q, 212 < 11, q2 constant: nowhere 75. open holes 122, 22, 12, 22 closed holes 122, 12, 12, 42 domain: 12q, q2 range: 12q, 22 < 34, q2 increasing: 12q, 222 < 10, 22 < 12, q2 decreasing: 122, 02 constant: nowhere 77. open hole 11, 12 domain: 12q, 12 < 11, q2 range: 12q, 12 < 11, q2 increasing: 12q, 12 < 11, q2 decreasing: nowhere constant: nowhere 79. Profit is increasing from October through December and decreasing from January through October. 81. C1x2 5 • 10x, 0 # x # 50 9x, 50 , x # 100 8x, x . 100 83. C1x2 5 e 250x, 0 # x # 10 175x 1 750, x . 10 85. C1x2 5 e 1000 1 35x, 0 # x # 100 2000 1 25x, x . 100 87. R1x2 5 e 50,000 1 3x, 0 # x # 100,000 4x 2 50,000, x . 100,000 89. P1x2 5 65x 2 800 91. ƒ1x2 5 0.98 1 0.2233x44, x $ 0 93. ƒ1t2 5 3121233t44, t $ 0 95. a. 20 million tons per year b. 110 million tons per year 97. 0 ft/sec 99. Demand for product is increasing at an approximate rate of 236 units over the first quarter. 101. The domain is incorrect. It should be 12q, 02 < 10, q2. The range is also incorrect and should be 10, q2. The graph is also incorrect. It should contain an open circle at 10, 02. 103. The portion of C1x2 for x . 30 should be: 15 1 x 2 30 105. true 107. false 109. yes, if a 5 2b 111. odd 113. odd 115. domain: all real numbers range: set of integers Section 3.3 1. l 3. a 5. b 7. i 9. c 11. g 13. y 5 0 x 0 1 3 15. y 5 02x 0 5 0x0 17. y 5 3 0 x 0 19. y 5 x3 2 4 21. y 5 1x 1 123 1 3 23. y 5 12x23 25. 27. 29. 31. 33. 35. Number miles beyond first 30 e Answers to Odd-Numbered Exercises 1189 37. 39. 41. 43. 45. 47. 49. 51. 53. 55. 57. 59. 61. 63. 65. 67. 69. 71. 73. 75. ƒ1x2 5 1x 2 322 1 2 77. ƒ1x2 5 21x 1 122 1 1 79. ƒ1x2 5 21x 2 222 2 5 81. S 1x2 5 10x and S 1x2 5 10x 1 50 83. T 1x2 5 0.331x 2 63002 85. a. BSA 1w2 5 Å 9w 200; b. BSA 1w 2 32 5 Ç 91w 2 32 200 87. (b) shifted to the right 3 units 89. (b) should be deleted since 0 3 2 x0 5 0 x 2 30 . The correct sequence of steps would be: (a) S (c) S (d ), where (c): Shift to the right 3. 91. true 93. true 95. 1a 1 3, b 1 22 97. Any part of the graph of ƒ 1x2 that is below the x-axis is reflected above it for 0 ƒ1x20 . a. b. 1190 Answers to Odd-Numbered Exercises 99. If 0 , a , 1, you have a horizontal stretch. If a . 1, the graph is a horizontal compression. a. b. 101. Each horizontal line in the graph of y 5 Œ xœ is stretched twice in length. There is a vertical shift of one unit up. Section 3.4 1. ƒ1x2 1 g1x2 5 x 1 2 ƒ1x2 2 g1x2 5 3x ƒ1x2⋅g1x2 5 22x2 1 x 1 1 s domain: 12q, q2 ƒ1x2 g1x2 5 2x 1 1 1 2 x domain: 12q, 12 ∪ 11, q2 3. ƒ1x2 1 g1x2 5 3x2 2 x 2 4 ƒ1x2 2 g1x2 5 x2 2 x 1 4 ƒ1x2 ⋅ g1x2 5 2x4 2 x3 2 8x2 1 4x s domain: 12q, q2 ƒ1x2 g1x2 5 2x2 2 x x2 2 4 domain: 12q, 222 ∪ 122, 22 ∪ 12, q2 5. ƒ1x2 1 g1x2 5 1 1 x2 x ƒ1x2 2 g1x2 5 1 2 x2 x ƒ1x2 ⋅ g1x2 5 1 ƒ1x2 g1x2 5 1 x2 w domain: 12q, 02 ∪ 10, q2 7. ƒ1x2 1 g1x2 5 3!x ƒ1x2 2 g1x2 5 2!x ƒ1x2 ⋅ g1x2 5 2x s domain: 30, q2 ƒ1x2 g1x2 5 1 2 domain: 10, q2 9. ƒ1x2 1 g1x2 5 !4 2 x 1 !x 1 3 ƒ1x2 2 g1x2 5 !4 2 x 2 !x 1 3 ƒ1x2⋅g1x2 5 !4 2 x⋅!x 1 3 s domain: 323, 44 ƒ1x2 g1x2 5 !4 2 x !x 1 3 x 1 3 domain: 123, 44 11. 1ƒ + g21x2 5 2x2 2 5 domain: 12q, q2 1g + ƒ 21x2 5 4x2 1 4x 2 2 domain: 12q, q2 13. 1ƒ + g21x2 5 1 x 1 1 domain: 12q, 212 ∪ 121, q2 1g + ƒ 21x2 5 1 x 2 1 1 2 domain: 12q, 12 ∪ 11, q2 15. 1ƒ + g21x2 5 1 0 x 2 1 0 domain: 12q, 12 ∪ 11, q2 1g + ƒ 21x2 5 1 0 x 0 2 1 domain: 12q, 212 ∪ 121, 12 ∪ 11, q2 17. 1ƒ + g21x2 5 !x 1 4 domain: 324, q2 1g + ƒ 21x2 5 !x 2 1 1 5 domain: 31, q2 19. 1ƒ + g21x2 5 x domain: 12q, q2 1g + ƒ 21x2 5 x domain: 12q, q2 21. 15 23. 13 25. 26!3 27. 110 3 29. 11 31. 3!2 33. undefined 35. undefined 37. 13 39. ƒ1g1122 5 1 3 g1ƒ1222 5 2 41. ƒ1g1122 5 undefined g 1 ƒ1222 5 undefined 43. ƒ1g1122 5 1 3 g1ƒ1222 5 4 45. ƒ1g1122 5 !5 g1ƒ1222 5 6 47. ƒ1g1122 5 undefined g1ƒ1222 5 undefined 49. ƒ1g1122 5 3 "3 g1ƒ1222 5 4 51. ƒ1g1x22 5 2ax 2 1 2 b 1 1 5 x 2 1 1 1 5 x g1ƒ1x22 5 12x 1 1221 2 5 2x 2 5 x 53. ƒ1g1x22 5 "1x2 1 12 2 1 5 "x2 5 0x0 5 x g1ƒ1x22 5 A !x 2 1B2 1 1 5 1x 2 12 1 1 5 x 55. ƒ1g1x22 5 1 1 x 5 x g1ƒ1x22 5 1 1 x 5 x 57. ƒ1g1x22 5 4a !x 1 9 2 b 2 29 5 4ax 1 9 4 b29 5 x g1ƒ1x22 5 "14x2 2 92 1 9 2 5 "4x2 2 5 2x 2 5 x 59. ƒ 1g1x22 5 1 x 1 1 x 2 1 5 1 x 1 1 2 x x 5 1 1 x 5 x g 1ƒ1x22 5 1 x 2 1 1 1 1 x 2 1 5 1 1 x 2 1 x 2 1 1 x 2 1 5 x x 2 1 1 x 2 1 5 x 61. ƒ1x2 5 2x2 1 5x g1x2 5 3x 2 1 63. ƒ1x2 5 2 0 x 0 g1x2 5 x 2 3 65. ƒ1x2 5 3 !x 2 2 g1x2 5 x 1 1 Since x $ 1 e Answers to Odd-Numbered Exercises 1191 67. F1C1K22 5 9 5 1K 2 273.152 1 32 69. a. A1x2 5 Ax 4B2 b. A11002 5 625 ft2 c. A12002 5 2500 ft2 71. a. C1 p2 5 62,000 2 20p b. R1 p2 5 600,000 2 200p c. P1 p2 5 538,000 2 180p 73. a. C1n1t22 5 210t2 1 500t 1 1375 b. C1n11622 5 6815 The cost of production on a day when the assembly line was running for 16 hours is $6,815,000. 75. a. A1r1t22 5 p110t 2 0.2t222 b. 11,385 square miles 77. A1t2 5 pC150!tD2 5 22,500pt ft2 79. d1h2 5 "h2 1 4 81. Must exclude 22 from the domain. 83. The operation is composition, not multiplication. 85. Function notation, not multiplication. 87. false 89. true 91. 1g + ƒ21x2 5 1 x domain: x 2 0. 93. 1g + ƒ21x2 5 x domain: C2a, qB 95. domain: 327, 94 97. domain: 12q, 32 ∪ 123, 214 ∪ 34, 62 ∪ 16, q2 Section 3.5 1. function, not one-to-one 3. function, one-to-one 5. function, one-to-one 7. not a function 9. function, not one-to-one 11. function, not one-to-one 13. function, one-to-one 15. function, not one-to-one 17. not one-to-one function 19. one-to-one function 21. not one-to-one function 23. one-to-one function 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. ƒ211x2 5 x 1 1 domain ƒ: 12q, q2 domain ƒ21: 12q, q2 range ƒ: 12q, q2 range ƒ21: 12q, q2 45. ƒ211x2 5 21 3 x 1 2 3 domain ƒ: 12q, q2 domain ƒ21: 12q, q2 range ƒ: 12q, q2 range ƒ21: 12q, q2 47. ƒ211x2 5 ! 3 x 2 1 domain ƒ: 12q, q2 domain ƒ21: 12q, q2 range ƒ: 12q, q2 range ƒ21: 12q, q2 49. ƒ211x2 5 x2 1 3 domain ƒ: 33, q2 domain ƒ21: 30, q2 range ƒ: 30, q2 range ƒ21: 33, q2 51. ƒ211x2 5 !x 1 1 domain ƒ: 30, q2 domain ƒ21: 321, q2 range ƒ: 321, q2 range ƒ21: 30, q2 1192 Answers to Odd-Numbered Exercises 53. ƒ211x2 5 22 1 !x 1 3 domain ƒ: 322, q2 domain ƒ21: 323, q2 range ƒ: 323, q2 range ƒ21: 322, q2 55. ƒ211x2 5 2 x domain ƒ: 12q, 02 ∪ 10, q2 range ƒ: 12q, 02 ∪ 10, q2 domain ƒ21: 12q, 02 ∪ 10, q2 range ƒ21: 12q, 02 ∪ 10, q2 57. ƒ211x2 5 3x 2 2 x 5 322 x domain ƒ: 12q, 32 ∪ 13, q2 range ƒ: 12q, 02 ∪ 10, q2 domain ƒ21: 12q, 02 ∪ 10, q2 range ƒ21: 12q, 32 ∪ 13, q2 59. ƒ211x2 5 5x 2 1 x 1 7 domain ƒ: 12q, 52 ∪ 15, q2 range ƒ: 12q, 272 ∪ 127, q2 domain ƒ21: 12q, 272 ∪ 127, q2 range ƒ21: 12q, 52 ∪ 15, q2 61. not one-to-one 63. one-to-one ƒ 211x2 5 c x x # 21 ! 3 x 21 , x , 1 x x $ 1 65. ƒ211x2 5 5 9 1x 2 322; this now represents degrees Fahrenheit being turned into degrees Celsius. 67. C1x2 5 b 250x, 0 # x # 10 175x 1 750, x . 10 C 211x2 5 b x 250, 0 # x # 2500 x 2 750 175 , x . 2500 69. E1x2 5 7.5x, E 211x2 5 x 7.5, x $ 0 tells you how many hours the student will have to work to bring home x dollars. 71. domain: 30, 244 range: 397.5528, 101.704 73. domain: 397.5528, 101.704 range: 30, 244 75. No, it’s not a function because it fails the vertical line test. 77. ƒ is not one-to-one, so it does not have an inverse function. 79. false 81. false 83. 1b, 02 85. ƒ1x2 5 "1 2 x2, 0 # x # 1, 0 # y # 1, ƒ211x2 5 "1 2 x2, 0 # x # 1, 0 # y # 1, domain and range of both are 30, 14 87. m 2 0 89. not one-to-one 91. not one-to-one 93. No, the functions are not inverses of each other. Had we restricted the domain of g1x2, then they would be inverses. 95. Yes, the functions are inverses of each other. Section 3.6 1. y 5 kx 3. V 5 kx3 5. z 5 km 7. ƒ 5 k l 9. F 5 kw L 11. v 5 kgt 13. R 5 k PT 15. y 5 k!x 17. d 5 rt 19. V 5 lwh 21. A 5 pr2 23. V 5 p 16 hr 2 25. V 5 400,000 P 27. F 5 2p lL 29. t 5 19.2 s 31. R 5 4.9 I 2 33. R 5 0.01L A 35. F 5 0.025m1m2 d 2 37. W 5 7.5H 39. 1292 mph 41. F 5 1.618H 43. 24 cm 45. $37.50 47. 20,000 49. 600 w/m2 51. Bank of America: 1.5% Navy Federal Credit Union: 3% 53. 11 12 or 0.92 atm 55. Should be y is inversely proportional to x. 57. true 59. b 61. s2 pl 5 1.23C 2 n k7/6 L 11/6 63. a. The least-squares regression line is y 5 2.93x 1 201.72. b. The variation constant is 120.07, and the equation of the direct variation is y 5 120.074x0.259. c. When the oil price is $72.70 per barrel in September 2006, the predicted stock index from the least-squares regression line is 415 and from the equation of direct variation is 364. The least-squares regression line gives a closer approximation to the actual value, 417. 65. a. y 5 2141.73x 1 2,419.35 b. 3,217.69, y 5 3217.69 x0.41 Answers to Odd-Numbered Exercises 1193 c. When the 5-year maturity rate is 5.02% in September 2006, the predicted number of housing units from the least-squares regression line is 1708, and from the equation of inverse variation is 1661. The equation of the least-squares regression line gives a closer approximation to the actual value, 1861. 67. a. y 5 0.123x 1 2.391 b. About $3.38 per gallon. Yes. c. $3.74 Review Exercises 1. yes 3. yes 5. no 7. yes 9. no 11. a. 2 b. 4 c. x 5 23, 4 13. a. 0 b. 22 c. x 5 25, 2 15. 5 17. 2665 19. 22 21. 4 23. 12q, q2 25. 12q, 242 ∪ 124, q2 27. 34, q2 29. D 5 18 31. neither 33. odd 35. neither 37. odd 39. a. 324, 74 b. 322, 44 c. increasing: 13, 72, decreasing: 10, 32, constant: 124, 02 41. 22 43. domain: 12q, q2 range: 10, q2 open hole 10, 02, closed hole 10, 22 45. domain: 12q, q2 range: 321, q2 open hole 11, 32 closed hole 11, 212 47. C1x2 5 b 25, x # 2, 25 1 10.501x 2 22, x . 2 49. 51. 53. 55. 57. 59. y 5 !x 1 3 domain: 323, q2 61. y 5 !x 2 2 1 3 domain: 32, q2 63. y 5 5!x 2 6 domain: 30, q2 65. y 5 1x 1 222 2 12 67. g1x2 1 h1x2 5 22x 2 7 g1x2 2 h1x2 5 24x 2 1 g1x2 ? h1x2 5 23x2 1 5x 1 12 domain: 12q, q2 g1x2 h1x2 5 23x 2 4 x 2 3 domain: 12q, 32 ∪ 13, q2 69. g1x2 1 h1x2 5 1 x2 1 !x g1x2 2 h1x2 5 1 x2 2 !x g1x2⋅h1x2 5 1 x3/2 domain: 10, q2 g1x2 h1x2 5 1 x5/2 71. g1x2 1 h1x2 5 !x 2 4 1 !2x 1 1 g1x2 2 h1x2 5 !x 2 4 2 !2x 1 1 domain: 34, q2 g1x2 ⋅h1x2 5 !x 2 4 ⋅!2x 1 1 g1x2 h1x2 5 !x 2 4 !2x 1 1 domain: 34, q2 73. 1ƒ + g21x2 5 6x 2 1 domain: 12q, q2 1g + ƒ21x2 5 6x 2 7 domain: 12q, q2 75. 1ƒ + g21x2 5 8 2 2x 13 2 3x domain: 12q, 42 ∪ Q4, 13 3 R ∪ Q13 3 , qR 1g + ƒ21x2 5 x 1 3 4x 1 10 domain: 12q, 232 ∪ Q23, 25 2R ∪ Q25 2, qR 77. 1ƒ + g21x2 5 "x2 2 9 domain: 12q, 234 ∪ 33, q2 1g + ƒ21x2 5 x 2 9 domain: 35, q2 79. ƒ1g 1322 5 857, g 1ƒ12122 5 51 81. ƒ1g1322 5 17 31, g 1ƒ12122 5 1 83. ƒ1g 1322 5 12, g 1ƒ12122 5 2 85. ƒ1x2 5 3x2 1 4x 1 7, g 1x2 5 x 2 2 87. ƒ1x2 5 1 !x, g 1x2 5 x2 1 7 89. A 1t2 5 625p 1t 1 22 in2 91. yes 93. no 95. yes 97. yes 99. yes t w t 1194 Answers to Odd-Numbered Exercises 101. 103. 105. ƒ211x2 5 1 21x 2 12 5 x 2 1 2 domain ƒ: 12q, q2 domain ƒ21: 12q, q2 range ƒ: 12q, q2 range ƒ21: 12q, q2 107. ƒ211x2 5 x2 2 4 domain ƒ: 324, q2 domain ƒ21: 30, q2 range ƒ: 30, q2 range ƒ21: 324, q2 109. ƒ211x2 5 6 2 3x x 2 1 domain ƒ: 12q, 232 ∪ 123, q2 range ƒ: 12q, 12 ∪ 11, q2 domain ƒ21: 12q, 12 ∪ 11, q2 range ƒ 21: 12q, 232 ∪ 123, q2 111. S1x2 5 22,000 1 0.08 x, S 211x2 5 x 2 22,000 0.08 , sales required to earn a desired income 113. C 5 2pr 115. A 5 pr2 117. W 5 8.5H 119. domain: 12q, 212 ∪ 13, q2 121. a. 12q, 22 ∪ 12, q2 b. 521, 0, 16 ∪ 12, q2 c. increasing: 12, q2, decreasing: 12q, 212, constant: 121, 02 ∪ 10, 12 ∪ 11, 22 1 2 3 4 –1 –2 –3 –3 –2 –1 1 2 3 123. The graph of ƒ can be obtained by shifting the graph of g two units to the left. That is, ƒ1x2 5 g1x 1 22. 125. domain: 321.5, 42 127. yes 129. a. y 5 64.67x 1 805 b. No, it is not close to the actual price at $517.20. c. $1581.05 Practice Test 1. b 3. c 5. !x 2 2 x2 1 11 domain: 32, q2 7. x 1 9 domain: 12, q2 9. 4 11. neither 13. domain: 33, q2 range: 12q, 24 15. –5 –3 –1 3 2 1 4 5 –2 3 2 4 5 6 8 7 x y domain: 12q, 212 ∪ 121, q2 range: 31, q2 17. a. 22 b. 4 c. 23 d. x 5 23, 2 19. 6x 1 3h 2 4 21. 232 23. ƒ211x2 5 x2 1 5 domain ƒ: 35, q2 domain ƒ21: 30, q2 range ƒ: 30, q2 range ƒ21: 35, q2 25. ƒ211x2 5 5x 2 1 x 1 2 domain ƒ: 12q, 52 ∪ 15, q2 range ƒ: 12q, 222 ∪ 122, q2 domain ƒ21: 12q, 222 ∪ 122, q2 range ƒ21: 12q, 52 ∪ 15, q2 27. 30, q2 29. S1x2 5 0.42 x where x is the original price of a suit 31. quadrant III, “quarter of unit circle” 33. C 1x2 5 e 15, 0 # x # 30 x 2 15, x . 30 35. F 5 30 m P Answers to Odd-Numbered Exercises 1195 37. yes Cumulative Test 1. 3 1 !5 2 3. x 1x 2 22, x 2 22 5. 145 1 0i 7. 40% 9. x 5 1 6 !35 5 11. x 5 62, 6i!3 13. 10, 52 15. d 5 11.03, 11.25, 2.452 17. y 5 23 19. 1x 1 222 1 1y 1 122 5 20 21. 12q, 12 ∪ 11, q2 23. 46 25. r 5 135 t 27. h1x2 5 x 2 2 CH A P T E R 4 Section 4.1 1. b 3. a 5. b 7. c 9. 11. 13. 15. 17. 19. 21. 23. ƒ1x2 5 1x 1 322 2 12 25. ƒ1x2 5 21x 1 522 1 28 27. ƒ1x2 5 21x 1 222 2 10 29. ƒ1x2 5 241x 2 222 1 9 31. ƒ1x2 5 1 21x 2 422 2 5 33. 35. 37. 39. 41. A 1 33, 494 33 B 43. A7, 2 39 2 B 45. 1275,12.952 47. A15 28, 829 392B 49. y 5 221x 1 122 1 4 51. y 5 251x 2 222 1 5 53. y 5 5 9 1x 1 122 2 3 55. y 5 12A x 2 1 2B2 23 4 57. y 5 5 4 1x 2 2.522 2 3.5 59. a. 350,000 units b. $12,262,500 61. Gaining: January 2010; Losing: February 2010–June 2011 63. a. 120 feet b. 50 yards 65. 2,083,333 sq. ft 67. a. 1 second, 116 feet b. 3.69 seconds 69. a. 26,000 feet b. 8944 feet 71. a. 100 boards b. $24,000 73. 15 to 16 units to break even or 64 to 65 units to break even. 75. a. ƒ1t2 5 349 162 t2 1 16 b. 878 million 77. a. y5 20.011t 2 22522 1 400 b. 425 minutes 79. The corrections that need to be made are vertex 123, 212 and x-intercepts 122, 02 124, 02. 81. ƒ1x2 5 21x 2 122 1 4. The negative must be factored out of the x2 and x terms. 83. true 85. false 87. ƒ1x2 5 a Ax 1 b 2aB2 1 4ac 2 b2 4a 89. a. The maximum area of the rectangular pasture is 62,500 square feet. b. The maximum area of the circular fence is approximately 79,577 square feet. 91. a. 11425, 4038.252 b. 10, 2232 c. 14.04, 02, 12845.96, 02 d. x 5 1425 1196 Answers to Odd-Numbered Exercises 93. a. y 5 22x2 1 12.8x 1 4.32 b. y 5 221x 2 3.222 1 24.8, 13.2, 24.82 c. Yes 95. a. b. The equation of the best fit parabola is y 5 21.0589x2 1 12.268x 1 4.3042 with r2 5 0.9814. This is shown below on the scatterplot: c. i. Using the equation from part 1b2, the initial height from which the pumpkin was thrown is about 4.3 feet, and the initial velocity is about 12.3 feet per second. ii. The maximum height occurs at the vertex and is approximately 39.8 feet; this occurs about 5.85 seconds into the flight. iii. The pumpkin lands approximately 12 seconds after it is launched. Section 4.2 1. polynomial; degree 2 3. polynomial; degree 5 5. not a polynomial 7. not a polynomial 9. not a polynomial 11. h 13. b 15. e 17. c 19. 21. 23. 25. 27. 3 1multiplicity 12, 24 1multiplicity 32 29. 0 1multiplicity 22, 7 1multiplicity 22, 24 1multiplicity 12 31. 0 1multiplicity 22, 1 1multiplicity 22 33. 0 1multiplicity 12, 3 2 1multiplicity 12, 29 4 1multiplicity 12 35. 0 1multiplicity 22, 23 1multiplicity 12 37. 0 1multiplicity 42 39. P1x2 5 x 1x 1 32 1x 2 12 1x 2 22 41. P1x2 5 x 1x 1 52 1x 1 32 1x 2 2 2 1x 2 62 43. P1x2 5 12x 1 12 13x 2 22 14x 2 32 45. P1x2 5 x2 2 2x 2 1 47. P1x2 5 x2 1x 1 223 49. P1x2 5 1x 1 322 1x 2 725 51. P1x2 5 x21x 1 12Ax 1 !3B2 Ax 2 !3B2 53. ƒ1x2 5 21x 1 322 a. 23 1multiplicity 22 b. touches at 23 c. 10, 292 d. falls left and right, without bound e. 55. ƒ1x2 5 1x 2 223 a. 2 1multiplicity 32 b. crosses at 2 c. 10, 282 d. falls left, rises right e. 57. ƒ1x2 5 x 1x 2 32 1x 1 32 a. 0, 3, 23 1multiplicity 12 b. crosses at each zero c. 10, 02 d. falls left, rises right e. 59. ƒ1x2 5 2x 1x 2 22 1x 1 12 a. 0, 2, –1 1multiplicity 12 b. crosses at each zero c. 10, 02 d. falls right, rises left e. Ben’s Data 0 5 10 15 20 25 30 35 40 45 0 2 4 6 8 Time (in seconds) Vertical Distance (in feet) Ben’s Data 0 5 10 15 20 25 30 35 40 45 0 2 4 6 8 Time (in seconds) Vertical Distance (in feet) Answers to Odd-Numbered Exercises 1197 61. ƒ1x2 5 2x31x 1 32 a. 0 1multiplicity 32, 23 1multiplicity 12 b. crosses at both 0 and 23 c. 10, 02 d. falls left and right, without bound e. 63. ƒ1x2 5 12x41x 2 42 1x 1 12 a. 0 1multiplicity 42, 4 1multiplicity 12, 21 1multiplicity 12 b. touches at 0 and crosses at 4 and 21 c. 10, 02 d. rises left and right, without bound e. 65. ƒ1x2 5 2x31x 2 42 1x 1 12 a. 0 1multiplicity 32, 4 1multiplicity 12, 21 1multiplicity 12 b. crosses at each zero c. 10, 02 d. falls left, rises right e. 67. ƒ1x2 51x 2 22 1x 1 22 1x 2 12 a. 1, 2, 221multiplicity 12 b. crosses at each zero c. 10, 42 d. falls left, rises right e. 69. ƒ1x2 521x 1 2221x 2 122 a. 22 1multiplicity 22, 1 1multiplicity 22 b. touches at both 22 and 1 c. 10, 242 d. falls left and right, without bound e. 71. ƒ1x2 5 x21x 2 2231x 1 322 a. 0 1multiplicity 22, 2 1multiplicity 32, 23 1multiplicity 22 b. touches at both 0 and 23, and crosses at 2 c. 10, 02 d. falls left, and rises right e. 73. a. 23 1multiplicity 12, 21 1multiplicity 22, 2 1multiplicity 12 b. even c. negative d. 10, 62 e. ƒ1x2 5 21x 1 1221x 2 22 1x 1 32 75. a. 0 1multiplicity 22, 22 1multiplicity 22, 3 2 1multiplicity 12 b. odd c. positive d. 10, 02 e. ƒ1x2 5 x212x 2 32 1x 1 222 77. a. Revenue for the company is increasing when advertising costs are less than $400,000. Revenue for the company is decreasing when advertising costs are between $400,000 and $600,000. b. The zeros of the revenue function occur when $0 and $600,000 are spent on advertising. When either $0 or $600,000 is spent on advertising, the company’s revenue is $0. 79. The velocity of air in the trachea is increasing when the radius of the trachea is between 0 and 0.45 cm and decreasing when the radius of the trachea is between 0.45 cm and 0.65 cm. 81. Sixth-degree polynomial because there are five known turning points. 83. down 85. 6th degree 87. If h is a zero, then 1x 2 h2 is a factor. So, in this case the function would be: P1x2 5 1x 1 22 1x 1 12 1x 2 32 1x 2 42 89. ƒ1x2 5 1x 2 1221x 1 223 Yes, the zeros are 22 and 1. But you must remember that this is a fifth-degree polynomial. At the 22 zero the graph crosses, but it should be noted that at 1 it only touches at this value. The correct graph would be the following: 91. false 93. true 95. n 97. ƒ1x2 5 1x 1 1221x 2 325 , g1x2 5 1x 1 1241x 2 323 h1x2 5 1x 1 1261x 2 32 99. The zeros of the polynomial are 0, a, and 2b. 101. no x-intercepts 103. y 5 22x5 , yes 0.2 0.4 0.6 1 2 3 4 5 r v 1198 Answers to Odd-Numbered Exercises 105. x-intercepts: 122.25, 02, 16.2, 02, 114.2, 02 zeros: 22.25 1multiplicity 22, 6.2 1multiplicity12, 14.2 1multiplicity 12 107. 122.56, 217.122, 120.58, 12.592, 11.27, 211.732 Section 4.3 1. Q1x2 5 2x 2 1, r1x2 5 0 3. Q1x2 5 x 2 3, r1x2 5 0 5. Q1x2 5 3x 2 3, r1x2 5 211 7. Q1x2 5 3x 2 28, r1x2 5 130 9. Q1x2 5 x 2 4, r1x2 5 12 11. Q1x2 5 3x 1 5, r1x2 5 0 13. Q1x2 5 2x 2 3, r1x2 5 0 15. Q1x2 5 4x2 1 4x 1 1, r1x2 5 0 17. Q1x2 5 2x2 2 x 2 1 2, r1x2 5 15 2 19. Q1x2 5 4x2 2 10x 2 6, r1x2 5 0 21. Q1x2 5 22x2 2 3x 2 9, r1x2 5 227x2 1 3x 1 9 23. Q1x2 5 x2 1 1, r1x2 5 0 25. Q1x2 5 x2 1 x 1 1 6, r1x2 5 2 121 6 x 1 121 3 27. Q1x2 5 23x3 1 5.2x2 1 3.12x 2 0.128, r1x2 5 0.9232 29. Q1x2 5 x2 2 0.6x 1 0.09, r1x2 5 0 31. Q1x2 5 3x 1 1, r1x2 5 0 33. Q1x2 5 7x 2 10, r1x2 5 15 35. Q1x2 5 2x3 1 3x 2 2, r1x2 5 0 37. Q1x2 5 x3 2 x2 1 x 2 1, r1x2 5 2 39. Q1x2 5 x3 2 2x2 1 4x 2 8, r1x2 5 0 41. Q1x2 5 2x2 2 6x 1 2, r1x2 5 0 43. Q1x2 5 2x3 2 5 3x2 1 53 9 x 1 106 27 , r1x2 5 2 112 81 45. Q1x2 5 2x3 1 6x2 2 18x 2 54, r1x2 5 0 47. Q1x2 5 x6 1 x5 1 x4 2 7x3 2 7x2 2 4x 2 4, r1x2 5 23 49. Q1x2 5 x5 1 !5x4 2 44x3 2 44!5x2 2 245x 2 245!5, r1x2 5 0 51. Q1x2 5 2x 2 7, r1x2 5 0 53. Q1x2 5 x2 2 9, r1x2 5 0 55. Q1x2 5 x4 1 2x3 1 8x2 1 18x 1 36, r1x2 5 71 57. Q1x2 5 x2 1 1, r1x2 5 224 59. Q1x2 5 x6 1 x5 1 x4 1 x3 1 x2 1 x 1 1, r1x2 5 0 61. The width is 3x2 1 2x 1 1 feet. 63. The trip takes x2 1 1 hours. 65. In long division, you must subtract 1not add2 each term. 67. Forgot the “0” placeholder. 69. true 71. false 73. yes 75. Q1x2 5 x2n 1 2x n 1 1, r 1x2 5 0 77. 2x 2 1; linear 79. x3 2 1; cubic 81. 26x2 1 16x 2 10; quadratic function Section 4.4 1. ƒ112 5 0 3. g112 5 4 5. ƒ1222 5 84 7. Yes, the given is a zero of the polynomial. 9. Yes, the given is a zero of the polynomial. 11. 24, 1, 3; P1x2 5 1x 2 12 1x 1 42 1x 2 32 13. 23, 1 2, 2; P1x2 5 12x 2 12 1x 1 32 1x 2 22 15. 23, 5; P1x2 5 1x2 1 42 1x 2 52 1x 1 32 17. 23, 1; P1x2 5 1x 2 12 1x 1 32 1x2 2 2x 1 22 19. 22, 21 1both multiplicity 22; P1x2 5 1x 1 222 1x 1 122 21. 61, 62, 64 23. 61, 62, 63, 64, 66, 612 25. 61 2, 61, 62, 64, 68 27. 61, 62, 64, 65, 610, 620, 61 5, 62 5, 64 5 29. 61, 62, 64, 68; rational zeros: 24, 21, 2, 1 31. 61, 63, 61 2, 63 2; rational zeros: 1 2, 1, 3 33. POSITIVE REAL ZEROS NEGATIVE REAL ZEROS 1 1 35. POSITIVE REAL ZEROS NEGATIVE REAL ZEROS 1 0 37. POSITIVE REAL ZEROS NEGATIVE REAL ZEROS 2 1 0 1 Answers to Odd-Numbered Exercises 1199 39. POSITIVE REAL ZEROS NEGATIVE REAL ZEROS 1 1 41. POSITIVE REAL ZEROS NEGATIVE REAL ZEROS 2 2 0 2 2 0 0 0 43. POSITIVE REAL ZEROS NEGATIVE REAL ZEROS 4 0 2 0 0 0 45. a. Number of sign variations for P1x2: 0 Number of sign variations for P12x2: 3 POSITIVE REAL ZEROS NEGATIVE REAL ZEROS 0 3 0 1 b. possible rational zeros: 61, 62, 63, 66 c. rational zeros: 21, 22, 23 d. P1x2 5 1x 1 12 1x 1 22 1x 1 32 47. a. Number of sign variations for P1x2: 2 Number of sign variations for P12x2: 1 POSITIVE REAL ZEROS NEGATIVE REAL ZEROS 2 1 0 1 b. possible rational zeros: 61, 67 c. rational zeros: 21, 1, 7 d. P1x2 5 1x 1 12 1x 2 12 1x 2 72 49. a. Number of sign variations for P1x2: 1 Number of sign variations for P12x2: 2 POSITIVE REAL ZEROS NEGATIVE REAL ZEROS 1 2 1 0 b. possible rational zeros: 61, 62, 65, 610 c. rational zeros: 0, 1, 22, 25 d. P1x2 5 x 1 x 2 12 1x 1 22 1x 1 52 51. a. Number of sign variations for P1x2: 4 Number of sign variations for P12x2: 0 POSITIVE REAL ZEROS NEGATIVE REAL ZEROS 4 0 2 0 0 0 b. possible rational zeros: 61, 62, 613, 626 c. rational zeros: 1, 2 d. P1x2 5 1x 2 12 1x 2 22 1x2 2 4x 1 132 53. a. Number of sign variations for P1x2: 2 Number of sign variations for P12x2: 1 POSITIVE REAL ZEROS NEGATIVE REAL ZEROS 2 1 0 1 b. possible rational zeros: 61, 61 2, 61 5, 6 1 10 c. rational zeros: 21, 21 2, 1 5 d. P1x2 5 1x 2 12 1 2x 1 12 15x 2 12 55. a. Number of sign variations for P1x2: 1 Number of sign variations for P12x2: 2 POSITIVE REAL ZEROS NEGATIVE REAL ZEROS 1 2 1 0 b. possible rational zeros: 61, 62, 65, 610, 6 1 2, 6 1 3, 6 1 6, 62 3, 65 2, 65 3, 65 6, 610 3 c. rational zeros: 21, 25 2, 2 3 d. P1x2 5 61x 1 12ax 1 5 2b ax 2 2 3b 57. a. Number of sign variations for P1x2: 4 Number of sign variations for P12x2: 0 POSITIVE REAL ZEROS NEGATIVE REAL ZEROS 4 0 2 0 0 0 b. possible rational zeros: 61, 62, 64 c. rational zeros: 1 d. P1x2 5 1x 2 122 1x2 1 42 1200 Answers to Odd-Numbered Exercises 59. a. Number of sign variations for P1x2: 1 Number of sign variations for P12x2: 1 POSITIVE REAL ZEROS NEGATIVE REAL ZEROS 1 1 b. possible rational zeros: 61, 62, 63, 64, 66, 69, 612, 618, 636 c. rational zeros: 21, 1 d. P1x2 5 1x 1 12 1x 2 12 1x2 1 92 1x2 1 42 61. a. Number of sign variations for P1x2: 4 Number of sign variations for P12x2: 0 POSITIVE REAL ZEROS NEGATIVE REAL ZEROS 4 0 2 0 0 0 b. possible rational zeros: 61, 65, 61 2, 61 4, 6 5 2, 65 4 c. rational zeros: 1 2 d. P1x2 5 4Ax 2 1 2B2 1x2 2 4x 1 52 63. 65. 67. x 5 1.34 69. x 5 0.22 71. x 5 20.43 73. a. P1x2 5 23x2 2 2x 1 26, x $ 0 b. 263 subscribers 75. P1x2 5 20.0002x2 1 8x 2 1500; 0 or 2 positive real zeros. 77. 18 hours 79. It is true that one can get 5 negative zeros here, but there may be just 1 or 3. POSITIVE REAL ZEROS NEGATIVE REAL ZEROS 0 5 0 3 0 1 81. true 83. false 85. b, c 87. possible rational zeros: 61, 62, 64, 68, 616, 632 zeros: 2 89. a. 23 4, 2 3 b. P1x2 5 13x 2 2214x 1 321x2 1 2x 1 52 Section 4.5 1. x 5 62i; P1x2 51x 1 2i2 1x 2 2i2 3. x 5 1 6 i; P1x2 5 1x 2 11 2 i22 1x 2 11 1 i22 5. x 5 62, 62i; P1x2 5 1x 2 22 1x 1 22 1x 2 2i2 1x 1 2i 2 7. x 5 6!5, 6!5 i; P1x2 5 Ax 2 !5BAx 1 !5BAx 2 !5 iBAx 1 !5 iB 9. 2i 11. 22i, 3 1 i 13. 1 1 3i, 2 2 5i 15. i, 1 1 i 17. P1x2 5 x3 2 2x2 1 5x 19. P1x2 5 x3 2 3x2 1 28x 2 26 21. P1x2 5 x4 2 2x3 1 11x2 2 18x 1 18 23. 62i, 23, 5; P1x2 5 1x 2 2i 2 1x 1 2i 2 1x 2 52 1x 1 32 25. 6i, 1, 3; P1x2 51x 2 i2 1x 1 i2 1x 2 32 1x 2 12 27. 63i, 1 1multiplicity 22; P1x2 51x 2 3i2 1x 1 3i2 1x 2 122 29. 1 6 i, 21 6 2!2; P1x2 5 1x 2 11 1 i221x 2112 i221x2A2122!2BBAx2A2112!2BB 31. 3 6 i, 62; P1x2 5 1x 2 13 1 i22 1x 2 13 2 i22 1x 2 22 1x 1 22 33. 2 6 i, 1, 4; P1x2 5 1x – 12 1 i22 1x 2 12 2 i22 1x 2 12 1x 2 42 35. P1x2 5 1x 1 3i2 1x 2 3i2 1x 2 12 37. P1x2 5 1x 1 i2 1x 2 i2 1x 2 52 39. P1x2 5 1x 1 2i 2 1x 2 2i 2 1x 1 12 41. P1x2 5 1x 2 321x 2 121 1 !5i221x 2 121 2 !5i22 43. P1x2 5 1x 1 32 1x 2 52 1x 1 2i2 1x 2 2i2 45. P1x2 5 1x 1 12 1x 2 52 1x 1 2i2 1x 2 2i2 47. P1x2 5 1x 2 12 1x 2 22 1x 2 12 2 3i22 1x 2 12 1 3i22 Answers to Odd-Numbered Exercises 1201 49. P1x2 5 21x 1 12 1x 2 22 1x 2 12 2 i22 1x 2 12 1 i22 51. P1x2 5 1x 2 1221x 1 2i2 1x 2 2i2 53. P1x2 5 1x 2 12 1x 1 12 1x 2 2i2 1x 1 2i2 1x 2 3i2 1x 1 3i2 55. P1x2 5 12x 2 1221x 2 12 2 i22 1x 2 12 1 i22 57. P1x2 5 1x 2 121x 1 1213x 2 221x 2 2i21x 1 2i2 59. Yes—since there are no real zeros, this function is either always positive or always negative; the end behavior is similar to an even power function with a positive leading coefficient 1rising in both ends2. We expect profit to eventually rise 1without bound!2. 61. No—since there is only one real zero 1and it corresponds to producing a positive number of units2, then that is the “break-even point”; the end behavior is similar to an odd power function with a negative leading coefficient 1falling to the right2. We expect profit to fall 1without bound!2 after the break-even point. 63. Since the profit function is a third-degree polynomial, we know that the function has three zeros and at most two turning points. Looking at the graph, we can see there is one real zero where t # 0. There are no real zeros when t . 0; therefore, the other two zeros must be complex conjugates. Therefore, the company always has a profit greater than approximately $5.1 million and, in fact, the profit will increase towards infinity as t increases. 65. Since the concentration function is a third-degree polynomial, we know the function has three zeros and at most two turning points. Looking at the graph, we can see there will be one real zero where t $ 8. The remaining zeros are a pair of complex conjugates. Therefore, the concentration of the drug in the bloodstream will decrease to zero as the hours go by. Note that the concentration will not approach negative infinity since concentration is a nonnegative quantity. 67. Just because 1 is a zero does not mean 21 is a zero 1holds only for complex conjugates2. Should have used synthetic division with 1. 69. false 71. true 73. No—if there is an imaginary zero, then its conjugate is also a zero. Therefore, all imaginary zeros can only correspond to an even-degree polynomial. 75. P1x2 5 x6 1 3b2 x4 1 3b4 x2 1 b6 77. all roots are complex REAL ZEROS COMPLEX ZEROS 0 4 2 2 4 0 79. 3 5, 6i, 62i; P1x2 5 251x 2 0.62 1x 2 2i2 1x 1 2i2 1x 2 i2 1x 1 i2 Section 4.6 1. 12q, 232 ∪ 123, q2 3. A 2q, 21 @ 3B ∪ A 21 @ 3, 1 @ 2B ∪ A1 @ 2, qB 5. 12q, 242 ∪ 124, 32 ∪ 13, q2 7. 12q, q2 9. 12q, 222 ∪ 122, 32 ∪ 13, q2 11. HA: y 5 0 VA: x 5 22 13. HA: none VA: x 5 25 15. HA: none VA: x 5 1 2, x 5 24 3 17. HA: y 5 1 3 VA: none 19. HA: y 5 12 7 VA: x 5 0.5, x 5 21.5 21. y 5 x 1 6 23. y 5 2x 1 24 25. y 5 4x 1 11 2 27. b 29. a 31. e 33. 35. 37. 39. 41. 43. 45. 47. 49. 51. 53. 55. –5 –3 1 2 3 4 5 1 –1 –3 –4 –5 2 3 4 5 x y 1202 Answers to Odd-Numbered Exercises 57. 59. a. x-intercept: 12, 02; y-intercept: 10, 0.52 b. HA: y 5 0 VA: x 5 21, x 5 4 c. f 1x2 5 x 2 2 1x 1 121x 2 42 61. a. x-intercept: 10, 02; y-intercept: 10, 02 b. HA: y 5 23 VA: x 5 24, x 5 4 c. f 1x2 5 23x2 1x 1 421x 2 42 63. a. C112 > 0.0198 b. C1602 > 0.0324 c. C13002 > 0 d. y 5 0; after several days, C1t2 > 0 65. a. N102 5 52 wpm b. N1122 > 107 wpm c. N1362 > 120 wpm d. y 5 130; 130 wpm 67. y 5 10, 10 ounces of food 69. 2w2 1 1000 w 71. 2000 or 8000 units; average profit of $16 per unit. 73. The concentration of the drug in the bloodstream 15 hours after taking the dose is approximately 25.4 µg/mL. There are two times, 1 hour and 15 hours, after taking the medication at which the concentration of the drug in the bloodstream is approximately 25.4 µg/mL. The first time, approximately 1 hour, occurs as the concentration of the drug is increasing to a level high enough that the body will be able to maintain a concentration of approximately 25 µg/mL throughout the day. The second time, approximately 15 hours, occurs many hours later in the day as the concentration of the medication in the bloodstream drops. 75. There is a common factor in the numerator and denominator. There is a hold at x 5 1. 77. The horizontal asymptote was found incorrectly. The correct horizontal asymptote is y 5 21. 79. true 81. false 83. HA: y 5 1 VA: x 5 c, x 5 2d 85. Two possibilities: y 5 4x2 1x 1 321x 2 12 and y 5 4x5 1x 1 3231x 2 122 87. VA: x 5 22, yes 89. HA: y 5 0 VA: x 5 0, x 5 21 3 Intercepts: A22 5, 0B, yes 91. a. f: HA: y 5 0 VA: x 5 3 g: HA: y 5 2 VA: x 5 3 h: HA: y 5 23 VA: x 5 3 b. graphs of ƒ and g: as x S 6q, f 1x2 S 0 and g1x2 S 2 c. graphs of g and h below: as x S 6q, g1x2 S 2 and h1x2 S 23 d. g1x2 5 2x 2 5 x 2 3 , h1x2 5 23x 1 10 x 2 3 Yes, if the degree of the numerator is the same as the degree of the denominator, then the horizontal asymptote is the ratio of the leading coefficients for both g and h. Review Exercises 1. b 3. a 5. 7. 9. f 1x2 5 A x 2 3 2B2 2 49 4 11. ƒ1x2 5 41x 1 122 2 11 13. 15. 17. A 5 26, 599 52 B 19. A 2 2 15, 451 125B 21. y 5 1 91x 1 222 1 3 23. y 5 5.61x 2 2.722 1 3.4 25. a. P1x2 5 22x2 1 35 3 x 2 14 b. x > 4.1442433, 1.68909 c. d. 11.6891, 4.1442 or 1689 to 4144 Answers to Odd-Numbered Exercises 1203 27. A1x2 5 21 21x 2 122 1 9 2, maximum at x 5 1 base: 3 units, height: 3 units 29. yes, 6 31. no 33. d 35. a 37. 39. 41. 6 1multiplicity 52, 24 1multiplicity 22 43. 0, 22, 2, 3, 23, all multiplicity 1 45. ƒ1x2 5 x 1x 1 32 1x 2 42 47. ƒ1x2 5 x 15x 1 22 14x 2 32 49. ƒ1x2 5 x4 2 2x3 2 11x2 1 12x 1 36 51. ƒ1x2 5 1x 2 72 1x 1 22 a. 22, 7 1both multiplicity 12 b. crosses at 22, 7 c. 10, 2142 d. rises right and left e. 53. ƒ1x2 5 6x7 1 3x5 2 x2 1 x 2 4 a. 10.8748, 02 with multiplicity 1 b. crosses at its only real zero c. 10, 242 d. falls left and rises right e. 55. a. b. 1, 3, 7 1all with multiplicity 12 c. between 1 and 3 hours, and more than 7 hours is financially beneficial 57. Q1x2 5 x 1 4, r1x2 5 2 59. Q1x2 5 2x3 2 4x2 2 2x 27 2, r 1x2 5 223 61. Q1x2 5 x3 1 2x2 1 x 2 4, r1x2 5 0 63. Q1x2 5 x5 2 8x4 1 64x3 2 512x2 1 4096x 2 32,768, r 1x2 5 262,080 65. Q1x2 5 x 1 3, r1x2 5 24x 2 8 67. Q1x2 5 x2 2 5x 1 7, r1x2 5 215 69. 3x3 1 2x2 2 x 1 4 feet 71. ƒ1222 5 2207 73. g112 5 0 75. no 77. yes 79. P1x2 5 x 1x 1 22 1x 2 422 81. P1x2 5 x21x 1 32 1x 2 222 83. POSITIVE REAL ZEROS NEGATIVE REAL ZEROS 1 1 85. POSITIVE REAL ZEROS NEGATIVE REAL ZEROS 5 2 5 0 3 2 3 0 1 2 1 0 87. possible rational zeros: 61, 62, 63, 66 89. possible rational zeros: 61, 62, 64, 68, 616, 632, 664, 61 2 91. possible rational zeros: 61, 61 2; zeros: 1 2 93. possible rational zeros: 61, 62, 64, 68, 616; zeros: 1, 2, 4, 22 95. a. POSITIVE REAL ZEROS NEGATIVE REAL ZEROS 1 0 b. 61, 65 c. 21 is a lower bound, 5 is an upper bound d. none e. not possible f. 97. a. POSITIVE REAL ZEROS NEGATIVE REAL ZEROS 3 0 1 0 b. 61, 62, 63, 64, 66, 612 c. 24 is a lower bound, 12 is upper bound d. 1, 2, 6 e. P1x2 5 1x 2 12 1x 2 62 1x 2 22 f. 99. a. POSITIVE REAL ZEROS NEGATIVE REAL ZEROS 0 0 0 2 2 2 2 0 b. 61, 62, 63, 64, 66, 68, 612, 624 1204 Answers to Odd-Numbered Exercises c. 24 is a lower bound, 8 is upper bound d. 22, 21, 1, 6 e. P1x2 5 1x 2 22 1x 1 12 1x 1 22 1x 2 62 f. 101. P1x2 5 1x 2 5i2 1x 1 5i2 103. P1x2 5 1x 2 11 2 2i22 1x 2 11 1 2i22 105. 2i, 3 2 i 107. 2i, 2 1 I 109. 2i, 4, 21; P1x2 5 1x 2 i2 1x 1 i2 1x 2 42 1x 1 12 111. 3i, 1 6 i; P1x2 5 1x 2 3i2 1x 1 3i2 1x 2 11 1 i22 1x 2 11 2 i22 113. P1x2 5 1x 2 32 1x 1 32 1x 2 3i2 1x 1 3i2 115. P1x2 51x 2 2i2 1x 1 2i2 1x 2 12 117. HA: y 5 21 VA: x 5 22 119. HA: none VA: x 5 21 Slant: y 5 4x 2 4 121. HA: y 5 2 VA: none 123. 125. 127. 129. a. 1480, 212112 b. 10, 2592 c. 1212.14, 02, 1972.14, 02 d. x 5 480 131. x-intercepts: 121, 02, 10.4, 02, 12.8, 02; zeros: 21, 0.4, 2.8, each with multiplicity 1 133. linear function 135. a. 22, 1multiplicity 22, 3, 4 b. P1x2 5 1x 1 2221x 2 32 1x 2 42 137. 7 2, 22 6 3i; P1x2 5 12x 2 721x 1 2 2 3i21x 1 2 1 3i2 139. a. yes, one-to-one b. f 211x2 5 x 1 3 2 2 x c. Practice Test 1. 3. 13, 1 22 5. ƒ1x2 5 x 1x 2 2231x 2 122 7. Q1x2 5 22x2 2 2x 211 2 , r1x2 5 219 2 x 1 7 2 9. yes, complex zeros. 11. P1x2 5 1x 2 72 1x 1 22 1x 2 12 13. yes, complex zeros. 15. possible rational zeros: 61, 62, 63, 64, 66, 612, 61 3, 62 3, 64 3 17. 3 2, 62i 19. Given the points 10, 3002, 12, 2852, 110, 3152, 152, 3002, you can have a polynomial of degree 3 because there are 2 turning points. 21. Given the points 11970, 0.082, 11988, 0.132, 12002, 0.042, 12005, 0.062, the lowest-degree polynomial that can be represented is a third-degree polynomial. 23. a. x-intercept: 10, 02, y-intercept: 10, 02 b. x 5 62 c. y 5 0 d. none e. Answers to Odd-Numbered Exercises 1205 25. a. x-intercept: 13, 02, y-intercept: A0, 3 8B b. x 5 22, x 5 4 c. y 5 0 d. none e. 27. a. y 5 x2 2 3x 2 7.99 b. y 5 1x 2 1.522 2 10.24 c. 121.7, 02 and 14.7, 02 d. yes Cumulative Test 1. 5 4x7y10 3. 4x 2 12 5. 331 3 min. 7. x 5 3 9. y 5 1 3 x 2 1 3 11. x2 1 1 y 2 622 5 2 13. neither 15. right 1 unit and then up 3 units 17. g1ƒ12122 5 0 19. ƒ1x2 5 1x 1 222 1 3 21. Q1x2 5 4x2 1 4x 1 1, r 1x2 5 28 23. possible rational zeros: 61, 62, 63, 64, 66, 61 2, 61 3, 61 4, 61 6,6 1 12, 62 3, 63 2, 63 4, 61 4 zeros: 22, 23 4, 1 3 25. HA: y 5 0 VA: x 5 62 27. f 1x2 5 31x 1 12 x12x 2 32 x-intercept: 121, 02 HA: y 5 0 VA: x 5 0, x 5 3 2 yes C H A P T E R 5 Section 5.1 1. 16 3. 1 25 5. 4 7. 27 9. 21 11. 5.2780 13. 9.7385 15. 7.3891 17. 0.0432 19. 27 21. 16 23. 4 25. 19.81 27. f 29. e 31. b 33. y-intercept: 10, 12 HA: y 5 0 domain: 12q, q2 range: 10, q2 other points: A21, 1 6B, 11, 62 35. y-intercept: 10, 12 HA: y 5 0 domain: 12q, q2 range: 10, q2 other points: 11, 0.12, 121, 102 37. y-intercept: 10, 12 HA: y 5 0 domain: 12q, q2 range: 10, q2 other points: A1, 1 eB, 121, e2 39. y-intercept: 10, 02 HA: y 5 21 domain: 12q, q2 range: 121, q2 other points: 12, 32, 11, 12 41. y-intercept: 10, 12 HA: y 5 2 domain: 12q, q2 range: 12q, 22 other points: 11, 2 2 e2, A21, 2 2 1 eB 43. y-intercept: 10, e 2 42 HA: y 5 24 domain: 12q, q2 range: 124, q2 other points: 121, 232, 11, e2 242 1206 Answers to Odd-Numbered Exercises 45. y-intercept: 10, 32 HA: y 5 0 domain: 12q, q2 range: 10, q2 other points: 12, 3e2, 11, 3!e2 47. y-intercept: 10, 52 HA: y 5 1 domain: 12q, q2 range: 11, q2 other points: 10, 52, 12, 22 49. 10.4 million 51. P1302 5 150012 30⁄52 > 96,000 53. 168 mg 55. $3031 57. $3,448.42 59. $13,011.03 61. $4319.55 63. $13,979.42 65. 3.4 mg/L 67. p 1price per unit2 D 1p2—approximate demand for product in units 1.00 1,955,000 5.00 1,020,500 10.00 452,810 20.00 89,147 40.00 3455 60.00 134 80.00 5 90.00 1 69. The mistake is that 421 @ 2 2 42. Rather, 421 @ 2 5 1 4 1 @ 2 5 1 2. 71. 2.5% needs to be converted to a decimal, 0.025. 73. false 75. true 77. 79. 81. y-intercept: 10, be 2 a2 HA: y 5 2a 83. Domain: 12q, q2 85. 87. close on the interval 123, 32 89. as x increases, ƒ1x2 S e, g1x2 S e2, h1x2 S e4 91. a. 210 200 190 180 170 160 150 140 0 2 Time (in minutes) Temperature (in degrees Fahrenheit) 4 6 b. The best fit exponential curve is y 5 228.34 10.91732x with r2 5 0.9628. This best fit curve is shown on the following scatterplot. The fit is very good, as evidenced by the fact that the square of the correlation coefficient is very close to 1. 210 220 200 190 180 170 160 150 140 0 2 Time (in minutes) Temperature (in degrees Fahrenheit) 4 6 c. i. Compute the y-value when x 5 6 to obtain about 136 degrees Fahrenheit. ii. The temperature of the soup the moment it was taken out of the microwave is the y-value at x 5 0, namely, about 228 degrees Fahrenheit. d. The shortcoming of this model for large values of x is that the curve approaches the x-axis, not 72 degrees. As such, it is no longer useful for describing the temperature beyond the x-value at which the temperature is 72 degrees. Section 5.2 1. 53 5 125 3. 81 1⁄ 4 5 3 5. 225 5 1 32 7. 1022 5 0.01 9. 104 5 10,000 11. A1 4B23 5 64 13. e21 5 1 e 15. e0 5 1 17. ex 5 5 19. xz 5 y 21. log8 15122 5 3 23. log10.000012 5 25 25. log225 1152 5 1 2 27. log2 @ 5A 8 125B 5 3 29. log 1 @ 27 132 5 21 3 Answers to Odd-Numbered Exercises 1207 31. ln 6 5 x 33. logy x 5 z 35. 0 37. 5 39. 7 41. 26 43. undefined 45. undefined 47. 1.46 49. 5.94 51. undefined 53. 28.11 55. 125, q2 57. A2q, 5 2B 59. A2q, 7 2B 61. 12q, 02 ∪ 10, q2 63. R 65. b 67. c 69. d 71. domain: 11, q2 range: 12q, q2 73. domain: 10, q2 range: 12q, q2 75. domain: 122, q2 range: 12q, q2 77. domain: 10, q2 range: 12q, q2 79. domain: 124, q2 range: 12q, q2 81. domain: 10, q2 range: 12q, q2 83. 60 decibels 85. 117 decibels 87. 8.5 89. 6.6 91. 3.3 93. Normal rainwater: 5.6 Acid rain/tomato juice: 4 95. 3.6 97. 13,236 years 99. 25 dB loss 101. a. USAGE WAVELENGTH FREQUENCY Super Low Frequency— Communication with Submarines 10,000,000 m 30 Hz Ultra Low Frequency— Communication within Mines 1,000,000 m 300 Hz Very Low Frequency— Avalanche Beacons 100,000 m 3000 Hz Low Frequency— Navigation, AM Longwave Broadcasting 10,000 m 30,000 Hz Medium Frequency—AM Broadcasts, Amateur Radio 1,000 m 300,000 Hz High Frequency— Shortwave broadcasts, Citizens Band Radio 100 m 3,000,000 Hz Very High Frequency— FM Radio, Television 10 m 30,000,000 Hz Ultra High Frequency— Television, Mobile Phones 0.050 m 6,000,000,000 Hz b. 103. log2 4 5 x is equivalent to 2x 5 4 1not x 5 242. 105. The domain is the set of all real numbers such that x 1 5 . 0, which is written as 125, q2. 107. false 109. true 111. domain: 1a, q2 range: 12q, q2 x-intercept: 1a 1 eb, 02 113. 115. y 5 x 117. x-intercept: 11, 02 VA: x 5 0 119. 10, q2 121. a. S b. A reasonable estimate for Vmax is about 156 µmol/min. c. Km is the value of [S] that results in the velocity being half of its maximum value, which by 1b2 is about 156. So, we need the value of [S] that corresponds to v 5 78. From the graph, this is very difficult to ascertain because of the very small units. We can simply say that it occurs between 0.0001 and 0.0002. A more accurate estimate can be obtained if a best fit curve is known. 1208 Answers to Odd-Numbered Exercises d. 1i2 v 5 33.70 ln13S42 1 395.80 with r2 5 0.9984. It is shown on the scatterplot below. 1ii2 Using the equation, we must solve the following equation for [S]: 100 5 33.70 ln13S42 1 395.80 2295.80 5 33.70 ln3S4 28.77745 5 ln3S4 e28.77745 5 S S < 0.000154171 Section 5.3 1. 0 3. 1 5. 8 7. 23 9. 3 2 11. 5 13. x 1 5 15. 8 17. 1 9 19. 3 logb 1x2 1 5 logb 1 y2 21. 1 2 logb 1x2 1 1 3 logb 1y2 23. 1 3 logb 1r2 2 1 2 logb 1s2 25. logb 1x2 2 logb 1y2 2 logb 1z2 27. 2 log x 1 1 2 log 1x 1 52 29. 3 ln x 1 2 ln1x 2 22 2 1 2 ln1x2 1 52 31. 2 log 1x 2 12 2 log 1x 2 32 2 log 1x 1 32 33. logb 1x3y52 35. logb a u5 v2 b 37. logb 1x1/2 y2/32 39. log a u2 v3z2b 41. ln a x2 2 1 1x2 1 322b 43. ln a 1x 1 321/2 x1x 1 221/3b 45. 1.2091 47. 22.3219 49. 1.6599 51. 2.0115 53. 3.7856 55. 110 decibels 57. 5.5 59. 3 log 5 2 log 52 5 3 log 5 2 2 log 5 5 log 5 61. Cannot apply the product and quotient properties to logarithms with different bases. So, you cannot reduce the given expression further without using the change of base formula. 63. true 65. false 67. Proof: Let u 5 logb M, v 5 logb N. Then, bu 5 M, bv 5 N. Observe that logb AM NB 5 logbAb b B 5 logb Abu2vB 5 u 2 v 5 logb M 2 logb N 69. 6 logb x 2 9 logb y 1 15 logb z 71. yes 73. no 75. no 77. yes Section 5.4 1. x 5 4 3. x 5 22 5. x 5 62 7. x 5 24 9. x 5 2 3 2 11. x 5 21 13. x 5 3, 4 15. x 5 0, 6 17. x 5 1, 4 19. x 5 3 1 log 1812 2 < 2.454 21. x 5 log3 152 2 1 < 0.465 23. x 5 log2 1272 1 1 3 < 1.918 25. x 5 ln 5 < 1.609 27. x 5 10 ln4 < 13.863 29. x 5 log3 1102 < 2.096 31. x 5 ln 1222 2 4 3 < 20.303 33. x 5 ln 6 2 < 0.896 35. x 5 ln a27 1 !61 2 b < 20.904 37. x 5 0 39. x 5 ln 7 < 1.946 41. x 5 0 43. x 5 log10192 2 < 0.477 45. x 5 50 47. x 5 40 49. x 5 9 32 51. x 5 63 53. x 5 5 55. x 5 6 57. x 5 21 59. no solution 61. x 5 25 8 63. x 5 6"e5 < 612.182 65. x 5 47.5 67. x 5 6"e4 2 1 < 6 7.321 69. x 5 1 2 123 1 e222 < 21.432 71. x < 21.25 73. x 5 2 1 "4 1 4e4 2 < 8.456 75. x 5 23 1 "13 2 < 0.303 77. x 5 1 1 !7 < 3.646 79. a. 151 beats per minute b. 7 minutes c. 66 beats per minute 81. 31.9 years 83. 19.74 years 85. 3.16 3 1015 joules 87. 1 W @ m2 89. 4.61 hours 91. 15.89 years 93. 6.2 v u Answers to Odd-Numbered Exercises 1209 95. ln14ex2 2 4x. Should first divide both sides by 4, then take the natural log: 4ex 5 9 ex 5 9 4 ln1ex2 5 lnA9 4B x 5 lnA9 4B 97. The correction that needs to be made is that x 5 25 needs to be removed from the list of solutions. The domain of the logs cannot include a negative solution. 99. true 101. false 103. x 5 1 1 "1 1 4b2 2 105. t 5 25 lna 3000 2 y 2y b 107. f 211x2 5 lnAx 1 "x2 2 1B; x $ 1 109. x 5 36 !5 2 111. –5 –3 –1 2 3 4 5 12 14 16 –4 2 4 6 8 10 x y x = 2.40 x = –0.25 113. domain: 12q, q2 y-axis symmetry Section 5.5 1. c 1iv2 3. a 1iii2 5. f 1i2 7. 119 million 9. 7.7 years, 2010 11. 2151.9 subscribers 13. $6,770,673 15. 332 million 17. 1.53 million 19. 13.53 ml 21. a. k 5 2ln a 8 15b > 0.6286 b. 636,000 mp3 players 23. 7575 years 25. 131,158,556 years old 27. 1058F 29. 3.8 hours before 7 am 31. $19,100 33. a. 84,520 b. 100,000 c. 100,000 35. 202.422 cases 37. 1.89 years 39. r 5 0 41. a. b. 75 c. 4 d. 4 43. a. 18 years b. 10 years 45. a. 30 years b. $328, 120 47. r 5 0.07, not 7 49. true 51. false 53. less time 55. a. For the same periodic payment, it will take Wing Shan fewer years to pay off the loan if she can afford to pay biweekly. b. 11.58 years c. 10.33 years d. 8.54 years, 7.69 years, respectively Review Exercises 1. 17,559.94 3. 5.52 5. 24.53 7. 5.89 9. 73.52 11. 6.25 13. b 15. c 17. y-intercept: 10, 212. HA: y 5 0 19. y-intercept: 10, 22 HA: y 5 1 21. y-intercept: 10, 12 HA: y 5 0 23. y-intercept: 10, 3.22 HA: y 5 0 25. $6144.68 27. $23,080.29 29. 43 5 64 31. 1022 5 1 100 33. log6 216 5 3 35. log 2@13A 4 169B 5 2 37. 0 39. 24 41. 1.51 43. 22.08 45. 122, q2 47. 12q, q2 49. b 51. d 1210 Answers to Odd-Numbered Exercises 53. 55. 57. 6.5 59. 50 decibels 61. 1 63. 6 65. a logc 1x2 1 b logc 1 y 2 67. logj 1r2 1 logj 1s2 2 3 logj t 69. 1 2 log 1a2 2 3 2 log 1b2 2 2 5 log 1c2 71. 0.5283 73. 0.2939 75. x 5 24 77. x 5 4 3 79. x 5 26 81. x < 20.218 83. no solution 85. x 5 0 87. x 5 100 3 89. x 5 128!2 91. x < 63.004 93. x < 0.449 95. $28,536.88 97. 16.6 years 99. 4.59 million 101. 6250 bacteria 103. 56 years 105. 16 fish 107. 343 mice 109. HA: y 5 e!2 < 4.11 111. 12.376, 2.0712 113. 10, q2 115. domain: 12q, q2, symmetric origin HA: y 5 21 1as x S 2q2, y 5 1 1as x S q2 117. a. N 5 4e20.038508t < 410.96222t b. N 5 410.96222t c. yes Practice Test 1. x3 3. 24 5. x 5 6!1 1 ln 42 < 62.177 7. x 5 21 1 lnA300 27 B 0.2 > 7.04 9. x 5 4 1 e2 < 11.389 11. x 5 ee < 15.154 13. x 5 9 15. x 5 23 6 !9 1 4e 2 < 0.729 17. x 5 lnA1 2B 5 2ln122 < 20.693 19. 121, 02 ∪ 11, q2 21. x-intercept: none y-intercept: 10, 22 HA: y 5 1 23. x-intercept: a 3 1 1 @ e 2 , 0b y-intercept: none VA: x 5 3 2 25. $8051.62 27. 90 decibels 29. 7.9 3 1011 , E , 2.5 3 1013 joules 31. 7800 bacteria 33. 3 days 35. domain: 12q, q2 symmetric origin Cumulative Test 1. x 5⁄6 y2 3. x 5 2 6 !19 5 5. 12q, 219D 7. y 5 3 4x 1 3 4 9. a. 1 b. 5 c. 1 d. undefined e. domain: 122, q2 range: 10, q2 f. increasing: 14, q2, decreasing: 10, 42, constant: 122, 02 11. yes 13. 11, 212 15. Q1x2 5 2x3 1 3x 2 5, r1x2 5 0 17. HA: none VA: x 5 3 Slant: y 5 x 1 3 19. 125 21. 5 23. x 5 0.5 25. 8.62 years 27. a. N 5 6e20.247553t < 610.97554864212t b. 2.72 grams C H A P T E R 6 Section 6.1 1. 180° 3. 2120° 5. 300° 7. 2288° 9. 270° 11. a. 72° b. 162° 13. a. 48° b. 138° 15. a. 1° b. 91° 17. 54°/36° 19. 120°/60° Answers to Odd-Numbered Exercises 1211 21. g 5 30° 23. a 5 120°, b 5 30°, g 5 30° 25. a 5 18°, b 5 108°, g 5 54° 27. c 5 5 29. b 5 8 31. c 5 !89 33. c 5 25 35. 10!2 < 14.14 in. 37. 2 cm 39. Other leg: 5!3 < 8.66 m; Hypotenuse: 10 m 41. Other leg: 4!3 < 6.93 yd; Hypotenuse: 8!3 < 13.9 yd 43. Short leg: 5 in.; Long leg: 5!3 < 8.66 in. 45. ƒ 5 3 47. a 5 15 49. c 5 11.55 km 51. ƒ 5 1 3 in. 53. 120° 55. 144° 57. 60 min or 1 h 59. !7300 < 85 ft 61. 241 ft 63. 9.8 ft 65. 17 ft 67. 48 ft 3 28 ft 69. 38 ft 71. 225 ft 73. 20 points 75. 50.004 ft 77. 7.8 cm 79. 4 cm 81. The length opposite the 60° angle is !3 times (not twice) the length opposite 30° angle. 83. false 85. true 87. true 89. 110° 91. DC 5 3 93. 25 95. The triangles are isosceles right triangles 45°245°290°. 97. 8 99. 2!3 101. x 5 2 103. 28.89 ft, 33.36 ft Section 6.2 1. 4 5 3. 5 4 5. 4 3 7. !5 5 9. !5 11. 2 13. 2!10 7 15. 7 3 17. 3!10 20 19. 30° 21. 90° 2 x 23. 60° 25. cos190° 2 x 2 y2 27. sin170° 2 A2 29. tan145° 1 x2 31. sec130° 1 u2 33. a 35. b 37. c 39. !3 3 41. !3 43. 2!3 3 45. 2!3 3 47. !3 3 49. !2 51. 0.6018 53. 0.1392 55. 1.3764 57. 1.0098 59. 1.0002 61. 0.7002 63. 0.4337 65. 30° 67. a. 2/5 b. 5/2 69. 10 mi 71. 2.405 73. 1.335 75. 0 points 77. 50 points (a bullseye) 79. 1.09 angstroms 81. The opposite of angle y is 3 (not 4). 83. Secant is the reciprocal of cosine (not sine). 85. true 87. sin 30° 5 1 2 and cos 30° 5 !3 2 89. tan 30° 5 !3 3 and tan 60° 5 !3 91. sec 45° 5 !2 and csc 45° 5 !2 93. 0 95. 0 97. 3!5 5 99. 2!13 13 101. 13 9 103. 97 65 105. a. 2.92398 b. 2.92380; (b) is more accurate. 107. a. 0.70274 b. 0.70281; (b) is more accurate. Section 6.3 1. three 3. two 5. 47° 7. 55° 9. 83° 11. a < 14 in. 13. a < 18 ft 15. a < 5.50 mi 17. c < 12 km 19. c < 20.60 cm 21. a < 50° 23. a < 62° 25. a < 82.12 yd 27. c < 10.6 km 29. c < 19,293 km 31. b < 58°, a < 6.4 ft, b < 10 ft 33. a < 18°, a < 3.0 mm, b < 9.2 mm 35. b < 35.8°, b < 80.1 mi, c < 137 mi 37. b < 61.62°, a < 936.9 ft, c < 1971 ft 39. a < 56.0°, b < 34.0°, c < 51.3 ft 41. a < 55.480°, b < 34.520°, b < 24,235 km 43. c 5 27.0 in., a 5 24.4 in., a 5 64.6° 45. c 5 4.16 cm, b 5 1.15 cm, a 5 73.90° 1212 Answers to Odd-Numbered Exercises 47. 286 ft 49. a 5 88 ft 51. 260 ft (262 rounded to two significant digits) 53. 11° (she is too low) 55. 80 ft 57. 170 m 59. 0.000016° 61. 26 ft 63. 4400 ft 65. 8° 67. 136.7° 69. 91.99 ft 71. 27 ft 73. 0 point ( just misses the entire target) 75. 121 ft 77. 24 ft 79. d < 3.5 ft 81. 4.7 in. 83. Tan21 should have been used (not tan). 85. true 87. false 89. false 91. 2 mi 93. 0.5 95. 2 7 97. 40° 99. 0.8 101. u 103. 35°; u 105. 0.785 Section 6.4 1. QI 3. QII 5. QIV 7. negative y-axis 9. x-axis 11. QIII 13. QI 15. QIII 17. QII 19. QII 21. QIV 23. 25. 27. 29. 31. 33. 35. 37. c 39. e 41. f 43. 52° 45. 268° 47. 330° 49. 150° 51. 320.0001° sin u cos u tan u cot u sec u csc u 53. 2!5 5 !5 5 2 1 2 !5 !5 2 55. 2!5 5 !5 5 2 1 2 !5 !5 2 57. 4!41 41 5!41 41 4 5 5 4 !41 5 !41 4 59. 2!5 5 2 !5 5 2 2 2 1 2 2!5 !5 2 Answers to Odd-Numbered Exercises 1213 95. 21200° 97. 23240° 99. Don and Ron end up in the same place, 180° from where they started. 101. 1440° 103. tan u 5 7 24 105. 120 ft 107. profit 109. 1.4 cm 111. r 5 !5 (not 5) 113. false 115. true 117. true 119. false 121. 2 3 5 123. m 5 tan u 125. y 5 2!2 127. y 5 21tanu21x 2 a2 129. cos 270° 5 0 131. 0 133. 0 135. Does not exist because we are dividing by 0 137. 21 Section 6.5 1. QIV 3. QII 5. QI 7. QI 9. QIII 11. sin u 5 2 4 5 13. tanu 5 2 60 11 15. sinu 5 2 84 85 17. tanu 5 !3 19. tan u 5 2 !11 5 21. 2 !3 3 23. 2!3 3 25. 1 27. 21 29. 0 31. 1 33. 1 35. not possible 37. possible 39. not possible 41. possible 43. possible 45. possible 47. 21 2 49. 2 !3 2 51. !3 3 53. 1 55. 22 57. 1 59. 30° and 330° 61. 210° and 330° 63. 90° and 270° 65. 270° 67. 20.8387 69. 211.4301 71. 2.0627 73. 21.0711 75. 22.6051 77. 110° 79. 143° 81. 322° 83. 140° 85. 340° 87. 1° 89. 335° 91. 1.3 93. 12° 95. Lower leg is bent at knee at an angle of 15°. 97. 75.5° 99. The reference angle is made with the terminal side and the negative x-axis (not the y-axis). 101. true 103. false 105. false 107. 2 a "a2 1 b2 109. 2"a2 2 b2 b 111. QI 113. QI 115. QI Section 6.6 1. 1 5 or 0.2 3. 2 11 or < 0.18 5. 1 50 or 0.02 7. 1 8 or 0.125 9. 1 5 or 0.2 11. 1 24 or 0.042 13. p 6 15. p 4 17. 7p 4 19. 5p 12 21. 17p 18 23. 13p 3 25. 2 7p 6 27. 220p 29. 30° 31. 135° 33. 67.5° 35. 75° 37. 1620° 39. 171° 41. 284° 43. 229.18° 45. 48.70° 47. 2160.37° 61. 27!65 65 24!65 65 7 4 4 7 2!65 4 2!65 7 63. !15 5 2 !10 5 2 !6 2 2 !6 3 2 !10 2 !15 3 65. 2 !6 4 2 !10 4 !15 5 !15 3 2 2!10 5 2 2!6 3 67. 2 2!29 29 2 5!29 29 2 5 5 2 2 !29 5 2 !29 2 69. 2 !5 5 2!5 5 2 1 2 2 2 !5 2 2 !5 71. !5 5 2 2!5 5 2 1 2 2 2 2 !5 2 !5 73. 2!5 5 !5 5 2 1 2 !5 !5 2 75. !5 5 2!5 5 1 2 2 !5 2 !5 77. 2 !10 10 3!10 10 2 1 3 2 3 !10 3 2 !10 79. 2!13 13 2 3!13 13 2 2 3 2 3 2 2 !13 3 !13 2 81. 1 0 Undefined 0 Undefined 1 83. 21 0 Undefined 0 Undefined 21 85. 1 0 Undefined 0 Undefined 1 87. 21 0 Undefined 0 Undefined 21 89. 21 0 Undefined 0 Undefined 21 91. 1 0 Undefined 0 Undefined 1 93. 2!3 2 2 1 2 !3 !3 3 2 2 22!3 3 1214 Answers to Odd-Numbered Exercises 49. 198.48° 51. 0.820 53. 1.95 55. 0.986 57. 12 mm 59. 2p 3 ft 61. 5 2 in. 63. 11p 5 mm 65. 200p 3 km 67. 12.8 sq ft 69. 2.85 sq km 71. 5 sq yd 73. 8.62 sq cm 75. 0.0236 sq ft 77. 2 5 m/sec 79. 272 km/hr 81. 9.8 m 83. 1.5 mi 85. 5p 2 rad sec 87. 2p 9 rad sec 89. 6p in./sec 91. p 4 mm/sec 93. 26.2 cm 95. 653 in. (or 54.5 ft) 97. 5262 km 99. 37 ft 101. 200p < 628 ft 103. 50° 105. 60 in. (5 ft) 107. 911 mi 109. 157 sq ft 111. 78 sq in. 113. 74 sq mi 115. 1.4 nm 117. 70 mph 119. 1037.51 mph; 1040.35 mph 121. 16 ft/sec 123. 66 2 3p rad/min 125. 12.05 mph 127. 640 rev/min 129. 10.11 rad/sec or 1.6 rotations per second 131. 17.59 m/s 133. 40p rad/min 135. 32p ft 137. Angular speed needs to be in radians (not degrees) per second. 139. true 141. true 143. 2p < 6.3 cm/sec 145. tripled 147. 15 2 p square units 149. 68°45r44s Section 6.7 1. 2 !3 2 3. 2 !3 2 5. !2 2 7. 21 9. 2!2 11. !3 13. 2 15. 2 !3 2 17. 2 !3 2 19. 2 !2 2 21. 2 !3 2 23. !2 2 25. 1 27. !2 2 29. 0 31. 22 33. !3 3 35. p 6, 11p 6 37. 4p 3 , 5p 3 39. 0, p, 2p, 3p, 4p 41. p, 3p 43. 3p 4 , 7p 4 45. 3p 4 , 5p 4 47. 0, p, 2p 49. p 2, 3p 2 51. 7p 6 , 11p 6 53. p 6, 11p 6 55. 22.9°F 57. 99.05°F 59. 2.6 ft 61. 135 lb 63. 13,000 guests 65. 10.7 mcg/mL 67. 35°C 69. Used the x-coordinate for sine and the y-coordinate for cosine; should have done the opposite 71. true 73. false 75. true 77. odd 79. p 4, 5p 4 83. sec u 85. Yes, u 5 p 4, 3p 4 , 5p 4 , 7p 4 87. sin1423°2 < 0.891 and sin12423°2 < 20.891 89. 6.314, 26.314 91. 0.5 93. 0.8660 Section 6.8 1. c 3. a 5. h 7. b 9. e 11. A 5 3 2, p 5 2p 3 13. A 5 1, p 5 2p 5 15. A 5 2 3, p 5 4p 3 17. A 5 3, p 5 2 19. A 5 5, p 5 6 21. 23. 25. Answers to Odd-Numbered Exercises 1215 27. 29. 31. 33. 35. 37. 39. 41. y 5 2sin 2x 43. y 5 cos px 45. y 5 22sin ap 2 xb 47. y 5 sin 8px 49. Amplitude: 2; period: 2; phase shift: 1 p (right) 51. Amplitude: 5; period: 2p 3 ; phase shift: 2 2 3 (left) 53. Amplitude: 6; period: 2; phase shift: 22 (left) 55. Amplitude: 3; period: p; phase shift: 2p 2 (left) 57. Amplitude: 1 4; period: 8p; phase shift: 2p (right) 59. Amplitude: 2; period: 4; phase shift: 4 (right) 61. 63. 65. 67. 1216 Answers to Odd-Numbered Exercises 69. 71. 73. 75. 77. 79. 81. 83. 85. 87. 89. 91. Answers to Odd-Numbered Exercises 1217 93. 95. 3500 widgets 97. Amplitude: 1 mg/L; period 8 weeks 99. Amplitude: 4 cm; Mass: 4 g 101. 1 4p cycles per second 103. Amplitude: 0.005 cm; Frequency: 256 hertz 105. Amplitude: 0.008 cm; Frequency: 375 hertz 107. 660 m/sec 109. 660 m/sec 111. y 5 25 1 25 sin c 2p 4 1t 2 12d or y 5 25 2 25 cos apt 2 b 113. Forgot to reflect about the x-axis 115. true 117. false 119. 10, A2 121. x 5 np B , where n is an integer 123. a0, 2 A 2 b 125. a p13 1 4n2 2B , 0 b, where n an integer. 127. c25A 2 , 3A 2 d 129. no 131. coincide 133. a. Y2 is Y1 shifted to the left p 3. b. Y2 is Y1 shifted to the right p 3. 135. As t increases, the amplitude goes to zero for Y3. 137. a. Y2 is Y1 shifted upward by 1 unit. b. Y2 is Y2 shifted downward by 1 unit. 139. 5 Section 6.9 1. b 3. h 5. c 7. d 9. 11. 13. 15. 1218 Answers to Odd-Numbered Exercises 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. Answers to Odd-Numbered Exercises 1219 41. 43. 45. 47. 49. 51. 53. 55. 57. Domain: all real numbers x, such that x 2 n, where n is an integer Range: all real numbers 59. Domain: all real numbers x, such that x 2 2n 1 1 10 p, where n is an integer Range: 12q, 224 ∪ 32, q2 61. Domain: all real numbers x, such that x 2 2np, where n is an integer Range: 12q, 14 ∪ 33, q2 63. Domain: 5x : x 2 4n 1 6, n an integer6 Range: all real numbers 65. Domain: 5x : x 2 n, n an integer6 Range: A2q, 2 5 2D ∪ C2 3 2, qB 67. 3 mi 69. 55.4 sq in. 71. Forgot the amplitude of 3 73. true 75. n 5 integer 77. x 5 2p, 2 p 2, 0, p 2, p 79. anp 2 C B , 0b, where n an integer 81. Infinitely many. The tangent function is an increasing, periodic function and y 5 x is increasing. 83. A 5 !2 85. p Review Exercises 1. a. 62° b. 152° 3. a. 55° b. 145° 5. a. 0.99° b. 90.99° 7. g 5 25° 9. a 5 140°, b 5 20°, g 5 20° 11. b 5 !128 5 8!2 13. c 5 !65 15. 12!2 yd 17. Leg: 3!3 ft; Hypotenuse: 6 ft 19. F 5 4 21. C 5 147.6 km 23. 10!3 in. 25. 150° 1220 Answers to Odd-Numbered Exercises 27. 32 m 29. 2!13 13 31. !13 2 33. 3 2 35. cos 60° 37. cot 45° 39. csc 60° 41. b 43. b 45. c 47. 0.6691 49. 0.9548 51. 1.5399 53. 1.5477 55. 75 ft sin u cos u tan u cot u sec u csc u 57. 24 5 3 5 2 4 3 2 3 4 5 3 2 5 4 59. !10 10 23!10 10 21 3 23 2 !10 3 !10 61. 1 2 !3 2 !3 3 !3 2!3 3 2 63. 2!5 5 2!5 5 2 1 2 2 2 !5 2 2!5 65. 2!7.2 3 2!7.2 6 2 1 2 2!7.2 1.2 2!7.2 2.4 67. 2 1 2 69. 2 !3 3 71. 2 2!3 3 73. 20.2419 75. 1.0355 77. 22.7904 79. 20.6494 81. 3p 4 83. 11p 6 85. 6p 5 87. 9p 89. 60° 91. 225° 93. 100° 95. 1800° 97. 240p in./min < 754 in./min 99. 2 !3 3 101. 2 1 2 103. 1 105. 21 107. 21 109. 1 2 111. 2p 113. y 5 4cosx 115. 5 117. Amplitude: 2; Period: 1 119. Amplitude: 1 5; Period: 2p 3 121. 123. Amplitude Period Phase Shift Vertical Shift 125. 3 2p p 2 (right) 12 127. 4 2p 3 2p 4 (left) 22 129. 1 3 2 1 2p (right) 1 2 (down) 131. 133. 135. Domain: all real numbers such that x 2 np, where n is an integer; Range: all real numbers 137. Domain: all real numbers such that x 2 2n 1 1 4 p, where n is an integer; Range: 12q, 234 ∪ 33, q2 139. Domain: e x : x 2 6n 1 7 6 , n an integerf Range: A2q, 2 3 4D ∪ C2 1 4, qB. 141. Answers to Odd-Numbered Exercises 1221 143. 145. 147. 71.57 ft, 82.64 ft 149. a. 1.02041 b. 1.02085, (b) is more accurate. 151. 2.612 153. 21 155. QIV 157. 100°46r45s 159. 20.9659 161. a. Y2 is Y1 shifted to the left by p 6 unit. b. Y2 is Y1 shifted to the right p 6 unit. 163. 5 Practice Test 1. 6000 ft 3. The first is an exact value of the cosine function; the second is an approximation. 5. QIV 7. 585° 9. 15p 4 sq in. 11. Amplitude: 5; Period: 2p 3 13. 15. x 5 np 2 or anp 2 , 0b, where n is an integer. 17. 12q, 244 ∪ 32, q2 19. 21. true 23. y 5 4 sin 321x 1 3 22421 2 25. 27. a. 1.34409 b. 1.34352, (b) is more accurate. Cumulative Test 1. x 2 1 5 3. x 2 21, 0; no solution 5. 12q, 04 ∪ 11, 34 7. y 5 0.5x 1 3.4 9. 2 5 8 11. 1 13. ƒ1x2 5 22x2 1 7 15. Factors of a0 5 10; 61, 62, 65, 610; Factors of an 5 2; 61, 62 Possible rational zeros: 61, 62, 65, 610, 61 2, 65 2 Testing the zeros: P1252 5 0, P1212 5 0, P A1 2B 5 0, P122 5 0 17. y-intercept: 10, 02; domain: 12q, q2; range: 121, q2; horizontal asymptote: y 5 21 19. 21. 0.435 23. sin u 5 3!13 13 , cos u 5 2 2!13 13 , tan u 5 2 3 2, cot u 5 2 2 3, sec u 5 2 !13 2 , csc u 5 !13 3 25. Amplitude: 4; Period: p; Phase shift: 2 p 2 (left) 1222 Answers to Odd-Numbered Exercises C H A P TE R 7 Section 7.1 1. 8 7 3. 25 3 5. 21 5 7. 2!5 5 9. 25!7 7 11. 21 4 13. 2 !3 3 15. 4 3 17. !11 5 19. b a 1a 2 02 21. 5 64 23. 28 25. 25 27. 2 !3 2 29. 2 !21 5 31. 2!17 33. 2!5 35. 28!161 161 37. 215!11 44 39. 22!13 13 41. !6 2 43. !6 45. 2 !62 8 47. cos u 5 23 5, sin u 5 4 5 49. cos u 5 2 !5 5 , sin u 5 22!5 5 51. cos u 5 25!34 34 , sin u 5 23!34 34 53. sin u 5 22!29 29 , cos u 5 5!29 29 55. sin u 5 4!17 17 , cos u 5 !17 17 57. 1 sin u 59. 21 61. cos2 u sin u 63. 1 sin u 65. 1 1 2 sin u cos u 67. 1 1 2 sin u cos u cos2 u 69. 1 71. r 5 8 sin u11 2 sin2 u2 5 8 sin u 2 8 sin3 u u 5 30°, r 5 3 u 5 60°, r 5 !3 u 5 90°, r 5 0 73. Each dollar spent on costs produces $1.33 in revenue. 75. Cosine is negative in quadrant III. 77. false 79. 90° 81. cot u 5 6"1 2 sin2 u sin u 83. 8 k cos u k 85. 5 k sec u k 87. true 89. yes a. 0.3746 b. 0.9272 c. 0.4040 d. 0.4040 (c) and (d) are the same. 91. (a) and (c) Section 7.2 1. 1 3. csc x 5. 21 7. sec2 x 9. 1 11. sin2 x 2 cos2 x 13. sec x 15. 1 17. sin2 x 19. csc2 x 21. 2cos x 23. 1 24.– 50. See Instructor’s Solution Manual. 51. conditional 53. identity 55. conditional 57. conditional 59. conditional 61. identity 63. conditional 65. a cos u 67. The cos x and sin x terms do not cancel. The numerators become cos2 x and sin2 x, respectively. 69. This is a conditional equation. Just because the equation is true for p 4 does not mean it is true for all x. 71. false 73. QI, QIV 75. QIII and QIV 77. a2 1 b2 79. No, let A 5 308 and B 5 608. 81. No, take A 5 p 4 . 83. cos1A 1 B2 5 cos A cos B 2 sin A sin B 85. sin1A 1 B2 5 sin A cos B 1 cos A sin B 87. cos2 A 5 1 1 cos12A2 2 89. Y1 5 Y3 Section 7.3 1. !6 2 !2 4 3. !6 2 !2 4 5. 22 1 !3 7. !2 1 !6 4 9. 2 1 !3 11. 2 1 !3 13. !2 2 !6 15. 4 !211 1 !32 5 !6 2 !2 17. cos x 19. 2sin x 21. 0 23. 22 cos1A 2 B2 25. 22 sin1A 1 B2 27. tan126°2 29. 1 1 2!30 12 31. 26!6 1 4 25 33. 3 2 4!15 4 1 3!15 5 192 2 25!15 2119 35. identity 37. conditional 39. identity 41. identity 43. identity 45. conditional 47. identity 49. identity 51. conditional Answers to Odd-Numbered Exercises 1223 53. y 5 sinax 1 p 3 b 55. y 5 cosax 2 p 4 b 57. y 5 2sin14x2 59. y 5 tanax 1 p 4 b 61. y 5 tanap 6 1 x b 67. cos1kz 2 ct2 5 cos1kz2cos1ct2 1 sin1kz2sin1ct2; when z l 5 integer, then kz 5 2pn and the sin kz term goes to zero. 69. T1t2 5 38 2 2.5 sin ap 6 tb 71. Tangent of a sum is not the sum of the tangents. Needed to use the tangent of a sum identity. 73. false 75. false 79. A 5 np and B 5 mp, where n and m are integers, or A and B, where B 5 A 6 2np. 81. a. b. c. The difference quotients of y 5 sin x better approximate y 5 cos x as h goes to zero. 83. a. b. c. The graphs better approximate y 5 2 cos 2x as h goes to zero. 1224 Answers to Odd-Numbered Exercises Section 7.4 1. 24 5 3. 120 119 5. 120 169 7. 24 3 9. !19 10 11. 119 120 13. tan130°2 5 !3 3 15. 1 2 sinap 4 b 5 !2 4 17. cos14x2 19. 2 !3 3 21. 2 !3 2 23. 2 !3 2 41. y 5 cot x 43. y 5 sec12x2 45. y 5 1 2 sin14x2 47. y 5 121 2 sin12x2 49. y 5 2 sin x 2 3 cos12x2 51. C1t2 5 2 1 10 cos 2t 53. 22,565,385 lb 55. !2 ft 57. Sine is negative (not positive). 59. false 61. false 63. true 65. true 67. false 69. tan124x2 5 4 tan x1tan2 x 2 12 11 2 tan2 x22 2 4 tan2 x 71. no 72.–76. See Instructor’s Solution Manual. 77. Y1 and Y3 Section 7.5 1. "2 2 !3 2 3. 2"2 1 !3 2 5. "2 2 !3 2 7. "3 1 2!2 9. 2 2 "2 1 !2 11. 1 2 !2 or 21 "3 1 2!2 13. 2 2 "2 2 !2 15. 1 17. 2!13 13 19. 3!13 13 21. !5 2 1 2 or Å 3 2 !5 2 23. Å 3 1 2!2 6 or 1 1 !2 !6 25. 2 !15 5 27. Å 1 2 24/!601 2 29. 2Å 1 2 !0.91 1 1 !0.91 31. Å 7 3 33. cosa5p 12 b 35. tan175°2 37. tana5p 8 b 51. y 5 2 1 2 cos x Answers to Odd-Numbered Exercises 1225 53. y 5 cos x 55. y 5 1 2 2 cos x 57. y 5 tan x 59. 1 3 60.– 62. See Instructor’s Solution Manual. 63. sinax 2b is positive, not negative, in this case. 65. false 67. false 68.–72. See Instructor’s Solution Manual. 73. 0 , x , p 75. yes 77. identity 79. Y1 and Y3 Section 7.6 1. 1 2 3sin13x2 1 sin x4 3. 5 2 3cos12x2 2 cos110x24 5. 23cos x 1 cos13x24 7. 1 2 3cos x 2 cos14x24 9. 1 2 ccosa2x 3 b 1 cos12x2d 11. 23 2 3cos11.9x2 1 cos11.1x24 13. 23sin12!3x2 2 sin14!3x24 15. 2 cos14x2cos x 17. 2 sin x cos12x2 19. 22 sin x cosA3 2 xB 21. 2 cosA3 2 xBcosA5 6 xB 23. 2 sin10. 5x2cos10.1x2 25. 22 sinA !5xBcosA2!5xB 27. 2 cosa p 24 xbcosa5p 24 xb 29. 2tan x 31. tan12x2 33. cota3x 2 b 43. R1t2 5 !3 cosap 6 t 1 4p 3 b 45. Average frequency: 443 Hz; Beat frequency: 102 Hz 47. 2 sinc 2pct 2 a 1 1.55 1 1 0.63b106d cos c 2pct 2 a 1 1.55 2 1 0.63b106d 49. 2 sinc 2p119792t 2 d cos c 2p14392t 2 d 5 2 sin11979pt2cos1439pt2 51. 251!6 2 !32 3 < 5.98 ft2 53. cos A cos B 2 cos1AB2 and sin A sin B 2 sin1AB2. Should have used the product-to-sum identity. 55. false 57. true 59. Answers will vary. Here is one approach. sin A sin B sin C 1 2 3cos1A 2 B22cos 1A 1 B24 5 1 2 ≥ cos1A 2 B2sin C 2 cos1A 1 B2sin C ¥ 1 2 3sin1A2B1C21sin1C2A1B24 1 2 3sin1A2B1C21sin1C2A1B24 sin A sin B sin C 5 1 4 3sin1A 2 B 1 C2 1 sin1C 2 A 1 B24 2sin1A1 B 1 C2 2 sin1C2 A 2 B24 60.– 62. See Instructor’s Solution Manual. 63. y 5 1 2 3 2 cosa7p 6 xb 1 3 2 cosa5p 6 xb 65. y 5 21 2 cosa3p 2 xb 2 1 2 cosap 6 xb f f f 1226 Answers to Odd-Numbered Exercises 67. y 5 sin14x2 69. Y1 and Y3 Section 7.7 1. p 4 3. 2p 3 5. 3p 4 7. p 6 9. p 6 11. 2p 3 13. 0 15. p 17. 60° 19. 45° 21. 120° 23. 30° 25. 230° 27. 135° 29. 290° 31. 90° 33. 57.10° 35. 62.18° 37. 48.10° 39. 215.30° 41. 166.70° 43. 20.63 45. 1.43 47. 0.92 49. 2.09 51. 0.31 53. 5p 12 55. not possible 57. p 6 59. 2p 3 61. !3 63. p 3 65. not possible 67. 0 69. 2p 4 71. not possible 73. 2p 3 75. 2p 4 77. !7 4 79. 12 13 81. 3 4 83. 5!23 23 85. 4!15 15 87. 11 60 89. 24 25 91. 56 65 93. 24 25 95. 120 119 97. "1 2 u2 99. "1 2 u2 u 101. April or October 103. in month 3 105. t < 0.026476 sec or 26 ms 107. 173.4; June 22–23 109. 11.3 < 11 years 111. tan u 5 7 x 1 1 x 1 2 7 x ⋅1 x 5 8x x2 2 7 113. 0.70 m; 0.24 m 115. u 5 p 2 tan21a150 x b 2 tan21a 300 200 2 xb 117. The wrong interval for the identity was used. The correct domain is c2p 2, p 2 d . 119. cot21 x 2 1 tan21 x 121. false 123. false 125. 1 2 is not in the domain of the inverse secant function. 127. "x2 2 1 x 129. 0 131. a. ap 4, 5p 4 b b. ƒ211x2 5 cos211x 2 32 1 p 4 Domain: 32, 44 133. a. a p 12, 7p 12 b b. ƒ211x2 5 p 12 1 1 2 arccot 14x 2 82 Domain: All real numbers 135. The identity sin1sin21 x2 5 x only holds for 21 # x # 1. 137. c2p 2, 0b ∪ a0, p 2 d 139. a. 720 1681 b. 0.42832 c. yes Section 7.8 1. 3p 4 , 5p 4 3. 7p 6 , 11p 6 , 19p 6 , 23p 6 5. np, where n is an integer 7. 7p 12, 11p 12 , 19p 12 , 23p 12 9. 7p 3 1 4np or 11p 3 1 4np, where n is an integer 11. 24p 3 , 25p 6 , 211p 6 , 2p 3 , p 6, 2p 3 , 7p 6 , 5p 3 13. 24p 3 , 22p 3 Answers to Odd-Numbered Exercises 1227 15. p 6 1 np 4 , where n is an integer 17. 2p, 25p 3 , 2p 3 19. 2p 2, 27p 6 , 211p 6 21. p 6, p 3, 7p 6 , 4p 3 23. p 12, 7p 12, 13p 12 , 19p 12 25. p 3, 2p 3 , 4p 3 , 5p 3 27. 2p 3 29. p 4, 3p 4 , 5p 4 , 7p 4 31. p 2, 3p 2 , p 3, 5p 3 33. 3p 2 , 7p 6 , 11p 6 35. p 2 37. 0, p 39. p 12, 5p 12, 7p 12, 11p 12 , 13p 12 , 17p 12 , 19p 12 , 23p 12 41. 115.83°, 154.17°, 295.83°, 334.17° 43. 333.63° 45. 29.05°, 209.05° 47. 200.70°, 339.30° 49. 41.41°, 318.59° 51. 56.31°, 126.87°, 236.31°, 306.87° 53. 9.74°, 80.26°,101.79°,168.21°,189.74°, 260.26°, 281.79°, 348.21° 55. 80.12°, 279.88° 57. 64.93°, 121.41°, 244.93°, 301.41° 59. 15°, 45°, 75°,105°,135°,165°,195°, 225°, 255°, 285°, 315°, 345° 61. p 4, 5p 4 63. p 65. p 6 67. p 3 69. p 4, 3p 4 , 5p 4 , 7p 4 71. p 2, 3p 2 73. 0, p 4, p, 7p 4 75. p 6, 5p 6 , 7p 6 , 11p 6 , p 2, 3p 2 77. p 6, p 3, 7p 6 , 4p 3 79. p 6, 5p 6 , 7p 6 , 11p 6 81. 3p 2 83. 2p 3 , 4p 3 85. p 3, 5p 3 , p 87. p 24, 5p 24, 13p 24 , 17p 24 , 25p 24 , 29p 24 , 37p 24 , 41p 24 89. 57.47°, 122.53°, 323.62°, 216.38° 91. 30°, 150°, 199.47°, 340.53° 93. 14.48°, 165.52°, 270° 95. 111.47°, 248.53° 97. p 3, 5p 3 99. Fourth quarter of 2017, second quarter of 2018, and fourth quarter of 2019 101. around 9 pm 103. March 105. A 5 1 2 h1b1 1 b22 5 1 21x sin u21x 1 1x cos u 1 x 1 x cos u22 5 x2 sin u11 1 cos u2 107. 2016 109. 24° 111. 3 4 sec 113. 10, 12, ap 3, 3 2b, 1p, 232, a5p 3 , 3 2b 115. March and September 117. Around 1 am and 11 am 119. Extraneous solution. Forgot to check. 121. Can’t divide by cos x. Must factor. 123. false 125. true 127. p 6, 5p 6 , 7p 6 , 11p 6 129. p 6 or 30° 131. no solution 133. 5 1 2n, where n is an integer 135. x 5 p 6 < 0.524 and x 5 5p 6 < 2.618 137. no solution 139. no solution 141. 1228 Answers to Odd-Numbered Exercises 143. x < 1.3 145. 2.21911 Review Exercises 1. 211 7 3. 2 8 15 5. 4 27 7. 3!3 9. 2 7 24 11. 2 5 12 13. sin u 5 2 !3 2 , cos u 5 21 2 15. cos u sin2 u 17. sin u 19. 1 sin u 1 cos u 21. sec2 x 23. sec2 u 25. cos2 x 27. 214 1 2 csc x 1 csc2 x2 28.–34. See Instructor’s Solution Manual. 35. identity 37. conditional 39. identity 41. !2 2 !6 4 43. !3 2 3 3 1 !3 5 !3 2 2 45. sin x 47. tan x 49. 117 44 51. 2 897 1025 53. identity 55. y 5 cosax 1 p 2 b 57. y 5 tana2x 3 b 59. 7 25 61. 671 1800 63. 336 625 65. !3 2 67. 3 2 68.–74. See Instructor’s Solution Manual. 75. 2"2 2 !2 2 77. !2 2 1 79. 2 2 "2 1 !3 81. 7!2 10 83. 25 4 85. sina p 12b 86.–90. See Instructor’s Solution Manual. 91. y 5 sina p 24 xb 93. y 5 2tanax 2b 95. 33sin17x2 1 sin13x24 97. 22 sin14x2sin x 99. 2 sinax 3bcos x 101. cot13x2 102.–106. See Instructor’s Solution Manual. 107. p 4 109. p 2 111. p 6 113. 290° 115. 60° 117. 260° 119. 237.50° 121. 22.50° 123. 1.75 125. 20.10 127. 2p 4 129. 2!3 131. p 3 133. 60 61 135. 7 6 137. 6!35 35 139. July 141. 2p 3 , 5p 6 , 5p 3 , 11p 6 143. 23p 2 , 2p 2 145. 9p 4 , 21p 4 Answers to Odd-Numbered Exercises 1229 147. p 3, 2p 3 , 4p 3 , 5p 3 149. 3p 8 , 7p 8 , 11p 8 , 15p 8 151. 0, p, 3p 4 , 7p 4 153. 80.46°, 170.46°, 260.46°, 350.46° 155. 90°, 138.59°, 221.41°, 270° 157. 17.62°, 162.38° 159. p 4, 5p 4 161. p 3, p 163. 0, p 6, p, 11p 6 165. 3p 2 167. p 169. p 2, 3p 2 171. 90°, 135°, 270°, 315° 173. 0° 175. 60°, 90°, 270°, 300° 177. (a) and (c) 179. Y1 5 Y2 181. 3 cos13x2 183. Y1 and Y3 185. Y1 and Y3 187. Y1 and Y3 189. a. 23 5 b. 20.6 c. yes 191. 0.5787 Practice Test 1. x 5 12n 1 12 2 p, where n is an integer 3. 2"2 2 !2 2 5. Å 3 10 5 !30 10 7. cos110x2 9. cosaa 1 b 2 b 11. 20 cos x cos 3 13. µ 4p 3 1 2np 5p 3 1 2np , where n is an integer 15. 14.48°, 90°, 165.52°, 270° 17. conditional 19. 2 !26 26 21. cotap 6 x 1 p 8 b 23. One-to-one on the interval cc 2 1 2, cb ∪ ac, c 1 1 2d . f 211x2 5 1 p csc21ax 2 a b b 2 c p . 25. 8 3 1 8n, 16 3 1 8n, where n is an integer 27. p 2 1 6np, 7p 2 1 6np, where n is an integer 29. a. 3!10 10 b. 0.94868 c. yes Cumulative Test 1. x 5 3 6 2!5 3. 1x 1 3221 y 1 122 5 25 5. even 7. ƒ + g 5 1 x3 2 1; domain is x 2 0 9. a26 5, 23 5b 11. Q1x2 5 5x 2 4, r 1x2 5 25x 1 7 13. HA: y 5 0.7; VA: x 5 22, x 5 3 15. 123, q2 17. 4 19. 0.4695 21. 27p 12 23. conditional 25. 5 12 27. 1.3994 C H A P T E R 8 Section 8.1 1. SSA 3. SSS 5. ASA 7. g 5 75°, b 5 12.2 m, c 5 13.7 m 9. b 5 62°, a 5 163 cm, c 5 215 cm 11. b 5 116.1°, a 5 80.2 yd, b 5 256.6 yd 13. g 5 120°, a 5 7 m, b 5 7 m 15. a 5 97°, a 5 118 yd, b 5 52 yd 17. b1 5 20°, g1 5 144°, c1 5 9; b2 5 160°, g2 5 4°, c2 5 1 19. a 5 40°, b 5 100°, b 5 18 21. no triangle 23. b 5 90°, g 5 60°, c 5 16 25. b 5 23°, g 5 123°, c 5 15 27. b1 5 21.9°, g1 5 136.8°, c1 5 11.36 b2 5 158.1°, g1 5 0.6°, c1 5 0.17 29. b 5 62°, g 5 2°, c 5 0.3 31. b1 5 77°, a1 5 63°, a 5 457 b2 5 103°, a2 5 37°, a 5 309 33. a 5 31°, g 5 43°, c 5 2 35. 1246 ft 37. 1.7 mi 39. 1.3 mi 41. 26 ft 43. 270 ft 45. 63.83° 47. 76.5 ft 49. 60.14 ft 51. 1.2 cm 53. No solution because b . a and a , 90° 55. false 57. true 59. true 61.–70. See Instructor’s Solution Manual. Section 8.2 1. C 3. S 5. S 7. C 9. b 5 5, a 5 47°, g 5 33° 11. a 5 5, b 5 158°, g 5 6° 1230 Answers to Odd-Numbered Exercises 13. a 5 2, b 5 80°, g 5 80° 15. b 5 5, a 5 43°, g 5 114° 17. b 5 7, a 5 30°, g 5 90° 19. a 5 93°, b 5 39°, g 5 48° 21. a 5 51.3°, b 5 51.3°, g 5 77.4° 23. a 5 75°, b 5 57°, g 5 48° 25. no triangle 27. a 5 67°, b 5 23°, g 5 90° 29. g 5 105°, b 5 5, c 5 9 31. b 5 12°, g 5 137°, c 5 16 33. a 5 66°, b 5 77°, g 5 37° 35. g 5 2°, a 5 168°, a 5 13 37. b 5 11.16, a 5 42.40°, g 5 85.91° 39. a 5 46.76°, b 5 58.45°, g 5 74.79° 41. a 5 1.09, b 5 61.327°, g 5 47.460° 43. 2710 mi 45. 1280 mi 47. 63.7 ft 49. 16 ft 51. 0.8 mi 53. 26° 55. 21.67° 57. /QRP 5 151.03°, /RQP 5 3.97° 59. 0.5° 61. about 83° 63. Should have used the smaller angle b in Step 2 65. false 67. true 69. true 70.–72. See Instructor’s Solution Manual. 73. Ç 1 2 cosA2 cos21A1 4BB 2 75.–80. See Instructor’s Solution Manual. Section 8.3 1. 55.4 3. 0.5 5. 23.6 7. 6.4 9. 4408.4 11. 9.6 13. 97.4 15. 25.0 17. 26.7 19. 111.64 21. 111,632,076 23. no triangle 25. 13.15 27. 174.76 29. 11.98 31. 19.21 33. 312,297 nm2 35. 10,591 ft2 37. 16° or 164° 39. 312.4 mi2 41. a. 41,842 sq ft b. $89,123 43. 47,128 sq ft 45. 23.38 ft2 47. Area of PQ1Q2 5 12 Area of PQ2Q3 5 22.55 Area of Q1Q2Q3 5 15.21 Area of PQ1Q3 5 49.76 The sum of the areas of the smaller three triangles does not equal the area of the outer triangle. 49. 10.86 51. Semiperimeter is half the perimeter. 53. true 55. true 57. false 58.–60. See Instructor’s Solution Manual. 61. 0.69 63.–70. See Instructor’s Solution Manual. Section 8.4 1. !13 3. 5!2 5. 25 7. !73; u 5 69.4° 9. !26; u 5 348.7° 11. !17; u 5 166.0° 13. 8; u 5 180° 15. 2!3; u 5 60° 17. 822, 229 19. 8212, 99 21. 80, 2149 23. 8236, 489 25. 86.3, 3.09 27. 822.8, 15.89 29. 82.6, 23.19 31. 88.2, 23.89 33. 821, 1.79 35. X2 5 13, 212 13Y 37. X60 11, 11 61Y 39. X24 25, 2 7 25Y 41. X23 5, 24 5Y 43. h !10 10 , 3!10 10 i 45. 7i 1 3j 47. 5i 2 3j 49. 2i 1 0j 51. 2i 1 0j 53. 25i 1 5j 55. 7i 1 0j 57. Vertical: 1100 ft /sec Horizontal: 1905 ft /sec 59. 2801 lb 61. 11.7 mph; 31° west of due north 63. 52.41° east of north; 303 mph 65. 250 lb 67. Vertical: 51.4 ft /sec Horizontal: 61.3 ft /sec 69. 29.93 yd 71. 10.9° 73. 1156 lb 75. Magnitude: 351.16; Angle: 23.75° from 180° N force 77. Tension: 5!2; k u S k 5 5 79. a. 0 S b. 10 1 3!2 1 !10 units 81. 8.97 Nm 83. 31.95 Nm 85. Magnitude: 8.67; Direction: 18.05° counterclockwise of S 87. 19° 89. Magnitude is never negative. Should not have factored out the negative but instead squared it in finding the magnitude. 91. false 93. true 95. vector 97. "a2 1 b2 98.–100. See Instructor’s Solution Manual. Answers to Odd-Numbered Exercises 1231 101. 826, 49 5 1 2828, 49 2281, 219 102.–104. See Instructor’s Solution Manual. 105. Directly 107. 109. 183, 100.4° Section 8.5 1. 2 3. 23 5. 42 7. 11 9. 213a 11. 21.4 13. 98° 15. 109° 17. 3° 19. 30° 21. 105° 23. 180° 25. no 27. yes 29. no 31. yes 33. yes 35. yes 37. 400 ft-lb 39. 80,000 ft-lb 41. 1299 ft-lb 43. 148 ft-lb 45. 1607 lb 47. 694,593 ft-lb 49. u S⋅v S 5 49,300, and it represents the total cost of buying the prescribed number of 10-packs of both types of battery. 50.–54. See Instructor’s Solution Manual. 55. a. v1 5 214.72 b. 64.57° 57. n S 5 8r, 09; u S⋅n S 5 r k u S k 59. 22 61. The dot product of two vectors is a scalar (not a vector). Should have summed the products of components. 63. false 65. true 67. 17 68.–74. See Instructor’s Solution Manual. 75. a. 22u S b. 2cu S 77. Any vector that is perpendicular to u S 79. 26 is minimum and 6 is maximum. 81. 21083 83. 31.43° 85. 47° Section 8.6 The answers to Exercises 1 to 8 are all plotted on the same graph, below. 9. !2ccosa7p 4 b 1 i sina7p 4 b d 5 !21cos 315° 1 i sin 315°2 11. 2ccosap 3 b 1 i sinap 3 b d 5 21cos 60° 1 i sin 60°2 13. 4!2ccosa3p 4 b 1 i sina3p 4 b d 5 4!21cos 135° 1 i sin 135°2 15. 2!3ccosa5p 3 b 1 i sina5p 3 b d 5 2!31cos 300° 1 i sin 300°2 17. 31cos 0 1 i sin 02 5 31cos 0° 1 i sin 0°2 19. !2 2 ccosa5p 4 b 1 i sina5p 4 b d 5 !2 2 1cos 225° 1 i sin 225°2 21. 5.321cos 0 1 i sin 02 23. !581cos 293.2° 1 i sin 293.2°2 25. !611cos 140.2° 1 i sin 140.2°2 27. 131cos 112.6° 1 i sin 112.6°2 29. 101cos 323.1° 1 i sin 323.1°2 31. !13 4 1cos 123.7° 1 i sin 123.7°2 33. 5.591cos 24.27° 1 i sin 24.27°2 35. !171cos 212.84° 1 i sin 212.84°2 37. 4.541cos 332.31° 1 i sin 332.31°2 39. 25 41. !2 2 !2i 43. 22 2 2!3i 45. 23 2 1 !3 2 i 47. 1 1 i 49. 2.1131 2 4.5315i 51. 20.5209 1 2.9544i 53. 5.3623 2 4.4995i 55. 22.8978 1 0.7765i 57. 0.6180 2 1.9021i 59. 1001cos 0° 1 i sin0°2 1 1201cos 30° 1 i sin30°2; 212.6 lb 61. 801cos 0° 1 i sin0°2 1 1501cos 30° 1 i sin30°2; 19.7° 63. u S 5 81cos 150° 1 i sin150°2; v S 5 61cos 45° 1 i sin45°2; w S 5 k w S k 1cos u 1 i sinu2, where w S 5 "14!3 2 3!222 1 124 2 3!222; u 5 tan21a 24 2 3!2 4!3 2 3!2b 1232 Answers to Odd-Numbered Exercises 65. R S 5 k R S k 1cos u 1 i sinu2, where R S 5 "1100!2 1 18022 1 1100!222; u 5 tan21a 100!2 100!2 1 180b R S 5 351.161cos 123.75°2 1 i sin123.75°22 67. The point is in QIII (not QI). 69. true 71. true 73. 0° 75. k b k 77. a!51cos 296.6° 1 i sin 296.6°2 79. pa1 2 2 !3 2 ib 81. z 5 8.791cos28° 1 i sin28°2 83. 1.414, 45°, 1.4141cos45° 1 i sin45°2 85. 2.2361cos 26.6° 1 i sin 26.6°2 87. 35,323° Section 8.7 1. 26 1 6!3i 3. 24!2 2 4!2i 5. 0 1 8i 7. 9!2 2 1 9!2 2 i 9. 0 1 12i 11. 3 2 1 3!3 2 i 13. 2!2 1 !2i 15. 0 2 2i 17. 3 2 1 3!3 2 i 19. 25 2 2 5!3 2 i 21. 4 2 4i 23. 264 1 0i 25. 28 1 8!3i 27. 1,048,576 1 0i 29. 21,048,576!3 2 1,048,576i 31. 2 1cos 150° 1 i sin 150°2 and 2 1cos 330° 1 i sin 330°2 33. !6 1cos 157.5° 1 i sin 157.5°2, !6 1cos 337.5° 1 i sin 337.5°2 35. 21cos 20° 1 i sin 20°2, 21cos 140° 1 i sin 140°2, 21cos 260° 1 i sin 260°2 37. ! 3 2 1cos 110° 1 i sin 110°2, ! 3 2 1cos 230° 1 i sin 230°2, ! 3 2 1cos350° 1 i sin 350°2 39. 2 1cos 78.75° 1 i sin 78.75°2, 2 1cos 168.75° 1 i sin 168.75°2, 2 1cos 258.75° 1 i sin 258.75°2, 2 1cos 348.75° 1 i sin 348.75°2 41. x 5 62, x 5 62i 43. 22, 1 2 !3i, 1 1 !3i 45. !2 2 !2i, !2 1 !2i, 2!2 2 !2i, 2!2 1 !2i 47. 1, 21, 1 2 1 !3 2 i, 1 2 2 !3 2 i, 21 2 1 !3 2 i, 2 1 2 2 !3 2 i 49. 2 !2 2 1 !2 2 i, !2 2 2 !2 2 i 51. ! 4 2ccosap 8 1 pk 2 b 1 i sinap 8 1 pk 2 b d , k 5 0, 1, 2, 3 53. 2ccosap 5 1 2pk 5 b 1 i sinap 5 1 2pk 5 b d , k 5 0, 1, 2, 3, 4 55. p2ccosa p 14 1 2pk 7 b 1 i sina p 14 1 2pk 7 b d , k 5 0, 1, 2, 3, 4, 5, 6 Answers to Odd-Numbered Exercises 1233 57. 1cos 45° 1 i sin 45°2, 1cos 117° 1 i sin 117°2, 1cos 189° 1 i sin 189°2, 1cos 261° 1 i sin 261°2, 1cos 333° 1 i sin 333°2 59. ccosa5 p 18 1 pk 3 b 1 i sin a 5 p 18 1 pk 3 b d , k 5 0, 1, 2, 3, 4, 5 61. Reversed the order of angles being subtracted 63. Use De Moivre’s theorem; in general, 1a 1 b26 2 a6 1 b6. 65. true 67. false 69. true 70.–78. See Instructor’s Solution Manual. 79. 2 n1m 2 e p 4 1m2n2i 81. cos 66° 1 i sin 66°, cos 138° 1 i sin 138°, cos 210° 1 i sin 210°, cos 282° 1 i sin 282°, cos 354° 1 i sin 354° 83. cos 40° 1 i sin 40°, cos 100° 1 i sin 100°, cos 160° 1 i sin 160°, cos 220° 1 i sin 220°, cos 280° 1 i sin 280°, cos 340° 1 i sin 340° 85. 3 1cos15° 1 sin15°2, 3 1cos135° 1 sin135°2, 3 1cos 255° 1 sin 255°2 Section 8.8 The answers to Exercises 1 to 10 are all plotted on the same graph, below. 3 4 6 12 0 1 4 3 5 9 7 12 17 12 11 6 11 12 13 12 19 12 23 4 7 3 5 12 5 12 7 2 3 3 4 4 5 6 7 6 5 4 3 2 3 2 6 8 2 10 11. a4, p 3 b 13. a2, 4p 3 b 15. a4!2, 3p 4 b 17. 13, 02 19. a2, 7p 6 b 21. 12, 22!32 23. a !3 2 , 21 2b 25. 10, 02 27. 121, 2!32 29. a !2 2 , 2 !2 2 b 31. d 33. a 35. 37. 39. 1234 Answers to Odd-Numbered Exercises 41. 43. 45. 47. 49. 51. Line: y 5 22x 1 1 53. Circle: 1x 2 122 1 y2 5 9 55. 57. 59. 61. 63. Answers to Odd-Numbered Exercises 1235 65. 67. 69. For (a)–(c), all three graphs generate the same set of points, as seen below: Note that all three graphs are figure eights. Extending the domain in (b) results in twice as fast movement, while doing so in (c) results in movement that is four times as fast. 71. 6 times 73. a. r 5 8 cos 3u b. 50 times because the period is 2p 75. The point is in QIII; the angle found was the reference angle (needed to add p). 77. true 79. r 5 a cos u 81. 12a, u 6 180°2 83. The graphs intersect when u 5 p 3, 2p 3 , 4p 3 , 5p 3 . 85. x3 2 y3 2 2axy 5 0 87. It is a circle of radius a and center 1a cos b, a sin b2. 89. The inner loop is generated beginning with u 5 p 2 and ending with 3p 2 . 91. The very tip of the inner loop begins with u 5 p 2; then it crosses the origin (the first time) at u 5 cos21A21 3B, winds around, and eventually ends with u 5 3p 2 . 93. 67.5°, 157.5°, 247.5°, 337.5° Review Exercises 1. g 5 150°, c 5 12, b 5 8 3. g 5 130°, a 5 1, b 5 9 5. b 5 158°, a 5 11, b 5 22 7. b 5 90°, a 5 !2, c 5 !2 9. b 5 146°, b 5 266, c 5 178 11. b 5 26°, g 5 134°, c 5 15 or b 5 154°, g 5 6°, c 5 2 13. b 5 127°, g 5 29°, b 5 20 or b 5 5°, g 5 151°, b 5 2 15. no triangle 17. b 5 15°, g 5 155°, c 5 10 or b 5 165°, g 5 5°, c 5 2 19. 12.2 mi 21. a 5 42°, b 5 88°, c 5 46 23. a 5 51°, b 5 54°, g 5 75° 25. a 5 42°, b 5 48°, g 5 90° 27. b 5 28°, g 5 138°, a 5 4 29. b 5 68°, g 5 22°, a 5 11 31. a 5 51°, b 5 59°, g 5 70° 33. b 5 37°, g 5 43°, a 5 26 35. b 5 4°, g 5 166°, a 5 28 37. no triangle 39. b 5 10°, g 5 155°, c 5 10.3 41. 141.8 43. 51.5 45. 89.8 47. 41.7 49. 5.2 in. 51. 13 53. 13 55. 26; 112.6° 57. 20; 323.1° 59. 82, 119 61. 838, 279 63. 82.6, 9.79 65. 823.1, 11.69 67. h !2 2 , 2 !2 2 i 69. (5i 1 j) 71. 26 73. 29 75. 16 77. 59° 79. 49° 81. 166° 83. no 85. yes 87. no 89. no 91. Real axis Imaginary axis –5i –4i –3i –2i –i i 2i –6 + 2i 3i 4i 5i –2 –6 –4 –8 1 2 1236 Answers to Odd-Numbered Exercises 93. 2 1cos 315° 1 i sin 315°2 95. 8 1cos 270° 1 i sin 270°2 97. 61 1cos 169.6° 1 i sin 169.6°2 99. 17 1cos 28.1° 1 i sin 28.1°2 101. 3 2 3!3i 103. 21 1 i 105. 23.7588 2 1.3681i 107. 212i 109. 221 2 2 21!3 2 i 111. 2 !3 2 1 1 2 i 113. 26 115. 2324 117. 16 2 16!3i 119. 2 1cos 30° 1 i sin 30°2, 2 1cos 210° 1 i sin 210°2 121. 4 1cos 45° 1 i sin 45°2, 4 1cos 135° 1 i sin 135°2, 4 1cos 225° 1 i sin 225°2, 4 1cos 315° 1 i sin 315°2 123. 26, 3 1 3!3i, 3 2 3!3i 125. !2 2 1 !2 2 i, !2 2 2 !2 2 i, 2 !2 2 1 !2 2 i, 2 !2 2 2 !2 2 i, All the points from Exercises 127–132 are graphed on the single graph below. 127. a2!2, 3p 4 b 129. a10, 7p 6 b 131. a2, 3p 2 b 133. a23 2, 3!3 2 b 135. 11, !32 137. a21 2, 2 !3 2 b 139. 141. 143. Z 5 30.4°, Y 5 107.6°, B 5 45.0 145. C 5 4.0, X 5 74.1°, Y 5 64.3° 149. 65, 293° 151. 140° 153. 12,246° 155. 2 1cos 30° 1 sin 30°2, 2 1cos 120° 1 sin 120°2, 2 1cos 210° 1 sin 210°2, 2 1cos 300° 1 sin 300°2 157. 10°, 50°, 130°, 170°, 250°, 290° Practice Test 1. a 5 7.8, c 5 14.6, and g 5 110° 3. a 5 35.4°, b 5 48.2°, and g 5 96.4° 5. no triangle 7. b 5 1.82, c 5 4.08, g 5 50° 9. 57 11. magnitude 5 13, u 5 112.6° 13. a. 8214, 59; b. 216 15. 32,768C21 1 i!3 D 17. a23!3 2 , 23 2b 3 4 6 12 0 127 129 131 12 17 12 11 6 11 12 13 12 19 12 23 4 7 3 5 12 5 12 7 2 3 3 4 4 5 6 7 6 5 4 3 2 3 2 130 128 132 Answers to Odd-Numbered Exercises 1237 19. 20.–22. See Instructor’s Solution Manual. 23. 220 25. !2 27. a 5 5.49, b 5 4.48, c 5 4.23, d 5 1.16 29. 4ccosa3p 8 1 pk 2 b 1 i sina3p 8 1 pk 2 b d , k 5 0, 1, 2, 3 31. 112° Cumulative Test 1. x 5 2 6 4i 3. 2 5. 22x 1 h 7. undefined 9. x 5 22, multiplicity 3; x 5 0, multiplicity 2 11. VA: x 5 2; slant asymptote: y 5 x 1 2 13. $37,250 15. 0.001 17. 15!2 ft < 21.21 ft 19. 21. 23 1 8!6 25 23. c 5 6, a 5 132°, b 5 27° 25. 22 1 2i 27. 12°, 78°, 192°, 248° CH A P T E R 9 Section 9.1 1. 11, 02 3. 18, 212 5. 11, 212 7. 11, 22 9. u 5 32 17, v 5 11 17 11. no solution 13. infinitely many solutions: 1a, 4a212 15. infinitely many solutions: aa, 5a 2 15 3 b 17. 11, 32 19. 16, 82 21. 11.2, 0.042 23. 13, 12 25. 12, 52 27. 123, 42 29. A1, 22 7B 31. A19 7 , 11 35B 33. infinitely many solutions: aa, 5 2 2a 5 b 35. 14, 02 37. 122, 12 39. infinitely many solutions: aa, 1.25 2 0.02a 0.05 b 41. A75 32, 7 16B 43. c 45. d 47. 10, 02 49. 121, 212 51. 10, 262 53. no solution 55. 6 AusPens per kit 57. 5 Montblanc pens, 64 Cross pens 59. 15.86 ml of 8% HCl, 21.14 ml of 15% HCl 61. $300,000 of sales 63. 169 highway miles, 180.5 city miles 65. plane speed: 450 mph, wind speed: 50 mph 67. 10% stock: $3500, 14% stock: $6500 69. 8 CD players 71. Every term in the first equation is not multiplied by 21 correctly. The equation should be 22x 2 y 5 3, and the resulting solution should be x 5 11, y 5 225. 73. Did not distribute 21 correctly. In Step 3, the calculation should be 2123y 2 42 5 3y 1 4. 75. false 77. false 79. A 5 24, B 5 7 81. 2% drink: 8 cups, 4% drink: 96 cups 83. 18.9, 6.42 85. infinitely many solutions: aa, 7 2 23a 5 b 1238 Answers to Odd-Numbered Exercises 87. 12.426, 20.0592 Section 9.2 1. x 5 23 2, y 5 23, z 5 9 2 3. x 5 22, y 5 9 2, z 5 1 2 5. x 5 5, y 5 3, z 5 21 7. x 5 90 31, y 5 103 31 , z 5 9 31 9. x 5 213 4 , y 5 1 2, z 5 25 2 11. x 5 22, y 5 21, z 5 0 13. x 5 2, y 5 5, z 5 21 15. no solution 17. no solution 19. x 5 1 2 a, y 5 21a 1 1 22, z 5 a 21. x 5 41 1 4a, y 5 31 1 3a, z 5 a 23. x1 5 21 2, x2 5 7 4, x3 5 23 4 25. no solution 27. x1 5 1, x2 5 21 1 a, x3 5 a 29. x 5 2 3a 1 8 3, y 5 21 3a 2 10 3 , z 5 a 31. 100 basic widgets, 100 midprice widgets, and 100 top-of-the-line widgets produced. 33. 8 touchdowns, 6 extra points, 4 field goals 35. 6 Sriracha Chicken sandwiches 3 Tuna Salad sandwiches 5 Roast Beef sandwiches 37. h0 5 0, v0 5 52, a 5 232 39. y 5 20.0625x2 1 5.25x 2 50 41. money market: $10,000, mutual fund: $4000, stock: $6000 43. 33 regular model skis, 72 trick skis, 5 slalom skis 45. game 1: 885 points, game 2: 823 points, game 3: 883 points 47. Equation 122 and Equation 132 must be added correctly—should be 2x 2 y 1 z 5 2. Also, should begin by eliminating one variable from Equation 112. 49. true 51. a 5 4, b 5 22, c 5 24 53. a 5 255 24, b 5 21 4, c 5 223 24 , d 5 1 4, e 5 44 55. no solution 57. x1 5 22, x2 5 1, x3 5 24, x4 5 5 59. x 5 41 1 4a, y 5 31 1 3a, z 5 a 61. same as answer in Exercise 59 63. A 280 7 , 280 7 , 48 7 B Section 9.3 1. d 3. a 5. b 7. A x 2 5 1 B x 1 4 9. A x 2 4 1 B x 1 C x2 11. 2x 2 6 1 3x 1 33 x2 1 x 1 5 13. Ax 1 B x2 1 10 1 Cx 1 D Ax2 1 10B 2 15. 1 x 2 1 x 1 1 17. 1 x 2 1 19. 2 x 2 3 1 7 x 1 5 21. 3 x 2 1 1 4 1x 2 122 23. 4 x 1 3 2 15 1x 1 322 25. 3 x 1 1 1 1 x 2 5 1 2 1x 2 522 27. 22 x 1 4 1 7x x2 1 3 29. 22 x 2 7 1 4x 2 3 3x2 2 7x 1 5 31. x x2 1 9 2 9x 1x2 1 922 33. 2x 2 3 x2 1 1 1 5x 1 1 1x2 1 122 35. 1 x 2 1 1 1 21x 1 12 1 23x 2 1 21x2 1 12 37. 3 x 2 1 1 2x 1 5 x2 1 2x 2 1 39. 1 x 2 1 1 1 2 x x2 1 x 1 1 41. 1 d0 1 1 di 5 1 f 43. The form of the decomposition is incorrect. It should be A x 1 Bx 1 C x2 1 1 . Once this correction is made, the correct decomposition is 1 x 1 2x 1 3 x2 1 1. 45. false 47. 1 x 2 1 2 1 x 1 2 1 1 x 2 2 49. 1 x 1 1 x 1 1 2 1 x3 51. x x2 1 1 2 2x 1x2 1 122 1 x 1 2 1x2 1 123 53. yes 55. no 57. yes Section 9.4 1. d 3. b 5. 7. 9. 11. Answers to Odd-Numbered Exercises 1239 13. 15. 17. 19. 21. 23. no solution 25. –5 –3 –1 2 1 3 4 5 –5 –4 –3 2 1 3 4 5 x y 27. 29. 31. 33. 35. 37. 39. 41. 43. 45. no solution 47. 49. 4 units2 51. 7.5 units2 53. x $ 0, y $ 0 x 1 20y # 2400 25x 1 150y # 6000 55. a. 275 # 10x 1 20y 125 # 15x 1 10y 200 # 20x 1 15y x $ 0, y $ 0 b. 2 4 6 8 10 12 14 16 18 2 4 6 8 10 12 14 x y c. Two possible diet combinations are 2 ounces of food A and 14 ounces of food B or 10 ounces of food A and 10 ounces of food B. 57. a. x $ 2y x 1 y $ 1000 x $ 0, y $ 0 b. 100 200 300 400 500 600 700 800 900 100 200 300 400 500 600 700 800 900 1000 x y c. Two possible solutions would be for the manufacturer to produce 700 USB wireless mice and 300 Bluetooth mice or 800 USB wireless mice and 300 Bluetooth mice. 59. • P # 80 2 0.01x P $ 60 x $ 0 61. 20,000 63. The shading should be above the line. 65. true 67. false 69. shaded rectangle d 1240 Answers to Odd-Numbered Exercises 71. 73. 75. Section 9.5 1. ƒ1x, y2 5 z 5 2x 1 3y ƒ121, 42 5 10 ƒ12, 42 5 16 1MAX2 ƒ122, 212 5 27 1MIN2 ƒ11, 212 5 21 3. ƒ1x, y2 5 z 5 1.5x 1 4.5y ƒ121, 42 5 16.5 ƒ12, 42 5 21 1MAX2 ƒ122, 212 5 27.5 1MIN2 ƒ11, 212 5 23 5. minimize at ƒ10, 02 5 0 7. no maximum 9. minimize at ƒ10, 02 5 0 11. maximize at ƒ11, 62 5 53 20 5 2.65 13. Frances T-shirts: 130 Charley T-shirts: 50 1 profit $9502 15. laptops: 25, desktops: 0 1 profit $75002 17. first-class cars: 3, second-class cars: 27 19. 200 of each type of ski 21. Should compare the values of the objective function at the vertices rather than comparing the y-values of the vertices 23. false 25. maximum at 10, a2 and is a 27. minimum at 10, 02 and is 0 29. maximum at 16.7, 4.52 and is 176.9 31. maximum at A4 9, 115 9 B and is approximately 25 Review Exercises 1. 13, 02 3. A13 4 , 8B 5. 12, 12 7. A19 8 , 13 8 B 9. 122, 12 11. A12, 35 6 B 13. 13, 222 15. 121, 22 17. c 19. d 21. 6% NaCl: 10.5 ml, 18% NaCl: 31.5 ml 23. x 5 21, y 5 2a 1 2, z 5 a 25. no solution 27. y 5 20.0050x2 1 0.4486x 2 3.8884 29. A x 2 1 1 B 1x 2 122 1 C x 1 3 1 D x 2 5 31. A x 1 B 4x 1 5 1 C 2x 1 1 1 D 12x 1 122 33. A x 2 3 1 B x 1 4 35. Ax 1 B x2 1 17 1 Cx 1 D Ax2 1 17B 2 37. 4 x 2 1 1 5 x 1 7 39. 1 2x 1 15 21x 2 52 2 3 21x 1 52 41. 22 x 1 1 1 2 x 43. 5 x 1 2 2 27 1x 1 222 45. 47. 49. 51. 53. no solution 55. 57. 59. minimum value of z is 0, occurs at 10, 02 61. maximum value of z is 25.6, occurs at 10, 82 63. minimum value of z is 230, occurs at 10, 62 65. ocean watercolor: 10 geometric shape: 30 1 profit $3902 67. 12, 232 69. 13.6, 3, 0.82 71. yes Answers to Odd-Numbered Exercises 1241 73. 75. maximum is 12.06, occurs at 11.8, 0.62 Practice Test 1. 17, 32 3. x 5 a, y 5 a 2 2 5. x 5 1, y 5 25, z 5 3 7. x 5 21, y 5 3, z 5 4 9. 5 x 2 3 x 1 1 11. 7 x 1 2 2 9 1x 1 222 13. 1 3x 1 2 31x 2 32 2 1 x 1 3 15. 17. 19. 21. minimum value of z is 7, occurs at 10, 12 23. money market: $14,000, aggressive stock: $8500, conservative stock: $7500 25. 111, 19, 12 Cumulative Test 1. x 2 1 5 , x 2 5 6 3. no solution 5. 12q, 04 ∪ 11, 34 7. y 5 0.5x 1 3.4 9. 2 5 8 11. 1 13. ƒ1x2 5 22x2 1 7 15. possible rational zeros: 61, 62, 65, 610, 61 2, 65 2 rational zeros: 25, 21, 1 2, 2 17. domain: 12q, q2 range: 121, q2 y-intercept: 10, 02 HA: y 5 21 19. Shift the graph of y 5 ln x left one unit, and then down three units. 21. 0.435 23. x 5 2, y 5 0, z 5 5 25. 27. yes C H A P T E R 1 0 Section 10 . 1 1. 2 3 3 3. 1 3 4 5. 1 3 1 7. c 3 22 24 6 7 23d 9. £ 2 23 4 21 1 2 5 22 23 3 23 1 7 § 11. £ 1 1 0 1 0 21 0 1 1 3 3 2 5 § 13. £ 24 3 5 2 23 22 22 4 3 3 2 23 1 § 15. e 23x 1 7y 5 2 x 1 5y 5 8 17. • 2x 5 4 7x 1 9y 1 3z 5 23 4x 1 6y 2 5z 5 8 19. e x 5 a y 5 b 21. not reduced form 23. reduced form 25. not reduced form 27. reduced form 29. row-echelon form 31. c1 22 0 7 23 5d 33. £ 1 22 21 0 5 21 3 22 5 3 3 0 8 § 35. D 1 22 5 21 0 1 1 23 0 22 1 22 0 0 1 21 4 2 3 5 26 T 37. D 1 0 5 210 0 1 2 23 0 0 27 6 0 0 8 210 4 25 22 3 29 T 1242 Answers to Odd-Numbered Exercises 39. D 1 0 4 0 0 1 2 0 0 0 1 0 0 0 0 1 4 27 211 21 23 T 41. c1 0 0 1 28 6d 43. £ 1 0 0 0 1 0 0 0 1 3 22 21 0 § 45. £ 1 0 0 0 1 0 0 0 1 3 2 5 21 § 47. c1 0 22 0 1 22 1 2d 49. C 1 0 1 0 1 1 0 0 0 3 1 21 2 0 S 51. x 5 27, y 5 5 53. x 2 2y 5 23 or x 5 2a 2 3, y 5 a 55. no solution 57. x 5 4a 1 41, y 5 3a 1 31, z 5 a 59. x1 5 21 2, x2 5 7 4, x3 5 23 4 61. no solution 63. x1 5 1, x2 5 a 2 1, x3 5 a 65. x 5 2 3a 1 8 3, y 5 21 3a 2 10 3 , z 5 a 67. no solution 69. x1 5 22, x2 5 1, x3 5 24, x4 5 5 71. 11, 222 73. no solution 75. 122, 1, 32 77. 13, 22, 22 79. no solution 81. x 5 a 4 1 3, y 5 7a 4 1 1, z 5 a 83. x 5 72 2 11a 14 , y 5 13a 1 4 14 , z 5 a 85. x 5 1, y 5 2, z 5 23, w 5 1 87. 8 touchdowns, 5 extra points, 1 two-point conversion, and 2 field goals 89. 2 Egg Salad, 2 Club, 8 Turkey, 2 Roast Beef 91. initial height: 0 ft, initial velocity: 50 ft/sec, acceleration: 232 ft/sec2 93. y 5 20.053x2 1 4.58x 2 34.76 95. about 88 ml of 1.5% solution and 12 ml of the 30% solution 97. 200 basic widgets, 100 midprice widgets, and 75 top-of-the-line widgets produced 99. money market: $5500, mutual fund: $2500, stock: $2000 101. 25 units product x, 40 units product y, 6 units product z 103. general admission: 25, reserved: 30, end zone: 45 105. a 5 222 17, b 5 244 17, c 5 2280 17 107. Need to line up a single variable in a given column before forming the augmented matrix. The correct matrix is £ 21 1 1 1 1 22 1 1 1 3 2 23 6 § , after reducing, £ 1 0 0 0 1 0 0 0 1 3 2 1 3 § 109. Row 3 is not inconsistent. It implies z 5 0. 111. false 113. true 115. ƒ1x2 5 211 6 x4 1 44 3 x3 2 223 6 x2 1 94 3 x 1 44 117. 119. a. y 5 20.24x2 1 0.93x 1 6.09 b. b. y 5 20.24x2 1 0.93x 1 6.09 Section 10.2 1. 2 3 3 3. 2 3 2 5. 3 3 3 7. 1 3 1 9. 4 3 4 11. x 5 25, y 5 1 13. x 5 23, y 5 22, z 5 3 15. x 5 6, y 5 3 17. c21 5 1 5 2 5d 19. £ 22 4 2 22 21 3 § 21. not defined 23. not defined 25. c22 6 0 4 8 2d 27. £ 0 25 210 5 215 25 § 29. c22 12 3 13 2 14d 31. c 8 3 11 5d 33. £ 23 21 6 24 7 1 13 14 9 § 35. £ 3 6 22 22 17 24 § 37. not defined 39. 30 604 41. 326 1 294 43. £ 7 10 28 0 15 5 23 0 27 § 45. c12 20 30 42d 47. £ 24 24 216 § 49. not defined 51. A 5 c0.70 0.30d , B 5 c0.89 0.84d a. 46A 5 c32.2 13.8d , out of 46 million people, 32.2 million said that they had tried to quit smoking, while 13.8 million said that they had not. b. 46B 5 c40.94 38.64d , out of 46 million people, 40.94 million believed that smoking would increase the chance of getting lung cancer, and 38.64 million believed that smoking would shorten their lives. 53. A 5 c0.693 0.729 0.597 0.637d , B 5 c100M 110Md AB 5 c149.49M 129.77Md 149.49 million registered voters, of those 129.77 million actually vote. Answers to Odd-Numbered Exercises 1243 55. A 5 30.575 0.5 1.004, B 5 £ 7523 2700 15200 § AB 5 $20,875.73 57. A 5 D 230 3 44 9 430 19 46 20 290 5 45 19 330 5 47 24 T 2A 5 D 460 6 88 18 860 38 92 40 580 10 90 38 660 10 94 48 T, nutritional information corresponding to 2 sandwiches. 0.5A 5 D 115 1.5 22 4.5 215 9.5 23 10 145 2.5 22.5 9.5 165 2.5 23.5 12 T, nutritional information corresponding to one-half of a sandwich. 59. AB 5 £ $0.228 $0.081 $0.015 § , total cost per mile to run each type of automobile 61. N 5 £ 2 1 0 § XN 5 £ 10 16 20 § The nutritional content of the meal is 10 g of carbohydrates, 16 g of protein, and 20 g of fat. 63. N 5 £ 200 25 0 § XN 5 £ 9.25 13.25 15.75 § Company 1 would charge $9.25, Company 2 would charge $13.25, and Company 3 would charge $15.75, respectively, for 200 minutes of talking and 25 text messages. The better cell phone provider for this employee would be Company 1. 65. Not multiplying correctly. It should be: c3 2 1 4d # c21 3 22 5d 5 c27 19 29 23d 67. false 69. true 71. 5 c a2 11 1 a12a21 a11a12 1 a12a22 a21a11 1 a22a21 a2 22 1 a21a12 d 73. A 5 c1 1 1 1d , A2 5 c2 2 2 2d , An 5 2n21 A, n $ 1 75. must have m 5 p 77. D 33 35 296 282 31 19 146 138 T 79. not defined 81. £ 5 24 4 2 215 23 26 4 28 § Section 10 . 3 1. c22 5 7 22d cx yd 5 c 10 24d 3. c 1 22 23 1d cx yd 5 c8 6d 5. £ 3 5 21 1 0 2 21 1 21 § £ x y z § 5 £ 2 17 4 § 7. £ 3 0 1 0 1 22 1 2 0 § £ x y z § 5 £ 10 4 6 § 9. yes 11. yes 13. no 15. yes 17. no 19. c0 21 1 2d 21. c2 1 13 8 39 20 39 2 4 117 d 23. c20.1618 0.2284 0.5043 20.1237d 25. £ 1 2 1 2 0 1 2 0 1 2 0 21 2 21 2 § 27. A21 does not exist. 29. £ 21 2 21 2 5 2 1 2 1 2 23 2 0 21 1 § 31. £ 1 2 1 2 0 3 4 1 4 21 2 1 4 3 4 21 2 § 33. x 5 2, y 5 21 35. x 5 1 2, y 5 1 3 37. x 5 4, y 5 23 39. x 5 0, y 5 0, z 5 1 41. A21 does not exist 43. x 5 21, y 5 1, z 5 27 45. x 5 3, y 5 5, z 5 4 47. a. c$50 $20d b. sweatshirt: $50, t-shirt: $20 49. JAW 51. LEG 53. EYE 55. X 5 £ 8 4 6 6 10 5 10 4 8 § 21 £ 18 21 22 § 5 £ 1 1 1 § The combination of one serving each of foods A, B, and C will create a meal of 18 g of carbohydrates, 21 g of protein, and 22 g of fat. 57. X 5 £ 0.03 0.06 0.15 0.04 0.05 0.18 0.05 0.07 0.13 § 21 £ 49.50 52.00 58.50 § 5 £ 350 400 100 § The employee’s normal monthly usage is 350 minutes talking, 400 text messages, and 100 megabytes of data usage. 59. A is not invertible because the identity matrix was not reached. 61. false 63. x 5 9 65. A⋅A21 5 ca b c dd ⋅a 1 ad 2 bc c d 2b 2c ad b 5 1 ad 2 bc a ca b c dd ⋅c d 2b 2c adb 5 1 ad 2 bc cad 2 bc 0 0 ad 2 bcd c ad 2 bc ad 2 bc 0 0 ad 2 bc ad 2 bc d 5 c1 0 0 1d 5 I 67. ad 2 bc 5 0 1244 Answers to Odd-Numbered Exercises 69. E 2 115 6008 431 6008 21067 6008 103 751 411 6008 2391 6008 731 6008 2 22 751 57 751 28 751 2 85 751 3 751 2 429 6008 145 6008 1035 6008 12 751 U 71. x 5 1.8, y 5 21.6 73. 13.7, 22.4, 9.32 Section 10 . 4 1. 22 3. 31 5. 228 7. 20.6 9. 0 11. x 5 5, y 5 26 13. x 5 22, y 5 1 15. x 5 23, y 5 24 17. x 5 22, y 5 5 19. x 5 2, y 5 2 21. D 5 0, inconsistent or dependent system 23. D 5 0, inconsistent or dependent system 25. x 5 1 2, y 5 21 27. x 5 1.5, y 5 2.1 29. x 5 0, y 5 7 31. x 5 1 3, y 5 3 4 33. 7 35. 225 37. 2180 39. 0 41. 238 43. 0 45. 95.7 47. x 5 2, y 5 3, z 5 5 49. x 5 22, y 5 3, z 5 5 51. x 5 2, y 5 23, z 5 1 53. D 5 0, inconsistent or dependent system 55. D 5 0, inconsistent or dependent system 57. x 5 23, y 5 1, z 5 4 59. x 5 2, y 5 23, z 5 5 61. 6 units2 63. 6 units2 65. y 5 2x 67. I1 5 7 2, I2 5 5 2, I3 5 1 69. The second determinant should be subtracted; that is, it should be 21 23 2 1 21 . 71. In Dx and Dy the column c 6 23 d should replace the column corresponding to the variable that is being solved for in each case. Precisely, Dx should be 6 3 23 21 and Dy should be 2 6 21 23 . 73. true 75. false 77. abc 79. 2419 81. 2b1 a2 c2 a3 c3 1 b2 a1 c1 a3 c3 2 b3 a1 c1 a2 c2 5 2b1 C1a221c32 2 1a321c22D 1 b2 C1a121c32 2 1a321c1 2 b32 C1a121c22 2 1a221c12D 5 2a2b1c3 1 a3b1c2 1 a1b2c3 2 a3b2c1 2 a1b3c2 1 a2b3c1 83. 2180 85. 21019 87. x 5 26.4, y 5 1.5, z 5 3.4 Review Exercises 1. c5 7 3 24 2 22d 3. £ 2 0 21 0 1 23 1 0 4 3 3 22 23 § 5. no 7. no 9. c1 22 0 1 1 21d 11. £ 1 24 3 0 22 3 0 1 24 3 21 22 8 § 13. c1 0 0 1 3 5 21 5 d 15. £ 1 0 0 0 1 0 0 0 1 3 24 8 24 § 17. x 5 5 4, y 5 7 8 19. x 5 2, y 5 1 21. x 5 274 21, y 5 273 21, z 5 23 7 23. x 5 1, y 5 3, z 5 25 25. x 5 23 7a 2 2, y 5 2 7a 1 2, z 5 a 27. y 5 20.005x2 1 0.45x 2 3.89 29. not defined 31. c3 5 2 7 8 1d 33. c9 24 9 9d 35. c 4 13 18 11d 37. c 0 219 218 29d 39. c27 211 28 3 7 2d 41. c10 213 18 220d 43. c17 28 18 33 0 42d 45. c10 9 20 22 24 2d 47. yes 49. yes 51. c 2 5 21 5 3 10 1 10 d 53. c0 21 2 1 0d 55. £ 21 6 7 12 2 1 12 1 2 21 4 21 4 1 6 2 1 12 2 5 12 § 57. £ 0 22 5 1 5 1 22 5 1 5 21 2 3 10 1 10 § 59. x 5 5, y 5 4 61. x 5 8, y 5 12 63. x 5 1, y 5 2, z 5 3 65. 28 67. 5.4 69. x 5 3, y 5 1 71. x 5 6, y 5 0 73. x 5 90, y 5 155 75. 11 77. 2abd 79. x 5 1, y 5 1, z 5 2 81. x 5 215 7 , y 5 225 7 , z 5 19 14 83. 1 unit2 85. a. y 5 20.16x2 2 0.05x 2 4.10 b. y 5 0.16x2 2 0.05x 2 4.10 87. £ 2238 206 50 2113 159 135 40 230 0 § 89. x 5 2.25, y 5 24.35 91. x 5 29.5, y 5 3.4 Answers to Odd-Numbered Exercises 1245 Practice Test 1. c 1 22 22 3 1 2d 3. £ 6 9 1 2 23 1 10 12 2 3 5 3 9 § 5. £ 1 3 5 0 1 211 0 7 15 § 7. x 5 21 3a 1 7 6, y 5 1 9a 2 2 9, z 5 a 9. c211 19 26 8d 11. c 1 19 3 19 5 19 2 4 19 d 13. no inverse 15. x 5 23, y 5 1, z 5 7 17. 231 19. x 5 1, y 5 21, z 5 2 21. money market: $3500, conservative stock: $4500, aggressive stock: $7000 23. 112.5, 26.42 Cumulative Test 1. x 5 3 6 2!5 3. 1x 1 322 1 1y 1 122 5 25 5. even 7. ƒ1g1x22 5 A1 xB3 2 1 5 1 x 2 1 domain: 12q, 02 ∪ 10, q2 9. A26 5, 23 5B 11. Q1x2 5 5x 2 4, r1x2 5 25x 1 7 13. VA: x 5 22, x 5 3 HA: y 5 0.7 15. 123, q2 17. x 5 4 19. no solution 21. Maximum: zA10 7 , 10 7 B 5 120 7 23. c78 210 40 26 0 14d 25. x 5 3 11, y 5 2 2 11 27. c 35 12 219 214d , c 7 131 6 131 2 19 262 2 35 262 d C H A P T E R 1 1 Section 11 . 1 1. hyperbola 3. circle 5. hyperbola 7. ellipse 9. parabola 11. circle Section 11 . 2 1. c 3. d 5. c 7. a 9. x2 5 12y 11. y2 5 220x 13. 1x 2 322 5 81y 2 52 15. 1y 2 422 5 281x 2 22 17. 1x 2 222 5 41321y 2 12 5 121y 2 12 19. 1y 1 122 5 41121x 2 22 5 41x 2 22 21. 1y 2 222 5 81x 1 12 23. 1x 2 222 5 281y 1 12 25. vertex: 10, 02 focus: 10, 22 directrix: y 5 22 length of latus rectum: 8 27. vertex: 10, 02 focus: A 21 2, 0B directrix: x 5 1 2 length of latus rectum: 2 29. vertex: 10, 02 focus: 10, 42 directrix: y 5 24 length of latus rectum: 16 31. vertex: 10, 02 focus: 11, 02 directrix: x 5 21 length of latus rectum: 4 33. vertex: 123, 22 35. vertex: 13, 212 37. vertex: 125, 02 39. vertex: 10, 22 41. vertex: 123, 212 43. vertex: A1 2, 5 4B 45. 10, 22, receiver placed 2 feet from vertex 47. opens up: y 5 1 8x2, for any x in [22.5, 2.5] opens right: x 5 1 8y2, for any y in [22.5, 2.5] 49. x2 5 41402y 5 160y 51. yes, opening height 18.75 feet, mast 17 feet 53. 374.25 feet, x2 5 1497y 55. The maximum profit of $400,000 is achieved when 3000 units are produced. 57. If the vertex is at the origin and the focus is at 13, 02, then the parabola must open to the right. So, the general equation is y2 5 4px, for some p . 0. 59. true 61. false 3 1246 Answers to Odd-Numbered Exercises 63. Equate d1 and d2 and simplify: "1x 2 022 1 1y 2 p22 5 0y 1 p0 x2 5 4py 65. 67. vertex: 12.5, 23.52 69. vertex: 11.8, 1.52 opens right opens left Section 11.3 1. d 3. a 5. center: 10, 02 vertices: 165, 02, 10, 642 7. center: 10, 02 vertices: 164, 02, 10, 682 9. center: 10, 02 vertices: 1610, 02, 10, 612 11. center: 10, 02 vertices: A63 2, 0B, A0, 61 9B 13. center: 10, 02 vertices: 162, 02, 10, 642 15. center: 10, 02 vertices: 162, 02, 10, 6!22 17. x2 36 1 y2 20 5 1 19. x2 7 1 y2 16 5 1 21. x2 4 1 y2 16 5 1 23. x2 9 1 y2 49 5 1 25. c 27. b 29. center: 11, 22 vertices: 123, 22, 15, 22, 11, 02, 11, 42 31. center: 123, 42 vertices: A22!2 2 3, 4B, A2!2 2 3, 4B, A23, 4 1 4!5B, A23, 4 2 4!5B 33. center: 10, 32 vertices: 122, 32, 12, 32, 10, 32, 10, 42 35. center: 11, 12 vertices: A162!2, 12, 11, 32, 11, 212 37. center: 122, 232 vertices: A226!10, 232, A22, 2365!2B 39. 1x 2 222 25 1 1y 2 522 9 5 1 41. 1x 2 422 7 1 1y 1 422 16 5 1 43. 1x 2 322 4 1 1y 2 222 16 5 1 45. 1x 1 122 9 1 1y 1 422 25 5 1 Answers to Odd-Numbered Exercises 1247 47. x2 225 1 y2 5625 5 1 49. a. x2 5625 1 y2 400 5 1 b. The width of the track at the end of the field is 24 yards, so it will NOT encompass the football field, which is 30 yards wide. 51. x2 5,914,000,0002 1 y2 5,729,000,0002 5 1 53. x2 150,000,0002 1 y2 146,000,0002 5 1 55. straight line 57. a. x2 64 1 y2 25 5 1 b. 42 inches c. 1509 steps 59. It should be a2 5 6, b2 5 4, so that a 5 6!6, b 5 62. 61. false 63. true 65. Pluto: e > 0.25 Earth: e > 0.02 67. As c increases, ellipse becomes more elongated. 69. As c increases, circle gets smaller. 71. As c decreases, the major axis becomes longer. Section 11.4 1. b 3. d 5. 7. 9. 11. 13. 15. 17. x2 16 2 y2 20 5 1 19. y2 9 2 x2 7 5 1 21. x2 2 y2 5 a2 23. y2 4 2 x2 5 b2 25. c 27. b 29. 31. 33. 35. 37. 39. 1x 2 222 16 2 1y 2 522 9 5 1 41. 1y 1 422 9 2 1x 2 422 7 5 1 43. Ship will come ashore between the two stations: 28.5 miles from one and 121.5 miles from the other. 45. 0.000484 seconds 47. y2 2 4 5x2 5 1 49. 275 feet 51. The transverse axis should be vertical. The points are 13, 02, 123, 02, and the vertices are 10, 22, 10, 222. 53. false 55. true 57. x2 2 y2 5 a2 or y2 2 x2 5 a2 59. As c increases, the graphs become more squeezed down toward the x-axis. 1248 Answers to Odd-Numbered Exercises 61. As c decreases, the vertices are located at A61 c, 0B and are moving away from the origin. Section 11.5 1. 12, 62, 121, 32 3. 11, 02 5. no solution 7. 10, 12 9. 10.63, 21.612, 120.63, 21.612 11. no solution 13. 11, 12 15. A2!2, !2B, A22!2, 2!2B, A !2, 2!2B, A2!2, 22!2B 17. 126, 332, 12, 12 19. 13, 42, 122, 212 21. 10, 232, A2 5, 211 5 B 23. 121, 212, A1 4, 3 2B 25. 121, 242, 14, 12 27. 11, 32, 121, 232 29. 12, 42 31. A1 2, 1 3B, A1 2,21 3B 33. no solution 35. 37. 39. 3 and 7 41. 8 and 9, 28 and 29 43. 8 cm 3 10 cm 45. 400 ft 3 500 ft or 1000 3 ft 3 600 ft 47. professor: 2 m/sec, Kirani James: 10 m/sec 49. In general, y2 2 y 2 0. Must solve this system using substitution. 51. false 53. false 55. 2n 57. Consider e y 5 x2 1 1 y 5 1 . Any system in which the linear equation is the tangent line to the parabola at its vertex will have only one solution. 59. no solution 61. 121.57, 21.642 63. 11.067, 4.1192, 11.986, 0.6382, 121.017, 24.7572 Section 11.6 1. b 3. j 5. h 7. c 9. d 11. k 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. ( 1, 1) (1,2) Answers to Odd-Numbered Exercises 1249 33. 35. 37. 39. 41. 43. 45. 47. 49. 51. 9 2p units2 53. There is no common region here—it is empty, as is seen in the graph below: 55. false 57. 0 # a # b 59. 61. 63. 65. Section 11.7 1. A3!2, !2B 3. a23!3 2 1 1, 3 2 1 !3b 5. a21 2 2 3!3 2 , !3 2 2 3 2b 7. a3!3 2 , 3 2b 9. hyperbola; X2 2 2 Y2 2 5 1 11. parabola; 2X2 2 2Y 2 1 5 0 13. hyperbola; X2 6 2 Y2 2 5 1 15. ellipse; X2 2 1 Y2 1 5 1 1250 Answers to Odd-Numbered Exercises 17. parabola; 2X2 2 2Y 2 1 5 0 19. ellipse; X2 1 1 Y2 9 5 1 21. hyperbola; X2 3 2 Y2 2 5 1 23. parabola; Y2 2 X 2 4 5 0 25. 45° 27. 60° 29. 30° 31. 45° 33. 15° 35. < 40.3° 37. < 50.7° 39. X2 4 1 Y2 9 5 1; rotation of 60° 41. X2 9 2 Y2 1 5 1; rotation of 45° 43. X2 2 Y 2 3 5 0; rotation of 30° 45. X2 25 1 Y2 4 5 1; rotation of 30° 47. X2 1 4X 2 Y 5 0; rotation of 45° 49. true 51. true 53. a. For 90°, the new equation is x2 b2 1 y2 a2 5 1. b. For 180°, the original equation results. 55. a , 0 hyperbola; a 5 0 parabola; a . 0, a 2 1 ellipse; a 5 1 circle. 57. Amount of rotation is 30°. a. Answers to Odd-Numbered Exercises 1251 b. 59. The amount of rotation is about 36°. a. b. 61. a. The amount of rotation is about 63°. It is a parabola. b. The amount of rotation is about 19°. It is a hyperbola. c. The amount of rotation is about 26.5°. Section 11.8 1. r 5 5 2 2 sin u 3. r 5 8 1 1 2 sin u 5. r 5 1 1 1 cos u 7. r 5 6 4 1 3 cos u 9. r 5 12 3 2 4 cos u 11. r 5 3 1 2 sin u 13. r 5 18 5 1 3 sin u 15. parabola 17. ellipse 19. hyperbola 21. ellipse 23. parabola 25. hyperbola 27. parabola; e 5 1 29. hyperbola; e 5 2 31. ellipse; e 5 1 2 33. parabola; e 5 1 35. ellipse; e 5 1 3 1252 Answers to Odd-Numbered Exercises 37. hyperbola; e 5 3 2 39. parabola; e 5 1 41. r 5 57,934,50811 2 0.20622 1 2 0.206 cos u 43. r 5 75,000,00011 2 0.22322 1 2 0.223 cos u 45. Conic becomes more elliptic as e S 1 and more circular as e S 0. 46.– 48. See Instructor’s Solution Manual. 49. 2ep 1 2 e2 51. ± ep 1 2 e 2 ep 1 1 e 2 ,p ≤ 52.– 60. See Instructor’s Solution Manual. 61. a. With u step 5 p/3, plot points 12, 02, 114.93, p/32, 114.93, 2p/32, 12, p2, 11.07, 4p/32, 11.07, 5p/32, and 12, 2p2. b. With u step 5 0.8p, plot points 12, 02, 14.85, 0.8p2, 11.03, 1.6p2, 140.86, 2.4p2, 11.26, 3.2p2, and 12, 4p2. Section 11.9 1. 3. 5. 7. 9. 11. 13. 15. Answers to Odd-Numbered Exercises 1253 17. 19. 21. 23. Arrow in different directions, depending on t 25. 27. 29. 31. y 5 1 x2 33. y 5 x 2 2 35. y 5 "x2 1 1 37. x 1 y 5 2 39. x 1 4y 5 8 41. 17.7 sec 43. yes 45. Height: 5742 ft; Horizontal distance: 13,261 ft 47. 125 sec 49. 51. 53. The original domain must be t $ 0; therefore, only the part of the parabola where y $ 0 is part of the plane curve. 55. false 57. Quarter circle in QI 58.– 60. See Instructor’s Solution Manual. 61. x 1 y 2a 1 y 2 x 2b 5 1 63. y 5 b a !x 65. 67. 69. a 5 2, 0 # t , 2p a 5 3, 0 # t , 2p t x y 0 A 1 B 0 p 2 0 A 1 B p 2A 2B 0 3p 2 0 2A 2B 2p A 1 B 0 1254 Answers to Odd-Numbered Exercises Review Exercises 1. false 3. true 5. y2 5 12x 7. y2 5 220x 9. 1x 2 222 5 81y 2 32 11. 1x 2 122 5 412121y 2 62 5 241y 2 62 13. vertex: 10, 02 focus: 10, 232 directrix: y 5 3 latus rectum: 12 15. vertex: 10, 02 focus: A1 4, 0B directrix: x 5 21 4 latus rectum: 1 17. vertex: 12, 222 focus: 13, 222 directrix: x 5 1 latus rectum: 4 19. vertex: 123, 12 focus: 123, 212 directrix: y 5 3 latus rectum: 8 21. vertex: A25 2, 275 8 B focus: A25 2, 279 8 B directrix: y 5 271 8 latus rectum: 2 23. 25 8 5 3.125 feet from the center 25. 27. 29. x2 25 1 y2 16 5 1 31. x2 9 1 y2 64 5 1 33. 35. 37. 1x 2 322 25 1 1y 2 322 9 5 1 39. x2 778,300,0002 1 y2 777,400,0002 5 1 41. 43. 45. x2 9 2 y2 16 5 1 47. y2 9 2 x2 5 1 49. 51. 53. 1x 2 422 16 2 1y 2 322 9 5 1 55. Ship will come ashore between the two stations 65.36 miles from one and 154.64 miles from the other. 57. 122, 272, 11, 242 59. 11, 22, 121, 22 61. no solution 63. no solution 65. 12, 32, 123, 222 67. A1 2, 1 !7B, A21 2, 1 !7B, A1 2, 2 1 !7B, A21 2, 2 1 !7B 69. 71. 73. 75. Answers to Odd-Numbered Exercises 1255 77. 79. 81. a23 2 1 !3, 23!3 2 1 1b 83. X2 4 2 Y2 4 5 1 85. 60° 87. X2 1 4 5 Y 89. r 5 21 7 2 3 sin u 91. hyperbola 93. e 5 1 2; vertices A4 3, 0B, 14, p2 or in rectangular form A4 3, 0B, 124, 02 95. 97. 99. x 5 4 2 y2 101. y 5 2x 1 4 103. The vertex is located at 10.6, 21.22. The parabola opens down. 105. a. y 5 21.46!8.81 2 3x b. Vertex at 12.94, 21.42, opens to the left. c. Yes, (a) and (b) agree with each other. 107. As c increases, the minor axis along the x-axis decreases. 109. As c increases, the vertices of the hyperbolas located at a6 1 2c, 0b are moving toward the origin. 111. 10.635, 2.4802, 120.635, 2.4802, 121.245, 0.6452, 11.245, 0.6452 1256 Answers to Odd-Numbered Exercises 113. 115. a. The amount of rotation is about 22.5°. It is an ellipse. b. The amount of rotation is about 11.6°. It is a hyperbola. 117. With u step 5 p 4, points 12, 02, a1.06, p 4 b, a0.89, p 2 b, a1.06, 3p 4 b, 12, p2, a17.22, 5p 4 b, a28, 3p 2 b, a17.22, 7p 4 b, and 12, 2p2 are plotted. 119. a 5 2, b 5 3, t in 30, 2p4 a 5 3, b 5 2, t in 30, 2p4 Practice Test 1. c 3. d 5. f 7. y2 5 216x 9. 1x 1 122 5 2121y 2 52 11. x2 7 1 y2 16 5 1 13. 1x 2 222 20 1 y2 36 5 1 15. x2 2 y2 4 5 1 17. y2 16 2 1x 2 222 20 5 1 19. 21. 23. x2 5 6y; 2 2 # x # 2 25. 27. 29. Ellipse: e 5 2 3 31. 5.3 sec, 450 ft 33. 35. a. b. The vertex is located at 122.1, 1.22. The parabola opens upward. c. Yes Answers to Odd-Numbered Exercises 1257 Cumulative Test 1. 26, 3 3. 27 5. ƒ1x2 5 1 31x 2 722 1 7 7. 24, 4 9. 21.9626 11. 1 cos u 1 sin u 13. p 3, 2p 3 , 4p 3 , 5p 3 15. 825.1, 14.19 17. Soda: $1.29; Soft pretzel: $1.45 19. 21. c7 216 33 18 23 217d 23. 1x 2 622 9 1 1y 1 222 25 5 1 25. 12, 42 14, 22 27. CH A P T E R 1 2 Section 12 . 1 1. 1, 2, 3, 4 3. 1, 3, 5, 7 5. 1 2, 2 3, 3 4, 4 5 7. 2, 2, 4 3, 2 3 9. 2x2, x3, 2x4, x5 11. 21 6, 1 12, 2 1 20, 1 30 13. 1 512 15. 2 1 420 17. 10201 10000 5 1.0201 19. 23 21. an 5 2n 23. an 5 1 1n 1 12n 25. an 5 1212n2n 3n 27. an 5 1212n11 29. 72 31. 812 33. 1 5852 35. 83, 156, 160 37. 1 n1n 1 12 39. 12n 1 3212n 1 22 41. 7, 10, 13, 16 43. 1, 2, 6, 24 45. 100, 50, 25 3 , 25 72 47. 1, 2, 2, 4 49. 1, 21, 22, 5 51. 10 53. 30 55. 36 57. 5 59. 1 2 x 1 x2 2 x3 61. 109 15 63. 1 1 x 1 x2 2 1 x3 6 1 x4 24 65. 20 9 67. not possible 69. a 6 n50 1212n 2n 71. a q n51 1212n21n 73. a 6 n51 1n 1 12! 1n 2 12! 5 a 5 n51 n1n 1 12 75. a q n51 1212n21 xn21 1n 2 12! 5 a q n50 1212n xn n! 77. $28,640.89; total balance in account after 6 years (or 72 months) 79. sn 5 20 1 2n; a paralegal with 20 years experience would make $60 per hour 81. an 5 1.03an21; a0 5 30,000 83. an11 5 1000 2 75n; approximately 10.7 years 85. A1 5 100, A2 5 200.10, A3 5 300.30, A4 5 400.60, A36 5 3663.72 87. 7; 7.38906 89. 0.095310; 0.095310 91. The mistake is that 6! 2 3!2!, but rather 6! 5 6⋅5⋅4⋅3⋅2⋅1. 93. 1212n11 5 e 1, n 5 1, 3, 5, . . . 21, n 5 2, 4, 6, . . . So, the terms should all be the opposite sign. 95. true 97. false 99. C, C 1 D, C 1 2D, C 1 3D 101. 1 and 1 103. < 2.705; < 2.717; < 2.718; 105. 109 15 Section 12.2 1. arithmetic, d 5 3 3. not arithmetic 5. arithmetic, d 5 20.03 7. arithmetic, d 5 2 3 9. not arithmetic 11. 3, 1, 21, 23; arithmetic; d 5 22 13. 1, 4, 9, 16; not arithmetic 15. 2, 7, 12, 17; arithmetic; d 5 5 17. 0, 10, 20, 30; arithmetic; d 5 10 19. 21, 2, 23, 4; not arithmetic 21. an 5 11 1 1n 2 125 5 5n 1 6 23. an 5 24 1 1n 2 12122 5 26 1 2n 25. an 5 0 1 1n 2 12 2 3 5 2 3n 2 2 3 27. an 5 0 1 1n 2 12 e 5 en 2 e 29. 124 31. 2684 33. 16 3 35. a5 5 44, a17 5 152; an 5 8 1 1n 2 129 5 9n 2 1 37. a7 5 21, a17 5 241; an 5 23 1 1n 2 121242 5 24n 1 27 39. a4 5 3, a22 5 15; an 5 1 1 1n 2 12 2 3 5 2 3n 1 1 3 41. 552 43. 2780 45. 51 47. 416 49. 3875 51. 21 2 C1 6 2 13 2 D 5 21 2 A1 2 39 6 B 5 2133 2 53. 630 55. S43 5 43 4 15 2 432 5 2817 2 57. 1368 59. Colin: $347,500; Camden: $340,000 61. 850 seats 1258 Answers to Odd-Numbered Exercises 63. 1101 glasses on the bottom row, each row had 20 fewer glasses than the one before 65. 1600 feet 67. 210 oranges 69. a. 23 seats in the first row b. 1125 seats 71. an 5 a1 1 1n 2 12 d, not a1 1 nd 73. There are 11 terms, not 10. So, n 5 11, and thus, S11 5 11 2 11 1 212 5 121. 75. false 77. true 79. 1n 1 1212a 1 nb2 2 81. 27,420 83. 5050 85. 2500 87. 18,850 Section 12.3 1. yes, r 5 3 3. no 5. yes, r 5 1 2 7. yes, r 5 1.7 9. 6, 18, 54, 162, 486 11. 1, 24, 16, 264, 256 13. 10,000, 10,600, 11,236, 11,910.16, 12,624.77 15. 2 3, 1 3, 1 6, 1 12, 1 24 17. an 5 5122n21 19. an 5 11232n21 21. an 5 100011.072n21 23. an 5 16 3 A21 4Bn21 25. a7 5 2128 27. a13 5 4096 3 29. a15 5 6.10 3 10216 31. 8191 3 33. 59,048 35. 2.2 37. 6560 39. 16,383 41. 2 43. 21 4 45. not possible, diverges 47. 227 2 49. 10,526 51. 2 3 53. 100 55. $44,610.95 57. an 5 200010.52n; a4 5 125, a7 5 16 59. 17 feet 61. 58,640 students 63. 67 days; $9618 65. $3877.64 67. 26 weeks: $13,196.88 52 weeks: $26,811.75 69. $367,987 71. 1 2 1 2 1 2 5 1 73. should be r 5 21 3 75. should use r 5 23 all the way through the calculation; also, a1 2 12 (not 4) 77. false 79. true 81. 0 b 0 , 1, a 1 2 b 83. 47 99 85. 237, 529, 996, 894, 754 87. a q n50 xn 5 1 1 2 x for 0 x 0 # 1 89. 1 1 2 2x for 0 x 0 , 1 2 Section 12.4 1.–24. See Instructor’s Solution Manual. 25. 7 steps 27. 31 steps 29.–30. See Instructor’s Solution Manual. 31. false 33.–37. See Instructor’s Solution Manual. 39. 255 256, yes Section 12.5 1. 35 3. 45 5. 1 7. 1 9. 17,296 11. x4 1 8x3 1 24x2 1 32x 1 16 13. y5 2 15y4 1 90y3 2 270y2 1 405y 2 243 15. x5 1 5x4y 1 10x3y2 1 10x2y3 1 5xy4 1 y5 17. x3 1 9x2y 1 27xy2 1 27y3 19. 125x3 2 150x2 1 60x 2 8 21. 1 x4 1 20y x3 1 150y2 x2 1 500y3 x 1 625y4 23. x8 1 4x6y2 1 6x4y4 1 4x2y6 1 y8 25. a5x5 1 5a4bx4y 1 10a3b2x3y2 1 10a2b2x2y3 1 5ab4xy4 1 b5y5 27. x3 1 12x5@2 1 60x2 1 160x3@2 1 240x 1 192x1@2 1 64 29. a3 1 4a9@4b1@4 1 6a3@2b1@2 1 4a3@4b3@4 1 b 31. x 1 8x3@4y1@2 1 24x1@2y 1 32x1@4y3@2 1 16y2 33. r4 2 4r3s 1 6r2s2 2 4rs3 1 s4 35. a6x6 1 6a5bx5y 1 15a4b2x4y2 1 20a3b3x3y3 1 15a2b4x2y4 1 6ab5xy5 1 b6y6 37. 3360 39. 5670 41. 22,680 43. 70 45. 3,838,380 47. 2,598,960 49. a7 5b 2 7! 5!, but rather 7! 5! 2! 5 7⋅6⋅5! 5!12⋅12 5 21 51. false 53. true 55.–56. See Instructor’s Solution Manual. 57. 1 2 3x 1 3x2 2 x3 Answers to Odd-Numbered Exercises 1259 59. Graphs of the respective functions get closer to the graph of y4 5 11 2 x23 when 1 , x , 2; when x . 1, no longer true. 61. graph of the curve better approximation to the graph of y 5 A1 1 1 xB3, for 1 , x , 2; no, does not get closer to this graph if 0 , x , 1 Section 12.6 1. 360 3. 15,120 5. 40,320 7. 1716 9. 252 11. 15,890,700 13. 1 15. 27,405 17. 215,553,195 19. 24 21. 12 23. 10,000 25. 32,760 27. Each of 20 questions has four answer choices. So, there are 420 < 1.1 3 1012 possible ways to answer the questions on the exam. 29. 100,000; 81,000 31. 2.65 3 1032 33. 59,280 35. 997,002,000 37. 22,957,480 39. 2,598,960 41. 1326 43. 4.9 3 1014 45. 256 47. 15,625 49. 400 51. The combination formula nCr should be used instead. 53. true 55. false 57. nCr nCr11 5 r 1 1 n 2 r 59. C1n, r2⋅r! 5 n! 1n 2 r2! 61. answers the same 63. a. 5,040 b. 5,040 c. yes d. nP r 5 r!nCr Section 12.7 1. 52, 3, . . . , 126 3. 5BBBB, BBBG, BBGB, BBGG, BGBB, BGBG, BGGB, BGGG, GBBB, GBBG, GBGB, GBGG, GGBB, GGBG, GGGB, GGGG6 5. 5RR, RB, RW, BB, BR, BW, WR, WB6 7. 1 8 9. 7 8 11. 1 18 13. 1 2 15. 5 12 17. 10 13 19. 4 13 21. 3 4 23. 3 4 25. 0 27. a. 270,725 b. 715 270,725 > 0.26% c. 13 270,725 < 0.005% 29. 8 52 5 2 13 < 15.4% 31. 4 663 < 0.6% 33. 1 32 < 3.1% 35. 31 32 < 96.9% 37. A18 38B4 < 5.03% 39. 20% 41. 25% 43. 48 1326 < 3.6% 45. 0.001526% 47. a. {(Brown, Blue), (Brown, Brown), (Blue, Brown), (Blue, Blue)} b. 1 @ 4 c. 3 @ 4 49. 2,598,960 51. 1287 2,598,960 53. The events aren’t mutually exculsive. So, probability 5 4 52 1 13 52 2 1 52 5 16 52 5 4 13. 55. true 57. false 59. 0.0027 61. 1 3 < 0.333 63.–64. Use random number generator and compare. 65. 0.2907 Review Exercises 1. 1, 8, 27, 64 3. 5, 8, 11, 14 5. a5 5 32 243 < 0.13 7. a15 5 2 1 3600 9. an 5 1212n113n 11. an 5 1212n 13. 56 15. 1 n 1 1 17. 5, 3, 1, 21 19. 1, 2, 4, 32 21. 15 23. 69 25. a 7 n51 1212n 2n21 27. a q n50 xn n! 29. $36,639.90; amount in the account after 5 years 31. arithmetic, d 5 22 33. arithmetic, d 5 1 2 35. arithmetic, d 5 1 37. an 5 24 1 1n 2 12152 5 5n 2 9 39. an 5 1 1 1n 2 12A22 3B 5 22 3n 1 5 3 41. a1 5 5, d 5 2, an 5 5 1 1n 2 12122 5 2n 1 3 43. a1 5 10, d 5 6, an 5 10 1 1n 2 126 5 6n 1 4 45. 630 47. 420 49. Bob: $885,000 Tania: $990,000 51. geometric, r 5 22 53. geometric, r 5 1 2 55. 3, 6, 12, 24, 48 57. 100, 2400, 1600, 26400, 25,600 59. an 5 a1rn21 5 7⋅2n21 61. an 5 1222n21 63. a25 5 33,554,432 65. a12 5 22.048 3 1026 67. 4920.50 69. 16,400 71. 3 73. $60,875.61 75.–78. See Instructor’s Solution Manual. 79. 165 81. 1 83. x4 2 20x3 1 150x2 2 500x 1 625 85. 8x3 2 60x2 1 150x 2 125 87. x5@2 1 5x2 1 10x3@2 1 10x 1 5x1@2 1 1 89. r5 2 5r4s 1 10r3s2 2 10r2s3 1 5rs4 2 s5 91. 112 93. 37,500 95. 22,957,480 97. 840 99. 95,040 f 2 of spades 1260 Answers to Odd-Numbered Exercises 101. 792 103. 1 105. 30 107. 5040 109. 120 seating arrangements, 15 years 111. 94,109,400 113. 20,358,520 115. 1 16 < 6.25% 117. 30 36 < 83.3% 119. 2 3 < 66.7% 121. 7 12 < 58.3% 123. 2 13 < 15.4% 125. 31 32 < 96.88% 127. 5369 3600 129. 34,875 14 131. 1 1 1 2x 133. 99,900, yes 135. Graphs become better approximations of the graph of y 5 11 1 2x24 for 20.1 , x , 0.1; does not get closer for 0.1 , x , 1. 137. a. 11,440 b. 11,440 c. yes d. nP r r! 5 nCr 139. 0.0722 Practice Test 1. xn21 3. Sn 5 1 2 xn 1 2 x 5. 0x0 , 1 7. 1 2 A1 4B10 < 1 9. 24,950 11. 2520 13. 455 15. 2184 17. x10 1 5x7 1 10x4 1 10x 1 5 x2 1 1 x5 19. more permutations than combinations since order is taken into account when determining the number of permutations 21. 18 38 < 0.47 23. 18 38 > 0.47 25. 4 13 < 0.308 27. 184,756 Cumulative Test 1. 2 3. 5 6 3!7i 4 5. x 5 28 7. 2x 1 h 2 3 9. 115, 62 11. VA: x 5 3 HA: y 5 25 13. 2.585 15. no solution 17. z11, 42 5 24 19. c99 29 18 4 29 7d 21. 2161x 2 32 5 1y 2 522 23. 3 25. 210 5 1024 27. 23432 Agriculture pasture areas, 345, 924 Animals bird wings, 853 cat food amounts, 412 deer population, 751 dog runs, 295, 1048 dog training, 136 gray wolf population, 472 horse paddock, 1048 Archaeology dating of fossils, 453 Art/Music/Theater/Entertainment American Idol, 1149 candy costs, 104 chord frequencies, 105 Dancing with the Stars, 1149 frequencies of sound, 448–450 murals, 795 musical tones and frequencies, 711–712, 714, 747–748 New York Philharmonic sound levels, 472 phonographs, 582 The Price is Right, 583 rock concert sound levels, 447, 473 Survivor, 1149 television sizes, 118 theater seating, 1115, 1116 viewing paintings in a museum, 734 Automotive antifreeze concentration, 104 braking, 819 car depreciation, 1125 car options, 1164 car repair costs, 233 car value, 148, 158, 200, 482, 489 driving under a bridge, 1013, 1089 gas and electricity costs, 959 gas mileage, 346, 879, 922 gas/oil mixtures, 879 gasoline mixtures, 99–100 gasoline prices, 99, 104, 171–172, 248, 962 hybrid cars, 962 license plates, 1149 NASCAR race, 530 NASCAR revenue, 171 new car models, 472–473, 482 production levels, 890 rental car costs, 91, 135, 199 service costs, 89 speed of a car, 576 tire size and speedometer readings, 578, 581, 582 towing power, 819 traffic, 534 trip speeds, 370 van depreciation, 472 van value, 472 windshield wipers, 581 Aviation airplane slides, 788 bearing, 534, 787 distance between planes, 787, 1095 elevation of a rocket, 537 glide path, 533–534 heading and airspeed, 810 length of an aircraft wing, 787 locating airplanes, 532 midair refueling, 533 NASA astronaut weights, 136 NASA centrifuge, 582 NASA “vomit comet,” 346 sonic booms, 620 wind speed, 101, 105, 879–880 Biology/Life Sciences Archimedes spiral, 853 average height, 200 bacterial growth, 481, 489, 490, 1107 dating of fossils, 424, 453 DNA structure, 582 eye color, 1156, 1158 human anatomy, 318, 1166 human body temperature, 155, 310, 752 human body weight, 136, 200, 345, 361 sex of offspring, 1152, 1158, 1164 volume of air in the lungs, 751 yo-yo dieting, 591 Budgeting business, 103 car repair costs, 233 cost of flower seed, 795 costs, 270 event planning, 103, 252, 270, 879, 909 long-distance call costs, 252, 310 monthly driving costs, 199 personal, 104 race entry fee, 270, 310 salaries, 1116, 1123 service charges, 196–197 Business break-even point, 45, 118, 386, 751–752, 880 budgeting, 103 cash flow of a stock fund, 695 cell phone plan, 318 cell phone provision, 962, 974 computer assembly, 919 computer sales, 695 costs, 89, 91, 104, 127, 135, 137, 200–201, 270, 295–296, 341, 345, 714, 879 deduction for business expenses, 961 defective products, 1158 demand, 619 job applicants, 961 job offers, 876 leadership positions, 1164 losses, 550 markups, 103, 137 NASCAR revenue, 171 prices, 103, 104, 118, 148, 171, 270, 295, 318, 328, 889 production levels, 890, 912, 919, 923, 948 profits, 34, 45, 118, 136, 148, 160, 184, 270, 285, 345, 346, 386, 394–395, 413, 418, 422, 482, 550, 1013 real estate, 148, 437, 1107, 1148 revenues, 136, 155, 171, 347, 361, 714, 795–796, 819 salaries, 285, 310, 325, 326, 437, 879, 1106–1107, 1115, 1116, 1123, 1125, 1126, 1162, 1163 sales, 270, 318, 481–482, 522, 591, 663, 734, 751, 948, 1104, 1107 stock values, 115–116, 118, 127, 146, 148, 361, 488, 887 supply and demand, 127 time to pay off debt, 483 tipping, 961–962 typing speed, 412 van depreciation, 472 wages, 318, 327 working together, 105 Chemistry acidity/basicity of a solution, 452–453 enzyme reactions, 455 mixtures, 99–100, 104, 158, 879, 881, 922, 948 pH of a solution, 452–453, 488 structure of molecules, 535 temperature and pressure, 319 water molecules, 565 Communications cell phone costs, 91, 137, 318, 962, 974 cell phone coverage, 207 cell phone plans, 481, 879 cell phone provision by business, 962, 974 cell phone purchases, 482 cell phone towers, 171, 207 cell phone triangulation, 788 cell phone use, 25, 347 communication with astronauts, 531 laser communications, 25, 104, 137, 735 Applications Index [ [ 1261 1262 Applications Index Communications (continued ) long-distance call cost, 270 monthly phone costs, 200 parabolic antennas, 1010–1011 postage rates, 270–271 satellite dish receivers, 1010, 1013, 1089 telephone infrastructure, 1132 touch-tone dialing, 714 10-digit dialing, 1142 Computers building a system, 1148 computer business, 919 depreciation, 437, 1125 e-mail passwords, 1164 human brain vs., 890 Internet costs, 91, 137, 328 passwords, 1148 production levels, 912 Scrabble game, 890 website hits, 1126 Construction and Home Improvement amount of fencing to buy, 1045–1046, 1048 area of a floor, 525 bathroom remodeling, 199 budgeting, 104 building a treehouse, 536 dog runs, 207 gardening, 119, 413 garden maze, 853 great room dimensions, 796 housecleaning, 102, 119 house painting, 105 hurricane preparations, 918–919 ladder size, 104 landscaping, 104, 105 pitch of a roof, 500 rain gutters, 751 sprinkler system, 270, 328 sprinkling grass, 184 window height bathroom remodeling, 788 Consumer Situations baseball card values, 252 cell phone costs, 91, 137, 318, 962, 974 cell phone plans, 481, 879 cell phone purchases, 482 cell phone use, 25, 347 comparative shopping, 135 depreciation of goods, 437 discount prices, 103, 236 food costs, 103, 104 revolving restaurant, 507 sales tax, 318, 327 search and rescue, 534 television sizes, 118 wedding invitations, 1148 Demographics (US/World) adult smokers in the United States, 961 average marriage age, 196, 951 cell phones in United States, 25, 347 HIV/AIDS infection rate, 481 population growth, 433, 437, 475, 481, 488, 489, 490 registered voters, 961 time to register for a class, 819 underage smoking, 347 U.S. population, 15 women in science, 961 Design champagne fountain, 1115 field of tulips, 1115 home décor, 789 new car models, 472–473, 482 nuclear cooling towers, 1037 suspension bridge, 1013 university campus, 207 Earth Sciences earthquakes, 448, 452, 463, 472, 473, 490, 735 Economics cash flow of a stock fund, 695 exchange rates, 252 gasoline prices, 171–172 house prices, 137, 296, 481 market price, 295 oil consumption, 490 price increases, 148 prices, 118 profit function, 346–347 stock prices, 115–116, 118, 127, 146, 148, 361, 488, 887 supply and demand, 127, 184, 271, 318, 438, 913 U.S. national debt, 15 Education/Learning band uniforms, 270 bookstore markups, 103 class seating, 1149 fraternity elections, 1149 grades, 89, 105, 127, 137, 158, 160, 233, 483 leadership, 1148 memorization by professor, 412 multiple-choice tests, 1149 number of students on college campus, 478 paper margins, 118 research grants, 104 salaries, 879 sleep hours, 103 sorority financials, 119 sorority T-shirt costs, 270 sound level in college stadium, 452 sound level of teacher speaking, 488 travel time, 105 tutoring costs, 325 university population growth, 1126 Electricity/Electronics/Optics alternating current, 734 band-pass filters, 155 cardioid microphones, 853 cell phone towers, 171, 207 circuit theory, 988 dish TV, 534 electrical circuits, 68 electrical fields, 35, 45 electromagnetic spectrum, 453 electromagnetic wave propagation, 687 eyeglass lens, 1013, 1093 focal length of a lens, 58, 91, 105–106 gas and electricity costs, 959 laser beams, 25, 184 laser communication, 25, 104, 137 lens law, 891 light cast by a lamp, 1037 optical signals, 714 radio waves, 184 refraction of light, 522, 747, 751 resistance in circuits, 58 speed of a lighthouse beacon, 576–577 speed of an electric saw, 582 square waves, 271 television sizes, 118 Engineering Archimedes spiral, 853 bridge construction, 778 cell phone towers, 207 Christmas lights, 508 electrical fields, 35, 45 gears, 581 height of a projectile, 45 nuclear cooling towers, 1037 solar cooker, 1013 solar furnace, 1013 Environment carrying capacity, 482–483 dry erase markers, 879 envelope waste, 253 fires, 207 forestry, 550 gas mileage, 346 gasoline and electricity consumption, 962 global climate change, 253 height of a mountain, 639 hours of daylight, 734 insect infestation, 795 oil spill, 296 oxygen level fluctuations, 620 paper use, 118 plastic bag use, 200 pollen levels, 734, 750 sprinkler coverage, 581 tides, 591 toxic fume emissions, 581 Finance average credit card debt, 200 depreciation, 437, 472, 1125 federal funds rate, 253, 361–362 federal income tax, 137, 285, 961 insurance costs, 127 investments. See Investments loan payments, 58 mortgage rates, 422 paying off credit cards, 479–480 PIN number combinations, 1148 raffles, 1149 sorority financials, 119 time to pay off debt, 483 Applications Index 1263 Food/Nutrition amount of meat to buy, 877 for cats, 412 coffee costs, 104 costs, 103, 104 diets, 103, 158, 889–890, 912, 942, 947 dinner combinations, 1148 donut store products, 919 energy drinks, 881 exercise and nutrition, 948 nutrient levels in diet, 962, 974 oranges display, 1116 pasta making, 583 pizza slices, 574 sandwich consumption, 962 solar cooker, 1013 Forensic Science blood splatter, 521–522 time of death, 482 Geography average temperatures, 134, 155, 160, 200, 252 Bermuda Triangle, 764, 785, 795 Earth’s circumference, 25 surveying glaciers, 789 Zip codes, 1149 Geometry angle between two lines, 687 angle traced by hands of a clock, 507, 550 areas, 34, 96, 119, 148, 326, 341–342, 345, 370, 413, 418, 419, 575, 639, 714, 795, 796, 912, 987 circles, 96, 103, 295, 326, 328, 891, 948, 1048 circumference, 96 cylinders, 34, 252 decagons, 796 depth of treasure, 537 distance traveled by hands of a clock, 580–581 graphs of quadratic function, 948 height of a flagpole, 508 height of a lighthouse, 508 height of a man, 508 height of a tree, 505, 508, 651 hexagons, 796, 839 lines, 987 octagons, 839 parallelograms, 796 pentagons, 839 perimeters, 34, 45, 91, 96, 103, 118, 158, 413, 924 polygon angles, 1132 quadrilaterals, 687 rectangles, 45, 96–97, 103, 119, 148, 158, 249, 326, 345, 370, 413, 418, 419, 924, 1048 rotation of combination lock dial, 550 semicircles, 104 squares, 119, 295, 839 surface areas, 34 tangent and slope related, 34 Tower of Pisa, 775 triangles, 96, 103, 119, 158, 418, 550, 714, 795, 796, 987 vectors, 819 volumes, 34, 45, 119, 252, 296, 326, 419 Government DUI levels, 160 federal funds rate, 361–362 FEMA hurricane response, 864, 917–918 gun laws, 951 income taxes, 137, 285, 961 law enforcement, 788 postage rates, 270–271 registered voters, 961 sales tax, 318, 327 U.S. national debt, 15 Health/Medicine anesthesia, 472, 481 average height, 200 blood alcohol level, 160 body surface area, 127, 286 body temperature, 155, 310, 591, 592, 752 body weight, 136 bunions, 550–551 diets, 103, 158, 889–890, 912, 942, 947 drug concentrations, 592, 687 drug conversions/dosages, 91, 285, 347, 386–387, 395, 412, 413, 438 drug mixtures, 879 elbow joint torque, 788 elliptical trainers, 1026 exercise heart rate, 137, 472 flu outbreak model, 118 health care costs, 253 herd immunity, 228 HIV/AIDS, 481, 1107 insurance costs, 127 IV solutions, 104–105 malaria outbreak, 252–253 muscle force, 811 orthotic knee braces, 565 pollen levels, 734, 750 radiation therapy, 509 sleep hours, 103 smoking, 347, 961 traction, 536 velocity of air in trachea, 361 viral spread, 482, 490 weight loss/gain, 345, 361 Investments allocating principal, 104, 158, 880, 890, 922, 924, 948, 993 annuities, 1107, 1126 compound interest, 434–435, 437–438, 470–471, 472, 487, 488, 490, 1106, 1107, 1124, 1126, 1162 debt accrual, 734 future value, 1107 interest rates, 104, 158, 319 investments, 98–99 multiple, 98–99 real estate, 148, 437 saving, 734 simple interest, 97–98 stock values, 115–116, 118, 127, 146, 148, 361, 488, 887 Math and Numbers approximating ex, 1107 approximating functions, 1107 biofilium, 663 cryptography, 926, 972 data curve-fitting, 890, 922, 943, 947–948, 993 difference quotients, 687 Fibonacci sequence, 1108 finding numbers, 1048 linear regression, 158 lottery numbers, 1139, 1149, 1164 optimization, 819 partial-fraction decomposition, 891 pursuit theory, 820 raffles, 1164 random number generator, 1158 safe combinations, 1164 sum and product, 103, 118 temperature conversions, 91, 285, 347, 386–387, 395, 412, 413, 438 Mental Exercises memorization by professor, 412 memorization of cards, 413 Personal finance average credit card debt, 200 budgeting, 104 car repair costs, 233 federal income tax, 137, 285, 961 investments. See Investments paying off credit cards, 479–480 safes, 1149 Physics and Astronomy asteroids, 1026, 1076 braking, 819 bullet speed, 148, 809 bungee jumping, 1126 calibrating recording devices, 1037 closing doors, 819 communication with astronauts, 531 distance traveled by a bullet, 1084 Earth’s orbit, 234 Earth’s rotation, 582 electromagnetic spectrum, 453 falling objects and gravity, 34, 68, 119, 252, 271, 345, 418, 890, 947, 1115–1116, 1126, 1163 fireworks, 296 flight of a projectile, 1084 frequencies of sound, 448–450 frequency of oscillations, 620 Halley’s Comet orbit, 853, 1026 height of a hot-air balloon, 777–778 interplanetary distances, 25 ISS orbit, 573 Jupiter’s rotation, 582 laser beams, 483 lifting weights, 819 low earth orbit satellite orbits, 580 1264 Applications Index Physics and Astronomy (continued ) missile fired from a ship, 809 Moon’s orbit, 25 musical tones and frequencies, 711 NASA astronaut weights, 136 NASA centrifuge, 582 NASA “vomit comet,” 346 Newton’s Law of Heating and Cooling, 439, 482 parabolic telescope, 1013 path of a projectile, 809 path of a punted football, 342–343, 345 pendulums, 68, 127, 853–854 planetary orbits, 68, 1025–1026, 1075, 1089, 1093 Pluto’s orbit, 853 pre-crusher wheels, 695–696 radioactive decay, 434, 437, 476–477, 482, 489, 490 raising wrecks, 819 resultant force, 810, 811, 827 rocket tracking, 778 satellite orbits, 1021–1022 sliding box, 810 solar cooker, 1013 solar furnace, 1013 solar radiation and distance, 319 sound levels, 447, 452, 453, 463, 472, 473, 488, 490 sound wave amplitude and frequency, 620 Space Shuttle escape basket, 777 speed of light, 91 speed of sound, 104, 127, 318 spring height, 620 spring stretch and force, 318 temperature conversions, 91, 200, 292–293, 295, 310, 328 theory of relativity, 127–128 torque, 811 towing power, 819 work, 819 x-ray crystallography, 523 Puzzles cryptography, 974 distance puzzles, 103 sleep hours, 103 Sports and Recreation AKC field trials, 783 amusement park rides, 507, 581, 582, 583, 1025, 1084 archery, 508–509, 523, 535, 778 aviation, 105 balancing a seesaw, 105 baseball, 119, 252, 787, 810, 1084 basketball, 270, 889, 947, 974, 1149 bicycle gears, 581 bicycle speed, 582 blackjack, 1096, 1149, 1158, 1164 boating, 105, 1163 bowling, 778, 788–789 bungee jumping, 1126 canasta, 1140, 1164 cards, 1149, 1152, 1155, 1158, 1159, 1164 child’s swing, 611–613 club sign-up, 811 dog field trials, 507–508 dog training, 136 drawing cards, 1164, 1166 exercise and nutrition, 948 exercise equipment, 810 firecrackers, 252 fireworks, 296 football, 34, 105, 171, 232, 330, 342–343, 345, 370, 422, 452, 810, 889, 947, 948, 1022, 1025, 1106, 1149, 1158, 1162, 1164 footraces, 1048 golf, 82, 137, 155, 533, 1082–1083 Ironman Triathlon, 1149 jogging, 105 lifeguard posts, 778 Little League football, 477–478 marbles, 1166 marching band formation, 1113–1114 memorization of cards, 413 motorcycle manufacture, 341 NASCAR, 171 NASCAR race, 530 Olympic Decathlon, 209–210, 228 party tent, 508 peg game, 1147 poker, 1140, 1159 rock climbing, 778, 788 rolling dice, 1153 roulette, 1158, 1166 rowing, 270, 310 running, 200 Samurai sword artistry, 853 Scrabble game, 890 seasonal sales at Magic Kingdom, 591 season tickets, 1149 skiing, 534 soccer, 346 swimming pool dimensions, 249 swimming pool volume, 296 tennis, 534–535, 778, 819 tossing a coin, 1126 Tower of Hanoi game, 1131–1132 track, 550, 1025 video games, 853 volleyball, 1145 walking, 105 weightlifting, 809 Zim-Zam, 550 Transportation aviation. See Aviation boating, 105, 805–806, 807–808, 809–810, 1013 bypass around town, 232 ship navigation, 1034, 1037, 1090 shortcuts, 522 train, 919 trip speeds, 370 Travel blood alcohol level and DUI, 160 distance traveled, 158, 171 travel time, 105 trip lengths, 94, 105 trip speeds, 370 Weather atmospheric temperature, 591 average rainfall, 200 average temperatures, 134, 155, 160, 200, 252, 760 FEMA hurricane response, 864, 917–918 global climate change, 253 global warming, 271 humidity, 137 hurricanes, 912, 917–919 raindrops, 326 staking down trees before a storm, 502, 508 temperature conversions, 200 Zoology alligator length, 104 bacterial growth, 481, 489, 490, 1107 carrying capacity, 482–483, 489 deer population, 751 firefly bioluminescence, 620–621 fish population, 489 gray wolf population, 472 hamster on a wheel, 583 insect infestation, 795 phytoplankton growth, 481 snake length, 104 Abscissa, 166 Absolute value, 11 definition of, 149 properties of, 150 Absolute value equations, 150–152 definition of, 150 quadratic, 152 Absolute value function, 257 Absolute value inequalities, 152–154 properties of, 152 Accuracy, 525 Actual velocity, 804 Acute angles, 495 calculating trigonometric function values for, 545–547 Acute triangles, 766 Addition. See also Sum(s) associative property of, 9 commutative property of, 9 of complex numbers, 71 of functions, 287–288 of matrices, 952–954 order of operations and, 6–9 of ordinates, 614–616 of polynomials, 28 of rational expressions, 51–53 of real numbers, 9–10 of vectors, 799, 801–802 Addition method. See Elimination method Additive identity, for a matrix, 954 Additive identity property definition of, 9 for matrices, 954 of real numbers, 9 Additive inverse definition of, 9 for a matrix, 954 Additive inverse property for matrices, 954 of real numbers, 9 Algebraic expressions definition of, 8, 84 evaluating, 8–9 Algebraic properties of vectors, 802 Algebraic signs, of trigonometric functions, 552–554 Algebraic techniques, solving trigonometric equations using, 741–742 Alternating sequences, 1099 Amplitude, of sinusoidal functions, 600 Angle(s), 494–496 acute, 495 acute, calculating trigonometric function values for, 545–547 central, 567 complementary, 496 coterminal, 541–542 degree measure of, 494–496 of depression, 529, 530 of elevation, 529 of inclination, 530 negative, 494 nonacute. See Nonacute angles obtuse, 495 positive, 494 quadrantal, 539 quadrantal, calculating trigonometric function values for, 547–548 radian measure of, 567–569 rays of, 494 reference, 556–557, 558 right, 495 of rotation, transforming a general second-degree equation into an equation of a conic and, 1060–1064 in standard position, 538–541 straight, 495 supplementary, 496 of a triangle, finding, 497 between vectors, 814–818 vertex of, 494 Angle of rotation formula, 1061 Angular speed, 576–577, 578 Apparent velocity, 804 Applications. See also Applications index involving composite functions, 289, 292–293 involving ellipses, 1021–1022 involving exponential functions, 432–435 involving finite arithmetic sequences, 1113–1114 involving functions, 248–249 involving geometric series, 1123–1124 involving hyperbolas, 1034 involving inverses of square matrices, 972 involving linear equations, 93–103, 195–197 involving logarithmic equations, 469–471 involving logarithms, 446–450 involving matrices, 959, 972 involving matrix multiplication, 959 involving parabolas, 1010–1011 involving quadratic equations, 115–116 involving quadratic functions, 340–343 involving series, 1104 involving systems of linear equations in two variables, 876–877 involving systems of linear inequalities in two variables, 909–911 involving systems of nonlinear equations, 1045–1046 involving trigonometric equations, 747–748 solving application problems using modeling, 93–95 Approximations of decimals, 6 of exponential functions, 426 of real zeros, bisection method for, 382–383 Arc length, 572–573 Area of a circular sector, 573–575 of a triangle, 791–794 Arguments, variable, evaluating functions with, 244 Arithmetic operations. See also Addition; Division; Multiplication; Subtraction order of, 6–9 Arithmetic sequences, 1109–1114 common difference in, 1109–1110 definition of, 1109 finite. See Finite arithmetic sequences general term of, 1110–1111 sum of, 1111–1116 Association, direction of, in scatterplots, 213–214 Associative property of addition for matrices, 954 of real numbers, 9 Associative property of multiplication, of real numbers, 9 Asymptotes, of rational functions, 398–403, 406–408 Augmented matrices, 930–932 row operations on, 931–932 Average rate of change, 260–263 Axis(es) of an ellipse, 1015 of a graph, 166 imaginary, 821 polar, 841 real, 821 rotation of, 1057–1064 transverse, of a hyperbola, 1028 Axis of symmetry of a parabola, 1003 of a quadratic function, 333 Back substitution, Gaussian elimination with, 933–935 Base(s) b, exponential functions with, 931–932 of an exponential function, 426 exponents and, 17 Subject Index [ [ 1265 Base(s) (continued ) of a logarithmic function, 440 natural (e), 431–432 Basic properties of real numbers, 9–13 Basic rules of algebra, 9 Bearing, 531, 532 Bell-shaped curve, 477 Best fit line. See Linear regression Binomial(s), 27, 1133 cubing, 32 FOIL method for multiplying, 29–33 Binomial coefficients, 1133–1134, 1135 definition of, 1134 Pascal’s triangle and, 1136–1137 Binomial expansions finding a particular term of, 1138 Pascal’s triangle in, 1137 patterns of, 1133–1134 Binomial theorem, 1133–1138 binomial coefficients in, 1133–1134, 1135 binomial expansion and. See Binomial expansions Bisection method, for approximating real zeros, 382–383 Bounded graphs, 911 Bounded regions, 908 Brackets ([]) for grouping, 8 in inequalities, 129 Branches, of hyperbolas, 1028 Calculator use. See TI-83+/TI-84 calculators Cardioids, 847–848 Carrying capacity, 478 Cartesian coordinate system, 166, 538 Cartesian plane, 166, 543–548 quadrants of, 538–539 Center of a circle, 202, 203–204, 205 of an ellipse, 1015 of a hyperbola, 1028 Central angle, 567 Change-of-base formula, 460–461 Circles, 202–205 area of, formula for, 96 center of, 202, 203–204, 205 circumference of, formula for, 96 defined, 1000 equations of, transforming equations to standard form, 205 general form of equation of, 204 graphing, 202–204 radius of, 202, 203, 205 standard equation of, 202–204, 205 unit, 202, 584–589 Circular functions, 585–589 domains of, 588 even, 588, 589 odd, 588, 589 properties of, 587–589 ranges of, 588 translations of, 633–636 Circular sectors, area of, 573–575 Closed intervals, inequalities and, 129–130 Coefficients binomial, 1134 correlation (r), 216–219 definition of, 8 leading, of a polynomial, 27 leading, of a polynomial function, 332 of a monomial, 27 Cofactor of a square matrix, 977–978 definition of, 977 Cofunction(s), trigonometric, 515–516 Cofunction identities, 516, 680 Cofunction theorem, 516 Column index, of a matrix, 929 Column matrices, 930 Combinations, 1145–1146 definition of, 1146–1147 permutations distinguished from, 1143 Combined inequalities, 129. See also Double inequalities Combined variation, 315, 316 Combining like radicals, 63 Common difference, in arithmetic sequences, 1109–1110 Common logarithm(s) (base 10), properties of, 457 Common logarithmic function, 442 Common ratio, of a geometric sequence, 1117–1118 Commutative property of addition for matrices, 954 of real numbers, 9 Commutative property of multiplication, of real numbers, 9 Complement(s), of events, 1153 Complementary angles, 496 Completing the square quadratic equations and, 110–112 quadratic functions and, 336–337 transforming equations of circles to standard form by, 205 Complex conjugate(s), 72–73 Complex conjugate pairs, 389–392 Complex conjugate zeros theorem, 389–392 Complex numbers, 70–74 adding, 71 definition of, 71 dividing, 72–73 equality of, 71 imaginary unit i and, 70–71 modulus of, 821–822, 823 multiplying, 72 polar (trigonometric) form of, 822–825 powers of, 832–833 products of, 829–830 quotients of, 830–831 raising to integer powers, 73–74 real part of, 71 in rectangular form, 821–822 roots of, 833–837 subtracting, 71 Complex plane, 821 Complex rational expressions, 54–56 definition of, 54 simplifying, 54–56 Complex zeros in conjugate pairs, 389–392 factoring polynomials with, 390–391 Composite functions, applications involving, 289, 292–293 Composition of functions, 236, 289–293 Compounding, 434 Compound interest, 434–435 definition of, 97 Cones, double, 1000 Conic sections, 1000–1035. See also Circle; Circles; Ellipses; Hyperbolas; Parabolas alternative definition of, 1067–1069 angle of rotation necessary to transform a general second-degree equation into an equation of, 1060–1064 definitions of, 1000–1001 determining type of, 1001–1002 graphing by point-plotting, 1073 graphing from its equation, 1071–1072 names of, 1000 polar equations of, 1067–1074 rotated, graphing, 1063–1064 Conjugates complex, 72–73 of the denominator, 63–64 Constant(s), 8 of variation (proportionality), 312, 313, 314 Constant functions, 255–256, 259 Constant term, of a polynomials, 27 Constraints, in linear programming, 914 Consumer surplus, 909–911 Continuous compounding, 435 Continuous graphs, 351 Converging series, 1104 Corner points, in graphs, 911 Correlation coefficient (r), 216–219 computing using Excel 2010, 217–218 computing using TI-83+/TI-84 calculators, 216–217 Cosecant function graphing, 627–628, 631, 632–633 inverse, 725–727 Cosine(s), Law of, 780–786 Cosine function, 513 cofunction identity for, 680 graphing, 597–598 graphing sums of, 616, 628 inverse, 720–722 sum and difference identities for, 676–680 summary of, 598 Cotangent function graphing, 625, 628, 629, 630–631 inverse, 725–727, 728 Coterminal angles, 541–542 Cramer’s rule, 980–985 for solving systems of linear equations in three variables, 982–985 for solving systems of linear equations in two variables, 981–982 1266 Subject Index Cryptography, matrices in, 926 Cube(s) of binomials, 32 difference of, factoring, 39 sum of, factoring, 38 Cube function, 256 Cube root(s), 62 Cube root function, 257 Current velocity, 804 Curves, plane, 1078 Cycloids, 1080–1081 Damped harmonic motion, 609, 612–613 Decay, exponential, 474, 476–477 Decibels (dBs), 446–447 definition of, 447 Decimals approximation of, 6 converting between scientific notation and, 27–33 rational and irrational, 4, 5 Decreasing functions, 259–260 Degree of a monomial, 27 of a polynomial, 27 Degree measure, 494–496 Degrees (of angles), converting between radians and, 569–571 De Moivre’s theorem, 832–833 Denominators, 4 conjugates of, 63–64 least common, 12 rationalizing, 63–65 Dependent systems of equations, 866, 884–885 matrices and, 939, 940, 941 Dependent variables, 209, 240 Descartes’ rule of signs, 375, 378–379 Determinant of a matrix, 968 Cramer’s rule and, 980–985 definition of, 978 expanding by the first row, 978 finding, 978–980 of an n 3 n matrix, 977–980 of a 2 3 2 matrix, 976–977 Difference(s). See also Subtraction common, in arithmetic sequences, 1109–1110 of two cubes, factoring, 39 of two squares, factoring, 37–38 Difference function, 287–288 Difference identities, trigonometric, 675–685 Difference quotient, 246 calculating, 263 definition of, 263 Directed line segments, 798 Direction along a curve, 1078 of vectors, 798–801 Direction angle of a vector, 800–801 of vectors, 800–801 Directrix, of a parabola, 1003 Direct variation, 312–314 with powers, 313–314 Discriminants, 115 Distance formula, definition of, 168 Distance-rate-time problems, 100–102 graphing, 167–168 Distinguishable permutations, 1146–1147 Distributive property, of real numbers, 9, 10, 11 Diverging series, 1104 Dividend, 363 Division. See also Quotient(s) of complex numbers, 72–73 of functions, 287–289 long, of polynomials, 363–367 order of operations and, 6–9 of rational expressions, 49–51 synthetic, of polynomials, 367–369 Division algorithm, 366 Division sign (/), 6, 7 Divisor, 363 Domains of circular functions, 588 explicit, 247 of functions, 239, 247–249, 288–289, 291, 301–302 implicit, 247 of logarithmic functions, finding, 443–444 of rational expressions, 46–47 of rational functions, 396–397 of a relation, 238 Dot product, 802, 813–818 Double-angle identities, 689–694 applying, 691–693 derivation of, 690 Double cones, 1000 Double inequalities, 129 linear, solving, 133–134 Doubling time growth model, 432–434 Eccentricity alternative definition of conics and, 1067–1069 of ellipses, 1015, 1017 Elements of matrices, 929 of a set, 4 Elimination method Gaussian, with back substitution, 933–935 Gauss-Jordan, 935–938 solving systems of linear equations in three variables using, 934–937 solving systems of linear equations in two variables using, 869–872 solving systems of nonlinear equations using, 1040–1043 Ellipses, 1015–1023 applications involving, 1021–1022 centered at point (h,k), 1019–1021 centered at the origin, 1015–1019 center of, 1015 defined, 1000, 1015 eccentricity of, 1015, 1017 equations of, 1016–1017, 1019, 1020 foci of, 1015 major axis of, 1015 minor axis of, 1015 rotating, 1059–1060 vertices of, 1015 Ellipsis (punctuation), 4 Empty set, 4 e (natural base), 431–432 End behavior, of a polynomial function, 357 Endpoints in interval notation, 129 of a ray, 494 Equality of complex numbers, 71 of matrices, 950–952 Equal sign (5), 7 Equal vectors, 798–799, 801 Equations absolute value, 150–152 of a circle, 202–204, 205 of a conic, angle of rotation necessary to transform a general second-degree equation into, 1060–1064 definition of, 84 dependent systems of. See Dependent systems of equations of an ellipse, 1016–1017, 1019, 1020 equivalent, generating, 84 exponential. See Exponential equations first-degree, 85 functions defined by, 240–242 general form. See General form graphs of, 174–175 of a hyperbola, 1029–1030, 1033, 1034 inconsistent systems of. See Inconsistent systems of equations independent systems of, 866 linear. See Linear equations; Systems of linear equations in three variables; Systems of linear equations in two variables logarithmic. See Logarithmic equations matrix, 964–965 nonlinear, systems of. See Systems of nonlinear equations with no solution, 84 of a parabola, 1004, 1007 parametric. See Parametric equations polar, of conic sections, 1067–1074 polynomial. See Polynomial equations quadratic. See Quadratic equations quadratic in form, 123–125 radical, 121–123 rational. See Rational equations solution set of, 84 solutions (roots) of, 84 solving, 84 standard form. See Standard form trigonometric. See Trigonometric equations using complex roots, solving, 836–837 Subject Index 1267 1268 Subject Index Equilateral triangles, 497 Equilibrants, 807 Equilibrium point, 909 Equivalent equations, generating, 84 Evaluation of algebraic expressions, 8–9 Evaluation of functions difference quotient and, 246 exponential, 426–427 logarithmic, 440–442 with negatives, 245 with quotients, 245–246 by substitution, 243 with sums, 245 with variable arguments, 244 Even functions, 257–259, 588, 589 Even trigonometric identities, 666 Events complements of, 1153 definition of, 1151 independent, 1156–1157 mutually exclusive, 1154–1156 not occurring, probability of, 1153–1154 probability of, 1152–1153 union of, probability of, 1154–1156 Excel 2010 computing correlation coefficients using, 217–218 creating scatterplots using, 212–213 finding line of best fit using, 222–224 Experiments, 1151 Explicit domains, 247 Exponent(s) fractional, equations quadratic in form with, solving, 125 integer, 17–19 natural-number, 17 negative, equations quadratic in form with, solving, 124 properties of, 456 rational. See Rational exponents Exponential decay models, 474, 476–477 Exponential equations, 465–471 inverse properties and, 465 one-to-one properties and, 465 strategies for solving, 465 Exponential expressions, simplification of, 20–21 Exponential functions, 426–436 applications of, 432–435 definition of, 426 evaluating, 426–427 graphs of, 427–430 natural base e and, 431–432 Exponential growth models, 474, 475 Expressions algebraic, 8–9, 84 exponential, simplification of, 20–21 radical, simplified form of, 65 rational. See Rational expressions Extraneous solutions, 121, 122 definition of, 87 Factor(s). See also Partial-fraction decomposition linear, in partial fraction decomposition, 894–897 of polynomials, 36 Factorial(s), definition of, 1100 Factorial notation, 1100–1101 Factoring, 36 of expressions with rational exponents, 67 of polynomial functions, 377–378 of polynomials. See Factoring polynomials of quadratic equations, 107–109 solving equations with, 125–126 of trinomials, as product of two binomials, 39–42 Factoring polynomials, 36–44, 390–393 with complex zeros, 390–391 factor theorem for, 373–375 greatest common factor and, 36–37 by grouping, 42–43 over the integers, 36 real zeros and, 380–381 special forms of, 37–39 strategy for, 43 trinomials as product of two binomials and, 39–42 upper and lower bound rules and, 381 Factor theorem, 373–375 Feasible solutions, linear programming and, 914 Fibonacci sequence, 1101–1102 Finite arithmetic sequences, 1111–1114 application involving, 1113–1114 definition of, 1112 Finite geometric series, 1119–1120 Finite sequences, 1098. See also Finite arithmetic sequences Finite series, 1102, 1103–1104. See also Finite geometric series First-degree equations, 85. See also Linear equations Focus/foci of ellipses, 1015 of a parabola, 1003 FOIL method, for multiplying binomials, 29–30 Force vectors, 804 Formulas angle of rotation, 1061 for area of a circle, 96 for area of a rectangle, 96 for area of a triangle, 96 change-of-base, 460–461 distance, definition of, 168 geometric, 96 Heron’s, 793–794 midpoint, definition of, 169 or circumference of a circle, 96 partial-sum, proving with mathematical induction, 1130 for perimeter of a rectangle, 96 for perimeter of a triangle, 96 quadratic, 112–115 recursion, 1101–1102 rotation of axes, 1057–1060 458 angles evaluating trigonometric functions exactly for, 518–519 in standard position, 540 458-458-908 triangles, 499–500 Four-leaved rose, 847 Fraction(s) linear inequalities with, solving, 132–133 partial, 893. See also Partial fraction decomposition properties of, 12–13 Fractional exponents, equations quadratic in form with, solving, 125 Function(s), 238–249 absolute value, 257 adding, 287–288 applications involving, 248–249 argument of, 243 average rate of change and, 260–263 common logarithmic, 442 composite, 289, 292–293 composition of, 236, 289–293 constant, 255–256, 259 cube, 256 cube root, 257 decreasing, 259–260 defined by equations, 240–242 definition of, 238–239 difference, 287–288 dividing, 287–289 domain of, 239, 247–249, 288–289, 291, 301–302 evaluating. See Evaluation of functions even, 257–259 exponential. See Exponential functions expressing, 242 function notation and, 242–246 graphing, 243–244 greatest integer, 266 identity, 256 increasing, 259–260 inverse. See Inverse functions linear, 255 logarithmic. See Logarithmic functions multiplying, 287–288 objective, in linear programming, 914–916 odd, 257–259 one-to-one, 298–300 periodic, 594 piecewise-defined. See Piecewise-defined functions polynomial. See Polynomial functions power. See Power functions product, 287–288 quadratic. See Quadratic functions quotient, 287–289 range of, 239 rational. See Rational functions reciprocal, 257 relations and, 238–240 sinusoidal. See Sinusoidal functions Subject Index 1269 square, 256 square root. See Square root function step, 266 subtracting, 287–288 sum, 287–288 trigonometric. See Trigonometric functions vertical line test for, 241–242 Function notation, 242–246, 288 Fundamental counting theorem, 1141–1143 Fundamental period, 594 Fundamental theorem of algebra, 388–389 Gaussian distribution models, 474, 477–478 Gaussian elimination, with back substitution, 933–935 Gauss-Jordan elimination, 935–938 General form of the equation of a circle, 204 of the equation of a quadratic function, 336–339 of the equation of a straight line, 186 General terms of an arithmetic sequence, 1110–1111 of a geometric sequence, 1118–1119 of a sequence, 1098–1099 Geometric formula, 96 Geometric sequences, 1117–1119 common ratio of, 1117–1118 definition of, 1117 general term of, 1118–1119 Geometric series, 1119–1124 applications involving, 1123–1124 finite, 1119–1120 infinite, 1121–1123 Geometry problems, 95–97. See also Applications index Geostationary satellites, 1021 Geosynchronous (GEO) orbit, 1021 Graphs/graphing, 164–231 asymptotes of rational functions and, 398–403, 406–408 best fit line and. See Linear regression bounded, 911 Cartesian coordinate system and, 166 of a circle, 202–204 compression of, vertical and horizontal, 280–282 of a conic from its equation, 1071–1072 continuous, 351 of curves defined by parametric equations, 1079–1080 of a cycloid, 1081 distance between two points and, 167–168 of ellipses, 1017–1018, 1020–1021 of an equation, 174–175 of a function, 243–244 of a hyperbola, 1031–1032, 1033 intercepts and, 176–177, 180–181, 186–188 of inverse functions, 303–304 of a line, 186–197 of linear equations, 190–197 of linear inequalities, 130–131, 904–905 linear regression and. See Linear regression of logarithmic functions, 443–446 with logarithmic scales, 449–450 midpoint of a line segment joining two points and, 168–169 of parabolas with vertex at (h,k), 1007–1009 of parabolas with vertex at the origin, 1005–1007 of piecewise functions, 264–266 point-plotting and, 166, 173–175 by point-plotting in polar coordinates, 1073 point-slope form and, 192–193 of polar equations, 843–850 of polynomial functions, 351–352, 355–359, 383–384 of polynomial inequalities, 139 of projectile motion, 1081–1083 of quadratic functions, 332–343 of rational functions, 404–409 reflection about the axes in, 278–280 of a rotated conic, 1063–1064 of scatterplots. See Scatterplots of shifted sinusoidal functions, 606–608 shifts of, horizontal and vertical, 274–278 of sinusoidal functions, 594–606 slope and, 188–190 slope-intercept form and, 190–191 smooth, 351 solving systems of linear equations in two variables using, 873–875 stretching of, vertical and horizontal, 280–282 of sums of functions, 614–616 symmetry and, 177–181 of systems of linear inequalities in two variables, 906–908 of systems of nonlinear inequalities, 1051–1052 transformations of, 274–283 unbounded, 911 Greatest common factor (GCF), 36–37 Greatest integer function, 266 Grouping factoring a polynomial by, 42–43 factoring by, solving polynomial equations using, 125–126 notations for, 8 Growth, logistic, 474, 478 Growth models doubling time, 432–434 exponential, 474, 475 logistic, 474, 478 Half-angle identities, 697–704 applying, 700–704 derivation of, 698–699 Half-life, in exponential decay, 476 Half open intervals, inequalities and, 129–130 Half-planes, 903 Harmonic motion, 609–614 damped, 609, 612–613 examples of, 610 resonance and, 609, 614 simple, 609, 610–612 Heading, 531 Heaviside step function, 266 Heron’s formula, 793–794 Horizontal asymptotes, of rational functions, 398, 400–403, 406–407 Horizontal components, of vectors, 803 Horizontal line(s), finding equation of, 193 Horizontal line test, 299–300 definition of, 299 Horizontal shifts graphing exponential functions using, 429–430 graphing logarithmic functions using, 445 Hyperbolas, 1028–1035 applications involving, 1034 branches of, 1028 centered at point (h,k), 1032–1034 centered at the origin, 1028–1032 center of, 1028 defined, 1000 equation of, 1029–1030, 1033, 1034 graphs of, 1031–1032, 1033 transverse axis of, 1028 vertices of, 1028 Hypotenuse, of a right triangle, 497 Identities cofunctions, 516 defined, 654 definition of, 84 reciprocal, 513–515 trigonometric. See Trigonometric identities Identity function, 256 Imaginary axis, 821 Imaginary numbers, 70 pure, 71 Imaginary unit (i), of complex numbers, 70–71 Immediate value theorem, definition of, 354 Implicit domains, 247 Improper rational expressions, 893 Improper rational functions, 401 Inconsistent systems of equations, 866, 885–886 matrices and, 939 Increasing functions, 259–260 Independent events, 1156–1157 Independent systems of equations, 866 Independent variables, 209, 240 Index(es) of a root, 62 row and column, of a matrix, 929 of summation, 1102–1103 Inequalities absolute value. See Absolute value inequalities double (combined). See Double inequalities equations compared with, 129 linear. See Linear inequalities; Systems of linear inequalities in two variables nonlinear. See Nonlinear inequalities; Systems of nonlinear inequalities 1270 Subject Index Inequalities (continued ) polynomial, 139–143 properties of, 132 quadratic, 140–143 rational, 144–146 strict, 129 Infinite geometric series, 1121–1123 Infinite sequences, 1098 Infinite series, 1102, 1104 Infinity (∞), 130 Initial point, of a vector, 798 Initial ray (side) of an angle, 494 Inspection, solving trigonometric equations by, 737–740 Integer(s), 4, 5 Integer exponents, 17–19 complex numbers and, 73–74 properties of, 19–21 Intercepts as graphing aid, 180–181 in graphs, 176–177, 180–181, 186–188 slope-intercept form and, 190–191 Interest. See also Applications index compound, 97, 434–435 simple, 97–98 solving problems involving, 97–99, 434–435 Interest rates, 434 Intermediate value theorem, 382–383 Intersection of sets, 131 Interval(s) closed and open, 129–130 half open, 129–130 test, on real number line, 139 Interval notation, linear inequalities and, 129–131 Inverse(s) additive, 9 additive, for a matrix, 954 multiplicative, 9 of square matrices. See Inverses of square matrices Inverse functions, 301–307 definition of, 301 finding, 304–307 graphical interpretation of, 303–304 trigonometric equations requiring the use of, solving, 742–744 Inverse properties, exponential equations and, 465 Inverses of square matrices, 926, 965–970 applications involving, 972 definition of, 966 solving systems of linear equations using, 970–972 Inverse trigonometric functions, 716–732 finding exact values for expressions involving, 728–731 Inverse variation, 314–315 Irrational numbers, 4, 5 Irreducible quadratic factors, in partial fraction decomposition, 897–899 Isosceles triangles, 497 Joint variation, 315, 316 Latus rectum, 1005–1007 Law of Cosines, 780–786 Law of Sines, 769–771 Leading coefficient of a polynomial, 27 of a polynomial function, 332 Least common denominator (LCD), 12 adding and subtracting rational expressions using, 53 Least common multiple (LCM), 12 Legs, of a right triangle, 497 Like radicals, combining, 63 Like terms, combining, 28 Limaçons, 847 Line(s) best fit. See Linear regression defined, 494 equations of. See Linear equations graphing, 186–197 horizontal, finding equation of, 193 joining two points, midpoint of, 168–169 parallel, 194 parametric representation of, 885, 886 perpendicular, 195 slope of, 188–190 Linear equations, 84–90, 190–197 applications involving, 93–103, 195–197 definition of, 85 general form of, 186 in one variable, solving, 84–86 parallel, 194–195 perpendicular, 194–195 point-slope form of, 192–193 rational equations reducible to, solving, 87–89 slope-intercept form of, 190–191 solving using mathematical models, 93–95 systems of. See Systems of linear equations in three variables; Systems of linear equations in two variables of a vertical line, finding, 193 Linear factors, in partial fraction decomposition, 894–897 Linear functions, 255 Linear inequalities, 129–135 graphing, 130–131 interval notation for, 129–131 solving, 131–135 systems of. See Systems of linear inequalities in two variables in two variables. See Linear inequalities in two variables; Systems of linear inequalities in two variables Linear inequalities in two variables, 903–905 graphing, 904–905 systems of. See Systems of linear inequalities in two variables Linearity, in scatterplots, 214–216 Linear programming, 914–918 constraints and, 914 feasible solutions and, 914 objective function and, 914–916 solving an optimization problem using, 914–918 Linear regression, 219–225 determining best fit line and, 219–224 prediction using best fit line and, 224–225 Linear relationships, 214–215 strength of, 216–219 Linear speed, 575–576, 577–578 Line segments, 494 directed, 798 Logarithm(s), 440 applications of, 446–450 change-of-base formula and, 460–461 common (base 10), properties of, 457 evaluating, 440–442 finding value of, 441–442 natural (base e), properties of, 457 properties of, 456–460 Logarithmic equations, 468–471 applications of, 469–471 strategy for solving, 468 Logarithmic functions, 440–450 common, 442 definition of, 440 exponential form of, 440–441 finding domains of, 443–444 graphs of, 443–446 logarithmic form of, 440–441 natural, 442 properties of, 456–460 Logarithmic models, 474, 479–480 Logarithmic scales, 448–450 graphing using, 449–450 Logistic growth models, 474, 478 Long division, of polynomials, 363–367 Loran, 1034 Lower/ upper bound rules, for real zeros, 381 Lowest terms rational functions in, 399 reducing to. See Simplification Magnitude, of vectors, 798–801 Main diagonal entries, of matrices, 929 Major axis, of an ellipse, 1015 Mathematical induction, 1128–1130 principle pf, 1128 Matrix(ces), 926–992 applications involving, 959, 972 augmented, 930–931 column, 930 column index of, 929 in cryptography, 926 dependent systems of equations and, 939, 940, 941 determinants of. See Determinant of a matrix elements of, 929 equality of, 950–952 finding the order of, 929 Gaussian elimination with back substitution and, 933–935 Gauss-Jordan elimination and, 935–938 Subject Index 1271 inconsistent systems of equations and, 939 inverses of, 926, 965–970 main diagonal entries of, 929 nonsingular, 967 reduced row-echelon form of, 932 row-echelon form of, 932–933 row index of, 929, 930 row operations on, 931–932 singular, 967 square. See Inverses of square matrices; Square matrices zero, 953–954 Matrix addition, 952–954 definition of, 952 properties of, 954 Matrix algebra, 951 solving systems of linear equations using, 970–972 Matrix equations, 964–965 Matrix multiplication, 955–959 application involving, 959 definition of, 956 by inverse, 966 properties of, 958–959 Matrix subtraction, 952–953 definition of, 952 Members, of a set, 4 Midpoint, of a line segment joining two points, 168–169 Midpoint formula, definition of, 169 Minor axis, of an ellipse, 1015 Minor of a square matrix, 977–978 definition of, 977 Mixture(s), definition of, 99 Mixture problems, 97 Modeling of doubling time growth, 432–434 of exponential decay, 474, 476–477 exponential decay models and, 474, 476–477 of exponential growth, 474, 475 exponential growth models and, 474, 475 Gaussian (normal) distribution models and, 474, 477–478 logarithmic models and, 474, 479–480 logistic growth models and, 474, 478 normal distribution models and, 474, 477–478 solving application problems using, 93–95 with a system of three linear equations, 887–888 with systems of linear equations in three variables, 887–888 using variation, 312–317 Modulus, of a complex number, 821–822, 923 Monomials, 27 coefficient of, 27 degree of, 27 multiplying with a polynomial, 29 Motion, projectile, 1081–1083 Multiplication. See also Product(s) associative property of, 9 of binomials, FOIL method for, 29–33 commutative property of, 9 of complex numbers, 72 of functions, 287–288 of matrices. See Matrix multiplication order of operations and, 6–9 of polynomials, 28–33 of rational expressions, 49–51 scalar, of vectors, 802 of vectors, 813–818 Multiplication sign (?), 6, 7 Multiplicative identity matrix, 965–966 Multiplicative identity property, of real numbers, 9 Multiplicative inverse, 9 Multiplicative inverse property, of real numbers, 9 Multiplicity of a zero, 354–355, 357 definition of, 354 Mutually exclusive events, 1154–1156 Natural base (e), 431–432 Natural exponential function, 431 Natural logarithm(s) (base e), properties of, 457 Natural logarithmic function, 442 Natural number(s), 4, 5 Natural-number exponents, 17 Negative(s) evaluating functions with, 245 properties of, 10–11 Negative angles, 494 Negative association, in a scatterplot, 213, 214 Negative exponents, equations quadratic in form with, solving, 124 Negative infinity (2∞), 130 Negative-integer exponent property, 18 Nonacute angles calculating trigonometric function values for, 545–547 trigonometric functions of, 552–563 Nondistinguishable permutations, 1146–1147 Nonlinear equations, systems of. See Systems of nonlinear equations Nonlinear inequalities systems of, 1051–1054 in two variables, 1049–1051 Nonlinear relationships, 214, 215 Nonsingular matrices, 967 Normal distribution models, 474, 477–478 Notation factorial, 1100–1101 function, 242–246, 288 interval. See Interval notation scientific, 21–23 sigma (S) (summation), 1102–1103 nth partial sum of an arithmetic sequence, 1111–1114 definition of, 1102 nth roots, 62–65 principal, 62 nth root theorem, 833–834 Null set, 4 Number(s) complex. See Complex numbers imaginary. See Imaginary numbers irrational, 4, 5 natural, 4, 5 rational, 4, 5. See also Fraction(s) real. See Real number(s) sets and subsets of, 4 whole, 4, 5 Number line. See Real number line Numerators, 4 n zeros theorem, 388–389 Objective functions, in linear programming, 914–916 Oblique triangles acute, 766 Law of Cosines to solve, 780–786 Law of Sines to solve, 769–771 obtuse, 766 solving, 766–776 Obtuse angles, 495 Obtuse triangles, 766 Odd functions, 257–259, 588, 589 Odd trigonometric identities, 666 One-to-one functions, 298–300 One-to-one properties, exponential equations and, 465 Open intervals, inequalities and, 129–130 Operations. See also Addition; Division; Multiplication; Subtraction order of, 6–9 Optimization problem, solving using linear programming, 914–918 Order of matrices, finding, 929 of operations, 6–9 Ordered pairs, 166 Ordinates, 166 addition of, 614–616 Orientation, along a curve, 1078 Origin of a graph, 166 in polar coordinate system, 841 Orthogonal vectors, 815–816 Outcomes, 1151 Parabolas, 330, 332–343, 1003–1011 applications involving, 1010–1011 axis of symmetry of, 1003 defined, 1000, 1003 directrix of, 1003 equation of, 1004, 1007 finding the equation of, 340–343 focus of, 1003 with vertex at the origin, 1003–1007 with vertex at the point (h,k), 1007–1009 vertex of, 1003 Parallel lines, definition of, 194 Parallel vectors, 815 Parameter(s), defined, 1078 1272 Subject Index Parametric equations, 1078–1083 applications of, 1080–1083 defined, 1078 graphing curves defined by, 1079–1080 Parametric representation, of a line, 885, 886 Parentheses for grouping, 8 in inequalities, 129 Partial fraction(s), 893 Partial fraction decomposition, 892–901 with combinations of factors, 900 with distinct irreducible quadratic factors, 897–898 with distinct linear factors, 894–895 with repeated irreducible quadratic factors, 899 with repeated linear factors, 895–897 Partial-sum formulas, proving with mathematical induction, 1130 Pascal’s triangle, 1136–1137 Perfect squares, 111 factoring, 38 square roots of, 60–61 Periodic functions, 594 of sinusoidal functions, 602 Permutations, 1143–1145 combinations distinguished from, 1143 definition of, 1143 distinguishable, 1146–1147 with repetition (nondistinguishable), 1146–1147 Perpendicular lines, definition of, 195 Perpendicular vectors, 815–816 Piecewise-defined functions, 263–266 finding inverse of, 307 Plane Cartesian, 166 division into half-planes, 903 graph of linear equation as, 882 Plane curves, 1078 Point-plotting, 166, 173–175 graphic conics by, 1073 of polar coordinates, 841–842 Point-slope form, of linear equations, 192–193 Polar axis, 841 Polar coordinates, 841–842 converting between rectangular coordinates and, 842–843 equations of conic in, 1067–1074 point-plotting, 841–842 Polar equations converting between rectangular form equations and, 849–850 graphs of, 843–850 Polar form, of a complex number, 822–825 Pole, in polar coordinate system, 841 Polynomial(s), 27–33. See also Complex rational expressions; Rational expressions adding, 28 complex zeros of. See Complex zeros degree of, 27 dividing, 363–367 factored completely, 36 factoring. See Factoring polynomials factors of, 36 greatest common factor of, 36–37 multiplying, 28–33 prime, 36, 42 real zeros of. See Real zeros special products of, 29–33 in standard form, 27 subtracting, 28 synthetic division of, 367–369 zeros of, 139. See also Complex zeros; Real zeros Polynomial equations, factoring by grouping to solve, 125–126 Polynomial functions, 349–359 definition of, 332, 349 factoring, 377–378 graphing, 351–352, 355–359, 383–384 identifying, 349–351 leading coefficient of, 332 quadratic. See Quadratic functions ratios of. See Rational functions real zeros of, 353–355 Polynomial inequalities, 139–143 Position vectors, 799 Positive angles, 494 Positive association, in a scatterplot, 213 Power(s). See also Exponent(s) of complex numbers, 832–833 direct variation with, 313–314 Power functions characteristics of, 352 definition of, 351 graphing polynomial functions using transformations of, 351–352 Power property, of integer exponents, 19 Prediction, using line of best fit, 224–225 Predictor variables, 209, 240 Prime polynomials, 36, 42 Principal (investment), 434 Principal nth root, 62 Principal square root, 60, 70, 109 Probability, 1151–1157 of an event, 1152–1153 of an event not occurring, 1153–1154 independent events and, 1156–1157 mutually exclusive events and, 1154–1156 sample space and, 1151 Producer surplus, 909–911 Product(s). See also Multiplication of complex numbers, 829–830 dot, 802, 813–818 special, 29–33 Product function, 287–288 Product property, of integer exponents, 19 Product to a power property, of integer exponents, 19 Product-to-sum identities, 708–709 Projectile motion, 1081–1083 Proper rational expressions, 893 Proper rational functions, 401 Proportionality. See Variation Pure imaginary numbers, 71 Pythagorean identities, trigonometric, 657–660, 666 Pythagorean theorem, 498–499 Quadrant(s), of Cartesian plane, 166, 538–539 Quadrantal angles, 539 calculating trigonometric function values for, 547–548 Quadratic equations, 107–109 absolute value, 152 applications involving, 115–116 completing the square to solve, 110–112 factoring, 107–109 quadratic formula and, 112–115 square root method to solve, 109–110 Quadratic factors, irreducible, in partial fraction decomposition, 897–899 Quadratic formula, 112–115 Quadratic functions, 330, 332–343 application problems involving, 340–343 axis of symmetry of, 333 definition of, 333 in general form, graphing, 336–339 graphs of, 333–343 in standard form, graphing, 333–336 vertex of, 333 Quadratic inequalities, 140–143 Quadratic in form equations, 123–125 Quotient(s), 363. See also Division of complex numbers, 830–831 difference. See Difference quotient evaluating functions with, 245–246 Quotient function, 287–289 Quotient identities, trigonometric, 656–657, 666 Quotient property, of integer exponents, 19 Quotient to a power property, of integer exponents, 19 Radian(s), 567 converting between degrees and, 569–571 Radian measure, of an angle, 567–569 Radical(s), 62 like, combining, 63 properties of, 62 Pythagorean theorem with, 499 simplifying, 62 Radical equations, 121–123 Radical expressions, simplified form of, 65 Radical sign A" B, 60 Radicand, 60 Radius, of a circle, 202, 203, 205 Ranges of circular functions, 588 of a function, 239 of a relation, 238 of trigonometric functions, 555–556 Rate of change average, 260–263 slope as, 195–196 Subject Index 1273 Ratio(s) right triangle. See Right triangle ratios setting up trigonometric functions as, 515 trigonometric, 512. See also Trigonometric functions Rational equations definition of, 87 reducible to linear equations, solving, 87–89 Rational exponents, 66–67 equations with, solving by factoring, 125 Rational expressions, 46–53 adding, 51–53 complex. See Complex rational expressions dividing, 49–51 domains of, 46–47 improper, 893 multiplying, 49–51 proper, 893 simplifying, 48–49 subtracting, 51–53 Rational functions, 396–410 asymptotes of, 398–403 definition of, 396 domain of, 396–397 graphing, 404–409 improper, 401 proper, 401 Rational inequalities, 144–146 Rationalizing denominators, 63–65 Rational numbers, 4, 5. See also Fraction(s) Rational root test, 375–378 Rational zero theorem, 375–378 Rays, 494 Real axis, 821 Real number(s) properties of, 9–13 set of, 4–5, 84 trichotomy property of, 129 Real number line absolute value and, 11 definition of, 5 distance between two points on, 150 test intervals on, 139 Real part, of a complex number, 71 Real zeros, 372–385 approximating, 382–383 Descartes’ rule of signs and, 375, 378–379 factoring polynomials and, 380–381 factor theorem and, 373–375 finding, 375 graphing polynomial functions and, 383–384 intermediate value theorem and, 382–383 number of, 375 of a polynomial function, 353–355 rational zero theorem and, 375–378 remainder theorem and, 372–373 upper and lower bound rules for, 381 Reciprocal(s), 9, 654 Reciprocal function, 257 Reciprocal identities, 513–515 trigonometric, 654–656, 665 Rectangles area of, formula for, 96 perimeter of, formula for, 96 Rectangular coordinates, converting between polar coordinates and, 842–843 Rectangular coordinate system, 166, 538, 821 Rectangular form, of complex numbers, 821–822 Rectangular form equations, converting between polar equations and, 849–850 Recursion formulas, 1101–1102 Reduced row-echelon form, of a matrix, 932 Reducing to lowest terms. See Simplification Reference angles, 556–557, 558 Reference right triangles, 556, 557 Relations definition of, 238 domain of, 238 functions and, 238–240 range of, 238 Remainder(s), 363 Remainder theorem, 372–373 Repetition, permutations with, 1146–1147 Resonance, harmonic motion and, 609, 614 Response variables, 209, 240 Resultant vectors, 804–808 Resultant velocity, 804 Richter scale, definition of, 448 Right angles, 495 Right triangle(s), 497 458-458-908, 499–500 hypotenuse of, 497 legs of, 497 Pythagorean theorem and, 498–499 reference, 556, 557 solving, 525–532 308-608-908, 501–502 Right triangle ratios, 511–520 cofunctions and, 515–516 evaluating trigonometric functions exactly for, 517–519 reciprocal identities and, 513–515 Roots of complex numbers, 833–837 cube, 62 of equations, 84 square. See Square root(s) Rose, 847 Rotation of axes formulas, 1057–1060 Rounding, of decimals, 6 Row-echelon form, of a matrix, 932–933 Row index, of a matrix, 929 Row matrices, 930 Row operations, on matrices, 931–932 Sample space, 1151 Scalar(s), 798 Scalar multiplication, 954–955 definition of, 954 of vectors, 802 Scatterplots, 209–225 creating using Excel 2010, 212–213 creating using TI-83+ or TI-84 calculators, 211 linear regression and. See Linear regression patterns in, 213–219 Scientific notation, 21–23 Secant function graphing, 626–627, 628, 631, 632 inverse, 725–727 Sectors, circular, area of, 573–575 Sequences, 1098–1102 alternating, 1099 arithmetic. See Arithmetic sequences; Finite arithmetic sequences definition of, 1098 factorials and, 1101–1102 Fibonacci, 1101–1102 finite and infinite, 1098. See also Finite arithmetic sequences geometric. See Geometric sequences terms of, 1098–1099 Series, 1102–1104 application involving, 1104 converging and diverging, 1104 definition of, 1102 finite and infinite, 1102, 1103–1104. See also Finite geometric series geometric. See Geometric series sigma (summation) notation and, 1102–1103 Sets, 4 empty (null), 4 intersection of, 131 of real numbers, 4–5, 84 union of, 131 Shifts graphing exponential functions using, 429–430 graphing logarithmic functions using, 445 Sigma (S) notation, 1102–1103 Sign(s), algebraic, of trigonometric functions, 552–554 Significant digits, 525–526 Similar triangles, applications involving, 505 Simple harmonic motion, 609, 610–612 Simple interest, 97–98 definition of, 97 Simplification of a complex rational expression, 54–56 of exponential expressions, 20–21 of expressions with rational exponents, 66–67 order of operations and, 7–8 of radical expressions, 65 of radicals using imaginary numbers, 70 of rational expressions, 48–49, 54–56 of square roots, 61 Simplified form, of radical expressions, 65 Sine(s), Law of, 769–771 Sine function, 513 cofunction identity for, 680 graphing, 594–596, 615–616, 628 inverse, 716–719 sum and difference identities for, 680–683 summary of, 596 1274 Subject Index Singular matrices, 967 Sinusoidal functions amplitude and period of graphs of, 599–600 graphs of, 594–606 shifted, graphing, 606–608 608 angles evaluating trigonometric functions exactly for, 519 in standard position, 540 Slant asymptotes, of rational functions, 403, 408 Slope of a line, 188–190 as rate of change, 195–196 of the secant line, 260–262 Slope-intercept form, of linear equations, 190–191 Smooth graphs, 351 Software. See Excel 2010 Solutions of equations, 84 extraneous, 87, 121, 122 feasible, linear programming and, 914 Solution sets of equations, 84 of a system of inequalities, 905 Special products, 29–33 Speed angular, 576–577, 578 defined, 575 linear, 575–576, 577–578 Spirals, 848 Square(s) completing, quadratic equations and, 110–112 difference of, factoring, 37–38 perfect. See Perfect squares in special products, 31 Square function, 256 Square matrices, 929 Cramer’s rule and, 980–985 inverse of, finding, 967–970 inverses of. See Inverses of square matrices Square root(s), 60–61 principal, 60, 70, 109 properties of, 61 simplifying, 61 Square root function, 257 inverse of, 305–307 Square root method, quadratic equations and, 109–110 Square root property, 109–110 Standard form of a complex number, 71 of complex numbers, 821–822 of equation of a hyperbola, 1029–1030, 1034 of equation of a circle, 202–204, 205 of equation of a straight line, 186 of polynomials, 27 quadratic functions in, graphing, 333–336 Standard position, angles in, 538–541 Step functions, 266 Straight angles, 495 Stretching, vertical and horizontal, of graphs, 280–282 Strict inequalities, 129 Subsets, 4 Substitution evaluating functions by, 243 solving systems of linear equations in two variables using, 867–869 solving systems of nonlinear equations using, 1044–1045 u-substitution, 123–125 Substitution principle, evaluating algebraic expressions using, 8 Subtraction. See also Difference(s) of complex numbers, 71 of functions, 287–288 of matrices, 952–953 order of operations and, 6–9 of polynomials, 28 of rational expressions, 51–53 of real numbers, 9–10 Sum(s). See also Addition of an arithmetic sequence, 1111–1114 evaluating functions with, 245 of an infinite geometric series, 1121 nth partial sum, 1102 of two cubes, factoring, 38 Sum function, 287–288 Sum identities, trigonometric, 675–685 Summation notation, 1102–1103 Sum-to-product identities, 710–712 Supplementary angles, 496 Surplus, consumer and producer, 909–911 Symmetry of graphs, 177–179 as graphing aid, 180–181 tests for, 178–179 types of, 178 Synthetic division, of polynomials, 367–369 Systems of linear equations Cramer’s rule for solving, 981–985 in four variables, Gauss-Jordan elimination to solve, 937–938 matrices to write, 930 solving using matrix algebra and inverses of square matrices, 970–972 in three variables. See Systems of linear equations in three variables in two variables. See Systems of linear equations in two variables writing as matrix equations, 964–965 Systems of linear equations in three variables, 881–888 combining elimination and substitution methods to solve, 882–883 Cramer’s rule for solving, 982–985 dependent, 884–885 Gauss-Jordan elimination to solve, 936–937 inconsistent, 885–886 modeling with, 887–888 solving, 881–884 Systems of linear equations in two variables, 866–878 applications involving, 876–877 Cramer’s rule for solving, 981–982 dependent, 866 elimination method for solving, 869–872 Gaussian elimination with back-substitution to solve, 934 graphing method for solving, 873–875 identifying method to use for solving, 875–876 inconsistent, 866 independent, 866 solving, 866–875 substitution method for solving, 867–869 Systems of linear inequalities in two variables, 905–911 applications involving, 909–911 graphing, 906–908 solution set of, 905 Systems of nonlinear equations, 1039–1047 applications involving, 1045–1046 solving using elimination, 1040–1043 solving using substitution, 1044–1045 Systems of nonlinear inequalities, 1051–1054 graphs of, 1051–1052 solving, 1052–1054 Tangent function, 513 graphing, 623–625, 628, 629–630 inverse, 723–725 sum and difference identities for, 683–685 Technology use. See Excel 2010; TI-83+/ TI-84 calculators Term(s) of an algebraic expression, 8 of a binomial expansion, finding, 1138 constant, of a polynomial, 27 general. See General terms like, combining, 28 lowest. See Lowest terms; Simplification; Substitution of a polynomial, 27 of a sequence, 1098–1099 Terminal point, of a vector, 798 Terminal ray (side) of an angle, 494 Test intervals, on real number line, 139 30° angles evaluating trigonometric functions exactly for, 517–518, 519 in standard position, 540 30°-60°-90° triangles, 501–502 TI-83+/TI-84 calculators computing correlation coefficients using, 216–217 creating scatterplots using, 211 finding line of best fit using, 221–222 Transformations graphing logarithmic functions using, 446 of power functions, graphing polynomial functions using, 351–352 Transverse axis, of a hyperbola, 1028 Triangles, 497–505 angle sum of, 497 area of, 791–794 area of, formula for, 96 congruent, 503 equilateral, 497 finding angles of, 497 isosceles, 497 oblique. See Oblique triangles perimeter of, formula for, 96 right. See Right triangle(s); Right triangle ratios similar, 502–505 Trichotomy property, of real numbers, 129 Trigonometric equations, 737–749 applications involving, 747–748 linear, solving, 741 quadratic, solving, 741–742 requiring the use of inverse functions, solving, 742–744 solving by inspection, 737–740 solving using algebraic techniques, 741–742 solving using trigonometric identities, 744–748 Trigonometric form, of a complex number, 822–825 Trigonometric functions, 492–651. See also Cosine function; Sine function; Tangent function algebraic signs of, 552–554 angles and degree measure and, 494–496 angles in standard position and, 538–541 angular speed and, 576–578 arc length and, 572–573 area of a circular sector and, 573–575 cartesian plane and, 543–548 cofunctions and, 515–516 coterminal angles and, 541–542 evaluating exactly for special angles measures, 517–519 evaluating using calculators, 519–520 graphing, 623–636 graphs of sinusoidal functions and, 594–608 harmonic motion and, 609–614 inverse, 716–732 linear speed and, 575–576, 577–578 of nonacute angles, 552–563 radian measure and, 567–579 ranges of, 555–556 reciprocal identities and, 513–515 reference angles and reference right triangles and, 556–559 right triangle ratios and, 511–513 setting up as ratios, 515 solving right triangles and, 525–532 sums, of graphing, 614–616 translations of, 633–636 triangles and, 497–505 unit circle approach and, 584–589 Trigonometric identities, 654–661 basic, 665–666 cofunction, 680 double-angle, 689–694 even-odd, 666 half-angle, 697–704 product-to-sum, 708–709 Pythagorean, 657–660, 666 quotient, 656–657, 666 reciprocal, 654–656, 665 solving trigonometric equations using, 744–748 sum and difference, 675–685 sum-to-product, 710–712 verifying, 668–671 Trigonometric ratios, 512. See also Trigonometric functions Trinomials, 27 factoring as product of two binomials, 39–42 Truncation, of decimals, 6 Unbounded graphs, 911 Union of sets, 131 of events, probability of, 1154–1156 Unit circle, 202, 584–589 Unit step function, 266 Unit vectors, 803–804 Upper bound rule, for real zeros, 381 Upper/lower bound rules, for real zeros, 381 u-substitution, 123–125 Variables, 8 dependent (response), 209, 240 independent (predictor), 209, 240 Variation, 312–317 combined, 315, 316 constant of, 312, 313, 314 direct, 312–314 inverse, 314–315 joint, 315, 316 Vectors, 798–808 addition of, 799, 801–802 algebraic interpretation of, 799–800 algebraic properties of, 802 angle between, 814–818 components of, 799 direction angle of, 800–801 direction of, 798–801 equal, 798–799, 801 force, 804 geometric interpretation of, 798–799 horizontal and vertical components of, 803 magnitude of, 798–801 multiplication of, 813–818 parallel, 815 perpendicular (orthogonal), 815–816 position, 799 resultant, 804–808 scalar multiplication of, 802 in standard position, 799 unit, 803–804 velocity, 804 work problems and, 816–818 Velocity actual, 804 apparent, 804 current, 804 defined, 575 resultant, 804 Velocity vectors, 804 Vertex(ices) of an ellipse, 1015 in graphs, 911 of a hyperbola, 1028 of a parabola, 1003 of a quadratic function, 333 Vertical asymptotes, of rational functions, 398, 399–400 Vertical components, of vectors, 803 Vertical line(s), finding equation of, 193 Vertical line test, 241–242 Vertical shifts graphing exponential functions using, 429–430 graphing logarithmic functions using, 445 Whole numbers, 4, 5 Word problems, procedure for solving, 93–95 Work problems, vectors and, 816–818 x-axis, 166 x-coordinate, 166 x-intercepts, 176–177, 186–188 y-axis, 166 y-coordinate, 166 y-intercepts, 176–177, 186–188 Zero, properties of, 12 Zero(s) complex, of a polynomial. See Complex zeros multiplicity of. See Multiplicity of a zero of a polynomial, 139. See also Complex zeros; Real zeros real. See Real zeros Zero-exponent property, 18–19 Zero matrices, 953–954 Zero product property, 12 factoring quadratic equations and, 107–109 Zero row, in Pascal’s triangle, 1137 Subject Index 1275 Exponents and Radicals a0 5 1, a 2 0 ax ay 5 ax2y aa bb x 5 ax bx " n am 5 am/n 5 a" n ab m a2x 5 1 ax 1ax2y 5 axy "a 5 a1/2 " n ab 5 " n a" n b axay 5 ax1y 1ab2x 5 axbx " n a 5 a1/n Å n aa bb 5 " n a " n b Absolute Value 1. 0 x 0 5 e x if x $ 0 2x if x , 0 2. If 0 x 0 5 c, then x 5 c or x 5 2c. 1c . 02 3. If 0 x 0 , c, then 2c , x , c. 1c . 02 4. If 0 x 0 . c, then x , 2c or x . c. 1c . 02 Properties of Logarithms 1. logb 1MN2 5 logb M 1 logb N 2. logb aM N b 5 logb M 2 logb N 3. logb M p 5 p logb M 4. logb M 5 loga M loga b 5 ln M ln b 5 log M log b 5. logb bx 5 x; ln ex 5 x 6. blogb x 5 x; eln x 5 x, x . 0 Special Factorizations 1. Difference of two squares: A2 2 B2 5 1A 1 B21A 2 B2 2. Perfect square trinomials: A2 1 2AB 1 B2 5 1A 1 B22 A2 2 2AB 1 B2 5 1A 2 B22 3. Sum of two cubes: A3 1 B3 5 1A 1 B21A2 2 AB 1 B22 4. Difference of two cubes: A3 2 B3 5 1A 2 B21A2 1 AB 1 B22 Arithmetic Operations ab 1 ac 5 a1b 1 c2 a b 1 c d 5 ad 1 bc bd a 1 b c 5 a c 1 b c aa bb ac db 5 ad bc a ab cb 5 ab c a 2 b c 2 d 5 b 2 a d 2 c ab 1 ac a 5 b 1 c, a 2 0 aa bb c 5 a bc a ab cb 5 ac b Formulas/Equations The distance from 1x1, y12 to 1x2, y22 is "1x2 2 x122 1 1y2 2 y122. The midpoint of the line segment with endpoints 1x1, y12 and 1x2, y22 is ax1 1 x2 2 , y1 1 y2 2 b. The standard equation of a circle of radius r with center at 1h, k2 is 1x 2 h22 1 1y 2 k22 5 r2. The slope m of the line containing the points 1x1, y12 and 1x2, y22 is slope 1m2 5 change in y change in x 5 y2 2 y1 x2 2 x1 1x1 2 x22 m is undefined if x1 5 x2. The equation of a line with slope m and y-intercept 10, b2 is y 5 mx 1 b. The equation of a line with slope m containing the point 1x1, y12 is y 2 y1 5 m1x 2 x12. The solutions of the equation ax2 1 bx 1 c 5 0, a 2 0, are x 5 2b 6 "b2 2 4ac 2a . If b2 2 4ac . 0, there are two unequal real solutions. If b2 2 4ac 5 0, there is a repeated real solution. If b2 2 4ac , 0, there are two complex solutions that are not real. Distance Formula Midpoint Formula Standard Equation of a Circle Slope Formula Slope-Intercept Equation of a Line Point-Slope Equation of a Line Quadratic Formula Functions Constant Function Linear Function Quadratic Function Polynomial Function Rational Function Exponential Function Logarithmic Function ƒ1x2 5 b ƒ1x2 5 mx 1 b, where m is the slope and b is the y-intercept ƒ1x2 5 ax2 1 bx 1 c, a 2 0 or ƒ1x2 5 a1x 2 h22 1 k parabola vertex 1h, k2 ƒ1x2 5 anxn 1 an21xn21 1 c1 a1x 1 a0 R1x2 5 n1x2 d1x2 5 anxn 1 an21xn21 1 c1 a1x 1 a0 bmxm 1 am21xm21 1 c1 b1x 1 b0 ƒ1x2 5 bx, b . 0, b 2 1 ƒ1x2 5 logbx, b . 0, b 2 1 Transformations In each case, c represents a positive real number. Function Draw the graph of f and: Vertical translations Horizontal translations Reflections Shift f upward c units. Shift f downward c units. Shift f to the right c units. Shift f to the left c units. Reflect f about the x-axis. Reflect f about the y-axis. ey 5 ƒ1x2 1 c y 5 ƒ1x2 2 c ey 5 ƒ1x 2 c2 y 5 ƒ1x 1 c2 ey 5 2ƒ1x2 y 5 ƒ12x2 Sequences 1. Infinite Sequence: 5an6 5 a1, a2, a3, c, an, c 2. Summation Notation: a n i51 ai 5 a1 1 a2 1 a3 1 c1 an 3. nth Term of an Arithmetic Sequence: an 5 a1 1 1n 2 12d 4. Sum of First n Terms of an Arithmetic Sequence: Sn 5 n 2 1a1 1 an2 5. nth Term of a Geometric Sequence: an 5 a1rn21 6. Sum of First n Terms of a Geometric Sequence: Sn 5 a111 2 rn2 1 2 r 1r 2 12 7. Sum of an Infinite Geometric Series with r , 1: S 5 a1 1 2 r The Binomial Theorem 1. n! 5 n1n 2 121n 2 22 c 3 ⋅ 2 ⋅ 1; 1! 5 1; 0! 5 1 2. an rb 5 n! r!1n 2 r2! 3. Binomial theorem: 1a 1 b2n 5 an 0ban 1 an 1ban21b 1 an 2ban22b2 1 c1 an nbbn Permutations, Combinations, and Probability 1. nP r, the number of permutations of n elements taken r at a time, is given by nP r 5 n! 1n 2 r2!. 2. nCr, the number of combinations of n elements taken r at a time, is given by nCr 5 n! 1n 2 r2!r!. 3. Probability of an Event: P1E2 5 n1E2 n1S2 , where n1E2 5 the number of outcomes in event E and n1S2 5 the number of outcomes in the sample space. Conic Sections Directrix x = p Focus (p, 0) x y y2 = 4px Vertex x2 = 4py Directrix y = p Focus (0, p) x y Vertex Parabola x y Vertex (a, 0) Vertex (a, 0) Foci Center Major axis x2 a2 y2 b2 + = 1 x y Vertex (0, a) Vertex (0, a) Foci Center Major axis x2 b2 y2 a2 + = 1 Ellipse x y Focus Focus Center Vertex (a, 0) Vertex x2 a2 y2 b2 = 1 (a, 0) x y Focus Vertex Focus Vertex (0, a) Center y2 a2 x2 b2 = 1 (0, a) Hyperbola 3.1 1289 Exact Values of Trigonometric Functions x degrees x radians sin x cos x tan x 0° 0 0 1 0 30° p 6 1 2 !3 2 !3 3 45° p 4 !2 2 !2 2 1 60° p 3 !3 2 1 2 !3 90° p 2 1 0 — Right Triangle Trigonometry sin u 5 opp hyp csc u 5 hyp opp cos u 5 adj hyp sec u 5 hyp adj tan u 5 opp adj cot u 5 adj opp b a θ c Adjacent Opposite Hypotenuse Trigonometric Functions in the Cartesian Plane sin u 5 y r csc u 5 r y cos u 5 x r sec u 5 r x tan u 5 y x cot u 5 x y (x, y) θ x y r x y Angle Measurement p radians 5 180° s 5 ru A 5 1 2r2u 1u in radians2 To convert from degrees to radians, multiply by p 180°. To convert from radians to degrees, multiply by 180° p . s θ r r Oblique Triangle Law of Sines In any triangle, sin a a 5 sin b b 5 sin g c . Law of Cosines a2 5 b2 1 c2 2 2bc cos a b2 5 a2 1 c2 2 2ac cos b c2 5 a2 1 b2 2 2ab cos g α β γ a b c Circular Functions (cos u, sin u) x y 45º 4 π 360º 2π 0º 0 0 315º 4 7π 270º 2 3π 225º 4 5π π 180º 135º 4 3π 90º 2 π (0, 1) (0, –1) (1, 0) (–1, 0) 2 √2 2 √2 ( , – ) 2 √2 2 √2 ( , ) 2 √2 2 √2 (– , – ) 2 √2 2 √2 (– , ) x y 60º 3 π 30º 6 π 360º 2π 0º 0 0 330º 6 11π 300º 3 5π 270º 2 3π 240º 3 4π 210º 6 7π π 180º 150º 6 5π 90º 2 π 120º 3 2π (0, 1) (0, –1) (1, 0) (–1, 0) 2 √3 2 1 ( , – ) 2 √3 2 1 ( , ) 2 √3 2 1 ( , – ) 2 √3 2 1 (– , – ) 2 √3 2 1 (– , – ) 2 √3 2 1 (– , ) 2 √3 2 1 (– , ) 2 √3 2 1 ( , ) Special Right Triangles 1 1 45º 45º √2 60º 30º 1 2 √3 1290 CHAPTER 3 Identities Identities for Negatives sin12x2 5 2sinx cos12x2 5 cos x tan12x2 5 2tanx Pythagorean Identities sin2 x 1 cos2 x 5 1 tan2 x 1 1 5 sec2 x 1 1 cot2 x 5 csc2 x Cofunction Identities aReplace p 2 with 90° if x is in degree measure.b sinap 2 2 xb 5 cos x cosap 2 2 xb 5 sin x tanap 2 2 xb 5 cot x cotap 2 2 xb 5 tan x secap 2 2 xb 5 csc x cscap 2 2 xb 5 sec x Product-to-Sum Identities sin x cos y 5 1 23sin 1x 1 y2 1 sin 1x 2 y24 cos x sin y 5 1 23sin 1x 1 y2 2 sin 1x 2 y24 sin x sin y 5 1 23cos 1x 2 y2 2 cos 1x 1 y24 cos x cos y 5 1 23cos 1x 1 y2 1 cos 1x 2 y24 Sum-to-Product Identities sin x 1 sin y 5 2 sin ax 1 y 2 b cos ax 2 y 2 b sin x 2 sin y 5 2 cos ax 1 y 2 b sin ax 2 y 2 b cos x 1 cos y 5 2 cos ax 1 y 2 b cos ax 2 y 2 b cos x 2 cos y 5 22 sin ax 1 y 2 b sin ax 2 y 2 b Reciprocal Identities csc x 5 1 sin x sec x 5 1 cos x cot x 5 1 tan x Quotient Identities tan x 5 sin x cos x cot x 5 cos x sin x Sum Identities sin1x 1 y2 5 sin x cos y 1 cos x sin y cos1x 1 y2 5 cos x cos y 2 sin x sin y tan1x 1 y2 5 tan x 1 tan y 1 2 tan x tan y Difference Identities sin1x 2 y2 5 sin x cos y 2 cos x sin y cos1x 2 y2 5 cos x cos y 1 sin x sin y tan1x 2 y2 5 tan x 2 tan y 1 1 tan x tan y Double-Angle Identities sin12x2 5 2 sin x cos x cos12x2 5 • cos2 x 2 sin2 x 1 2 2 sin2 x 2 cos2 x 2 1 tan12x2 5 2 tan x 1 2 tan2 x 5 2 cot x cot2 x 2 1 5 2 cot x 2 tan x Half-Angle Identities sin ax 2b 5 6Å 1 2 cos x 2 cos ax 2b 5 6Å 1 1 cos x 2 tan ax 2b 5 1 2 cos x sin x 5 sin x 1 1 cos x 5 6Å 1 2 cos x 1 1 cos x Identities for Reducing Powers sin2 x 5 1 2 cos12x2 2 cos2 x 5 1 1 cos12x2 2 tan2 x 5 1 2 cos12x2 1 1 cos12x2 Sign 11/22 is determined by quadrant in which x /2 lies 3.1 1291 Amplitude, Period, and Phase Shift y 5 A sin 1Bx 1 C 2 y 5 A cos 1Bx 1 C 2 Amplitude 5 A Period 5 2p B Phase shift 5 C B µ left if C B . 0 right if C B , 0 y 5 A tan 1Bx 1 C 2 y 5 A cot 1Bx 1 C 2 Period 5 p B Phase shift 5 C B µ left if C B . 0 right if C B , 0 Powers and Roots of Complex Numbers zn 5 1x 1 iy 2 n 5 3r 1cos u 1 i sin u24 n 5 r n 3cos 1nu2 1 i sin 1nu24 n 5 1, 2, . . . ! n z 5 1x 1 iy 2 1/n 5 3r 1cos u 1 i sin u24 1/n 5 r 1/n ccosau 1 2kp n b 1 i sinau 1 2kp n b d k 5 0, 1, 2, . . . , n 2 1 Graphs of the Trigonometric Functions –1 1 x y y = sinx 2π π –π –2π –1 1 x y 2π π –π –2π y = cosx 2π π –π –2π –5 –3 –1 1 3 5 x y y = tanx 2π π –π –2π –5 –3 –1 1 3 5 x y y = cscx 2π π –π –2π –5 –3 –1 1 3 5 x y y = secx 2π π –π –2π –5 –3 –1 1 3 5 x y y = cotx Polar Coordinates x 5 r cos u y 5 r sin u r 2 5 x 2 1 y2 tan u 5 y x Complex Numbers x 1 iy 5 r 1cos u 1 i sin u2 Rectangular Trigonometric form (polar) form x y x θ y r (x, y) (r, θ) Heron’s Formula for Area If the semiperimeter s of a triangle is s 5 a 1 b 1 c 2 then the area of that triangle is A 5 "s1s 2 a21s 2 b21s 2 c2 1292 CHAPTER 3 Inverse Trigonometric Functions y 5 sin21 x x 5 sin y 2p 2 # y # p 2 21 # x # 1 y 5 cos21 x x 5 cos y 0 # y # p 21 # x # 1 y 5 tan21 x x 5 tan y 2p 2 , y , p 2 x is any real number y 5 cot21 x x 5 cot y 0 , y , p x is any real number y 5 sec21 x x 5 sec y 0 # y # p, y 2 p 2 x # 21 or x $ 1 y 5 csc21 x x 5 csc y 2p 2 # y # p 2, y 2 0 x # 21 or x $ 1 Graphs of the Inverse Trigonometric Functions π –1 –0.5 0.5 1 x y y = sin–1x or y = arcsinx π 4 π 2 4 π 2 – – –1 –0.5 0.5 1 x y π 4 3π 4 π y = cos–1 x or y = arccos x x y π 4 π 2 y = tan–1 x or y = arctan x 3 2 1 –3 –2 –1 π 4 π 2 – – Vectors B A Vector v = AB → u v u + v u v u + v ku (k > 0) (k < 0) ku u x y j i (0, 1) (1, 0) v u θ Vector Addition Scalar Multiplication For vectors u 5 8a, b9 and v 5 8c, d9, and real number k, u 5 ai 1 bj u 5 "a2 1 b2 u 1 v 5 8a 1 c, b 1 d9 ku 5 8ka, kb9 u # v 5 ac 1 bd cos u 5 u # v u v Compv u 5 ucos u 5 u # v u
4821	https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/pages/lecture-notes/	Browse Course Material Course Info Instructor Prof. Anantha Chandrakasan Departments Electrical Engineering and Computer Science As Taught In Spring 2006 Level Undergraduate Topics Engineering Computer Science Computer Design and Engineering Electrical Engineering Digital Systems Electronics Learning Resource Types assignment Problem Sets grading Exams with Solutions notes Lecture Notes group_work Projects with Examples Download Course search GIVE NOW about ocw help & faqs contact us 6.111 \| Spring 2006 \| Undergraduate Introductory Digital Systems Laboratory Lecture Notes \| SES # \| TOPICS \| --- \| \| L1 \| Introduction (PDF) \| \| L2 \| Combinational logic (PDF) \| \| L3 \| Introduction to Verilog® - combinational logic (PDF) \| \| L4 \| Sequential building blocks (PDF) \| \| L5 \| Simple sequential circuits and Verilog® (PDF) \| \| L6 \| Finite-state machines and synchronization (PDF) \| \| L7 \| Memory basics and timing (PDF) \| \| L8-L9 \| Arithmetic structures (PDF) \| \| L10 \| Analog building blocks (PDF) \| \| L11 \| System integration issues and major/minor FSM (PDF) \| \| L12 \| Reconfigurable logic architecture (PDF - 1.3 MB) \| \| L13 \| Video (PDF - 4.0 MB) \| \| L14 \| Project kickoff (PDF) \| \| L15 \| LSI integration and performance transformations (PDF - 1.1 MB) \| \| L16 \| Power dissipation in digital systems (PDF) \| Course Info Instructor Prof. Anantha Chandrakasan Departments Electrical Engineering and Computer Science As Taught In Spring 2006 Level Undergraduate Topics Engineering Computer Science Computer Design and Engineering Electrical Engineering Digital Systems Electronics Learning Resource Types assignment Problem Sets grading Exams with Solutions notes Lecture Notes group_work Projects with Examples Download Course Over 2,500 courses & materials Freely sharing knowledge with learners and educators around the world. Learn more © 2001–2025 Massachusetts Institute of Technology Creative Commons License Terms and Conditions Proud member of: © 2001–2025 Massachusetts Institute of Technology You are leaving MIT OpenCourseWare Please be advised that external sites may have terms and conditions, including license rights, that differ from ours. MIT OCW is not responsible for any content on third party sites, nor does a link suggest an endorsement of those sites and/or their content. Continue
4822	https://or.stackexchange.com/questions/6207/how-to-linearize-inequalities-having-max-or-min	linear programming - How to linearize inequalities having max or min? - Operations Research Stack Exchange Join Operations Research By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How to linearize inequalities having max or min? Ask Question Asked 4 years, 5 months ago Modified4 years, 5 months ago Viewed 264 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. I'm modeling an LP problem in which I have to maximize an objective function. Two of the constraints are the following, where k i k i are constants and x i x i decision variables (continuous). Could anyone help me on how to linearize these constraints? max[(x 1+k 1),(x 2+x 3+x 4)]min[(x 1+k 1),(x 2+x 3+x 4)]≤k 2≥−k 2 max[(x 1+k 1),(x 2+x 3+x 4)]≤k 2 min[(x 1+k 1),(x 2+x 3+x 4)]≥−k 2 linear-programming linearization Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications edited May 1, 2021 at 9:36 TheSimpliFire♦ 5,523 5 5 gold badges 25 25 silver badges 58 58 bronze badges asked Apr 30, 2021 at 21:30 TR FernandesTR Fernandes 59 4 4 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 8 Save this answer. Show activity on this post. max(y,z)≤b max(y,z)≤b is equivalent to y z≤b≤b y≤b z≤b The min min constraint is similar. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Apr 30, 2021 at 22:03 RobPrattRobPratt 36.2k 2 2 gold badges 51 51 silver badges 94 94 bronze badges Add a comment\| Your Answer Thanks for contributing an answer to Operations Research Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. 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4824	https://www.themathdoctors.org/the-art-of-proving-trig-identities/	Skip to content The Art of Proving Trig Identities August 22, 2019 January 26, 2024 / NQOTW, Trigonometry / Intuition, Pedagogy, Proof / By Dave Peterson (A new question of the week) Last week we looked at a recent question about basic trigonometric equations. That discussion continued into the subject of identities, which we’ll look at here. We’ll be sitting in on an extended chat about many important aspects of this kind of work. It’s still very long, even after extensive editing, but worth reading through. How do I get there? Practice! Picking up the thread, here is Sarah’s new question: How do you prove an identity? Doctor Rick picked up the discussion: Finding a way to prove an identity is something of an art form, learned by practice. Here is an example of a more or less general approach to proving trig identities, which won’t always give the shortest or most elegant proof, but at least it’s a backup: Proving Trigonometric Identities Again, you can search the Archive for the words “trig identity proof” to get suggestions for many specific proofs. If you would like to present a specific example of a trig identity to be proved, we will be happy to work through the details with you. Sarah answered: My problem is that after l see the answer, l get it, but I’m not capable of getting to the answer myself, and I’ve been getting a lot of practice. Usually, my mathematical intuition and recognition is quite good, not sure where that went. I’m attaching some examples: If you are not familiar with the “triple-equal” symbol “≡”, it means “is identically equal to”, that is, “is equal to, for all values of the variable“. This is a way to distinguish identities from equations that are to be solved for specific values. Not all textbooks use it, but it is a good practice. Problem 1: Aim toward the goal Doctor Rick replied, using the first example to illustrate how he thinks: Thanks. I’d like to walk through one problem at a time with you. Let’s start with the separate image: Show that -1 + cos α 1 + cos α ------------- + ------------- ≡ 2(1 + tan α) sec α + tan α sec α - tan α The first thing I would choose to do here is to carry out the addition of fractions on the left. This looks like a really good first step because I can see what will happen with the common denominator: it will be a difference of squares. Do you see that? It’s very possible that we’ll be able to apply a “Pythagorean identity” to that denominator. Give that a try, and show me your work as far as you can take it. Here we have a step that is both conventional, and supported by a look ahead. Sarah showed how one attempt got her tangled up: I’ve worked out all of the ones l sent already, but only with some help here and there, once l realised what was the first step, the rest followed easily. I’m attaching one of my original attempts. I’m ending up taking a much longer way which sometimes ends up leading to nowhere. I usually used to manage something like this, but when l don’t manage all the ones l try, I’m finding it off-putting instead of fun. Doctor Rick wrote back, again focusing on the thinking: Hi, Sarah. Proving trig identities is a lot like solving puzzles. As you know, they can be a lot of fun when you succeed, but really tedious when we can’t find our way through them. I like to do Sudoku, but now and then there is a supposedly easy puzzle that I get stuck on. If I put the puzzle aside for later, I may then find one thing I’d missed, and everything comes together after that. Since Sudoku are just puzzles, I can just skip one if it gets boring. We ought to be able to do that with trig identities, too; it’s OK not to solve them all. We all have a “blind spot” now and then. But you’d like to get better at this. Looking at your work, I wonder why you tried replacing cos α by (sin α)/(tan α). Did you have some other identity in which this actually helped? It isn’t something I’d think of. As Doctor Rob said in the Archive answer I showed you, a good first step (when you don’t know what else to do) is to replace all tan, cot, sec, and csc with expressions using only sin and cos. Then you only need to consider the identities involving sin and cos — that’s simpler to think about. Introducing tan when it wasn’t there, doesn’t make things simpler. If you had done what I’m suggesting (which, of course, is different from what I suggested last time), you’d have: -1 + cos α 1 + cos α sin α ------------- + ------------- ≡ 2(1 + ----- ) 1 sin α 1 sin α cos α ----- + ----- ----- - ----- cos α cos α cos α cos α This certainly doesn’t look simpler yet, but the fact that we only have sin and cos should help soon. Now there’s something I see to do next: multiply the numerator and denominator of each fraction on the left by cos α. Why? Because this will make the expression simpler, by getting rid of the “fractions within fractions”. Now, eventually I’m going to come back around to my earlier suggestion: carrying out the addition of fractions (common denominator and all that). Asking myself why I’d do this, I’d say that it’s because I don’t know any (basic, memorable) identities involving a sum of fractions. I’m aiming to make things look more like what I know. As with putting the expression in terms of sin and cos, I’m trying to increase my chance of recognizing a pattern I can use. Does this help at all? We can keep discussing your thinking on various problems and see what we discover. Let me just say this (which you probably already know, at least in theory): One of the most overlooked steps in problem-solving is the last — to look over your work to see what lessons you can learn from it. Did you see a pattern that turned out to be useful? Or, looking back with the benefit of hindsight, do you see something you could have done that would have gotten you to the goal more quickly? When I asked myself and you why we did certain things above, it was with the idea of abstracting some possibly useful principles for future problems. The key idea here is that we are trying to make an expression (a) simpler, (b) more familiar, and (c) more like the other side. Each of these short-term goals makes it likely we will be able to move forward. After some discussion, he demonstrated both methods for this problem, which I’ll include here. The first, which Sarah said worked better for her, started with combining the fractions: Here’s my work with the first method: -1 + cos α 1 + cos α ------------- + ------------- sec α + tan α sec α - tan α (-1 + cos α)(sec α - tan α) + (1 + cos α)(sec α + tan α) ≡ -------------------------------------------------------- (sec α + tan α)(sec α - tan α) -sec α + tan α + cos α sec α - cos α tan α + sec α + tan α + cos α sec α + cos α tan α ≡ ---------------------------------------------------------- sec2 α - tan2 α 2 tan α + 2 ≡ --------------- 1 ≡ 2(1 + tan α) I did more than one thing at a time on some steps to condense the proof somewhat. This method depends on knowing the identity sec2 x - tan2 x = 1 and a few other identities involving more than just sin and cos. For those who can’t see these things so readily, converting to sin and cos can make the steps easier to see, though it does take longer. The second method, as mentioned, is a good fallback when you don’t see what else to do, as it allows you to use the more familiar identities, even though it may get more complicated: Here is the rest of the sin-and-cos-first method. I hadn’t tried following that method through to the end, but I will now: To show that -1 + cos α 1 + cos α ------------- + ------------- ≡ 2(1 + tan α) : sec α + tan α sec α - tan α -1 + cos α 1 + cos α LHS ≡ ------------- + ------------- 1 sin α 1 sin α ----- + ----- ----- - ----- cos α cos α cos α cos α cos α (-1 + cos α) cos α (1 + cos α) ≡ ------------------ + ----------------- 1 + sin α 1 - sin α (-1 + cos α)(1 - sin α) + (1 + cos α)(1 + sin α) ≡ cos α ------------------------------------------------ (1 + sin α)(1 - sin α) (-1 + cos α + sin α - sin α cos α + 1 + sin α + cos α + sin α cos α) ≡ cos α ----------------------------------- 1 - sin2 α 2(sin α + cos α) ≡ cos α ---------------- cos2 α 2 (sin α + cos α) ≡ ----------------- cos α ≡ 2(tan α + 1) ≡ 2(1 + tan α), RHS of identity to be proved. Back in the original discussion, Sarah responded: l don’t know why l changed cos to sin/tan, but l probably said if you can change tan to sin/cos then it works that way too. l now realise it makes it much longer. Also, you substituted on both the left and right hand side. I was told that you can’t do that, because you need to start working on one side, and get to the other, or first work on one side and reach a dead end, do the same on the other side, and have both dead ends equal. I’ve tried a couple of others and l seem to be getting the hang of it slowly. Doctor Rob’s article is helpful. I think l should practice easy ones to get the technique and confidence, then move to harder ones, seeing them as a challenge. If you want to discuss some others, and we can both show our methods, l’m open to that – we can both analyse after the different methods (if they’re different) Problem 6 (a,b): Train your intuition Doctor Rick answered her question, then suggested the next problems to try: Working on both sides is something we have discussed in Ask Dr. Math, and Dr. Peterson looked at some of these discussions in our Blog: Different Ways to Prove a Trigonometric Identity (The first thing he looks at is Dr. Rob’s general method that I showed you earlier. Then he gets into your question about whether we have to start at one end and go through to the other.) For practice, let’s look at the four-part Question 6 that you showed me earlier. I consider part (a) to be much simpler than the one we’ve been discussing. Part (b) is also relatively easy. (a) (cos x + sin x)(cos x – sin x) ≡ 1 – 2 sin2 x (b) sin2 x – sin4 x ≡ cos2 x – cos4 x What did you do with these identities? Did you notice any patterns or things to try right away? What happened when you tried them? The first thing that occurs to me in each case works out well, but it may be that my intuition has been trained by lots of practice so that I see things that you don’t notice yet – or I don’t even notice things that you do, that don’t work out. Our goal, at least in part, is to train your intuition to focus on truly useful things. A teacher may consider some things “obvious” that a student can’t yet see at all; and may totally miss the distracting obstacles the student needs to avoid. This is why we want to see it through the student’s eyes in order to help! Sarah showed her work for each problem: l tried these two again; this time, intuition worked both times and am attaching my method. Doctor Rick commented on the thinking behind each solution: Regarding the first problem, (cos x + sin x)(cos x - sin x) ≡ 1 - 2 sin2 x you have done well. I see that you multiplied out the left side and then canceled the “cross terms”. Were you able to see, without doing that all out, that you’d get a difference of squares? If you can see this ahead of time, it can help you recognize that multiplying out the product is definitely a simplifying step. I’d probably multiply out anyway in this situation, but getting a glimpse of what’s coming next is part of the skill set that makes proving identities easier. On my part, I also recognize that the difference of squares that you get, cos2 x – sin2 x, is the double-angle identity — it’s cos(2x). This isn’t useful in proving the identity, but I also know that 1 – 2sin2 x (or something like that — I may not remember it exactly) is an alternate form of that identity. Thus I saw something coming — namely, that this proof will be something I’ve seen before, when the alternate form was proved. The more you’ve seen before, the easier it is to recognize useful patterns. Your method on the second problem is not what I did, but I can see that you used a good technique. I’m supposing that you noticed that the right-hand side of the identity to be proved had only cosines, so you chose to rewrite the left-hand side in terms of cosines as well. That’s good thinking, using the form of the “target” expression to guide you in choosing your steps. Here’s my method. I worked on both sides at once in my exploration. Where one side is distinctly simpler than the other, it’s good to start work on the more complicated side, on the theory that it will be easier to recognize when you’re heading toward the simpler expression. In this problem, though, both sides are equally complicated. So I began like this, factoring on both sides: sin2 x - sin4 x cos2 x - cos4 x siin2 x (1 - sin2 x) cos2 x (1 - cos2 x) sin2 x (cos2 x) ≡ cos2 x (sin2 x) Now I had identically equal expressions on both sides, so I could rework my exploration into a standard-style proof. This is the idea of “building the bridge from both sides toward the middle, then crossing the bridge from one side to the other.” sin2 x - sin4 x ≡ sin2 x (1 - sin2 x) ≡ sin2 x (cos2 x) ≡ (1 - cos2 x)(cos2 x) ≡ cos2 x - cos4 x This looks a little simpler than your proof, but yours is perfectly good, and it has the advantage that the reader can make a good guess at how you thought it through. “Elegant” proofs (not that mine is particularly elegant) tend to hide the thought process that produced the proof, so that you may think, “I would never have thought of that.” Thus they aren’t so much help in learning to do your own proofs! What is intuition? Sarah expanded an earlier thought: I just realised something from what you said: sometimes the more you know, the more it can interfere and make it harder – how do you know what is useful and what is not? The more you know, the more you need to sift to see what you need and what will actually work! Also, thank you – especially for getting my interest into this, highlighting the whys, the thought process, explaining things so clearly, the advantages of certain methods and for everything 🙂 Doctor Rick responded: To tell the truth, as I was writing about “the more you’ve seen before,” I had a contrary thought similar, in a way, to yours: As I get older, I find that more of the people I see remind me of someone I once knew — just because I’ve known a lot of people! That doesn’t do me much good, though; it just confuses me — could this actually bethe person I’m reminded of? (If you’ve ever read Agatha Christie’s Miss Marple mystery novels, you’ll recognize that this is how the elderly Miss Marple solved crimes — they reminded her of people and situations she’d known in her little village, and that gave her insight into likely motives. But that’s fiction, and in any case, my mind doesn’t work that way.) Getting back to math, all I can say is that we develop our intuition by noting what steps have proved useful in the problems we have worked. Yes, the number of things we might do at a given step can grow, but we find that not all of them are equally likely to move us forward in proving an identity. In this way, we develop heuristics such as those listed by Dr. Rob — techniques that are more likely to help solve a problem. We can’t be sure these things will help, but we try them first. Mentioning heuristics reminds me that another of Dr. Peterson’s blogs examined a discussion I had with a student about a challenging identity proof. You might be interested in this one too: Proving an Identity in Different Ways Sarah added: A bit of a digression here, but you got me thinking; what would you say mathematical intuition is? And how do you train your intuition (other than by practice)? Do you think it is innate or do you develop it? I personally think it is both, because not everyone can “see” certain things, but l think you can develop and work on what you have. Doctor Rick answered: Regarding the nature of intuition, I haven’t studied psychology, I’m just writing as a layman regarding that field. To me, intuition is essentially subconscious pattern-matching — making associations between a current situation and things we’ve experienced before, so as to reach a conclusion without formal logical reasoning. Of course, in mathematical proof, we can’t forget about formal logical reasoning! Rather, we use the intuitive process as an aid in the exploration phase as we search for a line of formal logical reasoning. Under this understanding, intuition is learned, as it is built up of our experiences. Thus practice is the key to mathematical intuition, or problem-solving skill. But as I have already said, in order for that practice to develop our skill effectively, we need to do that final step in problem-solving: to think about what we have done, looking for lessons to be learned. A harder one? Sarah was ready for a challenge: Do you have a good one for me, which requires some easy stuff, but which may be harder to see please? I’m willing to try out a more challenging one. If you do have one, l’m very ready to try it out. Or any one really, which might teach me the techniques required. Your part about intuition – very interesting; agreed. Doctor Rick offered a problem to try: It’s hard to tell how difficult you might find a particular identity proof problem, just as the difficulty ratings of Sudoku puzzles don’t always match my experience with them. But I have some problems that at least look interesting. Let’s see how you do with this one: 1 Prove that ------------- ≡ sec x – tan x sec x + tan x Sarah was up to the task: Thank you for the question. l managed this one on my own. I did LHS = 1/((1/cosx) + sinx/cosx) = cosx/ (1+sinx) Then l multiplied by (1-sinx) /(1-sinx) cosx – cosx sinx / (1-sin^2x) = cosx(1-sinx) / cos^2 x = (1-sinx)/cosx = sec x – tan x Getting the hang of them now, thanks for the help! Doctor Rick responded: Your work is good. You used the fall-back idea of writing everything in terms of sin and cos. Then, I suppose, you may have recognized the useful pattern that multiplying (1 ± sin x) or (1 ± cos x) by what we might call its trig conjugate (I just made that up) results in a difference of squares to which the basic Pythagorean identity can be applied. If you were more familiar with the other Pythagorean identities, you could have done the same with fewer steps. (I’ve told you that I don’t have these identities fully memorized myself, I have to derive them in my head to be sure.) Here is what this approach would look like: 1 sec x - tan x sec x - tan x ------------- × ------------- ≡ --------------- sec x + tan x sec x - tan x sec2 x - tan2 x sec x - tan x ≡ ------------- ≡ sec x - tan x 1 Again, you did well. Let me also show you how one might approach this as if we were solving an equation – something that I know you have been trained not to do: 1 ------------- = sec x – tan x sec x + tan x 1 = (sec x - tan x)(sec x + tan x) 1 = sec2 x - tan2 x But this is always true (if necessary, you can do this to verify it:) 1 = 1/cos2 x - sin2 x / cos2 x 1 = (1 - sin2 x)/cos2 x cos2 x = 1 - sin2 x This you definitely recognize as an identity. Now, this “solution of the equation” has a clear relationship to my proof of the identity, Rather than multiplying both sides of the equation by (sec x + tan x), we work with the LHS alone, multiplying it by 1 in the form (sec x + tan x)/(sec x + tan x). This is the idea presented in Dr. Peterson’s blog “Different Ways to Prove a Trigonometric Identity“, in the section “Can you work on both sides?”. This was the end of the discussion. I’m sure the subject will come up again. Leave a Comment Cancel Reply This site uses Akismet to reduce spam. Learn how your comment data is processed.
4825	https://www.ck12.org/flexi/chemistry/Reaction-Order/derive-an-expression-to-calculate-the-time-required-for-completion-of-a-zero-order-reaction/	Flexi answers - Derive an expression to calculate the time required for completion of a zero order reaction. \| CK-12 Foundation All Subjects Chemistry Reaction Order Question Derive an expression to calculate the time required for completion of a zero order reaction. Flexi Says: A zero-order reaction is one in which the rate of reaction is independent of the concentration of the reactants. The rate law for a zero-order reaction is given by: \begin{align}\frac{d[A]}{dt} = -k\end{align} where [A] is the concentration of the reactant, t is time, and k is the rate constant. To find the time required for the reaction to complete, we need to integrate this equation from the initial concentration [A]0 to the final concentration [A] (which is 0 for a complete reaction), and from the initial time (0) to the final time t: \begin{align}\int_{[A]0}^{0} d[A] = -k \int{0}^{t} dt\end{align} The integral of d[A] from [A]0 to 0 is -[A]0, and the integral of dt from 0 to t is t. So we get: \begin{align}-[A]_0 = -kt\end{align} Solving for t gives the time required for the reaction to complete: \begin{align}t = \frac{[A]_0}{k}\end{align} Analogy / Example Try Asking: Is the decomposition of hydrogen peroxide a first order reaction?A molecule of ethyl alcohol is converted to acetaldehyde in one's body by zero order kinetics. If the concentration of alcohol is 0.015 mol/L and the rate constant = 6.410^-5 mol/L·min, what is the concentration of alcohol after 3.5 hours? a. 0.15 mol/L b. 0.0032 mol/L c. 0.0016 mol/L d. 4.310^-3 mol/L e. 9.610^7 mol/LHow can you tell if a reaction is first or second order? How can Flexi help? By messaging Flexi, you agree to our Terms and Privacy Policy
4826	https://math.stackexchange.com/questions/130773/find-bound-for-sum-of-square-roots	Skip to main content Find bound for sum of square roots Ask Question Asked Modified 12 years, 5 months ago Viewed 9k times This question shows research effort; it is useful and clear 16 Save this question. Show activity on this post. Let a1,...,an be real numbers, such that a1+...+an=A. What can we say about a1−−√+...+an−−√? I would like to bound from above thus sum in terms of A. calculus combinatorics number-theory approximation Share CC BY-SA 3.0 Follow this question to receive notifications edited Mar 26, 2013 at 21:59 azimut 24.4k1111 gold badges7878 silver badges142142 bronze badges asked Apr 12, 2012 at 6:23 Nick G.H.Nick G.H. 42533 silver badges1414 bronze badges 2 ai are all nonnegative? – Paul Commented Apr 12, 2012 at 6:26 Yes, we can assume that they are nonnegative. – Nick G.H. Commented Apr 12, 2012 at 6:48 Add a comment \| 4 Answers 4 Reset to default This answer is useful 19 Save this answer. Show activity on this post. Assuming ai≥0, you could consider the vector x=(a1−−√,...,an−−√), then you have \|\|x\|\|22=A, and Hölder's inequality gives \|\|x\|\|1≤n−−√\|\|x\|\|2, from which you get the bound a1−−√+...+an−−√≤nA−−−√ Share CC BY-SA 3.0 Follow this answer to receive notifications edited Mar 26, 2013 at 21:58 azimut 24.4k1111 gold badges7878 silver badges142142 bronze badges answered Apr 12, 2012 at 6:35 copper.hatcopper.hat 178k1010 gold badges127127 silver badges270270 bronze badges 1 3 This bound is 'tight' in the sense that if you take all ai equal, then it is an equality. – copper.hat Commented Apr 12, 2012 at 6:39 Add a comment \| This answer is useful 14 Save this answer. Show activity on this post. Hint: Use the Cauchy-Schwarz Inequality. We have (∑ai−−√)2≤n∑ai. (In the notation of the article linked to, xi=ai−−√ and yi=1.) Share CC BY-SA 3.0 Follow this answer to receive notifications edited Apr 12, 2012 at 6:41 answered Apr 12, 2012 at 6:35 André NicolasAndré Nicolas 515k4747 gold badges583583 silver badges1k1k bronze badges Add a comment \| This answer is useful 8 Save this answer. Show activity on this post. Since we are taking the square root of each ai, we will assume that ai≥0. Using Jensen's Inequality yields (1n∑i=1nai−−√)2≤1n∑i=1n(ai−−√)2=1n∑i=1nai(1) Rearranging (1) gives ∑i=1nai−−√≤n∑i=1nai−−−−−−√(2) Share CC BY-SA 3.0 Follow this answer to receive notifications answered Apr 12, 2012 at 8:44 robjohn♦robjohn 354k3838 gold badges497497 silver badges890890 bronze badges Add a comment \| This answer is useful 4 Save this answer. Show activity on this post. Just for fun, you can also look at it in terms of elementary statistics. For k=1,…,n let xk=ak−−√. Let x¯ be the mean of the xk. For fixed x¯, the variance of the xk is minimized when x1=…=xn, when it is 0. But the variance is 1n∑kx2k−x¯2, so the minimum value of ∑kx2k is nx¯2, occurring when x1=…=xn. For fixed A=∑kx2k, therefore, the maximum value of x¯ is A/n−−−−√, and hence ∑kxk≤nA−−−√, with equality when x1=…=xn. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Apr 12, 2012 at 7:08 Brian M. ScottBrian M. Scott 632k5757 gold badges822822 silver badges1.4k1.4k bronze badges Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions calculus combinatorics number-theory approximation See similar questions with these tags. 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4827	https://arxiv.org/pdf/2412.01668	H´ ENON MAPS WITH MANY RATIONAL PERIODIC POINTS HYEONGGEUN KIM, HOLLY KRIEGER, MARA-IOANA POSTOLACHE, AND VIVIAN SZETO Abstract. In this paper we study the dynamical uniform boundedness conjecture in the context of H´ enon maps. Building on work of Doyle and Hyde on polynomial maps in one variable, we construct for each odd integer d≥3 a H´ enon map of degree ddefined over Qwith at least ( d−4) 2integral periodic points and an integral cycle of length at least 8d+10 3. This provides a quadratic lower bound on any conjectural uniform bound for periodic rational points of H´ enon maps, and answers in the affir-mative a question of Ingram concerning the existence of many rational periodic points and long rational cycles for H´ enon maps. Introduction Fix an integer d ≥ 2 and let f : Pn → Pn be a morphism of degree d defined over a number field K. We say a point P is preperiodic for f if the forward orbit {P, f (P ), f ◦2(P ), . . . } of P is a finite set. By a theorem of Northcott [Nor50], the set Preper( f, K ) of K-rational preperiodic points of f is finite. The dynamical uniform boundedness conjecture proposed by Morton and Silverman [MS94] asserts that the size of Preper( f, K ) is bounded by a quantity depending only the dimension n, the degree d, and the extension degree [ K : Q]. Very little is known about this conjecture. By work of Fakhruddin [Fak03], the Morton-Silverman conjecture implies the analogous uniform boundedness conjecture on rational torsion points of abelian varieties, currently only known in the case of di-mension 1 [Maz77; Mer96]. However, computational and conditional evidence suggests that it is indeed difficult to produce maps with large sets of rational preperiodic points (see [Ben07] and [Loo21] for some of the strongest known evidence). In the case when n = 1 and K = Q, a basic interpolation argument allows the construction of rational maps f : P1 → P1 over Q with at least d + 1 rational points. Recently, Doyle and Hyde [DH] provided in each degree d ≥ 2 an explicit example rd(x)of a polynomial of degree at most d which has at least d + 6 rational preperiodic points, and gave a non-constructive proof that there exist examples in degree at most d with d + ⌊log 2(d)⌋ rational preperiodic points, providing the best known lower bound on any uniform constant arising in the Morton-Silverman conjecture. Surprisingly, DeMarco and Mavraki [DM24] have shown that the examples of Doyle-Hyde have applications to uniform unlikely intersections in complex dynamics. Date : July 9, 2025. 1 arXiv:2412.01668v2 [math.DS] 8 Jul 2025 2KIM, KRIEGER, POSTOLACHE, AND SZETO H´ enon maps appear as a natural generalization of quadratic polynomials and are a fruitful entry point to studying polynomial dynamics in dimension greater than one, as initiated in [HOV94]. Given an integer d ≥ 2, H´ enon map of degree d is a map of the form (x, y ) 7 → (y, −δx + p(y)) where p is a degree d polynomial of one variable and δ a non-zero constant: note that a H´ enon map is a birational map of the plane with constant Jacobian determinant δ. A H´ enon map does not extend to a morphism of the projective plane, as the natural extension has [1 , 0, 0] as a point of indeterminacy; however, any H´ enon map nonetheless admits a dynamical height function and so the periodic points satisfy a Northcott finiteness property [Sil94; Kaw06]. That is, for any degree d H´ enon map f defined over a number field K, the set Per( f, K ) of periodic points with coordinates in K is a finite set. In analogy with the conjecture of Morton and Silverman, one might hope that the size of the set of K-rational periodic points depends only on the degree of the H´ enon map and the extension degree [ K : Q]; Ingram [Ing14] has provided evidence for this in the case that d = 2 and K = Q by computing the set of rational periodic points for H´ enon maps of the form ( y, x + y2 + b) where b ∈ Q has small height. In this article, we study the polynomials constructed by Doyle and Hyde and use them to construct an explicit family of H´ enon maps {hd(x, y )} over Q with degree d → ∞ and a large number of rational periodic points and long rational cycles. This answers in the affirmative the first two parts of the following question of Ingram [H´ e24]: Question (Ingram) . Over a number field K, is it possible to construct infinite families of generalized H´ enon maps of algebraic degree d, and K-rational cycles of length at least d + 3? Can one construct maps with Nd periodic points, where Nd − d → ∞ , or even Nd/d → ∞ as d → ∞ ? Theorem A. For each odd degree d > 2 there exists an explicitly constructed polyno-mial sd of degree at most d with rational coefficients such that the H´ enon map hd(x, y ) = ( y, −x + sd(y)) has at least (d − 4) 2 rational periodic points. If d = 2 k + 1 , the constructed polynomials sd are integer-valued polynomials which satisfy lim k→∞ (−1) ks2k+1 (x) = 2 √3 sin πx 3 =: s∞(x, y ). A direct computation shows (see section 5.2) that any point of C2 with integer coordi-nates is periodic under iteration of the limiting H´ enon map h∞(x, y ) = ( y, −x + s∞(y), with period n ∈ { 1, 4, 5, 6, 12 , 20 } determined by the class of ( x, y ) modulo 6. Though the proof of Theorem A is not asymptotic, each map hd preserves the lattice Z2 and so one expects a large set of periodic integral points within the plane region where sd is a good approximation of s∞. Interestingly, the dynamics near the boundary of this region allows for integer cycles of hd with long period, despite the bound of 20 on the period of an integer cycle for the limiting map h∞: see Figure 1. H´ENON MAPS WITH MANY RATIONAL PERIODIC POINTS 3 Figure 1. Periodic integer cycles for hd with d = 43 plotted in Z2 and colored by period length. There are a large number of periodic points with cycle length at most 20, but the boundary of this region admits a cycle of length 118. Theorem B. Let d ≡ 1 mod 6 and hd(x, y ) = ( y, −x + sd(y)) as above. Then hd has a cycle of integer points of length 8d+10 3 . One can produce similar results in other odd degrees by introducing a shift to the polynomial sd. This is in striking contrast with the one-dimensional results of Doyle and Hyde, which are many-to-one on their sets of preperiodic integer points and hence produce only short cycles. The family of degree d H´ enon maps has dimension d + 2 and so this theorem also improves substantially on what is obtained by interpolation to produce an integer cycle (this is still true in the larger family of generalized H´ enon maps, which can have mixed dimension (see [FM89]) if d is a composite integer, but has maximal component dimension d + 6). The organization of the paper is as follows. In Section 2 we introduce the polynomials sd used to construct the H´ enon maps hd, and in Section 3 we prove various auxiliary estimates which will allow us to understand hd, as well as noting as an application an optimality result for Doyle-Hyde. Section 4 provides the proof of Theorem A, and Theorem B is proved in Section 5, which also includes an analysis of the limiting map h∞.4 KIM, KRIEGER, POSTOLACHE, AND SZETO Acknowledgements. The work for this paper was done during a 2023 Summer Re-search In Mathematics (SRIM) programme in Cambridge. The first author was sup-ported by SRIM and the third and fourth authors by the Philippa Fawcett Internship Programme. The fourth author was also supported by the Trinity College Summer Studentship Scheme. We are grateful to the Faculty of Mathematics and the SRIM organizer Dhruv Ranganathan for their support. We also thank Romain Dujardin, Trevor Hyde, and Patrick Ingram for helpful conversations related to this work. 2. Background In this section, we introduce the polynomials sd as integer-valued approximations of discrete trigonometric functions. We also review the notion of dynamical compression and the constructions given in [DH]. 2.1. Compressing polynomials. Let K be a number field, f ∈ K[x] be a polynomial and f n := f ◦n be its n-fold composition. Recall that α ∈ K is preperiodic if the forward orbit of α under f is finite. Let Preper( f, K ) denote the set of preperiodic points in K.Let [ m] := {1, 2, · · · , m } for positive integer m. A polynomial g ∈ C[x] is said to exhibit dynamical compression if it is linearly conjugate to some polynomial f ∈ C[x]which satisfies f ([ m]) ⊆ [m] for some m > d + 1. 2.2. Discrete approximations of sine and cosine. Define the following functions of one variable: c2j (x) := 1(2 j)! j Y i=1 x2 − 2i − 12 2! , c 2j+1 (x) := 1(2 j + 1)! x k Y i=1 (x2 − i2). We then define for any k ≥ 0 the functions s2k(x) := k X j=0 (−1) k−j c2j (x)and s2k+1 (x) := k X j=0 (−1) k−j c2j+1 (x). The degree d polynomial sd ∈ Q[x] is integer-valued on Z + d+1 2 and takes values in {− 1, 0, 1} for integers in the interval [ −d+1 2 , d+1 2 ]. In [DH], Doyle and Hyde show that by applying the following transformation: rd(x) = ( sd(x − 3 − d+1 2 ) + 2 if d even sd(x − 3 − d+1 2 ) − x + d + 6 if d odd one obtains a family of integer-valued polynomials rd ∈ Q[x] that exhibit dynamical compression, namely: ( rd([ d + 6]) ⊆ [d + 5] if d even rd([ d + 6]) ⊆ [d + 4] if d odd .H´ENON MAPS WITH MANY RATIONAL PERIODIC POINTS 5 The polynomial family sd is alternatively characterized by initial values together with the inductive relation δs d+1 (x) = sd(x), s 1(x) = x where δ is the central difference operator : δf (x) = f x + 12 − f x − 12 . For d ≡ 1 mod 4 , s d can be interpreted as a discrete analogue of the sine func-tion on some finite interval; indeed, the construction of sd boils down to being a polynomial interpolation on the set −d+1 2 , −d−12 , · · · , d+1 2 of a 6-periodic sequence −1, −1, 0, 1, 1, 0, · · · . Applying the discrete δ, see that each sd corresponds to ± sin or ± cos. We formalize this observation, which summarizes the computation of subsections 4.2 and 4.4 in [DH]: Lemma 2.1. Let σ : Z → Z be a 6-periodic function defined by σ(0) = 0, σ(1) = 1, σ(2) = 1 and σ(m + 3) = −σ(m) for all m. Then sd(m) agrees with σ(m + 32 (d − 1)) on values m = −d+1 2 , −d−12 , · · · , d+1 2 for all d ≥ 1. Proof. This proof rephrases the definition of sd, which was given in terms of the doubly periodic sequence ρ(m, d ) in [DH], in a more intuitive way. Consider the following table, where each row is 6-periodic and each column is 4-periodic: dm · · · −3 −52 −2 −32 −1 −12 0 12 1 32 2 52 3 · · · 1 0 −1 −1 0 1 1 02 1 0 −1 −1 0 13 0 1 1 0 −1 −1 04 −1 0 1 1 0 −15 0 1 1 0 −1 −1 0Note that applying the central difference operator δ to the ( d + 1)-th row gives the d-th row. Unpacking the definition given in [DH], it is straightforward to check that ρ(m, d ) = σ(m + d − 2). Since sd is defined by the interpolation sd(m − d+1 2 ) = ρ(m, d )on integers 0 ≤ m ≤ d + 1, and the d-th row is given by values of the function σ(m + 32 (d − 1)), it follows that sd interpolates the centremost d + 2 values of the d-th row above. □ The family sd hence serves as an example of a family of polynomials with small range (i.e. within −1, 0, 1) over an interval of size d + 2; in fact, sd has a linearly (in d)bounded range over an interval of size d + 6. The transformation from sd to rd ensures that rd is integer-valued on Z and dynamically compresses the interval [ d + 6]. 3. Estimates for sd In this section we prove a series of estimates of sd which allow us to study the H´ enon maps of Section 4. As an application of these estimates to one-dimensional dynamics, 6 KIM, KRIEGER, POSTOLACHE, AND SZETO we will also see in Subsection 3.3 that the dynamically compressing maps rd have no rational preperiodic points outside the set of integers they compress; that is, the bounds given for the explicit examples of Section 4 of [DH] are sharp. 3.1. Local uniform convergence of sd. In this subsection we establish the local uniform convergence of subsequences of sd to the trigonometric functions they approx-imate. Proposition 3.1. (−1) ds2d+1 (z) locally uniformly converges to the sine function 2√3 sin πz 3 as d → ∞ on the complex plane; that is, ∞ X j=0 (−1) j c2j+1 (z) = 2 √3 sin πz 3 locally uniformly on z ∈ C. Proof. Observe that (−1) j c2j+1 (z) = ( −1) j z (z + 1)( z + 2) · · · (z + j)( z − 1)( z − 2) · · · (z − j)(2 j + 1)! = z (1 − z)(2 − z) · · · (j − z)(1 + z)(2 + z) · · · (j + z)(32 )( 52 ) · · · (2j+1 2 )14j j!= z (1 − z)(j)(1 + z)(j) (3 /2) (j) 14j j!, where q(j) := q(q + 1) · · · (q + j − 1) denotes the rising factorial. Hence the infinite sum of these functions are ∞ X j=0 (−1) j c2j+1 (z) = z ∞ X j=0 (1 − z)(j)(1 + z)(j) (3 /2) (j) 14j j! = z2F1 1 − z, 1 + z; 32; 14 , where 2F1 is the Gaussian hypergeometric function defined by 2 F1(a, b ; c; z) = ∞ X n=0 a(n)b(n) c(n) zn n!for c̸ ∈ Z≤0 and \|z\| < 1. Note that the ratios between consecutive terms converge to (a + n)( b + n)z (c + n)n → z, which implies that the series converges uniformly on compact subsets of (a, b, c, z ) ∈ C4 : c̸ ∈ Z≤0, \|z\| < 1 . We use formula 15.1.16 of [AS72]: 2 F1 a, 2 − a; 32; sin 2 w = sin((2 a − 2) w)(a − 1) sin(2 w).H´ENON MAPS WITH MANY RATIONAL PERIODIC POINTS 7 Putting a = 1 − z and w = π/ 6, we get the desired result of ∞ X j=0 (−1) j c2j+1 (z) = z2F1 1 − z, 1 + z; 32; 14 = 2 √3 sin πz 3 , where the convergence is locally uniform in z ∈ C. □ Since all sd and the sine function are holomorphic, it follows from standard results in complex analysis that the derivative of ( −1) ds2d+1 (z) also locally converges to the de-rivative of the sine wave. A similar argument shows that ( −1) ds2d(z) and its derivative locally uniformly converge to 2√3 cos( πz 3 ) and its derivative on the complex plane. 3.2. Tail behaviour of sd. We now work with sd as functions of a real (rather than complex) variable and show that sd is strictly increasing on the real interval [ d+3 2 , ∞); this is achieved by first showing that sd is strictly increasing on [ d+3 2 , d+5 2 ] and then applying a double induction argument. For our analysis, we need some explicit auxiliary bounds on the derivative of cd on the real line. For convenience, we put cd(x) in the same setting for both even and odd d by use of the generalized binomial coefficient, which we denote ˜cd(x) := 1 d!x(x − 1)( x − 2) · · · (x − d + 1) , noting that cd(x) = ˜ cd(x + d−12 ). Lemma 3.2. \|c′ d (x)\| ≤ 1 for \|x\| ≤ d−12 . Proof. We prove that \|˜c′ d (x)\| ≤ 1 on 0 ≤ x ≤ d − 1. By the product rule, we have the following expression: ˜c′ d (x) = 1 d! X 0≤j<d Y 0≤i<d,i ̸=j (x − i). By the triangle inequality, \|˜c′ d (x)\| ≤ 1 d! X 0≤j<d Y 0≤i<d,i ̸=j \|x − i\| ≤ 1 d! X 0≤j<d (d − 1)! = 1 . □ Lemma 3.3. \|c′ d (x)\| ≤ 6+4 log dd(d−1) for \|x\| ≤ d−32 . Proof. We prove that \|˜c′ d (x)\| ≤ 6+4 log dd(d−1) on 1 ≤ x ≤ d − 2. By continuity of ˜ c′ d , it suffices to consider non-integer x. We carry on from the expression above: \|˜c′ d (x)\| ≤ 1 d! X 0≤j<d Y 0≤i<d,i ̸=j \|x − i\|. Fix an integer 1 ≤ N < d − 2 and take N < x < N + 1. Note that for j < N , Y i̸=j \|x − i\| ≤ (N + 1)!( d − N − 1)! x − j ≤ (N + 1)!( d − N − 1)! N − j .8 KIM, KRIEGER, POSTOLACHE, AND SZETO Similarly for j > N + 1, Y i̸=j \|x − i\| ≤ (N + 1)!( d − N − 1)! j − x ≤ (N + 1)!( d − N − 1)! j − N − 1 . For j = N and N + 1, we simply get X j=N,N +1 Y i̸=j \|x − i\| ≤ (( x − N ) + ( N + 1 − x)) · Y i̸=N,N +1 \|x − i\| ≤ (N + 1)!( d − N − 1)! . Combining these bounds, we obtain: \|˜c′ d (x)\| ≤ 1 d!(N + 1)!( d − N − 1)!(1 + HN −1 + Hd−N ) ≤ 1 + 2 HddCN +1 ≤ 1 + 2(log d + 1) d(d − 1) /2 = 6 + 4 log dd(d − 1) , where Hn = 1 + 1 /2 + · · · + 1 /n < log n + 1. □ Lemma 3.4. sd(x) is strictly increasing on d+3 2 ≤ x ≤ d+5 2 for all d ≥ 1. Proof. For d ≤ 44, this is verified by numerical calculation, so it suffices to prove that s′ d (x) > 0 on [ d+3 2 , d+5 2 ] for all d > 44. We established that s′ d (x) − c′ d+2 (x) + c′ d+4 (x) − c′ d+6 (x) + · · · = ( 2π 3√3 cos πx 3 d odd − 2π 3√3 sin πx 3 d even where the sum converges uniformly on d+3 2 ≤ x ≤ d+5 2 . Moreover note that \|c′ d+6 (x) − c′ d+8 (x) + · · · \| ≤ \| c′ d+6 (x)\| + ∞ X k=4 \|c′ d+2 k (x)\| < 1 + ∞ X k=4 6 + 4 log( d + 2 k)(d + 2 k)( d + 2 k − 1) =: Ed by Lemmas 3.2 and 3.3. Since each of the summands are decreasing in d and numerical calculation gives Ed < 2 for d = 44, it follows that Ed < 2 for all d > 44. Hence it suffices to show that c′ d+2 (x) − c′ d+4 (x) > 2π 3√3 + 2 on d+3 2 ≤ x ≤ d+5 2 . Equivalently, it suffices to show that ˜c′ d+2 (x) − ˜c′ d+4 (x + 1) > 2π 3√3 + 2 on d + 3 ≤ x ≤ d + 4. Expanding the LHS expression with the product rule, H´ENON MAPS WITH MANY RATIONAL PERIODIC POINTS 9 ˜c′ d+2 (x) − ˜c′ d+4 (x + 1) = ddx 1(d + 2)! x(x − 1) · · · (x − d − 1) 1 − (x + 1)( x − d − 2) (d + 3)( d + 4) = 1(d + 2)! 1 − (x + 1)( x − d − 2) (d + 3)( d + 4) d+1 X j=0 Y i̸=j, 0≤i≤d+1 (x − i) − 1(d + 2)! x(x − 1) · · · (x − d − 1) 2x − d − 1(d + 3)( d + 4) . We can bound the expressions above on d + 3 ≤ x ≤ d + 4 as follows: • By putting x = d + 4, 1 − (x + 1)( x − d − 2) (d + 3)( d + 4) ≥ 1 − 2( d + 5) (d + 3)( d + 4) ≥ dd + 3 . • By putting x = d + 3, d+1 X j=0 Y i̸=j, 0≤i≤d+1 (x − i) ≥ (d + 3)! 12 + 13 + · · · + 1 d + 3 (d + 3)! (log d − 1/2) , where we use the lower bound Hd > log d + 1 /2. • By putting x = d + 4, 1(d + 2)! x(x − 1) · · · (x − d − 1) ≤ (d + 4)( d + 3) 2 . • By putting x = d + 4, 2x − d − 1(d + 3)( d + 4) ≤ d + 7 (d + 3)( d + 4) < 2 d + 3 . Combining all of these bounds, we get the following bound: ˜c′ d+2 (x) − ˜c′ d+4 (x + 1) > d (log d − 1/2) − d − 4on d + 3 ≤ x ≤ d + 4. Since d (log d − 1/2) − d − 4 > 2π 3√3 2 for all d > 44, this proves the lemma. □ Proposition 3.5. sd(x) is strictly increasing on x ≥ d+3 2 for all d ≥ 1. Proof. We proceed by induction on d. Claim is true for d = 1 since sd(x) = x. Suppose the claim is true for d − 1. We already established that sd(x) is increasing on [ d+3 2 , d+5 2 ]by Lemma 3.4. For d+5 2 ≤ x < y ≤ d+7 2 , observe that sd(y) = sd(y − 1) + sd−1(y − 1/2) s d(x − 1) + sd−1(x − 1/2) = sd(x), since δs d = sd−1 holds by definition, and sd−1 is strictly increasing on [ d+2 2 , ∞) by in-duction hypothesis. Thus sd is strictly increasing on [ d+5 2 , d+7 2 ]. Repeating this process 10 KIM, KRIEGER, POSTOLACHE, AND SZETO for [ d+7 2 , d+9 2 ] and so on, we conclude that sd is strictly increasing on [ d+3 2 , ∞). Hence the claim is true for d. By induction, the claim is true for all d. □ We conclude this subsection with a lower bound of sd on some interval in the integers. This result will be useful in proving the boundedness of K(rd) and understanding the preperiodic points of rd, and will also be used in Section 4 when analyzing the periodic points of the H´ enon map hd. Lemma 3.6. For all d ≥ 3 and x ∈ Z + d+1 2 , we have sd(x) ≥ 3x for all x ≥ d+7 2 . Proof. We proceed by induction on d. The claim is true for s3(x) = x3−7x 6 . Now take d ≥ 4 and suppose claim is true for some d − 1. Take x = d+7+2 k 2 with integer k ≥ 0, then by sd−1 = δs d and the induction hypothesis we obtain sd(x) = sd d + 5 2 + k X j=0 sd−1 d + 6 + 2 j 2 ≥ sd d + 5 2 + k X j=0 32(d + 6 + 2 j) = sd d + 5 2 32(k + 1)( d + 6 + k). Since sd(d+5 2 ) ≥ d + 1 by Lemma 4.2(5) of [DH], we have sd(x) − 3x ≥ d + 1 + 32(k + 1)( d + 6 + k) − 32(d + 7 + 2 k)= d − 12 + 32k(k + d + 5) ≥ 92 + 32k(k + 9) > 0for all k ≥ 0 and d ≥ 4. Therefore the claim is true for d. □ 3.3. Rational preperiodic points of rd. As a consequence of our estimates in the previous subsections, we will show that the effective examples rd of [DH] have no ‘extra’ rational preperiodic points. Proposition 3.7. Let K(f ) ⊂ C denote the filled Julia set of polynomial f . We have that K(rd) ∩ R ⊂ (0 , d + 7) for all d ≥ 2. Proof. It suffices to show that \|rd(x)\| > \|x\| for x ≥ d + 7 and x ≤ 0. This is true for d = 2, so wlog assume d ≥ 3. • d even: Under conjugation by translation, we must show that sd(x) − d + 3 2 > \|x\| for \|x\| ≥ d+7 2 . By symmetry of sd, it suffices to show this for x ≥ d+7 2 only. Recall from Lemma 3.6 that sd(n) ≥ 3n for all n ∈ Z + 1 /2 with n ≥ d+7 2 .Taking any real number n ≤ x < n + 1, we have sd(x) ≥ sd(n) ≥ 3n > x + d + 3 2 . Note that we use Proposition 3.5 for sd(x) ≥ sd(n). H´ENON MAPS WITH MANY RATIONAL PERIODIC POINTS 11 • d odd: Under conjugation by translation, we must show that sd(x) − x − 1 > x for x ≥ d+7 2 and sd(x) − x − 1 < x for x ≤ − d+7 2 . By symmetry of sd, it suffices to show that sd(x) > 2x + 1 . for x ≥ d+7 2 . Noting that sd(n) ≥ 3n for all integers n ≥ d+7 2 , for any real number n ≤ x < n + 1 we have sd(x) ≥ sd(n) ≥ 3n > 2x + 1 . Again note that we use Proposition 3.5 for sd(x) ≥ sd(n). □ We now consider the non-Archimedean behavior of rd(x) to complete our study of the set of rational preperiodic points for rd(x). From the closed form for sd(x) we will show that for any prime p, the p-adic escape radius of rd is bounded above by 1. Proposition 3.8. Let d ≥ 2. Let p be a prime, with associated p-adic absolute value \| · \| p. If x ∈ Q with \|x\|p > 1 then \|rd(x)\|p > \|x\|p. Proof. We first consider the case when d = 2 n is even. Recall then that rd(x) = sd x − 3 − d + 1 2 2 , where sd(x) = n X k=0 (−1) n−kc2k(x)and c2k(x) = 1(2 k)! k Y j=1 x2 − 2j − 12 2! . Suppose that \|x\|p > \|1/2\|p, and write y = x − 3 − d+1 2 . Then \|y\|p = \|x\|p > \|1/2\|p and so y2 − 2j−12 2 p = \|y\|2 p , so \|c2k(y)\|p = \|y\|2kp \|(2 k)! \|−1 p . Since \|x\|p > 1, this is strictly increasing in k and so we see that \|rd(x)\|p = \|sd(y)\|p = \|c2n(x)\|p ≥ \| y\|2np = \|x\|2np > \|x\|p as claimed. It remains to deal with the case when p = 2 and \|x\|2 = 2 . In that case, x − 3 − (d + 1) /2 is an integer, and we have \|c2k(x)\|2 = \|1/4\|k 2 \|(2 k)! \|−12 . Again this increases in k and so \|rd(x)\|2 = \|sd(x − 3 − (d + 1) /2) \|2 = \|c2n(x − 3 − (d + 1) /2) \|2 = \|1/4\|n 2 \|(2 n)! \|−12 > \|x\|2, as desired. The case when d is odd is a similar computation, noting that we use d ≥ 2 to ensure that \|sd(x − 3 − (d + 1) /2) \|p > \|x\|p whenever \|x\|p > 1.12 KIM, KRIEGER, POSTOLACHE, AND SZETO □ Corollary 3.9. Let d ≥ 2 and rd(x) as defined above. Then the set of rational preperiodic points of rd(x) is precisely 1 , 2, . . . , d + 6. Proof. By the previous proposition, any rational preperiodic point is a p-adic integer for all primes p, thus a rational integer. Since the filled Julia set satisfies K(rd) ∩ R ⊂ (0 , d + 7), the only possible rational preperiodic points lie in the set 1 , 2, . . . , d + 6 , and as rd(x) maps this set to itself (Theorem 4.4, [DH]), this is precisely the set of rational preperiodic points for rd(x). □ 3.4. Optimality among dynamically compressing polynomials. In Section 5 of [DH], the authors list integer-valued polynomials of degree 2 ≤ d ≤ 20 that exhibit dy-namical compression of large intervals. The example provided for d = 2 coincides with the polynomial r2 and compresses r2() ⊆ . By elementary means, one can show that this is the optimal choice for dynamical compression in degree 2 in the following sense: if an integer-valued quadratic polynomial f of degree 2 satisfies f ([ m]) ⊂ [m], then m ≤ 8. In fact, m = 8 is attained by only r2(x) = x2−9x+22 2 and r2(x) + 1. In contrast, r3 is does not compress the maximal possible range in degree 3 since it compresses r3() ⊆ , while f (x) = x3−18 x2+89 x−66 6 , which is the example listed in [DH], compresses f () ⊆ . Nevertheless, it is possible by similar elementary means to show that f compresses the largest interval for degree 3 and is the unique such polynomial. Unfortunately, the elementary approach does not generalize well even to other small degrees. 4. H´ enon maps with many periodic points In this section, we study the dynamical behaviour of the following family of H´ enon maps: hd(x, y ) = ( y, −x + sd(y)) for odd d ≥ 3. Recall that sd is integer-valued for odd d, so hd (and its inverse) preserves the integral lattice Z2. This family exhibits many interesting properties, namely having a large number of periodic integer points near the origin with long periodic cycles. We are interested in asymptotic results so focus on the case of d odd to simplify the estimates and exposition: similar results will likely hold for some suitably chosen shift in the case of d even. Note that hd is conjugate to its inverse: h−1 d (x, y ) = ( −y + sd(x), x ) = ( r ◦ hd ◦ r−1)( x, y ), where r(x, y ) = ( y, x ) is a map that reflects the x and y coordinates. 4.1. Escape radius. In this subsection, we investigate the (archimedean and non-archimedean) filled Julia sets of the H´ enon map hd. The filled Julia set K(f ) of a plane polynomial automorphism f : C2 → C2 is defined to be K(f ) = K+(f ) ∩ K−(f )where K±(f ) = z ∈ C2 : sup n≥1 ∥f ±n(z)∥ < ∞ .H´ENON MAPS WITH MANY RATIONAL PERIODIC POINTS 13 In particular, the filled Julia set K(f ) contains all periodic points of f .We prove that there exists some Rv > 0 for each rational place v such that every periodic point of hd must satisfy max {\| x\|v, \|y\|v} < R v for all v. The proof involves cut-ting up Q2 into subsets between which the dynamics of the points are well-understood; it is a similar approach to studying the filtration properties of H´ enon maps as demon-strated in [BS91]. The escape radii results will imply that all periodic points of hd are integer points within a certain box and leads to an upper bound on the number of periodic points on hd as a quadratic function of d.For each place v of Q, we define the following norm on Q2: ∥(x, y )∥v = max {\| x\|v, \|y\|v} . Let s ∈ Q[x] be a non-zero odd polynomial of degree at least two, and define h(x, y ) = ( y, −x + s(y)) . For each place v, choose Rv > 0 such that \|s(y)\|v > 2\|y\|v for all \|y\|v ≥ Rv. Lemma 4.1. For each rational place v, consider the sets: Dv = (x, y ) ∈ Q2 : \|y\|v ≥ \| x\|v ≥ Rv , Tv = (x, y ) ∈ Q2 : \|x\|v < R v ≤ \| y\|v For all ( x, y ) ∈ D v ∪ T v, we have h(x, y ) ∈ D v and ∥h(x, y )∥v > ∥(x, y )∥v. Proof. For point ( x, y ) ∈ D v ∪ T v, note that \|s(y)\|v > 2\|y\|v and ∥(x, y )∥v = \|y\|v.Applying the triangle inequality, we obtain: \| − x + s(y)\|v ≥ 2\|y\|v − \| x\|v > \|y\|v ≥ Rv. Hence h(x, y ) ∈ D v and ∥h(x, y )∥v > ∥(x, y )∥v. □ The lemma above shows that the ∥ · ∥ v-norm of ( x, y ) ∈ D v ∪ T v strictly increases under iterations of h, which implies that all points ( x, y ) ∈ D v ∪ T v are not periodic points of h. The following diagram describes the dynamics of points under h: \|x\|v \|y\|v Rv Rv ⟳ Tv Dv Since h−1 = r ◦ h ◦ r for r(x, y ) = ( y, x ), it follows that the ∥ · ∥ v-norm of ( x, y ) ∈ r(Dv ∪Tv) strictly increases under iterations of h−1. Therefore any point in Dv ∪T v ∪r(Dv ∪T v)cannot be a periodic point of h. Since this set is precisely the complement of the open ball of radius Rv under the ∥ · ∥ v-norm, we reach the following conclusion: 14 KIM, KRIEGER, POSTOLACHE, AND SZETO Proposition 4.2. If ( x, y ) ∈ Q2 is a periodic point of h, then ∥(x, y )∥v < R v for all places v of Q.We now apply this to the case of s = sd and h = hd for odd d ≥ 3. We wish our bounds to depend explicitly on d, which amounts to finding for each place some Rv > 0such that \|sd(y)\|v > 2\|y\|v for all \|y\|v ≥ Rv.For v = ∞, recall from the proof of Theorem 3.7 that sd(y) > 2y + 1 for x ≥ d+7 2 . It follows from sd being an odd function that \|sd(y)\| > 2\|y\| for all \|x\| ≥ d+7 2 . Hence we can take R∞ = d+7 2 .For v = p prime, observe that the leading coefficient of sd is precisely 1 /d ! and that all coefficients are integer multiplies of 1 /d !. Take \|y\|p > 1 + 2 \|d!\|p, then by applying the ultrametric inequality on \|sd(y)\|p, we obtain: \|sd(y)\|p − 2\|y\|p ≥ 1 d! p \|y\|dp − 1 d! p \|y\|d−1 p − 2\|y\|d−1 p = \|y\|d−1 p \|y\|p − 1 − 2\|d!\|p \|d!\|p 0. Hence we can take Rp = 1 + 3 \|d!\|p. Using these results, we conclude: Corollary 4.3. For odd d ≥ 3, every rational periodic point ( x, y ) of hd is an integer point satisfying max {\| x\|, \|y\|} ≤ d+5 2 . Proof. By R∞ = d+7 2 , every rational periodic point ( x, y ) of hd satisfies max {\| x\|, \|y\|} < d+7 2 . It remains to show that ( x, y ) is an integer point. Observe that it suffices to prove Rp < p , since ∥(x, y )∥p < R p < p for all p implies that both x and y are integers. This inequality is true for all ( p, d ) with the exception of ( p, d ) = (2 , 3). For this case, we have ∥(x, y )∥2 < 1 + 3 · \| 3! \|2 < 22 and thus an exhaustive check of all half-integer points with max {\| x\|, \|y\|} < 3+7 2 = 5 is sufficient. □ 4.2. Counting periodic points. In the previous section, we showed that every ra-tional periodic point of hd must be an integer point inside the box ∥(x, y )∥∞ ≤ d+5 2 .We now turn to show that most integer points inside this box are periodic points of hd. Theorem 4.4. Define R to be as follows: R :=  d+1 2 if d ≡ 1 mod 6 , d+1 2 − 2 if d ≡ 3 mod 6 , d+1 2 − 3 if d ≡ 5 mod 6 . Any integer point ( x, y ) ∈ Z2 with ∥(x, y )∥∞ ≤ R is periodic under hd. Proof. Define the box B = {(x, y ) ∈ Z2 : ∥(x, y )∥∞ ≤ R}. It is sufficient to prove for (x, y ) ∈ B that the iterates eventually return to B. Recall that \|sd(y)\| ≤ 1 for \|y\| ≤ d+1 2 ,hence for ( x, y ) ∈ B we have ∥hd(x, y )∥∞ ≤ \| x\| + \|sd(y)\| ≤ R + 1 . Here equality holds iff x = R, sd(y) = −1 or x = −R, sd(y) = 1. In any other case, we are done since we immediately get hd(x, y ) ∈ B.For the two cases above, observe that hd(−x, −y) = −hd(x, y ) and so it suffices to consider the case of x = R, sd(y) = −1. The computation is tedious, but since values of H´ENON MAPS WITH MANY RATIONAL PERIODIC POINTS 15 sd over the integers −d+5 2 , · · · , d+5 2 are completely understood given d mod 6, we can divide into cases and manually iterate hd starting from the point hd(x, y ) = ( y, −R−1). As an example, suppose that d ≡ 1 mod 6 and y ≡ 4 mod 6. Recall by Lemma 2.1 that sd is 6-periodic from −R to R, with sd(R−1) = 0 , s d(R) = 1 and that sd(R+1) = 2 . We have (y, −R − 1) 7 → (−R − 1, −y − 2) since sd is odd and sd(R + 1) = 2 . If \| − y − 2\| > R and \|y\| ≤ R we must have y ∈ { R − 1, R } which contradicts sd(y) = −1. So \| − y − 2\| ≤ R and the value of sd(−y − 2) = 0 . So (−R − 1, −y − 2) 7 → (−y − 2, R + 1) 7 → (R + 1 , y + 4) . If \|y + 4 \| ≤ R then ( R + 1 , y + 4) 7 → (y + 4 , −R) and we have returned to the box B.If \|y + 4 \| > R then we must have y = R − 3 as y ≡ 4 mod 6 . But we compute directly the orbit of ( R − 3, −R − 1): (R − 3, −R − 1) 7 → (−R − 1, −R + 1) 7 → (−R + 1 , R + 1) 7 → (R + 1 , R + 1) 7→ (R + 1 , −R + 1) 7 → (−R + 1 , −R − 1) 7 → (−R − 1, R − 3) 7 → (R − 3, R ), and ( R − 3, R ) ∈ B. Proceeding in this way for all cases of d and y, one sees that the iterates return to B after finitely many times. □ From the two results above, we deduce the following. Corollary 4.5. The number of rational periodic points of hd satisfies: (d − 4) 2 ≤ Per( hd, Q) ≤ (d + 6) 2. Remark. Computations suggest that the actual number of integer periodic points of hd is: Per( hd, Q) =  d2 − 8d 3 56 3 if d ≡ 1 mod 6 ,d2 + 8 if d ≡ 3 mod 6 ,d2 − 8d 3 40 3 if d ≡ 5 mod 6 . Long cycles In this section, we investigate long periodic cycles present within the integer periodic points of hd. This is the most surprising aspect of this family of H´ enon maps, as it turns out that for d ≡ 1 mod 6 the length of the longest cycle is 8d+10 3 , substantially better than interpolation in the degree d H´ enon family provides. With small modifications, similar results can be obtained in any odd degree, though we do not provide proof. Theorem 5.1. For d ≡ 1 mod 6, there exists a periodic cycle of length 8d+10 3 . Proof. For d ≡ 1 mod 6 recall that R = d+1 2 . We first tabulate the tail values of sd as follows: y R − 2 R − 1 R R + 1 R + 2 sd(y) −1 0 1 2 d + 2 16 KIM, KRIEGER, POSTOLACHE, AND SZETO We also recall that sd is an odd function and that it is 6-periodic on −R, · · · , R with values −1, −1, 0, 1, 1, 0. This determines the values of sd on integers −R − 2 to R + 2. We claim that the orbit of ( R + 1 , −R + 1) under iterations of hd take the following form: (−R − 1, −R + 1) (−R + 1 , R + 1) (R + 1 , R + 1) (R + 1 , −R + 1) (−R − 1, −R + 7) (R + 1 , −R + 7) ... ...(−R − 1, R − 7) (R + 1 , R − 7) (−R − 1, R − 1) (−R − 1, −R − 1) (R − 1, −R − 1) (R + 1 , R − 1) 8 8 8 8 111 8 8 8 8 111 Here the numbers on the arrows denote the number of iterations of hd involved. The horizontal arrows are clear. For the right-hand side vertical arrows, it suffices to prove that h◦8 d (R + 1 , y ) = ( R + 1 , y + 6) for integer y satisfying −R + 1 ≤ y ≤ R − 7 and y ≡ R − 1 mod 6. This is a simple check: (R + 1 , y ) 7 → (y, −R − 1) 7 → (−R − 1, −y − 2) 7 → (−y − 2, −R) 7 → (R, y + 3) 7→ (y + 3 , −R) 7 → (−R, −y − 4) 7 → (−y − 4, R + 1) 7 → (R + 1 , y + 6) The left-hand side vertical arrows follow from hd(−x, −y) = −hd(x, y ). Therefore the period of ( R + 1 , −R + 1) is 6 + 2 · 8 · 2R − 26 = 8d + 10 3 . □ As noted in the introduction, the family of H´ enon maps is defined by d + 2 param-eters, so we see that these cycles are much longer than one can hope for from a naive interpolation argument (see Section 11 of [H´ e24] for details of this argument). 5.1. Effects of shifting. In the case when d ≡ 3, 5 mod 6, computations suggest that the longest periodic cycle of hd has length ≤ 20 for all d. This resonates with the fact, which will be discussed later in the last section, that the longest cycle of the limiting H´ enon map h∞ has length 20. It is an interesting feature of the dynamics on the boundary of the escape region in the case d ≡ 1 mod 6 that hd is a good but not too good approximation of the map h∞. This suggests that for d ≡ 3, 5 mod 6 similar long cycles might be obtained by minor (integer-valued) perturbations of hd. Indeed, here we list some experimental results that arise from shifting our H´ enon maps. Define hd,c (x, y ) to be hd,c (x, y ) = ( −y, x + sd(y + c)) where c ∈ Z. We find that such H´ enon maps may have even more integer periodic points and even longer cycles for a suitably chosen shift c. The following tables contain H´ENON MAPS WITH MANY RATIONAL PERIODIC POINTS 17 polynomial interpolations of computational results obtained for 15 ≤ d ≤ 299; we do not provide any proof. • d ≡ 1 mod 6: c Number of integer periodic points Longest cycle 0 d2 − 8d 3 56 38d+10 3 ±1 d2 + 2 d + 4 10 d−73 ±2 d2 − 6d + 18 16 d−61 3 • d ≡ 3 mod 6: c Number of integer periodic points Longest cycle 0 d2 + 8 20 −1 d2 + 4 d 8d − 39 1 d2 + 4 d + 1 8d − 39 ±2 d2 − 4d + 7 60 • d ≡ 5 mod 6: c Number of integer periodic points Longest cycle 0 d2 − 8d 3 40 3 20 ±1 d2 − 2d + 29 14 d−31 3 ±2 d2 − 22 d 3 161 328 d−185 3 5.2. Transcendental H´ enon maps. Recall from Proposition 3.1 that ( −1) ds2d+1 (x)locally uniformly converges to 2√3 sin( πx 3 ) on the real line as d tends to infinity. We conclude this article by studying the dynamical behaviour of the limiting H´ enon map: h∞(x, y ) = y, −x + 2 √3 sin πy 3 . As each hd is integer-valued for each odd d, the rapid convergence of sd on a neighbor-hood of 0 ensures that the integer periodic points of hd will have dynamics described by h∞ for points of sufficiently small norm compared to d. Proposition 5.2. Any integer point ( x, y ) ∈ Z2 is periodic under iteration of h∞ with periods 1,4,5,6,12 or 20. The period is determined by the values of x, y modulo 6 as follows: 18 KIM, KRIEGER, POSTOLACHE, AND SZETO y mod 6 x mod 6 0 1 2 3 4 55 12 20 20 20 20 12 4 20 20 12 12 20 20 3 4 20 12 4 12 20 2 20 20 20 12 12 20 1 12 12 20 20 20 20 0 4 12 20 4 20 12 There are the following exceptions: period of ( x, y ) =  1 (x, y ) = (0 , 0) , 5 (x, y ) = (2 , 0) , (2 , 1) , (1 , 2) , (0 , 2) , (−1, 1) , (−2, 0) , (−2, −1) , (−1, −2) , (0 , −2) , (1 , −1) , 6 (x, y ) = (1 , 0) , (1 , 1) , (0 , 1) , (−1, 0) , (−1, −1) , (0 , −1) . Proof. One can check this manually by iterating the point (6 x + a, 6y + b) for integers 0 ≤ a, b < 6. Here we show only one instance of this. Take a = 1, b = 0. The forward orbit of (6 x + 1 , 6y) over h∞ is as follows: (6 x + 1 , 6y) 7 → (6 y, −6x − 1) 7 → (−6x − 1, 6y − 1) 7 → (−6y − 1, 6x) 7 → (6 x, 6y + 1) 7→ (6 y + 1 , −6x + 1) 7 → (−6x + 1 , −6y) 7 → (−6y, 6x − 1) 7 → (6 x − 1, 6y − 1) 7→ (6 y − 1, −6x) 7 → (−6x, −6y + 1) 7 → (−6y + 1 , 6x + 1) 7 → (6 x + 1 , 6y). Note that all of these points are distinct except for the case of x = y = 0, when the sixth iterate is equal to the initial point, hence the point (1 , 0) (and its iterates) having a period of six. □ We now analyse the local behaviour of hd near the integer points. Observe that the Jacobian matrix of h∞ is 6-periodic in both x and y: Dh ∞(x, y ) = 0 1 −1 2π 3√3 cos( πx 3 ) . Note also from the proof of Proposition 5.2 that the forward orbit modulo 6 only depend on the values of x and y modulo 6, excluding some ( x, y ) that are close to the origin. The two facts combined makes it possible to explicitly calculate the multiplier and its eigenvalues for all integer points. Here we instead take a visual approach: We give a small perturbation on each integer point and plot its forward orbit. Each points in the forward orbit is coloured as follows. Forward orbits of points near: • period 1 points are coloured in bright red; • period 4 points are coloured in dark red/black; • period 6 points are coloured in bright blue; • period 12 points are coloured in navy/cyan; H´ENON MAPS WITH MANY RATIONAL PERIODIC POINTS 19 • period 5 points are coloured in green/dark purple; • period 20 points are coloured in lime/bright purple. After applying 2 × 10 5 iterations of h∞ to randomly perturbed points near integer points, we obtain the following picture: In the same way as hd, all integer points have multipliers with determinant 1. Hence the eigenvalues are either (i) real numbers whose product is 1 (of saddle type ), or (ii) complex conjugates with modulus 1. Looking at the picture above, we can categorize the integer points as follows: • Period 6, 12 points and period 4 points ( x, y ) ∈ 3Z × 3Z with x − y ≡ 0 mod 6 are not of saddle type (the orbits near these period 4 points are coloured in dark red). The small closed loops formed around those points suggest that the perturbed points are indeed being rotated. • Period 5, 20 points are of saddle type. The orbits of perturbations of these points converge to two distinct limit sets, each coloured in green and purple. 20 KIM, KRIEGER, POSTOLACHE, AND SZETO • Period 4 points ( x, y ) ∈ 3Z × 3Z with x − y ≡ 3 mod 6 are of saddle type, and the orbits near these points are coloured in black. The following picture depicts the forward orbit of a perturbation near (3 , 0) after 5 × 10 5 iterations of h∞: One might naturally ask whether a better choice of limiting map might produce longer cycles; however, it may be that any similar construction of a family of H´ enon maps arising from a convergent sequence of integer-valued polynomials (analogous to sd) can only admit long cycles on the boundary of the escape region, which would imply a linear bound in cycle length. It would also be interesting to explore the family of H´ enon maps of the form ( x, y ) 7 → (y, −x+fd(y)) where fd is the (not explicit) family of polynomials with d + ⌊log 2(d)⌋ rational preperiodic points whose existence is known by Doyle and Hyde. REFERENCES 21 References [AS72] M. Abramowitz and I. A. Stegun. Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. National Bureau of Standards Applied Mathematics Series 55. Tenth Printing. USA: ERIC, 1972. [Ben07] R. Benedetto. “Preperiodic points of polynomials over global fields”. J. Reine Angew. Math. 608 (2007), pp. 123–153. [BS91] E. Bedford and J. Smillie. “Polynomial diffeomorphisms of C2: currents, equilibrium measure and hyperbolicity.” Inventiones mathematicae 103 .1 (1991), pp. 69–100. [DH] J. R. Doyle and T. Hyde. “Polynomials with many rational preperiodic points”. To appear in Trans. Amer. Math. Soc. (). doi : 10.1090/tran/9406 .[DM24] L. DeMarco and N. M. Mavraki. “Dynamics on P1: preperiodic points and pairwise stability”. Compositio Mathematica 160 .2 (2024), 356––387. [Fak03] N. Fakhruddin. “Questions on self-maps of algebraic varieties”. J. Ramanu-jan Math. Soc. 18 .2 (2003), pp. 109–122. [FM89] S. Friedland and J. Milnor. “Dynamical properties of plane polynomial au-tomorphisms”. Ergodic Theory Dyn. Syst. 9.1 (1989), pp. 67–99. [HOV94] J. H. Hubbard and R. W. Oberste-Vorth. “H´ enon mappings in the complex domain I : the global topology of dynamical space”. Publications Math´ ematiques de l’IH ´ES 79 (1994), pp. 5–46. [H´ e24] J. X. de H´ enon. “H´ enon Maps: A List of Open Problems”. Arnold Mathe-matical Journal 10 (2024), pp. 585–620. [Ing14] P. Ingram. “Canonical heights for H´ enon maps”. Proceedings of the London Mathematical Society 108 .3 (2014), pp. 780–808. [Kaw06] S. Kawaguchi. “Canonical height functions for affine plane automorphisms”. Mathematische Annalen 335 (2006), pp. 285–310. [Loo21] N. R. Looper. The Uniform Boundedness and Dynamical Lang Conjectures for polynomials . 2021. arXiv: 2105.05240 [math.NT] .[Maz77] B. Mazur. “Modular curves and the Eisenstein ideal”. Publications Math´ ematiques de l’IH ´ES 47 (1977), pp. 33–186. [Mer96] L. Merel. “Bornes pour la torsiondes courbes elliptiques sur les corps de nombres”. Inventiones mathematicae 124 .1 (1996), pp. 437–449. [MS94] P. Morton and J. H. Silverman. “Rational periodic points of rational func-tions”. International Mathematics Research Notices 2 (1994), pp. 97–110. [Nor50] D. G. Northcott. “Periodic Points on an Algebraic Variety”. Annals of Math-ematics 51 .1 (1950), pp. 167–177. [Sil94] J. H. Silverman. “Geometric and arithmetic properties of the H´ enon map”. Mathematische Zeitschrift 215 .2 (1994), pp. 237–250. 22 REFERENCES Email address : hk523@cam.ac.uk Email address : hkrieger@dpmms.cam.ac.uk Email address : mip32@cam.ac.uk Email address : ws397@cam.ac.uk
4828	https://www.wallstreetmojo.com/deadweight-loss/	All BlogsEconomics ResourcesDeadweight Loss Deadweight Loss Publication Date : 25 Jun, 2022 Blog Author : Priya Choubey Edited by : Collins Enosh Reviewed by : Dheeraj Vaidya, CFA, FRM Share Table Of Contents Deadweight Loss Definition Deadweight loss refers to the cost borne by society when there is an imbalance between the demand and supply. It is a market inefficiency that is caused by the improper allocation of resources. In a free market scenario, the price of goods and services depends majorly on their demand and supply. But sometimes, market inefficiency is caused by an external force—government laws, taxation, subsidies, monopoly, price floors, or price ceilings. In such a market, commodities are either overvalued or undervalued. Table of contents Deadweight Loss Definition Deadweight Loss Explained Causes of Deadweight Loss #1 - Price Ceiling #2 - Price Floor #3 - Monopoly #4 - Taxation #5 - Subsidy #6 - Product Surplus Calculate Deadweight Loss Example Deadweight Loss Graph Frequently Asked Questions (FAQs) Recommended Articles Deadweight loss is the economic cost borne by society. It is a market inefficiency caused by an imbalance between consumption and allocation of resources. The deadweight inefficiency of a product can never be negative; it can be zero. Deadweight loss is zero when the demand is perfectly elastic or when the supply is perfectly inelastic. This market inefficiency is represented by the following formula:Deadweight Loss = ½ (New Price – Original Price) (Original Quantity – New Quantity) Deadweight Loss Explained A deadweight loss is a market inefficiency caused by a mismatch between goods consumption and demand. Due to the inefficiency, products are either overvalued or undervalued. In such scenarios, demand and supply are not driven by market forces. Instead, demand and supply are moved artificially—by factors like taxation, subsidies, product surplus, consumer surplus, monopoly, oligopoly, price ceiling, and price floor. Highly elastic commodities are prone to such inefficiencies. Price changes significantly impact the demand for a highly elastic commodity. Deadweight losses also arise when there is a positive externality. In such scenarios, the marginal benefit from a product is higher than the marginal social cost. Deadweight losses are not seen in an efficient market—where the market is run by fair competition. While the value of deadweight loss of a product can never be negative, it can be zero. Deadweight loss is zero when the demand is perfectly elastic or when the supply is perfectly inelastic. Causes of Deadweight Loss A deadweight inefficiency occurs when the market is unnaturally controlled by governments or external forces. Deadweight market inefficiency is caused by the following causes: #1 - Price Ceiling The government ascertains a maximum price for products—to prevent overcharging. However, price ceilings discourage sellers, as it curtails the possibility of earning high returns. Thus, price ceilings bring down goods supply. #2 - Price Floor Often, the government fixes a minimum selling price for goods. This increases product prices. Similarly, governments often fix a minimum wage for laborers and employees. But high wages result in job loss for incompetent employees. #3 - Monopoly When a single market player has a monopoly, the regulation of goods price and supply is unnatural. The selling price set by the monopolist is significantly higher than the marginal cost—the market becomes inefficient. Further, if customers are unable to afford the product or service—demand falls. #4 - Taxation When the government raises the taxes on certain goods or services, it influences the price and demand for that product. When taxes raise a product’s price, its demand starts falling. But this cuts into producers' profit margin. #5 - Subsidy Governments provide subsidies on certain goods or services—bringing the price down. As a result, the product demand rises. However, this artificially created demand drives consumers to buy a particular commodity in more quantity. #6 - Product Surplus When the market is flooded with excessive goods and the demand is low, a product surplus is created. When demand is low, the commodity’s price falls. Manufacturers incur losses due to the gap between supply and demand. Calculate Deadweight Loss For calculations, deadweight loss is half of the price change multiplied by the change in demand. It is computed using the following formula: Or, Here, ∆P is the price difference. ∆Q is the difference in the quantity demanded. P1 is the original price. Q1 is the original quantity. P2 is the new price. Q2 is the new quantity. Example Let us assume that economic equilibrium will be achieved for a product at the price of $8.The demand at this price is 8000 units. The government then imposes a price floor; the price is increased to $10. At this price, the expected demand falls to 7000 units. Based on the given data, calculate the deadweight loss. Solution:Dead weight = 0.5 (P2-P1) (Q1-Q2) = 0.5 (10-8) (8000-7000) = $1000 Thus, due to the price floor, manufacturers incur a loss of $1000. Deadweight Loss Graph The deadweight loss is the gap between the demand and supply of goods. Graphically is it represented as follows: In the above graph, the demand curve intersects with the supply curve at point' E,' i.e., equilibrium. Equilibrium is a scenario where the consumption and the allocation of goods are equal. However, due to the price ceiling, the demand curve shifts to the left—P2 is the new price. Similarly, Q2 is the new demanded quantity. Subsidies also shift the demand curve to the left. In contrast, price floors and taxes shift the demand curve towards the right. Frequently Asked Questions (FAQs) What is deadweight loss? Deadweight inefficiency is the economic cost incurred by society when there is an imbalance of demand and supply. This could be an inefficient resource allocation caused by government intervention, monopoly, collusion, product surplus, or product deficit. How to calculate deadweight loss? It is computed as half of the value acquired by multiplying the product's price change and the difference in quantity demanded. If ∆P is the price difference and ∆Q is the difference in the quantity demanded, deadweight inefficiency is computed using the following formula:Deadweight Loss = ½ (New Price – Original Price) (Original Quantity – New Quantity) Why does a monopoly cause a deadweight loss? When a single market player enjoys a monopoly, the monopolist regulates goods prices and supply. When supply is low, consumers are charged exorbitantly—significantly higher than the marginal cost. When consumers lose purchasing power, demand falls. Can deadweight loss be negative? It cannot be a negative value. But, it can be zero. This occurs when the demand is perfectly elastic or when the supply is perfectly inelastic. Recommended Articles This is a guide to what is Deadweight Loss and its Definition. We explain deadweight loss in economics, its meaning, calculation, graphs, & causes like monopoly, tax, price floor & price-ceiling. You can learn more about it from the following articles – Deadweight Loss Formula Antitrust Acts Supply-Side Economics FINANCIAL MODELING RESOURCES Create a Full Dynamic Financial Model in 2 Days (6 hours) \| Any Graduate Or Professional is eligible \| Build & Forecast IS, BS, CF from Scratch. Join WallStreetMojo YouTube INVESTMENT BANKING RESOURCES The Exact Training Used by Top Investment Banks \| FM, DCF, LBO, M&A, Accounting, Derivatives & More \| $2400+ in Exclusive Benefits \| 100+ Wall Street-Level Skills. Join WallStreetMojo Instagram CHATGPT RESOURCES Boost Productivity 10X with AI-Powered Excel \| Save Hours, Eliminate Errors \| $300+ in Exclusive Bonuses \| Advanced Data Analysis & Reporting with AI. Join WallStreetMojo LinkedIn EXCEL RESOURCES Master Excel, VBA, PowerBI Like a Pro \| 70+ Hours of Expert Training \| Real-world Excel applications \| Earn Your Certification & Land High-Paying Roles! Join WallStreetMojo Facebook Join WallStreetMojo X
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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Optimizing the algorithm for brute force numbers in the problem Ask Question Asked 4 years, 9 months ago Modified4 years, 9 months ago Viewed 138 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. How many pairs (i, j) exist such that 1 <= i <= j <= n, j - i <= a? 'n' and 'a' input numbers. The problem is my algorithm is too slow when increasing 'n' or 'a'. I cannot think of a correct algorithm. Execution time must be less than 10 seconds. Tests: n = 3 a = 1 Number of pairs = 5 n = 5 a = 4 Number of pairs = 15 n = 10 a = 5 Number of pairs = 45 n = 100 a = 0 Number of pairs = 100 n = 1000000 a = 333333 Number of pairs = 277778388889 n = 100000 a = 555555 Number of pairs = 5000050000 n = 1000000 a = 1000000 Number of pairs = 500000500000 n = 998999 a = 400000 Number of pairs = 319600398999 n = 898982 a = 40000 Number of pairs = 35160158982 ```python n, a = input().split() i = 1 j = 1 answer = 0 while True: if n >= j: if a >= (j-i): answer += 1 j += 1 else: i += 1 j = i if j > n: break else: i += 1 j = i if j > n: break print(answer) ``` python algorithm Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications edited Dec 27, 2020 at 13:51 Red 27.7k 8 8 gold badges 44 44 silver badges 63 63 bronze badges asked Dec 27, 2020 at 13:48 Alexander KarpenkoAlexander Karpenko 23 4 4 bronze badges Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. One can derive a direct formula to solve this problem. python ans = ((a+1)a)/2 + (a+1) + (a+1)(n-a-1) Thus the time complexity is O(1). This is the fastest way to solve this problem. Derivation: The first a number can have pairs of (a+1)C2 + (a+1). Every additional number has 'a+1' options to pair with. So, therefore, there are n-a-1 number remaining and have (a+1) options, (a+1)(n-a-1) Therefore the final answer is (a+1)C2 + (a+1) + (a+1)(n-a-1) implies ((a+1)a)/2 + (a+1) + (a+1)(n-a-1). Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Dec 27, 2020 at 16:03 Red 27.7k 8 8 gold badges 44 44 silver badges 63 63 bronze badges answered Dec 27, 2020 at 14:25 YogeshYogesh 800 1 1 gold badge 13 13 silver badges 23 23 bronze badges 1 Comment Add a comment John Coleman John ColemanOver a year ago Nice, I was in the process of deriving a closed-form formula from my answer when I saw how your approach was able to get the formula with much less fuss. 2020-12-27T14:35:51.477Z+00:00 0 Reply Copy link This answer is useful 3 Save this answer. Show activity on this post. You are using a quadratic algorithm but you should be able to get it to linear (or perhaps even better). The main problem is to determine how many pairs, given i and j. It is natural to split that off into a function. A key point is that, for i fixed, the j which go with that i are in the range i to min(n,i+a). This is since j-i <= a is equivalent to j <= i+a. There are min(n,i+a) - i + 1 valid j for a given i. This leads to: python def count_pairs(n,a): return sum(min(n,i+a) - i + 1 for i in range(1,n+1)) count_pairs(898982,40000) evaluates to 35160158982 in about a second. If that is still to slow, do some more mathematical analysis. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Dec 27, 2020 at 14:20 answered Dec 27, 2020 at 14:11 John ColemanJohn Coleman 52.1k 7 7 gold badges 59 59 silver badges 127 127 bronze badges 5 Comments Add a comment Red RedOver a year ago Oh,that explains it. Impressive. 2020-12-27T14:17:21.5Z+00:00 0 Reply Copy link John Coleman John ColemanOver a year ago @AnnZen n,a would actually be a bit more idiomatic (more like the range() function) and mesh better with OP's code, so I will tweak it. Thanks for noticing that my parameter order was slightly unnatural 2020-12-27T14:19:56.723Z+00:00 1 Reply Copy link Red RedOver a year ago You can improve its efficiency by eliminating the evaluation of +1 in each iteration: return sum(min(n,i+a) - i for i in range(2,n+2)) 2020-12-27T14:27:47.92Z+00:00 0 Reply Copy link Red RedOver a year ago And by evaluating the -i inside the min function: return sum(min(n-i,a) for i in range(2,n+2)) 2020-12-27T14:35:57.07Z+00:00 0 Reply Copy link John Coleman John ColemanOver a year ago @AnnZen Doubtless there are tweaks, but compared with the accepted answer it would be like rearranging deck furniture on the Titanic. 2020-12-27T14:37:11.957Z+00:00 1 Reply Copy link Add a comment This answer is useful 1 Save this answer. Show activity on this post. Here is an improvement: ```python n, a = map(int, input().split()) i = 1 j = 1 answer = 0 while True: if n >= j <= a + i: answer += 1 j += 1 continue i = j = i + 1 if j > n: break print(answer) ``` Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Dec 27, 2020 at 14:11 answered Dec 27, 2020 at 13:58 RedRed 27.7k 8 8 gold badges 44 44 silver badges 63 63 bronze badges 1 Comment Add a comment Red RedOver a year ago @selfAlex Updated. continue is faster then else. 2020-12-27T14:12:18.73Z+00:00 1 Reply Copy link Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. To learn more, see our tips on writing great answers. 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4830	https://en.wikipedia.org/wiki/Gastric_acid	Jump to content Gastric acid Afrikaans العربية 閩南語 / Bn-lm-gí Беларуская Български Català Čeština Dansk Deutsch ދިވެހިބަސް Eesti Ελληνικά Español Esperanto Euskara فارسی Français Galego 한국어 Հայերեն हिन्दी Bahasa Indonesia Íslenska Italiano ქართული Қазақша Lëtzebuergesch Македонски Nederlands 日本語 Norsk bokmål Occitan Oʻzbekcha / ўзбекча Polski Português Romnă Русский Shqip Српски / srpski Srpskohrvatski / српскохрватски Suomi Svenska తెలుగు ไทย Türkçe Українська Tiếng Việt 吴语粵語中文 Edit links From Wikipedia, the free encyclopedia \| \| \| --- \| \| \| This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. Find sources: "Gastric acid" – news · newspapers · books · scholar · JSTOR (March 2022) (Learn how and when to remove this message) \| Digestive fluid formed in the stomach Gastric acid or stomach acid is the acidic component – hydrochloric acid – of gastric juice, produced by parietal cells in the gastric glands of the stomach lining. In humans, the pH is between one and three, much lower than most other animals, but is very similar to that of carrion-eating carnivores that need protection from ingesting pathogens. With this higher acidity, gastric acid plays a key protective role against pathogens. It is also key in the digestion of proteins by activating digestive enzymes, which together break down the long chains of amino acids. Gastric acid is regulated in feedback systems to increase production when needed, such as after a meal. Other cells in the stomach produce bicarbonate, a base, to buffer the fluid, ensuring a regulated pH. These cells also produce mucus – a viscous barrier to prevent gastric acid from damaging the stomach. The pancreas further produces large amounts of bicarbonate, secreting this through the pancreatic duct to the duodenum to neutralize gastric acid passing into the digestive tract. The secretion is a complex and relatively energetically expensive process. Parietal cells contain an extensive secretory network (called canaliculi) from which the hydrochloric acid is secreted into the lumen of the stomach. The pH level is maintained by the proton pump H+/K+ ATPase. The parietal cell releases bicarbonate into the bloodstream in the process, which causes a temporary rise of pH in the blood, known as an alkaline tide. The gastric juice also contains digestive enzymes produced by other cells in the gastric glands – gastric chief cells. Gastric chief cells secrete an inactivated pepsinogen. Once in the stomach lumen gastric acid activates the proenzyme to pepsin. Secretion [edit] A typical adult human stomach will secrete about 1.5 liters of gastric juice daily. Gastric juice is the combination of gastric gland secretions including the main component of hydrochloric acid (gastric acid), gastric lipase and pepsinogen. Once in the stomach pepsinogen is changed by gastric acid to the digestive enzyme pepsin adding this enzyme to the gastric juice. In humans, the pH of gastric acid is between one and three, much lower than most other animals, but is very similar to that of carrion eating carnivores, needing extra protection from ingesting pathogens. Gastric acid secretion is produced in several steps. Chloride and hydrogen ions are secreted separately from the cytoplasm of parietal cells and mixed in the canaliculi. This creates a negative potential of between −40 and −70 mV across the parietal cell membrane that causes potassium ions and a small number of sodium ions to diffuse from the cytoplasm into the parietal cell canaliculi. Gastric acid is then secreted along with other gland secretions into the gastric pit for release into the stomach lumen. The enzyme carbonic anhydrase catalyses the reaction between carbon dioxide and water to form carbonic acid. This acid immediately dissociates into hydrogen and bicarbonate ions. The hydrogen ions leave the cell through H+/K+ ATPase antiporter pumps. At the same time, sodium ions are actively reabsorbed[citation needed] . This means that the majority of secreted K+ (potassium) and Na+ (sodium) ions return to the cytoplasm. In the canaliculus, secreted hydrogen and chloride ions mix and are secreted into the lumen of the oxyntic gland. The highest concentration that gastric acid reaches in the stomach is 160 mM in the canaliculi. This is about 3 million times that of arterial blood, but almost exactly isotonic with other bodily fluids. The lowest pH of the secreted acid is 0.8, but the acid is diluted in the stomach lumen to a pH of between 1 and 3. There is a small continuous basal secretion of gastric acid between meals of usually less than 10 mEq/hour. There are three phases in the secretion of gastric acid which increase the secretion rate in order to digest a meal: The cephalic phase: Thirty percent of the total gastric acid secretions to be produced is stimulated by anticipation of eating and the smell or taste of food. This signalling occurs from higher centres in the brain through the vagus nerve (Cranial Nerve X). It activates parietal cells to release acid and ECL cells to release histamine. The vagus nerve (CN X) also releases gastrin releasing peptide onto G cells. Finally, it also inhibits somatostatin release from D cells. The gastric phase: About sixty percent of the total acid for a meal is secreted in this phase. Acid secretion is stimulated by distension of the stomach and by amino acids present in the food. The intestinal phase: The remaining 10% of acid is secreted when chyme enters the small intestine, and is stimulated by small intestine distension and by amino acids. The duodenal cells release entero-oxyntin which acts on parietal cells without affecting gastrin. Regulation of secretion [edit] See also: Phases of digestion Gastric acid production is regulated by both the autonomic nervous system and several hormones. The parasympathetic nervous system, via the vagus nerve, and the hormone gastrin stimulate the parietal cell to produce gastric acid, both directly acting on parietal cells and indirectly, through the stimulation of the secretion of the hormone histamine from enterochromaffin-like cells (ECLs). Vasoactive intestinal peptide, cholecystokinin, and secretin all inhibit production. The production of gastric acid in the stomach is tightly regulated by positive regulators and negative feedback mechanisms. Four types of cells are involved in this process: parietal cells, G cells, D cells and enterochromaffin-like cells. Beside this, the endings of the vagus nerve (CN X) and the intramural nervous plexus in the digestive tract influence the secretion significantly. Nerve endings in the stomach secrete two stimulatory neurotransmitters: acetylcholine and gastrin-releasing peptide. Their action is both direct on parietal cells and mediated through the secretion of gastrin from G cells and histamine from enterochromaffin-like cells. Gastrin acts on parietal cells directly and indirectly too, by stimulating the release of histamine. The release of histamine is the most important positive regulation mechanism of the secretion of gastric acid in the stomach. Its release is stimulated by gastrin and acetylcholine and inhibited by somatostatin. Neutralization [edit] In the duodenum, gastric acid is neutralized by bicarbonate. This also blocks gastric enzymes that function optimally in the acid range of pH. The secretion of bicarbonate from the pancreas is stimulated by secretin. This polypeptide hormone gets activated and secreted from so-called S cells in the mucosa of the duodenum and jejunum when the pH in the duodenum falls below 4.5 to 5.0. The neutralization is described by the equation: : HCl + NaHCO3 → NaCl + H2CO3 The carbonic acid rapidly equilibrates with carbon dioxide and water through catalysis by carbonic anhydrase enzymes bound to the gut epithelial lining, leading to a net release of carbon dioxide gas within the lumen associated with neutralisation. In the absorptive upper intestine, such as the duodenum, both the dissolved carbon dioxide and carbonic acid will tend to equilibrate with the blood, leading to most of the gas produced on neutralisation being exhaled through the lungs. Clinical significance [edit] Gastroesophageal reflux disease (GERD) is a common disorder that occurs when stomach acid repeatedly flows back into the esophagus, this backwash of acid (reflux) also known as heartburn can irritate the lining of the esophagus. Most people are able to manage the discomfort of GERD with lifestyle changes and medications, notably proton pump inhibitors, and H2 blockers. Antacids may also be used to neutralise gastric acid. Sometimes, surgery may be needed to ease symptoms. Chronic inflammation of the gastric mucosa can lead to atrophic gastritis resulting in a decreased secretion of gastric acid, and consequent digestive problems. In hypochlorhydria and achlorhydria, gastric acid is either low or absent, respectively. This can potentially lead to less protection against ingested pathogens such as Vibrio or Helicobacter bacteria. In Zollinger–Ellison syndrome gastrin levels are increased, leading to excess gastric acid production, which can cause gastric ulcers. Hypercalcemia also increases gastrin and gastric acid and can cause ulcers. In diseases featuring excess vomiting, hypochloremic metabolic alkalosis (decreased blood acidity by H+ and chlorine depletion) may develop. History [edit] \| \| \| --- \| \| \| This section needs expansion. You can help by adding to it. (November 2010) \| The role of gastric acid in digestion was established in the 1820s and 1830s by William Beaumont on Alexis St. Martin, who, as a result of an accident, had a fistula (hole) in his stomach, which allowed Beaumont to observe the process of digestion and to extract gastric acid, verifying that acid played a crucial role in digestion. See also [edit] Discovery and development of proton pump inhibitors References [edit] ^ Jump up to: a b Fujimori, S (21 November 2020). "Gastric acid level of humans must decrease in the future". World Journal of Gastroenterology. 26 (43): 6706–6709. doi:10.3748/wjg.v26.i43.6706. PMC 7684463. PMID 33268958. ^ Marieb EN, Hoehn K (2018). Human Anatomy and Physiology (11th ed.). Pearson Education. p. 1264. ISBN 978-0-13-458099-9. ^ Jump up to: a b c Dworken HJ (2016). Human digestive system: gastric secretion. Encyclopædia Britannica Inc. ^ Martinsen TC, Fossmark R, Waldum HL (November 2019). "The Phylogeny and Biological Function of Gastric Juice-Microbiological Consequences of Removing Gastric Acid". Int J Mol Sci. 20 (23): 6031. doi:10.3390/ijms20236031. PMC 6928904. PMID 31795477. ^ Sehgal, Shalini; Saji, Hephzibah; Banik, Samudra Prosad (1 January 2022). "Chapter 11 - Role of food structure in digestion and health". Nutrition and Functional Foods in Boosting Digestion, Metabolism and Immune Health: 151–165. doi:10.1016/B978-0-12-821232-5.00019-7. ISBN 978-0-12-821232-5. ^ Beasley DE, Koltz AM, Lambert JE, Fierer N, Dunn RR (2015-07-29). "The Evolution of Stomach Acidity and Its Relevance to the Human Microbiome". PLOS ONE. 10 (7): e0134116. Bibcode:2015PLoSO..1034116B. doi:10.1371/journal.pone.0134116. PMC 4519257. PMID 26222383.{{cite journal}}: CS1 maint: article number as page number (link) ^ Guyton AC, Hall JE (2006). Textbook of Medical Physiology (11 ed.). Philadelphia: Elsevier Saunders. p. 797. ISBN 0-7216-0240-1. ^ Page 192 in: Agabegi ED, Agabegi SS (2008). Step-Up to Medicine (Step-Up Series). Hagerstwon, MD: Lippincott Williams & Wilkins. ISBN 978-0-7817-7153-5. ^ Jump up to: a b Lecture, "Function of the Stomach and Small Intestine" Deakin University School of Medicine. October 15, 2012 ^ "acetylcholine \| Definition, Function, & Facts \| Britannica". www.britannica.com. Retrieved 2021-12-13. ^ "Somatostatin". www.hormone.org. Retrieved 2021-12-13. ^ Lönnerholm G, Knutson L, Wistrand PJ, Flemström G (June 1989). "Carbonic anhydrase in the normal rat stomach and duodenum and after treatment with omeprazole and ranitidine". Acta Physiologica Scandinavica. 136 (2): 253–262. doi:10.1111/j.1748-1716.1989.tb08659.x. PMID 2506730. ^ "Gastroesophageal reflux disease (GERD) - Symptoms and causes". Mayo Clinic. Retrieved 2023-09-10. ^ "Zollinger-Ellison syndrome". Mayo Clinic. Retrieved 10 January 2025. ^ Sadiq, Nazia M.; Anastasopoulou, Catherine; Patel, Goonja; Badireddy, Madhu (2025), "Hypercalcemia", StatPearls, Treasure Island (FL): StatPearls Publishing, PMID 28613465, retrieved 2025-05-21 ^ Harré R (1981). Great Scientific Experiments. Phaidon (Oxford). pp. 39–47. ISBN 0-7148-2096-2. External links [edit] Wikimedia Commons has media related to Gastric acid. The Parietal Cell: Mechanism of Acid Secretion; Colorado State University Archived 2021-05-02 at the Wayback Machine \| v t e Physiology of the gastrointestinal system \| \| --- \| \| GI tract \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| Upper \| \| \| \| --- \| \| Exocrine \| Chief cells + Pepsinogen Parietal cells + Gastric acid + Intrinsic factor Foveolar cells + HCO3− + Mucus \| \| Processes \| Swallowing Vomiting Gastric emptying \| \| Fluids \| Saliva Gastric acid \| \| Gastric acid secretion \| Gastrin + G cells Histamine + ECL cells Somatostatin + D cells \| \| \| Lower \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| Endocrine/paracrine \| \| \| \| --- \| \| Bile and pancreatic secretion \| Enterogastrone Cholecystokinin + I cells Secretin + S cells \| \| Glucose homeostasis (incretins) \| GIP + K cells GLP-1 + L cells \| \| Endocrine cell types \| Enteroendocrine cells Enterochromaffin cell APUD cell \| \| Exocrine cell types \| Goblet cells \| \| \| Fluids \| Intestinal juice \| \| Processes \| Segmentation contractions Migrating motor complex Borborygmus Defecation \| \| \| Enteric nervous system \| Submucous plexus Myenteric plexus \| \| Either/both \| \| \| \| --- \| \| Processes \| Peristalsis (Interstitial cell of Cajal Basal electrical rhythm) Gastrocolic reflex Digestion Enterocyte \| \| \| \| Accessory \| \| \| \| --- \| \| Fluids \| Bile Pancreatic juice \| \| Processes \| Enterohepatic circulation \| \| \| Abdominopelvic \| Peritoneal fluid \| \| \| \| --- \| \| Authority control databases: National \| \| Retrieved from " Categories: Acids Body fluids Stomach Hidden categories: CS1 maint: article number as page number Articles needing additional references from March 2022 All articles needing additional references Articles with short description Short description matches Wikidata All articles with unsourced statements Articles with unsourced statements from June 2023 Articles to be expanded from November 2010 All articles to be expanded Commons category link is on Wikidata Webarchive template wayback links
4831	https://pmc.ncbi.nlm.nih.gov/articles/PMC2781829/	Human population growth and the demographic transition - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Philos Trans R Soc Lond B Biol Sci . 2009 Oct 27;364(1532):2985–2990. doi: 10.1098/rstb.2009.0137 Search in PMC Search in PubMed View in NLM Catalog Add to search Human population growth and the demographic transition John Bongaarts John Bongaarts 1 The Population Council, One Dag Hammarskjold Plaza, New York, NY 10017, USA Find articles by John Bongaarts 1, Author information Copyright and License information 1 The Population Council, One Dag Hammarskjold Plaza, New York, NY 10017, USA Email: jbongaarts@popcouncil.org © 2009 The Royal Society This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. PMC Copyright notice PMCID: PMC2781829 PMID: 19770150 Abstract The world and most regions and countries are experiencing unprecedentedly rapid demographic change. The most obvious example of this change is the huge expansion of human numbers: four billion have been added since 1950. Projections for the next half century expect a highly divergent world, with stagnation or potential decline in parts of the developed world and continued rapid growth in the least developed regions. Other demographic processes are also undergoing extraordinary change: women's fertility has dropped rapidly and life expectancy has risen to new highs. Past trends in fertility and mortality have led to very young populations in high fertility countries in the developing world and to increasingly older populations in the developed world. Contemporary societies are now at very different stages of their demographic transitions. This paper summarizes key trends in population size, fertility and mortality, and age structures during these transitions. The focus is on the century from 1950 to 2050, which covers the period of most rapid global demographic transformation. Keywords: population growth, demographic transition, fertility, mortality, age structure 1. Introduction After centuries of very slow and uneven growth, the world population reached one billion in 1800. The modern expansion of human numbers started then, rising at a slow but more steady pace over the next 150 years to 2.5 billion in 1950. During the second half of the twentieth century, however, growth rates accelerated to historically unprecedented levels. As a result, world population more than doubled to 6.5 billion in 2005 (United Nations 1962, 1973, 2007). This population expansion is expected to continue for several more decades before peaking near 10 billion later in the twenty-first century. Around 2070, the world's population will be 10 times larger than in 1800. The recent period of very rapid demographic change in most countries around the world is characteristic of the central phases of a secular process called the demographic transition. Over the course of this transition, declines in birth rates followed by declines in death rates bring about an era of rapid population growth. This transition usually accompanies the development process that transforms an agricultural society into an industrial one. Before the transition's onset, population growth (which equals the difference between the birth and death rate in the absence of migration) is near zero as high death rates more or less offset the high birth rates typical of agrarian societies before the industrial revolution. Population growth is again near zero after the completion of the transition as birth and death rates both reach low levels in the most developed societies. During the intervening transition period, rapid demographic change occurs, characterized by two distinct phases. During the first phase, the population growth rate rises as the death rate declines while the birth rate remains high. In the second phase, the growth rate declines (but remains positive) due to a decline in the birth rate. The entire transition typically takes more than a century to complete and ends with a much larger population size. The plot of world population size over time in figure 1 (top solid line) shows the typical S-shaped pattern of estimated and projected population size over the course of the transition. Population growth accelerated for most of the twentieth century reaching the transition's midpoint in the 1980s and has recently begun to decelerate slightly. Today, we are still on the steepest part of this growth curve with additions to world population exceeding 75 million per year between 1971 and 2016. Figure 1. Open in a new tab Population size estimates, 1900–2005 and projections 2005–2050. High, medium and low variants. Contemporary societies are at very different stages of their demographic transitions. Key trends in population size, fertility and mortality during these transitions are summarized below. The focus is on the century from 1950 to 2050, covering the period of most rapid global demographic change. The main source of data is the United Nation's 2006 world population assessment, which provides estimates for 1950–2005 and projections from 2005 to 2050 (United Nations 2007). 2. Future population trends The projected rise in world population to 9.2 billion in 2050 represents an increase of 2.7 billion over the 2005 population of 6.5 billion. Nearly all of this future growth will occur in the ‘South’—i.e. Africa, Asia (excluding Japan, Australia and New Zealand), and Latin America—where population size is projected to increase from 5.3 to 7.9 billion between 2005 and 2050 (table 1). In contrast, in the ‘North’ (Europe, Northern America, Japan and Australia/New Zealand), population size is forecast to remain virtually stable, growing slightly from 1.22 to 1.25 billion between 2005 and 2050. The difference in trends between these two world regions reflects the later stage of the transition in the North compared with the South. Table 1. Population estimates (1950–2005) and projections (2005–2050), by region. Adapted from United Nations (2007). \| \| population (billions) \| % increase \| :--- \| \| 1950 \| 2005 \| 2050 \| 1950–2005 \| 2005–2050 \| \| Africa \| 0.22 \| 0.92 \| 2.00 \| 311 \| 117 \| \| Sub-Saharan \| 0.18 \| 0.77 \| 1.76 \| 327 \| 129 \| \| Asia \| 1.41 \| 3.94 \| 5.27 \| 179 \| 34 \| \| Latin America \| 0.17 \| 0.56 \| 0.77 \| 233 \| 38 \| \| Europe \| 0.55 \| 0.73 \| 0.66 \| 33 \| −9 \| \| Northern America \| 0.17 \| 0.33 \| 0.45 \| 94 \| 34 \| \| South \| 1.72 \| 5.30 \| 7.95 \| 208 \| 50 \| \| North \| 0.81 \| 1.22 \| 1.25 \| 49 \| 2 \| \| World \| 2.54 \| 6.51 \| 9.19 \| 157 \| 41 \| Open in a new tab The global demographic transition began in the nineteenth century in the now economically developed parts of the world (the North) with declines in death rates. Large reductions in birth rates followed in the early part of the twentieth century. These transitions are now more or less complete. But, as shown in table 1, trends for the two principal regions in the North are expected to diverge between 2005 and 2050: an increase from 0.33 to 0.45 billion in Northern America, and a decline from 0.73 to 0.66 billion in Europe. In fact, several countries in Europe (e.g. Russia) and East Asia (e.g. Japan) face significant population declines as birth rates have fallen below death rates. The demographic transitions in Africa, Asia and Latin America started later and are still underway. In 2005, Asia had a population of 3.94 billion, more than half of the world total, and its population is expected to grow by 34 per cent to 5.27 billion by 2050. Africa, with 0.92 billion inhabitants in 2005, is likely to experience by far the most rapid relative expansion, more than doubling to 2.0 billion by 2050. Latin America, with 0.56 billion in 2005, is the smallest of the regions of the South; its projected growth trend is similar to that of Asia. It may seem surprising that population growth continues at a rapid pace in sub-Saharan Africa, where the AIDS epidemic is most severe. This epidemic has indeed caused many deaths, but population growth continues because the epidemic is no longer expanding and the birth rate is expected to remain higher than the elevated death rate in the future (UNAIDS 2007; Bongaarts et al. 2008). The epidemic's demographic impact can be assessed by comparing the standard UN population projection (which includes the epidemic's effect) with a separate hypothetical projection in which AIDS mortality is excluded (United Nations 2007). In sub-Saharan Africa, the former projects a 2050 population of 1.76 billion and the latter a population of 1.95 billion. The difference of 0.2 billion in 2050 between these projections with and without the epidemic is due to deaths from AIDS as well as the absence of the descendents from people who died from AIDS. According to these projections, the population of sub-Saharan Africa will grow by one billion between 2005 and 2050 despite the substantial impact of the AIDS epidemic. In fact, no country is expected to see a decline in its population size between 2005 and 2050 due to high AIDS mortality. Most populations in sub-Saharan Africa will more than double in size, several will triple and Niger is expected to quadruple by 2050 (United Nations 2007). Transitions in the developing world have generally produced more rapid population growth rates in mid-transition than historically observed in the North. In some developing countries (e.g. Kenya and Uganda), peak growth rates approached four per cent per year in recent decades (implying a doubling of population size in two decades), levels that were very rarely observed in developed countries except with massive immigration. Two factors account for this very rapid expansion of population in these still largely traditional societies: the spread of medical technology (e.g. immunization, antibiotics) after World War II, which led to extremely rapid declines in death rates, and a lag in declines in birth rates. Population sizes for the 10 largest countries in 2005 and in 2050 are presented in table 2. In 2005, China (1.31 billion) and India (1.13 billion) were by far the largest countries, together accounting for nearly half the South's total. The top 10 include six Asian countries and only one country each in Latin America and Africa. By 2050, the ranking is expected to have shifted substantially, with India's population exceeding China's, and with Ethiopia and DR Congo rising to the top 10, replacing Japan and the Russian Federation. Table 2. Ten largest countries by population size in 1995 (estimate) and 2050 (medium projection). Adapted from United Nations (2007). \| rank \| 1995 \| 2050 \| :--- \| country \| population size (millions) \| country \| projected population (millions) \| \| 1 \| China \| 1313 \| India \| 165 \| \| 2 \| India \| 1134 \| China \| 140 \| \| 3 \| United States \| 300 \| United States \| 40 \| \| 4 \| Indonesia \| 226 \| Indonesia \| 29 \| \| 5 \| Brazil \| 187 \| Pakistan \| 29 \| \| 6 \| Pakistan \| 158 \| Nigeria \| 28 \| \| 7 \| Bangladesh \| 153 \| Brazil \| 25 \| \| 8 \| Russian Federation \| 144 \| Bangladesh \| 25 \| \| 9 \| Nigeria \| 141 \| D.R. Congo \| 18 \| \| 10 \| Japan \| 128 \| Ethiopia \| 18 \| Open in a new tab To simplify the presentation of results, all projections discussed in this study are taken from the medium variant of the UN projections (United Nations 2007). The UN has a good record of making relatively accurate projections (National Research Council 2000), but the future is of course uncertain and actual population trends over the next half century will likely diverge to some extent from current projections. The UN makes an effort to capture this uncertainty by publishing separate high and low projections. For the world, the high and low variants reach 7.8 and 10.8 billion, respectively, in 2050, indicating a rather wide range of possible outcomes (see dashed lines in figure 1). 3. Drivers of population growth: fertility and mortality The world's population increases every year because the global birth rate exceeds the death rate. For example, in 2000–2005 population size increased at a rate of 1.17 per cent per year, which equals the difference between a birth rate of 2.03 per cent and a death rate of 0.86 per cent. At the country level, population growth is also affected by migration, but for the regional aggregates of population used in this analysis, migration is usually a minor factor, and it will therefore not be discussed in detail. The annual birth and death rates of populations are in turn primarily determined by levels of fertility and mortality experienced by individuals. The most widely used fertility indicator is the total fertility rate (TFR), which equals the number of births a woman would have by the end of her reproductive years if she experienced the age-specific fertility rates prevailing in a given year. Mortality is often measured by the life expectancy (LE) at birth, which equals the average number of years a newborn would live if subjected to age-specific mortality rates observed in a given year. (a). Fertility The UN's past estimates and future projections of fertility levels by region for the period 1950–2050 are presented in figure 2. In the 1950s, the TFR in the South was high and virtually stable at around six births per woman on average. This high level of fertility reflects a near absence of birth control, a condition that has prevailed for centuries before the middle of the twentieth century. In the late 1960s, a rapid decline in fertility started nearly simultaneously in Asia and Latin America. In contrast, Africa has experienced only limited reproductive change. As a result of these divergent past trends, fertility levels in 2000–2005 differed widely among regions from as high as 5 births per woman (bpw) in Africa, to 2.5 bpw in Asia and Latin America. Average fertility in the North was already low in the early 1950s and has since declined to 2.0 bpw in Northern America and to 1.4 bpw in Europe. Figure 2. Open in a new tab Trends in the total fertility rate by region. The decline in the average fertility in the South from 6 to 3 bpw over the past half century has been very rapid by historical standards. This reproductive revolution is mainly due to two factors. First, desired family size of parents has declined as the cost of children rose and child survival increased. Second, government intervention played a key role. In China this took the form of a coercive and unpopular one-child policy, but most other countries implemented voluntary family planning programmes. The aim of these programmes is to provide information about and access to contraceptives at subsidized prices so that women who want to limit their childbearing can more readily do so. UN projections for the South assume that the TFR will eventually reach and then fall slightly below the so-called ‘replacement’ level in all regions. Replacement fertility is just above 2 bpw and it represents the level at which each generation just replaces the previous one, thus leading to zero population growth (in the absence of mortality change and migration). Below-replacement fertility produces, in the long run, population decline. As is evident from figure 2, the TFRs in Asia and Latin America are expected to reach the replacement level around 2020. Africa is assumed to be on a much slower trajectory towards replacement fertility because of its lower level of socio-economic development. High fertility therefore remains a key cause of future population growth in this region. In contrast, the already low fertility of the North is expected to remain below replacement and is no longer driving population growth. (b). Mortality and life expectancy Mortality levels have also changed rapidly over the past several decades (figure 3). The South experienced exceptional improvements in LE from an average of 41 years in 1950–1955 to 64 years in 2000–2005. By the early 2000, Latin America reached mortality levels similar to those prevailing in the North in the 1970s, and Asia was just a few years behind. Africa experienced the highest mortality and improvements in LE stalled in the 1990s due to the AIDS epidemic. As a result, Africa's LE, at 52 years in 2000–2005, was still substantially below that of Asia (68) and Latin America (72). As expected, Europe and Northern America already achieved relatively low levels of mortality by 1950, but they have nevertheless seen significant further improvements since then. Europe's LE (74) is now lower than North America's (78) because of a rise in mortality in Eastern Europe after the break-up of the Soviet Union. Figure 3. Open in a new tab Trends in LE by region. Projections of future LEs by the UN assume continued improvements over time in all regions. The North is expected to reach 82 years in 2050 despite the increasing difficulty in achieving increments as countries reach ever higher levels of LE. Asia and Latin America are expected to continue to close the gap with the North, and Africa will continue to lag, in part because the continent remains affected by the AIDS epidemic. It should be noted that the assumptions made by the UN about future trends in fertility and mortality are not based on a firm theoretical basis. Instead, the UN relies on empirical regularities in past trends in countries that have completed their transitions, mostly in the North, where fertility declined to approximately the replacement level, and increases in LE became smaller over time. This is a plausible approach that unfortunately leaves room for potential inaccuracies in projection results. 4. Changing population age composition Over the course of the demographic transition, declines in fertility and mortality cause important changes in a population's age composition. In general, countries in the early stages of the transition have a younger age structure than countries in the later stages. Figure 4 presents the distribution of the 2005 population in four broad age groups: 0–14, 15–24, 25–64 and 65+ by region. Most of the regions in the South—Africa, Latin America, South Asia and West Asia—have very young age structures with about half of the population under age 25 (62% in Africa). The exception is East Asia (mostly China) where this proportion is 37 per cent. In the North, the population under 25 is still smaller: 35 per cent in North America and just 30 per cent in Europe. The reverse pattern is observed for the proportion 65+, which is much higher in the North than in the South, ranging from as high as 15 per cent in Europe to as low as just 3 per cent in Africa. Figure 4. Open in a new tab Distribution of population by age, by region, 2005. (a). The age-dependency ratio A changing age distribution has significant social and economic consequences, e.g. for the allocation of education, healthcare and social security resources to the young and old. Assessments of this impact often rely on the so-called age-dependency ratio (DR) that summarizes key changes in the age structure. The DR at a given point in time equals the ratio of population aged below 15 and over 65 to the population of age 15–64. This ratio aims to measure how many ‘dependents’ there are for each person in the ‘productive’ age group. Obviously, not every person below 15 and over 65 is a dependent and not every person between ages 15 and 65 is productive. Despite its crudeness, this indicator is widely used to document broad trends in the age composition. Over the course of a demographic transition, the DR shows a characteristic pattern of change. Figure 5 presents this pattern as observed in the South from 1950 to 2005 and projected from 2005 to 2050. Early in the transition, the DR typically first rises slightly as improvements in survival chances of children raise the number of young people. Next, the DR falls sharply as declines in fertility reduce the proportion of the population under age 15. This decline has important economic consequences because it creates a so-called ‘demographic dividend’, which boosts economic growth by increasing the size of the labour force relative to dependents and by stimulating savings (Birdsall et al. 2001). Finally, at the end of the transition, the DR increases again as the proportion of the population over age 65 rises. Figure 5 also plots the DR of the North from 1950–2050. From 1950 to 2010 it showed a slight decline, but after 2010 it rises steeply as very low fertility and increasing longevity increases the proportion 65+. This ageing of the North poses serious challenges to support systems for the elderly (OECD 1998, 2001). Figure 5. Open in a new tab Dependency ratio estimates, 1950–2005. (b). Population momentum At the end of the demographic transition natural population growth reaches zero once three conditions are met: Fertility levels-off at the replacement level of about 2.1 bpw (more precisely, the net reproduction rate should be 1). If fertility remains above replacement, population growth continues. Mortality stops declining. In practice, this is not likely to happen because improvements in medical technology and healthcare as well as changes in lifestyles, etc. will probably ensure continued increases in LE. The age structure has adjusted to the post-transitional levels of fertility and mortality. The adjustment of the age structure at the end of the transition takes many decades to complete. A key implication of this slow adjustment process is that population growth continues for many years after replacement fertility is reached if, as is often the case, the population is still relatively young when fertility reaches the replacement level. The tendency of population size to increase after a two-child family size has been reached is referred to as population momentum; it is the consequence of a young population age structure (‘young’ is defined relative to the age structure in the current life table) (Bongaarts & Bulatao 1999). The population momentum inherent in the age structure of a particular population at a given point in time can be estimated with a hypothetical population projection in which future fertility is set instantly to the replacement level, mortality is held constant and migration is set to zero. Since such a variant is not directly available from UN projections, it will not be presented here. However, the UN does provide ‘instant replacement’ projections in which mortality and migration trends are the same as in the standard projection. This projection gives an approximation of the combined effect on future growth of population momentum and declining mortality in the South because the role of migration is small. The difference between this hypothetical projection and the standard medium UN projection is a measure of the impact of high fertility on future population growth. Results of these two projections are presented in figure 6, which compares the per cent growth between 2005 and 2050 for regions in the South. The black bars give the growth in the standard (medium variant) projection and the grey bars give the growth in the ‘instant replacement’ projection. Three results are noteworthy. First, the two projections differ most in Africa (+117% versus +50%) which is as expected because fertility is still very high in this region. Second, in all regions of the South outside China, populations would be expected to rise by 50 per cent (62% in West Asia) if fertility were set to replacement in 2005. This implies that momentum and declining mortality are responsible for nearly half of the projected future population growth in Africa and for the large majority of growth in Latin America, and South and West Asia. Third, in East Asia and in Latin America the replacement projection exceeds the medium UN projection. This finding is explained by the fact that fertility in these regions is assumed to average below the replacement level over the next half century. Figure 6. Open in a new tab Percentage increase in population 2005–2050, by region, alternative projections. Black bars denote medium UN projection; grey bars denote instant replacement projection (hypothetical). 5. Conclusion The world and most countries are going through a period of unprecedentedly rapid demographic change. The most obvious example of this change is the huge expansion of human numbers: four billion have been added since 1950. Other demographic processes are also experiencing extraordinary change: women are having fewer births and LEs have risen to new highs. Past trends in fertility and/or mortality have led to very young populations in high fertility countries in the South and to increasingly older populations in the North. Still other important demographic changes which were not reviewed here include rapid urbanization, international migration, and changes in family and household structure. Global population growth will continue for decades, reaching around 9.2 billion in 2050 and peaking still higher later in the century. The demographic drivers of this growth are high fertility in parts of the South, as well as declining mortality and momentum. This large expansion in human numbers and of the accompanying changes in the age structure will have multiple consequences for society, the economy and the environment as discussed in the subsequent chapters in this issue. Footnotes One contribution of 14 to a Theme Issue ‘The impact of population growth on tomorrow's world’. References Birdsall N., Kelley A., Sinding S.2001Population matters: demographic change, economic growth and poverty in the developing world Oxford, UK: Oxford University Press [Google Scholar] Bongaarts J., Bulatao R.1999Completing the demographic transition. Popul. Dev. Rev. 25, 515–529 (doi:10.1111/j.1728-4457.1999.00515.x) [Google Scholar] Bongaarts J., Buettner J., Heilig G., Pelletier F.2008Has the AIDS epidemic peaked? Popul. Dev. Rev. 34, 199–224 (doi:10.1111/j.1728-4457.2008.00217.x) [Google Scholar] National Research Council 2000Beyond six billion: forecasting the world's population (eds Bongaarts J., Bulatao R.). Washington, DC: National Academy Press [Google Scholar] OECD 1998Maintaining prosperity in an ageing society Paris: OECD Publications [Google Scholar] OECD 2001The fiscal implications of ageing: projections of age-related spending. OECD Economic Outlook 69, 145–167 [Google Scholar] UNAIDS 2007AIDS Epidemic Update Geneva: UNAIDS [Google Scholar] United Nations 1962Demographic yearbook New York, NY: United Nations [Google Scholar] United Nations 1973The determinants and consequences of population trends New York, NY: Department of Economic and Social Affairs, Population Studies 50, United Nations [Google Scholar] United Nations 2007World population prospects: the 2006 revision New York, NY: United Nations Population Division [Google Scholar] Articles from Philosophical Transactions of the Royal Society B: Biological Sciences are provided here courtesy of The Royal Society ACTIONS View on publisher site PDF (303.8 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract 1. Introduction 2. Future population trends 3. Drivers of population growth: fertility and mortality 4. Changing population age composition 5. Conclusion Footnotes References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
4832	https://www.youtube.com/watch?v=jVzmP_XLggE	Discrete Structures: Recurrence Relations -- Solving Recurrences, Part 1 Darin Brezeale 843 subscribers 28 likes Description 769 views Posted: 16 Aug 2021 I use the Expand, Guess, and Verify approach to solve first-order recurrence relations of the form S(n) = aS(n-1) or S(n) = S(n-1) + b This assumes that you are familiar with Recursive Definition: Mathematical Induction: 4 comments Transcript: now let's solve a recurrence relationship if you don't know what a recursive definition is or if you don't know what mathematical induction is then you should watch my videos on those two things and i'll link to those in the description below but here we have a recursive definition so our basis is s of one is equal to three and then each additional value because when n is greater than equal to this is true s of n is equal to s of n minus one plus 2. before we see how to solve this which is where we produce a closed form solution let's sketch out a few things that would be helpful to understand the process you wouldn't normally do this but i'm doing this so what my process is will appear less magical so if we have s of n equals s of n minus 1 plus 2 what this means in english is that the current value is equal to previous value plus 2 which means that s of n minus 1 is equal to its previous value assuming we can go farther back so let's say for example n was 10. well sub 10 is equal to s of 9 plus 2 and s of 9 would be equal s of 8 plus 2 and so forth because this term is the previous one to this and the term that's previous to s of n minus 2 would be s of n minus 3 plus 2 and so forth we're going to use this here in a moment so first we're going to do the expand part so we have s of n is equal to s of n minus 1 plus 2 that's what we were given right here and now we're going to replace s of n minus 1 by the right hand side so this says let me use square brackets to make it clear what i'm inserting s of n minus 1 is equal to s of n minus 2 plus 2 plus our original 2. this two right here simply comes down to here and now i'm going to simplify a little bit this becomes s of n minus two plus there's two twos but i'm gonna write it as two times two instead of four so why am i doing that well ultimately when we get to the guess step we need to recognize a pattern that shows us how the value of n relates to the overall value and in lots of cases you'll have something like this you might have a term that is you know 7 times n you might have a term that was 2 to the nth power you might have a term that was 5 to the n minus 1 power something like that if i replace this with 4 i may not be able to see what the pattern is all right so that's the reason why i did this let's do this again now i'm going to replace s of n minus 2 with s of n minus 3 plus 2 plus what we had right here 2 times 2 and now if i simplify this it becomes s of n minus three plus there's a two we have two more twos so it's really three twos and i think i'm ready to recognize a pattern right here i've got a three and that's a three there's a 2 and there's a 2. and in fact if we look at this right here i can say this is s of n minus 1 plus 1 2. so there's a 1 there's a 1. so what i think this eventually becomes is s of n minus k plus k 2's which i will rewrite in standard form as s of n minus k plus 2 times k right where this we normally have the actual number on the left now what are we doing here we're actually starting at some value of n and this part right here is working its way back to the basis so imagine n was 4 to begin with well s of 4 is equal to s of 3 plus a single 2 right because this definition but it's also true that s 4 is equal to s of 2 plus a 2 plus another 2. and s of 4 is equal to s of 1 plus a 2 plus a 2 plus a 2. so that means that eventually n minus k will be equal to one and if we use algebraic manipulations on this we get that k is equal to n minus one so now i'm gonna substitute these things into this relationship right here and this right here becomes s of one plus two times n minus one which is equal to s of one is three plus two times n minus two which is equal to two n plus one and so my guess this was the verify step excuse me the expand step right here the guess step is that s of n is equal to two n plus one right that's my guess at a closed form solution so how do we verify that right now just as a way to spot check my work sometimes i will iterate using the recursive definition through several values and then also use the formula for some of those values just to sort of get a feeling that it seems to work but i have found occasions where i made a mistake in the verif in the expansion steps excuse me and if i only check say a couple of values from iterating i would find that like s of 1 and s of 2 were the same values when i used the iterative approach as when i used my close form solution even though i had the wrong closed form solution eventually if i iterated enough i would find the closed form solution at some point didn't match but it is possible to come up with a mistake here that still can generate the first couple values okay so really what we needed to verify is to use mathematical induction and that's what we're looking at now so remember we have a basis step and so i like to think it was the left side and the right side the left side is what i was given and we were told that s of 1 was equal to 3. where does that come from well that's just what we were told it is right so if that turns out to be wrong there's i really have no way of dealing with that so that's just what we're given but our formula for the closed form solution should still work for this and so we have s of n is equal to s of n or excuse me s of n excuse me what was it is equal to n plus one sorry which is equal to or we have s of one it's equal to then two times one plus one which is two plus one which equals three those are the same as they should be because if we can't even get this nothing else will matter the next step of induction is we'll make an assumption so we assume that s of k is equal to two times k plus one right basically it's our closed form solution but using k and now we're on check and see can we use the formula to get future values so s of k plus 1 is equal to s of k plus 2. now where is this this right here is based upon the given recurrence relationship basically the previous value to this so if this is k plus one the previous value is k plus two that's what we really have here it's just that everything's been shifted over by one but it's still the formula but if our closed form solution is correct then we can replace s of k by the closed form solution to get two of k plus one plus two right this right here which is equal to two k plus three now i want to on the right hand side use the formula for the whole thing so this says that s of k plus 1 is equal to 2 of k plus 1 because that's what the input is plus one which is two k plus two plus one which is equal to k plus three and those are the same therefore according to mathematical induction we've proven it or verified it now if you weren't sure if this was correct then or if you just want to feel comfortable with it you could always use the formula to produce several values and then use the original definition to produce several values by iterating and make sure you get the same thing but i don't know that i've ever produced or used induction to show something was really true that turned out not really to be true i don't know that i've ever been able to make that mistake because i would have to make mistakes here and here that somehow matched up i don't know that i've ever done that all right so that's one example let's look at a second example so now we have that the basis for our recursive definition is s of 1 is equal to 2 and the recurrence relationship is that s of n is equal 3 times the previous value or 3 times s of n minus 1 and that's true and n is greater than or equal to 2. okay if you need to see this in terms of s of n minus 1 and 2 and so forth you could expand this but we're just going to jump right down to the expand part so this says that s of n is equal to 3 s of n minus 1 is equal 3 times s or shoot me three times s of n minus two maybe i should have expanded it let's just do that real fast s of n is equal to three s of n minus one s of n minus one is equal to three times s of n minus 2 s of n minus 3 0 3 s of n minus excuse me this is 2 3 and so forth so this is the original 3 that we had from right here and then s of n minus 1 is replaced by 3 of sn s of n minus 2 which is equal to now once again i could just say 9 but that's don't make it more difficult to find the pattern i'm going to say 3 squared times s of n minus 2. now we expand again this would be 3 squared times replace this with s of n minus 2 is equal to 3 times s of n minus 3 which is 3 cubed times s of n minus 3 dot dot which is equal to what's the pattern here well i could say this right here just right here 3 to the first power times s of n minus 1. there's a 1 there's a 1. there's a 2 there's a 2. there's a 3 there's a 3. so it looks like we have 3 to the k power times s of n minus k now we know that n minus k eventually becomes 1 and from that i can get the k is equal to n minus 1. so now i'm going to guess that s of n is equal to 3 to the n minus 1 power okay times s of 1 which is 3 to the n minus 1 power times 2. so to make this little neater we'll say this right here is equal to 2 dot 3 to the n minus 1 power that's my guess at a closed form solution all right let's verify using induction so the basis step so s of 1 was equal to 2. we were given that the formula so this is given the formula says s of 1 is equal to 3 of 1 minus one the left one is the input right that's n times two so three the zero power just be one so it's just equal to two so let's check out we assume s sub k is equal to times three to the k minus one power and now we have s of k plus one is equal to according to the recurrence relationship 3 times s of k and now we replace this s of k with our closed form solution two times three excuse me two times sorry three times two times three to the k minus one power which is really equal to two times three which is really three to the first power times three to the k minus one power when we combine these we add the the powers here which is equal to times three to the k power all right what would the formula have told us we just went through directly well this is s of k plus 1 is equal to 3 to the k plus 1 minus 1 power times two or i've sort of swapped the sides of these which equal two times k plus one minus one is really just three to the k power and once again those are the same and this is a little easier to be comfortable with because i look at it and say okay you just take two multiply it by three multiply by another three multiply by another three and so forth so two multiply by three multiply by another three and so forth so there are two examples what i'll do in my next video is sort of combine the approach where we'll have something like this s of n is equal to say a where it's not equal to one s of n minus one plus a b well that's not equal to zero so you could say this right here has a is three and b is zero and in the previous example we had a was one and b was two because we were adding two each time i'll combine those into the next video because it's a much more challenging process
4833	https://www.linkedin.com/posts/unicefusa_how-the-muac-tape-helps-save-lives-activity-7365481813038747650-jWnv	How the MUAC Tape Helps Save Lives \| UNICEF USA Agree & Join LinkedIn By clicking Continue to join or sign in, you agree to LinkedIn’s User Agreement, Privacy Policy, and Cookie Policy. Skip to main contentLinkedIn Top Content People Learning Jobs Games Sign inJoin now UNICEF USA’s Post UNICEF USA 54,831 followers 1mo Report this post This simple tape? It helps save lives. The mid-upper arm circumference (MUAC) tape is a powerful tool that helps health workers quickly assess a child’s nutritional status and ensure those who are malnourished get the treatment they need. In and out of emergencies, and around the world, UNICEF works every day to detect, treat and prevent malnutrition. Swipe through to learn more. 142 Comments LikeComment Share Copy LinkedIn Facebook Twitter Viktor Zhokhov Senior/Lead Fullstack Engineer · React, Node.js, NestJS · 14+ yrs · Scalable Architectures · B2B/Remote 1mo Report this comment Да зж. Би из блюдЮ LikeReply 1 Reaction 📚WM SMILE GENERATION UGANDA LIMITED🌿 1mo Report this comment Such a simple yet powerful tool! At 📚WM SMILE GENERATION UGANDA LIMITED🌿, we value initiatives like UNICEF’s use of the MUAC tape that make a real difference in detecting and treating malnutrition, helping children get the care they need to survive and thrive. LikeReply 1 Reaction See more comments To view or add a comment, sign in 54,831 followers View ProfileFollow More from this author ### Embracing Partnership and Purpose at the Parliament of the World’s Religions UNICEF USA 2y Explore content categories Career Productivity Finance Soft Skills & Emotional Intelligence Project Management Education Technology Leadership Ecommerce Show more Show less LinkedIn© 2025 About Accessibility User Agreement Privacy Policy Your California Privacy Choices Cookie Policy Copyright Policy Brand Policy Guest Controls Community Guidelines العربية (Arabic) বাংলা (Bangla) Čeština (Czech) Dansk (Danish) Deutsch (German) Ελληνικά (Greek) English (English) Español (Spanish) فارسی (Persian) Suomi (Finnish) Français (French) हिंदी (Hindi) Magyar (Hungarian) Bahasa Indonesia (Indonesian) Italiano (Italian) עברית (Hebrew) 日本語 (Japanese) 한국어 (Korean) मराठी (Marathi) Bahasa Malaysia (Malay) Nederlands (Dutch) Norsk (Norwegian) ਪੰਜਾਬੀ (Punjabi) Polski (Polish) Português (Portuguese) Română (Romanian) Русский (Russian) Svenska (Swedish) తెలుగు (Telugu) ภาษาไทย (Thai) Tagalog (Tagalog) Türkçe (Turkish) Українська (Ukrainian) Tiếng Việt (Vietnamese) 简体中文 (Chinese (Simplified)) 正體中文 (Chinese (Traditional)) Language Sign in to view more content Create your free account or sign in to continue your search Sign in Welcome back Email or phone Password Show Forgot password? Sign in or By clicking Continue to join or sign in, you agree to LinkedIn’s User Agreement, Privacy Policy, and Cookie Policy. New to LinkedIn? Join now or New to LinkedIn? Join now By clicking Continue to join or sign in, you agree to LinkedIn’s User Agreement, Privacy Policy, and Cookie Policy.
4834	https://www.homeschoolmath.net/teaching/math-stars.php	Math Stars Worksheets Free worksheets By Grades 1st grade 2nd grade 3rd grade 4th grade 5th grade 6th grade 7th grade Elementary Number Charts Addition Multiplication Division Long division Basic operations Measuring Telling time Place value Rounding Roman numerals Money US Money Canadian Australian British European S. 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Good problems can inspire curiosity about number relationships and geometric properties. It is hoped that in accepting the challenge of mathematical problem solving, students, their parents, and their teachers will be led to explore new mathematical horizons. Math Stars are in sets for Grades 1-8 and include commentaries for teachers. All Math Stars Newsletters are ready for classroom use and available for downloading as PDF files. Grade 1 Math Stars Worksheet PDF Grade 2 Math Stars Worksheet PDF Grade 3 Math Stars Worksheet PDF Grade 4 Math Stars Worksheet PDF Grade 5 Math Stars Worksheet PDF Grade 6 Math Stars Worksheet PDF Grade 7 Math Stars Worksheet PDF Grade 8 Math Stars Worksheet PDF (This content was originally published by the North Carolina State Board of Education on mathlearnnc.sharpschool.com, which is no longer available.) 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Hint: it has to do with a "recipe" that many math lessons follow. The do's and don'ts of teaching problem solving in math Advice on how you can teach problem solving in elementary, middle, and high school math. How to set up algebraic equations to match word problems Students often have problems setting up an equation for a word problem in algebra. To do that, they need to see the RELATIONSHIP between the different quantities in the problem. This article explains some of those relationships. Seven reasons behind math anxiety and how to prevent it Mental math "mathemagic" with Arthur Benjamin (video) Keeping math skills sharp in the summer Geometric vanish puzzles Science resources Short reviews of the various science resources and curricula I have used with my own children. Copyright 2003- 2025 HomeschoolMath.net AboutContactPrivacy policyEdit Cookie Consent Powered By 00:00/00:53 10 Sec 1.5M 12 Welcome to home school math Next Stay
4835	https://www.old.goldbook.iupac.org/html/R/R05271.html	IUPAC Gold Book - relative molecular mass, M_r IUPAC>Gold Book>alphabetical index>R>relative molecular mass, Related index: IUPAC>Gold Book>math/physics>quantities Indexes alphabetical chemistry math/physics general source documents Download Gold Book PDF FAQ about sitemap PREVIOUSR05271NEXT relative molar massrelative permeability, relative molecular mass, Ratio of the mass of a molecule to the unified atomic mass unit. Sometimes called the molecular weight or relative molar mass. Source: Green Book, 2nd ed., p. 41 PAC, 1996, 68, 957(Glossary of terms in quantities and units in Clinical Chemistry (IUPAC-IFCC Recommendations 1996)) on page 990 Related index: IUPAC>Gold Book>math/physics>quantities Interactive Link Maps First Level Second Level Third Level Cite as: IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). XML on-line corrected version: (2006-) created by M. Nic, J. Jirat, B. Kosata; updates compiled by A. Jenkins. ISBN 0-9678550-9-8. Last update: 2014-02-24; version: 2.3.3. DOI of this term: Original PDF version: The PDF version is out of date and is provided for reference purposes only. For some entries, the PDF version may be unavailable. Current PDF version \| Version for print \| History of this term
4836	https://www.youtube.com/watch?v=U-N5n4yVIE4	Find Vector Between Two Points Study Force 57300 subscribers 32 likes Description 8853 views Posted: 7 Jul 2021 ✔ ✔ ✔ Ask questions here: Follow us: ▶ Facebook: ▶ Instagram: ▶ Twitter: Q. Consider the points A(2, 3), B(-1, 4) and C(0, -3). a. Find (AB), the vector from point A to point B. b. Find (AC), the vector from point A to point C. c. Find (BC), the vector from point B to point C. Draw the three vectors that you found above as well as (OA), (OB), and (OC). 3 comments Transcript: in this quick tutorial i'll show you how to find the vector between two points consider the points a which has the coordinates two and three b negative one and four and c being zero negative three find the vector a b which is the vector between points a and b i want to start by creating a cartesian plane to represent these three points here's my cartesian plane we have the point a at two and three which is over here and the point b at negative one and four and the point c being at zero and negative three so they want the vector that spans from a over to b the order in which this is written is important because the way that this is written tells us that we start at a and end at b so to find the components of the vector that represents this arrow and we'll call that vector a b to find the components of a b we have to subtract the coordinates of a from that of b not the opposite way around so it's very important that you follow the order in which these letters are written so we subtract the coordinates of a from b for the x coordinate we'll say negative 1 minus 2 and for the y component 4 minus 3. negative 1 minus 2 is negative 3 and 4 minus 3 is 1. so the components of the vector that spans a to b is negative three and one now had you ignored the fact that we start at a and end at b your components would be the opposite where it would be positive three and negative one and that would not be correct let's try the next one find the vector that spans between a and c so let's go ahead and create that vector there it is we start at a and end at c i'll do the answer so this is a and this is b we'll subtract the coordinates of a from that of c i'll write down the vector has the components zero minus two and negative three minus three zero minus 2 is negative 2 and negative 3 minus 3 is negative 6. that represents the components of that vector and finally a vector that spans between b and c so we start at b and end at c i'll write the answer down over here we'll subtract the coordinates of b from that of c starting with the x components zero minus negative one zero minus negative one and negative three minus four negative three minus four we should end up with positive one and negative seven now the nice thing about doing this is that if you wanted to center your vectors at the origin you can do so by plotting where these points are for instance if i plotted negative 3 and 1 that would be over here that would represent one of my vectors and it would mean the exact same thing as if it were up here if i plotted this vector we would be at negative 2 and negative 6. so that's 1 2 3 4 5 6. so the vector ac is the one that i am graphing right now that's the exact same thing as that and the same applies if i were to graph bc bc would have the coordinates 1 and negative 7 which would mean somewhere over here there it is so let me go ahead and erase those to prepare for question d which says draw the three vectors that you found above and i've done that already notice that we have these black arrows that represent those vectors as well as three other vectors that go from the o to a o represents origin so i will do that first one o to a is right there o to b is over here and o to c is that one and there you have it now you know how to find vectors between two points
4837	https://nrich.maths.org/secondary-interactive-games-and-challenges	Skip to main content List Secondary interactive games and challenges This is our collection of games specially chosen for use in the computer room or on the whiteboard. Each explores a strategic or mathematical theme; they will work well as teacher-led class activities but are also suitable for students to play directly, as they can all be played by individuals against the computer and none require direct teacher mediation. As with all of our games, players should be armed with a pen and paper to help to plan their strategies and record their findings as they play. The games are listed in approximate order of difficulty. problem Matching fractions, decimals and percentages Age 7 to 14 Challenge level Can you match pairs of fractions, decimals and percentages, and beat your previous scores? problem Missing multipliers Age 7 to 14 Challenge level What is the smallest number of answers you need to reveal in order to work out the missing headers? problem Got it Age 7 to 14 Challenge level A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target. problem Fruity totals Age 7 to 16 Challenge level In this interactivity each fruit has a hidden value. Can you deduce what each one is worth? problem The Number Jumbler Age 7 to 14 Challenge level The Number Jumbler can always work out your chosen symbol. Can you work out how? game Favourite Estimating angles Age 7 to 14 Challenge level How good are you at estimating angles? problem Latin lilies Age 7 to 14 Challenge level In this game you are challenged to gain more columns of lily pads than your opponent. game Favourite Factors and multiples game Age 7 to 16 Challenge level A game in which players take it in turns to choose a number. Can you block your opponent? game Favourite The remainders game Age 7 to 14 Challenge level Play this game and see if you can figure out the computer's chosen number. game Favourite Countdown Age 7 to 14 Challenge level Here is a chance to play a version of the classic Countdown Game. problem Treasure hunt Age 7 to 14 Challenge level Can you find a reliable strategy for choosing coordinates that will locate the treasure in the minimum number of guesses? problem Dozens Age 7 to 14 Challenge level Can you select the missing digit(s) to find the largest multiple? problem Square it Age 11 to 16 Challenge level Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square. problem Polygon rings Age 11 to 14 Challenge level Join pentagons together edge to edge. Will they form a ring? problem Arithmagons Age 11 to 16 Challenge level Can you find the values at the vertices when you know the values on the edges? problem Fractions and percentages card game Age 11 to 16 Challenge level Can you find the pairs that represent the same amount of money? game Last biscuit Age 11 to 18 Challenge level Can you find a strategy that ensures you get to take the last biscuit in this game? problem Magic potting sheds Age 11 to 16 Challenge level Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it? problem Connect three Age 11 to 16 Challenge level In this game the winner is the first to complete a row of three. Are some squares easier to land on than others? game Favourite Diamond collector Age 11 to 16 Challenge level Collect as many diamonds as you can by drawing three straight lines. problem Crossing the bridge Age 11 to 18 Challenge level Four friends must cross a bridge. How can they all cross it in just 17 minutes? problem All in a jumble Age 11 to 14 Challenge level My measurements have got all jumbled up! Swap them around and see if you can find a combination where every measurement is valid. problem In the bag Age 11 to 14 Challenge level Can you guess the colours of the 10 marbles in the bag? Can you develop an effective strategy for reaching 1000 points in the least number of rounds? game Favourite Vector racer Age 11 to 16 Challenge level The classic vector racing game. game Favourite Up, down, flying around Age 11 to 14 Challenge level Play this game to learn about adding and subtracting positive and negative numbers problem Countdown fractions Age 11 to 16 Challenge level Here is a chance to play a fractions version of the classic Countdown Game. problem Shifting times tables Age 11 to 14 Challenge level Can you find a way to identify times tables after they have been shifted up or down? problem Nine colours Age 11 to 16 Challenge level Can you use small coloured cubes to make a 3 by 3 by 3 cube so that each face of the bigger cube contains one of each colour? problem Mixing lemonade Age 11 to 14 Challenge level Can you work out which drink has the stronger flavour? problem Frogs Age 11 to 14 Challenge level How many moves does it take to swap over some red and blue frogs? Do you have a method? problem Charlie's delightful machine Age 11 to 16 Challenge level Here is a machine with four coloured lights. Can you develop a strategy to work out the rules controlling each light? problem Multiplication arithmagons Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons? problem Finding factors Age 14 to 16 Challenge level Can you find the hidden factors which multiply together to produce each quadratic expression? problem 3D treasure hunt Some treasure has been hidden in a three-dimensional grid! Can you work out a strategy to find it as efficiently as possible? problem Standard index form matching Age 14 to 16 Challenge level Can you match these calculations in Standard Index Form with their answers?
4838	https://phys.libretexts.org/Courses/Georgia_State_University/GSU-TM-Physics_II_(2212)/02%3A_Electrostatics_-_Charges_Forces_and_Fields	2: Electrostatics - Charges, Forces and Fields - Physics LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 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"GSU-TM-Physics_II_(2212)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "GSU-TM-Physics_I_(2211)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Sat, 19 Aug 2023 01:17:54 GMT 2: Electrostatics - Charges, Forces and Fields 76510 76510 Joshua Halpern { } Anonymous Anonymous 2 false false [ "article:topic-guide", "authorname:openstax", "license:ccby", "showtoc:no", "program:openstax", "source-phys-4378", "licenseversion:40", "source@ ] [ "article:topic-guide", "authorname:openstax", "license:ccby", "showtoc:no", "program:openstax", "source-phys-4378", "licenseversion:40", "source@ ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Campus Bookshelves 3. Georgia State University 4. GSU-TM-Physics II (2212) 5. 2: Electrostatics - Charges, Forces and Fields Expand/collapse global location GSU-TM-Physics II (2212) Front Matter 1: Introduction to Physics, Measurements and Mathematics Tools 2: Electrostatics - Charges, Forces and Fields 3: Electric Potential and Capacitance 4: Current and Resistance 5: Resistive Networks 6: Sources of Magnetism, Magnetic Forces and Fields 7: Electromagnetic Induction 8: Electromagnetic Waves 9: Geometrical Optics 10: Physical Optics 11: Nuclear Physics 12: Atomic Structure Back Matter 2: Electrostatics - Charges, Forces and Fields Last updated Aug 19, 2023 Save as PDF 1.9.27: Math-vector basics and diffrential equations 2.1: Unit Learning Outcomes Page ID 76510 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents No headers 2.1: Unit Learning Outcomes 2.2: Electrical Charge 2.3: Conductors, Insulators, and Charging by Induction 2.4: Electrostatic Force - Coulomb's Law 2.5: Electric Field 2.6: Calculating Electric Fields of Charge Distributions 2.7: Electric Flux 2.8: Gauss’s Law 2.9: Applying Gauss’s Law 2.10: Conductors in Electrostatic Equilibrium 2.11: Summary 2.12: Practice 2.13: Electric Charges and Fields (Answer) Thumbnail: The eight source charges each apply a force on the single test charge Q. Each force can be calculated independently of the other seven forces. This is the essence of the superposition principle. This page titled 2: Electrostatics - Charges, Forces and Fields is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. Back to top 1.9.27: Math-vector basics and diffrential equations 2.1: Unit Learning Outcomes Was this article helpful? Yes No Recommended articles 1: Electric Charges and FieldsIn this chapter, we begin the study of the electric force, which acts on all objects with a property called charge. The electric force is much stronge... 1: Electric Charges and FieldsIn this chapter, we begin the study of the electric force, which acts on all objects with a property called charge. 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4839	https://math.stackexchange.com/questions/1214067/question-about-necessary-and-sufficient-conditions	number theory - Question about necessary and sufficient conditions? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Question about necessary and sufficient conditions? Ask Question Asked 10 years, 6 months ago Modified10 years, 6 months ago Viewed 173 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. \begingroup I am working on a question which begins with The number \alpha is a common root of the equations x^2+ax+b=0 and x^2+cx+d=0. Given that a\neq c, show that \alpha=-\frac{b-d}{a-c} This was easy enough to show algebraically by assuming that alpha does satisfy the above two equations. Am I correct in saying that, because I made this assumption, it is sufficient that the root alpha takes on the above form, but it is not necessary? This doesn't seem to fit with the question though, because that would imply that if alpha takes on that value, then the two equations automatically have a common root, however it is not necessary for alpha to take on this value for them to have a common root. In that case, I have not shown that alpha being a common root of the equations implies that alpha must take on this value, I have shown that if alpha does take. On this value, then the two equations have a common root... I'm sorry if this is very confusing! I am trying to explain myself as best as I can... The second part of the question is Hence, or otherwise, show that the equations have at least one common root if and only if (b-d)^2-a(b-d)(a-c)+b(a-c)^2=0 Since I am not sure whether the first statement for the value of alpha is necessary or sufficient, I am having trouble deciding how far I can deal with the second part of the question using the result from the first part... Thank you in advance for any guidance. If anyone has a link to any particularly good resource that will help with using necessary/sufficient and if/iff I would be very grateful. number-theory functions notation Share Cite Follow Follow this question to receive notifications edited Mar 31, 2015 at 14:13 N. F. Taussig 79.3k 14 14 gold badges 62 62 silver badges 77 77 bronze badges asked Mar 31, 2015 at 8:11 MeepMeep 3,287 4 4 gold badges 34 34 silver badges 59 59 bronze badges \endgroup 3 \begingroup Is it a=-\frac{b-d}{a-c} or \alpha=-\frac{b-d}{a-c} ?\endgroup Claude Leibovici –Claude Leibovici 2015-03-31 08:32:58 +00:00 Commented Mar 31, 2015 at 8:32 \begingroup Apologies- should be alpha. Will change it.\endgroup Meep –Meep 2015-03-31 09:13:10 +00:00 Commented Mar 31, 2015 at 9:13 \begingroupsimilar question\endgroup Bumblebee –Bumblebee 2015-03-31 09:29:55 +00:00 Commented Mar 31, 2015 at 9:29 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. \begingroup Assume for now that a\ne c. Then you’ve shown that if if the two equations have a common root \alpha, then that common root must be -\frac{b-d}{a-c}\;.\tag{1} You’ve not shown, however, that the expression in (1) always is a common root of the two equations. That is, you’ve shown that having \alpha equal to (1) is necessary, but not that it’s sufficient. And in fact it’s not sufficient. Consider the equations x^2=0 and x^2+x+1=0, with a=b=0 and c=d=1: then -\frac{b-d}{a-c}=-\frac{0-1}{0-1}=-1\;, but -1 isn’t a root of either equation. To find a sufficient condition, substitute the expression (1) for x in each of the equations; in the first you get \frac{(b-d)^2}{(a-c)^2}-\frac{a(b-d)}{a-c}+b on the left-hand side, so in order for (1) to be a root of x^2+ax+b=0, we must have \frac{(b-d)^2}{(a-c)^2}-\frac{a(b-d)}{a-c}+b=0\;.\tag{2} Similarly, from the second equation we find that we must have \frac{(b-d)^2}{(a-c)^2}-\frac{c(b-d)}{a-c}+d=0\;.\tag{3} Multiplying (2) through by (a-c)^2 yields (b-d)^2-a(b-d)(a-c)+b(a-c)^2=0\;,\tag{4} which should look familiar. Doing the same thing to (3) yields (b-d)^2-c(b-d)(a-c)+d(a-c)^2=0\;;\tag{5} with a bit of algebra you can show that (4) and (5) are equivalent, in the sense that each is true if and only if the other is true. (HINT: Solve them for \frac{(b-d)^2}{a-c}.) Thus, we need only consider (4). If (4) holds, and a\ne c, then (2) and (3) hold, and (1) is a root of the first equation. What does (4) imply when a=c? Does it still ensure that the two equations have a common root? Share Cite Follow Follow this answer to receive notifications answered Mar 31, 2015 at 8:58 Brian M. ScottBrian M. Scott 633k 57 57 gold badges 824 824 silver badges 1.4k 1.4k bronze badges \endgroup 10 \begingroup Thank you for your reply! I have been thinking over this for some time now but I am still a bit confused. My first difficulty is with double headed arrows in the first part of the argument.\endgroup Meep –Meep 2015-03-31 09:56:44 +00:00 Commented Mar 31, 2015 at 9:56 \begingroup@21joanna12: The first part of the problem is just a one-way implication: if there is a common root it must be (1). In your discussion you exactly inverted what you'd proved. You don't have to deal with a two-way implication until the second part of question.\endgroup Brian M. Scott –Brian M. Scott 2015-03-31 10:05:48 +00:00 Commented Mar 31, 2015 at 10:05 \begingroup Apologies- I accidentally submitted my comment before I fully write out what I was having trouble with. I wrote: "Assume f(x) and g(x) (the two functions) have a common roots alpha \leftrightarrow\alpha^2+a\alpha+b=0 and \alpha^2+c\alpha+d..." and the rest was just algebraic manippulation so it had double headed arrows all the way through. I thought that if you can use double headed arrows all the way through your argument, then what you have at each step is necessary and sufficient?\endgroup Meep –Meep 2015-03-31 10:10:09 +00:00 Commented Mar 31, 2015 at 10:10 \begingroup I am just unsure of how II used the double headed arrows wrong, because using them still allows me to go back up the argument from (1) to 'assume f(x) and g(x) have a common root alpha' which to me seems to say (1) is sufficient?\endgroup Meep –Meep 2015-03-31 10:14:29 +00:00 Commented Mar 31, 2015 at 10:14 \begingroup My second difficulty is with understanding why (4) is both necessary and sufficient for at least one common root alpha. I understand that if the common root alpha exists, it is necessary that it takes the form (1) and that is alpha exists and a is not equal to c, then (4) follows. But since (1) is necessary, but not sufficient, and I have used it to get (4), then it seems to me that (4) is also necessary, but not sufficient, for there to be at least one common root alpha. And II haven't shown any 'if' because I cannot say that (4) implies the common root in any way...\endgroup Meep –Meep 2015-03-31 10:20:56 +00:00 Commented Mar 31, 2015 at 10:20 \|Show 5 more comments You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions number-theory functions notation See similar questions with these tags. 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4840	https://www.cnblogs.com/liuning8023/p/3525920.html	Published Time: 2014-01-19T14:59:00.0000000+08:00 算术平均、几何平均、调和平均、平方平均和移动平均 - 船长&CAP - 博客园会员周边众包新闻博问闪存赞助商 Chat2DB 所有博客当前博客我的博客我的园子账号设置会员中心简洁模式 ... 退出登录注册登录船长&CAP “0 + 1 = The World, 我们既愚蠢/也聪明/愚蠢的是/我们世界只有0和1/聪明的是/我们却用0和1描述了这个世界” 博客园首页新随笔联系订阅管理算术平均、几何平均、调和平均、平方平均和移动平均迁移到：本文内容算术平均几何平均调和平均平方平均移动平均参考资料算术平均、几何平均、调和平均、平方平均和移动平均跟计算编程有什么关系：Just One Word，不能只会算术平均数，还有其他很多选择，以及不同场景使用不同的平均数。算术平均算术平均（Arithmetic mean）是最基本、最常用的一种平均指标，描述数据集中趋势的一个统计指标。计算公式为：即，n 个数据相加后除以 n。0 也记入。统计学上，算术平均较中位数和众数更少受到随机因素影响，但缺点是它极易受到极大极小值的影响。例如，有数组 (5, 7, 5, 4, 6, 7, 8, 5, 4, 7, 8, 6, 20)，平均值是 7.1，但实际上大部分数据（10个）都不超过7，如果去掉 20，平均数为 6。上面是简单算术平均，它只是加权算术平均的一种特殊形式。若原始数据，被分成 k 组，各组的值为 (x1，x2，...，xk)，各组频率分别为 (f1，f2，...，fk)，则加权算术平均数的计算公式为：由公式可以看出，加权算术平均数同时受到两个因素的影响，一个是各组数值的大小 xi，另一个是各组分布频数 fi。在数值不变的情况下，某组的频数越多，该组数值对平均数的作用就大，反之，越小。算术平均可以用来反映一组数据的一般情况，也可以对不同组的数据进行比较。平均数可以直观、简明的表示一组数据，所以，在日常生活中经常用到，如平均速度、平均身高、平均产量、平均成绩等等。算术平均主要适用于数值型数据，不适用于品质数据。几何平均几何平均（Geometric mean），是另一种计算平均值的方法。对几何平均，也可以像算术平均一样，做加权的几何平均。简单几何平均的计算公式为：即，n 个数据相乘后开 n 次方。其中，xi 都是正实数。几何平均适用于对比率、指数等进行平均，主要用于平均增长（变化）率，对数正态分布。算术-几何平均数若有两个正实数 x 和 y，则它们的算术-几何平均数为，先计算这两个数的算术平均数，称为 a1；再计算它们的几何平均数，称为 g1。重复这个步骤，便得到了两个数列 (a__n) 和 (g__n)：这两个数列都收敛于一个相同的数，这个数称为 x 和 y 的算术-几何平均数，记为 M(x, y) 或 agm(x, y)。示例：计算 a0 = 24和 g0 \= 6的算术-几何平均数 M(24, 6) 如下表所示： nangn024611521213.513.41640786500...313.45820393250...13.45813903099...413.45817148175...13.45817148171... a0 = 24和 g0 \= 6的算术-几何平均数。 1 和的算术-几何平均数的倒数，称为高斯常数。调和平均调和平均（Harmonic Mean），也分简单和加权的形式。加权调和平均数是加权算术平均数的变形。多数多情况下，我们只掌握每组某个标志的数值总和（m），而缺少总体单位数（f）的资料，因此，不能直接采用加权算术平均数法计算平均数，而则采用加权调和平均数。先由加权算术平均数公式推到加权调和平均公式，最后推到简单调和平均公式，它是加权调和公式的特殊形式。加权算术平均的计算公式为：即，加权调和平均公式为：当 mi\=1 时，则公式退化成简单调和平均公式：即，n 个数据的倒数取算术平均，再取倒数。调和平均一般用于计算平均速率。示例：某工厂购进材料三批，每批价格及采购金额资料如下表：价格x（元/千克）采购金额 m（元）采购数量 m/x（千克）第一批3510000286第二批4020000500第三批4515000330合计——450001116 每千克 40.32 元。二个数的调和平均数最常用的是二个正数值 x1 和 x2 的调和平均数 H：而 x1 和 x2 的算术平均数 A 与几何平均数 G 分别为：那么，它们存在如下关系：应用可以用在相同距离，但速度不同的平均速度，如一段路，前半段时速 60 公里，后半段时速 30 公里〔两段距离相等〕，则其平均速度为两者的调和平均数 40 公里。两个电阻 R1 和 R2 并联后的等效电阻 Req 为调和平均数的一半。物理学中的减缩质量也为调和平均数的一半。毕达哥拉斯平均是算术平均数（A）、几何平均数（G）及调和平均数（H），这三种平均数的总称。平方平均平方平均（Quadratic mean），简称方均根（Root Mean Square，RMS），是平方根的广义平均（generalized mean），计算公式为：即，n 个数据的平方取算数平均，再开平方根。利用柯西不等式，平方平均与算术平均的关系是：平方平均不小于算术平均。应用平方平均数常用来计算一组数据和某个数据的“平均差”。像交流电的电压、电流数值以及均匀加速直线运动的位移中点平均速度，都是以其实际数值的方均根表示。例如，交流电 220V 表示电压信号的均方根（又称为有效值），即 220V，为交流电瞬时值（瞬时值又称暂态值）的最大值的。统计中的标准差 s：即，所有数据与算术平均值相减，取它们的平方平均数。移动平均移动平均（Moving Average，MA），又称“移动平均线”简称均线，是一种简单平滑预测技术，它的基本思想是：根据时间序列资料、逐项推移，依次计算包含一定项数的序时平均值，以反映长期趋势的方法。因此，当时间序列的数值由于受周期变动和随机波动的影响，起伏较大，不易显示出事件的发展趋势时，使用移动平均法可以消除这些因素的影响，显示出事件的发展方向与趋势（即趋势线），然后依趋势线分析预测序列的长期趋势。移动平均法适用于即期预测。当产品需求既不快速增长也不快速下降，且不存在季节性因素时，移动平均法能有效地消除预测中的随机波动，是非常有用的。移动平均可抚平短期波动，反映出长期趋势或周期。最常见的是利用股价、回报或交易量等变量计算出移动平均。数学上，移动平均可视为一种卷积（卷积是通过两个函数 f 和 g 生成第三个函数的一种数学算子，表征函数 f 与经过翻转和平移的 g 的重叠部分的累积。如果将参加卷积的一个函数看作区间的指示函数，卷积还可以被看作是“移动平均”的推广）。移动平均法可以分为：简单移动平均和加权移动平均。参考资料 Wiki 算术-几何平均 M(x,y) Wiki 广义平均（幂平均） Wiki 毕达哥拉斯平均 Wiki 移动平均 Wiki MBAlib 移动平均 Wiki 卷积 posted @ 2014-01-19 14:59 船长&CAP 阅读(52551) 评论(1) 编辑收藏举报努力加载评论中... 刷新页面返回顶部公告 Copyright © 2025 船长&CAP Powered by .NET 9.0 on Kubernetes
4841	https://www.osmosis.org/learn/Mixed_platelet_and_coagulation_disorders:_Pathology_review	Mixed platelet and coagulation disorders: Pathology review: Video, Causes, & Meaning \| Osmosis Plans Medicine (USMLE)Medicine (non-USMLE)Medicine (Osteopathy)Physician AssistantNurse PractitionerRegistered NursingLicensed Practical NursingDentistryPharmacyHealth Professional Library Medicine (USMLE)Medicine (non-USMLE)Medicine (Osteopathy)Physician AssistantNurse PractitionerRegistered NursingLicensed Practical NursingDentistryPharmacyHealth Professional Resources EventsBlogQuestions & answer topicsRaise the Line PodcastSpreadjoy e-cardsOsmosis around the world InstitutionsSign inTry it free Fall in love with Osmosis at 20% off! Save now until September 30 at 11:59 PM PT. Learn more Skip to the video Mixed platelet and coagulation disorders: Pathology review Foundational Sciences Pathology Blood and lymphoreticular system Blood and lymphoreticular system pathology review Coagulation disorders (hypocoagulable and hypercoagulable conditions) Also appears in 10,134 views Add to playlist 00:00 / 00:00 Videos Notes Mixed platelet and coagulation disorders: Pathology review Blood and lymphoreticular system Anemia, cytopenias, and polycythemia anemias Acute intermittent porphyriaPorphyria cutanea tardaAutoimmune hemolytic anemiaGlucose-6-phosphate dehydrogenase (G6PD) deficiencyHemolytic disease of the newbornHereditary spherocytosisParoxysmal nocturnal hemoglobinuriaPyruvate kinase deficiencySickle cell disease (NORD)Fanconi anemiaFolate (Vitamin B9) deficiencyMegaloblastic anemiaVitamin B12 deficiencyAlpha-thalassemiaAnemia of chronic diseaseBeta-thalassemiaIron deficiency anemiaLead poisoningSideroblastic anemiaAnemia of chronic diseaseAplastic anemiaDiamond-Blackfan anemiaFanconi anemiaAplastic anemiaImmune thrombocytopeniaLeukemoid reactionPolycythemia vera (NORD) Coagulation disorders (hypercoaguable and hypocoaguable conditions) Antiphospholipid syndromeAntithrombin III deficiencyFactor V LeidenProtein C deficiencyProtein S deficiencyHemophiliaVitamin K deficiencyDisseminated intravascular coagulationHeparin-induced thrombocytopeniaVon Willebrand diseaseBernard-Soulier syndromeGlanzmann's thrombastheniaHemolytic-uremic syndromeImmune thrombocytopeniaThrombotic thrombocytopenic purpura Infectious and immunologic disorders Autoimmune hemolytic anemiaHemolytic-uremic syndromeParoxysmal nocturnal hemoglobinuriaThrombotic thrombocytopenic purpura Neoplasms Langerhans cell histiocytosisMastocytosis (NORD)Essential thrombocythemia (NORD)Myelodysplastic syndromesMyelofibrosis (NORD)Polycythemia vera (NORD)Acute leukemiaChronic leukemiaHodgkin lymphomaNon-Hodgkin lymphomaMonoclonal gammopathy of undetermined significanceMultiple myelomaWaldenstrom macroglobulinemia Traumatic, mechanical, and vascular disorders AspleniaRuptured spleen Blood and lymphoreticular system pathology review Extrinsic hemolytic normocytic anemia: Pathology reviewHeme synthesis disorders: Pathology reviewIntrinsic hemolytic normocytic anemia: Pathology reviewMacrocytic anemia: Pathology reviewMicrocytic anemia: Pathology reviewNon-hemolytic normocytic anemia: Pathology reviewCoagulation disorders: Pathology reviewMixed platelet and coagulation disorders: Pathology reviewPlatelet disorders: Pathology reviewThrombosis syndromes (hypercoagulability): Pathology reviewLeukemias: Pathology reviewLymphomas: Pathology reviewMyeloproliferative disorders: Pathology reviewPlasma cell disorders: Pathology review Assessments USMLE® Step 1 questions Start 0 / 2 complete PANCE® questions Start 0 / 3 complete Questions USMLE® Step 1 style questions USMLE 0 of 2 complete Start Save to Queue Preview A 4-year-old girl presents to the emergency department with abdominal pain, diarrhea, and fatigue. The patient was in her usual state of health until yesterday afternoon, when her family returned home from a barbeque. The patient is otherwise healthy and takes no medications aside from amoxicillin for a recent episode of otitis media. According to her parents, the patient’s urine has looked “darker” than usual. Temperature is 39.0°C (102.2°F), pulse is 115/min, respirations are 22/min, and blood pressure is 100/70 mmHg. Physical examination demonstrates a pale appearing girl with diffuse abdominal tenderness to palpation, delayed capillary refill, and gingival bleeding. Multiple tiny, brownish-purple, blanchable spots are noted under the skin. Which of the following laboratory findings are most consistent with this patient’s disease process? Transcript Watch video only Content Reviewers Yifan Xiao, MD,Antonia Syrnioti, MD Contributors Samantha McBundy, MFA, CMI,Robyn Hughes, MScBMC,Maria Emfietzoglou, MD,Sam Gillespie, BSc At the emergency department, a 70 year old male named Max is admitted because of high fever with chills, and hypotension. He complains of having urinary urgency, frequency and dysuria, or painful urination, for the last few days. A few hours after admission, he rapidly deteriorates and starts to bleed from venipuncture sites. Urine and blood cultures are ordered and are both positive for gram negative rods. Lab tests show low platelet count, and bleeding time, PT and PTT are prolonged, fibrinogen is decreased and d-dimers are elevated. Peripheral blood smear shows schistocytes. Now, there’s also an 18 year old female, named Sylvia, that came in with recurrent severe nose bleeds. She also complains of heavy menstrual periods. Family history reveals her father also suffered from bleeding diathesis. Lab tests show normal platelet count, prolonged bleeding time and PTT, and normal PT. Both Max and Sylvia are suffering from a hemostasis disorder. Hemostasis disorders, also known as bleeding disorders, can be broadly divided into three groups. The first includes problems with primary hemostasis, which is the formation of the weak platelet plug, and so, they’re referred to as platelet disorders. Now, the second group includes problems with secondary hemostasis, which is making a strong fibrin clot through activation of the intrinsic, extrinsic and common coagulation pathways, and are also known as coagulation disorders. And the last group includes disorders that affect both primary and secondary hemostasis and are known as mixed platelet and coagulation disorders. Okay, in this video, we will focus on mixed platelet and coagulation disorders, that include disseminated intravascular coagulation, or DIC, and von Willebrand disease. Alright, so let’s take a closer look at these disorders, starting with DIC, which is a massive overactivation of the coagulation system including both platelets and clotting factors. For your exams, it’s important to know that DIC can occur in response to serious conditions including gram negative bacterial sepsis, trauma, and obstetric complications such as abruptio placenta and retained dead fetus in utero, acute pancreatitis, malignancies such as adenocarcinomas and promyelocytic leukemia, nephrotic syndrome, snakebites, and transfusion reactions. Okay, whatever the cause, there is a release of a procoagulant that tips the scales in favor of clot formation. Procoagulants could be enzymes that help to proteolytically cleave and activate clotting factors or proteins like bacterial components such as lipopolysaccharide or tissue factor also known as thromboplastin. For your test, remember that the release of tissue factor from abruptio placenta into the maternal circulation, is, in fact, the most common cause of DIC in pregnancy. DIC leads to widespread clotting, which can block off small arteries leading to organ ischemia. These clots also act like jagged rocks in a river and damage the red blood cells floating by, causing microangiopathic hemolytic anemia. These damaged RBCs can be seen on a blood smear as schistocytes but sometimes they get destroyed completely. At the same time, excessive clot formation depletes platelets and clotting factors, which paradoxically, leads to increased bleeding. Now, let’s move onto von Willebrand disease, the most common inherited bleeding disorder. It’s usually caused by autosomal dominant mutations of von Willebrand factor. These proteins normally serves as the glue between the platelet receptor Gp1b and the collagen underneath the endothelial cells. So, for the test remember that if there’s a problem with von Willebrand factor, it’s hard for platelets to adhere to collagen in damaged blood vessels, leading to impaired platelet function. Inherited von Willebrand disease is subclassified into type 1, which is a decrease in the quantity of von Willebrand factor, and type 2, which is a decrease in the function of von Willebrand factor. Meanwhile, von Willebrand factor also stabilize factor 8 of the intrinsic coagulation pathway. So without von Willebrand factor, there’s less functioning factor 8 around, leading to decreased activation of the coagulation cascade. So mixed platelet and coagulation disorders affect both primary and secondary hemostasis, and as a result they can present with symptoms caused by dysfunctions in both pathways. Primary hemostatic, or platelet, problems usually present with petechiae, which are pinpoint superficial skin bleeds, anterior epistaxis, which are usually mild nosebleeds, immediate bleeding after surgical procedures, like tooth extraction, or bleeding from mucosal surfaces, like gingival, gastrointestinal, or vaginal bleeding. In contrast, secondary hemostatic, or coagulation, problems can present with large bruises after minor trauma, like bumping into a door. They also suffer from ecchymoses, which is discoloration caused by bleeding under the skin, deep tissue hematomas, hemarthrosis, which is bleeding inside the joint space, posterior epistaxis, which causes a severe nosebleed, GI bleeding, urinary bleeding, and persistent bleeding after surgical procedures.Now, a dangerous complication is intracerebral hemorrhage, or bleeding into the brain, which can cause a stroke or increased intracranial pressure. Sources "Robbins Basic Pathology" Elsevier (2017) "Harrison's Principles of Internal Medicine, Twentieth Edition (Vol.1 & Vol.2)" McGraw-Hill Education / Medical (2018) "Diagnosis and Treatment of Benign Bleeding Disorders" Journal of the Advanced Practitioner in Oncology (2016) "Bleeding and Coagulopathies in Critical Care" New England Journal of Medicine (2014) "Disseminated intravascular coagulation" Nature Reviews Disease Primers (2016) "von Willebrand disease (VWD): evidence-based diagnosis and management guidelines, the National Heart, Lung, and Blood Institute (NHLBI) Expert Panel report (USA)" Haemophilia (2008) "The diagnosis and management of von Willebrand disease: a United Kingdom Haemophilia Centre Doctors Organization guideline approved by the British Committee for Standards in Haematology" British Journal of Haematology (2014) Company About usCareersLibraryEditorial BoardBlogBecome An Osmosis Student Leader Support Help centerContact usFAQ For Institutions Medicine (DO)Medicine (MD)Nurse Practitioner (NP)Physician Assistant (PA)DentistryNursing (RN) More PlansiOS appAndroid appSwagCreate custom contentRaise the Line Podcast Copyright © 2025 Elsevier, its licensors, and contributors. 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4842	https://www.genevaenvironmentnetwork.org/events/world-wetlands-day-2023/	World Wetlands Day 2023 – Geneva Environment Network Skip to content Menu English Français Environment in Geneva GEN Secretariat Key Areas and SDGs Organizations Experts History Environment House Organizations Conference Centre Visits Staff Association and Intern Board Directions Events GEN Events Plastics Treaty INC-5.2 World Environment Day GEN 25th Anniversary Events Archive Jobs Resources Newsletter Geneva Green Guide Learning Publications Glossary Links Updates Menu Event World Day World Wetlands Day 2023 02 Feb 2023 Nature SDG6\|SDG14\|SDG15 "It's time for wetland restoration" is the 2023 theme for World Wetlands Day, celebrated each year on 2 February. #GenerationRestoration #ForWetlands About World Wetlands Day, celebrated each year on 2 February, marks the date of the adoption of theConvention on Wetlands (Ramsar Convention)on 2 February 1971 in Ramsar, Iran. Wetlands are among the world’s most diverse productive environments; cradles of biological diversity that provide the water and productivity upon which countless species of plants and animals depend for survival. Wetlands are indispensable for the countless benefits or “ecosystem services” that they provide humanity, ranging from freshwater supply, food and building materials, and biodiversity, to flood control, groundwater recharge, and climate change mitigation. Wetlands are also essential for biodiversity: though wetlands cover only around 6% of the Earth’s land surface, 40% of all plant and animal species live or breed in wetlands. However, wetlands as the most threatened ecosystem. According to the World Heritage and Ramsar Conventions, 64% of the world’s wetlands have disappeared since the beginning of the last century. In most regions across the world, wetlands continue to decline compromising the benefits that wetlands provide to people. According to theGlobal Wetland Outlook, we lose wetlands three times faster than natural forests. Therefore, the conservation of wetlands is a vital task of humanity, which can help achieving the Sustainable Development Goals by 2030. Official website 2023 Theme The theme for the 2023 edition is “It’s time for wetland restoration”. Pitch for the Grant 2023: Call for project ideas in wetlands On occasion of World Wetlands Day 2023, the Convention on Wetlands is inviting individuals, community organizations and social enterprises to submit pitches of project ideas or initiatives, which aim to contribute to healthy wetlands by either promoting conservation, restoration or improved management. → Submit your applications from 2 February to 2 March 2023 UN Decade on Restoration (2021-2030) TheUN Decade on Ecosystem Restoration calls for the protection and revival of ecosystems around the world for the benefit of people and nature. It aims to halt the degradation of ecosystems, and restore them to achieve global goals. Only with healthy ecosystems, such as wetlands, can we enhance people’s livelihoods, counteract climate change, and stop the collapse of biodiversity. The UN Decade runs from 2021 through 2030, which is also the deadline for the Sustainable Development Goals and the timeline scientists have identified as the last chance to prevent catastrophic climate change. Led by theUnited Nations Environment Programme (UNEP)and theFood and Agriculture Organization (FAO)of the United Nations, the UN Decade is building a strong, broad-based global movement to ramp up restoration and put the world on track for a sustainable future. That will include building political momentum for restoration as well as thousands of initiatives on the ground. More on the UN Decade on Restoration Events Activities in Geneva for World Wetlands Day will be listed here. For more information on events organized by local actors, you may consult dans ma Nature Genève. Ciné Environment House \| Chaos in the climate of the Sundarbans 25 January 2023 \| 12:45 – 13:45 CET \| IEH1 & Online Visite de presse \| By invitation 31 January 2023 \| 8:30 – 12:30 CET Time to Fast-Track Restoration of Wetlands: Seven Priority Actions 1 February 2023 \| 9:00 – 10:15 CET & 16:00 -17:15 CET \| Secretariat of the Convention on Wetlands Pygargue à queue blanche : un géant plane dorénavant sur le lac ! 5 February 2023 \| 8:30 – 12:30 CET \| Lagune des Eaux-vives et ses environs \| État de Genève Foulque macroule : la gardienne des roseaux 5 February 2023 \| 9:00 – 11:00 CET \| Lagune des Eaux-vives et ses environs \| Rdv devant Baby Plage \| NARIES Fuligule nyroca : à la recherche de l'oiseau rare ! 5 February 2023 \| 9:30 – 15:30 CET \| Lagune des Eaux-vives et ses environs \|Muséum d’histoire naturelle de la Ville de Genève Grèbe huppé : l’oiseau sous-marin à la conquête des ports 5 February 2023 \| 11:00-12:30 CET \| Lagune des Eaux-vives et ses environs \| Groupe Ornithologique du Bassin Genevois La Nette rousse et les Characées 5 February 2023 \| 13:00 – 14:00 CET \| Rdv au niveau du panneau didactique sur le quai Gustave-Ador \| Association pour la Sauvegarde du Léman Ephémère : les dessous d’une vie pas si brève 5 February 2023 \| 11:30-12:30 CET \| Lagune des Eaux-vives et ses environs \| HEPIA Découverte du Fuligule Morillon 5 February 2023 \| 13:30 – 14:00 CET \| Rdv à l’embarcadère « Port Noir » \| Pro Natura Genève Les coulisses d’une roselière lacustre 5 February 2023 \| 10:30 – 12:00 CET \| Centre Nature de la Pointe à la Bise \| Pro Natura Genève Les canards de la Rade 5 February 2023 \| 13:00 – 17:00 CET \| Pavillon Plantamour – Parc Mon Repos \| La Libellule The Role of Geneva As a global hub for environment governance and the host region of the Secretariat of the Convention on Wetlands, Geneva is key place to protect wetlands and reverse the trend of global loss of wetlands. Many organizations are active in the region to conserve wetland areas locally and globally. Learn more about the importance of wetlands and the role of Geneva in our update below. Wetlands and the Role of Geneva Share this page Share on Facebook Share on Bluesky Share on X (Twitter) Share on LinkedIn Share by Email Receive updates Newsletter Contact Contact Search the website Search for: Search Follow us Facebook Bluesky X(Twitter) LinkedIn YouTube Instagram Network coordinated by With support from Terms of Use
4843	https://www.geogebra.org/m/hQW8KxXz	Home Resources Profile Classroom App Downloads The Pythagorean Theorem and The Circle Author:kmack7 Topic:Circle In the below activity grab point P and drag it around the the graph. Notice the trace it creates. We see that when moving around the point P that we create a circle. A circle is the locus of points in a plane equidistant from a given point in the same plane. The given point is called the center and the given distance is the radius. See the activity below and observe that when point P moves around the circle, the equation shown doesn't change. So, we have a right triangle. We know side a is the y component of point P and side b is the x component. The Pythagorean Theorem is where . This is used to find the length of the hypotenuse, h. The equation of a circle is , look familiar?? The sides of the right triangle and the hypotenuse are used in building the circle and thus its equation. Since P lies on the circle, what must be true about its coordinates? Pick a point and verify that it satisfies your answer. Check my answer In the activity below, change the radius of circle A by dragging the slider back and forth. Notice what changes in the equation of the circle. So, we see that by changing the radius of the circle, the only thing in the equation that changes is what the equation equals. Much like in the pythagorean theorem, when c changes, the hypotenuse changes, so when the radius changes, the circle gets bigger/smaller. From this, we can conclude that the hypotenuse of the right triangle = the radius of a circle. Now we can relate the Pythagorean Theorem to the Equation of a circle: New Resources רישום חופשי Set Operations (2 Sets) Dividing a 3-digit number by a 1-digit number (1) [z`]]](/m/akq95bkx) Untitled Discover Resources inversion Latice Demo: Shuffle without the Shuffle() command สามเหลี่ยมพีทาโกรัส2 Colored circle Discover Topics Square Sphere Probability Variance Linear Functions
4844	https://www.reddit.com/r/calculus/comments/8up9ii/definite_integral_of_an_absolute_value_function/	Definite Integral of an absolute value function : r/calculus Skip to main contentDefinite Integral of an absolute value function : r/calculus Open menu Open navigationGo to Reddit Home r/calculus A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to calculus r/calculus r/calculus Welcome to r/calculus - a space for learning calculus and related disciplines. Remember to read the rules before posting and flair your posts appropriately. 168K Members Online •7 yr. ago [deleted] Definite Integral of an absolute value function Hey, im a bit rusty on calculus , but im studying for my final exam , and i found something a bit weird, im asked to find the area bounded by the curve f(x)=\|xx-4x+3\| , x=0 , x=5 , and the x axis, but the points x=1 and x=3 are sharp points, which means that f(x) is not differentiable at those points. can i still integrate? bc i cant undo something that cant be done. "si" means if in this case, and sorry 4 my broken english , im not a native speaker Read more Share Related Answers Section Related Answers Methods to integrate absolute value functions Antiderivative of 1/x function Properties of the natural logarithm function Integral of negative sine function Explanation why integral of 1/x is ln x New to Reddit? Create your account and connect with a world of communities. Continue with Google Continue with Google. Opens in new tab Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community 0 0 Top Posts Reddit reReddit: Top posts of June 29, 2018 Reddit reReddit: Top posts of June 2018 Reddit reReddit: Top posts of 2018 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
4845	https://www.studocu.com/en-us/messages/question/3025165/a-trough-has-ends-shaped-like-isosceles-triangles-with-width-3-m-and-height-4-m-and-the-trough-is	[Solved] A trough has ends shaped like isosceles triangles with width 3 m - Adv Calculus (MATH 472) - Studocu Skip to main content Teachers University High School Discovery Sign in Welcome to Studocu Sign in to access study resources Sign in Register Guest user Add your university or school 0 followers 0 Uploads 0 upvotes Home My Library AI Notes Ask AI AI Quiz New Recent A trough has ends shaped like isosceles triangles, with width 3 m and height 4 m, and the trough is 20 m long. Water is being pumped into the trough at a rate of9 m3/min.At what rate (in m/min) does the height of the water change when the water is 1 m deep? Adv Calculus (MATH 472) My Library Courses You don't have any courses yet. Add Courses Studylists You don't have any Studylists yet. Create a Studylist University of Idaho Adv Calculus Question A trough has ends shaped like isosceles triangles with width 3 m University of Idaho Adv Calculus Question Anonymous Student 2 years ago A trough has ends shaped like isosceles triangles, with width 3 m and height 4 m, and the trough is 20 m long. Water is being pumped into the trough at a rate of 9 m3/min. At what rate (in m/min) does the height of the water change when the water is 1 m deep? Like 0 Related documents (12) 2.8 - Related Rates Analytic Geometry And Calculus I Lecture notes None (12) 2.8, Related Rates - SP23 MATH170-05, section 05, Spring 2023 Web Assign Analytic Geometry And Calculus I Assignments None (02) 1.4, The Tangent and Velocity Problems - SP23 MATH170-05, section 05, Spring 2023 Web Assign Analytic Geometry And Calculus I Assignments None (06) 2.1, Derivatives and Rates of Change - SP23 MATH170-05, section 05, Spring 2023 Web Assign Analytic Geometry And Calculus I Assignments None (09) 2.4, Derivatives of Trig Functions - SP23 MATH170-05, section 05, Spring 2023 Web Assign Analytic Geometry And Calculus I Assignments None 2.1 - rate of change - Description is required Please give as much additional information as possible,Finite Mathematics Assignments None Annotated-Written Homework%237 corrections Analytic Geometry And Calculus I Assignments None Answer Created with AI 2 years ago Solution To solve this problem, we need to use the concept of related rates in calculus. First, let's denote the following variables: V: volume of the water in the trough h: height of the water in the trough r: rate at which water is being pumped into the trough dh/dt: rate at which the height of the water is changing The volume of the water in the trough can be represented as the volume of a triangular prism, which is given by the formula: js V = 1/2 base height length In this case, the base of the triangle is the width of the water surface, which changes as the height of the water changes. Since the ends of the trough are isosceles triangles, the width of the water surface is proportional to its height. So, we can write the base as 3h/4. Substituting this into the volume formula, we get: js V = 1/2 (3h/4) h 20 Simplifying this, we get: js V = 15h^2 Differentiating both sides with respect to time (t), we get: js dV/dt = 30h dh/dt We know that dV/dt = r = 9 m³/min. Substituting this and h = 1 m into the equation, we can solve for dh/dt: js 9 = 30 1 dh/dt Solving for dh/dt, we get: js dh/dt = 9/30 = 0.3 m/min So, the height of the water changes at a rate of 0.3 m/min when the water is 1 m deep. Like 0 AI answers may contain errors. Please double check important information and use responsibly. Ask a new question Related Answered Questions 2 years ago Sandra is a 25-year-old woman who weighs 120 lb. She burns300 − 50t cal/hr while walking on her treadmill. Her caloric intake from drinking Gatorade is90t calories during the t th hour. What is her net decrease in calories after walking for 2 hours? Adv Calculus (MATH 472) 2 years ago Consider a limousine that gets m(v) = (120-2x)/4 mpg at speed v (in mph), the chauffeur costs $16/h, and gas is $3.50/gal.(a) Find the cost (in dollars) per mile at speed v.C(v) =(b) Find the cheapest driving speed (in mph). You may enter an exact answer using the CalcPad, or use a rounded answer that is accurate to at least four decimal places. Adv Calculus (MATH 472) 2 years ago You are constructing a box for your cat to sleep in. The plush material for the square bottom of the box costs 7 dollars per square foot and the material for the sides costs 2 dollars per square foot. You need a box with volume4 ft3. Use x to represent the length of the square base, and use h to represent the height of the box.(a) Determine a formula for the height h in terms of x.h = (b) Determine a formula for the total cost C(x) of the materials for the box.C(x) = (c) Find the dimensions (in ft) of the box that minimize cost.length of the base=height= Adv Calculus (MATH 472) 2 years ago Consider a wire 2 ft long cut into two pieces. One piece forms a circle with radius r and the other forms a square with side length x.(a) Determine a formula for the radius r in terms of x.r = (b) Determine a function A(x) representing the sum of the areas (in square feet) of the two shapes, as a function of xA(x)= (c) Determine the value of x (in ft) which maximizes the sum of the areas of the two shapes. Adv Calculus (MATH 472) 2 years ago Two poles are connected by a wire that is also connected to the ground. The first pole is 60 ft tall and the second pole is 30 ft tall. There is a distance of 90 ft between the two poles. Let x denote the distance from the shorter pole to the anchor. (a) Determine the total length T(x) (in feet) of wire used as a function of xT(x)= (b) Where should the wire be anchored to the ground (in ft, from the shorter pole) to minimize the amount of wire needed? Adv Calculus (MATH 472) 2 years ago Consider the function f(x) = 3x∛4-x on the interval [0,5] Find all critical values of f(x) and then determine the maximum & minimum values of f(x) on [0,5] Adv Calculus (MATH 472) Ask AI Home My Library Discovery Discovery Universities High Schools High School Levels Teaching resources Lesson plan generator Test generator Live quiz generator Ask AI English (US) United States Company About us Studocu Premium Academic Integrity Jobs Blog Dutch Website Study Tools All Tools Ask AI AI Notes AI Quiz Generator Notes to Quiz Videos Notes to Audio Infographic Generator Contact & Help F.A.Q. Contact Newsroom Legal Terms Privacy policy Cookie Settings Cookie Statement Copyright & DSA View our reviews on Trustpilot English (US) United States Studocu is not affiliated to or endorsed by any school, college or university. 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4846	https://math.stackexchange.com/questions/750709/proof-that-b-log-bx-x	logarithms - Proof that $b^{\log_b(x)} = x$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Proof that b log b(x)=x b log b⁡(x)=x Ask Question Asked 11 years, 5 months ago Modified11 years, 5 months ago Viewed 2k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. I understand that the exponential functions are inverses, and would therefore map x x when formed as a composition, but I cannot find any formal mathmatical proofs. My thought process is: log b(b log b(x))=log b(x)→log b(x)=log b(x)→x=x log b⁡(b log b⁡(x))=log b⁡(x)→log b⁡(x)=log b⁡(x)→x=x Is that the only way of going about it or are there other formal proofs? Thanks! logarithms exponential-function Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Apr 12, 2014 at 15:04 Cortizol 3,796 25 25 silver badges 37 37 bronze badges asked Apr 12, 2014 at 14:36 OpieDopeeOpieDopee 1,169 1 1 gold badge 13 13 silver badges 31 31 bronze badges 2 2 How do you define the exponential and the logarithm?Jonathan –Jonathan 2014-04-12 14:37:23 +00:00 Commented Apr 12, 2014 at 14:37 Thank you so much question provider and answer providers. This question and the following answers were very valuable for my article, and I appreciate your insight very much.Stephen Elliott –Stephen Elliott 2024-03-17 08:39:56 +00:00 Commented Mar 17, 2024 at 8:39 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. This is a tautology since log b(x)log b⁡(x) is defined as the exponent m m such that b m=x b m=x. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Apr 12, 2014 at 14:55 user141421user141421 2,566 13 13 silver badges 11 11 bronze badges Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. By definition, if A x=B A x=B, then x=log A B x=log A⁡B. Now, replace the value of x from the latter identity into the former. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Apr 12, 2014 at 14:39 LucianLucian 49.4k 2 2 gold badges 86 86 silver badges 157 157 bronze badges 1 Isn't that what I did already?OpieDopee –OpieDopee 2014-04-12 15:08:01 +00:00 Commented Apr 12, 2014 at 15:08 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. What exponent a a can you put on b b so that b a=x b a=x? log b(x)log b⁡(x) is just the name for an a a that makes b a b a be equal to x x. So by definition b log b(x)=x b log b⁡(x)=x. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Apr 12, 2014 at 14:52 patiencepatience 829 1 1 gold badge 10 10 silver badges 23 23 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions logarithms exponential-function See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 0How Can Solving (2 n)(log b 2)=(5 n)(log b 5)(2 n)(log b⁡2)=(5 n)(log b⁡5) be Generalized for any Base of log b n log b⁡n? 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4847	https://www.symbolab.com/study-guides/collegealgebra1/simplifying-algebraic-expressions.html	Study Guide - Simplifying Algebraic Expressions Upgrade to Pro Continue to site We've updated our Privacy Policy effective December 15. Please read our updated Privacy Policy and tap Continue SolutionsIntegral CalculatorDerivative CalculatorAlgebra CalculatorMatrix CalculatorMore... GraphingLine Graph CalculatorExponential Graph CalculatorQuadratic Graph CalculatorSine Graph CalculatorMore... CalculatorsBMI CalculatorCompound Interest CalculatorPercentage CalculatorAcceleration CalculatorMore... GeometryPythagorean Theorem CalculatorCircle Area CalculatorIsosceles Triangle CalculatorTriangles CalculatorMore... ToolsNotebookGroupsCheat SheetsWorksheetsStudy GuidesPracticeVerify Solution en EnglishEspañolPortuguêsFrançaisDeutschItalianoРусский中文(简体)한국어日本語Tiếng Việtעבריתالعربية Upgrade × Symbolab for Chrome Snip & solve on any website Add to Chrome Study Guides>College Algebra Simplifying Algebraic Expressions Sometimes we can simplify an algebraic expression to make it easier to evaluate or to use in some other way. To do so, we use the properties of real numbers. We can use the same properties in formulas because they contain algebraic expressions. Example 12: Simplifying Algebraic Expressions Simplify each algebraic expression. 1. 3 x−2 y+x−3 y−7 3x - 2y+x - 3y - 7 3 x−2 y+x−3 y−7 2. 2 r−5(3−r)+4 2r - 5\left(3-r\right)+4 2 r−5(3−r)+4 3. (4 t−5 4 s)−(2 3 t+2 s)\left(4t-\frac{5}{4}s\right)-\left(\frac{2}{3}t+2s\right)(4 t−4 5s)−(3 2t+2 s) 4. 2 m n−5 m+3 m n+n 2mn - 5m+3mn+n 2 mn−5 m+3 mn+n Solution \begin{array}\text{ }3x-2y+x-3y-7 \hfill& =3x+x-2y-3y-7 \hfill& \text{Commutative property of addition} \ \hfill& =4x-5y-7 \hfill& \text{Simplify}\end{array} \begin{array}2r-5\left(3-r\right)+4 \hfill& =2r-15+5r+4 \hfill& \text{Distributive property} \ \hfill& =2r+5y-15+4 \hfill& \text{Commutative property of addition} \ \hfill& =7r-11 \hfill& \text{Simplify}\end{array} \begin{array}4t-4\left(t-\frac{5}{4}s\right)-\left(\frac{2}{3}t+2s\right) \hfill& =4t-\frac{5}{4}s-\frac{2}{3}t-2s \hfill& \text{Distributive property} \ \hfill& =4t-\frac{2}{3}t-\frac{5}{4}s-2s \hfill& \text{Commutative property of addition} \ \hfill& =\text{10}{3}t-\frac{13}{4}s \hfill& \text{Simplify}\end{array} \begin{array}\text{ }mn-5m+3mn+n \hfill& =2mn+3mn-5m+n \hfill& \text{Commutative property of addition} \ \hfill& =5mn-5m+n \hfill& \text{Simplify}\end{array} Try It 12 Simplify each algebraic expression. 1. 2 3 y−2(4 3 y+z)\frac{2}{3}y - 2\left(\frac{4}{3}y+z\right)3 2y−2(3 4y+z) 2. 5 t−2−3 t+1\frac{5}{t}-2-\frac{3}{t}+1 t 5−2−t 3+1 3. 4 p(q−1)+q(1−p)4p\left(q - 1\right)+q\left(1-p\right)4 p(q−1)+q(1−p) 4. 9 r−(s+2 r)+(6−s)9r-\left(s+2r\right)+\left(6-s\right)9 r−(s+2 r)+(6−s) Solution Example 13: Simplifying a Formula A rectangle with length L L L and width W W W has a perimeter P P P given by P=L+W+L+W P=L+W+L+W P=L+W+L+W. Simplify this expression. Solution \begin{array}\text{ }P=L+W+L+W \ P=L+L+W+W \hfill& \text{Commutative property of addition} \ P=2L+2W \hfill& \text{Simplify} \ P=2\left(L+W\right) \hfill& \text{Distributive property}\end{array} Try It 13 If the amount P P P is deposited into an account paying simple interest r r r for time t t t, the total value of the deposit A A A is given by A=P+P r t A=P+Prt A=P+P r t. Simplify the expression. (This formula will be explored in more detail later in the course.) Solution Licenses & Attributions CC licensed content, Specific attribution College Algebra.Provided by:OpenStaxAuthored by:OpenStax College Algebra.Located at: BY: Attribution. 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4848	https://www.gauthmath.com/solution/1839009639061506/Range-of-y-4cos-x-pi-4	Solved: Range of y=4cos(-x+pi/4) [Math] Drag Image or Click Here to upload Command+to paste Upgrade Sign in Homework Homework Assignment Solver Assignment Calculator Calculator Resources Resources Blog Blog App App Gauth Unlimited answers Gauth AI Pro Start Free Trial Homework Helper Study Resources Math Trigonometry Questions Question Range of y=4cos(-x+pi/4) Gauth AI Solution 100%(1 rated) Answer The answer is [-4, 4] Explanation Analyze the properties of the cosine function The cosine function, $$\cos(x)$$cos(x), has a range of $$[-1, 1]$$[−1,1]. This means that the value of $$\cos(x)$$cos(x) is always between -1 and 1, inclusive. Consider the transformation of the cosine function The given function is $$y = 4\cos(-x + \frac{\pi}{4})$$y=4 cos(−x+4 π). The argument of the cosine function is $$-x + \frac{\pi}{4}$$−x+4 π. The coefficient 4 outside the cosine function affects the amplitude and thus the range. Determine the range of the transformed function Since the range of $$\cos(u)$$cos(u) is $$[-1, 1]$$[−1,1] for any $$u$$u, the range of $$4\cos(-x + \frac{\pi}{4})$$4 cos(−x+4 π) is $$4 \times [-1, 1] = [-4, 4]$$4×[−1,1]=[−4,4]. The term $$-x + \frac{\pi}{4}$$−x+4 π only affects the phase and period of the cosine function, not the range. Helpful Not Helpful Explain Simplify this solution Gauth AI Pro Back-to-School 3 Day Free Trial Limited offer! Enjoy unlimited answers for free. Join Gauth PLUS for $0 Previous questionNext question Related Sine and Identify the domain, range, a 1. y=4cos 3 θ _ Domain: _ Range: _ Period: Verified Answer How may different arrangements are there of the letters in The number of possible arrangements is MISSISSIPPI？ 100% (2 rated) Which equation represents a circle with center -5,-6 and radius of 5? x+52+y+62= square root of 5 x+52+y+62=25 x+52+y+62=5 x-52+y-62=25 75% (4 rated) What is the range of y=4cos x-1 ? 100% (19 rated) What is the range of y=4cos x-1 ? 97% (409 rated) The product of eight and seven when multiplied by F is less than the product of four and seven plus ten. a. 8+7F<4+7+10 b. 87F>47+10 C. 87F ≤ 47+10 d. 87F<47+10 100% (5 rated) Tell the range of the function y=4cos 2x+3 100% (6 rated) Which of the following lists only contains shapes that fall under the category of parallelogram? A square, rectangle, triangle B trapezoid, square, rectangle C quadrilateral, square, rectangle D rhombus, rectangle, square 100% (3 rated) What is the range of the function y=4cos x ? 98% (731 rated) This quiz: 10 points possibl Question 1 of 10 This question: 1 points pos: Determine the amplitude, range, period, and phase shift and then sketch the graph of the function using the quarter points. y=4cos -x+frac π 4 Which of the following functions is equivalent to y=4cos -x+frac π 4 ? A. y=-4cos x-frac π 4 B. y=4cos x-frac π 4 C. y=4cos x+frac π 4 D. y=-4cos x+frac π 4 The amplitude of y=4cos -x+frac π 4 is square Type an exact answer, using π as needed. Use integers or fractions for any numbers in the expression. The range of y=4 cos -x+frac π 4 is square . Type your answer in interval notation. Type an exact answer, using π as needed. Use integers or fractions for any numbers in the expression. The period of y=4 cos -x+frac π 4 is square . Simplify your answer. Type an exact answer, using π as needed. Use integers or fractions for any numbers in the expression. The phase shift of y=4cos -x+frac π 4 is square . Simplify your answer. Type an exact answer, using π as needed. Use integers or fractions for any numbers in the expression. Use the coordinates of the five quarter points of y= cos x to determine the corresponding quarter points on the graph of y=4cos -x+frac π 4. Simplify your answers. Type ordered pairs. Type exact answers, using π as needed. Use integers or fractions for any numbers in the expressions. Choose the correct graph of y=4c 05 -x+frac π 4 below. 100% (4 rated) Gauth it, Ace it! contact@gauthmath.com Company About UsExpertsWriting Examples Legal Honor CodePrivacy PolicyTerms of Service Download App
4849	https://web.mit.edu/deweck/www/PDF_archive/2%20Refereed%20Journal/2_12_SMO_AWSMOO1_deWeck_Kim.pdf	Submitted to Structural and Multidisciplinary Optimization for Review 1 Adaptive Weighted Sum Method for Multiobjective Optimization Il Yong Kim Queen’s University, Kingston, Ontario, K7L 3N6, Canada Olivier de Weck† Massachusetts Institute of Technology, Cambridge, Massachusetts, 02139, USA This paper presents an adaptive weighted sum method for multiobjective optimization problems. The authors developed the bi-objective adaptive weighted sum method, which determines uniformly-spaced Pareto optimal solutions, finds solutions on non-convex regions, and neglects non-Pareto optimal solutions. However, the method could solve only problems with two objective functions. In this work, the bi-objective adaptive weighted sum method is extended to problems with more than two objective functions. In the first phase, the usual weighted sum method is performed to approximate Pareto surfaces quickly, and a mesh of Pareto front patches is identified. Each Pareto front patch is then refined by imposing additional equality constraints that connect the pseudo nadir point and the expected Pareto optimal solutions on a piecewise planar surface in the objective space. It is demonstrated that the method produces a well-distributed Pareto front mesh for effective visualization and finds solutions in non-convex regions. Two numerical examples and a simple structural optimization problem are solved as case studies. Nomenclature J = objective function vector x = design vector p = vector of fixed parameters g = inequality constraint vector h = equality constraint vector m = number of objectives i α = th i weighting factor i J = normalized objective function Utopia J = utopia point Nadir J = nadir point i J = th i anchor point i J j P = position vector of the th j expected solution on the piecewise linearized plane I. Introduction Design optimization is to seek the best design that minimizes the objective function by changing design variables while satisfying design constraints. During design optimization one often needs to consider several design criteria or objective functions simultaneously. For example, we may want to maximize range and payload mass while trying to minimize lifecycle cost for an airplane design. When more than one design objective is associated, the design Assistant Professor, Department of Mechanical and Materials Engineering, McLaughlin Hall 305, Queen’s University. † Assistant Professor, Department of Aeronautics and Astronautics and Engineering Systems Division, Room 33-410, MIT, AIAA senior member. Submitted to Structural and Multidisciplinary Optimization for Review 2 problem becomes multiobjective, in which case the usual design optimization for a scalar objective function cannot be used. Multiobjective optimization can be stated as follows: ( ) 1,..., ) , , , min , s.t. , 0 , =0 (i n i LB i i UB x x x = ≤ ≤ ≤ J x p g(x p) h(x p) (1) where the objective function vector J is a function of design vector x and a fixed parameter vector p ; g and h are inequality and equality constraints; and , i LB x and , i UB x are the lower and upper bounds for the th i design variable, respectively. Stadler1,2 applied the notion of Pareto optimality to the fields of engineering and science in the 1970s. The most widely-used method for multiobjective optimization is the weighted sum method. The method transforms multiple objectives into an aggregated objective function by multiplying each objective function by a weighting factor and summing up all weighted objective functions: weighted sum 1 1 2 2 m m J w J w J w J = + + + ! (2) where ( 1, , ) i w i m = ! is a weighting factor for the th i objective function (potentially also dividing each objective by a scaling factor, i.e. i i i w sf α = ). If 1 1 m i i w = = ∑ and 0 1 i w ≤ ≤ , the weighted sum is said to be a convex combination of objectives. Each single objective optimization determines one particular optimal solution point on the Pareto front. The weighted sum method then changes weights systemically, and each different single objective optimization determines a different optimal solution. The solutions obtained approximate the Pareto front. Initial work on the weighted sum method can be found in Zadeh3. Koski4 applied the weighted sum method to structural optimization. Marglin5 developed the -constraint ε method, and Lin6 developed the equality constraint method. Heuristic methods are also used for multiobjective optimization: Suppapitnarm7 applied simulated annealing to multiobjective optimization, and multiobjective optimization by Genetic Algorithms can be found in Goldberg8, and Fonseca and Fleming9 among others. Das and Dennis10 developed the NBI (Normal Boundary Intersection) method, in which sub-optimizations are performed on normal lines to the utopia hyperplane that is defined and bounded by all anchor points. The NBI method produces well-distributed solutions, and it is easily scalable to -dimensional n problems. The method can also determine Pareto optimal solutions in non-convex regions, which the weighed sum method misses. The weak points of the method are (1) in highly nonlinear problems, it is hard to obtain optimal solutions due to equality constraints, (2) non-Pareto optimal solutions (dominated solutions) are also obtained, and a Pareto filtering must be used to filter out those solutions, and (3) in high dimensional problems (more than two objective functions), the projection of the utopia plane does not cover the entire Pareto front, and some Pareto front regions are not discovered by this method. Messac and Mattson11,12 used physical programming for generating Pareto fronts for concept selection. They also developed the normal constraint method13, which generates uniformly distributed solutions along the Pareto front without missing any Pareto front regions. The method can be extended to -dimensional n problems. The well-known drawbacks of the weighted sum method, as discussed in a number of studies11,14,15, are that (1) often the optimal solution distribution is not uniform, and that (2) more seriously, optimal solutions in non-convex regions are not detected. The adaptive weighted sum (AWS) method16 was developed recently by the authors to address these two drawbacks. By imposing additional inequality constraints in the usual weighted sum method, the optimization is performed only in a newly-defined feasible region where more exploration is needed. The adaptive weighted sum method successfully solves multiobjective optimization problems: the AWS method produces well-distributed solutions, finds Pareto optimal solutions in non-convex regions, and neglects non-Pareto optimal solutions. The AWS method, however, was previously only applicable to bi-objective optimization problems. Therefore, we will refer to the previous technique as the “bi-objective adaptive weighted sum method” to differentiate it from the generalized multiobjective adaptive weighted sum method presented here. Submitted to Structural and Multidisciplinary Optimization for Review 3 In this work, the bi-objective adaptive weighted sum method is extended to multiobjective optimization problems with more than two objective functions. Unfortunately, the additional inequality constraints that are used in the bi-objective adaptive weighted sum method are not suitable for higher-dimensional multiobjective optimization. The reason is that the Pareto-front patches (extension of segments in two dimensions) to be constructed by additional constraints can be arbitrarily-shaped hyperplanes in multidimensional problems, and it is difficult to define feasible regions for refinement by inequality constraints alone. In the multiobjective adaptive weighted sum method, additional equality constraints are introduced that connect the pseudo-nadir point and the expected locations of Pareto optimal solutions on the piecewise linearized plane in the objective space. Sub-optimizations for further refinement are conducted along these lines (equality constraints) determining solutions near desired positions, which leads to a well-distributed mesh representation of the Pareto. II. Multiobjective Adaptive Weighted Sum Method: Fundamental Concepts The fundamental philosophy of the adaptive weighted sum method is to adaptively refine the Pareto front. In the first stage, the method determines a rough profile of the Pareto front. By estimating the size of each Pareto patch (line segment in the case of two-dimensional problems), the regions for further refinement in the objective space are determined. In the subsequent stage, only these regions are specified as feasible domains for sub-optimization by assigning additional constraints. In the bi-objective adaptive weighted sum method, the feasible domain for further exploration is determined by specifying two inequality constraints. The usual weighted sum method is then performed as sub-optimization in these feasible domains obtaining more Pareto optimal solutions. When a new set of Pareto optimal solutions are determined, the Pareto patch size estimation is again performed to determine the regions for further refinement. These steps are repeated until a termination criterion is met. Figure 1 compares the typical weighted sum method and the bi-objective adaptive weighted sum method for a sample problem that has a relatively flat region and a non-convex region. We found that the inequality constraints as boundaries for constructing feasible regions are not suitable for optimization problems with more than two objective functions. Feasible regions for further refinement in the two-Figure 1. The concept and procedure of the adaptive weighted sum method. J1 Feasible region J2 1 δ 1 δ 2 δ 2 δ Feasible region J1 J2 1st region for more refinement 2nd region for more refinement J2 J1 (a) Usual weighted sum method J1 J2 True Pareto front Utopia point (b) Procedure of the bi-objective adaptive weighted sum method Submitted to Structural and Multidisciplinary Optimization for Review 4 dimensional case can be defined easily by laying two inequality constraints that are parallel to each of the axes with prescribed offset distances from the end points, because the Pareto front is a two-dimensional curve, and there are always only two end points for each Pareto-front segment. In higher-dimensional cases, however, the Pareto front becomes a surface (for three objective functions) or a hypersurface (for more than three objective functions), and it becomes difficult to impose constraints such that sub-optimizations are performed only in a selected Pareto-front patch and to adaptively refine the patches. This is because Pareto-front patches may have arbitrary shapes, and the number of edges for each Pareto-front patch may vary. In addition, when the number of vertices is larger than the dimension of the objective space, all vertices or their connecting edges may not lie in the same (hyper-) plane, and it becomes even more difficult to impose constraints for sub-optimization of further refinement and to perform adaptive refinement in the following stages. Indeed, the problems encountered then resemble adaptive remeshing in the Finite Element Method, but in higher dimensions. There has been very extensive research conducted in this field for decades, and it remains to be seen whether the sophisticated and sometimes complicated adaptive remeshing techniques of the FEM are applicable to Pareto front generation. It is also important to note that the FEM techniques can be applied only to three-dimensional problems. In this work, we adopt equality constraints to define feasible regions for further refinement, which is more robust for obtaining well-distributed solutions in multidimensional problems than inequality constraints. Adding equality constraints increases the likelihood of entrapment in local minima, but also facilitates the adaptive procedure by simplifying patch refinement. Although Pareto front patches of any shape can be used, we demonstrate the method with quadrilateral patches with applications to three-dimensional problems in this paper. In the first stage, the approximate shape of the Pareto front is determined by using the usual weighted sum method. Pareto front patches are then identified, and patches for further refinement are selected on the basis of the patch size. Sub-optimization is performed only in the selected patches by specifying additional equality constraints. Figure 2 shows the concept of the multiobjective adaptive weighted sum method with equality constraints for multiobjective optimization. In the bi-objective adaptive weighted sum method, feasible regions for further refinement are defined by two inequality constraints (Fig. 1), but in the multiobjective adaptive weighted sum method, one or several equality constraints are specified that connect the pseudo nadir point and expected solutions on the piecewise linearized Pareto front. Actual solutions obtained will be on the equality constraint line, but they may be located in different positions from the expected solutions as shown in the figure. Figure 2. Adaptive weighted sum method for multidimensional problems (2-D representation). Expected solution (Pseudo) Nadir Point Actual solution J1 J2 Piecewise linearized Pareto front Expected solution (Pseudo) Nadir Point Actual solution J1 J2 Piecewise linearized Pareto front Figure 3. Adaptive weighted sum method for multidimensional problems (3-D representation). Expected solution Actual solution Pseudo Nadir Point J2 J1 J3 Utopia Point 1 N 2 N 4 N 3 N Utopia J Nadir J Piecewise linearized Pareto front Expected solution Actual solution Pseudo Nadir Point J2 J1 J3 J2 J1 J3 Utopia Point 1 N 2 N 4 N 3 N Utopia J Nadir J Piecewise linearized Pareto front Submitted to Structural and Multidisciplinary Optimization for Review 5 The equality constraint, which is represented by a line, allows us to have great control over the position of new Pareto optimal solutions obtained, and this simplifies adaptive refinement. In the three-dimensional case, the Pareto front becomes a surface, and the linearized Pareto front patch is represented by four line segments that connect four vertices, as shown in Fig. 3. When solutions are obtained in all patches for further refinement, a new set of Pareto front patches are identified, and a patch-size evaluation is performed to determine where to further refine. These steps are repeated until a termination criterion is met. The complete and detailed procedure is presented in the following section. III. Multiobjective Adaptive Weighted Sum Method: Procedures In this section, we detail the procedure for implementing the multiobjective adaptive weighted sum method. [Step 1] Stage = 1. Normalize the objective functions. When i x is the optimal solution vector for the single objective optimization of the th i objective function i J , the utopia point U J is defined as 1 2 1 2 [ ( ) ( ) ( )] Utopia m m J J J = J x x x ! , (3) and the pseudo nadir point Nadir J is defined as 1 2 [ ] Nadir Nadir Nadir Nadir m J J J = J ! (4) where m is the number of objective functions or the dimension of the objective space, and each component Nadir i J is determined by 1 2 max[ ( ) ( ) ( )] Nadir m i i i i J J J J = x x x ! . (5) The th i anchor point i J is defined as 1 2 [ ( ) ( ) ( )] i i i i m J J J = J x x x ! . (6) Now the normalized objective function i J is obtained as Utopia i i i Nadir Utopia i i J J J J J − = − . (7) [Step 2] Perform multiobjective optimization using the usual weighted sum approach with a small number of divisions, initial n . For three objective functions, the weighted single objective function Total J is obtained as ( ) ( ) ( ) ( ) 2 1 1 1 2 2 3 1 2 1 1 2 2 2 3 1 1 1 1 , [0,1] Total i J J J J J J J α α α α α α α α α α = + − + −     = + − + − ∈ (8) where i α is the th i weighting factor. As a general form, the weighted single objective function of m objective functions, m Total J , is determined by ( ) 1 1 1 1 , 2 m m Total m Total m m J J J m α α − − − = + − ≥ (9) where 1 1 Total J J ≡ .Note that m-1 weighting factors are needed to explore an m-dimensional objective space. Submitted to Structural and Multidisciplinary Optimization for Review 6 The uniform step size of the th i weighting factor i α is determined by the number of initial divisions along the th i objective dimension: initial, i 1 , 1, , 1 i i m n α ∆ = = − ! (10) where m is the number of objective functions. In this work, we use the same step size for all weighting factors. There is a scheme that systemically determines each weighting factor and helps produce well-distributed solutions10. In the adaptive weighted sum method, however, the usual step size strategy Eq. (10) can be used because this approximate multiobjective optimization is conducted only once, after which adaptive refinement is conducted. [Step 3] Delete nearly overlapping solutions. It occurs often that several nearly identical solutions are obtained when the weighted sum method is used. The Euclidian distances between these solutions are nearly zero, and among these, only one solution is needed to represent the Pareto front. In the computer implementation, if the distance among solutions in the objective space is less than a predetermined distance (ε ), then all solutions except one are deleted. [Step 4] Identify Pareto-front patches. Patches of any shape can be used, but in this work we use quadrilateral patches in three-dimensional problems. Four Pareto-optimal solutions become the four nodes of each patch, and edges are line segments that connect two neighboring nodes of each patch. Constructing and maintaining meshes on the Pareto front may be tedious, but there are two advantages of using a mesh, which will be discussed in detail in the following sections: (1) patches play the role of primitives for further refinement for subsequent stages, as will be seen in Step 5, and (2) if only non-dominated solution points are displayed, it is difficult to visualize and interpret the shape of the Pareto front. A mesh representation makes it very easy to visualize the Pareto surface as in the case of finite element meshes. [Step 5] Stage = Stage + 1. Determine the layout for further refinements in each of the Pareto-front patches. The larger the patch is, the more it needs to be refined. Figure 4 shows an example of refinement, in which a patch is composed of four nodes in three dimensional objective space, as will be studied in this paper. Because the lower patch is larger, it is refined more than the upper one. In each mesh, the locations of expected solutions are determined by interpolation, and sub-optimizations are conducted along the lines that connect the pseudo nadir point and the expected solutions. The actual solutions may be different from expected solutions, and there can be dominated solutions, which must be deleted by a Pareto filter. The position vector of the th j expected solution on the piecewise linearized plane ( j P ) is obtained as the weighed sum of the four vectors of the nodal solutions as 3 1 2 4 1 2 3 4 , [0,1] j i β β β β β = + + + ∈ P N N N N (11) where i N is the position vector the th i node of a Pareto-front patch (Fig. 4(b)), and i β is a weighting factor for interpolation. (a) Original patches (b) Refined patches (Expected solutions) (c) Refined patches (Actual solutions) Figure 4. Adaptive refinement procedure. j P 1 N 2 N 4 N 3 N 1 N 2 N 4 N 3 N j P 1 N 2 N 4 N 3 N 1 N 2 N 4 N 3 N Dominated solutions Submitted to Structural and Multidisciplinary Optimization for Review 7 The normalized vector of j P is obtained as j Utopia j i i i Nadir Utopia i i P J P J J − = − (12) where j i P is the th i coordinate of the th j expected solution on the piecewise linearized (hyper-)plane. The refinement level, which is represented by the step size of weighting factors, is determined based on the relative average length of the patch in each direction. [Step 6] Impose an additional equality constraint for each expected solution and conduct a sub-optimization with the weighted sum method. For the th j normalized expected solution, j P , the sub-optimization problem is defined as minimize ( ) ( ) ( ( ) ) subject to 1 ( ) ( ) 0 ( ) 0 i j Nadir Nadir j Nadir Nadir ⋅ − ⋅ − = − − = ≤ w J x P J J x J P J J x J h x g x (13) where ( ) j Nadir i = − − w P J is a vector of weighting factors, and ( ) h x and ( ) g x are normalized equality and inequality constraint vectors. Note that the normalized nadir point Nadir J is a vector whose components are one, i.e. (1,1, ,1) Nadir = J ! . Utopia Point 1 J 2 J 3 J (0, 0, 0) Utopia = J Expected solution Actual solution Pseudo Nadir Point (1,1,1) Nadir = J j P j Nadir − P J j J Utopia Point 1 J 2 J 3 J (0, 0, 0) Utopia = J Expected solution Actual solution Pseudo Nadir Point (1,1,1) Nadir = J j P j Nadir − P J j J Figure 5. Configuration of an additional equality constraint for refinement (3-D representation). Submitted to Structural and Multidisciplinary Optimization for Review 8 The equality constraint ( ) ( ) ( ) ( ( ) ) / ( ) 1 j Nadir Nadir j Nadir Nadir − ⋅ − − − = P J J x J P J J x J makes the two vectors and ( ) j Nadir Nadir − − P J J x J be collinear in the objective space. This constraint therefore ensures that the solution is obtained only along the line j Nadir − P J , which connects the expected solution on the piecewise linearized plane and the pseudo nadir point. The objective function ( ) ( ) j Nadir − − ⋅ P J J x is a scalar function to be minimized determining the solution that is nearest to the utopia point in the direction of ( ) j Nadir − − P J . The actual solution obtained for the th j normalized expected solution, j P , would be different from the expected solution. In Figure 5, the origin of the vector j Nadir − P J is actually (0,0, 0) but moved for better visualization. [Step 7] Perform Pareto filtering. In the bi-objective adaptive weighted sum method, non-Pareto optimal solutions are automatically rejected, so the filtering is not needed. In the multiobjective adaptive weighted sum method, however, any solution that lies on the equality constraint is feasible, and non-Pareto optimal solutions may be obtained. In each step, it is necessary to perform Pareto filtering to obtain the true Pareto front. [Step 8] Delete overlapping solutions. Identify Pareto-front patches with all Pareto optimal solutions including newly obtained solutions in the previous steps. If a termination criterion is met, stop; otherwise go to Step 5. Several types of termination criteria may be used: (1) the number of stages reaches a prescribed number; (2) the size of largest Pareto-front patch falls below a prescribed value; (3) the standard deviation among the sizes of all Pareto-front patches falls below a prescribed value. In this work, the maximum number of stages is used as the termination criterion. IV. Numerical Examples Three numerical examples are presented in this section to demonstrate the performance of the multiobjective adaptive weighted sum method. All examples are three-dimensional problems. Sequential Quadratic Programming (SQP) in MATLAB is used for every optimization. A. Example 1: Convex Pareto Front The first example is a multiobjective maximization problem whose Pareto front is convex. The problem statement is 1 2 3 4 3 2 1 2 3 1 1 2 2 3 3 maximize [ ] subject to 2 5 1 0 ( 1,2,3). T i J J J x x x J x J x J x x i + + ≤ = = = ≥ = (14) The Pareto front of this problem is convex, but the curvatures are different in three axes. Figure 6 shows the Pareto front obtained by the usual weighted sum method. The step sizes for the two weighting factors in Eq. (8), 1 2 and α α ∆ ∆ , are 1/9. The number of Pareto optimal solutions on the front is 100 with ten solutions coincident on Figure 6. Pareto front obtained by the usual weighted sum method for Example 1. 1 J 2 J 3 J 2 J 3 J 1 J 1 J 2 J 3 J 2 J 3 J 1 J Submitted to Structural and Multidisciplinary Optimization for Review 9 the top vertex. These solutions represent a 9 9 × Pareto-front mesh. The patch size varies greatly according to the position: the meshes on the top are slender, and the meshes near to the two anchor points 1 J and 3 J on the bottom are relatively large and slender while the patches in the middle region of the front are small and nearly square. Figure 7 shows the three stages of the Pareto front evolution produced by the multiobjective adaptive weighted sum method. In the first stage, the usual three-dimensional weighted sum method with 1 2 0.5 α α ∆ = ∆ = is performed obtaining a coarse 2 2 × Pareto-front mesh. Based on this initial Pareto front, the relative size of each Pareto-front patch is estimated, and adaptive refinement is performed. Note that the Pareto front is not symmetric in any direction, and this is the reason that mesh refinement in the second stage is asymmetric. The number of Pareto front patches is 72 in the third stage. Contrary to the Pareto front representation by the usual weighted sum method in Fig. 6, the mesh shape and size are quite uniform. In this work, the goal of adaptive refinement is to have not only uniformly distributed solutions but also solutions that form a well-shaped mesh layout. Multiobjective optimization is conducted in order to present trade-off information to engineers such that best decisions can be made. In the two-dimensional case, it is not difficult to interpret a Pareto front that is represented only by solution points. In higher dimensions, however, the point-based representation is often hard to interpret. By maintaining meshes as in this work, further adaptive refinement considering the mesh size can be performed systematically, and it is very easy to visualize the Pareto front obtained – at least in three objective dimensions at-a-time. B. Example 2: Non-convex Pareto Front with Dominated Solutions In the previous example, the Pareto front was convex, and the problem associated with the usual weighted sum method was only that we could not obtain evenly-distributed solutions, or the mesh layout was not uniform. In this example, we solve a multiobjective problem whose Pareto front has non-convex regions and is disconnected due to dominated solutions. The problem statement is 2 1 2 3 1 3 1 1 2 2 3 3 1 2 3 maximize [ ] subject to cos 0 0 0 1.2 . T x J J J x e x J x J x J x x x x π − − − + ≤ = = = ≤ ≤ ≥ ≥ (15) Figure 8 shows the Pareto front of this problem including the dominated solution region from two different viewpoints, which was generated with MATLAB executing a full factorial evaluation (which would not be feasible for higher dimensional design spaces). The boundary of the Pareto front is composed of three curves: the curve between 1 J and 3 J is convex with a gap due to a dominated solution region, but the other two curves are not 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J Figure 7. Pareto front obtained by the multiobjective adaptive weighted sum method for Example 1. (a) Stage 1 (b) Stage 2 (c) Stage 3 Submitted to Structural and Multidisciplinary Optimization for Review 10 convex. There is a dominated region in the middle, which looks like a valley. The usual weighted sum method is used to determine the Pareto front with a 20 20 × mesh. As appears in Fig. 9, solutions are obtained only on the convex curve between 1 J and 3 J and on the short convex curve segment near 2 J . Dominated solutions are not obtained, however most of the Pareto front, which is non-convex, is not revealed by this method. The multiobjective adaptive weighted sum method is performed to find the Pareto front adaptively. In the first stage, the usual weighted sum method with 1 2 1 α α ∆ = ∆ = determines the three anchor points forming an approximate Pareto front (Fig. 10). In the second stage, the overall shape of the non-convex Pareto front is found; a result which cannot be obtained no matter how many solutions are found by the usual weighted sum method. Because equality constraints are used, dominated solutions (non-Pareto optimal solutions) are obtained by the multiobjective adaptive weighted sum method. A Pareto filtering step is conducted in each stage. In the final stage, the Pareto front in its “entirety” is determined. We can clearly see a dominated region in the middle, and the mesh representation makes it easy to interpret the surface. 1 J 2 J 3 J 2 J 3 J 1 J 1 J 2 J 3 J 1 J 2 J 3 J 2 J 3 J 1 J 1 J 2 J 3 J 2 J 3 J 1 J 1 J 2 J 3 J 1 J 2 J 3 J Figure 8. Pareto front profile (including the dominated region) for Example 2. Figure 9. Pareto front obtained by the usual weighted sum method for Example 2. 1 J 2 J 3 J 2 J 3 J 1 J 1 J 2 J 3 J 2 J 3 J 1 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J 1 J 2 J 3 J Figure 10. Pareto front obtained by the multiobjective adaptive weighted sum method for Example 2. (a) Stage 1 (b) Stage 2 (c) Stage 3 (d) Stage 4 (e) Stage 5 Submitted to Structural and Multidisciplinary Optimization for Review 11 C. Example 3: Three-bar Problem Finally, the multiobjective adaptive weighted sum method is applied to the three-bar problem15. The geometry and material properties are presented in Fig. 11. The upper end of each bar is fixed, and a horizontal and vertical load are applied at Point P. The objective functions to be minimized are the total volume, stress in Truss 1, and stress in Truss 3. The mathematical problem statement is 1 3 minimize [Volume( ) ( ) ( )] subject to 200 MPa ( ) 200 MPa ( 1,2,3) 0.1cm 2 cm ( 1,2,3) T i i i A i σ σ σ − ≤ ≤ = ≤ ≤ = A A A A (16) where the design variable i A is the cross-sectional area of the th i truss. The Pareto front of this problem is non-convex, and some part of the objective space is dominated. Figure 12 shows the results obtained with the multiobjective adaptive weighted sum method. For better visualization, the graph is inverted, i.e. all three objective functions are multiplied by minus one (it is often easier to interpret the Pareto surface when we view it from outside the feasible range rather than from the inside). The pseudo nadir point is the origin where three reference planes in the figure meet. In the first stage, the approximate Pareto front is represented by 6 patches. The multiobjective adaptive weighted sum method is then applied determining more refined Pareto front meshes until convergence is reached in three stages. L L 3 L F F 1 δ 2 δ 1 A 2 A 3 A P Design variables: , 1,2,3 i A i = :displancement of the point in the th direction i P i δ upper limit lower limit 2 2 upper limit lower limit 20 , 100 , 200 200 , 200 2 , 0.1 F kN L cm E GPa MPa MPa A cm A cm σ σ = = = = = − = = L L 3 L F F 1 δ 2 δ 1 δ 2 δ 1 A 2 A 3 A P Design variables: , 1,2,3 i A i = :displancement of the point in the th direction i P i δ upper limit lower limit 2 2 upper limit lower limit 20 , 100 , 200 200 , 200 2 , 0.1 F kN L cm E GPa MPa MPa A cm A cm σ σ = = = = = − = = Pseudo Nadir Point Stress 1 Stress 3 Volume Pseudo Nadir Point Stress 1 Stress 3 Volume Stress 1 Stress 3 Volume Stress 1 Stress 3 Volume Stress 1 Stress 3 Volume Stress 1 Stress 3 Volume Figure 12. Pareto front obtained by the multiobjective adaptive weighted sum method for Example 3. (a) Stage 1 (b) Stage 2 (c) Stage 3 Stress 1 Stress 3 Volume Stress 1 Stress 3 Volume Stress 1 Stress 3 Volume Figure 11. Three-bar problem. Submitted to Structural and Multidisciplinary Optimization for Review 12 V. Discussions The multiobjective adaptive weighted sum method effectively solves multiobjective optimization problems with more than two objective functions. In the bi-objective adaptive weighted sum method, which is applicable to only optimizations with two objective functions, inequality constraints are used to specify regions for further refinement. In the multiobjective adaptive weighted sum method, on the other hand, equality constraints are used and the method is scalable to -dimensional n problems. The equality constraints allow us to decide where to obtain additional solutions, and this makes the Pareto front mesh well conditioned. There are two important issues in using the multiobjective adaptive weighted sum method. First, adaptive refinement is conducted only within the Pareto front that is determined by the usual weighted sum method in the first stage. If the first stage optimization misses some regions on the Pareto front, they are not discovered in the following adaptive refinement stages. In general, the usual weighted sum method finds most parts of the Pareto front reliably and quickly. This is, in addition to its adaptivity, a distinctive advantage of the adaptive weighted sum method, which utilizes the usual weighted sum method in the first stage, over the NBI method. Figure 13 shows a Pareto surface that is obtained by the adaptive weighted sum method (based on the first stage usual weighted sum method) and the utopia plane for the NBI method. The view-direction is rotated such that it is normal to the utopia plane. As can be seen in the figure, the NBI method cannot determine the three regions that are not covered by the normal projection of the utopia plane. The main difference between AWS and NBI, aside from adaptive refinement, is the fact that equality constraints in AWS are imposed radially from the pseudo-Nadir point rather than normal to the utopia plane defined by the anchor points as is done in NBI. Second, equality constraints are generally difficult to be satisfied. Optimizers are often good at finding a (local) optimal solution if it starts from within a feasible region, but optimization may fail to find a feasible solution if the initial design is far from the feasible region. When a line constraint (equality constraint) is specified in the objective space, we should make sure the initial design lies on or near the line. It is often difficult to find such an initial design, however, and we have to try many initial designs, which is computationally expensive. A convex Pareto front example, non-convex Pareto front with dominated regions, and a three-bar problem are solved successfully by the multiobjective adaptive weighted sum method. The advantages over the usual weighted sum method – uniform distribution and the ability to determine non-convex Pareto front – are presented by the examples. Adaptivity and the ability to find the entire Pareto front are the merits of this method that the NBI method does not have. This method will be applied to problems with practical applications and complicated Pareto fronts as further work. This includes benchmarking AWS against NBI and other methods in terms of uniformity, completeness and computational effort for Pareto front generation. One typical case in which the usual weighted sum method fails to discover all parts of the Pareto front is when the anchor points are not unique (in the two-dimensional case, an anchor solution may be a line segment, and in the three-dimensional case, it can be a plane). The usual weighted sum method for the first stage will also be improved such that problems with non-unique anchor solutions can be treated. Furthermore, visualization challenges for patch representations in cases with more than three objectives (m>3) will be investigated. References 1 Stadler, W., “A Survey of Multicriteria Optimization, or the Vector Maximum Problem,” Journal of Optimization Theory and Applications, Vol. 29, 1979, pp. 1-52. 2 Stadler, W., “Applications of Multicriteria Optimization in Engineering and the Sciences (A Survey),” Multiple Criteria Decision Making – Past Decade and Future Trends, edited by M. Zeleny, JAI Press, Greenwich, Connecticut, 1984. 3 Zadeh, L., “Optimality and Non-Scalar-Valued Performance Criteria,” IEEE Transactions on Automatic Control, AC-8, 59, 1963. Utopia plane for the NBI Utopia plane for the NBI Figure 13. Comparison of the AWS method and the NBI method. Submitted to Structural and Multidisciplinary Optimization for Review 13 4 Koski, J., “Multicriteria truss optimization,” Multicriteria Optimization in Engineering and in the Sciences, edited by Stadler, New York, Plenum Press, 1988. 5 Marglin, S. Public Investment Criteria, MIT Press, Cambridge, Massachusetts, 1967. 6 Lin, J., “Multiple objective problems: Pareto-optimal solutions by method of proper equality constraints,” IEEE Transactions on Automatic Control, Vol. 21, 1976, pp. 641-650. 7 Suppapitnarm, A., Seffen, K. A., Parks, G. T., and Clarkson, P. J., “A simulated annealing algorithm for multiobjective optimization,” Engineering Optimization, Vol. 33, No. 1, 2000. pp. 59-85. 8 Goldberg, D. E., Genetic Algorithms in Search, Optimization and Machine Learning, Addison Wesley, 1989. 9 Fonseca, C., and Fleming, P., “An overview of evolutionary algorithms in multiobjective optimization,” Evolutionary Computation, Vol. 3, 1995, pp. 1-18. 10 Das, I., and Dennis, J. E., “Normal-Boundary Intersection: A New Method for Generating Pareto Optimal Points in Multicriteria Optimization Problems,” SIAM Journal on Optimization, Vol. 8, No. 3, 1998, pp. 631-657. 11 Messac, A., and Mattson, C. A., “Generating Well-Distributed Sets of Pareto Points for Engineering Design using Physical Programming,” Optimization and Engineering, Kluwer Publishers, Vol. 3, 2002, pp. 431-450. 12 Mattson, C. A., and Messac, A., “Concept Selection Using s-Pareto Frontiers,” AIAA Journal. Vol. 41, 2003, pp. 1190-1204. 13 Messac, A. and Mattson, C. A., “Normal Constraint Method with Guarantee of Even Representation of Complete Pareto Frontier,” AIAA Journal, Vol. 42, 2004. 14 Das, I., and Dennis, J. E., “A closer look at drawbacks of minimizing weighted sums of objectives for Pareto set generation in multicriteria optimization problems,” Structural Optimization, Vol. 14, 1997, pp. 63-69. 15 Koski, J., “Defectiveness of weighting method in multicriterion optimization of structures,” Communications in Applied Numerical Methods, Vol. 1, 1985, pp. 333-337. 16 Kim, I.Y., and de Weck, O., “Adaptive weighted-sum method for bi-objective optimization: Pareto front generation,” Structural and Multidisciplinary Optimization, 2004 (in press).
4850	https://medicoapps.org/m-derivatives-of-mesoderm/	Derivatives Of Mesoderm - New Skip to content Home Daily Updates Store Free PDF DownloadsFree PDF Downloads Log in / Signup Main Menu Home Daily Updates Store Log in / Signup Derivatives Of Mesoderm Leave a Comment / Module / By Renu Maurya Derivatives Of Mesoderm DERIVATIVES OF MESODERM Mesoderm is divided into three parts:- 1) Paraxial mesoderm 2) Intermediate mesoderm 3) Lateral plate mesoderm 1. PARAXIAL MESODERM: It is organised intosomiteswhich in turn gives rise to:- Sclerotomes:Forms axial skeleton includingvertebrae, ribsand parts of neurocranium. Myotomes:Form allvoluntary (skeletal) muscles of head, trunk and limbs. Dermatomes:Form dermis of skin, especially over dorsal regions. 2. INTERMEDIATE MESODERM: It gives rise to major portion ofurogenital system:- Urinary organs:Kidney,ureters,trigone of bladder,posterior wall female urethra, posterior wall of upper half of prostatic part of male urethra, inner glandular zone of prostate. Reproductive organs:Gonads(testis and ovary),epididymis,ducts deferens, seminal vesicle, ejaculatory duct,uterus,uterine tube and upper part of vagina. 3. LATERAL PLATE MESODERM: It forms:- i) Somatopleuric mesoderm (parietal layer) All connective tissues including specialised connective tissue likebone,cartilage, adipose tissue. Dermis of skin over ventrolateral body wall and limbs. Superficial and deep fascia. Ligaments, tendons, aponeurosis. Parietal pleura, parietal peritoneum and tunica vaginalis of testis. Dura-mater. Lid muscles, extraocular muscles. Sclera, choroid,vitreous. Corneal stroma, iris and ciliary body (except epithelium). ii) Splanchnopleuric mesoderm (visceral layer) Smooth muscle and connective tissue of respiratory tract, gut, blood vessels and heart. Adrenal cortex. Mesothelium (visceral layer) of pleural, peritoneal and pericardeal cavities. Mesenchyme surrounding pericardial coelom gives rise to myocardium and serous pericardium. Spleen and lymph nodes. iii) Spetum transversum Diaphragm Fibrous pericardium iv) Angiogenic mesoderm Endocardium of heart Endothelium of blood and lymphatic vessels. Microglia, tissue macrophage. Circulating blood vessels. Exam Question DERIVATIVES OF MESODERM Mesoderm is divided into three parts:- 1) Paraxial mesoderm 2) Intermediate mesoderm 3) Lateral plate mesoderm 1. PARAXIAL MESODERM: It is organised into somites which in turn gives rise to:- Sclerotomes:Forms axial skeleton including vertebrae, ribs and parts of neurocranium. Myotomes:Form all voluntary (skeletal) muscles of head, trunk and limbs. Dermatomes:Form dermis of skin, especially over dorsal regions. 2. INTERMEDIATE MESODERM: It gives rise to major portion of urogenital system:- Urinary organs:Kidney,ureters,trigone of bladder,posterior wall female urethra, posterior wall of upper half of prostatic part of male urethra, inner glandular zone of prostate. Reproductive organs:Gonads(testis and ovary),epididymis,ducts deferens, seminal vesicle, ejaculatory duct,uterus,uterine tube and upper part of vagina. 3. LATERAL PLATE MESODERM: It forms:- i) Somatopleuric mesoderm (parietal layer) All connective tissues including specialised connective tissue like bone,cartilage, adipose tissue. Dermis of skin over ventrolateral body wall and limbs. Superficial and deep fascia. Ligaments, tendons, aponeurosis. Parietal pleura, parietal peritoneum and tunica vaginalis of testis. Dura-mater. Lid muscles, extraocular muscles. Sclera, choroid,vitreous. Corneal stroma, iris and ciliary body (except epithelium). ii) Splanchnopleuric mesoderm (visceral layer) Smooth muscle and connective tissue of respiratory tract, gut, blood vessels and heart. Adrenal cortex. Mesothelium (visceral layer) of pleural, peritoneal and pericardeal cavities. Mesenchyme surrounding pericardial coelom gives rise to myocardium and serous pericardium. Spleen and lymph nodes. iii) Spetum transversum Diaphragm Fibrous pericardium iv) Angiogenic mesoderm Endocardium of heart Endothelium of blood and lymphatic vessels. Microglia, tissue macrophage. Circulating blood vessels. Don’t Forget to Solve all the previous Year Question asked on Derivatives Of Mesoderm Click Here to Start Quiz Share this: Click to share on X (Opens in new window)X Click to share on Facebook (Opens in new window)Facebook Like this: Like Loading... 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4851	https://brightchamps.com/en-us/math/numbers/representation-of-real-numbers-on-number-line	Our Programs Learn More About us Login Table Of Contents What are Real Numbers? What is the Real Number Line? How to Represent Real Numbers on a Number Line? How to Represent the Ordering of Real Numbers on a Number Line? Absolute Value of a Real Number on a Number Line Common Mistakes and How to Avoid Them in Representation of Real Numbers on Number Line Real-life Applications of Representation of Real Numbers on Number Line Solved Examples of Representation of Real Numbers on Number Line FAQs of Representation of Real Numbers on Number Line Explore More numbers 128 Learners Last updated on July 22nd, 2025 Representation of Real Numbers on Number Line Real numbers include natural numbers, whole numbers, as well as rational and irrational numbers. On a straight number line, each integer is placed at equal intervals. The number line extends infinitely in both directions. To organize and compare numbers, we can use a number line. In this article, we will learn about the representation of real numbers on the number line in detail. Representation of Real Numbers on Number Line for US Students What are Real Numbers? The set of rational numbers Q and the set of irrational numbers Q’ together make up the set of real numbers. It is denoted as R. Subsets of whole numbers, natural numbers, integers, and rational and irrational numbers are all considered real numbers. Therefore, R = Q ∪ Q’ Struggling with Math? Get 1:1 Coaching to Boost Grades Fast ! What is the Real Number Line? On a number line, each number has a unique point known as a coordinate and has a distinct position. For instance, the real number 3 is positioned between 2 and 4. On a number line, two numbers cannot share the same position. The origin of a number line is at 0. On the right side of the origin are the positive numbers, while on the left side are the negative numbers. The visual representation of a real number line is: How to Represent Real Numbers on a Number Line? Following the given steps will help us to easily indicate real numbers on a number line using graphs and coordinates. Step 1: Draw a straight line, mark the origin at 0, and draw arrows on both sides of the origin point. Step 2: Use a fixed scale for marking real numbers. Place real numbers on both sides of the origin at equal intervals. Step 3: Mark positive numbers to the right of the origin, and negative numbers to the left. Step 4: By identifying the correct positions of natural numbers, whole numbers, and integers, we can easily place them on a number line. If we have a large number, such as 100, then we can use a larger scale, for example, marking each unit 1 as unit 20. Through this, we can reach 100 after taking 5 steps. Step 5: First, convert the rational or irrational numbers to a decimal form, to mark them on a number line. Now we can mark real numbers such as -7/2, -2, 0, 1/2, and 3 on a number line. How to Represent the Ordering of Real Numbers on a Number Line? We use the number line to compare and arrange real numbers. Symbols such as greater than (>), less than (<), and equal to (=) are used to compare numbers. Step 1: On a number line, the larger numbers are placed to the right and the smaller numbers are positioned to the left. Step 2: On the number line, negative numbers are always on the left side of the zero. The negative numbers closer to zero are considered greater. For example, -3 is greater than -13 because -3 is closer to the origin point than -13. Take a look at the given image to understand the comparison of real numbers on a number line. Here -3 > -13. Absolute Value of a Real Number on a Number Line The absolute value is the distance between a real number and the origin of the number line. It is denoted \|x\|, where x is the real number. Since distance is always positive, the absolute value will also be positive. For instance, 3 is a real number, then the absolute value will be \|3\| = 3. Take a look at the given image: Here, the real number (3) is 3 units away from the origin. The distance between a negative number and the origin point is the same as its equivalent positive number. For example, \|-3\| = 3, here also -3 is 3 units far from the origin. Common Mistakes and How to Avoid Them in Representation of Real Numbers on Number Line When students represent real numbers on a number line, they sometimes make some mistakes that lead to incorrect comparisons and arrangement of numbers. Here are some common mistakes and helpful solutions to avoid errors when representing real numbers on a number line. Mistake 1 Forgetting the Origin as a Reference Students should use the origin (0) as a reference when they mark real numbers on the number line. If they ignore the origin, they will mark real numbers at inconsistent intervals. They should place positive numbers to the right, while negative numbers to the left side of the origin. Mistake 2 Ignoring the Equal Distance Between Numbers Always remember to keep an equal interval between numbers when drawing a number line. They can use a ruler or a fixed scale to keep an equal spacing between numbers. If they mistakenly mark 3 closer to 4, the number line will be confusing and lead to incorrect conclusions. Mistake 3 Placing Negative Numbers on the Right Side of the Origin Remember to place negative numbers to the left of the origin (0). Only positive numbers should be marked to the right of the number line. Keep in mind that greater negative numbers are closer to the origin. For example, -2 is greater than -4 because it is closer to the origin (0). Mistake 4 Confusing Positive and Negative Decimals Students should remember that positive decimals must be placed between positive numbers and negative decimals marked between negative numbers. For example, it is incorrect when kids mistakenly mark -0.5 between 0.5 and 1. It should be placed on the left side of the origin, between -0.6 and -1 since -0.5 is a negative number. Mistake 5 Forgetting to Convert Fractions into Decimals Keep in mind to convert fractions to decimal form when marking fractions on a number line. By dividing the numerator by the denominator, students can change the fractions to decimals. Place the decimal number carefully on the number line between the whole numbers. For example, the fraction 1/2 can be converted to 0.5 and carefully marked between 0 and 1 on the number line. Level Up with a Math Certification! 2X Faster Learning (Grades 1-12) Real-life Applications of Representation of Real Numbers on Number Line The real number line plays an important role in various real-life situations by helping to compare and arrange numbers in order. The real-life applications of the real number line are as follows: Weather forecasters and geologists use the real number line to represent the temperatures and their variations. For example, the temperature in Los Angeles is 17℃, and in New York it is -7℃. By using a number line, they can easily compare temperatures and find the temperature differences between these two cities. In sports, scores are labeled on a number line, the positive scores of a team are marked on the right side, and the negative scores are on the left. For example, in a football match, team A scored + 3, which means they achieved 3 goals, whereas team B lost 2 points, bringing their score to -2. The savings and debts of an account holder can be represented using a number line. It helps the bank to understand the person’s balance and loans accurately. For example, if a person has $300 in his account and withdraws $500, now he has a balance of -$200. When cooking, we can use a real number line to mark the quantity of items used for a recipe. For instance, to bake a cake we need 2/5 cups of flour and 3/4 cups of milk. These fractions can be easily marked on a number line which helps visualize and measure ingredients easily. Hey! Solved Examples of Representation of Real Numbers on Number Line Problem 1 Represent -3.5 on the number line. Okay, lets begin Explanation To begin, draw a straight line and mark 0 as its origin. Then find -3 on the number line. Thus, -3.5 will fall exactly in the middle of -3 and -4. Divide the space between -3 and -4 into 10 equal parts and mark the given number on the number line. Well explained 👍 Problem 2 Find and plot a rational number between 1/10 and 3/8. Okay, lets begin Explanation We can convert 1/10 and 3/8 to a common denominator. The common denominator of the unlike fractions 1/10 and 3/8 is 40. 1/10 = 1 × 4 / 10 × 4 = 4/40 Thus, 1/10 = 4/40. Next, we can convert 3/8 to a denominator of 40. 3/8 = 3 × 5 / 8 × 5 = 15/40 Thus, 3/8 = 15/40. Now, find the midpoint: (4/40 + 15/40) ÷ 2 = 19/40 ÷ 2 = 19/80. So, 19/80 is a rational number between 1/10 and 3/8, and plot the number on the number line. Well explained 👍 Problem 3 Represent 5/6 on the number line. Okay, lets begin Explanation To begin, we can convert the fraction into its decimal form. 5 ÷ 6 = 0.8333... 0.8333… is a non-terminating decimal, so we can round it to 0.83, which is an approximate value. This means, 5/6 is slightly greater than 0.8 and lower than 0.85. Now we can draw a number line and mark 0 and 1. To represent tenths, divide the space between 0 and 1 into 10 equal parts (0.1, 0.2,... 09, 1.0). 5/6 can be marked between 0.8 and 0.9. Plot the points between 0.8 and 0.9 on the number line. Again divide the space between 0.8 and 0.9 into 10 equal parts to represent hundredths (0.01, 0.02,...) Now count 3 steps after 0.8 because: 0.83 = 0.80 + 0.03. On the number line, mark the point as 5/6. In the image, the black dot stands for the 5/6. Well explained 👍 Problem 4 Find and plot a rational number between -5/6 and -1/3. Okay, lets begin Explanation First, we need to convert -5/6 and -1/3 to like fractions. For that, we must find a common denominator. 12 is the common denominator of 6 and 3 (because it is the least common denominator). -5/6 =(-5 × 2) / (6 × 2) = -10/12 -1/3 = (-1 × 4) / (3 × 4) = -4/12 Now find the midpoint of -10/12 and -4/12: (-10/12 + -4/12) ÷ 2 = -14/12 ÷ 2 = -7/12 Next, mark the spot -7/12 on the number line. Therefore, -7/12 is a rational number between -5/6 and -1/3. Well explained 👍 Problem 5 Represent 4.5 on the number line. Okay, lets begin Explanation Draw a number line and mark 4 and 5. Divide the space between 4 and 5 into 10 equal parts. Spot 4.5 between 4 and 5. Mark 4.5 on the number line. Well explained 👍 Turn your child into a math star! 1 Math Hack Schools Won't Teach! FAQs of Representation of Real Numbers on Number Line 1. Define real numbers. 2.Can we represent a fraction on a number line? 3.Can we represent a decimal number on a number line? 4.How can we represent negative and positive numbers on a number line? 5.Can we represent irrational numbers on a number line? 6.How can children in United States use numbers in everyday life to understand Representation of Real Numbers on Number Line? 7.What are some fun ways kids in United States can practice Representation of Real Numbers on Number Line with numbers? 8.What role do numbers and Representation of Real Numbers on Number Line play in helping children in United States develop problem-solving skills? 9.How can families in United States create number-rich environments to improve Representation of Real Numbers on Number Line skills? Struggling with Math? Get 1:1 Coaching to Boost Grades Fast ! Explore More numbers Previous to Representation of Real Numbers on Number Line Commutative Property\|Division of Fractions \|Prime Factorization\|Conjugates and Rationalization\|Operations on Rational Numbers\|Quotient\|Reciprocal of Fractions\|Properties of Integers \|Multiplier \|Binary Number System \| Binary Division\|Greatest Common Divisor (GCD)\|International Number System\|Equivalent Ratios\|Indian Place Value Chart\|Counting In Words\|Significant Figures\|Numerator And Denominator\|Divisor\|Cardinal Numbers Next to Representation of Real Numbers on Number Line Multiplication on Number Line \|Multiplying Mixed Fractions\|Rational Numbers on Number Line\|Dividend\|Difference Between Place Value And Face Value \|Euler’s Number\|Number Theory\|Multiplying Fractions with Whole Numbers\|Remainder\|Multiples of 415\|Divisibility Rule of 207\|Square Root of 23\|Square root of 117 \|Square Root of 245\|Square root of 38 \|Square root of 3600 \|Square Root of 87\|LCM of 6,9 and 12\|Divisibility Rule of 249\|117 in Roman Numerals Hiralee Lalitkumar Makwana About the Author Hiralee Lalitkumar Makwana has almost two years of teaching experience. She is a number ninja as she loves numbers. Her interest in numbers can be seen in the way she cracks math puzzles and hidden patterns. Fun Fact : She loves to read number jokes and games. Brightchamps Follow Us Email us at care@brightchamps.com Math Topics Numbers Multiplication Tables Algebra Geometry Calculus Measurement Math Formulas Data Commercial Math Explore by Country United States United Kingdom Saudi Arabia United Arab Emirates Our Programs CodeCHAMPS FinCHAMPS LingoCHAMPS RoboCHAMPS Math Questions Sitemap \| © Copyright 2025 BrightCHAMPS INDONESIA - Axa Tower 45th floor, JL prof. Dr Satrio Kav. 18, Kel. Karet Kuningan, Kec. Setiabudi, Kota Adm. Jakarta Selatan, Prov. 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4852	https://pubmed.ncbi.nlm.nih.gov/9468482/	Platelet adhesion to native type I collagen fibrils. Role of GPVI in divalent cation-dependent and -independent adhesion and thromboxane A2 generation - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Role of GPVI in divalent cation-dependent and -independent adhesion and thromboxane A2 generation T Nakamura1,G A Jamieson,M Okuma,J Kambayashi,N N Tandon Affiliations Expand Affiliation 1 Otsuka America Pharmaceutical Inc., Rockville, Maryland 20850, USA. PMID: 9468482 DOI: 10.1074/jbc.273.8.4338 Free article Item in Clipboard Platelet adhesion to native type I collagen fibrils. Role of GPVI in divalent cation-dependent and -independent adhesion and thromboxane A2 generation T Nakamura et al. J Biol Chem.1998. Free article Show details Display options Display options Format J Biol Chem Actions Search in PubMed Search in NLM Catalog Add to Search . 1998 Feb 20;273(8):4338-44. doi: 10.1074/jbc.273.8.4338. Authors T Nakamura1,G A Jamieson,M Okuma,J Kambayashi,N N Tandon Affiliation 1 Otsuka America Pharmaceutical Inc., Rockville, Maryland 20850, USA. PMID: 9468482 DOI: 10.1074/jbc.273.8.4338 Item in Clipboard Full text links Cite Display options Display options Format Abstract Three glycoproteins (GPs), namely GPIa-IIa, GPVI, and GPIV, have been recently implicated in platelet-collagen adhesive interactions. We have employed antibodies to these GPs to investigate further their role in platelet adhesion to immobilized monomeric and polymeric fibrillar collagen under static conditions in the presence and the absence of Mg2+. In the presence of Mg2+, each antibody inhibited platelet adhesion to fibrillar collagen from 70 to 85%, especially during the early phase (<15 min), but the inhibitory effects diminished dramatically to 25% or less by 60 min. Combination of anti-GPVI with anti-GPIa-IIa antibodies completely inhibited platelet adhesion at 60 min. Anti-GPIV and anti-GPIa-IIa or anti-GPVI antibodies in combinations were more effective in inhibiting adhesion than was anti-GPIa-IIa or anti-GPVI alone. In the absence of Mg2+, anti-GPVI completely inhibited adhesion at 60 min, while anti-GPIV antibody inhibited adhesion by about 50% and minimal effects were seen with anti-GPIa-IIa, suggesting that GPIa-IIa does not play a significant role in the divalent cation-independent platelet adhesion to immobilized fibrillar collagen. Under either divalent cation-dependent or -independent conditions, platelets adhered to fibrillar collagen were able to secrete contents of both alpha-granules and dense granules and generate thromboxane A2 (TXA2), but platelets adhering to acid soluble monomeric collagen neither secreted their granular contents nor generated TXA2. Although anti-GPVI antibodies were not able to inhibit Mg2+-dependent adhesion, they completely inhibited TXA2 generation under both divalent cation-dependent and -independent conditions. With the other antibodies, TXA2 generation corresponded with the amount of adhesion observed. These results suggest that GPVI is directly associated with the TXA2 generating system during platelet-collagen interaction. PubMed Disclaimer Similar articles Activation of the GP IIb-IIIa complex induced by platelet adhesion to collagen is mediated by both alpha2beta1 integrin and GP VI.Nakamura T, Kambayashi J, Okuma M, Tandon NN.Nakamura T, et al.J Biol Chem. 1999 Apr 23;274(17):11897-903. doi: 10.1074/jbc.274.17.11897.J Biol Chem. 1999.PMID: 10207010 Signal transduction pathways mediated by glycoprotein Ia/IIa in human platelets: comparison with those of glycoprotein VI.Inoue K, Ozaki Y, Satoh K, Wu Y, Yatomi Y, Shin Y, Morita T.Inoue K, et al.Biochem Biophys Res Commun. 1999 Mar 5;256(1):114-20. doi: 10.1006/bbrc.1999.0295.Biochem Biophys Res Commun. 1999.PMID: 10066433 Collagen-platelet interactions: evidence for a direct interaction of collagen with platelet GPIa/IIa and an indirect interaction with platelet GPIIb/IIIa mediated by adhesive proteins.Coller BS, Beer JH, Scudder LE, Steinberg MH.Coller BS, et al.Blood. 1989 Jul;74(1):182-92.Blood. 1989.PMID: 2546619 Structure-activity relationships of snake toxins targeting platelet receptors, glycoprotein Ib-IX-V and glycoprotein VI.Andrews RK, Gardiner EE, Shen Y, Berndt MC.Andrews RK, et al.Curr Med Chem Cardiovasc Hematol Agents. 2003 Jun;1(2):143-9. doi: 10.2174/1568016033477559.Curr Med Chem Cardiovasc Hematol Agents. 2003.PMID: 15320694 Review. Platelet-collagen interaction: is GPVI the central receptor?Nieswandt B, Watson SP.Nieswandt B, et al.Blood. 2003 Jul 15;102(2):449-61. doi: 10.1182/blood-2002-12-3882. Epub 2003 Mar 20.Blood. 2003.PMID: 12649139 Review. See all similar articles Cited by GPVI and alpha2beta1 play independent critical roles during platelet adhesion and aggregate formation to collagen under flow.Sarratt KL, Chen H, Zutter MM, Santoro SA, Hammer DA, Kahn ML.Sarratt KL, et al.Blood. 2005 Aug 15;106(4):1268-77. doi: 10.1182/blood-2004-11-4434. Epub 2005 May 10.Blood. 2005.PMID: 15886326 Free PMC article. Impaired activation of platelets lacking protein kinase C-theta isoform.Nagy B Jr, Bhavaraju K, Getz T, Bynagari YS, Kim S, Kunapuli SP.Nagy B Jr, et al.Blood. 2009 Mar 12;113(11):2557-67. doi: 10.1182/blood-2008-07-169268. Epub 2009 Jan 22.Blood. 2009.PMID: 19164598 Free PMC article. Matrix protein microarrays for spatially and compositionally controlled microspot thrombosis under laminar flow.Okorie UM, Diamond SL.Okorie UM, et al.Biophys J. 2006 Nov 1;91(9):3474-81. doi: 10.1529/biophysj.106.083287. Epub 2006 Aug 11.Biophys J. 2006.PMID: 16905604 Free PMC article. Aegyptin, a novel mosquito salivary gland protein, specifically binds to collagen and prevents its interaction with platelet glycoprotein VI, integrin alpha2beta1, and von Willebrand factor.Calvo E, Tokumasu F, Marinotti O, Villeval JL, Ribeiro JMC, Francischetti IMB.Calvo E, et al.J Biol Chem. 2007 Sep 14;282(37):26928-26938. doi: 10.1074/jbc.M705669200. Epub 2007 Jul 24.J Biol Chem. 2007.PMID: 17650501 Free PMC article. Salivary Thromboxane A2-Binding Proteins from Triatomine Vectors of Chagas Disease Inhibit Platelet-Mediated Neutrophil Extracellular Traps (NETs) Formation and Arterial Thrombosis.Mizurini DM, Aslan JS, Gomes T, Ma D, Francischetti IM, Monteiro RQ.Mizurini DM, et al.PLoS Negl Trop Dis. 2015 Jun 25;9(6):e0003869. doi: 10.1371/journal.pntd.0003869. eCollection 2015.PLoS Negl Trop Dis. 2015.PMID: 26110417 Free PMC article. See all "Cited by" articles MeSH terms Animals Actions Search in PubMed Search in MeSH Add to Search Antibodies / immunology Actions Search in PubMed Search in MeSH Add to Search Blood Platelets / metabolism Actions Search in PubMed Search in MeSH Add to Search Cations, Divalent Actions Search in PubMed Search in MeSH Add to Search Cell Adhesion / immunology Actions Search in PubMed Search in MeSH Add to Search Collagen / metabolism Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Platelet Factor 4 / metabolism Actions Search in PubMed Search in MeSH Add to Search Platelet Membrane Glycoproteins / immunology Actions Search in PubMed Search in MeSH Add to Search Rats Actions Search in PubMed Search in MeSH Add to Search Serotonin / metabolism Actions Search in PubMed Search in MeSH Add to Search Thromboxane A2 / biosynthesis Actions Search in PubMed Search in MeSH Add to Search beta-Thromboglobulin / metabolism Actions Search in PubMed Search in MeSH Add to Search Substances Antibodies Actions Search in PubMed Search in MeSH Add to Search Cations, Divalent Actions Search in PubMed Search in MeSH Add to Search Platelet Membrane Glycoproteins Actions Search in PubMed Search in MeSH Add to Search beta-Thromboglobulin Actions Search in PubMed Search in MeSH Add to Search Serotonin Actions Search in PubMed Search in MeSH Add to Search Platelet Factor 4 Actions Search in PubMed Search in MeSH Add to Search Thromboxane A2 Actions Search in PubMed Search in MeSH Add to Search Collagen Actions Search in PubMed Search in MeSH Add to Search Related information PubChem Compound (MeSH Keyword) LinkOut - more resources Full Text Sources Elsevier Science Other Literature Sources The Lens - Patent Citations Database Full text links[x] Elsevier Science [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). 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4853	https://www.youtube.com/watch?v=fMVOwML7zTY	Understand Fraction Division w/ Pictures and Models - [6-2-11] Math and Science 1630000 subscribers 1394 likes Description 90487 views Posted: 16 Sep 2021 More Lessons: Twitter: In this lesson, we will learn what it means to divide two fractions. First, we draw a picture of each fraction, then we can see that dividing one fraction into another is just seeing how many fractions fit inside of the other fraction. To perform the division without a picture, we just change the division to multiplication, then multiply by the reciprocal fractions. Dividing fractions is central to math, and we use this concept in algebra, calculus, and every branch of science and engineering. 80 comments Transcript: Intro hello welcome back the title of this lesson is called dividing fractions with models uh part one so another title would be understanding what fraction division really means right a lot of students can calculate the answer when we're dividing fractions but really don't remember or understand what it actually means to divide fractions Fraction Division so for instance our first problem will be the following let's say we take the fraction one-half and want to divide that fraction by another fraction which is one-fourth what would this be actually equal to so it's helpful to remember what fractions what fraction division or what division in general really means what it means is we take the fraction one-half and we want to see how many times can this fraction or whatever it is fit inside of the one half before we tackle this problem let's go back and revision and revise or remember what division actually means in general let's take the simpler problem six and let's divide it by 2. now you all know 6 divided by 2 is 3 right because 3 times 2 is 6. but let's try to remember what does it mean to actually divide these numbers the 6 we're going to represent by these little circles or 1 2 3 here's 4 here's five here's six of them and we're going to divide it by two so one and i can kind of put a little division symbol here six divided by two right what does it actually mean to take and divide these here there's a couple of ways to think about it but one way is to say well how many times can this fit inside of the six that we are given we're dividing by two so it can fit one time here and then another time here and then a third time here so we say that six divided by 2 is 3. so it's helpful in division problems to think of it as we're starting with an amount and we're dividing by another amount how many times can that second thing fit in so we're going to be taking one half and dividing by one-fourth and doing the same thing we just need to represent one-half and one-fourth to figure out the answer so let's go over here and do that we have the same problem one-half divided by one-fourth when we represent one-half what do we do we have a whole in this case it's like a candy bar a whole candy bar think of it but one half is really only this amount and one fourth was what we're dividing by if we take the same size candy bar and cut it into four equal pieces and only have one fourth then we have this guy right here so the first amount the one half is only this amount right here and then the second amount the 1 4 is the amount here we really want to see how many times can this fit into there and so we've redrawn it here in the second drawing down below so here you can see we've redrawn it as we have before the one-half is here and the one-fourth is here but we can now see that two of them exactly fit inside whereas before we have the one-half and the one-fourth and just kind of showing you the fractions here down below you can see that one of them fit in there and then the second one-fourth is exactly the correct size to fit another another time so how many times can this one-fourth fit in there it can fit exactly 2 times so we say the fraction one half divided by 1 4 is exactly equal to 2 because the fraction 1 4 fits into this fraction one half two whole times and that is basically the concept of fraction division it's the same thing as regular division that we already understand and know and love but uh applied to fractions so let's take another example what about the fraction two-thirds and divided by 1 6. we want to see how many times can this fraction fit inside of this one so one-third would be taking a candy bar into three equal pieces and only having one of them and one-sixth would be taking a candy bar and cutting it into six one two three four five six pieces and only having one of them how many times can it fit into the one-third well actually this is just one-third in my drawing here but in our original fraction really we're taking two-thirds and dividing by 1 6. so if this is one-third then this must be two-thirds right here and then we're dividing by 1 6 which is down below so how many times can one-sixth fit into two-thirds well it can fit one time two times three times four times so this is a third this is another third for two thirds and dividing by 1 6 means it goes exactly four times exactly four times and that's the final answer all right let's take a look at the next problem let's say we have the fraction one-fifth and we're going to divide it by one-tenth so here we have the fraction one-fifth because we cut the candy bar into five pieces and only have one of them here we cut the candy bar into one two three four five six seven eight nine ten pieces and only have one of them how many times can this fit in there well we have the one fifth we have the one tenth and it can fit two whole times so one fifth divided by one tenth means it goes two whole times all right for our next problem what about the fraction one-half divided by one-eighth well one-half is represented by this half of the candy bar and then one-eighth is cutting it into eight equal pieces and only having one of them how many times can this fit into the one half so if we have the one half here it'll fit one time two times three times four times exactly so when we divide these we get an answer of four and that is the final answer all right now here's our very last problem it is taking the fraction one third and dividing it by one-ninth one-third is taking candy bar cutting into three pieces only having one of them ninth is cutting it into nine pieces one two three four five six seven eight nine and only having one of them how many times can it go in i think you can see that into one third one ninth can fit one two three times exactly so the fractional answer is three three times exactly all right now Summary in all of these problems the answers were whole numbers we took and divided these fractions we got a whole number of three we took and divided these fractions got a whole number of four in all of the other examples uh earlier in the lesson they also divided a whole number of times much like when you divide regular numbers 6 divided by 3 is 2 right 6 divided by 2 is 3 but if you take 6 divided by 5 it doesn't divide evenly there's a fractional answer same thing is true of fractions i've constructed these to all divide a whole number of times but if i change the problem a little bit it won't go in a whole number of times they'll have a fractional leftover if i change this fraction instead of 1 9 to make it 1 8 or 1 7 or 1 6 i'm going to change the width of the fraction i'm going to change the size of it and then it might go in and there might be a little bit of spillage left over or it might not go in an exact number of times so in general when we divide fractions we will not get answers that are whole numbers because in this case it lines up exactly but if i divide by a different size fraction we might have a little bit of left over extra and that won't divide a whole number of times and so you may have a fractional answer i just want you to keep that in mind i'd like you to go through all of these try to understand the main concept behind fraction division follow me on to the next lesson we'll continue building your skills with visualizing how to divide fractions
4854	https://www.droracle.ai/articles/7544/contraindications-for-reglan	What are the contraindications for Reglan (Metoclopramide)? Select Language▼ What are the contraindications for Reglan (Metoclopramide)? Medical Advisory BoardAll articles are reviewed for accuracy by our Medical Advisory Board Educational purpose only • Exercise caution as content is pending human review Article Review Status Submitted Under Review Approved Last updated: February 14, 2025 • View editorial policy From the Guidelines Metoclopramide (Reglan) is contraindicated in patients with pheochromocytoma, seizure disorder, GI bleeding, and GI obstruction. Contraindications The contraindications for metoclopramide are based on its potential to exacerbate certain conditions, such as: + Pheochromocytoma: metoclopramide may cause a hypertensive crisis in patients with this condition 1 + Seizure disorder: metoclopramide may lower the seizure threshold and worsen seizure control 1 + GI bleeding: metoclopramide may increase the risk of bleeding due to its effects on gastric motility and acid secretion 1 + GI obstruction: metoclopramide may worsen intestinal obstruction due to its prokinetic effects 1 ### Additional Considerations Metoclopramide should be used with caution in patients with a history of extrapyramidal disorders, such as Parkinson's disease, due to the risk of worsening symptoms 2 The European Medicines Agency (EMA) has recommended restricting metoclopramide to short-term use (up to 5 days) to minimize the risk of neurological side effects, such as extrapyramidal disorders 3 The maximum recommended dose of metoclopramide in adults has been restricted to 30 mg/day to minimize the risk of adverse effects 3 From the FDA Drug Label Metoclopramide should not be used whenever stimulation of gastrointestinal motility might be dangerous, e.g., in the presence of gastrointestinal hemorrhage, mechanical obstruction or perforation. Metoclopramide is contraindicated in patients with pheochromocytoma because the drug may cause a hypertensive crisis, probably due to release of catecholamines from the tumor. Such hypertensive crises may be controlled by phentolamine. Metoclopramide is contraindicated in patients with known sensitivity or intolerance to the drug. Metoclopramide should not be used in epileptics or patients receiving other drugs which are likely to cause extrapyramidal reactions, since the frequency and severity of seizures or extrapyramidal reactions may be increased. Do not receive metoclopramide if you: • have stomach or intestine problems that could get worse with metoclopramide, such as bleeding, blockage or a tear in your stomach or bowel wall • have an adrenal gland tumor called pheochromocytoma • are allergic to metoclopramide or anything in it. • take medicines that can cause uncontrolled movements, such as medicines for mental illness • have seizures The contraindications for Reglan (Metoclopramide) are: Gastrointestinal hemorrhage Mechanical obstruction Perforation Pheochromocytoma Known sensitivity or intolerance to the drug Epileptics Patients receiving other drugs which are likely to cause extrapyramidal reactions Stomach or intestine problems that could get worse with metoclopramide Allergic to metoclopramide or anything in it Take medicines that can cause uncontrolled movements Seizures 45 From the Research Contraindications for Reglan (Metoclopramide) The following are contraindications for Reglan (Metoclopramide) as stated in the studies: + Combination with MAO inhibitors, tricyclic antidepressants, sympathomimetic amines, or to patients with pheochromocytoma 6 + GI hemorrhage, obstruction, or perforation 6 + Long-term use (>12 weeks) due to the risk of tardive dyskinesia, a movement disorder that may be irreversible 7, 8, 9 High-risk groups for tardive dyskinesia include: + Elderly females + Diabetics + Patients with liver or kidney failure + Patients with concomitant antipsychotic drug therapy 8, 9 It is recommended to limit the use of parenteral metoclopramide to one or two days and oral preparations for four to 12 weeks of therapy 10 References 1 Guideline Guideline Directed Topic Overview Dr.Oracle Medical Advisory Board & Editors, 2025 2 Guideline Guideline Directed Topic Overview Dr.Oracle Medical Advisory Board & Editors, 2025 3 Guideline Guideline Directed Topic Overview Dr.Oracle Medical Advisory Board & Editors, 2025 4 Drug Official FDA Drug Label For metoclopramide (PO) FDA, 2025 5 Drug Official FDA Drug Label For metoclopramide (PO) FDA, 2025 6 Research Review of a new gastrointestinal drug--metoclopramide. American journal of hospital pharmacy, 1981 7 Research Metoclopramide for the treatment of diabetic gastroparesis. Expert review of gastroenterology & hepatology, 2019 8 Research Gastroparesis, metoclopramide, and tardive dyskinesia: Risk revisited. Neurogastroenterology and motility, 2019 9 Research Review article: metoclopramide and tardive dyskinesia. Alimentary pharmacology & therapeutics, 2010 10 Research Metoclopramide: a dopamine receptor antagonist. American family physician, 1990 Related Questions What is Metoclopramide (Reglan) used for?What are the contraindications for Reglan (Metoclopramide) IV?Can I use droperidol and Reglan (metoclopramide) for treating gastroparesis?Can Reglan (metoclopramide) be used daily?Can I use Reglan (Metoclopramide) for gastroparesis?Can medications be taken periprocedurally?What is the differential diagnosis for a 61-year-old male, 17 years post-gastric bypass, with a body mass index (BMI) of 24, gout, and macrocytosis, who is taking allopurinol, Cialis (tadalafil), and tamsulosin, with normal comprehensive metabolic panel (CMP) and complete blood count (CBC) aside from macrocytosis?What does hypotension with a blood pressure of 78/49 and a mean arterial pressure (MAP) of 48 indicate and how is it treated?What is an amyloid spell?What are the bolus and drip amounts for Protonix (pantoprazole)?What is the dose and frequency for Reglan (Metoclopramide)? 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4855	https://mathbitsnotebook.com/Geometry/3DShapes/3DLinesPlanes.html	\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- \| \| \| \| \| --- \| \| GeometrySmallLogo \| Lines and PlanesMathBitsNotebook.com Topical Outline \| Geometry Outline \| MathBits' Teacher Resources Terms of Use Contact Person: Donna Roberts divider \| The intersection of a line and a plane can be the empty set, a point, or a line. Consider the following theorems relating lines and planes. The diagrams supplied for each theorem represent one possible depiction of the situation. \| \| \| --- \| \| theorem If a line is perpendicular to each of two intersecting lines at their point of intersection, then the line is perpendicular to the plane determined by the two intersecting lines. picture1mbpicmath1 \| theorem Through a given point there passes one and only one plane perpendicular to a given line. picture3mbpicmath3a \| \| theorem Through a given point there passes one and only one line perpendicular to a given plane. picture3mb picmath4a \| theorem If two planes are perpendicular to the same line, the planes are parallel. picture7mb picture7math \| \| theorem Two lines perpendicular to the same plane are coplanar. picture2mb picmath2 \| theorem If a line is perpendicular to a plane, then any line perpendicular to the given line at its point of intersection with the given plane is in the given plane. picture5mb pic6math \| \| theorem Two planes are perpendicular to each other if and only if one plane contains a line perpendicular to the second plane. picture5mb pic5math \| theorem If a line is perpendicular to a plane, then every plane containing the line is perpendicular to the given plane. picture5mb picmath7 \| \| theorem If a plane intersects two parallel planes, then the intersection is two parallel lines. picture8mb pic8math \| The angle where two planes meet is called a dihedral angle. Example: Carpenters and construction workers deal with dihedral angles when planning the construction of the trusses in a roof. ArmyRoof \| divider \| \| \| NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use". \| Topical Outline \| Geometry Outline \| MathBitsNotebook.com \| MathBits' Teacher Resources Terms of UseContact Person: Donna Roberts \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- \| \| \| \| \| --- \| \| GeometrySmallLogo \| Lines and PlanesMathBitsNotebook.com Topical Outline \| Geometry Outline \| MathBits' Teacher Resources Terms of Use Contact Person: Donna Roberts divider \| The intersection of a line and a plane can be the empty set, a point, or a line. Consider the following theorems relating lines and planes. The diagrams supplied for each theorem represent one possible depiction of the situation. \| \| \| --- \| \| theorem If a line is perpendicular to each of two intersecting lines at their point of intersection, then the line is perpendicular to the plane determined by the two intersecting lines. picture1mbpicmath1 \| theorem Through a given point there passes one and only one plane perpendicular to a given line. picture3mbpicmath3a \| \| theorem Through a given point there passes one and only one line perpendicular to a given plane. picture3mb picmath4a \| theorem If two planes are perpendicular to the same line, the planes are parallel. picture7mb picture7math \| \| theorem Two lines perpendicular to the same plane are coplanar. picture2mb picmath2 \| theorem If a line is perpendicular to a plane, then any line perpendicular to the given line at its point of intersection with the given plane is in the given plane. picture5mb pic6math \| \| theorem Two planes are perpendicular to each other if and only if one plane contains a line perpendicular to the second plane. picture5mb pic5math \| theorem If a line is perpendicular to a plane, then every plane containing the line is perpendicular to the given plane. picture5mb picmath7 \| \| theorem If a plane intersects two parallel planes, then the intersection is two parallel lines. picture8mb pic8math \| The angle where two planes meet is called a dihedral angle. Example: Carpenters and construction workers deal with dihedral angles when planning the construction of the trusses in a roof. ArmyRoof \| divider \| \| \| NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use". \| Topical Outline \| Geometry Outline \| MathBitsNotebook.com \| MathBits' Teacher Resources Terms of UseContact Person: Donna Roberts \| \| \| \| --- \| \| GeometrySmallLogo \| Lines and PlanesMathBitsNotebook.com Topical Outline \| Geometry Outline \| MathBits' Teacher Resources Terms of Use Contact Person: Donna Roberts divider \| Lines and PlanesMathBitsNotebook.com Topical Outline \| Geometry Outline \| MathBits' Teacher Resources Terms of Use Contact Person: Donna Roberts The intersection of a line and a plane can be the empty set, a point, or a line. Consider the following theorems relating lines and planes. The diagrams supplied for each theorem represent one possible depiction of the situation. \| \| \| --- \| \| theorem If a line is perpendicular to each of two intersecting lines at their point of intersection, then the line is perpendicular to the plane determined by the two intersecting lines. picture1mbpicmath1 \| theorem Through a given point there passes one and only one plane perpendicular to a given line. picture3mbpicmath3a \| \| theorem Through a given point there passes one and only one line perpendicular to a given plane. picture3mb picmath4a \| theorem If two planes are perpendicular to the same line, the planes are parallel. picture7mb picture7math \| \| theorem Two lines perpendicular to the same plane are coplanar. picture2mb picmath2 \| theorem If a line is perpendicular to a plane, then any line perpendicular to the given line at its point of intersection with the given plane is in the given plane. picture5mb pic6math \| \| theorem Two planes are perpendicular to each other if and only if one plane contains a line perpendicular to the second plane. picture5mb pic5math \| theorem If a line is perpendicular to a plane, then every plane containing the line is perpendicular to the given plane. picture5mb picmath7 \| \| theorem If a plane intersects two parallel planes, then the intersection is two parallel lines. picture8mb pic8math \| The angle where two planes meet is called a dihedral angle. Example: Carpenters and construction workers deal with dihedral angles when planning the construction of the trusses in a roof. ArmyRoof \| Example: Carpenters and construction workers deal with dihedral angles when planning the construction of the trusses in a roof. \| \| \| NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use". \| Topical Outline \| Geometry Outline \| MathBitsNotebook.com \| MathBits' Teacher Resources Terms of UseContact Person: Donna Roberts
4856	https://www.khanacademy.org/standards/CCSS.Math/5.MD	Standards Mapping - Common Core Math \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Math: Pre-K - 8th grade Pre-K through grade 2 (Khan Kids) 2nd grade 3rd grade 4th grade 5th grade 6th grade 7th grade 8th grade Basic geometry and measurement See Pre-K - 8th grade Math Math: Illustrative Math-aligned 3rd grade math 4th grade math 5th grade math 6th grade math 7th grade math 8th grade math Math: Eureka Math-aligned 3rd grade math 4th grade math 5th grade math 6th grade math 7th grade math 8th grade math Math: Get ready courses Get ready for 3rd grade Get ready for 4th grade Get ready for 5th grade Get ready for 6th grade Get ready for 7th grade Get ready for 8th grade Get ready for Algebra 1 Get ready for Geometry Get ready for Algebra 2 Get ready for Precalculus Get ready for AP® Calculus Get ready for AP® Statistics Math: high school & college Algebra 1 Geometry Algebra 2 Integrated math 1 Integrated math 2 Integrated math 3 Trigonometry Precalculus High school statistics Statistics & probability College algebra AP®︎/College Calculus AB AP®︎/College Calculus BC AP®︎/College Statistics Multivariable calculus Differential equations Linear algebra See all Math Math: Multiple grades Early math review Arithmetic Basic geometry and measurement Pre-algebra Algebra basics Test prep SAT Math SAT Reading and Writing Get ready for SAT Prep: Math NEW Get Ready for SAT Prep: Reading and writing NEW LSAT MCAT Science Middle school biology Middle school Earth and space science Middle school chemistry NEW Middle school physics NEW High school biology High school chemistry High school physics Hands-on science activities NEW Teacher resources (NGSS) NEW AP®︎/College Biology AP®︎/College Chemistry AP®︎/College Environmental Science AP®︎/College Physics 1 AP®︎/College Physics 2 Organic chemistry Cosmology and astronomy Electrical engineering See all Science Economics Macroeconomics AP®︎/College Macroeconomics Microeconomics AP®︎/College Microeconomics Finance and capital markets See all Economics Reading & language arts Up to 2nd grade (Khan Kids) 2nd grade 3rd grade 4th grade reading and vocab NEW 5th grade reading and vocab NEW 6th grade reading and vocab 7th grade reading and vocab NEW 8th grade reading and vocab NEW 9th grade reading and vocab NEW 10th grade reading and vocab NEW Grammar See all Reading & Language Arts Computing Intro to CS - 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Convert units (metrics) Convert units (US customary) Convert units multi-step word problems (metric) Convert units multi-step word problems (US customary) Convert units of time Convert units word problems (metrics) Convert units word problems (US customary) Converting units of time review (seconds, minutes, & hours) Converting units: centimeters to meters Converting units: metric distance Converting units: minutes to hours Converting units: US volume Measurement word problem: blood drive Measurement word problem: distance home Measurement word problem: elevator Measurement word problem: tea party Metric units of length review (mm, cm, m, & km) Metric units of mass review (g and kg) Metric units of volume review (L and mL) Same length in different units Time word problem: Susan's break U.S. customary and metric units US Customary units of length review (in, ft, yd, & mi) US Customary units of volume review (c, pt, qt, & gal) US Customary units of weight review (oz & lb) 5.MD.B ------ Represent and interpret data. ----------------------------- 5.MD.B.2 Fully covered Make a line plot to display a data set of measurements in fractions of a unit (1/2, 1/4, 1/8). Use operations on fractions for this grade to solve problems involving information presented in line plots. Graph data on line plots (through 1/8 of a unit) Interpret dot plots with fraction operations Line plot distribution: trail mix 5.MD.C ------ Geometric measurement: understand concepts of volume and relate volume to multiplication and to addition. --------------------------------------------------------------------------------------------------------- ### 5.MD.C.3 ### Recognize volume as an attribute of solid figures and understand concepts of volume measurement. 5.MD.C.3.a Fully covered A cube with side length 1 unit, called a “unit cube,” is said to have “one cubic unit” of volume, and can be used to measure volume. Volume intro Volume with unit cubes 5.MD.C.3.b Fully covered A solid figure which can be packed without gaps or overlaps using n unit cubes is said to have a volume of n cubic units. Measuring volume with unit cubes Volume intro 5.MD.C.4 Mostly covered Measure volumes by counting unit cubes, using cubic cm, cubic in, cubic ft, and improvised units. Compare volumes with unit cubes Measuring volume with unit cubes Volume of rectangular prisms review Volume of rectangular prisms with unit cubes Volume with unit cubes ### 5.MD.C.5 ### Relate volume to the operations of multiplication and addition and solve real world and mathematical problems involving volume. 5.MD.C.5.a Fully covered Find the volume of a right rectangular prism with whole-number side lengths by packing it with unit cubes, and show that the volume is the same as would be found by multiplying the edge lengths, equivalently by multiplying the height by the area of the base. Represent threefold whole-number products as volumes, e.g., to represent the associative property of multiplication. Compare volumes with unit cubes Create volume expressions Creating volume expressions Measuring volume as area times length Volume as area of base times height Volume formula intuition 5.MD.C.5.b Fully covered Apply the formulas V = l × w × h and V = b × h for rectangular prisms to find volumes of right rectangular prisms with whole-number edge lengths in the context of solving real world and mathematical problems. Compare volumes with unit cubes Create volume expressions Creating volume expressions Measuring volume as area times length Volume as area of base times height Volume formula intuition Volume of a rectangular prism Volume of rectangular prisms Volume word problem: water tank Volume word problems 5.MD.C.5.c Fully covered Recognize volume as additive. Find volumes of solid figures composed of two non-overlapping right rectangular prisms by adding the volumes of the non-overlapping parts, applying this technique to solve real world problems. 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4857	https://testbook.com/maths/profit-and-loss	Typesetting math: 100% Test Series Profit and loss terms are used to identify whether a sale is advantageous or not. We all are somewhat familiar with the concepts of profit and loss, when a person runs a business, he or she either faces loss or earns profits. When a person sells a product at a higher rate than the cost price, the difference between both amounts is called profit while when a person sells a product at a lower rate than the cost price, then the difference between both amounts is called loss. Profit and Loss Profit is the difference amount when a person sells a product at a higher rate than cost price & loss is the difference amount when a person sells a product at a lower rate than cost price. Every commodity, product or item has a cost price and selling price and depending on the values of these prices, we compute the profit gained or the loss incurred for an individual product. 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Explore this Supercoaching Maths Notes Free PDFs \| Topic \| PDF Link \| --- \| \| General and Middle Term in Binomial Free Notes PDF \| Download PDF \| \| Circle Study Notes \| Download PDF \| \| Tangents and Normal to Conics \| Download PDF \| \| Increasing and Decreasing Function in Maths \| Download PDF \| \| Wheatstone Bridge Notes \| Download PDF \| \| Alternating Current Notes \| Download PDF \| \| Friction in Physics \| Download PDF \| \| Drift Velocity Notes \| Download PDF \| \| Chemical Equilibrium Notes \| Download PDF \| \| Quantum Number in Chemistry Notes \| Download PDF \| Profit: When a person sells a product at a higher rate than the cost price, then the difference between both amounts. Profit Formula = Selling price – Cost price Loss: When a person sells a product at a lower rate than the cost price, then the difference between both amounts. Loss = Cost Price – Selling Price Profit and Loss Made Easy Profit and Loss is a basic concept in math used to check if we earned or lost money in a deal. It is all about comparing the cost price (CP) — the amount we spent to buy something — and the selling price (SP) — the amount we sold it for. If SP > CP, it means you made a profit. If SP < CP, it means you had a loss. These concepts are used in shops, businesses, and even in our daily purchases. Knowing how to calculate profit or loss helps us make smart financial decisions. For example, if you bought a book for ₹200 and sold it for ₹250, your profit is ₹50. Test Series 138.7k Users NCERT XI-XII Physics Foundation Pack Mock Test 323 Total Tests \| 3 Free Tests English,Hindi 3 Live Test 163 Class XI Chapter Tests 157 Class XII Chapter Tests View Test Series 93.3k Users Physics for Medical Exams Mock Test 74 Total Tests \| 2 Free Tests English 74 Previous Year Chapter Test View Test Series 64.5k Users NCERT XI-XII Math Foundation Pack Mock Test 117 Total Tests \| 2 Free Tests English,Hindi 3 AIM IIT 🎯 49 Class XI 65 Class XII View Test Series Terms used in Profit and Loss Cost Price: Cost price is the price at which a person purchases a product. For example, if Ahana purchased a book for 250 rupees, this is the cost price for that particular book. Cost price is abbreviated as C.P. Selling Price: Selling price is the price at which a person sells a product. For example, if Ankur sold a book for 350 rupees, then this is thought to be the selling price of the book. The selling price is abbreviated as S.P. Market Price: It is the price that is marked on an article or commodity. It is also known as list price or tag price. If there is no discount on the marked price, then the selling price is equal to the marked price. Markup: It is the amount by which cost price is increased to reach market price. Markup = market price – cost price Profit: Profit is the amount earned when something is sold for more than it cost to buy or make. It is calculated using the formula: Profit = Selling Price - Cost Price Loss: Loss is the amount lost when something is sold for less than it cost to buy or make. It is calculated using the formula: Loss = Cost Price - Selling Price Discount: The reduction offered by a merchant on a marked price is called a discount. Successive Discount: If an article is sold at two discounts then it is said that it is sold after two successive discounts. Dishonest Dealing: In it, a person/shopkeeper sells any product at the wrong weight and earns a profit. This can be done either by using false weight or by false reading. 1) A shopkeeper claims to sell rice at a cost price but uses a false weight of 900gm instead of 1000gm. 2) A person sells cloth to a customer but uses false reading and gives 90 meters of cloth instead of 100 meters. Successive Selling: In it, a product is sold for more than one time from one person to another person at some profit or loss. For example – A sold a pen to B at 10% profit and then B sold the pen to C at 20% profit. Sales Tax: When purchasing any product we have to give a certain tax to the government. This additional payment is known as sales tax. Tax is always calculated on the selling price of a product. Profit and Loss Formulas Profit and loss formula is employed in maths to determine the price of an entity in the market and comprehend how advantageous a business is. If the selling price > cost price, then the difference between the S.P. and C.P. is called profit. Similarly, if the selling price < cost price, then the difference between the C.P. and the S.P. is called loss. \| \| \| \| --- \| Profit and Loss Terms \| Meaning \| Formulas \| \| Profit or Gain \| The selling price of the object > than its cost price \| Profit=Selling price(SP) – Cost Price(CP) \| \| Loss \| The cost price of the object > than its selling price \| Loss=Cost Price(CP) – Selling Price(SP) \| \| Selling Price \| The piece for which a commodity is sold is said to be the selling price for that particular item denoted as SP. \| ᠎᠎᠎᠎SP=(100+Profit%100)×CP OR SP=(100−Loss%100)×CP \| \| Cost Price \| The expense at which an object is bought is termed as the cost price for that object, abbreviated as C.P. \| CP=(100100+Profit%)×SP OR CP=(100100−Loss%)×SP \| \| Discount \| To manage the competitors in the industry and promote the sale of goods, vendors offer discounts to consumers. \| Discount= MP – SP(Marked Price – Selling Price) \| Profit Percentage and Loss Percentage The profit percentage (%), as well as the loss percentage (%), is obtained with the help of the below-mentioned formulas. Along with the profit percentage (%) and loss percentage (%) other percentage-related formulas are also discussed below: \| \| \| --- \| \| Profit and Loss Terms \| Formulas in Percentage \| \| Profit percentage (%) \| Profit=(SP) – (CP) Profit percentage%=(ProfitCost Price)×100 \| \| Loss percentage (%) \| Loss= (CP) – (SP) Loss percentage%=(LossCost Price)×100 \| \| Discount (%) \| (DiscountMarked Price)×100 \| \| Markup (%) \| (markupcost price)×100 Where Markup = Selling Price – Cost \| For false weight, the profit percentage can be determined by the formula:Gain%=ErrorTrueValue−Error×100%. Profit and Loss Tricks Students can find different tips and tricks on profit and loss below for solving the questions. Tip # 1: Candidates need to make sure that they know all the important formulas related to profit and loss which are discussed above and some are mentioned below. If ath part of items are sold at x% loss, then for making no profit no loss, Required gain percentage in selling the rest items = ax / (1 - a) If two objects are sold at same selling price, one at x% profit and other at x% loss, then Loss % = (x² / 100) If the cost price of x articles is equal to selling price of y articles, then Profit percentage = {(x - y) / y} × 100 Tip # 2: If there are two successive profits or losses at x% and y% respectively, then the resultant profit or loss% = (x + y + xy)/100 For profit, we take x and/or y as +ve value For loss, we take x and/or y as −ve value Tip # 3: Profit percentage and loss percentage are always calculated on C.P. unless stated otherwise. Also, Check out the Latest Railway Exams with Maths as a Part of its Syllabus Profit and Loss Shortcuts Now that you know how to calculate profit, loss, and their percentages, let’s go over some useful formulas and tricks to solve problems related to profit and loss quickly: Profit (P) = Selling Price (SP) – Cost Price (CP); when SP > CP Loss (L) = Cost Price (CP) – Selling Price (SP); when CP > SP Profit Percentage (P%) = (P ÷ CP) × 100 Loss Percentage (L%) = (L ÷ CP) × 100 SP when there’s profit = [(100 + P%) ÷ 100] × CP SP when there’s loss = [(100 – L%) ÷ 100] × CP CP when there's profit = [100 ÷ (100 + P%)] × SP CP when there's loss = [100 ÷ (100 – L%)] × SP Discount = Marked Price (MP) – Selling Price (SP) SP with discount = MP – Discount Special Cases: If there is a false weight, then Profit% = [(True weight – False weight) ÷ False weight] × 100 If two successive profits are given as m% and n%, then Net Profit% = m + n + (mn ÷ 100) If one transaction leads to a profit of m% and another to a loss of n%, then Net % profit/loss = m – n – (mn ÷ 100) If an item is sold at m% profit and then again at n% profit, the original cost price is: CP = [100 × 100 × P ÷ (100 + m)(100 + n)] (For loss, replace "+" with "–") When profit% and loss% are the same, Net Loss% = (P² ÷ 100) Profit and Loss Solved Examples: Example 1: Marked price of a cricket bat is Rs 1000 and it is sold at Rs 800. Find the discount percentage. Solution: Discount = MP – SP = 1000 – 800 = Rs 200 Discount Percentage = (D/MP) × 100 = (200/1000) × 100 = 20%. Example 2: Marked price of a product is Rs 240 and 25% discount is provided on it. Find the selling price. Solution: Discount = SP × 25% = 240 × (25/100) = Rs 60 Selling price = MP – Discount = 240 – 60 = Rs 180. Alternate Method: Selling Price = (100 – D %) × MP/100 = (100 – 25) × 240/100 = Rs 180. Example 3: A T-shirt is sold after providing two successive discounts of 20%. If the marked price of a T-shirt is Rs 200 then find the selling price. Solution: Discount 1 = 200 × 20/100 = Rs 40 Selling price after 1st discount = 200 – 40 = Rs 160 Discount 2 = 160 × 20/100 = Rs 32 Selling price after 2nd discount = 160 – 32 = Rs 128 Alternate Method: Effective discount = 20 + 20 – (20 × 20)/100 = 36% Discount = 200 × 36/100 = Rs 72 Selling price = 200 – 72 = Rs 128. Example 4: A man gains 30% by selling an article for a certain price. If he sells it at double the current selling price, then what will be the profit percentage? Solution: Let the cost price be Rs. x. ∴ Selling price = Rs. 1.3x Now, new SP = Rs. 2.6x ∴ Profit % = [(2.6x− x )] × 100 = 160% Example 5: If A bought an article at Rs.200 and sold it to B at 20% profit. Again B sold the article at 10% profit to C. Find the amount paid by C. Solution: Price paid by B = 200 + (200/100 × 20) = 200 + 40 = Rs. 240 ∴ Price paid by C = 240 + (240/100 × 10) = 240 + 24 = Rs. 264 Alternate Method: Net profit = 20 + 10 + 20 × 10/100 = 32% Hence, amount paid by C = 200 + (200/100 × 32) = Rs. 264. Example 6: A man sold 2 bicycles at the same selling price. One at 20% loss and other at 20% profit. Find overall profit and loss percentage. Solution: Let the selling price be 300x Then, CP for 1st bicycle = 250x Then, CP of 2nd bicycle = 375x Hence, Net CP = 625x and net SP = 600x ∴ Net loss % = (25x/625x) × 100 = 4% Example 7: If the cost price of 5 oranges is equal to the selling price of 4 oranges, then find a profit percentage. Solution: Let cost price of an orange is Rs. 4 and selling price of an orange is Rs. 5 (we can assume it as it satisfies the given condition of the cost price of 5 oranges is equal to selling price of 4 oranges) Hence, profit percentage = [(5 – 4)/4] × 100 = 25% Example 8: 10 pens costs Rs. 100 each. If half of the pens are sold at 10% loss then find at what price remaining each pen should be sold for making no loss and no profit. Solution: Total cost price of 10 pens = 10 × 100 = Rs. 1000 Selling price of 1 pen = 100 – (100 × 10%) = Rs. 90 Hence, selling price of 5 pens = Rs. 450 Now, selling price of remaining 5 pens = 1000 – 450 = Rs. 550 Hence, selling price of 1 pen = Rs. 110 ∴ Profit % = [(110 – 100)/100] = 10% Example 9: Ram purchased a bicycle for Rs. 5954. He had paid a VAT of 14.5%. Find the list price of the bicycle. Solution: Let the list price be Rs. a. VAT = 14.5% So, a × (114.5/100) = 5954 ⇒ a = (5954 × 100)/114.5 ⇒ a = 5200 ∴ The list price of the bicycle was Rs. 5200. Example 10: Rajesh bought accessories worth Rs. 150. Out of the amount spent for buying accessories, Rs. 10 was spent on sales tax due to taxable purchases. If the tax rate was 10%, calculate the price of the tax-free items. Solution: Total Price = 150 Tax paid = Rs. 10 Tax = 10% Let the taxable purchases = Rs x ⇒ 10% of x = 10 ⇒ 0.1x = 10 ∴ x = 100 ∴ Tax free items = 150 – 100 – 10 = Rs.40 Important Notes Points from the Article Here are a few important points to remember about profit and loss. They will help you quickly understand and solve related questions. Profit = SP - CP Loss = CP - SP Profit (%) = {Profit/CP} × 100 Loss (%) = {Loss/CP} × 100 Discount = Marked Price - Selling Price Discount (%) = (Discount/MP) × 100 \| \| \| --- \| \| If you are checking Profit and Loss article, also check related maths articles: \| \| \| Profit Formula \| Interest \| \| Simple Interest \| Average \| \| Ratio and Proportion \| Percentage Increase & Decrease \| We hope you found this article regarding Profit and Loss was informative and helpful, and please do not hesitate to contact us for any doubts or queries regarding the same. 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Profit is amount when a person sells a product at a higher rate than cost price & loss is amount when a person sells a product at a lower rate than cost price. What is the formula of profit and loss? The formula for the profit and loss is:Profit percentage = (Profit /Cost Price) x 100Loss percentage = (Loss / Cost price) x 100 What is the SP formula? S.P. is the abbreviation for selling price. It is basically the price at which an article is traded. If the vendor gains profit then S.P. = C.P. + Profit. What do MP and CP stand for? Mp stands for market price and CP stands for cost price. What is the cost price formula? The amount spent for a product or entity to buy it is called a cost price.Cost price formula = Selling Price + Loss What is the discount formula? The formula to estimate the discount in terms of SP and MP is:Discount = Marked Price – Selling Price Why are Profit and Loss questions important in exams? 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4858	https://pmc.ncbi.nlm.nih.gov/articles/PMC11864215/	Phenotype of sickle cell disease. Correlation of haplotypes and polymorphisms in cluster β, BCL11A, and HBS1L−MYB. Pilot study - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Advanced Search Journal List User Guide New Try this search in PMC Beta Search View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Front Med (Lausanne) . 2025 Feb 12;12:1347026. doi: 10.3389/fmed.2025.1347026 Search in PMC Search in PubMed View in NLM Catalog Add to search Phenotype of sickle cell disease. Correlation of haplotypes and polymorphisms in cluster β, BCL11A, and HBS1L−MYB. Pilot study Paloma Ropero Paloma Ropero 1 Servicio de Hematología y Hemoterapia, Hospital Clínico San Carlos, Madrid, Spain 2 Instituto de Investigación Sanitaria Hospital Clínico San Carlos, Madrid, Spain Find articles by Paloma Ropero 1,2,, Miriam Peral Miriam Peral 3 Instituto de Biología y Genética Molecular, Valladolid, Spain Find articles by Miriam Peral 3, Luis Javier Sánchez-Martínez Luis Javier Sánchez-Martínez 4 Departamento de Biodiversidad, Ecología y Evolución, Facultad de Biología, Universidad Complutense de Madrid, Madrid, Spain Find articles by Luis Javier Sánchez-Martínez 4, Sara Rochas Sara Rochas 1 Servicio de Hematología y Hemoterapia, Hospital Clínico San Carlos, Madrid, Spain 2 Instituto de Investigación Sanitaria Hospital Clínico San Carlos, Madrid, Spain Find articles by Sara Rochas 1,2, Miguel Gómez-Álvarez Miguel Gómez-Álvarez 1 Servicio de Hematología y Hemoterapia, Hospital Clínico San Carlos, Madrid, Spain Find articles by Miguel Gómez-Álvarez 1, Jorge M Nieto Jorge M Nieto 1 Servicio de Hematología y Hemoterapia, Hospital Clínico San Carlos, Madrid, Spain 2 Instituto de Investigación Sanitaria Hospital Clínico San Carlos, Madrid, Spain Find articles by Jorge M Nieto 1,2, Fernando A González Fernando A González 1 Servicio de Hematología y Hemoterapia, Hospital Clínico San Carlos, Madrid, Spain 2 Instituto de Investigación Sanitaria Hospital Clínico San Carlos, Madrid, Spain Find articles by Fernando A González 1,2, Ana Villegas Ana Villegas 1 Servicio de Hematología y Hemoterapia, Hospital Clínico San Carlos, Madrid, Spain Find articles by Ana Villegas 1, Celina Benavente Celina Benavente 1 Servicio de Hematología y Hemoterapia, Hospital Clínico San Carlos, Madrid, Spain 2 Instituto de Investigación Sanitaria Hospital Clínico San Carlos, Madrid, Spain Find articles by Celina Benavente 1,2 Author information Article notes Copyright and License information 1 Servicio de Hematología y Hemoterapia, Hospital Clínico San Carlos, Madrid, Spain 2 Instituto de Investigación Sanitaria Hospital Clínico San Carlos, Madrid, Spain 3 Instituto de Biología y Genética Molecular, Valladolid, Spain 4 Departamento de Biodiversidad, Ecología y Evolución, Facultad de Biología, Universidad Complutense de Madrid, Madrid, Spain Edited by: Robert W. Maitta, Case Western Reserve University, United States Reviewed by: Parikipandla Sridevi, Central Tribal University of Andhra Pradesh, India Mohamed A. Yassin, Qatar University, Qatar ✉ Correspondence: Paloma Ropero, paloma.ropero@salud.madrid.org Received 2023 Nov 30; Accepted 2025 Jan 14; Collection date 2025. Copyright © 2025 Ropero, Peral, Sánchez-Martínez, Rochas, Gómez-Álvarez, Nieto, González, Villegas and Benavente. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms. PMC Copyright notice PMCID: PMC11864215 PMID: 40012971 Abstract Objective/Background Sickle cell disease (SCD) is a monogenic disease with a highly variable phenotype depending on the amount of fetal hemoglobin (HbF), the main modulator. Variation of HbF levels among patients is genetically regulated. HbF determines both the phenotype of the disease and the response to treatment with the main drug used, hydroxyurea. The efforts of the researchers have focused on discovering the genetic factors responsible for HbF variation, mainly describing the haplotypes of the β cluster and single nucleotide polymorphisms (SNPs) at three different loci: BCL11A, HBS1L-MYB, and the β-globin cluster. This study aimed to determine the possible correlation between the number of SNPs and haplotypes with higher HbF levels in a cohort of patients with SCD. A positive association could explain why certain haplotypes, such as Senegal or Arab-Indian, show higher HbF levels and less severe disease. Methods To test this hypothesis, the characterization of haplotypes was performed using the PCR-RFLP technique and genotyping of three SNPs representative of the three loci with the greatest association with HbF variation: Xmn I (rs7482144), BCL11A (rs4671393), and HBS1L-MYB (rs9376092). Results We found more SNPs in haplotypes related to higher HbF than those with less HbF, although only the SNP Xmn I (rs7482144) showed a statistically significant association. Conclusion We found a direct correlation between haplotypes and the number of SNPs. Haplotypes with higher levels of HbF and less severe phenotypes showed a higher number of SNPs. Thus, the Benin and Bantu haplotypes traditionally associated with poor prognosis showed the fewest mutated SNPs. Keywords: sickle cell disease, cluster haplotypes β, Xmn I, BCL11A, HBS1L-MYB, fetal hemoglobin 1 Introduction Sickle cell disease (SCD) is characterized by complex pathophysiology largely driven by vaso-occlusion and hemolytic anemia. Patients with SCD may experience a wide range of symptoms and complications, including acute chest syndrome, infections, pulmonary hypertension, stroke, and painful vaso-occlusive crises (1). The disease is caused by a single nucleotide transversion at codon 6 GAG > GTG (HBB: c.20A > T) (NM_000518.4) of the β-globin (HBB) gene, thus leading to the production of the most common hemoglobin variant worldwide, HbS, characterized by the substitution of the amino acid Glu → Val at the β6(A3) position (2). In addition, at least five different haplotypes have been characterized in the β cluster, suggesting distinct geographical origins of the same β S gene (Senegal, Benin, Bantu, Arab-Indian, and Cameroon), with haplotype differences in fetal Hb (HbF) levels (3, 4) having been documented as well. SCD (OMIM 603903) is one of the most common autosomal recessive disorders worldwide, with more than 300,000 newborns affected each year, with an expected increase to more than 400,000 by 2050 (5). In Europe, the prevalence of SCD in the 27 member states is estimated to be approximately 1 in 150, while the SCD registry in Spain shows 1,142 registered cases. Early detection, prophylactic treatments with penicillin, and vaccines have improved the quality of life and increased the life expectancy of patients with SCD, although the only curative treatment is allogeneic transplantation from a human leucocyte antigens-compatible donor (6). Although all patients homozygous for the HbS allele have the same genotype (β S/β S), the severity of the disease may be highly variable among affected subjects, from patients with severe clinical symptoms to cases with milder symptoms. The phenotypic heterogeneity is due to both genetic and environmental determinants. The main determinants are the presence of HbF (3, 7), a modulator of the clinical and hematological features of SCD (1, 2). Only one drug approved by the Food and Drug Administration (FDA) and European Medicines Agency (EMA) induces the production of HbF, hydroxyurea (HU); however, not all patients manage to increase HbF levels and improve. Although most patients treated with this drug respond adequately, between 10 and 20% of adults show a minimal response (8, 9). This variability is probably due, among other causes, to baseline HbF levels varying between the haplotypes of the β cluster, heterogeneity among genes responsible for HU metabolism, and quantitative trait locus that affect the expression of γ-globin (HBG) genes, including Xmn I of the β-globin locus on chromosome 11p15, BCL11A on chromosome 2p15, and the intergenic region HBS1L-MYB on chromosome 6q23. The Xmn I variant (rs7482144) exerts a direct effect on the expression of the HBG2 gene (10), while BCL11A (rs4671393) and HBS1L-MYB intergenic region (rs9376092) variants increase HbF levels by decreasing the expression of the transcriptional repressors of γ-globin chain synthesis BCL11A and MYB, respectively (7). The main objective of this work has been to assess whether there is a direct correlation between SNPs Xmn I (rs7482144) located in HBG2, rs4671393 located in BCL11A, and rs9376092 located in HBS1L-MYB and haplotypes and whether haplotypes related to lower severity, having increased values of HbF, show a greater number of these SNPs. To meet this objective, we first analyzed the statistical association of the percentage of HbF for each SNP and then the number and frequency of SNPs in each haplotype. 2 Materials and methods 2.1 Sample collection A total of 28 patients diagnosed with SCD (14 women and 14 men) without treatment with HU and older than 6 years (mean age 11 years) were studied. The samples were received from different Spanish regions at the San Carlos Clinical Hospital in Madrid between 2019 and 2020. 2.2 Hematological measurements All patients underwent a hematometry study with reticulocyte count (Coulter LH750 Analyzer; Beckman Coulter, Brea, CA, USA) and red blood cell morphology. HbA 2 and HbF levels were measured using high-performance liquid ion exchange chromatography (HPLC-CE; VARIANT™; Bio-Rad Laboratories, Hercules, CA, USA). Hemoglobins were studied through capillary zone electrophoresis [Sebia Capillarys Flext (Sebia, Norcross, GA)] and HPLC-CE using the short software for Bio-Rad β-thalassemia (Bio-Rad, Hercules, CA) according to manufacturer’s instructions. 2.3 Molecular analysis and genotyping SNPs After genomic DNA isolation (Biorobot® EZ1; Qiagen GmbH, Hilden, Germany), DNA was quantified on an Invitrogen Qubit 4 fluorometer (Thermo Scientific, Wilmington, DE, USA). The association with α-thalassemia was ruled out by screening for the most common α-thalassemia point mutations and deletions worldwide (21 overall) through multiplex PCR followed by reverse hybridization using the commercial Alpha-Globin StripAssay kit (ViennaLab Diagnostic GmbH, Vienna, Austria) with a clinical sensitivity > 90%. The molecular characterization of HbS was performed with the β-Globin StripAssay MED (ViennaLab Diagnostic GmbH, Vienna, Austria) commercial kit, and its confirmation using automatic Sanger sequencing of the β-globin gene following the previously described protocol (11). The haplotypes of the β cluster were obtained through amplification and digestion with restriction enzymes (PCR-RFLP) according to the protocol described by Rahimi et al. (12). The genotyping of SNPs located in the Gγ (rs7482144) and BCL11A (rs4671393) genes and the HBS1L-MYB intergenic region (rs9376092) was performed through automatic Sanger sequencing using the primers shown in Table 1. TABLE 1. Primers used in SNP genotyping. Sequence Gγ - XmnI (rs7482144) F: 5′ AAC TGT TGC TTT ATA GGA TTT T 3′ R: 5′ AGG AGC TTA TTG ATA ACC TCA GAC 3′ BCL11A (rs4671393) F: 5′ ATG GGA AGA GAC CCC AAA AC 3′ R: 5′ CCT TCT GCT TCC TGT TCA CC 3′ HBS1L-MYB (rs9376092) F: 5′ GAT CAC CCA TCC ATT CAT CC 3′ R: 5′ TCA CCT TCT GAT GTG AAG GAC T 3′ Open in a new tab F = forward primer (5′→3′) and R = reverse primer (3′→5′). 2.4 Statistical analysis In the descriptive study of the data, the qualitative variables are shown alongside their frequency distribution. Quantitative variables are summarized with their mean and standard deviation (SD). Quantitative variables showing an asymmetric distribution are summarized with median and interquartile range (IQR). In comparing parameters between the study groups, the association is assessed using the non-parametric Fisher test because the groups have a small sample size. A significance value of 5% is accepted for all tests. Data processing and analysis are performed using the statistical software IBM SPSS Statistics v.2°. 2.5 Ethical and legal aspects All hematological indices and clinical findings were performed with the prior informed consent of the patients, and the study was approved by the Ethics Committee of the San Carlos Clinical Hospital, Madrid, Spain. All experiments were conducted in accordance with the Declaration of Helsinki. 3 Results 3.1 Hematological data The mean value of the hematological data is shown in Table 2. The values obtained from the study of hemoglobins were: HbA2 (2.62 ± 0.52%), HbF (14.95 ± 9.13%), and HbS (81.86 ± 8.46%). TABLE 2. Hematological data. Mean ± SD Reference values RBC (x 10 6/mL)3.11 ± 0.72 M: 4.52–5.90 F: 4.10–5.10 Hb (g/dL)8.78 ± 1.27 M: 13.5–17.0 F: 11.6–15.0 PCV (%)25.87 ± 4.11 M: 38.3–48.6 F: 36–45 MCV (fL)86.36 ± 12.06 80–100.1 MCH (pg)29.24 ± 4.72 27–32 MCHC (g/dL)33.93 ± 1.40 33.4–35.5 RDW (%)22.40 ± 4.06< 15.00 Reticulocytes (%)9.13 ± 3.60 0.5–2.5 Open in a new tab The number of red cells (RBCs), hemoglobin (Hb), and packet cell volume (PCV) [25.87 ± 4.11%] parameters are decreased compared with normal values. Red cell distribution width (RDW) and reticulocyte count are increased. The rest of the magnitudes, mean corpuscular volume (MCV), mean corpuscular hemoglobin (MCH), and mean corpuscular hemoglobin concentration (MCHC), show values within the reference range. 3.2 Molecular analysis and genotyping SNPs Haplotypes have been inferred based on the presence (+) or absence (−) of cutting at polymorphic sites by specific restriction enzymes (5′ε-Hin cII; 5′Gγ-Xmn I; GγIVSII-HicIII; AγIVSII-HicIII; 3′ψβ-Hin cII; and β-Ava II). In the study population, the African haplotypes Benin (−−−−++), Bantu (−−+−−+), Senegal (−++−++) and Cameroon (−−+++++) have been reported. The largest haplotype in the sample was Benin (70%), followed by Bantu (15%), Senegal (11%), and Cameroon (4%). The haplotypes with the highest HbF values were Benin: 16.59 ± 9.44% and Senegal: 14.52 ± 4.76%, while the Bantu haplotype showed the lowest HbF values: 5.94 ± 2.42%. The frequency of the SNPs studied is shown in Table 3. The most frequent were the wild-type homozygous state for Xmn I (rs7482144) and HBS1L-MYB (rs9376092), with 88.9 and 77.8%, respectively. TABLE 3. Frequency of SNP genotype in the study population. Xmn I (rs7482144)BCL11A (rs4671393)HBS1L-MYB (rs9376092) CCCTTTGGGAAACCCAAA Frequency 24 (88.9%)2 (7.4%)1 (3.7%)11 (40.7%)11 (40.7%)5 (18.5%)21 (77.8%)6 (22.2%)0 (0%) Open in a new tab For Xmn I (rs7482144): C is the wild-type allele, and T is the mutated allele. For BCL11A (rs4671393): G is the wild-type allele, and A is the mutated allele. For HBS1L-MYB (rs9376092): C is the wild-type allele, and A is the mutated allele. In all cases, the coexistence of alpha thalassemia as well as any other hemoglobinopathy was ruled out. 3.3 Statistical analysis The variation of HbF associated with the three SNPs in the three possible genotypes (homozygous for the wild-type allele, homozygous for the mutated allele, and heterozygous) is shown in Figure 1. Only the SNPs HBS1L-MYB (rs9376092) showed a statistically significant association in the study population (p = 0.002). FIGURE 1. Open in a new tab For Xmn I (rs7482144), there is no statistically significant association (p = 0.98). For the wild-type CC genotype, the mean ± SD of HbF is [15.00 ± 9.60%]; in CT heterozygous, it is [15.28 ± 6.47%], while in homozygous for the mutated T allele, it is [13.00 ± 0.00%]. The distribution of HbF among the genotypes of the SNP Xmn I (rs7482144) shows that HbF values below 10% are only found in homozygous for the wild-type allele (CC genotype), although HbF among these individuals has a very heterogeneous distribution. When the mutated T allele is present, both in heterozygous and homozygous, HbF levels keep above 10%. For BCL11A (rs4671393), in homozygous for the wild-type G allele, the mean ± SD of HbF is [13.5 ± 9.76%]; in GA heterozygous [14.24 ± 8.77%] and in homozygous for the mutated A allele [20.09 ± 8.36%]. The mean HbF values between the SNP genotypes show no statistically significant association (p = 0.38). The highest values of HbF among the genotypes of the SNP BCL11A (rs4671393) are reported when the mutated allele A is present, both in GA heterozygosity and in AA homozygosity, with these being much higher in the latter case. In HBS1L-MYB (rs9376092), for the homozygous for the wild-type C allele, the mean ± SD of HbF is [12.8 ± 8.06%], and in the CA heterozygous, it is [14.95 ± 9.13%]. The correlation of HbF between the genotypes found of the SNP HBS1L-MYB (rs9376092) shows a statistically significant association in the study population (p< 0.05). No homozygous individual has been found for the mutated allele. The distribution of HbF between homozygous for the wild-type allele (CC genotype) and heterozygous (CA genotype) is different, being higher in the latter situation. The number and frequency of SNPs for each haplotype are shown in Table 4. The distribution of SNPs in BCL11A and HBS1L-MYB among the different haplotypes did not produce statistically significant results. However, a statistically significant association with the SNP Xmn I (rs7482144) (p< 0.05) was observed. The distribution of HbF among SNPs in haplotypes is shown in Figure 2. TABLE 4. Genotype frequency of SNPs in Bantu, Benin, and Senegal haplotypes. Xmn I (rs7482144)BCL11A (rs4671393)HBS1L-MYB (rs9376092)HbF (%)Number of chromosomes Number of mutations CC (−/−)CT (±)TT (+/+)GG (−/−)GA (±)AA (+/+)CC (−/−)CA (±)AA (+/+) Bantu 4 (100%)0 0 1 (25%)3 (75%)0 4 (100%)0 0 5.9 8 3 (12.5%) Benin 20 (100%)0 0 10 (50%)6 (30%)4 (20%)14 (70%)6 (30%)0 16.9 40 20 (16.7%) Senegal 0 2 (66.7%)1 (33.3%)0 2 (66.7%)1 (33.3%)3 (100%)0 0 13 6 8 (44%) Frequency 24 (88.9%)2 (7.4%)1 (3.7%)11 (40.7%)11 (40.7%)5 (18.4%)21 (77.8%)6 (22.8%)0 p-value< 0.001 0.39 0.57 Open in a new tab In the Bantu haplotype, the homozygous state for the wild-type allele of the three SNPs is the majority, comprising 37.5% of all possible genotypes reported; 12.5% corresponds to the heterozygous state and 0% to the homozygous state for the mutated allele; 12.5% of alleles were mutated. Therefore, the low levels of HbF of this haplotype can be due to its small number of SNPs, and the fact that none of them is homozygous for the mutated allele associated with increased HbF. In the Benin haplotype, characterized by intermediate HbF values and severity, 73.3% have the wild-type genotype, 20% of the possible genotypes have at least one copy of the mutated allele of the SNPs, and the homozygous state for the mutated allele of the three SNPs comprises 6.6% of all reported genotypes. Overall, 16.7% of the alleles were mutated. For the Senegal haplotype, in which the homozygous state for the wild-type allele of the three SNPs comprises 33.3%, the heterozygous 44.4%, and the mutated homozygous state 22.22%, 44% of alleles for these SNPs were mutated. Number of mutated chromosomes. FIGURE 2. Open in a new tab The Bantu haplotype is characterized by showing the Xmn I (rs7482144) and HBS1L-MYB (rs9376092) polymorphisms in homozygosity for the wild-type allele in 100% of cases, while the BCL11A polymorphism (rs4671393) is mostly (75%) in heterozygous individuals. HbF levels do not exceed 10%. In the Benin haplotype, the homozygous state for the wild-type allele is the majority in all three SNPs, accounting for 100%, 47.4%, and 68.4% of Xmn I (rs7482144), BCL11A (rs4671393), and HBS1L-MYB (rs9376092), respectively. Only the homozygous status for the mutated allele of SNP BCL11A (rs4671393) is reported. HbF levels are highly heterogeneous. The Senegal haplotype only shows homozygosity for the wild-type allele of SNP HBS1L-MYB (rs9376092). The heterozygous state is found in 66.7% of cases of both SNP Xmn I (rs7482144) and BCL11A (rs4671393). The homozygous genotype for the mutated allele of both SNPs constitutes 33.3%. HbF levels lie in the 10–20% range. 4 Discussion Sickle cell disease (SCD) has been studied for many years owing to the unexpected phenotypic heterogeneity of a disease caused by a single mutation. Several factors have been related to an improvement in complications and severity of the disease. The main and most studied ones are HbF levels, which are also subject to high heterogeneity, although there are others, such as the simultaneous inheritance of α-thalassemia or environmental factors. For most of the time, SCD has been restricted mainly to sub-Saharan Africa, where approximately 80% of such births occur every year (5). However, it is currently found in most countries owing, among other causes, to adoptions, economic, political, or wartime migrations, thus becoming a global public health issue (3, 13). Most treatments aim to increase HbF levels to achieve a more favorable phenotype, with HU being the main drug used. Although these treatments are only palliative and the only curative treatment is the transplantation of hematopoietic precursors, the vast majority of patients cannot access the latter because of the shortage of compatible donors, low incomes, and inadequate sanitation in countries with the highest incidence, usually coinciding with developing countries (1). Therefore, there is a growing interest in glimpsing the genetic factors responsible for the increase in HbF levels and their variability among patients and trying to genetically or pharmacologically manipulate them to achieve a less severe disease phenotype. In this regard, this study aimed to determine whether haplotypes related to higher levels of HbF and, therefore, to a disease with fewer complications show a greater number of SNPs previously described as positive modulators of HbF synthesis. The presence of these SNPs could partially explain the heterogeneity of HbF levels between haplotypes. HbF is a highly variable parameter among individuals with SCD. In the study cohort, the values ranged from 1.9 to 32%. Most of this variation (89%) is controlled by genetic factors identified over the last few years, among which the haplotypes of the β cluster and SNPs in the three quantitative trait locus studied stand out (14). In this study, the analysis of the haplotypes of the β cluster showed four different patterns, identifying only African haplotypes, the majority being Benin, which is the most frequent in nearby countries such as Algeria or Tunisia (15). Thus, the patients studied came from or had ancestry from countries in sub-Saharan Africa. No cases have been identified for the Arab-Indian haplotype, which is more restricted to the Saudi population. Individuals with the Bantu haplotype showed the lowest HbF levels, approximately 5%, while individuals with the highest levels were Benin and Senegal, reaching approximately 15%. HbF values in Bantu and Senegal haplotypes are consistent with levels described in other populations, while HbF values in Benin vary widely among individuals (16). The mean HbF of the Cameroon haplotype could not be calculated as it was only present in one individual (1.85%). In our study population, probably due to sample size, the impact of β cluster haplotypes on HbF levels is not statistically significant. Although suggesting a trend toward higher levels when the mutated allele is present, the distribution of HbF among the three genotypes of the SNP Xmn I (rs7482144) did not show any statistical significance since the mutated allele was identified only in two heterozygous (7.4%) subjects and one (3.7%) homozygous individual, with most individuals being homozygous for the wild-type allele (88.9%). Recent studies indicate that other SNPs within the β-globin cluster are more significantly associated with HbF variation than the one used in this study (17). Regarding the genotypes of the SNP BCL11A (rs4671393) located in intron 2 of the BCL11A oncogene, greater variability was reported than in the previous case. The mutated allele was present in 11 heterozygous individuals (40.7%) and five (18.4%) homozygous individuals. The high frequency of the mutated allele in the sample may be because the individuals comprising it were African or of African descent, where the frequency of the mutated allele is much higher than in the rest of the populations. This SNP is associated with higher levels of HbF, and some studies have described it as the most influential, with 13% of the variability being attributed to it (17–19). In our study, genotypes containing this mutated allele show higher mean HbF levels than those homozygous for the wild-type allele. There is an increase in HbF in homozygous individuals for the mutated allele, where the mean HbF level is greater than 20%; however, owing to the restricted sample size, no statistically significant association could be established. Of the SNP HBS1L-MYB (rs9376092), only two of the three possible genotypes have been reported, and no homozygous individuals for the mutated allele were reported. Despite this obstacle, it is the only variable showing a statistically significant association with HbF levels. This SNP is also associated with HbF variation in patient cohorts from other populations and healthy populations (20). It is located in block 2 of the HBS1L-MYB intergenic region, where the strongest association with HbF levels has been reported. Several studies have concluded that other SNPs (rs9399137 and rs9402686) within this block could be more related to this HbF variability (7). There are other SNPs at cluster loci β-globin, BCL11A, and HBS1L-MYB related to variation in HbF levels. They could be used in future research to assess whether these are also found more frequently in haplotypes with higher HbF levels. The main hypothesis of the study was that haplotypes with higher HbF levels have a higher number of mutated SNPs. In the Bantu haplotype, no individuals with two copies of the mutated allele for any of the three SNPs were found. Therefore, the low levels of HbF of this haplotype could be due to the small number of SNPs and the fact that none of them is homozygous for the mutated allele associated with the increased HbF. In the Benin haplotype, characterized by intermediate HbF values and severity, the homozygous state for the mutated allele of the three SNPs together constitutes 6.7% of all reported genotypes, while the Senegal haplotype is the one with the highest relative frequency of SNPs in the homozygous state for the mutated allele (22.22%) from the reported genotypes. Based on the results obtained and according to the hypothesis of the study, a direct correlation is observed between the number of SNPs homozygous for the mutated allele and haplotypes with higher levels of HbF. This distribution of SNPs could be responsible for the increase in HbF levels. Although only the SNP Xmn I (rs7482144) showed a statistically significant association, this result is probably due to the small sample size, so it would be necessary to expand the sample to extract more statistically robust results. All these genetic modulators of HbF levels, and therefore of the clinic and severity of SCD, could be used as biomarkers to stratify patients based on their ability to produce HbF, with the objective of a more customized clinical and pharmacological management according to the expected phenotype. This could have implications in genetic counseling and prenatal diagnosis of patients. The genetic SNPs described may become potential pharmacological targets to devise novel therapies that increase the level of HbF in individuals with a more severe phenotype and improve their clinical development. In this regard, strategies using BCL11A as a genetic target, such as silencing this gene, are being devised since this is a repressor of the synthesis of γ-globin chains. Silencing increases HbF production and corrects the disease phenotype in mouse models of SCD without affecting erythropoiesis or the expression of other genes (21). Other strategies rely on interference with BCL11A enhancers through genetic engineering to decrease their synthesis and thus enhance HbF production (22). The global burden of the disease is expected to increase in the coming years, owing to improved treatment and migration to countries with higher incomes that allow for increased patient survival (5). These estimates highlight the importance of finding and exploiting genetic or pharmacological targets to improve quality of life and decrease mortality in patients with SCD. The high heterogeneity of this disease, not yet fully explained by the genetic factors described here, implies that additional genes are involved in HbF production to be discovered. Therefore, we will better understand the genetic mechanisms underlying this disease and new candidate genes to study. The advancements that are being performed allow us to improve the quality of life of patients, and every day we are closer to achieving a customized therapy that adjusts to the characteristics of each individual. 5 Conclusion In conclusion, our study affirms that individuals exhibiting elevated HbF levels manifest a milder phenotype. Additionally, the mutated alleles of the identified SNPs are linked to an inclination for increased HbF production. The correlation between haplotypes, the quantity of SNPs, and higher HbF levels indicates that a less severe phenotype is associated with a greater number of SNPs. Notably, the Benin and Bantu haplotypes, conventionally associated with a poorer prognosis, harbor the fewest mutated SNPs. To further validate these findings, more comprehensive studies involving larger patient cohorts are warranted, especially considering the limited representation of individuals with Arab-Indian or Cameroon haplotypes in our current analysis. Acknowledgments We would like to express their gratitude for the permission to include the translated study, titled “FENOTIPO DE LA ENFERMEDAD DE CÉLULAS FALCIFORME. RELACIÓN DE LOS HAPLOTIPOS Y LOS POLIMORFISMOS EN EL CLÚSTER B, BCL11A Y HBS1L-MYB. ESTUDIO PILOTO.” Ropero et al. (23). Funding Statement The authors declare that no financial support was received for the research, authorship, and/or publication of this article. Data availability statement The datasets presented in this study can be found in online repositories. The names of the repository/repositories and accession number(s) can be found in this article/supplementary material. Ethics statement The studies involving humans were approved by the Comité de Ética del Hospital Clínico San Carlos de Madrid. The studies were conducted in accordance with the local legislation and institutional requirements. The participants provided their written informed consent to participate in this study. Written informed consent was not obtained from the individual(s) for the publication of any potentially identifiable images or data included in this article because no pictures of the participants have been taken. Author contributions PR: Writing – review & editing, Conceptualization, Supervision, Validation. MP: Formal Analysis, Investigation, Methodology, Writing – original draft. LS-M: Data curation, Formal Analysis, Writing – original draft. SR: Writing – original draft. MG-Á: Writing – original draft. JN: Writing – original draft. 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The names of the repository/repositories and accession number(s) can be found in this article/supplementary material. 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4859	http://aleph0.clarku.edu/~djoyce/elements/bookI/bookI.html	Euclid's Elements, Book I Book I Table of contents Definitions (23) Postulates (5) Common Notions (5) Propositions (48) Guide to Book I Definitions Definition 1.A point is that which has no part. Definition 2.A line is breadthless length. Definition 3.The ends of a line are points. Definition 4.A straight line is a line which lies evenly with the points on itself. Definition 5.A surface is that which has length and breadth only. Definition 6.The edges of a surface are lines. Definition 7.A plane surface is a surface which lies evenly with the straight lines on itself. Definition 8.A plane angle is the inclination to one another of two lines in a plane which meet one another and do not lie in a straight line. Definition 9.And when the lines containing the angle are straight, the angle is called rectilinear.Definition 10.When a straight line standing on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands. Definition 11.An obtuse angle is an angle greater than a right angle. Definition 12.An acute angle is an angle less than a right angle. Definition 13.A boundary is that which is an extremity of anything. Definition 14.A figure is that which is contained by any boundary or boundaries. Definition 15.A circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among those lying within the figure equal one another. Definition 16.And the point is called the center of the circle. Definition 17.A diameter of the circle is any straight line drawn through the center and terminated in both directions by the circumference of the circle, and such a straight line also bisects the circle. Definition 18.A semicircle is the figure contained by the diameter and the circumference cut off by it. And the center of the semicircle is the same as that of the circle. Definition 19.Rectilinear figures are those which are contained by straight lines, trilateral figures being those contained by three, quadrilateral those contained by four, and multilateral those contained by more than four straight lines. Definition 20.Of trilateral figures, an equilateral triangle is that which has its three sides equal, an isosceles triangle that which has two of its sides alone equal, and a scalene triangle that which has its three sides unequal. Definition 21.Further, of trilateral figures, a right-angled triangle is that which has a right angle, an obtuse-angled triangle that which has an obtuse angle, and an acute-angled triangle that which has its three angles acute. Definition 22.Of quadrilateral figures, a square is that which is both equilateral and right-angled; an oblong that which is right-angled but not equilateral; a rhombus that which is equilateral but not right-angled; and a rhomboid that which has its opposite sides and angles equal to one another but is neither equilateral nor right-angled. And let quadrilaterals other than these be called trapezia.Definition 23Parallel straight lines are straight lines which, being in the same plane and being produced indefinitely in both directions, do not meet one another in either direction. Postulates Let the following be postulated: Postulate 1.To draw a straight line from any point to any point.Postulate 2.To produce a finite straight line continuously in a straight line.Postulate 3.To describe a circle with any center and radius.Postulate 4.That all right angles equal one another. Postulate 5.That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles. Common Notions Common notion 1.Things which equal the same thing also equal one another. Common notion 2.If equals are added to equals, then the wholes are equal. Common notion 3.If equals are subtracted from equals, then the remainders are equal. Common notion 4.Things which coincide with one another equal one another. Common notion 5.The whole is greater than the part. Propositions Proposition 1.To construct an equilateral triangle on a given finite straight line.Proposition 2.To place a straight line equal to a given straight line with one end at a given point.Proposition 3.To cut off from the greater of two given unequal straight lines a straight line equal to the less. Proposition 4.If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, then they also have the base equal to the base, the triangle equals the triangle, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides. Proposition 5.In isosceles triangles the angles at the base equal one another, and, if the equal straight lines are produced further, then the angles under the base equal one another. Proposition 6.If in a triangle two angles equal one another, then the sides opposite the equal angles also equal one another. Proposition 7.Given two straight lines constructed from the ends of a straight line and meeting in a point, there cannot be constructed from the ends of the same straight line, and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each equal to that from the same end. Proposition 8.If two triangles have the two sides equal to two sides respectively, and also have the base equal to the base, then they also have the angles equal which are contained by the equal straight lines. Proposition 9.To bisect a given rectilinear angle. Proposition 10.To bisect a given finite straight line. Proposition 11.To draw a straight line at right angles to a given straight line from a given point on it. Proposition 12.To draw a straight line perpendicular to a given infinite straight line from a given point not on it. Proposition 13.If a straight line stands on a straight line, then it makes either two right angles or angles whose sum equals two right angles. Proposition 14.If with any straight line, and at a point on it, two straight lines not lying on the same side make the sum of the adjacent angles equal to two right angles, then the two straight lines are in a straight line with one another. Proposition 15.If two straight lines cut one another, then they make the vertical angles equal to one another. Corollary. If two straight lines cut one another, then they will make the angles at the point of section equal to four right angles. Proposition 16.In any triangle, if one of the sides is produced, then the exterior angle is greater than either of the interior and opposite angles. Proposition 17.In any triangle the sum of any two angles is less than two right angles. Proposition 18.In any triangle the angle opposite the greater side is greater. Proposition 19.In any triangle the side opposite the greater angle is greater. Proposition 20.In any triangle the sum of any two sides is greater than the remaining one. Proposition 21.If from the ends of one of the sides of a triangle two straight lines are constructed meeting within the triangle, then the sum of the straight lines so constructed is less than the sum of the remaining two sides of the triangle, but the constructed straight lines contain a greater angle than the angle contained by the remaining two sides. Proposition 22.To construct a triangle out of three straight lines which equal three given straight lines: thus it is necessary that the sum of any two of the straight lines should be greater than the remaining one. Proposition 23.To construct a rectilinear angle equal to a given rectilinear angle on a given straight line and at a point on it. Proposition 24.If two triangles have two sides equal to two sides respectively, but have one of the angles contained by the equal straight lines greater than the other, then they also have the base greater than the base. Proposition 25.If two triangles have two sides equal to two sides respectively, but have the base greater than the base, then they also have the one of the angles contained by the equal straight lines greater than the other. Proposition 26.If two triangles have two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that opposite one of the equal angles, then the remaining sides equal the remaining sides and the remaining angle equals the remaining angle. Proposition 27.If a straight line falling on two straight lines makes the alternate angles equal to one another, then the straight lines are parallel to one another. Proposition 28.If a straight line falling on two straight lines makes the exterior angle equal to the interior and opposite angle on the same side, or the sum of the interior angles on the same side equal to two right angles, then the straight lines are parallel to one another. Proposition 29.A straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle equal to the interior and opposite angle, and the sum of the interior angles on the same side equal to two right angles. Proposition 30.Straight lines parallel to the same straight line are also parallel to one another. Proposition 31.To draw a straight line through a given point parallel to a given straight line. Proposition 32.In any triangle, if one of the sides is produced, then the exterior angle equals the sum of the two interior and opposite angles, and the sum of the three interior angles of the triangle equals two right angles. Proposition 33.Straight lines which join the ends of equal and parallel straight lines in the same directions are themselves equal and parallel. Proposition 34.In parallelogrammic areas the opposite sides and angles equal one another, and the diameter bisects the areas. Proposition 35.Parallelograms which are on the same base and in the same parallels equal one another. Proposition 36.Parallelograms which are on equal bases and in the same parallels equal one another. Proposition 37.Triangles which are on the same base and in the same parallels equal one another. Proposition 38.Triangles which are on equal bases and in the same parallels equal one another. Proposition 39.Equal triangles which are on the same base and on the same side are also in the same parallels. Proposition 40.Equal triangles which are on equal bases and on the same side are also in the same parallels. Proposition 41.If a parallelogram has the same base with a triangle and is in the same parallels, then the parallelogram is double the triangle. Proposition 42.To construct a parallelogram equal to a given triangle in a given rectilinear angle. Proposition 43.In any parallelogram the complements of the parallelograms about the diameter equal one another. Proposition 44.To a given straight line in a given rectilinear angle, to apply a parallelogram equal to a given triangle. Proposition 45.To construct a parallelogram equal to a given rectilinear figure in a given rectilinear angle. Proposition 46.To describe a square on a given straight line. Proposition 47.In right-angled triangles the square on the side opposite the right angle equals the sum of the squares on the sides containing the right angle. Proposition 48.If in a triangle the square on one of the sides equals the sum of the squares on the remaining two sides of the triangle, then the angle contained by the remaining two sides of the triangle is right. Guide About the Definitions The Elements begins with a list of definitions. Some of these indicate little more than certain concepts will be discussed, such as Def.I.1, Def.I.2, and Def.I.5, which introduce the terms point, line, and surface. (Note that for Euclid, the concept of line includes curved lines.) Others are substantial definitions which actually describe new concepts in terms of old ones. For example, Def.I.10 defines a right angle as one of two equal adjacent angles made when one straight line meets another. Other definitions look like they’re substantial, but actually are not. For instance, Def.I.4 says a straight line “is a line which lies evenly with the points on itself.” No where in the Elements is the defining phrase “which lies evenly with the points on itself” applicable. Thus, this definition indicates, at most, that some lines under discussion will be straight lines. It has been suggested that the definitions were added to the Elements sometime after Euclid wrote them. Another possibility is that they are actually from a different work, perhaps older. In Def.I.22 special kinds of quadrilaterals are defined including square, oblong (a rectangle that are not squares), rhombus (equilateral but not a square), and rhomboid (parallelogram but not a rhombus). Except for squares, these other shapes are not mentioned in the Elements. Euclid does use parallelograms, but they’re not defined in this definition. Also, the exclusive nature of some of these terms—the part that indicates not a square—is contrary to Euclid’s practice of accepting squares and rectangles as kinds of parallelograms. About the Postulates Following the list of definitions is a list of postulates. Each postulate is an axiom—which means a statement which is accepted without proof— specific to the subject matter, in this case, plane geometry. Most of them are constructions. For instance, Post.I.1 says a straight line can be drawn between two points, and Post.I.3 says a circle can be drawn given a specified point to be the center and another point to be on the circumference. The fourth postulate, Post.I.4, is not a construction, but says that all right angles are equal. About magnitudes and the Common Notions The Common Notions are also axioms, but they refer to magnitudes of various kinds. The kind of magnitude that appears most frequently is that of straight line. Other important kinds are angles, plane figures, and solid figures In proposition III.16 (but nowhere else) angles with curved sides are compared with rectilinear angles which shows that rectilinear angles are to be considered as a special kind of plane angle. That agrees with Euclid’s definition of them in I.Def.9 and I.Def.8. Also in Book III, parts of circumferences of circles, that is, arcs, appear as magnitudes. Only arcs of equal circles can be compared or added, so arcs of equal circles comprise a kind of magnitude, while arcs of unequal circles are magnitudes of different kinds. These kinds are all different from straight lines. Whereas areas of figures are comparable, different kinds of curves are not. Book V includes the general theory of ratios. No particular kind of magnitude is specified in that book. It may come as a surprise that ratios do not themselves form a kind of magnitude since they can be compared, but they cannot be added. See the guide on Book V for more information. Number theory is treated in Books VII through IX. It could be considered that numbers form a kind of magnitude as pointed out by Aristotle. Beginning in Book XI, solids are considered, and they form the last kind of magnitude discussed in the Elements. The propositions Following the definitions, postulates, and common notions, there are 48 propositions. Each of these propositions includes a statement followed by a proof of the statement. Each statement of the proof is logically justified by a definition, postulate, common notion, or an earlier proposition that has already been proven. There are gaps in the logic of some of the proofs, and these are mentioned in the commentaries after the propositions. Also included in the proof is a diagram illustrating the proof. Some of the propositions are constructions. A construction depends, ultimately, on the constructive postulates about drawing lines and circles. The first part of a proof for a constructive proposition is how to perform the construction. The rest of the proof (usually the longer part), shows that the proposed construction actually satisfies the goal of the proposition. In the list of propositions in each book, the constructions are displayed in red. Most of the propositions, however, are not constructions. Their statements say that under certain conditions, certain other conditions logically follow. For example, Prop.I.5 says that if a triangle has the property that two of its sides are equal, then it follows that the angles opposite these sides (called the “base angles”) are also equal. Even the propositions that are not constructions may have constructions included in their proofs since auxiliary lines or circles may be needed in the explanation. But the bulk of the proof is, as for the constructive propositions, a sequence of statements that are logically justified and which culminates in the statement of the proposition. Logical structure of Book I The various postulates and common notions are frequently used in Book I. Only two of the propositions rely solely on the postulates and axioms, namely, I.1 and I.4. The logical chains of propositions in Book I are longer than in the other books; there are long sequences of propositions each relying on the previous. Dependencies within Book I 123 3, 45, 6 578 1, 3, 89, 11 1, 4, 910 8, 1012 111314, 15 3, 4, 10, 151627 13, 1617 3, 5, 1618 5, 1819 3, 5, 1920 16, 2021 3, 2022 8, 2223 3, 4, 5, 19, 2324 4, 2425 3, 4, 1626 13, 15, 2728, 29 2930 23, 2731 13, 29, 3132 4, 27, 2933 4, 26, 293443 4, 29, 3435 33, 34, 3536 31, 34, 3537 31, 34, 3638 31, 3739 31, 3840 34, 3741 10, 23, 31, 38, 4142 15, 29, 31, 42, 4344 14, 29, 30, 33, 34, 42, 4445 3, 11, 29, 31, 3446 4, 14, 31, 41, 4647 3, 8, 11, 4748 First page: I.Def.1 Next: Book II Previous: Introduction ©1996 David E. Joyce Department of Mathematics and Computer Science Clark University Worcester, MA 01610
4860	https://www.youtube.com/watch?v=G224KyLE9qQ	Sum of Even Numbers Between 50 and 100 inclusive Math With Cal 294 subscribers 41 likes Description 7610 views Posted: 26 Nov 2022 Hi everyone, in order to find the sum of the even numbers between 50 and 100 inclusive, we can employ the Sum of the first n terms formula of the arithmetic progression. This video gives a step by step approach on explaining the formula and how to get the sum of the even numbers between 50 and 100 inclusive. ✏️ S_n: Denotes the sum of the first n terms of the progression ✏️ n: Denotes the number of terms in the progression ✏️ a: Denotes the first term of the progression ✏️ d: Denotes the common difference between each term in the progression 👇 Thank you so much for subscribing to this channel ☺️ 🚨Charge up your skills in Arithmetic and Geometric Progression in this playlist! 🚀 Want to learn more H2 Math? 🎞 🎞 🎞 🎞 🎞 Want to learn more Additional Math? 🎞 🎞 🎞 arithmeticprogression #sumofnterms #h2math 4 comments Transcript: hello everyone today we are going to find the sum of even numbers from 50 to 100 inclusive we could do this easily by employing the sum of the first n terms formula of an arithmetic progression now if you are new to this formula this represents the sum of the first n terms while n represents the number of terms in the arithmetic progression that we are looking at also a represents the first term of the progression well D is the common difference between each term in the progression now coming back to this example to visualize things better we are going to list just a few terms in the progression starting with 50 51 52 53 54 and so on until 100. now the key idea is to find what are the exact values of a d and n so that we could use this formula right here so these are the even terms that we want and the first term is actually 50. now this means that a is equals to 50. now between the terms 50 and 52 the common difference is simply 2 because if we were to add 2 continuously to 52 it gives us an even number 54 56 58 and so on until 100. now the next part is to determine the number of even terms between 50 and 100 inclusive now this can be a bit tricky if we are not sure how to do that let's imagine the numbers 1 to 10. now if you see over here there are five even numbers or by taking 10 divided by 2 which is 5. now similarly for the number of even terms between 50 to 100 it is 100 minus 50 divided by 2. however do remember to include the term 50 as required by the question so the total number of even terms is 26 instead of 25 so all we need to do is to add a 1 over here now with these values of a d and n simply plug them into the sum of the first n terms over here so we could write s 26 of the even terms is equals to 26 over 2 multiplied by 2 times 50 plus n minus 1 which is 26 minus 1 times 2 and this will be equals to 26 multiplied by 50 plus 25 which gives us a value of 1950. thus the sum of all the even numbers between 50 to 100 inclusive is 1950.
4861	https://www.youtube.com/watch?v=K63p4fz4MEg	Ortho Para Ratio - Aromatic Nitration of Toluene The Organic Chemistry Tutor 9640000 subscribers 815 likes Description 60514 views Posted: 8 May 2018 This organic chemistry video tutorial provides a basic introduction into the ortho para ratio of the aromatic nitration of toluene. Bulky groups such as the tert-butyl group favor the para product while the small substituents such as the methyl group favor the ortho product. Subscribe: Access to Premium Videos: New Organic Chemistry Playlist 32 comments Transcript: now what products will we get if we mix toluene with nitric acid and sulfuric acid so here's toluene is basically a benzene ring with a methyl group and so in this reaction this is going to be the nitration of toluene so what products can we get in this reaction now the methyl group is a weakly activating group it's also an ortho para director so it's going to direct the incoming no.2 group either to the ortho position or to the para position and so what we're going to get is a mixture of products and so we can get the ortho product and we can also get the para product so which of these two products will be the major product is it going to be the ortho product or the para product what would you say well it turns out that you get a good mixture of both products but it's not an equal 50/50 a mixture rather the ortho product will exceed the para product in this example and the ratio is about 60% of the ortho product and 40% of the para product so the methyl group because it's not bulky it's gonna favour the ortho product over the para product but you still get a good mix of both now what's gonna happen if we replace the methyl group with a bulkier let's say fo Group how will that affect the ortho para ratio for the nitration of ethyl benzene so we're still going to get a good mixture of both products just like before we're going to get the ortho product and we're also going to get the para product so which of these two products do you think will be the major product is it ortho para or do you think they will be the same in this example turns out that for this reaction there is no major or minor product the ortho and the para product they existed about the same yield 50/50 and so for ethyl benzene we get a good mixture of both so notice what happened when we replaced the methyl group with an ethyl group so by replacing the methyl group with four more bulkier or a group that's more sterically hindered the orthopod ik its yield went down and the yield of the para product went up and so keep that in mind if you use a very bulky substituent then the para product will be favored but if you use a substituent that is not bulky like the methyl group then you're going to get a good yield for the auto product as well so the next example will illustrate this so this time we're gonna use a bulky tert-butyl group and we're gonna be acting with nitric acid and sulfuric acid and so in this case it's very clear to see which product will be the major product what do you think the numbers will be for this reaction so we know it's not going to be a 50/50 ratio anymore what do you think the yields would be in this example so this is the ortho product and here is the para product so without knowing the numbers you know that the para product will be the major product because this site is more accessible whereas this site is 2 sterically hindered and still they yield for the ortho site will be low so the ortho product is going to be the minor product in this example and according to one textbook the ortho ratio in this example is roughly about 20% and a pair of ratio is 80% so when using bulky groups the para product will be the major product and if you use group that's not bulky like a methyl group then the ortho product will be slightly above the para product but you still get a good mixture of both now let's go over some example problems so for these problems I want you to draw all the possible products that can form in this reaction and identify which one is the major product and which one is the minor product so here we have a benzene ring with a tert-butyl group and a methyl group so where will the no.2 go now the methyl group wants the no.2 group to go ortho or para with respect to it so these two positions are ortho with respect to the methyl group and this is para but para is occupied by the tert-butyl group so it won't go there now the left side and the right side would lead to the same product so we're not going to worry about the left side we're just going to focus on the right side so the methyl group wants the no.2 group to go there now the tert-butyl group is also an alkyl group so it's an ortho para director so the tert-butyl group wants the no.2 group to go here but it can't go there and so these two they compete with each other and they're both weakly activating groups so the no.2 can go in both of those two positions and so we're going to get a mixture of products so here's the first product where the no.2 group is very close to the syrup you and then for the second product let's put it over here we're gonna put the no.2 group or fill with respect to the methyl group so these are the two products that we can get in this reaction now which one is going to be the major product and which one is going to be the minor product well this carbon highlighted in red is more accessible because the methyl group is less sterically hindered than the tert-butyl group so therefore this is going to be the major product and this is going to be the minor product we can still get it however not in good deal because this tert-butyl group is very bulky we have a carbon with three methyl groups on that carbon and so it's very difficult for the nitronium ion to approach at this position so this is going to be the major product due to steric effects now let's consider one more example so this time the tert-butyl group and the methyl group will be ortho with respect to each other compared to the last example where they were para with respect to each other and so let's reactor with nitric acid and sulfuric acid again so let's focus on the methyl group using the same color red now the methyl group which is right here it wants to direct the income in no.2 group to the ortho position so it can be on either side because those two sides will lead to a different product or the para position with respect to the methyl group now the tert-butyl group also wants the no.2 group to go in the same position ortho with respect to it or para so in this case the methyl group and the tert-butyl group they're working with each other and rather than against each other so now the only thing we'll need to consider is steric effects now let's call this site a and this is going to be site B and C which of these three sites is most accessible to the incoming no.2 group or the nitronium ion C is the least accessible because it's between a bulky tert-butyl group and a methyl group now a is the most accessible because it's very far away from the tert-butyl group b is close to the tert-butyl group but it's not between the tert-butyl group and the methyl group so B is better than seeing therefore a is the most accessible carbon to the electrophile B is in the middle and C is the worse so the major product will be this one where we place the no.2 group at position a because that is the most accessible carbon and then B is going to give us the minor product which let me put this here now ask for products see because it's so sterically hindered this product will be negligible if it does form the yield will be very very low so we don't need to write it so therefore product a is the major product product B is the minor product so these two products they will be significant in the reaction the product C will be insignificant so it's not going to be observed and so that's it for this video hopefully it gave you a good understanding of the ortho/para ratio and how to tell which product will be the major product looking at steric effects thanks for watching
4862	https://asm.matweb.com/search/specificmaterial.asp?bassnum=mq316a	\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| AISI Type 316 Stainless Steel, annealed sheet Subcategory: Ferrous Metal; Metal; Stainless Steel; T 300 Series Stainless Steel Key Words: UNS S31600, SS316, 316SS, AISI 316, DIN 1.4401, DIN 1.4408, DIN X5CrNiMo17122, TGL 39672 X5CrNiMo1911, TGL 7143X5CrNiMo1811, ISO 2604-1 F62, ISO 2604-2 TS60, ISO 2604-2 TS61, ISO 2604-4 P60, ISO 2604-4 P61, ISO 4954 X5CrNiMo17122E, ISO 683/13 20, ISO 683/13 20a, ISO 6931 X5CrNiMo17122, JIS SUS 316 \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| --- \| Component \| Wt. % \| \| \| \| \| \| \| \| \| \| C \| 0.08 \| \| \| Cr \| Max 18 \| \| \| Fe \| 62 \| \| \| \| \| \| \| --- \| Component \| Wt. % \| \| \| \| \| \| \| \| \| \| Mn \| 2 \| \| \| Mo \| Max 3 \| \| \| Ni \| Max 14 \| \| \| \| \| \| \| --- \| Component \| Wt. % \| \| \| \| \| \| \| \| \| \| P \| 0.045 \| \| \| S \| 0.03 \| \| \| Si \| 1 \| \| \| Material Notes: Molybdenum content increases resistance to marine environments. High creep strength at elevated temperatures and good heat resistance. Biocompatible. Fabrication characteristics similar to Types 302 and 304. Applications: food and pharmaceutical processing equipment, marine exterior trim, surgical implants, and industrial equipment that handles the corrosive process chemicals used to produce inks, rayons, photographic chemicals, paper, textiles, bleaches, and rubber. Corrosion Resistance: better corrosion resistance than 302 and 304; resists sodium and calcium brines; hypochlorite solutions, phosphoric acid; and the sulfite liquors and sulfurous acids used in the paper pulp industry. \| \| \| \| \| --- --- \| \| Physical Properties \| Metric \| English \| Comments \| \| \| \| \| \| \| \| \| \| \| Density \| 8 g/cc \| 0.289 lb/in³ \| \| \| Mechanical Properties \| \| \| \| \| \| \| \| \| \| \| \| \| \| Hardness, Rockwell B \| 79 \| 79 \| \| \| Tensile Strength, Ultimate \| 580 MPa \| 84100 psi \| \| \| Tensile Strength, Yield \| 290 MPa \| 42100 psi \| \| \| Elongation at Break \| 50 % \| 50 % \| in 50 mm \| \| Modulus of Elasticity \| 193 GPa \| 28000 ksi \| in tension \| \| Charpy Impact \| 105 J \| 77.4 ft-lb \| V-notch \| \| Izod Impact \| 129 J \| 95.1 ft-lb \| \| \| Electrical Properties \| \| \| \| \| \| \| \| \| \| \| \| \| \| Electrical Resistivity \| 7.4e-005 ohm-cm \| 7.4e-005 ohm-cm \| at 20ºC \| \| Magnetic Permeability \| 1.008 \| 1.008 \| at RT \| \| Thermal Properties \| \| \| \| \| \| \| \| \| \| \| \| \| \| CTE, linear 20°C \| 16 µm/m-°C \| 8.89 µin/in-°F \| 0 - 100ºC \| \| CTE, linear 250°C \| 16.2 µm/m-°C \| 9 µin/in-°F \| at 0-315°C (32-600°F) \| \| CTE, linear 500°C \| 17.5 µm/m-°C \| 9.72 µin/in-°F \| 0 - 540ºC \| \| Specific Heat Capacity \| 0.5 J/g-°C \| 0.12 BTU/lb-°F \| from 0-100°C (32-212°F) \| \| Thermal Conductivity at Elevated Temperature \| 16.3 W/m-K \| 113 BTU-in/hr-ft²-°F \| 100ºC \| \| Melting Point \| 1370 - 1400 °C \| 2500 - 2550 °F \| \| \| Solidus \| 1370 °C \| 2500 °F \| \| \| Liquidus \| 1400 °C \| 2550 °F \| \| \| Maximum Service Temperature, Air \| 870 °C \| 1600 °F \| Intermittent Service \| \| Maximum Service Temperature, Air \| 925 °C \| 1700 °F \| Continuous Service \| References for this datasheet. Some of the values displayed above may have been converted from their original units and/or rounded in order to display the information in a consistant format. Users requiring more precise data for scientific or engineering calculations can click on the property value to see the original value as well as raw conversions to equivalent units. We advise that you only use the original value or one of its raw conversions in your calculations to minimize rounding error. We also ask that you refer to MatWeb's disclaimer and terms of use regarding this information. MatWeb data and tools provided by MatWeb, LLC. \| \| \| \|
4863	https://study.com/academy/lesson/simplifying-complex-rational-expressions.html	Simplifying Complex Rational Expressions \| Steps & Examples - Lesson \| Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Copyright Math Courses Algebra I: High School Simplifying Complex Rational Expressions \| Steps & Examples Contributors: Antonette Dela Cruz, Yuanxin (Amy) Yang Alcocer Author Author: Antonette Dela Cruz Antonette Dela Cruz is a veteran teacher of Mathematics with 25 years of teaching experience. She has a bachelor’s degree in Chemical Engineering (cum laude) and a graduate degree in Business Administration (magna cum laude) from the University of the Philippines. She’s currently teaching Analysis of Functions and Trigonometry Honors at Volusia County Schools in Florida. Instructor Instructor: Yuanxin (Amy) Yang Alcocer Amy has a master's degree in secondary education and has been teaching math for over 9 years. Amy has worked with students at all levels from those with special needs to those that are gifted. Understand the steps for simplifying complex rational expressions. Explore the instances where rational fractions and complex rational expressions are relevant. Updated: 11/21/2023 Table of Contents Complex Rational Expressions Examples and Applications of Complex Rational Expressions Simplifying Complex Rational Expressions Examples of Simplifying Complex Rational Fractions Lesson Summary Show FAQ What are the 3 steps on simplifying rational expression? STEP 1: Simplify the numerator and the denominator separately by converting each into one fraction using LCD (least common denominator). STEP 2: Turn the division of fractions into a multiplication operation to simplify them into one single fraction. (Keep the numerator fraction and multiply to the reciprocal of the denominator fraction) STEP 3: Remove common factors from the numerator and denominator to get to the simplest terms. How do you simplify complex rational expressions? Simplifying a complex rational expression turns the two fractions of the numerator and denominator into one single fraction. This is by converting the division of two fractions into a multiplication operation by getting the product of the numerator fraction and the reciprocal (flipped) of the denominator fraction. The last step is removing the resulting numerator and denominator's common factors. Create an account LessonTranscript VideoQuizCourse An error occurred trying to load this video. 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Your next lesson will play in 10 seconds 0:03 What Is a Complex… 0:46 Rewriting the Problem 1:36 Simplifying the Problem 2:22 Another Example 3:57 Lesson Summary QuizCourseView Video OnlySaveTimeline 44K views Recommended lessons and courses for you Related LessonsRelated Courses ##### Dividing Functions: Examples & Overview 6:16 ##### How to Subtract Fractions with Variables \| Examples 5:55 ##### How to Rationalize the Denominator with a Radical Expression 3:52 ##### How to Solve Complex Fractions 5:20 ##### Rationalizing the Numerator \| Steps & Examples 5:30 ##### How to Find an Unknown in a Proportion 6:06 ##### Partial Fractions \| Formula, Solution & Examples 4:44 ##### Partial Fraction Decomposition \| Equations & Examples 14:09 ##### Least Common Multiple \| LCM Overview & Examples 10:56 ##### Solving Word Problems with Algebraic Division Expressions 2:54 ##### Division and Reciprocals of Radical Expressions 5:53 ##### Converting Repeating Decimals to Fractions \| Overview & Examples 7:06 ##### Multiplying then Simplifying Radical Expressions 3:57 ##### Simplifying Square Roots \| Overview & Examples 4:49 ##### Simplifying Square Root Expressions \| Steps & Examples 7:03 ##### How to Divide Radicals, Square Roots & Rational Expressions 7:07 ##### Factoring Polynomial Expressions \| Definition, Methods & Examples 6:18 ##### Rational Equations \| Definition, Formula & Examples 7:58 ##### How to Solve and Graph And & Or Compound Inequalities 7:36 ##### Direct & Inverse Variation \| Equations, Relationships & Problems 6:27 ##### Algebra II: High School ##### CUNY Assessment Test in Math: Practice & Study Guide ##### MHT-CET: Practice & Study Guide ##### GED Math: Quantitative, Arithmetic & Algebraic Problem Solving ##### NY Regents Exam - Integrated Algebra: Test Prep & Practice ##### Prentice Hall Algebra 2: Online Textbook Help ##### Glencoe Algebra 1: Online Textbook Help ##### McDougal Littell Pre-Algebra: Online Textbook Help ##### McDougal Littell Algebra 2: Online Textbook Help ##### McDougal Littell Algebra 1: Online Textbook Help ##### Glencoe Pre-Algebra: Online Textbook Help ##### Linear Algebra: Help & Tutorials ##### Prentice Hall Pre-Algebra: Online Textbook Help ##### Accuplacer Math: Quantitative Reasoning, Algebra, and Statistics Placement Study Guide and Test Prep ##### Algebra 2 and Trig Textbook ##### College 101: College Readiness ##### Saxon Algebra 2 Homeschool: Online Textbook Help ##### Saxon Algebra 1 Homeschool: Online Textbook Help ##### Saxon Algebra 1/2 Homeschool: Online Textbook Help ##### Test for Admission into Catholic High Schools (TACHS) Study Guide and Exam Prep Complex Rational Expressions ---------------------------- Math is like a building block and depends on a solid foundation for the basic concepts. In the number system, a rational number is a ratio of two integers with a numerator (top portion) and a denominator (bottom part). Rational numbers are also called fractions. When the numerator and denominator are polynomials, the fraction is also called a rational fraction or rational expression. It becomes a complex rational expression when it is a fraction of fractions. That is, a complex rational expression is a fraction of rational expressions. In real-life applications, there are formulas used that have the same format as complex rational expressions and require the same algebraic handling. The finance formula for solving monthly payments for a loan is a complex rational expression. Also, the formula for solving combined resistances connected in parallel has fractions in the denominator and is another application. To unlock this lesson you must be a Study.com memberCreate an account Examples and Applications of Complex Rational Expressions --------------------------------------------------------- Fig. 1 Example of rational numbers, rational expressions, and complex rational expressions Shown in Fig. 1 are examples of rational functions, rational expressions, and complex rational expressions. The basic format of all three types is based on a core fraction. Thus, the mathematics of simplifying will use the tools in algebraic manipulations of fractions. The difference between having unknown variables and not having them lies in the conditions set for the values allowed for the unknown variable. Since a fraction can never have zero in the denominator, conditions are set for the unknown variable. The possible values of the unknown variable in the denominator are all real numbers that will not result in a denominator equal to zero. That is, if the final result is {eq}\frac{3}{x+1} {/eq} Applications using complex rational expressions are abundant, especially in finance and physics. Among these real-life applications are the following: Application 1 Using of formula to calculate annuity to get monthly payments on a loan: {eq}A=\frac{pi}{1-\frac{1}{(1+i)}^{n}} {/eq}, where {eq}p {/eq} is the principal (the amount borrowed), and {eq}i =\frac{r}{n} {/eq}, where {eq}r {/eq} is the interest rate expressed as a decimal and {eq}n {/eq} is the number of payments per year. Application 2 Calculation of combined resistances connected in parallel circuits in physics: Combined Resistance = {eq}=\frac{1}{\frac{1}{R_{1}}+\frac{1}{R_{2}}} {/eq}, where {eq}R_{1} {/eq} and {eq}R_{2} {/eq} are the individual resistances. Application 3 Determining the focal length of lenses: {eq}f =\frac{1}{\frac{1}{p}+\frac{1}{q}} {/eq}, where {eq}p {/eq} is the distance that the object is from the lens and {eq}q {/eq} is the distance that the image is from the lens. To unlock this lesson you must be a Study.com memberCreate an account Simplifying Complex Rational Expressions ---------------------------------------- In general, listed are the three main steps in simplifying complex rational expressions: STEP 1: Convert the numerator and denominator to one single fraction. If there are mixed fractions ( whole number and fraction), convert the whole number into a fraction with the same denominator as the other fraction. Combine the two fractions. STEP 2: Carry out the division of the two fractions. Multiply the numerator fraction by the reciprocal of the denominator fraction. STEP 3: Divide both numerator and denominator by the common factors until the result is at its simplest terms. Be sure to add variable conditions so that the denominator will never be zero. Example 1 Simplifying complex rational expression without variables To simplify the following: {eq}\frac{2 +\frac{2}{3}}{\frac{3}{5}} {/eq}, the steps are as follows: STEP 1: The numerator has to be rewritten as a single fraction. {eq}2+\frac{2}{3}= 2\cdot \frac{3}{3}+\frac{2}{3}=\frac{6}{3}+\frac{2}{3}=\frac{8}{3} {/eq} STEP 2: The division of the fractions will be carried out. {eq}\frac{\frac{8}{3}}{\frac{3}{5}}=\frac{8}{3}\cdot \frac{5}{3}=\frac{40}{9} {/eq} STEP 3: Check for common factors and simplify. The {eq}\frac{40}{9} {/eq} is in its simplest terms, so this is the final answer in fraction form. Example 2 Simplifying complex rational expressions with variables. The next example has unknown variables: {eq}\frac{x+y}{\frac{4}{7}} {/eq} STEP 1: Both numerator and denominator are already in fraction form. The numerator is rewritten as a fraction by putting 1 in the denominator. {eq}x+y = \frac{x+y}{1} {/eq} STEP 2: Division of the two fractions is carried out. Division of the fractions is multiplying the numerator fraction by the reciprocal of the denominator fraction. {eq}\frac{(x+y)}{1} \div \frac{4}{7} = \frac{(x+y)}{1} \cdot \frac{7}{4}=\frac{7x+7y}{4} {/eq} STEP 3: Check for common factors for the numerator and denominator. In this example, there is no common factor, and the fraction is already in its simplest form. There are also no limitations or conditions for the variables since they will not result in a zero in the denominator. Final answer will be {eq}\frac{7x+7y}{4} {/eq}, where {eq}x {/eq} and {eq}y {/eq} are all real numbers. To unlock this lesson you must be a Study.com memberCreate an account Examples of Simplifying Complex Rational Fractions -------------------------------------------------- Additional examples of complex rational expressions are shown, simplified completely: Example 3 {eq}\frac{\frac{2}{7}}{1 +\frac{2}{3}} {/eq} STEP 1: The denominator has a mixed fraction that needs to be rewritten: {eq}1+\frac{2}{3}= 1\cdot \frac{3}{3}+\frac{2}{3}=\frac{3}{3}+\frac{2}{3}=\frac{5}{3} {/eq} STEP 2: Division is carried out on the numerator and the denominator fractions. {eq}\frac{\frac{2}{7}}{\frac{5}{3}}=\frac{2}{7}\cdot \frac{3}{5}=\frac{6}{35} {/eq} STEP 3: There are no common factors to simplify the fraction further. Example 4 {eq}\frac{\frac{x^{2}}{y}}{\frac{x}{y^{2}}} {/eq} STEP 1: Both numerator and denominator are already in fraction form. STEP 2: The Division of the fractions is calculated by multiplying the reciprocal of the denominator fraction by the numerator fraction. {eq}\frac{x^{2}}{y}\cdot \frac{y^{2}}{x} {/eq} STEP 3: There are common factors in the numerator and denominator that may be simplified. {eq}\frac{x^{2}}{y}\cdot \frac{y^{2}}{x}=x\cdot y= xy {/eq} Example 5 {eq}\frac{1+\frac{x}{y}}{y+\frac{2x^{2}}{3y}} {/eq} STEP 1: Convert the numerator and the denominator into single fractions. for the numerator: {eq}1+\frac{x}{y}=1\cdot \frac{y}{y}+\frac{x}{y}=\frac{y+x}{y} {/eq} for the denominator: {eq}y+\frac{2x^{2}}{3y}=y\cdot \frac{3y}{3y}+\frac{2x^{2}}{3y}=\frac{3y^{2}}{3y}+\frac{2x^{2}}{3y}=\frac{3y^{2}+2x^{2}}{3y} {/eq} STEP 2: Divide the two fractions by multiplying the reciprocal of the denominator by the numerator fraction. {eq}\frac{y+x}{y}\cdot \frac{3y}{3y^{2}+2x^{2}} {/eq} STEP 3: Remove common factors of numerators and denominators {eq}\frac{y+x}{1}\cdot \frac{3}{3y^{2}+2x^{2}}=\frac{3(y+x)}{3y^{2}+2x^{2}}=\frac{3y+3x}{3y^{2}+2x^{2}} {/eq} To unlock this lesson you must be a Study.com memberCreate an account Lesson Summary -------------- Complex rational expressions are fractions of fractions. The numerator and the denominator are also fractions. Simplifying complex rational expressions follow these three steps: (1) converting the numerator and denominator into single fractions, (2) performing division on the fractions, and (3) simplifying the numerator and denominator by removing common factors. To do the first step, the terms must be in the same denominator. This is carried out using the least common denominator or multiplying a whole number by a fraction equal to one and giving the number a denominator. The second step is multiplying the numerator fraction by the reciprocal of the denominator fraction. Getting the reciprocal is simply flipping the fraction upside down. The third step is finding common factors in the numerator and denominator so the fraction can be simplified to its lowest terms. Various applications in real life use complex rational expressions. In physics, the formula of combined resistances of a parallel circuit is a complex rational expression. The formula for annuity in finance also is an example of a complex rational expression. Knowing the proper way to simplify these rational expressions may be very beneficial in determining variables needed for real-life applications. To unlock this lesson you must be a Study.com memberCreate an account Video Transcript What Is a Complex Rational Expression? If you thought fractions were scary before, a complex rational expression will scare you even more! What is it? A complex rational expression is a fraction of fractions. So we have a fraction in our numerator and a fraction in our denominator. That is a complex rational expression. This is what we will be talking about in this video lesson, but don't run away just yet! I will show you that it's not quite as scary as you think. I will show you just how easy it can be to simplify them and get them reduced to something that you can easily work with. The process that we will be using to tame these fraction monsters is a two-step process that involves rewriting our problem and then simplifying. Are you ready to get started? Let's go! Rewriting the Problem Let's start with a problem that includes only numbers, so we can see how the process works and how easy it is to use. First problem. Don't get scared. This fraction monster won't bite. It might look tough and mean, but it's actually quite soft on the inside. Let's open this fraction up and see what's on the inside. We begin by rewriting our fraction. We know that fractions are actually division problems. So we actually have 4/5 divided by 10/6. We also know that when we divide by a fraction, we can actually turn it into a multiplication problem by flipping the second fraction. So our 4/5 divided by 10/6 becomes 4/5 times 6/10. We have flipped our second fraction. So this is our rewritten problem: Rewritten problem. Simplifying the Problem Ah, multiplication! We can do multiplication easily with two fractions, can't we? Yes, we simply multiply across our numerator and multiply across our denominator. But before we do so, can we simplify any of these numbers? Is there a number in the numerator and the denominator that share a common factor? Yes, there is. The 4 in the numerator and the 10 in the denominator both can be divided by 2. So we can simplify the 4 to 2 and the 10 to 5 by dividing both of these numbers by 2. So, now our problem is 2/5 times 6/5. Now we can perform our multiplication to get 12/25. This is our final answer, and we are done. That wasn't so bad, was it? Another Example Now, let's look at another example. This time we will see a problem with variables because many problems you will see will involve variables. Second problem. Okay, so this problem is a little bit more complicated, since our denominator isn't a nice-looking fraction now. So, we need to turn our denominator into a fraction. We do this by adding our y/x and our 3. We recall that we need a common denominator before adding numerators. We see that our common denominator needs to be x, so the 3 needs to be turned into 3 x/x. Now, doing the addition, we get (y + 3 x)/x. We can rewrite this for our denominator in our problem. Now we can go ahead and rewrite our problem as the division of two fractions. We remember that we can turn the division into a multiplication by flipping the second fraction. So our rewritten problem is 3/x times x/(y + 3 x). Now, we can simplify by canceling anything that both the numerator and denominator have in common. We see that there is an x in the numerator and an x in the denominator. So we can cancel out the x's. Now we can go ahead and multiply the rest to get 3/(y + 3 x). This is our answer. We can't simplify any more. So, this problem is a bit more complicated, but in the end, it's not so bad when we take it step by step. Last Example Now, let's look at an example that has more than one term in both the numerator and denominator. Third problem. We need to turn both the numerator and denominator into fractions. To turn the numerator into a fraction, we need to rewrite the 4 as 4 x/x so that we can subtract our fractions. Now, doing the subtraction, we get (4 x - 3)/x. To turn the denominator into a fraction, we need to rewrite the 9 x as 9 x 2/x so that we can add our fractions. Now, doing the addition, we get (y + 9 x 2)/x. Now we can go ahead and rewrite our problem as the division of two fractions. Writing as Division of Two Fractions. We remember that we can turn the division into a multiplication by flipping the second fraction. So our rewritten problem is (4 x - 3)/x times x/(y + 9 x 2) We can simplify by canceling anything that both the numerator and denominator have in common. The denominator of the first fraction is x and the numerator of the second fraction is x, so we can cancel out the x's. Now we can go ahead and multiply the rest to get (4 x - 3) / (y + 9 x 2). This cannot be simplified any more and so it is our answer. Lesson Summary We are now done. So, what have we learned? We've learned that a complex rational expression is a fraction of fractions. To simplify them, we first turn the problem into a division problem. We rewrite the top fraction divided by the bottom fraction. Next, we turn the division into a multiplication by flipping the second fraction. Then we look for common terms we can cancel in both the numerator and denominator. Then we perform the multiplication, and we are done! Learning Outcomes Once you've completed this lesson, you should be able to: Define 'complex rational expression' Explain how to simplify complex rational expressions Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Unlock your education See for yourself why 30 million people use Study.com Become a Study.com member and start learning now. 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Try it now Algebra I: High School 19 chapters 163 lessons 1 flashcard sets Chapter 1 High School Algebra: Solving Math Word Problems Solving Word Problems: Steps & Examples 5:54 minSolving Word Problems with Multiple Steps 5:30 minRestating Word Problems Using Words or Images 4:32 minPersonalizing a Word Problem to Increase Understanding 4:50 min Chapter 2 High School Algebra: Percent Notation Percentage \| Definition & Calculation 4:20 minConverting Percent to Decimal \| Overview & Examples 2:37 minConverting Percent Notation to Fraction Notation 3:16 minChanging Between Decimals and Percents 4:53 minDecimal to Fraction \| Conversion & Examples 7:32 minConverting Fractions to Percents 3:43 min Chapter 3 High School Algebra: Calculations, Ratios, Percent & Proportions Ratios & Rates \| Differences & Examples 6:37 minHow to Solve Problems with Money 8:29 minProportion \| Definition, Formula & Types 6:05 minCalculations with Ratios and Proportions 5:35 minPercents: Definition, Application & Examples 6:20 minHow to Solve Word Problems That Use Percents 6:30 minCombination in Mathematics \| Definition, Formula & Examples 7:14 minPermutation Definition, Formula & Examples 6:58 minHow to Solve Problems with Time 6:18 minDistance Equations \| Formula, Calculation & Examples 6:31 min Chapter 4 High School Algebra: Real Numbers Types of Numbers & Its Classifications 6:56 minGraphing Rational Numbers on a Number Line \| Chart & Examples 5:02 minNotation for Rational Numbers, Fractions & Decimals 6:16 minThe Order of Real Numbers: Inequalities 4:36 minFinding the Absolute Value of a Real Number 3:11 min Chapter 5 High School Algebra: Exponents and Exponential Expressions How to Use Exponential Notation 2:44 minScientific Notation \| Definition, Conversion & Examples 6:49 minSimplifying and Solving Exponential Expressions 7:27 minExponential Expressions & The Order of Operations 4:36 minMultiplying Exponents \| Overview, Methods & Rules 4:07 minDividing Exponential Expressions 4:43 minThe Power of Zero: Simplifying Exponential Expressions 5:11 minNegative Exponents: Writing Powers of Fractions and Decimals 3:55 minPower of Powers: Simplifying Exponential Expressions 3:33 min Chapter 6 High School Algebra: Properties of Exponents Properties of Exponents \| Formula & Examples 5:26 minHow to Define a Zero and Negative Exponent 3:13 minSimplifying Expressions with Exponents \| Overview & Examples 4:52 minRational Exponents \| Definition, Calculation & Examples 3:22 minSimplifying Algebraic Expressions with Rational Exponent 7:41 min Chapter 7 High School Algebra: Radical Expressions Square Root \| Definition, Formula & Examples 7:05 minEstimating Square Roots \| Overview & Examples 5:10 minSimplifying Square Roots When not a Perfect Square 4:45 minSimplifying Square Root Expressions \| Steps & Examples 7:03 minDivision and Reciprocals of Radical Expressions 5:53 minRadicands and Radical Expressions 4:29 minEvaluating Square Roots of Perfect Squares 5:12 minFactoring Radical Expressions 4:45 minSimplifying Square Roots of Powers in Radical Expressions 3:51 minMultiplying then Simplifying Radical Expressions 3:57 minHow to Divide Radicals, Square Roots & Rational Expressions 7:07 minSimplifying Square Roots \| Overview & Examples 4:49 minRationalizing the Denominator \| Overview & Examples 7:01 minAddition and Subtraction Using Radical Notation 3:08 minHow to Multiply Radical Expressions 6:35 minSolving Radical Equations \| Overview & Examples 6:48 minSolving Radical Equations with Two Radical Terms 6:00 min Chapter 8 High School Algebra: Algebraic Expressions and Equations What is the Correct Setup to Solve Math Problems?: Writing Arithmetic Expressions 5:50 minCommon Math Formulas \| Overview, Uses & Importance 7:08 minAlgebraic Expression \| Definition, Operations & Examples 5:12 minEvaluating Algebraic Expressions \| Rules & Examples 7:27 minCombining Like Terms in Algebraic Expressions 7:04 minPractice Simplifying Algebraic Expressions 8:27 minNegative Signs and Simplifying Algebraic Expressions 9:38 minWriting Equations with Inequalities: Open Sentences and True/False Statements 4:22 minAlgebra Equations \| Formula, Types & Examples 7:28 minDefining, Translating, & Solving One-Step Equations 6:15 minSolving Equations Using the Addition Principle 5:20 minMultiplication Principle \| Definition, Equations & Examples 4:03 minSolving Equations Using Both Addition and Multiplication Principles 6:21 minCollecting Like Terms On One Side of an Equation 6:28 minSolving Equations Containing Parentheses 6:50 minSolutions to Systems of Equations \| Overview & Examples 4:45 minTranslating Words into Algebraic Expressions \| Phrases & Examples 6:31 minHow to Solve One-Step Algebra Equations in Word Problems 5:05 minSolving Multiple Step Equations \| Explanation, Steps & Examples 5:44 minSolving Algebra Word Problems \| Multi-Step Equations & Examples 6:16 minAlgebra Terms Flashcards Chapter 9 High School Algebra: Algebraic Distribution Distribution in Math \| Definition, Process & Examples 5:39 minDistributing First vs. Adding First: Differences & Examples 6:44 minDistributing Positive and Negative Signs 5:56 minDistributing Algebraic Expressions with Numbers and Variables 7:57 minChanging Negative Exponents to Fractions 6:24 minFractional Exponents \| Definition, Rules & Examples 6:38 minDistribution of More Than One Term in Algebra 6:12 min Chapter 10 High School Algebra: Properties of Functions Function in Math \| Definition & Examples 7:57 minTransformations: How to Shift Graphs on a Plane 7:12 minHow to Add, Subtract, Multiply and Divide Functions 6:43 minDomain & Range of a Function \| Definition, Equation & Examples 8:32 minHow to Compose Functions 6:52 minInverse Functions \| Definition, Methods & Calculation 6:05 minApplying Function Operations Practice Problems 5:17 min Chapter 11 High School Algebra: Working With Inequalities Inequality Signs in Math \| Symbols, Examples & Variation 7:09 minGraphing Inequalities \| Definition, Rules & Examples 7:59 minInequality Notation \| Overview & Examples 8:16 minGraphing Inequalities \| Overview & Examples 12:06 minSolve & Graph an Absolute Value Inequality \| Formula & Examples 8:02 minAbsolute Value Inequalities \| Definition, Calculation & Examples 9:06 minTranslating Math Sentences to Inequalities 5:36 min Chapter 12 High School Algebra: Linear Equations Linear Equations \| Definition, Formula & Solution 7:28 minApplying the Distributive Property to Linear Equations 4:18 minForms of a Linear Equation \| Overview, Graphs & Conversion 6:38 minAbstract Algebraic Examples and Going from a Graph to a Rule 10:37 minUndefined & Zero Slope Graph \| Definition & Examples 4:23 minParallel vs Perpendicular vs Transverse Lines Overview & Examples 6:06 minParallel & Perpendicular Lines \| Equation, Graph & Examples 6:07 minLinear Equation \| Parts, Writing & Examples 8:58 minSystem of Equations in Algebra \| Overview, Methods & Examples 8:39 minHow Do I Use a System of Equations? 9:47 minNonlinear Function \| Definition, Examples & Graphs 6:03 min Chapter 13 High School Algebra: Factoring Factoring in Algebra \| Definition, Equations & Examples 5:32 minFinding the Prime Factorization of a Number \| Meaning & Examples 5:36 minUsing Prime Factorizations to Find the Least Common Multiples 7:28 minEquivalent Expressions and Fraction Notation 5:46 minUsing Fraction Notation: Addition, Subtraction, Multiplication & Division 6:12 minFactoring Out Variables: Instructions & Examples 6:46 minCombining Numbers and Variables When Factoring 6:35 minTransforming Factoring Into A Division Problem 5:11 minFactoring by Grouping \| Definition, Steps & Examples 7:46 min Chapter 14 High School Algebra: Quadratic Equations Quadratic Equation \| Definition, Formula & Examples 5:13 minSolving Quadratics: Assigning the Greatest Common Factor and Multiplication Property of Zero 5:24 minQuadratic Function \| Formula, Equations & Examples 9:20 minHow to Solve Quadratics That Are Not in Standard Form 6:14 minSolving Quadratic Inequalities Using Two Binomials 5:36 min Chapter 15 High School Algebra: Graphing and Factoring Quadratic Equations Tables & Graphs in the Real World \| Uses & Examples 5:50 minScatter Plot Graph \| Overview, Uses & Examples 7:17 minParabola \| Definition & Parabolic Shape Equation 4:36 minTypes of Parabolas \| Overview, Graphs & Examples 6:15 minMultiplying Binomials Using FOIL and the Area Method 7:26 minMultiplying Binomials \| Overview, Methods & Examples 5:46 minFactoring Quadratic Equations Using Reverse Foil Method 8:50 minFactoring Quadratic Equations \| Solution & Examples 7:35 minQuadratic Trinomial \| Definition, Factorization & Examples 7:53 minHow to Complete the Square \| Method & Examples 8:43 minCompleting the Square Practice Problems 7:31 minHow to Solve a Quadratic Equation by Factoring 7:53 min Chapter 16 High School Algebra: Properties of Polynomial Functions Cubic, Quartic & Quintic Equations \| Graphs & Examples 11:14 minAdding, Subtracting & Multiplying Polynomials \| Steps & Examples 6:53 minPolynomial Long Division \| Overview & Examples 8:05 minSynthetic Division of Polynomials \| Method & Examples 6:51 minDividing Polynomials with Long and Synthetic Division: Practice Problems 10:11 minOperations with Polynomials in Several Variables 6:09 min Chapter 17 High School Algebra: Rational Expressions How to Multiply and Divide Rational Expressions 8:07 minMultiplying and Dividing Rational Expressions: Practice Problems 4:40 minAdding & Subtracting Rational Expressions \| Overview & Examples 8:02 minPractice Adding and Subtracting Rational Expressions 9:12 minRational Equations \| Definition, Formula & Examples 7:58 minRational Equations: Practice Problems 13:15 minDivision and Reciprocals of Rational Expressions 5:09 minViewing now Simplifying Complex Rational Expressions \| Steps & Examples 4:37 minUp next Solving Direct Variation \| Equation, Problems & Examples 5:12 min Watch next lessonSolving Equations of Inverse Variation 5:13 min Chapter 18 High School Algebra: Matrices and Absolute Value Matrix in Math \| Definition, Properties & Rules 5:39 minFinding the Determinant of a Matrix \| Properties, Rules & Formula 7:02 minAbsolute Value \| Explanation & Examples 4:42 minAbsolute Value Expression \| Evaluation, Simplification & Examples 5:28 minSolving Absolute Value Functions & Equations \| Rules & Examples 5:26 minAbsolute Value \| Overview & Practice Problems 7:09 minAbsolute Value \| Graph & Transformations 8:14 minGraphing Absolute Value Functions \| Definition & Translation 6:08 min Chapter 19 High School Algebra: Data, Statistics, and Probability Independent & Dependent Events \| Overview, Probability & Examples 12:06 minMeasures of Central Tendency \| Definition, Formula & Examples 8:30 minOrganizing and Understanding Data with Tables & Schedules 6:33 minPie Chart vs. Bar Graph \| Overview, Uses & Examples 9:36 min 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4864	https://therassolution.kleinschmidtgroup.com/ras-post/critical-depth-primer/	Critical Depth Primer - Kleinschmidt Blog Book Forum Videos Critical Depth Primer Written by Chris Goodell \| June 24, 2014 Critical depth is an important hydraulic parameter when evaluating hydraulic modeling results. As we know from our college hydraulics 101 class, water flowing at depths less than critical depth is supercritical and water flowing at depths greater than critical depth is subcritical. Supercritical flow is characterized by relatively shallow depths and high velocities. Subcritical flow is characterized by relatively deep depths and slower velocities. The forces dominating the movement of supercritical flow are inertial, while the forces dominating the movement of subcritical flow are gravitational. The flow regime (subcritical or supercritical) a particular cross section, or series of cross sections is in, can be determined by the Froude Number, , where F = Froude Number, V = Velocity, g = gravitational constant, and d = depth. A Froude Number greater than 1 indicates supercritical flow, a Froude Number less than 1 indicates subcritical. A Froude Number = 1 is considered “critical” and possesses the minimum amount of specific energy (potential energy plus kinematic energy, per unit mass). This is considered an unstable condition in nature and is very rare. If you ever get a Froude Number = 1 in your results, most likely, there is a problem with the computations (i.e. in steady flow, RAS could not come up with a valid solution, so it defaults to critical depth). For those of you who have run steady flow HEC-RAS models, you know that there are three flow regime options for computing a steady flow run: subcritical, supercritical, and mixed flow (both sub and supercritical). If you select RAS to run in subcritical, and somewhere in your system RAS is not able to compute a valid subcritical answer, then it defaults to critical depth and moves on. If you select RAS to run in supercritical, and somewhere in your system RAS is not able to compute a valid supercritical answer, then it also defaults to critical depth and moves on. If you select mixed flow, RAS will compute both a subcritical and supercritical profile and anywhere there is a valid solution for both regimes, RAS will select the one that has the higher specific force value. If, in mixed flow, there are any cross sections that “default to critical depth”, that means there was a problem with RAS obtaining a solution. Usually, this means your cross section spacing is too far apart, you are in an area of rapidly varied flow (with not enough cross sections), or just bad input data. In unsteady flow, contrary to intuition, checking the “Mixed Flow” box in the unsteady flow analysis window does NOT tell RAS to evaluate both sub and supercritical solutions. This is done anyway in unsteady flow-whether “Mixed Flow Regime” is checked or not. Let me repeat: HEC-RAS can compute solutions in both subcritical and supercritical in unsteady flow regardless of whether “Mixed Flow Regime” is checked or not. Checking the “Mixed Flow Regime” box in the unsteady flow analysis window simply uses a stabilizing scheme for situations near critical depth and with large changes in velocity with respect to time (the local acceleration term in the St. Venant Equation). This is described further in It’s very helpful to understand the solution you’ve obtained after running RAS by viewing the water surface profile plot with the “critical depth” variable turned on. This allows you to gage how close you are to critical depth, and more importantly, allows you to quickly evaluate if your solution has defaulted to critical depth anywhere, indicating a problem with the solution. The critical depth variable can be turned on in any of the graphical plots by selecting Options…Variables. Then check the box next to Critical Depth Elevation. When you do this, RAS will plot critical depth, but ONLY at certain locations. Notice the plot from the Single Bridge HEC-RAS example data set. Critical depth (in red) is only plotted downstream of the bridge, and at a couple of cross sections upstream of the bridge. That is because RAS will only compute critical depth if your answer is supercritical, close to critical depth, or RAS is not able to come up with a valid solution (defaulting to critical depth) and at the boundaries. If you would like HEC-RAS to compute critical depth everywhere for you, go to the steady flow analysis window, and select Options…Critical Depth Output Option. Then check the box next to “Critical Always Calculated. Comments Ghanim Khalaf on June 24, 2014 Thank you very much. Maria K on June 26, 2014 Thank you very much for this post! bahnisikha.das513@gmail.com on March 21, 2016 can u please tell where do I find the details of how critical depth is calculated by parabolic method in steady flow analysis Chris Goodell on March 28, 2016 The hydraulic reference manual is a good place to start. Adam S. Kamara on May 24, 2017 Thanks Prof. Goodell. Just completed a project using RAS. Hydraulic is nice to study. Sunday Adebayo on June 9, 2017 Thank You… But have being looking for solution to any "Incomplete data error" when running a steady flow analysis in subcritical flow regime… the errors includes: Station data contains a wall with zero width, found the elevation goes up and then down (or down and then up) without moving over. Please remove wall (take out the middle point) or add some width to the wall. At point(s): 2,3,4,5,6,7,8,…26. No manning n data or friction heights K set Right bank station must be greater than left bank Main Channel length is less than or equal to zero, HEC-RAS requires a positive reach length. Right over bank length is less than or equal to zero, HEC- RAS requires a positive reach. NB: Am working with two rivers that met at a junction and continues flowing. PLS AM NEW TO HEC-RAS, but i need this to do my Masters Thesis. THANKS Mirko on June 9, 2017 Regarding The hydraulic reference manual (v4.1. page2-13), which value of Froude number in Cross Section Output Table is for critical depth, Fr=1 or Fr=0.94? Chris Goodell on June 12, 2017 FR=1. FR = 0.94 is just a quick threshold check for when to determine critical depth by more accurate methods (specific energy method). Richard Smith on June 23, 2017 Great blog. I work a lot with FEMA flood studies where a "no-rise" condition means no rise to the 0.00' comparing pre- and post-project modeling when determining if a conditional letter of map revision (CLOMR) is required. Frequently the post project model will have an identical cross section as the pre-project but will have a different critical depth and this can show as a 0.01 or greater rise which of course is not allowed if trying to achieve a no-rise. The models are all run in subcritical flow mode (a FEMA requirement). I have set both models to parabolic and also multiple methods but although I get different answers the rise still occurs. Adjusting computation tolerances courser does not help with matching the critical depth and adjusting finer appears to make the differences greater. Other identical pre- and post-project cross sections flowing at critical depth match elevations very well. Any thoughts on what is causing this would really be helpful. Thank you. Richard Smith on June 23, 2017 Chris, my bad, I found the top of bank was set differently between the two models. I do see this a lot and usually can find a tiny error between the cross sections and correcting that solved the problem. Chris Goodell on June 23, 2017 Not sure what would cause that other than some difference between your cross sections. You might try copying the cross section from one geometry to the other geometry just to double check there is nothing different. Chris Goodell on June 23, 2017 Glad you found it! morgan omale on August 7, 2017 Hi chris, its morgan from Nigeria… i just succeeded in running an unsteady flow simulation..using my dam breach hydrograph output data as initial boundary condition..(flow hydro graph…but my water level in the profile was constant even the energy line was also constant…How do i manipulate my way through the energy line and water level..although the discharge Q was reducing from upstream downstream because of the dam breach over topping flood..PLEASE HELP ASAP… Chris Goodell on August 7, 2017 Sounds like maybe you have a downstream boundary or control in your model that is keeping the water high, level, and slow. Could be an error too. Check your htab parameters and make sure they cover the full range of ws elevations computed in the simulation. Especially the bridge and culvert htabs. Also make sure your computational interval is small enough. For a dam breach model, it should typically be 30 seconds or less. Nate on August 28, 2017 Chris, do you know the basis for why HEC-RAS (and hydraulic modeling in general) defaults to the flow profile with the higher specific force value at a particular cross-section? Why can't the lower specific force value be just as correct? I've tried searching the HEC-RAS manual and other references but haven't had any luck finding an explanation for why the higher specific force governs. Chris Goodell on August 29, 2017 You might try reading the Specific Force section in Chow's "Open Channel Hydraulics". That's Chapter 3-7 in my version of Chow. It might help explain. Unknown on September 20, 2017 Hi Chris- I am running a Hec-Ras model which I am having a divided flow computed for my cross sections and at one of my sections, just downstream of a box culvert my flow hits super critical and the divided flow is not computed. Can you think of any reason or way to fix this? The program wants to push the entire flow back through the channel rather than use the storage as defined outside of the banks. Any help as quick as possible would be greatly appreciated. Chris Deepak Garggottri Acharya on August 12, 2018 Hi Chris How can I analyze the river with Caving (Over hanging) section by HEC-RAS? Sharif on July 30, 2021 Hi, I have a project in unsteady conditions. I have a flow hydrograph for upstream as the boundary condition. But there is a drop downstream, and the only boundary condition is that critical depth occurs there. How can I model this? Chris Goodell on July 30, 2021 Sharif, I would just use normal depth and adjust your slope until you more or less land on critical depth there. Sharif on July 31, 2021 Thanks a lot. it works. Add Your Comment Leave a Reply Cancel reply You must Register or Login to post a comment. Username or Email Address Password Please enter an answer in digits: seven + 16 = [x] Remember Me Search for Posts Search Welcome to the RAS Solution You’ve arrived at The RAS Solution. The best help site for all things HEC-RAS. Use the Search bar above to find topics you are interested in. Whether you are new to HEC-RAS or a seasoned expert, I think you’ll find a lot of great stuff in here. Check out the HEC-RAS Vodcast(video podcast) I do with Ben Cary called Full Momentum. The Author Chris Goodell, Principal Consultant for H&H at Kleinschmidt Associates, is a former HEC-RAS Development Team member and teaches HEC-RAS courses around the world. He is the author of the popular book “Breaking the HEC-RAS Code” and co-host of the Full Momentum vodcast. 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4865	https://www.gauthmath.com/solution/Locate-point-R-so-that-triangle-CRL-is-an-isosceles-triangle-Can-you-find-more-t-1704594047525893	Solved: Locate point R so that △ CRL is an isosceles triangle. Can you find more than one answer? [Math] Drag Image or Click Here to upload Command+to paste Upgrade Sign in Homework Homework Assignment Solver Assignment Calculator Calculator Resources Resources Blog Blog App App Gauth Unlimited answers Gauth AI Pro Start Free Trial Homework Helper Study Resources Math Questions Question Locate point R so that △ CRL is an isosceles triangle. Can you find more than one answer? Illustrate your answer(s). (3) Show transcript Expert Verified Solution 93%(383 rated) Answer Yes, there are multiple answers. Explanation Recognize that for $$\triangle CRL$$△CR L to be an isosceles triangle, point $$R$$R must be equidistant from points $$C$$C and $$L$$L Understand that the locus of points equidistant from two given points is the perpendicular bisector of the line segment joining those points. Conclude that if point $$R$$R is on the perpendicular bisector of line segment $$CL$$C L, then $$\triangle CRL$$△CR L will be an isosceles triangle. Note that there are infinitely many points on the perpendicular bisector that satisfy the condition, so there are multiple answers. So, point $$R$$R can be located on the perpendicular bisector of line segment $$CL$$C L to form an isosceles triangle $$\triangle CRL$$△CR L. There are multiple answers, as there are infinitely many points on the perpendicular bisector. Helpful Not Helpful Explain Simplify this solution Related For Exercises 12-14, use the graphs below. Can you find more than one answer? 12. Locate a point L so that △ LRY is an isosceles triangle. 13. Locate a point O so that △ MOE is an isosceles right triangle. 14. Locate a point R so that △ CRL is an isosceles right triangle. ㊈ 100% (4 rated) Locate point O so that △ MOE is an isosceles right triangle. List all possible answers. 100% (1 rated) Locate point L so that △ LRY is an isosceles triangle. List all possible answers. _ 100% (1 rated) 7 -7 15. Locate point L so that 16 △ LRY is an isosceles triangle. List all possible answers. tr a 100% (5 rated) What is the relationship between an isosceles triangle and the location of it ry It! Locate the Orthocenter orthocenter? Explain your answer 100% (4 rated) OF MI ture If you ar ngle, explain how What is the relationship between an isosceles triangle and the location of ry It! Locate the Orthocente orthocenter? Explain your answer ices at each of th f a Triangle 100% (2 rated) XAMPLL A Locate the Orthocenter = = -------=- - Try It! 4. What is the relationship between an isosceles triangle and the location of its orthosenter? Explain your answer. 100% (2 rated) AMPLE 4 Try It! Locate the Orthocenter 4. What is the relationship between an isosceles triangle and the location of its orthocenter? Explain your answer. MPLE 5 Try It! Find the Orthocenter 100% (3 rated) Which of the following illustrates an isosceles right triangle? Answer A 56.3 ° 11 2 100% (5 rated) Simplify 6+7i/1-4i giving your answer in the form a+bi. 75% (8 rated) Gauth it, Ace it! contact@gauthmath.com Company About UsExpertsWriting Examples Legal Honor CodePrivacy PolicyTerms of Service Download App
4866	https://ximera.osu.edu/math/calc1Book/calcBook/evt/evt	3.2 Extreme Value Theorem - Ximera Statistics Get HelpContact my instructor Report error to authorsRequest help using XimeraReport bug to programmers Another Math Editor Failed Saved! Saving… Reconnecting…Save Update Erase Edit MeProfileSuperviseLogout Sign InSign In with GoogleSign In with TwitterSign In with GitHub Sign In Warning × You are about to erase your work on this activity. Are you sure you want to do this? No, keep my work.Yes, delete my work. Updated Version Available × There is an updated version of this activity. If you update to the most recent version of this activity, then your current progress on this activity will be erased. Regardless, your record of completion will remain. How would you like to proceed? Keep the old version.Delete my work and update to the new version. Mathematical Expression Editor × +–× ÷ x ⁿ √ⁿ√ π θ φ ρ ()\|\| sin cos tan arcsin arccos arctan e ˣ ln log [?]\blue{[?]}[?] Cancel OK Chapter 1: Limits 1.1 Velocity We compute average velocity to estimate instantaneous velocity. 1.2 Numerical Limits We find limits using numerical information. 1.3 Graphical Limits In this section we use the graph of a function to find limits. 1.4 Computing Limits Compute limits using algebraic techniques. 1.5 Infinite Limits Determine when a limit is infinite. 1.6 End Behavior Find limits at infinity. 1.7 Squeeze Theorem Find limits using the Squeeze Theorem. 1.8 Continuity In this section we learn the definition of continuity and we study the types of discontinuities. 1.9 Intermediate Value Theorem In this section we apply a theoretically important existence theorem called the Intermediate Value Theorem. 1.10 Definition of Derivative In this section we learn the definition of the derivative and we use it to solve the tangent line problem. 1.11 Differentiability We determine differentiability at a point Chapter 2: Derivatives 2.1 Power Rule We learn how to find the derivative of a power function. 2.2 Basic Differentiation Rules The following rules allow us the find the derivative of multiples, sums and differences of functions whose derivatives are already known. 2.3 Derivatives of Natural Exponential and Log In this section we compute derivatives involving and . 2.4 Derivatives of Sine and Cosine In this section we compute derivatives involving and . 2.5 Product Rule We compute the derivative of a product. 2.6 Quotient Rule We compute the derivative of a quotient. 2.7 Chain Rule We compute the derivative of a composition. 2.8 Derivatives of General Exponential and Log In this section we compute derivatives involving and . 2.9 Derivatives of Inverse Trig Functions In this section we compute derivatives involving and . 2.10 Table of Derivatives We learn the derivatives of many familiar functions. 2.11 More Problems Use the differentiation rules to compute derivatives 2.12 Logarithmic Differentiation We use the logarithm to compute the derivative of a function. 2.13 Linear Approximation In this lesson we will use the tangent line to approximate the value of a function near the point of tangency. 2.14 Rectilinear Motion In this section we analyze the motion of a particle moving in a straight line. Our analysis includes the position, velocity and acceleration of the particle. 2.15 Rates of Change In this section we interpret the derivative as an instantaneous rate of change. 2.16 Implicit Differentiation We learn to compute the derivative of an implicit function. 2.17 Related Rates In this section we discover the relationship between the rates of change of two or more related quantities. Chapter 3: Applications 3.1 Critical Numbers In this section we learn to find the critical numbers of a function. 3.2 Extreme Value Theorem In this section we learn the Extreme Value Theorem and we find the extremes of a function. 3.3 Mean Value Theorem We apply the Mean Value Theorem. 3.4 Increasing and Decreasing Functions In this section, we use the derivative to determine intervals on which a given function is increasing or decreasing. We will also determine the local extremes of the function. 3.5 Concavity In this section we learn about the two types of curvature and determine the curvature of a function. 3.6 L’Hopital’s Rule In this section we compute limits using L’Hopital’s Rule which requires our knowledge of derivatives. 3.7 Optimization We find extremes of functions which model real world situations. Chapter 4: Integrals 4.1 Indefinite Integrals In this section we learn to compute general anti-derivatives, also known as indefinite integrals. 4.2 Properties of Indefinite Integrals In this section we examine several properties of the indefinite integral. 4.3 Substitution In this section we learn to reverse the chain rule by making a substitution. 4.4 Riemann Sums We compute Riemann Sums to approximate the area under a curve. 4.5 Definite Integrals In this section we learn to compute the value of a definite integral using the fundamental theorem of calculus. 4.6 The Fundamental Theorem of Calculus In this section we learn to compute the value of a definite integral using the fundamental theorem of calculus. 4.7 Properties of Definite Integrals In this section we use properties of definite integrals to compute and interpret them. 4.8 Applications of Definite Integrals In this section we use definite integrals to study rectilinear motion and compute average value. 4.9 FTC, part II In this section we learn the second part of the fundamental theorem and we use it to compute the derivative of an area function. Review Final Exam Review In this section we prepare for the final exam. math Calculus I, by Andrew Incognito 3.2 Extreme Value Theorem In this section we learn the Extreme Value Theorem and we find the extremes of a function. 1 The Extreme Value Theorem In this section we will solve the problem of finding the maximum and minimum values of a continuous function on a closed interval. Extreme Value Theorem If is continuous on the closed interval , then has both an absolute maximum and an absolute minimum on the interval. It is important to note that the theorem contains two hypothesis. The first is that is continuous and the second is that the interval is closed. If either of these conditions fails to hold, then might fail to have either an absolute max or an absolute min (or both). It is also important to note that the theorem tells us that the max and the min occur in the interval, but it does not tell us how to find them. The main idea is finding the location of the absolute max and absolute min of a continuous function on a closed interval is contained in the following theorem. Fermat’s Theorem Suppose is defined on the open interval . If has an absolute maximum or absolute minimum at a point in the interval, then is a critical number for . As a result of Fermat’s Theorem, we can conclude that the absolute extremes of a continuous function on a closed interval must occur at the endpoints of the interval or at a critical numbers inside the interval. 2 Closed Interval Method Finding the absolute extremes of a continuous function, , on a closed interval is a three step process. Find the critical numbers of inside the interval . 2. Compute the values of at the critical numbers and at the endpoints. 3. The largest of the values from step 2 is the absolute maximum of on the closed interval , and the smallest of these values is the absolute minimum. example 1 Find the absolute maximum and the absolute minimum values of on the closed interval . Solution: The function is a polynomial, so it is continuous, and the interval is closed, so by the Extreme Value Theorem, we know that this function has an absolute maximum and an absolute minimum on the interval . First, we find the critical numbers of in the interval . The derivative is which exists for all values of . Solving the equation gives as the only critical number of the function. Moreover, this critical number is in the interval . By the closed interval method, we know that the absolute extremes occur at either the endpoints, and , or the critical number . Plugging these values into the original function yields: The largest of these values is 5, corresponding to , and the smallest is 1, corresponding to . Thus, we conclude that the absolute maximum of on the closed interval is occurring at the left endpoint and the absolute minimum of in the interval is occurring at the critical number . (problem 1) Find the absolute maximum and the absolute minimum of on the closed interval . The absolute maximum is and it occurs at The absolute minimum is and it occurs at example 2 Find the absolute max and the absolute min of on the closed interval . Solution: First, we find the critical numbers of in the interval . The function is a polynomial, so it is differentiable everywhere. We solve the equation . This becomes which has two solutions and (verify). Hence has two critical numbers in the interval. The absolute extremes occur at either the endpoints, or the critical numbers . Plugging these special values into the original function yields: From this data we conclude that the absolute maximum of on the interval is occurring at both the critical number and the right endpoint and the absolute minimum of in the interval is occurring at the critical number . (problem 2a) Find the absolute maximum and the absolute minimum of on the closed interval . The absolute maximum is and it occurs at The absolute minimum is and it occurs at (problem 2b) Find the absolute maximum and the absolute minimum of on the closed interval . (If the max/min occurs in more than one place, list them in ascending order). The absolute maximum is and it occurs at and The absolute minimum is and it occurs at and example 3 Find the absolute max and the absolute min of on the closed interval . Solution: First, we find the critical numbers of in the interval . The function is a polynomial, so it is differentiable everywhere. We solve the equation . This becomes and the solutions are and (verify). Noting that is not in the interval we see that has two critical numbers in the interval, namely and . The absolute extremes occur at either the endpoints, or the critical numbers . Plugging these special values into the original function yields: From this data we conclude that the absolute maximum of on the interval is occurring at the endpoint and the absolute minimum of in the interval is occurring at the critical number . (problem 3) Find the absolute maximum and the absolute minimum of on the closed interval . The absolute maximum is and it occurs at The absolute minimum is and it occurs at example 4 Find the absolute max and the absolute min of on the closed interval . Solution: First, we find the critical numbers of in the interval . We solve the equation . Using the product rule and the chain rule, we have which simplifies to . Thus we need to solve (verify) and the only solution is (verify). Hence has one critical number in the interval and it occurs at . The absolute extremes occur at either the endpoints, or the critical number . Plugging these special values into the original function yields: From this data we conclude that the absolute maximum of on the interval is occurring at the right endpoint and the absolute minimum of in the interval is occurring at the critical number . Reveal Hint (1 of 1) (1)(problem 4a) Find the absolute maximum and the absolute minimum of on the closed interval . The absolute maximum is and it occurs at The absolute minimum is and it occurs at Reveal Hint (1 of 2) (1)(problem 4b) Find the absolute maximum and the absolute minimum of on the closed interval . use the quotient rule to find The absolute maximum is and it occurs at The absolute minimum is and it occurs at Note that the critical number is not in the interval . Here is a detailed, lecture style video on the Extreme Value Theorem: 3 Proof of Fermat’s Theorem Suppose that is defined on the open interval and that has an absolute max at . Thus, for all in . If is undefined then, is a critical number for . If is defined, then from the definition of the derivative we have The difference quotient in the left-hand limit is positive (or zero) since the numerator is negative (or zero) and the denominator is negative. Thus, . But the difference quotient in the right-hand limit is negative (or zero) and so . Hence, and the theorem is proved. 2025-03-20 20:05:54 ← Previous Next → Courses Calculus OneCalculus TwoCalculus Three About FAQDevelopment TeamWorkshopContact Us Social FacebookTwitterGoogle PlusGitHub Built at The Ohio State University OSU with support fromNSF Grant DUE-1245433, theShuttleworth Foundation, the Department of Mathematics, and the Affordable Learning ExchangeALX. © 2013–2025, The Ohio State University — Ximera team 100 Math Tower, 231 West 18th Avenue, Columbus OH, 43210–1174 Phone: (773) 809–5659 \| Contact If you have trouble accessing this page and need to request an alternate format, contact ximera@math.osu.edu. Start typing the name of a mathematical function to automatically insert it. (For example, "sqrt" for root, "mat" for matrix, or "defi" for definite integral.) 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Symbols Type......to get norm∣∣[?]∣∣\|\|\blue{[?]}\|\|∣∣[?]∣∣ text[?]\text{\blue{[?]}}[?] sym_name[?]\backslash\texttt{\blue{[?]}}[?] abs∣[?]∣\left\|\blue{[?]}\right\|∣[?]∣ sqrt[?]\sqrt{\blue{[?]}}[?] paren([?])\left(\blue{[?]}\right)([?]) floor⌊[?]⌋\lfloor \blue{[?]} \rfloor⌊[?]⌋ factorial[?]!\blue{[?]}![?]! exp[?][?]{\blue{[?]}}^{\blue{[?]}}[?][?] sub[?][?]{\blue{[?]}}{\blue{[?]}}[?][?] frac[?][?]\dfrac{\blue{[?]}}{\blue{[?]}}[?][?] int∫[?]d[?]\displaystyle\int{\blue{[?]}}d\blue{[?]}∫[?]d[?] defi∫[?][?][?]d[?]\displaystyle\int{\blue{[?]}}^{\blue{[?]}}\blue{[?]}d\blue{[?]}∫[?][?][?]d[?] deriv d d[?][?]\displaystyle\frac{d}{d\blue{[?]}}\blue{[?]}d[?]d[?] sum∑[?][?][?]\displaystyle\sum_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}[?]∑[?][?] prod∏[?][?][?]\displaystyle\prod_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}[?]∏[?][?] root[?][?]\sqrt[\blue{[?]}]{\blue{[?]}}[?][?] vec⟨[?]⟩\left\langle \blue{[?]} \right\rangle⟨[?]⟩ mat([?])\left(\begin{matrix} \blue{[?]} \end{matrix}\right)([?]) ⋅\cdot⋅ infinity∞\infty∞ arcsin arcsin⁡([?])\arcsin\left(\blue{[?]}\right)arcsin([?]) arccos arccos⁡([?])\arccos\left(\blue{[?]}\right)arccos([?]) arctan arctan⁡([?])\arctan\left(\blue{[?]}\right)arctan([?]) sin sin⁡([?])\sin\left(\blue{[?]}\right)sin([?]) cos cos⁡([?])\cos\left(\blue{[?]}\right)cos([?]) tan tan⁡([?])\tan\left(\blue{[?]}\right)tan([?]) sec sec⁡([?])\sec\left(\blue{[?]}\right)sec([?]) csc csc⁡([?])\csc\left(\blue{[?]}\right)csc([?]) cot cot⁡([?])\cot\left(\blue{[?]}\right)cot([?]) log log⁡([?])\log\left(\blue{[?]}\right)lo g([?]) ln ln⁡([?])\ln\left(\blue{[?]}\right)ln([?]) alpha α\alpha α beta β\beta β gamma γ\gamma γ delta δ\delta δ epsilon ϵ\epsilon ϵ zeta ζ\zeta ζ eta η\eta η theta θ\theta θ iota ι\iota ι kappa κ\kappa κ lambda λ\lambda λ mu μ\mu μ nu ν\nu ν xi ξ\xi ξ omicron ο\omicron ο pi π\pi π rho ρ\rho ρ sigma σ\sigma σ tau τ\tau τ upsilon υ\upsilon υ phi ϕ\phi ϕ chi χ\chi χ psi ψ\psi ψ omega ω\omega ω Gamma Γ\Gamma Γ Delta Δ\Delta Δ Theta Θ\Theta Θ Lambda Λ\Lambda Λ Xi Ξ\Xi Ξ Pi Π\Pi Π Sigma Σ\Sigma Σ Phi Φ\Phi Φ Psi Ψ\Psi Ψ Omega Ω\Omega Ω × Global settings: Settings ×
4867	https://artofproblemsolving.com/wiki/index.php/1978_IMO_Problems/Problem_2?srsltid=AfmBOoqIBg9gHTOTIRprx-ordkkM4353kVeIasJi4Oryz9bEgsKRfFhP	Art of Problem Solving 1978 IMO Problems/Problem 2 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 1978 IMO Problems/Problem 2 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 1978 IMO Problems/Problem 2 Problem We consider a fixed point in the interior of a fixed sphere We construct three segments , perpendicular two by two with the vertexes on the sphere We consider the vertex which is opposite to in the parallelepiped (with right angles) with as edges Find the locus of the point when take all the positions compatible with our problem. Solution Let be the radius of the given fixed sphere. Let point be the center of the sphere. Let point be the 4th vertex of the face of the parallelepiped that contains points , , and . Let point be the point where the line that passes through intersects the circle on the side nearest to point Let We start the calculations as follows: Therefore, [Equation 1] Using law of cosines: [Equation 2] Using law of cosines again we also get: Since , then [Equation 3] Substituting [Equation 2] and [Equation 3] into [Equation 1] we get: [Equation 4] Now we apply the law of cosines again: Since, and then, [Equation 5] Substituting [Equation 4] into [Equation 5] we get: Notice that all of the terms with cancel and thus we're left with: regardless of . [Equation 6] Now we need to find Since points , , and are on the plane perpendicular to the plane with points , , and , then these points lie on the big circle of the sphere. Therefore the distance can be found using the formula: Solving for we get: [Equation 7] Now we need to get which will be using the formula: [Equation 8] Substituting [Equation 6] and [Equation 7] into [Equation 8] we get: This results in: which is constant regardless of and constant regardless of where points , , and are located as long as they're still perpendicular to each other. In space, this is a sphere with radius which is equal to Therefore, the locus of vertex is a sphere of radius with center at , where is the radius of the given sphere and the distance from the center of the given sphere to point ~ Tomas Diaz. orders@tomasdiaz.com Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page. See Also 1978 IMO (Problems) • Resources Preceded by Problem 11•2•3•4•5•6Followed by Problem 3 All IMO Problems and Solutions Retrieved from " Categories: Geometry Problems Olympiad Geometry Problems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
4868	https://mathworld.wolfram.com/FocalParameter.html	TOPICS Focal Parameter The distance (sometimes also denoted ) from the focus to the conic section directrix of a conic section. The following table gives the focal parameter for the different types of conics, where is the semimajor axis, is the distances from the origin to the focus, and is the eccentricity. \| \| \| conic \| \| ellipse \| \| parabola \| \| hyperbola \| See also Conic Section, Conic Section Directrix, Eccentricity, Focus, Semilatus Rectum, Universal Parabolic Constant Explore with Wolfram\|Alpha More things to try: conic sections 141(2^141) + 1 eigenvectors {{1,0,0},{0,0,1},{0,1,0}} Cite this as: Weisstein, Eric W. "Focal Parameter." From MathWorld--A Wolfram Resource. Subject classifications
4869	https://scienceworld.wolfram.com/physics/LloydsMirror.html	Lloyd's Mirror -- from Eric Weisstein's World of Physics OpticsInterferenceInterferometryGeneral Interferometry Lloyd's Mirror An interferometer consisting of a flat piece of dielectric or metal which serves as a mirror from which one portion of the wave is reflected. The other portion proceeds directly to the screen. The separation of maxima is given by (1) where is the wavelength, but because upon reflection from the mirror, the intensity will be the inverse of that for a double slit interference, namely (2) Beamsplitter, Double Slit Interference © 1996-2007 Eric W. Weisstein
4870	https://www.cuemath.com/calculators/central-angle-calculator/	Central Angle Calculator - How to Calculate Central Angle? We use cookies to improve your experience. Learn more OK Sign up Central Angle Calculator 'Cuemath's Central Angle Calculator' is an online tool that helps to calculate the central angle for a given arc length and radius. What is Central Angle Calculator? Cuemath's online Central Angle calculator helps you to calculate the central angle in a few seconds. How to Use Central Angle Calculator? Please follow the below steps to find the central angle: Step 1: Enter the arc length and radius in the given input box. Step 2:Click on the"Calculate"button to find the central angle for a given arc length and radius. Step 3:Click on the"Reset" button to find the central angle for different values. How to Find Central Angle? A central angle is defined as the angle subtended by an arc at the center of a circle. The radius vectors form the arms of the angle. A central angle is calculated using the formula: Central Angle = Arc length(AB) / Radius(OA) = (s× 360°) /2πr,where 's' is arc length, and 'r' is radius of the circle. Want to find complex math solutions within seconds? Use our free online calculator to solve challenging questions. With Cuemath, find solutions in simple and easy steps. Book a Free Trial Class Solved Example: Find Central Angle if the arc length is 5 units and radius is 6 units? Solution: Central Angle = Arc length(AB) / Radius(OA) = (s× 360°) /2πr =(5× 360°) /2× 3.14× 6 = 1800 / 37.68 = 47.77 Similarly, you can try the calculator to find the central angle for 1) arc length = 11units and radius = 9units 2) arc length = 7units and radius = 8units Related Articles: Central angle Angles Explore math program Math worksheets and visual curriculum Sign up FOLLOW CUEMATH Facebook Youtube Instagram Twitter LinkedIn Tiktok MATH PROGRAM Online math classes Online Math Courses online math tutoring Online Math Program After School Tutoring Private math tutor Summer Math Programs Math Tutors Near Me Math Tuition Homeschool Math Online Solve Math Online Curriculum NEW OFFERINGS Coding SAT Science English MATH ONLINE CLASSES 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math ABOUT US Our Mission Our Journey Our Team QUICK LINKS Maths Games Maths Puzzles Our Pricing Math Questions Blogs Events FAQs MATH TOPICS Algebra 1 Algebra 2 Geometry Calculus math Pre-calculus math Math olympiad Numbers Measurement MATH TEST CAASPP CogAT STAAR NJSLA SBAC Math Kangaroo AMC 8 MATH CURRICULUM 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math FOLLOW CUEMATH Facebook Youtube Instagram Twitter LinkedIn Tiktok MATH PROGRAM Online math classes Online Math Courses online math tutoring Online Math Program After School Tutoring Private math tutor Summer Math Programs Math Tutors Near Me Math Tuition Homeschool Math Online Solve Math Online Curriculum NEW OFFERINGS Coding SAT Science English MATH CURRICULUM 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math MATH TEST CAASPP CogAT STAAR NJSLA SBAC Math Kangaroo AMC 8 ABOUT US Our Mission Our Journey Our Team MATH TOPICS Algebra 1 Algebra 2 Geometry Calculus math Pre-calculus math Math olympiad Numbers Measurement QUICK LINKS Maths Games Maths Puzzles Our Pricing Math Questions Blogs Events FAQs MATH ONLINE CLASSES 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math Terms and ConditionsPrivacy Policy
4871	https://math.stackexchange.com/questions/2264593/minimal-touching-subgraph	graph theory - Minimal 'touching' subgraph - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Minimal 'touching' subgraph Ask Question Asked 8 years, 5 months ago Modified8 years, 5 months ago Viewed 58 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. Whilst playing Rimworld it dawned to me that a part of the game (at least how I play it) was finding an optimal solution to the following problem: Given a finite connected graph V V (just denoting the set of vertices), find a subset W⊂V W⊂V such that W W is connected Every point of V∖W V∖W is adjacent to some point of W W \|W\|\|W\| is minimal with respect to the above two requirements This sounds like it could be a well-studied problem, but I know very little about graph theory and would not know where to look. If it is too hard to approach in general, one could first limit to the case where W W is an n×m n×m grid, with two vertices being adjacent if they lie next to each other vertically or horizontally. From playing with it, I would suspect some cauliflower-like shape appearing. graph-theory Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked May 3, 2017 at 20:53 Bib-lostBib-lost 4,050 16 16 silver badges 44 44 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. Unfortunately, in general this is the connected dominating set problem (Wikipedia link), for which no efficient algorithm is known. As the Wikipedia link mentions, this is related to finding a spanning tree with the maximum number of leaves. More precisely, given a connected dominating set W W, construct a spanning tree T T for the graph by first taking a spanning tree for W W, and then joining every vertex V∖W V∖W to one of its neighbors in W W. Every vertex in V∖W V∖W is a leaf, and no vertex of W W should be a leaf in an optimal solution: that means it's not used to connect to any vertex of V∖W V∖W, so we should remove it. So if we want to minimize \|W\|\|W\|, we want to maximize the number of leaves in T T. For grid graphs, we can get an approximately-optimal solution very easily. In an m×n m×n grid graph, there are m n m n vertices. If k k of them are leaves and m n−k m n−k are non-leaves, then the non-leaf vertices can have degree at most 4 4, so the sum of the degrees is at most k+4(m n−k)=4 m n−3 k k+4(m n−k)=4 m n−3 k. But the sum of the degrees is twice the number of edges, which must be 2 m n−2 2 m n−2. So 2 m n−2≤4 m n−3 k⟹3 k≤2 m n+2⟹k≤2 3 m n+2 3.2 m n−2≤4 m n−3 k⟹3 k≤2 m n+2⟹k≤2 3 m n+2 3. Since W W consists of the nonleaf vertices, \|W\|≥m n−k≥1 3 m n−2 3\|W\|≥m n−k≥1 3 m n−2 3: we must include at least 1 3 1 3 of the vertices in W W. Here's a construction that achieves \|W\|<1 3 m n+m+n\|W\|<1 3 m n+m+n: Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered May 3, 2017 at 21:51 Misha LavrovMisha Lavrov 160k 11 11 gold badges 167 167 silver badges 304 304 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions graph-theory See similar questions with these tags. 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4872	https://www.noaa.gov/sites/default/files/legacy/document/2020/Oct/07354626675.pdf	Ecological Modelling 186 (2005) 196–211 Mean free-path length theory of predator–prey interactions: Application to juvenile salmon migration James J. Andersona,∗, Eliezer Gurariea, Richard W. Zabelb a School of Aquatic and Fishery Sciences, University of Washington, Box 358218, Seattle, WA 98195, USA b Northwest Fisheries Science Center, National Marine Fisheries Service, National Oceanic and Atmospheric Administration, 2725 Montlake Blvd. East Seattle, WA 98112, USA Received 3 February 2004; received in revised form 30 November 2004; accepted 10 January 2005 Available online 12 February 2005 Abstract Ecological theory traditionally describes predator–prey interactions in terms of a law of mass action in which the prey mortality rate depends on the density of predators and prey. This simplifying assumption makes population-based models more tractable but ignores potentially important behaviors that characterize predator–prey dynamics. Here, we expand traditional predator–prey models by incorporating directed and random movements of both predators and prey. The model is based on theory originally developed to predict collision rates of molecules. The temporal and spatial dimensions of predators–prey encounters are determined by deﬁning movement rules and the predator’s ﬁeld of vision. These biologically meaningful parameters can accommodate a broad range of behaviors within an analytically tractable framework suitable for population-based models. We apply the model to prey (juvenile salmon) migrating through a ﬁeld of predators (piscivores) and ﬁnd that traditional predator–prey models were not adequate to describe observations. Model parameters estimated from the survival of juvenile chinook salmon migrating through the Snake River in the northwestern United States are similar to estimates derived from independent approaches and data. For this system, we conclude that survival depends more on travel distance than travel time or migration velocity. Crown Copyright © 2004 Published by Elsevier B.V. All rights reserved. Keywords: Predator–prey; Juvenile salmon; Migration; Survival; Mean free-path length 1. Introduction The ﬁrst predator–prey models (Lotka, 1925; Volter-ra, 1926), and nearly all subsequent predator–prey ∗ Corresponding author. Present address: Columbia Basin Re-search Puget Sound Plaza, 1325 4th Avenue, Ste 1820, Seattle, WA 98101-2509, USA. Tel.: +1 206 543 4772; fax: +1 206 616 7452. E-mail address: jim@cbr.washington.edu (J.J. Anderson). models formulated in terms of differential equations, apply a law of mass action, which has its origins in chemical reaction theory ﬁrst proposed by Waage and Guldberg (1864). In predator–prey models predation events are typically assumed analogous to the com-bining of molecules, such that in its simplest form the predation rate is equal to the product of predator and prey densities and a constant expressing a rate of in-teraction. The models have since evolved (Berryman, 0304-3800/$ – see front matter. Crown Copyright © 2004 Published by Elsevier B.V. All rights reserved. doi:10.1016/j.ecolmodel.2005.01.014 J.J. Anderson et al. / Ecological Modelling 186 (2005) 196–211 197 1992), incorporating such reﬁnements as predator sati-ation (Hollings, 1959), multi-predator and prey species (May, 1974), and dispersion and diffusion (Okubo and Levin, 2001). These improvements characterize the ef-fective densities of predators and prey more realisti-cally, but they ultimately use the Lotka–Volterra model mass action assumption. Models that go beyond the mass action assumption, e.g. ratio-dependent models (Arditi and Ginzburg, 1989) do not allow for variabil-ity in the behavior of predators and prey. It has been noted that the spatial scales of interac-tions between organisms, sometimes referred to as the “ecological neighborhood,” is important in determin-ing population dynamics and in particular predator prey interactions (Addicott et al., 1987; Murdoch et al., 1992; Pascual and Levin, 1999; Hosseini, 2003). Within these local scales of interaction, the movements and distributions of predators and prey in response to each other can signiﬁcantly complicate the interactions and are likely to invalidate many assumptions in simple predator–prey models (Lima, 2002). Peterson and DeAngelis (2000) address this issue by developing an individual-based model (IBM) of juvenile salmon mi-grating through a ﬁeld of piscivores. Simulated pulses of ﬁsh move through a series of cells representing a stretch of river and within each cell there is some probability of encountering predators. The authors ﬁnd that the estimated intensity of predation depends on the size of the cells (DeAngelis and Peterson, 2001). This result is useful in illustrating the importance of spatial effects and the effects of scale issues on a model. From a practical perspective, however, the appropriate cell size is not immediately evident and therefore the system dynamics depend on a free parameter with an uncertain ecological meaning. In general, IBMs can incorporate detailed behaviors and show great promise in exploring predator–prey reactions, but they are limited by difﬁculties in obtaining general results that can be scaled upward to the population level (Pascual and Levin, 1999; Murdoch et al., 1992). The problem of understanding the role of local scales of interaction on population dynamics can be ap-proached by identifying individual level behaviors that are important for determining population dynamics and incorporating these behaviors into population models. Gerritsen and Strickler, 1977 developed a spatially ex-plicit model of predator–prey encounter rates in terms of relative velocities and densities of zooplankton and their prey. Here, we consider this problem from a per-spective somewhat similar to Gerritsen and Strickler’s model, henceforth labeled GS, in which we deﬁne an interaction length scale in terms of fundamental behav-iors of the interacting species. Because we are primarily interested in the survival of migrating prey, we pay par-ticular attention to the interaction between directed and random components of movement. Our approach is motivated by the “mean free-path” theory, which lies at the foundations of kinetic theory as conceived by Maxwell and Boltzmann in the late nine-teenth century. In the kinematic theory of gases, the col-lision rate between molecules depends not only on the relative magnitude of velocities but also on the degree of randomness they exhibit. An emergent property is the mean free-path length between molecular collisions (Feynman et al., 1963). We apply this theory to predator and prey populations by describing their movements in terms of both directed and random components. This allows us to distinguish predator tactics such as “cruis-ing search” and “sit-and-wait” (Hart, 1997). As free-path length theory characterizes the mean features of a gas in terms of the small-scale properties of molecules, the theory applied to animals characterizes large-scale predator–prey dynamics in terms of the small-scale in-teractions of the predators and prey. To demonstrate the practical capabilities of a mean free-path length theory (MFL), we apply the model to data on the mortality of juvenile salmon migrating through the Snake River. During their seaward migra-tion, millions of juvenile salmon are consumed by in-digenous and non-indigenous predators (Ward et al., 1995). This predation is one of several factors that determine the dynamics of salmon populations in the Columbia River Basin (NRC, 1996). Understanding the details of the predator–prey interactions is therefore of practical as well as theoretical interest. We utilize an extensive data set where several hundred thousand spring chinook salmon (Oncorhynchus tshawytscha) were tagged and released at several hatcheries in the Snake River Basin (Fig. 1). After a migration rang-ing from approximately 100 to 800 km, the ﬁsh were detected at a downstream site and mortality for each release group was estimated. We test the results of the MFL model, by comparing the ﬁtted model coefﬁcients to independently estimated values. Finally, we note that the GS model, which described interactions among zooplankton in explicit spheri- 198 J.J. Anderson et al. / Ecological Modelling 186 (2005) 196–211 Fig. 1. Map showing Snake River tributaries and the hatcheries from which ﬁsh were released for survival studies. cal coordinates, yields similar dynamics to the MFL model. This conclusion corroborates a result obtained by Evans (1989). 2. Model of predator–prey interactions The assumptions of the model are as follows: (1) a prey can be described as a point in space that takes a path with mean velocity U and random component u; (2) predators are likewise points with mean velocity V and random component v∗. The directed components of velocity have magnitudes U and V; (3) predators are randomly distributed with a uniform probability dis-tribution in space with density ρ; (4) the scale of the predator–prey encounter volume is described by a con-stant cross-sectional area of interaction α. A predation event, which eliminates the prey, occurs when a preda-tor and prey are within a distance r = √α/π. Over an ensemble of prey and predator–prey inter-actions, the average time between encounters is τ and the probability of a single encounter follows a Poisson distribution such that probability of predator encounter in time = exp(−t/τ) (1) J.J. Anderson et al. / Ecological Modelling 186 (2005) 196–211 199 Fig. 2. Illustration of a prey (arrow) moving a distance x through a unit area A, of predator habitat. Predators (cylinders) have a cross-sectional area α. The probability of encountering a predator while traveling x is equal to the fraction of the unit area covered by predators. where t is the total exposure time. This is also an ex-pression of the survival over time. To derive an expres-sion for τ, ﬁrst deﬁne the chance a prey encounters a predator in traveling a short distance as probability of predator encounter in distance = x λ (2) where λ is the prey’s path length and x is the small distance in the direction the prey travels. Similar to the mean free-path length in molecular collisions (Feynman et al., 1963), the chance of en-counter can also be expressed in terms of the fraction of a unit encounter area occupied by predators. Con-sider a plane of unit area, Aunit = 1, that is perpendicu-lar to the relative average movement between the prey and its predators (Fig. 2). Over a short distance of its path x, the prey sweeps out a small volume Aunitx. This volume contains n predators where n = ρAunitx. The area of the plane covered by predators is Apredator = αn = αρAunitx where α is the encounter area of a predator–prey interaction. The encounter area can be expressed in terms of the encounter distance, r, at which a predation event occurs, so α = πr2. The chance of encountering a predator traveling the small distance x is equivalent to the area occupied by the predators divided by the unit area of the plane, which is simply probability of predator encounter in x = πr2ρx (3) Equating Eqs. (2) and (3) the path length is λ = 1 ρπr2 (4) The encounter time, or mean time between encoun-ters, and path length are related by w, which is the mag-nitude of the relative speed of the prey with respect to the predators. Thus, λ = wτ and the characteristic en-counter time is τ = 1 ρπr2w (5) The relative speed, which we designate the en-counter speed, is expressed as a root-mean-square (rms) speed between the predator and prey as w = E[(v − u)2] (6) Next, represent prey and predator velocity vectors in terms of their average and random parts as u = U + u∗ and v = V + v∗ (7) where U and V are the mean prey and predator velocity vectors and u and v∗ are the associated random or ﬂuc-tuating velocity vectors about the mean vectors. This decomposition into mean and random parts is com-monly used in kinematic studies and in hydrodynamics (Sverdrup et al., 1942). By our deﬁnition, the random components are uncorrelated and have zero means, and so the expected value of the square of the difference be-tween the velocity vectors is E[(v − u)2] = E[U2] − E[2UV] + E[V2] +E[u∗2] + E[v∗2] (8) The rms encounter speed deﬁned by Eq. (6) reduces to w = W2 + ω2 (9) where the squared mean encounter speed is W2 = V2 + U2 − 2UV cos θ (10) where V and U are the magnitudes of vectors V and U, respectively, θ the angle between the mean paths of the predator and prey, and the mean squared random 200 J.J. Anderson et al. / Ecological Modelling 186 (2005) 196–211 encounter speed is described in terms of the variances of v and u as ω2 = E[v∗2] + E[u∗2] (11) Note that ω2 is the sum of the variances of the random parts of the predator and prey speeds and it admits any distribution for the random velocities as long as the mean values are zero. Using Eq. (9) in Eq. (5) to deﬁne τ in Eq. (1), the probability of a prey encountering a predator over time, which is also equivalent to the prey survival, S, over time, is S = exp −t λ W2 + ω2 (12) From Eq. (12) we can express the rate of predation in the form of traditional predator–prey terminology. Noting ρ = P/H where P is the predator population in the habitat of volume H, then the rate of mortality ex-pressed in a Lotka–Volterra type predator–prey equa-tion is dS dt = − πr2 H W2 + ω2PS (13) Noting the predator’s encounter area, r2, has a dimension of area/predator, the combined terms, πr2√ W2 + ω2/H, has as dimension of t−1 predator−1 and is thus equivalent to the time constant in a Lotka–Volterra type predator–prey equation. However, here the rate term is speciﬁed in terms of the preda-tor’s ecological neighborhood through the encounter length scale r, and the predator–prey mean and ran-dom encounter speeds, W and ω. With visual preda-tors, r should depend on the range of visibility in the environment, the sensory ability of the predators, and the near ﬁeld probability of a predator’s chance of cap-turing a prey. The encounter speeds characterize the probabilities of predator–prey encounters over space and time. The model, in the form of Eq. (12) or Eq. (13) has characteristics not found in a formulation in which the rate coefﬁcient is simply a constant. For example, because the predation rate depends on the mean and random speeds of the predator and prey, the predator’s foraging rate is explicitly coupled to its energy expenditure, and so the optimal foraging strategy for the predator may depend on the character of the prey’s speed. This issue was explored by Gerritsen (1984) for the case of random free-swimming aquatic animals. In addition, prey survival over time and space depends on the relative motions of predators and prey. Some special cases illustrating the effect of behaviors on the survival equation are discusses below. 2.1. Special cases of prey survival The general MFL model can be further simpliﬁed to a number of cases in which the predators and prey have different mean and random motions that are germane to ecological situations. If prey migrate through a corridor or ﬁeld of resident predators, then by deﬁnition the predator mean speed is zero (V = 0), and by deﬁnition of the prey’s migration behavior, its mean speed, is simply its average migra-tion velocity U. Then the average prey migration time t, distance x, and velocity are related by x = Ut, and Eq. (12) simpliﬁes to S = exp − 1 λ x 2 + ω 2 t 2 (14) Eq. (14) will be referred to as the XT model because survival depends on both the distance traveled and the travel time. The relative importance of migration dis-tance and time on prey survival depends on the random encounter velocity ω. The length scale λ is a measure of distance between predator–prey encounters and from Eq. (4) it depends on the encounter length scale and the predator density as λ = 1 πr2ρ (15) Whereas, the x term in the equation could equally be replaced with Ut, the dependence on x is appropri-ate for characterizing the survival of animals as they migrate over ﬁxed distances. In this case, the impor-tance of migration velocity, i.e. travel time over a ﬁxed distance, in determining the prey’s survival depends on the magnitude of the random encounter speed relative to the average migration velocity. To illustrate the im-portance of migration velocity, we rewrite Eq. (14) as S = exp − x λ 1 + ω U 2 ߙ (16) and plot S vs. U/ω for migration over an arbitrary dis-tance x = (Fig. 3). The result is a hockey stick like J.J. Anderson et al. / Ecological Modelling 186 (2005) 196–211 201 Fig. 3. XT model survival vs. average migration velocity U, divided by the random component of the velocity ω. survival response versus prey migration rate expressed in terms of the random velocity relative to the directed velocity. The break, at which survival becomes propor-tional to migration velocity, occurs when the migration velocity drops below the random encounter velocity, i.e. ω/U ≥1. If the prey migration is rapid, so the predators are essentially stationary relative to the prey’s migration velocity, then U2 ∼W2  ω2. In this case, prey that move with different average migration velocities expe-rience the same survival probability over a ﬁxed migra-tion distance even though their migration times may be different over the distance. Under these conditions, the asymptotic form of prey survival depends on distance as S = exp −x λ (17) Eq. (17) will be referred to as the gauntlet model and over a ﬁxed migration distance survival is independent of the prey’s velocity. Although the average travel time is related to survival through the average velocity, the equation S = exp(−Ut/λ) is not a suitable description of survival in a gauntlet model since both velocity and travel time are required to characterize survival and their product is simply the migration length x. In this case, survival simply decreases exponentially with dis-tance traveled and is independent of the amount of time it takes to travel the distance. If predators and prey are both mobile and resident within the habitat, then W = 0 and we obtain a tradi-tional exposure model S = exp − ω λ t (18) Note the survival over time depends on the predation length scale characterized by λ, and the intensity of the predator–prey interactions as characterized by ω. In this case, survival declines exponentially with exposure time t. In these special cases, the speciﬁc rate of prey mor-tality, deﬁned (1/S) dS/dt = f ( •)/λ, depends on the re-ciprocal of the predator length scale λ and a rate func-tion f (•) that characterizes the rate of predator–prey encounters. The length scale depends on the number of predators in the habitat and the predator–prey en-counter distance r. The encounter rate function char-acterizes the rate at which predators and prey interact, which in turn depends on their behaviors. In the gaunt-let model, each prey encounters an individual predator at most once while in the exposure model multiple en-counters are possible. 2.2. Comparison with the Gerritsen–Strickler model The Gerritsen and Strickler (1977) predator–prey encounter model for zooplankton is similar to the MFL model in that both characterize a relative speed between predators and prey and describe predation in terms of an encounter distance. Here, we illustrate the similarity between the GS model and Eq. (18) and corroborate a result obtained by Evans (1989). The GS model assumes uniform distributions of predator and prey moving randomly in a three-dimensional ﬁeld. The model characterizes the fre-quency of interaction using a spherical polar coordinate reference frame. The predator and prey have swimming velocities v and u, respectively and their relative speed is w = u2 + v2 − 2uv cos θ (19) where θ is the angle of prey relative to the angle of the vector v of the predator’s motion. Assuming the prey and predators are randomly distributed, the encounter rate of a prey to predators in the GS model is 1 S dS dt = πr2ρ 6 ( ¯ u + ¯ v)3 −\| ¯ u − ¯ v\|3 ¯ u · ¯ v (20) where ¯ u and ¯ v are the mean speeds of the predator and prey populations. The equivalent encounter rate in the 202 J.J. Anderson et al. / Ecological Modelling 186 (2005) 196–211 Fig. 4. Comparison of encounter rate of predators by prey as function of prey and predator speeds for the GS model (—) and the MFL model (– –). MFL model is 1 S dS dt = πr2ρ E (u∗ 2) + E(v∗ 2) (21) Note that the mean speeds in the GS model are equivalent to the rms speeds in the MFL model such that ¯ u = E(u∗2) = urms. Thus, the formulations of the encounter rate by the Eq. (20) for the GS model and (21) for the MFL model are different. Furthermore, in the GS formulation the mean speeds are deterministic, while in the MFL model speeds have probability distributions with intensities characterized by the rms speeds. The encounter rates for the case with probability distributions in the animals’ speeds were numerically calculated by Gerritsen and Stricker. On average, Eq. (20) underestimates the encounter rates by about 1–10% depending on the distributions assumed for the predator and prey speeds. However, even though the formulations are different, their char-acteristics are virtually identical. This is illustrated in Fig. 4, which compares the encounter rates of the two equations for differing predator and prey speeds. Note that in the deterministic form of the GS model, encounter rates are between 0 and 5% lower than the rates of the MFL model. This is the same level of under prediction that Gerritsen and Strickler determined when comparing the deterministic speeds form of the GS model to a numerical form in which the predator and prey speeds were normally distributed. We conclude that the MFL model is a simpler ex-pression than the GS model and that it provides a close ﬁt to the deterministic form of the GS model. The MFL model agrees even more closely with the numer-ical form of the GS model using randomly distributed predator and prey speeds. In a brief discussion of the GS model, Evans (1989) simpliﬁes the GS relative encounter rates and includes a turbulent term that is functionally equivalent to the random velocities with the important distinction of be-ing for both predator and prey. He concludes that the values obtained for encounter rates differ from the GS model by no more than 6%, corroborating the result obtained above. 3. A case study with migrating juvenile salmon We apply the model to data on the survival of juve-nile salmon during their seaward migration. Each year during their migration through the Snake and Columbia Rivers and their tributaries, predators consumed mil-lions of juvenile salmonids (Rieman et al., 1991). The major piscivores are the native northern pikeminnow (Ptychocheilus oregonensis) and three non-indigenous species—smallmouth bass (Micropterus dolomieu), walleye (Stizostedion vitreum) and channel catﬁsh (Ic-talurus punctatus) (Poe et al., 1991). In addition, birds consume migrating juvenile ﬁsh. Notably predation by the Caspian tern (Sterna caspia) has increased over the past decade as a result of a substantial increase in its population (Roby et al., 1998). These predators are generalists that feed seasonally and may employ different tactics so our application of the model to the survival data reﬂects a mixture of dif-ferent foraging strategies by predators. However, the model is ﬂexible and accommodates a variety of be-haviors. Ultimately, the fact that we know the fate of thousands of prey after migrating past a ﬁeld of preda-tors yields a rich data set for our analysis. We use mark recapture studies conducted on Snake River system. Between 1993 and 2003, 287 tagged groups of spring chinook were released from 17 hatcheries located in the tributaries of the Snake River Basin (Fig. 1). The ﬁsh, tagged with passive integrated transponders (PIT) (Prentice et al., 1990), were re-leased from locations in the Snake River tributaries ranging from 31 to 772 km upstream of the detec-tion site at Lower Granite Dam. Release sample sizes ranged from 6 to 51,196 ﬁsh with a median release size of 796 ﬁsh. Fish were released between days of J.J. Anderson et al. / Ecological Modelling 186 (2005) 196–211 203 Fig. 5. Tag group release day of year vs. distance of migration to Lower Granite Dam. the year 71 through 130 and, in general, ﬁsh released further upstream tended to be released earlier (Fig. 5). Survivals to Lower Granite Dam were estimated with the multiple-recapture model for single-release groups (Muir et al., 2001). The relationship between log survival and the mi-gration travel time was variable from one year to the next (Fig. 6). Individually, only 1996–1998, 2001 and 2002 exhibited signiﬁcant (p < 0.01) linear regressions of log(S) versus t, which were weighted by the standard error of S. Correlations were typically low with only 1993 and 1995 exhibiting r2 > 0.5. No relationship be-tween log survival and travel time was evident for all years combined (r2 = 0.08, p < 0.0001). A stronger and consistent relationship was evident between log sur-vival and migration distance (Fig. 7). Weighted regres-sions of log(S) versus x were signiﬁcant (p < 0.005) for all years except 1997. The r2 correlations were above 0.7 for 6 of the 10 years and the regression of all years combined was very signiﬁcant (r2 = 0.65, p < 0.0001). Migration velocities increased with release site dis-tance from Lower Granite Dam. An unweighted linear regression gave u (cm/s) = 1.4 + 0.026x (km) (r2 = 0.70, p < 0.0001). However, there was no relationship be-tween release day and migration velocity (r2 = 0.03, p < 0.0019). Table 1 Regressions coefﬁcients for Eq. (22) coefﬁcients for spring chinook hatchery releases and migration to Lower Granite Dam over the years 1993–2003 Year n r2 a (×10−6 km−2) b (×10−5 day−2) Estimate S.E. Estimate S.E. 1993 17 0.94 3.11 0.34 −3.21 13.78 1994 38 0.97 4.76 0.21 6.22 11.79 1995 43 0.80 7.25 1.00 24.35 28.08 1996 15 0.97 7.96 0.44 −18.04 4.60 1997 12 0.68 1.56 1.46 20.66 9.28 1998 15 0.88 1.32 0.27 9.45 3.70 1999 22 0.54 1.19 0.40 4.76 2.85 2000 43 0.60 1.34 0.39 6.73 2.39 2001 26 0.81 1.93 0.43 6.21 2.27 2002 26 0.86 5.19 0.87 −2.97 5.43 2003 25 0.78 6. 91 1.00 0.20 5.07 All years 287 0.65 4.85 0.25 −1.85 2.12 “n” is the number of release groups. 204 J.J. Anderson et al. / Ecological Modelling 186 (2005) 196–211 Fig. 6. Regression of log survival, log(S) vs. migration travel time, T(d). Regressions are weighted by 1/(S.E.)2 of S. Similar results were found in other studies. Muir et al. (2001), analyzing a subset (1992 through 1998) of the data used in our analysis, noted that the esti-mated survival from the hatcheries was inversely cor-related with migration distance to Lower Granite Dam (r2 = 0.64, p < 0.001). In a linear regression, survival also had a weak inverse relationship with travel time to Lower Granite Dam (r2 = 0.17, p > 0.07). Studies have also noted that survivals of juvenile salmon mi-grating through the Columbia and Snake Rivers were signiﬁcantly related to the distance traveled but not to travel time (Bickford and Skalski, 2000; Smith et al., 2002). To apply the MFL model to the PIT tag data, we note that the juvenile salmon rapidly migrate through the tributaries while the predators are resident within the tributaries. Therefore, Eq. (14) is an appropriate form of the MFL model for ﬁtting the juvenile salmon survival data. To estimate the model parameters, we write Eq. (14) in a multiple-linear form (log S)2 = ax2 + bt2 (22) where the parameters are deﬁned λ = 1 √ a and ω = b a (23) Because of the large difference in sample sizes, we weighted the individual survival estimates by one over the square of the standard error of the survival and ﬁt Eq. (22) to each year and to the combined years (Table 1). Eq. (23) requires that a > 0 and b ≥0, otherwise the model coefﬁcients are imaginary. However, the regres-sion puts no constraints on the values of a and b. This is somewhat problematic if the random speed ω, is near zero. In that case, b is also near zero and may take on a negative value as a result of estimation error. Therefore, J.J. Anderson et al. / Ecological Modelling 186 (2005) 196–211 205 Fig. 7. Regression of log survival, log(S) vs. migration distance, X (km). Regressions are weighted by 1/(S.E.)2 of S. we also calculate λ and ω using regression coefﬁcients a and b plus and minus their standard errors (S.E.). We pair the estimates of a plus its standard error with es-timates of b minus its standard error because a and b vary inversely in Eq. (22). That is, if due to estimation error the point estimate of a is low, then we expect that the point estimate of b will be high. In Table 2, NA indicates imaginary estimates of ω. 4. Results The r2 of the XT model ﬁt to data over the years 1993–2003 range from 0.54 to 0.97; the ﬁt to all years is 0.65, and the p-values are signiﬁcant at the 0.005 level or greater for all years (Table 1). Estimates of λ range between 354 and 917 km and estimates of ω range between 0.5 and 12 cm/s. Including all years to-gether, the encounter velocity is less than 0.6 cm/s and Table 2 Model parameters, λ in km, ω in cm/s Year λmax λave λmin ωmax ωave ωmin 1993 601 567 539 5.0 NA NA 1994 469 458 449 5.3 3.1 NA 1995 400 371 348 7.3 5.0 NA 1996 365 354 345 NA NA NA 1997 3131 800 575 12.0 9.9 7.4 1998 975 871 794 8.6 7.3 5.7 1999 1127 917 793 6.9 5.5 3.5 2000 1026 864 761 7.1 6.1 4.9 2001 815 719 651 5.7 4.9 3.9 2002 481 439 406 1.9 NA NA 2003 412 381 356 2.4 0.5 NA All years 466 454 443 0.6 NA NA Subscripts min and max are deﬁned by standard errors on the param-eters. NA indicates estimate not computable because of a negative b estimate. 206 J.J. Anderson et al. / Ecological Modelling 186 (2005) 196–211 the mean free-path length is 454 km. In some years, because b is negative, the random encounter velocity ω cannot be calculated, but in all years except 1996 the upper estimate of ω can be calculated. An estimate of ω is not possible in 1996 because b is negative and the standard error is less than the absolute value of b. However, λ is not affected by this negative b estimate in 1996. Regressing log S against x gives λ = 354, which is identical to λ obtained by Eq. (22) that includes the negative b. We conclude that in 1996 ω = 0, which im-plies survival is best characterized by a gauntlet process of Eq. (17). 5. Comparison with independent predictions To evaluate the model, we compare parameters es-timated from the XT model to estimates derived from other methods using independent observations. 5.1. Encounter distance From Eq. (4), the predation encounter distance, r, characterizing the average distance at which a preda-tion event occurs can be written 1 √πρλ (24) To estimate r, the predator density over the migra-tion path is required. To derive a very approximate estimate of ρ, we use population estimates of northern pikeminnow and smallmouth bass, which are the major predators of juvenile salmon in the river (Poe et al., 1991; Knutsen and Ward, 1999). Estimated populations for Lower Granite Reservoir are 26,000 northern pikeminnow larger than 250 mm (NMFS, 2000) and 20,911 smallmouth bass larger than 174 mm (Bennett et al., 1997). Dividing the combined pop-ulations by the volume of Lower Granite Reservoir, 597 × 106 m3, the predator density approximation is ρ = 7.9 × 10− 5 predators m− 3. Using the range of the mean estimates of λ, from Table 2, the predator–prey encounter distance varies between 6.6 and 10.7 cm with the estimate for the combined years of r = 9.4 cm. Because λ is derived from survival estimates of ﬁsh migrating for several weeks, the encounter distance represents an average of day and night conditions over the migratory period. For an independent estimate of the encounter distance, consider observations of predator reaction distance, which should be somewhat greater than the encounter distance because reaction distance identiﬁes the distance at which a predator ﬁrst reacts to a prey while the encounter distance, by deﬁnition, is the dis-tance within which on average a predation event occurs. Reaction distance depends on water clarity and light level (Vogel and Beauchamp, 1999). In 1997, the water clarity based on horizontal secchi disk readings in the Grande Ronde, a tributary of the Snake River, ranged between 20 and 100 cm, with a mean of about 50 cm (Steel, 1999). This equates to a turbidity reading of about 40 NTU (Steel and Neuhauser, 2002). Addition-ally, secchi disk readings in Lower Granite Reservoir typically vary between 10 and 50 cm (CRDART, 1998). Using the Vogel and Beauchamp (1999) reaction dis-tance formula for the response of lake trout to rainbow and cutthroat trout prey, the reaction distance is 37 cm under midday conditions (100 lx) and a turbidity of 40 NTU, while in a midcrepuscular period (0.17 lx) the reaction distance is 5 cm. Additionally, note that laboratory studies on brook trout (Sweka and Hartman, 2001) and rainbow trout (Barrett et al., 1992) found reaction distances less than 20 cm for turbidity levels greater than 30 NTUs. Furthermore, because the reaction distance is zero at night, the reaction distance averaged over the day should be about half the midday values and thus between 10 and 20 cm. Thus, the 9 cm encounter distance derived from the XT model is reasonable because it is close to the reac-tion distances estimated above. Consequentially as has been found in other systems (Gregory and Levings, 1998), we may expect that, in the Snake River tribu-taries, water visibility may be an important determinant of predator–prey encounter distance and therefore of juvenile salmon survival. 5.2. Random encounter velocity The predator–prey random encounter velocity de-pends on the random prey velocity u, and the random predator velocity v∗, according to Eq. (11). An upper estimate of u can be derived from Zabel (2002) in which the distribution of smolt migration travel times was modeled with an advection-diffusion equation con-trolled by two parameters: migration velocity u, and a spread term σ2. The travel time distribution is inversely J.J. Anderson et al. / Ecological Modelling 186 (2005) 196–211 207 related to the migration velocity distribution and, based on Tweedie (1957), the variance in the migration ve-locity is Var x t = u x σ2 + 2 x2 σ 4 (25) where x is the migration distance. The square root of Eq. (25) provides an upper estimate of u. From 4 years of data, wild Snake River spring chinook migrating 233 km from the Salmon River to Lower Granite Dam √ Var(x/t) ranged between 9.0 and 16.1 cm/s. For a second measure of the encounter velocity, consider predator velocities determined from radio-tagged Northern pikeminnow in John Day Reservoir. Using ﬁsh positions determined several times per day in May 1993 and 1994, an average velocity over the ground was 7 cm/s in the tailrace and 1 cm/s in mid-reservoir (Martinelli and Shively, 1997; Martinelli et al., 1993). Additionally, the coefﬁcients of variation were about one in both areas so the random velocity about equals the average velocity, and to a ﬁrst order the predator random velocity is expected to be a few cm/s. In comparison, the XT model estimated encounter velocity for the Snake River tributaries ranges between 0 and 12 cm/s. Although these estimates are relatively close, a direct comparison between them is problem-atic. First, uncertainty in all estimates is large and it is not possible to put meaningful conﬁdence intervals on any of them. Second, the estimates derived from prey and predator movements only represent compo-nents of the combined estimate of the XT model. Third, the prey random velocity estimated from Eq. (25) con-tains additional elements other than the actual random swimming velocity. Eq. (25) is derived from a differ-ence in arrival times of ﬁsh after traveling the distance x. Because early in migration juvenile salmon mostly migrate at night and are presumed to hold station near the bottom during the day (Zabel, 2002), the spread in a release group’s arrival time at a downstream location is the result of both their actual random swimming ve-locity and their diel station holding behavior. Thus, the velocity derived from Eq. (25) should represent an up-per limit of u. Considering these caveats, XT model’s estimate of ω on the order of a few cm/s is reasonable. 6. Discussion The MFL model was motivated by the ﬁnding that the survival of migrating juvenile salmon was indepen-dent of their migration time (Muir et al., 2001; Smith et al., 2002). While many predator–prey models do not easily explain this puzzling result, the mean free-path perspective provides an intuitive explanation that is valuable for exploring how behavior affects the spatial-temporal characteristics of predator–prey encounters. Furthermore, while the MFL model does not provide spatial details of predator–prey encounters, it implic-itly contains spatial scales and directionality and so provides a way to characterize the dynamics within homogeneous cells frequently used in individual based models (IBMs). Below we elaborate on these ideas. The contrast in the way the MFL model treats spatial scales compared to IBMs is illustrated by comparing our model to an IBM of salmon predator–prey inter-actions (DeAngelis and Peterson, 2001). One problem with IBMs is that model response is often sensitive to the cell size. DeAngelis and Peterson (2001) explored this problem in their model of salmon smolts migrat-ing through a river described by a series of cells and suggested a sharp increase in smolt mortality as cell size increased from 4 to 50 km was an artifact of im-properly characterizing the ecological neighborhood of the predator–prey interaction. The MFL model is con-tinuous and overcomes this problem by avoiding the need of setting the correct cell size; in effect, the data determine the scale of the model. Further comparisons between the MFL model and the Peterson and DeAngelis model emphasize the con-sequences of how predator–prey interactions are spec-iﬁed. The MFL model encounter rate depends on the predator and prey random speeds and the prey migra-tion rate, and so if fast migrating prey pass a gauntlet of slow predators, survival depends on the gauntlet length but not the rate of passage since a predator only gets one chance at a prey. In the Peterson and DeAngelis model, encounter rate is characterized by the GS equation, which depends on predator and prey random speeds, but not on the mean prey velocity (Peterson and De Angelis, 2000). Consequentially, a prey can encounter a predator multiple times and so mortality depends on the migration rate since it determines cell residence time. These alternative encounter representations pro-duce diametrically opposed survival relationships. At 208 J.J. Anderson et al. / Ecological Modelling 186 (2005) 196–211 low smolt density, cell-based survival increases in an exponential-like manner with increasing ﬂow, while MFL-based survival asymptotically approaches a con-stant as ﬂow increases (Fig. 3). These differences il-lustrate the importance of properly characterizing how animals move between cells and what they do within a cell. We emphasis this issue because cells and grids are powerful constructs for modeling animal dynamics in complex landscapes (Tischendorf, 1997; Nestler et al., 1997; Bian, 2003; Booker et al., 2004). Further-more, the MFL model can readily be incorporated into a Peterson/DeAngelis-type cell-based model. How animals move and interact with their environ-ment is fundamental in determining predator–prey en-counter rates and therefore population dynamics. For example, in an individual-based model, Mitchell and Lima (2002) found that random search predators are more efﬁcient at ﬁnding stationary prey than randomly moving prey while the opposite is true if the preda-tors have a spatial memory of the prey distribution. From studies on sit-and-wait invertebrate predators in streams, Woodward and Hildrew (2002) suggested that prey availability, effectively the product of density and encounter rate, may be more important in determin-ing predator impact than prey density itself. They also noted that the impact of sit-and-wait predators might be different from the impact of search predators. Luttbeg and Schmitz (2000) found that differing assumptions about prey behavior and risk perception could produce similar or different community dynamics, community persistence and resource use. They emphasized the importance of better understanding and representing the behavioral and cognitive capabilities of the prey. Blaine and DeAngelis (1997), analyzing an herbivore-periphyton model, found that the conditions in which the herbivore moved between patches affected how spa-tial heterogeneity and biomass responded to nutrient in-puts. These studies and others illustrate the importance of animal behavior, memory, and adaptive responses to population dynamics. However, the speciﬁc behaviors are not readily represented by the traditional sit-and-wait and searching categories of predator–prey interac-tions (Hart, 1997). In contract the MFL approach, offers a simple and unique quantiﬁcation of fundamental be-haviors in terms of the random and directed motions of both predators and prey. Furthermore, because encoun-ters are described by speeds, the energetic costs of the predator–prey strategies can be quantiﬁed and ranked. Undoubtedly, behavior and environmental covari-ates can add formable layers of complexity to a model. However, by linking covariates to the effective preda-tor density and predator encounter distance, which to-gether determine the mean free-path length, we can potentially reduce the complexity. For example, An-derson and Van Holmes (manuscript in review) relate encounter distance to transparency based on a study of Gregory and Levings (1998) and effective predator density to temperature based on how predator digestion changes with temperature (Vigg and Burley, 1991). In principle, the effect of the prey consumption rate on predator satiation could also be formulated in terms of the effective predator density, although a simple ana-lytical solution is unlikely. However, instead of incor-porating more complex predator dynamics into a prey survival model, it is feasible to use the MFL theory to deﬁne prey encounters in a detailed predator foraging model, for example as in Jeschke et al. (2002). In the present model, the effects of predator satiation are re-ﬂected in the effective predator density, which in turn affects the mean free-path length. Therefore, we expect a longer mean-free path length as predators become satiated. While the statistically derived mean-free path char-acterization of animal movement is relatively unique in ecology, which typically considers individual-based animal movements, the two approaches have similari-ties. Matthiopoulos (2003) studied an animal’s use of space limited by preference and accessibility in terms of the total path length (l) and the maximum distance (ρ) achieved over the path. The MFL model contains similar concepts: l is equivalent to λ, and ρ is equiva-lent to Ut. Cuddington and Yodzis (2002), considering individual random walks through a spatially complex fractal environment, characterized how environmental constraints, such as a river valley, limit animal mobil-ity. The number of new sites a walker visits depends on the fractal, or spectral dimension, of the environment. On a one-dimensional landscape, the number of new sites visited is proportional to the square-root of time while the number of sites visited on a two-dimensional surface is proportional to time. Thus, reduced fractal dimensionality of the environment reduces encounters with new sites. The MFL model has an analogous in-terpretation. In the form, S = exp −ω λ t 1 + (W /ω)2 , the term W/ω is a fractal-like dimension. When W/ω > 1 J.J. Anderson et al. / Ecological Modelling 186 (2005) 196–211 209 predators and prey tend to move at some ﬁxed angle to each other and prey survival over a ﬁxed migration distance is greatest and independent of time. As the ratio decreases, movements of predator and prey take on higher dimensions, while mortality increases and becomes dependent on time. Thus, we see that fractal characteristics of movement can affect predator–prey dynamics in both restricted and unrestricted environ-ments. Furthermore, the equation’s rate coefﬁcient, ω/λ, characterizes the strength of mortality per unit time. We suggest that the mean ﬁree-path length con-cept, quantiﬁed by these two ratios, provide a very ba-sic way to compare and contrast diverse predator–prey systems. Returning to the issue that motivated our work, we ﬁnd that the model ﬁt to smolt survival data pro-duced random encounter velocities of a few cm/s with within-year variations of about 50% and an across-year variations of about 100% (Table 2). These rather large variations may be associated with environmental prop-erties such as water transparency or river hydraulics that affect the parameters in ways not explicit in the model. The large random encounter velocity in 1997, which was a year of high ﬂow and low transparency, supports this possibility. On the other hand, our data represent ﬁsh survival from 17 hatcheries located on different tributaries between 31 and 772 km above Lower Granite Dam, and so we may expect variations in ﬁtting such a diverse set of data with a two-parameter model. In any case, with random encounter and migra-tion velocities about 5 and 15 cm/s respectively, then U/ω is about 3, and from Fig. 3 we see that survival is largely independent of migration velocity. In other words, ﬁsh pass a gauntlet of predators so that migra-tion distance, not migration velocity, is a major factor determining the survival of juvenile spring chinook migrating through the tributaries of the Snake River. Finally, as a check on the model, we estimated the parameters using independent methods and data. The predator encounter length scale, derived from predator density and the prey’s mean free-path length, was close to the predator’s visual ﬁeld during the migration. 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4873	https://www.quora.com/How-does-one-calculate-the-terminal-velocity-of-a-falling-object-with-a-given-mass-and-surface-area-from-a-given-height-on-Earth-with-respect-to-both-variable-air-resistance-and-gravitational-pull	How does one calculate the terminal velocity of a falling object with a given mass and surface area from a given height on Earth with respect to both variable air resistance and gravitational pull? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Physics Coefficient of Air Resist... Mass (people) Falling Bodies Surface Area Scientific Calculations Gravity (physics) Mechanics (general) Terminal Velocity 5 How does one calculate the terminal velocity of a falling object with a given mass and surface area from a given height on Earth with respect to both variable air resistance and gravitational pull? All related (39) Sort Recommended Arthur Clarke Former Chemical Engineer · Author has 1.2K answers and 1.7M answer views ·6y Originally Answered: How does one calculate the terminal velocity of a falling object with a given mass and surface area from a given height on earth with respect both variable air resistance and gravitational pull? · I do not intend to give chapter and verse on this. Instead a bit of background on the theory and you can do the rest. Isaac Newton studied the motion of objects subject to air and water resistance under gravity and published his findings in the celebrated work “Philosophiæ Naturalis Principia Mathematica” or Principia for short. It was written in Latin because English was too vulgar for such high-brow philosophy. His work involved dropping inflated hog’s bladders, natural balloons, from the top of the newly built St Paul’s Cathedral, on the inside that is. He timed the rate of descent and dete Continue Reading I do not intend to give chapter and verse on this. Instead a bit of background on the theory and you can do the rest. Isaac Newton studied the motion of objects subject to air and water resistance under gravity and published his findings in the celebrated work “Philosophiæ Naturalis Principia Mathematica” or Principia for short. It was written in Latin because English was too vulgar for such high-brow philosophy. His work involved dropping inflated hog’s bladders, natural balloons, from the top of the newly built St Paul’s Cathedral, on the inside that is. He timed the rate of descent and determined the terminal velocities subject to several variables, bladder diameter and overall weight being the key ones. Using correlations he drew lines of best fit and determined the constants that described the experimental results. He came up with what we now call Newton’s equation for the passage of spheres through viscous media (i.e. air and also water). This equation has no viscosity term. Instead it has a Drag Coefficient (Cd) as viscosity was not defined in Newton’s day. From this you can estimate the terminal velocity of a sphere falling through air for example. Now nearly every high school physics student knows Stokes equation (published 1851) for the terminal velocity of a sphere falling under gravity. But not every student knows that this equation is only for laminar flow conditions, when Reynold’s number (Re) is less than or equal to 2. For most cases of say macroscopic objects falling through the air, Newton’s equation is the one to use. For both spherical and non-spherical objects there are plots of Re versus Cd which can be used iteratively to solve the problem. For those who wish to know more I suggest personal research enough to solve a few numeral problems. Getting the right answer is a reward unto itself and will leave a lasting impression. Upvote · 9 1 Sponsored by Avnet Silica We're at the Pulse of the Market. Explore growth data, key markets and technologies, and the startups we’ve scaled from concept to market. Learn More 9 1 Related questions More answers below How can I calculate the terminal velocity of a falling object? What is the terminal velocity of a ball dropped from a height? How much air resistance acts on a falling 100 N box of nails when it reaches terminal velocity? Does mass affect the speed of a falling object on Earth (with air resistance)? How does the terminal velocity of a falling object change when the object has more air resistance? Ojvind Bernander Ph.D. in Computational Neuroscience&Physics, California Institute of Technology (Caltech) (Graduated 1993) · Author has 746 answers and 308.1K answer views ·6y Originally Answered: How does one calculate the terminal velocity of a falling object with a given mass and surface area from a given height on earth with respect both variable air resistance and gravitational pull? · v = sqrt(m g/k rho A) k depends on the exact geometry. As a first approximation, set it to 1. The object has area A, mass m. Far above the Earth’s surface, adjust g, dividing it by (R/Rsurf)^2. The density rho decreases exponentially with height h according to the Barometric formula, 101325 exp(−0.00012 h), metric units. Continue Reading v = sqrt(m g/k rho A) k depends on the exact geometry. As a first approximation, set it to 1. The object has area A, mass m. Far above the Earth’s surface, adjust g, dividing it by (R/Rsurf)^2. The density rho decreases exponentially with height h according to the Barometric formula, 101325 exp(−0.00012 h), metric units. Upvote · Pranav Prabhu I Do Some Internships and Community Service (2019–present) · Author has 1.3K answers and 2.7M answer views ·7y Originally Answered: How can I calculate the terminal velocity of a falling object? · Use the terminal velocity formula, v = the square root of ((2mg)/(ρAC)). m = mass of the falling object. g = the acceleration due to gravity. ρ = the density of the fluid the object is falling through. A = the projected area of the object. C = the drag coefficient. Find the mass of the falling object. This should be measured in grams or kilograms, in the metric system. Know the acceleration due to the gravity of the Earth. Close enough to the earth to encounter air resistance, this acceleration is 9.8 meters per second squared, or 32 feet per second squared. Calculate the downward pull of Continue Reading Use the terminal velocity formula, v = the square root of ((2mg)/(ρAC)). m = mass of the falling object. g = the acceleration due to gravity. ρ = the density of the fluid the object is falling through. A = the projected area of the object. C = the drag coefficient. Find the mass of the falling object. This should be measured in grams or kilograms, in the metric system. Know the acceleration due to the gravity of the Earth. Close enough to the earth to encounter air resistance, this acceleration is 9.8 meters per second squared, or 32 feet per second squared. Calculate the downward pull of gravity. The force with which the falling object is being pulled down equals the object's mass times acceleration due to gravity, or F = MA. This number, multiplied by two, goes in the top of the terminal velocity formula. How to Calculate Terminal Velocity Upvote · 9 3 Sponsored by All Out Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out. Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association. Shop Now 999 621 Simon Bridge Scientist · Author has 86.2K answers and 38.8M answer views ·5y Originally Answered: How can I calculate the terminal velocity of a falling object? · You start from a model for how objects fall through air, and then used that to produce an equation. These models can get arbitrarily complicated, depending on how much you want to just model from scratch and how much you are prepared to measure. For a start … the force on the object would be gravity and drag… acting in opposite directions. So you write, using Newton’s second law: m g−F d r a g=m a m g−F d r a g=m a … for an object just going straight down. So what will the drag for depend on? A common starter model says drag depends on the square of the falling speed, and the cross-section area. F d r a g=C d A v 2 F d r a g=C d A v 2 … w Continue Reading You start from a model for how objects fall through air, and then used that to produce an equation. These models can get arbitrarily complicated, depending on how much you want to just model from scratch and how much you are prepared to measure. For a start … the force on the object would be gravity and drag… acting in opposite directions. So you write, using Newton’s second law: m g−F d r a g=m a m g−F d r a g=m a … for an object just going straight down. So what will the drag for depend on? A common starter model says drag depends on the square of the falling speed, and the cross-section area. F d r a g=C d A v 2 F d r a g=C d A v 2 … where C d C d is the drag coefficient, which will depend on t=what the object is made out of. You’d look this up for common objects… otherwise you need some way to model how the material properties affect the drag coefficient. You can look for v so that a = 0 (constant speed). That would mean that mg = C_dAv^2 … solve for v. Upvote · 9 1 Assistant Bot · Sep 6 Terminal velocity is the steady speed reached when downward gravity equals upward aerodynamic drag (and buoyancy if significant). For an object falling through air from rest, with mass m, frontal area A, drag coefficient Cd, and starting at height h, compute the fall including variable air density and gravity if needed. Two approaches: closed-form approximation (constant parameters) and full variable model (numerical). Steps, equations, and practical recipe follow. 1) Definitions and parameters m: mass (kg) 2) Terminal velocity at a fixed altitude (algebraic) Assuming constant ρ and g, termina Continue Reading Terminal velocity is the steady speed reached when downward gravity equals upward aerodynamic drag (and buoyancy if significant). For an object falling through air from rest, with mass m, frontal area A, drag coefficient Cd, and starting at height h, compute the fall including variable air density and gravity if needed. Two approaches: closed-form approximation (constant parameters) and full variable model (numerical). Steps, equations, and practical recipe follow. 1) Definitions and parameters m: mass (kg) 2) Terminal velocity at a fixed altitude (algebraic) Assuming constant ρ and g, terminal velocity vt solves mg = D(vt), so vt = sqrt( (2 m g) / (ρ Cd A) ). This is the standard closed-form result and holds when drag ∝ v² (high Reynolds number). Use ρ evaluated at the altitude where vt is achieved (often near surface: ρ ≈ 1.225 kg/m³). 3) When air density and gravity vary with altitude (numerical model) Governing ODE (downward positive): m dv/dt = m g(z) − 0.5 ρ(z) Cd A v \|v\|. ρ(z): use an atmosphere model. For troposphere (approx up to 11 km) with temperature lapse rate L ≈ 0.0065 K/m, ρ(z) = ρ0 · (T(z)/T0)^( (g0 / (R L)) - 1 ), or directly compute pressure with barometric formula then ρ = p/(R_specific T). For many problems near ground, ρ(z) ≈ ρ0 is adequate. g(z) = g0 · (R_earth/(R_earth + z))^2; usually negligible change <0.1% for z ≪ R_earth. Numerical procedure: Choose time step Δt small enough (e.g., 0.001–0.01 s depending on stiffness). At each time step compute ρ(z) and g(z), then acceleration a = g(z) − (0.5 ρ(z) Cd A v \|v\|)/m. Integrate with a stable method (explicit RK4, or velocity-Verlet); update v and z until v converges to steady value or until z=0 (impact). Terminal velocity is the asymptotic v when dv/dt → 0; practically, take vt ≈ v after dv/dt falls below a small tolerance. 4) Analytical solution for v(z) when ρ and g constant (closed-form time/position relations) If ρ and g constant and drag quadratic, integration yields: dv/dt = g − (k/m) v², with k = 0.5 ρ Cd A. Solution for velocity in time: v(t) = vt tanh( (g t)/vt ), where vt = sqrt( m g / k ) = sqrt( (2 m g)/(ρ Cd A) ). Velocity as function of distance fallen: v(z) = vt sqrt(1 − exp(−2 g z / vt²)) / sqrt(1 + exp(−2 g z / vt²)) (equivalently from integrating dz/dt). These closed forms let you estimate how far is required to reach, say, 95% of vt: distance to reach fraction f of vt ≈ (vt² / (2 g)) · ln( (1+f)/(1−f) ). 5) Practical notes and recommended workflow If h is small (< a few hundred meters) and object is dense (large m/A), use constant ρ and closed-form vt formula; compute vt and check the distance needed to approach vt. If required distance ≪ h, vt will be reached before impact. If h is very large (skydiver, > several km) or precision matters, integrate numerically using altitude-dependent ρ and optionally g(z). For low speeds / low Reynolds number (Stokes regime), drag is linear (F = 6πμrv) and vt = mg/(6πμr). Use quadratic drag only when Re ≳ 1000; check Re = v L / ν after a preliminary vt estimate. Use conservative estimates for Cd and A if orientation or tumbling is unknown; tumbling increases effective A and Cd. 6) Example (quick calculation) Given m=80 kg, A=0.7 m², Cd=1.0, ρ=1.225 kg/m³, g=9.81 m/s²: vt = sqrt( (2·80·9.81) / (1.225·1.0·0.7) ) ≈ sqrt( 1569.6 / 0.8575 ) ≈ sqrt(1830) ≈ 42.8 m/s (~154 km/h). Summary Use vt = sqrt((2mg)/(ρCdA)) for constant-air-density estimate. For altitude-dependent air density or long falls, integrate m dv/dt = m g(z) − 0.5 ρ(z) Cd A v\|v\| with dz/dt = v numerically. Verify drag regime (quadratic vs linear) and use appropriate Cd, atmosphere model, and a stable ODE integrator for accurate results. Upvote · Related questions More answers below What is the terminal velocity of a falling object on Earth? What is the minimum height needed for an object to reach Earth's surface without bouncing back up at its terminal velocity? Does Earth's mass affect its gravitational pull? What is the overall energy transfer for an object falling at a terminal velocity taking into account air resistances? Would a heavier object reach a faster terminal velocity than a lighter object with the same surface area when you drop them from sky? Why? Jeff Obunga Student at Portsmouth University. · Author has 1.9K answers and 2.3M answer views ·8y Originally Answered: How can I calculate the terminal velocity of a falling object? · Have a a look at the following equation : V = (2mg/pAc)^1/2 Where: V= Terminal velocity m = mass A = projected area g = Gravity p = density C = drag coefficient Hope this helps. Upvote · 9 2 Promoted by Grammarly Grammarly Great Writing, Simplified ·Aug 18 Which are the best AI tools for students? There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Do Continue Reading There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Docs – Your all-in-one writing surface Think of docs as your smart notebook meets your favorite editor. It’s a writing surface where you can brainstorm, draft, organize your thoughts, and edit—all in one place. It comes with a panel of smart tools to help you refine your work at every step of the writing process and even includes AI Chat to help you get started or unstuck. Expert Review – Your built-in subject expert Need to make sure your ideas land with credibility? Expert Review gives you tailored, discipline-aware feedback grounded in your field—whether you're writing about a specific topic, looking for historical context, or looking for some extra back-up on a point. It’s like having the leading expert on the topic read your paper before you submit it. AI Grader – Your predictive professor preview Curious what your instructor might think? Now, you can get a better idea before you hit send. 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All five tools work together inside Grammarly’s document editor to help you grow your skills and get your writing across the finish line—whether you’re just starting out or fine-tuning your final draft. The best part? It’s built for school, and it’s ready when you are. Try these features and more for free at Grammarly.com and get started today! Upvote · 999 200 99 34 9 3 Les McLean Ph.D. in Engineering · Author has 4.9K answers and 12.3M answer views ·6y Originally Answered: How can I calculate the terminal velocity of a falling object? · Terminal velocity is: Vt = √[2mg/(ρACd)] where m is the mass of the object, ρ is density of the fluid (nominally 1.225kg/m^3 for air), A is the projected frontal area and Cd is the drag coefficient which depends on the shape and the surface roughness of the object. For a smooth bullet, falling nose first, Cd = 0.30 EXAMPLE: Say that a bullet has a mass of 10g = 0.01kg and a projected frontal area of 0.001m^2. Then Vt = √{2(0.01)(9.81)/[1.225(0.001)(0.30)]} Vt = √533.88 = 23.11m/s Upvote · 9 1 Aaron Brown Author has 25.5K answers and 104.1M answer views ·8y Originally Answered: How can I calculate the terminal velocity of a falling object? · It’s easy if you ignore air resistance. On earth, gravity accelerates objects at 9.8 meters per second per second. That means if an object falls for T seconds, it falls 4.9T^2 meters. After T seconds the object is falling at 9.8T meters per second. So if an object falls D meters, it takes SQRT(D/4.9) seconds, and it is going 9.8SQRT(D/4.9) meters per second at that time. With air resistance it depends on the shape and weight of the object. Most things reach a maximum velocity at which friction cancels out air resistance. For a person that’s something like 50 meters per second, which you reach Continue Reading It’s easy if you ignore air resistance. On earth, gravity accelerates objects at 9.8 meters per second per second. That means if an object falls for T seconds, it falls 4.9T^2 meters. After T seconds the object is falling at 9.8T meters per second. So if an object falls D meters, it takes SQRT(D/4.9) seconds, and it is going 9.8SQRT(D/4.9) meters per second at that time. With air resistance it depends on the shape and weight of the object. Most things reach a maximum velocity at which friction cancels out air resistance. For a person that’s something like 50 meters per second, which you reach in about six seconds, during which you fall about 150 meters. So after the first 150 meters, you won’t be speeding up anymore, but you will lose speed when you hit the ground. Upvote · 9 2 9 3 John Gompers Author has 382 answers and 113.1K answer views ·3y Related If we neglect air resistance, what would be the terminal velocity of a falling object near Earth’s surface? Why or why not? If we neglect air resistance, a falling object doesn't have a terminal velocity. In terms of velocity, an object will continue doing the same thing unless a force acts on it. For a falling object, the main force is Gravity, which makes the object accelerate downwards. Trying to stop it (without much success) is Air Resistance. An objects Terminal Velocity is the velocity where the air resistance ‘upwards' against its fall matches the force of gravity pulling it down. When the two forces match, they cancel out, and the object will continue doing what it was doing (in this case, falling down at a Continue Reading If we neglect air resistance, a falling object doesn't have a terminal velocity. In terms of velocity, an object will continue doing the same thing unless a force acts on it. For a falling object, the main force is Gravity, which makes the object accelerate downwards. Trying to stop it (without much success) is Air Resistance. An objects Terminal Velocity is the velocity where the air resistance ‘upwards' against its fall matches the force of gravity pulling it down. When the two forces match, they cancel out, and the object will continue doing what it was doing (in this case, falling down at a constant rate). Take away the air resistance, you take away the force opposing gravity, and the object will continue accelerating until it hits something solid (like the ground). This is only dependent on how high the object is dropped from, and there is no theoretical maximum velocity. Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Upvote · 9 2 9 2 Martin Franěk Ing. in Software and Applications&Physics, FJFI ČVUT V Praze (Graduated 2013) · Author has 456 answers and 273.3K answer views ·3y Related What is the maximum possible terminal velocity of a falling body on Earth, and how do we know this limit exists? Well, the terminal velocity is determined by shape (air resistance) and mass. Theoretically speaking - if you have a really good aerodynamic shape and take something really dense (like neutron-star dense or just-below-collapsing-to-black-hole dense), then you may achieve just about any velocity you want (as terminal velocity). A nice example would be “needle” just few atoms thick - and long as needed - then it will have really small resistance (small area and no turbulence) and weight can be increased nearly infinitely - by having the rod “long enough”. BUT… (there is always a but… and when it i Continue Reading Well, the terminal velocity is determined by shape (air resistance) and mass. Theoretically speaking - if you have a really good aerodynamic shape and take something really dense (like neutron-star dense or just-below-collapsing-to-black-hole dense), then you may achieve just about any velocity you want (as terminal velocity). A nice example would be “needle” just few atoms thick - and long as needed - then it will have really small resistance (small area and no turbulence) and weight can be increased nearly infinitely - by having the rod “long enough”. BUT… (there is always a but… and when it is not, then there is butt in the way) In the real world you will be limited in achieving high velocities by the fact, that the atmosphere itself is not infinite - you have just about 100km or so of something you may call atmosphere. And even there many would tell you, you are already outside of atmosphere… And even with zero resistance you would by free-fall from 100km get “just” to little over 5000km/h - and even that is computed with constant sea-level gravity - wich would be slightly lower 100km from surface. Back to theory limitations though - the highest velocity you can achieve by simple free-fall will be equal to escape velocity - about 11186 km/s. This velocity is defined, that object - thrown at this velocity - will have just enough energy to counter the effect of gravity - so it will slow down to zero just as it is out of reach of the effect of gravity - in infinity. The velocity and energy are (via symmetry) equal to object, wich would start at infinity and free-falls to Earth. To sum it - theoretical limit is speed of light. Nothing goes faster (that we know). For just about any velocity, you can create a body, wich will have terminal velocity that high. But practically speaking, nothing will free-fall to earth faster, than 11186km/s - even if the object has higher terminal velocity. Upvote · Gene M.S. in Physics, University of Minnesota - Twin Cities (Graduated 1971) · Author has 9K answers and 2.7M answer views ·2y Related How does the surface area of the parachute affect the terminal velocity of the falling object? You point out one of the distinctions between friction and air/fluid resistance. The force of air resistance depends on the cross sectional area in the direction of motion. For a falling object, parachute & passenger, the area and speed identifies a volume of air that must be moved/accelerated out of the way for passage. That takes a force! Newton explained if you push air out of the way downward, it pushes back upward. Now you have gravity down and air up. The faster you’re moving, the quicker you must move air. “Quicker” means more acceleration/force so that at some speed the air resistance wi Continue Reading You point out one of the distinctions between friction and air/fluid resistance. The force of air resistance depends on the cross sectional area in the direction of motion. For a falling object, parachute & passenger, the area and speed identifies a volume of air that must be moved/accelerated out of the way for passage. That takes a force! Newton explained if you push air out of the way downward, it pushes back upward. Now you have gravity down and air up. The faster you’re moving, the quicker you must move air. “Quicker” means more acceleration/force so that at some speed the air resistance will equal your weight giving you no net force/acceleration. That’s “terminal” in the sense that you’ve reached your destination speed … as in a train/bus terminal/destination. WWII chutes were fixed and hard to control. Modern chutes are closer to gliders where both speed and direction can be controlled by shaping the canapé. Pre-A/C in cars, kids & adults played with the wind with their hands out the windows. Changing the area of your hand in the direction of motion gave good feedback to this. Inclination also explained how lift is produced on wings. My best lesson was taking a sheet of plywood out of a box store on a windy day. The best video I’ve seen was a raptor grabbing a large salmon from a western river. It was taken sideways by both feet. The bird was having trouble staying airborne by the length x width of the fish hitting the waves and air. Without pausing in its attempt to take off, it “walked” its way to a better mid-point; and then reversed its grip (one foot at a time) until the fish was pointed foreword. At that point, it soared off to home or at least a safe place to eat. Practical physics - 101. Upvote · Jess H. Brewer Physics professor since 1977. · Author has 22.9K answers and 59.6M answer views ·7y Related What is the maximum possible terminal velocity speed in which an object can fall from the highest point in the sky under earth's gravitational pull? “The highest point in the sky” would be, as Joseph pointed out, infinitely far from the Earth. Assuming that the object could avoid being pulled in by other planets or stars on the way, neglecting the gravitational attraction of more important masses like the Sun or the Milky Way or… you know, let’s just pretend there is nothing else in the universe besides this object and the Earth, and that we don’t mind waiting literally forever for it to arrive. Subject to those caveats, we can get its velocity at a distance of r r from the center of the Earth by setting its kinetic energy plus its gravitatio Continue Reading “The highest point in the sky” would be, as Joseph pointed out, infinitely far from the Earth. Assuming that the object could avoid being pulled in by other planets or stars on the way, neglecting the gravitational attraction of more important masses like the Sun or the Milky Way or… you know, let’s just pretend there is nothing else in the universe besides this object and the Earth, and that we don’t mind waiting literally forever for it to arrive. Subject to those caveats, we can get its velocity at a distance of r r from the center of the Earth by setting its kinetic energy plus its gravitational potential energy to zero, which was true at the outset and must still be true now: 1 2 m v 2−G M m/r=0 1 2 m v 2−G M m/r=0 or v=√2 G M r v=2 G M r. Now, this object is going to burn up in the atmosphere once it gets below about 10 km altitude — trust me on this. So its maximum velocity will be just before that, at r=6,388,164 r=6,388,164 m from the center of the Earth. Plugging into the equation gives v=11,173 v=11,173 m/s or 40,222 40,222 kph. Definitely burns up. Upvote · 99 16 9 3 Andy Ross Primary Educator (2001–present) · Author has 224 answers and 424.1K answer views ·1y Related How do you calculate what altitude something falls from so that its terminal velocity matches orbital velocity? What a strange but interesting question. You have to assume that there is no atmosphere otherwise there’s a complex time-varying terminal velocity which depends on the density of the atmosphere at varying altitudes. It’s perhaps more useful to consider the energy of the body, both in orbit and when it is plummeting, than the velocity. That energy is what it requires to lift the object to great height AND give it the velocity required for orbit. Consider a theoretical planet with no atmosphere and a giant hole that extends right down to the middle of its core, the place where the local pull of g Continue Reading What a strange but interesting question. You have to assume that there is no atmosphere otherwise there’s a complex time-varying terminal velocity which depends on the density of the atmosphere at varying altitudes. It’s perhaps more useful to consider the energy of the body, both in orbit and when it is plummeting, than the velocity. That energy is what it requires to lift the object to great height AND give it the velocity required for orbit. Consider a theoretical planet with no atmosphere and a giant hole that extends right down to the middle of its core, the place where the local pull of gravity originates from. If falling under the planet’s gravity the gravitational potential energy that the object had in orbit will be converted into kinetic energy as it falls and in the absence of an atmosphere it will continue to accelerate until impact. So if we take your terminal velocity reference then we are assuming that there’s an atmosphere, so acceleration will eventually stop as the object falls. In which case it is losing energy to friction with the air. You can hopefully see a problem here: we can’t compute it’s terminal velocity without knowing how much atmosphere is slowing it down. But, if we ignore air resistance (and your terminal velocity limitation) what you can quickly deduce is that its maximum vertical speed would be determined by the purely gravitational attraction. And because we can’t magically create energy during its fall, the object will at the moment of impact have the speed that’s determined by its gravitational potential energy gain in flying to space in the first place and the work done against the pull of the planet’s mass that took. What its fall won’t do is convert the energy put in to to place the object into an orbital motion. That’s the energy that stops it from simply falling back down to Earth (satellites have to be sent around the Earth, not just parked in space because gravity will just pull them down no matter how high you send them). So your question is interesting but not well constructed. We can convert orbital energy into kinetic, of course (that’s what returning space craft do), but the speed of terminal velocity will be determined by the altitude of the returning object, its speed across the ground, the air resistance and the trajectory it takes. Because re-entry causes the object to slow down (drag) it will always have a speed that’s less than its on-orbit speed because its gain in velocity as it drops will cause a huge increase in drag - there’s a nonlinear relationship there - although the size of this drag depends on the density of the atmosphere. Terminal velocity is a vector quantity because it’s based on vertical and horizontal motion. So although there is an increase in vertical as it falls there’s a reduction in horizontal due to drag. Damn this is complex. With a bit of time this could be an interesting question for students but it’s not well framed at the moment and there are too many wooly elements to it. You might be interested in Earth-grazing asteroids, rocks from space that hit our atmosphere at a shallow angle, lose a few km per second in speed but have sufficient speed and trajectory to skim out of the atmosphere and back into the vacuum. Upvote · Related questions How can I calculate the terminal velocity of a falling object? What is the terminal velocity of a ball dropped from a height? How much air resistance acts on a falling 100 N box of nails when it reaches terminal velocity? Does mass affect the speed of a falling object on Earth (with air resistance)? How does the terminal velocity of a falling object change when the object has more air resistance? What is the terminal velocity of a falling object on Earth? What is the minimum height needed for an object to reach Earth's surface without bouncing back up at its terminal velocity? Does Earth's mass affect its gravitational pull? What is the overall energy transfer for an object falling at a terminal velocity taking into account air resistances? Would a heavier object reach a faster terminal velocity than a lighter object with the same surface area when you drop them from sky? Why? What is the relationship between mass and the time it takes for an object to reach terminal velocity while falling freely? With no air resistance, can I reach terminal velocity? When does a falling object reach terminal velocity? What is the height of a ball falling from rest at terminal velocity? What are all the factors needed for a falling object to have the highest terminal velocity under planet earth's gravitational pull? Related questions How can I calculate the terminal velocity of a falling object? What is the terminal velocity of a ball dropped from a height? How much air resistance acts on a falling 100 N box of nails when it reaches terminal velocity? Does mass affect the speed of a falling object on Earth (with air resistance)? How does the terminal velocity of a falling object change when the object has more air resistance? What is the terminal velocity of a falling object on Earth? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025 Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. 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4874	https://www.edulyte.com/maths/volume-of-a-cone/	Unlocking the Secrets of Cone Volume: Formulas and Calculations Skip to content Features AI Quiz Builder Build and Share Smarter Quizzes Faster with AI. Online Course Builder Quickly Create and Sell Your Online Courses – No Coding Needed. Appointment Scheduler Create and Share Your Personal Booking Link with Ease. Custom Events Create Your Event: Online or Offline, Free or Paid Easily. Custom Quiz Builder Custom Quiz Builder: Create Your Own Quizzes Digital Downloads Maximise Your Earnings with Digital Downloads Tutor Search Carefully chosen experts await to guide your learning journey. One-on-One Mastery Direct & Driven: sessions crafted just for you by educators you can trust! Group Classes Learn Together, Achieve More. guided by our top educators. Courses English English mastery made easy: resources, courses & expert tips! PTE Wave goodbye to PTE worries: master PTE, bypass the stress! 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Grammar Free notes, worksheets, and quizzes at your fingertips!Explore All Topics Menu Blog Learner sign up Educator sign up Log in 2D Shapes2cosacosb Formula30-60-90 Formulas3D ShapesAbsolute Value FormulaAcute AngleAcute Angle triangleAdditionAlgebra FormulasAlgebra of MatricesAlgebraic EquationsAlgebraic ExpressionsAngle FormulaAnnulusAnova FormulaAnti-derivative FormulaAntiderivative FormulaApplication of DerivativesApplications of IntegrationArc Length FormulaArccot FormulaArctan FormulaArea Formula for QuadrilateralsArea FormulasArea Of A Sector Of A Circle FormulaArea Of An Octagon FormulaArea Of Isosceles TriangleArea Of ShapesArea Under the Curve FormulaArea of RectangleArea of Regular Polygon FormulaArea of TriangleArea of a Circle FormulaArea of a Pentagon FormulaArea of a Square FormulaArea of a Trapezoid FormulaArithmetic Mean FormulaArithmetic ProgressionsArithmetic Sequence Recursive FormulaArithmetic and Geometric ProgressionAscending OrderAssociative Property FormulaAsymptote FormulaAverage Deviation FormulaAverage Rate of Change FormulaAveragesAxioms Of ProbabilityAxis of Symmetry FormulaBasic Math FormulasBasics Of AlgebraBinary FormulaBinomial Probability FormulaBinomial Theorem FormulaBinomial distributionBodmas RuleBoolean AlgebraBusiness MathematicsCalculusCelsius FormulaCentral Angle of a Circle FormulaCentral Limit Theorem FormulaCentroid of a Trapezoid FormulaChain RuleChain Rule FormulaChange of Base FormulaChi Square FormulaCirclesCircumference FormulaCoefficient of Determination FormulaCoefficient of Variation FormulaCofactor FormulaComplete the square formulaComplex numbersCompound Interest FormulaConditional Probability FormulaConeConfidence Interval FormulaCongruence of TrianglesCorrelation Coefficient FormulaCos Double Angle FormulaCos Square theta FormulaCos Theta FormulaCosec Cot FormulaCosecant FormulaCosine FormulaCovariance FormulaCubeCurated Maths Resources for Teachers – EdulyteCylinderDecimalsDifferential calculusDiscover the world of MathsEllipseEquilateral triangleEuler’s formulaEven numbersExponentsFibonacci TheoryFractionFraction to decimalGeometric sequenceHeptagonHyperbolaIntegersIntegrationIntegration by partsLinesLocusMatricesNatural numbersNumber lineOdd numbersParallelogramPercentage formulaPerimeterPolygonPolynomialsPrismProbabilityPyramidPythagoras theoremRoman NumeralsScalene triangleSetsShapes NamesSimple interest formulaSlope formulaSolid shapesSphereSquareStandard deviation formulaSubtractionSymmetryTimeTrianglesTrigonometry formulaTypes of anglesValue of PiVariance formulaVectorVolume formulasVolume of a coneVolume of sphere formulaWhole numbers Home » Maths » Volume of a cone Volume of a cone Edulyte Maths Lessons Learn Maths anytime, anywhere Sign Up Unlocking the Secrets of Cone Volume: Formulas and Calculations Comprehensive Definition, Description, Examples & Rules Includes a Free Worksheet Edulyte Maths Lessons Learn Maths anytime, anywhere Sign Up What will you learn Introduction Conеs arе cool shapеs, likе icе crеam conеs or party hats. Thеy’rе not just fun to look at; they’re essential in many things wе usе еvеry day. One crucial thing wе want to know about conеs is how much stuff thеy can hold insidе. Wе call this thеir “volumе.” Think about it likе this: whеn you fill up a cup with juicе, you’rе finding its volumе. Understanding how much space is inside a conе helps us in different ways. Knowing thе volume of a conе hеlps us dеsign containеrs that can hold a lot of things. It’s like figuring out how much soda can fit in a conе-shapеd cup at a party. Engineers use cone volume to ensure large structures can hold desired amounts. This applies to water tanks and roller coasters. Today, we’re going to sее why understanding thе volumе of a conе is important. It’s not just about math; it’s about making things work bеttеr in our еvеryday livеs! What is thе Volumе of a Conе? Thе volumе of a conе is thе amount of spacе insidе it. To find this volumе, we use a simple formula. Imaginе thе conе as an icе crеam conе – widе at thе bottom and coming to a point at thе top. The volume is the amount of space the cone occupies in a 3-dimensional space. The formula for thе volumе of a conе is V = (1/3) π r^2 h. Volumе of a Conе Formula The formula for finding the volumе of a conе is V = (1/3) π r^2 h. Let’s break down this formula to understand еach part: V (Volumе): This is what we’re trying to find – how much spacе is insidе thе conе. π (Pi): It’s a special numbеr (approximatеly 3.14) used in many math formulas. For thе conе volumе, wе multiply π by thе radius squarеd and thе height. r (Radius): Imagine a circlе at thе bottom of thе conе. Thе radius is thе distancе from the cеntеr of this circlе to its еdgе. Wе squarе this valuе (multiply it by itsеlf). h (Hеight): This is how tall thе conе is, measured from thе tip to thе basе. Undеrstanding this formula hеlps us calculatе how much spacе diffеrеnt conеs can hold. Whеthеr it’s a traffic conе, an icе crеam conе, or a mountain-shapеd conе in math problems, wе can find out thеir volumеs using this formula. Step-by-Step Guide: How to Calculate the Volume of a Cone First, we need to understand the formula we have to use. Ensure you have a table with formulas and labels ready. This table will help you measure different parts of the cone. First, make a rough diagram of the problem to visualize the cone. Mark the measurements given and those you need to find. Use proper notations. Remember to halve the given Diameter. You can use the formula ( L = r^2 + h^2 ) to calculate the height or radius when you have the Slant Height. Note down all the respective values for r, h, and L, and now plug in the formula. Proceed with calculations and name the diagram once before proceeding. If you need clarification on the calculation, note this: you can use the volume of a cylinder with the same height. Divide it by 3, and you will get a value roughly equal to the volume of the cone. Remember to mention all unit lengths during operations. Review the entire answer after a problem is complete. Example: Given, R = 10 cm H = 10 cm V = (1/3) π r^2 h => (1 / 3) 3.1415 10^2 10 => 1047.197 cm^3 Real-world Applications We know cones have unique geometrical properties. Engineers today are implementing cones in their designs. Some modern applications of cones are Liquid Dispеnsеrs: Crеating еfficiеnt liquid dispеnsеrs rеliеs on knowlеdgе of conе volumе. It guarantееs еffеctivе storagе and dispеnsing of prеcisе bеvеragе amounts at sеlf-sеrvе stations. Pеtrolеum Industry: In thе oil sеctor, еnginееrs еmploy conе volumе computations whеn constructing tanks with conе-shapеd basеs. This еnsurеs prеcisе planning of storagе capacity within oil rеfinеriеs. 3D Printing: Mastеry of conе volumе provеs critical in thе world of 3D printing. It aids in optimizing matеrial usagе, minimizing wastе, and boosting thе ovеrall еfficiеncy of thе 3D printing procеss. Horticulturе Plantеrs: Crafting tapеrеd plantеrs or flowеrpots dеmands an undеrstanding of conе volumе. Determining the correct soil amount is crucial information for gardening. It promotes successful gardening and provides optimal space for plant growth. Firеwork Crafting: Pyrotеchnicians utilizе conе volumе calculations for firеwork shеlls. Crafting mesmerizing displays with specific burst patterns is essential. It adds an artistic touch to pyrotechnic shows. Comparison with Other Geometric Shapes Everybody is familiar with cubes and spheres. These geometrical shapes are unique in their own way. Knowing these concepts is crucial in engineering designs, especially when used with cones. Torus: Thе torus, rеsеmbling a doughnut or ring shapе, has a volumе formula of V=(πr^2)(2πR).Hеrе, R rеprеsеnts thе distancе from thе tubе’s cеntеr to thе torus cеntеr, and r is thе tubе radius. Unlikе common shapеs, thе torus formula involvеs thе multiplication of two radii and a constant (2 pi squarеd). This еquation dеfinеs thе spacе еnclosеd by thе torus surfacе. Frustum: Thе frustum, a truncatеd part of a conе or pyramid, is rеprеsеntеd by V=1/3 π h [(R^3-r^3)/ r]. In this formula, h is thе height, R is thе largеr basе radius, and r is thе smallеr top radius. Thе frustum formula incorporatеs hеights and radii of both basеs, including thеir sum and product. It еffеctivеly addrеssеs thе tapеrеd naturе of thе frustum. Common Mistakes to Avoid People often make errors when calculating the volume of a cone. These mistakes happen because they are in a hurry or unsure. Forgetting to convert Diameter to Radius in Volume Calculation Using slant height L as the Height of the cone Forgetting to square the radius in the formula Not using proper cubic units for the final answer Converting all values to a single unit of measurement (e.g., all should be in cm,m, or inches) Tips and Tricks Here are some tips and tricks to solve faster. Make a given column in the beginning and note down all values from the question/that are available to you. Convert all measurement units to a single unit, which should all be in either cm, m, inches, etc. Try to use a calculator for all your calculations. To remember the digits of pi, remember the sentence – “May I have a large container of coffee, please?”. Here, the number of letters in each word is the value of the special number. Which is approximately ‘3.1415926….’. Step Up Your Math Game Today! Free sign-up for a personalised dashboard, learning tools, and unlimited possibilities! Sign up Now Key Takeaways Working with cones is crucial in Engineering and CAD visualization. To apply these concepts, one must understand geometric shapes. The unit of measurement of the volume of a cone is Cubic The slant Height is not the same as the height of the cylinder The base of a cone is a 2-dimensional circle It has one vertice and can roll on one of its faces. Quiz Check your score in the end Download Free Worksheet Quiz Check your score in the end Question of Question comes here Answer Option Answer Option Answer Option Answer Option Submit Frequently Asked Questions Can I Usе a Diffеrеnt Unit of Mеasurеmеnt for radius and height in thе Formula? Yes, but make sure you write the final value’s correct dimensions. Arе Thеrе Altеrnativе Formulas for Calculating Conе Volumе? Re-interpreting the volume of a cone by writing the slant height L instead of r and h. Thеrе arе yet other variations for specific conе typеs. What Arе Common Mistakеs to Avoid when Calculating Conе Volumе? Some common errors arise from people using the incorrect value for height. The L value represents the slant height. Another mistake occurs when people don’t use the radius. The question provides the diameter. (R = D/2) Is Thеrе a Quick Mеthod for Estimating Conе Volumе? The volume of a cylinder relates to the formula. We can estimate the volume of an approximate cone by dividing the volume of the cylinder by 3. 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4875	https://www.teachoo.com/4500/599/Ex-6.1--10---Solve--x-3---x-2---1---Chapter-6-Linear-Inequalities/category/Ex-6.1/	Ex 5.1, 10 - Solve: x/3 > x/2 + 1 - Chapter 6 Linear Inequalities Everything Class 6 Class 6 Maths (Ganita Prakash) Class 6 Maths (old NCERT) Class 6 Science (Curiosity) Class 6 Science (old NCERT) Class 6 Social Science Class 6 English Class 7 Class 7 Maths (Ganita Prakash) Class 7 Maths (old NCERT) Class 7 Science (Curiosity) Class 7 Science (old NCERT) Class 7 Social Science Class 7 English Class 8 Class 8 Maths (Ganita Prakash) Class 8 Maths (old NCERT) Class 8 Science (Curiosity) Class 8 Science (old NCERT) Class 8 Social Science Class 8 English Class 9 Class 9 Maths Class 9 Science Class 9 Social Science Class 9 English Class 10 Class 10 Maths Class 10 Science Class 10 Social Science Class 10 English Class 11 Class 11 Maths Class 11 English Class 11 Computer Science (Python) Class 12 Class 12 Maths Class 12 English Class 12 Economics Class 12 Accountancy Class 12 Physics Class 12 Chemistry Class 12 Biology Class 12 Computer Science (Python) Class 12 Physical Education Courses GST and Accounting Course Excel Course Tally Course Finance and CMA Data Course Payroll Course Interesting Learn English Learn Excel Learn Tally Learn GST (Goods and Services Tax) Learn Accounting and Finance Popular GST Demo GST Tax Invoice Format Accounts Tax Practical Tally Ledger List Maths Updated 2025-26 Class 6 Maths (Ganita Prakash) Class 6 Maths (old NCERT) Class 7 Maths (Ganita Prakash) Class 7 Maths (old NCERT) Class 8 Maths (Ganita Prakash) Class 8 Maths (old NCERT) Class 9 Maths Class 10 Maths Class 11 Maths Class 12 Maths Sample Paper Class 10 Maths Sample Paper Class 12 Maths Accounts & Finance Learn Excel free Learn Tally free Learn GST (Goods and Services Tax) Learn Accounting and Finance GST and Income Tax Return Filing Course Tally Ledger List Popular Class 6 Science (Curiosity) Class 6 Science (old NCERT) Class 7 Science (Curiosity) Class 7 Science (old NCERT) Class 8 Science (Curiosity) Class 8 Science (old NCERT) Class 9 Science Class 10 Science Remove Ads Login Maths Remove ads Science GST Accounts Tax Englishtan Excel Social Science Class 6 Class 6 Maths (Ganita Prakash) Class 6 Maths (old NCERT) Class 6 Science (Curiosity) Class 6 Science (old NCERT) Class 6 Social Science Class 6 English Class 7 Class 7 Maths (Ganita Prakash) Class 7 Maths (old NCERT) Class 7 Science (Curiosity) Class 7 Science (old NCERT) Class 7 Social Science Class 7 English Class 8 Class 8 Maths (Ganita Prakash) Class 8 Maths (old NCERT) Class 8 Science (Curiosity) Class 8 Science (old NCERT) Class 8 Social Science Class 8 English Class 9 Class 9 Maths Class 9 Science Class 9 Social Science Class 9 English Class 10 Class 10 Maths Class 10 Science Class 10 Social Science Class 10 English Class 11 Class 11 Maths Class 11 Computer Science (Python) Class 11 English Class 12 Class 12 Maths Class 12 English Class 12 Economics Class 12 Accountancy Class 12 Physics Class 12 Chemistry Class 12 Biology Class 12 Computer Science (Python) Class 12 Physical Education Courses GST and Accounting Course Excel Course Tally Course Finance and CMA Data Course Payroll Course Interesting Learn English Learn Excel Learn Tally Learn GST (Goods and Services Tax) Learn Accounting and Finance Popular GST Demo GST Tax Invoice Format Accounts Tax Practical Tally Ledger List Remove Ads Login Chapter 5 Class 11 Linear Inequalities Serial order wise Ex 5.1 Ex 5.1 Ex 5.1, 1 Ex 5.1, 2 Ex 5.1, 3 Ex 5.1, 4 Important Ex 5.1, 5 Ex 5.1, 6 Important Ex 5.1, 7 Ex 5.1, 8 Important Ex 5.1, 9 Ex 5.1, 10 Important You are here Ex 5.1, 11 Important Ex 5.1, 12 Ex 5.1, 13 Ex 5.1, 14 Ex 5.1, 15 Ex 5.1, 16 Important Ex 5.1, 17 Important Ex 5.1, 18 Ex 5.1, 19 Ex 5.1, 20 Important Ex 5.1, 21 Ex 5.1, 22 Important Ex 5.1, 23 Important Ex 5.1, 24 Ex 5.1, 25 Important Ex 5.1, 26 Examples→ Chapter 5 Class 11 Linear Inequalities Serial order wise Ex 5.1 Examples Miscellaneous Solving Linear Inequality Graphically Solving Pair of Linear Inequalities Ex 5.1, 10 - Chapter 5 Class 11 Linear Inequalities Last updated at December 16, 2024 by Teachoo ADVERTISEMENT Powered by VidCrunch Next Stay Playback speed 1x Normal Quality Auto Back 720p 360p 240p 144p Auto Back 0.25x 0.5x 1x Normal 1.5x 2x / Skip Ads by Next: Ex 5.1, 11 Important →Remove AdsShare on WhatsApp Transcript Ex 5.1, 10 Solve the given inequality for real x: 𝑥/3 > 𝑥/2 + 1 𝑥/3 > 𝑥/2 + 1 𝑥/3 −𝑥/2 > 1 (2𝑥 − 3𝑥)/6 > 1 (−𝑥)/6 > 1 −x > 6 Since x is negative, we multiply both sides by -1 & change the signs (– 1) × (–x) < (– 1) × (6) x < –6 C Hence x is a real number which is less than –6 Thus , solution is (–∞, –6). Show More Chapter 5 Class 11 Linear Inequalities Serial order wise Ex 5.1 Ex 5.1, 1 Ex 5.1, 2 Ex 5.1, 3 Ex 5.1, 4 Important Ex 5.1, 5 Ex 5.1, 6 Important Ex 5.1, 7 Ex 5.1, 8 Important Ex 5.1, 9 Ex 5.1, 10 Important You are here Ex 5.1, 11 Important Ex 5.1, 12 Ex 5.1, 13 Ex 5.1, 14 Ex 5.1, 15 Ex 5.1, 16 Important Ex 5.1, 17 Important Ex 5.1, 18 Ex 5.1, 19 Ex 5.1, 20 Important Ex 5.1, 21 Ex 5.1, 22 Important Ex 5.1, 23 Important Ex 5.1, 24 Ex 5.1, 25 Important Ex 5.1, 26 Examples→ Made by Davneet Singh Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo Hi, it looks like you're using AdBlock :( Displaying ads are our only source of revenue. To help Teachoo create more content, and view the ad-free version of Teachooo... please purchase Teachoo Black subscription. Join Teachoo Black at ₹39 only Please login to view more pages. 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4876	https://pdfcoffee.com/coordinate-geometry-loney-sl-pdf-free.html	Email: [email protected] Login Register English Deutsch Español Français Português Coordinate Geometry Loney SL Author / Uploaded DivyaGoel1 Categories Ellipse Circle Cartesian Coordinate System Line (Geometry) Coordinate System uTd-nu y: /o- — — BY THE SAME AUTHOR. A Treatise on Elementary Dynamics. Third Edition, Revised and Enlarged. Cro Views 16,678 Downloads 3,065 File size 21MB Report DMCA / Copyright DOWNLOAD FILE Recommend Stories ###### Coordinate Geometry by Sl Loney 876 166 16MB Read more ###### The Elements of Coordinate Geometry - S.L. 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Third Edition, Revised and Enlarged.Crown7s.8vo.6 c?.Solutions of the Examples in the Elementary Dynamics. Crown 8vo. 75. Qd.The Elementsof Statics and Dynamics. Fcap. 8vo. Elements of Statics. Fourth Edition, 4s. 6d. II. Elements of Dynamics. Fourth Ed., 3s. Qd. two Parts bound in one Volume. The 7s. Qd.Part PartI."Mr Loney shows that he knows how to combine perspicuity with brevity in a remarkable degree. One feature of both books is that the author points out the portions that are adapted for a first reading, and also those that are required for particular examinations." GUdTd'tQjTtPARTELEMENTS OFI.STATICS."Students reading for the different examinations at Cambridge, for the London University Matriculation, and intermediate Science, for the Woolwich Entrance Examinations, will fi^nd a statement of the part of the book to be read on this subject. Mr Loney deserves much praise for the uniformly high standard he has shown in this his Glasgow Herald. latest effort."andSolutions of the Examples Staticsand Dynamics.inthe Elements ofEx. Fcap. Svo.7s. Qd.Mechanics and Hydrostatics for Beginners. Fcap. 8vo.Third Edition.Ex.3s. Qd.Parts. Crown 8vo. Second and including the Solution ofPlane Trigonometry. In two Edition. Part Triangles. 5s.PartII.tions.DeI.uptoMoivre's Theorem and the higher por-3s. 6d.The two Parts boundin oneYolume.7s. Qd.NATUREsays ; " ...It would be difficult to find a better introduction to Plane Trigonometry."SCHOOL GUARDIANsays; " ...It is a model of method, and the simplicity of its explanations, its numerous illustrative diagrams, and its large and varied collection of exercises and examples combine to render it especially suitable for the use of young pupils and private students.... The typography of the volume is in every respect admirable...."Theclearness,and accuracy;""The SCHOOLMASTER says; " ...Mr Loney, using the soundest judgment in his choice of matter, presents it with inimitable brevity and clearness ; the publishers, too, have vied with the author in their efforts to excel, and, as a result, the book is in every way worthy of commendation." The SCOTSMAN says " ...It expounds the subject with a skill and fulness which give evidence of a peculiar experience in teaching and of a special appreciation of the needs of students. ;ThePRACTICAL TEACHERcompetitorsitcannotsays; "...Among its numerous a deservedly high and popularfail to attainposition."The NATIONAL TEACHER (Ireland) says '• ...Seldom do such books come under the notice of the reviewer. For conciseness and clearness it has few compeers. It is deep without dullness ; comprehensive without wearisomeness. It comes before us with a newness and freshness almost amounting to novelty...." ;The SPEAKER says ; "Mr Loney as a writer of elementary mathematical treatises maintains a high standard. His Elementary Dynamics is marked by its brevity and clearness and deserves its success.... The Cambridge Press has every reason to be proud of its achievement."The made aEDUCATIONAL REVIEWsays;"...The author has notspecial point of developing any one particular branch of the subject, but his work is of uniform character throughout, and, we may add, of uniformly good quality.... The arrangement of sections is excellent ; the attention is clearly directed to important points, and the style in which the book is produced may be fairly described asluxurious."GLASGOW HERALDsays ; " ...Mr Loney's text-book is sure The to be widely used in the several mathematical classes where sound work The publisher's part of the work is also most admirably is expected.done.CORRESPONDENTsays; "...It is well The UNIVERSITY written and the chapters relating to the changes of sign and magnitude of trigonometrical ratios in different quadrants are especially deserving of praise...."SCIENCE AND ARTsays; "...In the analytical part of the subject, considerable attention has been given to complex quantities, and the author is to be congratulated on the lucid way in which he has treated them.... The book may be strongly recommended as a first-rate text-book."aontion:C.J.CLAY and SONS,CAMBRIDGE UNIVERSITY PRESS WAREHOUSE, AVE MARIA LANE. (ffilagfloia:263,ARGTLE STREET.THE ELEMENTS OPCOOEDINATE GEOMETRY.THE ELEMENTS OFCOOEDINATE aEOMETRYBYS.L.LONEY,M.A.,LATE FELLOW OF SIDNEY SUSSEX COLLEGE, CAMBRIDGE, PROFESSOR AT THE ROYAL HOLLOWAY COLLEGE.^^^l^^mU. MASS. MATH, DEPTiMACMILLAN AND AND NEW YOEK. 1895 [All Bights reserved.']CO.CTambrtlJse:PRINTED BYJ.&C.F.CLAY,AT THE UNIVEBSITY PRESS.150553PKEFACE. "TNthe following work I have tried to present theelements of Coordinate Geometry in a manner suitablepresentJunior Students.Thebook only deals with Cartesian andPolarBeginners andforWithin theseCoordinates.that the bookis fairlylimits I venture tocomplete, and that no proposi-tions of very great importance haveThe Straight Line and morefullysinceitishopebeen omitted.Circle have been treatedthan the other portions of the subject, generally in theelementary conceptionsthat beginners find great difficulties.There are a large number of Examples, over 1100 inall,and theycharacter.are,inThe examplesgeneral, of an are especiallythe earlier parts of the book.elementarynumerous inPREFACE.viforreadingportions of the proof sheets, but especially toMr W.IJ.am muchindebted to several friendsDobbs, M.A. who has kindly read the whole of thebook and made many valuable suggestions. For any shallbecriticisms,suggestions, orgrateful. S.EoTAIicorrections, IHOLLOWAY COLLEGE,Egham, Surbey. July4, 1895.L.LONEY.CONTENTS. CHAP. I.II.Introduction....Algebraic KesultsCoordinates. Lengths of Straight Lines and Areas of Triangles Polar CoordinatesEquation to a LocusLocus.IV.The Straight Line. Eect angular Coordinates.... ......Length of a perpendicular...Line.51......Equations representing two or more Straight Lines lines givenby one equationGreneral equation of the second degreeVII.66.lociAngle between twoTransformation of Coordinates Invariants42Polar Equations andEquations involving an arbitrary constantVI.3958Oblique Coordinates Examples of31.44..".Bisectors of anglesThe Straight824Straight line through two points Angle between two given straight lines Conditions that they may be parallel and per-V119III.pendicularPAGE.....73 8088 90 94 109 115.CONTENTS.VliiPAGECHAP.VIII.The Circle118Equation to a tangent126 137Pole and polarEquation to a circle in polar coordinates Equation referred to oblique axes Equations in terms of one variable ..150.160Systems of Circles Orthogonal circlesX.148...IX..145....,..160.Kadical axis161Coaxal circles166The ParabolaConic Sections.174.180Equation to a tangentSomeproperties of the parabola.187..190Pole and polar195Diameters Equations in terms of one variable XI.The Parabola{continued')..198.....Loci connected with the parabola Three normals passing through a given point Parabola referred to two tangents as axes ...211The Ellipse225Auxiliary circle and eccentric angleEquation to a tangentSomeConjugate diameters237......249.Pour normals through any point Examples of loci XIII..231.......properties of the ellipsePole and polar206.217.XII.206......254 265266The Hyperbola Asymptotes Equation referred to the asymptotes as axes One variable. Examples242271284 .296 299.CONTENTS. CHAP.XIV.Polar Equation to, a ConicIX....Polar equation to a tangent, polar, and normalXV.General Equation.Tracing of CurvesParticular cases of conic sectionsPAGE306 313,322.... .322Transformation of equation to centre as origin Equation to asymptotes Tracing a parabola...... ...... ......Tracing a central conic......Eccentricity and foci of general conicXVI.General Equation Tangent Conjugate diameters326.329 332 338 342349 349 352Conies through the intersections of two conies356The equation S=Xuv358...... .......General equation to thej)airof tangentsdrawnfrom any pointThe director The foci The axescircle367 369Lengths of straight lines drawn in given directions to meet the conic Conies passing through four 23oints ...Conies touching four linesTheLM=B?conicXVII. Miscellaneous PropositionsOn364 365....370 378 380 382 385the four normals from any point to a central conicConfocal conies.......Circles of curvatureand contact of the third order385.EnvelopesAnswers392 398 407.i—xiiiERKATA. Page „ ,,,,87,Ex.27, line 4.235, Ex. 18, line 3.„,,282, Ex.,,3.line 5.For "JR" read " S."For "odd" read "even." Dele"and Page37,Ex. 15."For "transverse" read "conjugate."CHAPTERI.INTRODUCTION.SOME ALGEBRAIC RESULTS. 1.Quadratic Equations.Theroots of the quad-ratic equation a'3^mayeasily+ 6x + c =be shewn to be-&+•JlP'— 4ac1-b- s/b^ — 4:aG'^"'^•2^2i.They areand unequal,equal, or imaginary, according as the quantity b^—iac is positive, zero, or negative,therefore reali.e.2.andaccording asb^=4:ac.Relations between the roots of any algebraic equation of the terms of the equation.the coejicientsIf any equation be written so that the coefficient of the highest term is unity, it is shewn in any treatise on Algebra thatthe sum of the roots is equal to the coefficient of (1) the second term with its sign changed, (2)theat a time,issum of the products of the roots, taken two equal to the coefficient of the third term,the sum of their products, taken three at a time, (3) equal to the coefficient of the fourth term with its sign changed, isand so L.on.e1,.COORDINATE GEOMETRY. Ex.If1.a and/3ax'^2.If a,b+ bx + c = 0,i.e.x^c+ - x + ~ = 0, a— ba + p=we haveEx.be the roots of the equationac^and a^ = -.and 7 be the roots of the cubic equationj8,ax^ + bx^ + cx + d=0, i.e.x^+-x^ofawe have+-x + - = 0, aa+ p + y:a^y + ya + a^=:ando-Pl-3.shewn that the solutionIt can easily beof theequations+a^xh^y+G^z= 0,a^ + h^y + c^z = 0,andX ISy~ ^2^1^1^2^1^2 ~ ^2^1'^1^2~ ^2^1Determinant Notation. 4.Theis calledquantity-a determinant of thesecond order and stands for the quantityExs.(1)d-yfd^^1,ha-})^— aj)^,so that= Ob^^ — 6»2&i\%^\ = 2x5-4x3 = 10-12=-2; ;'\|!4, 5i3, (ii)-7,-4\|-6 = 3X(-6)-{-7)X(- 4) = 18 - 28 = -10.DETERMINANTS. «!,5.The quantityCOORDINATE GEOMETRY.8.The quantity j(h.1^2>%J^461,&2)hih^11^25^1) ^2^3> 5^3) ^4called a determinant of the fourth order and stands for the quantityisK «i X^2» h, ^3^4J— Clo^3> ^4Xi^lJ^35C-,^3}\11&i,+ 6^3h 433562JX^4! C^1?254&1, cCj_X.ELIMINATION.5Elimination. Suppose11.we havethe two equations+ a^y =aj^x(1),\x +b^y ^0 two unknown quantities x andbetween the be some relation holding between the four ^or, from (1), we have bi, and 63(2),y.There mustcoefficients6i, ctaj•and, from(2),we havey~%'-=— yKEquating these two valuesi.e.a-J)^=-^Xof -we have— ajb^ =(3).The result (3) is the condition that both the equations and (2) should be true for the same values of x and y. The process of finding this condition is called the eliminating of X and y from the equations (1) and (2), and the result (3) is often called the eliminant of (1) and (2). Using the notation of Art. 4, the result (3) may be (1)written in the form1)'^0.is obtained from (1) and (2) by taking the x and y in the order in which they occur in the equations, placing them in this order to form a determinant, and equating it to zero.This resultcoefficients of12.we have a-^x + a^y + a^^ = \x+ h^y^ h^z = G^x + G^y + C3S = unknown quantitiesSuppose, again, thatand between the threethe three equations (1),(2),(3), x, y,andz.COORDINATE GEOMETRY.6Bydividing each equation by zwe havethree equationsXbetween the two unknown quantities — and yz%,zTwoofELIMINATION. 14.andIf againwe havethe four equationsa-^x+dil/+cf'zZ+ a^u =0,h^x+h^y+b^z+b^u=0,Ci«;+c^i/+G^z+c^u=0,djX + d^y+d.^z+d^ — 0,could be shewn that the result of eliminating the four quantities cc, y, z^ and u is the determinantit«1JCHAPTER COORDINATES.II.LENGTHS OF STRAIGHT LINES AND AREAS OF TRIANGLES.OX07and 15. Coordinates. Let be two fixed The line is straight lines in the plane of the paper. the axis of y, whilst the called the axis of cc, the line two together are called the axes of coordinates.OXOYThe pointiscalled the origin of coordinates or,moreshortly, the origin.FFrom any point in the plane draw a straight lineOF to meet OX M. The distance OM is called the Abscissa, and the distanceparallel toinMP the Ordinate of the point P, whilst the abscissa and the ordinate together are called its Coordinates.OXDistances measured parallel to or without a suffix, {e.g.Xj, x.-^... x\ measured parallel to OY are called suffix, (e.g.2/i, 2/2,---If the distancesthe coordinates of2/'.y",---)-OM and MPP are,are called a?, with x",...), and distances y, with or without abe respectively x and ?/, by the symbolfor brevity, denoted{x, y).Conversely, when we are given that the coordinates of For from we a point are (x, y) we know its position. {—x) along and have only to measure a distancePOMOXCOORDINATES. then from 21 measure a distance9MPOY{=y) parallel to arrive at the position of the point P. For example be equal to the unit of length and in the figure, ifand weMP= WM,OMP is the point (1, 2). Produce XO backwards to form thenthe line OX' and backwards to become OY'. In Analytical Geometry we have the same rule as to signs that the student has already met with in Trigonometry. Lines measured parallel to OX are positive whilst those measured parallel to OX' are negative ; lines measured parallel to OY are positive and those parallel to OY' are16.YOnegative.quadrant YOX' and P^M^, drawn y, meet OX' in M^^ and if the numerical values of the quantities OM^ and J/aPg be a and h, the coordinates of P are {-a and h) and the position of Pg is given by the symbol (—a, h). IfP2 b® i^li®parallel to the axis ofSimilarly, if P3 be in the third quadrant X'OY', both of coordinates are negative, and, if the numerical lengths of Oi/3 and J/3P3 be c and d, then P3 is denoted by theitssymbol (—c,Finally,positive 17.and Ex. (i)—d).in the fourth quadrant its abscissa is its ordinate is negative.ifP4lieLay down on "paper (2,-1),(ii)(-3,the position of the points 2),and(iii)(-2, -3).To get the first point we measure a distance 2 along OX and then a distance 1 parallel to OF'; we thus arrive at the required point. To get the second point, we measure a distance 3 along OX', and then 2 parallel to OY. To get the third point, we measure 2 along OX' and then 3 parallel to OT. These three points are respectively the points P4 P., and Pg in ,,the figure of Art. 15.18. When the axes of coordinates are as in the figure of Art. 15, not at right angles, they are said to be Oblique Axes, and the angle between their two positive directions and 07, i.e. the angle XOY, is generally denoted byOXthe Greek letterw.COORDINATE GEOMETRY.10 In general,it ishowever foundtoOXbe more convenient to They are thenand OZat right angles. take the axes said to be Rectangular Axes.It may always be assumed throughout this book that the axes are rectangular unless it is otherwise stated.The systemspoken of in the last System of CoordiIt is so called because this system was first intronates. duced by the philosopher Des Cartes. There are other systems of coordinates in use, but the Cartesian system is by far the most important. 19.fewarticles isTo find20.of coordinatesknownas the Cartesianthe distance betweentwo points whose co-ordinates are given.Let Pi and P^ be the two given points, and let their coordinates be respectively {x^ y^) ,and(a^sj 2/2)-Draw rallelJ/jtotoPji/i and P^M^ paOY, to meet OX inand M^.Draw P^R parallelOX to meet M-^P^ in R.'qMjvTThenP^R = M^Mt^ = OM^ - OMc^ = oi^-X2, RP, = M,P,-M,P, = y,~y,, andz P^i^Pi= z6>ifiPa-l 80° -PiJfiX^l 80° -^^ ^^^^^^^^^"^^ '/^^^let>/yyC ""-., ''-•.,'pM"^(ii) For the second point, the radius vector revolves from OX through 150° and is then in the position OP^ ; measuring a distance 3 along it we arrive at Pg2—2COORDINATE GEOMETKY.20(iii) For the third point, let the radius vector revolve from OX through 45° into the position OL. We have now to measure along OL a distance - 2, i.e. we have to measure a distance 2 not along OL but in the opposite direction. Producing iO to Pg, so that OP3 is 2 units of length, we have the required point P3.Toget the fourth point, we let the radius vector rotate from 330° into the position and measure on it a distance -3, i.e. 3 in the direction produced. thus have the point P^y which is the same as the point given by (ii). (iv)OX throughOMMOWe(v) If the radius vector rotate through - 210°, position OP2, and the point required is Pg.itbe in thewillFor the sixthpoint, the radius vector, after rotating through in the position OM: then measure - 3 along it, i.e. 3 in the direction produced, and once more arrive at the point Pg. (vi)^- 30°,WeisMO32. It will be observed that in the previous example the same point P^ is denoted by each of the four sets of polar coordinates (3, 150°),-210°) and (-3, -30°). be found that the same point is given(-3, 330°),(3,In general it v^ill by each of the polar coordinates (r, 0), (- r, 180° + 6), {r, - (360° or,and {- r, - (180° expressing the angles in radians, by each of the 6)]6% co-ordinates (r,e\ {-r,7r +6), {r,- (27r - 0)} and{- r,-(tt-$)}.It is also clear that adding 360° (or any multiple of 360°) to the vectorial angle does not alter the final position of the revolving line, so that {r, 6) is always the same point as (r, ^ + ?i 360°), where n is an integer. .So, adding 180° or any odd multiple of 180° to the vectorial angle and changing the sign of the radius vectorgives thesame point asbefore.[-r, ^ isthe same point as [—r,Thus the point+ (2n + 1)180°] 6+180°],i.e. isthe point[r,6\33. To find the length of the straight line joining two points whose polar coordinates are given. LetA and Bcoordinates be(r^,be the two points and let their polar 6y) and (r^, 6^ respectively, so thatOA^r^, OB = r^, lXOA^O^, andlX0B = 6^,POLAR COORDINATES.21Then (Trigonometry, Art. 164) AB" - OA'' + OB'' -20 A. OB cos = r-^ + r^ - 2r-^r^ cos {0^ - 6^.AOB34. To find the area of a triangle the coordinates of whose angular points are given. Let ABC be the triangle and be the polar coordinates of angular points. We havelet(r-^,0^),(r^,62),and(rg, ^3)itsAABO=AOBC+aOCA -AOBA(1).NowA0BC = i0B,0C sin BOC [Trigonometry, Art. 198] 'Soand= ^r^r^ sin (^3 - $^). A OCA = \0G OA sin CO A = ^r^r, sin (6, - 6,), AOAB^^OA. OB sin AOB = ^r^r^ sin {6^ - 6.^ = - Jn^2 sin (^2 - ^1). .Hence(1) givesA ABC = J\r\x = J'I\2/=2r2/=x/3r2/=V2x = -'2,\ 2/^0x^-.^?>,\ /' 2/^--ix== 0, 2/ = -2jx=l,r53\ 'yx^l x=^iJ\V3/'r-J2A3/=-v2r\y=i x = -l,r \y=-si^)'x=J2, }L and x=J3) -./2/'-'"% = -l/2/ =EQUATION TO A LOCUS.127•COORDINATE GEOMETRY.28a large number of values of x and the corresponding values of ?/, the points thus obtained would be found all to lie on the curve in the figure. Ifwe tookBothof its brancheswould be foundto stretchawaytoinfinity towards the right of the figure.took any point on this curve and measured accuracy its x and y the values thus obtained would be found to satisfy equation (1). Also we should not be able to find any point, not lying on the curve, whose coordinates would satisfy (1). In the language of Analytical Geometry the equation This curve is called (1) is the equation to the above curve. a Parabola and will be fully discussed in Chapter X. Also,withweifsufficientIf a point move so as to satisfy any given condition describe some definite curve, or locus, and there can always be found an equation between the x and y of any42.it willpoint on the path.This equationiscalled the equation to the locus orHencecurve.Def.Equation to a curve.The equationtoacurve is the relation which exists between the coordinates of any foint on the curve^ and which holds for no other points except those lying on the curve.43. will beConversely to every equation between x and y it is, in general, a definite geometricalfound that therelocus.Thus in Art. 39 the equationisx + y=\, and the P^P^ (produceddefinite path, or locus, is the straight lineindefinitelyboth ways).In Art. 40 the equation path, or locus,isthe dottedisx'^+y'^^ 4,and thedefinitecircle.Again the equation 2/ = 1 states that the moving point is such that its ordinate is always unity, i.e. that it is always at a distance 1 from the axis of x. The definite path, or locus, is therefore a straight line parallel to and at a distance unity from it.OXEQUATION TO A LOCUS.29In the next chapter it will be found that if the equation be of the first degree {i.e. if it contain no products, squares, or higher powers of x and y) the locus44.corresponding is always a straight line. If the equation be of the second or higher degree, the corresponding locus is, in general, a curved line.We append45.a few simple examples of the forma-tion of the equation to a locus.Ex. 1. A point moves so that from tioo given perpendicular axes find the equationthe algebraic isequaltosum ofitsdistancesa constant quantity a;to its locus.Take the two straight lines as the axes of coordinates. Let {x, y) be any point satisfying the given condition. We then ha,wex + y = a. This being the relation connecting the coordinates of any point on the locus is the equation to the locus. It will be found in the next chapter that this equation represents a straightline.Ex. 2. The sum of the squares of the distances of a moving point from the tioo fixed points {a, 0) and {-a, 0) is equal to a constant quantity 2c^. Find the equation to its locus. Let (a;, y) be any position of the moving point. Then, by Art. 20, the condition of the question gives {[x- af + /} +{ (a;+ af + if] = 2c\x^ + y'^ = c^- a~.i.CiThis being the relation between the coordinates of any, and every, point that satisfies the given condition is, by Art. 42, the equation to the required locus. This equation tells us that the square of the distance of the point {x, y) from the origin is constant and equal to c^ - a^, and therefore the locus of the point is a circle whose centre is the origin.isEx. 3. A point moves so that its distance from the point (-1,0) always three times its distance from the point (0, 2).Let {x, y) be any point which then havesatisfiesthe given condition.WeJ{x + iy' + {y-0)^=Bj{x - 0)2+ {y - 2)2, so that,on squaring, x'^i.e.+ 2x + l + y'^=9{x'^ + y'^-4:y + 4), 8(a;2 + y2)_2a;-36?/ + 35 = 0.This being the relation between the coordinates of each, and that satisfies the given relation is, by Art. 42, theevery, pointrequired equation. It will be found, in a later chapter, that this equation represents a circle.COOEDINATE GEOMETRY.30EXAMPLES.IV.By taking a number of solutions, as in Arts. 39 the loci of the following equations—41,sketch:1.2x + dy = l0.4.a;2-4aa; + ?/2 + 3a22.^x-y = l.= 0.5.x'^-2ax-Vy'^ = Q.3. y'^= x.6.^x = y^-^.^' + ^'=1. '4^97A and B being the fixed points (a, 0) and ( obtain the equations giving the locus of P, when 8.is- P52 _ a constant quantity = 2fc2.PA"^10.PA = nPB, n being constant. P^+PjB = c, a constant quantity.11.PB^ + PC^=2PA^, C9.a, 0) respectively,being the point(c, 0).12. Find the locus of a point whose distance from the point equal to its distance from the axis of y.whichFind the equation to the locus of a point distant from the points whose coordinates are 13.(1, 0)15.{a+b,and(0,-2).14.(2, 3)isand(1, 2)always equi(4, 5).a-h) and {a-b, a + b).Find the equation to the locus of a point which moves so that 16. its distance from the axis of x the axis of y.17. its distance from the point tance from the axis of y. 18. thesumisthree times(a, 0)isitsdistance fromalways four timesof the squares of its distancesfrom the axesits dis-isequalto 3.19. the square of20.from 21.itsitsdistance from the pointdistance from the point(3, 0) is(0, 2) isequal to4.three times its distance(0, 2).itsfrom thedistance from the axis of xisalways one halfitsdistanceorigin.Afixed point is at a perpendicular distance a from a fixed 22. straight line and a point moves so that its distance from the fixed point is always equal to its distance from the fixed line. Find the equation to its locus, the axes of coordinates being drawn through the fixed point and being parallel and perpendicular to the given line.23. In the previous question if the first distance be (1), always half, (2), always twice, the second distance, find the equations to theandrespective loci.CHAPTER THE STRAIGHTLINE.IV.RECTANGULAR COORDINATES.46. To find the equation to a straight line which is parallel to one of the coordinate axes. Let CL be any line parallel to the axis of y and passing through a point C on the axis of x such that OG = c. Let F be any point on X and y.Then the alwaysc,this lineabscissa of the pointwhose coordinates areFisso thatx=c(1).This being true for every point on the line CL (produced indefinitely both ways), and for no other point, is, by Art. 42, the equation to the line.XIt will be noted that the equation does not contain thecoordinatey.Similarly the equation to a straight line parallel to the axis oi X is y — d.Cor. The equation to the axis of a? is The equation to the axis oi y is x — 0.2/= 0.47. To find the equation to a st7'aight line which cuts off a given intercept on the axis of y and is inclined at a given angle to the axis of x. Let the given intercept becandletthe given angle be a.COORDINATE GEOMETRY.32Let C be a point on the axis Through C draw a straight line Z(7Z' inclined at an angle a (= tan~^ m) to the axis of x^ so that tan aOCof y such thatisc.— m.^^^^ OThe straight line LCL' is therefore the straight line required, and we have to -'l find the relation between the lying on coordinates of any pointPDraw PM perpendicular G parallel to OX.toMXit.OXtomeet in^alinethroughLet the coordinates = y. and Then MP = NP +ofPbecuand?/»so thatOM=xMPMN =C]Srt^iia + 00 = m.x +c,y = mx+c.i.e.This relation being true for any point on the given straight line is, by Art. 42, the equation to the straight line.[In this, and other similar cases, it could be shewn, is only true for points lyingconversely, that the equation on the given straight line.]to any straight line passing through which cuts off a zero intercept from the axis found by putting c — O and hence is 3/ = mx.The equationCor.the origin, of2/,isi.e.48. The angle a which is used in the previous article is the angle through which a straight line, originally parallel to OZ, would have to turn in order to coincide with the given direction, the rotation being always in the positive direction. Also m is always the tangent In the case of such a straight line as AB, in the figure of this angle. is equal to the tangent of the angle PAX (not of the of Art. 50, angle PAO). In this case therefore wi, being the tangent of an obtuse angle, is a negative quantity.mThe student should verify the truth of the equation on the straight line LCL', and alsoof the last for straightfor such a straight line asA^B^ in thearticle for all pointsHnes in other positions, e.g. figure of Art. 59. In thislatter casebothmandc are negativequantities.Acareful consideration of all the possible cases of a few propositions will soon satisfy him that this verification is not always necessary, but that it is sufficient to consider the standard figure.THE STRAIGHT33LINE.49. Ex. The equation to the straight line cutting off an intercept 3 from the negative direction of the axis of y, and inclined at 120° to the axis of a;, is= a;tanl20° + (-3), y= -x^S-S, y + x^S + S = 0. ?/i.e. i.e.50.1^0off giventhe equation to the straight linefinda andi7iterceptsOXLet A and B be on such that OA = a and OB =ABJoinand producedefinitely both ways.which cutsh from the axes.andOYrespectively,and beh.it in-PLetbeany point (x, y) on this straight perpendicular line, and drawPMtoOX.We require therelation thatalways holds between x and long as P lies on AB.ByEuc. YI.3/,sowe have4,OM_PBMP _AP ^"""^OA~AB''OB~ABOM MP PB + AP = + ^.e.ThisABOBOAXyaD1,^therefore the required equation ; for it is the between the coordinates of any point the given straight line. isrelation that holdslying on 51.The equationin the preceding articlemayhe also obtainedby expressing the fact that the sum of the areas of the triangles and OPB is equal to OAB, so thatOP A\axy + \hy.x = \ax'b, and hencea52. Ex. 1. Find the equation to through the -point (3, - 4) and cutting opposite signs,fromstraightline passing equal but ofthe tioo axes.Let the intercepts cut— a.theoff intercepts,offfrom the two axes be of lengths a andCOORDINATE GEOMETRY.34 The equationto the straight line is then-aax-y = ai.e.(3,(1).Since, in addition, the straight line is to go through the point -4), these coordinates must satisfy (1), so that3-(-4) = a, and therefore The required equationa = l. isthereforex-y = 7. Ex. 2. Find the equation to the straight line lohich passes through the point (-5, 4) and is such that the portion of it between the axes is divided by the point in the ratio ofl 2. :+ t = 1. bLet the required straight line be a in the points whose coordinates are {a,and0)The coordinatesThis meets the axes (0, &).theof the point dividing points in the ratio 1 2, are (Art. 22)joining theseline:2.a+1.0 If thisbe the point,(-2.0 + 1.&2(1.i.e,-^b,and -.we have5, 4)2a „ b -5:=and 4=-, so thatThe,,a=--Y- andbrequired straight lineisthereforeX y -i^^l2 oyI.e.53.To find= 12.8a;'= 60. a straightthe equation toline in tennsofperpendicular let fall upon it from the origin and the angle that this perpendicular makes with the axis of x. theLet be jo.ORbe the perpendicular fromLet a be the angle that with OX.LetPORmakesbe any point, whoseco-ABordinates are x and y, lying on draw the ordinate PM, and also perpendicular to OR and perpendicular to ML.PN',MLandlet itslength85THE STRAIGHT LINE.OL = OMco^aThen(1),LR = NP = MF&inNMP.andlNMP^W - lNMO= iMOL^a.ButLR = MP&m.a Hence, adding Oil/ cos a(2).and (2), we have a=OL + LR=OR ifPsin + (1)=79,X COS a + y sin a = p.i.e.Thisisthe required equation.—54. In Arts. 47 53 we have found that the corresponding equations are only of the first degree in x and y. shall now prove thatAnyWeequation of the first degreei7ix and y always repre-a straight line. For the most general form of such an equation is Ax + By^C = ^ (1), which do where A^ B, and C are constants, i.e. quantities not contain x and y and which remain the same for all points on the locus. Let (cCi, 2/1), (a?2) 2/2)) ^iicl (rt's, 2/3) be any three points on sentsthe locus of the equation(1).Since the point {x-^, y^) lies on the locus, its coordinates when substituted for x and y in (1) must satisfy it.HenceAx^ +Ry^+C-=0(2).C^O(3),Ax^ + Ry^ +So andAxs +£ys+C =(4).Since these three equations hold between the three quantities A, B, and C, we can, as in Art. 12, eliminate them.Theresult is= ^352/35(5).-•-But, by Art. 25, the relation (5) states that the area of the triangle whose vertices are (x^, y^), (x^, 3/2)5 ^^^ (^3> 2/3) is zero.Also these are any three points on thelocus.3—2,COORDINATE GEOMETRY.36must therefore be a straight line ; for a curved not be such that the triangle obtained by joining any three points on it should be zero.Thelocusline couldThe proposition55.of the preceding article ^a;may be andmay alsobe deducedFor the equationfrom Art. 47.% + (7=0A C y=- — x-^,writtenthis is the+sameas the straight liney = mx + c,AC^?3i=-— andifc=-—.isx>But in Art. 47 it was shewn that y = mx + c was the equation to a straight line cutting off an intercept c from the axis of y and inclined at an angle tan~^m to the axis of x.Ax + By + C=0The equationtherefore represents a straight line cutting offCan intercept - — from x>the axis of y and inclined at an angle tan~^We56.(-—\|to the axis of x.can reduce the general equation of theAx + By + C =degreefirst(1)to the form of Art. 53. For, if p be the perpendicular from the origin on (1) and a the angle it makes with the axis, the equation to the straight line must beX cos a 4- 2/ sin a - /» = This equation must therefore be the same(2).asABCcos aHence pcos asin aC-A-B(1).—psin a\/cos^a + sin^ aJa^ + B'1sJa^'+ B^Hence cos a --A s/A^-B.-sm a =+ B^',\fA-',Cand^ p =+ B''The equation (1) may by dividing it by JA^ + B^ and arranging constant termisnegative.+ B^ form (2)sfA^therefore be reduced to the itso that theTHE STRAIGHT Ex.57.Reduceto tlie37LINE.perpendicular form the equation^ + 2/\/3 + 7 = + JA'' B^= ^TTs = sJ4:=2.Here Dividingby(1)we have2,i.e.^(-i)+y(--^)-i=o,i.e.X cos 240° + y sin 240° - 1 = 0.To58.(1).trace the straight line given hyan equation ofthe first degree.Let the equation beAx + By + G =(1).This can be written in the form(a)A ComparingthisBwith the result of Art. 50,wesee that it—represents a straightHne which cuts——Its positionisequationreduces to the formfrom the axes.off intercepts(J -^andtherefore known.jOIfGbezero, the(1)A and thus (by Art.47, Cor.)represents a straightpassing through the origin inchned at an angle tan~^ to the axis of (^)Thex.Its position is thereforemaystraight lineIfwe put y —in (1)therefore liesonGwe have x — —-r.it.byfirndingit.JLi-'i-')~ r)known.also be tracedthe coordinates of any two points onIhneThe pointCOORDINATE GEOMETRY.38 Ifwe put G^(»-.)ojonlies= 0, we haveG2/=— ^so that the point,it.Hence, as before, we have the position of the straight line.Ex.69.Trace the straight (1)3a;-4i/(3)%y = x',+lines7 = 0; (4)(2)x = ^i7a;+ 8y + 9 = 0j(5)Putting 2/ = 0, we have rc= -\|, (1) and putting x = Q, we have y = ^. Measuring 0A-^{= -^) along the axis2/= -2.ofx we have one point onthe Hne.Measuring OB^ (=t) along the axis of y we have another point. A-^B^ produced both ways, is the required line,Hence,Putting in succession y and x equal to zero, we have the (2) intercepts on the axes equal to - f and - f. If then 0-42= -f and 0^2= - \|, we have A^B^, the required line. (3)The point(0, 0) satisfiesthe equation so that the originisonthe line.Also the point therefore OC3. (4)(3,The line ic = 21),is,byi.e.C.^,liesonit.The requiredArt. 46, parallel to the axis of yline isand passesthrough the point A^ on the axis of x such that 0A^ = 2.The line y= - 2 is parallel to the axis of x and passes through (5) the point B^ on the axis of y, such that 0B^= - 2.60.Straight Line at Infinity. We have seen Ax + By + (7 = represents a straight linethat the equationSTRAIGHT LINE JOINING TWO POINTS. which cutsoiF interceptsccJi.Jj— - and — — from39the axes ofcoordinates. IfXofAvanish, but notBor C, the intercept on the axisThe equation of the straight line the form y = constant, and hence, as inis infinitely great.then reduces to Art. 46, represents a straight line parallel to Ox.Bvanish, but not A or C, the straight line meets So if the axis of y at an infinite distance and is therefore parallel toit.BIf A and both vanish, but not C, these two intercepts are both infinite and therefore the straight line Q .x + .y + C = is altogether at infinity.The multiplication of an equation by a constant Thus the equations it. and 10a;- 152/+ 25 2a;-32/+5 = represent the same straight line. Conversely, if two equations of the first degree represent the same straight line, one equation must be equal to the other multiplied by a constant quantity, so that the ratios of the corresponding coefficients must be the same. For example, if the equations and A-^^x + B^y + Cj = a^x + \y + Ci = we must have 61.does not alter\«!CjL62. To jind the equation to the straight line which passes through the two given points {x\ y') and (x", y").ByArt. 47, the equation to y--Byanystraight lineismx -VG(1).mand properly determining the quantities (1) represent any straight line we please.cwe canmakeIf (1) pass through the point 2/'Substituting forcfrom(a;',y')^we have= mas' + c(2),(2).the equation (1) becomesy-y' = m(x-x')(3).XCOOKDINATE GEOMETRY.40This is the equation to the line going through (x\ y') making an angle tan~^ with OX. If in addition (3) passes through the point {x", y"), thenm—y=m{xy -yX'-Substituting this value in equation63.Ex.Findtherwe(3),get as the requiredV" — v' X" — x^^through the points (-1,— x),ytiequation 3)andto(4,'the straightlinewhich passes-2).Let the required equation bey=mx + c Since(1)goes through the3=-m + Hencefirst c,point,(1).we haveso that c =m + S.becomes(1)y = mx + m + S If in addition the line-2 = 47?i + m + 3, Hence(2)(2).goes through the second point, so thatm=we have-1.becomesy=-x + 2,i.e.x + y = 2.Or, again, using the result of the last article the equation isy-B = ^-^^^{x + l)=-x-l, y + x-=2.i.e.64.Tofix definitelythe position of a straight linewemust have always two quantities given. Thus one point on the straight line and the direction of the straight line will determine it; or again two points lying on the straight line will determineit.Analytically, the general equation to a straight line two arbitrary constants, which will have to be determined so that the general equation may represent any particular straight line. will containmThus, in Art. 47, the quantities and c which remain the same, so long as we are considering the same straigld line, are the two constants for the straight line.41EXAMPLES.Similarly, in Art. 50, the quantities a and h are the constants for the straight line.65. In any equation to a locus the quantities x and y, which are the coordinates of any point on the locus, are called Current Coordinates;traced out by a point whichthe curvemaybe conceived as" runs " along the locus.EXAMPLES.V.Find the equation to the straight line 1. cutting off an intercept unity from the positive direction of the axis of y and inclined at 45° to the axis of x. 2. cutting off an intercept - 5 from the axis of y and being equally inclined to the axes. 3. cutting off an intercept 2 from the negative direction of the axis of y and inclined at 30° to OX.4.cutting off an intercept - 3 from the axis of y to the axis of x.and inclinedatan angle tan~i fFind the equationto the straight line5.cutting off intercepts 3 and 2 from the axes.6.cutting off intercepts- 5 and6from the axes.Find the equation to the straight line which passes through the point (5, 6) and has intercepts on the axes equal in magnitude and both positive, (1) equal in magnitude but opposite in sign. (2) 7.8.Find the equations to the straight lines which pass through (1, - 2) and cut off equal distances from the two axes.the point9. Find the equation to the straight line which passes through the given point {x\ y') and is such that the given point bisects the part intercepted between the axes.10. Find the equation to the straight line which passes through the point ( - 4, 3) and is such that the portion of it between the axes is divided by the point in the ratio 5 3. :Trace the straight lines whose equations are+ 2?/+3 = 0. + 7r/ = 0.11.a;13.3a;12.5a--7//-914.2a;-3?/Find the equations to the straight lines passing following pairs of points. 15.(0, 0)17.(-1,and 3)(2,and-2).(6,-7).16.(3, 4)18.(0,= 0.+ 4 = 0.andthrough the (5, 6).-a) and(&, 0).COORDINATE GEOMETRY.42 and{a+ h, a-h).19.(a, &)20.{at^, 2at-^)22.(« cos 0123.(acos0jLJ & sin 0j)24.( sec 01, 6,anda sin(at^^ 2at;).^sin^g)Find the equations to the sides of the whose angular points are respectively 25.[Exs. v.]02).triangles the coordinates ofand (-1,-2).and (-1, -2).27. Find the equations to the diagonals of the rectangle the equations of whose sides are x = a, x = a\ y = b, and y = b\28. Find the equation to the straight line which bisects the distance between the points {a, b) and {a', b') and also bisects the distance between the points ( - a, b) and (a', - b'). 29. Find the equations to the straight lines which go through the origin and trisect the portion of the straight line 3a; + 2/ = 12 which is intercepted between the axes of coordinates.Angles between straightlines.To find the angle between two given straight lines. Let the two straight lines be AL^ and AL^j meeting the axis of X in L^ and L^, 66.I.Let their equations bey — m^x^-G-^ and y ~ in.j,x ^r c.^ By Art. 47 we therefore have tan^ZjA'^mi, and td^Vi. AL.^X^Wj.^,.Now— L AL^X — L AL.2.X. tan L^AL^ — tan \AL^X — AL^X\ LL-^AL^^AL^X— tan AL^X 1 + tan AL^X. tan AL^X ta,nrn^ 1— n^+mi«i2(1).43ANGLES BETWEEN STRAIGHT LINES. Hence the required angle — lL^AL= tan-i"'^•""'^l[In any numerical example, if the quantity tity it is the tangent of the acute angle it is the tangent of the obtuse angle.]II.and(2).+ mim2 (2)Let the equations of the straight ^i£c + ^i2/ + Ci = 0,+ G^^O. By dividing the equations by B^ andwrittenandbe a positive quan-between the lineslines;ifnegative,beA^^x^- B^^yB^, theymay beCOORDINATE GEOMETRY.44To find67.the condition thattwo straight linesmayhe parallel.Twostraight lines are parallel when the angle between therefore the tangent of this angle is zero.them is zero and The equation(2) of the last articlethen givesTwostraight lines whose equations are given in the "m" form are therefore parallel when their "7?i's" are the same, or, in other words, if their equations differ only in the constant term.The straight line Ax + By + G' = parallel to the straight line Ax + By two equations are the same.isany straight+ C = 0. Forthelinewhich"m's"isof theAgain the equation A {x-x')+B {y-y') = clearly represents the straight line which passes through the point {x', y') and is parallel toAx + By + C=0.Theresult (3) of the last article gives, as the conditionfor parallel lines,Ex.68.Findthrough the point(4,the equation to the straight line, which passes - 5), and which is parallel to the straight line 3:cAnystraight linewhich+ 4r/ + 5--=0(1).is parallel to (1)hasitsequation of theform 3a;[For the"w"of both (1)and+ 4^/ + (7=0 (2) is(2).the same.]This straight line will pass through the point(4,- 5)if3x4 + 4x(-5) + C = 0, (7=20-12 = 8.i.e. ifThe equation(2)then becomes 3a;+42/69.To findthe condition thatequations are given,Let the straightand+ 8 = 0.may linestwost^'aight linesjhe 'perpendicular.bey — m^x y — m.^x-i-Ci, -- G.2_.whoseCONDITIONS OF PERPENDICULARITY.45be the angle between them we have, by Art. 66,Iftan^^ r^""^^ 1 +mim2(1).If the lines be perpendicular, then ^= 00tanbe=90°,and therefore.The right-hand member of equation (1) must therefore and this can only happen when its denominatorinfinite,is zero.The condition 1The to+of perpendicularity is therefore that m^TTi^straight line yy = »...H-.c.,— O, —Tn-^Tn2i.e.tu^x+c.^is=—I.'therefore perpendicular«, = -!.ify/c'-tIt follows that the straight linesA^x +B^y + C^ = which m^ = —for^andAA a)m^^ — ^ /,are at right anglesifA,,_ V A^A^+B^B^ = 0.ai.e.and A^x + B^y + 0^ = 0,From70.Athe preceding articleitfollows that thetwostraight linesand are at right angles;A^x + B,y + Ci = Q(1),B,x-A,y+C^ =(2),for the product of their m'sderived from (1) by interchanging the coefficients y, changing the sign of one of them, and changing the constant into any other constant.Also ofa;(2) isandEx. whereThestraight line throughB^x'(x', y')perpendicular to- A-^y' + 62= 0, so that Cg = A^y'- B^x'.This straight lineisthereforeB,{x-x')-A^{y-y') = 0.(1) is (2)COORDINATE GEOMETRY.4671. Ex. 1. Find the equation to the straight line which passes through the point (4, —5) and is perpendicular to the straight lineSx + 4ij + 5 =AnyFirst Method.(1).straight line perpendicular to (1) is by thelast article4:X-Sij+ C=0(2).[We should expect an arbitrary constant in (2) because an infinite number of straight lines perpendicular to (1).] The straight line (2) passes through the point (4, - 5) if i.e.there are4x4-3x(-5) + C = 0, (7= -16-15= -31.aThe required equationistherefore4:X-Sy = 31.AnySecond Method. point isstraight line passing throughthe giveny -{-5)=m{x~4:). This straight line m'sis-perpendicular tois1,mXi.e. if(- 1) = -(1) ifthe product of their1,m=\|.i.e. ifThe required equationisthereforey + 5=i{x-4), 4:X-'6yi.e.AnyThird Method. the point It is(4,- 5),= Sl.straight line isy=mx + c.Itpasses throughif-5 = 4m + cperpendicular to(3).(1) ifmx{-i)=-l(4).Hence m = f and then (3) gives c = —V. The required equation is therefore y = '^x-^-^,4x-By = Sl.i.e.[In the first method, we start with any straight line which is perpendicular to the given straight line and pick out that particular straight line which goes through the given point. In the second method, we start with any straight line passing through the given point and pick out that particular one which is perpendicular to the given straight hne. In the third method, we start with any straight line whatever and determine its constants, so that it may satisfy the two given conditions.The student shouldillustratebyfigures. ]Ex. 2. Find the equation to the straight line which passes through the point (x', y') and is perpendicular to the given straight line yy' = 2a {x+ x').THE STRAIGHT The givenstraight line isyy'Any47LINE.- 2ax - 2ax' = 0.straight line perpendicular to it is (Art. 70)2ay + xy'+G=0 This will pass through the point straight line requiredthe coordinates x'if2ai/ +i.eAtxY+C = 0,G=-2ay' -x'y'. G the required equationi.e. ifSubstituting in(1).and therefore will be the and y' satisfy it,(x', y')(1) for2a{y-y')istherefore+ y'{x-x') = 0.72. To find the equations to the straight lines which pass through a given point (x', t/') and make a given angle a with the given straight line y — nix + c.LetP bethe given point and let the given straight linebe LMJSf, making an angle with the axis of x such that= m.tan(i.e. except when a right angle or zero) thereIn general aistwo straight lines PMR and making an angle a withareFNSthe givenline.Let theselinesthe axis ofof x in R and S and let with the positive direction ofmeet the axisthem make angles ^ and Our partners will collect data and use cookies for ad personalization and measurement. Learn how we and our ad partner Google, collect and use data. Agree & close
4877	https://www.radiacode.com/isotope/co-60?lang=en	We value your privacy We use cookies to enhance your browsing experience, serve personalised ads or content, and analyse our traffic. By clicking "Accept All", you consent to our use of cookies. Cookie Policy Radiacode — Pocket-Sized Next-Generation Geiger Counter and Gamma Spectrometer Ultra-Sensitive Nuclear Radiation Detector & Scintillation Spectrometry with Real-Time Isotope visualisation DOUBLE DEALDiscounted Price + Free Shipping Ga-67 Next Isotope: Ga-67 Cobalt-60 (Co-60) is a radioactive isotope of cobalt with a half-life of approximately 5.27 years. It decays by beta emission to nickel-60, accompanied by the emission of high-energy gamma rays. These gamma rays, with characteristic energies of 1.17 MeV and 1.33 MeV, make Co-60 a valuable source of gamma radiation for a wide range of applications. Co-60 is produced artificially in nuclear reactors by neutron activation of stable cobalt-59. Co-60 is widely used in various industries and medical applications. In medicine, it is a key isotope for radiotherapy, particularly for treating cancer, where its gamma rays are used to target and destroy malignant cells. It is also employed in sterilization of medical equipment and pharmaceuticals, ensuring the elimination of bacteria and other pathogens. In industry, Co-60 is used for non-destructive testing (NDT) to inspect materials and structures, such as pipelines and aircraft components, for internal defects. Additionally, it is used in food irradiation to extend shelf life and eliminate harmful microorganisms. Co-60 does not occur naturally and is produced artificially in nuclear reactors by exposing cobalt-59 to neutron radiation. It is encountered in controlled environments, such as hospitals, industrial facilities, and research laboratories, where it is used in radiation-based applications. Co-60 may also be present in nuclear waste and requires careful handling and disposal due to its strong gamma emissions and radiological hazards. Its use is strictly regulated to ensure safety and minimize environmental impact. Co-60 Cobalt-60 Technogenic β, γ radiation Half-life: 5,3 years Main emission lines: 1173, 1332 keV Cobalt-60 (Co-60) is a radioactive isotope of cobalt with a half-life of approximately 5.27 years. It decays by beta emission to nickel-60, accompanied by the emission of high-energy gamma rays. These gamma rays, with characteristic energies of 1.17 MeV and 1.33 MeV, make Co-60 a valuable source of gamma radiation for a wide range of applications. Co-60 is produced artificially in nuclear reactors by neutron activation of stable cobalt-59. Co-60 is widely used in various industries and medical applications. In medicine, it is a key isotope for radiotherapy, particularly for treating cancer, where its gamma rays are used to target and destroy malignant cells. It is also employed in sterilization of medical equipment and pharmaceuticals, ensuring the elimination of bacteria and other pathogens. In industry, Co-60 is used for non-destructive testing (NDT) to inspect materials and structures, such as pipelines and aircraft components, for internal defects. Additionally, it is used in food irradiation to extend shelf life and eliminate harmful microorganisms. Co-60 does not occur naturally and is produced artificially in nuclear reactors by exposing cobalt-59 to neutron radiation. It is encountered in controlled environments, such as hospitals, industrial facilities, and research laboratories, where it is used in radiation-based applications. Co-60 may also be present in nuclear waste and requires careful handling and disposal due to its strong gamma emissions and radiological hazards. Its use is strictly regulated to ensure safety and minimize environmental impact. Videos Findings Control source Co-60
4878	https://www.mathopenref.com/isosceles.html	Math Open Reference Home Contact About Subject Index Isosceles Triangle A triangle which has two of its sides equal in length. Try this Drag the orange dots on each vertex to reshape the triangle. Notice it always remains an isosceles triangle, the sides AB and AC always remain equal in length Options Hide \|< >\| RESET The word isosceles is pronounced "eye-sos-ell-ease" with the emphasis on the 'sos'. It is any triangle that has two sides the same length. If all three sides are the same length it is called an equilateral triangle. Obviously all equilateral triangles also have all the properties of an isosceles triangle. Properties The unequal side of an isosceles triangle is usually referred to as the 'base' of the triangle. The base angles of an isosceles triangle are always equal. In the figure above, the angles ∠ABC and ∠ACB are always the same When the 3rd angle is a right angle, it is called a "right isosceles triangle". The altitude is a perpendicular distance from the base to the topmost vertex. Constructing an Isosceles Triangle It is possible to construct an isosceles triangle of given dimensions using just a compass and straightedge. See these three constructions: Isosceles triangle, given base and side Isosceles triangle, given base and altitude Isosceles triangle, given leg and apex angle Solving an isosceles triangle The base, leg or altitude of an isosceles triangle can be found if you know the other two. A perpendicular bisector of the base forms an altitude of the triangle as shown on the right. This forms two congruent right triangles that can be solved using Pythagoras' Theorem as shown below. Finding the base To find the base given the leg and altitude, use the formula: Base = 2 √ L 2 − A 2 where: L is the length of a leg A is the altitude Finding the leg To find the leg length given the base and altitude, use the formula: Leg = √ A 2 + B 2 2 where: B is the length of the base A is the altitude Altitude To find the altitude given the base and leg, use the formula: Altitude = √ L 2 − B 2 2 where: L is the length of a leg B is the base Interior angles If you are given one interior angle of an isosceles triangle you can find the other two. For example, We are given the angle at the apex as shown on the right of 40°. We know that the interior angles of all triangles add to 180°. So the two base angles must add up to 180-40, or 140°. Since the two base angles are congruent (same measure), they are each 70°. If we are given a base angle of say 45°, we know the base angles are congruent (same measure) and the interior angles of any triangle always add to 180°. So the apex angle must be 180-45-45 or 90°. Other triangle topics General Triangle definition Hypotenuse Triangle interior angles Triangle exterior angles Triangle exterior angle theorem Pythagorean Theorem Proof of the Pythagorean Theorem Pythagorean triples Triangle circumcircle Triangle incircle Triangle medians Triangle altitudes Midsegment of a triangle Triangle inequality Side / angle relationship Perimeter / Area Perimeter of a triangle Area of a triangle Heron's formula Area of an equilateral triangle Area by the "side angle side" method Area of a triangle with fixed perimeter Triangle types Right triangle Isosceles triangle Scalene triangle Equilateral triangle Equiangular triangle Obtuse triangle Acute triangle 3-4-5 triangle 30-60-90 triangle 45-45-90 triangle Triangle centers Incenter of a triangle Circumcenter of a triangle Centroid of a triangle Orthocenter of a triangle Euler line Congruence and Similarity Congruent triangles Solving triangles Solving the Triangle Law of sines Law of cosines Triangle quizzes and exercises Triangle type quiz Ball Box problem How Many Triangles? Satellite Orbits (C) 2011 Copyright Math Open Reference. All rights reserved
4879	https://linear.axler.net/Norms.pdf	Inner Products and Norms, part 2: Norms Norms Deﬁnition: norm, ∥v∥ For v ∈V, the norm of v, denoted ∥v∥, is deﬁned by ∥v∥= p ⟨v, v⟩. Norms Deﬁnition: norm, ∥v∥ For v ∈V, the norm of v, denoted ∥v∥, is deﬁned by ∥v∥= p ⟨v, v⟩. If (z1, . . . , zn) ∈Fn, then ∥(z1, . . . , zn)∥= q \|z1\|2 + · · · + \|zn\|2. Norms Deﬁnition: norm, ∥v∥ For v ∈V, the norm of v, denoted ∥v∥, is deﬁned by ∥v∥= p ⟨v, v⟩. If (z1, . . . , zn) ∈Fn, then ∥(z1, . . . , zn)∥= q \|z1\|2 + · · · + \|zn\|2. In the vector space of continuous real-valued functions on [−1, 1] with inner product ⟨f, g⟩= R 1 −1 f(x)g(x) dx, we have ∥f∥= sZ 1 −1 f(x) 2 dx. Norms Deﬁnition: norm, ∥v∥ For v ∈V, the norm of v, denoted ∥v∥, is deﬁned by ∥v∥= p ⟨v, v⟩. If (z1, . . . , zn) ∈Fn, then ∥(z1, . . . , zn)∥= q \|z1\|2 + · · · + \|zn\|2. In the vector space of continuous real-valued functions on [−1, 1] with inner product ⟨f, g⟩= R 1 −1 f(x)g(x) dx, we have ∥f∥= sZ 1 −1 f(x) 2 dx. Basic properties of the norm Suppose v ∈V. ∥v∥= 0 if and only if v = 0. ∥λv∥= \|λ\| ∥v∥for all λ ∈F. Norms Deﬁnition: norm, ∥v∥ For v ∈V, the norm of v, denoted ∥v∥, is deﬁned by ∥v∥= p ⟨v, v⟩. If (z1, . . . , zn) ∈Fn, then ∥(z1, . . . , zn)∥= q \|z1\|2 + · · · + \|zn\|2. In the vector space of continuous real-valued functions on [−1, 1] with inner product ⟨f, g⟩= R 1 −1 f(x)g(x) dx, we have ∥f∥= sZ 1 −1 f(x) 2 dx. Basic properties of the norm Suppose v ∈V. ∥v∥= 0 if and only if v = 0. ∥λv∥= \|λ\| ∥v∥for all λ ∈F. Proof:Suppose λ ∈F. Then ∥λv∥2 = ⟨λv, λv⟩ Norms Deﬁnition: norm, ∥v∥ For v ∈V, the norm of v, denoted ∥v∥, is deﬁned by ∥v∥= p ⟨v, v⟩. If (z1, . . . , zn) ∈Fn, then ∥(z1, . . . , zn)∥= q \|z1\|2 + · · · + \|zn\|2. In the vector space of continuous real-valued functions on [−1, 1] with inner product ⟨f, g⟩= R 1 −1 f(x)g(x) dx, we have ∥f∥= sZ 1 −1 f(x) 2 dx. Basic properties of the norm Suppose v ∈V. ∥v∥= 0 if and only if v = 0. ∥λv∥= \|λ\| ∥v∥for all λ ∈F. Proof:Suppose λ ∈F. Then ∥λv∥2 = ⟨λv, λv⟩ = λ⟨v, λv⟩ Norms Deﬁnition: norm, ∥v∥ For v ∈V, the norm of v, denoted ∥v∥, is deﬁned by ∥v∥= p ⟨v, v⟩. If (z1, . . . , zn) ∈Fn, then ∥(z1, . . . , zn)∥= q \|z1\|2 + · · · + \|zn\|2. In the vector space of continuous real-valued functions on [−1, 1] with inner product ⟨f, g⟩= R 1 −1 f(x)g(x) dx, we have ∥f∥= sZ 1 −1 f(x) 2 dx. Basic properties of the norm Suppose v ∈V. ∥v∥= 0 if and only if v = 0. ∥λv∥= \|λ\| ∥v∥for all λ ∈F. Proof:Suppose λ ∈F. Then ∥λv∥2 = ⟨λv, λv⟩ = λ⟨v, λv⟩ = λ¯ λ⟨v, v⟩ Norms Deﬁnition: norm, ∥v∥ For v ∈V, the norm of v, denoted ∥v∥, is deﬁned by ∥v∥= p ⟨v, v⟩. If (z1, . . . , zn) ∈Fn, then ∥(z1, . . . , zn)∥= q \|z1\|2 + · · · + \|zn\|2. In the vector space of continuous real-valued functions on [−1, 1] with inner product ⟨f, g⟩= R 1 −1 f(x)g(x) dx, we have ∥f∥= sZ 1 −1 f(x) 2 dx. Basic properties of the norm Suppose v ∈V. ∥v∥= 0 if and only if v = 0. ∥λv∥= \|λ\| ∥v∥for all λ ∈F. Proof:Suppose λ ∈F. Then ∥λv∥2 = ⟨λv, λv⟩ = λ⟨v, λv⟩ = λ¯ λ⟨v, v⟩ = \|λ\|2∥v∥2. Norms Deﬁnition: norm, ∥v∥ For v ∈V, the norm of v, denoted ∥v∥, is deﬁned by ∥v∥= p ⟨v, v⟩. If (z1, . . . , zn) ∈Fn, then ∥(z1, . . . , zn)∥= q \|z1\|2 + · · · + \|zn\|2. In the vector space of continuous real-valued functions on [−1, 1] with inner product ⟨f, g⟩= R 1 −1 f(x)g(x) dx, we have ∥f∥= sZ 1 −1 f(x) 2 dx. Basic properties of the norm Suppose v ∈V. ∥v∥= 0 if and only if v = 0. ∥λv∥= \|λ\| ∥v∥for all λ ∈F. Proof:Suppose λ ∈F. Then ∥λv∥2 = ⟨λv, λv⟩ = λ⟨v, λv⟩ = λ¯ λ⟨v, v⟩ = \|λ\|2∥v∥2. Taking square roots now gives the desired equality. Orthogonality Deﬁnition: orthogonal Two vectors u, v ∈V are called orthogonal if ⟨u, v⟩= 0. Orthogonality Deﬁnition: orthogonal Two vectors u, v ∈V are called orthogonal if ⟨u, v⟩= 0. Orthogonality and 0 0 is orthogonal to every vector in V. 0 is the only vector in V that is orthogonal to itself. Orthogonality Deﬁnition: orthogonal Two vectors u, v ∈V are called orthogonal if ⟨u, v⟩= 0. Orthogonality and 0 0 is orthogonal to every vector in V. 0 is the only vector in V that is orthogonal to itself. Pythagorean Theorem Suppose u and v are orthogonal vectors in V. Then ∥u + v∥2 = ∥u∥2 + ∥v∥2. Orthogonality Deﬁnition: orthogonal Two vectors u, v ∈V are called orthogonal if ⟨u, v⟩= 0. Orthogonality and 0 0 is orthogonal to every vector in V. 0 is the only vector in V that is orthogonal to itself. Pythagorean Theorem Suppose u and v are orthogonal vectors in V. Then ∥u + v∥2 = ∥u∥2 + ∥v∥2. Proof ∥u + v∥2 = ⟨u + v, u + v⟩ Orthogonality Deﬁnition: orthogonal Two vectors u, v ∈V are called orthogonal if ⟨u, v⟩= 0. Orthogonality and 0 0 is orthogonal to every vector in V. 0 is the only vector in V that is orthogonal to itself. Pythagorean Theorem Suppose u and v are orthogonal vectors in V. Then ∥u + v∥2 = ∥u∥2 + ∥v∥2. Proof ∥u + v∥2 = ⟨u + v, u + v⟩ = ⟨u, u⟩+ ⟨u, v⟩+ ⟨v, u⟩+ ⟨v, v⟩ Orthogonality Deﬁnition: orthogonal Two vectors u, v ∈V are called orthogonal if ⟨u, v⟩= 0. Orthogonality and 0 0 is orthogonal to every vector in V. 0 is the only vector in V that is orthogonal to itself. Pythagorean Theorem Suppose u and v are orthogonal vectors in V. Then ∥u + v∥2 = ∥u∥2 + ∥v∥2. Proof ∥u + v∥2 = ⟨u + v, u + v⟩ = ⟨u, u⟩+ ⟨u, v⟩+ ⟨v, u⟩+ ⟨v, v⟩ = ∥u∥2 + ∥v∥2 Woman Teaching Geometry Cauchy–Schwarz Inequality Cauchy–Schwarz Inequality Suppose u, v ∈V. Then \|⟨u, v⟩\| ≤∥u∥∥v∥. This inequality is an equality if and only if one of u, v is a scalar multiple of the other. Cauchy–Schwarz Inequality Cauchy–Schwarz Inequality Suppose u, v ∈V. Then \|⟨u, v⟩\| ≤∥u∥∥v∥. This inequality is an equality if and only if one of u, v is a scalar multiple of the other. Examples: If x1, . . . , xn, y1, . . . , yn ∈R, then \|x1y1 + · · · + xnyn\|2 ≤(x12 + · · · + xn2)(y12 + · · · + yn2). Cauchy–Schwarz Inequality Cauchy–Schwarz Inequality Suppose u, v ∈V. Then \|⟨u, v⟩\| ≤∥u∥∥v∥. This inequality is an equality if and only if one of u, v is a scalar multiple of the other. Examples: If x1, . . . , xn, y1, . . . , yn ∈R, then \|x1y1 + · · · + xnyn\|2 ≤(x12 + · · · + xn2)(y12 + · · · + yn2). If f, g are continuous real-valued functions on [−1, 1], then Z 1 −1 f(x)g(x) dx 2 ≤ Z 1 −1 f(x) 2 dx Z 1 −1 g(x) 2 dx . Triangle Inequality Triangle Inequality Suppose u, v ∈V. Then ∥u + v∥≤∥u∥+ ∥v∥. This inequality is an equality if and only if one of u, v is a nonnegative multiple of the other. Triangle Inequality Triangle Inequality Suppose u, v ∈V. Then ∥u + v∥≤∥u∥+ ∥v∥. This inequality is an equality if and only if one of u, v is a nonnegative multiple of the other. u + v v u Triangle Inequality Triangle Inequality Suppose u, v ∈V. Then ∥u + v∥≤∥u∥+ ∥v∥. This inequality is an equality if and only if one of u, v is a nonnegative multiple of the other. u + v v u Proof ∥u + v∥2 = ⟨u + v, u + v⟩ Triangle Inequality Triangle Inequality Suppose u, v ∈V. Then ∥u + v∥≤∥u∥+ ∥v∥. This inequality is an equality if and only if one of u, v is a nonnegative multiple of the other. u + v v u Proof ∥u + v∥2 = ⟨u + v, u + v⟩ = ⟨u, u⟩+ ⟨v, v⟩+ ⟨u, v⟩+ ⟨v, u⟩ Triangle Inequality Triangle Inequality Suppose u, v ∈V. Then ∥u + v∥≤∥u∥+ ∥v∥. This inequality is an equality if and only if one of u, v is a nonnegative multiple of the other. u + v v u Proof ∥u + v∥2 = ⟨u + v, u + v⟩ = ⟨u, u⟩+ ⟨v, v⟩+ ⟨u, v⟩+ ⟨v, u⟩ = ⟨u, u⟩+ ⟨v, v⟩+ ⟨u, v⟩+ ⟨u, v⟩ Triangle Inequality Triangle Inequality Suppose u, v ∈V. Then ∥u + v∥≤∥u∥+ ∥v∥. This inequality is an equality if and only if one of u, v is a nonnegative multiple of the other. u + v v u Proof ∥u + v∥2 = ⟨u + v, u + v⟩ = ⟨u, u⟩+ ⟨v, v⟩+ ⟨u, v⟩+ ⟨v, u⟩ = ⟨u, u⟩+ ⟨v, v⟩+ ⟨u, v⟩+ ⟨u, v⟩ = ∥u∥2 + ∥v∥2 + 2 Re⟨u, v⟩ Triangle Inequality Triangle Inequality Suppose u, v ∈V. Then ∥u + v∥≤∥u∥+ ∥v∥. This inequality is an equality if and only if one of u, v is a nonnegative multiple of the other. u + v v u Proof ∥u + v∥2 = ⟨u + v, u + v⟩ = ⟨u, u⟩+ ⟨v, v⟩+ ⟨u, v⟩+ ⟨v, u⟩ = ⟨u, u⟩+ ⟨v, v⟩+ ⟨u, v⟩+ ⟨u, v⟩ = ∥u∥2 + ∥v∥2 + 2 Re⟨u, v⟩ ≤∥u∥2 + ∥v∥2 + 2\|⟨u, v⟩\| Triangle Inequality Triangle Inequality Suppose u, v ∈V. Then ∥u + v∥≤∥u∥+ ∥v∥. This inequality is an equality if and only if one of u, v is a nonnegative multiple of the other. u + v v u Proof ∥u + v∥2 = ⟨u + v, u + v⟩ = ⟨u, u⟩+ ⟨v, v⟩+ ⟨u, v⟩+ ⟨v, u⟩ = ⟨u, u⟩+ ⟨v, v⟩+ ⟨u, v⟩+ ⟨u, v⟩ = ∥u∥2 + ∥v∥2 + 2 Re⟨u, v⟩ ≤∥u∥2 + ∥v∥2 + 2\|⟨u, v⟩\| ≤∥u∥2 + ∥v∥2 + 2∥u∥∥v∥ Triangle Inequality Triangle Inequality Suppose u, v ∈V. Then ∥u + v∥≤∥u∥+ ∥v∥. This inequality is an equality if and only if one of u, v is a nonnegative multiple of the other. u + v v u Proof ∥u + v∥2 = ⟨u + v, u + v⟩ = ⟨u, u⟩+ ⟨v, v⟩+ ⟨u, v⟩+ ⟨v, u⟩ = ⟨u, u⟩+ ⟨v, v⟩+ ⟨u, v⟩+ ⟨u, v⟩ = ∥u∥2 + ∥v∥2 + 2 Re⟨u, v⟩ ≤∥u∥2 + ∥v∥2 + 2\|⟨u, v⟩\| ≤∥u∥2 + ∥v∥2 + 2∥u∥∥v∥ = (∥u∥+ ∥v∥)2 Parallelogram Equality Parallelogram Equality Suppose u, v ∈V. Then ∥u + v∥2 + ∥u −v∥2 = 2(∥u∥2 + ∥v∥2). Parallelogram Equality Parallelogram Equality Suppose u, v ∈V. Then ∥u + v∥2 + ∥u −v∥2 = 2(∥u∥2 + ∥v∥2). u + v u - v u u v v Parallelogram Equality Parallelogram Equality Suppose u, v ∈V. Then ∥u + v∥2 + ∥u −v∥2 = 2(∥u∥2 + ∥v∥2). u + v u - v u u v v Proof ∥u + v∥2 + ∥u −v∥2 Parallelogram Equality Parallelogram Equality Suppose u, v ∈V. Then ∥u + v∥2 + ∥u −v∥2 = 2(∥u∥2 + ∥v∥2). u + v u - v u u v v Proof ∥u + v∥2 + ∥u −v∥2 = ⟨u + v, u + v⟩+ ⟨u −v, u −v⟩ Parallelogram Equality Parallelogram Equality Suppose u, v ∈V. Then ∥u + v∥2 + ∥u −v∥2 = 2(∥u∥2 + ∥v∥2). u + v u - v u u v v Proof ∥u + v∥2 + ∥u −v∥2 = ⟨u + v, u + v⟩+ ⟨u −v, u −v⟩ = ∥u∥2 + ∥v∥2 + ⟨u, v⟩+ ⟨v, u⟩ + ∥u∥2 + ∥v∥2 −⟨u, v⟩−⟨v, u⟩ Parallelogram Equality Parallelogram Equality Suppose u, v ∈V. Then ∥u + v∥2 + ∥u −v∥2 = 2(∥u∥2 + ∥v∥2). u + v u - v u u v v Proof ∥u + v∥2 + ∥u −v∥2 = ⟨u + v, u + v⟩+ ⟨u −v, u −v⟩ = ∥u∥2 + ∥v∥2 + ⟨u, v⟩+ ⟨v, u⟩ + ∥u∥2 + ∥v∥2 −⟨u, v⟩−⟨v, u⟩ = 2(∥u∥2 + ∥v∥2) Linear Algebra Done Right, by Sheldon Axler
4880	https://arxiv.org/pdf/1808.08486	Published Time: Sun, 22 Jan 2023 22:18:57 GMT arXiv:1808.08486v1 [math.CO] 26 Aug 2018 MODIFIED ERD ¨ OS–GINZBURG–ZIV CONSTANTS FOR Z/n Z AND (Z/n Z)2 AARON BERGER AND DANIELLE WANG Abstract. For an abelian group G and an integer t > 0, the modified Erd¨ os– Ginzburg–Ziv constant s′ t (G) is the smallest integer ℓ such that any zero-sum sequence of length at least ℓ with elements in G contains a zero-sum sub-sequence (not necessarily consecutive) of length t. We compute s′ t (G) for G = Z/n Z and for t = n, G = ( Z/n Z)2 . Keywords: Zero-sum sequence, Zero-sum subsequence, Erd¨ os–Ginzburg–Ziv Constant. Introduction In 1961, Erd¨ os, Ginzburg, and Ziv proved the following classical theorem. Theorem 1.1 (Erd¨ os–Ginzburg–Ziv [ 6]) . Any sequence of length 2n − 1 in Z/n Z contains a zero-sum subsequence of length n. Here, a subsequence need not be consecutive, and a sequence is zero-sum if its elements sum to 0. This theorem has lead to many problems involving zero-sum sequences over groups. In general, let G be an abelian group, and let G0 ⊆ G be a susbset. Let L ⊆ N.Then sL(G0) is defined to be the minimal ℓ such that any sequence of length ℓ with elements in G0 contains a zero-sum subsequence whose length is in L. When G0 = G and L = {exp( G)}, this constant is called the Erd¨ os–Ginzburg–Ziv constant. When G = Z, this problem turns out to be not very interesting — if G0 con-tains a nonzero element, then sL(G0) = ∞. This has lead to [ 2] the study of the modified Erd¨ os–Ginzburg–Ziv constant s′L(G0), defined as the smallest ℓ such that any zero-sum sequence of length at least ℓ with elements in G0 contains a zero-sum subsequence whose length is in L. When L = {t} is a single element, we omit the set brackets for convenience. In [ 3], the first author determined modified EGZ constants in the infinite cyclic case. Here we treat the finite cyclic case and extensions. Problem 1.2 ([ 3, Problem 2]) . Compute s′ t (G) for G = Z/n Z and ( Z/n Z)2.In this paper, we answer Problem 1.2 for G = Z/n Z and for t = n, G = ( Z/n Z)2.Note that in both cases, when n does not divide t, the quantity st(G) is infinite. Theorem 1.3. The modified EGZ constant of Z/n Z is given by s′ nt (Z/n Z) = (t + 1) n − ℓ + 1 , where ℓ is the smallest integer such that ℓ ∤ n. Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA 02139-4307 E-mail addresses : bergera@mit.edu, diwang@mit.edu . 12AARON BERGER AND DANIELLE WANG Theorem 1.4. We have s′ n (( Z/n Z)2) = 4 n − ℓ + 1 where ℓ is the smallest integer such that d ≥ 4 and ℓ ∤ n. The cyclic case In this section we give the proof of Theorem 1.3 . As in [ 10 ], if J is a sequence of elements of Z/n Z or ( Z/n Z)2, we use ( k \| J) to denote the number of zero-sum subsequences of J of size k. Proposition 2.1. If d \| n and J is a zero-sum sequence in Z/n Z of length 2n − d,then (n \| J) > 0.Proof. By Theorem 1.1 , we can break off subsequences of J of size d with sum 0 (mod d) until we have fewer than 2 d − 1 remaining. In fact, since d \| n, we will have exactly d remaining. But since the sum was zero-sum to begin with, the last d must also sum to zero, so we have 2( n/d ) − 1 blocks of size d with sums dx 1, . . . , dx 2( n/d )−1 for some xi. By Theorem 1.1 , some n/d of these must sum to 0 in Z/(n/d )Z, so the union of these blocks gives a subsequence of length n whose sum is zero in Z/n Z. Corollary 2.2. Let ℓ be the smallest positive integer such that ℓ ∤ n, and let t ≥ 1. If J is a zero-sum sequence in Z/n Z of length at least (t+1) n−ℓ+1 , then (nt \| J) > 0.Proof. We induct on t. The case t = 1 follows from Proposition 2.1 since ℓ−1, . . . , 1all divide n. Suppose the result is true for positive integers less than t > 1. Then J contains a zero-sum subsequence of length ( t − 1) n. Remove these elements from J. We are left with a zero-sum sequence of length 2 n − ℓ + 1. This is the t = 1 case, so we can find another zero-sum subsequence of length n. Combine this with the ( t − 1) n to get the desired subsequence of length nt . Proposition 2.3. Suppose ℓ ∤ n and t ≥ 1. Then there exists a zero-sum subse-quence in Z/n Z of length (1 + t)n − ℓ which contains no zero-sum subsequence of length nt .Proof. Consider a sequence of 0’s and 1’s with multiplicities a ≤ tn − 1, b ≤ n − 1respectively where a + b = ( t + 1) n − ℓ. Such a sequence will have no zero-sum subsequence of length nt . It suffices to find a, b such that g = gcd( n, ℓ ) \| b, because then we can add some constant to every term of the sequence to make it zero-sum. Note that adding a constant to every term does not introduce any new zero-sum subsequences. It suffices to take b = tn − g and a = n − ℓ + g ≤ n − ℓ/ 2 ≤ n − 1. Corollary 2.2 and Proposition 2.3 together imply Theorem 1.3 . The case (Z/n Z)2 In this section we prove Theorem 1.4 . We first prove some preliminary lemmas. The following results from [ 10 ] are key. Lemma 3.1 ([ 10 , Corollary 2.4]) . Let p be a prime, and let J be a sequence of elements in (Z/p Z)2. If \|J\| = 3 p − 2 or \|J\| = 3 p − 1, then (p \| J) = 0 implies (2 p \| J) ≡ − 1 (mod p). Lemma 3.2 ([ 10 , Corollary 2.5]) . Let p and J be as in Lemma 3.1 . If \|J\| is a zero-sum sequence with exactly 3p elements, then (p \| J) > 0.MODIFIED ERD ¨OS–GINZBURG–ZIV CONSTANTS FOR Z/n Z AND ( Z/n Z)2 3 Theorem 3.3 ([ 10 , Theorem 3.2]) . If J is a sequence of length 4n − 3 in (Z/n Z)2 then (n \| J) > 0. We generalize Lemma 3.2 to non-prime n. Lemma 3.4. If J is a zero-sum sequence of length 3n in (Z/n Z)2, then (n \| J) > 0.Proof. We induct on n. The base case n = 1 is clear. Assume the the lemma is true for all positive integers less than n. Let n = pm with p prime and m < n .Since 3 n > 4m − 3, we can find some m elements of J whose sum is 0 (mod m). Say their sum is mx 1 and remove these m elements. We can continue doing this until there remain only 3 m elements. But since J was a zero-sum sequence, the remaining 3 m elements must sum to 0 (mod m), so by the induction hypothesis, we can remove another m with sum a multiple of m. This gives us 3 p − 2 blocks of size m whose sums are mx 1, . . . , mx 3p−2 for some xi.If some p of the xi sum to 0 (mod p), then combining the blocks would give us n elements whose sum is 0 (mod n), as desired. If not, by Lemma 3.1 , we must have some 2 p of the xi summing to 0 (mod p), so we have 2 n elements whose sum is 0 (mod n). But since J itself is zero-sum and has size 3 n, the complement is zero-sum as well and has size n. Proposition 3.5. If d \| n, and J is a zero-sum sequence in (Z/n Z)2 of length 4n − d, then (n \| J) > 0.Proof. Note that 4 n − d ≥ 3m. By Theorem 3.3 , we can break off subsequences of size d with sum 0 (mod d) until we have only 3 d elements remaining. Then by Lemma 3.4 we can break off another d elements, to obtain 4( n/d ) − 3 blocks of size d, with sums dx 1, . . . , dx 4( n/d )−3 for some xi. By Theorem 3.3 , some n/d of the xi must sum to 0 in ( Z/(n/d )Z)2. Combining the corresponding blocks gives a subsequence of length n whose sum is zero in ( Z/n Z)2. The following corollary is clear from Proposition 3.5 and Theorem 3.3 . Corollary 3.6. Let ℓ be the smallest integer greater than or equal to 4 such that ℓ ∤ n. If J is a zero-sum sequence in (Z/n Z)2 of length at least 4n − ℓ + 1 , then (n \| J) > 0. Proposition 3.7. Suppose 4 ≤ ℓ ∤ n. There exists a zero-sum sequence in (Z/n Z)2 of length 4n − ℓ which contains no zero-sum subsequences of length n.Proof. First, consider a sequence of the form (0 , 0) a ≤ n − 1(0 , 1) b ≤ n − 1(1 , 0) c ≤ n − 1(1 , 1) d ≤ n − 1, where a denotes the number of (0 , 0)’s, etc., and a + b + c + d = 4 n − ℓ. It is easy to check that this sequence contains no zero-sum subsequence of length n. Now, we claim that there exists ( r, s ) ∈ (Z/n Z)2 such that adding ( r, s ) to each term of the above sequence will result in a zero-sum sequence. Note that adding ( r, s ) to each term does not change the fact that there is no zero-sum subsequence of length n.In fact, all we need is g := gcd( n, ℓ ) \| c + d, b + d. 4 AARON BERGER AND DANIELLE WANG We claim that the following a, b, c, d work. a = n − ℓ + g + 1 (or n − ℓ + 2 g + 1 if g = 1) b = n − 1 c = n − 1 d = n − g + 1 (or n − 2g + 1 if g = 1) . Note that g ≤ ℓ/ 2 because ℓ ∤ n, so a ≤ n − ℓ/ 2 + 1 ≤ n − 1 if g 6 = 1, and a = n − ℓ + 3 ≤ n − 1 if g = 1. It is easy to show that we always have a, d ≥ 0 and d ≤ n − 1, and that these a, b, c, d satisfy the divisibility relation. Now, Corollary 3.6 and Proposition 3.7 imply Theorem 1.4 . Open problems Harborth [ 8] first considered the problem of computing sn(( Z/n Z)d) for higher dimensions. He proved the following bounds. Theorem 4.1 (Harborth [ 8]) . We have (n − 1)2 d + 1 ≤ sn(( Z/n Z)d) ≤ (n − 1) nd + 1 . For d > 2 the precise value of sn(( Z/n Z)d) is not known. See [ 4, 5] for some better lower bounds and [ 1, 9] for some better upper bounds. In general the lower bound in Theorem 4.1 is not tight, but Harborth showed that it is an equality for n = 2 k a power of 2. Conjecture 4.2. If n = 2 k and d ≥ 1, we have s′ n (( Z/n Z)d) = 2 dn − ℓ + 1 , where ℓ is the smallest integer such that ℓ ≥ 2d and ℓ ∤ n. By an argument similar to the ( Z/n Z)2 case, we can reduce this conjecture to the case n = 2 d, in which case ℓ = 2 d +1. We also have not determined the modified EGZ constants for ( Z/n Z)2 for subseqences of length greater than n. Problem 4.3. Compute s′ nt (( Z/n Z)2) for t > 1. The constant sn(Z/m Z×Z/n Z) is known to be 2 m+2 n−3 for m \| n [7, Theorem 5.8.3]. Problem 4.4. Compute s′ nt (Z/m Z × Z/n Z) for t ≥ 1 and m \| n. Acknowledgements This research was conducted at the University of Minnesota Duluth REU and was supported by NSF / DMS grant 1650947 and NSA grant H98230-18-1-0010. We would like to thank Joe Gallian for running the program. References Alon, N., and Dubiner, M. Zero-sum sets of prescribed size. Combinatorics, Paul Erd¨ os is Eighty 1 (1993), 33–50. Augspurger, C., Minter, M., Shoukry, K., Sissokho, P., and Voss, K. Avoiding zero-sum subsequences of prescribed length over the integers. arXiv preprint arXiv:1603.03978 (2016). MODIFIED ERD ¨OS–GINZBURG–ZIV CONSTANTS FOR Z/n ZAND ( Z/n Z)25 Berger, A. An analogue of the Erd¨ os-Ginzburg-Ziv theorem over Z. arXiv preprint arXiv:1608.04125 (2016). Edel, Y., Elsholtz, C., Geroldinger, A., Kubertin, S., and Rackham, L. Zero-sum problems in finite abelian groups and affine caps. Quarterly Journal of Mathematics 58 , 2 (2007), 159–186. Elsholtz, C. Lower bounds for multidimensional zero sums. Combinatorica 24 , 3 (2004), 351–358. Erd¨ os, P., Ginzburg, A., and Ziv, A. Theorem in the additive number theory. Bull. Res. Council Israel F 10 (1961), 41–43. Halter-Koch, F., and Geroldinger, A. Non-unique factorizations: Algebraic, Combina-torial and Analytic Theory . Chapman and Hall/CRC, 2006. Harborth, H. Ein extremalproblem f¨ ur Gitterpunkte. J. Reine Angew. Math. (1973). Meshulam, R. On subsets of finite abelian groups with no 3-term arithmetic progressions. Journal of Combinatorial Theory, Series A 71 , 1 (1995), 168–172. Reiher, C. On Kemnitz’ conjecture concerning lattice-points in the plane. The Ramanujan Journal 13 , 1-3 (2007), 333–337.
4881	https://www.biointeractive.org/sites/default/files/Mouse_HardyWeinberg_Teacher.pdf	Allele and Phenotype Frequencies in Rock Pocket Mouse Populations Published August 2012 Revised August 2015 www.BioInteractive.org Page 1 of 4 LESSON TEACHER MATERIALS The Making of the Fittest: Natural Selection and Adaptation ALLELE AND PHENOTYPE FREQUENCIES IN ROCK POCKET MOUSE POPULATIONS OVERVIEW This lesson serves as a supplement to the short film The Making of the Fittest: Natural Selection and Adaptation ( by reinforcing the concepts of variation and natural selection. If your class does not cover Hardy-Weinberg equilibrium, you may wish to use another lesson, “Color Variation over Time in Rock Pocket Mouse Populations” ( KEY CONCEPTS AND LEARNING OBJECTIVES • Mutations that increase fitness of an organism increase in frequency in a population. • Evolution can happen quickly (hundreds of years, or even less); advantageous genetic mutations can increase in frequency in a population quite rapidly, even if the fitness advantage to the organism is small. Students will be able to • describe how variation, selection, and time fuel the process of evolution; and • manipulate and analyze data. CURRICULUM CONNECTIONS Curriculum Standards NGSS (April 2013) HS-LS1-1, HS-LS2-1, HS-LS2-2, HS-LS2-6, HS-LS2-7, HS-LS3-1, HS-LS3-3, HS-LS4-2, HS-LS4-3, HS-LS4-4, HS-LS4-5 HS.LS1.A, HS.LS2.C, HS.LS4.B, HS.LS4.C Common Core (2010) CCSS.ELA-Literacy.RST.9-10.3, CCSS.ELA-Literacy.RST.9-10.4, CCSS.ELA-Literacy.RST.9-10.7 CCSS.Math.Practice.MP.2, CCSS.Math.Practice.MP.3, CCSS.Math.Practice.MP.4, CCSS.Math.Practice.MP.5 AP Biology (2012–13) 1.A.1, 1.A.2, 1.A.4, 1.C.3, 4.C.3 IB Biology (2016) 3.1, 5.1, 5.2, 10.3 KEY TERMS adaptation, allele, allele frequency, Hardy-Weinberg, melanism, mutation, natural selection, variation TIME REQUIREMENTS This lesson was designed to be completed within one or two 50-minute class periods. SUGGESTED AUDIENCE This lesson is appropriate for high school biology (all levels including AP and IB) and introductory college biology. PRIOR KNOWLEDGE Students should have a basic understanding of evolution, adaptation, and algebra. MATERIALS calculator computer and the Selection Coefficient spreadsheet file found under the “Survival of the Fittest—Battling Beetles” activity at Allele and Phenotype Frequencies in Rock Pocket Mouse Populations www.BioInteractive.org Page 2 of 4 LESSON TEACHER MATERIALS The Making of the Fittest: Natural Selection and Adaptation TEACHING TIPS • Fill a few plastic sandwich bags with 15 grams of paper clips and pass them around so that students have an idea of how much a rock pocket mouse weighs. • You may want to show the film more than once so students can take notes. Encourage them to write down questions they have about the film’s content. • Students should understand that individuals do not evolve but that populations evolve and that variations may be favorable, neutral, or negative. Variations occur randomly and not “as needed.” • Explain that the Hardy-Weinberg theorem describes conditions under which evolution will not occur. This informs us about the conditions under which it will occur, such as small population size, migration, nonrandom mating, occurrence of mutation, and, finally, selection. Selection is influenced by factors such as environmental change, predation, and disease. These actively influence the competitive advantage of specific traits present in a population. Point out that a population will almost always meet one or more of these conditions. In other words, populations are always evolving. • Reinforce with students that dominant alleles are not always more common. The allele for dark-colored fur in rock pocket mice is dominant, but it is only common in populations living on dark substrates. • If your students are not comfortable with the math involved in solving the Hardy-Weinberg questions, it may be helpful to go over the problems in Part 1 as a class. We provide the stepped-out math in the answer key for Part 1. • We provide information about how to mathematically derive the selection coefficient in the “Survival of the Fittest—Battling Beetles” activity located at The spreadsheet is also at that location. • Part 3 requires access to a computer. You may want to assign this part as homework and e-mail the spreadsheet file to your students. ANSWER KEY QUESTIONS TO ANSWER WHILE WATCHING THE FILM 1. Watch the short film The Making of the Fittest: Natural Selection and Adaptation. As you watch, record the following information. a. What specific trait did researchers study in this investigation? Fur color, specifically melanism, is the trait they studied. b. How does this trait affect the survival of the mice in different environments? Depending on the color of the substrate, fur color may or may not help rock pocket mice blend in with their environment. On a light-colored substrate, mice with fur that is light in color are camouflaged and are not very obvious to predators. On a dark substrate, mice with dark-colored fur blend in and are better able to avoid predation. Predators readily spot mice with light-colored fur. Mice that survive predation are more likely to live, reproduce, and pass on their favorable trait for fur color. c. What is the genetic basis of the trait? Mutations in the Mc1r gene are responsible for the appearance of dark fur color in this particular population of rock pocket mice. (Note: Mutations of other genes in the pigment pathway may play a role in other mouse populations. See the lesson “The Biochemistry and Cell Signaling Pathway of the Mc1r Gene” for more information.) PART 1: REVIEWING THE PRINCIPLES OF THE HARDY-WEINBERG THEOREM 1. If there are 12 rock pocket mice with dark-colored fur and 4 with light-colored fur in a population, what is the value of q? Remember that light-colored fur is recessive. q = 0.5 Explanation: q2 = (4/16), or 0.25; therefore, q = the √0.25, or 0.5 2. If the frequency of p in a population is 60% (0.6), what is the frequency of q? q = .4 Explanation: If the frequency of p in a population is .6 then the frequency of q is .4 since p + q = 1. 3. In a population of 1,000 rock pocket mice, 360 have dark-colored fur. The others have light-colored fur. If the population is at Hardy-Weinberg equilibrium, what percentage of mice in the population are homozygous dominant, dark-colored mice? Allele and Phenotype Frequencies in Rock Pocket Mouse Populations www.BioInteractive.org Page 3 of 4 LESSON TEACHER MATERIALS The Making of the Fittest: Natural Selection and Adaptation p2 = 0.04, or 4% Explanation: q2 = 640/1,000 = 0.64, so, q = 0.8; because p + q = 1, p = 0.2 and p2 = (0.2)(0.2) = 0.04 PART 2: APPLYING HARDY-WEINBERG TO POCKET MOUSE FIELD DATA 1. Calculate the overall frequencies of light-colored mice and dark-colored mice caught on light-colored substrates. frequency = number of mice of one color/total number of mice Frequency of light-colored mice = 120/168 = 71%; Frequency of dark-colored mice = 48/168 = 29% 2. Calculate the overall frequencies of light-colored mice and dark-colored mice caught on dark-colored substrates. frequency = number of mice of one color/total number of mice Frequency of light-colored mice = 3/57 = 5%; Frequency of dark-colored mice = 54/57 = 95% 3. Using the Hardy-Weinberg equation and data from the table above, determine the number of mice with the DD and Dd genotypes on the light, rocky, granite substrate. Frequency of mice with the dd genotype on light-colored substrate = 71% Frequency of mice with the DD genotype on light-colored substrate = 3% Frequency of mice with the Dd genotype on light-colored substrate = 26% 4. Using the Hardy-Weinberg equation and data from the table above, determine the number of mice with the DD and Dd genotypes on the darky, rocky lava substrate. Frequency of mice with the dd genotype on dark-colored substrate = 5% Frequency of mice with the DD genotype on dark-colored substrate = 61% Frequency of mice with the Dd genotype on dark-colored substrate = 34% 5. Which fur color seems to have the greatest overall selective advantage? Use data collected from both dark-colored and light-colored substrates to support your answer. Dark fur color seems to have the greatest selective advantage. On the light-colored substrate, 29% of the mice have dark fur, while only 5% of the mice on the dark-colored substrate have light fur. Also, at collecting site no. 6, where there is a light-colored, rocky substrate, 43 out of 77 mice collected had dark-colored fur—over half of the sampled population. Dark-colored fur seems to have a selective advantage over light fur color. 6. According to the film, what environmental change gave a selective advantage for one coat color over another? The color of the landscape changed so that some members of the population were more visible to predators than other members were. That is what happened in the film. When sections of the landscape became dark, the light-colored mice were at a selective disadvantage. 7. In a separate study, 76 rock pocket mice were collected from four different, widely separated areas of dark lava rock. One collecting site was in Arizona. The other three were in New Mexico. Dr. Nachman and colleagues observed no significant differences in the color of the rocks in the four locations sampled. However, the dark-colored mice from the three New Mexico locations were slightly darker than the dark-colored mice from the Arizona population. The entire Mc1r gene was sequenced in all 76 of the mice collected. The mutations responsible for the dark fur color in the Arizona mice were absent from the three different populations of New Mexico mice. No Mc1r mutations were associated with dark fur color in the New Mexico populations. These findings suggest that adaptive dark coloration has occurred at least twice in the rock pocket mouse and that these similar phenotypic changes have different genetic bases. How does this study support the concept that natural selection is not random? Evidence that natural selection is not random is the fact that when different genetic mutations produce the same phenotypic results in different areas, these similar adaptations are favored under similar conditions. An example provided in the film is the different populations of rock pocket mice with mutations that result in dark fur color. Allele and Phenotype Frequencies in Rock Pocket Mouse Populations www.BioInteractive.org Page 4 of 4 LESSON TEACHER MATERIALS The Making of the Fittest: Natural Selection and Adaptation Dr. Carroll summed it up in the statement: “Evolution can and does repeats itself.” This is evidence that natural selection is not random. 8. To determine if the rock pocket mouse population is evolving, explain why it is necessary to collect fur color frequency data over a period of many years. The data collected represent only one moment in time. If the population is evolving, the frequency of the two alleles for fur color will change over time. If the population is not under selective pressure, or is not evolving, the frequencies will remain approximately the same. PART 3: HARDY-WEINBERG EXTENDED 1. Use the spreadsheet to determine how the selection coefficient (s) influences the phenotype of future generations. Substitute increasingly large numbers for s. Record each new value and describe what happens to the frequencies of p and q over the next five generations. When the value of s increases, the value of p increases while the value of q decreases in each generation. Students should record the values of s and either record or print the values for p and q. During this exercise, the frequency of the dominant phenotype increases as the frequency of the recessive phenotype decreases. 2. Explain how the selection coefficient and natural selection are related. The selection coefficient is a numerical representation of how much advantage or disadvantage a particular variation or trait provides an organism. It provides a way to mathematically model and predict evolutionary change. 3. In areas with primarily dark-colored substrate, dark-colored mice have a selective advantage over light-colored mice. Therefore, mice with one or more copies of the dominant Mc1r D allele have a selective advantage over mice with two copies of the Mc1r d allele. In the film, Dr. Carroll says that with a 1% selection advantage, it takes 1,000 years for 95% of the mice to have the dominant phenotype. With a 10% selection advantage, it would take just 100 years. Use the spreadsheet to verify these facts. a. Find out how many generations following the first appearance of a dark-colored mutation it would take for 95% of the mice to express the dominant dark-colored phenotype given a 1% advantage (s = 0.01). Rock pocket mice have approximately one litter of pups a year, so the number of generations will be equal to the number of years. You will not be able to use the graph on the Main Page tab since it only goes up to 100 generations. So, you will need to look at column D of the worksheet called Main Worksheet. Scroll down until the value is greater than 0.95. Record your answer below. It would take about 936 generations for 95% of the mice to express the dominant dark-colored phenotype. b. Repeat the process for a 10% advantage (s = 0.1). It would take about 100 generations. c. What would the selection coefficient need to be for 95% of the mice to have the dominant phenotype in just 50 years? Record your answer below. The coefficient would need to be about 0.22, a 22% advantage. AUTHOR Mary Colvard, Cobleskill-Richmondville High School (retired), New York FIELD TESTERS Kim Burley, LTCHS; Christina McCoy-Crawford, First Baptist School; Sherri Story, Kings Fork High School
4882	https://international-maths-challenge.com/a-family-of-sequences-and-number-triangles/	A Family of Sequences and Number Triangles - International Maths Challenge QOTD Academic Board Login Signup Signup Home About our Challenges Schools Fees & Registration FAQs Sample Questions A Family of Sequences and Number Triangles The triangular numbers (and higher triangular numbers) can be generated using this recurrence relationship: These will form Pascal’s triangle (if we shift the variables n->n+d-1 and d->r, we get the familiar C(n,r) indexing for the Pascal Triangle). The d=2 case gives the usual “flat” 2d triangular numbers, and other d values provide triangular numbers of different dimensions. It turns out that recurrence relation can be generalized to generate a family of sequences and triangles. Consider this more general relation: Doing some initial exploring reveals four interesting cases: The triangular numbers With all these additional parameters set to 1, we get our original relation, the familiar triangular numbers, and Pascal’s triangle. The k-polygonal numbers If we set the “zero dimension” to k-2, we end up with the k-polygonal numbers. The triangular numbers arise in the special case where k=3. Except in the k=3 case, the triangles that are generated are not symmetrical. Below is the triangle generated by setting k=5. The symmetrically shifted k-polygonal numbers As far as I know, there is not a standard name for these. Each k value will generate a triangle that is symmetrical about its center and whose edge values are equal to k-2. For a given k value, if you enter sequences generated by particular values of d,you’ll find that some are well known. The codes in the diagrams correspond to the sequence ids from the Encyclopedia. Here is the triangle generated by k=4: And here is the triangle generated for k=5: The Eulerian numbers (Euler’s number triangle) This is a particularly nice way to generate the Eulerian numbers, which have a nice connection to the triangular numbers. There is a little inconsistency in the way the Eulerian numbers are indexed, however. For this formula to work, it should be altered slightly so that d>0. The resulting formula looks like this: And the triangle looks like this: It is surprising that so many interesting and well known sequences and triangles can be generated from such a simple formula, and that they can be interpreted as being part of a single family. For more such insights, log into www.international-maths-challenge.com. Credit for article given to dan.mackinnon Blog Math-A-Maze Why Take the Challenge? Privacy Policy Terms & Conditions Contact Us × Updates at the IMC! The merchandise and certificate form is now closed. All winners will receive an email by Thursday 9 AM EDT with updates on their certificate status, merchandise, communication with their school, and other relevant details. ×
4883	https://www.khanacademy.org/science/up-class-11-physics/x3a9a44f124d01cf7:work-energy-and-power/x3a9a44f124d01cf7:work-energy-problems-involving-friction/v/work-energy-problem-with-friction	Work/energy problem with friction (video) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. 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Skip to lesson content UP Physics Grade 11 Course: UP Physics Grade 11>Unit 8 Lesson 6: Work-energy problems involving friction Energy dissipated by friction Work/energy problem with friction Science> UP Physics Grade 11> Work, energy and power> Work-energy problems involving friction © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Work/energy problem with friction Google Classroom Microsoft Teams About About this video Transcript Explore the concept of conservation of energy with a twist - introducing friction! Learn how energy isn't always fully conserved due to nonconservative forces like friction. Understand how potential energy converts into kinetic energy, and how friction impacts this transformation.Created by Sal Khan. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted Aryan Purohit 10 years ago Posted 10 years ago. Direct link to Aryan Purohit's post “So to find out the work d...” more So to find out the work done by frictional force, it doesn't matter if that force is acting along an incline(diagonal) or in a straight line the method of finding will always be the same, force distance? I mean thats not the case with other energies, if work done is along a straight line we can safely use forcedistance, but if its acting diagonally we have to use forcedistancecostheta. Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Teacher Mackenzie (UK) 10 years ago Posted 10 years ago. Direct link to Teacher Mackenzie (UK)'s post “Always use force x dista...” more Always use force x distance x cos theta Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Harshini Vijayajeevan 12 years ago Posted 12 years ago. Direct link to Harshini Vijayajeevan's post “what if there's a elastic...” more what if there's a elastic collision involved ? How can we relate work-energy theroem to that application ? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer keya2sweet 13 years ago Posted 13 years ago. Direct link to keya2sweet's post “how did he get 8,455 when...” more how did he get 8,455 when he minus 30.000 from 38,500? At 8:14 Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Shreyas Minocha 10 years ago Posted 10 years ago. Direct link to Shreyas Minocha's post “What would the downsides ...” more What would the downsides of lack of friction be in terms of physics? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Ravin 10 years ago Posted 10 years ago. Direct link to Ravin's post “Car tyres would not be ab...” more Car tyres would not be able to grip the road so they will slip our shoes wont grip the ground and this would make movement harder Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Emily 8 years ago Posted 8 years ago. Direct link to Emily's post “Why doesn't force due to ...” more Why doesn't force due to gravity in the x direction put work in? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Mark Geary 8 years ago Posted 8 years ago. Direct link to Mark Geary's post “Gravity does not act in t...” more Gravity does not act in the x direction, therefore, does not perform work in that direction. Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Mayank 12 years ago Posted 12 years ago. Direct link to Mayank's post “shouldn't we have calcula...” more shouldn't we have calculated force that acts in system by subtracting 60 from 90 x 9.8 cos 5 .And then calculating velocity. Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Jason Wang 6 years ago Posted 6 years ago. Direct link to Jason Wang's post “Hi, I was trying to answ...” more Hi, I was trying to answer this question using the work energy principle (Wnet = final kinetic energy - initial kinetic energy). Wnet = Fnet x distance = (force of gravity parallel to ramp - force of friction) x distance = (mgsin5 - 60)x500. And then using Wnet to solve for velocity. The answer is obviously wrong, but I can't seem to figure out why. Any help would be greatly appreciated. Jason Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Vitor 10 years ago Posted 10 years ago. Direct link to Vitor's post “Is this problem correct? ...” more Is this problem correct? I can'T reach that value with the fórmula of conservation of mechanic energy Answer Button navigates to signup page •Comment Button navigates to signup page (0 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Sabbarish Govindarajan 10 years ago Posted 10 years ago. Direct link to Sabbarish Govindarajan's post “The surface is rough, or,...” more The surface is rough, or, there's some friction. So, some energy will be lost because of friction. So, Mechanical Energy won't be conserved in this case. But for solving these type of problems, we can use Initial Energy= Final Energy+ Work Done by friction. That is some of the energy will be lost as heat due to friction.(Like while rubbing palms, you can feel the heat) And the energy lost is the Work done by friction. Remember, Use Momentum Conservation, when there's no external forces acting on it. Use Energy Conservation, when there's no energy loss due to friction or some other resistance(Viscosity, Air resistance). 6 comments Comment on Sabbarish Govindarajan's post “The surface is rough, or,...” (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more richard 11 years ago Posted 11 years ago. Direct link to richard's post “if he going down ice and ...” more if he going down ice and the friction is inside the gearbox (not due to air resistance or heat), why would that slow him down? Is that the same thing as friction due to air resistance? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Akhilesh Pandey 10 years ago Posted 10 years ago. Direct link to Akhilesh Pandey's post “At 8:09 , Why dosent he u...” more At 8:09 , Why dosent he use cos (theta) while calculating the frictional force on the biker?? Thank You Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Evan Lewis 9 years ago Posted 9 years ago. Direct link to Evan Lewis's post “We don't use cos(theta) b...” more We don't use cos(theta) because the force of friction is already parallel to the direction of displacement. You only have to do that if the force is not parallel to the direction of the displacement, because only parallel forces can do work on an object. Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Video transcript Welcome back. Welcome back. Welcome back. I'll now do another conservation of energy problem, and this time I'll add another twist. So far, everything we've been doing, energy was conserved by the law of conservation. But that's because all of the forces that were acting in these systems were conservative forces. And now I'll introduce you to a problem that has a little bit of friction, and we'll see that some of that energy gets lost to friction. And we can think about it a little bit. Well where does that energy go? And I'm getting this problem from the University of Oregon's zebu.uoregon.edu. And they seem to have some nice physics problems, so I'll use theirs. And I just want to make sure they get credit. So let's see. They say a 90 kilogram bike and rider. So the bike and rider combined are 90 kilograms. So let's just say the mass is 90 kilograms. Start at rest from the top of a 500 meter long hill. OK, so I think they mean that the hill is something like this. So if this is the hill, that the hypotenuse here is 500 hundred meters long. So the length of that, this is 500 meters. A 500 meter long hill with a 5 degree incline. So this is 5 degrees. And we can kind of just view it like a wedge, like we've done in other problems. There you go. That's pretty straight. OK. Assuming an average friction force of 60 newtons. OK, so they're not telling us the coefficient of friction and then we have to figure out the normal force and all of that. They're just telling us, what is the drag of friction? Or how much is actually friction acting against this rider's motion? We could think a little bit about where that friction is coming from. So the force of friction is equal to 60 newtons And of course, this is going to be going against his motion or her motion. And the question asks us, find the speed of the biker at the bottom of the hill. So the biker starts up here, stationary. That's the biker. My very artful rendition of the biker. And we need to figure out the velocity at the bottom. This to some degree is a potential energy problem. It's definitely a conservation of mechanical energy problem. So let's figure out what the energy of the system is when the rider starts off. So the rider starts off at the top of this hill. So definitely some potential energy. And is stationary, so there's no kinetic energy. So all of the energy is potential, and what is the potential energy? Well potential energy is equal to mass times the acceleration of gravity times height, right? Well that's equal to, if the mass is 90, the acceleration of gravity is 9.8 meters per second squared. And then what's the height? Well here we're going to have to break out a little trigonometry. We need to figure out this side of this triangle, if you consider this whole thing a triangle. Let's see. We want to figure out the opposite. We know the hypotenuse and we know this angle here. So the sine of this angle is equal to opposite over hypotenuse. So, SOH. Sine is opposite over hypotenuse. So we know that the height-- so let me do a little work here-- we know that sine of 5 degrees is equal to the height over 500. Or that the height is equal to 500 sine of 5 degrees. And I calculated the sine of 5 degrees ahead of time. Let me make sure I still have it. That's cause I didn't have my calculator with me today. But you could do this on your own. So this is equal to 500, and the sine of 5 degrees is 0.087. So when you multiply these out, what do I get? I'm using the calculator on Google actually. 500 times sine. You get 43.6. So this is equal to 43.6. So the height of the hill is 43.6 meters. So going back to the potential energy, we have the mass times the acceleration of gravity times the height. Times 43.6. And this is equal to, and then I can use just my regular calculator since I don't have to figure out trig functions anymore. So 90-- so you can see the whole thing-- times 9.8 times 43.6 is equal to, let's see, roughly 38,455. So this is equal to 38,455 joules or newton meters. And that's a lot of potential energy. So what happens? At the bottom of the hill-- sorry, I have to readjust my chair-- at the bottom of the hill, all of this gets converted to, or maybe I should pose that as a question. Does all of it get converted to kinetic energy? Almost. We have a force of friction here. And friction, you can kind of view friction as something that eats up mechanical energy. These are also called nonconservative forces because when you have these forces at play, all of the force is not conserved. So a way to think about it is, is that the energy, let's just call it total energy. So let's say total energy initial, well let me just write initial energy is equal to the energy wasted in friction-- I should have written just letters-- from friction plus final energy. So we know what the initial energy is in this system. That's the potential energy of this bicyclist and this roughly 38 and 1/2 kilojoules or 38,500 joules, roughly. And now let's figure out the energy wasted from friction, and the energy wasted from friction is the negative work that friction does. And what does negative work mean? Well the bicyclist is moving 500 meters in this direction. So distance is 500 meters. But friction isn't acting along the same direction as distance. The whole time, friction is acting against the distance. So when the force is going in the opposite direction as the distance, your work is negative. So another way of thinking of this problem is energy initial is equal to, or you could say the energy initial plus the negative work of friction, right? If we say that this is a negative quantity, then this is equal to the final energy. And here, I took the friction and put it on the other side because I said this is going to be a negative quantity in the system. And so you should always just make sure that if you have friction in the system, just as a reality check, that your final energy is less than your initial energy. Our initial energy is, let's just say 38.5 kilojoules. What is the negative work that friction is doing? Well it's 500 meters. And the entire 500 meters, it's always pushing back on the rider with a force of 60 newtons. So force times distance. So it's minus 60 newtons, cause it's going in the opposite direction of the motion, times 500. And this is going to equal the ending, oh, no. This is going to equal the final energy, right? And what is this? 60 times 500, that's 3,000. No, 30,000, right. So let's subtract 30,000 from 38,500. So let's see. Minus 30. I didn't have to do that. I could have done that in my head. So that gives us 8,455 joules is equal to the final energy. And what is all the final energy? Well by this time, the rider's gotten back to, I guess we could call the sea level. So all of the energy is now going to be kinetic energy, right? What's the formula for kinetic energy? It's 1/2 mv squared. And we know what m is. The mass of the rider is 90. So we have this is 90. So if we divide both sides. So the 1/2 times 90. That's 45. So 8,455 divided by 45. So we get v squared is equal to 187.9. And let's take the square root of that and we get the velocity, 13.7. So if we take the square root of both sides of this, so the final velocity is 13.7. I know you can't read that. 13.7 meters per second. And this was a slightly more interesting problem because here we had the energy wasn't completely conserved. Some of the energy, you could say, was eaten by friction. And actually that energy just didn't disappear into a vacuum. It was actually generated into heat. And it makes sense. If you slid down a slide of sandpaper, your pants would feel very warm by the time you got to the bottom of that. But the friction of this, they weren't specific on where the friction came from, but it could have come from the gearing within the bike. It could have come from the wind. Maybe the bike actually skidded a little bit on the way down. I don't know. But hopefully you found that a little bit interesting. And now you can not only work with conservation of mechanical energy, but you can work problems where there's a little bit of friction involved as well. Anyway, I'll see you in the next video. 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4884	https://nhigham.com/wp-content/uploads/2021/07/cohen_an21.pdf	Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-D Neil Calkin, Nick Cohen Clemson University, University of California, Irvine July 23, 2021 Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 1 / 34 Thanks to... Rob Corless Neil Calkin NSF Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 2 / 34 How to count configurations of non-attacking kings on a chess board K K K K Figure: The five possible 2 × 2 boards. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 3 / 34 How many ways can you place non-attacking kings on a chessboard? How many configurations for 1 × 1 chess boards? Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 4 / 34 How many ways can you place non-attacking kings on a chessboard? How many configurations for 1 × 1 chess boards? 2. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 4 / 34 How many ways can you place non-attacking kings on a chessboard? How many configurations for 1 × 1 chess boards? 2. 2 × 2 boards? Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 4 / 34 How many ways can you place non-attacking kings on a chessboard? How many configurations for 1 × 1 chess boards? 2. 2 × 2 boards? 5. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 4 / 34 How many ways can you place non-attacking kings on a chessboard? How many configurations for 1 × 1 chess boards? 2. 2 × 2 boards? 5. 2 × 2 × 2 boards? Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 4 / 34 How many ways can you place non-attacking kings on a chessboard? How many configurations for 1 × 1 chess boards? 2. 2 × 2 boards? 5. 2 × 2 × 2 boards? 9. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 4 / 34 How many ways can you place non-attacking kings on a chessboard? How many configurations for 1 × 1 chess boards? 2. 2 × 2 boards? 5. 2 × 2 × 2 boards? 9. 3 × 3 boards? Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 4 / 34 How many ways can you place non-attacking kings on a chessboard? How many configurations for 1 × 1 chess boards? 2. 2 × 2 boards? 5. 2 × 2 × 2 boards? 9. 3 × 3 boards? 35. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 4 / 34 How many ways can you place non-attacking kings on a chessboard? How many configurations for 1 × 1 chess boards? 2. 2 × 2 boards? 5. 2 × 2 × 2 boards? 9. 3 × 3 boards? 35. 3 × 3 × 3 boards? Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 4 / 34 How many ways can you place non-attacking kings on a chessboard? How many configurations for 1 × 1 chess boards? 2. 2 × 2 boards? 5. 2 × 2 × 2 boards? 9. 3 × 3 boards? 35. 3 × 3 × 3 boards? 2,089. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 4 / 34 How many ways can you place non-attacking kings on a chessboard? How many configurations for 1 × 1 chess boards? 2. 2 × 2 boards? 5. 2 × 2 × 2 boards? 9. 3 × 3 boards? 35. 3 × 3 × 3 boards? 2,089. 4 × 4 × 4 boards? Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 4 / 34 How many ways can you place non-attacking kings on a chessboard? How many configurations for 1 × 1 chess boards? 2. 2 × 2 boards? 5. 2 × 2 × 2 boards? 9. 3 × 3 boards? 35. 3 × 3 × 3 boards? 2,089. 4 × 4 × 4 boards? 3,144,692. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 4 / 34 In the limit One question we can ask is: How does the number of boards increase as we increase the number of squares in the board? multiplicity2D = lim m,n→∞F(m, n)1/mn Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 5 / 34 In the limit One question we can ask is: How does the number of boards increase as we increase the number of squares in the board? multiplicity2D = lim m,n→∞F(m, n)1/mn ≈1.3426 . . . Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 5 / 34 In the limit One question we can ask is: How does the number of boards increase as we increase the number of squares in the board? multiplicity2D = lim m,n→∞F(m, n)1/mn ≈1.3426 . . . How much information can be stored per square? capacity2D = log2 multiplicity2D ≈0.42507 . . . Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 5 / 34 In the limit One question we can ask is: How does the number of boards increase as we increase the number of squares in the board? multiplicity2D = lim m,n→∞F(m, n)1/mn ≈1.3426 . . . How much information can be stored per square? capacity2D = log2 multiplicity2D ≈0.42507 . . . What about for three dimensions? Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 5 / 34 Why is this problem important? Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 6 / 34 Why is this problem important? Statistical mechanics (Entropy) Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 6 / 34 Why is this problem important? Statistical mechanics (Entropy) Information Theory (Channel Capacity) Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 6 / 34 Why is this problem important? Statistical mechanics (Entropy) Information Theory (Channel Capacity) Dynamical Systems (Subshifts of Finite Type) Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 6 / 34 Why is this problem interesting? Look at these beautiful matrices! We can solve many types of recurrence relations exactly. It feels like an exact solution should be right around the corner. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 7 / 34 Why is this problem interesting? Look at these beautiful matrices! We can solve many types of recurrence relations exactly. It feels like an exact solution should be right around the corner. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 7 / 34 Why is this problem interesting? Look at these beautiful matrices! We can solve many types of recurrence relations exactly. It feels like an exact solution should be right around the corner. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 7 / 34 Look at these beautiful eigenvectors! Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 8 / 34 The 2-D problem The 2-D problem is relatively easy. 1 Identify the 1-D slices 2 Build an adjacency matrix Ak of all possible slices of height k with 1s identifying slices that are allowed to be placed next to each other. 3 Compute eT 1 An+1 k e1 to calculate the number of configurations of kings on a board of dimension k × n. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 9 / 34 The 2-D problem: identifying the 1-D slices The 1-D slices can be generated using the simple recursive relationship shown below. It is easy to see that the number of 1-D boards are Fibonacci numbers. K Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 10 / 34 The 2-D problem: building an adjacency matrix An adjacency matrix is indexed by the previously shown ordering of 1-D slices and shows us which slices may be adjacent. If we index as indicated in the previous slides, it is easy to see that Ak−1 appears in the top left corner and copies of Ak−2 appear in the bottom left and top right corners. Ak−1 Ak−2 Ak−2 0 Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 11 / 34 The 2-D problem: building an adjacency matrix A =       1 1 1 1 1 1 0 0 1 0 1 0 0 0 0 1 1 0 0 0 1 0 0 0 0       = A K K K K K Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 12 / 34 The 2-D Problem: Interpreting the Adjacency Matrix It is easy to see (and well known) that the (i, j) entry of An+1 k counts the number of configurations of n slices sandwiched between slice i and slice j. To simplify computation, we can sum the entries of An−1 k to count the number of k × n configurations of kings like so: v ←(1, 1, . . . , 1) for i ←1 to n −1: v ←Ak v return P vi Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 13 / 34 The 2-D Problem: Interpreting the Adjacency Matrix An 1 approaches a multiple of the Perron eigenvector, and its growth rate, the Perron eigenvalue, tells us what the asymptotic growth rate of the number of boards is as we increase the thickness. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 14 / 34 The 2-D Problem: Interpreting the Adjacency Matrix An 1 approaches a multiple of the Perron eigenvector, and its growth rate, the Perron eigenvalue, tells us what the asymptotic growth rate of the number of boards is as we increase the thickness. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 14 / 34 The 2-D Problem: Interpreting the Adjacency Matrix An 1 approaches a multiple of the Perron eigenvector, and its growth rate, the Perron eigenvalue, tells us what the asymptotic growth rate of the number of boards is as we increase the thickness. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 14 / 34 The 2-D Problem: Interpreting the Adjacency Matrix The dimension of Ak grows like Fib(k) × Fib(k). E.g. A30 is 2, 178, 309 × 2, 178, 309. The number of non-zero entries in these matrices grow like 2k (A30 has 1, 431, 655, 765 nonzero entries), but it turns out, you only need to store the vector, since the matrix operation on the vector can be coded without the need to hold the matrix in computer memory, so the memory requirements only grow like ϕk, where ϕ ≈1.618 is the golden ratio. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 15 / 34 The 3-D Problem is Hard Just as we create an adjacency matrix of 1-D slices for the 2-D problem, we can also create an adjacency matrix of 2-D slices for the 3-D problem. But there is no nice way of generating these 2-D slices that yields easy to store matrices. Unlike for the 2-D problem, we need to store the entire matrix in memory. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 16 / 34 The 3-D Problem is Hard Just as we create an adjacency matrix of 1-D slices for the 2-D problem, we can also create an adjacency matrix of 2-D slices for the 3-D problem. But there is no nice way of generating these 2-D slices that yields easy to store matrices. Unlike for the 2-D problem, we need to store the entire matrix in memory... sort of. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 16 / 34 Matrix Compression Studying the Perron eigenvector, we noticed that many entries were repeated. These repeated entries usually corresponded with symmetries of slices such as flips or rotations. K K x= 1.0 0.5513875 0.3554157 0.5513875 0.3554157 Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 17 / 34 Matrix Compression We then realized that we could use these repeated eigenvector entries to reduce the dimension of the matrices used in our calculations by summing the rows corresponding to identical eigenvector entries and eliminating the redundant column index. A =       1 1 1 1 1 1 0 0 1 0 1 0 0 0 0 1 1 0 0 0 1 0 0 0 0       − →     1 1 1 1 2 1 0 0 1 0 0 0 1 0 0 0     Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 18 / 34 Matrix Compression In fact, we can do all the compression in one process as follows: Ac = L A R Ac is the compressed matrix A is a d × d matrix while Ac is a d′ × d′ matrix with d′ < d. R is a d × d′ matrix that has one column for every unique eigenvector entry, and the column is the indicator function for that class of slices. L is a d′ × d matrix that has one row for every unique eigenvector entry. For each row and its corresponding class, L has a 1 at the very first instance of an index corresponding to that class and 0s elsewhere. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 19 / 34 Matrix Compression L =   1 0 0 0 0 0 1 0 0 0 0 0 1 0 0   A =       1 1 1 1 1 1 0 0 1 0 1 0 0 0 0 1 1 0 0 0 1 0 0 0 0       R =       1 0 0 0 1 0 0 0 1 0 1 0 0 0 1       Ac =   1 2 2 1 1 0 1 0 0   L, A, and R for 3 × 1 slices. If x is a Perron eigenvector of A and λ is the Perron eigenvalue of A, then L x is a Perron eigenvector of L A R with L A R L x = λL x. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 20 / 34 Matrix Compression L x =   1 0 0 0 0 0 1 0 0 0 0 0 1 0 0         xa xb xc xb xc       =   xa xb xc   R(L x) =       1 0 0 0 1 0 0 0 1 0 1 0 0 0 1         xa xb xc  =       xa xb xc xb xc       Ac (L x) = L A R L x = L Ax = λ L x Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 21 / 34                                 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 1 1 1 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 1 1 1 0 0 0 0 0 0 1 0 0 1 1 1 1 0 0 1 0 0 1 0 0 1 0 0 1 1 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 1 1 0 0 1 0 0 1 0 0 1 0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1                                 − →   0 0 1 0 3 1 12 6 1   The adjacency matrix for 3 × 3 slices with one opposite edge pair glued together: uncompressed and compressed. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 22 / 34 Matrix Compression We searched the literature for others using matrix compression and found: "Exact and Approximate Compression of Transfer Matrices for Graph Homomorphisms" by Lundow and Markström "Compression of Transfer Matrices" by Lundow. These authors describe matrix compression via graph homomorphisms. In our case, the vertices are slices and edges are possible adjacencies between slices. The key step in using this matrix compression technique is identifying the graph homomorphisms. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 23 / 34 Graph Homomorphisms for Non-Attacking Kings K K K K K K K K K K K K K We find graph homomorphisms by identifying the regions where kings can be placed in adjacent slices. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 24 / 34 Graph Homomorphisms for Non-Attacking Kings We call the adjacent region of possible kings positions to a slice a PKP . To check whether two slices were equivalent, we compared their PKP patterns under rotation, reflection, and translation (when applicable). We converted each PKP to an integer, and for each class of equivalent PKPs, we determined the least such integer, which we called the min PKP . K K K K 27 216 432 54 Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 25 / 34 Graph homomorphisms for non-attacking kings We partitioned all slices into classes that share the same min PKP . This gives us an effective method for computing the graph homomorphisms for our problem. K K K K 27 216 432 54 Graph homomorphisms for non-attacking kings We partitioned all slices into classes that share the same min PKP . This gives us an effective method for computing the graph homomorphisms for our problem. K K K K 27 216 432 54 All four slices go in bucket labeled 27. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 26 / 34 Computation The end result compressed matrices we created had entries ci,j whose value is the number of adjacencies between some representative of the ith slice class and all slices in class j. If slice class i has 1000 members, we needed to check roughly 1000 times fewer adjacencies than in a non-compressed adjacency matrix. Counting these adjacencies was the primary time constraint we ran into. The largest slice dimensions we were able to work with had roughly 1000 members per class, and so matrix compression sped up our code by roughly a factor of 1000. The compressed matrix also decreased the storage requirements by roughly 1000 fold. Both uncompressed and compressed matrices could be stored as sparse matrices. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 27 / 34 PSA: GPUs are Your Friend Originally we used multiple CPUs to find slice adjacencies with Java. Later we used the PyTorch Python library to move this operation to GPUs and achieved roughly another 1000 times speedup. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 28 / 34 Results 1.1722475193 ≤multiplicity3D ≤1.1798420399 0.2292772260 ≤capacity3D ≤0.2385937211, Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 29 / 34 Results n F(n, n, n) F ′(n, n, n) F ′′(n, n, n) F ′′′(n, n, n) 1 2 1 1 1 2 9 9 9 9 3 2,089 469 109 28 4 3,144,692 955,597 285,457 86,409 5 2,748, 141,446, 7,797, 442, 613,397,101 194,951 443,501 888,551 6 107,008,949, 3,540,028,254, 126,286,208, 868,167,431,857 720,734,235 383,726,353 7 13,894,384, 73,142,142, 421,725,200, 033,156,308,816, 037,998,950, 626,057,564, 935,906,058,416 249,305,520,745 456,468,571 Table: The number of n × n × n solids of various sizes with 0, 1, 2, and 3 pairs of attached ends. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 30 / 34 Open Questions Can we create an indexing scheme for the 2-D slices that allow better storage of uncompressed adjacency matrices (much like what is done for 1-D slices)? Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 31 / 34 Open Questions Can we create an indexing scheme for the 2-D slices that allow better storage of uncompressed adjacency matrices (much like what is done for 1-D slices)? Can we bound the relative sizes of the other eigenvalues for an adjacency matrix? Doing so may greatly improve the lower bound on multiplicity3D and capacity3D. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 31 / 34 Open Questions Can we create an indexing scheme for the 2-D slices that allow better storage of uncompressed adjacency matrices (much like what is done for 1-D slices)? Can we bound the relative sizes of the other eigenvalues for an adjacency matrix? Doing so may greatly improve the lower bound on multiplicity3D and capacity3D. Can we develop a technique that scales to 4-D and beyond? Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 31 / 34 Open Questions Can we create an indexing scheme for the 2-D slices that allow better storage of uncompressed adjacency matrices (much like what is done for 1-D slices)? Can we bound the relative sizes of the other eigenvalues for an adjacency matrix? Doing so may greatly improve the lower bound on multiplicity3D and capacity3D. Can we develop a technique that scales to 4-D and beyond? Compressed matrices are not always full rank. Why? How much more compression is possible? Is it practical? Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 31 / 34 Open Questions Can we create an indexing scheme for the 2-D slices that allow better storage of uncompressed adjacency matrices (much like what is done for 1-D slices)? Can we bound the relative sizes of the other eigenvalues for an adjacency matrix? Doing so may greatly improve the lower bound on multiplicity3D and capacity3D. Can we develop a technique that scales to 4-D and beyond? Compressed matrices are not always full rank. Why? How much more compression is possible? Is it practical? Email any questions or ideas to cohenn1@uci.edu Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 31 / 34 Open Questions Can we create an indexing scheme for the 2-D slices that allow better storage of uncompressed adjacency matrices (much like what is done for 1-D slices)? Can we bound the relative sizes of the other eigenvalues for an adjacency matrix? Doing so may greatly improve the lower bound on multiplicity3D and capacity3D. Can we develop a technique that scales to 4-D and beyond? Compressed matrices are not always full rank. Why? How much more compression is possible? Is it practical? Email any questions or ideas to cohenn1@uci.edu Thanks for your attention! Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 31 / 34 Lower bound Maxminum principle Switch indices and apply maximum principle twice each. multiplicity3D ≥    λp+2q+1,t+2u+1 λp+2q+1,2u+1 1/t λ′ 2s,2q+1 1/2s    1/p Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 32 / 34 Upper bound The trace of powers of adjacency matrices counts cylindrical boards; i.e. the number of boards that both begin and end with the same slice The trace is sums of powers of the eigenvalues. Switch indices. η3 ≤λ′′ 2p,2q 1/4pq Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 33 / 34 References R. J. Baxter, I. G. Enting, and S. K. Tsang, “Hard-square lattice gas,” J. Statist. Phys., vol. 22, no. 4, pp. 465–489, 1980. N. J. Calkin, K. James, S. Purvis, S. Race, K. Schneider, and M. Yancey, “Counting kings: As easy as λ1, λ2, λ3 . . . ,” in Proceedings of the Thirty-Seventh Southeastern International Conference on Combinatorics, Graph Theory and Computing, vol. 183, 2006, pp. 83–95. N. J. Calkin and H. S. Wilf, “The number of independent sets in a grid graph,” SIAM Journal on Discrete Mathematics, vol. 11, no. 1, pp. 54–60, 1998. S. Friedland, “On the entropy of Zd subshifts of finite type,” Linear Algebra Appl., vol. 252, pp. 199–220, 1997. S. Friedland, P . H. k. Lundow, and K. Markström, “The 1-vertex transfer matrix and accurate estimation of channel capacity,” IEEE Trans. Inform. Theory, vol. 56, no. 8, pp. 3692–3699, 2010. P . H. k. Lundow, “Compression of transfer matrices,” in 1-3, vol. 231, 17th British Combinatorial Conference (Canterbury, 1999), 2001, pp. 321–329. P . H. k. Lundow and K. Markström, “Exact and approximate compression of transfer matrices for graph homomorphisms,” LMS J. Comput. Math., vol. 11, pp. 1–14, 2008. H. C. Marques Fernandes, Y. Levin, and J. J. Arenzon, “Equation of state for hard-square lattice gases,” Physical Review E, vol. 75, no. 5, 2007. Z. Nagy and K. Zeger, “Capacity bounds for the 3-dimensional (0, 1) runlength limited channel,” in Applied algebra, algebraic algorithms and error-correcting codes (Honolulu, HI, 1999), ser. Lecture Notes in Comput. Sci. Vol. 1719, Springer, Berlin, 1999, pp. 245–251. Neil Calkin, Nick Cohen (Clemson University, University of California, Irvine) Calculating the 3-D Kings Multiplicity Constant: Configurations of Non-Attacking Kings in 3-July 23, 2021 34 / 34
4885	https://www.geeksforgeeks.org/maths/tangent-addition-formula/	Tangent Addition Formula Last Updated : 23 Jul, 2025 Suggest changes Like Article Trigonometric identities are equalities using trigonometric functions that hold true for any value of the variables involved, hence defining both sides of the equality. These are equations that relate to various trigonometric functions and are true for every variable value in the domain. The formulae sin(A+B), cos(A-B), and tan(A+B) are some of the sum and difference identities. Table of Content Tangent Trigonometric Ratio Tangent Addition Formula Derivation of Tangent Addition Related Articles Sample Problems on Tangent Addition Sample Problems on Tangent Addition Formula Tangent Trigonometric Ratio The ratio of any two right triangle sides is called a trigonometric ratio. The tangent ratio is defined as the ratio of the length of the opposite side of an angle divided by the length of the adjacent side. If θ is the angle created by the base and hypotenuse of a right-angled triangle then, tan θ = Perpendicular/Base = sin θ/ cos θ Here, perpendicular is the side opposite to the angle and base is the side adjacent to it. Tangent Addition Formula In trigonometry, the tangent addition formula is referred to as the tan(A + B) formula for the compound angle (A+B). It is used when the angle for which the tangent function value is to be determined is supplied as the sum of any two angles. It may alternatively be written as tan(A + B) = sin (A + B)/cos (A + B) since the tangent function is a ratio of the sine and cosine functions. tan(A + B) = (tan A + tan B)/(1 - tan A tan B) Derivation of Tangent Addition The formula for tangent addition is derived by using the formulas for expansion of sum angle for sine and cosine ratios. Now, we know that, tan (A + B) = sin (A + B)/cos (A + B) ...... (1) Substitute sin (A + B) = sin A cos B + cos A sin B and cos (A + B) = cos A cos B - sin A sin B in the equation (1). tan (A + B) = (sin A cos B + cos A sin B)/(cos A cos B - sin A sin B) Dividing the numerator and denominator by cos A cos B, we get tan (A + B) = [(sin A cos B + cos A sin B)/(cos A cos B)]/[(cos A cos B - sin A sin B)/(cos A cos B)] tan (A + B) = [(sin A cos B)/(cos A cos B) + (cos A sin B)/(cos A cos B)]/[(cos A cos B)/(cos A cos B) - (sin A sin B)/(cos A cos B)] tan (A + B) = (tan A + tan B)/(1 - tan A tan B) This derives the formula for tangent addition of any two angles, A and B. Related Articles: Sin Cos Tan Formula Law of Sines Trigonometric Functions Trigonometric Chart Sample Problems on Tangent Addition Problem 1. If tan A = 1/2 and tan B = 1/3, find the value of tan (A+B) using the formula. Solution: We have, tan A = 1/2 and tan B = 1/3. Using the formula we get, tan (A + B) = (tan A + tan B)/(1 - tan A tan B) = (1/2 + 1/3)/(1 - (1/2)(1/3)) = (5/6)/(1 - 1/6) = (5/6)/(5/6) = 1 Problem 2. If tan A = 2/3 and tan B = 4/7, find the value of tan (A+B) using the formula. Solution: We have, tan A = 2/3 and tan B = 4/7. Using the formula we get, tan (A + B) = (tan A + tan B)/(1 - tan A tan B) = (2/3 + 4/7)/(1 - (2/3)(4/7)) = (16/21)/(1 - 8/21) = (16/21)/(13/21) = 16/13 Problem 3. If tan (A+B) = 15/11 and tan A = 2/11, find the value of tan B using the formula. Solution: We have, tan (A+B) = 15/11 and tan A = 2/11. Let tan B = x. Using the formula we get, tan (A + B) = (tan A + tan B)/(1 - tan A tan B) => 15/11 = (2/11 + x)/(1 - (2/11)(x)) => 15/11 = ((2 + 11x)/11)/((11 - 2x)/11) => 15/11 = (11x + 2)/(11 - 2x) => 165 - 30x = 121x + 22 => 151x = 143 => x = 143/151 => tan B = 143/151 Problem 4. If tan B = 6/13 and tan (A+B) = 9/13, find the value of tan A using the formula. Solution: We have, tan (A+B) = 9/13 and tan B = 6/13. Let tan A = x. Using the formula we get, tan (A + B) = (tan A + tan B)/(1 - tan A tan B) => 9/13 = (x + 6/13)/(1 - (x)(6/13)) => 9/13 = ((13x + 6)/13)/((13 - 6x)/13) => 9/13 = (13x + 6)/(13 - 6x) => 117 - 54x = 169x + 78 => 223x = 39 => x = 39/223 => tan A = 39/223 Problem 5. If sin A = 4/5 and cos B = 5/13, find the value of tan (A+B) using the formula. Solution: We have, sin A = 4/5. It means, cos A = 3/5. So, tan A = 4/3. Also, cos B = 5/13. It means, sin B = 12/13. So, tan B = 12/5. Using the formula we get, tan (A + B) = (tan A + tan B)/(1 - tan A tan B) = (4/3 + 12/5)/(1 - (4/3)(12/5)) = (56/15)/(1 - 48/15) = (56/15)/(-33/15) = -56/33 Sample Problems on Tangent Addition Formula Using the tangent addition formula, find tan(45° + 30°) If tan A = 2 and tan B = 3, Calculate tan(A+B)using tangent addition formula. Solve for x in the equation tan(𝑥+15°)=√3 If sin A = 1/√2 and cos B = 1/√2, find the value of tan (A+B) using the formula. If tan (A+B) = 15/11 and tan A = 2/11, find the value of tan B using the formula. 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4886	https://www.splashlearn.com/math-vocabulary/identity-function	Identity Function: Definition, Graph, Examples, Facts Parents Explore by Grade Preschool (Age 2-5)KindergartenGrade 1Grade 2Grade 3Grade 4Grade 5 Explore by Subject Math ProgramEnglish Program More Programs Homeschool ProgramSummer ProgramMonthly Mash-up Helpful Links Parenting BlogSuccess StoriesSupportGifting Also available on Educators Teach with Us For TeachersFor Schools and DistrictsData Protection Addendum Impact Success Stories Resources Lesson PlansClassroom ToolsTeacher BlogHelp & Support More Programs SpringBoardSummer Learning Our Library All By Grade PreschoolKindergartenGrade 1Grade 2Grade 3Grade 4Grade 5 By Subject MathEnglish By Topic CountingAdditionSubtractionMultiplicationPhonicsAlphabetVowels One stop for learning fun! Games, activities, lessons - it's all here! 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Domain and Range of Identity Function Properties of Identity Function Solved Examples on Identity Function Practice Problems on Identity Function Frequently Asked Questions about Identity Function What Is Identity Function? An identity function is a polynomial function that maps every element to itself. So, the image of any element is the element itself. It is a function in which the output is the same as the input. Identity function is denoted by “I” and it has the form I(x) = x for all real numbers. An identity function is also known as an identity map,identity relation or identity transformation. It is known as an identity function because the image of an element in the domain is the same as the output in the range. The identity of the element is maintained. Recommended Games Identity Quarter Hours Game Play More Games Identity Function Definition Identity function is a real-valued functionf R→Rsuch thatf(x)=xfor allx∈R. An identity function maps each real number to itself. The preimage and the image are equal for identity functions. It is noted from the above image that the function f is an identity function as each element of A is mapped onto itself. Hence, the function f is one-one and onto. Domain and Range of Identity Function The domain of the identity function is R, the set of real numbers. The range of the identity function is also R. The co-domain and the range of an identity function are equal sets. Hence, the identity function is onto. Inverse of Identity Function When we find the inverse of any function, the domain and range of that function gets exchanged. This implies that the identity function is invertible and is its own inverse. Identity function takes an element and maps it to itself. Inverse of the identity function (which is itself) again maps the same element to itself. Identity Function Graph To graph the identity function, we plot the different points as shown below. x-4-3-2-1 0 1 2 3 I(x)-4-4-2-1 0 1 2 3 The graph of the identity function is a straight line that passes through the origin. The straight line makes an angle of 45° with the positive x-axis. Slope of a straight line = tan θ where θ is the angle between line and positive x-axis. So, slope of an identity function graph = tan 45° = 1 Properties of Identity Function It is a real-valued linear function. The graph of an identity function is a straight line that makes an angle of 45° with both x-axis and y-axis. Identity function is bijective. The inverse of the identity function is the identity function itself. Identity function is an odd function since I(x)/n e q I(−x). Solved Examples on Identity Function 1.If g is an identity function, then find g(0), g(1), g(100), g(6.4). Solution: g is an identity function. So, it will return the same value. g(0) = 0 g(1) = 1 g(100) = 100 g(6.4) = 6.4 2. What is the domain of the identity function? Solution: Domain of the identity function is the set of real numbers. Identity function maps every real number to itself. 3. Define an identity function. What is the graph of identity function? Solution: An identity function is defined as a real-valued function f : R → R such that f(x) = x for all x in R. Its graph is a straight line passing through the origin. Practice Problems on Identity Function Identity Function: Definition, Graph, Examples, FAQs Attend this quiz & Test your knowledge. 1 The identity function will map the real number 0 to 1 0 10 x Correct Incorrect Correct answer is: 0 The identity function returns the same value. Thus, I(0) = 0 2 Which of the following is the domain of an identity function? Complex Numbers Integers Real Numbers Natural Numbers Correct Incorrect Correct answer is: Real Numbers The identity function is defined as f: R → R such that f(x) = x. 3 The identity function has the form f(x) = 1 f(x) = x 2 f(x) = c f(x) = x Correct Incorrect Correct answer is: f(x) = x The identity function is defined as f: R → R such that f(x) = x. 4 Which of the following is a slope of the graph of an identity function? 0 1 2 not defined Correct Incorrect Correct answer is: 1 θ = 45° Slope (m) = tan 45° = 1 5 The identity function is only onto only one-one one-one and onto many to one Correct Incorrect Correct answer is: one-one and onto The identity function is bijective. Thus, it is one-one and onto. Frequently Asked Questions about Identity Function What is the difference between a constant function and an identity function? A constant function is a function whose output remains the same irrespective of the input. It is defined as f(x) = c, where x ∈ R. On the other hand, an identity function is a function that returns the same value. It is defined as f(x) = x, where x ∈ R. What is a many-to-one function? According to the many-to-one function, the two or more different elements are mapped to the same image. One example of this function is: Is identity function even or odd? Identity function is an odd function. f(x) = x is an identity function. f( – x) = – x RELATED POSTS Converting Fractions into Decimals – Methods, Facts, Examples Math Symbols – Definition with Examples Math Glossary Terms beginning with Z Factor Tree – Definition with Examples LEARN & PLAY Pre-Kindergarten Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade Explore 15,000+ Games and Worksheets Try for free Latest Videos POPULAR POSTS Feet to CM (ft to cm) Conversion – Formula,… Prime Numbers – Definition, Chart, Examples,… Place Value – Definition with Examples AM and PM – Definition, Examples, FAQs,… Three Dimensional Shapes (3D Shapes)- Definition, Examples Math & ELA \| PreK To Grade 5 Kids see fun. You see real learning outcomes. Watch your kids fall in love with math & reading through our scientifically designed curriculum. 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4887	https://ocw.mit.edu/courses/18-02sc-multivariable-calculus-fall-2010/0768ecb704c18ede11fae15923c160f8_MIT18_02SC_notes_6.pdf	Distances to planes and lines In this note we will look at distances to planes and lines. Our approach is geometric. Very broadly, we will draw a sketch and use vector techniques. Please note is that our sketches are not oriented, drawn to scale or drawn in perspective. Rather they are a simple ’cartoon’ which shows the important features of the problem. 1. Distance: point to plane: ii) A plane with normal − → N and containing a point Q. − → N Ingredients: i) A point P , −→ PQ −→ PQ The distance from P to the plane is d = cos θ = \| \| . · \|N\| We will explain this formula by way of the following example. Example 1: Let P = (1, 3, 2). Find the distance from P to the plane x + 2y = 3. Answer: First we gather our ingredients. Q = (3, 0, 0) is a point on the plane (it is easy to ﬁnd such a point). N = normal to plane = i + 2j. R = point on plane closest to P (this is point unknown and we do not need to ﬁnd it to ﬁnd the distance). The ﬁgure shows that P Q R N θ cos θ = −→ PQ N · \|N\| . −→ PQ distance = \|PR\| = Computing −→ PQ = 2i − 3j − 2k gives distance = −→ PQ N · \|N\| = h2, −3, −2i · h1, √ 2 5 , 0i 4 √ 5 = . 2. Distance: point to line: Ingredients: i) A point P , ii) A line with direction vector v and containing a point Q. −→ v QP × The distance from P to the line is d = \|QP\| sin θ = . \|v\| We will explain this formula by way of the following example. Example 2: Let P = (1, 3, 2), ﬁnd the distance from the point P to the line through (1, 0, 0) and (1, 2, 0). Answer: First we gather our ingredients. P Q R v θ Q = (1, 0, 0) (this is easy to ﬁnd). v = h1, 2, 0i −h1, 0, 0i = 2j is parallel to the line. R = point on line closest to P (this is point is unknown). Using the relation A × B −→ PQ A\|\| v B\| sin θ, the ﬁgure shows that \| \| \| = − − → QP × distance = PR \| sin θ = = \| . \|v\| Computing: −→ PQ = 3j + 2k, −→ QP × v which implies = \|(3j + 2k) × j\| = \| − 2i\| = 2. \|v\| 3. Distance between parallel planes: The trick here is to reduce it to the distance from a point to a plane. Example 3: Find the distance between the planes x + 2y − z = 4 and x + 2y − z = 3. Both planes have normal N = i + 2j − k so they are parallel. Take any point on the ﬁrst plane, say, P = (4, 0, 0). Distance between planes = distance from P to second plane. Choose Q = (1, 0, 0) = point on second plane ⇒ d = \|−→· \|N\| \| = \|3i · (i + 2j − k)\|/ √ 6 = √ 6/2. QP N 4. Distance between skew lines: We place the lines in parallel planes and ﬁnd the distance between the planes as in the previous example As usual it’s easy to ﬁnd a point on each line. Thus, to ﬁnd the parallel planes we only need to ﬁnd the normal. N = v1 × v2, where v1 and v2 are the direction vectors of the lines. MIT OpenCourseWare 18.02SC Multivariable Calculus Fall 2010 For information about citing these materials or our Terms of Use, visit:
4888	https://mathoverflow.net/questions/56509/minimizing-a-sum-of-functions	Skip to main content Minimizing a sum of functions Ask Question Asked Modified 14 years, 6 months ago Viewed 2k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. Hello, I have an expression that I have to minimize. The expression is a sum of piecewise functions each function depending on 3 variables. Let's say Si(ai,bi,ci) is my function, i goes from 1 to N. The sum that I try to minimize is The sum of all Si functions (i from 1 to N). Is the first derivative an option in this case? We're talking about 3N variables (ai,bi,ci where i is from 1 to N). Thanks, Iulian oc.optimization-and-control Share CC BY-SA 2.5 Improve this question Follow this question to receive notifications asked Feb 24, 2011 at 9:53 Iulian SerbanoiuIulian Serbanoiu 39133 silver badges1010 bronze badges 1 It sounds like the variables are independent, and so the minimum of the sum is the sum of the minima, in which case find the minimum for each function i, and then sum the N minima. Gerhard "Sometimes The Obvious Is Simple" Paseman, 2011.02.24 – Gerhard Paseman Commented Feb 24, 2011 at 10:21 Add a comment \| 1 Answer 1 Reset to default This answer is useful 1 Save this answer. Show activity on this post. Unless your functions have a very special structure (convex? independent as in @Gerhard's comment? smooth?) I (a) don't see how you can differentiate [you say your functions are piecewise], and (b) don't see what good it would do you if you could (you might find some local minimum, if the gradient had some very simple form, but would not know it was a global minimum). Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered Feb 25, 2011 at 4:43 Igor RivinIgor Rivin 1 0 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions oc.optimization-and-control See similar questions with these tags. Featured on Meta Upcoming initiatives on Stack Overflow and across the Stack Exchange network... Related 2 Condition number for Ellipsoid method matrix 5 Least sum squares given constraints on subcomponents 3 A lower bound of a particular convex function 7 A Lagrangian problem with a countable family of local extrema ? 2 Functions that are easy to compare to a norm 4 Find distribution that minimises a function of its moments Question feed By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
4889	https://circuitdigest.com/electronic-circuits/positive-and-negative-feedback-in-op-amp-circuits	Positive and Negative Feedback in Op-Amps Circuits and their Practical Applications Skip to main content Continue to site >>> About \| Contact\| Advertise User account menu Log in Search Search News New Products Industry News Community News Articles Product Reviews Tech Articles Tutorials Interviews Case Studies DigiKey Store Cables and Wires LED/Optoelectronics Connectors and Interconnect Discrete Electromechanical Integrated Circuits (ICs) Embedded Computers Enclosures, Hardware and Office Isolators RF and Wireless Power and Circuit Protection Programmers Passives Sensors and Transducers Semiconductors Test Products Tools Projects 555 Timer Circuits Op-amp Circuits Audio Circuits Power Supply Circuits VLSI Projects All Electronic Circuits Arduino Projects Raspberry Pi Projects Raspberry Pi Pico Projects Raspberry Pi Zero Projects STM32 Projects ESP8266 Projects IoT Projects AI Projects PIC Projects ESP32 Projects NodeMCU Projects EV Projects PCB Projects Arduino ESP8266 Projects Home Automation Projects Robotics Projects All Microcontroller Projects Contests IoT & Edge AI Project Challenge 2024 World Energy Challenge 2023 India Automation Challenge 2021 Contact Advertise About Us More DigiKey Videos Partner Events and Conferences EE Calculators Home Electronic Circuits Positive and Negative Feedback in Op-Amps Circuits Positive and Negative Feedback in Op-Amps Circuits PublishedJune 11, 2021 0 VVivekanand Author Positive Feedback Vs Negative Feedback in Op-Amps Circuits Operational Amplifiers or op-amps for short are perhaps the most widely used component amongst allanalog electronics. Because of their versatility, only a few external components are needed to configure them to perform a wide range of tasks like amplification, addition, subtraction, multiplication, integration and so on, hence the name operational amplifier, since it performs mathematical functions. This functionality comes from the fact that they use feedback, which means sampling part of the output and adding or subtracting it from the input to achieve the desired result. There are two types of feedback, positive feedback and negative feedback in op-amp, both of which will be covered in this article in detail. Negative Feedback in Op-Amp Negative feedback takes a part of the output and subtracts it from the input in such a way that the output is in equilibrium with the input. This means that any change in the input is followed by a similar change in the output. The simplest example of negative feedback is the op-amp follower. In this case, the inverting input is connected to the output and the non-inverting input serves as the signal input. Following the rules of op-amp behaviour where the op-amp will try to maintain a 0V difference in voltage across the inverting and non-inverting inputs, we can understand that the output follows the input to maintain this 0V difference, hence the name follower. If the input to this circuit was 1V, then the output would also be 1V, since the output is directly connected to the inverting input, hence making the voltage difference between the inverting and non-inverting pins 0V. If you noticed carefully, you’ll realize that the gain of the circuit described is exactly 1, since the ratio of the input voltage and the output voltage is 1. For the purposes of demonstration, this circuit was constructed using an LM741 op-amp, powered by a ±12V rail with a triangle wave input (from a triangle wave generator made in a previous article). The above figure shows the waveforms of the circuit – the yellow waveform is the input, and the blue waveform is the output. The output is a replica of the input, so we know the follower works. Note the same vertical scale on both channels. What if we want a gain other than 1? This can be done by adding a voltage divider to the output and connecting the inverting input to the middle of the divider. The non-inverting input serves as the signal input as usual. In this case, both resistors are of equal value. If the input signal again is 1V, then the op-amp will try to change the output in such a way as to make the inverting input 1V in order to maintain a 0V differential across its input. To do that, the output must go to 2V, so that the voltage divider output (and hence the inverting input) is at 1V. This circuit has a gain of 2 – it multiplies the input voltage by a factor of 2. It is clear that the output maintains equilibrium with the input – the output responds linearly to changes in the input, so this circuit is used as an amplifier, and this configuration is the classic non-inverting amplifier. The previous follower circuit was modified by adding two resistors, and it is clearly seen that the output of the circuit is twice the input voltage. The scope waveforms shown in the above figure illustrate how the output, which is the blue waveform, is twice the amplitude of the input, which is the yellow waveform. Note how the output is distorted due to the slew rate limitation of the op-amp. The gain of both the circuits described is much less than the open-loop gain of the op-amp itself, so it can be said that negative feedback reduces the overall gain of the system in exchange for stability. Negative feedback op-amp applications: Op-amp Negative feedback finds use mainly in amplifiers, where the input is multiplied by a factor called gain, and the output should be linear and stable with changes in input. Positive Feedback in Op-Amp The non-inverting amplifier circuit can be modified a little bit to create a circuit that has positive feedback. The inverting and the non-inverting inputs of the op-amps are switched so that the inverting input becomes the signal input and the non-inverting input becomes the pin that receives feedback from the output through the voltage divider. Now, when the voltage on the input becomes higher than the voltage at the non-inverting input, the output goes low. Since the op-amp is powered from a ±12V rail, the output sits at -12V and therefore the non-inverting input at -6V. The output now stays latched at -12V till the input goes below -6V, at which point the output goes high to 12V, putting 6V at the non-inverting input. Now the input has to cross 6V to make the output change state again. Unlike the non-inverting amplifier configuration, this output of this circuit does not maintain equilibrium with the input, instead, it saturates to either supply rail in a non-linear fashion. Positive feedback op-amp applications: From this, we can conclude that positive feedback increases the gain of the system drastically, but is not stable and has only two states. Therefore, positive feedback cannot be used to create an amplifier since feedback is highly non-linear. The best way to demonstrate an op-amp with positive feedback is a positive feedback oscillator. If we modified the previous circuit, adding a capacitor between inverting input and ground and a resistor between the inverting input and output, we can make a simple relaxation oscillator. The scope waveform in the above figure shows the output of the oscillator on the yellow channel and the voltage at the non-inverting input on the blue channel. As you can see, the threshold voltage at the non-inverting input changes with every cycle of the oscillator, as described in the text above, between +6V and -6V. We have tried to cover the positive and negative feedback in the article along with how they work, simple circuits to demonstrate them, and practical applications are also described. Analog op amp Have any question related to this Article? 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4890	https://www.omnicalculator.com/math/subset	Subset Calculator This subset calculator can generate all the subsets of a given set, as well as find the total number of subsets. It can also count the number of proper subsets based on the number of elements your set has, or maybe you need to know how many subsets there are with a specific number of elements? No problem! Our subset calculator is here to help you. What is a subset of a set? And what is a proper subset? If you want to learn what these terms mean, read the article below, where we give the subset and proper subset definitions. We also explain the subset vs. proper subset distinction and show how to find subsets and proper subsets of a set. As a bonus, we will then tell you what a power set is, as well as present to you all the required formulas 😊 Subsets play an important role in statistics whenever you need to find the probability of a certain event. You might need it when working with combinations or permutations. What is a subset of a set? Let A and B be two sets. We say that A is a subset of B if every element of A is also an element of B. In other words, A consists of some (possibly all) of the elements of B but doesn't have any elements that B doesn't have. If A is a subset of B, we can also say that B is a superset of A. Examples: The empty set ∅ is a subset of any set; {1,2} is a subset of {1,2,3,4}; ∅, {1} and {1,2} are three different subsets of {1,2}; and Prime numbers and odd numbers are both subsets of the set of integers. Power set definition The set of all possible subsets of a set (including the empty set and the set itself!) is called the power set of a set. We usually denote the power set of any set A by P(A). Note that the power set consists of sets; in particular, the elements of A are NOT the elements of P(A)! Examples: If A = {1,2}, then P(A) = {∅, {1}, {2}, {1,2}}; and P(∅) = {∅}. As you can see in the examples, the power set always has more elements than the original set. How many? Check the section below. And if you'd like to learn even more about this type of set, the power set calculator may satisfy your curiosity! What is a proper subset? A is a proper subset of B if A is a subset of B and A isn't equal to B. In other words, A has some but not all of the elements of B, and A doesn't have any elements that don't belong to B. We can also say that B is a proper superset of A. Examples: {1} and {2} are proper subsets of {1,2}; The empty set ∅ is a proper subset of {1,2}; But {1,2} is NOT a proper subset of {1,2}; and Prime numbers and odd numbers are two distinct proper subsets of the set of all integers. Subset vs. proper subset facts There's no set without a subset. Each set has at least one subset: the empty set ∅; For each set, there is only one subset that is NOT a proper subset: the set itself; There is exactly one set with no proper subsets: the empty set; and Every non-empty set has at least two subsets (itself and the empty set) and at least one proper subset (the empty set). As a consequence, each set has one more subset than it has proper subsets. How many exactly? Check below. Notation issue Some people use the symbol ⊆ to indicate a subset and ⊂ to indicate a proper subset: A ⊆ B we read as A is a subset of B; and C ⊂ B we read as C is a proper subset of B Others, however, use ⊂ for subsets and ⊊ for proper subsets: A ⊂ B we read as A is a subset of B; and C ⊊ B we read as C is a proper subset of B Best stick to the convention introduced by your teacher. If you're unsure and want to be on the safe side, use ⊆ for subsets and ⊊ for proper subsets: the tiny equal/unequal sign at the bottom of the symbol indicates that the subset can/cannot be equal to the set, which leaves no space for any ambiguity. How to use this subset calculator Our subset calculator is here for you whenever you wonder how to find subsets and need to generate the list of subsets of a given set. Alternatively, you can use it to determine the number of subsets based on the number of elements in your set. Here's a quick set of instructions on how to use it: The subset calculator has two modes: set elements mode and set cardinality mode. For set elements mode: enter the elements of your set. Initially, you will see three fields, but more will pop up when you need them. You may enter up to 10 elements. We then count the subsets and proper subsets of your set. You can also display the list of subsets by ticking on the "Display subsets" checkbox after entering at least 1 element. You can only enter numbers as elements. If your set consists of letters, or any other elements, don't worry – replace them with any numbers you want. For readability, we recommend picking smaller numbers rather than larger ones, but, in the end, it's up to your creativity. Just remember to map the distinct elements of your set to distinct numbers! 3. For set cardinality mode: "set cardinality" is the number of elements in a set. Once you tell us how many elements your set has, we count the number of (proper) subsets and: For smaller sets (up to ten elements), the calculator displays the number of subsets with all possible cardinalities; and For larger sets (more than ten elements), you need to enter the cardinality for which you want the subsets counted. Tip: Make sure to check out the union and intersection calculator for further study of set operations. Example of how to find subsets and proper subsets Let us list all subsets of A = {a, b, c, d}. The subset of A containing no elements: ∅ (can also be represented by {}) The subsets of A containing one element: {a}; {b}; {c}; {d} The subsets of A containing two elements: {a, b}; {a, c}; {a, d}; {b, c}; {b, d}; {c, d} The subsets of A containing three elements: {a, b, c}; {a, b, d}; {a, c, d}; {b, c, d} The subset of A containing four elements: {a, b, c, d} There can't be a subset with more than four elements, as A itself has only four elements (a subset of A must not contain any element which is not in A). So, we listed all possible subsets of A: there are 16 of them. Among them, there is one subset of A, which is NOT a proper subset of A: A itself. Therefore, apart from {a, b, c, d}, the subsets listed above are all possible proper subsets of A. There are 15 of them. It's not hard, is it? But our set had just 4 elements. What if we were to find all the subsets of the set {a, b, c, ..., z} containing all twenty-six letters from the English alphabet? In the next section, we explain how to calculate how many subsets there are in a set without writing them all out! Number of subsets and proper subsets of a set Formula to find the number of subsets: If a set contains n elements, then the number of subsets of this set is equal to 2ⁿ. To understand this formula, let's follow this train of thought. Note that to construct a subset for each element of the original set, you have to decide whether this element will be included in the subset or not. Therefore, you have two possibilities for a given element. So, in total, you have 2 × 2 × ... × 2 possibilities, where the number of twos corresponds to the number of elements in the set, so there are n of them. 2. Formula to find the number of proper subsets: If a set contains n elements, then the number of subsets of this set is equal to 2ⁿ − 1. The only subset that is not proper is the set itself. So, to get the number of proper subsets, you just need to subtract one from the total number of subsets. 3. Formula to find the number of subsets with a given cardinality: Recall that "set cardinality" is the number of elements in a set. If a set contains n elements, then its subsets can have between 0 and n elements. The number of subsets with k elements, where 0 ≤ k ≤ n, is given by the binomial coefficient: (nk)=k!(n−k)!n! The symbol on the left-hand side is read "n choose k". The exclamation mark on the right-hand side is a factorial. This number, sometimes denoted by C(n,k) or nCk, is the number of k-combinations of an n-element set. That is, this is the number of ways in which k distinct elements can be chosen from a larger set of n distinguishable objects, where order doesn't matter. To learn more, check our combinations calculator. Example of how to find the number of subsets Example 1. Assume we have a set A with 4 elements. First, let's calculate the number of subsets and the number of proper subsets of A: Number of subsets of A: 2⁴ = 16 Number of proper subsets of A: 2⁴ − 1 = 15 2. Next, we find the number of subsets of A with a given number of elements: Number of subsets of A with 0 elements: 4! / (0! × 4!) = 1 Number of subsets of A with 1 element: 4! / (1! × 3!) = 4 / 1 = 4 Number of subsets of A with 2 elements: 4! / (2! × 2!) = 3 × 4 / 2 = 6 Number of subsets of A with 3 elements: 4! / (3! × 1!) = 4 / 1 = 4 Number of subsets of A with 4 elements: 4! / (4! × 0!) = 1 Take a look at those numbers: 1 4 6 4 1. Maybe you have recognized them as the fourth row of Pascal's triangle. Indeed, for a set of n elements, the n-th row of Pascal's triangle lists how many subsets with 0, 1, ..., n elements the set has! Example 2. Now we can finally get back to the set {a, b, c, ..., z} of all the letters of the English alphabet. As it has 26 elements, we use the Pascal's triangle calculator to generate the 26-th row of the Pascal's triangle: 1 26 325 2600 14950 65780 230230 657800 1562275 3124550 5311735 7726160 9657700 10400600 9657700 7726160 5311735 3124550 1562275 657800 230230 65780 14950 2600 325 26 1 From this we immediately see that {a, b, ..., z} has 1 subset with 0 elements 26 subsets with 1 element 325 subsets with 2 elements 2600 subsets with 3 elements ... 10400600 subsets with 13 elements! ... In total, there are 67108864 subsets! FAQs How do I calculate the number of subsets? Given a set A with cardinality n, there are 2ⁿ subsets, and 2ⁿ − 1 proper subsets. Here's why: Consider a set with one element, {a}. There are two subsets: ∅ and {a}. Now consider a set with two elements, {a, b}. There are four subsets: ∅, {a}, {b}, and {a, b}. This pattern goes on for any n, to deliver 2ⁿ. However, for the number of proper subsets, we subtract one subset (representing the original set, A): 2ⁿ − 1. Is the empty set a subset of every set? Yes, the empty set is a subset of every set. It seems paradoxical but consider the following. If A is a subset of B, then all elements of A must be in B — inversely, no element of A may be outside B. This is true for the empty set: all its elements (of which there are none) are in a given set B, and none of its elements are outside B. Interval Notation Calculator Partial Fraction Decomposition Calculator Set elements (enter up to 10 numbers) Results Expect your resulting subset details here upon entering your set elements. Share result Reload calculatorClear all changes Did we solve your problem today? Check out 41 similar algebra calculators 🔡 Absolute value equation Absolute value inequalities Adding and subtracting polynomials
4891	https://oar.marine.ie/bitstream/10793/760/1/Maths%20Lesson%20Plan%20and%20Worksheets-Sym	Explorer Education Programme Lesson Plan: Line Symmetry Class: Third / Fourth Class Strand: Shape and Space Strand Units: Symmetry www.explorers.ie TITLE: SYMMETRY AND THE SEA Aim / Description: The aim of this lesson plan is to teach students about line symmetry in the marine environment and familiarise them with drawing lines of symmetry in two dimensional shapes. The students will begin by identifying symmetrical pictures and work up to drawing symmetrical images by themselves. The use of drawing and identification will enable students to communicate their understanding of line symmetry in different shapes. For more information on sea animals and their habitats see the following teaching materials found at www.explorers.ie : “Explorer Species Information Book”, “What will I see on the Seashore - Anecdotes about the Seashore”, “Seashore Ecology - Seashore Survey looking at the Zones” and “Illustration of the Seashore Zones”. Materials • Worksheet-Maths 14: Cut and Fold Shapes • Worksheet-Maths 15: Pictures and Symmetry • Worksheet-Maths 16: The Other Half • Worksheet-Maths 17: Sketching Symmetry • Selection of pencils, coloured pencils and markers • Optional: Selection of symmetrical (have a line / lines of symmetry) and asymmetrical (have no line / lines of symmetry) specimens for students to observe Activity: Step 1: • Put children into groups and provide with Worksheet-Maths 14. • Have them cut out the shape and then fold it to identify the lines of symmetry. • Ask the students how many lines of symmetry they can find on each shape. • Extension for Fourth Class: Identify the lines of symmetry as vertical, horizontal or diagonal. • In groups, the students can talk about why the lines of symmetry are correct using visual and oral communication skills. Explorer Education Programme Lesson Plan: Line Symmetry Class: Third / Fourth Class Strand: Shape and Space Strand Units: Symmetry www.explorers.ie Step 2: • Provide the Worksheet-Maths 15 to each student • Ask students to study each image to identify if they have any line / lines of symmetry. The lines of symmetry may be vertical, horizontal or diagonal. Explain how images with lines of symmetry present are called symmetrical. Images with no lines of symmetry are called asymmetrical. • Ask students to circle the correct answer for each image by identifying the images which portray symmetry. • Discuss the results with students. (Sample questions: Was it easy to identify the symmetrical images? Why or why not. Can they think of other examples of symmetry in nature or real life? i.e. Humans, animals, leaf, window. Extension for Fourth Class Step 3: • Provide the Worksheet-Maths 16 to each student. • Ask students to complete the missing half of the pictures provided. • Discuss lines of symmetry in each picture Step 4: • Provide Worksheet-Maths 17 to each student • Allow students to draw a symmetrical sketch from the list provided. • Have students mark the line / lines of symmetry on their sketch. Outcome / Objective: The children in the class should have developed an understanding for concepts of line symmetry in Shape & Space (Geometry) through the use of: • Identification • Exploration • Drawing The children in the class should have developed and applied skills in the following: • Communication and expression • Understanding and recalling • Reasoning and implementing Explorer Education Programme Worksheet-Maths 15: Pictures and Symmetry Class: Third / Fourth Class Strand: Shape and Space www.explorers.ie Aim / Description: Ask students to identify if the pictures shown are symmetrical (have lines of symmetry) or not and circle the correct answer for each one. Yes No Yes No Yes No Yes No Explorer Education Programme Worksheet-Maths 14: Cut and Fold Shapes Class: Third / Fourth Class Strand: Shape and Space www.explorers.ie Aim / Description: The aim of this worksheet is to provide students with different shapes to cut and fold. Once the shapes have been cut out, students can fold them to show the line/ lines of symmetry. Ask students how many lines of symmetry they can find on each shape. Shapes: Rectangle Explorer Education Programme Worksheet-Maths 14: Cut and Fold Shapes Class: Third / Fourth Class Strand: Shape and Space www.explorers.ie Circle Square Explorer Education Programme Worksheet-Maths 14: Cut and Fold Shapes Class: Third / Fourth Class Strand: Shape and Space www.explorers.ie Parallelogram Heart Explorer Education Programme Worksheet-Maths 14: Cut and Fold Shapes Class: Third / Fourth Class Strand: Shape and Space www.explorers.ie Octagon 5 pointed star Explorer Education Programme Worksheet-Maths 14: Cut and Fold Shapes Class: Third / Fourth Class Strand: Shape and Space www.explorers.ie Moon Hexagon Explorer Education Programme Worksheet-Maths 14: Cut and Fold Shapes Class: Third / Fourth Class Strand: Shape and Space www.explorers.ie Left and right arrow Diamond Explorer Education Programme Worksheet-Maths 16: The Other Half Class: Fourth Class Strand: Shape and Space www.explorers.ie Aim / Description: Ask students to draw the other half of each picture in a symmetrical way. Explorer Education Programme Worksheet-Maths 17: Sketching Symmetry Class: Fourth Class Strand: Shape and Space www.explorers.ie Aim / Description: Make a symmetrical drawing of one of the following: • Jellyfish • Crab • Turtle • Fish • Person
4892	https://www.geeksforgeeks.org/python/initialize-matrix-in-python/	Initialize Matrix in Python - GeeksforGeeks Skip to content Tutorials Python Java DSA ML & Data Science Interview Corner Programming Languages Web Development CS Subjects DevOps Software and Tools School Learning Practice Coding Problems Courses DSA / Placements ML & Data Science Development Cloud / DevOps Programming Languages All Courses Tracks Languages Python C C++ Java Advanced Java SQL JavaScript Interview Preparation GfG 160 GfG 360 System Design Core Subjects Interview Questions Interview Puzzles Aptitude and Reasoning Data Science Python Data Analytics Complete Data Science Dev Skills Full-Stack Web Dev DevOps Software Testing CyberSecurity Tools Computer Fundamentals AI Tools MS Excel & Google Sheets MS Word & Google Docs Maths Maths For Computer Science Engineering Mathematics Switch to Dark Mode Sign In Free Python 3 Tutorial Data Types Control Flow Functions List String Set Tuple Dictionary Oops Exception Handling Python Programs Python Projects Python Interview Questions Python MCQ NumPy Pandas Python Database Data Science With Python Machine Learning with Python Django Flask R Sign In ▲ Open In App Initialize Matrix in Python Last Updated : 11 Jul, 2025 Comments Improve Suggest changes 8 Likes Like Report Initializing a matrix inPythonrefers to creating a 2D structure (list of lists) to store data in rows and columns. For example, a 3x2 matrix can be represented as three rows, each containing two elements. Let’s explore different methods to do this efficientely. Using list comprehension List comprehension creates the matrix by building each element separately in nested loops, ensuring each row is independent. Very flexible for initializing with any values or expressions. Slightly longer but avoids shared references. Python ```python3 rows = 3 cols = 2 res = [[0 for _ in range(cols)] for _ in range(rows)] print(res) ``` rows = 3 cols = 2res = [[0 for _ in range(cols)] for _ in range(rows)]print(res) Output Explanation: (for _ in range(rows)) runs three times to create three rows. For each row, the inner loop ([0 for _ in range(cols)]) runs twice to generate a list with two zeros. Using Multiplication inside list comprehension Builds each row by repeating zeros and replicates rows with a list comprehension, producing independent rows. It’s shorter and faster for simple zero matrices. Best for uniform initialization. Python ```python3 rows = 3 cols = 2 res = [cols for _ in range(rows)] print(res) ``` rows = 3 cols = 2res = [cols for _ in range(rows)]print(res) Output Explanation: (for _ in range(rows))runs three times to create three rows. For each row, cols creates a new list with two zeros by multiplying the value 0 by the number of columns. Using numpy zeros Generates a fast, memory-efficient 2D array of zeros using NumPy, ideal for numeric data and large matrices. Returns a NumPy array instead of a list. Great for scientific or vectorized operations. Python ```python3 import numpy as np rows, cols = 3, 2 res = np.zeros((rows, cols), dtype=int) print(res) ``` import numpy as np rows, cols = 3, 2 res = np.zeros((rows, cols), dtype=int)print(res) Output[[0 0] [0 0] [0 0]] Explanation: np.zeros((rows, cols), dtype=int) function generates a matrix where each element is initialized to 0 and dtype=intensures the values are integers. Using with deep copy Creates one zero-filled row and deep copies it multiple times to prevent shared references between rows. Useful when matrix elements are mutable objects. Slightly slower due to deep copying overhead. Python ```python3 import copy rows, cols = 3, 2 row = cols res = [copy.deepcopy(row) for _ in range(rows)] print(res) ``` import copy rows, cols = 3, 2row = cols res = [copy.deepcopy(row) for _ in range(rows)]print(res) Output Explanation: This code first creates a single row with two zeros using cols.It then usescopy.deepcopy(row) in a list comprehension to create three separate copies of that row. 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4893	https://www.doe.mass.edu/frameworks/search/Map.aspx?ST_CODE=Mathematics.8.EE.A.01	Standards Navigator - Curriculum Map Standards Map Mathematics > Grade 8 > Expressions and Equations Accessibility Mode:- [x] Note: You are viewing this information in accessibility mode. To view the map, enlarge your window or use a larger device. Note: Click any standard to move it to the center of the map. Mathematics \| Grade : 8 Domain - Expressions and Equations Cluster - Work with radicals and integer exponents. [8.EE.A.1] - Know and apply the properties of integer exponents to generate equivalent numerical expressions.For example, 3² x 3-5 = 3-3 = 1/3 3 = 1/27. Resources Exponent The number that indicates how many times the base is used as a factor, e.g., in 4 3 = 4 x 4 x 4 = 64, the exponent is 3, indicating that 4 is repeated as a factor three times. Expression A mathematical phrase that combines operations, numbers, and/or variables (e.g., 3 2 ÷ a). Integer All positive and negative whole numbers, including zero. MCAS Items: 2025 Spring Release - Mathematics - Grade 8 - Item 3 2023 Spring Release - Mathematics - Grade 8 - Item 8 2022 Spring Release - Mathematics - Grade 8 - Item 10 2019 Spring Release - Mathematics - Grade 8 - Item 3 2019 Spring Release - Mathematics - Grade 8 - Item 8 2018 Spring Release - Mathematics - Grade 8 - Item 20 [8.EE.A.3] - Use numbers expressed in the form of a single digit times an integer power of 10 to estimate very large or very small quantities, and to express how many times as much one is than the other.For example, estimate the population of the United States as 3 x 10 8 and the population of the world as 7 x 10 9, and determine that the world population is more than 20 times larger. [8.EE.A.4] - Perform operations with numbers expressed in scientific notation, including problems where both decimal and scientific notation are used. Use scientific notation and choose units of appropriate size for measurements of very large or very small quantities (e.g., use millimeters per year for seafloor spreading). Interpret scientific notation that has been generated by technology. [AI.N-RN.A.1] - Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents. For example, we define 5 1/3 to be the cube root of 5 because we want (5 1/3)3 = 5(1/3)3 to hold, so (5 1/3)3 must equal 5. [AI.N-RN.A.2] - Rewrite expressions involving radicals and rational exponents using the properties of exponents. [AI.A-SSE.A.2] - Use the structure of an expression to identify ways to rewrite it. For example, see (x + 2)2 – 9 as a difference of squares that can be factored as ((x + 2) + 3)((x + 2 ) – 3). [AI.A-APR.A.1] - Understand that polynomials form a system analogous to the integers, namely, they are closed under certain operations. [AI.A-APR.A.1.a] - Perform operations on polynomial expressions (addition, subtraction, multiplication), and compare the system of polynomials to the system of integers when performing operations. [AI.A-APR.A.1.b] - Factor and/or expand polynomial expressions, identify and combine like terms, and apply the Distributive property. [MII.N-RN.A.1] - Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents. For example, we define 5 1/3 to be the cube root of 5 because we want (5 1/3)3 = 5(1/3)3 to hold, so (5 1/3)3 must equal 5. [MII.N-RN.A.2] - Rewrite expressions involving radicals and rational exponents using the properties of exponents. [MII.A-SSE.A.2] - Use the structure of an expression to identify ways to rewrite it. For example, see (x + 2)2 – 9 as a difference of squares that can be factored as ((x + 2) + 3)((x + 2) – 3). [MII.A-APR.A.1.a] - Perform operations on polynomial expressions (addition, subtraction, multiplication), and compare the system of polynomials to the system of integers when performing operations. [MII.A-APR.A.1.b] - Factor and/or expand polynomial expressions; identify and combine like terms; and apply the Distributive property. [PC.F-BF.B.5] - (+) Understand the inverse relationship between exponents and logarithms and use this relationship to solve problems involving logarithms and exponents. Mathematics \| Grade : 8 Domain - Expressions and Equations Cluster - Work with radicals and integer exponents. [8.EE.A.1] - Know and apply the properties of integer exponents to generate equivalent numerical expressions.For example, 3² x 3-5 = 3-3 = 1/3 3 = 1/27. Resources: Exponent The number that indicates how many times the base is used as a factor, e.g., in 4 3 = 4 x 4 x 4 = 64, the exponent is 3, indicating that 4 is repeated as a factor three times. Expression A mathematical phrase that combines operations, numbers, and/or variables (e.g., 3 2 ÷ a). Integer All positive and negative whole numbers, including zero. MCAS Items: 2025 Spring Release - Mathematics - Grade 8 - Item 3 2023 Spring Release - Mathematics - Grade 8 - Item 8 2022 Spring Release - Mathematics - Grade 8 - Item 10 2019 Spring Release - Mathematics - Grade 8 - Item 3 2019 Spring Release - Mathematics - Grade 8 - Item 8 2018 Spring Release - Mathematics - Grade 8 - Item 20 Predecessor Standards: No Predecessor Standards found. Successor Standards: AI.N-RN.A.1 Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents. For example, we define 5 1/3 to be the cube root of 5 because we want (5 1/3)3 = 5(1/3)3 to hold, so (5 1/3)3 must equal 5. AI.N-RN.A.2 Rewrite expressions involving radicals and rational exponents using the properties of exponents. AI.A-SSE.A.2 Use the structure of an expression to identify ways to rewrite it. For example, see (x + 2)2 – 9 as a difference of squares that can be factored as ((x + 2) + 3)((x + 2 ) – 3). AI.A-APR.A.1 Understand that polynomials form a system analogous to the integers, namely, they are closed under certain operations. AI.A-APR.A.1.a Perform operations on polynomial expressions (addition, subtraction, multiplication), and compare the system of polynomials to the system of integers when performing operations. AI.A-APR.A.1.b Factor and/or expand polynomial expressions, identify and combine like terms, and apply the Distributive property. MII.N-RN.A.1 Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents. For example, we define 5 1/3 to be the cube root of 5 because we want (5 1/3)3 = 5(1/3)3 to hold, so (5 1/3)3 must equal 5. MII.N-RN.A.2 Rewrite expressions involving radicals and rational exponents using the properties of exponents. MII.A-SSE.A.2 Use the structure of an expression to identify ways to rewrite it. For example, see (x + 2)2 – 9 as a difference of squares that can be factored as ((x + 2) + 3)((x + 2) – 3). MII.A-APR.A.1.a Perform operations on polynomial expressions (addition, subtraction, multiplication), and compare the system of polynomials to the system of integers when performing operations. MII.A-APR.A.1.b Factor and/or expand polynomial expressions; identify and combine like terms; and apply the Distributive property. PC.F-BF.B.5 (+) Understand the inverse relationship between exponents and logarithms and use this relationship to solve problems involving logarithms and exponents. Same Level Standards: 8.EE.A.3 Use numbers expressed in the form of a single digit times an integer power of 10 to estimate very large or very small quantities, and to express how many times as much one is than the other.For example, estimate the population of the United States as 3 x 10 8 and the population of the world as 7 x 10 9, and determine that the world population is more than 20 times larger. 8.EE.A.4 Perform operations with numbers expressed in scientific notation, including problems where both decimal and scientific notation are used. Use scientific notation and choose units of appropriate size for measurements of very large or very small quantities (e.g., use millimeters per year for seafloor spreading). Interpret scientific notation that has been generated by technology.
4894	https://www.math.unl.edu/~sdunbar1/ExperimentationCR/Lessons/Fibonacci/FibonacciII/fibonacci2.shtml	[an error occurred while processing this directive] Math 896 Section 700 Experimentation, Conjecture and Reasoning Spring 2005 Question of the Day What mathematical identities do you know? Do you know any identities or relationships that are satisfied by special sets of integers, such as the sequence of odd numbers and the sets of square numbers? What is the difference between an identity and a formula? Key Concepts The Fibonacci sequence satisfies a number of identities among it members. The Fibonacci numbers have some remarkable divisibility properties. Vocabulary An identity is an equation that is satisfied by any number that replaces the letter for which the equation is defined, that is, the equation is true for all values of the variable. Mathematical Ideas Identities about Fibonacci numbers Cassini’s Identity This section is adapted from: Concrete Mathematics Second Edition, R. L Graham, D. E. Knuth, O. Patashnik, Addison-Wesley, 1994, pages 290-301. One of the oldest theorems about Fibonacci numbers is due to the French astronomer Jean-Dominique Cassini in 1680. Cassini also discovered the dark gap between the rings A and B of Saturn, now known as the Cassini division. Cassini deduced that Try it with a few of the Fibonacci numbers. For example, . In some ways this is inconvenient, because it relates Fibonacci numbers which are two indices apart. However, because , we can write . Then substituting into Cassini’s identity, we get Try this also with a few of the Fibonacci numbers to test it out. For example, Divisibility of Fibonacci numbers In this section, we will do a mathematical experiment in numerical pattern finding and investigation. We will make a conjecture about the pattern and try to prove it with the identities we developed above. Consider the following table which I constructed so that you could study some Fibonacci relationships easily without having to do large amounts of computation: \| n \| F_n \| “Divisors of Fn” \| \| 1 \| 1 \| \| \| 2 \| 1 \| \| \| 3 \| 2 \| \| \| 4 \| 3 \| \| \| 5 \| 5 \| \| \| 6 \| 8 \| \| \| 7 \| 13 \| \| \| 8 \| 21 \| \| \| 9 \| 34 \| \| \| 10 \| 55 \| \| \| 11 \| 89 \| \| \| 12 \| 144 \| \| \| 13 \| 233 \| \| \| 14 \| 377 \| \| \| 15 \| 610 \| \| \| 16 \| 987 \| \| \| 17 \| 1597 \| \| \| 18 \| 2584 \| \| \| 19 \| 4181 \| \| \| 20 \| 6765 \| \| \| \| The first column is just index numbers. The second column is the Fibonacci number corresponding to that index. The third column is the set of positive integers that divide evenly into the Fibonacci number. Start with a simple observation that you may notice after examining the table for a while. Among the divisors of is . Among the divisors of is . Among the divisors of is . Among the divisors of is . Are we ready to make a conjecture? We might conjecture that evenly divides . Can we test our conjecture? For instance, certainly divides . And divides . Can you test a few more? This can be proved in the following way. First observe by definition, and then by a quick substitution: Now use the definition of Fibonacci numbers again, and the previous information, and continue. Notice that from the 3rd line on, each line is the sum of the two previous lines, and so the coefficients are the Fibonacci numbers! Therefore we have proved: (Check this out by substituting a few numbers for to make sure it matches what you see above.) Now substitute to get So does indeed divide , indeed with quotient ! This proof can be generalized to show that is a multiple of , and even that . Binet’s Formula for the Fibonacci numbers Let be the symbol for the Golden Ratio. Then recall that also appears in so many formulas along with the Golden Ratio that we give it a special symbol . And finally, we need one more symbol . Then using either solutions of difference equations or generating functions, (both are ideas form more advanced mathematics) we can prove that Spell that out a little more explicitly: This is an amazing formula, since it does not look like it is even a rational number, let alone an integer. Yet it is! Try out some small values of to test it, for example, , and . With the idea that as (more precisely , we can make another interesting formula for the Fibonacci numbers: Fibonacci number bases and a cute conversion You are familiar with number bases, for example base 2 number systems or the base 10 numbers using place notation with powers of 2 or 10 respectively. One can do the same thing with Fibonacci numbers: That is, every natural number is either a Fibonacci number or it is expressible uniquely as the sum of Fibonacci numbers. Here is how to express a number in the Fibonacci number system: Write down a natural number, say 38. Find the largest Fibonacci number that does not exceed the given number, in this case, is the largest Fibonacci number that does not exceed 38. Subtract the Fibonacci number from the given number and look at the new number, in this case, 4 Now find the largest number that does not exceed this new number, for the example, is the largest Fibonacci number not exceeding 4. Continue this process, in the example we are down to 1, and so stop Now we can put this number base idea together with the single--term Binet formula above to make a cute number trick. By pure coincidence is also very nearly the number of kilometers in a mile. (The exact number of kilometers in a mile is ) This will give a handy way to convert mentally from kilometers to miles or vice-versa. Suppose we want to convert, say 30 kilometers to its equivalent in miles. Just use the Fibonacci representation . That is: Then the number of miles in 30 kilometers would be approximately The actual number of miles in 30 kilometers is , so this is close enough for mental work. SO the idea is represent the number of kilometers in the Fibonacci number system and then “shift down a notch” and add up again. This is because by the Binet formula, shifting down divides by , more or less. Similarly, to go from miles to kilometers, “shift up a notch”. For example, 30 miles , so in kilometers, this is . The actual conversion is kilometers, pretty close. Sample Worked Problems Problem 13, Page 59, Even More Fibonacci Relationships Compute the first few terms of the new combined sequence to be (for n = 3, 4,..., 10 respectively) Now that can be compared to the Lucas sequence: L1 = 2, L2 = 1, L3 = 3, L4 = 4, L5 = 7, L6 = 11, L7 = 18, L8 = 29, L9 = 47, L10 = 76. The coincidence is too similar to be ignored! It certainly appears that Fn+2 -Fn-2 = Ln+1. Let’s try it with an integer larger than we have tried yet: say n = 14. Then we would check to see if F16 - F12 = L15. This translates into 987 - 144 = 843 which is true. It certainly appears as if we have a relationship figured out. We could prove this relationship in several possible ways, including mathematical induction, and Binet’s formula as well as some other ways that are simpler but use concepts introduced in advanced courses (namely the linearity of the recursion formula and superposition of solutions.) The problem does not call for it, but for another example, here is a proof using mathematical induction: First consider the initial case with n = 3, or the assertion that F5 + F1 = L4. This we demonstrate directly by substituting the values: 5 - 1 = 4. Now assume that we know that Fn+2 - Fn-2 = Ln+1 is true for all values of 3 < n < N and we wish to show that FN+2 - Fn-2 = Ln+1 is true. (Incidentally, this is called the Strong Induction Method, which is an equivalent formulation of the more familiar simple induction method.) We apply the definition of the Fibonacci numbers to the equation whose truth we are trying to establish. We find that if we can establish the truth of then we will be done. But this last equation can be rearranged into Each parenthesized term on the left is equivalent to the corresponding term on the right by the Strong Induction hypothesis. We are done. Problems 34 and 35, A big fib and Decomposing naturals The problem statement is “Suppose we have a natural number that is not a Fibonacci number, let’s call it N. Suppose that F is the largest Fibonacci number that does not exceed N. Show that the number N - F must be smaller than the Fibonacci number that comes right before F. In order to show this, it helps to be slight more definite with the notation. Say that the largest Fibonacci number smaller than N is Fn. Then the Fibonacci number that comes right before Fn is Fn-1. Also, the Fibonacci number that comes right after is Fn+1. But since N is not a Fibonacci number, and Fn is the largest Fibonacci number less than N, we know that N < Fn+1. Since Fn+1 = Fn + Fn-1, then N < Fn - Fn-1. Rearranging, N - Fn < Fn-1. Now use mathematical induction in the strong form to show that every natural number can be written as a sum of distinct non-consecutive Fibonacci numbers. First, 1 can be written as the trivial sum of the first Fibonacci number by itself: 1 = F1. Let N be given. Now assume that every number less than N can be written as the sum of distinct non-consecutive Fibonacci numbers. Then let F be the largest Fibonacci number less than N, so N = F + (N - F). But we just showed that N - F is less than the immediately previous Fibonacci number. By the strong induction hypothesis, N - F can be written as the sum of distinct non-consecutive Fibonacci numbers. The proof is done. (This is what the book problem alludes to as systematically reducing the problem to a smaller problem.) Problem 39, Generalized Sums Consider first the generalized Fibonacci relation Fn+1 = Fn + Fn-1, with starting terms F0 = 1 and F1 = 1. Then as in Problem 38, the first 15 terms in the sequence are 1, 1, 3, 7, 17, 41, 99, 239, 577, 1393, 3363, 8119, 19601, 47321, 114243. We observe that these grow quite rapidly,w with large gaps between the terms. Now we ask if every natural number can be expressed as the sum of distinct non-consecutive generalized Fibonacci numbers? The answer is no, and we can illustrate why not with several counter-examples: 11 can be written as the sum of several generalized Fibonacci numbers, but they are not non-consecutive, 11 = 7 + 3 + 1. Likewise for 12: 12 = 7 + 3 + 1 + 1. But 13 cannot be expressed as the sum of distinct generalized Fibonacci numbers, we would have to repeat one 13 = 7 + 3 + 3, or another 13 = 7 + 3 + 1 + 1 + 1. Similarly for 14, 15 and 16. It gets worse, since the largest number we can express even with all 5 of the first 5 terms of the generalized Fibonacci sequence is 1 + 1 + 3 + 7 + 17 = 22. So there is no way to reproduce the natural numbers 23 < n < 40 as a sum of distinct generalized Fibonacci numbers. We would have to use multiple copies of some of the generalized Fibonacci numbers. Problems to Work for Understanding Solidifying Ideas: pages 59: 9,10,11,12,14 New Ideas: page 61: 29, 30 Habits of Mind: page 62: number 40 Reading Suggestion: Concrete Mathematics Second Edition, R. L Graham, D. E. Knuth, O. Patahsnik, Addison-Wesley, 1994, pages 290-301. [an error occurred while processing this directive] [an error occurred while processing this directive] Last modified: [an error occurred while processing this directive]
4895	https://www.sciencedirect.com/science/article/pii/S1059131118305405	Skip to article My account Sign in View PDF Seizure Volume 65, February 2019, Pages 111-117 Comparing the efficacy of sodium valproate and levetiracetam following initial lorazepam in elderly patients with generalized convulsive status epilepticus (GCSE): A prospective randomized controlled pilot study Author links open overlay panel, , , , , , , , , , , rights and content Under an Elsevier user license Open archive Highlights ¢ First randomized controlled study in the elderly patients with GCSE. ¢ SVP & LEV following initial lorazepam could control SE in 68&74% respectively. ¢ SVP & LEV following initial lorazepam had statistically comparable efficacy. ¢ Overall seizure control of 77.1% was achieved by adding second AED. ¢ Lorazepam & LEV/SVP combination was safe & higher STESS had poor treatment response. Abstract Purpose: This randomized control study was conducted to compare the efficacy of sodium valproate (SVP) and levetiracetam (LEV) following initial intravenous lorazepam in elderly patients (age: >60years) with generalized convulsive status epilepticus (GCSE) and to identify predictors of poor seizure control. Methods: A total of 118 patients (mean age: 67.5±7.5 years, M:F=1.6:1), who had presented with GCSE were randomized into the SVP or LEV treatment arms. All patients received initial intravenous lorazepam (0.1mg/kg) followed by one of the two antiepileptic drugs (AEDs), parenteral SVP (2025mg/kg) or LEV (2025mg/kg). Those who failed to achieve control with the initial AED, were crossed over to receive the other AED. One-hundred patients (SVP=50; LEV=50) completed the study. Results: SE could be controlled with lorazepam and one of the AEDs (SVP or LEV) in 71.18% (84/118). Intention-to-treat analysis showed that the two groups did not differ significantly in terms of seizure control [SVP: 41/60 (68.3%); LEV: 43/58 (74.1%), p=0.486]. Of 100 patients who completed the study, seizure control was achieved in 38/50(76%) in the SVP and 43/50(86%) in the LEV group (p=0.202). After crossing over to the second AED, SE could be controlled in an additional in 50% (6/12) in SVP (+LEV) group and in 14.3% (1/7) in LEV (+SVP) group. Overall, after the second AED, seizure control was achieved in 77.1% (91/118). Higher STESS was associated with poor therapeutic response (p=0.049). Conclusions: The efficacy of SVP and LEV following initial lorazepam in controlling GCSE in elderly population was comparable, hence the choice of AED could be individualized. Keywords Elderly seizures Generalised convulsive status epilepticus (GCSE) Levetiracetam (LEV) Sodium valproate (SVP) Status epilepticus (SE) Treatment of SE Cited by (0) © 2019 British Epilepsy Association. Published by Elsevier Ltd.
4896	https://www.webelements.com/periodicity/group_number/	WebElements Periodic Table » Periodicity » Group numbers » Periodic table gallery Full property list Abundances Atomic numbers Atomic radii Atomic spectra Atomic weights Block in periodic table Boiling point Bond enthalpies Bond length in element Bulk modulus CAS Registry ID Classification Colour Covalent radii Critical temperature Crystal structure Density of solid Description Discovery Effective nucl. charges Electron bind. energies Electrical resistivity Electron affinity Electron shell structure Electronegativity Electronic configuration Enthalpy: atomization Enthalpy of fusion Expansion coefficient Enthalpy: vaporization Geology Group Name Group numbers Hardness Health hazards History of the element Hydration enthalpies Hydrolysis constants Ionization energies Isolation Isotopes Lattice energies Liquid Range Mass in human Meaning of name Melting points Molar volume Names and symbols NMR Orbital_properties Oxidation numbers Period numbers Poisson's ratio Ionic radii Radii - ionic, Pauling Radii - metallic (12) Radii - valence orbital Radii - Van der Waals Reactions of elements Reduction potentials Reduction potentials Reflectivity Refractive index Rigidity modulus Standard state Superconductivity Term symbol Thermal conductivity Thermodynamics Uses Velocity of sound X-ray mass abs. coeff. Young's modulus Group numbers Group numbers Group numbers Gallery of images Periodic table gallery Heatscapes Group 1 Group 2 Group 3 Group 4 Group 5 Group 6 Group 7 Group 8 Group 9 Group 10 Group 11 Group 12 Group 13 Group 14 Group 15 Group 16 Group 17 Group 18 Period 2 Period 3 Period 4sp Period 4 Period 4d Period 5sp Period 5spd Period 5d Period 6sp Period 6spd Period 6 Period 6d Period 7sp Period 7spd Period 7 Period 7d s- and p-Block d-Block The group number is an identifier used to describe the column of the standard periodic table in which the element appears. Groups 1-2 termed s-block elements. Groups 1-2 (except hydrogen) and 13-18 are termed main group elements. Groups 3-12 are termed d-block elements. Groups 3-11 are termed transition elements. Transition elements are those whose atoms have an incomplete d-subshell or whose cations have an incomplete d-subshell. Groups 13-18 termed p-block elements. Main group elements in the first two rows of the table are called typical elements. The first row of the f-block elements are called lanthanoids (or, less desirably, lanthanides. The second row of the f-block elements are called actanoids (or, less desirably, actanides. The following names for specific groups in the periodic table are in common use: Group 1: alkali metals Group 2: alkaline earth metals Group 11: coinage metals (not an IUPAC approved name) Group 15: pnictogens (not an IUPAC approved name) Group 16: chalcogens Group 17: halogens Group 18: noble gases In addition, groups may be idenitifed by the first element in each group - so the Group 16 set of elements is sometimes called the oxygen group. Units None Notes There is considerable confusion surrounding the Group labels. The scheme used in WebElements is numeric and is the current IUPAC convention. The other two systems are less desirable since they are confusing, but still in common usage. The designations A and B are completely arbitrary. The first of these (A left, B right) is based upon older IUPAC recommendations and frequently used in Europe. The last set (main group elements A, transition elements B) was in common use in America. IUPAC, European, and American Group labelling schemes\| Group \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 \| 10 \| 11 \| 12 \| 13 \| 14 \| 15 \| 16 \| 17 \| 18 \| --- --- --- --- --- --- --- --- --- \| European \| IA \| IIA \| IIIA \| IVA \| VA \| VIA \| VIIA \| VIIIA \| VIIIA \| VIIIA \| IB \| IIB \| IIIB \| IVB \| VB \| VIB \| VIIB \| VIIIB \| \| American \| IA \| IIA \| IIIB \| IVAB \| VB \| VIB \| VIIB \| VIIIB \| VIIIB \| VIIIB \| IB \| IIB \| IIIA \| IVA \| VA \| VIA \| VIIA \| VIIIA \| Literature sources International Union of Pure & Applied Chemistry: Commission on the Nomeclature of Inorganic Chemistry, Nomenclature of Inorganic Chemistry Recommendations 2005, Eds. N.G. Connelly, T. Damhus, R.M. Hartshorn, and A.T. Hutton, Royal Sociery of Chemistry, 2005. ISBN-13: 978-0854044382. W.C. Fernelius, and W.H. Powell, "Confusion in the periodic table of the elements", J. Chem. Ed., 1982, 59, 504-508. Periodic table in any volume of the Inorganic Oxford Chemistry Primers, Oxford University Press, Oxford, UK. Explore the element of your choice through this periodic table.\| 1 \| 2 \| \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 \| 10 \| 11 \| 12 \| 13 \| 14 \| 15 \| 16 \| 17 \| 18 \| --- --- --- --- --- --- --- --- --- \| 1 H \| \| \| 2 He \| \| 3 Li \| 4 Be \| \| 5 B \| 6 C \| 7 N \| 8 O \| 9 F \| 10 Ne \| \| 11 Na \| 12 Mg \| \| 13 Al \| 14 Si \| 15 P \| 16 S \| 17 Cl \| 18 Ar \| \| 19 K \| 20 Ca \| \| 21 Sc \| 22 Ti \| 23 V \| 24 Cr \| 25 Mn \| 26 Fe \| 27 Co \| 28 Ni \| 29 Cu \| 30 Zn \| 31 Ga \| 32 Ge \| 33 As \| 34 Se \| 35 Br \| 36 Kr \| \| 37 Rb \| 38 Sr \| \| 39 Y \| 40 Zr \| 41 Nb \| 42 Mo \| 43 Tc \| 44 Ru \| 45 Rh \| 46 Pd \| 47 Ag \| 48 Cd \| 49 In \| 50 Sn \| 51 Sb \| 52 Te \| 53 I \| 54 Xe \| \| 55 Cs \| 56 Ba \| \| 71 Lu \| 72 Hf \| 73 Ta \| 74 W \| 75 Re \| 76 Os \| 77 Ir \| 78 Pt \| 79 Au \| 80 Hg \| 81 Tl \| 82 Pb \| 83 Bi \| 84 Po \| 85 At \| 86 Rn \| \| 87 Fr \| 88 Ra \| \| 103 Lr \| 104 Rf \| 105 Db \| 106 Sg \| 107 Bh \| 108 Hs \| 109 Mt \| 110 Ds \| 111 Rg \| 112 Cn \| 113 Nh \| 114 Fl \| 115 Mc \| 116 Lv \| 117 Ts \| 118 Og \| \| \| \| Lanthanoids \| \| 57 La \| 58 Ce \| 59 Pr \| 60 Nd \| 61 Pm \| 62 Sm \| 63 Eu \| 64 Gd \| 65 Tb \| 66 Dy \| 67 Ho \| 68 Er \| 69 Tm \| 70 Yb \| \| \| Actinoids \| \| 89 Ac \| 90 Th \| 91 Pa \| 92 U \| 93 Np \| 94 Pu \| 95 Am \| 96 Cm \| 97 Bk \| 98 Cf \| 99 Es \| 100 Fm \| 101 Md \| 102 No \| \| Key properties Electronegativity Melting point Density Molar volume Ionization energy (1st) Covalent radius Metallic radius Abundance: earth's crust Shop Sorry, shop closed Social Facebook Twitter Pinterest Administration Privacy About Copyright Thanks... WebElements: THE periodic table on the WWW [www.webelements.com] Copyright 1993-2025 Mark Winter [ The University of Sheffield and WebElements Ltd, UK]. All rights reserved. You can reference the WebElements periodic table as follows: "WebElements, accessed September 2025."
4897	https://math.stackexchange.com/questions/522077/complete-and-elementary-proof-that-ax-1-x-converges-as-x-goes-to-0	calculus - Complete and elementary proof that $(a^x - 1)/x $ converges as x goes to 0 - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Complete and elementary proof that (a x−1)/x(a x−1)/x converges as x goes to 0 Ask Question Asked 11 years, 11 months ago Modified11 years, 11 months ago Viewed 2k times This question shows research effort; it is useful and clear 7 Save this question. Show activity on this post. Anybody who has taken a calculus course knows that lim x→0 a x−1 x lim x→0 a x−1 x exists for any positive real number a a, simply because the limit is by definition the derivative of the function a x a x at x=0 x=0. However, for this argument to be non-circular one must have an independent technique for proving that a x a x is differentiable. The standard approach involves the following two steps: 1) Calculate the derivative of log a x log a⁡x by reducing it to the calculation of lim h→0(1+h)1 h lim h→0(1+h)1 h 2) Apply the inverse function theorem. I find this unsatisfying for two reasons. First, the inverse function theorem is not entirely trivial, even in one variable. Second, the limit in step 1 is quite difficult; in books it is often calculated along the sequence h=1 n h=1 n where n n runs over the positive integers, but doing the full calculation seems to be quite a bit more difficult (if one hopes to avoid circular reasoning). So I would like a different argument which uses only the elementary theory of limits and whatever algebra is needed. For instance, I would like to avoid logarithms if their use involves an appeal to the inverse function theorem. Is this possible? calculus limits Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Oct 11, 2013 at 1:22 Alex 597 2 2 gold badges 6 6 silver badges 14 14 bronze badges asked Oct 11, 2013 at 1:09 Paul SiegelPaul Siegel 9,537 36 36 silver badges 63 63 bronze badges 12 6 When you really get down to it, as I think you are trying to do here, what is your definition of a x a x? If you want to dig that deep, you will need to recognize that a x a x is generally a pretty tough thing to define without calculus when x x is a general real number.2'5 9'2 –2'5 9'2 2013-10-11 01:28:45 +00:00 Commented Oct 11, 2013 at 1:28 1 You can define it as a x=inf{a r:r>x,r∈Q}a x=inf{a r:r>x,r∈Q}.abnry –abnry 2013-10-11 01:31:24 +00:00 Commented Oct 11, 2013 at 1:31 Yes, that's how I want to define it: give an algebraic definition on Q Q, prove that the function so defined is continuous on Q Q with the standard metric, and define a x a x to be the unique continuous extension to R R. None of this involves anything all that much more sophisticated than the definition of a limit.Paul Siegel –Paul Siegel 2013-10-11 01:40:34 +00:00 Commented Oct 11, 2013 at 1:40 Of course, I don't know an answer even with most other definitions of a x a x. One way is to first define logarithms using integrals, but this ends up using the fundamental theorem of calculus and the inverse function theorem. Another way is to define e x e x as the unique solution to the ODE y′=y y′=y, but this uses the existence/uniqueness theorem for ODE's which is just as hard as the inverse function theorem. The only definition I can think of which makes this limit transparent is via power series.Paul Siegel –Paul Siegel 2013-10-11 01:46:17 +00:00 Commented Oct 11, 2013 at 1:46 OK, but what do you mean by "the unique continuous extension to R R"? Do you want to get down to the business of establishing that there is a unique continuous extension? Separate comment: using power series would also be circular. Either you are relying on knowing the derivative of a x a x to get the power series, or you are defining the a x a x by its power series but then what is the meaning of the ln(a)ln⁡(a)'s in the coefficients?2'5 9'2 –2'5 9'2 2013-10-11 02:28:51 +00:00 Commented Oct 11, 2013 at 2:28 \|Show 7 more comments 5 Answers 5 Sorted by: Reset to default This answer is useful 5 Save this answer. Show activity on this post. Prove that lim n(1+1 n)n=lim n(1+1 n)n+1 lim n(1+1 n)n=lim n(1+1 n)n+1 exists, and call this limit e e. The reason why these two limit exists is because it can be proven with the Bernoulli inequality that (1+1 n)n(1+1 n)n is increasing and (1+1 n)n+1(1+1 n)n+1 is decreasing. It follows that they are both convergent, and their ratio converges to 1 1, thus the limit exists. From here, we also get immediately that Now, for each x∈(0,∞)x∈(0,∞) we have ⌊x⌋≤x<⌊x⌋+1⌊x⌋≤x<⌊x⌋+1. For simplicity I will denote n:=⌊x⌋n:=⌊x⌋. Then, we get (1+1 n+1)n≤(1+1 x)x≤(1+1 n)n+1(1+1 n+1)n≤(1+1 x)x≤(1+1 n)n+1 As n→∞n→∞ when x→∞x→∞, by a squeeze type argument we get lim x→∞(1+1 x)x=e lim x→∞(1+1 x)x=e Using 1−1 x=x−1 x=1 1+1 x−1 1−1 x=x−1 x=1 1+1 x−1 we also get lim x→∞(1−1 x)x=e−1 lim x→∞(1−1 x)x=e−1 and then lim y→0(1+y)1 y=e lim y→0(1+y)1 y=e Let y=a x−1 y=a x−1 [Note: no logarithms are used here, we just use the fact that if the limit exists for all y→0 y→0, it also exists for this articular choice of y y.] Then, we get lim x→0(a x)1 a x−1=e lim x→0(a x)1 a x−1=e Thus, we proved that lim x→0 a x a x−1=e,lim x→0 a x a x−1=e, exists. At this point logarithms would solve the problem, but you can probably finish the argument without using logarithms. For example, you can prove that a y a y is strictly increasing/decreasing (which reduces to x>y⇒a x−y>1 x>y⇒a x−y>1) and prove the following lemma: Lemma If f f is strictly monotonic and continuous, and for some c c the limit lim x→c f(g(x))lim x→c f(g(x)) exists, then lim x→c g(x)lim x→c g(x) exists. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 11, 2013 at 1:34 N. S.N. S. 135k 12 12 gold badges 150 150 silver badges 271 271 bronze badges 1 This looks very nice - I even don't mind using logarithms in the last step since all that is really used is the continuity and invertibility of a x a x.Paul Siegel –Paul Siegel 2013-10-11 02:21:44 +00:00 Commented Oct 11, 2013 at 2:21 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. The most common definition of e e is e:=lim x→0(1+x)1/x e:=lim x→0(1+x)1/x although you often see it with n=1/x n=1/x and as n→∞n→∞. Now lim x→0(e x−1 x)=lim x→0⎛⎝⎜(lim y→0(1+y)1/y)x−1 x⎞⎠⎟=lim x→0(lim y→0(1+y)x/y−1 x)=lim x→0 lim y→0((1+y)x/y−1 x)lim x→0(e x−1 x)=lim x→0((lim y→0(1+y)1/y)x−1 x)=lim x→0(lim y→0(1+y)x/y−1 x)=lim x→0 lim y→0((1+y)x/y−1 x) Now are you willing to believe that lim(x,y)→(0,0)((1+y)x/y−1 x)lim(x,y)→(0,0)((1+y)x/y−1 x) exists? If so it equals the last line above, and it also equals lim x→0((1+x)x/x−1 x)lim x→0((1+x)x/x−1 x) by tracking along the line y=x y=x. This last quantity is clearly 1 1. From here, you can establish that the derivative of e x e x is e x e x, from which the Chain Rule gives that the derivative of a x a x is a x ln(a)a x ln⁡(a), from which you can get that lim x→0(a x−1 x)=ln(a)lim x→0(a x−1 x)=ln⁡(a). See if you can find justification for the existence of that limit as (x,y)→(0,0)(x,y)→(0,0). Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 11, 2013 at 2:21 2'5 9'22'5 9'2 57.2k 8 8 gold badges 89 89 silver badges 162 162 bronze badges 1 Your argument seems to exploit the continuity of the function ((1+y)x/y−1)/x((1+y)x/y−1)/x which, as far as I can tell, is no easier than the original problem.Paul Siegel –Paul Siegel 2013-10-11 03:58:46 +00:00 Commented Oct 11, 2013 at 3:58 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. From the identity a x−1 x=a x/n−1 x/n⋅1 n⋅(1+a x/n+a 2 x/n+⋯+a(n−1)x/n)a x−1 x=a x/n−1 x/n⋅1 n⋅(1+a x/n+a 2 x/n+⋯+a(n−1)x/n) it shouldn't be terribly hard to show that the limit exists (although it might be a bit tedious to work through all of the detail). I can't say how illuminating this approach would be, though. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 11, 2013 at 2:42 user14972 user14972 2 This is a compelling observation, but it is not clear how exactly to exploit it.Paul Siegel –Paul Siegel 2013-10-11 04:09:10 +00:00 Commented Oct 11, 2013 at 4:09 @Paul: The proof sketch I have in mind is: (1) prove that it converges when x x is restricted to the sequence x n=1/n!x n=1/n!. (2) prove it for x x restricted to positive rationals. (3) prove it for x x restricted to nonzero rationals. (4) prove the limit. Point (2), I think, is the only tricky part, and I think it can be done by using the fact we can make a x a x arbitrarily close to 1 by forcing x x sufficiently small.user14972 –user14972 2013-10-11 05:48:11 +00:00 Commented Oct 11, 2013 at 5:48 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. If we know, from the properties of exponents, that f(x)=a x f(x)=a x satisfies f(x+y)=f(x)f(y)f(x+y)=f(x)f(y), then f(x+h)−f(x)=a x+h−a x=a x a h−a x=a x(a h−1)f(x+h)−f(x)=a x+h−a x=a x a h−a x=a x(a h−1) so f(x+h)−f(x)h=a x a h−1 h=a x f(h)−f(0)h f(x+h)−f(x)h=a x a h−1 h=a x f(h)−f(0)h. All we need, therefore, is that a x a x is differentiable at 0 0. Then, as has been said in other solutions, we have to look at some foundation stuff. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 11, 2013 at 20:05 marty cohenmarty cohen 111k 10 10 gold badges 88 88 silver badges 186 186 bronze badges 1 I'm not sure this is helpful: my question could be rephrased as "how can one prove that a x a x is differentiable at 0 0 using only elementary techniques?" But I don't think foundational issues are all that relevant here: there is no problem defining a x a x on Q Q, and it can be extended to R R by continuity. If you like, I would be happy with an argument which calculates the limit along all sequences of rational numbers.Paul Siegel –Paul Siegel 2013-10-12 11:49:43 +00:00 Commented Oct 12, 2013 at 11:49 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. This question is ultimately related to conceptual foundation of exponentiation and logarithm. The limit of (a x−1)/x(a x−1)/x as x→0 x→0 requires that we first define the symbol a x a x. This itself is a tough problem (for a beginner in calculus) and there are many approaches possible. Based on the question given here, one possible approach goes like this. 1) Let f(x)=lim n→∞n(x 1/n−1)f(x)=lim n→∞n(x 1/n−1). Then it can be shown (using various theorems on limits and nothing else) that this limit exists for all x>0 x>0 and hence defines a function f(x)f(x) which has following properties: f(1)=0 f(1)=0 f(x y)=f(x)+f(y)f(x y)=f(x)+f(y) f(x)f(x) is strictly increasing function of x x and maps the domain (0,∞)(0,∞) into range (−∞,∞)(−∞,∞) 2) Next we can use the inverse function theorem to define a new function g(x)g(x) such that y=g(x)y=g(x) and x=f(y)x=f(y) are equivalent. Let e=g(1)e=g(1) so that f(e)=1 f(e)=1 and then we define function a x=g(x f(a))a x=g(x f(a)). 3) It can then be proved that lim x→0(a x−1)/x=f(a)lim x→0(a x−1)/x=f(a) and lim h→0(1+h)1/h=e=g(1)lim h→0(1+h)1/h=e=g(1) In traditional notation f(x)f(x) is represented by log x log⁡x and g(x)g(x) is represented by e x e x or exp(x)exp⁡(x). The above presentation does not involve any circular reasoning, but does require the deep conceptual framework of real numbers and limits. Even at the beginning of this presentation we need to know the meaning of x 1/n x 1/n which can not be given without the completeness property of real numbers. UPDATE: After looking at OP's comments on my answer I think what is needed here is to show that the limit lim x→0(a x−1)/x lim x→0(a x−1)/x exists under the following definition of a x a x: If x x is rational then a x a x can be defined to some extent via algebra as is done in courses of elementary algebra. If x x is irrational then let x n x n be a sequence of rationals tending to x x as n→∞n→∞ and then we define a x=lim n→∞a x n a x=lim n→∞a x n. This definition is intuitive but has technical challenges namely a) to show that lim n→∞a x n lim n→∞a x n exists and also show that b) if x n,y n x n,y n are rational and x n→x,y n→x x n→x,y n→x then lim n→∞a x n=lim n→∞a y n lim n→∞a x n=lim n→∞a y n Using this definition we can proceed as follows: 1) Show that if x n x n is a sequence of reals tending to x x then a x=lim n→∞a x n a x=lim n→∞a x n 2) Show that lim x→0 a x=1 lim x→0 a x=1 3) Using 2) above show that it is sufficient to consider x→0+x→0+ when calculating lim(a x−1)/x lim(a x−1)/x 4) Establish that lim n→∞n(a 1/n−1)=L lim n→∞n(a 1/n−1)=L exists. Use this to find bounds for a 1/n a 1/n like a 1/n=1+{(L+ϕ(n))/n}a 1/n=1+{(L+ϕ(n))/n} where ϕ(n)ϕ(n) to zero as n→∞n→∞ 5) Using 4) above estimate a x a x for small x x i.e. for 1/(n+1)<x<1/n 1/(n+1)<x<1/n and then show that (a x−1)/x(a x−1)/x also tends to L L as x→0+x→0+. All this can be done rigorously with epsilon-delta type arguments. Proving 4) will require the theorem that bounded and monotone sequences are convergent. Thus we can provide a rigorous proof that lim x→0(a x−1)/x lim x→0(a x−1)/x exists for all a>0 a>0 and use this limit as definition of log a log⁡a if one so desires. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Oct 13, 2013 at 9:40 answered Oct 11, 2013 at 18:33 Paramanand Singh♦Paramanand Singh 92.6k 15 15 gold badges 159 159 silver badges 349 349 bronze badges 3 Well, this is an attempt to solve the problem by defining it away. As I wrote in the comments above, I want to use the most naive definition of a x a x: first define it on rational numbers using algebra, then extend to the reals by continuity.Paul Siegel –Paul Siegel 2013-10-12 11:29:24 +00:00 Commented Oct 12, 2013 at 11:29 To Paul Siegel: The approach you mention for defining a x a x is intuitive, but difficult technically. What I presented was one way of doing things. I still prefer to start the definition of these stuff with log x=∫x 1(1/t)d t log⁡x=∫1 x(1/t)d t as this is the easiest approach with very simple proofs of almost all properties of logarithm and exponents.Paramanand Singh –Paramanand Singh♦ 2013-10-12 17:58:23 +00:00 Commented Oct 12, 2013 at 17:58 To Paul Siegel: In case you define a x a x first for rational x x and then for irrational x x by continuity, then you need to also define the function log x log⁡x suitably. If you provide a definition of log x log⁡x in your approach then I think it should be possible to supply an easy proof of lim x→0(a x−1)/x=log a lim x→0(a x−1)/x=log⁡a Paramanand Singh –Paramanand Singh♦ 2013-10-12 18:00:11 +00:00 Commented Oct 12, 2013 at 18:00 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. 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4898	https://3d.bk.tudelft.nl/courses/geo5017/books/Linear_algebra.pdf	SCHAUM’S outlines Linear Algebra Fourth Edition Seymour Lipschutz, Ph.D. Temple University Marc Lars Lipson, Ph.D. University of Virginia Schaum’s Outline Series New York Chicago San Francisco Lisbon London Madrid Mexico City Milan New Delhi San Juan Seoul Singapore Sydney Toronto SCHAUM’S outlines Copyright © 2009, 2001, 1991, 1968 by The McGraw-Hill Companies, Inc. All rights reserved. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior writ-ten permission of the publisher. ISBN: 978-0-07-154353-8 MHID: 0-07-154353-8 The material in this eBook also appears in the print version of this title: ISBN: 978-0-07-154352-1, MHID: 0-07-154352-X. All trademarks are trademarks of their respective owners. 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Preface Linear algebra has in recent years become an essential part of the mathematical background required by mathematicians and mathematics teachers, engineers, computer scientists, physicists, economists, and statisticians, among others. This requirement reﬂects the importance and wide applications of the subject matter. This book is designed for use as a textbook for a formal course in linear algebra or as a supplement to all current standard texts. It aims to present an introduction to linear algebra which will be found helpful to all readers regardless of their ﬁelds of speciﬁcation. More material has been included than can be covered in most ﬁrst courses. This has been done to make the book more ﬂexible, to provide a useful book of reference, and to stimulate further interest in the subject. Each chapter begins with clear statements of pertinent deﬁnitions, principles, and theorems together with illustrative and other descriptive material. This is followed by graded sets of solved and supplementary problems. The solved problems serve to illustrate and amplify the theory, and to provide the repetition of basic principles so vital to effective learning. Numerous proofs, especially those of all essential theorems, are included among the solved problems. The supplementary problems serve as a complete review of the material of each chapter. The ﬁrst three chapters treat vectors in Euclidean space, matrix algebra, and systems of linear equations. These chapters provide the motivation and basic computational tools for the abstract investigations of vector spaces and linear mappings which follow. After chapters on inner product spaces and orthogonality and on determinants, there is a detailed discussion of eigenvalues and eigenvectors giving conditions for representing a linear operator by a diagonal matrix. This naturally leads to the study of various canonical forms, speciﬁcally, the triangular, Jordan, and rational canonical forms. Later chapters cover linear functions and the dual space V, and bilinear, quadratic, and Hermitian forms. The last chapter treats linear operators on inner product spaces. The main changes in the fourth edition have been in the appendices. First of all, we have expanded Appendix A on the tensor and exterior products of vector spaces where we have now included proofs on the existence and uniqueness of such products. We also added appendices covering algebraic structures, including modules, and polynomials over a ﬁeld. Appendix D, ‘‘Odds and Ends,’’ includes the Moore–Penrose generalized inverse which appears in various applications, such as statistics. There are also many additional solved and supplementary problems. Finally, we wish to thank the staff of the McGraw-Hill Schaum’s Outline Series, especially Charles Wall, for their unfailing cooperation. SEYMOUR LIPSCHUTZ MARC LARS LIPSON iii This page intentionally left blank Contents CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors 1 1.1 Introduction 1.2 Vectors in Rn 1.3 Vector Addition and Scalar Multi-plication 1.4 Dot (Inner) Product 1.5 Located Vectors, Hyperplanes, Lines, Curves in Rn 1.6 Vectors in R3 (Spatial Vectors), ijk Notation 1.7 Complex Numbers 1.8 Vectors in Cn CHAPTER 2 Algebra of Matrices 27 2.1 Introduction 2.2 Matrices 2.3 Matrix Addition and Scalar Multiplica-tion 2.4 Summation Symbol 2.5 Matrix Multiplication 2.6 Transpose of a Matrix 2.7 Square Matrices 2.8 Powers of Matrices, Polynomials in Matrices 2.9 Invertible (Nonsingular) Matrices 2.10 Special Types of Square Matrices 2.11 Complex Matrices 2.12 Block Matrices CHAPTER 3 Systems of Linear Equations 57 3.1 Introduction 3.2 Basic Deﬁnitions, Solutions 3.3 Equivalent Systems, Elementary Operations 3.4 Small Square Systems of Linear Equations 3.5 Systems in Triangular and Echelon Forms 3.6 Gaussian Elimination 3.7 Echelon Matrices, Row Canonical Form, Row Equivalence 3.8 Gaussian Elimination, Matrix Formulation 3.9 Matrix Equation of a System of Linear Equations 3.10 Systems of Linear Equations and Linear Combinations of Vectors 3.11 Homogeneous Systems of Linear Equations 3.12 Elementary Matrices 3.13 LU Decomposition CHAPTER 4 Vector Spaces 112 4.1 Introduction 4.2 Vector Spaces 4.3 Examples of Vector Spaces 4.4 Linear Combinations, Spanning Sets 4.5 Subspaces 4.6 Linear Spans, Row Space of a Matrix 4.7 Linear Dependence and Independence 4.8 Basis and Dimension 4.9 Application to Matrices, Rank of a Matrix 4.10 Sums and Direct Sums 4.11 Coordinates CHAPTER 5 Linear Mappings 164 5.1 Introduction 5.2 Mappings, Functions 5.3 Linear Mappings (Linear Transformations) 5.4 Kernel and Image of a Linear Mapping 5.5 Singular and Nonsingular Linear Mappings, Isomorphisms 5.6 Operations with Linear Mappings 5.7 Algebra A(V ) of Linear Operators CHAPTER 6 Linear Mappings and Matrices 195 6.1 Introduction 6.2 Matrix Representation of a Linear Operator 6.3 Change of Basis 6.4 Similarity 6.5 Matrices and General Linear Mappings CHAPTER 7 Inner Product Spaces, Orthogonality 226 7.1 Introduction 7.2 Inner Product Spaces 7.3 Examples of Inner Product Spaces 7.4 Cauchy–Schwarz Inequality, Applications 7.5 Orthogonal-ity 7.6 Orthogonal Sets and Bases 7.7 Gram–Schmidt Orthogonalization Process 7.8 Orthogonal and Positive Deﬁnite Matrices 7.9 Complex Inner Product Spaces 7.10 Normed Vector Spaces (Optional) v CHAPTER 8 Determinants 264 8.1 Introduction 8.2 Determinants of Orders 1 and 2 8.3 Determinants of Order 3 8.4 Permutations 8.5 Determinants of Arbitrary Order 8.6 Proper-ties of Determinants 8.7 Minors and Cofactors 8.8 Evaluation of Determi-nants 8.9 Classical Adjoint 8.10 Applications to Linear Equations, Cramer’s Rule 8.11 Submatrices, Minors, Principal Minors 8.12 Block Matrices and Determinants 8.13 Determinants and Volume 8.14 Determi-nant of a Linear Operator 8.15 Multilinearity and Determinants CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 292 9.1 Introduction 9.2 Polynomials of Matrices 9.3 Characteristic Polyno-mial, Cayley–Hamilton Theorem 9.4 Diagonalization, Eigenvalues and Eigenvectors 9.5 Computing Eigenvalues and Eigenvectors, Diagonalizing Matrices 9.6 Diagonalizing Real Symmetric Matrices and Quadratic Forms 9.7 Minimal Polynomial 9.8 Characteristic and Minimal Polyno-mials of Block Matrices CHAPTER 10 Canonical Forms 325 10.1 Introduction 10.2 Triangular Form 10.3 Invariance 10.4 Invariant Direct-Sum Decompositions 10.5 Primary Decomposition 10.6 Nilpotent Operators 10.7 Jordan Canonical Form 10.8 Cyclic Subspaces 10.9 Rational Canonical Form 10.10 Quotient Spaces CHAPTER 11 Linear Functionals and the Dual Space 349 11.1 Introduction 11.2 Linear Functionals and the Dual Space 11.3 Dual Basis 11.4 Second Dual Space 11.5 Annihilators 11.6 Transpose of a Linear Mapping CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms 359 12.1 Introduction 12.2 Bilinear Forms 12.3 Bilinear Forms and Matrices 12.4 Alternating Bilinear Forms 12.5 Symmetric Bilinear Forms, Quadratic Forms 12.6 Real Symmetric Bilinear Forms, Law of Inertia 12.7 Hermitian Forms CHAPTER 13 Linear Operators on Inner Product Spaces 377 13.1 Introduction 13.2 Adjoint Operators 13.3 Analogy Between A(V ) and C, Special Linear Operators 13.4 Self-Adjoint Operators 13.5 Orthogonal and Unitary Operators 13.6 Orthogonal and Unitary Matrices 13.7 Change of Orthonormal Basis 13.8 Positive Deﬁnite and Positive Operators 13.9 Diagonalization and Canonical Forms in Inner Product Spaces 13.10 Spectral Theorem APPENDIX A Multilinear Products 396 APPENDIX B Algebraic Structures 403 APPENDIX C Polynomials over a Field 411 APPENDIX D Odds and Ends 415 List of Symbols 420 Index 421 vi Contents CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors 1.1 Introduction There are two ways to motivate the notion of a vector: one is by means of lists of numbers and subscripts, and the other is by means of certain objects in physics. We discuss these two ways below. Here we assume the reader is familiar with the elementary properties of the ﬁeld of real numbers, denoted by R. On the other hand, we will review properties of the ﬁeld of complex numbers, denoted by C. In the context of vectors, the elements of our number ﬁelds are called scalars. Although we will restrict ourselves in this chapter to vectors whose elements come from R and then from C, many of our operations also apply to vectors whose entries come from some arbitrary ﬁeld K. Lists of Numbers Suppose the weights (in pounds) of eight students are listed as follows: 156; 125; 145; 134; 178; 145; 162; 193 One can denote all the values in the list using only one symbol, say w, but with different subscripts; that is, w1; w2; w3; w4; w5; w6; w7; w8 Observe that each subscript denotes the position of the value in the list. For example, w1 ¼ 156; the first number; w2 ¼ 125; the second number; . . . Such a list of values, w ¼ ðw1; w2; w3; . . . ; w8Þ is called a linear array or vector. Vectors in Physics Many physical quantities, such as temperature and speed, possess only ‘‘magnitude.’’ These quantities can be represented by real numbers and are called scalars. On the other hand, there are also quantities, such as force and velocity, that possess both ‘‘magnitude’’ and ‘‘direction.’’ These quantities, which can be represented by arrows having appropriate lengths and directions and emanating from some given reference point O, are called vectors. Now we assume the reader is familiar with the space R3 where all the points in space are represented by ordered triples of real numbers. Suppose the origin of the axes in R3 is chosen as the reference point O for the vectors discussed above. Then every vector is uniquely determined by the coordinates of its endpoint, and vice versa. There are two important operations, vector addition and scalar multiplication, associated with vectors in physics. The deﬁnition of these operations and the relationship between these operations and the endpoints of the vectors are as follows. 1 CHAPTER 1 (i) Vector Addition: The resultant u þ v of two vectors u and v is obtained by the parallelogram law; that is, u þ v is the diagonal of the parallelogram formed by u and v. Furthermore, if ða; b; cÞ and ða0; b0; c0Þ are the endpoints of the vectors u and v, then ða þ a0; b þ b0; c þ c0Þ is the endpoint of the vector u þ v. These properties are pictured in Fig. 1-1(a). (ii) Scalar Multiplication: The product ku of a vector u by a real number k is obtained by multiplying the magnitude of u by k and retaining the same direction if k > 0 or the opposite direction if k < 0. Also, if ða; b; cÞ is the endpoint of the vector u, then ðka; kb; kcÞ is the endpoint of the vector ku. These properties are pictured in Fig. 1-1(b). Mathematically, we identify the vector u with its ða; b; cÞ and write u ¼ ða; b; cÞ. Moreover, we call the ordered triple ða; b; cÞ of real numbers a point or vector depending upon its interpretation. We generalize this notion and call an n-tuple ða1; a2; . . . ; anÞ of real numbers a vector. However, special notation may be used for the vectors in R3 called spatial vectors (Section 1.6). 1.2 Vectors in Rn The set of all n-tuples of real numbers, denoted by Rn, is called n-space. A particular n-tuple in Rn, say u ¼ ða1; a2; . . . ; anÞ is called a point or vector. The numbers ai are called the coordinates, components, entries, or elements of u. Moreover, when discussing the space Rn, we use the term scalar for the elements of R. Two vectors, u and v, are equal, written u ¼ v, if they have the same number of components and if the corresponding components are equal. Although the vectors ð1; 2; 3Þ and ð2; 3; 1Þ contain the same three numbers, these vectors are not equal because corresponding entries are not equal. The vector ð0; 0; . . . ; 0Þ whose entries are all 0 is called the zero vector and is usually denoted by 0. EXAMPLE 1.1 (a) The following are vectors: ð2; 5Þ; ð7; 9Þ; ð0; 0; 0Þ; ð3; 4; 5Þ The ﬁrst two vectors belong to R2, whereas the last two belong to R3. The third is the zero vector in R3. (b) Find x; y; z such that ðx y; x þ y; z 1Þ ¼ ð4; 2; 3Þ. By deﬁnition of equality of vectors, corresponding entries must be equal. Thus, x y ¼ 4; x þ y ¼ 2; z 1 ¼ 3 Solving the above system of equations yields x ¼ 3, y ¼ 1, z ¼ 4. Figure 1-1 2 CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors Column Vectors Sometimes a vector in n-space Rn is written vertically rather than horizontally. Such a vector is called a column vector, and, in this context, the horizontally written vectors in Example 1.1 are called row vectors. For example, the following are column vectors with 2; 2; 3, and 3 components, respectively: 1 2 ; 3 4 ; 1 5 6 2 4 3 5; 1:5 2 3 15 2 6 4 3 7 5 We also note that any operation deﬁned for row vectors is deﬁned analogously for column vectors. 1.3 Vector Addition and Scalar Multiplication Consider two vectors u and v in Rn, say u ¼ ða1; a2; . . . ; anÞ and v ¼ ðb1; b2; . . . ; bnÞ Their sum, written u þ v, is the vector obtained by adding corresponding components from u and v. That is, u þ v ¼ ða1 þ b1; a2 þ b2; . . . ; an þ bnÞ The scalar product or, simply, product, of the vector u by a real number k, written ku, is the vector obtained by multiplying each component of u by k. That is, ku ¼ kða1; a2; . . . ; anÞ ¼ ðka1; ka2; . . . ; kanÞ Observe that u þ v and ku are also vectors in Rn. The sum of vectors with different numbers of components is not deﬁned. Negatives and subtraction are deﬁned in Rn as follows: u ¼ ð1Þu and u v ¼ u þ ðvÞ The vector u is called the negative of u, and u v is called the difference of u and v. Now suppose we are given vectors u1; u2; . . . ; um in Rn and scalars k1; k2; . . . ; km in R. We can multiply the vectors by the corresponding scalars and then add the resultant scalar products to form the vector v ¼ k1u1 þ k2u2 þ k3u3 þ þ kmum Such a vector v is called a linear combination of the vectors u1; u2; . . . ; um. EXAMPLE 1.2 (a) Let u ¼ ð2; 4; 5Þ and v ¼ ð1; 6; 9Þ. Then u þ v ¼ ð2 þ 1; 4 þ ð5Þ; 5 þ 9Þ ¼ ð3; 1; 4Þ 7u ¼ ð7ð2Þ; 7ð4Þ; 7ð5ÞÞ ¼ ð14; 28; 35Þ v ¼ ð1Þð1; 6; 9Þ ¼ ð1; 6; 9Þ 3u 5v ¼ ð6; 12; 15Þ þ ð5; 30; 45Þ ¼ ð1; 42; 60Þ (b) The zero vector 0 ¼ ð0; 0; . . . ; 0Þ in Rn is similar to the scalar 0 in that, for any vector u ¼ ða1; a2; . . . ; anÞ. u þ 0 ¼ ða1 þ 0; a2 þ 0; . . . ; an þ 0Þ ¼ ða1; a2; . . . ; anÞ ¼ u (c) Let u ¼ 2 3 4 2 4 3 5 and v ¼ 3 1 2 2 4 3 5. Then 2u 3v ¼ 4 6 8 2 4 3 5 þ 9 3 6 2 4 3 5 ¼ 5 9 2 2 4 3 5. CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors 3 Basic properties of vectors under the operations of vector addition and scalar multiplication are described in the following theorem. THEOREM 1.1: For any vectors u; v; w in Rn and any scalars k; k0 in R, (i) ðu þ vÞ þ w ¼ u þ ðv þ wÞ, (v) kðu þ vÞ ¼ ku þ kv, (ii) u þ 0 ¼ u; (vi) ðk þ k0Þu ¼ ku þ k0u, (iii) u þ ðuÞ ¼ 0; (vii) (kk’)u=k(k’u); (iv) u þ v ¼ v þ u, (viii) 1u ¼ u. We postpone the proof of Theorem 1.1 until Chapter 2, where it appears in the context of matrices (Problem 2.3). Suppose u and v are vectors in Rn for which u ¼ kv for some nonzero scalar k in R. Then u is called a multiple of v. Also, u is said to be in the same or opposite direction as v according to whether k > 0 or k < 0. 1.4 Dot (Inner) Product Consider arbitrary vectors u and v in Rn; say, u ¼ ða1; a2; . . . ; anÞ and v ¼ ðb1; b2; . . . ; bnÞ The dot product or inner product or scalar product of u and v is denoted and deﬁned by u v ¼ a1b1 þ a2b2 þ þ anbn That is, u v is obtained by multiplying corresponding components and adding the resulting products. The vectors u and v are said to be orthogonal (or perpendicular) if their dot product is zero—that is, if u v ¼ 0. EXAMPLE 1.3 (a) Let u ¼ ð1; 2; 3Þ, v ¼ ð4; 5; 1Þ, w ¼ ð2; 7; 4Þ. Then, u v ¼ 1ð4Þ 2ð5Þ þ 3ð1Þ ¼ 4 10 3 ¼ 9 u w ¼ 2 14 þ 12 ¼ 0; v w ¼ 8 þ 35 4 ¼ 39 Thus, u and w are orthogonal. (b) Let u ¼ 2 3 4 2 4 3 5 and v ¼ 3 1 2 2 4 3 5. Then u v ¼ 6 3 þ 8 ¼ 11. (c) Suppose u ¼ ð1; 2; 3; 4Þ and v ¼ ð6; k; 8; 2Þ. Find k so that u and v are orthogonal. First obtain u v ¼ 6 þ 2k 24 þ 8 ¼ 10 þ 2k. Then set u v ¼ 0 and solve for k: 10 þ 2k ¼ 0 or 2k ¼ 10 or k ¼ 5 Basic properties of the dot product in Rn (proved in Problem 1.13) follow. THEOREM 1.2: For any vectors u; v; w in Rn and any scalar k in R: (i) ðu þ vÞ w ¼ u w þ v w; (iii) u v ¼ v u, (ii) ðkuÞ v ¼ kðu vÞ, (iv) u u 0; and u u ¼ 0 iff u ¼ 0. Note that (ii) says that we can ‘‘take k out’’ from the ﬁrst position in an inner product. By (iii) and (ii), u ðkvÞ ¼ ðkvÞ u ¼ kðv uÞ ¼ kðu vÞ 4 CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors That is, we can also ‘‘take k out’’ from the second position in an inner product. The space Rn with the above operations of vector addition, scalar multiplication, and dot product is usually called Euclidean n-space. Norm (Length) of a Vector The norm or length of a vector u in Rn, denoted by kuk, is deﬁned to be the nonnegative square root of u u. In particular, if u ¼ ða1; a2; . . . ; anÞ, then kuk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ u u p ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ a2 1 þ a2 2 þ þ a2 n q That is, kuk is the square root of the sum of the squares of the components of u. Thus, kuk 0, and kuk ¼ 0 if and only if u ¼ 0. A vector u is called a unit vector if kuk ¼ 1 or, equivalently, if u u ¼ 1. For any nonzero vector v in Rn, the vector ^ v ¼ 1 kvk v ¼ v kvk is the unique unit vector in the same direction as v. The process of ﬁnding ^ v from v is called normalizing v. EXAMPLE 1.4 (a) Suppose u ¼ ð1; 2; 4; 5; 3Þ. To ﬁnd kuk, we can ﬁrst ﬁnd kuk2 ¼ u u by squaring each component of u and adding, as follows: kuk2 ¼ 12 þ ð2Þ2 þ ð4Þ2 þ 52 þ 32 ¼ 1 þ 4 þ 16 þ 25 þ 9 ¼ 55 Then kuk ¼ ﬃﬃﬃﬃﬃ 55 p . (b) Let v ¼ ð1; 3; 4; 2Þ and w ¼ ð1 2 ; 1 6 ; 5 6 ; 1 6Þ. Then kvk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1 þ 9 þ 16 þ 4 p ¼ ﬃﬃﬃﬃﬃ 30 p and kwk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 9 36 þ 1 36 þ 25 36 þ 1 36 r ¼ ﬃﬃﬃﬃﬃ 36 36 r ¼ ﬃﬃﬃ 1 p ¼ 1 Thus w is a unit vector, but v is not a unit vector. However, we can normalize v as follows: ^ v ¼ v kvk ¼ 1 ﬃﬃﬃﬃﬃ 30 p ; 3 ﬃﬃﬃﬃﬃ 30 p ; 4 ﬃﬃﬃﬃﬃ 30 p ; 2 ﬃﬃﬃﬃﬃ 30 p This is the unique unit vector in the same direction as v. The following formula (proved in Problem 1.14) is known as the Schwarz inequality or Cauchy– Schwarz inequality. It is used in many branches of mathematics. THEOREM 1.3 (Schwarz): For any vectors u; v in Rn, ju vj kukkvk. Using the above inequality, we also prove (Problem 1.15) the following result known as the ‘‘triangle inequality’’ or Minkowski’s inequality. THEOREM 1.4 (Minkowski): For any vectors u; v in Rn, ku þ vk kuk þ kvk. Distance, Angles, Projections The distance between vectors u ¼ ða1; a2; . . . ; anÞ and v ¼ ðb1; b2; . . . ; bnÞ in Rn is denoted and deﬁned by dðu; vÞ ¼ ku vk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ ða1 b1Þ2 þ ða2 b2Þ2 þ þ ðan bnÞ2 q One can show that this deﬁnition agrees with the usual notion of distance in the Euclidean plane R2 or space R3. CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors 5 The angle y between nonzero vectors u; v in Rn is deﬁned by cos y ¼ u v kukkvk This deﬁnition is well deﬁned, because, by the Schwarz inequality (Theorem 1.3), 1 u v kukkvk 1 Note that if u v ¼ 0, then y ¼ 90 (or y ¼ p=2). This then agrees with our previous deﬁnition of orthogonality. The projection of a vector u onto a nonzero vector v is the vector denoted and deﬁned by projðu; vÞ ¼ u v kvk2 v ¼ u v v v v We show below that this agrees with the usual notion of vector projection in physics. EXAMPLE 1.5 (a) Suppose u ¼ ð1; 2; 3Þ and v ¼ ð2; 4; 5Þ. Then dðu; vÞ ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ ð1 2Þ2 þ ð2 4Þ2 þ ð3 5Þ2 q ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ 1 þ 36 þ 4 p ¼ ﬃﬃﬃﬃﬃ 41 p To ﬁnd cos y, where y is the angle between u and v, we ﬁrst ﬁnd u v ¼ 2 8 þ 15 ¼ 9; kuk2 ¼ 1 þ 4 þ 9 ¼ 14; kvk2 ¼ 4 þ 16 þ 25 ¼ 45 Then cos y ¼ u v kukkvk ¼ 9 ﬃﬃﬃﬃﬃ 14 p ﬃﬃﬃﬃﬃ 45 p Also, projðu; vÞ ¼ u v kvk2 v ¼ 9 45 ð2; 4; 5Þ ¼ 1 5 ð2; 4; 5Þ ¼ 2 5 ; 4 5 ; 1 (b) Consider the vectors u and v in Fig. 1-2(a) (with respective endpoints A and B). The (perpendicular) projection of u onto v is the vector u with magnitude kuk ¼ kuk cos y ¼ kuk u v kukvk ¼ u v kvk To obtain u, we multiply its magnitude by the unit vector in the direction of v, obtaining u ¼ kuk v kvk ¼ u v kvk v kvk ¼ u v kvk2 v This is the same as the above deﬁnition of projðu; vÞ. Figure 1-2 z y x 0 u ( ) b B b b b ( , , ) 1 2 3 u = B – A A a a a ( , , ) 1 2 3 P b a b a b a ( – , – , – ) 1 1 2 2 3 3 0 u ( ) a Projection of onto u u A u B C θ 6 CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors 1.5 Located Vectors, Hyperplanes, Lines, Curves in Rn This section distinguishes between an n-tuple PðaiÞ Pða1; a2; . . . ; anÞ viewed as a point in Rn and an n-tuple u ¼ ½c1; c2; . . . ; cn viewed as a vector (arrow) from the origin O to the point Cðc1; c2; . . . ; cnÞ. Located Vectors Any pair of points AðaiÞ and BðbiÞ in Rn deﬁnes the located vector or directed line segment from A to B, written AB !. We identify AB ! with the vector u ¼ B A ¼ ½b1 a1; b2 a2; . . . ; bn an because AB ! and u have the same magnitude and direction. This is pictured in Fig. 1-2(b) for the points Aða1; a2; a3Þ and Bðb1; b2; b3Þ in R3 and the vector u ¼ B A which has the endpoint Pðb1 a1, b2 a2, b3 a3Þ. Hyperplanes A hyperplane H in Rn is the set of points ðx1; x2; . . . ; xnÞ that satisfy a linear equation a1x1 þ a2x2 þ þ anxn ¼ b where the vector u ¼ ½a1; a2; . . . ; an of coefﬁcients is not zero. Thus a hyperplane H in R2 is a line, and a hyperplane H in R3 is a plane. We show below, as pictured in Fig. 1-3(a) for R3, that u is orthogonal to any directed line segment PQ !, where Pð piÞ and QðqiÞ are points in H: [For this reason, we say that u is normal to H and that H is normal to u:] Because Pð piÞ and QðqiÞ belong to H; they satisfy the above hyperplane equation—that is, a1 p1 þ a2 p2 þ þ an pn ¼ b and a1q1 þ a2q2 þ þ anqn ¼ b v ¼ PQ ! ¼ Q P ¼ ½q1 p1; q2 p2; . . . ; qn pn Let Then u v ¼ a1ðq1 p1Þ þ a2ðq2 p2Þ þ þ anðqn pnÞ ¼ ða1q1 þ a2q2 þ þ anqnÞ ða1 p1 þ a2 p2 þ þ an pnÞ ¼ b b ¼ 0 Thus v ¼ PQ ! is orthogonal to u; as claimed. Figure 1-3 CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors 7 Lines in Rn The line L in Rn passing through the point Pðb1; b2; . . . ; bnÞ and in the direction of a nonzero vector u ¼ ½a1; a2; . . . ; an consists of the points Xðx1; x2; . . . ; xnÞ that satisfy X ¼ P þ tu or x1 ¼ a1t þ b1 x2 ¼ a2t þ b2 :::::::::::::::::::: xn ¼ ant þ bn or LðtÞ ¼ ðait þ biÞ 8 > > < > > : where the parameter t takes on all real values. Such a line L in R3 is pictured in Fig. 1-3(b). EXAMPLE 1.6 (a) Let H be the plane in R3 corresponding to the linear equation 2x 5y þ 7z ¼ 4. Observe that Pð1; 1; 1Þ and Qð5; 4; 2Þ are solutions of the equation. Thus P and Q and the directed line segment v ¼ PQ ! ¼ Q P ¼ ½5 1; 4 1; 2 1 ¼ ½4; 3; 1 lie on the plane H. The vector u ¼ ½2; 5; 7 is normal to H, and, as expected, u v ¼ ½2; 5; 7 ½4; 3; 1 ¼ 8 15 þ 7 ¼ 0 That is, u is orthogonal to v. (b) Find an equation of the hyperplane H in R4 that passes through the point Pð1; 3; 4; 2Þ and is normal to the vector u ¼ ½4; 2; 5; 6. The coefﬁcients of the unknowns of an equation of H are the components of the normal vector u; hence, the equation of H must be of the form 4x1 2x2 þ 5x3 þ 6x4 ¼ k Substituting P into this equation, we obtain 4ð1Þ 2ð3Þ þ 5ð4Þ þ 6ð2Þ ¼ k or 4 6 20 þ 12 ¼ k or k ¼ 10 Thus, 4x1 2x2 þ 5x3 þ 6x4 ¼ 10 is the equation of H. (c) Find the parametric representation of the line L in R4 passing through the point Pð1; 2; 3; 4Þ and in the direction of u ¼ ½5; 6; 7; 8. Also, ﬁnd the point Q on L when t ¼ 1. Substitution in the above equation for L yields the following parametric representation: x1 ¼ 5t þ 1; x2 ¼ 6t þ 2; x3 ¼ 7t þ 3; x4 ¼ 8t 4 or, equivalently, LðtÞ ¼ ð5t þ 1; 6t þ 2; 7t þ 3; 8t 4Þ Note that t ¼ 0 yields the point P on L. Substitution of t ¼ 1 yields the point Qð6; 8; 4; 4Þ on L. Curves in Rn Let D be an interval (ﬁnite or inﬁnite) on the real line R. A continuous function F: D ! Rn is a curve in Rn. Thus, to each point t 2 D there is assigned the following point in Rn: FðtÞ ¼ ½F1ðtÞ; F2ðtÞ; . . . ; FnðtÞ Moreover, the derivative (if it exists) of FðtÞ yields the vector VðtÞ ¼ dFðtÞ dt ¼ dF1ðtÞ dt ; dF2ðtÞ dt ; . . . ; dFnðtÞ dt 8 CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors which is tangent to the curve. Normalizing VðtÞ yields TðtÞ ¼ VðtÞ kVðtÞk Thus, TðtÞ is the unit tangent vector to the curve. (Unit vectors with geometrical signiﬁcance are often presented in bold type.) EXAMPLE 1.7 Consider the curve FðtÞ ¼ ½sin t; cos t; t in R3. Taking the derivative of FðtÞ [or each component of FðtÞ] yields VðtÞ ¼ ½cos t; sin t; 1 which is a vector tangent to the curve. We normalize VðtÞ. First we obtain kVðtÞk2 ¼ cos2 t þ sin2 t þ 1 ¼ 1 þ 1 ¼ 2 Then the unit tangent vection TðtÞ to the curve follows: TðtÞ ¼ VðtÞ kVðtÞk ¼ cos t ﬃﬃﬃ 2 p ; sin t ﬃﬃﬃ 2 p ; 1ﬃﬃﬃ 2 p 1.6 Vectors in R3 (Spatial Vectors), ijk Notation Vectors in R3, called spatial vectors, appear in many applications, especially in physics. In fact, a special notation is frequently used for such vectors as follows: i ¼ ½1; 0; 0 denotes the unit vector in the x direction: j ¼ ½0; 1; 0 denotes the unit vector in the y direction: k ¼ ½0; 0; 1 denotes the unit vector in the z direction: Then any vector u ¼ ½a; b; c in R3 can be expressed uniquely in the form u ¼ ½a; b; c ¼ ai þ bj þ cj Because the vectors i; j; k are unit vectors and are mutually orthogonal, we obtain the following dot products: i i ¼ 1; j j ¼ 1; k k ¼ 1 and i j ¼ 0; i k ¼ 0; j k ¼ 0 Furthermore, the vector operations discussed above may be expressed in the ijk notation as follows. Suppose u ¼ a1i þ a2j þ a3k and v ¼ b1i þ b2j þ b3k Then u þ v ¼ ða1 þ b1Þi þ ða2 þ b2Þj þ ða3 þ b3Þk and cu ¼ ca1i þ ca2j þ ca3k where c is a scalar. Also, u v ¼ a1b1 þ a2b2 þ a3b3 and kuk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ u u p ¼ a2 1 þ a2 2 þ a2 3 EXAMPLE 1.8 Suppose u ¼ 3i þ 5j 2k and v ¼ 4i 8j þ 7k. (a) To ﬁnd u þ v, add corresponding components, obtaining u þ v ¼ 7i 3j þ 5k (b) To ﬁnd 3u 2v, ﬁrst multiply by the scalars and then add: 3u 2v ¼ ð9i þ 13j 6kÞ þ ð8i þ 16j 14kÞ ¼ i þ 29j 20k CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors 9 (c) To ﬁnd u v, multiply corresponding components and then add: u v ¼ 12 40 14 ¼ 42 (d) To ﬁnd kuk, take the square root of the sum of the squares of the components: kuk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ 9 þ 25 þ 4 p ¼ ﬃﬃﬃﬃﬃ 38 p Cross Product There is a special operation for vectors u and v in R3 that is not deﬁned in Rn for n 6¼ 3. This operation is called the cross product and is denoted by u v. One way to easily remember the formula for u v is to use the determinant (of order two) and its negative, which are denoted and deﬁned as follows: a b c d ¼ ad bc and a b c d ¼ bc ad Here a and d are called the diagonal elements and b and c are the nondiagonal elements. Thus, the determinant is the product ad of the diagonal elements minus the product bc of the nondiagonal elements, but vice versa for the negative of the determinant. Now suppose u ¼ a1i þ a2j þ a3k and v ¼ b1i þ b2j þ b3k. Then u v ¼ ða2b3 a3b2Þi þ ða3b1 a1b3Þj þ ða1b2 a2b1Þk ¼ a1 a2 a3 b1 b2 b3 i a1 a2 a3 b1 b2 b3 j þ a1 a2 a3 b1 b2 b3 i That is, the three components of u v are obtained from the array a1 a2 a3 b1 b2 b3 (which contain the components of u above the component of v) as follows: (1) Cover the ﬁrst column and take the determinant. (2) Cover the second column and take the negative of the determinant. (3) Cover the third column and take the determinant. Note that u v is a vector; hence, u v is also called the vector product or outer product of u and v. EXAMPLE 1.9 Find u v where: (a) u ¼ 4i þ 3j þ 6k, v ¼ 2i þ 5j 3k, (b) u ¼ ½2; 1; 5, v ¼ ½3; 7; 6. (a) Use 4 3 6 2 5 3 to get u v ¼ ð9 30Þi þ ð12 þ 12Þj þ ð20 6Þk ¼ 39i þ 24j þ 14k (b) Use 2 1 5 3 7 6 to get u v ¼ ½6 35; 15 12; 14 þ 3 ¼ ½41; 3; 17 Remark: The cross products of the vectors i; j; k are as follows: i j ¼ k; j k ¼ i; k i ¼ j j i ¼ k; k j ¼ i; i k ¼ j Thus, if we view the triple ði; j; kÞ as a cyclic permutation, where i follows k and hence k precedes i, then the product of two of them in the given direction is the third one, but the product of two of them in the opposite direction is the negative of the third one. Two important properties of the cross product are contained in the following theorem. 10 CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors THEOREM 1.5: Let u; v; w be vectors in R3. (a) The vector u v is orthogonal to both u and v. (b) The absolute value of the ‘‘triple product’’ u v w represents the volume of the parallelopiped formed by the vectors u; v, w. [See Fig. 1-4(a).] We note that the vectors u; v, u v form a right-handed system, and that the following formula gives the magnitude of u v: ku vk ¼ kukkvk sin y where y is the angle between u and v. 1.7 Complex Numbers The set of complex numbers is denoted by C. Formally, a complex number is an ordered pair ða; bÞ of real numbers where equality, addition, and multiplication are deﬁned as follows: ða; bÞ ¼ ðc; dÞ if and only if a ¼ c and b ¼ d ða; bÞ þ ðc; dÞ ¼ ða þ c; b þ dÞ ða; bÞ ðc; dÞ ¼ ðac bd; ad þ bcÞ We identify the real number a with the complex number ða; 0Þ; that is, a $ ða; 0Þ This is possible because the operations of addition and multiplication of real numbers are preserved under the correspondence; that is, ða; 0Þ þ ðb; 0Þ ¼ ða þ b; 0Þ and ða; 0Þ ðb; 0Þ ¼ ðab; 0Þ Thus we view R as a subset of C, and replace ða; 0Þ by a whenever convenient and possible. We note that the set C of complex numbers with the above operations of addition and multiplication is a field of numbers, like the set R of real numbers and the set Q of rational numbers. Figure 1-4 CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors 11 The complex number ð0; 1Þ is denoted by i. It has the important property that i2 ¼ ii ¼ ð0; 1Þð0; 1Þ ¼ ð1; 0Þ ¼ 1 or i ¼ ﬃﬃﬃﬃﬃﬃ ﬃ 1 p Accordingly, any complex number z ¼ ða; bÞ can be written in the form z ¼ ða; bÞ ¼ ða; 0Þ þ ð0; bÞ ¼ ða; 0Þ þ ðb; 0Þ ð0; 1Þ ¼ a þ bi The above notation z ¼ a þ bi, where a Re z and b Im z are called, respectively, the real and imaginary parts of z, is more convenient than ða; bÞ. In fact, the sum and product of complex numbers z ¼ a þ bi and w ¼ c þ di can be derived by simply using the commutative and distributive laws and i2 ¼ 1: z þ w ¼ ða þ biÞ þ ðc þ diÞ ¼ a þ c þ bi þ di ¼ ða þ bÞ þ ðc þ dÞi zw ¼ ða þ biÞðc þ diÞ ¼ ac þ bci þ adi þ bdi2 ¼ ðac bdÞ þ ðbc þ adÞi We also deﬁne the negative of z and subtraction in C by z ¼ 1z and w z ¼ w þ ðzÞ Warning: The letter i representing ﬃﬃﬃﬃﬃﬃ ﬃ 1 p has no relationship whatsoever to the vector i ¼ ½1; 0; 0 in Section 1.6. Complex Conjugate, Absolute Value Consider a complex number z ¼ a þ bi. The conjugate of z is denoted and deﬁned by z ¼ a þ bi ¼ a bi Then z z ¼ ða þ biÞða biÞ ¼ a2 b2i2 ¼ a2 þ b2. Note that z is real if and only if z ¼ z. The absolute value of z, denoted by jzj, is deﬁned to be the nonnegative square root of z z. Namely, jzj ¼ ﬃﬃﬃﬃ z z p ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ a2 þ b2 p Note that jzj is equal to the norm of the vector ða; bÞ in R2. Suppose z 6¼ 0. Then the inverse z1 of z and division in C of w by z are given, respectively, by z1 ¼ z z z ¼ a a2 þ b2 b a2 þ b2 i and w z w z z z ¼ wz1 EXAMPLE 1.10 Suppose z ¼ 2 þ 3i and w ¼ 5 2i. Then z þ w ¼ ð2 þ 3iÞ þ ð5 2iÞ ¼ 2 þ 5 þ 3i 2i ¼ 7 þ i zw ¼ ð2 þ 3iÞð5 2iÞ ¼ 10 þ 15i 4i 6i2 ¼ 16 þ 11i z ¼ 2 þ 3i ¼ 2 3i and w ¼ 5 2i ¼ 5 þ 2i w z ¼ 5 2i 2 þ 3i ¼ ð5 2iÞð2 3iÞ ð2 þ 3iÞð2 3iÞ ¼ 4 19i 13 ¼ 4 13 19 13 i jzj ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 4 þ 9 p ¼ ﬃﬃﬃﬃﬃ 13 p and jwj ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ 25 þ 4 p ¼ ﬃﬃﬃﬃﬃ 29 p Complex Plane Recall that the real numbers R can be represented by points on a line. Analogously, the complex numbers C can be represented by points in the plane. Speciﬁcally, we let the point ða; bÞ in the plane represent the complex number a þ bi as shown in Fig. 1-4(b). In such a case, jzj is the distance from the origin O to the point z. The plane with this representation is called the complex plane, just like the line representing R is called the real line. 12 CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors 1.8 Vectors in Cn The set of all n-tuples of complex numbers, denoted by Cn, is called complex n-space. Just as in the real case, the elements of Cn are called points or vectors, the elements of C are called scalars, and vector addition in Cn and scalar multiplication on Cn are given by ½z1; z2; . . . ; zn þ ½w1; w2; . . . ; wn ¼ ½z1 þ w1; z2 þ w2; . . . ; zn þ wn z½z1; z2; . . . ; zn ¼ ½zz1; zz2; . . . ; zzn where the zi, wi, and z belong to C. EXAMPLE 1.11 Consider vectors u ¼ ½2 þ 3i; 4 i; 3 and v ¼ ½3 2i; 5i; 4 6i in C3. Then u þ v ¼ ½2 þ 3i; 4 i; 3 þ ½3 2i; 5i; 4 6i ¼ ½5 þ i; 4 þ 4i; 7 6i ð5 2iÞu ¼ ½ð5 2iÞð2 þ 3iÞ; ð5 2iÞð4 iÞ; ð5 2iÞð3Þ ¼ ½16 þ 11i; 18 13i; 15 6i Dot (Inner) Product in Cn Consider vectors u ¼ ½z1; z2; . . . ; zn and v ¼ ½w1; w2; . . . ; wn in Cn. The dot or inner product of u and v is denoted and deﬁned by u v ¼ z1 w1 þ z2 w2 þ þ zn wn This deﬁnition reduces to the real case because wi ¼ wi when wi is real. The norm of u is deﬁned by kuk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ u u p ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ z1 z1 þ z2 z2 þ þ zn zn p ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ jz1j2 þ jz2j2 þ þ jvnj2 q We emphasize that u u and so kuk are real and positive when u 6¼ 0 and 0 when u ¼ 0. EXAMPLE 1.12 Consider vectors u ¼ ½2 þ 3i; 4 i; 3 þ 5i and v ¼ ½3 4i; 5i; 4 2i in C3. Then u v ¼ ð2 þ 3iÞð3 4iÞ þ ð4 iÞð5iÞ þ ð3 þ 5iÞð4 2iÞ ¼ ð2 þ 3iÞð3 þ 4iÞ þ ð4 iÞð5iÞ þ ð3 þ 5iÞð4 þ 2iÞ ¼ ð6 þ 13iÞ þ ð5 20iÞ þ ð2 þ 26iÞ ¼ 9 þ 19i u u ¼ j2 þ 3ij2 þ j4 ij2 þ j3 þ 5ij2 ¼ 4 þ 9 þ 16 þ 1 þ 9 þ 25 ¼ 64 kuk ¼ ﬃﬃﬃﬃﬃ 64 p ¼ 8 The space Cn with the above operations of vector addition, scalar multiplication, and dot product, is called complex Euclidean n-space. Theorem 1.2 for Rn also holds for Cn if we replace u v ¼ v u by u v ¼ u v On the other hand, the Schwarz inequality (Theorem 1.3) and Minkowski’s inequality (Theorem 1.4) are true for Cn with no changes. SOLVED PROBLEMS Vectors in Rn 1.1. Determine which of the following vectors are equal: u1 ¼ ð1; 2; 3Þ; u2 ¼ ð2; 3; 1Þ; u3 ¼ ð1; 3; 2Þ; u4 ¼ ð2; 3; 1Þ Vectors are equal only when corresponding entries are equal; hence, only u2 ¼ u4. CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors 13 1.2. Let u ¼ ð2; 7; 1Þ, v ¼ ð3; 0; 4Þ, w ¼ ð0; 5; 8Þ. Find: (a) 3u 4v, (b) 2u þ 3v 5w. First perform the scalar multiplication and then the vector addition. (a) 3u 4v ¼ 3ð2; 7; 1Þ 4ð3; 0; 4Þ ¼ ð6; 21; 3Þ þ ð12; 0; 16Þ ¼ ð18; 21; 13Þ (b) 2u þ 3v 5w ¼ ð4; 14; 2Þ þ ð9; 0; 12Þ þ ð0; 25; 40Þ ¼ ð5; 39; 54Þ 1.3. Let u ¼ 5 3 4 2 4 3 5; v ¼ 1 5 2 2 4 3 5; w ¼ 3 1 2 2 4 3 5. Find: (a) 5u 2v, (b) 2u þ 4v 3w. First perform the scalar multiplication and then the vector addition: (a) 5u 2v ¼ 5 5 3 4 2 4 3 5 2 1 5 2 2 4 3 5 ¼ 25 15 20 2 4 3 5 þ 2 10 4 2 4 3 5 ¼ 27 5 24 2 4 3 5 (b) 2u þ 4v 3w ¼ 10 6 8 2 4 3 5 þ 4 20 8 2 4 3 5 þ 9 3 6 2 4 3 5 ¼ 23 17 22 2 4 3 5 1.4. Find x and y, where: (a) ðx; 3Þ ¼ ð2; x þ yÞ, (b) ð4; yÞ ¼ xð2; 3Þ. (a) Because the vectors are equal, set the corresponding entries equal to each other, yielding x ¼ 2; 3 ¼ x þ y Solve the linear equations, obtaining x ¼ 2; y ¼ 1: (b) First multiply by the scalar x to obtain ð4; yÞ ¼ ð2x; 3xÞ. Then set corresponding entries equal to each other to obtain 4 ¼ 2x; y ¼ 3x Solve the equations to yield x ¼ 2, y ¼ 6. 1.5. Write the vector v ¼ ð1; 2; 5Þ as a linear combination of the vectors u1 ¼ ð1; 1; 1Þ, u2 ¼ ð1; 2; 3Þ, u3 ¼ ð2; 1; 1Þ. We want to express v in the form v ¼ xu1 þ yu2 þ zu3 with x; y; z as yet unknown. First we have 1 2 5 2 4 3 5 ¼ x 1 1 1 2 4 3 5 þ y 1 2 3 2 4 3 5 þ z 2 1 1 2 4 3 5 ¼ x þ y þ 2z x þ 2y z x þ 3y þ z 2 4 3 5 (It is more convenient to write vectors as columns than as rows when forming linear combinations.) Set corresponding entries equal to each other to obtain x þ y þ 2z ¼ 1 x þ 2y z ¼ 2 x þ 3y þ z ¼ 5 or x þ y þ 2z ¼ 1 y 3z ¼ 3 2y z ¼ 4 or x þ y þ 2z ¼ 1 y 3z ¼ 3 5z ¼ 10 This unique solution of the triangular system is x ¼ 6, y ¼ 3, z ¼ 2. Thus, v ¼ 6u1 þ 3u2 þ 2u3. 14 CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors 1.6. Write v ¼ ð2; 5; 3Þ as a linear combination of u1 ¼ ð1; 3; 2Þ; u2 ¼ ð2; 4; 1Þ; u3 ¼ ð1; 5; 7Þ: Find the equivalent system of linear equations and then solve. First, 2 5 3 2 4 3 5 ¼ x 1 3 2 2 4 3 5 þ y 2 4 1 2 4 3 5 þ z 1 5 7 2 4 3 5 ¼ x þ 2y þ z 3x 4y 5z 2x y þ 7z 2 4 3 5 Set the corresponding entries equal to each other to obtain x þ 2y þ z ¼ 2 3x 4y 5z ¼ 5 2x y þ 7z ¼ 3 or x þ 2y þ z ¼ 2 2y 2z ¼ 1 5y þ 5z ¼ 1 or x þ 2y þ z ¼ 2 2y 2z ¼ 1 0 ¼ 3 The third equation, 0x þ 0y þ 0z ¼ 3, indicates that the system has no solution. Thus, v cannot be written as a linear combination of the vectors u1, u2, u3. Dot (Inner) Product, Orthogonality, Norm in Rn 1.7. Find u v where: (a) u ¼ ð2; 5; 6Þ and v ¼ ð8; 2; 3Þ, (b) u ¼ ð4; 2; 3; 5; 1Þ and v ¼ ð2; 6; 1; 4; 8Þ. Multiply the corresponding components and add: (a) u v ¼ 2ð8Þ 5ð2Þ þ 6ð3Þ ¼ 16 10 18 ¼ 12 (b) u v ¼ 8 þ 12 þ 3 20 8 ¼ 5 1.8. Let u ¼ ð5; 4; 1Þ, v ¼ ð3; 4; 1Þ, w ¼ ð1; 2; 3Þ. Which pair of vectors, if any, are perpendicular (orthogonal)? Find the dot product of each pair of vectors: u v ¼ 15 16 þ 1 ¼ 0; v w ¼ 3 þ 8 þ 3 ¼ 14; u w ¼ 5 8 þ 3 ¼ 0 Thus, u and v are orthogonal, u and w are orthogonal, but v and w are not. 1.9. Find k so that u and v are orthogonal, where: (a) u ¼ ð1; k; 3Þ and v ¼ ð2; 5; 4Þ, (b) u ¼ ð2; 3k; 4; 1; 5Þ and v ¼ ð6; 1; 3; 7; 2kÞ. Compute u v, set u v equal to 0, and then solve for k: (a) u v ¼ 1ð2Þ þ kð5Þ 3ð4Þ ¼ 5k 10. Then 5k 10 ¼ 0, or k ¼ 2. (b) u v ¼ 12 3k 12 þ 7 þ 10k ¼ 7k þ 7. Then 7k þ 7 ¼ 0, or k ¼ 1. 1.10. Find kuk, where: (a) u ¼ ð3; 12; 4Þ, (b) u ¼ ð2; 3; 8; 7Þ. First ﬁnd kuk2 ¼ u u by squaring the entries and adding. Then kuk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ kuk2 q . (a) kuk2 ¼ ð3Þ2 þ ð12Þ2 þ ð4Þ2 ¼ 9 þ 144 þ 16 ¼ 169. Then kuk ¼ ﬃﬃﬃﬃﬃﬃﬃ ﬃ 169 p ¼ 13. (b) kuk2 ¼ 4 þ 9 þ 64 þ 49 ¼ 126. Then kuk ¼ ﬃﬃﬃﬃﬃﬃﬃ ﬃ 126 p . CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors 15 1.11. Recall that normalizing a nonzero vector v means ﬁnding the unique unit vector ^ v in the same direction as v, where ^ v ¼ 1 kvk v Normalize: (a) u ¼ ð3; 4Þ, (b) v ¼ ð4; 2; 3; 8Þ, (c) w ¼ ð1 2, 2 3, 1 4). (a) First ﬁnd kuk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ 9 þ 16 p ¼ ﬃﬃﬃﬃﬃ 25 p ¼ 5. Then divide each entry of u by 5, obtaining ^ u ¼ ð3 5, 4 5). (b) Here kvk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 16 þ 4 þ 9 þ 64 p ¼ ﬃﬃﬃﬃﬃ 93 p . Then ^ v ¼ 4 ﬃﬃﬃﬃﬃ 93 p ; 2 ﬃﬃﬃﬃﬃ 93 p ; 3 ﬃﬃﬃﬃﬃ 93 p ; 8 ﬃﬃﬃﬃﬃ 93 p (c) Note that w and any positive multiple of w will have the same normalized form. Hence, ﬁrst multiply w by 12 to ‘‘clear fractions’’—that is, ﬁrst ﬁnd w0 ¼ 12w ¼ ð6; 8; 3Þ. Then kw0k ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 36 þ 64 þ 9 p ¼ ﬃﬃﬃﬃﬃﬃﬃ ﬃ 109 p and ^ w ¼ b w0 ¼ 6 ﬃﬃﬃﬃﬃﬃﬃ ﬃ 109 p ; 8 ﬃﬃﬃﬃﬃﬃﬃ ﬃ 109 p ; 3 ﬃﬃﬃﬃﬃﬃﬃ ﬃ 109 p 1.12. Let u ¼ ð1; 3; 4Þ and v ¼ ð3; 4; 7Þ. Find: (a) cos y, where y is the angle between u and v; (b) projðu; vÞ, the projection of u onto v; (c) dðu; vÞ, the distance between u and v. First ﬁnd u v ¼ 3 12 þ 28 ¼ 19, kuk2 ¼ 1 þ 9 þ 16 ¼ 26, kvk2 ¼ 9 þ 16 þ 49 ¼ 74. Then (a) cos y ¼ u v kukkvk ¼ 19 ﬃﬃﬃﬃﬃ 26 p ﬃﬃﬃﬃﬃ 74 p , (b) projðu; vÞ ¼ u v kvk2 v ¼ 19 74 ð3; 4; 7Þ ¼ 57 74 ; 76 74 ; 133 74 ¼ 57 74 ; 38 37 ; 133 74 ; (c) dðu; vÞ ¼ ku vk ¼ kð2; 7 3Þk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 4 þ 49 þ 9 p ¼ ﬃﬃﬃﬃﬃ 62 p : 1.13. Prove Theorem 1.2: For any u; v; w in Rn and k in R: (i) ðu þ vÞ w ¼ u w þ v w, (ii) ðkuÞ v ¼ kðu vÞ, (iii) u v ¼ v u, (iv) u u 0, and u u ¼ 0 iff u ¼ 0. Let u ¼ ðu1; u2; . . . ; unÞ, v ¼ ðv1; v2; . . . ; vnÞ, w ¼ ðw1; w2; . . . ; wnÞ. (i) Because u þ v ¼ ðu1 þ v1; u2 þ v2; . . . ; un þ vnÞ, ðu þ vÞ w ¼ ðu1 þ v1Þw1 þ ðu2 þ v2Þw2 þ þ ðun þ vnÞwn ¼ u1w1 þ v1w1 þ u2w2 þ þ unwn þ vnwn ¼ ðu1w1 þ u2w2 þ þ unwnÞ þ ðv1w1 þ v2w2 þ þ vnwnÞ ¼ u w þ v w (ii) Because ku ¼ ðku1; ku2; . . . ; kunÞ, ðkuÞ v ¼ ku1v1 þ ku2v2 þ þ kunvn ¼ kðu1v1 þ u2v2 þ þ unvnÞ ¼ kðu vÞ (iii) u v ¼ u1v1 þ u2v2 þ þ unvn ¼ v1u1 þ v2u2 þ þ vnun ¼ v u (iv) Because u2 i is nonnegative for each i, and because the sum of nonnegative real numbers is nonnegative, u u ¼ u2 1 þ u2 2 þ þ u2 n 0 Furthermore, u u ¼ 0 iff ui ¼ 0 for each i, that is, iff u ¼ 0. 16 CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors 1.14. Prove Theorem 1.3 (Schwarz): ju vj kukkvk. For any real number t, and using Theorem 1.2, we have 0 ðtu þ vÞ ðtu þ vÞ ¼ t2ðu uÞ þ 2tðu vÞ þ ðv vÞ ¼ kuk2t2 þ 2ðu vÞt þ kvk2 Let a ¼ kuk2, b ¼ 2ðu vÞ, c ¼ kvk2. Then, for every value of t, at2 þ bt þ c 0. This means that the quadratic polynomial cannot have two real roots. This implies that the discriminant D ¼ b2 4ac 0 or, equivalently, b2 4ac. Thus, 4ðu vÞ2 4kuk2kvk2 Dividing by 4 gives us our result. 1.15. Prove Theorem 1.4 (Minkowski): ku þ vk kuk þ kvk. By the Schwarz inequality and other properties of the dot product, ku þ vk2 ¼ ðu þ vÞ ðu þ vÞ ¼ ðu uÞ þ 2ðu vÞ þ ðv vÞ kuk2 þ 2kukkvk þ kvk2 ¼ ðkuk þ kvkÞ2 Taking the square root of both sides yields the desired inequality. Points, Lines, Hyperplanes in Rn Here we distinguish between an n-tuple Pða1; a2; . . . ; anÞ viewed as a point in Rn and an n-tuple u ¼ ½c1; c2; . . . ; cn viewed as a vector (arrow) from the origin O to the point Cðc1; c2; . . . ; cnÞ. 1.16. Find the vector u identiﬁed with the directed line segment PQ ! for the points: (a) Pð1; 2; 4Þ and Qð6; 1; 5Þ in R3, (b) Pð2; 3; 6; 5Þ and Qð7; 1; 4; 8Þ in R4. (a) u ¼ PQ ! ¼ Q P ¼ ½6 1; 1 ð2Þ; 5 4 ¼ ½5; 3; 9 (b) u ¼ PQ ! ¼ Q P ¼ ½7 2; 1 3; 4 þ 6; 8 5 ¼ ½5; 2; 10; 13 1.17. Find an equation of the hyperplane H in R4 that passes through Pð3; 4; 1; 2Þ and is normal to u ¼ ½2; 5; 6; 3. The coefﬁcients of the unknowns of an equation of H are the components of the normal vector u. Thus, an equation of H is of the form 2x1 þ 5x2 6x3 3x4 ¼ k. Substitute P into this equation to obtain k ¼ 26. Thus, an equation of H is 2x1 þ 5x2 6x3 3x4 ¼ 26. 1.18. Find an equation of the plane H in R3 that contains Pð1; 3; 4Þ and is parallel to the plane H0 determined by the equation 3x 6y þ 5z ¼ 2. The planes H and H0 are parallel if and only if their normal directions are parallel or antiparallel (opposite direction). Hence, an equation of H is of the form 3x 6y þ 5z ¼ k. Substitute P into this equation to obtain k ¼ 1. Then an equation of H is 3x 6y þ 5z ¼ 1. 1.19. Find a parametric representation of the line L in R4 passing through Pð4; 2; 3; 1Þ in the direction of u ¼ ½2; 5; 7; 8. Here L consists of the points XðxiÞ that satisfy X ¼ P þ tu or xi ¼ ait þ bi or LðtÞ ¼ ðait þ biÞ where the parameter t takes on all real values. Thus we obtain x1 ¼ 4 þ 2t; x2 ¼ 2 þ 2t; x3 ¼ 3 7t; x4 ¼ 1 þ 8t or LðtÞ ¼ ð4 þ 2t; 2 þ 2t; 3 7t; 1 þ 8tÞ CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors 17 1.20. Let C be the curve FðtÞ ¼ ðt2; 3t 2; t3; t2 þ 5Þ in R4, where 0 t 4. (a) Find the point P on C corresponding to t ¼ 2. (b) Find the initial point Q and terminal point Q 0 of C. (c) Find the unit tangent vector T to the curve C when t ¼ 2. (a) Substitute t ¼ 2 into FðtÞ to get P ¼ f ð2Þ ¼ ð4; 4; 8; 9Þ. (b) The parameter t ranges from t ¼ 0 to t ¼ 4. Hence, Q ¼ f ð0Þ ¼ ð0; 2; 0; 5Þ and Q 0 ¼ Fð4Þ ¼ ð16; 10; 64; 21Þ. (c) Take the derivative of FðtÞ—that is, of each component of FðtÞ—to obtain a vector V that is tangent to the curve: VðtÞ ¼ dFðtÞ dt ¼ ½2t; 3; 3t2; 2t Now ﬁnd V when t ¼ 2; that is, substitute t ¼ 2 in the equation for VðtÞ to obtain V ¼ Vð2Þ ¼ ½4; 3; 12; 4. Then normalize V to obtain the desired unit tangent vector T. We have kVk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 16 þ 9 þ 144 þ 16 p ¼ ﬃﬃﬃﬃﬃﬃﬃ ﬃ 185 p and T ¼ 4 ﬃﬃﬃﬃﬃﬃﬃ ﬃ 185 p ; 3 ﬃﬃﬃﬃﬃﬃﬃ ﬃ 185 p ; 12 ﬃﬃﬃﬃﬃﬃﬃ ﬃ 185 p ; 4 ﬃﬃﬃﬃﬃﬃﬃ ﬃ 185 p Spatial Vectors (Vectors in R3), ijk Notation, Cross Product 1.21. Let u ¼ 2i 3j þ 4k, v ¼ 3i þ j 2k, w ¼ i þ 5j þ 3k. Find: (a) u þ v, (b) 2u 3v þ 4w, (c) u v and u w, (d) kuk and kvk. Treat the coefﬁcients of i, j, k just like the components of a vector in R3. (a) Add corresponding coefﬁcients to get u þ v ¼ 5i 2j 2k. (b) First perform the scalar multiplication and then the vector addition: 2u 3v þ 4w ¼ ð4i 6j þ 8kÞ þ ð9i þ 3j þ 6kÞ þ ð4i þ 20j þ 12kÞ ¼ i þ 17j þ 26k (c) Multiply corresponding coefﬁcients and then add: u v ¼ 6 3 8 ¼ 5 and u w ¼ 2 15 þ 12 ¼ 1 (d) The norm is the square root of the sum of the squares of the coefﬁcients: kuk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 4 þ 9 þ 16 p ¼ ﬃﬃﬃﬃﬃ 29 p and kvk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 9 þ 1 þ 4 p ¼ ﬃﬃﬃﬃﬃ 14 p 1.22. Find the (parametric) equation of the line L: (a) through the points Pð1; 3; 2Þ and Qð2; 5; 6Þ; (b) containing the point Pð1; 2; 4Þ and perpendicular to the plane H given by the equation 3x þ 5y þ 7z ¼ 15: (a) First ﬁnd v ¼ PQ ! ¼ Q P ¼ ½1; 2; 8 ¼ i þ 2j 8k. Then LðtÞ ¼ ðt þ 1; 2t þ 3; 8t þ 2Þ ¼ ðt þ 1Þi þ ð2t þ 3Þj þ ð8t þ 2Þk (b) Because L is perpendicular to H, the line L is in the same direction as the normal vector N ¼ 3i þ 5j þ 7k to H. Thus, LðtÞ ¼ ð3t þ 1; 5t 2; 7t þ 4Þ ¼ ð3t þ 1Þi þ ð5t 2Þj þ ð7t þ 4Þk 1.23. Let S be the surface xy2 þ 2yz ¼ 16 in R3. (a) Find the normal vector Nðx; y; zÞ to the surface S. (b) Find the tangent plane H to S at the point Pð1; 2; 3Þ. 18 CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors (a) The formula for the normal vector to a surface Fðx; y; zÞ ¼ 0 is Nðx; y; zÞ ¼ Fxi þ Fyj þ Fzk where Fx, Fy, Fz are the partial derivatives. Using Fðx; y; zÞ ¼ xy2 þ 2yz 16, we obtain Fx ¼ y2; Fy ¼ 2xy þ 2z; Fz ¼ 2y Thus, Nðx; y; zÞ ¼ y2i þ ð2xy þ 2zÞj þ 2yk. (b) The normal to the surface S at the point P is NðPÞ ¼ Nð1; 2; 3Þ ¼ 4i þ 10j þ 4k Hence, N ¼ 2i þ 5j þ 2k is also normal to S at P. Thus an equation of H has the form 2x þ 5y þ 2z ¼ c. Substitute P in this equation to obtain c ¼ 18. Thus the tangent plane H to S at P is 2x þ 5y þ 2z ¼ 18. 1.24. Evaluate the following determinants and negative of determinants of order two: (a) (i) 3 4 5 9 , (ii) 2 1 4 3 , (iii) 4 5 3 2 (b) (i) 3 6 4 2 , (ii) 7 5 3 2 , (iii) 4 1 8 3 Use a b c d ¼ ad bc and a b c d ¼ bc ad. Thus, (a) (i) 27 20 ¼ 7, (ii) 6 þ 4 ¼ 10, (iii) 8 þ 15 ¼ 7: (b) (i) 24 6 ¼ 18, (ii) 15 14 ¼ 29, (iii) 8 þ 12 ¼ 4: 1.25. Let u ¼ 2i 3j þ 4k, v ¼ 3i þ j 2k, w ¼ i þ 5j þ 3k. Find: (a) u v, (b) u w (a) Use 2 3 4 3 1 2 to get u v ¼ ð6 4Þi þ ð12 þ 4Þj þ ð2 þ 9Þk ¼ 2i þ 16j þ 11k: (b) Use 2 3 4 1 5 3 to get u w ¼ ð9 20Þi þ ð4 6Þj þ ð10 þ 3Þk ¼ 29i 2j þ 13k: 1.26. Find u v, where: (a) u ¼ ð1; 2; 3Þ, v ¼ ð4; 5; 6Þ; (b) u ¼ ð4; 7; 3Þ, v ¼ ð6; 5; 2Þ. (a) Use 1 2 3 4 5 6 to get u v ¼ ½12 15; 12 6; 5 8 ¼ ½3; 6; 3: (b) Use 4 7 3 6 5 2 to get u v ¼ ½14 þ 15; 18 þ 8; 20 42 ¼ ½29; 26; 22: 1.27. Find a unit vector u orthogonal to v ¼ ½1; 3; 4 and w ¼ ½2; 6; 5. First ﬁnd v w, which is orthogonal to v and w. The array 1 3 4 2 6 5 gives v w ¼ ½15 þ 24; 8 þ 5; 6 61 ¼ ½9; 13; 12: Normalize v w to get u ¼ ½9= ﬃﬃﬃﬃﬃﬃﬃ ﬃ 394 p , 13= ﬃﬃﬃﬃﬃﬃﬃ ﬃ 394 p , 12= ﬃﬃﬃﬃﬃﬃﬃ ﬃ 394 p : 1.28. Let u ¼ ða1; a2; a3Þ and v ¼ ðb1; b2; b3Þ so u v ¼ ða2b3 a3b2; a3b1 a1b3; a1b2 a2b1Þ. Prove: (a) u v is orthogonal to u and v [Theorem 1.5(a)]. (b) ku vk2 ¼ ðu uÞðv vÞ ðu vÞ2 (Lagrange’s identity). CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors 19 (a) We have u ðu vÞ ¼ a1ða2b3 a3b2Þ þ a2ða3b1 a1b3Þ þ a3ða1b2 a2b1Þ ¼ a1a2b3 a1a3b2 þ a2a3b1 a1a2b3 þ a1a3b2 a2a3b1 ¼ 0 Thus, u v is orthogonal to u. Similarly, u v is orthogonal to v. (b) We have ku vk2 ¼ ða2b3 a3b2Þ2 þ ða3b1 a1b3Þ2 þ ða1b2 a2b1Þ2 ð1Þ ðu uÞðv vÞ ðu vÞ2 ¼ ða2 1 þ a2 2 þ a2 3Þðb2 1 þ b2 2 þ b2 3Þ ða1b1 þ a2b2 þ a3b3Þ2 ð2Þ Expansion of the right-hand sides of (1) and (2) establishes the identity. Complex Numbers, Vectors in Cn 1.29. Suppose z ¼ 5 þ 3i and w ¼ 2 4i. Find: (a) z þ w, (b) z w, (c) zw. Use the ordinary rules of algebra together with i2 ¼ 1 to obtain a result in the standard form a þ bi. (a) z þ w ¼ ð5 þ 3iÞ þ ð2 4iÞ ¼ 7 i (b) z w ¼ ð5 þ 3iÞ ð2 4iÞ ¼ 5 þ 3i 2 þ 4i ¼ 3 þ 7i (c) zw ¼ ð5 þ 3iÞð2 4iÞ ¼ 10 14i 12i2 ¼ 10 14i þ 12 ¼ 22 14i 1.30. Simplify: (a) ð5 þ 3iÞð2 7iÞ, (b) ð4 3iÞ2, (c) ð1 þ 2iÞ3. (a) ð5 þ 3iÞð2 7iÞ ¼ 10 þ 6i 35i 21i2 ¼ 31 29i (b) ð4 3iÞ2 ¼ 16 24i þ 9i2 ¼ 7 24i (c) ð1 þ 2iÞ3 ¼ 1 þ 6i þ 12i2 þ 8i3 ¼ 1 þ 6i 12 8i ¼ 11 2i 1.31. Simplify: (a) i0; i3; i4, (b) i5; i6; i7; i8, (c) i39; i174, i252, i317: (a) i0 ¼ 1, i3 ¼ i2ðiÞ ¼ ð1ÞðiÞ ¼ i; i4 ¼ ði2Þði2Þ ¼ ð1Þð1Þ ¼ 1 (b) i5 ¼ ði4ÞðiÞ ¼ ð1ÞðiÞ ¼ i, i6 ¼ ði4Þði2Þ ¼ ð1Þði2Þ ¼ i2 ¼ 1, i7 ¼ i3 ¼ i, i8 ¼ i4 ¼ 1 (c) Using i4 ¼ 1 and in ¼ i4qþr ¼ ði4Þqir ¼ 1qir ¼ ir, divide the exponent n by 4 to obtain the remainder r: i39 ¼ i4ð9Þþ3 ¼ ði4Þ9i3 ¼ 19i3 ¼ i3 ¼ i; i174 ¼ i2 ¼ 1; i252 ¼ i0 ¼ 1; i317 ¼ i1 ¼ i 1.32. Find the complex conjugate of each of the following: (a) 6 þ 4i, 7 5i, 4 þ i, 3 i, (b) 6, 3, 4i, 9i. (a) 6 þ 4i ¼ 6 4i, 7 5i ¼ 7 þ 5i, 4 þ i ¼ 4 i, 3 i ¼ 3 þ i (b) 6 ¼ 6, 3 ¼ 3, 4i ¼ 4i, 9i ¼ 9i (Note that the conjugate of a real number is the original number, but the conjugate of a pure imaginary number is the negative of the original number.) 1.33. Find z z and jzj when z ¼ 3 þ 4i. For z ¼ a þ bi, use z z ¼ a2 þ b2 and z ¼ ﬃﬃﬃﬃ z z p ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ a2 þ b2 p . z z ¼ 9 þ 16 ¼ 25; jzj ¼ ﬃﬃﬃﬃﬃ 25 p ¼ 5 1.34. Simpify 2 7i 5 þ 3i : To simplify a fraction z=w of complex numbers, multiply both numerator and denominator by w, the conjugate of the denominator: 2 7i 5 þ 3i ¼ ð2 7iÞð5 3iÞ ð5 þ 3iÞð5 3iÞ ¼ 11 41i 34 ¼ 11 34 41 34 i 20 CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors 1.35. Prove: For any complex numbers z, w 2 C, (i) z þ w ¼ z þ w, (ii) zw ¼ z w, (iii) z ¼ z. Suppose z ¼ a þ bi and w ¼ c þ di where a; b; c; d 2 R. (i) z þ w ¼ ða þ biÞ þ ðc þ diÞ ¼ ða þ cÞ þ ðb þ dÞi ¼ ða þ cÞ ðb þ dÞi ¼ a þ c bi di ¼ ða biÞ þ ðc diÞ ¼ z þ w (ii) zw ¼ ða þ biÞðc þ diÞ ¼ ðac bdÞ þ ðad þ bcÞi ¼ ðac bdÞ ðad þ bcÞi ¼ ða biÞðc diÞ ¼ z w (iii) z ¼ a þ bi ¼ a bi ¼ a ðbÞi ¼ a þ bi ¼ z 1.36. Prove: For any complex numbers z; w 2 C, jzwj ¼ jzjjwj. By (ii) of Problem 1.35, jzwj2 ¼ ðzwÞðzwÞ ¼ ðzwÞð z wÞ ¼ ðz zÞðw wÞ ¼ jzj2jwj2 The square root of both sides gives us the desired result. 1.37. Prove: For any complex numbers z; w 2 C, jz þ wj jzj þ jwj. Suppose z ¼ a þ bi and w ¼ c þ di where a; b; c; d 2 R. Consider the vectors u ¼ ða; bÞ and v ¼ ðc; dÞ in R2. Note that jzj ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ a2 þ b2 p ¼ kuk; jwj ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ c2 þ d2 p ¼ kvk and jz þ wj ¼ jða þ cÞ þ ðb þ dÞij ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ða þ cÞ2 þ ðb þ dÞ2 q ¼ kða þ c; b þ dÞk ¼ ku þ vk By Minkowski’s inequality (Problem 1.15), ku þ vk kuk þ kvk, and so jz þ wj ¼ ku þ vk kuk þ kvk ¼ jzj þ jwj 1.38. Find the dot products u v and v u where: (a) u ¼ ð1 2i; 3 þ iÞ, v ¼ ð4 þ 2i; 5 6iÞ; (b) u ¼ ð3 2i; 4i; 1 þ 6iÞ, v ¼ ð5 þ i; 2 3i; 7 þ 2iÞ. Recall that conjugates of the second vector appear in the dot product ðz1; . . . ; znÞ ðw1; . . . ; wnÞ ¼ z1 w1 þ þ zn wn (a) u v ¼ ð1 2iÞð4 þ 2iÞ þ ð3 þ iÞð5 6iÞ ¼ ð1 2iÞð4 2iÞ þ ð3 þ iÞð5 þ 6iÞ ¼ 10i þ 9 þ 23i ¼ 9 þ 13i v u ¼ ð4 þ 2iÞð1 2iÞ þ ð5 6iÞð3 þ iÞ ¼ ð4 þ 2iÞð1 þ 2iÞ þ ð5 6iÞð3 iÞ ¼ 10i þ 9 23i ¼ 9 13i (b) u v ¼ ð3 2iÞð5 þ iÞ þ ð4iÞð2 3iÞ þ ð1 þ 6iÞð7 þ 2iÞ ¼ ð3 2iÞð5 iÞ þ ð4iÞð2 þ 3iÞ þ ð1 þ 6iÞð7 2iÞ ¼ 20 þ 35i v u ¼ ð5 þ iÞð3 2iÞ þ ð2 3iÞð4iÞ þ ð7 þ 2iÞð1 þ 6iÞ ¼ ð5 þ iÞð3 þ 2iÞ þ ð2 3iÞð4iÞ þ ð7 þ 2iÞð1 6iÞ ¼ 20 35i In both cases, v u ¼ u v. This holds true in general, as seen in Problem 1.40. 1.39. Let u ¼ ð7 2i; 2 þ 5iÞ and v ¼ ð1 þ i; 3 6iÞ. Find: (a) u þ v, (b) 2iu, (c) ð3 iÞv, (d) u v, (e) kuk and kvk. (a) u þ v ¼ ð7 2i þ 1 þ i; 2 þ 5i 3 6iÞ ¼ ð8 i; 1 iÞ (b) 2iu ¼ ð14i 4i2; 4i þ 10i2Þ ¼ ð4 þ 14i; 10 þ 4iÞ (c) ð3 iÞv ¼ ð3 þ 3i i i2; 9 18i þ 3i þ 6i2Þ ¼ ð4 þ 2i; 15 15iÞ CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors 21 (d) u v ¼ ð7 2iÞð1 þ iÞ þ ð2 þ 5iÞð3 6iÞ ¼ ð7 2iÞð1 iÞ þ ð2 þ 5iÞð3 þ 6iÞ ¼ 5 9i 36 3i ¼ 31 12i (e) kuk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ 72 þ ð2Þ2 þ 22 þ 52 q ¼ ﬃﬃﬃﬃﬃ 82 p and kvk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 12 þ 12 þ ð3Þ2 þ ð6Þ2 q ¼ ﬃﬃﬃﬃﬃ 47 p 1.40. Prove: For any vectors u; v 2 Cn and any scalar z 2 C, (i) u v ¼ v u, (ii) ðzuÞ v ¼ zðu vÞ, (iii) u ðzvÞ ¼ zðu vÞ. Suppose u ¼ ðz1; z2; . . . ; znÞ and v ¼ ðw1; w2; . . . ; wnÞ. (i) Using the properties of the conjugate, v u ¼ w1 z1 þ w2 z2 þ þ wn zn ¼ w1 z1 þ w2 z2 þ þ wn zn ¼ w1z1 þ w2z2 þ þ wnzn ¼ z1 w1 þ z2 w2 þ þ zn wn ¼ u v (ii) Because zu ¼ ðzz1; zz2; . . . ; zznÞ, ðzuÞ v ¼ zz1 w1 þ zz2 w2 þ þ zzn wn ¼ zðz1 w1 þ z2 w2 þ þ zn wnÞ ¼ zðu vÞ (Compare with Theorem 1.2 on vectors in Rn.) (iii) Using (i) and (ii), u ðzvÞ ¼ ðzvÞ u ¼ zðv uÞ ¼ zðv uÞ ¼ zðu vÞ SUPPLEMENTARY PROBLEMS Vectors in Rn 1.41. Let u ¼ ð1; 2; 4Þ, v ¼ ð3; 5; 1Þ, w ¼ ð2; 1; 3Þ. Find: (a) 3u 2v; (b) 5u þ 3v 4w; (c) u v, u w, v w; (d) kuk, kvk; (e) cos y, where y is the angle between u and v; (f ) dðu; vÞ; (g) projðu; vÞ. 1.42. Repeat Problem 1.41 for vectors u ¼ 1 3 4 2 4 3 5, v ¼ 2 1 5 2 4 3 5, w ¼ 3 2 6 2 4 3 5. 1.43. Let u ¼ ð2; 5; 4; 6; 3Þ and v ¼ ð5; 2; 1; 7; 4Þ. Find: (a) 4u 3v; (b) 5u þ 2v; (c) u v; (d) kuk and kvk; (e) projðu; vÞ; ( f ) dðu; vÞ. 1.44. Normalize each vector: (a) u ¼ ð5; 7Þ; (b) v ¼ ð1; 2; 2; 4Þ; (c) w ¼ 1 2 ; 1 3 ; 3 4 . 1.45. Let u ¼ ð1; 2; 2Þ, v ¼ ð3; 12; 4Þ, and k ¼ 3. (a) Find kuk, kvk, ku þ vk, kkuk: (b) Verify that kkuk ¼ jkjkuk and ku þ vk kuk þ kvk. 1.46. Find x and y where: (a) ðx; y þ 1Þ ¼ ðy 2; 6Þ; (b) xð2; yÞ ¼ yð1; 2Þ. 1.47. Find x; y; z where ðx; y þ 1; y þ zÞ ¼ ð2x þ y; 4; 3zÞ. 22 CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors 1.48. Write v ¼ ð2; 5Þ as a linear combination of u1 and u2, where: (a) u1 ¼ ð1; 2Þ and u2 ¼ ð3; 5Þ; (b) u1 ¼ ð3; 4Þ and u2 ¼ ð2; 3Þ. 1.49. Write v ¼ 9 3 16 2 4 3 5 as a linear combination of u1 ¼ 1 3 3 2 4 3 5, u2 ¼ 2 5 1 2 4 3 5, u3 ¼ 4 2 3 2 4 3 5. 1.50. Find k so that u and v are orthogonal, where: (a) u ¼ ð3; k; 2Þ, v ¼ ð6; 4; 3Þ; (b) u ¼ ð5; k; 4; 2Þ, v ¼ ð1; 3; 2; 2kÞ; (c) u ¼ ð1; 7; k þ 2; 2Þ, v ¼ ð3; k; 3; kÞ. Located Vectors, Hyperplanes, Lines in Rn 1.51. Find the vector v identiﬁed with the directed line segment PQ ! for the points: (a) Pð2; 3; 7Þ and Qð1; 6; 5Þ in R3; (b) Pð1; 8; 4; 6Þ and Qð3; 5; 2; 4Þ in R4. 1.52. Find an equation of the hyperplane H in R4 that: (a) contains Pð1; 2; 3; 2Þ and is normal to u ¼ ½2; 3; 5; 6; (b) contains Pð3; 1; 2; 5Þ and is parallel to 2x1 3x2 þ 5x3 7x4 ¼ 4. 1.53. Find a parametric representation of the line in R4 that: (a) passes through the points Pð1; 2; 1; 2Þ and Qð3; 5; 7; 9Þ; (b) passes through Pð1; 1; 3; 3Þ and is perpendicular to the hyperplane 2x1 þ 4x2 þ 6x3 8x4 ¼ 5. Spatial Vectors (Vectors in R3), ijk Notation 1.54. Given u ¼ 3i 4j þ 2k, v ¼ 2i þ 5j 3k, w ¼ 4i þ 7j þ 2k. Find: (a) 2u 3v; (b) 3u þ 4v 2w; (c) u v, u w, v w; (d) kuk, kvk, kwk. 1.55. Find the equation of the plane H: (a) with normal N ¼ 3i 4j þ 5k and containing the point Pð1; 2; 3Þ; (b) parallel to 4x þ 3y 2z ¼ 11 and containing the point Qð2; 1; 3Þ. 1.56. Find the (parametric) equation of the line L: (a) through the point Pð2; 5; 3Þ and in the direction of v ¼ 4i 5j þ 7k; (b) perpendicular to the plane 2x 3y þ 7z ¼ 4 and containing Pð1; 5; 7Þ. 1.57. Consider the following curve C in R3 where 0 t 5: FðtÞ ¼ t3i t2j þ ð2t 3Þk (a) Find the point P on C corresponding to t ¼ 2. (b) Find the initial point Q and the terminal point Q 0. (c) Find the unit tangent vector T to the curve C when t ¼ 2. 1.58. Consider a moving body B whose position at time t is given by RðtÞ ¼ t2i þ t3j þ 3tk. [Then VðtÞ ¼ dRðtÞ=dt and AðtÞ ¼ dVðtÞ=dt denote, respectively, the velocity and acceleration of B.] When t ¼ 1, ﬁnd for the body B: (a) position; (b) velocity v; (c) speed s; (d) acceleration a. CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors 23 1.59. Find a normal vector N and the tangent plane H to each surface at the given point: (a) surface x2y þ 3yz ¼ 20 and point Pð1; 3; 2Þ; (b) surface x2 þ 3y2 5z2 ¼ 160 and point Pð3; 2; 1Þ: Cross Product 1.60. Evaluate the following determinants and negative of determinants of order two: (a) 2 5 3 6 ; 3 6 1 4 ; 4 2 7 3 (b) 6 4 7 5 ; 1 3 2 4 ; 8 3 6 2 1.61. Given u ¼ 3i 4j þ 2k, v ¼ 2i þ 5j 3k, w ¼ 4i þ 7j þ 2k, ﬁnd: (a) u v, (b) u w, (c) v w. 1.62. Given u ¼ ½2; 1; 3, v ¼ ½4; 2; 2, w ¼ ½1; 1; 5, ﬁnd: (a) u v, (b) u w, (c) v w. 1.63. Find the volume V of the parallelopiped formed by the vectors u; v; w appearing in: (a) Problem 1.60 (b) Problem 1.61. 1.64. Find a unit vector u orthogonal to: (a) v ¼ ½1; 2; 3 and w ¼ ½1; 1; 2; (b) v ¼ 3i j þ 2k and w ¼ 4i 2j k. 1.65. Prove the following properties of the cross product: (a) u v ¼ ðv uÞ (d) u ðv þ wÞ ¼ ðu vÞ þ ðu wÞ (b) u u ¼ 0 for any vector u (e) ðv þ wÞ u ¼ ðv uÞ þ ðw uÞ (c) ðkuÞ v ¼ kðu vÞ ¼ u ðkvÞ ( f ) ðu vÞ w ¼ ðu wÞv ðv wÞu Complex Numbers 1.66. Simplify: (a) ð4 7iÞð9 þ 2iÞ; (b) ð3 5iÞ2; (c) 1 4 7i; (d) 9 þ 2i 3 5i; (e) ð1 iÞ3. 1.67. Simplify: (a) 1 2i ; (b) 2 þ 3i 7 3i ; (c) i15; i25; i34; (d) 1 3 i 2 . 1.68. Let z ¼ 2 5i and w ¼ 7 þ 3i. Find: (a) v þ w; (b) zw; (c) z=w; (d) z; w; (e) jzj, jwj. 1.69. Show that for complex numbers z and w: (a) Re z ¼ 1 2 ðz þ zÞ, (b) Im z ¼ 1 2 ðz z), (c) zw ¼ 0 implies z ¼ 0 or w ¼ 0. Vectors in Cn 1.70. Let u ¼ ð1 þ 7i; 2 6iÞ and v ¼ ð5 2i; 3 4iÞ. Find: (a) u þ v (b) ð3 þ iÞu (c) 2iu þ ð4 þ 7iÞv (d) u v (e) kuk and kvk. 24 CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors 1.71. Prove: For any vectors u; v; w in Cn: (a) ðu þ vÞ w ¼ u w þ v w, (b) w ðu þ vÞ ¼ w u þ w v. 1.72. Prove that the norm in Cn satisﬁes the following laws: ½N1 For any vector u, kuk 0; and kuk ¼ 0 if and only if u ¼ 0. ½N2 For any vector u and complex number z, kzuk ¼ jzjkuk. ½N3 For any vectors u and v, ku þ vk kuk þ kvk. ANSWERS TO SUPPLEMENTARY PROBLEMS 1.41. (a) ð3; 16; 4Þ; (b) (6,1,35); (c) 3; 12; 8; (d) ﬃﬃﬃﬃﬃ 21 p , ﬃﬃﬃﬃﬃ 35 p , ﬃﬃﬃﬃﬃ 14 p ; (e) 3= ﬃﬃﬃﬃﬃ 21 p ﬃﬃﬃﬃﬃ 35 p ; ( f ) ﬃﬃﬃﬃﬃ 62 p ; (g) 3 35 ð3; 5; 1Þ ¼ ð 9 35, 15 35, 3 35) 1.42. (Column vectors) (a) ð1; 7; 22Þ; (b) ð1; 26; 29Þ; (c) 15; 27; 34; (d) ﬃﬃﬃﬃﬃ 26 p , ﬃﬃﬃﬃﬃ 30 p ; (e) 15=ð ﬃﬃﬃﬃﬃ 26 p ﬃﬃﬃﬃﬃ 30 p Þ; ( f ) ﬃﬃﬃﬃﬃ 86 p ; (g) 15 30 v ¼ ð1; 1 2 ; 5 2Þ 1.43. (a) ð13; 14; 13; 45; 0Þ; (b) ð20; 29; 22; 16; 23Þ; (c) 6; (d) ﬃﬃﬃﬃﬃ 90 p ; ﬃﬃﬃﬃﬃ 95 p ; (e) 6 95 v; ( f ) ﬃﬃﬃﬃﬃﬃﬃ ﬃ 167 p 1.44. (a) ð5= ﬃﬃﬃﬃﬃ 76 p ; 9= ﬃﬃﬃﬃﬃ 76 p Þ; (b) ð1 5 ; 2 5 ; 2 5 ; 4 5Þ; (c) ð6= ﬃﬃﬃﬃﬃﬃﬃ ﬃ 133 p ; 4 ﬃﬃﬃﬃﬃﬃﬃ ﬃ 133 p ; 9 ﬃﬃﬃﬃﬃﬃﬃ ﬃ 133 p Þ 1.45. (a) 3; 13; ﬃﬃﬃﬃﬃﬃﬃ ﬃ 120 p ; 9 1.46. (a) x ¼ 3; y ¼ 5; (b) x ¼ 0; y ¼ 0, and x ¼ 1; y ¼ 2 1.47. x ¼ 3; y ¼ 3; z ¼ 3 2 1.48. (a) v ¼ 5u1 u2; (b) v ¼ 16u1 23u2 1.49. v ¼ 3u1 u2 þ 2u3 1.50. (a) 6; (b) 3; (c) 3 2 1.51. (a) v ¼ ½1; 9; 2; (b) [2; 3; 6; 10] 1.52. (a) 2x1 þ 3x2 5x3 þ 6x4 ¼ 35; (b) 2x1 3x2 þ 5x3 7x4 ¼ 16 1.53. (a) ½2t þ 1; 7t þ 2; 6t þ 1; 11t þ 2; (b) ½2t þ 1; 4t þ 1; 6t þ 3; 8t þ 3 1.54. (a) 23j þ 13k; (b) 9i 6j 10k; (c) 20; 12; 37; (d) ﬃﬃﬃﬃﬃ 29 p ; ﬃﬃﬃﬃﬃ 38 p ; ﬃﬃﬃﬃﬃ 69 p 1.55. (a) 3x 4y þ 5z ¼ 20; (b) 4x þ 3y 2z ¼ 1 1.56. (a) ½4t þ 2; 5t þ 5; 7t 3; (b) ½2t þ 1; 3t 5; 7t þ 7 1.57. (a) P ¼ Fð2Þ ¼ 8i 4j þ k; (b) Q ¼ Fð0Þ ¼ 3k, Q0 ¼ Fð5Þ ¼ 125i 25j þ 7k; (c) T ¼ ð6i 2j þ kÞ= ﬃﬃﬃﬃﬃ 41 p 1.58. (a) i þ j þ 2k; (b) 2i þ 3j þ 2k; (c) ﬃﬃﬃﬃﬃ 17 p ; (d) 2i þ 6j 1.59. (a) N ¼ 6i þ 7j þ 9k, 6x þ 7y þ 9z ¼ 45; (b) N ¼ 6i 12j 10k, 3x 6y 5z ¼ 16 CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors 25 1.60. (a) 3; 6; 26; (b) 2; 10; 34 1.61. (a) 2i þ 13j þ 23k; (b) 22i þ 2j þ 37k; (c) 31i 16j 6k 1.62. (a) ½5; 8; 6; (b) ½2; 7; 1; (c) ½7; 18; 5 1.63. (a) 143; (b) 17 1.64. (a) ð7; 1; 3Þ= ﬃﬃﬃﬃﬃ 59 p ; (b) ð5i þ 11j 2kÞ= ﬃﬃﬃﬃﬃﬃﬃ ﬃ 150 p 1.66. (a) 50 55i; (b) 16 30i; (c) 1 65 ð4 þ 7iÞ; (d) 1 2 ð1 þ 3iÞ; (e) 2 2i 1.67. (a) 1 2 i; (b) 1 58 ð5 þ 27iÞ; (c) 1; i; 1; (d) 1 50 ð4 þ 3iÞ 1.68. (a) 9 2i; (b) 29 29i; (c) 1 61 ð1 41iÞ; (d) 2 þ 5i, 7 3i; (e) ﬃﬃﬃﬃﬃ 29 p , ﬃﬃﬃﬃﬃ 58 p 1.69. (c) Hint: If zw ¼ 0, then jzwj ¼ jzjjwj ¼ j0j ¼ 0 1.70. (a) ð6 þ 5i, 5 10iÞ; (b) ð4 þ 22i, 12 16iÞ; (c) ð8 41i, 4 33iÞ; (d) 12 þ 2i; (e) ﬃﬃﬃﬃﬃ 90 p , ﬃﬃﬃﬃﬃ 54 p 26 CHAPTER 1 Vectors in Rn and Cn, Spatial Vectors Algebra of Matrices 2.1 Introduction This chapter investigates matrices and algebraic operations deﬁned on them. These matrices may be viewed as rectangular arrays of elements where each entry depends on two subscripts (as compared with vectors, where each entry depended on only one subscript). Systems of linear equations and their solutions (Chapter 3) may be efﬁciently investigated using the language of matrices. Furthermore, certain abstract objects introduced in later chapters, such as ‘‘change of basis,’’ ‘‘linear transformations,’’ and ‘‘quadratic forms,’’ can be represented by these matrices (rectangular arrays). On the other hand, the abstract treatment of linear algebra presented later on will give us new insight into the structure of these matrices. The entries in our matrices will come from some arbitrary, but ﬁxed, ﬁeld K. The elements of K are called numbers or scalars. Nothing essential is lost if the reader assumes that K is the real ﬁeld R. 2.2 Matrices A matrix A over a ﬁeld K or, simply, a matrix A (when K is implicit) is a rectangular array of scalars usually presented in the following form: A ¼ a11 a12 . . . a1n a21 a22 . . . a2n am1 am2 . . . amn 2 6 6 4 3 7 7 5 The rows of such a matrix A are the m horizontal lists of scalars: ða11; a12; . . . ; a1nÞ; ða21; a22; . . . ; a2nÞ; . . . ; ðam1; am2; . . . ; amnÞ and the columns of A are the n vertical lists of scalars: a11 a21 . . . am1 2 6 6 4 3 7 7 5; a12 a22 . . . am2 2 6 6 4 3 7 7 5; . . . ; a1n a2n . . . amn 2 6 6 4 3 7 7 5 Note that the element aij, called the ij-entry or ij-element, appears in row i and column j. We frequently denote such a matrix by simply writing A ¼ ½aij. A matrix with m rows and n columns is called an m by n matrix, written m n. The pair of numbers m and n is called the size of the matrix. Two matrices A and B are equal, written A ¼ B, if they have the same size and if corresponding elements are equal. Thus, the equality of two m n matrices is equivalent to a system of mn equalities, one for each corresponding pair of elements. A matrix with only one row is called a row matrix or row vector, and a matrix with only one column is called a column matrix or column vector. A matrix whose entries are all zero is called a zero matrix and will usually be denoted by 0. 27 CHAPTER 2 Matrices whose entries are all real numbers are called real matrices and are said to be matrices over R. Analogously, matrices whose entries are all complex numbers are called complex matrices and are said to be matrices over C. This text will be mainly concerned with such real and complex matrices. EXAMPLE 2.1 (a) The rectangular array A ¼ 1 4 5 0 3 2 is a 2 3 matrix. Its rows are ð1; 4; 5Þ and ð0; 3; 2Þ, and its columns are 1 0 ; 4 3 ; 5 2 (b) The 2 4 zero matrix is the matrix 0 ¼ 0 0 0 0 0 0 0 0 . (c) Find x; y; z; t such that x þ y 2z þ t x y z t ¼ 3 7 1 5 By deﬁnition of equality of matrices, the four corresponding entries must be equal. Thus, x þ y ¼ 3; x y ¼ 1; 2z þ t ¼ 7; z t ¼ 5 Solving the above system of equations yields x ¼ 2, y ¼ 1, z ¼ 4, t ¼ 1. 2.3 Matrix Addition and Scalar Multiplication Let A ¼ ½aij and B ¼ ½bij be two matrices with the same size, say m n matrices. The sum of A and B, written A þ B, is the matrix obtained by adding corresponding elements from A and B. That is, A þ B ¼ a11 þ b11 a12 þ b12 . . . a1n þ b1n a21 þ b21 a22 þ b22 . . . a2n þ b2n am1 þ bm1 am2 þ bm2 . . . amn þ bmn 2 6 6 4 3 7 7 5 The product of the matrix A by a scalar k, written k A or simply kA, is the matrix obtained by multiplying each element of A by k. That is, kA ¼ ka11 ka12 . . . ka1n ka21 ka22 . . . ka2n kam1 kam2 . . . kamn 2 6 6 4 3 7 7 5 Observe that A þ B and kA are also m n matrices. We also deﬁne A ¼ ð1ÞA and A B ¼ A þ ðBÞ The matrix A is called the negative of the matrix A, and the matrix A B is called the difference of A and B. The sum of matrices with different sizes is not deﬁned. 28 CHAPTER 2 Algebra of Matrices EXAMPLE 2.2 Let A ¼ 1 2 3 0 4 5 and B ¼ 4 6 8 1 3 7 . Then A þ B ¼ 1 þ 4 2 þ 6 3 þ 8 0 þ 1 4 þ ð3Þ 5 þ ð7Þ " # ¼ 5 4 11 1 1 2 " # 3A ¼ 3ð1Þ 3ð2Þ 3ð3Þ 3ð0Þ 3ð4Þ 3ð5Þ " # ¼ 3 6 9 0 12 15 " # 2A 3B ¼ 2 4 6 0 8 10 " # þ 12 18 24 3 9 21 " # ¼ 10 22 18 3 17 31 " # The matrix 2A 3B is called a linear combination of A and B. Basic properties of matrices under the operations of matrix addition and scalar multiplication follow. THEOREM 2.1: Consider any matrices A; B; C (with the same size) and any scalars k and k0. Then (i) ðA þ BÞ þ C ¼ A þ ðB þ CÞ, (v) kðA þ BÞ ¼ kA þ kB, (ii) A þ 0 ¼ 0 þ A ¼ A, (vi) ðk þ k0ÞA ¼ kA þ k0A, (iii) A þ ðAÞ ¼ ðAÞ þ A ¼ 0; (vii) ðkk0ÞA ¼ kðk0AÞ, (iv) A þ B ¼ B þ A, (viii) 1 A ¼ A. Note ﬁrst that the 0 in (ii) and (iii) refers to the zero matrix. Also, by (i) and (iv), any sum of matrices A1 þ A2 þ þ An requires no parentheses, and the sum does not depend on the order of the matrices. Furthermore, using (vi) and (viii), we also have A þ A ¼ 2A; A þ A þ A ¼ 3A; . . . and so on. The proof of Theorem 2.1 reduces to showing that the ij-entries on both sides of each matrix equation are equal. (See Problem 2.3.) Observe the similarity between Theorem 2.1 for matrices and Theorem 1.1 for vectors. In fact, the above operations for matrices may be viewed as generalizations of the corresponding operations for vectors. 2.4 Summation Symbol Before we deﬁne matrix multiplication, it will be instructive to ﬁrst introduce the summation symbol S (the Greek capital letter sigma). Suppose f ðkÞ is an algebraic expression involving the letter k. Then the expression P n k¼1 f ðkÞ or equivalently Pn k¼1 f ðkÞ has the following meaning. First we set k ¼ 1 in f ðkÞ, obtaining f ð1Þ Then we set k ¼ 2 in f ðkÞ, obtaining f ð2Þ, and add this to f ð1Þ, obtaining f ð1Þ þ f ð2Þ CHAPTER 2 Algebra of Matrices 29 Then we set k ¼ 3 in f ðkÞ, obtaining f ð3Þ, and add this to the previous sum, obtaining f ð1Þ þ f ð2Þ þ f ð3Þ We continue this process until we obtain the sum f ð1Þ þ f ð2Þ þ þ f ðnÞ Observe that at each step we increase the value of k by 1 until we reach n. The letter k is called the index, and 1 and n are called, respectively, the lower and upper limits. Other letters frequently used as indices are i and j. We also generalize our deﬁnition by allowing the sum to range from any integer n1 to any integer n2. That is, we deﬁne P n2 k¼n1 f ðkÞ ¼ f ðn1Þ þ f ðn1 þ 1Þ þ f ðn1 þ 2Þ þ þ f ðn2Þ EXAMPLE 2.3 (a) P 5 k¼1 xk ¼ x1 þ x2 þ x3 þ x4 þ x5 and P n i¼1 aibi ¼ a1b1 þ a2b2 þ þ anbn (b) P 5 j¼2 j2 ¼ 22 þ 32 þ 42 þ 52 ¼ 54 and P n i¼0 aixi ¼ a0 þ a1x þ a2x2 þ þ anxn (c) P p k¼1 aikbkj ¼ ai1b1j þ ai2b2j þ ai3b3j þ þ aipbpj 2.5 Matrix Multiplication The product of matrices A and B, written AB, is somewhat complicated. For this reason, we ﬁrst begin with a special case. The product AB of a row matrix A ¼ ½ai and a column matrix B ¼ ½bi with the same number of elements is deﬁned to be the scalar (or 1 1 matrix) obtained by multiplying corresponding entries and adding; that is, AB ¼ ½a1; a2; . . . ; an b1 b2 . . . bn 2 6 6 4 3 7 7 5 ¼ a1b1 þ a2b2 þ þ anbn ¼ P n k¼1 akbk We emphasize that AB is a scalar (or a 1 1 matrix). The product AB is not deﬁned when A and B have different numbers of elements. EXAMPLE 2.4 (a) ½7; 4; 5 3 2 1 2 4 3 5 ¼ 7ð3Þ þ ð4Þð2Þ þ 5ð1Þ ¼ 21 8 5 ¼ 8 (b) ½6; 1; 8; 3 4 9 2 5 2 6 6 4 3 7 7 5 ¼ 24 þ 9 16 þ 15 ¼ 32 We are now ready to deﬁne matrix multiplication in general. 30 CHAPTER 2 Algebra of Matrices DEFINITION: Suppose A ¼ ½aik and B ¼ ½bkj are matrices such that the number of columns of A is equal to the number of rows of B; say, A is an m p matrix and B is a p n matrix. Then the product AB is the m n matrix whose ij-entry is obtained by multiplying the ith row of A by the jth column of B. That is, a11 . . . a1p : . . . : ai1 . . . aip : . . . : am1 . . . amp 2 6 6 6 6 4 3 7 7 7 7 5 b11 . . . b1j . . . b1n : . . . : . . . : : . . . : . . . : : . . . : . . . : bp1 . . . bpj . . . bpn 2 6 6 6 6 4 3 7 7 7 7 5 ¼ c11 . . . c1n : . . . : : cij : : . . . : cm1 . . . cmn 2 6 6 6 6 4 3 7 7 7 7 5 where cij ¼ ai1b1j þ ai2b2j þ þ aipbpj ¼ P p k¼1 aikbkj The product AB is not deﬁned if A is an m p matrix and B is a q n matrix, where p 6¼ q. EXAMPLE 2.5 (a) Find AB where A ¼ 1 3 2 1 and B ¼ 2 0 4 5 2 6 . Because A is 2 2 and B is 2 3, the product AB is deﬁned and AB is a 2 3 matrix. To obtain the ﬁrst row of the product matrix AB, multiply the ﬁrst row [1, 3] of A by each column of B, 2 5 ; 0 2 ; 4 6 respectively. That is, AB ¼ 2 þ 15 0 6 4 þ 18 ¼ 17 6 14 To obtain the second row of AB, multiply the second row ½2; 1 of A by each column of B. Thus, AB ¼ 17 6 14 4 5 0 þ 2 8 6 ¼ 17 6 14 1 2 14 (b) Suppose A ¼ 1 2 3 4 and B ¼ 5 6 0 2 . Then AB ¼ 5 þ 0 6 4 15 þ 0 18 8 ¼ 5 2 15 10 and BA ¼ 5 þ 18 10 þ 24 0 6 0 8 ¼ 23 34 6 8 The above example shows that matrix multiplication is not commutative—that is, in general, AB 6¼ BA. However, matrix multiplication does satisfy the following properties. THEOREM 2.2: Let A; B; C be matrices. Then, whenever the products and sums are deﬁned, (i) ðABÞC ¼ AðBCÞ (associative law), (ii) AðB þ CÞ ¼ AB þ AC (left distributive law), (iii) ðB þ CÞA ¼ BA þ CA (right distributive law), (iv) kðABÞ ¼ ðkAÞB ¼ AðkBÞ, where k is a scalar. We note that 0A ¼ 0 and B0 ¼ 0, where 0 is the zero matrix. CHAPTER 2 Algebra of Matrices 31 2.6 Transpose of a Matrix The transpose of a matrix A, written AT, is the matrix obtained by writing the columns of A, in order, as rows. For example, 1 2 3 4 5 6 T ¼ 1 4 2 5 3 6 2 4 3 5 and ½1; 3; 5T ¼ 1 3 5 2 4 3 5 In other words, if A ¼ ½aij is an m n matrix, then AT ¼ ½bij is the n m matrix where bij ¼ aji. Observe that the tranpose of a row vector is a column vector. Similarly, the transpose of a column vector is a row vector. The next theorem lists basic properties of the transpose operation. THEOREM 2.3: Let A and B be matrices and let k be a scalar. Then, whenever the sum and product are deﬁned, (i) ðA þ BÞT ¼ AT þ BT, (iii) ðkAÞT ¼ kAT, (ii) ðATÞT ¼ A; (iv) ðABÞT ¼ BTAT. We emphasize that, by (iv), the transpose of a product is the product of the transposes, but in the reverse order. 2.7 Square Matrices A square matrix is a matrix with the same number of rows as columns. An n n square matrix is said to be of order n and is sometimes called an n-square matrix. Recall that not every two matrices can be added or multiplied. However, if we only consider square matrices of some given order n, then this inconvenience disappears. Speciﬁcally, the operations of addition, multiplication, scalar multiplication, and transpose can be performed on any n n matrices, and the result is again an n n matrix. EXAMPLE 2.6 The following are square matrices of order 3: A ¼ 1 2 3 4 4 4 5 6 7 2 4 3 5 and B ¼ 2 5 1 0 3 2 1 2 4 2 4 3 5 The following are also matrices of order 3: A þ B ¼ 3 3 4 4 1 6 6 8 3 2 6 4 3 7 5; 2A ¼ 2 4 6 8 8 8 10 12 14 2 6 4 3 7 5; AT ¼ 1 4 5 2 4 6 3 4 7 2 6 4 3 7 5 AB ¼ 5 7 15 12 0 20 17 7 35 2 6 4 3 7 5; BA ¼ 27 30 33 22 24 26 27 30 33 2 6 4 3 7 5 Diagonal and Trace Let A ¼ ½aij be an n-square matrix. The diagonal or main diagonal of A consists of the elements with the same subscripts—that is, a11; a22; a33; . . . ; ann 32 CHAPTER 2 Algebra of Matrices The trace of A, written trðAÞ, is the sum of the diagonal elements. Namely, trðAÞ ¼ a11 þ a22 þ a33 þ þ ann The following theorem applies. THEOREM 2.4: Suppose A ¼ ½aij and B ¼ ½bij are n-square matrices and k is a scalar. Then (i) trðA þ BÞ ¼ trðAÞ þ trðBÞ, (iii) trðATÞ ¼ trðAÞ, (ii) trðkAÞ ¼ k trðAÞ, (iv) trðABÞ ¼ trðBAÞ. EXAMPLE 2.7 Let A and B be the matrices A and B in Example 2.6. Then diagonal of A ¼ f1; 4; 7g and trðAÞ ¼ 1 4 þ 7 ¼ 4 diagonal of B ¼ f2; 3; 4g and trðBÞ ¼ 2 þ 3 4 ¼ 1 Moreover, trðA þ BÞ ¼ 3 1 þ 3 ¼ 5; trð2AÞ ¼ 2 8 þ 14 ¼ 8; trðATÞ ¼ 1 4 þ 7 ¼ 4 trðABÞ ¼ 5 þ 0 35 ¼ 30; trðBAÞ ¼ 27 24 33 ¼ 30 As expected from Theorem 2.4, trðA þ BÞ ¼ trðAÞ þ trðBÞ; trðATÞ ¼ trðAÞ; trð2AÞ ¼ 2 trðAÞ Furthermore, although AB 6¼ BA, the traces are equal. Identity Matrix, Scalar Matrices The n-square identity or unit matrix, denoted by In, or simply I, is the n-square matrix with 1’s on the diagonal and 0’s elsewhere. The identity matrix I is similar to the scalar 1 in that, for any n-square matrix A, AI ¼ IA ¼ A More generally, if B is an m n matrix, then BIn ¼ ImB ¼ B. For any scalar k, the matrix kI that contains k’s on the diagonal and 0’s elsewhere is called the scalar matrix corresponding to the scalar k. Observe that ðkIÞA ¼ kðIAÞ ¼ kA That is, multiplying a matrix A by the scalar matrix kI is equivalent to multiplying A by the scalar k. EXAMPLE 2.8 The following are the identity matrices of orders 3 and 4 and the corresponding scalar matrices for k ¼ 5: 1 0 0 0 1 0 0 0 1 2 4 3 5; 1 1 1 1 2 6 6 4 3 7 7 5; 5 0 0 0 5 0 0 0 5 2 4 3 5; 5 5 5 5 2 6 6 4 3 7 7 5 Remark 1: It is common practice to omit blocks or patterns of 0’s when there is no ambiguity, as in the above second and fourth matrices. Remark 2: The Kronecker delta function dij is deﬁned by dij ¼ 0 if i 6¼ j 1 if i ¼ j Thus, the identity matrix may be deﬁned by I ¼ ½dij. CHAPTER 2 Algebra of Matrices 33 2.8 Powers of Matrices, Polynomials in Matrices Let A be an n-square matrix over a ﬁeld K. Powers of A are deﬁned as follows: A2 ¼ AA; A3 ¼ A2A; . . . ; Anþ1 ¼ AnA; . . . ; and A0 ¼ I Polynomials in the matrix A are also deﬁned. Speciﬁcally, for any polynomial f ðxÞ ¼ a0 þ a1x þ a2x2 þ þ anxn where the ai are scalars in K, f ðAÞ is deﬁned to be the following matrix: f ðAÞ ¼ a0I þ a1A þ a2A2 þ þ anAn [Note that f ðAÞ is obtained from f ðxÞ by substituting the matrix A for the variable x and substituting the scalar matrix a0I for the scalar a0.] If f ðAÞ is the zero matrix, then A is called a zero or root of f ðxÞ. EXAMPLE 2.9 Suppose A ¼ 1 2 3 4 . Then A2 ¼ 1 2 3 4 1 2 3 4 ¼ 7 6 9 22 and A3 ¼ A2A ¼ 7 6 9 22 1 2 3 4 ¼ 11 38 57 106 Suppose f ðxÞ ¼ 2x2 3x þ 5 and gðxÞ ¼ x2 þ 3x 10. Then f ðAÞ ¼ 2 7 6 9 22 3 1 2 3 4 þ 5 1 0 0 1 ¼ 16 18 27 61 gðAÞ ¼ 7 6 9 22 þ 3 1 2 3 4 10 1 0 0 1 ¼ 0 0 0 0 Thus, A is a zero of the polynomial gðxÞ. 2.9 Invertible (Nonsingular) Matrices A square matrix A is said to be invertible or nonsingular if there exists a matrix B such that AB ¼ BA ¼ I where I is the identity matrix. Such a matrix B is unique. That is, if AB1 ¼ B1A ¼ I and AB2 ¼ B2A ¼ I, then B1 ¼ B1I ¼ B1ðAB2Þ ¼ ðB1AÞB2 ¼ IB2 ¼ B2 We call such a matrix B the inverse of A and denote it by A1. Observe that the above relation is symmetric; that is, if B is the inverse of A, then A is the inverse of B. EXAMPLE 2.10 Suppose that A ¼ 2 5 1 3 and B ¼ 3 5 1 2 . Then AB ¼ 6 5 10 þ 10 3 3 5 þ 6 ¼ 1 0 0 1 and BA ¼ 6 5 15 15 2 þ 2 5 þ 6 ¼ 1 0 0 1 Thus, A and B are inverses. It is known (Theorem 3.16) that AB ¼ I if and only if BA ¼ I. Thus, it is necessary to test only one product to determine whether or not two given matrices are inverses. (See Problem 2.17.) Now suppose A and B are invertible. Then AB is invertible and ðABÞ1 ¼ B1A1. More generally, if A1; A2; . . . ; Ak are invertible, then their product is invertible and ðA1A2 . . . AkÞ1 ¼ A1 k . . . A1 2 A1 1 the product of the inverses in the reverse order. 34 CHAPTER 2 Algebra of Matrices Inverse of a 2 2 Matrix Let A be an arbitrary 2 2 matrix, say A ¼ a b c d . We want to derive a formula for A1, the inverse of A. Speciﬁcally, we seek 22 ¼ 4 scalars, say x1, y1, x2, y2, such that a b c d x1 x2 y1 y2 ¼ 1 0 0 1 or ax1 þ by1 ax2 þ by2 cx1 þ dy1 cx2 þ dy2 ¼ 1 0 0 1 Setting the four entries equal to the corresponding entries in the identity matrix yields four equations, which can be partitioned into two 2 2 systems as follows: ax1 þ by1 ¼ 1; ax2 þ by2 ¼ 0 cx1 þ dy1 ¼ 0; cx2 þ dy2 ¼ 1 Suppose we let jAj ¼ ab bc (called the determinant of A). Assuming jAj 6¼ 0, we can solve uniquely for the above unknowns x1, y1, x2, y2, obtaining x1 ¼ d jAj ; y1 ¼ c jAj ; x2 ¼ b jAj ; y2 ¼ a jAj Accordingly, A1 ¼ a b c d 1 ¼ d=jAj b=jAj c=jAj a=jAj ¼ 1 jAj d b c a In other words, when jAj 6¼ 0, the inverse of a 2 2 matrix A may be obtained from A as follows: (1) Interchange the two elements on the diagonal. (2) Take the negatives of the other two elements. (3) Multiply the resulting matrix by 1=jAj or, equivalently, divide each element by jAj. In case jAj ¼ 0, the matrix A is not invertible. EXAMPLE 2.11 Find the inverse of A ¼ 2 3 4 5 and B ¼ 1 3 2 6 . First evaluate jAj ¼ 2ð5Þ 3ð4Þ ¼ 10 12 ¼ 2. Because jAj 6¼ 0, the matrix A is invertible and A1 ¼ 1 2 5 3 4 2 ¼ 5 2 3 2 2 1 Now evaluate jBj ¼ 1ð6Þ 3ð2Þ ¼ 6 6 ¼ 0. Because jBj ¼ 0, the matrix B has no inverse. Remark: The above property that a matrix is invertible if and only if A has a nonzero determinant is true for square matrices of any order. (See Chapter 8.) Inverse of an n n Matrix Suppose A is an arbitrary n-square matrix. Finding its inverse A1 reduces, as above, to ﬁnding the solution of a collection of n n systems of linear equations. The solution of such systems and an efﬁcient way of solving such a collection of systems is treated in Chapter 3. 2.10 Special Types of Square Matrices This section describes a number of special kinds of square matrices. Diagonal and Triangular Matrices A square matrix D ¼ ½dij is diagonal if its nondiagonal entries are all zero. Such a matrix is sometimes denoted by D ¼ diagðd11; d22; . . . ; dnnÞ CHAPTER 2 Algebra of Matrices 35 where some or all the dii may be zero. For example, 3 0 0 0 7 0 0 0 2 2 4 3 5; 4 0 0 5 ; 6 0 9 8 2 6 6 4 3 7 7 5 are diagonal matrices, which may be represented, respectively, by diagð3; 7; 2Þ; diagð4; 5Þ; diagð6; 0; 9; 8Þ (Observe that patterns of 0’s in the third matrix have been omitted.) A square matrix A ¼ ½aij is upper triangular or simply triangular if all entries below the (main) diagonal are equal to 0—that is, if aij ¼ 0 for i > j. Generic upper triangular matrices of orders 2, 3, 4 are as follows: a11 a12 0 a22 ; b11 b12 b13 b22 b23 b33 2 4 3 5; c11 c12 c13 c14 c22 c23 c24 c33 c34 c44 2 6 6 4 3 7 7 5 (As with diagonal matrices, it is common practice to omit patterns of 0’s.) The following theorem applies. THEOREM 2.5: Suppose A ¼ ½aij and B ¼ ½bij are n n (upper) triangular matrices. Then (i) A þ B, kA, AB are triangular with respective diagonals: ða11 þ b11; . . . ; ann þ bnnÞ; ðka11; . . . ; kannÞ; ða11b11; . . . ; annbnnÞ (ii) For any polynomial f ðxÞ, the matrix f ðAÞ is triangular with diagonal ð f ða11Þ; f ða22Þ; . . . ; f ðannÞÞ (iii) A is invertible if and only if each diagonal element aii 6¼ 0, and when A1 exists it is also triangular. A lower triangular matrix is a square matrix whose entries above the diagonal are all zero. We note that Theorem 2.5 is true if we replace ‘‘triangular’’ by either ‘‘lower triangular’’ or ‘‘diagonal.’’ Remark: A nonempty collection A of matrices is called an algebra (of matrices) if A is closed under the operations of matrix addition, scalar multiplication, and matrix multiplication. Clearly, the square matrices with a given order form an algebra of matrices, but so do the scalar, diagonal, triangular, and lower triangular matrices. Special Real Square Matrices: Symmetric, Orthogonal, Normal [Optional until Chapter 12] Suppose now A is a square matrix with real entries—that is, a real square matrix. The relationship between A and its transpose AT yields important kinds of matrices. (a) Symmetric Matrices A matrix A is symmetric if AT ¼ A. Equivalently, A ¼ ½aij is symmetric if symmetric elements (mirror elements with respect to the diagonal) are equal—that is, if each aij ¼ aji. A matrix A is skew-symmetric if AT ¼ A or, equivalently, if each aij ¼ aji. Clearly, the diagonal elements of such a matrix must be zero, because aii ¼ aii implies aii ¼ 0. (Note that a matrix A must be square if AT ¼ A or AT ¼ A.) 36 CHAPTER 2 Algebra of Matrices EXAMPLE 2.12 Let A ¼ 2 3 5 3 6 7 5 7 8 2 4 3 5; B ¼ 0 3 4 3 0 5 4 5 0 2 4 3 5; C ¼ 1 0 0 0 0 1 : (a) By inspection, the symmetric elements in A are equal, or AT ¼ A. Thus, A is symmetric. (b) The diagonal elements of B are 0 and symmetric elements are negatives of each other, or BT ¼ B. Thus, B is skew-symmetric. (c) Because C is not square, C is neither symmetric nor skew-symmetric. (b) Orthogonal Matrices A real matrix A is orthogonal if AT ¼ A1—that is, if AAT ¼ ATA ¼ I. Thus, A must necessarily be square and invertible. EXAMPLE 2.13 Let A ¼ 1 9 8 9 4 9 4 9 4 9 7 9 8 9 1 9 4 9 2 6 4 3 7 5. Multiplying A by AT yields I; that is, AAT ¼ I. This means ATA ¼ I, as well. Thus, AT ¼ A1; that is, A is orthogonal. Now suppose A is a real orthogonal 3 3 matrix with rows u1 ¼ ða1; a2; a3Þ; u2 ¼ ðb1; b2; b3Þ; u3 ¼ ðc1; c2; c3Þ Because A is orthogonal, we must have AAT ¼ I. Namely, AAT ¼ a1 a2 a3 b1 b2 b3 c1 c2 c3 2 4 3 5 a1 b1 c1 a2 b2 c2 a3 b3 c3 2 4 3 5 ¼ 1 0 0 0 1 0 0 0 1 2 4 3 5 ¼ I Multiplying A by AT and setting each entry equal to the corresponding entry in I yields the following nine equations: a2 1 þ a2 2 þ a2 3 ¼ 1; a1b1 þ a2b2 þ a3b3 ¼ 0; a1c1 þ a2c2 þ a3c3 ¼ 0 b1a1 þ b2a2 þ b3a3 ¼ 0; b2 1 þ b2 2 þ b2 3 ¼ 1; b1c1 þ b2c2 þ b3c3 ¼ 0 c1a1 þ c2a2 þ c3a3 ¼ 0; c1b1 þ c2b2 þ c3b3 ¼ 0; c2 1 þ c2 2 þ c2 3 ¼ 1 Accordingly, u1 u1 ¼ 1, u2 u2 ¼ 1, u3 u3 ¼ 1, and ui uj ¼ 0 for i 6¼ j. Thus, the rows u1, u2, u3 are unit vectors and are orthogonal to each other. Generally speaking, vectors u1, u2; . . . ; um in Rn are said to form an orthonormal set of vectors if the vectors are unit vectors and are orthogonal to each other; that is, ui uj ¼ 0 if i 6¼ j 1 if i ¼ j In other words, ui uj ¼ dij where dij is the Kronecker delta function: We have shown that the condition AAT ¼ I implies that the rows of A form an orthonormal set of vectors. The condition ATA ¼ I similarly implies that the columns of A also form an orthonormal set of vectors. Furthermore, because each step is reversible, the converse is true. The above results for 3 3 matrices are true in general. That is, the following theorem holds. THEOREM 2.6: Let A be a real matrix. Then the following are equivalent: (a) A is orthogonal. (b) The rows of A form an orthonormal set. (c) The columns of A form an orthonormal set. For n ¼ 2, we have the following result (proved in Problem 2.28). CHAPTER 2 Algebra of Matrices 37 THEOREM 2.7: Let A be a real 2 2 orthogonal matrix. Then, for some real number y, A ¼ cos y sin y sin y cos y or A ¼ cos y sin y sin y cos y (c) Normal Matrices A real matrix A is normal if it commutes with its transpose AT—that is, if AAT ¼ ATA. If A is symmetric, orthogonal, or skew-symmetric, then A is normal. There are also other normal matrices. EXAMPLE 2.14 Let A ¼ 6 3 3 6 . Then AAT ¼ 6 3 3 6 6 3 3 6 ¼ 45 0 0 45 and ATA ¼ 6 3 3 6 6 3 3 6 ¼ 45 0 0 45 Because AAT ¼ ATA, the matrix A is normal. 2.11 Complex Matrices Let A be a complex matrix—that is, a matrix with complex entries. Recall (Section 1.7) that if z ¼ a þ bi is a complex number, then z ¼ a bi is its conjugate. The conjugate of a complex matrix A, written A, is the matrix obtained from A by taking the conjugate of each entry in A. That is, if A ¼ ½aij, then A ¼ ½bij, where bij ¼ aij. (We denote this fact by writing A ¼ ½ aij.) The two operations of transpose and conjugation commute for any complex matrix A, and the special notation AH is used for the conjugate transpose of A. That is, AH ¼ ð AÞT ¼ ðATÞ Note that if A is real, then AH ¼ AT. [Some texts use A instead of AH:] EXAMPLE 2.15 Let A ¼ 2 þ 8i 5 3i 4 7i 6i 1 4i 3 þ 2i . Then AH ¼ 2 8i 6i 5 þ 3i 1 þ 4i 4 þ 7i 3 2i 2 4 3 5. Special Complex Matrices: Hermitian, Unitary, Normal [Optional until Chapter 12] Consider a complex matrix A. The relationship between A and its conjugate transpose AH yields important kinds of complex matrices (which are analogous to the kinds of real matrices described above). A complex matrix A is said to be Hermitian or skew-Hermitian according as to whether AH ¼ A or AH ¼ A: Clearly, A ¼ ½aij is Hermitian if and only if symmetric elements are conjugate—that is, if each aij ¼ aji—in which case each diagonal element aii must be real. Similarly, if A is skew-symmetric, then each diagonal element aii ¼ 0. (Note that A must be square if AH ¼ A or AH ¼ A.) A complex matrix A is unitary if AHA1 ¼ A1AH ¼ I—that is, if AH ¼ A1: Thus, A must necessarily be square and invertible. We note that a complex matrix A is unitary if and only if its rows (columns) form an orthonormal set relative to the dot product of complex vectors. A complex matrix A is said to be normal if it commutes with AH—that is, if AAH ¼ AHA 38 CHAPTER 2 Algebra of Matrices (Thus, A must be a square matrix.) This deﬁnition reduces to that for real matrices when A is real. EXAMPLE 2.16 Consider the following complex matrices: A ¼ 3 1 2i 4 þ 7i 1 þ 2i 4 2i 4 7i 2i 5 2 4 3 5 B ¼ 1 2 1 i 1 þ i i 1 1 þ i 1 þ i 1 þ i 0 2 4 3 5 C ¼ 2 þ 3i 1 i 1 þ 2i (a) By inspection, the diagonal elements of A are real, and the symmetric elements 1 2i and 1 þ 2i are conjugate, 4 þ 7i and 4 7i are conjugate, and 2i and 2i are conjugate. Thus, A is Hermitian. (b) Multiplying B by BH yields I; that is, BBH ¼ I. This implies BHB ¼ I, as well. Thus, BH ¼ B1, which means B is unitary. (c) To show C is normal, we evaluate CCH and CHC: CCH ¼ 2 þ 3i 1 i 1 þ 2i 2 3i i 1 1 2i ¼ 14 4 4i 4 þ 4i 6 and similarly CHC ¼ 14 4 4i 4 þ 4i 6 . Because CCH ¼ CHC, the complex matrix C is normal. We note that when a matrix A is real, Hermitian is the same as symmetric, and unitary is the same as orthogonal. 2.12 Block Matrices Using a system of horizontal and vertical (dashed) lines, we can partition a matrix A into submatrices called blocks (or cells) of A. Clearly a given matrix may be divided into blocks in different ways. For example, 1 2 0 1 3 2 3 5 7 2 3 1 4 5 9 4 6 3 1 8 2 6 6 4 3 7 7 5; 1 2 0 1 3 2 3 5 7 2 3 1 4 5 9 4 6 3 1 8 2 6 6 4 3 7 7 5; 1 2 0 1 3 2 3 5 7 2 3 1 4 5 9 4 6 3 1 8 2 6 6 4 3 7 7 5 The convenience of the partition of matrices, say A and B, into blocks is that the result of operations on A and B can be obtained by carrying out the computation with the blocks, just as if they were the actual elements of the matrices. This is illustrated below, where the notation A ¼ ½Aij will be used for a block matrix A with blocks Aij. Suppose that A ¼ ½Aij and B ¼ ½Bij are block matrices with the same numbers of row and column blocks, and suppose that corresponding blocks have the same size. Then adding the corresponding blocks of A and B also adds the corresponding elements of A and B, and multiplying each block of A by a scalar k multiplies each element of A by k. Thus, A þ B ¼ A11 þ B11 A12 þ B12 . . . A1n þ B1n A21 þ B21 A22 þ B22 . . . A2n þ B2n . . . . . . . . . . . . Am1 þ Bm1 Am2 þ Bm2 . . . Amn þ Bmn 2 6 6 6 4 3 7 7 7 5 and kA ¼ kA11 kA12 . . . kA1n kA21 kA22 . . . kA2n . . . . . . . . . . . . kAm1 kAm2 . . . kAmn 2 6 6 4 3 7 7 5 CHAPTER 2 Algebra of Matrices 39 The case of matrix multiplication is less obvious, but still true. That is, suppose that U ¼ ½Uik and V ¼ ½Vkj are block matrices such that the number of columns of each block Uik is equal to the number of rows of each block Vkj. (Thus, each product UikVkj is deﬁned.) Then UV ¼ W11 W12 . . . W1n W21 W22 . . . W2n . . . . . . . . . . . . Wm1 Wm2 . . . Wmn 2 6 6 4 3 7 7 5; where Wij ¼ Ui1V1j þ Ui2V2j þ þ UipVpj The proof of the above formula for UV is straightforward but detailed and lengthy. It is left as an exercise (Problem 2.85). Square Block Matrices Let M be a block matrix. Then M is called a square block matrix if (i) M is a square matrix. (ii) The blocks form a square matrix. (iii) The diagonal blocks are also square matrices. The latter two conditions will occur if and only if there are the same number of horizontal and vertical lines and they are placed symmetrically. Consider the following two block matrices: A ¼ 1 2 3 4 5 1 1 1 1 1 9 8 7 6 5 4 4 4 4 4 3 5 3 5 3 2 6 6 6 6 4 3 7 7 7 7 5 and B ¼ 1 2 3 4 5 1 1 1 1 1 9 8 7 6 5 4 4 4 4 4 3 5 3 5 3 2 6 6 6 6 4 3 7 7 7 7 5 The block matrix A is not a square block matrix, because the second and third diagonal blocks are not square. On the other hand, the block matrix B is a square block matrix. Block Diagonal Matrices Let M ¼ ½Aij be a square block matrix such that the nondiagonal blocks are all zero matrices; that is, Aij ¼ 0 when i 6¼ j. Then M is called a block diagonal matrix. We sometimes denote such a block diagonal matrix by writing M ¼ diagðA11; A22; . . . ; ArrÞ or M ¼ A11 A22 Arr The importance of block diagonal matrices is that the algebra of the block matrix is frequently reduced to the algebra of the individual blocks. Speciﬁcally, suppose f ðxÞ is a polynomial and M is the above block diagonal matrix. Then f ðMÞ is a block diagonal matrix, and f ðMÞ ¼ diagð f ðA11Þ; f ðA22Þ; . . . ; f ðArrÞÞ Also, M is invertible if and only if each Aii is invertible, and, in such a case, M1 is a block diagonal matrix, and M1 ¼ diagðA1 11 ; A1 22 ; . . . ; A1 rr Þ Analogously, a square block matrix is called a block upper triangular matrix if the blocks below the diagonal are zero matrices and a block lower triangular matrix if the blocks above the diagonal are zero matrices. 40 CHAPTER 2 Algebra of Matrices EXAMPLE 2.17 Determine which of the following square block matrices are upper diagonal, lower diagonal, or diagonal: A ¼ 1 2 0 3 4 5 0 0 6 2 4 3 5; B ¼ 1 0 0 0 2 3 4 0 5 0 6 0 0 7 8 9 2 6 6 4 3 7 7 5; C ¼ 1 0 0 0 2 3 0 4 5 2 4 3 5; D ¼ 1 2 0 3 4 5 0 6 7 2 4 3 5 (a) A is upper triangular because the block below the diagonal is a zero block. (b) B is lower triangular because all blocks above the diagonal are zero blocks. (c) C is diagonal because the blocks above and below the diagonal are zero blocks. (d) D is neither upper triangular nor lower triangular. Also, no other partitioning of D will make it into either a block upper triangular matrix or a block lower triangular matrix. SOLVED PROBLEMS Matrix Addition and Scalar Multiplication 2.1 Given A ¼ 1 2 3 4 5 6 and B ¼ 3 0 2 7 1 8 , ﬁnd: (a) A þ B, (b) 2A 3B. (a) Add the corresponding elements: A þ B ¼ 1 þ 3 2 þ 0 3 þ 2 4 7 5 þ 1 6 þ 8 ¼ 4 2 5 3 6 2 (b) First perform the scalar multiplication and then a matrix addition: 2A 3B ¼ 2 4 6 8 10 12 þ 9 0 6 21 3 24 ¼ 7 4 0 29 7 36 (Note that we multiply B by 3 and then add, rather than multiplying B by 3 and subtracting. This usually prevents errors.) 2.2. Find x; y; z; t where 3 x y z t ¼ x 6 1 2t þ 4 x þ y z þ t 3 : Write each side as a single equation: 3x 3y 3z 3t ¼ x þ 4 x þ y þ 6 z þ t 1 2t þ 3 Set corresponding entries equal to each other to obtain the following system of four equations: 3x ¼ x þ 4; 3y ¼ x þ y þ 6; 3z ¼ z þ t 1; 3t ¼ 2t þ 3 or 2x ¼ 4; 2y ¼ 6 þ x; 2z ¼ t 1; t ¼ 3 The solution is x ¼ 2, y ¼ 4, z ¼ 1, t ¼ 3. 2.3. Prove Theorem 2.1 (i) and (v): (i) ðA þ BÞ þ C ¼ A þ ðB þ CÞ, (v) kðA þ BÞ ¼ kA þ kB. Suppose A ¼ ½aij, B ¼ ½bij, C ¼ ½cij. The proof reduces to showing that corresponding ij-entries in each side of each matrix equation are equal. [We prove only (i) and (v), because the other parts of Theorem 2.1 are proved similarly.] CHAPTER 2 Algebra of Matrices 41 (i) The ij-entry of A þ B is aij þ bij; hence, the ij-entry of ðA þ BÞ þ C is ðaij þ bijÞ þ cij. On the other hand, the ij-entry of B þ C is bij þ cij; hence, the ij-entry of A þ ðB þ CÞ is aij þ ðbij þ cijÞ. However, for scalars in K, ðaij þ bijÞ þ cij ¼ aij þ ðbij þ cijÞ Thus, ðA þ BÞ þ C and A þ ðB þ CÞ have identical ij-entries. Therefore, ðA þ BÞ þ C ¼ A þ ðB þ CÞ. (v) The ij-entry of A þ B is aij þ bij; hence, kðaij þ bijÞ is the ij-entry of kðA þ BÞ. On the other hand, the ij-entries of kA and kB are kaij and kbij, respectively. Thus, kaij þ kbij is the ij-entry of kA þ kB. However, for scalars in K, kðaij þ bijÞ ¼ kaij þ kbij Thus, kðA þ BÞ and kA þ kB have identical ij-entries. Therefore, kðA þ BÞ ¼ kA þ kB. Matrix Multiplication 2.4. Calculate: (a) ½8; 4; 5 3 2 1 2 4 3 5, (b) ½6; 1; 7; 5 4 9 3 2 2 6 6 4 3 7 7 5, (c) ½3; 8; 2; 4 5 1 6 2 4 3 5 (a) Multiply the corresponding entries and add: ½8; 4; 5 3 2 1 2 4 3 5 ¼ 8ð3Þ þ ð4Þð2Þ þ 5ð1Þ ¼ 24 8 5 ¼ 11 (b) Multiply the corresponding entries and add: ½6; 1; 7; 5 4 9 3 2 2 6 6 6 4 3 7 7 7 5 ¼ 24 þ 9 21 þ 10 ¼ 22 (c) The product is not deﬁned when the row matrix and the column matrix have different numbers of elements. 2.5. Let ðr sÞ denote an r s matrix. Find the sizes of those matrix products that are deﬁned: (a) ð2 3Þð3 4Þ; (c) ð1 2Þð3 1Þ; (e) ð4 4Þð3 3Þ (b) ð4 1Þð1 2Þ, (d) ð5 2Þð2 3Þ, (f) ð2 2Þð2 4Þ In each case, the product is deﬁned if the inner numbers are equal, and then the product will have the size of the outer numbers in the given order. (a) 2 4, (c) not deﬁned, (e) not deﬁned (b) 4 2, (d) 5 3, (f) 2 4 2.6. Let A ¼ 1 3 2 1 and B ¼ 2 0 4 3 2 6 . Find: (a) AB, (b) BA. (a) Because A is a 2 2 matrix and B a 2 3 matrix, the product AB is deﬁned and is a 2 3 matrix. To obtain the entries in the ﬁrst row of AB, multiply the ﬁrst row ½1; 3 of A by the columns 2 3 ; 0 2 ; 4 6 of B, respectively, as follows: AB ¼ 1 3 2 1 2 0 4 3 2 6 ¼ 2 þ 9 0 6 4 þ 18 ¼ 11 6 14 42 CHAPTER 2 Algebra of Matrices To obtain the entries in the second row of AB, multiply the second row ½2; 1 of A by the columns of B: AB ¼ 1 3 2 1 2 0 4 3 2 6 ¼ 11 6 14 4 3 0 þ 2 8 6 Thus, AB ¼ 11 6 14 1 2 14 : (b) The size of B is 2 3 and that of A is 2 2. The inner numbers 3 and 2 are not equal; hence, the product BA is not deﬁned. 2.7. Find AB, where A ¼ 2 3 1 4 2 5 and B ¼ 2 1 0 6 1 3 5 1 4 1 2 2 2 4 3 5. Because A is a 2 3 matrix and B a 3 4 matrix, the product AB is deﬁned and is a 2 4 matrix. Multiply the rows of A by the columns of B to obtain AB ¼ 4 þ 3 4 2 þ 9 1 0 15 þ 2 12 þ 3 2 8 2 þ 20 4 6 þ 5 0 þ 10 10 24 2 þ 10 ¼ 3 6 13 13 26 5 0 32 : 2.8. Find: (a) 1 6 3 5 2 7 , (b) 2 7 1 6 3 5 , (c) ½2; 7 1 6 3 5 . (a) The ﬁrst factor is 2 2 and the second is 2 1, so the product is deﬁned as a 2 1 matrix: 1 6 3 5 2 7 ¼ 2 42 6 35 ¼ 40 41 (b) The product is not deﬁned, because the ﬁrst factor is 2 1 and the second factor is 2 2. (c) The ﬁrst factor is 1 2 and the second factor is 2 2, so the product is deﬁned as a 1 2 (row) matrix: ½2; 7 1 6 3 5 ¼ ½2 þ 21; 12 35 ¼ ½23; 23 2.9. Clearly, 0A ¼ 0 and A0 ¼ 0, where the 0’s are zero matrices (with possibly different sizes). Find matrices A and B with no zero entries such that AB ¼ 0. Let A ¼ 1 2 2 4 and B ¼ 6 2 3 1 . Then AB ¼ 0 0 0 0 . 2.10. Prove Theorem 2.2(i): ðABÞC ¼ AðBCÞ. Let A ¼ ½aij, B ¼ ½bjk, C ¼ ½ckl, and let AB ¼ S ¼ ½sik, BC ¼ T ¼ ½tjl. Then sik ¼ P m j¼1 aijbjk and tjl ¼ P n k¼1 bjkckl Multiplying S ¼ AB by C, the il-entry of ðABÞC is si1c1l þ si2c2l þ þ sincnl ¼ P n k¼1 sikckl ¼ P n k¼1 P m j¼1 ðaijbjkÞckl On the other hand, multiplying A by T ¼ BC, the il-entry of AðBCÞ is ai1t1l þ ai2t2l þ þ aintnl ¼ P m j¼1 aijtjl ¼ P m j¼1 P n k¼1 aijðbjkcklÞ The above sums are equal; that is, corresponding elements in ðABÞC and AðBCÞ are equal. Thus, ðABÞC ¼ AðBCÞ. CHAPTER 2 Algebra of Matrices 43 2.11. Prove Theorem 2.2(ii): AðB þ CÞ ¼ AB þ AC. Let A ¼ ½aij, B ¼ ½bjk, C ¼ ½cjk, and let D ¼ B þ C ¼ ½djk, E ¼ AB ¼ ½eik, F ¼ AC ¼ ½ fik. Then djk ¼ bjk þ cjk; eik ¼ P m j¼1 aijbjk; fik ¼ P m j¼1 aijcjk Thus, the ik-entry of the matrix AB þ AC is eik þ fik ¼ P m j¼1 aijbjk þ P m j¼1 aijcjk ¼ P m j¼1 aijðbjk þ cjkÞ On the other hand, the ik-entry of the matrix AD ¼ AðB þ CÞ is ai1d1k þ ai2d2k þ þ aimdmk ¼ P m j¼1 aijdjk ¼ P m j¼1 aijðbjk þ cjkÞ Thus, AðB þ CÞ ¼ AB þ AC, because the corresponding elements are equal. Transpose 2.12. Find the transpose of each matrix: A ¼ 1 2 3 7 8 9 ; B ¼ 1 2 3 2 4 5 3 5 6 2 4 3 5; C ¼ ½1; 3; 5; 7; D ¼ 2 4 6 2 4 3 5 Rewrite the rows of each matrix as columns to obtain the transpose of the matrix: AT ¼ 1 7 2 8 3 9 2 4 3 5; BT ¼ 1 2 3 2 4 5 3 5 6 2 4 3 5; CT ¼ 1 3 5 7 2 6 6 4 3 7 7 5; DT ¼ ½2; 4; 6 (Note that BT ¼ B; such a matrix is said to be symmetric. Note also that the transpose of the row vector C is a column vector, and the transpose of the column vector D is a row vector.) 2.13. Prove Theorem 2.3(iv): ðABÞT ¼ BTAT. Let A ¼ ½aik and B ¼ ½bkj. Then the ij-entry of AB is ai1b1j þ ai2b2j þ þ aimbmj This is the ji-entry (reverse order) of ðABÞT. Now column j of B becomes row j of BT, and row i of A becomes column i of AT. Thus, the ij-entry of BTAT is ½b1j; b2j; . . . ; bmj½ai1; ai2; . . . ; aimT ¼ b1jai1 þ b2jai2 þ þ bmjaim Thus, ðABÞT ¼ BTAT on because the corresponding entries are equal. Square Matrices 2.14. Find the diagonal and trace of each matrix: (a) A ¼ 1 3 6 2 5 8 4 2 9 2 4 3 5, (b) B ¼ 2 4 8 3 7 9 5 0 2 2 4 3 5, (c) C ¼ 1 2 3 4 5 6 . (a) The diagonal of A consists of the elements from the upper left corner of A to the lower right corner of A or, in other words, the elements a11, a22, a33. Thus, the diagonal of A consists of the numbers 1; 5, and 9. The trace of A is the sum of the diagonal elements. Thus, trðAÞ ¼ 1 5 þ 9 ¼ 5 (b) The diagonal of B consists of the numbers 2; 7, and 2. Hence, trðBÞ ¼ 2 7 þ 2 ¼ 3 (c) The diagonal and trace are only deﬁned for square matrices. 44 CHAPTER 2 Algebra of Matrices 2.15. Let A ¼ 1 2 4 3 , and let f ðxÞ ¼ 2x3 4x þ 5 and gðxÞ ¼ x2 þ 2x þ 11. Find (a) A2, (b) A3, (c) f ðAÞ, (d) gðAÞ. (a) A2 ¼ AA ¼ 1 2 4 3 1 2 4 3 ¼ 1 þ 8 2 6 4 12 8 þ 9 ¼ 9 4 8 17 (b) A3 ¼ AA2 ¼ 1 2 4 3 9 4 8 17 ¼ 9 16 4 þ 34 36 þ 24 16 51 ¼ 7 30 60 67 (c) First substitute A for x and 5I for the constant in f ðxÞ, obtaining f ðAÞ ¼ 2A3 4A þ 5I ¼ 2 7 30 60 67 4 1 2 4 3 þ 5 1 0 0 1 Now perform the scalar multiplication and then the matrix addition: f ðAÞ ¼ 14 60 120 134 þ 4 8 16 12 þ 5 0 0 5 ¼ 13 52 104 117 (d) Substitute A for x and 11I for the constant in gðxÞ, and then calculate as follows: gðAÞ ¼ A2 þ 2A 11I ¼ 9 4 8 17 þ 2 1 2 4 3 11 1 0 0 1 ¼ 9 4 8 17 þ 2 4 8 6 þ 11 0 0 11 ¼ 0 0 0 0 Because gðAÞ is the zero matrix, A is a root of the polynomial gðxÞ. 2.16. Let A ¼ 1 3 4 3 . (a) Find a nonzero column vector u ¼ x y such that Au ¼ 3u. (b) Describe all such vectors. (a) First set up the matrix equation Au ¼ 3u, and then write each side as a single matrix (column vector) as follows: 1 3 4 3 x y ¼ 3 x y ; and then x þ 3y 4x 3y ¼ 3x 3y Set the corresponding elements equal to each other to obtain a system of equations: x þ 3y ¼ 3x 4x 3y ¼ 3y or 2x 3y ¼ 0 4x 6y ¼ 0 or 2x 3y ¼ 0 The system reduces to one nondegenerate linear equation in two unknowns, and so has an inﬁnite number of solutions. To obtain a nonzero solution, let, say, y ¼ 2; then x ¼ 3. Thus, u ¼ ð3; 2ÞT is a desired nonzero vector. (b) To ﬁnd the general solution, set y ¼ a, where a is a parameter. Substitute y ¼ a into 2x 3y ¼ 0 to obtain x ¼ 3 2 a. Thus, u ¼ ð3 2 a; aÞT represents all such solutions. Invertible Matrices, Inverses 2.17. Show that A ¼ 1 0 2 2 1 3 4 1 8 2 4 3 5 and B ¼ 11 2 2 4 0 1 6 1 1 2 4 3 5 are inverses. Compute the product AB, obtaining AB ¼ 11 þ 0 þ 12 2 þ 0 2 2 þ 0 2 22 þ 4 þ 18 4 þ 0 3 4 1 3 44 4 þ 48 8 þ 0 8 8 þ 1 8 2 4 3 5 ¼ 1 0 0 0 1 0 0 0 1 2 4 3 5 ¼ I Because AB ¼ I, we can conclude (Theorem 3.16) that BA ¼ I. Accordingly, A and B are inverses. CHAPTER 2 Algebra of Matrices 45 2.18. Find the inverse, if possible, of each matrix: (a) A ¼ 5 3 4 2 ; (b) B ¼ 2 3 1 3 ; (c) 2 6 3 9 : Use the formula for the inverse of a 2 2 matrix appearing in Section 2.9. (a) First ﬁnd jAj ¼ 5ð2Þ 3ð4Þ ¼ 10 12 ¼ 2. Next interchange the diagonal elements, take the negatives of the nondiagonal elements, and multiply by 1=jAj: A1 ¼ 1 2 2 3 4 5 ¼ 1 3 2 2 5 2 " # (b) First ﬁnd jBj ¼ 2ð3Þ ð3Þð1Þ ¼ 6 þ 3 ¼ 9. Next interchange the diagonal elements, take the negatives of the nondiagonal elements, and multiply by 1=jBj: B1 ¼ 1 9 3 3 1 2 ¼ 1 3 1 3 1 9 2 9 " # (c) First ﬁnd jCj ¼ 2ð9Þ 6ð3Þ ¼ 18 18 ¼ 0. Because jCj ¼ 0; C has no inverse. 2.19. Let A ¼ 1 1 1 0 1 2 1 2 4 2 6 6 4 3 7 7 5. Find A1 ¼ x1 x2 x3 y1 y2 y3 z1 z2 z3 2 4 3 5. Multiplying A by A1 and setting the nine entries equal to the nine entries of the identity matrix I yields the following three systems of three equations in three of the unknowns: x1 þ y1 þ z1 ¼ 1 x2 þ y2 þ z2 ¼ 0 x3 þ y3 þ z3 ¼ 0 y1 þ 2z1 ¼ 0 y2 þ 2z2 ¼ 1 y3 þ 2z3 ¼ 0 x1 þ 2y1 þ 4z1 ¼ 0 x2 þ 2y2 þ 4z2 ¼ 0 x3 þ 2y3 þ 4z3 ¼ 1 [Note that A is the coefﬁcient matrix for all three systems.] Solving the three systems for the nine unknowns yields x1 ¼ 0; y1 ¼ 2; z1 ¼ 1; x2 ¼ 2; y2 ¼ 3; z2 ¼ 1; x3 ¼ 1; y3 ¼ 2; z3 ¼ 1 Thus; A1 ¼ 0 2 1 2 3 2 1 1 1 2 6 4 3 7 5 (Remark: Chapter 3 gives an efﬁcient way to solve the three systems.) 2.20. Let A and B be invertible matrices (with the same size). Show that AB is also invertible and ðABÞ1 ¼ B1A1. [Thus, by induction, ðA1A2 . . . AmÞ1 ¼ A1 m . . . A1 2 A1 1 .] Using the associativity of matrix multiplication, we get ðABÞðB1A1Þ ¼ AðBB1ÞA1 ¼ AIA1 ¼ AA1 ¼ I ðB1A1ÞðABÞ ¼ B1ðA1AÞB ¼ A1IB ¼ B1B ¼ I Thus, ðABÞ1 ¼ B1A1. 46 CHAPTER 2 Algebra of Matrices Diagonal and Triangular Matrices 2.21. Write out the diagonal matrices A ¼ diagð4; 3; 7Þ, B ¼ diagð2; 6Þ, C ¼ diagð3; 8; 0; 5Þ. Put the given scalars on the diagonal and 0’s elsewhere: A ¼ 4 0 0 0 3 0 0 0 7 2 4 3 5; B ¼ 2 0 0 6 ; C ¼ 3 8 0 5 2 6 6 4 3 7 7 5 2.22. Let A ¼ diagð2; 3; 5Þ and B ¼ diagð7; 0; 4Þ. Find (a) AB, A2, B2; (b) f ðAÞ, where f ðxÞ ¼ x2 þ 3x 2; (c) A1 and B1. (a) The product matrix AB is a diagonal matrix obtained by multiplying corresponding diagonal entries; hence, AB ¼ diagð2ð7Þ; 3ð0Þ; 5ð4ÞÞ ¼ diagð14; 0; 20Þ Thus, the squares A2 and B2 are obtained by squaring each diagonal entry; hence, A2 ¼ diagð22; 32; 52Þ ¼ diagð4; 9; 25Þ and B2 ¼ diagð49; 0; 16Þ (b) f ðAÞ is a diagonal matrix obtained by evaluating f ðxÞ at each diagonal entry. We have f ð2Þ ¼ 4 þ 6 2 ¼ 8; f ð3Þ ¼ 9 þ 9 2 ¼ 16; f ð5Þ ¼ 25 þ 15 2 ¼ 38 Thus, f ðAÞ ¼ diagð8; 16; 38Þ. (c) The inverse of a diagonal matrix is a diagonal matrix obtained by taking the inverse (reciprocal) of each diagonal entry. Thus, A1 ¼ diagð1 2 ; 1 3 ; 1 5Þ, but B has no inverse because there is a 0 on the diagonal. 2.23. Find a 2 2 matrix A such that A2 is diagonal but not A. Let A ¼ 1 2 3 1 . Then A2 ¼ 7 0 0 7 , which is diagonal. 2.24. Find an upper triangular matrix A such that A3 ¼ 8 57 0 27 . Set A ¼ x y 0 z . Then x3 ¼ 8, so x ¼ 2; and z3 ¼ 27, so z ¼ 3. Next calculate A3 using x ¼ 2 and y ¼ 3: A2 ¼ 2 y 0 3 2 y 0 3 ¼ 4 5y 0 9 and A3 ¼ 2 y 0 3 4 5y 0 9 ¼ 8 19y 0 27 Thus, 19y ¼ 57, or y ¼ 3. Accordingly, A ¼ 2 3 0 3 . 2.25. Let A ¼ ½aij and B ¼ ½bij be upper triangular matrices. Prove that AB is upper triangular with diagonal a11b11, a22b22; . . . ; annbnn. Let AB ¼ ½cij. Then cij ¼ Pn k¼1 aikbkj and cii ¼ Pn k¼1 aikbki. Suppose i > j. Then, for any k, either i > k or k > j, so that either aik ¼ 0 or bkj ¼ 0. Thus, cij ¼ 0, and AB is upper triangular. Suppose i ¼ j. Then, for k < i, we have aik ¼ 0; and, for k > i, we have bki ¼ 0. Hence, cii ¼ aiibii, as claimed. [This proves one part of Theorem 2.5(i); the statements for A þ B and kA are left as exercises.] CHAPTER 2 Algebra of Matrices 47 Special Real Matrices: Symmetric and Orthogonal 2.26. Determine whether or not each of the following matrices is symmetric—that is, AT ¼ A—or skew-symmetric—that is, AT ¼ A: (a) A ¼ 5 7 1 7 8 2 1 2 4 2 4 3 5; (b) B ¼ 0 4 3 4 0 5 3 5 0 2 4 3 5; (c) C ¼ 0 0 0 0 0 0 (a) By inspection, the symmetric elements (mirror images in the diagonal) are 7 and 7, 1 and 1, 2 and 2. Thus, A is symmetric, because symmetric elements are equal. (b) By inspection, the diagonal elements are all 0, and the symmetric elements, 4 and 4, 3 and 3, and 5 and 5, are negatives of each other. Hence, B is skew-symmetric. (c) Because C is not square, C is neither symmetric nor skew-symmetric. 2.27. Suppose B ¼ 4 x þ 2 2x 3 x þ 1 is symmetric. Find x and B. Set the symmetric elements x þ 2 and 2x 3 equal to each other, obtaining 2x 3 ¼ x þ 2 or x ¼ 5. Hence, B ¼ 4 7 7 6 . 2.28. Let A be an arbitrary 2 2 (real) orthogonal matrix. (a) Prove: If ða; bÞ is the ﬁrst row of A, then a2 þ b2 ¼ 1 and A ¼ a b b a or A ¼ a b b a : (b) Prove Theorem 2.7: For some real number y, A ¼ cos y sin y sin y cos y or A ¼ cos y sin y sin y cos y (a) Suppose ðx; yÞ is the second row of A. Because the rows of A form an orthonormal set, we get a2 þ b2 ¼ 1; x2 þ y2 ¼ 1; ax þ by ¼ 0 Similarly, the columns form an orthogonal set, so a2 þ x2 ¼ 1; b2 þ y2 ¼ 1; ab þ xy ¼ 0 Therefore, x2 ¼ 1 a2 ¼ b2, whence x ¼ b: Case (i): x ¼ b. Then bða þ yÞ ¼ 0, so y ¼ a. Case (ii): x ¼ b. Then bðy aÞ ¼ 0, so y ¼ a. This means, as claimed, A ¼ a b b a or A ¼ a b b a (b) Because a2 þ b2 ¼ 1, we have 1 a 1. Let a ¼ cos y. Then b2 ¼ 1 cos2 y, so b ¼ sin y. This proves the theorem. 2.29. Find a 2 2 orthogonal matrix A whose ﬁrst row is a (positive) multiple of ð3; 4Þ. Normalize ð3; 4Þ to get ð3 5 ; 4 5Þ. Then, by Problem 2.28, A ¼ 3 5 4 5 4 5 3 5 " # or A ¼ 3 5 4 5 4 5 3 5 " # : 2.30. Find a 3 3 orthogonal matrix P whose ﬁrst two rows are multiples of u1 ¼ ð1; 1; 1Þ and u2 ¼ ð0; 1; 1Þ, respectively. (Note that, as required, u1 and u2 are orthogonal.) 48 CHAPTER 2 Algebra of Matrices First ﬁnd a nonzero vector u3 orthogonal to u1 and u2; say (cross product) u3 ¼ u1 u2 ¼ ð2; 1; 1Þ. Let A be the matrix whose rows are u1; u2; u3; and let P be the matrix obtained from A by normalizing the rows of A. Thus, A ¼ 1 1 1 0 1 1 2 1 1 2 6 4 3 7 5 and P ¼ 1= ﬃﬃﬃ 3 p 1= ﬃﬃﬃ 3 p 1= ﬃﬃﬃ 3 p 0 1= ﬃﬃﬃ 2 p 1= ﬃﬃﬃ 2 p 2= ﬃﬃﬃ 6 p 1= ﬃﬃﬃ 6 p 1= ﬃﬃﬃ 6 p 2 6 6 6 4 3 7 7 7 5 Complex Matrices: Hermitian and Unitary Matrices 2.31. Find AH where (a) A ¼ 3 5i 2 þ 4i 6 þ 7i 1 þ 8i , (b) A ¼ 2 3i 5 þ 8i 4 3 7i 6 i 5i 2 4 3 5 Recall that AH ¼ AT, the conjugate tranpose of A. Thus, (a) AH ¼ 3 þ 5i 6 7i 2 4i 1 8i , (b) AH ¼ 2 þ 3i 4 6 þ i 5 8i 3 þ 7i 5i 2.32. Show that A ¼ 1 3 2 3 i 2 3 i 2 3 i 1 3 2 3 i " # is unitary. The rows of A form an orthonormal set: 1 3 2 3 i; 2 3 i 1 3 2 3 i; 2 3 i ¼ 1 9 þ 4 9 þ 4 9 ¼ 1 1 3 2 3 i; 2 3 i 2 3 i; 1 3 2 3 i ¼ 2 9 i þ 4 9 þ 2 9 i 4 9 ¼ 0 2 3 i; 1 3 2 3 i 2 3 i; 1 3 2 3 i ¼ 4 9 þ 1 9 þ 4 9 ¼ 1 Thus, A is unitary. 2.33. Prove the complex analogue of Theorem 2.6: Let A be a complex matrix. Then the following are equivalent: (i) A is unitary. (ii) The rows of A form an orthonormal set. (iii) The columns of A form an orthonormal set. (The proof is almost identical to the proof on page 37 for the case when A is a 3 3 real matrix.) First recall that the vectors u1; u2; . . . ; un in Cn form an orthonormal set if they are unit vectors and are orthogonal to each other, where the dot product in Cn is deﬁned by ða1; a2; . . . ; anÞ ðb1; b2; . . . ; bnÞ ¼ a1 b1 þ a2 b2 þ þ an bn Suppose A is unitary, and R1; R2; . . . ; Rn are its rows. Then RT 1 ; RT 2 ; . . . ; RT n are the columns of AH. Let AAH ¼ ½cij. By matrix multiplication, cij ¼ Ri RT j ¼ Ri Rj. Because A is unitary, we have AAH ¼ I. Multi-plying A by AH and setting each entry cij equal to the corresponding entry in I yields the following n2 equations: R1 R1 ¼ 1; R2 R2 ¼ 1; . . . ; Rn Rn ¼ 1; and Ri Rj ¼ 0; for i 6¼ j Thus, the rows of A are unit vectors and are orthogonal to each other; hence, they form an orthonormal set of vectors. The condition ATA ¼ I similarly shows that the columns of A also form an orthonormal set of vectors. Furthermore, because each step is reversible, the converse is true. This proves the theorem. Block Matrices 2.34. Consider the following block matrices (which are partitions of the same matrix): (a) 1 2 0 1 3 2 3 5 7 2 3 1 4 5 9 2 4 3 5, (b) 1 2 0 1 3 2 3 5 7 2 3 1 4 5 9 2 4 3 5 CHAPTER 2 Algebra of Matrices 49 Find the size of each block matrix and also the size of each block. (a) The block matrix has two rows of matrices and three columns of matrices; hence, its size is 2 3. The block sizes are 2 2, 2 2, and 2 1 for the ﬁrst row; and 1 2, 1 2, and 1 1 for the second row. (b) The size of the block matrix is 3 2; and the block sizes are 1 3 and 1 2 for each of the three rows. 2.35. Compute AB using block multiplication, where A ¼ 1 2 1 3 4 0 0 0 2 2 4 3 5 and B ¼ 1 2 3 1 4 5 6 1 0 0 0 1 2 4 3 5 Here A ¼ E F 01 2 G and B ¼ R S 01 3 T , where E; F; G; R; S; T are the given blocks, and 01 2 and 01 3 are zero matrices of the indicated sites. Hence, AB ¼ ER ES þ FT 01 3 GT ¼ ½ 0 0 0 9 12 15 19 26 33 2 3 7 þ 1 0 2 6 4 3 7 5 ¼ 9 12 15 4 19 26 33 7 0 0 0 2 2 4 3 5 2.36. Let M ¼ diagðA; B; CÞ, where A ¼ 1 2 3 4 , B ¼ ½5, C ¼ 1 3 5 7 . Find M2. Because M is block diagonal, square each block: A2 ¼ 7 10 15 22 ; B2 ¼ ½25; C2 ¼ 16 24 40 64 ; so M2 ¼ 7 10 15 22 25 16 24 40 64 2 6 6 6 6 4 3 7 7 7 7 5 Miscellaneous Problem 2.37. Let f ðxÞ and gðxÞ be polynomials and let A be a square matrix. Prove (a) ð f þ gÞðAÞ ¼ f ðAÞ þ gðAÞ, (b) ð f gÞðAÞ ¼ f ðAÞgðAÞ, (c) f ðAÞgðAÞ ¼ gðAÞ f ðAÞ. Suppose f ðxÞ ¼ Pr i¼1 aixi and gðxÞ ¼ Ps j¼1 bjxj. (a) We can assume r ¼ s ¼ n by adding powers of x with 0 as their coefﬁcients. Then f ðxÞ þ gðxÞ ¼ P n i¼1 ðai þ biÞxi Hence, ð f þ gÞðAÞ ¼ P n i¼1 ðai þ biÞAi ¼ P n i¼1 aiAi þ P n i¼1 biAi ¼ f ðAÞ þ gðAÞ (b) We have f ðxÞgðxÞ ¼ P i;j aibjxiþj. Then f ðAÞgðAÞ ¼ P i aiAi ! P j bjAj ! ¼ P i;j aibjAiþj ¼ ð fgÞðAÞ (c) Using f ðxÞgðxÞ ¼ gðxÞf ðxÞ, we have f ðAÞgðAÞ ¼ ð fgÞðAÞ ¼ ðg f ÞðAÞ ¼ gðAÞ f ðAÞ 50 CHAPTER 2 Algebra of Matrices SUPPLEMENTARY PROBLEMS Algebra of Matrices Problems 2.38–2.41 refer to the following matrices: A ¼ 1 2 3 4 ; B ¼ 5 0 6 7 ; C ¼ 1 3 4 2 6 5 ; D ¼ 3 7 1 4 8 9 2.38. Find (a) 5A 2B, (b) 2A þ 3B, (c) 2C 3D. 2.39. Find (a) AB and ðABÞC, (b) BC and AðBCÞ. [Note that ðABÞC ¼ AðBCÞ.] 2.40. Find (a) A2 and A3, (b) AD and BD, (c) CD. 2.41. Find (a) AT, (b) BT, (c) ðABÞT, (d) ATBT. [Note that ATBT 6¼ ðABÞT.] Problems 2.42 and 2.43 refer to the following matrices: A ¼ 1 1 2 0 3 4 ; B ¼ 4 0 3 1 2 3 ; C ¼ 2 3 0 1 5 1 4 2 1 0 0 3 2 4 3 5; D ¼ 2 1 3 2 4 3 5: 2.42. Find (a) 3A 4B, (b) AC, (c) BC, (d) AD, (e) BD, ( f ) CD. 2.43. Find (a) AT, (b) ATB, (c) ATC. 2.44. Let A ¼ 1 2 3 6 . Find a 2 3 matrix B with distinct nonzero entries such that AB ¼ 0. 2.45 Let e1 ¼ ½1; 0; 0, e2 ¼ ½0; 1; 0, e3 ¼ ½0; 0; 1, and A ¼ a1 a2 a3 a4 b1 b2 b3 b4 c1 c2 c3 c4 2 4 3 5. Find e1A, e2A, e3A. 2.46. Let ei ¼ ½0; . . . ; 0; 1; 0; . . . ; 0, where 1 is the ith entry. Show (a) eiA ¼ Ai, ith row of A. (c) If eiA ¼ eiB, for each i, then A ¼ B. (b) BeT j ¼ Bj, jth column of B. (d) If AeT j ¼ BeT j , for each j, then A ¼ B. 2.47. Prove Theorem 2.2(iii) and (iv): (iii) ðB þ CÞA ¼ BA þ CA, (iv) kðABÞ ¼ ðkAÞB ¼ AðkBÞ. 2.48. Prove Theorem 2.3: (i) ðA þ BÞT ¼ AT þ BT, (ii) ðATÞT ¼ A, (iii) ðkAÞT ¼ kAT. 2.49. Show (a) If A has a zero row, then AB has a zero row. (b) If B has a zero column, then AB has a zero column. Square Matrices, Inverses 2.50. Find the diagonal and trace of each of the following matrices: (a) A ¼ 2 5 8 3 6 7 4 0 1 2 4 3 5, (b) B ¼ 1 3 4 6 1 7 2 5 1 2 4 3 5, (c) C ¼ 4 3 6 2 5 0 Problems 2.51–2.53 refer to A ¼ 2 5 3 1 , B ¼ 4 2 1 6 , C ¼ 6 4 3 2 . 2.51. Find (a) A2 and A3, (b) f ðAÞ and gðAÞ, where f ðxÞ ¼ x3 2x2 5; gðxÞ ¼ x2 3x þ 17: CHAPTER 2 Algebra of Matrices 51 2.52. Find (a) B2 and B3, (b) f ðBÞ and gðBÞ, where f ðxÞ ¼ x2 þ 2x 22; gðxÞ ¼ x2 3x 6: 2.53. Find a nonzero column vector u such that Cu ¼ 4u. 2.54. Find the inverse of each of the following matrices (if it exists): A ¼ 7 4 5 3 ; B ¼ 2 3 4 5 ; C ¼ 4 6 2 3 ; D ¼ 5 2 6 3 2.55. Find the inverses of A ¼ 1 1 2 1 2 5 1 3 7 2 4 3 5 and B ¼ 1 1 1 0 1 1 1 3 2 2 4 3 5. [Hint: See Problem 2.19.] 2.56. Suppose A is invertible. Show that if AB ¼ AC, then B ¼ C. Give an example of a nonzero matrix A such that AB ¼ AC but B 6¼ C. 2.57. Find 2 2 invertible matrices A and B such that A þ B 6¼ 0 and A þ B is not invertible. 2.58. Show (a) A is invertible if and only if AT is invertible. (b) The operations of inversion and transpose commute; that is, ðATÞ1 ¼ ðA1ÞT. (c) If A has a zero row or zero column, then A is not invertible. Diagonal and triangular matrices 2.59. Let A ¼ diagð1; 2; 3Þ and B ¼ diagð2; 5; 0Þ. Find (a) AB, A2, B2; (b) f ðAÞ, where f ðxÞ ¼ x2 þ 4x 3; (c) A1 and B1. 2.60. Let A ¼ 1 2 0 1 and B ¼ 1 1 0 0 1 1 0 0 1 2 4 3 5. (a) Find An. (b) Find Bn. 2.61. Find all real triangular matrices A such that A2 ¼ B, where (a) B ¼ 4 21 0 25 , (b) B ¼ 1 4 0 9 . 2.62. Let A ¼ 5 2 0 k . Find all numbers k for which A is a root of the polynomial: (a) f ðxÞ ¼ x2 7x þ 10, (b) gðxÞ ¼ x2 25, (c) hðxÞ ¼ x2 4. 2.63. Let B ¼ 1 0 26 27 : Find a matrix A such that A3 ¼ B. 2.64. Let B ¼ 1 8 5 0 9 5 0 0 4 2 4 3 5. Find a triangular matrix A with positive diagonal entries such that A2 ¼ B. 2.65. Using only the elements 0 and 1, ﬁnd the number of 3 3 matrices that are (a) diagonal, (b) upper triangular, (c) nonsingular and upper triangular. Generalize to n n matrices. 2.66. Let Dk ¼ kI, the scalar matrix belonging to the scalar k. Show (a) DkA ¼ kA, (b) BDk ¼ kB, (c) Dk þ Dk0 ¼ Dkþk0, (d) DkDk0 ¼ Dkk0 2.67. Suppose AB ¼ C, where A and C are upper triangular. (a) Find 2 2 nonzero matrices A; B; C, where B is not upper triangular. (b) Suppose A is also invertible. Show that B must also be upper triangular. 52 CHAPTER 2 Algebra of Matrices Special Types of Real Matrices 2.68. Find x; y; z such that A is symmetric, where (a) A ¼ 2 x 3 4 5 y z 1 7 2 4 3 5, (b) A ¼ 7 6 2x y z 2 x 2 5 2 4 3 5. 2.69. Suppose A is a square matrix. Show (a) A þ AT is symmetric, (b) A AT is skew-symmetric, (c) A ¼ B þ C, where B is symmetric and C is skew-symmetric. 2.70. Write A ¼ 4 5 1 3 as the sum of a symmetric matrix B and a skew-symmetric matrix C. 2.71. Suppose A and B are symmetric. Show that the following are also symmetric: (a) A þ B; (b) kA, for any scalar k; (c) A2; (d) An, for n > 0; (e) f ðAÞ, for any polynomial f ðxÞ. 2.72. Find a 2 2 orthogonal matrix P whose ﬁrst row is a multiple of (a) ð3; 4Þ, (b) ð1; 2Þ. 2.73. Find a 3 3 orthogonal matrix P whose ﬁrst two rows are multiples of (a) ð1; 2; 3Þ and ð0; 2; 3Þ, (b) ð1; 3; 1Þ and ð1; 0; 1Þ. 2.74. Suppose A and B are orthogonal matrices. Show that AT, A1, AB are also orthogonal. 2.75. Which of the following matrices are normal? A ¼ 3 4 4 3 , B ¼ 1 2 2 3 , C ¼ 1 1 1 0 1 1 0 0 1 2 4 3 5. Complex Matrices 2.76. Find real numbers x; y; z such that A is Hermitian, where A ¼ 3 x þ 2i yi 3 2i 0 1 þ zi yi 1 xi 1 2 4 3 5: 2.77. Suppose A is a complex matrix. Show that AAH and AHA are Hermitian. 2.78. Let A be a square matrix. Show that (a) A þ AH is Hermitian, (b) A AH is skew-Hermitian, (c) A ¼ B þ C, where B is Hermitian and C is skew-Hermitian. 2.79. Determine which of the following matrices are unitary: A ¼ i=2 ﬃﬃﬃ 3 p =2 ﬃﬃﬃ 3 p =2 i=2 ; B ¼ 1 2 1 þ i 1 i 1 i 1 þ i ; C ¼ 1 2 1 i 1 þ i i 1 1 þ i 1 þ i 1 þ i 0 2 4 3 5 2.80. Suppose A and B are unitary. Show that AH, A1, AB are unitary. 2.81. Determine which of the following matrices are normal: A ¼ 3 þ 4i 1 i 2 þ 3i and B ¼ 1 0 1 i i . CHAPTER 2 Algebra of Matrices 53 Block Matrices 2.82. Let U ¼ 1 2 0 0 0 3 4 0 0 0 0 0 5 1 2 0 0 3 4 1 2 6 6 4 3 7 7 5 and V ¼ 3 2 0 0 2 4 0 0 0 0 1 2 0 0 2 3 0 0 4 1 2 6 6 6 6 4 3 7 7 7 7 5 . (a) Find UV using block multiplication. (b) Are U and V block diagonal matrices? (c) Is UV block diagonal? 2.83. Partition each of the following matrices so that it becomes a square block matrix with as many diagonal blocks as possible: A ¼ 1 0 0 0 0 2 0 0 3 2 4 3 5; B ¼ 1 2 0 0 0 3 0 0 0 0 0 0 4 0 0 0 0 5 0 0 0 0 0 0 6 2 6 6 6 6 4 3 7 7 7 7 5 ; C ¼ 0 1 0 0 0 0 2 0 0 2 4 3 5 2.84. Find M2 and M3 for (a) M ¼ 2 0 0 0 0 1 4 0 0 2 1 0 0 0 0 3 2 6 6 4 3 7 7 5, (b) M ¼ 1 1 0 0 2 3 0 0 0 0 1 2 0 0 4 5 2 6 6 4 3 7 7 5. 2.85. For each matrix M in Problem 2.84, ﬁnd f ðMÞ where f ðxÞ ¼ x2 þ 4x 5. 2.86. Suppose U ¼ ½Uik and V ¼ ½Vkj are block matrices for which UV is deﬁned and the number of columns of each block Uik is equal to the number of rows of each block Vkj. Show that UV ¼ ½Wij, where Wij ¼ P k UikVkj. 2.87. Suppose M and N are block diagonal matrices where corresponding blocks have the same size, say M ¼ diagðAiÞ and N ¼ diagðBiÞ. Show (i) M þ N ¼ diagðAi þ BiÞ, (iii) MN ¼ diagðAiBiÞ, (ii) kM ¼ diagðkAiÞ, (iv) f ðMÞ ¼ diagð f ðAiÞÞ for any polynomial f ðxÞ. ANSWERS TO SUPPLEMENTARY PROBLEMS Notation: A ¼ ½R1; R2; . . . denotes a matrix A with rows R1; R2; . . . . 2.38. (a) ½5; 10; 27; 34, (b) ½17; 4; 12; 13, (c) ½7; 27; 11; 8; 36; 37 2.39. (a) ½7; 14; 39; 28, ½21; 105; 98; 17; 285; 296 (b) ½5; 15; 20; 8; 60; 59, ½21; 105; 98; 17; 285; 296 2.40. (a) ½7; 6; 9; 22, ½11; 38; 57; 106; (b) ½11; 9; 17; 7; 53; 39, ½15; 35; 5; 10; 98; 69; (c) not deﬁned 2.41. (a) ½1; 3; 2; 4, (b) ½5; 6; 0; 7, (c) ½7; 39; 14; 28; (d) ½5; 15; 10; 40 2.42. (a) ½13; 3; 18; 4; 17; 0, (b) ½5; 2; 4; 5; 11; 3; 12; 18, (c) ½11; 12; 0; 5; 15; 5; 8; 4, (d) ½9; 9, (e) ½1; 9, (f ) not deﬁned 54 CHAPTER 2 Algebra of Matrices 2.43. (a) ½1; 0; 1; 3; 2; 4, (b) ½4; 0; 3; 7; 6; 12; 4; 8; 6], (c) not deﬁned 2.44. ½2; 4; 6; 1; 2; 3 2.45. ½a1; a2; a3; a4, ½b1; b2; b3; b4, ½c1; c2; c3; c4 2.50. (a) 2; 6; 1; trðAÞ ¼ 5, (b) 1; 1; 1; trðBÞ ¼ 1, (c) not deﬁned 2.51. (a) ½11; 15; 9; 14, ½67; 40; 24; 59, (b) ½50; 70; 42; 36, gðAÞ ¼ 0 2.52. (a) ½14; 4; 2; 34, ½60; 52; 26; 200, (b) f ðBÞ ¼ 0, ½4; 10; 5; 46 2.53. u ¼ ½2a; aT 2.54. ½3; 4; 5; 7, ½ 5 2 ; 3 2; 2; 1, not deﬁned, ½1; 2 3; 2; 5 3 2.55. ½1; 1; 1; 2; 5; 3; 1; 2; 1, ½1; 1; 0; 1; 3; 1; 1; 4; 1 2.56. A ¼ ½1; 2; 1; 2, B ¼ ½0; 0; 1; 1, C ¼ ½2; 2; 0; 0 2.57. A ¼ ½1; 2; 0; 3; B ¼ ½4; 3; 3; 0 2.58. (c) Hint: Use Problem 2.48 2.59. (a) AB ¼ diagð2; 10; 0Þ, A2 ¼ diagð1; 4; 9Þ, B2 ¼ diagð4; 25; 0Þ; (b) f ðAÞ ¼ diagð2; 9; 6Þ; (c) A1 ¼ diagð1; 1 2 ; 1 3Þ, C1 does not exist 2.60. (a) ½1; 2n; 0; 1, (b) ½1; n; 1 2 nðn 1Þ; 0; 1; n; 0; 0; 1 2.61. (a) ½2; 3; 0; 5, ½2; 3; 0; 5, ½2; 7; 0; 5, ½2; 7; 0; 5, (b) none 2.62. (a) k ¼ 2, (b) k ¼ 5, (c) none 2.63. ½1; 0; 2; 3 2.64. ½1; 2; 1; 0; 3; 1; 0; 0; 2 2.65. All entries below the diagonal must be 0 to be upper triangular, and all diagonal entries must be 1 to be nonsingular. (a) 8 ð2nÞ, (b) 26 ð2nðnþ1Þ=2Þ, (c) 23 ð2nðn1Þ=2Þ 2.67. (a) A ¼ ½1; 1; 0; 0, B ¼ ½1; 2; 3; 4, C ¼ ½4; 6; 0; 0 2.68. (a) x ¼ 4, y ¼ 1, z ¼ 3; (b) x ¼ 0, y ¼ 6, z any real number 2.69. (c) Hint: Let B ¼ 1 2 ðA þ ATÞ and C ¼ 1 2 ðA ATÞ: 2.70. B ¼ ½4; 3; 3; 3, C ¼ ½0; 2; 2; 0 2.72. (a) ½3 5, 4 5; 4 5, 3 5], (b) ½1= ﬃﬃﬃ 5 p , 2= ﬃﬃﬃ 5 p ; 2= ﬃﬃﬃ 5 p , 1= ﬃﬃﬃ 5 p 2.73. (a) ½1= ﬃﬃﬃﬃﬃ 14 p , 2= ﬃﬃﬃﬃﬃ 14 p , 3= ﬃﬃﬃﬃﬃ 14 p ; 0; 2= ﬃﬃﬃﬃﬃ 13 p , 3= ﬃﬃﬃﬃﬃ 13 p ; 12= ﬃﬃﬃﬃﬃﬃﬃ ﬃ 157 p , 3= ﬃﬃﬃﬃﬃﬃﬃ ﬃ 157 p , 2= ﬃﬃﬃﬃﬃﬃﬃ ﬃ 157 p (b) ½1= ﬃﬃﬃﬃﬃ 11 p , 3= ﬃﬃﬃﬃﬃ 11 p , 1= ﬃﬃﬃﬃﬃ 11 p ; 1= ﬃﬃﬃ 2 p , 0; 1= ﬃﬃﬃ 2 p ; 3= ﬃﬃﬃﬃﬃ 22 p , 2= ﬃﬃﬃﬃﬃ 22 p , 3= ﬃﬃﬃﬃﬃ 22 p 2.75. A; C CHAPTER 2 Algebra of Matrices 55 2.76. x ¼ 3, y ¼ 0, z ¼ 3 2.78. (c) Hint: Let B ¼ 1 2 ðA þ AHÞ and C ¼ 1 2 ðA AHÞ. 2.79. A; B; C 2.81. A 2.82. (a) UV ¼ diagð½7; 6; 17; 10; ½1; 9; 7; 5); (b) no; (c) yes 2.83. A: line between ﬁrst and second rows (columns); B: line between second and third rows (columns) and between fourth and ﬁfth rows (columns); C: C itself—no further partitioning of C is possible. 2.84. (a) M2 ¼ diagð½4, ½9; 8; 4; 9, ½9Þ, M3 ¼ diagð½8; ½25; 44; 22; 25, ½27Þ (b) M2 ¼ diagð½3; 4; 8; 11, ½9; 12; 24; 33Þ M3 ¼ diagð½11; 15; 30; 41, ½57; 78; 156; 213Þ 2.85. (a) diagð½7, ½8; 24; 12; 8, ½16Þ, (b) diagð½2; 8; 16; 181], ½8; 20; 40; 48Þ 56 CHAPTER 2 Algebra of Matrices Systems of Linear Equations 3.1 Introduction Systems of linear equations play an important and motivating role in the subject of linear algebra. In fact, many problems in linear algebra reduce to ﬁnding the solution of a system of linear equations. Thus, the techniques introduced in this chapter will be applicable to abstract ideas introduced later. On the other hand, some of the abstract results will give us new insights into the structure and properties of systems of linear equations. All our systems of linear equations involve scalars as both coefﬁcients and constants, and such scalars may come from any number ﬁeld K. There is almost no loss in generality if the reader assumes that all our scalars are real numbers—that is, that they come from the real ﬁeld R. 3.2 Basic Deﬁnitions, Solutions This section gives basic deﬁnitions connected with the solutions of systems of linear equations. The actual algorithms for ﬁnding such solutions will be treated later. Linear Equation and Solutions A linear equation in unknowns x1; x2; . . . ; xn is an equation that can be put in the standard form a1x1 þ a2x2 þ þ anxn ¼ b ð3:1Þ where a1; a2; . . . ; an, and b are constants. The constant ak is called the coefﬁcient of xk, and b is called the constant term of the equation. A solution of the linear equation (3.1) is a list of values for the unknowns or, equivalently, a vector u in Kn, say x1 ¼ k1; x2 ¼ k2; . . . ; xn ¼ kn or u ¼ ðk1; k2; . . . ; knÞ such that the following statement (obtained by substituting ki for xi in the equation) is true: a1k1 þ a2k2 þ þ ankn ¼ b In such a case we say that u satisﬁes the equation. Remark: Equation (3.1) implicitly assumes there is an ordering of the unknowns. In order to avoid subscripts, we will usually use x; y for two unknowns; x; y; z for three unknowns; and x; y; z; t for four unknowns; they will be ordered as shown. 57 CHAPTER 3 EXAMPLE 3.1 Consider the following linear equation in three unknowns x; y; z: x þ 2y 3z ¼ 6 We note that x ¼ 5; y ¼ 2; z ¼ 1, or, equivalently, the vector u ¼ ð5; 2; 1Þ is a solution of the equation. That is, 5 þ 2ð2Þ 3ð1Þ ¼ 6 or 5 þ 4 3 ¼ 6 or 6 ¼ 6 On the other hand, w ¼ ð1; 2; 3Þ is not a solution, because on substitution, we do not get a true statement: 1 þ 2ð2Þ 3ð3Þ ¼ 6 or 1 þ 4 9 ¼ 6 or 4 ¼ 6 System of Linear Equations A system of linear equations is a list of linear equations with the same unknowns. In particular, a system of m linear equations L1; L2; . . . ; Lm in n unknowns x1; x2; . . . ; xn can be put in the standard form a11x1 þ a12x2 þ þ a1nxn ¼ b1 a21x1 þ a22x2 þ þ a2nxn ¼ b2 ð3:2Þ ::::::::::::::::::::::::::::::::::::::::::::::::::: am1x1 þ am2x2 þ þ amnxn ¼ bm where the aij and bi are constants. The number aij is the coefﬁcient of the unknown xj in the equation Li, and the number bi is the constant of the equation Li. The system (3.2) is called an m n (read: m by n) system. It is called a square system if m ¼ n—that is, if the number m of equations is equal to the number n of unknowns. The system (3.2) is said to be homogeneous if all the constant terms are zero—that is, if b1 ¼ 0, b2 ¼ 0; . . . ; bm ¼ 0. Otherwise the system is said to be nonhomogeneous. A solution (or a particular solution) of the system (3.2) is a list of values for the unknowns or, equivalently, a vector u in Kn, which is a solution of each of the equations in the system. The set of all solutions of the system is called the solution set or the general solution of the system. EXAMPLE 3.2 Consider the following system of linear equations: x1 þ x2 þ 4x3 þ 3x4 ¼ 5 2x1 þ 3x2 þ x3 2x4 ¼ 1 x1 þ 2x2 5x3 þ 4x4 ¼ 3 It is a 3 4 system because it has three equations in four unknowns. Determine whether (a) u ¼ ð8; 6; 1; 1Þ and (b) v ¼ ð10; 5; 1; 2Þ are solutions of the system. (a) Substitute the values of u in each equation, obtaining 8 þ 6 þ 4ð1Þ þ 3ð1Þ ¼ 5 or 8 þ 6 þ 4 þ 3 ¼ 5 or 5 ¼ 5 2ð8Þ þ 3ð6Þ þ 1 2ð1Þ ¼ 1 or 16 þ 18 þ 1 2 ¼ 1 or 1 ¼ 1 8 þ 2ð6Þ 5ð1Þ þ 4ð1Þ ¼ 3 or 8 þ 12 5 þ 4 ¼ 3 or 3 ¼ 3 Yes, u is a solution of the system because it is a solution of each equation. (b) Substitute the values of v into each successive equation, obtaining 10 þ 5 þ 4ð1Þ þ 3ð2Þ ¼ 5 or 10 þ 5 þ 4 þ 6 ¼ 5 or 5 ¼ 5 2ð10Þ þ 3ð5Þ þ 1 2ð2Þ ¼ 1 or 20 þ 15 þ 1 4 ¼ 1 or 8 ¼ 1 No, v is not a solution of the system, because it is not a solution of the second equation. (We do not need to substitute v into the third equation.) 58 CHAPTER 3 Systems of Linear Equations The system (3.2) of linear equations is said to be consistent if it has one or more solutions, and it is said to be inconsistent if it has no solution. If the ﬁeld K of scalars is inﬁnite, such as when K is the real ﬁeld R or the complex ﬁeld C, then we have the following important result. THEOREM 3.1: Suppose the ﬁeld K is inﬁnite. Then any system l of linear equations has (i) a unique solution, (ii) no solution, or (iii) an inﬁnite number of solutions. This situation is pictured in Fig. 3-1. The three cases have a geometrical description when the system l consists of two equations in two unknowns (Section 3.4). Augmented and Coefﬁcient Matrices of a System Consider again the general system (3.2) of m equations in n unknowns. Such a system has associated with it the following two matrices: M ¼ a11 a12 . . . a1n b1 a21 a22 . . . a2n b2 ::::::::::::::::::::::::::::::::::::::: am1 am2 . . . amn bn 2 6 6 4 3 7 7 5 and A ¼ a11 a12 . . . a1n a21 a22 . . . a2n ::::::::::::::::::::::::::::::: am1 am2 . . . amn 2 6 6 4 3 7 7 5 The ﬁrst matrix M is called the augmented matrix of the system, and the second matrix A is called the coefﬁcient matrix. The coefﬁcient matrix A is simply the matrix of coefﬁcients, which is the augmented matrix M without the last column of constants. Some texts write M ¼ ½A; B to emphasize the two parts of M, where B denotes the column vector of constants. The augmented matrix M and the coefﬁcient matrix A of the system in Example 3.2 are as follows: M ¼ 1 1 4 3 5 2 3 1 2 1 1 2 5 4 3 2 4 3 5 and A ¼ 1 1 4 3 2 3 1 2 1 2 5 4 2 4 3 5 As expected, A consists of all the columns of M except the last, which is the column of constants. Clearly, a system of linear equations is completely determined by its augmented matrix M, and vice versa. Speciﬁcally, each row of M corresponds to an equation of the system, and each column of M corresponds to the coefﬁcients of an unknown, except for the last column, which corresponds to the constants of the system. Degenerate Linear Equations A linear equation is said to be degenerate if all the coefﬁcients are zero—that is, if it has the form 0x1 þ 0x2 þ þ 0xn ¼ b ð3:3Þ Figure 3-1 CHAPTER 3 Systems of Linear Equations 59 The solution of such an equation depends only on the value of the constant b. Speciﬁcally, (i) If b 6¼ 0, then the equation has no solution. (ii) If b ¼ 0, then every vector u ¼ ðk1; k2; . . . ; knÞ in Kn is a solution. The following theorem applies. THEOREM 3.2: Let l be a system of linear equations that contains a degenerate equation L, say with constant b. (i) If b 6¼ 0, then the system l has no solution. (ii) If b ¼ 0, then L may be deleted from the system without changing the solution set of the system. Part (i) comes from the fact that the degenerate equation has no solution, so the system has no solution. Part (ii) comes from the fact that every element in Kn is a solution of the degenerate equation. Leading Unknown in a Nondegenerate Linear Equation Now let L be a nondegenerate linear equation. This means one or more of the coefﬁcients of L are not zero. By the leading unknown of L, we mean the ﬁrst unknown in L with a nonzero coefﬁcient. For example, x3 and y are the leading unknowns, respectively, in the equations 0x1 þ 0x2 þ 5x3 þ 6x4 þ 0x5 þ 8x6 ¼ 7 and 0x þ 2y 4z ¼ 5 We frequently omit terms with zero coefﬁcients, so the above equations would be written as 5x3 þ 6x4 þ 8x6 ¼ 7 and 2y 4z ¼ 5 In such a case, the leading unknown appears ﬁrst. 3.3 Equivalent Systems, Elementary Operations Consider the system (3.2) of m linear equations in n unknowns. Let L be the linear equation obtained by multiplying the m equations by constants c1; c2; . . . ; cm, respectively, and then adding the resulting equations. Speciﬁcally, let L be the following linear equation: ðc1a11 þ þ cmam1Þx1 þ þ ðc1a1n þ þ cmamnÞxn ¼ c1b1 þ þ cmbm Then L is called a linear combination of the equations in the system. One can easily show (Problem 3.43) that any solution of the system (3.2) is also a solution of the linear combination L. EXAMPLE 3.3 Let L1, L2, L3 denote, respectively, the three equations in Example 3.2. Let L be the equation obtained by multiplying L1, L2, L3 by 3; 2; 4, respectively, and then adding. Namely, 3L1: 3x1 þ 3x2 þ 12x3 þ 9x4 ¼ 15 2L2: 4x1 6x2 2x3 þ 4x4 ¼ 2 4L1: 4x1 þ 8x2 20x3 þ 16x4 ¼ 12 ðSumÞ L: 3x1 þ 5x2 10x3 þ 29x4 ¼ 25 60 CHAPTER 3 Systems of Linear Equations Then L is a linear combination of L1, L2, L3. As expected, the solution u ¼ ð8; 6; 1; 1Þ of the system is also a solution of L. That is, substituting u in L, we obtain a true statement: 3ð8Þ þ 5ð6Þ 10ð1Þ þ 29ð1Þ ¼ 25 or 24 þ 30 10 þ 29 ¼ 25 or 9 ¼ 9 The following theorem holds. THEOREM 3.3: Two systems of linear equations have the same solutions if and only if each equation in each system is a linear combination of the equations in the other system. Two systems of linear equations are said to be equivalent if they have the same solutions. The next subsection shows one way to obtain equivalent systems of linear equations. Elementary Operations The following operations on a system of linear equations L1; L2; . . . ; Lm are called elementary operations. ½E1 Interchange two of the equations. We indicate that the equations Li and Lj are interchanged by writing: ‘‘Interchange Li and Lj’’ or ‘‘Li ! Lj’’ ½E2 Replace an equation by a nonzero multiple of itself. We indicate that equation Li is replaced by kLi (where k 6¼ 0) by writing ‘‘Replace Li by kLi’’ or ‘‘kLi ! Li’’ ½E3 Replace an equation by the sum of a multiple of another equation and itself. We indicate that equation Lj is replaced by the sum of kLi and Lj by writing ‘‘Replace Lj by kLi þ Lj’’ or ‘‘kLi þ Lj ! Lj’’ The arrow ! in ½E2 and ½E3 may be read as ‘‘replaces.’’ The main property of the above elementary operations is contained in the following theorem (proved in Problem 3.45). THEOREM 3.4: Suppose a system of m of linear equations is obtained from a system l of linear equations by a ﬁnite sequence of elementary operations. Then m and l have the same solutions. Remark: Sometimes (say to avoid fractions when all the given scalars are integers) we may apply ½E2 and ½E3 in one step; that is, we may apply the following operation: ½E Replace equation Lj by the sum of kLi and k0Lj (where k0 6¼ 0), written ‘‘Replace Lj by kLi þ k0Lj’’ or ‘‘kLi þ k0Lj ! Lj’’ We emphasize that in operations ½E3 and [E], only equation Lj is changed. Gaussian elimination, our main method for ﬁnding the solution of a given system of linear equations, consists of using the above operations to transform a given system into an equivalent system whose solution can be easily obtained. The details of Gaussian elimination are discussed in subsequent sections. 3.4 Small Square Systems of Linear Equations This section considers the special case of one equation in one unknown, and two equations in two unknowns. These simple systems are treated separately because their solution sets can be described geometrically, and their properties motivate the general case. CHAPTER 3 Systems of Linear Equations 61 Linear Equation in One Unknown The following simple basic result is proved in Problem 3.5. THEOREM 3.5: Consider the linear equation ax ¼ b. (i) If a 6¼ 0, then x ¼ b=a is a unique solution of ax ¼ b. (ii) If a ¼ 0, but b 6¼ 0, then ax ¼ b has no solution. (iii) If a ¼ 0 and b ¼ 0, then every scalar k is a solution of ax ¼ b. EXAMPLE 3.4 Solve (a) 4x 1 ¼ x þ 6, (b) 2x 5 x ¼ x þ 3, (c) 4 þ x 3 ¼ 2x þ 1 x. (a) Rewrite the equation in standard form obtaining 3x ¼ 7. Then x ¼ 7 3 is the unique solution [Theorem 3.5(i)]. (b) Rewrite the equation in standard form, obtaining 0x ¼ 8. The equation has no solution [Theorem 3.5(ii)]. (c) Rewrite the equation in standard form, obtaining 0x ¼ 0. Then every scalar k is a solution [Theorem 3.5(iii)]. System of Two Linear Equations in Two Unknowns (2 2 System) Consider a system of two nondegenerate linear equations in two unknowns x and y, which can be put in the standard form A1x þ B1y ¼ C1 A2x þ B2y ¼ C2 ð3:4Þ Because the equations are nondegenerate, A1 and B1 are not both zero, and A2 and B2 are not both zero. The general solution of the system (3.4) belongs to one of three types as indicated in Fig. 3-1. If R is the ﬁeld of scalars, then the graph of each equation is a line in the plane R2 and the three types may be described geometrically as pictured in Fig. 3-2. Speciﬁcally, (1) The system has exactly one solution. Here the two lines intersect in one point [Fig. 3-2(a)]. This occurs when the lines have distinct slopes or, equivalently, when the coefﬁcients of x and y are not proportional: A1 A2 6¼ B1 B2 or; equivalently; A1B2 A2B1 6¼ 0 For example, in Fig. 3-2(a), 1=3 6¼ 1=2. y L1 x L2 0 –3 3 –3 3 L x y L x y 1 2 : – = – 1 : 3 + 2 = 12 6 (a) y (b) L1 x L2 0 3 –3 3 L x y L x y 1 2 : + 3 = 3 : 2 + 6 = – 8 6 –3 y (c) L L 1 2 and x 0 3 –3 3 L x y L x y 1 2 : + 2 = 4 : 2 + 4 = 8 6 –3 Figure 3-2 62 CHAPTER 3 Systems of Linear Equations (2) The system has no solution. Here the two lines are parallel [Fig. 3-2(b)]. This occurs when the lines have the same slopes but different y intercepts, or when A1 A2 ¼ B1 B2 6¼ C1 C2 For example, in Fig. 3-2(b), 1=2 ¼ 3=6 6¼ 3=8. (3) The system has an inﬁnite number of solutions. Here the two lines coincide [Fig. 3-2(c)]. This occurs when the lines have the same slopes and same y intercepts, or when the coefﬁcients and constants are proportional, A1 A2 ¼ B1 B2 ¼ C1 C2 For example, in Fig. 3-2(c), 1=2 ¼ 2=4 ¼ 4=8. Remark: The following expression and its value is called a determinant of order two: A1 B1 A2 B2 ¼ A1B2 A2B1 Determinants will be studied in Chapter 8. Thus, the system (3.4) has a unique solution if and only if the determinant of its coefﬁcients is not zero. (We show later that this statement is true for any square system of linear equations.) Elimination Algorithm The solution to system (3.4) can be obtained by the process of elimination, whereby we reduce the system to a single equation in only one unknown. Assuming the system has a unique solution, this elimination algorithm has two parts. ALGORITHM 3.1: The input consists of two nondegenerate linear equations L1 and L2 in two unknowns with a unique solution. Part A. (Forward Elimination) Multiply each equation by a constant so that the resulting coefﬁcients of one unknown are negatives of each other, and then add the two equations to obtain a new equation L that has only one unknown. Part B. (Back-Substitution) Solve for the unknown in the new equation L (which contains only one unknown), substitute this value of the unknown into one of the original equations, and then solve to obtain the value of the other unknown. Part A of Algorithm 3.1 can be applied to any system even if the system does not have a unique solution. In such a case, the new equation L will be degenerate and Part B will not apply. EXAMPLE 3.5 (Unique Case). Solve the system L1: 2x 3y ¼ 8 L2: 3x þ 4y ¼ 5 The unknown x is eliminated from the equations by forming the new equation L ¼ 3L1 þ 2L2. That is, we multiply L1 by 3 and L2 by 2 and add the resulting equations as follows: 3L1: 6x þ 9y ¼ 24 2L2: 6x þ 8y ¼ 10 Addition : 17y ¼ 34 CHAPTER 3 Systems of Linear Equations 63 We now solve the new equation for y, obtaining y ¼ 2. We substitute y ¼ 2 into one of the original equations, say L1, and solve for the other unknown x, obtaining 2x 3ð2Þ ¼ 8 or 2x 6 ¼ 8 or 2x ¼ 2 or x ¼ 1 Thus, x ¼ 1, y ¼ 2, or the pair u ¼ ð1; 2Þ is the unique solution of the system. The unique solution is expected, because 2=3 6¼ 3=4. [Geometrically, the lines corresponding to the equations intersect at the point ð1; 2Þ.] EXAMPLE 3.6 (Nonunique Cases) (a) Solve the system L1: x 3y ¼ 4 L2: 2x þ 6y ¼ 5 We eliminated x from the equations by multiplying L1 by 2 and adding it to L2—that is, by forming the new equation L ¼ 2L1 þ L2. This yields the degenerate equation 0x þ 0y ¼ 13 which has a nonzero constant b ¼ 13. Thus, this equation and the system have no solution. This is expected, because 1=ð2Þ ¼ 3=6 6¼ 4=5. (Geometrically, the lines corresponding to the equations are parallel.) (b) Solve the system L1: x 3y ¼ 4 L2: 2x þ 6y ¼ 8 We eliminated x from the equations by multiplying L1 by 2 and adding it to L2—that is, by forming the new equation L ¼ 2L1 þ L2. This yields the degenerate equation 0x þ 0y ¼ 0 where the constant term is also zero. Thus, the system has an inﬁnite number of solutions, which correspond to the solutions of either equation. This is expected, because 1=ð2Þ ¼ 3=6 ¼ 4=ð8Þ. (Geometrically, the lines corresponding to the equations coincide.) To ﬁnd the general solution, let y ¼ a, and substitute into L1 to obtain x 3a ¼ 4 or x ¼ 3a þ 4 Thus, the general solution of the system is x ¼ 3a þ 4; y ¼ a or u ¼ ð3a þ 4; aÞ where a (called a parameter) is any scalar. 3.5 Systems in Triangular and Echelon Forms The main method for solving systems of linear equations, Gaussian elimination, is treated in Section 3.6. Here we consider two simple types of systems of linear equations: systems in triangular form and the more general systems in echelon form. Triangular Form Consider the following system of linear equations, which is in triangular form: 2x1 3x2 þ 5x3 2x4 ¼ 9 5x2 x3 þ 3x4 ¼ 1 7x3 x4 ¼ 3 2x4 ¼ 8 64 CHAPTER 3 Systems of Linear Equations That is, the ﬁrst unknown x1 is the leading unknown in the ﬁrst equation, the second unknown x2 is the leading unknown in the second equation, and so on. Thus, in particular, the system is square and each leading unknown is directly to the right of the leading unknown in the preceding equation. Such a triangular system always has a unique solution, which may be obtained by back-substitution. That is, (1) First solve the last equation for the last unknown to get x4 ¼ 4. (2) Then substitute this value x4 ¼ 4 in the next-to-last equation, and solve for the next-to-last unknown x3 as follows: 7x3 4 ¼ 3 or 7x3 ¼ 7 or x3 ¼ 1 (3) Now substitute x3 ¼ 1 and x4 ¼ 4 in the second equation, and solve for the second unknown x2 as follows: 5x2 1 þ 12 ¼ 1 or 5x2 þ 11 ¼ 1 or 5x2 ¼ 10 or x2 ¼ 2 (4) Finally, substitute x2 ¼ 2, x3 ¼ 1, x4 ¼ 4 in the ﬁrst equation, and solve for the ﬁrst unknown x1 as follows: 2x1 þ 6 þ 5 8 ¼ 9 or 2x1 þ 3 ¼ 9 or 2x1 ¼ 6 or x1 ¼ 3 Thus, x1 ¼ 3 , x2 ¼ 2, x3 ¼ 1, x4 ¼ 4, or, equivalently, the vector u ¼ ð3; 2; 1; 4Þ is the unique solution of the system. Remark: There is an alternative form for back-substitution (which will be used when solving a system using the matrix format). Namely, after ﬁrst ﬁnding the value of the last unknown, we substitute this value for the last unknown in all the preceding equations before solving for the next-to-last unknown. This yields a triangular system with one less equation and one less unknown. For example, in the above triangular system, we substitute x4 ¼ 4 in all the preceding equations to obtain the triangular system 2x1 3x2 þ 5x3 ¼ 17 5x2 x3 ¼ 1 7x3 ¼ 7 We then repeat the process using the new last equation. And so on. Echelon Form, Pivot and Free Variables The following system of linear equations is said to be in echelon form: 2x1 þ 6x2 x3 þ 4x4 2x5 ¼ 15 x3 þ 2x4 þ 2x5 ¼ 5 3x4 9x5 ¼ 6 That is, no equation is degenerate and the leading unknown in each equation other than the ﬁrst is to the right of the leading unknown in the preceding equation. The leading unknowns in the system, x1, x3, x4, are called pivot variables, and the other unknowns, x2 and x5, are called free variables. Generally speaking, an echelon system or a system in echelon form has the following form: a11x1 þ a12x2 þ a13x3 þ a14x4 þ þ a1nxn ¼ b1 a2j2xj2 þ a2;j2þ1xj2þ1 þ þ a2nxn ¼ b2 :::::::::::::::::::::::::::::::::::::::::::::: arjrxjr þ þ arnxn ¼ br ð3:5Þ where 1 < j2 < < jr and a11, a2j2; . . . ; arjr are not zero. The pivot variables are x1, xj2; . . . ; xjr. Note that r n. The solution set of any echelon system is described in the following theorem (proved in Problem 3.10). CHAPTER 3 Systems of Linear Equations 65 THEOREM 3.6: Consider a system of linear equations in echelon form, say with r equations in n unknowns. There are two cases: (i) r ¼ n. That is, there are as many equations as unknowns (triangular form). Then the system has a unique solution. (ii) r < n. That is, there are more unknowns than equations. Then we can arbitrarily assign values to the n r free variables and solve uniquely for the r pivot variables, obtaining a solution of the system. Suppose an echelon system contains more unknowns than equations. Assuming the ﬁeld K is inﬁnite, the system has an inﬁnite number of solutions, because each of the n r free variables may be assigned any scalar. The general solution of a system with free variables may be described in either of two equivalent ways, which we illustrate using the above echelon system where there are r ¼ 3 equations and n ¼ 5 unknowns. One description is called the ‘‘Parametric Form’’ of the solution, and the other description is called the ‘‘Free-Variable Form.’’ Parametric Form Assign arbitrary values, called parameters, to the free variables x2 and x5, say x2 ¼ a and x5 ¼ b, and then use back-substitution to obtain values for the pivot variables x1, x3, x5 in terms of the parameters a and b. Speciﬁcally, (1) Substitute x5 ¼ b in the last equation, and solve for x4: 3x4 9b ¼ 6 or 3x4 ¼ 6 þ 9b or x4 ¼ 2 þ 3b (2) Substitute x4 ¼ 2 þ 3b and x5 ¼ b into the second equation, and solve for x3: x3 þ 2ð2 þ 3bÞ þ 2b ¼ 5 or x3 þ 4 þ 8b ¼ 5 or x3 ¼ 1 8b (3) Substitute x2 ¼ a, x3 ¼ 1 8b, x4 ¼ 2 þ 3b, x5 ¼ b into the ﬁrst equation, and solve for x1: 2x1 þ 6a ð1 8bÞ þ 4ð2 þ 3bÞ 2b ¼ 15 or x1 ¼ 4 3a 9b Accordingly, the general solution in parametric form is x1 ¼ 4 3a 9b; x2 ¼ a; x3 ¼ 1 8b; x4 ¼ 2 þ 3b; x5 ¼ b or, equivalently, v ¼ ð4 3a 9b; a; 1 8b; 2 þ 3b; bÞ where a and b are arbitrary numbers. Free-Variable Form Use back-substitution to solve for the pivot variables x1, x3, x4 directly in terms of the free variables x2 and x5. That is, the last equation gives x4 ¼ 2 þ 3x5. Substitution in the second equation yields x3 ¼ 1 8x5, and then substitution in the ﬁrst equation yields x1 ¼ 4 3x2 9x5. Accordingly, x1 ¼ 4 3x2 9x5; x2 ¼ free variable; x3 ¼ 1 8x5; x4 ¼ 2 þ 3x5; x5 ¼ free variable or, equivalently, v ¼ ð4 3x2 9x5; x2; 1 8x5; 2 þ 3x5; x5Þ is the free-variable form for the general solution of the system. We emphasize that there is no difference between the above two forms of the general solution, and the use of one or the other to represent the general solution is simply a matter of taste. Remark: A particular solution of the above system can be found by assigning any values to the free variables and then solving for the pivot variables by back-substitution. For example, setting x2 ¼ 1 and x5 ¼ 1, we obtain x4 ¼ 2 þ 3 ¼ 5; x3 ¼ 1 8 ¼ 7; x1 ¼ 4 3 9 ¼ 8 Thus, u ¼ ð8; 1; 7; 5; 1Þ is the particular solution corresponding to x2 ¼ 1 and x5 ¼ 1. 66 CHAPTER 3 Systems of Linear Equations 3.6 Gaussian Elimination The main method for solving the general system (3.2) of linear equations is called Gaussian elimination. It essentially consists of two parts: Part A. (Forward Elimination) Step-by-step reduction of the system yielding either a degenerate equation with no solution (which indicates the system has no solution) or an equivalent simpler system in triangular or echelon form. Part B. (Backward Elimination) Step-by-step back-substitution to ﬁnd the solution of the simpler system. Part B has already been investigated in Section 3.4. Accordingly, we need only give the algorithm for Part A, which is as follows. ALGORITHM 3.2 for (Part A): Input: The m n system (3.2) of linear equations. ELIMINATION STEP: Find the ﬁrst unknown in the system with a nonzero coefﬁcient (which now must be x1). (a) Arrange so that a11 6¼ 0. That is, if necessary, interchange equations so that the ﬁrst unknown x1 appears with a nonzero coefﬁcient in the ﬁrst equation. (b) Use a11 as a pivot to eliminate x1 from all equations except the ﬁrst equation. That is, for i > 1: (1) Set m ¼ ai1=a11; (2) Replace Li by mL1 þ Li The system now has the following form: a11x1 þ a12x2 þ a13x3 þ þ a1nxn ¼ b1 a2j2xj2 þ þ a2nxn ¼ b2 ::::::::::::::::::::::::::::::::::::::: amj2xj2 þ þ amnxn ¼ bn where x1 does not appear in any equation except the ﬁrst, a11 6¼ 0, and xj2 denotes the ﬁrst unknown with a nonzero coefﬁcient in any equation other than the ﬁrst. (c) Examine each new equation L. (1) If L has the form 0x1 þ 0x2 þ þ 0xn ¼ b with b 6¼ 0, then STOP The system is inconsistent and has no solution. (2) If L has the form 0x1 þ 0x2 þ þ 0xn ¼ 0 or if L is a multiple of another equation, then delete L from the system. RECURSION STEP: Repeat the Elimination Step with each new ‘‘smaller’’ subsystem formed by all the equations excluding the ﬁrst equation. OUTPUT: Finally, the system is reduced to triangular or echelon form, or a degenerate equation with no solution is obtained indicating an inconsistent system. The next remarks refer to the Elimination Step in Algorithm 3.2. (1) The following number m in (b) is called the multiplier: m ¼ ai1 a11 ¼ coefficient to be deleted pivot (2) One could alternatively apply the following operation in (b): Replace Li by ai1L1 þ a11Li This would avoid fractions if all the scalars were originally integers. CHAPTER 3 Systems of Linear Equations 67 Gaussian Elimination Example Here we illustrate in detail Gaussian elimination using the following system of linear equations: L1: x 3y 2z ¼ 6 L2: 2x 4y 3z ¼ 8 L3: 3x þ 6y þ 8z ¼ 5 Part A. We use the coefﬁcient 1 of x in the ﬁrst equation L1 as the pivot in order to eliminate x from the second equation L2 and from the third equation L3. This is accomplished as follows: (1) Multiply L1 by the multiplier m ¼ 2 and add it to L2; that is, ‘‘Replace L2 by 2L1 þ L2.’’ (2) Multiply L1 by the multiplier m ¼ 3 and add it to L3; that is, ‘‘Replace L3 by 3L1 þ L3.’’ These steps yield ð2ÞL1: 2x þ 6y þ 4z ¼ 12 L2: 2x 4y 3z ¼ 8 New L2: 2y þ z ¼ 4 3L1: 3x 9y 6z ¼ 18 L3: 3x þ 6y þ 8z ¼ 5 New L3: 3y þ 2z ¼ 13 Thus, the original system is replaced by the following system: L1: x 3y 2z ¼ 6 L2: 2y þ z ¼ 4 L3: 3y þ 2z ¼ 13 (Note that the equations L2 and L3 form a subsystem with one less equation and one less unknown than the original system.) Next we use the coefﬁcient 2 of y in the (new) second equation L2 as the pivot in order to eliminate y from the (new) third equation L3. This is accomplished as follows: (3) Multiply L2 by the multiplier m ¼ 3 2 and add it to L3; that is, ‘‘Replace L3 by 3 2 L2 þ L3:’’ (Alternately, ‘‘Replace L3 by 3L2 þ 2L3,’’ which will avoid fractions.) This step yields 3 2 L2: 3y þ 3 2 z ¼ 6 L3: 3y þ 2z ¼ 13 New L3: 7 2 z ¼ 7 or 3L2: 6y þ 3z ¼ 12 2L3: 6y þ 4z ¼ 26 New L3: 7z ¼ 14 Thus, our system is replaced by the following system: L1: x 3y 2z ¼ 6 L2: 2y þ z ¼ 4 L3: 7z ¼ 14 ðor 7 2 z ¼ 7Þ The system is now in triangular form, so Part A is completed. Part B. The values for the unknowns are obtained in reverse order, z; y; x, by back-substitution. Speciﬁcally, (1) Solve for z in L3 to get z ¼ 2. (2) Substitute z ¼ 2 in L2, and solve for y to get y ¼ 3. (3) Substitute y ¼ 3 and z ¼ 2 in L1, and solve for x to get x ¼ 1. Thus, the solution of the triangular system and hence the original system is as follows: x ¼ 1; y ¼ 3; z ¼ 2 or; equivalently; u ¼ ð1; 3; 2Þ: 68 CHAPTER 3 Systems of Linear Equations Condensed Format The Gaussian elimination algorithm involves rewriting systems of linear equations. Sometimes we can avoid excessive recopying of some of the equations by adopting a ‘‘condensed format.’’ This format for the solution of the above system follows: Number Equation Operation ð1Þ x 3y 2z ¼ 6 ð2Þ 2x 4y 3z ¼ 8 ð3Þ 3x þ 6y þ 8z ¼ 5 ð20Þ 2y þ z ¼ 4 Replace L2 by 2L1 þ L2 ð30Þ 3y þ 2z ¼ 13 Replace L3 by 3L1 þ L3 ð300Þ 7z ¼ 14 Replace L3 by 3L2 þ 2L3 That is, ﬁrst we write down the number of each of the original equations. As we apply the Gaussian elimination algorithm to the system, we only write down the new equations, and we label each new equation using the same number as the original corresponding equation, but with an added prime. (After each new equation, we will indicate, for instructional purposes, the elementary operation that yielded the new equation.) The system in triangular form consists of equations (1), ð20Þ, and ð300Þ, the numbers with the largest number of primes. Applying back-substitution to these equations again yields x ¼ 1, y ¼ 3, z ¼ 2. Remark: If two equations need to be interchanged, say to obtain a nonzero coefﬁcient as a pivot, then this is easily accomplished in the format by simply renumbering the two equations rather than changing their positions. EXAMPLE 3.7 Solve the following system: x þ 2y 3z ¼ 1 2x þ 5y 8z ¼ 4 3x þ 8y 13z ¼ 7 We solve the system by Gaussian elimination. Part A. (Forward Elimination) We use the coefﬁcient 1 of x in the ﬁrst equation L1 as the pivot in order to eliminate x from the second equation L2 and from the third equation L3. This is accomplished as follows: (1) Multiply L1 by the multiplier m ¼ 2 and add it to L2; that is, ‘‘Replace L2 by 2L1 þ L2.’’ (2) Multiply L1 by the multiplier m ¼ 3 and add it to L3; that is, ‘‘Replace L3 by 3L1 þ L3.’’ The two steps yield x þ 2y 3z ¼ 1 y 2z ¼ 2 2y 4z ¼ 4 or x þ 2y 3z ¼ 1 y 2z ¼ 2 (The third equation is deleted, because it is a multiple of the second equation.) The system is now in echelon form with free variable z. Part B. (Backward Elimination) To obtain the general solution, let the free variable z ¼ a, and solve for x and y by back-substitution. Substitute z ¼ a in the second equation to obtain y ¼ 2 þ 2a. Then substitute z ¼ a and y ¼ 2 þ 2a into the ﬁrst equation to obtain x þ 2ð2 þ 2aÞ 3a ¼ 1 or x þ 4 þ 4a 3a ¼ 1 or x ¼ 3 a Thus, the following is the general solution where a is a parameter: x ¼ 3 a; y ¼ 2 þ 2a; z ¼ a or u ¼ ð3 a; 2 þ 2a; aÞ CHAPTER 3 Systems of Linear Equations 69 EXAMPLE 3.8 Solve the following system: x1 þ 3x2 2x3 þ 5x4 ¼ 4 2x1 þ 8x2 x3 þ 9x4 ¼ 9 3x1 þ 5x2 12x3 þ 17x4 ¼ 7 We use Gaussian elimination. Part A. (Forward Elimination) We use the coefﬁcient 1 of x1 in the ﬁrst equation L1 as the pivot in order to eliminate x1 from the second equation L2 and from the third equation L3. This is accomplished by the following operations: (1) ‘‘Replace L2 by 2L1 þ L2’’ and (2) ‘‘Replace L3 by 3L1 þ L3’’ These yield: x1 þ 3x2 2x3 þ 5x4 ¼ 4 2x2 þ 3x3 x4 ¼ 1 4x2 6x3 þ 2x4 ¼ 5 We now use the coefﬁcient 2 of x2 in the second equation L2 as the pivot and the multiplier m ¼ 2 in order to eliminate x2 from the third equation L3. This is accomplished by the operation ‘‘Replace L3 by 2L2 þ L3,’’ which then yields the degenerate equation 0x1 þ 0x2 þ 0x3 þ 0x4 ¼ 3 This equation and, hence, the original system have no solution: DO NOT CONTINUE Remark 1: As in the above examples, Part A of Gaussian elimination tells us whether or not the system has a solution—that is, whether or not the system is consistent. Accordingly, Part B need never be applied when a system has no solution. Remark 2: If a system of linear equations has more than four unknowns and four equations, then it may be more convenient to use the matrix format for solving the system. This matrix format is discussed later. 3.7 Echelon Matrices, Row Canonical Form, Row Equivalence One way to solve a system of linear equations is by working with its augmented matrix M rather than the system itself. This section introduces the necessary matrix concepts for such a discussion. These concepts, such as echelon matrices and elementary row operations, are also of independent interest. Echelon Matrices A matrix A is called an echelon matrix, or is said to be in echelon form, if the following two conditions hold (where a leading nonzero element of a row of A is the ﬁrst nonzero element in the row): (1) All zero rows, if any, are at the bottom of the matrix. (2) Each leading nonzero entry in a row is to the right of the leading nonzero entry in the preceding row. That is, A ¼ ½aij is an echelon matrix if there exist nonzero entries a1j1; a2j2; . . . ; arjr; where j1 < j2 < < jr 70 CHAPTER 3 Systems of Linear Equations with the property that aij ¼ 0 for ðiÞ i r; j < ji ðiiÞ i > r The entries a1j1, a2j2; . . . ; arjr, which are the leading nonzero elements in their respective rows, are called the pivots of the echelon matrix. EXAMPLE 3.9 The following is an echelon matrix whose pivots have been circled: A ¼ 0 2 3 4 5 9 0 7 0 0 0 3 4 1 2 5 0 0 0 0 0 5 7 2 0 0 0 0 0 0 8 6 0 0 0 0 0 0 0 0 2 6 6 6 6 4 3 7 7 7 7 5 Observe that the pivots are in columns C2; C4; C6; C7, and each is to the right of the one above. Using the above notation, the pivots are a1j1 ¼ 2; a2j2 ¼ 3; a3j3 ¼ 5; a4j4 ¼ 8 where j1 ¼ 2, j2 ¼ 4, j3 ¼ 6, j4 ¼ 7. Here r ¼ 4. Row Canonical Form A matrix A is said to be in row canonical form (or row-reduced echelon form) if it is an echelon matrix— that is, if it satisﬁes the above properties (1) and (2), and if it satisﬁes the following additional two properties: (3) Each pivot (leading nonzero entry) is equal to 1. (4) Each pivot is the only nonzero entry in its column. The major difference between an echelon matrix and a matrix in row canonical form is that in an echelon matrix there must be zeros below the pivots [Properties (1) and (2)], but in a matrix in row canonical form, each pivot must also equal 1 [Property (3)] and there must also be zeros above the pivots [Property (4)]. The zero matrix 0 of any size and the identity matrix I of any size are important special examples of matrices in row canonical form. EXAMPLE 3.10 The following are echelon matrices whose pivots have been circled: 2 3 2 0 4 5 6 0 0 0 1 3 2 0 0 0 0 0 0 6 2 0 0 0 0 0 0 0 2 6 6 4 3 7 7 5; 1 2 3 0 0 1 0 0 0 2 4 3 5; 0 1 3 0 0 4 0 0 0 1 0 3 0 0 0 0 1 2 2 4 3 5 The third matrix is also an example of a matrix in row canonical form. The second matrix is not in row canonical form, because it does not satisfy property (4); that is, there is a nonzero entry above the second pivot in the third column. The ﬁrst matrix is not in row canonical form, because it satisﬁes neither property (3) nor property (4); that is, some pivots are not equal to 1 and there are nonzero entries above the pivots. CHAPTER 3 Systems of Linear Equations 71 Elementary Row Operations Suppose A is a matrix with rows R1; R2; . . . ; Rm. The following operations on A are called elementary row operations. ½E1 (Row Interchange): Interchange rows Ri and Rj. This may be written as ‘‘Interchange Ri and Rj’’ or ‘‘Ri ! Rj’’ ½E2 (Row Scaling): Replace row Ri by a nonzero multiple kRi of itself. This may be written as ‘‘Replace Ri by kRi ðk 6¼ 0Þ’’ or ‘‘kRi ! Ri’’ ½E3 (Row Addition): Replace row Rj by the sum of a multiple kRi of a row Ri and itself. This may be written as ‘‘Replace Rj by kRi þ Rj’’ or ‘‘kRi þ Rj ! Rj’’ The arrow ! in E2 and E3 may be read as ‘‘replaces.’’ Sometimes (say to avoid fractions when all the given scalars are integers) we may apply ½E2 and ½E3 in one step; that is, we may apply the following operation: ½E Replace Rj by the sum of a multiple kRi of a row Ri and a nonzero multiple k0Rj of itself. This may be written as ‘‘Replace Rj by kRi þ k0Rj ðk0 6¼ 0Þ’’ or ‘‘kRi þ k0Rj ! Rj’’ We emphasize that in operations ½E3 and ½E only row Rj is changed. Row Equivalence, Rank of a Matrix A matrix A is said to be row equivalent to a matrix B, written A B if B can be obtained from A by a sequence of elementary row operations. In the case that B is also an echelon matrix, B is called an echelon form of A. The following are two basic results on row equivalence. THEOREM 3.7: Suppose A ¼ ½aij and B ¼ ½bij are row equivalent echelon matrices with respective pivot entries a1j1; a2j2; . . . arjr and b1k1; b2k2; . . . bsks Then A and B have the same number of nonzero rows—that is, r ¼ s—and the pivot entries are in the same positions—that is, j1 ¼ k1, j2 ¼ k2; . . . ; jr ¼ kr. THEOREM 3.8: Every matrix A is row equivalent to a unique matrix in row canonical form. The proofs of the above theorems will be postponed to Chapter 4. The unique matrix in Theorem 3.8 is called the row canonical form of A. Using the above theorems, we can now give our ﬁrst deﬁnition of the rank of a matrix. DEFINITION: The rank of a matrix A, written rankðAÞ, is equal to the number of pivots in an echelon form of A. The rank is a very important property of a matrix and, depending on the context in which the matrix is used, it will be deﬁned in many different ways. Of course, all the deﬁnitions lead to the same number. The next section gives the matrix format of Gaussian elimination, which ﬁnds an echelon form of any matrix A (and hence the rank of A), and also ﬁnds the row canonical form of A. 72 CHAPTER 3 Systems of Linear Equations One can show that row equivalence is an equivalence relation. That is, (1) A A for any matrix A. (2) If A B, then B A. (3) If A B and B C, then A C. Property (2) comes from the fact that each elementary row operation has an inverse operation of the same type. Namely, (i) ‘‘Interchange Ri and Rj’’ is its own inverse. (ii) ‘‘Replace Ri by kRi’’ and ‘‘Replace Ri by ð1=kÞRi’’ are inverses. (iii) ‘‘Replace Rj by kRi þ Rj’’ and ‘‘Replace Rj by kRi þ Rj’’ are inverses. There is a similar result for operation [E] (Problem 3.73). 3.8 Gaussian Elimination, Matrix Formulation This section gives two matrix algorithms that accomplish the following: (1) Algorithm 3.3 transforms any matrix A into an echelon form. (2) Algorithm 3.4 transforms the echelon matrix into its row canonical form. These algorithms, which use the elementary row operations, are simply restatements of Gaussian elimination as applied to matrices rather than to linear equations. (The term ‘‘row reduce’’ or simply ‘‘reduce’’ will mean to transform a matrix by the elementary row operations.) ALGORITHM 3.3 (Forward Elimination): The input is any matrix A. (The algorithm puts 0’s below each pivot, working from the ‘‘top-down.’’) The output is an echelon form of A. Step 1. Find the ﬁrst column with a nonzero entry. Let j1 denote this column. (a) Arrange so that a1j1 6¼ 0. That is, if necessary, interchange rows so that a nonzero entry appears in the ﬁrst row in column j1. (b) Use a1j1 as a pivot to obtain 0’s below a1j1. Speciﬁcally, for i > 1: ð1Þ Set m ¼ aij1=a1j1; ð2Þ Replace Ri by mR1 þ Ri [That is, apply the operation ðaij1=a1j1ÞR1 þ Ri ! Ri:] Step 2. Repeat Step 1 with the submatrix formed by all the rows excluding the ﬁrst row. Here we let j2 denote the ﬁrst column in the subsystem with a nonzero entry. Hence, at the end of Step 2, we have a2j2 6¼ 0. Steps 3 to r. Continue the above process until a submatrix has only zero rows. We emphasize that at the end of the algorithm, the pivots will be a1j1; a2j2; . . . ; arjr where r denotes the number of nonzero rows in the ﬁnal echelon matrix. Remark 1: The following number m in Step 1(b) is called the multiplier: m ¼ aij1 a1j1 ¼ entry to be deleted pivot CHAPTER 3 Systems of Linear Equations 73 Remark 2: One could replace the operation in Step 1(b) by the following which would avoid fractions if all the scalars were originally integers. Replace Ri by aij1R1 þ a1j1Ri: ALGORITHM 3.4 (Backward Elimination): The input is a matrix A ¼ ½aij in echelon form with pivot entries a1j1; a2j2; . . . ; arjr The output is the row canonical form of A. Step 1. (a) (Use row scaling so the last pivot equals 1.) Multiply the last nonzero row Rr by 1=arjr. (b) (Use arjr ¼ 1 to obtain 0’s above the pivot.) For i ¼ r 1; r 2; . . . ; 2; 1: ð1Þ Set m ¼ aijr; ð2Þ Replace Ri by mRr þ Ri (That is, apply the operations aijrRr þ Ri ! Ri.) Steps 2 to r1. Repeat Step 1 for rows Rr1, Rr2; . . . ; R2. Step r. (Use row scaling so the ﬁrst pivot equals 1.) Multiply R1 by 1=a1j1. There is an alternative form of Algorithm 3.4, which we describe here in words. The formal description of this algorithm is left to the reader as a supplementary problem. ALTERNATIVE ALGORITHM 3.4 Puts 0’s above the pivots row by row from the bottom up (rather than column by column from right to left). The alternative algorithm, when applied to an augmented matrix M of a system of linear equations, is essentially the same as solving for the pivot unknowns one after the other from the bottom up. Remark: We emphasize that Gaussian elimination is a two-stage process. Speciﬁcally, Stage A (Algorithm 3.3). Puts 0’s below each pivot, working from the top row R1 down. Stage B (Algorithm 3.4). Puts 0’s above each pivot, working from the bottom row Rr up. There is another algorithm, called Gauss–Jordan, that also row reduces a matrix to its row canonical form. The difference is that Gauss–Jordan puts 0’s both below and above each pivot as it works its way from the top row R1 down. Although Gauss–Jordan may be easier to state and understand, it is much less efﬁcient than the two-stage Gaussian elimination algorithm. EXAMPLE 3.11 Consider the matrix A ¼ 1 2 3 1 2 2 4 4 6 10 3 6 6 9 13 2 4 3 5. (a) Use Algorithm 3.3 to reduce A to an echelon form. (b) Use Algorithm 3.4 to further reduce A to its row canonical form. (a) First use a11 ¼ 1 as a pivot to obtain 0’s below a11; that is, apply the operations ‘‘Replace R2 by 2R1 þ R2’’ and ‘‘Replace R3 by 3R1 þ R3.’’ Then use a23 ¼ 2 as a pivot to obtain 0 below a23; that is, apply the operation ‘‘Replace R3 by 3 2 R2 þ R3.’’ This yields A 1 2 3 1 2 0 0 2 4 6 0 0 3 6 7 2 4 3 5 1 2 3 1 2 0 0 2 4 6 0 0 0 0 2 2 4 3 5 The matrix is now in echelon form. 74 CHAPTER 3 Systems of Linear Equations (b) Multiply R3 by 1 2 so the pivot entry a35 ¼ 1, and then use a35 ¼ 1 as a pivot to obtain 0’s above it by the operations ‘‘Replace R2 by 6R3 þ R2’’ and then ‘‘Replace R1 by 2R3 þ R1.’’ This yields A 1 2 3 1 2 0 0 2 4 6 0 0 0 0 1 2 4 3 5 1 2 3 1 0 0 0 2 4 0 0 0 0 0 1 2 4 3 5: Multiply R2 by 1 2 so the pivot entry a23 ¼ 1, and then use a23 ¼ 1 as a pivot to obtain 0’s above it by the operation ‘‘Replace R1 by 3R2 þ R1.’’ This yields A 1 2 3 1 0 0 0 1 2 0 0 0 0 0 1 2 4 3 5 1 2 0 7 0 0 0 1 2 0 0 0 0 0 1 2 4 3 5: The last matrix is the row canonical form of A. Application to Systems of Linear Equations One way to solve a system of linear equations is by working with its augmented matrix M rather than the equations themselves. Speciﬁcally, we reduce M to echelon form (which tells us whether the system has a solution), and then further reduce M to its row canonical form (which essentially gives the solution of the original system of linear equations). The justiﬁcation for this process comes from the following facts: (1) Any elementary row operation on the augmented matrix M of the system is equivalent to applying the corresponding operation on the system itself. (2) The system has a solution if and only if the echelon form of the augmented matrix M does not have a row of the form ð0; 0; . . . ; 0; bÞ with b 6¼ 0. (3) In the row canonical form of the augmented matrix M (excluding zero rows), the coefﬁcient of each basic variable is a pivot entry equal to 1, and it is the only nonzero entry in its respective column; hence, the free-variable form of the solution of the system of linear equations is obtained by simply transferring the free variables to the other side. This process is illustrated below. EXAMPLE 3.12 Solve each of the following systems: (a) x1 þ x2 2x3 þ 4x4 ¼ 5 2x1 þ 2x2 3x3 þ x4 ¼ 3 3x1 þ 3x2 4x3 2x4 ¼ 1 (b) x1 þ x2 2x3 þ 3x4 ¼ 4 2x1 þ 3x2 þ 3x3 x4 ¼ 3 5x1 þ 7x2 þ 4x3 þ x4 ¼ 5 (c) x þ 2y þ z ¼ 3 2x þ 5y z ¼ 4 3x 2y z ¼ 5 (a) Reduce its augmented matrix M to echelon form and then to row canonical form as follows: M ¼ 1 1 2 4 5 2 2 3 1 3 3 3 4 2 1 2 4 3 5 1 1 2 4 5 0 0 1 7 7 0 0 2 14 14 2 4 3 5 1 1 0 10 9 0 0 1 7 7 0 0 0 0 0 2 4 3 5 Rewrite the row canonical form in terms of a system of linear equations to obtain the free variable form of the solution. That is, x1 þ x2 10x4 ¼ 9 x3 7x4 ¼ 7 or x1 ¼ 9 x2 þ 10x4 x3 ¼ 7 þ 7x4 (The zero row is omitted in the solution.) Observe that x1 and x3 are the pivot variables, and x2 and x4 are the free variables. CHAPTER 3 Systems of Linear Equations 75 (b) First reduce its augmented matrix M to echelon form as follows: M ¼ 1 1 2 3 4 2 3 3 1 3 5 7 4 1 5 2 4 3 5 1 1 2 3 4 0 1 7 7 5 0 2 14 14 15 2 4 3 5 1 1 2 3 4 0 1 7 7 5 0 0 0 0 5 2 4 3 5 There is no need to continue to ﬁnd the row canonical form of M, because the echelon form already tells us that the system has no solution. Speciﬁcally, the third row of the echelon matrix corresponds to the degenerate equation 0x1 þ 0x2 þ 0x3 þ 0x4 ¼ 5 which has no solution. Thus, the system has no solution. (c) Reduce its augmented matrix M to echelon form and then to row canonical form as follows: M ¼ 1 2 1 3 2 5 1 4 3 2 1 5 2 6 4 3 7 5 1 2 1 3 0 1 3 10 0 8 4 4 2 6 4 3 7 5 1 2 1 3 0 1 3 10 0 0 28 84 2 6 4 3 7 5 1 2 1 3 0 1 3 10 0 0 1 3 2 6 4 3 7 5 1 2 0 0 0 1 0 1 0 0 1 3 2 6 4 3 7 5 1 0 0 2 0 1 0 1 0 0 1 3 2 6 4 3 7 5 Thus, the system has the unique solution x ¼ 2, y ¼ 1, z ¼ 3, or, equivalently, the vector u ¼ ð2; 1; 3Þ. We note that the echelon form of M already indicated that the solution was unique, because it corresponded to a triangular system. Application to Existence and Uniqueness Theorems This subsection gives theoretical conditions for the existence and uniqueness of a solution of a system of linear equations using the notion of the rank of a matrix. THEOREM 3.9: Consider a system of linear equations in n unknowns with augmented matrix M ¼ ½A; B. Then, (a) The system has a solution if and only if rankðAÞ ¼ rankðMÞ. (b) The solution is unique if and only if rankðAÞ ¼ rankðMÞ ¼ n. Proof of (a). The system has a solution if and only if an echelon form of M ¼ ½A; B does not have a row of the form ð0; 0; . . . ; 0; bÞ; with b 6¼ 0 If an echelon form of M does have such a row, then b is a pivot of M but not of A, and hence, rankðMÞ > rankðAÞ. Otherwise, the echelon forms of A and M have the same pivots, and hence, rankðAÞ ¼ rankðMÞ. This proves (a). Proof of (b). The system has a unique solution if and only if an echelon form has no free variable. This means there is a pivot for each unknown. Accordingly, n ¼ rankðAÞ ¼ rankðMÞ. This proves (b). The above proof uses the fact (Problem 3.74) that an echelon form of the augmented matrix M ¼ ½A; B also automatically yields an echelon form of A. 76 CHAPTER 3 Systems of Linear Equations 3.9 Matrix Equation of a System of Linear Equations The general system (3.2) of m linear equations in n unknowns is equivalent to the matrix equation a11 a12 . . . a1n a21 a22 . . . a2n ::::::::::::::::::::::::::::::: am1 am2 . . . amn 2 6 6 4 3 7 7 5 x1 x2 x3 . . . xn 2 6 6 6 6 4 3 7 7 7 7 5 ¼ b1 b2 . . . bm 2 6 6 4 3 7 7 5 or AX ¼ B where A ¼ ½aij is the coefﬁcient matrix, X ¼ ½xj is the column vector of unknowns, and B ¼ ½bi is the column vector of constants. (Some texts write Ax ¼ b rather than AX ¼ B, in order to emphasize that x and b are simply column vectors.) The statement that the system of linear equations and the matrix equation are equivalent means that any vector solution of the system is a solution of the matrix equation, and vice versa. EXAMPLE 3.13 The following system of linear equations and matrix equation are equivalent: x1 þ 2x2 4x3 þ 7x4 ¼ 4 3x1 5x2 þ 6x3 8x4 ¼ 8 4x1 3x2 2x3 þ 6x4 ¼ 11 and 1 2 4 7 3 5 6 8 4 3 2 6 2 4 3 5 x1 x2 x3 x4 2 6 6 4 3 7 7 5 ¼ 4 8 11 2 4 3 5 We note that x1 ¼ 3, x2 ¼ 1, x3 ¼ 2, x4 ¼ 1, or, in other words, the vector u ¼ ½3; 1; 2; 1 is a solution of the system. Thus, the (column) vector u is also a solution of the matrix equation. The matrix form AX ¼ B of a system of linear equations is notationally very convenient when discussing and proving properties of systems of linear equations. This is illustrated with our ﬁrst theorem (described in Fig. 3-1), which we restate for easy reference. THEOREM 3.1: Suppose the ﬁeld K is inﬁnite. Then the system AX ¼ B has: (a) a unique solution, (b) no solution, or (c) an inﬁnite number of solutions. Proof. It sufﬁces to show that if AX ¼ B has more than one solution, then it has inﬁnitely many. Suppose u and v are distinct solutions of AX ¼ B; that is, Au ¼ B and Av ¼ B. Then, for any k 2 K, A½u þ kðu vÞ ¼ Au þ kðAu AvÞ ¼ B þ kðB BÞ ¼ B Thus, for each k 2 K, the vector u þ kðu vÞ is a solution of AX ¼ B. Because all such solutions are distinct (Problem 3.47), AX ¼ B has an inﬁnite number of solutions. Observe that the above theorem is true when K is the real ﬁeld R (or the complex ﬁeld C). Section 3.3 shows that the theorem has a geometrical description when the system consists of two equations in two unknowns, where each equation represents a line in R2. The theorem also has a geometrical description when the system consists of three nondegenerate equations in three unknowns, where the three equations correspond to planes H1, H2, H3 in R3. That is, (a) Unique solution: Here the three planes intersect in exactly one point. (b) No solution: Here the planes may intersect pairwise but with no common point of intersection, or two of the planes may be parallel. (c) Inﬁnite number of solutions: Here the three planes may intersect in a line (one free variable), or they may coincide (two free variables). These three cases are pictured in Fig. 3-3. Matrix Equation of a Square System of Linear Equations A system AX ¼ B of linear equations is square if and only if the matrix A of coefﬁcients is square. In such a case, we have the following important result. CHAPTER 3 Systems of Linear Equations 77 THEOREM 3.10: A square system AX ¼ B of linear equations has a unique solution if and only if the matrix A is invertible. In such a case, A1B is the unique solution of the system. We only prove here that if A is invertible, then A1B is a unique solution. If A is invertible, then AðA1BÞ ¼ ðAA1ÞB ¼ IB ¼ B and hence, A1B is a solution. Now suppose v is any solution, so Av ¼ B. Then v ¼ Iv ¼ ðA1AÞv ¼ A1ðAvÞ ¼ A1B Thus, the solution A1B is unique. EXAMPLE 3.14 Consider the following system of linear equations, whose coefﬁcient matrix A and inverse A1 are also given: x þ 2y þ 3z ¼ 1 x þ 3y þ 6z ¼ 3 2x þ 6y þ 13z ¼ 5 ; A ¼ 1 2 3 1 3 6 2 6 13 2 4 3 5; A1 ¼ 3 8 3 1 7 3 0 2 1 2 4 3 5 By Theorem 3.10, the unique solution of the system is A1B ¼ 3 8 3 1 7 3 0 2 1 2 4 3 5 1 3 5 2 4 3 5 ¼ 6 5 1 2 4 3 5 That is, x ¼ 6, y ¼ 5, z ¼ 1. Remark: We emphasize that Theorem 3.10 does not usually help us to ﬁnd the solution of a square system. That is, ﬁnding the inverse of a coefﬁcient matrix A is not usually any easier than solving the system directly. Thus, unless we are given the inverse of a coefﬁcient matrix A, as in Example 3.14, we usually solve a square system by Gaussian elimination (or some iterative method whose discussion lies beyond the scope of this text). ( ) Unique solution a H2 H3 H1 H1 H2 H3 ( ) Infinite number of solutions c H3 H H 1 2 and (i) (ii) (iii) H H H 1 2 3 , , and (i) ( ) No solutions b H3 H2 H1 (ii) (iii) (i ) v H1 H2 H3 H2 H3 H1 H3 Figure 3-3 78 CHAPTER 3 Systems of Linear Equations 3.10 Systems of Linear Equations and Linear Combinations of Vectors The general system (3.2) of linear equations may be rewritten as the following vector equation: x1 a11 a21 . . . am1 2 6 6 4 3 7 7 5 þ x2 a12 a22 . . . am2 2 6 6 4 3 7 7 5 þ þ xn a1n a2n . . . amn 2 6 6 4 3 7 7 5 ¼ b1 b2 . . . bm 2 6 6 4 3 7 7 5 Recall that a vector v in Kn is said to be a linear combination of vectors u1; u2; . . . ; um in Kn if there exist scalars a1; a2; . . . ; am in K such that v ¼ a1u1 þ a2u2 þ þ amum Accordingly, the general system (3.2) of linear equations and the above equivalent vector equation have a solution if and only if the column vector of constants is a linear combination of the columns of the coefﬁcient matrix. We state this observation formally. THEOREM 3.11: A system AX ¼ B of linear equations has a solution if and only if B is a linear combination of the columns of the coefﬁcient matrix A. Thus, the answer to the problem of expressing a given vector v in Kn as a linear combination of vectors u1; u2; . . . ; um in Kn reduces to solving a system of linear equations. Linear Combination Example Suppose we want to write the vector v ¼ ð1; 2; 5Þ as a linear combination of the vectors u1 ¼ ð1; 1; 1Þ; u2 ¼ ð1; 2; 3Þ; u3 ¼ ð2; 1; 1Þ First we write v ¼ xu1 þ yu2 þ zu3 with unknowns x; y; z, and then we ﬁnd the equivalent system of linear equations which we solve. Speciﬁcally, we ﬁrst write 1 2 5 2 4 3 5 ¼ x 1 1 1 2 4 3 5 þ y 1 2 3 2 4 3 5 þ z 2 1 1 2 4 3 5 ðÞ Then 1 2 5 2 4 3 5 ¼ x x x 2 4 3 5 þ y 2y 3y 2 4 3 5 þ 2z z z 2 4 3 5 ¼ x þ y þ 2z x þ 2y z x þ 3y þ z 2 4 3 5 Setting corresponding entries equal to each other yields the following equivalent system: x þ y þ 2z ¼ 1 x þ 2y z ¼ 2 x þ 3y þ z ¼ 5 ðÞ For notational convenience, we have written the vectors in Rn as columns, because it is then easier to ﬁnd the equivalent system of linear equations. In fact, one can easily go from the vector equation () directly to the system (). Now we solve the equivalent system of linear equations by reducing the system to echelon form. This yields x þ y þ 2z ¼ 1 y 3z ¼ 3 2y z ¼ 4 and then x þ y þ 2z ¼ 1 y 3z ¼ 3 5z ¼ 10 Back-substitution yields the solution x ¼ 6, y ¼ 3, z ¼ 2. Thus, v ¼ 6u1 þ 3u2 þ 2u3. CHAPTER 3 Systems of Linear Equations 79 EXAMPLE 3.15 (a) Write the vector v ¼ ð4; 9; 19Þ as a linear combination of u1 ¼ ð1; 2; 3Þ; u2 ¼ ð3; 7; 10Þ; u3 ¼ ð2; 1; 9Þ: Find the equivalent system of linear equations by writing v ¼ xu1 þ yu2 þ zu3, and reduce the system to an echelon form. We have x þ 3y þ 2z ¼ 4 2x 7y þ z ¼ 9 3x þ 10y þ 9z ¼ 19 or x þ 3y þ 2z ¼ 4 y þ 5z ¼ 17 y þ 3z ¼ 7 or x þ 3y þ 2z ¼ 4 y þ 5z ¼ 17 8z ¼ 24 Back-substitution yields the solution x ¼ 4, y ¼ 2, z ¼ 3. Thus, v is a linear combination of u1; u2; u3. Speciﬁcally, v ¼ 4u1 2u2 þ 3u3. (b) Write the vector v ¼ ð2; 3; 5Þ as a linear combination of u1 ¼ ð1; 2; 3Þ; u2 ¼ ð2; 3; 4Þ; u3 ¼ ð1; 3; 5Þ Find the equivalent system of linear equations by writing v ¼ xu1 þ yu2 þ zu3, and reduce the system to an echelon form. We have x þ 2y þ z ¼ 2 2x þ 3y þ 3z ¼ 3 3x 4y 5z ¼ 5 or x þ 2y þ z ¼ 2 y þ z ¼ 1 2y 2z ¼ 1 or x þ 2y þ z ¼ 2 5y þ 5z ¼ 1 0 ¼ 3 The system has no solution. Thus, it is impossible to write v as a linear combination of u1; u2; u3. Linear Combinations of Orthogonal Vectors, Fourier Coefﬁcients Recall ﬁrst (Section 1.4) that the dot (inner) product u v of vectors u ¼ ða1; . . . ; anÞ and v ¼ ðb1; . . . ; bnÞ in Rn is deﬁned by u v ¼ a1b1 þ a2b2 þ þ anbn Furthermore, vectors u and v are said to be orthogonal if their dot product u v ¼ 0. Suppose that u1; u2; . . . ; un in Rn are n nonzero pairwise orthogonal vectors. This means ðiÞ ui uj ¼ 0 for i 6¼ j and ðiiÞ ui ui 6¼ 0 for each i Then, for any vector v in Rn, there is an easy way to write v as a linear combination of u1; u2; . . . ; un, which is illustrated in the next example. EXAMPLE 3.16 Consider the following three vectors in R3: u1 ¼ ð1; 1; 1Þ; u2 ¼ ð1; 3; 2Þ; u3 ¼ ð5; 1; 4Þ These vectors are pairwise orthogonal; that is, u1 u2 ¼ 1 3 þ 2 ¼ 0; u1 u3 ¼ 5 1 4 ¼ 0; u2 u3 ¼ 5 þ 3 8 ¼ 0 Suppose we want to write v ¼ ð4; 14; 9Þ as a linear combination of u1; u2; u3. Method 1. Find the equivalent system of linear equations as in Example 3.14 and then solve, obtaining v ¼ 3u1 4u2 þ u3. Method 2. (This method uses the fact that the vectors u1; u2; u3 are mutually orthogonal, and hence, the arithmetic is much simpler.) Set v as a linear combination of u1; u2; u3 using unknown scalars x; y; z as follows: ð4; 14; 9Þ ¼ xð1; 1; 1Þ þ yð1; 3; 2Þ þ zð5; 1; 4Þ ðÞ 80 CHAPTER 3 Systems of Linear Equations Take the dot product of () with respect to u1 to get ð4; 14; 9Þ ð1; 1; 1Þ ¼ xð1; 1; 1Þ ð1; 1; 1Þ or 9 ¼ 3x or x ¼ 3 (The last two terms drop out, because u1 is orthogonal to u2 and to u3.) Next take the dot product of () with respect to u2 to obtain ð4; 14; 9Þ ð1; 3; 2Þ ¼ yð1; 3; 2Þ ð1; 3; 2Þ or 56 ¼ 14y or y ¼ 4 Finally, take the dot product of () with respect to u3 to get ð4; 14; 9Þ ð5; 1; 4Þ ¼ zð5; 1; 4Þ ð5; 1; 4Þ or 42 ¼ 42z or z ¼ 1 Thus, v ¼ 3u1 4u2 þ u3. The procedure in Method 2 in Example 3.16 is valid in general. Namely, THEOREM 3.12: Suppose u1; u2; . . . ; un are nonzero mutually orthogonal vectors in Rn. Then, for any vector v in Rn, v ¼ v u1 u1 u1 u1 þ v u2 u2 u2 u2 þ þ v un un un un We emphasize that there must be n such orthogonal vectors ui in Rn for the formula to be used. Note also that each ui ui 6¼ 0, because each ui is a nonzero vector. Remark: The following scalar ki (appearing in Theorem 3.12) is called the Fourier coefﬁcient of v with respect to ui: ki ¼ v ui ui ui ¼ v ui kuik2 It is analogous to a coefﬁcient in the celebrated Fourier series of a function. 3.11 Homogeneous Systems of Linear Equations A system of linear equations is said to be homogeneous if all the constant terms are zero. Thus, a homogeneous system has the form AX ¼ 0. Clearly, such a system always has the zero vector 0 ¼ ð0; 0; . . . ; 0Þ as a solution, called the zero or trivial solution. Accordingly, we are usually interested in whether or not the system has a nonzero solution. Because a homogeneous system AX ¼ 0 has at least the zero solution, it can always be put in an echelon form, say a11x1 þ a12x2 þ a13x3 þ a14x4 þ þ a1nxn ¼ 0 a2j2xj2 þ a2;j2þ1xj2þ1 þ þ a2nxn ¼ 0 :::::::::::::::::::::::::::::::::::::::::::: arjrxjr þ þ arnxn ¼ 0 Here r denotes the number of equations in echelon form and n denotes the number of unknowns. Thus, the echelon system has n r free variables. The question of nonzero solutions reduces to the following two cases: (i) r ¼ n. The system has only the zero solution. (ii) r < n. The system has a nonzero solution. Accordingly, if we begin with fewer equations than unknowns, then, in echelon form, r < n, and the system has a nonzero solution. This proves the following important result. THEOREM 3.13: A homogeneous system AX ¼ 0 with more unknowns than equations has a nonzero solution. CHAPTER 3 Systems of Linear Equations 81 EXAMPLE 3.17 Determine whether or not each of the following homogeneous systems has a nonzero solution: (a) x þ y z ¼ 0 2x 3y þ z ¼ 0 x 4y þ 2z ¼ 0 (b) x þ y z ¼ 0 2x þ 4y z ¼ 0 3x þ 2y þ 2z ¼ 0 (c) x1 þ 2x2 3x3 þ 4x4 ¼ 0 2x1 3x2 þ 5x3 7x4 ¼ 0 5x1 þ 6x2 9x3 þ 8x4 ¼ 0 (a) Reduce the system to echelon form as follows: x þ y z ¼ 0 5y þ 3z ¼ 0 5y þ 3z ¼ 0 and then x þ y z ¼ 0 5y þ 3z ¼ 0 The system has a nonzero solution, because there are only two equations in the three unknowns in echelon form. Here z is a free variable. Let us, say, set z ¼ 5. Then, by back-substitution, y ¼ 3 and x ¼ 2. Thus, the vector u ¼ ð2; 3; 5Þ is a particular nonzero solution. (b) Reduce the system to echelon form as follows: x þ y z ¼ 0 2y þ z ¼ 0 y þ 5z ¼ 0 and then x þ y z ¼ 0 2y þ z ¼ 0 11z ¼ 0 In echelon form, there are three equations in three unknowns. Thus, the system has only the zero solution. (c) The system must have a nonzero solution (Theorem 3.13), because there are four unknowns but only three equations. (Here we do not need to reduce the system to echelon form.) Basis for the General Solution of a Homogeneous System Let W denote the general solution of a homogeneous system AX ¼ 0. A list of nonzero solution vectors u1; u2; . . . ; us of the system is said to be a basis for W if each solution vector w 2 W can be expressed uniquely as a linear combination of the vectors u1; u2; . . . ; us; that is, there exist unique scalars a1; a2; . . . ; as such that w ¼ a1u1 þ a2u2 þ þ asus The number s of such basis vectors is equal to the number of free variables. This number s is called the dimension of W, written as dim W ¼ s. When W ¼ f0g—that is, the system has only the zero solution— we deﬁne dim W ¼ 0. The following theorem, proved in Chapter 5, page 171, tells us how to ﬁnd such a basis. THEOREM 3.14: Let W be the general solution of a homogeneous system AX ¼ 0, and suppose that the echelon form of the homogeneous system has s free variables. Let u1; u2; . . . ; us be the solutions obtained by setting one of the free variables equal to 1 (or any nonzero constant) and the remaining free variables equal to 0. Then dim W ¼ s, and the vectors u1; u2; . . . ; us form a basis of W. We emphasize that the general solution W may have many bases, and that Theorem 3.12 only gives us one such basis. EXAMPLE 3.18 Find the dimension and a basis for the general solution W of the homogeneous system x1 þ 2x2 3x3 þ 2x4 4x5 ¼ 0 2x1 þ 4x2 5x3 þ x4 6x5 ¼ 0 5x1 þ 10x2 13x3 þ 4x4 16x5 ¼ 0 82 CHAPTER 3 Systems of Linear Equations First reduce the system to echelon form. Apply the following operations: ‘‘Replace L2 by 2L1 þ L2’’ and ‘‘Replace L3 by 5L1 þ L3’’ and then ‘‘Replace L3 by 2L2 þ L3’’ These operations yield x1 þ 2x2 3x3 þ 2x4 4x5 ¼ 0 x3 3x4 þ 2x5 ¼ 0 2x3 6x4 þ 4x5 ¼ 0 and x1 þ 2x2 3x3 þ 2x4 4x5 ¼ 0 x3 3x4 þ 2x5 ¼ 0 The system in echelon form has three free variables, x2; x4; x5; hence, dim W ¼ 3. Three solution vectors that form a basis for W are obtained as follows: (1) Set x2 ¼ 1, x4 ¼ 0, x5 ¼ 0. Back-substitution yields the solution u1 ¼ ð2; 1; 0; 0; 0Þ. (2) Set x2 ¼ 0, x4 ¼ 1, x5 ¼ 0. Back-substitution yields the solution u2 ¼ ð7; 0; 3; 1; 0Þ. (3) Set x2 ¼ 0, x4 ¼ 0, x5 ¼ 1. Back-substitution yields the solution u3 ¼ ð2; 0; 2; 0; 1Þ. The vectors u1 ¼ ð2; 1; 0; 0; 0Þ, u2 ¼ ð7; 0; 3; 1; 0Þ, u3 ¼ ð2; 0; 2; 0; 1Þ form a basis for W. Remark: Any solution of the system in Example 3.18 can be written in the form au1 þ bu2 þ cu3 ¼ að2; 1; 0; 0; 0Þ þ bð7; 0; 3; 1; 0Þ þ cð2; 0; 2; 0; 1Þ ¼ ð2a þ 7b 2c; a; 3b 2c; b; cÞ or x1 ¼ 2a þ 7b 2c; x2 ¼ a; x3 ¼ 3b 2c; x4 ¼ b; x5 ¼ c where a; b; c are arbitrary constants. Observe that this representation is nothing more than the parametric form of the general solution under the choice of parameters x2 ¼ a, x4 ¼ b, x5 ¼ c. Nonhomogeneous and Associated Homogeneous Systems Let AX ¼ B be a nonhomogeneous system of linear equations. Then AX ¼ 0 is called the associated homogeneous system. For example, x þ 2y 4z ¼ 7 3x 5y þ 6z ¼ 8 and x þ 2y 4z ¼ 0 3x 5y þ 6z ¼ 0 show a nonhomogeneous system and its associated homogeneous system. The relationship between the solution U of a nonhomogeneous system AX ¼ B and the solution W of its associated homogeneous system AX ¼ 0 is contained in the following theorem. THEOREM 3.15: Let v0 be a particular solution of AX ¼ B and let W be the general solution of AX ¼ 0. Then the following is the general solution of AX ¼ B: U ¼ v0 þ W ¼ fv0 þ w : w 2 Wg That is, U ¼ v0 þ W is obtained by adding v0 to each element in W. We note that this theorem has a geometrical interpretation in R3. Speciﬁcally, suppose W is a line through the origin O. Then, as pictured in Fig. 3-4, U ¼ v0 þ W is the line parallel to W obtained by adding v0 to each element of W. Similarly, whenever W is a plane through the origin O, then U ¼ v0 þ W is a plane parallel to W. CHAPTER 3 Systems of Linear Equations 83 3.12 Elementary Matrices Let e denote an elementary row operation and let eðAÞ denote the results of applying the operation e to a matrix A. Now let E be the matrix obtained by applying e to the identity matrix I; that is, E ¼ eðIÞ Then E is called the elementary matrix corresponding to the elementary row operation e. Note that E is always a square matrix. EXAMPLE 3.19 Consider the following three elementary row operations: ð1Þ Interchange R2 and R3: ð2Þ Replace R2 by 6R2: ð3Þ Replace R3 by 4R1 þ R3: The 3 3 elementary matrices corresponding to the above elementary row operations are as follows: E1 ¼ 1 0 0 0 0 1 0 1 0 2 4 3 5; E2 ¼ 1 0 0 0 6 0 0 0 1 2 4 3 5; E3 ¼ 1 0 0 0 1 0 4 0 1 2 4 3 5 The following theorem, proved in Problem 3.34, holds. THEOREM 3.16: Let e be an elementary row operation and let E be the corresponding m m elementary matrix. Then eðAÞ ¼ EA where A is any m n matrix. In other words, the result of applying an elementary row operation e to a matrix A can be obtained by premultiplying A by the corresponding elementary matrix E. Now suppose e0 is the inverse of an elementary row operation e, and let E0 and E be the corresponding matrices. We note (Problem 3.33) that E is invertible and E0 is its inverse. This means, in particular, that any product P ¼ Ek . . . E2E1 of elementary matrices is invertible. Figure 3-4 84 CHAPTER 3 Systems of Linear Equations Applications of Elementary Matrices Using Theorem 3.16, we are able to prove (Problem 3.35) the following important properties of matrices. THEOREM 3.17: Let A be a square matrix. Then the following are equivalent: (a) A is invertible (nonsingular). (b) A is row equivalent to the identity matrix I. (c) A is a product of elementary matrices. Recall that square matrices A and B are inverses if AB ¼ BA ¼ I. The next theorem (proved in Problem 3.36) demonstrates that we need only show that one of the products is true, say AB ¼ I, to prove that matrices are inverses. THEOREM 3.18: Suppose AB ¼ I. Then BA ¼ I, and hence, B ¼ A1. Row equivalence can also be deﬁned in terms of matrix multiplication. Speciﬁcally, we will prove (Problem 3.37) the following. THEOREM 3.19: B is row equivalent to A if and only if there exists a nonsingular matrix P such that B ¼ PA. Application to Finding the Inverse of an n n Matrix The following algorithm ﬁnds the inverse of a matrix. ALGORITHM 3.5: The input is a square matrix A. The output is the inverse of A or that the inverse does not exist. Step 1. Form the n 2n (block) matrix M ¼ ½A; I, where A is the left half of M and the identity matrix I is the right half of M. Step 2. Row reduce M to echelon form. If the process generates a zero row in the A half of M, then STOP A has no inverse. (Otherwise A is in triangular form.) Step 3. Further row reduce M to its row canonical form M ½I; B where the identity matrix I has replaced A in the left half of M. Step 4. Set A1 ¼ B, the matrix that is now in the right half of M. The justiﬁcation for the above algorithm is as follows. Suppose A is invertible and, say, the sequence of elementary row operations e1; e2; . . . ; eq applied to M ¼ ½A; I reduces the left half of M, which is A, to the identity matrix I. Let Ei be the elementary matrix corresponding to the operation ei. Then, by applying Theorem 3.16. we get Eq . . . E2E1A ¼ I or ðEq . . . E2E1IÞA ¼ I; so A1 ¼ Eq . . . E2E1I That is, A1 can be obtained by applying the elementary row operations e1; e2; . . . ; eq to the identity matrix I, which appears in the right half of M. Thus, B ¼ A1, as claimed. EXAMPLE 3.20 Find the inverse of the matrix A ¼ 1 0 2 2 1 3 4 1 8 2 4 3 5. CHAPTER 3 Systems of Linear Equations 85 First form the (block) matrix M ¼ ½A; I and row reduce M to an echelon form: M ¼ 1 0 2 1 0 0 2 1 3 0 1 0 4 1 8 0 0 1 2 4 3 5 1 0 2 1 0 0 0 1 1 2 1 0 0 1 0 4 0 1 2 4 3 5 1 0 2 1 0 0 0 1 1 2 1 0 0 0 1 6 1 1 2 4 3 5 In echelon form, the left half of M is in triangular form; hence, A has an inverse. Next we further row reduce M to its row canonical form: M 1 0 0 11 2 2 0 1 0 4 0 1 0 0 1 6 1 1 2 4 3 5 1 0 0 11 2 2 0 1 0 4 0 1 0 0 1 6 1 1 2 4 3 5 The identity matrix is now in the left half of the ﬁnal matrix; hence, the right half is A1. In other words, A1 ¼ 11 2 2 4 0 1 6 1 1 2 4 3 5 Elementary Column Operations Now let A be a matrix with columns C1; C2; . . . ; Cn. The following operations on A, analogous to the elementary row operations, are called elementary column operations: ½F1 (Column Interchange): Interchange columns Ci and Cj. ½F2 (Column Scaling): Replace Ci by kCi (where k 6¼ 0). ½F3 (Column Addition): Replace Cj by kCi þ Cj. We may indicate each of the column operations by writing, respectively, ð1Þ Ci $ Cj; ð2Þ kCi ! Ci; ð3Þ ðkCi þ CjÞ ! Cj Moreover, each column operation has an inverse operation of the same type, just like the corresponding row operation. Now let f denote an elementary column operation, and let F be the matrix obtained by applying f to the identity matrix I; that is, F ¼ f ðIÞ Then F is called the elementary matrix corresponding to the elementary column operation f . Note that F is always a square matrix. EXAMPLE 3.21 Consider the following elementary column operations: ð1Þ Interchange C1 and C3; ð2Þ Replace C3 by 2C3; ð3Þ Replace C3 by 3C2 þ C3 The corresponding three 3 3 elementary matrices are as follows: F1 ¼ 0 0 1 0 1 0 1 0 0 2 4 3 5; F2 ¼ 1 0 0 0 1 0 0 0 2 2 4 3 5; F3 ¼ 1 0 0 0 1 3 0 0 1 2 4 3 5 The following theorem is analogous to Theorem 3.16 for the elementary row operations. THEOREM 3.20: For any matrix A; f ðAÞ ¼ AF. That is, the result of applying an elementary column operation f on a matrix A can be obtained by postmultiplying A by the corresponding elementary matrix F. 86 CHAPTER 3 Systems of Linear Equations Matrix Equivalence A matrix B is equivalent to a matrix A if B can be obtained from A by a sequence of row and column operations. Alternatively, B is equivalent to A, if there exist nonsingular matrices P and Q such that B ¼ PAQ. Just like row equivalence, equivalence of matrices is an equivalence relation. The main result of this subsection (proved in Problem 3.38) is as follows. THEOREM 3.21: Every m n matrix A is equivalent to a unique block matrix of the form Ir 0 0 0 where Ir is the r-square identity matrix. The following deﬁnition applies. DEFINITION: The nonnegative integer r in Theorem 3.18 is called the rank of A, written rankðAÞ. Note that this deﬁnition agrees with the previous deﬁnition of the rank of a matrix. 3.13 LU DECOMPOSITION Suppose A is a nonsingular matrix that can be brought into (upper) triangular form U using only row-addition operations; that is, suppose A can be triangularized by the following algorithm, which we write using computer notation. ALGORITHM 3.6: The input is a matrix A and the output is a triangular matrix U. Step 1. Repeat for i ¼ 1; 2; . . . ; n 1: Step 2. Repeat for j ¼ i þ 1, i þ 2; . . . ; n (a) Set mij : ¼ aij=aii. (b) Set Rj : ¼ mijRi þ Rj [End of Step 2 inner loop.] [End of Step 1 outer loop.] The numbers mij are called multipliers. Sometimes we keep track of these multipliers by means of the following lower triangular matrix L: L ¼ 1 0 0 . . . 0 0 m21 1 0 . . . 0 0 m31 m32 1 . . . 0 0 mn1 mn2 mn3 . . . mn;n1 1 2 6 6 6 6 4 3 7 7 7 7 5 That is, L has 1’s on the diagonal, 0’s above the diagonal, and the negative of the multiplier mij as its ij-entry below the diagonal. The above matrix L and the triangular matrix U obtained in Algorithm 3.6 give us the classical LU factorization of such a matrix A. Namely, THEOREM 3.22: Let A be a nonsingular matrix that can be brought into triangular form U using only row-addition operations. Then A ¼ LU, where L is the above lower triangular matrix with 1’s on the diagonal, and U is an upper triangular matrix with no 0’s on the diagonal. ......................................................... CHAPTER 3 Systems of Linear Equations 87 EXAMPLE 3.22 SupposeA ¼ 1 2 3 3 4 13 2 1 5 2 4 3 5.WenotethatA maybereducedtotriangularformbytheoperations ‘‘Replace R2 by 3R1 þ R2’’; ‘‘Replace R3 by 2R1 þ R3’’; and then ‘‘Replace R3 by 3 2 R2 þ R3’’ That is, A 1 2 3 0 2 4 0 3 1 2 4 3 5 1 2 3 0 2 4 0 0 7 2 4 3 5 This gives us the classical factorization A ¼ LU, where L ¼ 1 0 0 3 1 0 2 3 2 1 2 6 4 3 7 5 and U ¼ 1 2 3 0 2 4 0 0 7 2 6 4 3 7 5 We emphasize: (1) The entries 3; 2; 3 2 in L are the negatives of the multipliers in the above elementary row operations. (2) U is the triangular form of A. Application to Systems of Linear Equations Consider a computer algorithm M. Let CðnÞ denote the running time of the algorithm as a function of the size n of the input data. [The function CðnÞ is sometimes called the time complexity or simply the complexity of the algorithm M.] Frequently, CðnÞ simply counts the number of multiplications and divisions executed by M, but does not count the number of additions and subtractions because they take much less time to execute. Now consider a square system of linear equations AX ¼ B, where A ¼ ½aij; X ¼ ½x1; . . . ; xnT; B ¼ ½b1; . . . ; bnT and suppose A has an LU factorization. Then the system can be brought into triangular form (in order to apply back-substitution) by applying Algorithm 3.6 to the augmented matrix M ¼ ½A; B of the system. The time complexity of Algorithm 3.6 and back-substitution are, respectively, CðnÞ 1 2 n3 and CðnÞ 1 2 n2 where n is the number of equations. On the other hand, suppose we already have the factorization A ¼ LU. Then, to triangularize the system, we need only apply the row operations in the algorithm (retained by the matrix L) to the column vector B. In this case, the time complexity is CðnÞ 1 2 n2 Of course, to obtain the factorization A ¼ LU requires the original algorithm where CðnÞ 1 2 n3. Thus, nothing may be gained by ﬁrst ﬁnding the LU factorization when a single system is involved. However, there are situations, illustrated below, where the LU factorization is useful. Suppose, for a given matrix A, we need to solve the system AX ¼ B 88 CHAPTER 3 Systems of Linear Equations repeatedly for a sequence of different constant vectors, say B1; B2; . . . ; Bk. Also, suppose some of the Bi depend upon the solution of the system obtained while using preceding vectors Bj. In such a case, it is more efﬁcient to ﬁrst ﬁnd the LU factorization of A, and then to use this factorization to solve the system for each new B. EXAMPLE 3.23 Consider the following system of linear equations: x þ 2y þ z ¼ k1 2x þ 3y þ 3z ¼ k2 3x þ 10y þ 2z ¼ k3 or AX ¼ B; where A ¼ 1 2 1 2 3 3 3 10 2 2 4 3 5 and B ¼ k1 k2 k3 2 4 3 5 Suppose we want to solve the system three times where B is equal, say, to B1; B2; B3. Furthermore, suppose B1 ¼ ½1; 1; 1T, and suppose Bjþ1 ¼ Bj þ Xj ðfor j ¼ 1; 2Þ where Xj is the solution of AX ¼ Bj. Here it is more efﬁcient to ﬁrst obtain the LU factorization of A and then use the LU factorization to solve the system for each of the B’s. (This is done in Problem 3.42.) SOLVED PROBLEMS Linear Equations, Solutions, 2 2 Systems 3.1. Determine whether each of the following equations is linear: (a) 5x þ 7y 8yz ¼ 16, (b) x þ py þ ez ¼ log 5, (c) 3x þ ky 8z ¼ 16 (a) No, because the product yz of two unknowns is of second degree. (b) Yes, because p; e, and log 5 are constants. (c) As it stands, there are four unknowns: x; y; z; k. Because of the term ky it is not a linear equation. However, assuming k is a constant, the equation is linear in the unknowns x; y; z. 3.2. Determine whether the following vectors are solutions of x1 þ 2x2 4x3 þ 3x4 ¼ 15: (a) u ¼ ð3; 2; 1; 4Þ and (b) v ¼ ð1; 2; 4; 5Þ: (a) Substitute to obtain 3 þ 2ð2Þ 4ð1Þ þ 3ð4Þ ¼ 15, or 15 ¼ 15; yes, it is a solution. (b) Substitute to obtain 1 þ 2ð2Þ 4ð4Þ þ 3ð5Þ ¼ 15, or 4 ¼ 15; no, it is not a solution. 3.3. Solve (a) ex ¼ p, (b) 3x 4 x ¼ 2x þ 3, (c) 7 þ 2x 4 ¼ 3x þ 3 x (a) Because e 6¼ 0, multiply by 1=e to obtain x ¼ p=e. (b) Rewrite in standard form, obtaining 0x ¼ 7. The equation has no solution. (c) Rewrite in standard form, obtaining 0x ¼ 0. Every scalar k is a solution. 3.4. Prove Theorem 3.4: Consider the equation ax ¼ b. (i) If a 6¼ 0, then x ¼ b=a is a unique solution of ax ¼ b. (ii) If a ¼ 0 but b 6¼ 0, then ax ¼ b has no solution. (iii) If a ¼ 0 and b ¼ 0, then every scalar k is a solution of ax ¼ b. Suppose a 6¼ 0. Then the scalar b=a exists. Substituting b=a in ax ¼ b yields aðb=aÞ ¼ b, or b ¼ b; hence, b=a is a solution. On the other hand, suppose x0 is a solution to ax ¼ b, so that ax0 ¼ b. Multiplying both sides by 1=a yields x0 ¼ b=a. Hence, b=a is the unique solution of ax ¼ b. Thus, (i) is proved. On the other hand, suppose a ¼ 0. Then, for any scalar k, we have ak ¼ 0k ¼ 0. If b 6¼ 0, then ak 6¼ b. Accordingly, k is not a solution of ax ¼ b, and so (ii) is proved. If b ¼ 0, then ak ¼ b. That is, any scalar k is a solution of ax ¼ b, and so (iii) is proved. CHAPTER 3 Systems of Linear Equations 89 3.5. Solve each of the following systems: (a) 2x 5y ¼ 11 3x þ 4y ¼ 5 (b) 2x 3y ¼ 8 6x þ 9y ¼ 6 (c) 2x 3y ¼ 8 4x þ 6y ¼ 16 (a) Eliminate x from the equations by forming the new equation L ¼ 3L1 þ 2L2. This yields the equation 23y ¼ 23; and so y ¼ 1 Substitute y ¼ 1 in one of the original equations, say L1, to get 2x 5ð1Þ ¼ 11 or 2x þ 5 ¼ 11 or 2x ¼ 6 or x ¼ 3 Thus, x ¼ 3, y ¼ 1 or the pair u ¼ ð3; 1Þ is the unique solution of the system. (b) Eliminate x from the equations by forming the new equation L ¼ 3L1 þ L2. This yields the equation 0x þ 0y ¼ 30 This is a degenerate equation with a nonzero constant; hence, this equation and the system have no solution. (Geometrically, the lines corresponding to the equations are parallel.) (c) Eliminate x from the equations by forming the new equation L ¼ 2L1 þ L2. This yields the equation 0x þ 0y ¼ 0 This is a degenerate equation where the constant term is also zero. Thus, the system has an inﬁnite number of solutions, which correspond to the solution of either equation. (Geometrically, the lines corresponding to the equations coincide.) To ﬁnd the general solution, set y ¼ a and substitute in L1 to obtain 2x 3a ¼ 8 or 2x ¼ 3a þ 8 or x ¼ 3 2 a þ 4 Thus, the general solution is x ¼ 3 2 a þ 4; y ¼ a or u ¼ 3 2 a þ 4; a where a is any scalar. 3.6. Consider the system x þ ay ¼ 4 ax þ 9y ¼ b (a) For which values of a does the system have a unique solution? (b) Find those pairs of values (a; b) for which the system has more than one solution. (a) Eliminate x from the equations by forming the new equation L ¼ aL1 þ L2. This yields the equation ð9 a2Þy ¼ b 4a ð1Þ The system has a unique solution if and only if the coefﬁcient of y in (1) is not zero—that is, if 9 a2 6¼ 0 or if a 6¼ 3. (b) The system has more than one solution if both sides of (1) are zero. The left-hand side is zero when a ¼ 3. When a ¼ 3, the right-hand side is zero when b 12 ¼ 0 or b ¼ 12. When a ¼ 3, the right-hand side is zero when b þ 12 0 or b ¼ 12. Thus, (3; 12) and ð3; 12Þ are the pairs for which the system has more than one solution. Systems in Triangular and Echelon Form 3.7. Determine the pivot and free variables in each of the following systems: 2x1 3x2 6x3 5x4 þ 2x5 ¼ 7 x3 þ 3x4 7x5 ¼ 6 x4 2x5 ¼ 1 (a) 2x 6y þ 7z ¼ 1 4y þ 3z ¼ 8 2z ¼ 4 (b) x þ 2y 3z ¼ 2 2x þ 3y þ z ¼ 4 3x þ 4y þ 5z ¼ 8 (c) (a) In echelon form, the leading unknowns are the pivot variables, and the others are the free variables. Here x1, x3, x4 are the pivot variables, and x2 and x5 are the free variables. 90 CHAPTER 3 Systems of Linear Equations (b) The leading unknowns are x; y; z, so they are the pivot variables. There are no free variables (as in any triangular system). (c) The notion of pivot and free variables applies only to a system in echelon form. 3.8. Solve the triangular system in Problem 3.7(b). Because it is a triangular system, solve by back-substitution. (i) The last equation gives z ¼ 2. (ii) Substitute z ¼ 2 in the second equation to get 4y þ 6 ¼ 8 or y ¼ 1 2. (iii) Substitute z ¼ 2 and y ¼ 1 2 in the ﬁrst equation to get 2x 6 1 2 þ 7ð2Þ ¼ 1 or 2x þ 11 ¼ 1 or x ¼ 5 Thus, x ¼ 5, y ¼ 1 2, z ¼ 2 or u ¼ ð5; 1 2 ; 2Þ is the unique solution to the system. 3.9. Solve the echelon system in Problem 3.7(a). Assign parameters to the free variables, say x2 ¼ a and x5 ¼ b, and solve for the pivot variables by back-substitution. (i) Substitute x5 ¼ b in the last equation to get x4 2b ¼ 1 or x4 ¼ 2b þ 1. (ii) Substitute x5 ¼ b and x4 ¼ 2b þ 1 in the second equation to get x3 þ 3ð2b þ 1Þ 7b ¼ 6 or x3 b þ 3 ¼ 6 or x3 ¼ b þ 3 (iii) Substitute x5 ¼ b, x4 ¼ 2b þ 1, x3 ¼ b þ 3, x2 ¼ a in the ﬁrst equation to get 2x1 3a 6ðb þ 3Þ 5ð2b þ 1Þ þ 2b ¼ 7 or 2x1 3a 14b 23 ¼ 7 or x1 ¼ 3 2 a þ 7b þ 15 Thus, x1 ¼ 3 2 a þ 7b þ 15; x2 ¼ a; x3 ¼ b þ 3; x4 ¼ 2b þ 1; x5 ¼ b or u ¼ 3 2 a þ 7b þ 15; a; b þ 3; 2b þ 1; b is the parametric form of the general solution. Alternatively, solving for the pivot variable x1; x3; x4 in terms of the free variables x2 and x5 yields the following free-variable form of the general solution: x1 ¼ 3 2 x2 þ 7x5 þ 15; x3 ¼ x5 þ 3; x4 ¼ 2x5 þ 1 3.10. Prove Theorem 3.6. Consider the system (3.4) of linear equations in echelon form with r equations and n unknowns. (i) If r ¼ n, then the system has a unique solution. (ii) If r < n, then we can arbitrarily assign values to the n r free variable and solve uniquely for the r pivot variables, obtaining a solution of the system. (i) Suppose r ¼ n. Then we have a square system AX ¼ B where the matrix A of coefﬁcients is (upper) triangular with nonzero diagonal elements. Thus, A is invertible. By Theorem 3.10, the system has a unique solution. (ii) Assigning values to the n r free variables yields a triangular system in the pivot variables, which, by (i), has a unique solution. CHAPTER 3 Systems of Linear Equations 91 Gaussian Elimination 3.11. Solve each of the following systems: x þ 2y 4z ¼ 4 2x þ 5y 9z ¼ 10 3x 2y þ 3z ¼ 11 (a) x þ 2y 3z ¼ 1 3x þ y 2z ¼ 7 5x þ 3y 4z ¼ 2 (b) x þ 2y 3z ¼ 1 2x þ 5y 8z ¼ 4 3x þ 8y 13z ¼ 7 (c) Reduce each system to triangular or echelon form using Gaussian elimination: (a) Apply ‘‘Replace L2 by 2L1 þ L2’’ and ‘‘Replace L3 by 3L1 þ L3’’ to eliminate x from the second and third equations, and then apply ‘‘Replace L3 by 8L2 þ L3’’ to eliminate y from the third equation. These operations yield x þ 2y 4z ¼ 4 y z ¼ 2 8y þ 15z ¼ 23 and then x þ 2y 4z ¼ 4 y z ¼ 2 7z ¼ 7 The system is in triangular form. Solve by back-substitution to obtain the unique solution u ¼ ð2; 1; 1Þ. (b) Eliminate x from the second and third equations by the operations ‘‘Replace L2 by 3L1 þ L2’’ and ‘‘Replace L3 by 5L1 þ L3.’’ This gives the equivalent system x þ 2y 3z ¼ 1 7y 11z ¼ 10 7y þ 11z ¼ 7 The operation ‘‘Replace L3 by L2 þ L3’’ yields the following degenerate equation with a nonzero constant: 0x þ 0y þ 0z ¼ 3 This equation and hence the system have no solution. (c) Eliminate x from the second and third equations by the operations ‘‘Replace L2 by 2L1 þ L2’’ and ‘‘Replace L3 by 3L1 þ L3.’’ This yields the new system x þ 2y 3z ¼ 1 y 2z ¼ 2 2y 4z ¼ 4 or x þ 2y 3z ¼ 1 y 2z ¼ 2 (The third equation is deleted, because it is a multiple of the second equation.) The system is in echelon form with pivot variables x and y and free variable z. To ﬁnd the parametric form of the general solution, set z ¼ a and solve for x and y by back-substitution. Substitute z ¼ a in the second equation to get y ¼ 2 þ 2a. Then substitute z ¼ a and y ¼ 2 þ 2a in the ﬁrst equation to get x þ 2ð2 þ 2aÞ 3a ¼ 1 or x þ 4 þ a ¼ 1 or x ¼ 3 a Thus, the general solution is x ¼ 3 a; y ¼ 2 þ 2a; z ¼ a or u ¼ ð3 a; 2 þ 2a; aÞ where a is a parameter. 3.12. Solve each of the following systems: x1 3x2 þ 2x3 x4 þ 2x5 ¼ 2 3x1 9x2 þ 7x3 x4 þ 3x5 ¼ 7 2x1 6x2 þ 7x3 þ 4x4 5x5 ¼ 7 (a) x1 þ 2x2 3x3 þ 4x4 ¼ 2 2x1 þ 5x2 2x3 þ x4 ¼ 1 5x1 þ 12x2 7x3 þ 6x4 ¼ 3 (b) Reduce each system to echelon form using Gaussian elimination: 92 CHAPTER 3 Systems of Linear Equations (a) Apply ‘‘Replace L2 by 3L1 þ L2’’ and ‘‘Replace L3 by 2L1 þ L3’’ to eliminate x from the second and third equations. This yields x1 3x2 þ 2x3 x4 þ 2x5 ¼ 2 x3 þ 2x4 3x5 ¼ 1 3x3 þ 6x4 9x5 ¼ 3 or x1 3x2 þ 2x3 x4 þ 2x5 ¼ 2 x3 þ 2x4 3x5 ¼ 1 (We delete L3, because it is a multiple of L2.) The system is in echelon form with pivot variables x1 and x3 and free variables x2; x4; x5. To ﬁnd the parametric form of the general solution, set x2 ¼ a, x4 ¼ b, x5 ¼ c, where a; b; c are parameters. Back-substitution yields x3 ¼ 1 2b þ 3c and x1 ¼ 3a þ 5b 8c. The general solution is x1 ¼ 3a þ 5b 8c; x2 ¼ a; x3 ¼ 1 2b þ 3c; x4 ¼ b; x5 ¼ c or, equivalently, u ¼ ð3a þ 5b 8c; a; 1 2b þ 3c; b; cÞ. (b) Eliminate x1 from the second and third equations by the operations ‘‘Replace L2 by 2L1 þ L2’’ and ‘‘Replace L3 by 5L1 þ L3.’’ This yields the system x1 þ 2x2 3x3 þ 4x4 ¼ 2 x2 þ 4x3 7x4 ¼ 3 2x2 þ 8x3 14x4 ¼ 7 The operation ‘‘Replace L3 by 2L2 þ L3’’ yields the degenerate equation 0 ¼ 1. Thus, the system has no solution (even though the system has more unknowns than equations). 3.13. Solve using the condensed format: 2y þ 3z ¼ 3 x þ y þ z ¼ 4 4x þ 8y 3z ¼ 35 The condensed format follows: Number Equation Operation ð2Þ ð 1 =Þ 2y þ 3z ¼ 3 L1 $ L2 ð1Þ ð 2 =Þ x þ y þ z ¼ 4 L1 $ L2 ð3Þ 4x þ 8y 3z ¼ 35 ð30Þ 4y 7z ¼ 19 Replace L3 by 4L1 þ L3 ð300Þ 13z ¼ 13 Replace L3 by 2L2 þ L3 Here (1), (2), and (300) form a triangular system. (We emphasize that the interchange of L1 and L2 is accomplished by simply renumbering L1 and L2 as above.) Using back-substitution with the triangular system yields z ¼ 1 from L3, y ¼ 3 from L2, and x ¼ 2 from L1. Thus, the unique solution of the system is x ¼ 2, y ¼ 3, z ¼ 1 or the triple u ¼ ð2; 3; 1Þ. 3.14. Consider the system x þ 2y þ z ¼ 3 ay þ 5z ¼ 10 2x þ 7y þ az ¼ b (a) Find those values of a for which the system has a unique solution. (b) Find those pairs of values ða; bÞ for which the system has more than one solution. Reduce the system to echelon form. That is, eliminate x from the third equation by the operation ‘‘Replace L3 by 2L1 þ L3’’ and then eliminate y from the third equation by the operation CHAPTER 3 Systems of Linear Equations 93 ‘‘Replace L3 by 3L2 þ aL3.’’ This yields x þ 2y þ z ¼ 3 ay þ 5z ¼ 10 3y þ ða 2Þz ¼ b 6 and then x þ 2y þ z ¼ 3 ay þ 5z ¼ 10 ða2 2a 15Þz ¼ ab 6a 30 Examine the last equation ða2 2a 15Þz ¼ ab 6a 30. (a) The system has a unique solution if and only if the coefﬁcient of z is not zero; that is, if a2 2a 15 ¼ ða 5Þða þ 3Þ 6¼ 0 or a 6¼ 5 and a 6¼ 3: (b) The system has more than one solution if both sides are zero. The left-hand side is zero when a ¼ 5 or a ¼ 3. When a ¼ 5, the right-hand side is zero when 5b 60 ¼ 0, or b ¼ 12. When a ¼ 3, the right-hand side is zero when 3b 12 ¼ 0, or b ¼ 4. Thus, ð5; 12Þ and ð3; 4Þ are the pairs for which the system has more than one solution. Echelon Matrices, Row Equivalence, Row Canonical Form 3.15. Row reduce each of the following matrices to echelon form: (a) A ¼ 1 2 3 0 2 4 2 2 3 6 4 3 2 4 3 5; (b) B ¼ 4 1 6 1 2 5 6 3 4 2 4 3 5 (a) Use a11 ¼ 1 as a pivot to obtain 0’s below a11; that is, apply the row operations ‘‘Replace R2 by 2R1 þ R2’’ and ‘‘Replace R3 by 3R1 þ R3:’’ Then use a23 ¼ 4 as a pivot to obtain a 0 below a23; that is, apply the row operation ‘‘Replace R3 by 5R2 þ 4R3.’’ These operations yield A 1 2 3 0 0 0 4 2 0 0 5 3 2 4 3 5 1 2 3 0 0 0 4 2 0 0 0 2 2 4 3 5 The matrix is now in echelon form. (b) Hand calculations are usually simpler if the pivot element equals 1. Therefore, ﬁrst interchange R1 and R2. Next apply the operations ‘‘Replace R2 by 4R1 þ R2’’ and ‘‘Replace R3 by 6R1 þ R3’’; and then apply the operation ‘‘Replace R3 by R2 þ R3.’’ These operations yield B 1 2 5 4 1 6 6 3 4 2 4 3 5 1 2 5 0 9 26 0 9 26 2 4 3 5 1 2 5 0 9 26 0 0 0 2 4 3 5 The matrix is now in echelon form. 3.16. Describe the pivoting row-reduction algorithm. Also describe the advantages, if any, of using this pivoting algorithm. The row-reduction algorithm becomes a pivoting algorithm if the entry in column j of greatest absolute value is chosen as the pivot a1j1 and if one uses the row operation ðaij1=a1j1ÞR1 þ Ri ! Ri The main advantage of the pivoting algorithm is that the above row operation involves division by the (current) pivot a1j1, and, on the computer, roundoff errors may be substantially reduced when one divides by a number as large in absolute value as possible. 3.17. Let A ¼ 2 2 2 1 3 6 0 1 1 7 10 2 2 4 3 5. Reduce A to echelon form using the pivoting algorithm. 94 CHAPTER 3 Systems of Linear Equations First interchange R1 and R2 so that 3 can be used as the pivot, and then apply the operations ‘‘Replace R2 by 2 3 R1 þ R2’’ and ‘‘Replace R3 by 1 3 R1 þ R3.’’ These operations yield A 3 6 0 1 2 2 2 1 1 7 10 2 2 4 3 5 3 6 0 1 0 2 2 1 3 0 5 10 5 3 2 6 4 3 7 5 Now interchange R2 and R3 so that 5 can be used as the pivot, and then apply the operation ‘‘Replace R3 by 2 5 R2 þ R3.’’ We obtain A 3 6 0 1 0 5 10 5 3 0 2 2 1 3 2 4 3 5 3 6 0 1 0 5 10 5 3 0 0 6 1 2 4 3 5 The matrix has been brought to echelon form using partial pivoting. 3.18. Reduce each of the following matrices to row canonical form: (a) A ¼ 2 2 1 6 4 4 4 1 10 13 8 8 1 26 23 2 4 3 5; (b) B ¼ 5 9 6 0 2 3 0 0 7 2 4 3 5 (a) First reduce A to echelon form by applying the operations ‘‘Replace R2 by 2R1 þ R2’’ and ‘‘Replace R3 by 4R1 þ R3,’’ and then applying the operation ‘‘Replace R3 by R2 þ R3.’’ These operations yield A 2 2 1 6 4 0 0 3 2 5 0 0 3 2 7 2 4 3 5 2 2 1 6 4 0 0 3 2 5 0 0 0 4 2 2 4 3 5 Now use back-substitution on the echelon matrix to obtain the row canonical form of A. Speciﬁcally, ﬁrst multiply R3 by 1 4 to obtain the pivot a34 ¼ 1, and then apply the operations ‘‘Replace R2 by 2R3 þ R2’’ and ‘‘Replace R1 by 6R3 þ R1.’’ These operations yield A 2 2 1 6 4 0 0 3 2 5 0 0 0 1 1 2 2 4 3 5 2 2 1 0 1 0 0 3 0 6 0 0 0 1 1 2 2 4 3 5 Now multiply R2 by 1 3 , making the pivot a23 ¼ 1, and then apply ‘‘Replace R1 by R2 þ R1,’’ yielding A 2 2 1 0 1 0 0 1 0 2 0 0 0 1 1 2 2 4 3 5 2 2 0 0 3 0 0 1 0 2 0 0 0 1 1 2 2 4 3 5 Finally, multiply R1 by 1 2 , so the pivot a11 ¼ 1. Thus, we obtain the following row canonical form of A: A 1 1 0 0 3 2 0 0 1 0 2 0 0 0 1 1 2 2 4 3 5 (b) Because B is in echelon form, use back-substitution to obtain B 5 9 6 0 2 3 0 0 1 2 6 4 3 7 5 5 9 0 0 2 0 0 0 1 2 6 4 3 7 5 5 9 0 0 1 0 0 0 1 2 6 4 3 7 5 5 0 0 0 1 0 0 0 1 2 6 4 3 7 5 1 0 0 0 1 0 0 0 1 2 6 4 3 7 5 The last matrix, which is the identity matrix I, is the row canonical form of B. (This is expected, because B is invertible, and so its row canonical form must be I.) 3.19. Describe the Gauss–Jordan elimination algorithm, which also row reduces an arbitrary matrix A to its row canonical form. CHAPTER 3 Systems of Linear Equations 95 The Gauss–Jordan algorithm is similar in some ways to the Gaussian elimination algorithm, except that here each pivot is used to place 0’s both below and above the pivot, not just below the pivot, before working with the next pivot. Also, one variation of the algorithm ﬁrst normalizes each row—that is, obtains a unit pivot—before it is used to produce 0’s in the other rows, rather than normalizing the rows at the end of the algorithm. 3.20. Let A ¼ 1 2 3 1 2 1 1 4 1 3 2 5 9 2 8 2 4 3 5. Use Gauss–Jordan to ﬁnd the row canonical form of A. Use a11 ¼ 1 as a pivot to obtain 0’s below a11 by applying the operations ‘‘Replace R2 by R1 þ R2’’ and ‘‘Replace R3 by 2R1 þ R3.’’ This yields A 1 2 3 1 2 0 3 1 2 1 0 9 3 4 4 2 4 3 5 Multiply R2 by 1 3 to make the pivot a22 ¼ 1, and then produce 0’s below and above a22 by applying the operations ‘‘Replace R3 by 9R2 þ R3’’ and ‘‘Replace R1 by 2R2 þ R1.’’ These operations yield A 1 2 3 1 2 0 1 1 3 2 3 1 3 0 9 3 4 4 2 6 6 4 3 7 7 5 1 0 11 3 1 3 8 3 0 1 1 3 2 3 1 3 0 0 0 2 1 2 6 6 4 3 7 7 5 Finally, multiply R3 by 1 2 to make the pivot a34 ¼ 1, and then produce 0’s above a34 by applying the operations ‘‘Replace R2 by 2 3 R3 þ R2’’ and ‘‘Replace R1 by 1 3 R3 þ R1.’’ These operations yield A 1 0 11 3 1 3 8 3 0 1 1 3 2 3 1 3 0 0 0 1 1 2 2 6 6 4 3 7 7 5 1 0 11 3 0 17 6 0 1 1 3 0 2 3 0 0 0 1 1 2 2 6 6 4 3 7 7 5 which is the row canonical form of A. Systems of Linear Equations in Matrix Form 3.21. Find the augmented matrix M and the coefﬁcient matrix A of the following system: x þ 2y 3z ¼ 4 3y 4z þ 7x ¼ 5 6z þ 8x 9y ¼ 1 First align the unknowns in the system, and then use the aligned system to obtain M and A. We have x þ 2y 3z ¼ 4 7x þ 3y 4z ¼ 5 8x 9y þ 6z ¼ 1 ; then M ¼ 1 2 3 4 7 3 4 5 8 9 6 1 2 4 3 5 and A ¼ 1 2 3 7 3 4 8 9 6 2 4 3 5 3.22. Solve each of the following systems using its augmented matrix M: x þ 2y z ¼ 3 x þ 3y þ z ¼ 5 3x þ 8y þ 4z ¼ 17 (a) x 2y þ 4z ¼ 2 2x 3y þ 5z ¼ 3 3x 4y þ 6z ¼ 7 (b) x þ y þ 3z ¼ 1 2x þ 3y z ¼ 3 5x þ 7y þ z ¼ 7 (c) (a) Reduce the augmented matrix M to echelon form as follows: M ¼ 1 2 1 3 1 3 1 5 3 8 4 17 2 4 3 5 1 2 1 3 0 1 2 2 0 2 7 8 2 4 3 5 1 2 1 3 0 1 2 2 0 0 3 4 2 4 3 5 96 CHAPTER 3 Systems of Linear Equations Now write down the corresponding triangular system x þ 2y z ¼ 3 y þ 2z ¼ 2 3z ¼ 4 and solve by back-substitution to obtain the unique solution x ¼ 17 3 ; y ¼ 2 3 ; z ¼ 4 3 or u ¼ ð17 3 ; 2 3 ; 4 3Þ Alternately, reduce the echelon form of M to row canonical form, obtaining M 1 2 1 3 0 1 2 2 0 0 1 4 3 2 6 6 4 3 7 7 5 1 2 0 13 3 0 1 0 2 3 0 0 1 4 3 2 6 6 4 3 7 7 5 1 0 0 17 3 0 1 0 2 3 0 0 1 4 3 2 6 6 4 3 7 7 5 This also corresponds to the above solution. (b) First reduce the augmented matrix M to echelon form as follows: M ¼ 1 2 4 2 2 3 5 3 3 4 6 7 2 4 3 5 1 2 4 2 0 1 3 1 0 2 6 1 2 4 3 5 1 2 4 2 0 1 3 1 0 0 0 3 2 4 3 5 The third row corresponds to the degenerate equation 0x þ 0y þ 0z ¼ 3, which has no solution. Thus, ‘‘DO NOT CONTINUE.’’ The original system also has no solution. (Note that the echelon form indicates whether or not the system has a solution.) (c) Reduce the augmented matrix M to echelon form and then to row canonical form: M ¼ 1 1 3 1 2 3 1 3 5 7 1 7 2 4 3 5 1 1 3 1 0 1 7 1 0 2 14 2 2 4 3 5 1 0 10 0 0 1 7 1 (The third row of the second matrix is deleted, because it is a multiple of the second row and will result in a zero row.) Write down the system corresponding to the row canonical form of M and then transfer the free variables to the other side to obtain the free-variable form of the solution: x þ 10z ¼ 0 y 7z ¼ 1 and x ¼ 10z y ¼ 1 þ 7z Here z is the only free variable. The parametric solution, using z ¼ a, is as follows: x ¼ 10a; y ¼ 1 þ 7a; z ¼ a or u ¼ ð10a; 1 þ 7a; aÞ 3.23. Solve the following system using its augmented matrix M: x1 þ 2x2 3x3 2x4 þ 4x5 ¼ 1 2x1 þ 5x2 8x3 x4 þ 6x5 ¼ 4 x1 þ 4x2 7x3 þ 5x4 þ 2x5 ¼ 8 Reduce the augmented matrix M to echelon form and then to row canonical form: M ¼ 1 2 3 2 4 1 2 5 8 1 6 4 1 4 7 5 2 8 2 6 4 3 7 5 1 2 3 2 4 1 0 1 2 3 2 2 0 2 4 7 2 7 2 6 4 3 7 5 1 2 3 2 4 1 0 1 2 3 2 2 0 0 0 1 2 3 2 6 4 3 7 5 1 2 3 0 8 7 0 1 2 0 8 7 0 0 0 1 2 3 2 6 4 3 7 5 1 0 1 0 24 21 0 1 2 0 8 7 0 0 0 1 2 3 2 6 4 3 7 5 Write down the system corresponding to the row canonical form of M and then transfer the free variables to the other side to obtain the free-variable form of the solution: x1 þ x3 þ 24x5 ¼ 21 x2 2x3 8x5 ¼ 7 x4 þ 2x5 ¼ 3 and x1 ¼ 21 x3 24x5 x2 ¼ 7 þ 2x3 þ 8x5 x4 ¼ 3 2x5 CHAPTER 3 Systems of Linear Equations 97 Here x1; x2; x4 are the pivot variables and x3 and x5 are the free variables. Recall that the parametric form of the solution can be obtained from the free-variable form of the solution by simply setting the free variables equal to parameters, say x3 ¼ a, x5 ¼ b. This process yields x1 ¼ 21 a 24b; x2 ¼ 7 þ 2a þ 8b; x3 ¼ a; x4 ¼ 3 2b; x5 ¼ b or u ¼ ð21 a 24b; 7 þ 2a þ 8b; a; 3 2b; bÞ which is another form of the solution. Linear Combinations, Homogeneous Systems 3.24. Write v as a linear combination of u1; u2; u3, where (a) v ¼ ð3; 10; 7Þ and u1 ¼ ð1; 3; 2Þ; u2 ¼ ð1; 4; 2Þ; u3 ¼ ð2; 8; 1Þ; (b) v ¼ ð2; 7; 10Þ and u1 ¼ ð1; 2; 3Þ, u2 ¼ ð1; 3; 5Þ, u3 ¼ ð1; 5; 9Þ; (c) v ¼ ð1; 5; 4Þ and u1 ¼ ð1; 3; 2Þ, u2 ¼ ð2; 7; 1Þ, u3 ¼ ð1; 6; 7Þ. Find the equivalent system of linear equations by writing v ¼ xu1 þ yu2 þ zu3. Alternatively, use the augmented matrix M of the equivalent system, where M ¼ ½u1; u2; u3; v. (Here u1; u2; u3; v are the columns of M.) (a) The vector equation v ¼ xu1 þ yu2 þ zu3 for the given vectors is as follows: 3 10 7 2 4 3 5 ¼ x 1 3 2 2 4 3 5 þ y 1 4 2 2 4 3 5 þ z 2 8 1 2 4 3 5 ¼ x þ y þ 2z 3x þ 4y þ 8z 2x þ 2y þ z 2 4 3 5 Form the equivalent system of linear equations by setting corresponding entries equal to each other, and then reduce the system to echelon form: x þ y þ 2z ¼ 3 3x þ 4y þ 8z ¼ 10 2x þ 2y þ z ¼ 7 or x þ y þ 2z ¼ 3 y þ 2z ¼ 1 4y þ 5z ¼ 13 or x þ y þ 2z ¼ 3 y þ 2z ¼ 1 3z ¼ 9 The system is in triangular form. Back-substitution yields the unique solution x ¼ 2, y ¼ 7, z ¼ 3. Thus, v ¼ 2u1 þ 7u2 3u3. Alternatively, form the augmented matrix M ¼ [u1; u2; u3; v] of the equivalent system, and reduce M to echelon form: M ¼ 1 1 2 3 3 4 8 10 2 2 1 7 2 4 3 5 1 1 2 3 0 1 2 1 0 4 5 13 2 4 3 5 1 1 2 3 0 1 2 1 0 0 3 9 2 4 3 5 The last matrix corresponds to a triangular system that has a unique solution. Back-substitution yields the solution x ¼ 2, y ¼ 7, z ¼ 3. Thus, v ¼ 2u1 þ 7u2 3u3. (b) Form the augmented matrix M ¼ ½u1; u2; u3; v of the equivalent system, and reduce M to the echelon form: M ¼ 1 1 1 2 2 3 5 7 3 5 9 10 2 4 3 5 1 1 1 2 0 1 3 3 0 2 6 4 2 4 3 5 1 1 1 2 0 1 3 3 0 0 0 2 2 4 3 5 The third row corresponds to the degenerate equation 0x þ 0y þ 0z ¼ 2, which has no solution. Thus, the system also has no solution, and v cannot be written as a linear combination of u1; u2; u3. (c) Form the augmented matrix M ¼ ½u1; u2; u3; v of the equivalent system, and reduce M to echelon form: M ¼ 1 2 1 1 3 7 6 5 2 1 7 4 2 4 3 5 1 2 1 1 0 1 3 2 0 3 9 6 2 4 3 5 1 2 1 1 0 1 3 2 0 0 0 0 2 4 3 5 98 CHAPTER 3 Systems of Linear Equations The last matrix corresponds to the following system with free variable z: x þ 2y þ z ¼ 1 y þ 3z ¼ 2 Thus, v can be written as a linear combination of u1; u2; u3 in many ways. For example, let the free variable z ¼ 1, and, by back-substitution, we get y ¼ 2 and x ¼ 2. Thus, v ¼ 2u1 2u2 þ u3. 3.25. Let u1 ¼ ð1; 2; 4Þ, u2 ¼ ð2; 3; 1Þ, u3 ¼ ð2; 1; 1Þ in R3. Show that u1; u2; u3 are orthogonal, and write v as a linear combination of u1; u2; u3, where (a) v ¼ ð7; 16; 6Þ, (b) v ¼ ð3; 5; 2Þ. Take the dot product of pairs of vectors to get u1 u2 ¼ 2 6 þ 4 ¼ 0; u1 u3 ¼ 2 þ 2 4 ¼ 0; u2 u3 ¼ 4 3 1 ¼ 0 Thus, the three vectors in R3 are orthogonal, and hence Fourier coefﬁcients can be used. That is, v ¼ xu1 þ yu2 þ zu3, where x ¼ v u1 u1 u1 ; y ¼ v u2 u2 u2 ; z ¼ v u3 u3 u3 (a) We have x ¼ 7 þ 32 þ 24 1 þ 4 þ 16 ¼ 63 21 ¼ 3; y ¼ 14 48 þ 6 4 þ 9 þ 1 ¼ 28 14 ¼ 2; z ¼ 14 þ 16 6 4 þ 1 þ 1 ¼ 24 6 ¼ 4 Thus, v ¼ 3u1 2u2 þ 4u3. (b) We have x ¼ 3 þ 10 þ 8 1 þ 4 þ 16 ¼ 21 21 ¼ 1; y ¼ 6 15 þ 2 4 þ 9 þ 1 ¼ 7 14 ¼ 1 2 ; z ¼ 6 þ 5 2 4 þ 1 þ 1 ¼ 9 6 ¼ 3 2 Thus, v ¼ u1 1 2 u2 þ 3 2 u3. 3.26. Find the dimension and a basis for the general solution W of each of the following homogeneous systems: 2x1 þ 4x2 5x3 þ 3x4 ¼ 0 3x1 þ 6x2 7x3 þ 4x4 ¼ 0 5x1 þ 10x2 11x3 þ 6x4 ¼ 0 (a) x 2y 3z ¼ 0 2x þ y þ 3z ¼ 0 3x 4y 2z ¼ 0 (b) (a) Reduce the system to echelon form using the operations ‘‘Replace L2 by 3L1 þ 2L2,’’ ‘‘Replace L3 by 5L1 þ 2L3,’’ and then ‘‘Replace L3 by 2L2 þ L3.’’ These operations yield 2x1 þ 4x2 5x3 þ 3x4 ¼ 0 x3 x4 ¼ 0 3x3 3x4 ¼ 0 and 2x1 þ 4x2 5x3 þ 3x4 ¼ 0 x3 x4 ¼ 0 The system in echelon form has two free variables, x2 and x4, so dim W ¼ 2. A basis ½u1; u2 for W may be obtained as follows: (1) Set x2 ¼ 1, x4 ¼ 0. Back-substitution yields x3 ¼ 0, and then x1 ¼ 2. Thus, u1 ¼ ð2; 1; 0; 0Þ. (2) Set x2 ¼ 0, x4 ¼ 1. Back-substitution yields x3 ¼ 1, and then x1 ¼ 1. Thus, u2 ¼ ð1; 0; 1; 1Þ. (b) Reduce the system to echelon form, obtaining x 2y 3z ¼ 0 5y þ 9z ¼ 0 2y þ 7z ¼ 0 and x 2y 3z ¼ 0 5y þ 9z ¼ 0 17z ¼ 0 There are no free variables (the system is in triangular form). Hence, dim W ¼ 0, and W has no basis. Speciﬁcally, W consists only of the zero solution; that is, W ¼ f0g. 3.27. Find the dimension and a basis for the general solution W of the following homogeneous system using matrix notation: x1 þ 2x2 þ 3x3 2x4 þ 4x5 ¼ 0 2x1 þ 4x2 þ 8x3 þ x4 þ 9x5 ¼ 0 3x1 þ 6x2 þ 13x3 þ 4x4 þ 14x5 ¼ 0 Show how the basis gives the parametric form of the general solution of the system. When a system is homogeneous, we represent the system by its coefﬁcient matrix A rather than by its CHAPTER 3 Systems of Linear Equations 99 augmented matrix M, because the last column of the augmented matrix M is a zero column, and it will remain a zero column during any row-reduction process. Reduce the coefﬁcient matrix A to echelon form, obtaining A ¼ 1 2 3 2 4 2 4 8 1 9 3 6 13 4 14 2 4 3 5 1 2 3 2 4 0 0 2 5 1 0 0 4 10 2 2 4 3 5 1 2 3 2 4 0 0 2 5 1 (The third row of the second matrix is deleted, because it is a multiple of the second row and will result in a zero row.) We can now proceed in one of two ways. (a) Write down the corresponding homogeneous system in echelon form: x1 þ 2x2 þ 3x3 2x4 þ 4x5 ¼ 0 2x3 þ 5x4 þ x5 ¼ 0 The system in echelon form has three free variables, x2; x4; x5, so dim W ¼ 3. A basis ½u1; u2; u3 for W may be obtained as follows: (1) Set x2 ¼ 1, x4 ¼ 0, x5 ¼ 0. Back-substitution yields x3 ¼ 0, and then x1 ¼ 2. Thus, u1 ¼ ð2; 1; 0; 0; 0Þ. (2) Set x2 ¼ 0, x4 ¼ 1, x5 ¼ 0. Back-substitution yields x3 ¼ 5 2, and then x1 ¼ 19 2 . Thus, u2 ¼ ð19 2 ; 0; 5 2 ; 1; 0Þ. (3) Set x2 ¼ 0, x4 ¼ 0, x5 ¼ 1. Back-substitution yields x3 ¼ 1 2, and then x1 ¼ 5 2. Thus, u3 ¼ ð 5 2, 0, 1 2 ; 0; 1Þ. [One could avoid fractions in the basis by choosing x4 ¼ 2 in (2) and x5 ¼ 2 in (3), which yields multiples of u2 and u3.] The parametric form of the general solution is obtained from the following linear combination of the basis vectors using parameters a; b; c: au1 þ bu2 þ cu3 ¼ ð2a þ 19 2 b 5 2 c; a; 5 2 b 1 2 c; b; cÞ (b) Reduce the echelon form of A to row canonical form: A 1 2 3 2 4 0 0 1 5 2 1 2 " # 1 2 3 19 2 5 2 0 0 1 5 2 1 2 " # Write down the corresponding free-variable solution: x1 ¼ 2x2 þ 19 2 x4 5 2 x5 x3 ¼ 5 2 x4 1 2 x5 Using these equations for the pivot variables x1 and x3, repeat the above process to obtain a basis ½u1; u2; u3 for W. That is, set x2 ¼ 1, x4 ¼ 0, x5 ¼ 0 to get u1; set x2 ¼ 0, x4 ¼ 1, x5 ¼ 0 to get u2; and set x2 ¼ 0, x4 ¼ 0, x5 ¼ 1 to get u3. 3.28. Prove Theorem 3.15. Let v0 be a particular solution of AX ¼ B, and let W be the general solution of AX ¼ 0. Then U ¼ v0 þ W ¼ fv0 þ w : w 2 Wg is the general solution of AX ¼ B. Let w be a solution of AX ¼ 0. Then Aðv0 þ wÞ ¼ Av0 þ Aw ¼ B þ 0 ¼ B Thus, the sum v0 þ w is a solution of AX ¼ B. On the other hand, suppose v is also a solution of AX ¼ B. Then Aðv v0Þ ¼ Av Av0 ¼ B B ¼ 0 Therefore, v v0 belongs to W. Because v ¼ v0 þ ðv v0Þ, we ﬁnd that any solution of AX ¼ B can be obtained by adding a solution of AX ¼ 0 to a solution of AX ¼ B. Thus, the theorem is proved. 100 CHAPTER 3 Systems of Linear Equations Elementary Matrices, Applications 3.29. Let e1; e2; e3 denote, respectively, the elementary row operations ‘‘Interchange rows R1 and R2; ’’ ‘‘Replace R3 by 7R3; ’’ ‘‘Replace R2 by 3R1 þ R2’’ Find the corresponding three-square elementary matrices E1; E2; E3. Apply each operation to the 3 3 identity matrix I3 to obtain E1 ¼ 0 1 0 1 0 0 0 0 1 2 4 3 5; E2 ¼ 1 0 0 0 1 0 0 0 7 2 4 3 5; E3 ¼ 1 0 0 3 1 0 0 0 1 2 4 3 5 3.30. Consider the elementary row operations in Problem 3.29. (a) Describe the inverse operations e1 1 , e1 2 , e1 3 . (b) Find the corresponding three-square elementary matrices E0 1, E0 2, E0 3. (c) What is the relationship between the matrices E0 1, E0 2, E0 3 and the matrices E1, E2, E3? (a) The inverses of e1, e2, e3 are, respectively, ‘‘Interchange rows R1 and R2; ’’ ‘‘Replace R3 by 1 7 R3; ’’ ‘‘Replace R2 by 3R1 þ R2:’’ (b) Apply each inverse operation to the 3 3 identity matrix I3 to obtain E0 1 ¼ 0 1 0 1 0 0 0 0 1 2 4 3 5; E0 2 ¼ 1 0 0 0 1 0 0 0 1 7 2 4 3 5; E0 3 ¼ 1 0 0 3 1 0 0 0 1 2 4 3 5 (c) The matrices E0 1, E0 2, E0 3 are, respectively, the inverses of the matrices E1, E2, E3. 3.31. Write each of the following matrices as a product of elementary matrices: (a) A ¼ 1 3 2 4 ; (b) B ¼ 1 2 3 0 1 4 0 0 1 2 4 3 5; (c) C ¼ 1 1 2 2 3 8 3 1 2 2 4 3 5 The following three steps write a matrix M as a product of elementary matrices: Step 1. Row reduce M to the identity matrix I, keeping track of the elementary row operations. Step 2. Write down the inverse row operations. Step 3. Write M as the product of the elementary matrices corresponding to the inverse operations. This gives the desired result. If a zero row appears in Step 1, then M is not row equivalent to the identity matrix I, and M cannot be written as a product of elementary matrices. (a) (1) We have A ¼ 1 3 2 4 1 3 0 2 1 3 0 1 1 0 0 1 ¼ I where the row operations are, respectively, ‘‘Replace R2 by 2R1 þ R2; ’’ ‘‘Replace R2 by 1 2 R2; ’’ ‘‘Replace R1 by 3R2 þ R1’’ (2) Inverse operations: ‘‘Replace R2 by 2R1 þ R2;’’ ‘‘Replace R2 by 2R2;’’ ‘‘Replace R1 by 3R2 þ R1’’ (3) A ¼ 1 0 2 1 1 0 0 2 1 3 0 1 CHAPTER 3 Systems of Linear Equations 101 (b) (1) We have B ¼ 1 2 3 0 1 4 0 0 1 2 4 3 5 1 2 0 0 1 0 0 0 1 2 4 3 5 1 0 0 0 1 0 0 0 1 2 4 3 5 ¼ I where the row operations are, respectively, ‘‘Replace R2 by 4R3 þ R2; ’’ ‘‘Replace R1 by 3R3 þ R1; ’’ ‘‘Replace R1 by 2R2 þ R1’’ (2) Inverse operations: ‘‘Replace R2 by 4R3 þ R2; ’’ ‘‘Replace R1 by 3R3 þ R1; ’’ ‘‘Replace R1 by 2R2 þ R1’’ (3) B ¼ 1 0 0 0 1 4 0 0 1 2 4 3 5 1 0 3 0 1 0 0 0 1 2 4 3 5 1 2 0 0 1 0 0 0 1 2 4 3 5 (c) (1) First row reduce C to echelon form. We have C ¼ 1 1 2 2 3 8 3 1 2 2 4 3 5 1 1 2 0 1 4 0 2 8 2 4 3 5 1 1 2 0 1 4 0 0 0 2 4 3 5 In echelon form, C has a zero row. ‘‘STOP.’’ The matrix C cannot be row reduced to the identity matrix I, and C cannot be written as a product of elementary matrices. (We note, in particular, that C has no inverse.) 3.32. Find the inverse of (a) A ¼ 1 2 4 1 1 5 2 7 3 2 4 3 5; (b) B ¼ 1 3 4 1 5 1 3 13 6 2 4 3 5. (a) Form the matrix M ¼ [A; I] and row reduce M to echelon form: M ¼ 1 2 4 1 0 0 1 1 5 0 1 0 2 7 3 0 0 1 2 6 4 3 7 5 1 2 4 1 0 0 0 1 1 1 1 0 0 3 5 2 0 1 2 6 4 3 7 5 1 2 4 1 0 0 0 1 1 1 1 0 0 0 2 5 3 1 2 6 4 3 7 5 In echelon form, the left half of M is in triangular form; hence, A has an inverse. Further reduce M to row canonical form: M 1 2 0 9 6 2 0 1 0 7 2 5 2 1 2 0 0 1 5 2 3 2 1 2 2 6 6 4 3 7 7 5 1 0 0 16 11 3 0 1 0 7 2 5 2 1 2 0 0 1 5 2 3 2 1 2 2 6 6 4 3 7 7 5 The ﬁnal matrix has the form ½I; A1; that is, A1 is the right half of the last matrix. Thus, A1 ¼ 16 11 3 7 2 5 2 1 2 5 2 3 2 1 2 2 6 6 4 3 7 7 5 (b) Form the matrix M ¼ ½B; I and row reduce M to echelon form: M ¼ 1 3 4 1 0 0 1 5 1 0 1 0 3 13 6 0 0 1 2 4 3 5 1 3 4 1 0 0 0 2 3 1 1 0 0 4 6 3 0 1 2 4 3 5 1 3 4 1 0 0 0 2 3 1 1 0 0 0 0 1 2 1 2 4 3 5 In echelon form, M has a zero row in its left half; that is, B is not row reducible to triangular form. Accordingly, B has no inverse. 102 CHAPTER 3 Systems of Linear Equations 3.33. Show that every elementary matrix E is invertible, and its inverse is an elementary matrix. Let E be the elementary matrix corresponding to the elementary operation e; that is, eðIÞ ¼ E. Let e0 be the inverse operation of e and let E0 be the corresponding elementary matrix; that is, e0ðIÞ ¼ E0. Then I ¼ e0ðeðIÞÞ ¼ e0ðEÞ ¼ E0E and I ¼ eðe0ðIÞÞ ¼ eðE0Þ ¼ EE0 Therefore, E0 is the inverse of E. 3.34. Prove Theorem 3.16: Let e be an elementary row operation and let E be the corresponding m-square elementary matrix; that is, E ¼ eðIÞ. Then eðAÞ ¼ EA, where A is any m n matrix. Let Ri be the row i of A; we denote this by writing A ¼ ½R1; . . . ; Rm. If B is a matrix for which AB is deﬁned then AB ¼ ½R1B; . . . ; RmB. We also let ei ¼ ð0; . . . ; 0; ^ 1; 0; . . . ; 0Þ; ^¼ i Here ^¼ i means 1 is the ith entry. One can show (Problem 2.45) that eiA ¼ Ri. We also note that I ¼ ½e1; e2; . . . ; em is the identity matrix. (i) Let e be the elementary row operation ‘‘Interchange rows Ri and Rj.’’ Then, for ^¼ i and ^ ^ ¼ j, E ¼ eðIÞ ¼ ½e1; . . . ; b ej; . . . ; b b ei; . . . ; em and eðAÞ ¼ ½R1; . . . ; b Rj; . . . ; b b Ri; . . . ; Rm Thus, EA ¼ ½e1A; . . . ; c ejA; . . . ; c c eiA; . . . ; emA ¼ ½R1; . . . ; b Rj; . . . ; b b Ri; . . . ; Rm ¼ eðAÞ (ii) Let e be the elementary row operation ‘‘Replace Ri by kRi ðk 6¼ 0Þ.’’ Then, for^¼ i, E ¼ eðIÞ ¼ ½e1; . . . ; b kei; . . . ; em and eðAÞ ¼ ½R1; . . . ; c kRi; . . . ; Rm Thus, EA ¼ ½e1A; . . . ; d keiA; . . . ; emA ¼ ½R1; . . . ; c kRi; . . . ; Rm ¼ eðAÞ (iii) Let e be the elementary row operation ‘‘Replace Ri by kRj þ Ri.’’ Then, for^¼ i, E ¼ eðIÞ ¼ ½e1; . . . ; d kej þ ei; . . . ; em and eðAÞ ¼ ½R1; . . . ; d kRj þ Ri; . . . ; Rm Using ðkej þ eiÞA ¼ kðejAÞ þ eiA ¼ kRj þ Ri, we have EA ¼ ½e1A; . . . ; ðkej þ eiÞA; . . . ; emA ¼ ½R1; . . . ; d kRj þ Ri; . . . ; Rm ¼ eðAÞ 3.35. Prove Theorem 3.17: Let A be a square matrix. Then the following are equivalent: (a) A is invertible (nonsingular). (b) A is row equivalent to the identity matrix I. (c) A is a product of elementary matrices. Suppose A is invertible and suppose A is row equivalent to matrix B in row canonical form. Then there exist elementary matrices E1; E2; . . . ; Es such that Es . . . E2E1A ¼ B. Because A is invertible and each elementary matrix is invertible, B is also invertible. But if B 6¼ I, then B has a zero row; whence B is not invertible. Thus, B ¼ I, and (a) implies (b). CHAPTER 3 Systems of Linear Equations 103 If (b) holds, then there exist elementary matrices E1; E2; . . . ; Es such that Es . . . E2E1A ¼ I. Hence, A ¼ ðEs . . . E2E1Þ1 ¼ E1 1 E1 2 . . . ; E1 s . But the E1 i are also elementary matrices. Thus (b) implies (c). If (c) holds, then A ¼ E1E2 . . . Es. The Ei are invertible matrices; hence, their product A is also invertible. Thus, (c) implies (a). Accordingly, the theorem is proved. 3.36. Prove Theorem 3.18: If AB ¼ I, then BA ¼ I, and hence B ¼ A1. Suppose A is not invertible. Then A is not row equivalent to the identity matrix I, and so A is row equivalent to a matrix with a zero row. In other words, there exist elementary matrices E1; . . . ; Es such that Es . . . E2E1A has a zero row. Hence, Es . . . E2E1AB ¼ Es . . . E2E1, an invertible matrix, also has a zero row. But invertible matrices cannot have zero rows; hence A is invertible, with inverse A1. Then also, B ¼ IB ¼ ðA1AÞB ¼ A1ðABÞ ¼ A1I ¼ A1 3.37. Prove Theorem 3.19: B is row equivalent to A (written B AÞ if and only if there exists a nonsingular matrix P such that B ¼ PA. If B A, then B ¼ esð. . . ðe2ðe1ðAÞÞÞ . . .Þ ¼ Es . . . E2E1A ¼ PA where P ¼ Es . . . E2E1 is nonsingular. Conversely, suppose B ¼ PA, where P is nonsingular. By Theorem 3.17, P is a product of elementary matrices, and so B can be obtained from A by a sequence of elementary row operations; that is, B A. Thus, the theorem is proved. 3.38. Prove Theorem 3.21: Every m n matrix A is equivalent to a unique block matrix of the form Ir 0 0 0 , where Ir is the r r identity matrix. The proof is constructive, in the form of an algorithm. Step 1. Row reduce A to row canonical form, with leading nonzero entries a1j1, a2j2; . . . ; arjr. Step 2. Interchange C1 and C1j1, interchange C2 and C2j2; . . . , and interchange Cr and Cjr. This gives a matrix in the form Ir B 0 0 , with leading nonzero entries a11; a22; . . . ; arr. Step 3. Use column operations, with the aii as pivots, to replace each entry in B with a zero; that is, for i ¼ 1; 2; . . . ; r and j ¼ r þ 1, r þ 2; . . . ; n, apply the operation bijCi þ Cj ! Cj. The ﬁnal matrix has the desired form Ir 0 0 0 . Lu Factorization 3.39. Find the LU factorization of (a) A ¼ 1 3 5 2 4 7 1 2 1 2 4 3 5; (b) B ¼ 1 4 3 2 8 1 5 9 7 2 4 3 5: (a) Reduce A to triangular form by the following operations: ‘‘Replace R2 by 2R1 þ R2; ’’ ‘‘Replace R3 by R1 þ R3; ’’ and then ‘‘Replace R3 by 5 2 R2 þ R3’’ These operations yield the following, where the triangular form is U: A 1 3 5 0 2 3 0 5 6 2 4 3 5 1 3 5 0 2 3 0 0 3 2 2 4 3 5 ¼ U and L ¼ 1 0 0 2 1 0 1 5 2 1 2 4 3 5 The entries 2; 1; 5 2 in L are the negatives of the multipliers 2; 1; 5 2 in the above row operations. (As a check, multiply L and U to verify A ¼ LU.) 104 CHAPTER 3 Systems of Linear Equations (b) Reduce B to triangular form by ﬁrst applying the operations ‘‘Replace R2 by 2R1 þ R2’’ and ‘‘Replace R3 by 5R1 þ R3.’’ These operations yield B 1 4 3 0 0 7 0 11 8 2 4 3 5: Observe that the second diagonal entry is 0. Thus, B cannot be brought into triangular form without row interchange operations. Accordingly, B is not LU-factorable. (There does exist a PLU factorization of such a matrix B, where P is a permutation matrix, but such a factorization lies beyond the scope of this text.) 3.40. Find the LDU factorization of the matrix A in Problem 3.39. The A ¼ LDU factorization refers to the situation where L is a lower triangular matrix with 1’s on the diagonal (as in the LU factorization of A), D is a diagonal matrix, and U is an upper triangular matrix with 1’s on the diagonal. Thus, simply factor out the diagonal entries in the matrix U in the above LU factorization of A to obtain D and L. That is, L ¼ 1 0 0 2 1 0 1 5 2 1 2 4 3 5; D ¼ 1 0 0 0 2 0 0 0 3 2 2 4 3 5; U ¼ 1 3 5 0 1 3 0 0 1 2 4 3 5 3.41. Find the LU factorization of the matrix A ¼ 1 2 1 2 3 3 3 10 2 2 4 3 5. Reduce A to triangular form by the following operations: ð1Þ ‘‘Replace R2 by 2R1 þ R2; ’’ ð2Þ ‘‘Replace R3 by 3R1 þ R3; ’’ ð3Þ ‘‘Replace R3 by 4R2 þ R3’’ These operations yield the following, where the triangular form is U: A 1 2 1 0 1 1 0 4 5 2 4 3 5 1 2 1 0 1 1 0 0 1 2 4 3 5 ¼ U and L ¼ 1 0 0 2 1 0 3 4 1 2 4 3 5 The entries 2; 3; 4 in L are the negatives of the multipliers 2; 3; 4 in the above row operations. (As a check, multiply L and U to verify A ¼ LU.) 3.42. Let A be the matrix in Problem 3.41. Find X1; X2; X3, where Xi is the solution of AX ¼ Bi for (a) B1 ¼ ð1; 1; 1Þ, (b) B2 ¼ B1 þ X1, (c) B3 ¼ B2 þ X2. (a) Find L1B1 by applying the row operations (1), (2), and then (3) in Problem 3.41 to B1: B1 ¼ 1 1 1 2 4 3 5 ! ð1Þ and ð2Þ 1 1 4 2 4 3 5 ! ð3Þ 1 1 8 2 4 3 5 Solve UX ¼ B for B ¼ ð1; 1; 8Þ by back-substitution to obtain X1 ¼ ð25; 9; 8Þ. (b) First ﬁnd B2 ¼ B1 þ X1 ¼ ð1; 1; 1Þ þ ð25; 9; 8Þ ¼ ð24; 10; 9Þ. Then as above B2 ¼ ½24; 10; 9T ! ð1Þ and ð2Þ ½24; 58; 63T ! ð3Þ ½24; 58; 295T Solve UX ¼ B for B ¼ ð24; 58; 295Þ by back-substitution to obtain X2 ¼ ð943; 353; 295Þ. (c) First ﬁnd B3 ¼ B2 þ X2 ¼ ð24; 10; 9Þ þ ð943; 353; 295Þ ¼ ð919; 343; 286Þ. Then, as above B3 ¼ ½943; 353; 295T ! ð1Þ and ð2Þ ½919; 2181; 2671T ! ð3Þ ½919; 2181; 11 395T Solve UX ¼ B for B ¼ ð919; 2181; 11 395Þ by back-substitution to obtain X3 ¼ ð37 628; 13 576; 11 395Þ. CHAPTER 3 Systems of Linear Equations 105 Miscellaneous Problems 3.43. Let L be a linear combination of the m equations in n unknowns in the system (3.2). Say L is the equation ðc1a11 þ þ cmam1Þx1 þ þ ðc1a1n þ þ cmamnÞxn ¼ c1b1 þ þ cmbm ð1Þ Show that any solution of the system (3.2) is also a solution of L. Let u ¼ ðk1; . . . ; knÞ be a solution of (3.2). Then ai1k1 þ ai2k2 þ þ ainkn ¼ bi ði ¼ 1; 2; . . . ; mÞ ð2Þ Substituting u in the left-hand side of (1) and using (2), we get ðc1a11 þ þ cmam1Þk1 þ þ ðc1a1n þ þ cmamnÞkn ¼ c1ða11k1 þ þ a1nknÞ þ þ cmðam1k1 þ þ amnknÞ ¼ c1b1 þ þ cmbm This is the right-hand side of (1); hence, u is a solution of (1). 3.44. Suppose a system m of linear equations is obtained from a system l by applying an elementary operation (page 64). Show that m and l have the same solutions. Each equation L in m is a linear combination of equations in l. Hence, by Problem 3.43, any solution of l will also be a solution of m. On the other hand, each elementary operation has an inverse elementary operation, so l can be obtained from m by an elementary operation. This means that any solution of m is a solution of l. Thus, l and m have the same solutions. 3.45. Prove Theorem 3.4: Suppose a system m of linear equations is obtained from a system l by a sequence of elementary operations. Then m and l have the same solutions. Each step of the sequence does not change the solution set (Problem 3.44). Thus, the original system l and the ﬁnal system m (and any system in between) have the same solutions. 3.46. A system l of linear equations is said to be consistent if no linear combination of its equations is a degenerate equation L with a nonzero constant. Show that l is consistent if and only if l is reducible to echelon form. Suppose l is reducible to echelon form. Then l has a solution, which must also be a solution of every linear combination of its equations. Thus, L, which has no solution, cannot be a linear combination of the equations in l. Thus, l is consistent. On the other hand, suppose l is not reducible to echelon form. Then, in the reduction process, it must yield a degenerate equation L with a nonzero constant, which is a linear combination of the equations in l. Therefore, l is not consistent; that is, l is inconsistent. 3.47. Suppose u and v are distinct vectors. Show that, for distinct scalars k, the vectors u þ kðu vÞ are distinct. Suppose u þ k1ðu vÞ ¼ u þ k2ðu vÞ: We need only show that k1 ¼ k2. We have k1ðu vÞ ¼ k2ðu vÞ; and so ðk1 k2Þðu vÞ ¼ 0 Because u and v are distinct, u v 6¼ 0. Hence, k1 k2 ¼ 0, and so k1 ¼ k2. 3.48. Suppose AB is deﬁned. Prove (a) Suppose A has a zero row. Then AB has a zero row. (b) Suppose B has a zero column. Then AB has a zero column. 106 CHAPTER 3 Systems of Linear Equations (a) Let Ri be the zero row of A, and C1; . . . ; Cn the columns of B. Then the ith row of AB is ðRiC1; RiC2; . . . ; RiCnÞ ¼ ð0; 0; 0; . . . ; 0Þ (b) BT has a zero row, and so BTAT ¼ ðABÞT has a zero row. Hence, AB has a zero column. SUPPLEMENTARY PROBLEMS Linear Equations, 2 2 Systems 3.49. Determine whether each of the following systems is linear: (a) 3x 4y þ 2yz ¼ 8, (b) ex þ 3y ¼ p, (c) 2x 3y þ kz ¼ 4 3.50. Solve (a) px ¼ 2, (b) 3x þ 2 ¼ 5x þ 7 2x, (c) 6x þ 2 4x ¼ 5 þ 2x 3 3.51. Solve each of the following systems: (a) 2x þ 3y ¼ 1 5x þ 7y ¼ 3 (b) 4x 2y ¼ 5 6x þ 3y ¼ 1 (c) 2x 4 ¼ 3y 5y x ¼ 5 (d) 2x 4y ¼ 10 3x 6y ¼ 15 3.52. Consider each of the following systems in unknowns x and y: (a) x ay ¼ 1 ax 4y ¼ b (b) ax þ 3y ¼ 2 12x þ ay ¼ b (c) x þ ay ¼ 3 2x þ 5y ¼ b For which values of a does each system have a unique solution, and for which pairs of values ða; bÞ does each system have more than one solution? General Systems of Linear Equations 3.53. Solve (a) x þ y þ 2z ¼ 4 2x þ 3y þ 6z ¼ 10 3x þ 6y þ 10z ¼ 17 (b) x 2y þ 3z ¼ 2 2x 3y þ 8z ¼ 7 3x 4y þ 13z ¼ 8 (c) x þ 2y þ 3z ¼ 3 2x þ 3y þ 8z ¼ 4 5x þ 8y þ 19z ¼ 11 3.54. Solve (a) x 2y ¼ 5 2x þ 3y ¼ 3 3x þ 2y ¼ 7 (b) x þ 2y 3z þ 2t ¼ 2 2x þ 5y 8z þ 6t ¼ 5 3x þ 4y 5z þ 2t ¼ 4 (c) x þ 2y þ 4z 5t ¼ 3 3x y þ 5z þ 2t ¼ 4 5x 4y þ 6z þ 9t ¼ 2 3.55. Solve (a) 2x y 4z ¼ 2 4x 2y 6z ¼ 5 6x 3y 8z ¼ 8 (b) x þ 2y z þ 3t ¼ 3 2x þ 4y þ 4z þ 3t ¼ 9 3x þ 6y z þ 8t ¼ 10 3.56. Consider each of the following systems in unknowns x; y; z: (a) x 2y ¼ 1 x y þ az ¼ 2 ay þ 9z ¼ b (b) x þ 2y þ 2z ¼ 1 x þ ay þ 3z ¼ 3 x þ 11y þ az ¼ b (c) x þ y þ az ¼ 1 x þ ay þ z ¼ 4 ax þ y þ z ¼ b For which values of a does the system have a unique solution, and for which pairs of values ða; bÞ does the system have more than one solution? The value of b does not have any effect on whether the system has a unique solution. Why? CHAPTER 3 Systems of Linear Equations 107 Linear Combinations, Homogeneous Systems 3.57. Write v as a linear combination of u1; u2; u3, where (a) v ¼ ð4; 9; 2Þ, u1 ¼ ð1; 2; 1Þ, u2 ¼ ð1; 4; 2Þ, u3 ¼ ð1; 3; 2Þ; (b) v ¼ ð1; 3; 2Þ, u1 ¼ ð1; 2; 1Þ, u2 ¼ ð2; 6; 5Þ, u3 ¼ ð1; 7; 8Þ; (c) v ¼ ð1; 4; 6Þ, u1 ¼ ð1; 1; 2Þ, u2 ¼ ð2; 3; 5Þ, u3 ¼ ð3; 5; 8Þ. 3.58. Let u1 ¼ ð1; 1; 2Þ, u2 ¼ ð1; 3; 2Þ, u3 ¼ ð4; 2; 1Þ in R3. Show that u1; u2; u3 are orthogonal, and write v as a linear combination of u1; u2; u3, where (a) v ¼ ð5; 5; 9Þ, (b) v ¼ ð1; 3; 3Þ, (c) v ¼ ð1; 1; 1Þ. (Hint: Use Fourier coefﬁcients.) 3.59. Find the dimension and a basis of the general solution W of each of the following homogeneous systems: (a) x y þ 2z ¼ 0 2x þ y þ z ¼ 0 5x þ y þ 4z ¼ 0 (b) x þ 2y 3z ¼ 0 2x þ 5y þ 2z ¼ 0 3x y 4z ¼ 0 (c) x þ 2y þ 3z þ t ¼ 0 2x þ 4y þ 7z þ 4t ¼ 0 3x þ 6y þ 10z þ 5t ¼ 0 3.60. Find the dimension and a basis of the general solution W of each of the following systems: (a) x1 þ 3x2 þ 2x3 x4 x5 ¼ 0 2x1 þ 6x2 þ 5x3 þ x4 x5 ¼ 0 5x1 þ 15x2 þ 12x3 þ x4 3x5 ¼ 0 (b) 2x1 4x2 þ 3x3 x4 þ 2x5 ¼ 0 3x1 6x2 þ 5x3 2x4 þ 4x5 ¼ 0 5x1 10x2 þ 7x3 3x4 þ 18x5 ¼ 0 Echelon Matrices, Row Canonical Form 3.61. Reduce each of the following matrices to echelon form and then to row canonical form: (a) 1 1 2 2 4 9 1 5 12 2 4 3 5; (b) 1 2 1 2 1 2 4 1 2 5 3 6 3 7 7 2 4 3 5; (c) 2 4 2 2 5 1 3 6 2 2 0 4 4 8 2 6 5 7 2 4 3 5 3.62. Reduce each of the following matrices to echelon form and then to row canonical form: (a) 1 2 1 2 1 2 2 4 3 5 5 7 3 6 4 9 10 11 1 2 4 3 6 9 2 6 6 4 3 7 7 5; (b) 0 1 2 3 0 3 8 12 0 0 4 6 0 2 7 10 2 6 6 4 3 7 7 5; (c) 1 3 1 3 2 8 5 10 1 7 7 11 3 11 7 15 2 6 6 4 3 7 7 5 3.63. Using only 0’s and 1’s, list all possible 2 2 matrices in row canonical form. 3.64. Using only 0’s and 1’s, ﬁnd the number n of possible 3 3 matrices in row canonical form. Elementary Matrices, Applications 3.65. Let e1; e2; e3 denote, respectively, the following elementary row operations: ‘‘Interchange R2 and R3; ’’ ‘‘Replace R2 by 3R2; ’’ ‘‘Replace R1 by 2R3 þ R1’’ (a) Find the corresponding elementary matrices E1; E2; E3. (b) Find the inverse operations e1 1 , e1 2 , e1 3 ; their corresponding elementary matrices E0 1, E0 2, E0 3; and the relationship between them and E1; E2; E3. (c) Describe the corresponding elementary column operations f1; f2; f3. (d) Find elementary matrices F1; F2; F3 corresponding to f1; f2; f3, and the relationship between them and E1; E2; E3. 108 CHAPTER 3 Systems of Linear Equations 3.66. Express each of the following matrices as a product of elementary matrices: A ¼ 1 2 3 4 ; B ¼ 3 6 2 4 ; C ¼ 2 6 3 7 ; D ¼ 1 2 0 0 1 3 3 8 7 2 4 3 5 3.67. Find the inverse of each of the following matrices (if it exists): A ¼ 1 2 1 2 3 1 3 4 4 2 4 3 5; B ¼ 1 2 3 2 6 1 3 10 1 2 4 3 5; C ¼ 1 3 2 2 8 3 1 7 1 2 4 3 5; D ¼ 2 1 1 5 2 3 0 2 1 2 4 3 5 3.68. Find the inverse of each of the following n n matrices: (a) A has 1’s on the diagonal and superdiagonal (entries directly above the diagonal) and 0’s elsewhere. (b) B has 1’s on and above the diagonal, and 0’s below the diagonal. Lu Factorization 3.69. Find the LU factorization of each of the following matrices: (a) 1 1 1 3 4 2 2 3 2 2 4 3 5, (b) 1 3 1 2 5 1 3 4 2 2 4 3 5, (c) 2 3 6 4 7 9 3 5 4 2 4 3 5, (d) 1 2 3 2 4 7 3 7 10 2 4 3 5 3.70. Let A be the matrix in Problem 3.69(a). Find X1; X2; X3; X4, where (a) X1 is the solution of AX ¼ B1, where B1 ¼ ð1; 1; 1ÞT. (b) For k > 1, Xk is the solution of AX ¼ Bk, where Bk ¼ Bk1 þ Xk1. 3.71. Let B be the matrix in Problem 3.69(b). Find the LDU factorization of B. Miscellaneous Problems 3.72. Consider the following systems in unknowns x and y: ðaÞ ax þ by ¼ 1 cx þ dy ¼ 0 ðbÞ ax þ by ¼ 0 cx þ dy ¼ 1 Suppose D ¼ ad bc 6¼ 0. Show that each system has the unique solution: (a) x ¼ d=D, y ¼ c=D, (b) x ¼ b=D, y ¼ a=D. 3.73. Find the inverse of the row operation ‘‘Replace Ri by kRj þ k0Ri ðk0 6¼ 0Þ.’’ 3.74. Prove that deleting the last column of an echelon form (respectively, the row canonical form) of an augmented matrix M ¼ ½A; B yields an echelon form (respectively, the row canonical form) of A. 3.75. Let e be an elementary row operation and E its elementary matrix, and let f be the corresponding elementary column operation and F its elementary matrix. Prove (a) f ðAÞ ¼ ðeðATÞÞT, (b) F ¼ ET, (c) f ðAÞ ¼ AF. 3.76. Matrix A is equivalent to matrix B, written A B, if there exist nonsingular matrices P and Q such that B ¼ PAQ. Prove that is an equivalence relation; that is, (a) A A, (b) If A B, then B A, (c) If A B and B C, then A C. CHAPTER 3 Systems of Linear Equations 109 ANSWERS TO SUPPLEMENTARY PROBLEMS Notation: A ¼ ½R1; R2; . . . denotes the matrix A with rows R1; R2; . . . . The elements in each row are separated by commas (which may be omitted with single digits), the rows are separated by semicolons, and 0 denotes a zero row. For example, A ¼ ½1; 2; 3; 4; 5; 6; 7; 8; 0 ¼ 1 2 3 4 5 6 7 8 0 0 0 0 2 4 3 5 3.49. (a) no, (b) yes, (c) linear in x; y; z, not linear in x; y; z; k 3.50. (a) x ¼ 2=p, (b) no solution, (c) every scalar k is a solution 3.51. (a) ð2; 1Þ, (b) no solution, (c) ð5; 2Þ, (d) ð5 2a; aÞ 3.52. (a) a 6¼ 2; ð2; 2Þ; ð2; 2Þ, (b) a 6¼ 6; ð6; 4Þ; ð6; 4Þ, (c) a 6¼ 5 2 ; ð5 2 ; 6Þ 3.53. (a) ð2; 1; 1 2Þ, (b) no solution, (c) u ¼ ð7a 1; 2a þ 2; aÞ. 3.54. (a) ð3; 1Þ, (b) u ¼ ða þ 2b; 1 þ 2a 2b; a; bÞ, (c) no solution 3.55. (a) u ¼ ð1 2 a þ 2; a; 1 2Þ, (b) u ¼ ð1 2 ð7 5b 4aÞ; a; 1 2 ð1 þ bÞ; bÞ 3.56. (a) a 6¼ 3; ð3; 3Þ; ð3; 3Þ, (b) a 6¼ 5 and a 6¼ 1; ð5; 7Þ; ð1; 5Þ, (c) a 6¼ 1 and a 6¼ 2; ð2; 5Þ 3.57. (a) 2; 1; 3, (b) 6; 3; 1, (c) not possible 3.58. (a) 3; 2; 1, (b) 2 3 ; 1; 1 3, (c) 2 3 ; 1 7 ; 1 21 3.59. (a) dim W ¼ 1; u1 ¼ ð1; 1; 1Þ, (b) dim W ¼ 0, no basis, (c) dim W ¼ 2; u1 ¼ ð2; 1; 0; 0Þ; u2 ¼ ð5; 0; 2; 1Þ 3.60. (a) dim W ¼ 3; u1 ¼ ð3; 1; 0; 0; 0Þ, u2 ¼ ð7; 0; 3; 1; 0Þ, u3 ¼ ð3; 0; 1; 0; 1Þ, (b) dim W ¼ 2, u1 ¼ ð2; 1; 0; 0; 0Þ, u2 ¼ ð5; 0; 5; 3; 1Þ 3.61. (a) ½1; 0; 1 2 ; 0; 1; 5 2 ; 0, (b) ½1; 2; 0; 0; 2; 0; 0; 1; 0; 5; 0; 0; 0; 1; 2, (c) ½1; 2; 0; 4; 5; 3; 0; 0; 1; 5; 15 2 ; 5 2 ; 0 3.62. (a) ½1; 2; 0; 0; 4; 2; 0; 0; 1; 0; 1; 2; 0; 0; 0; 1; 2; 1; 0, (b) ½0; 1; 0; 0; 0; 0; 1; 0; 0; 0; 0; 1; 0, (c) ½1; 0; 0; 4; 0; 1; 0; 1; 0; 0; 1; 2; 0 3.63. 5: ½1; 0; 0; 1, ½1; 1; 0; 0, ½1; 0; 0; 0, ½0; 1; 0; 0; 0 3.64. 16 3.65. (a) ½1; 0; 0; 0; 0; 1; 0; 1; 0, ½1; 0; 0; 0; 3; 0; 0; 0; 1, ½1; 0; 2; 0; 1; 0; 0; 0; 1, (b) R2 $ R3; 1 3 R2 ! R2; 2R3 þ R1 ! R1; each E0 i ¼ E1 i , (c) C2 $ C3; 3C2 ! C2; 2C3 þ C1 ! C1, (d) each Fi ¼ ET i . 3.66. A ¼ ½1; 0; 3; 1½1; 0; 0; 2½1; 2; 0; 1, B is not invertible, C ¼ ½1; 0; 3 2 ; 1½1; 0; 0; 2½1; 6; 0; 1½2; 0; 0; 1, D ¼ ½100; 010; 301½100; 010; 021½100; 013; 001½120; 010; 001 3.67. A1 ¼ ½8; 12; 5; 5; 7; 3; 1; 2; 1, B has no inverse, C1 ¼ ½29 2 ; 17 2 ; 7 2 ; 5 2 ; 3 2 ; 1 2 ; 3; 2; 1; D1 ¼ ½8; 3; 1; 5; 2; 1; 10; 4; 1 110 CHAPTER 3 Systems of Linear Equations 3.68. A1 ¼ ½1; 1; 1; 1; . . . ; 0; 1; 1; 1; 1; . . . ; 0; 0; 1; 1; 1; 1; 1; . . . ; . . . ; . . . ; 0; . . . 0; 1 B1 has 1’s on diagonal, 1’s on superdiagonal, and 0’s elsewhere. 3.69. (a) ½100; 310; 211½1; 1; 1; 0; 1; 1; 0; 0; 1, (b) ½100; 210; 351½1; 3; 1; 0; 1; 3; 0; 0; 10, (c) ½100; 210; 3 2 ; 1 2 ; 1½2; 3; 6; 0; 1; 3; 0; 0; 7 2, (d) There is no LU decomposition. 3.70. X1 ¼ ½1; 1; 1T; B2 ¼ ½2; 2; 0T, X2 ¼ ½6; 4; 0T, B3 ¼ ½8; 6; 0T, X3 ¼ ½22; 16; 2T, B4 ¼ ½30; 22; 2T, X4 ¼ ½86; 62; 6T 3.71. B ¼ ½100; 210; 351 diagð1; 1; 10Þ ½1; 3; 1; 0; 1; 3; 0; 0; 1 3.73. Replace Ri by kRj þ ð1=k0ÞRi. 3.75. (c) f ðAÞ ¼ ðeðATÞÞT ¼ ðEATÞT ¼ ðATÞTET ¼ AF 3.76. (a) A ¼ IAI: (b) If A ¼ PBQ, then B ¼ P1AQ1. (c) If A ¼ PBQ and B ¼ P0CQ0, then A ¼ ðPP0ÞCðQ 0QÞ. CHAPTER 3 Systems of Linear Equations 111 Vector Spaces 4.1 Introduction This chapter introduces the underlying structure of linear algebra, that of a ﬁnite-dimensional vector space. The deﬁnition of a vector space V, whose elements are called vectors, involves an arbitrary ﬁeld K, whose elements are called scalars. The following notation will be used (unless otherwise stated or implied): V the given vector space u; v; w vectors in V K the given number field a; b; c; or k scalars in K Almost nothing essential is lost if the reader assumes that K is the real ﬁeld R or the complex ﬁeld C. The reader might suspect that the real line R has ‘‘dimension’’ one, the cartesian plane R2 has ‘‘dimension’’ two, and the space R3 has ‘‘dimension’’ three. This chapter formalizes the notion of ‘‘dimension,’’ and this deﬁnition will agree with the reader’s intuition. Throughout this text, we will use the following set notation: a 2 A Element a belongs to set A a; b 2 A Elements a and b belong to A 8x 2 A For every x in A 9x 2 A There exists an x in A A B A is a subset of B A \ B Intersection of A and B A [ B Union of A and B ; Empty set 4.2 Vector Spaces The following deﬁnes the notion of a vector space V where K is the ﬁeld of scalars. DEFINITION: Let V be a nonempty set with two operations: (i) Vector Addition: This assigns to any u; v 2 V a sum u þ v in V. (ii) Scalar Multiplication: This assigns to any u 2 V, k 2 K a product ku 2 V. Then V is called a vector space (over the ﬁeld K) if the following axioms hold for any vectors u; v; w 2 V: 112 CHAPTER 4 [A1] ðu þ vÞ þ w ¼ u þ ðv þ wÞ [A2] There is a vector in V, denoted by 0 and called the zero vector, such that, for any u 2 V; u þ 0 ¼ 0 þ u ¼ u [A3] For each u 2 V; there is a vector in V, denoted by u, and called the negative of u, such that u þ ðuÞ ¼ ðuÞ þ u ¼ 0. [A4] u þ v ¼ v þ u. [M1] kðu þ vÞ ¼ ku þ kv, for any scalar k 2 K: [M2] ða þ bÞu ¼ au þ bu; for any scalars a; b 2 K. [M3] ðabÞu ¼ aðbuÞ; for any scalars a; b 2 K. [M4] 1u ¼ u, for the unit scalar 1 2 K. The above axioms naturally split into two sets (as indicated by the labeling of the axioms). The ﬁrst four are concerned only with the additive structure of V and can be summarized by saying V is a commutative group under addition. This means (a) Any sum v1 þ v2 þ þ vm of vectors requires no parentheses and does not depend on the order of the summands. (b) The zero vector 0 is unique, and the negative u of a vector u is unique. (c) (Cancellation Law) If u þ w ¼ v þ w, then u ¼ v. Also, subtraction in V is deﬁned by u v ¼ u þ ðvÞ, where v is the unique negative of v. On the other hand, the remaining four axioms are concerned with the ‘‘action’’ of the ﬁeld K of scalars on the vector space V. Using these additional axioms, we prove (Problem 4.2) the following simple properties of a vector space. THEOREM 4.1: Let V be a vector space over a ﬁeld K. (i) For any scalar k 2 K and 0 2 V; k0 ¼ 0. (ii) For 0 2 K and any vector u 2 V; 0u ¼ 0. (iii) If ku ¼ 0, where k 2 K and u 2 V, then k ¼ 0 or u ¼ 0. (iv) For any k 2 K and any u 2 V; ðkÞu ¼ kðuÞ ¼ ku. 4.3 Examples of Vector Spaces This section lists important examples of vector spaces that will be used throughout the text. Space Kn Let K be an arbitrary ﬁeld. The notation Kn is frequently used to denote the set of all n-tuples of elements in K. Here Kn is a vector space over K using the following operations: (i) Vector Addition: ða1; a2; . . . ; anÞ þ ðb1; b2; . . . ; bnÞ ¼ ða1 þ b1; a2 þ b2; . . . ; an þ bnÞ (ii) Scalar Multiplication: kða1; a2; . . . ; anÞ ¼ ðka1; ka2; . . . ; kanÞ The zero vector in Kn is the n-tuple of zeros, 0 ¼ ð0; 0; . . . ; 0Þ and the negative of a vector is deﬁned by ða1; a2; . . . ; anÞ ¼ ða1; a2; . . . ; anÞ Observe that these are the same as the operations deﬁned for Rn in Chapter 1. The proof that Kn is a vector space is identical to the proof of Theorem 1.1, which we now regard as stating that Rn with the operations deﬁned there is a vector space over R. CHAPTER 4 Vector Spaces 113 Polynomial Space PðtÞ Let PðtÞ denote the set of all polynomials of the form pðtÞ ¼ a0 þ a1t þ a2t2 þ þ asts ðs ¼ 1; 2; . . .Þ where the coefﬁcients ai belong to a ﬁeld K. Then PðtÞ is a vector space over K using the following operations: (i) Vector Addition: Here pðtÞ þ qðtÞ in PðtÞ is the usual operation of addition of polynomials. (ii) Scalar Multiplication: Here kpðtÞ in PðtÞ is the usual operation of the product of a scalar k and a polynomial pðtÞ. The zero polynomial 0 is the zero vector in PðtÞ. Polynomial Space PnðtÞ Let PnðtÞ denote the set of all polynomials pðtÞ over a ﬁeld K, where the degree of pðtÞ is less than or equal to n; that is, pðtÞ ¼ a0 þ a1t þ a2t2 þ þ asts where s n. Then PnðtÞ is a vector space over K with respect to the usual operations of addition of polynomials and of multiplication of a polynomial by a constant (just like the vector space PðtÞ above). We include the zero polynomial 0 as an element of PnðtÞ, even though its degree is undeﬁned. Matrix Space Mm;n The notation Mm;n, or simply M; will be used to denote the set of all m n matrices with entries in a ﬁeld K. Then Mm;n is a vector space over K with respect to the usual operations of matrix addition and scalar multiplication of matrices, as indicated by Theorem 2.1. Function Space FðXÞ Let X be a nonempty set and let K be an arbitrary ﬁeld. Let FðXÞ denote the set of all functions of X into K. [Note that FðXÞ is nonempty, because X is nonempty.] Then FðXÞ is a vector space over K with respect to the following operations: (i) Vector Addition: The sum of two functions f and g in FðXÞ is the function f þ g in FðXÞ deﬁned by ð f þ gÞðxÞ ¼ f ðxÞ þ gðxÞ 8x 2 X (ii) Scalar Multiplication: The product of a scalar k 2 K and a function f in FðXÞ is the function kf in FðXÞ deﬁned by ðkf ÞðxÞ ¼ kf ðxÞ 8x 2 X The zero vector in FðXÞ is the zero function 0, which maps every x 2 X into the zero element 0 2 K; 0ðxÞ ¼ 0 8x 2 X Also, for any function f in FðXÞ, negative of f is the function f in FðXÞ deﬁned by ðf ÞðxÞ ¼ f ðxÞ 8x 2 X Fields and Subﬁelds Suppose a ﬁeld E is an extension of a ﬁeld K; that is, suppose E is a ﬁeld that contains K as a subﬁeld. Then E may be viewed as a vector space over K using the following operations: (i) Vector Addition: Here u þ v in E is the usual addition in E. (ii) Scalar Multiplication: Here ku in E, where k 2 K and u 2 E, is the usual product of k and u as elements of E. That is, the eight axioms of a vector space are satisﬁed by E and its subﬁeld K with respect to the above two operations. 114 CHAPTER 4 Vector Spaces 4.4 Linear Combinations, Spanning Sets Let V be a vector space over a ﬁeld K. A vector v in V is a linear combination of vectors u1; u2; . . . ; um in V if there exist scalars a1; a2; . . . ; am in K such that v ¼ a1u1 þ a2u2 þ þ amum Alternatively, v is a linear combination of u1; u2; . . . ; um if there is a solution to the vector equation v ¼ x1u1 þ x2u2 þ þ xmum where x1; x2; . . . ; xm are unknown scalars. EXAMPLE 4.1 (Linear Combinations in Rn) Suppose we want to express v ¼ ð3; 7; 4Þ in R3 as a linear combination of the vectors u1 ¼ ð1; 2; 3Þ; u2 ¼ ð2; 3; 7Þ; u3 ¼ ð3; 5; 6Þ We seek scalars x, y, z such that v ¼ xu1 þ yu2 þ zu3; that is, 3 3 4 2 4 3 5 ¼ x 1 2 3 2 4 3 5 þ y 2 3 7 2 4 3 5 þ z 3 5 6 2 4 3 5 or x þ 2y þ 3z ¼ 3 2x þ 3y þ 5z ¼ 7 3x þ 7y þ 6z ¼ 4 (For notational convenience, we have written the vectors in R3 as columns, because it is then easier to ﬁnd the equivalent system of linear equations.) Reducing the system to echelon form yields x þ 2y þ 3z ¼ 3 y z ¼ 1 y 3z ¼ 13 and then x þ 2y þ 3z ¼ 3 y z ¼ 1 4z ¼ 12 Back-substitution yields the solution x ¼ 2, y ¼ 4, z ¼ 3. Thus, v ¼ 2u1 4u2 þ 3u3. Remark: Generally speaking, the question of expressing a given vector v in Kn as a linear combination of vectors u1; u2; . . . ; um in Kn is equivalent to solving a system AX ¼ B of linear equations, where v is the column B of constants, and the u’s are the columns of the coefﬁcient matrix A. Such a system may have a unique solution (as above), many solutions, or no solution. The last case—no solution—means that v cannot be written as a linear combination of the u’s. EXAMPLE 4.2 (Linear combinations in PðtÞ) Suppose we want to express the polynomial v ¼ 3t2 þ 5t 5 as a linear combination of the polynomials p1 ¼ t2 þ 2t þ 1; p2 ¼ 2t2 þ 5t þ 4; p3 ¼ t2 þ 3t þ 6 We seek scalars x, y, z such that v ¼ xp1 þ yp2 þ zp3; that is, 3t2 þ 5t 5 ¼ xðt2 þ 2t þ 1Þ þ yð2t2 þ 5t þ 4Þ þ zðt2 þ 3t þ 6Þ ðÞ There are two ways to proceed from here. (1) Expand the right-hand side of () obtaining: 3t2 þ 5t 5 ¼ xt2 þ 2xt þ x þ 2yt2 þ 5yt þ 4y þ zt2 þ 3zt þ 6z ¼ ðx þ 2y þ zÞt2 þ ð2x þ 5y þ 3zÞt þ ðx þ 4y þ 6zÞ Set coefﬁcients of the same powers of t equal to each other, and reduce the system to echelon form: x þ 2y þ z ¼ 3 2x þ 5y þ 3z ¼ 5 x þ 4y þ 6z ¼ 5 or x þ 2y þ z ¼ 3 y þ z ¼ 1 2y þ 5z ¼ 8 or x þ 2y þ z ¼ 3 y þ z ¼ 1 3z ¼ 6 CHAPTER 4 Vector Spaces 115 The system is in triangular form and has a solution. Back-substitution yields the solution x ¼ 3, y ¼ 1, z ¼ 2. Thus, v ¼ 3p1 þ p2 2p3 (2) The equation () is actually an identity in the variable t; that is, the equation holds for any value of t. We can obtain three equations in the unknowns x, y, z by setting t equal to any three values. For example, Set t ¼ 0 in ð1Þ to obtain: x þ 4y þ 6z ¼ 5 Set t ¼ 1 in ð1Þ to obtain: 4x þ 11y þ 10z ¼ 3 Set t ¼ 1 in ð1Þ to obtain: y þ 4z ¼ 7 Reducing this system to echelon form and solving by back-substitution again yields the solution x ¼ 3, y ¼ 1, z ¼ 2. Thus (again), v ¼ 3p1 þ p2 2p3. Spanning Sets Let V be a vector space over K. Vectors u1; u2; . . . ; um in V are said to span V or to form a spanning set of V if every v in V is a linear combination of the vectors u1; u2; . . . ; um—that is, if there exist scalars a1; a2; . . . ; am in K such that v ¼ a1u1 þ a2u2 þ þ amum The following remarks follow directly from the deﬁnition. Remark 1: Suppose u1; u2; . . . ; um span V. Then, for any vector w, the set w; u1; u2; . . . ; um also spans V. Remark 2: Suppose u1; u2; . . . ; um span V and suppose uk is a linear combination of some of the other u’s. Then the u’s without uk also span V. Remark 3: Suppose u1; u2; . . . ; um span V and suppose one of the u’s is the zero vector. Then the u’s without the zero vector also span V. EXAMPLE 4.3 Consider the vector space V ¼ R3. (a) We claim that the following vectors form a spanning set of R3: e1 ¼ ð1; 0; 0Þ; e2 ¼ ð0; 1; 0Þ; e3 ¼ ð0; 0; 1Þ Speciﬁcally, if v ¼ ða; b; cÞ is any vector in R3, then v ¼ ae1 þ be2 þ ce3 For example, v ¼ ð5; 6; 2Þ ¼ 5e1 6e2 þ 2e3. (b) We claim that the following vectors also form a spanning set of R3: w1 ¼ ð1; 1; 1Þ; w2 ¼ ð1; 1; 0Þ; w3 ¼ ð1; 0; 0Þ Speciﬁcally, if v ¼ ða; b; cÞ is any vector in R3, then (Problem 4.62) v ¼ ða; b; cÞ ¼ cw1 þ ðb cÞw2 þ ða bÞw3 For example, v ¼ ð5; 6; 2Þ ¼ 2w1 8w2 þ 11w3. (c) One can show (Problem 3.24) that v ¼ ð2; 7; 8Þ cannot be written as a linear combination of the vectors u1 ¼ ð1; 2; 3Þ; u2 ¼ ð1; 3; 5Þ; u3 ¼ ð1; 5; 9Þ Accordingly, u1, u2, u3 do not span R3. 116 CHAPTER 4 Vector Spaces EXAMPLE 4.4 Consider the vector space V ¼ PnðtÞ consisting of all polynomials of degree n. (a) Clearly every polynomial in PnðtÞ can be expressed as a linear combination of the n þ 1 polynomials 1; t; t2; t3; . . . ; tn Thus, these powers of t (where 1 ¼ t0) form a spanning set for PnðtÞ. (b) One can also show that, for any scalar c, the following n þ 1 powers of t c, 1; t c; ðt cÞ2; ðt cÞ3; . . . ; ðt cÞn (where ðt cÞ0 ¼ 1), also form a spanning set for PnðtÞ. EXAMPLE 4.5 Consider the vector space M ¼ M2;2 consisting of all 2 2 matrices, and consider the following four matrices in M: E11 ¼ 1 0 0 0 ; E12 ¼ 0 1 0 0 ; E21 ¼ 0 0 1 0 ; E22 ¼ 0 0 0 1 Then clearly any matrix A in M can be written as a linear combination of the four matrices. For example, A ¼ 5 6 7 8 ¼ 5E11 6E12 þ 7E21 þ 8E22 Accordingly, the four matrices E11, E12, E21, E22 span M. 4.5 Subspaces This section introduces the important notion of a subspace. DEFINITION: Let V be a vector space over a ﬁeld K and let W be a subset of V. Then W is a subspace of V if W is itself a vector space over K with respect to the operations of vector addition and scalar multiplication on V. The way in which one shows that any set W is a vector space is to show that W satisﬁes the eight axioms of a vector space. However, if W is a subset of a vector space V, then some of the axioms automatically hold in W, because they already hold in V. Simple criteria for identifying subspaces follow. THEOREM 4.2: Suppose W is a subset of a vector space V. Then W is a subspace of V if the following two conditions hold: (a) The zero vector 0 belongs to W. (b) For every u; v 2 W; k 2 K: (i) The sum u þ v 2 W. (ii) The multiple ku 2 W. Property (i) in (b) states that W is closed under vector addition, and property (ii) in (b) states that W is closed under scalar multiplication. Both properties may be combined into the following equivalent single statement: (b0) For every u; v 2 W; a; b 2 K, the linear combination au þ bv 2 W. Now let V be any vector space. Then V automatically contains two subspaces: the set {0} consisting of the zero vector alone and the whole space V itself. These are sometimes called the trivial subspaces of V. Examples of nontrivial subspaces follow. EXAMPLE 4.6 Consider the vector space V ¼ R3. (a) Let U consist of all vectors in R3 whose entries are equal; that is, U ¼ fða; b; cÞ : a ¼ b ¼ cg For example, (1, 1, 1), (73, 73, 73), (7, 7, 7), (72, 72, 72) are vectors in U. Geometrically, U is the line through the origin O and the point (1, 1, 1) as shown in Fig. 4-1(a). Clearly 0 ¼ ð0; 0; 0Þ belongs to U, because CHAPTER 4 Vector Spaces 117 all entries in 0 are equal. Further, suppose u and v are arbitrary vectors in U, say, u ¼ ða; a; aÞ and v ¼ ðb; b; bÞ. Then, for any scalar k 2 R, the following are also vectors in U: u þ v ¼ ða þ b; a þ b; a þ bÞ and ku ¼ ðka; ka; kaÞ Thus, U is a subspace of R3. (b) Let W be any plane in R3 passing through the origin, as pictured in Fig. 4-1(b). Then 0 ¼ ð0; 0; 0Þ belongs to W, because we assumed W passes through, the origin O. Further, suppose u and v are vectors in W. Then u and v may be viewed as arrows in the plane W emanating from the origin O, as in Fig. 4-1(b). The sum u þ v and any multiple ku of u also lie in the plane W. Thus, W is a subspace of R3. EXAMPLE 4.7 (a) Let V ¼ Mn;n, the vector space of n n matrices. Let W1 be the subset of all (upper) triangular matrices and let W2 be the subset of all symmetric matrices. Then W1 is a subspace of V, because W1 contains the zero matrix 0 and W1 is closed under matrix addition and scalar multiplication; that is, the sum and scalar multiple of such triangular matrices are also triangular. Similarly, W2 is a subspace of V. (b) Let V ¼ PðtÞ, the vector space PðtÞ of polynomials. Then the space PnðtÞ of polynomials of degree at most n may be viewed as a subspace of PðtÞ. Let QðtÞ be the collection of polynomials with only even powers of t. For example, the following are polynomials in QðtÞ: p1 ¼ 3 þ 4t2 5t6 and p2 ¼ 6 7t4 þ 9t6 þ 3t12 (We assume that any constant k ¼ kt0 is an even power of t.) Then QðtÞ is a subspace of PðtÞ. (c) Let V be the vector space of real-valued functions. Then the collection W1 of continuous functions and the collection W2 of differentiable functions are subspaces of V. Intersection of Subspaces Let U and W be subspaces of a vector space V. We show that the intersection U \ W is also a subspace of V. Clearly, 0 2 U and 0 2 W, because U and W are subspaces; whence 0 2 U \ W. Now suppose u and v belong to the intersection U \ W. Then u; v 2 U and u; v 2 W. Further, because U and W are subspaces, for any scalars a; b 2 K, au þ bv 2 U and au þ bv 2 W Thus, au þ bv 2 U \ W. Therefore, U \ W is a subspace of V. The above result generalizes as follows. THEOREM 4.3: The intersection of any number of subspaces of a vector space V is a subspace of V. Figure 4-1 118 CHAPTER 4 Vector Spaces Solution Space of a Homogeneous System Consider a system AX ¼ B of linear equations in n unknowns. Then every solution u may be viewed as a vector in Kn. Thus, the solution set of such a system is a subset of Kn. Now suppose the system is homogeneous; that is, suppose the system has the form AX ¼ 0. Let W be its solution set. Because A0 ¼ 0, the zero vector 0 2 W. Moreover, suppose u and v belong to W. Then u and v are solutions of AX ¼ 0, or, in other words, Au ¼ 0 and Av ¼ 0. Therefore, for any scalars a and b, we have Aðau þ bvÞ ¼ aAu þ bAv ¼ a0 þ b0 ¼ 0 þ 0 ¼ 0 Thus, au þ bv belongs to W, because it is a solution of AX ¼ 0. Accordingly, W is a subspace of Kn. We state the above result formally. THEOREM 4.4: The solution set W of a homogeneous system AX ¼ 0 in n unknowns is a subspace of Kn. We emphasize that the solution set of a nonhomogeneous system AX ¼ B is not a subspace of Kn. In fact, the zero vector 0 does not belong to its solution set. 4.6 Linear Spans, Row Space of a Matrix Suppose u1; u2; . . . ; um are any vectors in a vector space V. Recall (Section 4.4) that any vector of the form a1u1 þ a2u2 þ þ amum, where the ai are scalars, is called a linear combination of u1; u2; . . . ; um. The collection of all such linear combinations, denoted by spanðu1; u2; . . . ; umÞ or spanðuiÞ is called the linear span of u1; u2; . . . ; um. Clearly the zero vector 0 belongs to spanðuiÞ, because 0 ¼ 0u1 þ 0u2 þ þ 0um Furthermore, suppose v and v0 belong to spanðuiÞ, say, v ¼ a1u1 þ a2u2 þ þ amum and v0 ¼ b1u1 þ b2u2 þ þ bmum Then, v þ v0 ¼ ða1 þ b1Þu1 þ ða2 þ b2Þu2 þ þ ðam þ bmÞum and, for any scalar k 2 K, kv ¼ ka1u1 þ ka2u2 þ þ kamum Thus, v þ v0 and kv also belong to spanðuiÞ. Accordingly, spanðuiÞ is a subspace of V. More generally, for any subset S of V, spanðSÞ consists of all linear combinations of vectors in S or, when S ¼ f, span(S) ¼ f0g. Thus, in particular, S is a spanning set (Section 4.4) of spanðSÞ. The following theorem, which was partially proved above, holds. THEOREM 4.5: Let S be a subset of a vector space V. (i) Then spanðSÞ is a subspace of V that contains S. (ii) If W is a subspace of V containing S, then spanðSÞ W. Condition (ii) in theorem 4.5 may be interpreted as saying that spanðSÞ is the ‘‘smallest’’ subspace of V containing S. EXAMPLE 4.8 Consider the vector space V ¼ R3. (a) Let u be any nonzero vector in R3. Then spanðuÞ consists of all scalar multiples of u. Geometrically, spanðuÞ is the line through the origin O and the endpoint of u, as shown in Fig. 4-2(a). CHAPTER 4 Vector Spaces 119 (b) Let u and v be vectors in R3 that are not multiples of each other. Then spanðu; vÞ is the plane through the origin O and the endpoints of u and v as shown in Fig. 4-2(b). (c) Consider the vectors e1 ¼ ð1; 0; 0Þ, e2 ¼ ð0; 1; 0Þ, e3 ¼ ð0; 0; 1Þ in R3. Recall [Example 4.1(a)] that every vector in R3 is a linear combination of e1, e2, e3. That is, e1, e2, e3 form a spanning set of R3. Accordingly, spanðe1; e2; e3Þ ¼ R3. Row Space of a Matrix Let A ¼ ½aij be an arbitrary m n matrix over a ﬁeld K. The rows of A, R1 ¼ ða11; a12; . . . ; a1nÞ; R2 ¼ ða21; a22; . . . ; a2nÞ; . . . ; Rm ¼ ðam1; am2; . . . ; amnÞ may be viewed as vectors in Kn; hence, they span a subspace of Kn called the row space of A and denoted by rowsp(A). That is, rowspðAÞ ¼ spanðR1; R2; . . . ; RmÞ Analagously, the columns of A may be viewed as vectors in Km called the column space of A and denoted by colsp(A). Observe that colspðAÞ ¼ rowspðATÞ. Recall that matrices A and B are row equivalent, written A B, if B can be obtained from A by a sequence of elementary row operations. Now suppose M is the matrix obtained by applying one of the following elementary row operations on a matrix A: ð1Þ Interchange Ri and Rj; ð2Þ Replace Ri by kRi; ð3Þ Replace Rj by kRi þ Rj Then each row of M is a row of A or a linear combination of rows of A. Hence, the row space of M is contained in the row space of A. On the other hand, we can apply the inverse elementary row operation on M to obtain A; hence, the row space of A is contained in the row space of M. Accordingly, A and M have the same row space. This will be true each time we apply an elementary row operation. Thus, we have proved the following theorem. THEOREM 4.6: Row equivalent matrices have the same row space. We are now able to prove (Problems 4.45–4.47) basic results on row equivalence (which ﬁrst appeared as Theorems 3.7 and 3.8 in Chapter 3). THEOREM 4.7: Suppose A ¼ ½aij and B ¼ ½bij are row equivalent echelon matrices with respective pivot entries a1j1; a2j2; . . . ; arjr and b1k1; b2k2; . . . ; bsks Then A and B have the same number of nonzero rows—that is, r ¼ s—and their pivot entries are in the same positions—that is, j1 ¼ k1; j2 ¼ k2; . . . ; jr ¼ kr. THEOREM 4.8: Suppose A and B are row canonical matrices. Then A and B have the same row space if and only if they have the same nonzero rows. 0 (a) u Figure 4-2 0 (b) u 120 CHAPTER 4 Vector Spaces COROLLARY 4.9: Every matrix A is row equivalent to a unique matrix in row canonical form. We apply the above results in the next example. EXAMPLE 4.9 Consider the following two sets of vectors in R4: u1 ¼ ð1; 2; 1; 3Þ; u2 ¼ ð2; 4; 1; 2Þ; u3 ¼ ð3; 6; 3; 7Þ w1 ¼ ð1; 2; 4; 11Þ; w2 ¼ ð2; 4; 5; 14Þ Let U ¼ spanðuiÞ and W ¼ spanðwiÞ. There are two ways to show that U ¼ W. (a) Show that each ui is a linear combination of w1 and w2, and show that each wi is a linear combination of u1, u2, u3. Observe that we have to show that six systems of linear equations are consistent. (b) Form the matrix A whose rows are u1, u2, u3 and row reduce A to row canonical form, and form the matrix B whose rows are w1 and w2 and row reduce B to row canonical form: A ¼ 1 2 1 3 2 4 1 2 3 6 3 7 2 6 4 3 7 5 1 2 1 3 0 0 3 8 0 0 6 16 2 6 4 3 7 5 1 2 0 1 3 0 0 1 8 3 0 0 0 0 2 6 4 3 7 5 B ¼ 1 2 4 11 2 4 5 14 1 2 4 11 0 0 3 8 1 2 0 1 3 0 0 1 8 3 " # Because the nonzero rows of the matrices in row canonical form are identical, the row spaces of A and B are equal. Therefore, U ¼ W. Clearly, the method in (b) is more efﬁcient than the method in (a). 4.7 Linear Dependence and Independence Let V be a vector space over a ﬁeld K. The following deﬁnes the notion of linear dependence and independence of vectors over K. (One usually suppresses mentioning K when the ﬁeld is understood.) This concept plays an essential role in the theory of linear algebra and in mathematics in general. DEFINITION: We say that the vectors v1; v2; . . . ; vm in V are linearly dependent if there exist scalars a1; a2; . . . ; am in K, not all of them 0, such that a1v1 þ a2v2 þ þ amvm ¼ 0 Otherwise, we say that the vectors are linearly independent. The above deﬁnition may be restated as follows. Consider the vector equation x1v1 þ x2v2 þ þ xmvm ¼ 0 ðÞ where the x’s are unknown scalars. This equation always has the zero solution x1 ¼ 0; x2 ¼ 0; . . . ; xm ¼ 0. Suppose this is the only solution; that is, suppose we can show: x1v1 þ x2v2 þ þ xmvm ¼ 0 implies x1 ¼ 0; x2 ¼ 0; . . . ; xm ¼ 0 Then the vectors v1; v2; . . . ; vm are linearly independent, On the other hand, suppose the equation () has a nonzero solution; then the vectors are linearly dependent. A set S ¼ fv1; v2; . . . ; vmg of vectors in V is linearly dependent or independent according to whether the vectors v1; v2; . . . ; vm are linearly dependent or independent. An inﬁnite set S of vectors is linearly dependent or independent according to whether there do or do not exist vectors v1; v2; . . . ; vk in S that are linearly dependent. Warning: The set S ¼ fv1; v2; . . . ; vmg above represents a list or, in other words, a ﬁnite sequence of vectors where the vectors are ordered and repetition is permitted. CHAPTER 4 Vector Spaces 121 The following remarks follow directly from the above deﬁnition. Remark 1: Suppose 0 is one of the vectors v1; v2; . . . ; vm, say v1 ¼ 0. Then the vectors must be linearly dependent, because we have the following linear combination where the coefﬁcient of v1 6¼ 0: 1v1 þ 0v2 þ þ 0vm ¼ 1 0 þ 0 þ þ 0 ¼ 0 Remark 2: Suppose v is a nonzero vector. Then v, by itself, is linearly independent, because kv ¼ 0; v 6¼ 0 implies k ¼ 0 Remark 3: Suppose two of the vectors v1; v2; . . . ; vm are equal or one is a scalar multiple of the other, say v1 ¼ kv2. Then the vectors must be linearly dependent, because we have the following linear combination where the coefﬁcient of v1 6¼ 0: v1 kv2 þ 0v3 þ þ 0vm ¼ 0 Remark 4: Two vectors v1 and v2 are linearly dependent if and only if one of them is a multiple of the other. Remark 5: If the set fv1; . . . ; vmg is linearly independent, then any rearrangement of the vectors fvi1; vi2; . . . ; vimg is also linearly independent. Remark 6: If a set S of vectors is linearly independent, then any subset of S is linearly independent. Alternatively, if S contains a linearly dependent subset, then S is linearly dependent. EXAMPLE 4.10 (a) Let u ¼ ð1; 1; 0Þ, v ¼ ð1; 3; 2Þ, w ¼ ð4; 9; 5Þ. Then u, v, w are linearly dependent, because 3u þ 5v 2w ¼ 3ð1; 1; 0Þ þ 5ð1; 3; 2Þ 2ð4; 9; 5Þ ¼ ð0; 0; 0Þ ¼ 0 (b) We show that the vectors u ¼ ð1; 2; 3Þ, v ¼ ð2; 5; 7Þ, w ¼ ð1; 3; 5Þ are linearly independent. We form the vector equation xu þ yv þ zw ¼ 0, where x, y, z are unknown scalars. This yields x 1 2 3 2 4 3 5 þ y 2 5 7 2 4 3 5 þ z 1 3 5 2 4 3 5 ¼ 0 0 0 2 4 3 5 or x þ 2y þ z ¼ 0 2x þ 5y þ 3z ¼ 0 3x þ 7y þ 5z ¼ 0 or x þ 2y þ z ¼ 0 y þ z ¼ 0 2z ¼ 0 Back-substitution yields x ¼ 0, y ¼ 0, z ¼ 0. We have shown that xu þ yv þ zw ¼ 0 implies x ¼ 0; y ¼ 0; z ¼ 0 Accordingly, u, v, w are linearly independent. (c) Let V be the vector space of functions from R into R. We show that the functions f ðtÞ ¼ sin t, gðtÞ ¼ et, hðtÞ ¼ t2 are linearly independent. We form the vector (function) equation xf þ yg þ zh ¼ 0, where x, y, z are unknown scalars. This function equation means that, for every value of t, x sin t þ yet þ zt2 ¼ 0 Thus, in this equation, we choose appropriate values of t to easily get x ¼ 0, y ¼ 0, z ¼ 0. For example, ðiÞ Substitute t ¼ 0 ðiiÞ Substitute t ¼ p ðiiiÞ Substitute t ¼ p=2 to obtain xð0Þ þ yð1Þ þ zð0Þ ¼ 0 to obtain xð0Þ þ 0ðepÞ þ zðp2Þ ¼ 0 to obtain xð1Þ þ 0ðep=2Þ þ 0ðp2=4Þ ¼ 0 or or or y ¼ 0 z ¼ 0 x ¼ 0 We have shown xf þ yg þ zf ¼ 0 implies x ¼ 0; y ¼ 0; z ¼ 0 Accordingly, u, v, w are linearly independent. 122 CHAPTER 4 Vector Spaces Linear Dependence in R3 Linear dependence in the vector space V ¼ R3 can be described geometrically as follows: (a) Any two vectors u and v in R3 are linearly dependent if and only if they lie on the same line through the origin O, as shown in Fig. 4-3(a). (b) Any three vectors u, v, w in R3 are linearly dependent if and only if they lie on the same plane through the origin O, as shown in Fig. 4-3(b). Later, we will be able to show that any four or more vectors in R3 are automatically linearly dependent. Linear Dependence and Linear Combinations The notions of linear dependence and linear combinations are closely related. Speciﬁcally, for more than one vector, we show that the vectors v1; v2; . . . ; vm are linearly dependent if and only if one of them is a linear combination of the others. Suppose, say, vi is a linear combination of the others, vi ¼ a1v1 þ þ ai1vi1 þ aiþ1viþ1 þ þ amvm Then by adding vi to both sides, we obtain a1v1 þ þ ai1vi1 vi þ aiþ1viþ1 þ þ amvm ¼ 0 where the coefﬁcient of vi is not 0. Hence, the vectors are linearly dependent. Conversely, suppose the vectors are linearly dependent, say, b1v1 þ þ bjvj þ þ bmvm ¼ 0; where bj 6¼ 0 Then we can solve for vj obtaining vj ¼ b1 j b1v1 b1 j bj1vj1 b1 j bjþ1vjþ1 b1 j bmvm and so vj is a linear combination of the other vectors. We now state a slightly stronger statement than the one above. This result has many important consequences. LEMMA 4.10: Suppose two or more nonzero vectors v1; v2; . . . ; vm are linearly dependent. Then one of the vectors is a linear combination of the preceding vectors; that is, there exists k > 1 such that vk ¼ c1v1 þ c2v2 þ þ ck1vk1 Figure 4-3 CHAPTER 4 Vector Spaces 123 Linear Dependence and Echelon Matrices Consider the following echelon matrix A, whose pivots have been circled: A ¼ 0 2 3 4 5 6 7 0 0 4 3 2 3 4 0 0 0 0 7 8 9 0 0 0 0 0 6 7 0 0 0 0 0 0 0 2 6 6 6 6 4 3 7 7 7 7 5 Observe that the rows R2, R3, R4 have 0’s in the second column below the nonzero pivot in R1, and hence any linear combination of R2, R3, R4 must have 0 as its second entry. Thus, R1 cannot be a linear combination of the rows below it. Similarly, the rows R3 and R4 have 0’s in the third column below the nonzero pivot in R2, and hence R2 cannot be a linear combination of the rows below it. Finally, R3 cannot be a multiple of R4, because R4 has a 0 in the ﬁfth column below the nonzero pivot in R3. Viewing the nonzero rows from the bottom up, R4, R3, R2, R1, no row is a linear combination of the preceding rows. Thus, the rows are linearly independent by Lemma 4.10. The argument used with the above echelon matrix A can be used for the nonzero rows of any echelon matrix. Thus, we have the following very useful result. THEOREM 4.11: The nonzero rows of a matrix in echelon form are linearly independent. 4.8 Basis and Dimension First we state two equivalent ways to deﬁne a basis of a vector space V. (The equivalence is proved in Problem 4.28.) DEFINITION A: A set S ¼ fu1; u2; . . . ; ung of vectors is a basis of V if it has the following two properties: (1) S is linearly independent. (2) S spans V. DEFINITION B: A set S ¼ fu1; u2; . . . ; ung of vectors is a basis of V if every v 2 V can be written uniquely as a linear combination of the basis vectors. The following is a fundamental result in linear algebra. THEOREM 4.12: Let V be a vector space such that one basis has m elements and another basis has n elements. Then m ¼ n. A vector space V is said to be of ﬁnite dimension n or n-dimensional, written dim V ¼ n if V has a basis with n elements. Theorem 4.12 tells us that all bases of V have the same number of elements, so this deﬁnition is well deﬁned. The vector space {0} is deﬁned to have dimension 0. Suppose a vector space V does not have a ﬁnite basis. Then V is said to be of inﬁnite dimension or to be inﬁnite-dimensional. The above fundamental Theorem 4.12 is a consequence of the following ‘‘replacement lemma’’ (proved in Problem 4.35). LEMMA 4.13: Suppose fv1; v2; . . . ; vng spans V, and suppose fw1; w2; . . . ; wmg is linearly indepen-dent. Then m n, and V is spanned by a set of the form fw1; w2; . . . ; wm; vi1; vi2; . . . ; vinmg Thus, in particular, n þ 1 or more vectors in V are linearly dependent. Observe in the above lemma that we have replaced m of the vectors in the spanning set of V by the m independent vectors and still retained a spanning set. 124 CHAPTER 4 Vector Spaces Examples of Bases This subsection presents important examples of bases of some of the main vector spaces appearing in this text. (a) Vector space Kn: Consider the following n vectors in Kn: e1 ¼ ð1; 0; 0; 0; . . . ; 0; 0Þ; e2 ¼ ð0; 1; 0; 0; . . . ; 0; 0Þ; . . . ; en ¼ ð0; 0; 0; 0; . . . ; 0; 1Þ These vectors are linearly independent. (For example, they form a matrix in echelon form.) Furthermore, any vector u ¼ ða1; a2; . . . ; anÞ in Kn can be written as a linear combination of the above vectors. Speciﬁcally, v ¼ a1e1 þ a2e2 þ þ anen Accordingly, the vectors form a basis of Kn called the usual or standard basis of Kn. Thus (as one might expect), Kn has dimension n. In particular, any other basis of Kn has n elements. (b) Vector space M ¼ Mr;s of all r s matrices: The following six matrices form a basis of the vector space M2;3 of all 2 3 matrices over K: 1 0 0 0 0 0 ; 0 1 0 0 0 0 ; 0 0 1 0 0 0 ; 0 0 0 1 0 0 ; 0 0 0 0 1 0 ; 0 0 0 0 0 1 More generally, in the vector space M ¼ Mr;s of all r s matrices, let Eij be the matrix with ij-entry 1 and 0’s elsewhere. Then all such matrices form a basis of Mr;s called the usual or standard basis of Mr;s. Accordingly, dim Mr;s ¼ rs. (c) Vector space PnðtÞ of all polynomials of degree n: The set S ¼ f1; t; t2; t3; . . . ; tng of n þ 1 polynomials is a basis of PnðtÞ. Speciﬁcally, any polynomial f ðtÞ of degree n can be expessed as a linear combination of these powers of t, and one can show that these polynomials are linearly independent. Therefore, dim PnðtÞ ¼ n þ 1. (d) Vector space PðtÞ of all polynomials: Consider any ﬁnite set S ¼ ff1ðtÞ; f2ðtÞ; . . . ; fmðtÞg of polynomials in PðtÞ, and let m denote the largest of the degrees of the polynomials. Then any polynomial gðtÞ of degree exceeding m cannot be expressed as a linear combination of the elements of S. Thus, S cannot be a basis of PðtÞ. This means that the dimension of PðtÞ is inﬁnite. We note that the inﬁnite set S0 ¼ f1; t; t2; t3; . . .g, consisting of all the powers of t, spans PðtÞ and is linearly independent. Accordingly, S0 is an inﬁnite basis of PðtÞ. Theorems on Bases The following three theorems (proved in Problems 4.37, 4.38, and 4.39) will be used frequently. THEOREM 4.14: Let V be a vector space of ﬁnite dimension n. Then: (i) Any n þ 1 or more vectors in V are linearly dependent. (ii) Any linearly independent set S ¼ fu1; u2; . . . ; ung with n elements is a basis of V. (iii) Any spanning set T ¼ fv1; v2; . . . ; vng of V with n elements is a basis of V. THEOREM 4.15: Suppose S spans a vector space V. Then: (i) Any maximum number of linearly independent vectors in S form a basis of V. (ii) Suppose one deletes from S every vector that is a linear combination of preceding vectors in S. Then the remaining vectors form a basis of V. CHAPTER 4 Vector Spaces 125 THEOREM 4.16: Let V be a vector space of ﬁnite dimension and let S ¼ fu1; u2; . . . ; urg be a set of linearly independent vectors in V. Then S is part of a basis of V; that is, S may be extended to a basis of V. EXAMPLE 4.11 (a) The following four vectors in R4 form a matrix in echelon form: ð1; 1; 1; 1Þ; ð0; 1; 1; 1Þ; ð0; 0; 1; 1Þ; ð0; 0; 0; 1Þ Thus, the vectors are linearly independent, and, because dim R4 ¼ 4, the four vectors form a basis of R4. (b) The following n þ 1 polynomials in PnðtÞ are of increasing degree: 1; t 1; ðt 1Þ2; . . . ; ðt 1Þn Therefore, no polynomial is a linear combination of preceding polynomials; hence, the polynomials are linear independent. Furthermore, they form a basis of PnðtÞ, because dim PnðtÞ ¼ n þ 1. (c) Consider any four vectors in R3, say ð257; 132; 58Þ; ð43; 0; 17Þ; ð521; 317; 94Þ; ð328; 512; 731Þ By Theorem 4.14(i), the four vectors must be linearly dependent, because they come from the three-dimensional vector space R3. Dimension and Subspaces The following theorem (proved in Problem 4.40) gives the basic relationship between the dimension of a vector space and the dimension of a subspace. THEOREM 4.17: Let W be a subspace of an n-dimensional vector space V. Then dim W n. In particular, if dim W ¼ n, then W ¼ V. EXAMPLE 4.12 Let W be a subspace of the real space R3. Note that dim R3 ¼ 3. Theorem 4.17 tells us that the dimension of W can only be 0, 1, 2, or 3. The following cases apply: (a) If dim W ¼ 0, then W ¼ f0g, a point. (b) If dim W ¼ 1, then W is a line through the origin 0. (c) If dim W ¼ 2, then W is a plane through the origin 0. (d) If dim W ¼ 3, then W is the entire space R3. 4.9 Application to Matrices, Rank of a Matrix Let A be any m n matrix over a ﬁeld K. Recall that the rows of A may be viewed as vectors in Kn and that the row space of A, written rowsp(A), is the subspace of Kn spanned by the rows of A. The following deﬁnition applies. DEFINITION: The rank of a matrix A, written rank(A), is equal to the maximum number of linearly independent rows of A or, equivalently, the dimension of the row space of A. Recall, on the other hand, that the columns of an m n matrix A may be viewed as vectors in Km and that the column space of A, written colsp(A), is the subspace of Km spanned by the columns of A. Although m may not be equal to n—that is, the rows and columns of A may belong to different vector spaces—we have the following fundamental result. THEOREM 4.18: The maximum number of linearly independent rows of any matrix A is equal to the maximum number of linearly independent columns of A. Thus, the dimension of the row space of A is equal to the dimension of the column space of A. Accordingly, one could restate the above deﬁnition of the rank of A using columns instead of rows. 126 CHAPTER 4 Vector Spaces Basis-Finding Problems This subsection shows how an echelon form of any matrix A gives us the solution to certain problems about A itself. Speciﬁcally, let A and B be the following matrices, where the echelon matrix B (whose pivots are circled) is an echelon form of A: A ¼ 1 2 1 3 1 2 2 5 5 6 4 5 3 7 6 11 6 9 1 5 10 8 9 9 2 6 8 11 9 12 2 6 6 6 6 4 3 7 7 7 7 5 and B ¼ 1 2 1 3 1 2 0 1 3 1 2 1 0 0 0 1 1 2 0 0 0 0 0 0 0 0 0 0 0 0 2 6 6 6 6 4 3 7 7 7 7 5 We solve the following four problems about the matrix A, where C1; C2; . . . ; C6 denote its columns: (a) Find a basis of the row space of A. (b) Find each column Ck of A that is a linear combination of preceding columns of A. (c) Find a basis of the column space of A. (d) Find the rank of A. (a) We are given that A and B are row equivalent, so they have the same row space. Moreover, B is in echelon form, so its nonzero rows are linearly independent and hence form a basis of the row space of B. Thus, they also form a basis of the row space of A. That is, basis of rowspðAÞ: ð1; 2; 1; 3; 1; 2Þ; ð0; 1; 3; 1; 2; 1Þ; ð0; 0; 0; 1; 1; 2Þ (b) Let Mk ¼ ½C1; C2; . . . ; Ck, the submatrix of A consisting of the ﬁrst k columns of A. Then Mk1 and Mk are, respectively, the coefﬁcient matrix and augmented matrix of the vector equation x1C1 þ x2C2 þ þ xk1Ck1 ¼ Ck Theorem 3.9 tells us that the system has a solution, or, equivalently, Ck is a linear combination of the preceding columns of A if and only if rankðMkÞ ¼ rankðMk1Þ, where rankðMkÞ means the number of pivots in an echelon form of Mk. Now the ﬁrst k column of the echelon matrix B is also an echelon form of Mk. Accordingly, rankðM2Þ ¼ rankðM3Þ ¼ 2 and rankðM4Þ ¼ rankðM5Þ ¼ rankðM6Þ ¼ 3 Thus, C3, C5, C6 are each a linear combination of the preceding columns of A. (c) The fact that the remaining columns C1, C2, C4 are not linear combinations of their respective preceding columns also tells us that they are linearly independent. Thus, they form a basis of the column space of A. That is, basis of colspðAÞ: ½1; 2; 3; 1; 2T; ½2; 5; 7; 5; 6T; ½3; 6; 11; 8; 11T Observe that C1, C2, C4 may also be characterized as those columns of A that contain the pivots in any echelon form of A. (d) Here we see that three possible deﬁnitions of the rank of A yield the same value. (i) There are three pivots in B, which is an echelon form of A. (ii) The three pivots in B correspond to the nonzero rows of B, which form a basis of the row space of A. (iii) The three pivots in B correspond to the columns of A, which form a basis of the column space of A. Thus, rankðAÞ ¼ 3. CHAPTER 4 Vector Spaces 127 Application to Finding a Basis for W ¼ spanðu1; u2; . . . ; urÞ Frequently, we are given a list S ¼ fu1; u2; . . . ; urg of vectors in Kn and we want to ﬁnd a basis for the subspace W of Kn spanned by the given vectors—that is, a basis of W ¼ spanðSÞ ¼ spanðu1; u2; . . . ; urÞ The following two algorithms, which are essentially described in the above subsection, ﬁnd such a basis (and hence the dimension) of W. Algorithm 4.1 (Row space algorithm) Step 1. Form the matrix M whose rows are the given vectors. Step 2. Row reduce M to echelon form. Step 3. Output the nonzero rows of the echelon matrix. Sometimes we want to ﬁnd a basis that only comes from the original given vectors. The next algorithm accomplishes this task. Algorithm 4.2 (Casting-out algorithm) Step 1. Form the matrix M whose columns are the given vectors. Step 2. Row reduce M to echelon form. Step 3. For each column Ck in the echelon matrix without a pivot, delete (cast out) the vector uk from the list S of given vectors. Step 4. Output the remaining vectors in S (which correspond to columns with pivots). We emphasize that in the ﬁrst algorithm we form a matrix whose rows are the given vectors, whereas in the second algorithm we form a matrix whose columns are the given vectors. EXAMPLE 4.13 Let W be the subspace of R5 spanned by the following vectors: u1 ¼ ð1; 2; 1; 3; 2Þ; u2 ¼ ð1; 3; 3; 5; 3Þ; u3 ¼ ð3; 8; 7; 13; 8Þ u4 ¼ ð1; 4; 6; 9; 7Þ; u5 ¼ ð5; 13; 13; 25; 19Þ Find a basis of W consisting of the original given vectors, and ﬁnd dim W. Form the matrix M whose columns are the given vectors, and reduce M to echelon form: M ¼ 1 1 3 1 5 2 3 8 4 13 1 3 7 6 13 3 5 13 9 25 2 3 8 7 19 2 6 6 6 6 4 3 7 7 7 7 5 1 1 3 1 5 0 1 2 2 3 0 0 0 1 2 0 0 0 0 0 0 0 0 0 0 2 6 6 6 6 4 3 7 7 7 7 5 The pivots in the echelon matrix appear in columns C1, C2, C4. Accordingly, we ‘‘cast out’’ the vectors u3 and u5 from the original ﬁve vectors. The remaining vectors u1, u2, u4, which correspond to the columns in the echelon matrix with pivots, form a basis of W. Thus, in particular, dim W ¼ 3. Remark: The justiﬁcation of the casting-out algorithm is essentially described above, but we repeat it again here for emphasis. The fact that column C3 in the echelon matrix in Example 4.13 does not have a pivot means that the vector equation xu1 þ yu2 ¼ u3 has a solution, and hence u3 is a linear combination of u1 and u2. Similarly, the fact that C5 does not have a pivot means that u5 is a linear combination of the preceding vectors. We have deleted each vector in the original spanning set that is a linear combination of preceding vectors. Thus, the remaining vectors are linearly independent and form a basis of W. 128 CHAPTER 4 Vector Spaces Application to Homogeneous Systems of Linear Equations Consider again a homogeneous system AX ¼ 0 of linear equations over K with n unknowns. By Theorem 4.4, the solution set W of such a system is a subspace of Kn, and hence W has a dimension. The following theorem, whose proof is postponed until Chapter 5, holds. THEOREM 4.19: The dimension of the solution space W of a homogeneous system AX ¼ 0 is n r, where n is the number of unknowns and r is the rank of the coefﬁcient matrix A. In the case where the system AX ¼ 0 is in echelon form, it has precisely n r free variables, say xi1; xi2; . . . ; xinr. Let vj be the solution obtained by setting xij ¼ 1 (or any nonzero constant) and the remaining free variables equal to 0. We show (Problem 4.50) that the solutions v1; v2; . . . ; vnr are linearly independent; hence, they form a basis of the solution space W. We have already used the above process to ﬁnd a basis of the solution space W of a homogeneous system AX ¼ 0 in Section 3.11. Problem 4.48 gives three other examples. 4.10 Sums and Direct Sums Let U and W be subsets of a vector space V. The sum of U and W, written U þ W, consists of all sums u þ w where u 2 U and w 2 W. That is, U þ W ¼ fv : v ¼ u þ w; where u 2 U and w 2 Wg Now suppose U and W are subspaces of V. Then one can easily show (Problem 4.53) that U þ W is a subspace of V. Recall that U \ W is also a subspace of V. The following theorem (proved in Problem 4.58) relates the dimensions of these subspaces. THEOREM 4.20: Suppose U and W are ﬁnite-dimensional subspaces of a vector space V. Then U þ W has ﬁnite dimension and dimðU þ WÞ ¼ dim U þ dim W dimðU \ WÞ EXAMPLE 4.14 Let V ¼ M2;2, the vector space of 2 2 matrices. Let U consist of those matrices whose second row is zero, and let W consist of those matrices whose second column is zero. Then U ¼ a b 0 0 ; W ¼ a 0 c 0 and U þ W ¼ a b c 0 ; U \ W ¼ a 0 0 0 That is, U þ W consists of those matrices whose lower right entry is 0, and U \ W consists of those matrices whose second row and second column are zero. Note that dim U ¼ 2, dim W ¼ 2, dimðU \ WÞ ¼ 1. Also, dimðU þ WÞ ¼ 3, which is expected from Theorem 4.20. That is, dimðU þ WÞ ¼ dim U þ dim V dimðU \ WÞ ¼ 2 þ 2 1 ¼ 3 Direct Sums The vector space V is said to be the direct sum of its subspaces U and W, denoted by V ¼ U W if every v 2 V can be written in one and only one way as v ¼ u þ w where u 2 U and w 2 W. The following theorem (proved in Problem 4.59) characterizes such a decomposition. THEOREM 4.21: The vector space V is the direct sum of its subspaces U and W if and only if: (i) V ¼ U þ W, (ii) U \ W ¼ f0g. CHAPTER 4 Vector Spaces 129 EXAMPLE 4.15 Consider the vector space V ¼ R3: (a) Let U be the xy-plane and let W be the yz-plane; that is, U ¼ fða; b; 0Þ : a; b 2 Rg and W ¼ fð0; b; cÞ : b; c 2 Rg Then R3 ¼ U þ W, because every vector in R3 is the sum of a vector in U and a vector in W. However, R3 is not the direct sum of U and W, because such sums are not unique. For example, ð3; 5; 7Þ ¼ ð3; 1; 0Þ þ ð0; 4; 7Þ and also ð3; 5; 7Þ ¼ ð3; 4; 0Þ þ ð0; 9; 7Þ (b) Let U be the xy-plane and let W be the z-axis; that is, U ¼ fða; b; 0Þ : a; b 2 Rg and W ¼ fð0; 0; cÞ : c 2 Rg Now any vector ða; b; cÞ 2 R3 can be written as the sum of a vector in U and a vector in V in one and only one way: ða; b; cÞ ¼ ða; b; 0Þ þ ð0; 0; cÞ Accordingly, R3 is the direct sum of U and W; that is, R3 ¼ U W. General Direct Sums The notion of a direct sum is extended to more than one factor in the obvious way. That is, V is the direct sum of subspaces W1; W2; . . . ; Wr, written V ¼ W1 W2 Wr if every vector v 2 V can be written in one and only one way as v ¼ w1 þ w2 þ þ wr where w1 2 W1; w2 2 W2; . . . ; wr 2 Wr. The following theorems hold. THEOREM 4.22: Suppose V ¼ W1 W2 Wr. Also, for each k, suppose Sk is a linearly independent subset of Wk. Then (a) The union S ¼ S k Sk is linearly independent in V. (b) If each Sk is a basis of Wk, then S k Sk is a basis of V. (c) dim V ¼ dim W1 þ dim W2 þ þ dim Wr. THEOREM 4.23: Suppose V ¼ W1 þ W2 þ þ Wr and dim V ¼ P k dim Wk. Then V ¼ W1 W2 Wr: 4.11 Coordinates Let V be an n-dimensional vector space over K with basis S ¼ fu1; u2; . . . ; ung. Then any vector v 2 V can be expressed uniquely as a linear combination of the basis vectors in S, say v ¼ a1u1 þ a2u2 þ þ anun These n scalars a1; a2; . . . ; an are called the coordinates of v relative to the basis S, and they form a vector [a1; a2; . . . ; an] in Kn called the coordinate vector of v relative to S. We denote this vector by ½vS, or simply ½v; when S is understood. Thus, ½vS ¼ ½a1; a2; . . . ; an For notational convenience, brackets ½. . ., rather than parentheses ð. . .Þ, are used to denote the coordinate vector. 130 CHAPTER 4 Vector Spaces Remark: The above n scalars a1; a2; . . . ; an also form the coordinate column vector ½a1; a2; . . . ; anT of v relative to S. The choice of the column vector rather than the row vector to represent v depends on the context in which it is used. The use of such column vectors will become clear later in Chapter 6. EXAMPLE 4.16 Consider the vector space P2ðtÞ of polynomials of degree 2. The polynomials p1 ¼ t þ 1; p2 ¼ t 1; p3 ¼ ðt 1Þ2 ¼ t2 2t þ 1 form a basis S of P2ðtÞ. The coordinate vector [v] of v ¼ 2t2 5t þ 9 relative to S is obtained as follows. Set v ¼ xp1 þ yp2 þ zp3 using unknown scalars x, y, z, and simplify: 2t2 5t þ 9 ¼ xðt þ 1Þ þ yðt 1Þ þ zðt2 2t þ 1Þ ¼ xt þ x þ yt y þ zt2 2zt þ z ¼ zt2 þ ðx þ y 2zÞt þ ðx y þ zÞ Then set the coefﬁcients of the same powers of t equal to each other to obtain the system z ¼ 2; x þ y 2z ¼ 5; x y þ z ¼ 9 The solution of the system is x ¼ 3, y ¼ 4, z ¼ 2. Thus, v ¼ 3p1 4p2 þ 2p3; and hence; ½v ¼ ½3; 4; 2 EXAMPLE 4.17 Consider real space R3. The following vectors form a basis S of R3: u1 ¼ ð1; 1; 0Þ; u2 ¼ ð1; 1; 0Þ; u3 ¼ ð0; 1; 1Þ The coordinates of v ¼ ð5; 3; 4Þ relative to the basis S are obtained as follows. Set v ¼ xv1 þ yv2 þ zv3; that is, set v as a linear combination of the basis vectors using unknown scalars x, y, z. This yields 5 3 4 2 4 3 5 ¼ x 1 1 0 2 4 3 5 þ y 1 1 0 2 4 3 5 þ z 0 1 1 2 4 3 5 The equivalent system of linear equations is as follows: x þ y ¼ 5; x þ y þ z ¼ 3; z ¼ 4 The solution of the system is x ¼ 3, y ¼ 2, z ¼ 4. Thus, v ¼ 3u1 þ 2u2 þ 4u3; and so ½vs ¼ ½3; 2; 4 Remark 1: There is a geometrical interpretation of the coordinates of a vector v relative to a basis S for the real space Rn, which we illustrate using the basis S of R3 in Example 4.17. First consider the space R3 with the usual x, y, z axes. Then the basis vectors determine a new coordinate system of R3, say with x0, y0, z0 axes, as shown in Fig. 4-4. That is, (1) The x0-axis is in the direction of u1 with unit length ku1k. (2) The y0-axis is in the direction of u2 with unit length ku2k. (3) The z0-axis is in the direction of u3 with unit length ku3k. Then each vector v ¼ ða; b; cÞ or, equivalently, the point Pða; b; cÞ in R3 will have new coordinates with respect to the new x0, y0, z0 axes. These new coordinates are precisely ½vS, the coordinates of v with respect to the basis S. Thus, as shown in Example 4.17, the coordinates of the point Pð5; 3; 4Þ with the new axes form the vector [3, 2, 4]. Remark 2: Consider the usual basis E ¼ fe1; e2; . . . ; eng of Kn deﬁned by e1 ¼ ð1; 0; 0; . . . ; 0; 0Þ; e2 ¼ ð0; 1; 0; . . . ; 0; 0Þ; . . . ; en ¼ ð0; 0; 0; . . . ; 0; 1Þ CHAPTER 4 Vector Spaces 131 Let v ¼ ða1; a2; . . . ; anÞ be any vector in Kn. Then one can easily show that v ¼ a1e1 þ a2e2 þ þ anen; and so ½vE ¼ ½a1; a2; . . . ; an That is, the coordinate vector ½vE of any vector v relative to the usual basis E of Kn is identical to the original vector v. Isomorphism of V and Kn Let V be a vector space of dimension n over K, and suppose S ¼ fu1; u2; . . . ; ung is a basis of V. Then each vector v 2 V corresponds to a unique n-tuple ½vS in Kn. On the other hand, each n-tuple [c1; c2; . . . ; cn] in Kn corresponds to a unique vector c1u1 þ c2u2 þ þ cnun in V. Thus, the basis S induces a one-to-one correspondence between V and Kn. Furthermore, suppose v ¼ a1u1 þ a2u2 þ þ anun and w ¼ b1u1 þ b2u2 þ þ bnun Then v þ w ¼ ða1 þ b1Þu1 þ ða2 þ b2Þu2 þ þ ðan þ bnÞun kv ¼ ðka1Þu1 þ ðka2Þu2 þ þ ðkanÞun where k is a scalar. Accordingly, ½v þ wS ¼ ½a1 þ b1; . . . ; an þ bn ¼ ½a1; . . . ; an þ ½b1; . . . ; bn ¼ ½vS þ ½wS ½kvS ¼ ½ka1; ka2; . . . ; kan ¼ k½a1; a2; . . . ; an ¼ k½vS Thus, the above one-to-one correspondence between V and Kn preserves the vector space operations of vector addition and scalar multiplication. We then say that V and Kn are isomorphic, written V ﬃKn We state this result formally. Figure 4-4 132 CHAPTER 4 Vector Spaces THEOREM 4.24: Let V be an n-dimensional vector space over a ﬁeld K. Then V and Kn are isomorphic. The next example gives a practical application of the above result. EXAMPLE 4.18 Suppose we want to determine whether or not the following matrices in V ¼ M2;3 are linearly dependent: A ¼ 1 2 3 4 0 1 ; B ¼ 1 3 4 6 5 4 ; C ¼ 3 8 11 16 10 9 The coordinate vectors of the matrices in the usual basis of M2;3 are as follows: ½A ¼ ½1; 2; 3; 4; 0; 1; ½B ¼ ½1; 3; 4; 6; 5; 4; ½C ¼ ½3; 8; 11; 16; 10; 9 Form the matrix M whose rows are the above coordinate vectors and reduce M to an echelon form: M ¼ 1 2 3 4 0 1 1 3 4 6 5 4 3 8 11 16 10 9 2 4 3 5 1 2 3 4 0 1 0 1 1 2 5 3 0 2 2 4 10 6 2 4 3 5 1 2 3 4 0 1 0 1 1 2 5 3 0 0 0 0 0 0 2 4 3 5 Because the echelon matrix has only two nonzero rows, the coordinate vectors [A], [B], [C] span a subspace of dimension 2 and so are linearly dependent. Accordingly, the original matrices A, B, C are linearly dependent. SOLVED PROBLEMS Vector Spaces, Linear Combinations 4.1. Suppose u and v belong to a vector space V. Simplify each of the following expressions: (a) E1 ¼ 3ð2u 4vÞ þ 5u þ 7v, (c) E3 ¼ 2uv þ 3ð2u þ 4vÞ (b) E2 ¼ 3u 6ð3u 5vÞ þ 7u, (d) E4 ¼ 5u 3 v þ 5u Multiply out and collect terms: (a) E1 ¼ 6u 12v þ 5u þ 7v ¼ 11u 5v (b) E2 ¼ 3u 18u þ 30v þ 7u ¼ 8u þ 30v (c) E3 is not deﬁned because the product uv of vectors is not deﬁned. (d) E4 is not deﬁned because division by a vector is not deﬁned. 4.2. Prove Theorem 4.1: Let V be a vector space over a ﬁeld K. (i) k0 ¼ 0. (ii) 0u ¼ 0. (iii) If ku ¼ 0, then k ¼ 0 or u ¼ 0. (iv) ðkÞu ¼ kðuÞ ¼ ku. (i) By Axiom [A2] with u ¼ 0, we have 0 þ 0 ¼ 0. Hence, by Axiom [M1], we have k0 ¼ kð0 þ 0Þ ¼ k0 þ k0 Adding k0 to both sides gives the desired result. (ii) For scalars, 0 þ 0 ¼ 0. Hence, by Axiom [M2], we have 0u ¼ ð0 þ 0Þu ¼ 0u þ 0u Adding 0u to both sides gives the desired result. (iii) Suppose ku ¼ 0 and k 6¼ 0. Then there exists a scalar k1 such that k1k ¼ 1. Thus, u ¼ 1u ¼ ðk1kÞu ¼ k1ðkuÞ ¼ k10 ¼ 0 (iv) Using u þ ðuÞ ¼ 0 and k þ ðkÞ ¼ 0 yields 0 ¼ k0 ¼ k½u þ ðuÞ ¼ ku þ kðuÞ and 0 ¼ 0u ¼ ½k þ ðkÞu ¼ ku þ ðkÞu Adding ku to both sides of the ﬁrst equation gives ku ¼ kðuÞ; and adding ku to both sides of the second equation gives ku ¼ ðkÞu. Thus, ðkÞu ¼ kðuÞ ¼ ku. CHAPTER 4 Vector Spaces 133 4.3. Show that (a) kðu vÞ ¼ ku kv, (b) u þ u ¼ 2u. (a) Using the deﬁnition of subtraction, that u v ¼ u þ ðvÞ, and Theorem 4.1(iv), that kðvÞ ¼ kv, we have kðu vÞ ¼ k½u þ ðvÞ ¼ ku þ kðvÞ ¼ ku þ ðkvÞ ¼ ku kv (b) Using Axiom [M4] and then Axiom [M2], we have u þ u ¼ 1u þ 1u ¼ ð1 þ 1Þu ¼ 2u 4.4. Express v ¼ ð1; 2; 5Þ in R3 as a linear combination of the vectors u1 ¼ ð1; 1; 1Þ; u2 ¼ ð1; 2; 3Þ; u3 ¼ ð2; 1; 1Þ We seek scalars x, y, z, as yet unknown, such that v ¼ xu1 þ yu2 þ zu3. Thus, we require 1 2 5 2 4 3 5 ¼ x 1 1 1 2 4 3 5 þ y 1 2 3 2 4 3 5 þ z 2 1 1 2 4 3 5 or x þ y þ 2z ¼ 1 x þ 2y z ¼ 2 x þ 3y þ z ¼ 5 (For notational convenience, we write the vectors in R3 as columns, because it is then easier to ﬁnd the equivalent system of linear equations.) Reducing the system to echelon form yields the triangular system x þ y þ 2z ¼ 1; y 3z ¼ 3; 5z ¼ 10 The system is consistent and has a solution. Solving by back-substitution yields the solution x ¼ 6, y ¼ 3, z ¼ 2. Thus, v ¼ 6u1 þ 3u2 þ 2u3. Alternatively, write down the augmented matrix M of the equivalent system of linear equations, where u1, u2, u3 are the ﬁrst three columns of M and v is the last column, and then reduce M to echelon form: M ¼ 1 1 2 1 1 2 1 2 1 3 1 5 2 4 3 5 1 1 2 1 0 1 3 3 0 2 1 4 2 4 3 5 1 1 2 1 0 1 3 3 0 0 5 10 2 4 3 5 The last matrix corresponds to a triangular system, which has a solution. Solving the triangular system by back-substitution yields the solution x ¼ 6, y ¼ 3, z ¼ 2. Thus, v ¼ 6u1 þ 3u2 þ 2u3. 4.5. Express v ¼ ð2; 5; 3Þ in R3 as a linear combination of the vectors u1 ¼ ð1; 3; 2Þ; u2 ¼ ð2; 4; 1Þ; u3 ¼ ð1; 5; 7Þ We seek scalars x, y, z, as yet unknown, such that v ¼ xu1 þ yu2 þ zu3. Thus, we require 2 5 3 2 4 3 5 ¼ x 1 3 2 2 4 3 5 þ y 2 4 1 2 4 3 5 þ z 1 5 7 2 4 3 5 or x þ 2y þ z ¼ 2 3x 4y 5z ¼ 5 2x y þ 7z ¼ 3 Reducing the system to echelon form yields the system x þ 2y þ z ¼ 2; 2y 2z ¼ 1; 0 ¼ 3 The system is inconsistent and so has no solution. Thus, v cannot be written as a linear combination of u1, u2, u3. 4.6. Express the polynomial v ¼ t2 þ 4t 3 in PðtÞ as a linear combination of the polynomials p1 ¼ t2 2t þ 5; p2 ¼ 2t2 3t; p3 ¼ t þ 1 Set v as a linear combination of p1, p2, p3 using unknowns x, y, z to obtain t2 þ 4t 3 ¼ xðt2 2t þ 5Þ þ yð2t2 3tÞ þ zðt þ 1Þ ðÞ We can proceed in two ways. 134 CHAPTER 4 Vector Spaces Method 1. Expand the right side of () and express it in terms of powers of t as follows: t2 þ 4t 3 ¼ xt2 2xt þ 5x þ 2yt2 3yt þ zt þ z ¼ ðx þ 2yÞt2 þ ð2x 3y þ zÞt þ ð5x þ 3zÞ Set coefﬁcients of the same powers of t equal to each other, and reduce the system to echelon form. This yields x þ 2y ¼ 1 2x 3y þ z ¼ 4 5x þ 3z ¼ 3 or x þ 2y ¼ 1 y þ z ¼ 6 10y þ 3z ¼ 8 or x þ 2y ¼ 1 y þ z ¼ 6 13z ¼ 52 The system is consistent and has a solution. Solving by back-substitution yields the solution x ¼ 3, y ¼ 2, z ¼ 4. Thus, v ¼ 3p1 þ 2p2 þ 4p2. Method 2. The equation () is an identity in t; that is, the equation holds for any value of t. Thus, we can set t equal to any numbers to obtain equations in the unknowns. (a) Set t ¼ 0 in () to obtain the equation 3 ¼ 5x þ z. (b) Set t ¼ 1 in () to obtain the equation 2 ¼ 4x y þ 2z. (c) Set t ¼ 1 in () to obtain the equation 6 ¼ 8x þ 5y. Solve the system of the three equations to again obtain the solution x ¼ 3, y ¼ 2, z ¼ 4. Thus, v ¼ 3p1 þ 2p2 þ 4p3. 4.7. Express M as a linear combination of the matrices A, B, C, where M ¼ 4 7 7 9 ; and A ¼ 1 1 1 1 ; B ¼ 1 2 3 4 ; C ¼ 1 1 4 5 Set M as a linear combination of A, B, C using unknown scalars x, y, z; that is, set M ¼ xA þ yB þ zC. This yields 4 7 7 9 ¼ x 1 1 1 1 þ y 1 2 3 4 þ z 1 1 4 5 ¼ x þ y þ z x þ 2y þ z x þ 3y þ 4z x þ 4y þ 5z Form the equivalent system of equations by setting corresponding entries equal to each other: x þ y þ z ¼ 4; x þ 2y þ z ¼ 7; x þ 3y þ 4z ¼ 7; x þ 4y þ 5z ¼ 9 Reducing the system to echelon form yields x þ y þ z ¼ 4; y ¼ 3; 3z ¼ 3; 4z ¼ 4 The last equation drops out. Solving the system by back-substitution yields z ¼ 1, y ¼ 3, x ¼ 2. Thus, M ¼ 2A þ 3B C. Subspaces 4.8. Prove Theorem 4.2: W is a subspace of V if the following two conditions hold: (a) 0 2 W. (b) If u; v 2 W, then u þ v, ku 2 W. By (a), W is nonempty, and, by (b), the operations of vector addition and scalar multiplication are well deﬁned for W. Axioms [A1], [A4], [M1], [M2], [M3], [M4] hold in W because the vectors in W belong to V. Thus, we need only show that [A2] and [A3] also hold in W. Now [A2] holds because the zero vector in V belongs to W by (a). Finally, if v 2 W, then ð1Þv ¼ v 2 W, and v þ ðvÞ ¼ 0. Thus [A3] holds. 4.9. Let V ¼ R3. Show that W is not a subspace of V, where (a) W ¼ fða; b; cÞ : a 0g, (b) W ¼ fða; b; cÞ : a2 þ b2 þ c2 1g. In each case, show that Theorem 4.2 does not hold. CHAPTER 4 Vector Spaces 135 (a) W consists of those vectors whose ﬁrst entry is nonnegative. Thus, v ¼ ð1; 2; 3Þ belongs to W. Let k ¼ 3. Then kv ¼ ð3; 6; 9Þ does not belong to W, because 3 is negative. Thus, W is not a subspace of V. (b) W consists of vectors whose length does not exceed 1. Hence, u ¼ ð1; 0; 0Þ and v ¼ ð0; 1; 0Þ belong to W, but u þ v ¼ ð1; 1; 0Þ does not belong to W, because 12 þ 12 þ 02 ¼ 2 > 1. Thus, W is not a subspace of V. 4.10. Let V ¼ PðtÞ, the vector space of real polynomials. Determine whether or not W is a subspace of V, where (a) W consists of all polynomials with integral coefﬁcients. (b) W consists of all polynomials with degree 6 and the zero polynomial. (c) W consists of all polynomials with only even powers of t. (a) No, because scalar multiples of polynomials in W do not always belong to W. For example, f ðtÞ ¼ 3 þ 6t þ 7t2 2 W but 1 2 f ðtÞ ¼ 3 2 þ 3t þ 7 2 t2 62 W (b and c) Yes. In each case, W contains the zero polynomial, and sums and scalar multiples of polynomials in W belong to W. 4.11. Let V be the vector space of functions f : R ! R. Show that W is a subspace of V, where (a) W ¼ f f ðxÞ : f ð1Þ ¼ 0g, all functions whose value at 1 is 0. (b) W ¼ f f ðxÞ : f ð3Þ ¼ f ð1Þg, all functions assigning the same value to 3 and 1. (c) W ¼ f f ðtÞ : f ðxÞ ¼ f ðxÞg, all odd functions. Let ^ 0 denote the zero function, so ^ 0ðxÞ ¼ 0 for every value of x. (a) ^ 0 2 W, because ^ 0ð1Þ ¼ 0. Suppose f ; g 2 W. Then f ð1Þ ¼ 0 and gð1Þ ¼ 0. Also, for scalars a and b, we have ðaf þ bgÞð1Þ ¼ af ð1Þ þ bgð1Þ ¼ a0 þ b0 ¼ 0 Thus, af þ bg 2 W, and hence W is a subspace. (b) ^ 0 2 W, because ^ 0ð3Þ ¼ 0 ¼ ^ 0ð1Þ. Suppose f; g 2 W. Then f ð3Þ ¼ f ð1Þ and gð3Þ ¼ gð1Þ. Thus, for any scalars a and b, we have ðaf þ bgÞð3Þ ¼ af ð3Þ þ bgð3Þ ¼ af ð1Þ þ bgð1Þ ¼ ðaf þ bgÞð1Þ Thus, af þ bg 2 W, and hence W is a subspace. (c) ^ 0 2 W, because ^ 0ðxÞ ¼ 0 ¼ 0 ¼ ^ 0ðxÞ. Suppose f; g 2 W. Then f ðxÞ ¼ f ðxÞ and gðxÞ ¼ gðxÞ. Also, for scalars a and b, ðaf þ bgÞðxÞ ¼ af ðxÞ þ bgðxÞ ¼ af ðxÞ bgðxÞ ¼ ðaf þ bgÞðxÞ Thus, ab þ gf 2 W, and hence W is a subspace of V. 4.12. Prove Theorem 4.3: The intersection of any number of subspaces of V is a subspace of V. Let fWi : i 2 Ig be a collection of subspaces of V and let W ¼ \ðWi : i 2 IÞ. Because each Wi is a subspace of V, we have 0 2 Wi, for every i 2 I. Hence, 0 2 W. Suppose u; v 2 W. Then u; v 2 Wi, for every i 2 I. Because each Wi is a subspace, au þ bv 2 Wi , for every i 2 I. Hence, au þ bv 2 W. Thus, W is a subspace of V. Linear Spans 4.13. Show that the vectors u1 ¼ ð1; 1; 1Þ, u2 ¼ ð1; 2; 3Þ, u3 ¼ ð1; 5; 8Þ span R3. We need to show that an arbitrary vector v ¼ ða; b; cÞ in R3 is a linear combination of u1, u2, u3. Set v ¼ xu1 þ yu2 þ zu3; that is, set ða; b; cÞ ¼ xð1; 1; 1Þ þ yð1; 2; 3Þ þ zð1; 5; 8Þ ¼ ðx þ y þ z; x þ 2y þ 5z; x þ 3y þ 8zÞ 136 CHAPTER 4 Vector Spaces Form the equivalent system and reduce it to echelon form: x þ y þ z ¼ a x þ 2y þ 5z ¼ b x þ 3y þ 8z ¼ c or x þ y þ z ¼ a y þ 4z ¼ b a 2y þ 7c ¼ c a or x þ y þ z ¼ a y þ 4z ¼ b a z ¼ c 2b þ a The above system is in echelon form and is consistent; in fact, x ¼ a þ 5b 3c; y ¼ 3a 7b þ 4c; z ¼ a þ 2b c is a solution. Thus, u1, u2, u3 span R3. 4.14. Find conditions on a, b, c so that v ¼ ða; b; cÞ in R3 belongs to W ¼ spanðu1; u2; u3Þ; where u1 ¼ ð1; 2; 0Þ; u2 ¼ ð1; 1; 2Þ; u3 ¼ ð3; 0; 4Þ Set v as a linear combination of u1, u2, u3 using unknowns x, y, z; that is, set v ¼ xu1 þ yu2 þ zu3: This yields ða; b; cÞ ¼ xð1; 2; 0Þ þ yð1; 1; 2Þ þ zð3; 0; 4Þ ¼ ðx y þ 3z; 2x þ y; 2y 4zÞ Form the equivalent system of linear equations and reduce it to echelon form: x y þ 3z ¼ a 2x þ y ¼ b 2y 4z ¼ c or x y þ 3z ¼ a 3y 6z ¼ b 2a 2y 4z ¼ c or x y þ 3z ¼ a 3y 6z ¼ b 2a 0 ¼ 4a 2b þ 3c The vector v ¼ ða; b; cÞ belongs to W if and only if the system is consistent, and it is consistent if and only if 4a 2b þ 3c ¼ 0. Note, in particular, that u1, u2, u3 do not span the whole space R3. 4.15. Show that the vector space V ¼ PðtÞ of real polynomials cannot be spanned by a ﬁnite number of polynomials. Any ﬁnite set S of polynomials contains a polynomial of maximum degree, say m. Then the linear span span(S) of S cannot contain a polynomial of degree greater than m. Thus, spanðSÞ 6¼ V, for any ﬁnite set S. 4.16. Prove Theorem 4.5: Let S be a subset of V. (i) Then span(S) is a subspace of V containing S. (ii) If W is a subspace of V containing S, then spanðSÞ W. (i) Suppose S is empty. By deﬁnition, spanðSÞ ¼ f0g. Hence spanðSÞ ¼ f0g is a subspace of V and S spanðSÞ. Suppose S is not empty and v 2 S. Then v ¼ 1v 2 spanðSÞ; hence, S spanðSÞ. Also 0 ¼ 0v 2 spanðSÞ. Now suppose u; w 2 spanðSÞ, say u ¼ a1u1 þ þ arur ¼ P i aiui and w ¼ b1w1 þ þ bsws ¼ P j bjwj where ui, wj 2 S and ai; bj 2 K. Then u þ v ¼ P i aiui þ P j bjwj and ku ¼ k P i aiui ¼ P i kaiui belong to span(S) because each is a linear combination of vectors in S. Thus, span(S) is a subspace of V. (ii) Suppose u1; u2; . . . ; ur 2 S. Then all the ui belong to W. Thus, all multiples a1u1; a2u2; . . . ; arur 2 W, and so the sum a1u1 þ a2u2 þ þ arur 2 W. That is, W contains all linear combinations of elements in S, or, in other words, spanðSÞ W, as claimed. Linear Dependence 4.17. Determine whether or not u and v are linearly dependent, where (a) u ¼ ð1; 2Þ, v ¼ ð3; 5Þ, (c) u ¼ ð1; 2; 3Þ, v ¼ ð4; 5; 6Þ (b) u ¼ ð1; 3Þ, v ¼ ð2; 6Þ, (d) u ¼ ð2; 4; 8Þ, v ¼ ð3; 6; 12Þ Two vectors u and v are linearly dependent if and only if one is a multiple of the other. (a) No. (b) Yes; for v ¼ 2u. (c) No. (d) Yes, for v ¼ 3 2 u. CHAPTER 4 Vector Spaces 137 4.18. Determine whether or not u and v are linearly dependent, where (a) u ¼ 2t2 þ 4t 3, v ¼ 4t2 þ 8t 6, (b) u ¼ 2t2 3t þ 4, v ¼ 4t2 3t þ 2, (c) u ¼ 1 3 4 5 0 1 ; v ¼ 4 12 16 20 0 4 , (d) u ¼ 1 1 1 2 2 2 ; v ¼ 2 2 2 3 3 3 Two vectors u and v are linearly dependent if and only if one is a multiple of the other. (a) Yes; for v ¼ 2u. (b) No. (c) Yes, for v ¼ 4u. (d) No. 4.19. Determine whether or not the vectors u ¼ ð1; 1; 2Þ, v ¼ ð2; 3; 1Þ, w ¼ ð4; 5; 5Þ in R3 are linearly dependent. Method 1. Set a linear combination of u, v, w equal to the zero vector using unknowns x, y, z to obtain the equivalent homogeneous system of linear equations and then reduce the system to echelon form. This yields x 1 1 1 2 4 3 5 þ y 2 3 1 2 4 3 5 þ z 4 5 5 2 4 3 5 ¼ 0 0 0 2 4 3 5 or x þ 2y þ 4z ¼ 0 x þ 3y þ 5z ¼ 0 2x þ y þ 5z ¼ 0 or x þ 2y þ 4z ¼ 0 y þ z ¼ 0 The echelon system has only two nonzero equations in three unknowns; hence, it has a free variable and a nonzero solution. Thus, u, v, w are linearly dependent. Method 2. Form the matrix A whose columns are u, v, w and reduce to echelon form: A ¼ 1 2 4 1 3 5 2 1 5 2 4 3 5 1 2 4 0 1 1 0 3 3 2 4 3 5 1 2 4 0 1 1 0 0 0 2 4 3 5 The third column does not have a pivot; hence, the third vector w is a linear combination of the ﬁrst two vectors u and v. Thus, the vectors are linearly dependent. (Observe that the matrix A is also the coefﬁcient matrix in Method 1. In other words, this method is essentially the same as the ﬁrst method.) Method 3. Form the matrix B whose rows are u, v, w, and reduce to echelon form: B ¼ 1 1 2 2 3 1 4 5 5 2 4 3 5 0 1 2 0 1 3 0 1 3 2 4 3 5 1 1 2 0 1 3 0 0 0 2 4 3 5 Because the echelon matrix has only two nonzero rows, the three vectors are linearly dependent. (The three given vectors span a space of dimension 2.) 4.20. Determine whether or not each of the following lists of vectors in R3 is linearly dependent: (a) u1 ¼ ð1; 2; 5Þ, u2 ¼ ð1; 3; 1Þ, u3 ¼ ð2; 5; 7Þ, u4 ¼ ð3; 1; 4Þ, (b) u ¼ ð1; 2; 5Þ, v ¼ ð2; 5; 1Þ, w ¼ ð1; 5; 2Þ, (c) u ¼ ð1; 2; 3Þ, v ¼ ð0; 0; 0Þ, w ¼ ð1; 5; 6Þ. (a) Yes, because any four vectors in R3 are linearly dependent. (b) Use Method 2 above; that is, form the matrix A whose columns are the given vectors, and reduce the matrix to echelon form: A ¼ 1 2 1 2 5 5 5 1 2 2 4 3 5 1 2 1 0 1 3 0 9 3 2 4 3 5 1 2 1 0 1 3 0 0 24 2 4 3 5 Every column has a pivot entry; hence, no vector is a linear combination of the previous vectors. Thus, the vectors are linearly independent. (c) Because 0 ¼ ð0; 0; 0Þ is one of the vectors, the vectors are linearly dependent. 138 CHAPTER 4 Vector Spaces 4.21. Show that the functions f ðtÞ ¼ sin t, gðtÞ cos t, hðtÞ ¼ t from R into R are linearly independent. Set a linear combination of the functions equal to the zero function 0 using unknown scalars x, y, z; that is, set xf þ yg þ zh ¼ 0. Then show x ¼ 0, y ¼ 0, z ¼ 0. We emphasize that xf þ yg þ zh ¼ 0 means that, for every value of t, we have xf ðtÞ þ ygðtÞ þ zhðtÞ ¼ 0. Thus, in the equation x sin t þ y cos t þ zt ¼ 0: ðiÞ Set t ¼ 0 ðiiÞ Set t ¼ p=2 ðiiiÞ Set t ¼ p to obtain to obtain to obtain xð0Þ þ yð1Þ þ zð0Þ ¼ 0 xð1Þ þ yð0Þ þ zp=2 ¼ 0 xð0Þ þ yð1Þ þ zðpÞ ¼ 0 or or or y ¼ 0: x þ pz=2 ¼ 0: y þ pz ¼ 0: The three equations have only the zero solution; that is, x ¼ 0, y ¼ 0, z ¼ 0. Thus, f , g, h are linearly independent. 4.22. Suppose the vectors u, v, w are linearly independent. Show that the vectors u þ v, u v, u 2v þ w are also linearly independent. Suppose xðu þ vÞ þ yðu vÞ þ zðu 2v þ wÞ ¼ 0. Then xu þ xv þ yu yv þ zu 2zv þ zw ¼ 0 or ðx þ y þ zÞu þ ðx y 2zÞv þ zw ¼ 0 Because u, v, w are linearly independent, the coefﬁcients in the above equation are each 0; hence, x þ y þ z ¼ 0; x y 2z ¼ 0; z ¼ 0 The only solution to the above homogeneous system is x ¼ 0, y ¼ 0, z ¼ 0. Thus, u þ v, u v, u 2v þ w are linearly independent. 4.23. Show that the vectors u ¼ ð1 þ i; 2iÞ and w ¼ ð1; 1 þ iÞ in C2 are linearly dependent over the complex ﬁeld C but linearly independent over the real ﬁeld R. Recall that two vectors are linearly dependent (over a ﬁeld K) if and only if one of them is a multiple of the other (by an element in K). Because ð1 þ iÞw ¼ ð1 þ iÞð1; 1 þ iÞ ¼ ð1 þ i; 2iÞ ¼ u u and w are linearly dependent over C. On the other hand, u and w are linearly independent over R, as no real multiple of w can equal u. Speciﬁcally, when k is real, the ﬁrst component of kw ¼ ðk; k þ kiÞ must be real, and it can never equal the ﬁrst component 1 þ i of u, which is complex. Basis and Dimension 4.24. Determine whether or not each of the following form a basis of R3: (a) (1, 1, 1), (1, 0, 1); (c) (1, 1, 1), (1, 2, 3), ð2; 1; 1Þ; (b) (1, 2, 3), (1, 3, 5), (1, 0, 1), (2, 3, 0); (d) (1, 1, 2), (1, 2, 5), (5, 3, 4). (a and b) No, because a basis of R3 must contain exactly three elements because dim R3 ¼ 3. (c) The three vectors form a basis if and only if they are linearly independent. Thus, form the matrix whose rows are the given vectors, and row reduce the matrix to echelon form: 1 1 1 1 2 3 2 1 1 2 4 3 5 1 1 1 0 1 2 0 3 1 2 4 3 5 1 1 1 0 1 2 0 0 5 2 4 3 5 The echelon matrix has no zero rows; hence, the three vectors are linearly independent, and so they do form a basis of R3. CHAPTER 4 Vector Spaces 139 (d) Form the matrix whose rows are the given vectors, and row reduce the matrix to echelon form: 1 1 2 1 2 5 5 3 4 2 4 3 5 1 1 2 0 1 3 0 2 6 2 4 3 5 1 1 2 0 1 3 0 0 0 2 4 3 5 The echelon matrix has a zero row; hence, the three vectors are linearly dependent, and so they do not form a basis of R3. 4.25. Determine whether (1, 1, 1, 1), (1, 2, 3, 2), (2, 5, 6, 4), (2, 6, 8, 5) form a basis of R4. If not, ﬁnd the dimension of the subspace they span. Form the matrix whose rows are the given vectors, and row reduce to echelon form: B ¼ 1 1 1 1 1 2 3 2 2 5 6 4 2 6 8 5 2 6 6 4 3 7 7 5 1 1 1 1 0 1 2 1 0 3 4 2 0 4 6 3 2 6 6 4 3 7 7 5 1 1 1 1 0 1 2 1 0 0 2 1 0 0 2 1 2 6 6 4 3 7 7 5 1 1 1 1 0 1 2 1 0 0 2 1 0 0 0 0 2 6 6 4 3 7 7 5 The echelon matrix has a zero row. Hence, the four vectors are linearly dependent and do not form a basis of R4. Because the echelon matrix has three nonzero rows, the four vectors span a subspace of dimension 3. 4.26. Extend fu1 ¼ ð1; 1; 1; 1Þ; u2 ¼ ð2; 2; 3; 4Þg to a basis of R4. First form the matrix with rows u1 and u2, and reduce to echelon form: 1 1 1 1 2 2 3 4 1 1 1 1 0 0 1 2 Then w1 ¼ ð1; 1; 1; 1Þ and w2 ¼ ð0; 0; 1; 2Þ span the same set of vectors as spanned by u1 and u2. Let u3 ¼ ð0; 1; 0; 0Þ and u4 ¼ ð0; 0; 0; 1Þ. Then w1, u3, w2, u4 form a matrix in echelon form. Thus, they are linearly independent, and they form a basis of R4. Hence, u1, u2, u3, u4 also form a basis of R4. 4.27. Consider the complex ﬁeld C, which contains the real ﬁeld R, which contains the rational ﬁeld Q. (Thus, C is a vector space over R, and R is a vector space over Q.) (a) Show that f1; ig is a basis of C over R; hence, C is a vector space of dimension 2 over R. (b) Show that R is a vector space of inﬁnite dimension over Q. (a) For any v 2 C, we have v ¼ a þ bi ¼ að1Þ þ bðiÞ, where a; b 2 R. Hence, f1; ig spans C over R. Furthermore, if xð1Þ þ yðiÞ ¼ 0 or x þ yi ¼ 0, where x, y 2 R, then x ¼ 0 and y ¼ 0. Hence, f1; ig is linearly independent over R. Thus, f1; ig is a basis for C over R. (b) It can be shown that p is a transcendental number; that is, p is not a root of any polynomial over Q. Thus, for any n, the n þ 1 real numbers 1; p; p2; . . . ; pn are linearly independent over Q. R cannot be of dimension n over Q. Accordingly, R is of inﬁnite dimension over Q. 4.28. Suppose S ¼ fu1; u2; . . . ; ung is a subset of V. Show that the following Deﬁnitions A and B of a basis of V are equivalent: (A) S is linearly independent and spans V. (B) Every v 2 V is a unique linear combination of vectors in S. Suppose (A) holds. Because S spans V, the vector v is a linear combination of the ui, say u ¼ a1u1 þ a2u2 þ þ anun and u ¼ b1u1 þ b2u2 þ þ bnun Subtracting, we get 0 ¼ v v ¼ ða1 b1Þu1 þ ða2 b2Þu2 þ þ ðan bnÞun 140 CHAPTER 4 Vector Spaces But the ui are linearly independent. Hence, the coefﬁcients in the above relation are each 0: a1 b1 ¼ 0; a2 b2 ¼ 0; . . . ; an bn ¼ 0 Therefore, a1 ¼ b1; a2 ¼ b2; . . . ; an ¼ bn. Hence, the representation of v as a linear combination of the ui is unique. Thus, (A) implies (B). Suppose (B) holds. Then S spans V. Suppose 0 ¼ c1u1 þ c2u2 þ þ cnun However, we do have 0 ¼ 0u1 þ 0u2 þ þ 0un By hypothesis, the representation of 0 as a linear combination of the ui is unique. Hence, each ci ¼ 0 and the ui are linearly independent. Thus, (B) implies (A). Dimension and Subspaces 4.29. Find a basis and dimension of the subspace W of R3 where (a) W ¼ fða; b; cÞ : a þ b þ c ¼ 0g, (b) W ¼ fða; b; cÞ : ða ¼ b ¼ cÞg (a) Note that W 6¼ R3, because, for example, ð1; 2; 3Þ 62 W. Thus, dim W < 3. Note that u1 ¼ ð1; 0; 1Þ and u2 ¼ ð0; 1; 1Þ are two independent vectors in W. Thus, dim W ¼ 2, and so u1 and u2 form a basis of W. (b) The vector u ¼ ð1; 1; 1Þ 2 W. Any vector w 2 W has the form w ¼ ðk; k; kÞ. Hence, w ¼ ku. Thus, u spans W and dim W ¼ 1. 4.30. Let W be the subspace of R4 spanned by the vectors u1 ¼ ð1; 2; 5; 3Þ; u2 ¼ ð2; 3; 1; 4Þ; u3 ¼ ð3; 8; 3; 5Þ (a) Find a basis and dimension of W. (b) Extend the basis of W to a basis of R4. (a) Apply Algorithm 4.1, the row space algorithm. Form the matrix whose rows are the given vectors, and reduce it to echelon form: A ¼ 1 2 5 3 2 3 1 4 3 8 3 5 2 4 3 5 1 2 5 3 0 7 9 2 0 14 18 4 2 4 3 5 1 2 5 3 0 7 9 2 0 0 0 0 2 4 3 5 The nonzero rows ð1; 2; 5; 3Þ and ð0; 7; 9; 2Þ of the echelon matrix form a basis of the row space of A and hence of W. Thus, in particular, dim W ¼ 2. (b) We seek four linearly independent vectors, which include the above two vectors. The four vectors ð1; 2; 5; 3Þ, ð0; 7; 9; 2Þ, (0, 0, 1, 0), and (0, 0, 0, 1) are linearly independent (because they form an echelon matrix), and so they form a basis of R4, which is an extension of the basis of W. 4.31. Let W be the subspace of R5 spanned by u1 ¼ ð1; 2; 1; 3; 4Þ, u2 ¼ ð2; 4; 2; 6; 8Þ, u3 ¼ ð1; 3; 2; 2; 6Þ, u4 ¼ ð1; 4; 5; 1; 8Þ, u5 ¼ ð2; 7; 3; 3; 9Þ. Find a subset of the vectors that form a basis of W. Here we use Algorithm 4.2, the casting-out algorithm. Form the matrix M whose columns (not rows) are the given vectors, and reduce it to echelon form: M ¼ 1 2 1 1 2 2 4 3 4 7 1 2 2 5 3 3 6 2 1 3 4 8 6 8 9 2 6 6 6 6 4 3 7 7 7 7 5 1 2 1 1 2 0 0 1 2 3 0 0 3 6 5 0 0 1 2 3 0 0 2 4 1 2 6 6 6 6 4 3 7 7 7 7 5 1 2 1 1 2 0 0 1 2 3 0 0 0 0 4 0 0 0 0 0 0 0 0 0 0 2 6 6 6 6 4 3 7 7 7 7 5 The pivot positions are in columns C1, C3, C5. Hence, the corresponding vectors u1, u3, u5 form a basis of W, and dim W ¼ 3. CHAPTER 4 Vector Spaces 141 4.32. Let V be the vector space of 2 2 matrices over K. Let W be the subspace of symmetric matrices. Show that dim W ¼ 3, by ﬁnding a basis of W. Recall that a matrix A ¼ ½aij is symmetric if AT ¼ A, or, equivalently, each aij ¼ aji. Thus, A ¼ a b b d denotes an arbitrary 2 2 symmetric matrix. Setting (i) a ¼ 1, b ¼ 0, d ¼ 0; (ii) a ¼ 0, b ¼ 1, d ¼ 0; (iii) a ¼ 0, b ¼ 0, d ¼ 1, we obtain the respective matrices: E1 ¼ 1 0 0 0 ; E2 ¼ 0 1 1 0 ; E3 ¼ 0 0 0 1 We claim that S ¼ fE1; E2; E3g is a basis of W; that is, (a) S spans W and (b) S is linearly independent. (a) The above matrix A ¼ a b b d ¼ aE1 þ bE2 þ dE3. Thus, S spans W. (b) Suppose xE1 þ yE2 þ zE3 ¼ 0, where x, y, z are unknown scalars. That is, suppose x 1 0 0 0 þ y 0 1 1 0 þ z 0 0 0 1 ¼ 0 0 0 0 or x y y z ¼ 0 0 0 0 Setting corresponding entries equal to each other yields x ¼ 0, y ¼ 0, z ¼ 0. Thus, S is linearly independent. Therefore, S is a basis of W, as claimed. Theorems on Linear Dependence, Basis, and Dimension 4.33. Prove Lemma 4.10: Suppose two or more nonzero vectors v1; v2; . . . ; vm are linearly dependent. Then one of them is a linear combination of the preceding vectors. Because the vi are linearly dependent, there exist scalars a1; . . . ; am, not all 0, such that a1v1 þ þ amvm ¼ 0. Let k be the largest integer such that ak 6¼ 0. Then a1v1 þ þ akvk þ 0vkþ1 þ þ 0vm ¼ 0 or a1v1 þ þ akvk ¼ 0 Suppose k ¼ 1; then a1v1 ¼ 0, a1 6¼ 0, and so v1 ¼ 0. But the vi are nonzero vectors. Hence, k > 1 and vk ¼ a1 k a1v1 a1 k ak1vk1 That is, vk is a linear combination of the preceding vectors. 4.34. Suppose S ¼ fv1; v2; . . . ; vmg spans a vector space V. (a) If w 2 V, then fw; v1; . . . ; vmg is linearly dependent and spans V. (b) If vi is a linear combination of v1; . . . ; vi1, then S without vi spans V. (a) The vector w is a linear combination of the vi, because fvig spans V. Accordingly, fw; v1; . . . ; vmg is linearly dependent. Clearly, w with the vi span V, as the vi by themselves span V; that is, fw; v1; . . . ; vmg spans V. (b) Suppose vi ¼ k1v1 þ þ ki1vi1. Let u 2 V. Because fvig spans V, u is a linear combination of the vj’s, say u ¼ a1v1 þ þ amvm: Substituting for vi, we obtain u ¼ a1v1 þ þ ai1vi1 þ aiðk1v1 þ þ ki1vi1Þ þ aiþ1viþ1 þ þ amvm ¼ ða1 þ aik1Þv1 þ þ ðai1 þ aiki1Þvi1 þ aiþ1viþ1 þ þ amvm Thus, fv1; . . . ; vi1; viþ1; . . . ; vmg spans V. In other words, we can delete vi from the spanning set and still retain a spanning set. 4.35. Prove Lemma 4.13: Suppose fv1; v2; . . . ; vng spans V, and suppose fw1; w2; . . . ; wmg is linearly independent. Then m n, and V is spanned by a set of the form fw1; w2; . . . ; wm; vi1; vi2; . . . ; vinmg Thus, any n þ 1 or more vectors in V are linearly dependent. 142 CHAPTER 4 Vector Spaces It sufﬁces to prove the lemma in the case that the vi are all not 0. (Prove!) Because fvig spans V, we have by Problem 4.34 that fw1; v1; . . . ; vng ð1Þ is linearly dependent and also spans V. By Lemma 4.10, one of the vectors in (1) is a linear combination of the preceding vectors. This vector cannot be w1, so it must be one of the v’s, say vj: Thus by Problem 4.34, we can delete vj from the spanning set (1) and obtain the spanning set fw1; v1; . . . ; vj1; vjþ1; . . . ; vng ð2Þ Now we repeat the argument with the vector w2. That is, because (2) spans V, the set fw1; w2; v1; . . . ; vj1; vjþ1; . . . ; vng ð3Þ is linearly dependent and also spans V. Again by Lemma 4.10, one of the vectors in (3) is a linear combination of the preceding vectors. We emphasize that this vector cannot be w1 or w2, because fw1; . . . ; wmg is independent; hence, it must be one of the v’s, say vk. Thus, by Problem 4.34, we can delete vk from the spanning set (3) and obtain the spanning set fw1; w2; v1; . . . ; vj1; vjþ1; . . . ; vk1; vkþ1; . . . ; vng We repeat the argument with w3, and so forth. At each step, we are able to add one of the w’s and delete one of the v’s in the spanning set. If m n, then we ﬁnally obtain a spanning set of the required form: fw1; . . . ; wm; vi1; . . . ; vinmg Finally, we show that m > n is not possible. Otherwise, after n of the above steps, we obtain the spanning set fw1; . . . ; wng. This implies that wnþ1 is a linear combination of w1; . . . ; wn, which contradicts the hypothesis that fwig is linearly independent. 4.36. Prove Theorem 4.12: Every basis of a vector space V has the same number of elements. Suppose fu1; u2; . . . ; ung is a basis of V, and suppose fv1; v2; . . .g is another basis of V. Because fuig spans V, the basis fv1; v2; . . .g must contain n or less vectors, or else it is linearly dependent by Problem 4.35—Lemma 4.13. On the other hand, if the basis fv1; v2; . . .g contains less than n elements, then fu1; u2; . . . ; ung is linearly dependent by Problem 4.35. Thus, the basis fv1; v2; . . .g contains exactly n vectors, and so the theorem is true. 4.37. Prove Theorem 4.14: Let V be a vector space of ﬁnite dimension n. Then (i) Any n þ 1 or more vectors must be linearly dependent. (ii) Any linearly independent set S ¼ fu1; u2; . . . ung with n elements is a basis of V. (iii) Any spanning set T ¼ fv1; v2; . . . ; vng of V with n elements is a basis of V. Suppose B ¼ fw1; w2; . . . ; wng is a basis of V. (i) Because B spans V, any n þ 1 or more vectors are linearly dependent by Lemma 4.13. (ii) By Lemma 4.13, elements from B can be adjoined to S to form a spanning set of V with n elements. Because S already has n elements, S itself is a spanning set of V. Thus, S is a basis of V. (iii) Suppose T is linearly dependent. Then some vi is a linear combination of the preceding vectors. By Problem 4.34, V is spanned by the vectors in T without vi and there are n 1 of them. By Lemma 4.13, the independent set B cannot have more than n 1 elements. This contradicts the fact that B has n elements. Thus, T is linearly independent, and hence T is a basis of V. 4.38. Prove Theorem 4.15: Suppose S spans a vector space V. Then (i) Any maximum number of linearly independent vectors in S form a basis of V. (ii) Suppose one deletes from S every vector that is a linear combination of preceding vectors in S. Then the remaining vectors form a basis of V. (i) Suppose fv1; . . . ; vmg is a maximum linearly independent subset of S, and suppose w 2 S. Accord-ingly, fv1; . . . ; vm; wg is linearly dependent. No vk can be a linear combination of preceding vectors. CHAPTER 4 Vector Spaces 143 Hence, w is a linear combination of the vi. Thus, w 2 spanðviÞ, and hence S spanðviÞ. This leads to V ¼ spanðSÞ spanðviÞ V Thus, fvig spans V, and, as it is linearly independent, it is a basis of V. (ii) The remaining vectors form a maximum linearly independent subset of S; hence, by (i), it is a basis of V. 4.39. Prove Theorem 4.16: Let V be a vector space of ﬁnite dimension and let S ¼ fu1; u2; . . . ; urg be a set of linearly independent vectors in V. Then S is part of a basis of V; that is, S may be extended to a basis of V. Suppose B ¼ fw1; w2; . . . ; wng is a basis of V. Then B spans V, and hence V is spanned by S [ B ¼ fu1; u2; . . . ; ur; w1; w2; . . . ; wng By Theorem 4.15, we can delete from S [ B each vector that is a linear combination of preceding vectors to obtain a basis B0 for V. Because S is linearly independent, no uk is a linear combination of preceding vectors. Thus, B0 contains every vector in S, and S is part of the basis B0 for V. 4.40. Prove Theorem 4.17: Let W be a subspace of an n-dimensional vector space V. Then dim W n. In particular, if dim W ¼ n, then W ¼ V. Because V is of dimension n, any n þ 1 or more vectors are linearly dependent. Furthermore, because a basis of W consists of linearly independent vectors, it cannot contain more than n elements. Accordingly, dim W n. In particular, if fw1; . . . ; wng is a basis of W, then, because it is an independent set with n elements, it is also a basis of V. Thus, W ¼ V when dim W ¼ n. Rank of a Matrix, Row and Column Spaces 4.41. Find the rank and basis of the row space of each of the following matrices: (a) A ¼ 1 2 0 1 2 6 3 3 3 10 6 5 2 4 3 5, (b) B ¼ 1 3 1 2 3 1 4 3 1 4 2 3 4 7 3 3 8 1 7 8 2 6 6 4 3 7 7 5. (a) Row reduce A to echelon form: A 1 2 0 1 0 2 3 1 0 4 6 2 2 4 3 5 1 2 0 1 0 2 3 1 0 0 0 0 2 4 3 5 The two nonzero rows ð1; 2; 0; 1Þ and ð0; 2; 3; 1Þ of the echelon form of A form a basis for rowsp(A). In particular, rankðAÞ ¼ 2. (b) Row reduce B to echelon form: B 1 3 1 2 3 0 1 2 1 1 0 3 6 3 3 0 1 2 1 1 2 6 6 4 3 7 7 5 1 3 1 2 3 0 1 2 1 1 0 0 0 0 0 0 0 0 0 0 2 6 6 4 3 7 7 5 The two nonzero rows ð1; 3; 1; 2; 3Þ and ð0; 1; 2; 1; 1Þ of the echelon form of B form a basis for rowsp(B). In particular, rankðBÞ ¼ 2. 4.42. Show that U ¼ W, where U and W are the following subspaces of R3: U ¼ spanðu1; u2; u3Þ ¼ spanð1; 1; 1Þ; ð2; 3; 1Þ; ð3; 1; 5Þg W ¼ spanðw1; w2; w3Þ ¼ spanð1; 1; 3Þ; ð3; 2; 8Þ; ð2; 1; 3Þg 144 CHAPTER 4 Vector Spaces Form the matrix A whose rows are the ui, and row reduce A to row canonical form: A ¼ 1 1 1 2 3 1 3 1 5 2 4 3 5 1 1 1 0 1 1 0 2 2 2 4 3 5 1 0 2 0 1 1 0 0 0 2 4 3 5 Next form the matrix B whose rows are the wj, and row reduce B to row canonical form: B ¼ 1 1 3 3 2 8 2 1 3 2 4 3 5 1 1 3 0 1 1 0 3 3 2 4 3 5 1 0 2 0 1 1 0 0 0 2 4 3 5 Because A and B have the same row canonical form, the row spaces of A and B are equal, and so U ¼ W. 4.43. Let A ¼ 1 2 1 2 3 1 2 4 3 7 7 4 1 2 2 5 5 6 3 6 6 15 14 15 2 6 6 4 3 7 7 5. (a) Find rankðMkÞ, for k ¼ 1; 2; . . . ; 6, where Mk is the submatrix of A consisting of the ﬁrst k columns C1; C2; . . . ; Ck of A. (b) Which columns Ckþ1 are linear combinations of preceding columns C1; . . . ; Ck? (c) Find columns of A that form a basis for the column space of A. (d) Express column C4 as a linear combination of the columns in part (c). (a) Row reduce A to echelon form: A 1 2 1 2 3 1 0 0 1 3 1 2 0 0 1 3 2 5 0 0 3 9 5 12 2 6 6 4 3 7 7 5 1 2 1 2 3 1 0 0 1 3 1 2 0 0 0 0 1 3 0 0 0 0 0 0 2 6 6 4 3 7 7 5 Observe that this simultaneously reduces all the matrices Mk to echelon form; for example, the ﬁrst four columns of the echelon form of A are an echelon form of M4. We know that rankðMkÞ is equal to the number of pivots or, equivalently, the number of nonzero rows in an echelon form of Mk. Thus, rankðM1Þ ¼ rankðM2Þ ¼ 1; rankðM3Þ ¼ rankðM4Þ ¼ 2 rankðM5Þ ¼ rankðM6Þ ¼ 3 (b) The vector equation x1C1 þ x2C2 þ þ xkCk ¼ Ckþ1 yields the system with coefﬁcient matrix Mk and augmented Mkþ1. Thus, Ckþ1 is a linear combination of C1; . . . ; Ck if and only if rankðMkÞ ¼ rankðMkþ1Þ or, equivalently, if Ckþ1 does not contain a pivot. Thus, each of C2, C4, C6 is a linear combination of preceding columns. (c) In the echelon form of A, the pivots are in the ﬁrst, third, and ﬁfth columns. Thus, columns C1, C3, C5 of A form a basis for the columns space of A. Alternatively, deleting columns C2, C4, C6 from the spanning set of columns (they are linear combinations of other columns), we obtain, again, C1, C3, C5. (d) The echelon matrix tells us that C4 is a linear combination of columns C1 and C3. The augmented matrix M of the vector equation C4 ¼ xC1 þ yC2 consists of the columns C1, C3, C4 of A which, when reduced to echelon form, yields the matrix (omitting zero rows) 1 1 2 0 1 3 or x þ y ¼ 2 y ¼ 3 or x ¼ 1; y ¼ 3 Thus, C4 ¼ C1 þ 3C3 ¼ C1 þ 3C3 þ 0C5. 4.44. Suppose u ¼ ða1; a2; . . . ; anÞ is a linear combination of the rows R1; R2; . . . ; Rm of a matrix B ¼ ½bij, say u ¼ k1R1 þ k2R2 þ þ kmRm: Prove that ai ¼ k1b1i þ k2b2i þ þ kmbmi; i ¼ 1; 2; . . . ; n where b1i; b2i; . . . ; bmi are the entries in the ith column of B. CHAPTER 4 Vector Spaces 145 We are given that u ¼ k1R1 þ k2R2 þ þ kmRm. Hence, ða1; a2; . . . ; anÞ ¼ k1ðb11; . . . ; b1nÞ þ þ kmðbm1; . . . ; bmnÞ ¼ ðk1b11 þ þ kmbm1; . . . ; k1b1n þ þ kmbmnÞ Setting corresponding components equal to each other, we obtain the desired result. 4.45. Prove Theorem 4.7: Suppose A ¼ ½aij and B ¼ ½bij are row equivalent echelon matrices with respective pivot entries a1j1; a2j2; . . . ; arjr and b1k1; b2k2; . . . ; bsks (pictured in Fig. 4-5). Then A and B have the same number of nonzero rows—that is, r ¼ s—and their pivot entries are in the same positions; that is, j1 ¼ k1; j2 ¼ k2; . . . ; jr ¼ kr. Clearly A ¼ 0 if and only if B ¼ 0, and so we need only prove the theorem when r 1 and s 1. We ﬁrst show that j1 ¼ k1. Suppose j1 < k1. Then the j1th column of B is zero. Because the ﬁrst row R of A is in the row space of B, we have R ¼ c1R1 þ c1R2 þ þ cmRm, where the Ri are the rows of B. Because the j1th column of B is zero, we have a1j1 ¼ c10 þ c20 þ þ cm0 ¼ 0 But this contradicts the fact that the pivot entry a1j1 6¼ 0. Hence, j1 k1 and, similarly, k1 j1. Thus j1 ¼ k1. Now let A0 be the submatrix of A obtained by deleting the ﬁrst row of A, and let B0 be the submatrix of B obtained by deleting the ﬁrst row of B. We prove that A0 and B0 have the same row space. The theorem will then follow by induction, because A0 and B0 are also echelon matrices. Let R ¼ ða1; a2; ... ; anÞ be any row of A0 and let R1; ... ; Rm be the rows of B. Because R is in the row space of B, there exist scalars d1; ... ; dm such that R ¼ d1R1 þ d2R2 þ þ dmRm. Because A is in echelon form and R is not the ﬁrst row of A, the j1th entry of R is zero: ai ¼ 0 for i ¼ j1 ¼ k1. Furthermore, because B is in echelon form, all the entries in the k1th column of B are 0 except the ﬁrst: b1k1 6¼ 0, but b2k1 ¼ 0; ... ; bmk1 ¼ 0. Thus, 0 ¼ ak1 ¼ d1b1k1 þ d20 þ þ dm0 ¼ d1b1k1 Now b1k1 6¼ 0 and so d1 ¼ 0. Thus, R is a linear combination of R2; . . . ; Rm and so is in the row space of B0. Because R was any row of A0, the row space of A0 is contained in the row space of B0. Similarly, the row space of B0 is contained in the row space of A0. Thus, A0 and B0 have the same row space, and so the theorem is proved. 4.46. Prove Theorem 4.8: Suppose A and B are row canonical matrices. Then A and B have the same row space if and only if they have the same nonzero rows. Obviously, if A and B have the same nonzero rows, then they have the same row space. Thus we only have to prove the converse. Suppose A and B have the same row space, and suppose R 6¼ 0 is the ith row of A. Then there exist scalars c1; . . . ; cs such that R ¼ c1R1 þ c2R2 þ þ csRs ð1Þ where the Ri are the nonzero rows of B. The theorem is proved if we show that R ¼ Ri; that is, that ci ¼ 1 but ck ¼ 0 for k 6¼ i. A ¼ a1j1 a2j2 :::::::::::::::::::::::::::::::::::::: arjr 2 6 6 4 3 7 7 5; b ¼ b1k1 b2k2 :::::::::::::::::::::::::::::::::::::: bsks 2 6 6 4 3 7 7 5 Figure 4-5 146 CHAPTER 4 Vector Spaces Let aij, be the pivot entry in R—that is, the ﬁrst nonzero entry of R. By (1) and Problem 4.44, aiji ¼ c1b1ji þ c2b2ji þ þ csbsji ð2Þ But, by Problem 4.45, biji is a pivot entry of B, and, as B is row reduced, it is the only nonzero entry in the jth column of B. Thus, from (2), we obtain aiji ¼ cibiji. However, aiji ¼ 1 and biji ¼ 1, because A and B are row reduced; hence, ci ¼ 1. Now suppose k 6¼ i, and bkjk is the pivot entry in Rk. By (1) and Problem 4.44, aijk ¼ c1b1jk þ c2b2jk þ þ csbsjk ð3Þ Because B is row reduced, bkjk is the only nonzero entry in the jth column of B. Hence, by (3), aijk ¼ ckbkjk. Furthermore, by Problem 4.45, akjk is a pivot entry of A, and because A is row reduced, aijk ¼ 0. Thus, ckbkjk ¼ 0, and as bkjk ¼ 1, ck ¼ 0. Accordingly R ¼ Ri; and the theorem is proved. 4.47. Prove Corollary 4.9: Every matrix A is row equivalent to a unique matrix in row canonical form. Suppose A is row equivalent to matrices A1 and A2, where A1 and A2 are in row canonical form. Then rowspðAÞ ¼ rowspðA1Þ and rowspðAÞ ¼ rowspðA2Þ. Hence, rowspðA1Þ ¼ rowspðA2Þ. Because A1 and A2 are in row canonical form, A1 ¼ A2 by Theorem 4.8. Thus, the corollary is proved. 4.48. Suppose RB and AB are deﬁned, where R is a row vector and A and B are matrices. Prove (a) RB is a linear combination of the rows of B. (b) The row space of AB is contained in the row space of B. (c) The column space of AB is contained in the column space of A. (d) If C is a column vector and AC is deﬁned, then AC is a linear combination of the columns of A: (e) rankðABÞ rankðBÞ and rankðABÞ rankðAÞ. (a) Suppose R ¼ ða1; a2; . . . ; amÞ and B ¼ ½bij. Let B1; . . . ; Bm denote the rows of B and B1; . . . ; Bn its columns. Then RB ¼ ðRB1; RB2; . . . ; RBnÞ ¼ ða1b11 þ a2b21 þ þ ambm1; . . . ; a1b1n þ a2b2n þ þ ambmnÞ ¼ a1ðb11; b12; . . . ; b1nÞ þ a2ðb21; b22; . . . ; b2nÞ þ þ amðbm1; bm2; . . . ; bmnÞ ¼ a1B1 þ a2B2 þ þ amBm Thus, RB is a linear combination of the rows of B, as claimed. (b) The rows of AB are RiB, where Ri is the ith row of A. Thus, by part (a), each row of AB is in the row space of B. Thus, rowspðABÞ rowspðBÞ, as claimed. (c) Using part (b), we have colspðABÞ ¼ rowspðABÞT ¼ rowspðBTATÞ rowspðATÞ ¼ colspðAÞ: (d) Follows from ðcÞ where C replaces B: (e) The row space of AB is contained in the row space of B; hence, rankðABÞ rankðBÞ. Furthermore, the column space of AB is contained in the column space of A; hence, rankðABÞ rankðAÞ. 4.49. Let A be an n-square matrix. Show that A is invertible if and only if rankðAÞ ¼ n. Note that the rows of the n-square identity matrix In are linearly independent, because In is in echelon form; hence, rankðInÞ ¼ n. Now if A is invertible, then A is row equivalent to In; hence, rankðAÞ ¼ n. But if A is not invertible, then A is row equivalent to a matrix with a zero row; hence, rankðAÞ < n; that is, A is invertible if and only if rankðAÞ ¼ n. CHAPTER 4 Vector Spaces 147 Applications to Linear Equations 4.50. Find the dimension and a basis of the solution space W of each homogeneous system: x þ 2y þ 2z s þ 3t ¼ 0 x þ 2y þ 3z þ s þ t ¼ 0 3x þ 6y þ 8z þ s þ 5t ¼ 0 (a) x þ 2y þ z 2t ¼ 0 2x þ 4y þ 4z 3t ¼ 0 3x þ 6y þ 7z 4t ¼ 0 (b) x þ y þ 2z ¼ 0 2x þ 3y þ 3z ¼ 0 x þ 3y þ 5z ¼ 0 (c) (a) Reduce the system to echelon form: x þ 2y þ 2z s þ 3t ¼ 0 z þ 2s 2t ¼ 0 2z þ 4s 4t ¼ 0 or x þ 2y þ 2z s þ 3t ¼ 0 z þ 2s 2t ¼ 0 The system in echelon form has two (nonzero) equations in ﬁve unknowns. Hence, the system has 5 2 ¼ 3 free variables, which are y, s, t. Thus, dim W ¼ 3. We obtain a basis for W: ð1Þ Set y ¼ 1; s ¼ 0; t ¼ 0 to obtain the solution v1 ¼ ð2; 1; 0; 0; 0Þ: ð2Þ Set y ¼ 0; s ¼ 1; t ¼ 0 to obtain the solution v2 ¼ ð5; 0; 2; 1; 0Þ: ð3Þ Set y ¼ 0; s ¼ 0; t ¼ 1 to obtain the solution v3 ¼ ð7; 0; 2; 0; 1Þ: The set fv1; v2; v3g is a basis of the solution space W. (b) (Here we use the matrix format of our homogeneous system.) Reduce the coefﬁcient matrix A to echelon form: A ¼ 1 2 1 2 2 4 4 3 3 6 7 4 2 4 3 5 1 2 1 2 0 0 2 1 0 0 4 2 2 4 3 5 1 2 1 2 0 0 2 1 0 0 0 0 2 4 3 5 This corresponds to the system x þ 2y þ 2z 2t ¼ 0 2z þ t ¼ 0 The free variables are y and t, and dim W ¼ 2. (i) Set y ¼ 1, z ¼ 0 to obtain the solution u1 ¼ ð2; 1; 0; 0Þ. (ii) Set y ¼ 0, z ¼ 2 to obtain the solution u2 ¼ ð6; 0; 1; 2Þ. Then fu1; u2g is a basis of W. (c) Reduce the coefﬁcient matrix A to echelon form: A ¼ 1 1 2 2 3 3 1 3 5 2 4 3 5 1 1 2 0 1 1 0 2 3 2 4 3 5 1 1 2 0 1 1 0 0 5 2 4 3 5 This corresponds to a triangular system with no free variables. Thus, 0 is the only solution; that is, W ¼ f0g. Hence, dim W ¼ 0. 4.51. Find a homogeneous system whose solution set W is spanned by fu1; u2; u3g ¼ fð1; 2; 0; 3Þ; ð1; 1; 1; 4Þ; ð1; 0; 2; 5Þg Let v ¼ ðx; y; z; tÞ. Then v 2 W if and only if v is a linear combination of the vectors u1, u2, u3 that span W. Thus, form the matrix M whose ﬁrst columns are u1, u2, u3 and whose last column is v, and then row reduce M to echelon form. This yields M ¼ 1 1 1 x 2 1 0 y 0 1 2 z 3 4 5 t 2 6 6 4 3 7 7 5 1 1 1 x 0 1 2 2x þ y 0 1 2 z 0 1 2 3x þ t 2 6 6 4 3 7 7 5 1 1 1 x 0 1 2 2x þ y 0 0 0 2x þ y þ z 0 0 0 5x y þ t 2 6 6 4 3 7 7 5 148 CHAPTER 4 Vector Spaces Then v is a linear combination of u1, u2, u3 if rankðMÞ ¼ rankðAÞ, where A is the submatrix without column v. Thus, set the last two entries in the fourth column on the right equal to zero to obtain the required homogeneous system: 2x þ y þ z ¼ 0 5x þ y t ¼ 0 4.52. Let xi1; xi2; . . . ; xik be the free variables of a homogeneous system of linear equations with n unknowns. Let vj be the solution for which xij ¼ 1, and all other free variables equal 0. Show that the solutions v1; v2; . . . ; vk are linearly independent. Let A be the matrix whose rows are the vi. We interchange column 1 and column i1, then column 2 and column i2; . . . ; then column k and column ik, and we obtain the k n matrix B ¼ ½I; C ¼ 1 0 0 . . . 0 0 c1;kþ1 . . . c1n 0 1 0 . . . 0 0 c2;kþ1 . . . c2n ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 0 0 0 . . . 0 1 ck;kþ1 . . . ckn 2 6 6 4 3 7 7 5 The above matrix B is in echelon form, and so its rows are independent; hence, rankðBÞ ¼ k. Because A and B are column equivalent, they have the same rank—rankðAÞ ¼ k. But A has k rows; hence, these rows (i.e., the vi) are linearly independent, as claimed. Sums, Direct Sums, Intersections 4.53. Let U and W be subspaces of a vector space V. Show that (a) U þ V is a subspace of V. (b) U and W are contained in U þ W. (c) U þ W is the smallest subspace containing U and W; that is, U þ W ¼ spanðU; WÞ. (d) W þ W ¼ W. (a) Because U and W are subspaces, 0 2 U and 0 2 W. Hence, 0 ¼ 0 þ 0 belongs to U þ W. Now suppose v; v0 2 U þ W. Then v ¼ u þ w and v0 ¼ u0 þ v0, where u; u0 2 U and w; w0 2 W. Then av þ bv0 ¼ ðau þ bu0Þ þ ðaw þ bw0Þ 2 U þ W Thus, U þ W is a subspace of V. (b) Let u 2 U. Because W is a subspace, 0 2 W. Hence, u ¼ u þ 0 belongs to U þ W. Thus, U U þ W. Similarly, W U þ W. (c) Because U þ W is a subspace of V containing U and W, it must also contain the linear span of U and W. That is, spanðU; WÞ U þ W. On the other hand, if v 2 U þ W, then v ¼ u þ w ¼ 1u þ 1w, where u 2 U and w 2 W. Thus, v is a linear combination of elements in U [ W, and so v 2 spanðU; WÞ. Hence, U þ W spanðU; WÞ. The two inclusion relations give the desired result. (d) Because W is a subspace of V, we have that W is closed under vector addition; hence, W þ W W. By part (a), W W þ W. Hence, W þ W ¼ W. 4.54. Consider the following subspaces of R5: U ¼ spanðu1; u2; u3Þ ¼ spanfð1; 3; 2; 2; 3Þ; ð1; 4; 3; 4; 2Þ; ð2; 3; 1; 2; 9Þg W ¼ spanðw1; w2; w3Þ ¼ spanfð1; 3; 0; 2; 1Þ; ð1; 5; 6; 6; 3Þ; ð2; 5; 3; 2; 1Þg Find a basis and the dimension of (a) U þ W, (b) U \ W. CHAPTER 4 Vector Spaces 149 (a) U þ W is the space spanned by all six vectors. Hence, form the matrix whose rows are the given six vectors, and then row reduce to echelon form: 1 3 2 2 3 1 4 3 4 2 2 3 1 2 9 1 3 0 2 1 1 5 6 6 3 2 5 3 2 1 2 6 6 6 6 6 6 4 3 7 7 7 7 7 7 5 1 3 2 2 3 0 1 1 2 1 0 3 3 6 3 0 0 2 0 2 0 2 4 4 0 0 1 7 2 5 2 6 6 6 6 6 6 4 3 7 7 7 7 7 7 5 1 3 2 2 3 0 1 1 2 1 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 6 6 6 6 6 6 4 3 7 7 7 7 7 7 5 The following three nonzero rows of the echelon matrix form a basis of U \ W: ð1; 3; 2; 2; 2; 3Þ; ð0; 1; 1; 2; 1Þ; ð0; 0; 1; 0; 1Þ Thus, dimðU þ WÞ ¼ 3. (b) Let v ¼ ðx; y; z; s; tÞ denote an arbitrary element in R5. First ﬁnd, say as in Problem 4.49, homogeneous systems whose solution sets are U and W, respectively. Let M be the matrix whose columns are the ui and v, and reduce M to echelon form: M ¼ 1 1 2 x 3 4 3 y 2 3 1 z 2 4 2 s 3 2 9 t 2 6 6 6 6 4 3 7 7 7 7 5 1 1 2 x 0 1 3 3x þ y 0 0 0 x þ y þ z 0 0 0 4x 2y þ s 0 0 0 6x þ y þ t 2 6 6 6 6 4 3 7 7 7 7 5 Set the last three entries in the last column equal to zero to obtain the following homogeneous system whose solution set is U: x þ y þ z ¼ 0; 4x 2y þ s ¼ 0; 6x þ y þ t ¼ 0 Now let M0 be the matrix whose columns are the wi and v, and reduce M0 to echelon form: M0 ¼ 1 1 2 x 3 5 5 y 0 6 3 z 2 6 2 s 1 3 1 t 2 6 6 6 6 4 3 7 7 7 7 5 1 1 2 x 0 2 1 3x þ y 0 0 0 9x þ 3y þ z 0 0 0 4x 2y þ s 0 0 0 2x y þ t 2 6 6 6 6 4 3 7 7 7 7 5 Again set the last three entries in the last column equal to zero to obtain the following homogeneous system whose solution set is W: 9 þ 3 þ z ¼ 0; 4x 2y þ s ¼ 0; 2x y þ t ¼ 0 Combine both of the above systems to obtain a homogeneous system, whose solution space is U \ W, and reduce the system to echelon form, yielding x þ y þ z ¼ 0 2y þ 4z þ s ¼ 0 8z þ 5s þ 2t ¼ 0 s 2t ¼ 0 There is one free variable, which is t; hence, dimðU \ WÞ ¼ 1. Setting t ¼ 2, we obtain the solution u ¼ ð1; 4; 3; 4; 2Þ, which forms our required basis of U \ W. 4.55. Suppose U and W are distinct four-dimensional subspaces of a vector space V, where dim V ¼ 6. Find the possible dimensions of U \ W. Because U and W are distinct, U þ W properly contains U and W; consequently, dimðU þ WÞ > 4. But dimðU þ WÞ cannot be greater than 6, as dim V ¼ 6. Hence, we have two possibilities: (a) dimðU þ WÞ ¼ 5 or (b) dimðU þ WÞ ¼ 6. By Theorem 4.20, dimðU \ WÞ ¼ dim U þ dim W dimðU þ WÞ ¼ 8 dimðU þ WÞ Thus (a) dimðU \ WÞ ¼ 3 or (b) dimðU \ WÞ ¼ 2. 150 CHAPTER 4 Vector Spaces 4.56. Let U and W be the following subspaces of R3: U ¼ fða; b; cÞ : a ¼ b ¼ cg and W ¼ fð0; b; cÞg (Note that W is the yz-plane.) Show that R3 ¼ U W. First we show that U \ W ¼ f0g. Suppose v ¼ ða; b; cÞ 2 U \ W. Then a ¼ b ¼ c and a ¼ 0. Hence, a ¼ 0, b ¼ 0, c ¼ 0. Thus, v ¼ 0 ¼ ð0; 0; 0Þ. Next we show that R3 ¼ U þ W. For, if v ¼ ða; b; cÞ 2 R3, then v ¼ ða; a; aÞ þ ð0; b a; c aÞ where ða; a; aÞ 2 U and ð0; b a; c aÞ 2 W Both conditions U \ W ¼ f0g and U þ W ¼ R3 imply that R3 ¼ U W. 4.57. Suppose that U and W are subspaces of a vector space V and that S ¼ fuig spans U and S0 ¼ fwjg spans W. Show that S [ S0 spans U þ W. (Accordingly, by induction, if Si spans Wi, for i ¼ 1; 2; . . . ; n, then S1 [ . . . [ Sn spans W1 þ þ Wn.) Let v 2 U þ W. Then v ¼ u þ w, where u 2 U and w 2 W. Because S spans U, u is a linear combination of ui, and as S0 spans W, w is a linear combination of wj; say u ¼ a1ui1 þ a2ui2 þ þ aruir and v ¼ b1wj1 þ b2wj2 þ þ bswjs where ai; bj 2 K. Then v ¼ u þ w ¼ a1ui1 þ a2ui2 þ þ aruir þ b1wj1 þ b2wj2 þ þ bswjs Accordingly, S [ S0 ¼ fui; wjg spans U þ W. 4.58. Prove Theorem 4.20: Suppose U and V are ﬁnite-dimensional subspaces of a vector space V. Then U þ W has ﬁnite dimension and dimðU þ WÞ ¼ dim U þ dim W dimðU \ WÞ Observe that U \ W is a subspace of both U and W. Suppose dim U ¼ m, dim W ¼ n, dimðU \ WÞ ¼ r. Suppose fv1; . . . ; vrg is a basis of U \ W. By Theorem 4.16, we can extend fvig to a basis of U and to a basis of W; say fv1; . . . ; vr; u1; . . . ; umrg and fv1; . . . ; vr; w1; . . . ; wnrg are bases of U and W, respectively. Let B ¼ fv1; . . . ; vr; u1; . . . ; umr; w1; . . . ; wnrg Note that B has exactly m þ n r elements. Thus, the theorem is proved if we can show that B is a basis of U þ W. Because fvi; ujg spans U and fvi; wkg spans W, the union B ¼ fvi; uj; wkg spans U þ W. Thus, it sufﬁces to show that B is independent. Suppose a1v1 þ þ arvr þ b1u1 þ þ bmrumr þ c1w1 þ þ cnrwnr ¼ 0 ð1Þ where ai, bj, ck are scalars. Let v ¼ a1v1 þ þ arvr þ b1u1 þ þ bmrumr ð2Þ By (1), we also have v ¼ c1w1 cnrwnr ð3Þ Because fvi; ujg U, v 2 U by (2); and as fwkg W, v 2 W by (3). Accordingly, v 2 U \ W. Now fvig is a basis of U \ W, and so there exist scalars d1; . . . ; dr for which v ¼ d1v1 þ þ drvr. Thus, by (3), we have d1v1 þ þ drvr þ c1w1 þ þ cnrwnr ¼ 0 But fvi; wkg is a basis of W, and so is independent. Hence, the above equation forces c1 ¼ 0; . . . ; cnr ¼ 0. Substituting this into (1), we obtain a1v1 þ þ arvr þ b1u1 þ þ bmrumr ¼ 0 But fvi; ujg is a basis of U, and so is independent. Hence, the above equation forces a1 ¼ 0; . . . ; ar ¼ 0; b1 ¼ 0; . . . ; bmr ¼ 0. Because (1) implies that the ai, bj, ck are all 0, B ¼ fvi; uj; wkg is independent, and the theorem is proved. CHAPTER 4 Vector Spaces 151 4.59. Prove Theorem 4.21: V ¼ U W if and only if (i) V ¼ U þ W, (ii) U \ W ¼ f0g. Suppose V ¼ U W. Then any v 2 V can be uniquely written in the form v ¼ u þ w, where u 2 U and w 2 W. Thus, in particular, V ¼ U þ W. Now suppose v 2 U \ W. Then ð1Þ v ¼ v þ 0; where v 2 U; 0 2 W; ð2Þ v ¼ 0 þ v; where 0 2 U; v 2 W: Thus, v ¼ 0 þ 0 ¼ 0 and U \ W ¼ f0g. On the other hand, suppose V ¼ U þ W and U \ W ¼ f0g. Let v 2 V. Because V ¼ U þ W, there exist u 2 U and w 2 W such that v ¼ u þ w. We need to show that such a sum is unique. Suppose also that v ¼ u0 þ w0, where u0 2 U and w0 2 W. Then u þ w ¼ u0 þ w0; and so u u0 ¼ w0 w But u u0 2 U and w0 w 2 W; hence, by U \ W ¼ f0g, u u0 ¼ 0; w0 w ¼ 0; and so u ¼ u0; w ¼ w0 Thus, such a sum for v 2 V is unique, and V ¼ U W. 4.60. Prove Theorem 4.22 (for two factors): Suppose V ¼ U W. Also, suppose S ¼ fu1; . . . ; umg and S0 ¼ fw1; . . . ; wng are linearly independent subsets of U and W, respectively. Then (a) The union S [ S0 is linearly independent in V. (b) If S and S0 are bases of U and W, respectively, then S [ S0 is a basis of V. (c) dim V ¼ dim U þ dim W. (a) Suppose a1u1 þ þ amum þ b1w1 þ þ bnwn ¼ 0, where ai, bj are scalars. Then ða1u1 þ þ amumÞ þ ðb1w1 þ þ bnwnÞ ¼ 0 ¼ 0 þ 0 where 0; a1u1 þ þ amum 2 U and 0; b1w1 þ þ bnwn 2 W. Because such a sum for 0 is unique, this leads to a1u1 þ þ amum ¼ 0 and b1w1 þ þ bnwn ¼ 0 Because S1 is linearly independent, each ai ¼ 0, and because S2 is linearly independent, each bj ¼ 0. Thus, S ¼ S1 [ S2 is linearly independent. (b) By part (a), S ¼ S1 [ S2 is linearly independent, and, by Problem 4.55, S ¼ S1 [ S2 spans V ¼ U þ W. Thus, S ¼ S1 [ S2 is a basis of V. (c) This follows directly from part (b). Coordinates 4.61. Relative to the basis S ¼ fu1; u2g ¼ fð1; 1Þ; ð2; 3Þg of R2, ﬁnd the coordinate vector of v, where (a) v ¼ ð4; 3Þ, (b) v ¼ ða; bÞ. In each case, set v ¼ xu1 þ yu2 ¼ xð1; 1Þ þ yð2; 3Þ ¼ ðx þ 2y; x þ 3yÞ and then solve for x and y. (a) We have ð4; 3Þ ¼ ðx þ 2y; x þ 3yÞ or x þ 2y ¼ 4 x þ 3y ¼ 3 The solution is x ¼ 18, y ¼ 7. Hence, ½v ¼ ½18; 7. (b) We have ða; bÞ ¼ ðx þ 2y; x þ 3yÞ or x þ 2y ¼ a x þ 3y ¼ b The solution is x ¼ 3a 2b, y ¼ a þ b. Hence, ½v ¼ ½3a 2b; a þ b. 152 CHAPTER 4 Vector Spaces 4.62. Find the coordinate vector of v ¼ ða; b; cÞ in R3 relative to (a) the usual basis E ¼ fð1; 0; 0Þ; ð0; 1; 0Þ; ð0; 0; 1Þg, (b) the basis S ¼ fu1; u2; u3g ¼ fð1; 1; 1Þ; ð1; 1; 0Þ; ð1; 0; 0Þg. (a) Relative to the usual basis E, the coordinates of ½vE are the same as v. That is, ½vE ¼ ½a; b; c. (b) Set v as a linear combination of u1, u2, u3 using unknown scalars x, y, z. This yields a b c 2 4 3 5 ¼ x 1 1 1 2 4 3 5 þ y 1 1 0 2 4 3 5 þ z 1 0 0 2 4 3 5 or x þ y þ z ¼ a x þ y ¼ b x ¼ c Solving the system yields x ¼ c, y ¼ b c, z ¼ a b. Thus, ½vS ¼ ½c; b c; a b. 4.63. Consider the vector space P3ðtÞ of polynomials of degree 3. (a) Show that S ¼ fðt 1Þ3; ðt 1Þ2; t 1; 1g is a basis of P3ðtÞ. (b) Find the coordinate vector ½v of v ¼ 3t3 4t2 þ 2t 5 relative to S. (a) The degree of ðt 1Þk is k; writing the polynomials of S in reverse order, we see that no polynomial is a linear combination of preceding polynomials. Thus, the polynomials are linearly independent, and, because dim P3ðtÞ ¼ 4, they form a basis of P3ðtÞ. (b) Set v as a linear combination of the basis vectors using unknown scalars x, y, z, s. We have v ¼ 3t3 þ 4t2 þ 2t 5 ¼ xðt 1Þ3 þ yðt 1Þ2 þ zðt 1Þ þ sð1Þ ¼ xðt3 3t2 þ 3t 1Þ þ yðt2 2t þ 1Þ þ zðt 1Þ þ sð1Þ ¼ xt3 3xt2 þ 3xt x þ yt2 2yt þ y þ zt z þ s ¼ xt3 þ ð3x þ yÞt2 þ ð3x 2y þ zÞt þ ðx þ y z þ sÞ Then set coefﬁcients of the same powers of t equal to each other to obtain x ¼ 3; 3x þ y ¼ 4; 3x 2y þ z ¼ 2; x þ y z þ s ¼ 5 Solving the system yields x ¼ 3, y ¼ 13, z ¼ 19, s ¼ 4. Thus, ½v ¼ ½3; 13; 19; 4. 4.64. Find the coordinate vector of A ¼ 2 3 4 7 in the real vector space M ¼ M2;2 relative to (a) the basis S ¼ 1 1 1 1 ; 1 1 1 0 ; 1 1 0 0 ; 1 0 0 0 , (b) the usual basis E ¼ 1 0 0 0 ; 0 1 0 0 ; 0 0 1 0 ; 0 0 0 1 (a) Set A as a linear combination of the basis vectors using unknown scalars x, y, z, t as follows: A ¼ 2 3 4 7 ¼ x 1 1 1 1 þ y 1 1 1 0 þ z 1 1 0 0 þ t 1 0 0 0 ¼ x þ z þ t x y z x þ y x Set corresponding entries equal to each other to obtain the system x þ z þ t ¼ 2; x y z ¼ 3; x þ y ¼ 4; x ¼ 7 Solving the system yields x ¼ 7, y ¼ 11, z ¼ 21, t ¼ 30. Thus, ½AS ¼ ½7; 11; 21; 30. (Note that the coordinate vector of A is a vector in R4, because dim M ¼ 4.) (b) Expressing A as a linear combination of the basis matrices yields 2 3 4 7 ¼ x 1 0 0 0 þ y 0 1 0 0 þ z 0 0 1 0 þ t 0 0 0 1 ¼ x y z t Thus, x ¼ 2, y ¼ 3, z ¼ 4, t ¼ 7. Hence, ½A ¼ ½2; 3; 4; 7, whose components are the elements of A written row by row. CHAPTER 4 Vector Spaces 153 Remark: This result is true in general; that is, if A is any m n matrix in M ¼ Mm;n, then the coordinates of A relative to the usual basis of M are the elements of A written row by row. 4.65. In the space M ¼ M2;3, determine whether or not the following matrices are linearly dependent: A ¼ 1 2 3 4 0 5 ; B ¼ 2 4 7 10 1 13 ; C ¼ 1 2 5 8 2 11 If the matrices are linearly dependent, ﬁnd the dimension and a basis of the subspace W of M spanned by the matrices. The coordinate vectors of the above matrices relative to the usual basis of M are as follows: ½A ¼ ½1; 2; 3; 4; 0; 5; ½B ¼ ½2; 4; 7; 10; 1; 13; ½C ¼ ½1; 2; 5; 8; 2; 11 Form the matrix M whose rows are the above coordinate vectors, and reduce M to echelon form: M ¼ 1 2 3 4 0 5 2 4 7 10 1 13 1 2 5 8 2 11 2 4 3 5 1 2 3 4 0 5 0 0 1 2 1 3 0 0 0 0 0 0 2 4 3 5 Because the echelon matrix has only two nonzero rows, the coordinate vectors ½A, ½B, ½C span a space of dimension two, and so they are linearly dependent. Thus, A, B, C are linearly dependent. Furthermore, dim W ¼ 2, and the matrices w1 ¼ 1 2 3 4 0 5 and w2 ¼ 0 0 1 2 1 3 corresponding to the nonzero rows of the echelon matrix form a basis of W. Miscellaneous Problems 4.66. Consider a ﬁnite sequence of vectors S ¼ fv1; v2; . . . ; vng. Let T be the sequence of vectors obtained from S by one of the following ‘‘elementary operations’’: (i) interchange two vectors, (ii) multiply a vector by a nonzero scalar, (iii) add a multiple of one vector to another. Show that S and T span the same space W. Also show that T is independent if and only if S is independent. Observe that, for each operation, the vectors in T are linear combinations of vectors in S. On the other hand, each operation has an inverse of the same type (Prove!); hence, the vectors in S are linear combinations of vectors in T. Thus S and T span the same space W. Also, T is independent if and only if dim W ¼ n, and this is true if and only if S is also independent. 4.67. Let A ¼ ½aij and B ¼ ½bij be row equivalent m n matrices over a ﬁeld K, and let v1; . . . ; vn be any vectors in a vector space V over K. Let u1 ¼ a11v1 þ a12v2 þ þ a1nvn u2 ¼ a21v1 þ a22v2 þ þ a2nvn um ¼ am1v1 þ am2v2 þ þ amnvn w1 ¼ b11v1 þ b12v2 þ þ b1nvn w2 ¼ b21v1 þ b22v2 þ þ b2nvn ::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::::::::::::::::::::::::::: wm ¼ bm1v1 þ bm2v2 þ þ bmnvn Show that fuig and fwig span the same space. Applying an ‘‘elementary operation’’ of Problem 4.66 to fuig is equivalent to applying an elementary row operation to the matrix A. Because A and B are row equivalent, B can be obtained from A by a sequence of elementary row operations; hence, fwig can be obtained from fuig by the corresponding sequence of operations. Accordingly, fuig and fwig span the same space. 4.68. Let v1; . . . ; vn belong to a vector space V over K, and let P ¼ ½aij be an n-square matrix over K. Let w1 ¼ a11v1 þ a12v2 þ þ a1nvn; . . . ; wn ¼ an1v1 þ an2v2 þ þ annvn (a) Suppose P is invertible. Show that fwig and fvig span the same space; hence, fwig is independent if and only if fvig is independent. (b) Suppose P is not invertible. Show that fwig is dependent. (c) Suppose fwig is independent. Show that P is invertible. 154 CHAPTER 4 Vector Spaces (a) Because P is invertible, it is row equivalent to the identity matrix I. Hence, by Problem 4.67, fwig and fvig span the same space. Thus, one is independent if and only if the other is. (b) Because P is not invertible, it is row equivalent to a matrix with a zero row. This means that fwig spans a space that has a spanning set of less than n elements. Thus, fwig is dependent. (c) This is the contrapositive of the statement of (b), and so it follows from (b). 4.69. Suppose that A1; A2; . . . are linearly independent sets of vectors, and that A1 A2 . . .. Show that the union A ¼ A1 [ A2 [ . . . is also linearly independent. Suppose A is linearly dependent. Then there exist vectors v1; . . . ; vn 2 A and scalars a1; . . . ; an 2 K, not all of them 0, such that a1v1 þ a2v2 þ þ anvn ¼ 0 ð1Þ Because A ¼ [ Ai and the vi 2 A, there exist sets Ai1; . . . ; Ain such that v1 2 Ai1; v2 2 Ai2; . . . ; vn 2 Ain Let k be the maximum index of the sets Aij : k ¼ maxði1; . . . ; inÞ. It follows then, as A1 A2 . . . ; that each Aij is contained in Ak. Hence, v1; v2; . . . ; vn 2 Ak, and so, by (1), Ak is linearly dependent, which contradicts our hypothesis. Thus, A is linearly independent. 4.70. Let K be a subﬁeld of a ﬁeld L, and let L be a subﬁeld of a ﬁeld E. (Thus, K L E, and K is a subﬁeld of E.) Suppose E is of dimension n over L, and L is of dimension m over K. Show that E is of dimension mn over K. Suppose fv1; . . . ; vng is a basis of E over L and fa1; . . . ; amg is a basis of L over K. We claim that faivj : i ¼ 1; . . . ; m; j ¼ 1; . . . ; ng is a basis of E over K. Note that faivjg contains mn elements. Let w be any arbitrary element in E. Because fv1; . . . ; vng spans E over L, w is a linear combination of the vi with coefﬁcients in L: w ¼ b1v1 þ b2v2 þ þ bnvn; bi 2 L ð1Þ Because fa1; . . . ; amg spans L over K, each bi 2 L is a linear combination of the aj with coefﬁcients in K: b1 ¼ k11a1 þ k12a2 þ þ k1mam b2 ¼ k21a1 þ k22a2 þ þ k2mam :::::::::::::::::::::::::::::::::::::::::::::::::: bn ¼ kn1a1 þ kn2a2 þ þ kmnam where kij 2 K. Substituting in (1), we obtain w ¼ ðk11a1 þ þ k1mamÞv1 þ ðk21a1 þ þ k2mamÞv2 þ þ ðkn1a1 þ þ knmamÞvn ¼ k11a1v1 þ þ k1mamv1 þ k21a1v2 þ þ k2mamv2 þ þ kn1a1vn þ þ knmamvn ¼ P i;j kjiðaivjÞ where kji 2 K. Thus, w is a linear combination of the aivj with coefﬁcients in K; hence, faivjg spans E over K. The proof is complete if we show that faivjg is linearly independent over K. Suppose, for scalars xji 2 K; we have P i;j xjiðaivjÞ ¼ 0; that is, ðx11a1v1 þ x12a2v1 þ þ x1mamv1Þ þ þ ðxn1a1vn þ xn2a2vn þ þ xnmamvmÞ ¼ 0 or ðx11a1 þ x12a2 þ þ x1mamÞv1 þ þ ðxn1a1 þ xn2a2 þ þ xnmamÞvn ¼ 0 Because fv1; . . . ; vng is linearly independent over L and the above coefﬁcients of the vi belong to L, each coefﬁcient must be 0: x11a1 þ x12a2 þ þ x1mam ¼ 0; . . . ; xn1a1 þ xn2a2 þ þ xnmam ¼ 0 CHAPTER 4 Vector Spaces 155 But fa1; . . . ; amg is linearly independent over K; hence, because the xji 2 K, x11 ¼ 0; x12 ¼ 0; . . . ; x1m ¼ 0; . . . ; xn1 ¼ 0; xn2 ¼ 0; . . . ; xnm ¼ 0 Accordingly, faivjg is linearly independent over K, and the theorem is proved. SUPPLEMENTARY PROBLEMS Vector Spaces 4.71. Suppose u and v belong to a vector space V. Simplify each of the following expressions: (a) E1 ¼ 4ð5u 6vÞ þ 2ð3u þ vÞ, (c) E3 ¼ 6ð3u þ 2vÞ þ 5u 7v, (b) E2 ¼ 5ð2u 3vÞ þ 4ð7v þ 8Þ, (d) E4 ¼ 3ð5u þ 2=vÞ: 4.72. Let V be the set of ordered pairs (a; b) of real numbers with addition in V and scalar multiplication on V deﬁned by ða; bÞ þ ðc; dÞ ¼ ða þ c; b þ dÞ and kða; bÞ ¼ ðka; 0Þ Show that V satisﬁes all the axioms of a vector space except [M4]—that is, except 1u ¼ u. Hence, [M4] is not a consequence of the other axioms. 4.73. Show that Axiom [A4] of a vector space V (that u þ v ¼ v þ u) can be derived from the other axioms for V. 4.74. Let V be the set of ordered pairs (a; b) of real numbers. Show that V is not a vector space over R with addition and scalar multiplication deﬁned by (i) ða; bÞ þ ðc; dÞ ¼ ða þ d; b þ cÞ and kða; bÞ ¼ ðka; kbÞ, (ii) ða; bÞ þ ðc; dÞ ¼ ða þ c; b þ dÞ and kða; bÞ ¼ ða; bÞ, (iii) ða; bÞ þ ðc; dÞ ¼ ð0; 0Þ and kða; bÞ ¼ ðka; kbÞ, (iv) ða; bÞ þ ðc; dÞ ¼ ðac; bdÞ and kða; bÞ ¼ ðka; kbÞ. 4.75. Let V be the set of inﬁnite sequences (a1; a2; . . .) in a ﬁeld K. Show that V is a vector space over K with addition and scalar multiplication deﬁned by ða1; a2; . . .Þ þ ðb1; b2; . . .Þ ¼ ða1 þ b1; a2 þ b2; . . .Þ and kða1; a2; . . .Þ ¼ ðka1; ka2; . . .Þ 4.76. Let U and W be vector spaces over a ﬁeld K. Let V be the set of ordered pairs (u; w) where u 2 U and w 2 W. Show that V is a vector space over K with addition in V and scalar multiplication on V deﬁned by ðu; wÞ þ ðu0; w0Þ ¼ ðu þ u0; w þ w0Þ and kðu; wÞ ¼ ðku; kwÞ (This space V is called the external direct product of U and W.) Subspaces 4.77. Determine whether or not W is a subspace of R3 where W consists of all vectors (a; b; c) in R3 such that (a) a ¼ 3b, (b) a b c, (c) ab ¼ 0, (d) a þ b þ c ¼ 0, (e) b ¼ a2, ( f ) a ¼ 2b ¼ 3c. 4.78. Let V be the vector space of n-square matrices over a ﬁeld K. Show that W is a subspace of V if W consists of all matrices A ¼ ½aij that are (a) symmetric (AT ¼ A or aij ¼ aji), (b) (upper) triangular, (c) diagonal, (d) scalar. 4.79. Let AX ¼ B be a nonhomogeneous system of linear equations in n unknowns; that is, B 6¼ 0. Show that the solution set is not a subspace of Kn. 4.80. Suppose U and W are subspaces of V for which U [ W is a subspace. Show that U W or W U. 4.81. Let V be the vector space of all functions from the real ﬁeld R into R. Show that W is a subspace of V where W consists of all: (a) bounded functions, (b) even functions. [Recall that f : R ! R is bounded if 9M 2 R such that 8x 2 R, we have j f ðxÞj M; and f ðxÞ is even if f ðxÞ ¼ f ðxÞ; 8x 2 R.] 156 CHAPTER 4 Vector Spaces 4.82. Let V be the vector space (Problem 4.75) of inﬁnite sequences (a1; a2; . . .) in a ﬁeld K. Show that W is a subspace of V if W consists of all sequences with (a) 0 as the ﬁrst element, (b) only a ﬁnite number of nonzero elements. Linear Combinations, Linear Spans 4.83. Consider the vectors u ¼ ð1; 2; 3Þ and v ¼ ð2; 3; 1Þ in R3. (a) Write w ¼ ð1; 3; 8Þ as a linear combination of u and v. (b) Write w ¼ ð2; 4; 5Þ as a linear combination of u and v. (c) Find k so that w ¼ ð1; k; 4Þ is a linear combination of u and v. (d) Find conditions on a, b, c so that w ¼ ða; b; cÞ is a linear combination of u and v. 4.84. Write the polynomial f ðtÞ ¼ at2 þ bt þ c as a linear combination of the polynomials p1 ¼ ðt 1Þ2, p2 ¼ t 1, p3 ¼ 1. [Thus, p1, p2, p3 span the space P2ðtÞ of polynomials of degree 2.] 4.85. Find one vector in R3 that spans the intersection of U and W where U is the xy-plane—that is, U ¼ fða; b; 0Þg—and W is the space spanned by the vectors (1, 1, 1) and (1, 2, 3). 4.86. Prove that span(S) is the intersection of all subspaces of V containing S. 4.87. Show that spanðSÞ ¼ spanðS [ f0gÞ. That is, by joining or deleting the zero vector from a set, we do not change the space spanned by the set. 4.88. Show that (a) If S T, then spanðSÞ spanðTÞ. (b) span½spanðSÞ ¼ spanðSÞ. Linear Dependence and Linear Independence 4.89. Determine whether the following vectors in R4 are linearly dependent or independent: (a) ð1; 2; 3; 1Þ, ð3; 7; 1; 2Þ, ð1; 3; 7; 4Þ; (b) ð1; 3; 1; 2Þ, ð2; 5; 1; 3Þ, ð1; 3; 7; 2Þ. 4.90. Determine whether the following polynomials u, v, w in PðtÞ are linearly dependent or independent: (a) u ¼ t3 4t2 þ 3t þ 3, v ¼ t3 þ 2t2 þ 4t 1, w ¼ 2t3 t2 3t þ 5; (b) u ¼ t3 5t2 2t þ 3, v ¼ t3 4t2 3t þ 4, w ¼ 2t3 17t2 7t þ 9. 4.91. Show that the following functions f , g, h are linearly independent: (a) f ðtÞ ¼ et, gðtÞ ¼ sin t, hðtÞ ¼ t2; (b) f ðtÞ ¼ et, gðtÞ ¼ e2t, hðtÞ ¼ t. 4.92. Show that u ¼ ða; bÞ and v ¼ ðc; dÞ in K2 are linearly dependent if and only if ad bc ¼ 0. 4.93. Suppose u, v, w are linearly independent vectors. Prove that S is linearly independent where (a) S ¼ fu þ v 2w; u v w; u þ wg; (b) S ¼ fu þ v 3w; u þ 3v w; v þ wg. 4.94. Suppose fu1; . . . ; ur; w1; . . . ; wsg is a linearly independent subset of V. Show that spanðuiÞ \ spanðwjÞ ¼ f0g 4.95. Suppose v1; v2; . . . ; vn are linearly independent. Prove that S is linearly independent where (a) S ¼ fa1v1; a2v2; . . . ; anvng and each ai 6¼ 0. (b) S ¼ fv1; . . . ; vk1; w; vkþ1; . . . ; vng and w ¼ P i bivi and bk 6¼ 0. 4.96. Suppose ða11; . . . ; a1nÞ; ða21; . . . ; a2nÞ; . . . ; ðam1; . . . ; amnÞ are linearly independent vectors in Kn, and suppose v1; v2; . . . ; vn are linearly independent vectors in a vector space V over K. Show that the following CHAPTER 4 Vector Spaces 157 vectors are also linearly independent: w1 ¼ a11v1 þ þ a1nvn; w2 ¼ a21v1 þ þ a2nvn; . . . ; wm ¼ am1v1 þ þ amnvn Basis and Dimension 4.97. Find a subset of u1, u2, u3, u4 that gives a basis for W ¼ spanðuiÞ of R5, where (a) u1 ¼ ð1; 1; 1; 2; 3Þ, u2 ¼ ð1; 2; 1; 2; 1Þ, u3 ¼ ð3; 5; 1; 2; 5Þ, u4 ¼ ð1; 2; 1; 1; 4Þ (b) u1 ¼ ð1; 2; 1; 3; 1Þ, u2 ¼ ð2; 4; 2; 6; 2Þ, u3 ¼ ð1; 3; 1; 2; 1Þ, u4 ¼ ð3; 7; 3; 8; 1Þ (c) u1 ¼ ð1; 0; 1; 0; 1Þ, u2 ¼ ð1; 1; 2; 1; 0Þ, u3 ¼ ð2; 1; 3; 1; 1Þ, u4 ¼ ð1; 2; 1; 1; 1Þ (d) u1 ¼ ð1; 0; 1; 1; 1Þ, u2 ¼ ð2; 1; 2; 0; 1Þ, u3 ¼ ð1; 1; 2; 3; 4Þ, u4 ¼ ð4; 2; 5; 4; 6Þ 4.98. Consider the subspaces U ¼ fða; b; c; dÞ : b 2c þ d ¼ 0g and W ¼ fða; b; c; dÞ : a ¼ d; b ¼ 2cg of R4. Find a basis and the dimension of (a) U, (b) W, (c) U \ W. 4.99. Find a basis and the dimension of the solution space W of each of the following homogeneous systems: ðaÞ x þ 2y 2z þ 2s t ¼ 0 x þ 2y z þ 3s 2t ¼ 0 2x þ 4y 7z þ s þ t ¼ 0 ðbÞ x þ 2y z þ 3s 4t ¼ 0 2x þ 4y 2z s þ 5t ¼ 0 2x þ 4y 2z þ 4s 2t ¼ 0 4.100. Find a homogeneous system whose solution space is spanned by the following sets of three vectors: (a) ð1; 2; 0; 3; 1Þ, ð2; 3; 2; 5; 3Þ, ð1; 2; 1; 2; 2Þ; (b) (1, 1, 2, 1, 1), (1, 2, 1, 4, 3), (3, 5, 4, 9, 7). 4.101. Determine whether each of the following is a basis of the vector space PnðtÞ: (a) f1; 1 þ t; 1 þ t þ t2; 1 þ t þ t2 þ t3; . . . ; 1 þ t þ t2 þ þ tn1 þ tng; (b) f1 þ t; t þ t2; t2 þ t3; . . . ; tn2 þ tn1; tn1 þ tng: 4.102. Find a basis and the dimension of the subspace W of PðtÞ spanned by (a) u ¼ t3 þ 2t2 2t þ 1, v ¼ t3 þ 3t2 3t þ 4, w ¼ 2t3 þ t2 7t 7, (b) u ¼ t3 þ t2 3t þ 2, v ¼ 2t3 þ t2 þ t 4, w ¼ 4t3 þ 3t2 5t þ 2. 4.103. Find a basis and the dimension of the subspace W of V ¼ M2;2 spanned by A ¼ 1 5 4 2 ; B ¼ 1 1 1 5 ; C ¼ 2 4 5 7 ; D ¼ 1 7 5 1 Rank of a Matrix, Row and Column Spaces 4.104. Find the rank of each of the following matrices: (a) 1 3 2 5 4 1 4 1 3 5 1 4 2 4 3 2 7 3 6 13 2 6 6 4 3 7 7 5, (b) 1 2 3 2 1 3 2 0 3 8 7 2 2 1 9 10 2 6 6 4 3 7 7 5, (c) 1 1 2 4 5 5 5 8 1 1 2 2 2 6 6 4 3 7 7 5 4.105. For k ¼ 1; 2; . . . ; 5, ﬁnd the number nk of linearly independent subsets consisting of k columns for each of the following matrices: (a) A ¼ 1 1 0 2 3 1 2 0 2 5 1 3 0 2 7 2 4 3 5, (b) B ¼ 1 2 1 0 2 1 2 3 0 4 1 1 5 0 6 2 4 3 5 158 CHAPTER 4 Vector Spaces 4.106. Let (a) A ¼ 1 2 1 3 1 6 2 4 3 8 3 15 1 2 2 5 3 11 4 8 6 16 7 32 2 6 6 4 3 7 7 5, (b) B ¼ 1 2 2 1 2 1 2 4 5 4 5 5 1 2 3 4 4 6 3 6 7 7 9 10 2 6 6 4 3 7 7 5 For each matrix (where C1; . . . ; C6 denote its columns): (i) Find its row canonical form M. (ii) Find the columns that are linear combinations of preceding columns. (iii) Find columns (excluding C6) that form a basis for the column space. (iv) Express C6 as a linear combination of the basis vectors obtained in (iii). 4.107. Determine which of the following matrices have the same row space: A ¼ 1 2 1 3 4 5 ; B ¼ 1 1 2 2 3 1 ; C ¼ 1 1 3 2 1 10 3 5 1 2 4 3 5 4.108. Determine which of the following subspaces of R3 are identical: U1 ¼ span½ð1; 1; 1Þ; ð2; 3; 1Þ; ð3; 1; 5Þ; U2 ¼ span½ð1; 1; 3Þ; ð3; 2; 8Þ; ð2; 1; 3Þ U3 ¼ span½ð1; 1; 1Þ; ð1; 1; 3Þ; ð3; 1; 7Þ 4.109. Determine which of the following subspaces of R4 are identical: U1 ¼ span½ð1; 2; 1; 4Þ; ð2; 4; 1; 5Þ; ð3; 6; 2; 9Þ; U2 ¼ span½ð1; 2; 1; 2Þ; ð2; 4; 1; 3Þ; U3 ¼ span½ð1; 2; 3; 10Þ; ð2; 4; 3; 11Þ 4.110. Find a basis for (i) the row space and (ii) the column space of each matrix M: (a) M ¼ 0 0 3 1 4 1 3 1 2 1 3 9 4 5 2 4 12 8 8 7 2 6 6 4 3 7 7 5, (b) M ¼ 1 2 1 0 1 1 2 2 1 3 3 6 5 2 7 2 4 1 1 0 2 6 6 4 3 7 7 5. 4.111. Show that if any row is deleted from a matrix in echelon (respectively, row canonical) form, then the resulting matrix is still in echelon (respectively, row canonical) form. 4.112. Let A and B be arbitrary m n matrices. Show that rankðA þ BÞ rankðAÞ þ rankðBÞ. 4.113. Let r ¼ rankðA þ BÞ. Find 2 2 matrices A and B such that (a) r < rankðAÞ, rank(B); (b) r ¼ rankðAÞ ¼ rankðBÞ; (c) r > rankðAÞ, rank(B). Sums, Direct Sums, Intersections 4.114. Suppose U and W are two-dimensional subspaces of K3. Show that U \ W 6¼ f0g. 4.115. Suppose U and W are subspaces of V such that dim U ¼ 4, dim W ¼ 5, and dim V ¼ 7. Find the possible dimensions of U \ W. 4.116. Let U and W be subspaces of R3 for which dim U ¼ 1, dim W ¼ 2, and U 6 W. Show that R3 ¼ U W. 4.117. Consider the following subspaces of R5: U ¼ span½ð1; 1; 1; 2; 0Þ; ð1; 2; 2; 0; 3Þ; ð1; 1; 2; 2; 1Þ W ¼ span½ð1; 2; 3; 0; 2Þ; ð1; 1; 3; 2; 4Þ; ð1; 1; 2; 2; 5Þ CHAPTER 4 Vector Spaces 159 (a) Find two homogeneous systems whose solution spaces are U and W, respectively. (b) Find a basis and the dimension of U \ W. 4.118. Let U1, U2, U3 be the following subspaces of R3: U1 ¼ fða; b; cÞ : a ¼ cg; U2 ¼ fða; b; cÞ : a þ b þ c ¼ 0g; U3 ¼ fð0; 0; cÞg Show that (a) R3 ¼ U1 þ U2, (b) R3 ¼ U2 þ U3, (c) R3 ¼ U1 þ U3. When is the sum direct? 4.119. Suppose U, W1, W2 are subspaces of a vector space V. Show that ðU \ W1Þ þ ðU \ W2Þ U \ ðW1 þ W2Þ Find subspaces of R2 for which equality does not hold. 4.120. Suppose W1; W2; . . . ; Wr are subspaces of a vector space V. Show that (a) spanðW1; W2; . . . ; WrÞ ¼ W1 þ W2 þ þ Wr. (b) If Si spans Wi for i ¼ 1; . . . ; r, then S1 [ S2 [ [ Sr spans W1 þ W2 þ þ Wr. 4.121. Suppose V ¼ U W. Show that dim V ¼ dim U þ dim W. 4.122. Let S and T be arbitrary nonempty subsets (not necessarily subspaces) of a vector space V and let k be a scalar. The sum S þ T and the scalar product kS are deﬁned by S þ T ¼ ðu þ v : u 2 S; v 2 Tg; kS ¼ fku : u 2 Sg [We also write w þ S for fwg þ S.] Let S ¼ fð1; 2Þ; ð2; 3Þg; T ¼ fð1; 4Þ; ð1; 5Þ; ð2; 5Þg; w ¼ ð1; 1Þ; k ¼ 3 Find: (a) S þ T, (b) w þ S, (c) kS, (d) kT, (e) kS þ kT, (f) kðS þ TÞ. 4.123. Show that the above operations of S þ T and kS satisfy (a) Commutative law: S þ T ¼ T þ S. (b) Associative law: ðS1 þ S2Þ þ S3 ¼ S1 þ ðS2 þ S3Þ. (c) Distributive law: kðS þ TÞ ¼ kS þ kT. (d) S þ f0g ¼ f0g þ S ¼ S and S þ V ¼ V þ S ¼ V. 4.124. Let V be the vector space of n-square matrices. Let U be the subspace of upper triangular matrices, and let W be the subspace of lower triangular matrices. Find (a) U \ W, (b) U þ W. 4.125. Let V be the external direct sum of vector spaces U and W over a ﬁeld K. (See Problem 4.76.) Let ^ U ¼ fðu; 0Þ : u 2 Ug and ^ W ¼ fð0; wÞ : w 2 Wg Show that (a) ^ U and ^ W are subspaces of V, (b) V ¼ ^ U ^ W. 4.126. Suppose V ¼ U þ W. Let ^ V be the external direct sum of U and W. Show that V is isomorphic to ^ V under the correspondence v ¼ u þ w $ ðu; wÞ. 4.127. Use induction to prove (a) Theorem 4.22, (b) Theorem 4.23. Coordinates 4.128. The vectors u1 ¼ ð1; 2Þ and u2 ¼ ð4; 7Þ form a basis S of R2. Find the coordinate vector ½v of v relative to S where (a) v ¼ ð5; 3Þ, (b) v ¼ ða; bÞ. 4.129. The vectors u1 ¼ ð1; 2; 0Þ, u2 ¼ ð1; 3; 2Þ, u3 ¼ ð0; 1; 3Þ form a basis S of R3. Find the coordinate vector ½v of v relative to S where (a) v ¼ ð2; 7; 4Þ, (b) v ¼ ða; b; cÞ. 160 CHAPTER 4 Vector Spaces 4.130. S ¼ ft3 þ t2; t2 þ t; t þ 1; 1g is a basis of P3ðtÞ. Find the coordinate vector ½v of v relative to S where (a) v ¼ 2t3 þ t2 4t þ 2, (b) v ¼ at3 þ bt2 þ ct þ d. 4.131. Let V ¼ M2;2. Find the coordinate vector [A] of A relative to S where S ¼ 1 1 1 1 ; 1 1 1 0 ; 1 1 0 0 ; 1 0 0 0 and ðaÞ A ¼ 3 5 6 7 ; ðbÞ A ¼ a b c d 4.132. Find the dimension and a basis of the subspace W of P3ðtÞ spanned by u ¼ t3 þ 2t2 3t þ 4; v ¼ 2t3 þ 5t2 4t þ 7; w ¼ t3 þ 4t2 þ t þ 2 4.133. Find the dimension and a basis of the subspace W of M ¼ M2;3 spanned by A ¼ 1 2 1 3 1 2 ; B ¼ 2 4 3 7 5 6 ; C ¼ 1 2 3 5 7 6 Miscellaneous Problems 4.134. Answer true or false. If false, prove it with a counterexample. (a) If u1, u2, u3 span V, then dim V ¼ 3. (b) If A is a 4 8 matrix, then any six columns are linearly dependent. (c) If u1, u2, u3 are linearly independent, then u1, u2, u3, w are linearly dependent. (d) If u1, u2, u3, u4 are linearly independent, then dim V 4. (e) If u1, u2, u3 span V, then w, u1, u2, u3 span V. (f) If u1, u2, u3, u4 are linearly independent, then u1, u2, u3 are linearly independent. 4.135. Answer true or false. If false, prove it with a counterexample. (a) If any column is deleted from a matrix in echelon form, then the resulting matrix is still in echelon form. (b) If any column is deleted from a matrix in row canonical form, then the resulting matrix is still in row canonical form. (c) If any column without a pivot is deleted from a matrix in row canonical form, then the resulting matrix is in row canonical form. 4.136. Determine the dimension of the vector space W of the following n-square matrices: (a) symmetric matrices, (b) antisymmetric matrices, (d) diagonal matrices, (c) scalar matrices. 4.137. Let t1; t2; . . . ; tn be symbols, and let K be any ﬁeld. Let V be the following set of expressions where ai 2 K: a1t1 þ a2t2 þ þ antn Deﬁne addition in V and scalar multiplication on V by ða1t1 þ þ antnÞ þ ðb1t1 þ þ bntnÞ ¼ ða1 þ b1Þt1 þ þ ðanbnmÞtn kða1t1 þ a2t2 þ þ antnÞ ¼ ka1t1 þ ka2t2 þ þ kantn Show that V is a vector space over K with the above operations. Also, show that ft1; . . . ; tng is a basis of V, where tj ¼ 0t1 þ þ 0tj1 þ 1tj þ 0tjþ1 þ þ 0tn CHAPTER 4 Vector Spaces 161 ANSWERS TO SUPPLEMENTARY PROBLEMS [Some answers, such as bases, need not be unique.] 4.71. (a) E1 ¼ 26u 22v; (b) The sum 7v þ 8 is not deﬁned, so E2 is not deﬁned; (c) E3 ¼ 23u þ 5v; (d) Division by v is not deﬁned, so E4 is not deﬁned. 4.77. (a) Yes; (b) No; e.g., ð1; 2; 3Þ 2 W but 2ð1; 2; 3Þ 62 W; (c) No; e.g., ð1; 0; 0Þ; ð0; 1; 0Þ 2 W, but not their sum; (d) Yes; (e) No; e.g., ð1; 1; 1Þ 2 W, but 2ð1; 1; 1Þ 62 W; (f) Yes 4.79. The zero vector 0 is not a solution. 4.83. (a) w ¼ 3u1 u2, (b) Impossible, (c) k ¼ 11 5 , (d) 7a 5b þ c ¼ 0 4.84. Using f ¼ xp1 þ yp2 þ zp3, we get x ¼ a, y ¼ 2a þ b, z ¼ a þ b þ c 4.85. v ¼ ð2; 1; 0Þ 4.89. (a) Dependent, (b) Independent 4.90. (a) Independent, (b) Dependent 4.97. (a) u1, u2, u4; (b) u1, u2, u3; (c) u1, u2, u4; (d) u1, u2, u3 4.98. (a) dim U ¼ 3, (b) dim W ¼ 2, (c) dimðU \ WÞ ¼ 1 4.99. (a) Basis: fð2; 1; 0; 0; 0Þ; ð4; 0; 1; 1; 0Þ; ð3; 0; 1; 0; 1Þg; dim W ¼ 3; (b) Basis: fð2; 1; 0; 0; 0Þ; ð1; 0; 1; 0; 0Þg; dim W ¼ 2 4.100. (a) 5x þ y z s ¼ 0; x þ y z t ¼ 0; (b) 3x y z ¼ 0; 2x 3y þ s ¼ 0; x 2y þ t ¼ 0 4.101. (a) Yes, (b) No, because dim PnðtÞ ¼ n þ 1, but the set contains only n elements. 4.102. (a) dim W ¼ 2, (b) dim W ¼ 3 4.103. dim W ¼ 2 4.104. (a) 3, (b) 2, (c) 3 4.105. (a) n1 ¼ 4; n2 ¼ 5; n3 ¼ n4 ¼ n5 ¼ 0; (b) n1 ¼ 4; n2 ¼ 6; n3 ¼ 3; n4 ¼ n5 ¼ 0 4.106. (a) (i) M ¼ ½1; 2; 0; 1; 0; 3; 0; 0; 1; 2; 0; 1; 0; 0; 0; 0; 1; 2; 0; (ii) C2, C4, C6; (iii) C1, C3, C5; (iv) C6 ¼ 3C1 þ C3 þ 2C5. (b) (i) M ¼ ½1; 2; 0; 0; 3; 1; 0; 0; 1; 0; 1; 1; 0; 0; 0; 1; 1; 2; 0; (ii) C2, C5, C6; (iii) C1, C3, C4; (iv) C6 ¼ C1 C3 þ 2C4 4.107. A and C are row equivalent to 1 0 7 0 1 4 , but not B 4.108. U1 and U2 are row equivalent to 1 0 2 0 1 1 , but not U3 4.109. U1 and U3 are row equivalent to 1 2 0 1 0 0 1 3 ; but not U2 4.110. (a) (i) ð1; 3; 1; 2; 1Þ, ð0; 0; 1; 1; 1Þ, ð0; 0; 0; 4; 7Þ; (ii) C1, C3, C4; (b) (i) ð1; 2; 1; 0; 1Þ, ð0; 0; 1; 1; 2Þ; (ii) C1, C3 162 CHAPTER 4 Vector Spaces 4.113. (a) A ¼ 1 1 0 0 ; B ¼ 1 1 0 0 ; (b) A ¼ 1 0 0 0 ; B ¼ 0 2 0 0 ; (c) A ¼ 1 0 0 0 ; B ¼ 0 0 0 1 4.115. dimðU \ WÞ ¼ 2, 3, or 4 4.117. (a) (i) 3x þ 4y z t ¼ 0 4x þ 2y þ s ¼ 0 (ii) 4x þ 2y s ¼ 0 9x þ 2y þ z þ t ¼ 0 ; (b) Basis: fð1; 2; 5; 0; 0Þ; ð0; 0; 1; 0; 1Þg; dimðU \ WÞ ¼ 2 4.118. The sum is direct in (b) and (c). 4.119. In R2, let U, V, W be, respectively, the line y ¼ x, the x-axis, the y-axis. 4.122. (a) fð2; 6Þ; ð2; 7Þ; ð3; 7Þ; ð3; 8Þ; ð4; 8Þg; (b) fð2; 3Þ; ð3; 4Þg; (c) fð3; 6Þ; ð6; 9Þg; (d) fð3; 12Þ; ð3; 15Þ; ð6; 15Þg; (e and f) fð6; 18Þ; ð6; 21Þ; ð9; 21Þ; ð9; 24Þ; ð12; 24Þg 4.124. (a) Diagonal matrices, (b) V 4.128. (a) [41; 11], (b) [7a 4b; 2a þ b] 4.129. (a) [11; 13; 10], (b) [c 3b þ 7a; c þ 3b 6a; c 2b þ 4a] 4.130. (a) [2; 1; 2; 2], (b) [a; b c; c b þ a; d c þ b a] 4.131. (a) [7; 1; 13; 10], (b) [d; c d; b þ c 2d; a b 2c þ 2d] 4.132. dim W ¼ 2; basis: ft3 þ 2t2 3t þ 4; t2 þ 2t 1g 4.133. dim W ¼ 2; basis: f½1; 2; 1; 3; 1; 2; ½0; 0; 1; 1; 3; 2g 4.134. (a) False; (1, 1), (1, 2), (2, 1) span R2; (b) True; (c) False; (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), w ¼ ð0; 0; 0; 1Þ; (d) True; (e) True; (f) True 4.135. (a) True; (b) False; e.g. delete C2 from 1 0 3 0 1 2 ; (c) True 4.136. (a) 1 2 nðn þ 1Þ, (b) 1 2 nðn 1Þ, (c) n, (d) 1 CHAPTER 4 Vector Spaces 163 Linear Mappings 5.1 Introduction The main subject matter of linear algebra is the study of linear mappings and their representation by means of matrices. This chapter introduces us to these linear maps and Chapter 6 shows how they can be represented by matrices. First, however, we begin with a study of mappings in general. 5.2 Mappings, Functions Let A and B be arbitrary nonempty sets. Suppose to each element in a 2 A there is assigned a unique element of B; called the image of a. The collection f of such assignments is called a mapping (or map) from A into B, and it is denoted by f : A ! B The set A is called the domain of the mapping, and B is called the target set. We write f ðaÞ, read ‘‘f of a;’’ for the unique element of B that f assigns to a 2 A. One may also view a mapping f : A ! B as a computer that, for each input value a 2 A, produces a unique output f ðaÞ 2 B. Remark: The term function is used synonymously with the word mapping, although some texts reserve the word ‘‘function’’ for a real-valued or complex-valued mapping. Consider a mapping f : A ! B. If A0 is any subset of A, then f ðA0Þ denotes the set of images of elements of A0; and if B0 is any subset of B, then f 1ðB0Þ denotes the set of elements of A; each of whose image lies in B. That is, f ðA0Þ ¼ f f ðaÞ : a 2 A0g and f 1ðB0Þ ¼ fa 2 A : f ðaÞ 2 B0g We call f ðA0) the image of A0 and f 1ðB0Þ the inverse image or preimage of B0. In particular, the set of all images (i.e., f ðAÞ) is called the image or range of f. To each mapping f : A ! B there corresponds the subset of A B given by fða; f ðaÞÞ : a 2 Ag. We call this set the graph of f . Two mappings f : A ! B and g : A ! B are deﬁned to be equal, written f ¼ g, if f ðaÞ ¼ gðaÞ for every a 2 A—that is, if they have the same graph. Thus, we do not distinguish between a function and its graph. The negation of f ¼ g is written f 6¼ g and is the statement: There exists an a 2 A for which f ðaÞ 6¼ gðaÞ: Sometimes the ‘‘barred’’ arrow 7! is used to denote the image of an arbitrary element x 2 A under a mapping f : A ! B by writing x 7! f ðxÞ This is illustrated in the following example. 164 CHAPTER 5 EXAMPLE 5.1 (a) Let f : R ! R be the function that assigns to each real number x its square x2. We can denote this function by writing f ðxÞ ¼ x2 or x 7! x2 Here the image of 3 is 9, so we may write f ð3Þ ¼ 9. However, f 1ð9Þ ¼ f3; 3g. Also, f ðRÞ ¼ ½0; 1Þ ¼ fx : x 0g is the image of f. (b) Let A ¼ fa; b; c; dg and B ¼ fx; y; z; tg. Then the following deﬁnes a mapping f : A ! B: f ðaÞ ¼ y; f ðbÞ ¼ x; f ðcÞ ¼ z; f ðdÞ ¼ y or f ¼ fða; yÞ; ðb; xÞ; ðc; zÞ; ðd; yÞg The ﬁrst deﬁnes the mapping explicitly, and the second deﬁnes the mapping by its graph. Here, f ðfa; b; dgÞ ¼ f f ðaÞ; f ðbÞ; f ðdÞg ¼ fy; x; yg ¼ fx; yg Furthermore, f ðAÞ ¼ fx; y; zg is the image of f. EXAMPLE 5.2 Let V be the vector space of polynomials over R, and let pðtÞ ¼ 3t2 5t þ 2. (a) The derivative deﬁnes a mapping D : V ! V where, for any polynomials f ðtÞ, we have Dð f Þ ¼ df =dt. Thus, DðpÞ ¼ Dð3t2 5t þ 2Þ ¼ 6t 5 (b) The integral, say from 0 to 1, deﬁnes a mapping J : V ! R. That is, for any polynomial f ðtÞ, Jð f Þ ¼ ð1 0 f ðtÞ dt; and so JðpÞ ¼ ð1 0 ð3t2 5t þ 2Þ ¼ 1 2 Observe that the mapping in (b) is from the vector space V into the scalar ﬁeld R, whereas the mapping in (a) is from the vector space V into itself. Matrix Mappings Let A be any m n matrix over K. Then A determines a mapping FA : Kn ! Km by FAðuÞ ¼ Au where the vectors in Kn and Km are written as columns. For example, suppose A ¼ 1 4 5 2 3 6 and u ¼ 1 3 5 2 4 3 5 then FAðuÞ ¼ Au ¼ 1 4 5 2 3 6 1 3 5 2 4 3 5 ¼ 36 41 Remark: For notational convenience, we will frequently denote the mapping FA by the letter A, the same symbol as used for the matrix. Composition of Mappings Consider two mappings f : A ! B and g : B ! C, illustrated below: A ! f B ! g C The composition of f and g, denoted by g f , is the mapping g f : A ! C deﬁned by ðg f ÞðaÞ gð f ðaÞÞ CHAPTER 5 Linear Mappings 165 That is, ﬁrst we apply f to a 2 A, and then we apply g to f ðaÞ 2 B to get gð f ðaÞÞ 2 C. Viewing f and g as ‘‘computers,’’ the composition means we ﬁrst input a 2 A to get the output f ðaÞ 2 B using f , and then we input f ðaÞ to get the output gð f ðaÞÞ 2 C using g. Our ﬁrst theorem tells us that the composition of mappings satisﬁes the associative law. THEOREM 5.1: Let f : A ! B, g : B ! C, h : C ! D. Then h ðg f Þ ¼ ðh gÞ f We prove this theorem here. Let a 2 A. Then ðh ðg f ÞÞðaÞ ¼ hððg f ÞðaÞÞ ¼ hðgð f ðaÞÞÞ ððh gÞ f ÞðaÞ ¼ ðh gÞð f ðaÞÞ ¼ hðgð f ðaÞÞÞ Thus, ðh ðg f ÞÞðaÞ ¼ ððh gÞ f ÞðaÞ for every a 2 A, and so h ðg f Þ ¼ ðh gÞ f. One-to-One and Onto Mappings We formally introduce some special types of mappings. DEFINITION: A mapping f : A ! B is said to be one-to-one (or 1-1 or injective) if different elements of A have distinct images; that is, If f ðaÞ ¼ f ða0Þ; then a ¼ a0: DEFINITION: A mapping f : A ! B is said to be onto (or f maps A onto B or surjective) if every b 2 B is the image of at least one a 2 A. DEFINITION: A mapping f : A ! B is said to be a one-to-one correspondence between A and B (or bijective) if f is both one-to-one and onto. EXAMPLE 5.3 Let f : R ! R, g : R ! R, h : R ! R be deﬁned by f ðxÞ ¼ 2x; gðxÞ ¼ x3 x; hðxÞ ¼ x2 The graphs of these functions are shown in Fig. 5-1. The function f is one-to-one. Geometrically, this means that each horizontal line does not contain more than one point of f. The function g is onto. Geometrically, this means that each horizontal line contains at least one point of g. The function h is neither one-to-one nor onto. For example, both 2 and 2 have the same image 4, and 16 has no preimage. Identity and Inverse Mappings Let A be any nonempty set. The mapping f : A ! A deﬁned by f ðaÞ ¼ a—that is, the function that assigns to each element in A itself—is called identity mapping. It is usually denoted by 1A or 1 or I. Thus, for any a 2 A, we have 1AðaÞ ¼ a. Figure 5-1 166 CHAPTER 5 Linear Mappings Now let f : A ! B. We call g : B ! A the inverse of f, written f 1, if f g ¼ 1B and g f ¼ 1A We emphasize that f has an inverse if and only if f is a one-to-one correspondence between A and B; that is, f is one-to-one and onto (Problem 5.7). Also, if b 2 B, then f 1ðbÞ ¼ a, where a is the unique element of A for which f ðaÞ ¼ b 5.3 Linear Mappings (Linear Transformations) We begin with a deﬁnition. DEFINITION: Let V and U be vector spaces over the same ﬁeld K. A mapping F : V ! U is called a linear mapping or linear transformation if it satisﬁes the following two conditions: (1) For any vectors v; w 2 V, Fðv þ wÞ ¼ FðvÞ þ FðwÞ. (2) For any scalar k and vector v 2 V, FðkvÞ ¼ kFðvÞ. Namely, F : V ! U is linear if it ‘‘preserves’’ the two basic operations of a vector space, that of vector addition and that of scalar multiplication. Substituting k ¼ 0 into condition (2), we obtain Fð0Þ ¼ 0. Thus, every linear mapping takes the zero vector into the zero vector. Now for any scalars a; b 2 K and any vector v; w 2 V, we obtain Fðav þ bwÞ ¼ FðavÞ þ FðbwÞ ¼ aFðvÞ þ bFðwÞ More generally, for any scalars ai 2 K and any vectors vi 2 V, we obtain the following basic property of linear mappings: Fða1v1 þ a2v2 þ þ amvmÞ ¼ a1Fðv1Þ þ a2Fðv2Þ þ þ amFðvmÞ Remark 1: A linear mapping F : V ! U is completely characterized by the condition Fðav þ bwÞ ¼ aFðvÞ þ bFðwÞ ðÞ and so this condition is sometimes used as its deﬁntion. Remark 2: The term linear transformation rather than linear mapping is frequently used for linear mappings of the form F : Rn ! Rm. EXAMPLE 5.4 (a) Let F : R3 ! R3 be the ‘‘projection’’ mapping into the xy-plane; that is, F is the mapping deﬁned by Fðx; y; zÞ ¼ ðx; y; 0Þ. We show that F is linear. Let v ¼ ða; b; cÞ and w ¼ ða0; b0; c0Þ. Then Fðv þ wÞ ¼ Fða þ a0; b þ b0; c þ c0Þ ¼ ða þ a0; b þ b0; 0Þ ¼ ða; b; 0Þ þ ða0; b0; 0Þ ¼ FðvÞ þ FðwÞ and, for any scalar k, FðkvÞ ¼ Fðka; kb; kcÞ ¼ ðka; kb; 0Þ ¼ kða; b; 0Þ ¼ kFðvÞ Thus, F is linear. (b) Let G : R2 ! R2 be the ‘‘translation’’ mapping deﬁned by Gðx; yÞ ¼ ðx þ 1; y þ 2Þ. [That is, G adds the vector (1, 2) to any vector v ¼ ðx; yÞ in R2.] Note that Gð0Þ ¼ Gð0; 0Þ ¼ ð1; 2Þ 6¼ 0 Thus, the zero vector is not mapped into the zero vector. Hence, G is not linear. CHAPTER 5 Linear Mappings 167 EXAMPLE 5.5 (Derivative and Integral Mappings) Consider the vector space V ¼ PðtÞ of polynomials over the real ﬁeld R. Let uðtÞ and vðtÞ be any polynomials in V and let k be any scalar. (a) Let D : V ! V be the derivative mapping. One proves in calculus that dðu þ vÞ dt ¼ du dt þ dv dt and dðkuÞ dt ¼ k du dt That is, Dðu þ vÞ ¼ DðuÞ þ DðvÞ and DðkuÞ ¼ kDðuÞ. Thus, the derivative mapping is linear. (b) Let J : V ! R be an integral mapping, say Jð f ðtÞÞ ¼ ð1 0 f ðtÞ dt One also proves in calculus that, ð1 0 ½uðtÞ þ vðtÞdt ¼ ð1 0 uðtÞ dt þ ð1 0 vðtÞ dt and ð1 0 kuðtÞ dt ¼ k ð1 0 uðtÞ dt That is, Jðu þ vÞ ¼ JðuÞ þ JðvÞ and JðkuÞ ¼ kJðuÞ. Thus, the integral mapping is linear. EXAMPLE 5.6 (Zero and Identity Mappings) (a) Let F : V ! U be the mapping that assigns the zero vector 0 2 U to every vector v 2 V. Then, for any vectors v; w 2 V and any scalar k 2 K, we have Fðv þ wÞ ¼ 0 ¼ 0 þ 0 ¼ FðvÞ þ FðwÞ and FðkvÞ ¼ 0 ¼ k0 ¼ kFðvÞ Thus, F is linear. We call F the zero mapping, and we usually denote it by 0. (b) Consider the identity mapping I : V ! V, which maps each v 2 V into itself. Then, for any vectors v; w 2 V and any scalars a; b 2 K, we have Iðav þ bwÞ ¼ av þ bw ¼ aIðvÞ þ bIðwÞ Thus, I is linear. Our next theorem (proved in Problem 5.13) gives us an abundance of examples of linear mappings. In particular, it tells us that a linear mapping is completely determined by its values on the elements of a basis. THEOREM 5.2: Let V and U be vector spaces over a ﬁeld K. Let fv1; v2; . . . ; vng be a basis of V and let u1; u2; . . . ; un be any vectors in U. Then there exists a unique linear mapping F : V ! U such that Fðv1Þ ¼ u1; Fðv2Þ ¼ u2; . . . ; FðvnÞ ¼ un. We emphasize that the vectors u1; u2; . . . ; un in Theorem 5.2 are completely arbitrary; they may be linearly dependent or they may even be equal to each other. Matrices as Linear Mappings Let A be any real m n matrix. Recall that A determines a mapping FA : Kn ! Km by FAðuÞ ¼ Au (where the vectors in Kn and Km are written as columns). We show FA is linear. By matrix multiplication, FAðv þ wÞ ¼ Aðv þ wÞ ¼ Av þ Aw ¼ FAðvÞ þ FAðwÞ FAðkvÞ ¼ AðkvÞ ¼ kðAvÞ ¼ kFAðvÞ In other words, using A to represent the mapping, we have Aðv þ wÞ ¼ Av þ Aw and AðkvÞ ¼ kðAvÞ Thus, the matrix mapping A is linear. 168 CHAPTER 5 Linear Mappings Vector Space Isomorphism The notion of two vector spaces being isomorphic was deﬁned in Chapter 4 when we investigated the coordinates of a vector relative to a basis. We now redeﬁne this concept. DEFINITION: Two vector spaces V and U over K are isomorphic, written V ﬃU, if there exists a bijective (one-to-one and onto) linear mapping F : V ! U. The mapping F is then called an isomorphism between V and U. Consider any vector space V of dimension n and let S be any basis of V. Then the mapping v 7! ½vS which maps each vector v 2 V into its coordinate vector ½vS, is an isomorphism between V and Kn. 5.4 Kernel and Image of a Linear Mapping We begin by deﬁning two concepts. DEFINITION: Let F : V ! U be a linear mapping. The kernel of F, written Ker F, is the set of elements in V that map into the zero vector 0 in U; that is, Ker F ¼ fv 2 V : FðvÞ ¼ 0g The image (or range) of F, written Im F, is the set of image points in U; that is, Im F ¼ fu 2 U : there exists v 2 V for which FðvÞ ¼ ug The following theorem is easily proved (Problem 5.22). THEOREM 5.3: Let F : V ! U be a linear mapping. Then the kernel of F is a subspace of V and the image of F is a subspace of U. Now suppose that v1; v2; . . . ; vm span a vector space V and that F : V ! U is linear. We show that Fðv1Þ; Fðv2Þ; . . . ; FðvmÞ span Im F. Let u 2 Im F. Then there exists v 2 V such that FðvÞ ¼ u. Because the vi’s span V and v 2 V, there exist scalars a1; a2; . . . ; am for which v ¼ a1v1 þ a2v2 þ þ amvm Therefore, u ¼ FðvÞ ¼ Fða1v1 þ a2v2 þ þ amvmÞ ¼ a1Fðv1Þ þ a2Fðv2Þ þ þ amFðvmÞ Thus, the vectors Fðv1Þ; Fðv2Þ; . . . ; FðvmÞ span Im F. We formally state the above result. PROPOSITION 5.4: Suppose v1; v2; . . . ; vm span a vector space V, and suppose F : V ! U is linear. Then Fðv1Þ; Fðv2Þ; . . . ; FðvmÞ span Im F. EXAMPLE 5.7 (a) Let F : R3 ! R3 be the projection of a vector v into the xy-plane [as pictured in Fig. 5-2(a)]; that is, Fðx; y; zÞ ¼ ðx; y; 0Þ Clearly the image of F is the entire xy-plane—that is, points of the form (x; y; 0). Moreover, the kernel of F is the z-axis—that is, points of the form (0; 0; c). That is, Im F ¼ fða; b; cÞ : c ¼ 0g ¼ xy-plane and Ker F ¼ fða; b; cÞ : a ¼ 0; b ¼ 0g ¼ z-axis (b) Let G : R3 ! R3 be the linear mapping that rotates a vector v about the z-axis through an angle y [as pictured in Fig. 5-2(b)]; that is, Gðx; y; zÞ ¼ ðx cos y y sin y; x sin y þ y cos y; zÞ CHAPTER 5 Linear Mappings 169 Observe that the distance of a vector v from the origin O does not change under the rotation, and so only the zero vector 0 is mapped into the zero vector 0. Thus, Ker G ¼ f0g. On the other hand, every vector u in R3 is the image of a vector v in R3 that can be obtained by rotating u back by an angle of y. Thus, Im G ¼ R3, the entire space. EXAMPLE 5.8 Consider the vector space V ¼ PðtÞ of polynomials over the real ﬁeld R, and let H : V ! V be the third-derivative operator; that is, H½ f ðtÞ ¼ d3f =dt3. [Sometimes the notation D3 is used for H, where D is the derivative operator.] We claim that Ker H ¼ fpolynomials of degree 2g ¼ P2ðtÞ and Im H ¼ V The ﬁrst comes from the fact that Hðat2 þ bt þ cÞ ¼ 0 but HðtnÞ 6¼ 0 for n 3. The second comes from that fact that every polynomial gðtÞ in V is the third derivative of some polynomial f ðtÞ (which can be obtained by taking the antiderivative of gðtÞ three times). Kernel and Image of Matrix Mappings Consider, say, a 3 4 matrix A and the usual basis fe1; e2; e3; e4g of K4 (written as columns): A ¼ a1 a2 a3 a4 b1 b2 b3 b4 c1 c2 c3 c4 2 4 3 5; e1 ¼ 1 0 0 0 2 6 6 4 3 7 7 5; e2 ¼ 1 0 0 0 2 6 6 4 3 7 7 5; e3 ¼ 1 0 0 0 2 6 6 4 3 7 7 5; e4 ¼ 1 0 0 0 2 6 6 4 3 7 7 5 Recall that A may be viewed as a linear mapping A : K4 ! K3, where the vectors in K4 and K3 are viewed as column vectors. Now the usual basis vectors span K4, so their images Ae1, Ae2, Ae3, Ae4 span the image of A. But the vectors Ae1, Ae2, Ae3, Ae4 are precisely the columns of A: Ae1 ¼ ½a1; b1; c1T; Ae2 ¼ ½a2; b2; c2T; Ae3 ¼ ½a3; b3; c3T; Ae4 ¼ ½a4; b4; c4T Thus, the image of A is precisely the column space of A. On the other hand, the kernel of A consists of all vectors v for which Av ¼ 0. This means that the kernel of A is the solution space of the homogeneous system AX ¼ 0, called the null space of A. We state the above results formally. PROPOSITION 5.5: Let A be any m n matrix over a ﬁeld K viewed as a linear map A : Kn ! Km. Then Ker A ¼ nullspðAÞ and Im A ¼ colspðAÞ Here colsp(A) denotes the column space of A, and nullsp(A) denotes the null space of A. Figure 5-2 170 CHAPTER 5 Linear Mappings Rank and Nullity of a Linear Mapping Let F : V ! U be a linear mapping. The rank of F is deﬁned to be the dimension of its image, and the nullity of F is deﬁned to be the dimension of its kernel; namely, rankðFÞ ¼ dimðIm FÞ and nullityðFÞ ¼ dimðKer FÞ The following important theorem (proved in Problem 5.23) holds. THEOREM 5.6 Let V be of ﬁnite dimension, and let F : V ! U be linear. Then dim V ¼ dimðKer FÞ þ dimðIm FÞ ¼ nullityðFÞ þ rankðFÞ Recall that the rank of a matrix A was also deﬁned to be the dimension of its column space and row space. If we now view A as a linear mapping, then both deﬁnitions correspond, because the image of A is precisely its column space. EXAMPLE 5.9 Let F : R4 ! R3 be the linear mapping deﬁned by Fðx; y; z; tÞ ¼ ðx y þ z þ t; 2x 2y þ 3z þ 4t; 3x 3y þ 4z þ 5tÞ (a) Find a basis and the dimension of the image of F . First ﬁnd the image of the usual basis vectors of R4, Fð1; 0; 0; 0Þ ¼ ð1; 2; 3Þ; Fð0; 0; 1; 0Þ ¼ ð1; 3; 4Þ Fð0; 1; 0; 0Þ ¼ ð1; 2; 3Þ; Fð0; 0; 0; 1Þ ¼ ð1; 4; 5Þ By Proposition 5.4, the image vectors span Im F. Hence, form the matrix M whose rows are these image vectors and row reduce to echelon form: M ¼ 1 2 3 1 2 3 1 3 4 1 4 5 2 6 6 4 3 7 7 5 1 2 3 0 0 0 0 1 1 0 2 2 2 6 6 4 3 7 7 5 1 2 3 0 1 1 0 0 0 0 0 0 2 6 6 4 3 7 7 5 Thus, (1, 2, 3) and (0, 1, 1) form a basis of Im F. Hence, dimðIm FÞ ¼ 2 and rankðFÞ ¼ 2. (b) Find a basis and the dimension of the kernel of the map F. Set FðvÞ ¼ 0, where v ¼ ðx; y; z; tÞ, Fðx; y; z; tÞ ¼ ðx y þ z þ t; 2x 2y þ 3z þ 4t; 3x 3y þ 4z þ 5tÞ ¼ ð0; 0; 0Þ Set corresponding components equal to each other to form the following homogeneous system whose solution space is Ker F: x y þ z þ t ¼ 0 2x 2y þ 3z þ 4t ¼ 0 3x 3y þ 4z þ 5t ¼ 0 or x y þ z þ t ¼ 0 z þ 2t ¼ 0 z þ 2t ¼ 0 or x y þ z þ t ¼ 0 z þ 2t ¼ 0 The free variables are y and t. Hence, dimðKer FÞ ¼ 2 or nullityðFÞ ¼ 2. (i) Set y ¼ 1, t ¼ 0 to obtain the solution (1; 1; 0; 0Þ, (ii) Set y ¼ 0, t ¼ 1 to obtain the solution (1; 0; 2; 1Þ. Thus, (1; 1; 0; 0) and (1; 0; 2; 1) form a basis for Ker F. As expected from Theorem 5.6, dimðIm FÞ þ dimðKer FÞ ¼ 4 ¼ dim R4. Application to Systems of Linear Equations Let AX ¼ B denote the matrix form of a system of m linear equations in n unknowns. Now the matrix A may be viewed as a linear mapping A : Kn ! Km CHAPTER 5 Linear Mappings 171 Thus, the solution of the equation AX ¼ B may be viewed as the preimage of the vector B 2 Km under the linear mapping A. Furthermore, the solution of the associated homogeneous system AX ¼ 0 may be viewed as the kernel of the linear mapping A. Applying Theorem 5.6 to this homogeneous system yields dimðKer AÞ ¼ dim Kn dimðIm AÞ ¼ n rank A But n is exactly the number of unknowns in the homogeneous system AX ¼ 0. Thus, we have proved the following theorem of Chapter 4. THEOREM 4.19: The dimension of the solution space W of a homogenous system AX ¼ 0 of linear equations is s ¼ n r, where n is the number of unknowns and r is the rank of the coefﬁcient matrix A. Observe that r is also the number of pivot variables in an echelon form of AX ¼ 0, so s ¼ n r is also the number of free variables. Furthermore, the s solution vectors of AX ¼ 0 described in Theorem 3.14 are linearly independent (Problem 4.52). Accordingly, because dim W ¼ s, they form a basis for the solution space W. Thus, we have also proved Theorem 3.14. 5.5 Singular and Nonsingular Linear Mappings, Isomorphisms Let F : V ! U be a linear mapping. Recall that Fð0Þ ¼ 0. F is said to be singular if the image of some nonzero vector v is 0—that is, if there exists v 6¼ 0 such that FðvÞ ¼ 0. Thus, F : V ! U is nonsingular if the zero vector 0 is the only vector whose image under F is 0 or, in other words, if Ker F ¼ f0g. EXAMPLE 5.10 Consider the projection map F : R3 ! R3 and the rotation map G : R3 ! R3 appearing in Fig. 5-2. (See Example 5.7.) Because the kernel of F is the z-axis, F is singular. On the other hand, the kernel of G consists only of the zero vector 0. Thus, G is nonsingular. Nonsingular linear mappings may also be characterized as those mappings that carry independent sets into independent sets. Speciﬁcally, we prove (Problem 5.28) the following theorem. THEOREM 5.7: Let F : V ! U be a nonsingular linear mapping. Then the image of any linearly independent set is linearly independent. Isomorphisms Suppose a linear mapping F : V ! U is one-to-one. Then only 0 2 V can map into 0 2 U, and so F is nonsingular. The converse is also true. For suppose F is nonsingular and FðvÞ ¼ FðwÞ, then Fðv wÞ ¼ FðvÞ FðwÞ ¼ 0, and hence, v w ¼ 0 or v ¼ w. Thus, FðvÞ ¼ FðwÞ implies v ¼ w— that is, F is one-to-one. We have proved the following proposition. PROPOSITION 5.8: A linear mapping F : V ! U is one-to-one if and only if F is nonsingular. Recall that a mapping F : V ! U is called an isomorphism if F is linear and if F is bijective (i.e., if F is one-to-one and onto). Also, recall that a vector space V is said to be isomorphic to a vector space U, written V ﬃU, if there is an isomorphism F : V ! U. The following theorem (proved in Problem 5.29) applies. THEOREM 5.9: Suppose V has ﬁnite dimension and dim V ¼ dim U. Suppose F : V ! U is linear. Then F is an isomorphism if and only if F is nonsingular. 172 CHAPTER 5 Linear Mappings 5.6 Operations with Linear Mappings We are able to combine linear mappings in various ways to obtain new linear mappings. These operations are very important and will be used throughout the text. Let F : V ! U and G : V ! U be linear mappings over a ﬁeld K. The sum F þ G and the scalar product kF, where k 2 K, are deﬁned to be the following mappings from V into U: ðF þ GÞðvÞ FðvÞ þ GðvÞ and ðkFÞðvÞ kFðvÞ We now show that if F and G are linear, then F þ G and kF are also linear. Speciﬁcally, for any vectors v; w 2 V and any scalars a; b 2 K, ðF þ GÞðav þ bwÞ ¼ Fðav þ bwÞ þ Gðav þ bwÞ ¼ aFðvÞ þ bFðwÞ þ aGðvÞ þ bGðwÞ ¼ a½FðvÞ þ GðvÞ þ b½FðwÞ þ GðwÞ ¼ aðF þ GÞðvÞ þ bðF þ GÞðwÞ and ðkFÞðav þ bwÞ ¼ kFðav þ bwÞ ¼ k½aFðvÞ þ bFðwÞ ¼ akFðvÞ þ bkFðwÞ ¼ aðkFÞðvÞ þ bðkFÞðwÞ Thus, F þ G and kF are linear. The following theorem holds. THEOREM 5.10: Let V and U be vector spaces over a ﬁeld K. Then the collection of all linear mappings from V into U with the above operations of addition and scalar multi-plication forms a vector space over K. The vector space of linear mappings in Theorem 5.10 is usually denoted by HomðV; UÞ Here Hom comes from the word ‘‘homomorphism.’’ We emphasize that the proof of Theorem 5.10 reduces to showing that HomðV; UÞ does satisfy the eight axioms of a vector space. The zero element of HomðV; UÞ is the zero mapping from V into U, denoted by 0 and deﬁned by 0ðvÞ ¼ 0 for every vector v 2 V. Suppose V and U are of ﬁnite dimension. Then we have the following theorem. THEOREM 5.11: Suppose dim V ¼ m and dim U ¼ n. Then dim½HomðV; UÞ ¼ mn. Composition of Linear Mappings Now suppose V, U, and W are vector spaces over the same ﬁeld K, and suppose F : V ! U and G : U ! W are linear mappings. We picture these mappings as follows: V ! F U ! G W Recall that the composition function G F is the mapping from V into W deﬁned by ðG FÞðvÞ ¼ GðFðvÞÞ. We show that G F is linear whenever F and G are linear. Speciﬁcally, for any vectors v; w 2 V and any scalars a; b 2 K, we have ðG FÞðav þ bwÞ ¼ GðFðav þ bwÞÞ ¼ GðaFðvÞ þ bFðwÞÞ ¼ aGðFðvÞÞ þ bGðFðwÞÞ ¼ aðG FÞðvÞ þ bðG FÞðwÞ Thus, G F is linear. The composition of linear mappings and the operations of addition and scalar multiplication are related as follows. CHAPTER 5 Linear Mappings 173 THEOREM 5.12: Let V, U, W be vector spaces over K. Suppose the following mappings are linear: F : V ! U; F0 : V ! U and G : U ! W; G0 : U ! W Then, for any scalar k 2 K: (i) G ðF þ F0Þ ¼ G F þ G F0. (ii) ðG þ G0Þ F ¼ G F þ G0 F. (iii) kðG FÞ ¼ ðkGÞ F ¼ G ðkFÞ. 5.7 Algebra AðVÞ of Linear Operators Let V be a vector space over a ﬁeld K. This section considers the special case of linear mappings from the vector space V into itself—that is, linear mappings of the form F : V ! V. They are also called linear operators or linear transformations on V. We will write AðVÞ, instead of HomðV; VÞ, for the space of all such mappings. Now AðVÞ is a vector space over K (Theorem 5.8), and, if dim V ¼ n, then dim AðVÞ ¼ n2. Moreover, for any mappings F; G 2 AðVÞ, the composition G F exists and also belongs to AðVÞ. Thus, we have a ‘‘multiplication’’ deﬁned in AðVÞ. [We sometimes write FG instead of G F in the space AðVÞ.] Remark: An algebra A over a ﬁeld K is a vector space over K in which an operation of multiplication is deﬁned satisfying, for every F; G; H 2 A and every k 2 K: (i) FðG þ HÞ ¼ FG þ FH, (ii) ðG þ HÞF ¼ GF þ HF, (iii) kðGFÞ ¼ ðkGÞF ¼ GðkFÞ. The algebra is said to be associative if, in addition, ðFGÞH ¼ FðGHÞ. The above deﬁnition of an algebra and previous theorems give us the following result. THEOREM 5.13: Let V be a vector space over K. Then AðVÞ is an associative algebra over K with respect to composition of mappings. If dim V ¼ n, then dim AðVÞ ¼ n2. This is why AðVÞ is called the algebra of linear operators on V. Polynomials and Linear Operators Observe that the identity mapping I : V ! V belongs to AðVÞ. Also, for any linear operator F in AðVÞ, we have FI ¼ IF ¼ F. We can also form ‘‘powers’’ of F. Namely, we deﬁne F0 ¼ I; F2 ¼ F F; F3 ¼ F2 F ¼ F F F; F4 ¼ F3 F; . . . Furthermore, for any polynomial pðtÞ over K, say, pðtÞ ¼ a0 þ a1t þ a2t2 þ þ ast2 we can form the linear operator pðFÞ deﬁned by pðFÞ ¼ a0I þ a1F þ a2F2 þ þ asFs (For any scalar k, the operator kI is sometimes denoted simply by k.) In particular, we say F is a zero of the polynomial pðtÞ if pðFÞ ¼ 0. EXAMPLE 5.11 Let F : K3 ! K3 be deﬁned by Fðx; y; zÞ ¼ ð0; x; yÞ. For any ða; b; cÞ 2 K3, ðF þ IÞða; b; cÞ ¼ ð0; a; bÞ þ ða; b; cÞ ¼ ða; a þ b; b þ cÞ F3ða; b; cÞ ¼ F2ð0; a; bÞ ¼ Fð0; 0; aÞ ¼ ð0; 0; 0Þ Thus, F3 ¼ 0, the zero mapping in AðVÞ. This means F is a zero of the polynomial pðtÞ ¼ t3. 174 CHAPTER 5 Linear Mappings Square Matrices as Linear Operators Let M ¼ Mn;n be the vector space of all square n n matrices over K. Then any matrix A in M deﬁnes a linear mapping FA : Kn ! Kn by FAðuÞ ¼ Au (where the vectors in Kn are written as columns). Because the mapping is from Kn into itself, the square matrix A is a linear operator, not simply a linear mapping. Suppose A and B are matrices in M. Then the matrix product AB is deﬁned. Furthermore, for any (column) vector u in Kn, FABðuÞ ¼ ðABÞu ¼ AðBuÞ ¼ AðFBðUÞÞ ¼ FAðFBðuÞÞ ¼ ðFA FBÞðuÞ In other words, the matrix product AB corresponds to the composition of A and B as linear mappings. Similarly, the matrix sum A þ B corresponds to the sum of A and B as linear mappings, and the scalar product kA corresponds to the scalar product of A as a linear mapping. Invertible Operators in AðVÞ Let F : V ! V be a linear operator. F is said to be invertible if it has an inverse—that is, if there exists F1 in AðVÞ such that FF1 ¼ F1F ¼ I. On the other hand, F is invertible as a mapping if F is both one-to-one and onto. In such a case, F1 is also linear and F1 is the inverse of F as a linear operator (proved in Problem 5.15). Suppose F is invertible. Then only 0 2 V can map into itself, and so F is nonsingular. The converse is not true, as seen by the following example. EXAMPLE 5.12 Let V ¼ PðtÞ, the vector space of polynomials over K. Let F be the mapping on V that increases by 1 the exponent of t in each term of a polynomial; that is, Fða0 þ a1t þ a2t2 þ þ astsÞ ¼ a0t þ a1t2 þ a2t3 þ þ astsþ1 Then F is a linear mapping and F is nonsingular. However, F is not onto, and so F is not invertible. The vector space V ¼ PðtÞ in the above example has inﬁnite dimension. The situation changes signiﬁcantly when V has ﬁnite dimension. Namely, the following theorem applies. THEOREM 5.14: Let F be a linear operator on a ﬁnite-dimensional vector space V. Then the following four conditions are equivalent. (i) F is nonsingular: Ker F ¼ f0g. (iii) F is an onto mapping. (ii) F is one-to-one. (iv) F is invertible. The proof of the above theorem mainly follows from Theorem 5.6, which tells us that dim V ¼ dimðKer FÞ þ dimðIm FÞ By Proposition 5.8, (i) and (ii) are equivalent. Note that (iv) is equivalent to (ii) and (iii). Thus, to prove the theorem, we need only show that (i) and (iii) are equivalent. This we do below. (a) Suppose (i) holds. Then dimðKer FÞ ¼ 0, and so the above equation tells us that dim V ¼ dimðIm FÞ. This means V ¼ Im F or, in other words, F is an onto mapping. Thus, (i) implies (iii). (b) Suppose (iii) holds. Then V ¼ Im F, and so dim V ¼ dimðIm FÞ. Therefore, the above equation tells us that dimðKer FÞ ¼ 0, and so F is nonsingular. Therefore, (iii) implies (i). Accordingly, all four conditions are equivalent. Remark: Suppose A is a square n n matrix over K. Then A may be viewed as a linear operator on Kn. Because Kn has ﬁnite dimension, Theorem 5.14 holds for the square matrix A. This is why the terms ‘‘nonsingular’’ and ‘‘invertible’’ are used interchangeably when applied to square matrices. EXAMPLE 5.13 Let F be the linear operator on R2 deﬁned by Fðx; yÞ ¼ ð2x þ y; 3x þ 2yÞ. (a) To show that F is invertible, we need only show that F is nonsingular. Set Fðx; yÞ ¼ ð0; 0Þ to obtain the homogeneous system 2x þ y ¼ 0 and 3x þ 2y ¼ 0 CHAPTER 5 Linear Mappings 175 Solve for x and y to get x ¼ 0, y ¼ 0. Hence, F is nonsingular and so invertible. (b) To ﬁnd a formula for F1, we set Fðx; yÞ ¼ ðs; tÞ and so F1ðs; tÞ ¼ ðx; yÞ. We have ð2x þ y; 3x þ 2yÞ ¼ ðs; tÞ or 2x þ y ¼ s 3x þ 2y ¼ t Solve for x and y in terms of s and t to obtain x ¼ 2s t, y ¼ 3s þ 2t. Thus, F1ðs; tÞ ¼ ð2s t; 3s þ 2tÞ or F1ðx; yÞ ¼ ð2x y; 3x þ 2yÞ where we rewrite the formula for F1 using x and y instead of s and t. SOLVED PROBLEMS Mappings 5.1. State whether each diagram in Fig. 5-3 deﬁnes a mapping from A ¼ fa; b; cg into B ¼ fx; y; zg. (a) No. There is nothing assigned to the element b 2 A. (b) No. Two elements, x and z, are assigned to c 2 A. (c) Yes. 5.2. Let f : A ! B and g : B ! C be deﬁned by Fig. 5-4. (a) Find the composition mapping ðg f Þ : A ! C. (b) Find the images of the mappings f , g, g f . (a) Use the deﬁnition of the composition mapping to compute ðg f Þ ðaÞ ¼ gð f ðaÞÞ ¼ gðyÞ ¼ t; ðg f Þ ðbÞ ¼ gð f ðbÞÞ ¼ gðxÞ ¼ s ðg f Þ ðcÞ ¼ gð f ðcÞÞ ¼ gðyÞ ¼ t Observe that we arrive at the same answer if we ‘‘follow the arrows’’ in Fig. 5-4: a ! y ! t; b ! x ! s; c ! y ! t (b) By Fig. 5-4, the image values under the mapping f are x and y, and the image values under g are r, s, t. Figure 5-3 Figure 5-4 176 CHAPTER 5 Linear Mappings Hence, Im f ¼ fx; yg and Im g ¼ fr; s; tg Also, by part (a), the image values under the composition mapping g f are t and s; accordingly, Im g f ¼ fs; tg. Note that the images of g and g f are different. 5.3. Consider the mapping F : R3 ! R2 deﬁned by Fðx; y; zÞ ¼ ðyz; x2Þ. Find (a) Fð2; 3; 4Þ; (b) Fð5; 2; 7Þ; (c) F1ð0; 0Þ, that is, all v 2 R3 such that FðvÞ ¼ 0. (a) Substitute in the formula for F to get Fð2; 3; 4Þ ¼ ð3 4; 22Þ ¼ ð12; 4Þ. (b) Fð5; 2; 7Þ ¼ ð2 7; 52Þ ¼ ð14; 25Þ. (c) Set FðvÞ ¼ 0, where v ¼ ðx; y; zÞ, and then solve for x, y, z: Fðx; y; zÞ ¼ ðyz; x2Þ ¼ ð0; 0Þ or yz ¼ 0; x2 ¼ 0 Thus, x ¼ 0 and either y ¼ 0 or z ¼ 0. In other words, x ¼ 0, y ¼ 0 or x ¼ 0; z ¼ 0—that is, the z-axis and the y-axis. 5.4. Consider the mapping F : R2 ! R2 deﬁned by Fðx; yÞ ¼ ð3y; 2xÞ. Let S be the unit circle in R2, that is, the solution set of x2 þ y2 ¼ 1. (a) Describe FðSÞ. (b) Find F1ðSÞ. (a) Let (a; b) be an element of FðSÞ. Then there exists ðx; yÞ 2 S such that Fðx; yÞ ¼ ða; bÞ. Hence, ð3y; 2xÞ ¼ ða; bÞ or 3y ¼ a; 2x ¼ b or y ¼ a 3 ; x ¼ b 2 Because ðx; yÞ 2 S—that is, x2 þ y2 ¼ 1—we have b 2 2 þ a 3 2 ¼ 1 or a2 9 þ b2 4 ¼ 1 Thus, FðSÞ is an ellipse. (b) Let Fðx; yÞ ¼ ða; bÞ, where ða; bÞ 2 S. Then ð3y; 2xÞ ¼ ða; bÞ or 3y ¼ a, 2x ¼ b. Because ða; bÞ 2 S, we have a2 þ b2 ¼ 1. Thus, ð3yÞ2 þ ð2xÞ2 ¼ 1. Accordingly, F1ðSÞ is the ellipse 4x2 þ 9y2 ¼ 1. 5.5. Let the mappings f : A ! B, g : B ! C, h : C ! D be deﬁned by Fig. 5-5. Determine whether or not each function is (a) one-to-one; (b) onto; (c) invertible (i.e., has an inverse). (a) The mapping f : A ! B is one-to-one, as each element of A has a different image. The mapping g : B ! C is not one-to one, because x and z both have the same image 4. The mapping h : C ! D is one-to-one. (b) The mapping f : A ! B is not onto, because z 2 B is not the image of any element of A. The mapping g : B ! C is onto, as each element of C is the image of some element of B. The mapping h : C ! D is also onto. (c) A mapping has an inverse if and only if it is one-to-one and onto. Hence, only h has an inverse. z y x w B g f C h 5 6 4 1 a D b c 2 3 A Figure 5-5 CHAPTER 5 Linear Mappings 177 5.6. Suppose f : A ! B and g : B ! C. Hence, ðg f Þ : A ! C exists. Prove (a) If f and g are one-to-one, then g f is one-to-one. (b) If f and g are onto mappings, then g f is an onto mapping. (c) If g f is one-to-one, then f is one-to-one. (d) If g f is an onto mapping, then g is an onto mapping. (a) Suppose ðg f ÞðxÞ ¼ ðg f ÞðyÞ. Then gð f ðxÞÞ ¼ gð f ðyÞÞ. Because g is one-to-one, f ðxÞ ¼ f ðyÞ. Because f is one-to-one, x ¼ y. We have proven that ðg f ÞðxÞ ¼ ðg f ÞðyÞ implies x ¼ y; hence g f is one-to-one. (b) Suppose c 2 C. Because g is onto, there exists b 2 B for which gðbÞ ¼ c. Because f is onto, there exists a 2 A for which f ðaÞ ¼ b. Thus, ðg f ÞðaÞ ¼ gð f ðaÞÞ ¼ gðbÞ ¼ c. Hence, g f is onto. (c) Suppose f is not one-to-one. Then there exist distinct elements x; y 2 A for which f ðxÞ ¼ f ðyÞ. Thus, ðg f ÞðxÞ ¼ gð f ðxÞÞ ¼ gð f ðyÞÞ ¼ ðg f ÞðyÞ. Hence, g f is not one-to-one. Therefore, if g f is one-to-one, then f must be one-to-one. (d) If a 2 A, then ðg f ÞðaÞ ¼ gð f ðaÞÞ 2 gðBÞ. Hence, ðg f ÞðAÞ gðBÞ. Suppose g is not onto. Then gðBÞ is properly contained in C and so ðg f ÞðAÞ is properly contained in C; thus, g f is not onto. Accordingly, if g f is onto, then g must be onto. 5.7. Prove that f : A ! B has an inverse if and only if f is one-to-one and onto. Suppose f has an inverse—that is, there exists a function f 1 : B ! A for which f 1 f ¼ 1A and f f 1 ¼ 1B. Because 1A is one-to-one, f is one-to-one by Problem 5.6(c), and because 1B is onto, f is onto by Problem 5.6(d); that is, f is both one-to-one and onto. Now suppose f is both one-to-one and onto. Then each b 2 B is the image of a unique element in A, say b. Thus, if f ðaÞ ¼ b, then a ¼ b; hence, f ðbÞ ¼ b. Now let g denote the mapping from B to A deﬁned by b 7! b. We have (i) ðg f ÞðaÞ ¼ gð f ðaÞÞ ¼ gðbÞ ¼ b ¼ a for every a 2 A; hence, g f ¼ 1A. (ii) ð f gÞðbÞ ¼ f ðgðbÞÞ ¼ f ðbÞ ¼ b for every b 2 B; hence, f g ¼ 1B. Accordingly, f has an inverse. Its inverse is the mapping g. 5.8. Let f : R ! R be deﬁned by f ðxÞ ¼ 2x 3. Now f is one-to-one and onto; hence, f has an inverse mapping f 1. Find a formula for f 1. Let y be the image of x under the mapping f ; that is, y ¼ f ðxÞ ¼ 2x 3. Hence, x will be the image of y under the inverse mapping f 1. Thus, solve for x in terms of y in the above equation to obtain x ¼ 1 2 ðy þ 3Þ. Then the formula deﬁning the inverse function is f 1ðyÞ ¼ 1 2 ðy þ 3Þ, or, using x instead of y, f 1ðxÞ ¼ 1 2 ðx þ 3Þ. Linear Mappings 5.9. Suppose the mapping F : R2 ! R2 is deﬁned by Fðx; yÞ ¼ ðx þ y; xÞ. Show that F is linear. We need to show that Fðv þ wÞ ¼ FðvÞ þ FðwÞ and FðkvÞ ¼ kFðvÞ, where u and v are any elements of R2 and k is any scalar. Let v ¼ ða; bÞ and w ¼ ða0; b0Þ. Then v þ w ¼ ða þ a0; b þ b0Þ and kv ¼ ðka; kbÞ We have FðvÞ ¼ ða þ b; aÞ and FðwÞ ¼ ða0 þ b0; a0Þ. Thus, Fðv þ wÞ ¼ Fða þ a0; b þ b0Þ ¼ ða þ a0 þ b þ b0; a þ a0Þ ¼ ða þ b; aÞ þ ða0 þ b0; a0Þ ¼ FðvÞ þ FðwÞ and FðkvÞ ¼ Fðka; kbÞ ¼ ðka þ kb; kaÞ ¼ kða þ b; aÞ ¼ kFðvÞ Because v, w, k were arbitrary, F is linear. 178 CHAPTER 5 Linear Mappings 5.10. Suppose F : R3 ! R2 is deﬁned by Fðx; y; zÞ ¼ ðx þ y þ z; 2x 3y þ 4zÞ. Show that F is linear. We argue via matrices. Writing vectors as columns, the mapping F may be written in the form FðvÞ ¼ Av, where v ¼ ½x; y; zT and A ¼ 1 1 1 2 3 4 Then, using properties of matrices, we have Fðv þ wÞ ¼ Aðv þ wÞ ¼ Av þ Aw ¼ FðvÞ þ FðwÞ FðkvÞ ¼ AðkvÞ ¼ kðAvÞ ¼ kFðvÞ and Thus, F is linear. 5.11. Show that the following mappings are not linear: (a) F : R2 ! R2 deﬁned by Fðx; yÞ ¼ ðxy; xÞ (b) F : R2 ! R3 deﬁned by Fðx; yÞ ¼ ðx þ 3; 2y; x þ yÞ (c) F : R3 ! R2 deﬁned by Fðx; y; zÞ ¼ ðjxj; y þ zÞ (a) Let v ¼ ð1; 2Þ and w ¼ ð3; 4Þ; then v þ w ¼ ð4; 6Þ. Also, FðvÞ ¼ ð1ð2Þ; 1Þ ¼ ð2; 1Þ and FðwÞ ¼ ð3ð4Þ; 3Þ ¼ ð12; 3Þ Hence, Fðv þ wÞ ¼ ð4ð6Þ; 4Þ ¼ ð24; 6Þ 6¼ FðvÞ þ FðwÞ (b) Because Fð0; 0Þ ¼ ð3; 0; 0Þ 6¼ ð0; 0; 0Þ, F cannot be linear. (c) Let v ¼ ð1; 2; 3Þ and k ¼ 3. Then kv ¼ ð3; 6; 9Þ. We have FðvÞ ¼ ð1; 5Þ and kFðvÞ ¼ 3ð1; 5Þ ¼ ð3; 15Þ: Thus, FðkvÞ ¼ Fð3; 6; 9Þ ¼ ð3; 15Þ 6¼ kFðvÞ Accordingly, F is not linear. 5.12. Let V be the vector space of n-square real matrices. Let M be an arbitrary but ﬁxed matrix in V. Let F : V ! V be deﬁned by FðAÞ ¼ AM þ MA, where A is any matrix in V. Show that F is linear. For any matrices A and B in V and any scalar k, we have FðA þ BÞ ¼ ðA þ BÞM þ MðA þ BÞ ¼ AM þ BM þ MA þ MB ¼ ðAM þ MAÞ ¼ ðBM þ MBÞ ¼ FðAÞ þ FðBÞ and FðkAÞ ¼ ðkAÞM þ MðkAÞ ¼ kðAMÞ þ kðMAÞ ¼ kðAM þ MAÞ ¼ kFðAÞ Thus, F is linear. 5.13. Prove Theorem 5.2: Let V and U be vector spaces over a ﬁeld K. Let fv1; v2; . . . ; vng be a basis of V and let u1; u2; . . . ; un be any vectors in U. Then there exists a unique linear mapping F : V ! U such that Fðv1Þ ¼ u1; Fðv2Þ ¼ u2; . . . ; FðvnÞ ¼ un. There are three steps to the proof of the theorem: (1) Deﬁne the mapping F : V ! U such that FðviÞ ¼ ui; i ¼ 1; . . . ; n. (2) Show that F is linear. (3) Show that F is unique. Step 1. Let v 2 V. Because fv1; . . . ; vng is a basis of V, there exist unique scalars a1; . . . ; an 2 K for which v ¼ a1v1 þ a2v2 þ þ anvn. We deﬁne F : V ! U by FðvÞ ¼ a1u1 þ a2u2 þ þ anun CHAPTER 5 Linear Mappings 179 (Because the ai are unique, the mapping F is well deﬁned.) Now, for i ¼ 1; . . . ; n, vi ¼ 0v1 þ þ 1vi þ þ 0vn Hence, FðviÞ ¼ 0u1 þ þ 1ui þ þ 0un ¼ ui Thus, the ﬁrst step of the proof is complete. Step 2. Suppose v ¼ a1v1 þ a2v2 þ þ anvn and w ¼ b1v1 þ b2v2 þ þ bnvn. Then v þ w ¼ ða1 þ b1Þv1 þ ða2 þ b2Þv2 þ þ ðan þ bnÞvn and, for any k 2 K, kv ¼ ka1v1 þ ka2v2 þ þ kanvn. By deﬁnition of the mapping F, FðvÞ ¼ a1u1 þ a2u2 þ þ anvn and FðwÞ ¼ b1u1 þ b2u2 þ þ bnun Hence, Fðv þ wÞ ¼ ða1 þ b1Þu1 þ ða2 þ b2Þu2 þ þ ðan þ bnÞun ¼ ða1u1 þ a2u2 þ þ anunÞ þ ðb1u1 þ b2u2 þ þ bnunÞ ¼ FðvÞ þ FðwÞ and FðkvÞ ¼ kða1u1 þ a2u2 þ þ anunÞ ¼ kFðvÞ Thus, F is linear. Step 3. Suppose G : V ! U is linear and Gðv1Þ ¼ ui; i ¼ 1; . . . ; n. Let v ¼ a1v1 þ a2v2 þ þ anvn Then GðvÞ ¼ Gða1v1 þ a2v2 þ þ anvnÞ ¼ a1Gðv1Þ þ a2Gðv2Þ þ þ anGðvnÞ ¼ a1u1 þ a2u2 þ þ anun ¼ FðvÞ Because GðvÞ ¼ FðvÞ for every v 2 V; G ¼ F. Thus, F is unique and the theorem is proved. 5.14. Let F : R2 ! R2 be the linear mapping for which Fð1; 2Þ ¼ ð2; 3Þ and Fð0; 1Þ ¼ ð1; 4Þ. [Note that fð1; 2Þ; ð0; 1Þg is a basis of R2, so such a linear map F exists and is unique by Theorem 5.2.] Find a formula for F; that is, ﬁnd Fða; bÞ. Write ða; bÞ as a linear combination of (1, 2) and (0, 1) using unknowns x and y, ða; bÞ ¼ xð1; 2Þ þ yð0; 1Þ ¼ ðx; 2x þ yÞ; so a ¼ x; b ¼ 2x þ y Solve for x and y in terms of a and b to get x ¼ a, y ¼ 2a þ b. Then Fða; bÞ ¼ xFð1; 2Þ þ yFð0; 1Þ ¼ að2; 3Þ þ ð2a þ bÞð1; 4Þ ¼ ðb; 5a þ 4bÞ 5.15. Suppose a linear mapping F : V ! U is one-to-one and onto. Show that the inverse mapping F1 : U ! V is also linear. Suppose u; u0 2 U. Because F is one-to-one and onto, there exist unique vectors v; v0 2 V for which FðvÞ ¼ u and Fðv0Þ ¼ u0. Because F is linear, we also have Fðv þ v0Þ ¼ FðvÞ þ Fðv0Þ ¼ u þ u0 and FðkvÞ ¼ kFðvÞ ¼ ku By deﬁnition of the inverse mapping, F1ðuÞ ¼ v; F1ðu0Þ ¼ v0; F1ðu þ u0Þ ¼ v þ v0; F1ðkuÞ ¼ kv: Then F1ðu þ u0Þ ¼ v þ v0 ¼ F1ðuÞ þ F1ðu0Þ and F1ðkuÞ ¼ kv ¼ kF1ðuÞ Thus, F1 is linear. 180 CHAPTER 5 Linear Mappings Kernel and Image of Linear Mappings 5.16. Let F : R4 ! R3 be the linear mapping deﬁned by Fðx; y; z; tÞ ¼ ðx y þ z þ t; x þ 2z t; x þ y þ 3z 3tÞ Find a basis and the dimension of (a) the image of F; (b) the kernel of F. (a) Find the images of the usual basis of R4: Fð1; 0; 0; 0Þ ¼ ð1; 1; 1Þ; Fð0; 0; 1; 0Þ ¼ ð1; 2; 3Þ Fð0; 1; 0; 0Þ ¼ ð1; 0; 1Þ; Fð0; 0; 0; 1Þ ¼ ð1; 1; 3Þ By Proposition 5.4, the image vectors span Im F. Hence, form the matrix whose rows are these image vectors, and row reduce to echelon form: 1 1 1 1 0 1 1 2 3 1 1 3 2 6 6 6 4 3 7 7 7 5 1 1 1 0 1 2 0 1 2 0 2 4 2 6 6 6 4 3 7 7 7 5 1 1 1 0 1 2 0 0 0 0 0 0 2 6 6 6 4 3 7 7 7 5 Thus, (1, 1, 1) and (0, 1, 2) form a basis for Im F; hence, dimðIm FÞ ¼ 2. (b) Set FðvÞ ¼ 0, where v ¼ ðx; y; z; tÞ; that is, set Fðx; y; z; tÞ ¼ ðx y þ z þ t; x þ 2z t; x þ y þ 3z 3tÞ ¼ ð0; 0; 0Þ Set corresponding entries equal to each other to form the following homogeneous system whose solution space is Ker F: x y þ z þ t ¼ 0 x þ 2z t ¼ 0 x þ y þ 3z 3t ¼ 0 or x y þ z þ t ¼ 0 y þ z 2t ¼ 0 2y þ 2z 4t ¼ 0 or x y þ z þ t ¼ 0 y þ z 2t ¼ 0 The free variables are z and t. Hence, dimðKer FÞ ¼ 2. (i) Set z ¼ 1, t ¼ 0 to obtain the solution (2; 1; 1; 0). (ii) Set z ¼ 0, t ¼ 1 to obtain the solution (1, 2, 0, 1). Thus, (2; 1; 1; 0) and (1, 2, 0, 1) form a basis of Ker F. [As expected, dimðIm FÞ þ dimðKer FÞ ¼ 2 þ 2 ¼ 4 ¼ dim R4, the domain of F.] 5.17. Let G : R3 ! R3 be the linear mapping deﬁned by Gðx; y; zÞ ¼ ðx þ 2y z; y þ z; x þ y 2zÞ Find a basis and the dimension of (a) the image of G, (b) the kernel of G. (a) Find the images of the usual basis of R3: Gð1; 0; 0Þ ¼ ð1; 0; 1Þ; Gð0; 1; 0Þ ¼ ð2; 1; 1Þ; Gð0; 0; 1Þ ¼ ð1; 1; 2Þ By Proposition 5.4, the image vectors span Im G. Hence, form the matrix M whose rows are these image vectors, and row reduce to echelon form: M ¼ 1 0 1 2 1 1 1 1 2 2 4 3 5 1 0 1 0 1 1 0 1 1 2 4 3 5 1 0 1 0 1 1 0 0 0 2 4 3 5 Thus, (1, 0, 1) and (0; 1; 1) form a basis for Im G; hence, dimðIm GÞ ¼ 2. (b) Set GðvÞ ¼ 0, where v ¼ ðx; y; zÞ; that is, Gðx; y; zÞ ¼ ðx þ 2y z; y þ z; x þ y 2zÞ ¼ ð0; 0; 0Þ CHAPTER 5 Linear Mappings 181 Set corresponding entries equal to each other to form the following homogeneous system whose solution space is Ker G: x þ 2y z ¼ 0 y þ z ¼ 0 x þ y 2z ¼ 0 or x þ 2y z ¼ 0 y þ z ¼ 0 y z ¼ 0 or x þ 2y z ¼ 0 y þ z ¼ 0 The only free variable is z; hence, dimðKer GÞ ¼ 1. Set z ¼ 1; then y ¼ 1 and x ¼ 3. Thus, (3; 1; 1) forms a basis of Ker G. [As expected, dimðIm GÞ þ dimðKer GÞ ¼ 2 þ 1 ¼ 3 ¼ dim R3, the domain of G.] 5.18. Consider the matrix mapping A : R4 ! R3, where A ¼ 1 2 3 1 1 3 5 2 3 8 13 3 2 4 3 5. Find a basis and the dimension of (a) the image of A, (b) the kernel of A. (a) The column space of A is equal to Im A. Now reduce AT to echelon form: AT ¼ 1 1 3 2 3 8 3 5 13 1 2 3 2 6 6 4 3 7 7 5 1 1 3 0 1 2 0 2 4 0 3 6 2 6 6 4 3 7 7 5 1 1 3 0 1 2 0 0 0 0 0 0 2 6 6 4 3 7 7 5 Thus, fð1; 1; 3Þ; ð0; 1; 2Þg is a basis of Im A, and dimðIm AÞ ¼ 2. (b) Here Ker A is the solution space of the homogeneous system AX ¼ 0, where X ¼ fx; y; z; tÞT. Thus, reduce the matrix A of coefﬁcients to echelon form: 1 2 3 1 0 1 2 3 0 2 4 6 2 4 3 5 1 2 3 1 0 1 2 3 0 0 0 0 2 4 3 5 or x þ 2y þ 3z þ t ¼ 0 y þ 2z 3t ¼ 0 The free variables are z and t. Thus, dimðKer AÞ ¼ 2. (i) Set z ¼ 1, t ¼ 0 to get the solution (1; 2; 1; 0). (ii) Set z ¼ 0, t ¼ 1 to get the solution (7; 3; 0; 1). Thus, (1; 2; 1; 0) and (7; 3; 0; 1) form a basis for Ker A. 5.19. Find a linear map F : R3 ! R4 whose image is spanned by (1; 2; 0; 4) and (2; 0; 1; 3). Form a 4 3 matrix whose columns consist only of the given vectors, say A ¼ 1 2 2 2 0 0 0 1 1 4 3 3 2 6 6 4 3 7 7 5 Recall that A determines a linear map A : R3 ! R4 whose image is spanned by the columns of A. Thus, A satisﬁes the required condition. 5.20. Suppose f : V ! U is linear with kernel W, and that f ðvÞ ¼ u. Show that the ‘‘coset’’ v þ W ¼ fv þ w : w 2 Wg is the preimage of u; that is, f 1ðuÞ ¼ v þ W. We must prove that (i) f 1ðuÞ v þ W and (ii) v þ W f 1ðuÞ. We ﬁrst prove (i). Suppose v0 2 f 1ðuÞ. Then f ðv0Þ ¼ u, and so f ðv0 vÞ ¼ f ðv0Þ f ðvÞ ¼ u u ¼ 0 that is, v0 v 2 W. Thus, v0 ¼ v þ ðv0 vÞ 2 v þ W, and hence f 1ðuÞ v þ W. 182 CHAPTER 5 Linear Mappings Now we prove (ii). Suppose v0 2 v þ W. Then v0 ¼ v þ w, where w 2 W. Because W is the kernel of f ; we have f ðwÞ ¼ 0. Accordingly, f ðv0Þ ¼ f ðv þ wÞ þ f ðvÞ þ f ðwÞ ¼ f ðvÞ þ 0 ¼ f ðvÞ ¼ u Thus, v0 2 f 1ðuÞ, and so v þ W f 1ðuÞ. Both inclusions imply f 1ðuÞ ¼ v þ W. 5.21. Suppose F : V ! U and G : U ! W are linear. Prove (a) rankðG FÞ rankðGÞ, (b) rankðG FÞ rankðFÞ. (a) Because FðVÞ U, we also have GðFðVÞÞ GðUÞ, and so dim½GðFðVÞÞ dim½GðUÞ. Then rankðG FÞ ¼ dim½ðG FÞðVÞ ¼ dim½GðFðVÞÞ dim½GðUÞ ¼ rankðGÞ. (b) We have dim½GðFðVÞÞ dim½FðVÞ. Hence, rankðG FÞ ¼ dim½ðG FÞðVÞ ¼ dim½GðFðVÞÞ dim½FðVÞ ¼ rankðFÞ 5.22. Prove Theorem 5.3: Let F : V ! U be linear. Then, (a) Im F is a subspace of U, (b) Ker F is a subspace of V. (a) Because Fð0Þ ¼ 0; we have 0 2 Im F. Now suppose u; u0 2 Im F and a; b 2 K. Because u and u0 belong to the image of F, there exist vectors v; v0 2 V such that FðvÞ ¼ u and Fðv0Þ ¼ u0. Then Fðav þ bv0Þ ¼ aFðvÞ þ bFðv0Þ ¼ au þ bu0 2 Im F Thus, the image of F is a subspace of U. (b) Because Fð0Þ ¼ 0; we have 0 2 Ker F . Now suppose v; w 2 Ker F and a; b 2 K. Because v and w belong to the kernel of F, FðvÞ ¼ 0 and FðwÞ ¼ 0. Thus, Fðav þ bwÞ ¼ aFðvÞ þ bFðwÞ ¼ a0 þ b0 ¼ 0 þ 0 ¼ 0; and so av þ bw 2 Ker F Thus, the kernel of F is a subspace of V. 5.23. Prove Theorem 5.6: Suppose V has ﬁnite dimension and F : V ! U is linear. Then dim V ¼ dimðKer FÞ þ dimðIm FÞ ¼ nullityðFÞ þ rankðFÞ Suppose dimðKer FÞ ¼ r and fw1; . . . ; wrg is a basis of Ker F , and suppose dimðIm FÞ ¼ s and fu1; . . . ; usg is a basis of Im F. (By Proposition 5.4, Im F has ﬁnite dimension.) Because every uj 2 Im F, there exist vectors v1; . . . ; vs in V such that Fðv1Þ ¼ u1; . . . ; FðvsÞ ¼ us. We claim that the set B ¼ fw1; . . . ; wr; v1; . . . ; vsg is a basis of V; that is, (i) B spans V, and (ii) B is linearly independent. Once we prove (i) and (ii), then dim V ¼ r þ s ¼ dimðKer FÞ þ dimðIm FÞ. (i) B spans V. Let v 2 V. Then FðvÞ 2 Im F. Because the uj span Im F, there exist scalars a1; . . . ; as such that FðvÞ ¼ a1u1 þ þ asus. Set ^ v ¼ a1v1 þ þ asvs v. Then Fð^ vÞ ¼ Fða1v1 þ þ asvs vÞ ¼ a1Fðv1Þ þ þ asFðvsÞ FðvÞ ¼ a1u1 þ þ asus FðvÞ ¼ 0 Thus, ^ v 2 Ker F. Because the wi span Ker F, there exist scalars b1; . . . ; br, such that ^ v ¼ b1w1 þ þ brwr ¼ a1v1 þ þ asvs v Accordingly, v ¼ a1v1 þ þ asvs b1w1 brwr Thus, B spans V. CHAPTER 5 Linear Mappings 183 (ii) B is linearly independent. Suppose x1w1 þ þ xrwr þ y1v1 þ þ ysvs ¼ 0 ð1Þ where xi; yj 2 K. Then 0 ¼ Fð0Þ ¼ Fðx1w1 þ þ xrwr þ y1v1 þ þ ysvsÞ ¼ x1Fðw1Þ þ þ xrFðwrÞ þ y1Fðv1Þ þ þ ysFðvsÞ ð2Þ But FðwiÞ ¼ 0, since wi 2 Ker F, and FðvjÞ ¼ uj. Substituting into (2), we will obtain y1u1 þ þ ysus ¼ 0. Since the uj are linearly independent, each yj ¼ 0. Substitution into (1) gives x1w1 þ þ xrwr ¼ 0. Since the wi are linearly independent, each xi ¼ 0. Thus B is linearly independent. Singular and Nonsingular Linear Maps, Isomorphisms 5.24. Determine whether or not each of the following linear maps is nonsingular. If not, ﬁnd a nonzero vector v whose image is 0. (a) F : R2 ! R2 deﬁned by Fðx; yÞ ¼ ðx y; x 2yÞ. (b) G : R2 ! R2 deﬁned by Gðx; yÞ ¼ ð2x 4y; 3x 6yÞ. (a) Find Ker F by setting FðvÞ ¼ 0, where v ¼ ðx; yÞ, ðx y; x 2yÞ ¼ ð0; 0Þ or x y ¼ 0 x 2y ¼ 0 or x y ¼ 0 y ¼ 0 The only solution is x ¼ 0, y ¼ 0. Hence, F is nonsingular. (b) Set Gðx; yÞ ¼ ð0; 0Þ to ﬁnd Ker G: ð2x 4y; 3x 6yÞ ¼ ð0; 0Þ or 2x 4y ¼ 0 3x 6y ¼ 0 or x 2y ¼ 0 The system has nonzero solutions, because y is a free variable. Hence, G is singular. Let y ¼ 1 to obtain the solution v ¼ ð2; 1Þ, which is a nonzero vector, such that GðvÞ ¼ 0. 5.25. The linear map F : R2 ! R2 deﬁned by Fðx; yÞ ¼ ðx y; x 2yÞ is nonsingular by the previous Problem 5.24. Find a formula for F1. Set Fðx; yÞ ¼ ða; bÞ, so that F1ða; bÞ ¼ ðx; yÞ. We have ðx y; x 2yÞ ¼ ða; bÞ or x y ¼ a x 2y ¼ b or x y ¼ a y ¼ a b Solve for x and y in terms of a and b to get x ¼ 2a b, y ¼ a b. Thus, F1ða; bÞ ¼ ð2a b; a bÞ or F1ðx; yÞ ¼ ð2x y; x yÞ (The second equation is obtained by replacing a and b by x and y, respectively.) 5.26. Let G : R2 ! R3 be deﬁned by Gðx; yÞ ¼ ðx þ y; x 2y; 3x þ yÞ. (a) Show that G is nonsingular. (b) Find a formula for G1. (a) Set Gðx; yÞ ¼ ð0; 0; 0Þ to ﬁnd Ker G. We have ðx þ y; x 2y; 3x þ yÞ ¼ ð0; 0; 0Þ or x þ y ¼ 0; x 2y ¼ 0; 3x þ y ¼ 0 The only solution is x ¼ 0, y ¼ 0; hence, G is nonsingular. (b) Although G is nonsingular, it is not invertible, because R2 and R3 have different dimensions. (Thus, Theorem 5.9 does not apply.) Accordingly, G1 does not exist. 184 CHAPTER 5 Linear Mappings 5.27. Suppose that F : V ! U is linear and that V is of ﬁnite dimension. Show that V and the image of F have the same dimension if and only if F is nonsingular. Determine all nonsingular linear mappings T : R4 ! R3. By Theorem 5.6, dim V ¼ dimðIm FÞ þ dimðKer FÞ. Hence, V and Im F have the same dimension if and only if dimðKer FÞ ¼ 0 or Ker F ¼ f0g (i.e., if and only if F is nonsingular). Because dim R3 is less than dim R4, we have that dimðIm TÞ is less than the dimension of the domain R4 of T. Accordingly no linear mapping T : R4 ! R3 can be nonsingular. 5.28. Prove Theorem 5.7: Let F : V ! U be a nonsingular linear mapping. Then the image of any linearly independent set is linearly independent. Suppose v1; v2; . . . ; vn are linearly independent vectors in V. We claim that Fðv1Þ; Fðv2Þ; . . . ; FðvnÞ are also linearly independent. Suppose a1Fðv1Þ þ a2Fðv2Þ þ þ anFðvnÞ ¼ 0, where ai 2 K. Because F is linear, Fða1v1 þ a2v2 þ þ anvnÞ ¼ 0. Hence, a1v1 þ a2v2 þ þ anvn 2 Ker F But F is nonsingular—that is, Ker F ¼ f0g. Hence, a1v1 þ a2v2 þ þ anvn ¼ 0. Because the vi are linearly independent, all the ai are 0. Accordingly, the FðviÞ are linearly independent. Thus, the theorem is proved. 5.29. Prove Theorem 5.9: Suppose V has ﬁnite dimension and dim V ¼ dim U. Suppose F : V ! U is linear. Then F is an isomorphism if and only if F is nonsingular. If F is an isomorphism, then only 0 maps to 0; hence, F is nonsingular. Conversely, suppose F is nonsingular. Then dimðKer FÞ ¼ 0. By Theorem 5.6, dim V ¼ dimðKer FÞ þ dimðIm FÞ. Thus, dim U ¼ dim V ¼ dimðIm FÞ Because U has ﬁnite dimension, Im F ¼ U. This means F maps V onto U. Thus, F is one-to-one and onto; that is, F is an isomorphism. Operations with Linear Maps 5.30. Deﬁne F : R3 ! R2 and G : R3 ! R2 by Fðx; y; zÞ ¼ ð2x; y þ zÞ and Gðx; y; zÞ ¼ ðx z; yÞ. Find formulas deﬁning the maps: (a) F þ G, (b) 3F, (c) 2F 5G. (a) ðF þ GÞðx; y; zÞ ¼ Fðx; y; zÞ þ Gðx; y; zÞ ¼ ð2x; y þ zÞ þ ðx z; yÞ ¼ ð3x z; 2y þ zÞ (b) ð3FÞðx; y; zÞ ¼ 3Fðx; y; zÞ ¼ 3ð2x; y þ zÞ ¼ ð6x; 3y þ 3zÞ (c) ð2F 5GÞðx; y; zÞ ¼ 2Fðx; y; zÞ 5Gðx; y; zÞ ¼ 2ð2x; y þ zÞ 5ðx z; yÞ ¼ ð4x; 2y þ 2zÞ þ ð5x þ 5z; 5yÞ ¼ ðx þ 5z; 3y þ 2zÞ 5.31. Let F : R3 ! R2 and G : R2 ! R2 be deﬁned by Fðx; y; zÞ ¼ ð2x; y þ zÞ and Gðx; yÞ ¼ ðy; xÞ. Derive formulas deﬁning the mappings: (a) G F, (b) F G. (a) ðG FÞðx; y; zÞ ¼ GðFðx; y; zÞÞ ¼ Gð2x; y þ zÞ ¼ ðy þ z; 2xÞ (b) The mapping F G is not deﬁned, because the image of G is not contained in the domain of F. 5.32. Prove: (a) The zero mapping 0, deﬁned by 0ðvÞ ¼ 0 2 U for every v 2 V, is the zero element of HomðV; UÞ. (b) The negative of F 2 HomðV; UÞ is the mapping ð1ÞF, that is, F ¼ ð1ÞF. Let F 2 HomðV; UÞ. Then, for every v 2 V: ðF þ 0ÞðvÞ ¼ FðvÞ þ 0ðvÞ ¼ FðvÞ þ 0 ¼ FðvÞ ðaÞ Because ðF þ 0ÞðvÞ ¼ FðvÞ for every v 2 V, we have F þ 0 ¼ F . Similarly, 0 þ F ¼ F: ðF þ ð1ÞFÞðvÞ ¼ FðvÞ þ ð1ÞFðvÞ ¼ FðvÞ FðvÞ ¼ 0 ¼ 0ðvÞ ðbÞ Thus, F þ ð1ÞF ¼ 0: Similarly ð1ÞF þ F ¼ 0: Hence, F ¼ ð1ÞF: CHAPTER 5 Linear Mappings 185 5.33. Suppose F1; F2; . . . ; Fn are linear maps from V into U. Show that, for any scalars a1; a2; . . . ; an, and for any v 2 V, ða1F1 þ a2F2 þ þ anFnÞðvÞ ¼ a1F1ðvÞ þ a2F2ðvÞ þ þ anFnðvÞ The mapping a1F1 is deﬁned by ða1F1ÞðvÞ ¼ a1FðvÞ. Hence, the theorem holds for n ¼ 1. Accordingly, by induction, ða1F1 þ a2F2 þ þ anFnÞðvÞ ¼ ða1F1ÞðvÞ þ ða2F2 þ þ anFnÞðvÞ ¼ a1F1ðvÞ þ a2F2ðvÞ þ þ anFnðvÞ 5.34. Consider linear mappings F : R3 ! R2, G : R3 ! R2, H : R3 ! R2 deﬁned by Fðx; y; zÞ ¼ ðx þ y þ z; x þ yÞ; Gðx; y; zÞ ¼ ð2x þ z; x þ yÞ; Hðx; y; zÞ ¼ ð2y; xÞ Show that F, G, H are linearly independent [as elements of HomðR3; R2Þ]. Suppose, for scalars a; b; c 2 K, aF þ bG þ cH ¼ 0 ð1Þ (Here 0 is the zero mapping.) For e1 ¼ ð1; 0; 0Þ 2 R3, we have 0ðe1Þ ¼ ð0; 0Þ and ðaF þ bG þ cHÞðe1Þ ¼ aFð1; 0; 0Þ þ bGð1; 0; 0Þ þ cHð1; 0; 0Þ ¼ að1; 1Þ þ bð2; 1Þ þ cð0; 1Þ ¼ ða þ 2b; a þ b þ cÞ Thus by (1), ða þ 2b; a þ b þ cÞ ¼ ð0; 0Þ and so a þ 2b ¼ 0 and a þ b þ c ¼ 0 ð2Þ Similarly for e2 ¼ ð0; 1; 0Þ 2 R3, we have 0ðe2Þ ¼ ð0; 0Þ and ðaF þ bG þ cHÞðe2Þ ¼ aFð0; 1; 0Þ þ bGð0; 1; 0Þ þ cHð0; 1; 0Þ ¼ að1; 1Þ þ bð0; 1Þ þ cð2; 0Þ ¼ ða þ 2c; a þ bÞ Thus, a þ 2c ¼ 0 and a þ b ¼ 0 ð3Þ Using (2) and (3), we obtain a ¼ 0; b ¼ 0; c ¼ 0 ð4Þ Because (1) implies (4), the mappings F, G, H are linearly independent. 5.35. Let k be a nonzero scalar. Show that a linear map T is singular if and only if kT is singular. Hence, T is singular if and only if T is singular. Suppose T is singular. Then TðvÞ ¼ 0 for some vector v 6¼ 0. Hence, ðkTÞðvÞ ¼ kTðvÞ ¼ k0 ¼ 0 and so kT is singular. Now suppose kT is singular. Then ðkTÞðwÞ ¼ 0 for some vector w 6¼ 0. Hence, TðkwÞ ¼ kTðwÞ ¼ ðkTÞðwÞ ¼ 0 But k 6¼ 0 and w 6¼ 0 implies kw 6¼ 0. Thus, T is also singular. 5.36. Find the dimension d of: (a) HomðR3; R4Þ, (b) HomðR5; R3Þ, (c) HomðP3ðtÞ; R2Þ, (d) HomðM2;3; R4Þ. Use dim½HomðV; UÞ ¼ mn, where dim V ¼ m and dim U ¼ n. (a) d ¼ 3ð4Þ ¼ 12. (c) Because dim P3ðtÞ ¼ 4, d ¼ 4ð2Þ ¼ 8. (b) d ¼ 5ð3Þ ¼ 15. (d) Because dim M2;3 ¼ 6, d ¼ 6ð4Þ ¼ 24. 186 CHAPTER 5 Linear Mappings 5.37. Prove Theorem 5.11. Suppose dim V ¼ m and dim U ¼ n. Then dim½HomðV; UÞ ¼ mn. Suppose fv1; . . . ; vmg is a basis of V and fu1; . . . ; ung is a basis of U. By Theorem 5.2, a linear mapping in HomðV; UÞ is uniquely determined by arbitrarily assigning elements of U to the basis elements vi of V. We deﬁne Fij 2 HomðV; UÞ; i ¼ 1; . . . ; m; j ¼ 1; . . . ; n to be the linear mapping for which FijðviÞ ¼ uj, and FijðvkÞ ¼ 0 for k 6¼ i. That is, Fij maps vi into uj and the other v’s into 0. Observe that fFijg contains exactly mn elements; hence, the theorem is proved if we show that it is a basis of HomðV; UÞ. Proof that fFijg generates HomðV; UÞ. Consider an arbitrary function F 2 HomðV; UÞ. Suppose Fðv1Þ ¼ w1; Fðv2Þ ¼ w2; . . . ; FðvmÞ ¼ wm. Because wk 2 U, it is a linear combination of the u’s; say, wk ¼ ak1u1 þ ak2u2 þ þ aknun; k ¼ 1; . . . ; m; aij 2 K ð1Þ Consider the linear mapping G ¼ Pm i¼1 Pn j¼1 aijFij. Because G is a linear combination of the Fij, the proof that fFijg generates HomðV; UÞ is complete if we show that F ¼ G. We now compute GðvkÞ; k ¼ 1; . . . ; m. Because FijðvkÞ ¼ 0 for k 6¼ i and FkiðvkÞ ¼ ui; GðvkÞ ¼ P m i¼1 P n j¼1 aijFijðvkÞ ¼ P n j¼1 akjFkjðvkÞ ¼ P n j¼1 akjuj ¼ ak1u1 þ ak2u2 þ þ aknun Thus, by (1), GðvkÞ ¼ wk for each k. But FðvkÞ ¼ wk for each k. Accordingly, by Theorem 5.2, F ¼ G; hence, fFijg generates HomðV; UÞ. Proof that fFijg is linearly independent. Suppose, for scalars cij 2 K, P m i¼1 P n j¼1 cijFij ¼ 0 For vk; k ¼ 1; . . . ; m, 0 ¼ 0ðvkÞ ¼ P m i¼1 P n j¼1 cijFijðvkÞ ¼ P n j¼1 ckjFkjðvkÞ ¼ P n j¼1 ckjuj ¼ ck1u1 þ ck2u2 þ þ cknun But the ui are linearly independent; hence, for k ¼ 1; . . . ; m, we have ck1 ¼ 0; ck2 ¼ 0; . . . ; ckn ¼ 0. In other words, all the cij ¼ 0, and so fFijg is linearly independent. 5.38. Prove Theorem 5.12: (i) G ðF þ F0Þ ¼ G F þ G F0. (ii) ðG þ G0Þ F ¼ G F þ G0 F. (iii) kðG FÞ ¼ ðkGÞ F ¼ G ðkFÞ. (i) For every v 2 V, ðG ðF þ F0ÞÞðvÞ ¼ GððF þ F0ÞðvÞÞ ¼ GðFðvÞ þ F0ðvÞÞ ¼ GðFðvÞÞ þ GðF0ðvÞÞ ¼ ðG FÞðvÞ þ ðG F0ÞðvÞ ¼ ðG F þ G F0ÞðvÞ Thus, G ðF þ F0Þ ¼ G F þ G F0. (ii) For every v 2 V, ððG þ G0Þ FÞðvÞ ¼ ðG þ G0ÞðFðvÞÞ ¼ GðFðvÞÞ þ G0ðFðvÞÞ ¼ ðG FÞðvÞ þ ðG0 FÞðvÞ ¼ ðG F þ G0 FÞðvÞ Thus, ðG þ G0Þ F ¼ G F þ G0 F. CHAPTER 5 Linear Mappings 187 (iii) For every v 2 V, ðkðG FÞÞðvÞ ¼ kðG FÞðvÞ ¼ kðGðFðvÞÞÞ ¼ ðkGÞðFðvÞÞ ¼ ðkG FÞðvÞ and ðkðG FÞÞðvÞ ¼ kðG FÞðvÞ ¼ kðGðFðvÞÞÞ ¼ GðkFðvÞÞ ¼ GððkFÞðvÞÞ ¼ ðG kFÞðvÞ Accordingly, kðG FÞ ¼ ðkGÞ F ¼ G ðkFÞ. (We emphasize that two mappings are shown to be equal by showing that each of them assigns the same image to each point in the domain.) Algebra of Linear Maps 5.39. Let F and G be the linear operators on R2 deﬁned by Fðx; yÞ ¼ ðy; xÞ and Gðx; yÞ ¼ ð0; xÞ. Find formulas deﬁning the following operators: (a) F þ G, (b) 2F 3G, (c) FG, (d) GF, (e) F2, (f) G2. (a) ðF þ GÞðx; yÞ ¼ Fðx; yÞ þ Gðx; yÞ ¼ ðy; xÞ þ ð0; xÞ ¼ ðy; 2xÞ. (b) ð2F 3GÞðx; yÞ ¼ 2Fðx; yÞ 3Gðx; yÞ ¼ 2ðy; xÞ 3ð0; xÞ ¼ ð2y; xÞ. (c) ðFGÞðx; yÞ ¼ FðGðx; yÞÞ ¼ Fð0; xÞ ¼ ðx; 0Þ. (d) ðGFÞðx; yÞ ¼ GðFðx; yÞÞ ¼ Gðy; xÞ ¼ ð0; yÞ. (e) F2ðx; yÞ ¼ FðFðx; yÞÞ ¼ Fðy; xÞ ¼ ðx; yÞ. (Note that F2 ¼ I, the identity mapping.) (f) G2ðx; yÞ ¼ GðGðx; yÞÞ ¼ Gð0; xÞ ¼ ð0; 0Þ. (Note that G2 ¼ 0, the zero mapping.) 5.40. Consider the linear operator T on R3 deﬁned by Tðx; y; zÞ ¼ ð2x; 4x y; 2x þ 3y zÞ. (a) Show that T is invertible. Find formulas for (b) T1, (c) T2, (d) T2. (a) Let W ¼ Ker T. We need only show that T is nonsingular (i.e., that W ¼ f0g). Set Tðx; y; zÞ ¼ ð0; 0; 0Þ, which yields Tðx; y; zÞ ¼ ð2x; 4x y; 2x þ 3y zÞ ¼ ð0; 0; 0Þ Thus, W is the solution space of the homogeneous system 2x ¼ 0; 4x y ¼ 0; 2x þ 3y z ¼ 0 which has only the trivial solution (0, 0, 0). Thus, W ¼ f0g. Hence, T is nonsingular, and so T is invertible. (b) Set Tðx; y; zÞ ¼ ðr; s; tÞ [and so T 1ðr; s; tÞ ¼ ðx; y; zÞ]. We have ð2x; 4x y; 2x þ 3y zÞ ¼ ðr; s; tÞ or 2x ¼ r; 4x y ¼ s; 2x þ 3y z ¼ t Solve for x, y, z in terms of r, s, t to get x ¼ 1 2 r, y ¼ 2r s, z ¼ 7r 3s t. Thus, T1ðr; s; tÞ ¼ ð1 2 r; 2r s; 7r 3s tÞ or T1ðx; y; zÞ ¼ ð1 2 x; 2x y; 7x 3y zÞ (c) Apply T twice to get T2ðx; y; zÞ ¼ Tð2x; 4x y; 2x þ 3y zÞ ¼ ½4x; 4ð2xÞ ð4x yÞ; 2ð2xÞ þ 3ð4x yÞ ð2x þ 3y zÞ ¼ ð4x; 4x þ y; 14x 6y þ zÞ (d) Apply T1 twice to get T2ðx; y; zÞ ¼ T2ð1 2 x; 2x y; 7x 3y zÞ ¼ ½1 4 x; 2ð1 2 xÞ ð2x yÞ; 7ð1 2 xÞ 3ð2x yÞ ð7x 3y zÞ ¼ ð1 4 x; x þ y; 19 2 x þ 6y þ zÞ 188 CHAPTER 5 Linear Mappings 5.41. Let V be of ﬁnite dimension and let T be a linear operator on V for which TR ¼ I, for some operator R on V. (We call R a right inverse of T.) (a) Show that T is invertible. (b) Show that R ¼ T1. (c) Give an example showing that the above need not hold if V is of inﬁnite dimension. (a) Let dim V ¼ n. By Theorem 5.14, T is invertible if and only if T is onto; hence, T is invertible if and only if rankðTÞ ¼ n. We have n ¼ rankðIÞ ¼ rankðTRÞ rankðTÞ n. Hence, rankðTÞ ¼ n and T is invertible. (b) TT1 ¼ T1T ¼ I. Then R ¼ IR ¼ ðT1TÞR ¼ T1ðTRÞ ¼ T 1I ¼ T 1. (c) Let V be the space of polynomials in t over K; say, pðtÞ ¼ a0 þ a1t þ a2t2 þ þ asts. Let T and R be the operators on V deﬁned by Tð pðtÞÞ ¼ 0 þ a1 þ a2t þ þ asts1 and Rð pðtÞÞ ¼ a0t þ a1t2 þ þ astsþ1 We have ðTRÞð pðtÞÞ ¼ TðRð pðtÞÞÞ ¼ Tða0t þ a1t2 þ þ astsþ1Þ ¼ a0 þ a1t þ þ asts ¼ pðtÞ and so TR ¼ I, the identity mapping. On the other hand, if k 2 K and k 6¼ 0, then ðRTÞðkÞ ¼ RðTðkÞÞ ¼ Rð0Þ ¼ 0 6¼ k Accordingly, RT 6¼ I. 5.42. Let F and G be linear operators on R2 deﬁned by Fðx; yÞ ¼ ð0; xÞ and Gðx; yÞ ¼ ðx; 0Þ. Show that (a) GF ¼ 0, the zero mapping, but FG 6¼ 0. (b) G2 ¼ G. (a) ðGFÞðx; yÞ ¼ GðFðx; yÞÞ ¼ Gð0; xÞ ¼ ð0; 0Þ. Because GF assigns 0 ¼ ð0; 0Þ to every vector (x; y) in R2, it is the zero mapping; that is, GF ¼ 0. On the other hand, ðFGÞðx; yÞ ¼ FðGðx; yÞÞ ¼ Fðx; 0Þ ¼ ð0; xÞ. For example, ðFGÞð2; 3Þ ¼ ð0; 2Þ. Thus, FG 6¼ 0, as it does not assign 0 ¼ ð0; 0Þ to every vector in R2. (b) For any vector (x; y) in R2, we have G2ðx; yÞ ¼ GðGðx; yÞÞ ¼ Gðx; 0Þ ¼ ðx; 0Þ ¼ Gðx; yÞ. Hence, G2 ¼ G. 5.43. Find the dimension of (a) AðR4Þ, (b) AðP2ðtÞÞ, (c) AðM2;3). Use dim½AðVÞ ¼ n2 where dim V ¼ n. Hence, (a) dim½AðR4Þ ¼ 42 ¼ 16, (b) dim½AðP2ðtÞÞ ¼ 32 ¼ 9, (c) dim½AðM2;3Þ ¼ 62 ¼ 36. 5.44. Let E be a linear operator on V for which E2 ¼ E. (Such an operator is called a projection.) Let U be the image of E, and let W be the kernel. Prove (a) If u 2 U, then EðuÞ ¼ u (i.e., E is the identity mapping on U ). (b) If E 6¼ I, then E is singular—that is, EðvÞ ¼ 0 for some v 6¼ 0. (c) V ¼ U W. (a) If u 2 U, the image of E, then EðvÞ ¼ u for some v 2 V. Hence, using E2 ¼ E, we have u ¼ EðvÞ ¼ E2ðvÞ ¼ EðEðvÞÞ ¼ EðuÞ (b) If E 6¼ I, then for some v 2 V, EðvÞ ¼ u, where v 6¼ u. By (i), EðuÞ ¼ u. Thus, Eðv uÞ ¼ EðvÞ EðuÞ ¼ u u ¼ 0; where v u 6¼ 0 (c) We ﬁrst show that V ¼ U þ W. Let v 2 V. Set u ¼ EðvÞ and w ¼ v EðvÞ. Then v ¼ EðvÞ þ v EðvÞ ¼ u þ w By deﬂnition, u ¼ EðvÞ 2 U, the image of E. We now show that w 2 W, the kernel of E, EðwÞ ¼ Eðv EðvÞÞ ¼ EðvÞ E2ðvÞ ¼ EðvÞ EðvÞ ¼ 0 and thus w 2 W. Hence, V ¼ U þ W. We next show that U \ W ¼ f0g. Let v 2 U \ W. Because v 2 U, EðvÞ ¼ v by part (a). Because v 2 W, EðvÞ ¼ 0. Thus, v ¼ EðvÞ ¼ 0 and so U \ W ¼ f0g. The above two properties imply that V ¼ U W. CHAPTER 5 Linear Mappings 189 SUPPLEMENTARY PROBLEMS Mappings 5.45. Determine the number of different mappings from ðaÞ f1; 2g into f1; 2; 3g; ðbÞ f1; 2; ... ; rg into f1; 2; ... ; sg: 5.46. Let f : R ! R and g : R ! R be deﬁned by f ðxÞ ¼ x2 þ 3x þ 1 and gðxÞ ¼ 2x 3. Find formulas deﬁning the composition mappings: (a) f g; (b) g f ; (c) g g; (d) f f. 5.47. For each mappings f : R ! R ﬁnd a formula for its inverse: (a) f ðxÞ ¼ 3x 7, (b) f ðxÞ ¼ x3 þ 2. 5.48. For any mapping f : A ! B, show that 1B f ¼ f ¼ f 1A. Linear Mappings 5.49. Show that the following mappings are linear: (a) F : R3 ! R2 deﬁned by Fðx; y; zÞ ¼ ðx þ 2y 3z; 4x 5y þ 6zÞ. (b) F : R2 ! R2 deﬁned by Fðx; yÞ ¼ ðax þ by; cx þ dyÞ, where a, b, c, d belong to R. 5.50. Show that the following mappings are not linear: (a) F : R2 ! R2 deﬁned by Fðx; yÞ ¼ ðx2; y2Þ. (b) F : R3 ! R2 deﬁned by Fðx; y; zÞ ¼ ðx þ 1; y þ zÞ. (c) F : R2 ! R2 deﬁned by Fðx; yÞ ¼ ðxy; yÞ. (d) F : R3 ! R2 deﬁned by Fðx; y; zÞ ¼ ðjxj; y þ zÞ. 5.51. Find Fða; bÞ, where the linear map F : R2 ! R2 is deﬁned by Fð1; 2Þ ¼ ð3; 1Þ and Fð0; 1Þ ¼ ð2; 1Þ. 5.52. Find a 2 2 matrix A that maps (a) ð1; 3ÞT and ð1; 4ÞT into ð2; 5ÞT and ð3; 1ÞT, respectively. (b) ð2; 4ÞT and ð1; 2ÞT into ð1; 1ÞT and ð1; 3ÞT, respectively. 5.53. Find a 2 2 singular matrix B that maps ð1; 1ÞT into ð1; 3ÞT. 5.54. Let V be the vector space of real n-square matrices, and let M be a ﬁxed nonzero matrix in V. Show that the ﬁrst two of the following mappings T : V ! V are linear, but the third is not: (a) TðAÞ ¼ MA, (b) TðAÞ ¼ AM þ MA, (c) TðAÞ ¼ M þ A. 5.55. Give an example of a nonlinear map F : R2 ! R2 such that F1ð0Þ ¼ f0g but F is not one-to-one. 5.56. Let F : R2 ! R2 be deﬁned by Fðx; yÞ ¼ ð3x þ 5y; 2x þ 3yÞ, and let S be the unit circle in R2. (S consists of all points satisfying x2 þ y2 ¼ 1.) Find (a) the image FðSÞ, (b) the preimage F1ðSÞ. 5.57. Consider the linear map G : R3 ! R3 deﬁned by Gðx; y; zÞ ¼ ðx þ y þ z; y 2z; y 3zÞ and the unit sphere S2 in R3, which consists of the points satisfying x2 þ y2 þ z2 ¼ 1. Find (a) GðS2Þ, (b) G1ðS2Þ. 5.58. Let H be the plane x þ 2y 3z ¼ 4 in R3 and let G be the linear map in Problem 5.57. Find (a) GðHÞ, (b) G1ðHÞ. 5.59. Let W be a subspace of V. The inclusion map, denoted by i : W , ! V, is deﬁned by iðwÞ ¼ w for every w 2 W. Show that the inclusion map is linear. 5.60. Suppose F : V ! U is linear. Show that FðvÞ ¼ FðvÞ. Kernel and Image of Linear Mappings 5.61. For each linear map F ﬁnd a basis and the dimension of the kernel and the image of F: (a) F : R3 ! R3 deﬁned by Fðx; y; zÞ ¼ ðx þ 2y 3z; 2x þ 5y 4z; x þ 4y þ zÞ, (b) F : R4 ! R3 deﬁned by Fðx; y; z; tÞ ¼ ðx þ 2y þ 3z þ 2t; 2x þ 4y þ 7z þ 5t; x þ 2y þ 6z þ 5tÞ. 190 CHAPTER 5 Linear Mappings 5.62. For each linear map G, ﬁnd a basis and the dimension of the kernel and the image of G: (a) G : R3 ! R2 deﬁned by Gðx; y; zÞ ¼ ðx þ y þ z; 2x þ 2y þ 2zÞ, (b) G : R3 ! R2 deﬁned by Gðx; y; zÞ ¼ ðx þ y; y þ zÞ, (c) G : R5 ! R3 deﬁned by Gðx; y; z; s; tÞ ¼ ðx þ 2y þ 2z þ s þ t; x þ 2y þ 3z þ 2s t; 3x þ 6y þ 8z þ 5s tÞ: 5.63. Each of the following matrices determines a linear map from R4 into R3: (a) A ¼ 1 2 0 1 2 1 2 1 1 3 2 2 2 4 3 5, (b) B ¼ 1 0 2 1 2 3 1 1 2 0 5 3 2 4 3 5. Find a basis as well as the dimension of the kernel and the image of each linear map. 5.64. Find a linear mapping F : R3 ! R3 whose image is spanned by (1, 2, 3) and (4, 5, 6). 5.65. Find a linear mapping G : R4 ! R3 whose kernel is spanned by (1, 2, 3, 4) and (0, 1, 1, 1). 5.66. Let V ¼ P10ðtÞ, the vector space of polynomials of degree 10. Consider the linear map D4 : V ! V, where D4 denotes the fourth derivative d4ð f Þ=dt4. Find a basis and the dimension of (a) the image of D4; (b) the kernel of D4. 5.67. Suppose F : V ! U is linear. Show that (a) the image of any subspace of V is a subspace of U; (b) the preimage of any subspace of U is a subspace of V. 5.68. Show that if F : V ! U is onto, then dim U dim V. Determine all linear maps F : R3 ! R4 that are onto. 5.69. Consider the zero mapping 0 : V ! U deﬁned by 0ðvÞ ¼ 0; 8 v 2 V. Find the kernel and the image of 0. Operations with linear Mappings 5.70. Let F : R3 ! R2 and G : R3 ! R2 be deﬁned by Fðx; y; zÞ ¼ ðy; x þ zÞ and Gðx; y; zÞ ¼ ð2z; x yÞ. Find formulas deﬁning the mappings F þ G and 3F 2G. 5.71. Let H : R2 ! R2 be deﬁned by Hðx; yÞ ¼ ðy; 2xÞ. Using the maps F and G in Problem 5.70, ﬁnd formulas deﬁning the mappings: (a) H F and H G, (b) F H and G H, (c) H ðF þ GÞ and H F þ H G. 5.72. Show that the following mappings F, G, H are linearly independent: (a) F; G; H 2 HomðR2; R2Þ deﬁned by Fðx; yÞ ¼ ðx; 2yÞ, Gðx; yÞ ¼ ðy; x þ yÞ, Hðx; yÞ ¼ ð0; xÞ, (b) F; G; H 2 HomðR3; RÞ deﬁned by Fðx; y; zÞ ¼ x þ y þ z, Gðx; y; zÞ ¼ y þ z, Hðx; y; zÞ ¼ x z. 5.73. For F; G 2 HomðV; UÞ, show that rankðF þ GÞ rankðFÞ þ rankðGÞ. (Here V has ﬁnite dimension.) 5.74. Let F : V ! U and G : U ! V be linear. Show that if F and G are nonsingular, then G F is nonsingular. Give an example where G F is nonsingular but G is not. [Hint: Let dim V < dim U: 5.75. Find the dimension d of (a) HomðR2; R8Þ, (b) HomðP4ðtÞ; R3Þ, (c) HomðM2;4; P2ðtÞÞ. 5.76. Determine whether or not each of the following linear maps is nonsingular. If not, ﬁnd a nonzero vector v whose image is 0; otherwise ﬁnd a formula for the inverse map: (a) F : R3 ! R3 deﬁned by Fðx; y; zÞ ¼ ðx þ y þ z; 2x þ 3y þ 5z; x þ 3y þ 7zÞ, (b) G : R3 ! P2ðtÞ deﬁned by Gðx; y; zÞ ¼ ðx þ yÞt2 þ ðx þ 2y þ 2zÞt þ y þ z, (c) H : R2 ! P2ðtÞ deﬁned by Hðx; yÞ ¼ ðx þ 2yÞt2 þ ðx yÞt þ x þ y. 5.77. When can dim ½HomðV; UÞ ¼ dim V? CHAPTER 5 Linear Mappings 191 Algebra of Linear Operators 5.78. Let F and G be the linear operators on R2 deﬁned by Fðx; yÞ ¼ ðx þ y; 0Þ and Gðx; yÞ ¼ ðy; xÞ. Find formulas deﬁning the linear operators: (a) F þ G, (b) 5F 3G, (c) FG, (d) GF, (e) F2, ( f ) G2. 5.79. Show that each linear operator T on R2 is nonsingular and ﬁnd a formula for T1 , where (a) Tðx; yÞ ¼ ðx þ 2y; 2x þ 3yÞ, (b) Tðx; yÞ ¼ ð2x 3y; 3x 4yÞ. 5.80. Show that each of the following linear operators T on R3 is nonsingular and ﬁnd a formula for T1, where (a) Tðx; y; zÞ ¼ ðx 3y 2z; y 4z; zÞ; (b) Tðx; y; zÞ ¼ ðx þ z; x y; yÞ. 5.81. Find the dimension of AðVÞ, where (a) V ¼ R7, (b) V ¼ P5ðtÞ, (c) V ¼ M3;4. 5.82. Which of the following integers can be the dimension of an algebra AðVÞ of linear maps: 5, 9, 12, 25, 28, 36, 45, 64, 88, 100? 5.83. Let T be the linear operator on R2 deﬁned by Tðx; yÞ ¼ ðx þ 2y; 3x þ 4yÞ. Find a formula for f ðTÞ, where (a) f ðtÞ ¼ t2 þ 2t 3, (b) f ðtÞ ¼ t2 5t 2. Miscellaneous Problems 5.84. Suppose F : V ! U is linear and k is a nonzero scalar. Prove that the maps F and kF have the same kernel and the same image. 5.85. Suppose F and G are linear operators on V and that F is nonsingular. Assume that V has ﬁnite dimension. Show that rankðFGÞ ¼ rankðGFÞ ¼ rankðGÞ. 5.86. Suppose V has ﬁnite dimension. Suppose T is a linear operator on V such that rankðT2Þ ¼ rankðTÞ. Show that Ker T \ Im T ¼ f0g. 5.87. Suppose V ¼ U W. Let E1 and E2 be the linear operators on V deﬁned by E1ðvÞ ¼ u, E2ðvÞ ¼ w, where v ¼ u þ w, u 2 U, w 2 W. Show that (a) E2 1 ¼ E1 and E2 2 ¼ E2 (i.e., that E1 and E2 are projections); (b) E1 þ E2 ¼ I, the identity mapping; (c) E1E2 ¼ 0 and E2E1 ¼ 0. 5.88. Let E1 and E2 be linear operators on V satisfying parts (a), (b), (c) of Problem 5.88. Prove V ¼ Im E1 Im E2 5.89. Let v and w be elements of a real vector space V. The line segment L from v to v þ w is deﬁned to be the set of vectors v þ tw for 0 t 1. (See Fig. 5.6.) (a) Show that the line segment L between vectors v and u consists of the points: (i) ð1 tÞv þ tu for 0 t 1, (ii) t1v þ t2u for t1 þ t2 ¼ 1, t1 0, t2 0. (b) Let F : V ! U be linear. Show that the image FðLÞ of a line segment L in V is a line segment in U. Figure 5-6 192 CHAPTER 5 Linear Mappings 5.90. Let F : V ! U be linear and let W be a subspace of V. The restriction of F to W is the map FjW : W ! U deﬁned by FjWðvÞ ¼ FðvÞ for every v in W. Prove the following: (a) FjW is linear; (b) KerðFjWÞ ¼ ðKer FÞ \ W; (c) ImðFjWÞ ¼ FðWÞ. 5.91. A subset X of a vector space V is said to be convex if the line segment L between any two points (vectors) P; Q 2 X is contained in X. (a) Show that the intersection of convex sets is convex; (b) suppose F : V ! U is linear and X is convex. Show that FðXÞ is convex. ANSWERS TO SUPPLEMENTARY PROBLEMS 5.45. ðaÞ 32 ¼ 9; ðbÞ sr 5.46. (a) ð f gÞðxÞ ¼ 4x2 þ 1, (b) ðg f ÞðxÞ ¼ 2x2 þ 6x 1, (c) ðg gÞðxÞ ¼ 4x 9, (d) ð f f ÞðxÞ ¼ x4 þ 6x3 þ 14x2 þ 15x þ 5 5.47. (a) f 1ðxÞ ¼ 1 3 ðx þ 7Þ, (b) f 1ðxÞ ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ x 2 3 p 5.49. Fðx; y; zÞ ¼ Aðx; y; zÞT, where (a) A ¼ 1 2 3 4 5 6 , (b) A ¼ a b c d 5.50. (a) u ¼ ð2; 2Þ, k ¼ 3; then FðkuÞ ¼ ð36; 36Þ but kFðuÞ ¼ ð12; 12Þ; (b) Fð0Þ 6¼ 0; (c) u ¼ ð1; 2Þ, v ¼ ð3; 4Þ; then Fðu þ vÞ ¼ ð24; 6Þ but FðuÞ þ FðvÞ ¼ ð14; 6Þ; (d) u ¼ ð1; 2; 3Þ, k ¼ 2; then FðkuÞ ¼ ð2; 10Þ but kFðuÞ ¼ ð2; 10Þ. 5.51. Fða; bÞ ¼ ða þ 2b; 3a þ bÞ 5.52. (a) A ¼ 17 5 23 6 ; (b) None. (2; 4) and (1; 2) are linearly dependent but not (1, 1) and (1, 3). 5.53. B ¼ 1 0 3 0 [Hint: Send ð0; 1ÞT into ð0; 0ÞT.] 5.55. Fðx; yÞ ¼ ðx2; y2Þ 5.56. (a) 13x2 42xy þ 34y2 ¼ 1, (b) 13x2 þ 42xy þ 34y2 ¼ 1 5.57. (a) x2 8xy þ 26y2 þ 6xz 38yz þ 14z2 ¼ 1, (b) x2 þ 2xy þ 3y2 þ 2xz 8yz þ 14z2 ¼ 1 5.58. (a) x y þ 2z ¼ 4, (b) x þ 6z ¼ 4 5.61. (a) dimðKer FÞ ¼ 1, fð7; 2; 1Þg; dimðIm FÞ ¼ 2, fð1; 2; 1Þ; ð0; 1; 2Þg; (b) dimðKer FÞ ¼ 2, fð2; 1; 0; 0Þ; ð1; 0; 1; 1Þg; dimðIm FÞ ¼ 2, fð1; 2; 1Þ; ð0; 1; 3Þg 5.62. (a) dimðKer GÞ ¼ 2, fð1; 0; 1Þ; ð1; 1; 0Þg; dimðIm GÞ ¼ 1, fð1; 2Þg; (b) dimðKer GÞ ¼ 1, fð1; 1; 1Þg; Im G ¼ R2, fð1; 0Þ; ð0; 1Þg; (c) dimðKer GÞ ¼ 3, fð2; 1; 0; 0; 0Þ; ð1; 0; 1; 1; 0Þ; ð5; 0; 2; 0; 1Þg; dimðIm GÞ ¼ 2, fð1; 1; 3Þ; ð0; 1; 2Þg 5.63. (a) dimðKer AÞ ¼ 2, fð4; 2; 5; 0Þ; ð1; 3; 0; 5Þg; dimðIm AÞ ¼ 2, fð1; 2; 1Þ; ð0; 1; 1Þg; (b) dimðKer BÞ ¼ 1, fð1; 2 3 ; 1; 1Þg; Im B ¼ R3 5.64. Fðx; y; zÞ ¼ ðx þ 4y; 2x þ 5y; 3x þ 6yÞ CHAPTER 5 Linear Mappings 193 5.65. Fðx; y; z; tÞ ¼ ðx þ y z; 2x þ y t; 0Þ 5.66. (a) f1; t; t2; . . . ; t6g, (b) f1; t; t2; t3g 5.68. None, because dim R4 > dim R3: 5.69. Ker 0 ¼ V, Im 0 ¼ f0g 5.70. ðF þ GÞðx; y; zÞ ¼ ðy þ 2z; 2x y þ zÞ, ð3F 2GÞðx; y; zÞ ¼ ð3y 4z; x þ 2y þ 3zÞ 5.71. (a) ðH FÞðx; y; zÞ ¼ ðx þ z; 2yÞ, ðH GÞðx; y; zÞ ¼ ðx y; 4zÞ; (b) not deﬁned; (c) ðH ðF þ GÞÞðx; y; zÞ ¼ ðH F þ H GÞðx; y; zÞ ¼ ð2x y þ z; 2y þ 4zÞ 5.74. Fðx; yÞ ¼ ðx; y; yÞ; Gðx; y; zÞ ¼ ðx; yÞ 5.75. (a) 16, (b) 15, (c) 24 5.76. (a) v ¼ ð2; 3; 1Þ; (b) G1ðat2 þ bt þ cÞ ¼ ðb 2c; a b þ 2c; a þ b cÞ; (c) H is nonsingular, but not invertible, because dim P2ðtÞ > dim R2. 5.77. dim U ¼ 1; that is, U ¼ K. 5.78. (a) ðF þ GÞðx; yÞ ¼ ðx; xÞ; (b) ð5F 3GÞðx; yÞ ¼ ð5x þ 8y; 3xÞ; (c) ðFGÞðx; yÞ ¼ ðx y; 0Þ; (d) ðGFÞðx; yÞ ¼ ð0; x þ yÞ; (e) F2ðx; yÞ ¼ ðx þ y; 0Þ (note that F2 ¼ F); ( f ) G2ðx; yÞ ¼ ðx; yÞ. [Note that G2 þ I ¼ 0; hence, G is a zero of f ðtÞ ¼ t2 þ 1.] 5.79. (a) T1ðx; yÞ ¼ ð3x þ 2y; 2x yÞ, (b) T1ðx; yÞ ¼ ð4x þ 3y; 3x þ 2yÞ 5.80. (a) T1ðx; y; zÞ ¼ ðx þ 3y þ 14z; y 4z; zÞ, (b) T1ðx; y; zÞ ¼ ðy þ z; y; x y zÞ 5.81. (a) 49, (b) 36, (c) 144 5.82. Squares: 9, 25, 36, 64, 100 5.83. (a) Tðx; yÞ ¼ ð6x þ 14y; 21x þ 27yÞ; (b) Tðx; yÞ ¼ ð0; 0Þ—that is, f ðTÞ ¼ 0 194 CHAPTER 5 Linear Mappings Linear Mappings and Matrices 6.1 Introduction Consider a basis S ¼ fu1; u2; . . . ; ung of a vector space V over a ﬁeld K. For any vector v 2 V, suppose v ¼ a1u1 þ a2u2 þ þ anun Then the coordinate vector of v relative to the basis S, which we assume to be a column vector (unless otherwise stated or implied), is denoted and deﬁned by ½vS ¼ ½a1; a2; . . . ; anT Recall (Section 4.11) that the mapping v7!½vS, determined by the basis S, is an isomorphism between V and Kn. This chapter shows that there is also an isomorphism, determined by the basis S, between the algebra AðVÞ of linear operators on V and the algebra M of n-square matrices over K. Thus, every linear mapping F: V ! V will correspond to an n-square matrix ½FS determined by the basis S. We will also show how our matrix representation changes when we choose another basis. 6.2 Matrix Representation of a Linear Operator Let T be a linear operator (transformation) from a vector space V into itself, and suppose S ¼ fu1; u2; . . . ; ung is a basis of V. Now Tðu1Þ, Tðu2Þ; . . . ; TðunÞ are vectors in V, and so each is a linear combination of the vectors in the basis S; say, Tðu1Þ ¼ a11u1 þ a12u2 þ þ a1nun Tðu2Þ ¼ a21u1 þ a22u2 þ þ a2nun :::::::::::::::::::::::::::::::::::::::::::::::::::::: TðunÞ ¼ an1u1 þ an2u2 þ þ annun The following deﬁnition applies. DEFINITION: The transpose of the above matrix of coefﬁcients, denoted by mSðTÞ or ½TS, is called the matrix representation of T relative to the basis S, or simply the matrix of T in the basis S. (The subscript S may be omitted if the basis S is understood.) Using the coordinate (column) vector notation, the matrix representation of T may be written in the form mSðTÞ ¼ ½TS ¼ ½Tðu1ÞS; ½Tðu2ÞS; . . . ; ½Tðu1ÞS That is, the columns of mðTÞ are the coordinate vectors of Tðu1Þ, Tðu2Þ; . . . ; TðunÞ, respectively. CHAPTER 6 195 EXAMPLE 6.1 Let F: R2 ! R2 be the linear operator deﬁned by Fðx; yÞ ¼ ð2x þ 3y; 4x 5yÞ. (a) Find the matrix representation of F relative to the basis S ¼ fu1; u2g ¼ fð1; 2Þ; ð2; 5Þg. (1) First ﬁnd Fðu1Þ, and then write it as a linear combination of the basis vectors u1 and u2. (For notational convenience, we use column vectors.) We have Fðu1Þ ¼ F 1 2 ¼ 8 6 ¼ x 1 2 þ y 2 5 and x þ 2y ¼ 8 2x þ 5y ¼ 6 Solve the system to obtain x ¼ 52, y ¼ 22. Hence, Fðu1Þ ¼ 52u1 22u2. (2) Next ﬁnd Fðu2Þ, and then write it as a linear combination of u1 and u2: Fðu2Þ ¼ F 2 5 ¼ 19 17 ¼ x 1 2 þ y 2 5 and x þ 2y ¼ 19 2x þ 5y ¼ 17 Solve the system to get x ¼ 129, y ¼ 55. Thus, Fðu2Þ ¼ 129u1 55u2. Now write the coordinates of Fðu1Þ and Fðu2Þ as columns to obtain the matrix ½FS ¼ 52 129 22 55 (b) Find the matrix representation of F relative to the (usual) basis E ¼ fe1; e2g ¼ fð1; 0Þ; ð0; 1Þg. Find Fðe1Þ and write it as a linear combination of the usual basis vectors e1 and e2, and then ﬁnd Fðe2Þ and write it as a linear combination of e1 and e2. We have Fðe1Þ ¼ Fð1; 0Þ ¼ ð2; 2Þ ¼ 2e1 þ 4e2 Fðe2Þ ¼ Fð0; 1Þ ¼ ð3; 5Þ ¼ 3e1 5e2 and so ½FE ¼ 2 3 4 5 Note that the coordinates of Fðe1Þ and Fðe2Þ form the columns, not the rows, of ½FE. Also, note that the arithmetic is much simpler using the usual basis of R2. EXAMPLE 6.2 Let V be the vector space of functions with basis S ¼ fsin t; cos t; e3tg, and let D: V ! V be the differential operator deﬁned by Dð f ðtÞÞ ¼ dð f ðtÞÞ=dt. We compute the matrix representing D in the basis S: Dðsin tÞ ¼ cos t ¼ 0ðsin tÞ þ 1ðcos tÞ þ 0ðe3tÞ Dðcos tÞ ¼ sin t ¼ 1ðsin tÞ þ 0ðcos tÞ þ 0ðe3tÞ Dðe3tÞ ¼ 3e3t ¼ 0ðsin tÞ þ 0ðcos tÞ þ 3ðe3tÞ and so ½D ¼ 0 1 0 1 0 0 0 0 3 2 6 4 3 7 5 Note that the coordinates of Dðsin tÞ, Dðcos tÞ, Dðe3tÞ form the columns, not the rows, of ½D. Matrix Mappings and Their Matrix Representation Consider the following matrix A, which may be viewed as a linear operator on R2, and basis S of R2: A ¼ 3 2 4 5 and S ¼ fu1; u2g ¼ 1 2 ; 2 5 (We write vectors as columns, because our map is a matrix.) We ﬁnd the matrix representation of A relative to the basis S. 196 CHAPTER 6 Linear Mappings and Matrices (1) First we write Aðu1Þ as a linear combination of u1 and u2. We have Aðu1Þ ¼ 3 2 4 5 1 2 ¼ 1 6 ¼ x 1 2 þ y 2 5 and so x þ 2y ¼ 1 2x þ 5y ¼ 6 Solving the system yields x ¼ 7, y ¼ 4. Thus, Aðu1Þ ¼ 7u1 4u2. (2) Next we write Aðu2Þ as a linear combination of u1 and u2. We have Aðu2Þ ¼ 3 2 4 5 2 5 ¼ 4 7 ¼ x 1 2 þ y 2 5 and so x þ 2y ¼ 4 2x þ 5y ¼ 7 Solving the system yields x ¼ 6, y ¼ 1. Thus, Aðu2Þ ¼ 6u1 þ u2. Writing the coordinates of Aðu1Þ and Aðu2Þ as columns gives us the following matrix representation of A: ½AS ¼ 7 6 4 1 Remark: Suppose we want to ﬁnd the matrix representation of A relative to the usual basis E ¼ fe1; e2g ¼ f½1; 0T; ½0; 1Tg of R2: We have Aðe1Þ ¼ 3 2 4 5 1 0 ¼ 3 4 ¼ 3e1 þ 4e2 Aðe2Þ ¼ 3 2 4 5 0 1 ¼ 2 5 ¼ 2e1 5e2 and so ½AE ¼ 3 2 4 5 Note that ½AE is the original matrix A. This result is true in general: The matrix representation of any n n square matrix A over a field K relative to the usual basis E of Kn is the matrix A itself; that is; ½AE ¼ A Algorithm for Finding Matrix Representations Next follows an algorithm for ﬁnding matrix representations. The ﬁrst Step 0 is optional. It may be useful to use it in Step 1(b), which is repeated for each basis vector. ALGORITHM 6.1: The input is a linear operator T on a vector space V and a basis S ¼ fu1; u2; . . . ; ung of V. The output is the matrix representation ½TS. Step 0. Find a formula for the coordinates of an arbitrary vector v relative to the basis S. Step 1. Repeat for each basis vector uk in S: (a) Find TðukÞ. (b) Write TðukÞ as a linear combination of the basis vectors u1; u2; . . . ; un. Step 2. Form the matrix ½TS whose columns are the coordinate vectors in Step 1(b). EXAMPLE 6.3 Let F: R2 ! R2 be deﬁned by Fðx; yÞ ¼ ð2x þ 3y; 4x 5yÞ. Find the matrix representa-tion ½FS of F relative to the basis S ¼ fu1; u2g ¼ fð1; 2Þ; ð2; 5Þg. (Step 0) First ﬁnd the coordinates of ða; bÞ 2 R2 relative to the basis S. We have a b ¼ x 1 2 þ y 2 5 or x þ 2y ¼ a 2x 5y ¼ b or x þ 2y ¼ a y ¼ 2a þ b CHAPTER 6 Linear Mappings and Matrices 197 Solving for x and y in terms of a and b yields x ¼ 5a þ 2b, y ¼ 2a b. Thus, ða; bÞ ¼ ð5a þ 2bÞu1 þ ð2a bÞu2 (Step 1) Now we ﬁnd Fðu1Þ and write it as a linear combination of u1 and u2 using the above formula for ða; bÞ, and then we repeat the process for Fðu2Þ. We have Fðu1Þ ¼ Fð1; 2Þ ¼ ð4; 14Þ ¼ 8u1 6u2 Fðu2Þ ¼ Fð2; 5Þ ¼ ð11; 33Þ ¼ 11u1 11u2 (Step 2) Finally, we write the coordinates of Fðu1Þ and Fðu2Þ as columns to obtain the required matrix: ½FS ¼ 8 11 6 11 Properties of Matrix Representations This subsection gives the main properties of the matrix representations of linear operators T on a vector space V. We emphasize that we are always given a particular basis S of V. Our ﬁrst theorem, proved in Problem 6.9, tells us that the ‘‘action’’ of a linear operator T on a vector v is preserved by its matrix representation. THEOREM 6.1: Let T: V ! V be a linear operator, and let S be a (ﬁnite) basis of V. Then, for any vector v in V, ½TS½vS ¼ ½TðvÞS. EXAMPLE 6.4 Consider the linear operator F on R2 and the basis S of Example 6.3; that is, Fðx; yÞ ¼ ð2x þ 3y; 4x 5yÞ and S ¼ fu1; u2g ¼ fð1; 2Þ; ð2; 5Þg Let v ¼ ð5; 7Þ; and so FðvÞ ¼ ð11; 55Þ Using the formula from Example 6.3, we get ½v ¼ ½11; 3T and ½FðvÞ ¼ ½55; 33T We verify Theorem 6.1 for this vector v (where ½F is obtained from Example 6.3): ½F½v ¼ 8 11 6 11 11 3 ¼ 55 33 ¼ ½FðvÞ Given a basis S of a vector space V, we have associated a matrix ½T to each linear operator T in the algebra AðVÞ of linear operators on V. Theorem 6.1 tells us that the ‘‘action’’ of an individual linear operator T is preserved by this representation. The next two theorems (proved in Problems 6.10 and 6.11) tell us that the three basic operations in AðVÞ with these operators—namely (i) addition, (ii) scalar multiplication, and (iii) composition—are also preserved. THEOREM 6.2: Let V be an n-dimensional vector space over K, let S be a basis of V, and let M be the algebra of n n matrices over K. Then the mapping m: AðVÞ ! M defined by mðTÞ ¼ ½TS is a vector space isomorphism. That is, for any F; G 2 AðVÞ and any k 2 K, (i) mðF þ GÞ ¼ mðFÞ þ mðGÞ or ½F þ G ¼ ½F þ ½G (ii) mðkFÞ ¼ kmðFÞ or ½kF ¼ k½F (iii) m is bijective (one-to-one and onto). 198 CHAPTER 6 Linear Mappings and Matrices THEOREM 6.3: For any linear operators F; G 2 AðVÞ, mðG FÞ ¼ mðGÞmðFÞ or ½G F ¼ ½G½F (Here G F denotes the composition of the maps G and F.) 6.3 Change of Basis Let V be an n-dimensional vector space over a ﬁeld K. We have shown that once we have selected a basis S of V, every vector v 2 V can be represented by means of an n-tuple ½vS in Kn, and every linear operator T in AðVÞ can be represented by an n n matrix over K. We ask the following natural question: How do our representations change if we select another basis? In order to answer this question, we ﬁrst need a deﬁnition. DEFINITION: Let S ¼ fu1; u2; . . . ; ung be a basis of a vector space V; and let S0 ¼ fv1; v2; . . . ; vng be another basis. (For reference, we will call S the ‘‘old’’ basis and S0 the ‘‘new’’ basis.) Because S is a basis, each vector in the ‘‘new’’ basis S0 can be written uniquely as a linear combination of the vectors in S; say, v1 ¼ a11u1 þ a12u2 þ þ a1nun v2 ¼ a21u1 þ a22u2 þ þ a2nun ::::::::::::::::::::::::::::::::::::::::::::::::: vn ¼ an1u1 þ an2u2 þ þ annun Let P be the transpose of the above matrix of coefﬁcients; that is, let P ¼ ½pij, where pij ¼ aji. Then P is called the change-of-basis matrix (or transition matrix) from the ‘‘old’’ basis S to the ‘‘new’’ basis S0. The following remarks are in order. Remark 1: The above change-of-basis matrix P may also be viewed as the matrix whose columns are, respectively, the coordinate column vectors of the ‘‘new’’ basis vectors vi relative to the ‘‘old’’ basis S; namely, P ¼ ½v1S; ½v2S; . . . ; ½vnS Remark 2: Analogously, there is a change-of-basis matrix Q from the ‘‘new’’ basis S0 to the ‘‘old’’ basis S. Similarly, Q may be viewed as the matrix whose columns are, respectively, the coordinate column vectors of the ‘‘old’’ basis vectors ui relative to the ‘‘new’’ basis S0; namely, Q ¼ ½u1S0; ½u2S0; . . . ; ½unS0 Remark 3: Because the vectors v1; v2; . . . ; vn in the new basis S0 are linearly independent, the matrix P is invertible (Problem 6.18). Similarly, Q is invertible. In fact, we have the following proposition (proved in Problem 6.18). PROPOSITION 6.4: Let P and Q be the above change-of-basis matrices. Then Q ¼ P1. Now suppose S ¼ fu1; u2; . . . ; ung is a basis of a vector space V, and suppose P ¼ ½pij is any nonsingular matrix. Then the n vectors vi ¼ p1iui þ p2iu2 þ þ pniun; i ¼ 1; 2; . . . ; n corresponding to the columns of P, are linearly independent [Problem 6.21(a)]. Thus, they form another basis S0 of V. Moreover, P will be the change-of-basis matrix from S to the new basis S0. CHAPTER 6 Linear Mappings and Matrices 199 EXAMPLE 6.5 Consider the following two bases of R2: S ¼ fu1; u2g ¼ fð1; 2Þ; ð3; 5Þg and S0 ¼ fv1; v2g ¼ fð1; 1Þ; ð1; 2Þg (a) Find the change-of-basis matrix P from S to the ‘‘new’’ basis S0. Write each of the new basis vectors of S0 as a linear combination of the original basis vectors u1 and u2 of S. We have 1 1 ¼ x 1 2 þ y 3 5 or x þ 3y ¼ 1 2x þ 5y ¼ 1 yielding x ¼ 8; y ¼ 3 1 1 ¼ x 1 2 þ y 3 5 or x þ 3y ¼ 1 2x þ 5y ¼ 1 yielding x ¼ 11; y ¼ 4 Thus, v1 ¼ 8u1 þ 3u2 v2 ¼ 11u1 þ 4u2 and hence; P ¼ 8 11 3 4 : Note that the coordinates of v1 and v2 are the columns, not rows, of the change-of-basis matrix P. (b) Find the change-of-basis matrix Q from the ‘‘new’’ basis S0 back to the ‘‘old’’ basis S. Here we write each of the ‘‘old’’ basis vectors u1 and u2 of S0 as a linear combination of the ‘‘new’’ basis vectors v1 and v2 of S0. This yields u1 ¼ 4v1 3v2 u2 ¼ 11v1 8v2 and hence; Q ¼ 4 11 3 8 As expected from Proposition 6.4, Q ¼ P1. (In fact, we could have obtained Q by simply ﬁnding P1.) EXAMPLE 6.6 Consider the following two bases of R3: E ¼ fe1; e2; e3g ¼ fð1; 0; 0Þ; ð0; 1; 0Þ; ð0; 0; 1Þg and S ¼ fu1; u2; u3g ¼ fð1; 0; 1Þ; ð2; 1; 2Þ; ð1; 2; 2Þg (a) Find the change-of-basis matrix P from the basis E to the basis S. Because E is the usual basis, we can immediately write each basis element of S as a linear combination of the basis elements of E. Speciﬁcally, u1 ¼ ð1; 0; 1Þ ¼ e1 þ e3 u2 ¼ ð2; 1; 2Þ ¼ 2e1 þ e2 þ 2e3 u3 ¼ ð1; 2; 2Þ ¼ e1 þ 2e2 þ 2e3 and hence; P ¼ 1 2 1 0 1 2 1 2 2 2 6 4 3 7 5 Again, the coordinates of u1; u2; u3 appear as the columns in P. Observe that P is simply the matrix whose columns are the basis vectors of S. This is true only because the original basis was the usual basis E. (b) Find the change-of-basis matrix Q from the basis S to the basis E. The deﬁnition of the change-of-basis matrix Q tells us to write each of the (usual) basis vectors in E as a linear combination of the basis elements of S. This yields e1 ¼ ð1; 0; 0Þ ¼ 2u1 þ 2u2 u3 e2 ¼ ð0; 1; 0Þ ¼ 2u1 þ u2 e3 ¼ ð0; 0; 1Þ ¼ 3u1 2u2 þ u3 and hence; Q ¼ 2 2 3 2 1 2 1 0 1 2 6 4 3 7 5 We emphasize that to ﬁnd Q, we need to solve three 3 3 systems of linear equations—one 3 3 system for each of e1; e2; e3. 200 CHAPTER 6 Linear Mappings and Matrices Alternatively, we can ﬁnd Q ¼ P1 by forming the matrix M ¼ ½P; I and row reducing M to row canonical form: M ¼ 1 2 1 1 0 0 0 1 2 0 1 0 1 2 2 0 0 1 2 6 4 3 7 5 1 0 0 2 2 3 0 1 0 2 1 2 0 0 1 1 0 1 2 6 4 3 7 5 ¼ ½I; P1 thus; Q ¼ P1 ¼ 2 2 3 2 1 2 1 0 1 2 6 4 3 7 5 (Here we have used the fact that Q is the inverse of P.) The result in Example 6.6(a) is true in general. We state this result formally, because it occurs often. PROPOSITION 6.5: The change-of-basis matrix from the usual basis E of Kn to any basis S of Kn is the matrix P whose columns are, respectively, the basis vectors of S. Applications of Change-of-Basis Matrix First we show how a change of basis affects the coordinates of a vector in a vector space V. The following theorem is proved in Problem 6.22. THEOREM 6.6: Let P be the change-of-basis matrix from a basis S to a basis S0 in a vector space V. Then, for any vector v 2 V, we have P½vS0 ¼ ½vS and hence; P1½vS ¼ ½vS0 Namely, if we multiply the coordinates of v in the original basis S by P1, we get the coordinates of v in the new basis S0. Remark 1: Although P is called the change-of-basis matrix from the old basis S to the new basis S0, we emphasize that P1 transforms the coordinates of v in the original basis S into the coordinates of v in the new basis S0. Remark 2: Because of the above theorem, many texts call Q ¼ P1, not P, the transition matrix from the old basis S to the new basis S0. Some texts also refer to Q as the change-of-coordinates matrix. We now give the proof of the above theorem for the special case that dim V ¼ 3. Suppose P is the change-of-basis matrix from the basis S ¼ fu1; u2; u3g to the basis S0 ¼ fv1; v2; v3g; say, v1 ¼ a1u1 þ a2u2 þ a3a3 v2 ¼ b1u1 þ b2u2 þ b3u3 v3 ¼ c1u1 þ c2u2 þ c3u3 and hence; P ¼ a1 b1 c1 a2 b2 c2 a3 b3 c3 2 4 3 5 Now suppose v 2 V and, say, v ¼ k1v1 þ k2v2 þ k3v3. Then, substituting for v1; v2; v3 from above, we obtain v ¼ k1ða1u1 þ a2u2 þ a3u3Þ þ k2ðb1u1 þ b2u2 þ b3u3Þ þ k3ðc1u1 þ c2u2 þ c3u3Þ ¼ ða1k1 þ b1k2 þ c1k3Þu1 þ ða2k1 þ b2k2 þ c2k3Þu2 þ ða3k1 þ b3k2 þ c3k3Þu3 CHAPTER 6 Linear Mappings and Matrices 201 Thus, ½vS0 ¼ k1 k2 k3 2 4 3 5 and ½vS ¼ a1k1 þ b1k2 þ c1k3 a2k1 þ b2k2 þ c2k3 a3k1 þ b3k2 þ c3k3 2 4 3 5 Accordingly, P½vS0 ¼ a1 b1 c1 a2 b2 c2 a3 b3 c3 2 4 3 5 k1 k2 k3 2 4 3 5 ¼ a1k1 þ b1k2 þ c1k3 a2k1 þ b2k2 þ c2k3 a3k1 þ b3k2 þ c3k3 2 4 3 5 ¼ ½vS Finally, multiplying the equation ½vS ¼ P½vS, by P1, we get P1½vS ¼ P1P½vS0 ¼ I½vS0 ¼ ½vS0 The next theorem (proved in Problem 6.26) shows how a change of basis affects the matrix representation of a linear operator. THEOREM 6.7: Let P be the change-of-basis matrix from a basis S to a basis S0 in a vector space V. Then, for any linear operator T on V, ½TS0 ¼ P1½TSP That is, if A and B are the matrix representations of T relative, respectively, to S and S0, then B ¼ P1AP EXAMPLE 6.7 Consider the following two bases of R3: E ¼ fe1; e2; e3g ¼ fð1; 0; 0Þ; ð0; 1; 0Þ; ð0; 0; 1Þg and S ¼ fu1; u2; u3g ¼ fð1; 0; 1Þ; ð2; 1; 2Þ; ð1; 2; 2Þg The change-of-basis matrix P from E to S and its inverse P1 were obtained in Example 6.6. (a) Write v ¼ ð1; 3; 5Þ as a linear combination of u1; u2; u3, or, equivalently, ﬁnd ½vS. One way to do this is to directly solve the vector equation v ¼ xu1 þ yu2 þ zu3; that is, 1 3 5 2 4 3 5 ¼ x 1 0 1 2 4 3 5 þ y 2 1 2 2 4 3 5 þ z 1 2 2 2 4 3 5 or x þ 2y þ z ¼ 1 y þ 2z ¼ 3 x þ 2y þ 2z ¼ 5 The solution is x ¼ 7, y ¼ 5, z ¼ 4, so v ¼ 7u1 5u2 þ 4u3. On the other hand, we know that ½vE ¼ ½1; 3; 5T, because E is the usual basis, and we already know P1. Therefore, by Theorem 6.6, ½vS ¼ P1½vE ¼ 2 2 3 2 1 2 1 0 1 2 4 3 5 1 3 5 2 4 3 5 ¼ 7 5 4 2 4 3 5 Thus, again, v ¼ 7u1 5u2 þ 4u3. (b) Let A ¼ 1 3 2 2 4 1 3 1 2 2 4 3 5, which may be viewed as a linear operator on R3. Find the matrix B that represents A relative to the basis S. 202 CHAPTER 6 Linear Mappings and Matrices The deﬁnition of the matrix representation of A relative to the basis S tells us to write each of Aðu1Þ, Aðu2Þ, Aðu3Þ as a linear combination of the basis vectors u1; u2; u3 of S. This yields Aðu1Þ ¼ ð1; 3; 5Þ ¼ 11u1 5u2 þ 6u3 Aðu2Þ ¼ ð1; 2; 9Þ ¼ 21u1 14u2 þ 8u3 Aðu3Þ ¼ ð3; 4; 5Þ ¼ 17u1 8e2 þ 2u3 and hence; B ¼ 11 21 17 5 14 8 6 8 2 2 6 4 3 7 5 We emphasize that to ﬁnd B, we need to solve three 3 3 systems of linear equations—one 3 3 system for each of Aðu1Þ, Aðu2Þ, Aðu3Þ. On the other hand, because we know P and P1, we can use Theorem 6.7. That is, B ¼ P1AP ¼ 2 2 3 2 1 2 1 0 1 2 4 3 5 1 3 2 2 4 1 3 1 2 2 4 3 5 1 2 1 0 1 2 1 2 2 2 4 3 5 ¼ 11 21 17 5 14 8 6 8 2 2 4 3 5 This, as expected, gives the same result. 6.4 Similarity Suppose A and B are square matrices for which there exists an invertible matrix P such that B ¼ P1AP; then B is said to be similar to A, or B is said to be obtained from A by a similarity transformation. We show (Problem 6.29) that similarity of matrices is an equivalence relation. By Theorem 6.7 and the above remark, we have the following basic result. THEOREM 6.8: Two matrices represent the same linear operator if and only if the matrices are similar. That is, all the matrix representations of a linear operator T form an equivalence class of similar matrices. A linear operator T is said to be diagonalizable if there exists a basis S of V such that T is represented by a diagonal matrix; the basis S is then said to diagonalize T. The preceding theorem gives us the following result. THEOREM 6.9: Let A be the matrix representation of a linear operator T. Then T is diagonalizable if and only if there exists an invertible matrix P such that P1AP is a diagonal matrix. That is, T is diagonalizable if and only if its matrix representation can be diagonalized by a similarity transformation. We emphasize that not every operator is diagonalizable. However, we will show (Chapter 10) that every linear operator can be represented by certain ‘‘standard’’ matrices called its normal or canonical forms. Such a discussion will require some theory of ﬁelds, polynomials, and determinants. Functions and Similar Matrices Suppose f is a function on square matrices that assigns the same value to similar matrices; that is, f ðAÞ ¼ f ðBÞ whenever A is similar to B. Then f induces a function, also denoted by f , on linear operators T in the following natural way. We deﬁne f ðTÞ ¼ f ð½TSÞ where S is any basis. By Theorem 6.8, the function is well deﬁned. The determinant (Chapter 8) is perhaps the most important example of such a function. The trace (Section 2.7) is another important example of such a function. CHAPTER 6 Linear Mappings and Matrices 203 EXAMPLE 6.8 Consider the following linear operator F and bases E and S of R2: Fðx; yÞ ¼ ð2x þ 3y; 4x 5yÞ; E ¼ fð1; 0Þ; ð0; 1Þg; S ¼ fð1; 2Þ; ð2; 5Þg By Example 6.1, the matrix representations of F relative to the bases E and S are, respectively, A ¼ 2 3 4 5 and B ¼ 52 129 22 55 Using matrix A, we have (i) Determinant of F ¼ detðAÞ ¼ 10 12 ¼ 22; (ii) Trace of F ¼ trðAÞ ¼ 2 5 ¼ 3: On the other hand, using matrix B, we have (i) Determinant of F ¼ detðBÞ ¼ 2860 þ 2838 ¼ 22; (ii) Trace of F ¼ trðBÞ ¼ 52 55 ¼ 3. As expected, both matrices yield the same result. 6.5 Matrices and General Linear Mappings Last, we consider the general case of linear mappings from one vector space into another. Suppose V and U are vector spaces over the same ﬁeld K and, say, dim V ¼ m and dim U ¼ n. Furthermore, suppose S ¼ fv1; v2; . . . ; vmg and S0 ¼ fu1; u2; . . . ; ung are arbitrary but ﬁxed bases, respectively, of V and U. Suppose F: V ! U is a linear mapping. Then the vectors Fðv1Þ, Fðv2Þ; . . . ; FðvmÞ belong to U, and so each is a linear combination of the basis vectors in S0; say, Fðv1Þ ¼ a11u1 þ a12u2 þ þ a1nun Fðv2Þ ¼ a21u1 þ a22u2 þ þ a2nun ::::::::::::::::::::::::::::::::::::::::::::::::::::::: FðvmÞ ¼ am1u1 þ am2u2 þ þ amnun DEFINITION: The transpose of the above matrix of coefﬁcients, denoted by mS;S0ðFÞ or ½FS;S0, is called the matrix representation of F relative to the bases S and S0. [We will use the simple notation mðFÞ and ½F when the bases are understood.] The following theorem is analogous to Theorem 6.1 for linear operators (Problem 6.67). THEOREM 6.10: For any vector v 2 V, ½FS;S0½vS ¼ ½FðvÞS0. That is, multiplying the coordinates of v in the basis S of V by ½F, we obtain the coordinates of FðvÞ in the basis S0 of U. Recall that for any vector spaces V and U, the collection of all linear mappings from V into U is a vector space and is denoted by HomðV; UÞ. The following theorem is analogous to Theorem 6.2 for linear operators, where now we let M ¼ Mm;n denote the vector space of all m n matrices (Problem 6.67). THEOREM 6.11: The mapping m: HomðV; UÞ ! M deﬁned by mðFÞ ¼ ½F is a vector space isomorphism. That is, for any F; G 2 HomðV; UÞ and any scalar k, (i) mðF þ GÞ ¼ mðFÞ þ mðGÞ or ½F þ G ¼ ½F þ ½G (ii) mðkFÞ ¼ kmðFÞ or ½kF ¼ k½F (iii) m is bijective (one-to-one and onto). 204 CHAPTER 6 Linear Mappings and Matrices Our next theorem is analogous to Theorem 6.3 for linear operators (Problem 6.67). THEOREM 6.12: Let S; S0; S00 be bases of vector spaces V; U; W, respectively. Let F: V ! U and G U ! W be linear mappings. Then ½G FS;S00 ¼ ½GS0;S00½FS;S0 That is, relative to the appropriate bases, the matrix representation of the composition of two mappings is the matrix product of the matrix representations of the individual mappings. Next we show how the matrix representation of a linear mapping F: V ! U is affected when new bases are selected (Problem 6.67). THEOREM 6.13: Let P be the change-of-basis matrix from a basis e to a basis e0 in V, and let Q be the change-of-basis matrix from a basis f to a basis f 0 in U. Then, for any linear map F: V ! U, ½Fe0; f 0 ¼ Q1½Fe; f P In other words, if A is the matrix representation of a linear mapping F relative to the bases e and f , and B is the matrix representation of F relative to the bases e0 and f 0, then B ¼ Q1AP Our last theorem, proved in Problem 6.36, shows that any linear mapping from one vector space V into another vector space U can be represented by a very simple matrix. We note that this theorem is analogous to Theorem 3.18 for m n matrices. THEOREM 6.14: Let F: V ! U be linear and, say, rankðFÞ ¼ r. Then there exist bases of V and U such that the matrix representation of F has the form A ¼ Ir 0 0 0 where Ir is the r-square identity matrix. The above matrix A is called the normal or canonical form of the linear map F. SOLVED PROBLEMS Matrix Representation of Linear Operators 6.1. Consider the linear mapping F: R2 ! R2 deﬁned by Fðx; yÞ ¼ ð3x þ 4y; 2x 5yÞ and the following bases of R2: E ¼ fe1; e2g ¼ fð1; 0Þ; ð0; 1Þg and S ¼ fu1; u2g ¼ fð1; 2Þ; ð2; 3Þg (a) Find the matrix A representing F relative to the basis E. (b) Find the matrix B representing F relative to the basis S. (a) Because E is the usual basis, the rows of A are simply the coefﬁcients in the components of Fðx; yÞ; that is, using ða; bÞ ¼ ae1 þ be2, we have Fðe1Þ ¼ Fð1; 0Þ ¼ ð3; 2Þ ¼ 3e1 þ 2e2 Fðe2Þ ¼ Fð0; 1Þ ¼ ð4; 5Þ ¼ 4e1 5e2 and so A ¼ 3 4 2 5 Note that the coefﬁcients of the basis vectors are written as columns in the matrix representation. CHAPTER 6 Linear Mappings and Matrices 205 (b) First ﬁnd Fðu1Þ and write it as a linear combination of the basis vectors u1 and u2. We have Fðu1Þ ¼ Fð1; 2Þ ¼ ð11; 8Þ ¼ xð1; 2Þ þ yð2; 3Þ; and so x þ 2y ¼ 11 2x þ 3y ¼ 8 Solve the system to obtain x ¼ 49, y ¼ 30. Therefore, Fðu1Þ ¼ 49u1 þ 30u2 Next ﬁnd Fðu2Þ and write it as a linear combination of the basis vectors u1 and u2. We have Fðu2Þ ¼ Fð2; 3Þ ¼ ð18; 11Þ ¼ xð1; 2Þ þ yð2; 3Þ; and so x þ 2y ¼ 18 2x þ 3y ¼ 11 Solve for x and y to obtain x ¼ 76, y ¼ 47. Hence, Fðu2Þ ¼ 76u1 þ 47u2 Write the coefﬁcients of u1 and u2 as columns to obtain B ¼ 49 76 30 47 (b0) Alternatively, one can ﬁrst ﬁnd the coordinates of an arbitrary vector ða; bÞ in R2 relative to the basis S. We have ða; bÞ ¼ xð1; 2Þ þ yð2; 3Þ ¼ ðx þ 2y; 2x þ 3yÞ; and so x þ 2y ¼ a 2x þ 3y ¼ b Solve for x and y in terms of a and b to get x ¼ 3a þ 2b, y ¼ 2a b. Thus, ða; bÞ ¼ ð3a þ 2bÞu1 þ ð2a bÞu2 Then use the formula for ða; bÞ to ﬁnd the coordinates of Fðu1Þ and Fðu2Þ relative to S: Fðu1Þ ¼ Fð1; 2Þ ¼ ð11; 8Þ ¼ 49u1 þ 30u2 Fðu2Þ ¼ Fð2; 3Þ ¼ ð18; 11Þ ¼ 76u1 þ 47u2 and so B ¼ 49 76 30 47 6.2. Consider the following linear operator G on R2 and basis S: Gðx; yÞ ¼ ð2x 7y; 4x þ 3yÞ and S ¼ fu1; u2g ¼ fð1; 3Þ; ð2; 5Þg (a) Find the matrix representation ½GS of G relative to S. (b) Verify ½GS½vS ¼ ½GðvÞS for the vector v ¼ ð4; 3Þ in R2. First ﬁnd the coordinates of an arbitrary vector v ¼ ða; bÞ in R2 relative to the basis S. We have a b ¼ x 1 3 þ y 2 5 ; and so x þ 2y ¼ a 3x þ 5y ¼ b Solve for x and y in terms of a and b to get x ¼ 5a þ 2b, y ¼ 3a b. Thus, ða; bÞ ¼ ð5a þ 2bÞu1 þ ð3a bÞu2; and so ½v ¼ ½5a þ 2b; 3a bT (a) Using the formula for ða; bÞ and Gðx; yÞ ¼ ð2x 7y; 4x þ 3yÞ, we have Gðu1Þ ¼ Gð1; 3Þ ¼ ð19; 13Þ ¼ 121u1 70u2 Gðu2Þ ¼ Gð2; 5Þ ¼ ð31; 23Þ ¼ 201u1 116u2 and so ½GS ¼ 121 201 70 116 (We emphasize that the coefﬁcients of u1 and u2 are written as columns, not rows, in the matrix representation.) (b) Use the formula ða; bÞ ¼ ð5a þ 2bÞu1 þ ð3a bÞu2 to get v ¼ ð4; 3Þ ¼ 26u1 þ 15u2 GðvÞ ¼ Gð4; 3Þ ¼ ð20; 7Þ ¼ 131u1 þ 80u2 Then ½vS ¼ ½26; 15T and ½GðvÞS ¼ ½131; 80T 206 CHAPTER 6 Linear Mappings and Matrices Accordingly, ½GS½vS ¼ 121 201 70 116 26 15 ¼ 131 80 ¼ ½GðvÞS (This is expected from Theorem 6.1.) 6.3. Consider the following 2 2 matrix A and basis S of R2: A ¼ 2 4 5 6 and S ¼ fu1; u2g ¼ 1 2 ; 3 7 The matrix A deﬁnes a linear operator on R2. Find the matrix B that represents the mapping A relative to the basis S. First ﬁnd the coordinates of an arbitrary vector ða; bÞT with respect to the basis S. We have a b ¼ x 1 2 þ y 3 7 or x þ 3y ¼ a 2x 7y ¼ b Solve for x and y in terms of a and b to obtain x ¼ 7a þ 3b, y ¼ 2a b. Thus, ða; bÞT ¼ ð7a þ 3bÞu1 þ ð2a bÞu2 Then use the formula for ða; bÞT to ﬁnd the coordinates of Au1 and Au2 relative to the basis S: Au1 ¼ 2 4 5 6 1 2 ¼ 6 7 ¼ 63u1 þ 19u2 Au2 ¼ 2 4 5 6 3 7 ¼ 22 27 ¼ 235u1 þ 71u2 Writing the coordinates as columns yields B ¼ 63 235 19 71 6.4. Find the matrix representation of each of the following linear operators F on R3 relative to the usual basis E ¼ fe1; e2; e3g of R3; that is, ﬁnd ½F ¼ ½FE: (a) F deﬁned by Fðx; y; zÞ ¼ ðx þ 2y 3z; 4x 5y 6z; 7x þ 8y þ 9z). (b) F deﬁned by the 3 3 matrix A ¼ 1 1 1 2 3 4 5 5 5 2 4 3 5. (c) F deﬁned by Fðe1Þ ¼ ð1; 3; 5Þ; Fðe2Þ ¼ ð2; 4; 6Þ, Fðe3Þ ¼ ð7; 7; 7Þ. (Theorem 5.2 states that a linear map is completely deﬁned by its action on the vectors in a basis.) (a) Because E is the usual basis, simply write the coefﬁcients of the components of Fðx; y; zÞ as rows: ½F ¼ 1 2 3 4 5 6 7 8 9 2 4 3 5 (b) Because E is the usual basis, ½F ¼ A, the matrix A itself. (c) Here Fðe1Þ ¼ ð1; 3; 5Þ ¼ e1 þ 3e2 þ 5e3 Fðe2Þ ¼ ð2; 4; 6Þ ¼ 2e1 þ 4e2 þ 6e3 Fðe3Þ ¼ ð7; 7; 7Þ ¼ 7e1 þ 7e2 þ 7e3 and so ½F ¼ 1 2 7 3 4 7 5 6 7 2 4 3 5 That is, the columns of ½F are the images of the usual basis vectors. 6.5. Let G be the linear operator on R3 deﬁned by Gðx; y; zÞ ¼ ð2y þ z; x 4y; 3xÞ. (a) Find the matrix representation of G relative to the basis S ¼ fw1; w2; w3g ¼ fð1; 1; 1Þ; ð1; 1; 0Þ; ð1; 0; 0Þg (b) Verify that ½G½v ¼ ½GðvÞ for any vector v in R3. CHAPTER 6 Linear Mappings and Matrices 207 First ﬁnd the coordinates of an arbitrary vector ða; b; cÞ 2 R3 with respect to the basis S. Write ða; b; cÞ as a linear combination of w1; w2; w3 using unknown scalars x; y, and z: ða; b; cÞ ¼ xð1; 1; 1Þ þ yð1; 1; 0Þ þ zð1; 0; 0Þ ¼ ðx þ y þ z; x þ y; xÞ Set corresponding components equal to each other to obtain the system of equations x þ y þ z ¼ a; x þ y ¼ b; x ¼ c Solve the system for x; y, z in terms of a; b, c to ﬁnd x ¼ c, y ¼ b c, z ¼ a b. Thus, ða; b; cÞ ¼ cw1 þ ðb cÞw2 þ ða bÞw3, or equivalently, ½ða; b; cÞ ¼ ½c; b c; a bT (a) Because Gðx; y; zÞ ¼ ð2y þ z; x 4y; 3xÞ, Gðw1Þ ¼ Gð1; 1; 1Þ ¼ ð3; 3; 3Þ ¼ 3w1 6x2 þ 6x3 Gðw2Þ ¼ Gð1; 1; 0Þ ¼ ð2; 3; 3Þ ¼ 3w1 6w2 þ 5w3 Gðw3Þ ¼ Gð1; 0; 0Þ ¼ ð0; 1; 3Þ ¼ 3w1 2w2 w3 Write the coordinates Gðw1Þ, Gðw2Þ, Gðw3Þ as columns to get ½G ¼ 3 3 3 6 6 2 6 5 1 2 4 3 5 (b) Write GðvÞ as a linear combination of w1; w2; w3, where v ¼ ða; b; cÞ is an arbitrary vector in R3, GðvÞ ¼ Gða; b; cÞ ¼ ð2b þ c; a 4b; 3aÞ ¼ 3aw1 þ ð2a 4bÞw2 þ ða þ 6b þ cÞw3 or equivalently, ½GðvÞ ¼ ½3a; 2a 4b; a þ 6b þ cT Accordingly, ½G½v ¼ 3 3 3 6 6 2 6 5 1 2 4 3 5 c b c a b 2 4 3 5 ¼ 3a 2a 4b a þ 6b þ c 2 4 3 5 ¼ ½GðvÞ 6.6. Consider the following 3 3 matrix A and basis S of R3: A ¼ 1 2 1 3 1 0 1 4 2 2 4 3 5 and S ¼ fu1; u2; u3g ¼ 1 1 1 2 4 3 5; 0 1 1 2 4 3 5; 1 2 3 2 4 3 5 8 < : 9 = ; The matrix A deﬁnes a linear operator on R3. Find the matrix B that represents the mapping A relative to the basis S. (Recall that A represents itself relative to the usual basis of R3.) First ﬁnd the coordinates of an arbitrary vector ða; b; cÞ in R3 with respect to the basis S. We have a b c 2 4 3 5 ¼ x 1 1 1 2 4 3 5 þ y 0 1 1 2 4 3 5 þ z 1 2 3 2 4 3 5 or x þ z ¼ a x þ y þ 2z ¼ b x þ y þ 3z ¼ c Solve for x; y; z in terms of a; b; c to get x ¼ a þ b c; y ¼ a þ 2b c; z ¼ c b thus; ða; b; cÞT ¼ ða þ b cÞu1 þ ða þ 2b cÞu2 þ ðc bÞu3 208 CHAPTER 6 Linear Mappings and Matrices Then use the formula for ða; b; cÞT to ﬁnd the coordinates of Au1, Au2, Au3 relative to the basis S: Aðu1Þ ¼ Að1; 1; 1ÞT ¼ ð0; 2; 3ÞT ¼ u1 þ u2 þ u3 Aðu2Þ ¼ Að1; 1; 0ÞT ¼ ð1; 1; 2ÞT ¼ 4u1 3u2 þ 3u3 Aðu3Þ ¼ Að1; 2; 3ÞT ¼ ð0; 1; 3ÞT ¼ 2u1 u2 þ 2u3 so B ¼ 1 4 2 1 3 1 1 3 2 2 4 3 5 6.7. For each of the following linear transformations (operators) L on R2, ﬁnd the matrix A that represents L (relative to the usual basis of R2): (a) L is deﬁned by Lð1; 0Þ ¼ ð2; 4Þ and Lð0; 1Þ ¼ ð5; 8Þ. (b) L is the rotation in R2 counterclockwise by 90. (c) L is the reﬂection in R2 about the line y ¼ x. (a) Because fð1; 0Þ; ð0; 1Þg is the usual basis of R2, write their images under L as columns to get A ¼ 2 5 4 8 (b) Under the rotation L, we have Lð1; 0Þ ¼ ð0; 1Þ and Lð0; 1Þ ¼ ð1; 0Þ. Thus, A ¼ 0 1 1 0 (c) Under the reﬂection L, we have Lð1; 0Þ ¼ ð0; 1Þ and Lð0; 1Þ ¼ ð1; 0Þ. Thus, A ¼ 0 1 1 0 6.8. The set S ¼ fe3t, te3t, t2e3tg is a basis of a vector space V of functions f : R ! R. Let D be the differential operator on V; that is, Dð f Þ ¼ df =dt. Find the matrix representation of D relative to the basis S. Find the image of each basis function: Dðe3tÞ ¼ 3e3t Dðte3tÞ ¼ e3t þ 3te3t Dðt2e3tÞ ¼ 2te3t þ 3t2e3t ¼ 3ðe3tÞ þ 0ðte3tÞ þ 0ðt2e3tÞ ¼ 1ðe3tÞ þ 3ðte3tÞ þ 0ðt2e3tÞ ¼ 0ðe3tÞ þ 2ðte3tÞ þ 3ðt2e3tÞ and thus; ½D ¼ 3 1 0 0 3 2 0 0 3 2 4 3 5 6.9. Prove Theorem 6.1: Let T: V ! V be a linear operator, and let S be a (ﬁnite) basis of V. Then, for any vector v in V, ½TS½vS ¼ ½TðvÞS. Suppose S ¼ fu1; u2; . . . ; ung, and suppose, for i ¼ 1; . . . ; n, TðuiÞ ¼ ai1u1 þ ai2u2 þ þ ainun ¼ P n j¼1 aijuj Then ½TS is the n-square matrix whose jth row is ða1j; a2j; . . . ; anjÞ ð1Þ Now suppose v ¼ k1u1 þ k2u2 þ þ knun ¼ P n i¼1 kiui Writing a column vector as the transpose of a row vector, we have ½vS ¼ ½k1; k2; . . . ; knT ð2Þ CHAPTER 6 Linear Mappings and Matrices 209 Furthermore, using the linearity of T, TðvÞ ¼ T P n i¼1 kiui ¼ P n i¼1 kiTðuiÞ ¼ P n i¼1 ki P n j¼1 aijuj ¼ P n j¼1 P n i¼1 aijki uj ¼ P n j¼1 ða1jk1 þ a2jk2 þ þ anjknÞuj Thus, ½TðvÞS is the column vector whose jth entry is a1jk1 þ a2jk2 þ þ anjkn ð3Þ On the other hand, the jth entry of ½TS½vS is obtained by multiplying the jth row of ½TS by ½vS—that is (1) by (2). But the product of (1) and (2) is (3). Hence, ½TS½vS and ½TðvÞS have the same entries. Thus, ½TS½vS ¼ ½TðvÞS. 6.10. Prove Theorem 6.2: Let S ¼ fu1; u2; . . . ; ung be a basis for V over K, and let M be the algebra of n-square matrices over K. Then the mapping m: AðVÞ ! M deﬁned by mðTÞ ¼ ½TS is a vector space isomorphism. That is, for any F; G 2 AðVÞ and any k 2 K, we have (i) ½F þ G ¼ ½F þ ½G, (ii) ½kF ¼ k½F, (iii) m is one-to-one and onto. (i) Suppose, for i ¼ 1; . . . ; n, FðuiÞ ¼ P n j¼1 aijuj and GðuiÞ ¼ P n j¼1 bijuj Consider the matrices A ¼ ½aij and B ¼ ½bij. Then ½F ¼ AT and ½G ¼ BT. We have, for i ¼ 1; . . . ; n, ðF þ GÞðuiÞ ¼ FðuiÞ þ GðuiÞ ¼ P n j¼1 ðaij þ bijÞuj Because A þ B is the matrix ðaij þ bijÞ, we have ½F þ G ¼ ðA þ BÞT ¼ AT þ BT ¼ ½F þ ½G (ii) Also, for i ¼ 1; . . . ; n; ðkFÞðuiÞ ¼ kFðuiÞ ¼ k P n j¼1 aijuj ¼ P n j¼1 ðkaijÞuj Because kA is the matrix ðkaijÞ, we have ½kF ¼ ðkAÞT ¼ kAT ¼ k½F (iii) Finally, m is one-to-one, because a linear mapping is completely determined by its values on a basis. Also, m is onto, because matrix A ¼ ½aij in M is the image of the linear operator, FðuiÞ ¼ P n j¼1 aijuj; i ¼ 1; . . . ; n Thus, the theorem is proved. 6.11. Prove Theorem 6.3: For any linear operators G; F 2 AðVÞ, ½G F ¼ ½G½F. Using the notation in Problem 6.10, we have ðG FÞðuiÞ ¼ GðFðuiÞÞ ¼ G P n j¼1 aijuj ¼ P n j¼1 aijGðujÞ ¼ P n j¼1 aij P n k¼1 bjkuk ¼ P n k¼1 P n j¼1 aijbjk uk Recall that AB is the matrix AB ¼ ½cik, where cik ¼ Pn j¼1 aijbjk. Accordingly, ½G F ¼ ðABÞT ¼ BTAT ¼ ½G½F The theorem is proved. 210 CHAPTER 6 Linear Mappings and Matrices 6.12. Let A be the matrix representation of a linear operator T. Prove that, for any polynomial f ðtÞ, we have that f ðAÞ is the matrix representation of f ðTÞ. [Thus, f ðTÞ ¼ 0 if and only if f ðAÞ ¼ 0.] Let f be the mapping that sends an operator T into its matrix representation A. We need to prove that fð f ðTÞÞ ¼ f ðAÞ. Suppose f ðtÞ ¼ antn þ þ a1t þ a0. The proof is by induction on n, the degree of f ðtÞ. Suppose n ¼ 0. Recall that fðI0Þ ¼ I, where I0 is the identity mapping and I is the identity matrix. Thus, fð f ðTÞÞ ¼ fða0I0Þ ¼ a0fðI0Þ ¼ a0I ¼ f ðAÞ and so the theorem holds for n ¼ 0. Now assume the theorem holds for polynomials of degree less than n. Then, because f is an algebra isomorphism, fð f ðTÞÞ ¼ fðanT n þ an1Tn1 þ þ a1T þ a0I0Þ ¼ anfðTÞfðT n1Þ þ fðan1Tn1 þ þ a1T þ a0I0Þ ¼ anAAn1 þ ðan1An1 þ þ a1A þ a0IÞ ¼ f ðAÞ and the theorem is proved. Change of Basis The coordinate vector ½vS in this section will always denote a column vector; that is, ½vS ¼ ½a1; a2; . . . ; anT 6.13. Consider the following bases of R2: E ¼ fe1; e2g ¼ fð1; 0Þ; ð0; 1Þg and S ¼ fu1; u2g ¼ fð1; 3Þ; ð1; 4Þg (a) Find the change-of-basis matrix P from the usual basis E to S. (b) Find the change-of-basis matrix Q from S back to E. (c) Find the coordinate vector ½v of v ¼ ð5; 3Þ relative to S. (a) Because E is the usual basis, simply write the basis vectors in S as columns: P ¼ 1 1 3 4 (b) Method 1. Use the deﬁnition of the change-of-basis matrix. That is, express each vector in E as a linear combination of the vectors in S. We do this by ﬁrst ﬁnding the coordinates of an arbitrary vector v ¼ ða; bÞ relative to S. We have ða; bÞ ¼ xð1; 3Þ þ yð1; 4Þ ¼ ðx þ y; 3x þ 4yÞ or x þ y ¼ a 3x þ 4y ¼ b Solve for x and y to obtain x ¼ 4a b, y ¼ 3a þ b. Thus, v ¼ ð4a bÞu1 þ ð3a þ bÞu2 and ½vS ¼ ½ða; bÞS ¼ ½4a b; 3a þ bT Using the above formula for ½vS and writing the coordinates of the ei as columns yields e1 ¼ ð1; 0Þ ¼ 4u1 3u2 e2 ¼ ð0; 1Þ ¼ u1 þ u2 and Q ¼ 4 1 3 1 Method 2. Because Q ¼ P1; ﬁnd P1, say by using the formula for the inverse of a 2 2 matrix. Thus, P1 ¼ 4 1 3 1 (c) Method 1. Write v as a linear combination of the vectors in S, say by using the above formula for v ¼ ða; bÞ. We have v ¼ ð5; 3Þ ¼ 23u1 18u2; and so ½vS ¼ ½23; 18T. Method 2. Use, from Theorem 6.6, the fact that ½vS ¼ P1½vE and the fact that ½vE ¼ ½5; 3T: ½vS ¼ P1½vE ¼ 4 1 3 1 5 3 ¼ 23 18 CHAPTER 6 Linear Mappings and Matrices 211 6.14. The vectors u1 ¼ ð1; 2; 0Þ, u2 ¼ ð1; 3; 2Þ, u3 ¼ ð0; 1; 3Þ form a basis S of R3. Find (a) The change-of-basis matrix P from the usual basis E ¼ fe1; e2; e3g to S. (b) The change-of-basis matrix Q from S back to E. (a) Because E is the usual basis, simply write the basis vectors of S as columns: P ¼ 1 1 0 2 3 1 0 2 3 2 4 3 5 (b) Method 1. Express each basis vector of E as a linear combination of the basis vectors of S by ﬁrst ﬁnding the coordinates of an arbitrary vector v ¼ ða; b; cÞ relative to the basis S. We have a b c 2 4 3 5 ¼ x 1 2 0 2 4 3 5 þ y 1 3 2 2 4 3 5 þ z 0 1 3 2 4 3 5 or x þ y ¼ a 2x þ 3y þ z ¼ b 2y þ 3z ¼ c Solve for x; y; z to get x ¼ 7a 3b þ c, y ¼ 6a þ 3b c, z ¼ 4a 2b þ c. Thus, v ¼ ða; b; cÞ ¼ ð7a 3b þ cÞu1 þ ð6a þ 3b cÞu2 þ ð4a 2b þ cÞu3 or ½vS ¼ ½ða; b; cÞS ¼ ½7a 3b þ c; 6a þ 3b c; 4a 2b þ cT Using the above formula for ½vS and then writing the coordinates of the ei as columns yields e1 ¼ ð1; 0; 0Þ ¼ 7u1 6u2 þ 4u3 e2 ¼ ð0; 1; 0Þ ¼ 3u1 þ 3u2 2u3 e3 ¼ ð0; 0; 1Þ ¼ u1 u2 þ u3 and Q ¼ 7 3 1 6 3 1 4 2 1 2 4 3 5 Method 2. Find P1 by row reducing M ¼ ½P; I to the form ½I; P1: M ¼ 1 1 0 1 0 0 2 3 1 0 1 0 0 2 3 0 0 1 2 6 4 3 7 5 1 1 0 1 0 0 0 1 1 2 1 0 0 2 3 0 0 1 2 6 4 3 7 5 1 1 0 1 0 0 0 1 1 2 1 0 0 0 1 4 2 1 2 6 4 3 7 5 1 0 0 7 3 1 0 1 0 6 3 1 0 0 1 4 2 1 2 6 4 3 7 5 ¼ ½I; P1 Thus, Q ¼ P1 ¼ 7 3 1 6 3 1 4 2 1 2 4 3 5. 6.15. Suppose the x-axis and y-axis in the plane R2 are rotated counterclockwise 45 so that the new x 0-axis and y 0-axis are along the line y ¼ x and the line y ¼ x, respectively. (a) Find the change-of-basis matrix P. (b) Find the coordinates of the point Að5; 6Þ under the given rotation. (a) The unit vectors in the direction of the new x 0- and y 0-axes are u1 ¼ ð1 2 ﬃﬃﬃ 2 p ; 1 2 ﬃﬃﬃ 2 p Þ and u2 ¼ ð 1 2 ﬃﬃﬃ 2 p ; 1 2 ﬃﬃﬃ 2 p Þ (The unit vectors in the direction of the original x and y axes are the usual basis of R2.) Thus, write the coordinates of u1 and u2 as columns to obtain P ¼ 1 2 ﬃﬃﬃ 2 p 1 2 ﬃﬃﬃ 2 p 1 2 ﬃﬃﬃ 2 p 1 2 ﬃﬃﬃ 2 p " # (b) Multiply the coordinates of the point by P1: 1 2 ﬃﬃﬃ 2 p 1 2 ﬃﬃﬃ 2 p 1 2 ﬃﬃﬃ 2 p 1 2 ﬃﬃﬃ 2 p " # 5 6 ¼ 11 2 ﬃﬃﬃ 2 p 1 2 ﬃﬃﬃ 2 p " # (Because P is orthogonal, P1 is simply the transpose of P.) 212 CHAPTER 6 Linear Mappings and Matrices 6.16. The vectors u1 ¼ ð1; 1; 0Þ, u2 ¼ ð0; 1; 1Þ, u3 ¼ ð1; 2; 2Þ form a basis S of R3. Find the coordinates of an arbitrary vector v ¼ ða; b; cÞ relative to the basis S. Method 1. Express v as a linear combination of u1; u2; u3 using unknowns x; y; z. We have ða; b; cÞ ¼ xð1; 1; 0Þ þ yð0; 1; 1Þ þ zð1; 2; 2Þ ¼ ðx þ z; x þ y þ 2z; y þ 2zÞ this yields the system x þ z ¼ a x þ y þ 2z ¼ b y þ 2z ¼ c or x þ z ¼ a y þ z ¼ a þ b y þ 2z ¼ c or x þ z ¼ a y þ z ¼ a þ b z ¼ a b þ c Solving by back-substitution yields x ¼ b c, y ¼ 2a þ 2b c, z ¼ a b þ c. Thus, ½vS ¼ ½b c; 2a þ 2b c; a b þ cT Method 2. Find P1 by row reducing M ¼ ½P; I to the form ½I; P1, where P is the change-of-basis matrix from the usual basis E to S or, in other words, the matrix whose columns are the basis vectors of S. We have M ¼ 1 0 1 1 0 0 1 1 2 0 1 0 0 1 2 0 0 1 2 6 4 3 7 5 1 0 1 1 0 0 0 1 1 1 1 0 0 1 2 0 0 1 2 6 4 3 7 5 1 0 1 1 0 0 0 1 1 1 1 0 0 0 1 1 1 1 2 6 4 3 7 5 1 0 0 0 1 1 0 1 0 2 2 1 0 0 1 1 1 1 2 6 4 3 7 5 ¼ ½I; P1 Thus; P1 ¼ 0 1 1 2 2 1 1 1 1 2 6 4 3 7 5 and ½vS ¼ P1½vE ¼ 0 1 1 2 2 1 1 1 1 2 6 4 3 7 5 a b c 2 6 4 3 7 5 ¼ b c 2a þ 2b c a b þ c 2 6 4 3 7 5 6.17. Consider the following bases of R2: S ¼ fu1; u2g ¼ fð1; 2Þ; ð3; 4Þg and S0 ¼ fv1; v2g ¼ fð1; 3Þ; ð3; 8Þg (a) Find the coordinates of v ¼ ða; bÞ relative to the basis S. (b) Find the change-of-basis matrix P from S to S0. (c) Find the coordinates of v ¼ ða; bÞ relative to the basis S0. (d) Find the change-of-basis matrix Q from S0 back to S. (e) Verify Q ¼ P1. (f ) Show that, for any vector v ¼ ða; bÞ in R2, P1½vS ¼ ½vS0. (See Theorem 6.6.) (a) Let v ¼ xu1 þ yu2 for unknowns x and y; that is, a b ¼ x 1 2 þ y 3 4 or x þ 3y ¼ a 2x 4y ¼ b or x þ 3y ¼ a 2y ¼ 2a þ b Solve for x and y in terms of a and b to get x ¼ 2a 3 2 b and y ¼ a þ 1 2 b. Thus, ða; bÞ ¼ ð2a 3 2Þu1 þ ða þ 1 2 bÞu2 or ½ða; bÞS ¼ ½2a 3 2 b; a þ 1 2 bT (b) Use part (a) to write each of the basis vectors v1 and v2 of S0 as a linear combination of the basis vectors u1 and u2 of S; that is, v1 ¼ ð1; 3Þ ¼ ð2 9 2Þu1 þ ð1 þ 3 2Þu2 ¼ 13 2 u1 þ 5 2 u2 v2 ¼ ð3; 8Þ ¼ ð6 12Þu1 þ ð3 þ 4Þu2 ¼ 18u1 þ 7u2 CHAPTER 6 Linear Mappings and Matrices 213 Then P is the matrix whose columns are the coordinates of v1 and v2 relative to the basis S; that is, P ¼ 13 2 18 5 2 7 " # (c) Let v ¼ xv1 þ yv2 for unknown scalars x and y: a b ¼ x 1 3 þ y 3 8 or x þ 3y ¼ a 3x þ 8y ¼ b or x þ 3y ¼ a y ¼ b 3a Solve for x and y to get x ¼ 8a þ 3b and y ¼ 3a b. Thus, ða; bÞ ¼ ð8a þ 3bÞv1 þ ð3a bÞv2 or ½ða; bÞS0 ¼ ½8a þ 3b; 3a bT (d) Use part (c) to express each of the basis vectors u1 and u2 of S as a linear combination of the basis vectors v1 and v2 of S0: u1 ¼ ð1; 2Þ ¼ ð8 6Þv1 þ ð3 þ 2Þv2 ¼ 14v1 þ 5v2 u2 ¼ ð3; 4Þ ¼ ð24 12Þv1 þ ð9 þ 4Þv2 ¼ 36v1 þ 13v2 Write the coordinates of u1 and u2 relative to S0 as columns to obtain Q ¼ 14 36 5 13 . (e) QP ¼ 14 36 5 13 13 2 18 5 2 7 " # ¼ 1 0 0 1 ¼ I (f ) Use parts (a), (c), and (d) to obtain P1½vS ¼ Q½vS ¼ 14 36 5 13 2a 3 2 b a þ 1 2 b " # ¼ 8a þ 3b 3a b ¼ ½vS0 6.18. Suppose P is the change-of-basis matrix from a basis fuig to a basis fwig, and suppose Q is the change-of-basis matrix from the basis fwig back to fuig. Prove that P is invertible and that Q ¼ P1. Suppose, for i ¼ 1; 2; . . . ; n, that wi ¼ ai1u1 þ ai2u2 þ . . . þ ainun ¼ P n j¼1 aijuj ð1Þ and, for j ¼ 1; 2; . . . ; n, uj ¼ bj1w1 þ bj2w2 þ þ bjnwn ¼ P n k¼1 bjkwk ð2Þ Let A ¼ ½aij and B ¼ ½bjk. Then P ¼ AT and Q ¼ BT. Substituting (2) into (1) yields wi ¼ P n j¼1 aij P n k¼1 bjkwk ¼ P n k¼1 P n j¼1 aijbjk wk Because fwig is a basis, P aijbjk ¼ dik, where dik is the Kronecker delta; that is, dik ¼ 1 if i ¼ k but dik ¼ 0 if i 6¼ k. Suppose AB ¼ ½cik. Then cik ¼ dik. Accordingly, AB ¼ I, and so QP ¼ BTAT ¼ ðABÞT ¼ IT ¼ I Thus, Q ¼ P1. 6.19. Consider a ﬁnite sequence of vectors S ¼ fu1; u2; . . . ; ung. Let S0 be the sequence of vectors obtained from S by one of the following ‘‘elementary operations’’: (1) Interchange two vectors. (2) Multiply a vector by a nonzero scalar. (3) Add a multiple of one vector to another vector. Show that S and S0 span the same subspace W. Also, show that S0 is linearly independent if and only if S is linearly independent. 214 CHAPTER 6 Linear Mappings and Matrices Observe that, for each operation, the vectors S0 are linear combinations of vectors in S. Also, because each operation has an inverse of the same type, each vector in S is a linear combination of vectors in S0. Thus, S and S0 span the same subspace W. Moreover, S0 is linearly independent if and only if dim W ¼ n, and this is true if and only if S is linearly independent. 6.20. Let A ¼ ½aij and B ¼ ½bij be row equivalent m n matrices over a ﬁeld K, and let v1; v2; . . . ; vn be any vectors in a vector space V over K. For i ¼ 1; 2; . . . ; m, let ui and wi be deﬁned by ui ¼ ai1v1 þ ai2v2 þ þ ainvn and wi ¼ bi1v1 þ bi2v2 þ þ binvn Show that fuig and fwig span the same subspace of V. Applying an ‘‘elementary operation’’ of Problem 6.19 to fuig is equivalent to applying an elementary row operation to the matrix A. Because A and B are row equivalent, B can be obtained from A by a sequence of elementary row operations. Hence, fwig can be obtained from fuig by the corresponding sequence of operations. Accordingly, fuig and fwig span the same space. 6.21. Suppose u1; u2; . . . ; un belong to a vector space V over a ﬁeld K, and suppose P ¼ ½aij is an n-square matrix over K. For i ¼ 1; 2; . . . ; n, let vi ¼ ai1u1 þ ai2u2 þ þ ainun. (a) Suppose P is invertible. Show that fuig and fvig span the same subspace of V. Hence, fuig is linearly independent if and only if fvig is linearly independent. (b) Suppose P is singular (not invertible). Show that fvig is linearly dependent. (c) Suppose fvig is linearly independent. Show that P is invertible. (a) Because P is invertible, it is row equivalent to the identity matrix I. Hence, by Problem 6.19, fvig and fuig span the same subspace of V. Thus, one is linearly independent if and only if the other is linearly independent. (b) Because P is not invertible, it is row equivalent to a matrix with a zero row. This means fvig spans a substance that has a spanning set with less than n elements. Thus, fvig is linearly dependent. (c) This is the contrapositive of the statement of part (b), and so it follows from part (b). 6.22. Prove Theorem 6.6: Let P be the change-of-basis matrix from a basis S to a basis S0 in a vector space V. Then, for any vector v 2 V, we have P½vS0 ¼ ½vS, and hence, P1½vS ¼ ½vS0. Suppose S ¼ fu1; . . . ; ung and S0 ¼ fw1; . . . ; wng, and suppose, for i ¼ 1; . . . ; n, wi ¼ ai1u1 þ ai2u2 þ þ ainun ¼ P n j¼1 aijuj Then P is the n-square matrix whose jth row is ða1j; a2j; . . . ; anjÞ ð1Þ Also suppose v ¼ k1w1 þ k2w2 þ þ knwn ¼ Pn i¼1 kiwi. Then ½vS0 ¼ ½k1; k2; . . . ; knT ð2Þ Substituting for wi in the equation for v, we obtain v ¼ P n i¼1 kiwi ¼ P n i¼1 ki P n j¼1 aijuj ¼ P n j¼1 P n i¼1 aijki uj ¼ P n j¼1 ða1jk1 þ a2jk2 þ þ anjknÞuj Accordingly, ½vS is the column vector whose jth entry is a1jk1 þ a2jk2 þ þ anjkn ð3Þ On the other hand, the jth entry of P½vS0 is obtained by multiplying the jth row of P by ½vS0—that is, (1) by (2). However, the product of (1) and (2) is (3). Hence, P½vS0 and ½vS have the same entries. Thus, P½vS0 ¼ ½vS0, as claimed. Furthermore, multiplying the above by P1 gives P1½vS ¼ P1P½vS0 ¼ ½vS0. CHAPTER 6 Linear Mappings and Matrices 215 Linear Operators and Change of Basis 6.23. Consider the linear transformation F on R2 deﬁned by Fðx; yÞ ¼ ð5x y; 2x þ yÞ and the following bases of R2: E ¼ fe1; e2g ¼ fð1; 0Þ; ð0; 1Þg and S ¼ fu1; u2g ¼ fð1; 4Þ; ð2; 7Þg (a) Find the change-of-basis matrix P from E to S and the change-of-basis matrix Q from S back to E. (b) Find the matrix A that represents F in the basis E. (c) Find the matrix B that represents F in the basis S. (a) Because E is the usual basis, simply write the vectors in S as columns to obtain the change-of-basis matrix P. Recall, also, that Q ¼ P1. Thus, P ¼ 1 2 4 7 and Q ¼ P1 ¼ 7 2 4 1 (b) Write the coefﬁcients of x and y in Fðx; yÞ ¼ ð5x y; 2x þ yÞ as rows to get A ¼ 5 1 2 1 (c) Method 1. Find the coordinates of Fðu1Þ and Fðu2Þ relative to the basis S. This may be done by ﬁrst ﬁnding the coordinates of an arbitrary vector ða; bÞ in R2 relative to the basis S. We have ða; bÞ ¼ xð1; 4Þ þ yð2; 7Þ ¼ ðx þ 2y; 4x þ 7yÞ; and so x þ 2y ¼ a 4x þ 7y ¼ b Solve for x and y in terms of a and b to get x ¼ 7a þ 2b, y ¼ 4a b. Then ða; bÞ ¼ ð7a þ 2bÞu1 þ ð4a bÞu2 Now use the formula for ða; bÞ to obtain Fðu1Þ ¼ Fð1; 4Þ ¼ ð1; 6Þ ¼ 5u1 2u2 Fðu2Þ ¼ Fð2; 7Þ ¼ ð3; 11Þ ¼ u1 þ u2 and so B ¼ 5 1 2 1 Method 2. By Theorem 6.7, B ¼ P1AP. Thus, B ¼ P1AP ¼ 7 2 4 1 5 1 2 1 1 2 4 7 ¼ 5 1 2 1 6.24. Let A ¼ 2 3 4 1 . Find the matrix B that represents the linear operator A relative to the basis S ¼ fu1; u2g ¼ f½1; 3T; ½2; 5Tg. [Recall A deﬁnes a linear operator A: R2 ! R2 relative to the usual basis E of R2]. Method 1. Find the coordinates of Aðu1Þ and Aðu2Þ relative to the basis S by ﬁrst ﬁnding the coordinates of an arbitrary vector ½a; bT in R2 relative to the basis S. By Problem 6.2, ½a; bT ¼ ð5a þ 2bÞu1 þ ð3a bÞu2 Using the formula for ½a; bT, we obtain Aðu1Þ ¼ 2 3 4 1 1 3 ¼ 11 1 ¼ 53u1 þ 32u2 and Aðu2Þ ¼ 2 3 4 1 2 5 ¼ 19 3 ¼ 89u1 þ 54u2 Thus; B ¼ 53 89 32 54 Method 2. Use B ¼ P1AP, where P is the change-of-basis matrix from the usual basis E to S. Thus, simply write the vectors in S (as columns) to obtain the change-of-basis matrix P and then use the formula 216 CHAPTER 6 Linear Mappings and Matrices for P1. This gives P ¼ 1 2 3 5 and P1 ¼ 5 2 3 1 Then B ¼ P1AP ¼ 1 2 3 5 2 3 4 1 5 2 3 1 ¼ 53 89 32 54 6.25. Let A ¼ 1 3 1 2 5 4 1 2 2 2 4 3 5: Find the matrix B that represents the linear operator A relative to the basis S ¼ fu1; u2; u3g ¼ f½1; 1; 0T; ½0; 1; 1T; ½1; 2; 2Tg [Recall A that deﬁnes a linear operator A: R3 ! R3 relative to the usual basis E of R3.] Method 1. Find the coordinates of Aðu1Þ, Aðu2Þ, Aðu3Þ relative to the basis S by ﬁrst ﬁnding the coordinates of an arbitrary vector v ¼ ða; b; cÞ in R3 relative to the basis S. By Problem 6.16, ½vS ¼ ðb cÞu1 þ ð2a þ 2b cÞu2 þ ða b þ cÞu3 Using this formula for ½a; b; cT, we obtain Aðu1Þ ¼ ½4; 7; 1T ¼ 8u1 þ 7u2 5u3; Aðu2Þ ¼ ½4; 1; 0T ¼ u1 6u2 þ 3u3 Aðu3Þ ¼ ½9; 4; 1T ¼ 3u1 11u2 þ 6u3 Writing the coefﬁcients of u1; u2; u3 as columns yields B ¼ 8 1 3 7 6 11 5 3 6 2 4 3 5 Method 2. Use B ¼ P1AP, where P is the change-of-basis matrix from the usual basis E to S. The matrix P (whose columns are simply the vectors in S) and P1 appear in Problem 6.16. Thus, B ¼ P1AP ¼ 0 1 1 2 2 1 1 1 1 2 4 3 5 1 3 1 2 5 4 1 2 2 2 4 3 5 1 0 1 1 1 2 0 1 2 2 4 3 5 ¼ 8 1 3 7 6 11 5 3 6 2 4 3 5 6.26. Prove Theorem 6.7: Let P be the change-of-basis matrix from a basis S to a basis S0 in a vector space V. Then, for any linear operator T on V, ½TS0 ¼ P1½TSP. Let v be a vector in V. Then, by Theorem 6.6, P½vS0 ¼ ½vS. Therefore, P1½TSP½vS0 ¼ P1½TS½vS ¼ P1½TðvÞS ¼ ½TðvÞS0 But ½TS0½vS0 ¼ ½TðvÞS0. Hence, P1½TSP½vS0 ¼ ½TS0½vS0 Because the mapping v 7! ½vS0 is onto Kn, we have P1½TSPX ¼ ½TS0X for every X 2 Kn. Thus, P1½TSP ¼ ½TS0, as claimed. Similarity of Matrices 6.27. Let A ¼ 4 2 3 6 and P ¼ 1 2 3 4 . (a) Find B ¼ P1AP. (b) Verify trðBÞ ¼ trðAÞ: (c) Verify detðBÞ ¼ detðAÞ: (a) First ﬁnd P1 using the formula for the inverse of a 2 2 matrix. We have P1 ¼ 2 1 3 2 1 2 " # CHAPTER 6 Linear Mappings and Matrices 217 Then B ¼ P1AP ¼ 2 1 3 2 1 2 4 2 3 6 1 2 3 4 ¼ 25 30 27 2 15 (b) trðAÞ ¼ 4 þ 6 ¼ 10 and trðBÞ ¼ 25 15 ¼ 10. Hence, trðBÞ ¼ trðAÞ. (c) detðAÞ ¼ 24 þ 6 ¼ 30 and detðBÞ ¼ 375 þ 405 ¼ 30. Hence, detðBÞ ¼ detðAÞ. 6.28. Find the trace of each of the linear transformations F on R3 in Problem 6.4. Find the trace (sum of the diagonal elements) of any matrix representation of F such as the matrix representation ½F ¼ ½FE of F relative to the usual basis E given in Problem 6.4. (a) trðFÞ ¼ trð½FÞ ¼ 1 5 þ 9 ¼ 5. (b) trðFÞ ¼ trð½FÞ ¼ 1 þ 3 þ 5 ¼ 9. (c) trðFÞ ¼ trð½FÞ ¼ 1 þ 4 þ 7 ¼ 12. 6.29. Write A B if A is similar to B—that is, if there exists an invertible matrix P such that A ¼ P1BP. Prove that is an equivalence relation (on square matrices); that is, (a) A A, for every A. (b) If A B, then B A. (c) If A B and B C, then A C. (a) The identity matrix I is invertible, and I1 ¼ I. Because A ¼ I1AI, we have A A. (b) Because A B, there exists an invertible matrix P such that A ¼ P1BP. Hence, B ¼ PAP1 ¼ ðP1Þ1AP and P1 is also invertible. Thus, B A. (c) Because A B, there exists an invertible matrix P such that A ¼ P1BP, and as B C, there exists an invertible matrix Q such that B ¼ Q1CQ. Thus, A ¼ P1BP ¼ P1ðQ1CQÞP ¼ ðP1Q1ÞCðQPÞ ¼ ðQPÞ1CðQPÞ and QP is also invertible. Thus, A C. 6.30. Suppose B is similar to A, say B ¼ P1AP. Prove (a) Bn ¼ P1AnP, and so Bn is similar to An. (b) f ðBÞ ¼ P1f ðAÞP, for any polynomial f ðxÞ, and so f ðBÞ is similar to f ðAÞ: (c) B is a root of a polynomial gðxÞ if and only if A is a root of gðxÞ. (a) The proof is by induction on n. The result holds for n ¼ 1 by hypothesis. Suppose n > 1 and the result holds for n 1. Then Bn ¼ BBn1 ¼ ðP1APÞðP1An1PÞ ¼ P1AnP (b) Suppose f ðxÞ ¼ anxn þ þ a1x þ a0. Using the left and right distributive laws and part (a), we have P1f ðAÞP ¼ P1ðanAn þ þ a1A þ a0IÞP ¼ P1ðanAnÞP þ þ P1ða1AÞP þ P1ða0IÞP ¼ anðP1AnPÞ þ þ a1ðP1APÞ þ a0ðP1IPÞ ¼ anBn þ þ a1B þ a0I ¼ f ðBÞ (c) By part (b), gðBÞ ¼ 0 if and only if P1gðAÞP ¼ 0 if and only if gðAÞ ¼ P0P1 ¼ 0. Matrix Representations of General Linear Mappings 6.31. Let F: R3 ! R2 be the linear map deﬁned by Fðx; y; zÞ ¼ ð3x þ 2y 4z; x 5y þ 3zÞ. (a) Find the matrix of F in the following bases of R3 and R2: S ¼ fw1; w2; w3g ¼ fð1; 1; 1Þ; ð1; 1; 0Þ; ð1; 0; 0Þg and S0 ¼ fu1; u2g ¼ fð1; 3Þ; ð2; 5Þg 218 CHAPTER 6 Linear Mappings and Matrices (b) Verify Theorem 6.10: The action of F is preserved by its matrix representation; that is, for any v in R3, we have ½FS;S0½vS ¼ ½FðvÞS0. (a) From Problem 6.2, ða; bÞ ¼ ð5a þ 2bÞu1 þ ð3a bÞu2. Thus, Fðw1Þ ¼ Fð1; 1; 1Þ ¼ ð1; 1Þ ¼ 7u1 þ 4u2 Fðw2Þ ¼ Fð1; 1; 0Þ ¼ ð5; 4Þ ¼ 33u1 þ 19u2 Fðw3Þ ¼ Fð1; 0; 0Þ ¼ ð3; 1Þ ¼ 13u1 þ 8u2 Write the coordinates of Fðw1Þ, Fðw2Þ; Fðw3Þ as columns to get ½FS;S0 ¼ 7 33 13 4 19 8 (b) If v ¼ ðx; y; zÞ, then, by Problem 6.5, v ¼ zw1 þ ðy zÞw2 þ ðx yÞw3. Also, FðvÞ ¼ ð3x þ 2y 4z; x 5y þ 3zÞ ¼ ð13x 20y þ 26zÞu1 þ ð8x þ 11y 15zÞu2 Hence; ½vS ¼ ðz; y z; x yÞT and ½FðvÞS0 ¼ 13x 20y þ 26z 8x þ 11y 15z Thus, ½FS;S0½vS ¼ 7 33 13 4 19 8 z y x x y 2 4 3 5 ¼ 13x 20y þ 26z 8x þ 11y 15z ¼ ½FðvÞS0 6.32. Let F: Rn ! Rm be the linear mapping deﬁned as follows: Fðx1; x2; . . . ; xnÞ ¼ ða11x1 þ þ a1nxn, a21x1 þ þ a2nxn; . . . ; am1x1 þ þ amnxnÞ (a) Show that the rows of the matrix ½F representing F relative to the usual bases of Rn and Rm are the coefﬁcients of the xi in the components of Fðx1; . . . ; xnÞ. (b) Find the matrix representation of each of the following linear mappings relative to the usual basis of Rn: (i) F: R2 ! R3 deﬁned by Fðx; yÞ ¼ ð3x y; 2x þ 4y; 5x 6yÞ. (ii) F: R4 ! R2 deﬁned by Fðx; y; s; tÞ ¼ ð3x 4y þ 2s 5t; 5x þ 7y s 2tÞ. (iii) F: R3 ! R4 deﬁned by Fðx; y; zÞ ¼ ð2x þ 3y 8z; x þ y þ z; 4x 5z; 6yÞ. (a) We have Fð1; 0; . . . ; 0Þ ¼ ða11; a21; . . . ; am1Þ Fð0; 1; . . . ; 0Þ ¼ ða12; a22; . . . ; am2Þ ::::::::::::::::::::::::::::::::::::::::::::::::::::: Fð0; 0; . . . ; 1Þ ¼ ða1n; a2n; . . . ; amnÞ and thus; ½F ¼ a11 a12 . . . a1n a21 a22 . . . a2n ::::::::::::::::::::::::::::::::: am1 am2 . . . amn 2 6 6 4 3 7 7 5 (b) By part (a), we need only look at the coefﬁcients of the unknown x; y; . . . in Fðx; y; . . .Þ. Thus, ðiÞ ½F ¼ 3 1 2 4 5 6 2 4 3 5; ðiiÞ ½F ¼ 3 4 2 5 5 7 1 2 ; ðiiiÞ ½F ¼ 2 3 8 1 1 1 4 0 5 0 6 0 2 6 6 4 3 7 7 5 6.33. Let A ¼ 2 5 3 1 4 7 . Recall that A determines a mapping F: R3 ! R2 deﬁned by FðvÞ ¼ Av, where vectors are written as columns. Find the matrix ½F that represents the mapping relative to the following bases of R3 and R2: (a) The usual bases of R3 and of R2. (b) S ¼ fw1; w2; w3g ¼ fð1; 1; 1Þ; ð1; 1; 0Þ; ð1; 0; 0Þg and S0 ¼ fu1; u2g ¼ fð1; 3Þ; ð2; 5Þg. (a) Relative to the usual bases, ½F is the matrix A. CHAPTER 6 Linear Mappings and Matrices 219 (b) From Problem 9.2, ða; bÞ ¼ ð5a þ 2bÞu1 þ ð3a bÞu2. Thus, Fðw1Þ ¼ 2 5 3 1 4 7 1 1 1 2 6 4 3 7 5 ¼ 4 4 ¼ 12u1 þ 8u2 Fðw2Þ ¼ 2 5 3 1 4 7 1 1 0 2 6 4 3 7 5 ¼ 7 3 ¼ 41u1 þ 24u2 Fðw3Þ ¼ 2 5 3 1 4 7 1 0 0 2 6 4 3 7 5 ¼ 2 1 ¼ 8u1 þ 5u2 Writing the coefﬁcients of Fðw1Þ, Fðw2Þ, Fðw3Þ as columns yields ½F ¼ 12 41 8 8 24 5 . 6.34. Consider the linear transformation T on R2 deﬁned by Tðx; yÞ ¼ ð2x 3y; x þ 4yÞ and the following bases of R2: E ¼ fe1; e2g ¼ fð1; 0Þ; ð0; 1Þg and S ¼ fu1; u2g ¼ fð1; 3Þ; ð2; 5Þg (a) Find the matrix A representing T relative to the bases E and S. (b) Find the matrix B representing T relative to the bases S and E. (We can view T as a linear mapping from one space into another, each having its own basis.) (a) From Problem 6.2, ða; bÞ ¼ ð5a þ 2bÞu1 þ ð3a bÞu2. Hence, Tðe1Þ ¼ Tð1; 0Þ ¼ ð2; 1Þ ¼ 8u1 þ 5u2 Tðe2Þ ¼ Tð0; 1Þ ¼ ð3; 4Þ ¼ 23u1 13u2 and so A ¼ 8 23 5 13 (b) We have Tðu1Þ ¼ Tð1; 3Þ ¼ ð7; 13Þ ¼ 7e1 þ 13e2 Tðu2Þ ¼ Tð2; 5Þ ¼ ð11; 22Þ ¼ 11e1 þ 22e2 and so B ¼ 7 11 13 22 6.35. How are the matrices A and B in Problem 6.34 related? By Theorem 6.12, the matrices A and B are equivalent to each other; that is, there exist nonsingular matrices P and Q such that B ¼ Q1AP, where P is the change-of-basis matrix from S to E, and Q is the change-of-basis matrix from E to S. Thus, P ¼ 1 2 3 5 ; Q ¼ 5 2 3 1 ; Q1 ¼ 1 2 3 5 and Q1AP ¼ 1 2 3 5 8 23 5 13 1 2 3 5 ¼ 7 11 13 22 ¼ B 6.36. Prove Theorem 6.14: Let F: V ! U be linear and, say, rankðFÞ ¼ r. Then there exist bases V and of U such that the matrix representation of F has the following form, where Ir is the r-square identity matrix: A ¼ Ir 0 0 0 Suppose dim V ¼ m and dim U ¼ n. Let W be the kernel of F and U0 the image of F. We are given that rank ðFÞ ¼ r. Hence, the dimension of the kernel of F is m r. Let fw1; . . . ; wmrg be a basis of the kernel of F and extend this to a basis of V: fv1; . . . ; vr; w1; . . . ; wmrg Set u1 ¼ Fðv1Þ; u2 ¼ Fðv2Þ; . . . ; ur ¼ FðvrÞ 220 CHAPTER 6 Linear Mappings and Matrices Then fu1; . . . ; urg is a basis of U0, the image of F. Extend this to a basis of U, say fu1; . . . ; ur; urþ1; . . . ; ung Observe that Fðv1Þ ¼ u1 ¼ 1u1 þ 0u2 þ þ 0ur þ 0urþ1 þ þ 0un Fðv2Þ ¼ u2 ¼ 0u1 þ 1u2 þ þ 0ur þ 0urþ1 þ þ 0un :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: FðvrÞ ¼ ur ¼ 0u1 þ 0u2 þ þ 1ur þ 0urþ1 þ þ 0un Fðw1Þ ¼ 0 ¼ 0u1 þ 0u2 þ þ 0ur þ 0urþ1 þ þ 0un :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: FðwmrÞ ¼ 0 ¼ 0u1 þ 0u2 þ þ 0ur þ 0urþ1 þ þ 0un Thus, the matrix of F in the above bases has the required form. SUPPLEMENTARY PROBLEMS Matrices and Linear Operators 6.37. Let F: R2 ! R2 be deﬁned by Fðx; yÞ ¼ ð4x þ 5y; 2x yÞ. (a) Find the matrix A representing F in the usual basis E. (b) Find the matrix B representing F in the basis S ¼ fu1; u2g ¼ fð1; 4Þ; ð2; 9Þg. (c) Find P such that B ¼ P1AP. (d) For v ¼ ða; bÞ, ﬁnd ½vS and ½FðvÞS. Verify that ½FS½vS ¼ ½FðvÞS. 6.38. Let A: R2 ! R2 be deﬁned by the matrix A ¼ 5 1 2 4 . (a) Find the matrix B representing A relative to the basis S ¼ fu1; u2g ¼ fð1; 3Þ; ð2; 8Þg. (Recall that A represents the mapping A relative to the usual basis E.) (b) For v ¼ ða; bÞ, ﬁnd ½vS and ½AðvÞS. 6.39. For each linear transformation L on R2, ﬁnd the matrix A representing L (relative to the usual basis of R2): (a) L is the rotation in R2 counterclockwise by 45. (b) L is the reﬂection in R2 about the line y ¼ x. (c) L is deﬁned by Lð1; 0Þ ¼ ð3; 5Þ and Lð0; 1Þ ¼ ð7; 2Þ. (d) L is deﬁned by Lð1; 1Þ ¼ ð3; 7Þ and Lð1; 2Þ ¼ ð5; 4Þ. 6.40. Find the matrix representing each linear transformation T on R3 relative to the usual basis of R3: (a) Tðx; y; zÞ ¼ ðx; y; 0Þ. (b) Tðx; y; zÞ ¼ ðz; y þ z; x þ y þ zÞ. (c) Tðx; y; zÞ ¼ ð2x 7y 4z; 3x þ y þ 4z; 6x 8y þ zÞ. 6.41. Repeat Problem 6.40 using the basis S ¼ fu1; u2; u3g ¼ fð1; 1; 0Þ; ð1; 2; 3Þ; ð1; 3; 5Þg. 6.42. Let L be the linear transformation on R3 deﬁned by Lð1; 0; 0Þ ¼ ð1; 1; 1Þ; Lð0; 1; 0Þ ¼ ð1; 3; 5Þ; Lð0; 0; 1Þ ¼ ð2; 2; 2Þ (a) Find the matrix A representing L relative to the usual basis of R3. (b) Find the matrix B representing L relative to the basis S in Problem 6.41. 6.43. Let D denote the differential operator; that is, Dð f ðtÞÞ ¼ df =dt. Each of the following sets is a basis of a vector space V of functions. Find the matrix representing D in each basis: (a) fet; e2t; te2tg. (b) f1; t; sin 3t; cos 3tg. (c) fe5t; te5t; t2e5tg. CHAPTER 6 Linear Mappings and Matrices 221 6.44. Let D denote the differential operator on the vector space V of functions with basis S ¼ fsin y, cos yg. (a) Find the matrix A ¼ ½DS. (b) Use A to show that D is a zero of f ðtÞ ¼ t2 þ 1. 6.45. Let V be the vector space of 2 2 matrices. Consider the following matrix M and usual basis E of V: M ¼ a b c d and E ¼ 1 0 0 0 ; 0 1 0 0 ; 0 0 1 0 ; 0 0 0 1 Find the matrix representing each of the following linear operators T on V relative to E: (a) TðAÞ ¼ MA. (b) TðAÞ ¼ AM. (c) TðAÞ ¼ MA AM. 6.46. Let 1V and 0V denote the identity and zero operators, respectively, on a vector space V. Show that, for any basis S of V, (a) ½1VS ¼ I, the identity matrix. (b) ½0VS ¼ 0, the zero matrix. Change of Basis 6.47. Find the change-of-basis matrix P from the usual basis E of R2 to a basis S, the change-of-basis matrix Q from S back to E, and the coordinates of v ¼ ða; bÞ relative to S, for the following bases S: (a) S ¼ fð1; 2Þ; ð3; 5Þg. (c) S ¼ fð2; 5Þ; ð3; 7Þg. (b) S ¼ fð1; 3Þ; ð3; 8Þg. (d) S ¼ fð2; 3Þ; ð4; 5Þg. 6.48. Consider the bases S ¼ fð1; 2Þ; ð2; 3Þg and S0 ¼ fð1; 3Þ; ð1; 4Þg of R2. Find the change-of-basis matrix: (a) P from S to S0. (b) Q from S0 back to S. 6.49. Suppose that the x-axis and y-axis in the plane R2 are rotated counterclockwise 30 to yield new x 0-axis and y 0-axis for the plane. Find (a) The unit vectors in the direction of the new x 0-axis and y 0-axis. (b) The change-of-basis matrix P for the new coordinate system. (c) The new coordinates of the points Að1; 3Þ, Bð2; 5Þ, Cða; bÞ. 6.50. Find the change-of-basis matrix P from the usual basis E of R3 to a basis S, the change-of-basis matrix Q from S back to E, and the coordinates of v ¼ ða; b; cÞ relative to S, where S consists of the vectors: (a) u1 ¼ ð1; 1; 0Þ; u2 ¼ ð0; 1; 2Þ; u3 ¼ ð0; 1; 1Þ. (b) u1 ¼ ð1; 0; 1Þ; u2 ¼ ð1; 1; 2Þ; u3 ¼ ð1; 2; 4Þ. (c) u1 ¼ ð1; 2; 1Þ; u2 ¼ ð1; 3; 4Þ; u3 ¼ ð2; 5; 6Þ. 6.51. Suppose S1; S2; S3 are bases of V. Let P and Q be the change-of-basis matrices, respectively, from S1 to S2 and from S2 to S3. Prove that PQ is the change-of-basis matrix from S1 to S3. Linear Operators and Change of Basis 6.52. Consider the linear operator F on R2 deﬁned by Fðx; yÞ ¼ ð5x þ y; 3x 2yÞ and the following bases of R2: S ¼ fð1; 2Þ; ð2; 3Þg and S0 ¼ fð1; 3Þ; ð1; 4Þg (a) Find the matrix A representing F relative to the basis S. (b) Find the matrix B representing F relative to the basis S0. (c) Find the change-of-basis matrix P from S to S0. (d) How are A and B related? 6.53. Let A: R2 ! R2 be deﬁned by the matrix A ¼ 1 1 3 2 . Find the matrix B that represents the linear operator A relative to each of the following bases: (a) S ¼ fð1; 3ÞT; ð2; 5ÞTg. (b) S ¼ fð1; 3ÞT; ð2; 4ÞTg. 222 CHAPTER 6 Linear Mappings and Matrices 6.54. Let F: R2 ! R2 be deﬁned by Fðx; yÞ ¼ ðx 3y; 2x 4yÞ. Find the matrix A that represents F relative to each of the following bases: (a) S ¼ fð2; 5Þ; ð3; 7Þg. (b) S ¼ fð2; 3Þ; ð4; 5Þg. 6.55. Let A: R3 ! R3 be deﬁned by the matrix A ¼ 1 3 1 2 7 4 1 4 3 2 4 3 5. Find the matrix B that represents the linear operator A relative to the basis S ¼ fð1; 1; 1ÞT; ð0; 1; 1ÞT; ð1; 2; 3ÞTg. Similarity of Matrices 6.56. Let A ¼ 1 1 2 3 and P ¼ 1 2 3 5 . (a) Find B ¼ P1AP. (b) Verify that trðBÞ ¼ trðAÞ: (c) Verify that detðBÞ ¼ detðAÞ. 6.57. Find the trace and determinant of each of the following linear maps on R2: (a) Fðx; yÞ ¼ ð2x 3y; 5x þ 4yÞ. (b) Gðx; yÞ ¼ ðax þ by; cx þ dyÞ. 6.58. Find the trace and determinant of each of the following linear maps on R3: (a) Fðx; y; zÞ ¼ ðx þ 3y; 3x 2z; x 4y 3zÞ. (b) Gðx; y; zÞ ¼ ðy þ 3z; 2x 4z; 5x þ 7yÞ. 6.59. Suppose S ¼ fu1; u2g is a basis of V, and T: V ! V is deﬁned by Tðu1Þ ¼ 3u1 2u2 and Tðu2Þ ¼ u1 þ 4u2. Suppose S0 ¼ fw1; w2g is a basis of V for which w1 ¼ u1 þ u2 and w2 ¼ 2u1 þ 3u2. (a) Find the matrices A and B representing T relative to the bases S and S0, respectively. (b) Find the matrix P such that B ¼ P1AP. 6.60. Let A be a 2 2 matrix such that only A is similar to itself. Show that A is a scalar matrix, that is, that A ¼ a 0 0 a . 6.61. Show that all matrices similar to an invertible matrix are invertible. More generally, show that similar matrices have the same rank. Matrix Representation of General Linear Mappings 6.62. Find the matrix representation of each of the following linear maps relative to the usual basis for Rn: (a) F: R3 ! R2 deﬁned by Fðx; y; zÞ ¼ ð2x 4y þ 9z; 5x þ 3y 2zÞ. (b) F: R2 ! R4 deﬁned by Fðx; yÞ ¼ ð3x þ 4y; 5x 2y; x þ 7y; 4xÞ: (c) F: R4 ! R deﬁned by Fðx1; x2; x3; x4Þ ¼ 2x1 þ x2 7x3 x4. 6.63. Let G: R3 ! R2 be deﬁned by Gðx; y; zÞ ¼ ð2x þ 3y z; 4x y þ 2zÞ. (a) Find the matrix A representing G relative to the bases S ¼ fð1; 1; 0Þ; ð1; 2; 3Þ; ð1; 3; 5Þg and S0 ¼ fð1; 2Þ; ð2; 3Þg (b) For any v ¼ ða; b; cÞ in R3, ﬁnd ½vS and ½GðvÞS0. (c) Verify that A½vS ¼ ½GðvÞS0. 6.64. Let H: R2 ! R2 be deﬁned by Hðx; yÞ ¼ ð2x þ 7y; x 3yÞ and consider the following bases of R2: S ¼ fð1; 1Þ; ð1; 2Þg and S0 ¼ fð1; 4Þ; ð1; 5Þg (a) Find the matrix A representing H relative to the bases S and S0. (b) Find the matrix B representing H relative to the bases S0 and S. CHAPTER 6 Linear Mappings and Matrices 223 6.65. Let F: R3 ! R2 be deﬁned by Fðx; y; zÞ ¼ ð2x þ y z; 3x 2y þ 4zÞ. (a) Find the matrix A representing F relative to the bases S ¼ fð1; 1; 1Þ; ð1; 1; 0Þ; ð1; 0; 0Þg and S0 ¼ ð1; 3Þ; ð1; 4Þg (b) Verify that, for any v ¼ ða; b; cÞ in R3, A½vS ¼ ½FðvÞS0. 6.66. Let S and S0 be bases of V, and let 1V be the identity mapping on V. Show that the matrix A representing 1V relative to the bases S and S0 is the inverse of the change-of-basis matrix P from S to S0; that is, A ¼ P1. 6.67. Prove (a) Theorem 6.10, (b) Theorem 6.11, (c) Theorem 6.12, (d) Theorem 6.13. [Hint: See the proofs of the analogous Theorems 6.1 (Problem 6.9), 6.2 (Problem 6.10), 6.3 (Problem 6.11), and 6.7 (Problem 6.26).] Miscellaneous Problems 6.68. Suppose F: V ! V is linear. A subspace W of V is said to be invariant under F if FðWÞ W. Suppose W is invariant under F and dim W ¼ r. Show that F has a block triangular matrix representation M ¼ A B 0 C where A is an r r submatrix. 6.69. Suppose V ¼ U þ W, and suppose U and V are each invariant under a linear operator F: V ! V. Also, suppose dim U ¼ r and dim W ¼ S. Show that F has a block diagonal matrix representation M ¼ A 0 0 B where A and B are r r and s s submatrices. 6.70. Two linear operators F and G on V are said to be similar if there exists an invertible linear operator T on V such that G ¼ T 1 F T. Prove (a) F and G are similar if and only if, for any basis S of V, ½FS and ½GS are similar matrices. (b) If F is diagonalizable (similar to a diagonal matrix), then any similar matrix G is also diagonalizable. ANSWERS TO SUPPLEMENTARY PROBLEMS Notation: M ¼ ½R1; R2; . . . represents a matrix M with rows R1; R2; . . . : 6.37. (a) A ¼ ½4; 5; 2; 1; (b) B ¼ ½220; 487; 98; 217; (c) P ¼ ½1; 2; 4; 9; (d) ½vS ¼ ½9a 2b; 4a þ bT and ½FðvÞS ¼ ½32a þ 47b; 14a 21bT 6.38. (a) B ¼ ½6; 28; 4; 15; (b) ½vS ¼ ½4a b; 3 2 a þ 1 2 bT and ½AðvÞS ¼ ½18a 8b; 1 2 ð13a þ 7bÞ 6.39. (a) ½ ﬃﬃﬃ 2 p ; ﬃﬃﬃ 2 p ; ﬃﬃﬃ 2 p ; ﬃﬃﬃ 2 p ; (b) ½0; 1; 1; 0; (c) ½3; 7; 5; 2; (d) ½1; 2; 18; 11 6.40. (a) ½1; 0; 0; 0; 1; 0; 0; 0; 0; (b) ½0; 0; 1; 0; 1; 1; 1; 1; 1; (c) ½2; 7; 4; 3; 1; 4; 6; 8; 1 6.41. (a) ½1; 3; 5; 0; 5; 10; 0; 3; 6; (b) ½0; 1; 2; 1; 2; 3; 1; 0; 0; (c) ½15; 65; 104; 49; 219; 351; 29; 130; 208 6.42. (a) ½1; 1; 2; 1; 3; 2; 1; 5; 2; (b) ½0; 2; 14; 22; 0; 5; 8 6.43. (a) ½1; 0; 0; 0; 2; 1; 0; 0; 2; (b) ½0; 1; 0; 0; 0; 0; 0; 0; 3; 0; 0; 3; 0; (c) ½5; 1; 0; 0; 5; 2; 0; 0; 5 224 CHAPTER 6 Linear Mappings and Matrices 6.44. (a) A ¼ ½0; 1; 1; 0; (b) A2 þ I ¼ 0 6.45. (a) ½a; 0; b; 0; 0; a; 0; b; c; 0; d; 0; 0; c; 0; d; (b) ½a; c; 0; 0; b; d; 0; 0; 0; 0; a; c; 0; 0; b; d; (c) ½0; c; b; 0; b; a d; 0; b; c; 0; d a; c; 0; c; b; 0 6.47. (a) ½1; 3; 2; 5; ½5; 3; 2; 1; ½v ¼ ½5a þ 3b; 2a bT; (b) ½1; 3; 3; 8; ½8; 3; 3; 1; ½v ¼ ½8a 3b; 3a þ bT; (c) ½2; 3; 5; 7; ½7; 3; 5; 2; ½v ¼ ½7a þ 3b; 5a 2bT; (d) ½2; 4; 3; 5; ½ 5 2 ; 2; 3 2 ; 1; ½v ¼ ½ 5 2 a þ 2b; 3 2 a bT 6.48. (a) P ¼ ½3; 5; 1; 2; (b) Q ¼ ½2; 5; 1; 3 6.49. Here K ¼ ﬃﬃﬃ 3 p : (a) 1 2 ðK; 1Þ; 1 2 ð1; KÞ; ðbÞ P ¼ 1 2 ½K; 1; 1; K; ðcÞ 1 2 ½K þ 3; 3K 1T; 1 2 ½2K 5; 5K 2T; 1 2 ½aK þ b; bK aT 6.50. P is the matrix whose columns are u1; u2; u3; Q ¼ P1; ½v ¼ Q½a; b; cT: (a) Q ¼ ½1; 0; 0; 1; 1; 1; 2; 2; 1; ½v ¼ ½a; a b þ c; 2a þ 2b cT; (b) Q ¼ ½0; 2; 1; 2; 3; 2; 1; 1; 1; ½v ¼ ½2b þ c; 2a þ 3b 2c; a b þ cT; (c) Q ¼ ½2; 2; 1; 7; 4; 1; 5; 3; 1; ½v ¼ ½2a þ 2b c; 7a þ 4b c; 5a 3b þ cT 6.52. (a) ½23; 39; 15; 26; (b) ½35; 41; 27; 32; (c) ½3; 5; 1; 2; (d) B ¼ P1AP 6.53. (a) ½28; 47; 15; 25; (b) ½13; 18; 15 2 ; 10 6.54. (a) ½43; 60; 33; 46; (b) 1 2 ½3; 7; 5; 9 6.55. ½10; 8; 20; 13; 11; 28; 5; 4; 10 6.56. (a) ½34; 57; 19; 32; (b) trðBÞ ¼ trðAÞ ¼ 2; (c) detðBÞ ¼ detðAÞ ¼ 5 6.57. (a) trðFÞ ¼ 6; detðFÞ ¼ 23; (b) trðGÞ ¼ a þ d; detðGÞ ¼ ad bc 6.58. (a) trðFÞ ¼ 2; detðFÞ ¼ 13; (b) trðGÞ ¼ 0; detðGÞ ¼ 22 6.59. (a) A ¼ ½3; 1; 2; 4; B ¼ ½8; 11; 2; 1; (b) P ¼ ½1; 2; 1; 3 6.62. (a) ½2; 4; 9; 5; 3; 2; (b) ½3; 5; 1; 4; 4; 2; 7; 0; (c) ½2; 1; 7; 1 6.63. (a) ½9; 1; 4; 7; 2; 1; (b) ½vS ¼ ½a þ 2b c; 5a 5b þ 2c; 3a þ 3b cT, and ½GðvÞS0 ¼ ½2a 11b þ 7c; 7b 4cT 6.64. (a) A ¼ ½47; 85; 38; 69; (b) B ¼ ½71; 88; 41; 51 6.65. A ¼ ½3; 11; 5; 1; 8; 3 CHAPTER 6 Linear Mappings and Matrices 225 Inner Product Spaces, Orthogonality 7.1 Introduction The deﬁnition of a vector space V involves an arbitrary ﬁeld K. Here we ﬁrst restrict K to be the real ﬁeld R, in which case V is called a real vector space; in the last sections of this chapter, we extend our results to the case where K is the complex ﬁeld C, in which case V is called a complex vector space. Also, we adopt the previous notation that u; v; w are vectors in V a; b; c; k are scalars in K Furthermore, the vector spaces V in this chapter have ﬁnite dimension unless otherwise stated or implied. Recall that the concepts of ‘‘length’’ and ‘‘orthogonality’’ did not appear in the investigation of arbitrary vector spaces V (although they did appear in Section 1.4 on the spaces Rn and Cn). Here we place an additional structure on a vector space V to obtain an inner product space, and in this context these concepts are deﬁned. 7.2 Inner Product Spaces We begin with a deﬁnition. DEFINITION: Let V be a real vector space. Suppose to each pair of vectors u; v 2 V there is assigned a real number, denoted by hu; vi. This function is called a (real) inner product on V if it satisﬁes the following axioms: ½I1 (Linear Property): hau1 þ bu2; vi ¼ ahu1; vi þ bhu2; vi. ½I2 (Symmetric Property): hu; vi ¼ hv; ui. ½I3 (Positive Deﬁnite Property): hu; ui 0.; and hu; ui ¼ 0 if and only if u ¼ 0. The vector space V with an inner product is called a (real) inner product space. Axiom ½I1 states that an inner product function is linear in the ﬁrst position. Using ½I1 and the symmetry axiom ½I2, we obtain hu; cv1 þ dv2i ¼ hcv1 þ dv2; ui ¼ chv1; ui þ dhv2; ui ¼ chu; v1i þ dhu; v2i That is, the inner product function is also linear in its second position. Combining these two properties and using induction yields the following general formula: P i aiui; P j bjvj ¼ P i P j aibjhui; vji CHAPTER 7 226 That is, an inner product of linear combinations of vectors is equal to a linear combination of the inner products of the vectors. EXAMPLE 7.1 Let V be a real inner product space. Then, by linearity, h3u1 4u2; 2v1 5v2 þ 6v3i ¼ 6hu1; v1i 15hu1; v2i þ 18hu1; v3i 8hu2; v1i þ 20hu2; v2i 24hu2; v3i h2u 5v; 4u þ 6vi ¼ 8hu; ui þ 12hu; vi 20hv; ui 30hv; vi ¼ 8hu; ui 8hv; ui 30hv; vi Observe that in the last equation we have used the symmetry property that hu; vi ¼ hv; ui. Remark: Axiom ½I1 by itself implies h0; 0i ¼ h0v; 0i ¼ 0hv; 0i ¼ 0: Thus, ½I1, ½I2, ½I3 are equivalent to ½I1, ½I2, and the following axiom: ½I0 3 If u 6¼ 0; then hu; ui is positive: That is, a function satisfying ½I1, ½I2, ½I0 3 is an inner product. Norm of a Vector By the third axiom ½I3 of an inner product, hu; ui is nonnegative for any vector u. Thus, its positive square root exists. We use the notation kuk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ hu; ui p This nonnegative number is called the norm or length of u. The relation kuk2 ¼ hu; ui will be used frequently. Remark: If kuk ¼ 1 or, equivalently, if hu; ui ¼ 1, then u is called a unit vector and it is said to be normalized. Every nonzero vector v in V can be multiplied by the reciprocal of its length to obtain the unit vector ^ v ¼ 1 kvk v which is a positive multiple of v. This process is called normalizing v. 7.3 Examples of Inner Product Spaces This section lists the main examples of inner product spaces used in this text. Euclidean n-Space Rn Consider the vector space Rn. The dot product or scalar product in Rn is deﬁned by u v ¼ a1b1 þ a2b2 þ þ anbn where u ¼ ðaiÞ and v ¼ ðbiÞ. This function deﬁnes an inner product on Rn. The norm kuk of the vector u ¼ ðaiÞ in this space is as follows: kuk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ u u p ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ a2 1 þ a2 2 þ þ a2 n q On the other hand, by the Pythagorean theorem, the distance from the origin O in R3 to a point Pða; b; cÞ is given by ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ a2 þ b2 þ c2 p . This is precisely the same as the above-deﬁned norm of the vector v ¼ ða; b; cÞ in R3. Because the Pythagorean theorem is a consequence of the axioms of CHAPTER 7 Inner Product Spaces, Orthogonality 227 Euclidean geometry, the vector space Rn with the above inner product and norm is called Euclidean n-space. Although there are many ways to deﬁne an inner product on Rn, we shall assume this inner product unless otherwise stated or implied. It is called the usual (or standard) inner product on Rn. Remark: Frequently the vectors in Rn will be represented by column vectors—that is, by n 1 column matrices. In such a case, the formula hu; vi ¼ uTv deﬁnes the usual inner product on Rn. EXAMPLE 7.2 Let u ¼ ð1; 3; 4; 2Þ, v ¼ ð4; 2; 2; 1Þ, w ¼ ð5; 1; 2; 6Þ in R4. (a) Show h3u 2v; wi ¼ 3hu; wi 2hv; wi: By deﬁnition, hu; wi ¼ 5 3 þ 8 þ 12 ¼ 22 and hv; wi ¼ 20 þ 2 4 þ 6 ¼ 24 Note that 3u 2v ¼ ð5; 13; 16; 4Þ. Thus, h3u 2v; wi ¼ 25 13 þ 32 þ 24 ¼ 18 As expected, 3hu; wi 2hv; wi ¼ 3ð22Þ 2ð24Þ ¼ 18 ¼ h3u 2v; wi. (b) Normalize u and v: By deﬁnition, kuk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1 þ 9 þ 16 þ 4 p ¼ ﬃﬃﬃﬃﬃ 30 p and kvk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 16 þ 4 þ 4 þ 1 p ¼ 5 We normalize u and v to obtain the following unit vectors in the directions of u and v, respectively: ^ u ¼ 1 kuk u ¼ 1 ﬃﬃﬃﬃﬃ 30 p ; 3 ﬃﬃﬃﬃﬃ 30 p ; 4 ﬃﬃﬃﬃﬃ 30 p ; 2 ﬃﬃﬃﬃﬃ 30 p and ^ v ¼ 1 kvk v ¼ 4 5 ; 2 5 ; 2 5 ; 1 5 Function Space C½a; b and Polynomial Space PðtÞ The notation C½a; b is used to denote the vector space of all continuous functions on the closed interval ½a; b—that is, where a t b. The following deﬁnes an inner product on C½a; b, where f ðtÞ and gðtÞ are functions in C½a; b: h f ; gi ¼ ðb a f ðtÞgðtÞ dt It is called the usual inner product on C½a; b. The vector space PðtÞ of all polynomials is a subspace of C½a; b for any interval ½a; b, and hence, the above is also an inner product on PðtÞ. EXAMPLE 7.3 Consider f ðtÞ ¼ 3t 5 and gðtÞ ¼ t2 in the polynomial space PðtÞ with inner product h f ; gi ¼ ð1 0 f ðtÞgðtÞ dt: (a) Find h f ; gi. We have f ðtÞgðtÞ ¼ 3t3 5t2. Hence, h f ; gi ¼ ð1 0 ð3t3 5t2Þ dt ¼ 3 4 t4 5 3 t3 1 0 ¼ 3 4 5 3 ¼ 11 12 228 CHAPTER 7 Inner Product Spaces, Orthogonality (b) Find k f k and kgk. We have ½ f ðtÞ2 ¼ f ðtÞ f ðtÞ ¼ 9t2 30t þ 25 and ½gðtÞ2 ¼ t4. Then k f k2 ¼ h f ; f i ¼ ð1 0 ð9t2 30t þ 25Þ dt ¼ 3t3 15t2 þ 25t 1 0 ¼ 13 kgk2 ¼ hg; gi ¼ ð1 0 t4 dt ¼ 1 5 t5 1 0 ¼ 1 5 Therefore, k f k ¼ ﬃﬃﬃﬃﬃ 13 p and kgk ¼ ﬃﬃ 1 5 q ¼ 1 5 ﬃﬃﬃ 5 p . Matrix Space M ¼ Mm;n Let M ¼ Mm;n, the vector space of all real m n matrices. An inner product is deﬁned on M by hA; Bi ¼ trðBTAÞ where, as usual, trð Þ is the trace—the sum of the diagonal elements. If A ¼ ½aij and B ¼ ½bij, then hA; Bi ¼ trðBTAÞ ¼ P m i¼1 P n j¼1 aijbij and kAk2 ¼ hA; Ai ¼ P m i¼1 P n j¼1 a2 ij That is, hA; Bi is the sum of the products of the corresponding entries in A and B and, in particular, hA; Ai is the sum of the squares of the entries of A. Hilbert Space Let V be the vector space of all inﬁnite sequences of real numbers ða1; a2; a3; . . .Þ satisfying P 1 i¼1 a2 i ¼ a2 1 þ a2 2 þ < 1 that is, the sum converges. Addition and scalar multiplication are deﬁned in V componentwise; that is, if u ¼ ða1; a2; . . .Þ and v ¼ ðb1; b2; . . .Þ then u þ v ¼ ða1 þ b1; a2 þ b2; . . .Þ and ku ¼ ðka1; ka2; . . .Þ An inner product is deﬁned in v by hu; vi ¼ a1b1 þ a2b2 þ The above sum converges absolutely for any pair of points in V. Hence, the inner product is well deﬁned. This inner product space is called l2-space or Hilbert space. 7.4 Cauchy–Schwarz Inequality, Applications The following formula (proved in Problem 7.8) is called the Cauchy–Schwarz inequality or Schwarz inequality. It is used in many branches of mathematics. THEOREM 7.1: (Cauchy–Schwarz) For any vectors u and v in an inner product space V, hu; vi2 hu; uihv; vi or jhu; vij kukkvk Next we examine this inequality in speciﬁc cases. EXAMPLE 7.4 (a) Consider any real numbers a1; . . . ; an, b1; . . . ; bn. Then, by the Cauchy–Schwarz inequality, ða1b1 þ a2b2 þ þ anbnÞ2 ða2 1 þ þ a2 nÞðb2 1 þ þ b2 nÞ That is, ðu vÞ2 kuk2kvk2, where u ¼ ðaiÞ and v ¼ ðbiÞ. CHAPTER 7 Inner Product Spaces, Orthogonality 229 (b) Let f and g be continuous functions on the unit interval ½0; 1. Then, by the Cauchy–Schwarz inequality, ð1 0 f ðtÞgðtÞ dt 2 ð1 0 f 2ðtÞ dt ð1 0 g2ðtÞ dt That is, ðh f ; giÞ2 k f k2kvk2. Here V is the inner product space C½0; 1. The next theorem (proved in Problem 7.9) gives the basic properties of a norm. The proof of the third property requires the Cauchy–Schwarz inequality. THEOREM 7.2: Let V be an inner product space. Then the norm in V satisﬁes the following properties: ½N1 kvk 0; and kvk ¼ 0 if and only if v ¼ 0. ½N2 kkvk ¼ jkjkvk. ½N3 ku þ vk kuk þ kvk. The property ½N3 is called the triangle inequality, because if we view u þ v as the side of the triangle formed with sides u and v (as shown in Fig. 7-1), then ½N3 states that the length of one side of a triangle cannot be greater than the sum of the lengths of the other two sides. Angle Between Vectors For any nonzero vectors u and v in an inner product space V, the angle between u and v is deﬁned to be the angle y such that 0 y p and cos y ¼ hu; vi kukkvk By the Cauchy–Schwartz inequality, 1 cos y 1, and so the angle exists and is unique. EXAMPLE 7.5 (a) Consider vectors u ¼ ð2; 3; 5Þ and v ¼ ð1; 4; 3Þ in R3. Then hu; vi ¼ 2 12 þ 15 ¼ 5; kuk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ 4 þ 9 þ 25 p ¼ ﬃﬃﬃﬃﬃ 38 p ; kvk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1 þ 16 þ 9 p ¼ ﬃﬃﬃﬃﬃ 26 p Then the angle y between u and v is given by cos y ¼ 5 ﬃﬃﬃﬃﬃ 38 p ﬃﬃﬃﬃﬃ 26 p Note that y is an acute angle, because cos y is positive. (b) Let f ðtÞ ¼ 3t 5 and gðtÞ ¼ t2 in the polynomial space PðtÞ with inner product h f ; gi ¼ Ð 1 0 f ðtÞgðtÞ dt. By Example 7.3, h f ; gi ¼ 11 12 ; k f k ¼ ﬃﬃﬃﬃﬃ 13 p ; kgk ¼ 1 5 ﬃﬃﬃ 5 p Then the ‘‘angle’’ y between f and g is given by cos y ¼ 11 12 ð ﬃﬃﬃﬃﬃ 13 p Þ 1 5 ﬃﬃﬃ 5 p ¼ 55 12 ﬃﬃﬃﬃﬃ 13 p ﬃﬃﬃ 5 p Note that y is an obtuse angle, because cos y is negative. Figure 7-1 230 CHAPTER 7 Inner Product Spaces, Orthogonality 7.5 Orthogonality Let V be an inner product space. The vectors u; v 2 V are said to be orthogonal and u is said to be orthogonal to v if hu; vi ¼ 0 The relation is clearly symmetric—if u is orthogonal to v, then hv; ui ¼ 0, and so v is orthogonal to u. We note that 0 2 V is orthogonal to every v 2 V, because h0; vi ¼ h0v; vi ¼ 0hv; vi ¼ 0 Conversely, if u is orthogonal to every v 2 V, then hu; ui ¼ 0 and hence u ¼ 0 by ½I3: Observe that u and v are orthogonal if and only if cos y ¼ 0, where y is the angle between u and v. Also, this is true if and only if u and v are ‘‘perpendicular’’—that is, y ¼ p=2 (or y ¼ 90). EXAMPLE 7.6 (a) Consider the vectors u ¼ ð1; 1; 1Þ, v ¼ ð1; 2; 3Þ, w ¼ ð1; 4; 3Þ in R3. Then hu; vi ¼ 1 þ 2 3 ¼ 0; hu; wi ¼ 1 4 þ 3 ¼ 0; hv; wi ¼ 1 8 9 ¼ 16 Thus, u is orthogonal to v and w, but v and w are not orthogonal. (b) Consider the functions sin t and cos t in the vector space C½p; p of continuous functions on the closed interval ½p; p. Then hsin t; cos ti ¼ ðp p sin t cos t dt ¼ 1 2 sin2 tjp p ¼ 0 0 ¼ 0 Thus, sin t and cos t are orthogonal functions in the vector space C½p; p. Remark: A vector w ¼ ðx1; x2; . . . ; xnÞ is orthogonal to u ¼ ða1; a2; . . . ; anÞ in Rn if hu; wi ¼ a1x1 þ a2x2 þ þ anxn ¼ 0 That is, w is orthogonal to u if w satisﬁes a homogeneous equation whose coefﬁcients are the elements of u. EXAMPLE 7.7 Find a nonzero vector w that is orthogonal to u1 ¼ ð1; 2; 1Þ and u2 ¼ ð2; 5; 4Þ in R3. Let w ¼ ðx; y; zÞ. Then we want hu1; wi ¼ 0 and hu2; wi ¼ 0. This yields the homogeneous system x þ 2y þ z ¼ 0 2x þ 5y þ 4z ¼ 0 or x þ 2y þ z ¼ 0 y þ 2z ¼ 0 Here z is the only free variable in the echelon system. Set z ¼ 1 to obtain y ¼ 2 and x ¼ 3. Thus, w ¼ ð3; 2; 1Þ is a desired nonzero vector orthogonal to u1 and u2. Any multiple of w will also be orthogonal to u1 and u2. Normalizing w, we obtain the following unit vector orthogonal to u1 and u2: ^ w ¼ w kwk ¼ 3 ﬃﬃﬃﬃﬃ 14 p ; 2 ﬃﬃﬃﬃﬃ 14 p ; 1 ﬃﬃﬃﬃﬃ 14 p Orthogonal Complements Let S be a subset of an inner product space V. The orthogonal complement of S, denoted by S? (read ‘‘S perp’’) consists of those vectors in V that are orthogonal to every vector u 2 S; that is, S? ¼ fv 2 V : hv; ui ¼ 0 for every u 2 Sg CHAPTER 7 Inner Product Spaces, Orthogonality 231 In particular, for a given vector u in V, we have u? ¼ fv 2 V : hv; ui ¼ 0g that is, u? consists of all vectors in V that are orthogonal to the given vector u. We show that S? is a subspace of V. Clearly 0 2 S?, because 0 is orthogonal to every vector in V. Now suppose v, w 2 S?. Then, for any scalars a and b and any vector u 2 S, we have hav þ bw; ui ¼ ahv; ui þ bhw; ui ¼ a 0 þ b 0 ¼ 0 Thus, av þ bw 2 S?, and therefore S? is a subspace of V. We state this result formally. PROPOSITION 7.3: Let S be a subset of a vector space V. Then S? is a subspace of V. Remark 1: Suppose u is a nonzero vector in R3. Then there is a geometrical description of u?. Speciﬁcally, u? is the plane in R3 through the origin O and perpendicular to the vector u. This is shown in Fig. 7-2. Remark 2: Let W be the solution space of an m n homogeneous system AX ¼ 0, where A ¼ ½aij and X ¼ ½xi. Recall that W may be viewed as the kernel of the linear mapping A: Rn ! Rm. Now we can give another interpretation of W using the notion of orthogonality. Speciﬁcally, each solution vector w ¼ ðx1; x2; . . . ; xnÞ is orthogonal to each row of A; hence, W is the orthogonal complement of the row space of A. EXAMPLE 7.8 Find a basis for the subspace u? of R3, where u ¼ ð1; 3; 4Þ. Note that u? consists of all vectors w ¼ ðx; y; zÞ such that hu; wi ¼ 0, or x þ 3y 4z ¼ 0. The free variables are y and z. (1) Set y ¼ 1, z ¼ 0 to obtain the solution w1 ¼ ð3; 1; 0Þ. (2) Set y ¼ 0, z ¼ 1 to obtain the solution w1 ¼ ð4; 0; 1Þ. The vectors w1 and w2 form a basis for the solution space of the equation, and hence a basis for u?. Suppose W is a subspace of V. Then both W and W ? are subspaces of V. The next theorem, whose proof (Problem 7.28) requires results of later sections, is a basic result in linear algebra. THEOREM 7.4: Let W be a subspace of V. Then V is the direct sum of W and W ?; that is, V ¼ W W ?. Figure 7-2 232 CHAPTER 7 Inner Product Spaces, Orthogonality 7.6 Orthogonal Sets and Bases Consider a set S ¼ fu1; u2; . . . ; urg of nonzero vectors in an inner product space V. S is called orthogonal if each pair of vectors in S are orthogonal, and S is called orthonormal if S is orthogonal and each vector in S has unit length. That is, (i) Orthogonal: hui; uji ¼ 0 for i 6¼ j (ii) Orthonormal: hui; uji ¼ 0 for i 6¼ j 1 for i ¼ j Normalizing an orthogonal set S refers to the process of multiplying each vector in S by the reciprocal of its length in order to transform S into an orthonormal set of vectors. The following theorems apply. THEOREM 7.5: Suppose S is an orthogonal set of nonzero vectors. Then S is linearly independent. THEOREM 7.6: (Pythagoras) Suppose fu1; u2; . . . ; urg is an orthogonal set of vectors. Then ku1 þ u2 þ þ urk2 ¼ ku1k2 þ ku2k2 þ þ kurk2 These theorems are proved in Problems 7.15 and 7.16, respectively. Here we prove the Pythagorean theorem in the special and familiar case for two vectors. Speciﬁcally, suppose hu; vi ¼ 0. Then ku þ vk2 ¼ hu þ v; u þ vi ¼ hu; ui þ 2hu; vi þ hv; vi ¼ hu; ui þ hv; vi ¼ kuk2 þ kvk2 which gives our result. EXAMPLE 7.9 (a) Let E ¼ fe1; e2; e3g ¼ fð1; 0; 0Þ; ð0; 1; 0Þ; ð0; 0; 1Þg be the usual basis of Euclidean space R3. It is clear that he1; e2i ¼ he1; e3i ¼ he2; e3i ¼ 0 and he1; e1i ¼ he2; e2i ¼ he3; e3i ¼ 1 Namely, E is an orthonormal basis of R3. More generally, the usual basis of Rn is orthonormal for every n. (b) Let V ¼ C½p; p be the vector space of continuous functions on the interval p t p with inner product deﬁned by h f ; gi ¼ Ð p p f ðtÞgðtÞ dt. Then the following is a classical example of an orthogonal set in V: f1; cos t; cos 2t; cos 3t; . . . ; sin t; sin 2t; sin 3t; . . .g This orthogonal set plays a fundamental role in the theory of Fourier series. Orthogonal Basis and Linear Combinations, Fourier Coefﬁcients Let S consist of the following three vectors in R3: u1 ¼ ð1; 2; 1Þ; u2 ¼ ð2; 1; 4Þ; u3 ¼ ð3; 2; 1Þ The reader can verify that the vectors are orthogonal; hence, they are linearly independent. Thus, S is an orthogonal basis of R3. Suppose we want to write v ¼ ð7; 1; 9Þ as a linear combination of u1; u2; u3. First we set v as a linear combination of u1; u2; u3 using unknowns x1; x2; x3 as follows: v ¼ x1u1 þ x2u2 þ x3u3 or ð7; 1; 9Þ ¼ x1ð1; 2; 1Þ þ x2ð2; 1; 4Þ þ x3ð3; 2; 1Þ ðÞ We can proceed in two ways. METHOD 1: Expand ðÞ (as in Chapter 3) to obtain the system x1 þ 2x2 þ 3x3 ¼ 7; 2x1 þ x2 2x3 ¼ 1; x1 4x2 þ x3 ¼ 7 Solve the system by Gaussian elimination to obtain x1 ¼ 3, x2 ¼ 1, x3 ¼ 2. Thus, v ¼ 3u1 u2 þ 2u3. CHAPTER 7 Inner Product Spaces, Orthogonality 233 METHOD 2: (This method uses the fact that the basis vectors are orthogonal, and the arithmetic is much simpler.) If we take the inner product of each side of ðÞ with respect to ui, we get hv; uii ¼ hx1u2 þ x2u2 þ x3u3; uii or hv; uii ¼ xihui; uii or xi ¼ hv; uii hui; uii Here two terms drop out, because u1; u2; u3 are orthogonal. Accordingly, x1 ¼ hv; u1i hu1; u1i ¼ 7 þ 2 þ 9 1 þ 4 þ 1 ¼ 18 6 ¼ 3; x2 ¼ hv; u2i hu2; u2i ¼ 14 þ 1 36 4 þ 1 þ 16 ¼ 21 21 ¼ 1 x3 ¼ hv; u3i hu3; u3i ¼ 21 2 þ 9 9 þ 4 þ 1 ¼ 28 14 ¼ 2 Thus, again, we get v ¼ 3u1 u2 þ 2u3. The procedure in Method 2 is true in general. Namely, we have the following theorem (proved in Problem 7.17). THEOREM 7.7: Let fu1; u2; . . . ; ung be an orthogonal basis of V. Then, for any v 2 V, v ¼ hv; u1i hu1; u1i u1 þ hv; u2i hu2; u2i u2 þ þ hv; uni hun; uni un Remark: The scalar ki hv; uii hui; uii is called the Fourier coefﬁcient of v with respect to ui, because it is analogous to a coefﬁcient in the Fourier series of a function. This scalar also has a geometric interpretation, which is discussed below. Projections Let V be an inner product space. Suppose w is a given nonzero vector in V, and suppose v is another vector. We seek the ‘‘projection of v along w,’’ which, as indicated in Fig. 7-3(a), will be the multiple cw of w such that v0 ¼ v cw is orthogonal to w. This means hv cw; wi ¼ 0 or hv; wi chw; wi ¼ 0 or c ¼ hv; wi hw; wi Accordingly, the projection of v along w is denoted and deﬁned by projðv; wÞ ¼ cw ¼ hv; wi hw; wi w Such a scalar c is unique, and it is called the Fourier coefﬁcient of v with respect to w or the component of v along w. The above notion is generalized as follows (see Problem 7.25). Figure 7-3 234 CHAPTER 7 Inner Product Spaces, Orthogonality THEOREM 7.8: Suppose w1; w2; . . . ; wr form an orthogonal set of nonzero vectors in V. Let v be any vector in V. Deﬁne v0 ¼ v ðc1w1 þ c2w2 þ þ crwrÞ where c1 ¼ hv; w1i hw1; w1i ; c2 ¼ hv; w2i hw2; w2i ; . . . ; cr ¼ hv; wri hwr; wri Then v0 is orthogonal to w1; w2; . . . ; wr. Note that each ci in the above theorem is the component (Fourier coefﬁcient) of v along the given wi. Remark: The notion of the projection of a vector v 2 V along a subspace W of V is deﬁned as follows. By Theorem 7.4, V ¼ W W ?. Hence, v may be expressed uniquely in the form v ¼ w þ w0; where w 2 W and w0 2 W ? We deﬁne w to be the projection of v along W, and denote it by projðv; WÞ, as pictured in Fig. 7-2(b). In particular, if W ¼ spanðw1; w2; . . . ; wrÞ, where the wi form an orthogonal set, then projðv; WÞ ¼ c1w1 þ c2w2 þ þ crwr Here ci is the component of v along wi, as above. 7.7 Gram–Schmidt Orthogonalization Process Suppose fv1; v2; . . . ; vng is a basis of an inner product space V. One can use this basis to construct an orthogonal basis fw1; w2; . . . ; wng of V as follows. Set w1 ¼ v1 w2 ¼ v2 hv2; w1i hw1; w1i w1 w3 ¼ v3 hv3; w1i hw1; w1i w1 hv3; w2i hw2; w2i w2 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: wn ¼ vn hvn; w1i hw1; w1i w1 hvn; w2i hw2; w2i w2 hvn; wn1i hwn1; wn1i wn1 In other words, for k ¼ 2; 3; . . . ; n, we deﬁne wk ¼ vk ck1w1 ck2w2 ck;k1wk1 where cki ¼ hvk; wii=hwi; wii is the component of vk along wi. By Theorem 7.8, each wk is orthogonal to the preceeding w’s. Thus, w1; w2; . . . ; wn form an orthogonal basis for V as claimed. Normalizing each wi will then yield an orthonormal basis for V. The above construction is known as the Gram–Schmidt orthogonalization process. The following remarks are in order. Remark 1: Each vector wk is a linear combination of vk and the preceding w’s. Hence, one can easily show, by induction, that each wk is a linear combination of v1; v2; . . . ; vn. Remark 2: Because taking multiples of vectors does not affect orthogonality, it may be simpler in hand calculations to clear fractions in any new wk, by multiplying wk by an appropriate scalar, before obtaining the next wkþ1. CHAPTER 7 Inner Product Spaces, Orthogonality 235 Remark 3: Suppose u1; u2; . . . ; ur are linearly independent, and so they form a basis for U ¼ spanðuiÞ. Applying the Gram–Schmidt orthogonalization process to the u’s yields an orthogonal basis for U. The following theorems (proved in Problems 7.26 and 7.27) use the above algorithm and remarks. THEOREM 7.9: Let fv1; v2; . . . ; vng be any basis of an inner product space V. Then there exists an orthonormal basis fu1; u2; . . . ; ung of V such that the change-of-basis matrix from fvig to fuig is triangular; that is, for k ¼ 1; . . . ; n, uk ¼ ak1v1 þ ak2v2 þ þ akkvk THEOREM 7.10: Suppose S ¼ fw1; w2; . . . ; wrg is an orthogonal basis for a subspace W of a vector space V. Then one may extend S to an orthogonal basis for V; that is, one may ﬁnd vectors wrþ1; . . . ; wn such that fw1; w2; . . . ; wng is an orthogonal basis for V. EXAMPLE 7.10 Apply the Gram–Schmidt orthogonalization process to ﬁnd an orthogonal basis and then an orthonormal basis for the subspace U of R4 spanned by v1 ¼ ð1; 1; 1; 1Þ; v2 ¼ ð1; 2; 4; 5Þ; v3 ¼ ð1; 3; 4; 2Þ (1) First set w1 ¼ v1 ¼ ð1; 1; 1; 1Þ. (2) Compute v2 hv2; w1i hw1; w1i w1 ¼ v2 12 4 w1 ¼ ð2; 1; 1; 2Þ Set w2 ¼ ð2; 1; 1; 2Þ. (3) Compute v3 hv3; w1i hw1; w1i w1 hv3; w2i hw2; w2i w2 ¼ v3 ð8Þ 4 w1 ð7Þ 10 w2 ¼ 8 5 ; 17 10 ; 13 10 ; 7 5 Clear fractions to obtain w3 ¼ ð6; 17; 13; 14Þ. Thus, w1; w2; w3 form an orthogonal basis for U. Normalize these vectors to obtain an orthonormal basis fu1; u2; u3g of U. We have kw1k2 ¼ 4, kw2k2 ¼ 10, kw3k2 ¼ 910, so u1 ¼ 1 2 ð1; 1; 1; 1Þ; u2 ¼ 1 ﬃﬃﬃﬃﬃ 10 p ð2; 1; 1; 2Þ; u3 ¼ 1 ﬃﬃﬃﬃﬃﬃﬃ ﬃ 910 p ð16; 17; 13; 14Þ EXAMPLE 7.11 Let V be the vector space of polynomials f ðtÞ with inner product h f ; gi ¼ Ð 1 1 f ðtÞgðtÞ dt. Apply the Gram–Schmidt orthogonalization process to f1; t; t2; t3g to ﬁnd an orthogonal basis f f0; f1; f2; f3g with integer coefﬁcients for P3ðtÞ. Here we use the fact that, for r þ s ¼ n, htr; tsi ¼ ð1 1 tn dt ¼ tnþ1 n þ 1 1 1 ¼ 2=ðn þ 1Þ when n is even 0 when n is odd (1) First set f0 ¼ 1. (2) Compute t ¼ ht; 1i h1; 1i ð1Þ ¼ t 0 ¼ t. Set f1 ¼ t. (3) Compute t2 ht2; 1i h1; 1i ð1Þ ht2; ti ht; ti ðtÞ ¼ t2 2 3 2 ð1Þ þ 0ðtÞ ¼ t2 1 3 Multiply by 3 to obtain f2 ¼ 3t2 ¼ 1. 236 CHAPTER 7 Inner Product Spaces, Orthogonality (4) Compute t3 ht3; 1i h1; 1i ð1Þ ht3; ti ht; ti ðtÞ ht3; 3t2 1i h3t2 1; 3t2 1i ð3t2 1Þ ¼ t3 0ð1Þ 2 5 2 3 ðtÞ 0ð3t2 1Þ ¼ t3 3 5 t Multiply by 5 to obtain f3 ¼ 5t3 3t. Thus, f1; t; 3t2 1; 5t3 3tg is the required orthogonal basis. Remark: Normalizing the polynomials in Example 7.11 so that pð1Þ ¼ 1 yields the polynomials 1; t; 1 2 ð3t2 1Þ; 1 2 ð5t3 3tÞ These are the ﬁrst four Legendre polynomials, which appear in the study of differential equations. 7.8 Orthogonal and Positive Deﬁnite Matrices This section discusses two types of matrices that are closely related to real inner product spaces V. Here vectors in Rn will be represented by column vectors. Thus, hu; vi ¼ uTv denotes the inner product in Euclidean space Rn. Orthogonal Matrices A real matrix P is orthogonal if P is nonsingular and P1 ¼ PT, or, in other words, if PPT ¼ PTP ¼ I. First we recall (Theorem 2.6) an important characterization of such matrices. THEOREM 7.11: Let P be a real matrix. Then the following are equivalent: (a) P is orthogonal; (b) the rows of P form an orthonormal set; (c) the columns of P form an orthonormal set. (This theorem is true only using the usual inner product on Rn. It is not true if Rn is given any other inner product.) EXAMPLE 7.12 (a) Let P ¼ 1= ﬃﬃﬃ 3 p 1= ﬃﬃﬃ 3 p 1= ﬃﬃﬃ 3 p 0 1= ﬃﬃﬃ 2 p 1= ﬃﬃﬃ 2 p 2= ﬃﬃﬃ 6 p 1= ﬃﬃﬃ 6 p 1= ﬃﬃﬃ 6 p 2 4 3 5: The rows of P are orthogonal to each other and are unit vectors. Thus P is an orthogonal matrix. (b) Let P be a 2 2 orthogonal matrix. Then, for some real number y, we have P ¼ cos y sin y sin y cos y or P ¼ cos y sin y sin y cos y The following two theorems (proved in Problems 7.37 and 7.38) show important relationships between orthogonal matrices and orthonormal bases of a real inner product space V. THEOREM 7.12: Suppose E ¼ feig and E0 ¼ fe0 ig are orthonormal bases of V. Let P be the change-of-basis matrix from the basis E to the basis E0. Then P is orthogonal. THEOREM 7.13: Let fe1; . . . ; eng be an orthonormal basis of an inner product space V. Let P ¼ ½aij be an orthogonal matrix. Then the following n vectors form an orthonormal basis for V: e0 i ¼ a1ie1 þ a2ie2 þ þ anien; i ¼ 1; 2; . . . ; n CHAPTER 7 Inner Product Spaces, Orthogonality 237 Positive Deﬁnite Matrices Let A be a real symmetric matrix; that is, AT ¼ A. Then A is said to be positive deﬁnite if, for every nonzero vector u in Rn, hu; Aui ¼ uTAu > 0 Algorithms to decide whether or not a matrix A is positive deﬁnite will be given in Chapter 12. However, for 2 2 matrices, we have simple criteria that we state formally in the following theorem (proved in Problem 7.43). THEOREM 7.14: A 2 2 real symmetric matrix A ¼ a b c d ¼ a b b d is positive deﬁnite if and only if the diagonal entries a and d are positive and the determinant jAj ¼ ad bc ¼ ad b2 is positive. EXAMPLE 7.13 Consider the following symmetric matrices: A ¼ 1 3 3 4 ; B ¼ 1 2 2 3 ; C ¼ 1 2 2 5 A is not positive deﬁnite, because jAj ¼ 4 9 ¼ 5 is negative. B is not positive deﬁnite, because the diagonal entry 3 is negative. However, C is positive deﬁnite, because the diagonal entries 1 and 5 are positive, and the determinant jCj ¼ 5 4 ¼ 1 is also positive. The following theorem (proved in Problem 7.44) holds. THEOREM 7.15: Let A be a real positive deﬁnite matrix. Then the function hu; vi ¼ uTAv is an inner product on Rn. Matrix Representation of an Inner Product (Optional) Theorem 7.15 says that every positive deﬁnite matrix A determines an inner product on Rn. This subsection may be viewed as giving the converse of this result. Let V be a real inner product space with basis S ¼ fu1; u2; . . . ; ung. The matrix A ¼ ½aij; where aij ¼ hui; uji is called the matrix representation of the inner product on V relative to the basis S. Observe that A is symmetric, because the inner product is symmetric; that is, hui; uji ¼ huj; uii. Also, A depends on both the inner product on V and the basis S for V. Moreover, if S is an orthogonal basis, then A is diagonal, and if S is an orthonormal basis, then A is the identity matrix. EXAMPLE 7.14 The vectors u1 ¼ ð1; 1; 0Þ, u2 ¼ ð1; 2; 3Þ, u3 ¼ ð1; 3; 5Þ form a basis S for Euclidean space R3. Find the matrix A that represents the inner product in R3 relative to this basis S. First compute each hui; uji to obtain hu1; u1i ¼ 1 þ 1 þ 0 ¼ 2; hu2; u2i ¼ 1 þ 4 þ 9 ¼ 14; hu1; u2i ¼ 1 þ 2 þ 0 ¼ 3; hu2; u3i ¼ 1 þ 6 þ 15 ¼ 22; hu1; u3i ¼ 1 þ 3 þ 0 ¼ 4 hu3; u3i ¼ 1 þ 9 þ 25 ¼ 35 Then A ¼ 2 3 4 3 14 22 4 22 35 2 4 3 5. As expected, A is symmetric. The following theorems (proved in Problems 7.45 and 7.46, respectively) hold. THEOREM 7.16: Let A be the matrix representation of an inner product relative to basis S for V. Then, for any vectors u; v 2 V, we have hu; vi ¼ ½uTA½v where ½u and ½v denote the (column) coordinate vectors relative to the basis S. 238 CHAPTER 7 Inner Product Spaces, Orthogonality THEOREM 7.17: Let A be the matrix representation of any inner product on V. Then A is a positive deﬁnite matrix. 7.9 Complex Inner Product Spaces This section considers vector spaces over the complex ﬁeld C. First we recall some properties of the complex numbers (Section 1.7), especially the relations between a complex number z ¼ a þ bi; where a; b 2 R; and its complex conjugate z ¼ a bi: z z ¼ a2 þ b2; jzj ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ a2 þ b2 p ; z1 þ z2 ¼ z1 þ z2 z1z2 ¼ z1z2; z ¼ z Also, z is real if and only if z ¼ z. The following deﬁnition applies. DEFINITION: Let V be a vector space over C. Suppose to each pair of vectors, u; v 2 V there is assigned a complex number, denoted by hu; vi. This function is called a (complex) inner product on V if it satisﬁes the following axioms: ½I1 (Linear Property) hau1 þ bu2; vi ¼ ahu1; vi þ bhu2; vi ½I2 (Conjugate Symmetric Property) hu; vi ¼ hv; ui ½I3 (Positive Deﬁnite Property) hu; ui 0; and hu; ui ¼ 0 if and only if u ¼ 0. The vector space V over C with an inner product is called a (complex) inner product space. Observe that a complex inner product differs from the real case only in the second axiom ½I 2 : Axiom ½I1 (Linear Property) is equivalent to the two conditions: ðaÞ hu1 þ u2; vi ¼ hu1; vi þ hu2; vi; ðbÞ hku; vi ¼ khu; vi On the other hand, applying ½I1 and ½I2 , we obtain hu; kvi ¼ hkv; ui ¼ khv; ui ¼ khv; ui ¼ khu; vi That is, we must take the conjugate of a complex number when it is taken out of the second position of a complex inner product. In fact (Problem 7.47), the inner product is conjugate linear in the second position; that is, hu; av1 þ bv2i ¼ ahu; v1i þ bhu; v2i Combining linear in the ﬁrst position and conjugate linear in the second position, we obtain, by induction, P i aiui; P j bjvj + ¼ P i;j aibjhui; vji The following remarks are in order. Remark 1: Axiom ½I1 by itself implies that h0; 0i ¼ h0v; 0i ¼ 0hv; 0i ¼ 0. Accordingly, ½I1 , ½I2 , and ½I3 are equivalent to ½I1 , ½I2 , and the following axiom: ½I3 0 If u 6¼ 0; then hu; ui > 0: That is, a function satisfying ½I1, ½I2 , and ½I3 0 is a (complex) inner product on V. Remark 2: By ½I2 ; hu; ui ¼ hu; ui. Thus, hu; ui must be real. By ½I3 ; hu; ui must be nonnegative, and hence, its positive real square root exists. As with real inner product spaces, we deﬁne kuk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ hu; ui p to be the norm or length of u. Remark 3: In addition to the norm, we deﬁne the notions of orthogonality, orthogonal comple-ment, and orthogonal and orthonormal sets as before. In fact, the deﬁnitions of distance and Fourier coefﬁcient and projections are the same as in the real case. CHAPTER 7 Inner Product Spaces, Orthogonality 239 EXAMPLE 7.15 (Complex Euclidean Space Cn). Let V ¼ Cn, and let u ¼ ðziÞ and v ¼ ðwiÞ be vectors in Cn. Then hu; vi ¼ P k zkwk ¼ z1w1 þ z2w2 þ þ znwn is an inner product on V, called the usual or standard inner product on Cn. V with this inner product is called Complex Euclidean Space. We assume this inner product on Cn unless otherwise stated or implied. Assuming u and v are column vectors, the above inner product may be deﬁned by hu; vi ¼ uT v where, as with matrices, v means the conjugate of each element of v. If u and v are real, we have wi ¼ wi. In this case, the inner product reduced to the analogous one on Rn. EXAMPLE 7.16 (a) Let V be the vector space of complex continuous functions on the (real) interval a t b. Then the following is the usual inner product on V: h f ; gi ¼ ðb a f ðtÞgðtÞ dt (b) Let U be the vector space of m n matrices over C. Suppose A ¼ ðzijÞ and B ¼ ðwijÞ are elements of U. Then the following is the usual inner product on U: hA; Bi ¼ trðBHAÞ ¼ P m i¼1 P n j¼1 wijzij As usual, BH ¼ BT; that is, BH is the conjugate transpose of B. The following is a list of theorems for complex inner product spaces that are analogous to those for the real case. Here a Hermitian matrix A (i.e., one where AH ¼ AT ¼ AÞ plays the same role that a symmetric matrix A (i.e., one where AT ¼ A) plays in the real case. (Theorem 7.18 is proved in Problem 7.50.) THEOREM 7.18: (Cauchy–Schwarz) Let V be a complex inner product space. Then jhu; vij kukkvk THEOREM 7.19: Let W be a subspace of a complex inner product space V. Then V ¼ W W ?. THEOREM 7.20: Suppose fu1; u2; . . . ; ung is a basis for a complex inner product space V. Then, for any v 2 V, v ¼ hv; u1i hu1; u1i u1 þ hv; u2i hu2; u2i u2 þ þ hv; uni hun; uni un THEOREM 7.21: Suppose fu1; u2; . . . ; ung is a basis for a complex inner product space V. Let A ¼ ½aij be the complex matrix deﬁned by aij ¼ hui; uji. Then, for any u; v 2 V, hu; vi ¼ ½uTA½v where ½u and ½v are the coordinate column vectors in the given basis fuig. (Remark: This matrix A is said to represent the inner product on V.) THEOREM 7.22: Let A be a Hermitian matrix (i.e., AH ¼ AT ¼ AÞ such that X TA X is real and positive for every nonzero vector X 2 Cn. Then hu; vi ¼ uTA v is an inner product on Cn. THEOREM 7.23: Let A be the matrix that represents an inner product on V. Then A is Hermitian, and X TAX is real and positive for any nonzero vector in Cn. 240 CHAPTER 7 Inner Product Spaces, Orthogonality 7.10 Normed Vector Spaces (Optional) We begin with a deﬁnition. DEFINITION: Let V be a real or complex vector space. Suppose to each v 2 V there is assigned a real number, denoted by kvk. This function k k is called a norm on V if it satisﬁes the following axioms: ½N1 kvk 0; and kvk ¼ 0 if and only if v ¼ 0. ½N2 kkvk ¼ jkjkvk. ½N3 ku þ vk kuk þ kvk. A vector space V with a norm is called a normed vector space. Suppose V is a normed vector space. The distance between two vectors u and v in V is denoted and deﬁned by dðu; vÞ ¼ ku vk The following theorem (proved in Problem 7.56) is the main reason why dðu; vÞ is called the distance between u and v. THEOREM 7.24: Let V be a normed vector space. Then the function dðu; vÞ ¼ ku vk satisﬁes the following three axioms of a metric space: ½M1 dðu; vÞ 0; and dðu; vÞ ¼ 0 if and only if u ¼ v. ½M2 dðu; vÞ ¼ dðv; uÞ. ½M3 dðu; vÞ dðu; wÞ þ dðw; vÞ. Normed Vector Spaces and Inner Product Spaces Suppose V is an inner product space. Recall that the norm of a vector v in V is deﬁned by kvk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ hv; vi p One can prove (Theorem 7.2) that this norm satisﬁes ½N1, ½N2, and ½N3. Thus, every inner product space V is a normed vector space. On the other hand, there may be norms on a vector space V that do not come from an inner product on V, as shown below. Norms on Rn and Cn The following deﬁne three important norms on Rn and Cn: kða1; . . . ; anÞk1 ¼ maxðjaijÞ kða1; . . . ; anÞk1 ¼ ja1j þ ja2j þ þ janj kða1; . . . ; anÞk2 ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ja1j2 þ ja2j2 þ þ janj2 q (Note that subscripts are used to distinguish between the three norms.) The norms k k1, k k1, and k k2 are called the inﬁnity-norm, one-norm, and two-norm, respectively. Observe that k k2 is the norm on Rn (respectively, Cn) induced by the usual inner product on Rn (respectively, Cn). We will let d1, d1, d2 denote the corresponding distance functions. EXAMPLE 7.17 Consider vectors u ¼ ð1; 5; 3Þ and v ¼ ð4; 2; 3Þ in R3. (a) The inﬁnity norm chooses the maximum of the absolute values of the components. Hence, kuk1 ¼ 5 and kvk1 ¼ 4 CHAPTER 7 Inner Product Spaces, Orthogonality 241 (b) The one-norm adds the absolute values of the components. Thus, kuk1 ¼ 1 þ 5 þ 3 ¼ 9 and kvk1 ¼ 4 þ 2 þ 3 ¼ 9 (c) The two-norm is equal to the square root of the sum of the squares of the components (i.e., the norm induced by the usual inner product on R3). Thus, kuk2 ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ 1 þ 25 þ 9 p ¼ ﬃﬃﬃﬃﬃ 35 p and kvk2 ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 16 þ 4 þ 9 p ¼ ﬃﬃﬃﬃﬃ 29 p (d) Because u v ¼ ð1 4; 5 2; 3 þ 3Þ ¼ ð3; 7; 6Þ, we have d1ðu; vÞ ¼ 7; d1ðu; vÞ ¼ 3 þ 7 þ 6 ¼ 16; d2ðu; vÞ ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 9 þ 49 þ 36 p ¼ ﬃﬃﬃﬃﬃ 94 p EXAMPLE 7.18 Consider the Cartesian plane R2 shown in Fig. 7-4. (a) Let D1 be the set of points u ¼ ðx; yÞ in R2 such that kuk2 ¼ 1. Then D1 consists of the points ðx; yÞ such that kuk2 2 ¼ x2 þ y2 ¼ 1. Thus, D1 is the unit circle, as shown in Fig. 7-4. (b) Let D2 be the set of points u ¼ ðx; yÞ in R2 such that kuk1 ¼ 1. Then D1 consists of the points ðx; yÞ such that kuk1 ¼ jxj þ jyj ¼ 1. Thus, D2 is the diamond inside the unit circle, as shown in Fig. 7-4. (c) Let D3 be the set of points u ¼ ðx; yÞ in R2 such that kuk1 ¼ 1. Then D3 consists of the points ðx; yÞ such that kuk1 ¼ maxðjxj, jyjÞ ¼ 1. Thus, D3 is the square circumscribing the unit circle, as shown in Fig. 7-4. Norms on C½a; b Consider the vector space V ¼ C½a; b of real continuous functions on the interval a t b. Recall that the following deﬁnes an inner product on V: h f ; gi ¼ ðb a f ðtÞgðtÞ dt Accordingly, the above inner product deﬁnes the following norm on V ¼ C½a; b (which is analogous to the k k2 norm on Rn): k f k2 ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðb a ½ f ðtÞ2 dt s Figure 7-4 242 CHAPTER 7 Inner Product Spaces, Orthogonality The following deﬁne the other norms on V ¼ C½a; b: k f k1 ¼ ðb a j f ðtÞj dt and k f k1 ¼ maxðj f ðtÞjÞ There are geometrical descriptions of these two norms and their corresponding distance functions, which are described below. The ﬁrst norm is pictured in Fig. 7-5. Here k f k1 ¼ area between the function j f j and the t-axis d1ð f ; gÞ ¼ area between the functions f and g This norm is analogous to the norm k k1 on Rn. The second norm is pictured in Fig. 7-6. Here k f k1 ¼ maximum distance between f and the t-axis d1ð f ; gÞ ¼ maximum distance between f and g This norm is analogous to the norms k k1 on Rn. SOLVED PROBLEMS Inner Products 7.1. Expand: (a) h5u1 þ 8u2; 6v1 7v2i, (b) h3u þ 5v; 4u 6vi, (c) k2u 3vk2 Use linearity in both positions and, when possible, symmetry, hu; vi ¼ hv; ui. Figure 7-5 Figure 7-6 CHAPTER 7 Inner Product Spaces, Orthogonality 243 (a) Take the inner product of each term on the left with each term on the right: h5u1 þ 8u2; 6v1 7v2i ¼ h5u1; 6v1i þ h5u1; 7v2i þ h8u2; 6v1i þ h8u2; 7v2i ¼ 30hu1; v1i 35hu1; v2i þ 48hu2; v1i 56hu2; v2i [Remark: Observe the similarity between the above expansion and the expansion (5a–8b)(6c–7d ) in ordinary algebra.] (b) h3u þ 5v; 4u 6vi ¼ 12hu; ui 18hu; vi þ 20hv; ui 30hv; vi ¼ 12hu; ui þ 2hu; vi 30hv; vi (c) k2u 3vk2 ¼ h2u 3v; 2u 3vi ¼ 4hu; ui 6hu; vi 6hv; ui þ 9hv; vi ¼ 4kuk2 12ðu; vÞ þ 9kvk2 7.2. Consider vectors u ¼ ð1; 2; 4Þ; v ¼ ð2; 3; 5Þ; w ¼ ð4; 2; 3Þ in R3. Find (a) u v, (b) u w; (c) v w, (d) ðu þ vÞ w, (e) kuk, (f ) kvk. (a) Multiply corresponding components and add to get u v ¼ 2 6 þ 20 ¼ 16: (b) u w ¼ 4 þ 4 12 ¼ 4. (c) v w ¼ 8 6 15 ¼ 13. (d) First ﬁnd u þ v ¼ ð3; 1; 9Þ. Then ðu þ vÞ w ¼ 12 2 27 ¼ 17. Alternatively, using ½I1, ðu þ vÞ w ¼ u w þ v w ¼ 4 13 ¼ 17. (e) First ﬁnd kuk2 by squaring the components of u and adding: kuk2 ¼ 12 þ 22 þ 42 ¼ 1 þ 4 þ 16 ¼ 21; and so kuk ¼ ﬃﬃﬃﬃﬃ 21 p (f ) kvk2 ¼ 4 þ 9 þ 25 ¼ 38, and so kvk ¼ ﬃﬃﬃﬃﬃ 38 p . 7.3. Verify that the following deﬁnes an inner product in R2: hu; vi ¼ x1y1 x1y2 x2y1 þ 3x2y2; where u ¼ ðx1; x2Þ; v ¼ ðy1; y2Þ We argue via matrices. We can write hu; vi in matrix notation as follows: hu; vi ¼ uTAv ¼ ½x1; x2 1 1 1 3 y1 y2 Because A is real and symmetric, we need only show that A is positive deﬁnite. The diagonal elements 1 and 3 are positive, and the determinant kAk ¼ 3 1 ¼ 2 is positive. Thus, by Theorem 7.14, A is positive deﬁnite. Accordingly, by Theorem 7.15, hu; vi is an inner product. 7.4. Consider the vectors u ¼ ð1; 5Þ and v ¼ ð3; 4Þ in R2. Find (a) hu; vi with respect to the usual inner product in R2. (b) hu; vi with respect to the inner product in R2 in Problem 7.3. (c) kvk using the usual inner product in R2. (d) kvk using the inner product in R2 in Problem 7.3. (a) hu; vi ¼ 3 þ 20 ¼ 23. (b) hu; vi ¼ 1 3 1 4 5 3 þ 3 5 4 ¼ 3 4 15 þ 60 ¼ 44. (c) kvk2 ¼ hv; vi ¼ hð3; 4Þ; ð3; 4Þi ¼ 9 þ 16 ¼ 25; hence, jvk ¼ 5. (d) kvk2 ¼ hv; vi ¼ hð3; 4Þ; ð3; 4Þi ¼ 9 12 12 þ 48 ¼ 33; hence, kvk ¼ ﬃﬃﬃﬃﬃ 33 p . 7.5. Consider the following polynomials in PðtÞ with the inner product h f ; gi ¼ Ð 1 0 f ðtÞgðtÞ dt: f ðtÞ ¼ t þ 2; gðtÞ ¼ 3t 2; hðtÞ ¼ t2 2t 3 (a) Find h f ; gi and h f ; hi. (b) Find k f k and kgk. (c) Normalize f and g. 244 CHAPTER 7 Inner Product Spaces, Orthogonality (a) Integrate as follows: h f ; gi ¼ ð1 0 ðt þ 2Þð3t 2Þ dt ¼ ð1 0 ð3t2 þ 4t 4Þ dt ¼ t3 þ 2t2 4t 1 0 ¼ 1 h f ; hi ¼ ð1 0 ðt þ 2Þðt2 2t 3Þ dt ¼ t4 4 7t2 2 6t 1 0 ¼ 37 4 (b) h f ; f i ¼ Ð 1 0 ðt þ 2Þðt þ 2Þ dt ¼ 19 3 ; hence, k f k ¼ ﬃﬃﬃﬃ 19 3 q ¼ 1 3 ﬃﬃﬃﬃﬃ 57 p hg; gi ¼ ð1 0 ð3t 2Þð3t 2Þ ¼ 1; hence; kgk ¼ ﬃﬃﬃ 1 p ¼ 1 (c) Because k f k ¼ 1 3 ﬃﬃﬃﬃﬃ 57 p and g is already a unit vector, we have ^ f ¼ 1 k f k f ¼ 3 ﬃﬃﬃﬃﬃ 57 p ðt þ 2Þ and ^ g ¼ g ¼ 3t 2 7.6. Find cos y where y is the angle between: (a) u ¼ ð1; 3; 5; 4Þ and v ¼ ð2; 3; 4; 1Þ in R4, (b) A ¼ 9 8 7 6 5 4 and B ¼ 1 2 3 4 5 6 , where hA; Bi ¼ trðBTAÞ: Use cos y ¼ hu; vi kukkvk (a) Compute: hu; vi ¼ 2 9 20 þ 4 ¼ 23; kuk2 ¼ 1 þ 9 þ 25 þ 16 ¼ 51; kvk2 ¼ 4 þ 9 þ 16 þ 1 ¼ 30 Thus; cos y ¼ 23 ﬃﬃﬃﬃﬃ 51 p ﬃﬃﬃﬃﬃ 30 p ¼ 23 3 ﬃﬃﬃﬃﬃﬃﬃ ﬃ 170 p (b) Use hA; Bi ¼ trðBTAÞ ¼ Pm i¼1 Pn j¼1 aijbij, the sum of the products of corresponding entries. hA; Bi ¼ 9 þ 16 þ 21 þ 24 þ 25 þ 24 ¼ 119 Use kAk2 ¼ hA; Ai ¼ Pm i¼1 Pn j¼1 a2 ij ; the sum of the squares of all the elements of A. kAk2 ¼ hA; Ai ¼ 92 þ 82 þ 72 þ 62 þ 52 þ 42 ¼ 271; kBk2 ¼ hB; Bi ¼ 12 þ 22 þ 32 þ 42 þ 52 þ 62 ¼ 91; and so and so kAk ¼ ﬃﬃﬃﬃﬃﬃﬃ ﬃ 271 p kBk ¼ ﬃﬃﬃﬃﬃ 91 p Thus; cos y ¼ 119 ﬃﬃﬃﬃﬃﬃﬃ ﬃ 271 p ﬃﬃﬃﬃﬃ 91 p 7.7. Verify each of the following: (a) Parallelogram Law (Fig. 7-7): ku þ vk2 þ ku vk2 ¼ 2kuk2 þ 2kvk2. (b) Polar form for hu; vi (which shows the inner product can be obtained from the norm function): hu; vi ¼ 1 4 ðku þ vk2 ku vk2Þ: Expand as follows to obtain ku þ vk2 ¼ hu þ v; u þ vi ¼ kuk2 þ 2hu; vi þ kvk2 ð1Þ ku vk2 ¼ hu v; u vi ¼ kuk2 2hu; vi þ kvk2 ð2Þ Add (1) and (2) to get the Parallelogram Law (a). Subtract (2) from (1) to obtain ku þ vk2 ku vk2 ¼ 4hu; vi Divide by 4 to obtain the (real) polar form (b). CHAPTER 7 Inner Product Spaces, Orthogonality 245 7.8. Prove Theorem 7.1 (Cauchy–Schwarz): For u and v in a real inner product space V; hu; ui2 hu; uihv; vi or jhu; vij kukkvk: For any real number t, htu þ v; tu þ vi ¼ t2hu; ui þ 2thu; vi þ hv; vi ¼ t2kuk2 þ 2thu; vi þ kvk2 Let a ¼ kuk2, b ¼ 2hu; vÞ, c ¼ kvk2. Because ktu þ vk2 0, we have at2 þ bt þ c 0 for every value of t. This means that the quadratic polynomial cannot have two real roots, which implies that b2 4ac 0 or b2 4ac. Thus, 4hu; vi2 4kuk2kvk2 Dividing by 4 gives our result. 7.9. Prove Theorem 7.2: The norm in an inner product space V satisﬁes (a) ½N1 kvk 0; and kvk ¼ 0 if and only if v ¼ 0. (b) ½N2 kkvk ¼ jkjkvk. (c) ½N3 ku þ vk kuk þ kvk. (a) If v 6¼ 0, then hv; vi > 0, and hence, kvk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ hv; vi p > 0. If v ¼ 0, then h0; 0i ¼ 0. Consequently, k0k ¼ ﬃﬃﬃ 0 p ¼ 0. Thus, ½N1 is true. (b) We have kkvk2 ¼ hkv; kvi ¼ k2hv; vi ¼ k2kvk2. Taking the square root of both sides gives ½N2. (c) Using the Cauchy–Schwarz inequality, we obtain ku þ vk2 ¼ hu þ v; u þ vi ¼ hu; ui þ hu; vi þ hu; vi þ hv; vi kuk2 þ 2kukkvk þ kvk2 ¼ ðkuk þ kvkÞ2 Taking the square root of both sides yields ½N3. Orthogonality, Orthonormal Complements, Orthogonal Sets 7.10. Find k so that u ¼ ð1; 2; k; 3Þ and v ¼ ð3; k; 7; 5Þ in R4 are orthogonal. First ﬁnd hu; vi ¼ ð1; 2; k; 3Þ ð3; k; 7; 5Þ ¼ 3 þ 2k þ 7k 15 ¼ 9k 12 Then set hu; vi ¼ 9k 12 ¼ 0 to obtain k ¼ 4 3. 7.11. Let W be the subspace of R5 spanned by u ¼ ð1; 2; 3; 1; 2Þ and v ¼ ð2; 4; 7; 2; 1Þ. Find a basis of the orthogonal complement W ? of W. We seek all vectors w ¼ ðx; y; z; s; tÞ such that hw; ui ¼ x þ 2y þ 3z s þ 2t ¼ 0 hw; vi ¼ 2x þ 4y þ 7z þ 2s t ¼ 0 Eliminating x from the second equation, we ﬁnd the equivalent system x þ 2y þ 3z s þ 2t ¼ 0 z þ 4s 5t ¼ 0 Figure 7-7 246 CHAPTER 7 Inner Product Spaces, Orthogonality The free variables are y; s, and t. Therefore, (1) Set y ¼ 1, s ¼ 0, t ¼ 0 to obtain the solution w1 ¼ ð2; 1; 0; 0; 0Þ. (2) Set y ¼ 0, s ¼ 1, t ¼ 0 to ﬁnd the solution w2 ¼ ð13; 0; 4; 1; 0Þ. (3) Set y ¼ 0, s ¼ 0, t ¼ 1 to obtain the solution w3 ¼ ð17; 0; 5; 0; 1Þ. The set fw1; w2; w3g is a basis of W ?. 7.12. Let w ¼ ð1; 2; 3; 1Þ be a vector in R4. Find an orthogonal basis for w?. Find a nonzero solution of x þ 2y þ 3z þ t ¼ 0, say v1 ¼ ð0; 0; 1; 3Þ. Now ﬁnd a nonzero solution of the system x þ 2y þ 3z þ t ¼ 0; z 3t ¼ 0 say v2 ¼ ð0; 5; 3; 1Þ. Last, ﬁnd a nonzero solution of the system x þ 2y þ 3z þ t ¼ 0; 5y þ 3z þ t ¼ 0; z 3t ¼ 0 say v3 ¼ ð14; 2; 3; 1Þ. Thus, v1, v2, v3 form an orthogonal basis for w?. 7.13. Let S consist of the following vectors in R4: u1 ¼ ð1; 1; 0; 1Þ; u2 ¼ ð1; 2; 1; 3Þ; u3 ¼ ð1; 1; 9; 2Þ; u4 ¼ ð16; 13; 1; 3Þ (a) Show that S is orthogonal and a basis of R4. (b) Find the coordinates of an arbitrary vector v ¼ ða; b; c; dÞ in R4 relative to the basis S. (a) Compute u1 u2 ¼ 1 þ 2 þ 0 3 ¼ 0; u2 u3 ¼ 1 þ 2 9 þ 6 ¼ 0; u1 u3 ¼ 1 þ 1 þ 0 2 ¼ 0; u2 u4 ¼ 16 26 þ 1 þ 9 ¼ 0; u1 u4 ¼ 16 13 þ 0 3 ¼ 0 u3 u4 ¼ 16 13 9 þ 6 ¼ 0 Thus, S is orthogonal, and S is linearly independent. Accordingly, S is a basis for R4 because any four linearly independent vectors form a basis of R4. (b) Because S is orthogonal, we need only ﬁnd the Fourier coefﬁcients of v with respect to the basis vectors, as in Theorem 7.7. Thus, k1 ¼ hv; u1i hu1; u1i ¼ a þ b d 3 ; k2 ¼ hv; u2i hu2; u2i ¼ a þ 2b þ c þ 3d 15 ; k3 ¼ hv; u3i hu3; u3i ¼ a þ b 9c þ 2d 87 k4 ¼ hv; u4i hu4; u4i ¼ 16a 13b þ c þ 3d 435 are the coordinates of v with respect to the basis S. 7.14. Suppose S, S1, S2 are the subsets of V. Prove the following: (a) S S??. (b) If S1 S2, then S? 2 S? 1 . (c) S? ¼ span ðSÞ?. (a) Let w 2 S. Then hw; vi ¼ 0 for every v 2 S?; hence, w 2 S??. Accordingly, S S??. (b) Let w 2 S? 2 . Then hw; vi ¼ 0 for every v 2 S2. Because S1 S2, hw; vi ¼ 0 for every v ¼ S1. Thus, w 2 S? 1 , and hence, S? 2 S? 1 . (c) Because S spanðSÞ, part (b) gives us spanðSÞ? S?. Suppose u 2 S? and v 2 spanðSÞ. Then there exist w1; w2; . . . ; wk in S such that v ¼ a1w1 þ a2w2 þ þ akwk. Then, using u 2 S?, we have hu; vi ¼ hu; a1w1 þ a2w2 þ þ akwki ¼ a1hu; w1i þ a2hu; w2i þ þ akhu; wki ¼ a1ð0Þ þ a2ð0Þ þ þ akð0Þ ¼ 0 Thus, u 2 spanðSÞ?. Accordingly, S? spanðSÞ?. Both inclusions give S? ¼ spanðSÞ?. 7.15. Prove Theorem 7.5: Suppose S is an orthogonal set of nonzero vectors. Then S is linearly independent. CHAPTER 7 Inner Product Spaces, Orthogonality 247 Suppose S ¼ fu1; u2; . . . ; urg and suppose a1u1 þ a2u2 þ þ arur ¼ 0 ð1Þ Taking the inner product of (1) with u1, we get 0 ¼ h0; u1i ¼ ha1u1 þ a2u2 þ þ arur; u1i ¼ a1hu1; u1i þ a2hu2; u1i þ þ arhur; u1i ¼ a1hu1; u1i þ a2 0 þ þ ar 0 ¼ a1hu1; u1i Because u1 6¼ 0, we have hu1; u1i 6¼ 0. Thus, a1 ¼ 0. Similarly, for i ¼ 2; . . . ; r, taking the inner product of (1) with ui, 0 ¼ h0; uii ¼ ha1u1 þ þ arur; uii ¼ a1hu1; uii þ þ aihui; uii þ þ arhur; uii ¼ aihui; uii But hui; uii 6¼ 0, and hence, every ai ¼ 0. Thus, S is linearly independent. 7.16. Prove Theorem 7.6 (Pythagoras): Suppose fu1; u2; . . . ; urg is an orthogonal set of vectors. Then ku1 þ u2 þ þ urk2 ¼ ku1k2 þ ku2k2 þ þ kurk2 Expanding the inner product, we have ku1 þ u2 þ þ urk2 ¼ hu1 þ u2 þ þ ur; u1 þ u2 þ þ uri ¼ hu1; u1i þ hu2; u2i þ þ hur; uri þ P i6¼j hui; uji The theorem follows from the fact that hui; uii ¼ kuik2 and hui; uji ¼ 0 for i 6¼ j. 7.17. Prove Theorem 7.7: Let fu1; u2; . . . ; ung be an orthogonal basis of V. Then for any v 2 V, v ¼ hv; u1i hu1; u1i u1 þ hv; u2i hu2; u2i u2 þ þ hv; uni hun; uni un Suppose v ¼ k1u1 þ k2u2 þ þ knun. Taking the inner product of both sides with u1 yields hv; u1i ¼ hk1u2 þ k2u2 þ þ knun; u1i ¼ k1hu1; u1i þ k2hu2; u1i þ þ knhun; u1i ¼ k1hu1; u1i þ k2 0 þ þ kn 0 ¼ k1hu1; u1i Thus, k1 ¼ hv; u1i hu1; u1i. Similarly, for i ¼ 2; . . . ; n, hv; uii ¼ hk1ui þ k2u2 þ þ knun; uii ¼ k1hu1; uii þ k2hu2; uii þ þ knhun; uii ¼ k1 0 þ þ kihui; uii þ þ kn 0 ¼ kihui; uii Thus, ki ¼ hv; uii hu1; uii. Substituting for ki in the equation v ¼ k1u1 þ þ knun, we obtain the desired result. 7.18. Suppose E ¼ fe1; e2; . . . ; eng is an orthonormal basis of V. Prove (a) For any u 2 V, we have u ¼ hu; e1ie1 þ hu; e2ie2 þ þ hu; enien. (b) ha1e1 þ þ anen; b1e1 þ þ bneni ¼ a1b1 þ a2b2 þ þ anbn. (c) For any u; v 2 V, we have hu; vi ¼ hu; e1ihv; e1i þ þ hu; enihv; eni. (a) Suppose u ¼ k1e1 þ k2e2 þ þ knen. Taking the inner product of u with e1, hu; e1i ¼ hk1e1 þ k2e2 þ þ knen; e1i ¼ k1he1; e1i þ k2he2; e1i þ þ knhen; e1i ¼ k1ð1Þ þ k2ð0Þ þ þ knð0Þ ¼ k1 248 CHAPTER 7 Inner Product Spaces, Orthogonality Similarly, for i ¼ 2; . . . ; n, hu; eii ¼ hk1e1 þ þ kiei þ þ knen; eii ¼ k1he1; eii þ þ kihei; eii þ þ knhen; eii ¼ k1ð0Þ þ þ kið1Þ þ þ knð0Þ ¼ ki Substituting hu; eii for ki in the equation u ¼ k1e1 þ þ knen, we obtain the desired result. (b) We have P n i¼1 aiei; P n j¼1 bjej + ¼ P n i;j¼1 aibjhei; eji ¼ P n i¼1 aibihei; eii þ P i6¼j aibjhei; eji But hei; eji ¼ 0 for i 6¼ j, and hei; eji ¼ 1 for i ¼ j. Hence, as required, P n i¼1 aiei; P n j¼1 bjej + ¼ P n i¼1 aibi ¼ a1b1 þ a2b2 þ þ anbn (c) By part (a), we have u ¼ hu; e1ie1 þ þ hu; enien and v ¼ hv; e1ie1 þ þ hv; enien Thus, by part (b), hu; vi ¼ hu; e1ihv; e1i þ hu; e2ihv; e2i þ þ hu; enihv; eni Projections, Gram–Schmidt Algorithm, Applications 7.19. Suppose w 6¼ 0. Let v be any vector in V. Show that c ¼ hv; wi hw; wi ¼ hv; wi kwk2 is the unique scalar such that v0 ¼ v cw is orthogonal to w. In order for v0 to be orthogonal to w we must have hv cw; wi ¼ 0 or hv; wi chw; wi ¼ 0 or hv; wi ¼ chw; wi Thus, c hv; wi hw; wi. Conversely, suppose c ¼ hv; wi hw; wi. Then hv cw; wi ¼ hv; wi chw; wi ¼ hv; wi hv; wi hw; wi hw; wi ¼ 0 7.20. Find the Fourier coefﬁcient c and the projection of v ¼ ð1; 2; 3; 4Þ along w ¼ ð1; 2; 1; 2Þ in R4. Compute hv; wi ¼ 1 4 þ 3 8 ¼ 8 and kwk2 ¼ 1 þ 4 þ 1 þ 4 ¼ 10. Then c ¼ 8 10 ¼ 4 5 and projðv; wÞ ¼ cw ¼ ð 4 5 ; 8 5 ; 4 5 ; 8 5Þ 7.21. Consider the subspace U of R4 spanned by the vectors: v1 ¼ ð1; 1; 1; 1Þ; v2 ¼ ð1; 1; 2; 4Þ; v3 ¼ ð1; 2; 4; 3Þ Find (a) an orthogonal basis of U; (b) an orthonormal basis of U. (a) Use the Gram–Schmidt algorithm. Begin by setting w1 ¼ u ¼ ð1; 1; 1; 1Þ. Next ﬁnd v2 hv2; w1i hw1; w1i w1 ¼ ð1; 1; 2; 4Þ 8 4 ð1; 1; 1; 1Þ ¼ ð1; 1; 0; 2Þ Set w2 ¼ ð1; 1; 0; 2Þ. Then ﬁnd v3 hv3; w1i hw1; w1i w1 hv3; w2i hw2; w2i w2 ¼ ð1; 2; 4; 3Þ ð4Þ 4 ð1; 1; 1; 1Þ ð9Þ 6 ð1; 1; 0; 2Þ ¼ ð1 2 ; 3 2 ; 3; 1Þ Clear fractions to obtain w3 ¼ ð1; 3; 6; 2Þ. Then w1; w2; w3 form an orthogonal basis of U. CHAPTER 7 Inner Product Spaces, Orthogonality 249 (b) Normalize the orthogonal basis consisting of w1; w2; w3. Because kw1k2 ¼ 4, kw2k2 ¼ 6, and kw3k2 ¼ 50, the following vectors form an orthonormal basis of U: u1 ¼ 1 2 ð1; 1; 1; 1Þ; u2 ¼ 1ﬃﬃﬃ 6 p ð1; 1; 0; 2Þ; u3 ¼ 1 5 ﬃﬃﬃ 2 p ð1; 3; 6; 2Þ 7.22. Consider the vector space PðtÞ with inner product h f ; gi ¼ Ð 1 0 f ðtÞgðtÞ dt. Apply the Gram– Schmidt algorithm to the set f1; t; t2g to obtain an orthogonal set f f0; f1; f2g with integer coefﬁcients. First set f0 ¼ 1. Then ﬁnd t ht; 1i h1; 1i 1 ¼ t 1 2 1 1 ¼ t 1 2 Clear fractions to obtain f1 ¼ 2t 1. Then ﬁnd t2 ht2; 1i h1; 1i ð1Þ ht2; 2t 1i h2t 1; 2t 1i ð2t 1Þ ¼ t2 1 3 1 ð1Þ 1 6 1 3 ð2t 1Þ ¼ t2 t þ 1 6 Clear fractions to obtain f2 ¼ 6t2 6t þ 1. Thus, f1; 2t 1; 6t2 6t þ 1g is the required orthogonal set. 7.23. Suppose v ¼ ð1; 3; 5; 7Þ. Find the projection of v onto W or, in other words, ﬁnd w 2 W that minimizes kv wk, where W is the subspance of R4 spanned by (a) u1 ¼ ð1; 1; 1; 1Þ and u2 ¼ ð1; 3; 4; 2Þ, (b) v1 ¼ ð1; 1; 1; 1Þ and v2 ¼ ð1; 2; 3; 2Þ. (a) Because u1 and u2 are orthogonal, we need only compute the Fourier coefﬁcients: c1 ¼ hv; u1i hu1; u1i ¼ 1 þ 3 þ 5 þ 7 1 þ 1 þ 1 þ 1 ¼ 16 4 ¼ 4 c2 ¼ hv; u2i hu2; u2i ¼ 1 9 þ 20 14 1 þ 9 þ 16 þ 4 ¼ 2 30 ¼ 1 15 Then w ¼ projðv; WÞ ¼ c1u1 þ c2u2 ¼ 4ð1; 1; 1; 1Þ 1 15 ð1; 3; 4; 2Þ ¼ ð59 15 ; 63 5 ; 56 15 ; 62 15Þ: (b) Because v1 and v2 are not orthogonal, ﬁrst apply the Gram–Schmidt algorithm to ﬁnd an orthogonal basis for W. Set w1 ¼ v1 ¼ ð1; 1; 1; 1Þ. Then ﬁnd v2 hv2; w1i hw1; w1i w1 ¼ ð1; 2; 3; 2Þ 8 4 ð1; 1; 1; 1Þ ¼ ð1; 0; 1; 0Þ Set w2 ¼ ð1; 0; 1; 0Þ. Now compute c1 ¼ hv; w1i hw1; w1i ¼ 1 þ 3 þ 5 þ 7 1 þ 1 þ 1 þ 1 ¼ 16 4 ¼ 4 c2 ¼ hv; w2i hw2; w2i 1 þ 0 þ 5 þ 0 1 þ 0 þ 1 þ 0 ¼ 6 2 ¼ 3 Then w ¼ projðv; WÞ ¼ c1w1 þ c2w2 ¼ 4ð1; 1; 1; 1Þ 3ð1; 0; 1; 0Þ ¼ ð7; 4; 1; 4Þ. 7.24. Suppose w1 and w2 are nonzero orthogonal vectors. Let v be any vector in V. Find c1 and c2 so that v0 is orthogonal to w1 and w2, where v0 ¼ v c1w1 c2w2. If v0 is orthogonal to w1, then 0 ¼ hv c1w1 c2w2; w1i ¼ hv; w1i c1hw1; w1i c2hw2; w1i ¼ hv; w1i c1hw1; w1i c20 ¼ hv; w1i c1hw1; w1i Thus, c1 ¼ hv; w1i=hw1; w1i. (That is, c1 is the component of v along w1.) Similarly, if v0 is orthogonal to w2, then 0 ¼ hv c1w1 c2w2; w2i ¼ hv; w2i c2hw2; w2i Thus, c2 ¼ hv; w2i=hw2; w2i. (That is, c2 is the component of v along w2.) 250 CHAPTER 7 Inner Product Spaces, Orthogonality 7.25. Prove Theorem 7.8: Suppose w1; w2; . . . ; wr form an orthogonal set of nonzero vectors in V. Let v 2 V. Deﬁne v0 ¼ v ðc1w1 þ c2w2 þ þ crwrÞ; where ci ¼ hv; wii hwi; wii Then v0 is orthogonal to w1; w2; . . . ; wr. For i ¼ 1; 2; . . . ; r and using hwi; wji ¼ 0 for i 6¼ j, we have hv c1w1 c2x2 crwr; wii ¼ hv; wii c1hw1; wii cihwi; wii crhwr; wii ¼ hv; wii c1 0 cihwi; wii cr 0 ¼ hv; wii cihwi; wii ¼ hv; wii hv; wii hwi; wii hwi; wii ¼ 0 The theorem is proved. 7.26. Prove Theorem 7.9: Let fv1; v2; . . . ; vng be any basis of an inner product space V. Then there exists an orthonormal basis fu1; u2; . . . ; ung of V such that the change-of-basis matrix from fvig to fuig is triangular; that is, for k ¼ 1; 2; . . . ; n, uk ¼ ak1v1 þ ak2v2 þ þ akkvk The proof uses the Gram–Schmidt algorithm and Remarks 1 and 3 of Section 7.7. That is, apply the algorithm to fvig to obtain an orthogonal basis fwi; . . . ; wng, and then normalize fwig to obtain an orthonormal basis fuig of V. The speciﬁc algorithm guarantees that each wk is a linear combination of v1; . . . ; vk, and hence, each uk is a linear combination of v1; . . . ; vk. 7.27. Prove Theorem 7.10: Suppose S ¼ fw1; w2; . . . ; wrg, is an orthogonal basis for a subspace W of V. Then one may extend S to an orthogonal basis for V; that is, one may ﬁnd vectors wrþ1; . . . ; wr such that fw1; w2; . . . ; wng is an orthogonal basis for V. Extend S to a basis S0 ¼ fw1; . . . ; wr; vrþ1; . . . ; vng for V. Applying the Gram–Schmidt algorithm to S0, we ﬁrst obtain w1; w2; . . . ; wr because S is orthogonal, and then we obtain vectors wrþ1; . . . ; wn, where fw1; w2; . . . ; wng is an orthogonal basis for V. Thus, the theorem is proved. 7.28. Prove Theorem 7.4: Let W be a subspace of V. Then V ¼ W W ?. By Theorem 7.9, there exists an orthogonal basis fu1; . . . ; urg of W, and by Theorem 7.10 we can extend it to an orthogonal basis fu1; u2; . . . ; ung of V. Hence, urþ1; . . . ; un 2 W ?. If v 2 V, then v ¼ a1u1 þ þ anun; where a1u1 þ þ arur 2 W and arþ1urþ1 þ þ anun 2 W ? Accordingly, V ¼ W þ W ?. On the other hand, if w 2 W \ W ?, then hw; wi ¼ 0. This yields w ¼ 0. Hence, W \ W ? ¼ f0g. The two conditions V ¼ W þ W ? and W \ W ? ¼ f0g give the desired result V ¼ W W ?. Remark: Note that we have proved the theorem for the case that V has ﬁnite dimension. We remark that the theorem also holds for spaces of arbitrary dimension. 7.29. Suppose W is a subspace of a ﬁnite-dimensional space V. Prove that W ¼ W ??. By Theorem 7.4, V ¼ W W ?, and also V ¼ W ? W ??. Hence, dim W ¼ dim V dim W ? and dim W ?? ¼ dim V dim W ? This yields dim W ¼ dim W ??. But W W ?? (see Problem 7.14). Hence, W ¼ W ??, as required. 7.30. Prove the following: Suppose w1; w2; . . . ; wr form an orthogonal set of nonzero vectors in V. Let v be any vector in V and let ci be the component of v along wi. Then, for any scalars a1; . . . ; ar, we have v P r k¼1 ckwk v P r k¼1 akwk That is, P ciwi is the closest approximation to v as a linear combination of w1; . . . ; wr. CHAPTER 7 Inner Product Spaces, Orthogonality 251 By Theorem 7.8, v P ckwk is orthogonal to every wi and hence orthogonal to any linear combination of w1; w2; . . . ; wr. Therefore, using the Pythagorean theorem and summing from k ¼ 1 to r, v P akwk k k2 ¼ v P ckwk þ P ðck akÞwk k k2¼ v P ckwk k k2þ Pðck akÞwk k k2 v P ckwk k k2 The square root of both sides gives our theorem. 7.31. Suppose fe1; e2; . . . ; erg is an orthonormal set of vectors in V. Let v be any vector in V and let ci be the Fourier coefﬁcient of v with respect to ui. Prove Bessel’s inequality: P r k¼1 c2 k kvk2 Note that ci ¼ hv; eii, because keik ¼ 1. Then, using hei; eji ¼ 0 for i 6¼ j and summing from k ¼ 1 to r, we get 0 v P ckek; v P ck; ek h i ¼ hv; vi 2 v; P ckeki þ P c2 k ¼ hv; vi P 2ckhv; eki þ P c2 k ¼ hv; vi P 2c2 k þ P c2 k ¼ hv; vi P c2 k This gives us our inequality. Orthogonal Matrices 7.32. Find an orthogonal matrix P whose ﬁrst row is u1 ¼ ð1 3 ; 2 3 ; 2 3Þ. First ﬁnd a nonzero vector w2 ¼ ðx; y; zÞ that is orthogonal to u1—that is, for which 0 ¼ hu1; w2i ¼ x 3 þ 2y 3 þ 2z 3 ¼ 0 or x þ 2y þ 2z ¼ 0 One such solution is w2 ¼ ð0; 1; 1Þ. Normalize w2 to obtain the second row of P: u2 ¼ ð0; 1= ﬃﬃﬃ 2 p ; 1= ﬃﬃﬃ 2 p Þ Next ﬁnd a nonzero vector w3 ¼ ðx; y; zÞ that is orthogonal to both u1 and u2—that is, for which 0 ¼ hu1; w3i ¼ x 3 þ 2y 3 þ 2z 3 ¼ 0 or x þ 2y þ 2z ¼ 0 0 ¼ hu2; w3i ¼ yﬃﬃﬃ 2 p yﬃﬃﬃ 2 p ¼ 0 or y z ¼ 0 Set z ¼ 1 and ﬁnd the solution w3 ¼ ð4; 1; 1Þ. Normalize w3 and obtain the third row of P; that is, u3 ¼ ð4= ﬃﬃﬃﬃﬃ 18 p ; 1= ﬃﬃﬃﬃﬃ 18 p ; 1= ﬃﬃﬃﬃﬃ 18 p Þ: P ¼ 1 3 2 3 2 3 0 1= ﬃﬃﬃ 2 p 1= ﬃﬃﬃ 2 p 4=3 ﬃﬃﬃ 2 p 1=3 ﬃﬃﬃ 2 p 1=3 ﬃﬃﬃ 2 p 2 4 3 5 Thus; We emphasize that the above matrix P is not unique. 7.33. Let A ¼ 1 1 1 1 3 4 7 5 2 2 4 3 5. Determine whether or not: (a) the rows of A are orthogonal; (b) A is an orthogonal matrix; (c) the columns of A are orthogonal. (a) Yes, because ð1; 1; 1Þ ð1; 3; 4Þ ¼ 1 þ 3 4 ¼ 0, ð1; 1 1Þ ð7; 5; 2Þ ¼ 7 5 2 ¼ 0, and ð1; 3; 4Þ ð7; 5; 2Þ ¼ 7 15 þ 8 ¼ 0. (b) No, because the rows of A are not unit vectors, for example, ð1; 1; 1Þ2 ¼ 1 þ 1 þ 1 ¼ 3. (c) No; for example, ð1; 1; 7Þ ð1; 3; 5Þ ¼ 1 þ 3 35 ¼ 31 6¼ 0. 7.34. Let B be the matrix obtained by normalizing each row of A in Problem 7.33. (a) Find B. (b) Is B an orthogonal matrix? (c) Are the columns of B orthogonal? 252 CHAPTER 7 Inner Product Spaces, Orthogonality (a) We have kð1; 1; 1Þk2 ¼ 1 þ 1 þ 1 ¼ 3; kð1; 3; 4Þk2 ¼ 1 þ 9 þ 16 ¼ 26 kð7; 5; 2Þk2 ¼ 49 þ 25 þ 4 ¼ 78 Thus; B ¼ 1= ﬃﬃﬃ 3 p 1= ﬃﬃﬃ 3 p 1= ﬃﬃﬃ 3 p 1= ﬃﬃﬃﬃﬃ 26 p 3= ﬃﬃﬃﬃﬃ 26 p 4= ﬃﬃﬃﬃﬃ 26 p 7= ﬃﬃﬃﬃﬃ 78 p 5= ﬃﬃﬃﬃﬃ 78 p 2= ﬃﬃﬃﬃﬃ 78 p 2 6 4 3 7 5 (b) Yes, because the rows of B are still orthogonal and are now unit vectors. (c) Yes, because the rows of B form an orthonormal set of vectors. Then, by Theorem 7.11, the columns of B must automatically form an orthonormal set. 7.35. Prove each of the following: (a) P is orthogonal if and only if PT is orthogonal. (b) If P is orthogonal, then P1 is orthogonal. (c) If P and Q are orthogonal, then PQ is orthogonal. (a) We have ðPTÞT ¼ P. Thus, P is orthogonal if and only if PPT ¼ I if and only if PTTPT ¼ I if and only if PT is orthogonal. (b) We have PT ¼ P1, because P is orthogonal. Thus, by part (a), P1 is orthogonal. (c) We have PT ¼ P1 and QT ¼ Q1. Thus, ðPQÞðPQÞT ¼ PQQTPT ¼ PQQ1P1 ¼ I. Therefore, ðPQÞT ¼ ðPQÞ1, and so PQ is orthogonal. 7.36. Suppose P is an orthogonal matrix. Show that (a) hPu; Pvi ¼ hu; vi for any u; v 2 V; (b) kPuk ¼ kuk for every u 2 V. Use PTP ¼ I and hu; vi ¼ uTv. (a) hPu; Pvi ¼ ðPuÞTðPvÞ ¼ uTPTPv ¼ uTv ¼ hu; vi. (b) We have kPuk2 ¼ hPu; Pui ¼ uTPTPu ¼ uTu ¼ hu; ui ¼ kuk2 Taking the square root of both sides gives our result. 7.37. Prove Theorem 7.12: Suppose E ¼ feig and E0 ¼ fe0 ig are orthonormal bases of V. Let P be the change-of-basis matrix from E to E0. Then P is orthogonal. Suppose e0 i ¼ bi1e1 þ bi2e2 þ þ binen; i ¼ 1; . . . ; n ð1Þ Using Problem 7.18(b) and the fact that E0 is orthonormal, we get dij ¼ he0 i; e0 ji ¼ bi1bj1 þ bi2bj2 þ þ binbjn ð2Þ Let B ¼ ½bij be the matrix of the coefﬁcients in (1). (Then P ¼ BT.) Suppose BBT ¼ ½cij. Then cij ¼ bi1bj1 þ bi2bj2 þ þ binbjn ð3Þ By (2) and (3), we have cij ¼ dij. Thus, BBT ¼ I. Accordingly, B is orthogonal, and hence, P ¼ BT is orthogonal. 7.38. Prove Theorem 7.13: Let fe1; . . . ; eng be an orthonormal basis of an inner product space V. Let P ¼ ½aij be an orthogonal matrix. Then the following n vectors form an orthonormal basis for V: e0 i ¼ a1ie1 þ a2ie2 þ þ anien; i ¼ 1; 2; . . . ; n CHAPTER 7 Inner Product Spaces, Orthogonality 253 Because feig is orthonormal, we get, by Problem 7.18(b), he0 i; e0 ji ¼ a1ia1j þ a2ia2j þ þ anianj ¼ hCi; Cji where Ci denotes the ith column of the orthogonal matrix P ¼ ½aij: Because P is orthogonal, its columns form an orthonormal set. This implies he0 i; e0 ji ¼ hCi; Cji ¼ dij: Thus, fe0 ig is an orthonormal basis. Inner Products And Positive Deﬁnite Matrices 7.39. Which of the following symmetric matrices are positive deﬁnite? (a) A ¼ 3 4 4 5 , (b) B ¼ 8 3 3 2 , (c) C ¼ 2 1 1 3 , (d) D ¼ 3 5 5 9 Use Theorem 7.14 that a 2 2 real symmetric matrix is positive deﬁnite if and only if its diagonal entries are positive and if its determinant is positive. (a) No, because jAj ¼ 15 16 ¼ 1 is negative. (b) Yes. (c) No, because the diagonal entry 3 is negative. (d) Yes. 7.40. Find the values of k that make each of the following matrices positive deﬁnite: (a) A ¼ 2 4 4 k , (b) B ¼ 4 k k 9 , (c) C ¼ k 5 5 2 (a) First, k must be positive. Also, jAj ¼ 2k 16 must be positive; that is, 2k 16 > 0. Hence, k > 8. (b) We need jBj ¼ 36 k2 positive; that is, 36 k2 > 0. Hence, k2 < 36 or 6 < k < 6. (c) C can never be positive deﬁnite, because C has a negative diagonal entry 2. 7.41. Find the matrix A that represents the usual inner product on R2 relative to each of the following bases of R2: ðaÞ fv1 ¼ ð1; 3Þ; v2 ¼ ð2; 5Þg; ðbÞ fw1 ¼ ð1; 2Þ; w2 ¼ ð4; 2Þg: (a) Compute hv1; v1i ¼ 1 þ 9 ¼ 10, hv1; v2i ¼ 2 þ 15 ¼ 17, hv2; v2i ¼ 4 þ 25 ¼ 29. Thus, A ¼ 10 17 17 29 . (b) Compute hw1; w1i ¼ 1 þ 4 ¼ 5, hw1; w2i ¼ 4 4 ¼ 0, hw2; w2i ¼ 16 þ 4 ¼ 20. Thus, A ¼ 5 0 0 20 . (Because the basis vectors are orthogonal, the matrix A is diagonal.) 7.42. Consider the vector space P2ðtÞ with inner product h f ; gi ¼ Ð 1 1 f ðtÞgðtÞ dt. (a) Find h f ; gi, where f ðtÞ ¼ t þ 2 and gðtÞ ¼ t2 3t þ 4. (b) Find the matrix A of the inner product with respect to the basis f1; t; t2g of V. (c) Verify Theorem 7.16 by showing that h f ; gi ¼ ½ f TA½g with respect to the basis f1; t; t2g. (a) h f ; gi ¼ ð1 1 ðt þ 2Þðt2 3t þ 4Þ dt ¼ ð1 1 ðt3 t2 2t þ 8Þ dt ¼ t4 4 t3 3 t2 þ 8t 1 1 ¼ 46 3 (b) Here we use the fact that if r þ s ¼ n, htr; tri ¼ ð1 1 tn dt ¼ tnþ1 n þ 1 1 1 ¼ 2=ðn þ 1Þ if n is even; 0 if n is odd: Then h1; 1i ¼ 2, h1; ti ¼ 0, h1; t2i ¼ 2 3, ht; ti ¼ 2 3, ht; t2i ¼ 0, ht2; t2i ¼ 2 5. Thus, A ¼ 2 0 2 3 0 2 3 0 2 3 0 2 5 2 4 3 5 254 CHAPTER 7 Inner Product Spaces, Orthogonality (c) We have ½ f T ¼ ð2; 1; 0Þ and ½gT ¼ ð4; 3; 1Þ relative to the given basis. Then ½ f TA½g ¼ ð2; 1; 0Þ 2 0 2 3 0 2 3 0 2 3 0 2 5 2 4 3 5 4 3 1 2 4 3 5 ¼ ð4; 2 3 ; 4 3Þ 4 3 1 2 4 3 5 ¼ 46 3 ¼ h f ; gi 7.43. Prove Theorem 7.14: A ¼ a b b c is positive deﬁnite if and only if a and d are positive and jAj ¼ ad b2 is positive. Let u ¼ ½x; yT. Then f ðuÞ ¼ uTAu ¼ ½x; y a b b d x y ¼ ax2 þ 2bxy þ dy2 Suppose f ðuÞ > 0 for every u 6¼ 0. Then f ð1; 0Þ ¼ a > 0 and f ð0; 1Þ ¼ d > 0. Also, we have f ðb; aÞ ¼ aðad b2Þ > 0. Because a > 0, we get ad b2 > 0. Conversely, suppose a > 0, b ¼ 0, ad b2 > 0. Completing the square gives us f ðuÞ ¼ a x2 þ 2b a xy þ b2 a2 y2 þ dy2 b2 a y2 ¼ a x þ by a 2 þ ad b2 a y2 Accordingly, f ðuÞ > 0 for every u 6¼ 0. 7.44. Prove Theorem 7.15: Let A be a real positive deﬁnite matrix. Then the function hu; vi ¼ uTAv is an inner product on Rn. For any vectors u1; u2, and v, hu1 þ u2; vi ¼ ðu1 þ u2ÞTAv ¼ ðuT 1 þ uT 2 ÞAv ¼ uT 1 Av þ uT 2 Av ¼ hu1; vi þ hu2; vi and, for any scalar k and vectors u; v, hku; vi ¼ ðkuÞTAv ¼ kuTAv ¼ khu; vi Thus ½I1 is satisﬁed. Because uTAv is a scalar, ðuTAvÞT ¼ uTAv. Also, AT ¼ A because A is symmetric. Therefore, hu; vi ¼ uTAv ¼ ðuTAvÞT ¼ vTATuTT ¼ vTAu ¼ hv; ui Thus, ½I2 is satisﬁed. Last, because A is positive deﬁnite, X TAX > 0 for any nonzero X 2 Rn. Thus, for any nonzero vector v; hv; vi ¼ vTAv > 0. Also, h0; 0i ¼ 0TA0 ¼ 0. Thus, ½I3 is satisﬁed. Accordingly, the function hu; vi ¼ Av is an inner product. 7.45. Prove Theorem 7.16: Let A be the matrix representation of an inner product relative to a basis S of V. Then, for any vectors u; v 2 V, we have hu; vi ¼ ½uTA½v Suppose S ¼ fw1; w2; . . . ; wng and A ¼ ½kij. Hence, kij ¼ hwi; wji. Suppose u ¼ a1w1 þ a2w2 þ þ anwn and v ¼ b1w1 þ b2w2 þ þ bnwn Then hu; vi ¼ P n i¼1 P n j¼1 aibjhwi; wji ð1Þ On the other hand, ½uTA½v ¼ ða1; a2; . . . ; anÞ k11 k12 . . . k1n k21 k22 . . . k2n :::::::::::::::::::::::::::::: kn1 kn2 . . . knn 2 6 6 6 4 3 7 7 7 5 b1 b2 . . . bn 2 6 6 6 6 4 3 7 7 7 7 5 ¼ P n i¼1 aiki1; P n i¼1 aiki2; . . . ; P n i¼1 aikin b1 b2 . . . bn 2 6 6 6 6 4 3 7 7 7 7 5 ¼ P n j¼1 P n i¼1 aibjkij ð2Þ Equations ð1Þ and (2) give us our result. CHAPTER 7 Inner Product Spaces, Orthogonality 255 7.46. Prove Theorem 7.17: Let A be the matrix representation of any inner product on V. Then A is a positive deﬁnite matrix. Because hwi; wji ¼ hwj; wii for any basis vectors wi and wj, the matrix A is symmetric. Let X be any nonzero vector in Rn. Then ½u ¼ X for some nonzero vector u 2 V. Theorem 7.16 tells us that X TAX ¼ ½uTA½u ¼ hu; ui > 0. Thus, A is positive deﬁnite. Complex Inner Product Spaces 7.47. Let V be a complex inner product space. Verify the relation hu; av1 þ bv2i ¼ ahu; v1i þ bhu; v2i Using ½I2 , ½I1 , and then ½I2 , we ﬁnd hu; av1 þ bv2i ¼ hav1 þ bv2; ui ¼ ahv1; ui þ bhv2; ui ¼ ahv1; ui þ bhv2; ui ¼ ahu; v1i þ bhu; v2i 7.48. Suppose hu; vi ¼ 3 þ 2i in a complex inner product space V. Find (a) hð2 4iÞu; vi; (b) hu; ð4 þ 3iÞvi; (c) hð3 6iÞu; ð5 2iÞvi: (a) hð2 4iÞu; vi ¼ ð2 4iÞhu; vi ¼ ð2 4iÞð3 þ 2iÞ ¼ 14 8i (b) hu; ð4 þ 3iÞvi ¼ ð4 þ 3iÞhu; vi ¼ ð4 3iÞð3 þ 2iÞ ¼ 18 i (c) hð3 6iÞu; ð5 2iÞvi ¼ ð3 6iÞð5 2iÞhu; vi ¼ ð3 6iÞð5 þ 2iÞð3 þ 2iÞ ¼ 129 18i 7.49. Find the Fourier coefﬁcient (component) c and the projection cw of v ¼ ð3 þ 4i; 2 3iÞ along w ¼ ð5 þ i; 2iÞ in C2. Recall that c ¼ hv; wi=hw; wi. Compute hv; wi ¼ ð3 þ 4iÞð5 þ iÞ þ ð2 3iÞð2iÞ ¼ ð3 þ 4iÞð5 iÞ þ ð2 3iÞð2iÞ ¼ 19 þ 17i 6 4i ¼ 13 þ 13i hw; wi ¼ 25 þ 1 þ 4 ¼ 30 Thus, c ¼ ð13 þ 13iÞ=30 ¼ 13 30 þ 13 30 i: Accordingly, projðv; wÞ ¼ cw ¼ ð26 15 þ 39 15 i; 13 15 þ 1 15 iÞ 7.50. Prove Theorem 7.18 (Cauchy–Schwarz): Let V be a complex inner product space. Then jhu; vij kukkvk. If v ¼ 0, the inequality reduces to 0 0 and hence is valid. Now suppose v 6¼ 0. Using z z ¼ jzj2 (for any complex number z) and hv; ui ¼ hu; vi, we expand ku hu; vitvk2 0, where t is any real value: 0 ku hu; vitvk2 ¼ hu hu; vitv; u hu; vitvi ¼ hu; ui hu; vithu; vi hu; vÞthv; ui þ hu; vihu; vit2hv; vi ¼ kuk2 2tjhu; vij2 þ jhu; vij2t2kvk2 Set t ¼ 1=kvk2 to ﬁnd 0 kuk2 jhu; vij2 kvk2 , from which jhu; vij2 kvk2kvk2. Taking the square root of both sides, we obtain the required inequality. 7.51. Find an orthogonal basis for u? in C3 where u ¼ ð1; i; 1 þ iÞ. Here u? consists of all vectors s ¼ ðx; y; zÞ such that hw; ui ¼ x iy þ ð1 iÞz ¼ 0 Find one solution, say w1 ¼ ð0; 1 i; iÞ. Then ﬁnd a solution of the system x iy þ ð1 iÞz ¼ 0; ð1 þ iÞy iz ¼ 0 Here z is a free variable. Set z ¼ 1 to obtain y ¼ i=ð1 þ iÞ ¼ ð1 þ iÞ=2 and x ¼ ð3i 3Þ2. Multiplying by 2 yields the solution w2 ¼ ð3i 3, 1 þ i, 2). The vectors w1 and w2 form an orthogonal basis for u?. 256 CHAPTER 7 Inner Product Spaces, Orthogonality 7.52. Find an orthonormal basis of the subspace W of C3 spanned by v1 ¼ ð1; i; 0Þ and v2 ¼ ð1; 2; 1 iÞ: Apply the Gram–Schmidt algorithm. Set w1 ¼ v1 ¼ ð1; i; 0Þ. Compute v2 hv2; w1i hw1; w1i w1 ¼ ð1; 2; 1 iÞ 1 2i 2 ð1; i; 0Þ ¼ ð1 2 þ i; 1 1 2 i; 1 iÞ Multiply by 2 to clear fractions, obtaining w2 ¼ ð1 þ 2i; 2 i; 2 2iÞ. Next ﬁnd kw1k ¼ ﬃﬃﬃ 2 p and then kw2k ¼ ﬃﬃﬃﬃﬃ 18 p . Normalizing fw1; w2g, we obtain the following orthonormal basis of W: u1 ¼ 1ﬃﬃﬃ 2 p ; iﬃﬃﬃ 2 p ; 0 ; u2 ¼ 1 þ 2i ﬃﬃﬃﬃﬃ 18 p ; 2 i ﬃﬃﬃﬃﬃ 18 p ; 2 2i ﬃﬃﬃﬃﬃ 18 p 7.53. Find the matrix P that represents the usual inner product on C3 relative to the basis f1; i; 1 ig. Compute the following six inner products: h1; 1i ¼ 1; hi; ii ¼ i i ¼ 1; h1; ii ¼ i ¼ i; hi; 1 ii ¼ ið1 iÞ ¼ 1 þ i; h1; 1 ii ¼ 1 i ¼ 1 þ i h1 i; 1 ii ¼ 2 Then, using ðu; vÞ ¼ hv; ui, we obtain P ¼ 1 i 1 þ i i 1 1 þ i 1 i 1 i 2 2 4 3 5 (As expected, P is Hermitian; that is, PH ¼ P.) Normed Vector Spaces 7.54. Consider vectors u ¼ ð1; 3; 6; 4Þ and v ¼ ð3; 5; 1; 2Þ in R4. Find (a) kuk1 and kvj1, (b) kuk1 and kvk1, (c) kuk2 and kvk2, (d) d1ðu; vÞ; d1ðu; vÞ, d2ðu; vÞ. (a) The inﬁnity norm chooses the maximum of the absolute values of the components. Hence, kuk1 ¼ 6 and kvk1 ¼ 5 (b) The one-norm adds the absolute values of the components. Thus, kuk1 ¼ 1 þ 3 þ 6 þ 4 ¼ 14 and kvk1 ¼ 3 þ 5 þ 1 þ 2 ¼ 11 (c) The two-norm is equal to the square root of the sum of the squares of the components (i.e., the norm induced by the usual inner product on R3). Thus, kuk2 ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1 þ 9 þ 36 þ 16 p ¼ ﬃﬃﬃﬃﬃ 62 p and kvk2 ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 9 þ 25 þ 1 þ 4 p ¼ ﬃﬃﬃﬃﬃ 39 p (d) First ﬁnd u v ¼ ð2; 8; 7; 6Þ. Then d1ðu; vÞ ¼ ku vk1 ¼ 8 d1ðu; vÞ ¼ ku vk1 ¼ 2 þ 8 þ 7 þ 6 ¼ 23 d2ðu; vÞ ¼ ku vk2 ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 4 þ 64 þ 49 þ 36 p ¼ ﬃﬃﬃﬃﬃﬃﬃ ﬃ 153 p 7.55. Consider the function f ðtÞ ¼ t2 4t in C½0; 3. (a) Find k f k1, (b) Plot f ðtÞ in the plane R2, (c) Find k f k1, (d) Find k f k2. (a) We seek k f k1 ¼ maxðj f ðtÞjÞ. Because f ðtÞ is differentiable on ½0; 3, j f ðtÞj has a maximum at a critical point of f ðtÞ (i.e., when the derivative f 0ðtÞ ¼ 0), or at an endpoint of ½0; 3. Because f 0ðtÞ ¼ 2t 4, we set 2t 4 ¼ 0 and obtain t ¼ 2 as a critical point. Compute f ð2Þ ¼ 4 8 ¼ 4; f ð0Þ ¼ 0 0 ¼ 0; f ð3Þ ¼ 9 12 ¼ 3 Thus, k f k1 ¼ j f ð2Þj ¼ j 4j ¼ 4. CHAPTER 7 Inner Product Spaces, Orthogonality 257 (b) Compute f ðtÞ for various values of t in ½0; 3, for example, t 0 1 2 3 f ðtÞ 0 3 4 3 Plot the points in R2 and then draw a continuous curve through the points, as shown in Fig. 7-8. (c) We seek k f k1 ¼ Ð 3 0 j f ðtÞj dt. As indicated in Fig. 7-3, f ðtÞ is negative in ½0; 3; hence, j f ðtÞj ¼ ðt2 4tÞ ¼ 4t t2 k f k1 ¼ ð3 0 ð4t t2Þ dt ¼ 2t2 t3 3 3 0 ¼ 18 9 ¼ 9 Thus; (d) k f k2 2 ¼ ð3 0 f ðtÞ2 dt ¼ ð3 0 ðt4 8t3 þ 16t2Þ dt ¼ t5 5 2t4 þ 16t3 3 3 0 ¼ 153 5 . Thus, k f k2 ¼ ﬃﬃﬃﬃﬃﬃﬃ ﬃ 153 5 r . 7.56. Prove Theorem 7.24: Let V be a normed vector space. Then the function dðu; vÞ ¼ ku vk satisﬁes the following three axioms of a metric space: ½M1 dðu; vÞ 0; and dðu; vÞ ¼ 0 iff u ¼ v. ½M2 dðu; vÞ ¼ dðv; uÞ. ½M3 dðu; vÞ dðu; wÞ þ dðw; vÞ. If u 6¼ v, then u v 6¼ 0, and hence, dðu; vÞ ¼ ku vk > 0. Also, dðu; uÞ ¼ ku uk ¼ k0k ¼ 0. Thus, ½M1 is satisﬁed. We also have dðu; vÞ ¼ ku vk ¼ k 1ðv uÞk ¼ j 1jkv uk ¼ kv uk ¼ dðv; uÞ and dðu; vÞ ¼ ku vk ¼ kðu wÞ þ ðw vÞk ku wk þ kw vk ¼ dðu; wÞ þ dðw; vÞ Thus, ½M2 and ½M3 are satisﬁed. SUPPLEMENTARY PROBLEMS Inner Products 7.57. Verify that the following is an inner product on R2, where u ¼ ðx1; x2Þ and v ¼ ðy1; y2Þ: f ðu; vÞ ¼ x1y1 2x1 y2 2x2 y1 þ 5x2 y2 7.58. Find the values of k so that the following is an inner product on R2, where u ¼ ðx1; x2Þ and v ¼ ðy1; y2Þ: f ðu; vÞ ¼ x1y1 3x1 y2 3x2 y1 þ kx2 y2 Figure 7-8 258 CHAPTER 7 Inner Product Spaces, Orthogonality 7.59. Consider the vectors u ¼ ð1; 3Þ and v ¼ ð2; 5Þ in R2. Find (a) hu; vi with respect to the usual inner product in R2. (b) hu; vi with respect to the inner product in R2 in Problem 7.57. (c) kvk using the usual inner product in R2. (d) kvk using the inner product in R2 in Problem 7.57. 7.60. Show that each of the following is not an inner product on R3, where u ¼ ðx1; x2; x3Þ and v ¼ ðy1; y2; y3Þ: (a) hu; vi ¼ x1y1 þ x2y2; (b) hu; vi ¼ x1y2x3 þ y1x2y3. 7.61. Let V be the vector space of m n matrices over R. Show that hA; Bi ¼ trðBTAÞ deﬁnes an inner product in V. 7.62. Suppose jhu; vij ¼ kukkvk. (That is, the Cauchy–Schwarz inequality reduces to an equality.) Show that u and v are linearly dependent. 7.63. Suppose f ðu; vÞ and gðu; vÞ are inner products on a vector space V over R. Prove (a) The sum f þ g is an inner product on V, where ð f þ gÞðu; vÞ ¼ f ðu; vÞ þ gðu; vÞ. (b) The scalar product kf , for k > 0, is an inner product on V, where ðkf Þðu; vÞ ¼ kf ðu; vÞ. Orthogonality, Orthogonal Complements, Orthogonal Sets 7.64. Let V be the vector space of polynomials over R of degree 2 with inner product deﬁned by h f ; gi ¼ Ð 1 0 f ðtÞgðtÞ dt. Find a basis of the subspace W orthogonal to hðtÞ ¼ 2t þ 1. 7.65. Find a basis of the subspace W of R4 orthogonal to u1 ¼ ð1; 2; 3; 4Þ and u2 ¼ ð3; 5; 7; 8Þ. 7.66. Find a basis for the subspace W of R5 orthogonal to the vectors u1 ¼ ð1; 1; 3; 4; 1Þ and u2 ¼ ð1; 2; 1; 2; 1Þ. 7.67. Let w ¼ ð1; 2; 1; 3Þ be a vector in R4. Find (a) an orthogonal basis for w?; (b) an orthonormal basis for w?. 7.68. Let W be the subspace of R4 orthogonal to u1 ¼ ð1; 1; 2; 2Þ and u2 ¼ ð0; 1; 2; 1Þ. Find (a) an orthogonal basis for W; (b) an orthonormal basis for W. (Compare with Problem 7.65.) 7.69. Let S consist of the following vectors in R4: u1 ¼ ð1; 1; 1; 1Þ; u2 ¼ ð1; 1; 1; 1Þ; u3 ¼ ð1; 1; 1; 1Þ; u4 ¼ ð1; 1; 1; 1Þ (a) Show that S is orthogonal and a basis of R4. (b) Write v ¼ ð1; 3; 5; 6Þ as a linear combination of u1; u2; u3; u4. (c) Find the coordinates of an arbitrary vector v ¼ ða; b; c; dÞ in R4 relative to the basis S. (d) Normalize S to obtain an orthonormal basis of R4. 7.70. Let M ¼ M2;2 with inner product hA; Bi ¼ trðBTAÞ. Show that the following is an orthonormal basis for M: 1 0 0 0 ; 0 1 0 0 ; 0 0 1 0 ; 0 0 0 1 7.71. Let M ¼ M2;2 with inner product hA; Bi ¼ trðBTAÞ. Find an orthogonal basis for the orthogonal complement of (a) diagonal matrices, (b) symmetric matrices. CHAPTER 7 Inner Product Spaces, Orthogonality 259 7.72. Suppose fu1; u2; . . . ; urg is an orthogonal set of vectors. Show that fk1u1; k2u2; . . . ; krurg is an orthogonal set for any scalars k1; k2; . . . ; kr. 7.73. Let U and W be subspaces of a ﬁnite-dimensional inner product space V. Show that (a) ðU þ WÞ? ¼ U ? \ W ?; (b) ðU \ WÞ? ¼ U? þ W ?. Projections, Gram–Schmidt Algorithm, Applications 7.74. Find the Fourier coefﬁcient c and projection cw of v along w, where (a) v ¼ ð2; 3; 5Þ and w ¼ ð1; 5; 2Þ in R3: (b) v ¼ ð1; 3; 1; 2Þ and w ¼ ð1; 2; 7; 4Þ in R4: (c) v ¼ t2 and w ¼ t þ 3 in PðtÞ; with inner product h f ; gi ¼ Ð 1 0 f ðtÞgðtÞ dt (d) v ¼ 1 2 3 4 and w ¼ 1 1 5 5 in M ¼ M2;2; with inner product hA; Bi ¼ trðBTAÞ: 7.75. Let U be the subspace of R4 spanned by v1 ¼ ð1; 1; 1; 1Þ; v2 ¼ ð1; 1; 2; 2Þ; v3 ¼ ð1; 2; 3; 4Þ (a) Apply the Gram–Schmidt algorithm to ﬁnd an orthogonal and an orthonormal basis for U. (b) Find the projection of v ¼ ð1; 2; 3; 4Þ onto U. 7.76. Suppose v ¼ ð1; 2; 3; 4; 6Þ. Find the projection of v onto W, or, in other words, ﬁnd w 2 W that minimizes kv wk, where W is the subspace of R5 spanned by (a) u1 ¼ ð1; 2; 1; 2; 1Þ and u2 ¼ ð1; 1; 2; 1; 1Þ, (b) v1 ¼ ð1; 2; 1; 2; 1Þ and v2 ¼ ð1; 0; 1; 5; 1Þ. 7.77. Consider the subspace W ¼ P2ðtÞ of PðtÞ with inner product h f ; gi ¼ Ð 1 0 f ðtÞgðtÞ dt. Find the projection of f ðtÞ ¼ t3 onto W. (Hint: Use the orthogonal polynomials 1; 2t 1, 6t2 6t þ 1 obtained in Problem 7.22.) 7.78. Consider PðtÞ with inner product h f ; gi ¼ Ð 1 1 f ðtÞgðtÞ dt and the subspace W ¼ P3ðtÞ: (a) Find an orthogonal basis for W by applying the Gram–Schmidt algorithm to f1; t; t2; t3g. (b) Find the projection of f ðtÞ ¼ t5 onto W. Orthogonal Matrices 7.79. Find the number and exhibit all 2 2 orthogonal matrices of the form 1 3 x y z . 7.80. Find a 3 3 orthogonal matrix P whose ﬁrst two rows are multiples of u ¼ ð1; 1; 1Þ and v ¼ ð1; 2; 3Þ, respectively. 7.81. Find a symmetric orthogonal matrix P whose ﬁrst row is ð1 3 ; 2 3 ; 2 3Þ. (Compare with Problem 7.32.) 7.82. Real matrices A and B are said to be orthogonally equivalent if there exists an orthogonal matrix P such that B ¼ PTAP. Show that this relation is an equivalence relation. Positive Deﬁnite Matrices and Inner Products 7.83. Find the matrix A that represents the usual inner product on R2 relative to each of the following bases: (a) fv1 ¼ ð1; 4Þ; v2 ¼ ð2; 3Þg, (b) fw1 ¼ ð1; 3Þ; w2 ¼ ð6; 2Þg. 7.84. Consider the following inner product on R2: f ðu; vÞ ¼ x1y1 2x1y2 2x2y1 þ 5x2y2; where u ¼ ðx1; x2Þ v ¼ ðy1; y2Þ Find the matrix B that represents this inner product on R2 relative to each basis in Problem 7.83. 260 CHAPTER 7 Inner Product Spaces, Orthogonality 7.85. Find the matrix C that represents the usual basis on R3 relative to the basis S of R3 consisting of the vectors u1 ¼ ð1; 1; 1Þ, u2 ¼ ð1; 2; 1Þ, u3 ¼ ð1; 1; 3Þ. 7.86. Let V ¼ P2ðtÞ with inner product h f ; gi ¼ Ð 1 0 f ðtÞgðtÞ dt. (a) Find h f ; gi, where f ðtÞ ¼ t þ 2 and gðtÞ ¼ t2 3t þ 4. (b) Find the matrix A of the inner product with respect to the basis f1; t; t2g of V. (c) Verify Theorem 7.16 that h f ; gi ¼ ½ f TA½g with respect to the basis f1; t; t2g. 7.87. Determine which of the following matrices are positive deﬁnite: (a) 1 3 3 5 , (b) 3 4 4 7 , (c) 4 2 2 1 , (d) 6 7 7 9 . 7.88. Suppose A and B are positive deﬁnite matrices. Show that: (a) A þ B is positive deﬁnite and (b) kA is positive deﬁnite for k > 0. 7.89. Suppose B is a real nonsingular matrix. Show that: (a) BTB is symmetric and (b) BTB is positive deﬁnite. Complex Inner Product Spaces 7.90. Verify that ha1u1 þ a2u2 b1v1 þ b2v2i ¼ a1 b1hu1; v1i þ a1 b2hu1; v2i þ a2 b1hu2; v1i þ a2 b2hu2; v2i More generally, prove that hPm i¼1 aiui; Pn j¼1 bjvji ¼ P i;j ai bjhui; vii. 7.91. Consider u ¼ ð1 þ i; 3; 4 iÞ and v ¼ ð3 4i; 1 þ i; 2iÞ in C3. Find (a) hu; vi, (b) hv; ui, (c) kuk, (d) kvk, (e) dðu; vÞ. 7.92. Find the Fourier coefﬁcient c and the projection cw of (a) u ¼ ð3 þ i; 5 2iÞ along w ¼ ð5 þ i; 1 þ iÞ in C2, (b) u ¼ ð1 i; 3i; 1 þ iÞ along w ¼ ð1; 2 i; 3 þ 2iÞ in C3. 7.93. Let u ¼ ðz1; z2Þ and v ¼ ðw1; w2Þ belong to C2. Verify that the following is an inner product of C2: f ðu; vÞ ¼ z1 w1 þ ð1 þ iÞz1 w2 þ ð1 iÞz2 w1 þ 3z2 w2 7.94. Find an orthogonal basis and an orthonormal basis for the subspace W of C3 spanned by u1 ¼ ð1; i; 1Þ and u2 ¼ ð1 þ i; 0; 2Þ. 7.95. Let u ¼ ðz1; z2Þ and v ¼ ðw1; w2Þ belong to C2. For what values of a; b; c; d 2 C is the following an inner product on C2? f ðu; vÞ ¼ az1 w1 þ bz1 w2 þ cz2 w1 þ dz2 w2 7.96. Prove the following form for an inner product in a complex space V: hu; vi ¼ 1 4 ku þ vk2 1 4 ku vk2 þ 1 4 ku þ ivk2 1 4 ku ivk2 [Compare with Problem 7.7(b).] 7.97. Let V be a real inner product space. Show that (i) kuk ¼ kvk if and only if hu þ v; u vi ¼ 0; (ii) ku þ vk2 ¼ kuk2 þ kvk2 if and only if hu; vi ¼ 0. Show by counterexamples that the above statements are not true for, say, C2. 7.98. Find the matrix P that represents the usual inner product on C3 relative to the basis f1; 1 þ i; 1 2ig. CHAPTER 7 Inner Product Spaces, Orthogonality 261 7.99. A complex matrix A is unitary if it is invertible and A1 ¼ AH. Alternatively, A is unitary if its rows (columns) form an orthonormal set of vectors (relative to the usual inner product of Cn). Find a unitary matrix whose ﬁrst row is: (a) a multiple of ð1; 1 iÞ; (b) a multiple of ð1 2 ; 1 2 i; 1 2 1 2 iÞ. Normed Vector Spaces 7.100. Consider vectors u ¼ ð1; 3; 4; 1; 2Þ and v ¼ ð3; 1; 2; 3; 1Þ in R5. Find (a) kuk1 and kvk1 , (b) kuk1 and kvk1, (c) kuk2 and kvk2, (d) d1ðu; vÞ; d1ðu; vÞ, d2ðu; vÞ 7.101. Repeat Problem 7.100 for u ¼ ð1 þ i; 2 4iÞ and v ¼ ð1 i; 2 þ 3iÞ in C2. 7.102. Consider the functions f ðtÞ ¼ 5t t2 and gðtÞ ¼ 3t t2 in C½0; 4. Find (a) d1ð f ; gÞ, (b) d1ð f ; gÞ, (c) d2ð f ; gÞ 7.103. Prove (a) k k1 is a norm on Rn. (b) k k1 is a norm on Rn. 7.104. Prove (a) k k1 is a norm on C½a; b. (b) k k1 is a norm on C½a; b. ANSWERS TO SUPPLEMENTARY PROBLEMS Notation: M ¼ ½R1; R2; . . . denotes a matrix M with rows R1; R2; : . . . Also, basis need not be unique. 7.58. k > 9 7.59. (a) 13, (b) 71, (c) ﬃﬃﬃﬃﬃ 29 p , (d) ﬃﬃﬃﬃﬃ 89 p 7.60. Let u ¼ ð0; 0; 1Þ; then hu; ui ¼ 0 in both cases 7.64. f7t2 5t; 12t2 5g 7.65. fð1; 2; 1; 0Þ; ð4; 4; 0; 1Þg 7.66. ð1; 0; 0; 0; 1Þ; ð6; 2; 0; 1; 0Þ; ð5; 2; 1; 0; 0Þ 7.67. (a) u1 ¼ ð0; 0; 3; 1Þ; u2 ¼ ð0; 5; 1; 3Þ; u3 ¼ ð14; 2; 1; 3Þ; (b) u1= ﬃﬃﬃﬃﬃ 10 p ; u2= ﬃﬃﬃﬃﬃ 35 p ; u3= ﬃﬃﬃﬃﬃﬃﬃ ﬃ 210 p 7.68. (a) ð0; 2; 1; 0Þ; ð15; 1; 2; 5Þ, (b) ð0; 2; 1; 0Þ= ﬃﬃﬃ 5 p ; ð15; 1; 2; 5Þ= ﬃﬃﬃﬃﬃﬃﬃ ﬃ 255 p 7.69. (b) v ¼ 1 4 ð5u1 þ 3u2 13u3 þ 9u4Þ, (c) ½v ¼ 1 4 ½a þ b þ c þ d; a þ b c d; a b þ c d; a b c þ d 7.71. (a) ½0; 1; 0; 0; ½0; 0; 1; 0, (b) ½0; 1; 1; 0 7.74. (a) c ¼ 23 30, (b) c ¼ 1 7, (c) c ¼ 15 148, (d) c ¼ 19 26 7.75. (a) w1 ¼ ð1; 1; 1; 1Þ; w2 ¼ ð0; 2; 1; 1Þ; w3 ¼ ð12; 4; 1; 7Þ, (b) projðv; UÞ ¼ 1 5 ð1; 12; 3; 6Þ 7.76. (a) projðv; WÞ ¼ 1 8 ð23; 25; 30; 25; 23Þ, (b) First ﬁnd an orthogonal basis for W; say, w1 ¼ ð1; 2; 1; 2; 1Þ and w2 ¼ ð0; 2; 0; 3; 2Þ. Then projðv; WÞ ¼ 1 17 ð34; 76; 34; 56; 42Þ 7.77. projð f ; WÞ ¼ 3 2 t2 3 5 t þ 1 20 262 CHAPTER 7 Inner Product Spaces, Orthogonality 7.78. (a) f1; t; 3t2 1; 5t3 3tg, projð f ; WÞ ¼ 10 9 t3 5 21 t 7.79. Four: ½a; b; b; a, ½a; b; b; a, ½a; b; b; a, ½a; b; b; a, where a ¼ 1 3 and b ¼ 1 3 ﬃﬃﬃ 8 p 7.80. P ¼ ½1=a; 1=a; 1=a; 1=b; 2=b; 3=b; 5=c; 2=c; 3=c, where a ¼ ﬃﬃﬃ 3 p ; b ¼ ﬃﬃﬃﬃﬃ 14 p ; c ¼ ﬃﬃﬃﬃﬃ 38 p 7.81. 1 3 ½1; 2; 2; 2; 2; 1; 2; 1; 2 7.83. (a) ½17; 10; 10; 13, (b) ½10; 0; 0; 40 7.84. (a) ½65; 68; 68; 73, (b) ½58; 8; 8; 8 7.85. ½3; 4; 3; 4; 6; 2; 3; 2; 11 7.86. (a) 83 12, (b) ½1; a; b; a; b; c; b; c; d, where a ¼ 1 2, b ¼ 1 3, c ¼ 1 4, d ¼ 1 5 7.87. (a) No, (b) Yes, (c) No, (d) Yes 7.91. (a) 4i, (b) 4i, (c) ﬃﬃﬃﬃﬃ 28 p , (d) ﬃﬃﬃﬃﬃ 31 p , (e) ﬃﬃﬃﬃﬃ 59 p 7.92. (a) c ¼ 1 28 ð19 5iÞ, (b) c ¼ 1 19 ð3 þ 6iÞ 7.94. fv1 ¼ ð1; i; 1Þ= ﬃﬃﬃ 3 p ; v2 ¼ ð2i; 1 3i; 3 iÞ= ﬃﬃﬃﬃﬃ 24 p g 7.95. a and d real and positive, c ¼ b and ad bc positive. 7.97. u ¼ ð1; 2Þ; v ¼ ði; 2iÞ 7.98. P ¼ ½1; 1 i; 1 þ 2i; 1 þ i; 2; 1 þ 3i; 1 2i; 1 3i; 5 7.99. (a) ð1= ﬃﬃﬃ 3 p Þ½1; 1 i; 1 þ i; 1, (b) ½a; ai; a ai; bi; b; 0; a; ai; a ai, where a ¼ 1 2 and b ¼ 1= ﬃﬃﬃ 2 p . 7.100. (a) 4 and 3, (b) 11 and 10, (c) ﬃﬃﬃﬃﬃ 31 p and ﬃﬃﬃﬃﬃ 24 p , (d) 6; 19; 9 7.101. (a) ﬃﬃﬃﬃﬃ 20 p and ﬃﬃﬃﬃﬃ 13 p , (b) ﬃﬃﬃ 2 p þ ﬃﬃﬃﬃﬃ 20 p and ﬃﬃﬃ 2 p þ ﬃﬃﬃﬃﬃ 13 p , (c) ﬃﬃﬃﬃﬃ 22 p and ﬃﬃﬃﬃﬃ 15 p , (d) 7; 9; ﬃﬃﬃﬃﬃ 53 p 7.102. (a) 8, (b) 16, (c) 16= ﬃﬃﬃ 3 p CHAPTER 7 Inner Product Spaces, Orthogonality 263 Determinants 8.1 Introduction Each n-square matrix A ¼ ½aij is assigned a special scalar called the determinant of A, denoted by detðAÞ or jAj or a11 a12 . . . a1n a21 a22 . . . a2n ::::::::::::::::::::::::::::: an1 an2 . . . ann We emphasize that an n n array of scalars enclosed by straight lines, called a determinant of order n, is not a matrix but denotes the determinant of the enclosed array of scalars (i.e., the enclosed matrix). The determinant function was ﬁrst discovered during the investigation of systems of linear equations. We shall see that the determinant is an indispensable tool in investigating and obtaining properties of square matrices. The deﬁnition of the determinant and most of its properties also apply in the case where the entries of a matrix come from a commutative ring. We begin with a special case of determinants of orders 1, 2, and 3. Then we deﬁne a determinant of arbitrary order. This general deﬁnition is preceded by a discussion of permutations, which is necessary for our general deﬁnition of the determinant. 8.2 Determinants of Orders 1 and 2 Determinants of orders 1 and 2 are deﬁned as follows: ja11j ¼ a11 and a11 a12 a21 a22 ¼ a11a22 a12a21 Thus, the determinant of a 1 1 matrix A ¼ ½a11 is the scalar a11; that is, detðAÞ ¼ ja11j ¼ a11. The determinant of order two may easily be remembered by using the following diagram: a11 a12 a21 a22 That, is, the determinant is equal to the product of the elements along the plus-labeled arrow minus the product of the elements along the minus-labeled arrow. (There is an analogous diagram for determinants of order 3, but not for higher-order determinants.) EXAMPLE 8.1 (a) Because the determinant of order 1 is the scalar itself, we have: detð27Þ ¼ 27; detð7Þ ¼ 7; detðt 3Þ ¼ t 3 (b) 5 3 4 6 ¼ 5ð6Þ 3ð4Þ ¼ 30 12 ¼ 18; 3 2 5 7 ¼ 21 þ 10 ¼ 31 ! ! þ CHAPTER 8 264 Application to Linear Equations Consider two linear equations in two unknowns, say a1z þ b1y ¼ c1 a2x þ b2y ¼ c2 Let D ¼ a1b2 a2b1, the determinant of the matrix of coefﬁcients. Then the system has a unique solution if and only if D 6¼ 0. In such a case, the unique solution may be expressed completely in terms of determinants as follows: x ¼ Nx D ¼ b2c1 b1c2 a1b2 a2b1 ¼ c1 b1 c2 b2 a1 b1 a2 b2 ; y ¼ Ny D ¼ a1c2 a2c1 a1b2 a2b1 ¼ a1 c1 a2 c2 a1 b1 a2 b2 Here D appears in the denominator of both quotients. The numerators Nx and Ny of the quotients for x and y, respectively, can be obtained by substituting the column of constant terms in place of the column of coefﬁcients of the given unknown in the matrix of coefﬁcients. On the other hand, if D ¼ 0, then the system may have no solution or more than one solution. EXAMPLE 8.2 Solve by determinants the system 4x 3y ¼ 15 2x þ 5y ¼ 1 First ﬁnd the determinant D of the matrix of coefﬁcients: D ¼ 4 3 2 5 ¼ 4ð5Þ ð3Þð2Þ ¼ 20 þ 6 ¼ 26 Because D 6¼ 0, the system has a unique solution. To obtain the numerators Nx and Ny, simply replace, in the matrix of coefﬁcients, the coefﬁcients of x and y, respectively, by the constant terms, and then take their determinants: Nx ¼ 15 3 1 5 ¼ 75 þ 3 ¼ 78 Ny ¼ 4 15 2 1 ¼ 4 30 ¼ 26 Then the unique solution of the system is x ¼ Nx D ¼ 78 26 ¼ 3; y ¼ Ny D ¼ 26 26 ¼ 1 8.3 Determinants of Order 3 Consider an arbitrary 3 3 matrix A ¼ ½aij. The determinant of A is deﬁned as follows: detðAÞ ¼ a11 a12 a13 a21 a22 a23 a31 a32 a33 ¼ a11a22a33 þ a12a23a31 þ a13a21a32 a13a22a31 a12a21a33 a11a23a32 Observe that there are six products, each product consisting of three elements of the original matrix. Three of the products are plus-labeled (keep their sign) and three of the products are minus-labeled (change their sign). The diagrams in Fig. 8-1 may help us to remember the above six products in detðAÞ. That is, the determinant is equal to the sum of the products of the elements along the three plus-labeled arrows in CHAPTER 8 Determinants 265 Fig. 8-1 plus the sum of the negatives of the products of the elements along the three minus-labeled arrows. We emphasize that there are no such diagrammatic devices with which to remember determinants of higher order. EXAMPLE 8.3 Let A ¼ 2 1 1 0 5 2 1 3 4 2 4 3 5 and B ¼ 3 2 1 4 5 1 2 3 4 2 4 3 5. Find detðAÞ and detðBÞ. Use the diagrams in Fig. 8-1: detðAÞ ¼ 2ð5Þð4Þ þ 1ð2Þð1Þ þ 1ð3Þð0Þ 1ð5Þð1Þ ð3Þð2Þð2Þ 4ð1Þð0Þ ¼ 40 2 þ 0 5 12 0 ¼ 21 detðBÞ ¼ 60 4 þ 12 10 9 þ 32 ¼ 81 Alternative Form for a Determinant of Order 3 The determinant of the 3 3 matrix A ¼ ½aij may be rewritten as follows: detðAÞ ¼ a11ða22a23 a23a32Þ a12ða21a33 a23a31Þ þ a13ða21a32 a22a31Þ ¼ a11 a22 a23 a32 a33 a12 a21 a23 a31 a33 þ a13 a21 a22 a31 a32 which is a linear combination of three determinants of order 2 whose coefﬁcients (with alternating signs) form the ﬁrst row of the given matrix. This linear combination may be indicated in the form a11 a11 a12 a13 a21 a22 a23 a31 a32 a33 a12 a11 a12 a13 a21 a22 a23 a31 a32 a33 þ a13 a11 a12 a13 a21 a22 a23 a31 a32 a33 Note that each 2 2 matrix can be obtained by deleting, in the original matrix, the row and column containing its coefﬁcient. EXAMPLE 8.4 1 2 3 4 2 3 0 5 1 ¼ 1 1 2 3 4 2 3 0 5 1 2 1 2 3 4 2 3 0 5 1 þ 3 1 2 3 4 2 3 0 5 1 ¼ 1 2 3 5 1 2 4 3 0 1 þ 3 4 2 0 5 ¼ 1ð2 15Þ 2ð4 þ 0Þ þ 3ð20 þ 0Þ ¼ 13 þ 8 þ 60 ¼ 55 Figure 8-1 266 CHAPTER 8 Determinants 8.4 Permutations A permutation s of the set f1; 2; . . . ; ng is a one-to-one mapping of the set onto itself or, equivalently, a rearrangement of the numbers 1; 2; . . . ; n. Such a permutation s is denoted by s ¼ 1 2 . . . n j1 j2 . . . jn or s ¼ j1j2 jn; where ji ¼ sðiÞ The set of all such permutations is denoted by Sn, and the number of such permutations is n!. If s 2 Sn; then the inverse mapping s1 2 Sn; and if s; t 2 Sn, then the composition mapping s t 2 Sn. Also, the identity mapping e ¼ s s1 2 Sn. (In fact, e ¼ 123 . . . n.) EXAMPLE 8.5 (a) There are 2! ¼ 2 1 ¼ 2 permutations in S2; they are 12 and 21. (b) There are 3! ¼ 3 2 1 ¼ 6 permutations in S3; they are 123, 132, 213, 231, 312, 321. Sign (Parity) of a Permutation Consider an arbitrary permutation s in Sn, say s ¼ j1j2 jn: We say s is an even or odd permutation according to whether there is an even or odd number of inversions in s. By an inversion in s we mean a pair of integers ði; kÞ such that i > k, but i precedes k in s. We then deﬁne the sign or parity of s, written sgn s, by sgn s ¼ 1 if s is even 1 if s is odd EXAMPLE 8.6 (a) Find the sign of s ¼ 35142 in S5. For each element k, we count the number of elements i such that i > k and i precedes k in s. There are 2 numbers ð3 and 5Þ greater than and preceding 1; 3 numbers ð3; 5; and 4Þ greater than and preceding 2; 1 number ð5Þ greater than and preceding 4: (There are no numbers greater than and preceding either 3 or 5.) Because there are, in all, six inversions, s is even and sgn s ¼ 1. (b) The identity permutation e ¼ 123 . . . n is even because there are no inversions in e. (c) In S2, the permutation 12 is even and 21 is odd. In S3, the permutations 123, 231, 312 are even and the permutations 132, 213, 321 are odd. (d) Let t be the permutation that interchanges two numbers i and j and leaves the other numbers ﬁxed. That is, tðiÞ ¼ j; tðjÞ ¼ i; tðkÞ ¼ k; where k 6¼ i; j We call t a transposition. If i < j, then there are 2ð j iÞ 1 inversions in t, and hence, the transposition t is odd. Remark: One can show that, for any n, half of the permutations in Sn are even and half of them are odd. For example, 3 of the 6 permutations in S3 are even, and 3 are odd. 8.5. Determinants of Arbitrary Order Let A ¼ ½aij be a square matrix of order n over a ﬁeld K. Consider a product of n elements of A such that one and only one element comes from each row and one and only one element comes from each column. Such a product can be written in the form a1j1a2j2 anjn CHAPTER 8 Determinants 267 that is, where the factors come from successive rows, and so the ﬁrst subscripts are in the natural order 1; 2; . . . ; n. Now because the factors come from different columns, the sequence of second subscripts forms a permutation s ¼ j1 j2 jn in Sn. Conversely, each permutation in Sn determines a product of the above form. Thus, the matrix A contains n! such products. DEFINITION: The determinant of A ¼ ½aij, denoted by detðAÞ or jAj, is the sum of all the above n! products, where each such product is multiplied by sgn s. That is, jAj ¼ P s ðsgn sÞa1j1a2j2 anjn or jAj ¼ P s2Sn ðsgn sÞa1sð1Þa2sð2Þ ansðnÞ The determinant of the n-square matrix A is said to be of order n. The next example shows that the above deﬁnition agrees with the previous deﬁnition of determinants of orders 1, 2, and 3. EXAMPLE 8.7 (a) Let A ¼ ½a11 be a 1 1 matrix. Because S1 has only one permutation, which is even, detðAÞ ¼ a11, the number itself. (b) Let A ¼ ½aij be a 2 2 matrix. In S2, the permutation 12 is even and the permutation 21 is odd. Hence, detðAÞ ¼ a11 a12 a21 a22 ¼ a11a22 a12a21 (c) Let A ¼ ½aij be a 3 3 matrix. In S3, the permutations 123, 231, 312 are even, and the permutations 321, 213, 132 are odd. Hence, detðAÞ ¼ a11 a12 a13 a21 a22 a23 a31 a32 a33 ¼ a11a22a33 þ a12a23a31 þ a13a21a32 a13a22a31 a12a21a33 a11a23a32 Remark: As n increases, the number of terms in the determinant becomes astronomical. Accordingly, we use indirect methods to evaluate determinants rather than the deﬁnition of the determinant. In fact, we prove a number of properties about determinants that will permit us to shorten the computation considerably. In particular, we show that a determinant of order n is equal to a linear combination of determinants of order n 1, as in the case n ¼ 3 above. 8.6 Properties of Determinants We now list basic properties of the determinant. THEOREM 8.1: The determinant of a matrix A and its transpose AT are equal; that is, jAj ¼ jATj. By this theorem (proved in Problem 8.22), any theorem about the determinant of a matrix A that concerns the rows of A will have an analogous theorem concerning the columns of A. The next theorem (proved in Problem 8.24) gives certain cases for which the determinant can be obtained immediately. THEOREM 8.2: Let A be a square matrix. (i) If A has a row (column) of zeros, then jAj ¼ 0. (ii) If A has two identical rows (columns), then jAj ¼ 0. 268 CHAPTER 8 Determinants (iii) If A is triangular (i.e., A has zeros above or below the diagonal), then jAj ¼ product of diagonal elements. Thus, in particular, jIj ¼ 1, where I is the identity matrix. The next theorem (proved in Problems 8.23 and 8.25) shows how the determinant of a matrix is affected by the elementary row and column operations. THEOREM 8.3: Suppose B is obtained from A by an elementary row (column) operation. (i) If two rows (columns) of A were interchanged, then jBj ¼ jAj. (ii) If a row (column) of A were multiplied by a scalar k, then jBj ¼ kjAj. (iii) If a multiple of a row (column) of A were added to another row (column) of A, then jBj ¼ jAj. Major Properties of Determinants We now state two of the most important and useful theorems on determinants. THEOREM 8.4: The determinant of a product of two matrices A and B is the product of their determinants; that is, detðABÞ ¼ detðAÞ detðBÞ The above theorem says that the determinant is a multiplicative function. THEOREM 8.5: Let A be a square matrix. Then the following are equivalent: (i) A is invertible; that is, A has an inverse A1. (ii) AX ¼ 0 has only the zero solution. (iii) The determinant of A is not zero; that is, detðAÞ 6¼ 0. Remark: Depending on the author and the text, a nonsingular matrix A is deﬁned to be an invertible matrix A, or a matrix A for which jAj 6¼ 0, or a matrix A for which AX ¼ 0 has only the zero solution. The above theorem shows that all such deﬁnitions are equivalent. We will prove Theorems 8.4 and 8.5 (in Problems 8.29 and 8.28, respectively) using the theory of elementary matrices and the following lemma (proved in Problem 8.26), which is a special case of Theorem 8.4. LEMMA 8.6: Let E be an elementary matrix. Then, for any matrix A; jEAj ¼ jEjjAj. Recall that matrices A and B are similar if there exists a nonsingular matrix P such that B ¼ P1AP. Using the multiplicative property of the determinant (Theorem 8.4), one can easily prove (Problem 8.31) the following theorem. THEOREM 8.7: Suppose A and B are similar matrices. Then jAj ¼ jBj. 8.7 Minors and Cofactors Consider an n-square matrix A ¼ ½aij. Let Mij denote the ðn 1Þ-square submatrix of A obtained by deleting its ith row and jth column. The determinant jMijj is called the minor of the element aij of A, and we deﬁne the cofactor of aij, denoted by Aij; to be the ‘‘signed’’ minor: Aij ¼ ð1ÞiþjjMijj CHAPTER 8 Determinants 269 Note that the ‘‘signs’’ ð1Þiþj accompanying the minors form a chessboard pattern with þ’s on the main diagonal: þ þ . . . þ þ . . . þ þ . . . ::::::::::::::::::::::::::::::: 2 6 6 4 3 7 7 5 We emphasize that Mij denotes a matrix, whereas Aij denotes a scalar. Remark: The sign ð1Þiþj of the cofactor Aij is frequently obtained using the checkerboard pattern. Speciﬁcally, beginning with þ and alternating signs: þ; ; þ; ; . . . ; count from the main diagonal to the appropriate square. EXAMPLE 8.8 Let A ¼ 1 2 3 4 5 6 7 8 9 2 4 3 5. Find the following minors and cofactors: (a) jM23j and A23, (b) jM31j and A31. (a) jM23j ¼ 1 2 3 4 5 6 7 8 9 ¼ 1 2 7 8 ¼ 8 14 ¼ 6, and so A23 ¼ ð1Þ2þ3jM23j ¼ ð6Þ ¼ 6 (b) jM31j ¼ 1 2 3 4 5 6 7 8 9 ¼ 2 3 5 6 ¼ 12 15 ¼ 3, and so A31 ¼ ð1Þ1þ3jM31j ¼ þð3Þ ¼ 3 Laplace Expansion The following theorem (proved in Problem 8.32) holds. THEOREM 8.8: (Laplace) The determinant of a square matrix A ¼ ½aij is equal to the sum of the products obtained by multiplying the elements of any row (column) by their respective cofactors: jAj ¼ ai1Ai1 þ ai2Ai2 þ þ ainAin ¼ P n j¼1 aijAij jAj ¼ a1jA1j þ a2jA2j þ þ anjAnj ¼ P n i¼1 aijAij The above formulas for jAj are called the Laplace expansions of the determinant of A by the ith row and the jth column. Together with the elementary row (column) operations, they offer a method of simplifying the computation of jAj, as described below. 8.8 Evaluation of Determinants The following algorithm reduces the evaluation of a determinant of order n to the evaluation of a determinant of order n 1. ALGORITHM 8.1: (Reduction of the order of a determinant) The input is a nonzero n-square matrix A ¼ ½aij with n > 1. Step 1. Choose an element aij ¼ 1 or, if lacking, aij 6¼ 0. Step 2. Using aij as a pivot, apply elementary row (column) operations to put 0’s in all the other positions in the column (row) containing aij. Step 3. Expand the determinant by the column (row) containing aij. 270 CHAPTER 8 Determinants The following remarks are in order. Remark 1: Algorithm 8.1 is usually used for determinants of order 4 or more. With determinants of order less than 4, one uses the speciﬁc formulas for the determinant. Remark 2: Gaussian elimination or, equivalently, repeated use of Algorithm 8.1 together with row interchanges can be used to transform a matrix A into an upper triangular matrix whose determinant is the product of its diagonal entries. However, one must keep track of the number of row interchanges, because each row interchange changes the sign of the determinant. EXAMPLE 8.9 Use Algorithm 8.1 to ﬁnd the determinant of A ¼ 5 4 2 1 2 3 1 2 5 7 3 9 1 2 1 4 2 6 6 4 3 7 7 5. Use a23 ¼ 1 as a pivot to put 0’s in the other positions of the third column; that is, apply the row operations ‘‘Replace R1 by 2R2 þ R1,’’ ‘‘Replace R3 by 3R2 þ R3,’’ and ‘‘Replace R4 by R2 þ R4.’’ By Theorem 8.3(iii), the value of the determinant does not change under these operations. Thus, jAj ¼ 5 4 2 1 2 3 1 2 5 7 3 9 1 2 1 4 ¼ 1 2 0 5 2 3 1 2 1 2 0 3 3 1 0 2 Now expand by the third column. Speciﬁcally, neglect all terms that contain 0 and use the fact that the sign of the minor M23 is ð1Þ2þ3 ¼ 1. Thus, jAj ¼ 1 2 0 5 2 3 1 2 1 2 0 3 3 1 0 2 ¼ 1 2 5 1 2 3 3 1 2 ¼ ð4 18 þ 5 30 3 þ 4Þ ¼ ð38Þ ¼ 38 8.9 Classical Adjoint Let A ¼ ½aij be an n n matrix over a ﬁeld K and let Aij denote the cofactor of aij. The classical adjoint of A, denoted by adj A, is the transpose of the matrix of cofactors of A. Namely, adj A ¼ ½AijT We say ‘‘classical adjoint’’ instead of simply ‘‘adjoint’’ because the term ‘‘adjoint’’ is currently used for an entirely different concept. EXAMPLE 8.10 Let A ¼ 2 3 4 0 4 2 1 1 5 2 4 3 5. The cofactors of the nine elements of A follow: A11 ¼ þ 4 2 1 5 ¼ 18; A21 ¼ 3 4 1 5 ¼ 11; A31 ¼ þ 3 4 4 2 ¼ 10; A12 ¼ 0 2 1 5 ¼ 2; A22 ¼ þ 2 4 1 5 ¼ 14; A32 ¼ 2 4 0 2 ¼ 4; A13 ¼ þ 0 4 1 1 ¼ 4 A23 ¼ 2 3 1 1 ¼ 5 A33 ¼ þ 2 3 0 4 ¼ 8 CHAPTER 8 Determinants 271 The transpose of the above matrix of cofactors yields the classical adjoint of A; that is, adj A ¼ 18 11 10 2 14 4 4 5 8 2 4 3 5 The following theorem (proved in Problem 8.34) holds. THEOREM 8.9: Let A be any square matrix. Then Aðadj AÞ ¼ ðadj AÞA ¼ jAjI where I is the identity matrix. Thus, if jAj 6¼ 0, A1 ¼ 1 jAj ðadj AÞ EXAMPLE 8.11 Let A be the matrix in Example 8.10. We have detðAÞ ¼ 40 þ 6 þ 0 16 þ 4 þ 0 ¼ 46 Thus, A does have an inverse, and, by Theorem 8.9, A1 ¼ 1 jAj ðadj AÞ ¼ 1 46 18 11 10 2 14 4 4 5 8 2 6 4 3 7 5 ¼ 9 23 11 46 5 23 1 23 7 23 2 23 2 23 5 46 4 23 2 6 4 3 7 5 8.10 Applications to Linear Equations, Cramer’s Rule Consider a system AX ¼ B of n linear equations in n unknowns. Here A ¼ ½aij is the (square) matrix of coefﬁcients and B ¼ ½bi is the column vector of constants. Let Ai be the matrix obtained from A by replacing the ith column of A by the column vector B. Furthermore, let D ¼ detðAÞ; N1 ¼ detðA1Þ; N2 ¼ detðA2Þ; . . . ; Nn ¼ detðAnÞ The fundamental relationship between determinants and the solution of the system AX ¼ B follows. THEOREM 8.10: The (square) system AX ¼ B has a solution if and only if D 6¼ 0. In this case, the unique solution is given by x1 ¼ N1 D ; x2 ¼ N2 D ; . . . ; xn ¼ Nn D The above theorem (proved in Problem 8.10) is known as Cramer’s rule for solving systems of linear equations. We emphasize that the theorem only refers to a system with the same number of equations as unknowns, and that it only gives the solution when D 6¼ 0. In fact, if D ¼ 0, the theorem does not tell us whether or not the system has a solution. However, in the case of a homogeneous system, we have the following useful result (to be proved in Problem 8.54). THEOREM 8.11: A square homogeneous system AX ¼ 0 has a nonzero solution if and only if D ¼ jAj ¼ 0. 272 CHAPTER 8 Determinants EXAMPLE 8.12 Solve the system using determinants x þ y þ z ¼ 5 x 2y 3z ¼ 1 2x þ y z ¼ 3 8 < : First compute the determinant D of the matrix of coefﬁcients: D ¼ 1 1 1 1 2 3 2 1 1 ¼ 2 6 þ 1 þ 4 þ 3 þ 1 ¼ 5 Because D 6¼ 0, the system has a unique solution. To compute Nx, Ny, Nz, we replace, respectively, the coefﬁcients of x; y; z in the matrix of coefﬁcients by the constant terms. This yields Nx ¼ 5 1 1 1 2 3 3 1 1 ¼ 20; Ny ¼ 1 5 1 1 1 3 2 3 1 ¼ 10; Nz ¼ 1 1 5 1 2 1 2 1 3 ¼ 15 Thus, the unique solution of the system is x ¼ Nx=D ¼ 4, y ¼ Ny=D ¼ 2, z ¼ Nz=D ¼ 3; that is, the vector u ¼ ð4; 2; 3Þ. 8.11 Submatrices, Minors, Principal Minors Let A ¼ ½aij be a square matrix of order n. Consider any r rows and r columns of A. That is, consider any set I ¼ ði1; i2; . . . ; irÞ of r row indices and any set J ¼ ðj1; j2; . . . ; jrÞ of r column indices. Then I and J deﬁne an r r submatrix of A, denoted by AðI; JÞ, obtained by deleting the rows and columns of A whose subscripts do not belong to I or J, respectively. That is, AðI; JÞ ¼ ½ast : s 2 I; t 2 J The determinant jAðI; JÞj is called a minor of A of order r and ð1Þi1þi2þþirþj1þj2þþjrjAðI; JÞj is the corresponding signed minor. (Note that a minor of order n 1 is a minor in the sense of Section 8.7, and the corresponding signed minor is a cofactor.) Furthermore, if I0 and J0 denote, respectively, the remaining row and column indices, then jAðI0; J 0Þj denotes the complementary minor, and its sign (Problem 8.74) is the same sign as the minor. EXAMPLE 8.13 Let A ¼ ½aij be a 5-square matrix, and let I ¼ f1; 2; 4g and J ¼ f2; 3; 5g. Then I0 ¼ f3; 5g and J 0 ¼ f1; 4g, and the corresponding minor jMj and complementary minor jM0j are as follows: jMj ¼ jAðI; JÞj ¼ a12 a13 a15 a22 a23 a25 a42 a43 a45 and jM0j ¼ jAðI0; J 0Þj ¼ a31 a34 a51 a54 Because 1 þ 2 þ 4 þ 2 þ 3 þ 5 ¼ 17 is odd, jMj is the signed minor, and jM0j is the signed complementary minor. Principal Minors A minor is principal if the row and column indices are the same, or equivalently, if the diagonal elements of the minor come from the diagonal of the matrix. We note that the sign of a principal minor is always þ1, because the sum of the row and identical column subscripts must always be even. CHAPTER 8 Determinants 273 EXAMPLE 8.14 Let A ¼ 1 2 1 3 5 4 3 1 2 2 4 3 5. Find the sums C1, C2, and C3 of the principal minors of A of orders 1, 2, and 3, respectively. (a) There are three principal minors of order 1. These are j1j ¼ 1; j5j ¼ 5; j 2j ¼ 2; and so C1 ¼ 1 þ 5 2 ¼ 4 Note that C1 is simply the trace of A. Namely, C1 ¼ trðAÞ: (b) There are three ways to choose two of the three diagonal elements, and each choice gives a minor of order 2. These are 1 2 3 5 ¼ 1; 1 1 3 2 ¼ 1; 5 4 1 2 ¼ 14 (Note that these minors of order 2 are the cofactors A33, A22, and A11 of A, respectively.) Thus, C2 ¼ 1 þ 1 14 ¼ 14 (c) There is only one way to choose three of the three diagonal elements. Thus, the only minor of order 3 is the determinant of A itself. Thus, C3 ¼ jAj ¼ 10 24 3 15 4 þ 12 ¼ 44 8.12 Block Matrices and Determinants The following theorem (proved in Problem 8.36) is the main result of this section. THEOREM 8.12: Suppose M is an upper (lower) triangular block matrix with the diagonal blocks A1; A2; . . . ; An. Then detðMÞ ¼ detðA1Þ detðA2Þ . . . detðAnÞ EXAMPLE 8.15 Find jMj where M ¼ 2 3 4 7 8 1 5 3 2 1 0 0 2 1 5 0 0 3 1 4 0 0 5 2 6 2 6 6 6 6 4 3 7 7 7 7 5 Note that M is an upper triangular block matrix. Evaluate the determinant of each diagonal block: 2 3 1 5 ¼ 10 þ 3 ¼ 13; 2 1 5 3 1 4 5 2 6 ¼ 12 þ 20 þ 30 þ 25 16 18 ¼ 29 Then jMj ¼ 13ð29Þ ¼ 377. Remark: Suppose M ¼ A B C D , where A; B; C; D are square matrices. Then it is not generally true that jMj ¼ jAjjDj jBjjCj. (See Problem 8.68.) 8.13 Determinants and Volume Determinants are related to the notions of area and volume as follows. Let u1; u2; . . . ; un be vectors in Rn. Let S be the (solid) parallelopiped determined by the vectors; that is, S ¼ fa1u1 þ a2u2 þ þ anun : 0 ai 1 for i ¼ 1; . . . ; ng (When n ¼ 2; S is a parallelogram.) Let VðSÞ denote the volume of S (or area of S when n ¼ 2Þ. Then VðSÞ ¼ absolute value of det ðAÞ 274 CHAPTER 8 Determinants where A is the matrix with rows u1; u2; . . . ; un. In general, VðSÞ ¼ 0 if and only if the vectors u1; . . . ; un do not form a coordinate system for Rn (i.e., if and only if the vectors are linearly dependent). EXAMPLE 8.16 Let u1 ¼ ð1; 1; 0Þ, u2 ¼ ð1; 1; 1Þ, u3 ¼ ð0; 2; 3Þ. Find the volume VðSÞ of the parallelo-piped S in R3 (Fig. 8-2) determined by the three vectors. Evaluate the determinant of the matrix whose rows are u1; u2; u3: 1 1 0 1 1 1 0 2 3 ¼ 3 þ 0 þ 0 0 2 3 ¼ 2 Hence, VðSÞ ¼ j 2j ¼ 2. 8.14 Determinant of a Linear Operator Let F be a linear operator on a vector space V with ﬁnite dimension. Let A be the matrix representation of F relative to some basis S of V. Then we deﬁne the determinant of F, written detðFÞ, by detðFÞ ¼ jAj If B were another matrix representation of F relative to another basis S0 of V, then A and B are similar matrices (Theorem 6.7) and jBj ¼ jAj (Theorem 8.7). In other words, the above deﬁnition detðFÞ is independent of the particular basis S of V. (We say that the deﬁnition is well deﬁned.) The next theorem (to be proved in Problem 8.62) follows from analogous theorems on matrices. THEOREM 8.13: Let F and G be linear operators on a vector space V. Then (i) detðF GÞ ¼ detðFÞ detðGÞ. (ii) F is invertible if and only if detðFÞ 6¼ 0. EXAMPLE 8.17 Let F be the following linear operator on R3 and let A be the matrix that represents F relative to the usual basis of R3: Fðx; y; zÞ ¼ ð2x 4y þ z; x 2y þ 3z; 5x þ y zÞ and A ¼ 2 4 1 1 2 3 5 1 1 2 4 3 5 Then detðFÞ ¼ jAj ¼ 4 60 þ 1 þ 10 6 4 ¼ 55 z y x 0 u3 u2 u1 Figure 8-2 CHAPTER 8 Determinants 275 8.15 Multilinearity and Determinants Let V be a vector space over a ﬁeld K. Let a ¼ V n; that is, a consists of all the n-tuples A ¼ ðA1; A2; . . . ; AnÞ where the Ai are vectors in V. The following deﬁnitions apply. DEFINITION: A function D: a ! K is said to be multilinear if it is linear in each component: (i) If Ai ¼ B þ C, then DðAÞ ¼ Dð. . . ; B þ C; . . .Þ ¼ Dð. . . ; B; . . . ; Þ þ Dð. . . ; C; . . .Þ (ii) If Ai ¼ kB, where k 2 K, then DðAÞ ¼ Dð. . . ; kB; . . .Þ ¼ kDð. . . ; B; . . .Þ We also say n-linear for multilinear if there are n components. DEFINITION: A function D: a ! K is said to be alternating if DðAÞ ¼ 0 whenever A has two identical elements: DðA1; A2; . . . ; AnÞ ¼ 0 whenever Ai ¼ Aj; i 6¼ j Now let M denote the set of all n-square matrices A over a ﬁeld K. We may view A as an n-tuple consisting of its row vectors A1; A2; . . . ; An; that is, we may view A in the form A ¼ ðA1; A2; . . . ; AnÞ. The following theorem (proved in Problem 8.37) characterizes the determinant function. THEOREM 8.14: There exists a unique function D: M ! K such that (i) D is multilinear, (ii) D is alternating, (iii) DðIÞ ¼ 1. This function D is the determinant function; that is, DðAÞ ¼ jAj; for any matrix A 2 M. SOLVED PROBLEMS Computation of Determinants 8.1. Evaluate the determinant of each of the following matrices: (a) A ¼ 6 5 2 3 , (b) B ¼ 2 3 4 7 ; (c) C ¼ 4 5 1 2 ; (d) D ¼ t 5 6 3 t þ 2 Use the formula a b c d ¼ ad bc: (a) jAj ¼ 6ð3Þ 5ð2Þ ¼ 18 10 ¼ 8 (b) jBj ¼ 14 þ 12 ¼ 26 (c) jCj ¼ 8 5 ¼ 13 (d) jDj ¼ ðt 5Þðt þ 2Þ 18 ¼ t2 3t 10 18 ¼ t2 10t 28 8.2. Evaluate the determinant of each of the following matrices: (a) A ¼ 2 3 4 5 4 3 1 2 1 2 4 3 5, (b) B ¼ 1 2 3 2 4 1 1 5 2 2 4 3 5, (c) C ¼ 1 3 5 3 1 2 1 2 1 2 4 3 5 276 CHAPTER 8 Determinants Use the diagram in Fig. 8-1 to obtain the six products: (a) jAj ¼ 2ð4Þð1Þ þ 3ð3Þð1Þ þ 4ð2Þð5Þ 1ð4Þð4Þ 2ð3Þð2Þ 1ð3Þð5Þ ¼ 8 þ 9 þ 40 16 12 15 ¼ 14 (b) jBj ¼ 8 þ 2 þ 30 12 þ 5 8 ¼ 9 (c) jCj ¼ 1 þ 6 þ 30 5 þ 4 9 ¼ 25 8.3. Compute the determinant of each of the following matrices: (a) A ¼ 2 3 4 5 6 7 8 9 1 2 4 3 5, (b) B ¼ 4 6 8 9 0 2 7 3 0 0 5 6 0 0 0 3 2 6 6 4 3 7 7 5, (c) C ¼ 1 2 1 1 3 3 4 1 2 1 1 4 1 2 6 4 3 7 5: (a) One can simplify the entries by ﬁrst subtracting twice the ﬁrst row from the second row—that is, by applying the row operation ‘‘Replace R2 by 21 þ R2.’’ Then jAj ¼ 2 3 4 5 6 7 8 9 1 ¼ 2 3 4 1 0 1 8 9 1 ¼ 0 24 þ 36 0 þ 18 3 ¼ 27 (b) B is triangular, so jBj ¼ product of the diagonal entries ¼ 120. (c) The arithmetic is simpler if fractions are ﬁrst eliminated. Hence, multiply the ﬁrst row R1 by 6 and the second row R2 by 4. Then j24Cj ¼ 3 6 2 3 2 4 1 4 1 ¼ 6 þ 24 þ 24 þ 4 48 þ 18 ¼ 28; so jCj ¼ 28 24 ¼ 7 6 8.4. Compute the determinant of each of the following matrices: (a) A ¼ 2 5 3 2 2 3 2 5 1 3 2 2 1 6 4 3 2 6 6 4 3 7 7 5, (b) B ¼ 6 2 1 0 5 2 1 1 2 1 1 1 2 2 3 3 0 2 3 1 1 1 3 4 2 2 6 6 6 6 4 3 7 7 7 7 5 (a) Use a31 ¼ 1 as a pivot to put 0’s in the ﬁrst column, by applying the row operations ‘‘Replace R1 by 2R3 þ R1,’’ ‘‘Replace R2 by 2R3 þ R2,’’ and ‘‘Replace R4 by R3 þ R4.’’ Then jAj ¼ 2 5 3 2 2 3 2 5 1 3 2 2 1 6 4 3 ¼ 0 1 1 6 0 3 2 1 1 3 2 2 0 3 2 5 ¼ 1 1 6 3 2 1 3 2 5 ¼ 10 þ 3 36 þ 36 2 15 ¼ 4 (b) First reduce jBj to a determinant of order 4, and then to a determinant of order 3, for which we can use Fig. 8-1. First use c22 ¼ 1 as a pivot to put 0’s in the second column, by applying the row operations ‘‘Replace R1 by 2R2 þ R1,’’ ‘‘Replace R3 by R2 þ R3,’’ and ‘‘Replace R5 by R2 þ R5.’’ Then jBj ¼ 2 0 1 4 3 2 1 1 2 1 1 0 1 0 2 3 0 2 3 1 1 0 2 2 3 ¼ 2 1 4 3 1 1 0 2 3 2 3 1 1 2 2 3 ¼ 1 1 4 5 0 1 0 0 5 2 3 5 1 2 2 7 ¼ 1 4 5 5 3 5 1 2 7 ¼ 21 þ 20 þ 50 þ 15 þ 10 140 ¼ 34 CHAPTER 8 Determinants 277 Cofactors, Classical Adjoints, Minors, Principal Minors 8.5. Let A ¼ 2 1 3 4 5 4 7 2 4 0 6 3 3 2 5 2 2 6 6 4 3 7 7 5: (a) Find A23, the cofactor (signed minor) of 7 in A. (b) Find the minor and the signed minor of the submatrix M ¼ Að2; 4; 2; 3Þ. (c) Find the principal minor determined by the ﬁrst and third diagonal entries—that is, by M ¼ Að1; 3; 1; 3Þ. (a) Take the determinant of the submatrix of A obtained by deleting row 2 and column 3 (those which contain the 7), and multiply the determinant by ð1Þ2þ3: A23 ¼ 2 1 4 4 0 3 3 2 2 ¼ ð61Þ ¼ 61 The exponent 2 þ 3 comes from the subscripts of A23—that is, from the fact that 7 appears in row 2 and column 3. (b) The row subscripts are 2 and 4 and the column subscripts are 2 and 3. Hence, the minor is the determinant jMj ¼ a22 a23 a42 a43 ¼ 4 7 2 5 ¼ 20 þ 14 ¼ 6 and the signed minor is ð1Þ2þ4þ2þ3jMj ¼ jMj ¼ ð6Þ ¼ 6. (c) The principal minor is the determinant jMj ¼ a11 a13 a31 a33 ¼ 2 3 4 6 ¼ 12 þ 12 ¼ 24 Note that now the diagonal entries of the submatrix are diagonal entries of the original matrix. Also, the sign of the principal minor is positive. 8.6. Let B ¼ 1 1 1 2 3 4 5 8 9 2 4 3 5. Find: (a) jBj, (b) adj B, (c) B1 using adj B. (a) jBj ¼ 27 þ 20 þ 16 15 32 18 ¼ 2 (b) Take the transpose of the matrix of cofactors: adj B ¼ 3 4 8 9 2 4 5 9 2 3 5 8 1 1 8 9 1 1 5 9 1 1 5 8 1 1 3 4 1 1 2 4 1 1 2 3 2 6 6 6 6 6 6 6 6 4 3 7 7 7 7 7 7 7 7 5 T ¼ 5 2 1 1 4 3 1 2 1 2 6 4 3 7 5 T ¼ 5 1 1 2 4 2 1 3 1 2 6 4 3 7 5 (c) Because jBj 6¼ 0, B1 ¼ 1 jBj ðadj BÞ ¼ 1 2 5 1 1 2 4 2 1 3 1 2 4 3 5 ¼ 5 2 1 2 1 2 1 2 1 1 2 3 2 1 2 2 6 4 3 7 5 8.7. Let A ¼ 1 2 3 4 5 6 0 7 8 2 4 3 5, and let Sk denote the sum of its principal minors of order k. Find Sk for (a) k ¼ 1, (b) k ¼ 2, (c) k ¼ 3. 278 CHAPTER 8 Determinants (a) The principal minors of order 1 are the diagonal elements. Thus, S1 is the trace of A; that is, S1 ¼ trðAÞ ¼ 1 þ 5 þ 8 ¼ 14 (b) The principal minors of order 2 are the cofactors of the diagonal elements. Thus, S2 ¼ A11 þ A22 þ A33 ¼ 5 6 7 8 þ 1 3 0 8 þ 1 2 4 5 ¼ 2 þ 8 3 ¼ 3 (c) There is only one principal minor of order 3, the determinant of A. Then S3 ¼ jAj ¼ 40 þ 0 þ 84 0 42 64 ¼ 18 8.8. Let A ¼ 1 3 0 1 4 2 5 1 1 0 3 2 3 2 1 4 2 6 6 4 3 7 7 5. Find the number Nk and sum Sk of principal minors of order: (a) k ¼ 1, (b) k ¼ 2, (c) k ¼ 3, (d) k ¼ 4. Each (nonempty) subset of the diagonal (or equivalently, each nonempty subset of f1; 2; 3; 4gÞ determines a principal minor of A, and Nk ¼ n k ¼ n! k!ðn kÞ! of them are of order k. Thus; N1 ¼ 4 1 ¼ 4; N2 ¼ 4 2 ¼ 6; N3 ¼ 4 3 ¼ 4; N4 ¼ 4 4 ¼ 1 (a) S1 ¼ j1j þ j2j þ j3j þ j4j ¼ 1 þ 2 þ 3 þ 4 ¼ 10 (b) S2 ¼ 1 3 4 2 þ 1 0 1 3 þ 1 1 3 4 þ 2 5 0 3 þ 2 1 2 4 þ 3 2 1 4 ¼ 14 þ 3 þ 7 þ 6 þ 10 þ 14 ¼ 54 (c) S3 ¼ 1 3 0 4 2 5 1 0 3 þ 1 3 1 4 2 1 3 2 4 þ 1 0 1 1 3 2 3 1 4 þ 2 5 1 0 3 2 2 1 4 ¼ 57 þ 65 þ 22 þ 54 ¼ 198 (d) S4 ¼ detðAÞ ¼ 378 Determinants and Systems of Linear Equations 8.9. Use determinants to solve the system 3y þ 2x ¼ z þ 1 3x þ 2z ¼ 8 5y 3z 1 ¼ x 2y : 8 < : First arrange the equation in standard form, then compute the determinant D of the matrix of coefﬁcients: 2x þ 3y z ¼ 1 3x þ 5y þ 2z ¼ 8 x 2y 3z ¼ 1 and D ¼ 2 3 1 3 5 2 1 2 3 ¼ 30 þ 6 þ 6 þ 5 þ 8 þ 27 ¼ 22 Because D 6¼ 0, the system has a unique solution. To compute Nx; Ny; Nz, we replace, respectively, the coefﬁcients of x; y; z in the matrix of coefﬁcients by the constant terms. Then Nx ¼ 1 3 1 8 5 2 1 2 1 ¼ 66; Ny ¼ 2 1 1 3 8 2 1 1 3 ¼ 22; Nz ¼ 2 3 1 3 5 8 1 2 1 ¼ 44 CHAPTER 8 Determinants 279 Thus, x ¼ Nx D ¼ 66 22 ¼ 3; y ¼ Ny D ¼ 22 22 ¼ 1; z ¼ Nz D ¼ 44 22 ¼ 2 8.10. Consider the system kx þ y þ z ¼ 1 x þ ky þ z ¼ 1 x þ y þ kz ¼ 1 8 < : Use determinants to ﬁnd those values of k for which the system has (a) a unique solution, (b) more than one solution, (c) no solution. (a) The system has a unique solution when D 6¼ 0, where D is the determinant of the matrix of coefﬁcients. Compute D ¼ k 1 1 1 k 1 1 1 k ¼ k3 þ 1 þ 1 k k k ¼ k3 3k þ 2 ¼ ðk 1Þ2ðk þ 2Þ Thus, the system has a unique solution when ðk 1Þ2ðk þ 2Þ 6¼ 0; when k 6¼ 1 and k 6¼ 2 (b and c) Gaussian elimination shows that the system has more than one solution when k ¼ 1, and the system has no solution when k ¼ 2. Miscellaneous Problems 8.11. Find the volume VðSÞ of the parallelepiped S in R3 determined by the vectors: (a) u1 ¼ ð1; 1; 1Þ; u2 ¼ ð1; 3; 4Þ; u3 ¼ ð1; 2; 5Þ. (b) u1 ¼ ð1; 2; 4Þ; u2 ¼ ð2; 1; 3Þ; u3 ¼ ð5; 7; 9Þ. VðSÞ is the absolute value of the determinant of the matrix M whose rows are the given vectors. Thus, (a) jMj ¼ 1 1 1 1 3 4 1 2 5 ¼ 15 4 þ 2 3 þ 8 þ 5 ¼ 7. Hence, VðSÞ ¼ j 7j ¼ 7. (b) jMj ¼ 1 2 4 2 1 3 5 7 9 ¼ 9 30 þ 56 20 þ 21 36 ¼ 0. Thus, VðSÞ ¼ 0, or, in other words, u1; u2; u3 lie in a plane and are linearly dependent. 8.12. Find detðMÞ where M ¼ 3 4 0 0 0 2 5 0 0 0 0 9 2 0 0 0 5 0 6 7 0 0 4 3 4 2 6 6 6 6 4 3 7 7 7 7 5 ¼ 3 4 0 0 0 2 5 0 0 0 0 9 2 0 0 0 5 0 6 7 0 0 4 3 4 2 6 6 6 6 4 3 7 7 7 7 5 M is a (lower) triangular block matrix; hence, evaluate the determinant of each diagonal block: 3 4 2 5 ¼ 15 8 ¼ 7; j2j ¼ 2; 6 7 3 4 ¼ 24 21 ¼ 3 Thus, jMj ¼ 7ð2Þð3Þ ¼ 42. 8.13. Find the determinant of F: R3 ! R3 deﬁned by Fðx; y; zÞ ¼ ðx þ 3y 4z; 2y þ 7z; x þ 5y 3zÞ 280 CHAPTER 8 Determinants The determinant of a linear operator F is equal to the determinant of any matrix that represents F. Thus ﬁrst ﬁnd the matrix A representing F in the usual basis (whose rows, respectively, consist of the coefﬁcients of x; y; z). Then A ¼ 1 3 4 0 2 7 1 5 3 2 4 3 5; and so detðFÞ ¼ jAj ¼ 6 þ 21 þ 0 þ 8 35 0 ¼ 8 8.14. Write out g ¼ gðx1; x2; x3; x4Þ explicitly where gðx1; x2; . . . ; xnÞ ¼ Q i<j ðxi xjÞ: The symbol Q is used for a product of terms in the same way that the symbol P is used for a sum of terms. That is, Q i<j ðxi xjÞ means the product of all terms ðxi xjÞ for which i < j. Hence, g ¼ gðx1; . . . ; x4Þ ¼ ðx1 x2Þðx1 x3Þðx1 x4Þðx2 x3Þðx2 x4Þðx3 x4Þ 8.15. Let D be a 2-linear, alternating function. Show that DðA; BÞ ¼ DðB; AÞ. Because D is alternating, DðA; AÞ ¼ 0, DðB; BÞ ¼ 0. Hence, DðA þ B; A þ BÞ ¼ DðA; AÞ þ DðA; BÞ þ DðB; AÞ þ DðB; BÞ ¼ DðA; BÞ þ DðB; AÞ However, DðA þ B; A þ BÞ ¼ 0. Hence, DðA; BÞ ¼ DðB; AÞ, as required. Permutations 8.16. Determine the parity (sign) of the permutation s ¼ 364152. Count the number of inversions. That is, for each element k, count the number of elements i in s such that i > k and i precedes k in s. Namely, k ¼ 1: 3 numbers ð3; 6; 4Þ k ¼ 2: 4 numbers ð3; 6; 4; 5Þ k ¼ 3: 0 numbers k ¼ 4: 1 number ð6Þ k ¼ 5: 1 number ð6Þ k ¼ 6: 0 numbers Because 3 þ 4 þ 0 þ 1 þ 1 þ 0 ¼ 9 is odd, s is an odd permutation, and sgn s ¼ 1. 8.17. Let s ¼ 24513 and t ¼ 41352 be permutations in S5. Find (a) t s, (b) s1. Recall that s ¼ 24513 and t ¼ 41352 are short ways of writing s ¼ 1 2 3 4 5 2 4 5 1 3 or sð1Þ ¼ 2; sð2Þ ¼ 4; sð3Þ ¼ 5; sð4Þ ¼ 1; sð5Þ ¼ 3 t ¼ 1 2 3 4 5 4 1 3 5 2 c or tð1Þ ¼ 4; tð2Þ ¼ 1; tð3Þ ¼ 3; tð4Þ ¼ 5; tð5Þ ¼ 2 (a) The effects of s and then t on 1; 2; 3; 4; 5 are as follows: 1 ! 2 ! 1; 2 ! 4 ! 5; 3 ! 5 ! 2; 4 ! 1 ! 4; 5 ! 3 ! 3 [That is, for example, ðt sÞð1Þ ¼ tðsð1ÞÞ ¼ tð2Þ ¼ 1: Thus, t s ¼ 15243. (b) By deﬁnition, s1ð jÞ ¼ k if and only if sðkÞ ¼ j. Hence, s1 ¼ 2 4 5 1 3 1 2 3 4 5 ¼ 1 2 3 4 5 4 1 5 2 3 or s1 ¼ 41523 8.18. Let s ¼ j1 j2 . . . jn be any permutation in Sn. Show that, for each inversion ði; kÞ where i > k but i precedes k in s, there is a pair ði; jÞ such that i < k and sðiÞ > sð jÞ ð1Þ and vice versa. Thus, s is even or odd according to whether there is an even or an odd number of pairs satisfying (1). CHAPTER 8 Determinants 281 Choose i and k so that sðiÞ ¼ i and sðkÞ ¼ k. Then i > k if and only if sðiÞ > sðkÞ, and i precedes k in s if and only if i < k. 8.19. Consider the polynomials g ¼ gðx1; . . . ; xnÞ and sðgÞ, deﬁned by g ¼ gðx1; . . . ; xnÞ ¼ Q i<j ðxi xjÞ and sðgÞ ¼ Q i<j ðxsðiÞ xsðjÞÞ (See Problem 8.14.) Show that sðgÞ ¼ g when s is an even permutation, and sðgÞ ¼ g when s is an odd permutation. That is, sðgÞ ¼ ðsgn sÞg. Because s is one-to-one and onto, sðgÞ ¼ Q ij ðxi xjÞ Thus, sðgÞ or sðgÞ ¼ g according to whether there is an even or an odd number of terms of the form xi xj, where i > j. Note that for each pair ði; jÞ for which i < j and sðiÞ > sð jÞ there is a term ðxsðiÞ xsð jÞÞ in sðgÞ for which sðiÞ > sð jÞ. Because s is even if and only if there is an even number of pairs satisfying (1), we have sðgÞ ¼ g if and only if s is even. Hence, sðgÞ ¼ g if and only if s is odd. 8.20. Let s; t 2 Sn. Show that sgnðt sÞ ¼ ðsgn tÞðsgn sÞ. Thus, the product of two even or two odd permutations is even, and the product of an odd and an even permutation is odd. Using Problem 8.19, we have sgnðt sÞ g ¼ ðt sÞðgÞ ¼ tðsðgÞÞ ¼ tððsgn sÞgÞ ¼ ðsgn tÞðsgn sÞg Accordingly, sgn ðt sÞ ¼ ðsgn tÞðsgn sÞ. 8.21. Consider the permutation s ¼ j1 j2 jn. Show that sgn s1 ¼ sgn s and, for scalars aij, show that aj11aj22 ajnn ¼ a1k1a2k2 ankn where s1 ¼ k1k2 kn. We have s1 s ¼ e, the identity permutation. Because e is even, s1 and s are both even or both odd. Hence sgn s1 ¼ sgn s. Because s ¼ j1j2 jn is a permutation, aj11aj22 ajnn ¼ a1k1a2k2 ankn. Then k1; k2; . . . ; kn have the property that sðk1Þ ¼ 1; sðk2Þ ¼ 2; . . . ; sðknÞ ¼ n Let t ¼ k1k2 kn. Then, for i ¼ 1; . . . ; n, ðs tÞðiÞ ¼ sðtðiÞÞ ¼ sðkiÞ ¼ i Thus, s t ¼ e, the identity permutation. Hence, t ¼ s1. Proofs of Theorems 8.22. Prove Theorem 8.1: jATj ¼ jAj. If A ¼ ½aij, then AT ¼ ½bij, with bij ¼ aji. Hence, jATj ¼ P s2Sn ðsgn sÞb1sð1Þb2sð2Þ bnsðnÞ ¼ P s2Sn ðsgn sÞasð1Þ;1asð2Þ;2 asðnÞ;n Let t ¼ s1. By Problem 8.21 sgn t ¼ sgn s, and asð1Þ;1asð2Þ;2 asðnÞ;n ¼ a1tð1Þa2tð2Þ antðnÞ. Hence, jATj ¼ P s2Sn ðsgn tÞa1tð1Þa2tð2Þ antðnÞ 282 CHAPTER 8 Determinants However, as s runs through all the elements of Sn; t ¼ s1 also runs through all the elements of Sn. Thus, jATj ¼ jAj. 8.23. Prove Theorem 8.3(i): If two rows (columns) of A are interchanged, then jBj ¼ jAj. We prove the theorem for the case that two columns are interchanged. Let t be the transposition that interchanges the two numbers corresponding to the two columns of A that are interchanged. If A ¼ ½aij and B ¼ ½bij, then bij ¼ aitðjÞ. Hence, for any permutation s, b1sð1Þb2sð2Þ bnsðnÞ ¼ a1ðt sÞð1Þa2ðt sÞð2Þ anðt sÞðnÞ Thus, jBj ¼ P s2Sn ðsgn sÞb1sð1Þb2sð2Þ bnsðnÞ ¼ P s2Sn ðsgn sÞa1ðt sÞð1Þa2ðt sÞð2Þ anðt sÞðnÞ Because the transposition t is an odd permutation, sgnðt sÞ ¼ ðsgn tÞðsgn sÞ ¼ sgn s. Accordingly, sgn s ¼ sgn ðt sÞ; and so jBj ¼ P s2Sn ½sgnðt sÞa1ðt sÞð1Þa2ðt sÞð2Þ anðt sÞðnÞ But as s runs through all the elements of Sn; t s also runs through all the elements of Sn: Hence, jBj ¼ jAj. 8.24. Prove Theorem 8.2. (i) If A has a row (column) of zeros, then jAj ¼ 0. (ii) If A has two identical rows (columns), then jAj ¼ 0. (iii) If A is triangular, then jAj ¼ product of diagonal elements. Thus, jIj ¼ 1. (i) Each term in jAj contains a factor from every row, and so from the row of zeros. Thus, each term of jAj is zero, and so jAj ¼ 0. (ii) Suppose 1 þ 1 6¼ 0 in K. If we interchange the two identical rows of A, we still obtain the matrix A. Hence, by Problem 8.23, jAj ¼ jAj, and so jAj ¼ 0. Now suppose 1 þ 1 ¼ 0 in K. Then sgn s ¼ 1 for every s 2 Sn: Because A has two identical rows, we can arrange the terms of A into pairs of equal terms. Because each pair is 0, the determinant of A is zero. (iii) Suppose A ¼ ½aij is lower triangular; that is, the entries above the diagonal are all zero: aij ¼ 0 whenever i < j. Consider a term t of the determinant of A: t ¼ ðsgn sÞa1i1a2i2 anin; where s ¼ i1i2 in Suppose i1 6¼ 1. Then 1 < i1 and so a1i1 ¼ 0; hence, t ¼ 0: That is, each term for which i1 6¼ 1 is zero. Now suppose i1 ¼ 1 but i2 6¼ 2. Then 2 < i2, and so a2i2 ¼ 0; hence, t ¼ 0. Thus, each term for which i1 6¼ 1 or i2 6¼ 2 is zero. Similarly, we obtain that each term for which i1 6¼ 1 or i2 6¼ 2 or . . . or in 6¼ n is zero. Accordingly, jAj ¼ a11a22 ann ¼ product of diagonal elements. 8.25. Prove Theorem 8.3: B is obtained from A by an elementary operation. (i) If two rows (columns) of A were interchanged, then jBj ¼ jAj. (ii) If a row (column) of A were multiplied by a scalar k, then jBj ¼ kjAj. (iii) If a multiple of a row (column) of A were added to another row (column) of A; then jBj ¼ jAj. (i) This result was proved in Problem 8.23. (ii) If the jth row of A is multiplied by k, then every term in jAj is multiplied by k, and so jBj ¼ kjAj. That is, jBj ¼ P s ðsgn sÞa1i1a2i2 ðkajijÞ anin ¼ k P s ðsgn sÞa1i1a2i2 anin ¼ kjAj CHAPTER 8 Determinants 283 (iii) Suppose c times the kth row is added to the jth row of A. Using the symbol ^ to denote the jth position in a determinant term, we have jBj ¼ P s ðsgn sÞa1i1a2i2 ðcakik þ ajijÞ . . . anin ¼ c P s ðsgn sÞa1i1a2i2 c akik anin þ P s ðsgn sÞa1i1a2i2 ajij anin The ﬁrst sum is the determinant of a matrix whose kth and jth rows are identical. Accordingly, by Theorem 8.2(ii), the sum is zero. The second sum is the determinant of A. Thus, jBj ¼ c 0 þ jAj ¼ jAj. 8.26. Prove Lemma 8.6: Let E be an elementary matrix. Then jEAj ¼ jEjjAj. Consider the elementary row operations: (i) Multiply a row by a constant k 6¼ 0, (ii) Interchange two rows, (iii) Add a multiple of one row to another. Let E1; E2; E3 be the corresponding elementary matrices That is, E1; E2; E3 are obtained by applying the above operations to the identity matrix I. By Problem 8.25, jE1j ¼ kjIj ¼ k; jE2j ¼ jIj ¼ 1; jE3j ¼ jIj ¼ 1 Recall (Theorem 3.11) that EiA is identical to the matrix obtained by applying the corresponding operation to A. Thus, by Theorem 8.3, we obtain the following which proves our lemma: jE1Aj ¼ kjAj ¼ jE1jjAj; jE2Aj ¼ jAj ¼ jE2jjAj; jE3Aj ¼ jAj ¼ 1jAj ¼ jE3jjAj 8.27. Suppose B is row equivalent to a square matrix A. Prove that jBj ¼ 0 if and only if jAj ¼ 0. By Theorem 8.3, the effect of an elementary row operation is to change the sign of the determinant or to multiply the determinant by a nonzero scalar. Hence, jBj ¼ 0 if and only if jAj ¼ 0. 8.28. Prove Theorem 8.5: Let A be an n-square matrix. Then the following are equivalent: (i) A is invertible, (ii) AX ¼ 0 has only the zero solution, (iii) detðAÞ 6¼ 0. The proof is by the Gaussian algorithm. If A is invertible, it is row equivalent to I. But jIj 6¼ 0. Hence, by Problem 8.27, jAj 6¼ 0. If A is not invertible, it is row equivalent to a matrix with a zero row. Hence, detðAÞ ¼ 0. Thus, (i) and (iii) are equivalent. If AX ¼ 0 has only the solution X ¼ 0, then A is row equivalent to I and A is invertible. Conversely, if A is invertible with inverse A1, then X ¼ IX ¼ ðA1AÞX ¼ A1ðAXÞ ¼ A10 ¼ 0 is the only solution of AX ¼ 0. Thus, (i) and (ii) are equivalent. 8.29. Prove Theorem 8.4: jABj ¼ jAjjBj. If A is singular, then AB is also singular, and so jABj ¼ 0 ¼ jAjjBj. On the other hand, if A is nonsingular, then A ¼ En E2E1, a product of elementary matrices. Then, Lemma 8.6 and induction yields jABj ¼ jEn E2E1Bj ¼ jEnj jE2jjE1jjBj ¼ jAjjBj 8.30. Suppose P is invertible. Prove that jP1j ¼ jPj1. P1P ¼ I: Hence; 1 ¼ jIj ¼ jP1Pj ¼ jP1jjPj; and so jP1j ¼ jPj1: 8.31. Prove Theorem 8.7: Suppose A and B are similar matrices. Then jAj ¼ jBj. Because A and B are similar, there exists an invertible matrix P such that B ¼ P1AP. Therefore, using Problem 8.30, we get jBj ¼ jP1APj ¼ jP1jjAjjPj ¼ jAjjP1jjP ¼ jAj. We remark that although the matrices P1 and A may not commute, their determinants jP1j and jAj do commute, because they are scalars in the ﬁeld K. 8.32. Prove Theorem 8.8 (Laplace): Let A ¼ ½aij, and let Aij denote the cofactor of aij. Then, for any i or j jAj ¼ ai1Ai1 þ þ ainAin and jAj ¼ a1jA1j þ þ anjAnj d 284 CHAPTER 8 Determinants Because jAj ¼ jATj, we need only prove one of the expansions, say, the ﬁrst one in terms of rows of A. Each term in jAj contains one and only one entry of the ith row ðai1; ai2; . . . ; ainÞ of A. Hence, we can write jAj in the form jAj ¼ ai1A i1 þ ai2A i2 þ þ ainA in (Note that A ij is a sum of terms involving no entry of the ith row of A.) Thus, the theorem is proved if we can show that A ij ¼ Aij ¼ ð1ÞiþjjMijj where Mij is the matrix obtained by deleting the row and column containing the entry aij: (Historically, the expression A ij was deﬁned as the cofactor of aij, and so the theorem reduces to showing that the two deﬁnitions of the cofactor are equivalent.) First we consider the case that i ¼ n, j ¼ n. Then the sum of terms in jAj containing ann is annA nn ¼ ann P s ðsgn sÞa1sð1Þa2sð2Þ an1;sðn1Þ where we sum over all permutations s 2 Sn for which sðnÞ ¼ n. However, this is equivalent (Prove!) to summing over all permutations of f1; . . . ; n 1g. Thus, A nn ¼ jMnnj ¼ ð1ÞnþnjMnnj. Now we consider any i and j. We interchange the ith row with each succeeding row until it is last, and we interchange the jth column with each succeeding column until it is last. Note that the determinant jMijj is not affected, because the relative positions of the other rows and columns are not affected by these interchanges. However, the ‘‘sign’’ of jAj and of A ij is changed n 1 and then n j times. Accordingly, A ij ¼ ð1ÞniþnjjMijj ¼ ð1ÞiþjjMijj 8.33. Let A ¼ ½aij and let B be the matrix obtained from A by replacing the ith row of A by the row vector ðbi1; . . . ; binÞ. Show that jBj ¼ bi1Ai1 þ bi2Ai2 þ þ binAin Furthermore, show that, for j 6¼ i, aj1Ai1 þ aj2Ai2 þ þ ajnAin ¼ 0 and a1jA1i þ a2jA2i þ þ anjAni ¼ 0 Let B ¼ ½bij. By Theorem 8.8, jBj ¼ bi1Bi1 þ bi2Bi2 þ þ binBin Because Bij does not depend on the ith row of B; we get Bij ¼ Aij for j ¼ 1; . . . ; n. Hence, jBj ¼ bi1Ai1 þ bi2Ai2 þ þ binAin Now let A0 be obtained from A by replacing the ith row of A by the jth row of A. Because A0 has two identical rows, jA0j ¼ 0. Thus, by the above result, jA0j ¼ aj1Ai1 þ aj2Ai2 þ þ ajnAin ¼ 0 Using jATj ¼ jAj, we also obtain that a1jA1i þ a2jA2i þ þ anjAni ¼ 0. 8.34. Prove Theorem 8.9: Aðadj AÞ ¼ ðadj AÞA ¼ jAjI. Let A ¼ ½aij and let Aðadj AÞ ¼ ½bij. The ith row of A is ðai1; ai2; . . . ; ainÞ ð1Þ Because adj A is the transpose of the matrix of cofactors, the jth column of adj A is the tranpose of the cofactors of the jth row of A: ðAj; Aj2; . . . ; AjnÞT ð2Þ Now bij; the ij entry in Aðadj AÞ, is obtained by multiplying expressions (1) and (2): bij ¼ ai1Aj1 þ ai2Aj2 þ þ ainAjn CHAPTER 8 Determinants 285 By Theorem 8.8 and Problem 8.33, bij ¼ jAj if i ¼ j 0 if i 6¼ j Accordingly, Aðadj AÞ is the diagonal matrix with each diagonal element jAj. In other words, Aðadj AÞ ¼ jAjI. Similarly, ðadj AÞA ¼ jAjI. 8.35. Prove Theorem 8.10 (Cramer’s rule): The (square) system AX ¼ B has a unique solution if and only if D 6¼ 0. In this case, xi ¼ Ni=D for each i. By previous results, AX ¼ B has a unique solution if and only if A is invertible, and A is invertible if and only if D ¼ jAj 6¼ 0. Now suppose D 6¼ 0. By Theorem 8.9, A1 ¼ ð1=DÞðadj AÞ. Multiplying AX ¼ B by A1, we obtain X ¼ A1AX ¼ ð1=DÞðadj AÞB ð1Þ Note that the ith row of ð1=DÞðadj AÞ is ð1=DÞðA1i; A2i; . . . ; AniÞ. If B ¼ ðb1; b2; . . . ; bnÞT, then, by (1), xi ¼ ð1=DÞðb1A1i þ b2A2i þ þ bnAniÞ However, as in Problem 8.33, b1A1i þ b2A2i þ þ bnAni ¼ Ni, the determinant of the matrix obtained by replacing the ith column of A by the column vector B. Thus, xi ¼ ð1=DÞNi, as required. 8.36. Prove Theorem 8.12: Suppose M is an upper (lower) triangular block matrix with diagonal blocks A1; A2; . . . ; An. Then detðMÞ ¼ detðA1Þ detðA2Þ detðAnÞ We need only prove the theorem for n ¼ 2—that is, when M is a square matrix of the form M ¼ A C 0 B . The proof of the general theorem follows easily by induction. Suppose A ¼ ½aij is r-square, B ¼ ½bij is s-square, and M ¼ ½mij is n-square, where n ¼ r þ s. By deﬁnition, detðMÞ ¼ P s2Sn ðsgn sÞm1sð1Þm2sð2Þ mnsðnÞ If i > r and j r, then mij ¼ 0. Thus, we need only consider those permutations s such that sfr þ 1; r þ 2; . . . ; r þ sg ¼ fr þ 1; r þ 2; . . . ; r þ sg and sf1; 2; . . . ; rg ¼ f1; 2; . . . ; rg Let s1ðkÞ ¼ sðkÞ for k r, and let s2ðkÞ ¼ sðr þ kÞ r for k s. Then ðsgn sÞm1sð1Þm2sð2Þ mnsðnÞ ¼ ðsgn s1Þa1s1ð1Þa2s1ð2Þ ars1ðrÞðsgn s2Þb1s2ð1Þb2s2ð2Þ bss2ðsÞ which implies detðMÞ ¼ detðAÞ detðBÞ. 8.37. Prove Theorem 8.14: There exists a unique function D : M ! K such that (i) D is multilinear, (ii) D is alternating, (iii) DðIÞ ¼ 1. This function D is the determinant function; that is, DðAÞ ¼ jAj. Let D be the determinant function, DðAÞ ¼ jAj. We must show that D satisﬁes (i), (ii), and (iii), and that D is the only function satisfying (i), (ii), and (iii). By Theorem 8.2, D satisﬁes (ii) and (iii). Hence, we show that it is multilinear. Suppose the ith row of A ¼ ½aij has the form ðbi1 þ ci1; bi2 þ ci2; . . . ; bin þ cinÞ. Then DðAÞ ¼ DðA1; . . . ; Bi þ Ci; . . . ; AnÞ ¼ P Sn ðsgn sÞa1sð1Þ ai1;sði1ÞðbisðiÞ þ cisðiÞÞ ansðnÞ ¼ P Sn ðsgn sÞa1sð1Þ bisðiÞ ansðnÞ þ P Sn ðsgn sÞa1sð1Þ cisðiÞ ansðnÞ ¼ DðA1; . . . ; Bi; . . . ; AnÞ þ DðA1; . . . ; Ci; . . . ; AnÞ 286 CHAPTER 8 Determinants Also, by Theorem 8.3(ii), DðA1; . . . ; kAi; . . . ; AnÞ ¼ kDðA1; . . . ; Ai; . . . ; AnÞ Thus, D is multilinear—D satisﬁes (i). We next must prove the uniqueness of D. Suppose D satisﬁes (i), (ii), and (iii). If fe1; . . . ; eng is the usual basis of Kn, then, by (iii), Dðe1; e2; . . . ; enÞ ¼ DðIÞ ¼ 1. Using (ii), we also have that Dðei1; ei2; . . . ; einÞ ¼ sgn s; where s ¼ i1i2 in ð1Þ Now suppose A ¼ ½aij. Observe that the kth row Ak of A is Ak ¼ ðak1; ak2; . . . ; aknÞ ¼ ak1e1 þ ak2e2 þ þ aknen Thus, DðAÞ ¼ Dða11e1 þ þ a1nen; a21e1 þ þ a2nen; . . . ; an1e1 þ þ annenÞ Using the multilinearity of D, we can write DðAÞ as a sum of terms of the form DðAÞ ¼ P Dða1i1ei1; a2i2ei2; . . . ; anineinÞ ¼ Pða1i1a2i2 aninÞDðei1; ei2; . . . ; einÞ ð2Þ where the sum is summed over all sequences i1i2 . . . in, where ik 2 f1; . . . ; ng. If two of the indices are equal, say ij ¼ ik but j 6¼ k, then, by (ii), Dðei1; ei2; . . . ; einÞ ¼ 0 Accordingly, the sum in (2) need only be summed over all permutations s ¼ i1i2 in. Using (1), we ﬁnally have that DðAÞ ¼ P s ða1i1a2i2 aninÞDðei1; ei2; . . . ; einÞ ¼ P s ðsgn sÞa1i1a2i2 anin; where s ¼ i1i2 in Hence, D is the determinant function, and so the theorem is proved. SUPPLEMENTARY PROBLEMS Computation of Determinants 8.38. Evaluate: (a) 2 6 4 1 , (b) 5 1 3 2 , (c) 2 8 5 3 , (d) 4 9 1 3 , (e) a þ b a b a þ b 8.39. Find all t such that (a) t 4 3 2 t 9 ¼ 0, (b) t 1 4 3 t 2 ¼ 0 8.40. Compute the determinant of each of the following matrices: (a) 2 1 1 0 5 2 1 3 4 2 4 3 5, (b) 3 2 4 2 5 1 0 6 1 2 4 3 5, (c) 2 1 4 6 3 2 4 1 2 2 4 3 5, (d) 7 6 5 1 2 1 3 2 1 2 4 3 5 CHAPTER 8 Determinants 287 8.41. Find the determinant of each of the following matrices: (a) 1 2 2 3 1 0 2 0 3 1 1 2 4 3 0 2 2 6 6 4 3 7 7 5, (b) 2 1 3 2 3 0 1 2 1 1 4 3 2 2 1 1 2 6 6 4 3 7 7 5 8.42. Evaluate: (a) 2 1 3 4 2 1 2 1 3 3 5 4 5 2 1 4 , (b) 2 1 4 3 1 1 0 2 3 2 3 1 1 2 2 3 , (c) 1 2 3 1 1 1 2 0 2 0 4 5 1 4 4 6 8.43. Evaluate each of the following determinants: (a) 1 2 1 3 1 2 1 1 2 3 3 1 0 2 1 5 1 2 3 4 2 3 1 1 2 , (b) 1 3 5 7 9 2 4 2 4 2 0 0 1 2 3 0 0 5 6 2 0 0 2 3 1 , (c) 1 2 3 4 5 5 4 3 2 1 0 0 6 5 1 0 0 0 7 4 0 0 0 2 3 Cofactors, Classical Adjoints, Inverses 8.44. Find detðAÞ, adj A, and A1, where (a) A ¼ 1 1 0 1 1 1 0 2 1 2 4 3 5, (b) A ¼ 1 2 2 3 1 0 1 1 1 2 4 3 5 8.45. Find the classical adjoint of each matrix in Problem 8.41. 8.46. Let A ¼ a b c d . (a) Find adj A, (b) Show that adjðadj AÞ ¼ A, (c) When does A ¼ adj A? 8.47. Show that if A is diagonal (triangular) then adj A is diagonal (triangular). 8.48. Suppose A ¼ ½aij is triangular. Show that (a) A is invertible if and only if each diagonal element aii 6¼ 0. (b) The diagonal elements of A1 (if it exists) are a1 ii , the reciprocals of the diagonal elements of A. Minors, Principal Minors 8.49. Let A ¼ 1 2 3 2 1 0 2 3 3 1 2 5 4 3 0 1 2 6 6 4 3 7 7 5 and B ¼ 1 3 1 5 2 3 1 4 0 5 2 1 3 0 5 2 2 6 6 4 3 7 7 5. Find the minor and the signed minor corresponding to the following submatrices: (a) Að1; 4; 3; 4Þ, (b) Bð1; 4; 3; 4Þ, (c) Að2; 3; 2; 4Þ, (d) Bð2; 3; 2; 4Þ. 8.50. For k ¼ 1; 2; 3, ﬁnd the sum Sk of all principal minors of order k for (a) A ¼ 1 3 2 2 4 3 5 2 1 2 4 3 5, (b) B ¼ 1 5 4 2 6 1 3 2 0 2 4 3 5, (c) C ¼ 1 4 3 2 1 5 4 7 11 2 4 3 5 288 CHAPTER 8 Determinants 8.51. For k ¼ 1; 2; 3; 4, ﬁnd the sum Sk of all principal minors of order k for (a) A ¼ 1 2 3 1 1 2 0 5 0 1 2 2 4 0 1 3 2 6 6 4 3 7 7 5, (b) B ¼ 1 2 1 2 0 1 2 3 1 3 0 4 2 7 4 5 2 6 6 4 3 7 7 5 Determinants and Linear Equations 8.52. Solve the following systems by determinants: (a) 3x þ 5y ¼ 8 4x 2y ¼ 1 , (b) 2x 3y ¼ 1 4x þ 7y ¼ 1 , (c) ax 2by ¼ c 3ax 5by ¼ 2c ðab 6¼ 0Þ 8.53. Solve the following systems by determinants: (a) 2x 5y þ 2z ¼ 2 x þ 2y 4z ¼ 5 3x 4y 6z ¼ 1 8 < : , (b) 2z þ 3 ¼ y þ 3x x 3z ¼ 2y þ 1 3y þ z ¼ 2 2x 8 < : 8.54. Prove Theorem 8.11: The system AX ¼ 0 has a nonzero solution if and only if D ¼ jAj ¼ 0. Permutations 8.55. Find the parity of the permutations s ¼ 32154, t ¼ 13524, p ¼ 42531 in S5. 8.56. For the permutations in Problem 8.55, ﬁnd (a) t s, (b) p s, (c) s1, (d) t1. 8.57. Let t 2 Sn: Show that t s runs through Sn as s runs through Sn; that is, Sn ¼ ft s : s 2 Sng: 8.58. Let s 2 Sn have the property that sðnÞ ¼ n. Let s 2 Sn1 be deﬁned by sðxÞ ¼ sðxÞ. (a) Show that sgn s ¼ sgn s, (b) Show that as s runs through Sn, where sðnÞ ¼ n, s runs through Sn1; that is, Sn1 ¼ fs : s 2 Sn; sðnÞ ¼ ng: 8.59. Consider a permutation s ¼ j1 j2 . . . jn. Let feig be the usual basis of Kn, and let A be the matrix whose ith row is eji [i.e., A ¼ ðej1, ej2; . . . ; ejnÞ]. Show that jAj ¼ sgn s. Determinant of Linear Operators 8.60. Find the determinant of each of the following linear transformations: (a) T:R2 ! R2 deﬁned by Tðx; yÞ ¼ ð2x 9y; 3x 5yÞ, (b) T:R3 ! R3 deﬁned by Tðx; y; zÞ ¼ ð3x 2z; 5y þ 7z; x þ y þ zÞ, (c) T:R3 ! R2 deﬁned by Tðx; y; zÞ ¼ ð2x þ 7y 4z; 4x 6y þ 2zÞ. 8.61. Let D:V ! V be the differential operator; that is, Dð f ðtÞÞ ¼ df =dt. Find detðDÞ if V is the vector space of functions with the following bases: (a) f1; t; . . . ; t5g, (b) fet; e2t; e3tg, (c) fsin t; cos tg. 8.62. Prove Theorem 8.13: Let F and G be linear operators on a vector space V. Then (i) detðF GÞ ¼ detðFÞ detðGÞ, (ii) F is invertible if and only if detðFÞ 6¼ 0. 8.63. Prove (a) detð1VÞ ¼ 1, where 1V is the identity operator, (b) -detðT 1Þ ¼ detðTÞ1 when T is invertible. CHAPTER 8 Determinants 289 Miscellaneous Problems 8.64. Find the volume VðSÞ of the parallelopiped S in R3 determined by the following vectors: (a) u1 ¼ ð1; 2; 3Þ, u2 ¼ ð3; 4; 1Þ, u3 ¼ ð2; 1; 5Þ, (b) u1 ¼ ð1; 1; 3Þ, u2 ¼ ð1; 2; 4Þ, u3 ¼ ð4; 1; 5Þ. 8.65. Find the volume VðSÞ of the parallelepiped S in R4 determined by the following vectors: u1 ¼ ð1; 2; 5; 1Þ; u2 ¼ ð2; 1; 2; 1Þ; u3 ¼ ð3; 0; 1 2Þ; u4 ¼ ð1; 1; 4; 1Þ 8.66. Let V be the space of 2 2 matrices M ¼ a b c d over R. Determine whether D:V ! R is 2-linear (with respect to the rows), where ðaÞ DðMÞ ¼ a þ d; ðbÞ DðMÞ ¼ ad; ðcÞ DðMÞ ¼ ac bd; ðdÞ DðMÞ ¼ ab cd; ðeÞ DðMÞ ¼ 0 ðfÞ DðMÞ ¼ 1 8.67. Let A be an n-square matrix. Prove jkAj ¼ knjAj. 8.68. Let A; B; C; D be commuting n-square matrices. Consider the 2n-square block matrix M ¼ A B C D . Prove that jMj ¼ jAjjDj jBjjCj. Show that the result may not be true if the matrices do not commute. 8.69. Suppose A is orthogonal; that is, ATA ¼ I. Show that detðAÞ ¼ 1. 8.70. Let V be the space of m-square matrices viewed as m-tuples of row vectors. Suppose D:V ! K is m-linear and alternating. Show that (a) Dð. . . ; A; . . . ; B; . . .Þ ¼ Dð. . . ; B; . . . ; A; . . .Þ; sign changed when two rows are interchanged. (b) If A1; A2; . . . ; Am are linearly dependent, then DðA1; A2; . . . ; AmÞ ¼ 0. 8.71. Let V be the space of m-square matrices (as above), and suppose D: V ! K. Show that the following weaker statement is equivalent to D being alternating: DðA1; A2; . . . ; AnÞ ¼ 0 whenever Ai ¼ Aiþ1 for some i Let V be the space of n-square matrices over K. Suppose B 2 V is invertible and so detðBÞ 6¼ 0. Deﬁne D: V ! K by DðAÞ ¼ detðABÞ=detðBÞ, where A 2 V. Hence, DðA1; A2; . . . ; AnÞ ¼ detðA1B; A2B; . . . ; AnBÞ=detðBÞ where Ai is the ith row of A, and so AiB is the ith row of AB. Show that D is multilinear and alternating, and that DðIÞ ¼ 1. (This method is used by some texts to prove that jABj ¼ jAjjBj.) 8.72. Show that g ¼ gðx1; . . . ; xnÞ ¼ ð1ÞnVn1ðxÞ where g ¼ gðxiÞ is the difference product in Problem 8.19, x ¼ xn, and Vn1 is the Vandermonde determinant deﬁned by Vn1ðxÞ 1 1 . . . 1 1 x1 x2 . . . xn1 x x2 1 x2 2 . . . x2 n1 x2 :::::::::::::::::::::::::::::::::::::::::::: xn1 1 xn1 2 . . . xn1 n1 xn1 2 6 6 6 6 6 6 6 4 8.73. Let A be any matrix. Show that the signs of a minor A½I; J and its complementary minor A½I0; J0 are equal. 290 CHAPTER 8 Determinants 8.74. Let A be an n-square matrix. The determinantal rank of A is the order of the largest square submatrix of A (obtained by deleting rows and columns of A) whose determinant is not zero. Show that the determinantal rank of A is equal to its rank—the maximum number of linearly independent rows (or columns). ANSWERS TO SUPPLEMENTARY PROBLEMS Notation: M ¼ ½R1; R2; . . . denotes a matrix with rows R1; R2; : . . . 8.38. (a) 22, (b) 13, (c) 46, (d) 21, (e) a2 þ ab þ b2 8.39. (a) 3; 10; (b) 5; 2 8.40. (a) 21, (b) 11, (c) 100, (d) 0 8.41. (a) 131, (b) 55 8.42. (a) 33, (b) 0, (c) 45 8.43. (a) 32, (b) 14, (c) 468 8.44. (a) jAj ¼ 2; adj A ¼ ½1; 1; 1; 1; 1; 1; 2; 2; 0, (b) jAj ¼ 1; adj A ¼ ½1; 0; 2; 3; 1; 6; 2; 1; 5. Also, A1 ¼ ðadj AÞ=jAj 8.45. (a) ½16; 29; 26; 2; 30; 38; 16; 29; 8; 51; 13; 1; 13; 1; 28; 18, (b) ½21; 14; 17; 19; 44; 11; 33; 11; 29; 1; 13; 21; 17; 7; 19; 18 8.46. (a) adj A ¼ ½d; b; c; a, (c) A ¼ kI 8.49. (a) 3; 3, (b) 23; 23, (c) 3; 3, (d) 17; 17 8.50. (a) 2; 17; 73, (b) 7; 10; 105, (c) 13; 54; 0 8.51. (a) 6; 13; 62; 219; (b) 7; 37; 30; 20 8.52. (a) x ¼ 21 26 ; y ¼ 29 26; (b) x ¼ 5 13 ; y ¼ 1 13; (c) x ¼ c a ; y ¼ c b 8.53. (a) x ¼ 5; y ¼ 2; z ¼ 1, (b) Because D ¼ 0, the system cannot be solved by determinants. 8.55. (a) sgn s ¼ 1; sgn t ¼ 1; sgn p ¼ 1 8.56. (a) t s ¼ 53142, (b) p s ¼ 52413, (c) s1 ¼ 32154, (d) t1 ¼ 14253 8.60. (a) detðTÞ ¼ 17, (b) detðTÞ ¼ 4, (c) not deﬁned 8.61. (a) 0, (b) 6, (c) 1 8.64. (a) 18, (b) 0 8.65. 17 8.66. (a) no, (b) yes, (c) yes, (d) no, (e) yes, (f ) no CHAPTER 8 Determinants 291 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 9.1 Introduction The ideas in this chapter can be discussed from two points of view. Matrix Point of View Suppose an n-square matrix A is given. The matrix A is said to be diagonalizable if there exists a nonsingular matrix P such that B ¼ P1AP is diagonal. This chapter discusses the diagonalization of a matrix A. In particular, an algorithm is given to ﬁnd the matrix P when it exists. Linear Operator Point of View Suppose a linear operator T: V ! V is given. The linear operator T is said to be diagonalizable if there exists a basis S of V such that the matrix representation of T relative to the basis S is a diagonal matrix D. This chapter discusses conditions under which the linear operator T is diagonalizable. Equivalence of the Two Points of View The above two concepts are essentially the same. Speciﬁcally, a square matrix A may be viewed as a linear operator F deﬁned by FðXÞ ¼ AX where X is a column vector, and B ¼ P1AP represents F relative to a new coordinate system (basis) S whose elements are the columns of P. On the other hand, any linear operator T can be represented by a matrix A relative to one basis and, when a second basis is chosen, T is represented by the matrix B ¼ P1AP where P is the change-of-basis matrix. Most theorems will be stated in two ways: one in terms of matrices A and again in terms of linear mappings T. Role of Underlying Field K The underlying number ﬁeld K did not play any special role in our previous discussions on vector spaces and linear mappings. However, the diagonalization of a matrix A or a linear operator T will depend on the CHAPTER 9 292 roots of a polynomial DðtÞ over K, and these roots do depend on K. For example, suppose DðtÞ ¼ t2 þ 1. Then DðtÞ has no roots if K ¼ R, the real ﬁeld; but DðtÞ has roots i if K ¼ C, the complex ﬁeld. Furthermore, ﬁnding the roots of a polynomial with degree greater than two is a subject unto itself (frequently discussed in numerical analysis courses). Accordingly, our examples will usually lead to those polynomials DðtÞ whose roots can be easily determined. 9.2 Polynomials of Matrices Consider a polynomial f ðtÞ ¼ antn þ þ a1t þ a0 over a ﬁeld K. Recall (Section 2.8) that if A is any square matrix, then we deﬁne f ðAÞ ¼ anAn þ þ a1A þ a0I where I is the identity matrix. In particular, we say that A is a root of f ðtÞ if f ðAÞ ¼ 0, the zero matrix. EXAMPLE 9.1 Let A ¼ 1 2 3 4 . Then A2 ¼ 7 10 15 22 . Let f ðtÞ ¼ 2t2 3t þ 5 and gðtÞ ¼ t2 5t 2 Then f ðAÞ ¼ 2A2 3A þ 5I ¼ 14 20 30 44 þ 3 6 9 12 þ 5 0 0 5 ¼ 16 14 21 37 and gðAÞ ¼ A2 5A 2I ¼ 7 10 15 22 þ 5 10 15 20 þ 2 0 0 2 ¼ 0 0 0 0 Thus, A is a zero of gðtÞ. The following theorem (proved in Problem 9.7) applies. THEOREM 9.1: Let f and g be polynomials. For any square matrix A and scalar k, (i) ð f þ gÞðAÞ ¼ f ðAÞ þ gðAÞ (iii) ðkf ÞðAÞ ¼ kf ðAÞ (ii) ð fgÞðAÞ ¼ f ðAÞgðAÞ (iv) f ðAÞgðAÞ ¼ gðAÞ f ðAÞ: Observe that (iv) tells us that any two polynomials in A commute. Matrices and Linear Operators Now suppose that T: V ! V is a linear operator on a vector space V. Powers of T are deﬁned by the composition operation: T2 ¼ T T; T3 ¼ T2 T; . . . Also, for any polynomial f ðtÞ ¼ antn þ þ a1t þ a0, we deﬁne f ðTÞ in the same way as we did for matrices: f ðTÞ ¼ anTn þ þ a1T þ a0I where I is now the identity mapping. We also say that T is a zero or root of f ðtÞ if f ðTÞ ¼ 0; the zero mapping. We note that the relations in Theorem 9.1 hold for linear operators as they do for matrices. Remark: Suppose A is a matrix representation of a linear operator T. Then f ðAÞ is the matrix representation of f ðTÞ, and, in particular, f ðTÞ ¼ 0 if and only if f ðAÞ ¼ 0. CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 293 9.3 Characteristic Polynomial, Cayley–Hamilton Theorem Let A ¼ ½aij be an n-square matrix. The matrix M ¼ A tIn, where In is the n-square identity matrix and t is an indeterminate, may be obtained by subtracting t down the diagonal of A. The negative of M is the matrix tIn A, and its determinant DðtÞ ¼ detðtIn AÞ ¼ ð1Þn detðA tInÞ which is a polynomial in t of degree n and is called the characteristic polynomial of A. We state an important theorem in linear algebra (proved in Problem 9.8). THEOREM 9.2: (Cayley–Hamilton) Every matrix A is a root of its characteristic polynomial. Remark: Suppose A ¼ ½aij is a triangular matrix. Then tI A is a triangular matrix with diagonal entries t aii; hence, DðtÞ ¼ detðtI AÞ ¼ ðt a11Þðt a22Þ ðt annÞ Observe that the roots of DðtÞ are the diagonal elements of A. EXAMPLE 9.2 Let A ¼ 1 3 4 5 . Its characteristic polynomial is DðtÞ ¼ jtI Aj ¼ t 1 3 4 t 5 ¼ ðt 1Þðt 5Þ 12 ¼ t2 6t 7 As expected from the Cayley–Hamilton theorem, A is a root of DðtÞ; that is, DðAÞ ¼ A2 6A 7I ¼ 13 18 24 37 þ 6 18 24 30 þ 7 0 0 7 ¼ 0 0 0 0 Now suppose A and B are similar matrices, say B ¼ P1AP, where P is invertible. We show that A and B have the same characteristic polynomial. Using tI ¼ P1tIP, we have DBðtÞ ¼ detðtI BÞ ¼ detðtI P1APÞ ¼ detðP1tIP P1APÞ ¼ det½P1ðtI AÞP ¼ detðP1Þ detðtI AÞ detðPÞ Using the fact that determinants are scalars and commute and that detðP1Þ detðPÞ ¼ 1, we ﬁnally obtain DBðtÞ ¼ detðtI AÞ ¼ DAðtÞ Thus, we have proved the following theorem. THEOREM 9.3: Similar matrices have the same characteristic polynomial. Characteristic Polynomials of Degrees 2 and 3 There are simple formulas for the characteristic polynomials of matrices of orders 2 and 3. (a) Suppose A ¼ a11 a12 a21 a22 . Then DðtÞ ¼ t2 ða11 þ a22Þt þ detðAÞ ¼ t2 trðAÞ t þ detðAÞ Here trðAÞ denotes the trace of A—that is, the sum of the diagonal elements of A. (b) Suppose A ¼ a11 a12 a13 a21 a22 a23 a31 a32 a33 2 4 3 5. Then DðtÞ ¼ t3 trðAÞ t2 þ ðA11 þ A22 þ A33Þt detðAÞ (Here A11, A22, A33 denote, respectively, the cofactors of a11, a22, a33.) 294 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors EXAMPLE 9.3 Find the characteristic polynomial of each of the following matrices: (a) A ¼ 5 3 2 10 , (b) B ¼ 7 1 6 2 , (c) C ¼ 5 2 4 4 . (a) We have trðAÞ ¼ 5 þ 10 ¼ 15 and jAj ¼ 50 6 ¼ 44; hence, DðtÞ þ t2 15t þ 44. (b) We have trðBÞ ¼ 7 þ 2 ¼ 9 and jBj ¼ 14 þ 6 ¼ 20; hence, DðtÞ ¼ t2 9t þ 20. (c) We have trðCÞ ¼ 5 4 ¼ 1 and jCj ¼ 20 þ 8 ¼ 12; hence, DðtÞ ¼ t2 t 12. EXAMPLE 9.4 Find the characteristic polynomial of A ¼ 1 1 2 0 3 2 1 3 9 2 4 3 5. We have trðAÞ ¼ 1 þ 3 þ 9 ¼ 13. The cofactors of the diagonal elements are as follows: A11 ¼ 3 2 3 9 ¼ 21; A22 ¼ 1 2 1 9 ¼ 7; A33 ¼ 1 1 0 3 ¼ 3 Thus, A11 þ A22 þ A33 ¼ 31. Also, jAj ¼ 27 þ 2 þ 0 6 6 0 ¼ 17. Accordingly, DðtÞ ¼ t3 13t2 þ 31t 17 Remark: The coefﬁcients of the characteristic polynomial DðtÞ of the 3-square matrix A are, with alternating signs, as follows: S1 ¼ trðAÞ; S2 ¼ A11 þ A22 þ A33; S3 ¼ detðAÞ We note that each Sk is the sum of all principal minors of A of order k. The next theorem, whose proof lies beyond the scope of this text, tells us that this result is true in general. THEOREM 9.4: Let A be an n-square matrix. Then its characteristic polynomial is DðtÞ ¼ tn S1tn1 þ S2tn2 þ þ ð1ÞnSn where Sk is the sum of the principal minors of order k. Characteristic Polynomial of a Linear Operator Now suppose T: V ! V is a linear operator on a vector space V of ﬁnite dimension. We deﬁne the characteristic polynomial DðtÞ of T to be the characteristic polynomial of any matrix representation of T. Recall that if A and B are matrix representations of T, then B ¼ P1AP, where P is a change-of-basis matrix. Thus, A and B are similar, and by Theorem 9.3, A and B have the same characteristic polynomial. Accordingly, the characteristic polynomial of T is independent of the particular basis in which the matrix representation of T is computed. Because f ðTÞ ¼ 0 if and only if f ðAÞ ¼ 0, where f ðtÞ is any polynomial and A is any matrix representation of T, we have the following analogous theorem for linear operators. THEOREM 9.20: (Cayley–Hamilton) A linear operator T is a zero of its characteristic polynomial. CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 295 9.4 Diagonalization, Eigenvalues and Eigenvectors Let A be any n-square matrix. Then A can be represented by (or is similar to) a diagonal matrix D ¼ diagðk1; k2; . . . ; knÞ if and only if there exists a basis S consisting of (column) vectors u1; u2; . . . ; un such that Au1 ¼ k1u1 Au2 ¼ k2u2 :::::::::::::::::::::::::::::::::::: Aun ¼ knun In such a case, A is said to be diagonizable. Furthermore, D ¼ P1AP, where P is the nonsingular matrix whose columns are, respectively, the basis vectors u1; u2; . . . ; un. The above observation leads us to the following deﬁnition. DEFINITION: Let A be any square matrix. A scalar l is called an eigenvalue of A if there exists a nonzero (column) vector v such that Av ¼ lv Any vector satisfying this relation is called an eigenvector of A belonging to the eigenvalue l. We note that each scalar multiple kv of an eigenvector v belonging to l is also such an eigenvector, because AðkvÞ ¼ kðAvÞ ¼ kðlvÞ ¼ lðkvÞ The set El of all such eigenvectors is a subspace of V (Problem 9.19), called the eigenspace of l. (If dim El ¼ 1, then El is called an eigenline and l is called a scaling factor.) The terms characteristic value and characteristic vector (or proper value and proper vector) are sometimes used instead of eigenvalue and eigenvector. The above observation and deﬁnitions give us the following theorem. THEOREM 9.5: An n-square matrix A is similar to a diagonal matrix D if and only if A has n linearly independent eigenvectors. In this case, the diagonal elements of D are the corresponding eigenvalues and D ¼ P1AP, where P is the matrix whose columns are the eigenvectors. Suppose a matrix A can be diagonalized as above, say P1AP ¼ D, where D is diagonal. Then A has the extremely useful diagonal factorization: A ¼ PDP1 Using this factorization, the algebra of A reduces to the algebra of the diagonal matrix D, which can be easily calculated. Speciﬁcally, suppose D ¼ diagðk1; k2; . . . ; knÞ. Then Am ¼ ðPDP1Þm ¼ PDmP1 ¼ P diagðkm 1 ; . . . ; km n ÞP1 More generally, for any polynomial f ðtÞ, f ðAÞ ¼ f ðPDP1Þ ¼ Pf ðDÞP1 ¼ P diagð f ðk1Þ; f ðk2Þ; . . . ; f ðknÞÞP1 Furthermore, if the diagonal entries of D are nonnegative, let B ¼ P diagð ﬃﬃﬃﬃ ﬃ k1 p ; ﬃﬃﬃﬃ ﬃ k2 p ; . . . ; ﬃﬃﬃﬃ ﬃ kn p Þ P1 Then B is a nonnegative square root of A; that is, B2 ¼ A and the eigenvalues of B are nonnegative. 296 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors EXAMPLE 9.5 Let A ¼ 3 1 2 2 and let v1 ¼ 1 2 and v2 ¼ 1 1 . Then Av1 ¼ 3 1 2 2 1 2 ¼ 1 2 ¼ v1 and Av2 ¼ 3 1 2 2 1 1 ¼ 4 4 ¼ 4v2 Thus, v1 and v2 are eigenvectors of A belonging, respectively, to the eigenvalues l1 ¼ 1 and l2 ¼ 4. Observe that v1 and v2 are linearly independent and hence form a basis of R2. Accordingly, A is diagonalizable. Furthermore, let P be the matrix whose columns are the eigenvectors v1 and v2. That is, let P ¼ " 1 1 2 1 # ; and so P1 ¼ 1 3 1 3 2 3 1 3 " # Then A is similar to the diagonal matrix D ¼ P1AP ¼ 1 3 1 3 2 3 1 3 " #" 3 1 2 2 #" 1 1 2 1 # ¼ " 1 0 0 4 # As expected, the diagonal elements 1 and 4 in D are the eigenvalues corresponding, respectively, to the eigenvectors v1 and v2, which are the columns of P. In particular, A has the factorization A ¼ PDP1 ¼ " 1 1 2 1 #" 1 0 0 4 # 1 3 1 3 2 3 1 3 " # Accordingly, A4 ¼ " 1 1 2 1 #" 1 0 0 256 # 1 3 1 3 2 3 1 3 " # ¼ " 171 85 170 86 # Moreover, suppose f ðtÞ ¼ t3 5t2 þ 3t þ 6; hence, f ð1Þ ¼ 5 and f ð4Þ ¼ 2. Then f ðAÞ ¼ Pf ðDÞP1 ¼ 1 1 2 1 5 0 0 2 1 3 1 3 2 3 1 3 " # ¼ 3 1 2 4 Last, we obtain a ‘‘positive square root’’ of A. Speciﬁcally, using ﬃﬃﬃ 1 p ¼ 1 and ﬃﬃﬃ 4 p ¼ 2, we obtain the matrix B ¼ P ﬃﬃﬃﬃ D p P1 ¼ 1 1 2 1 1 0 0 2 1 3 1 3 2 3 1 3 " # ¼ 5 3 1 3 2 3 4 3 " # where B2 ¼ A and where B has positive eigenvalues 1 and 2. Remark: Throughout this chapter, we use the following fact: If P ¼ a b c d ; then P1 ¼ d=jPj b=jPj c=jPj a=jPj : That is, P1 is obtained by interchanging the diagonal elements a and d of P, taking the negatives of the nondiagonal elements b and c, and dividing each element by the determinant jPj. Properties of Eigenvalues and Eigenvectors Example 9.5 indicates the advantages of a diagonal representation (factorization) of a square matrix. In the following theorem (proved in Problem 9.20), we list properties that help us to ﬁnd such a representation. CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 297 THEOREM 9.6: Let A be a square matrix. Then the following are equivalent. (i) A scalar l is an eigenvalue of A. (ii) The matrix M ¼ A lI is singular. (iii) The scalar l is a root of the characteristic polynomial DðtÞ of A. The eigenspace El of an eigenvalue l is the solution space of the homogeneous system MX ¼ 0, where M ¼ A lI; that is, M is obtained by subtracting l down the diagonal of A. Some matrices have no eigenvalues and hence no eigenvectors. However, using Theorem 9.6 and the Fundamental Theorem of Algebra (every polynomial over the complex ﬁeld C has a root), we obtain the following result. THEOREM 9.7: Let A be a square matrix over the complex ﬁeld C. Then A has at least one eigenvalue. The following theorems will be used subsequently. (The theorem equivalent to Theorem 9.8 for linear operators is proved in Problem 9.21, and Theorem 9.9 is proved in Problem 9.22.) THEOREM 9.8: Suppose v1; v2; . . . ; vn are nonzero eigenvectors of a matrix A belonging to distinct eigenvalues l1; l2; . . . ; ln. Then v1; v2; . . . ; vn are linearly independent. THEOREM 9.9: Suppose the characteristic polynomial DðtÞ of an n-square matrix A is a product of n distinct factors, say, DðtÞ ¼ ðt a1Þðt a2Þ ðt anÞ. Then A is similar to the diagonal matrix D ¼ diagða1; a2; . . . ; anÞ. If l is an eigenvalue of a matrix A, then the algebraic multiplicity of l is deﬁned to be the multiplicity of l as a root of the characteristic polynomial of A, and the geometric multiplicity of l is deﬁned to be the dimension of its eigenspace, dim El. The following theorem (whose equivalent for linear operators is proved in Problem 9.23) holds. THEOREM 9.10: The geometric multiplicity of an eigenvalue l of a matrix A does not exceed its algebraic multiplicity. Diagonalization of Linear Operators Consider a linear operator T: V ! V. Then T is said to be diagonalizable if it can be represented by a diagonal matrix D. Thus, T is diagonalizable if and only if there exists a basis S ¼ fu1; u2; . . . ; ung of V for which Tðu1Þ ¼ k1u1 Tðu2Þ ¼ k2u2 ::::::::::::::::::::::::::::::::::::::: TðunÞ ¼ knun In such a case, T is represented by the diagonal matrix D ¼ diagðk1; k2; . . . ; knÞ relative to the basis S. The above observation leads us to the following deﬁnitions and theorems, which are analogous to the deﬁnitions and theorems for matrices discussed above. DEFINITION: Let T be a linear operator. A scalar l is called an eigenvalue of T if there exists a nonzero vector v such that TðvÞ ¼ lv. Every vector satisfying this relation is called an eigenvector of T belonging to the eigenvalue l. 298 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors The set El of all eigenvectors belonging to an eigenvalue l is a subspace of V, called the eigenspace of l. (Alternatively, l is an eigenvalue of T if lI T is singular, and, in this case, El is the kernel of lI T.) The algebraic and geometric multiplicities of an eigenvalue l of a linear operator T are deﬁned in the same way as those of an eigenvalue of a matrix A. The following theorems apply to a linear operator T on a vector space V of ﬁnite dimension. THEOREM 9.50: T can be represented by a diagonal matrix D if and only if there exists a basis S of V consisting of eigenvectors of T. In this case, the diagonal elements of D are the corresponding eigenvalues. THEOREM 9.60: Let T be a linear operator. Then the following are equivalent: (i) A scalar l is an eigenvalue of T. (ii) The linear operator lI T is singular. (iii) The scalar l is a root of the characteristic polynomial DðtÞ of T. THEOREM 9.70: Suppose V is a complex vector space. Then T has at least one eigenvalue. THEOREM 9.80: Suppose v1; v2; . . . ; vn are nonzero eigenvectors of a linear operator T belonging to distinct eigenvalues l1; l2; . . . ; ln. Then v1; v2; . . . ; vn are linearly independent. THEOREM 9.90: Suppose the characteristic polynomial DðtÞ of T is a product of n distinct factors, say, DðtÞ ¼ ðt a1Þðt a2Þ ðt anÞ. Then T can be represented by the diagonal matrix D ¼ diagða1; a2; . . . ; anÞ. THEOREM 9.100: The geometric multiplicity of an eigenvalue l of T does not exceed its algebraic multiplicity. Remark: The following theorem reduces the investigation of the diagonalization of a linear operator T to the diagonalization of a matrix A. THEOREM 9.11: Suppose A is a matrix representation of T. Then T is diagonalizable if and only if A is diagonalizable. 9.5 Computing Eigenvalues and Eigenvectors, Diagonalizing Matrices This section gives an algorithm for computing eigenvalues and eigenvectors for a given square matrix A and for determining whether or not a nonsingular matrix P exists such that P1AP is diagonal. ALGORITHM 9.1: (Diagonalization Algorithm) The input is an n-square matrix A. Step 1. Find the characteristic polynomial DðtÞ of A. Step 2. Find the roots of DðtÞ to obtain the eigenvalues of A. Step 3. Repeat (a) and (b) for each eigenvalue l of A. (a) Form the matrix M ¼ A lI by subtracting l down the diagonal of A. (b) Find a basis for the solution space of the homogeneous system MX ¼ 0. (These basis vectors are linearly independent eigenvectors of A belonging to l.) CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 299 Step 4. Consider the collection S ¼ fv1; v2; . . . ; vmg of all eigenvectors obtained in Step 3. (a) If m 6¼ n, then A is not diagonalizable. (b) If m ¼ n, then A is diagonalizable. Speciﬁcally, let P be the matrix whose columns are the eigenvectors v1; v2; . . . ; vn. Then D ¼ P1AP ¼ diagðl1; l2; . . . ; lnÞ where li is the eigenvalue corresponding to the eigenvector vi. EXAMPLE 9.6 The diagonalizable algorithm is applied to A ¼ 4 2 3 1 . (1) The characteristic polynomial DðtÞ of A is computed. We have trðAÞ ¼ 4 1 ¼ 3; jAj ¼ 4 6 ¼ 10; hence, DðtÞ ¼ t2 3t 10 ¼ ðt 5Þðt þ 2Þ (2) Set DðtÞ ¼ ðt 5Þðt þ 2Þ ¼ 0. The roots l1 ¼ 5 and l2 ¼ 2 are the eigenvalues of A. (3) (i) We ﬁnd an eigenvector v1 of A belonging to the eigenvalue l1 ¼ 5. Subtract l1 ¼ 5 down the diagonal of A to obtain the matrix M ¼ 1 2 3 6 . The eigenvectors belonging to l1 ¼ 5 form the solution of the homogeneous system MX ¼ 0; that is, 1 2 3 6 x y ¼ 0 0 or x þ 2y ¼ 0 3x 6y ¼ 0 or x þ 2y ¼ 0 The system has only one free variable. Thus, a nonzero solution, for example, v1 ¼ ð2; 1Þ, is an eigenvector that spans the eigenspace of l1 ¼ 5. (ii) We ﬁnd an eigenvector v2 of A belonging to the eigenvalue l2 ¼ 2. Subtract 2 (or add 2) down the diagonal of A to obtain the matrix M ¼ 6 2 3 1 and the homogenous system 6x þ 2y ¼ 0 3x þ y ¼ 0 or 3x þ y ¼ 0: The system has only one independent solution. Thus, a nonzero solution, say v2 ¼ ð1; 3Þ; is an eigenvector that spans the eigenspace of l2 ¼ 2: (4) Let P be the matrix whose columns are the eigenvectors v1 and v2. Then P ¼ 2 1 1 3 ; and so P1 ¼ 3 7 1 7 1 7 2 7 " # Accordingly, D ¼ P1AP is the diagonal matrix whose diagonal entries are the corresponding eigenvalues; that is, D ¼ P1AP ¼ 3 7 1 7 1 7 2 7 " # 4 2 3 1 2 1 1 3 ¼ 5 0 0 2 EXAMPLE 9.7 Consider the matrix B ¼ 5 1 1 3 . We have trðBÞ ¼ 5 þ 3 ¼ 8; jBj ¼ 15 þ 1 ¼ 16; so DðtÞ ¼ t2 8t þ 16 ¼ ðt 4Þ2 Accordingly, l ¼ 4 is the only eigenvalue of B. 300 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors Subtract l ¼ 4 down the diagonal of B to obtain the matrix M ¼ 1 1 1 1 and the homogeneous system x y ¼ 0 x y ¼ 0 or x y ¼ 0 The system has only one independent solution; for example, x ¼ 1; y ¼ 1. Thus, v ¼ ð1; 1Þ and its multiples are the only eigenvectors of B. Accordingly, B is not diagonalizable, because there does not exist a basis consisting of eigenvectors of B. EXAMPLE 9.8 Consider the matrix A ¼ 3 5 2 3 . Here trðAÞ ¼ 3 3 ¼ 0 and jAj ¼ 9 þ 10 ¼ 1. Thus, DðtÞ ¼ t2 þ 1 is the characteristic polynomial of A. We consider two cases: (a) A is a matrix over the real ﬁeld R. Then DðtÞ has no (real) roots. Thus, A has no eigenvalues and no eigenvectors, and so A is not diagonalizable. (b) A is a matrix over the complex ﬁeld C. Then DðtÞ ¼ ðt iÞðt þ iÞ has two roots, i and i. Thus, A has two distinct eigenvalues i and i, and hence, A has two independent eigenvectors. Accordingly there exists a nonsingular matrix P over the complex ﬁeld C for which P1AP ¼ i 0 0 i Therefore, A is diagonalizable (over C). 9.6 Diagonalizing Real Symmetric Matrices and Quadratic Forms There are many real matrices A that are not diagonalizable. In fact, some real matrices may not have any (real) eigenvalues. However, if A is a real symmetric matrix, then these problems do not exist. Namely, we have the following theorems. THEOREM 9.12: Let A be a real symmetric matrix. Then each root l of its characteristic polynomial is real. THEOREM 9.13: Let A be a real symmetric matrix. Suppose u and v are eigenvectors of A belonging to distinct eigenvalues l1 and l2. Then u and v are orthogonal, that; is, hu; vi ¼ 0. The above two theorems give us the following fundamental result. THEOREM 9.14: Let A be a real symmetric matrix. Then there exists an orthogonal matrix P such that D ¼ P1AP is diagonal. The orthogonal matrix P is obtained by normalizing a basis of orthogonal eigenvectors of A as illustrated below. In such a case, we say that A is ‘‘orthogonally diagonalizable.’’ EXAMPLE 9.9 Let A ¼ 2 2 2 5 , a real symmetric matrix. Find an orthogonal matrix P such that P1AP is diagonal. First we ﬁnd the characteristic polynomial DðtÞ of A. We have trðAÞ ¼ 2 þ 5 ¼ 7; jAj ¼ 10 4 ¼ 6; so DðtÞ ¼ t2 7t þ 6 ¼ ðt 6Þðt 1Þ Accordingly, l1 ¼ 6 and l2 ¼ 1 are the eigenvalues of A. (a) Subtracting l1 ¼ 6 down the diagonal of A yields the matrix M ¼ 4 2 2 1 and the homogeneous system 4x 2y ¼ 0 2x y ¼ 0 or 2x þ y ¼ 0 A nonzero solution is u1 ¼ ð1; 2Þ. CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 301 (b) Subtracting l2 ¼ 1 down the diagonal of A yields the matrix M ¼ 1 2 2 4 and the homogeneous system x 2y ¼ 0 (The second equation drops out, because it is a multiple of the ﬁrst equation.) A nonzero solution is u2 ¼ ð2; 1Þ. As expected from Theorem 9.13, u1 and u2 are orthogonal. Normalizing u1 and u2 yields the orthonormal vectors ^ u1 ¼ ð1= ﬃﬃﬃ 5 p ; 2= ﬃﬃﬃ 5 p Þ and ^ u2 ¼ ð2= ﬃﬃﬃ 5 p ; 1= ﬃﬃﬃ 5 p Þ Finally, let P be the matrix whose columns are ^ u1 and ^ u2, respectively. Then P ¼ 1= ﬃﬃﬃ 5 p 2= ﬃﬃﬃ 5 p 2= ﬃﬃﬃ 5 p 1= ﬃﬃﬃ 5 p and P1AP ¼ 6 0 0 1 As expected, the diagonal entries of P1AP are the eigenvalues corresponding to the columns of P. The procedure in the above Example 9.9 is formalized in the following algorithm, which ﬁnds an orthogonal matrix P such that P1AP is diagonal. ALGORITHM 9.2: (Orthogonal Diagonalization Algorithm) The input is a real symmetric matrix A. Step 1. Find the characteristic polynomial DðtÞ of A. Step 2. Find the eigenvalues of A, which are the roots of DðtÞ. Step 3. For each eigenvalue l of A in Step 2, ﬁnd an orthogonal basis of its eigenspace. Step 4. Normalize all eigenvectors in Step 3, which then forms an orthonormal basis of Rn. Step 5. Let P be the matrix whose columns are the normalized eigenvectors in Step 4. Application to Quadratic Forms Let q be a real polynomial in variables x1; x2; . . . ; xn such that every term in q has degree two; that is, qðx1; x2; . . . ; xnÞ ¼ P i cix2 i þ P i m. Then f þ g ¼ ðan þ bnÞtn þ þ ða1 þ b1Þt þ ða0 þ b0Þ Hence, ð f þ gÞðAÞ ¼ ðan þ bnÞAn þ þ ða1 þ b1ÞA þ ða0 þ b0ÞI ¼ anAn þ bnAn þ þ a1A þ b1A þ a0I þ b0I ¼ f ðAÞ þ gðAÞ (ii) By deﬁnition, fg ¼ cnþmtnþm þ þ c1t þ c0 ¼ P nþm k¼0 cktk, where ck ¼ a0bk þ a1bk1 þ þ akb0 ¼ P k i¼0 aibki Hence, ð fgÞðAÞ ¼ P nþm k¼0 ckAk and f ðAÞgðAÞ ¼ P n i¼0 aiAi P m j¼0 bjAj ¼ P n i¼0 P m j¼0 aibjAiþj ¼ P nþm k¼0 ckAk ¼ ð fgÞðAÞ (iii) By deﬁnition, kf ¼ kantn þ þ ka1t þ ka0, and so ðkf ÞðAÞ ¼ kanAn þ þ ka1A þ ka0I ¼ kðanAn þ þ a1A þ a0IÞ ¼ kf ðAÞ (iv) By (ii), gðAÞf ðAÞ ¼ ðgf ÞðAÞ ¼ ð fgÞðAÞ ¼ f ðAÞgðAÞ. 9.8. Prove the Cayley–Hamilton Theorem 9.2: Every matrix A is a root of its characterstic polynomial DðtÞ. Let A be an arbitrary n-square matrix and let DðtÞ be its characteristic polynomial, say, DðtÞ ¼ jtI Aj ¼ tn þ an1tn1 þ þ a1t þ a0 Now let BðtÞ denote the classical adjoint of the matrix tI A. The elements of BðtÞ are cofactors of the matrix tI A and hence are polynomials in t of degree not exceeding n 1. Thus, BðtÞ ¼ Bn1tn1 þ þ B1t þ B0 where the Bi are n-square matrices over K which are independent of t. By the fundamental property of the classical adjoint (Theorem 8.9), ðtI AÞBðtÞ ¼ jtI AjI, or ðtI AÞðBn1tn1 þ þ B1t þ B0Þ ¼ ðtn þ an1tn1 þ þ a1t þ a0ÞI Removing the parentheses and equating corresponding powers of t yields Bn1 ¼ I; Bn2 ABn1 ¼ an1I; . . . ; B0 AB1 ¼ a1I; AB0 ¼ a0I Multiplying the above equations by An; An1; . . . ; A; I, respectively, yields AnBn1 ¼ AnI; An1Bn2 AnBn1 ¼ an1An1; . . . ; AB0 A2B1 ¼ a1A; AB0 ¼ a0I Adding the above matrix equations yields 0 on the left-hand side and DðAÞ on the right-hand side; that is, 0 ¼ An þ an1An1 þ þ a1A þ a0I Therefore, DðAÞ ¼ 0, which is the Cayley–Hamilton theorem. 308 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors Eigenvalues and Eigenvectors of 2 2 Matrices 9.9. Let A ¼ 3 4 2 6 . (a) Find all eigenvalues and corresponding eigenvectors. (b) Find matrices P and D such that P is nonsingular and D ¼ P1AP is diagonal. (a) First ﬁnd the characteristic polynomial DðtÞ of A: DðtÞ ¼ t2 trðAÞ t þ jAj ¼ t2 þ 3t 10 ¼ ðt 2Þðt þ 5Þ The roots l ¼ 2 and l ¼ 5 of DðtÞ are the eigenvalues of A. We ﬁnd corresponding eigenvectors. (i) Subtract l ¼ 2 down the diagonal of A to obtain the matrix M ¼ A 2I, where the corresponding homogeneous system MX ¼ 0 yields the eigenvectors corresponding to l ¼ 2. We have M ¼ 1 4 2 8 ; corresponding to x 4y ¼ 0 2x 8y ¼ 0 or x 4y ¼ 0 The system has only one free variable, and v1 ¼ ð4; 1Þ is a nonzero solution. Thus, v1 ¼ ð4; 1Þ is an eigenvector belonging to (and spanning the eigenspace of) l ¼ 2. (ii) Subtract l ¼ 5 (or, equivalently, add 5) down the diagonal of A to obtain M ¼ 8 4 2 1 ; corresponding to 8x 4y ¼ 0 2x y ¼ 0 or 2x y ¼ 0 The system has only one free variable, and v2 ¼ ð1; 2Þ is a nonzero solution. Thus, v2 ¼ ð1; 2Þ is an eigenvector belonging to l ¼ 5. (b) Let P be the matrix whose columns are v1 and v2. Then P ¼ 4 1 1 2 and D ¼ P1AP ¼ 2 0 0 5 Note that D is the diagonal matrix whose diagonal entries are the eigenvalues of A corresponding to the eigenvectors appearing in P. Remark: Here P is the change-of-basis matrix from the usual basis of R2 to the basis S ¼ fv1; v2g, and D is the matrix that represents (the matrix function) A relative to the new basis S. 9.10. Let A ¼ 2 2 1 3 . (a) Find all eigenvalues and corresponding eigenvectors. (b) Find a nonsingular matrix P such that D ¼ P1AP is diagonal, and P1. (c) Find A6 and f ðAÞ, where t4 3t3 6t2 þ 7t þ 3. (d) Find a ‘‘real cube root’’ of B—that is, a matrix B such that B3 ¼ A and B has real eigenvalues. (a) First ﬁnd the characteristic polynomial DðtÞ of A: DðtÞ ¼ t2 trðAÞ t þ jAj ¼ t2 5t þ 4 ¼ ðt 1Þðt 4Þ The roots l ¼ 1 and l ¼ 4 of DðtÞ are the eigenvalues of A. We ﬁnd corresponding eigenvectors. (i) Subtract l ¼ 1 down the diagonal of A to obtain the matrix M ¼ A lI, where the corresponding homogeneous system MX ¼ 0 yields the eigenvectors belonging to l ¼ 1. We have M ¼ 1 2 1 2 ; corresponding to x þ 2y ¼ 0 x þ 2y ¼ 0 or x þ 2y ¼ 0 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 309 The system has only one independent solution; for example, x ¼ 2, y ¼ 1. Thus, v1 ¼ ð2; 1Þ is an eigenvector belonging to (and spanning the eigenspace of) l ¼ 1. (ii) Subtract l ¼ 4 down the diagonal of A to obtain M ¼ 2 2 1 1 ; corresponding to 2x þ 2y ¼ 0 x y ¼ 0 or x y ¼ 0 The system has only one independent solution; for example, x ¼ 1, y ¼ 1. Thus, v2 ¼ ð1; 1Þ is an eigenvector belonging to l ¼ 4. (b) Let P be the matrix whose columns are v1 and v2. Then P ¼ 2 1 1 1 and D ¼ P1AP ¼ 1 0 0 4 ; where P1 ¼ 1 3 1 3 1 3 2 3 " # (c) Using the diagonal factorization A ¼ PDP1, and 16 ¼ 1 and 46 ¼ 4096, we get A6 ¼ PD6P1 ¼ 2 1 1 1 " # 1 0 0 4096 " # 1 3 1 3 1 3 2 3 " # ¼ 1366 2230 1365 2731 " # Also, f ð1Þ ¼ 2 and f ð4Þ ¼ 1. Hence, f ðAÞ ¼ Pf ðDÞP1 ¼ 2 1 1 1 " # 2 0 0 1 " # 1 3 1 3 1 3 2 3 " # ¼ 1 2 1 0 " # (d) Here 1 0 0 ﬃﬃﬃ 4 3 p is the real cube root of D. Hence the real cube root of A is B ¼ P ﬃﬃﬃﬃ D 3 p P1 ¼ 2 1 1 1 " # 1 0 0 ﬃﬃﬃ 4 3 p " # 1 3 1 3 1 3 2 3 " # ¼ 1 3 2 þ ﬃﬃﬃ 4 3 p 2 þ 2 ﬃﬃﬃ 4 3 p 1 þ ﬃﬃﬃ 4 3 p 1 þ 2 ﬃﬃﬃ 4 3 p " # 9.11. Each of the following real matrices deﬁnes a linear transformation on R2: (a) A ¼ 5 6 3 2 , (b) B ¼ 1 1 2 1 , (c) C ¼ 5 1 1 3 Find, for each matrix, all eigenvalues and a maximum set S of linearly independent eigenvectors. Which of these linear operators are diagonalizable—that is, which can be represented by a diagonal matrix? (a) First ﬁnd DðtÞ ¼ t2 3t 28 ¼ ðt 7Þðt þ 4Þ. The roots l ¼ 7 and l ¼ 4 are the eigenvalues of A. We ﬁnd corresponding eigenvectors. (i) Subtract l ¼ 7 down the diagonal of A to obtain M ¼ 2 6 3 9 ; corresponding to 2x þ 6y ¼ 0 3x 9y ¼ 0 or x 3y ¼ 0 Here v1 ¼ ð3; 1Þ is a nonzero solution. (ii) Subtract l ¼ 4 (or add 4) down the diagonal of A to obtain M ¼ 9 6 3 2 ; corresponding to 9x þ 6y ¼ 0 3x þ 2y ¼ 0 or 3x þ 2y ¼ 0 Here v2 ¼ ð2; 3Þ is a nonzero solution. Then S ¼ fv1; v2g ¼ fð3; 1Þ; ð2; 3Þg is a maximal set of linearly independent eigenvectors. Because S is a basis of R2, A is diagonalizable. Using the basis S, A is represented by the diagonal matrix D ¼ diagð7; 4Þ. (b) First ﬁnd the characteristic polynomial DðtÞ ¼ t2 þ 1. There are no real roots. Thus B, a real matrix representing a linear transformation on R2, has no eigenvalues and no eigenvectors. Hence, in particular, B is not diagonalizable. 310 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors (c) First ﬁnd DðtÞ ¼ t2 8t þ 16 ¼ ðt 4Þ2. Thus, l ¼ 4 is the only eigenvalue of C. Subtract l ¼ 4 down the diagonal of C to obtain M ¼ 1 1 1 1 ; corresponding to x y ¼ 0 The homogeneous system has only one independent solution; for example, x ¼ 1, y ¼ 1. Thus, v ¼ ð1; 1Þ is an eigenvector of C. Furthermore, as there are no other eigenvalues, the singleton set S ¼ fvg ¼ fð1; 1Þg is a maximal set of linearly independent eigenvectors of C. Furthermore, because S is not a basis of R2, C is not diagonalizable. 9.12. Suppose the matrix B in Problem 9.11 represents a linear operator on complex space C2. Show that, in this case, B is diagonalizable by ﬁnding a basis S of C2 consisting of eigenvectors of B. The characteristic polynomial of B is still DðtÞ ¼ t2 þ 1. As a polynomial over C, DðtÞ does factor; speciﬁcally, DðtÞ ¼ ðt iÞðt þ iÞ. Thus, l ¼ i and l ¼ i are the eigenvalues of B. (i) Subtract l ¼ i down the diagonal of B to obtain the homogeneous system ð1 iÞx y ¼ 0 2x þ ð1 iÞy ¼ 0 or ð1 iÞx y ¼ 0 The system has only one independent solution; for example, x ¼ 1, y ¼ 1 i. Thus, v1 ¼ ð1; 1 iÞ is an eigenvector that spans the eigenspace of l ¼ i. (ii) Subtract l ¼ i (or add i) down the diagonal of B to obtain the homogeneous system ð1 þ iÞx y ¼ 0 2x þ ð1 þ iÞy ¼ 0 or ð1 þ iÞx y ¼ 0 The system has only one independent solution; for example, x ¼ 1, y ¼ 1 þ i. Thus, v2 ¼ ð1; 1 þ iÞ is an eigenvector that spans the eigenspace of l ¼ i. As a complex matrix, B is diagonalizable. Speciﬁcally, S ¼ fv1; v2g ¼ fð1; 1 iÞ; ð1; 1 þ iÞg is a basis of C2 consisting of eigenvectors of B. Using this basis S, B is represented by the diagonal matrix D ¼ diagði; iÞ. 9.13. Let L be the linear transformation on R2 that reﬂects each point P across the line y ¼ kx, where k > 0. (See Fig. 9-1.) (a) Show that v1 ¼ ðk; 1Þ and v2 ¼ ð1; kÞ are eigenvectors of L. (b) Show that L is diagonalizable, and ﬁnd a diagonal representation D. (a) The vector v1 ¼ ðk; 1Þ lies on the line y ¼ kx, and hence is left ﬁxed by L; that is, Lðv1Þ ¼ v1. Thus, v1 is an eigenvector of L belonging to the eigenvalue l1 ¼ 1. The vector v2 ¼ ð1; kÞ is perpendicular to the line y ¼ kx, and hence, L reﬂects v2 into its negative; that is, Lðv2Þ ¼ v2. Thus, v2 is an eigenvector of L belonging to the eigenvalue l2 ¼ 1. y x 0 L P ( ) P L( ) v2 v2 y k = x Figure 9-1 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 311 (b) Here S ¼ fv1; v2g is a basis of R2 consisting of eigenvectors of L. Thus, L is diagonalizable, with the diagonal representation D ¼ 1 0 0 1 (relative to the basis S). Eigenvalues and Eigenvectors 9.14. Let A ¼ 4 1 1 2 5 2 1 1 2 2 4 3 5: (a) Find all eigenvalues of A. (b) Find a maximum set S of linearly independent eigenvectors of A. (c) Is A diagonalizable? If yes, ﬁnd P such that D ¼ P1AP is diagonal. (a) First ﬁnd the characteristic polynomial DðtÞ of A. We have trðAÞ ¼ 4 þ 5 þ 2 ¼ 11 and jAj ¼ 40 2 2 þ 5 þ 8 4 ¼ 45 Also, ﬁnd each cofactor Aii of aii in A: A11 ¼ 5 2 1 2 ¼ 12; A22 ¼ 4 1 1 2 ¼ 9; A33 ¼ 4 1 2 5 ¼ 18 Hence; DðtÞ ¼ t3 trðAÞ t2 þ ðA11 þ A22 þ A33Þt jAj ¼ t3 11t2 þ 39t 45 Assuming Dt has a rational root, it must be among 1, 3, 5, 9, 15, 45. Testing, by synthetic division, we get 3 1 11 þ 39 45 3 24 þ 45 1 8 þ 15 þ 0 Thus, t ¼ 3 is a root of DðtÞ. Also, t 3 is a factor and t2 8t þ 15 is a factor. Hence, DðtÞ ¼ ðt 3Þðt2 8t þ 15Þ ¼ ðt 3Þðt 5Þðt 3Þ ¼ ðt 3Þ2ðt 5Þ Accordingly, l ¼ 3 and l ¼ 5 are eigenvalues of A. (b) Find linearly independent eigenvectors for each eigenvalue of A. (i) Subtract l ¼ 3 down the diagonal of A to obtain the matrix M ¼ 1 1 1 2 2 2 1 1 1 2 4 3 5; corresponding to x þ y z ¼ 0 Here u ¼ ð1; 1; 0Þ and v ¼ ð1; 0; 1Þ are linearly independent solutions. (ii) Subtract l ¼ 5 down the diagonal of A to obtain the matrix M ¼ 1 1 1 2 0 2 1 1 3 2 4 3 5; corresponding to x þ y z ¼ 0 2x 2z ¼ 0 x þ y 3z ¼ 0 or x z ¼ 0 y 2z ¼ 0 Only z is a free variable. Here w ¼ ð1; 2; 1Þ is a solution. Thus, S ¼ fu; v; wg ¼ fð1; 1; 0Þ; ð1; 0; 1Þ; ð1; 2; 1Þg is a maximal set of linearly independent eigenvectors of A. Remark: The vectors u and v were chosen so that they were independent solutions of the system x þ y z ¼ 0. On the other hand, w is automatically independent of u and v because w belongs to a different eigenvalue of A. Thus, the three vectors are linearly independent. 312 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors (c) A is diagonalizable, because it has three linearly independent eigenvectors. Let P be the matrix with columns u; v; w. Then P ¼ 1 1 1 1 0 2 0 1 1 2 4 3 5 and D ¼ P1AP ¼ 3 3 5 2 4 3 5 9.15. Repeat Problem 9.14 for the matrix B ¼ 3 1 1 7 5 1 6 6 2 2 4 3 5. (a) First ﬁnd the characteristic polynomial DðtÞ of B. We have trðBÞ ¼ 0; jBj ¼ 16; B11 ¼ 4; B22 ¼ 0; B33 ¼ 8; so P i Bii ¼ 12 Therefore, DðtÞ ¼ t3 12t þ 16 ¼ ðt 2Þ2ðt þ 4Þ. Thus, l1 ¼ 2 and l2 ¼ 4 are the eigen-values of B. (b) Find a basis for the eigenspace of each eigenvalue of B. (i) Subtract l1 ¼ 2 down the diagonal of B to obtain M ¼ 1 1 1 7 7 1 6 6 0 2 4 3 5; corresponding to x y þ z ¼ 0 7x 7y þ z ¼ 0 6x 6y ¼ 0 or x y þ z ¼ 0 z ¼ 0 The system has only one independent solution; for example, x ¼ 1, y ¼ 1, z ¼ 0. Thus, u ¼ ð1; 1; 0Þ forms a basis for the eigenspace of l1 ¼ 2. (ii) Subtract l2 ¼ 4 (or add 4) down the diagonal of B to obtain M ¼ 7 1 1 7 1 1 6 6 6 2 4 3 5; corresponding to 7x y þ z ¼ 0 7x y þ z ¼ 0 6x 6y þ 6z ¼ 0 or x y þ z ¼ 0 6y 6z ¼ 0 The system has only one independent solution; for example, x ¼ 0, y ¼ 1, z ¼ 1. Thus, v ¼ ð0; 1; 1Þ forms a basis for the eigenspace of l2 ¼ 4. Thus S ¼ fu; vg is a maximal set of linearly independent eigenvectors of B. (c) Because B has at most two linearly independent eigenvectors, B is not similar to a diagonal matrix; that is, B is not diagonalizable. 9.16. Find the algebraic and geometric multiplicities of the eigenvalue l1 ¼ 2 of the matrix B in Problem 9.15. The algebraic multiplicity of l1 ¼ 2 is 2, because t 2 appears with exponent 2 in DðtÞ. However, the geometric multiplicity of l1 ¼ 2 is 1, because dim El1 ¼ 1 (where El1 is the eigenspace of l1). 9.17. Let T: R3 ! R3 be deﬁned by Tðx; y; zÞ ¼ ð2x þ y 2z; 2x þ 3y 4z; x þ y zÞ. Find all eigenvalues of T, and ﬁnd a basis of each eigenspace. Is T diagonalizable? If so, ﬁnd the basis S of R3 that diagonalizes T; and ﬁnd its diagonal representation D. First ﬁnd the matrix A that represents T relative to the usual basis of R3 by writing down the coefﬁcients of x; y; z as rows, and then ﬁnd the characteristic polynomial of A (and T). We have A ¼ ½T ¼ 2 1 2 2 3 4 1 1 1 2 4 3 5 and trðAÞ ¼ 4; jAj ¼ 2 A11 ¼ 1; A22 ¼ 0; A33 ¼ 4 P i Aii ¼ 5 Therefore, DðtÞ ¼ t3 4t2 þ 5t 2 ¼ ðt 1Þ2ðt 2Þ, and so l ¼ 1 and l ¼ 2 are the eigenvalues of A (and T). We next ﬁnd linearly independent eigenvectors for each eigenvalue of A. CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 313 (i) Subtract l ¼ 1 down the diagonal of A to obtain the matrix M ¼ 1 1 2 2 2 4 1 1 2 2 4 3 5; corresponding to x þ y 2z ¼ 0 Here y and z are free variables, and so there are two linearly independent eigenvectors belonging to l ¼ 1. For example, u ¼ ð1; 1; 0Þ and v ¼ ð2; 0; 1Þ are two such eigenvectors. (ii) Subtract l ¼ 2 down the diagonal of A to obtain M ¼ 0 1 2 2 1 4 1 1 3 2 4 3 5; corresponding to y 2z ¼ 0 2x þ y 4z ¼ 0 x þ y 3z ¼ 0 or x þ y 3z ¼ 0 y 2z ¼ 0 Only z is a free variable. Here w ¼ ð1; 2; 1Þ is a solution. Thus, T is diagonalizable, because it has three independent eigenvectors. Speciﬁcally, choosing S ¼ fu; v; wg ¼ fð1; 1; 0Þ; ð2; 0; 1Þ; ð1; 2; 1Þg as a basis, T is represented by the diagonal matrix D ¼ diagð1; 1; 2Þ. 9.18. Prove the following for a linear operator (matrix) T: (a) The scalar 0 is an eigenvalue of T if and only if T is singular. (b) If l is an eigenvalue of T, where T is invertible, then l1 is an eigenvalue of T1. (a) We have that 0 is an eigenvalue of T if and only if there is a vector v 6¼ 0 such that TðvÞ ¼ 0v—that is, if and only if T is singular. (b) Because T is invertible, it is nonsingular; hence, by (a), l 6¼ 0. By deﬁnition of an eigenvalue, there exists v 6¼ 0 such that TðvÞ ¼ lv. Applying T 1 to both sides, we obtain v ¼ T1ðlvÞ ¼ lT 1ðvÞ; and so T 1ðvÞ ¼ l1v Therefore, l1 is an eigenvalue of T1. 9.19. Let l be an eigenvalue of a linear operator T: V ! V, and let El consists of all the eigenvectors belonging to l (called the eigenspace of l). Prove that El is a subspace of V. That is, prove (a) If u 2 El, then ku 2 El for any scalar k. (b) If u; v; 2 El, then u þ v 2 El. (a) Because u 2 El, we have TðuÞ ¼ lu. Then TðkuÞ ¼ kTðuÞ ¼ kðluÞ ¼ lðkuÞ; and so ku 2 El: (We view the zero vector 0 2 V as an ‘‘eigenvector’’ of l in order for El to be a subspace of V.) (b) As u; v 2 El, we have TðuÞ ¼ lu and TðvÞ ¼ lv. Then Tðu þ vÞ ¼ TðuÞ þ TðvÞ ¼ lu þ lv ¼ lðu þ vÞ; and so u þ v 2 El 9.20. Prove Theorem 9.6: The following are equivalent: (i) The scalar l is an eigenvalue of A. (ii) The matrix lI A is singular. (iii) The scalar l is a root of the characteristic polynomial DðtÞ of A. The scalar l is an eigenvalue of A if and only if there exists a nonzero vector v such that Av ¼ lv or ðlIÞv Av ¼ 0 or ðlI AÞv ¼ 0 or lI A is singular. In such a case, l is a root of DðtÞ ¼ jtI Aj. Also, v is in the eigenspace El of l if and only if the above relations hold. Hence, v is a solution of ðlI AÞX ¼ 0. 9.21. Prove Theorem 9.80: Suppose v1; v2; . . . ; vn are nonzero eigenvectors of T belonging to distinct eigenvalues l1; l2; . . . ; ln. Then v1; v2; . . . ; vn are linearly independent. 314 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors Suppose the theorem is not true. Let v1; v2; . . . ; vs be a minimal set of vectors for which the theorem is not true. We have s > 1, because v1 6¼ 0. Also, by the minimality condition, v2; . . . ; vs are linearly independent. Thus, v1 is a linear combination of v2; . . . ; vs, say, v1 ¼ a2v2 þ a3v3 þ þ asvs ð1Þ (where some ak 6¼ 0Þ. Applying T to (1) and using the linearity of T yields Tðv1Þ ¼ Tða2v2 þ a3v3 þ þ asvsÞ ¼ a2Tðv2Þ þ a3Tðv3Þ þ þ asTðvsÞ ð2Þ Because vj is an eigenvector of T belonging to lj, we have TðvjÞ ¼ ljvj. Substituting in (2) yields l1v1 ¼ a2l2v2 þ a3l3v3 þ þ aslsvs ð3Þ Multiplying (1) by l1 yields l1v1 ¼ a2l1v2 þ a3l1v3 þ þ asl1vs ð4Þ Setting the right-hand sides of (3) and (4) equal to each other, or subtracting (3) from (4) yields a2ðl1 l2Þv2 þ a3ðl1 l3Þv3 þ þ asðl1 lsÞvs ¼ 0 ð5Þ Because v2; v3; . . . ; vs are linearly independent, the coefﬁcients in (5) must all be zero. That is, a2ðl1 l2Þ ¼ 0; a3ðl1 l3Þ ¼ 0; . . . ; asðl1 lsÞ ¼ 0 However, the li are distinct. Hence l1 lj 6¼ 0 for j > 1. Hence, a2 ¼ 0, a3 ¼ 0; . . . ; as ¼ 0. This contradicts the fact that some ak 6¼ 0. The theorem is proved. 9.22. Prove Theorem 9.9. Suppose DðtÞ ¼ ðt a1Þðt a2Þ . . . ðt anÞ is the characteristic polynomial of an n-square matrix A, and suppose the n roots ai are distinct. Then A is similar to the diagonal matrix D ¼ diagða1; a2; . . . ; anÞ. Let v1; v2; . . . ; vn be (nonzero) eigenvectors corresponding to the eigenvalues ai. Then the n eigenvectors vi are linearly independent (Theorem 9.8), and hence form a basis of Kn. Accordingly, A is diagonalizable (i.e., A is similar to a diagonal matrix D), and the diagonal elements of D are the eigenvalues ai. 9.23. Prove Theorem 9.100: The geometric multiplicity of an eigenvalue l of T does not exceed its algebraic multiplicity. Suppose the geometric multiplicity of l is r. Then its eigenspace El contains r linearly independent eigenvectors v1; . . . ; vr. Extend the set fvig to a basis of V, say, fvi; . . . ; vr; w1; . . . ; wsg. We have Tðv1Þ ¼ lv1; Tðv2Þ ¼ lv2; . . . ; TðvrÞ ¼ lvr; Tðw1Þ ¼ a11v1 þ þ a1rvr þ b11w1 þ þ b1sws Tðw2Þ ¼ a21v1 þ þ a2rvr þ b21w1 þ þ b2sws :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: TðwsÞ ¼ as1v1 þ þ asrvr þ bs1w1 þ þ bssws Then M ¼ lIr A 0 B is the matrix of T in the above basis, where A ¼ ½aijT and B ¼ ½bijT: Because M is block diagonal, the characteristic polynomial ðt lÞr of the block lIr must divide the characteristic polynomial of M and hence of T. Thus, the algebraic multiplicity of l for T is at least r, as required. Diagonalizing Real Symmetric Matrices and Quadratic Forms 9.24. Let A ¼ 7 3 3 1 . Find an orthogonal matrix P such that D ¼ P1AP is diagonal. CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 315 First ﬁnd the characteristic polynomial DðtÞ of A. We have DðtÞ ¼ t2 trðAÞ t þ jAj ¼ t2 6t 16 ¼ ðt 8Þðt þ 2Þ Thus, the eigenvalues of A are l ¼ 8 and l ¼ 2. We next ﬁnd corresponding eigenvectors. Subtract l ¼ 8 down the diagonal of A to obtain the matrix M ¼ 1 3 3 9 ; corresponding to x þ 3y ¼ 0 3x 9y ¼ 0 or x 3y ¼ 0 A nonzero solution is u1 ¼ ð3; 1Þ. Subtract l ¼ 2 (or add 2) down the diagonal of A to obtain the matrix M ¼ 9 3 3 1 ; corresponding to 9x þ 3y ¼ 0 3x þ y ¼ 0 or 3x þ y ¼ 0 A nonzero solution is u2 ¼ ð1; 3Þ. As expected, because A is symmetric, the eigenvectors u1 and u2 are orthogonal. Normalize u1 and u2 to obtain, respectively, the unit vectors ^ u1 ¼ ð3= ﬃﬃﬃﬃﬃ 10 p ; 1= ﬃﬃﬃﬃﬃ 10 p Þ and ^ u2 ¼ ð1= ﬃﬃﬃﬃﬃ 10 p ; 3= ﬃﬃﬃﬃﬃ 10 p Þ: Finally, let P be the matrix whose columns are the unit vectors ^ u1 and ^ u2, respectively. Then P ¼ 3= ﬃﬃﬃﬃﬃ 10 p 1= ﬃﬃﬃﬃﬃ 10 p 1= ﬃﬃﬃﬃﬃ 10 p 3= ﬃﬃﬃﬃﬃ 10 p " # and D ¼ P1AP ¼ 8 0 0 2 As expected, the diagonal entries in D are the eigenvalues of A. 9.25. Let B ¼ 11 8 4 8 1 2 4 2 4 2 4 3 5. (a) Find all eigenvalues of B. (b) Find a maximal set S of nonzero orthogonal eigenvectors of B. (c) Find an orthogonal matrix P such that D ¼ P1BP is diagonal. (a) First ﬁnd the characteristic polynomial of B. We have trðBÞ ¼ 6; jBj ¼ 400; B11 ¼ 0; B22 ¼ 60; B33 ¼ 75; so P i Bii ¼ 135 Hence, DðtÞ ¼ t3 6t2 135t 400. If DðtÞ has an integer root it must divide 400. Testing t ¼ 5, by synthetic division, yields 5 1 6 135 400 5 þ 55 þ 400 1 11 80 þ 0 Thus, t þ 5 is a factor of DðtÞ, and t2 11t 80 is a factor. Thus, DðtÞ ¼ ðt þ 5Þðt2 11t 80Þ ¼ ðt þ 5Þ2ðt 16Þ The eigenvalues of B are l ¼ 5 (multiplicity 2), and l ¼ 16 (multiplicity 1). (b) Find an orthogonal basis for each eigenspace. Subtract l ¼ 5 (or, add 5) down the diagonal of B to obtain the homogeneous system 16x 8y þ 4z ¼ 0; 8x þ 4y 2z ¼ 0; 4x 2y þ z ¼ 0 That is, 4x 2y þ z ¼ 0. The system has two independent solutions. One solution is v1 ¼ ð0; 1; 2Þ. We seek a second solution v2 ¼ ða; b; cÞ, which is orthogonal to v1, such that 4a 2b þ c ¼ 0; and also b 2c ¼ 0 316 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors One such solution is v2 ¼ ð5; 8; 4Þ. Subtract l ¼ 16 down the diagonal of B to obtain the homogeneous system 5x 8y þ 4z ¼ 0; 8x 17y 2z ¼ 0; 4x 2y 20z ¼ 0 This system yields a nonzero solution v3 ¼ ð4; 2; 1Þ. (As expected from Theorem 9.13, the eigenvector v3 is orthogonal to v1 and v2.) Then v1; v2; v3 form a maximal set of nonzero orthogonal eigenvectors of B. (c) Normalize v1; v2; v3 to obtain the orthonormal basis: ^ v1 ¼ v1= ﬃﬃﬃ 5 p ; ^ v2 ¼ v2= ﬃﬃﬃﬃﬃﬃﬃ ﬃ 105 p ; ^ v3 ¼ v3= ﬃﬃﬃﬃﬃ 21 p Then P is the matrix whose columns are ^ v1; ^ v2; ^ v3. Thus, P ¼ 0 5= ﬃﬃﬃﬃﬃﬃﬃ ﬃ 105 p 4= ﬃﬃﬃﬃﬃ 21 p 1= ﬃﬃﬃ 5 p 8= ﬃﬃﬃﬃﬃﬃﬃ ﬃ 105 p 2= ﬃﬃﬃﬃﬃ 21 p 2= ﬃﬃﬃ 5 p 4= ﬃﬃﬃﬃﬃﬃﬃ ﬃ 105 p 1= ﬃﬃﬃﬃﬃ 21 p 2 6 4 3 7 5 and D ¼ P1BP ¼ 5 5 16 2 6 4 3 7 5 9.26. Let qðx; yÞ ¼ x2 þ 6xy 7y2. Find an orthogonal substitution that diagonalizes q. Find the symmetric matrix A that represents q and its characteristic polynomial DðtÞ. We have A ¼ 1 3 3 7 and DðtÞ ¼ t2 þ 6t 16 ¼ ðt 2Þðt þ 8Þ The eigenvalues of A are l ¼ 2 and l ¼ 8. Thus, using s and t as new variables, a diagonal form of q is qðs; tÞ ¼ 2s2 8t2 The corresponding orthogonal substitution is obtained by ﬁnding an orthogonal set of eigenvectors of A. (i) Subtract l ¼ 2 down the diagonal of A to obtain the matrix M ¼ 1 3 3 9 ; corresponding to x þ 3y ¼ 0 3x 9y ¼ 0 or x þ 3y ¼ 0 A nonzero solution is u1 ¼ ð3; 1Þ. (ii) Subtract l ¼ 8 (or add 8) down the diagonal of A to obtain the matrix M ¼ 9 3 3 1 ; corresponding to 9x þ 3y ¼ 0 3x þ y ¼ 0 or 3x þ y ¼ 0 A nonzero solution is u2 ¼ ð1; 3Þ. As expected, because A is symmetric, the eigenvectors u1 and u2 are orthogonal. Now normalize u1 and u2 to obtain, respectively, the unit vectors ^ u1 ¼ ð3= ﬃﬃﬃﬃﬃ 10 p ; 1= ﬃﬃﬃﬃﬃ 10 p Þ and ^ u2 ¼ ð1= ﬃﬃﬃﬃﬃ 10 p ; 3= ﬃﬃﬃﬃﬃ 10 p Þ: Finally, let P be the matrix whose columns are the unit vectors ^ u1 and ^ u2, respectively, and then ½x; yT ¼ P½s; tT is the required orthogonal change of coordinates. That is, P ¼ 3= ﬃﬃﬃﬃﬃ 10 p 1= ﬃﬃﬃﬃﬃ 10 p 1= ﬃﬃﬃﬃﬃ 10 p 3= ﬃﬃﬃﬃﬃ 10 p # and x ¼ 3s t ﬃﬃﬃﬃﬃ 10 p ; y ¼ s þ 3t ﬃﬃﬃﬃﬃ 10 p One can also express s and t in terms of x and y by using P1 ¼ PT. That is, s ¼ 3x þ y ﬃﬃﬃﬃﬃ 10 p ; t ¼ x þ 3t ﬃﬃﬃﬃﬃ 10 p CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 317 Minimal Polynomial 9.27. Let A ¼ 4 2 2 6 3 4 3 2 3 2 4 3 5 and B ¼ 3 2 2 4 4 6 2 3 5 2 4 3 5. The characteristic polynomial of both matrices is DðtÞ ¼ ðt 2Þðt 1Þ2. Find the minimal polynomial mðtÞ of each matrix. The minimal polynomial mðtÞ must divide DðtÞ. Also, each factor of DðtÞ (i.e., t 2 and t 1) must also be a factor of mðtÞ. Thus, mðtÞ must be exactly one of the following: f ðtÞ ¼ ðt 2Þðt 1Þ or gðtÞ ¼ ðt 2Þðt 1Þ2 (a) By the Cayley–Hamilton theorem, gðAÞ ¼ DðAÞ ¼ 0, so we need only test f ðtÞ. We have f ðAÞ ¼ ðA 2IÞðA IÞ ¼ 2 2 2 6 5 4 3 2 1 2 4 3 5 3 2 2 6 4 4 3 2 2 2 4 3 5 ¼ 0 0 0 0 0 0 0 0 0 2 4 3 5 Thus, mðtÞ ¼ f ðtÞ ¼ ðt 2Þðt 1Þ ¼ t2 3t þ 2 is the minimal polynomial of A. (b) Again gðBÞ ¼ DðBÞ ¼ 0, so we need only test f ðtÞ. We get f ðBÞ ¼ ðB 2IÞðB IÞ ¼ 1 2 2 4 6 6 2 3 3 2 4 3 5 2 2 2 4 5 6 2 3 4 2 4 3 5 ¼ 2 2 2 4 4 4 2 2 2 2 4 3 5 6¼ 0 Thus, mðtÞ 6¼ f ðtÞ. Accordingly, mðtÞ ¼ gðtÞ ¼ ðt 2Þðt 1Þ2 is the minimal polynomial of B. [We emphasize that we do not need to compute gðBÞ; we know gðBÞ ¼ 0 from the Cayley–Hamilton theorem.] 9.28. Find the minimal polynomial mðtÞ of each of the following matrices: (a) A ¼ 5 1 3 7 , (b) B ¼ 1 2 3 0 2 3 0 0 3 2 4 3 5, (c) C ¼ 4 1 1 2 (a) The characteristic polynomial of A is DðtÞ ¼ t2 12t þ 32 ¼ ðt 4Þðt 8Þ. Because DðtÞ has distinct factors, the minimal polynomial mðtÞ ¼ DðtÞ ¼ t2 12t þ 32. (b) Because B is triangular, its eigenvalues are the diagonal elements 1; 2; 3; and so its characteristic polynomial is DðtÞ ¼ ðt 1Þðt 2Þðt 3Þ. Because DðtÞ has distinct factors, mðtÞ ¼ DðtÞ. (c) The characteristic polynomial of C is DðtÞ ¼ t2 6t þ 9 ¼ ðt 3Þ2. Hence the minimal polynomial of C is f ðtÞ ¼ t 3 or gðtÞ ¼ ðt 3Þ2. However, f ðCÞ 6¼ 0; that is, C 3I 6¼ 0. Hence, mðtÞ ¼ gðtÞ ¼ DðtÞ ¼ ðt 3Þ2: 9.29. Suppose S ¼ fu1; u2; . . . ; ung is a basis of V, and suppose F and G are linear operators on V such that ½F has 0’s on and below the diagonal, and ½G has a 6¼ 0 on the superdiagonal and 0’s elsewhere. That is, ½F ¼ 0 a21 a31 . . . an1 0 0 a32 . . . an2 :::::::::::::::::::::::::::::::::::::::: 0 0 0 . . . an;n1 0 0 0 . . . 0 2 6 6 6 6 4 3 7 7 7 7 5 ; ½G ¼ 0 a 0 . . . 0 0 0 a . . . 0 ::::::::::::::::::::::::::: 0 0 0 . . . a 0 0 0 . . . 0 2 6 6 6 6 4 3 7 7 7 7 5 Show that (a) Fn ¼ 0, (b) Gn1 6¼ 0, but Gn ¼ 0. (These conditions also hold for ½F and ½G.) (a) We have Fðu1Þ ¼ 0 and, for r > 1, FðurÞ is a linear combination of vectors preceding ur in S. That is, FðurÞ ¼ ar1u1 þ ar2u2 þ þ ar;r1ur1 318 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors Hence, F2ðurÞ ¼ FðFðurÞÞ is a linear combination of vectors preceding ur1, and so on. Hence, FrðurÞ ¼ 0 for each r. Thus, for each r, FnðurÞ ¼ Fnrð0Þ ¼ 0, and so Fn ¼ 0, as claimed. (b) We have Gðu1Þ ¼ 0 and, for each k > 1, GðukÞ ¼ auk1. Hence, GrðukÞ ¼ arukr for r < k. Because a 6¼ 0, an1 6¼ 0. Therefore, Gn1ðunÞ ¼ an1u1 6¼ 0, and so Gn1 6¼ 0. On the other hand, by (a), Gn ¼ 0. 9.30. Let B be the matrix in Example 9.12(a) that has 1’s on the diagonal, a’s on the superdiagonal, where a 6¼ 0, and 0’s elsewhere. Show that f ðtÞ ¼ ðt lÞn is both the characteristic polynomial DðtÞ and the minimum polynomial mðtÞ of A. Because A is triangular with l’s on the diagonal, DðtÞ ¼ f ðtÞ ¼ ðt lÞn is its characteristic polynomial. Thus, mðtÞ is a power of t l. By Problem 9.29, ðA lIÞr1 6¼ 0. Hence, mðtÞ ¼ DðtÞ ¼ ðt lÞn. 9.31. Find the characteristic polynomial DðtÞ and minimal polynomial mðtÞ of each matrix: (a) M ¼ 4 1 0 0 0 0 4 1 0 0 0 0 4 0 0 0 0 0 4 1 0 0 0 0 4 2 6 6 6 6 4 3 7 7 7 7 5 , (b) M0 ¼ 2 7 0 0 0 2 0 0 0 0 1 1 0 0 2 4 2 6 6 4 3 7 7 5 (a) M is block diagonal with diagonal blocks A ¼ 4 1 0 0 4 1 0 0 4 2 4 3 5 and B ¼ 4 1 0 4 The characteristic and minimal polynomial of A is f ðtÞ ¼ ðt 4Þ3 and the characteristic and minimal polynomial of B is gðtÞ ¼ ðt 4Þ2. Then DðtÞ ¼ f ðtÞgðtÞ ¼ ðt 4Þ5 but mðtÞ ¼ LCM½ f ðtÞ; gðtÞ ¼ ðt 4Þ3 (where LCM means least common multiple). We emphasize that the exponent in mðtÞ is the size of the largest block. (b) Here M0 is block diagonal with diagonal blocks A0 ¼ 2 7 0 2 and B0 ¼ 1 1 2 4 The char-acteristic and minimal polynomial of A0 is f ðtÞ ¼ ðt 2Þ2. The characteristic polynomial of B0 is gðtÞ ¼ t2 5t þ 6 ¼ ðt 2Þðt 3Þ, which has distinct factors. Hence, gðtÞ is also the minimal polynomial of B. Accordingly, DðtÞ ¼ f ðtÞgðtÞ ¼ ðt 2Þ3ðt 3Þ but mðtÞ ¼ LCM½ f ðtÞ; gðtÞ ¼ ðt 2Þ2ðt 3Þ 9.32. Find a matrix A whose minimal polynomial is f ðtÞ ¼ t3 8t2 þ 5t þ 7. Simply let A ¼ 0 0 7 1 0 5 0 1 8 2 4 3 5, the companion matrix of f ðtÞ [deﬁned in Example 9.12(b)]. 9.33. Prove Theorem 9.15: The minimal polynomial mðtÞ of a matrix (linear operator) A divides every polynomial that has A as a zero. In particular (by the Cayley–Hamilton theorem), mðtÞ divides the characteristic polynomial DðtÞ of A. Suppose f ðtÞ is a polynomial for which f ðAÞ ¼ 0. By the division algorithm, there exist polynomials qðtÞ and rðtÞ for which f ðtÞ ¼ mðtÞqðtÞ þ rðtÞ and rðtÞ ¼ 0 or deg rðtÞ < deg mðtÞ. Substituting t ¼ A in this equation, and using that f ðAÞ ¼ 0 and mðAÞ ¼ 0, we obtain rðAÞ ¼ 0. If rðtÞ 6¼ 0, then rðtÞ is a polynomial of degree less than mðtÞ that has A as a zero. This contradicts the deﬁnition of the minimal polynomial. Thus, rðtÞ ¼ 0, and so f ðtÞ ¼ mðtÞqðtÞ; that is, mðtÞ divides f ðtÞ. CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 319 9.34. Let mðtÞ be the minimal polynomial of an n-square matrix A. Prove that the characteristic polynomial DðtÞ of A divides ½mðtÞn. Suppose mðtÞ ¼ tr þ c1tr1 þ þ cr1t þ cr. Deﬁne matrices Bj as follows: B0 ¼ I B1 ¼ A þ c1I B2 ¼ A2 þ c1A þ c2I Br1 ¼ Ar1 þ c1Ar2 þ þ cr1I so so so so I ¼ B0 c1I ¼ B1 A ¼ B1 AB0 c2I ¼ B2 AðA þ c1IÞ ¼ B2 AB1 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: cr1I ¼ Br1 ABr2 Then ABr1 ¼ crI ðAr þ c1Ar1 þ þ cr1A þ crIÞ ¼ crI mðAÞ ¼ crI Set BðtÞ ¼ tr1B0 þ tr2B1 þ þ tBr2 þ Br1 Then ðtI AÞBðtÞ ¼ ðtrB0 þ tr1B1 þ þ tBr1Þ ðtr1AB0 þ tr2AB1 þ þ ABr1Þ ¼ trB0 þ tr1ðB1 AB0Þ þ tr2ðB2 AB1Þ þ þ tðBr1 ABr2Þ ABr1 ¼ trI þ c1tr1I þ c2tr2I þ þ cr1tI þ crI ¼ mðtÞI Taking the determinant of both sides gives jtI AjjBðtÞj ¼ jmðtÞIj ¼ ½mðtÞn. Because jBðtÞj is a poly-nomial, jtI Aj divides ½mðtÞn; that is, the characteristic polynomial of A divides ½mðtÞn. 9.35. Prove Theorem 9.16: The characteristic polynomial DðtÞ and the minimal polynomial mðtÞ of A have the same irreducible factors. Suppose f ðtÞ is an irreducible polynomial. If f ðtÞ divides mðtÞ, then f ðtÞ also divides DðtÞ [because mðtÞ divides DðtÞ. On the other hand, if f ðtÞ divides DðtÞ, then by Problem 9.34, f ðtÞ also divides ½mðtÞn. But f ðtÞ is irreducible; hence, f ðtÞ also divides mðtÞ. Thus, mðtÞ and DðtÞ have the same irreducible factors. 9.36. Prove Theorem 9.19: The minimal polynomial mðtÞ of a block diagonal matrix M with diagonal blocks Ai is equal to the least common multiple (LCM) of the minimal polynomials of the diagonal blocks Ai. We prove the theorem for the case r ¼ 2. The general theorem follows easily by induction. Suppose M ¼ A 0 0 B , where A and B are square matrices. We need to show that the minimal polynomial mðtÞ of M is the LCM of the minimal polynomials gðtÞ and hðtÞ of A and B, respectively. Because mðtÞ is the minimal polynomial of M; mðMÞ ¼ mðAÞ 0 0 mðBÞ ¼ 0, and mðAÞ ¼ 0 and mðBÞ ¼ 0. Because gðtÞ is the minimal polynomial of A, gðtÞ divides mðtÞ. Similarly, hðtÞ divides mðtÞ. Thus mðtÞ is a multiple of gðtÞ and hðtÞ. Now let f ðtÞ be another multiple of gðtÞ and hðtÞ. Then f ðMÞ ¼ f ðAÞ 0 0 f ðBÞ ¼ 0 0 0 0 ¼ 0. But mðtÞ is the minimal polynomial of M; hence, mðtÞ divides f ðtÞ. Thus, mðtÞ is the LCM of gðtÞ and hðtÞ. 9.37. Suppose mðtÞ ¼ tr þ ar1tr1 þ þ a1t þ a0 is the minimal polynomial of an n-square matrix A. Prove the following: (a) A is nonsingular if and only if the constant term a0 6¼ 0. (b) If A is nonsingular, then A1 is a polynomial in A of degree r 1 < n. (a) The following are equivalent: (i) A is nonsingular, (ii) 0 is not a root of mðtÞ, (iii) a0 6¼ 0. Thus, the statement is true. 320 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors (b) Because A is nonsingular, a0 6¼ 0 by (a). We have mðAÞ ¼ Ar þ ar1Ar1 þ þ a1A þ a0I ¼ 0 Thus; 1 a0 ðAr1 þ ar1Ar2 þ þ a1IÞA ¼ I Accordingly; A1 ¼ 1 a0 ðAr1 þ ar1Ar2 þ þ a1IÞ SUPPLEMENTARY PROBLEMS Polynomials of Matrices 9.38. Let A ¼ 2 3 5 1 and B ¼ 1 2 0 3 . Find f ðAÞ, gðAÞ, f ðBÞ, gðBÞ, where f ðtÞ ¼ 2t2 5t þ 6 and gðtÞ ¼ t3 2t2 þ t þ 3. 9.39. Let A ¼ 1 2 0 1 . Find A2, A3, An, where n > 3, and A1. 9.40. Let B ¼ 8 12 0 0 8 12 0 0 8 2 4 3 5. Find a real matrix A such that B ¼ A3. 9.41. For each matrix, ﬁnd a polynomial having the following matrix as a root: (a) A ¼ 2 5 1 3 , (b) B ¼ 2 3 7 4 , (c) C ¼ 1 1 2 1 2 3 2 1 4 2 4 3 5 9.42. Let A be any square matrix and let f ðtÞ be any polynomial. Prove (a) ðP1APÞn ¼ P1AnP. (b) f ðP1APÞ ¼ P1f ðAÞP. (c) f ðATÞ ¼ ½ f ðAÞT. (d) If A is symmetric, then f ðAÞ is symmetric. 9.43. Let M ¼ diag½A1; . . . ; Ar be a block diagonal matrix, and let f ðtÞ be any polynomial. Show that f ðMÞ is block diagonal and f ðMÞ ¼ diag½ f ðA1Þ; . . . ; f ðArÞ: 9.44. Let M be a block triangular matrix with diagonal blocks A1; . . . ; Ar, and let f ðtÞ be any polynomial. Show that f ðMÞ is also a block triangular matrix, with diagonal blocks f ðA1Þ; . . . ; f ðArÞ. Eigenvalues and Eigenvectors 9.45. For each of the following matrices, ﬁnd all eigenvalues and corresponding linearly independent eigen-vectors: (a) A ¼ 2 3 2 5 , (b) B ¼ 2 4 1 6 , (c) C ¼ 1 4 3 7 When possible, ﬁnd the nonsingular matrix P that diagonalizes the matrix. 9.46. Let A ¼ 2 1 2 3 . (a) Find eigenvalues and corresponding eigenvectors. (b) Find a nonsingular matrix P such that D ¼ P1AP is diagonal. (c) Find A8 and f ðAÞ where f ðtÞ ¼ t4 5t3 þ 7t2 2t þ 5. (d) Find a matrix B such that B2 ¼ A. CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 321 9.47. Repeat Problem 9.46 for A ¼ 5 6 2 2 . 9.48. For each of the following matrices, ﬁnd all eigenvalues and a maximum set S of linearly independent eigenvectors: (a) A ¼ 1 3 3 3 5 3 6 6 4 2 4 3 5, (b) B ¼ 3 1 1 7 5 1 6 6 2 2 4 3 5, (c) C ¼ 1 2 2 1 2 1 1 1 4 2 4 3 5 Which matrices can be diagonalized, and why? 9.49. For each of the following linear operators T: R2 ! R2, ﬁnd all eigenvalues and a basis for each eigenspace: (a) Tðx; yÞ ¼ ð3x þ 3y; x þ 5yÞ, (b) Tðx; yÞ ¼ ð3x 13y; x 3yÞ. 9.50. Let A ¼ a b c d be a real matrix. Find necessary and sufﬁcient conditions on a; b; c; d so that A is diagonalizable—that is, so that A has two (real) linearly independent eigenvectors. 9.51. Show that matrices A and AT have the same eigenvalues. Give an example of a 2 2 matrix A where A and AT have different eigenvectors. 9.52. Suppose v is an eigenvector of linear operators F and G. Show that v is also an eigenvector of the linear operator kF þ k0G, where k and k0 are scalars. 9.53. Suppose v is an eigenvector of a linear operator T belonging to the eigenvalue l. Prove (a) For n > 0; v is an eigenvector of Tn belonging to ln. (b) f ðlÞ is an eigenvalue of f ðTÞ for any polynomial f ðtÞ. 9.54. Suppose l 6¼ 0 is an eigenvalue of the composition F G of linear operators F and G. Show that l is also an eigenvalue of the composition G F. [Hint: Show that GðvÞ is an eigenvector of G F.] 9.55. Let E: V ! V be a projection mapping; that is, E2 ¼ E. Show that E is diagonalizable and, in fact, can be represented by the diagonal matrix M ¼ Ir 0 0 0 , where r is the rank of E. Diagonalizing Real Symmetric Matrices and Quadratic Forms 9.56. For each of the following symmetric matrices A, ﬁnd an orthogonal matrix P and a diagonal matrix D such that D ¼ P1AP: (a) A ¼ 5 4 4 1 , (b) A ¼ 4 1 1 4 , (c) A ¼ 7 3 3 1 9.57. For each of the following symmetric matrices B, ﬁnd its eigenvalues, a maximal orthogonal set S of eigenvectors, and an orthogonal matrix P such that D ¼ P1BP is diagonal: (a) B ¼ 0 1 1 1 0 1 1 1 0 2 4 3 5, (b) B ¼ 2 2 4 2 5 8 4 8 17 2 4 3 5 9.58. Using variables s and t, ﬁnd an orthogonal substitution that diagonalizes each of the following quadratic forms: (a) qðx; yÞ ¼ 4x2 þ 8xy 11y2, (b) qðx; yÞ ¼ 2x2 6xy þ 10y2 9.59. For each of the following quadratic forms qðx; y; zÞ, ﬁnd an orthogonal substitution expressing x; y; z in terms of variables r; s; t, and ﬁnd qðr; s; tÞ: (a) qðx; y; zÞ ¼ 5x2 þ 3y2 þ 12xz; (b) qðx; y; zÞ ¼ 3x2 4xy þ 6y2 þ 2xz 4yz þ 3z2 322 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 9.60. Find a real 2 2 symmetric matrix A with eigenvalues: (a) l ¼ 1 and l ¼ 4 and eigenvector u ¼ ð1; 1Þ belonging to l ¼ 1; (b) l ¼ 2 and l ¼ 3 and eigenvector u ¼ ð1; 2Þ belonging to l ¼ 2. In each case, ﬁnd a matrix B for which B2 ¼ A. Characteristic and Minimal Polynomials 9.61. Find the characteristic and minimal polynomials of each of the following matrices: (a) A ¼ 3 1 1 2 4 2 1 1 3 2 4 3 5, (b) B ¼ 3 2 1 3 8 3 3 6 1 2 4 3 5 9.62. Find the characteristic and minimal polynomials of each of the following matrices: (a) A ¼ 2 5 0 0 0 0 2 0 0 0 0 0 4 2 0 0 0 3 5 0 0 0 0 0 7 2 6 6 6 6 4 3 7 7 7 7 5 , (b) B ¼ 4 1 0 0 0 1 2 0 0 0 0 0 3 1 0 0 0 0 3 1 0 0 0 0 3 2 6 6 6 6 4 3 7 7 7 7 5 , (c) C ¼ 3 2 0 0 0 1 4 0 0 0 0 0 3 1 0 0 0 1 3 0 0 0 0 0 4 2 6 6 6 6 4 3 7 7 7 7 5 9.63. Let A ¼ 1 1 0 0 2 0 0 0 1 2 4 3 5 and B ¼ 2 0 0 0 2 2 0 0 1 2 4 3 5. Show that A and B have different characteristic polynomials (and so are not similar) but have the same minimal polynomial. Thus, nonsimilar matrices may have the same minimal polynomial. 9.64. Let A be an n-square matrix for which Ak ¼ 0 for some k > n. Show that An ¼ 0. 9.65. Show that a matrix A and its transpose AT have the same minimal polynomial. 9.66. Suppose f ðtÞ is an irreducible monic polynomial for which f ðAÞ ¼ 0 for a matrix A. Show that f ðtÞ is the minimal polynomial of A. 9.67. Show that A is a scalar matrix kI if and only if the minimal polynomial of A is mðtÞ ¼ t k. 9.68. Find a matrix A whose minimal polynomial is (a) t3 5t2 þ 6t þ 8, (b) t4 5t3 2t þ 7t þ 4. 9.69. Let f ðtÞ and gðtÞ be monic polynomials (leading coefﬁcient one) of minimal degree for which A is a root. Show f ðtÞ ¼ gðtÞ: [Thus, the minimal polynomial of A is unique.] ANSWERS TO SUPPLEMENTARY PROBLEMS Notation: M ¼ ½R1; R2; . . . denotes a matrix M with rows R1; R2; . . . : 9.38. f ðAÞ ¼ ½26; 3; 5; 27, gðAÞ ¼ ½40; 39; 65; 27, f ðBÞ ¼ ½3; 6; 0; 9, gðBÞ ¼ ½3; 12; 0; 15 9.39. A2 ¼ ½1; 4; 0; 1, A3 ¼ ½1; 6; 0; 1, An ¼ ½1; 2n; 0; 1, A1 ¼ ½1; 2; 0; 1 9.40. Let A ¼ ½2; a; b; 0; 2; c; 0; 0; 2. Set B ¼ A3 and then a ¼ 1, b ¼ 1 2, c ¼ 1 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 323 9.41. Find DðtÞ: (a) t2 þ t 11, (b) t2 þ 2t þ 13, (c) t3 7t2 þ 6t 1 9.45. (a) l ¼ 1; u ¼ ð3; 1Þ; l ¼ 4; v ¼ ð1; 2Þ, (b) l ¼ 4; u ¼ ð2; 1Þ, (c) l ¼ 1; u ¼ ð2; 1Þ; l ¼ 5; v ¼ ð2; 3Þ. Only A and C can be diagonalized; use P ¼ ½u; v. 9.46. (a) l ¼ 1; u ¼ ð1; 1Þ; l ¼ 4; v ¼ ð1; 2Þ, (b) P ¼ ½u; v, (c) f ðAÞ ¼ ½3; 1; 2; 1; A8 ¼ ½21 846; 21 845; 43 690; 43 691, (d) B ¼ 4 3 ; 1 3 ; 2 3 ; 5 3 9.47. (a) l ¼ 1; u ¼ ð3; 2Þ; l ¼ 2; v ¼ ð2; 1Þ, (b) P ¼ ½u; v, (c) f ðAÞ ¼ ½2; 6; 2; 9; A8 ¼ ½1021; 1530; 510; 764, (d) B ¼ ½3 þ 4 ﬃﬃﬃ 2 p ; 6 þ 6 ﬃﬃﬃ 2 p ; 2 2 ﬃﬃﬃ 2 p ; 4 3 ﬃﬃﬃ 2 p 9.48. (a) l ¼ 2; u ¼ ð1; 1; 0Þ; v ¼ ð1; 0; 1Þ; l ¼ 4; w ¼ ð1; 1; 2Þ, (b) l ¼ 2; u ¼ ð1; 1; 0Þ; l ¼ 4; v ¼ ð0; 1; 1Þ, (c) l ¼ 3; u ¼ ð1; 1; 0Þ; v ¼ ð1; 0; 1Þ; l ¼ 1; w ¼ ð2; 1; 1Þ. Only A and C can be diagonalized; use P ¼ ½u; v; w: 9.49. (a) l ¼ 2; u ¼ ð3; 1Þ; l ¼ 6; v ¼ ð1; 1Þ, (b) No real eigenvalues 9.50. We need ½trðAÞ2 4½detðAÞ 0 or ða dÞ2 þ 4bc 0. 9.51. A ¼ ½1; 1; 0; 1 9.56. (a) P ¼ ½2; 1; 1; 2= ﬃﬃﬃ 5 p , D ¼ ½7; 0; 0; 3, (b) P ¼ ½1; 1; 1; 1= ﬃﬃﬃ 2 p , D ¼ ½3; 0; 0; 5, (c) P ¼ ½3; 1; 1; 3= ﬃﬃﬃﬃﬃ 10 p , D ¼ ½8; 0; 0; 2 9.57. (a) l ¼ 1; u ¼ ð1; 1; 0Þ; v ¼ ð1; 1; 2Þ; l ¼ 2; w ¼ ð1; 1; 1Þ, (b) l ¼ 1; u ¼ ð2; 1; 1Þ; v ¼ ð2; 3; 1Þ; l ¼ 22; w ¼ ð1; 2; 4Þ; Normalize u; v; w, obtaining ^ u; ^ v; ^ w, and set P ¼ ½^ u; ^ v; ^ w. (Remark: u and v are not unique.) 9.58. (a) x ¼ ð4s þ tÞ= ﬃﬃﬃﬃﬃ 17 p ; y ¼ ðs þ 4tÞ= ﬃﬃﬃﬃﬃ 17 p ; qðs; tÞ ¼ 5s2 12t2, (b) x ¼ ð3s tÞ= ﬃﬃﬃﬃﬃ 10 p ; y ¼ ðs þ 3tÞ= ﬃﬃﬃﬃﬃ 10 p ; qðs; tÞ ¼ s2 þ 11t2 9.59. (a) x ¼ ð3s þ 2tÞ= ﬃﬃﬃﬃﬃ 13 p ; y ¼ r; z ¼ ð2s 3tÞ= ﬃﬃﬃﬃﬃ 13 p ; qðr; s; tÞ ¼ 3r2 þ 9s2 4t2, (b) x ¼ 5Ks þ Lt; y ¼ Jr þ 2Ks 2Lt; z ¼ 2Jr Ks Lt, where J ¼ 1= ﬃﬃﬃ 5 p , K ¼ 1= ﬃﬃﬃﬃﬃ 30 p , L ¼ 1= ﬃﬃﬃ 6 p ; qðr; s; tÞ ¼ 2r2 þ 2s2 þ 8t2 9.60. (a) A ¼ 1 2 ½5; 3; 3; 5; B ¼ 1 2 ½3; 1; 1; 3, (b) A ¼ 1 5 ½14; 2; 2; 11, B ¼ 1 5 ½ ﬃﬃﬃ 2 p þ 4 ﬃﬃﬃ 3 p ; 2 ﬃﬃﬃ 2 p 2 ﬃﬃﬃ 3 p ; 2 ﬃﬃﬃ 2 p 2 ﬃﬃﬃ 3 p ; 4 ﬃﬃﬃ 2 p þ ﬃﬃﬃ 3 p 9.61. (a) DðtÞ ¼ mðtÞ ¼ ðt 2Þ2ðt 6Þ, (b) DðtÞ ¼ ðt 2Þ2ðt 6Þ; mðtÞ ¼ ðt 2Þðt 6Þ 9.62. (a) DðtÞ ¼ ðt 2Þ3ðt 7Þ2; mðtÞ ¼ ðt 2Þ2ðt 7Þ, (b) DðtÞ ¼ ðt 3Þ5; mðtÞ ¼ ðt 3Þ3, (c) DðtÞ ¼ ðt 2Þ2ðt 4Þ2ðt 5Þ; mðtÞ ¼ ðt 2Þðt 4Þðt 5Þ 9.68. Let A be the companion matrix [Example 9.12(b)] with last column: (a) ½8; 6; 5T, (b) ½4; 7; 2; 5T 9.69. Hint: A is a root of hðtÞ ¼ f ðtÞ gðtÞ, where hðtÞ 0 or the degree of hðtÞ is less than the degree of f ðtÞ: 324 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors Canonical Forms 10.1 Introduction Let T be a linear operator on a vector space of ﬁnite dimension. As seen in Chapter 6, T may not have a diagonal matrix representation. However, it is still possible to ‘‘simplify’’ the matrix representation of T in a number of ways. This is the main topic of this chapter. In particular, we obtain the primary decomposition theorem, and the triangular, Jordan, and rational canonical forms. We comment that the triangular and Jordan canonical forms exist for T if and only if the characteristic polynomial DðtÞ of T has all its roots in the base ﬁeld K. This is always true if K is the complex ﬁeld C but may not be true if K is the real ﬁeld R. We also introduce the idea of a quotient space. This is a very powerful tool, and it will be used in the proof of the existence of the triangular and rational canonical forms. 10.2 Triangular Form Let T be a linear operator on an n-dimensional vector space V. Suppose T can be represented by the triangular matrix A ¼ a11 a12 . . . a1n a22 . . . a2n . . . . . . ann 2 6 6 4 3 7 7 5 Then the characteristic polynomial DðtÞ of T is a product of linear factors; that is, DðtÞ ¼ detðtI AÞ ¼ ðt a11Þðt a22Þ ðt annÞ The converse is also true and is an important theorem (proved in Problem 10.28). THEOREM 10.1: Let T:V ! V be a linear operator whose characteristic polynomial factors into linear polynomials. Then there exists a basis of V in which T is represented by a triangular matrix. THEOREM 10.1: (Alternative Form) Let A be a square matrix whose characteristic polynomial factors into linear polynomials. Then A is similar to a triangular matrix—that is, there exists an invertible matrix P such that P1AP is triangular. We say that an operator T can be brought into triangular form if it can be represented by a triangular matrix. Note that in this case, the eigenvalues of T are precisely those entries appearing on the main diagonal. We give an application of this remark. CHAPTER 10 325 EXAMPLE 10.1 Let A be a square matrix over the complex ﬁeld C. Suppose l is an eigenvalue of A2. Show that ﬃﬃﬃ l p or ﬃﬃﬃ l p is an eigenvalue of A. By Theorem 10.1, A and A2 are similar, respectively, to triangular matrices of the form B ¼ m1 . . . m2 . . . . . . . . . mn 2 6 6 4 3 7 7 5 and B2 ¼ m2 1 . . . m2 2 . . . . . . . . . m2 n 2 6 6 4 3 7 7 5 Because similar matrices have the same eigenvalues, l ¼ m2 i for some i. Hence, mi ¼ ﬃﬃﬃ l p or mi ¼ ﬃﬃﬃ l p is an eigenvalue of A. 10.3 Invariance Let T:V ! V be linear. A subspace W of V is said to be invariant under T or T-invariant if T maps W into itself—that is, if v 2 W implies TðvÞ 2 W. In this case, T restricted to W deﬁnes a linear operator on W; that is, T induces a linear operator ^ T:W ! W deﬁned by ^ TðwÞ ¼ TðwÞ for every w 2 W. EXAMPLE 10.2 (a) Let T: R3 ! R3 be the following linear operator, which rotates each vector v about the z-axis by an angle y (shown in Fig. 10-1): Tðx; y; zÞ ¼ ðx cos y y sin y; x sin y þ y cos y; zÞ Observe that each vector w ¼ ða; b; 0Þ in the xy-plane W remains in W under the mapping T; hence, W is T-invariant. Observe also that the z-axis U is invariant under T. Furthermore, the restriction of T to W rotates each vector about the origin O, and the restriction of T to U is the identity mapping of U. (b) Nonzero eigenvectors of a linear operator T:V ! V may be characterized as generators of T-invariant one-dimensional subspaces. Suppose TðvÞ ¼ lv, v 6¼ 0. Then W ¼ fkv; k 2 Kg, the one-dimensional subspace generated by v, is invariant under T because TðkvÞ ¼ kTðvÞ ¼ kðlvÞ ¼ klv 2 W Conversely, suppose dim U ¼ 1 and u 6¼ 0 spans U, and U is invariant under T. Then TðuÞ 2 U and so TðuÞ is a multiple of u—that is, TðuÞ ¼ mu. Hence, u is an eigenvector of T. The next theorem (proved in Problem 10.3) gives us an important class of invariant subspaces. THEOREM 10.2: Let T:V ! V be any linear operator, and let f ðtÞ be any polynomial. Then the kernel of f ðTÞ is invariant under T. The notion of invariance is related to matrix representations (Problem 10.5) as follows. THEOREM 10.3: Suppose W is an invariant subspace of T:V ! V. Then T has a block matrix repre-sentation A B 0 C , where A is a matrix representation of the restriction ^ T of T to W. 0 W y z x U T( ) v T w ( ) θ θ v w Figure 10-1 326 CHAPTER 10 Canonical Forms 10.4 Invariant Direct-Sum Decompositions A vector space V is termed the direct sum of subspaces W1; . . . ; Wr, written V ¼ W1 W2 . . . Wr if every vector v 2 V can be written uniquely in the form v ¼ w1 þ w2 þ . . . þ wr; with wi 2 Wi The following theorem (proved in Problem 10.7) holds. THEOREM 10.4: Suppose W1; W2; . . . ; Wr are subspaces of V, and suppose B1 ¼ fw11; w12; . . . ; w1n1g; . . . ; Br ¼ fwr1; wr2; . . . ; wrnrg are bases of W1; W2; . . . ; Wr, respectively. Then V is the direct sum of the Wi if and only if the union B ¼ B1 [ . . . [ Br is a basis of V. Now suppose T:V ! V is linear and V is the direct sum of (nonzero) T-invariant subspaces W1; W2; . . . ; Wr; that is, V ¼ W1 . . . Wr and TðWiÞ Wi; i ¼ 1; . . . ; r Let Ti denote the restriction of T to Wi. Then T is said to be decomposable into the operators Ti or T is said to be the direct sum of the Ti; written T ¼ T1 . . . Tr: Also, the subspaces W1; . . . ; Wr are said to reduce T or to form a T-invariant direct-sum decomposition of V. Consider the special case where two subspaces U and W reduce an operator T:V ! V; say dim U ¼ 2 and dim W ¼ 3, and suppose fu1; u2g and fw1; w2; w3g are bases of U and W, respectively. If T1 and T2 denote the restrictions of T to U and W, respectively, then T1ðu1Þ ¼ a11u1 þ a12u2 T1ðu2Þ ¼ a21u1 þ a22u2 T2ðw1Þ ¼ b11w1 þ b12w2 þ b13w3 T2ðw2Þ ¼ b21w1 þ b22w2 þ b23w3 T2ðw3Þ ¼ b31w1 þ b32w2 þ b33w3 Accordingly, the following matrices A; B; M are the matrix representations of T1, T2, T, respectively, A ¼ a11 a21 a12 a22 ; B ¼ b11 b21 b31 b12 b22 b32 b13 b23 b33 2 4 3 5; M ¼ A 0 0 B The block diagonal matrix M results from the fact that fu1; u2; w1; w2; w3g is a basis of V (Theorem 10.4), and that TðuiÞ ¼ T1ðuiÞ and TðwjÞ ¼ T2ðwjÞ. A generalization of the above argument gives us the following theorem. THEOREM 10.5: Suppose T:V ! V is linear and suppose V is the direct sum of T-invariant subspaces, say, W1; . . . ; Wr. If Ai is a matrix representation of the restriction of T to Wi, then T can be represented by the block diagonal matrix: M ¼ diagðA1; A2; . . . ; ArÞ 10.5 Primary Decomposition The following theorem shows that any operator T:V ! V is decomposable into operators whose minimum polynomials are powers of irreducible polynomials. This is the ﬁrst step in obtaining a canonical form for T. CHAPTER 10 Canonical Forms 327 THEOREM 10.6: (Primary Decomposition Theorem) Let T:V ! V be a linear operator with minimal polynomial mðtÞ ¼ f1ðtÞn1f2ðtÞn2 frðtÞnr where the fiðtÞ are distinct monic irreducible polynomials. Then V is the direct sum of T-invariant subspaces W1; . . . ; Wr, where Wi is the kernel of fiðTÞni. Moreover, fiðtÞni is the minimal polynomial of the restriction of T to Wi. The above polynomials fiðtÞni are relatively prime. Therefore, the above fundamental theorem follows (Problem 10.11) from the next two theorems (proved in Problems 10.9 and 10.10, respectively). THEOREM 10.7: Suppose T:V ! V is linear, and suppose f ðtÞ ¼ gðtÞhðtÞ are polynomials such that f ðTÞ ¼ 0 and gðtÞ and hðtÞ are relatively prime. Then V is the direct sum of the T-invariant subspace U and W, where U ¼ Ker gðTÞ and W ¼ Ker hðTÞ. THEOREM 10.8: In Theorem 10.7, if f ðtÞ is the minimal polynomial of T [and gðtÞ and hðtÞ are monic], then gðtÞ and hðtÞ are the minimal polynomials of the restrictions of T to U and W, respectively. We will also use the primary decomposition theorem to prove the following useful characterization of diagonalizable operators (see Problem 10.12 for the proof). THEOREM 10.9: A linear operator T:V ! V is diagonalizable if and only if its minimal polynomial mðtÞ is a product of distinct linear polynomials. THEOREM 10.9: (Alternative Form) A matrix A is similar to a diagonal matrix if and only if its minimal polynomial is a product of distinct linear polynomials. EXAMPLE 10.3 Suppose A 6¼ I is a square matrix for which A3 ¼ I. Determine whether or not A is similar to a diagonal matrix if A is a matrix over: (i) the real ﬁeld R, (ii) the complex ﬁeld C. Because A3 ¼ I, A is a zero of the polynomial f ðtÞ ¼ t3 1 ¼ ðt 1Þðt2 þ t þ 1Þ: The minimal polynomial mðtÞ of A cannot be t 1, because A 6¼ I. Hence, mðtÞ ¼ t2 þ t þ 1 or mðtÞ ¼ t3 1 Because neither polynomial is a product of linear polynomials over R, A is not diagonalizable over R. On the other hand, each of the polynomials is a product of distinct linear polynomials over C. Hence, A is diagonalizable over C. 10.6 Nilpotent Operators A linear operator T:V ! V is termed nilpotent if Tn ¼ 0 for some positive integer n; we call k the index of nilpotency of T if Tk ¼ 0 but Tk1 6¼ 0: Analogously, a square matrix A is termed nilpotent if An ¼ 0 for some positive integer n, and of index k if Ak ¼ 0 but Ak1 6¼ 0. Clearly the minimum polynomial of a nilpotent operator (matrix) of index k is mðtÞ ¼ tk; hence, 0 is its only eigenvalue. EXAMPLE 10.4 The following two r-square matrices will be used throughout the chapter: N ¼ NðrÞ ¼ 0 1 0 . . . 0 0 0 0 1 . . . 0 0 :::::::::::::::::::::::::::::::: 0 0 0 . . . 0 1 0 0 0 . . . 0 0 2 6 6 6 6 4 3 7 7 7 7 5 and JðlÞ ¼ l 1 0 . . . 0 0 0 l 1 . . . 0 0 :::::::::::::::::::::::::::::::: 0 0 0 . . . l 1 0 0 0 . . . 0 l 2 6 6 6 6 4 3 7 7 7 7 5 328 CHAPTER 10 Canonical Forms The ﬁrst matrix N, called a Jordan nilpotent block, consists of 1’s above the diagonal (called the super-diagonal), and 0’s elsewhere. It is a nilpotent matrix of index r. (The matrix N of order 1 is just the 1 1 zero matrix .) The second matrix JðlÞ, called a Jordan block belonging to the eigenvalue l, consists of l’s on the diagonal, 1’s on the superdiagonal, and 0’s elsewhere. Observe that JðlÞ ¼ lI þ N In fact, we will prove that any linear operator T can be decomposed into operators, each of which is the sum of a scalar operator and a nilpotent operator. The following (proved in Problem 10.16) is a fundamental result on nilpotent operators. THEOREM 10.10: Let T:V ! V be a nilpotent operator of index k. Then T has a block diagonal matrix representation in which each diagonal entry is a Jordan nilpotent block N. There is at least one N of order k, and all other N are of orders k. The number of N of each possible order is uniquely determined by T. The total number of N of all orders is equal to the nullity of T. The proof of Theorem 10.10 shows that the number of N of order i is equal to 2mi miþ1 mi1, where mi is the nullity of Ti. 10.7 Jordan Canonical Form An operator T can be put into Jordan canonical form if its characteristic and minimal polynomials factor into linear polynomials. This is always true if K is the complex ﬁeld C. In any case, we can always extend the base ﬁeld K to a ﬁeld in which the characteristic and minimal polynomials do factor into linear factors; thus, in a broad sense, every operator has a Jordan canonical form. Analogously, every matrix is similar to a matrix in Jordan canonical form. The following theorem (proved in Problem 10.18) describes the Jordan canonical form J of a linear operator T. THEOREM 10.11: Let T:V ! V be a linear operator whose characteristic and minimal polynomials are, respectively, DðtÞ ¼ ðt l1Þn1 ðt lrÞnr and mðtÞ ¼ ðt l1Þm1 ðt lrÞmr where the li are distinct scalars. Then T has a block diagonal matrix representa-tion J in which each diagonal entry is a Jordan block Jij ¼ JðliÞ. For each lij, the corresponding Jij have the following properties: (i) There is at least one Jij of order mi; all other Jij are of order mi. (ii) The sum of the orders of the Jij is ni. (iii) The number of Jij equals the geometric multiplicity of li. (iv) The number of Jij of each possible order is uniquely determined by T. EXAMPLE 10.5 Suppose the characteristic and minimal polynomials of an operator T are, respec-tively, DðtÞ ¼ ðt 2Þ4ðt 5Þ3 and mðtÞ ¼ ðt 2Þ2ðt 5Þ3 CHAPTER 10 Canonical Forms 329 Then the Jordan canonical form of T is one of the following block diagonal matrices: diag 2 1 0 2 ; 2 1 0 2 ; 5 1 0 0 5 1 0 0 5 2 4 3 5 0 @ 1 A or diag 2 1 0 2 ; ½2; ½2; 5 1 0 0 5 1 0 0 5 2 4 3 5 0 @ 1 A The ﬁrst matrix occurs if T has two independent eigenvectors belonging to the eigenvalue 2; and the second matrix occurs if T has three independent eigenvectors belonging to the eigenvalue 2. 10.8 Cyclic Subspaces Let T be a linear operator on a vector space V of ﬁnite dimension over K. Suppose v 2 V and v 6¼ 0. The set of all vectors of the form f ðTÞðvÞ, where f ðtÞ ranges over all polynomials over K, is a T-invariant subspace of V called the T-cyclic subspace of V generated by v; we denote it by Zðv; TÞ and denote the restriction of T to Zðv; TÞ by Tv: By Problem 10.56, we could equivalently deﬁne Zðv; TÞ as the intersection of all T-invariant subspaces of V containing v. Now consider the sequence v; TðvÞ; T2ðvÞ; T3ðvÞ; . . . of powers of T acting on v. Let k be the least integer such that TkðvÞ is a linear combination of those vectors that precede it in the sequence, say, TkðvÞ ¼ ak1Tk1ðvÞ a1TðvÞ a0v mvðtÞ ¼ tk þ ak1tk1 þ þ a1t þ a0 Then is the unique monic polynomial of lowest degree for which mvðTÞðvÞ ¼ 0. We call mvðtÞ the T-annihilator of v and Zðv; TÞ. The following theorem (proved in Problem 10.29) holds. THEOREM 10.12: Let Zðv; TÞ, Tv, mvðtÞ be deﬁned as above. Then (i) The set fv; TðvÞ; . . . ; Tk1ðvÞg is a basis of Zðv; TÞ; hence, dim Zðv; TÞ ¼ k. (ii) The minimal polynomial of Tv is mvðtÞ. (iii) The matrix representation of Tv in the above basis is just the companion matrix CðmvÞ of mvðtÞ; that is, CðmvÞ ¼ 0 0 0 . . . 0 a0 1 0 0 . . . 0 a1 0 1 0 . . . 0 a2 :::::::::::::::::::::::::::::::::::::::: 0 0 0 . . . 0 ak2 0 0 0 . . . 1 ak1 2 6 6 6 6 6 6 4 3 7 7 7 7 7 7 5 10.9 Rational Canonical Form In this section, we present the rational canonical form for a linear operator T:V ! V. We emphasize that this form exists even when the minimal polynomial cannot be factored into linear polynomials. (Recall that this is not the case for the Jordan canonical form.) 330 CHAPTER 10 Canonical Forms LEMMA 10.13: Let T:V ! V be a linear operator whose minimal polynomial is f ðtÞn, where f ðtÞ is a monic irreducible polynomial. Then V is the direct sum V ¼ Zðv1; TÞ Zðvr; TÞ of T-cyclic subspaces Zðvi; TÞ with corresponding T-annihilators f ðtÞn1; f ðtÞn2; . . . ; f ðtÞnr; n ¼ n1 n2 . . . nr Any other decomposition of V into T-cyclic subspaces has the same number of components and the same set of T-annihilators. We emphasize that the above lemma (proved in Problem 10.31) does not say that the vectors vi or other T-cyclic subspaces Zðvi; TÞ are uniquely determined by T, but it does say that the set of T-annihilators is uniquely determined by T. Thus, T has a unique block diagonal matrix representation: M ¼ diagðC1; C2; . . . ; CrÞ where the Ci are companion matrices. In fact, the Ci are the companion matrices of the polynomials f ðtÞni. Using the Primary Decomposition Theorem and Lemma 10.13, we obtain the following result. THEOREM 10.14: Let T:V ! V be a linear operator with minimal polynomial mðtÞ ¼ f1ðtÞm1f2ðtÞm2 fsðtÞms where the fiðtÞ are distinct monic irreducible polynomials. Then T has a unique block diagonal matrix representation: M ¼ diagðC11; C12; . . . ; C1r1; . . . ; Cs1; Cs2; . . . ; CsrsÞ where the Cij are companion matrices. In particular, the Cij are the companion matrices of the polynomials fiðtÞnij, where m1 ¼ n11 n12 n1r1; . . . ; ms ¼ ns1 ns2 nsrs The above matrix representation of T is called its rational canonical form. The polynomials fiðtÞnij are called the elementary divisors of T. EXAMPLE 10.6 Let V be a vector space of dimension 8 over the rational ﬁeld Q, and let T be a linear operator on V whose minimal polynomial is mðtÞ ¼ f1ðtÞf2ðtÞ2 ¼ ðt4 4t3 þ 6t2 4t 7Þðt 3Þ2 Thus, because dim V ¼ 8; the characteristic polynomial DðtÞ ¼ f1ðtÞ f2ðtÞ4: Also, the rational canonical form M of T must have one block the companion matrix of f1ðtÞ and one block the companion matrix of f2ðtÞ2. There are two possibilities: (a) diag½Cðt4 4t3 þ 6t2 4t 7Þ, Cððt 3Þ2Þ, Cððt 3Þ2Þ (b) diag½Cðt4 4t3 þ 6t2 4t 7Þ, Cððt 3Þ2Þ, Cðt 3Þ; Cðt 3Þ That is, (a) diag 0 0 0 7 1 0 0 4 0 1 0 6 0 0 1 4 2 6 6 4 3 7 7 5; 0 9 1 6 ; 0 9 1 6 0 B B @ 1 C C A; (b) diag 0 0 0 7 1 0 0 4 0 1 0 6 0 0 1 4 2 6 6 4 3 7 7 5; 0 9 1 6 ; ½3; ½3 0 B B @ 1 C C A 10.10 Quotient Spaces Let V be a vector space over a ﬁeld K and let W be a subspace of V. If v is any vector in V, we write v þ W for the set of sums v þ w with w 2 W; that is, v þ W ¼ fv þ w : w 2 Wg CHAPTER 10 Canonical Forms 331 These sets are called the cosets of W in V. We show (Problem 10.22) that these cosets partition V into mutually disjoint subsets. EXAMPLE 10.7 Let W be the subspace of R2 deﬁned by W ¼ fða; bÞ : a ¼ bg; that is, W is the line given by the equation x y ¼ 0. We can view v þ W as a translation of the line obtained by adding the vector v to each point in W. As shown in Fig. 10-2, the coset v þ W is also a line, and it is parallel to W. Thus, the cosets of W in R2 are precisely all the lines parallel to W. In the following theorem, we use the cosets of a subspace W of a vector space V to deﬁne a new vector space; it is called the quotient space of V by W and is denoted by V=W. THEOREM 10.15: Let W be a subspace of a vector space over a ﬁeld K. Then the cosets of W in V form a vector space over K with the following operations of addition and scalar multiplication: ðiÞ ðu þ wÞ þ ðv þ WÞ ¼ ðu þ vÞ þ W; ðiiÞ kðu þ WÞ ¼ ku þ W; where k 2 K We note that, in the proof of Theorem 10.15 (Problem 10.24), it is ﬁrst necessary to show that the operations are well deﬁned; that is, whenever u þ W ¼ u0 þ W and v þ W ¼ v0 þ W, then ðiÞ ðu þ vÞ þ W ¼ ðu0 þ v0Þ þ W and ðiiÞ ku þ W ¼ ku0 þ W for any k 2 K In the case of an invariant subspace, we have the following useful result (proved in Problem 10.27). THEOREM 10.16: Suppose W is a subspace invariant under a linear operator T:V ! V. Then T induces a linear operator T on V=W deﬁned by Tðv þ WÞ ¼ TðvÞ þ W. Moreover, if T is a zero of any polynomial, then so is T. Thus, the minimal polynomial of T divides the minimal polynomial of T. SOLVED PROBLEMS Invariant Subspaces 10.1. Suppose T:V ! V is linear. Show that each of the following is invariant under T: (a) f0g, (b) V, (c) kernel of T, (d) image of T. (a) We have Tð0Þ ¼ 0 2 f0g; hence, f0g is invariant under T. (b) For every v 2 V , TðvÞ 2 V; hence, V is invariant under T. (c) Let u 2 Ker T. Then TðuÞ ¼ 0 2 Ker T because the kernel of T is a subspace of V. Thus, Ker T is invariant under T. (d) Because TðvÞ 2 Im T for every v 2 V, it is certainly true when v 2 Im T. Hence, the image of T is invariant under T. 10.2. Suppose fWig is a collection of T-invariant subspaces of a vector space V. Show that the intersection W ¼ T i Wi is also T-invariant. Suppose v 2 W; then v 2 Wi for every i. Because Wi is T-invariant, TðvÞ 2 Wi for every i. Thus, TðvÞ 2 W and so W is T-invariant. Figure 10-2 332 CHAPTER 10 Canonical Forms 10.3. Prove Theorem 10.2: Let T:V ! V be linear. For any polynomial f ðtÞ, the kernel of f ðTÞ is invariant under T. Suppose v 2 Ker f ðTÞ—that is, f ðTÞðvÞ ¼ 0. We need to show that TðvÞ also belongs to the kernel of f ðTÞ—that is, f ðTÞðTðvÞÞ ¼ ð f ðTÞ TÞðvÞ ¼ 0. Because f ðtÞt ¼ tf ðtÞ, we have f ðTÞ T ¼ T f ðTÞ. Thus, as required, ð f ðTÞ TÞðvÞ ¼ ðT f ðTÞÞðvÞ ¼ Tð f ðTÞðvÞÞ ¼ Tð0Þ ¼ 0 10.4. Find all invariant subspaces of A ¼ 2 5 1 2 viewed as an operator on R2. By Problem 10.1, R2 and f0g are invariant under A. Now if A has any other invariant subspace, it must be one-dimensional. However, the characteristic polynomial of A is DðtÞ ¼ t2 trðAÞ t þ jAj ¼ t2 þ 1 Hence, A has no eigenvalues (in R) and so A has no eigenvectors. But the one-dimensional invariant subspaces correspond to the eigenvectors; thus, R2 and f0g are the only subspaces invariant under A. 10.5. Prove Theorem 10.3: Suppose W is T-invariant. Then T has a triangular block representation A B 0 C , where A is the matrix representation of the restriction ^ T of T to W. We choose a basis fw1; . . . ; wrg of W and extend it to a basis fw1; . . . ; wr; v1; . . . ; vsg of V. We have ^ Tðw1Þ ¼ Tðw1Þ ¼ a11w1 þ þ a1rwr ^ Tðw2Þ ¼ Tðw2Þ ¼ a21w1 þ þ a2rwr :::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ^ TðwrÞ ¼ TðwrÞ ¼ ar1w1 þ þ arrwr Tðv1Þ ¼ b11w1 þ þ b1rwr þ c11v1 þ þ c1svs Tðv2Þ ¼ b21w1 þ þ b2rwr þ c21v1 þ þ c2svs :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: TðvsÞ ¼ bs1w1 þ þ bsrwr þ cs1v1 þ þ cssvs But the matrix of T in this basis is the transpose of the matrix of coefﬁcients in the above system of equations (Section 6.2). Therefore, it has the form A B 0 C , where A is the transpose of the matrix of coefﬁcients for the obvious subsystem. By the same argument, A is the matrix of ^ T relative to the basis fwig of W. 10.6. Let ^ T denote the restriction of an operator T to an invariant subspace W. Prove (a) For any polynomial f ðtÞ, f ð ^ TÞðwÞ ¼ f ðTÞðwÞ. (b) The minimal polynomial of ^ T divides the minimal polynomial of T. (a) If f ðtÞ ¼ 0 or if f ðtÞ is a constant (i.e., of degree 1), then the result clearly holds. Assume deg f ¼ n > 1 and that the result holds for polynomials of degree less than n. Suppose that f ðtÞ ¼ antn þ an1tn1 þ þ a1t þ a0 f ð ^ TÞðwÞ ¼ ðan ^ Tn þ an1 ^ T n1 þ þ a0IÞðwÞ ¼ ðan ^ Tn1Þð ^ TðwÞÞ þ ðan1 ^ T n1 þ þ a0IÞðwÞ ¼ ðanTn1ÞðTðwÞÞ þ ðan1T n1 þ þ a0IÞðwÞ ¼ f ðTÞðwÞ Then (b) Let mðtÞ denote the minimal polynomial of T. Then by (a), mð ^ TÞðwÞ ¼ mðTÞðwÞ ¼ 0ðwÞ ¼ 0 for every w 2 W; that is, ^ T is a zero of the polynomial mðtÞ. Hence, the minimal polynomial of ^ T divides mðtÞ. CHAPTER 10 Canonical Forms 333 Invariant Direct-Sum Decompositions 10.7. Prove Theorem 10.4: Suppose W1; W2; . . . ; Wr are subspaces of V with respective bases B1 ¼ fw11; w12; . . . ; w1n1g; . . . ; Br ¼ fwr1; wr2; . . . ; wrnrg Then V is the direct sum of the Wi if and only if the union B ¼ S i Bi is a basis of V. Suppose B is a basis of V. Then, for any v 2 V, v ¼ a11w11 þ þ a1n1w1n1 þ þ ar1wr1 þ þ arnrwrnr ¼ w1 þ w2 þ þ wr where wi ¼ ai1wi1 þ þ ainiwini 2 Wi. We next show that such a sum is unique. Suppose v ¼ w0 1 þ w0 2 þ þ w0 r; where w0 i 2 Wi Because fwi1; . . . ; winig is a basis of Wi, w0 i ¼ bi1wi1 þ þ biniwini, and so v ¼ b11w11 þ þ b1n1w1n1 þ þ br1wr1 þ þ brnrwrnr Because B is a basis of V; aij ¼ bij, for each i and each j. Hence, wi ¼ w0 i, and so the sum for v is unique. Accordingly, V is the direct sum of the Wi. Conversely, suppose V is the direct sum of the Wi. Then for any v 2 V, v ¼ w1 þ þ wr, where wi 2 Wi. Because fwijig is a basis of Wi, each wi is a linear combination of the wiji, and so v is a linear combination of the elements of B. Thus, B spans V. We now show that B is linearly independent. Suppose a11w11 þ þ a1n1w1n1 þ þ ar1wr1 þ þ arnrwrnr ¼ 0 Note that ai1wi1 þ þ ainiwini 2 Wi. We also have that 0 ¼ 0 þ 0 0 2 Wi. Because such a sum for 0 is unique, ai1wi1 þ þ ainiwini ¼ 0 for i ¼ 1; . . . ; r The independence of the bases fwijig implies that all the a’s are 0. Thus, B is linearly independent and is a basis of V. 10.8. Suppose T:V ! V is linear and suppose T ¼ T1 T2 with respect to a T-invariant direct-sum decomposition V ¼ U W. Show that (a) mðtÞ is the least common multiple of m1ðtÞ and m2ðtÞ, where mðtÞ, m1ðtÞ, m2ðtÞ are the minimum polynomials of T; T1; T2, respectively. (b) DðtÞ ¼ D1ðtÞD2ðtÞ, where DðtÞ; D1ðtÞ, D2ðtÞ are the characteristic polynomials of T; T1; T2, respectively. (a) By Problem 10.6, each of m1ðtÞ and m2ðtÞ divides mðtÞ. Now suppose f ðtÞ is a multiple of both m1ðtÞ and m2ðtÞ, then f ðT1ÞðUÞ ¼ 0 and f ðT2ÞðWÞ ¼ 0. Let v 2 V, then v ¼ u þ w with u 2 U and w 2 W. Now f ðTÞv ¼ f ðTÞu þ f ðTÞw ¼ f ðT1Þu þ f ðT2Þw ¼ 0 þ 0 ¼ 0 That is, T is a zero of f ðtÞ. Hence, mðtÞ divides f ðtÞ, and so mðtÞ is the least common multiple of m1ðtÞ and m2ðtÞ. (b) By Theorem 10.5, T has a matrix representation M ¼ A 0 0 B , where A and B are matrix representations of T1 and T2, respectively. Then, as required, DðtÞ ¼ jtI Mj ¼ tI A 0 0 tI B ¼ jtI AjjtI Bj ¼ D1ðtÞD2ðtÞ 10.9. Prove Theorem 10.7: Suppose T:V ! V is linear, and suppose f ðtÞ ¼ gðtÞhðtÞ are polynomials such that f ðTÞ ¼ 0 and gðtÞ and hðtÞ are relatively prime. Then V is the direct sum of the T-invariant subspaces U and W where U ¼ Ker gðTÞ and W ¼ Ker hðTÞ. 334 CHAPTER 10 Canonical Forms Note ﬁrst that U and W are T-invariant by Theorem 10.2. Now, because gðtÞ and hðtÞ are relatively prime, there exist polynomials rðtÞ and sðtÞ such that rðtÞgðtÞ þ sðtÞhðtÞ ¼ 1 Hence; for the operator T; rðTÞgðTÞ þ sðTÞhðTÞ ¼ I ðÞ Let v 2 V; then; by ðÞ; v ¼ rðTÞgðTÞv þ sðTÞhðTÞv But the ﬁrst term in this sum belongs to W ¼ Ker hðTÞ, because hðTÞrðTÞgðTÞv ¼ rðTÞgðTÞhðTÞv ¼ rðTÞf ðTÞv ¼ rðTÞ0v ¼ 0 Similarly, the second term belongs to U. Hence, V is the sum of U and W. To prove that V ¼ U W, we must show that a sum v ¼ u þ w with u 2 U, w 2 W, is uniquely determined by v. Applying the operator rðTÞgðTÞ to v ¼ u þ w and using gðTÞu ¼ 0, we obtain rðTÞgðTÞv ¼ rðTÞgðTÞu þ rðTÞgðTÞw ¼ rðTÞgðTÞw Also, applying ðÞ to w alone and using hðTÞw ¼ 0, we obtain w ¼ rðTÞgðTÞw þ sðTÞhðTÞw ¼ rðTÞgðTÞw Both of the above formulas give us w ¼ rðTÞgðTÞv, and so w is uniquely determined by v. Similarly u is uniquely determined by v. Hence, V ¼ U W, as required. 10.10. Prove Theorem 10.8: In Theorem 10.7 (Problem 10.9), if f ðtÞ is the minimal polynomial of T (and gðtÞ and hðtÞ are monic), then gðtÞ is the minimal polynomial of the restriction T1 of T to U and hðtÞ is the minimal polynomial of the restriction T2 of T to W. Let m1ðtÞ and m2ðtÞ be the minimal polynomials of T1 and T2, respectively. Note that gðT1Þ ¼ 0 and hðT2Þ ¼ 0 because U ¼ Ker gðTÞ and W ¼ Ker hðTÞ. Thus, m1ðtÞ divides gðtÞ and m2ðtÞ divides hðtÞ ð1Þ By Problem 10.9, f ðtÞ is the least common multiple of m1ðtÞ and m2ðtÞ. But m1ðtÞ and m2ðtÞ are relatively prime because gðtÞ and hðtÞ are relatively prime. Accordingly, f ðtÞ ¼ m1ðtÞm2ðtÞ. We also have that f ðtÞ ¼ gðtÞhðtÞ. These two equations together with (1) and the fact that all the polynomials are monic imply that gðtÞ ¼ m1ðtÞ and hðtÞ ¼ m2ðtÞ, as required. 10.11. Prove the Primary Decomposition Theorem 10.6: Let T:V ! V be a linear operator with minimal polynomial mðtÞ ¼ f1ðtÞn1f2ðtÞn2 . . . frðtÞnr where the fiðtÞ are distinct monic irreducible polynomials. Then V is the direct sum of T-invariant subspaces W1; . . . ; Wr where Wi is the kernel of fiðTÞni. Moreover, fiðtÞni is the minimal polynomial of the restriction of T to Wi. The proof is by induction on r. The case r ¼ 1 is trivial. Suppose that the theorem has been proved for r 1. By Theorem 10.7, we can write V as the direct sum of T-invariant subspaces W1 and V1, where W1 is the kernel of f1ðTÞn1 and where V1 is the kernel of f2ðTÞn2 frðTÞnr. By Theorem 10.8, the minimal polynomials of the restrictions of T to W1 and V1 are f1ðtÞn1 and f2ðtÞn2 frðtÞnr, respectively. Denote the restriction of T to V1 by ^ T1. By the inductive hypothesis, V1 is the direct sum of subspaces W2; . . . ; Wr such that Wi is the kernel of fiðT1Þni and such that fiðtÞni is the minimal polynomial for the restriction of ^ T1 to Wi. But the kernel of fiðTÞni, for i ¼ 2; . . . ; r is necessarily contained in V1, because fiðtÞni divides f2ðtÞn2 frðtÞnr. Thus, the kernel of fiðTÞni is the same as the kernel of fiðT1Þni, which is Wi. Also, the restriction of T to Wi is the same as the restriction of ^ T1 to Wi (for i ¼ 2; . . . ; r); hence, fiðtÞni is also the minimal polynomial for the restriction of T to Wi. Thus, V ¼ W1 W2 Wr is the desired decomposition of T. 10.12. Prove Theorem 10.9: A linear operator T:V ! V has a diagonal matrix representation if and only if its minimal polynomal mðtÞ is a product of distinct linear polynomials. CHAPTER 10 Canonical Forms 335 Suppose mðtÞ is a product of distinct linear polynomials, say, mðtÞ ¼ ðt l1Þðt l2Þ ðt lrÞ where the li are distinct scalars. By the Primary Decomposition Theorem, V is the direct sum of subspaces W1; . . . ; Wr, where Wi ¼ KerðT liIÞ. Thus, if v 2 Wi, then ðT liIÞðvÞ ¼ 0 or TðvÞ ¼ liv. In other words, every vector in Wi is an eigenvector belonging to the eigenvalue li. By Theorem 10.4, the union of bases for W1; . . . ; Wr is a basis of V. This basis consists of eigenvectors, and so T is diagonalizable. Conversely, suppose T is diagonalizable (i.e., V has a basis consisting of eigenvectors of T ). Let l1; . . . ; ls be the distinct eigenvalues of T. Then the operator f ðTÞ ¼ ðT l1IÞðT l2IÞ ðT lsIÞ maps each basis vector into 0. Thus, f ðTÞ ¼ 0, and hence, the minimal polynomial mðtÞ of T divides the polynomial f ðtÞ ¼ ðt l1Þðt l2Þ ðt lsIÞ Accordingly, mðtÞ is a product of distinct linear polynomials. Nilpotent Operators, Jordan Canonical Form 10.13. Let T:V be linear. Suppose, for v 2 V, TkðvÞ ¼ 0 but Tk1ðvÞ 6¼ 0. Prove (a) The set S ¼ fv; TðvÞ; . . . ; Tk1ðvÞg is linearly independent. (b) The subspace W generated by S is T-invariant. (c) The restriction ^ T of T to W is nilpotent of index k. (d) Relative to the basis fTk1ðvÞ; . . . ; TðvÞ; vg of W, the matrix of T is the k-square Jordan nilpotent block Nk of index k (see Example 10.5). (a) Suppose av þ a1TðvÞ þ a2T2ðvÞ þ þ ak1Tk1ðvÞ ¼ 0 ðÞ Applying T k1 to ðÞ and using TkðvÞ ¼ 0, we obtain aTk1ðvÞ ¼ 0; because T k1ðvÞ 6¼ 0, a ¼ 0. Now applying Tk2 to ðÞ and using TkðvÞ ¼ 0 and a ¼ 0, we ﬁind a1Tk1ðvÞ ¼ 0; hence, a1 ¼ 0. Next applying Tk3 to ðÞ and using T kðvÞ ¼ 0 and a ¼ a1 ¼ 0, we obtain a2Tk1ðvÞ ¼ 0; hence, a2 ¼ 0. Continuing this process, we ﬁnd that all the a’s are 0; hence, S is independent. (b) Let v 2 W. Then v ¼ bv þ b1TðvÞ þ b2T 2ðvÞ þ þ bk1T k1ðvÞ Using TkðvÞ ¼ 0, we have TðvÞ ¼ bTðvÞ þ b1T2ðvÞ þ þ bk2Tk1ðvÞ 2 W Thus, W is T-invariant. (c) By hypothesis, TkðvÞ ¼ 0. Hence, for i ¼ 0; . . . ; k 1, ^ T kðTiðvÞÞ ¼ TkþiðvÞ ¼ 0 That is, applying ^ Tk to each generator of W, we obtain 0; hence, ^ Tk ¼ 0 and so ^ T is nilpotent of index at most k. On the other hand, ^ Tk1ðvÞ ¼ Tk1ðvÞ 6¼ 0; hence, T is nilpotent of index exactly k. (d) For the basis fTk1ðvÞ, Tk2ðvÞ; . . . ; TðvÞ; vg of W, ^ TðTk1ðvÞÞ ¼ TkðvÞ ¼ 0 ^ TðTk2ðvÞÞ ¼ Tk1ðvÞ ^ TðTk3ðvÞÞ ¼ Tk2ðvÞ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ^ TðTðvÞÞ ¼ T2ðvÞ ^ TðvÞ ¼ TðvÞ Hence, as required, the matrix of T in this basis is the k-square Jordan nilpotent block Nk. 336 CHAPTER 10 Canonical Forms 10.14. Let T:V ! V be linear. Let U ¼ Ker Ti and W ¼ Ker Tiþ1. Show that (a) U W, (b) TðWÞ U. (a) Suppose u 2 U ¼ Ker Ti. Then T iðuÞ ¼ 0 and so T iþ1ðuÞ ¼ TðTiðuÞÞ ¼ Tð0Þ ¼ 0. Thus, u 2 Ker T iþ1 ¼ W. But this is true for every u 2 U; hence, U W. (b) Similarly, if w 2 W ¼ Ker Tiþ1, then Tiþ1ðwÞ ¼ 0: Thus, T iþ1ðwÞ ¼ TiðTðwÞÞ ¼ Tið0Þ ¼ 0 and so TðWÞ U. 10.15. Let T:V be linear. Let X ¼ Ker Ti2, Y ¼ Ker Ti1, Z ¼ Ker Ti. Therefore (Problem 10.14), X Y Z. Suppose fu1; . . . ; urg; fu1; . . . ; ur; v1; . . . ; vsg; fu1; . . . ; ur; v1; . . . ; vs; w1; . . . ; wtg are bases of X; Y; Z, respectively. Show that S ¼ fu1; . . . ; ur; Tðw1Þ; . . . ; TðwtÞg is contained in Y and is linearly independent. By Problem 10.14, TðZÞ Y, and hence S Y. Now suppose S is linearly dependent. Then there exists a relation a1u1 þ þ arur þ b1Tðw1Þ þ þ btTðwtÞ ¼ 0 where at least one coefﬁcient is not zero. Furthermore, because fuig is independent, at least one of the bk must be nonzero. Transposing, we ﬁnd b1Tðw1Þ þ þ btTðwtÞ ¼ a1u1 arur 2 X ¼ Ker T i2 Hence; Ti2ðb1Tðw1Þ þ þ btTðwtÞÞ ¼ 0 Thus; Ti1ðb1w1 þ þ btwtÞ ¼ 0; and so b1w1 þ þ btwt 2 Y ¼ Ker Ti1 Because fui; vjg generates Y, we obtain a relation among the ui, vj, wk where one of the coefﬁcients (i.e., one of the bk) is not zero. This contradicts the fact that fui; vj; wkg is independent. Hence, S must also be independent. 10.16. Prove Theorem 10.10: Let T:V ! V be a nilpotent operator of index k. Then T has a unique block diagonal matrix representation consisting of Jordan nilpotent blocks N. There is at least one N of order k, and all other N are of orders k. The total number of N of all orders is equal to the nullity of T. Suppose dim V ¼ n. Let W1 ¼ Ker T, W2 ¼ Ker T 2; . . . ; Wk ¼ Ker Tk. Let us set mi ¼ dim Wi, for i ¼ 1; . . . ; k. Because T is of index k, Wk ¼ V and Wk1 6¼ V and so mk1 < mk ¼ n. By Problem 10.14, W1 W2 Wk ¼ V Thus, by induction, we can choose a basis fu1; . . . ; ung of V such that fu1; . . . ; umig is a basis of Wi. We now choose a new basis for V with respect to which T has the desired form. It will be convenient to label the members of this new basis by pairs of indices. We begin by setting vð1; kÞ ¼ umk1þ1; vð2; kÞ ¼ umk1þ2; . . . ; vðmk mk1; kÞ ¼ umk and setting vð1; k 1Þ ¼ Tvð1; kÞ; vð2; k 1Þ ¼ Tvð2; kÞ; . . . ; vðmk mk1; k 1Þ ¼ Tvðmk mk1; kÞ By the preceding problem, S1 ¼ fu1 . . . ; umk2; vð1; k 1Þ; . . . ; vðmk mk1; k 1Þg is a linearly independent subset of Wk1. We extend S1 to a basis of Wk1 by adjoining new elements (if necessary), which we denote by vðmk mk1 þ 1; k 1Þ; vðmk mk1 þ 2; k 1Þ; . . . ; vðmk1 mk2; k 1Þ Next we set vð1; k 2Þ ¼ Tvð1; k 1Þ; vð2; k 2Þ ¼ Tvð2; k 1Þ; . . . ; vðmk1 mk2; k 2Þ ¼ Tvðmk1 mk2; k 1Þ CHAPTER 10 Canonical Forms 337 Again by the preceding problem, S2 ¼ fu1; . . . ; umks; vð1; k 2Þ; . . . ; vðmk1 mk2; k 2Þg is a linearly independent subset of Wk2, which we can extend to a basis of Wk2 by adjoining elements vðmk1 mk2 þ 1; k 2Þ; vðmk1 mk2 þ 2; k 2Þ; . . . ; vðmk2 mk3; k 2Þ Continuing in this manner, we get a new basis for V, which for convenient reference we arrange as follows: vð1; kÞ . . . ; vðmk mk1; kÞ vð1; k 1Þ; . . . ; vðmk mk1; k 1Þ . . . ; vðmk1 mk2; k 1Þ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: vð1; 2Þ; . . . ; vðmk mk1; 2Þ; . . . ; vðmk1 mk2; 2Þ; . . . ; vðm2 m1; 2Þ vð1; 1Þ; . . . ; vðmk mk1; 1Þ; . . . ; vðmk1 mk2; 1Þ; . . . ; vðm2 m1; 1Þ; . . . ; vðm1; 1Þ The bottom row forms a basis of W1, the bottom two rows form a basis of W2, and so forth. But what is important for us is that T maps each vector into the vector immediately below it in the table or into 0 if the vector is in the bottom row. That is, Tvði; jÞ ¼ vði; j 1Þ for j > 1 0 for j ¼ 1 Now it is clear [see Problem 10.13(d)] that T will have the desired form if the vði; jÞ are ordered lexicographically: beginning with vð1; 1Þ and moving up the ﬁrst column to vð1; kÞ, then jumping to vð2; 1Þ and moving up the second column as far as possible. Moreover, there will be exactly mk mk1 diagonal entries of order k: Also, there will be ðmk1 mk2Þ ðmk mk1Þ ¼ 2mk1 mk mk2 diagonal entries of order k 1 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 2m2 m1 m3 diagonal entries of order 2 2m1 m2 diagonal entries of order 1 as can be read off directly from the table. In particular, because the numbers m1; . . . ; mk are uniquely determined by T, the number of diagonal entries of each order is uniquely determined by T. Finally, the identity m1 ¼ ðmk mk1Þ þ ð2mk1 mk mk2Þ þ þ ð2m2 m1 m3Þ þ ð2m1 m2Þ shows that the nullity m1 of T is the total number of diagonal entries of T. 10.17. Let A ¼ 0 1 1 0 1 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 6 6 6 6 4 3 7 7 7 7 5 and B ¼ 0 1 1 0 0 0 0 1 1 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 2 6 6 6 6 4 3 7 7 7 7 5 . The reader can verify that A and B are both nilpotent of index 3; that is, A3 ¼ 0 but A2 6¼ 0, and B3 ¼ 0 but B2 6¼ 0. Find the nilpotent matrices MA and MB in canonical form that are similar to A and B, respectively. Because A and B are nilpotent of index 3, MA and MB must each contain a Jordan nilpotent block of order 3, and none greater then 3. Note that rankðAÞ ¼ 2 and rankðBÞ ¼ 3, so nullityðAÞ ¼ 5 2 ¼ 3 and nullityðBÞ ¼ 5 3 ¼ 2. Thus, MA must contain three diagonal blocks, which must be one of order 3 and two of order 1; and MB must contain two diagonal blocks, which must be one of order 3 and one of order 2. Namely, MA ¼ 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 6 6 6 6 4 3 7 7 7 7 5 and MB ¼ 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 2 6 6 6 6 4 3 7 7 7 7 5 338 CHAPTER 10 Canonical Forms 10.18. Prove Theorem 10.11 on the Jordan canonical form for an operator T. By the primary decomposition theorem, T is decomposable into operators T1; . . . ; Tr; that is, T ¼ T1 Tr, where ðt liÞmi is the minimal polynomial of Ti. Thus, in particular, ðT1 l1IÞm1 ¼ 0; . . . ; ðTr lrIÞmr ¼ 0 Set Ni ¼ Ti liI. Then, for i ¼ 1; . . . ; r, Ti ¼ Ni þ liI; where Nmi i ¼ 0 That is, Ti is the sum of the scalar operator liI and a nilpotent operator Ni, which is of index mi because ðt liÞm i is the minimal polynomial of Ti. Now, by Theorem 10.10 on nilpotent operators, we can choose a basis so that Ni is in canonical form. In this basis, Ti ¼ Ni þ liI is represented by a block diagonal matrix Mi whose diagonal entries are the matrices Jij. The direct sum J of the matrices Mi is in Jordan canonical form and, by Theorem 10.5, is a matrix representation of T. Last, we must show that the blocks Jij satisfy the required properties. Property (i) follows from the fact that Ni is of index mi. Property (ii) is true because T and J have the same characteristic polynomial. Property (iii) is true because the nullity of Ni ¼ Ti liI is equal to the geometric multiplicity of the eigenvalue li. Property (iv) follows from the fact that the Ti and hence the Ni are uniquely determined by T. 10.19. Determine all possible Jordan canonical forms J for a linear operator T:V ! V whose characteristic polynomial DðtÞ ¼ ðt 2Þ5 and whose minimal polynomial mðtÞ ¼ ðt 2Þ2. J must be a 5 5 matrix, because DðtÞ has degree 5, and all diagonal elements must be 2, because 2 is the only eigenvalue. Moreover, because the exponent of t 2 in mðtÞ is 2, J must have one Jordan block of order 2, and the others must be of order 2 or 1. Thus, there are only two possibilities: J ¼ diag 2 1 2 ; 2 1 2 ; ½2 or J ¼ diag 2 1 2 ; ½2; ½2; ½2 10.20. Determine all possible Jordan canonical forms for a linear operator T:V ! V whose character-istic polynomial DðtÞ ¼ ðt 2Þ3ðt 5Þ2. In each case, ﬁnd the minimal polynomial mðtÞ. Because t 2 has exponent 3 in DðtÞ, 2 must appear three times on the diagonal. Similarly, 5 must appear twice. Thus, there are six possibilities: (a) diag 2 1 2 1 2 2 4 3 5; 5 1 5 0 @ 1 A, (b) diag 2 1 2 1 2 2 4 3 5; ½5; ½5 0 @ 1 A, (c) diag 2 1 2 ; ½2; 5 1 5 , (d) diag 2 1 2 ; ½2; ½5; ½5 , (e) diag ½2; ½2; ½2; 5 1 5 , (f ) diagð½2; ½2; ½2; ½5; ½5Þ The exponent in the minimal polynomial mðtÞ is equal to the size of the largest block. Thus, (a) mðtÞ ¼ ðt 2Þ3ðt 5Þ2, (b) mðtÞ ¼ ðt 2Þ3ðt 5Þ, (c) mðtÞ ¼ ðt 2Þ2ðt 5Þ2, (d) mðtÞ ¼ ðt 2Þ2ðt 5Þ, (e) mðtÞ ¼ ðt 2Þðt 5Þ2, (f ) mðtÞ ¼ ðt 2Þðt 5Þ Quotient Space and Triangular Form 10.21. Let W be a subspace of a vector space V. Show that the following are equivalent: (i) u 2 v þ W, (ii) u v 2 W, (iii) v 2 u þ W. Suppose u 2 v þ W. Then there exists w0 2 W such that u ¼ v þ w0. Hence, u v ¼ w0 2 W. Conversely, suppose u v 2 W. Then u v ¼ w0 where w0 2 W. Hence, u ¼ v þ w0 2 v þ W. Thus, (i) and (ii) are equivalent. We also have u v 2 W iff ðu vÞ ¼ v u 2 W iffv 2 u þ W. Thus, (ii) and (iii) are also equivalent. CHAPTER 10 Canonical Forms 339 10.22. Prove the following: The cosets of W in V partition V into mutually disjoint sets. That is, (a) Any two cosets u þ W and v þ W are either identical or disjoint. (b) Each v 2 V belongs to a coset; in fact, v 2 v þ W. Furthermore, u þ W ¼ v þ W if and only if u v 2 W, and so ðv þ wÞ þ W ¼ v þ W for any w 2 W. Let v 2 V. Because 0 2 W, we have v ¼ v þ 0 2 v þ W, which proves (b). Now suppose the cosets u þ W and v þ W are not disjoint; say, the vector x belongs to both u þ W and v þ W. Then u x 2 W and x v 2 W. The proof of (a) is complete if we show that u þ W ¼ v þ W. Let u þ w0 be any element in the coset u þ W. Because u x, x v, w0 belongs to W, ðu þ w0Þ v ¼ ðu xÞ þ ðx vÞ þ w0 2 W Thus, u þ w0 2 v þ W, and hence the cost u þ W is contained in the coset v þ W. Similarly, v þ W is contained in u þ W, and so u þ W ¼ v þ W. The last statement follows from the fact that u þ W ¼ v þ W if and only if u 2 v þ W, and, by Problem 10.21, this is equivalent to u v 2 W. 10.23. Let W be the solution space of the homogeneous equation 2x þ 3y þ 4z ¼ 0. Describe the cosets of W in R3. W is a plane through the origin O ¼ ð0; 0; 0Þ, and the cosets of W are the planes parallel to W. Equivalently, the cosets of W are the solution sets of the family of equations 2x þ 3y þ 4z ¼ k; k 2 R In fact, the coset v þ W, where v ¼ ða; b; cÞ, is the solution set of the linear equation 2x þ 3y þ 4z ¼ 2a þ 3b þ 4c or 2ðx aÞ þ 3ðy bÞ þ 4ðz cÞ ¼ 0 10.24. Suppose W is a subspace of a vector space V. Show that the operations in Theorem 10.15 are well deﬁned; namely, show that if u þ W ¼ u0 þ W and v þ W ¼ v0 þ W, then ðaÞ ðu þ vÞ þ W ¼ ðu0 þ v0Þ þ W and ðbÞ ku þ W ¼ ku0 þ W for any k 2 K (a) Because u þ W ¼ u0 þ W and v þ W ¼ v0 þ W, both u u0 and v v0 belong to W. But then ðu þ vÞ ðu0 þ v0Þ ¼ ðu u0Þ þ ðv v0Þ 2 W. Hence, ðu þ vÞ þ W ¼ ðu0 þ v0Þ þ W. (b) Also, because u u0 2 W implies kðu u0Þ 2 W, then ku ku0 ¼ kðu u0Þ 2 W; accordingly, ku þ W ¼ ku0 þ W. 10.25. Let V be a vector space and W a subspace of V. Show that the natural map Z: V ! V=W, deﬁned by ZðvÞ ¼ v þ W, is linear. For any u; v 2 V and any k 2 K, we have nðu þ vÞ ¼ u þ v þ W ¼ u þ W þ v þ W ¼ ZðuÞ þ ZðvÞ and ZðkvÞ ¼ kv þ W ¼ kðv þ WÞ ¼ kZðvÞ Accordingly, Z is linear. 10.26. Let W be a subspace of a vector space V. Suppose fw1; . . . ; wrg is a basis of W and the set of cosets f v1; . . . ; vsg, where vj ¼ vj þ W, is a basis of the quotient space. Show that the set of vectors B ¼ fv1; . . . ; vs, w1; . . . ; wrg is a basis of V. Thus, dim V ¼ dim W þ dimðV=WÞ. Suppose u 2 V. Because f vjg is a basis of V=W, u ¼ u þ W ¼ a1 v1 þ a2 v2 þ þ as vs Hence, u ¼ a1v1 þ þ asvs þ w, where w 2 W. Since fwig is a basis of W, u ¼ a1v1 þ þ asvs þ b1w1 þ þ brwr 340 CHAPTER 10 Canonical Forms Accordingly, B spans V. We now show that B is linearly independent. Suppose c1v1 þ þ csvs þ d1w1 þ þ drwr ¼ 0 ð1Þ Then c1 v1 þ þ cs vs ¼ 0 ¼ W Because f vjg is independent, the c’s are all 0. Substituting into (1), we ﬁnd d1w1 þ þ drwr ¼ 0. Because fwig is independent, the d’s are all 0. Thus, B is linearly independent and therefore a basis of V. 10.27. Prove Theorem 10.16: Suppose W is a subspace invariant under a linear operator T:V ! V. Then T induces a linear operator T on V=W deﬁned by Tðv þ WÞ ¼ TðvÞ þ W. Moreover, if T is a zero of any polynomial, then so is T. Thus, the minimal polynomial of T divides the minimal polynomial of T. We ﬁrst show that T is well deﬁned; that is, if u þ W ¼ v þ W, then Tðu þ WÞ ¼ Tðv þ WÞ. If u þ W ¼ v þ W, then u v 2 W, and, as W is T-invariant, Tðu vÞ ¼ TðuÞ TðvÞ 2 W. Accordingly, Tðu þ WÞ ¼ TðuÞ þ W ¼ TðvÞ þ W ¼ Tðv þ WÞ as required. We next show that T is linear. We have Tððu þ WÞ þ ðv þ WÞÞ ¼ Tðu þ v þ WÞ ¼ Tðu þ vÞ þ W ¼ TðuÞ þ TðvÞ þ W ¼ TðuÞ þ W þ TðvÞ þ W ¼ Tðu þ WÞ þ Tðv þ WÞ Furthermore, Tðkðu þ WÞÞ ¼ Tðku þ WÞ ¼ TðkuÞ þ W ¼ kTðuÞ þ W ¼ kðTðuÞ þ WÞ ¼ k ^ Tðu þ WÞ Thus, T is linear. Now, for any coset u þ W in V=W, T2ðu þ WÞ ¼ T 2ðuÞ þ W ¼ TðTðuÞÞ þ W ¼ TðTðuÞ þ WÞ ¼ Tð Tðu þ WÞÞ ¼ T2ðu þ WÞ Hence, T 2 ¼ T2. Similarly, T n ¼ Tn for any n. Thus, for any polynomial f ðtÞ ¼ antn þ þ a0 ¼ P aiti f ðTÞðu þ WÞ ¼ f ðTÞðuÞ þ W ¼ P aiTiðuÞ þ W ¼ P aiðTiðuÞ þ WÞ ¼ P aiTiðu þ WÞ ¼ P ai Tiðu þ WÞ ¼ ðP ai T iÞðu þ WÞ ¼ f ð TÞðu þ WÞ and so f ðTÞ ¼ f ð TÞ. Accordingly, if T is a root of f ðtÞ then f ðTÞ ¼ 0 ¼ W ¼ f ð TÞ; that is, T is also a root of f ðtÞ. The theorem is proved. 10.28. Prove Theorem 10.1: Let T:V ! V be a linear operator whose characteristic polynomial factors into linear polynomials. Then V has a basis in which T is represented by a triangular matrix. The proof is by induction on the dimension of V. If dim V ¼ 1, then every matrix representation of T is a 1 1 matrix, which is triangular. Now suppose dim V ¼ n > 1 and that the theorem holds for spaces of dimension less than n. Because the characteristic polynomial of T factors into linear polynomials, T has at least one eigenvalue and so at least one nonzero eigenvector v, say TðvÞ ¼ a11v. Let W be the one-dimensional subspace spanned by v. Set V ¼ V=W. Then (Problem 10.26) dim V ¼ dim V dim W ¼ n 1. Note also that W is invariant under T. By Theorem 10.16, T induces a linear operator T on V whose minimal polynomial divides the minimal polynomial of T. Because the characteristic polynomial of T is a product of linear polynomials, so is its minimal polynomial, and hence, so are the minimal and characteristic polynomials of T. Thus, V and T satisfy the hypothesis of the theorem. Hence, by induction, there exists a basis f v2; . . . ; vng of V such that Tð v2Þ ¼ a22 v2 Tð v3Þ ¼ a32 v2 þ a33 v3 ::::::::::::::::::::::::::::::::::::::::: Tð vnÞ ¼ an2 vn þ an3 v3 þ þ ann vn CHAPTER 10 Canonical Forms 341 Now let v2; . . . ; vn be elements of V that belong to the cosets v2; . . . ; vn, respectively. Then fv; v2; . . . ; vng is a basis of V (Problem 10.26). Because Tðv2Þ ¼ a22 v2, we have Tð v2Þ a22 v22 ¼ 0; and so Tðv2Þ a22v2 2 W But W is spanned by v; hence, Tðv2Þ a22v2 is a multiple of v, say, Tðv2Þ a22v2 ¼ a21v; and so Tðv2Þ ¼ a21v þ a22v2 Similarly, for i ¼ 3; . . . ; n TðviÞ ai2v2 ai3v3 aiivi 2 W; and so TðviÞ ¼ ai1v þ ai2v2 þ þ aiivi Thus, TðvÞ ¼ a11v Tðv2Þ ¼ a21v þ a22v2 :::::::::::::::::::::::::::::::::::::::: TðvnÞ ¼ an1v þ an2v2 þ þ annvn and hence the matrix of T in this basis is triangular. Cyclic Subspaces, Rational Canonical Form 10.29. Prove Theorem 10.12: Let Zðv; TÞ be a T-cyclic subspace, Tv the restriction of T to Zðv; TÞ, and mvðtÞ ¼ tk þ ak1tk1 þ þ a0 the T-annihilator of v. Then, (i) The set fv; TðvÞ; . . . ; Tk1ðvÞg is a basis of Zðv; TÞ; hence, dim Zðv; TÞ ¼ k. (ii) The minimal polynomial of Tv is mvðtÞ. (iii) The matrix of Tv in the above basis is the companion matrix C ¼ CðmvÞ of mvðtÞ [which has 1’s below the diagonal, the negative of the coefﬁcients a0; a1; . . . ; ak1 of mvðtÞ in the last column, and 0’s elsewhere]. (i) By deﬁnition of mvðtÞ, TkðvÞ is the ﬁrst vector in the sequence v, TðvÞ, T2ðvÞ; ... that, is a linear combination of those vectors that precede it in the sequence; hence, the set B ¼ fv; TðvÞ; ... ; Tk1ðvÞg is linearly independent. We now only have to show that Zðv; TÞ ¼ LðBÞ, the linear span of B. By the above, T kðvÞ 2 LðBÞ. We prove by induction that TnðvÞ 2 LðBÞ for every n. Suppose n > k and T n1ðvÞ 2 LðBÞ—that is, Tn1ðvÞ is a linear combination of v; ... ; Tk1ðvÞ. Then T nðvÞ ¼ TðTn1ðvÞÞ is a linear combination of TðvÞ; ... ; T kðvÞ. But TkðvÞ 2 LðBÞ; hence, T nðvÞ 2 LðBÞ for every n. Consequently, f ðTÞðvÞ 2 LðBÞ for any polynomial f ðtÞ. Thus, Zðv; TÞ ¼ LðBÞ, and so B is a basis, as claimed. (ii) Suppose mðtÞ ¼ ts þ bs1ts1 þ þ b0 is the minimal polynomial of Tv. Then, because v 2 Zðv; TÞ, 0 ¼ mðTvÞðvÞ ¼ mðTÞðvÞ ¼ TsðvÞ þ bs1Ts1ðvÞ þ þ b0v Thus, TsðvÞ is a linear combination of v, TðvÞ; . . . ; T s1ðvÞ, and therefore k s. However, mvðTÞ ¼ 0 and so mvðTvÞ ¼ 0: Then mðtÞ divides mvðtÞ; and so s k: Accordingly, k ¼ s and hence mvðtÞ ¼ mðtÞ. (iii) TvðvÞ ¼ TðvÞ TvðTðvÞÞ ¼ T2ðvÞ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: TvðTk2ðvÞÞ ¼ T k1ðvÞ TvðTk1ðvÞÞ ¼ T kðvÞ ¼ a0v a1TðvÞ a2T 2ðvÞ ak1T k1ðvÞ By deﬁnition, the matrix of Tv in this basis is the tranpose of the matrix of coefﬁcients of the above system of equations; hence, it is C, as required. 10.30. Let T:V ! V be linear. Let W be a T-invariant subspace of V and T the induced operator on V=W. Prove (a) The T-annihilator of v 2 V divides the minimal polynomial of T. (b) The T-annihilator of v 2 V=W divides the minimal polynomial of T. 342 CHAPTER 10 Canonical Forms (a) The T-annihilator of v 2 V is the minimal polynomial of the restriction of T to Zðv; TÞ; therefore, by Problem 10.6, it divides the minimal polynomial of T. (b) The T-annihilator of v 2 V=W divides the minimal polynomial of T, which divides the minimal polynomial of T by Theorem 10.16. Remark: In the case where the minimum polynomial of T is f ðtÞn, where f ðtÞ is a monic irreducible polynomial, then the T-annihilator of v 2 V and the T-annihilator of v 2 V=W are of the form f ðtÞm, where m n. 10.31. Prove Lemma 10.13: Let T:V ! V be a linear operator whose minimal polynomial is f ðtÞn, where f ðtÞ is a monic irreducible polynomial. Then V is the direct sum of T-cyclic subspaces Zi ¼ Zðvi; TÞ, i ¼ 1; . . . ; r, with corresponding T-annihilators f ðtÞn1; f ðtÞn2; . . . ; f ðtÞnr; n ¼ n1 n2 nr Any other decomposition of V into the direct sum of T-cyclic subspaces has the same number of components and the same set of T-annihilators. The proof is by induction on the dimension of V. If dim V ¼ 1, then V is T-cyclic and the lemma holds. Now suppose dim V > 1 and that the lemma holds for those vector spaces of dimension less than that of V. Because the minimal polynomial of T is f ðtÞn, there exists v1 2 V such that f ðTÞn1ðv1Þ 6¼ 0; hence, the T-annihilator of v1 is f ðtÞn. Let Z1 ¼ Zðv1; TÞ and recall that Z1 is T-invariant. Let V ¼ V=Z1 and let T be the linear operator on V induced by T. By Theorem 10.16, the minimal polynomial of T divides f ðtÞn; hence, the hypothesis holds for V and T. Consequently, by induction, V is the direct sum of T-cyclic subspaces; say, V ¼ Zð v2; TÞ Zð vr; T Þ where the corresponding T-annihilators are f ðtÞn2; . . . ; f ðtÞnr, n n2 nr. We claim that there is a vector v2 in the coset v2 whose T-annihilator is f ðtÞn2, the T-annihilator of v2. Let w be any vector in v2. Then f ðTÞn2ðwÞ 2 Z1. Hence, there exists a polynomial gðtÞ for which f ðTÞn2ðwÞ ¼ gðTÞðv1Þ ð1Þ Because f ðtÞn is the minimal polynomial of T, we have, by (1), 0 ¼ f ðTÞnðwÞ ¼ f ðTÞnn2gðTÞðv1Þ But f ðtÞn is the T-annihilator of v1; hence, f ðtÞn divides f ðtÞnn2gðtÞ, and so gðtÞ ¼ f ðtÞn2hðtÞ for some polynomial hðtÞ. We set v2 ¼ w hðTÞðv1Þ Because w v2 ¼ hðTÞðv1Þ 2 Z1, v2 also belongs to the coset v2. Thus, the T-annihilator of v2 is a multiple of the T-annihilator of v2. On the other hand, by (1), f ðTÞn2ðv2Þ ¼ f ðTÞnsðw hðTÞðv1ÞÞ ¼ f ðTÞn2ðwÞ gðTÞðv1Þ ¼ 0 Consequently, the T-annihilator of v2 is f ðtÞn2, as claimed. Similarly, there exist vectors v3; . . . ; vr 2 V such that vi 2 vi and that the T-annihilator of vi is f ðtÞni, the T-annihilator of vi. We set Z2 ¼ Zðv2; TÞ; . . . ; Zr ¼ Zðvr; TÞ Let d denote the degree of f ðtÞ, so that f ðtÞni has degree dni. Then, because f ðtÞni is both the T-annihilator of vi and the T-annihilator of vi, we know that fvi; TðviÞ; . . . ; Tdni1ðviÞg and f vi: TðviÞ; . . . ; Tdni1ðviÞg are bases for Zðvi; TÞ and Zðvi; TÞ, respectively, for i ¼ 2; . . . ; r. But V ¼ Zðv2; TÞ Zðvr; TÞ; hence, f v2; . . . ; T dn21ð v2Þ; . . . ; vr; . . . ; T dnr1ð vrÞg CHAPTER 10 Canonical Forms 343 is a basis for V. Therefore, by Problem 10.26 and the relation Tið vÞ ¼ TiðvÞ (see Problem 10.27), fv1; . . . ; Tdn11ðv1Þ; v2; . . . ; Ten21ðv2Þ; . . . ; vr; . . . ; Tdnr1ðvrÞg is a basis for V. Thus, by Theorem 10.4, V ¼ Zðv1; TÞ Zðvr; TÞ, as required. It remains to show that the exponents n1; . . . ; nr are uniquely determined by T. Because d ¼ degree of f ðtÞ; dim V ¼ dðn1 þ þ nrÞ and dim Zi ¼ dni; i ¼ 1; . . . ; r Also, if s is any positive integer, then (Problem 10.59) f ðTÞsðZiÞ is a cyclic subspace generated by f ðTÞsðviÞ, and it has dimension dðni sÞ if ni > s and dimension 0 if ni s. Now any vector v 2 V can be written uniquely in the form v ¼ w1 þ þ wr, where wi 2 Zi. Hence, any vector in f ðTÞsðVÞ can be written uniquely in the form f ðTÞsðvÞ ¼ f ðTÞsðw1Þ þ þ f ðTÞsðwrÞ where f ðTÞsðwiÞ 2 f ðTÞsðZiÞ. Let t be the integer, dependent on s, for which n1 > s; . . . ; nt > s; ntþ1 s Then f ðTÞsðVÞ ¼ f ðTÞsðZ1Þ f ðTÞsðZtÞ and so dim½ f ðTÞsðVÞ ¼ d½ðn1 sÞ þ þ ðnt sÞ ð2Þ The numbers on the left of (2) are uniquely determined by T. Set s ¼ n 1, and (2) determines the number of ni equal to n. Next set s ¼ n 2, and (2) determines the number of ni (if any) equal to n 1. We repeat the process until we set s ¼ 0 and determine the number of ni equal to 1. Thus, the ni are uniquely determined by T and V, and the lemma is proved. 10.32. Let V be a seven-dimensional vector space over R, and let T:V ! V be a linear operator with minimal polynomial mðtÞ ¼ ðt2 2t þ 5Þðt 3Þ3. Find all possible rational canonical forms M of T. Because dim V ¼ 7; there are only two possible characteristic polynomials, D1ðtÞ ¼ ðt2 2t þ 5Þ2 ðt 3Þ3 or D1ðtÞ ¼ ðt2 2t þ 5Þðt 3Þ5: Moreover, the sum of the orders of the companion matrices must add up to 7. Also, one companion matrix must be Cðt2 2t þ 5Þ and one must be Cððt 3Þ3Þ ¼ Cðt3 9t2 þ 27t 27Þ. Thus, M must be one of the following block diagonal matrices: (a) diag 0 5 1 2 ; 0 5 1 2 ; 0 0 27 1 0 27 0 1 9 2 4 3 5 0 @ 1 A; (b) diag 0 5 1 2 ; 0 0 27 1 0 27 0 1 9 2 4 3 5; 0 9 1 6 0 @ 1 A; (c) diag 0 5 1 2 ; 0 0 27 1 0 27 0 1 9 2 4 3 5; ½3; ½3 0 @ 1 A Projections 10.33. Suppose V ¼ W1 Wr. The projection of V into its subspace Wk is the mapping E: V ! V deﬁned by EðvÞ ¼ wk, where v ¼ w1 þ þ wr; wi 2 Wi. Show that (a) E is linear, (b) E2 ¼ E. (a) Because the sum v ¼ w1 þ þ wr, wi 2 W is uniquely determined by v, the mapping E is well deﬁned. Suppose, for u 2 V, u ¼ w0 1 þ þ w0 r, w0 i 2 Wi. Then v þ u ¼ ðw1 þ w0 1Þ þ þ ðwr þ w0 rÞ and kv ¼ kw1 þ þ kwr; kwi; wi þ w0 i 2 Wi are the unique sums corresponding to v þ u and kv. Hence, Eðv þ uÞ ¼ wk þ w0 k ¼ EðvÞ þ EðuÞ and EðkvÞ ¼ kwk þ kEðvÞ and therefore E is linear. 344 CHAPTER 10 Canonical Forms (b) We have that wk ¼ 0 þ þ 0 þ wk þ 0 þ þ 0 is the unique sum corresponding to wk 2 Wk; hence, EðwkÞ ¼ wk. Then, for any v 2 V, E2ðvÞ ¼ EðEðvÞÞ ¼ EðwkÞ ¼ wk ¼ EðvÞ Thus, E2 ¼ E, as required. 10.34. Suppose E:V ! V is linear and E2 ¼ E. Show that (a) EðuÞ ¼ u for any u 2 Im E (i.e., the restriction of E to its image is the identity mapping); (b) V is the direct sum of the image and kernel of E:V ¼ Im E Ker E; (c) E is the projection of V into Im E, its image. Thus, by the preceding problem, a linear mapping T:V ! V is a projection if and only if T2 ¼ T; this characterization of a projection is frequently used as its deﬁnition. (a) If u 2 Im E, then there exists v 2 V for which EðvÞ ¼ u; hence, as required, EðuÞ ¼ EðEðvÞÞ ¼ E2ðvÞ ¼ EðvÞ ¼ u (b) Let v 2 V. We can write v in the form v ¼ EðvÞ þ v EðvÞ. Now EðvÞ 2 Im E and, because Eðv EðvÞÞ ¼ EðvÞ E2ðvÞ ¼ EðvÞ EðvÞ ¼ 0 v EðvÞ 2 Ker E. Accordingly, V ¼ Im E þ Ker E. Now suppose w 2 Im E \ Ker E. By (i), EðwÞ ¼ w because w 2 Im E. On the other hand, EðwÞ ¼ 0 because w 2 Ker E. Thus, w ¼ 0, and so Im E \ Ker E ¼ f0g. These two conditions imply that V is the direct sum of the image and kernel of E. (c) Let v 2 V and suppose v ¼ u þ w, where u 2 Im E and w 2 Ker E. Note that EðuÞ ¼ u by (i), and EðwÞ ¼ 0 because w 2 Ker E. Hence, EðvÞ ¼ Eðu þ wÞ ¼ EðuÞ þ EðwÞ ¼ u þ 0 ¼ u That is, E is the projection of V into its image. 10.35. Suppose V ¼ U W and suppose T:V ! V is linear. Show that U and W are both T-invariant if and only if TE ¼ ET, where E is the projection of V into U. Observe that EðvÞ 2 U for every v 2 V, and that (i) EðvÞ ¼ v iff v 2 U, (ii) EðvÞ ¼ 0 iff v 2 W. Suppose ET ¼ TE. Let u 2 U. Because EðuÞ ¼ u, TðuÞ ¼ TðEðuÞÞ ¼ ðTEÞðuÞ ¼ ðETÞðuÞ ¼ EðTðuÞÞ 2 U Hence, U is T-invariant. Now let w 2 W. Because EðwÞ ¼ 0, EðTðwÞÞ ¼ ðETÞðwÞ ¼ ðTEÞðwÞ ¼ TðEðwÞÞ ¼ Tð0Þ ¼ 0; and so TðwÞ 2 W Hence, W is also T-invariant. Conversely, suppose U and W are both T-invariant. Let v 2 V and suppose v ¼ u þ w, where u 2 T and w 2 W. Then TðuÞ 2 U and TðwÞ 2 W; hence, EðTðuÞÞ ¼ TðuÞ and EðTðwÞÞ ¼ 0. Thus, ðETÞðvÞ ¼ ðETÞðu þ wÞ ¼ ðETÞðuÞ þ ðETÞðwÞ ¼ EðTðuÞÞ þ EðTðwÞÞ ¼ TðuÞ and ðTEÞðvÞ ¼ ðTEÞðu þ wÞ ¼ TðEðu þ wÞÞ ¼ TðuÞ That is, ðETÞðvÞ ¼ ðTEÞðvÞ for every v 2 V; therefore, ET ¼ TE, as required. SUPPLEMENTARY PROBLEMS Invariant Subspaces 10.36. Suppose W is invariant under T:V ! V. Show that W is invariant under f ðTÞ for any polynomial f ðtÞ. 10.37. Show that every subspace of V is invariant under I and 0, the identity and zero operators. CHAPTER 10 Canonical Forms 345 10.38. Let W be invariant under T1: V ! V and T2: V ! V. Prove W is also invariant under T1 þ T2 and T1T2. 10.39. Let T:V ! V be linear. Prove that any eigenspace, El is T-invariant. 10.40. Let V be a vector space of odd dimension (greater than 1) over the real ﬁeld R. Show that any linear operator on V has an invariant subspace other than V or f0g. 10.41. Determine the invariant subspace of A ¼ 2 4 5 2 viewed as a linear operator on (a) R2, (b) C2. 10.42. Suppose dim V ¼ n. Show that T:V ! V has a triangular matrix representation if and only if there exist T-invariant subspaces W1 W2 Wn ¼ V for which dim Wk ¼ k, k ¼ 1; . . . ; n. Invariant Direct Sums 10.43. The subspaces W1; . . . ; Wr are said to be independent if w1 þ þ wr ¼ 0, wi 2 Wi, implies that each wi ¼ 0. Show that spanðWiÞ ¼ W1 Wr if and only if the Wi are independent. [Here spanðWiÞ denotes the linear span of the Wi.] 10.44. Show that V ¼ W1 Wr if and only if (i) V ¼ spanðWiÞ and (ii) for k ¼ 1; 2; . . . ; r, Wk \ spanðW1; . . . ; Wk1; Wkþ1; . . . ; WrÞ ¼ f0g. 10.45. Show that spanðWiÞ ¼ W1 Wr if and only if dim ½spanðWiÞ ¼ dim W1 þ þ dim Wr. 10.46. Suppose the characteristic polynomial of T:V ! V is DðtÞ ¼ f1ðtÞn1f2ðtÞn2 frðtÞnr, where the fiðtÞ are distinct monic irreducible polynomials. Let V ¼ W1 Wr be the primary decomposition of V into T-invariant subspaces. Show that fiðtÞni is the characteristic polynomial of the restriction of T to Wi. Nilpotent Operators 10.47. Suppose T1 and T2 are nilpotent operators that commute (i.e., T1T2 ¼ T2T1). Show that T1 þ T2 and T1T2 are also nilpotent. 10.48. Suppose A is a supertriangular matrix (i.e., all entries on and below the main diagonal are 0). Show that A is nilpotent. 10.49. Let V be the vector space of polynomials of degree n. Show that the derivative operator on V is nilpotent of index n þ 1. 10.50. Show that any Jordan nilpotent block matrix N is similar to its transpose NT (the matrix with 1’s below the diagonal and 0’s elsewhere). 10.51. Show that two nilpotent matrices of order 3 are similar if and only if they have the same index of nilpotency. Show by example that the statement is not true for nilpotent matrices of order 4. Jordan Canonical Form 10.52. Find all possible Jordan canonical forms for those matrices whose characteristic polynomial DðtÞ and minimal polynomial mðtÞ are as follows: (a) DðtÞ ¼ ðt 2Þ4ðt 3Þ2; mðtÞ ¼ ðt 2Þ2ðt 3Þ2, (b) DðtÞ ¼ ðt 7Þ5; mðtÞ ¼ ðt 7Þ2, (c) DðtÞ ¼ ðt 2Þ7; mðtÞ ¼ ðt 2Þ3 10.53. Show that every complex matrix is similar to its transpose. (Hint: Use its Jordan canonical form.) 10.54. Show that all n n complex matrices A for which An ¼ I but Ak 6¼ I for k < n are similar. 10.55. Suppose A is a complex matrix with only real eigenvalues. Show that A is similar to a matrix with only real entries. 346 CHAPTER 10 Canonical Forms Cyclic Subspaces 10.56. Suppose T:V ! V is linear. Prove that Zðv; TÞ is the intersection of all T-invariant subspaces containing v. 10.57. Let f ðtÞ and gðtÞ be the T-annihilators of u and v, respectively. Show that if f ðtÞ and gðtÞ are relatively prime, then f ðtÞgðtÞ is the T-annihilator of u þ v. 10.58. Prove that Zðu; TÞ ¼ Zðv; TÞ if and only if gðTÞðuÞ ¼ v where gðtÞ is relatively prime to the T-annihilator of u. 10.59. Let W ¼ Zðv; TÞ, and suppose the T-annihilator of v is f ðtÞn, where f ðtÞ is a monic irreducible polynomial of degree d. Show that f ðTÞsðWÞ is a cyclic subspace generated by f ðTÞsðvÞ and that it has dimension dðn sÞ if n > s and dimension 0 if n s. Rational Canonical Form 10.60. Find all possible rational forms for a 6 6 matrix over R with minimal polynomial: (a) mðtÞ ¼ ðt2 2t þ 3Þðt þ 1Þ2, (b) mðtÞ ¼ ðt 2Þ3. 10.61. Let A be a 4 4 matrix with minimal polynomial mðtÞ ¼ ðt2 þ 1Þðt2 3Þ. Find the rational canonical form for A if A is a matrix over (a) the rational ﬁeld Q, (b) the real ﬁeld R, (c) the complex ﬁeld C. 10.62. Find the rational canonical form for the four-square Jordan block with l’s on the diagonal. 10.63. Prove that the characteristic polynomial of an operator T:V ! V is a product of its elementary divisors. 10.64. Prove that two 3 3 matrices with the same minimal and characteristic polynomials are similar. 10.65. Let Cð f ðtÞÞ denote the companion matrix to an arbitrary polynomial f ðtÞ. Show that f ðtÞ is the characteristic polynomial of Cð f ðtÞÞ. Projections 10.66. Suppose V ¼ W1 Wr. Let Ei denote the projection of V into Wi. Prove (i) EiEj ¼ 0, i 6¼ j; (ii) I ¼ E1 þ þ Er. 10.67. Let E1; . . . ; Er be linear operators on V such that (i) E2 i ¼ Ei (i.e., the Ei are projections); (ii) EiEj ¼ 0, i 6¼ j; (iii) I ¼ E1 þ þ Er Prove that V ¼ Im E1 Im Er. 10.68. Suppose E: V ! V is a projection (i.e., E2 ¼ E). Prove that E has a matrix representation of the form Ir 0 0 0 , where r is the rank of E and Ir is the r-square identity matrix. 10.69. Prove that any two projections of the same rank are similar. (Hint: Use the result of Problem 10.68.) 10.70. Suppose E: V ! V is a projection. Prove (i) I E is a projection and V ¼ Im E Im ðI EÞ, (ii) I þ E is invertible (if 1 þ 1 6¼ 0). Quotient Spaces 10.71. Let W be a subspace of V. Suppose the set of cosets fv1 þ W; v2 þ W; . . . ; vn þ Wg in V=W is linearly independent. Show that the set of vectors fv1; v2; . . . ; vng in V is also linearly independent. 10.72. Let W be a substance of V. Suppose the set of vectors fu1; u2; . . . ; ung in V is linearly independent, and that LðuiÞ \ W ¼ f0g. Show that the set of cosets fu1 þ W; . . . ; un þ Wg in V=W is also linearly independent. CHAPTER 10 Canonical Forms 347 10.73. Suppose V ¼ U W and that fu1; . . . ; ung is a basis of U. Show that fu1 þ W; . . . ; un þ Wg is a basis of the quotient spaces V=W. (Observe that no condition is placed on the dimensionality of V or W.) 10.74. Let W be the solution space of the linear equation a1x1 þ a2x2 þ þ anxn ¼ 0; ai 2 K and let v ¼ ðb1; b2; . . . ; bnÞ 2 Kn. Prove that the coset v þ W of W in Kn is the solution set of the linear equation a1x1 þ a2x2 þ þ anxn ¼ b; where b ¼ a1b1 þ þ anbn 10.75. Let V be the vector space of polynomials over R and let W be the subspace of polynomials divisible by t4 (i.e., of the form a0t4 þ a1t5 þ þ an4tn). Show that the quotient space V=W has dimension 4. 10.76. Let U and W be subspaces of V such that W U V. Note that any coset u þ W of W in U may also be viewed as a coset of W in V, because u 2 U implies u 2 V; hence, U=W is a subset of V=W. Prove that (i) U=W is a subspace of V=W, (ii) dimðV=WÞ dimðU=WÞ ¼ dimðV=UÞ. 10.77. Let U and W be subspaces of V. Show that the cosets of U \ W in V can be obtained by intersecting each of the cosets of U in V by each of the cosets of W in V: V=ðU \ WÞ ¼ fðv þ UÞ \ ðv0 þ WÞ : v; v0 2 Vg 10.78. Let T:V ! V 0 be linear with kernel W and image U. Show that the quotient space V=W is isomorphic to U under the mapping y:V=W ! U deﬁned by yðv þ WÞ ¼ TðvÞ. Furthermore, show that T ¼ i y Z, where Z:V ! V=W is the natural mapping of V into V=W (i.e., ZðvÞ ¼ v þ W), and i:U , ! V 0 is the inclusion mapping (i.e., iðuÞ ¼ u). (See diagram.) ANSWERS TO SUPPLEMENTARY PROBLEMS 10.41. (a) R2 and f0g, (b) C2; f0g; W1 ¼ spanð2; 1 2iÞ; W2 ¼ spanð2; 1 þ 2iÞ 10.52. (a) diag 2 1 2 ; 2 1 2 ; 3 1 3 ; diag 2 1 2 ; ½2: ½2; 3 1 3 ; (b) diag 7 1 7 ; 7 1 7 ; ½7 ; diag 7 1 7 ; ½7; ½7; ½7 ; (c) Let Mk denote a Jordan block with l ¼ 2 and order k. Then diagðM3; M3; M1Þ, diagðM3; M2; M2Þ, diagðM3; M2; M1; M1Þ, diagðM3; M1; M1; M1; M1Þ 10.60. Let A ¼ 0 3 1 2 ; B ¼ 0 1 1 2 ; C ¼ 0 0 8 1 0 12 0 1 6 2 4 3 5; D ¼ 0 4 1 4 . (a) diagðA; A; BÞ; diagðA; B; BÞ; diagðA; B; 1; 1Þ; (b) diagðC; CÞ; diagðC; D; 2Þ; diagðC; 2; 2; 2Þ 10.61. Let A ¼ 0 1 1 0 ; B ¼ 0 3 1 0 . (a) diagðA; BÞ, (b) diagðA; ﬃﬃﬃ 3 p ; ﬃﬃﬃ 3 p Þ, (c) diagði; i; ﬃﬃﬃ 3 p ; ﬃﬃﬃ 3 p Þ 10.62. Companion matrix with the last column ½l4; 4l3; 6l2; 4lT 348 CHAPTER 10 Canonical Forms CHAPTER 11 Linear Functionals and the Dual Space 11.1 Introduction In this chapter, we study linear mappings from a vector space V into its ﬁeld K of scalars. (Unless otherwise stated or implied, we view K as a vector space over itself.) Naturally all the theorems and results for arbitrary mappings on V hold for this special case. However, we treat these mappings separately because of their fundamental importance and because the special relationship of V to K gives rise to new notions and results that do not apply in the general case. 11.2 Linear Functionals and the Dual Space Let V be a vector space over a ﬁeld K. A mapping f:V ! K is termed a linear functional (or linear form) if, for every u; v 2 V and every a; b; 2 K, fðau þ bvÞ ¼ afðuÞ þ bfðvÞ In other words, a linear functional on V is a linear mapping from V into K. EXAMPLE 11.1 (a) Let pi:Kn ! K be the ith projection mapping; that is, piða1; a2; . . . anÞ ¼ ai. Then pi is linear and so it is a linear functional on Kn. (b) Let V be the vector space of polynomials in t over R. Let J:V ! R be the integral operator deﬁned by JðpðtÞÞ ¼ Ð 1 0 pðtÞ dt. Recall that J is linear; and hence, it is a linear functional on V. (c) Let V be the vector space of n-square matrices over K. Let T :V ! K be the trace mapping TðAÞ ¼ a11 þ a22 þ þ ann; where A ¼ ½aij That is, T assigns to a matrix A the sum of its diagonal elements. This map is linear (Problem 11.24), and so it is a linear functional on V. By Theorem 5.10, the set of linear functionals on a vector space V over a ﬁeld K is also a vector space over K, with addition and scalar multiplication deﬁned by ðf þ sÞðvÞ ¼ fðvÞ þ sðvÞ and ðkfÞðvÞ ¼ kfðvÞ where f and s are linear functionals on V and k 2 K. This space is called the dual space of V and is denoted by V. EXAMPLE 11.2 Let V ¼ Kn, the vector space of n-tuples, which we write as column vectors. Then the dual space V can be identiﬁed with the space of row vectors. In particular, any linear functional f ¼ ða1; ... ; anÞ in V has the representation fðx1; x2; . . . ; xnÞ ¼ ½a1; a2; . . . ; an½x2; x2; . . . ; xnT ¼ a1x1 þ a2x2 þ þ anxn Historically, the formal expression on the right was termed a linear form. CHAPTER 11 349 11.3 Dual Basis Suppose V is a vector space of dimension n over K. By Theorem 5.11, the dimension of the dual space V is also n (because K is of dimension 1 over itself). In fact, each basis of V determines a basis of V as follows (see Problem 11.3 for the proof). THEOREM 11.1: Suppose fv1; . . . ; vng is a basis of V over K. Let f1; . . . ; fn 2 V be the linear functionals as deﬁned by fiðvjÞ ¼ dij ¼ 1 if i ¼ j 0 if i 6¼ j Then ff1; . . . ; fng is a basis of V: The above basis ffig is termed the basis dual to fvig or the dual basis. The above formula, which uses the Kronecker delta dij, is a short way of writing f1ðv1Þ ¼ 1; f1ðv2Þ ¼ 0; f1ðv3Þ ¼ 0; . . . ; f1ðvnÞ ¼ 0 f2ðv1Þ ¼ 0; f2ðv2Þ ¼ 1; f2ðv3Þ ¼ 0; . . . ; f2ðvnÞ ¼ 0 :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: fnðv1Þ ¼ 0; fnðv2Þ ¼ 0; . . . ; fnðvn1Þ ¼ 0; fnðvnÞ ¼ 1 By Theorem 5.2, these linear mappings fi are unique and well deﬁned. EXAMPLE 11.3 Consider the basis fv1 ¼ ð2; 1Þ; v2 ¼ ð3; 1Þg of R2. Find the dual basis ff1; f2g. We seek linear functionals f1ðx; yÞ ¼ ax þ by and f2ðx; yÞ ¼ cx þ dy such that f1ðv1Þ ¼ 1; f1ðv2Þ ¼ 0; f2ðv2Þ ¼ 0; f2ðv2Þ ¼ 1 These four conditions lead to the following two systems of linear equations: f1ðv1Þ ¼ f1ð2; 1Þ ¼ 2a þ b ¼ 1 f1ðv2Þ ¼ f1ð3; 1Þ ¼ 3a þ b ¼ 0 and f2ðv1Þ ¼ f2ð2; 1Þ ¼ 2c þ d ¼ 0 f2ðv2Þ ¼ f2ð3; 1Þ ¼ 3c þ d ¼ 1 The solutions yield a ¼ 1, b ¼ 3 and c ¼ 1, d ¼ 2. Hence, f1ðx; yÞ ¼ x þ 3y and f2ðx; yÞ ¼ x 2y form the dual basis. The next two theorems (proved in Problems 11.4 and 11.5, respectively) give relationships between bases and their duals. THEOREM 11.2: Let fv1; . . . ; vng be a basis of V and let ff1; . . . ; fng be the dual basis in V. Then (i) For any vector u 2 V, u ¼ f1ðuÞv1 þ f2ðuÞv2 þ þ fnðuÞvn. (ii) For any linear functional s 2 V, s ¼ sðv1Þf1 þ sðv2Þf2 þ þ sðvnÞfn. THEOREM 11.3: Let fv1; . . . ; vng and fw1; . . . ; wng be bases of V and let ff1; . . . ; fng and fs1; . . . ; sng be the bases of V dual to fvig and fwig, respectively. Suppose P is the change-of-basis matrix from fvig to fwig. Then ðP1ÞT is the change-of-basis matrix from ffig to fsig. 11.4 Second Dual Space We repeat: Every vector space V has a dual space V, which consists of all the linear functionals on V. Thus, V has a dual space V, called the second dual of V, which consists of all the linear functionals on V. We now show that each v 2 V determines a speciﬁc element ^ v 2 V. First, for any f 2 V, we deﬁne ^ vðfÞ ¼ fðvÞ 350 CHAPTER 11 Linear Functionals and the Dual Space It remains to be shown that this map ^ v:V ! K is linear. For any scalars a; b 2 K and any linear functionals f; s 2 V, we have ^ vðaf þ bsÞ ¼ ðaf þ bsÞðvÞ ¼ afðvÞ þ bsðvÞ ¼ a^ vðfÞ þ b^ vðsÞ That is, ^ v is linear and so ^ v 2 V. The following theorem (proved in Problem 12.7) holds. THEOREM 11.4: If V has ﬁnite dimensions, then the mapping v 7! ^ v is an isomorphism of V onto V. The above mapping v 7! ^ v is called the natural mapping of V into V. We emphasize that this mapping is never onto V if V is not ﬁnite-dimensional. However, it is always linear, and moreover, it is always one-to-one. Now suppose V does have ﬁnite dimension. By Theorem 11.4, the natural mapping determines an isomorphism between V and V. Unless otherwise stated, we will identify V with V by this mapping. Accordingly, we will view V as the space of linear functionals on V and write V ¼ V. We remark that if ffig is the basis of V dual to a basis fvig of V, then fvig is the basis of V ¼ V that is dual to ffig. 11.5 Annihilators Let W be a subset (not necessarily a subspace) of a vector space V. A linear functional f 2 V is called an annihilator of W if fðwÞ ¼ 0 for every w 2 W—that is, if fðWÞ ¼ f0g. We show that the set of all such mappings, denoted by W 0 and called the annihilator of W, is a subspace of V. Clearly, 0 2 W 0: Now suppose f; s 2 W 0. Then, for any scalars a; b; 2 K and for any w 2 W, ðaf þ bsÞðwÞ ¼ afðwÞ þ bsðwÞ ¼ a0 þ b0 ¼ 0 Thus, af þ bs 2 W 0, and so W 0 is a subspace of V. In the case that W is a subspace of V, we have the following relationship between W and its annihilator W 0 (see Problem 11.11 for the proof). THEOREM 11.5: Suppose V has ﬁnite dimension and W is a subspace of V. Then ðiÞ dim W þ dim W 0 ¼ dim V and ðiiÞ W 00 ¼ W Here W 00 ¼ fv 2 V :fðvÞ ¼ 0 for every f 2 W 0g or, equivalently, W 00 ¼ ðW 0Þ0, where W 00 is viewed as a subspace of V under the identiﬁcation of V and V. 11.6 Transpose of a Linear Mapping Let T :V ! U be an arbitrary linear mapping from a vector space V into a vector space U. Now for any linear functional f 2 U, the composition f T is a linear mapping from V into K: That is, f T 2 V. Thus, the correspondence f 7! f T is a mapping from U into V; we denote it by Tt and call it the transpose of T. In other words, Tt:U ! V is deﬁned by TtðfÞ ¼ f T Thus, ðTtðfÞÞðvÞ ¼ fðTðvÞÞ for every v 2 V. CHAPTER 11 Linear Functionals and the Dual Space 351 THEOREM 11.6: The transpose mapping Tt deﬁned above is linear. Proof. For any scalars a; b 2 K and any linear functionals f; s 2 U, Ttðaf þ bsÞ ¼ ðaf þ bsÞ T ¼ aðf TÞ þ bðs TÞ ¼ aTtðfÞ þ bTtðsÞ That is, Tt is linear, as claimed. We emphasize that if T is a linear mapping from V into U, then Tt is a linear mapping from U into V. The same ‘‘transpose’’ for the mapping Tt no doubt derives from the following theorem (proved in Problem 11.16). THEOREM 11.7: Let T :V ! U be linear, and let A be the matrix representation of T relative to bases fvig of V and fuig of U. Then the transpose matrix AT is the matrix representation of Tt:U ! V relative to the bases dual to fuig and fvig. SOLVED PROBLEMS Dual Spaces and Dual Bases 11.1. Find the basis ff1; f2; f3g that is dual to the following basis of R3: fv1 ¼ ð1; 1; 3Þ; v2 ¼ ð0; 1; 1Þ; v3 ¼ ð0; 3; 2Þg The linear functionals may be expressed in the form f1ðx; y; zÞ ¼ a1x þ a2y þ a3z; f2ðx; y; zÞ ¼ b1x þ b2y þ b3z; f3ðx; y; zÞ ¼ c1x þ c2y þ c3z By deﬁnition of the dual basis, fiðvjÞ ¼ 0 for i 6¼ j, but fiðvjÞ ¼ 1 for i ¼ j. We ﬁnd f1 by setting f1ðv1Þ ¼ 1; f1ðv2Þ ¼ 0; f1ðv3Þ ¼ 0: This yields f1ð1; 1; 3Þ ¼ a1 a2 þ 3a3 ¼ 1; f1ð0; 1; 1Þ ¼ a2 a3 ¼ 0; f1ð0; 3; 2Þ ¼ 3a2 2a3 ¼ 0 Solving the system of equations yields a1 ¼ 1, a2 ¼ 0, a3 ¼ 0. Thus, f1ðx; y; zÞ ¼ x. We ﬁnd f2 by setting f2ðv1Þ ¼ 0, f2ðv2Þ ¼ 1, f2ðv3Þ ¼ 0. This yields f2ð1; 1; 3Þ ¼ b1 b2 þ 3b3 ¼ 0; f2ð0; 1; 1Þ ¼ b2 b3 ¼ 1; f2ð0; 3; 2Þ ¼ 3b2 2b3 ¼ 0 Solving the system of equations yields b1 ¼ 7, b2 ¼ 2, a3 ¼ 3. Thus, f2ðx; y; zÞ ¼ 7x 2y 3z. We ﬁnd f3 by setting f3ðv1Þ ¼ 0, f3ðv2Þ ¼ 0, f3ðv3Þ ¼ 1. This yields f3ð1; 1; 3Þ ¼ c1 c2 þ 3c3 ¼ 0; f3ð0; 1; 1Þ ¼ c2 c3 ¼ 0; f3ð0; 3; 2Þ ¼ 3c2 2c3 ¼ 1 Solving the system of equations yields c1 ¼ 2, c2 ¼ 1, c3 ¼ 1. Thus, f3ðx; y; zÞ ¼ 2x þ y þ z. 11.2. Let V ¼ fa þ bt : a; b 2 Rg, the vector space of real polynomials of degree 1. Find the basis fv1; v2g of V that is dual to the basis ff1; f2g of V deﬁned by f1ð f ðtÞÞ ¼ ð1 0 f ðtÞ dt and f2ð f ðtÞÞ ¼ ð2 0 f ðtÞ dt Let v1 ¼ a þ bt and v2 ¼ c þ dt. By deﬁnition of the dual basis, f1ðv1Þ ¼ 1; f1ðv2Þ ¼ 0 and f2ðv1Þ ¼ 0; fiðvjÞ ¼ 1 Thus, f1ðv1Þ ¼ Ð 1 0 ða þ btÞ dt ¼ a þ 1 2 b ¼ 1 f2ðv1Þ ¼ Ð 2 0 ða þ btÞ dt ¼ 2a þ 2b ¼ 0 ) and f1ðv2Þ ¼ Ð 1 0 ðc þ dtÞ dt ¼ c þ 1 2 d ¼ 0 f2ðv2Þ ¼ Ð 2 0 ðc þ dtÞ dt ¼ 2c þ 2d ¼ 1 ) Solving each system yields a ¼ 2, b ¼ 2 and c ¼ 1 2, d ¼ 1. Thus, fv1 ¼ 2 2t; v2 ¼ 1 2 þ tg is the basis of V that is dual to ff1; f2g. 352 CHAPTER 11 Linear Functionals and the Dual Space 11.3. Prove Theorem 11.1: Suppose fv1; . . . ; vng is a basis of V over K. Let f1; . . . ; fn 2 V be deﬁned by fiðvjÞ ¼ 0 for i 6¼ j, but fiðvjÞ ¼ 1 for i ¼ j. Then ff1; . . . ; fng is a basis of V. We ﬁrst show that ff1; . . . ; fng spans V. Let f be an arbitrary element of V, and suppose fðv1Þ ¼ k1; fðv2Þ ¼ k2; . . . ; fðvnÞ ¼ kn Set s ¼ k1f1 þ þ knfn. Then sðv1Þ ¼ ðk1f1 þ þ knfnÞðv1Þ ¼ k1f1ðv1Þ þ k2f2ðv1Þ þ þ knfnðv1Þ ¼ k1 1 þ k2 0 þ þ kn 0 ¼ k1 Similarly, for i ¼ 2; . . . ; n, sðviÞ ¼ ðk1f1 þ þ knfnÞðviÞ ¼ k1f1ðviÞ þ þ kifiðviÞ þ þ knfnðviÞ ¼ ki Thus, fðviÞ ¼ sðviÞ for i ¼ 1; ... ; n. Because f and s agree on the basis vectors, f ¼ s ¼ k1f1 þ þ knfn. Accordingly, ff1; ... ; fng spans V. It remains to be shown that ff1; . . . ; fng is linearly independent. Suppose a1f1 þ a2f2 þ þ anfn ¼ 0 Applying both sides to v1, we obtain 0 ¼ 0ðv1Þ ¼ ða1f1 þ þ anfnÞðv1Þ ¼ a1f1ðv1Þ þ a2f2ðv1Þ þ þ anfnðv1Þ ¼ a1 1 þ a2 0 þ þ an 0 ¼ a1 Similarly, for i ¼ 2; . . . ; n, 0 ¼ 0ðviÞ ¼ ða1f1 þ þ anfnÞðviÞ ¼ a1f1ðviÞ þ þ aifiðviÞ þ þ anfnðviÞ ¼ ai That is, a1 ¼ 0; . . . ; an ¼ 0. Hence, ff1; . . . ; fng is linearly independent, and so it is a basis of V. 11.4. Prove Theorem 11.2: Let fv1; . . . ; vng be a basis of V and let ff1; . . . ; fng be the dual basis in V. For any u 2 V and any s 2 V, (i) u ¼ P i fiðuÞvi. (ii) s ¼ P i fðviÞfi. Suppose u ¼ a1v1 þ a2v2 þ þ anvn ð1Þ Then f1ðuÞ ¼ a1f1ðv1Þ þ a2f1ðv2Þ þ þ anf1ðvnÞ ¼ a1 1 þ a2 0 þ þ an 0 ¼ a1 Similarly, for i ¼ 2; . . . ; n, fiðuÞ ¼ a1fiðv1Þ þ þ aifiðviÞ þ þ anfiðvnÞ ¼ ai That is, f1ðuÞ ¼ a1, f2ðuÞ ¼ a2; . . . ; fnðuÞ ¼ an. Substituting these results into (1), we obtain (i). Next we prove ðiiÞ. Applying the linear functional s to both sides of (i), sðuÞ ¼ f1ðuÞsðv1Þ þ f2ðuÞsðv2Þ þ þ fnðuÞsðvnÞ ¼ sðv1Þf1ðuÞ þ sðv2Þf2ðuÞ þ þ sðvnÞfnðuÞ ¼ ðsðv1Þf1 þ sðv2Þf2 þ þ sðvnÞfnÞðuÞ Because the above holds for every u 2 V, s ¼ sðv1Þf2 þ sðv2Þf2 þ þ sðvnÞfn, as claimed. 11.5. Prove Theorem 11.3. Let fvig and fwig be bases of V and let ffig and fsig be the respective dual bases in V. Let P be the change-of-basis matrix from fvig to fwig: Then ðP1ÞT is the change-of-basis matrix from ffig to fsig. Suppose, for i ¼ 1; . . . ; n, wi ¼ ai1v1 þ ai2v2 þ þ ainvn and si ¼ bi1f1 þ bi2f2 þ þ ainvn Then P ¼ ½aij and Q ¼ ½bij. We seek to prove that Q ¼ ðP1ÞT. Let Ri denote the ith row of Q and let Cj denote the jth column of PT. Then Ri ¼ ðbi1; bi2; . . . ; binÞ and Cj ¼ ðaj1; aj2; . . . ; ajnÞT CHAPTER 11 Linear Functionals and the Dual Space 353 By deﬁnition of the dual basis, siðwjÞ ¼ ðbi1f1 þ bi2f2 þ þ binfnÞðaj1v1 þ aj2v2 þ þ ajnvnÞ ¼ bi1aj1 þ bi2aj2 þ þ binajn ¼ RiCj ¼ dij where dij is the Kronecker delta. Thus, QPT ¼ ½RiCj ¼ ½dij ¼ I Therefore, Q ¼ ðPTÞ1 ¼ ðP1ÞT, as claimed. 11.6. Suppose v 2 V, v 6¼ 0, and dim V ¼ n. Show that there exists f 2 V such that fðvÞ 6¼ 0. We extend fvg to a basis fv; v2; . . . ; vng of V. By Theorem 5.2, there exists a unique linear mapping f:V ! K such that fðvÞ ¼ 1 and fðviÞ ¼ 0, i ¼ 2; . . . ; n. Hence, f has the desired property. 11.7. Prove Theorem 11.4: Suppose dim V ¼ n. Then the natural mapping v 7! ^ v is an isomorphism of V onto V. We ﬁrst prove that the map v 7! ^ v is linear—that is, for any vectors v; w 2 V and any scalars a; b 2 K, av þ bw ¼ a^ v þ b^ w. For any linear functional f 2 V, av þ bwðfÞ ¼ fðav þ bwÞ ¼ afðvÞ þ bfðwÞ ¼ a^ vðfÞ þ b ^ wðfÞ ¼ ða^ v þ b^ wÞðfÞ Because av þ bwðfÞ ¼ ða^ v þ b^ wÞðfÞ for every f 2 V, we have av þ bw ¼ a^ v þ b^ w. Thus, the map v 7! ^ v is linear. Now suppose v 2 V, v 6¼ 0. Then, by Problem 11.6, there exists f 2 V for which fðvÞ 6¼ 0. Hence, ^ vðfÞ ¼ fðvÞ 6¼ 0, and thus ^ v 6¼ 0. Because v 6¼ 0 implies ^ v 6¼ 0, the map v 7! ^ v is nonsingular and hence an isomorphism (Theorem 5.64). Now dim V ¼ dim V ¼ dim V, because V has ﬁnite dimension. Accordingly, the mapping v 7! ^ v is an isomorphism of V onto V. Annihilators 11.8. Show that if f 2 V annihilates a subset S of V, then f annihilates the linear span LðSÞ of S. Hence, S0 ¼ ½spanðSÞ0. Suppose v 2 spanðSÞ. Then there exists w1; . . . ; wr 2 S for which v ¼ a1w1 þ a2w2 þ þ arwr. fðvÞ ¼ a1fðw1Þ þ a2fðw2Þ þ þ arfðwrÞ ¼ a10 þ a20 þ þ ar0 ¼ 0 Because v was an arbitrary element of spanðSÞ; f annihilates spanðSÞ, as claimed. 11.9. Find a basis of the annihilator W 0 of the subspace W of R4 spanned by v1 ¼ ð1; 2; 3; 4Þ and v2 ¼ ð0; 1; 4; 1Þ By Problem 11.8, it sufﬁces to ﬁnd a basis of the set of linear functionals f such that fðv1Þ ¼ 0 and fðv2Þ ¼ 0, where fðx1; x2; x3; x4Þ ¼ ax1 þ bx2 þ cx3 þ dx4. Thus, fð1; 2; 3; 4Þ ¼ a þ 2b 3c þ 4d ¼ 0 and fð0; 1; 4; 1Þ ¼ b þ 4c d ¼ 0 The system of two equations in the unknowns a; b; c; d is in echelon form with free variables c and d. (1) Set c ¼ 1, d ¼ 0 to obtain the solution a ¼ 11, b ¼ 4, c ¼ 1, d ¼ 0. (2) Set c ¼ 0, d ¼ 1 to obtain the solution a ¼ 6, b ¼ 1, c ¼ 0, d ¼ 1. The linear functions f1ðxiÞ ¼ 11x1 4x2 þ x3 and f2ðxiÞ ¼ 6x1 x2 þ x4 form a basis of W 0. 11.10. Show that (a) For any subset S of V; S S00. (b) If S1 S2, then S0 2 S0 1. (a) Let v 2 S. Then for every linear functional f 2 S0, ^ vðfÞ ¼ fðvÞ ¼ 0. Hence, ^ v 2 ðS0Þ0. Therefore, under the identiﬁcation of V and V, v 2 S00. Accordingly, S S00. (b) Let f 2 S0 2. Then fðvÞ ¼ 0 for every v 2 S2. But S1 S2; hence, f annihilates every element of S1 (i.e., f 2 S0 1). Therefore, S0 2 S0 1. d d d d 354 CHAPTER 11 Linear Functionals and the Dual Space 11.11. Prove Theorem 11.5: Suppose V has ﬁnite dimension and W is a subspace of V. Then (i) dim W þ dim W 0 ¼ dim V, (ii) W 00 ¼ W. (i) Suppose dim V ¼ n and dim W ¼ r n. We want to show that dim W 0 ¼ n r. We choose a basis fw1; . . . ; wrg of W and extend it to a basis of V, say fw1; . . . ; wr; v1; . . . ; vnrg. Consider the dual basis ff1; . . . ; fr; s1; . . . ; snrg By deﬁnition of the dual basis, each of the above s’s annihilates each wi; hence, s1; . . . ; snr 2 W 0. We claim that fsig is a basis of W 0. Now fsjg is part of a basis of V, and so it is linearly independent. We next show that ffjg spans W 0. Let s 2 W 0. By Theorem 11.2, s ¼ sðw1Þf1 þ þ sðwrÞfr þ sðv1Þs1 þ þ sðvnrÞsnr ¼ 0f1 þ þ 0fr þ sðv1Þs1 þ þ sðvnrÞsnr ¼ sðv1Þs1 þ þ sðvnrÞsnr Consequently, fs1; . . . ; snrg spans W 0 and so it is a basis of W 0. Accordingly, as required dim W 0 ¼ n r ¼ dim V dim W: (ii) Suppose dim V ¼ n and dim W ¼ r. Then dim V ¼ n and, by (i), dim W 0 ¼ n r. Thus, by (i), dim W 00 ¼ n ðn rÞ ¼ r; therefore, dim W ¼ dim W 00. By Problem 11.10, W W 00. Accord-ingly, W ¼ W 00. 11.12. Let U and W be subspaces of V. Prove that ðU þ WÞ0 ¼ U 0 \ W 0. Let f 2 ðU þ WÞ0. Then f annihilates U þ W; and so, in particular, f annihilates U and W: That is, f 2 U0 and f 2 W 0; hence, f 2 U0 \ W 0: Thus, ðU þ WÞ0 U0 \ W 0: On the other hand, suppose s 2 U0 \ W 0: Then s annihilates U and also W. If v 2 U þ W, then v ¼ u þ w, where u 2 U and w 2 W. Hence, sðvÞ ¼ sðuÞ þ sðwÞ ¼ 0 þ 0 ¼ 0. Thus, s annihilates U þ W; that is, s 2 ðU þ WÞ0. Accordingly, U0 þ W 0 ðU þ WÞ0. The two inclusion relations together give us the desired equality. Remark: Observe that no dimension argument is employed in the proof; hence, the result holds for spaces of ﬁnite or inﬁnite dimension. Transpose of a Linear Mapping 11.13. Let f be the linear functional on R2 deﬁned by fðx; yÞ ¼ x 2y. For each of the following linear operators T on R2, ﬁnd ðTtðfÞÞðx; yÞ: (a) Tðx; yÞ ¼ ðx; 0Þ, (b) Tðx; yÞ ¼ ðy; x þ yÞ, (c) Tðx; yÞ ¼ ð2x 3y; 5x þ 2yÞ By deﬁnition, TtðfÞ ¼ f T; that is, ðTtðfÞÞðvÞ ¼ fðTðvÞÞ for every v. Hence, (a) ðTtðfÞÞðx; yÞ ¼ fðTðx; yÞÞ ¼ fðx; 0Þ ¼ x (b) ðTtðfÞÞðx; yÞ ¼ fðTðx; yÞÞ ¼ fðy; x þ yÞ ¼ y 2ðx þ yÞ ¼ 2x y (c) ðTtðfÞÞðx; yÞ ¼ fðTðx; yÞÞ ¼ fð2x 3y; 5x þ 2yÞ ¼ ð2x 3yÞ 2ð5x þ 2yÞ ¼ 8x 7y 11.14. Let T :V ! U be linear and let Tt:U ! V be its transpose. Show that the kernel of Tt is the annihilator of the image of T—that is, Ker Tt ¼ ðIm TÞ0. Suppose f 2 Ker Tt; that is, TtðfÞ ¼ f T ¼ 0. If u 2 Im T, then u ¼ TðvÞ for some v 2 V; hence, fðuÞ ¼ fðTðvÞÞ ¼ ðf TÞðvÞ ¼ 0ðvÞ ¼ 0 We have that fðuÞ ¼ 0 for every u 2 Im T; hence, f 2 ðIm TÞ0. Thus, Ker T t ðIm TÞ0. On the other hand, suppose s 2 ðIm TÞ0; that is, sðIm TÞ ¼ f0g . Then, for every v 2 V, ðTtðsÞÞðvÞ ¼ ðs TÞðvÞ ¼ sðTðvÞÞ ¼ 0 ¼ 0ðvÞ CHAPTER 11 Linear Functionals and the Dual Space 355 We have ðTtðsÞÞðvÞ ¼ 0ðvÞ for every v 2 V; hence, TtðsÞ ¼ 0. Thus, s 2 Ker Tt, and so ðIm TÞ0 Ker Tt. The two inclusion relations together give us the required equality. 11.15. Suppose V and U have ﬁnite dimension and T:V ! U is linear. Prove rankðTÞ ¼ rankðTtÞ. Suppose dim V ¼ n and dim U ¼ m, and suppose rankðTÞ ¼ r. By Theorem 11.5, dimðIm TÞ0 ¼ dim u dimðIm TÞ ¼ m rankðTÞ ¼ m r By Problem 11.14, Ker T t ¼ ðIm TÞ0. Hence, nullity ðTtÞ ¼ m r. It then follows that, as claimed, rankðTtÞ ¼ dim U nullityðTtÞ ¼ m ðm rÞ ¼ r ¼ rankðTÞ 11.16. Prove Theorem 11.7: Let T :V ! U be linear and let A be the matrix representation of T in the bases fvjg of V and fuig of U. Then the transpose matrix AT is the matrix representation of Tt:U ! V in the bases dual to fuig and fvjg. Suppose, for j ¼ 1; . . . ; m, TðvjÞ ¼ aj1u1 þ aj2u2 þ þ ajnun ð1Þ We want to prove that, for i ¼ 1; . . . ; n, T tðsiÞ ¼ a1if1 þ a2if2 þ þ amifm ð2Þ where fsig and ffjg are the bases dual to fuig and fvjg, respectively. Let v 2 V and suppose v ¼ k1v1 þ k2v2 þ þ kmvm. Then, by (1), TðvÞ ¼ k1Tðv1Þ þ k2Tðv2Þ þ þ kmTðvmÞ ¼ k1ða11u1 þ þ a1nunÞ þ k2ða21u1 þ þ a2nunÞ þ þ kmðam1u1 þ þ amnunÞ ¼ ðk1a11 þ k2a21 þ þ kmam1Þu1 þ þ ðk1a1n þ k2a2n þ þ kmamnÞun ¼ P n i¼1 ðk1a1i þ k2a2i þ þ kmamiÞui Hence, for j ¼ 1; . . . ; n. ðTtðsjÞðvÞÞ ¼ sjðTðvÞÞ ¼ sj P n i¼1 ðk1a1i þ k2a2i þ þ kmamiÞui ¼ k1a1j þ k2a2j þ þ kmamj ð3Þ On the other hand, for j ¼ 1; . . . ; n, ða1jf1 þ a2jf2 þ þ amjfmÞðvÞ ¼ ða1jf1 þ a2jf2 þ þ amjfmÞðk1v1 þ k2v2 þ þ kmvmÞ ¼ k1a1j þ k2a2j þ þ kmamj ð4Þ Because v 2 V was arbitrary, (3) and (4) imply that TtðsjÞ ¼ a1jf1 þ a2jf2 þ þ amjfm; j ¼ 1; . . . ; n which is (2). Thus, the theorem is proved. SUPPLEMENTARY PROBLEMS Dual Spaces and Dual Bases 11.17. Find (a) f þ s, (b) 3f, (c) 2f 5s, where f:R3 ! R and s:R3 ! R are deﬁned by fðx; y; zÞ ¼ 2x 3y þ z and sðx; y; zÞ ¼ 4x 2y þ 3z 11.18. Find the dual basis of each of the following bases of R3: (a) fð1; 0; 0Þ; ð0; 1; 0Þ; ð0; 0; 1Þg, (b) fð1; 2; 3Þ; ð1; 1; 1Þ; ð2; 4; 7Þg. 356 CHAPTER 11 Linear Functionals and the Dual Space 11.19. Let V be the vector space of polynomials over R of degree 2. Let f1; f2; f3 be the linear functionals on V deﬁned by f1ð f ðtÞÞ ¼ ð1 0 f ðtÞ dt; f2ð f ðtÞÞ ¼ f 0ð1Þ; f3ð f ðtÞÞ ¼ f ð0Þ Here f ðtÞ ¼ a þ bt þ ct2 2 V and f 0ðtÞ denotes the derivative of f ðtÞ. Find the basis f f1ðtÞ; f2ðtÞ; f3ðtÞg of V that is dual to ff1; f2; f3g. 11.20. Suppose u; v 2 V and that fðuÞ ¼ 0 implies fðvÞ ¼ 0 for all f 2 V. Show that v ¼ ku for some scalar k. 11.21. Suppose f; s 2 V and that fðvÞ ¼ 0 implies sðvÞ ¼ 0 for all v 2 V. Show that s ¼ kf for some scalar k. 11.22. Let V be the vector space of polynomials over K. For a 2 K, deﬁne fa:V ! K by fað f ðtÞÞ ¼ f ðaÞ. Show that (a) fa is linear; (b) if a 6¼ b, then fa 6¼ fb. 11.23. Let V be the vector space of polynomials of degree 2. Let a; b; c 2 K be distinct scalars. Let fa; fb; fc be the linear functionals deﬁned by fað f ðtÞÞ ¼ f ðaÞ, fbð f ðtÞÞ ¼ f ðbÞ, fcð f ðtÞÞ ¼ f ðcÞ. Show that ffa; fb; fcg is linearly independent, and ﬁnd the basis f f1ðtÞ; f2ðtÞ; f3ðtÞg of V that is its dual. 11.24. Let V be the vector space of square matrices of order n. Let T :V ! K be the trace mapping; that is, TðAÞ ¼ a11 þ a22 þ þ ann, where A ¼ ðaijÞ. Show that T is linear. 11.25. Let W be a subspace of V. For any linear functional f on W, show that there is a linear functional s on V such that sðwÞ ¼ fðwÞ for any w 2 W; that is, f is the restriction of s to W. 11.26. Let fe1; . . . ; eng be the usual basis of Kn. Show that the dual basis is fp1; . . . ; png where pi is the ith projection mapping; that is, piða1; . . . ; anÞ ¼ ai. 11.27. Let V be a vector space over R. Let f1; f2 2 V and suppose s:V ! R; deﬁned by sðvÞ ¼ f1ðvÞf2ðvÞ; also belongs to V. Show that either f1 ¼ 0 or f2 ¼ 0. Annihilators 11.28. Let W be the subspace of R4 spanned by ð1; 2; 3; 4Þ, ð1; 3; 2; 6Þ, ð1; 4; 1; 8Þ. Find a basis of the annihilator of W. 11.29. Let W be the subspace of R3 spanned by ð1; 1; 0Þ and ð0; 1; 1Þ. Find a basis of the annihilator of W. 11.30. Show that, for any subset S of V; spanðSÞ ¼ S00, where spanðSÞ is the linear span of S. 11.31. Let U and W be subspaces of a vector space V of ﬁnite dimension. Prove that ðU \ WÞ0 ¼ U0 þ W 0. 11.32. Suppose V ¼ U W. Prove that V 0 ¼ U0 W 0. Transpose of a Linear Mapping 11.33. Let f be the linear functional on R2 deﬁned by fðx; yÞ ¼ 3x 2y. For each of the following linear mappings T :R3 ! R2, ﬁnd ðT tðfÞÞðx; y; zÞ: (a) Tðx; y; zÞ ¼ ðx þ y; y þ zÞ, (b) Tðx; y; zÞ ¼ ðx þ y þ z; 2x yÞ 11.34. Suppose T1:U ! V and T2:V ! W are linear. Prove that ðT2 T1Þt ¼ T t 1 Tt 2. 11.35. Suppose T :V ! U is linear and V has ﬁnite dimension. Prove that Im Tt ¼ ðKer TÞ0. CHAPTER 11 Linear Functionals and the Dual Space 357 11.36. Suppose T :V ! U is linear and u 2 U. Prove that u 2 Im T or there exists f 2 V such that TtðfÞ ¼ 0 and fðuÞ ¼ 1. 11.37. Let V be of ﬁnite dimension. Show that the mapping T 7! Tt is an isomorphism from HomðV; VÞ onto HomðV; VÞ. (Here T is any linear operator on V.) Miscellaneous Problems 11.38. Let V be a vector space over R. The line segment uv joining points u; v 2 V is deﬁned by uv ¼ ftu þ ð1 tÞv:0 t 1g. A subset S of V is convex if u; v 2 S implies uv S. Let f 2 V. Deﬁne W þ ¼ fv 2 V : fðvÞ > 0g; W ¼ fv 2 V : fðvÞ ¼ 0g; W ¼ fv 2 V : fðvÞ < 0g Prove that W þ; W, and W are convex. 11.39. Let V be a vector space of ﬁnite dimension. A hyperplane H of V may be deﬁned as the kernel of a nonzero linear functional f on V. Show that every subspace of V is the intersection of a ﬁnite number of hyperplanes. ANSWERS TO SUPPLEMENTARY PROBLEMS 11.17. (a) 6x 5y þ 4z, (b) 6x 9y þ 3z, (c) 16x þ 4y 13z 11.18. (a) f1 ¼ x; f2 ¼ y; f3 ¼ z; (b) f1 ¼ 3x 5y 2z; f2 ¼ 2x þ y; f3 ¼ x þ 2y þ z 11.19. f1ðtÞ ¼ 3t 3 2 t2; f2ðtÞ ¼ 1 2 t þ 3 4 t2; f3ðtÞ ¼ 1 3t þ 3 2 t2 11.22. (b) Let f ðtÞ ¼ t. Then fað f ðtÞÞ ¼ a 6¼ b ¼ fbð f ðtÞÞ; and therefore, fa 6¼ fb 11.23. f1ðtÞ ¼ t2 ðb þ cÞt þ bc ða bÞða cÞ ; f2ðtÞ ¼ t2 ða þ cÞt þ ac ðb aÞðb cÞ ; f3ðtÞ ¼ t2 ða þ bÞt þ ab ðc aÞðc bÞ 11.28. ff1ðx; y; z; tÞ ¼ 5x y þ z; f2ðx; y; z; tÞ ¼ 2y tg 11.29. ffðx; y; zÞ ¼ x y þ zg 11.33. (a) ðT tðfÞÞðx; y; zÞ ¼ 3x þ y 2z, (b) ðTtðfÞÞðx; y; zÞ ¼ x þ 5y þ 3z 358 CHAPTER 11 Linear Functionals and the Dual Space Bilinear, Quadratic, and Hermitian Forms 12.1 Introduction This chapter generalizes the notions of linear mappings and linear functionals. Speciﬁcally, we introduce the notion of a bilinear form. These bilinear maps also give rise to quadratic and Hermitian forms. Although quadratic forms were discussed previously, this chapter is treated independently of the previous results. Although the ﬁeld K is arbitrary, we will later specialize to the cases K ¼ R and K ¼ C. Furthermore, we may sometimes need to divide by 2. In such cases, we must assume that 1 þ 1 6¼ 0, which is true when K ¼ R or K ¼ C. 12.2 Bilinear Forms Let V be a vector space of ﬁnite dimension over a ﬁeld K. A bilinear form on V is a mapping f :V V ! K such that, for all a; b 2 K and all ui; vi 2 V: (i) f ðau1 þ bu2; vÞ ¼ af ðu1; vÞ þ bf ðu2; vÞ, (ii) f ðu; av1 þ bv2Þ ¼ af ðu; v1Þ þ bf ðu; v2Þ We express condition (i) by saying f is linear in the ﬁrst variable, and condition (ii) by saying f is linear in the second variable. EXAMPLE 12.1 (a) Let f be the dot product on Rn; that is, for u ¼ ðaiÞ and v ¼ ðbiÞ, f ðu; vÞ ¼ u v ¼ a1b1 þ a2b2 þ þ anbn Then f is a bilinear form on Rn. (In fact, any inner product on a real vector space V is a bilinear form on V.) (b) Let f and s be arbitrarily linear functionals on V. Let f :V V ! K be deﬁned by f ðu; vÞ ¼ fðuÞsðvÞ. Then f is a bilinear form, because f and s are each linear. (c) Let A ¼ ½aij be any n n matrix over a ﬁeld K. Then A may be identiﬁed with the following bilinear form F on Kn, where X ¼ ½xi and Y ¼ ½yi are column vectors of variables: f ðX; YÞ ¼ X TAY ¼ P i;j aijxiyi ¼ a11x1y1 þ a12x1y2 þ þ annxnyn The above formal expression in the variables xi; yi is termed the bilinear polynomial corresponding to the matrix A. Equation (12.1) shows that, in a certain sense, every bilinear form is of this type. CHAPTER 12 359 Space of Bilinear Forms Let BðVÞ denote the set of all bilinear forms on V. A vector space structure is placed on BðVÞ, where for any f ; g 2 BðVÞ and any k 2 K, we deﬁne f þ g and kf as follows: ð f þ gÞðu; vÞ ¼ f ðu; vÞ þ gðu; vÞ and ðkf Þðu; vÞ ¼ kf ðu; vÞ The following theorem (proved in Problem 12.4) applies. THEOREM 12.1: Let V be a vector space of dimension n over K. Let ff1; . . . ; fng be any basis of the dual space V. Then f fij : i; j ¼ 1; . . . ; ng is a basis of BðVÞ, where fij is deﬁned by fijðu; vÞ ¼ fiðuÞfjðvÞ. Thus, in particular, dim BðVÞ ¼ n2. 12.3 Bilinear Forms and Matrices Let f be a bilinear form on V and let S ¼ fu1; . . . ; ung be a basis of V. Suppose u; v 2 V and u ¼ a1u1 þ þ anun and v ¼ b1u1 þ þ bnun Then f ðu; vÞ ¼ f ða1u1 þ þ anun; b1u1 þ þ bnunÞ ¼ P i;j aibjf ðui; ujÞ Thus, f is completely determined by the n2 values f ðui; ujÞ. The matrix A ¼ ½aij where aij ¼ f ðui; ujÞ is called the matrix representation of f relative to the basis S or, simply, the ‘‘matrix of f in S.’’ It ‘‘represents’’ f in the sense that, for all u; v 2 V, f ðu; vÞ ¼ P i;j aibj f ðui; ujÞ ¼ ½uT S A½vS ð12:1Þ [As usual, ½uS denotes the coordinate (column) vector of u in the basis S.] Change of Basis, Congruent Matrices We now ask, how does a matrix representing a bilinear form transform when a new basis is selected? The answer is given in the following theorem (proved in Problem 12.5). THEOREM 12.2: Let P be a change-of-basis matrix from one basis S to another basis S0. If A is the matrix representing a bilinear form f in the original basis S, then B ¼ PTAP is the matrix representing f in the new basis S0. The above theorem motivates the following deﬁnition. DEFINITION: A matrix B is congruent to a matrix A, written B ’ A, if there exists a nonsingular matrix P such that B ¼ PTAP. Thus, by Theorem 12.2, matrices representing the same bilinear form are congruent. We remark that congruent matrices have the same rank, because P and PT are nonsingular; hence, the following deﬁnition is well deﬁned. DEFINITION: The rank of a bilinear form f on V, written rankð f Þ, is the rank of any matrix representation of f . We say f is degenerate or nondegenerate according to whether rankð f Þ < dim V or rankð f Þ ¼ dim V. 12.4 Alternating Bilinear Forms Let f be a bilinear form on V. Then f is called (i) alternating if f ðv; vÞ ¼ 0 for every v 2 V; (ii) skew-symmetric if f ðu; vÞ ¼ f ðv; uÞ for every u; v 2 V. 360 CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms Now suppose (i) is true. Then (ii) is true, because, for any u; v; 2 V, 0 ¼ f ðu þ v; u þ vÞ ¼ f ðu; uÞ þ f ðu; vÞ þ f ðv; uÞ þ f ðv; vÞ ¼ f ðu; vÞ þ f ðv; uÞ On the other hand, suppose (ii) is true and also 1 þ 1 6¼ 0. Then (i) is true, because, for every v 2 V, we have f ðv; vÞ ¼ f ðv; vÞ. In other words, alternating and skew-symmetric are equivalent when 1 þ 1 6¼ 0. The main structure theorem of alternating bilinear forms (proved in Problem 12.23) is as follows. THEOREM 12.3: Let f be an alternating bilinear form on V. Then there exists a basis of V in which f is represented by a block diagonal matrix M of the form M ¼ diag 0 1 1 0 ; 0 1 1 0 ; . . . ; 0 1 1 0 ; ½0; ½0; . . . ½0 Moreover, the number of nonzero blocks is uniquely determined by f [because it is equal to 1 2 rankð f Þ. In particular, the above theorem shows that any alternating bilinear form must have even rank. 12.5 Symmetric Bilinear Forms, Quadratic Forms This section investigates the important notions of symmetric bilinear forms and quadratic forms and their representation by means of symmetric matrices. The only restriction on the ﬁeld K is that 1 þ 1 6¼ 0. In Section 12.6, we will restrict K to be the real ﬁeld R, which yields important special results. Symmetric Bilinear Forms Let f be a bilinear form on V. Then f is said to be symmetric if, for every u; v 2 V, f ðu; vÞ ¼ f ðv; uÞ One can easily show that f is symmetric if and only if any matrix representation A of f is a symmetric matrix. The main result for symmetric bilinear forms (proved in Problem 12.10) is as follows. (We emphasize that we are assuming that 1 þ 1 6¼ 0.) THEOREM 12.4: Let f be a symmetric bilinear form on V. Then V has a basis fv1; . . . ; vng in which f is represented by a diagonal matrix—that is, where f ðvi; vjÞ ¼ 0 for i 6¼ j. THEOREM 12.4: (Alternative Form) Let A be a symmetric matrix over K. Then A is congruent to a diagonal matrix; that is, there exists a nonsingular matrix P such that PTAP is diagonal. Diagonalization Algorithm Recall that a nonsingular matrix P is a product of elementary matrices. Accordingly, one way of obtaining the diagonal form D ¼ PTAP is by a sequence of elementary row operations and the same sequence of elementary column operations. This same sequence of elementary row operations on the identity matrix I will yield PT. This algorithm is formalized below. ALGORITHM 12.1: (Congruence Diagonalization of a Symmetric Matrix) The input is a symmetric matrix A ¼ ½aij of order n. Step 1. Form the n 2n (block) matrix M ¼ ½A1; I, where A1 ¼ A is the left half of M and the identity matrix I is the right half of M. Step 2. Examine the entry a11. There are three cases. CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms 361 Case I: a11 6¼ 0. (Use a11 as a pivot to put 0’s below a11 in M and to the right of a11 in A1:Þ For i ¼ 2; . . . ; n: (a) Apply the row operation ‘‘Replace Ri by ai1R1 þ a11Ri.’’ (b) Apply the corresponding column operation ‘‘Replace Ci by ai1C1 þ a11Ci.’’ These operations reduce the matrix M to the form M a11 0 0 A1 ðÞ Case II: a11 ¼ 0 but akk 6¼ 0, for some k > 1. (a) Apply the row operation ‘‘Interchange R1 and Rk.’’ (b) Apply the corresponding column operation ‘‘Interchange C1 and Ck.’’ (These operations bring akk into the ﬁrst diagonal position, which reduces the matrix to Case I.) Case III: All diagonal entries aii ¼ 0 but some aij 6¼ 0. (a) Apply the row operation ‘‘Replace Ri by Rj þ Ri.’’ (b) Apply the corresponding column operation ‘‘Replace Ci by Cj þ Ci.’’ (These operations bring 2aij into the ith diagonal position, which reduces the matrix to Case II.) Thus, M is ﬁnally reduced to the form ðÞ, where A2 is a symmetric matrix of order less than A. Step 3. Repeat Step 2 with each new matrix Ak (by neglecting the ﬁrst row and column of the preceding matrix) until A is diagonalized. Then M is transformed into the form M0 ¼ ½D; Q, where D is diagonal. Step 4. Set P ¼ QT. Then D ¼ PTAP. Remark 1: We emphasize that in Step 2, the row operations will change both sides of M, but the column operations will only change the left half of M. Remark 2: The condition 1 þ 1 6¼ 0 is used in Case III, where we assume that 2aij 6¼ 0 when aij 6¼ 0. The justiﬁcation for the above algorithm appears in Problem 12.9. EXAMPLE 12.2 Let A ¼ 1 2 3 2 5 4 3 4 8 2 4 3 5. Apply Algorithm 9.1 to ﬁnd a nonsingular matrix P such that D ¼ PTAP is diagonal. First form the block matrix M ¼ ½A; I; that is, let M ¼ ½A; I ¼ 1 2 3 1 0 0 2 5 4 0 1 0 3 4 8 0 0 1 2 4 3 5 Apply the row operations ‘‘Replace R2 by 2R1 þ R2’’ and ‘‘Replace R3 by 3R1 þ R3’’ to M, and then apply the corresponding column operations ‘‘Replace C2 by 2C1 þ C2’’ and ‘‘Replace C3 by 3C1 þ C3’’ to obtain 1 2 3 1 0 0 0 1 2 2 1 0 0 2 1 3 0 1 2 4 3 5 and then 1 0 0 1 0 0 0 1 2 2 1 0 0 2 1 3 0 1 2 4 3 5 362 CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms Next apply the row operation ‘‘Replace R3 by 2R2 þ R3’’ and then the corresponding column operation ‘‘Replace C3 by 2C2 þ C3’’ to obtain 1 0 0 1 0 0 0 1 2 2 1 0 0 0 5 7 2 1 2 4 3 5 and then 1 0 0 1 0 0 0 1 0 2 1 0 0 0 5 7 2 1 2 4 3 5 Now A has been diagonalized. Set P ¼ 1 2 7 0 1 2 0 0 1 2 4 3 5 and then D ¼ P1AP ¼ 1 0 0 0 1 0 0 0 5 2 4 3 5 We emphasize that P is the transpose of the right half of the ﬁnal matrix. Quadratic Forms We begin with a deﬁnition. DEFINITION A: A mapping q:V ! K is a quadratic form if qðvÞ ¼ f ðv; vÞ for some symmetric bilinear form f on V. If 1 þ 1 6¼ 0 in K, then the bilinear form f can be obtained from the quadratic form q by the following polar form of f : f ðu; vÞ ¼ 1 2 ½qðu þ vÞ qðuÞ qðvÞ Now suppose f is represented by a symmetric matrix A ¼ ½aij, and 1 þ 1 6¼ 0. Letting X ¼ ½xi denote a column vector of variables, q can be represented in the form qðXÞ ¼ f ðX; XÞ ¼ X TAX ¼ P i;j aijxixj ¼ P i aiix2 i þ 2 P i 0 for every v 6¼ 0, (ii) nonnegative semideﬁnite if qðvÞ ¼ f ðv; vÞ 0 for every v. EXAMPLE 12.3 Let f be the dot product on Rn. Recall that f is a symmetric bilinear form on Rn. We note that f is also positive deﬁnite. That is, for any u ¼ ðaiÞ 6¼ 0 in Rn, f ðu; uÞ ¼ a2 1 þ a2 2 þ þ a2 n > 0 Section 12.5 and Chapter 13 tell us how to diagonalize a real quadratic form q or, equivalently, a real symmetric matrix A by means of an orthogonal transition matrix P. If P is merely nonsingular, then q can be represented in diagonal form with only 1’s and 1’s as nonzero coefﬁcients. Namely, we have the following corollary. COROLLARY 12.6: Any real quadratic form q has a unique representation in the form qðx1; x2; . . . ; xnÞ ¼ x2 1 þ þ x2 p x2 pþ1 x2 r where r ¼ p þ n is the rank of the form. COROLLARY 12.6: (Alternative Form) Any real symmetric matrix A is congruent to the unique diagonal matrix D ¼ diagðIp; In; 0Þ where r ¼ p þ n is the rank of A. 12.7 Hermitian Forms Let V be a vector space of ﬁnite dimension over the complex ﬁeld C. A Hermitian form on V is a mapping f :V V ! C such that, for all a; b 2 C and all ui; v 2 V, (i) f ðau1 þ bu2; vÞ ¼ af ðu1; vÞ þ bf ðu2; vÞ, (ii) f ðu; vÞ ¼ f ðv; uÞ. (As usual, k denotes the complex conjugate of k 2 C.) Using (i) and (ii), we get f ðu; av1 þ bv2Þ ¼ f ðav1 þ bv2; uÞ ¼ af ðv1; uÞ þ bf ðv2; uÞ ¼ ^ af ðv1; uÞ þ bf ðv2; uÞ ¼ af ðu; v1Þ þ bf ðu; v2Þ That is, ðiiiÞ f ðu; av1 þ bv2Þ ¼ af ðu; v1Þ þ bf ðu; v2Þ: As before, we express condition (i) by saying f is linear in the ﬁrst variable. On the other hand, we express condition (iii) by saying f is ‘‘conjugate linear’’ in the second variable. Moreover, condition (ii) tells us that f ðv; vÞ ¼ f ðv; vÞ, and hence, f ðv; vÞ is real for every v 2 V. The results of Sections 12.5 and 12.6 for symmetric forms have their analogues for Hermitian forms. Thus, the mapping q:V ! R, deﬁned by qðvÞ ¼ f ðv; vÞ, is called the Hermitian quadratic form or complex quadratic form associated with the Hermitian form f . We can obtain f from q by the polar form f ðu; vÞ ¼ 1 4 ½qðu þ vÞ qðu vÞ þ 1 4 ½qðu þ ivÞ qðu ivÞ 364 CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms Now suppose S ¼ fu1; . . . ; ung is a basis of V. The matrix H ¼ ½hij where hij ¼ f ðui; ujÞ is called the matrix representation of f in the basis S. By (ii), f ðui; ujÞ ¼ f ðuj; uiÞ; hence, H is Hermitian and, in particular, the diagonal entries of H are real. Thus, any diagonal representation of f contains only real entries. The next theorem (to be proved in Problem 12.47) is the complex analog of Theorem 12.5 on real symmetric bilinear forms. THEOREM 12.7: Let f be a Hermitian form on V over C. Then there exists a basis of V in which f is represented by a diagonal matrix. Every other diagonal matrix representation of f has the same number p of positive entries and the same number n of negative entries. Again the rank and signature of the Hermitian form f are denoted and deﬁned by rankð f Þ ¼ p þ n and sigð f Þ ¼ p n These are uniquely deﬁned by Theorem 12.7. Analogously, a Hermitian form f is said to be (i) positive deﬁnite if qðvÞ ¼ f ðv; vÞ > 0 for every v 6¼ 0, (ii) nonnegative semideﬁnite if qðvÞ ¼ f ðv; vÞ 0 for every v. EXAMPLE 12.4 Let f be the dot product on Cn; that is, for any u ¼ ðziÞ and v ¼ ðwiÞ in Cn, f ðu; vÞ ¼ u v ¼ z1 w1 þ z2 w2 þ þ zn wn Then f is a Hermitian form on Cn. Moreover, f is also positive deﬁnite, because, for any u ¼ ðziÞ 6¼ 0 in Cn, f ðu; uÞ ¼ z1 z1 þ z2 z2 þ þ zn zn ¼ jz1j2 þ jz2j2 þ þ jznj2 > 0 SOLVED PROBLEMS Bilinear Forms 12.1. Let u ¼ ðx1; x2; x3Þ and v ¼ ðy1; y2; y3Þ. Express f in matrix notation, where f ðu; vÞ ¼ 3x1y1 2x1y3 þ 5x2y1 þ 7x2y2 8x2y3 þ 4x3y2 6x3y3 Let A ¼ ½aij, where aij is the coefﬁcient of xiyj. Then f ðu; vÞ ¼ X TAY ¼ ½x1; x2; x3 3 0 2 5 7 8 0 4 6 2 4 3 5 y1 y2 y3 2 4 3 5 12.2. Let A be an n n matrix over K. Show that the mapping f deﬁned by f ðX; YÞ ¼ X TAY is a bilinear form on Kn. For any a; b 2 K and any Xi; Yi 2 Kn, f ðaX1 þ bX2; YÞ ¼ ðaX1 þ bX2ÞTAY ¼ ðaX T 1 þ bX T 2 ÞAY ¼ aX T 1 AY þ bX T 2 AY ¼ af ðX1; YÞ þ bf ðX2; YÞ Hence, f is linear in the ﬁrst variable. Also, f ðX; aY1 þ bY2Þ ¼ X TAðaY1 þ bY2Þ ¼ aX TAY1 þ bX TAY2 ¼ af ðX; Y1Þ þ bf ðX; Y2Þ Hence, f is linear in the second variable, and so f is a bilinear form on Kn. CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms 365 12.3. Let f be the bilinear form on R2 deﬁned by f ½ðx1; x2Þ; ðy1; y2Þ ¼ 2x1y1 3x1y2 þ 4x2y2 (a) Find the matrix A of f in the basis fu1 ¼ ð1; 0Þ; u2 ¼ ð1; 1Þg. (b) Find the matrix B of f in the basis fv1 ¼ ð2; 1Þ; v2 ¼ ð1; 1Þg. (c) Find the change-of-basis matrix P from the basis fuig to the basis fvig, and verify that B ¼ PTAP. (a) Set A ¼ ½aij, where aij ¼ f ðui; ujÞ. This yields a11 ¼ f ½ð1; 0Þ; ð1; 0Þ ¼ 2 0 0 ¼ 2; a21 ¼ f ½ð1; 1Þ; ð1; 0Þ ¼ 2 0 þ 0 ¼ 2 a12 ¼ f ½ð1; 0Þ; ð1; 1Þ ¼ 2 3 0 ¼ 1; a22 ¼ f ½ð1; 1Þ; ð1; 1Þ ¼ 2 3 þ 4 ¼ 3 Thus, A ¼ 2 1 2 3 is the matrix of f in the basis fu1; u2g. (b) Set B ¼ ½bij, where bij ¼ f ðvi; vjÞ. This yields b11 ¼ f ½ð2; 1Þ; ð2; 1Þ ¼ 8 6 þ 4 ¼ 6; b21 ¼ f ½ð1; 1Þ; ð2; 1Þ ¼ 4 3 4 ¼ 3 b12 ¼ f ½ð2; 1Þ; ð1; 1Þ ¼ 4 þ 6 4 ¼ 6; b22 ¼ f ½ð1; 1Þ; ð1; 1Þ ¼ 2 þ 3 þ 4 ¼ 9 Thus, B ¼ 6 6 3 9 is the matrix of f in the basis fv1; v2g. (c) Writing v1 and v2 in terms of the ui yields v1 ¼ u1 þ u2 and v2 ¼ 2u1 u2. Then P ¼ 1 2 1 1 ; PT ¼ 1 1 2 1 PTAP ¼ 1 1 2 1 2 1 2 3 1 2 1 1 ¼ 6 6 3 9 ¼ B and 12.4. Prove Theorem 12.1: Let V be an n-dimensional vector space over K. Let ff1; . . . ; fng be any basis of the dual space V. Then f fij : i; j ¼ 1; . . . ; ng is a basis of BðVÞ, where fij is deﬁned by fijðu; vÞ ¼ fiðuÞfjðvÞ. Thus, dim BðVÞ ¼ n2. Let fu1; . . . ; ung be the basis of V dual to ffig. We ﬁrst show that f fijg spans BðVÞ. Let f 2 BðVÞ and suppose f ðui; ujÞ ¼ aij: We claim that f ¼ P i;j aij fij. It sufﬁces to show that f ðus; utÞ ¼ P aij fij ðus; utÞ for s; t ¼ 1; . . . ; n We have P aij fij ðus; utÞ ¼ P aij fijðus; utÞ ¼ P aijfiðusÞfjðutÞ ¼ P aijdisdjt ¼ ast ¼ f ðus; utÞ as required. Hence, ffijg spans BðVÞ. Next, suppose P aijfij ¼ 0. Then for s; t ¼ 1; . . . ; n, 0 ¼ 0ðus; utÞ ¼ ðP aij fijÞðus; utÞ ¼ ars The last step follows as above. Thus, f fijg is independent, and hence is a basis of BðVÞ. 12.5. Prove Theorem 12.2. Let P be the change-of-basis matrix from a basis S to a basis S0. Let A be the matrix representing a bilinear form in the basis S. Then B ¼ PTAP is the matrix representing f in the basis S0. Let u; v 2 V. Because P is the change-of-basis matrix from S to S0, we have P½uS0 ¼ ½uS and also P½vS0 ¼ ½vS; hence, ½uT S ¼ ½uT S0PT. Thus, f ðu; vÞ ¼ ½uT S A½vS ¼ ½uT S0PTAP½vS0 Because u and v are arbitrary elements of V, PTAP is the matrix of f in the basis S0. 366 CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms Symmetric Bilinear Forms, Quadratic Forms 12.6. Find the symmetric matrix that corresponds to each of the following quadratic forms: (a) qðx; y; zÞ ¼ 3x2 þ 4xy y2 þ 8xz 6yz þ z2, (b) q0ðx; y; zÞ ¼ 3x2 þ xz 2yz, (c) q00ðx; y; zÞ ¼ 2x2 5y2 7z2 The symmetric matrix A ¼ ½aij that represents qðx1; . . . ; xnÞ has the diagonal entry aii equal to the coefﬁcient of the square term x2 i and the nondiagonal entries aij and aji each equal to half of the coefﬁcient of the cross-product term xixj. Thus, (a) A ¼ 3 2 4 2 1 3 4 3 1 2 4 3 5, (b) A0 ¼ 3 0 1 2 0 0 1 1 2 1 0 2 4 3 5, (c) A00 ¼ 2 0 0 0 5 0 0 0 7 2 4 3 5 The third matrix A00 is diagonal, because the quadratic form q00 is diagonal; that is, q00 has no cross-product terms. 12.7. Find the quadratic form qðXÞ that corresponds to each of the following symmetric matrices: (a) A ¼ 5 3 3 8 ; (b) B ¼ 4 5 7 5 6 8 7 8 9 2 4 3 5, (c) C ¼ 2 4 1 5 4 7 6 8 1 6 3 9 5 8 9 1 2 6 6 4 3 7 7 5 The quadratic form qðXÞ that corresponds to a symmetric matrix M is deﬁned by qðXÞ ¼ X TMX, where X ¼ ½xi is the column vector of unknowns. (a) Compute as follows: qðx; yÞ ¼ X TAX ¼ ½x; y 5 3 3 8 x y ¼ ½5x 3y; 3x þ 8y x y ¼ 5x2 3xy 3xy þ 8y2 ¼ 5x2 6xy þ 8y2 As expected, the coefﬁcient 5 of the square term x2 and the coefﬁcient 8 of the square term y2 are the diagonal elements of A, and the coefﬁcient 6 of the cross-product term xy is the sum of the nondiagonal elements 3 and 3 of A (or twice the nondiagonal element 3, because A is symmetric). (b) Because B is a three-square matrix, there are three unknowns, say x; y; z or x1; x2; x3. Then qðx; y; zÞ ¼ 4x2 10xy 6y2 þ 14xz þ 16yz 9z2 qðx1; x2; x3Þ ¼ 4x2 1 10x1x2 6x2 2 þ 14x1x3 þ 16x2x3 9x2 3 or Here we use the fact that the coefﬁcients of the square terms x2 1; x2 2; x2 3 (or x2; y2; z2) are the respective diagonal elements 4; 6; 9 of B, and the coefﬁcient of the cross-product term xixj is the sum of the nondiagonal elements bij and bji (or twice bij, because bij ¼ bji). (c) Because C is a four-square matrix, there are four unknowns. Hence, qðx1; x2; x3; x4Þ ¼ 2x2 1 7x2 2 þ 3x2 3 þ x2 4 þ 8x1x2 2x1x3 þ 10x1x4 12x2x3 þ 16x2x4 þ 18x3x4 12.8. Let A ¼ 1 3 2 3 7 5 2 5 8 2 4 3 5. Apply Algorithm 12.1 to ﬁnd a nonsingular matrix P such that D ¼ PTAP is diagonal, and ﬁnd sigðAÞ, the signature of A. CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms 367 First form the block matrix M ¼ ½A; I: M ¼ ½A; I ¼ 1 3 2 1 0 0 3 7 5 0 1 0 2 5 8 0 0 1 2 4 3 5 Using a11 ¼ 1 as a pivot, apply the row operations ‘‘Replace R2 by 3R1 þ R2’’ and ‘‘Replace R3 by 2R1 þ R3’’ to M and then apply the corresponding column operations ‘‘Replace C2 by 3C1 þ C2’’ and ‘‘Replace C3 by 2C1 þ C3’’ to A to obtain 1 3 2 1 0 0 0 2 1 3 1 0 0 1 4 2 0 1 2 4 3 5 and then 1 0 0 1 0 0 0 2 1 3 1 0 0 1 4 2 0 1 2 4 3 5: Next apply the row operation ‘‘Replace R3 by R2 þ 2R3’’ and then the corresponding column operation ‘‘Replace C3 by C2 þ 2C3’’ to obtain 1 0 0 1 0 0 0 2 1 3 1 0 0 0 9 1 1 2 2 4 3 5 and then 1 0 0 1 0 0 0 2 0 3 1 0 0 0 18 1 1 2 2 4 3 5 Now A has been diagonalized and the transpose of P is in the right half of M. Thus, set P ¼ 1 3 1 0 1 1 0 0 2 2 4 3 5 and then D ¼ PTAP ¼ 1 0 0 0 2 0 0 0 18 2 4 3 5 Note D has p ¼ 2 positive and n ¼ 1 negative diagonal elements. Thus, the signature of A is sigðAÞ ¼ p n ¼ 2 1 ¼ 1. 12.9. Justify Algorithm 12.1, which diagonalizes (under congruence) a symmetric matrix A. Consider the block matrix M ¼ ½A; I. The algorithm applies a sequence of elementary row operations and the corresponding column operations to the left side of M, which is the matrix A. This is equivalent to premultiplying A by a sequence of elementary matrices, say, E1; E2; . . . ; Er, and postmultiplying A by the transposes of the Ei. Thus, when the algorithm ends, the diagonal matrix D on the left side of M is equal to D ¼ Er E2E1AET 1 ET 2 ET r ¼ QAQT; where Q ¼ Er E2E1 On the other hand, the algorithm only applies the elementary row operations to the identity matrix I on the right side of M. Thus, when the algorithm ends, the matrix on the right side of M is equal to Er E2E1I ¼ Er E2E1 ¼ Q Setting P ¼ QT, we get D ¼ PTAP, which is a diagonalization of A under congruence. 12.10. Prove Theorem 12.4: Let f be a symmetric bilinear form on V over K (where 1 þ 1 6¼ 0). Then V has a basis in which f is represented by a diagonal matrix. Algorithm 12.1 shows that every symmetric matrix over K is congruent to a diagonal matrix. This is equivalent to the statement that f has a diagonal representation. 12.11. Let q be the quadratic form associated with the symmetric bilinear form f . Verify the polar identity f ðu; vÞ ¼ 1 2 ½qðu þ vÞ qðuÞ qðvÞ. (Assume that 1 þ 1 6¼ 0.) We have qðu þ vÞ qðuÞ qðvÞ ¼ f ðu þ v; u þ vÞ f ðu; uÞ f ðv; vÞ ¼ f ðu; uÞ þ f ðu; vÞ þ f ðv; uÞ þ f ðv; vÞ f ðu; uÞ f ðv; vÞ ¼ 2f ðu; vÞ If 1 þ 1 6¼ 0, we can divide by 2 to obtain the required identity. 368 CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms 12.12. Consider the quadratic form qðx; yÞ ¼ 3x2 þ 2xy y2 and the linear substitution x ¼ s 3t; y ¼ 2s þ t (a) Rewrite qðx; yÞ in matrix notation, and ﬁnd the matrix A representing qðx; yÞ. (b) Rewrite the linear substitution using matrix notation, and ﬁnd the matrix P corresponding to the substitution. (c) Find qðs; tÞ using direct substitution. (d) Find qðs; tÞ using matrix notation. (a) Here qðx; yÞ ¼ ½x; y 3 1 1 1 x y . Thus, A ¼ 3 1 1 1 ; and qðXÞ ¼ X TAX, where X ¼ ½x; yT. (b) Here x y ¼ 1 3 2 1 s t . Thus, P ¼ 1 3 2 1 ; and X ¼ x y ; Y ¼ s t and X ¼ PY. (c) Substitute for x and y in q to obtain qðs; tÞ ¼ 3ðs 3tÞ2 þ 2ðs 3tÞð2s þ tÞ ð2s þ tÞ2 ¼ 3ðs2 6st þ 9t2Þ þ 2ð2s2 5st 3t2Þ ð4s2 þ 4st þ t2Þ ¼ 3s2 32st þ 20t2 (d) Here qðXÞ ¼ X TAX and X ¼ PY. Thus, X T ¼ Y TPT. Therefore, qðs; tÞ ¼ qðYÞ ¼ Y TPTAPY ¼ ½s; t 1 2 3 1 3 1 1 1 1 3 2 1 s t ¼ ½s; t 3 16 16 20 s t ¼ 3s2 32st þ 20t2 [As expected, the results in parts (c) and (d) are equal.] 12.13. Consider any diagonal matrix A ¼ diagða1; . . . ; anÞ over K. Show that for any nonzero scalars k1; . . . ; kn 2 K; A is congruent to a diagonal matrix D with diagonal entries a1k2 1; . . . ; ank2 n. Furthermore, show that (a) If K ¼ C, then we can choose D so that its diagonal entries are only 1’s and 0’s. (b) If K ¼ R, then we can choose D so that its diagonal entries are only 1’s, 1’s, and 0’s. Let P ¼ diagðk1; . . . ; knÞ. Then, as required, D ¼ PTAP ¼ diagðkiÞ diagðaiÞ diagðkiÞ ¼ diagða1k2 1; . . . ; ank2 nÞ (a) Let P ¼ diagðbiÞ, where bi ¼ 1= ﬃﬃﬃﬃ ai p if ai 6¼ 0 1 if ai ¼ 0 Then PTAP has the required form. (b) Let P ¼ diagðbiÞ, where bi ¼ 1= ﬃﬃﬃﬃﬃﬃ ﬃ jaij p if ai 6¼ 0 1 if ai ¼ 0 Then PTAP has the required form. Remark: We emphasize that (b) is no longer true if ‘‘congruence’’ is replaced by ‘‘Hermitian congruence.’’ 12.14. Prove Theorem 12.5: Let f be a symmetric bilinear form on V over R. Then there exists a basis of V in which f is represented by a diagonal matrix. Every other diagonal matrix representation of f has the same number p of positive entries and the same number n of negative entries. By Theorem 12.4, there is a basis fu1; . . . ; ung of V in which f is represented by a diagonal matrix with, say, p positive and n negative entries. Now suppose fw1; . . . ; wng is another basis of V, in which f is represented by a diagonal matrix with p0 positive and n0 negative entries. We can assume without loss of generality that the positive entries in each matrix appear ﬁrst. Because rankð f Þ ¼ p þ n ¼ p0 þ n0, it sufﬁces to prove that p ¼ p0. CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms 369 Let U be the linear span of u1; . . . ; up and let W be the linear span of wp0þ1; . . . ; wn. Then f ðv; vÞ > 0 for every nonzero v 2 U, and f ðv; vÞ 0 for every nonzero v 2 W. Hence, U \ W ¼ f0g. Note that dim U ¼ p and dim W ¼ n p0. Thus, dimðU þ WÞ ¼ dim U þ dimW dimðU \ WÞ ¼ p þ ðn p0Þ 0 ¼ p p0 þ n But dimðU þ WÞ dim V ¼ n; hence, p p0 þ n n or p p0. Similarly, p0 p and therefore p ¼ p0, as required. Remark: The above theorem and proof depend only on the concept of positivity. Thus, the theorem is true for any subﬁeld K of the real ﬁeld R such as the rational ﬁeld Q. Positive Deﬁnite Real Quadratic Forms 12.15. Prove that the following deﬁnitions of a positive deﬁnite quadratic form q are equivalent: (a) The diagonal entries are all positive in any diagonal representation of q. (b) qðYÞ > 0, for any nonzero vector Y in Rn. Suppose qðYÞ ¼ a1y2 1 þ a2y2 2 þ þ any2 n. If all the coefﬁcients are positive, then clearly qðYÞ > 0 whenever Y 6¼ 0. Thus, (a) implies (b). Conversely, suppose (a) is not true; that is, suppose some diagonal entry ak 0. Let ek ¼ ð0; . . . ; 1; . . . 0Þ be the vector whose entries are all 0 except 1 in the kth position. Then qðekÞ ¼ ak is not positive, and so (b) is not true. That is, (b) implies (a). Accordingly, (a) and (b) are equivalent. 12.16. Determine whether each of the following quadratic forms q is positive deﬁnite: (a) qðx; y; zÞ ¼ x2 þ 2y2 4xz 4yz þ 7z2 (b) qðx; y; zÞ ¼ x2 þ y2 þ 2xz þ 4yz þ 3z2 Diagonalize (under congruence) the symmetric matrix A corresponding to q. (a) Apply the operations ‘‘Replace R3 by 2R1 þ R3’’ and ‘‘Replace C3 by 2C1 þ C3,’’ and then ‘‘Replace R3 by R2 þ R3’’ and ‘‘Replace C3 by C2 þ C3.’’ These yield A ¼ 1 0 2 0 2 2 2 2 7 2 4 3 5 ’ 1 0 0 0 2 2 0 2 3 2 4 3 5 ’ 1 0 0 0 2 0 0 0 1 2 4 3 5 The diagonal representation of q only contains positive entries, 1; 2; 1, on the diagonal. Thus, q is positive deﬁnite. (b) We have A ¼ 1 0 1 0 1 2 1 2 3 2 4 3 5 ’ 1 0 0 0 1 2 0 2 2 2 4 3 5 ’ 1 0 0 0 1 0 0 0 2 2 4 3 5 There is a negative entry 2 on the diagonal representation of q. Thus, q is not positive deﬁnite. 12.17. Show that qðx; yÞ ¼ ax2 þ bxy þ cy2 is positive deﬁnite if and only if a > 0 and the discriminant D ¼ b2 4ac < 0. Suppose v ¼ ðx; yÞ 6¼ 0. Then either x 6¼ 0 or y 6¼ 0; say, y 6¼ 0. Let t ¼ x=y. Then qðvÞ ¼ y2½aðx=yÞ2 þ bðx=yÞ þ c ¼ y2ðat2 þ bt þ cÞ However, the following are equivalent: (i) s ¼ at2 þ bt þ c is positive for every value of t. (ii) s ¼ at2 þ bt þ c lies above the t-axis. (iii) a > 0 and D ¼ b2 4ac < 0. 370 CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms Thus, q is positive deﬁnite if and only if a > 0 and D < 0. [Remark: D < 0 is the same as detðAÞ > 0, where A is the symmetric matrix corresponding to q.] 12.18. Determine whether or not each of the following quadratic forms q is positive deﬁnite: (a) qðx; yÞ ¼ x2 4xy þ 7y2, (b) qðx; yÞ ¼ x2 þ 8xy þ 5y2, (c) qðx; yÞ ¼ 3x2 þ 2xy þ y2 Compute the discriminant D ¼ b2 4ac, and then use Problem 12.17. (a) D ¼ 16 28 ¼ 12. Because a ¼ 1 > 0 and D < 0; q is positive deﬁnite. (b) D ¼ 64 20 ¼ 44. Because D > 0; q is not positive deﬁnite. (c) D ¼ 4 12 ¼ 8. Because a ¼ 3 > 0 and D < 0; q is positive deﬁnite. Hermitian Forms 12.19. Determine whether the following matrices are Hermitian: (a) 2 2 þ 3i 4 5i 2 3i 5 6 þ 2i 4 þ 5i 6 2i 7 2 4 3 5, (b) 3 2 i 4 þ i 2 i 6 i 4 þ i i 7 2 4 3 5, (c) 4 3 5 3 2 1 5 1 6 2 4 3 5 A complex matrix A ¼ ½aij is Hermitian if A ¼ A—that is, if aij ¼ aji: (a) Yes, because it is equal to its conjugate transpose. (b) No, even though it is symmetric. (c) Yes. In fact, a real matrix is Hermitian if and only if it is symmetric. 12.20. Let A be a Hermitian matrix. Show that f is a Hermitian form on Cn where f is deﬁned by f ðX; YÞ ¼ X TA Y. For all a; b 2 C and all X1; X2; Y 2 Cn, f ðaX1 þ bX2; YÞ ¼ ðaX1 þ bX2ÞTA Y ¼ ðaX T 1 þ bX T 2 ÞA Y ¼ aX T 1 A Y þ bX T 2 A Y ¼ af ðX1; YÞ þ bf ðX2; YÞ Hence, f is linear in the ﬁrst variable. Also, f ðX; YÞ ¼ X TA Y ¼ ðX TA YÞT ¼ Y TATX ¼ Y TA X ¼ Y TA X ¼ f ðY; XÞ Hence, f is a Hermitian form on Cn. Remark: We use the fact that X TA Y is a scalar and so it is equal to its transpose. 12.21. Let f be a Hermitian form on V. Let H be the matrix of f in a basis S ¼ fuig of V. Prove the following: (a) f ðu; vÞ ¼ ½uT S H½vS for all u; v 2 V. (b) If P is the change-of-basis matrix from S to a new basis S0 of V, then B ¼ PTH P (or B ¼ QHQ, where Q ¼ PÞ is the matrix of f in the new basis S0. Note that (b) is the complex analog of Theorem 12.2. (a) Let u; v 2 V and suppose u ¼ a1u1 þ þ anun and v ¼ b1u1 þ þ bnun. Then, as required, f ðu; vÞ ¼ f ða1u1 þ þ anun; b1u1 þ þ bnunÞ ¼ P i;j ai bjf ðui; vjÞ ¼ ½a1; . . . ; anH½ b1; . . . ; bnT ¼ ½uT S H½vS CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms 371 (b) Because P is the change-of-basis matrix from S to S0, we have P½uS0 ¼ ½uS and P½vS0 ¼ ½vS; hence, ½uT S ¼ ½uT S0 PT and ½vS ¼ P½vS0: Thus, by (a), f ðu; vÞ ¼ ½uT S H½vS ¼ ½uT S0 PTH P½vS0 But u and v are arbitrary elements of V; hence, PTH P is the matrix of f in the basis S0: 12.22. Let H ¼ 1 1 þ i 2i 1 i 4 2 3i 2i 2 þ 3i 7 2 4 3 5, a Hermitian matrix. Find a nonsingular matrix P such that D ¼ PTH P is diagonal. Also, ﬁnd the signature of H. Use the modiﬁed Algorithm 12.1 that applies the same row operations but the corresponding conjugate column operations. Thus, ﬁrst form the block matrix M ¼ ½H; I: M ¼ 1 1 þ i 2i 1 0 0 1 i 4 2 3i 0 1 0 2i 2 þ 3i 7 0 0 1 2 4 3 5 Apply the row operations ‘‘Replace R2 by ð1 þ iÞR1 þ R2’’ and ‘‘Replace R3 by 2iR1 þ R3’’ and then the corresponding conjugate column operations ‘‘Replace C2 by ð1 iÞC1 þ C2’’ and ‘‘Replace C3 by 2iC1 þ C3’’ to obtain 1 1 þ i 2i 1 0 0 0 2 5i 1 þ i 1 0 0 5i 3 2i 0 1 2 4 3 5 and then 1 0 0 1 0 0 0 2 5i 1 þ i 1 0 0 5i 3 2i 0 1 2 4 3 5 Next apply the row operation ‘‘Replace R3 by 5iR2 þ 2R3’’ and the corresponding conjugate column operation ‘‘Replace C3 by 5iC2 þ 2C3’’ to obtain 1 0 0 1 0 0 0 2 5i 1 þ i 1 0 0 0 19 5 þ 9i 5i 2 2 4 3 5 and then 1 0 0 1 0 0 0 2 0 1 þ i 1 0 0 0 38 5 þ 9i 5i 2 2 4 3 5 Now H has been diagonalized, and the transpose of the right half of M is P. Thus, set P ¼ 1 1 þ i 5 þ 9i 0 1 5i 0 0 2 2 4 3 5; and then D ¼ PTH P ¼ 1 0 0 0 2 0 0 0 38 2 4 3 5: Note D has p ¼ 2 positive elements and n ¼ 1 negative elements. Thus, the signature of H is sigðHÞ ¼ 2 1 ¼ 1. Miscellaneous Problems 12.23. Prove Theorem 12.3: Let f be an alternating form on V. Then there exists a basis of V in which f is represented by a block diagonal matrix M with blocks of the form 0 1 1 0 or 0. The number of nonzero blocks is uniquely determined by f [because it is equal to 1 2 rankð f Þ. If f ¼ 0, then the theorem is obviously true. Also, if dim V ¼ 1, then f ðk1u; k2uÞ ¼ k1k2f ðu; uÞ ¼ 0 and so f ¼ 0. Accordingly, we can assume that dim V > 1 and f 6¼ 0. Because f 6¼ 0, there exist (nonzero) u1; u2 2 V such that f ðu1; u2Þ 6¼ 0. In fact, multiplying u1 by an appropriate factor, we can assume that f ðu1; u2Þ ¼ 1 and so f ðu2; u1Þ ¼ 1. Now u1 and u2 are linearly independent; because if, say, u2 ¼ ku1, then f ðu1; u2Þ ¼ f ðu1; ku1Þ ¼ kf ðu1; u1Þ ¼ 0. Let U ¼ spanðu1; u2Þ; then, 372 CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms (i) The matrix representation of the restriction of f to U in the basis fu1; u2g is 0 1 1 0 , (ii) If u 2 U, say u ¼ au1 þ bu2, then f ðu; u1Þ ¼ f ðau1 þ bu2; u1Þ ¼ b and f ðu; u2Þ ¼ f ðau1 þ bu2; u2Þ ¼ a Let W consists of those vectors w 2 V such that f ðw; u1Þ ¼ 0 and f ðw; u2Þ ¼ 0: Equivalently, W ¼ fw 2 V : f ðw; uÞ ¼ 0 for every u 2 Ug We claim that V ¼ U W. It is clear that U \ W ¼ f0g, and so it remains to show that V ¼ U þ W. Let v 2 V. Set u ¼ f ðv; u2Þu1 f ðv; u1Þu2 and w ¼ v u ð1Þ Because u is a linear combination of u1 and u2; u 2 U. We show next that w 2 W. By (1) and (ii), f ðu; u1Þ ¼ f ðv; u1Þ; hence, f ðw; u1Þ ¼ f ðv u; u1Þ ¼ f ðv; u1Þ f ðu; u1Þ ¼ 0 Similarly, f ðu; u2Þ ¼ f ðv; u2Þ and so f ðw; u2Þ þ f ðv u; u2Þ ¼ f ðv; u2Þ f ðu; u2Þ ¼ 0 Then w 2 W and so, by (1), v ¼ u þ w, where u 2 W. This shows that V ¼ U þ W; therefore, V ¼ U W. Now the restriction of f to W is an alternating bilinear form on W. By induction, there exists a basis u3; . . . ; un of W in which the matrix representing f restricted to W has the desired form. Accordingly, u1; u2; u3; . . . ; un is a basis of V in which the matrix representing f has the desired form. SUPPLEMENTARY PROBLEMS Bilinear Forms 12.24. Let u ¼ ðx1; x2Þ and v ¼ ðy1; y2Þ. Determine which of the following are bilinear forms on R2: (a) f ðu; vÞ ¼ 2x1y2 3x2y1, (c) f ðu; vÞ ¼ 3x2y2, (e) f ðu; vÞ ¼ 1, (b) f ðu; vÞ ¼ x1 þ y2, (d) f ðu; vÞ ¼ x1x2 þ y1y2, (f ) f ðu; vÞ ¼ 0 12.25. Let f be the bilinear form on R2 deﬁned by f ½ðx1; x2Þ; ðy1; y2Þ ¼ 3x1y1 2x1y2 þ 4x2y1 x2y2 (a) Find the matrix A of f in the basis fu1 ¼ ð1; 1Þ; u2 ¼ ð1; 2Þg. (b) Find the matrix B of f in the basis fv1 ¼ ð1; 1Þ; v2 ¼ ð3; 1Þg. (c) Find the change-of-basis matrix P from fuig to fvig, and verify that B ¼ PTAP. 12.26. Let V be the vector space of two-square matrices over R. Let M ¼ 1 2 3 5 , and let f ðA; BÞ ¼ trðATMBÞ, where A; B 2 V and ‘‘tr’’ denotes trace. (a) Show that f is a bilinear form on V. (b) Find the matrix of f in the basis 1 0 0 0 ; 0 1 0 0 ; 0 0 1 0 ; 0 0 0 1 12.27. Let BðVÞ be the set of bilinear forms on V over K. Prove the following: (a) If f ; g 2 BðVÞ, then f þ g, kg 2 BðVÞ for any k 2 K. (b) If f and s are linear functions on V, then f ðu; vÞ ¼ fðuÞsðvÞ belongs to BðVÞ. 12.28. Let ½ f denote the matrix representation of a bilinear form f on V relative to a basis fuig. Show that the mapping f 7! ½ f is an isomorphism of BðVÞ onto the vector space V of n-square matrices. CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms 373 12.29. Let f be a bilinear form on V. For any subset S of V, let S? ¼ fv 2 V : f ðu; vÞ ¼ 0 for every u 2 Sg and S> ¼ fv 2 V : f ðv; uÞ ¼ 0 for every u 2 Sg Show that: (a) S> and S> are subspaces of V; (b) S1 S2 implies S? 2 S? 1 and S> 2 S> 1 ; (c) f0g? ¼ f0g> ¼ V. 12.30. Suppose f is a bilinear form on V. Prove that: rankð f Þ ¼ dim V dim V ? ¼ dim V dim V >, and hence, dim V ? ¼ dim V >. 12.31. Let f be a bilinear form on V. For each u 2 V, let ^ u:V ! K and ~ u:V ! K be deﬁned by ^ uðxÞ ¼ f ðx; uÞ and ~ uðxÞ ¼ f ðu; xÞ. Prove the following: (a) ^ u and ~ u are each linear; i.e., ^ u; ~ u 2 V, (b) u 7! ^ u and u 7! ~ u are each linear mappings from V into V, (c) rankð f Þ ¼ rankðu 7! ^ uÞ ¼ rankðu 7! ~ uÞ. 12.32. Show that congruence of matrices (denoted by ’) is an equivalence relation; that is, (i) A ’ A; (ii) If A ’ B, then B ’ A; (iii) If A ’ B and B ’ C, then A ’ C. Symmetric Bilinear Forms, Quadratic Forms 12.33. Find the symmetric matrix A belonging to each of the following quadratic forms: (a) qðx; y; zÞ 2x2 8xy þ y2 16xz þ 14yz þ 5z2, (c) qðx; y; zÞ ¼ xy þ y2 þ 4xz þ z2 (b) qðx; y; zÞ ¼ x2 xz þ y2, (d) qðx; y; zÞ ¼ xy þ yz 12.34. For each of the following symmetric matrices A, ﬁnd a nonsingular matrix P such that D ¼ PTAP is diagonal: (a) A ¼ 1 0 2 0 3 6 2 6 7 2 4 3 5, (b) A ¼ 1 2 1 2 5 3 1 3 2 2 4 3 5, (c) A ¼ 1 1 0 2 1 2 1 0 0 1 1 2 2 0 2 1 2 6 6 4 3 7 7 5 12.35. Let qðx; yÞ ¼ 2x2 6xy 3y2 and x ¼ s þ 2t, y ¼ 3s t. (a) Rewrite qðx; yÞ in matrix notation, and ﬁnd the matrix A representing the quadratic form. (b) Rewrite the linear substitution using matrix notation, and ﬁnd the matrix P corresponding to the substitution. (c) Find qðs; tÞ using (i) direct substitution, (ii) matrix notation. 12.36. For each of the following quadratic forms qðx; y; zÞ, ﬁnd a nonsingular linear substitution expressing the variables x; y; z in terms of variables r; s; t such that qðr; s; tÞ is diagonal: (a) qðx; y; zÞ ¼ x2 þ 6xy þ 8y2 4xz þ 2yz 9z2, (b) qðx; y; zÞ ¼ 2x2 3y2 þ 8xz þ 12yz þ 25z2, (c) qðx; y; zÞ ¼ x2 þ 2xy þ 3y2 þ 4xz þ 8yz þ 6z2. In each case, ﬁnd the rank and signature. 12.37. Give an example of a quadratic form qðx; yÞ such that qðuÞ ¼ 0 and qðvÞ ¼ 0 but qðu þ vÞ 6¼ 0. 12.38. Let SðVÞ denote all symmetric bilinear forms on V. Show that (a) SðVÞ is a subspace of BðVÞ; (b) If dim V ¼ n, then dim SðVÞ ¼ 1 2 nðn þ 1Þ. 12.39. Consider a real quadratic polynomial qðx1; . . . ; xnÞ ¼ Pn i;j¼1 aijxixj; where aij ¼ aji. 374 CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms (a) If a11 6¼ 0, show that the substitution x1 ¼ y1 1 a11 ða12y2 þ þ a1nynÞ; x2 ¼ y2; . . . ; xn ¼ yn yields the equation qðx1; . . . ; xnÞ ¼ a11 y2 1 þ q0ðy2; . . . ; ynÞ, where q0 is also a quadratic polynomial. (b) If a11 ¼ 0 but, say, a12 6¼ 0, show that the substitution x1 ¼ y1 þ y2; x2 ¼ y1 y2; x3 ¼ y3; . . . ; xn ¼ yn yields the equation qðx1; . . . ; xnÞ ¼ P bij yi yj, where b11 6¼ 0, which reduces this case to case (a). Remark: This method of diagonalizing q is known as completing the square. Positive Deﬁnite Quadratic Forms 12.40. Determine whether or not each of the following quadratic forms is positive deﬁnite: (a) qðx; yÞ ¼ 4x2 þ 5xy þ 7y2, (c) qðx; y; zÞ ¼ x2 þ 4xy þ 5y2 þ 6xz þ 2yz þ 4z2 (b) qðx; yÞ ¼ 2x2 3xy y2; (d) qðx; y; zÞ ¼ x2 þ 2xy þ 2y2 þ 4xz þ 6yz þ 7z2 12.41. Find those values of k such that the given quadratic form is positive deﬁnite: (a) qðx; yÞ ¼ 2x2 5xy þ ky2, (b) qðx; yÞ ¼ 3x2 kxy þ 12y2 (c) qðx; y; zÞ ¼ x2 þ 2xy þ 2y2 þ 2xz þ 6yz þ kz2 12.42. Suppose A is a real symmetric positive deﬁnite matrix. Show that A ¼ PTP for some nonsingular matrix P. Hermitian Forms 12.43. Modify Algorithm 12.1 so that, for a given Hermitian matrix H, it ﬁnds a nonsingular matrix P for which D ¼ PTA P is diagonal. 12.44. For each Hermitian matrix H, ﬁnd a nonsingular matrix P such that D ¼ PTH P is diagonal: (a) H ¼ 1 i i 2 , (b) H ¼ 1 2 þ 3i 2 3i 1 , (c) H ¼ 1 i 2 þ i i 2 1 i 2 i 1 þ i 2 2 4 3 5 Find the rank and signature in each case. 12.45. Let A be a complex nonsingular matrix. Show that H ¼ AA is Hermitian and positive deﬁnite. 12.46. We say that B is Hermitian congruent to A if there exists a nonsingular matrix P such that B ¼ PTA P or, equivalently, if there exists a nonsingular matrix Q such that B ¼ QAQ. Show that Hermitian congruence is an equivalence relation. (Note: If P ¼ Q, then PTA P ¼ QAQ.) 12.47. Prove Theorem 12.7: Let f be a Hermitian form on V. Then there is a basis S of V in which f is represented by a diagonal matrix, and every such diagonal representation has the same number p of positive entries and the same number n of negative entries. Miscellaneous Problems 12.48. Let e denote an elementary row operation, and let f denote the corresponding conjugate column operation (where each scalar k in e is replaced by k in f ). Show that the elementary matrix corresponding to f is the conjugate transpose of the elementary matrix corresponding to e. 12.49. Let V and W be vector spaces over K. A mapping f :V W ! K is called a bilinear form on V and W if (i) f ðav1 þ bv2; wÞ ¼ af ðv1; wÞ þ bf ðv2; wÞ, (ii) f ðv; aw1 þ bw2Þ ¼ af ðv; w1Þ þ bf ðv; w2Þ for every a; b 2 K; vi 2 V; wj 2 W. Prove the following: CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms 375 (a) The set BðV; WÞ of bilinear forms on V and W is a subspace of the vector space of functions from V W into K. (b) If ff1; ... ; fmg is a basis of V and fs1; ... ; sng is a basis of W, then f fij : i ¼ 1; ... ; m; j ¼ 1; ... ; ng is a basis of BðV; WÞ, where fij is deﬁned by fijðv; wÞ ¼ fiðvÞsjðwÞ. Thus, dim BðV; WÞ ¼ dim V dim W. [Note that if V ¼ W, then we obtain the space BðVÞ investigated in this chapter.] 12.50. Let V be a vector space over K. A mapping f :V V . . . V zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ}\|ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{ m times ! K is called a multilinear (or m-linear) form on V if f is linear in each variable; that is, for i ¼ 1; . . . ; m, f ð. . . ; au þ bv; . . .Þ ¼ af ð. . . ; ^ u; . . .Þ þ bf ð. . . ; ^ v; . . .Þ where c . . . denotes the ith element, and other elements are held ﬁxed. An m-linear form f is said to be alternating if f ðv1; . . . vmÞ ¼ 0 whenever vi ¼ vj for i 6¼ j. Prove the following: (a) The set BmðVÞ of m-linear forms on V is a subspace of the vector space of functions from V V V into K. (b) The set AmðVÞ of alternating m-linear forms on V is a subspace of BmðVÞ. Remark 1: If m ¼ 2, then we obtain the space BðVÞ investigated in this chapter. Remark 2: If V ¼ Km, then the determinant function is an alternating m-linear form on V. ANSWERS TO SUPPLEMENTARY PROBLEMS Notation: M ¼ ½R1; R2; . . . denotes a matrix M with rows R1; R2; . . .. 12.24. (a) yes, (b) no, (c) yes, (d) no, (e) no, (f ) yes 12.25. (a) A ¼ ½4; 1; 7; 3, (b) B ¼ ½0; 4; 20; 32, (c) P ¼ ½3; 5; 2; 2 12.26. (b) ½1; 0; 2; 0; 0; 1; 0; 2; 3; 0; 5; 0; 0; 3; 0; 5 12.33. (a) ½2; 4; 8; 4; 1; 7; 8; 7; 5, (b) ½1; 0; 1 2 ; 0; 1; 0; 1 2 ; 0; 0, (c) ½0; 1 2 ; 2; 1 2 ; 1; 0; 2; 0; 1, (d) ½0; 1 2 ; 0; 1 2 ; 0; 1; 1 2 ; 0; 1 2 ; 0; 1 2 ; 0 12.34. (a) P ¼ ½1; 0; 2; 0; 1; 2; 0; 0; 1; D ¼ diagð1; 3; 9Þ; (b) P ¼ ½1; 2; 11; 0; 1; 5; 0; 0; 1; D ¼ diagð1; 1; 28Þ; (c) P ¼ ½1; 1; 1; 4; 0; 1; 1; 2; 0; 0; 1; 0; 0; 0; 0; 1; D ¼ diagð1; 1; 0; 9Þ 12.35. A ¼ ½2; 3; 3; 3, P ¼ ½1; 2; 3; 1, qðs; tÞ ¼ 43s2 4st þ 17t2 12.36. (a) x ¼ r 3s 19t, y ¼ s þ 7t, z ¼ t; qðr; s; tÞ ¼ r2 s2 þ 36t2; (b) x ¼ r 2t; y ¼ s þ 2t; z ¼ t; qðr; s; tÞ ¼ 2r2 3s2 þ 29t2; (c) x ¼ r s t; y ¼ s t; z ¼ t; qðr; s; tÞ ¼ r2 2s2 12.37. qðx; yÞ ¼ x2 y2, u ¼ ð1; 1Þ, v ¼ ð1; 1Þ 12.40. (a) yes, (b) no, (c) no, (d) yes 12.41. (a) k > 25 8 , (b) 12 < k < 12, (c) k > 5 12.44. (a) P ¼ ½1; i; 0; 1, D ¼ I; s ¼ 2; (b) P ¼ ½1; 2 þ 3i; 0; 1, D ¼ diagð1; 14Þ, s ¼ 0; (c) P ¼ ½1; i; 3 þ i; 0; 1; i; 0; 0; 1, D ¼ diagð1; 1; 4Þ; s ¼ 1 d 376 CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms Linear Operators on Inner Product Spaces 13.1 Introduction This chapter investigates the space AðVÞ of linear operators T on an inner product space V. (See Chapter 7.) Thus, the base ﬁeld K is either the real numbers R or the complex numbers C. In fact, different terminologies will be used for the real case and the complex case. We also use the fact that the inner products on real Euclidean space Rn and complex Euclidean space Cn may be deﬁned, respectively, by hu; vi ¼ uTv and hu; vi ¼ uT v where u and v are column vectors. The reader should review the material in Chapter 7 and be very familiar with the notions of norm (length), orthogonality, and orthonormal bases. We also note that Chapter 7 mainly dealt with real inner product spaces, whereas here we assume that V is a complex inner product space unless otherwise stated or implied. Lastly, we note that in Chapter 2, we used AH to denote the conjugate transpose of a complex matrix A; that is, AH ¼ AT. This notation is not standard. Many texts, expecially advanced texts, use A to denote such a matrix; we will use that notation in this chapter. That is, now A ¼ AT. 13.2 Adjoint Operators We begin with the following basic deﬁnition. DEFINITION: A linear operator T on an inner product space V is said to have an adjoint operator T on V if hTðuÞ; vi ¼ hu; TðvÞi for every u; v 2 V. The following example shows that the adjoint operator has a simple description within the context of matrix mappings. EXAMPLE 13.1 (a) Let A be a real n-square matrix viewed as a linear operator on Rn. Then, for every u; v 2 Rn; hAu; vi ¼ ðAuÞTv ¼ uTATv ¼ hu; ATvi Thus, the transpose AT of A is the adjoint of A. (b) Let B be a complex n-square matrix viewed as a linear operator on Cn. Then for every u; v; 2 Cn, hBu; vi ¼ ðBuÞT v ¼ uTBT v ¼ uTB v ¼ hu; Bvi Thus, the conjugate transpose B of B is the adjoint of B. CHAPTER 13 377 Remark: B may mean either the adjoint of B as a linear operator or the conjugate transpose of B as a matrix. By Example 13.1(b), the ambiguity makes no difference, because they denote the same object. The following theorem (proved in Problem 13.4) is the main result in this section. THEOREM 13.1: Let T be a linear operator on a ﬁnite-dimensional inner product space V over K. Then (i) There exists a unique linear operator T on V such that hTðuÞ; vi¼ hu; TðvÞi for every u; v 2 V. (That is, T has an adjoint T.) (ii) If A is the matrix representation T with respect to any orthonormal basis S ¼ fuig of V, then the matrix representation of T in the basis S is the conjugate transpose A of A (or the transpose AT of A when K is real). We emphasize that no such simple relationship exists between the matrices representing T and T if the basis is not orthonormal. Thus, we see one useful property of orthonormal bases. We also emphasize that this theorem is not valid if V has inﬁnite dimension (Problem 13.31). The following theorem (proved in Problem 13.5) summarizes some of the properties of the adjoint. THEOREM 13.2: Let T; T1; T2 be linear operators on V and let k 2 K. Then (i) ðT1 þ T2Þ ¼ T1 þ T2 , (iii) ðT1T2Þ ¼ T2 T1 , (ii) ðkTÞ ¼ kT, (iv) ðTÞ ¼ T. Observe the similarity between the above theorem and Theorem 2.3 on properties of the transpose operation on matrices. Linear Functionals and Inner Product Spaces Recall (Chapter 11) that a linear functional f on a vector space V is a linear mapping f:V ! K. This subsection contains an important result (Theorem 13.3) that is used in the proof of the above basic Theorem 13.1. Let V be an inner product space. Each u 2 V determines a mapping ^ u:V ! K deﬁned by ^ uðvÞ ¼ hv; ui Now, for any a; b 2 K and any v1; v2 2 V, ^ uðav1 þ bv2Þ ¼ hav1 þ bv2; ui ¼ ahv1; ui þ bhv2; ui ¼ a^ uðv1Þ þ b^ uðv2Þ That is, ^ u is a linear functional on V. The converse is also true for spaces of ﬁnite dimension and it is contained in the following important theorem (proved in Problem 13.3). THEOREM 13.3: Let f be a linear functional on a ﬁnite-dimensional inner product space V. Then there exists a unique vector u 2 V such that fðvÞ ¼ hv; ui for every v 2 V. We remark that the above theorem is not valid for spaces of inﬁnite dimension (Problem 13.24). 13.3 Analogy Between AðVÞ and C, Special Linear Operators Let AðVÞ denote the algebra of all linear operators on a ﬁnite-dimensional inner product space V. The adjoint mapping T 7! T on AðVÞ is quite analogous to the conjugation mapping z 7! z on the complex ﬁeld C. To illustrate this analogy we identify in Table 13-1 certain classes of operators T 2 AðVÞ whose behavior under the adjoint map imitates the behavior under conjugation of familiar classes of complex numbers. The analogy between these operators T and complex numbers z is reﬂected in the next theorem. 378 CHAPTER 13 Linear Operators on Inner Product Spaces THEOREM 13.4: Let l be an eigenvalue of a linear operator T on V. (i) If T ¼ T1 (i.e., T is orthogonal or unitary), then jlj ¼ 1. (ii) If T ¼ T (i.e., T is self-adjoint), then l is real. (iii) If T ¼ T (i.e., T is skew-adjoint), then l is pure imaginary. (iv) If T ¼ SS with S nonsingular (i.e., T is positive deﬁnite), then l is real and positive. Proof. In each case let v be a nonzero eigenvector of T belonging to l; that is, TðvÞ ¼ lv with v 6¼ 0. Hence, hv; vi is positive. Proof of (i). We show that l lhv; vi ¼ hv; vi: l lhv; vi ¼ hlv; lvi ¼ hTðvÞ; TðvÞi ¼ hv; TTðvÞi ¼ hv; IðvÞi ¼ hv; vi But hv; vi 6¼ 0; hence, l l ¼ 1 and so jlj ¼ 1. Proof of (ii). We show that lhv; vi ¼ lhv; vi: lhv; vi ¼ hlv; vi ¼ hTðvÞ; vi ¼ hv; TðvÞi ¼ hv; TðvÞi ¼ hv; lvi ¼ lhv; vi But hv; vi 6¼ 0; hence, l ¼ l and so l is real. Proof of (iii). We show that lhv; vi ¼ lhv; vi: lhv; vi ¼ hlv; vi ¼ hTðvÞ; vi ¼ hv; TðvÞi ¼ hv; TðvÞi ¼ hv; lvi ¼ lhv; vi But hv; vi 6¼ 0; hence, l ¼ l or l ¼ l, and so l is pure imaginary. Proof of (iv). Note ﬁrst that SðvÞ 6¼ 0 because S is nonsingular; hence, hSðvÞ, SðvÞi is positive. We show that lhv; vi ¼ hSðvÞ; SðvÞi: lhv; vi ¼ hlv; vi ¼ hTðvÞ; vi ¼ hSSðvÞ; vi ¼ hSðvÞ; SðvÞi But hv; vi and hSðvÞ; SðvÞi are positive; hence, l is positive. Table 13-1 Class of complex numbers Behavior under conjugation Class of operators in AðVÞ Behavior under the adjoint map Unit circle ðjzj ¼ 1Þ z ¼ 1=z Orthogonal operators (real case) T ¼ T 1 Unitary operators (complex case) Self-adjoint operators Also called: Real axis z ¼ z symmetric (real case) T ¼ T Hermitian (complex case) Skew-adjoint operators Also called: Imaginary axis z ¼ z skew-symmetric (real case) T ¼ T skew-Hermitian (complex case) Positive real axis z ¼ ww; w 6¼ 0 Positive deﬁnite operators T ¼ SS ð0; 1Þ with S nonsingular CHAPTER 13 Linear Operators on Inner Product Spaces 379 Remark: Each of the above operators T commutes with its adjoint; that is, TT ¼ TT. Such operators are called normal operators. 13.4 Self-Adjoint Operators Let T be a self-adjoint operator on an inner product space V; that is, suppose T ¼ T (If T is deﬁned by a matrix A, then A is symmetric or Hermitian according as A is real or complex.) By Theorem 13.4, the eigenvalues of T are real. The following is another important property of T. THEOREM 13.5: Let T be a self-adjoint operator on V. Suppose u and v are eigenvectors of T belonging to distinct eigenvalues. Then u and v are orthogonal; that is, hu; vi ¼ 0. Proof. Suppose TðuÞ ¼ l1u and TðvÞ ¼ l2v, where l1 6¼ l2. We show that l1hu; vi ¼ l2hu; vi: l1hu; vi ¼ hl1u; vi ¼ hTðuÞ; vi ¼ hu; TðvÞi ¼ hu; TðvÞi ¼ hu; l2vi ¼ l2hu; vi ¼ l2hu; vi (The fourth equality uses the fact that T ¼ T, and the last equality uses the fact that the eigenvalue l2 is real.) Because l1 6¼ l2, we get hu; vi ¼ 0. Thus, the theorem is proved. 13.5 Orthogonal and Unitary Operators Let U be a linear operator on a ﬁnite-dimensional inner product space V. Suppose U ¼ U 1 or equivalently UU ¼ UU ¼ I Recall that U is said to be orthogonal or unitary according as the underlying ﬁeld is real or complex. The next theorem (proved in Problem 13.10) gives alternative characterizations of these operators. THEOREM 13.6: The following conditions on an operator U are equivalent: (i) U ¼ U 1; that is, UU ¼ UU ¼ I. [U is unitary (orthogonal).] (ii) U preserves inner products; that is, for every v; w 2 V, hUðvÞ, UðwÞi ¼ hv; wi. (iii) U preserves lengths; that is, for every v 2 V, kUðvÞk ¼ kvk. EXAMPLE 13.2 (a) Let T :R3 ! R3 be the linear operator that rotates each vector v about the z-axis by a ﬁxed angle y as shown in Fig. 10-1 (Section 10.3). That is, T is deﬁned by Tðx; y; zÞ ¼ ðx cos y y sin y; x sin y þ y cos y; zÞ We note that lengths (distances from the origin) are preserved under T. Thus, T is an orthogonal operator. (b) Let V be l2-space (Hilbert space), deﬁned in Section 7.3. Let T :V ! V be the linear operator deﬁned by Tða1; a2; a3; . . .Þ ¼ ð0; a1; a2; a3; . . .Þ Clearly, T preserves inner products and lengths. However, T is not surjective, because, for example, ð1; 0; 0; . . .Þ does not belong to the image of T; hence, T is not invertible. Thus, we see that Theorem 13.6 is not valid for spaces of inﬁnite dimension. An isomorphism from one inner product space into another is a bijective mapping that preserves the three basic operations of an inner product space: vector addition, scalar multiplication, and inner 380 CHAPTER 13 Linear Operators on Inner Product Spaces products. Thus, the above mappings (orthogonal and unitary) may also be characterized as the isomorphisms of V into itself. Note that such a mapping U also preserves distances, because kUðvÞ UðwÞk ¼ kUðv wÞk ¼ kv wk Hence, U is called an isometry. 13.6 Orthogonal and Unitary Matrices Let U be a linear operator on an inner product space V. By Theorem 13.1, we obtain the following results. THEOREM 13.7A: A complex matrix A represents a unitary operator U (relative to an orthonormal basis) if and only if A ¼ A1. THEOREM 13.7B: A real matrix A represents an orthogonal operator U (relative to an orthonormal basis) if and only if AT ¼ A1. The above theorems motivate the following deﬁnitions (which appeared in Sections 2.10 and 2.11). DEFINITION: A complex matrix A for which A ¼ A1 is called a unitary matrix. DEFINITION: A real matrix A for which AT ¼ A1 is called an orthogonal matrix. We repeat Theorem 2.6, which characterizes the above matrices. THEOREM 13.8: The following conditions on a matrix A are equivalent: (i) A is unitary (orthogonal). (ii) The rows of A form an orthonormal set. (iii) The columns of A form an orthonormal set. 13.7 Change of Orthonormal Basis Orthonormal bases play a special role in the theory of inner product spaces V. Thus, we are naturally interested in the properties of the change-of-basis matrix from one such basis to another. The following theorem (proved in Problem 13.12) holds. THEOREM 13.9: Let fu1; . . . ; ung be an orthonormal basis of an inner product space V. Then the change-of-basis matrix from fuig into another orthonormal basis is unitary (orthogonal). Conversely, if P ¼ ½aij is a unitary (orthogonal) matrix, then the following is an orthonormal basis: fu0 i ¼ a1iu1 þ a2iu2 þ þ aniun : i ¼ 1; . . . ; ng Recall that matrices A and B representing the same linear operator T are similar; that is, B ¼ P1AP, where P is the (nonsingular) change-of-basis matrix. On the other hand, if V is an inner product space, we are usually interested in the case when P is unitary (or orthogonal) as suggested by Theorem 13.9. (Recall that P is unitary if the conjugate tranpose P ¼ P1, and P is orthogonal if the transpose PT ¼ P1.) This leads to the following deﬁnition. DEFINITION: Complex matrices A and B are unitarily equivalent if there exists a unitary matrix P for which B ¼ PAP. Analogously, real matrices A and B are orthogonally equivalent if there exists an orthogonal matrix P for which B ¼ PTAP. Note that orthogonally equivalent matrices are necessarily congruent. CHAPTER 13 Linear Operators on Inner Product Spaces 381 13.8 Positive Deﬁnite and Positive Operators Let P be a linear operator on an inner product space V. Then (i) P is said to be positive deﬁnite if P ¼ SS for some nonsingular operators S: (ii) P is said to be positive (or nonnegative or semideﬁnite) if P ¼ SS for some operator S: The following theorems give alternative characterizations of these operators. THEOREM 13.10A: The following conditions on an operator P are equivalent: (i) P ¼ T2 for some nonsingular self-adjoint operator T. (ii) P is positive deﬁnite. (iii) P is self-adjoint and hPðuÞ; ui > 0 for every u 6¼ 0 in V. The corresponding theorem for positive operators (proved in Problem 13.21) follows. THEOREM 13.10B: The following conditions on an operator P are equivalent: (i) P ¼ T2 for some self-adjoint operator T. (ii) P is positive; that is, P ¼ SS: (iii) P is self-adjoint and hPðuÞ; ui 0 for every u 2 V. 13.9 Diagonalization and Canonical Forms in Inner Product Spaces Let T be a linear operator on a ﬁnite-dimensional inner product space V over K. Representing T by a diagonal matrix depends upon the eigenvectors and eigenvalues of T, and hence, upon the roots of the characteristic polynomial DðtÞ of T. Now DðtÞ always factors into linear polynomials over the complex ﬁeld C but may not have any linear polynomials over the real ﬁeld R. Thus, the situation for real inner product spaces (sometimes called Euclidean spaces) is inherently different than the situation for complex inner product spaces (sometimes called unitary spaces). Thus, we treat them separately. Real Inner Product Spaces, Symmetric and Orthogonal Operators The following theorem (proved in Problem 13.14) holds. THEOREM 13.11: Let T be a symmetric (self-adjoint) operator on a real ﬁnite-dimensional product space V. Then there exists an orthonormal basis of V consisting of eigenvectors of T; that is, T can be represented by a diagonal matrix relative to an orthonormal basis. We give the corresponding statement for matrices. THEOREM 13.11: (Alternative Form) Let A be a real symmetric matrix. Then there exists an orthogonal matrix P such that B ¼ P1AP ¼ PTAP is diagonal. We can choose the columns of the above matrix P to be normalized orthogonal eigenvectors of A; then the diagonal entries of B are the corresponding eigenvalues. On the other hand, an orthogonal operator T need not be symmetric, and so it may not be represented by a diagonal matrix relative to an orthonormal matrix. However, such a matrix T does have a simple canonical representation, as described in the following theorem (proved in Problem 13.16). 382 CHAPTER 13 Linear Operators on Inner Product Spaces THEOREM 13.12: Let T be an orthogonal operator on a real inner product space V. Then there exists an orthonormal basis of V in which T is represented by a block diagonal matrix M of the form M ¼ diag Is; It; cos y1 sin y1 sin y1 cos y1 ; . . . ; cos yr sin yr sin yr cos yr The reader may recognize that each of the 2 2 diagonal blocks represents a rotation in the corresponding two-dimensional subspace, and each diagonal entry 1 represents a reﬂection in the corresponding one-dimensional subspace. Complex Inner Product Spaces, Normal and Triangular Operators A linear operator T is said to be normal if it commutes with its adjoint—that is, if TT ¼ TT. We note that normal operators include both self-adjoint and unitary operators. Analogously, a complex matrix A is said to be normal if it commutes with its conjugate transpose— that is, if AA ¼ AA. EXAMPLE 13.3 Let A ¼ 1 1 i 3 þ 2i . Then A ¼ 1 i 1 3 2i . Also AA ¼ 2 3 3i 3 þ 3i 14 ¼ AA. Thus, A is normal. The following theorem (proved in Problem 13.19) holds. THEOREM 13.13: Let T be a normal operator on a complex ﬁnite-dimensional inner product space V. Then there exists an orthonormal basis of V consisting of eigenvectors of T; that is, T can be represented by a diagonal matrix relative to an orthonormal basis. We give the corresponding statement for matrices. THEOREM 13.13: (Alternative Form) Let A be a normal matrix. Then there exists a unitary matrix P such that B ¼ P1AP ¼ PAP is diagonal. The following theorem (proved in Problem 13.20) shows that even nonnormal operators on unitary spaces have a relatively simple form. THEOREM 13.14: Let T be an arbitrary operator on a complex ﬁnite-dimensional inner product space V. Then T can be represented by a triangular matrix relative to an orthonormal basis of V. THEOREM 13.14: (Alternative Form) Let A be an arbitrary complex matrix. Then there exists a unitary matrix P such that B ¼ P1AP ¼ PAP is triangular. 13.10 Spectral Theorem The Spectral Theorem is a reformulation of the diagonalization Theorems 13.11 and 13.13. THEOREM 13.15: (Spectral Theorem) Let T be a normal (symmetric) operator on a complex (real) ﬁnite-dimensional inner product space V. Then there exists linear operators E1; . . . ; Er on V and scalars l1; . . . ; lr such that (i) T ¼ l1E1 þ l2E2 þ þ lrEr, (iii) E2 1 ¼ E1; E2 2 ¼ E2; . . . ; E2 r ¼ Er, (ii) E1 þ E2 þ þ Er ¼ I, (iv) EiEj ¼ 0 for i 6¼ j. CHAPTER 13 Linear Operators on Inner Product Spaces 383 The above linear operators E1; . . . ; Er are projections in the sense that E2 i ¼ Ei. Moreover, they are said to be orthogonal projections because they have the additional property that EiEj ¼ 0 for i 6¼ j. The following example shows the relationship between a diagonal matrix representation and the corresponding orthogonal projections. EXAMPLE 13.4 Consider the following diagonal matrices A; E1; E2; E3: A ¼ 2 3 3 5 2 6 6 4 3 7 7 5; E1 ¼ 1 0 0 0 2 6 6 4 3 7 7 5; E2 ¼ 0 1 1 0 2 6 6 4 3 7 7 5; E3 ¼ 0 0 0 1 2 6 6 4 3 7 7 5 The reader can verify that (i) A ¼ 2E1 þ 3E2 þ 5E3, (ii) E1 þ E2 þ E3 ¼ I, (iii) E2 i ¼ Ei, (iv) EiEj ¼ 0 for i 6¼ j. SOLVED PROBLEMS Adjoints 13.1. Find the adjoint of F:R3 ! R3 deﬁned by Fðx; y; zÞ ¼ ð3x þ 4y 5z; 2x 6y þ 7z; 5x 9y þ zÞ First ﬁnd the matrix A that represents F in the usual basis of R3—that is, the matrix A whose rows are the coefﬁcients of x; y; z—and then form the transpose AT of A. This yields A ¼ 3 4 5 2 6 7 5 9 1 2 4 3 5 and then AT ¼ 3 2 5 4 6 9 5 7 1 2 4 3 5 The adjoint F is represented by the transpose of A; hence, Fðx; y; zÞ ¼ ð3x þ 2y þ 5z; 4x 6y 9z; 5x þ 7y þ zÞ 13.2. Find the adjoint of G:C3 ! C3 deﬁned by Gðx; y; zÞ ¼ ½2x þ ð1 iÞy; ð3 þ 2iÞx 4iz; 2ix þ ð4 3iÞy 3z First ﬁnd the matrix B that represents G in the usual basis of C3, and then form the conjugate transpose B of B. This yields B ¼ 2 1 i 0 3 þ 2i 0 4i 2i 4 3i 3 2 4 3 5 and then B ¼ 2 3 2i 2i 1 þ i 0 4 þ 3i 0 4i 3 2 4 3 5 Then Gðx; y; zÞ ¼ ½2x þ ð3 2iÞy 2iz; ð1 þ iÞx þ ð4 þ 3iÞz; 4iy 3z: 13.3. Prove Theorem 13.3: Let f be a linear functional on an n-dimensional inner product space V. Then there exists a unique vector u 2 V such that fðvÞ ¼ hv; ui for every v 2 V. Let fw1; . . . ; wng be an orthonormal basis of V. Set u ¼ fðw1Þw1 þ fðw2Þw2 þ þ fðwnÞwn Let ^ u be the linear functional on V deﬁned by ^ uðvÞ ¼ hv; ui for every v 2 V. Then, for i ¼ 1; . . . ; n, ^ uðwiÞ ¼ hwi; ui ¼ hwi; fðw1Þw1 þ þ fðwnÞwni ¼ fðwiÞ 384 CHAPTER 13 Linear Operators on Inner Product Spaces Because ^ u and f agree on each basis vector, ^ u ¼ f. Now suppose u0 is another vector in V for which fðvÞ ¼ hv; u0i for every v 2 V. Then hv; ui ¼ hv; u0i or hv; u u0i ¼ 0. In particular, this is true for v ¼ u u0, and so hu u0; u u0i ¼ 0. This yields u u0 ¼ 0 and u ¼ u0. Thus, such a vector u is unique, as claimed. 13.4. Prove Theorem 13.1: Let T be a linear operator on an n-dimensional inner product space V. Then (a) There exists a unique linear operator T on V such that hTðuÞ; vi ¼ hu; TðvÞi for all u; v 2 V: (b) Let A be the matrix that represents T relative to an orthonormal basis S ¼ fuig. Then the conjugate transpose A of A represents T in the basis S. (a) We ﬁrst deﬁne the mapping T. Let v be an arbitrary but ﬁxed element of V. The map u 7! hTðuÞ; vi is a linear functional on V. Hence, by Theorem 13.3, there exists a unique element v0 2 V such that hTðuÞ; vi ¼ hu; v0i for every u 2 V. We deﬁne T : V ! V by TðvÞ ¼ v0. Then hTðuÞ; vi ¼ hu; TðvÞi for every u; v 2 V. We next show that T is linear. For any u; vi 2 V, and any a; b 2 K, hu; Tðav1 þ bv2Þi ¼ hTðuÞ; av1 þ bv2i ¼ ahTðuÞ; v1i þ bhTðuÞ; v2i ¼ ahu; Tðv1Þi þ bhu; Tðv2Þi ¼ hu; aTðv1Þ þ bTðv2Þi But this is true for every u 2 V; hence, Tðav1 þ bv2Þ ¼ aTðv1Þ þ bTðv2Þ. Thus, T is linear. (b) The matrices A ¼ ½aij and B ¼ ½bij that represent T and T, respectively, relative to the orthonormal basis S are given by aij ¼ hTðujÞ; uii and bij ¼ hTðujÞ; uii (Problem 13.67). Hence, bij ¼ hTðujÞ; uii ¼ hui; TðujÞi ¼ hTðuiÞ; uji ¼ aji Thus, B ¼ A, as claimed. 13.5. Prove Theorem 13.2: (i) ðT1 þ T2Þ ¼ T 1 þ T 2 , (iii) ðT1T2Þ ¼ T 2 T 1 , (ii) ðkTÞ ¼ kT, (iv) ðTÞ ¼ T. (i) For any u; v 2 V, hðT1 þ T2ÞðuÞ; vi ¼ hT1ðuÞ þ T2ðuÞ; vi ¼ hT1ðuÞ; vi þ hT2ðuÞ; vi ¼ hu; T 1 ðvÞi þ hu; T 2 ðvÞi ¼ hu; T 1 ðvÞ þ T 2 ðvÞi ¼ hu; ðT 1 þ T 2 ÞðvÞi The uniqueness of the adjoint implies ðT1 þ T2Þ ¼ T 1 þ T 2 . (ii) For any u; v 2 V, hðkTÞðuÞ; vi ¼ hkTðuÞ; vi ¼ khTðuÞ; vi ¼ khu; TðvÞi ¼ hu; kTðvÞi ¼ hu; ð kTÞðvÞi The uniqueness of the adjoint implies ðkTÞ ¼ kT. (iii) For any u; v 2 V, hðT1T2ÞðuÞ; vi ¼ hT1ðT2ðuÞÞ; vi ¼ hT2ðuÞ; T 1 ðvÞi ¼ hu; T 2 ðT 1 ðvÞÞi ¼ hu; ðT 2 T 1 ÞðvÞi The uniqueness of the adjoint implies ðT1T2Þ ¼ T 2 T 1 . (iv) For any u; v 2 V, hTðuÞ; vi ¼ hv; TðuÞi ¼ hTðvÞ; ui ¼ hu; TðvÞi The uniqueness of the adjoint implies ðTÞ ¼ T. CHAPTER 13 Linear Operators on Inner Product Spaces 385 13.6. Show that ðaÞ I ¼ I, and ðbÞ 0 ¼ 0. (a) For every u; v 2 V, hIðuÞ; vi ¼ hu; vi ¼ hu; IðvÞi; hence, I ¼ I. (b) For every u; v 2 V, h0ðuÞ; vi ¼ h0; vi ¼ 0 ¼ hu; 0i ¼ hu; 0ðvÞi; hence, 0 ¼ 0. 13.7. Suppose T is invertible. Show that ðT1Þ ¼ ðTÞ1. I ¼ I ¼ ðTT 1Þ ¼ ðT 1ÞT; hence; ðT 1Þ ¼ ðTÞ1: 13.8. Let T be a linear operator on V, and let W be a T-invariant subspace of V. Show that W ? is invariant under T. Let u 2 W ?. If w 2 W, then TðwÞ 2 W and so hw; TðuÞi ¼ hTðwÞ; ui ¼ 0. Thus, TðuÞ 2 W ? because it is orthogonal to every w 2 W. Hence, W ? is invariant under T. 13.9. Let T be a linear operator on V. Show that each of the following conditions implies T ¼ 0: (i) hTðuÞ; vi ¼ 0 for every u; v 2 V. (ii) V is a complex space, and hTðuÞ; ui ¼ 0 for every u 2 V. (iii) T is self-adjoint and hTðuÞ; ui ¼ 0 for every u 2 V. Give an example of an operator T on a real space V for which hTðuÞ; ui ¼ 0 for every u 2 V but T 6¼ 0. [Thus, (ii) need not hold for a real space V.] (i) Set v ¼ TðuÞ. Then hTðuÞ; TðuÞi ¼ 0, and hence, TðuÞ ¼ 0, for every u 2 V. Accordingly, T ¼ 0. (ii) By hypothesis, hTðv þ wÞ; v þ wi ¼ 0 for any v; w 2 V. Expanding and setting hTðvÞ; vi ¼ 0 and hTðwÞ; wi ¼ 0, we ﬁnd hTðvÞ; wi þ hTðwÞ; vi ¼ 0 ð1Þ Note w is arbitrary in (1). Substituting iw for w, and using hTðvÞ; iwi ¼ ihTðvÞ; wi ¼ ihTðvÞ; wi and hTðiwÞ; vi ¼ hiTðwÞ; vi ¼ ihTðwÞ; vi, we ﬁnd ihTðvÞ; wi þ ihTðwÞ; vi ¼ 0 Dividing through by i and adding to (1), we obtain hTðwÞ; vi ¼ 0 for any v; w; 2 V. By (i), T ¼ 0. (iii) By (ii), the result holds for the complex case; hence we need only consider the real case. Expanding hTðv þ wÞ; v þ wi ¼ 0, we again obtain (1). Because T is self-adjoint and as it is a real space, we have hTðwÞ; vi ¼ hw; TðvÞi ¼ hTðvÞ; wi. Substituting this into (1), we obtain hTðvÞ; wi ¼ 0 for any v; w 2 V. By (i), T ¼ 0. For an example, consider the linear operator T on R2 deﬁned by Tðx; yÞ ¼ ðy; xÞ. Then hTðuÞ; ui ¼ 0 for every u 2 V, but T 6¼ 0. Orthogonal and Unitary Operators and Matrices 13.10. Prove Theorem 13.6: The following conditions on an operator U are equivalent: (i) U ¼ U1; that is, U is unitary. (ii) hUðvÞ; UðwÞi ¼ hu; wi. (iii) kUðvÞk ¼ kvk. Suppose (i) holds. Then, for every v; w; 2 V, hUðvÞ; UðwÞi ¼ hv; UUðwÞi ¼ hv; IðwÞi ¼ hv; wi Thus, (i) implies (ii). Now if (ii) holds, then kUðvÞk ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ hUðvÞ; UðvÞi p ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ hv; vi p ¼ kvk Hence, (ii) implies (iii). It remains to show that (iii) implies (i). Suppose (iii) holds. Then for every v 2 V, hUUðvÞi ¼ hUðvÞ; UðvÞi ¼ hv; vi ¼ hIðvÞ; vi Hence, hðUU IÞðvÞ; vi ¼ 0 for every v 2 V. But UU I is self-adjoint (Prove!); then, by Problem 13.9, we have UU I ¼ 0 and so UU ¼ I. Thus, U ¼ U1, as claimed. 386 CHAPTER 13 Linear Operators on Inner Product Spaces 13.11. Let U be a unitary (orthogonal) operator on V, and let W be a subspace invariant under U. Show that W ? is also invariant under U. Because U is nonsingular, UðWÞ ¼ W; that is, for any w 2 W, there exists w0 2 W such that Uðw0Þ ¼ w. Now let v 2 W ?. Then, for any w 2 W, hUðvÞ; wi ¼ hUðvÞ; Uðw0Þi ¼ hv; w0i ¼ 0 Thus, UðvÞ belongs to W ?. Therefore, W ? is invariant under U. 13.12. Prove Theorem 13.9: The change-of-basis matrix from an orthonormal basis fu1; . . . ; ung into another orthonormal basis is unitary (orthogonal). Conversely, if P ¼ ½aij is a unitary (ortho-gonal) matrix, then the vectors ui0 ¼ P j ajiuj form an orthonormal basis. Suppose fvig is another orthonormal basis and suppose vi ¼ bi1u1 þ bi2u2 þ þ binun; i ¼ 1; . . . ; n ð1Þ Because fvig is orthonormal, dij ¼ hvi; vji ¼ bi1bj1 þ bi2bj2 þ þ binbjn ð2Þ Let B ¼ ½bij be the matrix of coefﬁcients in (1). (Then BT is the change-of-basis matrix from fuig to fvig.) Then BB ¼ ½cij, where cij ¼ bi1bj1 þ bi2bj2 þ þ binbjn. By (2), cij ¼ dij, and therefore BB ¼ I. Accordingly, B, and hence, BT, is unitary. It remains to prove that fu0 ig is orthonormal. By Problem 13.67, hu0 i; u0 ji ¼ a1ia1j þ a2ia2j þ þ anianj ¼ hCi; Cji where Ci denotes the ith column of the unitary (orthogonal) matrix P ¼ ½aij: Because P is unitary (orthogonal), its columns are orthonormal; hence, hu0 i; u0 ji ¼ hCi; Cji ¼ dij. Thus, fu0 ig is an orthonormal basis. Symmetric Operators and Canonical Forms in Euclidean Spaces 13.13. Let T be a symmetric operator. Show that (a) The characteristic polynomial DðtÞ of T is a product of linear polynomials (over R); (b) T has a nonzero eigenvector. (a) Let A be a matrix representing T relative to an orthonormal basis of V; then A ¼ AT. Let DðtÞ be the characteristic polynomial of A. Viewing A as a complex self-adjoint operator, A has only real eigenvalues by Theorem 13.4. Thus, DðtÞ ¼ ðt l1Þðt l2Þ ðt lnÞ where the li are all real. In other words, DðtÞ is a product of linear polynomials over R. (b) By (a), T has at least one (real) eigenvalue. Hence, T has a nonzero eigenvector. 13.14. Prove Theorem 13.11: Let T be a symmetric operator on a real n-dimensional inner product space V. Then there exists an orthonormal basis of V consisting of eigenvectors of T. (Hence, T can be represented by a diagonal matrix relative to an orthonormal basis.) The proof is by induction on the dimension of V. If dim V ¼ 1, the theorem trivially holds. Now suppose dim V ¼ n > 1. By Problem 13.13, there exists a nonzero eigenvector v1 of T. Let W be the space spanned by v1, and let u1 be a unit vector in W, e.g., let u1 ¼ v1=kv1k. Because v1 is an eigenvector of T, the subspace W of V is invariant under T. By Problem 13.8, W ? is invariant under T ¼ T. Thus, the restriction ^ T of T to W ? is a symmetric operator. By Theorem 7.4, V ¼ W W ?. Hence, dim W ? ¼ n 1, because dim W ¼ 1. By induction, there exists an orthonormal basis fu2; . . . ; ung of W ? consisting of eigenvectors of ^ T and hence of T. But hu1; uii ¼ 0 for i ¼ 2; . . . ; n because ui 2 W ?. Accordingly fu1; u2; . . . ; ung is an orthonormal set and consists of eigenvectors of T. Thus, the theorem is proved. CHAPTER 13 Linear Operators on Inner Product Spaces 387 13.15. Let qðx; yÞ ¼ 3x2 6xy þ 11y2. Find an orthonormal change of coordinates (linear substitution) that diagonalizes the quadratic form q. Find the symmetric matrix A representing q and its characteristic polynomial DðtÞ. We have A ¼ 3 3 3 11 and DðtÞ ¼ t2 trðAÞ t þ jAj ¼ t2 14t þ 24 ¼ ðt 2Þðt 12Þ The eigenvalues are l ¼ 2 and l ¼ 12. Hence, a diagonal form of q is qðs; tÞ ¼ 2s2 þ 12t2 (where we use s and t as new variables). The corresponding orthogonal change of coordinates is obtained by ﬁnding an orthogonal set of eigenvectors of A. Subtract l ¼ 2 down the diagonal of A to obtain the matrix M ¼ 1 3 3 9 corresponding to x 3y ¼ 0 3x þ 9y ¼ 0 or x 3y ¼ 0 A nonzero solution is u1 ¼ ð3; 1Þ. Next subtract l ¼ 12 down the diagonal of A to obtain the matrix M ¼ 9 3 3 1 corresponding to 9x 3y ¼ 0 3x y ¼ 0 or 3x y ¼ 0 A nonzero solution is u2 ¼ ð1; 3Þ. Normalize u1 and u2 to obtain the orthonormal basis ^ u1 ¼ ð3= ﬃﬃﬃﬃﬃ 10 p ; 1= ﬃﬃﬃﬃﬃ 10 p Þ; ^ u2 ¼ ð1= ﬃﬃﬃﬃﬃ 10 p ; 3= ﬃﬃﬃﬃﬃ 10 p Þ Now let P be the matrix whose columns are ^ u1 and ^ u2. Then P ¼ 3= ﬃﬃﬃﬃﬃ 10 p 1= ﬃﬃﬃﬃﬃ 10 p 1= ﬃﬃﬃﬃﬃ 10 p 3= ﬃﬃﬃﬃﬃ 10 p " # and D ¼ P1AP ¼ PTAP ¼ 2 0 0 12 Thus, the required orthogonal change of coordinates is x y ¼ P s t or x ¼ 3s t ﬃﬃﬃﬃﬃ 10 p ; y ¼ s þ 3t ﬃﬃﬃﬃﬃ 10 p One can also express s and t in terms of x and y by using P1 ¼ PT; that is, s ¼ 3x þ y ﬃﬃﬃﬃﬃ 10 p ; t ¼ x þ 3y ﬃﬃﬃﬃﬃ 10 p 13.16. Prove Theorem 13.12: Let T be an orthogonal operator on a real inner product space V. Then there exists an orthonormal basis of V in which T is represented by a block diagonal matrix M of the form M ¼ diag 1; . . . ; 1; 1; . . . ; 1; cos y1 sin y1 sin y1 cos y1 ; . . . ; cos yr sin yr sin yr cos yr Let S ¼ T þ T 1 ¼ T þ T. Then S ¼ ðT þ TÞ ¼ T þ T ¼ S. Thus, S is a symmetric operator on V. By Theorem 13.11, there exists an orthonormal basis of V consisting of eigenvectors of S. If l1; . . . ; lm denote the distinct eigenvalues of S, then V can be decomposed into the direct sum V ¼ V1 V2 Vm where the Vi consists of the eigenvectors of S belonging to li. We claim that each Vi is invariant under T. For suppose v 2 V; then SðvÞ ¼ liv and SðTðvÞÞ ¼ ðT þ T1ÞTðvÞ ¼ TðT þ T1ÞðvÞ ¼ TSðvÞ ¼ TðlivÞ ¼ liTðvÞ That is, TðvÞ 2 Vi. Hence, Vi is invariant under T. Because the Vi are orthogonal to each other, we can restrict our investigation to the way that T acts on each individual Vi. On a given Vi; we have ðT þ T 1Þv ¼ SðvÞ ¼ liv. Multiplying by T, we get ðT 2 liT þ IÞðvÞ ¼ 0 ð1Þ 388 CHAPTER 13 Linear Operators on Inner Product Spaces We consider the cases li ¼ 2 and li 6¼ 2 separately. If li ¼ 2, then ðT IÞ2ðvÞ ¼ 0, which leads to ðT IÞðvÞ ¼ 0 or TðvÞ ¼ v. Thus, T restricted to this Vi is either I or I. If li 6¼ 2, then T has no eigenvectors in Vi, because, by Theorem 13.4, the only eigenvalues of T are 1 or 1. Accordingly, for v 6¼ 0, the vectors v and TðvÞ are linearly independent. Let W be the subspace spanned by v and TðvÞ. Then W is invariant under T, because using (1) we get TðTðvÞÞ ¼ T2ðvÞ ¼ liTðvÞ v 2 W By Theorem 7.4, Vi ¼ W W ?. Furthermore, by Problem 13.8, W ? is also invariant under T. Thus, we can decompose Vi into the direct sum of two-dimensional subspaces Wj where the Wj are orthogonal to each other and each Wj is invariant under T. Thus, we can restrict our investigation to the way in which T acts on each individual Wj. Because T2 liT þ I ¼ 0, the characteristic polynomial DðtÞ of T acting on Wj is DðtÞ ¼ t2 lit þ 1. Thus, the determinant of T is 1, the constant term in DðtÞ. By Theorem 2.7, the matrix A representing T acting on Wj relative to any orthogonal basis of Wj must be of the form cos y sin y sin y cos y The union of the bases of the Wj gives an orthonormal basis of Vi, and the union of the bases of the Vi gives an orthonormal basis of V in which the matrix representing T is of the desired form. Normal Operators and Canonical Forms in Unitary Spaces 13.17. Determine which of the following matrices is normal: (a) A ¼ 1 i 0 1 , (b) B ¼ 1 i 1 2 þ i (a) AA ¼ 1 i 0 1 1 0 i 1 ¼ 2 i i 1 , AA ¼ 1 0 i 1 1 i 0 1 ¼ 1 i i 2 Because AA 6¼ AA, the matrix A is not normal. (b) BB 1 i 1 2 þ i 1 1 i 2 i ¼ 2 2 þ 2i 2 2i 6 ¼ 1 1 i 2 i 1 i 1 2 þ i ¼ BB Because BB ¼ BB, the matrix B is normal. 13.18. Let T be a normal operator. Prove the following: (a) TðvÞ ¼ 0 if and only if TðvÞ ¼ 0. (b) T lI is normal. (c) If TðvÞ ¼ lv, then TðvÞ ¼ lv; hence, any eigenvector of T is also an eigenvector of T. (d) If TðvÞ ¼ l1v and TðwÞ ¼ l2w where l1 6¼ l2, then hv; wi ¼ 0; that is, eigenvectors of T belonging to distinct eigenvalues are orthogonal. (a) We show that hTðvÞ; TðvÞi ¼ hTðvÞ; TðvÞi: hTðvÞ; TðvÞi ¼ hv; TTðvÞi ¼ hv; TTðvÞi ¼ hTðvÞ; TðvÞi Hence, by ½I3 in the deﬁnition of the inner product in Section 7.2, TðvÞ ¼ 0 if and only if TðvÞ ¼ 0. (b) We show that T lI commutes with its adjoint: ðT lIÞðT lIÞ ¼ ðT lIÞðT lIÞ ¼ TT lT lT þ l lI ¼ TT lT lT þ llI ¼ ðT lIÞðT lIÞ ¼ ðT lIÞðT lIÞ Thus, T lI is normal. CHAPTER 13 Linear Operators on Inner Product Spaces 389 (c) If TðvÞ ¼ lv, then ðT lIÞðvÞ ¼ 0. Now T lI is normal by (b); therefore, by (a), ðT lIÞðvÞ ¼ 0. That is, ðT lIÞðvÞ ¼ 0; hence, TðvÞ ¼ lv. (d) We show that l1hv; wi ¼ l2hv; wi: l1hv; wi ¼ hl1v; wi ¼ hTðvÞ; wi ¼ hv; TðwÞi ¼ hv; l2wi ¼ l2hv; wi But l1 6¼ l2; hence, hv; wi ¼ 0. 13.19. Prove Theorem 13.13: Let T be a normal operator on a complex ﬁnite-dimensional inner product space V. Then there exists an orthonormal basis of V consisting of eigenvectors of T. (Thus, T can be represented by a diagonal matrix relative to an orthonormal basis.) The proof is by induction on the dimension of V. If dim V ¼ 1, then the theorem trivially holds. Now suppose dim V ¼ n > 1. Because V is a complex vector space, T has at least one eigenvalue and hence a nonzero eigenvector v. Let W be the subspace of V spanned by v, and let u1 be a unit vector in W. Because v is an eigenvector of T, the subspace W is invariant under T. However, v is also an eigenvector of T by Problem 13.18; hence, W is also invariant under T. By Problem 13.8, W ? is invariant under T ¼ T. The remainder of the proof is identical with the latter part of the proof of Theorem 13.11 (Problem 13.14). 13.20. Prove Theorem 13.14: Let T be any operator on a complex ﬁnite-dimensional inner product space V. Then T can be represented by a triangular matrix relative to an orthonormal basis of V. The proof is by induction on the dimension of V. If dim V ¼ 1, then the theorem trivially holds. Now suppose dim V ¼ n > 1. Because V is a complex vector space, T has at least one eigenvalue and hence at least one nonzero eigenvector v. Let W be the subspace of V spanned by v, and let u1 be a unit vector in W. Then u1 is an eigenvector of T and, say, Tðu1Þ ¼ a11u1. By Theorem 7.4, V ¼ W W ?. Let E denote the orthogonal projection V into W ?. Clearly W ? is invariant under the operator ET. By induction, there exists an orthonormal basis fu2; . . . ; ung of W ? such that, for i ¼ 2; . . . ; n, ETðuiÞ ¼ ai2u2 þi3 u3 þ þ aiiui (Note that fu1; u2; . . . ; ung is an orthonormal basis of V.) But E is the orthogonal projection of V onto W ?; hence, we must have TðuiÞ ¼ ai1u1 þ ai2u2 þ þ aiiui for i ¼ 2; . . . ; n. This with Tðu1Þ ¼ a11u1 gives us the desired result. Miscellaneous Problems 13.21. Prove Theorem 13.10B: The following are equivalent: (i) P ¼ T2 for some self-adjoint operator T. (ii) P ¼ SS for some operator S; that is, P is positive. (iii) P is self-adjoint and hPðuÞ; ui 0 for every u 2 V. Suppose (i) holds; that is, P ¼ T2 where T ¼ T. Then P ¼ TT ¼ TT, and so (i) implies (ii). Now suppose (ii) holds. Then P ¼ ðSSÞ ¼ SS ¼ SS ¼ P, and so P is self-adjoint. Furthermore, hPðuÞ; ui ¼ hSSðuÞ; ui ¼ hSðuÞ; SðuÞi 0 Thus, (ii) implies (iii), and so it remains to prove that (iii) implies (i). Now suppose (iii) holds. Because P is self-adjoint, there exists an orthonormal basis fu1; . . . ; ung of V consisting of eigenvectors of P; say, PðuiÞ ¼ liui. By Theorem 13.4, the li are real. Using (iii), we show that the li are nonnegative. We have, for each i, 0 hPðuiÞ; uii ¼ hliui; uii ¼ lihui; uii Thus, hui; uii 0 forces li 0; as claimed. Accordingly, ﬃﬃﬃﬃ li p is a real number. Let T be the linear operator deﬁned by TðuiÞ ¼ ﬃﬃﬃﬃ li p ui for i ¼ 1; . . . ; n 390 CHAPTER 13 Linear Operators on Inner Product Spaces Because T is represented by a real diagonal matrix relative to the orthonormal basis fuig, T is self-adjoint. Moreover, for each i, T2ðuiÞ ¼ Tð ﬃﬃﬃﬃ li p uiÞ ¼ ﬃﬃﬃﬃ li p TðiiÞ ¼ ﬃﬃﬃﬃ li p ﬃﬃﬃﬃ li p ui ¼ liui ¼ PðuiÞ Because T2 and P agree on a basis of V; P ¼ T2. Thus, the theorem is proved. Remark: The above operator T is the unique positive operator such that P ¼ T 2; it is called the positive square root of P. 13.22. Show that any operator T is the sum of a self-adjoint operator and a skew-adjoint operator. Set S ¼ 1 2 ðT þ TÞ and U ¼ 1 2 ðT TÞ: Then T ¼ S þ U; where S ¼ ½1 2 ðT þ TÞ ¼ 1 2 ðT þ TÞ ¼ 1 2 ðT þ TÞ ¼ S U ¼ ½1 2 ðT TÞ ¼ 1 2 ðT TÞ ¼ 1 2 ðT TÞ ¼ U and that is, S is self-adjoint and U is skew-adjoint. 13.23. Prove: Let T be an arbitrary linear operator on a ﬁnite-dimensional inner product space V. Then T is a product of a unitary (orthogonal) operator U and a unique positive operator P; that is, T ¼ UP. Furthermore, if T is invertible, then U is also uniquely determined. By Theorem 13.10, TT is a positive operator; hence, there exists a (unique) positive operator P such that P2 ¼ TT (Problem 13.43). Observe that kPðvÞk2 ¼ hPðvÞ; PðvÞi ¼ hP2ðvÞ; vi ¼ hTTðvÞ; vi ¼ hTðvÞ; TðvÞi ¼ kTðvÞk2 ð1Þ We now consider separately the cases when T is invertible and noninvertible. If T is invertible, then we set ^ U ¼ PT1. We show that ^ U is unitary: ^ U ¼ ðPT1Þ ¼ T1P ¼ ðTÞ1P and ^ U ^ U ¼ ðTÞ1PPT1 ¼ ðTÞ1TTT1 ¼ I Thus, ^ U is unitary. We next set U ¼ ^ U1. Then U is also unitary, and T ¼ UP as required. To prove uniqueness, we assume T ¼ U0P0, where U0 is unitary and P0 is positive. Then TT ¼ P0 U0 U0P0 ¼ P0IP0 ¼ P2 0 But the positive square root of TT is unique (Problem 13.43); hence, P0 ¼ P. (Note that the invertibility of T is not used to prove the uniqueness of P.) Now if T is invertible, then P is also invertible by (1). Multiplying U0P ¼ UP on the right by P1 yields U0 ¼ U. Thus, U is also unique when T is invertible. Now suppose T is not invertible. Let W be the image of P; that is, W ¼ Im P. We deﬁne U1:W ! V by U1ðwÞ ¼ TðvÞ; where PðvÞ ¼ w ð2Þ We must show that U1 is well deﬁned; that is, that PðvÞ ¼ Pðv0Þ implies TðvÞ ¼ Tðv0Þ. This follows from the fact that Pðv v0Þ ¼ 0 is equivalent to kPðv v0Þk ¼ 0, which forces kTðv v0Þk ¼ 0 by (1). Thus, U1 is well deﬁned. We next deﬁne U2:W ! V. Note that, by (1), P and T have the same kernels. Hence, the images of P and T have the same dimension; that is, dimðIm PÞ ¼ dim W ¼ dimðIm TÞ. Consequently, W ? and ðIm TÞ? also have the same dimension. We let U2 be any isomorphism between W ? and ðIm TÞ?. We next set U ¼ U1 U2. [Here U is deﬁned as follows: If v 2 V and v ¼ w þ w0, where w 2 W, w0 2 W ?, then UðvÞ ¼ U1ðwÞ þ U2ðw0Þ.] Now U is linear (Problem 13.69), and, if v 2 V and PðvÞ ¼ w, then, by (2), TðvÞ ¼ U1ðwÞ ¼ UðwÞ ¼ UPðvÞ Thus, T ¼ UP, as required. It remains to show that U is unitary. Now every vector x 2 V can be written in the form x ¼ PðvÞ þ w0, where w0 2 W ?. Then UðxÞ ¼ UPðvÞ þ U2ðw0Þ ¼ TðvÞ þ U2ðw0Þ, where hTðvÞ; U2ðw0Þi ¼ 0 by deﬁnition CHAPTER 13 Linear Operators on Inner Product Spaces 391 of U2. Also, hTðvÞ; TðvÞi ¼ hPðvÞ; PðvÞi by (1). Thus, hUðxÞ; UðxÞi ¼ hTðvÞ þ U2ðw0Þ; TðvÞ þ U2ðw0Þi ¼ hTðvÞ; TðvÞi þ hU2ðw0Þ; U2ðw0Þi ¼ hPðvÞ; PðvÞi þ hw0; w0i ¼ hPðvÞ þ w0; PðvÞ þ w0Þ ¼ hx; xi [We also used the fact that hPðvÞ; w0i ¼ 0: Thus, U is unitary, and the theorem is proved. 13.24. Let V be the vector space of polynomials over R with inner product deﬁned by h f ; gi ¼ ð1 0 f ðtÞgðtÞ dt Give an example of a linear functional f on V for which Theorem 13.3 does not hold—that is, for which there is no polynomial hðtÞ such that fð f Þ ¼ h f ; hi for every f 2 V. Let f:V ! R be deﬁned by fð f Þ ¼ f ð0Þ; that is, f evaluates f ðtÞ at 0, and hence maps f ðtÞ into its constant term. Suppose a polynomial hðtÞ exists for which fðf Þ ¼ f ð0Þ ¼ ð1 0 f ðtÞhðtÞ dt ð1Þ for every polynomial f ðtÞ. Observe that f maps the polynomial tf ðtÞ into 0; hence, by (1), ð1 0 tf ðtÞhðtÞ dt ¼ 0 ð2Þ for every polynomial f ðtÞ. In particular (2) must hold for f ðtÞ ¼ thðtÞ; that is, ð1 0 t2h2ðtÞ dt ¼ 0 This integral forces hðtÞ to be the zero polynomial; hence, fð f Þ ¼ h f ; hi ¼ h f ; 0i ¼ 0 for every polynomial f ðtÞ. This contradicts the fact that f is not the zero functional; hence, the polynomial hðtÞ does not exist. SUPPLEMENTARY PROBLEMS Adjoint Operators 13.25. Find the adjoint of: (a) A ¼ 5 2i 3 þ 7i 4 6i 8 þ 3i ; (b) B ¼ 3 5i i 2i ; (c) C ¼ 1 1 2 3 13.26. Let T :R3 ! R3 be deﬁned by Tðx; y; zÞ ¼ ðx þ 2y; 3x 4z; yÞ: Find Tðx; y; zÞ: 13.27. Let T :C3 ! C3 be deﬁned by Tðx; y; zÞ ¼ ½ix þ ð2 þ 3iÞy; 3x þ ð3 iÞz; ð2 5iÞy þ iz: Find Tðx; y; zÞ: 13.28. For each linear function f on V; ﬁnd u 2 V such that fðvÞ ¼ hv; ui for every v 2 V: (a) f:R3 ! R deﬁned by fðx; y; zÞ ¼ x þ 2y 3z: (b) f:C3 ! C deﬁned by fðx; y; zÞ ¼ ix þ ð2 þ 3iÞy þ ð1 2iÞz: 13.29. Suppose V has ﬁnite dimension. Prove that the image of T is the orthogonal complement of the kernel of T; that is, Im T ¼ ðKer TÞ?: Hence, rankðTÞ ¼ rankðTÞ: 13.30. Show that TT ¼ 0 implies T ¼ 0: 392 CHAPTER 13 Linear Operators on Inner Product Spaces 13.31. Let V be the vector space of polynomials over R with inner product deﬁned by h f ; gi ¼ Ð 1 0 f ðtÞgðtÞ dt: Let D be the derivative operator on V; that is, Dð f Þ ¼ df =dt: Show that there is no operator D on V such that hDð f Þ; gi ¼ h f ; DðgÞi for every f ; g 2 V: That is, D has no adjoint. Unitary and Orthogonal Operators and Matrices 13.32. Find a unitary (orthogonal) matrix whose ﬁrst row is (a) ð2= ﬃﬃﬃﬃﬃ 13 p ; 3= ﬃﬃﬃﬃﬃ 13 p Þ, (b) a multiple of ð1; 1 iÞ, (c) a multiple of ð1; i; 1 iÞ: 13.33. Prove that the products and inverses of orthogonal matrices are orthogonal. (Thus, the orthogonal matrices form a group under multiplication, called the orthogonal group.) 13.34. Prove that the products and inverses of unitary matrices are unitary. (Thus, the unitary matrices form a group under multiplication, called the unitary group.) 13.35. Show that if an orthogonal (unitary) matrix is triangular, then it is diagonal. 13.36. Recall that the complex matrices A and B are unitarily equivalent if there exists a unitary matrix P such that B ¼ PAP. Show that this relation is an equivalence relation. 13.37. Recall that the real matrices A and B are orthogonally equivalent if there exists an orthogonal matrix P such that B ¼ PTAP. Show that this relation is an equivalence relation. 13.38. Let W be a subspace of V. For any v 2 V, let v ¼ w þ w0, where w 2 W, w0 2 W ?. (Such a sum is unique because V ¼ W W ?.) Let T :V ! V be deﬁned by TðvÞ ¼ w w0. Show that T is self-adjoint unitary operator on V. 13.39. Let V be an inner product space, and suppose U :V ! V (not assumed linear) is surjective (onto) and preserves inner products; that is, hUðvÞ; UðwÞi ¼ hu; wi for every v; w 2 V. Prove that U is linear and hence unitary. Positive and Positive Deﬁnite Operators 13.40. Show that the sum of two positive (positive deﬁnite) operators is positive (positive deﬁnite). 13.41. Let T be a linear operator on V and let f :V V ! K be deﬁned by f ðu; vÞ ¼ hTðuÞ; vi. Show that f is an inner product on V if and only if T is positive deﬁnite. 13.42. Suppose E is an orthogonal projection onto some subspace W of V. Prove that kI þ E is positive (positive deﬁnite) if k 0 ðk > 0Þ. 13.43. Consider the operator T deﬁned by TðuiÞ ¼ ﬃﬃﬃﬃ li p ui; i ¼ 1; . . . ; n, in the proof of Theorem 13.10A. Show that T is positive and that it is the only positive operator for which T2 ¼ P. 13.44. Suppose P is both positive and unitary. Prove that P ¼ I. 13.45. Determine which of the following matrices are positive (positive deﬁnite): ðiÞ 1 1 1 1 ; ðiiÞ 0 i i 0 ; ðiiiÞ 0 1 1 0 ; ðivÞ 1 1 0 1 ; ðvÞ 2 1 1 2 ; ðviÞ 1 2 2 1 13.46. Prove that a 2 2 complex matrix A ¼ a b c d is positive if and only if (i) A ¼ A, and (ii) a; d and jAj ¼ ad bc are nonnegative real numbers. CHAPTER 13 Linear Operators on Inner Product Spaces 393 13.47. Prove that a diagonal matrix A is positive (positive deﬁnite) if and only if every diagonal entry is a nonnegative (positive) real number. Self-adjoint and Symmetric Matrices 13.48. For any operator T, show that T þ T is self-adjoint and T T is skew-adjoint. 13.49. Suppose T is self-adjoint. Show that T2ðvÞ ¼ 0 implies TðvÞ ¼ 0. Using this to prove that TnðvÞ ¼ 0 also implies that TðvÞ ¼ 0 for n > 0. 13.50. Let V be a complex inner product space. Suppose hTðvÞ; vi is real for every v 2 V. Show that T is self-adjoint. 13.51. Suppose T1 and T2 are self-adjoint. Show that T1T2 is self-adjoint if and only if T1 and T1 commute; that is, T1T2 ¼ T2T1. 13.52. For each of the following symmetric matrices A, ﬁnd an orthogonal matrix P and a diagonal matrix D such that PTAP is diagonal: (a) A ¼ 1 2 2 2 ; (b) A ¼ 5 4 4 1 , (c) A ¼ 7 3 3 1 13.53. Find an orthogonal change of coordinates X ¼ PX 0 that diagonalizes each of the following quadratic forms and ﬁnd the corresponding diagonal quadratic form qðx0Þ: (a) qðx; yÞ ¼ 2x2 6xy þ 10y2, (b) qðx; yÞ ¼ x2 þ 8xy 5y2 (c) qðx; y; zÞ ¼ 2x2 4xy þ 5y2 þ 2xz 4yz þ 2z2 Normal Operators and Matrices 13.54. Let A ¼ 2 i i 2 . Verify that A is normal. Find a unitary matrix P such that PAP is diagonal. Find PAP. 13.55. Show that a triangular matrix is normal if and only if it is diagonal. 13.56. Prove that if T is normal on V, then kTðvÞk ¼ kTðvÞk for every v 2 V. Prove that the converse holds in complex inner product spaces. 13.57. Show that self-adjoint, skew-adjoint, and unitary (orthogonal) operators are normal. 13.58. Suppose T is normal. Prove that (a) T is self-adjoint if and only if its eigenvalues are real. (b) T is unitary if and only if its eigenvalues have absolute value 1. (c) T is positive if and only if its eigenvalues are nonnegative real numbers. 13.59. Show that if T is normal, then T and T have the same kernel and the same image. 13.60. Suppose T1 and T2 are normal and commute. Show that T1 þ T2 and T1T2 are also normal. 13.61. Suppose T1 is normal and commutes with T2. Show that T1 also commutes with T 2 . 13.62. Prove the following: Let T1 and T2 be normal operators on a complex ﬁnite-dimensional vector space V. Then there exists an orthonormal basis of V consisting of eigenvectors of both T1 and T2. (That is, T1 and T2 can be simultaneously diagonalized.) Isomorphism Problems for Inner Product Spaces 13.63. Let S ¼ fu1; . . . ; ung be an orthonormal basis of an inner product space V over K. Show that the mapping v 7! ½vs is an (inner product space) isomorphism between V and Kn. (Here ½vS denotes the coordinate vector of v in the basis S.) 394 CHAPTER 13 Linear Operators on Inner Product Spaces 13.64. Show that inner product spaces V and W over K are isomorphic if and only if V and W have the same dimension. 13.65. Suppose fu1; . . . ; ung and fu0 1; . . . ; u0 ng are orthonormal bases of V and W, respectively. Let T :V ! W be the linear map deﬁned by TðuiÞ ¼ u0 i for each i. Show that T is an isomorphism. 13.66. Let V be an inner product space. Recall that each u 2 V determines a linear functional ^ u in the dual space V by the deﬁnition ^ uðvÞ ¼ hv; ui for every v 2 V. (See the text immediately preceding Theorem 13.3.) Show that the map u 7! ^ u is linear and nonsingular, and hence an isomorphism from V onto V. Miscellaneous Problems 13.67. Suppose fu1; . . . ; ung is an orthonormal basis of V: Prove (a) ha1u1 þ a2u2 þ þ anun; b1u1 þ b2u2 þ þ bnuni ¼ a1 b1 þ a2 b2 þ . . . an bn (b) Let A ¼ ½aij be the matrix representing T: V ! V in the basis fuig: Then aij ¼ hTðuiÞ; uji: 13.68. Show that there exists an orthonormal basis fu1; . . . ; ung of V consisting of eigenvectors of T if and only if there exist orthogonal projections E1; . . . ; Er and scalars l1; . . . ; lr such that (i) T ¼ l1E1 þ þ lrEr, (ii) E1 þ þ Er ¼ I, (iii) EiEj ¼ 0 for i 6¼ j 13.69. Suppose V ¼ U W and suppose T1:U ! V and T2:W ! V are linear. Show that T ¼ T1 T2 is also linear. Here T is deﬁned as follows: If v 2 V and v ¼ u þ w where u 2 U, w 2 W, then TðvÞ ¼ T1ðuÞ þ T2ðwÞ ANSWERS TO SUPPLEMENTARY PROBLEMS Notation: ½R1; R2; . . . ; Rn denotes a matrix with rows R1; R2; . . . ; Rn. 13.25. (a) ½5 þ 2i; 4 þ 6i; 3 7i; 8 3i, (b) ½3; i; 5i; 2i, (c) ½1; 2; 1; 3 13.26. Tðx; y; zÞ ¼ ðx þ 3y; 2x þ z; 4yÞ 13.27. Tðx; y; zÞ ¼ ½ix þ 3y; ð2 3iÞx þ ð2 þ 5iÞz; ð3 þ iÞy iz 13.28. (a) u ¼ ð1; 2; 3Þ, (b) u ¼ ði; 2 3i; 1 þ 2iÞ 13.32. (a) ð1= ﬃﬃﬃﬃﬃ 13 p Þ½2; 3; 3; 2, (b) ð1= ﬃﬃﬃ 3 p Þ½1; 1 i; 1 þ i; 1, (c) 1 2 ½1; i; 1 i; ﬃﬃﬃ 2 p i; ﬃﬃﬃ 2 p ; 0; 1; i; 1 þ i 13.45. Only (i) and (v) are positive. Only (v) is positive deﬁnite. 13.52. (a and b) P ¼ ð1= ﬃﬃﬃ 5 p Þ½2; 1; 1; 2, (c) P ¼ ð1= ﬃﬃﬃﬃﬃ 10 p Þ½3; 1; 1; 3 (a) D ¼ ½2; 0; 0; 3; (b) D ¼ ½7; 0; 0; 3; (c) D ¼ ½8; 0; 0; 2 13.53. (a) x ¼ ð3x0 y0Þ= ﬃﬃﬃﬃﬃ 10 p ; y ¼ ðx0 þ 3y0Þ= ﬃﬃﬃﬃﬃ 10 p ; (b) x ¼ ð2x0 y0Þ= ﬃﬃﬃ 5 p ; y ¼ ðx0 þ 2y0Þ= ﬃﬃﬃ 5 p ; (c) x ¼ x0= ﬃﬃﬃ 3 p þ y0= ﬃﬃﬃ 2 p þ z0= ﬃﬃﬃ 6 p ; y ¼ x0= ﬃﬃﬃ 3 p 2z0= ﬃﬃﬃ 6 p ; z ¼ x0= ﬃﬃﬃ 3 p y0= ﬃﬃﬃ 2 p þ z0= ﬃﬃﬃ 6 p ; (a) qðx0Þ ¼ diagð1; 11Þ; (b) qðx0Þ ¼ diagð3; 7Þ; (c) qðx0Þ ¼ diagð1; 17Þ 13.54. (a) P ¼ ð1= ﬃﬃﬃ 2 p Þ½1; 1; 1; 1; PAP ¼ diagð2 þ i; 2 iÞ CHAPTER 13 Linear Operators on Inner Product Spaces 395 Multilinear Products A.1 Introduction The material in this appendix is much more abstract than that which has previously appeared. Accordingly, many of the proofs will be omitted. Also, we motivate the material with the following observation. Let S be a basis of a vector space V. Theorem 5.2 may be restated as follows. THEOREM 5.2: Let g:S ! V be the inclusion map of the basis S into V. Then, for any vector space U and any mapping f :S ! U; there exists a unique linear mapping f :V ! U such that f ¼ f g: Another way to state the fact that f ¼ f g is that the diagram in Fig. A-1(a) commutes. A.2 Bilinear Mapping and Tensor Products Let U, V, W be vector spaces over a ﬁeld K. Consider a map f : V W ! U Then f is said to be bilinear if, for each v 2 V; the map fv :W ! U deﬁned by fv w ð Þ ¼ f v; w ð Þ is linear; and, for each w 2 W; the map fw :V ! U deﬁned by fw v ð Þ ¼ f v; w ð Þ is linear. That is, f is linear in each of its two variables. Note that f is similar to a bilinear form except that the values of the map f are in a vector space U rather than the ﬁeld K. DEFINITION A.1: Let V and W be vector spaces over the same ﬁeld K. The tensor product of V and W is a vector space T over K together with a bilinear map g : V W ! T; denoted by g v; w ð Þ ¼ v w; with the following property: () For any vector space U over K and any bilinear map f :V W ! U there exists a unique linear map f :T ! U such that f g ¼ f : The tensor product (T, g) [or simply T when g is understood] of V and W is denoted by V W; and the element v w is called the tensor of v and w. Another way to state condition () is that the diagram in Fig. A-1(b) commutes. The fact that such a unique linear map f exists is called the ‘‘Universal Mapping Principle’’ (UMP). As illustrated in Fig. A-1(b), condition () also says that any bilinear map f :V W ! U ‘‘factors through’’ the tensor product T ¼ V W: The uniqueness in () implies that the image of g spans T; that is, span v w f g ð Þ ¼ T: APPENDIX A Figure A-1 396 THEOREM A.1: (Uniqueness of Tensor Products) Let (T, g) and T0; g0 ð Þ be tensor products of V and W. Then there exists a unique isomorphism h:T ! T0 such that hg ¼ g0: Proof. Because T is a tensor product, and g0 :V W ! T0 is bilinear, there exists a unique linear map h:T ! T0 such that hg ¼ g0: Similarly, because T0 is a tensor product, and g:V W ! T0 is bilinear, there exists a unique linear map h0 :T0 ! T such that h0g0 ¼ g: Using hg ¼ g0, we get h0hg ¼ g: Also, because T is a tensor product, and g:V W ! T is bilinear, there exists a unique linear map h :T ! T such that hg ¼ g: But 1Tg ¼ g: Thus, h0h ¼ h ¼ 1T. Similarly, hh0 ¼ 1T0: Therefore, h is an isomorphism from T to T0: THEOREM A.2: (Existence of Tensor Product) The tensor product T ¼ V W of vector spaces V and W over K exists. Let v1; . . . ; vm f g be a basis of V and let w1; . . . ; wn f g be a basis of W. Then the mn vectors vi wi i ¼ 1; . . . ; m; j ¼ 1; . . . ; n ð Þ form a basis of T. Thus, dim T ¼ mn ¼ dim V ð Þ dim W ð Þ: Outline of Proof. Suppose v1; . . . ; vm is a basis of V, and suppose w1; . . . ; wn f g is a basis of W. Consider the mn symbols tijji ¼ i; . . . ; m; j ¼ 1; . . . ; n . Let T be the vector space generated by the tij. That is, T consists of all linear combinations of the tij with coefﬁcients in K. [See Problem 4.137.] Let v 2 V and w 2 W. Say v ¼ a1v1 þ a2v2 þ þ amvm and w ¼ b1w1 þ b2w2 þ þ bmwm Let g:V W ! T be deﬁned by g v; w ð Þ ¼ X i X j aibjtij Then g is bilinear. [Proof left to reader.] Now let f :V W ! U be bilinear. Because the tij form a basis of T, Theorem 5.2 (stated above) tells us that there exists a unique linear map f :T ! U such that f tij ¼ f vi; wj . Then, for v ¼ P i aivi and w ¼ P j bjwj, we have f ðv; wÞ ¼ f X i aivi; X j bjwj ! ¼ X i X j aibj f vi; wj ¼ X i X j aibj tij ¼ f g v; w ð Þ ð Þ: Therefore, f ¼ f g where f is the required map in Deﬁnition A.1. Thus, T is a tensor product. Let fv0 1; . . . ; v0 mg be any basis of V and fw0 1; . . . ; w0 mg be any basis of W. Let v 2 V and w 2 W and say v ¼ a0 1v0 1 þ þ a0 mv0 m and w ¼ b0 1w0 1 þ þ b0 mw0 m Then v w ¼ g v; w ð Þ ¼ X i X j a0 ib0 i g v0 i; w0 i ð Þ ¼ X i X j a0 ib0 j v0 i w0 j Thus, the elements v0 i w0 j span T. There are mn such elements. They cannot be linearly dependent because tij is a basis of T, and hence, dim T ¼ mn. Thus, the v0 i w0 j form a basis of T. Next we give two concrete examples of tensor products. EXAMPLE A.1 Let V be the vector space of polynomials Pr1 x ð Þ and let W be the vector space of polynomials Ps1 y ð Þ. Thus, the following from bases of V and W, respectively, 1; x; x2; . . . ; xr1 and 1; y; y2; . . . ; y s1 In particular, dim V ¼ r and dim W ¼ s: Let T be the vector space of polynomials in variables x and y with basis xiy j where i ¼ 0; 1; . . . ; r 1; j ¼ 0; 1; . . . ; s 1 Appendix A Multilinear Products 397 Then T is the tensor product V W under the mapping xi y j ¼ xiyi For example, suppose v ¼ 2 5x þ 3x3 and w ¼ 7y þ 4y2. Then v w ¼ 14y þ 8y2 35xy 20xy2 þ 21x3y þ 12x3y2 Note, dim T ¼ rs ¼ dim V ð Þ dim W ð Þ: EXAMPLE A.2 Let V be the vector space of m n matrices over a ﬁeld K and let W be the vector space of p q matrices over K. Suppose A ¼ ½a11 belongs to V, and B belongs to W. Let T be the vector space of mp nq matrices over K. Then T is the tensor product of V and W where A B is the block matrix A B ¼ aijB ¼ a11B a12B a1nB a21B a22B a2nB am1B am2B amnB 2 6 6 4 3 7 7 5 For example, suppose A ¼ 1 2 3 4 and B ¼ 1 2 3 4 5 6 : Then A B ¼ 1 2 3 2 4 6 4 5 6 8 10 12 3 6 9 4 8 12 12 15 18 16 20 24 2 6 6 6 4 3 7 7 7 5 Isomorphisms of Tensor Products First we note that tensoring is associative in a cannonical way. Namely, THEOREM A.3: Let U, V, W be vector spaces over a ﬁeld K. Then there exists a unique isomorphism U V ð Þ W ! U V W ð Þ such that, for every u 2 U; v 2 V; w 2 W; u v ð Þ w 7! u v w ð Þ Accordingly, we may omit parenthesis when tensoring any number of factors. Speciﬁcally, given vectors spaces V1; V2; . . . ; Vm over a ﬁeld K, we may unambiguously form their tensor product V1 V2 . . . Vm and, for vectors vj in Vj, we may unambiguously form the tensor product v1 v2 . . . vm Moreover, given a vector space V over K, we may unambiguously deﬁne the following tensor product: rV ¼ V V . . . V r factors ð Þ Also, there is a canonical isomorphism rV ð Þ sV ð Þ ! rþsV Furthermore, viewing K as a vector space over itself, we have the canonical isomorphism K V ! V where we deﬁne a v ¼ av: 398 Appendix A Multilinear Products A.3 Alternating Multilinear Maps Let f :V r ! U where V and U are vector spaces over K. [Recall V r ¼ V V . . . V, r factors.] (1) The mapping f is said to be multilinear or r-linear if f v1; . . . ; vr ð Þ is linear as a function of each vj when the other vi’s are held ﬁxed. That is, f ð. . . ; vj þ v0 j; . . .Þ ¼ f ð. . . ; vj; . . .Þ þ f ð. . . ; v0 j; . . .Þ f ð. . . ; kvj; . . .Þ ¼ kf ð. . . ; vj; . . .Þ where only the jth position changes. (2) The mapping f is said to be alternating if f v1; . . . ; vr ð Þ ¼ 0 whenever vi ¼ vj with i 6¼ j One can easily show (Prove!) that if f is an alternating multilinear mapping on Vr, then f . . . ; vi; . . . ; vj; . . . ¼ f . . . ; vj; . . . ; vi; . . . That is, if two of the vectors are interchanged, then the associated value changes sign. EXAMPLE A.3 (Determinants) The determinant function D:M ! K on the space M of n n matrices may be viewed as an n-variable function D A ð Þ ¼ D R1; R2; . . . ; Rn ð Þ deﬁned on the rows R1; R2; . . . ; Rn of A. Recall (Chapter 8) that, in this context, D is both n-linear and alternating. We now need some additional notation. Let K ¼ k1; k2; . . . ; kr ½ denote an r-list (r-tuple) of elements from In ¼ 1; 2; . . . ; n ð Þ. We will then use the following notation where the vk’s denote vectors and the aik’s denote scalars: vK ¼ ðvk1; vk2; . . . ; vkrÞ and aK ¼ a1k1a2k2 . . . arkr Note vK is a list of r vectors, and aK is a product of r scalars. Now suppose the elements in K ¼ k1; k2; . . . ; kr ½ are distinct. Then K is a permutation sK of an r-list J ¼ i1; i2; . . . ; ir ½ in standard form, that is, where i1 < i2 < . . . < ir. The number of such standard-form r-lists J from In is the binomial coefﬁcient: n r ¼ n! r! n r ð Þ! [Recall sign sK ð Þ ¼ 1 ð ÞmK where mK is the number of interchanges that transforms K into J.] Now suppose A ¼ aij is an r n matrix. For a given ordered r-list J, we deﬁne DJ A ð Þ ¼ a1i1 a1i2 . . . a1ir a2i1 a2i2 . . . a2ir ari1 ari2 . . . arir That is, DJ(A) is the determinant of the r r submatrix of A whose column subscripts belong to J. Our main theorem below uses the following ‘‘shufﬂing’’ lemma. LEMMA A.4 Let V and U be vector spaces over K, and let f :V r ! U be an alternating r-linear mapping. Let v1; v2; . . . ; vn be vectors in V and let A ¼ aij be an r n matrix over K where r n. For i ¼ 1; 2; . . . ; r, let ui ¼ ai1vi þ ai2v2 þ þ ainvn ...................... Appendix A Multilinear Products 399 Then f u1; . . . ; ur ð Þ ¼ X f DJ A ð Þf ðvi1; vi2; . . . ; virÞ where the sum is over all standard-form r-lists J ¼ i1; i2; . . . ; ir f g. The proof is technical but straightforward. The linearity of f gives us the sum f u1; . . . ; ur ð Þ ¼ X K aKf vK ð Þ where the sum is over all r-lists K from 1; . . . ; n f g. The alternating property of f tells us that f vK ð Þ ¼ 0 when K does not contain distinct integers. The proof now mainly uses the fact that as we interchange the vj’s to transform f vK ð Þ ¼ f ðvk1; vk2; . . . ; vkrÞ to f vj ¼ f ðvi1; vi2; . . . ; virÞ so that i1 < < ir, the associated sign of aK, will change in the same way as the sign of the corresponding permutation sK changes when it is transformed to the identity permutation using transpositions. We illustrate the lemma below for r ¼ 2 and n ¼ 3. EXAMPLE A.4 Suppose f :V 2 ! U is an alternating multilinear function. Let v1; v2; v3 2 V and let u; w 2 V. Suppose u ¼ a1v1 þ a2v2 þ a3v3 and w ¼ b1v1 þ b2v2 þ b3v3 Consider f u; w ð Þ ¼ f a1v1 þ a2v2 þ a3v3; b1v1 þ b2v2 þ b3v3 ð Þ Using multilinearity, we get nine terms: f u; w ð Þ ¼ a1b1 f v1; vr ð Þ þ a1b2 f v1; v2 ð Þ þ a1b3 f v1; v3 ð Þ þ a2b1 f v2; v1 ð Þ þ a2b2 f v2; v2 ð Þ þ a2b3 f v2; v3 ð Þ þ a3b1 f v3; v1 ð Þ þ a3b2 f v3; v2 ð Þ þ a3b3 f v3; v3 ð Þ (Note that J ¼ 1; 2 ½ ; J0 ¼ 1; 3 ½ and J 00 ¼ 2; 3 ½ are the three standard-form 2-lists of I ¼ 1; 2; 3 ½ .) The alternating property of f tells us that each f vi; vi ð Þ ¼ 0; hence, three of the above nine terms are equal to 0. The alternating property also tells us that f vi; vf ¼ f vf ; vr . Thus, three of the terms can be transformed so their subscripts form a standard-form 2-list by a single interchange. Finally we obtain f u; w ð Þ ¼ a1b2 a2b1 ð Þ f v1; v2 ð Þ þ a1b3 a3b1 ð Þ f v1; v3 ð Þ þ a2b3 a3b2 ð Þ f v2; v3 ð Þ ¼ a1 a2 b1 b2 f v1; v2 ð Þ þ a1 a3 b1 b3 f v1; v3 ð Þ þ a2 a3 b2 b3 f v2; v3 ð Þ which is the content of Lemma A.4. A.4 Exterior Products The following deﬁnition applies. DEFINITION A.2: Let V be an n-dimensionmal vector space over a ﬁeld K, and let r be an integer such that 1 r n. The r-fold exterior product (or simply exterior product when r is understood) is a vector space E over K together with an alternating r-linear mapping g:V r ! E, denoted by g v1; . . . ; vr ð Þ ¼ v1 ^ .. . ^ vr, with the following property: () For any vector space U over K and any alternating r-linear map f :V r ! U there exists a unique linear map f :E ! U such that f g ¼ f . 400 Appendix A Multilinear Products The r-fold tensor product (E, g) (or simply E when g is understood) of V is denoted by ^r V, and the element v1 ^ ^ vr is called the exterior product or wedge product of the vi’s. Another way to state condition () is that the diagram in Fig. A-1(c) commutes. Again, the fact that such a unique linear map f exists is called the ‘‘Universal Mapping Principle (UMP)’’. As illustrated in Fig. A-1(c), condition () also says that any alternating r-linear map f :V r ! U ‘‘factors through’’ the exterior product E ¼ ^r V. Again, the uniqueness in () implies that the image of g spans E; that is, span v1 ^ ^ vr ð Þ ¼ E. THEOREM A.5: (Uniqueness of Exterior Products) Let (E, g) and E0; g0 ð Þ be r-fold exterior products of V. Then there exists a unique isomorphism h:E ! E0 such that hg ¼ g0. The proof is the same as the proof of Theorem A.1, which uses the UMP. THEOREM A.6: (Existence of Exterior Products) Let V be an n-dimensional vector space over K. Then the exterior product E ¼ ^r V exists. If r > n, then E ¼ 0 f g. If r n, then dim E ¼ n r . Moreover, if v1; . . . ; vn ½ is a basis of V, then the vectors vi1 ^ vi2 ^ ^ vir; where 1 i1 < i2 < < ir n, form a basis of E. We give a concrete example of an exterior product. EXAMPLE A.5 (Cross Product) Consider V ¼ R3 with the usual basis (i, j, k). Let E ¼ ^ 2V. Note dim V ¼ 3: Thus, dim E ¼ 3 with basis i ^ j; i ^ k; j ^ k: We identify E with R3 under the correspondence i ¼ j ^ k; j ¼ k ^ i ¼ i ^ k; k ¼ i ^ j Let u and w be arbitrary vectors in V ¼ R3, say u ¼ a1; a2; a3 ð Þ ¼ a1i þ a2j þ a3k and w ¼ b1; b2; b3 ð Þ ¼ b1i þ b2j þ b3k Then, as in Example A.3, u ^ w ¼ a1b2 a2b1 ð Þði ^ jÞ þ a1b3 a3b1 ð Þði ^ kÞ þ a2b3 a3b2 ð Þðj ^ kÞ Using the above identiﬁcation, we get u ^ w ¼ a2b3 a3b2 ð Þi a1b3 a3b1 ð Þj þ a1b2 a2b1 ð Þk ¼ a2 a3 b2 b3 i a1 a3 b1 b3 j þ a1 a2 b1 b2 k The reader may recognize that the above exterior product is precisely the well-known cross product in R3. Our last theorem tells us that we are actually able to ‘‘multiply’’ exterior products, which allows us to form an ‘‘exterior algebra’’ that is illustrated below. THEOREM A.7: Let V be a vector space over K. Let r and s be positive integers. Then there is a unique bilinear mapping ^ rV ^ sV ! ^ rþsV such that, for any vectors ui; wj in V, u1 ^ ^ ur ð Þ w1 ^ ^ ws ð Þ 7! u1 ^ ^ ur ^ w1 ^ ^ ws Appendix A Multilinear Products 401 EXAMPLE A.6 We form an exterior algebra A over a ﬁeld K using noncommuting variables x, y, z. Because it is an exterior algebra, our variables satisfy: x ^ x ¼ 0; y ^ y ¼ 0; z ^ z ¼ 0; and y ^ x ¼ x ^ y; z ^ x ¼ x ^ z; z ^ y ¼ y ^ z Every element of A is a linear combination of the eight elements 1; x; y; z; x ^ y; x ^ z; y ^ z; x ^ y ^ z We multiply two ‘‘polynomials’’ in A using the usual distributive law, but now we also use the above conditions. For example, 3 þ 4y 5x ^ y þ 6x ^ z ½ ^ 5x 2y ½ ¼ 15x 6y 20x ^ y þ 12x ^ y ^ z Observe we use the fact that 4y ½ ^ 5x ½ ¼ 20y ^ x ¼ 20x ^ y and 6x ^ z ½ ^ 2y ½ ¼ 12x ^ z ^ y ¼ 12x ^ y ^ z 402 Appendix A Multilinear Products Algebraic Structures B.1 Introduction We deﬁne here algebraic structures that occur in almost all branches of mathematics. In particular, we will deﬁne a ﬁeld that appears in the deﬁnition of a vector space. We begin with the deﬁnition of a group, which is a relatively simple algebraic structure with only one operation and is used as a building block for many other algebraic systems. B.2 Groups Let G be a nonempty set with a binary operation; that is, to each pair of elements a; b 2 G there is assigned an element ab 2 G. Then G is called a group if the following axioms hold: G1 ½ For any a; b; c 2 G, we have ab ð Þc ¼ a bc ð Þ (the associative law). G2 ½ There exists an element e 2 G, called the identity element, such that ae ¼ ea ¼ a for every a 2 G. G3 ½ For each a 2 G there exists an element a1 2 G, called the inverse of a, such that aa1 ¼ a1a ¼ e. A group G is said to be abelian (or: commutative) if the commutative law holds—that is, if ab ¼ ba for every a; b 2 G. When the binary operation is denoted by juxtaposition as above, the group G is said to be written multiplicatively. Sometimes, when G is abelian, the binary operation is denoted by + and G is said to be written additively. In such a case, the identity element is denoted by 0 and is called the zero element; the inverse is denoted by a and it is called the negative of a. If A and B are subsets of a group G, then we write AB ¼ abja 2 A; b 2 B f g or A þ B ¼ a þ bja 2 A; b 2 B f g We also write a for {a}. A subset H of a group G is called a subgroup of G if H forms a group under the operation of G. If H is a subgroup of G and a 2 G, then the set Ha is called a right coset of H and the set aH is called a left coset of H. DEFINITION: A subgroup H of G is called a normal subgroup if a1Ha H for every a 2 G. Equivalently, H is normal if aH ¼ Ha for every a 2 G—that is, if the right and left cosets of H coincide. Note that every subgroup of an abelian group is normal. THEOREM B.1: Let H be a normal subgroup of G. Then the cosets of H in G form a group under coset multiplication. This group is called the quotient group and is denoted by G/H. APPENDIX B 403 EXAMPLE B.1 The set Z of integers forms an abelian group under addition. (We remark that the even integers form a subgroup of Z but the odd integers do not.) Let H denote the set of multiples of 5; that is, H ¼ f. . . ; 10; 5; 0; 5; 10; . . .g. Then H is a subgroup (necessarily normal) of Z. The cosets of H in Z follow: 0 ¼ 0 þ H ¼ H ¼ . . . ; 10; 5; 0; 5; 10; . . . f g 1 ¼ 1 þ H ¼ f. . . ; 9; 4; 1; 6; 11; . . .g 2 ¼ 2 þ H ¼ . . . ; 8; 3; 2; 7; 12; . . . f g 3 ¼ 3 þ H ¼ . . . ; 7; 2; 3; 8; 13; . . . f g 4 ¼ 4 þ H ¼ . . . ; 6; 1; 4; 9; 14; . . . f g For any other integer n 2 Z, n ¼ n þ H coincides with one of the above cosets. Thus, by the above theorem, Z=H ¼ 0; 1; 2; 3; 4 f g forms a group under coset addition; its addition table follows: þ 0 1 2 3 4 0 0 1 2 3 4 1 1 2 3 4 0 2 2 3 4 0 1 3 3 4 0 1 2 4 4 0 1 2 3 This quotient group Z/H is referred to as the integers modulo 5 and is frequently denoted by Z5. Analogeusly, for any positive integer n, there exists the quotient group Zn called the integers modulo n. EXAMPLE B.2 The permutations of n symbols (see page 267) form a group under composition of mappings; it is called the symmetric group of degree n and is denoted by Sn. We investigate S3 here; its elements are E ¼ 1 2 3 1 2 3 s2 ¼ 1 2 3 3 2 1 f1 ¼ 1 2 3 2 3 1 s1 ¼ 1 2 3 1 3 2 s3 ¼ 1 2 3 2 1 3 f2 ¼ 1 2 3 3 1 2 Here 1 2 3 i j k is the permutation that maps 1 7! i; 2 7! j; 3 7! k. The multiplication table of S3 is E s1 s2 s3 f1 f2 E E s1 s2 s3 f1 f2 s1 s1 E f1 f2 s2 s3 s2 s2 f2 E f1 f3 s1 s3 s3 f1 f2 E s1 s2 f1 f1 s3 s1 s2 f2 E f2 f2 s2 s3 s1 E f1 (The element in the ath row and bth column is ab.) The set H ¼ E; s1 f g is a subgroup of S3; its right and left cosets are Right Cosets Left Cosets H ¼ E; s1 f g H ¼ E; s1 f g Hf1 ¼ f1; s2 f g f2H ¼ f1; s3 f g Hf2 ¼ f2; s3 f g f2H ¼ f2; s2 f g Observe that the right cosets and the left cosets are distinct; hence, H is not a normal subgroup of S3. A mapping f from a group G into a group G0 is called a homomorphism if f ab ð Þ ¼ f a ð Þf b ð Þ. For every a; b 2 G. (If f is also bijective, i.e., one-to-one and onto, then f is called an isomorphism and G and G0 are 404 Appendix B Algebraic Structures said to be isomorphic.) If f : G ! G0 is a homomorphism, then the kernel of f is the set of elements of G that map into the identity element e0 2 G0: kernel of f ¼ a 2 G j f a ð Þ ¼ e0 f g (As usual, f(G) is called the image of the mapping f : G ! G0.) The following theorem applies. THEOREM B.2: Let f: G ! G be a homomorphism with kernel K. Then K is a normal subgroup of G, and the quotient group G/K is isomorphic to the image of f. EXAMPLE B.3 Let G be the group of real numbers under addition, and let G0 be the group of positive real numbers under multiplication. The mapping f : G ! G0 deﬁned by f a ð Þ ¼ 2a is a homomorphism because f a þ b ð Þ ¼ 2aþb ¼ 2a2b ¼ f a ð Þf b ð Þ In particular, f is bijective, hence, G and G0 are isomorphic. EXAMPLE B.4 Let G be the group of nonzero complex numbers under multiplication, and let G0 be the group of nonzero real numbers under multiplication. The mapping f : G ! G0 deﬁned by f z ð Þ ¼ jzj is a homomorphism because f z1z2 ð Þ ¼ jz1z2j ¼ jz1jjz2j ¼ f z1 ð Þ f z2 ð Þ The kernel K of f consists of those complex numbers z on the unit circle—that is, for which jzj ¼ 1. Thus, G=K is isomorphic to the image of f—that is, to the group of positive real numbers under multiplication. B.3 Rings, Integral Domains, and Fields Let R be a nonempty set with two binary operations, an operation of addition (denoted by +) and an operation of multiplication (denoted by juxtaposition). Then R is called a ring if the following axioms are satisﬁed: R1 ½ For any a; b; c 2 R, we have a þ b ð Þ þ c ¼ a þ b þ c ð Þ. R2 ½ There exists an element 0 2 R; called the zero element, such that a þ 0 ¼ 0 þ a ¼ a for every a 2 R: R3 ½ For each a 2 R there exists an element a 2 R, called the negative of a, such that a þ a ð Þ ¼ a ð Þ þ a ¼ 0. R4 ½ For any a; b 2 R; we have a þ b ¼ b þ a: R5 ½ For any a; b; c 2 R; we have ab ð Þc ¼ a bc ð Þ: R6 ½ For any a; b; c 2 R; we have (i) a b þ c ð Þ ¼ ab þ ac; and (ii) b þ c ð Þa ¼ ba þ ca: Observe that the axioms R1 ½ through R4 ½ may be summarized by saying that R is an abelian group under addition. Subtraction is deﬁned in R by a b a þ b ð Þ. It can be shown (see Problem B.25) that a 0 ¼ 0 a ¼ 0 for every a 2 R: R is called a commutative ring if ab ¼ ba for every a; b 2 R: We also say that R is a ring with a unit element if there exists a nonzero element 1 2 R such that a 1 ¼ 1 a ¼ a for every a 2 R: A nonempty subset S of R is called a subring of R if S forms a ring under the operations of R. We note that S is a subring of R if and only if a; b 2 S implies a b 2 S and ab 2 S. A nonempty subset I of R is called a left ideal in R if (i) a b 2 I whenever a; b 2 I; and (ii) ra 2 I whenever r 2 R; a 2 I: Note that a left ideal I in R is also a subring of R. Similarly, we can deﬁne a right ideal and a two-sided ideal. Clearly all ideals in commutative rings are two sided. The term ideal shall mean two-sided ideal uniess otherwise speciﬁed. Appendix B Algebraic Structures 405 THEOREM B.3: Let I be a (two-sided) ideal in a ring R. Then the cosets a þ I j a 2 R f g form a ring under coset addition and coset multiplication. This ring is denoted by R=I and is called the quotient ring. Now let R be a commutative ring with a unit element. For any a 2 R, the set a ð Þ ¼ ra j r 2 R f g is an ideal; it is called the principal ideal generated by a. If every ideal in R is a principal ideal, then R is called a principal ideal ring. DEFINITION: A commutative ring R with a unit element is called an integral domain if R has no zero divisors—that is, if ab ¼ 0 implies a ¼ 0 or b ¼ 0. DEFINITION: A commutative ring R with a unit element is called a ﬁeld if every nonzero a 2 R has a multiplicative inverse; that is, there exists an element a1 2 R such that aa1 ¼ a1a ¼ 1: A ﬁeld is necessarily an integral domain; for if ab ¼ 0 and a 6¼ 0; then b ¼ 1 b ¼ a1ab ¼ a1 0 ¼ 0 We remark that a ﬁeld may also be viewed as a commutative ring in which the nonzero elements form a group under multiplication. EXAMPLE B.5 The set Z of integers with the usual operations of addition and multiplication is the classical example of an integral domain with a unit element. Every ideal I in Z is a principal ideal; that is, I ¼ n ð Þ for some integer n. The quotient ring Zn ¼ Z= n ð Þ is called the ring of integers module n. If n is prime, then Zn is a ﬁeld. On the other hand, if n is not prime then Zn has zero divisors. For example, in the ring Z6; 2 3 ¼ 0 and 2 6¼ 0 and 3 6¼ 0: EXAMPLE B.6 The rational numbers Q and the real numbers R each form a ﬁeld with respect to the usual operations of addition and multiplication. EXAMPLE B.7 Let C denote the set of ordered pairs of real numbers with addition and multiplication deﬁned by a; b ð Þ þ c; d ð Þ ¼ a þ c; b þ d ð Þ a; b ð Þ c; d ð Þ ¼ ac bd; ad þ bc ð Þ Then C satisﬁes all the required properties of a ﬁeld. In fact, C is just the ﬁeld of complex numbers (see page 4). EXAMPLE B.8 The set M of all 2 6 2 matrices with real entries forms a noncommutative ring with zero divisors under the operations of matrix addition and matrix multiplication. EXAMPLE B.9 Let R be any ring. Then the set R x ½ of all polynomials over R forms a ring with respect to the usual operations of addition and multiplication of polynomials. Moreover, if R is an integral domain then R x ½ is also an integral domain. Now let D be an integral domain. We say that b divides a in D if a ¼ bc for some c 2 D. An element u 2 D is called a unit if u divides 1—that is, if u has a multiplicative inverse. An element b 2 D is called an associate of a 2 D if b ¼ ua for some unit u 2 D. A nonunit p 2 D is said to be irreducible if p ¼ ab implies a or b is a unit. An integral domain D is called a unique factorization domain if every nonunit a 2 D can be written uniquely (up to associates and order) as a product of irreducible elements. EXAMPLE B.10 The ring Z of integers is the classical example of a unique factorization domain. The units of Z are 1 and 1. The only associates of n 2 Z are n and n. The irreducible elements of Z are the prime numbers. EXAMPLE B.11 The set D ¼ a þ b ﬃﬃﬃﬃﬃ 13 p j a; b integers is an integral domain. The units of D are 1; 18 5 ﬃﬃﬃﬃﬃ 13 p and 18 5 ﬃﬃﬃﬃﬃ 13 p . The elements 2; 3 ﬃﬃﬃﬃﬃ 13 p and 3 ﬃﬃﬃﬃﬃ 13 p are irreducible in D. Observe that 4 ¼ 2 2 ¼ 3 ﬃﬃﬃﬃﬃ 13 p 3 ﬃﬃﬃﬃﬃ 13 p : Thus, D is not a unique factorization domain. (See Problem B.40.) 406 Appendix B Algebraic Structures B.4 Modules Let M be an additive abelian group and let R be a ring with a unit element. Then M is said to be a (left) R-module if there exists a mapping R M ! M that satisﬁes the following axioms: M1 ½ r m1 þ m2 ð Þ ¼ rm1 þ rm2 M2 ½ r þ s ð Þm ¼ rm þ sm M3 ½ rs ð Þm ¼ r sm ð Þ M4 ½ 1 m ¼ m for any r; s 2 R and any mi 2 M. We emphasize that an R-module is a generalization of a vector space where we allow the scalars to come from a ring rather than a ﬁeld. EXAMPLE B.12 Let G be any additive abelian group. We make G into a module over the ring Z of integers by deﬁning ng ¼ g þ g þ þ g; zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ}\|ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{ n times 0g ¼ 0; n ð Þg ¼ ng where n is any positive integer. EXAMPLE B.13 Let R be a ring and let I be an ideal in R. Then I may be viewed as a module over R. EXAMPLE B.14 Let V be a vector space over a ﬁeld K and let T :V ! V be a linear mapping. We make V into a module over the ring K x ½ of polynomials over K by deﬁning f x ð Þv ¼ f T ð Þ v ð Þ: The reader should check that a scalar multiplication has been deﬁned. Let M be a module over R. An additive subgroup N of M is called a submodule of M if u 2 N and k 2 R imply ku 2 N: (Note that N is then a module over R.) Let M and M0 be R-modules. A mapping T :M ! M0 is called a homomorphism (or: R-homomorphism or R-linear) if (i) T u þ v ð Þ ¼ T u ð Þ þ T v ð Þ and (ii) T ku ð Þ ¼ kT u ð Þ for every u; v 2 M and every k 2 R. PROBLEMS Groups B.1. Determine whether each of the following systems forms a group G: (i) G ¼ set of integers; operation subtraction; (ii) G ¼ f1; 1g, operation multiplication; (iii) G ¼ set of nonzero rational numbers, operation division; (iv) G ¼ set of nonsingular n n matrices, operation matrix multiplication; (v) G ¼ fa þ bi : a; b 2 Zg, operation addition. B.2. Show that in a group G: (i) the identity element of G is unique; (ii) each a 2 G has a unique inverse a1 2 G; (iii) a1 ð Þ 1¼ a; and ab ð Þ1¼ b1a1; (iv) ab ¼ ac implies b ¼ c, and ba ¼ ca implies b ¼ c. Appendix B Algebraic Structures 407 B.3. In a group G, the powers of a 2 G are deﬁned by a0 ¼ e; an ¼ aan1; an ¼ an ð Þ1; where n 2 N Show that the following formulas hold for any integers r; s; t 2 Z: (i) aras ¼ arþs; (ii) ar ð Þs¼ ars; (iii) arþs ð Þt¼ arsþst. B.4. Show that if G is an abelian group, then ab ð Þn¼ anbn for any a; b 2 G and any integer n 2 Z: B.5. Suppose G is a group such that ab ð Þ2¼ a2b2 for every a; b 2 G. Show that G is abelian. B.6. Suppose H is a subset of a group G. Show that H is a subgroup of G if and only if (i) H is nonempty, and (ii) a; b 2 H implies ab1 2 H: B.7. Prove that the intersection of any number of subgroups of G is also a subgroup of G. B.8. Show that the set of all powers of a 2 G is a subgroup of G; it is called the cyclic group generated by a. B.9. A group G is said to be cyclic if G is generated by some a 2 G; that is, G ¼ an : n 2 Z ð Þ. Show that every subgroup of a cyclic group is cyclic. B.10. Suppose G is a cyclic subgroup. Show that G is isomorphic to the set Z of integers under addition or to the set Zn (of the integers module n) under addition. B.11. Let H be a subgroup of G. Show that the right (left) cosets of H partition G into mutually disjoint subsets. B.12. The order of a group G, denoted by jGj; is the number of elements of G. Prove Lagrange’s theorem: If H is a subgroup of a ﬁnite group G, then jHj divides jGj. B.13. Suppose jGj ¼ p where p is prime. Show that G is cyclic. B.14. Suppose H and N are subgroups of G with N normal. Show that (i) HN is a subgroup of G and (ii) H \ N is a normal subgroup of G. B.15. Let H be a subgroup of G with only two right (left) cosets. Show that H is a normal subgroup of G. B.16. Prove Theorem B.1: Let H be a normal subgroup of G. Then the cosets of H in G form a group G=H under coset multiplication. B.17. Suppose G is an abelian group. Show that any factor group G=H is also abelian. B.18. Let f : G ! G0 be a group homomorphism. Show that (i) f e ð Þ ¼ e0 where e and e0 are the identity elements of G and G0, respectively; (ii) f a1 ð Þ ¼ f a ð Þ1 for any a 2 G. B.19. Prove Theorem B.2: Let f : G ! G0 be a group homomorphism with kernel K. Then K is a normal subgroup of G, and the quotient group G=K is isomorphic to the image of f. B.20. Let G be the multiplicative group of complex numbers z such that jzj ¼ 1; and let R be the additive group of real numbers. Prove that G is isomorphic to R=Z: 408 Appendix B Algebraic Structures B.21. For a ﬁxed g 2 G, let ^ g : G ! G be deﬁned by ^ g a ð Þ ¼ g1ag: Show that G is an isomorphism of G onto G. B.22. Let G be the multiplicative group of n n nonsingular matrices over R. Show that the mapping A 7! jAj is a homomorphism of G into the multiplicative group of nonzero real numbers. B.23. Let G be an abelian group. For a ﬁxed n 2 Z; show that the map a 7! an is a homomorphism of G into G. B.24. Suppose H and N are subgroups of G with N normal. Prove that H \ N is normal in H and H= H \ N ð Þ is isomorphic to HN=N. Rings B.25. Show that in a ring R: (i) a 0 ¼ 0 a ¼ 0; (ii) a b ð Þ ¼ a ð Þb ¼ ab, (iii) a ð Þ b ð Þ ¼ ab: B.26. Show that in a ring R with a unit element: (i) 1 ð Þa ¼ a; (ii) 1 ð Þ 1 ð Þ ¼ 1. B.27. Let R be a ring. Suppose a2 ¼ a for every a 2 R: Prove that R is a commutative ring. (Such a ring is called a Boolean ring.) B.28. Let R be a ring with a unit element. We make R into another ring ^ R by deﬁning a b ¼ a þ b þ 1 and a b ¼ ab þ a þ b. (i) Verify that ^ R is a ring. (ii) Determine the 0-element and 1-element of ^ R. B.29. Let G be any (additive) abelian group. Deﬁne a multiplication in G by a b ¼ 0. Show that this makes G into a ring. B.30. Prove Theorem B.3: Let I be a (two-sided) ideal in a ring R. Then the cosets a þ I j a 2 R ð Þ form a ring under coset addition and coset multiplication. B.31. Let I1 and I2 be ideals in R. Prove that I1 þ I2 and I1 \ I2 are also ideals in R. B.32. Let R and R0 be rings. A mapping f : R ! R0 is called a homomorphism (or: ring homomorphism) if (i) f a þ b ð Þ ¼ f a ð Þ þ f b ð Þ and (ii) f ab ð Þ ¼ f a ð Þ f b ð Þ, for every a; b 2 R. Prove that if f : R ! R0 is a homomorphism, then the set K ¼ r 2 R j f r ð Þ ¼ 0 f g is an ideal in R. (The set K is called the kernel of f.) Integral Domains and Fields B.33. Prove that in an integral domain D, if ab ¼ ac; a 6¼ 0; then b ¼ c. B.34. Prove that F ¼ a þ b ﬃﬃﬃ 2 p j a; b rational is a ﬁeld. B.35. Prove that D ¼ a þ b ﬃﬃﬃ 2 p j a; b integers is an integral domain but not a ﬁeld. B.36. Prove that a ﬁnite integral domain D is a ﬁeld. B.37. Show that the only ideals in a ﬁeld K are 0 f g and K. B.38. A complex number a þ bi where a, b are integers is called a Gaussian integer. Show that the set G of Gaussian integers is an integral domain. Also show that the units in G are 1 and i. Appendix B Algebraic Structures 409 B.39. Let D be an integral domain and let I be an ideal in D. Prove that the factor ring D=I is an integral domain if and only if I is a prime ideal. (An ideal I is prime if ab 2 I implies a 2 I or b 2 I:) B.40. Consider the integral domain D ¼ a þ b ﬃﬃﬃﬃﬃ 13 p j a; b integers (see Example B.11). If a ¼ a þ b ﬃﬃﬃﬃﬃ 13 p , we deﬁne N a ð Þ ¼ a2 13b2. Prove: (i) N ab ð Þ ¼ N a ð ÞN b ð Þ; (ii) a is a unit if and only if N a ð Þ ¼ 1; (iii) the units of D are 1; 18 5 ﬃﬃﬃﬃﬃ 13 p and 18 5 ﬃﬃﬃﬃﬃ 13 p ; (iv) the numbers 2; 3 ﬃﬃﬃﬃﬃ 13 p and 3 ﬃﬃﬃﬃﬃ 13 p are irreducible. Modules B.41. Let M be an R-module and let A and B be submodules of M. Show that A þ B and A \ B are also submodules of M. B.42. Let M be an R-module with submodule N. Show that the cosets u þ N : u 2 M f g form an R-module under coset addition and scalar multiplication deﬁned by r u þ N ð Þ ¼ ru þ N. (This module is denoted by M=N and is called the quotient module.) B.43. Let M and M0 be R-modules and let f : M ! M0 be an R-homomorphism. Show that the set K ¼ u 2 M : f u ð Þ ¼ 0 f g is a submodule of f. (The set K is called the kernel of f.) B.44. Let M be an R-module and let E M ð Þ denote the set of all R-homomorphism of M into itself. Deﬁne the appropriate operations of addition and multiplication in E M ð Þ so that E M ð Þ becomes a ring. 410 Appendix B Algebraic Structures Polynomials over a Field C.1 Introduction We will investigate polynomials over a ﬁeld K and show that they have many properties that are analogous to properties of the integers. These results play an important role in obtaining canonical forms for a linear operator T on a vector space V over K. C.2 Ring of Polynomials Let K be a ﬁeld. Formally, a polynomial of f over K is an inﬁnite sequence of elements from K in which all except a ﬁnite number of them are 0: f ¼ . . . ; 0; an; . . . ; a1; a0 ð Þ (We write the sequence so that it extends to the left instead of to the right.) The entry ak is called the kth coefﬁcient of f. If n is the largest integer for which an 6¼ 0, then we say that the degree of f is n, written deg f ¼ n We also call an the leading coefﬁcient of f, and if an ¼ 1 we call f a monic polynomial. On the other hand, if every coefﬁcient of f is 0 then f is called the zero polynomial, written f ¼ 0. The degree of the zero polynomial is not deﬁned. Now if g is another polynomial over K, say g ¼ . . . ; 0; bm; . . . ; b1; b0 ð Þ then the sum f þ g is the polynomial obtained by adding corresponding coefﬁcients. That is, if m n, then f þ g ¼ . . . ; 0; an; . . . ; am þ bm; . . . ; a1 þ b1; a0 þ b0 ð Þ Furthermore, the product fg is the polynomial fg ¼ . . . ; 0; anbm; . . . ; a1b0 þ a0b1; a0b0 ð Þ that is, the kth coefﬁcient ck of fg is ck ¼ X k t¼0 a1bk1 ¼ a0bk þ a1bk1 þ þ akb0 The following theorem applies. THEOREM C.1: The set P of polynomials over a ﬁeld K under the above operations of addition and multiplication forms a commutative ring with a unit element and with no zero divisors—an integral domain. If f and g are nonzero polynomials in P, then deg fg ð Þ ¼ deg f ð Þ deg g ð Þ. APPENDIX C 411 Notation We identify the scalar a0 2 K with the polynomial a0 ¼ . . . ; 0; a0 ð Þ We also choose a symbol, say t, to denote the polynomial t ¼ . . . ; 0; 1; 0 ð Þ We call the symbol t an indeterminant. Multiplying t with itself, we obtain t2 ¼ . . . ; 0; 1; 0; 0 ð Þ; t3 ¼ . . . ; 0; 1; 0; 0; 0 ð Þ; . . . Thus, the above polynomial f can be written uniquely in the usual form f ¼ antn þ þ ast þ a0 When the symbol t is selected as the indeterminant, the ring of polynomials over K is denoted by K t ½ and a polynomial f is frequently denoted by f t ð Þ. We also view the ﬁeld K as a subset of K t ½ under the above identiﬁcation. This is possible because the operations of addition and multiplication of elements of K are preserved under this identiﬁcation: ð. . . ; 0; a0Þ þ ð. . . ; 0; b0Þ ¼ ð. . . ; 0; a0 þ b0Þ ð. . . ; 0; a0Þ ð. . . ; 0; b0Þ ¼ ð. . . ; 0; a0b0Þ We remark that the nonzero elements of K are the units of the ring K t ½ . We also remark that every nonzero polynomial is an associate of a unique monic polynomial. Hence, if d and d0 are monic polynomials for which d divides d0 and d0 divides d, then d ¼ d0. (A polynomial g divides a polynomial f if there is a polynomial h such that f ¼ hg:) C.3 Divisibility The following theorem formalizes the process known as ‘‘long division.’’ THEOREM C.2 (Division Algorithm): Let f and g be polynomials over a ﬁeld K with g 6¼ 0. Then there exist polynomials q and r such that f ¼ qg þ r where either r ¼ 0 or deg r < deg g. Proof: If f ¼ 0 or if deg f < deg g, then we have the required representation f ¼ 0g þ f Now suppose deg f deg g, say f ¼ antn þ þ a1t þ a0 and g ¼ bmtm þ þ b1t þ b0 where an; bm 6¼ 0 and n m. We form the polynomial f1 ¼ f an bm tnmg ð1Þ Then deg f1 < deg f . By induction, there exist polynomials q1 and r such that f1 ¼ q1g þ r 412 Appendix C Polynomials over a Field where either r ¼ 0 or deg r < deg g. Substituting this into (1) and solving for f, f ¼ q1 þ an bm tnm g þ r which is the desired representation. THEOREM C.3: The ring K t ½ of polynomials over a ﬁeld K is a principal ideal ring. If I is an ideal in K t ½ , then there exists a unique monic polynomial d that generates I, such that d divides every polynomial f 2 I. Proof. Let d be a polynomial of lowest degree in I. Because we can multiply d by a nonzero scalar and still remain in I, we can assume without loss in generality that d is a monic polynomial. Now suppose f 2 I. By Theorem C.2 there exist polynomials q and r such that f ¼ qd þ r where either r ¼ 0 or deg r < deg d Now f ; d 2 I implies qd 2 I; and hence, r ¼ f qd 2 I. But d is a polynomial of lowest degree in I. Accordingly, r ¼ 0 and f ¼ qd; that is, d divides f. It remains to show that d is unique. If d0 is another monic polynomial that generates I, then d divides d0 and d0 divides d. This implies that d ¼ d0, because d and d0 are monic. Thus, the theorem is proved. THEOREM C.4: Let f and g be nonzero polynomials in K t ½ . Then there exists a unique monic polynomial d such that (i) d divides f and g; and (ii) d0 divides f and g, then d0 divides d. DEFINITION: The above polynomial d is called the greatest common divisor of f and g. If d ¼ 1, then f and g are said to be relatively prime. Proof of Theorem C.4. The set I ¼ mf þ ng j m; n 2 K t ½ f g is an ideal. Let d be the monic polynomial that generates I. Note f ; g 2 I; hence, d divides f and g. Now suppose d0 divides f and g. Let J be the ideal generated by d0. Then f ; g 2 J, and hence, I J. Accordingly, d 2 J and so d0 divides d as claimed. It remains to show that d is unique. If d1 is another (monic) greatest common divisor of f and g, then d divides d1 and d1 divides d. This implies that d ¼ d1 because d and d1 are monic. Thus, the theorem is proved. COROLLARY C.5: Let d be the greatest common divisor of the polynomials f and g. Then there exist polynomials m and n such that d ¼ mf þ ng. In particular, if f and g are relatively prime, then there exist polynomials m and n such that mf þ ng ¼ 1. The corollary follows directly from the fact that d generates the ideal I ¼ mf þ ng j m; n 2 K t ½ f g C.4 Factorization A polynomial p 2 K t ½ of positive degree is said to be irreducible if p ¼ fg implies f or g is a scalar. LEMMA C.6: Suppose p 2 K t ½ is irreducible. If p divides the product fg of polynomials f ; g 2 K t ½ , then p divides f or p divides g. More generally, if p divides the product of n polynomials f1 f2 . . . fn, then p divides one of them. Proof. Suppose p divides fg but not f. Because p is irreducible, the polynomials f and p must then be relatively prime. Thus, there exist polynomials m; n 2 K t ½ such that mf þ np ¼ 1. Multiplying this Appendix C Polynomials over a Field 413 equation by g, we obtain mfg þ npg ¼ g. But p divides fg and so mfg, and p divides npg; hence, p divides the sum g ¼ mfg þ npg. Now suppose p divides f1 f2 fn: If p divides f1, then we are through. If not, then by the above result p divides the product f2 fn: By induction on n, p divides one of the polynomials f2; . . . fn: Thus, the lemma is proved. THEOREM C.7: (Unique Factorization Theorem) Let f be a nonzero polynomial in K t ½ : Then f can be written uniquely (except for order) as a product f ¼ kp1p2 pn where k 2 K and the pi are monic irreducible polynomials in K t ½ : Proof: We prove the existence of such a product ﬁrst. If f is irreducible or if f 2 K, then such a product clearly exists. On the other hand, suppose f ¼ gh where f and g are nonscalars. Then g and h have degrees less than that of f. By induction, we can assume g ¼ k1g1g2 gr and h ¼ k2h1h2 hs where k1; k2 2 K and the gi and hj are monic irreducible polynomials. Accordingly, f ¼ k1k2 ð Þg1g2 grk1h2 hs is our desired representation. We next prove uniqueness (except for order) of such a product for f. Suppose f ¼ kp1p2 pn ¼ k0q1q2 qm where k; k0 2 K and the p1; . . . ; pn; q1; . . . ; qm are monic irreducible polynomials. Now p1 divides k0q1 qm: Because p1 is irreducible, it must divide one of the qi by the above lemma. Say p1 divides q1. Because p1 and q1 are both irreducible and monic, p1 ¼ q1. Accordingly, kp2 pn ¼ k0q2 qm By induction, we have that n ¼ m and p2 ¼ q2; . . . ; pn ¼ qm for some rearrangement of the qi. We also have that k ¼ k0. Thus, the theorem is proved. If the ﬁeld K is the complex ﬁeld C, then we have the following result that is known as the fundamental theorem of algebra; its proof lies beyond the scope of this text. THEOREM C.8: (Fundamental Theorem of Algebra) Let f t ð Þ be a nonzero polynomial over the complex ﬁeld C. Then f t ð Þ can be written uniquely (except for order) as a product f t ð Þ ¼ k t r2 ð Þ t r2 ð Þ t rn ð Þ where k; ri 2 C—as a product of linear polynomials. In the case of the real ﬁeld R we have the following result. THEOREM C.9: Let f t ð Þ be a nonzero polynomial over the real ﬁeld R. Then f t ð Þ can be written uniquely (except for order) as a product f t ð Þ ¼ kp1 t ð Þp2 t ð Þ pm t ð Þ where k 2 R and the pi t ð Þ are monic irreducible polynomials of degree one or two. 414 Appendix C Polynomials over a Field Odds and Ends D.1 Introduction This appendix discusses various topics, such as equivalence relations, determinants and block matrices, and the generalized MP (Moore–Penrose) inverse. D.2 Relations and Equivalence Relations A binary relation or simply relation R from a set A to a set B assigns to each ordered pair a; b ð Þ 2 A B exactly one of the following statements: (i) ‘‘a is related to b,’’ written a R b, (ii) ‘‘a is not related to b’’ written a R b. A relation from a set A to the same set A is called a relation on A. Observe that any relation R from A to B uniquely deﬁnes a subset ^ R of A B as follows: ^ R ¼ a; b ð Þja R b f g Conversely, any subset ^ R of A B deﬁnes a relation from A to B as follows: a R b if and only if a; b ð Þ 2 R In view of the above correspondence between relations from A to B and subsets of A B, we redeﬁne a relation from A to B as follows: DEFINITION D.1: A relation R from A to B is a subset of A B. Equivalence Relations Consider a nonempty set S. A relation R on S is called an equivalence relation if R is reﬂexive, symmetric, and transitive; that is, if R satisﬁed the following three axioms: [E1] (Reﬂexivity) Every a 2 A is related to itself. That is, for every a 2 A, a R a. [E2] (Symmetry) If a is related to b, then b is related to a. That is, if a R b, then b R a. [E3] (Transitivity) If a is related to b and b is related to c, then a is related to c. That is, if a R b and b R c, then a R c: The general idea behind an equivalence relation is that it is a classiﬁcation of objects that are in some way ‘‘alike.’’ Clearly, the relation of equality is an equivalence relation. For this reason, one frequently uses ~ or to denote an equivalence relation. EXAMPLE D.1 (a) In Euclidean geometry, similarity of triangles is an equivalence relation. Speciﬁcally, suppose a; b; g are triangles. Then (i) a is similar to itself. (ii). If a is similar to b, then b is similar to a. (iii) If a is similar to b and b is similar to g, then a is similar to g. APPENDIX D 415 (b) The relation of set inclusion is not an equivalence relation. It is reﬂexive and transitive, but it is not symmetric because A B does not imply B A. Equivalence Relations and Partitions Let S be a nonempty set. Recall ﬁrst that a partition P of S is a subdivision of S into nonempty, nonoverlapping subsets; that is, a collection P ¼ fAjg of nonempty subsets of S such that (i) Each a 2 S belong to one of the Aj, (ii) The sets fAjg are mutually disjoint. The subsets in a partition P are called cells. Thus, each a 2 S belongs to exactly one of the cells. Also, any element b 2 Aj is called a representative of the cell Aj, and a subset B of S is called a system of representatives if B contains exactly one element in each of the cells in fAjg. Now suppose R is an equivalence relation on the nonempty set S. For each a 2 S, the equivalence class of a, denoted by [a], is the set of elements of S to which a is related: a ½ ¼ x j a Rx f g: The collection of equivalence classes, denoted by S=R, is called the quotient of S by R: S=R ¼ a ½ j a 2 S f g The fundamental property of an equivalence relation and its quotient set is contained in the following theorem: THEOREM D.1: Let R be an equivalence relation on a nonempty set S. Then the quotient set S=R is a partition of S. EXAMPLE D.2 Let be the relation on the set Z of integers deﬁned by x y mod 5 ð Þ which reads ‘‘x is congruent to y modulus 5’’ and which means that the difference x y is divisible by 5. Then is an equivalence relation on Z. Then there are exactly ﬁve equivalence classes in the quotient set Z= as follows: A0 ¼ . . . ; 10; 5; 0; 5; 10; . . . f g A1 ¼ . . . ; 9; 4; 1; 6; 11; . . . f g A2 ¼ . . . ; 8; 3; 2; 7; 12; . . . f g A3 ¼ . . . ; 7; 2; 3; 8; 13; . . . f g A4 ¼ . . . ; 6; 1; 4; 9; 14; . . . f g Note that any integer x, which can be expressed uniquely in the form x ¼ 5q þ r where 0 r < 5, is a member of the equivalence class Ar where r is the remainder. As expected, the equivalence classes are disjoint and their union is Z: Z ¼ A0 [ A1 [ A2 [ A3 [ A4 This quotient set Z= , called the integers modulo 5, is denoted Z=5Z or simply Z5: Usually one chooses 0; 1; 2; 3; 4 f g or 2; 1; 0; 1; 2 f g as a system of representatives of the equiva-lence classes. Analagously, for any positive integer m, there exists the congruence relation deﬁned by x y mod m ð Þ and the quotient set Z= is called the integers modulo m. 416 Appendix D Odds and Ends D.3 Determinants and Block Matrices Recall ﬁrst: THEOREM 8.12: Suppose M is an upper (lower) triangular block matrix with diagonal blocks Aj; A2; . . . ; An: Then det M ð Þ ¼ det Aj det A2 ð Þ . . . det An ð Þ: Accordingly, if M ¼ A B 0 D where A is r r and D is s s. Then det M ð Þ ¼ det A ð Þ det D ð Þ: THEOREM D.2: Consider the block matrix M ¼ A B C D where A is nonsingular, A is r r and D is s s: Then det M ð Þ ¼ det A ð Þ det D CA1B ð Þ Proof: Follows from the fact that M ¼ I 0 CA1 I A B 0 D CA1B and the above result. D.4 Full Rank Factorization A matrix B is said to have full row rank r if B has r rows that are linearly independent, and a matrix C is said to have full column rank r if C has r columns that are linearly independent. DEFINITION D.2: Let A be a m n matrix of rank r. Then A is said to have the full rank factorization A ¼ BC where B has full-column rank r and C has full-row rank r. THEOREM D.3: Every matrix A with rank r > 0 has a full rank factorization. There are many full rank factorizations of a matrix A. Fig. D-1 gives an algorithm to ﬁnd one such factorization. EXAMPLE D.3 Let A ¼ 1 1 1 2 2 2 1 3 1 1 2 3 2 4 3 5 where M ¼ 1 1 0 1 0 0 1 1 0 0 0 0 2 4 3 5 is the row cannonical form of A. We set B ¼ 1 1 2 1 1 2 2 4 3 5 and C ¼ 1 1 0 1 0 0 1 1 Then A ¼ BC is a full rank factorization of A. Algorithm D-1: The input is a matrix A of rank r > 0. The output is a full rank factorization of A. Step 1. Find the row cannonical form M of A. Step 2. Let B be the matrix whose columns are the columns of A corresponding to the columns of M with pivots. Step 3. Let C be the matrix whose rows are the nonzero rows of M. Then A ¼ BC is a full rank factorization of A. Figure D-1 Appendix D Odds and Ends 417 D.5 Generalized (Moore–Penrose) Inverse Here we assume that the ﬁeld of scalars is the complex ﬁeld C where the matrix AH is the conjugate transpose of a matrix A. [If A is a real matrix, then AH ¼ AT.] DEFINITION D.3: Let A be an m n matrix over C. A matrix, denoted by Aþ, is called the pseudoinverse or Morre–Penrose inverse or MP-inverse of A if A satisﬁes the following four equations: [MP1] AXA ¼ A; [MP3] AX ð ÞH¼ AX; [MP2] XAX ¼ X; [MP4] XA ð ÞH¼ XA; Clearly, Aþ is an n m matrix. Also, Aþ ¼ A1 if A is nonsingular. LEMMA D.4: Aþ is unique (when it exists). Proof. Suppose X and Y satisfy the four MP equations. Then AY ¼ AY ð ÞH¼ AXAY ð ÞH¼ AY ð ÞH AX ð ÞH¼ AYAX ¼ AYA ð ÞX ¼ AX The ﬁrst and fourth equations use [MP3], and the second and last equations use [MP1]. Similarly, YA ¼ XA (which uses [MP4] and [MP1]). Then, Y ¼ YAY ¼ YA ð ÞY ¼ XA ð ÞY ¼ X AY ð Þ ¼ X AX ð Þ ¼ X where the ﬁrst equation uses [MP2]. LEMMA D.5: Aþ exists for any matrix A. Fig. D-2 gives an algorithm that ﬁnds an MP-inverse for any matrix A. Combining the above two lemmas we obtain: THEOREM D.6: Every matrix A over C has a unique Moore–Penrose matrix Aþ. There are special cases when A has full-row rank or full-column rank. THEOREM D.7: Let A be a matrix over C. (a) If A has full column rank (columns are linearly independent), then Aþ ¼ AHA ð Þ 1AH: (b) If A has full row rank (rows are linearly independent), then Aþ ¼ AH AAH ð Þ 1: THEOREM D.8: Let A be a matrix over C. Suppose A ¼ BC is a full rank factorization of A. Then Aþ ¼ CþBþ ¼ CH CCH 1 BHB 1BH Moreover, AAþ ¼ BBþ and AþA ¼ CþC: Algorithm D-2. Input is an m n matrix A over C or rank r. Output is Aþ. Step 1. Interchange rows and columns of A so that PAQ ¼ A11 A12 A21 A22 where A11 is a nonsingular r r block. [Here P and Q are the products of elementary matrices corresponding to the interchanges of the rows and columns.] Step 2. Set B ¼ A11 A21 and C ¼ Ir; A1 11 A12 where Ir is the r r identity matrix. Step 3. Set Aþ ¼ Q CH CCH ð Þ 1 BHB ð Þ 1B11 h i P: Figure D-2 418 Appendix D Odds and Ends EXAMPLE D.4 Consider the full rank factorization A ¼ BC in Example D.1; that is, A ¼ 1 1 1 2 2 2 1 3 1 1 2 3 2 4 3 5 ¼ 1 1 2 1 1 2 2 4 3 5 1 1 0 1 0 0 1 1 ¼ BC Then CCH 1¼ 1 5 2 1 1 3 ; C CCH 1¼ 1 5 2 1 2 1 1 3 1 2 2 6 6 4 3 7 7 5; BHB 1¼ 1 11 6 5 5 6 ; B BHB 1¼ 1 11 1 7 4 1 4 7 Accordingly, the following is the Moore–Penrose inverse of A: Aþ ¼ 1 55 1 18 15 1 18 15 2 19 25 3 1 10 2 6 6 4 3 7 7 5 D.6 Least-Square Solution Consider a system AX ¼ B of linear equations. A least-square solution of AX ¼ B is the vector of smallest Euclidean norm that minimizes AX B k k2: That vector is X ¼ AþB [In case A is invertible, so Aþ ¼ A1, then X ¼ A1B, which is the unique solution of the system.] EXAMPLE D.5 Consider the following system AX ¼ B of linear equations: x þ y z þ 2t ¼ 1 2x þ 2y z þ 3t ¼ 3 x y þ 2z 3t ¼ 2 Then, using Example D.4, A ¼ 1 1 1 2 2 2 1 3 1 1 2 3 2 4 3 5; B ¼ 1 3 2 2 4 3 5; Aþ ¼ 1 55 1 18 15 1 18 15 2 19 25 3 1 10 2 6 6 4 3 7 7 5 Accordingly, X ¼ AþB ¼ 1=55 ð Þ 85; 85; 105; 20 ½ T¼ 17=11; 17=11; 21=11; 4=11 ½ T is the vector of smallest Euclidean norm which minimizes AX B k k2: Appendix D Odds and Ends 419 LIST OF SYMBOLS A ¼ ½aij, matrix, 27 A ¼ ½ aij, conjugate matrix, 38 jAj, determinant, 264, 268 A, adjoint, 377 AH, conjugate transpose, 38 AT, transpose, 33 Aþ, Moore–Penrose inverse, 418 Aij, minor, 269 AðI; JÞ, minor, 273 AðVÞ, linear operators, 174 adj A, adjoint (classical), 271 A B, row equivalence, 72 A ’ B, congruence, 360 C, complex numbers, 11 Cn, complex n-space, 13 C½a; b, continuous functions, 228 Cð f Þ, companion matrix, 304 colsp ðAÞ, column space, 120 dðu; vÞ, distance, 5, 241 diagða11; . . . ; annÞ, diagonal matrix, 35 diagðA11; . . . ; AnnÞ, block diagonal, 40 detðAÞ, determinant, 268 dim V, dimension, 124 fe1; . . . ; eng, usual basis, 125 Ek, projections, 384 f : A ! B, mapping, 164 FðXÞ, function space, 114 G F, composition, 173 HomðV; UÞ, homomorphisms, 174 i, j, k, 9 In, identity matrix, 33 Im F, image, 169 JðlÞ, Jordan block, 329 K, ﬁeld of scalars, 112 Ker F, kernel, 169 mðtÞ, minimal polynomial, 303 Mm;n; m n matrices, 114 n-space, 5, 13, 227, 240 P(t), polynomials, 114 PnðtÞ; polynomials, 114 projðu; vÞ, projection, 6, 234 projðu; VÞ, projection, 235 Q, rational numbers, 11 R, real numbers, 1 Rn, real n-space, 2 rowsp ðAÞ, row-space, 120 S?, orthogonal complement, 231 sgn s, sign, parity, 267 spanðSÞ, linear span, 119 trðAÞ, trace, 33 ½TS, matrix representation, 195 T, adjoint, 377 T-invariant, 327 Tt, transpose, 351 kuk, norm, 5, 13, 227, 241 ½uS, coordinate vector, 130 u v, dot product, 4, 13 hu; vi, inner product, 226, 238 u v, cross product, 10 u v, tensor product, 396 u ^ v, exterior product, 401 u v, direct sum, 129, 327 V ﬃU, isomorphism, 132, 169 V W, tensor product, 396 V, dual space, 349 V, second dual space, 350 Vr V, exterior product, 401 W 0, annihilator, 351 z, complex conjugate, 12 Zðv; TÞ, T-cyclic subspace, 330 dij, Kronecker delta, 37 DðtÞ, characteristic polynomial, 294 l, eigenvalue, 296 P , summation symbol, 29 L I S T O F S Y M B O L S 420 A Absolute value (complex), 12 Abelian group, 403 Adjoint, classical, 271 operator, 377, 384 Algebraic multiplicity, 298 Alternating mappings, 276, 360, 399 Angle between vectors, 6, 230 Annihilator, 330, 351, 354 Associate, 406 Associated homogeneous system, 83 Associative, 174, 403 Augmented matrix, 59 B Back-substitution, 63, 65, 67 Basis, 82, 124, 139 change of, 199, 211 dual, 350, 352 orthogonal, 243 orthonormal, 243 second dual, 367 standard, 125 usual, 125 Basis-ﬁnding algorithm, 127 Bessel inequality, 264 Bijective mapping, 166 Bilinear form, 359, 396 alternating, 276 matrix representation of, 360 polar form of, 363 real symmetric, 363 symmetric, 361 Bilinear mapping, 359, 396 Block matrix, 39, 50 determinants, 417 Jordan, 344 square, 40 Bounded, 156 C Cancellation law, 113 Canonical forms, 205, 325 Jordan, 329, 336 rational, 331 row, 74 triangular, 325 Casting-out algorithm, 128 Cauchy–Schwarz inequality, 5, 229, 240 Cayley–Hamilton theorem, 294, 308 Cells, 39, 415 Change of basis, 199, 211 Change-of-basis (transition) matrix, 199 Change-of-coordinate matrix, 221 Characteristic polynomial, 294, 305 value, 296 Classical adjoint, 271 Coefﬁcient, 57, 58, 411 Fourier, 233, 244 matrix, 59 Cofactor, 269 Column, 27 matrix, 27 operations, 89 space, 120 vector, 3 Colsp(A), column space, 126 Commutative law, 403 group, 113 Commuting (diagram), 396 Companion matrix, 304 Complement, orthogonal, 242 Complementary minor, 273 Completing the square, 393 Complex: conjugate, 13 inner product, 239 matrix, 38, 49 n-space, 13 numbers, 1, 11, 13 plane, 12 Complexity, 88 Components, 2 Composition of mappings, 165 Congruent matrices, 360 diagonalization, 61 Conjugate: complex, 12 linearity, 239 matrix, 38 symmetric, 239 Consistent system, 59 Constant term, 57, 58 Convex set, 193 Coordinates, 2, 130 vector, 130 Coset, 182, 332, 403 Cramer’s rule, 272 421 INDEX Cross product, 10 Curves, 8 Cyclic subspaces, 330, 342 group, 408 D dij, Kronecker delta function, 33 Decomposable, 327 Decomposition: direct-sum, 129 primary, 238 Degenerate, 360 bilinear form, 360 linear equations, 59 Dependence, linear, 133 Derivative, 168 Determinant, 63, 264, 267 computation of, 66, 270 linear operator, 275 order, 3, 266 Diagonal, 32 blocks, 40 matrix, 35, 47 quadratic form, 302 Diagonal (of a matrix), 10 Diagonalizable, 203, 292, 296 Diagonalization: algorithm, 299 in inner product space, 382 Dimension of solution spaces, 82 Dimension of vector spaces, 82, 139 ﬁnite, 124 inﬁnite, 124 subspaces, 126 Direct sum, 129, 327 decomposition, 327 Directed line segment, 7 Distance, 5, 241 Divides, 412 Division algorithm, 412 Domain, 164, 406 Dot product, 4 Dual: basis, 350, 352 space, 349, 352 E Echelon: form, 65, 72 matrices, 70 Eigenline, 296 Eigenspace, 299 Eigenvalue, 296, 298, 312 Eigenvector, 296, 298, 312 Elementary divisors, 331 Elementary matrix, 84 Elementary operations, 61 column, 86 row, 72, 120 Elimination, Gaussian, 67 Empty set, ;, 112 Equal: functions, 164 matrices, 27 vectors, 2 Equations (See Linear equations) Equivalence: classes, 416 matrix, 87 relation, 73, 415 row, 72 Equivalent systems, 61 Euclidean n-space, 5, 228 Exterior product, 401 F Field of scalars, 11, 406 Finite dimension, 124 Form: bilinear, 359 linear, 349 quadratic, 363 Forward elimination 63, 67, 73 Fourier coefﬁcient, 81, 233 series, 233 Free variable, 65, 66 Full rank, 41 factorization, 417 Function, 154 space F(X ), 114 Functional, linear, 349 Fundamental Theorem of Algebra, 414 G Gaussian elimination, 61, 67, 73 Gaussian integer, 409 Gauss–Jordan algorithm, 74 General solution, 58 Geometric multiplicity, 298 Gram–Schmidt orthogonalization, 235 Graph, 164 Greatest common divisor, 413 Group, 113, 403 H Hermitian: form, 364 matrix, 38, 49 quadratic form, 364 Hilbert space, 229 Homogeneous system, 58, 81 Homomorphism, 173, 404, 407 Hom(V, U ), 173 Hyperplane, 7, 358 I i, imaginary, 12 Ideal, 405 422 Index Identity: mapping, 166, 168 matrix, 33 ijk notation, 9 Image, 164, 169, 170 Imaginary part, 12 Im F, image, 169 Im z, imaginary part, 12 Inclusion mapping, 190 Inconsistent systems, 59 Independence, linear, 133 Index, 30 Index of nilpotency, 328 Inertia, Law of, 364 Inﬁnite dimension, 124 Inﬁnity-norm, 241 Injective mapping, 166 Inner product, 4 complex, 239 Inner product spaces, 226 linear operators on, 377 Integral, 168 domain, 406 Invariance, 224 Invariant subspaces, 224, 326, 332 direct-sum, 327 Inverse image, 164 Inverse mapping, 164 Inverse matrix, 34, 46, 85 computing, 85 inversion, 267 Invertible: matrices, 34, 46 Irreducible, 406 Isometry, 381 Isomorphic vector spaces, 169, 404 J Jordan: block, 304 canonical form, 329, 336 K Ker F, kernel, 169 Kernel, 169, 170 Kronecker delta function dij, 33 L l2-space, 229 Laplace expansion, 270 Law of inertia, 363 Leading: coefﬁcient, 60 nonzero element, 70 unknown, 60 Least square solution, 419 Legendre polynomial, 237 Length, 5, 227 Limits (summation), 30 Line, 8, 192 Linear: combination, 3, 29, 60, 79, 115 dependence, 121 form, 349 functional, 349 independence, 121 span, 119 Linear equation, 57 Linear equations (system), 58 consistent, 59 echelon form, 65 triangular form, 64 Linear mapping (function), 164, 167 image, 164, 169 kernel, 169 nullity, 171 rank, 171 Linear operator: adjoint, 377 characteristic polynomial, 304 determinant, 275 on inner product spaces, 377 invertible, 175 matrix representation, 195 Linear transformation (See linear mappings), 167 Located vectors, 7 LU decomposition, 87, 104 M Mm,n, matrix vector space, 114 Mappings (maps), 164 bilinear, 359, 396 composition of, 165 linear, 167 matrix, 168 Matrices: congruent, 360 equivalent, 87 similar, 203 Matrix, 27 augmented, 59 change-of-basis, 199 coefﬁcient, 59 companion, 304 diagonal, 35 echelon, 65, 70 elementary, 84 equivalence, 87 Hermitian, 38, 49 identity, 33 invertible, 34 nonsingular, 34 normal, 38 orthogonal, 237 positive deﬁnite, 238 rank, 72, 87 space, Mm,n, 114 square root, 296 triangular, 36 Index 423 Matrix mapping, 165 Matrix multiplication, 30 Matrix representation, 195, 238, 360 adjoint operator, 377, 384 bilinear form, 359 change of basis, 199 linear mapping, 195 Metric space, 241 Minimal polynomial, 303, 305 Minkowski’s inequality, 5 Minor, 269, 273 principle, 273 Module, 407 Monic polynomial 303, 411 Moore–Penrose inverse, 418 Multilinearity, 276, 399 Multiplicity, 298 Multiplier, 67, 73, 87 N n-linear, 276 n-space, 2 complex, 13 real, 2 Natural mapping, 351 New basis, 199 Nilpotent, 328, 336 Nonnegative semideﬂnite, 226 Nonsingular, 112 linear maps, 172 matrices, 34 Norm, 5, 227, 241 Normal, 7 matrix, 38 operator, 380, 383 Normalized, 227 Normalizing, 5, 227, 233 Normed vector space, 241 Nullity, 171 nullsp(A), 170 Null space, 170 O Old basis, 199 One-norm, 241 One-to-one: correspondence, 166 mapping, 166 Onto mapping (function), 166 Operators (See Linear operators) Order, n: determinant, 264 of a group, 408 Orthogonal, 4, 37, 80 basis, 231 complement, 231 matrix, 237 operator, 380 projection, 384 substitution, 302 Orthogonalization, Gram–Schmidt, 235 Orthogonally equivalent, 381 Orthonormal, 233 Outer product, 10 P Parameter, 64 form, 65 Particular solution, 58 Partition, 416 Permutations, 8, 267 Perpendicular, 4 Pivot, 67, 71 row reduction, 94 variables, 65 Pivoting (row reduction), 94 Polar form, 363 Polynomial, 411 characteristic, 294, 305 minimum, 303 space, Pn(t), 114 Positive deﬁnite, 226 matrices, 238 operators, 336, 382 Positive operators, 226 square root, 391 Primary decomposition theorem, 328 Prime ideal, 410 Principle ideal ring, 406 Principle minor, 273 Product: exterior, 401 inner, 4 tensor, 396 Projections, 167, 234, 344, 384 orthogonal, 384 Proper value, 296 vector, 296 Pythagorean theorem, 233 Q Q, rational numbers, 11 Quadratic form, 301, 315, 363 Quotient group, 403 ring, 406 spaces, 332, 416 R R, real numbers, 1, 12 Rn, real n-space, 2 Range, 164, 169 Rank, 72, 87, 126, 171, 364 Rational: canonical form, 331 numbers, Q, 11 Real: numbers, R, 1 part (complex number), 12 424 Index Real symmetric bilinear form, 363 Reduce, 73 Relation, 415 Representatives, 416 Restriction mapping, 192 Right-handed system, 11 Right inverse, 189 Ring, 405 quotient, 406 Root, 293 Rotation, 169 Row, 27 canonical form, 72 equivalence, 72 operations, 72 rank, 72 reduce, 73 reduced echelon form, 73 space, 120 S Sn, symmetric group, 267, 404 Scalar, 1, 12 matrix, 33 multiplication, 33 product, 27 Scaling factor, 296 Schwarz inequality, 5, 229, 240 (See Cauchy–Schwarz inequality) Second dual space, 350 Self-adjoint operator, 380 Sign of permutation, 267 Signature, 364 Similar, 203, 224 Similarity transformation, 203 Singular, 172 Size (matrix), 27 Skew-adjoint operator, 380 Skew-Hermitian, 38 Skew-symmetric, 360 matrix, 36, 48 Solution, (linear equations), 57 zero, 121 Spatial vectors, 9 Span, 116 Spanning sets, 116 Spectral theorem, 383 Square: matrix, 32, 44 system of linear equations, 58, 72 Square root of a matrix, 391 Standard: basis, 125 form, 57, 399 inner product, 228 Subdiagonal, 304 Subgroup, 403 Subset, 112 Subspace, 117, 133 Sum of vector spaces, 129 Summation symbol, 29 Superdiagonal, 304 Surjective map, 166 Sylvester’s theorem, 364 Symmetric: bilinear form, 361 matrices, 4, 36 Systems of linear equations, 58 T Tangent vector, T(t), 9 Target set, 164 Tensor product, 396 Time complexity, 88 Top-down, 73 Trace, 33 Transformation (linear), 167 Transition matrix, 199 Transpose: linear functional (dual space), 351 matrix, 32 Triangle inequality, 230 Triangular form, 64 Triangular matrix, 36, 47 block, 40 Triple product, 11 Two-norm, 241 U Unique factorization domain, 406, 414 Unit vector, 5, 227 matrix, 33 Unitary, 38, 49, 380 Universal mapping principle (UMP), 396 Usual: basis, 125 inner product, 228 V Vandermonde determinant, 290 Variable, free, 65 Vector, 2 coordinates, 130 located, 7 product, 10 spatial, 9 Vector space, 112, 226 basis, 124 dimension, 124 Volume, 274 W Wedge (exterior) product, 401 Z Z, integers, 406 Zero: mapping, 128, 168, 173 matrix, 27 polynomial, 411 solution, 121 vector, 2 Index 425
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Register now EN Deutsch Englisch Home Products Show all products Dehumidifier Fan heater Hygrometer Home fragrances Aroma diffuser Humidifier Air purifier Air washer 2in1 Humidifier accessories Air purifier accessories Fan Health Humidify the air Dehumidify the air Purify the air Cooling the flat Measure humidity Virus protection Alleviate allergy symptoms Neutralise odours Aromatherapy Rooms Office/home office Basement/hobby room Living room Bedroom Kitchen Children's room Dining room Service FAQ general Contact Wi-Fi Newsletter Blog overview Instruction manuals Give feedback About us Mission & values Team Designers History Contact Awards Distributors Retailers Media releases Image database media Vacancies Login EN Deutsch Englisch To contact Nadine Walder , 18 September 2024 Humidify the air How does humidity affect skin care? Are you no stranger to skin problems and blemishes that are influenced by the season and external factors? Do you often suffer from dry skin in winter and more oily skin in summer? The reasons for this may lie in the temperature and humidity, as the health and appearance of our skin is influenced not least by these two factors. In this article, we explain how humidity in particular affects your skin and how it should influence your skincare regime. Estimated reading time: 5 minutes Read on to find out: Factors that influence the skin and skin care Relationship between skin and humidity Skin care for high humidity in summer Low humidity: dry skin in winter Skin moisture: Ideal indoor climate for healthy skin all year round Factors that influence the skin and skin care The skin and its appearance are influenced by various factors. A basic distinction is made between internal and external influences. Internal factors include genes, hormonal balance and natural skin ageing. Genes can play a role in the development of neurodermatitis, for example, and skin prone to extreme dryness can also be hereditary. The appearance of the skin also interacts with our hormonal balance and is subject to its fluctuations, which women in particular often notice. If the hormonal balance becomes imbalanced, pimples, blackheads and the like can develop. Finally, the natural ageing process also plays a key role in the appearance of our skin and how we optimise our skin care. External influences on the skin and its appearance include free radicals (metabolic products that are very aggressive and can attack healthy cells), environmental pollution and UV radiation. These three factors contribute significantly to skin ageing, can impair the skin's appearance and favour or exacerbate problems such as acne or dry skin. Another often underestimated but decisive external factor is therelative humidity. It plays a key role in the skin's moisture content. Depending on the time of year and climatic conditions, humidity can pose different challenges for skin appearance, skin health and skin care. Relationship between skin and humidity Humidity plays a key role in regulating skin moisture. It influences the hydration of the skin, transepidermal water loss (TEWL) – a process in which water evaporates from the skin's surface into the environment –, the skin's moisture balance, the function of the skin barrier and how the skin feels in general. Astudy from 2016shows through a comprehensive literature review that epidemiological findings indicate that skin eczema, such as atopic dermatitis, increases with low humidity indoors. The study also indicates that the symptoms of dry skin occur more frequently at low humidity levels in air-conditioned indoor spaces. The results of the literature review also show that studies with mice have found that low humidity can cause a number of skin changes, including impairment of the desquamation process. Anotherstudyemphasises the link between low humidity and skin problems: Low humidity combined with low temperatures lead to a decrease in skin barrier function, increased susceptibility to mechanical stress and exacerbation of dermatitis. High humidity can lead to excess moisture and thus to problems such as oily skin or acne. The excess moisture mixes with sebum and dead skin cells and clogs the pores. Excessive humidity can also make the skin feel sticky and clammy. Fungal infections can also be a consequence of high relative humidity. Skin irritation and allergic reactions are further effects of excessive humidity. Sensitivity, redness and itching can develop or increase, especially in people with sensitive skin. For the skin to cope with different levels of humidity, it is crucial to adapt the skincare regime to the respective environmental conditions. Whether it's summer or winter, it's important to adapt your skincare regime to the humidity in order to achieve optimum results, strengthen the skin barrier and keep the skin healthy. Skin care for high humidity in summer In summer, the humidity indoors and outdoors is usually high. The skin can then produce an excessive amount of sebum, which can clog the pores and lead to blemishes. In the hotter months, it is advisable to choose lighter products and those that do not clog the pores. The skin should be cleansed daily with non-comedogenic cleanser. A 2-phase cleansing is recommended. In the first phase, you first do an oil-based cleansing and then use a foam or cream cleanser. It is also important to moisturise the skin sufficiently, even in high humidity. To do this, use creams that are oil-free, light and pore-friendly for your skin type. Moisturising ingredients such as glycerine and hyaluronic acid are beneficial. On the other hand, creams that are heavy or occlusive (protective layer-forming) products should be avoided. Heavy make-up should also be avoided in humid climates, as it can melt and further clog the pores. Regardless of the weather, you should also apply sun protection every day. In addition to these specific recommendations, there are several other measures that will help to support your skin health. These include drinking enough water, wearing breathable clothing and maintaining a healthy lifestyle in general. Low humidity:dry skin in winter Especially in heated rooms, the humidity often drops sharply in winter, which can dry out the skin. This can lead to redness, feelings of tightness and even eczema or worsen existing symptoms. In these conditions, you should rely on rich moisturisers and oils that strengthen the skin barrier and optimally lock in moisture (occlusive products). Use rich moisturisers with moisturising ingredients such as glycerin, ceramides and hyaluronic acid. You should apply these immediately after cleansing when the skin is still damp. You can also supplement your skincare routine with a moisturising serum. It is also advisable to exfoliate at least once a week to remove dead skin cells from the skin's surface. Chemical peels such as fruit acid peels are suitable here. Don't forget to use sun protection in winter too, as UV rays can damage the skin all year round. To keep your body and skin moisturised, you should drink plenty of water, just like in the hot months. A balanced diet with plenty of vitamins and antioxidants also supports skin health in the winter months. Skin moisture: Ideal indoor climate for healthy skin all year round To keep your skin healthy all year round, you can not only adapt your skincare to the respective conditions, but also optimise your indoor climate accordingly. A relative humidity of 40 % to 60 % is recommended indoors. In the winter months, when the humidity indoors drops significantly due to the heating, ahumidifiercan help you to achieve and maintain the ideal humidity level. Humidifiers release moisture into the room air and can therefore prevent the skin from losing moisture and drying out. A humidifier is a sensible investment, especially for sensitive skin or skin prone to dryness. In air-conditioned rooms, the air is often too dry, which is why a humidifier can also be helpful in summer. In environments with high humidity, on the other hand, adehumidifiercan be useful to support skin health. These reduce the humidity in the room and help to create a healthy skin climate. In addition and depending on the climatic conditions,air purifiersandair washerscan also be helpful in improving skin health. These devices filter pollutants, pollen and other particles from the air that can irritate the skin. For people with sensitive or allergy-prone skin, the use of an air purifier or air washer can be useful and very beneficial. Do you want to know more about humidity and what the difference is between absolute and relative humidity? More about humidity If you have questions related to indoor room climate,please get in touch with us. Or subscribe to ournewsletterto regularly get informed about current topics regarding indoor climate, experience reports or Stadler Form insights. 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