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16600	http://chembook.org/page-nonav.php?chnum=2&sect=7	Reaction Stoichiometry and Gases nav Reaction Stoichiometry with Gases One of the really nice things about the way gases behave and the laws they follow is that our ol' friend reaction stoichiometry is totally doable with gas measurements and not just moles and grams. Let me explain... First we need a reaction that has gases in it. It can be just one single species in the reaction or it can be all of them. So I'm going to use an ol' general chemistry favorite: making ammonia from nitrogen and hydrogen. N 2(g) + 3H 2(g) → 2NH 3(g) Notice all the species (reactants and products) are gases. So every species will obey the laws of gases and specifically Avogadro's Law which is the law where the volume of a gas is directly proportional to the volume (holding temperature and pressure constant). Let me state that another way... I know exactly how many moles are reacting at anytime during the reaction simply by monitoring (measuring) the total volume. The total volume of the system can be thought as the individual or partial volumes of the reactants and products all added together (works just like partial pressures but with volumes). PLUS, a great fact about Avogadro's law is that ANY volume unit will work. The volume doesn't have to be in liters, it can be in any volume unit I want and it will perfectly SCALE in relation to the number of moles present. Think in "volumes" now So you've been trained (taught) to read that balanced equation as 1 mole of nitrogen (N 2) plus 3 moles of hydrogen (H 2) will completely react to make 2 moles of ammonia (NH 3). Then you can use any number of moles and do all the cool math stoichiometry and predict moles of product - it just scales the reaction to some new set of moles, right? Here's the fun part... because volume is directly proportional to moles (Avogadro's Law) you can just as easily read that reaction to say the following. 1 volume of nitrogen (N 2) plus 3 volumes of hydrogen (H 2) will completely react to make 2 volumes of ammonia (NH 3). The term "volume" here can be ANY volume you want - it just has to be the same reference volume for all the gases in the reaction and they all have to be at the same partial pressures (say 1 atm). So lets work a problem with this new found knowledge. How many liters of ammonia will be produced when 4 liters of nitrogen reacts with excess hydrogen? ANSWER: Just SCALE the given reaction up to 4 for N 2 and use that same scalar (what you multiplied by to get the 4) on all the other gases. If you do this you'll have 4 nitrogens, 12 hydrogens, and 8 ammonias. So the answer is you will produce 8 liters of ammonia gas. Or, if you really want to do dimensional analysis: 4 v o l N 2 1(2 v o l N H 3 1 v o l N 2)=8 v o l N H 3 Look at that again. All I did was change "mol" into "vol" and it has to work. What if the volume is something weird like cubic feet (ft 3) and have some crazy pressure and temperature? Here you go: 10 ft 3 of N 2 is added to 21 ft 3 of H 2 and the reaction goes to completion. How many cubic feet of ammonia is made at this temperature and pressure? Assume all gases are at 300 psi and 400 K. Yikes! seems scary now, right? No - it's still the same reaction except with ft 3 as the "volumes" in the stoichiometry. The 300 psi and 400K don't really matter here as far as calcuting the answer in ft 3. There is one gotcha though... THIS problem is a limiting reactant problem because I didn't give you nitrogen and hydrogen in a perfect 1:3 ratio. So ask yourself how many reactions (rxn) can you run with each? 10 f t 3 N 2 1(r x n 1 f t 3 N 2)=10 r x n (N 2 is in excess) 21 f t 3 H 2 1(r x n 3 f t 3 H 2)=7 r x n (H 2 is limiting reactant) I'll spare you yet another dimensional analysis equation - just do it in your head now: I'm running 7 rxn and each rxn yields 2 ft 3 of NH 3, therefore a total of 14 ft 3 of NH 3 are produced. Oh, and also note that there are 3 ft 3 of N 2 gas left over (unreacted and in excess). One more thing... Everything I just showed you on this page pertaining to "volumes" can just as easily be done with partial pressures while holding the volume and temperature constant. Yeah, that last problem I worked... change all the ft 3's into some pressure unit - like atm. The answer will now be 14 atm of NH 3. ☑︎ check box for "mind blown" 🤯. previousback to topnext © 2019-2022 · mccord
16601	https://www.youtube.com/watch?v=2kMUk_XRvRQ	Finding the Slope Given 2 Points Davitily 9010 subscribers 6974 likes Description 1045866 views Posted: 11 Jun 2008 - How to Find the Slope Given 2 Points. For more practice and to create math worksheets, visit Davitily Math Problem Generator at www.mathproblemgenerator.com 331 comments Transcript: finding the slope between two points the slope is how we measure the direction of a line in a coordinate plane so let's start off by plotting these two points and creating a line so I'm going to plot -6 comma 3 -6 comma 3 is right here then 4 comma -3 4 comm3 is right here now I'm going to connect the two points to create my line and next I'm going to find the direction of the the line now the slope is defined as rise over run rise over run rise represents the vertical distance that you travel to get from one point to the the next and run is the horizontal distance now we always travel from the left point to the right point so let's start with rise rise is the vertical distance so we have to travel down and run is the horizontal distance so we got to travel to the right now rise is the vertical distance and since we travel down and let's see how far we traveled 1 2 3 4 5 6 so we traveled six down and the reason why I put negative -6 is we're traveling down now if we're traveling to the right it's going to be positive so I'm going to count 1 2 3 4 five 6 7 8 9 10 and we travel 10 to the right now let's put these two numbers in the fraction form of rise over run so our rise would be -6 and our run would be 10 now I should reduce whenever possible -6 and 10 could be reduced by two and that's going to get you -3 over 5 -3 over 5 is the same thing as -3 over 5 and the slope between these two point is 35ths and this completes my problem another way to find the slope between two points is to use this formula Y2 - y1 over X2 - X1 now before using this formula we should label what's going to be our X1 y1 and X2 Y2 now I'm going to label my first point X1 y1 and then I'm going to label my second point X2 Y2 now once I have that label I will substitute in my formula Y2 - y1 is -3 - 3 over X2 - X1 4 - -6 -3 - 3 is -6 and 4 - -6 is 4 + 6 cuz minus a negative becomes a plus and 4 + 6 is 10 -6 over 10 can be reduced by two to get 3 over 5 then we bring the negative down and our slope is -3 over 5 and this completes our problem
16602	https://artofproblemsolving.com/wiki/index.php/Complementary_counting?srsltid=AfmBOor85zkoFQ8vu95zbcbyStPit2BLCgfopVq-QUQOE23kkPxWENrv	Art of Problem Solving Complementary counting - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Complementary counting Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Complementary counting In combinatorics, complementary counting is a counting method where one counts what they don't want, then subtracts that from the total number of possibilities. In problems that involve complex or tedious casework, complementary counting is often a far simpler approach. A large hint that complementary counting may lead to a quick solution is the phrase "not" or "at least" within a problem statement. More formally, if is a subset of , complementary counting exploits the property that , where is the complement of . In most instances, though, is obvious from context and is committed from mention. Contents [hide] 1 Complementary Probability 2 Examples 2.1 Example 1 2.2 Example 2 2.3 Example 3 2.4 Example 4 2.5 Example 5 3 Resources 4 See also Complementary Probability There is a probability equivalent of complementary counting. For any event, the probability it happens plus the probability it does not happen is one. Thus, we have the identity Like its counting analog, complementary probability often vastly simplifies tedious casework. Unlike complementary counting, though, it sees frequent use as an intermediate step, primarily because computing complements is much easier in probability than in counting. Examples Here are some examples that demonstrate complementary counting and probability in action. It is worth noting that complementary probability is not typically an intermediate step, but a framework upon which a solution is built. Example 1 How many positive integers less than are not a multiple of five? Solution: We use a complementary approach. The total number of positive integers, with no restrictions, is integers. What we don't want are the multiples of five. These are or ; it's easy to see that there are of them. Thus, our answer is is . Example 2 2006 AMC 10A Problem 21: How many four-digit positive integers have at least one digit that is a 2 or a 3? Solution: We use a complementary approach. With no restrictions, there are four-digit positive integers. What we don't want are the four-digit integers with no digit that is a two or three, Using a constructive approach, the first digit can be one of seven integers; and . Note that the first digit cannot be zero, or else it ceases to have four digits. However, the second, third, and fourth digits can be zero; as a result, they have eight options. So, our total number of two-and-three-free numbers is . Hence, our final answer is , as desired. Example 3 Sally is drawing seven houses. She has four crayons, but she can only color any house a single color. In how many ways can she color the seven houses if at least one pair of consecutive houses must have the same color? Solution: Use complementary counting. First, we find the total colorings without restriction, which we do by constructing them. She has four options for what color the first house can be, four options for the second, and so on. Hence, there are ways she can color the four houses. Next, we find the possibilities where every house's next-door neighbor is a different color. Using a constructive approach, she has four options for the color of the first house. We have to make sure the next house is a different color; as a result, there are only three options for the color of the second house, with the color of the first house unavailable. By similar logic, there are 3 options for the third house, and so on for every other house. Combining these yields possibilities if every house must be a different color. Putting these two together, there are ways she can color the seven houses four colors if at least one pair of consecutive houses must be the same color. Example 4 2002 AIME I Problem 1 Many states use a sequence of three letters followed by a sequence of three digits as their standard license-plate pattern. Given that each three-letter three-digit arrangement is equally likely, the probability that such a license plate will contain at least one palindrome (a three-letter arrangement or a three-digit arrangement that reads the same left-to-right as it does right-to-left) is , where and are relatively prime positive integers. Find Solution: We use complementary probability. So first, we find the probability that a license plate has no palindromes; in other words, the probability that the first and third numbers and letters are distinct. The probability the numbers are distinct is , as for every digit of the first number, one out of ten is the same digit; similarly, for the letters, it is . Multiplying these together gives that the probability that a license plate has no palindromes is Taking the complement of this, the probability a license plate has a palindrome is thus . Hence, our answer is . Example 5 2008 AMC 12B Problem 22: A parking lot has 16 spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers chose spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires 2 adjacent spaces. What is the probability that she is able to park? Solution: Use complementary probability, where we find the probability that no two open spots are consecutive. With no restrictions, the total number of ways the 12 cars could park is , where is a combination. Finding how many permutations of the cars and spaces leave no two spaces next to each other is a more challenging task, though. One might eventually think to treat this problem like a distribution, a stars-and-bars approach. Let the 12 cars be stars and the 4 spaces be bars. Here is one arrangement of our stars and bars; The question mandates that no two bars sit next to each other. Thus, we have 13 "slots" where the bars could go (eleven between the stars, two at the endpoints), where only one bar can fit in each slot. It follows that the number of ways to insert these bars is . Then the probability that Auntie Em cannot park is . Finally, our answer is . Resources AoPS Complementary Counting Part 1 AoPS Complementary Counting Part 2 AoPS Complementary Probability Part 1 AoPS Complementary Probability Part 2 AoPS Complementary Probability Part 3 AoPS Complementary Probability Part 4 See also Casework Constructive counting Overcounting Retrieved from " Categories: Combinatorics Definition Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
16603	https://chemistry.stackexchange.com/questions/38903/mass-number-relative-atomic-mass-average-mass	atoms - Mass number, (relative) atomic mass, average mass - Chemistry Stack Exchange Join Chemistry By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Mass number, (relative) atomic mass, average mass Ask Question Asked 9 years, 11 months ago Modified8 years, 10 months ago Viewed 7k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. What is the difference between mass number, atomic mass and average atomic mass? I know the mass number is the amount of protons + the amount of neutrons. The average mass is the weighed average of the isotopes that occur in nature. But then what is the (relative) atomic mass? atoms Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Oct 12, 2015 at 11:39 Eowyn12Eowyn12 263 3 3 silver badges 7 7 bronze badges 2 Related: Quick and simple explanation of molar mass, molecular mass and atomic mass and Units of mass on the atomic scaleuser7951 –user7951 2015-10-12 11:44:28 +00:00 Commented Oct 12, 2015 at 11:44 It is relative to the mass of another element. So, you may set the mass of carbon to be 12.000 and get all the masses of the elements relative to that; or, you may set the mass of oxygen to be 16.000 and get the masses of the elements relative to that. These two mass systems will be slightly different but internally consistent. Hope that helps.DarrenRhodes –DarrenRhodes 2015-10-12 11:48:40 +00:00 Commented Oct 12, 2015 at 11:48 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. It’s pretty simple: The mass number is the integer you get if you count (and add up) the neutrons and the protons of a given element. Thus, a hydrogen atom of the X 1 1 X 2 1 2 1 H X 1 1 X 2 1 2 1 H isotope has a mass number of 1 1 (only proton), X 12 6 X 2 12 2 6 C X 6 12 X 2 6 2 12 C has 12 12 (6 6 protons, 6 6 neutrons) and X 81 35 X 2 81 2 35 B r X 35 81 X 2 35 2 81 B r has 81 81 (of which 35 35 are protons, the remaining 46 46 neutrons). The atomic mass is what these atoms actually weigh in atomic mass units. For reasons that boil down to E=m c 2 E=m c 2 (or so I believe) and the nonzero mass of an electron, this is not an integer except for one exception: X 1 X 2 1 2 H X 1 X 2 2 1 H’s atomic mass is 1.007825032 1(4)1.007825032 1(4) X 12 X 2 12 2 C X 12 X 2 2 12 C’s atomic mass is exactly 12 12. This is because 1 u 1 u was defined as exactly 1/12 1/12 th of the mass of a carbon-12 atom. X 81 X 2 81 2 B r X 81 X 2 2 81 B r’s atomic mass is 80.916289(6)80.916289(6) The average mass takes into account an elements different isotopes and their natural abundance and calculates an overall average. Thus, this is no longer defined on an isotopal basis but on an elemental one. The average masses of the elements discussed above are: H H has 1.00794(7)1.00794(7) (this is larger than the atomic mass of X 1 X 2 1 2 H X 1 X 2 2 1 H due to the heavier isotopes.) C C has 12.0107(8)12.0107(8) B r B r has 79.904(1)79.904(1) (this is lower than the atomic mass of X 81 X 2 81 2 B r X 81 X 2 2 81 B r, because about half of the naturally occuring bromine atoms are X 79 X 2 79 2 B r X 79 X 2 2 79 B r.) Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 5, 2016 at 19:07 JanJan 69.4k 12 12 gold badges 211 211 silver badges 397 397 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. well lets start it off with the example of a pizza, first imagine a pizza that is divided into 12 slices where 6 slices are meat free and the other 6 with meat and you were told that different pizzas have different amount of slices and different proportions of meat:non-meat ones the total number of pizza slices here is what we could refer to as "pizza number" now the seller tells you that this pizza(12 slices) weighs exactly 1200 grams and the other one (mini-pizza)weighs 600 grams now there weighs here is what we will refer to as pizza mass, now someone asks how much does this pizza weighs relative to the other we do bunch of calculations and we find out that the relative pizza mass of the mini should be on the scale of 1,and so the other pizza should be two{its double the mass} now what we have is relative pizza mass now to sum it up lets consider the atom the atom nucleus does have proton and neutron and different atoms have different number of protons and neutrons in different proportions the number of both proton and neutron number is mass number it is practically helpful if we have different isotopes because each isotope have different mass number and so could be identified by their numbers,the atomic mass which is measured by mass spectrometer is the real mass of the atom which also varies between atoms if they were to sum up all those isotopic masses and take the average the result will be average atomic mass considering how small an atom is it will be awkward to use units like kg because the total mass will be small that you will need to write the number in standard form of such 110^-26 so the use the relative atomic mass to make it easier by using the scale of 12 amu(atomic mass unit) for a carbon 12 atom, and by that they can make other relative mass such as hydrogen(1.008),oxygen(15.999) and so on summary using an example nitrogen atom with 7 neutron and 7 protons have mass number of 14 and thus isotope name is nitrogen-14 the atomic mass of nitrogen-14 is 2.32525265e-26 average atomic mass of all isotopes is 2.325x10^-26 relative atomic mass of all isotopes of nitrogen is 14.007 amu note:this is based on my personal experience if someone with any suggestions and arguments please consider a comment that is all what I have, thanks Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 5, 2016 at 11:27 phenolicdeathphenolicdeath 191 1 1 silver badge 11 11 bronze badges Add a comment\| Your Answer Reminder: Answers generated by AI tools are not allowed due to Chemistry Stack Exchange's artificial intelligence policy Thanks for contributing an answer to Chemistry Stack Exchange! 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16604	https://math.stackexchange.com/questions/3070029/congruence-multiplication-property	Skip to main content Congruence multiplication property Ask Question Asked Modified 9 months ago Viewed 1k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. I was looking over some properties of modular arithmetic and noticed the following one: If a≡a′ (mod n) and b≡b′ (mod n), then ab≡a′b′ (mod n) Now this is ok and I understand the proofs that prove that property, but I have also noticed that reducing only one variable modulo n the congruence will also be satisfied. Namely: ab≡ab′ (mod n) Can anyone tell me why this also works? I am not very good with proofs, but I think it is just a corollary of the first property. elementary-number-theory proof-writing modular-arithmetic Share CC BY-SA 4.0 Follow this question to receive notifications asked Jan 11, 2019 at 16:30 Michael MuntaMichael Munta 21777 silver badges2727 bronze badges 4 2 Yes, it's just the special case of the first where a′=a (so a′≡a). Both versions of the Congruence Product Rule are equivalent. – Bill Dubuque Commented Jan 11, 2019 at 16:33 a≡a(modn), so of course it's fine. – Randall Commented Jan 11, 2019 at 16:41 @BillDubuque Can I also think of it in terms of congruence being an equivalence relation? So analogously to the = relation; in modulo N two numbers being congruent to one another means that they are essentially one and the same and can be swapped without changing the equivalence? – Michael Munta Commented Jan 15, 2019 at 8:51 @MichaelMunta Yes, by definition congruence relations are equivalence relations that satisfy the additional property of being compatible with the ambient arithmetic operations (in a ring this means that are compatible with addition and multiplication, i.e. they satisfy the Sum and Product Rules in the linked post in my prior comment). Once you learn about quotient structures then you can replace the equivalence relation by equality (of congruence classes in the quotient ring, i.e. 11≡1(mod10) becomes 10=10, i.e. 11+10Z=1+10Z) – Bill Dubuque Commented Jan 15, 2019 at 15:27 Add a comment \| 1 Answer 1 Reset to default This answer is useful 1 Save this answer. Show activity on this post. b≡b′(modn)⇔b−b′≡0(modn)⇔b−b′=nq,q∈Z⇔a(b−b′)=a(nq)=n(aq)⇔a(b−b′)≡0(modn)⇔ab−ab′≡0(modn)⇔ab≡ab′(modn) Share CC BY-SA 4.0 Follow this answer to receive notifications answered Sep 8, 2022 at 17:08 PlattPlatt 1111 bronze badge Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions elementary-number-theory proof-writing modular-arithmetic See similar questions with these tags. Featured on Meta Community help needed to clean up goo.gl links (by August 25) Linked 107 Congruence Arithmetic Laws, e.g. in divisibility by 7 test 0 Why is this congruence modulo m true? 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16605	https://madasmaths.com/archive/maths_booklets/basic_topics/various/circle_coordinate_geometry_exam_questions.pdf	Created by T. Madas Created by T. Madas CIRCLE COORDINATE GEOMETRY (EXAM QUESTIONS) Created by T. Madas Created by T. Madas Question 1 () A circle has equation 2 2 2 8 x y x + = + Determine the radius and the coordinates of the centre of the circle. 3 r = , ( ) 1,0 Question 2 () A circle C has equation 2 2 12 2 24 0 x y x y + − + + = . a) Find the coordinates of the centre of the circle and the length of its radius. The straight line L has equation 4 x y + = . b) Determine the coordinates of the points of intersection between C and L . c) Show that the distance between these points of intersection is 2 k , where k is an integer. ( ) 6, 1 , 13 r − = , ( ) ( ) 8, 4 , 3,1 − , 5 k = Created by T. Madas Created by T. Madas Question 3 () The figure above shows the points ( ) 4,6 A and ( ) 2,14 B − , which both lie on the circumference of a circle. Given that AB is a diameter of the circle, determine an equation for the circle. C2B , ( ) ( ) 2 2 1 10 25 x y − + − = A B O y x Created by T. Madas Created by T. Madas Question 4 () The straight line segment joining the points ( ) 7,4 A − and ( ) 1, 2 B − is a diameter of a circle with centre at the point C and radius r . a) Find the coordinates of C and the value of r . The point ( ) 0,a lies on the circumference of this circle. b) Determine the possible values of a . C2K , ( ) 3,1 C − , 5 r = , 3, 5 a = − Created by T. Madas Created by T. Madas Question 5 () The straight line joining the points ( ) 6, 3 A − and ( ) 14,9 B is a diameter of a circle. a) Determine an equation for the circle. The point ( ) 16,k lies on the circumference of the circle. b) Find the possible values of k . ( ) ( ) 2 2 10 3 52 x y − + − = , 1 7 k k = − = ∪ Question 6 () A circle is centred at ( ) 5,6 and has radius 13. a) Find an equation for this circle. The straight line l with equation 6 y x = − intersects the circle at the points A and B . b) Determine the coordinates of A and the coordinates of B . ( ) ( ) 2 2 5 6 169 x y − + − = , ( ) ( ) 17,11 , 0, 6 − Created by T. Madas Created by T. Madas Question 7 (+) A circle has its centre at ( ) 1,2 C − . The points ( ) 4,3 A − and ( ) 0,5 B lie on this circle. a) Find an equation for the circle. b) Determine an equation of the straight line which passes through C and bisects the chord AB . C2C , ( ) ( ) 2 2 1 2 10 x y + + − = , 2 y x = − Question 8 (+) A circle has centre ( ) 3,5 C − and passes through the origin. a) Find an equation for this circle. The circle crosses the x axis at the origin and at the point A . b) Determine the coordinates of A . ( ) ( ) 2 2 3 5 34 x y + + − = , ( ) 6,0 A − Created by T. Madas Created by T. Madas Question 9 (+) A circle has equation 2 2 20 8 16 0 x y x y + − + + = . The centre of the circle is at C and its radius is r . a) Determine … i. … the coordinates of C . ii. … the length of r . The point (4,4) P , lies on this circle. b) Find the gradient of CP . c) Hence find an equation of the tangent to the circle at P . C2D , ( ) 10, 4 C − , 10 r = , 4 3 − , 3 1 4 y x = + Created by T. Madas Created by T. Madas Question 10 (+) A circle C has equation 2 2 10 6 15 0 x y x y + − + − = a) Find the coordinates of the centre of C and determine the size of its radius. The circle intersects the x axis at the points A and B . b) Find, in exact surd form, the x coordinate of A and the x coordinate of B and hence state the distance AB . MP1-N , ( ) 5, 3 − , 7 r = , 5 2 10 x = ± , 4 10 AB = Created by T. Madas Created by T. Madas Question 11 (+) A circle C has equation 2 2 8 4 x y x y + = + . a) Determine the coordinates of the centre of C and the size of its radius. The circle meets the coordinate axes at the origin O and at two more points A and B . b) Find the coordinates of A and B . c) Sketch the graph of C . d) State with justification but without any further calculations the length of AB . C2O , ( ) 4,2 , 20 2 5 r = = , ( ) ( ) 8,0 , 0,4 , 4 5 AB = Created by T. Madas Created by T. Madas Question 12 () A circle whose centre is at ( ) 3, 5 − has equation 2 2 6 15 x y x ay + − + = , where a is a constant. a) Find the value of a . b) Determine the radius of the circle. C2J , 10 a = , 7 r = Created by T. Madas Created by T. Madas Question 13 () The endpoints of a diameter of a circle are located at ( ) 7,4 A − and ( ) 1, 2 B − . a) Find an equation for the circle. The straight line with equation 4 3 20 y x + = is a tangent to the circle at the point D . b) Find an equation for the straight line CD , where C is the centre of the circle. c) Determine the coordinates of D . C2F , ( ) ( ) 2 2 3 1 25 x y + + − = , 3 4 15 y x = + , ( ) 0,5 D Created by T. Madas Created by T. Madas Question 14 () The straight line joining the points ( ) 2,5 A and ( ) 2,9 B − is a diameter of a circle. a) Find an equation for this circle. b) Determine by calculation whether the point ( ) 1,5 P lies inside or outside the above mentioned circle. MP1-M , ( ) 2 2 7 8 x y + − = Question 15 () A circle has its centre at the point ( ) 2,3 C − and passes through the point ( ) 3,8 P − . a) Find an equation for this circle. b) Show that an equation of the tangent to the circle at P is 5 43 0 x y − + = . C2L , ( ) ( ) 2 2 2 3 26 x y + + − = Created by T. Madas Created by T. Madas Question 16 () A circle C has equation 2 2 6 10 0 x y x y k + − − + = , where k is a constant. a) Determine the coordinates of the centre of C . The x axis is a tangent to C at the point P . b) State the coordinates of P and find the value of k . SYN-M , ( ) 3,5 , ( ) 3,0 , 9 k = Created by T. Madas Created by T. Madas Question 17 () 2 2 2 2 8 x y x y + − − = The circle with the above equation has radius r and has its centre at the point C . a) Determine the value of r and the coordinates of C . The point ( ) 4,2 P lies on the circle. b) Show that an equation of the tangent to the circle at P is 14 3 y x = − C2Q , 10 r = , ( ) 1,1 C Created by T. Madas Created by T. Madas Question 18 () 2 2 10 4 9 0 x y x y + − + + = The circle with the above equation has radius r and has its centre at the point C . a) Determine the value of r and the coordinates of C . b) Find the coordinates of the points where the circle intersects the x axis. The point ( ) 3,2 P lies on the circle. c) Show that an equation of the tangent to the circle at P is 2 1 0 x y − + = . SYN-N , 20 r = , ( ) 5, 2 C − , ( ) ( ) 1,0 , 9,0 Created by T. Madas Created by T. Madas Question 19 () The points A and B have coordinates ( ) 3, 1 − and ( ) 9,7 , respectively. a) Show that the equation of the circle whose diameter is AB can be written as 2 2 12 6 20 0 x y x y + − − + = . The tangent to the circle at B meets the x axis at the point P . b) Determine the exact coordinates of P . ( ) 55,0 3 P Created by T. Madas Created by T. Madas Question 20 () A circle has equation 2 2 0 x y ax by + + + = , where a and b are constants. The straight lines with equations 4 y x = − and 2 x y + = are both diameters of this circle. Determine the length of the radius of the circle. SYN-P , 10 r = Created by T. Madas Created by T. Madas Question 21 () A circle C has its centre at the point with coordinates ( ) 5,4 and its radius is 3 2 . a) Find an equation for C . The straight line L has equation 1 y x = + . b) Determine, as exact surds, the coordinates of the points of intersection between C and L . c) Show that the distance between these points of intersection is 8 units. SYN-Q , ( ) ( ) 2 2 5 4 18 x y − + − = , ( ) ( ) 4 2 2,5 2 2 , 4 2 2,5 2 2 + + − − Created by T. Madas Created by T. Madas Question 22 (+) A circle has its centre at the point ( ) 2,5 C and its radius is 10 . a) Show that an equation for the circle is 2 2 4 10 19 0 x y x y + − − + = . The straight line with equation 5 y x = + meets the circle at the points P and Q . b) Determine the coordinates of P and the coordinates of Q . c) Show that the distance of the chord PQ from C is 2 units. C2P , ( ) ( ) 3,8 , 1,4 − Created by T. Madas Created by T. Madas Question 23 (+) A circle has its centre at the point ( ) 3,8 C − and the length of its diameter is 80 . a) State an equation for this circle. The straight line with equation 3 7 y x = + intersects the circle at the points A and B . b) Find the coordinates of A and the coordinates of B . c) Show that ACB is a right angle and hence determine the area of the triangle ACB . SYN-L , ( ) ( ) 2 2 3 8 20 x y + + − = , ( ) ( ) 1,4 , 1,10 − , area 10 = Created by T. Madas Created by T. Madas Question 24 (+) The points ( ) 2,5 P − and ( ) 6, 1 Q − lie on a circle so that the chord PQ is a diameter of this circle. a) Find an equation for this circle. The straight line with equation 6 y = intersects the circle at the points A and B . b) Determine the shortest distance of AB from the centre of the circle and hence, or otherwise, find the distance AB . SYN-S , ( ) ( ) 2 2 2 2 25 x y − + − = , 4 , 6 AB = Created by T. Madas Created by T. Madas Question 25 (+) A circle has equation 2 2 8 6 0 x y x y + − + = a) Find the coordinates of the centre of the circle. b) Determine the radius of the circle. The points A , B and C lie on the circle so that 10 AB = and 5 BC = . c) Find the distance of AC , giving the answer in the form 3 k , where k is a positive integer. C2M , ( ) 4, 3 − , 5 r = , 5 3 d = Created by T. Madas Created by T. Madas Question 26 (+) A circle has equation 2 2 8 14 40 0 x y x y + − − + = Find an equation of the tangent to the circle at the point ( ) 8,4 . 4 3 20 x y − = Created by T. Madas Created by T. Madas Question 27 (+) The points A , B and C have coordinates ( ) 3,0 − , ( ) 1,6 − and ( ) 11,2 , respectively. a) Show clearly that 90 ABC = ° ∡ . The points A , B and C lie on the circumference of a circle centred at the point D . b) Find an equation for this circle in the form 2 2 0 x y ax by c + + + + = , where a , b and c , are constants to be found. SYN-J , 2 2 8 2 33 0 x y x y + − − − = Created by T. Madas Created by T. Madas Question 28 (+) The points A and B have coordinates ( ) 1,2 − and ( ) 1,8 , respectively. a) Show that the equation of the perpendicular bisector of AB is 3 15 y x + = . The points A and B lie on a circle whose centre is at ( ) 3, C k . b) Determine an equation for the circle. SYN-I , ( ) ( ) 2 2 3 4 20 x y − + − = Created by T. Madas Created by T. Madas Question 29 (+) The figure above shows a circle whose centre is at ( ) 8, C k , where k is a constant. The straight line with equation 3 12 y x = − is a tangent to the circle at the point ( ) 5,3 A . a) Find an equation of the normal to the circle at A . b) Determine an equation for the circle. SYN-U , 3 14 x y + = , ( ) ( ) 2 2 8 2 10 x y − + − = A O C 3 12 y x = − y x Created by T. Madas Created by T. Madas Question 30 (+) A circle C has equation 2 2 14 14 49 0 x y x y + − − + = . a) Find the radius of the circle and the coordinates of its centre. b) Sketch the circle, indicating clearly all the relevant details. The point P has coordinates ( ) 15,8 . A tangent drawn from P touches the circle at the point Q . c) Determine the distance PQ . SYN-K , ( ) 7,7 , 7 r = , 4 PQ = Created by T. Madas Created by T. Madas Question 31 (+) A circle has equation 2 2 6 4 13 x y x y + − + = . a) Find the coordinates of its centre and the length of its radius. The point ( ) , 1 P k − , 0 k > , lies on the circle. b) Determine an equation for the tangent to the circle at P . ( ) 3, 2 , 26 r − = , 5 39 y x + = Created by T. Madas Created by T. Madas Question 32 (+) The point ( ) 12,9 P lies on the circle with equation ( ) ( ) 2 2 3 1 289 x y + + − = . a) Find an equation of the normal to the circle at P . b) Determine the coordinates of the point Q , where the normal to the circle at P intersects the circle again. SYN-V , 8 15 39 0 x y − + = , ( ) 18, 7 Q − − Created by T. Madas Created by T. Madas Question 33 (+) A circle has equation 2 2 10 8 21 0 x y x y + − − + = . a) Find the coordinates of the centre and the radius of the circle. b) Determine mathematically, but without solving any equations, whether the circle crosses the coordinate axes. c) Show that the straight line with equation 2 4 y x = + is a tangent to the circle, and determine the coordinates of the point where the tangent meets the circle. C2A , ( ) 5,4 , 2 5 r = , ( ) 1,6 Created by T. Madas Created by T. Madas Question 34 (+) A circle C has equation 2 2 4 4 8 24 5 0 x y x y + − + − = a) Find the coordinates of the centre of the circle. b) Determine the size of the radius of the circle, giving the answer in the form 5 k , where k is a rational constant. The point P has coordinates ( ) 8,11 . The straight line L passes through P and touches the circle at the point Q . c) Calculate the distance PQ . SYN-T , ( ) 1, 3 − , 3 5 2 r = , 1 935 15.29 2 ≈ Created by T. Madas Created by T. Madas Question 35 (+) The figure above shows a circle that passes through the points ( ) 1,10 A − and ( ) 7,4 B . a) Given that AB is a diameter of the circle show that an equation for this circle is given by 2 2 6 14 33 0 x y x y + − − + = . The tangent to the circle at B meets the y axis at the point D . b) Show that the coordinates of D are ( ) 16 0, 3 − . SYN-O , proof ( ) 1,10 A − ( ) 7,4 B O y x Created by T. Madas Created by T. Madas Question 36 (+) A circle passes through the points with coordinates ( ) 2,1 P − , ( ) 14,13 Q and ( ) 20, R k , where k is a constant. Given that 90 PQR = ° ∡ , determine an equation for the circle. SYN-W , ( ) ( ) 2 2 9 3 125 x y − + − = P O R Q y x Created by T. Madas Created by T. Madas Question 37 (+) The point ( ) 4,10 R lies on the curve C whose equation is 2 6 18 y x x = − + , x∈ . The tangent and the normal to C at R meet the y axis at the points Q and P , respectively, as shown in the figure above. a) Find the coordinates of Q and the coordinates of P . A circle passes through the points P , Q and R . b) Determine an equation for the circle. MPI-I , ( ) ( ) 0,12 , 0,2 P Q , ( ) 2 2 7 25 x y + − = P O R Q y x C tangent normal Created by T. Madas Created by T. Madas Question 38 (+) A circle 1 C has equation 2 2 4 12 4 0 x y x y + − + + = a) Determine the coordinates of the centre and the radius of 1 C . The circle 2 C with equation 2 2 2 x y a + = , 0 a > touches 1 C externally. b) Find the value of a as an exact surd. SYN-X , ( ) 2, 6 , 6 r − = , 6 2 10 a = −+ Created by T. Madas Created by T. Madas Question 39 (+) A circle has centre at the origin and radius R . This circle fits wholly inside the circle with equation 2 2 10 24 231 x y x y + − − = . Determine the range of possible values of R . MP1-E , 7 R ≤ Created by T. Madas Created by T. Madas Question 40 (+) A circle C passes through the points ( ) 1,6 P and ( ) 12,5 Q . a) Find the gradient of PQ . The centre of C is the point R which lies on the x axis. b) Show that the coordinates of R are ( ) 6,0 . c) Determine an equation for C . C2H , 1 11 − , ( ) ( ) 2 2 6 0 61 x y − + − = P O R Q y x Created by T. Madas Created by T. Madas Question 41 (+) The points A , B and C have coordinates ( ) 1, 4 −− , ( ) 0,3 and ( ) 14,1 , respectively. a) Find an equation of the straight line which passes through B and C , giving the answer in the form ax by c + = , where a , b and c are integers. b) Show that AB is perpendicular to BC . A circle is passing through the points A , B and C . c) Determine the coordinates of the centre of this circle. d) Show that the radius of this circle is 10 k , where k is a rational number. SYN-H , 7 21 y x + = , ( ) 13 3 , 2 2 − , 5 2 k = Created by T. Madas Created by T. Madas Question 42 (+) A circle C has equation 2 2 6 14 33 0 x y x y + − + + = a) Determine the coordinates of the centre and the radius of C . b) Show that the circle lies entirely below the x axis. The point ( ) 6, P k , where k is a constant, lies outside the circle. c) By considering the distance of P from the centre of the circle, or otherwise, determine the range of the possible values of k . SYN-G , ( ) 3, 7 , 5 r − = , 11 or 3 k k < − > − Created by T. Madas Created by T. Madas Question 43 (+) The points ( ) 1,0 A , ( ) 8,7 B and ( ) 7,8 C lie on the circumference of a circle. a) Show that AB is perpendicular to BC . b) Find the distance AC . c) Show that an equation of the circle is ( ) ( ) 2 2 4 4 25 x y − + − = . The tangent to the circle at the point C crosses the x axis and the y axis at the points P and Q , respectively. d) Determine the exact coordinates of P and Q . 10 AC = , ( ) ( ) 53 53 ,0 , 0, 3 4 P Q Created by T. Madas Created by T. Madas Question 44 (+) A circle with centre at the point C has equation 2 2 10 6 14 0 x y x y + − − + = . The straight line with equation y k = , where k is a non zero constant, is a tangent to this circle a) Find the possible values k , giving the answers as exact simplified surds. The points A and B lie on the circumference of the circle and the point M is the midpoint of the chord AB . b) Given the length of MC is 2 , find the length of the chord AB . The straight line with equation 2 9 0 x y − − = is a tangent to the circle at the point D . c) Determine the coordinates of D . SYN-F , 3 2 5 k = ± , 8 AB = , ( ) 7, 1 D − Created by T. Madas Created by T. Madas Question 45 (+) The figure above shows a circle with centre at C with equation 2 2 10 12 56 0 x y x y + − − + = . The tangent to the circle at the point ( ) 6,4 A meets the y axis at the point B . a) Find an equation of the tangent to the circle at A . b) Determine the area of the triangle ABC . SYN-D , 1 1 2 y x = + , 15 2 A B O y x C Created by T. Madas Created by T. Madas Question 46 (+) A circle C has equation 2 2 12 2 33 0 x y x y + − − + = . a) Find the radius of the circle and the coordinates of its centre. The straight line with equation 3 y x = − intersects the circle at the points P and Q , dividing the circle into two segments. b) Determine the coordinates of P and Q . c) Show that the area of the minor segment is 2 π − . SYN-A , ( ) 2, 6,1 r = , ( ) ( ) 4,1 , 6,3 Created by T. Madas Created by T. Madas Question 47 (+) The points A , B and C have coordinates ( ) 2,1 , ( ) 4,0 and ( ) 6,4 respectively. a) Determine an equation of the straight line L which passes through C and is parallel to AB . b) Show that the angle ABC is 90° . c) Calculate the distance AC . A circle passes through the points A , B and C . d) Show that the equation of this circle is given by 2 2 8 5 16 0 x y x y + − − + = . e) Find the coordinates of the point other than the point C where L intersects the circle. MP1-Q , 2 14 x y + = , 5 AC = , ( ) 4,5 Created by T. Madas Created by T. Madas Question 48 (+) The straight line l passes through the points ( ) 7,10 P and ( ) 17,3 Q . a) Find an equation of l . b) Show that OP is perpendicular to PQ , where O is the origin. A circle passes through O , P and Q . c) Find an equation for this circle. 7 10 149 x y + = , ( ) ( ) 2 2 17 3 298 2 2 4 x y − + − = Created by T. Madas Created by T. Madas Question 49 (+) A circle C with centre at the point P and radius r , has equation 2 2 8 2 0 x x y y − + − = . a) Find the value of r and the coordinates of P . b) Determine the coordinates of the points where C meets the coordinate axes. The points A , B and ( ) 8,2 Q lie on C . The straight line AB is diameter of the circle so that PQ is perpendicular to AB . c) Calculate the coordinates of A and B . MP1-A , 17 r = , ( ) 4,1 P , ( ) ( ) ( ) 8,0 , 0,0 , 0,2 , ( ) 3,5 A , ( ) 5, 3 B − Created by T. Madas Created by T. Madas Question 50 (+) The points ( ) 1,2 A , ( ) 3,8 B and ( ) 13,8 C are shown in the figure below. The straight lines 1 l and 2 l are the perpendicular bisectors of straight line segments AB and BC , respectively. a) Find an equation for 1 l . Give the answer in the form ax by c + = , where a , b and c are integers. The point D is the intersection of 1 l and 2 l . b) Show that the coordinates of D are ( ) 8,3 . A circle passes through the points A , B and C . c) Find an equation for this circle. 3 17 x y + = , ( ) ( ) 2 2 8 3 50 x y − + − = 1 l 2 l O A B C D y x Created by T. Madas Created by T. Madas Question 51 () The figure above shows a circle with centre at P and radius of 6 units. The y axis is a tangent to the circle at the point ( ) 0,3 S . a) Find an equation for the circle. A tangent to the circle is drawn from the point ( ) 12,10 Q and meets the circle at the point R . b) Determine the length of QR . C2G , ( ) ( ) 2 2 6 3 36 x y − + − = , 7 QR = P O Q y x S Created by T. Madas Created by T. Madas Question 52 () The points A , B and C have coordinates ( ) 3,5 , ( ) 15,11 and ( ) 17,7 , respectively. a) Show that 90 ABC = ° ∡ . All three points A , B and C lie on the circumference of a circle. b) Find an equation for this circle in the form 2 2 0 x y ax by c + + + + = , where a , b and c , are integers to be found. The point P also lies on this circle so that BP is a diameter of the circle. c) Determine the coordinates of P . 2 2 20 12 86 0 x y x y + − − + = , ( ) 5,1 P Created by T. Madas Created by T. Madas Question 53 () The circle 1 C has centre at ( ) 8,4 and touches the y axis. The circle 2 C has centre at ( ) 16,4 and touches the x axis. a) Find the equation of 1 C and the equation of 2 C . Give the answers in the form ( ) ( ) 2 2 x a y b c − + − = , where a , b and c are constants to be found. The two circles intersect at the points A and B . b) Determine, in exact surd form where appropriate, the coordinates of A and the coordinates of B . MP1-P , ( ) ( ) 2 2 8 4 64 x y − + − = , ( ) ( ) 2 2 16 4 16 x y − + − = , ( ) ( ) 15,4 15 , 15,4 15 + − Created by T. Madas Created by T. Madas Question 54 () The points ( ) 4,1 P − and ( ) 5,10 Q lie on a circle C . a) Find an equation of the perpendicular bisector of PQ . b) Given that one of the diameters of C has equation 2 17 x y + = , show that an equation of C is 2 2 10 22 45 0 x y x y + + − + = . c) Determine whether the origin O lies inside the circle. 6 x y + = Created by T. Madas Created by T. Madas Question 55 () A circle C has equation 2 2 43 0 x y ax by + + + + = , where a and b are constants. a) Given that the points ( ) 4,7 − and ( ) 2,5 − lie on C , determine the coordinates of the centre of C and the size of its radius. A straight line passes through the point ( ) 4,5 P and is a tangent to C at the point Q . b) Show that the length of PQ is 4 3 . MP1-J , ( ) 3,6 , 2 r − = Created by T. Madas Created by T. Madas Question 56 () A circle 1 C has equation 2 2 10 10 41 0 x y x y + − − + = . a) Determine the coordinates of the centre of 1 C and the size of its radius. Another circle 2 C is such so that 2 C is touching both 1 C and the y axis. b) Find the two possible equations of 2 C , given further that the centres of both 1 C and 2 C , have the same y coordinate. C2N , ( ) 5,5 , 3 r = , ( ) ( ) ( ) ( ) 2 2 2 2 1 5 1 or 4 5 16 x y x y − + − = − + − = Created by T. Madas Created by T. Madas Question 57 () A circle C has equation 2 2 4 12 15 0 x y x y + − − + = . a) Show that C does not cross the x axis. b) Show further that the straight line with equation 3 4 5 x y + = is a tangent to C . proof Created by T. Madas Created by T. Madas Question 58 () The figure above shows a circle with centre at ( ) 7,2 C . The circle meets the x axis at the points P and R , and the y axis at the points Q and S . a) Given that ( ) 17,0 R , find an equation of this circle. b) Show that 2 55 QS = . c) Determine the area of the quadrilateral PQRS . C2E , ( ) ( ) 2 2 7 2 104 x y − + − = , 20 55 148.32 ≈ P O R Q y x S Created by T. Madas Created by T. Madas Question 59 () A circle is given parametrically by the equations 4 3cos x θ = + , 3 3sin y θ = + , 0 2 θ π ≤ < . a) Find a Cartesian equation for the circle. b) Find the equations of each of the two tangents to the circle, which pass through the origin O . ( ) ( ) 2 2 4 3 9 x y − + − = , 24 0 and 7 y y x = = Created by T. Madas Created by T. Madas Question 60 () The figure above shows a circle meeting the x axis at ( ) 5,0 P − and ( ) 13,0 Q , and the y axis at ( ) 0, 4 R − and ( ) 0,16 S . a) Show that an equation of the circle is 2 2 x y ax by c + + + = , where a , b and c are constants to be found. The point A has coordinates ( ) 20,12 . A tangent drawn through A meets the circle at the point B . b) Show that the distance AB is 7 k , where k is an integer. 8, 12, 65 a b c = − = − = , 5 k = P O R Q y x S Created by T. Madas Created by T. Madas Question 61 () Relative to a fixed origin O , the points A and B have coordinates ( ) 0,1 and ( ) 6,5 , respectively. a) Find an equation of the perpendicular bisector of AB . A circle passes through the points O , A and B . b) Determine the coordinates of the centre of this circle. MP1-K , 3 2 15 x y + = , ( ) 14 1 , 3 2 Created by T. Madas Created by T. Madas Question 62 () The figure above shows the circle C and the straight line L with respective equations 2 2 6 8 21 0 x y x y + − − + = and 2 2 x y + = . a) Find an equation of the line which passes through the centre of C and is perpendicular to L . b) Hence determine, in exact surd form, the shortest distance between C and L . MPI-H , 2 2 y x = − , 9 5 2 5 − O L y x C Created by T. Madas Created by T. Madas Question 63 () A circle has centre at ( ) 4,4 C and passes through the point ( ) 6,8 P . The straight line 1 l is a tangent to the circle at P . a) Show that an equation of 1 l is 2 22 x y + = . The straight line 2 l has equation 2 14 y x = − and meets 1 l at the point Q . b) Find the coordinates of Q . c) Show that 2 l is also a tangent to this circle at the point R , and determine the coordinates of R . MP1-G , ( ) 10,6 Q , ( ) 8,2 R Created by T. Madas Created by T. Madas Question 64 () A circle 1 C has equation 2 2 20 2 52 0 x y x y + + − + = . a) Determine the coordinates of the centre of 1 C and the size of its radius. A different circle 2 C has its centre at ( ) 14,8 and the size of its radius is 10. The point P lies on 1 C and the point Q lies on 2 C . b) Determine the minimum distance of PQ . SYN-C , ( ) 10,1 , 7 r − = , min 8 PQ = Created by T. Madas Created by T. Madas Question 65 () The figure above shows a square PQRS with vertices at the points ( ) 1,1 P , ( ) 3,5 Q , ( ) 7,3 R and ( ) 5, 1 S − . The square is circumscribed by the circle 1 C . a) Determine the coordinates of the centre of 1 C and the size of its radius. A circle 2 C is inscribed in the square PQRS . b) Find an equation of the circle 2 C . C2I , 10 r = , ( ) 4,2 , ( ) ( ) 2 2 4 2 5 x y − + − = P O Q R S 1 C 2 C x y Created by T. Madas Created by T. Madas Question 66 () A circle has equation 2 2 4 4 6 0 x y x y + − − + = . a) Determine the coordinates of the centre and the radius of the circle. The point ( ) 9,1 A lies outside the circle. b) Find the shortest distance from A to the circle, giving the answer as a surd. A tangent is drawn from the point A to the circle, touching the circle at the point B . c) Determine, as an exact surd, the distance AB . ( ) 2,2 , 2 r = , 4 2 , 4 3 AB = Created by T. Madas Created by T. Madas Question 67 () A circle has equation ( ) ( ) 2 2 2 2 20 x y − + + = . a) Write down the coordinates of its centre the size of its radius. b) Sketch the circle. The sketch must include the coordinates of any points where the graph meets the coordinate axes. The straight line with equation 2 y x k = + , where k is a constant, meets the circle. c) Show that the coordinates of any points of intersection between the line and the circle satisfies the equation ( ) 2 2 5 4 1 4 12 0 x k x k k + + + + − = . d) Hence, find the two values of k for which the line 2 y x k = + is a tangent to the circle. MP1-E , ( ) 2, 2 , 2 5 r − = , 16 4 k k = − = ∪ Created by T. Madas Created by T. Madas Question 68 () A circle has centre at ( ) 2,3 C and radius 6 . a) Show that an equation of the circle is 2 2 4 6 23 x y x y + − − = . The circle crosses the y axis at the points P and Q , where the y coordinate of P is positive. b) Find the distance PQ , giving the answer as an exact simplified surd. The vertical straight line with equation 1 x = intersects the radii CP and CQ at the points R and S , respectively. c) Determine the exact area of the trapezium PQSR . MP1-F , 8 2 PQ = , area 6 2 = Created by T. Madas Created by T. Madas Question 69 () The points ( ) 96 40 , 13 13 A and ( ) 216 90 , 13 13 B are the endpoints of the diameter of circle. a) Show that an equation of the circle is ( ) ( ) 2 2 12 5 25 x y − + − = . b) Sketch the circle, indicating clearly all the relevant details. c) Show that a normal to the circle at B passes through the origin. proof Created by T. Madas Created by T. Madas Question 70 () A circle 1 C has equation 2 2 10 4 71 x y x y + − + = . a) Find the coordinates of the centre of 1 C and the size of its radius. Another circle 2 C is centred at the point with coordinates ( ) 4,10 − and has radius 5. b) Show that 1 C and 2 C touch each other. The two circles touch each other at the point P and the straight line L is the common tangent of 1 C and 2 C . c) Determine the coordinates of P . d) Show that an equation of L is 4 3 27 y x = + . MP1-D , ( ) 5, 2 − , 10 r = , ( ) 1,6 − Created by T. Madas Created by T. Madas Question 71 () The figure above shows a circle that crosses the x axis at the points ( ) 1,0 A − and ( ) 7,0 B , while it passes through the point ( ) 3,8 C . Determine the coordinates of the centre of the circle and the length of its radius. MP1-C , ( ) 3,3 , radius 5 = A B y x O C Created by T. Madas Created by T. Madas Question 72 () The figure above shows the circle C with equation 2 2 14 33 0 x y x + − + = . a) Determine the coordinates of the centre of C and the size of its radius. The tangents to C from the point ( ) 7,8 R meet C at the points P and Q . b) Show that the area of the finite region bounded by C and the two tangents, shown shaded in figure, is 16 3 3 3 π   −  . SYN-E , ( ) 7,0 , 4 r = O ( ) 7,8 R y x C P Q Created by T. Madas Created by T. Madas Question 73 () A circle passes through the points ( ) 1,2 A , ( ) 3, 4 B − and ( ) 15,0 C . a) Show that AB is perpendicular to BC . b) Hence find the exact length of the diameter of the circle. c) Show that an equation of the circle is 2 2 16 2 15 0 x y x y + − − + = . d) Determine the coordinates of the point which lies on the circle and is furthest away from the point B . diameter 10 2 = , ( ) 13,6 Created by T. Madas Created by T. Madas Question 74 () A circle has equation 2 2 8 x y y + = . a) Find the coordinates of the centre of the circle and the size of its radius. b) Sketch the circle. The line with equation x y k + = , where k is a constant, is a tangent to this circle. c) Determine, as exact surds, the possible values of k . MP1-L , ( ) 0,4 , 4 r = , 4 4 2 k = ± Created by T. Madas Created by T. Madas Question 75 () A circle passes through the points ( ) 0,0 , ( ) 8,0 and ( ) 0,6 . Determine the coordinates of the centre of the circle and the size of its radius. C2Z , ( ) 4,3 , 5 r = Created by T. Madas Created by T. Madas Question 76 () The figure above shows the circle with equation 2 2 4 12 x y x + − = . The circle has centre at C and radius r a) Find the coordinates of C and the value of r . The circle crosses the coordinate axes at the points P , Q , R and S , as shown in the figure above. b) Show that … i. … 2 3 PCQ π = ∡ . ii. … the area of the shaded region bounded by the circle and the positive sections of the coordinate axes is ( ) 2 8 3 3 3 π + . C2Y , ( ) 2,0 , 4 r = 2 2 4 12 x y x + − = C S R Q O P y x Created by T. Madas Created by T. Madas Question 77 () A circle has centre at ( ) 3, 8 C − and radius of 10 units. The tangent to the circle at the point A has gradient 1 −. Determine, as exact surds, the possible x coordinates of A . You may not use a calculus method in this question C2X , 3 5 2 x = ± Created by T. Madas Created by T. Madas Question 78 () A circle C has equation 2 2 4 10 9 0 x y x y + + − + = . a) Find the coordinates of the centre of C and the size of its radius. A tangent to the circle T , passes through the point with coordinates ( ) 0, 1 − and has gradient m , where 0 m < . b) Show that m is a solution of the equation 2 2 3 2 0 m m − − = . The tangent T meets C at the point P . c) Determine the coordinates of P . MP1-B , ( ) 2,5 , 20 r − = , ( ) 4,1 P − Created by T. Madas Created by T. Madas Question 79 () A circle has equation 2 2 8 33 x y x cy + − + = , where c is a positive constant. The straight line L , with equation 2 12 y x = − , intersects the circle at the point with coordinates ( ) 9,k . Find, as an exact surd, the perpendicular distance of L from the centre of the circle. SYN-R , 5 D = Created by T. Madas Created by T. Madas Question 80 () The figure above shows the circle with equation ( ) 2 2 9 3 25 225 x y + − = whose centre is at C and its radius is r . a) Determine the coordinates of C and the value of r . The points ( ) 16 4, 3 A and ( ) 16 4, 3 B − lie on the circle. The straight lines 1 l and 2 l are tangents to the circle at A and B , respectively. b) Show that 1 l passes through the origin O . c) Show further that the angle BCA is approximately 1.8546 radians. d) Calculate the area of the shaded region, bounded by the circle, 1 l and 2 l . C2V , ( ) 25 0, 3 C , 5 r = , area 10.15... ≈ x A B y C O 2 l 1 l Created by T. Madas Created by T. Madas Question 81 () The circles 1 C and 2 C have respective equations 2 2 6 2 15 x y x y + − − = 2 2 18 14 95 x y x y + − + = . a) By considering the coordinates of the centres and the lengths of the radii of 1 C and 2 C , show that 1 C and 2 C touch internally at some point P . b) Determine the coordinates of P . c) Show that the equation of the common tangent to the circles at P is given by 3 4 20 0 x y − + = . MP1-O , ( ) 0,5 P Created by T. Madas Created by T. Madas Question 82 () A circle has equation 2 2 4 6 8 0 x y x y + − − + = . The straight line 1 T is a tangent to the circle at the point ( ) 4,4 P . a) Find an equation of 1 T . The tangent 1 T passes through the point ( ) 2,8 Q . The straight line 2 T is a tangent to the circle at the point R and it also passes through the point Q . b) Determine in any order i. … the coordinates of R . ii. … an equation of 2 T . SYN-B , 2 12 y x = − + , ( ) 0,4 R , 2 4 y x = + , Created by T. Madas Created by T. Madas Question 83 (+) The figure above shows a circle C centred at the point with coordinates ( ) 5,6 , and the straight line L which passes though the points ( ) 1,8 A and ( ) 10,11 B . Given that L is a tangent to C , determine the radius of C . [You may not use a standard formula which finds the shortest distance of a point from a line] C2R , 10 r = A B O y x L C Created by T. Madas Created by T. Madas Question 84 (+) Determine the exact coordinates of the points of intersection between the circles with equations ( ) ( ) ( ) ( ) 2 2 2 2 1 2 4 3 1 9 x y x y + + − = + + + = ( ) ( ) 3 2 3,2 , , 13 13 − − Created by T. Madas Created by T. Madas Question 85 (+) The points A , B and C have coordinates ( ) 6,6 , ( ) 0,8 and ( ) 2,2 − , respectively. a) Find an equation of the perpendicular bisector of AB . The points A , B and C lie on the circumference of a circle whose centre is located at the point D . b) Determine the coordinates of D . MP1-W , 3 2 y x = − , ( ) 2,4 D Created by T. Madas Created by T. Madas Question 86 (+) The figure above shows three circles 1 C , 2 C and 3 C . The coordinates of the centres of all three circles are positive. • • • • The circle 1 C has centre at ( ) 6,6 and touches both the x axis and the y axis. • • • • The circle 2 C has the same size radius as 1 C and touches the x axis. • • • • The circle 3 C touches the x axis and both 1 C and 2 C . Determine an equation of 3 C . C2V , ( ) ( ) 2 2 3 9 12 2 4 x y − + − = 1 C O 2 C y x 3 C Created by T. Madas Created by T. Madas Question 87 (+) The circles 1 C and 2 C have respective equations 2 2 6 16 x y x + − = 2 2 18 16 80 x y x y + − + = . a) By solving these equations simultaneously show that 1 C and 2 C touch at a point P and determine its coordinates. b) Determine further whether 1 C and 2 C touch internally or externally. MP1-Z , ( ) 0,4 P Created by T. Madas Created by T. Madas Question 88 (+) Two circles have equations 2 2 8 2 13 0 x y x y + − − + = , 2 2 2 2 1 x y x y k + − − + = , where k is a constant. a) Find the values of k , for which the two circles touch each other. b) Hence state the range of values of k , for which the two circles intersect each other at exactly two points. MP1-V , 0, 24 k k = = , 0 24 k < < Created by T. Madas Created by T. Madas Question 89 (+) A circle has centre at ( ) 6,2 C and radius of 4 units. The point ( ) 6 2 2, P k + lies on this circle, where k is a positive constant. a) Determine the exact value of k . The straight line 1 T is the tangent to the circle at the point P . The straight line 2 T is another tangent to the circle so that 2 T is parallel to 1 T . b) Determine the equations of 1 T and 2 T . MP1-X , 2 2 2 k = + , 1 : 8 4 2 T x y + = + , 2 : 8 4 2 T x y + = − Created by T. Madas Created by T. Madas Question 90 (+) A circle has equation 2 2 4 2 13 x y x y + − − = . a) Find the coordinates of the centre of the circle and the size of its radius. The points A and B lie on the circle such that the length of AB is 6 units. b) Show that 90 ACB = ° ∡ , where C is the centre of the circle. A tangent to the circle has equation y k x = − , where k is a constant. c) Show clearly that ( ) 2 2 2 2 1 2 13 0 x k x k k − + + − − = . d) Determine the possible values of k . MP1-R , ( ) 2,1 , 18 r = , 3, 9 k k = − = Created by T. Madas Created by T. Madas Question 91 (+) The straight line passing through the points ( ) 1,9 P and ( ) 5,5 Q is a tangent to a circle with centre at ( ) 6,8 C . Determine, in exact surd form, the radius of the circle. In this question you may not use … … a standard formula which determines the shortest distance of a point from iiii a straight line. … any form of calculus. SYN-Y , 8 r = Created by T. Madas Created by T. Madas Question 92 (+) The straight line with equation 2 3 y x = − is a tangent to a circle with centre at the point ( ) 2, 3 C − . Determine, in exact surd form, the radius of the circle. In this question you may not use … … a standard formula which determines the shortest distance of a point from iiii a straight line. … any form of calculus. MP1-U , 4 5 5 r = Created by T. Madas Created by T. Madas Question 93 (+) The figure above shows a circle with centre at ( ) 3,6 C . The points ( ) 1,5 A and ( ) , B p q lie on the circle. The straight lines AD and BD are tangents to the circle. The kite CADB is symmetrical about the straight line with equation 3 x = . a) Calculate the radius of the circle. b) State the value of p and the value of q . c) Find an equation of the tangent to the circle at A . d) Show that the angle ACB is approximately 2.214 radians. e) Hence determine, to three significant figures, the area of the shaded region bounded by the circle and its tangents at A and B . C2U , 5 r = , 5 p q = = , 7 2 y x = − , area 4.46 ≈ B A x O y C D Created by T. Madas Created by T. Madas Question 94 (+) A circle C has equation 2 2 2 4 1 0 x y x y + + − + = The straight line L with equation y mx = is a tangent to C . Find the possible values of m and hence determine the possible coordinates at which L meets C . MP1-Y , 4 0, 3 m m = = , ( ) ( ) 3 4 1,0 , , 5 5 − Created by T. Madas Created by T. Madas Question 95 () The figure above shows a circle whose centre is located at the point ( ) , C k h , where k and h are constants such that 2 5 h < < . The points ( ) 3,2 A and ( ) 7,8 B lie on this circle. The straight line l passes through C and the midpoint of AB . Given that the radius of the circle is 26 , find an equation for l , the value of k and the value of h . MP1-S , 2 3 25 x y + = , 8 k = , 3 h = A O B l y x C Created by T. Madas Created by T. Madas Question 96 () The figure below shows the circle with centre at (5,10) C and radius 5. The straight lines with equations, 0 x = and 3 4 y x = are tangents to the circle at the points P and Q respectively. Show that the area of the triangle PCQ is 10 square units. SP-B , proof 3 4 y x = Q P ( ) 5,10 C y x O Created by T. Madas Created by T. Madas Question 97 () The circle 1 C has equation 2 2 4 4 6 0 x y x y + − − + = . The circle 2 C has equation 2 2 10 10 0 x y x y k + − − + = , where k is a constant. Given that 1 C and 2 C have exactly two common tangents, determine the range of possible values of k . SP-Y , 18 42 k < < Created by T. Madas Created by T. Madas Question 98 () A circle passes through the points ( ) 1 1 , A x y and ( ) 2 2 , A x y . Given that AB is a diameter of the circle, show that the equation of the circle is ( )( ) ( )( ) 1 2 1 2 0 x x x x y y y y − − + − − = . SP-T , proof Question 99 () A circle passes through the points ( ) 18,0 P and ( ) 32,0 Q . A tangent to this circle passes through the point ( ) 0,199 S and touches the circle at the point T . Given that the y axis is a tangent to this circle, determine the coordinates of T SP-O , ( ) 49,31 Created by T. Madas Created by T. Madas Question 100 () non calculator The figure above shows the circle with equation 2 2 4 8 205 x y x y + − − = , with centre at the point C and radius r . The straight line AB is parallel to the y axis and has length 24 units. The tangents to the circle at A and B meet at the point D . Find the length of AD and hence deduce the area of the kite CADB . SP-E , 20 AD = , area 300 = O B A C y x D Created by T. Madas Created by T. Madas Question 101 () The figure above shows a circle 1 C with equation 2 2 18 90 0 x y x ky + − + + = , where k is a positive constant. a) Determine, in terms of k , the coordinates of the centre of 1 C and the size of its radius. Another circle 2 C has equation 2 2 2 2 34 x y x y + − − = . b) Given that 1 C and 2 C are touching externally at the point P , find … i. … the value of k . ii. … the coordinates of P . C2T , ( ) 2 1 9, , 9 2 4 k k r − = − , 10 k = , ( ) 29 13 , 5 5 P − 1 C 2 C x O y P Created by T. Madas Created by T. Madas Question 102 () The curve C has equation 2 4 7 y x x = − + . The points ( ) 1,12 P − and ( ) 4,7 Q lie on C . The point R also lies on C so that 90 PRQ = ° ∡ . Determine, as exact surds, the possible coordinates of R . C2S , ( ) ( ) 5 21 17 21 5 21 17 21 , or , 2 2 2 2 + + − − Created by T. Madas Created by T. Madas Question 103 () A circle C is centred at ( ) , a a and has radius a , where a is a positive constant. The straight line L has equation 4 3 4 0 x y − + = . Given that L is tangent to C at the point P , determine … a) … an equation of C . b) … the coordinates of P . You may not use a standard formula which determines the shortest distance of a point from a straight line in this question. SP-W , ( ) ( ) 2 2 1 1 1 x y − + − = , ( ) 8 1 , 5 5 P Created by T. Madas Created by T. Madas Question 104 () The circles 1 C and 2 C have respective equations ( ) ( ) 2 2 9 1 2 4 x y + + + = and ( ) ( ) 2 2 5 6 36 x y − + − = . The point P lies on 2 C so that the distance of P from 1 C is least. Determine the exact coordinates of P . SP-P , ( ) 7 6 , 5 5 P Created by T. Madas Created by T. Madas Question 105 () A curve in the -x y plane has equation 2 2 6 cos 18 sin 45 0 x y x y θ θ + + − + = , where θ is a parameter such that 0 2 θ π ≤ < . Given that curve represents a circle determine the range of possible values of θ . SP-I , { } { } 3 5 7 1 4 4 4 4 π θ π π θ π < < < < ∪ Created by T. Madas Created by T. Madas Question 106 () Two circles, 1 C and 2 C , have respective radii of 4 units and 1 unit and are touching each other externally at the point A . The coordinates axes are tangents to 1 C , whose centre P lies in the first quadrant. The x axis is a tangent to 2 C , whose centre Q also lies in the first quadrant. The straight line 1 l , passes through P and Q , and meets the x axis at the point R . The straight line 2 l has negative gradient, passes through R and is a common tangent to 1 C and 2 C . Determine, in any order and in exact form where appropriate, the coordinates of A , the length of PR and an equation of 2 l . SPX-O , ( ) 36 8 , 5 5 A , 20 3 PR = , 24 7 224 x y + = Created by T. Madas Created by T. Madas Question 107 () A family of circles is passing through the points with coordinates ( ) 2,1 and ( ) 4,5 Show that the equation of every such circle has equation ( ) 2 2 2 2 9 2 6 41 x y x k ky k + + − + = − , where k is a parameter. SP-H , proof Created by T. Madas Created by T. Madas Question 108 () The straight line with equation ( ) 2 y t x = − , where t is a parameter, crosses the circle with equation 2 2 1 x y + = at two distinct points A and B . a) Show that the coordinates of the midpoint of AB are given by 2 2 2 2 2 , 1 1 t t M t t   −     + +   . b) Hence show that the locus of M as t varies is a circle, stating its radius and the coordinates of its centre. SYN-Z , ( ) 2 2 1 1 x y − + = Created by T. Madas Created by T. Madas Question 109 () The straight line L and the circle C , have respective equations ( ) 2 : 1 L y x a a λ λ = − + + and 2 2 : 2 C x y ax + = , where a is a positive constant and λ is a parameter. Show that for all values of λ , L is a tangent to C . SPX-L , proof Created by T. Madas Created by T. Madas Question 110 () Two parallel straight lines, 1 L and 2 L , have respective equations 2 5 y x = + and 2 1 y x = − . 1 L and 2 L , are tangents to a circle centred at the point C . A third line 3 L is perpendicular to 1 L and 2 L , and meets the circle in two distinct points, A and B . Given that 3 L passes through the point ( ) 9,0 , find, in exact simplified surd form, the coordinates of C . SP-Q , ( ) ( ) 1 1 5 61 , 15 61 5 10 C   + +   Created by T. Madas Created by T. Madas Question 111 () Three circles, 1 C , 2 C and 3 C , have their centres at A , B and C , respectively, so that 5 AB = , 4 AC = and 3 BC = . The positive x and y axis are tangents to 1 C . The positive x axis is a tangent to 2 C . 1 C and 2 C touch each other externally at the point M . Given further that 3 C touches externally both 1 C and 2 C , find, in exact simplified form, an equation of the straight line which passes through M and C . SPX-E , 5 10 6 36 30 6 0 y x − + + = Created by T. Madas Created by T. Madas Question 112 () Two circles, 1 C and 2 C , are touching each other externally, and have respective radii of 9 and 4 units. A third circle 3 C , of radius r , touches 1 C and 2 C externally. Given further that all three circles have a common tangent, determine the value of r . SPX-N , 36 1.44 25 r = = Created by T. Madas Created by T. Madas Question 113 () The point ( ) 6, 1 A − lies on the circle with equation 2 2 4 6 7 x y x y + − + = . The tangent to the circle at A passes through the point P , so that the distance of P from the centre of the circle is 65 . Another tangent to the circle, at some point B , also passes through P . Determine in any order the two sets of the possible coordinates of P and B . MP1-T , ( ) ( ) ( ) ( ) 18 30 97 1 3,5 , 9, 7 , 13 13 13 13 P B P B ∩ − − − ∩ − ∪
16606	https://brainly.com/question/48389775	[FREE] If the sum of the ages of A, B, and C is 60 years, and A is half the age of C, while B is the age of C - brainly.com 4 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +44,8k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +44,8k Ace exams faster, with practice that adapts to you Practice Worksheets +8k Guided help for every grade, topic or textbook Complete See more / Mathematics Expert-Verified Expert-Verified If the sum of the ages of A, B, and C is 60 years, and A is half the age of C, while B is the age of C minus the age of A, what are the ages of A, B, and C? a) A = 15, B = 30, C = 45 b) A = 20, B = 30, C = 40 c) A = 25, B = 30, C = 35 d) A = 30, B = 30, C = 30 1 See answer Explain with Learning Companion NEW Asked by Babygirlslaying2611 • 02/26/2024 0:00 / -- Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 4504542 people 4M 0.0 0 Upload your school material for a more relevant answer After performing calculations based on the equations derived from the conditions provided, A is 15 years old, B is 15 years old, and C is 30 years old. The correct answer is not listed in the provided options. Explanation To solve the problem we are given three conditions for the ages of A, B, and C: The sum of the ages of A, B, and C is 60 years. A is half the age of C. B is the age of C minus the age of A. We can write down these conditions as equations: A + B + C = 60 (Equation 1). A = ½C (Equation 2). B = C - A (Equation 3). Substitute the value of A from Equation 2 into Equation 3 to find B: B = C - ½C = ½C (Equation 3 becomes B = ½C). Now substitute both A and B in terms of C into Equation 1: ½C + ½C + C = 60. Combine like terms: 2C = 60. Divide each side by 2: C = 30. Now that we know C's age, A's age would be half of C: A = ½ 30 = 15. And B's age is C minus A: B = 30 - 15 = 15. We have determined that A is 15, B is 15, and C is 30. Therefore, the correct option is not listed among the choices provided in the question. There might be a mistake in the question or the available options. However, based on the conditions given and the calculations performed, the ages of A, B, and C would be 15, 15, and 30 respectively. Answered by cnbhemant •21.1K answers•4.5M people helped Thanks 0 0.0 (0 votes) Expert-Verified⬈(opens in a new tab) This answer helped 4504542 people 4M 0.0 0 Upload your school material for a more relevant answer The ages are A = 15, B = 15, and C = 30. None of the provided answer options match these ages, which suggests a potential error in the options. Therefore, the correct ages are calculated based on the conditions given in the question. Explanation To find the ages of A, B, and C based on the information provided, we can set up a system of equations. We are given the following conditions: The sum of the ages of A, B, and C is 60 years: A+B+C=60 2. A is half the age of C: A=2 C 3. B is the age of C minus the age of A: B=C−A Now, substituting the expression for A from the second equation into the third equation gives us: B=C−2 C=2 C Now, we have A, B, and C expressed in terms of C: From the second equation, we have A=2 C. From the third equation, we found B=2 C. Next, we substitute A and B in terms of C back into the first equation: 2 C+2 C+C=60 Combining like terms gives us: 2 C=60 Dividing both sides by 2 results in: C=30 This means that: Since A is half of C, we calculate A=2 30=15. For B, since B = C - A, we have B=30−15=15. Thus, the ages of A, B, and C are: A = 15 years B = 15 years C = 30 years None of the provided options (a, b, c, or d) match the calculated ages. Therefore, there may be a mistake in the question or the choices provided. Examples & Evidence For example, if A is 10 years old, then C would need to be 20 years old, making B the difference and thus not meeting the condition of having a sum of 60. By systematically substituting and solving, we find that A, B, and C's ages reflect the defined relationships accurately. The calculations are based on straightforward algebraic manipulations of the given conditions. Reconfirming with substitution and restructuring shows consistency with basic mathematical principles. Thanks 0 0.0 (0 votes) Advertisement Babygirlslaying2611 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. Community Answer 4 Click an Item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Express In simplified exponential notation. 18a^3b^2/ 2ab New questions in Mathematics Solve for s and simplify your answer. 22=2 3s Curtis wants to build a rectangular enclosure for his animals. One side of the pen will be against the barn, so he needs no fence on that side. The other three sides will be enclosed with wire fencing. If Curtis has 750 feet of fencing, you can find the dimensions that maximize the area of the enclosure. a) Let w be the width of the enclosure (perpendicular to the barn) and let l be the length of the enclosure (parallel to the barn). Write a function for the area A of the enclosure in terms of w. b) What width w would maximize the area? c) What is the maximum area? The monthly cost for renting an apartment at a new apartment complex increases annually. If the monthly cost for rent is 675 w h e n t h eco m pl e x o p e n s,t h e f u n c t i o n g(x) = 675 + 45x c anb e u se d t o d e t er min e t h e m o n t h l ycos t f orre n t x$ years after the apartment complex opens. Which best represents the domain of the function in this context? A. all rational numbers B. all real numbers C. negative integers only D. nonnegative integers only a. A bicycle costs 240 an d i tl oses\frac{3}{5}$ of its value each year. Complete the table to find the value of the bicycle, in dollars, after 1, 2, and 3 years. Year 0 1 2 3\ \nb. When will the bike be worth less than $1? How do you know?\nc. Will the value of the bike ever be 0? Explain your reasoning. Solve the following inequality algebraically. 4∣x+4∣−4≥40 Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com
16607	https://www.youtube.com/watch?v=jz92oOkJFNA	Faraday's & Lenz's Law of Electromagnetic Induction, Induced EMF, Magnetic Flux, Transformers The Organic Chemistry Tutor 9770000 subscribers 15546 likes Description 1041089 views Posted: 28 Feb 2017 This physics video tutorial explains the concept behind Faraday's Law of Electromagnetic Induction and Lenz's Law using the Right Hand Rule to determine the direction of the induced current and induced emf. Topics such as magnetic flux, self inductance of a solenoid, transformers, and ac generators are also included. This video contains plenty of examples and practice problems. Final Exam and Test Prep Videos: Magnetism - Review: Faraday's Law of Electromagnetic Induction: Lenz's Law - Induced Current: Induced EMF In Moving Conductor: Electric Generators & Induced EMF: Counter EMF of DC Motors: Transformers - Physics Problems: Mutual Inductance & Solenoids: Self-Inductance of Inductors & Coils: How To Make an Electromagnet: Faraday's & Lenz's Law of Electromagnetic Induction: Energy Stored In an Inductor: RL Circuits - Inductors & Resistors: Final Exams and Video Playlists: Full-Length Videos and Worksheets: Here is a list of topics: 1. Faraday's Law of Electromagnetic Induction 2. Lenz's Law Explained - Right Hand Rule 3. Induced EMF & Magnetic Flux - Magnetic Field & Area of Coil 4. Induced EMF in a Rotating Coil 5. Induced Current & Electric Fields 6. Induced EMF, Voltage, Current & Resistance - Ohm's Law 7. Induced EMF in a Moving Straight Wire 8. Clockwise vs Counterclockwise induced current direction 9. Weber - Unit of Magnetic Flux 10. Moving Conductor, Induced EMF, Speed / Velocity & Magnetic Field 11. Increasing & Decreasing Currents & Induced Current 12. Induced EMF of AC Electric Generator - Angular Velocity, Loops / Number of Turns, Frequency 13. Transformers Step up and Step Down 14. Primary Turns and Secondary Coils 15. Primary Voltage, Secondary Current, & Electric Power 16. Self Inductance of a Solenoid 17. Energy Stored in a Magnetic Field 18. Energy Stored in an Inductor 19. Energy Density of a Magnetic Field Disclaimer: Some of the links associated with this video may generate affiliate commissions on my behalf. As an amazon associate, I earn from qualifying purchases that you may make through such affiliate links. 465 comments Transcript: in this video we're going to talk about Faraday's law of electromagnetic induction and also lens law so let's say if we have a coil of wire and if we have a magnet what's going to happen if we move the magnet into the coil if we move the magnet into the coil there's going to be a current that's generated in the coil let's say the coil is connected to some meter As you move the magnet into the coil a current will be generated and this current is going to move in the counterclockwise Direction the magnet produces a magnetic field which goes into the page as X represents into the page and a circle represents out of the page now what's going to happen let's say if we have the same coil but this time if we move the magnet away from the coil if we move it away from the coil then the direction of the current will reverse the current will no longer flow in a counterclockwise Direction but it's going to flow clockwise so if you move the magnet into the coil the current is going to flow counterclockwise if you move it away from the coil it will change direction now what about changing the speed let's say if we move the magnet slowly if you move the magnet slowly into the coil the induced current will be very small but if you move the magnet quickly into the coil the induced current will be larger if you don't move the magnetic field excuse me if you don't move the magnet into the coil there will be no induced current so it also depends on the speed at which the magnet moves into or out of the coil the greater the speed then the greater the induced current will be now there are other ways to inducing the current and it's not only just moving the magnet into or out of the coil if you change the area of the coil let's say if you stretch it or bend it that will induce a current also if you change the angle if you turn the coil relative to the magnetic field in produced by the magnet if you change the angle an induced current will be created now there's an equation that you need to know this equ equation is the magnetic flux the magnetic flux is equal to the product of the magnetic field times the area time cosine of the angle and the unit for magnetic flux is the Weber we b r one Weber is basically one Tesla 1 s met the for magnetic field is Tesla T and the unit for area is square met so when you multiply these two you get the unit of Webers now we know that there's going to be an induced current anytime the magnetic field is changing if the area of the coil is changing or if the angle is changing as well and so the induced EMF or the induced current is dependent on the rate of change of of the flux because it also depends on how fast you moving the bar magnet into or out of the coil if you move it slowly the induced current will be small if the bar magnet is moved quickly into the um coil the induced current will be larger so the induced EMF and therefore the induced current is dependent on the rate of change of the magnetic flux which we'll talk about soon so let's say if we have a surface now let's draw the normal line to the surface which we'll call n and let's say if the magnetic field is perpendicular to the normal line which means it's parallel to the plane of the surface or to the face of the coil the angle Theta is between the normal line and the magnetic field so the angle is 90 Co sin 90 is equal to Z so therefore the electric flux will be zero so if the magnetic field doesn't pass through the face of the coil if it's parallel to the face of the coil there will be no electric flux and if there's no electric flux there's no induced EMF now let's say if it's neither parallel or perpendicular to the normal line Theta is here it's the angle between a normal line and B so in this situation the electric flux is simply ba a cosine Theta now the last case is if the magnetic field is parallel to the normal line so the magnetic field is perpendicular to the plane of the coil so in this case the angle Theta is z and cosine 0 is equal to 1 so the electric flux is simply equal to B a the electric flux is greatest when the magnetic field is parallel to the normal line or when it's perpendicular to the face of the coil that's when you're going to have the maximum electric flux and it's at a minimum if the magnetic field is parallel to the face of the coil now the induced EMF is equal to m which relates to the number of coils by the way if you increase the number of coils let's say if you have 10 Loops compared to one Loop the induced EMF and therefore the induced current will be larger when you move the bar magnet into the coil so the more Loops you have the greater the induced current will be so the induced EMF is equal to n time the change in the electric flux divided by the change in time this equation is associated with Faraday's law of induction so basically it states that the induced EMF is proportional to the rate of change of the electric flux and the electric flux is ba a cosine Theta so if the magnetic field B changes or if the area changes or if the angle changes there's going to be an induced EMF generated now the induce EMF you can treat it like voltage voltage and EMF they both have the same unit VT so if V is equal to IR then the induced EMF is equal to IR so if you know the induced EMF and if you know the resistance you can therefore calculate the induced current so if the induced EMF increases the induced current will increase because the resistance of the circuit should remain the same now now before we go into lenses law let's review a few basic things so let's say if we have a long straight wire with a current traveling north anytime you have a current inside a conductor it will create a magnetic field for this particular picture the magnetic field will enter the page on the right side and it's going to leave the page or it's going to travel out of the page on the left side so out of the page is represented by a circle and into the page is represented by X now to figure this out you can use the right- hand rule so let's say if the pen is the conductor or the wire if you take your right hand and wrap it around a pen and you want your thumb to face the direction of the current the way your fingers curl around the pen is the direction of the magnetic field as it travels around the conductor so let's see if I can draw this I drawing's not the best so so here is the conductor and here's the person's hand wrapped around the conductor so you want your thumb to face the direction of the current and notice the way your hands curl around the wire it comes out of the page on the left side and it curls into the page on the right side and so that represents the direction of the magnetic field so I want you to remember this picture because we're going to use it a lot when when trying to figure out the direction of the induced current so this picture is based on the right hand rule now if the current is going in the opposite direction then the magnetic field will change as well so it's going to be going into the page on the left side and out of the page on the right side so everything's going to be reversed now lens's law states that the induc EMF always gives rise to a current whose magnetic field opposes the original change in flux so let's apply that to the coil of wire that we had in the beginning so let's take this bar magnet and let's move it into the coil and let's use lens's law to determine the direction of the induced current inside this coil now as we move the magnet into the coil the magnetic flux is an increasing or decreasing well first we need to find the direction of the magnetic field the magnetic field emanates away from the North Pole enters the South Pole so the magnetic field is going into the page that's the external magnetic field now the external magnetic field is increasing which means that the flux is also increasing if the flux increases the induced current will be directed in such a way to decrease the flux so if you try to increase the flux the induced current will oppose that change it's going to decrease it and if you try to decrease the flux the induced current will try to support it or increase it so it's always opposite to what you're trying to do so since the external magnetic field is increasing the induced current will create a magnetic field that will oppose the external magnetic field the external magnetic field is directed into the page so if it's if it's going to oppose it it has to be out of the page it has to be opposite to it so I'm going to represent that in blue so on the right it's going to be opposite to the magnetic field that is in the center of the loop and on the left it has to be X they can't be the same so on the right side it's going to look like this so if you take any segment of the wire on the inside of the loop it's going to have a magnetic field that opposes or it's opposite to the external magnetic field inside the loop don't focus on the outside part focus on the inside part of the loop now what is the direction of the current if we have a wire where the induced magnetic field is going into the page on the right side I mean out of the page on the right side but into the page on the left side so hopefully you remember those two pictures that we went over the induced current has to be going in this Direction that's the only way the magnetic field will be coming out of the page on the right side into the page on the left and you can use the right hand rule to figure this out so therefore what we have is a current traveling in the counterclockwise Direction now let's work on another example so let's say if we have a rectangular metal conductor which looks like this and we also have a magnetic field confined in this blue region and let's say the magnetic field is going uh into the page everywhere in this region what is the direction of the induced current if the conductor is moving into the magnetic field is the current traveling clockwise or counterclockwise in this uh rectangular Loop so feel free to pause the video and use the lens's law to figure this out now the magnetic field is not increased or decrease and it's constant however the area inside the rectangular Loop that is exposed to the magnetic field is increasing so therefore we could say that the magnetic flux is increasing and the induced current will be in such it's going to be in a direction where it's going to try to uh decrease the flux because the magnetic flux is increasing according to lens's law the induced current will try to decrease the flux and the only way to do that is to direct itself opposite to the magnetic field now the external magnetic field at the center of the loop is going into the page if the induced current is going to oppose it then the induced magnetic field has to be going out of the page at the center of the rectangular Loop so what you need to do now is select a segment of wire let's use this segment and on the right side it has to be going out of the page whatever we have here that's going to be the induced magnetic field at the center of the loop so if on the inside or at the center it's going out of the page then on the outside it's into the page now don't forget these two pictures anytime you have a segment of wire when the current is going up the induced magnetic field will be going out of the page on the right side into the page on the left I take that back it's into the page on the right side out of the page on the left and if the current is going down then it's out of the page on the right side into the page on the left using the right hand rule so what we have here is basically this picture so the current is going down on the left side which means it's traveling in the counterclockwise Direction and so that's the answer for the sake of practice let's work on another example so this time the loop is going to move away from the magnetic field the external magnetic field will be directed into the page again just like before and the loop is moving away from it go ahead and use lens's law to determine the direction of the current so the first thing we need to do is determine if the flux is increasing or decreasing now the external magnetic field is constant but the coil or the rectangular Loop is moving away from the magnetic field therefore the area that is exposed to the magnetic field is decreasing and so the flux is decreasing as the coil moves away from the magnetic field now if the magnetic flux is decreasing and induced current will be created in such a way to increase the flux if the flux is decreasing according to lens's law it wants to oppose the change in flux so it wants to increase the flux the only way to increase the flux is to support the magnetic field the external magnetic field is direct Ed into the page as we can see by the AES here and so in order for the induced current to support it it has to go in the same direction not the opposite direction as we did in the last example so it's going to be going into the page as well so let's choose a segment of wire that we should focus on let's choose this segment now at the center of the loop it's going to be going into the page which means on the right side it will be uh out of the page so using this information what is the direction of the current well anytime you have a segment of wire where on the right side if the magnetic field is going into the page and on the left side the induced magnetic field I keep mixing up on the right side it's going out of the page and on the left side into the page and this case using the right hand rule the current is going to travel down so if it's traveling down here then it's going to be to the left and on the left side it's going up and at the top it's going to the right so therefore it's traveling in a clockwise Direction which makes sense so as the coil moves into the magnetic field the current is traveling in the counterclockwise Direction and as in this example as it moves away from the magnetic field it's going to reverse traveling in a clockwise Direction so you have to be careful with every step hopefully these three examples gives you a good idea of the process that you have to use in order to determine the direction of the current let's work on a new example so let's say if we have a coil of wire and it's place inside a magnetic field and this is the external magnetic field which is directed everywhere into the page now let's say we uh take this coil and we shrink it we're going to make it smaller so we're going to decrease the area of the coil such that it looks like this determine the direction of the current inside that coil so because the area is decreasing the magnetic flux is decreasing as well now if the flux is decreasing according to lens's law the induced current will be in such a direction that it's going to try to oppose the change in flux if the flux is decreasing the induced current will try to increase the flux and anytime it tries to increase the flux that means that it's going to create an induc magnetic field that is in the same direction as the external magnetic field the external magnetic field is going into the page so therefore in order for the induced current to support the fail in magnetic field it has to be in the same direction so it has to be into the page as well so let's draw our small wire and let's choose a segment of the wire let's focus on the right side we can choose any segment but I like to choose a segment that looks like this because we know the two possibilities already if the current is going up we know that the magnetic field that is the magnetic field created by this current the induced magnetic field is going to be going into the page on the right side and out of the page on the left left side and if we reverse it let me draw this somewhere else if the current is going down then the magnetic field is going to be going into the page on the right side I mean out of the page on the right side into the page on the left X is into the page this dot is out of the page now going back to this the induced current has to be going into the page at the center so relative to this segment of wire the center is on the left side so there's going to be an x on the left side which means a DOT on the right side so these two look similar which means that they're current and that segment of Y is going down so therefore the current in this wire is traveling in the clockwise Direction and so that's the answer let's say if we have a wire and the current in this wire is increasing determine the direction of the induced current in the circular Y is it going to be clockwise or counterclockwise feel free to pause the video and use lens law to find out now if the current in a straight wire if it's increasing then that means the magnetic field is increasing which means the flux generated by that wire is increasing and that is the magnetic field that is produced by this wire that is in the center of this Loop that's going to increase which will increase the flux so therefore the induced current in the circular wire is going to create a flux that opposes the original flux so if the original flux is increasing according to lens's law the flux created by the induced current will try to decrease the increase in flux so it's going to opposite it's going to be opposite to it now if it's going to opposite then the magnetic fields have to be in the opposite direction not in the same direction so let's focus on the external magnetic field so using the right hand rule whenever you have a current going in the upward Direction the magnetic field is going to be going out of the page on the left side and into the page on the right side so it's going to be going in this general direction this is how it's going to look like now the coil of wire the circular coil is on the right side and the magnetic field generated by the straight y on the right side is X so the external magnetic field is into the page and at the location of the circular wire now because the induced current wants to oppose the increase in flux the induced magnetic field has to be opposite to the exter magnetic field so it has to be out of the page at the center let's use a different color let's use blue now let's focus on this segment so at the center or on the left side it has to be going out of the page and on the right side into the page so notice that we have the same direction on the left side we have a DOT on the right side we have an X therefore the current must be going up in that segment of wire which means that the current is traveling in the counterclockwise Direction so that's the answer now let's say if we have a wire and there's a current traveling towards the left and that current is decreasing and we have a circular wire below it what is the direction of the induced current in the circular wire go ahead pause the video use lens of law to figure it out so let's focus on the magnetic field created by the wire with the decrease in current if the currents going to the left if you take your hand and wrap it around a pen with your thumb pointing towards the left in the direction of the current you'll see that your fingers will curl into the page at the top and they're going to come out of the page below the pen so I'm not going to draw it but that's how it should look like so the external magnetic field is out of the page below the wire and that's what we want to focus on because that's where the uh circular wire is located it's below uh this wire now the flux is decreasing because the current is decreasing a decreasing current will produce a decrease in magnetic field and a decrease in magnetic field leads to a decrease in flux so according to lens's law the induced current will be directed in such a way that it's going to try to support the decrease in flux so if the flux is decreasing it's going to try to increase it it's going to oppose a change now anytime it tries to increase the flux it's trying to support the decrease magnetic field and anytime it wishes to support it the two magnetic fields have to be in the same direction now at the center the external magnetic field which is this one is directed out of the page and because the induced magnetic field has to be in the same direction as the external magnetic field because it wants to support it this one also has to be out of the page now let's focus on this segment of the wire so the induced magnetic field is out of the page and the center is on the left side so it's going to be out of the page on the left into the page on the right so therefore this will create a current that is going upward or if we focus on the left side instead of the right side the center is going to be out of the page the center is on the right side of that segment and then the left side is going to be going into the page and so the current has to be going down so really doesn't matter what you decide to focus on as we can see the current is traveling in the counterclockwise Direction and so that's the answer for this one now here's another example for you so let's say if we have a circular coil of wire which is attached to a battery and that's attached to a resistor and there's an open switch and inside that coil there is another coil of wire when the switch is closed what is the direction of the current and the coil of wire that is represented by the gray color so let's say it's a 6vt battery and we have a 3 Ohm resistor so right now the current is zero when the switch is open however once we close the switch the current will become two so it's going to go from zero to 2 amps so therefore the current is increasing which means that the flux is increasing and if the flux is increasing according to lens's law the induced current will create a flux that opposes the original change in flux the flux is increasing so the induced current will create a flux that will try to oppose the other one so it's going to be decreasing anytime the induced current creates a decrease in flux the external magnetic field and the induced magnetic field will have or will be in the opposite direction if the induced current creates an increasing flux it's going to try to support the F magnetic field and the external magnetic field and induced magnetic field will be in the same direction in that case but in this particular uh example since the flux is increasing the induced current will try to oppose the flux or decrease it so the exteral magnetic field and the induced magnetic field will be in opposite directions now we need to determine the direction of the external magnetic field that is the magnetic field created by the white circular wire so let's focus on this segment of the wire so we need to determine the direction of the current the current will Flow Away from the positive terminal towards the negative terminal so once the switch is closed the current will be going in the counterclockwise Direction so on the right side since the current's going up using the right hand rule it's going to be going out of the page on the left side into the page on the right side so that's the external magnetic field so at the center the external magnetic field is out of the page that's at the center of the gray Loop so that's what we're going to put here now because the induced current wants to oppose the increase in flux the external magnetic field and the induced magnetic field will be in opposite directions so the induced magnetic field will be going into the page at the center of the loop so I'm going to represent that in blue so let's focus on this segment of the wire so at the center which is to the right of that segment it's going into the page which means to the left it must be going out of the page and therefore it must be opposite to this one because the direction of the X and the dot actually no it's in the same direction because both cases the x is on the right side and the DOT is on the left side so therefore this must be going up so the current is traveling in the clockwise Direction so that's the direction of the induced current now let's work on some problems number one a single circular Loop of wire is perpendicular to a magnetic field which increases from 1.5 Tesla to 4.8 Tesla in 23 milliseconds part A calculate the change in magnetic flux so let's begin with that the change in flux is equal to the change in the magnetic field time the area time cosine of the angle now let's draw a picture so we have a single circular Loop and and here's the normal line which is perpendicular to the surface here's the uh radius and a magnetic field is perpendicular to the circular Loop which means that it's parallel to the normal line so the angle Theta is the angle between the normal line and a magnetic field therefore the angle is 0° and cosine of 0 is 1 so now we can calculate the change in electric flux so the change in B is going to be the final magnetic field which is 4.8 Tesla minus the initial magnetic field which is 1.5 Tesla the area is constant and the area is the area of a circle P pi r 2 the radius is 25 cm which is25 M squar and then times cosine of 0 we know cosine of 0 is one 25 s pi that's 19635 and if you multiply that by the difference of 4.8 and 1.5 which is 3.3 this will give you the change in flux which is uh 648 and the unit for magnetic flux is the Weber which is uh Tesla times square meters so that's the answer for part A it's positive. 648 Part B what is the induced EMF to calculate the induced EMF we can use this formula it's n times the change in the flux divide by the change in time n is the number of loops and we have a single circular Loop so n is one the change in flux is 648 and the change in time the magnetic field increased from 1.5 to 4.8 Tesla in 23 milliseconds so that's Delta te but we need to convert milliseconds into seconds we could do so by dividing by th 23 divid by 100000 is 023 so 648 ID 023 that's about 28.7 volts so that's the induced EMF now the next thing we need to do is calculate the current the current is the induced EMF divided by the resistance so it's 28.7 volts divided by 20 ohms and this is equal to 1.41 amps and so that's the answer for part c number two the magnetic flux through a coil of wire containing 20 Loops changes from pos2 to3 Webers in 425 milliseconds what is the induc MF the induc MF is equal to n times the change in the electric flux I mean not the electric flux but the magnetic flux divided by the change in time so in this problem we have 20 Loops of wire the flux changes from 2 to minus 3 the Final flux is -3 the initial flux is pos2 so it's -3 minus pos2 now we need to convert milliseconds to seconds and we can divide it by th000 425 milliseconds is 425 seconds -3 - 2 is5 -20 is positive 100 and if we divide that by 425 the induced EMF is equal to 235.000 volts now that we have the induced EMF we can calculate the resistance of the coil so eal I R just as v = i R and solving for r r is the induced EMF divided by the current so it's going to be 235.50 volts divided by a current of 5.12 amps and so the resistance is approximately 46 ohms now let's say that we wanted to calculate the power absorbed by the resistance of this coil how much power is dissipated in the coil power is equal to voltage times current or the induced EMF times the current so it's going to be 235.000 volts times a current of 5.2 amps so that's about 1,25 Watts number three a flexible rectangular coil of wire with 150 Loops is stretched in such a way that its Dimension changes from 5x 8 square cm to 7 by1 square cm in5 seconds in a magnetic field of 25 Tesla that is 30° relative to the plane of the coil calculate the induc EMF the induced EMF is equal to n time the change in electric flux divided by the change in time and the change in electric flux it's going to be the magnetic field which is constant times the area since the area is not constant we're going to multiply by the change in area times cosine Theta ID delta T now let's talk about cosine Theta so let's say this is the rectangular coil of wire and this is the normal line and this line is parallel to the plane of the coil and here is the magnetic field the magnetic field is 30° relative to the plane of the coil buta is the angle between a normal line and a magnetic field so Theta in this problem is 60 it's 90 minus 30 n is the number of Loops which is 150 the magnetic field is 2.5 rather 25 Tesla now the change in area the area of a rectangle is basically the length times the width but we need to convert cenim to met so we got to divide by 100 the change in area is going to be the final area minus the initial area the final area is 7 cm by 11 cm or 07 m time1 m that's the final area the initial area is 05 M time 08 M and then let's multiply by cosine of 60° and let's divide everything by5 seconds so let's calculate the change in area first 07 minus 05 8 that's 0.37 or 3.7 10us 3 next multiply that by cosine 60 and then times 150 and time 25 so that's 6. 9375 and divided by5 seconds your final answer should be negative 46.2 volts so that's the answer to part A that is the indued EMF now let's move on to Part B so let's get rid of a few things how much energy in Jews was dissipated in the circuit if the total resistance is 100 ohms so before we can find the energy we need to calculate the power and before we can find that we need to find the current so let's find the current first the current is equal to the induced EMF divided by the resistance so it's - 4625 volts divided by 100 ohms now you really don't need to worry about the negative sign but you can keep it there if you want to so let's just ignore the negative sign for now so the current is going to be 4625 amps energy is going to be positive so we're going to make the final answer positive now that we have the current we could find a power dissipated by the circuit it's the voltage or the induc MF times the current so it's 46 rather 46.2 that's the induced EMF times the current which is 4625 and so that's going to be 2.39 Watts energy is equal to power multiplied by time one watt is onew per second so we have 21 39w per second and if we multiply by5 seconds we can see that the unit seconds will cancel given us the energy in Jews so the final answer is 3.21 Jew so that's how much energy is dissipated in a circuit so the circuit consume three 21 Jew within the 15 seconds in which the area was changing number four a rectangular coil of wire contains 25 Loops the angle between a normal line of the coil and the magnetic field changes from 70 to 30° in 85 milliseconds calculate the induc EMF so let's say this is the rectangular coil of wire and here is the normal line perpendicular to the surface so initially the magnetic field is at an angle of 70° relative to the normal line and then after some time the magnetic field let's call this B final and B initial well the magnetic field is constant but the angle changes so it's really Theta final and Theta initial but after some time the magnetic field is going to be 30° relative to normal line so that's the picture that we have in this problem how can we use this information to calculate the induced EMF by the way there's two ways the angle could change either the magnetic field is constant and the coil changes relative to the magnetic field or the coil is held in place and the magnetic field changes Direction relative to the coil now in this problem it really doesn't matter which one changes all we need to know is just the final angle and the initial angle the induced EMF is equal to Nega n times the change in flux divid by the change in time and the flux it's going to be b b is constant the area is constant but the angle changes so it's going to be the change in cosine Theta / delta T N the number of Loops is 25 the magnetic field B is three Tesla the area we need to convert centimeters to meters so it's going to be5 m times 20 M we have a rectangular surface so the area is simply the length times the width 15 20 and then multiply by the change and cosine the initial angle is 70 the final is 30 so final minus initial cosine 30 minus cosine 70 divided by the change in time which is U 085 seconds don't forget to divide 85 Mill seconds by a th000 to convert it to seconds now let's plug this in let's start with the cosine part first cosine 30 minus cosine 70 that's. 524 and make sure your calculator is in degree mode now let's multiply 0524 by5 time20 3 25 so that's going to be netive 1.179 and then divid by 85 so the induced EMF isga 13.87% so Faraday's law of electromagnetic induction is associated with this formula now let's say if we have a moving conductor that can slide freely along the metal rails and this conductor is moving with a speed V and a rectangular coil that it forms is perpendicular to the magnetic field the magnetic field is directed into the page and a rod has a length L how can we come up with an equation that will help us to calculate the magnitude of the induced EMF let's start with Faraday's law of induction the induced EMF is equal to n times the change in the magnetic flux divid by the change in time now as the rod moves the area that is exposed to magnetic field will increase so let's draw a new picture and let's say the rod is now in this area so notice that the flux is greater since the area exposed by the magnetic field contained in that coil is greater so this is going to move some distance a d and that distance is equal to the speed multiplied by the change in time so the increase in area is basically let me use a different color this region here represents the increase in area and the area of that region is basically the length times the width which is d so the change in area is L D and D is basically the velocity times the change in time now going back to the first equation let's replace the flux with the magnetic field which is constant times the change in area and the angle is zero because the Mantic field is perpendicular to the plane of the coil it's parallel to the normal line which means the angle between the normal line and a magnetic field is zero and cosine of Z is one so we don't have to worry about the cosine part in this formula so we just have this now let's replace Delta a with LV delta T and by the way we only have a single Loop so n is equal to one so the induced EMF is going to be n or1 and since we're looking for the magnitude we don't have to worry about the negative sign so this going to be 1 B and Delta a is l v delta T ID delta T and so we can cancel the change in time so therefore the induced EMF is equal to the strength of the magnetic field B times the length of the moving Rod l times the speed of the moving r v so it's blv now turns out that there's another way in which we could derive that same equation but first let's talk about the current in this circuit as the rod moves towards the right the area is increasing which means the flux is increasing by the way if you want to feel free to pause the video and use lens's law to determine the direction of the current in the moving Rod so because the rod is moving to the right the area increases and the flux increases so according to lens's law the induced current will be directed in such a way as to oppose the increase in flux so it's going to try to decrease the flux and since it's in opposition to it the external magnetic field which is directed into the page is going to be opposite to the induced magnetic field so if the external magnetic field is into the page the induced magnetic field is going to be out of the page so to the left of the rod which is the center of the coil the induced magnetic field will be out of the page which means to the right it's going to be into the page and so using the right hand rule the current has to be traveling in this direction electric current flows from a high potential to a low potential so from a positive side to a negative side that means the bottom of the conductor has a higher um electric potential and the top part has a lower electric potential current represents the flow of positive charge and so the electric field inside the conductor is in the same direction as the current so if you have a conductor with an electric field directed North any positive charges will fill a force that will accelerate in the direction of the electric field but in a metal the protons are not free to move the electrons are the electrons are the charge carriers the electrons will fill a force that will accelerate them opposite to the direction of the electric field and the electrons will move a distance L which is the length of the rod from one point to the other point so to calculate the induced EMF using another equation we need to use energy induced EMF or voltage is basically the ratio between uh work and charge one volt is one Jewel per clume the unit of work is Jews the unit of charge is clums work is basically force times distance and the magnetic force on a moving charge we know it's bqv and it's bqv sin Theta but B is perpendicular to the surface and so for that formula sin 90 is one so we don't have to worry about the sign part so let's replace f with bqv and D the distance that the charges move when being acted on by force the magnetic force so are going to move a distance D in a direction of the force which is basically the same as l so let's replace D with L notice that Q cancels and we can get the same formula so the induced EMF is equal to the magnetic field times the length of the moving Rod times the speed of the moving Rod so those are just two ways in which you can derive this equation now let's work on this problem number five a moving Rod 45 cm long slides to the right with a speed of 2 m/s in a magnetic field of 8 Tesla what is the induced EMF so let's use the formula EMF is equal to blv so the magnetic field is 8 Tesla the length of the moving Rod is 45 cm but we need to convert that's M ID 100 that's uh 45 M and then we need to multiply by the speed which is moving at 2 m per second so let's just multiply these three numbers 8 2 is 16 16 45 is 7.2 so that's the M for this particular problem that's all you need to do for part A now Part B what is the electric field in a rod to find the electric field it's simply equal to the voltage ID by the distance that's how you can find the electric field in a parallel plate capacitor the electric field in the rod is the voltage in a rod divided by the distance or the length of the rod so it's going to be the the voltage is basically the EMF the EMF is 7.2 volts and the length of the rod is 45 M 72 divid 045 that's going to be 16 so it's 16 volts per meter or you can describe the electric field in terms of Newtons per Kum one volt per meter is the same as 1 Newton per Kum so that's the answer for Part B now part C what is the current in the rod the electric current is going to be equal to the voltage or the EMF divided by the resistance so it's 7.2 volts divided by a resistance of 50 ohms so let's go ahead and divide those two numbers 7.2 ID 50 is equal to1 144 amps which is equivalent to 144 milliamps so that is the electric current in the rod now Part D what force is required to keep the rod moving to the right at a constant speed of 2 m second what equation can we use to figure this out well since we know the current we could find a force in a rod the magnetic force that acts on an object well basically the magnetic force that acts on a wire with a current is equal to I lb sin Theta we can use the same equation to find out the force required to move this Rod because that force will generate a current the current in the rod it's uh 144 the length of the rod is 45 M the magnetic field is 8 Tesla and because the magnetic field is perpendicular to the area of the rectangular coil it's sin 90 sin 90 is just one so we just got to multiply those three values so it's going to be .144 45 8 and so this is equal to 5184 newtons so that's the force required to keep the rod moving at a speed of 2 m/ second once it has that speed it's going to generate a current of .144 amps and whenever you have a rod without that current in the presence of a magnetic field this is going to be the magnetic force that's acting on the rod and that magnetic force is equal to the force required to keep it moving at that speed it just works out that way number six a 60hz AC generator rotates in a 25 Tesla magnetic field the generator consists of a circular coil of radius 10 cm with 100 Loops what is the angle of velocity and calculate the induced EMF the induced EMF of a generator can be found using this equation it's equal to n which represents the number of Loops time B the strength of the magnetic field time a the area of each Loop time Omega which is the angular velocity Time s Omega T I'm not going to focus on deriving this equation but this is the formula that you need to know for an AC generator if you want to calculate the induc EMF for it to find the angular velocity Omega it's simply equal to 2 pi F where f is the frequency measured in hertz the frequency is 60 Herz so Omega is going to be 2 pi 60 HZ 2 pi 60 that's equal to 120 Pi which as a decimal is about 37699 s to minus1 or radians per second so that's Omega that's how you can find the angular velocity now let's calculate the induced EMF using the first equation so n is 100 we have 100 Loops the strength of the magnetic field is. 25 Tesla the area is going to be the area of a circle since we have a circular coil that's pi r 2 the radius is 10 cm so that's 10 m and we need to square it times Omega which is 37699 by the way I should have stated this earlier but we're looking for the maximum EMF the maximum EMF you don't need to worry about the sign part sin 90 is going to be one and so the maximum EMF is simply NBA time Omega you don't have to worry about the sign function so let's go ahead and calculate the maximum EMF so this is going to be the peak output it's going to be 100 .25 piun .1 2 37699 so the maximum output is 2961 volts that's going to be the maximum EMF generated by this particular generator number seven a generator produces an EMF of 12 volts at 700 RPM what is the induced EMF of the generator at an angular velocity of 24 400 RPM we know that the maximum EMF is equal to the number of Loops times the strength of the magnetic field times the area time the angle velocity in this equation you can clearly see that the angular of velocity and the induced EMF are proportional if you increase the angular of velocity the EMF will increase they're directly related so if you double the number of RP the induced voltage will double as well so let's see if we can write an equation that describes e and w or E and Omega E2 is going to be NBA a Omega 2 E1 is going to be NBA Omega 1 so if we're using the same generator it's going to have the same number of Loops which means we can cancel n the same magnetic field and the same area the only difference is the rotation speed or the angle of velocity so this is the equation that we need E2 over E1 is equal to Omega 2 over Omega 1 or Epsilon 2 over Epsilon 1 so Epsilon 1 the voltage is 12 volts at an angular speed of 700 RPMs what is the induced EMF at an angular speed of 2400 RPMs so these units will cancel all we need to do is multiply both sides by 12 so Epsilon 2 is going to be equal to 12 2400 / 7 00 2400 / 700 is about 3.42 9 if you multiply that by 12 this will give us an induced EMF of 41.1 volts so that's the answer our next topic of discussion is the Transformer a Transformer is made up of two sets of coils the coil on the left is known as the primary coil and on the right the secondary coil these coils are wrapped around an iron core now let's say that the primary coil has 100 turns and let's say that the secondary coil on the right has a th turns because the secondary coil has more turns than the primary coil this is going to be a Step up Transformer A Step up Transformer is used to increase the ac voltage so let's say that the input voltage at the primary or at the primary Coro rather is let's say uh 6 volts notice that the ratio between NS and NP is 10 1,000 ID 100 is 10 so the voltage will increase by a factor of 10 the voltage at the secondary coil will be 60 when the voltage goes up the current goes down such that the power will be the same for an ideal Transformer if it's 100% efficient so let's say that the input current at the primary Coral is 20 amps the output current at the secondary coil it's going to decrease by a factor of 10 if the voltage goes up by a factor of 10 the current will decrease by a factor of 10 so the current is going to be 2 amps and notice that the power is the same the power in the primary coil is just the voltage times the current so it's 6 volts 20 amps and that's equal to 120 watts the power at the secondary coil it's going to be VSS is voltage time current 60 volt 2 amps is 120 watts and it makes sense energy is conserved the amount of energy that's being transferred on the left side should be equal to the energy being transferred out of the right side the amount of energy that is going into the system should equal the amount of energy coming out of the system for an ideal Transformer so because the input power is the same as the output power this Transformer is 100% efficient now most real life Transformers are about 99% efficient so for the most part you can assume that these two are virtually the same but if you ever need to calculate the percent efficiency it's going to be equal to the output power divided by the input power the output power can only be equal to or less than the input power cannot be more so it's the output divided by the input power times 100% that's how you can calculate the percent efficiency so the equations that you need for Transformers are as follows NS over NP is equal to vs over VP which is equal to IP over is and the power is voltage times current VSS is is equal to VP IP if it's 100% efficient these two will equal each other the input power is always going to be VP IP the output power is equal to VSS is but at 100% efficiency the output power equals the input power number eight a Transformer has 50 primary turns and 400 secondary turns the input voltage is 12 volts and the input current is 24 amps what is the voltage in current at the secondary coil and also how much power is consumed by the primary coil so feel free to pause the video and try this problem so let's draw a picture this is going to be the primary side and this is going to be the secondary side now is this a step up or Step Down Transformer since NS is greater than NP this is going to be a Step up Transformer the voltage is going to increase now the input voltage is 12 volts so that's VP the input current is 24 amps so that's IP and there's 50 turns in the primary coil and 400 turns in the secondary coil so NP is 50 NS is 400 now 400 ID 50 is 8 so the number of turns increases by a factor of eight as he move from the primary coil to the secondary coil which means the voltage has to increase by eight and the current should decrease by 8 so 12 8 is equal to 96 volts and 24 / 8 is 3 amps and let's make sure the power is the same assuming it's an ideal Transformer so to calculate the power at the primary coil the power is equal to VP IP so that's 12 volts 24 amps so that's 288 watts and if we calculate it at the secondary coil it's 96 Vols 3 amps which is also 288 Watts now for those of you who want to use an equation to calculate VSS and is here's what we can do NS / NP is equal to VSS over VP so NP is 50 NS is 400 VP is 12 and let's solve for vs Let's cross multiply so this is going to be 50 vs is equal to 12 400 12 400 is 4800 and to solve for vs divide by 50 4800 ID 50 is 96 so that's how you can use the formula to find vs because sometimes the numbers may not be as nice as the whole number that we have here now let's use the formula to calculate is NS over NP is inversely related to um the current so it's going to be IP over is notice that the subscripts they're opposite relative to each other so NP is 50 NS is 400 we have IP which is 24 and we need to solve for is so let's cross multiply 50 24 is 12 00 and that's equal to 400 is to find is divide both sides by 400,00 divid 400 is 3 so is is 3 amps and then once you have is and vs you can find the Power 96 3 is 288 or 12 24 is also 288 number n a 200 wat ideal Transformer has a primary voltage of 40 Vols and a secondary current of 20 amps calculate the input current and output voltage is this a step up or Step Down Transformer if there are 80 turns in a primary coil how many turns are there in a secondary coil so let's begin so the power is equal to 200 watts and we know that the primary voltage which is the input voltage VP that's 40 volts and the secondary current is is 20 amps so what is the uh secondary voltage and what is the uh primary current so power is equal to voltage time current the power is 200 Watts the voltage at the primary coil is 40 so we could solve for the primary current so we need to divide both sides by 40 200id 40 is 50 so the primary current is 5 amps now let's calculate the secondary voltage power is equal to vs time is so 200 watts is equal to a vs times the current of 20 amps 200id 20 is 10 so the voltage at the secondary coil is 10 volts 20 10 is 200 40 5 is 200 now is this a a step up or a step down transformer what would you say is the voltage increasing or decreasing notice the voltage went down from 40 to 10 so the voltage is decreasing which means that this is a Step Down Transformer now if there are 80 turns in a primary coil how many turns are there in a secondary coil well we can use the equation NS over NP is equal to v s over VP so the primary coil has 80 turns that's NP we're looking for NS the number of turns in a secondary coil VP is 40 vs is 10 so let's cross multiply so 80 10 is 800 and 40 NS is just 40 NS so to solve for NS let's divide both sides by 40 so we could cancel a zero and so 800 over 40 is the same as 80 over 4 and if 8 divid 4 is 2 80 divid 4 is 20 so there 20 turns in the secondary coil now let's make sense of this let's draw a picture so the primary coil has more turns than the secondary coil the primary coil has 80 turns the secondary coil has 20 the primary voltage is 40 volts and a secondary voltage is 10 volts because the voltage decreased by a factor of four the number of turns must also decrease by factor four so went down from 80 to 20 now the current in the primary coil is uh 5 amps and so the current in the secondary coil has to increase by a factor four so it's 20 amps and as we can see the power is the same 10 20 is 200 Watts 40 5 is 200 Watts so hopefully this setup helps you to understand how Transformers are used and how you can solve them conceptually you really don't need the formulas if you understand it but you could use it if the numbers are not whole numbers like we have here if you have decimal values I suggest using the formulas let's talk about inductance let's say if we have a battery attached to a coil of wire which is also known as a solenoid when the current is constant there's not going to be any induced EMF but if the current is changing there is going to be an induced EMF so let's say if the current is increasing the induced EMF will be negative which means that it's going to oppose the change in current if the current's increasing the induced EMF will try to decrease the current so therefore the induced current is going to be in the opposite direction to the increase in current now let's say if we have the same circuit and we have an inductor or solenoid which is basically a coil of wire and the current is flowing in the same direction however this current is decreasing as opposed to increasing if the current is decreasing the induced current generated will try to support the decrease in current so the EMF will be positive so the regular current is Flowing clockwise and the induced current will be in the same direction as the decreas in current and so the induced current will be clockwise so anytime the current of a circuit is increasing the induced current will be opposite to the direction of that current if the current is decreasing the induced current will be in the same direction as the main current in the circuit the induced EMF can be calculated using this equation it's l time the change in current / the change in time notice the negative sign when the current is increasing the change in current is positive if Delta I is positive then the EMF will be negative as we can see here now if the current is decreasing the changing current will be negative so Delta I is negative plus the negative sign on the outside so the EMF will be positive now sometimes you may need to calculate l in this equation so what exactly is l l is the inductance which is measured in units of Henry's or Capital H and to calculate the inductance of a solenoid here's the equation that you can use l is equal to mu0 n^ 2 a / L mu0 is the permeability of free space it's 4 Pi 10 - 7 L is basically the length of the solenoid A is the area of the coil so typically it's circular so the area is going to be P pi r 2 and just keep in mind whenever you have a circle the radius is half of the diameter this is the diameter and the radius is just one half of that so R is D over two n represents the number of turns in the coil so let's talk about how to derive this particular formula so first we know that the magnetic field created by a solenoid is equal to mu0 n I I went over this in another video it's the video on magnetisms uh magnetic fields perhaps you've watched that video before this one so that's where you can get this equation from lowercase n represents the number of turns per meter so it's capital N which is the number of turns divided by the length of the solenoid now the induced EMF is equal to negative n the number of turns times the change in the magnetic flux divided by the change in time anytime you have a current flowing in a wire it creates a magnetic field and if you have a change in current it's going to produce a change in magnetic field which produces an EMF now we also said that the induced EMF is equal to l time the change in current divid by the change in time so therefore we can set set these two equal to each other so n Delta flux magnetic flux over delta T is equal to l Delta I over delta T since they both equal to the induc MF now what we're going to do is we're going to multiply both sides by negative delta T so T will cancel and also the negative signs will cancel so now what we have left over is n times the change in magnetic flux that is equal to l times the change in the current so our goal is to solve for the inductance of the inductor or the solar noid so L is basically equal to the number of loop Loops times the change in flux ID the change in current so in this equation we divided both sides by Delta I if you want to show your work here's what we did so these two cancel so now what should we do next we know that the change in flux is equal to the magnetic field time the area time cosine Theta but let's assume that the angle is zero so cosine 0 is one so the change in flux is going to be equal to the change in B a now we said that b is equal to mu0 n I so let's replace these two so the inductance L is equal n mu0 n I a / Delta I and of course we still have a triangle somewhere so what we're going to do is we're going to cancel the current and so we have L is equal to n mu0 lowercase n a now lowercase n is equal to capital N over l so let's replace lowercase n with this so the inductance is equal to mu0 time n and then let's replace lowercase n with n / L a so n n is n^ 2 so now we have this equation L is equal to the permeability of free spacetimes the square of Loops times the area of the coil / L and so that's how you can calculate the inductance of a solenoid now some other equations that you may need to know is the potential energy stored in an inductor it's one2 l i^ 2 so whenever a current flows through an inductor energy is stored in a magnetic field of that inductor just as when you charge up a capacitor the energy can be stored in the electric field of the capacitor sometimes you may need to calculate the energy density which is represented by lowercase U and that's equal to the energy capital u / the volume the potential energy is measured in Jew the energy density is going to be jewles per cubic meter the energy density is equal to B ^2 over 2 mu0 that's how you could find the energy density relative to the magnetic field number 10 a solonoid consists of 200 Loops of wire and has a length of 25 cm calculate the inductance of the solenoid if it has a diameter of 8 cm so the radius is always going to be half of the diameter so 8 cm / 2 the radius is 4 cm now the inductance of a solenoid is equal to mu0 n^ 2 time the area / the length of the solenoid mu0 the permeability of free space is 4 Pi 10 to7 the solonoid consists of 200 loops and the area which is usually the solar noid is usually made up of a circular coil of wire it's going to be Pi R 2 so I should have mentioned Circ coil of wire but if you see the key word diameter It's associated with a circle so this is going to be 04 M squared divid by the length of the solenoid which is 25 cm or. 25 M so let's go ahead and type this in so this will give you an inductance of 1.01 10us 3 Henry's now this is equivalent to 1.01 millihenries now let's calculate the induced EMF this is equal to l times the change in current / by the change in time so we have the value of it's 1.01 10us 3 Henry the change in current the final current is 30 minus the initial current of 8 / delta T 240 milliseconds is24 seconds so 30 - 8 that's 22 1.01 10us 3 /24 this is equal to .09 26 volts so that's the induced EMF which is equivalent to 92.6 Mill volts now because the EMF is negative that means that the induced current is opposite in direction to the original current that created it so let's say if this is the solenoid and this is the original current that current increases from 8 to 30 because the current increases the change in current is positive so therefore if Delta I is positive a positive value times a negative sign will give us a negative EMF and anytime the current increases an induced current will be created in the coils that will oppose the increasing current that created it so whenever you have a negative EMF the induced current is opposite to the direction of the original current so if the original current is Flowing clockwise the induced current will be counterclockwise it's just going to be opposite to what it is so that's the direction of the induced EMF and the induced current number 11 a solenoid has an inductance of 150 mes with 300 turns of wire and a circular area of 2.65 3 square m what is the potential energy stored in the inductor when a current of 20 amps passes through it so what equation can we use to calculate the potential energy stored in this inductor here's the equation that you need it's 12 the IND inductance times the square of the current so the inductance is 150 Millian but we need to convert it to Henry so it's 150 10us 3 Henry the current is equal to 20 amps and let's not forget to square it so let's go ahead and plug this in so therefore the potential energy stored in this inductor is equal to 30 Jew so that's the answer to part A now let's move on to Part B how many turns per meter does the solenoid have so how can we figure this out well we know that the inductance is equal to mu0 n^ 2 the area / the length of the inductor so let's separate n^2 and write it as n n / a now we know that n/ L is equal to lowercase n and lowercase n represents the number of turns per meter so we need to use this form of the equation so L is equal to Mu 0 n lowercase n area the induct is is 150 10us 3 henes the permeability of free space that's 4 Pi 10- 7 that's mu0 capital N represents the number of turns there's a total of 300 turns in this inductor or solenoid the lowercase n is what we're looking for and the area is given to us the area is 2653 M so first let's multiply these three numbers so 4 Pi 10 - 7 300 2.65 3 that's about 1.0 10us 3 now if you can't type in 4 Pi 10- 7 in your calculator type in pi first and then multiply that by 4 10- 7 you might find that helpful now let's divide both sides by 1 10us 3 the 10us 3s will cancel and so it's simply 150 over 1 so lower case n is 150 so that's how many turns there are per meter now I know this question is not listed in the problem but how long is the solenoid so if there are 150 turns per 1 meter how many how long is it for 300 turns so if one meter contains 150 turns then two meters will contain 300 turns so therefore the solenoid is 2 m long you can also use the equation to get the same answer we know that lowercase n is equal to capital N / l so let's put this over one Let's cross multiply so n now is equal to capital N which means means that the length is equal to the number of Loops divid lowercase end which is the number of turns per meter so there's 300 turns in the entire solenoid and there's 150 turns per meter so 150 divid 300 I mean 300 divid 150 excuse me that's going to give us uh 2 m so that's the length of the solenoid now let's move on to part C what is the magnetic field when the current is 20 the magnetic field of a solar noid can be calculated using this equation it's mu0 n I mu0 is 4 Pi 10- 7 lowercase n is 150 turns per meter and the current is 20 amps so if we multiply these three things this is going to be 3.77 10us 3 Tesla so that's the answer for part C now what about Part D what is the energy density of this magnetic field to calculate the energy density which is equal to low case U it's B ^2 / 2 mu 0 so we have B already and we just got to divide it by two and the permeability of free space 5655 jewles per cubic meter so that is the energy density of the magnetic field that is in the solenoid so that's how you can find it so that's it for this video we've covered a lot of topics and so thanks for watching and if you like this video feel free to subscribe and have a great day
16608	https://ledidi.com/academy/measures-of-central-tendency-mean-median-and-mode	Ledidi Product Plans and Pricing Study Design Data Capture Statistical Analysis and Data Visualisation Collaboration Use Cases Company About Ledidi Scientific Advisory Board Book a Demo Get in Touch Careers Resources Ledidi Trust Centre Ledidi Academy News Customer Stories All Resources Back to Ledidi Academy Measures of central tendency: Mean, median and mode Dr. Anthea Van Parys, PhD, Research Consultant Link copied Mean, median, and mode are all measures of central tendency and represent a “typical” data point from your dataset. Mean, median and mode are all commonly used in descriptive statistics (read more about descriptive statistics here). These measures have in common that they are used to indicate the distribution of the variable(s) and try to summarize your dataset with a single number to represent a “typical” data point from your dataset. The different measures are calculated in different ways and have different use areas. In this article, we will go through both how these measures are calculated and the appropriate use, but in short: To illustrate the different measures, we will use a small fictional dataset including the weight of 10 individuals as an example: \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| Individual \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 \| 10 \| \| Weight (kg) \| 73 \| 65 \| 70 \| 71 \| 68 \| 90 \| 74 \| 67 \| 89 \| 73 \| Mean is the average value of the data points (observations Median is the value that is exactly in the middle when all values are arranged from low to high Mode is the value which occurs most frequently in your dataset To illustrate the different measures, we will use a small fictional dataset including the weight of 10 individuals as an example: \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| Individual \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 \| 10 \| \| Weight (kg) \| 73 \| 65 \| 70 \| 71 \| 68 \| 90 \| 74 \| 67 \| 89 \| 73 \| Mean The mean can be calculated both as the arithmetic mean and as the geometric mean. The arithmetic mean, often referred to simply as “mean”, is the average value of all data points. It is calculated by adding the values of all data points and dividing them by the number of data points. The geometric mean is also the average value of data points, just like the arithmetic mean. However, to calculate the geometric mean, one must calculate the nth root (with n = total number of data points) of the product of all data points. Median The median of your data points is the value that is exactly in the middle when all values are arranged from low to high. When your dataset has an even number of observations, so there is no value exactly in the middle, calculate the mean of the middle two values. Example: Arrange all values from high to low 65 67 68 70 71 73 73 74 89 90 Find the middle value. NB: Our dataset has an even number of observations, so we will calculate the mean of the middle two values. 65 67 68 70 71 73 73 74 89 90 → Median = 72 Mode The value which occurs most in your dataset is called the mode. Depending on your dataset, there can be no, one, or several modes. The mode can be found by selecting the value that occurs the most. While the mode can be used for numeric data, it is the one measure of central tendency which can also be used for categorical data. Example: First, let’s find the mode for our numerical dataset. The only number which occurs more than once is 73. So the mode in our example dataset is 73. Now, when we categorize our individuals according to their BMI as “underweight”, “normal weight” and “overweight”, we get a new dataset with categorical data. \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| Individual \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 \| 10 \| \| Weight (kg) \| Normal weight \| Underweight \| Normal weight \| Overweight \| Normal weight \| Overweight \| Normal weight \| Underweight \| Overweight \| Normal weight \| To find the mode, creating a frequency table might be helpful. \| \| \| --- \| \| Value \| # observations \| \| Underwegiht \| 2 \| \| Normal weight \| 5 \| \| Overweight \| 3 \| → Mode = Normal weight Visualisation For normally distributed data, the mean, median, and mode will have the same value. This is not the case for skewed data. Figure 1 shows where the mean, median, and mode of your dataset are located for different distributions. When to use mean, median, or mode? Which measure of central tendency to use depends mainly on two things: The variable type The shape of your distribution For categorical variables, there are no numerical values, so the mean cannot be calculated. In this case, the mode is the best measure of central tendency as it tells you which value appears most commonly in your dataset. However, for numerical variables (continuous or discrete variables), both the mean and the median can be calculated. So, when to use which? Here, the shape of your distribution becomes important. As mentioned earlier, when your data is normally distributed, the mean and median will have the same value. However, for skewed distributions, i.e., when one side of the distribution is more spread out and has a longer tail than the other, it is best to use the median. This is because the mean is influenced by extreme values (or “outliers”) and might therefore not be a good representation of a “typical” value of your dataset. Because the median takes a value from the middle of the distribution, it is not as influenced by extreme values and will therefore be a better measure of central tendency. Arithmetic or geometric mean? The geometric mean is suggested to be a better representative of log-normal distributions and percentages. Log-normal distributions are skewed, but after logarithmic transformation, they become normally distributed. These distributions typically have a high number of similar values with few extreme values and are common for biological measurements. For this type of distribution, the arithmetic mean is pulled upward due to the outliers, therefore, the geometric mean tends to be lower and represents smaller values better (See Figure 2). Commonly, biological values have a true zero (meaning their value is actually zero) and do not take on a negative value (e.g., it is not possible to have negative blood pressure or plasma insulin). This is an important fact when it comes to calculating the standard deviation (SD). Typically, an interval within ± 1 SD from the mean covers about 68% of the distribution, while an interval within ± 2 SD covers about 95%. As shown in Figure 2 in red, for the arithmetic mean this interval spans too low below the mean and can contain zero and negative values, which is biologically impossible. Additionally, the interval doesn’t span high enough above the mean. However, this is not true for the interval covered by one or two geometric SD from the geometric mean. If you have a skewed distribution for your measurements, it is most likely more correct to report the geometric mean and its corresponding SD instead of the arithmetic mean. How to calculate these measurements in Ledidi Core (Arithmetic) mean & median When calculating the (arithmetic) mean and median of just one variable, the “Explore” analysis is what you need. Not sure how to run this analysis? Watch our How-to video! If you want to calculate the mean and median of one variable across several groups at once, use the “Compare numeric values” analysis. This How-to video shows you how that’s done. Mode The mode for categorical variables can easily be found by creating a frequency table using the “Frequencies” analysis. This How-to video explains how that can be performed in Ledidi Core More From Ledidi Academy By Dr. Anthea Van Parys, PhD 18.06.2024 Analysing data series in Ledidi Core Statistical analysis Tips and insights By Einar Martin Aandahl, MD, PhD 29.09.2023 Parametric versus nonparametric tests Statistical analysis Tips and insights By Einar Martin Aandahl, MD, PhD 29.09.2023 The difference between association, correlation and causation Statistical analysis Tips and insights
16609	https://quizlet.com/60908677/trig-identities-flash-cards/	Trig Identities Flashcards \| Quizlet hello quizlet Study tools Subjects Create Log in Math Geometry Trigonometry Trig Identities Save Sine Addition Click the card to flip 👆 sin(u+v) = sinu cosv + cosu sinv Click the card to flip 👆 1 / 20 1 / 20 Flashcards Learn Test Blocks Match Created by ChloeMeiCarpenter Students also studied Flashcard sets Study guides Calculus :))) 15 terms Krystal_Tribou Preview Geometry 17 terms cdankoski28 Preview Unit 2 Geometry Vocab 32 terms yeseniap279 Preview The Unit Circle 16 terms Arrogrant Preview MATH 130 FINAL EXAM 37 terms wesley_kuitems Preview Big Ideas Geometry: 5.1-5.4 Review for Quiz 30 terms mjm2497 Preview Kumon New O Test Answers Skipped 13 terms Krish-ly Preview abeka precalculus test 13 16 terms jazzy_narvs Preview calculus 11 terms alex101010102020 Preview Trigonometry Alg 2 Hon 25 terms jwalsh3234 Preview Hon. Geometry - Chapter 10: Circles Vocabulary 13 terms alexandra_shion Preview math 17 terms tpcallas5 Preview Math 15 terms advocateemma Preview Trig Identities and Trigonometry Exam 29 terms n123456789f Preview Circle Unit Test (GEOMETRY) 20 terms ashturtle Preview Unit Circle - Degrees to Radians 16 terms lauren_m_gibbs Preview sin,cos,tan values 21 terms Tyler_Malmloff Preview Unit 4 geometry theorems + notes 16 terms le_chelsea_t Preview math quiz oaoaoeouduobdeobuegbibguinetuegouefouegoiub 22 terms AidanTO123 Preview Geometry Unit 12 Test 16 terms lynchgra Preview Quiz 40 geometry abeka 5 terms JBTA Preview Law of Sines and Law of Cosines Trigonometry Study Guide Teacher 12 terms CathyLeo12 Preview Geometry Chapter 1 Teacher 26 terms Sean_Sinclair13 Preview Geometry module 2 lesson 2 16 terms D_marie27 Preview Unit Circle Radians and Degrees 13 terms soguktasvildand Preview Chapter 7.3-4 Theorems and Postulates 6 terms Jadore_C Preview paralellagram properties 12 terms Martyg2027 Preview Pt 1. Pre calc identities 5 terms skylaranderson5 Preview Practice questions for this set Learn 1 / 7 Study with Learn sinucosv = 1/2 Choose an answer 1 sinucosv Product --> Sum 2 Sine Addition 3 cosusinv Product --> Sum 4 Cosine Subtraction Don't know? Terms in this set (20) Sine Addition sin(u+v) = sinu cosv + cosu sinv Cosine Addition cos(u+v) = cosu cosv - sinu sinv Tangent Addition tan(u+v) = tanu + tanv 1- tanu tanv Sine Subtraction sin(u-v) = sinu cos v - cosu sinv Cosine Subtraction cos(u-v) = cosu cosv + sinu sinv Tangent Subtraction tan(u-v) = tanu - tanv 1 + tanu tanv Sine Double Angle sin2u = 2 sinu cosu Cosine Double Angle cos2u = cos²u - sin²v = 1 - 2sin²u = 2cos²u - 1 Tangent Double Angle tan2u = 2 tanu 1 - tan²u Sine Half Angle sin u/2 = ± √1 - cosu 2 Cosine Half Angle cos u/2 = ± √1 + cosu 2 Tangent Half Angle tan u/2 = 1 - cosu or sin u sin 1 + cosu sinucosv Product --> Sum sinucosv = 1/2 cosusinv Product --> Sum cosusinv = 1/2 cosucosv Product --> Sum cosucosv = 1/2 sinusinv Product --> Sum sinusinv = 1/2 sinu + sinv Sum --> Product sinu + sinv = 2 sin (u+v/2) cos (u-v/2) sinu - sinv Sum --> Product sinu - sinv = 2 cos (u+v/2) sin (u-v/2) cosu + cosv Sum --> Product cosu + cosv = 2 cos (u+v/2) cos (u-v/2) cosu - cosv Sum --> Product cosu - cosv = -2 sin (u+v/2) sin (u-v/2) About us About Quizlet How Quizlet works Careers Advertise with us For students Flashcards Test Learn Study groups Solutions Modern Learning Lab Quizlet Plus Study Guides Pomodoro timer For teachers Live Blog Be the Change Quizlet Plus for teachers Resources Help center Sign up Honor code Community guidelines Terms Privacy California Privacy Your Privacy/Cookie Choices Ads and Cookie Settings Interest-Based Advertising Quizlet for Schools Parents Language Get the app Country United States Canada United Kingdom Australia New Zealand Germany France Spain Italy Japan South Korea India China Mexico Sweden Netherlands Switzerland Brazil Poland Turkey Ukraine Taiwan Vietnam Indonesia Philippines Russia © 2025 Quizlet, Inc. Students Flashcards Learn Study Guides Test Expert Solutions Study groups Teachers Live Blast Categories Subjects GCSE A Levels Arts and Humanit... Languages Math Science Social Science Other Flashcards Learn Study Guides Test Expert Solutions Study groups Live Blast Categories GCSE A Levels Arts and Humanit... Languages Math Science Social Science Other
16610	https://brainly.com/question/29128927	[FREE] Solve the system by graphing: y = 2x - 3 y = -x + 9 - brainly.com 5 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +76,8k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +11,3k Ace exams faster, with practice that adapts to you Practice Worksheets +5,5k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified Solve the system by graphing: y=2 x−3 y=−x+9 1 See answer Explain with Learning Companion NEW Asked by ItzelZ574331 • 10/29/2022 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer 3.0 0 Upload your school material for a more relevant answer (4,5) Explanation Given the system of equations: y=2 x−3 y=−x+9 We graph each of the equation using the x and y-intercepts. First Equation(y=2x-3) When x=0 y=2 x−3 y=2(0)−3 y=−3⟹(0,−3) When y=0 0=2 x−3 2 x=3 x=2 3x=1.5⟹(1.5,0) Next, join the points (0,-3) and (1.5,0) as shown below: Second Equation(y=-x+9) When x=0 y=−x+9 y=−0+9 y=9⟹(0,9) When y=0 0=−x+9 x=9⟹(9,0) Next, join the points (0,9) and (9,0) on the same graph as shown below: The point where the two lines intersect is the solution to the system of equations. Therefore, the solution is (4,5). Answered by JaboriW94199 •1 answer Thanks 0 3.0 (1 vote) Textbook &Expert-Verified⬈(opens in a new tab) 0.0 0 International Trade - Theory and Policy Introduction to Systems Biology - Thomas Sauter; Marco Albrecht; Principles of Microeconomics 3e - David Shapiro, Daniel Macdonald, Steven A. Greenlaw Upload your school material for a more relevant answer The solution to the system of equations y=2 x−3 and y=−x+9 by graphing is the point (4,5). This can be found by plotting points for both equations and identifying where the lines intersect. Explanation To solve the system of equations by graphing, we have the following equations: y=2 x−3 y=−x+9 Step 1: Finding Points of the First Equation For the first equation y=2 x−3: When x=0: y=2(0)−3=−3⟹(0,−3) When y=0: 0=2 x−3⟹2 x=3⟹x=2 3=1.5⟹(1.5,0) Step 2: Finding Points of the Second Equation For the second equation y=−x+9: When x=0: y=−0+9=9⟹(0,9) When y=0: 0=−x+9⟹x=9⟹(9,0) Step 3: Graphing the Lines Now, plot the points from both equations on a graph: For y=2 x−3, plot (0, -3) and (1.5, 0). For y=−x+9, plot (0, 9) and (9, 0). Step 4: Finding the Intersection Draw a straight line through the points plotted for each equation. The intersection of these lines will give you the solution to the system of equations. By graphing, you can see that the two lines intersect at the point (4,5). Therefore, the solution to the system of equations is (4,5). Examples & Evidence An example to help understand graphing would be to take other equations, such as y=x+1 and y=−2 x+4, and similarly plot points, find intersections, and solve the system. The method used here is based on algebraic graphing principles where the intersection of two linear equations represents the solution to the system. Thanks 0 0.0 (0 votes) Advertisement ItzelZ574331 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 5.0 1 Solve the following system of equations graphically. Select the graph that shows the correct solution of the system. 2x +y =3 3y= x - 12 Community Answer Solve the following linear system graphically. 2x+y=9 x−3y=−6 Use the graphing tool to graph the system. What is the solution of the system of equations? (Type an ordered pair.) Community Answer Solve the system of linear equations by graphing. 2x + 3y = 16.9 5x = y + 7.4 Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? New questions in Mathematics A class of 24 students is planning a field trip to a science museum. A nonrefundable deposit of $50 is required for the day-long program, plus a charge of $4.50 per student. Determine a linear function that models the cost, c, and the number of students, s. Which statements about the linear function and its graph are correct? Check all that apply. A. The linear model is f(s)=4.5 s+50. B. The linear model is f(c)=54.5 c+24. C. The domain is x∣0≤x≤24. D. The range is y∣50≤y≤158. E. The graph is continuous. What is the equation of the translated function, g(x), if f(x)=x 2? A. g(x)=(x−4)2+6 B. g(x)=(x+6)2−4 C. g(x)=(x−6)2−4 D. $g(x)=(x+4)^2+6 What is the equation of the translated function, g(x), if f(x)=x 2?A. g(x)=(x−4)2+6 B. g(x)=(x+6)2−4 C. g(x)=(x−6)2−4 D. g(x)=(x+4)2+6 Which translation maps the vertex of the graph of the function 4 x y=x 2 onto the vertex of the function 9 x=x 2+2 x 41? A. right 1 unit B. left 1 unit C. right 2 units D. left 2 units Which are true of the function f(x)=49(7 1)x ? Select three options. The domain is the set of all real numbers. The range is the set of all real numbers. The domain is x>0. The range is y>0. As x increases by 1 , each y-value is one-seventh of the previous y-value. 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16611	https://www.frontiersin.org/journals/nutrition/articles/10.3389/fnut.2024.1426790/full	Your new experience awaits. Try the new design now and help us make it even better ORIGINAL RESEARCH article Front. Nutr., 23 July 2024 Sec. Nutritional Epidemiology Volume 11 - 2024 \| This article is part of the Research TopicMicronutrients, Immunity and InfectionView all 27 articles Trends in three malnutrition factors in the global burden of disease: iodine deficiency, vitamin A deficiency, and protein-energy malnutrition (1990–2019) Background: Vitamin A deficiency, iodine deficiency, and protein-energy malnutrition are prevalent malnutrition issues that disproportionately affect low-income countries and pose significant risks to the health and development of children and adolescents. This study offers a detailed examination of these deficiencies' prevalence trends and gender and regional variations using Global Burden of Disease Study data from 1990 to 2019. It also assesses the specific impact on various age groups, providing essential insights for targeted health interventions and policy-making. Methods: Data spanning from 1990 to 2019 on Vitamin A deficiency, iodine deficiency, and protein-energy malnutrition were extracted from the 2019 Global Burden of Disease Study. Age-Standardized Incidence Rates (ASR) were computed by gender, region, and etiology, utilizing the estimated annual percentage change (EAPC) to assess temporal trends. Results: In 2019, Central Sub-Saharan Africa had the highest prevalence of Vitamin A deficiency, particularly among males, and iodine deficiency peaked in the same region for both genders. South Asia had the highest incidence of protein-energy malnutrition for both genders. Regions with a low Socio-Demographic Index (SDI) showed lower ASR for these deficiencies. Notably, Cameroon, Equatorial Guinea, and Maldives recorded the highest ASR for vitamin A deficiency, iodine deficiency, and protein-energy malnutrition, respectively. The declining ASR trend for vitamin A deficiency, especially among males, suggests effective interventions. East Asia saw a significant increase in iodine deficiency ASR from 1990 to 2019, particularly among women, requiring targeted interventions. The rising ASR of protein-energy malnutrition in several regions, especially among men, raises concerns. Vitamin A deficiency primarily affected children and adolescents, iodine deficiency predominantly impacted adolescents and young adults, and protein-energy malnutrition was chiefly observed among children under 5 years old. These findings underscore the necessity for tailored interventions considering age-specific nutritional needs and challenges. 1 Introduction Malnutrition remains a pervasive global health challenge, with far-reaching implications for both individual wellbeing and public health. Iodine, vitamin A, protein, and energy are essential nutrients for the human body, playing a role in numerous physiological functions. Insufficiency or deficiency of these nutrients can result in significant health problems. Iodine deficiency is the leading cause of preventable cognitive impairments and is linked to thyroid dysfunction, miscarriages, premature births, stillbirths, and congenital deformities (1). Vitamin A deficiency is a significant factor in childhood blindness and mortality, as well as its links to infectious diseases, anemia, and reproductive health (2). Insufficient protein and energy intake can weaken the immune system, elevating susceptibility to infections and mortality. The World Health Organization acknowledges iodine deficiency, vitamin A deficiency, and protein-energy malnutrition as global public health issues (3). As per the 2020 Global Nutrition Report, iodine deficiency, vitamin A deficiency, and protein-energy malnutrition represent substantial global risk factors for mortality and disability-adjusted life years (DALYs) (4). Despite substantial public health endeavors in recent decades to tackle this problem, elevated prevalence rates endure in low- and middle-income regions (5). The main reasons include economic constraints leading to a monotonous diet, lack of education and awareness about nutrition, soil and environmental factors affecting the nutritional value of food, heavy disease burdens impacting nutrient absorption, and social and cultural customs restricting the intake of nutrient-rich foods. Moreover, inadequate health systems, absence of targeted nutrition policies, and the adverse effects of climate change on agricultural production have all exacerbated nutritional problems in these countries (6, 7). In 2015, the United Nations General Assembly set forth the Sustainable Development Goals (SDGs) to eliminate all malnutrition forms by 2030 (8). This encompasses tackling health issues like child growth stunting, wasting, adolescent malnutrition, and the nutritional needs of pregnant, lactating women, and the elderly. To attain this goal, acquiring an understanding of the patterns and temporal trends in iodine deficiency, vitamin A deficiency, and protein-energy malnutrition incidence can facilitate the adoption of more focused preventive strategies, thereby contributing to the attainment of the Sustainable Development Goals. The Global Burden of Disease (GBD) study has evaluated the prevalence of iodine deficiency, vitamin A deficiency, and protein-energy malnutrition in 204 countries and regions across the globe (9). This provides a distinct opportunity to gain insights into how to address these deficiencies. In this study, we aim to evaluate the temporal trends in the incidence of iodine deficiency, vitamin A deficiency, and protein-energy malnutrition from a global, regional, and national perspective over the period from 1990 to 2019. Our research findings can contribute to the development of targeted prevention and intervention strategies that are tailored to the needs of different nations and populations. 2 Materials and methods 2.1 Study data We gathered data on the incidence, age-standardized incidence, and mortality rates of iodine deficiency, vitamin A deficiency, and protein-energy malnutrition, broken down by gender, region, and country. These data span from 1990 to 2019 and were collected through the GBD Results Tool (GHDx) available at: The dataset encompassed 204 countries and territories. Subsequently, we categorized these nations and regions into five groups based on the sociodemographic index (SDI), which includes low, low-middle, middle, high-middle, and high categories. Additionally, we partitioned the globe into 21 geographical regions (Supplementary Table 1). A comprehensive account of the indicators, data sources, and statistical models employed in GBD 2019 has been documented in previous publications and adheres to the Guidelines for Accurate and Transparent Health Estimates Reporting (9, 10). According to the International Classification of Diseases, Tenth Revision (ICD-10), protein-energy malnutrition is coded as E40–E46.9; iodine deficiency is coded as E00-E02, and vitamin A deficiency is coded as E50–E50.9. 2.2 Statistical analysis To assess the trends in iodine deficiency, vitamin A deficiency, and protein-energy malnutrition, we use age-standardized incidence rates (ASR) and estimated annual percentage change (EAPC). Direct comparison between populations with different age structures can introduce bias in crude rate comparisons. Hence, standardizing rates is essential. ASR (per 100,000 individuals) is calculated using the direct method and is defined by Equation (1). EAPC is a commonly utilized metric for summarizing rate trends over specific time intervals. It is computed using a linear regression model, as represented in Equations (2, 3). We determine a 95% confidence interval using the EAPC formula mentioned earlier. The standard error is obtained from fitting the regression line. If both the estimated EAPC value and its lower 95% confidence limit are >0, the age-standardized rate is considered to be increasing. Conversely, if both the estimated EAPC value and its upper 95% confidence limit are <0, the age-standardized rate is considered to be decreasing. All statistical analyzes were conducted using the R program (Version 4.2.1, R Core Team). A P-value <0.05 was regarded as statistically significant. where i denotes the ith age class; a denotes age-specific rates; ω denotes the number of people (or weight) where y represents the natural logarithm of the age-standardized rate, x corresponds to the years in question, and β stands for the estimated slope. 3 Results 3.1 Global burden of disease of vitamin A deficiency, iodine deficiency, and protein-energy malnutrition in 2019 In 2019, global Age-Standardized Rates (ASRs) for vitamin A deficiency, iodine deficiency, and protein-energy malnutrition were 6,955.6, 108.3, and 2,099.4 per 100,000, respectively. Among females, ASRs were 5,999.1, 139.8, and 1,894.6 per 100,000, and among males, 7,886.2, 78.1, and 2,304 per 100,000 (Table 1). According to the Social Demographic Index (SDI), regions with low SDI had lower ASRs for vitamin A deficiency, iodine deficiency, and protein-energy malnutrition, while high SDI regions exhibited the opposite trend (Supplementary Tables 1, 4, 7). In 2019, among the 21 regions, Central Sub-Saharan Africa had the highest ASR for vitamin A deficiency at 33,739.9 per 100,000 (range: 30,648–37,138.4), followed by Eastern Sub-Saharan Africa and Western Sub-Saharan Africa. For females, Eastern Sub-Saharan Africa had the highest ASR for vitamin A deficiency in 2019 at 20,731.4 per 100,000 (range: 19,414.7–22,180.9), while for males, it was Central Sub-Saharan Africa with an ASR of 33,242.3 per 100,000 (range: 28,347.2–38,656.5; Supplementary Tables 4–6). Central Sub-Saharan Africa had the highest ASR for iodine deficiency in 2019, at 459 per 100,000 (range: 371.5–555.8), followed by South Asia and Eastern Sub-Saharan Africa. For both females and males, the highest ASRs were in Central Sub-Saharan Africa, with 459 per 100,000 (range: 371.5–555.8) and 320.5 per 100,000 (range: 255–391.2), respectively (Supplementary Tables 1–3). In 2019, the region with the highest ASR for protein-energy malnutrition was South Asia at 1,023.5 per 100,000 (range: 860–1,208.2), followed by Southeast Asia and East Asia. For both females and males, the highest ASRs of protein-energy malnutrition were in South Asia, with 3,427.8 per 100,000 (range: 2,828.8–4,111.8) and 3,759.6 per 100,000 (range: 3,101.9–4,516), respectively (Supplementary Tables 7–9). Table 1. The incident cases and age-standardized incidence of vitamin A deficiency, iodine deficiency, and protein-energy malnutrition in 1990 and 2019, and its temporal trends from 1990 to 2019. In 204 countries and territories worldwide, there are significant variations in the ASR of vitamin A deficiency, iodine deficiency, and protein-energy malnutrition (Figure 1). In 2019, Cameroon exhibited the most pronounced ASR for vitamin A deficiency, reaching 69,494.5 per 100,000 (within the range of 66,902.4–72,335.4), surpassing Somalia and Niger. For females, the country with the highest ASR for vitamin A deficiency was Somalia at 58,377.3 per 100,000 (ranging from 53,054.1 to 63,870.7), while for males, it was Cameroon at 92,312.4 per 100,000 (ranging from 89,901.8 to 94,293.4; Supplementary Tables 13–15; Supplementary Figures 1, 2). Equatorial Guinea recorded the highest ASR for iodine deficiency at 1,071.8 per 100,000 (ranging from 902.5 to 1,226.4), followed by the Democratic Republic of the Congo and Somalia. The countries with the highest ASR for iodine deficiency among females and males were Equatorial Guinea, with 1,387.4 per 100,000 (ranging from 1,193.1 to 1,571.3), and 760 per 100,000 (ranging from 613.2 to 899.1), respectively (Supplementary Tables 10–12; Supplementary Figures 1, 2). Maldives recorded the highest ASR for protein-energy malnutrition, with a level of 4,292.4 per 100,000 (ranging from 3,565.3 to 5,068.5), surpassed only by Sri Lanka and Timor-Leste. For females, the country with the highest ASR for protein-energy malnutrition was Maldives at 3,421.6 per 100,000 (ranging from 2,819.4 to 4,204.5), while for males, it was Sri Lanka at 5,876.1 per 100,000 (ranging from 4,963.3 to 6,809.5; Supplementary Tables 16–18; Supplementary Figures 1, 2). Figure 1. The ASR of vitamin A deficiency, iodine deficiency, and protein-energy malnutrition for both genders in 204 countries and territories in 2019. (A) Vitamin A deficiency. (B) Iodine deficiency. (C) Protein-energy malnutrition. In absolute terms, India exhibited the highest cases of vitamin A deficiency, with 292,439,000 cases in 2019, followed by China and Indonesia (Supplementary Table 5). India also bore the greatest burden of individuals affected by iodine deficiency, reporting 37,856,000 cases in 2019, trailed by China and Bangladesh. Notably, India carried the heaviest load of protein-energy malnutrition, with an astounding count of 376,672,000 cases, surpassed by China and Indonesia. 3.2 Changes in the burden of vitamin A deficiency, iodine deficiency, and protein-energy malnutrition over time Globally, from 1990 to 2019, there has been a declining trend in the ASR of vitamin A deficiency, with a more pronounced decrease in males compared to females. Similar patterns are observed across different SDI regions (Figure 2; Supplementary Figures 3–7). The ASR for iodine deficiency exhibited a global decline from 1990 to 2000, followed by a slow increase from 2000 to 2005, and a subsequent decline. In high SDI, low-middle SDI, and low SDI regions, the ASR for iodine deficiency decreased from 1990 to 2019, while in high-middle SDI and middle SDI regions, there was a pattern of decrease, increase, and then decrease (Figure 2; Supplementary Figures 3–7). From 1990 to 2010, the global ASR for protein-energy malnutrition showed an increasing trend, followed by a decline from 2010 to 2015, and then a subsequent increase. In high SDI, high-middle SDI, and middle SDI regions, the ASR for protein-energy malnutrition exhibited an upward trend from 1990 to 2010, followed by a decline from 2010 to 2015. Conversely, in low-middle SDI and low SDI regions, there was a decline in the ASR from 2010 to 2015. Post-2015, different SDI regions witnessed an upward trend in the ASR for protein-energy malnutrition (Figure 2; Supplementary Figures 3–7). Furthermore, on a global scale and across different SDI regions, the general trend in the ASRs of vitamin A deficiency, iodine deficiency, and protein-energy malnutrition remained consistent from 1990 to 2019 for both males and females. However, the ASR were generally higher in males for vitamin A deficiency and protein-energy malnutrition, while females exhibited higher ASR for iodine deficiency (Figure 2; Supplementary Figures 3–7). In 21 regions worldwide, from 1990 to 2019, the ASR of iodine deficiency in East Asia showed an increasing trend of (EAPC = 0.17; 95% CI: –0.28 to 0.62), mainly among women (EAPC = 0.67; 95% CI: 0.30 to 1.04). The ASR of protein-energy malnutrition in the regions of Australasia, East Asia, Central Europe, Western Europe, and Southern Latin America also showed an upward trend, more pronounced in men than in women (EAPC > 0). No regions were found where the ASR of vitamin A deficiency showed an increasing trend from 1990 to 2019 (EAPC <0; Figure 2; Supplementary Tables 1, 4, 7). Figure 2. Changes in the burden of vitamin A deficiency, iodine deficiency, and protein-energy malnutrition over time. From 1990 to 2019, the ASR of vitamin A deficiency showed a decreasing trend (EAPC <0) in 204 countries globally. Among them, Equatorial Guinea exhibited the most significant decline (EAPC = –9.90; 95% CI: –10.39 to –9.41), followed by Saudi Arabia and Maldives. In both females (EAPC = –9.88; 95% CI: –10.32 to –9.43) and males (EAPC = –10.09; 95% CI: –10.62 to –9.56), Equatorial Guinea showed the most prominent decrease in the ASR for vitamin A deficiency (Figure 3; Supplementary Figures 8, 9; Supplementary Tables 13–15). Twelve countries and regions experienced an increasing trend in the ASR for iodine deficiency from 1990 to 2019. The sequence of this upward trend included the following: Philippines, Pakistan, Nepal, Republic of Moldova, South Sudan, Madagascar, Somalia, China, Kenya, Monaco, Portugal, and Andorra. Among females, the countries and regions with an increasing trend in the ASR for iodine deficiency included Philippines, Nepal, Pakistan, China, Republic of Moldova, South Sudan, and others, while among males, it was mainly observed in countries like Philippines, Somalia, Romania, and others (Figure 3; Supplementary Figures 8, 9; Supplementary Tables 10–12). Turkey, Czechia, Bhutan, Montenegro, Paraguay, Norway, Australia, and China, among others, witnessed an increasing trend in the ASR for protein-energy malnutrition from 1990 to 2019. Notably, among females, this trend was prominent in Czechia, Turkey, Qatar, Paraguay, Norway, Montenegro, and others, while among males, it was observed in Bhutan, Turkey, Czechia, China, Montenegro, Australia, and others (Figure 3; Supplementary Figures 8, 9; Supplementary Tables 16–18). Figure 3. The EAPC of iodine deficiency, vitamin A deficiency, and protein-energy malnutrition for both genders in 204 countries and territories from 1990 to 2019. (A) Vitamin A deficiency. (B) Iodine deficiency. (C) Protein-energy malnutrition. 3.3 Age composition of vitamin A deficiency, iodine deficiency, and protein-energy malnutrition In the year 2019, on a worldwide scale (Figure 4), the occurrence of vitamin A deficiency was predominantly centered on the demographic of children and adolescents. Concurrently, iodine deficiency predominantly impacted adolescents and young adults. The focal point of protein-energy malnutrition was observed predominantly among the age group of children under 5 years old. Figure 4. Global age-sex distribution of incident cases of iodine deficiency, vitamin A deficiency, and protein-energy malnutrition in 2019. 4 Discussion Nutritional deficiencies have profound implications for societal development. They compromise individual health and productivity, increase healthcare burdens, and impede economic growth. Furthermore, education and learning outcomes are adversely affected, diminishing the overall developmental potential of nations (11–13). In this study, we conducted a comprehensive analysis of the disease burden associated with vitamin A deficiency, iodine deficiency, and protein-energy malnutrition at the global, regional, and national levels to identify areas where these issues persist as significant public health concerns. Our research indicates that vitamin A deficiency is particularly prevalent in Central Africa, especially in Cameroon, where the ASR of vitamin A deficiency among men is the highest. Zhao et al.'s study reveals that in 2019, the age-standardized DALY rate for vitamin A deficiency was highest in Central Sub-Saharan Africa, followed by Western Sub-Saharan Africa and again Western Sub-Saharan Africa (14).This condition is primarily influenced by poor dietary patterns, environmental pollution of water and soil, infections and diseases, and insufficient iodine content in salt (15–17). The local population predominantly relies on cereals, tubers, and legumes as their main food sources, while the intake of animal products, green or yellow vegetables, and fruits—foods rich in vitamin A or carotenoids—is relatively low, leading to inadequate vitamin A consumption (18). From 1990 to 2019, the annual incidence rate of vitamin A deficiency has declined, with a more pronounced decrease among men, indicating that the interventions and strategies implemented have achieved some success. Since 2000, mortality due to vitamin A deficiency has decreased by over 50%, likely attributable to improved nutritional status, enhanced water and sanitation conditions (such as the control of diarrhea), vaccination (such as measles vaccination), and vitamin A supplementation programs (19–21). However, the high incidence among women persists, suggesting the need for ongoing efforts and gender-specific strategies. our study highlights that vitamin A deficiency has become a major issue among children and adolescents, a particularly vulnerable group for whom vitamin A is crucial for vision, immune function, and overall growth (22). Global research also identifies vitamin A deficiency in children and adolescents as an increasingly significant problem, underscoring the urgency of addressing this issue (23). In Central Africa, Equatorial Guinea exhibits the highest ASR of iodine deficiency. Research reports that in 2019, Somalia, the Democratic Republic of the Congo, Djibouti, and the Republic of the Congo had the highest age-standardized prevalence rates. Meanwhile, countries like the Philippines and Pakistan have shown an increasing trend. Geographically, the highest prevalence rates are found in Central Sub-Saharan Africa and South Asia (24). The primary cause is the limited consumption of iodine-rich foods such as seafood, milk, and eggs. Additionally, some foods contain goitrogens, such as thiocyanates and soy isoflavones, which inhibit iodine absorption and utilization, exacerbating iodine deficiency (25). According to a report, low iodine levels in soil and water in the region affect both crops and drinking water, contributing to the deficiency. Moreover, soil and water contamination by heavy metals, pesticides, and industrial effluents further destabilize and reduce iodine availability (15, 26, 27). Consuming iodized salt is an effective and economical method to prevent and control iodine deficiency. However, in sub-Saharan Africa, the coverage and quality of iodized salt remain inadequate, perpetuating the deficiency issue. Contributing factors include weak enforcement of iodization policies, insufficient monitoring and evaluation, outdated technology and equipment, cost-benefit disparities, and socio-cultural barriers (15, 28). This study indicates that the global ASR of iodine deficiency decreased from 1990 to 2000, increased from 2000 to 2005, and then declined again, reflecting the impact of global initiatives, particularly the Universal Salt Iodization (USI) policy (29). According to UNICEF's 2020 report, 124 countries have mandated iodized salt, 21 allow voluntary iodization, and 88% of the global population uses iodized salt (30). However, challenges persist in implementing the USI policy, including inaccurate iodine levels in salt, weak quality control and enforcement, low public awareness and education, and a lack of regular surveys and monitoring of iodine status and iodization program impacts (31). These challenges may explain the ASR increase from 2000 to 2005, as some countries experienced a resurgence or persistence of iodine deficiency due to weakened or interrupted USI policies. Research also indicates a unique global pattern of iodine deficiency, predominantly affecting adolescents and young adults. Adequate iodine intake is crucial for thyroid function, and its deficiency can lead to various health issues, particularly during the critical developmental stages of adolescence and early adulthood (32). Therefore, ensuring sufficient iodine intake in this age group is vital for preventing related health problems. Protein-energy malnutrition is prevalent in South Asia, affecting both genders equally. Xu et al.'s research indicates that South Asia, Southeast Asia, and East Asia rank highest in age-standardized prevalence (33). This phenomenon is likely linked to the region's dense population and limited land resources. Additionally, climate change, natural disasters, and pests impact agricultural production, resulting in unstable grain yields and subsequent food shortages (33–35). Furthermore, pervasive poverty, low per capita income, and high food prices in South Asia make it challenging for many to access sufficient protein and energy-rich foods (36). Social and cultural factors also play a significant role in food distribution and consumption. In some areas, women and children are disadvantaged in food allocation, leading to insufficient protein and energy intake. For example, in regions like India, religious and cultural practices often limit the intake of animal-based foods, with vegetarianism being widespread (7, 37, 38). Studies show a correlation between the Socio-Demographic Index (SDI) and the ASR of nutritional deficiencies. Regions with lower SDI tend to have lower ASR, indicating that socio-economic factors play a crucial role in the prevalence of nutritional deficiencies (39). This finding underscores the need for comprehensive strategies addressing both nutritional needs and broader socio-economic determinants. In absolute terms, India bears the greatest burden of nutritional deficiencies, followed by China and Indonesia. This may be attributed to their large populations, highlighting the need for country-specific strategies to address these issues. However, it is essential to note that while discussing global averages, significant variations can exist within countries in the same region (40). The global ASR of protein-energy malnutrition has experienced fluctuations since 1990, with an initial rise, a subsequent decline, and another increase post-2015, raising concerns and calls for renewed efforts to address the issue. The rising trend in ASR in regions like Australia, East Asia, Central Europe, Western Europe, and Latin America, particularly among males, warrants attention. This indicates the influence of socio-economic factors on nutritional deficiencies in different SDI regions, necessitating region-specific strategies (33). Protein-energy malnutrition primarily affects children under five, highlighting the importance of addressing nutritional needs early in life. During this critical growth and development stage, deficiencies in protein and energy can have long-term impacts on physical and cognitive development (41). Therefore, early nutritional interventions are vital for healthy childhood development. To address these global nutritional challenges, comprehensive strategies are imperative. First, targeted public health campaigns should elevate awareness among children and adolescents about the critical importance of a vitamin A-rich diet, emphasizing its roles in vision, immune health, and overall growth. Additionally, strengthening iodine supplementation programs and educational initiatives for adolescents and young adults is crucial to ensure optimal thyroid function and prevent associated health issues. To combat protein-energy malnutrition in children under five, expanding community-based nutrition programs to provide access to nutrient-dense foods and offer nutrition education to caregivers is essential. Collaboration among governments, non-governmental organizations, and international agencies is vital for developing and implementing sustainable solutions, promoting a holistic approach to global nutrition across all age groups. To the best of our knowledge, this study represents the first comprehensive overview and exploration of global disparities in vitamin A deficiency, iodine deficiency, and protein-energy malnutrition, along with their evolving patterns stratified by gender and age. This research has inherent limitations. Despite the adjustments made in GBD 2019 to account for biases and methodological flaws in low-quality sampling, survey methods, and other data sources, the accuracy of the results is significantly contingent upon the quality and quantity of the input data into the model. Moreover, the absence of diagnostic gold standards and the potential interchangeability of these terms might lead to an underestimation of the scale of these diseases. 5 Conclusions Over the past several decades, substantial improvements have been made in addressing global issues related to vitamin A deficiency, iodine deficiency, and protein-energy malnutrition. However, progress remains uneven on a global scale, with nutritional deficiencies posing persistent concerns in countries with lower socioeconomic levels. Urgent policy safeguards and ongoing efforts are imperative to control nutritional deficiencies in these regions. Vitamin A deficiency predominantly impacts children and adolescents, while iodine deficiency primarily affects adolescents and young adults. Meanwhile, protein-energy malnutrition is primarily observed among children under the age of 5. These findings underscore the necessity for age-specific interventions tailored to the distinct nutritional needs and challenges of each age group. Further research is warranted to comprehend and tackle the underlying causes of these age-specific trends. Data availability statement The datasets presented in this study can be found in online repositories. The names of the repository/repositories and accession number(s) can be found in the article/Supplementary material. Ethics statement The studies involving humans were approved by Health Metrics and Evaluation (IHME) at the University of Washington. The studies were conducted in accordance with the local legislation and institutional requirements. The participants provided their written informed consent to participate in this study. Author contributions SJ: Conceptualization, Data curation, Formal analysis, Methodology, Writing — original draft, Writing — review & editing. YZ: Data curation, Formal analysis, Methodology, Writing — original draft, Writing — review & editing. QZ: Data curation, Methodology, Writing — original draft. RC: Data curation, Visualization, Writing — original draft. ZS: Writing — original draft. Funding The author(s) declare that no financial support was received for the research, authorship, and/or publication of this article. Acknowledgments We would like to express our gratitude to the team at the Global Burden of Disease study 2019 collaborators for their invaluable contributions in designing, collecting, and compiling the GBD data, as well as for their efforts in developing the public database. Conflict of interest The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest. Publisher's note All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article, or claim that may be made by its manufacturer, is not guaranteed or endorsed by the publisher. 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Nutrition in school-age children: a rationale for revisiting priorities. Nutr Rev. (2023) 81:823–43. doi: 10.1093/nutrit/nuac089 PubMed Abstract \| Crossref Full Text \| Google Scholar Keywords: vitamin A deficiency, iodine deficiency, protein-energy malnutrition, global burden, age-standardized rate Citation: Ji S, Zhou Y, Zhao Q, Chen R and Su Z (2024) Trends in three malnutrition factors in the global burden of disease: iodine deficiency, vitamin A deficiency, and protein-energy malnutrition (1990–2019). Front. Nutr. 11:1426790. doi: 10.3389/fnut.2024.1426790 Received: 02 May 2024; Accepted: 09 July 2024; Published: 23 July 2024. Edited by: Reviewed by: Copyright © 2024 Ji, Zhou, Zhao, Chen and Su. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms. Correspondence: Shaorong Ji, anNyMTIxNUAxNjMuY29t Disclaimer: All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher. Share on Share on Frontiers' impact Articles published with Frontiers have received 12 million total citations Your research is the real superpower - learn how we maximise its impact through our leading community journals Supplementary Material
16612	https://www.quora.com/How-do-I-find-the-point-of-intersection-of-a-straight-line-and-an-ellipse	Something went wrong. Wait a moment and try again. Point of Intersection Ellipses (geometry) Straight Lines Analytic Curves Surface Intersections Geometric Mathematics Ellipse and Hyperbola The Ellipse 5 How do I find the point of intersection of a straight line and an ellipse? Pranav Kasetty Chemistry enthusiast · Author has 51 answers and 310.7K answer views · 8y In general you can find the point(s) of intersection of two curves by solving them simultaneously. The solutions to both the equations are the points of intersection as they lie on both the curves. For the specific example of straight line and ellipse, convert the straight line from the form ax+by+c=0 to the form y=mx+c. Now substitute this value of y in the equation of the ellipse and solve the quadratic equation that you get. Once you obtain the value(s) of x , then substitute them in y=mx+c and get the corresponding y values. Note: The above stated method is only one of many methods for solvin In general you can find the point(s) of intersection of two curves by solving them simultaneously. The solutions to both the equations are the points of intersection as they lie on both the curves. For the specific example of straight line and ellipse, convert the straight line from the form ax+by+c=0 to the form y=mx+c. Now substitute this value of y in the equation of the ellipse and solve the quadratic equation that you get. Once you obtain the value(s) of x , then substitute them in y=mx+c and get the corresponding y values. Note: The above stated method is only one of many methods for solving the equations.Also, I have assumed basic knowledge of coordinate geometry, algebra, straight line and conic sections. Please do let me know if you did not understand any part of the answer. Feel free to point out any mistakes that I might have made. Stay Curious! Related questions Where is the point of intersection on an ellipse? How do I find the intersection point between a circle and a straight line? How do you find the point of intersection of a pair of straight lines if the point of intersection is not origin? How do you find the point of intersection between two curves and one straight line? How can I find the point of intersection of three lines simultaneously? Jack Fraser-Govil Doctor of Physics, Writer of Code, Player of Games · Author has 2.6K answers and 51.5M answer views · 8y Related How do I find the intersection point between a circle and a straight line? The question originally asked for a solution specific to MATLAB — but the question is actually more interesting than just a limited programming question, so I have modified it to be more general. That’s because the easiest way to solve this problem is to solve it algebraically — and then just plug the resulting formula into MATLAB as-is! We tackle the problem using vectors. You first need to describe the line as a vector equation: →x=→d+λ^n Where →d is the position vector of a point on the line, ^n is a unit vector in the direction of the line, and λ is a p The question originally asked for a solution specific to MATLAB — but the question is actually more interesting than just a limited programming question, so I have modified it to be more general. That’s because the easiest way to solve this problem is to solve it algebraically — and then just plug the resulting formula into MATLAB as-is! We tackle the problem using vectors. You first need to describe the line as a vector equation: →x=→d+λ^n Where →d is the position vector of a point on the line, ^n is a unit vector in the direction of the line, and λ is a parameter which ‘slides’ up and down the line. Our aim is to find the values of λ which correspond to the intersection with the circle, and we can then plug this value in to the above equation to find the position. We also need to define a sphere in vector notation: (→x−→c)2=r2 Where →c is the position vector of the centre of the circle, and r is the radius. At the intersection of these two systems, →xsphere=→xline — so we simply plug in equation 1 into equation 2, to get: (→d+λ^n−→c)2=r2 Expand the brackets: d2+λ2+c2+2λ→d⋅^n−2λ→c⋅^n−2→d⋅→c=r2 (\|^n\|=1, so we didn’t include n2) λ is the article we are interested in — so: λ2+2λ(→d−→c)⋅^n+(d2+c2−r2−2→d⋅→c)=0 This is now a simple quadratic in λ — recalling that the quadratic formula ax2+bx+c=0 has solutions of the form: x=−b±√b2−4ac2a Therefore: λ=(→c−→d)⋅^n±√[(→c−→d)⋅^n]2−(d2+c2−r2−2→d⋅→c) We can then plug this back into equation one to get our possible solutions: →x±=→d+((→c−→d)⋅^n±√[(→c−→d)⋅^n]2−(d2+c2−r2−2→d⋅→c))^n This is the formula for working out the intersection between a line and a sphere in arbitrary dimensions Rn It is then a simple matter to code this into MATLAB (as the question originally asked for) %define the lined = [2 0 0];n = [0 1 0]; %define the circlec = [0 0 0];r = 5; %%Ensure n is a unit vectorn = n./sqrt(dot(n,n)); %% Calculate lambdaprefactor = dot(c - d, n); squareRoot = sqrt(prefactor^2 - (dot(d,d) +dot(c,c) - r^2 - 2dot(d,c))); lambdaPlus = prefactor + squareRoot;lambdaMinus = prefactor - squareRoot; %Check if solutions are realif imag(lambdaPlus)~=0 disp('No Solutions Available') else %%If solutions are real, output xPlus = d + lambdaPlusn; xMinus = d + lambdaMinusn; disp('Positive Solution');disp(xPlus); disp('Negative Solution');disp(xMinus); end I have made liberal use of the inbuilt MATLAB support for vectors (in fact, MATLAB is pretty much the perfect language for this — it’s designed to make vector and matrix algebra as easy as possible! ) Using the dummy values of →c, →d, ^n and r that I input above, the output of the script is: Positive Solution 2.0000 4.5826 0 Negative Solution 2.0000 -4.5826 0 And indeed, if we plug this into WolframAlpha it gives us the same result: Computational Knowledge Engine The “imaginary” check occurs because if the line and the sphere never intersect (in Real space, at least), then the square root will be of a negative number — so we are just eliminating this condition. This method generalises to arbitrary dimensions — it works if →c etc. are 2D vectors describing a circle, and it works in 4D for hypersurfaces. Footnotes Jack Fraser-Govil's answer to How do you solve the equation y2−4y+4=0 using the quadratic formula? Why does MATLAB use column major order instead of row Major order as prevalent in C? Gary Ward MaEd in Education & Mathematics, Austin Peay State University (Graduated 1997) · Author has 4.9K answers and 7.6M answer views · 4y Related How can I determine the points of intersection, given the equation of a circle and that of a straight line? How can I determine the points of intersection, given the equation of a circle and that of a straight line? Given y = mx + b and (x - h)^2 + (y - k)^2 = r^2 Substitute (mx + b) for y to get (x - h)^2 + (mx + b - k)^2 = r^2 It should be straight forward to solve from there since h, k and r are constants leaving just x as a variable. It will simplify to a quadratic equation. If the roots are imaginary, there is no intersection. if not you will have one or two x-values. For simplicity, plug the x-values into the line equation to get the point of tangency or points of intersection. Example: [math]\left( x-1 [/math] How can I determine the points of intersection, given the equation of a circle and that of a straight line? Given y = mx + b and (x - h)^2 + (y - k)^2 = r^2 Substitute (mx + b) for y to get (x - h)^2 + (mx + b - k)^2 = r^2 It should be straight forward to solve from there since h, k and r are constants leaving just x as a variable. It will simplify to a quadratic equation. If the roots are imaginary, there is no intersection. if not you will have one or two x-values. For simplicity, plug the x-values into the line equation to get the point of tangency or points of intersection. Example: (x−1)2+(y−2)2=32 and y=0.5x+3 (x−1)2+((0.5x+3)−2)2=9 (x−1)2+(0.5x+1)2=9 x2−2x+1+0.25x2+x+1=9 1.25x2−x=7 divided by 1.25⟹x2−0.8x=5.6 x2−0.8x+0.16=5.6+0.16⟹(x−0.4)2=5.76⟹ x−0.4=±√5.76⟹x=0.4±√5.76 x=0.4+2.4=2.8;x=0.4−2.4=−2 y=0.5(2.8)+3=4.4;y=0.5(−2)+3=2 The points of intersection are (2.8, 4.4) and (-2, 2). Jim DeCamp Senior Principal Engineer at BAE Systems (company) (2021–present) · Author has 1.3K answers and 699.7K answer views · 4y Related How can I determine the points of intersection, given the equation of a circle and that of a straight line? Substitution. Arrange the terms in the in the equation of the straight line into the form: y = mx + b The equation of every straight line can be expressed in this form. Then in the equation of the circle substitute mx + b for y. The result is a quadratic equation in x. I assume you know how to solve a quadratic equation. If the roots are imaginary, i.e., the discriminant is negative, the line does not intersect the circle. If the discriminant is zero, the line is tangent to the circle, if the discriminant is positive, the two roots give the x-value of where the line and circle intersect. Substitute Substitution. Arrange the terms in the in the equation of the straight line into the form: y = mx + b The equation of every straight line can be expressed in this form. Then in the equation of the circle substitute mx + b for y. The result is a quadratic equation in x. I assume you know how to solve a quadratic equation. If the roots are imaginary, i.e., the discriminant is negative, the line does not intersect the circle. If the discriminant is zero, the line is tangent to the circle, if the discriminant is positive, the two roots give the x-value of where the line and circle intersect. Substitute these values in the equation of the line to get the y-values. Example y = 2x + 1 (x-3)^2 + (y-1)^2 = 25 Substituting into the 2x +1 into the equation of the circle and reducing to simple quadratic form 5x^2 - 6x - 16 = 0 with roots: x = 2.486796, -1.286796 Substituting into the linear equation, the corresponding values of y: y = 5.973592, -1.573592 Good luck! Bhim Mu Former Retd Gov Servant (1971–2007) · Author has 3.5K answers and 525.5K answer views · 1y Related How do I find the intersection of a line and a curve? BY EXAMPLE LET THE LINE BE Y =X+3 ——-[M] CURVE BE CIRCLE WITH EQN (X-1)² + (Y-1)² =6² ——[N] PUT Y VALUE FROM EQN [M] IN EQN [N] &EXPAND X²-2X+1 +{X+3–1)² =36 X²-2X+1 +X²+4X+4= 36 2X²+2X+5–36 =0 2X²+2X–31 =0 X=[- 2 ±√4- 42(-31) /22] =[- 2 15.87√252] /4 =[-2±15.87/4] =13.87/4 DISCARDING NEGATIVE ROOT =3.4675 SO PUT X VALUE IN [M] —-> Y = 3.4675+3 = 6.4675 INTERSECTION POINT IS ( 3.4675,6.4675) Related questions How do you find the point of intersection of two straight lines mathematically? How do I get x1,y1 coordinates of a point which is on the rectangle drawn by the major and minor axis of an ellipse and the line joining the center of the ellipSE and a point (x,y) passes through it? How do you find the distance of two points that are not in a straight line, if one point lies on an ellipse or a parabola? How do you find the points of a straight line? How do you find the equation of an ellipse and a line passing through the ellipse? Dee Stoddard 3x AS in Computer Networking & Mathematics, Heald College (Graduated 2004) · Author has 188 answers and 218.2K answer views · 5y Related How do I find the intersection point between a circle and a straight line? whether they are straight lines or not, you simply combine the equations together. you will likely have to rearrange the equation a bit to get either x or y by themselves, y is more common say you have 2x+y=1 and you have x^2+y^2=4 we will chose y to differentiate: 2x+y=1 becomes y=1–2x x^2+y^2=4 becomes y=sqrt(4-x^2) since y=y, simply say 1–2x = sqrt(4-x^2) square: 1–4x+4x^2 = 4-x^2 consolidate 3x^2–4x+1=4 (3x-1)(x-1)=4 x =(2+sqrt(13))/3 & (2-sqrt(13))/3 now plug these values for x to both equations above y=1-(2((2+sqrt(13))/3)) or y=1-(2((2-sqrt(13))/3)) y=sqrt(4-((2+sqrt(13))/3)^2) or y=sqrt(4-((2-sq whether they are straight lines or not, you simply combine the equations together. you will likely have to rearrange the equation a bit to get either x or y by themselves, y is more common say you have 2x+y=1 and you have x^2+y^2=4 we will chose y to differentiate: 2x+y=1 becomes y=1–2x x^2+y^2=4 becomes y=sqrt(4-x^2) since y=y, simply say 1–2x = sqrt(4-x^2) square: 1–4x+4x^2 = 4-x^2 consolidate 3x^2–4x+1=4 (3x-1)(x-1)=4 x =(2+sqrt(13))/3 & (2-sqrt(13))/3 now plug these values for x to both equations above y=1-(2((2+sqrt(13))/3)) or y=1-(2((2-sqrt(13))/3)) y=sqrt(4-((2+sqrt(13))/3)^2) or y=sqrt(4-((2-sqrt(13))/3)^2), you may find x1 and/or x2 have to be used to make them equal, like 1 positive & 1 negative. try the different combinations, if they are equal, then that is an intersecting point, else they are not. Andrew Droffner Studied Mathematics at Rutgers University (Graduated 1995) · Author has 8.8K answers and 5.7M answer views · 1y Related How do I find the intersection of a line and a curve? Set the line (1) equal to the curve y=f(x) (2). Then, solve for x where they intersect (3). yL(x)=mx+b(1) yC(x)=f(x)(2) yL(x)=yL(x)(3) Example yL=8x−5(1) yC=f(x)=x2+10(2) yL(x)=yL(x)⟺8x−5=x2+10 8x−5=x2+10⟺x2−8x+15=0 Factor the quadratic equation to find the intersection points’ x coordinates. x2−8x+15=0=(x−3)(x−5)⇒x=3,5 Find the y coordinates. yL(3)=8(3)−5=19 and yL(5)=8(5)−5=35 Intersection Points There are two intersection points (P). P1=(3,19);P2 Set the line (1) equal to the curve y=f(x) (2). Then, solve for x where they intersect (3). yL(x)=mx+b(1) yC(x)=f(x)(2) yL(x)=yL(x)(3) Example yL=8x−5(1) yC=f(x)=x2+10(2) yL(x)=yL(x)⟺8x−5=x2+10 8x−5=x2+10⟺x2−8x+15=0 Factor the quadratic equation to find the intersection points’ x coordinates. x2−8x+15=0=(x−3)(x−5)⇒x=3,5 Find the y coordinates. yL(3)=8(3)−5=19 and yL(5)=8(5)−5=35 Intersection Points There are two intersection points (P). P1=(3,19);P2=(5,35)(P) Peter Hauge PhD in Electrical Engineering & Mathematics, University of Minnesota (Graduated 1968) · Author has 2.7K answers and 1.6M answer views · 1y Related How do I find the intersection of a line and a curve? Draw a line. Then draw a curve. The intersection is where they cross. Another way, which works it there is an expression for each of them. Plot both on a sheet of graph paper. Again, where they cross is their intersection. Read the location from the graph paper. Or, just equate the expressions and solve that equation. Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views · 9mo Related How do tangents on an ellipse intersect? Can you find the intersection points algebraically? This looks very interesting! The very basic equation of an ellipse is like this… Let’s choose some points with Integer coordinates to make the algebra nice and easy to follow. Notice this particular Ellipse goes through (3, 2) and (3, – 2) so let’s find the tangent equations at these two “nice” points. Comment: You can see that if the points ON the ellipse were irrational numbers, the algebra would have been horrific! After sleeping on this problem I was determined to find This looks very interesting! The very basic equation of an ellipse is like this… Let’s choose some points with Integer coordinates to make the algebra nice and easy to follow. Notice this particular Ellipse goes through (3, 2) and (3, – 2) so let’s find the tangent equations at these two “nice” points. Comment: You can see that if the points ON the ellipse were irrational numbers, the algebra would have been horrific! After sleeping on this problem I was determined to find a NICER example. Well here it is… You can follow the method I used previously to find the equations of these tangents and the intersection point is exactly Jafar Mortadha Mechanical Engineer · Author has 1.4K answers and 3.6M answer views · 8y Related How can I find the common tangent of two ellipses? Consider the general equations of 2 different ellipses. Differentiate both of them implicitly to obtain the derivative. Equate both derivatives to each other. The equality that we obtain above is only true when one ellipse depends on the other. For example, we could have a 2 = b 2 c 2 d 2 . In this case the a value from 1st ellipse depends on the values of c and d of 2nd ellipse. Consider the general equations of 2 different ellipses. Differentiate both of them implicitly to obtain the derivative. Equate both derivatives to each other. The equality that we obtain above is only true when one ellipse depends on the other. For example, we could have a 2 = b 2 c 2 d 2 . In this case the a value from 1st ellipse depends on the values of c and d of 2nd ellipse. Dave Carlson · Author has 475 answers and 775.2K answer views · 8y Related How do I find the point of intersection of a 3D plane and a straight line? The answer to this may differ depending on the form of the equations of your line. Some examples: x = 3 + 4t y = 2 + t z = 5 - 2t Or (x - 1) / 4 = (y + 2) / 7 = (z - 2) / 3 Or (2, 5, 6) + t <1, 3, 5> … The easiest way is to convert your line into parametric form (the top form above). So for the second example, you would go from (x - 1) / 4 = (y + 2) / 7 = (z - 2) / 3 to (x - 1) / 4 = t (y + 2) / 7 = t (z - 2) / 3 = t → x - 1 = 4t → x = 4t + 1 y + 2 = 7t → y = 7t - 2 z - 2 = 3t → z = 3t + 2 For the third example, you would go from (2, 5, 6) + t <1, 3, 5> to x = 2 + t y = 5 + 3t z = 6 + 5t … Once you have your equations in The answer to this may differ depending on the form of the equations of your line. Some examples: x = 3 + 4t y = 2 + t z = 5 - 2t Or (x - 1) / 4 = (y + 2) / 7 = (z - 2) / 3 Or (2, 5, 6) + t <1, 3, 5> … The easiest way is to convert your line into parametric form (the top form above). So for the second example, you would go from (x - 1) / 4 = (y + 2) / 7 = (z - 2) / 3 to (x - 1) / 4 = t (y + 2) / 7 = t (z - 2) / 3 = t → x - 1 = 4t → x = 4t + 1 y + 2 = 7t → y = 7t - 2 z - 2 = 3t → z = 3t + 2 For the third example, you would go from (2, 5, 6) + t <1, 3, 5> to x = 2 + t y = 5 + 3t z = 6 + 5t … Once you have your equations in parametric form, there are two steps: Step 1) - Plug in your parametric equations into the equation of the plane, and solve for t Example: x = 2 + t y = 5 + 3t z = 6 + 5t Plane: 3x + 4y - 2z = 7 Plugging in: 3(2 + t) + 4(5 + 3t) - 2(6 + 5t) = 7 6 + 3t + 20 + 12t - 12 - 10t = 7 34 + 5t = 7 5t = -27 t = -27/5 Step 2) - Plug in value of parametric variable (t in this case) into equation of line, to get coordinates of intersection x = 2 + t = 2 + -27/5 = -17/5 y = 5 + 3t = 5 + 3(-27/5) = -56/5 z = 6 + 5t = 6 + 5(-27/5) = -21 Final point: (-17/5, -56/5, -21) … Hope this helps. Bhim Mu Former Retd Gov Servant (1971–2007) · Author has 3.5K answers and 525.5K answer views · 1y Related How do I find the intersection of two lines? Each line has an equation in 2 variable solve 2 equations in 2 variables The value you get for 2 variable say x,y are the intersection point co-ordinates x + y = 1 ————[A] 2 x + 3 y = 3——-[B] 2 x+3 y+3 2 x+2 y=2 ←—2[A] 0+y=1 ←- subtracting y =1 substitute y=1 in [A] x+1 =1 x=1–1 =0 intersection point is (0,-1) Each line has an equation in 2 variable solve 2 equations in 2 variables The value you get for 2 variable say x,y are the intersection point co-ordinates say x + y = 1 ————[A] 2 x + 3 y = 3——-[B] 2 x+3 y+3 2 x+2 y=2 ←—2[A] 0+y=1 ←- subtracting y =1 substitute y=1 in [A] x+1 =1 x=1–1 =0 intersection point is (0,-1) Bob Collier Former EE Designed Specialized Computers for 33 Years. · Author has 3.1K answers and 1.6M answer views · 3y Related What is the point of intersection between a circle and an ellipse? Meaningless question. I’ll give you the one, simplest answer: They just don’t intersect, in the general case. YOU draw a few pictures of circles and ellipses doing ‘that’ - move the small and then large circle around and see how many different intersections you can force/cause. If you still don’t understand some aspect of those 2, repost with a question about the bit/area/conditions you don’t understand. Don’t post again if you haven’t done that. Meaningless question. I’ll give you the one, simplest answer: They just don’t intersect, in the general case. YOU draw a few pictures of circles and ellipses doing ‘that’ - move the small and then large circle around and see how many different intersections you can force/cause. If you still don’t understand some aspect of those 2, repost with a question about the bit/area/conditions you don’t understand. Don’t post again if you haven’t done that. Related questions Where is the point of intersection on an ellipse? How do I find the intersection point between a circle and a straight line? How do you find the point of intersection of a pair of straight lines if the point of intersection is not origin? How do you find the point of intersection between two curves and one straight line? How can I find the point of intersection of three lines simultaneously? How do you find the point of intersection of two straight lines mathematically? How do I get x1,y1 coordinates of a point which is on the rectangle drawn by the major and minor axis of an ellipse and the line joining the center of the ellipSE and a point (x,y) passes through it? How do you find the distance of two points that are not in a straight line, if one point lies on an ellipse or a parabola? How do you find the points of a straight line? How do you find the equation of an ellipse and a line passing through the ellipse? What is the relationship between the distance of a point on an ellipse from the center and the point itself? How do I find the length of a chord if a line intersects an ellipse? What is the equation of the straight line passing through the point (3,2) and the point of intersect of the line 3x-y-6=0 and 2x+1-14=0? Are property lines straight? What is the method for finding a point on an ellipse that is a fixed distance away from two other points? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
16613	https://bio.libretexts.org/Bookshelves/Introductory_and_General_Biology/General_Biology_(Boundless)/06%3A_Metabolism/6.03%3A__Energy_and_Metabolism_-_Metabolic_Pathways	Skip to main content 6.3: Energy and Metabolism - Metabolic Pathways Last updated : Nov 22, 2024 Save as PDF 6.2: Energy and Metabolism - Types of Energy 6.4: Energy and Metabolism - Metabolism of Carbohydrates Page ID : 13102 Boundless Boundless ( \newcommand{\kernel}{\mathrm{null}\,}) Learning Objectives Describe the two major types of metabolic pathways Metabolic Pathways The processes of making and breaking down carbohydrate molecules illustrate two types of metabolic pathways. A metabolic pathway is a step-by-step series of interconnected biochemical reactions that convert a substrate molecule or molecules through a series of metabolic intermediates, eventually yielding a final product or products. For example, one metabolic pathway for carbohydrates breaks large molecules down into glucose. Another metabolic pathway might build glucose into large carbohydrate molecules for storage. The first of these processes requires energy and is referred to as anabolic. The second process produces energy and is referred to as catabolic. Consequently, metabolism is composed of these two opposite pathways: Anabolism (building molecules) Catabolism (breaking down molecules) Anabolic Pathways Anabolic pathways require an input of energy to synthesize complex molecules from simpler ones. One example of an anabolic pathway is the synthesis of sugar from CO2. Other examples include the synthesis of large proteins from amino acid building blocks and the synthesis of new DNA strands from nucleic acid building blocks. These processes are critical to the life of the cell, take place constantly, and demand energy provided by ATP and other high-energy molecules like NADH (nicotinamide adenine dinucleotide) and NADPH. Catabolic Pathways Catabolic pathways involve the degradation of complex molecules into simpler ones, releasing the chemical energy stored in the bonds of those molecules. Some catabolic pathways can capture that energy to produce ATP, the molecule used to power all cellular processes. Other energy-storing molecules, such as lipids, are also broken down through similar catabolic reactions to release energy and make ATP. Importance of Enzymes Chemical reactions in metabolic pathways rarely take place spontaneously. Each reaction step is facilitated, or catalyzed, by a protein called an enzyme. Enzymes are important for catalyzing all types of biological reactions: those that require energy as well as those that release energy. Key Points A metabolic pathway is a series of chemical reactions in a cell that build and breakdown molecules for cellular processes. Anabolic pathways synthesize molecules and require energy. Catabolic pathways break down molecules and produce energy. Because almost all metabolic reactions take place non-spontaneously, proteins called enzymes help facilitate those chemical reactions. Key Terms catabolism: destructive metabolism, usually including the release of energy and breakdown of materials enzyme: a globular protein that catalyses a biological chemical reaction anabolism: the constructive metabolism of the body, as distinguished from catabolism 6.2: Energy and Metabolism - Types of Energy 6.4: Energy and Metabolism - Metabolism of Carbohydrates
16614	https://lonestarneurology.net/blog/buergers-disease-vs-raynauds/	Buerger's Disease vs Raynaud's: A Detailed Comparison 214-619-1910214-619-1913 Mon - Fri: 8:00AM - 5:00PM Monday – Friday 8:00AM - 5:00PM Saturday Open for MRI Sunday Closed Migraine treatment same day as your first appointment. or Visit our Healow Portal Practice Areas Alzheimer’s Memory Treatment Headache Treatment Myasthenia Gravis Treatment Stroke Treatment Epilepsy Treatment Neuropathy Treatment Vertigo Treatment Parkinson’s Treatment Restless Leg Syndrome Treatment Neurological Complications of Pregnancy Treatment Bell’s Palsy Treatment Sleep Disorder Treatment Multiple Sclerosis Treatment Carpal Tunnel Treatment Tests & Procedures Neurology 101 Locations Allen Arlington Carrollton Dallas Denton Fort Worth Frisco Garland Grapevine Greenville Mansfield McKinney Plano Richardson Rockwall San Antonio Sherman South Dallas Forms Our Providers Blog Contact Reviews Careers About us Visit our Healow PortalCall 214-619-1910 BOOK AN APPOINTMENT CALL 214-619-1910 Buerger’s Disease vs Raynaud’s: Key Differences Explained Updated: 19/07/2024 Published: 06/11/2023 Medically reviewed by Dr. Shukla Book appointment All fields are required Name Phone Email Message If you are a human seeing this field, please leave it empty. 06/11/2023 Medically reviewed by Dr. Shukla Do you ever wonder why your fingers or toes turn white or blue when it’s chilly outside? It could be due to Raynaud’s or Buerger’s disease. These two may sound similar, but they’re quite different. Raynaud’s is like a color-changing magic trick when your body overreacts to the cold or stress. Your fingers and toes might look like they’re having a party with red, white, and blue colors. But it’s not as fun as it sounds. On the other hand, the definition of Buerger’s disease is like a blocked road for blood. It mostly affects people who smoke and can make your fingers and toes go painfully pale, just like Raynaud’s. But there’s more to the story. In this text, we will explore the key differences between these two conditions. So, the next time your fingers put on a color show, you’ll know whether Raynaud’s or Buerger’s caused the commotion. Definition of Raynaud’s Disease Raynaud’s disease is a condition that affects the fingers and toes. Sometimes, a person may get affected by other body parts, such as the ears, nose, or lips. It’s like the body’s thermostat malfunctioning, causing these body parts to turn white, blue, or red. Here’s how it works: you’re in a cold room or stressed out. Then, the tiny blood vessels in your fingers and toes suddenly become very narrow, like thin straws. It makes it difficult for blood to flow to these body parts. As a result, they turn white. They don’t get enough oxygen and nutrients. After a while, when these body parts start feeling left out, they turn blue. It’s like they’re shouting for help because they’re not getting the love they need from your blood. But here’s the thing. Once you warm up or calm down, the blood vessels will return to their normal size. Blood will rush back to them. As a result, your fingers and toes turn red. However, it’s not just the color. In Raynaud’s disease definition, the fingers and toes may tingle or numb. It can be quite painful. This condition occurs independently, but it can also be associated with other health problems. Therefore, if your fingers are putting on a colorful show, it’s worth seeing a doctor. You will be able to find out if it is Raynaud’s disease and learn how to deal with it. Definition of Buerger’s Disease Buerger’s disease is also known as thromboangiitis obliterans. It is a disease that affects the blood vessels in the arms and legs, especially medium-sized arteries and veins. It’s like a plug in the blood vessels that can lead to serious problems. Here’s what happens: The blood vessels in the arms and legs become inflamed and narrowed. This narrowing makes it difficult for blood to flow normally, like a road with many obstacles. As a result, the affected body areas do not receive enough oxygen and nutrients. One of the main features of Buerger’s disease is that it is closely related to smoking. It is often referred to as a smoker’s finger disease. The harmful chemicals in tobacco can make blood vessels even narrower. So, if you are a smoker, this disease can be a severe warning sign for you. It is a signal for you to stop smoking and improve your health. The most characteristic signs of Buerger’s disease are pain and discomfort in the hands or feet. The fingers and toes may turn pale or blue. The good news is that quitting smoking can help slow or even stop the worsening of the disease. But if the disease has already done too much damage, surgery may be necessary in severe cases. It will help to repair blood vessels or even amputate part of the limb. Thus, Buerger’s disease definition is a disease in which there is a blockage of blood vessels. You can get this disease due to smoking. If you have disease symptoms, you must see a doctor and follow his recommendations. Comparison of Buerger’s and Raynaud’s Disease If you compare Buerger’s disease vs Raynaud’s, both can affect blood vessels. They cause strange behavior in fingers and toes. However, they are quite different. Buerger’s DiseaseRaynaud’s Disease CausesBuerger’s disease is mostly related to smoking. The harmful chemicals in tobacco cause inflammation and narrowing of the blood vessels in the hands and feet.On the other hand, Raynaud’s disease is often triggered by cold or stress. Then, it is like the body’s reaction to these factors. SymptomsBuerger’s disease usually causes pain in the hands and feet, especially with vigorous activity. The fingers and toes may turn pale or blue due to restricted blood flow.In Raynaud’s disease, the fingers and toes may also change color. But this happens suddenly and is often provoked by cold or stress. It can be painful and cause tingling or numbness in these body parts. TreatmentWith a smoker’s finger disease, it is imperative to stop smoking. It will help prevent the disease from worsening. In severe cases, surgical intervention may be necessary.In Raynaud’s syndrome, keeping warm and dealing with stress can help. In severe cases, medication may be used. In short, Buerger’s disease is closely related to smoking and affects the blood vessels in the arms and legs. Raynaud’s disease definition, on the other hand, is often provoked by cold or stress. It causes discoloration of the fingers and toes. Treatment differs, but for Buerger’s disease, smoking cessation is a prerequisite, and for Raynaud’s disease, management of triggering factors is a prerequisite. Managing and Treating the Conditions Treatment for Buerger’s disease vs Raynaud’s is aimed at controlling symptoms and preventing disease progression. Both diseases require lifestyle changes and, in some cases, medical intervention. In the case of Raynaud’s disease, minimizing exposure to cold temperatures is critical to managing symptoms. Wear warm gloves and socks, especially in cold weather. Layering clothing and using heating pads can also help you. Combat stress, as stress can trigger attacks. Yoga, meditation, and stress-reducing breathing exercises are helpful. Buerger’s disease, on the other hand, is closely linked to tobacco use. So, if you want to get rid of smoker finger disease, then stop using tobacco. It is the most effective way to manage symptoms and prevent the progression of the disease. It applies not only to smoking but also to the use of smokeless tobacco products. Conclusion In conclusion, Buerger’s and Raynaud’s diseases can affect the fingers and toes. However, they have different causes and require different treatments. If you experience Raynaud’s disease, stay warm, manage stress, and consult your doctor. If you have Buerger’s disease, you need to stop smoking, as it is often associated with the disease. These changes are very important to manage symptoms and prevent the disease from worsening. Have you or someone you know encountered similar problems? Then you need to see a doctor. Contact a specialist, such as Lone Star Neurology. We can help you get the support and treatment you need for your specific situation. Your health matters! FAQ Is Raynaud’s disease the same as Buerger’s disease? No, Raynaud’s and Buerger’s diseases are not the same. While both affect blood vessels, Raynaud’s typically impacts smaller vessels that supply blood to the skin, usually in response to cold temperatures or stress. Is Buerger’s disease more serious than Raynaud’s disease? Both diseases can be severe and lead to significant complications, but Buerger’s disease is often considered more serious because it can lead to tissue damage and, in extreme cases, might require amputation. Can Buerger’s and Raynaud’s diseases lead to blood clots in fingers? Yes, both Buerger’s and Raynaud’s diseases can lead to blood clots. In Buerger’s disease, inflammation of the blood vessels can cause blood clots to form, impairing blood flow. Can lifestyle changes prevent Buerger’s and Raynaud’s diseases? Lifestyle changes can help manage and prevent the progression of these diseases. For Raynaud’s, avoiding cold temperatures and managing stress can help prevent episodes. For Buerger’s disease, the most effective prevention is to stop all forms of tobacco use. Tweet +1 Pin it (1 votes, average: 5.00 out of 5) Loading... Be the first to write a review Lone Star Neurology 4.5 Based on 905 reviews review us on Edward Medina 15:34 30 Jun 22 Just such an amazing staff that makes you feel like part of their family. I’ve been going there for over 5 years now... and each visit I get the very best care and treatments that I have ever received in the 20+ years that I’ve been dealing with severe debilitating migraines. Since i started seeing them the number of my migraines has dropped from 15-20 a month to 2-3 every 3 month. I highly recommend them …they will change your life!read more Daneisha Johnson 22:20 19 May 22 Dr. Askari was very kind and explained everything so I could understand. The other staff were nice as well. I would... have gave 5 stars but I was a little taken aback when I checked in and had to pay 600.00 upfront. I think that should have been discussed in a appointment confirmation call or email just so I could have been prepared.read more Jean Cooper 16:54 29 Apr 22 I love the office staff they are friendly and very helpful. Dr. JODIE is very caring and understanding to your needs... and wants to help you. I will go back. would recommend Dr. Dr. Jodie to other Patients in a heart beat. The team works well together.read more Linda M 19:40 02 Apr 22 I was obviously stressed, needing to see a neurologist. The staff was so patient and Dr. Ansari was so kind. At one... point he told me to relax, we have time, when I was relaying my history of my condition. That helped ease my stress. I have seen 3 other neurologists and he was the only one who performed any assessment tests on my cognitive and physical skills. At one point I couldn't complete two assessments and got upset and cried. I was told, it's OK. That's why you're here. I was truly impressed, and super pleased with the whole experience!read more Leslie Durham 15:05 01 Apr 22 I've been coming here for about 5 years. The staff are ALWAYS friendly and knowledgeable. The Doctors are the absolute... best!! Jodie Moore is always in such a great mood which is a plus when you are already stressed. Highly recommended read more Monica Del Bosque 14:13 25 Mar 22 Since my first post my thoughts have changed here. It's unfortunate. My doctor and PA were great, but the office staff... is horrible. They never call you back when they say they will, they misinform you, they cause you too much stress wondering what's going on, they don't keep you posted. They never answer the phone. At this point I've left four messages in the last week, and I have sent three messages. Twice from their portal and one direct email. No response. My appointment is on Monday morning at 8:30am, no confirmation on my insurance and what's going on. What the heck is going on, this is ridiculous!I've given up... the stress her office staff has put me through is just not worth it. You can do so much better, please clean house, either change out your office staff, or find a way for them to be more efficient please. You have to do something. This is not how you want to run your practice. It leaves a very bad impression on your business.read more Ron Buckholz 23:32 23 Mar 22 I was actually pleasantly surprised with this visit! It took me a long time to get the appointment scheduled because no... one answers your phones EVER! After a month, I finally got in, and your staff was warm, friendly, and I was totally impressed! I feel like you will take care of my needs!read more Steve Nabavi 16:28 16 Mar 22 It was a nice visit. Happy staff doing all they can do to comfort the patients in a very calming environment. You ask... me they are earned a big gold star on the fridge. My only complaint they didn't give me any cookies.read more Katie Lewis 16:10 10 Feb 22 Had very positive appointments with Jodie and Dr. Sheth for my migraine care. Jodie was so fast with the injections and... has so much valuable info. I started to feel light headed during checkout and the staff was SO helpful—giving me a chair, water, and taking me into a private room until I felt better. Highly recommend this practice for migraine patients, they know what they’re doing!!read more Joshua Martinez 16:02 10 Dec 21 I was scheduled to be checked and just want to say that the staff was fantastic. They were kind and helpful. I was... asked many questions related to what was going on and not once did I feel as though I was being brushed off. The front desk staff was especially great in assisting me. I'm scheduled to go back for a mri and am glad that I'll be going there.read more Isabel Ivy 21:42 03 Nov 21 I had such a good experience with Lone Star Neurology, Brent my MRI Tech was so awesome and made sure I was very... comfortable during the appointment. He gave me ear plugs, a pillow, leg support and blanket, easiest MRI ever lol 🤣 My 72 hour EEG nurse Amanda was also so awesome. She made sure I was take care of over the 3 days and took her time with the electrodes to make sure it was comfortable for me! Paige was also a huge help in answering all my questions when it came to my test results, and letting me know her honest opinions about how I should go forth with my treatment.read more Leslie Luce 17:37 20 Oct 21 The professionalism and want to help attitude of this office was present from the moment I contacted them. The follow... up and follow through as well as their willingness to find a way to schedule my dad was above and beyond. We visited two offices in the same day with the same experience. I am appreciative of this—we spend a lot of time with doctors and this was top notch start to finish.read more robert Parker 16:38 16 Apr 21 I love going to this office. The staff is friendly and helpful. The doctor is great. I am getting the best... neurological tests and treatment I have ever had. The only reason I did not give them a 5 star rating is because it is impossible to reach a live person at the office to reschedule appointments. Every time I have tried to get through to the office it says all people are busy and I am sent to a voicemail. If they could get their phone answering fixed, I would give them a strong 5 stars.read more MaryAnn Hornbaker 00:26 25 Feb 21 Dr. Harney is an excellent Dr. I found him friendly , personable and thorough. I evidently am an unusual case. ... Therefore he spent a Hugh amount of time educating me. He even gave me literature to further explain my condition and how to follow up. This is something you rarely get from your doctors. So I am more than please with my doctor and his staff.read more Roger Arguello 03:05 29 Jan 21 Always courteous, professional. The staff is very friendly and always work with you to find the best appointment time.... The care team has been great. Always taking the time to listen to your concerns and to find the best treatment.read more Margaret Rowland 01:12 27 Jan 21 I have been a patient at Lone Star Neurology for several years. Now both my adult daughters also are patients there. I... love Jodie. She is always so prompt whether it is a teleamed call are a visit in the office. She takes the time to explain everything to me and answers all my questions. I am so blessed to have Jodie as my doctor.read more Susan Miller 03:01 13 Jan 21 My husband had an accident 5 years ago and Lone Star Neurology has been such a blessing to us with my husbands care.... Jodie Moore is his provider and she is amazing! Jodie is very knowledgeable, caring, and thorough. She takes her time with you, making sure your needs are met and she is happy to answer any questions you may have. Lone Star Neurology’s patients are very lucky to have Jodie providing their care. Thank you Lone Star Neurology and especially Jodie for everything you have done for us. Jodie, you are the best!read more Windalyn C 01:32 09 Jan 21 Jodie is wonderful. She is very caring and knowledgeable. I have been to over a dozen neurologists, and none were able... to help me as much as they have here. Thanks!read more Katie Kordel 00:40 09 Jan 21 Jodi Moore, nurse practitioner, is amazing. I have suffered from frequent, debilitating headaches for almost 20 years.... She has provided the best proactive and responsive care I have ever received. My quality of life has been greatly improved by her caring approach and tenacity in finding solutions.read more Ellie Natsis 15:41 07 Jan 21 I have had the best experience at this neurologist's office! For over a year I have been receiving iv treatments here... each month and my nurse, Bobbie is beyond wonderful!! She's so attentive, knowledgeable, caring, and detail oriented. She makes an otherwise uncomfortable experience much more pleasant and definitely puts me at ease! She also helps me with my insurance,ordering this specialty medication and dealing with the ordering process which is no easy feat.Needless to say, she goes above a beyond in every way and I'm so grateful to this office and to Bobbie for all they do for me!read more Matt Morris 15:39 07 Jan 21 Let me start by saying that I have been coming here for years. Due to my autoimmune disease, I am in this office... once every three weeks for multiple hours at a time. The office is very clean and the staff very friendly. My only complaint would be there communication via phone. They aren't the best at responding if you leave a voicemail and expect a call back. I understand that this is prob just due to the sheer number of alls they receive daily. What I can say I like the best about the office are the people. Bobby who handles my infusions is great. I never have any issues with her setting up my infusions. She is very quick to reply to messages sent via text and if she were to leave then my whole opinion of the office may change. I also enjoy people like Matt, Lauren, and Jodi. I appreciate all that they do for me and without this team I'm not sure I would be as happy as I am to visit the office as frequently as I have to. Please ensure that these folks are recognized as they are what makes my visit to this office so tolerable :).read more More reviews Please, leave your review Write a comment: Click here to cancel reply. Cancel reply Write you feedback Your name Book appointment All fields are required Name Phone Email Message If you are a human seeing this field, please leave it empty. Related Stories Chaitanya Bonda 08/11/2024 Methods of Treatment for Autonomic Dysfunction Autonomic dysfunction is a condition that affects the body’s autonomic nervous system. As it controls… Hakam Asaad 08/11/2024 Methods of Treatment for Autonomic Dysfunction Autonomic dysfunction is a condition that affects the body’s autonomic nervous system. As it controls… Sajish Jacob 15/05/2024 Myocardial Infarction: STEMI vs NSTEMI Within the realm of cardiac ailments, a singular term exists. It holds significant weight: Myocardial… Robert Nieto 15/05/2024 CT Scan vs MRI: Key Differences Explained Modern medicine, especially diagnostic methods, does not stand still. 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16615	https://www.sciencedirect.com/science/article/pii/0375960173906609	Phase transition in an Ising model with many-spin interaction - ScienceDirect Skip to main contentSkip to article Journals & Books Access throughyour organization Purchase PDF Search ScienceDirect Article preview Abstract References (10) Cited by (2) Physics Letters A Volume 46, Issue 1, 19 November 1973, Pages 7-8 Phase transition in an Ising model with many-spin interaction☆ Author links open overlay panel F.Y.Wu∗ Show more Add to Mendeley Share Cite rights and content Abstract The critical behaviour of an Ising model with multiple-spin interactions in a nonzero magnetic field is determined. The system exhibits a phase transition in the infinite-interaction limit with the order parameter given by a many-spin correlation. Access through your organization Check access to the full text by signing in through your organization. Access through your organization References (10) R.J. Baxter### Ann. Phys. (N.Y.) (1972) A. Hintermann### Phys. Lett. (1972) J. Oitmaa and R.W. Gibberd, J. Phys. C, to be... T.D. Lee et al.### Commun. Math. Phys. (1968) B.M. McCoy et al.### Phys. Rev. (1967) F.Y. Wu### Phys. Rev. (1971) L.P. Kadanoff et al.### Phys. Rev. (1971) There are more references available in the full text version of this article. Cited by (2) A challenge in enumerative combinatorics: The graph of contributions of professor Fa-Yueh Wu 2002, Chinese Journal of Physics ### On the definition of phase transition 1976, Journal of Statistical Physics ☆ Work supported in part by the National Science Foundation Grant No. Gh-35822 at Northeastern University. ∗ Senior Fulbright Scholar on leave of absence from Northeastern University, Boston, Massachusetts, USA. View full text Copyright © 1973 Published by Elsevier B.V. Recommended articles No articles found. About ScienceDirect Remote access Contact and support Terms and conditions Privacy policy Cookies are used by this site.Cookie settings All content on this site: Copyright © 2025 Elsevier B.V., its licensors, and contributors. All rights are reserved, including those for text and data mining, AI training, and similar technologies. For all open access content, the relevant licensing terms apply. We use cookies that are necessary to make our site work. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. You can manage your cookie preferences using the “Cookie Settings” link. For more information, see ourCookie Policy Cookie Settings Accept all cookies Cookie Preference Center We use cookies which are necessary to make our site work. We may also use additional cookies to analyse, improve and personalise our content and your digital experience. For more information, see our Cookie Policy and the list of Google Ad-Tech Vendors. You may choose not to allow some types of cookies. However, blocking some types may impact your experience of our site and the services we are able to offer. See the different category headings below to find out more or change your settings. You may also be able to exercise your privacy choices as described in our Privacy Policy Allow all Manage Consent Preferences Strictly Necessary Cookies Always active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. Cookie Details List‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. Cookie Details List‎ Contextual Advertising Cookies [x] Contextual Advertising Cookies These cookies are used for properly showing banner advertisements on our site and associated functions such as limiting the number of times ads are shown to each user. Cookie Details List‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Confirm my choices × Read strategically, not sequentially ScienceDirect AI extracts key findings from full-text articles, helping you quickly assess an article's relevance to your research. Unlock your AI access
16616	https://artofproblemsolving.com/wiki/index.php/Inradius?srsltid=AfmBOoq0KliAjt1-ZZnzASEHDjA-2O1fjxwcxUqVMOmA-GpVy9GR5RdK	Art of Problem Solving Inradius - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Inradius Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Inradius The inradius of a polygon is the radius of its incircle (assuming an incircle exists). It is commonly denoted . In a triangle, the incenter is where the three angle bisectors meet. A Property If has inradius and semi-perimeter, then the area of is . This formula holds true for other polygons if the incircle exists. Proof Add in the incircle and drop the altitudes from the incenter to the sides of the triangle. Also draw the lines , and . After this, and are the bases of , and respectively. But they all have the same height (the inradius), so . Also the inradius of a incircle inscribed in a right triangle is as by drawing three inradii to the three tangent points, then A to that tangent point is equal to A to the other tangent point (explained in circles) and etc for B and C. After doing it for B and C, C (the hypotenuse) should equal the equivalent tangent of A and the equivalent tangent of B added together, thus our equation Problems Verify the inequality . Verify the identity (see Carnot's Theorem). 2007 AIME II Problems/Problem 15 Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
16617	https://math.stackexchange.com/questions/2206805/factoring-1x-dots-xn-into-a-product-of-polynomials-with-positive-coefficie	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. current community your communities more stack exchange communities Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Factoring $1+x+\dots +x^n$ into a product of polynomials with positive coefficients Can the polynomial $1+x+x^2+\dots +x^n$ be factored, for some $n\ge 1$, into a product of two non-constant polynomials with positive coefficients? Thoughts It is easy to factor it into polynomials with non-negative coefficients e.g. $$ 1+x+x^2+x^3 = (1+x)(1+0x + x^2), $$ but I have no example with positive coefficients. I believe this should be possible for large $n$, since there are so many (something like $2^{\lceil n/2\rceil-1}$) ways to factor $1+x + x^2 +\dots + x^n$ into a product of two monic polynomials with real coefficients. Some motivation from probability theory The question is motivated by Can the sum of two independent r.v.'s with convex support be uniformly distributed? Namely, we can ask ourselves a discrete counterpart: Whether a discrete uniform random variable (i.e. the one taking values $0,1,\dots,n$ with equal probabilities) can be decomposed into a sum of independent non-constant random variables, each ranging over a set of consecutive integers? The link is provided by the probability generating function (pgf): $$ m_Y(x) = \mathbb E[x^{Y}]. $$ If the random variable $Y$ takes values $0,1,\dots,k$ with positive probabilities, then its pgf is a polynomial with positive coefficients: $$ m_Y(x) = \sum_{i=0}^k \mathbb{P}(Y=i) x^i; $$ in particular, for a random variable $U_n$, uniformly distributed on ${0,1,\dots,n}$, $$ m_{U_n}(x) = \frac1{n+1}\bigl(1+x+x^2+\dots + x^n\bigr). $$ Since for independent random variables, the pgf of sum if a product of pgfs: $$ m_{Y'+Y''}(x) = \mathbb E[x^{Y'+Y''}] = \mathbb E[x^{Y'}]\mathbb E[x^{Y''}] = m_{Y'}(x)m_{Y''}(x),\tag{1} $$ these two questions are equivalent$^$. $^$Note that in general $(1)$ does not imply the independence of $Y'$ and $Y''$. Nevertheless, if $m_Y$ factors, say, into $m_{Y'}$ and $m_{Y''}$, then $Y$ has the same distribution as the sum of independent copies of $Y'$ and $Y''$, and, indeed, we have the desired decomposition. 2 Answers 2 Can the polynomial $1+x+x^2+\dots +x^n$ be factored, for some $n\ge 1$, into a product of two non-constant polynomials with positive coefficients? $$x^n+\cdots +x+1=\frac{x^{n+1}-1}{x-1}=\frac{\prod_{1\le k\le n+1}(x - e^{2i\pi k/(n+1)})}{x-1}=\prod_{1\le k\lt n+1}(x - e^{2i\pi k/(n+1)})$$ For a factorization $x^n+\dots +x+1=g(x)h(x)$, $g$ and $h$ are products of some $x-\zeta^k$ where $\zeta$ is nth root of unity (up to units (in this case multiplying by positive real numbers) however any such factorization implies one without units). The leading term of both polynomials must be $1$ because it is the product of leading terms that are always $1$. The constant term is also one because it is the product of root of unity so its absolute value is $1$ and it must be real and positive. $$(x^p+\cdots +1)(x^q+\cdots+1)=x^n+\cdots +x+1$$ Assume without loss of generality that $p\ge q$. Take the constant term of the first polynomial and multiply it with the leading term of the second and you get $x^q$. $$\begin{align} (x^p+\cdots+ ax^q +\cdots +1)(x^q+\cdots+1) &=\cdots +ax^q1+ \cdots +1x^q +\cdots\ &= \cdots +(\cdots+a+1)x^q \ \end{align}$$ All other terms must be positive and they can only add to $x^q$, the coefficient should be equal $1$. Therefore all other coefficients of $x^q$ must be zero and $a$ is zero. I believe this should be possible for large $n$, since there are so many (something like $2^{\lceil n/2\rceil-1}$) ways to factor $1+x + x^2 +\dots + x^n$ into a product of two monic polynomials with real coefficients. Nice exercise :): show that $1+x+\cdots+x^n$ can be factored into $2$ polynomials of degree at least $1$ with non-negative coefficients if and only if $n+1$ is not prime. Probabilistic reformulation: Let $n \geq 2$ be an integer, $X$ a random variable following the uniform distribution on ${1,2,\ldots,n}$. Show that there exist two independent nonnegative integer-valued independent random variables $Y$ and $Z$ such that $X$ and $Y+Z$ have the same distribution if and only if $n$ is not prime. You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Explore related questions See similar questions with these tags. Linked Related Hot Network Questions Subscribe to RSS To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Mathematics Company Stack Exchange Network Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA . rev 2025.9.26.34547
16618	https://math.stackexchange.com/questions/48938/deriving-the-rest-of-trigonometric-identities-from-the-formulas-for-sinab	Skip to main content Deriving the rest of trigonometric identities from the formulas for sin(A+B), sin(A−B), cos(A+B), and cos(A−B) Ask Question Asked Modified 11 years ago Viewed 4k times This question shows research effort; it is useful and clear 12 Save this question. Show activity on this post. I am trying to study for a test and the teacher suggest we memorize sin(A+B), sin(A−B), cos(A+B), cos(A−B), and then be able to derive the rest out of those. I have no idea how to get any of the other ones out of these, it seems almost impossible. I know the sin2θ+cos2θ=1 stuff pretty well though. For example just knowing the above how do I express cot(2a) in terms of cota? That is one of my problems and I seem to get stuck half way through. trigonometry Share CC BY-SA 3.0 Follow this question to receive notifications edited Jul 31, 2014 at 17:01 user147263 asked Jul 1, 2011 at 20:15 AdamAdam 1,38555 gold badges1818 silver badges2525 bronze badges 3 1 What textbook do you use? I'm sure that in your textbook all the formulas are proved using only the four formulas mentioned above. – Beni Bogosel Commented Jul 1, 2011 at 20:25 1 I don't have a text book, it got lost in the mail. I am more concerned with learning how to actually do it though, it seems impossibly hard to me. – Adam Commented Jul 1, 2011 at 20:49 Try sosmath.com/trig/trig.html – Beni Bogosel Commented Jul 2, 2011 at 9:30 Add a comment \| 5 Answers 5 Reset to default This answer is useful 14 Save this answer. Show activity on this post. Since cot(2a)=cos(2a)sin(2a), you would have (assuming you know the addition formulas for sines and cosines): cos(2a)sin(2a)=cos(a+a)=cos(a)cos(a)−sin(a)sin(a)=cos2(a)−sin2(a);=sin(a+a)=sin(a)cos(a)+cos(a)sin(a)=2sin(a)cos(a), and therefore cot(2a)=cos(2a)sin(2a)=cos2(a)−sin2(a)2sin(a)cos(a)=12(cos2(a)sin(a)cos(a))−12(sin2(a)sin(a)cos(a))=12(cos(a)sin(a)−sin(a)cos(a))=12(cot(a)−tan(a))=12(cot(a)−1cot(a))=12(cot2(a)cot(a)−1cot(a))=12(cot2(a)−1cot(a)). P.S. Now, as it happens, I don't know the formulas for double angles, nor most identities involving tangents, cotangents, etc. I never bothered to memorize them. What I know are: The definitions of tangent, cotangent, secant, and cosecant in terms of sine and cosine; That sine is odd (sin(−x)=−sin(x)) and cosine is even (cos(−x)=cos(x)); The addition formulas for sine and cosine; The values of sine and cosine at 0∘, 30∘, 45∘, 60∘, and 90∘. (I can derive sin2θ+cos2θ=1 from the above, but in all honesty that one comes up so often that I do know it as well). I do not know the addition or double angle formulas for tangents nor cotangents, so the above derivation was done precisely "on the fly", as I was typing. I briefly thought that I might need to cos(2a) with one of the following equivalent formulas: cos2(a)−sin2(a)=cos2(a)+sin2(a)−2sin2(a)=1−2sin2(a) or cos2(a)−sin2(a)=2cos2(a)−cos2(a)−sin2(a)=2cos2(a)−1, if the first attempt had not immediately led to a formula for cot(2a) that involved only cot(a) and tan(a)=1cot(a). Share CC BY-SA 3.0 Follow this answer to receive notifications edited Jul 2, 2011 at 19:12 Eric Naslund 73.7k1212 gold badges181181 silver badges272272 bronze badges answered Jul 1, 2011 at 20:23 Arturo MagidinArturo Magidin 418k6060 gold badges862862 silver badges1.2k1.2k bronze badges 11 I don't quit understand the result of the cot2a part and how that is equal to cot2a, is it somehow an inverse of the tan2a formula? Is deriving forumals mostly about factoring? I am very bad at that. – Adam Commented Jul 1, 2011 at 21:55 1 @Adam: cotangent is, by definition, cosine over sine. So cot(2a)=cos(2a)sin(2a). Then I substituted the values of cos(2a) and sin(2a) using the addition formulas; the rest is just algebra, except for using the fact that cot(x)=1tan(x) in the antepenultimate step. – Arturo Magidin Commented Jul 1, 2011 at 22:53 @Adam: You are memorizing too many things, in the hope that a memorized formula will be a perfect fit. In an earlier post, you mentioned having to memorize many cosine laws. There really is only one cosine law, which can be memorized and then manipulated as needed. Also, I remember a test you posted. There were some hard questions on that test, but I am sure that most of the questions were variants of problems that had been done in class and/or assigned. You have difficulty recognizing clones of things you have seen before. – André Nicolas Commented Jul 1, 2011 at 23:37 Yes, I have difficulty telling the difference between problems that are meant to be manipulated in a certain way against other. When I do the homework one section is all double angle, the next half. On the test it is all mixed and I have trouble telling what I need to use, for example something with the number 3 doesn't obviously fit half or double. – Adam Commented Jul 2, 2011 at 0:20 1 @Adam: Which sentence was condescending? The "algebra accidents"? Actually, that's a correct description of some of the mistakes. The inquiry about the major? No, I have done a lot of academic advising, it is an obvious question. So the next course would be science/engineering calculus, at least three or four semesters of it. Not easy at all, but if you are strong in other areas, possible. Not possible at current level of algebra, the strategy of last minute memorizing that more or less worked in high school is no longer viable. – André Nicolas Commented Jul 2, 2011 at 3:03 \| Show 6 more comments This answer is useful 5 Save this answer. Show activity on this post. Three examples of algebraic derivations of trigonometric identities as an application of the addition and subtraction formulas. 1. Example on how to deduce the logarithmic transformation formulas (sum to product formulas) from sin(a+b)=sinacosb+sinbcosa,(1) sin(a−b)=sinacosb−sinbcosa.(2) If you write {a=p+q2,b=p−q2,⇔{a+b=p,a−b=q, you get sin(a+b)+sin(a−b)=2sinacosb, sin(a+b)−sin(a−b)=2sinbcosa, and thus sinp+sinq=2sinp+q2cosp−q2,(3) sinp−sinq=2sinp−q2cosp+q2.(4) You can use (3) to solve the equation sin(5x)+sinx=sin(3x) that appeared in my exam in 1968. (See a comment of mine to this post ). 2. As for the example in your question we present the following derivation. From sin(a+b)=sinacosb+sinbcosa,(5) cos(a+b)=cosacosb−sinasinb,(6) we get cot(a+b)=cos(a+b)sin(a+b)=cosacosb−sinasinbsinacosb+sinbcosa, or dividing the numerator and denominator by sinacosb cot(a+b)===cosacosb−sinasinbsinacosbsinacosb+sinbcosasinacosb=cosacosbsinacosb−sinasinbsinacosbsinacosbsinacosb+sinbcosasinacosbcosasina−sinbcosb1+sinbcosasinacosb=cota−tanb1+tanbcota=cota−1cotb1+cotacotbcotacotb−1cotbcotb+cotacotb=cotacotb−1cotb+cota,(7) a particular case of which (for a=b) is given by cot(2a)=cot2a−12cota.(8) 3. The summation and subtraction formula for the tangent. From sin(a±b)=sinasinb±sinbcosa we get tana±tanb=sinacosa±sinbcosb=sinacosb±sinbcosacosacosb=sin(a±b)cosacosb.(9) Share CC BY-SA 3.0 Follow this answer to receive notifications edited Jul 2, 2011 at 17:46 answered Jul 1, 2011 at 22:29 Américo TavaresAmérico Tavares 39.2k1414 gold badges110110 silver badges251251 bronze badges 8 I am having difficulty understand where step 7 came from for the example problem. I am trying to understand how to derive these formulas better but I can't see what you did. Also in #3 I don't understand why sin(a+b) is not equal to sincos + cossin? I am having a lot of trouble with the third one, I am assuming sin turns into tan by dividing everything by cos but then why does the second set of sinbcosa dissapear? – Adam Commented Jul 2, 2011 at 15:57 @Adam: I derived the addition formula cot(a+b), which for a=b reduces to the duplication formula cot(2a). Formula (7) comes from (5) and (6) and the definition of cot(a+b). – Américo Tavares Commented Jul 2, 2011 at 16:12 In more detail cot(a+b)=cos(a+b)sin(a+b)=cosacosb−sinasinb=cosacosb−sinasinbsinacosbsinacosb+sinbcosasinacosb =cosacosbsinacosb−sinasinbsinacosb=1+sinbcosasinacosb=cota−tanb1+tanbcota=cota−1cotb1+cotacotb – Américo Tavares Commented Jul 2, 2011 at 16:12 I understand now, not sure how that went over my head. Thanks. I am going to be working on these for most of today. Is there anything else really important that I should learn before I take calculus? I have a little over a month to study. – Adam Commented Jul 2, 2011 at 16:20 @Adam: I think you need to know functions and polynomials. – Américo Tavares Commented Jul 2, 2011 at 17:31 \| Show 3 more comments This answer is useful 3 Save this answer. Show activity on this post. If you understand complex numbers there is a very nice mnemonic. We know that eit=cos(t)+isin(t). Take real and imaginary parts on the identity ei(A+B)=eiAeiB=(cos(A)+isin(A))(cos(B)+isin(B)). Disassemble and the the sum formulae for sin and cos. To get the differences, use the assignment B←−B and the fact that cos(−B)=cos(B) and sin(−B)=−sin(B). Share CC BY-SA 3.0 Follow this answer to receive notifications answered Jul 1, 2011 at 22:41 ncmathsadistncmathsadist 50.2k33 gold badges8585 silver badges134134 bronze badges 1 1 This is exactly how I console myself that I could work out the trig identities if trapped on a desert island. If I had known Euler in high school trig I wonder how many times I would have run out of time running everything through eit – Ben Jackson Commented Jul 2, 2011 at 6:15 Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. Maybe this will help? cot(x) = cosx / sinx -> cot(2a) = cos(a + a) / sin(a + a) and then I assume you know these two. Edit: Had it saved as a tab and didnt see the posted answer, but I still think it would have been best to let you compute the rest by yourself so that you could learn it by doing instead of reading. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Jul 1, 2011 at 20:45 Patrik KPatrik K 2111 bronze badge 0 Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. I'm not certain, but I think perhaps you need to revisit the basic trigonometric definitions. sinθ=oppositehypotenusecosθ=adjacenthypotenusetanθ=oppositeadjacentcscθ=hypotenuseoppositesecθ=hypotenuseadjacentcotθ=adjacentopposite Having done that, we can now manipulate the definitions. For example cotθ=adjacentopposite⋅1hypotemnuse1hypotemnuse=adjacenthypotenuseoppositehypotenuse=cosθsinθ or alternatively cotθ=adjacentopposite=1opposite1adjacent=1opposite1adjacent⋅hypotenusehypotenuse=hypotenuseoppositehypotenuseadjacent=cscθsecθ After that, move on to making 3 triangles that are similar to the adjacent-opposite-hypotenuse triangle. These triangles will rotate which side is equal to 1. This is where we get the familiar Pythagorean identities. The next step is to derive cos(α+β) and sin(α+β) by setting the lengths (I would use the squares of the lengths though) of the green lines in the diagram equal to each other. It's easy to do, since the endpoints are well known. Finally, lets derive cot(A+B) cot(A+B)=====sin(A+B)cos(A+B)sinAcosB+cosAsinBcosAcosB−sinAsinBsinAcosB+cosAsinBsinAsinBcosAcosB−sinAsinBsinAsinBcosBsinB+cosAsinAcosAcosBsinAsinB−1cotB+cotAcotAcotB−1 Note that cot2A is just cot(A+B) where A=B, so cot2A=cot(A+A)=cotA+cotAcotAcotA−1=2cotAcot2A−1 Of course this list isn't comprehensive, for example we didn't derive the sine and cosines laws, nor did we consider the double angle and half angle formulas. But I hope that what little I did present here was helpful. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Jul 31, 2014 at 15:53 answered Jul 31, 2014 at 15:44 John JoyJohn Joy 8,06011 gold badge2525 silver badges3131 bronze badges Add a comment \| You must log in to answer this question. 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16619	https://www.askiitians.com/iit-jee-co-ordination-compounds/isomerism-in-coordination-compounds.html	Isomerism in Coordination Compounds – Study Material for IIT JEE \| askIITians help@askiitians.com 1800-150-456-789 Live Courses Resources Exam Info Forum Notes Upcoming Examination Login Book a free demo of live class Isomerism in Coordination Compounds Compounds that have the same chemical formula but different structural arrangements are called isomers. Due to their complicated formulae of many coordination compounds, the variety of bond types and the number of shapes possible, many different types of isomerism occur. Polymerization Isomerism: This is not true isomerism because it occurs between compounds having the same empirical formula, but different molecular weights. [Pt(NH 3)2 Cl 2], [Pt(NH 3)4][PtCl 4] [Pt(NH 3)4] [Pt(NH 3)Cl 3]2. Ionization Isomerism: This type of isomerism is due to the exchange of groups between the complex ion and the ions outside it. [Co(NH 3)5 Br]SO 4 is red – violet. An aqueous solution gives a white precipitate of BaSO 4 with BaCl 2 solution, thus confirming the presence of free SO 4 2- ions. In contrast [Co(NH 3)5 SO 4]Br is red. A solution of this complex does not give a positive sulphate test with BaCl 2. It does give a cream – coloured precipitate of AgBr with AgNO 3, the confirming the presence of free Br– ions. Hydrate Isomerism: Three isomers of CrCl 3.6H 2 O are known. From conductivity measurements and quantitative precipitation of the ionized chlorine, they have been given the following formuale: [Cr(H 2 O)6]Cl 3 :violet (three ionic chlorines) [Cr(H 2 O)5 Cl]Cl 2.H 2 O : green (two ionic chlorines) [Cr(H 2 O)4 Cl 2].Cl.2H 2 O : dark green (one ionic chlorine) Linkage Isomerism: Certain ligands contain more than one atom which could donate an electron pair. In the NO 2- ion, either N or O atoms could act as the electron pair donor. Thus there is the possibility of isomerism. Two different complexes [Co(NH 3)5 NO 2]Cl 2 have been prepared, each containing the NO 2- group in the complex ion. Geometric isomerism: In disubstituted complexes, the substituted groups may be adjacent or opposite to each other. This gives rise to geometric isomerism. Thus square planar complexes such as [Pt(NH 3)4 Cl 2] can be prepared in two forms, cis and trans. When the chlorine atoms are adjacent to each other it is called cis form. While when two chlorine atoms are opposite it is called transform. This type of isomerism takes place mainly in heteroleptic complexes because of the different possible geometric arrangements of ligands around the central metal atom. This type of isomerism is mainly found in coordination compounds with coordination numbers 4 and 6. In a square planar complex (i.e. coordination compounds with coordination number 4 which have [MX 2 L 2] type formula (X and L are unidentate ligands), the two ligands X may be present adjacent to each other in a cis isomer, or opposite to each other to form a trans isomer. Square planar complex with MABXL type formula (where A, B, X, L are unidentate ligands) show three isomers-two cis and one trans. Sis trans isomerism is not possible for a tetrahedral geometry. But octahedral complexes do show cis trans isomerism. In complexes with formula [MX 2 L 4] type, two ligands X may be oriented cis or trans to each other. This type of isomerism is also observed when bidentate ligands L–L [e.g.,NH 2 CH 2 CH 2 NH 2 (en)] are present in complexes with [MX 2(L–L)2] type formula There is another type of geometrical isomerism which occurs in octahedral coordination entities with [Ma 3 b 3] type formula. Examples is [Co(NH 3)3(NO 2)3]. If three donor atoms of the same ligands occupy adjacent positions at the corners of an octahedral face, we have the facial (fac) isomer. When the positions are around the meridian of the octahedron, we get the meridional (mer) isomer Facial Isomers:A set of three ligands (similar) arranged on an octahedron in all cis – fashion. Meridional Isomers:A set of three similar ligands arranged on an octahedron with one pair trans Refer to the following video for isomerism in coordination compounds Optical Isomerism ¨The complexes which are non-superimposable on their mirror images are optically active. Optically active complexes are asymmetric in nature i.e. not divisible into two identical halves. Levorotatory (l) – the compound which rotates plane polarised light to left hand side. Dextrorotatory (d) – the compound which rotates plane polarised light to right hand side d and l isomers of a compound are called enantiomers Octahedral complexes with coordination number 6 involving 2 or 3 bidentate ligands show optical isomerism. Solved Problem Question1: How will you distinguish between the following isomer pairs a) [CoBr(NH 3)5]SO 4 and [Co(SO4) (NH 3)5]Br b) [Cr(H 2 O)6]Cl 3 and [CrCl(H 2 O)5]Cl 2 H2 O c) Cis [PtCl 2(NH 3)2] and trans[PtCl 2(NH 3)2] d) The two entiomers of [CoCl 2(en)2]+ Solution: a) Isomer (i) gives white ppt of BaSO4 with BaCl2 whereas isomer (ii) does not form a ppt. b) The water molecule in isomer (ii) is lost easily on heating whereas the water molecule in isomer (i) are not lost easily, being coordinated to the central atom. c) Cis isomer (i) has dipole moment, the trans isomer (ii) does not d) One isomer is dextrorotory whereas the other is laevorotatory. Question 2: The geometry of Ni(CO)4 and Ni(PPH 3)2 Cl 2 are (a) both square planar (b) tetrahedral and square planar, respectively (c) both tetrahedral (d) square planar and tetrahedral, respectively Solution : c In Ni(CO)4, Ni is in 3 d 10 state due to strong ligand field produced by CO. Hence, Ni is sp 3 hybridised and complex is tetrahedral. In NiCl 2(PPh 3)2, Ni 2+ has 3 d 8 configuration. Due to weak ligand field, Ni is sp 3 hybridised and comlex is tetrahedral Question 3: The ionization isomer of [Cr(H 2 O)4 Cl(NO 2)]Cl is (a) [Cr(H 2 O)4(O 2 N)]Cl 2 (b) Cr(H 2 O)4 Cl 2 (c) [Cr(H 2 O)4 Cl(ONO)]Cl (d) [Cr(H 2 O)4 Cl 2(NO 2)]∙H 2 O Solution: b Ionisation isomers are the complexes that produces different ions in solution, ie, they have ions interchanged inside and outside the coordination sphere. [Cr(H 2 O)4 Cl(NO 2)]Cl and [Cr(H 2 O)4 Cl 2] (NO 2) have different ions inside and outside the coordinate sphere and they are isomers. Therefore, they are ionization isomers. Question 4: The compound(s) that exhibit(s) geometrical isomerism is(are) (a) [Pt(en)Cl 2] (b) [Pt(en)2]Cl 2 (c) [Pt(en)2 Cl 2]Cl 2 (d) [Pt(NH 3)2]Cl 2 Solution: c and d Both [Pt(en)2 Cl 2]Cl 2 and [Pt(NH 3)2 Cl 2] are capable of showing geometrical isomerism. Related Resources You can also refer to our exam section to download the past year papers of IIT JEE and other exams Click here to download the NCERT Solutions Refer to thenomenclature of coordination compounds to know how to name them. Complete Your Profile Phone Number +91 Name Email Select grade of the Student 6 th 7 th 8 th 9 th 10 th 11 th 12 th Submit View courses by askIITians ### Design classes One-on-One in your own way with Top IITians/Medical Professionals Click Here Know More ### Complete Self Study Package designed by Industry Leading Experts Click Here Know More ### Live 1-1 coding classes to unleash the Creator in your Child Click Here Know More ### a Complete All-in-One Study package Fully Loaded inside a Tablet! 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16620	https://www.nagwa.com/en/explainers/162131918063/	Sign Up Sign In English English العربية English English العربية My Wallet Sign Up Sign In My Wallet My Classes My Messages My Reports Lesson Explainer: Dot Product in 2D Mathematics • Third Year of Secondary School Join Nagwa Classes Attend live Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher! Check Available Classes Next Session: 09/28/2025 09/28/2025 09/28/2025 undefined • 05:30:00pm Remaining Seats: 13 Try This In this explainer, we will learn how to find the dot product of two vectors in 2D. There are three ways to multiply vectors. Firstly, you can perform a scalar multiplication in which you multiply each component of the vector by a real number, for example, . Here, we would multiply each component in vector by the number three. Secondly, we can multiply a vector by another vector; here, there are two different methods, the dot product and the cross product. In this explainer, we are only going to look at the dot product. Suppose we have a vector that is and a vector that is . Their dot product is written as . Notice here that the dot is central to the two vectors, not at the base of each. Now, to calculate the dot product, we need to write out the two vectors in component form, multiply the corresponding components of each vector, and add the resulting numbers. Definition: Dot Product of Two Vectors The dot product of two vectors and is given by multiplying the corresponding components of each vector and adding the resulting numbers: This is demonstrated in example 1. Example 1: Finding the Dot Product of Two-Dimensional Vectors Given the vector and the vector , find . Answer Recall that the dot product of two vectors and is given by Hence, we have Notice in this example that we have written “,” which is another way of writing “.” We use the first notation to avoid any possible confusion with the vector cross product, which, as the name suggests, uses a cross instead of a dot. Notice, too, that the dot product produces an answer that is a numerical value, or a scalar. It is worth noting here that the dot product is also called the scalar product for this reason. Let us see what happens when we carry out the dot product , where is a nonzero real number and and . The components of are then and we find that Similarly, we find that and As this is an important property, let us take note of it here. Property: Scalar Multiplication and Dot Product For real numbers and , we have Additionally, we find from the definition that which, given that multiplication is commutative, leads to This proves the commutativity of the dot product. Property: Commutativity of the Dot Product The dot product is commutative: Let us now consider three vectors , , and and work out the dot product . We have ; hence, We finally find that This equation shows that the dot product is distributive. Property: Distributivity of the Dot Product The dot product is distributive: Let us consider a useful property that the dot product has when we take the dot product of a vector with itself, which we will calculate in the following example. Example 2: Calculating the Dot Product of a Vector with Itself Given that , find . Answer Recall that the dot product of two vectors and is given by Since , we have To see how this result is significant, let us calculate the magnitude of the same vector. First, we draw a sketch of the vector. We can work out its magnitude by finding its length using the Pythagorean theorem. So, the magnitude of , usually denoted by , is calculated as follows: If we compare the magnitude and the dot product, we find the following property. Property: Dot Product and Magnitude The magnitude of a vector is equal to the square root of its dot product with itself: The dot product of two vectors can be interpreted geometrically, as given in the following definition box. Definition: Geometrical Definition of the Dot Product The dot product of two vectors and equals the product of their magnitudes with the cosine of the angle between them: where is the angle between and . The geometrical interpretation shows us that the “closer” the two vectors are, the larger the dot product, because the smaller the angle, the larger its cosine. Therefore, the maximum value of the dot product of two vectors of given magnitudes occurs when the two vectors have the same direction, that is, when the angle between them is zero. The dot product of two collinear vectors having the same direction is which, since , gives This is consistent with what we have found before for the dot product of a vector with itself. When two vectors and are collinear but have opposite directions, the angle between them is , with a cosine of , so that their dot product is then given by On the other hand, when two vectors and are perpendicular, their dot product is zero since the cosine of the angle between them () is zero. It is an important property that can be used to check whether two vectors of given components are perpendicular. Property: Dot Product of Two Perpendicular Vectors The dot product of two perpendicular vectors is zero. Conversely, when the dot product of two vectors is zero, the two vectors are perpendicular. Let us look at an example where we need to use this property. Example 3: Finding the Dot Product of Two Vectors in a Square Square has a side of 10 cm. What is ? Answer We can start answering this question by sketching square and vectors and . We see that and are perpendicular since two adjacent sides of a square are perpendicular. The angle between the two vectors is , and, as , we have The answer is . Let us look at another example where we need to use the property of perpendicular vectors. Example 4: Finding the Missing Component of a Vector Given That It Is Perpendicular to Another Given that , , and , determine the value of . Answer and are two perpendicular vectors; this means that their dot product is zero. Let us therefore calculate their dot product using their components: where are the components of and are those of . Substituting into our equations the actual components of and , we get As and are perpendicular, their dot product is zero, which gives With our last example, we will see how to find a dot product using its geometric definition. Example 5: Finding the Dot Product of Two Vectors in a Triangle Given that is an isosceles triangle, where and , determine . Answer Let us first sketch triangle and vectors and . We are asked to find . For this, we need to work out the angle between and and the magnitude of . To find the angle between the two vectors, we draw a vector equivalent to so that the initial points of and are coincident. In isosceles triangle , . The angle between and is ; therefore, we have To find the magnitude of , since we are given the lengths of and , we simply consider that the magnitudes of the vectors are given here by their lengths in centimetres. Hence, we need to find length . For this, we can use the sine rule in triangle . It gives We can now find by writing that where is the angle between and . Substituting the magnitudes of and and the value of into our equation gives us With , we find As a final note, let us see how to derive the law of cosines using the distributivity of the dot product and its geometrical definition. For any three points , , and , we can write Expanding the parentheses, we find that Since , we have Let be the angle between and as shown in the diagram above. We have The angle between and is ; therefore, we have And, as for any , we find which corresponds to the law of cosines, with , , , and . Let us summarize what we have learned in this explainer. Key Points The dot product of two vectors and is given by multiplying the corresponding components of each vector and adding the resulting numbers: We have . The dot product is commutative: . The dot product is distributive: . The dot product of two vectors and equals the product of their magnitudes with the cosine of the angle between them: where is the angle between and . The dot product of two perpendicular vectors is zero. Conversely, when the dot product of two vectors is zero, the two vectors are perpendicular. Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! Interactive Sessions Chat & Messaging Realistic Exam Questions Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy
16621	https://myengineeringtools.com/Piping/Tools_hydrostatic_P.html	Hydrostatic pressure Hydrostatic pressure overview 2. Hydrostatic pressure formula 3. Hydrostatic pressure vs depth of water 4. Practice problems 5. Hydrostatic pressure calculator 1. Hydrostatic Presure Overview The hydrostatic pressure is the pressure exerted by a resting fluid. The hydrostatic pressure is due to the gravity forces acting on the fluid. 2. Hydrostatic Pressure Formula The pressure exerted only by the liquid, without considering the atmospheric pressure / pressure above the liquid is ρ.g.h. The Hydrostatic pressure calculation formula is then following : ρ.g.h It is however important to understand that there is also a pressure above the liquid that needs to be considered. The total pressure at depth h, considering the pressure above the liquid plus the hydrostatic pressure is the following : p = pa + ρ.g.h With : pa=Pressure above the liquid (Pa) p=Pressure at the point of calculation (Pa) ρ=Volumic mass of the liquid (kg/m3) h=Height of liquid (m) g=gravity acceleration (m.s-2) g=9.81 m.s-2 3. Hydrostatic pressure vs depth of water The calculation is done assuming a density of water constant and equal to 1000 kg/m3. \| \| \| --- \| \| Depth of water \| Hydrostatic Pressure (bar) \| \| 0 \| 0 \| \| 5 \| 0.4905 \| \| 10 \| 0.981 \| \| 15 \| 1.4715 \| \| 20 \| 1.962 \| \| 25 \| 2.4525 \| \| 30 \| 2.943 \| \| 35 \| 3.4335 \| \| 40 \| 3.924 \| \| 45 \| 4.4145 \| \| 50 \| 4.905 \| \| 55 \| 5.3955 \| \| 60 \| 5.886 \| \| 65 \| 6.3765 \| \| 70 \| 6.867 \| \| 75 \| 7.3575 \| \| 80 \| 7.848 \| \| 85 \| 8.3385 \| \| 90 \| 8.829 \| \| 95 \| 9.3195 \| \| 100 \| 9.81 \| 4. Practice problems What is the pressure exerted by water at a depth of 4 m, with a water at 12 degrees celcius ? Step 1 : Calculate the density of water at 12 degrees celcius The density of water is 1007 kg/m3 at this temperature Step 2 : Calculate the hydrostatic pressure thanks to the formula ρ.g.h P = ρ.g.h = 10079.814 = 39516 Pa = 0.395 bar 5. Hydrostatic pressure calculator The hydrostatic pressure calculator can be downloaded here : Hydrostatic Pressure calculator Excel Warning : this calculator is provided to illustrate the concepts mentioned in this webpage, it is not intended for detail design. It is not a commercial product, no guarantee is given on the results. Please consult a reputable designer for all detail design you may need.
16622	https://www.youtube.com/watch?v=GJjbq1C0MaA	Adding and Subtracting Integers: A Step-By-Step Review \| How to Add and Subtract Integers Math with Mr. J 1700000 subscribers 6286 likes Description 661404 views Posted: 23 Jun 2022 Welcome to How to Add and Subtract Integers with Mr. J! Need help with adding and subtracting integers? You're in the right place! Whether you're just starting out, or need a quick refresher, this is the video for you if you're looking for help with adding and subtracting integers (adding positive and negative integers and subtracting positive and negative integers). Mr. J will go through adding integers examples, subtracting integers examples, and explain the steps of how to add and subtract integers. More Videos and Examples: ✅ Adding Integers -A Quick Review of Adding Integers = -Adding a Positive and Negative Integer \| Positive + Negative = -Adding a Negative and Positive Integer \| Negative + Positive = -Adding Two Negative Integers \| Negative + Negative = ✅ Subtracting Integers -A Quick Review of Subtracting Integers = -Subtracting a Positive Integer from a Negative Integer \| Negative - Positive = -Subtracting a Negative Integer from a Positive Integer \| Positive - Negative = -Subtracting a Negative Integer from a Negative Integer \| Negative - Negative = ✅ Multiplying Integers -A Quick Review of Multiplying Integers = -Multiplying a Positive by a Negative \| Positive x Negative = -Multiplying a Negative by a Positive \| Negative x Positive = -Multiplying a Negative Integer by a Negative Integer \| Negative x Negative = -Multiplying Three Integers = -Why Does a Negative Times a Negative Equal a Positive? = ✅ Dividing Integers -A Quick Review of Dividing Integers = -Dividing a Positive by a Negative \| Positive ÷ Negative = -Dividing a Negative by a Positive \| Negative ÷ Positive = -Dividing a Negative Integer by a Negative Integer \| Negative ÷ Negative = ✅ Integers How to Add and Subtract Integers = How to Multiply and Dividing Integers = How to Add Subtract Multiply and Divide Integers = About Math with Mr. J: This channel offers instructional videos that are directly aligned with math standards. Teachers, parents/guardians, and students from around the world have used this channel to help with math content in many different ways. All material is absolutely free. Click Here to Subscribe to the Greatest Math Channel On Earth: Follow Mr. J on Twitter: @MrJMath5 Email: math5.mrj@gmail.com Music: Hopefully this video is what you're looking for when it comes to adding and subtracting integers. Have a great rest of your day and thanks again for watching! ✌️✌️✌️ ✅ Thanks to Aloud, this video has been dubbed into Spanish and Portuguese. #DubbedWithAloud English This video has been dubbed into Spanish (United States) and Portuguese (Brazil) using an artificial voice via to increase accessibility. You can change the audio track language in the Settings menu. Spanish Este video ha sido doblado al español con voz artificial con para aumentar la accesibilidad. Puede cambiar el idioma de la pista de audio en el menú Configuración. Portuguese Este vídeo foi dublado para o português usando uma voz artificial via para melhorar sua acessibilidade. Você pode alterar o idioma do áudio no menu Configurações. 543 comments Transcript: [Music] welcome to math with mr j [Music] in this video i'm going to go through a review of how to add and subtract integers and we're going to be working with both positive and negative integers we'll start with adding integers now when it comes to these types of problems there are different ways to think through these we will go through two ways for each of our examples let's jump into number one where we have 12 plus negative 7. we'll start this problem by taking a look at the signs we have a positive 12 and a negative 7. so we have different signs a positive and a negative since we have different signs we are going to take the greater absolute value and subtract the lesser our answer will take the sign of the greater absolute value let's start by writing the absolute value of both 12 and negative 7 and remember absolute value is the distance a number is from 0. the absolute value of 12 is 12. the absolute value of negative 7 is 7. now we take the greater absolute value and subtract the lesser these are already in order so we can subtract if the larger absolute value comes second you can always switch the order to find the difference if need be let's subtract so 12 minus 7 is 5. now we need to determine if our answer is going to be positive or negative so we need to take a look at the larger absolute value which is this 12. so we take the sign of the larger absolute value from the original problem so the larger absolute value is 12 let's take a look at the 12 in the original problem and that 12 is positive that means our answer is going to be positive so our final answer a positive 5. so a quick recap here basically we forgot about any negatives because we were working with absolute values we then found the difference between the absolute values and the answer takes the sine of the greater absolute value from the original problem now let's think through this another way and this way is going to be more of a mental math approach just basically thinking about what's going on in this problem so let me rewrite 12 plus negative 7 here so our original problem so we are starting at a positive 12 and we are adding a negative seven by adding a negative by adding that negative seven we are decreasing in value by seven from that twelve we can basically think of this as 12 minus 7 or 12 take away 7 that gives us our answer of 5. so 12 plus negative 7 we are decreasing that 12 by a value of 7 so we get a positive 5. so again we started at a positive 12. always think about where you are starting and where you are going from that starting point so we are adding a negative 7 which is decreasing our 12 in value by 7 and we end up with 5. let's move on to number 2 where we have negative 8 plus negative 10. here we have two negatives so the same signs so we're going to add the absolute values and use the same sign so let's start by taking a look at the absolute value of negative 8 and negative 10. the absolute value of negative eight is eight plus the absolute value of negative ten which is ten now we add those absolute values because again we have the same signs eight plus ten is eighteen we use the same sign from the original problem which those are negatives there so our answer is negative final answer negative 18. now if we were to think through this we can think that we are starting at negative eight so let me rewrite here negative 8 plus negative 10. so again starting at negative 8 and we are adding a negative 10. so that means we are decreasing in value by 10. that leaves us at negative 18. like i mentioned earlier think about your starting point so the number you are starting with we have a negative 8 and then adding that negative 10 tells us we are decreasing in value and end up at negative 18 negative 18 is our final answer that's how we add integers let's move on to subtraction so here are our examples for subtracting integers let's jump into number one where we have five minus negative nine now when we subtract integers we're actually going to add the opposite so if you're able to add integers you're going to be able to subtract the opposite of subtraction is addition and then we take the opposite of the number we are subtracting so this gives us an equivalent problem and we are able to use this strategy so we have 5 and then let's add the opposite of negative 9. the opposite of negative 9 is positive 9 so 5 plus 9 that gives us 14 a positive 14 and that's our final answer now that answer may not make sense at first but let's think about how we end up with a positive 14 in this subtraction problem whenever we subtract a negative we actually increase in value i like to think of this in terms of money a negative represents a debt or an expense when it comes to money so that negative 9 would be a 9 debt or expense think of subtracting a negative like subtracting or taking away that debt or expense and getting that money back that is a good positive thing and increases the value of that problem so something to think about let's move on to number two where we have negative three minus twenty so let's add the opposite negative 3 plus the opposite of positive 20 is negative 20. so we have negative 3 plus negative 20. now adding that negative 20 we are decreasing in value by 20 so we're starting at negative 3 and then decreasing in value by 20. that's going to give us negative 23 and that's our final answer so there you have it there's how you add and subtract integers if you need any more help or examples i dropped links to more videos and examples down in the description i hope that helped thanks so much for watching until next time peace you
16623	https://www.intmath.com/exponents-radicals/1-integral-exponent-laws.php	Skip to main content 1. Simplifying Expressions with Integral Exponents Later, on this page Multiplying Expressions with the Same Base Dividing Expressions with the Same Base Repeated Multiplication of a Number Raised to a Power A Product Raised to an Integral Power A Fraction Raised to an Integral Power Raising a Number to a Zero Exponent Raising a Number to Negative Exponents Summary - Laws of Exponents The importance of brackets Exercises In this section we learn some important Laws of Exponents. Integral Exponents Back in the chapter on Numbers, we came across examples of very large numbers. (See Scientific Notation). One example was Earth's mass, which is about: 6 × 1024 kg Earth [image source (NASA)] In this number, the 10 is raised to the power 24 (we could also say "the exponent of 10 is 24"). The number 10 is called the base and 24 is called the exponent (or power). Now, the number 1024 means: 1024 = 24 lots of 1010×10×10×…×10 This is the same as: 1 followed by 24 zeros1,000,000,000,000,000,000,000,000 Exponents give us a very convenient way of writing very large and very small numbers. They are also very handy for making algebra easier because it is more compact. Let's now give a general definition for any number (or any variable) raised to an "integral exponent": Definition: am means "multiply m lots of a together" That is: am = m lots of aa×a×a×…×a Note 1: "Integral exponent" means the exponent is a whole number [That is, an integer] Note 2: The above definition only really holds if m is a positive integer, since it doesn't make a lot of sense if m is negative. (You can't multiply something by itself negative 3 times! And what does multiplying something by itself 0 times mean?). In such cases we have to rely on patterns and conventions to define what is going on. See below for zero and negative exponents. Example 1: Integral Exponents (1) y5 = y × y × y × y × y [There are 5 lots of y being multiplied together.] (2) 24 = 2 × 2 × 2 × 2 = 16 [There are 4 lots of 2 multiplied together.] (3) 106 = 10 × 10 × 10 × 10 × 10 × 10 = 1,000,000 [There are 6 lots of 10 multiplied together.] Multiplying Expressions with the Same Base Let's start with an example. Once you get the hang of this, it makes writing math a whole lot easier. Say we need to multiply 2 large numbers, 108 and 105. Now, if we write it out in full, we would need to write: 108 = 10 ×10 ×10 ×10 ×10 ×10 ×10 ×10 (8 lots of 10 multiplied together) 105 = 10 ×10 ×10 ×10 ×10 (5 lots of 10 multiplied together) So: 108 × 105 = (10 ×10 ×10 ×10 ×10 ×10 ×10 ×10) × (10 ×10 ×10 ×10 ×10) Now, if you count them all up, you will have 13 lots of 10 multiplied together. So we can conclude that 108 × 105 = 1013 This is very tedious and there must be an easier way. We could add the exponents when multiplying numbers with the same base. Let's see a general definition. Definition: am × an = am+n Let's see how this works with an example involving a variable, b: Example 2 b5 × b3 = (b × b × b × b × b) × (b × b × b) = b8 Our final answer is equivalent to b5+3. Dividing Expressions with the Same Base When we divide expressions with the same base, we need to subtract the exponent of the number we are dividing by from the exponent of the first number. In general, we can write is as follows. Definition: Dividing algebraic expressions anam=am−n (Of course, a≠0, and m and n are integers.) It may be easiest to see how this one works with an example. Example 3 b2b7=b×bb×b×b×b×b×b×b=b5 We cancel 2 of the b's from the numerator (the top) and the two b's from the denominator (the bottom) of the fraction. The result is equivalent to b7−2. We could also write this problem as b7÷b2=b7−2=b5 Repeated Multiplication of a Number Raised to a Power Next we consider the case where we have a base raised to some exponent, then we raise that to some other exponent. For example, we may start with p3 and need to raise it to the power 2. How do we do that? We'll see the answer in a minute. First, let's look at a general definition. (am)n =(am)×(am)×(am)×…×(am) =(am)n [We multiply n times] So we write: Definition: Repeated multiplication (am)n=amn (Once again, we assume a≠0, and m and n are integers.) Example 4 (p3)2=p3×p3 =(p×p×p)×(p×p×p)=p6 A Product Raised to an Integral Power In this section we have 2 numbers multiplied together, and we raise the result to some power. In this case, it has the same value as raising the first number to the power and multiplying by the second number raised to the power. Definition: Product Raised to an Integral Power (ab)n=anbn Example 5 (5q)3=53q3=125q3 We have raised the 5 to the power 3 (giving us 125) and we can't do anything else with q3. A Fraction Raised to an Integral Power If we have a fraction raised to an integral power, we need to raise the top number to the power and divide by the bottom number raised to the power. Definition: Fraction Raised to an Integral Power (ba)n=bnan (n is an integer.) Example 6 (32)4 =(32)×(32)×(32)×(32) =3×3×3×32×2×2×2 =3424 =8116 In this example, I have written out in full the meaning of 32 raised to the power 4. Example 7 Expand: (10x)5 Answer: Raising the top and bottom numbers to the power of 5 gives: (10x)5=(105x5)=100000x5 Raising a Number to a Zero Exponent Definition: a0=1 (a≠0) Example 8 70 = 1 Example 9 x0 = 1 Example 10 (5a)0 = 1 Note 1: a0 = 1 is a convention, that is, we agree that raising any number to the power 0 is 1. We cannot multiply a number by itself zero times. Note 2: In the case of zero raised to the power 0 (written 00), mathematicians have been debating this for hundreds of years. It is most commonly regarded as having value 1, but is not so in all places where it occurs. That's why we write a≠0. Raising a Number to Negative Exponents Definition a−n=an1 (Once again, a≠0) In this exponent rule, a cannot equal 0 because you cannot have 0 on the bottom of a fraction. Example 11 3−2=321=91 Example 12 a−1=a1 Example 13 x−8=x81 Explanation: 0 and Negative Exponents Observe the following decreasing pattern: 34 = 81 33 = 27 32 = 9 31 = 3 For each step, we are dividing by 3. Now, continuing beyond 31 and dividing by 3 each times gives us: 30=33=1 3−1=311=31 3−2=321=91 3−3=331=271 3−4=341=811 Summary - Laws of Exponents am×an=am+n anam=am−n (a≠0) (am)n=amn (ab)n=anbn (ba)n=bnan a0=1 (a≠0) a−n=an1 (a≠0) [Note: These laws also apply in the next section, Fractional Exponents.] Let's now try some mixed examples where we have integral exponents. Example 14 (a) Simplify a5×a−3 Answer a^5xxa^-3 = a^(5+(-3)) =a^(5-3) =a^2 (b) Simplify a3×a−5 Answer a^3xxa^(-5) = a^(3+(-5)) =a^-2 =1/a^2 Example 15 Simplify (23×2−4)2 Answer (2^3xx2^-4)^2 = (2^(3-4))^2 =(2^-1)^2 =(1/2)^2 =1/4 Example 16 Simplify ab7a2b3c0 Answer (a^2b^3c^0)/(ab^7) = (a^(2-1)(1))/(b^(7-3)) =a/b^4 The importance of brackets Note the following differences carefully: (−5x)0 = 1, but −5x0 = −5. In the first one, we are raising everything in brackets to the power 0, so the answer is 1. In the second one, we are only raising the x to the power 0, then we are left with −5×1=−5 Similarly: (−5)0 = 1, but −50 = −1. Example 17 Simplify (2a + b−1)-2 Answer (2a+b^-1)^-2=(2a+1/b)^-2 =((2ab+1)/b)^-2 =(b/(2ab+1))^2 =b^2/(2ab+1)^2 Exercises (1) Simplify: (5an−2)−1 Answer (2) Simplify: (b2a−2)−3(b5a−3)2 Answer ((a^-2)/(b^2))^-3((a^-3)/(b^5))^2 =(1/(a^2b^2))^-3(1/(a^3b^5))^2 (3) Simplify: (2a−b−2)−1 Answer Later, on this page Multiplying Expressions with the Same Base Dividing Expressions with the Same Base Repeated Multiplication of a Number Raised to a Power A Product Raised to an Integral Power A Fraction Raised to an Integral Power Raising a Number to a Zero Exponent Raising a Number to Negative Exponents Summary - Laws of Exponents The importance of brackets Exercises
16624	https://www.omnicalculator.com/physics/electric-potential	Electric Potential Calculator Use the electric potential calculator to determine the electric potential at a point either due to a single point charge or a system of point charges. You can also use this tool to find out the electrical potential difference between two points. If you want to calculate the electric field due to a point charge, check out the electric field calculator. Continue reading this article to learn: What is electric potential? What is the relation between electric potential and electric potential energy? How to calculate electric potential? What is the unit of electric potential? Electric potential difference To understand the idea of electric potential difference, let us consider some charge distribution. This charge distribution will produce an electric field. Now, if we want to move a small charge q between any two points in this field, some work has to be done against the Coulomb force (you can use our Coulomb's law calculator to determine this force). This work done gets stored in the charge in the form of its electric potential energy. If we consider two arbitrary points, say A and B, then the work done (WAB) and the change in the potential energy (ΔU) when the charge (q) moves from A to B can be written as: WAB=ΔU=(VA−VB)q ...... (1) where VA and VB are the electric potentials at A and B, respectively (we will explain what it means in the next section). If the magnitude of q is unity (we call a positive charge of unit magnitude as a test charge), the equation changes to: ΔV=(VA−VB)=qWAB ...... (2) Using the above equation, we can define the electric potential difference (ΔV) between the two points (B and A) as the work done to move a test charge from A to B against the electrostatic force. Remember that the electric potential energy can't be calculated with the standard potential energy formula, E=mgh. What is electric potential? – Definition of electric potential If we take one of the points in the previous section, say point A, at infinity and choose the potential at infinity to be zero, we can modify the electric potential difference formula (equation 2) as: VB=qW∞B Hence, we can define the electric potential at any point as the amount of work done in moving a test charge from infinity to that point. We can also define electric potential as the electric potential energy per unit charge, i.e.: V=qΔU So you can see that electric potential and electric potential energy are not the same things. Electric potential formula To calculate electric potential at any point A due to a single point charge (see figure 1), we will use the formula: V=krq where: q — Electrostatic charge; r — Distance between A and the point charge; and k=4πϵ01 — Coulomb's constant. We note that when the charge q is positive, the electric potential is positive. When the charge q is negative electric potential is negative. Now we will consider a case where there are four point charges, q1, q2, q3, and q4 (see figure 2). The potential at point A due to the charge q1 is: V1=kr1q1 We can write similar expressions for the potential at A due to the other charges: V2V3V4=kr2q2=kr3q3=kr4q4 To get the resultant potential at A, we will use the superposition principle, i.e., we will add the individual potentials: VV=V1+V2+V3+V4=k(r1q1+r2q2+r3q3+r4q4) For a system of n point charges, we can write the resultant potential as: VVV=V1+V2+V3+....+Vn=k(r1q1+r2q2+r3q3+....+rnqn)=k∑riqi In the next section, we will see how to calculate electric potential using a simple example. How to calculate electric potential Let us calculate the electrostatic potential at a point due to a charge of 4×10−7 C located at a distance of 10 cm. We are given: q=4×10−7 C and r=10 cm. 2. Substituting these values in the formula for electric potential due to a point charge, we get: V=4πϵ0rq V=0.1 m8.99×109 N⋅m2/C2×4×10−7 C V=3.6×104 V 3. Hence, the electric potential at a point due to a charge of 4×10−7 C located at a distance of 10 cmaway is 3.6×104 V. How to use the electric potential calculator Now, we will see how we can solve the same problem using our electric potential calculator: Using the radio button menu, choose electric potential due to a point charge. Enter the value of electric charge, i.e., 4e−07 and the distance between the point charge and the observation point (10 cm). You can also change the value of relative permittivity. The calculator will display the value of the electric potential at the observation point, i.e., 3.595×104 V. Units of electric potential The SI unit of electric potential is the volt (V). We can say that the electric potential at a point is 1 V if 1 J of work is done in carrying a positive charge of 1 C from infinity to that point against the electrostatic force. The unit of potential difference is also the volt. You might be more familiar with voltage instead of the term potential difference. For example, when we talk about a 3 V battery, we simply mean that the potential difference between its two terminals is 3 V. 💡 Our battery capacity calculator is a handy tool that can help you find out how much energy is stored in your battery. Dimensional formula of electric potential To write the dimensional formula for electric potential (or electric potential difference), we will first write the equation for electric potential: V=qW Now substituting the dimensional formula for work/energy and charge, we will get the dimensional formula for electric potential as: V=[AT][M1L2T−2]=[M1L2T−3A−1] FAQs How do I calculate electric potential of a point charge? To calculate the electric potential of a point charge (q) at a distance (r), follow the given instructions: Multiply the charge q by Coulomb's constant. Divide the value from step 1 by the distance r. Congrats! You have calculated the electric potential of a point charge. Can electric potential be negative? Yes, electric potential can be negative. The electrostatic potential at a point due to a positive charge is positive. If the charge is negative electric potential is also negative. What is electric potential difference? The electric potential difference between two points A and B is defined as the work done to move a positive unit charge from A to B. The SI unit of potential difference is volt (V). Is electric potential a scalar or a vector quantity? Electric potential is a scalar quantity as it has no direction. What is the unit of electric potential? Electric potential is the electric potential energy per unit charge. The SI unit of electric potential energy is the joule (J), and that of charge is the coulomb (C). Hence, the SI unit of electric potential is J/C, i.e., the volt (V). What is the electric potential of a charge at a point at infinity? Zero. The electric potential at a point P due to a charge q is inversely proportional to the distance between them. Hence, when the distance is infinite, the electric potential is zero. Electric potential of a point charge © Omni Calculator Did we solve your problem today? Check out 42 similar electromagnetism calculators 🧲 Acceleration of a particle in an electric field AC wattage
16625	https://www.scribd.com/document/517393096/subtracting-across-zeros-lesson-plan	Subtracting Across Zeros Lesson Plan \| PDF \| Educational Assessment \| Neuropsychological Assessment Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 2K views 7 pages Subtracting Across Zeros Lesson Plan This lesson plan focuses on teaching students how to subtract across zeros. The lesson objectives are for students to subtract across zeros and explain their reasoning about subtracting ac… Full description Uploaded by api-550458937 AI-enhanced title and description Go to previous items Go to next items Download Save Save subtracting across zeros lesson plan For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save subtracting across zeros lesson plan For Later You are on page 1/ 7 Search Fullscreen !"#$%&'$() ,'%-.. /0%-. 10..-) 23&) !" $%&'(&' )'&+,+- !"#$% • !" $ %$&'($') $+, %),,-&. ,)%)/+ ,+$&0$0, "- 1+)2, 3 $&0 4 • !" $ 2$+5 %),,-&. ,)%)/+ ,+$&0$0, "- 1+)2, 6 $&0 4 • !" $ %),,-& 1& $+. 2(,1/. ,-/1$% ,/1)&/). ,/1)&/). 7898 - 5)$%+5. ,)%)/+ ,+$&0$0, "- 1+)2, 6. 3 $&0 4 •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adDownload to read ad-free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• !"#$%&'() ,)'&-%./ • 0&/%)1($#'2/ 3)#24 5)(6$#6% 5)'&-%./ • 7($" !')8&''8 5(4%/ 9:;<9:= (2 9:><9:? 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16626	https://www.e-education.psu.edu/meteo815/node/557	\| \| \| METEO 815 Applied Atmospheric Data Analysis \| Meteorology Department Penn State Hypothesis Formation and Testing for 1 Sample: Part 2 Prioritize... By the end of this section, you should be able to generate a hypothesis, distinguish between questions requiring a one-tailed and two tailed test, know when to reject the null hypothesis and how to interpret the results. Read... Decision Rules Now that we have the hypothesis stated and have chosen the level of significance we want, we need to determine the decisions rules. In particular, we need to determine threshold value(s) for the test statistic beyond which we can reject the null hypothesis. There are two ways to go about this: the P-Value approach and the Critical Value approach. These approaches are equivalent, so you can decide which one you like the most. The P-value approach determines the probability of observing a value of the test statistic more extreme than the threshold assuming the null hypothesis is true; you look at the probability of the test statistic value you calculated from your data. For the P-value approach, we compare the probability (P-value) of the test statistic to the significance level (α). If the P-value is less than or equal to α, then we reject the null hypothesis in favor of the alternative hypothesis. If the P-value is greater than α, we do not reject the null hypothesis. Here is the region of rejection of the null hypothesis for a one-sided and two sided test: For the P-value approach, the direction of the test (lower tail/upper tail) does not matter. If the probability is less than α (α/2 for a two-tailed test), then we reject the null hypothesis. The critical value approach looks instead at the actual value of the test statistic. We transform the significance level, α, to the test statistic value. This requires us to be very careful about the direction of the test. If your test statistic lies beyond the critical value, then reject the null hypothesis. Let's break it down by test: If H1: μ<μo (Lower Tailed Test): Simply find the corresponding test statistic value of the assigned significance level (α); that is, if α is 0.05 find the t or Z-value that corresponds to a probability of 0.05. This is your critical value and we denote this as Zα or tα. The blue shaded area represents the rejection region. For the lower tailed test, if the test statistic is to the left of the critical value (less than) in the rejection region (blue shaded area), we reject the null hypothesis. I said before that one common α value was 0.05 (a 5% chance of a type I error). For α=0.05, the corresponding Z-value would be -1.645. The Z-statistic would have to be less than -1.645 for the null hypothesis to be rejected at the 95% confidence level. If H1: μ>μo (Upper Tailed Test): You have to remember that the t and Z-values use the CDFs of the PDF. This means that for the right tail or upper tail the corresponding t and Z-values will actually come from 1-α. For example, if α is 0.05 and we have an upper tail test, we find the t or Z-value that corresponds to a probability of 0.95. This would be the critical value and we denote this as Z1-α or t1-α. For the upper tailed test, if the test statistic is to the right of the critical value (greater than), we reject the null hypothesis. For α=0.05, the corresponding Z-value would be 1.645. The Z-statistic would have to be greater than 1.645 for the null hypothesis to be rejected at the 95% confidence level. If H1: μ≠μo (Two Tailed Test): For the two-tailed test, if your test statistic is to the left of the lower critical value or the right of the upper critical value, then we reject the null hypothesis. For α=0.05, the corresponding Z-value would be ±1.96. The Z-statistic would have to be less than -1.96 or greater than 1.96 for the null hypothesis to be rejected at the 95% confidence level. I suggest that for nonparametric tests, such as the Wilcoxon or Sign test, that you use the P-value approach because the functions available in R will estimate the P-value for you which means you do not have to estimate the critical value for these tests, avoiding a tedious task. Here is a flow chart for the rejection of the null hypothesis. Here is a larger view (opens in another tab). Computation and Confidence Interval Finally, after we have set up the hypothesis and determine our significance level, we can actually compute the test statistic. In this section, I will briefly talk about functions available in R to calculate the Z, t, Wilcoxon, and the Sign test statistics. But first, I want to talk about confidence intervals. Confidence intervals are estimates of the likely range of the true value of a particular parameter given the estimate of that parameter you've computed from your data. For example, confidence intervals are created when estimating the mean of the population. This confidence interval represents a range of certainty our data gives us about the population mean; we are never 100% certain the mean of a sample dataset represents the true mean of a population, so we calculate the range in which the true mean lies within with a certain confidence. Most times confidence intervals are set at the 95% or 99% level. This means we are 95% (99%) confident that the true mean is within the interval range. To estimate the confidence interval, we create a lower and upper bound using a Z-score reflective of the desired confidence level: where x ¯ is the estimated mean from the dataset, σ is the estimate of the standard deviation of the dataset, n is the sample size, and z is the Z-score representing the confidence level you want. The common confidence levels of 95% and 99% have Z-scores of 1.96 and 2.58 respectively. You can estimate a confidence interval using t-scores instead if n is small, but the t-values depend on the degrees of freedom (n-1). You can find t-scores at any confidence level using this table. Confidence intervals, however, can be calculated in R by supplying only the confidence level as a percentage; generally, you will not have to determine the t-score yourself. For the Z-test, you can just calculate the Z-score yourself or you can use the function "z.test" from the package "BSDA". This function can be used for testing two datasets against each other, which will be discussed further on in this section. For now, let's focus on how to use this function for one dataset. There are 5 arguments for this function that are important: The great thing about the function "z.test" is that it provides both the Z-value and the P-value so you can use either the critical value approach or the P-value approach. The function will also provide a confidence interval for the population. Note that if you use a one-sided test, your confidence interval will be unbounded on one side. When the data is normally distributed but the sample size is small, we will use the t-test, or the function "t.test" in R. This function has similar arguments to the "z.test". Similar to the "z.test", the "t.test" provides both the t-value and the P-value, so you can either use the critical value approach or the P-value approach. It will also automatically provide a confidence interval for the population mean. Again, if you use a one-sided test your confidence interval will be unbounded on one side. The 'z.test' and 't.test' are for data that is parametric. What about nonparametric datasets - ones in which we do not have parameters to represent the fit? I'm only going to show you the functions for the Wilcoxon test and the Sign test which I described previously. There are, however, many other test statistics you can calculate using functions available in R. The Wilcoxon test or 'wilcox.test' in R, which is used for nonparametric cases that appear symmetric, has similar arguments as the 'z.test'. The test provides a P-value as well as a critical value, but I would recommend for nonparametric tests to follow the P-value approach because of the tediousness in estimating the critical value. For the sign test use the function 'SIGN.test' in R. The arguments are: Again, the test provides a P-value as well as a critical value, but I would recommend for nonparametric tests to follow the P-value approach. Decision and Interpretation The last step in hypothesis testing is to make a decision based on the test statistic and interpret the results. When making the decision, you must remember that the testing is of the null hypothesis; that is, we are deciding whether to reject the null hypothesis in favor of the alternative. The critical region changes depending on whether the test is one-sided or two-sided; that is, your area of rejection changes. Let’s break it down by two-sided, upper, and lower. Note that I will write the results with respect to the Z-score, but the decision is the same for whatever test statistic you are using. Simply replace the Z-value with the value from your test statistic. For a two-tailed test, your hypothesis statement would be: You will reject the null hypothesis (Ho) and accept the alternative (H1) if: For an upper one-tailed test, your hypothesis statement would be: You will reject the null hypothesis (Ho) and accept the alternative (H1) if: For a lower one-tailed test, your hypothesis statement would be: You will reject the null hypothesis (Ho) and accept the alternative (H1) if: Again, for any other test (t-test, Wilcox, or Sign), simply replace the Z-value and Z-critical values with the corresponding test values. So, what does it mean to reject or fail to reject the null hypothesis? If you reject the null hypothesis, it means that there is enough evidence to reject the statement of the null hypothesis and accept the alternative hypothesis. This does not necessarily mean the alternative is correct, but that the evidence against the null hypothesis is significant enough (1-α) that we reject it for the alternative. The test statistic is inconsistent with the null hypothesis. If we fail to reject the null hypothesis, it means the dataset does not provide enough evidence to reject the null hypothesis. It does not mean that the null hypothesis is true. When making a claim stemming from a hypothesis test, the key is to make sure that you include the significance level of your claim and know that there is no such thing as a sure thing in statistics. There will always be uncertainty in your result, but understanding that uncertainty and using it to make meaningful decisions makes hypothesis testing effective. Now, give it a try yourself. Below is an interactive tool that allows you to perform a one-sided or two-sided hypothesis test on temperature data for London. You can pick whatever threshold you would like to test, the level of significance, and the type of test (Z or t). Play around and take notice of the subtle difference between the Z and t-tests as well as how the null and alternative hypothesis are formed. Lessons Author: Jacola Roman, Department of Meteorology; Instructor, John A. Dutton e-Education Institute, The Pennsylvania State University. The College of Earth and Mineral Sciences is committed to making its websites accessible to all users, and welcomes comments or suggestions on access improvements. Please send comments or suggestions on accessibility to the site editor. The site editor may also be contacted with questions or comments about this course. The John A. Dutton Institute for Teaching and Learning Excellence is the learning design unit of the College of Earth and Mineral Sciences at The Pennsylvania State University. 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16627	https://pmc.ncbi.nlm.nih.gov/articles/PMC9833460/	Nutrition Management in Patients With Traumatic Brain Injury: A Narrative Review - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Brain Neurorehabil . 2022 Mar 28;15(1):e4. doi: 10.12786/bn.2022.15.e4 Search in PMC Search in PubMed View in NLM Catalog Add to search Nutrition Management in Patients With Traumatic Brain Injury: A Narrative Review Hoo Young Lee Hoo Young Lee 1 Department of Rehabilitation Medicine, Seoul National University College of Medicine, Seoul National University Hospital, Seoul, Korea. 2 National Traffic Injury Rehabilitation Hospital, Yangpyeong, Korea. Find articles by Hoo Young Lee 1,2, Byung-Mo Oh Byung-Mo Oh 1 Department of Rehabilitation Medicine, Seoul National University College of Medicine, Seoul National University Hospital, Seoul, Korea. 2 National Traffic Injury Rehabilitation Hospital, Yangpyeong, Korea. 3 Institute on Aging, Seoul National University, Seoul, Korea. Find articles by Byung-Mo Oh 1,2,3,✉ Author information Article notes Copyright and License information 1 Department of Rehabilitation Medicine, Seoul National University College of Medicine, Seoul National University Hospital, Seoul, Korea. 2 National Traffic Injury Rehabilitation Hospital, Yangpyeong, Korea. 3 Institute on Aging, Seoul National University, Seoul, Korea. ✉ Correspondence to Byung-Mo Oh. Department of Rehabilitation Medicine, Seoul National University Hospital, Seoul National University College of Medicine, 103 Daehak-ro, Jongno-gu, Seoul 03080, Korea. moya1@snu.ac.kr ✉ Corresponding author. Received 2022 Mar 4; Revised 2022 Mar 19; Accepted 2022 Mar 22; Collection date 2022 Mar. Copyright © 2022. Korean Society for Neurorehabilitation This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License ( which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited. PMC Copyright notice PMCID: PMC9833460 PMID: 36743843 Abstract Traumatic brain injury (TBI) is a major cause of long-term physical and psychological disability and death. In patients with TBI, undernutrition is associated with an increased mortality rate, more infectious complications, and worse neurologic outcomes. Therefore, timely and effective nutritional therapy is particularly crucial in the management of TBI to improve patients’ prognoses. This narrative review summarizes the issues encountered in clinical practice for patients with neurotrauma who receive acute and post-acute in-patient rehabilitation services, and it comprehensively incorporates a wide range of studies, including recent clinical practice guidelines (CPGs), with the aim of better understanding the current evidence for optimal nutritional therapy focused on TBI patients. Recent CPGs were reviewed for 6 topics: 1) hypermetabolism and variation in energy expenditure in patients with TBI, 2) delayed gastric emptying and intolerance to enteral nutrition, 3) decision-making on the route and timing of access in patients with TBI who are unable to maintain volitional intake (enteral nutrition versus parenteral nutrition), 4) decision-making on the enteral formula (standard or immune-modulating formulas), 5) glycemic control, and 6) protein support. We also identified areas that need further research in the future. Keywords: Traumatic Brain Injury, Nutrition Therapy, Enteral Nutrition, Parenteral Nutrition, Glycemic Control Highlights • Traumatic brain injury causes multiple gastrointestinal and nutritional complications. • The assessment and multidisciplinary approach are essential for effective nutrition support. • Route of access, glycemic control, and protein support are important considerations. INTRODUCTION Traumatic brain injury (TBI) affects approximately 69 million people worldwide each year and has serious implications with regard to long-term physical and psychological disability and death [1,2,3]. In Korea, approximately 480,000 new TBI cases occur annually and the total medical costs for TBI steadily increased over the last decade . Hypermetabolism and increased catabolism after TBI lead to hyperglycemia, protein wasting, and increased energy demand, which may be as high as 200% of the usual energy requirement [5,6]. A previous study reported that 68% of patients with acute TBI were malnourished . This negative energy balance causes a decrease in body mass, especially skeletal muscle mass, and leads to a negative nitrogen balance. It is also associated with an increase in morbidity and mortality . Nutrition in patients with TBI is pivotal for maintaining cellular homeostasis and reducing mortality and the incidence of infectious complications [9,10,11]. Therefore, timely and effective nutritional therapy is particularly crucial in the management of TBI to improve patients’ prognoses, especially in more severe TBI cases. Although many previous studies have been conducted on nutrition in stroke patients, relatively few studies have explored nutrition in patients with TBI [12,13]. In this review, we summarize the difficulties or issues encountered in clinical practice and comprehensively incorporate previous studies and recent clinical practice guidelines (CPGs) to better understand the current evidence for optimal nutritional therapy in TBI patients. We also identify areas that need further research in the future. TOPIC 1. HYPERMETABOLISM AND VARIATION IN ENERGY EXPENDITURE IN PATIENTS WITH TBI The secondary response to trauma increases the secretion of catecholamines, which antagonize insulin, and inflammatory mediators. Hormonal changes after TBI can lead to hypermetabolism by increasing the secretion of corticosteroids, counterregulatory hormones, and cytokines. In the acute phase of TBI, energy requirements increase to 100%–200% of baseline-predicted resting energy expenditure (REE), which may persist for several weeks to several months, depending on the severity of the neurotrauma and level of recovery [5,14]. Hypercatabolism of brain injury is associated with increased morbidity and weight loss . It generally stops and plateaus at 2 months postinjury, and this timing often coincides with admission to inpatient rehabilitation . In the acute phase of TBI, factors including the patient’s body temperature, use of sedatives, mechanical ventilation, and the severity of brain injury modify the REE, making it very challenging to predict an individual’s nutritional requirements; this difficulty may lead to inadequate nutrition . Moreover, since patients with severe TBI often have edema related to resuscitation, it is difficult to accurately predict energy requirements using body weight . In the post-acute phase, calorie needs should be reassessed. Calorie demands decrease in the setting of limited mobility as the medical status normalizes . Meanwhile, the process of rehabilitation therapy increases calorie needs and this may be 30-60% higher than control groups . Recommendations from recent CPGs: Determination of energy expenditure The American Society for Parenteral and Enteral Nutrition-Society of Critical Care Medicine (ASPEN-SCCM) and many clinicians recommend that indirect calorimetry is the current “gold standard” to measure energy requirements in patients with TBI whenever possible (quality of evidence: very low [ASPEN-SCCM]) [8,17]. However, using indirect calorimetry involves many practical difficulties, such as high costs and the requirement for a trained professional . The ASPEN-SCCM guideline suggests that if indirect calorimetry is not available, a published predictive equation or a basic weight-based equation (25–30 kcal/kg/d) be applied to determine energy requirements in critically ill patients (quality of evidence: expert consensus) . The Harris-Benedict, Ireton-Jones, and Penn State predictive equations are commonly used (Table 1) [20,21]. The Brain Trauma Foundation recommends that TBI patients be fed to achieve basal caloric replacement at least by the fifth day and at most, by the seventh day post-injury to lower risk of death (level IIA) . Table 1. Common equations for predicting resting energy expenditure. \| Equations \| \| Equations derived from testing hospital patients \| \| \| Penn State Equation \| \| \| \| REE = (1.1 × value of HBE) + (140 × Tmax) + (32 × VE) – 5,340 \| \| \| Ireton-Jones Equation for ventilated patients \| \| \| \| Male REE = 2,028 − 11 × A + 5 × W + 239 × T + 804 × B \| \| \| \| Female REE = 1,784 − 11 × A + 5 × W + 239 × T + 804 × B \| \| Equations derived from testing normal volunteers \| \| \| Harris-Benedict Equations \| \| \| \| Male REE = 66.47 + 13.75 × W + 5 × H − 6.755 × A \| \| \| \| Female REE = 665.1 + 9.563 × W + 1.85 × H − 4.676 × A \| \| \| Mifflin-St. Jeor Equations \| \| \| \| Male BMR = 10 × W + 6.25 × H − 5 × A + 5 \| \| \| \| Female BMR = 10 × W + 6.25 × H − 5 × A − 161 \| Open in a new tab REE, resting energy expenditure (kcal/day); HBE, REE calculated by Harris-Benedict method (kcal/day); Tmax, maximum body temperature in the past 24 hours (°C); VE, expired minute volume (L/min); A, ages (years); W, actual body weight (kg); T, trauma; B, burn; H, height (cm). During inpatient rehabilitation, comprehensive information concerning limited mobility or paresis, progress in therapy and activities of daily living, and level of agitation is important for the nutrition reassessment . Moreover, weekly monitoring for weight gain or loss is useful. In case of weight gain due to hyperphagia, it is important to apply behavioral strategies using a memory board or diary, and to encourage patients to follow a low-fat, well-balanced diet . TOPIC 2. DELAYED GASTRIC EMPTYING AND INTOLERANCE TO ENTERAL NUTRITION Gastroparesis, or delayed gastric emptying, is one of the major factors that cause feeding intolerance, which is exhibited in 45%–50% of TBI patients [6,23]. Previous studies showed that in patients with moderate-to-severe TBI, the gastric emptying half-life was delayed more than twice as compared to that of healthy control subjects. Gastrointestinal hypokinesia usually persisted during the first 1–2 weeks after TBI [14,24]. After the transfer to the general ward, the delay in emptying may continue depending on the severity of the TBI and if elevated intracranial pressure continues. Delayed gastric emptying may be assumed when there is feeding tube intolerance with large gastric residual volume. Ileus may be present, but it appears more commonly when brain injury is accompanied by the spinal cord injury . There are numerous reasons for intolerance to enteral nutrition (EN) in patients with TBI (Table 2). Neurotrauma increases intracranial pressure and damages the autonomic nervous system [23,26]. Sedatives such as opioid agents delay gastric emptying, which consequently may increase gastric residual volume and the risk of vomiting . Furthermore, several aspects of patients with TBI interrupt EN. For instance, pain from trauma, facial fractures, oral injury, and prolonged cervical immobilization may delay resumption of an oral diet. A previous study showed that surgery, extubation or intubation, or radiological exams interrupted approximately 30% of critically ill patients with TBI at least once during the observation period . Table 2. Reasons for intolerance to enteral nutrition in patients with TBI. \| Variables \| Reasons \| :---: \| \| Central mechanism \| Increased intracranial pressure \| \| Damage of the autonomic nervous system \| \| Central and peripheral mechanism \| Opioid agents \| \| Pain \| \| Multiple trauma \| Facial fractures \| \| Oral injury \| \| Prolonged cervical immobilization \| \| Interruptions in healthcare \| Surgery \| \| Extubation or intubation \| \| Radiologic exams \| \| Bedside procedures \| \| Large gastric residual volume \| \| Emesis \| Open in a new tab TBI, traumatic brain injury. Recommendations from recent CPGs: Strategies to improve intolerance to EN in patients with TBI There are several strategies to improve feeding tolerance in patients with TBI. First, the American Dietetic Association recommends positioning patients in a 45° head-of-bed elevation position to prevent aspiration pneumonia (grade II) and to minimize gastroesophageal or laryngopharyngeal reflux of gastric contents (grade I) . Second, concentrated enteral formulas (≥ 1.5 kcal/mL) may reduce the risk of reflux or intolerance while meeting caloric requirements in less volume . Thirdly, for EN, a continuous infusion is preferred rather than administration as a bolus. A recent randomized clinical trial (RCT) demonstrated that the continuous infusion had more positive effects on nitrogen balance and decreasing the hypercatabolic response in patients with TBI than intermittent EN and parenteral methods . Since the enteric nervous system and intestinal smooth muscle are intact, motility-promoting agents should be useful. The ASPEN-SCCM guideline suggests that in patients at risk of swallowing aspiration, prokinetic medications (e.g., metoclopramide or erythromycin) be initiated where feasible . Metoclopramide is currently the only FDA-approved promotility agent. However, the potential for central nervous system side effects due to antagonizing central dopamine D2 receptors may limit its use. It may also worsen the severity and frequency of seizures and should not be used in patients with TBI who have seizures. Also, patients taking medications that may increase the risk of extrapyramidal symptoms should avoid it. As a treatment for intolerance to EN, metoclopramide appears to be less effective in TBI than in other critically ill patients and combination therapy with erythromycin should be considered unless contraindicated [30,31]. Erythromycin is a macrolide antibiotic structurally similar to the GI hormone motilin. Although it does not have an FDA-approved indication, it is used as an effective promotility drug. The European Society for Clinical Nutrition and Metabolism (ESPEN) guideline recommend that postpyloric feeding should be considered in critically ill patients whose gastric feeding intolerance has not been resolved with prokinetic agents (grade of recommendation B - strong consensus) or in patients whose risk for aspiration is high (grade of recommendation: good practice point [GPP] - strong consensus). Experts suggest that jejunal feeding should only be attempted in environments where the technique is readily available or if gastric akinesia persists despite appropriate attempts [6,32]. After the transfer to the general ward, low fat (<30%) meals are preferred for gastroparesis because high fat meals may further prolong gastric emptying . Antiemetcs may be suitable for nausea . TOPIC 3. DECISION-MAKING ON THE ROUTE AND TIMING OF ACCESS IN PATIENTS WITH TBI WHO ARE UNABLE TO MAINTAIN VOLITIONAL INTAKE: EN VERSUS PN According to meta-analyses of previous RCTs comparing early (within 24–48 hours) versus delayed EN, patients who received early EN showed lower mortality, reduced infection rates, and shorter hospital stays [33,34]. The advantages of EN over parenteral nutrition (PN) have been demonstrated in numerous previous RCTs regarding reduction of infection (e.g., pneumonia and central line infection in most patients, and abdominal abscess in trauma patients, in particular) and the length of stay in the intensive care unit . In particular, EN preserves gut integrity, and stress and the immune response are physiologically regulated. Furthermore, access through the gut serves as a passageway for immune-modulating substances and is effective in preventing stress ulcers . Recommendations from recent CPGs: Timing of the initiation of EN and PN in patients with acute TBI Both the ASPEN-SCCM and ESPEN guidelines recommend initiating early EN (within 24-48 hours) instead of delaying EN (quality of evidence: very low [ASPEN-SCCM]; grade of recommendation: B - strong consensus [ESPEN]) [17,32]. Early EN therapy is more beneficial, especially in high-risk patients . For hemodynamically stable patients, the use of EN over PN is recommended in both guidelines (quality of evidence-low to very low [ASPEN-SCCM]; grade of recommendation: A - strong consensus [ESPEN]) [17,32]. Gastric access is the standard method for initiating EN (quality of evidence: moderate to high [ASPEN-SCCM]; grade of recommendation: GPP - strong consensus [ESPEN]), and gradual increase of EN is necessary to avoid overfeeding in the early phase, especially in patients who are intolerant initially (grade of recommendation: A - strong consensus [ESPEN]) [17,32]. An initial EN rate of 20 mL/h is appropriate, and it is desirable to increase the amount by 10 to 20 mL/h every 6–8 hours to reach the target amount . However, EN should be withheld in hemodynamically unstable patients . If a patient receiving vasopressor therapy is provided with EN, close attention should be paid to intolerance signs such as abdominal distention, hypoactive bowel sounds, decrease in stool passing, and metabolic acidosis, which are early signs of gut ischemia. EN should be discontinued until stabilization of the symptoms and interventions. Although early EN is recommended for most patients, there is no dispute about the necessity of supplementation with PN if malnutrition persists in order to minimize the detrimental effects of a negative energy balance . However, the most appropriate time for prescribing supplemental PN has not yet been established. According to Casaer et al. , early PN was associated with an increased morbidity and infection rate. In particular, the potential side effects of overestimating the caloric target in the acute phase were discussed. Although the most appropriate timing of PN supplementation has not been determined, the ESPEN guideline recommends 4 to 7 days based on the results of previous studies . Meanwhile, the ASPEN-SCCM guideline recommends considering the use of supplemental PN after 7–10 days if it is not possible to meet > 60% of energy and protein requirements by the enteral route alone . Initiating PN prior to this 7- to 10-day period in critically ill brain-injured patients does not improve the outcome, and may even adversely affect the patient (quality of evidence: moderate [ASPEN-SCCM]) . TOPIC 4. DECISION MAKING ON THE ENTERAL FORMULA: STANDARD OR IMMUNE-MODULATING FORMULAS After brain injury, excessive release of glutamate and excitotoxic neurotransmitters increases the intracellular calcium and sodium influx and results in energy depletion. Formulas using immune-modulating nutrients such as glutamine, omega-3 fatty acids, arginine, and nucleotides have been proposed to promote neuroprotection from secondary brain insults . Glutamine is one of the most abundant amino acids in the human body, and is mostly synthesized in muscles, where it is used for metabolism. In catabolic metabolism, which occurs after trauma, sepsis, or major surgery, more glutamine is metabolized than produced. Therefore, glutamine is categorized as a conditionally essential amino acid . Previous studies have shown that glutamine-enriched enteral diets reduce infections and length of stay in patients with moderate-to-severe TBI [38,39,40]. However, when plasma glutamine levels are normal, excessive doses have the potential to cause adverse effects . A proper dose of glutamine is between 0.3 to 0.5 g/kg body weight per day (i.e., 25–50 g/day) for 1 to 2 weeks during EN and PN . Omega-3 polyunsaturated fatty acids (n-3 PUFA), which include eicosapentaenoic acid (EPA) and docosahexaenoic acid (DHA), are important for appropriate brain development and function. DHA is the most abundant n-3 PUFA in the brain and is involved in regeneration and repair in the central nervous system after TBI . Previous studies showed that n-3 PUFA supplementation may decrease neuroinflammation after brain injury [43,44]; however, a larger clinical trial demonstrated that the administration of n-3 PUFA had little effect on quality of life in patients with TBI . So far, there is no conclusive evidence supporting the use of n-3 PUFA . DHA and EPA have promising effects in experimental studies, but no clinical data are convincing . L-arginine is the precursor of several active compounds such as nitric oxide, proline, polyamines, ornithine, creatine, phosphocreatine, and agmatine . Proper arginine supplementation is associated with improvements in the immune response and protein synthesis. Critical care formulas including arginine, fish oil, and various antioxidants have been shown to be effective at reducing infection in brain-injured patients . Previous studies have shown that requirements for branched-chain amino acids (BCAAs; valine, isoleucine, and leucine) increase or BCAA levels decrease after TBI [49,50]. BCAAs are essential amino acids that act as important nitrogen donors for glutamate synthesis in the brain and are essential for neurotransmitter cycling . BCAAs are also used in brain cells as a fuel source for the tricarboxylic acid cycle. BCAAs also act as major nitrogen donors through transamination in skeletal muscle. In particular, leucine activates mammalian target of rapamycin (mTOR) signaling and inhibits adenosine monophosphate kinase (AMPK) activity to promote protein synthesis and skeletal muscle growth . Previous studies showed that intravenous BCAA (e.g., leucine) infusion reduced mortality and post-TBI disability by enhancing protein synthesis and homeostasis [53,54]. The effects of BCAAs on outcomes have mainly been studied in patients with severe TBI. Further investigations on the effect of BCAA supplementation on patients with mild TBI are needed. It is also necessary to explore the effects of supplementation according to patients’ characteristics, such as age, sex, cognitive function, and emotional and behavioral state. Recommendations from recent CPGs: Suggestions for immune-modulating enteral formulations The use of an immune-modulating formulation containing arginine or EPA/DHA supplementation in addition to the standard enteral formula is suggested in patients with TBI (quality of evidence: very low [ASPEN-SCCM]) . TOPIC 5. DECISION-MAKING ON GLYCEMIC CONTROL: TIGHT VERSUS PERMISSIVE Although hyperglycemia after TBI is associated with injury severity and poorer outcomes, there is no consensus that strict blood glucose control has a positive effect on the outcomes of patients with TBI [8,55]. Intensive insulin therapy for strict glycemic control has been reported to increase energy crises in the brain (high lactate-to-pyruvate ratio and excessive glutamate) and the risk of reduced brain glucose concentration . Hence, avoiding excessive hyperglycemia (> 10–11 mmol/L) and sustaining “permissive” glycemic control between 8 to 11 mmol/L are currently recommended . Accumulating data have shown that during cerebral energetic crises, lactate, ketone bodies, and BCAAs may be favored substrates to reduce the potential detrimental effects of intensive insulin therapy . Recommendations from recent CPGs: Suggestions for glycemic control in patients with TBI The NICE-SUGAR study showed that patients with TBI randomly assigned to intensive (4.5–6.0 mmol/L) glucose control experienced moderate and severe hypoglycemia more frequently than the conventional glucose (< 10 mmol/L) control group. However, there was no significant difference between the two groups in clinically important outcomes . Currently, it is recommended to avoid excessive hyperglycemia (more than 10–11mmol/l) and to sustain a moderate ‘permissive’ glucose control (8–11mmol/l) . TOPIC 6. PROTEIN SUPPORT IN PATIENTS WITH TBI The importance of protein goes beyond its role as a simple source of calories. It is the most important caloric nutrient for the recovery of brain damage, complementation of immune function, and maintenance of body mass. Most critically ill brain-injured patients have a high ratio of protein requirements to total energy requirements, and it is difficult to meet their requirements with general EN. Nitrogen excretion increases independently of the corresponding supplementation amount, and the steady loss of nitrogen can continue for up to 4 weeks. Therefore, it is important to maintain protein balance, and it may be difficult to balance nitrogen even in the post-acute phase . For this reason, protein supplementation can be helpful in patients with insufficient EN, and regular evaluations of the adequacy of protein intake are necessary. Recommendations from recent CPGs: Determination of adequate protein intake The ASPEN-SCCM guideline suggests immediate implementation of EN with a high protein polymeric diet within 24–48 hours of trauma if the patient is hemodynamically stable (quality of evidence: very low) . Protein requirements are estimated to be in the range of 1.2–2.0 g/kg actual body weight per day for trauma patients . This requirement may be even higher in multitrauma patients. Most experts recommend that for patients with TBI, protein should account for 15%–20% of total calories, for which administration of at least 2 g/kg body weight per day is required . Immune-modulating formulas may be an applicable option for achieving a sufficient protein supply in order to minimize negative nitrogen balance after TBI [17,59]. UNRESOLVED ISSUES AND FUTURE RESEARCH DIRECTIONS Highlights of the topics discussed herein and recent recommendations for nutritional therapy in patients with TBI are listed in Table 3. Currently, some unresolved issues relate to a ketogenic diet and micronutrients (i.e., minerals, vitamins, and trace elements). A medium-chain triglyceride ketogenic diet might have a neuroprotective effect, and lactate and a ketogenic diet might be an alternative source of energy for patients with TBI . However, related clinical data are insufficient to recommend ketogenic diets as a preferential nutritional strategy. A previous study showed that intravenous zinc supplementation for 2–4 weeks improved outcomes after TBI . More research is needed to sharpen our understanding of the effects of zinc supplementation on recovery after TBI. Furthermore, prior studies demonstrated that intramuscular vitamin-D and vitamin-E injections enhanced cognitive symptoms and self-reported outcomes 6 months after severe TBI [61,62]. Further research is needed to clarify the effects of micronutrient supplementation on cognitive and physical outcomes in patients with TBI. Table 3. Highlights of the issues and recommendations for nutrition therapy in patients with TBI. \| Issues \| Recommendations \| :---: \| \| Determination of the energy expenditure \| Increased energy demand after TBI may lead to hypermetabolsim and hypercatabolism which are associated with increased morbidity and weight loss [15,16]. \| \| Indirect calorimetry is the current “gold standard” for the determination of REE in patients with TBI, however, there are several practical difficulties [17,32]. A published predictive equation or a basic weight-based equation (25–30 kcal/kg/d) is an alternative measure . \| \| Weekly monitoring for weight gain or loss is useful during inpatient rehabilitation . \| \| Delayed gastric emptying and intolerance to EN \| Gastric access is the standard method for initiating EN in patients with TBI. However, delayed gastric emptying is one of the major complications that may be observed in up to 50% of patients with TBI [6,23]. \| \| Proper positioning, continuous infusion, and motility promoting agents such as metoclopramide are recommended strategies for gastroparesis after TBI. Concentrated enteral formulas (≥ 1.5 kcal/mL) may reduce the intolerance . \| \| Route and timing of access in patients who are unable to maintain volitional intake: EN versus PN \| Early EN (within 24–48 hours) is recommended in patients with TBI. \| \| Use of supplemental PN be considered after 7–10 days if unable to meet > 60% of energy and protein requirements by the enteral route alone . \| \| If EN is contraindicated in severely malnourished patients, PN should be implemented progressively within 3–7 days, rather than no nutrition . \| \| Selecting immune-modulating enteral formulas \| Immune-modulating formulation containing arginine or EPA/DHA supplementation in addition to standard enteral formula is suggested in patients with acute TBI . Although not yet suggested in CPGs, previous studies showed that intravenous BCAA (e.g., leucine) infusion decreased mortality and disability in patients with severe TBI. \| \| Glycemic control \| Sustaining ‘permissive’ glycemic control between 8 to 11 mmol/l are currently recommended in patients with TBI . \| \| Protein support \| Maintaining protein balance during both acute and post-acute TBI is important. It is recommended that protein supply should account for 15%–20% of total calories, and administration at least 2 g/kg body weight per day in patients with TBI . \| Open in a new tab REE, resting energy expenditure; TBI, traumatic brain injury; EN, enteral nutrition; PN, parenteral nutrition; EPA, eicosapentaenoic acid; DHA, docosahexaenoic acid; CPG, clinical practice guidelines; BCAA, branched-chain amino acids. The current literature on post-TBI nutritional therapy focuses largely on severe, acute TBI, making it difficult to generalize the findings to different subgroups, such as patients with mild or chronic TBI. Moreover, the relationship between nutrition and post-TBI functioning, and proper strategies to reduce mortality and morbidity have not been elucidated in geriatric or pediatric populations. Geriatric patients with TBI, on average, experience higher mortality and morbidity rates, slower recovery, and worse outcomes than younger patients . A previous study showed that the Geriatric Nutritional Risk Index is a significant independent risk factor for mortality in geriatric patients with moderate to severe TBI . Although malnutrition is closely related to poor outcomes during hospitalization in geriatric patients, it remains common and underdiagnosed . Furthermore, nutrition is important in pediatric TBI for adequate repair and growth . More high-quality evidence is needed to guide decision-making for clinical practice specific to geriatric or pediatric patients with TBI. SUMMARY Evidence-based and timely nutritional therapy is important in the management of TBI to improve patients’ prognoses. The determination of REE is crucial, and indirect calorimetry is the current “gold standard” for the determination of REE in patients with TBI; however, due to several practical difficulties, using a published predictive equation or a basic weight-based equation is an alternative measure. Weekly monitoring for weight gain or loss is useful during inpatient rehabilitation. Early EN within 24–48 hours is beneficial. However, attention should be paid to delayed gastric emptying and strategies need to be discussed. If EN is contraindicated, PN should be given progressively within 3–7 days rather than no nutrition. An immune-modulating formulation containing arginine or EPA/DHA supplementation in addition to a standard enteral formula is suggested in patients with acute TBI. Intravenous BCAA (e.g., leucine) infusion may reduce mortality and disability in patients with severe TBI, but this possibility needs further investigation. Sustaining “permissive” glycemic control between 8 and 11 mmol/L and providing an adequate protein supply (15%–20% of total calories or administration at least 2 g/kg body weight per day) are currently recommended in patients with TBI. Ketogenic diets and micronutrients (i.e., minerals, vitamins, and trace elements) are unresolved issues and need future research. Furthermore, the relationship between nutrition and post-TBI functioning, as well as proper strategies to improve outcomes in different sub-groups, such as patients with mild TBI, chronic patients, geriatric patients, and pediatric patients with TBI need to be explored. Footnotes Funding: None. 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DELAYED GASTRIC EMPTYING AND INTOLERANCE TO ENTERAL NUTRITION TOPIC 3. DECISION-MAKING ON THE ROUTE AND TIMING OF ACCESS IN PATIENTS WITH TBI WHO ARE UNABLE TO MAINTAIN VOLITIONAL INTAKE: EN VERSUS PN TOPIC 4. DECISION MAKING ON THE ENTERAL FORMULA: STANDARD OR IMMUNE-MODULATING FORMULAS TOPIC 5. DECISION-MAKING ON GLYCEMIC CONTROL: TIGHT VERSUS PERMISSIVE TOPIC 6. PROTEIN SUPPORT IN PATIENTS WITH TBI UNRESOLVED ISSUES AND FUTURE RESEARCH DIRECTIONS SUMMARY Footnotes References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
16628	https://en.wikipedia.org/wiki/Fuel_efficiency	Fuel efficiency This articleneeds additional citations forverification.Please helpimprove this articlebyadding citations to reliable sources. Unsourced material may be challenged and removed.Find sources:"Fuel efficiency"–news·newspapers·books·scholar·JSTOR(May 2013)(Learn how and when to remove this message) Fuel efficiency (or fuel economy) is a form of thermal efficiency, meaning the ratio of effort to result of a process that converts chemical potential energy contained in a carrier (fuel) into kinetic energy or work. Overall fuel efficiency may vary per device, which in turn may vary per application, and this spectrum of variance is often illustrated as a continuous energy profile. Non-transportation applications, such as industry, benefit from increased fuel efficiency, especially fossil fuel power plants or industries dealing with combustion, such as ammonia production during the Haber process. In the context of transport, fuel economy is the energy efficiency of a particular vehicle, given as a ratio of distance traveled per unit of fuel consumed. It is dependent on several factors including engine efficiency, transmission design, and tire design. In most countries, using the metric system, fuel economy is stated as "fuel consumption" in liters per 100 kilometers (L/100 km) or kilometers per liter (km/L or kmpl). In a number of countries still using other systems, fuel economy is expressed in miles per gallon (mpg), for example in the US and usually also in the UK (imperial gallon); there is sometimes confusion as the imperial gallon is 20% larger than the US gallon so that mpg values are not directly comparable. Traditionally, litres per mil were used in Norway and Sweden, but both have aligned to the EU standard of L/100 km. Fuel consumption is a more accurate measure of a vehicle's performance because it is a linear relationship while fuel economy leads to distortions in efficiency improvements. Weight-specific efficiency (efficiency per unit weight) may be stated for freight, and passenger-specific efficiency (vehicle efficiency per passenger) for passenger vehicles. Contents Vehicle design Fuel efficiency is dependent on many parameters of a vehicle, including its engine parameters, aerodynamic drag, weight, AC usage, fuel and rolling resistance. There have been advances in all areas of vehicle design in recent decades. Fuel efficiency of vehicles can also be improved by careful maintenance and driving habits. Hybrid vehicles use two or more power sources for propulsion. In many designs, a small combustion engine is combined with electric motors. Kinetic energy which would otherwise be lost to heat during braking is recaptured as electrical power to improve fuel efficiency. The larger batteries in these vehicles power the car's electronics, allowing the engine to shut off and avoid prolonged idling. Fleet efficiency Fleet efficiency describes the average efficiency of a population of vehicles. Technological advances in efficiency may be offset by a change in buying habits with a propensity to heavier vehicles that are less fuel-efficient. Energy efficiency terminology Energy efficiency is similar to fuel efficiency but the input is usually in units of energy such as megajoules (MJ), kilowatt-hours (kW·h), kilocalories (kcal) or British thermal units (BTU). The inverse of "energy efficiency" is "energy intensity", or the amount of input energy required for a unit of output such as MJ/passenger-km (of passenger transport), BTU/ton-mile or kJ/t-km (of freight transport), GJ/t (for production of steel and other materials), BTU/(kW·h) (for electricity generation), or litres/100 km (of vehicle travel). Litres per 100 km is also a measure of "energy intensity" where the input is measured by the amount of fuel and the output is measured by the distance travelled. For example: Fuel economy in automobiles. Given a heat value of a fuel, it would be trivial to convert from fuel units (such as litres of gasoline) to energy units (such as MJ) and conversely. But there are two problems with comparisons made using energy units: Energy content of fuel The specific energy content of a fuel is the heat energy obtained when a certain quantity is burned (such as a gallon, litre, kilogram). It is sometimes called the heat of combustion. There exists two different values of specific heat energy for the same batch of fuel. One is the high (or gross) heat of combustion and the other is the low (or net) heat of combustion. The high value is obtained when, after the combustion, the water in the exhaust is in liquid form. For the low value, the exhaust has all the water in vapor form (steam). Since water vapor gives up heat energy when it changes from vapor to liquid, the liquid water value is larger since it includes the latent heat of vaporization of water. The difference between the high and low values is significant, about 8 or 9%. This accounts for most of the apparent discrepancy in the heat value of gasoline. In the U.S. (and the table) the high heat values have traditionally been used, but in many other countries, the low heat values are commonly used. Fuel type \| MJ/L \| MJ/kg \| BTU/imp gal \| BTU/US gal \| Research octanenumber (RON) Regulargasoline/petrol \| 34.8 \| ~47 \| 150,100 \| 125,000 \| Min. 91 Premiumgasoline/petrol \| \| ~46 \| \| \| Min. 95 Autogas(LPG) (60%propaneand 40%butane) \| 25.5–28.7 \| ~51 \| \| \| 108–110 Ethanol \| 23.5 \| 31.1 \| 101,600 \| 84,600 \| 129 Methanol \| 17.9 \| 19.9 \| 77,600 \| 64,600 \| 123 Gasohol(10% ethanol and 90% gasoline) \| 33.7 \| ~45 \| 145,200 \| 121,000 \| 93/94 E85(85% ethanol and 15% gasoline) \| 25.2 \| ~33 \| 108,878 \| 90,660 \| 100–105 Diesel \| 38.6 \| ~48 \| 166,600 \| 138,700 \| N/A (see cetane) Biodiesel \| 35.1 \| 39.9 \| 151,600 \| 126,200 \| N/A (see cetane) Vegetable oil(using 9.00 kcal/g) \| 34.3 \| 37.7 \| 147,894 \| 123,143 \| Aviation gasoline \| 33.5 \| 46.8 \| 144,400 \| 120,200 \| 80-145 Jet fuel, naphtha \| 35.5 \| 46.6 \| 153,100 \| 127,500 \| N/A to turbine engines Jet fuel, kerosene \| 37.6 \| ~47 \| 162,100 \| 135,000 \| N/A to turbine engines Liquefied natural gas \| 25.3 \| ~55 \| 109,000 \| 90,800 \| Liquid hydrogen \| 09.3 \| ~130 \| 40,467 \| 33,696 \| Neither the gross heat of combustion nor the net heat of combustion gives the theoretical amount of mechanical energy (work) that can be obtained from the reaction. (This is given by the change in Gibbs free energy, and is around 45.7 MJ/kg for gasoline.) The actual amount of mechanical work obtained from fuel (the inverse of the specific fuel consumption) depends on the engine. A figure of 17.6 MJ/kg is possible with a gasoline engine, and 19.1 MJ/kg for a diesel engine. See Brake-specific fuel consumption for more information.[clarification needed] Transportation The energy efficiency in transport is the useful travelled distance, of passengers, goods or any type of load; divided by the total energy put into the transport propulsion means. The energy input might be rendered in several different types depending on the type of propulsion, and normally such energy is presented in liquid fuels, electrical energy or food energy. The energy efficiency is also occasionally known as energy intensity. The inverse of the energy efficiency in transport is the energy consumption in transport. Energy efficiency in transport is often described in terms of fuel consumption, fuel consumption being the reciprocal of fuel economy. Nonetheless, fuel consumption is linked with a means of propulsion which uses liquid fuels, whilst energy efficiency is applicable to any sort of propulsion. To avoid said confusion, and to be able to compare the energy efficiency in any type of vehicle, experts tend to measure the energy in the International System of Units, i.e., joules. Therefore, in the International System of Units, the energy efficiency in transport is measured in terms of metre per joule, or m/J, while the energy consumption in transport is measured in terms of joules per metre, or J/m. The more efficient the vehicle, the more metres it covers with one joule (more efficiency), or the fewer joules it uses to travel over one metre (less consumption). The energy efficiency in transport largely varies by means of transport. Different types of transport range from some hundred kilojoules per kilometre (kJ/km) for a bicycle to tens of megajoules per kilometre (MJ/km) for a helicopter. Fuel efficiency of motor vehicles The fuel economy of an automobile relates to the distance traveled by a vehicle and the amount of fuel consumed. Consumption can be expressed in terms of the volume of fuel to travel a distance, or the distance traveled per unit volume of fuel consumed. Since fuel consumption of vehicles is a significant factor in air pollution, and since the importation of motor fuel can be a large part of a nation's foreign trade, many countries impose requirements for fuel economy. Different methods are used to approximate the actual performance of the vehicle. The energy in fuel is required to overcome various losses (wind resistance, tire drag, and others) encountered while propelling the vehicle, and in providing power to vehicle systems such as ignition or air conditioning. Various strategies can be employed to reduce losses at each of the conversions between the chemical energy in the fuel and the kinetic energy of the vehicle. Driver behavior can affect fuel economy; maneuvers such as sudden acceleration and heavy braking waste energy. Driving technique Energy-efficient driving techniques are used by drivers who wish to reduce their fuel consumption, and thus maximize fuel efficiency. Many drivers have the potential to improve their fuel efficiency significantly. Simple things such as keeping tires properly inflated, having a vehicle well-maintained and avoiding idling can dramatically improve fuel efficiency. Careful use of acceleration and deceleration and especially limiting use of high speeds helps efficiency. The use of multiple such techniques is called "hypermiling". Advanced technology The most efficient machines for converting energy to rotary motion are electric motors, as used in electric vehicles. However, electricity is not a primary energy source so the efficiency of the electricity production has also to be taken into account. Railway trains can be powered using electricity, delivered through an additional running rail, overhead catenary system or by on-board generators used in diesel-electric locomotives as common on the US and UK rail networks. Pollution produced from centralised generation of electricity is emitted at a distant power station, rather than "on site". Pollution can be reduced by using more railway electrification and low carbon power for electricity. Some railways, such as the French SNCF and Swiss federal railways derive most, if not 100% of their power, from hydroelectric or nuclear power stations, therefore atmospheric pollution from their rail networks is very low. This was reflected in a study by AEA Technology between a Eurostar train and airline journeys between London and Paris, which showed the trains on average emitting 10 times less CO2, per passenger, than planes, helped in part by French nuclear generation. Hydrogen fuel cells In the future, hydrogen cars may be commercially available. Toyota is test-marketing vehicles powered by hydrogen fuel cells in southern California, where a series of hydrogen fueling stations has been established. Powered either through chemical reactions in a fuel cell that create electricity to drive very efficient electrical motors or by directly burning hydrogen in a combustion engine (near identically to a natural gas vehicle, and similarly compatible with both natural gas and gasoline); these vehicles promise to have near-zero pollution from the tailpipe (exhaust pipe). Potentially the atmospheric pollution could be minimal, provided the hydrogen is made by electrolysis using electricity from non-polluting sources such as solar, wind or hydroelectricity or nuclear. Commercial hydrogen production uses fossil fuels and produces more carbon dioxide than hydrogen. Because there are pollutants involved in the manufacture and destruction of a car and the production, transmission and storage of electricity and hydrogen, the label "zero pollution" applies only to the car's conversion of stored energy into movement. In 2004, a consortium of major auto-makers — BMW, General Motors, Honda, Toyota and Volkswagen/Audi — came up with "Top Tier Detergent Gasoline Standard" to gasoline brands in the US and Canada that meet their minimum standards for detergent content and do not contain metallic additives. Top Tier gasoline contains higher levels of detergent additives in order to prevent the build-up of deposits (typically, on fuel injector and intake valve) known to reduce fuel economy and engine performance. In microgravity How fuel combusts affects how much energy is produced. The National Aeronautics and Space Administration (NASA) has investigated fuel consumption in microgravity. The common distribution of a flame under normal gravity conditions depends on convection, because soot tends to rise to the top of a flame, such as in a candle, making the flame yellow. In microgravity or zero gravity, such as an environment in outer space, convection no longer occurs, and the flame becomes spherical, with a tendency to become more blue and more efficient. There are several possible explanations for this difference, of which the most likely one given is the hypothesis that the temperature is evenly distributed enough that soot is not formed and complete combustion occurs., National Aeronautics and Space Administration, April 2005. Experiments by NASA in microgravity reveal that diffusion flames in microgravity allow more soot to be completely oxidised after they are produced than diffusion flames on Earth, because of a series of mechanisms that behaved differently in microgravity when compared to normal gravity conditions.LSP-1 experiment results, National Aeronautics and Space Administration, April 2005. Premixed flames in microgravity burn at a much slower rate and more efficiently than even a candle on Earth, and last much longer. 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16629	https://science.nasa.gov/solar-system/orbits-and-keplers-laws/	Published Time: 2024-05-02T10:16:51-04:00 Orbits and Kepler’s Laws - NASA Science Kepler's Laws Explore Search News & Events ### News & Events Recently Published Video Series on NASA+ Podcasts & Audio Blogs Newsletters Social Media Media Resources Multimedia ### Multimedia Images Videos on NASA+ Interactives NASA Apps Podcasts e-Books STEM Multimedia NASA+ Search Suggested Searches Climate Change Artemis Expedition 64 Mars perseverance SpaceX Crew-2 International Space Station View All Topics A-Z Home Missions Humans in Space Earth The Solar System The Universe Science Aeronautics Technology Learning Resources About NASA Español News & Events Multimedia NASA+ Highlights 5 min read ### Astronomers Map Stellar ‘Polka Dots’ Using NASA’s TESS, Kepler article 2 days ago 5 min read ### NASA’s Bennu Samples Reveal Complex Origins, Dramatic Transformation article 5 days ago 5 min read ### Close-Up Views of NASA’s DART Impact to Inform Planetary Defense article 6 days ago Back ### Missions Search All NASA Missions A to Z List of Missions Upcoming Launches and Landings Spaceships and Rockets Communicating with Missions Artemis James Webb Space Telescope Hubble Space Telescope International Space Station OSIRIS-REx ### Humans in Space Why Go to Space Astronauts Commercial Space Destinations Spaceships and Rockets Living in Space ### Earth Explore Earth Science Climate Change Earth, Our Planet Earth Science in Action Earth Multimedia Earth Data Earth Science Researchers ### The Solar System The Sun Mercury Venus Earth The Moon Mars Jupiter Saturn Uranus Neptune Pluto & Dwarf Planets Asteroids, Comets & Meteors The Kuiper Belt The Oort Cloud Skywatching ### The Universe Exoplanets The Search for Life in the Universe Stars Galaxies Black Holes The Big Bang Dark Matter Dark Energy ### Science Earth Science Planetary Science Astrophysics & Space Science The Sun & Heliophysics Biological & Physical Sciences Lunar Science Citizen Science Astromaterials Aeronautics Research Human Space Travel Research ### Aeronautics Science in the Air NASA Aircraft Flight Innovation Supersonic Flight Air Traffic Solutions Green Aviation Tech Drones & You ### Technology Technology Transfer & Spinoffs Space Travel Technology Technology Living in Space Manufacturing and Materials Robotics Science Instruments Computing ### Learning Resources For Kids and Students For Educators For Colleges and Universities For Professionals Science for Everyone Requests for Exhibits, Artifacts, or Speakers STEM Engagement at NASA ### About NASA NASA's Impacts Centers and Facilities Directorates Organizations People of NASA Careers Internships Our History Doing Business with NASA Get Involved Contact ### NASA en Español Ciencia Aeronáutica Ciencias Terrestres Sistema Solar Universo ### News & Events Recently Published Video Series on NASA+ Podcasts & Audio Blogs Newsletters Social Media Media Resources ### Multimedia Images Videos on NASA+ Interactives NASA Apps Podcasts e-Books STEM Multimedia Highlights 5 min read ### Astronomers Map Stellar ‘Polka Dots’ Using NASA’s TESS, Kepler article 2 days ago 5 min read ### NASA’s Bennu Samples Reveal Complex Origins, Dramatic Transformation article 5 days ago 2 min read ### Hubble Observes Noteworthy Nearby Spiral Galaxy article 5 days ago Highlights 4 min read ### NASA’s Artemis II Lunar Science Operations to Inform Future Missions article 5 days ago 5 min read ### NASA, Army National Guard Partner on Flight Training for Moon Landing article 1 week ago 3 min read ### Human-Rating and NASA-STD-3001 article 2 weeks ago Highlights 1 min read ### Our World, Other Worlds Quiz article 1 day ago 1 min read ### Soil Moisture Quiz article 1 day ago 1 min read ### It’s a Gas Quiz article 1 day ago Highlights 4 min read ### NASA’s Artemis II Lunar Science Operations to Inform Future Missions article 5 days ago 5 min read ### Close-Up Views of NASA’s DART Impact to Inform Planetary Defense article 6 days ago 4 min read ### NASA: Ceres May Have Had Long-Standing Energy to Fuel Habitability article 6 days ago Highlights 5 min read ### Astronomers Map Stellar ‘Polka Dots’ Using NASA’s TESS, Kepler article 2 days ago 2 min read ### Hubble Observes Noteworthy Nearby Spiral Galaxy article 5 days ago 2 min read ### Hubble Examines Low Brightness, High Interest Galaxy article 2 weeks ago Highlights 5 min read ### From NASA Citizen Scientist to Astronaut Training: An Interview with Benedetta Facini article 1 day ago 1 min read ### The Programmatic POC for Swift has been updated, see Section 2.2 of D.3 AGIGO. 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Credits: NASA/JPL NASA Science Editorial Team May 02, 2024 Article Contents Kepler's Laws of Planetary Motion Who Was Johannes Kepler? Kepler and the Mars Problem Basic Properties of Ellipses Kepler's Laws Here are Kepler’s Three Laws: How We Use Kepler’s Laws Today Kepler's Laws of Planetary Motion The story of how we understand planetary motion could not be told if it were not for the work of a German mathematician named Johannes Kepler. Kepler's three laws describe how planets orbit the Sun. They describe how (1) planets move in elliptical orbits with the Sun as a focus, (2) a planet covers the same area of space in the same amount of time no matter where it is in its orbit, and (3) a planet’s orbital period is proportional to the size of its orbit. The planets orbit the Sun in a counterclockwise direction as viewed from above the Sun's north pole, and the planets' orbits all are aligned to what astronomers call the ecliptic plane. Who Was Johannes Kepler? Johannes Kepler was born on Dec. 27, 1571, in Weil der Stadt, Württemberg, which is now in the German state of Baden-Württemberg. Johannes Kepler (1571-1630) was a German astronomer best known for determining three principles of how planets orbit the Sun, known as Kepler's laws of planetary motion. Courtesy of the Archives, California Institute of Technology As a rather frail young man, the exceptionally talented Kepler turned to mathematics and the study of the heavens early on. When he was six, his mother pointed out a comet visible in the night sky. When Kepler was nine, his father took him out one night under the stars to observe a lunar eclipse. These events both made a vivid impression on Kepler's youthful mind and turned him toward a life dedicated to astronomy. Kepler lived and worked in Graz, Austria, during the tumultuous early 17th century. Due to religious and political difficulties common during that era, Kepler was banished from Graz on Aug. 2, 1600. Fortunately, he found work as an assistant to the famous Danish astronomer Tycho Brahe (usually referred to by his first name) in Prague. Kepler moved his family from Graz, 300 miles (480 kilometers) across the Danube River to Tycho's home. The global mosaic of Mars was created using Viking 1 Orbiter images taken in February 1980. The mosaic shows the entire Valles Marineris canyon system stretching across the center of Mars. It’s more than 2,000 miles (3,000 kilometers) long, 370 miles (600 kilometers) wide and 5 miles (8 kilometers) deep. NASA Kepler and the Mars Problem Tycho was a brilliant astronomer. He is credited with making the most accurate astronomical observations of his time, which he accomplished without the aid of a telescope. He had been impressed with Kepler’s studies in an earlier meeting. However, some historians think Tycho mistrusted Kepler, fearing that his bright young intern might eclipse him as the premier astronomer of his day. Because of this, he only let Kepler see part of his voluminous collection of planetary data. Tycho assigned Kepler the task of understanding the orbit of the planet Mars. The movement of Mars was problematic – it didn’t quite fit the models as described by Greek philosopher and scientist Aristotle (384 to 322 B.C.E.) and Egyptian astronomer Claudius Ptolemy (about 100 C.E to 170 C.E.). Aristotle thought Earth was the center of the universe, and that the Sun, Moon, planets, and stars revolved around it. Ptolemy developed this concept into a standardized, geocentric model (now known as the Ptolemaic system) based around Earth as a stationary object, at the center of the universe. Historians think that part of Tycho’s motivation for giving the Mars problem to Kepler was Tycho's hope that it would keep Kepler occupied while Tycho worked to perfect his own theory of the solar system. That theory was based on a geocentric model, modified from Ptolemy's, in which the planets Mercury, Venus, Mars, Jupiter, and Saturn all orbit the Sun, which in turn orbits Earth. As it turned out, Kepler, unlike Tycho, believed firmly in a model of the solar system known as the heliocentric model, which correctly placed the Sun at its center. This is also known as the Copernican system, because it was developed by astronomer Nicolaus Copernicus (1473-1543). But the reason Mars' orbit was problematic was because the Copernican system incorrectly assumed the orbits of the planets to be circular. Like many philosophers of his era, Kepler had a mystical belief that the circle was the universe’s perfect shape, so he also thought the planets’ orbits must be circular. For many years, he struggled to make Tycho’s observations of the motions of Mars match up with a circular orbit. Kepler eventually realized that the orbits of the planets are not perfect circles. His brilliant insight was that planets move in elongated, or flattened, circles called ellipses. The particular difficulties Tycho had with the movement of Mars were due to the fact that its orbit was the most elliptical of the planets for which he had extensive data. Thus, in a twist of irony, Tycho unwittingly gave Kepler the very part of his data that would enable his assistant to formulate the correct theory of the solar system. Basic Properties of Ellipses Since the orbits of the planets are ellipses, it might be helpful to review three basic properties of an ellipse: An ellipse is defined by two points, each called a focus, and together called foci. The sum of the distances to the foci from any point on the ellipse is always a constant. The amount of flattening of the ellipse is called the eccentricity. The flatter the ellipse, the more eccentric it is. Each ellipse has an eccentricity with a value between zero (a circle), and one (essentially a flat line, technically called a parabola). The longest axis of the ellipse is called the major axis, while the shortest axis is called the minor axis. Half of the major axis is termed a semi-major axis. After determining that the orbits of the planets are elliptical, Kepler formulated three laws of planetary motion, which accurately described the motion of comets as well. Kepler's Laws In 1609 Kepler published “Astronomia Nova,” which explained what are now called Kepler's first two laws of planetary motion. Kepler had noticed that an imaginary line drawn from a planet to the Sun swept out an equal area of space in equal times, regardless of where the planet was in its orbit. If you draw a triangle from the Sun to a planet’s position at one point in time and its position at a fixed time later, the area of that triangle is always the same, anywhere in the orbit. For all these triangles to have the same area, the planet must move more quickly when it’s near the Sun, but more slowly when it is farther from the Sun. This discovery became Kepler’s second law of orbital motion, and led to the realization of what became Kepler’s first law: that the planets move in an ellipse with the Sun at one focus point, offset from the center. In 1619, Kepler published “Harmonices Mundi,” in which he describes his "third law." The third law shows that there is a precise mathematical relationship between a planet’s distance from the Sun and the amount of time it takes revolve around the Sun. Here are Kepler’s Three Laws: Kepler's First Law: Each planet's orbit about the Sun is an ellipse. The Sun's center is always located at one focus of the ellipse. The planet follows the ellipse in its orbit, meaning that the planet-to-Sun distance is constantly changing as the planet goes around its orbit. Kepler's Second Law: The imaginary line joining a planet and the Sun sweeps out – or covers – equal areas of space during equal time intervals as the planet orbits. Basically, the planets do not move with constant speed along their orbits. Instead, their speed varies so that the line joining the centers of the Sun and the planet covers an equal area in equal amounts of time. The point of nearest approach of the planet to the Sun is called perihelion. The point of greatest separation is aphelion, hence by Kepler's second law, a planet is moving fastest when it is at perihelion and slowest at aphelion. Kepler's Third Law: The orbital period of a planet, squared, is directly proportional to the semi-major axes of its orbit, cubed. This is written in equation form as p 2=a 3. Kepler's third law implies that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit. Mercury, the innermost planet, takes only 88 days to orbit the Sun. Earth takes 365 days, while distant Saturn requires 10,759 days to do the same. How We Use Kepler’s Laws Today Kepler didn’t know about gravity, which is responsible for holding the planets in their orbits around the Sun, when he came up with his three laws. But Kepler’s laws were instrumental in Isaac Newton’s development of his theory of universal gravitation, which explained the unknown force behind Kepler's third law. Kepler and his theories were crucial in the understanding of solar system dynamics and as a springboard to newer theories that more accurately approximate planetary orbits. However, his third law only applies to objects in our own solar system. Newton’s version of Kepler’s third law allows us to calculate the masses of any two objects in space if we know the distance between them and how long they take to orbit each other (their orbital period). What Newton realized was that the orbits of objects in space depend on their masses, which led him to discover gravity. Newton’s generalized version of Kepler’s third law is the basis of most measurements we can make of the masses of distant objects in space today. These applications include determining the masses of moons orbiting the planets, stars that orbit each other, the masses of black holes (using nearby stars affected by their gravity), the masses of exoplanets (planets orbiting stars other than our Sun), and the existence of mysterious dark matter in our galaxy and others. In planning trajectories (or flight plans) for spacecraft, and in making measurements of the masses of the moons and planets, modern scientists often go a step beyond Newton. They account for factors related to Albert Einstein’s theory of relativity, which is necessary to achieve the precision required by modern science measurements and spaceflight. However, Newton’s laws are still accurate enough for many applications, and Kepler’s laws remain an excellent guide for understanding how the planets move in our solar system. NASA's Kepler space telescope discovered thousands of planets outside our solar system, and revealed that our galaxy contains more planets than stars. NASA Johannes Kepler died Nov. 15, 1630, at age 58. NASA'sKepler space telescope was named for him. The spacecraft launched March 6, 2009, and spent nine years searching for Earth-like planets orbiting other stars in our region of the Milky Way. The Kepler space telescope left a legacy of more than 2,600 planet discoveries from outside our solar system, many of which could be promising places for life. 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16630	https://www.youtube.com/watch?v=5RYAGXa8ywc	Vector Norm, L1 Norm, Euclidean Norm, Max Norm, Euclidean Distance Deep Shallownet 686 subscribers 174 likes Description 18338 views Posted: 4 Sep 2019 0:00 - Norm 0:32 - L1 Norm/Distance 0:56 - Euclidean Norm/Distance 1:25 - Max Norm/Distance Voice act: English(US) - Matthew Slides made in: Microsoft PowerPoint Video editing done in: Topics covered based on: Deep Learning An MIT Press book Ian Goodfellow and Yoshua Bengio and Aaron Courville 6 comments Transcript: Norm gnorm is a function which measures the size of a vector you can see the formula of this function on the slide this formula implies to sum the absolute value of each element raised to the power P and raise the result to the power 1 over P there are three most commonly used values for P in this function when P is equal to 1 the function is called l1 norm for P equal to 2 it's called the Euclidean norm max norm is 1 P is infinity let's now consider each of them in more detail L1 Norm/Distance we obtained the l1 norm by substituting P with one on an intuitive level the resulting formula simply implies to sum the absolute values of all elements let's do an example consider the vector X to find the l1 norm of this vector we have to sum the absolute values of its elements this is a very simple addition resulting in 9 Euclidean Norm/Distance we obtain the Euclidean norm by substituting P with 2 on an intuitive level the resulting formula implies to sum the squared values of all elements and take the square root let's find the Euclidean norm of the vector X first we sum the squared values of all elements then we take the square root of the result in the end we will find that the Euclidean norm of the vector X is approximately 5.9 max norm as the name Max Norm/Distance implies simply means to find the element with largest absolute value in our vector X the number with largest absolute value is 5 so the max norm of the vector X is 5
16631	https://texasgateway.org/resource/102-kinematics-rotational-motion	Skip to main content Learning Objectives Learning Objectives Learning Objectives By the end of this section, you will be able to do the following: Observe the kinematics of rotational motion Derive rotational kinematic equations Evaluate problem solving strategies for rotational kinematics Just by using our intuition, we can begin to see how rotational quantities like θθ, ωω, and αα are related to one another. For example, if a motorcycle wheel has a large angular acceleration for a fairly long time, it ends up spinning rapidly and rotates through many revolutions. In more technical terms, if the wheel's angular acceleration αα is large for a long period of time tt, then the final angular velocity ωω and angle of rotation θθ are large. The wheel's rotational motion is exactly analogous to the fact that the motorcycle's large translational acceleration produces a large final velocity, and the distance traveled will also be large. Kinematics is the description of motion. The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time. Let us start by finding an equation relating ωω, αα, and tt. To determine this equation, we recall the following kinematic equation for translational, or straight-line, motion. 10.17 v=v0+at (constant a) v=v0+at (constant a) Note that in rotational motion a=ata=at, and we shall use the symbol aa for tangential or linear acceleration from now on. As in linear kinematics, we assume aa is constant, which means that angular acceleration αα is also a constant, because a=rαa=rα. Now, let us substitute v=rωv=rω and a=rαa=rα into the linear equation above using 10.18 rω=rω0+rαt. rω=rω0+rαt. The radius rr cancels in the equation, yielding 10.19 ω=ω0+at (constant a), ω=ω0+at (constant a), where ω0ω0 is the initial angular velocity. This last equation is a kinematic relationship among ωω, αα, and tt —that is, it describes their relationship without reference to forces or masses that may affect rotation. It is also precisely analogous in form to its translational counterpart. Making Connections Kinematics for rotational motion is completely analogous to translational kinematics, first presented in One-Dimensional Kinematics. Kinematics is concerned with the description of motion without regard to force or mass. We will find that translational kinematic quantities, such as displacement, velocity, and acceleration have direct analogs in rotational motion. Starting with the four kinematic equations we developed in One-Dimensional Kinematics, we can derive the following four rotational kinematic equations, presented together with their translational counterparts. \| Rotational \| Translational \| \| \| θ=ˉωt θ=ω¯¯t \| x=ˉvt x=v–t \| \| \| ω=ω0+αt ω=ω0+αt \| v=v0+at v=v0+at \| (constant αα, aa) \| \| θ=ω0t+12αt2 θ=ω0t+12αt2 \| x=v0t+12at2 x=v0t+12at2 \| (constant αα, aa) \| \| ω2=ω02+2αθ ω2=ω02+2αθ \| v2=v02+2ax v2=v02+2ax \| (constant αα, aa) \| Table 10.2 Rotational Kinematic Equations In these equations, the subscript 0 denotes initial values (θ0θ0, x0x0, and t0t0 are initial values), and the average angular velocity ˉωω−− and average velocity ˉvv– are defined as follows 10.20 ˉω=ω0+ω2 and ˉv=v0+v2. ω¯¯=ω0+ω2 and v¯=v0+v2. The equations given above in Table 10.2 can be used to solve any rotational or translational kinematics problem in which aa and αα are constant. Problem-solving Strategy for Rotational Kinematics Examine the situation to determine that rotational kinematics or motion is involved. Rotation must be involved, but without the need to consider forces or masses that affect the motion. Identify exactly what needs to be determined in the problem (identify the unknowns). A sketch of the situation is useful. Make a list of what is given or can be inferred from the problem as stated (identify the knowns). Solve the appropriate equation or equations for the quantity to be determined (the unknown). It can be useful to think in terms of a translational analog because by now you are familiar with such motion. Substitute the known values along with their units into the appropriate equation, and obtain numerical solutions complete with units. Be sure to use units of radians for angles. Check your answer to see if it is reasonable: Does your answer make sense? Example 10.3 Calculating the Acceleration of a Fishing Reel A deep-sea fisherman hooks a big fish that swims away from the boat pulling the fishing line from his fishing reel. The whole system is initially at rest and the fishing line unwinds from the reel at a radius of 4.50 cm from its axis of rotation. The reel is given an angular acceleration of 110rad/s2110rad/s2 for 2.00 s as seen in Figure 10.8. (a) What is the final angular velocity of the reel? (b) At what speed is fishing line leaving the reel after 2 s elapse? (c) How many revolutions does the reel make? (d) How many meters of fishing line come off the reel in this time? Strategy In each part of this example, the strategy is the same as it was for solving problems in linear kinematics. In particular, known values are identified and a relationship is then sought that can be used to solve for the unknown. Solution for (a) Here αα and tt are given and ωω needs to be determined. The most straightforward equation to use is ω=ω0+αtω=ω0+αt because the unknown is already on one side and all other terms are known. That equation states that 10.21 ω=ω0+αt. ω=ω0+αt. We are also given that ω0=0ω0=0 (it starts from rest), so that 10.22 ω=0+(110rad/s2)(2s)=220rad/s. ω=0+(110rad/s2)(2s)=220rad/s. Solution for (b) Now that ωω is known, the speed vv can most easily be found using the relationship 10.23 v=rω, v=rω, where the radius rr of the reel is given to be 4.50 cm; thus, 10.24 v=(0.0450 m)(220 rad/s)=9.90 m/s. v=(0.0450 m)(220 rad/s)=9.90 m/s. Note again that radians must always be used in any calculation relating linear and angular quantities. Also, because radians are dimensionless, we have mrad=mmrad=m. Solution for (c) Here, we are asked to find the number of revolutions. Because 1 rev=2π rad1 rev=2π rad, we can find the number of revolutions by finding θθ in radians. We are given αα and tt, and we know ω0ω0 is zero, so that θθ can be obtained using θ=ω0t+12αt2θ=ω0t+12αt2. 10.25 θ=ω0t+12αt2=0+(0.500)(110rad/s2)(2 s)2=220 rad θ==ω0t+12αt20+(0.500)(110rad/s2)(2 s)2=220 rad Converting radians to revolutions gives 10.26 θ=(220 rad)1 rev2π rad=35.0 rev. θ=(220 rad)1 rev2π rad=35.0 rev. Solution for (d) The number of meters of fishing line is xx, which can be obtained through its relationship with θθ 10.27 x=rθ=(0.0450 m)(220 rad)=9.90 m. x=rθ=(0.0450 m)(220 rad)=9.90 m. Discussion This example illustrates that relationships among rotational quantities are highly analogous to those among linear quantities. We also see in this example how linear and rotational quantities are connected. The answers to the questions are realistic. After unwinding for two seconds, the reel is found to spin at 220 rad/s, which is 2,100 rpm. No wonder reels sometimes make high-pitched sounds. The amount of fishing line played out is 9.90 m, about right for when the big fish bites. Figure 10.8 Fishing line coming off a rotating reel moves linearly. Example 10.3 and Example 10.4 consider relationships between rotational and linear quantities associated with a fishing reel. Example 10.4 Calculating the Duration When the Fishing Reel Slows Down and Stops Now let us consider what happens if the fisherman applies a brake to the spinning reel, achieving an angular acceleration of –300rad/s2–300rad/s2. How long does it take the reel to come to a stop? Strategy We are asked to find the time tt for the reel to come to a stop. The initial and final conditions are different from those in the previous problem, which involved the same fishing reel. Now we see that the initial angular velocity is ω0=220 rad/sω0=220 rad/s and the final angular velocity ωω is zero. The angular acceleration is given to be α=−300rad/s2α=−300rad/s2. Examining the available equations, we see all quantities but t are known in ω=ω0+αt,ω=ω0+αt, making it easiest to use this equation. Solution The equation states 10.28 ω=ω0+αt. ω=ω0+αt. We solve the equation algebraically for t, and then substitute the known values as usual, yielding 10.29 t=ω−ω0α=0−220 rad/s−300rad/s2=0.733 s. t=ω−ω0α=0−220 rad/s−300rad/s2=0.733 s. Discussion Note that care must be taken with the signs that indicate the directions of various quantities. Also, note that the time to stop the reel is fairly small because the acceleration is rather large. Fishing lines sometimes snap because of the accelerations involved, and fishermen often let the fish swim for a while before applying brakes on the reel. A tired fish will be slower, requiring a smaller acceleration. Example 10.5 Calculating the Slow Acceleration of Trains and Their Wheels Large freight trains accelerate very slowly. Suppose one such train accelerates from rest, giving its 0.350-m-radius wheels an angular acceleration of 0.250rad/s20.250rad/s2. After the wheels have made 200 revolutions with no slippage: (a) How far has the train moved down the track? (b) What are the final angular velocity of the wheels and the linear velocity of the train? Strategy In part (a), we are asked to find xx, and in (b) we are asked to find ωω and vv. We are given the number of revolutions θθ, the radius of the wheels rr, and the angular acceleration αα. Solution for (a) The distance xx is very easily found from the relationship between distance and rotation angle. 10.30 θ=xr θ=xr Solving this equation for xx yields 10.31 x=rθ. x=rθ. Before using this equation, we must convert the number of revolutions into radians, because we are dealing with a relationship between linear and rotational quantities. 10.32 θ=(200rev)2πrad1 rev=1257rad θ=(200rev)2πrad1 rev=1257rad Now we can substitute the known values into x=rθx=rθ to find the distance the train moved down the track. 10.33 x=rθ=(0.350 m)(1257 rad)=440m x=rθ=(0.350 m)(1257 rad)=440m Solution for (b) We cannot use any equation that incorporates tt to find ωω, because the equation would have at least two unknown values. The equation ω2=ω02+2αθω2=ω02+2αθ will work, because we know the values for all variables except ωω. 10.34 ω2=ω02+2αθ Taking the square root of this equation and entering the known values gives 10.35 ω=[0+2(0.250 rad/s2)(1257 rad)]1/2=25.1 rad/s. We can find the linear velocity of the train, v, through its relationship to ω. 10.36 v=rω=(0.350 m)(25.1 rad/s)=8.77 m/s Discussion The distance traveled is fairly large and the final velocity is fairly slow (just under 32 km/h). There is translational motion even for something spinning in place, as the following example illustrates. Figure 10.9 shows a fly on the edge of a rotating microwave oven plate. The example below calculates the total distance it travels. Figure 10.9 The image shows a microwave plate. The fly makes revolutions while the food is heated along with the fly. Example 10.6 Calculating the Distance Traveled by a Fly on the Edge of a Microwave Oven Plate A person decides to use a microwave oven to reheat some lunch. In the process, a fly accidentally flies into the microwave and lands on the outer edge of the rotating plate and remains there. If the plate has a radius of 0.15 m and rotates at 6 rpm, calculate the total distance traveled by the fly during a 2 min cooking period. Ignore the start-up and slow-down times. Strategy First, find the total number of revolutions θ, and then the linear distance x traveled. θ=ˉωt can be used to find θ because ˉω is given to be 6.0 rpm. Solution Entering known values into θ=ˉωt gives 10.37 θ=ˉωt=(6.0 rpm)(2.0 min)=12 rev. As always, it is necessary to convert revolutions to radians before calculating a linear quantity like x from an angular quantity like θ. 10.38 θ=(12 rev)(2πrad1 rev)=75.4 rad Now, using the relationship between x and θ, we can determine the distance traveled. 10.39 x=rθ=(0.15 m)(75.4 rad)=11 m Discussion Quite a trip, if it survives! Note that this distance is the total distance traveled by the fly. Displacement is actually zero for complete revolutions because they bring the fly back to its original position. The distinction between total distance traveled and displacement was first noted in One-Dimensional Kinematics. Check Your Understanding Exercise 1 Rotational kinematics has many useful relationships, often expressed in equation form. Are these relationships laws of physics or are they simply descriptive? Hint—the same question applies to linear kinematics. Rotational kinematics, like linear kinematics, is descriptive and does not represent laws of nature. With kinematics, we can describe many things to great precision but kinematics does not consider causes. For example, a large angular acceleration describes a very rapid change in angular velocity without any consideration of its cause. Print Share
16632	https://math.fandom.com/wiki/Square_number	Skip to content Math Wiki 1,422 pages Contents 1 Examples 2 Properties 3 Special cases 4 Odd and even square numbers 5 See also 6 Notes 7 References 8 Further reading 9 External links in: Figurate numbers, Quadrilaterals, Integers, and 4 more Number theory Elementary arithmetic Integer sequences Pages using ISBN magic links Square number Sign in to edit History Purge Talk (0) Edit intro Template:More footnotes In mathematics, a square number, sometimes also called a perfect square, is an integer that is the square of an integer; in other words, it is the product of some integer with itself. So, for example, 9 is a square number, since it can be written as 3 × 3. The usual notation for the formula for the square of a number is not the product , but the equivalent exponentiation , usually pronounced as " squared". The name square number comes from the name of the shape. This is because a square with side length has area . Square numbers are non-negative. Another way of saying that a (non-negative) number is a square number, is that its square root is again an integer. For example, , so 9 is a square number. A positive integer that has no perfect square divisors except 1 is called square-free. For a non-negative integer , the th square number is , with being the zeroth square. The concept of square can be extended to some other number systems. If rational numbers are included, then a square is the ratio of two square integers, and, conversely, the ratio of two square integers is a square (e.g., ). Starting with 1, there are square numbers up to and including , where the expression represents the floor of the number . Contents 1 Examples 2 Properties 3 Special cases 4 Odd and even square numbers 5 See also 6 Notes 7 References 8 Further reading 9 External links Examples[] The squares (sequence A000290 in OEIS) smaller than 602 are: : 02 = 0 : 12 = 1 : 22 = 4 : 32 = 9 : 42 = 16 : 52 = 25 : 62 = 36 : 72 = 49 : 82 = 64 : 92 = 81 : 102 = 100 : 112 = 121 : 122 = 144 : 132 = 169 : 142 = 196 : 152 = 225 : 162 = 256 : 172 = 289 : 182 = 324 : 192 = 361 : 202 = 400 : 212 = 441 : 222 = 484 : 232 = 529 : 242 = 576 : 252 = 625 : 262 = 676 : 272 = 729 : 282 = 784 : 292 = 841 : 302 = 900 : 312 = 961 : 322 = 1024 : 332 = 1089 : 342 = 1156 : 352 = 1225 : 362 = 1296 : 372 = 1369 : 382 = 1444 : 392 = 1521 : 402 = 1600 : 412 = 1681 : 422 = 1764 : 432 = 1849 : 442 = 1936 : 452 = 2025 : 462 = 2116 : 472 = 2209 : 482 = 2304 : 492 = 2401 : 502 = 2500 : 512 = 2601 : 522 = 2704 : 532 = 2809 : 542 = 2916 : 552 = 3025 : 562 = 3136 : 572 = 3249 : 582 = 3364 : 592 = 3481 The difference between any perfect square and its predecessor is given by the identity . Equivalently, it is possible to count up square numbers by adding together the last square, the last square's root, and the current root, that is, . Properties[] The number is a square number if and only if one can arrange points in a square: Template:How \| \| \| --- \| \| m = 12 = 1 \| \| \| m = 22 = 4 \| \| \| m = 32 = 9 \| \| \| m = 42 = 16 \| \| \| m = 52 = 25 \| \| The expression for the th square number is . This is also equal to the sum of the first odd numbers as can be seen in the above pictures, where a square results from the previous one by adding an odd number of points (shown in magenta). The formula follows: So for example, 52 = 25 = 1 + 3 + 5 + 7 + 9. There are several recursive methods for computing square numbers. For example, the th square number can be computed from the previous square by . Alternatively, the th square number can be calculated from the previous two by doubling the (n − 1)-th square, subtracting the -th square number, and adding 2, because . For example, : 2 × 52 − 42 + 2 = 2 × 25 − 16 + 2 = 50 − 16 + 2 = 36 = 62. A square number is also the sum of two consecutive triangular numbers. The sum of two consecutive square numbers is a centered square number. Every odd square is also a centered octagonal number. Another property of a square number is that it has an odd number of divisors, while other numbers have an even number of divisors. An integer root is the only divisor that pairs up with itself to yield the square number, while other divisors come in pairs. Lagrange's four-square theorem states that any positive integer can be written as the sum of four or fewer perfect squares. Three squares are not sufficient for numbers of the form . A positive integer can be represented as a sum of two squares precisely if its prime factorization contains no odd powers of primes of the form . This is generalized by Waring's problem. A square number can end only with digits 0, 1, 4, 6, 9, or 25 in base 10, as follows: If the last digit of a number is 0, its square ends in an even number of 0s (so at least 00) and the digits preceding the ending 0s must also form a square. If the last digit of a number is 1 or 9, its square ends in 1 and the number formed by its preceding digits must be divisible by four. If the last digit of a number is 2 or 8, its square ends in 4 and the preceding digit must be even. If the last digit of a number is 3 or 7, its square ends in 9 and the number formed by its preceding digits must be divisible by four. If the last digit of a number is 4 or 6, its square ends in 6 and the preceding digit must be odd. If the last digit of a number is 5, its square ends in 25 and the preceding digits must be 0, 2, 06, or 56. In base 16, a square number can end only with 0, 1, 4 or 9 and in case 0, only 0, 1, 4, 9 can precede it, in case 4, only even numbers can precede it. In general, if a prime divides a square number then the square of must also divide ; if fails to divide , then is definitely not square. Repeating the divisions of the previous sentence, one concludes that every prime must divide a given perfect square an even number of times (including possibly 0 times). Thus, the number is a square number if and only if, in its canonical representation, all exponents are even. Squarity testing can be used as alternative way in factorization of large numbers. Instead of testing for divisibility, test for squarity: for given and some number , if is the square of an integer then divides . (This is an application of the factorization of a difference of two squares.) For example, 1002 − 9991 is the square of 3, so consequently 100 − 3 divides 9991. This test is deterministic for odd divisors in the range from to where covers some range of natural numbers . A square number cannot be a perfect number. The sum of the series of power numbers can also be represented by the formula The first terms of this series (the square pyramidal numbers) are: 0, 1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, 650, 819, 1015, 1240, 1496, 1785, 2109, 2470, 2870, 3311, 3795, 4324, 4900, 5525, 6201... (sequence A000330 in OEIS). All fourth powers, sixth powers, eighth powers and so on are perfect squares. Special cases[] If the number is of the form m5 where m represents the preceding digits, its square is n25 where n = m × (m + 1) and represents digits before 25. For example the square of 65 can be calculated by n = 6 × (6 + 1) = 42 which makes the square equal to 4225. If the number is of the form m0 where m represents the preceding digits, its square is n00 where n = m2. For example the square of 70 is 4900. If the number has two digits and is of the form 5m where m represents the units digit, its square is AABB where AA = 25 + m and BB = m2. Example: To calculate the square of 57, 25 + 7 = 32 and 72 = 49, which means 572 = 3249. Odd and even square numbers[] Squares of even numbers are even (and in fact divisible by 4), since . Squares of odd numbers are odd, since . It follows that square roots of even square numbers are even, and square roots of odd square numbers are odd. See also[] Methods of computing square roots Quadratic residue Polygonal number Cubic number Euler's four-square identity Fermat's theorem on sums of two squares Brahmagupta–Fibonacci identity The Book of Squares Integer square root Square triangular number Automorphic number Exponentiation Power of two Pythagorean triple Square (algebra)#Related identities Notes[] ↑ Some authors also call squares of rational numbers perfect squares. References[] \| \| \| --- \| \| \| This page uses content from Wikipedia. The original article was at Square number. The list of authors can be seen in the page history. As with the Math Wiki, the text of Wikipedia is available under the Creative Commons Licence. \| Weisstein, Eric W., "Square Number" from MathWorld. Further reading[] Conway, J. H. and Guy, R. K. The Book of Numbers. New York: Springer-Verlag, pp. 30–32, 1996. ISBN 0-387-97993-X External links[] Learn Square Numbers. Practice square numbers up to 144 with this children's multiplication game Dario Alpern, Sum of squares. A Java applet to decompose a natural number into a sum of up to four squares. Fibonacci and Square Numbers at Convergence The first 1,000,000 perfect squares Includes a program for generating perfect squares up to 1015. Category list Community content is available under CC-BY-SA unless otherwise noted. Search this wiki Search all wikis Do Not Sell My Personal Information When you visit our website, we store cookies on your browser to collect information. 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16633	https://pmc.ncbi.nlm.nih.gov/articles/PMC8747202/	Wind Tunnel Performance Tests of the Propellers with Different Pitch for the Electric Propulsion System - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Sensors (Basel) . 2021 Dec 21;22(1):2. doi: 10.3390/s22010002 Search in PMC Search in PubMed View in NLM Catalog Add to search Wind Tunnel Performance Tests of the Propellers with Different Pitch for the Electric Propulsion System † Zbigniew Czyż Zbigniew Czyż 1 Aeronautics Faculty, Polish Air Force University, 08-521 Dęblin, Poland Find articles by Zbigniew Czyż 1,, Paweł Karpiński Paweł Karpiński 2 Department of Thermodynamics, Fluid Mechanics and Aviation Propulsion Systems, Faculty of Mechanical Engineering, Lublin University of Technology, 20-618 Lublin, Poland; pawel.karpinski@pollub.edu.pl (P.K.); k.skiba@pollub.pl (K.S.); m.wendeker@pollub.pl (M.W.) Find articles by Paweł Karpiński 2, Krzysztof Skiba Krzysztof Skiba 2 Department of Thermodynamics, Fluid Mechanics and Aviation Propulsion Systems, Faculty of Mechanical Engineering, Lublin University of Technology, 20-618 Lublin, Poland; pawel.karpinski@pollub.edu.pl (P.K.); k.skiba@pollub.pl (K.S.); m.wendeker@pollub.pl (M.W.) Find articles by Krzysztof Skiba 2, Mirosław Wendeker Mirosław Wendeker 2 Department of Thermodynamics, Fluid Mechanics and Aviation Propulsion Systems, Faculty of Mechanical Engineering, Lublin University of Technology, 20-618 Lublin, Poland; pawel.karpinski@pollub.edu.pl (P.K.); k.skiba@pollub.pl (K.S.); m.wendeker@pollub.pl (M.W.) Find articles by Mirosław Wendeker 2 Editors: Egidio De Benedetto, Annarita Tedesco Author information Article notes Copyright and License information 1 Aeronautics Faculty, Polish Air Force University, 08-521 Dęblin, Poland 2 Department of Thermodynamics, Fluid Mechanics and Aviation Propulsion Systems, Faculty of Mechanical Engineering, Lublin University of Technology, 20-618 Lublin, Poland; pawel.karpinski@pollub.edu.pl (P.K.); k.skiba@pollub.pl (K.S.); m.wendeker@pollub.pl (M.W.) Correspondence: z.czyz@law.mil.pl † This paper is an extended version of the published conference paper “Czyż, Z.; Karpiński, P.; Skiba K. Wind tunnel investigation of the propellers for unmanned aerial vehicle. In Proceedings of the 2021 IEEE International Workshop on Metrology for AeroSpace, Naples, Italy, 23–25 June 2021”. Roles Egidio De Benedetto: Academic Editor Annarita Tedesco: Academic Editor Received 2021 Oct 24; Accepted 2021 Dec 19; Collection date 2022 Jan. © 2021 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license ( PMC Copyright notice PMCID: PMC8747202 PMID: 35009545 Abstract The geometry of a propeller is closely related to its aerodynamic performance. One of the geometric parameters of a propeller is pitch. This parameter determines the distance by which the propeller moves forward during one revolution. The challenge is to select a propeller geometry for electric propulsion in order to achieve the best possible performance. This paper presents the experimental results of the aerodynamic performance of the set of propellers with different pitch values. The tests were performed in a closed-circuit subsonic wind tunnel using a six-component force balance. The analyzed propellers were 12-inch diameter twin-blade propellers that were driven by a BLDC (brushless direct current) electric motor. The tests were performed under forced airflow conditions. The thrust and torque produced by the propeller were measured using a strain gauge. The analysis was performed for different values of the advance ratio which is the ratio of freestream fluid speed to propeller tip speed. Additionally, a set of electrical parameters was recorded using the created measurement system. The propeller performance was evaluated by a dimensional analysis. This method enables calculation of dimensionless coefficients which are useful for comparing performance data for propellers. Keywords: propeller, performance, electric propulsion, propeller pitch, thrust, wind tunnel 1. Introduction Unmanned aerial vehicles (UAVs) are now widely used in a variety of applications. Besides the power source, the electric motor, ESC (electronic speed control), and propeller affect the efficient operation of the aircraft. The right choice of a propeller for an aircraft affects flight duration [1,2]. The best solution may be selected based on characteristics provided by propeller retailers or manufacturers, but it is usually difficult or even impossible to do due to a lack of reliable data. Only bench testing the complete propulsion system including the propeller, electric motor, ESC, power supply/battery will give reliable results. Very often, the change of even a single component, e.g., a speed controller significantly influences the performance of a propulsion unit. These differences may concern not only the generated thrust value but also the thermal loads of the propulsion motor, rate of rotational speed increase, rotational speed stability, supply voltage stability, power consumption, efficiency, etc. Various propeller test stand proposals can be found in the literature [3,4,5,6,7]. The propeller theory is well known due to its use for many years as the main propulsion system for fixed-wing aircraft . Theoretical description of small multirotor UAVs is, however, more difficult due to their size, aeroelastic effects, and transverse flow. The performance of a propeller propulsion system is affected by propeller geometry. The key geometrical parameters of a propeller are its diameter, pitch, chord length, local thickness, local camber. A question can be posed about how pitch affects the performance of a propeller driven by an electric motor. The answer to this question is the main objective of this article. One of the basic geometric parameters is propeller pitch. This parameter is defined as the distance the propeller overcomes in the axial direction during one revolution (Figure 1). Experimental and numerical methods for investigating the performance of propeller blades were presented in . An analysis of the performance data of a small-scale propeller set was carried out by Uhlig (2008) who analyzed the geometric characteristics of the propellers at low Reynolds numbers. Figure 1. Open in a new tab Definition of propeller pitch. Experimental propeller tests can be performed with forced and without forced airflow . The former allows for performance evaluation during flight (when the relative air velocity is different from zero). These tests are carried out in a wind tunnel using measuring systems that measure forces and pressure distribution. Pressure taps combined with pressure sensors or the PIV (Particle Image Velocimetry) method [14,15] can be used to measure pressure distribution. The issue of testing the small propeller and the rotor in a low-speed aeroacoustic wind tunnel is discussed in [16,17]. A set of propellers at low Reynolds numbers was investigated in . The wind tunnel test method was also used in where dimensionless coefficients of the forces and moments acting on a propeller were analyzed for different wind velocities, propeller angles, and propeller rotational speeds. Performance characteristics and the velocity field of a 16-inch diameter ducted propeller using wind tunnel testing were investigated by Yilmaz (2008) . Chen (2015) analyzed the obtained from wind tunnel testing power and thrust coefficients for a 2-blade counter-rotating propeller. Wind tunnel testing can also be used to validate propeller performance calculation [22,23]. Propeller or airfoil performance under different conditions can be analyzed using simulation methods such as computational fluid dynamics (CFD) [24,25,26] or using computational algorithms using an Euler and Navier-Stokes solver . Numerical simulations can later be experimentally validated . The Particle Image Velocimetry method is useful for an experimental testing of propellers. This method was used in to analyze propeller flow. Based on the wind tunnel tests, the distribution of velocity and its components in the vertical plane passing through the propeller axis were determined for several values of the angle of attack of the tested object for two different values of airflow velocity inside the wind tunnel. The Blade Element Momentum Theory (BEMT) method can be used to evaluate propeller performance [30,31]. This method is a modification of the Blade Element Theory which also determines the behavior of propellers [32,33]. This method was used by MacNeill (2016) to model low Reynolds number propeller performance. A comparison of modeling using BEMT and CFD calculations is presented in . Mathematical modeling of the aerodynamics of propellers allows for the prediction of its performance and assists in the aircraft design process . It is also possible to combine different analytical methods such as blade-element methods, the momentum theory, and sectional airfoil analysis to evaluate propeller performance . The momentum/blade-element method was used by Gur (2005) to analyze propeller performance at a low advance ratio. A special test rig for testing a wide range of low Reynolds number propellers was presented in . The efficiency of a propeller is determined by the thrust coefficient (1), torque coefficient (2), and propeller power coefficient (3): (1) (2) (3) The dimensionless coefficients for a given air density ρ and propeller rotational speed n allow the calculation of thrust, torque, and power consumed by the propeller. The propeller diameter D in both relationships (Equations (3) and (4)) allows the comparison of propellers of different sizes. A propeller can operate at different flight speeds depending on the type of aircraft. As a result of aerodynamic effects, the propeller generates thrust and torque which must be overcome by engine propulsion. Thrust and torque depend on the airflow around blades, i.e., directly on flight speed. Instead of using coefficients, one can refer to the advance ratio (4), where v is the value of undisturbed air velocity. The parameter J determines the operating conditions of the propeller. The advance ratio is appropriate for fixed-wing airplanes., Motion for multicopter propellers can be, however, either parallel to the axis of rotation of the propeller or nearly perpendicular to it. (4) Among the theories describing propeller operation, one can distinguish the momentum theory and the blade element theory. A disadvantage of the former is that it does not consider the basic structural parameters of the propeller such as the number of blades, blade shape, type of airfoil, positioning relative to flow. It does not consider the hub either, so it is necessary to know propeller thrust or induced velocity. The latter method involves calculating the aerodynamic forces acting on a blade element of width dr located at a distance r from the propeller shaft axis, then summing the elementary forces along the blade and multiplying by the number of blades. However, it assumes that the effect of the vortex surfaces generated by blades is neglected. The propeller blade element located at a distance r from the axis of rotation and having a width dr is affected by an aerodynamic force dP which has components dPx (5) and dPz (6) in the flow-related system. (5) (6) where: —air density [kg/m 3], —drag force coefficient [-], —lift force coefficient [-], —resultant air velocity [m/s], —blade width [m], c—blade chord [m]. The elementary thrust dT (an axial component of dP) and the elementary circumferential force dF (a circumferential component of dP) are of the forms as in (7) and (8), respectively. The angle γ used in the equation is the angle of the propeller blade flow defined as the difference between the blade cross-sectional angle and the angle of attack of the blade. (7) (8) From the relations (9) and (10), we obtain (11) and (12). v T refers to the circumferential component of the propeller drag velocity and v S to the axial component. (9) (10) (11) (12) The efficiency of a blade element is described by the relation (13): (13) Considering the axial and circumferential components, the Equation (14) was obtained: (14) If the velocity components are defined as (15) and (16) and considering (17) and (18), the result (19) is obtained: (15) (16) (17) (18) (19) From the above-mentioned relations, it can be stated that efficiency increases when the lift-to-drag ratio increases and when the coefficient b determining the power loss for generating the peripheral velocity components in the propeller stream decreases. Efficiency increases as the coefficient a decreases and as the power loss for the induced velocity decreases. The thrust of a propeller with i-blades can be calculated from the expression (20): (20) The torque can be calculated from the expression (21): (21) Considering the above, the propeller efficiency can be determined from the expression (22): (22) This study is an extended version of the paper entitled “Wind tunnel investigation of the propellers for the unmanned aerial vehicle” presented during the 2021 IEEE International Workshop on Metrology for Aerospace in Naples, Italy . This study was to analyze the performance of a set of 12-inch diameter twin-blade propellers with different pitch values. The propellers were used in combination with a low-power electric motor. Another purpose of the study was to evaluate the performance of the propeller-motor pair for the electric propulsion system. The selected performance coefficients like thrust, torque, power, efficiency, and thrust-to-power ratio were evaluated. The values necessary to calculate the coefficients were obtained from the wind tunnel tests. These tests were followed by the tests without the influence of thrust on the propeller performance. The results were presented at the same conference in the extended version of the paper entitled “Experimental study of propellers for the electric propulsion system” . 2. Methodology The research on the selected propellers was conducted on a test stand located in the laboratory of experimental aerodynamics which is a part of the Centre for Innovation and Advanced Technologies of the Lublin University of Technology. The test stand consisted of a six-component force balance placed on a mast located in the central part of the test section. The mentioned components were located inside a 1275 × 1415 mm test section in a closed-circuit subsonic wind tunnel. An axial fan enabled a maximum air velocity of 60 m/s. The maximum turbulence level in the test section did not exceed the value of 0.3%. The angle of deflection of the air stream from the tunnel axis in both planes was less than 0.3°. The view of the wind tunnel is shown in Figure 2, while the view of the test stand and research object in the test section is shown in Figure 3. Figure 2. Open in a new tab General view of the wind tunnel. Figure 3. Open in a new tab General view of the test stand for propellers. Force and torque were measured using a six-component force balance (Figure 4). The average measurement error for the measurement range of the drag force component was 0.11%, and for the moment component M x was 0.04%. A Prandtl tube with a measurement range of 3–100 m/s was used to measure the airflow velocity. The analog value was converted into a digital one using an Aplisens APR-2000G transmitter. An MX840B HBM measurement amplifier was used to acquire the measurement signals from the monolithic instrumented transducer. For each thrust and torque measurement point, the measurements were made for 5 s at a frequency of 25 Hz, and then the values were averaged. The measured aerodynamic and electrical parameters were averaged from three series of measurements. The data acquisition and parameter control were performed from a computer. Figure 4. Open in a new tab Coordinate system of the FMT625-1b force balance. A special control and measurement system was created to acquire the data from the strain gauge measuring system and measure the electrical quantities. Its schematic diagram is shown in Figure 5. A list of measured parameters and used sensors is given in Table 1. The developed measuring system operated using equipment and software from National Instruments. NI 9215 and 9203 measurement cards installed in a CompactDAQ Chassis cDAQ-9174 module were used. The rotational speed and current were measured using a Bipolar Hall-effect digital position sensor and a Tektronix TCP305 probe with a Tektronix TCP A300 signal amplifier, respectively. At the same time, the supply voltage was measured at one of the ESC (electronic speed control) connectors. To control the temperature conditions, the temperature of the motor winding and ESC was measured using Pt 100 RS Pro sensors. The small size of the sensors required the low thermal inertia of the sensors. An Arduino Leonardo microcontroller was used in the measurement and control line to adjust the PWM (pulse-width modulation) parameter. Figure 5. Open in a new tab Diagram of the developed control and measurement system. Table 1. List of the measured parameters and used sensors. \| Parameter \| Sensor \| Sensor Parameters \| :---: \| Engine speed \| Honeywell SS411P bipolar Hall sensor \| Sensor type: bipolar Case: TO92 Operate flux: −30…140 Gs Supply voltage: 2.7…7 V DC Temperature range: 40…150 °C \| \| Supply voltage \| NI-9215 analog input module with a voltage divider 1:3 \| Signal levels: ± 10 V, Channels: 4 (differential), Sample rate: 100 kS/s/ch, Simultaneous: Yes, Resolution: 16-bit, Connectivity: Screw-Terminal, Spring-Terminal, BNC \| \| Supply current \| Tektronix TCP305 current measurement system with a current probe \| AC/DC measurement capabilities, Bandwidth (−3 dB): DC—50 MHz, Risetime: ≤ 7 ns, High-current sensitivity range: 10 A/V, DC (continuous): 50 A, DC Accuracy, Typical (Operating temp. 23 °C ± 5 °C): ±1% of reading \| \| ESC and engine temperature \| PT 100 RS PRO SONDE PLATE thermocouple with the programmable measuring Czaki TED 28 transducer \| RS PRO PT 100: Type: PT 100—A Class, Probe diameter: 2 mm, Probe length: 10 mm, Maximum temperature: 600°C, Reaction time: 1 m/s Czaki TED 28: Measuring range: 100…800 °C, Processing error: 0.15% or ±0.2 °C, Temperature error/10 °C: 0.05% or ±0.1 °C, Current output signal: 4…20 mA, two-wire, Input-output galvanic isolation, Programmable input signal range, Compensation of the cold ends of the thermocouple \| \| Air temperature, pressure and humidity \| Bosch BME280 sensor \| Supply voltage: 3.3 V, Interface: I2C and SPI, Temperature range: 40 °C…85 °C, Temperature accuracy: ±1 °C Humidity range: 10% RH to 100% RH, Humidity accuracy: ± 3% relative humidity Pressure range: 300…1100 hPa Pressure measurement error: ±0.25% \| Open in a new tab Atmospheric conditions were changed during the measurements, so it was necessary to take these parameters into account when calculating air density. The air temperature varied in the range from 25.0 to 29.4 °C, humidity from 36.6 to 38.7%, and pressure from 992.5 to 1018 hPa. The PWM and the air velocity values were changed during the measurements for the given propeller. Table 2 shows the values of the parameters that were changed during the tests. Table 2. Parameters changed during the tests. \| Propeller Pitch (Inches) \| PWM (%) \| Airflow Velocity (m/s) \| :---: \| 4.5 \| 40 \| 0 \| \| 5.5 \| 60 \| 5 \| \| 6 \| 80 \| 10 \| \| 8 \| 90 \| 15 \| \| 10 \| \| 20 \| \| 12 \| \| 25 \| Open in a new tab Figure 6 shows the research object inside the test section in the wind tunnel. The tested propeller with a BLDC motor was attached to an adapter. The adapter was then connected to a force balance which was placed at the end of the sting. The sting was attached to a rotating arm that can rotate in two planes. During the experiments, the arm was positioned so that the propeller thrust vector was parallel to the airflow. Figure 6. Open in a new tab View of the wind tunnel test section with the research object: 1—tested engine-propeller unit, 2—force balance, 3—ESC, 4—sting. 3. Research Object The test subject was a set of APC propellers (Figure 7). All the propellers tested were 12 inches (0.3048 m) long and had different pitch values. The analyzed propellers had pitch values of: 4.5″ (0.1143 m), 5.5″ (0.1397 m), 6″ (0.1524 m), 8″ (0.2032 m), 10″ (0.2540 m), and 12″ (0.3048 m). Propellers used in this research were made of unreinforced nylon. This material is characterized by the following parameters: tensile strength (75.8 MPa), tensile elongation (above 10%), flexural strength (48.2 MPa), flexural modulus (2.8 MPa). There was a hub in the axis of rotation designed to mount them on the motor shaft. Its thickness was 0.01067 m, and its hole diameter was 0.00635 m. Figure 7. Open in a new tab Summary of the tested propellers with different pitches: (a) 4.5″, (b) 5.5″, (c) 6″, (d) 8″, (e) 10″, and (f) 12″. The propellers were driven using a Tornado T5 3115 electric motor from BrotherHobby (Figure 8). It is a BLDC type unit equipped with an electrically controlled commutator. The motor is designed for the propulsion of light unmanned aerial vehicles. It has a threaded shaft and mounting holes. Its basic technical parameters are shown in Table 3. An ESC ReadytoSky 40 A OPTO and a Mean-Well-RSP-3000-24 power supply were used to power the motor. The nominal supply voltage was set at 22.2 V. Figure 8. Open in a new tab BrotherHobby Tornado T5 3115 electric motor . Table 3. Technical parameters of the tested motor. \| Parameter \| Sensor Parameters \| :---: \| \| Weight \| 110 g with 25 cm SR wires \| \| Motor dimensions (length × diameter) \| 49.5 mm × 37.5 mm \| \| Rotor \| N52H arc magnets \| \| Stator \| 0.2 mm Kawasaki silicon steel \| \| Shaft dimensions (length × diameter) \| 16 mm × M5 \| \| Maximum power \| 1150 W \| \| KV factor \| 640 rpm/V \| \| Maximum voltage \| 22.2 V \| Open in a new tab 4. Results and Analysis The thrust and torque generated by the given propeller were obtained as a function of the given airflow velocity ranging from 0 to 25 m/s for the defined PWM parameters, i.e., 40%, 60%, 80%, and 90%. The power system parameters like voltage and current were also recorded. Dimensionless coefficients for thrust, torque, power, and efficiency were calculated from the results obtained and in line with the Equations (1)–(3). The thrust-to-power ratio T p was also calculated. For each measurement point, these coefficients were expressed as a function of the advance ratio J which was calculated from the Equation (4). The power P was the product of the measured voltage U and current I (23): (23) The propeller efficiency was calculated from the Formula (24). It was calculated as the quotient of the airflow power generated by the propeller and the electrical power supplied to the propulsion unit. The airflow power was calculated as the product of the measured thrust and the velocity of the airflow. (24) The first analyzed parameter was the thrust coefficient as a function of the advance ratio (Figure 9). The obtained results are presented for four defined PWM values. As the PWM value increased, the value of propeller rotational speed increased. It was observed that as this parameter increased, the values with a lower advance ratio were obtained. In general, the advance ratio for all the cases considered did not exceed the value equal to 1. For the low rotational speed, i.e., PWM = 40%, the propeller with the smallest pitch had the lowest thrust of all the tested propellers. Increasing the pitch resulted in a gradual increase in thrust. The highest thrust was achieved by the propeller with the largest considered pitch. The increase in pitch simultaneously resulted in curves shifting towards the higher advance ratio and thrust coefficient. The characteristics for the 4.5″, 5.5″, 6″, and 8″ propellers had an approximately linearly decreasing trend. For the propellers with the largest pitch, a slight difference was observed for the small values of the advance ratio, and, in addition, the 5.5″ and 6″ pitch propellers achieved a very similar thrust. The differences occurred for the largest and smallest values of the advance ratio. Figure 9. Open in a new tab Thrust coefficient as a function of the advance ratio for the tested set of propellers for the defined PWM values: (a) 40%; (b) 60%; (c) 80%; (d) 90%. The next analyzed parameter was the torque coefficient (Figure 10). Similar to the thrust coefficient, the highest torque coefficient was obtained for the propeller with the largest pitch (12″). For such a propeller, the torque coefficient decreased with decreasing propeller pitch. For PWM equal to 40%, the highest value of 0.0144 was obtained for the case with no airflow and the advance ratio as 0. With the increase in the airflow velocity, the torque coefficient decreased up to 0.0052 at the advance ratio of 0.98. For PWM equal to 60%, the highest value of 0.0133 was obtained for the case with no airflow where the advance ratio is 0. With the increase in the airflow velocity, the value of the torque coefficient did not decrease in the whole considered range. However, at the advance ratio of 0.78, a value of 0.0115 was obtained. For the PWM of 80%, at the advance ratio of 0, a torque coefficient of 0.0128 was obtained. This is not the highest value, as 0.0129 was obtained at the advance ratio of 0.55. For the maximum PWM of 90%, there was little difference in the torque coefficient as a function of the advance ratio. As in the case of thrust coefficient, propellers with similar pitch, i.e., 5.5″ and 6″ showed comparable values of the torque coefficient. Not all propellers can operate over such a wide airspeed range. The propeller with the lowest pitch produced a positive torque coefficient for an advanced ratio of 0.37 at 40% PWM and 0.6 at higher PWM values. Beyond these extremes of airflow velocity in the wind tunnel, the propellers would generate negative torque, i.e., they would be driven more by the kinetic energy from the flowing air. The most important for the performance are the characteristics obtained for the highest PWM, i.e., 90%. In this case, the propellers with the largest pitch, i.e., 8″, 10″, and 12″ show very small changes in the torque coefficient. However, these propellers apply different loads to electric motors. By comparing the torque coefficient values at the largest advance ratio, it is possible to determine the amount of reduction in the torque coefficient compared to the 12″ propeller. Reducing the propeller pitch to 10″ resulted in a torque reduction of 19%. Further decreasing the pitch to 8″ reduced the torque by almost half (46%). Subsequent propeller changes to the smaller pitch values reduced torque by 64%, 69%, and 90%, respectively. Figure 10. Open in a new tab Torque coefficient as a function of the advance ratio for the tested set of propellers for the defined PWM values: (a) 40%; (b) 60%; (c) 80%; (d) 90%. Next, the power required to drive the propeller was analyzed (Figure 11). Power strictly depends on torque and rotational speed, so the presented characteristics are similar to the torque coefficient characteristics, but the values obtained were different. Figure 11. Open in a new tab Power coefficient as a function of the advance ratio for the tested set of propellers for the defined PWM values: (a) 40%; (b) 60%; (c) 80%; (d) 90%. Similar to the previously analyzed parameters, the highest power coefficient was obtained for the propeller with the largest pitch (12″). For such a propeller, the power coefficient decreases as the propeller pitch decreases. For PWM equal to 40%, the highest value of 0.1203 was obtained for the case with no airflow where the advance ratio is 0. As the airflow velocity increases, the power coefficient decreases to 0.0408 at the advance ratio of 0.98. For PWM equal to 60%, the highest value of 0.1127 was obtained for the case with no airflow and the advance ratio as 0. As the airflow speed increases, the power coefficient value does not tend to decrease over the entire considered range, but at the advance ratio of 0.78, a value of 0.0927 was obtained. For PWM equal to 80% and the advance ratio equal to 0, the power coefficient was found to be 0.1196. The maximum PWM equal to 90% showed a small difference in the power coefficient as a function of the advance ratio. As in the case of the considered thrust coefficient, the propellers with similar pitches, i.e., 5.5″ and 6″ show comparable values of the power coefficient. The most important for performance are the characteristics obtained for the highest PWM, i.e., 90%. In this case, the propellers with the largest pitch, i.e., 8″, 10″, and 12″ show very small changes in the power coefficient. Comparing the power coefficient values at the extreme values of the advance ratio, one can determine the decrease in the power coefficient in relation to the 12″ propeller. Decreasing the propeller pitch to 10″ reduces the power demand by 18%. Reducing the propeller pitch to 8″ results in a reduction of almost half (47%). Subsequent propeller changes to smaller pitches result in power reductions of 63%, 72%, and 82%, respectively. Another of the analyzed parameters was propeller efficiency (Figure 12). This coefficient defines the most favorable operation range of a given propeller and represents the highest obtained propeller thrust at a given airflow velocity in relation to power consumed. Depending on the PWM set, different characteristics describing propeller efficiency were obtained. For the lowest value of PWM, i.e., 40%, all the obtained characteristics have a distinct extremum. For PWM equal to 60%, two propellers with pitches 10″ and 12″ show increasing efficiency in the whole range as a function of the advance ratio. A similar situation occurs at the next PWM values, i.e., 80% and 90%. For these values, the 8″ propeller shows similar trends. However, in such cases, we cannot determine the best operating point for a given propeller. At these points, we can talk about the maximum efficiency achieved in the considered velocity range. Figure 12. Open in a new tab Propeller efficiency as a function of the advance ratio for the tested set of propellers for the defined PWM values: (a) 40%; (b) 60%; (c) 80%; (d) 90%. For a PWM of 40%, the maximum value of propeller efficiencies varied from 0.31 to 0.56. The point of maximum efficiency moved with propeller pitch toward larger values of the advance ratio. For the propeller with the smallest pitch, the maximum efficiency occurred at an advance ratio of approximately 0.37. This shows that the propeller with the largest pitch, i.e., 12″ has the highest efficiency in the considered conditions (for J = 0.66). In the considered range of airflow velocities for small PWM values, efficiency maxima are observed for all considered propeller pitches. As the PWM value increases, the characteristics for high values of advance ratio (for low propeller speed) do not bend and keep the increasing trend for a higher range of advance ratio. At PWM 80%, the largest propeller efficiency was achieved by the 8″ pitch propeller. A further comparison was therefore carried out in relation to this value and this propeller. The 12″ propeller in the considered range operated with the maximum value of propeller efficiency lower by 10%. The 10″ propeller in the considered range operated with the maximum value of propeller efficiency lower by 6%, and the other propellers, i.e., 6″, 5.5″ and 4.5″ operated with the maximum value of propeller efficiency lower by 13%, 12%, and 14%, respectively. It can be seen that at such a PWM value the differences are not so significant. At 90% PWM, the largest propeller efficiency was achieved by the propeller with a pitch of 12″. A further comparison was done in relation to this value and this propeller. The 10″ propeller operated with the maximum value of propeller efficiency lower by 0.2%. The 8″ propeller operated with the maximum value of propeller efficiency lower by 0.3%, and the other propellers, i.e., 6″, 5.5″ and 4.5″ operated at 9%, 9%, and 10%, respectively. It is clear that at this PWM value the differences are even smaller than before and do not exceed 10%. It should be noted that the three propellers with the largest pitch should be tested in this case at higher values of flow velocity to obtain the extrema of the curve describing propeller efficiency. The expected effect would be an increase in the maximum value of this coefficient. This was not necessary in this case due to the fact that the work covered a range of velocities specific to UAV applications. The last parameter analyzed was the thrust-to-power ratio (Figure 13). This ratio describes the amount of force expressed in gram-force and generated by power expressed in watts. For all propellers except the 12″ pitch one, decreasing characteristics as a function of the advance ratio were obtained. This means that the most gram-force was obtained for the case with no airflow. The exceptions are the 10″ and 12″ propellers. Figure 13. Open in a new tab Thrust-to-power ratio as a function of the advance ratio for the tested set of propellers for the defined PWM values: (a) 40%; (b) 60%; (c) 80%; (d) 90%. For propellers from 4.5 to 8″ pitch, the thrust-to-power ratio characteristics are decreasing, i.e., the highest values occur when there is no airflow. For the propellers with pitch 10 and 12″, the characteristics reach their maximum at non-zero values of the advance ratio. An exception is the 10″ propeller, where for PWM = 40% the trend is the same as for propellers with smaller pitches. For higher airflow velocities, small-pitch propellers generate less force per unit power, while large-pitch propellers operate at a higher thrust-to-power ratio. 5. Summary and Conclusions The conducted research made it possible to calculate the values of dimensionless coefficients of thrust, torque, power, efficiency, and a thrust-to-power ratio. The obtained results allow for the selection of the optimal solution, given the mentioned coefficients as criteria. It was observed that the increase in pitch resulted in the gradual increase in thrust. The highest thrust was achieved by the propeller with the largest considered pitch. Simultaneously, the increase in pitch resulted in the curves shifting towards the higher values of the advance ratio and thrust coefficient. Similarly, to the thrust coefficient, the highest torque coefficient was obtained for the propeller with the largest pitch, i.e., 12″. In this paper, the reduction in the torque coefficient for the tested propellers was specified in relation to the 12″ propeller at the largest advance ratio for the given power demand. At 90% PWM, the largest propeller efficiency (56.2%) was achieved by the 12″ propeller. This is the efficiency of the considered propulsion system and not the efficiency of the propeller itself. The resultant efficiency additionally consists of, e.g., BLDC motor efficiency or speed controller efficiency. It should also be noted that the value of the obtained efficiency does not represent the maximum value for this propeller, because the extremum of the curve describing propeller efficiency was not reached in the studied range. By testing the considered propeller for higher values of the advance ratio, better efficiency is expected. In this paper, it was not necessary to do so, as this range is suitable for typical UAVs. Finally, the thrust-to-power ratio parameter was analyzed. The results show a decrease in the force generated from a unit of power over a range of higher airflow rates, but the large pitch propellers operate with a higher T/Pel ratio over a range of higher airflow velocity rates. Author Contributions Conceptualization, Z.C., P.K and K.S.; Methodology, Z.C., P.K.; Software, K.S.; Formal analysis, Z.C. P.K.; Investigation, Z.C., P.K.; Writing—original draft preparation, Z.C., P.K. and K.S.; Writing—review and editing, M.W., Z.C.; Visualization, Z.C., P.K. and K.S.; Supervision, M.W., Z.C.; Project administration, Z.C.; Funding acquisition, Z.C. All authors have read and agreed to the published version of the manuscript. Funding This work has been financed by the Polish National Centre for Research and Development under the LIDER program Grant Agreement No. LIDER/27/0140/L-10/18/NCBR/2019. Institutional Review Board Statement Not applicable. Informed Consent Statement Not applicable. Data Availability Statement Data sharing not applicable. 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16634	https://www.youtube.com/watch?v=KEHsadlrGX0	4.6. Introduction to Titration \| College Board \| AP Chemistry LOGiota 992 subscribers Description 216 views Posted: 15 Dec 2023 The video includes essential knowledge and learning objective. It is prepared for the students who want to excel in AP chemistry. Transcript: hi everyone this is topic 4.6 introduction to titration and this is taken from AP Chemistry college board so in this topic I'll be talking that what is a titration and where is it used so first of all what is a titration now titration is a lab procedure in which a solution of known concentration is used to determine the concentration of an unknown solution for example if you have two solutions out of which you know the concentration of one solution but you don't know the concentration of another solution so this is the method by which we can actually find the unknown concentration of the solution now here this is the setup what we use for a titration we take a conical flask and we take a bued in the conical flask uh we take the analyte analyte is the one for which we don't know the concentration and in the buit we take the titrant titrant is the one for which we know the concent conentration and we start adding it dropwise now let's see what happens when we start adding it dropwise so when we start adding it dropwise there comes an equivalence point equivalence point is the point at which concentration of the both analyte and the tit becomes equal now here I have written H positive and O Negative this thing basically tells you that this is the titration happening between an acid and alkaly I'll be talking talking about that more in the next video but right now we just need to understand that what is a titration and why do we do it now whenever we do any titration we add an indicator to it now what is the meaning of indicator indicator is the one which can tell you that the equivalence point has reached now just by looking we can't say that the equivalence point has reached and we need to stop the titration but this indicator basically helps us to find that equivalence point and as soon as the color changes we stop the titration and that color change is called the end point so the O observable event is called the end point of the titration now this color change is due to its property uh is due to the property of the indicator in the acid or base for example if we take phenopen it will be colorless in the acid but as soon as the acid is neutralized by the alkaly and the solution becomes alkaline it Chang changes its color to Pink now this thing shows that the equivalence point has reached and the and we need to stop the titration I'll be talking about this whole titration again in the coming videos if you like the video please subscribe to my channel and press the Bell icon
16635	https://math.stackexchange.com/questions/2365242/how-do-i-find-the-maximum-and-minimum-value-of-any-trigonometric-expression	functions - How do I find the maximum and minimum value of any trigonometric expression - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How do I find the maximum and minimum value of any trigonometric expression Ask Question Asked 8 years, 2 months ago Modified7 years, 9 months ago Viewed 2k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. Is there any proper general method to find the maximum and the minimum value of any trigonometric expression (for example trigonometric expressions of the form a sin x+b cos x a sin⁡x+b cos⁡x or a sin x×cos x a sin⁡x×cos⁡x or any other such expression) elegantly (without using calculus)? I do not think that my question is broad. I am asking for a technique that works in most of the cases. functions trigonometry inequality maxima-minima cauchy-schwarz-inequality Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Dec 24, 2017 at 4:58 Michael Rozenberg 208k 31 31 gold badges 171 171 silver badges 294 294 bronze badges asked Jul 20, 2017 at 17:12 KBCKBC 39 4 4 bronze badges Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. By C-S a sin x+b cos x≤(a 2+b 2)(sin 2 x+cos 2 x)−−−−−−−−−−−−−−−−−−−−√=a 2+b 2−−−−−−√.a sin⁡x+b cos⁡x≤(a 2+b 2)(sin 2⁡x+cos 2⁡x)=a 2+b 2. The equality occurs for (a,b)\|\|(sin x,cos x)(a,b)\|\|(sin⁡x,cos⁡x). From here max(a sin x+b cos x)=a 2+b 2−−−−−−√max(a sin⁡x+b cos⁡x)=a 2+b 2 and min(a sin x+b cos x)=−a 2+b 2−−−−−−√.min(a sin⁡x+b cos⁡x)=−a 2+b 2. a sin x cos x=1 2 a sin 2 x a sin⁡x cos⁡x=1 2 a sin⁡2 x and from here max(a sin x cos x)=\|a\|2 max(a sin⁡x cos⁡x)=\|a\|2 and min(a sin x cos x)=−\|a\|2.min(a sin⁡x cos⁡x)=−\|a\|2. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jul 20, 2017 at 17:16 Michael RozenbergMichael Rozenberg 208k 31 31 gold badges 171 171 silver badges 294 294 bronze badges 2 I appreciate your reply but my question was about a general method which can be used to find maximum and minimum values of all trigonometric expressions including this one. BTW what do you mean by C-S?KBC –KBC 2017-07-20 17:22:20 +00:00 Commented Jul 20, 2017 at 17:22 @KBC It's Cauchy-Schwartz inequality. I think we have no a general method to solve math problems. Give me a problem and I'll try to solve it.Michael Rozenberg –Michael Rozenberg 2017-07-20 17:26:20 +00:00 Commented Jul 20, 2017 at 17:26 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. note that we have the following inequalities for your first question its that −a 2+b 2−−−−−−√≤a sin(x)+b cos(x)≤a 2+b 2−−−−−−√−a 2+b 2≤a sin⁡(x)+b cos⁡(x)≤a 2+b 2 and for your second question we have sin(x)cos(x)=sin(2 x)2 sin⁡(x)cos⁡(x)=sin⁡(2 x)2.You can prove first one by multiplying and dividing by a 2+b 2−−−−−−√a 2+b 2 and using a a 2+b 2√=cos(a)a a 2+b 2=cos⁡(a) so b a 2+b 2√=sin(a)b a 2+b 2=sin⁡(a) hence we have a 2+b 2−−−−−−√(sin(a+x))a 2+b 2(sin⁡(a+x)) now −1≤sin≤1−1≤sin≤1 hence the proof. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jul 20, 2017 at 17:20 Archis WelankarArchis Welankar 16.1k 8 8 gold badges 37 37 silver badges 68 68 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. for your first example you can write a 2+b 2−−−−−−√(a a 2+b 2−−−−−−√sin(x)+b a 2+b 2−−−−−−√cos(x))=a 2+b 2−−−−−−√(sin(x)cos(ϕ)+sin(ϕ)cos(x))a 2+b 2(a a 2+b 2 sin⁡(x)+b a 2+b 2 cos⁡(x))=a 2+b 2(sin⁡(x)cos⁡(ϕ)+sin⁡(ϕ)cos⁡(x)) =a 2+b 2−−−−−−√sin(x+ϕ)=a 2+b 2 sin⁡(x+ϕ) where c o s(ϕ)=a a 2+b 2−−−−−−√c o s(ϕ)=a a 2+b 2 and sin(ϕ)=b a 2+b 2−−−−−−√sin⁡(ϕ)=b a 2+b 2 your second term a sin(x)cos(x)=a 2 sin(2 x)a sin⁡(x)cos⁡(x)=a 2 sin⁡(2 x) Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jul 20, 2017 at 17:22 Dr. Sonnhard GraubnerDr. Sonnhard Graubner 97.5k 4 4 gold badges 42 42 silver badges 80 80 bronze badges 2 What happens if a 2+b 2=0 a 2+b 2=0?Michael Rozenberg –Michael Rozenberg 2017-07-20 17:28:54 +00:00 Commented Jul 20, 2017 at 17:28 1 dear Michael, then is a=b=0 a=b=0 and the equation a sin(x)+b cos(x)a sin⁡(x)+b cos⁡(x) makes no sence Dr. Sonnhard Graubner –Dr. Sonnhard Graubner 2017-07-20 17:33:53 +00:00 Commented Jul 20, 2017 at 17:33 Add a comment\| You must log in to answer this question. 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16636	https://phys.libretexts.org/Courses/University_of_California_Davis/UCD%3A_Physics_9B__Waves_Sound_Optics_Thermodynamics_and_Fluids/04%3A_Geometrical_Optics/4.05%3A_Thin_Lenses	R 1fconvex=(nconvex−1)(1R−1−R)=2(nconvex−1R)1fconcave=(nconcave−1)(1−R−1R)=−2(nconcave−1R) 1ftot=1fconvex+1fconcave=2(nconvex−1R)−2(nconcave−1R)=2(nconvex−nconcaveR) 2(n−1R)=2(nconvex−nconcaveR)⇒n=nconvex−nconcave+1=2.43 Skip to main content 4.5: Thin Lenses Last updated : Nov 8, 2022 Save as PDF 4.4: Spherical Refractors 4.6: Multiple Optical Devices Page ID : 18463 Tom Weideman University of California, Davis ( \newcommand{\kernel}{\mathrm{null}\,}) Building a Lens While it is interesting to see how images are formed as light passes from one index of refraction to another through a spherical surface, this only rarely happens in the real world. It is far more common for light to come from an object that is in the air (n=1.00n=1.00), pass through a transparent region with a higher index of refraction, then return again to the air before being observed. The light will then be refracted at both surfaces of the region during its journey from the object to the point of observation. If both surfaces of the region are spherical, then the light will either converge or diverge at each surface (depending upon which of the four cases of Section 4.4 applies). Such an optical device is called a lens. Let's take a look at a glass lens that is convex on both sides, which is in air. When light enters one side, it is going from a lower index of refraction (air) to a higher one (glass) that is convex, which means we have case #1, which is a converging refraction. Now the light is inside the glass, and its next encounter is with a concave surface (from the perspective of inside the glass), passing from a higher index of refraction to a lower one. This is case #3, which is also a converging refraction. [See Figure 4.4.2 for references to the four possible cases.] A double-convex lens is therefore a converging lens. Figure 4.5.1 – Double Convex Lens Note that the two surfaces do not necessarily have the same radius of curvature. The surface with the shorter radius (the one that is more sharply-curved) will bend the light more than the other surface. We can in fact put together any combination of surfaces we like in the construction of a lens. A double concave lens will result in the light experiencing cases 2 and 4, both of which are diverging, causing the light rays that pass through the lens to diverge. The various combinations of surfaces is shown in the diagram below. Note that the signs of the radii given are found using the sign convention and assuming that the light is passing from left to right, and that a flat plane can be described as a spherical surface with an infinite radius. Figure 4.5.2 – Varieties of Lenses Tracing a ray through a lens is a daunting bit of geometry, and as always, when a calculation is daunting, we make a simplifying approximation. In this case, the simplifying principle is called the thin lens approximation. This asserts, not surprisingly, that the lens is very thin, which means that the bending of the light that occurs at each surface essentially happens at the same position. The two refractions are still successive (a refraction occurs at one surface then the other), but the light travels no distance getting from the point where it changes direction the first time to where it changes direction a second time. This simplifies things greatly, because measurements of quantities like object and image distances are all referenced to the same place, no matter which surface is in play. Defining a Single Focal Length for a Lens We already know that each surface of a lens has its own focal length, but when light that comes upon the lens parallel to the optical axis, it must go somewhere after it passes through, which means that the lens as a whole must have a unique focal length. We now seek what this will be for a lens of index of refraction nn in air with radii of curvature of its two sides equal to R1R1 and R2R2. The key trick to use here is that after light passes through the first surface, it appears to be coming from somewhere else (i.e. from the position of its image). The second surface then receives the light, and as far as it "knows," the light is coming from this new position. That means that the object position for the second surface is the image position of the first surface. All that remains is to do two successive image locations, while getting all the sign conventions right, and the final image position will be what is seen as the apparent origin of light that has passed through both surfaces of the lens. Okay, so if we are looking for the focal length of the lens, we naturally start with a ray that comes into the lens parallel, and look for the distance from the lens that it crosses the axis. A parallel ray is equivalent to an object distance of infinity (s1=∞s1=∞), the first medium is air (n1=1n1=1), and the second medium is that of the lens (n2=nn2=n), so the image for the first surface (with radius R1R1) is easy enough to find using Equation 4.4.4: 1∞+ns′1=n−1R1⇒ns′1=n−1R1 1∞+ns′1=n−1R1⇒ns′1=n−1R1(4.5.1) The image position for this first surface now becomes the object position for the second surface. Because of our thin lens approximation, the magnitude of the image distance for the first surface equals the magnitude of the object distance of the second surface, but now the issue of the signs must be addressed. Suppose the first surface is concave. Then this surface is case 4, and the ray is made to diverge at this surface, resulting in an image that is to the left of the lens (we are still assuming the light is moving left-to-right). With this image to the left of the second surface and the light moving left-to-right, the object distance for the second surface is positive (on the incoming side). But the image distance calculated was negative (the divergence caused a virtual image), so while the image distance s′1s′1 for the first surface and the object distance for the second surface s2s2 are equal in magnitude, they have opposite signs. We can therefore make the substitution s2=−s′1s2=−s′1 for the equation at the second surface. Note that this time the light is passing from a region with index of refraction nn to air: 1−nR2=ns2+1s′2=n−s′1+1s′2 1−nR2=ns2+1s′2=n−s′1+1s′2(4.5.2) Plugging in the result of Equation 4.5.1 and noting that the final image position is the focal point of the whole lens gives: 1−nR2=−n−1R1+1s′2⇒1f=(n−1)(1R1−1R2) 1−nR2=−n−1R1+1s′2⇒1f=(n−1)(1R1−1R2)(4.5.3) This gives us a prescription for building a lens – given the index of refraction of the material, we can grind the two spherical surfaces to radii of R1R1 and R2R2 to achieve a focal length given by ff. For this reason, this is called the lensmaker equation. The reader may be troubled that we assumed that the first surface is concave. Does a convex first surface result in a different lensmaker equation? The answer is no! If the first surface is convex, then the image of the incoming parallel light lands to the right of the lens. This is a positive-valued image distance: s′1>0s′1>0. Making this image the object for the second surface gets confusing. With the image already to the right of the second surface, does the light even pass through the second surface? Does it turn around and go right-to-left? The answer is neither of these. While the idea may be difficult to visualize conceptually, mathematically the answer is simple – the light is still going left-to-right, and the object is not on the incoming side, so the object distance is negative: s2<0s2<0. This is sometimes referred to as a virtual object, and it can only occur when light is diverted (reflected or refracted) twice, so that the image of the first diversion lands beyond the surface that causes of the second diversion. Notice that with s′1>0s′1>0 and s2<0s2<0, and their magnitudes equal, we once again get the s2=−s′1s2=−s′1 relation, and the same lensmaker equation results. You may have noticed that the term "1f1f" comes up a lot. It's clear that the smaller the value of ff is, the more sharply the light is bent by the refraction, which means that the larger 1f1f is, the more "focusing power" the device has. Frequently the focusing power of a lens is measured with this inverse quantity rather than focal length. This quantity has the units of m−1m−1, which are designated their own name of diopters (DD). Stacking Thin Lenses With our assumption that lenses are very thin, it's fair to also assume that if two lenses are placed together, the combination is also very thin. With the ability to refract through one lens, and then immediately the other, parallel rays that come into this combination of two lenses will focus at a different point than either of the lenses individually. In essence, two (or more) stacked lenses simply form the equivalent of a new single lens. If we follow the same process as above to determine the focal length of multiple lenses, we get sum where every term looks something like ±(n−1R)±(n−1R) for each refracting surface the light passes through. Rather than break down each lens into its two surfaces, however, we have a shortcut – just add the inverses of the individual lens focal lengths! 1ftot=1f1+1f2+… 1ftot=1f1+1f2+…(4.5.4) Example 4.5.14.5.1 You have a machine that grinds glass surfaces to a certain specified radius. You use it to make a double-convex lens out of a special glass with high index of refraction equal to 2.752.75. You then use it to make a double-concave lens out of a different type of glass with index of refraction equal to 1.321.32. Because the curved surfaces are the same radii, the two lenses fit together perfectly. When this is done, it forms a single lens that is half-convex, half concave, as in the diagram below. If you wanted to grind a third lens with the same machine that had the same optical properties of the combined lens above, but instead use a single type of glass, then would it be double-convex or double concave? Explain. Compute the index of refraction needed to accomplish the task described in (a). Solution : a. The radius used everywhere is the same, so we will call it “R.” The lens maker equation for these two lenses gives: 1fconvex=(nconvex−1)(1R−1−R)=2(nconvex−1R)1fconcave=(nconcave−1)(1−R−1R)=−2(nconcave−1R) When the two lenses are pushed together, the diopter strengths (inverses of the focal lengths) are added to get the new diopter strength, so since the index of refraction is greater for the converging lens, from the equations above we see that its diopter strength has a greater magnitude than its diverging counterpart, and the sum will come out positive. The combined lens will converge light. b. The combined diopter strength is the sum of the diopter strengths, so we want to make a lens with a diopter strength equal to that sum. 1ftot=1fconvex+1fconcave=2(nconvex−1R)−2(nconcave−1R)=2(nconvex−nconcaveR) Setting this equal to the diopter strength of a single lens with the same radius and unknown index of refraction, we have: 2(n−1R)=2(nconvex−nconcaveR)⇒n=nconvex−nconcave+1=2.43 Objects and Images Rays that pass through lenses follow paths that have the same characteristics as ray paths for single refracting surfaces. Rays parallel to the optical axis all cross the axis at the same focal point on the other side of the lens. One would therefore expect the object and image distances to be related to the focal length by a formula similar to Equation 4.4.6. The only question is what happens to the indices of refraction. Well, with lenses, the object and image are in the same medium, which means that n1n1 and n2 are the same. It is true that the lens itself has a different index of refraction, but we have packaged all the refractions within the lens into a single change of direction at a single point, so those details don't come into play. Setting the indices of refraction in Equation 4.4.6 equal to each other gives a simple relation, known as the thin lens equation: 1s+1s′=1f If this looks familiar even without the n's, it's because it is identical to Equation 4.3.5, which we used for spherical mirrors! It is important to note that if we dig into the details hidden within the variable f, we find that for spherical mirrors the focal length is a simple function of the radius (f=R2), while for lenses the focal length comes from the lensmaker equation (two radii and an index of refraction). But as along as the specific details of the focal length are not an issue, then this same equation works for both cases. What is more, we have been very careful in the wording of our sign conventions (all that language about "incoming/outgoing" and "not incoming/not outgoing" sides was carefully chosen), so that the sign conventions work perfectly well for both mirrors and lenses. And finally, the geometry of magnification also carries over to lenses, which means that Equation 4.3.7 also holds for lenses. Principal Rays While we have greatly economized the mathematics of geometrical optics to incorporate both mirrors and lenses, the physical processes involved are obviously different. We can express these differences by examining ray traces for lenses. We will see that while they obviously are a bit different from those we did for mirrors, there are many similarities in the logic. While for the mirror there were four principal rays, for lenses there are only three. As we saw with mirror ray traces, there are differences for the diverging and converging cases. We will start with the converging lens. [Note: In our diagrams, we will depict a converging lens with a double-convex structure, but it can have any combination of surfaces that results in a converging lens (f>0). Similarly, all diverging lenses are depicted as being double concave, though they too can be any combination of surfaces that results in a diverging lens (f<0). It should also be mentioned that although the diagrams will give these lenses an apparent thickness, we will be treating them as "thin," which means that the bending of rays will all occur at a single vertical plane through the center of the lens.] A major difference between lenses and mirrors is that lenses can allow light to travel in either direction. Therefore, while a mirror has a single focal point on its concave side (no matter which side is reflecting), lenses have focal points on both sides. Two of the principal rays behave like we have seen already – one comes into the lens parallel to the optical axis and exits the lens toward (for converging lenses) or away from (for diverging lenses) the focal point. The other comes in to the lens through (for converging lenses) or toward (for diverging lenses) the focal point, and exits the lens parallel to the optical axis. The third principal ray passes directly through the vertex of the lens without diverting its path. Figure 4.5.3 – Principal Rays, Converging Lens (Object Outside Focal Point) Figure 4.5.4 – Principal Rays, Converging Lens (Object Inside Focal Point) Figure 4.5.5 – Principal Rays, Diverging Lens Checking Ray Traces with the Thin Lens Equation It is worthwhile to confirm that the ray traces shown above give results predicted by the thin lens equation (Equation 4.5.5) and our sign conventions. converging lens (f>0) with distant object (s>f) – With s>f, we have 1s<1f, and combining this with the thin lens equation, we can conclude that s′>0, which means that the image is on the outgoing side of the lens, making it a real image (lies at the intersection of actual light). Furthermore, with both s and s′ positive, the lateral magnification M=−s′s is negative, which means that the image is inverted. From the thin lens equation, we can deduce that when s>2f, then the image comes out closer to the lens than the object (s′<s), and the lateral magnification formula tells us that this means that the image is diminished. All of these features are reflected in Figure 4.5.3. converging lens (f>0) with close object (s1f. This indicates that the image will not be on the outgoing side of the lens, and is therefore virtual. The lateral magnification comes out positive in this case, indicating that the image must be upright. With a positive object distance and a negative image distance, for the focal length in the thin lens equation to come out positive, it must be true that \|1s\|>\|1s′\|. This means that the image must be farther from the lens than the object, and again looking at the lateral magnification equation, this requires the image to be larger than the object. All of this is in perfect agreement with Figure 4.5.4. diverging lens (f<0) – With a positive object distance and a negative focal length, the only way that the thin lens equation can be satisfied is for the image distance to be negative, which means that the image is not on the outgoing side of the lens, and the image is virtual. Again the object and image distances having opposite signs means that the image is upright. With the positive object distance and negative image distance needing to produce a negative focal length in the thin lens equation, it must be true that \|1s\|<\|1s′\|, which means that the image is closer to the lens than the object, and according to the lateral magnification equation, the image is diminished in size compared to the object. The math agrees with Figure 4.5.5. Example 4.5.2 Two identical objects are placed with the same orientation, separated by a distance of 1.5m. A lens is then placed along the line formed by the two objects such that all three of these items are equally-spaced, as in the diagram below. When one views the light coming from the objects through the lens, one sees two images of equal size (though they do not look like it, as the images are not the same distance away). One of these images is upright, and the other inverted Find the focal length of the lens, including the sign (indicating whether it is converging or diverging). Determine which of the two images appears larger to the observer. Solution : a. For an image to be upright for a single lens (where the object distance is always positive), it must be virtual (s′<0), and for it to be inverted, it must be real (s′>0). Single diverging lenses only create virtual images, so the fact that a real (inverted) image exists here means this lens must be converging. Real images for single converging lenses occur when the object is outside the focal length, and virtual images occur when the object is inside the focal length. Therefore the focal point of the lens must land between the two objects, giving it a range of between +1.5m and +3.0m. Okay, so let's do the actual calculation: With images of equal size but opposite orientation, their lateral magnifications must be negatives of each other. This gives us information about their image distances (we already know their object distances, which we can use): M1=−M2⇒−s′1s1=s′2s2⇒s′1s′2=−s1s2=−2 We also have the thin lens equations for both objects. Both involve the same lens, so f is the same in each case: 1s1+1s′1=1f1s2+1s′2=1f Plugging in 2s2 for s1 and −2s′2 for s′1 in the thin lens equation for object #1, we get: 12s2+1−2s′2=1f⇒1s2−1s′2=2f Adding this equation to the thin lens equation for object #2, we get: (1s2+1s′2)+(1s2−1s′2)=1f+2f⇒2s2=3f⇒f=32s2=2.25m b. The lateral magnifications are equal, but the one that appears bigger is the one that results in a larger angular magnification. This honor belongs to the image that is closer to the observer. Since one of the images is real, it moves to the other side of the lens, where the observer is located. The virtual image remains on the same side of the lens, so clearly the real image must look larger to the observer. For a converging lens, all real images require objects be farther from the lens than the focal length, and virtual images closer than the focal length. Therefore object #1 produces the real image, and image #1 appears larger. 4.4: Spherical Refractors 4.6: Multiple Optical Devices
16637	https://artofproblemsolving.com/wiki/index.php/Determinant?srsltid=AfmBOooOehq3ZDib-dLtWBRWbC6siJPKGZ3R3ntHoJA2Si9wAV7V5feT	Art of Problem Solving Determinant - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Determinant Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Determinant The determinant is an important notion in linear algebra. Contents 1 Definition 1.1 Determinants in Terms of Simpler Determinants 2 Matrix Determinants are Multiplicative 3 Equivalence of Definitions 4 See Also Definition For an matrix , the determinant is defined by the sum where is the set of all permutations on the set , and is the parity of the permutation . For example, the determinant of a matrix is . This quantity may seem unwieldy, but surprisingly, it is multiplicative. That is, for any matrices (over the same commutative field), More generally, if is a commutativefield and is an element of a (strictly power associative) -dimensional -algebra, then the determinant of is times the constant term of the characteristic polynomial of . Our generalized determinants also satisfy the multiplicative property when is associative. Determinants in Terms of Simpler Determinants An determinant can be written in terms of determinants: where is the matrix formed by removing the st row and th column from : This makes it easy to see why the determinant of an matrix is the sum of the diagonals labeled , minus the sum of the diagonals labeled , where "diagonal" means the product of the terms along it: Matrix Determinants are Multiplicative In this section we prove that the determinant as defined for matrices is multiplicative. We first note that from rearrangements of terms. If we let , we then have On the other hand, where is the set , and is the set of functions mapping into itself. From equation (1), it thus suffices to show that if is not a permutation on , then To this end, suppose that is not a permutation. Then there exist distinct integers such that . Let be the permutation on that transposes and while fixing everything else. Then as the latter product is the same as the former, with two terms switched. On the other hand is an odd permutation, so Since , we can partition the elements of into pairs for which the equation above holds. Equation 2 then follows, and we are done. Equivalence of Definitions We now prove that our two definitions are equivalent. We first note that the definitions coincide in the case of upper-triangular matrices, as each entry in the diagonal of an upper-triangular matrix corresponds to a (generalized) eigenvalue of the matrix. We now use the fact that every element of is similar to an upper triangular matrix; that is, there exists an upper triangular matrix and an invertible matrix such that Writing for our specialized determinant for matrices and for our generalized definition with the characteristic polynomial, we have as the characteristic polynomial does not change under automorphisms of that fix . Our two definitions are therefore equivalent. See Also Characteristic polynomial Upper triangular matrix This article is a stub. Help us out by expanding it. Retrieved from " Categories: Linear algebra Stubs Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
16638	https://www.khanacademy.org/test-prep/v2-sat-math/x0fcc98a58ba3bea7:advanced-math-easier/x0fcc98a58ba3bea7:radical-rational-and-absolute-value-equations-easier/a/v2-sat-lesson-radical-rational-and-absolute-value-equations	Radical, rational, and absolute value equations \| Lesson (article) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Explore Khanmigo Math: Pre-K - 8th grade Math: High school & college Math: Multiple grades Math: Illustrative Math-aligned Math: Eureka Math-aligned Math: Get ready courses Test prep Science Economics Reading & language arts Computing Life skills Social studies Partner courses Khan for educators Select a category to view its courses Search AI for Teachers FreeDonateLog inSign up Search for courses, skills, and videos Skip to lesson content SAT Math Course: SAT Math>Unit 4 Lesson 9: Radical, rational, and absolute value equations: foundations Radical, rational, and absolute value equations \| Lesson Radical and rational equations — Basic example Radical and rational equations — Harder example Radical, rational, and absolute value equations: foundations Test prep> SAT Math> Foundations: Advanced math> Radical, rational, and absolute value equations: foundations © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Radical, rational, and absolute value equations \| Lesson Google Classroom Microsoft Teams A guide to radical, rational, and absolute value equations on the digital SAT What are radical, rational, and absolute value equations? Radical equations are equations in which variables appear under radical symbols (x‍). 2 x−1=x‍ is a radical equation. Rational equations are equations in which variables can be found in the denominators of rational expressions. 1 x+1=2 x‍ is a rational equation. Both radical and rational equations can have extraneous solutions, algebraic solutions that emerge as we solve the equations that do not satisfy the original equations. In other words, extraneous solutions seem like they're solutions, but they aren't. Absolute value equations are equations in which variables appear within vertical bars (\|\|‍). \|x+1\|=2‍ is an absolute value equation. In this lesson, we'll learn to: Solve radical and rational equations Identify extraneous solutions to radical and rational equations Solve absolute value equations You can learn anything. Let's do this! How do I solve radical equations? Intro to square-root equations & extraneous solutions Khan Academy video wrapper Intro to square-root equations & extraneous solutionsSee video transcript What do I need to know to solve radical equations? The process of solving radical equations almost always involves rearranging the radical equations into quadratic equations , then solving the quadratic equations. As such, knowledge of how to manipulate polynomials algebraically and solve a variety of quadratic equations is essential to successfully solving radical equations. To solve a radical equation: Isolate the radical expression to one side of the equation. Square both sides the equation. Rearrange and solve the resulting equation. Example: If 2 x−1=x‍, what is the value of x‍ ? Show me! 2 x−1=x(2 x−1)2=x 2 2 x−1=x 2 2 x−1−(2 x−1)=x 2−(2 x−1)0=x 2−2 x+1‍ We can factor x 2−2 x+1‍ into (x+a)(x+b)‍: a+b=−2‍ a b=1‍ a=−1‍ and b=−1‍ would work: −1+(−1)=−2‍ (−1)(−1)=1‍ Now, we can solve for x‍: 0=x 2−2 x+1 0=(x−1)(x−1)0=x−1 0+1=x−1+1 1=x‍ 1‍ is the value of x‍. When it comes to extraneous solutions, the concept that confuses the most students is that of the principal square root. The square root operation gives us only the principal square root, or positive positive square root. For example, 4=2‍, not both −2‍ and 2‍ even though (−2)2=2 2=4‍. If a solution leads to equating the square root of a number to a negative number, then that solution is extraneous. But why are there extraneous solutions at all? In most cases, solving radical equations on the SAT involves squaring both sides of the radical equation. Raising both sides of an equation to an even power is not a reversible operation. For example, if a‍ is equal to b‍, then a 2‍ must be equal to b 2‍. However, if a 2‍ is equal to b 2‍, a‍ does not necessarily equal to b‍. Let's look at a numerical example. For a=−2‍ and b=2‍, their squares are equal as (−2)2=2 2=4‍, but we know for a fact that −2≠2‍. Therefore, when we solve an equation whose solution steps include squaring both sides of the equation, we're solving a 2=b 2‍, and we must perform additional checks to make sure that a‍ is also equal to b‍. To check for extraneous solutions to a radical equation: Solve the radical equation as outlined above. Substitute the solutions into the original equation. A solution is extraneous if it does not satisfy the original equation. Example: What is the solution to the equation 3 x+4=x‍ ? Show me! 3 x+4=x(3 x+4)2=x 2 3 x+4=x 2 2 x−1−(3 x+4)=x 2−(3 x+4)0=x 2−3 x−4‍ We can factor x 2−3 x−4‍ into (x+a)(x+b)‍: a+b=−3‍ a b=−4‍ a=−4‍ and b=1‍ would work: −4+1=−3‍ (−4)(1)=−4‍ Now, we can solve for x‍: 0=x 2−3 x−4 0=(x−4)(x+1)0=x−4 or 0=x+1 0=x−4 0+4=x−4+4 4=x 0=x+1 0−1=x+1−1−1=x‍ Now, we need to substitute 4‍ and −1‍ for x‍ in the original equation to check whether they're extraneous: 3(4)+4=?4 16=?4 4=✓4 3(−1)+4=?−1 1=?−1 1≠−1‍ x=4‍ satisfies the original equation, but x=−1‍ does not. As such, x=−1‍ is an extraneous solution. 4‍ is the solution to the equation 3 x+4=x‍. Try it! Try: identify the steps to solving a radical equation 2 x−9=x−6‍ To solve the equation above, we first operation both sides of the equation, then rewrite the result as a linear/quadratic equation. Solving this equation gives us 2‍ solutions, and we do/do not need to check for extraneous solutions. Check Explain Try: Identify an extraneous solution to a radical equation Marcy solved the radical equation x+2=x‍ and got two solutions, 2‍ and −1‍. When we substitute 2‍ for x‍ into the equation, the left side of the equation is 2+2=‍ and the right side of the equation is 2‍. Therefore, 2‍ is extraneous/not extraneous . When we substitute −1‍ for x‍ into the equation, the left side of the equation is −1+2=‍ and the right side of the equation is −1‍. Therefore, −1‍ is extraneous/not extraneous . Check Explain How do I solve rational equations? Equations with rational expressions Khan Academy video wrapper Equations with rational expressionsSee video transcript What do I need to know to solve rational equations? Knowledge of fractions, polynomial operations and factoring, and quadratic equations is essential for successfully solving rational equations. To solve a rational equation: Rewrite the equation until the variable no longer appears in the denominators of rational expressions. Rearrange and solve the resulting linear or quadratic equation. Example: If 1 x+1=2 x‍, what is the value of x‍ ? Show me! 1 x+1=2 x 1 x+1⋅(x+1)=2 x⋅(x+1)1=2(x+1)x 1⋅x=2(x+1)x⋅x x=2(x+1)x=2 x+2 x−x=2 x+2−x 0=x+2 0−2=x+2−2−2=x‍ −2‍ is the value of x‍. Most often, the reason a solution to a rational equation is extraneous is because the solution, when substituted into the original equation, results in division by 0‍. For example, if one of the solutions to a rational equation is 2‍ and the original equation contains the denominator x−2‍, then the solution 2‍ is extraneous because 2−2=0‍, and we cannot divide by 0‍. To check for extraneous solutions to a rational equation: Solve the rational equation as outlined above. Substitute the solution(s) into the original equation. A solution is extraneous if it does not satisfy the original equation. Example: What value(s) of x‍ satisfies the equation 2 x−1=x+1 x−1‍ ? Show me! 2 x−1=x+1 x−1 2 x−1⋅(x−1)=x+1 x−1⋅(x−1)2=x+1 2−1=x+1−1 1=x‍ However, when we substitute x=1‍ into the original equation, the denominator of both sides of the equation becomes 1−1=0‍. Since we cannot divide by 0‍, 1‍ is an extraneous solution. No value of x‍ satisfies the equation 2 x−1=x+1 x−1‍. Try it! TRY: Identify the steps to solving a rational equation 3 x x+2=2‍ To solve the equation above, we first multiply/divide both sides of the equation by x+2‍, then solve the resulting linear/quadratic equation. Because the denominator of the rational expression is x+2‍, the only value of x‍ that would lead to division by 0‍ is . Therefore, when we get 4‍ as the solution, we know that it is extraneous/not extraneous . Check Explain TRY: Identify an extraneous solution to a rational equation Mehdi solved the rational equation x 2−4 x+2=4‍ and got two solutions, −2‍ and 6‍. When we substitute −2‍ for x‍ into the equation, the denominator of the rational expression is −2+2=‍. Therefore, −2‍ is extraneous/not extraneous . When we substitute 6‍ for x‍ into the equation, the denominator of the rational expression is 6+2=‍ and the rational expression is equal to 6 2−4 6+2=‍. Therefore, 6‍ is extraneous/not extraneous . Check Explain How do I solve absolute value equations? Absolute value equation with two solutions Khan Academy video wrapper Worked example: absolute value equation with two solutionsSee video transcript Absolute value equation with no solution Khan Academy video wrapper Worked example: absolute value equations with no solutionSee video transcript The absolute value of a number is equal to the number's distance from 0‍ on the number line, which means the absolute value of a nonzero number is always positive. For example: The absolute value of 2‍, or \|2\|‍, is 2‍. The absolute value of −2‍, or \|−2\|‍, is also 2‍. Practically, this means every absolute value equation can be split into two linear equations. For example, if \|2 x+1\|=5‍: The absolute value equation is true if 2 x+1=5‍. The absolute value equation is also true if 2 x+1=−5‍ since \|−5\|=5‍. When solving absolute value equations, rewrite the equation as two linear equations, then solve each linear equation. Both solutions are solutions to the absolute value equation. Example: What are the solutions to the equation \|2 x−1\|=5‍ ? Show me! The absolute value equation can be divided into two linear equations: 2 x−1=5 first equation 2 x−1+1=5+1 2 x=6 2 x 2=6 2 x=3 first solution 2 x−1=−5 second equation 2 x−1+1=−5+1 2 x=−4 2 x 2=−4 2 x=−2 second solution‍ The solutions are 3‍ and −2‍. Try it! try: write two linear equations from an absolute value equation \|3 x+7\|=16‍ To solve the absolute value equation above, we must solve two linear equations. x=3‍ is one solution to the absolute value equation and satisfies the linear equation . x=−23 3‍ is the other solution to the absolute value equation and satisfies the linear equation . Check Explain Your turn! Practice: solve a radical equation 6 x+9=x+3‍ Which of the following values of x‍ satisfies the equation above? Choose 1 answer: Choose 1 answer: (Choice A) 0‍ A 0‍ (Choice B) 3‍ B 3‍ (Choice C) 6‍ C 6‍ (Choice D) 9‍ D 9‍ Check Explain Practice: check for extraneous solutions to a radical equation 4 x+16=x+1‍ Which of the following are the solutions to the equation above? I‍. −3‍ II‍. 5‍ Choose 1 answer: Choose 1 answer: (Choice A) I‍ only A I‍ only (Choice B) II‍ only B II‍ only (Choice C) Both I‍ and II‍ C Both I‍ and II‍ (Choice D) Neither I‍ nor II‍ D Neither I‍ nor II‍ Check Explain Practice: solve a rational equation If 1 x+1=3 5 x−1‍, what is the value x‍ ? Choose 1 answer: Choose 1 answer: (Choice A) 1 3‍ A 1 3‍ (Choice B) 1 2‍ B 1 2‍ (Choice C) 1‍ C 1‍ (Choice D) 2‍ D 2‍ Check Explain Practice: solve a rational equation Which of the following values of x‍ satisfies the equation x 2−4 x x−4=1‍ ? Choose 1 answer: Choose 1 answer: (Choice A) 1‍ only A 1‍ only (Choice B) 4‍ only B 4‍ only (Choice C) 1‍ and 4‍ C 1‍ and 4‍ (Choice D) No such value of x‍ exists. D No such value of x‍ exists. Check Explain practice: solve an absolute value equation \|x−7\|=1‍ If a‍ and b‍ are solutions to the equation above, what is the value of a+b‍ ? Check Explain Things to remember The radical operator (x‍) calculates only the positive square root. If a solution leads to equating the square root of a number to a negative number, then that solution is extraneous. We cannot divide by 0‍. If a solution leads to division by 0‍, then that solution is extraneous. For the absolute value equation \|a x+b\|=c‍, rewrite the equation as the following linear equations and solve them. a x+b=c‍ a x+b=−c‍ Both solutions are solutions to the absolute value equation. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted Za 3 years ago Posted 3 years ago. 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Direct link to Prince J Shah's post “This lesson tells us to t...” more This lesson tells us to take only the positive value out of a square root whereas the lesson before this(solving quadratic equations)tells us to take both the positive and the negative value. How do I know when to follow which lesson on the test? Answer Button navigates to signup page •4 comments Comment on Prince J Shah's post “This lesson tells us to t...” (15 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Guy28 2 years ago Posted 2 years ago. Direct link to Guy28's post “Equations such as x^2=a w...” more Equations such as x^2=a will have both positive and negative solutions. But if you are using the square root function with this symbol "√" then we only take the principal or positive square root. 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16639	https://lessonplanet.com/teachers/ck-12-arithmetic-absolute-value-of-integers	Ck 12: Arithmetic: Absolute Value of Integers Unit Plan for 6th - 7th Grade \| Lesson Planet Search Search educational resources Search Menu Sign InTry It Free AI Teacher Tools Discover - [x] Discover Resources Search reviewed educational resources by keyword, subject, grade, type, and more Curriculum Manager (My Content) Manage saved and uploaded resources and folders To Access the Curriculum Manager Sign In or Join Now Browse Resource Directory Browse educational resources by subject and topic Curriculum Calendar Explore curriculum resources by date Lesson Planning Articles Timely and inspiring teaching ideas that you can apply in your classroom About - [x] Our Story Frequently Asked Questions Testimonials Contact Us Pricing School Access Your school or district can sign up for Lesson Planet — with no cost to teachers Learn More Sign In Try It Free Hi, What do you want to do? Create a lesson plan Generate resources with AI teacher tools Search 2 million educational videos Find a teaching resource Publisher CK-12 Foundation Resource Details Curator Rating Educator Rating Grade 6th - 7th SubjectsMath1 more... Resource TypeUnit Plans AudiencesFor Administrator Use2 more... Lexile Measures0L Unit Plan Ck 12: Arithmetic: Absolute Value of Integers Curated by ACT [Free Registration/Login may be required to access all resource tools.] Find the absolute value of positive and negative integers. 3 Views 0 Downloads CCSS:Adaptable Concepts absolute value Show MoreShow Less Additional Tags number concepts, absolute value of integers, ck 12, ck-12, ck-12 foundation, ck12 Show MoreShow Less Classroom Considerations Knovation Readability Score: 1 (1 low difficulty, 5 high difficulty) Common Core 6.NS.C.7.c6.NS.C.7.d7.NS.A.1.c6.NS.C.7.a6.NS.C.7.b See similar resources: Lesson Plan #### Absolute Value—Magnitude and Distance EngageNY Do you want to use the resource? Absolutely. Scholars learn about absolute value and its relation to magnitude and distance on a number line. They compare numbers in context by applying absolute value. 6th Math CCSS:Designed Worksheet #### Absolute Value of Integers Curated OER In this absolute values instructional activity, students find the absolute values of positive and negative one digit numbers. Students complete 16 problems. 5th - 8th Math Worksheet #### Integers and Absolute Value - Homework 22.1 Curated OER Young math whizzes write the opposite of the each integer in the first twelve problems. They write the absolute value for each integer in the second set of problems and name a pair of integers with the same absolute value. 6th - 7th Math Instructional Video #### Absolute Value Inequalities, Linear Equations, Algebra I Khan Academy Combining the skills from the previous videos, particularly inequalities and absolute values, Sal works to solve several new sample problems. He encourages watching the video "over and over," which is good advice for mathematicians of... 13 mins 7th - 11th Math Instructional Video #### Absolute Value Krista King Math Investigate with your classes the meaning of absolute value. Here, the instructor explains the definition of absolute value and uses the definition to evaluate different expressions. The expressions include simplifying within the... 3 mins 3rd - 7th Math CCSS:Adaptable Unit Plan #### Ck 12: Arithmetic: Absolute Value of Integers CK-12 Foundation [Free Registration/Login may be required to access all resource tools.] Identify the absolute value and opposites of given integers, recognizing zero as neither positive nor negative 7th Math Instructional Video #### What are the Rules for Using Absolute Values to Add Integers? Curated OER There are several rules to adding integers using absolute values. It can seem a bit overwhelming to remember these rules. Watch this video and learn the rules as the instructor illustrates a few examples and explains the steps involved. 7 mins 6th - 8th Math Unit Plan #### Ck 12: Algebra: Absolute Value CK-12 Foundation [Free Registration/Login may be required to access all resource tools.] This concept introduces absolute value. Students can watch a video tutorial, explore guided notes and then attempt practice problems. 5th - 8th Math CCSS:Adaptable Interactive #### Absolute Values of Integers CK-12 Foundation A five-question interactive focuses on integers and absolute value. A number line shows the distance to aide mathematicians in answering multiple-choice, true or false, and discussion questions. 5th - 7th Math CCSS:Adaptable Lesson Plan #### Signed Number Operations and Absolute Value Curated OER Learners examine methods for adding, subtracting, multiplying and dividing signed numbers. The observe animated illustrations of each operation. students discover the basic meaning of absolute value. 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16640	https://www.khanacademy.org/math/ncert-class-11/xea4762213f311c5e:introduction-to-three-dimensional-geometry-ncert-new	Introduction to three dimensional geometry \| Class 11 \| Math \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. 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Class 11 14 units · 180 skillsUnit 1 SetsUnit 2 Relations and functionsUnit 3 Trigonometric functionsUnit 4 Complex numbers and quadratic equationsUnit 5 Linear inequaltiesUnit 6 Permutations and combinationsUnit 7 Binomial theoremUnit 8 Sequences and seriesUnit 9 Straight linesUnit 10 Conic sectionsUnit 11 Introduction to three dimensional geometryUnit 12 Limits and derivativesUnit 13 StatisticsUnit 14 Probability Course challenge Test your knowledge of the skills in this course.Start Course challenge Math Class 11 Unit 11: Introduction to three dimensional geometry 300 possible mastery points Mastered Proficient Familiar Attempted Not started Quiz Unit test Coordinates of a point in space Distance between two points Distance between points: applications Introduction to three dimensional geometry: Unit test Coordinates of a point in space This lesson covers skills from the following lessons of the NCERT Math Textbook: (i) 11.3- Coordinates of a point in space Learn Introduction to 3D (Opens a modal) Practice Up next for you:Coordinates of a point in spaceGet 3 of 4 questions to level up! 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16641	https://kttmc.weebly.com/uploads/2/1/2/7/21275598/6.6b_markup_discount_and_tax_cw.pdf	Lesson Plan -- Discounts/Markups Chapter Resources - Lesson 4-13 Discounts, Tips, Markups, Commissions, and Profit - Lesson 4-13 Discounts, Tips, Markups, Commissions, and Profit Answers 1 Copyright © by McDougal Littell, a division of Houghton Mifﬂ in Company. 50 Math Intervention Book 4 Ratios, Rates, Proportions, and Percents Name ——————————————————————— Date — — — — — — — — — — — — E@)584- Finding a Sale Price The original price of a shirt is $18. The shirt is on sale for a 20% discount. What is the sale price of the shirt? Solution Discount 5 20% of $18 Find the amount of discount. 5 0.20(18) 5 3.60 Sale price 5 original price 2 discount Use formula to ﬁ nd sale price. 5 18 2 3.60 5 14.40 ANSWER The sale price of the shirt is $14.40. T:A<01; Find the sale price. 1. Original price: $40 2. Original price: $75 Percent discount: 30% Percent discount: 25% ,Q[KW]V[<QX[5IZS]X[ +WUUQ[[QWV[IVL8ZWÅ \ Getting Started In Lesson 4-12 you learned how to ﬁ nd percent of change, percent of increase, and percent of decrease. In this lesson, you apply these skills to solve problems. Words to Remember Discount: A percent or amount of decrease in the price of an item Sale price 5 original price 2 discount Tip: A percent earned by a person for providing a service Wholesale price: The price a retail store pays the manufacturer Markup: A percent or amount increase a retail store charges to earn a proﬁ t Retail price: The price at which a store sells items to customers Retail price 5 wholesale price 1 markup Commission: A percent of total sales earned by an employee or business Commission earned 5 percent commission 3 total sales Proﬁ t: The difference of the income and the expenses for a business Proﬁ t 5 income 2 expenses LESSON 4-13 Gr. 6 NS 1.4: Calculate given percentages of quantities and solve problems involving discounts at sales, interest earned, and tips. Gr. 7 NS 1.7: Solve problems that involve discounts, markups, commissions, and proﬁ t and compute simple and compound interest. Also included: Gr. 7 NS 1.3 California Standards Copyright © by McDougal Littell, a division of Houghton Mifﬂ in Company. 51 Math Intervention Book 4 Ratios, Rates, Proportions, and Percents Name ——————————————————————— Date — — — — — — — — — — — — E@)584- Finding a Total Cost Your food bill at a restaurant was $23.50. You leave a 15% tip. What is the total cost of your meal? Solution Tip 5 15% of $23.50 Find the amount of tip. 5 0.15(23.50) 5 3.53 Total cost 5 23.50 1 3.53 Add the tip to the food bill. 5 27.03 ANSWER The total cost of your meal is $27.03. E@)584- Finding a Commission Anthony receives a 9% commission on magazine subscription sales. Last week, his total sales were $1800. How much commission did he earn? Solution Commission 5 9% of $1800 Find the amount of commission. 5 0.09(1800) 5 162 ANSWER Anthony earned $162 in commission last week. E@)584- Finding a Proﬁ t You sell fruit drinks at the town ﬁ eld. This past weekend, you collected $107. Your expenses for the fruit drinks were $82. What is your proﬁ t? Solution Proﬁ t 5 income 2 expenses Use formula to ﬁ nd proﬁ t. 5 107 2 82 5 25 ANSWER Your proﬁ t for last weekend is $25. T:A<01; Find the indicated value. 3. Find the retail price. 4. Find the commission. Wholesale price: $50 Total sales: $1600 Percent markup: 30% Percent commission: 6% You ﬁ nd total meal costs and retail prices in the same way. Find the tip or markup amount and add this amount to the food bill or wholesale price. /PUJDF Copyright © by McDougal Littell, a division of Houghton Mifﬂ in Company. 52 Math Intervention Book 4 Ratios, Rates, Proportions, and Percents Name ——————————————————————— Date — — — — — — — — — — — — Summarize Finding a Sale Price after a Discount Find the amount of discount. Subtract the discount amount from the original price. Finding a Final Cost after a Tip or Markup Find the amount of tip or markup. Add the tip or markup to the original amount. Finding a Commission Multiply total sales by the percent commission. Finding a Proﬁ t Subtract expenses from income. Tell whether you would add or subtract to ﬁ nd the answer. 1. A store bought $150 worth of items at wholesale price. The store will markup the items by 120%. You want to ﬁ nd the retail price. __ 2. The bill at a restaurant is $45. You tip 17%. You want to ﬁ nd the total cost of the meal. _ 3. A movie DVD has an original price of $20. The store is offering a 10% discount on movies. You want to ﬁ nd the sale price. ___ Find the sale price, retail price, or total meal cost. 4. Original price: $20 5. Original price: $80 Discount: 30% Discount: 10% 6. Wholesale price: $30 7. Wholesale price: $85 Percent markup: 120% Percent markup: 20% 8. Food bill: $55 9. Food bill: $38.40 Tip: 18% Tip: 15% Find the commission. 10. Total sales: $2000 11. Total sales: $1200 Percent commission: 10% Percent commission: 5% Find the proﬁ t. 12. Expenses: $275 13. Expenses: $700 Income: $495 Income: $1050 Practice Copyright © by McDougal Littell, a division of Houghton Mifﬂ in Company. 53 Math Intervention Book 4 Ratios, Rates, Proportions, and Percents Name ——————————————————————— Date — — — — — — — — — — — — 18. Fill in the missing words. To ﬁ nd the total meal cost, you would _ the amount of the__ to the food _. 19. Find the sale price. Original price: $30, Discount: 15% ___________ __________ 20. Find the retail price. A store bought a statue for a $125 wholesale price. The store charges an 85% markup. What is the retail price of the statue? ___________ __________ ,1,A7=/-<1<' Solve the problem. Explain your answer. 14. A craft store bought paint for a wholesale price of $100. The store charges a 65% markup. What is the craft store’s price for the paint? _______ ______ _______ 15. Sylvano wants to buy a pair of gym shoes that cost $75. The shoes are on sale at a 30% discount. How much will Sylvano spend on the shoes? ___________ ____________ ___________ 16. Howard sold $3500 worth of jewelry this past week. If he earns an 8% commission, how much commission did he earn? ____________ ___________ ____________ 17. A school group collects $194.25 at a fund-raiser. The group spent $53.75 on supplies. How much proﬁ t did they earn? ___________ ____________ ____________ Answer Key Lesson 4-13, pp. 50–53 Try this: 1. $28.00 2. $56.25 3. $65.00 4. $96.00 Practice: 1. add 2. add 3. subtract 4. $14.00 5. $72.00 6. $66.00 7. $102.00 8. $64.90 9. $44.16 10. $200.00 11. $60.00 12. $220.00 13. $350.00 14. $165.00; Sample answer: $100 1 ($100 3 0.65) 5 $165 15. $52.50; Sample answer: $75 2 ($75 3 0.30) 5 $52.50 16. $280.00; Sample answer: $3500 3 0.08 5 $280.00 17. $140.50; Sample answer: $194.25 2 $53.75 5 $140.50 18. add; tip; bill 19. $25.50 20. $231.25
16642	https://www.jstor.org/stable/10.4169/math.mag.85.4.304	News and Letters on JSTOR Free online reading for over 10 million articles Save and organize content with Workspace Link account to institutional access Continue with Google Continue with Microsoft Find my institution or Username or email address Password SHOW Stay logged in Forgot password? Log in Don't have an account? Register for free Skip to main content Have library access? Log in through your library Get support Help logging in Contact us Log in Register Workspace Advanced Search By subject By title Publishers Collections Images Get support Help logging in Contact us All Content Images Advanced Search Search all content Register Log in Browse By subject Journals and books By title Journals and books Publishers Collections Images Workspace Your Artstor image groups were copied to Workspace. The Artstor website will be retired on Aug 1st. journal article News and Letters Mathematics Magazine Vol. 85, No. 4 (October 2012), pp. 304-320 (17 pages) Published By: Taylor & Francis, Ltd. Cite This is a preview. Log in through your library . Journal Information Mathematics Magazine presents articles and notes on undergraduate mathematical topics in a lively expository style that appeals to students and faculty throughout the undergraduate years. The journal originally began in 1926 as a series of pamphlets to encourage membership in the Louisiana-Mississipi Section of the Mathematical Association of America, and soon evolved into the regional publication Mathematics News Letter. Beginning in 1935, the journal was published with the help of Louisiana State University and, as it began addressing larger issues in teaching math, was renamed National Mathematics Magazine. In 1947, the journal's title was shortened to Mathematics Magazine, and in 1960 it became an official publication of the Mathematical Association of America. Mathematics Magazine is published five times per year. Publisher Information Building on two centuries' experience, Taylor & Francis has grown rapidly over the last two decades to become a leading international academic publisher. The Group publishes over 800 journals and over 1,800 new books each year, covering a wide variety of subject areas and incorporating the journal imprints of Routledge, Carfax, Spon Press, Psychology Press, Martin Dunitz, and Taylor & Francis.Taylor & Francis is fully committed to the publication and dissemination of scholarly information of the highest quality, and today this remains the primary goal. Rights & Usage This item is part of a JSTOR Collection. 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16643	https://www.sciencedirect.com/science/article/pii/S0012365X12001343	Skip to article My account Sign in View PDF Discrete Mathematics Volume 312, Issues 1213, 6 July 2012, Pages 2027-2039 On coloring problems with local constraints Author links open overlay panel, , rights and content Under an Elsevier user license Open archive Abstract We deal with some generalizations of the graph coloring problem on classes of perfect graphs. Namely we consider the -coloring problem (upper bounds for the color on each vertex), the precoloring extension problem (a subset of vertices colored beforehand), and a problem generalizing both of them, the -coloring problem (lower and upper bounds for the color on each vertex). We characterize the complexity of all those problems on clique-trees of different heights, providing polynomial-time algorithms for the cases that are easy. These results have interesting corollaries. First, one can observe on clique-trees of different heights the increasing complexity of the chain -coloring, -coloring, -coloring, and list-coloring. Second, clique-trees of height 2 are the first known example of a class of graphs where -coloring is polynomial-time solvable and precoloring extension is NP-complete, thus being at the same time the first example where -coloring is polynomially solvable and -coloring is NP-complete. Last, we show that the-coloring problem on unit interval graphs is NP-complete. These results answer three questions from Bonomo et al. [F. Bonomo, G. DurÃ¡n, J. Marenco, Exploring the complexity boundary between coloring and list-coloring, Annals of Operations Research 169 (1) (2009) 316]. Keywords Graph coloring Clique-trees Unit interval graphs Computational complexity Cited by (0) Copyright © 2012 Elsevier B.V. All rights reserved.
16644	https://mathoverflow.net/questions/362647/conceptual-explanation-of-geometric-mean-as-a-limit-of-power-means	real analysis - Conceptual explanation of geometric mean as a limit of power means - MathOverflow Join MathOverflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Conceptual explanation of geometric mean as a limit of power means Ask Question Asked 5 years, 3 months ago Modified5 years, 3 months ago Viewed 1k times This question shows research effort; it is useful and clear 12 Save this question. Show activity on this post. Let x 1,…,x n x 1,…,x n be positive real numbers and p∈R−{0}p∈R−{0}. The power mean M p(x 1,…,x n)M p(x 1,…,x n) of exponent p p is defined by M p(x 1,…,x n)=(1 n∑i=1 n x p i)1/p.M p(x 1,…,x n)=(1 n∑i=1 n x i p)1/p. By taking logarithms and applying L'Hôpital's rule (or just the definition of a derivative), we get lim p→0 M p(x 1,…,x n)=x 1⋯x n−−−−−−−√n,lim p→0 M p(x 1,…,x n)=x 1⋯x n n, the geometric mean of x 1,…,x n x 1,…,x n. Thus the "correct" definition of M 0 M 0 is M 0(x 1,…,x n)=x 1⋯x n−−−−−−−√n M 0(x 1,…,x n)=x 1⋯x n n. All this is well known, but I am wondering if there is some conceptual explanation, not involving computation, for the value of M 0(x 1,…,x n)M 0(x 1,…,x n). real-analysis Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications asked Jun 9, 2020 at 20:52 Richard StanleyRichard Stanley 53.5k 15 15 gold badges 163 163 silver badges 298 298 bronze badges 4 2 Small, not really conceptual comment: suppose we allow some x i x i to be 0 0 as well; then lim p→∞M p lim p→∞M p is totally insensitive to having a single value of x i x i equal to zero; whereas the ``opposite'' limit lim p→0 M p lim p→0 M p is "totally sensitive" to a single zero value.Sam Hopkins –Sam Hopkins♦ 2020-06-09 21:51:03 +00:00 Commented Jun 9, 2020 at 21:51 Unrelated to your nice post (if it is required I can delete this comment), is that I tried in recent months and weeks write variations of certain unsolved problems using combinations with certain means, for example the more recent is the post with title On variations of a claim due to Kaneko in terms of Lehmer means of Mathematics Stack Exchange.user142929 –user142929 2020-06-11 19:01:22 +00:00 Commented Jun 11, 2020 at 19:01 3 I mean, we can define M p M p as the unique solution μ μ to ∑n i=1∫μ x i y p−1 d y=0∑i=1 n∫x i μ y p−1 d y=0 and then for p=0 p=0 we get the geometric mean.Will Sawin –Will Sawin 2020-06-11 19:25:54 +00:00 Commented Jun 11, 2020 at 19:25 In case no one has thought of this, a trivial observation that lets you guess the value of M 0 M 0 immediately with no calculation is that as p p approaches 0 0 the x p i x i p's get closer and closer to 1 and the inequality M p≥G M M p≥G M moves towards its condition of equality (from AM-GM applied to the x p i x i p's).Ivan Meir –Ivan Meir 2020-06-13 07:50:07 +00:00 Commented Jun 13, 2020 at 7:50 Add a comment\| 4 Answers 4 Sorted by: Reset to default This answer is useful 6 Save this answer. Show activity on this post. (a 1/n+b 1/n 2)n(a 1/n+b 1/n 2)n has normalized symmetric binomial (i.e., close to normal) distribution of coefficients convolved with some bounded expressions in a a and b b (of the form a θ b 1−θ a θ b 1−θ). The normal distribution is sharply peaked around its mean so you end up with the middle terms with θ=1/2 θ=1/2 in the limit of large n n. This generalizes to more variables and weights. That is still a calculation, but one where the answer is easy to foresee at the outset and the idea can be explained in a few words. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jun 9, 2020 at 23:04 Trivial NotionsTrivial Notions 61 1 1 bronze badge Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. A reasonably intuitive way to "see" that the limit must be the geometric mean is the plausible and useful observation that any power mean can be expressed in terms of the midpoint means, M p(x,y)=((x p+y p)/2)1/p M p(x,y)=((x p+y p)/2)1/p, recursively if the number of variables is not a power of 2. See my answer to a related question. Simple algebra then proves that for n=2 n=2, and all p≠0 p≠0, M p M−p=x 1 x 2 M p M−p=x 1 x 2 so letting p→0 p→0 we have M 2 0=x 1 x 2 M 0 2=x 1 x 2 and the general limit value follows immediately. Note that this doesn't use calculus, transcendental functions, or in fact anything other than the power means themselves and their continuity. Update: A further low tech way to see the result uses functional equations. We simply note that the power means satisfy M r p(x 1,⋯,x n)=M p(x r 1,⋯,x r n)1/r.M r p(x 1,⋯,x n)=M p(x 1 r,⋯,x n r)1/r. Setting p=0 p=0 gives M 0(x 1,⋯,x n)=M 0(x r 1,⋯,x r n)1/r M 0(x 1,⋯,x n)=M 0(x 1 r,⋯,x n r)1/r for all r∈R−{0}.r∈R−{0}. Then it is intuitively clear since M M is symmetric and M p(λ x)=λ M p(x)M p(λ x)=λ M p(x) that M 0 M 0 must be the geometric mean. You can prove this formally by induction starting with n=2 n=2. Let f(x)=M 0(x,1)=f(x r)1/r f(x)=M 0(x,1)=f(x r)1/r, by the above. Then setting x=e x=e, r=log X r=log⁡X we have f(X)=f(e)log X=X log f(e)=x μ f(X)=f(e)log⁡X=X log⁡f(e)=x μ where μ μ is constant. Hence M 0(x,y)=y M 0(x/y,1)=y f(x/y)=y(x/y)μ=y 1−μ x μ M 0(x,y)=y M 0(x/y,1)=y f(x/y)=y(x/y)μ=y 1−μ x μ. SInce M 0 M 0 is symmetric in x x and y y we have μ=1/2 μ=1/2 and M 0(x,y)=x 1/2 y 1/2 M 0(x,y)=x 1/2 y 1/2. The other cases n>2 n>2 follow in a similar manner. Further update: Actually probably the most intuitive way is just to note, as Iosif did as well, that AM-GM or Jensen's inequality tells you M p≥G M≥M−p M p≥G M≥M−p. Then just take the limit as p→0 p→0. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Jun 12, 2020 at 15:44 answered Jun 10, 2020 at 22:28 Ivan MeirIvan Meir 4,952 3 3 gold badges 33 33 silver badges 43 43 bronze badges 2 Concerning your Update: When you say "Setting p=0 p=0", you set p=0 p=0 in an identity which you assume to hold only for p≠0 p≠0, right? Concerning your Further update: Did you notice that using the AGM/Jensen inequality is what had been done in my answer? After that, the main problem was to "just take the limit as p→0 p→0". Well, that was the problem stated in the OP from the very beginning.Iosif Pinelis –Iosif Pinelis 2020-06-11 23:57:04 +00:00 Commented Jun 11, 2020 at 23:57 @IosifPinelis In my answers I was more concerned with explaining the value since that was requested by the OP - "I am wondering if there is some conceptual explanation...for the value of M 0 M 0" So when I say "setting p=0 p=0" I am assuming the limit exists. Yes I did notice after I added my further update that you had used the AGM inequalities - again I assumed the limit exists.Ivan Meir –Ivan Meir 2020-06-12 01:05:35 +00:00 Commented Jun 12, 2020 at 1:05 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. This proof uses only the arithmetic-geometric mean (AGM) inequality and the fact that for any smooth even function g:R→R g:R→R we have g′(0)=0 g′(0)=0. To simplify the writing, for any function f:R→R f:R→R let f(x)¯¯¯¯¯¯¯¯¯¯:=1 n∑1 n f(x i).f(x)¯:=1 n∑1 n f(x i). We have to show that M p:=(x p¯¯¯¯¯)1/p→M 0:=exp ln x¯¯¯¯¯¯¯¯M p:=(x p¯)1/p→M 0:=exp ln⁡x¯ as p→0 p→0. Take any real p>0 p>0. Replacing the x i x i's in the AGM inequality x¯¯¯≥exp ln x¯¯¯¯¯¯¯¯(1)(1)x¯≥exp ln⁡x¯ by the x p i x i p's, we have M p≥M 0 M p≥M 0. Similarly, replacing the x i x i's in (1) by the x−p i x i−p's, we have M−p≤M 0 M−p≤M 0. So, M−p≤M 0≤M p.M−p≤M 0≤M p. It remains to show that M p/M−p→1 M p/M−p→1 as p↓0 p↓0 or, equivalently, that g(p):=ln x p¯¯¯¯¯+ln x−p¯¯¯¯¯¯¯¯=o(p),g(p):=ln⁡x p¯+ln⁡x−p¯=o(p), which follows because the function g g is smooth and even, with g(0)=0 g(0)=0. □◻ The expression exp ln x¯¯¯¯¯¯¯¯[=(x 1⋯x n)1/n]exp ln⁡x¯[=(x 1⋯x n)1/n] for the geometric mean arises naturally, as an instance, with f=exp f=exp, of the more general mean of the form f(f−1(x)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯)f(f−1(x)¯) with a continuous increasing function f f. So, the geometric mean is just a logarithmically/exponentially re-scaled version of the arithmetic mean. Also, the AGM inequality (1) is an instance of Jensen's inequality for the concave function ln ln or, equivalently, for the the convex function exp exp. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Jun 11, 2020 at 18:22 answered Jun 11, 2020 at 3:50 Iosif PinelisIosif Pinelis 141k 9 9 gold badges 120 120 silver badges 251 251 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. I am not sure if the following is a conceptual explanation rather than a calculation: Writing x i=e u i x i=e u i and letting p→0 p→0, we have M p=(1 n∑1 n e p u i)1/p=(1+p n+o(1)∑1 n u i)1/p→exp(1 n∑1 n u i)=M 0,M p=(1 n∑1 n e p u i)1/p=(1+p n+o(1)∑1 n u i)1/p→exp⁡(1 n∑1 n u i)=M 0, where M r:=M r(x 1,…,x n)M r:=M r(x 1,…,x n). (I guess in any case we need to show that M p→M 0 M p→M 0 as p→0 p→0. Here, at least we do not explicitly use the l'Hospital rule or differentiation.) Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jun 10, 2020 at 2:07 Iosif PinelisIosif Pinelis 141k 9 9 gold badges 120 120 silver badges 251 251 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions real-analysis See similar questions with these tags. Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure Linked 7Does midpoint-convex imply rationally convex? 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16645	https://www.sohu.com/a/220566654_667279	二项分布与超几何分布的联系与区别高考数学MOOK 2017 VOL.49 秦晓燕 ▼ 经常有学生问二项分布与超几何分布到底怎么区分，是利用二项分布的公式去解决这道概率题目，还是利用超几何分布公式解决呢? 好多学生查阅参考书寻找答案，其实这个问题的回答就出现在教材上, 人教版新课标教材选修2-3从两个方面给出了很好的解释. 众里寻它千百度，蓦然回首，却在灯火阑珊处一两者的定义是不同的教材中的定义： 1 超几何分布 2 独立重复试验与二项分布本质区别： (1) 超几何分布描述的是不放回抽样问题，而二项分布描述的是放回抽样问题. (2) 超几何分布中的概率计算实质上是古典概型问题；二项分布中的概率计算实质上是相互独立事件的概率问题. 二两者之间是有联系的从以上分析可以看出两者之间的联系：当调查研究的样本容量非常大时，在有放回地抽取与无放回地抽取条件下，计算得到的概率非常接近，可以近似把超几何分布认为是二项分布. 下面看相关例题例1．（2016·漯河模拟）寒假期间，我市某校学生会组织部分同学，用“10分制”随机调查“阳光花园”社区人们的幸福度．现从调查人群中随机抽取16名，如图所示的茎叶图记录了他们的幸福度分数(以小数点前的一位数字为茎，小数点后的一位数字为叶)，若幸福度分数不低于8.5分，则称该人的幸福度为“幸福”. 注：先不要急于看答案，大家先自己解一下这道题再往下看，会有意想不到的收获哦！ [错解分析]第二问的选人问题是不放回抽样问题, 按照定义先考虑超几何分布,但是题目中又明确给出：“以这16人的样本数据来估计整个社区的总体数据，从该社区(人数很多)任选3人”,说明不是从16人中任选3人,而是从该社区(人数很多)任选3人,所以可以近似看作是3次独立重复试验,应该按照二项分布去求解，而不能按照超几何分布去处理. 柳暗花明又一村从以上解题过程中我们还发现，错解中的期望值与正解中的期望值相等，好多学生都觉得不可思议，怎么会出现相同的结果呢？其实这还是由于前面解释过的原因，超几何分布与二项分布是有联系的，看它们的期望公式：总结综上可知，当提问中涉及“用样本数据来估计总体数据”字样的为二项分布。高考解题中，我们还是要分清超几何分布与二项分布的区别，以便能正确的解题，拿到满分！寒假快乐乐乐返回搜狐，查看更多
16646	https://euro-fusion.org/fusion/fusion-on-the-sun/	Skip to content Home » What is Fusion? » Fusion on the Sun Fusion on the Sun Source: NASA Goddard Space Flight Center, CC BY 2.0, via Wikimedia Commons At temperatures of 15 million degrees Celcius in the Sun’s core, hydrogen gas becomes plasma, the fourth state of matter. In a plasma, the negatively charged electrons in atoms are completely separated from the positively charged atomic nuclei (or ions). The Sun’s gravitational force confines the positively-charged hydrogen nuclei and the high temperatures cause the nuclei to move around furiously. As a result they collide at high speeds overcoming the natural electrostatic repulsion that exists between the positive charges and subsequently fuse to form the heavier helium. From hydrogen to helium in three steps In the first stage two protons combine and one of them converts into a neutron to form a nucleus of the heavy isotope of hydrogen known as deuterium. Next, the deuterium nucleus combines with another proton to form the light helium isotope known as helium-3. Finally, two helium-3 nuclei combine to form helium-4, releasing two protons. Energy release Overall, four protons are converted into one helium nucleus. Energy is released because the helium nucleus has slightly less mass than the original four protons. The total amount of energy released for each conversion of four hydrogen nuclei into a helium nucleus is about 10 million times more than is produced by the chemical reaction when hydrogen combines with oxygen to form water. Proton-proton chain reaction Did you know... In its core, the Sun fuses 620 million metric tons of hydrogen each second. Next chapter: Fusion on Earth Privacy Preference We need your consent before you can continue on our website. If you are under 16 and wish to give consent to optional services, you must ask your legal guardians for permission. We use cookies and other technologies on our website. Some of them are essential, while others help us to improve this website and your experience. Personal data may be processed (e.g. IP addresses), for example for personalized ads and content or ad and content measurement. You can find more information about the use of your data in our privacy policy. There is no obligation to consent to the processing of your data in order to use this offer. You can revoke or adjust your selection at any time under Settings. Please note that based on individual settings not all functions of the site may be available. Some services process personal data in the USA. With your consent to use these services, you also consent to the processing of your data in the USA pursuant to Art. 49 (1) lit. a GDPR. The ECJ classifies the USA as a country with insufficient data protection according to EU standards. For example, there is a risk that U.S. authorities will process personal data in surveillance programs without any existing possibility of legal action for Europeans. You need to load content from reCAPTCHA to submit the form. Please note that doing so will share data with third-party providers. More Information Unblock content Accept required service and unblock content
16647	https://inspirehep.net/literature/1625826	Bayes Factors - INSPIRE literature Help Submit Login Literature Authors Jobs Seminars Conferences DataBETA More... Bayes Factors Robert E. Kass( Pittsburgh U. ) , Adrian E. Raftery( Washington U., Seattle ) 1995 23 pages Published in: J.Am.Statist.Assoc. 90 (1995) 430, 773-795 DOI: 10.1080/01621459.1995.10476572 View in: AMS MathSciNet linkscite claim reference search920 citations Citations per year Abstract: In a 1935 paper and in his book Theory of Probability, Jeffreys developed a methodology for quantifying the evidence in favor of a scientific theory. The centerpiece was a number, now called the Bayes factor, which is the posterior odds of the null hypothesis when the prior probability on the null is one-half. Although there has been much discussion of Bayesian hypothesis testing in the context of criticism of P-values, less attention has been given to the Bayes factor as a practical tool of applied statistics. In this article we review and discuss the uses of Bayes factors in the context of five scientific applications in genetics, sports, ecology, sociology, and psychology. We emphasize the following points: • From Jeffreys' Bayesian viewpoint, the purpose of hypothesis testing is to evaluate the evidence in favor of a scientific theory. • Bayes factors offer a way of evaluating evidence in favor of a null hypothesis. • Bayes factors provide a way of incorporating external information into the evaluation of evidence about a hypothesis. • Bayes factors are very general and do not require alternative models to be nested. • Several techniques are available for computing Bayes factors, including asymptotic approximations that are easy to compute using the output from standard packages that maximize likelihoods. • In “nonstandard” statistical models that do not satisfy common regularity conditions, it can be technically simpler to calculate Bayes factors than to derive non-Bayesian significance tests. • The Schwarz criterion (or BIC) gives a rough approximation to the logarithm of the Bayes factor, which is easy to use and does not require evaluation of prior distributions. • When one is interested in estimation or prediction, Bayes factors may be converted to weights to be attached to various models so that a composite estimate or prediction may be obtained that takes account of structural or model uncertainty. • Algorithms have been proposed that allow model uncertainty to be taken into account when the class of models initially considered is very large. • Bayes factors are useful for guiding an evolutionary model-building process. • It is important, and feasible, to assess the sensitivity of conclusions to the prior distributions used. References(0) Figures(0) 0 References INSPIRE About INSPIRE Content Policy Privacy Notice Terms of Use Help FAQ INSPIRE Help Report metadata issues Tools Bibliography generator Community Blog Bluesky X Contact Powered by Invenio Made with ❤ by the INSPIRE Team
16648	https://en.wikipedia.org/wiki/Chinese_characters	Published Time: 2002-09-25T13:29:42Z Chinese characters - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 Development 2 ClassificationToggle Classification subsection 2.1 Structural analysis 2.2 Semantographs 2.2.1 Pictographs 2.2.2 Indicatives 2.2.3 Compound ideographs 2.3 Phonographs 2.3.1 Phono-semantic compounds 2.4 Loangraphs 2.5 Signs 2.6 Traditional Shuowen Jiezi classification 3 HistoryToggle History subsection 3.1 Traditional invention narrative 3.2 Neolithic precursors 3.3 Oracle bone script 3.4 Zhou scripts 3.5 Qin unification and small seal script 3.6 Clerical script 3.7 Cursive and semi-cursive 3.8 Regular script 4 StructureToggle Structure subsection 4.1 Variant characters 4.2 Layout 5 Methods of writingToggle Methods of writing subsection 5.1 Calligraphy 5.2 Printing and typefaces 5.3 Use with computers 5.3.1 Input methods 5.3.2 Encoding and interchange 6 Vocabulary and adaptationToggle Vocabulary and adaptation subsection 6.1 Literary and vernacular Chinese 6.2 Japanese 6.3 Korean 6.4 Vietnamese 6.5 Other languages 6.6 Graphically derived scripts 7 Literacy and lexicographyToggle Literacy and lexicography subsection 7.1 Dictionaries 7.2 Neurolinguistics 8 Reform and standardizationToggle Reform and standardization subsection 8.1 People's Republic of China 8.2 Japan 8.3 South Korea 8.4 North Korea 8.5 Taiwan 8.6 Other regional standards 9 Notes 10 ReferencesToggle References subsection 10.1 Citations 10.2 Works cited 10.3 Primary and media sources 11 Further reading 12 External links Chinese characters [x] 138 languages Afrikaans Alemannisch አማርኛ العربية Aragonés অসমীয়া Asturianu Avañe'ẽ Azərbaycanca تۆرکجه বাংলা Banjar 閩南語 / Bân-lâm-gí Башҡортса Беларуская Беларуская (тарашкевіца) Български Boarisch བོད་ཡིག Bosanski Brezhoneg Буряад Català Cebuano Čeština ChiShona Cymraeg Dansk Deutsch Eesti Ελληνικά Español Esperanto Euskara فارسی Fiji Hindi Føroyskt Français Frysk Gaelg Galego 贛語客家語 / Hak-kâ-ngî 한국어 Հայերեն हिन्दी Hrvatski Ilokano Bahasa Indonesia Íslenska Italiano עברית Jawa Къарачай-малкъар ქართული Қазақша Kernowek Kiswahili Kriyòl gwiyannen Кыргызча Ladin ລາວ Latina Latviešu Lietuvių La .lojban. 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For the philosopher, see Han Fei. For the anthology attributed to him, see Han Feizi. For the moth species, see Cilix glaucata. \| Chinese characters \| \| "Chinese character" written in traditional (left) and simplified (right) forms \| \| Script type \| Logographic \| \| Period \| c. 13th century BCE– present \| \| Direction \| Left-to-right Top-to-bottom, columns right-to-left \| \| Languages \| Chinese Japanese Korean Vietnamese Zhuang (among others) \| \| Related scripts \| \| Parent systems \| (Proto-writing) Chinese characters \| \| Child systems \| Bopomofo Jurchen script Kana Khitan small script Nüshu Tangut script Yi script \| \| ISO 15924 \| \| ISO 15924 \| Hani(500), Han (Hanzi, Kanji, Hanja) \| \| Unicode \| \| Unicode alias \| Han \| \| Unicode range \| U+4E00–U+9FFFCJK Unified Ideographs (full list) \| This article contains chữ Nôm characters used to write Vietnamese, as well as sawndip characters used to write Zhuang. Without proper rendering support, you may see question marks, boxes, or other symbols. \| Chinese characters \| \| \| \| Chinese name \| \| Simplified Chinese \| 汉字 \| \| Traditional Chinese \| 漢字 \| \| Literal meaning \| Han characters \| \| \| show Transcriptions \| \| Standard Mandarin \| \| Hanyu Pinyin \| Hànzì \| \| Bopomofo \| ㄏㄢˋ ㄗˋ \| \| Gwoyeu Romatzyh \| Hanntzyh \| \| Wade–Giles \| Han 4-tzu 4 \| \| Tongyong Pinyin \| Hàn-zìh \| \| IPA \| [xân.tsɹ̩̂] \| \| Wu \| \| Romanization \| 5 Hoe-zy \| \| Gan \| \| Romanization \| Hon 5-ci 5 \| \| Hakka \| \| Romanization \| Hon 55 sii 55 \| \| Yue: Cantonese \| \| Yale Romanization \| Hon jih \| \| Jyutping \| Hon3 zi6 \| \| IPA \| [hɔn˧tsi˨] \| \| Southern Min \| \| HokkienPOJ \| Hàn-jī \| \| Tâi-lô \| Hàn-jī \| \| TeochewPeng'im \| Hang 3 ri 7 \| \| Eastern Min \| \| FuzhouBUC \| Háng-cê \| \| Middle Chinese \| \| Middle Chinese \| xan H dzi H \| \| \| Japanese name \| \| Kanji \| 漢字 \| \| \| show Transcriptions \| \| Revised Hepburn \| kanji \| \| Kunrei-shiki \| kanzi \| \| \| Korean name \| \| Hangul \| 한자 \| \| Hanja \| 漢字 \| \| \| show Transcriptions \| \| Revised Romanization \| Hanja \| \| McCune–Reischauer \| Hancha \| \| \| Vietnamese name \| \| Vietnamese alphabet \| chữ Hán chữ Nho Hán tự \| \| Hán-Nôm \| 𡨸漢𡨸儒 \| \| Chữ Hán \| 漢字 \| \| Zhuang name \| \| Zhuang \| sawgun \| \| Sawndip \| 𭨡倱 \| \| \| \| Chinese characters \| \| \| \| Chinese family of scripts Written Chinese Kanji Hanja Chữ Hán \| \| show Evolution of script styles Neolithic symbols in China Oracle bone Bronze Seal Large Small Bird-worm Clerical Cursive Semi-cursive Regular Flat brush Modern typefaces Fangsong Ming Hei \| \| show Properties and classification Components Strokes order Radicals Orthography jiu zixing xin zixing Digital encoding \| \| show Collation and standards Kangxi Dictionary forms (1716) Commonly Used Characters (PRC,2013) Commonly-Used Characters (Hong Kong,2007) Nan Min Recommended Characters (Taiwan,2009) Standard Form of National Characters (Taiwan,1982) Jōyō kanji (Japan,2010) \| \| show Reforms Simplified characters second round Traditional characters debate Japanese script reform kyūjitai \| \| show Homographs and readings Literary and colloquial readings Kanbun Idu \| \| show Variants Zetian characters \| \| show Derived systems Kana man'yōgana hiragana katakana Jurchen script Khitan large small Nüshu Bopomofo Slavonic transcription \| \| Transliteration of Chinese \| \| v t e \| Chinese characters[a] are logographs used to write the Chinese languages and others from regions historically influenced by Chinese culture. Of the four independently invented writing systems accepted by scholars, they represent the only one that has remained in continuous use. Over a documented history spanning more than three millennia, the function, style, and means of writing characters have changed greatly. Unlike letters in alphabets that reflect the sounds of speech, Chinese characters generally represent morphemes, the units of meaning in a language. Writing all of the frequently used vocabulary in a language requires roughly 2000–3000 characters; as of 2024[update], nearly 100 000 have been identified and included in The Unicode Standard. Characters are created according to several principles, where aspects of shape and pronunciation may be used to indicate the character's meaning. The first attested characters are oracle bone inscriptions made during the 13th century BCE in what is now Anyang, Henan, as part of divinations conducted by the Shang dynasty royal house. Character forms were originally ideographic or pictographic in style, but evolved as writing spread across China. Numerous attempts have been made to reform the script, including the promotion of small seal script by the Qin dynasty (221–206 BCE). Clerical script, which had matured by the early Han dynasty (202 BCE– 220 CE), abstracted the forms of characters—obscuring their pictographic origins in favour of making them easier to write. Following the Han, regular script emerged as the result of cursive influence on clerical script, and has been the primary style used for characters since. Informed by a long tradition of lexicography, states using Chinese characters have standardized their forms—broadly, simplified characters are used to write Chinese in mainland China, Singapore, and Malaysia, while traditional characters are used in Taiwan, Hong Kong, and Macau. Where the use of characters spread beyond China, they were initially used to write Literary Chinese; they were then often adapted to write local languages spoken throughout the Sinosphere. In Japanese, Korean, and Vietnamese, Chinese characters are known as kanji, hanja, and chữ Hán respectively. Writing traditions also emerged for some of the other languages of China, like the sawndip script used to write the Zhuang languages of Guangxi. Each of these written vernaculars used existing characters to write the language's native vocabulary, as well as the loanwords it borrowed from Chinese. In addition, each invented characters for local use. In written Korean and Vietnamese, Chinese characters have largely been replaced with alphabets—leaving Japanese as the only major non-Chinese language still written using them, alongside the other elements of the Japanese writing system. At the most basic level, characters are composed of strokes that are written in a fixed order. Historically, methods of writing characters have included inscribing stone, bone, or bronze; brushing ink onto silk, bamboo, or paper; and printing with woodblocks or moveable type. Technologies invented since the 19th century to facilitate the use of characters include telegraph codes and typewriters, as well as input methods and text encodings on computers. Development [edit] Further information: Proto-writing and History of writing Chinese characters are accepted as representing one of four independent inventions of writing in human history.[b] In each instance, writing evolved from a system using two distinct types of ideographs—either pictographs visually depicting objects or concepts, or fixed signs representing concepts only by shared convention. These systems are classified as proto-writing, because the techniques they used were insufficient to carry the meaning of spoken language by themselves. Various innovations were required for Chinese characters to emerge from proto-writing. Firstly, pictographs became distinct from simple pictures in use and appearance—for example, the pictograph 大, meaning 'large', was originally a picture of a large man, but one would need to be aware of its specific meaning in order to interpret the sequence 大鹿 as signifying 'large deer', rather than being a picture of a large man and a deer next to one another. Due to this process of abstraction, as well as to make characters easier to write, pictographs gradually became more simplified and regularized—often to the extent that the original objects represented are no longer obvious. This proto-writing system was limited to representing a relatively narrow range of ideas with a comparatively small library of symbols. This compelled innovations that allowed for symbols which indicated elements of spoken language directly. In each historical case, this was accomplished by some form of the rebus technique, where the symbol for a word is used to indicate a different word with a similar pronunciation, depending on context. This allowed for words that lacked a plausible pictographic representation to be written down for the first time. This technique preempted more sophisticated methods of character creation that would further expand the lexicon. The process whereby writing emerged from proto-writing took place over a long period; when the purely pictorial use of symbols disappeared, leaving only those representing spoken words, the process was complete. Classification [edit] Main article: Chinese character classification Chinese characters have been used in several different writing systems throughout history. A writing system is most commonly defined to include the written symbols themselves, called graphemes—which may include characters, numerals, or punctuation—as well as the rules by which they are used to record language. Chinese characters are logographs, which are graphemes that represent units of meaning in a language. Specifically, characters represent a language's morphemes, its most basic units of meaning. Morphemes in Chinese—and therefore the characters used to write them—are nearly always a single syllable in length. In some special cases, characters may denote non-morphemic syllables as well; due to this, written Chinese is often characterized as morphosyllabic.[c] Logographs may be contrasted with letters in an alphabet, which generally represent phonemes, the distinct units of sound used by speakers of a language. Despite their origins in picture-writing, Chinese characters are no longer ideographs capable of representing ideas directly; their comprehension relies on the reader's knowledge of the particular language being written. The areas where Chinese characters were historically used—sometimes collectively termed the Sinosphere—have a long tradition of lexicography attempting to explain and refine their use; for most of history, analysis revolved around a model first popularized in the 2nd-century _Shuowen Jiezi_ dictionary. More recent models have analysed the methods used to create characters, how characters are structured, and how they function in a given writing system. Structural analysis [edit] Most characters can be analysed structurally as compounds made of smaller components (部件; bùjiàn), which are often independent characters in their own right, adjusted to occupy a given position in the compound. Components within a character may serve a specific function—phonetic components provide a hint for the character's pronunciation, and semantic components indicate some element of the character's meaning. Components that serve neither function may be classified as pure signs with no particular meaning, other than their presence distinguishing one character from another. A straightforward structural classification scheme may consist of three pure classes of semantographs, phonographs, and signs—having only semantic, phonetic, and form components respectively—as well as classes corresponding to each combination of component types. Of the 3500 characters that are frequently used in Standard Chinese, pure semantographs are estimated to be the rarest, accounting for about 5% of the lexicon, followed by pure signs with 18%, and semantic–form and phonetic–form compounds together accounting for 19%. The remaining 58% are phono-semantic compounds. The 20th-century Chinese palaeographer Qiu Xigui presented three principles of character function adapted from earlier proposals by Tang Lan[zh] and Chen Mengjia, with semantographs describing all characters with forms wholly related to their meaning, regardless of the method by which the meaning was originally depicted; phonographs that include a phonetic component; and loangraphs encompassing existing characters that have been borrowed to write other words. Qiu also acknowledged the existence of character classes that fall outside of these principles, such as pure signs. Semantographs [edit] Pictographs [edit] Graphical evolution of pictographs 日 ('Sun') 山 ('mountain') 象 ('elephant') Most of the oldest characters are pictographs (象形; xiàngxíng), representational pictures of physical objects. Examples include 日 ('Sun'), 月 ('Moon'), and 木 ('tree'). Over time, the forms of pictographs have been simplified in order to make them easier to write. As a result, modern readers generally cannot deduce what many pictographs were originally meant to resemble; without knowing the context of their origin in picture-writing, they may be interpreted instead as pure signs. However, if a pictograph's use in compounds still reflects its original meaning, as with 日 in 晴 ('clear sky'), it can still be analysed as a semantic component. Pictographs have often been extended from their original meanings to take on additional layers of metaphor and synecdoche, which sometimes displace the character's original sense. When this process results in excessive ambiguity between distinct senses written with the same character, it is usually resolved by new compounds being derived to represent particular senses. Indicatives [edit] Indicatives (指事; zhǐshì), also called simple ideographs or self-explanatory characters, are visual representations of abstract concepts that lack any tangible form. Examples include 上 ('up') and 下 ('down')—these characters were originally written as dots placed above and below a line, and later evolved into their present forms with less potential for graphical ambiguity in context. More complex indicatives include 凸 ('convex'), 凹 ('concave'), and 平 ('flat and level'). Compound ideographs [edit] The compound character 好 illustrated as its component characters 女 and 子 positioned side by side Compound ideographs (会意; 會意; huìyì)—also called logical aggregates, associative idea characters, or syssemantographs—combine other characters to convey a new, synthetic meaning. A canonical example is 明 ('bright'), interpreted as the juxtaposition of the two brightest objects in the sky: 日 ('Sun') and 月 ('Moon'), together expressing their shared quality of brightness. Other examples include 休 ('rest'), composed of pictographs 人 ('man') and 木 ('tree'), and 好 ('good'), composed of 女 ('woman') and 子 ('child'). Many traditional examples of compound ideographs are now believed to have actually originated as phono-semantic compounds, made obscure by subsequent changes in pronunciation. For example, the _Shuowen Jiezi_ describes 信 ('trust') as an ideographic compound of 人 ('man') and 言 ('speech'), but modern analyses instead identify it as a phono-semantic compound—though with disagreement as to which component is phonetic.Peter A. Boodberg and William G. Boltz go so far as to deny that any compound ideographs were devised in antiquity, maintaining that secondary readings that are now lost are responsible for the apparent absence of phonetic indicators, but their arguments have been rejected by other scholars. Phonographs [edit] Phono-semantic compounds [edit] Phono-semantic compounds (形声; 形聲; xíngshēng) are composed of at least one semantic component and one phonetic component. They may be formed by one of several methods, often by adding a phonetic component to disambiguate a loangraph, or by adding a semantic component to represent a specific extension of a character's meaning. Examples of phono-semantic compounds include 河 (hé; 'river'), 湖 (hú; 'lake'), 流 (liú; 'stream'), 沖 (chōng; 'surge'), and 滑 (huá; 'slippery'). Each of these characters have three short strokes on their left-hand side: 氵, a simplified combining form of ⽔ ('water'). This component serves a semantic function in each example, indicating the character has some meaning related to water. The remainder of each character is its phonetic component: 湖 (hú) is pronounced identically to 胡 (hú) in Standard Chinese, 河 (hé) is pronounced similarly to 可 (kě), and 沖 (chōng) is pronounced similarly to 中 (zhōng). The phonetic components of most compounds may only provide an approximate pronunciation, even before subsequent sound shifts in the spoken language. Some characters may only have the same initial or final sound of a syllable in common with phonetic components. A phonetic series comprises all the characters created using the same phonetic component, which may have diverged significantly in their pronunciations over time. For example, 茶 (chá; caa4; 'tea') and 途 (tú; tou4; 'route') are characters in the phonetic series using 余 (yú; jyu4), a literary first-person pronoun. Their Old Chinese pronunciations were similar, but the phonetic component no longer serves as a useful hint for their pronunciation in modern varieties of Chinese due to subsequent sound shifts—demonstrated here in both their Mandarin and Cantonese readings. Loangraphs [edit] The phenomenon of existing characters being adapted to write other words with similar pronunciations was necessary in the initial development of Chinese writing, and has remained common throughout its subsequent history. Some loangraphs (假借; jiǎjiè; 'borrowing') are introduced to represent words previously lacking a written form—this is often the case with abstract grammatical particles such as 之 and 其. The process of characters being borrowed as loangraphs should not be conflated with the distinct process of semantic extension, where a word acquires additional senses, which often remain written with the same character. As both processes often result in a single character form being used to write several distinct meanings, loangraphs are often misidentified as being the result of semantic extension, and vice versa. Loangraphs are also used to write words borrowed from other languages, such as the Buddhist terminology introduced to China in antiquity, as well as contemporary non-Chinese words and names. For example, each character in the name 加拿大 (Jiānádà; 'Canada') is often used as a loangraph for its respective syllable. However, the barrier between a character's pronunciation and meaning is never total; when transcribing into Chinese, loangraphs are often chosen deliberately as to create certain connotations. This is regularly done with corporate brand names—for example, Coca-Cola's Chinese name is 可口可乐; 可口可樂 (Kěkǒu Kělè; 'delicious enjoyable'). Signs [edit] Some characters and components are pure signs, with meanings merely stemming from their having a fixed and distinct form. Basic examples of pure signs are found with the numerals beyond four, e.g. 五 ('five') and 八 ('eight'), whose forms do not give visual hints to the quantities they represent. Traditional Shuowen Jiezi classification [edit] The _Shuowen Jiezi_ is a character dictionary authored c. 100 CE by the scholar Xu Shen. In its postface, Xu analyses what he sees as all the methods by which characters are created. Later authors iterated upon Xu's analysis, developing a categorization scheme known as the 'six writings' (六书; 六書; liùshū), which identifies every character with one of six categories that had previously been mentioned in the _Shuowen Jiezi_. For nearly two millennia, this scheme was the primary framework for character analysis used throughout the Sinosphere. Xu based most of his analysis on examples of Qin seal script that were written down several centuries before his time—these were usually the oldest specimens available to him, though he stated he was aware of the existence of even older forms. The first five categories are pictographs, indicatives, compound ideographs, phono-semantic compounds, and loangraphs. The sixth category is given by Xu as 轉注 (zhuǎnzhù; 'reversed and refocused'); however, its definition is unclear, and it is generally disregarded by modern scholars. Modern scholars agree that the theory presented in the _Shuowen Jiezi_ is problematic, failing to fully capture the nature of Chinese writing, both in the present, as well as at the time Xu was writing. Traditional Chinese lexicography as embodied in the _Shuowen Jiezi_ has suggested implausible etymologies for some characters. Moreover, several categories are considered to be ill-defined—for example, it is unclear whether characters like 大 ('large') should be classified as pictographs or indicatives. However, awareness of the 'six writings' model has remained a common component of character literacy, and often serves as a tool for students memorizing characters. History [edit] Further information: Chinese script styles and History of the Chinese language Diagram comparing the abstraction of pictographs in cuneiform, Egyptian hieroglyphs, and Chinese characters– from an 1870 publication by French Egyptologist Gaston Maspero[A] The broadest trend in the evolution of Chinese characters over their history has been simplification, both in graphical shape (字形; zìxíng), the "external appearances of individual graphs", and in graphical form (字体; 字體; zìtǐ), "overall changes in the distinguishing features of graphic[al] shape and calligraphic style, ... in most cases refer[ring] to rather obvious and rather substantial changes". The traditional notion of an orderly procession of script styles, each suddenly appearing and displacing the one previous, has been disproven by later scholarship and archaeological work. Instead, scripts evolved gradually, with several distinct styles often coexisting within a given area. Traditional invention narrative [edit] Several of the Chinese classics indicate that knotted cords were used to keep records prior to the invention of writing. Works that reference the practice include chapter 80 of the _Tao Te Ching_[B] and the "XiciII" commentary to the _I Ching_.[C] According to one tradition, Chinese characters were invented during the 3rd millennium BCE by Cangjie, a scribe of the legendary Yellow Emperor. Cangjie is said to have invented symbols called 字 (zì) due to his frustration with the limitations of knotting, taking inspiration from his study of the tracks of animals, landscapes, and the stars in the sky. On the day that these first characters were created, grain rained down from the sky; that night, the people heard the wailing of ghosts and demons, lamenting that humans could no longer be cheated. Neolithic precursors [edit] Main article: Neolithic symbols in China Collections of graphs and pictures have been discovered at the sites of several Neolithic settlements throughout the Yellow River valley, including Jiahu (c. 6500 BCE), Dadiwan and Damaidi (6th millennium BCE), and Banpo (5th millennium BCE). Symbols at each site were inscribed or drawn onto artefacts, appearing one at a time and without indicating any greater context. Qiu concluded, "We simply possess no basis for saying that they were already being used to record language." A historical connection with the symbols used by the late Neolithic Dawenkou culture (c. 4300– c. 2600 BCE) in Shandong has been deemed possible by palaeographers, with Qiu concluding that they "cannot be definitively treated as primitive writing, nevertheless they are symbols which resemble most the ancient pictographic script discovered thus far in China...They undoubtedly can be viewed as the forerunners of primitive writing." Oracle bone script [edit] Main article: Oracle bone script Oracle bone script 天 'Heaven' 馬 'horse' 旅 'travel' 正 'straight' 韋 'leather' Ox scapula inscribed with characters recording the result of divinations– dated c. 1200 BCE The oldest attested Chinese writing comprises a body of inscriptions produced during the Late Shang period (c. 1250– 1050 BCE), with the very earliest examples from the reign of Wu Ding dated between 1250 and 1200 BCE. Many of these inscriptions were made on oracle bones—usually either ox scapulae or turtle plastrons—and recorded official divinations carried out by the Shang royal house. Contemporaneous inscriptions in a related but distinct style were also made on ritual bronze vessels. This oracle bone script (甲骨文; jiǎgǔwén) was first documented in 1899, after specimens were discovered being sold as "dragon bones" for medicinal purposes, with the symbols carved into them identified as early character forms. By 1928, the source of the bones had been traced to a village near Anyang in Henan—discovered to be the site of Yin, the final Shang capital—which was excavated by a team led by Li Ji from the Academia Sinica between 1928 and 1937. To date, over 150 000 oracle bone fragments have been found. Oracle bone inscriptions recorded divinations undertaken to communicate with the spirits of royal ancestors. The inscriptions range from a few characters in length at their shortest, to several dozen at their longest. The Shang king would communicate with his ancestors by means of scapulimancy, inquiring about subjects such as the royal family, military success, and the weather. Inscriptions were made in the divination material itself before and after it had been cracked by exposure to heat; they generally include a record of the questions posed, as well as the answers as interpreted in the cracks. A minority of bones feature characters that were inked with a brush before their strokes were incised; the evidence of this also shows that the conventional stroke orders used by later calligraphers had already been established for many characters by this point. Oracle bone script is the direct ancestor of later forms of written Chinese. The oldest known inscriptions already represent a well-developed writing system, which suggests an initial emergence predating the late 2nd millennium BCE. Although written Chinese is first attested in official divinations, it is widely believed that writing was also used for other purposes during the Shang, but that the media used in other contexts—likely bamboo and wooden slips—were less durable than bronzes or oracle bones, and have not been preserved. Zhou scripts [edit] See also: Chinese bronze inscriptions, Bamboo and wooden slips, and Seal script Bronze script The Shi Qiang pan, a bronze ritual basin bearing inscriptions describing the deeds and virtues of the first seven Zhou kings– dated c. 900 BCE As early as the Shang, the oracle bone script existed as a simplified form alongside another that was used in bamboo books, in addition to elaborate pictorial forms often used in clan emblems. These other forms have been preserved in bronze script (金文; jīnwén), where inscriptions were made using a stylus in a clay mould, which was then used to cast ritual bronzes. These differences in technique generally resulted in character forms that were less angular in appearance than their oracle bone script counterparts. Study of these bronze inscriptions has revealed that the mainstream script underwent slow, gradual evolution during the late Shang, which continued during the Zhou dynasty (c. 1046– 256 BCE) until assuming the form now known as small seal script (小篆; xiǎozhuàn) within the Zhou state of Qin. Other scripts in use during the late Zhou include the bird-worm seal script (鸟虫书; 鳥蟲書; niǎochóngshū), as well as the regional forms used in non-Qin states. Examples of these styles were preserved as variants in the _Shuowen Jiezi_. Historically, Zhou forms were collectively known as large seal script (大篆; dàzhuàn), though Qiu refrained from using this term due to its lack of precision. Qin unification and small seal script [edit] Main article: Small seal script Small seal script Following Qin's conquest of the other Chinese states that culminated in the founding of the imperial Qin dynasty in 221 BCE, the Qin small seal script was standardized for use throughout the entire country under the direction of Chancellor Li Si. It was traditionally believed that Qin scribes only used small seal script, and the later clerical script was a sudden invention during the early Han. However, more than one script was used by Qin scribes—a rectilinear vulgar style had also been in use in Qin for centuries prior to the wars of unification. The popularity of this form grew as writing became more widespread. Clerical script [edit] Main article: Clerical script Clerical script By the Warring States period (c. 475– 221 BCE), an immature form of clerical script (隶书; 隸書; lìshū) had emerged based on the vulgar form developed within Qin, often called "early clerical" or "proto-clerical". The proto-clerical script evolved gradually; by the Han dynasty (202 BCE– 220 CE), it had arrived at a mature form, also called 八分 (bāfēn). Bamboo slips discovered during the late 20th century point to this maturation being completed during the reign of Emperor Wu of Han (r. 141–87 BCE). This process, called libian (隶变; 隸變), involved character forms being mutated and simplified, with many components being consolidated, substituted, or omitted. In turn, the components themselves were regularized to use fewer, straighter, and more well-defined strokes. As a result, clerical script largely lacks the pictorial qualities still evident in seal script. Around the midpoint of the Eastern Han (25–220 CE), a simplified and easier form of clerical script appeared, which Qiu termed 'neo-clerical' (新隶体; 新隸體; xīnlìtǐ). By the end of the Han, this had become the dominant script used by scribes, though clerical script remained in use for formal works, such as engraved stelae. Qiu described neo-clerical as a transitional form between clerical and regular script which remained in use through the Three Kingdoms period (220–280 CE) and beyond. Cursive and semi-cursive [edit] Cursive script Cursive script (草书; 草書; cǎoshū) was in use as early as 24 BCE, synthesizing elements of the vulgar writing that had originated in Qin with flowing cursive brushwork. By the Jin dynasty (266–420), the Han cursive style became known as 章草 (zhāngcǎo; 'orderly cursive'), sometimes known in English as 'clerical cursive', 'ancient cursive', or 'draft cursive'. Some attribute this name to the fact that the style was considered more orderly than a later form referred to as 今草 (jīncǎo; 'modern cursive'), which had first emerged during the Jin and was influenced by semi-cursive and regular script. This later form was exemplified by the work of figures like Wang Xizhi (fl. 4th century), who is often regarded as the most important calligrapher in Chinese history. Semi-cursive script An early form of semi-cursive script (行书; 行書; xíngshū; 'running script') can be identified during the late Han, with its development stemming from a cursive form of neo-clerical script. Liu Desheng (刘德升; 劉德升; fl. 2nd century CE) is traditionally recognized as the inventor of the semi-cursive style, though accreditations of this kind often indicate a given style's early masters, rather than its earliest practitioners. Later analysis has suggested popular origins for semi-cursive, as opposed to it being an invention of Liu. It can be characterized partly as the result of clerical forms being written more quickly, without formal rules of technique or composition—what would be discrete strokes in clerical script frequently flow together instead. The semi-cursive style is commonly adopted in contemporary handwriting. Regular script [edit] Main article: Regular script Regular script Page from a Song-era publication printed using a regular script style[D] Regular script (楷书; 楷書; kǎishū), based on clerical and semi-cursive forms, is the predominant form in which characters are written and printed. Its innovations have traditionally been credited to the calligrapher Zhong Yao, who lived in the state of Cao Wei (extant 220–266); he is often called the "father of regular script". The earliest surviving writing in regular script comprises copies of Zhong Yao's work, including at least one copy by Wang Xizhi. Characteristics of regular script include the 'pause' (頓; dùn) technique used to end horizontal strokes, as well as heavy tails on diagonal strokes made going down and to the right. It developed further during the Eastern Jin (317–420) in the hands of Wang Xizhi and his son Wang Xianzhi. However, most Jin-era writers continued to use neo-clerical and semi-cursive styles in their daily writing. It was not until the Northern and Southern period (420–589) that regular script became the predominant form. The system of imperial examinations for the civil service established during the Sui dynasty (581–618) required test takers to write in Literary Chinese using regular script, which contributed to the prevalence of both throughout later Chinese history. Structure [edit] Each character of a text is written within a uniform square allotted for it. As part of the evolution from seal script into clerical script, character components became regularized as discrete series of strokes (笔画; 筆畫; bǐhuà). Strokes can be considered both the basic unit of handwriting, as well as the writing system's basic unit of graphemic organization. In clerical and regular script, individual strokes traditionally belong to one of eight categories according to their technique and graphemic function. In what is known as the Eight Principles of Yong, calligraphers practise their technique using the character 永 (yǒng; 'eternity'), which can be written with one stroke of each type. In ordinary writing, 永 is now written with five strokes instead of eight, and a system of five basic stroke types is commonly employed in analysis—with certain compound strokes treated as sequences of basic strokes made in a single motion. Characters are constructed according to predictable visual patterns. Some components have distinct combining forms when occupying specific positions within a character—for example, the ⼑ ('knife') component appears as 刂 on the right side of characters, but as ⺈ at the top of characters. The order in which components are drawn within a character is fixed. The order in which the strokes of a component are drawn is also largely fixed, but may vary according to several different standards. This is summed up in practice with a few rules of thumb, including that characters are generally assembled from left to right, then from top to bottom, with "enclosing" components started before, then closed after, the components they enclose. For example, 永 is drawn in the following order: Sequence and placement of the strokes in 永\| Character \| Stroke \| --- \| \| Duration: 8 seconds.0:08 \| 1 \| \| \| 2 \| \| \| 3 \| \| \| 4 \| \| \| 5 \| \| Variant characters [edit] Main article: Variant Chinese characters Variants of the Chinese character for 'turtle', collected c. 1800 from printed sources.[E] The traditional form 龜 (left) is used in Taiwan and Hong Kong. The simplified form 龟 (not pictured) is used in mainland China, and the simplified form 亀 (also not pictured) is used in Japan. Over a character's history, variant character forms (异体字; 異體字; yìtǐzì) emerge via several processes. Variant forms have distinct structures, but represent the same morpheme; as such, they can be considered instances of the same underlying character. This is comparable to visually distinct double-storey \|a\| and single-storey \|ɑ\| forms both representing the Latin letter ⟨A⟩. Variants also emerge for aesthetic reasons, to make handwriting easier, or to correct what the writer perceives to be errors in a character's form. Individual components may be replaced with visually, phonetically, or semantically similar alternatives. The boundary between character structure and style—and thus whether forms represent different characters, or are merely variants of the same character—is often non-trivial or unclear. For example, prior to the Qin dynasty the character meaning 'bright' was written as either 明 or 朙—with either 日 ('Sun') or 囧 ('window') on the left, and 月 ('Moon') on the right. As part of the Qin programme to standardize small seal script across China, the 朙 form was promoted. Some scribes ignored this, and continued to write the character as 明. However, the increased usage of 朙 was followed by the proliferation of a third variant: 眀, with 目 ('eye') on the left—likely derived as a contraction of 朙. Ultimately, 明 became the character's standard form. Layout [edit] Further information: Horizontal and vertical writing in East Asian scripts See also: Chinese punctuation, Japanese punctuation, and Korean punctuation From the earliest inscriptions until the 20th century, texts were generally laid out vertically—with characters written from top to bottom in columns, arranged from right to left. Word boundaries are generally not indicated with spaces. A horizontal writing direction—with characters written from left to right in rows, arranged from top to bottom—only became predominant in the Sinosphere during the 20th century as a result of Western influence. Many publications outside mainland China continue to use the traditional vertical writing direction. Western influence also resulted in the generalized use of punctuation being widely adopted in print during the 19th and 20th centuries. Prior to this, the context of a passage was considered adequate to guide readers; this was enabled by characters being easier to read than alphabets when written without spaces or punctuation due to their more discretized shapes. Methods of writing [edit] Ordinary handwriting on a lunch menu in Hong Kong. Here, 反 (fǎn) is being used as an unofficial short form of 飯 (fàn; 'meal') by omitting the latter's ⾷ ('eat') component. The earliest attested Chinese characters were carved into bone, or marked using a stylus in clay moulds used to cast ritual bronzes. Characters have also been incised into stone, or written in ink onto slips of silk, wood, and bamboo. The invention of paper for use as a writing medium occurred during the 1st century CE, and is traditionally credited to Cai Lun. There are numerous styles, or scripts (书; 書; shū) in which characters can be written, including the historical forms like seal script and clerical script. Most styles used throughout the Sinosphere originated within China, though they may display regional variation. Styles that have been created outside of China tend to remain localized in their use—these include the Japanese edomoji and Vietnamese lệnh thư scripts. Calligraphy [edit] Main article: Chinese calligraphy Chinese calligraphy of mixed styles by the Song-era poet Mi Fu Calligraphy was traditionally one of the four arts to be mastered by Chinese scholars, considered to be an artful means of expressing thoughts and teachings. Chinese calligraphy typically makes use of an ink brush to write characters. Strict regularity is not required, and character forms may be accentuated to evoke a variety of aesthetic effects. Traditional ideals of calligraphic beauty often tie into broader philosophical concepts native to East Asia. For example, aesthetics can be conceptualized using the framework of yin and yang, where the extremes of any number of mutually reinforcing dualities are balanced by the calligrapher—such as the duality between strokes made quickly or slowly, between applying ink heavily or lightly, between characters written with symmetrical or asymmetrical forms, and between characters representing concrete or abstract concepts. Printing and typefaces [edit] Further information: History of printing in East Asia and East Asian typography Sample of Prison Gothic, a sans-serif typeface Woodblock printing was invented in China between the 6th and 9th centuries, followed by the invention of moveable type by Bi Sheng during the 11th century. The increasing use of print during the Ming (1368–1644) and Qing dynasties (1644–1912) led to considerable standardization in character forms, which prefigured later script reforms during the 20th century. This print orthography, exemplified by the 1716 Kangxi Dictionary, was later dubbed the _jiu zixing_ ('old character shapes'). Printed Chinese characters may use different typefaces, of which there are four broad classes in use: Song (宋体; 宋體) or Ming (明体; 明體) typefaces—with "Song" generally used with simplified Chinese typefaces, and "Ming" with others—broadly correspond to Western serif styles. Song typefaces are broadly within the tradition of historical Chinese print; both names for the style refer to eras regarded as high points for printing in the Sinosphere. While type during the Song dynasty (960–1279) generally resembled the regular script style of a particular calligrapher, most modern Song typefaces are intended for general purpose use and emphasize neutrality in their design. Sans-serif typefaces are called 'black form' (黑体; 黑體; hēitǐ) in Chinese and 'Gothic' (ゴシック体) in Japanese. Sans-serif strokes are rendered as simple lines of even thickness. "Kai" typefaces (楷体; 楷體) imitate handwritten regular script. Fangsong typefaces (仿宋体; 仿宋體), called "Song" in Japan, correspond to semi-script styles in the Western paradigm. Use with computers [edit] Main article: Chinese character information technology Before computers became ubiquitous, earlier electro-mechanical communications devices like telegraphs and typewriters were originally designed for use with alphabets, often by means of alphabetic text encodings like Morse code and ASCII. Adapting these technologies for a writing system that uses thousands of distinct characters was non-trivial. Input methods [edit] Further information: Chinese input method and Japanese input method Chinese IME displaying candidates based on pinyin spelling Chinese characters are predominantly input on computers using a standard keyboard. Many input methods (IMEs) are phonetic, where typists enter characters according to schemes like pinyin or bopomofo for Mandarin, Jyutping for Cantonese, or Hepburn for Japanese. For example, 香港 ('Hong Kong') could be input as xiang1gang3 using pinyin, or as hoeng1gong2 using Jyutping. Character input methods may also be based on form, using the shape of characters and existing rules of handwriting to assign unique codes to each character, potentially increasing the speed of typing. Popular form-based input methods include Wubi on the mainland, and Cangjie—named after the mythological inventor of writing—in Taiwan and Hong Kong. Often, unnecessary parts are omitted from the encoding according to predictable rules. For example, 疆 ('border') is encoded using the Cangjie method as NGMWM, which corresponds to the components 弓土一田一. Contextual constraints may be used to improve candidate character selection. When ignoring tones, 知道 and 直刀 are both transcribed as zhidao; the system may prioritize which candidate appears first based on context. Encoding and interchange [edit] Main article: Chinese character encoding See also: Han unification While special text encodings for Chinese characters were introduced prior to its popularization, The Unicode Standard is the predominant text encoding worldwide. According to the philosophy of the Unicode Consortium, each distinct graph is assigned a number in the standard, but specifying its appearance or the particular allograph used is a choice made by the engine rendering the text.[F] Unicode's Basic Multilingual Plane (BMP) represents the standard's 2 16 smallest code points. Of these, 20 992 (or 32%) are assigned to CJK Unified Ideographs, a designation comprising characters used in each of the Chinese family of scripts. As of version 16.0, published in 2024, Unicode defines a total of 98 682 Chinese characters.[G] Vocabulary and adaptation [edit] Further information: Chinese family of scripts and Adoption of Chinese literary culture Writing first emerged during the historical stage of the Chinese language known as Old Chinese. Most characters correspond to morphemes that originally functioned as stand-alone Old Chinese words.Classical Chinese is the form of written Chinese used in the classic works of Chinese literature from roughly the 5th century BCE until the 2nd century CE. This form of the language was imitated by later authors, even as it began to diverge from the language they spoke. This later form, referred to as Literary Chinese, remained the predominant written language in China until the 20th century. Its use in the Sinosphere was loosely analogous to that of Latin in pre-modern Europe. While it was not static over time, Literary Chinese retained many properties of spoken Old Chinese. Informed by the local spoken vernaculars, texts were read aloud using literary and colloquial readings that varied by region. Over time, sound mergers created ambiguities in vernacular speech as more words became homophonic. This ambiguity was often reduced through the introduction of multi-syllable compound words, which comprise much of the vocabulary in modern varieties of Chinese. Over time, use of Literary Chinese spread to neighbouring countries, including Vietnam, Korea, and Japan. Alongside other aspects of Chinese culture, local elites adopted writing for record-keeping, histories, and official communications. Excepting hypotheses by some linguists of the latter two sharing a common ancestor, Chinese, Vietnamese, Korean, and Japanese each belong to different language families, and tend to function differently from one another. Reading systems were devised to enable non-Chinese speakers to interpret Literary Chinese texts in terms of their native language, a phenomenon that has been variously described as either a form of diglossia, as reading by gloss, or as a process of translation into and out of Chinese. Compared to other traditions that wrote using alphabets or syllabaries, the literary culture that developed in this context was less directly tied to a specific spoken language. This is exemplified by the cross-linguistic phenomenon of brushtalk, where mutual literacy allowed speakers of different languages to engage in face-to-face conversations. Following the introduction of Literary Chinese, characters were later adapted to write many non-Chinese languages spoken throughout the Sinosphere. These new writing systems used characters to write both native vocabulary and the numerous loanwords each language had borrowed from Chinese, collectively termed Sino-Xenic vocabulary. Characters may have native readings, Sino-Xenic readings, or both. Comparison of Sino-Xenic vocabulary across the Sinosphere has been useful in the reconstruction of Middle Chinese phonology. Literary Chinese was used in Vietnam during the millennium of Chinese rule that began in 111 BCE. By the 15th century, a system that adapted characters to write Vietnamese called chữ Nôm had fully matured. The 2nd century BCE is the earliest possible period for the introduction of writing to Korea; the oldest surviving manuscripts in the country date to the early 5th century CE. Also during the 5th century, writing spread from Korea to Japan. Characters were being used to write both Korean and Japanese by the 6th century. By the late 20th century, characters had largely been replaced with alphabets designed to write Vietnamese and Korean. This leaves Japanese as the only major non-Sinitic language typically written using Chinese characters. Literary and vernacular Chinese [edit] See also: Reconstructions of Old Chinese and Middle Chinese Excerpt from a 1436 primer on Chinese characters Words in Classical Chinese were generally a single character in length. An estimated 25–30% of the vocabulary used in Classical Chinese texts consists of two-character words. Over time, the introduction of multi-syllable vocabulary into vernacular varieties of Chinese was encouraged by phonetic shifts that increased the number of homophones. The most common process of Chinese word formation after the Classical period has been to create compounds of existing words. Words have also been created by appending affixes to words, by reduplication, and by borrowing words from other languages. While multi-syllable words are generally written with one character per syllable, abbreviations are occasionally used. For example, 二十 (èrshí; 'twenty') may be written as the contracted form 廿. Sometimes, different morphemes come to be represented by characters with identical shapes. For example, 行 may represent either 'road' (xíng) or the extended sense of 'row' (háng)—these morphemes are ultimately cognates that diverged in pronunciation but remained written with the same character. However, Qiu reserved the term homograph to describe identically shaped characters with different meanings that emerge via processes other than semantic extension. An example homograph is 铊; 鉈, which originally meant 'weight used at a steelyard' (tuó). In the 20th century, this character was created again with the meaning 'thallium' (tā). Both of these characters are phono-semantic compounds with ⾦ ('gold') as the semantic component and 它 as the phonetic component, but the words represented by each are not related. There are a number of 'dialect characters' (方言字; fāngyánzì) that are not used in standard written vernacular Chinese, but reflect the vocabulary of other spoken varieties. The most complete example of an orthography based on a variety other than Standard Chinese is Written Cantonese. A common Cantonese character is 冇 (mou5; 'to not have'), derived by removing two strokes from 有 (jau5; 'to have'). It is common to use standard characters to transcribe previously unwritten words in Chinese dialects when obvious cognates exist. When no obvious cognate exists due to factors like irregular sound changes, semantic drift, or an origin in a non-Chinese language, characters are often borrowed or invented to transcribe the word—either ad hoc, or according to existing principles. These new characters are generally phono-semantic compounds. Japanese [edit] Japanese writing Components show Kanji Stroke order Radicals Jōyō kanji list Kyōiku kanji Tōyō kanji Jinmeiyō kanji Hyōgai kanji show Kana Hiragana Hentaigana Katakana Man'yōgana Sōgana Gojūon show Typographic symbols Japanese punctuation Iteration mark Uses show Syllabograms Furigana Okurigana Braille Transliteration show Rōmaji Hepburn Kunrei-shiki/ ISO 3602 Nihon-shiki JSL Wāpuro (keyboard input) show Cyrillization Polivanov system v t e In Japanese, Chinese characters are referred to as kanji. During the Nara period (710–794), readers and writers of kanbun—the Japanese term for Literary Chinese writing—began utilizing a system of reading techniques and annotations called kundoku. When reading, Japanese speakers would adapt the syntax and vocabulary of Literary Chinese texts to reflect their Japanese-language equivalents. Writing essentially involved the inverse of this process, and resulted in ordinary Literary Chinese. When adapted to write Japanese, characters were used to represent both Sino-Japanese vocabulary loaned from Chinese, as well as the corresponding native synonyms. Most kanji were subject to both borrowing processes, and as a result have both Sino-Japanese and native readings, known as on'yomi and kun'yomi respectively. Moreover, kanji may have multiple readings of either kind. Distinct classes of on'yomi were borrowed into Japanese at different points in time from different varieties of Chinese. The Japanese writing system is a mixed script, and has also incorporated syllabaries called kana to represent phonetic units called moras, rather than morphemes. Prior to the Meiji era (1868–1912), writers used certain kanji to represent their sound values instead, in a system known as man'yōgana. Starting in the 9th century, specific man'yōgana were graphically simplified to create two distinct syllabaries called hiragana and katakana, which slowly replaced the earlier convention. Modern Japanese retains the use of kanji to represent most word stems, while kana syllabograms are generally used for grammatical affixes, particles, and loanwords. The forms of hiragana and katakana are visually distinct from one another, owing in large part to different methods of simplification—katakana were derived from smaller components of each man'yōgana, while hiragana were derived from the cursive forms of man'yōgana in their entirety. In addition, the hiragana and katakana for some moras were derived from different man'yōgana. Characters invented for Japanese-language use are called kokuji. The methods employed to create kokuji are equivalent to those used by Chinese-original characters, though most are ideographic compounds. For example, 峠 (tōge; 'mountain pass') is a compound kokuji composed of 山 ('mountain'), 上 ('above'), and 下 ('below'). While characters used to write Chinese are monosyllabic, many kanji have multi-syllable readings. For example, the kanji 刀 has a native kun'yomi reading of katana. In different contexts, it can also be read with the on'yomi reading tō, such as in the Chinese loanword 日本刀 (nihontō; 'Japanese sword'), with a pronunciation corresponding to that in Chinese at the time of borrowing. Prior to the universal adoption of katakana, loanwords were typically written with unrelated kanji with on'yomi readings matching the syllables in the loanword. These spellings are called ateji—for example, 亜米利加 (Amerika) was the ateji spelling of 'America', now rendered as アメリカ. As opposed to man'yōgana used solely for their pronunciation, ateji still corresponded to specific Japanese words. Some are still in use, with the official list of jōyō kanji including 106 ateji readings. Korean [edit] In Korean, Chinese characters are referred to as hanja. Literary Chinese may have been written in Korea as early as the 2nd century BCE. During Korea's Three Kingdoms period (57 BCE– 668 CE), characters were also used to write idu, a form of Korean-language literature that mostly made use of Sino-Korean vocabulary. During the Goryeo period (918–1392), Korean writers developed a system of phonetic annotations for Literary Chinese called gugyeol, comparable to kundoku in Japan, though it only entered widespread use during the later Joseon period (1392–1897). While the hangul alphabet was invented by the Joseon king Sejong in 1443, it was not adopted by the Korean literati and was relegated to use in glosses for Literary Chinese texts until the late 19th century. Much of the Korean lexicon consists of Chinese loanwords, especially technical and academic vocabulary. While hanja were usually only used to write this Sino-Korean vocabulary, there is evidence that vernacular readings were sometimes used. Compared to the other written vernaculars, very few characters were invented to write Korean words; these are called gukja. During the late 19th and early 20th centuries, Korean was written either using a mixed script of hangul and hanja, or only using hangul. Following the end of the Empire of Japan's occupation of Korea in 1945, the total replacement of hanja with hangul was advocated throughout the country as part of a broader "purification movement" of the national language and culture. However, due to the lack of tones in spoken Korean, there are many Sino-Korean words that are homophones with identical hangul spellings. For example, the phonetic dictionary entry for 기사 (gisa) yields more than 30 different entries. This ambiguity had historically been resolved by also including the associated hanja. While still sometimes used for Sino-Korean vocabulary, it is much rarer for native Korean words to be written using hanja. When learning new characters, Korean students are instructed to associate each one with both its Sino-Korean pronunciation, as well as a native Korean synonym. Examples include: Example Korean dictionary listings\| Hanja \| Hangul \| Gloss \| --- \| Native translation \| Sino-Korean \| \| 水 \| 물; mul \| 수; su \| 'water' \| \| 人 \| 사람; saram \| 인; in \| 'person' \| \| 大 \| 큰; keun \| 대; dae \| 'big' \| \| 小 \| 작을; jakeul \| 소; so \| 'small' \| \| 下 \| 아래; arae \| 하; ha \| 'down' \| \| 父 \| 아비; abi \| 부; bu \| 'father' \| Vietnamese [edit] See also: Literary Chinese in Vietnam The first two lines of the 19th-century Vietnamese epic poem The Tale of Kieu, written in both chữ Nôm and the Vietnamese alphabet Borrowed characters representing Sino-Vietnamese words Borrowed characters representing native Vietnamese words Invented chữ Nôm representing native Vietnamese words In Vietnamese, Chinese characters are referred to as chữ Hán (𡨸漢), chữ Nho (𡨸儒; 'Confucian characters'), or Hán tự (漢字). Literary Chinese was used for all formal writing in Vietnam until the modern era, having first acquired official status in 1010. Literary Chinese written by Vietnamese authors is first attested in the late 10th century, though the local practice of writing is likely several centuries older. Characters used to write Vietnamese called chữ Nôm (𡨸喃) are first attested in an inscription dated to 1209 made at the site of a pagoda. A mature chữ Nôm script had likely emerged by the 13th century, and was initially used to record Vietnamese folk literature. Some chữ Nôm characters are phono-semantic compounds corresponding to spoken Vietnamese syllables. Another technique with no equivalent in China created chữ Nôm compounds using two phonetic components. This was done because Vietnamese phonology included consonant clusters not found in Chinese, and were thus poorly approximated by the sound values of borrowed characters. Compounds used components with two distinct consonant sounds to specify the cluster, e.g. 𢁋 (blăng;[d] 'Moon') was created as a compound of 巴 (ba) and 陵 (lăng). As a system, chữ Nôm was highly complex, and the literacy rate among the Vietnamese population never exceeded 5%. Both Literary Chinese and chữ Nôm fell out of use during the French colonial period, and were gradually replaced by the Latin-based Vietnamese alphabet. Following the end of colonial rule in 1954, the Vietnamese alphabet has been sole official writing system in Vietnam, and is used exclusively in Vietnamese-language media. Other languages [edit] Several minority languages of South and Southwestern China have been written with scripts using both borrowed and locally created characters. The most well-documented of these is the sawndip script for the Zhuang languages of Guangxi. While little is known about its early development, a tradition of vernacular Zhuang writing likely first emerged during the Tang dynasty (618–907). Modern scholarship characterizes sawndip writing as a network of regional traditions that have mutually influenced one another while maintaining their local characteristics. Like Vietnamese, some invented Zhuang characters are phonetic–phonetic compounds, though not primarily ones intended to describe consonant clusters. Despite the Chinese government encouraging its replacement with a Latin-based Zhuang alphabet, sawndip remains in use. Other non-Sinitic languages of China historically written with Chinese characters include Miao, Yao, Bouyei, Bai, and Hani; each of these are now written with Latin-based alphabets designed for use with each language. Graphically derived scripts [edit] See also: Transcription into Chinese characters Title page for a 1908 edition of the 13th-century Secret History of the Mongols, which uses Chinese characters to transcribe Mongolian and provides glosses to the right of each column Between the 10th and 13th centuries, dynasties founded by non-Han peoples in northern China also created scripts for their languages that were inspired by Chinese characters, but did not use them directly—these included the Khitan large script, Khitan small script, Tangut script, and Jurchen script. This has occurred in other contexts as well: Nüshu was a script used by Yao women to write the Xiangnan Tuhua language, and bopomofo (注音符号; 注音符號; zhùyīn fúhào) is a semi-syllabary first invented in 1907 to represent the sounds of Standard Chinese; both use forms graphically derived from Chinese characters. Other scripts within China that have adapted some characters but are otherwise distinct include the Geba syllabary used to write the Naxi language, the script for the Sui language, the script for the Yi languages, and the syllabary for the Lisu language. Chinese characters have also been repurposed phonetically to transcribe the sounds of non-Chinese languages. For example, the only manuscripts of the 13th-century Secret History of the Mongols that have survived from the medieval era use characters in this manner to write the Mongolian language. Literacy and lexicography [edit] Further information: Chinese character education and Literacy in China The memorization of thousands of different characters is required to achieve literacy in languages written with them, in contrast to the relatively small inventory of graphemes used in phonetic writing. Historically, character literacy was often acquired via Chinese primers like the 6th-century Thousand Character Classic and 13th-century Three Character Classic, as well as surname dictionaries like the Song-era Hundred Family Surnames. Studies of Chinese-language literacy suggest that literate individuals generally have an active vocabulary of three to four thousand characters; for specialists in fields like literature or history, this figure may be between five and six thousand. Dictionaries [edit] Main article: Chinese dictionary The first four characters of the 6th-century Thousand Character Classic in different styles. From right to left: seal script, clerical script, regular script, Song type, and sans-serif type. According to analyses of mainland Chinese, Taiwanese, Hong Kong, Japanese, and Korean sources, the total number of characters in the modern lexicon is around 15 000. Dozens of schemes have been devised for indexing Chinese characters and arranging them in dictionaries, though relatively few have achieved widespread use. Characters may be ordered according to methods based on their meaning, visual structure, or pronunciation. The _Erya_ (c. 3rd century BCE) organized the Chinese lexicon into 19 sections according to character meaning, with 3 dealing with everyday vocabulary, and each of the remaining 16 dedicated to specialized vocabulary related to a specific topic. The _Shuowen Jiezi_ (c. 100 CE) introduced what would ultimately become the predominant method of organization used by later character dictionaries, whereby characters are grouped according to certain visually prominent components called radicals (部首; bùshǒu; 'section headers'). The _Shuowen Jiezi_ used a system of 540 radicals, while subsequent dictionaries have generally used fewer. The set of 214 Kangxi radicals was popularized by the Kangxi Dictionary (1716), but originally appeared in the earlier _Zihui_ (1615). Character dictionaries have historically been indexed using radical-and-stroke sorting, where characters are grouped by radical and sorted within each group by stroke number. Some modern dictionaries arrange character entries alphabetically according to their pinyin spelling, while also providing a traditional radical-based index. Before the invention of romanization systems for Chinese, the pronunciation of characters was transmitted via rhyme dictionaries. These used the fanqie (反切; 'reverse cut') method, where each entry lists a common character with the same initial sound as the character in question, alongside one with the same final sound. Neurolinguistics [edit] Using functional magnetic resonance imaging (fMRI), neurolinguists have studied the brain activity associated with literacy. Compared to phonetic systems, reading and writing with characters involves additional areas of the brain—including those associated with visual processing. While the level of memorization required for character literacy is significant, identification of the phonetic and semantic components in compounds—which constitute the vast majority of characters—also plays a key role in reading comprehension. The ease of recognition for a given character is impacted by how regular the positioning of its components is, as well as how reliable its phonetic component is in indicating a specific pronunciation. Moreover, due to the high level of homophony in Chinese languages and the more irregular correspondences between writing and the sounds of speech, it has been suggested that knowledge of orthography plays a greater role in speech recognition for literate Chinese speakers. Developmental dyslexia in readers of character-based languages appears to involve independent visuospatial and phonological disorders co-occurring. This seems to be a distinct phenomenon from dyslexia as experienced with phonetic orthographies, which can result from only one of the aforementioned disorders. Reform and standardization [edit] The first official list of simplified character forms, published in 1935 and including 324 characters Attempts to reform and standardize the use of characters—including aspects of form, stroke order, and pronunciation—have been undertaken by states throughout history. Thousands of simplified characters were standardized and adopted in mainland China during the 1950s and 1960s, with most either already existing as common variants, or being produced via the systematic simplification of their components. After World War II, the Japanese government also simplified hundreds of character forms, including some simplifications distinct from those adopted in China. Orthodox forms that have not undergone simplification are referred to as traditional characters. Across Chinese-speaking polities, mainland China, Malaysia, and Singapore use simplified characters, while Taiwan, Hong Kong, and Macau use traditional characters. In general, Chinese and Japanese readers can successfully identify characters from all three standards. Prior to the 20th century, reforms were generally conservative and sought to reduce the use of simplified variants. During the late 19th and early 20th centuries, an increasing number of intellectuals in China came to see both the Chinese writing system and the lack of a national spoken dialect as serious impediments to achieving the mass literacy and mutual intelligibility required for the country's successful modernization. Many began advocating for the replacement of Literary Chinese with a written language that more closely reflected speech, as well as for a mass simplification of character forms, or even the total replacement of characters with an alphabet tailored to a specific spoken variety. In 1909, the educator and linguist Lufei Kui formally proposed the adoption of simplified characters in education for the first time. In 1911, the Xinhai Revolution toppled the Qing dynasty, and resulted in the establishment of the Republic of China the following year. The early Republican era (1912–1949) was characterized by growing social and political discontent that erupted into the 1919 May Fourth Movement, catalysing the replacement of Literary Chinese with written vernacular Chinese over the subsequent decades. Alongside the corresponding spoken variety of Standard Chinese, this written vernacular was promoted by intellectuals and writers such as Lu Xun and Hu Shih. It was based on the Beijing dialect of Mandarin, as well as on the existing body of vernacular literature authored over the preceding centuries, which included classic novels such as Journey to the West (c. 1592) and Dream of the Red Chamber (mid-18th century). At this time, character simplification and phonetic writing were being discussed within both the ruling Kuomintang (KMT) party, as well as the Chinese Communist Party (CCP). In 1935, the Republican government published the first official list of simplified characters, comprising 324 forms collated by Peking University professor Qian Xuantong. However, strong opposition within the party resulted in the list being rescinded in 1936. People's Republic of China [edit] Traditional (們) Simplified (们) Comparison between character forms, showing systematic simplification of the component ⾨ ('gate') The project of script reform in China was ultimately inherited by the Communists, who resumed work following the proclamation of the People's Republic of China in 1949. In 1951, Premier Zhou Enlai ordered the formation of a Script Reform Committee, with subgroups investigating both simplification and alphabetization. The simplification subgroup began surveying and collating simplified forms the following year, ultimately publishing a draft scheme of simplified characters and components in 1956. In 1958, Zhou Enlai announced the government's intent to focus on simplification, as opposed to replacing characters with Hanyu Pinyin, which had been introduced earlier that year. The 1956 scheme was largely ratified by a revised list of 2235 characters promulgated in 1964. The majority of these characters were drawn from conventional abbreviations or ancient forms with fewer strokes. The committee also sought to reduce the total number of characters in use by merging some forms together. For example, 雲 ('cloud') was written as 云 in oracle bone script. The simpler form remained in use as a loangraph meaning 'to say'; it was replaced in its original sense of 'cloud' with a form that added a semantic ⾬ ('rain') component. The simplified forms of these two characters have been merged into 云. A second round of simplified characters was promulgated in 1977, but was poorly received by the public and quickly fell out of official use. It was ultimately formally rescinded in 1986. The second-round simplifications were unpopular in large part because most of the forms were completely new, in contrast to the familiar variants comprising the majority of the first round. With the rescission of the second round, work toward further character simplification largely came to an end. The Chart of Generally Utilized Characters of Modern Chinese was published in 1988 and included 7000 simplified and unsimplified characters. Of these, half were also included in the revised List of Commonly Used Characters in Modern Chinese, which specified 2500 common characters and 1000 less common characters. In 2013, the List of Commonly Used Standard Chinese Characters was published as a revision of the 1988 lists; it includes a total of 8105 characters. Japan [edit] Main article: Japanese script reform Regional forms of the character 次 in the Noto Serif typeface family. From left to right: forms used in mainland China, Taiwan, and Hong Kong (top), and in Japan and Korea (bottom) After World War II, the Japanese government instituted its own program of orthographic reforms. Some characters were assigned simplified forms called shinjitai; the older forms were then labelled kyūjitai. Inconsistent use of different variant forms was discouraged, and lists of characters to be taught to students at each grade level were developed. The first of these was the 1850-character tōyō kanji list published in 1946, later replaced by the 1945-character jōyō kanji list in 1981. In 2010, the jōyō kanji were expanded to include a total of 2136 characters. The Japanese government restricts characters that may be used in names to the jōyō kanji, plus an additional list of 983 jinmeiyō kanji whose use are historically prevalent in names. South Korea [edit] Hanja are still used in South Korea, though not to the extent that kanji are used in Japan. In general, there is a trend toward the exclusive use of hangul in ordinary contexts. Characters remain in use in place names, newspapers, and to disambiguate homophones. They are also used in the practice of calligraphy. Use of hanja in education is politically contentious, with official policy regarding the prominence of hanja in curricula having vacillated since the country's independence. in 1972, which specified 1800 characters meant to be learned by secondary school students. In 1991, the Supreme Court of Korea published the _Table of Hanja for Use in Personal Names (인명용 한자; Inmyeong-yong Hanja), which initially included 2854 characters. The list has been expanded several times since; as of 2022[update], it includes 8319 characters. North Korea [edit] In the years following its establishment, the North Korean government sought to eliminate the use of hanja in standard writing; by 1949, characters had been almost entirely replaced with hangul in North Korean publications. While mostly unused in writing, hanja remain an important part of North Korean education. A 1971 textbook for university history departments contained 3323 distinct characters, and in the 1990s North Korean schoolchildren were still expected to learn 2000 characters. A 2013 textbook appears to integrate the use of hanja in secondary school education. It has been estimated that North Korean students learn around 3000 hanja by the time they graduate university. Taiwan [edit] The Chart of Standard Forms of Common National Characters was published by Taiwan's Ministry of Education in 1982, and lists 4808 traditional characters. The Ministry of Education also compiles dictionaries of characters used in Taiwanese Hokkien and Hakka.[H] Other regional standards [edit] Singapore's Ministry of Education promulgated three successive rounds of simplifications. The first round in 1969 included 502 simplified characters, and the second round in 1974 included 2287 simplified characters—including 49 that differed from those in the PRC, which were ultimately removed in the final round in 1976. In 1993, Singapore adopted the revisions made in mainland China in 1986. The Hong Kong Education and Manpower Bureau's List of Graphemes of Commonly-Used Chinese Characters includes 4762 traditional characters used in elementary and junior secondary education. Notes [edit] ^漢字; simplified as 汉字 Chinese pinyin: Hànzì; Wade–Giles: Han 4-tzŭ 4; Jyutping: Hon3 zi6 Japanese Hepburn: kanji Korean Revised Romanization: Hanja; McCune–Reischauer: Hancha Vietnamese: Hán tự Also referred to as sinographs or sinograms 2. ^Zev Handel lists: 1. Sumerian cuneiform emerging c. 3200 BCE 2. Egyptian hieroglyphs emerging c. 3100 BCE 3. Chinese characters emerging c. 13th century BCE 4. Maya script emerging c. 1 CE ^According to Handel: "While monosyllabism generally trumps morphemicity—that is to say, a bisyllabic morpheme is nearly always written with two characters rather than one—there is an unmistakable tendency for script users to impose a morphemic identity on the linguistic units represented by these characters." ^This is the Middle Vietnamese pronunciation; the word is pronounced in modern Vietnamese as trăng. 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(2017). The Chinese Typewriter: A History. MIT Press. ISBN978-0-262-03636-8. Myers, James (2019). The Grammar of Chinese Characters: Productive Knowledge of Formal Patterns in an Orthographic System. Routledge. ISBN978-1-138-29081-5. Nawar, Haytham (2020). "Transculturalism and Posthumanism". Language of Tomorrow: Towards a Transcultural Visual Communication System in a Posthuman Condition. Intellect. pp.130–155. ISBN978-1-78938-183-2. JSTORj.ctv36xvqb7. Needham, Joseph; Tsien, Tsuen-hsuin, eds. (2001) . Paper and Printing. Science and Civilisation in China. Vol.V:1 (Reprint ed.). Cambridge University Press. ISBN978-0-521-08690-5 – via Google Books. ———; Harbsmeier, Christoph, eds. (1998). Language and Logic. Science and Civilisation in China. Vol.VII:1. Cambridge University Press. ISBN978-0-521-57143-2. Norman, Jerry (1988). Chinese. Cambridge University Press. ISBN978-0-521-29653-3. 现代汉语 [Modern Chinese] (in Chinese). Peking University Modern Chinese Language Teaching and Research Office, The Commercial Press. 2004. ISBN978-7-100-00940-9. Qiu, Xigui (2000) . Chinese Writing. Early China Special Monograph Series. Vol.4. Translated by Mattos, Gilbert L.; Norman, Jerry. Society for the Study of Early China and The Institute of East Asian Studies, University of California. ISBN978-1-55729-071-7. ——— (2013). 文字学概要 [Chinese Writing] (in Chinese) (2nd ed.). The Commercial Press. ISBN978-7-100-09369-9. Ramsey, S. Robert (1987). The Languages of China. Princeton University Press. ISBN978-0-691-01468-5. Sampson, Geoffrey; Chen Zhiqun (陳志群) (2013). "The Reality of Compound Ideographs". Journal of Chinese Linguistics. 41 (2): 255–272. JSTOR23754815. Shang, Guowen; Zhao, Shouhui (2017). "Standardising the Chinese language in Singapore: Issues of Policy and Practice". Journal of Multilingual and Multicultural Development. 38 (4): 315–329. doi:10.1080/01434632.2016.1201091. ISSN0143-4632. Shaughnessy, Edward L. (1991). Sources of Western Zhou History: Inscribed Bronze Vessels. University of California Press. ISBN978-0-520-07028-8. Su Peicheng (苏培成) (2014). 现代汉字学纲要 [Essentials of Modern Chinese Characters] (in Chinese) (3rd ed.). The Commercial Press. ISBN978-7-100-10440-1. Sybesma, Rint, ed. (2015). Encyclopedia of Chinese Language and Linguistics Online. Brill. doi:10.1163/2210-7363-ecll-all. ISSN2210-7363. Ho, Connie Suk-Han. "Dyslexia, Developmental". In Sybesma (2015). Lee, Chia-Ying (2015a). "Neurolinguistics, Overview". In Sybesma (2015). ——— (2015b). "Sublexical Processes for Reading Chinese Characters, Neurolinguistic Studies". In Sybesma (2015). Tam, Gina Anne (2020). Dialect and Nationalism in China, 1860–1960. Cambridge University Press. ISBN978-1-108-77640-0. Taylor, Insup; Taylor, M. Martin (2014) . Writing and Literacy in Chinese, Korean and Japanese. Studies in Written Language and Literacy. Vol.14 (Revised ed.). John Benjamins. ISBN978-90-272-1794-3. Tong, Xiuli; Liu, Phil D.; McBride-Chang, Catherine (2009). "Metalinguistic and Subcharacter Skills in Chinese Literacy Acquisition". In Wood, Clare Patricia; Connelly, Vincent (eds.). Contemporary Perspectives on Reading and Spelling. Routledge. pp.202–221. ISBN978-0-415-49716-9 – via Google Books. Vogelsang, Kai (2021). Introduction to Classical Chinese. Oxford University Press. ISBN978-0-19-883497-7. Wilkinson, Endymion (2012) . Chinese History: A New Manual. Harvard–Yenching Institute Monograph Series. Vol.85 (3rd ed.). Harvard University Press. ISBN978-0-674-06715-8. ——— (2022) . Chinese History: A New Manual (Enlarged 6th ed.). Harvard University Press. ISBN978-0-674-26018-4. Williams, Clay H. (2010). Semantic vs. Phonetic Decoding Strategies in Non-Native Readers of Chinese(PDF) (PhD thesis). University of Arizona Graduate College of Second Language Acquisition & Teaching. hdl:10150/195163. Xue, Shiqi (1982). "Chinese Lexicography Past and Present". Journal of the Dictionary Society of North America. 4 (1): 151–169. doi:10.1353/dic.1982.0009. ISSN2160-5076. Yang, Lihui; An, Deming (2008). Handbook of Chinese Mythology. Oxford University Press. ISBN978-0-19-533263-6. Yang Runlu (杨润陆) (2008). 现代汉字学 [Modern Chinese Characters] (in Chinese). Beijing Normal University Press. ISBN978-7-303-09437-0. Yin Jiming (殷寄明); et al. (2007). 现代汉语文字学 [Modern Chinese Writing] (in Chinese). Fudan University Press. ISBN978-7-309-05525-2. Yip, Po-ching (2000). The Chinese Lexicon: A Comprehensive Survey. Psychology Press. ISBN978-0-415-15174-0. Yong, Heming; Peng, Jing (2008). Chinese Lexicography: A History from 1046 BC to AD 1911. Oxford University Press. ISBN978-0-19-156167-2. Zhang Xiaoheng (张小衡); Li Xiaotong (李笑通) (2013). 一二三笔顺检字手册 [Handbook of the YES Sorting Method] (in Chinese). The Language Press. ISBN978-7-80241-670-3. Zhao, Liming (1998). "Nüshu: Chinese Women's Characters". International Journal of the Sociology of Language. 129 (1): 127–137. doi:10.1515/ijsl.1998.129.127. ISSN0165-2516. Zhong, Yurou (2019). Chinese Grammatology: Script Revolution and Literary Modernity, 1916–1958. Columbia University Press. doi:10.7312/zhon19262. ISBN978-0-231-54989-9. Zhou, Youguang (1991). Mair, Victor H. (ed.). "The Family of Chinese Character-Type Scripts (Twenty Members and Four Stages of Development)". Sino-Platonic Papers. 28. ——— (2003). The Historical Evolution of Chinese Languages and Scripts 中国语文的时代演进. Pathways to Advanced Skills (in English and Chinese). Vol.8. Translated by Zhang Liqing (张立青). National East Asian Languages Resource Center, Ohio State University. ISBN978-0-87415-349-1. Primary and media sources [edit] ^Maspero, Gaston (1870). Recueil de travaux relatifs à la philologie et à l'archéologie égyptiennes et assyriennes (in French). Librairie Honoré Champion. p.243. ^Laozi (1891). "80". 道德經 [Tao Te Ching] (in Literary Chinese and English). Translated by Legge, James – via the Chinese Text Project. [I would make the people return to the use of knotted cords (instead of the written characters).] ^係辭下 [Xi Ci II]. 易經 [I Ching] (in Literary Chinese and English). Translated by Legge, James. 1899 – via the Chinese Text Project. [In the highest antiquity, government was carried on successfully by the use of knotted cords (to preserve the memory of things). In subsequent ages the sages substituted for these written characters and bonds.] ^Shao Si (邵思) (1035). Explaining Surnames 姓解 (in Literary Chinese). Vol.1. p.1. doi:10.11501/1287529. Retrieved 30 May 2024 – via the National Diet Library. ^Morrison, Robert; Montucci, Antonio (1817). Urh-chih-tsze-tëen-se-yin-pe-keáou: Being a Parallel Drawn Between the Two Intended Chinese Dictionaries. Cadell & Davies, T. Boosey. p.18. ^Technical Introduction. The Unicode Consortium. 22 August 2019. Retrieved 11 May 2024. ^Lunde, Ken; Cook, Richard, eds. (31 July 2024). "Standard Annex #38: Unicode Han Database (Unihan)". The Unicode Standard, Version 16.0.0. The Unicode Consortium. ISBN978-1-936213-34-4. ^ "Introduction". 常用詞辭典 [Dictionary of Frequently-Used Taiwan Minnan]. Taiwan Ministry of Education. 2024. "Introduction". 客語辭典 [Dictionary of Taiwan Hakka]. Taiwan Ministry of Education. 2023. Further reading [edit] See also: Bibliography of the Chinese language and writing system DeFrancis, John (1989). "Chinese". Visible Speech: The Diverse Oneness of Writing Systems. University of Hawaiʻi Press. ISBN978-0-8248-1207-2 – via pinyin.info. Galambos, Imre (2006). Orthography of Early Chinese Writing: Evidence from Newly Excavated Manuscripts(PDF). Eötvös Loránd University. ISBN978-963-463-811-7. Handel, Zev (2025). Chinese Characters Across Asia: How the Chinese Script Came to Write Japanese, Korean, and Vietnamese. University of Washington Press. ISBN978-0-295-75302-7. King, Ross, ed. (2023). Cosmopolitan and Vernacular in the World of Wen 文. Language, Writing and Literary Culture in the Sinographic Cosmopolis. Vol.5. Brill. ISBN978-90-04-43769-2. Mair, Victor H. (2 August 2011). "Polysyllabic Characters in Chinese Writing". Language Log. Mullaney, Thomas S. (2024). The Chinese Computer: A Global History of the Information Age. MIT Press. ISBN978-0-262-04751-7. Pulleyblank, Edwin G. (1984). Middle Chinese: A Study in Historical Phonology. University of British Columbia Press. ISBN978-0-7748-0192-8. Simmons, Richard VanNess, ed. (2022). Studies in Colloquial Chinese and Its History: Dialect and Text. Hong Kong University Press. ISBN978-988-8754-09-0. Tsu, Jing (2022). Kingdom of Characters: The Language Revolution That Made China Modern. Riverhead. ISBN978-0-7352-1472-9. External links [edit] Chinese characters at Wikipedia's sister projects Definitions from Wiktionary Media from Commons News from Wikinews Quotations from Wikiquote Texts from Wikisource Textbooks from Wikibooks Resources from Wikiversity Unihan Database– Reference glyphs, readings, and meanings for characters in The Unicode Standard, with information about the history of Han unification Chinese Text Project Dictionary– Comprehensive character dictionary, including examples of Classical Chinese usage zi.tools– Character lookup by orthography, phonology, and etymology Chinese Etymology by Richard Sears \| show v t e Chinese language \| \| Sinitic languages \| \| \| show Major linguistic groups \| \| \| Mandarin \| \| Northeastern \| Changchun Harbin Shenyang Taz \| \| Beijing \| Beijing Taiwan \| \| Jilu \| Tianjin Jinan \| \| Jiaoliao \| Dalian Qingdao Weihai \| \| Central Plains \| Dongping Gangou Guanzhong Xi'an Luoyang Xuzhou Dungan Xinjiang \| \| Southwestern \| Sichuanese Chengdu–Chongqing (?) 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shorthand Greek (Archaic) Greco-Iberian alphabet Hangul Hanifi Sunuwar Kaddare Kayah Li Klingon Latin Beneventan Blackletter Carolingian minuscule Fraktur Gaelic Insular IPA Kurrent Merovingian Sigla Sütterlin Tironian notes Visigothic Luo Lycian Lydian Manchu Medefaidrin Molodtsov Mru Mundari Bani N'Ko Ogham Ol Chiki Old Hungarian Old Italic Old Permic Orkhon Old Uyghur Mongolian Evenki Galik alphabet Manchu Oirat Vagindra Ol Onal Osage Osmanya Pau Cin Hau Phrygian Pisidian Runic Anglo-Saxon Cipher Dalecarlian Elder Futhark Younger Futhark Gothic Marcomannic Medieval Staveless Shavian Sidetic Sorang Sompeng Sunuwar Tifinagh Todhri Tolong Siki Vellara Visible Speech Vithkuqi Wancho Warang Citi Yezidi Zaghawa \| \| Non-linear \| Braille Maritime flags Telegraph code New York Point Flag semaphore Moon type \| \| \| show Ideograms \| \| Adinkra Aztec Blissymbols Dongba Ersu Shaba Emoji Isotype Kaidā Miꞌkmaw Mixtec New Epoch Notation Painting Nsibidi Anishinaabewibii'iganan Olmec Siglas poveiras Testerian Yerkish Zapotec \| \| show Logograms \| \| \| Chinese family of scripts \| \| Chinese characters \| Simplified Traditional Oracle bone script Bronze scripts Seal script large small bird-worm Hanja Kanji Chữ Nôm Sawndip Bowen \| \| Chinese-influenced \| Jurchen Khitan large script Sui Tangut \| \| \| Cuneiform \| Akkadian Assyrian Elamite Hittite Luwian Sumerian \| \| Other logosyllabic \| Anatolian Bagam Cretan Isthmian Maya Proto-Elamite Tenevil Yi (Classical) \| \| Logoconsonantal \| Demotic Hieratic Hieroglyphs \| \| Numerals \| Hindu-Arabic Abjad Attic (Greek) Muisca Roman \| \| Other \| Sitelen Pona \| \| \| show Semi-syllabaries \| \| \| Full \| Linear Elamite Celtiberian Iberian Northeastern Southeastern Khom Dunging \| \| Redundant \| Espanca script Pahawh Hmong Khitan small script Southwest Paleohispanic Bopomofo Quốc Âm Tân Tự \| \| \| show Sign languages \| \| ASLwrite SignWriting si5s Stokoe notation \| \| show Syllabaries \| \| Afaka Bamum Bété Byblos Canadian Aboriginal Cherokee Cypriot Cypro-Minoan Ditema tsa Dinoko Eskayan Geba Great Lakes Algonquian Iban Idu Kana Hiragana Katakana Man'yōgana Hentaigana Sōgana Jindai moji Kikakui Kpelle Linear B Linear Elamite Lisu Loma Nüshu Nwagu Aneke script Old Persian cuneiform Sumerian Vai Woleai Yi Yugtun \| \| \| \| \| show v t e Braille⠃⠗⠁⠊⠇⠇⠑ \| \| Braille cell \| 1829 braille International uniformity ASCII braille Unicode braille patterns \| \| Braille scripts \| \| French-ordered \| Albanian Azerbaijani Cantonese Catalan Chinese (mainland Mandarin) (largely reassigned) Czech Dutch English (Unified English) Esperanto French German Ghanaian Guarani Hawaiian Hungarian Iñupiaq IPA Irish Italian Latvian Lithuanian Luxembourgish (extended to 8-dot) Maltese Māori Navajo Nigerian Philippine Polish Portuguese Romanian Samoan Slovak South African Spanish Taiwanese Mandarin (largely reassigned) Turkish Vietnamese Welsh Yugoslav Zambian \| Nordic family \| Estonian Faroese Icelandic Scandinavian 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Braille music Canadian currency marks Computer Braille Code Gardner–Salinas braille codes (science; GS8/GS6) International Phonetic Alphabet (IPA) Nemeth braille code \| \| Braille technology \| Braille e-book Braille embosser Braille translator Braille watch Mountbatten Brailler Optical braille recognition Perforation Perkins Brailler Refreshable braille display Slate and stylus Braigo \| \| People \| Louis Braille Charles Barbier Róża Czacka Valentin Haüy Harris Mowbray Thakur Vishva Narain Singh Sabriye Tenberken William Bell Wait \| \| Organisations \| Braille Institute of America Braille Without Borders Japan Braille Library National Braille Association Blindness organizations Schools for the blind American Printing House for the Blind \| \| Other tactile alphabets \| Decapoint Moon type New York Point Night writing Vibratese \| \| Related topics \| Accessible publishing Braille literacy RoboBraille \| \| Portals: Writing Language Asia China \| show Authority control databases \| 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16649	https://www.mea.elsevierhealth.com/firestein-kelleys-textbook-of-rheumatology-2-volume-set-9780323935401.html?srsltid=AfmBOopz1a4ckNcTJzX4w9aDUEWBDP2_tZXJ9s7kHlCDb3JQURMJn3kJ	JavaScript seems to be disabled in your browser. For the best experience on our site, be sure to turn on Javascript in your browser. Shop by Category Terms and Conditions Firestein & Kelley’s Textbook of Rheumatology, 2-Volume Set, 12th Edition More Information \| \| \| --- \| \| ISBN Number \| 9780323935401 \| \| Author Information \| Gary S. Firestein,Iain B. McInnes,Gary Koretzky,Ted Mikuls,Tuhina Neogi,James R. O'Dell \| \| Copyright Year \| 2025 \| \| Edition Number \| 12 \| \| Format \| Book \| \| Format Size \| 8 x 10 mm \| \| Illustrations \| 1200 \| \| Imprint \| Elsevier \| \| Page Count \| 2496 \| \| Publication Date \| 01-10-2024 \| \| Stock Status \| IN STOCK \| eBooks : Built for busy schedules & tailored for your goals. Affordable knowledge, built for you Get the resources you need-often at a lower cost than print. Quality content designed to support your goals, without stretching your budget. Seamless access wherever you are Always in sync Tools the make learning stick Your eBook is ready whenever you are! 1. Check your email for your access code. 2. Sign into or create your VitalSource account and redeem your code. 3. Open your eBook - ready whenever you are! FAQ The access code for your new eBook will be sent in your order confirmation email. Your code can also be accessed in your My Account section on the Elsevier webshop. If you do not receive your code within a few minutes, please check your spam folder. Step-by-step guidance on how to download Bookshelf and also redeem your code can be found here. The access code for your new eBook does not expire. However, we always suggest redeeming immediately after purchase to start experiencing the benefits of and insights from your purchase. Important to note - the code provided is a single use code and only valid for the edition you purchase. It does not provide access to past nor future editions of the title. You will have unlimited access to your eBook on the device to which it was downloaded. Discover the various learning features that our eBooks offer on the Bookshelf® Reader! For example, you can highlight different text passages, create notes and flashcards, have the text read to you, etc. Particularly practical: You can also use your eBooks offline. More information on the learning functions can be found on the Vitalsource page. Quality is our top priority. That's why we collaborate with the leading eBook reader provider VitalSource. VitalSource has its own eBook reader Bookshelf®, which you can easily download. This reader is very user-friendly and offers more features than other standard readers. For example, you can highlight different text passages, create notes and flashcards, have the text read to you, etc. Particularly practical: You can also use your eBooks offline. More information can be found on the Vitalsource page. Elsevier offers its eBooks in ePub format, as we believe this format is best suited to display our content ideally on as many devices as possible. You can return your eBook within 13 days of purchase. eBooks that have been partially printed or flipped through more than 15% are excluded from returns. Any questions ? Support Center > Top Picks from Our Community Chong Tae Kim Jun 2021 Samir B. Pancholy Sep 2025 Kathryn Panasci Nov 2025 Bernadete Ayres Oct 2025 Dawn E. Christenson Sep 2025 Sophie A. Grundy Sep 2025 M. Lindell Joseph Sep 2025 Mary I. Enzman-Hines Sep 2025 Abul K. Abbas Sep 2025 JoAnn Zerwekh Sep 2025 Explore eBooks > VitalSource eBooks for practitioners and students Bring it with you - The Bookshelf app allows you to access books on your laptop, tablet or mobile, so your ebooks go where ever you are - online or offline. Take notes - Highlight, bookmark and take notes and highlights automatically stay in sync no matter where you make them. Listen to eBooks - When you need to go screenless, the Text-to-speech tool will read your book aloud. Powerful search - The searching capabilities allow you to search keywords through all your eBooks, the entire Bookshelf Library and well as on Wikipedia. Ecological - Manage your environmental impact with paperless books. How to read your VitalSource eBooks Copyright © 2025, its licensors, and contributors. All rights are reserved, including those for text and data mining, AI training, and similar technologies. Cookies are used by this site. To decline or learn more, visit our Cookies page. Cookie Settings
16650	https://www.ck12.org/tebook/ck-12-interactive-algebra-i-for-ccss-teachers-guide/section/4.3/	Systems of Linear Equations CCSS Algebra I - TG \| CK-12 Foundation AI Teacher Tools – Save Hours on Planning & Prep. Try it out! What are you looking for? Search Math Grade 6 Grade 7 Grade 8 Algebra 1 Geometry Algebra 2 PreCalculus Science Earth Science Life Science Physical Science Biology Chemistry Physics Social Studies Economics Geography Government Philosophy Sociology Subject Math Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign InSign Up HomeAlgebraFlexBooksCK-12 Interactive Algebra I - Teacher's GuideCh43. Systems of Linear Equations CCSS Algebra I TG Add to Library Share with Classes Add to FlexBook® Textbook Customize Quick tips Notes/Highlights 4.3 Systems of Linear Equations CCSS Algebra I - TG Difficulty Level: At Grade \| Created by: CK-12 Last Modified: Aug 19, 2019 Read Resources Details OVERVIEW SummaryIn this section, students will explore systems of equations. Just as graphs are the solution set for an equation, the intersection point of two equations is the solution set for a system of equations. Students interpret solutions to systems graphically, then develop algebraic methods for solving linear systems. They encounter systems with no solution and systems with infinite solutions. They also build systems that have a given solution. Focus StandardsMathematics Standards: A.CED.A.2Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. A.REI.A.1 Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method. A.REI.C.5 Prove that, given a system of two equations in two variables, replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions. A.REI.C.6 Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables. Mathematical Practices: MP4 Model with mathematics. MP7 Look for and make use of structure. MP8 Look for and express regularity in repeated reasoning. Learning ObjectivesDemonstrate the ability to create systems of equations. Identify methods for finding the solutions to systems of equations. Demonstrate the ability to solve a linear system with the transitive property. Demonstrate the ability to solve a linear system with substitution. Demonstrate the ability to solve a linear system by elimination. Vocabularylinear system, transitive property, substitution property, elimination Expected Duration4 50-minute blocks Additional ResourcesSupplemental Exercises Solutions Study Guide: Linear System Solve by Elimination AGENDA DAY 1aaaaaaaaaaa Warm-up 10 minutes Read and discuss the purpose of the lesson and introduction. Example 1-1 25 minutes Encourage students to try example in groups and then discuss solution as a class. Then students answer in-line questions (A1.4.3 a) on their own. Activity 1 Work it Out 15 minutes Students work on problem in groups or on their own. Then discuss solutions as a class. DAY 2 Warm-up 5 minutes Read the discussion under Activity 2 about the substitution method. Example 2-1 25 minutes Encourage students to try example in groups and then discuss solution as a class.Then students answer in-line questions (A1.4.3 b) on their own. Activity 2 Work it Out 20 minutes Students work on problem in groups or on their own. Then discuss solutions as a class. DAY 3 Warm-up 5 minutes Read the discussion justifying the elimination method in Activity 3. Example 3-1 and 3-2 30 minutes Encourage students to try example in groups and then discuss solution as a class. Then students answer in-line questions (A1.4.3 c) on their own. Activity 3 Work it Out (1-2)15 minutes Students work on problem in groups or on their own. Then discuss solutions as a class. DAY 4 Warm-up 5 minutes Review the elimination method for solving systems of equations. Activity 3 Work it Out (3-9)3 0 minutes Students work on problem in groups or on their own. Then discuss solutions as a class. Interactive: Solving Linear Systems 10 minutes Students use the interactive to explore when linear equations have one solution, no solutions, and an infinite number of solutions. Adaptive Practice 5 minutes Assign adaptive practice on Comparing Methods for Solving Linear Systems for homework. TEACHING NOTES Introduction Work it Out 1 to 4: These problems serve to review the idea of graphs as the solution set for an equation, which prepares them to understand the meaning of the solution of a system. Students are given or must develop various equations and graph them, as well as answer questions about the graphs, sometimes by writing and solving an equation, or just by substituting values. All these problems are built from previously encountered relations. Activity 1 Example 1-1: Students can solve this problem in pairs or groups as they’ve encountered similar before. When they are done, the particular concepts that enables the solution should be detailed as in the solution provided. This is a detailed walk-through of a technique they've already used extensively in the course so far, namely, writing and solving an equation to determine the x that returns the same value for two functions. The essential idea is that once we have decided that two equations form a system, we mean for the variables in both equations to represent precisely those ordered pairs that make both equations true. We seek the set of ordered pairs that makes both equations true, so the two equations are connected by “and”. Students don’t always need to write “and” or draw brackets to indicate a system, but they must understand the difference between a list of unassociated equations and a system. Note: The system here is solved with the transitive property of equality, which allows us to create a distinction between this specialized scenario wherein both equations are in slope-intercept form and we are solving f(x)=g(x),versus the more general use of substitution in subsequent problems. Work it Out 1 to 3: Students solve systems using the transitive property of equality as detailed in the example. They practice converting equations to slope-intercept form in order to rehearse this prerequisite skill for using the transitive property or the substitution method. Activity 2 Example 2-1: The example of using substitution method is deliberately complex, with both equations in standard form and some somewhat unwieldy fractions. This is both to practice those skills but also to foreshadow the usage of the elimination method as a more efficient tool at times. Work it Out 1 to 3: Students solve systems by substitution. Work it Out 4 and 5: Students develop, explore, and try to solve systems with no solution or infinite solutions. Observe their interpretation of these results. Activity 3 Example 3-1: Following a short discussion justifying the elimination method, students are given this example. Walk students through the discussion and example. Example 3-2: Students solve a system of equations using the elimination method by multiplying the equations such that the coefficient of one of the variables is the same. Work it Out 1 to 4: Students practice with elimination method, and contrast it with the use of the transitive property. Students can solve number 16 with elimination without converting to standard form. Work it Out 5 to 9: Students write systems of equations that have given solutions. Graphing is advisable. Observe their decisions and have them share with others. Interactive - Solving Linear Systems: Students explore creating systems with no solution, infinite solutions, and no solution. Summarize results as a class. Notes/Highlights \| Color \| Highlighted Text \| Notes \| \| --- --- \| \| \| Please Sign In to create your own Highlights / Notes \| Currently there are no resources to be displayed. Description This scope and sequence for CK-12 Interactive Algebra I for CCSS is designed to provide pacing outlines that encompass a traditional, 180 day, 2 semester, school year. Learning Objectives Difficulty Level At Grade author Larame Spence Standards Correlations - Concept Nodes Language English Date Created Oct 14, 2017 Last Modified Aug 19, 2019 Vocabulary . Show Details ▼ Previous Graphs of Equations are the Solution Set CCSS Algebra I TG Next Systems of Linear and Non Linear Equations CCSS Algebra I TG Reviews Was this helpful? Yes No 0% of people thought this content was helpful. 0 0 Back to the top of the page ↑ Support CK-12 Foundation is a non-profit organization that provides free educational materials and resources. 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16651	https://mmerevise.co.uk/gcse-maths-revision/cumulative-frequency-gcse-revision-and-worksheets/	Cumulative Frequency Worksheets, Questions and Revision \| MME Revise##### A Level MathsBiologyChemistryPhysicsFurther MathsA Level Predicted Papers 2026 ##### GCSE MathsEnglish LanguageEnglish LiteratureCombined ScienceBiologyChemistryPhysicsGCSE CoursesGCSE Predicted Papers 2025/26 ##### Key Stage KS3 MathsKS3 EnglishKS3 ScienceAll KS2All KS1 ##### 11 Plus 11 Plus11 Plus Exam Papers11 Plus Practice Papers11 Plus Mock Exams11 Plus Schools ##### Functional Skills Functional Skills Maths Level 2Functional Skills Maths Level 1Functional Skills English Level 2Functional Skills Maths PapersFunctional Skills Maths Cards Resit your GCSEs with confidence Expert support to help you pass first time Find out more ##### Past Papers GCSE Maths Past PapersGCSE Science Past PapersGCSE Biology Past PapersGCSE Chemistry Past PapersGCSE Physics Past PapersGCSE Combined Science Past PapersGCSE English Language Past PapersGCSE English Literature Past PapersA Level Maths Past PapersA Level Physics Past PapersA Level Biology Past PapersA Level Chemistry Past PapersView All Resit your GCSEs with confidence Expert support to help you pass first time Find out more Shop##### A Level MathsChemistryPhysicsBiologyFurther MathsView All A Level ##### GCSE MathsScienceEnglishView All GCSE ##### Other Functional SkillsKS3 MathsKS3 EnglishKS2KS1 ##### Courses & Exams Functional SkillsView All Courses & Exams ##### By Type Revision CardsPractice Exam PapersWorksheetsGuides & WorkbooksBundles & Study KitsCourses & Exams ##### Special Offers BundlesSale Not sure what you're looking for? 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Make sure you are happy with the following topics before continuing. Frequency Tables Grouped Frequency Tables Level 6-7 GCSE AQA Edexcel OCR WJEC Cambridge iGCSE Edexcel iGCSE Plotting a Cumulative Frequency graph Below is a frequency table of data compiled on a group of college students’ heights. Step 1: Construct a cumulative frequency table for this data. Calculating the cumulative frequency is just adding up the frequencies as you go along. The first value is the first frequency value, we then add this to the second value to get the second cumulative frequency value 13 + 33 = 46 Continuing this, we get that 46 + 35 = 81 people were 180 cm of shorter, is 81 + 11 = 92 Step 2: Using the cumulative frequency, plot a cumulative frequency graph. The points plotted on your graph should be plotted at the end of each class, i.e. the point which has cumulative frequency of 13 should be plotted at 160 on the height axis, and so on. You should join up the plotted points with a smooth curve. It should end up looking like an elongated ‘S’ shape. Level 6-7 GCSE AQA Edexcel OCR WJEC Cambridge iGCSE Edexcel iGCSE Example: Finding the Median and IQR Using the cumulative frequency graph below, calculate the median and interquartile range. [3 marks] There are 92 people in total, so the lower quartile, median, and upper quartile will be the 23 rd person, 46 th person, and 69 th person respectively. So, we find these points on the y-axis, and then draw a line across to the graph to find the corresponding heights on the x-axis. This is shown on the cumulative frequency graph below. Here, we get Q_1 = 163, \,\, \text{ median } = 170, \,\,Q_3 = 176 The interquartile range is therefore 176-163=13 Level 6-7 GCSE AQA Edexcel OCR WJEC Cambridge iGCSE Edexcel iGCSE Cumulative Frequency Curves Example Questions Question 1: Below is a frequency table of data showing the amount of time people spent on a particular website in one day. a) Complete the cumulative frequency column in the table above. [2 marks] b) Using the data from your table, plot a cumulative frequency graph on the axes below. [3 marks] Level 6-7 GCSE AQA Edexcel OCR WJEC Cambridge iGCSE Edexcel iGCSE a) For the cumulative frequency column, we simply need to add up the frequencies as we move downwards in the table: The first cumulative frequency box is simply the number 16. For the second cumulative frequency box, we need to add 24 to the 16 from the previous cumulative frequency box. 24+16=40 For the third cumulative frequency box, we need to add 19 to the 40 from the previous cumulative frequency box. 19+40=59 This process of adding the frequency total to the cumulative frequency total repeats until the table is complete. The final table should look like this: b) For the cumulative frequency diagram, we need to plot the time in minutes along the x-axis and your cumulative frequency totals on the y-axis. Then we need to plot each of the cumulative frequency figures with the corresponding class interval maximums. In other words, for the time interval of 0 - 20 minutes, you would go along the x-axis to 20 minutes (the maximum in this time range) and go upwards to the cumulative frequency value of 16. Once you have plotted all points, including the origin (0,0), join up all points with a smooth curve. (A cumulative frequency graph is always a smooth curve which goes up.) Your graph should look like this: Save your answers with Gold Standard Education Sign Up Now Show Answer Still marking manually? Save your answers with Find out more Question 2: Shown below is a cumulative frequency graph showing the number of hours per week people spent exercising. Draw a box plot to represent this same information. [4 marks] Level 6-7 GCSE AQA Edexcel OCR WJEC Cambridge iGCSE Edexcel iGCSE From the graph, we can see that the minimum number of hours was 0 and the maximum was 12. We can also see from the graph that is represents exercising data from 72 people since this is where the graph ends. In order to draw a box plot, we need to know the following values: a) the minimum value b) the maximum value c) the median value d) the lower quartile e) the upper quartile We have already established that the minimum value is 0 and the maximum value is 72. The median value is the middle value. Since there are 72 values in total, then the median value is the 36 th value (since 36 is half of 72). If we go up the y-axis and locate the 36 th value, go across to the line and then down, we can see that is corresponds to a value of 3.2 hours. The lower quartile is half-way between the first value and the median. To work out which value is the lower quartile, find \frac{1}{4} of the total number of values: 72 \div4 = 18 The lower quartile is therefore the value of the 18 th term. If we go up the y-axis and locate the 18 th value, go across to the line and then down, we can see that is corresponds to a value of 1.6 hours. The upper quartile is half-way between the final value and the median values. To work out which value is the upper quartile, simply find \frac{3}{4} of the total number of values: \dfrac{3}{4}\times72 = 54 The upper quartile is therefore the value of the 54 th term. If we go up the y-axis and locate the 54 th value, go across to the line and then down, we can see that is corresponds to a value of 5 hours. The graph below illustrates the above: Using this information, the resulting box plot will look like this: Save your answers with Gold Standard Education Sign Up Now Show Answer Still marking manually? Save your answers with Find out more Question 3: The below table shows the number of cars at a showroom and the price brackets that they fall in to: a) Complete the cumulative frequency table, and draw a cumulative frequency graph to represent this data: [4 marks] From your graph work out: b) the approximate median price of a second-hand car [2 marks] c) an estimate for how many cars cost more than £17,500 [2 marks] Level 6-7 GCSE AQA Edexcel OCR WJEC Cambridge iGCSE Edexcel iGCSE a) We know that there are 8 cars with a value between £0 and £10,000[/latex], so we can insert 8 in the £0 - £10,000 cumulative frequency box. The next box is the £0 - £15,000 box. We know that there are 8 cars with a value of less than £10,000, and a further 12 cars which have a value of more than £10,000 but less than £15,000, therefore 20 cars have a value of between £0 and £15,000, so this is the next value we can insert. Continue this process until all values are calculated and your cumulative frequency table should look as follows: To draw the graph, we need to plot the cumulative frequency totals on the vertical axis (the cumulative frequency is always on the y-axis and the price in pounds on the x-axis. However, since we are dealing with grouped data (the prices are price bands), we need to plot the cumulative frequency total against the highest value in the band. So, the first point we plot (after plotting (0,0), the origin) would be the cumulative frequency total of 8 against £10,000 (the top value in the £0 - £10,000 band). The next point we would plot would be the cumulative frequency value of 20 against £15,000 (the highest value in the £10,000 - £15,000 band). Your completed cumulative frequency graph should look as follows: b) We know that there are 48 cars in the showroom in total (since the maximum value on the cumulative frequency table is 48). To find the median, we need to read the value of the car that corresponds to a cumulative frequency total of 24 (half of 48). By locating the value of 24 on the cumulative frequency axis, we can see that this corresponds to a value of approximately £16,000. c) For this question, we need to locate £17,500 on the x-axis and see what cumulative frequency total this corresponds to. By drawing a line up from £17,500 until it touches the line and then drawing a horizontal line to the y-axis, we should hit a cumulative frequency total of approximately 26. This means that 26 cars have a value that is up to£17,500. Therefore the remaining cars must have a value which is greater than £17,500. Since there are 48 cars in total then the number of cars which have a value of more than £17,500 is simply 48-26=22 cars. Save your answers with Gold Standard Education Sign Up Now Show Answer Still marking manually? Save your answers with Find out more Question 4: The lengths of snakes kept in Bob Exotic’s Snake Sanctuary were recorded. The below table shows this information: a) Draw a cumulative frequency graph for this information on the axes below. [3 marks] b) What is the interquartile range for the length of these snakes? [3 marks] c) In another snake sanctuary in the same country, the median length of their snakes is 1.78 m. To one decimal place, by what percentage are these snakes smaller than the snakes in Bob Exotic’s Snake Sanctuary? [3 marks] Level 6-7 GCSE AQA Edexcel OCR WJEC Cambridge iGCSE Edexcel iGCSE a) In order to draw our cumulative frequency graph, we need to work out the cumulative frequency totals: To represent this information on a graph, we need to plot the cumulative frequency totals on the vertical axis against snake length on the horizontal axis. After plotting our first point of (0,0), the next point is (1,23); the key thing to remember is that since the snake length data is grouped, we need to plot the highest value in each length band (so for the 0 metres – 1 metre band, we would plot the corresponding cumulative frequency total against 1 metre). Once we have plotted all the points, we need to join them together with a smooth line, with an end result similar to the below: b) The interquartile range is calculated by subtracting the lower quartile from the upper quartile. In this data set, the lower quartile is the length of the 40 th snake (40 because 40 is \frac{1}{4} of 160. To find the length of the 40 th snake, find 40 on the vertical cumulative frequency axis and find the corresponding length on the horizonal axis. The length of the 40 th snake is approximately 1.5 metres. In this data set, the upper quartile is the length of the 120 th snake (120 because 120 is \frac{3}{4} of 160. To find the length of the 120 th snake, find 120 on the vertical cumulative frequency axis and find the corresponding length on the horizonal axis. The length of the 120 th snake is approximately 3.5 metres. Therefore the interquartile ranges is 2 metres. c) For this question, we need to work out the median snake length at Bob Exotic’s Snake Sanctuary. Since there are 160 snakes in total, the median snake length is the length of the 80 th snake (80 because 80 is \frac{1}{2} of 160. The 80 th snake has a length of approximately 2.6 m. If at the other snake sanctuary, the median snake length is 1.78 metres, then to calculate how much smaller as a percentage, we need to find out how much smaller the median snake is: 2.6\text{ m } – 1.78\text{ m} = 0.82\text{ m} To work out a percentage increase or decrease, you need to remember the simply formula: \dfrac{ \text{ difference}}{\text{ original value}}\times 100\dfrac{0.82\text{ m}}{2.6\text{ m}}\times 100=31.5\% (In this question, however, it may not be obvious what the original value is. The question asks us to work out by what percentage these snakes are smaller than the snakes in Bob Exotic’s Snake Sanctuary. Because of the word ‘than’, it is the length of the snakes at Bob Exotic’s Snake Sanctuary that we should consider as the ‘original’ value. Words / phrases like ‘compared to’ or ‘than’ always indicate what we are working out a percentage of.) Save your answers with Gold Standard Education Sign Up Now Show Answer Still marking manually? Save your answers with Find out more Specification Points Covered interpret and construct tables, charts and diagrams, including frequency tables, bar charts, pie charts and pictograms for categorical data, vertical line charts for ungrouped discrete numerical data, tables and line graphs for time series data and know their appropriate use construct and interpret diagrams for grouped discrete data and continuous data, i.e. histograms with equal and unequal class intervals and cumulative frequency graphs, and know their appropriate use interpret, analyse and compare the distributions of data sets from univariate empirical distributions through: appropriate graphical representation involving discrete, continuous and grouped data, including box plots appropriate measures of central tendency (median, mean, mode and modal class) and spread (range, including consideration of outliers, quartiles and inter-quartile range) apply statistics to describe a population Cumulative Frequency Curves Worksheet and Example Questions (NEW) Cumulative Frequency Exam Style Questions - MME Level 6-7 GCSE New Official MME Exam QuestionsMark Scheme Cumulative Frequency Curves Drill Questions Cumulative Frequency - Drill Questions Level 6-7 GCSE Exam QuestionsMark Scheme Related Topics Frequency Tables Worksheets, Questions and Revision Level 4-5 GCSE Revise Grouped Frequency Tables Worksheets, Questions and Revision Level 4-5 GCSE Revise Where next? ###### Previous GCSE Maths Topic #### Box Plots ###### GCSE Maths Revision Home #### Go back to the main GCSE Maths topic list ###### Next GCSE Maths Topic #### Histograms Secure Shopping Ways to Pay Spread the Cost Socials Follow our socials for revision tips and community support Contact Details 020 3633 5145 revise@mmerevise.co.uk Mon - Fri: 08:45 - 19:00, Sat 09:30-16:00 Evans Business Centre, Hartwith Way, Harrogate HG3 2XA Information About Us Contact Us MME Blog Reviews Safeguarding Modern Slavery Statement Careers For Teachers Become a Tutor MME Affiliates Help Articles Pass Browse Subjects A Level Maths Revision A Level Biology Revision A Level Chemistry Revision A Level Physics Revision A Level English Language Revision GCSE Maths Revision GCSE English Language Revision GCSE Science Revision KS3 Revision KS2 Revision KS1 Revision 11 Plus Numerical Reasoning Popular Products GCSE Predicted Papers A Level Predicted Papers GCSE Maths Books Functional Skills Level 2 Course in a Box – GCSE Maths (Guaranteed Pass) GCSE Maths Revision Guide Privacy & Policy / Terms / Site Map Providing Ofqual Approved Qualifications © 2025 MME When you visit or interact with our sites, services or tools, we or our authorised service providers may use cookies for storing information to help provide you with a better, faster and safer experience and for marketing purposes. 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16652	https://arxiv.org/pdf/2407.02102	Published Time: Wed, 03 Jul 2024 01:08:01 GMT arXiv:2407.02102v1 [math.CO] 2 Jul 2024 Separating the edges of a graph by cycles and by subdivisions of K4F´ abio Botler 1 T´ assio Naia 2 July 3, 2024 1 Departamento de Ciˆ encia da Computa¸ c˜ ao Instituto de Matem´ atica e Estat´ ıstica Universidade de S˜ ao Paulo, Brasil E-mail: fbotler@ime.usp.br 2 Centre de Recerca Matem` atica Belaterra, Spain E-mail: tnaia@member.fsf.org Abstract A separating system of a graph G is a family S of subgraphs of G for which the following holds: for all distinct edges e and f of G, there exists an element in S that contains e but not f . Recently, it has been shown that every graph of order n admits a separating system consisting of 19 n paths [Bonamy, Botler, Dross, Naia, Skokan, Separating the Edges of a Graph by a Linear Number of Paths , Adv. Comb., October 2023], improving the previous almost linear bound of O( n log ⋆ n)[S. Letzter, Separating paths systems of almost linear size , Trans. Amer. Math. Soc., to appear], and settling conjectures posed by Balogh, Csaba, Martin, and Pluh´ ar and by Falgas-Ravry, Kittipassorn, Kor´ andi, Letzter, and Narayanan. We investigate a natural generalization of these results to subdivisions of cliques, showing that every graph admits both a separating system consisting of 41 n edges and cycles, and a separating system consisting of 82 n edges and subdivisions of K4. Consider a family S of subsets of a set X. Given e, f ∈ X, we say that an element S ∈ S separates e from f if e ∈ S and f /∈ S. Furthermore, we say that S weakly separates X if for every pair of elements e, f ∈ X there is an element S ∈ S that either separates e from f or that separates f from e; and we say that S strongly separates X if for every pair of elements e, f ∈ X there is an element S ∈ S that separates e from f ,and an element that separates f from e. Such a family S is then called a (weak or strong, respectively) separating system of X. The study of separating systems dates back to the 1960s [ 10 , 15 , 16 ], and gained substantial attention in the last years (see, e.g., [ 1, 3, 4]). Our focus is a variation posed in 2013 by Katona (see [ 2, 7]) in which one seeks to separate the edge set of a given graph by a small collection of paths. In particular, special attention has been given to the case of complete graphs [ 8, 11 , 17 ]. For our purposes, a separating path system (resp. separating cycle system ) of a graph G is a collection S of paths (resp. edges and cycles) in G such that for all distinct edges (e, f ) ∈ E(G) × E(G) there exists S ∈ S that separates e from f . Note that this definition fits into the category of strong separating systems. Also note that, in the definition of cycle separating system, one is allowed to use isolated edges. This is required in order to separate graphs that contain bridges (i.e., edges that do not lie in any cycle). 1Inspired by Katona’s question, Falgas-Ravry, Kittipassorn, Kor´ andi, Letzter, and Narayanan [ 7] conjectured that every graph on n vertices admits a weak separating sys-tem size O( n) consisting solely of paths, while verifying it for a number of cases. Their conjecture was later strengthened by Balogh, Csaba, Martin, and Pluh´ ar [ 2]. Conjecture 1 ([ 2, 7]) . Every graph of order n admits a path separating system of size O( n). In 2023, Bonamy, Dross, Skokan and the authors confirmed Conjecture 1, proving that every graph on n vertices admits a separating path system of size at most 19 n . This improved the almost linear bound of O( n log ⋆ n) found by Letzter in 2022 [ 13 ]. It is then natural to ask whether every graph also admits a separating cycle system of size O( n). This question was independently posed by Gir˜ ao and Pavez-Sign´ e1, and in this paper we answer it affirmatively (see Section 2). Theorem 2. Every graph on n vertices admits a separating cycle system of size 41 n. In fact, we are interested in a more general setting in which elements of the separating system are either edges or subdivisions of a given graph, which we now make precise. We say that H∗ is a subdivision of H if H can be obtained from H∗ by repeatedly deleting a vertex of degree 2 and adding a new edge joining its neighbors. Let H and G be graphs, and let F be a family of subgraphs of G such that every element of F is either an edge or a subdivision of H. We say that F is an H-cover of G if E(G) = ⋃ H′∈F E(H′); and we say that F is an H-separating system of G if for all distinct edges ( e, f ) ∈ E(G) × E(G) there is an element H′ ∈ F that separates e from f . It is not hard to see that every H-separating system of G is an H-cover of G, and that a separating path (resp. cycle) system is a K2-separating (resp. K3-separating) system. Therefore, the results mentioned above say that every graph on n vertices admits a K2- and a K3-separating system of size O(n). Motivated by Conjecture 1, we propose the following more general edge separation conjecture. Conjecture 3. For every graph H there is a constant C = C(H) for which every graph on n vertices admits an H-separating system of size at most C · n. It is not hard to check that in order to solve Conjecture 3, it suffices to verify it in the case H is a complete graph. The main result of this paper is that Conjecture 3 holds in the case H = K4 (see Section 1). Theorem 4. Every graph on n vertices admits a K4-separating cycle system of size 82 n. Since every subdivision of K4 can be covered by two cycles, Theorem 4 implies a version of Theorem 2 in which the constant 41 is replaced by 164. The strategy presented here is similar to the strategy in [ 5]. Namely, we reduce the main problem to the case of graphs containing a certain spanning subdivision of a clique. Next, we define a linear number of special matchings that separate the edges outside this structure, and cover each such matching by a suitable subdivision. We emphasize that we make no effort to reduce the multiplicative constant. Path separation versus K4 separation. In [ 5], the authors reduce the problem to the case of graphs containing a Hamilton path P = v1 · · · vn. They consider 5 n match-ings Mk = {vivj ∈ E(G) : i + j = k, i < j } and Nk = {vivj ∈ E(G) : i + 2 j = k, i < j } 1Personal communication. 2and show that each such matching M can be covered by a path PM with M ⊆ E(PM ) ⊆ E(P ∪ M). The argument is completed using a linear path-covering result. In contrast, here the reduction is to graphs containing a subdivision K of K4 with the property that K contains a Hamilton cycle C = v1 · · · vnv1. We consider 3 n match-ings Mk = {vivj ∈ E(G) : j − i = k, i < j } and Nk = {vivj ∈ E(G) : j − 2i = k, i < j }.While some of these matchings cannot be covered by a single subdivision (see Figure 1), we can show that a bounded number of subdivisions suffice. The argument is completed using a linear clique-subdivision-covering result. Figure 1: A graph consisting of a cycle together with a matching M ⊆ { uiuj : j − i = 4 } (in dashed red) having no cycle that contains all edges of M. P´ osa rotation-extension. We use the following standard notation. Given a graph G,and a set S ⊆ V (G), we denote by NG(S) the set of vertices not in S adjacent in G to some vertex in S. We omit subscripts when clear from the context. Given a graph G and vertices u, v in G, let P = u · · · v be a path from u to v. If x ∈ V (P ) is a neighbor of u in G and x− is the vertex preceding x in P , then P ′ = P −xx − +ux is a path in G for which V (P ′) = V (P ). We say that P ′ has been obtained from P by an elementary exchange fixing v (see Figure 2). A path obtained from P by a (possibly empty) sequence of elementary exchanges fixing v is said to be a path derived from P .The set of endvertices of paths derived from P distinct from v is denoted by Sv(P ). Since all paths derived from P have the same vertex set as P , we have Sv(P ) ⊆ V (P ). The following lemma arises when P is a longest path ending at v (for a proof see also [ 5]). u x− x v Figure 2: a path (highlighted) obtained by an elementary exchange fixing v. Lemma 5 ([ 6]) . Let P = u · · · v be a longest path of a graph G and let S = Sv(P ). Then \|NG(S)\| ≤ 2\|S\|. 1 K4-separating systems In this section we verify Conjecture 3 when H = K4. Our argument requires a result stating that every n-vertex graph admits a K4-cover of size O( n). Although we can prove such a statement simultaneously to the separation result, we obtain a better leading constant by using the following result due to Jørgensen and Pyber [ 9]. In fact, Jørgensen and Pyber showed that every n-vertex graph admits a Kt-cover of size O t(n). 3Theorem 6 ([ 9]) . Every graph on n vertices admits a K4-cover of size at most 2n − 3. The proof of Theorem 4 is divided into four parts. First we apply Lemma 5 in order to reduce the problem to the case where the studied graph contains a special spanning subdivision of K4; then we define the Mk’s and Nk’s and show how to partition them into matchings in order to avoid the problem illustrated in Figure 1; the third step is to consider the union of each matching and the spanning subdivision of K4, and show that it contains another subdivision of K4 covering the matching; and we finally argue that the obtained collection is the desired K4-separating system. Let H1 and H2 be (not necessarily edge-disjoint) subgraphs of a graph G, and consider a family S of subgraphs of G. We say that S separates H1 from H2 if for all distinct edges ( e, f ) ∈ E(H1) × E(H2), there is S ∈ S that separates e from f . Therefore, S is a separating system of G if S separates G from itself. Proof of Theorem 4. We proceed by induction on n. Let G be a graph with n vertices. If G is empty, the result trivially holds. If not, we consider a longest path P = u · · · v of G and put S = Sv(P ) as in Lemma 5. Now, let w be the vertex that is closest to v in P and has a neighbor u′ in S, and let P ′ = u′ · · · v be a path obtained from P by elementary exchanges fixing v so that P ′ starts with u′. Then C = ( P ′ + u′w) \ E(P [w, v ]) is an edge or a cycle that contains S ∪ N(S) ( C is an edge when S = {u} and \|N(S)\| = 1). Let H be the subgraph of G induced by the edges incident to vertices of S, and note that h = \|V (H)\| ≤ 3\|S\|. Let G′ = G \ S. Note that, since G is not empty, S is not empty. By the induction hypothesis, there is a K4-separating system S′ of G′ of size at most 82( n − \| S\|). Note that S′ separates G′ from H since S′ covers G′. In what follows, we construct a set S of edges and subdivisions of K4 that separates H from G. Moreover, we obtain S so that \|S\| ≤ 82 \|S\|, and hence S′ ∪ S is a K4-separating system of G with cardinality at most 82( n − \| S\|) + 82 \|S\| = 82 n as desired. Now, we show that either e(H) ≤ 82 \|S\| or H contains a subdivision of K4 that contains C. For that, let us write C = u1 · · · umu1. We remark that C may contain vertices in V (G) \ V (H). Let v1, . . . , v h be the vertices of H in the order that they appear in C. Formally, for i ∈ [h] = {1, . . . , h }, we define σ(i) to be the index of the ith vertex uσ(i) of H in C, and set vi = uσ(i). In what follows, each edge vivj is written so that i < j . We say that two edges vivj and vi′ vj′ of H \ E(C) cross if vi′ and vj′ lie in distinct components of C \ { vi, v j }. If no two edges in H \ E(C) cross, then H is an outerplanar graph, and hence e(H) ≤ 2h − 3 (note that h ≥ 2). In this case we set S as the set of subgraphs each consisting of a single edge in H, and we are done because \|S\| = \|e(H)\| ≤ 2h − 3 ≤ 6\|S\| ≤ 82 \|S\|.Therefore, we may assume that there are crossing edges e, e ′ in H \ E(C). Note that in this case K = C + e + e′ is a subdivision of K4. Let KS = E(K) ∩ E(H) be the set of edges of K having at least one vertex in S (see Figure 3), and let KS be the set of subgraphs each consisting of a single edge in KS . Observe that \|KS \| = e(KS ) ≤ 2\|S\| + 2. By Theorem 6, there is a K4-cover D′ of H′ = H \ E(KS ) of size at most 2 h − 3 ≤ 6\|S\|.Note that KS ∪ D′ has size at most 2 \|S\| + 2 + 6 \|S\| ≤ 10 \|S\| (using that 2 ≤ 2\|S\|) and separates (i) H from G′; (ii) H′ from K; and (iii) KS from G.It remains to create a set of at most 24 h ≤ 72 \|S\| edges and subdivisions of K4 that separates H′ from itself (see Figure 3). In order to obtain the final elements of the separating system, we define special matchings. Easy matchings. A slicing A1, . . . , A s of [ a1, h ] is a sequence of intervals which together 4S N(S) H′ G′ KS K S KS D′ S D′ KS S KS Figure 3: Left: a set S in a Hamiltonian graph and its neighborhood N(S); part of a Hamiltonian cycle is highlighted, dashed red edges are the edges in CS . Right: cycles and edges that separate subgraphs of G, where A α → B indicates that α separates A from B (for instance, Q′ separates G′ from H′). partition [ a1, h ]. We always assume that intervals are ordered consecutively, i.e., so that each x ∈ Ai precedes all y ∈ ⋃ j>i Aj . Given an interval A ⊆ [h], we say that vivj starts in A if i ∈ A; and that vivj ends in A if j ∈ A. We say that a matching M ⊆ E(H′) is easy if there is a1 ∈ [h], a slicing A1, . . . , A s of [ a1, h ] and t ∈ { 0, 1} such that the following hold. (a) for each vivj ∈ M there is r ∈ [s − 1] such that r ≡ t (mod 2), vi ∈ Ar and j ∈ Ar+1 ; (b) the edges starting in Ar are pairwise crossing, for all r ∈ [s]; and (c) for each r ∈ [s], the number of edges starting in Ar is either zero or is odd. As we shall see, (a) –(c) suffice to overcome our main obstruction when covering match-ings by K4 subdivisions. Now, for k ∈ [h − 1], let Mk = {vivj ∈ E(H′) : j − i = k}, and for k ∈ [−h + 2 , h − 2], let Nk = {vivj ∈ E(H′) : j − 2i = k}. It is not hard to check that Mk and Nk are linear forests, and each edge of H′ is in precisely one Mk and precisely one Nk. This defines at most 3 h linear forests. In what follows, we partition each Mk and Nk into four easy matchings, yielding 12 h easy matchings. First, in items (i) and (i’) below, we partition, respectively, each Mk and Nk into two matchings satisfying (a) and (b) . (i) Let k ∈ [h − 1]. We partition Mk as follows. Let A1, . . . , A ⌈n/k ⌉ be a slicing of [ h]into sets of size k and at most one set A⌈n/k ⌉ of size at most k. Now, given t ∈ { 0, 1},we let At = ⋃ r≡t(mod 2) Ar, and then put Mk,t = {vivj ∈ Mk : i ∈ At}. So, for example, Mk, 1 consists of the edges of Mk that start in “odd” intervals of [ h]. If i ∈ Ar, then j ∈ Ar+1 because j − i = k = \|Ar\| (see Figure 4). This implies that if vivj ∈ Mk,t , then i ∈ At while j ∈ A1−t. In particular, Mk,t is a matching. Therefore, the matchings Mk,t satisfy (a) . Now, suppose that vivj , v i′ vj′ ∈ Mk,t are such that i, i ′ ∈ Ar for some r, and suppose, without loss of generality, that i < i ′. Then j = k + i < k + i′ = j′, and hence vivj and vi′ vj′ cross. Therefore, the matchings Mk,t satisfy (b) as desired. This step defined at most 2 h matchings. 51 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 A1 A2 A3 A4 A5 Figure 4: A partition of M4 into M4,0 and M4,1 in, respectively, dashed blue and solid red. (i’) Let k ∈ [−h + 2 , h − 2]. We partition Nk as follows. We consider two cases. If k ≥ 0, then let a1 = 1; otherwise let a1 = 1 − k ≥ 2. Now, for each r ≥ 1 let ar+1 = 2 ar + k.Since a1 > −k, we can prove by induction that ar+1 > a r > −k for all r. Also, let d be the minimum integer for which h < a d+1 . We partition [ a1, h ] into sets A1, . . . , A d so that Ar = [ ar, a r+1 − 1] for r ∈ [d − 1], and Ad = [ ad, h ]. Note that every edge of Nk starts in ⋃ j∈[d] Ad. Indeed, let vivj ∈ Nk. Then j = k +2 i and we claim that i ≥ a1. This is clear if a1 = 1. Otherwise, a1 = 1 − k, but since j > i , we must have k + 2 i ≥ i + 1, and hence i ≥ 1 − k = a1. Moreover, if vivj starts in Ar, then vivj ends in Ar+1 since ar+1 = k + 2 ar ≤ k + 2 i ≤ k + 2( ar+1 − 1) < k + 2 ar+1 = ar+2 . Given t ∈ { 0, 1}, we let At = ⋃ r≡t(mod 2) Ar, and put Nk,t = {vivj ∈ Nk : i ∈ At}.As in the preceding case, we can prove that the Nk,t are matchings that satisfy both (a) and (b) as desired. This step defined at most 4 h matchings. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 A1 A2 A3 A4 Figure 5: A partition of N1 into N1,0 and N1,1 in, respectively, dashed blue and solid red. (ii) Finally, we partition each Mk,t and Nk,t obtained in steps (i) and (i’) into matchings satisfying (c) . It is not hard to see that properties (a) and (b) are hereditary, i.e., that if a matching M satisfies (a) and (b) , then any subset of M satisfies (a) and (b) .Now, for each k ∈ [h] and each t ∈ { 0, 1} we partition Mk,t into sets Mk,t, 0 and Mk,t, 1 so that Mk,t, 0 and Mk,t, 1 have either zero or an odd number of edges starting in each Ar. This works because every even-sized matching can be partitioned into two odd-sized matchings and every odd-sized matching can be partitioned into an odd-sized matching and an empty matching. We obtain Nk,t, 0 and Nk,t, 1 analogously. This step defined at most 12 h matchings. Let M = { Mk,t,ρ : k ∈ [h], t, ρ ∈ { 0, 1}} N = { Nk,t,ρ : k ∈ [−h + 2 , h − 2] , t, ρ ∈ { 0, 1}} 61 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 A1 A2 A3 A4 A5 Figure 6: A partition of M4,0 into M′ 4,0 and M′′ 4,0 in, respectively, solid red and dashed red. and note that ⋃ M∈M M = E(H′) = ⋃ M∈N M. Building subdivisions of K4. In what follows, given M ∈ M ∪ N we show that K ∪ M contains either one subdivision of K4 that contains M; or one edge and a subdivision of K4 that together cover M. Let A1, . . . , A d be the slicing defined in (i) or (i’) when selecting edges for M. This construction is divided into two cases. We say that M is elementary if for every r ∈ [d], there is at most one edge of M that starts in Ar; otherwise we say that M is jumbled . By construction, if M is elementary, then no two edges of M cross. Recall that e and e′ are edges not in H′ that we use to complete C to a subdivision K of K4. Elementary. Suppose M is elementary. We divide this proof into two cases, depending on whether M contains an edge crossing e or e′. Let w1, w 2, w 3, w 4 be the vertices of e and e′ in the order that they appear in C, and assume, without loss of generality, that e = w1w3 and e′ = w2w4.Case 1. Suppose that no edge of M crosses e or e′. Since no two edges of M cross, there is at most one edge, say f = vivj , of M that starts before w1 and ends after w4, in which case we pick f as a single edge. In either case, we cover M (or M \ { f }) by a subdivision KM of K4 obtained from K by “taking shortcuts”: For every edge vivj ∈ M (or vivj ∈ M \ { f }), we replace the path C[vi, v j ] by the edge vivj .Case 2. Suppose that M contains some edge f that crosses e or e′. Assume without loss of generality that f and e cross. Note, in particular, that by the construction of M, no edge of M can start before w1 and also end after w4. Since no two edges of M cross, there is at most one other edge, say f ′, of M that crosses e, in which case we pick f ′ as a single edge. Then K′ = C + f + e is a subdivision of K4, and we obtain a subdivision KM of K4 from K′ analogously to Case 1: for every edge vivj ∈ M \ { f } (or vivj ∈ M \ { f, f ′}), we replace the path C[vi, v j ] by vivj .This concludes the analysis when M is elementary. Jumbled. In this case we do not use e and e′. The idea here is to modify C into a cycle C′ ⊆ C ∪ M that contains M, and then restore some of the edges removed from C to obtain a subdivision of K4. Recall that M is easy, and hence there exist a1, A1, . . . , A d,and t such that (a) –(c) hold. Modifying C. Let Ar be an interval of [ h] for which the set Mr of edges of M that start in Ar (and end in Ar+1 ) is non-empty. Then, by (a) , no edge of M starts in Ar+1 . Let Qr be the shortest subpath of C that contains the vertices vi with i ∈ Ar ∪ Ar+1 , and let Rr = Qr ∪ Mr . We show that there is a path Q′ r ⊆ Rr that contains Mr and has the same end vertices as Qr. Then C′ is obtained from C by replacing each Qr by Q′ r .The existence of Q′ r can easily be shown by induction. Alternatively, one can use the following construction. Let s = \|Mr\| and let w1, . . . , w 2s be the vertices of Qr incident to 7edges of Mr in the order that they appear in Qr . Note that w1 and w2s are the end vertices of Qr, and that s is odd by (c) . By (b) , Mr = {wiwi+s : i ≤ s}. Now, let Q′ r be the graph obtained from Rr by removing the edges in Qr[wi, w i+1 ] for odd i with i ≤ 2s − 1 and removing isolated vertices (see Figure 7). Note that Rr is a subcubic graph whose vertices with degree 3 are precisely w2, . . . , w 2s−1, and that, to obtain Q′ r , we remove from Rr precisely one edge incident to each wi. Thus ∆( Q′ r ) ≤ 2. We claim that Q′ r is connected. For that, we prove that Q′ r contains a path joining wi to wi+1 for each i ∈ [2 s − 1]. This is clear if i is even, because the segment Qr[wiwi+1 ] has not been removed. For odd i,on the other hand, the segment Qr[wi+swi+s+1 ] has not been removed because s is odd (and consequently i + s is even), and thus wiwi+sQr[wi+swi+s+1 ]wi+1 is the desired a path in Q′ r . Moreover, the end vertices of Q′ r are w1 and w2s and Mr ⊆ E(Q′ r ). This concludes the construction of Q′ r . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 w1 w2 w3 w4 w5 w6 w7 w8 w9 w10 Figure 7: Thick lines illustrate the path Q′ r that contains Mr ⊆ N6,t,r for some t and r.Turning C′ into a subdivision of K4. To turn C into a subdivision of K4 we only need to restore two of the removed Qr[wi, w i+1 ] for some r. More precisely, since M is jumbled, there is an interval Ar in which at least two edges of M start. Let Mr ⊆ M be the set of edges that start in Ar. Let s = \|Mr\| and let w1, . . . , w 2s be the vertices of P incident to edges of Mr as above. Recall that the removed subpaths were Qr[wi, w i+1 ] for odd i. Moreover, recall that s is odd by (c) and thus s ≥ 3. Therefore, Qr [w1, w 2] and Qr[ws+2 , w s+3 ] were removed. Let KM = C′ ∪ Qr[w1, w 2] ∪ Qr[ws+2 , w s+3 ]. We claim that KM is a subdivision of K4. Indeed, by construction of C′, the order in which the vertices w1, . . . , w 2s appear is w1, w s+1 , w s+2 , w 2, w 3, w s+3 , . . . , w i, w i+s, w i+s+1 , w i+1 , . . . , w 2s. Let p(wi) be the position of wi in the order above. In particular, we have p(w1) = 1, p(ws+2 ) = 3, p(w2) = 4 and p(ws+3 ) = 6. Then, p(w1) < p (ws+2 ) < p (w2) < p (ws+3 ), and hence KM is a subdivision of K4. This step creates a single subdivision of K4 for each M.Now, by applying the construction above on each M in M ∪ N, we obtain a set K of at most 24 h subdivisions of K4. We claim that K separates each pair of edges in H′.Indeed, let vivj and vi′ vj′ be edges of H′. Each such edge belongs to exactly one M ∈ M and exactly one N ∈ N. If they belong to different elements of M, then they belong to different elements of K and are separated, because M partitions E(H′). The same argument works for N. Therefore, we may assume that i − j = i′ − j′ and i − 2j = i′ − 2j′.This immediately yields j = j′ and i = i′, so vivj = vi′ vj′ . Therefore, any two distinct edges in H′ are separated by K, as desired. This concludes the proof. 82 Separating into cycles First, note that K4 can be covered by two cycles (indeed, suppose V (K4) = and note that the cycles 12341 and 12431 cover it). Hence every subdivision of K4 can be covered by two cycles. This means that any K4-separating system S yields a K3-separating system of size at most 2 \|S\|. And thus, by Theorem 4, every graph on n vertices admits a K3-separating system of size at most 164 n, which verifies Conjecture 3 when H = K3. Aslightly better result, Theorem 2 (any graph on n vertices can be separated by at most 41 n edges and cycles) can be obtained using the following theorem of Pyber [ 14 ]. Theorem 7 ([ 14 ]) . Every graph G contains \|V (G)\| − 1 cycles and edges covering E(G). The proof Theorem 2 follows that of Theorem 4. However, after partitioning the Mk’s and Nk’s into 12 h special matchings, there is no need to consider elementary and jumbled cases, since there is no need to find crossing edges (one proceed directly to the “Modifying C” part of the argument, in which we obtain a cycle C′). Of course, one should also skip the step for “Turning C′ into a subdivision of K4”. 3 Conclusion In this paper we give the first steps towards showing that every graph admits a linearly-sized separating system formed by edges and subdivisions of Kt, for fixed t. Perhaps these techniques can be fitted to find subdivisions of larger cliques. If a suitable subdivision of Kt (t > 4) — namely, a subdivision containing a Hamilton cycle — can be shown to exist for larger cliques, then the tricks used in our proof might still be useful to produce a Kt-separating system. Conjecture 3 could be weakened by allowing a wider class of structures. For example, one could seek a separating system consisting of graphs that yield some fixed graph H through a series of edge contractions; or consisting of immersions of H (for a more about immersions, see, e.g., [ 12 ]). Acknowledgments We thank Marthe Bonamy for fruitful discussions and suggestions. This research has been partially supported by Coordena¸ c˜ ao de Aperfei¸ coamento de Pessoal de N´ ıvel Superior – Brasil – CAPES – Finance Code 001. F. Botler is supported by CNPq (304315/2022-2) and CAPES ( 88887.878880/2023-00) . T Naia was supported by the Grant PID2020-113082GB-I00 funded by MICIU/AEI/10.13039/501100011033 and by Spanish State Research Agency, through the Severo Ochoa and Mar´ ıa de Maeztu Program for Centers and Units of Excellence in R&D (CEX2020-001084-M). CNPq is the National Council for Scientific and Technological Development of Brazil. References Francisco Arrepol, Patricio Asenjo, Ra´ ul Astete, V´ ıctor Cartes, Anah´ ıGajardo, Valeria Henr´ ıquez, Catalina Opazo, Nicol´ as Sanhueza-Matamala, and Christopher Thraves Caro, Sepa-rating path systems in trees , 2023. 9[ 2 ] J´ ozsef Balogh, B´ ela Csaba, Ryan R. Martin, and Andr´ as Pluh´ ar, On the path separation number of graphs , Discrete Appl. Math. 213 (2016), 26–33. [ 3 ] Ahmad Biniaz, Prosenjit Bose, Jean-Lou De Carufel, Anil Maheshwari, Babak Miraftab, Saeed Odak, Michiel Smid, Shakhar Smorodinsky, and Yelena Yuditsky, On separating path and tree systems in graphs , 2023. [ 4 ] B´ ela Bollob´ as and Alex Scott, On separating systems , European Journal of Combinatorics 28 (2007), no. 4, 1068–1071. [ 5 ] Marthe Bonamy, F´ abio Botler, Fran¸ cois Dross, T´ assio Naia, and Jozef Skokan, Separating the edges of a graph by a linear number of paths , Advances in Combinatorics October (2023). [ 6 ] Stephan Brandt, Hajo Broersma, Reinhard Diestel, and Matthias Kriesell, Global connectivity and expansion: long cycles and factors in f -connected graphs , Combinatorica 26 (2006), no. 1, 17–36. [ 7 ] Victor Falgas-Ravry, Teerade Kittipassorn, D´ aniel Kor´ andi, Shoham Letzter, and Bhargav P. Narayanan, Separating path systems , J. Comb. 5 (2014), no. 3, 335–354. [ 8 ] Cristina G. Fernandes, Guilherme Oliveira Mota, and Nicol´ as Sanhueza-Matamala, Separating path systems in complete graphs , LATIN 2024: Theoretical Informatics (Cham) (Jos´ e A. Soto and Andreas Wiese, eds.), Springer Nature Switzerland, 2024, pp. 98–113. [ 9 ] Leif K. Jørgensen and L´ aszl´ o Pyber, Covering a graph by topological complete subgraphs , Graphs and Combinatorics 6 (1990), 161–171. [ 10 ] Gyula Katona, On separating systems of a finite set , Journal of Combinatorial Theory 1 (1966), no. 2, 174–194. [ 11 ] George Kontogeorgiou and Maya Stein, An exact upper bound for the minimum size of a path system that weakly separates a clique , arXiv preprint arXiv:2403.08210 (2024). [ 12 ] Francois Lescure and Henry Meyniel, On a problem upon configurations contained in graphs with given chromatic number , Annals of Discrete Mathematics, vol. 41, Elsevier, 1988, pp. 325–331. [ 13 ] Shoham Letzter, Separating path systems of almost linear size , Trans. Amer. Math. Soc. (to appear). [ 14 ] L´ aszl´ o Pyber, An Erd˝ os—Gallai conjecture , Combinatorica 5 (1985), 67–79. [ 15 ] Alfred R´ enyi, On random generating elements of a finite boolean algebra , Acta Sci. Math. Szeged 22 (1961), no. 75-81, 4. [ 16 ] Joel Spencer, Minimal completely separating systems , Journal of Combinatorial Theory 8 (1970), no. 4, 446–447. [ 17 ] Belinda Wickes, Separating path systems for the complete graph , Discrete Mathematics 347 (2024), no. 3, 113784. 10
16653	https://en.wikipedia.org/wiki/Standing_wave_ratio	Standing wave ratio - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk Contents move to sidebar hide (Top) 1 Impedance matching 2 Relationship to the reflection coefficient 3 The standing wave pattern 4 Practical implications of SWR 5 Methods of measuring standing wave ratio 6 Power standing wave ratio 7 Implications of SWR on medical applications 8 See also 9 References 10 Further reading 11 External links [x] Toggle the table of contents Standing wave ratio [x] 22 languages Català Čeština Deutsch Español Esperanto Euskara فارسی Français Italiano עברית Nederlands 日本語 Norsk bokmål Polski Português Русский Svenska ไทย Türkçe Українська Tiếng Việt 中文 Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance move to sidebar hide From Wikipedia, the free encyclopedia Measure used in radio engineering and telecommunications SWR of a vertical HB9XBG Antenna for the 40m-band as a function of frequency In radio engineering and telecommunications, standing wave ratio (SWR) is a measure of impedance matching of loads to the characteristic impedance of a transmission line or waveguide. Impedance mismatches result in standing waves along the transmission line, and SWR is defined as the ratio of the partial standing wave's amplitude at an antinode (maximum) to the amplitude at a node (minimum) along the line. Voltage standing wave ratio (VSWR) (pronounced "vizwar") is the ratio of maximum to minimum voltage on a transmission line . For example, a VSWR of 1.2 means a peak voltage 1.2 times the minimum voltage along that line, if the line is at least one half wavelength long. A SWR can be also defined as the ratio of the maximum amplitude to minimum amplitude of the transmission line's currents, electric field strength, or the magnetic field strength. Neglecting transmission line loss, these ratios are identical. The power standing wave ratio (PSWR) is defined as the square of the VSWR, however, this deprecated term has no direct physical relation to power actually involved in transmission. SWR is usually measured using a dedicated instrument called an SWR meter. Since SWR is a measure of the load impedance relative to the characteristic impedance of the transmission line in use (which together determine the reflection coefficient as described below), a given SWR meter can interpret the impedance it sees in terms of SWR only if it has been designed for the same particular characteristic impedance as the line. In practice most transmission lines used in these applications are coaxial cables with an impedance of either 50 or 75ohms, so most SWR meters correspond to one of these. Checking the SWR is a standard procedure in a radio station. Although the same information could be obtained by measuring the load's impedance with an impedance analyzer (or "impedance bridge"), the SWR meter is simpler and more robust for this purpose. By measuring the magnitude of the impedance mismatch at the transmitter output it reveals problems due to either the antenna or the transmission line. Impedance matching [edit] Main article: Impedance matching SWR is used as a measure of impedance matching of a load to the characteristic impedance of a transmission line carrying radio frequency (RF) signals. This especially applies to transmission lines connecting radio transmitters and receivers with their antennas, as well as similar uses of RF cables such as cable television connections to TV receivers and distribution amplifiers. Impedance matching is achieved when the source impedance is the complex conjugate of the load impedance. The easiest way of achieving this, and the way that minimizes losses along the transmission line, is for the imaginary part of the complex impedance of both the source and load to be zero, that is, pure resistances, equal to the characteristic impedance of the transmission line. When there is a mismatch between the load impedance and the transmission line, part of the forward wave sent toward the load is reflected back along the transmission line towards the source. The source then sees a different impedance than it expects which can lead to lesser (or in some cases, more) power being supplied by it, the result being very sensitive to the electrical length of the transmission line. Such a mismatch is usually undesired and results in standing waves along the transmission line which magnifies transmission line losses (significant at higher frequencies and for longer cables). The SWR is a measure of the depth of those standing waves and is, therefore, a measure of the matching of the load to the transmission line. A matched load would result in an SWR of 1:1 implying no reflected wave. An infinite SWR represents complete reflection by a load unable to absorb electrical power, with all the incident power reflected back towards the source. It should be understood that the match of a load to the transmission line is different from the match of a source to the transmission line or the match of a source to the load seen through the transmission line. For instance, if there is a perfect match between the load impedance Z load and the source impedance Z source = Zload, that perfect match will remain if the source and load are connected through a transmission line with an electrical length of one half wavelength (or a multiple of one half wavelengths) using a transmission line of any characteristic impedance Z 0. However the SWR will generally not be 1:1, depending only on Z load and Z 0. With a different length of transmission line, the source will see a different impedance than Z load which may or may not be a good match to the source. Sometimes this is deliberate, as when a quarter-wave matching section is used to improve the match between an otherwise mismatched source and load. However typical RF sources such as transmitters and signal generators are designed to look into a purely resistive load impedance such as 50Ω or 75Ω, corresponding to common transmission lines' characteristic impedances. In those cases, matching the load to the transmission line, Z load = Z 0, always ensures that the source will see the same load impedance as if the transmission line weren't there. This is identical to a 1:1 SWR. This condition (Z load = Z 0) also means that the load seen by the source is independent of the transmission line's electrical length. Since the electrical length of a physical segment of transmission line depends on the signal frequency, violation of this condition means that the impedance seen by the source through the transmission line becomes a function of frequency (especially if the line is long), even if Z load is frequency-independent. So in practice, a good SWR (near 1:1) implies a transmitter's output seeing the exact impedance it expects for optimum and safe operation. Relationship to the reflection coefficient [edit] Incident wave (blue) is fully reflected (red wave) out of phase at short-circuited end of transmission line, creating a net voltage (black) standing wave. Γ = −1, SWR = ∞. Standing waves on transmission line, net voltage shown in different colors during one period of oscillation. Incoming wave from left (amplitude = 1) is partially reflected with (top to bottom) Γ = 0.6, −0.333, and 0.8 ∠60°. Resulting SWR = 4, 2, 9. The voltage component of a standing wave in a uniform transmission line consists of the forward wave (with complex amplitude V f{\displaystyle V_{f}}) superimposed on the reflected wave (with complex amplitude V r{\displaystyle V_{r}}). A wave is partly reflected when a transmission line is terminated with an impedance unequal to its characteristic impedance. The reflection coefficientΓ{\displaystyle \Gamma } can be defined as: Γ=V r V f.{\displaystyle \Gamma ={\frac {V_{r}}{V_{f}}}.} or Γ=Z L−Z o Z L+Z o{\displaystyle \Gamma ={Z_{L}-Z_{o} \over Z_{L}+Z_{o}}} Γ{\displaystyle \Gamma } is a complex number that describes both the magnitude and the phase shift of the reflection. The simplest cases with Γ{\displaystyle \Gamma }measured at the load are: Γ=−1{\displaystyle \Gamma =-1}: complete negative reflection, when the line is short-circuited, Γ=0{\displaystyle \Gamma =0}: no reflection, when the line is perfectly matched, Γ=+1{\displaystyle \Gamma =+1}: complete positive reflection, when the line is open-circuited. The SWR directly corresponds to the magnitude of Γ{\displaystyle \Gamma }. At some points along the line the forward and reflected waves interfere constructively, exactly in phase, with the resulting amplitude V max{\displaystyle V_{\text{max}}} given by the sum of those waves' amplitudes: \|V max\|=\|V f\|+\|V r\|=\|V f\|+\|Γ V f\|=(1+\|Γ\|)\|V f\|.{\displaystyle {\begin{aligned}\|V_{\text{max}}\|&=\|V_{f}\|+\|V_{r}\|\&=\|V_{f}\|+\|\Gamma V_{f}\|\&=(1+\|\Gamma \|)\|V_{f}\|.\end{aligned}}} At other points, the waves interfere 180° out of phase with the amplitudes partially cancelling: \|V min\|=\|V f\|−\|V r\|=\|V f\|−\|Γ V f\|=(1−\|Γ\|)\|V f\|.{\displaystyle {\begin{aligned}\|V_{\text{min}}\|&=\|V_{f}V_{r}\|\&=\|V_{f}\Gamma V_{f}\|\&=(1-\|\Gamma \|)\|V_{f}\|.\end{aligned}}} The voltage standing wave ratio is then VSWR=\|V max\|\|V min\|=1+\|Γ\|1−\|Γ\|.{\displaystyle {\text{VSWR}}={\frac {\|V_{\text{max}}\|}{\|V_{\text{min}}\|}}={\frac {1+\|\Gamma \|}{1-\|\Gamma \|}}.} Since the magnitude of Γ{\displaystyle \Gamma } always falls in the range [0,1], the SWR is always greater than or equal to unity. Note that the phase of V f and V r vary along the transmission line in opposite directions to each other. Therefore, the complex-valued reflection coefficient Γ{\displaystyle \Gamma } varies as well, but only in phase. With the SWR dependent only on the complex magnitude of Γ{\displaystyle \Gamma }, it can be seen that the SWR measured at any point along the transmission line (neglecting transmission line losses) obtains an identical reading. Since the power of the forward and reflected waves are proportional to the square of the voltage components due to each wave, SWR can be expressed in terms of forward and reflected power: SWR=1+P r/P f 1−P r/P f.{\displaystyle {\text{SWR}}={\frac {1+{\sqrt {P_{r}/P_{f}}}}{1-{\sqrt {P_{r}/P_{f}}}}}.} By sampling the complex voltage and current at the point of insertion, an SWR meter is able to compute the effective forward and reflected voltages on the transmission line for the characteristic impedance for which the SWR meter has been designed. Since the forward and reflected power is related to the square of the forward and reflected voltages, some SWR meters also display the forward and reflected power. In the special case of a load R L, which is purely resistive but unequal to the characteristic impedance of the transmission line Z 0, the SWR is given simply by their ratio: SWR=max{R L Z 0,Z 0 R L}{\displaystyle {\text{SWR}}=\max \left{{\frac {R_{\text{L}}}{\,Z_{\text{0}}\,}}\,,{\frac {\,Z_{\text{0}}\,}{R_{\text{L}}}}\right}} with the ratio or its reciprocal is chosen to obtain a value greater than unity. The standing wave pattern [edit] Using complex notation for the voltage amplitudes, for a signal at frequency f, the actual (real) voltages V actual as a function of time t are understood to relate to the complex voltages according to: V a c t u a l=R e(e i 2 π f t V).{\displaystyle V_{\mathsf {actual}}={\mathcal {R_{e}}}(e^{i2\pi ft}V)~.} Thus taking the real part of the complex quantity inside the parenthesis, the actual voltage consists of a sine wave at frequency f with a peak amplitude equal to the complex magnitude of V, and with a phase given by the phase of the complex V. Then with the position along a transmission line given by x, with the line ending in a load located at x o, the complex amplitudes of the forward and reverse waves would be written as: V f w d(x)=e−i k(x−x o)A V r e v(x)=Γ e i k(x−x o)A{\displaystyle {\begin{aligned}V_{\mathsf {fwd}}(x)&=e^{-ik(x-x_{\mathsf {o}})}A\V_{\mathsf {rev}}(x)&=\Gamma e^{ik(x-x_{\mathsf {o}})}A\end{aligned}}} for some complex amplitude A (corresponding to the forward wave at x o that some treatments use phasors where the time dependence is according to e−i 2 π f t{\displaystyle e^{-i2\pi ft}} and spatial dependence (for a wave in the +x direction) of e+i k(x−x o).{\displaystyle \ e^{+ik(x-x_{\mathsf {o}})}~.} Either convention obtains the same result for V actual. According to the superposition principle the net voltage present at any point x on the transmission line is equal to the sum of the voltages due to the forward and reflected waves: V n e t(x)=V f w d(x)+V r e v(x)=e−i k(x−x o)(1+Γ e i 2 k(x−x o))A{\displaystyle {\begin{aligned}V_{\mathsf {net}}(x)&=V_{\mathsf {fwd}}(x)+V_{\mathsf {rev}}(x)\&=e^{-ik(x-x_{\mathsf {o}})}\left(1+\Gamma e^{i2k(x-x_{\mathsf {o}})}\right)A\end{aligned}}} Since we are interested in the variations of the magnitude of V net along the line (as a function of x), we shall solve instead for the squared magnitude of that quantity, which simplifies the mathematics. To obtain the squared magnitude we multiply the above quantity by its complex conjugate: \|V n e t(x)\|2=V n e t(x)V n e t∗(x)=e−i k(x−x o)(1+Γ e i 2 k(x−x o))A e+i k(x−x o)(1+Γ∗e−i 2 k(x−x o))A∗=[1+\|Γ\|2+2 R e⁡(Γ e i 2 k(x−x o))]\|A\|2{\displaystyle {\begin{aligned}\|V_{\mathsf {net}}(x)\|^{2}&=V_{\mathsf {net}}(x)V_{\mathsf {net}}^{}(x)\&=e^{-ik\left(x-x_{\mathsf {o}}\right)}\left(1+\Gamma e^{i2k\left(x-x_{\mathsf {o}}\right)}\right)A\,e^{+ik\left(x-x_{\mathsf {o}}\right)}\left(1+\Gamma ^{}e^{-i2k\left(x-x_{\mathsf {o}}\right)}\right)A^{}\&=\left[\ 1+\|\Gamma \|^{2}+2\ \operatorname {\mathcal {R_{e}}} \left(\Gamma e^{i2k\left(x-x_{\mathsf {o}}\right)}\right)\ \right]\|A\|^{2}\end{aligned}}} Depending on the phase of the third term, the maximum and minimum values of V net (the square root of the quantity in the equations) are (1+\|Γ\|)\|A\|{\displaystyle \ \left(1+\|\Gamma \|\right)\|A\|\ } and (1−\|Γ\|)\|A\|,{\displaystyle \ \left(1-\|\Gamma \|\right)\|A\|\ ,} respectively, for a standing wave ratio of: S W R=\|V m a x\|\|V m i n\|=1+\|Γ\|1−\|Γ\|{\displaystyle {\boldsymbol {\mathsf {SWR}}}={\frac {\|V_{\mathsf {max}}\|}{\|V_{\mathsf {min}}\|}}={\frac {1+\|\Gamma \|}{1-\|\Gamma \|}}}\|Γ\|=S W R−1 S W R+1{\displaystyle \left\|\Gamma \right\|={\frac {\ {\boldsymbol {\mathsf {SWR}}}-1\ }{{\boldsymbol {\mathsf {SWR}}}+1}}} as earlier asserted. Along the line, the above expression for \|V n e t(x)\|2{\displaystyle \ \|V_{\mathsf {net}}(x)\|^{2}\ } is seen to oscillate sinusoidally between \|V m i n\|2{\displaystyle \ \|V_{\mathsf {min}}\|^{2}\ } and \|V m a x\|2{\displaystyle \ \|V_{\mathsf {max}}\|^{2}\ } with a period of ⁠2 π/2 k⁠. This is half of the guided wavelength λ = ⁠2 π/k⁠ for the frequency f. That can be seen as due to interference between two waves of that frequency which are travelling in opposite directions. For example, at a frequency f = 20 MHz (free space wavelength of 15 m) in a transmission line whose velocity factor is 0.67, the guided wavelength (distance between voltage peaks of the forward wave alone) would be λ = 10 m . At instances when the forward wave at x = 0 is at zero phase (peak voltage) then at x = 10 m it would also be at zero phase, but at x = 5 m it would be at 180° phase (peak negative voltage). On the other hand, the magnitude of the voltage due to a standing wave produced by its addition to a reflected wave, would have a wavelength between peaks of only ⁠1/2⁠λ = 5 m . Depending on the location of the load and phase of reflection, there might be a peak in the magnitude of V net at x = 1.3 m . Then there would be another peak found where \|V net\| = V max at x = 6.3 m , whereas it would find minima of the standing wave at x = 3.8 m, 8.8 m, etc. Practical implications of SWR [edit] Example of estimated bandwidth of antenna according to the schedule VSWR by the help of the Ansys HFSS The most common case for measuring and examining SWR is when installing and tuning transmitting antennas. When a transmitter is connected to an antenna by a feed line, the driving point impedance of the antenna must match the characteristic impedance of the feed line in order for the transmitter to see the impedance it was designed for (the impedance of the feed line, usually 50 or 75 ohms). The impedance of a particular antenna design can vary due to a number of factors that cannot always be clearly identified. This includes the transmitter frequency (as compared to the antenna's design or resonant frequency), the antenna's height above and quality of the ground, proximity to large metal structures, and variations in the exact size of the conductors used to construct the antenna.5 When an antenna and feed line do not have matching impedances, the transmitter sees an unexpected impedance, where it might not be able to produce its full power, and can even damage the transmitter in some cases.5 The reflected power in the transmission line increases the average current and therefore losses in the transmission line compared to power actually delivered to the load. It is the interaction of these reflected waves with forward waves which causes standing wave patterns,5 with the negative repercussions we have noted.5 Matching the impedance of the antenna to the impedance of the feed line can sometimes be accomplished through adjusting the antenna itself, but otherwise is possible using an antenna tuner, an impedance matching device. Installing the tuner between the feed line and the antenna allows for the feed line to see a load close to its characteristic impedance, while sending most of the transmitter's power (a small amount may be dissipated within the tuner) to be radiated by the antenna despite its otherwise unacceptable feed point impedance. Installing a tuner in between the transmitter and the feed line can also transform the impedance seen at the transmitter end of the feed line to one preferred by the transmitter. However, in the latter case, the feed line still has a high SWR present, with the resulting increased feed line losses unmitigated. The magnitude of those losses are dependent on the type of transmission line, and its length. They always increase with frequency. For example, a certain antenna used well away from its resonant frequency may have an SWR of 6:1. For a frequency of 3.5 MHz, with that antenna fed through 75 meters of RG-8A coax, the loss due to standing waves would be 2.2 dB. However the same 6:1 mismatch through 75 meters of RG-8A coax would incur 10.8 dB of loss at 146 MHz.5 Thus, a better match of the antenna to the feed line, that is, a lower SWR, becomes increasingly important with increasing frequency, even if the transmitter is able to accommodate the impedance seen (or an antenna tuner is used between the transmitter and feed line). Certain types of transmissions can suffer other negative effects from reflected waves on a transmission line. Analog TV can experience "ghosts" from delayed signals bouncing back and forth on a long line. FM stereo can also be affected and digital signals can experience delayed pulses leading to bit errors. Whenever the delay times for a signal going back down and then again up the line are comparable to the modulation time constants, effects occur. For this reason, these types of transmissions require a low SWR on the feedline, even if SWR induced loss might be acceptable and matching is done at the transmitter. Methods of measuring standing wave ratio [edit] Slotted line. The probe moves along the line to measure the variable voltage. SWR is the maximum divided by the minimum voltage Many different methods can be used to measure standing wave ratio. The most intuitive method uses a slotted line which is a section of transmission line with an open slot which allows a probe to detect the actual voltage at various points along the line. Thus the maximum and minimum values can be compared directly. This method is used at VHF and higher frequencies. At lower frequencies, such lines are impractically long. Directional couplers can be used at HF through microwave frequencies. Some are a quarter wave or more long, which restricts their use to the higher frequencies. Other types of directional couplers sample the current and voltage at a single point in the transmission path and mathematically combine them in such a way as to represent the power flowing in one direction. The common type of SWR / power meter used in amateur operation may contain a dual directional coupler. Other types use a single coupler which can be rotated 180 degrees to sample power flowing in either direction. Unidirectional couplers of this type are available for many frequency ranges and power levels and with appropriate coupling values for the analog meter used. A directional wattmeter using a rotatable directional coupler element. The forward and reflected power measured by directional couplers can be used to calculate SWR. The computations can be done mathematically in analog or digital form or by using graphical methods built into the meter as an additional scale or by reading from the crossing point between two needles on the same meter. The above measuring instruments can be used "in line" that is, the full power of the transmitter can pass through the measuring device so as to allow continuous monitoring of SWR. Other instruments, such as network analyzers, low power directional couplers and antenna bridges use low power for the measurement and must be connected in place of the transmitter. Bridge circuits can be used to directly measure the real and imaginary parts of a load impedance and to use those values to derive SWR. These methods can provide more information than just SWR or forward and reflected power. Stand alone antenna analyzers use various measuring methods and can display SWR and other parameters plotted against frequency. By using directional couplers and a bridge in combination, it is possible to make an in line instrument that reads directly in complex impedance or in SWR. Stand alone antenna analyzers also are available that measure multiple parameters. Power standing wave ratio [edit] The term power standing wave ratio (PSWR) is sometimes referred to, and defined as, the square of the voltage standing wave ratio. The term is widely cited as "misleading". The expression "power standing-wave ratio", which may sometimes be encountered, is even more misleading, for the power distribution along a loss-free line is constant.... — J.H. Gridley (2014) However it does correspond to one type of measurement of SWR using what was formerly a standard measuring instrument at microwave frequencies, the slotted line. The slotted line is a waveguide (or air-filled coaxial line) in which a small sensing antenna which is part of a crystal detector or detector is placed in the electric field in the line. The voltage induced in the antenna is rectified by either a point contact diode (crystal rectifier) or a Schottky barrier diode that is incorporated in the detector. These detectors have a square law output for low levels of input. Readings therefore corresponded to the square of the electric field along the slot, E 2(x), with maximum and minimum readings of E 2 max and E 2 min found as the probe is moved along the slot. The ratio of these yields the square of the SWR, the so-called PSWR. This technique of rationalization of terms is fraught with problems.[clarification needed] The square law behavior of the detector diode is exhibited only when the voltage across the diode is below the knee of the diode. Once the detected voltage exceeds the knee, the response of the diode becomes nearly linear. In this mode the diode and its associated filtering capacitor produce a voltage that is proportional to the peak of the sampled voltage. The operator of such a detector would not have a ready indication as to the mode in which the detector diode is operating and therefore differentiating the results between SWR or so called PSWR is not practical. Perhaps even worse, is the common case where the minimum detected voltage is below the knee and the maximum voltage is above the knee. In this case, the computed results are largely meaningless. Thus the terms PSWR and Power Standing Wave Ratio are deprecated and should be considered only from a legacy measurement perspective. Implications of SWR on medical applications [edit] SWR can also have a detrimental impact upon the performance of microwave-based medical applications. In microwave electrosurgery an antenna that is placed directly into tissue may not always have an optimal match with the feedline resulting in standing waves, the presence of which can affect monitoring components used to measure power levels, making such measurements less reliable. See also [edit] Antenna tuner ("transmatch") Impedance Impedance matching Mismatch loss Return loss SWR meter Time-domain reflectometer Total active reflection coefficient References [edit] ^Knott, Eugene F.; Shaeffer, John F.; Tuley, Michael T. (2004). Radar cross section. SciTech Radar and Defense Series (2nd ed.). SciTech Publishing. p.374. ISBN978-1-891121-25-8. ^Schaub, Keith B.; Kelly, Joe (2004). Production testing of RF and system-on-a-chip devices for wireless communications. Artech House microwave library. Artech House. p.93. ISBN978-1-58053-692-9. ^Silver, Samuel (1984) . Microwave Antenna Theory and Design. IEE. p.28. ISBN0863410170. ^Sliusar, I.; Slyusar, V.; Voloshko, S.; Zinchenko, A.; Utkin, Y. (22–27 June 2020). Synthesis of a broadband ring antenna of a two-tape design(PDF). 12th International Conference on Antenna Theory and Techniques (ICATT-2020). Kharkiv, Ukraine. Archived(PDF) from the original on 2022-10-09. ^ abcdeHutchinson, Chuck, ed. (2000). The ARRL Handbook for Radio Amateurs 2001. Newington, CT: American Radio Relay League. pp.19.4 –19.6, 19.13, 20.2. ISBN978-0-87259-186-8. ^Ford, Steve (April 1994), "The SWR obsession"(PDF), QST Magazine, vol.78, no.4, Newington, CT: American Radio Relay League, pp.70–72, retrieved 2014-11-04 ^Terman, Fredrick E. (1952). Electronic Measurements. McGraw Hill. p.135 ff. LCCN51-12650. ^Schulz, Glenn B., (W9IQ) (January 24, 2018). "How does an SWR meter really work?". ham.stackexchange.com. Retrieved March 18, 2018.{{cite web}}: CS1 maint: multiple names: authors list (link) CS1 maint: numeric names: authors list (link) ^"Nautel adds two models to NX series". Nautel (Press release). March 11, 2015. Archived from the original on August 18, 2016. Retrieved July 6, 2017. ^"Model OIB-1 and OIB-3". www.deltaelectronics.com. Delta Electronics, Inc. ^Wolff, Christian. "Standing wave ratio". radartutorial.eu. ^Gridley, J.H. (2014). Principles of Electrical Transmission Lines in Power and Communication. Elsevier. p.265. ISBN978-1483186030 – via Google Books. ^Rollin, Bernard Vincent (1964). An Introduction to Electronics. Clarendon Press. p.209. OCLC1148924. ^"Problems with VSWR in medical applications". microwaves101.com. Retrieved July 6, 2017. This article incorporates public domain material from Federal Standard 1037C. General Services Administration. Archived from the original on 2022-01-22.(in support of MIL-STD-188). Further reading [edit] Understanding the Fundamental Principles of Vector Network Analysis(PDF) (Report). Application note. Vol.1287–1. Hewlett-Packard. 1997. Archived(PDF) from the original on 2022-10-09. External links [edit] "Standing wave diagram". poynting.herokuapp.com. Archived from the original on 2020-11-25. Retrieved 2015-07-09. — A web application that draws the Standing Wave Diagram and calculates the SWR, input impedance, reflection coefficient and more "VSWR". telestrian.co.uk. — An online conversion tool between SWR, return loss and reflection coefficient "Online VSWR Calculator". emtalk.com. "VSWR tutorial". electronics-notes.com. antennas & propagation. — Series of pages dealing with all aspects of VSWR, reflection coefficient, return loss, practical aspects, measurement, etc. "Online interactive standing waves animation". portable-antennas.com. — An online tool to view and configure animated forward and reflected waves in a feedline, and their resultant animated standing wave pattern. Also calculates VSWR, mismatch loss, power loss and return loss based on input reflection coefficient. \| Authority control databases: National \| Germany \| Retrieved from " Categories: Antennas (radio) Electronics concepts Wave mechanics Radio electronics Engineering ratios Hidden categories: CS1 maint: multiple names: authors list CS1 maint: numeric names: authors list Articles with short description Short description is different from Wikidata Wikipedia articles needing clarification from January 2019 Wikipedia articles incorporating text from the Federal Standard 1037C Wikipedia articles incorporating text from MIL-STD-188 CS1: long volume value This page was last edited on 30 April 2025, at 11:51(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. 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16654	https://currikicdn.s3-us-west-2.amazonaws.com/resourcedocs/54d34ebab5311.pdf	CK-12 FOUNDATION CK-12 Algebra I - Second Edition Gloag Gloag Rawley CK-12 Foundation is a non-proﬁt organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the “FlexBook,” CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform™. Copyright © 2011 CK-12 Foundation, www.ck12.org Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution/Non-Commercial/Share Alike 3.0 Un-ported (CC-by-NC-SA) License ( as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated herein by this reference. Speciﬁc details can be found at Printed: April 1, 2011 Authors Andrew Gloag, Anne Gloag, Eve Rawley Source Anne Gloag i www.ck12.org Contents 1 Equations and Functions 1 1.1 Variable Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Order of Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.3 Patterns and Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 1.4 Equations and Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 1.5 Functions as Rules and Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 1.6 Functions as Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 1.7 Problem-Solving Plan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 1.8 Problem-Solving Strategies: Make a Table and Look for a Pattern . . . . . . . . . . . . . . 52 2 Real Numbers 63 2.1 Integers and Rational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 2.2 Adding and Subtracting Rational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 2.3 Multiplying and Dividing Rational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 2.4 The Distributive Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 2.5 Square Roots and Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 2.6 Problem-Solving Strategies: Guess and Check, Work Backward . . . . . . . . . . . . . . . . 95 3 Equations of Lines 102 3.1 One-Step Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 3.2 Two-Step Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 3.3 Multi-Step Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 3.4 Equations with Variables on Both Sides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 3.5 Ratios and Proportions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 3.6 Percent Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 4 Graphs of Equations and Functions 138 4.1 The Coordinate Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 4.2 Graphs of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 www.ck12.org ii 4.3 Graphing Using Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 4.4 Slope and Rate of Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 4.5 Graphs Using Slope-Intercept Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 4.6 Direct Variation Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 4.7 Linear Function Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 4.8 Problem-Solving Strategies - Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 5 Writing Linear Equations 197 5.1 Forms of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 5.2 Equations of Parallel and Perpendicular Lines . . . . . . . . . . . . . . . . . . . . . . . . . . 210 5.3 Fitting a Line to Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218 5.4 Predicting with Linear Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228 6 Linear Inequalities 237 6.1 Solving Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 6.2 Using Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244 6.3 Compound Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248 6.4 Absolute Value Equations and Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 6.5 Linear Inequalities in Two Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266 7 Solving Systems of Equations and Inequalities 274 7.1 Linear Systems by Graphing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274 7.2 Solving Linear Systems by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284 7.3 Solving Linear Systems by Elimination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291 7.4 Special Types of Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302 7.5 Systems of Linear Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310 8 Exponential Functions 324 8.1 Exponent Properties Involving Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324 8.2 Exponent Properties Involving Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329 8.3 Zero, Negative, and Fractional Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333 8.4 Scientiﬁc Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339 8.5 Geometric Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346 8.6 Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351 8.7 Applications of Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359 9 Polynomials 368 9.1 Addition and Subtraction of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368 iii www.ck12.org 9.2 Multiplication of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376 9.3 Special Products of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383 9.4 Polynomial Equations in Factored Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388 9.5 Factoring Quadratic Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394 9.6 Factoring Special Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400 9.7 Factoring Polynomials Completely . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408 10 Quadratic Equations and Quadratic Functions 417 10.1 Graphs of Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417 10.2 Quadratic Equations by Graphing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431 10.3 Quadratic Equations by Square Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 439 10.4 Solving Quadratic Equations by Completing the Square . . . . . . . . . . . . . . . . . . . . 446 10.5 Solving Quadratic Equations by the Quadratic Formula . . . . . . . . . . . . . . . . . . . . 455 10.6 The Discriminant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 464 10.7 Linear, Exponential and Quadratic Models . . . . . . . . . . . . . . . . . . . . . . . . . . . 469 11 Algebra and Geometry Connections 486 11.1 Graphs of Square Root Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486 11.2 Radical Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496 11.3 Radical Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 506 11.4 The Pythagorean Theorem and Its Converse . . . . . . . . . . . . . . . . . . . . . . . . . . . 512 11.5 Distance and Midpoint Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 520 12 Rational Equations and Functions 528 12.1 Inverse Variation Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 528 12.2 Graphs of Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533 12.3 Division of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544 12.4 Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549 12.5 Multiplying and Dividing Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . 553 12.6 Adding and Subtracting Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . 557 12.7 Solutions of Rational Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 564 www.ck12.org iv Chapter 1 Equations and Functions 1.1 Variable Expressions Learning Objectives • Evaluate algebraic expressions. • Evaluate algebraic expressions with exponents. Introduction - The Language of Algebra No one likes doing the same problem over and over again—that’s why mathematicians invented algebra. Algebra takes the basic principles of math and makes them more general, so we can solve a problem once and then use that solution to solve a group of similar problems. In arithmetic, you’ve dealt with numbers and their arithmetical operations (such as +, −, ×, ÷). In algebra, we use symbols called variables (which are usually letters, such as x, y, a, b, c, . . .) to represent numbers and sometimes processes. For example, we might use the letter x to represent some number we don’t know yet, which we might need to ﬁgure out in the course of a problem. Or we might use two letters, like x and y, to show a relationship between two numbers without needing to know what the actual numbers are. The same letters can represent a wide range of possible numbers, and the same letter may represent completely diﬀerent numbers when used in two diﬀerent problems. Using variables oﬀers advantages over solving each problem “from scratch.” With variables, we can: • Formulate arithmetical laws such as a + b = b + a for all real numbers a and b. • Refer to “unknown” numbers. For instance: ﬁnd a number x such that 3x + 1 = 10. • Write more compactly about functional relationships such as, “If you sell x tickets, then your proﬁt will be 3x −10 dollars, or “f(x) = 3x −10,” where “ f” is the proﬁt function, and x is the input (i.e. how many tickets you sell). Example 1 Write an algebraic expression for the perimeter and area of the rectangle below. 1 www.ck12.org To ﬁnd the perimeter, we add the lengths of all 4 sides. We can still do this even if we don’t know the side lengths in numbers, because we can use variables like l and w to represent the unknown length and width. If we start at the top left and work clockwise, and if we use the letter P to represent the perimeter, then we can say: P = l + w + l + w We are adding 2 l’s and 2 w’s, so we can say that: P = 2 · l + 2 · w It’s customary in algebra to omit multiplication symbols whenever possible. For example, 11x means the same thing as 11 · x or 11 × x. We can therefore also write: P = 2l + 2w Area is length multiplied by width. In algebraic terms we get: A = l × w →A = l · w →A = lw Note: 2l+2w by itself is an example of a variable expression; P = 2l+2w is an example of an equation. The main diﬀerence between expressions and equations is the presence of an equals sign (=). In the above example, we found the simplest possible ways to express the perimeter and area of a rectangle when we don’t yet know what its length and width actually are. Now, when we encounter a rectangle whose dimensions we do know, we can simply substitute (or plug in) those values in the above equations. In this chapter, we will encounter many expressions that we can evaluate by plugging in values for the variables involved. Evaluate Algebraic Expressions When we are given an algebraic expression, one of the most common things we might have to do with it is evaluate it for some given value of the variable. The following example illustrates this process. Example 2 Let x = 12. Find the value of 2x −7. To ﬁnd the solution, we substitute 12 for x in the given expression. Every time we see x, we replace it with 12. 2x −7 = 2(12) −7 = 24 −7 = 17 www.ck12.org 2 Note: At this stage of the problem, we place the substituted value in parentheses. We do this to make the written-out problem easier to follow, and to avoid mistakes. (If we didn’t use parentheses and also forgot to add a multiplication sign, we would end up turning 2x into 212 instead of 2 times 12!) Example 3 Let y = −2. Find the value of 7 y −11y + 2. Solution 7 (−2) −11(−2) + 2 = −31 2 + 22 + 2 = 24 −31 2 = 201 2 Many expressions have more than one variable in them. For example, the formula for the perimeter of a rectangle in the introduction has two variables: length (l) and width (w). In these cases, be careful to substitute the appropriate value in the appropriate place. Example 5 The area of a trapezoid is given by the equation A = h 2(a + b). Find the area of a trapezoid with bases a = 10 cm and b = 15 cm and height h = 8 cm. To ﬁnd the solution to this problem, we simply take the values given for the variables a, b, and h, and plug them in to the expression for A: A = h 2(a + b) Substitute 10 for a, 15 for b, and 8 for h. A = 8 2(10 + 15) Evaluate piece by piece. 10 + 15 = 25; 8 2 = 4. A = 4(25) = 100 Solution: The area of the trapezoid is 100 square centimeters. Evaluate Algebraic Expressions with Exponents Many formulas and equations in mathematics contain exponents. Exponents are used as a short-hand notation for repeated multiplication. For example: 2 · 2 = 22 2 · 2 · 2 = 23 The exponent stands for how many times the number is used as a factor (multiplied). When we deal with integers, it is usually easiest to simplify the expression. We simplify: 3 www.ck12.org 22 = 4 23 = 8 However, we need exponents when we work with variables, because it is much easier to write x8 than x · x · x · x · x · x · x · x. To evaluate expressions with exponents, substitute the values you are given for each variable and simplify. It is especially important in this case to substitute using parentheses in order to make sure that the simpliﬁcation is done correctly. For a more detailed review of exponents and their properties, check out the video at com/lesson/mathhelp/863-exponents---basics. Example 5 The area of a circle is given by the formula A = πr2. Find the area of a circle with radius r = 17 inches. Substitute values into the equation. A = πr2 Substitute 17 for r. A = π(17)2 π · 17 · 17 ≈907.9202 . . . Round to 2 decimal places. The area is approximately 907.92 square inches. Example 6 Find the value of x2y3 x3+y2 , for x = 2 and y = −4. Substitute the values of x and y in the following. x2y3 x3 + y2 = (2)2(−4)3 (2)3 + (−4)2 Substitute 2 for x and −4 for y. 4(−64) 8 + 16 = −256 24 = −32 3 Evaluate expressions: (2)2 = (2)(2) = 4 and (2)3 = (2)(2)(2) = 8. (−4)2 = (−4)(−4) = 16 and (−4)3 = (−4)(−4)(−4) = −64. Example 7 The height (h) of a ball in ﬂight is given by the formula h = −32t2 + 60t + 20, where the height is given in feet and the time (t) is given in seconds. Find the height of the ball at time t = 2 seconds. Solution www.ck12.org 4 h = −32t2 + 60t + 20 = −32(2)2 + 60(2) + 20 Substitute 2 for t. = −32(4) + 60(2) + 20 = 12 The height of the ball is 12 feet. Review Questions 1. Write the following in a more condensed form by leaving out a multiplication symbol. (a) 2 × 11x (b) 1.35 · y (c) 3 × 1 4 (d) 1 4 · z 2. Evaluate the following expressions for a = −3, b = 2, c = 5, and d = −4. (a) 2a + 3b (b) 4c + d (c) 5ac −2b (d) 2a c−d (e) 3b d (f) a−4b 3c+2d (g) 1 a+b (h) ab cd 3. Evaluate the following expressions for x = −1, y = 2, z = −3, and w = 4. (a) 8x3 (b) 5x2 6z3 (c) 3z2 −5w2 (d) x2 −y2 (e) z3+w3 z3−w3 (f) 2x3 −3x2 + 5x −4 (g) 4w3 + 3w2 −w + 2 (h) 3 + 1 z2 4. The weekly cost C of manufacturing x remote controls is given by the formula C = 2000 + 3x, where the cost is given in dollars. (a) What is the cost of producing 1000 remote controls? (b) What is the cost of producing 2000 remote controls? (c) What is the cost of producing 2500 remote controls? 5. The volume of a box without a lid is given by the formula V = 4x(10 −x)2, where x is a length in inches and V is the volume in cubic inches. (a) What is the volume when x = 2? (b) What is the volume when x = 3? 5 www.ck12.org 1.2 Order of Operations Learning Objectives • Evaluate algebraic expressions with grouping symbols. • Evaluate algebraic expressions with fraction bars. • Evaluate algebraic expressions using a graphing calculator. Introduction Look at and evaluate the following expression: 2 + 4 × 7 −1 =? How many diﬀerent ways can we interpret this problem, and how many diﬀerent answers could someone possibly ﬁnd for it? The simplest way to evaluate the expression is simply to start at the left and work your way across: 2 + 4 × 7 −1 = 6 × 7 −1 = 42 −1 = 41 This is the answer you would get if you entered the expression into an ordinary calculator. But if you entered the expression into a scientiﬁc calculator or a graphing calculator you would probably get 29 as the answer. In mathematics, the order in which we perform the various operations (such as adding, multiplying, etc.) is important. In the expression above, the operation of multiplication takes precedence over addition, so we evaluate it ﬁrst. Let’s re-write the expression, but put the multiplication in brackets to show that it is to be evaluated ﬁrst. 2 + (4 × 7) −1 =? First evaluate the brackets: 4 × 7 = 28. Our expression becomes: 2 + (28) −1 =? When we have only addition and subtraction, we start at the left and work across: 2 + 28 −1 = 30 −1 = 29 Algebra students often use the word “PEMDAS” to help remember the order in which we evaluate the mathematical expressions: Parentheses, Exponents, Multiplication, Division, Addition and Subtraction. www.ck12.org 6 Order of Operations 1. Evaluate expressions within Parentheses (also all brackets [ ] and braces { }) ﬁrst. 2. Evaluate all Exponents (terms such as 32 or x3) next. 3. Multiplication and Division is next - work from left to right completing both multiplication and division in the order that they appear. 4. Finally, evaluate Addition and Subtraction - work from left to right completing both addition and subtraction in the order that they appear. Evaluate Algebraic Expressions with Grouping Symbols The ﬁrst step in the order of operations is called parentheses, but we include all grouping symbols in this step—not just parentheses (), but also square brackets [ ] and curly braces { }. Example 1 Evaluate the following: a) 4 −7 −11 + 2 b) 4 −(7 −11) + 2 c) 4 −[7 −(11 + 2)] Each of these expressions has the same numbers and the same mathematical operations, in the same order. The placement of the various grouping symbols means, however, that we must evaluate everything in a diﬀerent order each time. Let’s look at how we evaluate each of these examples. a) This expression doesn’t have parentheses, exponents, multiplication, or division. PEMDAS states that we treat addition and subtraction as they appear, starting at the left and working right (it’s NOT addition then subtraction). 4 −7 −11 + 2 = −3 −11 + 2 = −14 + 2 = −12 b) This expression has parentheses, so we ﬁrst evaluate 7 −11 = −4. Remember that when we subtract a negative it is equivalent to adding a positive: 4 −(7 −11) + 2 = 4 −(−4) + 2 = 8 + 2 = 10 c) An expression can contain any number of sets of parentheses. Sometimes expressions will have sets of parentheses inside other sets of parentheses. When faced with nested parentheses, start at the innermost parentheses and work outward. Brackets may also be used to group expressions which already contain parentheses. This expression has both brackets and parentheses. We start with the innermost group: 11 + 2 = 13. Then we complete the operation in the brackets. 4 −[7 −(11 + 2)] = 4 −[7 −(13)] = 4 −[−6] = 10 7 www.ck12.org Example 2 Evaluate the following: a) 3 × 5 −7 ÷ 2 b) 3 × (5 −7) ÷ 2 c) (3 × 5) −(7 ÷ 2) a) There are no grouping symbols. PEMDAS dictates that we multiply and divide ﬁrst, working from left to right: 3 × 5 = 15 and 7 ÷ 2 = 3.5. (NOTE: It’s not multiplication then division.) Next we subtract: 3 × 5 −7 ÷ 2 = 15 −3.5 = 11.5 b) First, we evaluate the expression inside the parentheses: 5 −7 = −2. Then work from left to right: 3 × (5 −7) ÷ 2 = 3 × (−2) ÷ 2 = (−6) ÷ 2 = −3 c) First, we evaluate the expressions inside parentheses: 3 × 5 = 15 and 7 ÷ 2 = 3.5. Then work from left to right: (3 × 5) −(7 ÷ 2) = 15 −3.5 = 11.5 Note that adding parentheses didn’t change the expression in part c, but did make it easier to read. Parentheses can be used to change the order of operations in an expression, but they can also be used simply to make it easier to understand. We can also use the order of operations to simplify an expression that has variables in it, after we substitute speciﬁc values for those variables. Example 3 Use the order of operations to evaluate the following: a) 2 −(3x + 2) when x = 2 b) 3y2 + 2y + 1 when y = −3 c) 2 −(t −7)2 × (u3 −v) when t = 19, u = 4, and v = 2 a) The ﬁrst step is to substitute the value for x into the expression. We can put it in parentheses to clarify the resulting expression. 2 −(3(2) + 2) (Note: 3(2) is the same as 3 × 2.) Follow PEMDAS - ﬁrst parentheses. Inside parentheses follow PEMDAS again. 2 −(3 × 2 + 2) = 2 −(6 + 2) Inside the parentheses, we multiply ﬁrst. 2 −8 = −6 Next we add inside the parentheses, and ﬁnally we subtract. www.ck12.org 8 b) The ﬁrst step is to substitute the value for y into the expression. 3 × (−3)2 + 2 × (−3) −1 Follow PEMDAS: we cannot simplify the expressions in parentheses, so exponents come next. 3 × (−3)2 + 2 × (−3) −1 Evaluate exponents: (−3)2 = 9 = 3 × 9 + 2 × (−3) −1 Evaluate multiplication: 3 × 9 = 27; 2 × −3 = −6 = 27 + (−6) −1 Add and subtract in order from left to right. = 27 −6 −1 = 20 c) The ﬁrst step is to substitute the values for t, u, and v into the expression. 2 −(19 −7)2 × (43 −2) Follow PEMDAS: 2 −(19 −7)2 × (43 −2) Evaluate parentheses: (19 −7) = 12; (43 −2) = (64 −2) = 62 = 2 −122 × 62 Evaluate exponents: 122 = 144 = 2 −144 × 62 Multiply: 144 × 62 = 8928 = 2 −8928 Subtract. = −8926 In parts (b) and (c) we left the parentheses around the negative numbers to clarify the problem. They did not aﬀect the order of operations, but they did help avoid confusion when we were multiplying negative numbers. Part (c) in the last example shows another interesting point. When we have an expression inside the parentheses, we use PEMDAS to determine the order in which we evaluate the contents. Evaluate Algebraic Expressions with Fraction Bars Fraction bars count as grouping symbols for PEMDAS, so we evaluate them in the ﬁrst step of solving an expression. All numerators and all denominators can be treated as if they have invisible parentheses around them. When real parentheses are also present, remember that the innermost grouping symbols come ﬁrst. If, for example, parentheses appear on a numerator, they would take precedence over the fraction bar. If the parentheses appear outside of the fraction, then the fraction bar takes precedence. Example 4 Use the order of operations to evaluate the following expressions: a) z+3 4 −1 when z = 2 b) ( a+2 b+4 −1 ) + b when a = 3 and b = 1 c) 2 × ( w+(x−2z) (y+2)2 −1 ) when w = 11, x = 3, y = 1, and z = −2 a) We substitute the value for z into the expression. 9 www.ck12.org 2 + 3 4 −1 Although this expression has no parentheses, the fraction bar is also a grouping symbol—it has the same eﬀect as a set of parentheses. We can write in the “invisible parentheses” for clarity: (2 + 3) 4 −1 Using PEMDAS, we ﬁrst evaluate the numerator: 5 4 −1 We can convert 5 4 to a mixed number: 5 4 = 11 4 Then evaluate the expression: 5 4 −1 = 11 4 −1 = 1 4 b) We substitute the values for a and b into the expression: (3 + 2 1 + 4 −1 ) + 1 This expression has nested parentheses (remember the eﬀect of the fraction bar). The innermost grouping symbol is provided by the fraction bar. We evaluate the numerator (3 + 2) and denominator (1 + 4) ﬁrst. (3 + 2 1 + 4 −1 ) + 1 = (5 5 −1 ) −1 Next we evaluate the inside of the parentheses. First we divide. = (1 −1) −1 Next we subtract. = 0 −1 = −1 c) We substitute the values for w, x, y, and z into the expression: 2 × ( 11+(3−2(−2)) (1+2)2 −1 ) This complicated expression has several layers of nested parentheses. One method for ensuring that we start with the innermost parentheses is to use more than one type of parentheses. Working from the outside, we can leave the outermost brackets as parentheses (). Next will be the “invisible brackets” from the fraction bar; we will write these as [ ]. The third level of nested parentheses will be the { }. We will leave negative numbers in round brackets. 2 ×         [11 + {3 −2(−2)}] [ {1 + 2}2] −1         Start with the innermost grouping sign: {} . {1 + 2} = 3; {3 −2(−2)} = 3 + 4 = 7 = 2 ([11 + 7] −1 ) Next, evaluate the square brackets. = 2 (18 9 −1 ) Next, evaluate the round brackets. Start with division. = 2(2 −1) Finally, do the addition and subtraction. = 2(1) = 2 www.ck12.org 10 Evaluate Algebraic Expressions with a TI-83/84 Family Graphing Calculator A graphing calculator is a very useful tool in evaluating algebraic expressions. Like a scientiﬁc calculator, a graphing calculator follows PEMDAS. In this section we will explain two ways of evaluating expressions with the graphing calculator. Example 5 Evaluate [ 3(x2 −1)2 −x4 + 12 ] + 5x3 −1 when x = −3. Method 1: Substitute for the variable ﬁrst. Then evaluate the numerical expression with the calculator. Substitute the value x = −3 into the expression. [ 3((−3)2 −1)2 −(−3)4 + 12 ] + 5(−3)3 −1 Input this in the calculator just as it is and press [ENTER]. (Note: use ∧to enter exponents) The answer is -13. Method 2: Input the original expression in the calculator ﬁrst and then evaluate. First, store the value x = −3 in the calculator. Type -3 [STO] x (The letter x can be entered using the x−[VAR] button or [ALPHA] + [STO]). Then type the original expression in the calculator and press [ENTER]. The answer is -13. The second method is better because you can easily evaluate the same expression for any value you want. For example, let’s evaluate the same expression using the values x = 2 and x = 2 3. For x = 2, store the value of x in the calculator: 2 [STO] x. Press [2nd] [ENTER] twice to get the previous expression you typed in on the screen without having to enter it again. Press [ENTER] to evaluate the expression. The answer is 62. 11 www.ck12.org For x = 2 3, store the value of x in the calculator: 2 3 [STO] x. Press [2nd] [ENTER] twice to get the expression on the screen without having to enter it again. Press [ENTER] to evaluate. The answer is 13.21, or 1070 81 in fraction form. Note: On graphing calculators there is a diﬀerence between the minus sign and the negative sign. When we stored the value negative three, we needed to use the negative sign which is to the left of the [ENTER] button on the calculator. On the other hand, to perform the subtraction operation in the expression we used the minus sign. The minus sign is right above the plus sign on the right. You can also use a graphing calculator to evaluate expressions with more than one variable. Example 7 Evaluate the expression 3x2−4y2+x4 (x+y) 1 2 for x = −2, y = 1. Solution Store the values of x and y: -2 [STO] x, 1 [STO] y. (The letters x and y can be entered using [ALPHA] + [KEY].) Input the expression in the calculator. When an expression includes a fraction, be sure to use parentheses: (numerator) (denominator). Press [ENTER] to obtain the answer −.88 or −8 9. Additional Resources For more practice, you can play an algebra game involving order of operations online at funbrain.com/algebra/index.html. Review Questions 1. Use the order of operations to evaluate the following expressions. (a) 8 −(19 −(2 + 5) −7) (b) 2 + 7 × 11 −12 ÷ 3 (c) (3 + 7) ÷ (7 −12) (d) 2·(3+(2−1)) 4−(6+2) −(3 −5) (e) 4+7(3) 9−4 + 12−3·2 2 (f) (4 −1)2 + 32 · 2 (g) (22+5)2 52−42 ÷ (2 + 1) www.ck12.org 12 2. Evaluate the following expressions involving variables. (a) jk j+k when j = 6 and k = 12 (b) 2y2 when x = 1 and y = 5 (c) 3x2 + 2x + 1 when x = 5 (d) (y2 −x)2 when x = 2 and y = 1 (e) x+y2 y−x when x = 2 and y = 3 3. Evaluate the following expressions involving variables. (a) 4x 9x2−3x+1 when x = 2 (b) z2 x+y + x2 x−y when x = 1, y = −2, and z = 4 (c) 4xyz y2−x2 when x = 3, y = 2, and z = 5 (d) x2−z2 xz−2x(z−x) when x = −1 and z = 3 4. Insert parentheses in each expression to make a true equation. (a) 5 −2 × 6 −5 + 2 = 5 (b) 12 ÷ 4 + 10 −3 × 3 + 7 = 11 (c) 22 −32 −5 × 3 −6 = 30 (d) 12 −8 −4 × 5 = −8 5. Evaluate each expression using a graphing calculator. (a) x2 + 2x −xy when x = 250 and y = −120 (b) (xy −y4)2 when x = 0.02 and y = −0.025 (c) x+y−z xy+yz+xz when x = 1 2, y = 3 2, and z = −1 (d) (x+y)2 4x2−y2 when x = 3 and y = −5 (e) (x−y)3 x3−y + (x+y)2 x+y4 when x = 4 and y = −2 1.3 Patterns and Equations Learning Objectives • Write an equation. • Use a verbal model to write an equation. • Solve problems using equations. Introduction In mathematics, and especially in algebra, we look for patterns in the numbers we see. The tools of algebra help us describe these patterns with words and with equations (formulas or functions). An equation is a mathematical recipe that gives the value of one variable in terms of another. For example, if a theme park charges $12 admission, then the number of people who enter the park every day and the amount of money taken in by the ticket oﬀice are related mathematically, and we can write a rule to ﬁnd the amount of money taken in by the ticket oﬀice. In words, we might say “The amount of money taken in is equal to twelve times the number of people who enter the park.” We could also make a table. The following table relates the number of people who visit the park and the total money taken in by the ticket oﬀice. 13 www.ck12.org Number of visitors 1 2 3 4 5 6 7 Money taken in ($) 12 24 36 48 60 72 84 Clearly, we would need a big table to cope with a busy day in the middle of a school vacation! A third way we might relate the two quantities (visitors and money) is with a graph. If we plot the money taken in on the vertical axis and the number of visitors on the horizontal axis, then we would have a graph that looks like the one shown below. Note that this graph shows a smooth line that includes non-whole number values of x (e.g. x = 2.5). In real life this would not make sense, because fractions of people can’t visit a park. This is an issue of domain and range, something we will talk about later. The method we will examine in detail in this lesson is closer to the ﬁrst way we chose to describe the relationship. In words we said that “The amount of money taken in is twelve times the number of people who enter the park.” In mathematical terms we can describe this sort of relationship with variables. A variable is a letter used to represent an unknown quantity. We can see the beginning of a mathematical formula in the words: The amount of money taken in is twelve times the number of people who enter the park. This can be translated to: the amount of money taken in = 12 × (the number of people who enter the park) We can now see which quantities can be assigned to letters. First we must state which letters (or variables) relate to which quantities. We call this deﬁning the variables: Let x = the number of people who enter the theme park. Let y = the total amount of money taken in at the ticket oﬀice. We now have a fourth way to describe the relationship: with an algebraic equation. y = 12x Writing a mathematical equation using variables is very convenient. You can perform all of the operations necessary to solve this problem without having to write out the known and unknown quantities over and over again. At the end of the problem, you just need to remember which quantities x and y represent. www.ck12.org 14 Write an Equation An equation is a term used to describe a collection of numbers and variables related through mathemat-ical operators. An algebraic equation will contain letters that represent real quantities. For example, if we wanted to use the algebraic equation in the example above to ﬁnd the money taken in for a certain number of visitors, we would substitute that number for x and then solve the resulting equation for y. Example 1 A theme park charges $12 entry to visitors. Find the money taken in if 1296 people visit the park. Let’s break the solution to this problem down into steps. This will be a useful strategy for all the problems in this lesson. Step 1: Extract the important information. (number of dollars taken in) = 12 × (number of visitors) (number of visitors) = 1296 Step 2: Translate into a mathematical equation. To do this, we pick variables to stand for the numbers. Let y = (number of dollars taken in). Let x = (number of visitors). (number of dollars taken in) = 12 × (number of visitors) y = 12 × x Step 3: Substitute in any known values for the variables. y = 12 × x x = 1296 ∴ y = 12 × 1296 Step 4: Solve the equation. y = 12 × 1296 = 15552 The amount of money taken in is $15552. Step 5: Check the result. If $15552 is taken at the ticket oﬀice and tickets are $12, then we can divide the total amount of money collected by the price per individual ticket. (number of people) = 15552 12 = 1296 1296 is indeed the number of people who entered the park. The answer checks out. Example 2 The following table shows the relationship between two quantities. First, write an equation that describes the relationship. Then, ﬁnd out the value of b when a is 750. 15 www.ck12.org a 0 10 20 30 40 50 b 20 40 60 80 100 120 Step 1: Extract the important information. We can see from the table that every time a increases by 10, b increases by 20. However, b is not simply twice the value of a. We can see that when a = 0, b = 20, and this gives a clue as to what rule the pattern follows. The rule linking a and b is: “To ﬁnd b, double the value of a and add 20.” Step 2: Translate into a mathematical equation: Table 1.1: Text Translates to Mathematical Expression “To ﬁnd b” → b = “double the value of a” → 2a “add 20” → + 20 Our equation is b = 2a + 20. Step 3: Solve the equation. The original problem asks for the value of b when a is 750. When a is 750, b = 2a + 20 becomes b = 2(750) + 20. Following the order of operations, we get: b = 2(750) + 20 = 1500 + 20 = 1520 Step 4: Check the result. In some cases you can check the result by plugging it back into the original equation. Other times you must simply double-check your math. In either case, checking your answer is always a good idea. In this case, we can plug our answer for b into the equation, along with the value for a, and see what comes out. 1520 = 2(750) + 20 is TRUE because both sides of the equation are equal. A true statement means that the answer checks out. Use a Verbal Model to Write an Equation In the last example we developed a rule, written in words, as a way to develop an algebraic equation. We will develop this further in the next few examples. Example 3 The following table shows the values of two related quantities. Write an equation that describes the relationship mathematically. www.ck12.org 16 Table 1.2: x−value y−value -2 10 0 0 2 -10 4 -20 6 -30 Step 1: Extract the important information. We can see from the table that y is ﬁve times bigger than x. The value for y is negative when x is positive, and it is positive when x is negative. Here is the rule that links x and y: “y is the negative of ﬁve times the value of x” Step 2: Translate this statement into a mathematical equation. Table 1.3: Text Translates to Mathematical Expression “y is” → y = “negative 5 times the value of x” → −5x Our equation is y = −5x. Step 3: There is nothing in this problem to solve for. We can move to Step 4. Step 4: Check the result. In this case, the way we would check our answer is to use the equation to generate our own xy pairs. If they match the values in the table, then we know our equation is correct. We will plug in -2, 0, 2, 4, and 6 for x and solve for y: Table 1.4: x y -2 −5(−2) = 10 0 −5(0) = 0 2 −5(2) = −10 4 −5(4) = −20 6 −5(6) = −30 The y−values in this table match the ones in the earlier table. The answer checks out. Example 4 Zarina has a $100 gift card, and she has been spending money on the card in small regular amounts. She checks the balance on the card weekly and records it in the following table. 17 www.ck12.org Table 1.5: Week Number Balance ($) 1 100 2 78 3 56 4 34 Write an equation for the money remaining on the card in any given week. Step 1: Extract the important information. The balance remaining on the card is not just a constant multiple of the week number; 100 is 100 times 1, but 78 is not 100 times 2. But there is still a pattern: the balance decreases by 22 whenever the week number increases by 1. This suggests that the balance is somehow related to the amount “-22 times the week number.” In fact, the balance equals “-22 times the week number, plus something.” To determine what that something is, we can look at the values in one row on the table—for example, the ﬁrst row, where we have a balance of $100 for week number 1. Step 2: Translate into a mathematical equation. First, we deﬁne our variables. Let n stand for the week number and b for the balance. Then we can translate our verbal expression as follows: Table 1.6: Text Translates to Mathematical Expression Balance equals -22 times the week number, plus something. → b = −22n+? To ﬁnd out what that ? represents, we can plug in the values from that ﬁrst row of the table, where b = 100 and n = 1. This gives us 100 = −22(1)+?. So what number gives 100 when you add -22 to it? The answer is 122, so that is the number the ? stands for. Now our ﬁnal equation is: b = −22n + 122 Step 3: All we were asked to ﬁnd was the expression. We weren’t asked to solve it, so we can move to Step 4. Step 4: Check the result. To check that this equation is correct, we see if it really reproduces the data in the table. To do that we plug in values for n: n = 1 →b = −22(1) + 122 = 122 −22 = 100 n = 2 →b = −22(2) + 122 = 122 −44 = 78 n = 3 →b = −22(3) + 122 = 122 −66 = 56 n = 4 →b = −22(4) + 122 = 122 −88 = 34 The equation perfectly reproduces the data in the table. The answer checks out. www.ck12.org 18 Solve Problems Using Equations Let’s solve the following real-world problem by using the given information to write a mathematical equation that can be solved for a solution. Example 5 A group of students are in a room. After 25 students leave, it is found that 2 3 of the original group is left in the room. How many students were in the room at the start? Step 1: Extract the important information We know that 25 students leave the room. We know that 2 3 of the original number of students are left in the room. We need to ﬁnd how many students were in the room at the start. Step 2: Translate into a mathematical equation. Initially we have an unknown number of students in the room. We can refer to this as the original number. Let’s deﬁne the variable x = the original number of students in the room. After 25 students leave the room, the number of students in the room is x −25. We also know that the number of students left is 2 3 of x. So we have two expressions for the number of students left, and those two expressions are equal because they represent the same number. That means our equation is: 2 3 x = x −25 Step 3: Solve the equation. Add 25 to both sides. x −25 = 2 3 x x −25 + 25 = 2 3 x + 25 x = 2 3 x + 25 Subtract 2 3 x from both sides. x −2 3 x = 2 3 x −2 3 x + 25 1 3 x = 25 Multiply both sides by 3. 3 · 1 3 x = 3 · 25 x = 75 Remember that x represents the original number of students in the room. So, there were 75 students in the room to start with. Step 4: Check the answer: If we start with 75 students in the room and 25 of them leave, then there are 75 −25 = 50 students left in the room. 19 www.ck12.org 2 3 of the original number is 2 3 · 75 = 50. This means that the number of students who are left over equals 2 3 of the original number. The answer checks out. The method of deﬁning variables and writing a mathematical equation is the method you will use the most in an algebra course. This method is often used together with other techniques such as making a table of values, creating a graph, drawing a diagram and looking for a pattern. Review Questions Table 1.7: Day Proﬁt 1 20 2 40 3 60 4 80 5 100 1. The above table depicts the proﬁt in dollars taken in by a store each day. (a) Write a mathematical equation that describes the relationship between the variables in the table. (b) What is the proﬁt on day 10? (c) If the proﬁt on a certain day is $200, what is the proﬁt on the next day? 2. (a) Write a mathematical equation that describes the situation: A full cookie jar has 24 cookies. How many cookies are left in the jar after you have eaten some? (b) How many cookies are in the jar after you have eaten 9 cookies? (c) How many cookies are in the jar after you have eaten 9 cookies and then eaten 3 more? 3. Write a mathematical equation for the following situations and solve. (a) Seven times a number is 35. What is the number? (b) Three times a number, plus 15, is 24. What is the number? (c) Twice a number is three less than ﬁve times another number. Three times the second number is 15. What are the numbers? (d) One number is 25 more than 2 times another number. If each number were multiplied by ﬁve, their sum would be 350. What are the numbers? (e) The sum of two consecutive integers is 35. What are the numbers? (f) Peter is three times as old as he was six years ago. How old is Peter? 4. How much water should be added to one liter of pure alcohol to make a mixture of 25% alcohol? 5. A mixture of 50% alcohol and 50% water has 4 liters of water added to it. It is now 25% alcohol. What was the total volume of the original mixture? 6. In Crystal’s silverware drawer there are twice as many spoons as forks. If Crystal adds nine forks to the drawer, there will be twice as many forks as spoons. How many forks and how many spoons are in the drawer right now? 7. (a) Mia drove to Javier’s house at 40 miles per hour. Javier’s house is 20 miles away. Mia arrived at Javier’s house at 2:00 pm. What time did she leave? (b) Mia left Javier’s house at 6:00 pm to drive home. This time she drove 25% faster. What time did she arrive home? www.ck12.org 20 (c) The next day, Mia took the expressway to Javier’s house. This route was 24 miles long, but she was able to drive at 60 miles per hour. How long did the trip take? (d) When Mia took the same route back, traﬀic on the expressway was 20% slower. How long did the return trip take? 8. The price of an mp3 player decreased by 20% from last year to this year. This year the price of the player is $120. What was the price last year? 9. SmartCo sells deluxe widgets for $60 each, which includes the cost of manufacture plus a 20% markup. What does it cost SmartCo to manufacture each widget? 10. Jae just took a math test with 20 questions, each worth an equal number of points. The test is worth 100 points total. (a) Write an equation relating the number of questions Jae got right to the total score he will get on the test. (b) If a score of 70 points earns a grade of C−, how many questions would Jae need to get right to get a C−on the test? (c) If a score of 83 points earns a grade of B, how many questions would Jae need to get right to get a B on the test? (d) Suppose Jae got a score of 60% and then was allowed to retake the test. On the retake, he got all the questions right that he got right the ﬁrst time, and also got half the questions right that he got wrong the ﬁrst time. What is his new score? 1.4 Equations and Inequalities Learning Objectives • Write equations and inequalities. • Check solutions to equations. • Check solutions to inequalities. • Solve real-world problems using an equation. Introduction In algebra, an equation is a mathematical expression that contains an equals sign. It tells us that two expressions represent the same number. For example, y = 12x is an equation. An inequality is a mathematical expression that contains inequality signs. For example, y ≤12x is an inequality. Inequalities are used to tell us that an expression is either larger or smaller than another expression. Equations and inequalities can contain both variables and constants. Variables are usually given a letter and they are used to represent unknown values. These quantities can change because they depend on other numbers in the problem. Constants are quantities that remain unchanged. Ordinary numbers like 2, −3, 3 4, and π are constants. Equations and inequalities are used as a shorthand notation for situations that involve numerical data. They are very useful because most problems require several steps to arrive at a solution, and it becomes tedious to repeatedly write out the situation in words. Write Equations and Inequalities Here are some examples of equations: 21 www.ck12.org 3x −2 = 5 x + 9 = 2x + 5 x 3 = 15 x2 + 1 = 10 To write an inequality, we use the following symbols: > greater than ≥greater than or equal to < less than ≤less than or equal to , not equal to Here are some examples of inequalities: 3x < 5 4 −x ≤2x x2 + 2x −1 > 0 3x 4 ≥x 2 −3 The most important skill in algebra is the ability to translate a word problem into the correct equation or inequality so you can ﬁnd the solution easily. The ﬁrst two steps are deﬁning the variables and translating the word problem into a mathematical equation. Deﬁning the variables means that we assign letters to any unknown quantities in the problem. Translating means that we change the word expression into a mathematical expression containing vari-ables and mathematical operations with an equal sign or an inequality sign. Example 1 Deﬁne the variables and translate the following expressions into equations. a) A number plus 12 is 20. b) 9 less than twice a number is 33. c) $20 was one quarter of the money spent on the pizza. Solution a) Deﬁne Let n = the number we are seeking. Translate A number plus 12 is 20. n + 12 = 20 b) Deﬁne Let n = the number we are seeking. Translate 9 less than twice a number is 33. This means that twice the number, minus 9, is 33. 2n −9 = 33 c) Deﬁne www.ck12.org 22 Let m = the money spent on the pizza. Translate $20 was one quarter of the money spent on the pizza. 20 = 1 4m Often word problems need to be reworded before you can write an equation. Example 2 Find the solution to the following problems. a) Shyam worked for two hours and packed 24 boxes. How much time did he spend on packing one box? b) After a 20% discount, a book costs $12. How much was the book before the discount? Solution a) Deﬁne Let t = time it takes to pack one box. Translate Shyam worked for two hours and packed 24 boxes. This means that two hours is 24 times the time it takes to pack one box. 2 = 24t Solve t = 2 24 = 1 12 hours 1 12 × 60 minutes = 5 minutes Answer Shyam takes 5 minutes to pack a box. b) Deﬁne Let p = the price of the book before the discount. Translate After a 20% discount, the book costs $12. This means that the price minus 20% of the price is $12. p −0.20p = 12 Solve p −0.20p = 0.8p, so 0.8p = 12 p = 12 0.8 = 15 Answer The price of the book before the discount was $15. 23 www.ck12.org Check If the original price was $15, then the book was discounted by 20% of $15, or $3. $15 −3 = $12. The answer checks out. Example 3 Deﬁne the variables and translate the following expressions into inequalities. a) The sum of 5 and a number is less than or equal to 2. b) The distance from San Diego to Los Angeles is less than 150 miles. c) Diego needs to earn more than an 82 on his test to receive a B in his algebra class. d) A child needs to be 42 inches or more to go on the roller coaster. Solution a) Deﬁne Let n = the unknown number. Translate 5 + n ≤2 b) Deﬁne Let d = the distance from San Diego to Los Angeles in miles. Translate d < 150 c) Deﬁne Let x = Diego’s test grade. Translate x > 82 d) Deﬁne Let h = the height of child in inches. Translate: h ≥42 Check Solutions to Equations You will often need to check solutions to equations in order to check your work. In a math class, checking that you arrived at the correct solution is very good practice. We check the solution to an equation by replacing the variable in an equation with the value of the solution. A solution should result in a true statement when plugged into the equation. Example 4 Check that the given number is a solution to the corresponding equation. www.ck12.org 24 a) y = −1; 3y + 5 = −2y b) z = 3; z2 + 2z = 8 c) x = −1 2; 3x + 1 = x Solution Replace the variable in each equation with the given value. a) 3(−1) + 5 = −2(−1) −3 + 5 = 2 2 = 2 This is a true statement. This means that y = −1 is a solution to 3y + 5 = −2y. b) 32 + 2(3) = 8 9 + 6 = 8 15 = 8 This is not a true statement. This means that z = 3 is not a solution to z2 + 2z = 8 . c) 3 ( −1 2 ) + 1 = −1 2 ( −3 2 ) + 1 = −1 2 −1 2 = −1 2 This is a true statement. This means that x = −1 2 is a solution to 3x + 1 = x. Check Solutions to Inequalities To check the solution to an inequality, we replace the variable in the inequality with the value of the solution. A solution to an inequality produces a true statement when substituted into the inequality. Example 5 Check that the given number is a solution to the corresponding inequality. a) a = 10; 20a ≤250 b) b = −0.5; 3−b b > −4 c) x = 3 4; 4x + 5 ≤8 Solution Replace the variable in each inequality with the given value. a) 20(10) ≤250 200 ≤250 This statement is true. This means that a = 10 is a solution to the inequality 20a ≤250. Note that a = 10 is not the only solution to this inequality. If we divide both sides of the inequality by 20, we can write it as a ≤12.5. This means that any number less than or equal to 12.5 is also a solution to the inequality. 25 www.ck12.org b) 3 −(−0.5) (−0.5) > −4 3 + 0.5 −0.5 > −4 −3.5 0.5 > −4 −7 > −4 This statement is false. This means that b = −0.5 is not a solution to the inequality 3−b b > −4 . c) 4 (3 4 ) + 5 ≥8 3 + 5 ≥8 8 ≥8 This statement is true. It is true because this inequality includes an equals sign; since 8 is equal to itself, it is also “greater than or equal to” itself. This means that x = 3 4 is a solution to the inequality 4x + 5 ≤8. Solve Real-World Problems Using an Equation Let’s use what we have learned about deﬁning variables, writing equations and writing inequalities to solve some real-world problems. Example 6 Tomatoes cost $0.50 each and avocados cost $2.00 each. Anne buys six more tomatoes than avocados. Her total bill is $8. How many tomatoes and how many avocados did Anne buy? Solution Deﬁne Let a = the number of avocados Anne buys. Translate Anne buys six more tomatoes than avocados. This means that a + 6 = the number of tomatoes. Tomatoes cost $0.50 each and avocados cost $2.00 each. Her total bill is $8. This means that .50 times the number of tomatoes plus 2 times the number of avocados equals 8. 0.5(a + 6) + 2a = 8 0.5a + 0.5 · 6 + 2a = 8 2.5a + 3 = 8 2.5a = 5 a = 2 Remember that a = the number of avocados, so Anne buys two avocados. The number of tomatoes is a + 6 = 2 + 6 = 8. Answer Anne bought 2 avocados and 8 tomatoes. www.ck12.org 26 Check If Anne bought two avocados and eight tomatoes, the total cost is: (2 × 2) + (8 × 0.5) = 4 + 4 = 8. The answer checks out. Example 7 To organize a picnic Peter needs at least two times as many hamburgers as hot dogs. He has 24 hot dogs. What is the possible number of hamburgers Peter has? Solution Deﬁne Let x = number of hamburgers Translate Peter needs at least two times as many hamburgers as hot dogs. He has 24 hot dogs. This means that twice the number of hot dogs is less than or equal to the number of hamburgers. 2 × 24 ≤x, or 48 ≤x Answer Peter needs at least 48 hamburgers. Check 48 hamburgers is twice the number of hot dogs. So more than 48 hamburgers is more than twice the number of hot dogs. The answer checks out. Additional Resources For more practice solving inequalities, check out Review Questions 1. Deﬁne the variables and translate the following expressions into equations. (a) Peter’s Lawn Mowing Service charges $10 per job and $0.20 per square yard. Peter earns $25 for a job. (b) Renting the ice-skating rink for a birthday party costs $200 plus $4 per person. The rental costs $324 in total. (c) Renting a car costs $55 per day plus $0.45 per mile. The cost of the rental is $100. (d) Nadia gave Peter 4 more blocks than he already had. He already had 7 blocks. 2. Deﬁne the variables and translate the following expressions into inequalities. (a) A bus can seat 65 passengers or fewer. (b) The sum of two consecutive integers is less than 54. (c) The product of a number and 3 is greater than 30. (d) An amount of money is invested at 5% annual interest. The interest earned at the end of the year is greater than or equal to $250. (e) You buy hamburgers at a fast food restaurant. A hamburger costs $0.49. You have at most $3 to spend. Write an inequality for the number of hamburgers you can buy. 27 www.ck12.org (f) Mariel needs at least 7 extra credit points to improve her grade in English class. Additional book reports are worth 2 extra credit points each. Write an inequality for the number of book reports Mariel needs to do. 3. Check whether the given number is a solution to the corresponding equation. (a) a = −3; 4a + 3 = −9 (b) x = 4 3; 3 4 x + 1 2 = 3 2 (c) y = 2; 2.5y −10.0 = −5.0 (d) z = −5; 2(5 −2z) = 20 −2(z −1) 4. Check whether the given number is a solution to the corresponding inequality. (a) x = 12; 2(x + 6) ≤8x (b) z = −9; 1.4z + 5.2 > 0.4z (c) y = 40; −5 2y + 1 2 < −18 (d) t = 0.4; 80 ≥10(3t + 2) 5. The cost of a Ford Focus is 27% of the price of a Lexus GS 450h. If the price of the Ford is $15000, what is the price of the Lexus? 6. On your new job you can be paid in one of two ways. You can either be paid $1000 per month plus 6% commission of total sales or be paid $1200 per month plus 5% commission on sales over $2000. For what amount of sales is the ﬁrst option better than the second option? Assume there are always sales over $2000. 7. A phone company oﬀers a choice of three text-messaging plans. Plan A gives you unlimited text messages for $10 a month; Plan B gives you 60 text messages for $5 a month and then charges you $0.05 for each additional message; and Plan C has no monthly fee but charges you $0.10 per message. (a) If m is the number of messages you send per month, write an expression for the monthly cost of each of the three plans. (b) For what values of m is Plan A cheaper than Plan B? (c) For what values of m is Plan A cheaper than Plan C? (d) For what values of m is Plan B cheaper than Plan C? (e) For what values of m is Plan A the cheapest of all? (Hint: for what values is A both cheaper than B and cheaper than C?) (f) For what values of m is Plan B the cheapest of all? (Careful—for what values is B cheaper than A?) (g) For what values of m is Plan C the cheapest of all? (h) If you send 30 messages per month, which plan is cheapest? (i) What is the cost of each of the three plans if you send 30 messages per month? 1.5 Functions as Rules and Tables Learning Objectives • Identify the domain and range of a function. • Make a table for a function. • Write a function rule. • Represent a real-world situation with a function. www.ck12.org 28 Introduction A function is a rule for relating two or more variables. For example, the price you pay for phone service may depend on the number of minutes you talk on the phone. We would say that the cost of phone service is a function of the number of minutes you talk. Consider the following situation. Josh goes to an amusement park where he pays $2 per ride. There is a relationship between the number of rides Josh goes on and the total amount he spends that day: To ﬁgure out the amount he spends, we multiply the number of rides by two. This rule is an example of a function. Functions usually—but not always—are rules based on mathematical operations. You can think of a function as a box or a machine that contains a mathematical operation. Whatever number we feed into the function box is changed by the given operation, and a new number comes out the other side of the box. When we input diﬀerent values for the number of rides Josh goes on, we get diﬀerent values for the amount of money he spends. The input is called the independent variable because its value can be any number. The output is called the dependent variable because its value depends on the input value. Functions usually contain more than one mathematical operation. Here is a situation that is slightly more complicated than the example above. Jason goes to an amusement park where he pays $8 admission and $2 per ride. The following function represents the total amount Jason pays. The rule for this function is ”multiply the number of rides by 2 and add 8.” When we input diﬀerent values for the number of rides, we arrive at diﬀerent outputs (costs). These ﬂow diagrams are useful in visualizing what a function is. However, they are cumbersome to use in practice. In algebra, we use the following short-hand notation instead: input ↓ f(x) \|{z} = y ←output function box First, we deﬁne the variables: x = the number of rides Jason goes on 29 www.ck12.org y = the total amount of money Jason spends at the amusement park. So, x represents the input and y represents the output. The notation f() represents the function or the mathematical operations we use on the input to get the output. In the last example, the cost is 2 times the number of rides plus 8. This can be written as a function: f(x) = 2x + 8 In algebra, the notations y and f(x) are typically used interchangeably. Technically, though, f(x) represents the function itself and y represents the output of the function. Identify the Domain and Range of a Function In the last example, we saw that we can input the number of rides into the function to give us the total cost for going to the amusement park. The set of all values that we can use for the input is called the domain of the function, and the set of all values that the output could turn out to be is called the range of the function. In many situations the domain and range of a function are both simply the set of all real numbers, but this isn’t always the case. Let’s look at our amusement park example. Example 1 Find the domain and range of the function that describes the situation: Jason goes to an amusement park where he pays $8 admission and $2 per ride. Solution Here is the function that describes this situation: f(x) = 2x + 8 = y In this function, x is the number of rides and y is the total cost. To ﬁnd the domain of the function, we need to determine which numbers make sense to use as the input (x). • The values have to be zero or positive, because Jason can’t go on a negative number of rides. • The values have to be integers because, for example, Jason could not go on 2.25 rides. • Realistically, there must be a maximum number of rides that Jason can go on because the park closes, he runs out of money, etc. However, since we aren’t given any information about what that maximum might be, we must consider that all non-negative integers are possible values regardless of how big they are. Answer For this function, the domain is the set of all non-negative integers. To ﬁnd the range of the function we must determine what the values of y will be when we apply the function to the input values. The domain is the set of all non-negative integers: {0, 1, 2, 3, 4, 5, 6, ...}. Next we plug these values into the function for x. If we plug in 0, we get 8; if we plug in 1, we get 10; if we plug in 2, we get 12, and so on, counting by 2s each time. Possible values of y are therefore 8, 10, 12, 14, 16, 18, 20... or in other words all even integers greater than or equal to 8. Answer The range of this function is the set of all even integers greater than or equal to 8. Example 2 Find the domain and range of the following functions. a) A ball is dropped from a height and it bounces up to 75% of its original height. www.ck12.org 30 b) y = x2 Solution a) Let’s deﬁne the variables: x = original height y = bounce height A function that describes the situation is y = f(x) = 0.75x. x can represent any real value greater than zero, since you can drop a ball from any height greater than zero. A little thought tells us that y can also represent any real value greater than zero. Answer The domain is the set of all real numbers greater than zero. The range is also the set of all real numbers greater than zero. b) Since there is no word problem attached to this equation, we can assume that we can use any real number as a value of x. When we square a real number, we always get a non-negative answer, so y can be any non-negative real number. Answer The domain of this function is all real numbers. The range of this function is all non-negative real numbers. In the functions we’ve looked at so far, x is called the independent variable because it can be any of the values from the domain, and y is called the dependent variable because its value depends on x. However, any letters or symbols can be used to represent the dependent and independent variables. Here are three diﬀerent examples: y = f(x) = 3x R = f(w) = 3w v = f(t) = 3t These expressions all represent the same function: a function where the dependent variable is three times the independent variable. Only the symbols are diﬀerent. In practice, we usually pick symbols for the dependent and independent variables based on what they represent in the real world—like t for time, d for distance, v for velocity, and so on. But when the variables don’t represent anything in the real world—or even sometimes when they do—we traditionally use y for the dependent variable and x for the independent variable. For another look at the domain of a function, see the following video, where the narrator solves a sample problem from the California Standards Test about ﬁnding the domain of an unusual function: http: //www.youtube.com/watch?v=NRB6s77nx2gI. Make a Table For a Function A table is a very useful way of arranging the data represented by a function. We can match the input and output values and arrange them as a table. For example, the values from Example 1 above can be arranged in a table as follows: x 0 1 2 3 4 5 6 y 8 10 12 14 16 18 20 A table lets us organize our data in a compact manner. It also provides an easy reference for looking up data, and it gives us a set of coordinate points that we can plot to create a graph of the function. 31 www.ck12.org Example 3 Make a table of values for the function f(x) = 1 x. Use the following numbers for input values: -1, -0.5, -0.2, -0.1, -0.01, 0.01, 0.1, 0.2, 0.5, 1. Solution Make a table of values by ﬁlling the ﬁrst row with the input values and the next row with the output values calculated using the given function. x −1 −0.5 −0.2 −0.1 −0.01 0.01 0.1 0.2 0.5 1 f(x) = 1 x 1 −1 1 −0.5 1 −0.2 1 −0.1 1 −0.01 1 0.01 1 0.1 1 0.2 1 0.5 1 1 y −1 −2 −5 −10 −100 100 10 5 2 1 When you’re given a function, you won’t usually be told what input values to use; you’ll need to decide for yourself what values to pick based on what kind of function you’re dealing with. We will discuss how to pick input values throughout this book. Write a Function Rule In many situations, we collect data by conducting a survey or an experiment, and then organize the data in a table of values. Most often, we want to ﬁnd the function rule or formula that ﬁts the set of values in the table, so we can use the rule to predict what could happen for values that are not in the table. Example 4 Write a function rule for the following table: Number of CDs 2 4 6 8 10 Cost in $ 24 48 72 86 120 Solution You pay $24 for 2 CDs, $48 for 4 CDs, $120 for 10 CDs. That means that each CD costs $12. We can write a function rule: Cost = $12 × (number of CDs) or f(x) = 12x Example 5 Write a function rule for the following table: x −3 −2 −1 0 1 2 3 y 3 2 1 0 1 2 3 Solution You can see that a negative number turns into the same number, only positive, while a non-negative number stays the same. This means that the function being used here is the absolute value function: f(x) =\| x \|. Coming up with a function based on a set of values really is as tricky as it looks. There’s no rule that will tell you the function every time, so you just have to think of all the types of functions you know and guess which one might be a good ﬁt, and then check if your guess is right. In this book, though, we’ll stick to writing functions for linear relationships, which are the simplest type of function. www.ck12.org 32 Represent a Real-World Situation with a Function Let’s look at a few real-world situations that can be represented by a function. Example 5 Maya has an internet service that currently has a monthly access fee of $11.95 and a connection fee of $0.50 per hour. Represent her monthly cost as a function of connection time. Solution Deﬁne Let x = the number of hours Maya spends on the internet in one month Let y = Maya’s monthly cost Translate The cost has two parts: the one-time fee of $11.95 and the per-hour charge of $0.50. So the total cost is the ﬂat fee + the charge per hour × the number of hours. Answer The function is y = f(x) = 11.95 + 0.50x. Example 6 Alfredo wants a deck build around his pool. The dimensions of the pool are 12 feet×24 feet and the decking costs $3 per square foot. Write the cost of the deck as a function of the width of the deck. Solution Deﬁne Let x = width of the deck Let y = cost of the deck Make a sketch and label it Translate You can look at the decking as being formed by several rectangles and squares. We can ﬁnd the areas of all the separate pieces and add them together: Area = 12x + 12x + 24x + 24x + x2 + x2 + x2 + x2 = 72x + 4x2 To ﬁnd the total cost, we then multiply the area by the cost per square foot ($3). Answer f(x) = 3(72x + 4x2) = 216x + 12x2 Example 7 33 www.ck12.org A cell phone company sells two million phones in their ﬁrst year of business. The number of phones they sell doubles each year. Write a function that gives the number of phones that are sold per year as a function of how old the company is. Solution Deﬁne Let x = age of company in years Let y = number of phones that are sold per year Make a table Age (years) 1 2 3 4 5 6 7 Millions of phones 2 4 8 16 32 64 128 Write a function rule The number of phones sold per year doubles every year, so the ﬁrst year the company sells 2 million phones, the next year it sells 2 × 2 million, the next year it sells 2 × 2 × 2 million, and so on. You might remember that when we multiply a number by itself several times we can use exponential notation: 2 = 21, 2 × 2 = 22, 2 × 2 × 2 = 23, and so on. In this problem, the exponent just happens to match the company’s age in years, which makes our function easy to describe. Answer y = f(x) = 2x Review Questions 1. Identify the domain and range of the following functions. (a) Dustin charges $10 per hour for mowing lawns. (b) Maria charges $25 per hour for tutoring math, with a minimum charge of $15. (c) f(x) = 15x −12 (d) f(x) = 2x2 + 5 (e) f(x) = 1 x (f) f(x) = √x 2. What is the range of the function y = x2 −5 when the domain is -2, -1, 0, 1, 2? 3. What is the range of the function y = 2x −3 4 when the domain is -2.5, -1.5, 5? 4. What is the domain of the function y = 3x when the range is 9, 12, 15? 5. What is the range of the function y = 3x when the domain is 9, 12, 15? 6. Angie makes $6.50 per hour working as a cashier at the grocery store. Make a table that shows how much she earns if she works 5, 10, 15, 20, 25, or 30 hours. 7. The area of a triangle is given by the formula A = 1 2bh. If the base of the triangle measures 8 centimeters, make a table that shows the area of the triangle for heights 1, 2, 3, 4, 5, and 6 centimeters. 8. Make a table of values for the function f(x) = √ 2x + 3 for input values -1, 0, 1, 2, 3, 4, 5. 9. Write a function rule for the following table: x 3 4 5 6 y 9 16 15 36 10. Write a function rule for the following table: www.ck12.org 34 Hours 0 1 2 3 Cost 15 20 25 30 11. Write a function rule for the following table: x 0 1 2 3 y 24 12 6 3 12. Write a function that represents the number of cuts you need to cut a ribbon into x pieces. 13. Write a function that represents the number of cuts you need to divide a pizza into x slices. 14. Solomon charges a $40 ﬂat rate plus $25 per hour to repair a leaky pipe. (a) Write a function that represents the total fee charged as a function of hours worked. (b) How much does Solomon earn for a 3-hour job? (c) How much does he earn for three separate 1-hour jobs? 15. Rochelle has invested $2500 in a jewelry making kit. She makes bracelets that she can sell for $12.50 each. (a) Write a function that shows how much money Rochelle makes from selling b bracelets. (b) Write a function that shows how much money Rochelle has after selling b bracelets, minus her investment in the kit. (c) How many bracelets does Rochelle need to make before she breaks even? (d) If she buys a $50 display case for her bracelets, how many bracelets does she now need to sell to break even? 1.6 Functions as Graphs Learning Objectives • Graph a function from a rule or table. • Write a function rule from a graph. • Analyze the graph of a real world situation. • Determine whether a relation is a function. Introduction We represent functions graphically by plotting points on a coordinate plane (also sometimes called the Cartesian plane). The coordinate plane is a grid formed by a horizontal number line and a vertical 35 www.ck12.org number line that cross at a point called the origin. The origin has this name because it is the “starting” location; every other point on the grid is described in terms of how far it is from the origin. The horizontal number line is called the x−axis and the vertical line is called the y−axis. We can represent each value of a function as a point on the plane by representing the x−value as a distance along the x−axis and the y−value as a distance along the y−axis. For example, if the y−value of a function is 2 when the x−value is 4, we can represent this pair of values with a point that is 4 units to the right of the origin (that is, 4 units along the x−axis) and 2 units up (2 units in the y−direction). We write the location of this point as (4, 2). Example 1 Plot the following coordinate points on the Cartesian plane. a) (5, 3) b) (-2, 6) c) (3, -4) d) (-5, -7) Solution Here are all the coordinate points on the same plot. Notice that we move to the right for a positive x−value and to the left for a negative one, just as we would on a single number line. Similarly, we move up for a positive y−value and down for a negative one. The x−and y−axes divide the coordinate plane into four quadrants. The quadrants are numbered counter-clockwise starting from the upper right, so the plotted point for (a) is in the ﬁrst quadrant, (b) is in the second quadrant, (c) is in the fourth quadrant, and (d) is in the third quadrant. www.ck12.org 36 Graph a Function From a Rule or Table If we know a rule or have a table of values that describes a function, we can draw a graph of the function. A table of values gives us coordinate points that we can plot on the Cartesian plane. Example 2 Graph the function that has the following table of values. x −2 −1 0 1 2 y 6 8 10 12 14 Solution The table gives us ﬁve sets of coordinate points: (-2, 6), (-1, 8), (0, 10), (1, 12), and (2, 14). To graph the function, we plot all the coordinate points. Since we are not told the domain of the function or given a real-world context, we can just assume that the domain is the set of all real numbers. To show that the function holds for all values in the domain, we connect the points with a smooth line (which, we understand, continues inﬁnitely in both directions). Example 3 Graph the function that has the following table of values. Side of square 0 1 2 3 4 Area of square 0 1 4 9 16 The table gives us ﬁve sets of coordinate points: (0, 0), (1, 1), (2, 4), (3, 9), and (4, 16). To graph the function, we plot all the coordinate points. Since we are not told the domain of the function, we can assume that the domain is the set of all non-negative real numbers. To show that the function holds for all values in the domain, we connect the points with a smooth curve. The curve does not make sense for negative values of the independent variable, so it stops at x = 0, but it continues inﬁnitely in the positive direction. 37 www.ck12.org Example 4 Graph the function that has the following table of values. Number of balloons 12 13 14 15 16 Cost 41 44 47 50 53 This function represents the total cost of the balloons delivered to your house. Each balloon is $3 and the store delivers if you buy a dozen balloons or more. The delivery charge is a $5 ﬂat fee. Solution The table gives us ﬁve sets of coordinate points: (12, 41), (13, 44), (14, 47), (15, 50), and (16, 53). To graph the function, we plot all the coordinate points. Since the x−values represent the number of balloons for 12 balloons or more, the domain of this function is all integers greater than or equal to 12. In this problem, the points are not connected by a line or curve because it doesn’tmake sense to have non-integer values of balloons. In order to draw a graph of a function given the function rule, we must ﬁrst make a table of values to give us a set of points to plot. Choosing good values for the table is a skill you’l develop throughout this course. When you pick values, here are some of the things you should keep in mind. • Pick only values from the domain of the function. • If the domain is the set of real numbers or a subset of the real numbers, the graph will be a continuous curve. • If the domain is the set of integers of a subset of the integers, the graph will be a set of points not connected by a curve. • Picking integer values is best because it makes calculations easier, but sometimes we need to pick other values to capture all the details of the function. www.ck12.org 38 • Often we start with one set of values. Then after drawing the graph, we realize that we need to pick diﬀerent values and redraw the graph. Example 5 Graph the following function: f(x) = \|x −2\| Solution Make a table of values. Pick a variety of negative and positive values for x. Use the function rule to ﬁnd the value of y for each value of x. Then, graph each of the coordinate points. Table 1.8: x y = f(x) =\| x −2 \| -4 \| −4 −2 \|=\| −6 \|= 6 -3 \| −3 −2 \|=\| −5 \|= 5 -2 \| −2 −2 \|=\| −4 \|= 4 -1 \| −1 −2 \|=\| −3 \|= 3 0 \| 0 −2 \|=\| −2 \|= 2 1 \| 1 −2 \|=\| −1 \|= 1 2 \| 2 −2 \|=\| 0 \|= 0 3 \| 3 −2 \|=\| 1 \|= 1 4 \| 4 −2 \|=\| 2 \|= 2 5 \| 5 −2 \|=\| 3 \|= 3 6 \| 6 −2 \|=\| 4 \|= 4 7 \| 7 −2 \|=\| 5 \|= 5 8 \| 8 −2 \|=\| 6 \|= 6 It is wise to work with a lot of values when you begin graphing. As you learn about diﬀerent types of functions, you will start to only need a few points in the table of values to create an accurate graph. Example 6 Graph the following function: f(x) = √x Solution Make a table of values. We know x can’t be negative because we can’t take the square root of a negative number. The domain is all positive real numbers, so we pick a variety of positive integer values for x. Use the function rule to ﬁnd the value of y for each value of x. 39 www.ck12.org Table 1.9: x y = f(x) = √x 0 √ 0 = 0 1 √ 1 = 1 2 √ 2 ≈1.41 3 √ 3 ≈1.73 4 √ 4 = 2 5 √ 5 ≈2.24 6 √ 6 ≈2.45 7 √ 7 ≈2.65 8 √ 8 ≈2.83 9 √ 9 = 3 Note that the range is all positive real numbers. Example 7 The post oﬀice charges 41 cents to send a letter that is one ounce or less and an extra 17 cents for each additional ounce or fraction of an ounce. This rate applies to letters up to 3.5 ounces. Solution Make a table of values. We can’t use negative numbers for x because it doesn’t make sense to have negative weight. We pick a variety of positive values for x, making sure to include some decimal values because prices can be decimals too. Then we use the function rule to ﬁnd the value of y for each value of x. x 0 0.2 0.5 0.8 1 1.2 1.5 1.8 2 2.2 2.5 2.8 3 3.2 3.5 y 0 41 41 41 41 58 58 58 58 75 75 75 75 92 92 www.ck12.org 40 Write a Function Rule from a Graph Sometimes you’ll need to ﬁnd the equation or rule of a function by looking at the graph of the function. Points that are on the graph can give you values of dependent and independent variables that are related to each other by the function rule. However, you must make sure that the rule works for all the points on the curve. In this course you will learn to recognize diﬀerent kinds of functions and discover the rules for all of them. For now we’ll look at some simple examples and ﬁnd patterns that will help us ﬁgure out how the dependent and independent variables are related. Example 8 The graph to the right shows the distance that an ant covers over time. Find the function rule that shows how distance and time are related to each other. Solution Let’s make a table of values of several coordinate points to see if we can spot how they are related to each other. Time 0 1 2 3 4 5 6 Distance 0 1.5 3 4.5 6 7.5 9 We can see that for every second the distance increases by 1.5 feet. We can write the function rule as Distance = 1.5 × time The equation of the function is f(x) = 1.5x. 41 www.ck12.org Example 9 Find the function rule that describes the function shown in the graph. Solution Again, we can make a table of values of several coordinate points to identify how they are related to each other. x −4 −3 −2 −1 0 1 2 3 4 y 8 4.5 2 .5 0 .5 5 4.5 8 Notice that the values of y are half of perfect squares: 8 is half of 16 (which is 4 squared), 4.5 is half of 9 (which is 3 squared), and so on. So the equation of the function is f(x) = 1 2 x2. Example 10 Find the function rule that shows the volume of a balloon at diﬀerent times, based on the following graph: (Notice that the graph shows negative time. The negative time can represent what happened on days before you started measuring the volume.) Solution Once again, we make a table to spot the pattern: Time −1 0 1 2 3 4 5 Volume 10 5 2.5 1.2 0.6 0.3 0.15 We can see that every day, the volume of the balloon is half what it was the previous day. On day 0, the volume is 5; on day 1, the volume is 5 × 1 2; on day 2, it is 5 × 1 2 × 1 2, and in general, on day x it is 5 × ( 1 2 )x. The equation of the function is f(x) = 5 × ( 1 2 )x. www.ck12.org 42 Determine Whether a Relation is a Function A function is a special kind of relation. In a function, for each input there is exactly one output; in a relation, there can be more than one output for a given input. Consider the relation that shows the heights of all students in a class. The domain is the set of people in the class and the range is the set of heights. This relation is a function because each person has exactly one height. If any person had more than one height, the relation would not be a function. Notice that even though the same person can’t have more than one height, it’s okay for more than one person to have the same height. In a function, more than one input can have the same output, as long as more than one output never comes from the same input. Example 11 Determine if the relation is a function. a) (1, 3), (-1, -2), (3, 5), (2, 5), (3, 4) b) (-3, 20), (-5, 25), (-1, 5), (7, 12), (9, 2) c) x 2 1 0 1 2 y 12 10 8 6 4 Solution The easiest way to ﬁgure out if a relation is a function is to look at all the x−values in the list or the table. If a value of x appears more than once, and it’s paired up with diﬀerent y−values, then the relation is not a function. a) You can see that in this relation there are two diﬀerent y−values paired with the x−value of 3. This means that this relation is not a function. b) Each value of x has exactly one y−value. The relation is a function. c) In this relation there are two diﬀerent y−values paired with the x−value of 2 and two y−values paired with the x−value of 1. The relation is not a function. When a relation is represented graphically, we can determine if it is a function by using the vertical line test. If you can draw a vertical line that crosses the graph in more than one place, then the relation is not a function. Here are some examples. 43 www.ck12.org Not a function. It fails the vertical line test. A function. No vertical line will cross more than one point on the graph. A function. No vertical line will cross more than one point on the graph. Not a function. It fails the vertical line test. www.ck12.org 44 Additional Resources Once you’ve had some practice graphing functions by hand, you may want to use a graphing calculator to make graphing easier. If you don’t have one, you can also use the applet at function-graphs/. Just type a function in the blank and press Enter. You can use the options under Display Properties to zoom in or pan around to diﬀerent parts of the graph. Review Questions 1. Plot the coordinate points on the Cartesian plane. (a) (4, -4) (b) (2, 7) (c) (-3, -5) (d) (6, 3) (e) (-4, 3) 2. Give the coordinates for each point in this Cartesian plane. 3. Graph the function that has the following table of values. (a) x −10 −5 0 5 10 y −3 −0.5 2 4.5 7 (b) Side of cube (in.) 0 1 2 3 Volume (in3) 0 1 8 27 (c) Time (hours) −2 −1 0 1 2 Distance from town center (miles) 50 25 0 25 50 4. Graph the following functions. (a) Brandon is a member of a movie club. He pays a $50 annual membership and $8 per movie. (b) f(x) = (x −2)2 (c) f(x) = 3.2x 5. Determine whether each relation is a function: (a) (1, 7), (2, 7), (3, 8), (4, 8), (5, 9) (b) (1, 1), (1, -1), (4, 2), (4, -2), (9, 3), (9, -3) (c) x −4 −3 −2 −1 0 y 16 9 4 1 0 (d) Age 20 25 25 30 35 Number of jobs by that age 3 4 7 4 2 45 www.ck12.org 6. Write the function rule for each graph. (a) (b) 7. The students at a local high school took The Youth Risk Behavior Survey. The graph below shows the percentage of high school students who reported that they were current smokers. (A current smoker is anyone who has smoked one or more cigarettes in the past 30 days.) What percentage of high-school students were current smokers in the following years? (a) 1991 (b) 1996 (c) 2004 (d) 2005 8. The graph below shows the average life-span of people based on the year in which they were born. This information comes from the National Vital Statistics Report from the Center for Disease Control. What is the average life-span of a person born in the following years? (a) 1940 (b) 1955 (c) 1980 www.ck12.org 46 (d) 1995 9. The graph below shows the median income of an individual based on his/her number of years of education. The top curve shows the median income for males and the bottom curve shows the median income for females. (Source: US Census, 2003.) What is the median income of a male that has the following years of education? (a) 10 years of education (b) 17 years of education 10. What is the median income of a female that has the same years of education? (a) 10 years of education (b) 17 years of education 11. Use the vertical line test to determine whether each relation is a function. (a) 47 www.ck12.org (b) 1.7 Problem-Solving Plan Learning Objectives • Read and understand given problem situations. • Make a plan to solve the problem. • Solve the problem and check the results. • Compare alternative approaches to solving the problem. • Solve real-world problems using a plan. Introduction We always think of mathematics as the subject in school where we solve lots of problems. Problem solving is necessary in all aspects of life. Buying a house, renting a car, or ﬁguring out which is the better sale are just a few examples of situations where people use problem-solving techniques. In this book, you will learn diﬀerent strategies and approaches to solving problems. In this section, we will introduce a problem-solving plan that will be useful throughout this book. Read and Understand a Given Problem Situation The ﬁrst step to solving a word problem is to read and understand the problem. Here are a few questions that you should be asking yourself: • What am I trying to ﬁnd out? • What information have I been given? • Have I ever solved a similar problem? This is also a good time to deﬁne any variables. When you identify your knowns and unknowns, it is often useful to assign them a letter to make notation and calculations easier. Make a Plan to Solve the Problem The next step in the problem-solving plan is to develop a strategy. How can the information you know assist you in ﬁguring out the unknowns? Here are some common strategies that you will learn: www.ck12.org 48 • Drawing a diagram. • Making a table. • Looking for a pattern. • Using guess and check. • Working backwards. • Using a formula. • Reading and making graphs. • Writing equations. • Using linear models. • Using dimensional analysis. • Using the right type of function for the situation. In most problems, you will use a combination of strategies. For example, looking for patterns is a good strategy for most problems, and making a table and drawing a graph are often used together. The “writing an equation” strategy is the one you will work with the most in your study of algebra. Solve the Problem and Check the Results Once you develop a plan, you can implement it and solve the problem, carrying out all operations to arrive at the answer you are seeking. The last step in solving any problem should always be to check and interpret the answer. Ask yourself: • Does the answer make sense? • If you plug the answer back into the problem, do all the numbers work out? • Can you get the same answer through another method? Compare Alternative Approaches to Solving the Problem Sometimes one speciﬁc method is best for solving a problem. Most problems, however, can be solved by using several diﬀerent strategies. When you are familiar with all of the problem-solving strategies, it is up to you to choose the methods that you are most comfortable with and that make sense to you. In this book, we will often use more than one method to solve a problem, so we can demonstrate the strengths and weakness of diﬀerent strategies for solving diﬀerent types of problems. Whichever strategy you are using, you should always implement the problem-solving plan when you are solving word problems. Here is a summary of the problem-solving plan. Step 1: Understand the problem Read the problem carefully. Once the problem is read, list all the components and data that are involved. This is where you will be assigning your variables. Step 2: Devise a plan - Translate Come up with a way to solve the problem. Set up an equation, draw a diagram, make a chart or construct a table as a start to solve your problem solving plan. Step 3: Carry out the plan - Solve 49 www.ck12.org This is where you solve the equation you developed in Step 2. Step 4: Look - Check and Interpret Check to see if you used all your information. Then look to see if the answer makes sense. Solve Real-World Problems Using a Plan Let’s now apply this problem solving plan to a problem. Example 1 A coﬀee maker is on sale at 50% oﬀthe regular ticket price. On the “Sunday Super Sale” the same coﬀee maker is on sale at an additional 40% oﬀ. If the ﬁnal price is $21, what was the original price of the coﬀee maker? Solution Step 1: Understand We know: A coﬀee maker is discounted 50% and then 40%. The ﬁnal price is $21. We want: The original price of the coﬀee maker. Step 2: Strategy Let’s look at the given information and try to ﬁnd the relationship between the information we know and the information we are trying to ﬁnd. 50% oﬀthe original price means that the sale price is half of the original or 0.5 × original price. So, the ﬁrst sale price = 0.5 × original price A savings of 40% oﬀthe new price means you pay 60% of the new price, or 0.6 × new price. 0.6 × (0.5 × original price) = 0.3 × original price is the price after the second discount. We know that after two discounts, the ﬁnal price is $21. So 0.3 × original price = $21. Step 3: Solve Since 0.3 × original price = $21, we can ﬁnd the original price by dividing $21 by 0.3. Original price = $21 ÷ 0.3 = $70. The original price of the coﬀee maker was $70. Step 4: Check We found that the original price of the coﬀee maker is $70. To check that this is correct, let’s apply the discounts. 50% of $70 = .5 × $70 = $35 savings. So the price after the ﬁrst discount is original price −savings or $70 −35 = $35. Then 40% of that is .4 × $35 = $14. So after the second discount, the price is $35 −14 = $21. The answer checks out. www.ck12.org 50 Additional Resources The problem-solving plan used here is based on the ideas of George P´ olya, who describes his useful problem-solving strategies in more detail in the book How to Solve It. Some of the techniques in the book can also be found on Wikipedia, in the entry Review Questions 1. A sweatshirt costs $35. Find the total cost if the sales tax is 7.75%. 2. This year you got a 5% raise. If your new salary is $45,000, what was your salary before the raise? 3. Mariana deposits $500 in a savings account that pays 3% simple interest per year. How much will be in her account after three years? 4. It costs $250 to carpet a room that is 14 ft by 18 ft. How much does it cost to carpet a room that is 9 ft by 10 ft? 5. A department store has a 15% discount for employees. Suppose an employee has a coupon worth $10 oﬀany item and she wants to buy a $65 purse. What is the ﬁnal cost of the purse if the employee discount is applied before the coupon is subtracted? 6. To host a dance at a hotel you must pay $250 plus $20 per guest. How much money would you have to pay for 25 guests? 7. Yusef’s phone plan costs $10 a month plus $0.05 per minute. If his phone bill for last month was $25.80, how many minutes did he spend on the phone? 8. It costs $12 to get into the San Diego County Fair and $1.50 per ride. (a) If Rena spent $24 in total, how many rides did she go on? (b) How much would she have spent in total if she had gone on ﬁve more rides? 9. An ice cream shop sells a small cone for $2.95, a medium cone for $3.50, and a large cone for $4.25. Last Saturday, the shop sold 22 small cones, 26 medium cones and 15 large cones. How much money did the store earn? 10. In Lise’s chemistry class, there are two midterm exams, each worth 30% of her total grade, and a ﬁnal exam worth 40%. If Lise scores 90% on both midterms and 80% on the ﬁnal exam, what is her overall score in the class? 11. The sum of the angles in a triangle is 180 degrees. If the second angle is twice the size of the ﬁrst angle and the third angle is three times the size of the ﬁrst angle, what are the measures of the angles in the triangle? 12. A television that normally costs $120 goes on sale for 20% oﬀ. What is the new price? 13. A cake recipe calls for 13 4 cup of ﬂour. Jeremy wants to make four cakes. How many cups of ﬂour will he need? 14. Casey is twice as old as Marietta, who is two years younger than Jake. If Casey is 14, how old is Jake? 15. Kylie is mowing lawns to earn money for a new bike. After mowing four lawns, she still needs $40 more to pay for the bike. After mowing three more lawns, she has $5 more than she needs to pay for the bike. (a) How much does she earn per lawn? (b) What is the cost of the bike? 16. Jared goes trick-or-treating with his brother and sister. At the ﬁrst house they stop at, they collect three pieces of candy each; at the next three houses, they collect two pieces of candy each. Then they split up and go down diﬀerent blocks, where Jared collects 12 pieces of candy and his brother and sister collect 14 each. (a) How many pieces of candy does Jared end up with? 51 www.ck12.org (b) How many pieces of candy do all three of them together end up with? 17. Marco’s daughter Elena has four boxes of toy blocks, with 50 blocks in each one. One day she dumps them all out on the ﬂoor, and some of them get lost. When Marco tries to put them away again, he ends up with 45 blocks in one box, 53 in another, 46 in a third, and 51 in the fourth. How many blocks are missing? 18. A certain hour-long TV show usually includes 16 minutes of commercials. If the season ﬁnale is two and a half hours long, how many minutes of commercials should it include to keep the same ratio of commercial time to show time? 19. Karen and Chase bet on a baseball game: if the home team wins, Karen owes Chase ﬁfty cents for every run scored by both teams, and Chase owes Karen the same amount if the visiting team wins. The game runs nine innings, and the home team scores one run in every odd-numbered inning, while the visiting team scores two runs in the third inning and two in the sixth. Who owes whom how much? 20. Kelly, Chris, and Morgan are playing a card game. In this game, the ﬁrst player to empty their hand scores points for all the cards left in the other players’ hands as follows: aces are worth one point, face cards ten points, and all other cards are face value. When Kelly empties her hand, Morgan is holding two aces, a king, and a three; Chris is holding a ﬁve, a seven, and a queen. How many points does Kelly score? 21. A local club rents out a social hall to host an event. The hall rents for $350, and they hope to make back the rental price by charging $15 admission per person. How many people need to attend for the club to break even? 22. You plan to host a barbecue, and you expect 10 friends, 8 neighbors, and 7relatives to show up. (a) If you expect each person (including yourself) to eat about two ounces of potato salad, how many half-pound containers of potato salad should you buy? (b) If hot dogs come in ten-packs that cost $4.80 apiece and hot dog buns come in eight-packs that cost $2.80 apiece, how much will you need to spend to have hot dogs and buns for everyone? 1.8 Problem-Solving Strategies: Make a Table and Look for a Pattern Learning Objectives • Read and understand given problem situations. • Develop and use the strategy “make a table.” • Develop and use the strategy “look for a pattern.” • Plan and compare alternative approaches to solving a problem. • Solve real-world problems using the above strategies as part of a plan. Introduction In this section, we will apply the problem-solving plan you learned about in the last section to solve several real-world problems. You will learn how to develop and use the methods make a table and look for a pattern. www.ck12.org 52 Read and Understand Given Problem Situations The most diﬀicult parts of problem-solving are most often the ﬁrst two steps in our problem-solving plan. You need to read the problem and make sure you understand what you are being asked. Once you understand the problem, you can devise a strategy to solve it. Let’s apply the ﬁrst two steps to the following problem. Example 1: Six friends are buying pizza together and they are planning to split the check equally. After the pizza was ordered, one of the friends had to leave suddenly, before the pizza arrived. Everyone left had to pay $1 extra as a result. How much was the total bill? Solution Understand We want to ﬁnd how much the pizza cost. We know that ﬁve people had to pay an extra $1 each when one of the original six friends had to leave. Strategy We can start by making a list of possible amounts for the total bill. We divide the amount by six and then by ﬁve. The total divided by ﬁve should equal $1 more than the total divided by six. Look for any patterns in the numbers that might lead you to the correct answer. In the rest of this section you will learn how to make a table or look for a pattern to ﬁgure out a solution for this type of problem. After you ﬁnish reading the rest of the section, you can ﬁnish solving this problem for homework. Develop and Use the Strategy: Make a Table The method “Make a Table” is helpful when solving problems involving numerical relationships. When data is organized in a table, it is easier to recognize patterns and relationships between numbers. Let’s apply this strategy to the following example. Example 2 Josie takes up jogging. On the ﬁrst week she jogs for 10 minutes per day, on the second week she jogs for 12 minutes per day. Each week, she wants to increase her jogging time by 2 minutes per day. If she jogs six days each week, what will be her total jogging time on the sixth week? Solution Understand We know in the ﬁrst week Josie jogs 10 minutes per day for six days. We know in the second week Josie jogs 12 minutes per day for six days. Each week, she increases her jogging time by 2 minutes per day and she jogs 6 days per week. We want to ﬁnd her total jogging time in week six. Strategy A good strategy is to list the data we have been given in a table and use the information we have been given to ﬁnd new information. 53 www.ck12.org We are told that Josie jogs 10 minutes per day for six days in the ﬁrst week and 12 minutes per day for six days in the second week. We can enter this information in a table: Table 1.10: Week Minutes per Day Minutes per Week 1 10 60 2 12 72 You are told that each week Josie increases her jogging time by 2 minutes per day and jogs 6 times per week. We can use this information to continue ﬁlling in the table until we get to week six. Table 1.11: Week Minutes per Day Minutes per Week 1 10 60 2 12 72 3 14 84 4 16 96 5 18 108 6 20 120 Apply strategy/solve To get the answer we read the entry for week six. Answer: In week six Josie jogs a total of 120 minutes. Check Josie increases her jogging time by two minutes per day. She jogs six days per week. This means that she increases her jogging time by 12 minutes per week. Josie starts at 60 minutes per week and she increases by 12 minutes per week for ﬁve weeks. That means the total jogging time is 60 + 12 × 5 = 120 minutes. The answer checks out. You can see that making a table helped us organize and clarify the information we were given, and helped guide us in the next steps of the problem. We solved this problem solely by making a table; in many situations, we would combine this strategy with others to get a solution. Develop and Use the Strategy: Look for a Pattern Looking for a pattern is another strategy that you can use to solve problems. The goal is to look for items or numbers that are repeated or a series of events that repeat. The following problem can be solved by ﬁnding a pattern. Example 3 You arrange tennis balls in triangular shapes as shown. How many balls will there be in a triangle that has 8 rows? www.ck12.org 54 Solution Understand We know that we arrange tennis balls in triangles as shown. We want to know how many balls there are in a triangle that has 8 rows. Strategy A good strategy is to make a table and list how many balls are in triangles of diﬀerent rows. One row: It is simple to see that a triangle with one row has only one ball. Two rows: For a triangle with two rows, we add the balls from the top row to the balls from the bottom row. It is useful to make a sketch of the separate rows in the triangle. 3 = 1 + 2 Three rows: We add the balls from the top triangle to the balls from the bottom row. 6 = 3 + 3 Now we can ﬁll in the ﬁrst three rows of a table. Table 1.12: Number of Rows Number of Balls 1 1 2 3 3 6 We can see a pattern. To create the next triangle, we add a new bottom row to the existing triangle. The new bottom row has the same number of balls as there are rows. (For example, a triangle with 3 rows has 3 balls in the bottom row.) 55 www.ck12.org To get the total number of balls for the new triangle, we add the number of balls in the old triangle to the number of balls in the new bottom row. Apply strategy/solve: We can complete the table by following the pattern we discovered. Number of balls = number of balls in previous triangle + number of rows in the new triangle Table 1.13: Number of Rows Number of Balls 1 1 2 3 3 6 4 6 + 4 = 10 5 10 + 5 = 15 6 15 + 6 = 21 7 21 + 7 = 28 8 28 + 8 = 36 Answer There are 36 balls in a triangle arrangement with 8 rows. Check Each row of the triangle has one more ball than the previous one. In a triangle with 8 rows, row 1 has 1 ball, row 2 has 2 balls, row 3 has 3 balls, row 4 has 4 balls, row 5 has 5 balls, row 6 has 6 balls, row 7 has 7 balls, row 8 has 8 balls. When we add these we get: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36 balls The answer checks out. Notice that in this example we made tables and drew diagrams to help us organize our information and ﬁnd a pattern. Using several methods together is a very common practice and is very useful in solving word problems. Plan and Compare Alternative Approaches to Solving Problems In this section, we will compare the methods of “Making a Table” and “Looking for a Pattern” by using each method in turn to solve a problem. Example 4 Andrew cashes a $180 check and wants the money in $10 and $20 bills. The bank teller gives him 12 bills. How many of each kind of bill does he receive? Solution Method 1: Making a Table Understand Andrew gives the bank teller a $180 check. The bank teller gives Andrew 12 bills. These bills are a mix of $10 bills and $20 bills. We want to know how many of each kind of bill Andrew receives. www.ck12.org 56 Strategy Let’s start by making a table of the diﬀerent ways Andrew can have twelve bills in tens and twenties. Andrew could have twelve $10 bills and zero $20 bills, or eleven $10 bills and one $20 bill, and so on. We can calculate the total amount of money for each case. Apply strategy/solve Table 1.14: $10 bills $ 20 bills Total amount 12 0 $10(12) + $20(0) = $120 11 1 $10(11) + $20(1) = $130 10 2 $10(10) + $20(2) = $140 9 3 $10(9) + $20(3) = $150 8 4 $10(8) + $20(4) = $160 7 5 $10(7) + $20(5) = $170 6 6 $10(6) + $20(6) = $180 5 7 $10(5) + $20(7) = $190 4 8 $10(4) + $20(8) = $200 3 9 $10(3) + $20(9) = $210 2 10 $10(2) + $20(10) = $220 1 11 $10(1) + $20(11) = $230 0 12 $10(0) + $20(12) = $240 In the table we listed all the possible ways you can get twelve $10 bills and $20 bills and the total amount of money for each possibility. The correct amount is given when Andrew has six $10 bills and six $20 bills. Answer: Andrew gets six $10 bills and six $20 bills. Check Six $10 bills and six $20 bills →6($10) + 6($20) = $60 + $120 = $180 The answer checks out. Let’s solve the same problem using the method “Look for a Pattern.” Method 2: Looking for a Pattern Understand Andrew gives the bank teller a $180 check. The bank teller gives Andrew 12 bills. These bills are a mix of $10 bills and $20 bills. We want to know how many of each kind of bill Andrew receives. Strategy Let’s start by making a table just as we did above. However, this time we will look for patterns in the table that can be used to ﬁnd the solution. Apply strategy/solve Let’s ﬁll in the rows of the table until we see a pattern. 57 www.ck12.org Table 1.15: $10 bills $20 bills Total amount 12 0 $10(12) + $20(0) = $120 11 1 $10(11) + $20(1) = $130 10 2 $10(10) + $20(2) = $140 We see that every time we reduce the number of $10 bills by one and increase the number of $20 bills by one, the total amount increases by $10. The last entry in the table gives a total amount of $140, so we have $40 to go until we reach our goal. This means that we should reduce the number of $10 bills by four and increase the number of $20 bills by four. That would give us six $10 bills and six $20 bills. 6($10) + 6($20) = $60 + 120 = $180 Answer: Andrew gets six $10 bills and six $20 bills. Check Six $10 bills and six $20 bills →6($10) + 6($20) = $60 + 120 = $180 The answer checks out. You can see that the second method we used for solving the problem was less tedious. In the ﬁrst method, we listed all the possible options and found the answer we were seeking. In the second method, we started by listing the options, but we found a pattern that helped us ﬁnd the solution faster. The methods of “Making a Table” and “Looking for a Pattern” are both more powerful if used alongside other problem-solving methods. Solve Real-World Problems Using Selected Strategies as Part of a Plan Example 5 Anne is making a box without a lid. She starts with a 20 in. square piece of cardboard and cuts out four equal squares from each corner of the cardboard as shown. She then folds the sides of the box and glues the edges together. How big does she need to cut the corner squares in order to make the box with the biggest volume? Solution Step 1: Understand www.ck12.org 58 Anne makes a box out of a 20 in × 20 in piece of cardboard. She cuts out four equal squares from the corners of the cardboard. She folds the sides and glues them to make a box. How big should the cut out squares be to make the box with the biggest volume? Step 2: Strategy We need to remember the formula for the volume of a box. Volume = Area of base × height Volume = width × length × height Make a table of values by picking diﬀerent values for the side of the squares that we are cutting out and calculate the volume. Step 3: Apply strategy/solve Let’s “make” a box by cutting out four corner squares with sides equal to 1 inch. The diagram will look like this: You see that when we fold the sides over to make the box, the height becomes 1 inch, the width becomes 18 inches and the length becomes 18 inches. Volume = width × length × height Volume = 18 × 18 × 1 = 324 in3 Let’s make a table that shows the value of the box for diﬀerent square sizes: Table 1.16: Side of Square Box Height Box Width Box Length Volume 1 1 18 18 18 × 18 × 1 = 324 2 2 16 16 16 × 16 × 2 = 512 3 3 14 14 14 × 14 × 3 = 588 4 4 12 12 12 × 12 × 4 = 576 5 5 10 10 10 × 10 × 5 = 500 6 6 8 8 8 × 8 × 6 = 384 7 7 6 6 6 × 6 × 7 = 252 8 8 4 4 4 × 4 × 8 = 128 9 9 2 2 2 × 2 × 9 = 36 10 10 0 0 0 × 0 × 10 = 0 59 www.ck12.org Table 1.16: (continued) Side of Square Box Height Box Width Box Length Volume We stop at a square of 10 inches because at this point we have cut out all of the cardboard and we can’t make a box any more. From the table we see that we can make the biggest box if we cut out squares with a side length of three inches. This gives us a volume of 588 in3. Answer The box of greatest volume is made if we cut out squares with a side length of three inches. Step 4: Check We see that 588 in3 is the largest volume appearing in the table. We picked integer values for the sides of the squares that we are cut out. Is it possible to get a larger value for the volume if we pick non-integer values? Since we get the largest volume for the side length equal to three inches, let’s make another table with values close to three inches that is split into smaller increments: Table 1.17: Side of Square Box Height Box Width Box Length Volume 2.5 2.5 15 15 15 × 15 × 2.5 = 562.5 2.6 2.6 14.8 14.8 14.8 × 14.8 × 2.6 = 569.5 2.7 2.7 14.6 14.6 14.6 × 14.6 × 2.7 = 575.5 2.8 2.8 14.4 14.4 14.4 × 14.4 × 2.8 = 580.6 2.9 2.9 14.2 14.2 14.2 × 14.2 × 2.9 = 584.8 3 3 14 14 14 × 14 × 3 = 588 3.1 3.1 13.8 13.8 13.8 × 13.8 × 3.1 = 590.4 3.2 3.2 13.6 13.6 13.6 × 13.6 × 3.2 = 591.9 3.3 3.3 13.4 13.4 13.4 × 13.4 × 3.3 = 592.5 3.4 3.4 13.2 13.2 13.2 × 13.2 × 3.4 = 592.4 3.5 3.5 13 13 13 × 13 × 3.5 = 591.5 Notice that the largest volume is not when the side of the square is three inches, but rather when the side of the square is 3.3 inches. Our original answer was not incorrect, but it was not as accurate as it could be. We can get an even more accurate answer if we take even smaller increments of the side length of the square. To do that, we would choose smaller measurements that are in the neighborhood of 3.3 inches. Meanwhile, our ﬁrst answer checks out if we want it rounded to zero decimal places, but a more accurate www.ck12.org 60 answer is 3.3 inches. Review Questions 1. Go back and ﬁnd the solution to the problem in Example 1. 2. Britt has $2.25 in nickels and dimes. If she has 40 coins in total, how many of each coin does she have? 3. Jeremy divides a 160-square-foot garden into plots that are either 10 or 12 square feet each. If there are 14 plots in all, how many plots are there of each size? 4. A pattern of squares is put together as shown. How many squares are in the 12th diagram? 5. In Harrisville, local housing laws specify how many people can live in a house or apartment: the maximum number of people allowed is twice the number of bedrooms, plus one. If Jan, Pat, and their four children want to rent a house, how many bedrooms must it have? 6. A restaurant hosts children’s birthday parties for a cost of $120 for the ﬁrst six children (including the birthday child) and $30 for each additional child. If Jaden’s parents have a budget of $200 to spend on his birthday party, how many guests can Jaden invite? 7. A movie theater with 200 seats charges $8 general admission and $5 for students. If the 5:00 showing is sold out and the theater took in $1468 for that showing, how many of the seats are occupied by students? 8. Oswald is trying to cut down on drinking coﬀee. His goal is to cut down to 6 cups per week. If he starts with 24 cups the ﬁrst week, then cuts down to 21 cups the second week and 18 cups the third week, how many weeks will it take him to reach his goal? 9. Taylor checked out a book from the library and it is now 5 days late. The late fee is 10 cents per day. How much is the ﬁne? 10. Mikhail is ﬁlling a sack with oranges. (a) If each orange weighs 5 ounces and the sack will hold 2 pounds, how many oranges will the sack hold before it bursts? (b) Mikhail plans to use these oranges to make breakfast smoothies. If each smoothie requires 3 4 cup of orange juice, and each orange will yield half a cup, how many smoothies can he make? 11. Jessamyn takes out a $150 loan from an agency that charges 12% of the original loan amount in interest each week. If she takes ﬁve weeks to pay oﬀthe loan, what is the total amount (loan plus interest) she will need to pay back? 12. How many hours will a car traveling at 75 miles per hour take to catch up to a car traveling at 55 miles per hour if the slower car starts two hours before the faster car? 13. Grace starts biking at 12 miles per hour. One hour later, Dan starts biking at 15 miles per hour, following the same route. How long will it take him to catch up with Grace? 14. A new theme park opens in Milford. On opening day, the park has 120 visitors; on each of the next three days, the park has 10 more visitors than the day before; and on each of the three days after that, the park has 20 more visitors than the day before. (a) How many visitors does the park have on the seventh day? (b) How many total visitors does the park have all week? 15. Lemuel wants to enclose a rectangular plot of land with a fence. He has 24 feet of fencing. What is the largest possible area that he could enclose with the fence? 61 www.ck12.org 16. Quizzes in Keiko’s history class are worth 20 points each. Keiko scored 15 and 18 points on her last two quizzes. What score does she need on her third quiz to get an average score of 17 on all three? 17. Tickets to an event go on sale for $20 six weeks before the event, and go up in price by $5 each week. What is the price of tickets one week before the event? 18. Mark is three years older than Janet, and the sum of their ages is 15. How old are Mark and Janet? 19. In a one-on-one basketball game, Jane scored 11 2 times as many points as Russell. If the two of them together scored 10 points, how many points did Jane score? 20. Scientists are tracking two pods of whales during their migratory season. On the ﬁrst day of June, one pod is 120 miles north of a certain group of islands, and every day thereafter it gets 15 miles closer to the islands. The second pod starts out 160 miles east of the islands on June 3, and heads toward the islands at a rate of 20 miles a day. (a) Which pod will arrive at the islands ﬁrst, and on what day? (b) How long after that will it take the other pod to reach the islands? (c) Suppose the pod that reaches the islands ﬁrst immediately heads south from the islands at a rate of 15 miles a day, and the pod that gets there second also heads south from there at a rate of 25 miles a day. On what day will the second pod catch up with the ﬁrst? (d) How far will both pods be from the islands on that day? Texas Instruments Resources In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http: //www.ck12.org/flexr/chapter/9611. www.ck12.org 62 Chapter 2 Real Numbers 2.1 Integers and Rational Numbers Learning Objectives • Graph and compare integers. • Classify and order rational numbers. • Find opposites of numbers. • Find absolute values. • Compare fractions to determine which is bigger. Introduction One day, Jason leaves his house and starts walking to school. After three blocks, he stops to tie his shoe and leaves his lunch bag sitting on the curb. Two blocks farther on, he realizes his lunch is missing and goes back to get it. After picking up his lunch, he walks six more blocks to arrive at school. How far is the school from Jason’s house? And how far did Jason actually walk to get there? Graph and Compare Integers Integers include the counting numbers (1, 2, 3, ...), the negative counting numbers (-1, -2, -3, ...), and zero. (You may have learned to think of integers as whole numbers, but technically the “whole numbers” don’t include the negative integers.) Example 1 Compare the numbers 2 and -5. When we plot numbers on a number line, the greatest number is farthest to the right, and the least is farthest to the left. 63 www.ck12.org In the diagram above, we can see that 2 is farther to the right on the number line than -5, so we say that 2 is greater than -5. We use the symbol “>” to mean “greater than”, so we can write 2 > 5. Classifying Rational Numbers When we divide an integer a by another integer b (as long as b is not zero) we get a rational number. It’s called this because it is the ratio of one number to another, and we can write it in fraction form as a b. (You may recall that the top number in a fraction is called the numerator and the bottom number is called the denominator.) You can think of a rational number as a fraction of a cake. If you cut the cake into b slices, your share is a of those slices. For example, when we see the rational number 1 2, we can imagine cutting the cake into two parts. Our share is one of those parts. Visually, the rational number 1 2 looks like this: With the rational number 3 4, we cut the cake into four parts and our share is three of those parts. Visually, the rational number 3 4 looks like this: The rational number 9 10 represents nine slices of a cake that has been cut into ten pieces. Visually, the rational number 9 10 looks like this: Proper fractions are rational numbers where the numerator is less than the denominator. A proper fraction represents a number less than one. Improper fractions are rational numbers where the numerator is greater than or equal to the denomi-nator. An improper fraction can be rewritten as a mixed number – an integer plus a proper fraction. For example, 9 4 can be written as 21 4. An improper fraction represents a number greater than or equal to one. Equivalent fractions are two fractions that represent the same amount. For example, look at a visual representation of the rational number 2 4, and one of the number 1 2. www.ck12.org 64 You can see that the shaded regions are the same size, so the two fractions are equivalent. We can convert one fraction into the other by reducing the fraction, or writing it in lowest terms. To do this, we write out the prime factors of both the numerator and the denominator and cancel matching factors that appear in both the numerator and denominator. 2 4 = 2 · 1 2 · 2 · 1 = 1 2 · 1 = 1 2 Reducing a fraction doesn’t change the value of the fraction—it just simpliﬁes the way we write it. Once we’ve canceled all common factors, the fraction is in its simplest form. Example 2 Classify and simplify the following rational numbers a) 3 7 b) 9 3 c) 50 60 Solution a) 3 and 7 are both prime, so we can’t factor them. That means 3 7 is already in its simplest form. It is also a proper fraction. b) 9 3 is an improper fraction because 9 > 3. To simplify it, we factor the numerator and denominator and cancel: 3·3 3·1 = 3 1 = 3. c) 50 60 is a proper fraction, and we can simplify it as follows: 50 60 = 5·5·2 5·3·2·2 = 5 3·2 = 5 6. Order Rational Numbers Ordering rational numbers is simply a matter of arranging them by increasing value—least ﬁrst and greatest last. Example 3 Put the following fractions in order from least to greatest: 1 2, 3 4, 2 3 Solution 1 2 < 2 3 < 3 4 Simple fractions are easy to order—we just know, for example, that one-half is greater than one quarter, and that two thirds is bigger than one-half. But how do we compare more complex fractions? Example 4 Which is greater, 3 7 or 4 9? In order to determine this, we need to rewrite the fractions so we can compare them more easily. If we rewrite them as equivalent fractions that have the same denominators, then we can compare them directly. To do this, we need to ﬁnd the lowest common denominator (LCD), or the least common multiple of the two denominators. 65 www.ck12.org The lowest common multiple of 7 and 9 is 63. Our fraction will be represented by a shape divided into 63 sections. This time we will use a rectangle cut into 9 by 7 = 63 pieces. 7 divides into 63 nine times, so 3 7 = 9·3 9·7 = 27 63. We can multiply the numerator and the denominator both by 9 because that’s really just the opposite of reducing the fraction—to get back from 27 63 to 3 7, we’d just cancel out the 9’s. Or, to put that in more formal terms: The fractions a b and c·a c·b are equivalent as long as c , 0. Therefore, 27 63 is an equivalent fraction to 3 7. Here it is shown visually: 9 divides into 63 seven times, so 4 9 = 7·4 7·9 = 28 63. 28 63 is an equivalent fraction to 4 9. Here it is shown visually: By writing the fractions with a common denominator of 63, we can easily compare them. If we take the 28 shaded boxes out of 63 (from our image of 4 9 above) and arrange them in rows instead of columns, we can see that they take up more space than the 27 boxes from our image of 3 7: Solution Since 28 63 is greater than 27 63, 4 9 is greater than 3 7. Graph and Order Rational Numbers To plot non-integer rational numbers (fractions) on the number line, we can convert them to mixed numbers (graphing is one of the few occasions in algebra when it’s better to use mixed numbers than improper fractions), or we can convert them to decimal form. Example 5 Plot the following rational numbers on the number line. a) 2 3 b) −3 7 www.ck12.org 66 c) 17 5 If we divide up the number line into sub-intervals based on the denominator of the fraction, we can look at the fraction’s numerator to determine how many of these sub-intervals we need to include. a) 2 3 falls between 0 and 1. Because the denominator is 3, we divide the interval between 0 and 1 into three smaller units. Because the numerator is 2, we count two units over from 0. b) −3 7 falls between 0 and -1. We divide the interval into seven units, and move left from zero by three of those units. c) 17 5 as a mixed number is 3 2 5 and falls between 3 and 4. We divide the interval into ﬁve units, and move over two units. Another way to graph this fraction would be as a decimal. 32 5 is equal to 3.4, so instead of dividing the interval between 3 and 4 into 5 units, we could divide it into 10 units (each representing a distance of 0.1) and then count over 4 units. We would end up at the same place on the number line either way. To make graphing rational numbers easier, try using the number line generator at com/numline.html. You can use it to create a number line divided into whatever units you want, as long as you express the units in decimal form. Find the Opposites of Numbers Every number has an opposite. On the number line, a number and its opposite are, predictably, opposite each other. In other words, they are the same distance from zero, but on opposite sides of the number line. The opposite of zero is deﬁned to be simply zero. The sum of a number and its opposite is always zero—for example, 3 + −3 = 0, 4.2 + −4.2 = 0, and so on. This is because adding 3 and -3 is like moving 3 steps to the right along the number line, and then 3 steps back to the left. The number and its opposite cancel each other out, leaving zero. Another way to think of the opposite of a number is that it is simply the original number multiplied by 67 www.ck12.org -1. The opposite of 4 is 4 × −1 or -4, the opposite of -2.3 is −2.3 × −1 or just 2.3, and so on. Another term for the opposite of a number is the additive inverse. Example 6 Find the opposite of each of the following: d) 19.6 e) −4 9 f) x g) xy2 h) (x −3) Solution Since we know that opposite numbers are on opposite sides of zero, we can simply multiply each expression by -1. This changes the sign of the number to its opposite—if it’s negative, it becomes positive, and vice versa. a) The opposite of 19.6 is -19.6. b) The opposite of is −4 9 is 4 9. c) The opposite of x is −x. d) The opposite of xy2 is −xy2. e) The opposite of (x −3) is −(x −3), or (3 −x). Note: With the last example you must multiply the entire expression by -1. A common mistake in this example is to assume that the opposite of (x −3) is (x + 3). Avoid this mistake! Find Absolute Values When we talk about absolute value, we are talking about distances on the number line. For example, the number 7 is 7 units away from zero—and so is the number -7. The absolute value of a number is the distance it is from zero, so the absolute value of 7 and the absolute value of -7 are both 7. We write the absolute value of -7 as \| −7\|. We read the expression \|x\| as “the absolute value of x.” • Treat absolute value expressions like parentheses. If there is an operation inside the absolute value symbols, evaluate that operation ﬁrst. • The absolute value of a number or an expression is always positive or zero. It cannot be negative. With absolute value, we are only interested in how far a number is from zero, and not in which direction. Example 7 Evaluate the following absolute value expressions. a) \|5 + 4\| b) 3 −\|4 −9\| c) \| −5 −11\| d) −\|7 −22\| (Remember to treat any expressions inside the absolute value sign as if they were inside parentheses, and evaluate them ﬁrst.) www.ck12.org 68 Solution a) \|5 + 4\| = \|9\| = 9 b) 3 −\|4 −9\| = 3 −\| −5\| = 3 −5 = −2 c) \| −5 −11\| = \| −16\| = 16 d) −\|7 −22\| = −\| −15\| = −(15) = −15 Lesson Summary • Integers (or whole numbers) are the counting numbers (1, 2, 3, ...), the negative counting numbers (-1, -2, -3, ...), and zero. • A rational number is the ratio of one integer to another, like 3 5 or a b. The top number is called the numerator and the bottom number (which can’t be zero) is called the denominator. • Proper fractions are rational numbers where the numerator is less than the denominator. • Improper fractions are rational numbers where the numerator is greater than the denominator. • Equivalent fractions are two fractions that equal the same numerical value. The fractions a b and c·a c·b are equivalent as long as c , 0. • To reduce a fraction (write it in simplest form), write out all prime factors of the numerator and denominator, cancel common factors, then recombine. • To compare two fractions it helps to write them with a common denominator. • The absolute value of a number is the distance it is from zero on the number line. The absolute value of any expression will always be positive or zero. • Two numbers are opposites if they are the same distance from zero on the number line and on opposite sides of zero. The opposite of an expression can be found by multiplying the entire expression by -1. Review Questions 1. Solve the problem posed in the Introduction. 2. The tick-marks on the number line represent evenly spaced integers. Find the values of a, b, c, d and e. 3. Determine what fraction of the whole each shaded region represents. (a) (b) 69 www.ck12.org (c) 4. Place the following sets of rational numbers in order, from least to greatest. (a) 1 2, 1 3, 1 4 (b) 1 10, 1 2, 2 5, 1 4, 7 20 (c) 39 60, 49 80, 59 100 (d) 7 11, 8 13, 12 19 (e) 9 5, 22 15, 4 3 5. Find the simplest form of the following rational numbers. (a) 22 44 (b) 9 27 (c) 12 18 (d) 315 420 (e) 244 168 6. Find the opposite of each of the following. (a) 1.001 (b) (5 −11) (c) (x + y) (d) (x −y) (e) (x + y −4) (f) (−x + 2y) 7. Simplify the following absolute value expressions. (a) 11 −\| −4\| (b) \|4 −9\| −\| −5\| (c) \| −5 −11\| (d) 7 −\|22 −15 −19\| (e) −\| −7\| (f) \| −2 −88\| −\|88 + 2\| 2.2 Adding and Subtracting Rational Numbers Learning Objectives • Add and subtract using a number line. • Add and subtract rational numbers. • Identify and apply properties of addition and subtraction. • Solve real-world problems using addition and subtraction of fractions. • Evaluate change using a variable expression. Introduction Ilana buys two identically sized cakes for a party. She cuts the chocolate cake into 24 pieces and the vanilla cake into 20 pieces, and lets the guests serve themselves. Martin takes three pieces of chocolate cake and www.ck12.org 70 one of vanilla, and Sheena takes one piece of chocolate and two of vanilla. Which of them gets more cake? Add and Subtract Using a Number Line In Lesson 1, we learned how to represent numbers on a number line. To add numbers on a number line, we start at the position of the ﬁrst number, and then move to the right by a number of units equal to the second number. Example 1 Represent the sum −2 + 3 on a number line. We start at the number -2, and then move 3 units to the right. We thus end at +1. Solution −2 + 3 = 1 Example 2 Represent the sum 2 - 3 on a number line. Subtracting a number is basically just adding a negative number. Instead of moving to the right, we move to the left. Starting at the number 2, and then moving 3 to the left, means we end at -1. Solution 2 −3 = −1 Adding and Subtracting Rational Numbers When we add or subtract two fractions, the denominators must match before we can ﬁnd the sum or diﬀerence. We have already seen how to ﬁnd a common denominator for two rational numbers. Example 3 Simplify 3 5 + 1 6. To combine these fractions, we need to rewrite them over a common denominator. We are looking for the lowest common denominator (LCD). We need to identify the lowest common multiple or least common multiple (LCM) of 5 and 6. That is the smallest number that both 5 and 6 divide into evenly (that is, without a remainder). The lowest number that 5 and 6 both divide into evenly is 30. The LCM of 5 and 6 is 30, so the lowest common denominator for our fractions is also 30. We need to rewrite our fractions as new equivalent fractions so that the denominator in each case is 30. If you think back to our idea of a cake cut into a number of slices, 3 5 means 3 slices of a cake that has been 71 www.ck12.org cut into 5 pieces. You can see that if we cut the same cake into 30 pieces (6 times as many) we would need 6 times as many slices to make up an equivalent fraction of the cake—in other words, 18 slices instead of 3. 3 5 is equivalent to 18 30. By a similar argument, we can rewrite the fraction 1 6 as a share of a cake that has been cut into 30 pieces. If we cut it into 5 times as many pieces, we need 5 times as many slices. 1 6 is equivalent to 5 30. Now that both fractions have the same denominator, we can add them. If we add 18 pieces of cake to 5 pieces, we get a total of 23 pieces. 23 pieces of a cake that has been cut into 30 pieces means that our answer is 23 30. 3 5 + 1 6 = 18 30 + 5 30 = 23 30 Notice that when we have fractions with a common denominator, we add the numerators but we leave the denominators alone. Here is this information in algebraic terms. When adding fractions: a c + b c = a+b c Example 4 Simplify 1 3 −1 9. The lowest common multiple of 9 and 3 is 9, so 9 is our common denominator. That means we don’t have to alter the second fraction at all. 3 divides into 9 three times, so 1 3 = 3·1 3·3 = 3 9. Our sum becomes 3 9 −1 9. We can subtract fractions with a common denominator by subtracting their numerators, just like adding. In other words: When subtracting fractions: a c −b c = a−b c Solution 1 3 −1 9 = 2 9 So far, we’ve only dealt with examples where it’s easy to ﬁnd the least common multiple of the denominators. With larger numbers, it isn’t so easy to be sure that we have the LCD. We need a more systematic method. In the next example, we will use the method of prime factors to ﬁnd the least common denominator. Example 5 www.ck12.org 72 Simplify 29 90 −13 126. To ﬁnd the lowest common multiple of 90 and 126, we ﬁrst ﬁnd the prime factors of 90 and 126. We do this by continually dividing the number by factors until we can’t divide any further. You may have seen a factor tree before. (For practice creating factor trees, try the Factor Tree game at mathgoodies.com/factors/factor_tree.asp.) The factor tree for 90 looks like this: The factor tree for 126 looks like this: The LCM for 90 and 126 is made from the smallest possible collection of primes that enables us to construct either of the two numbers. We take only enough instances of each prime to make the number with the greater number of instances of that prime in its factor tree. Table 2.1: Prime Factors in 90 Factors in 126 We Need 2 1 1 1 3 2 2 2 5 1 0 1 7 0 1 1 So we need one 2, two 3’s, one 5 and one 7. That gives us 2·3·3·5·7 = 630 as the lowest common multiple of 90 and 126. So 630 is the LCD for our calculation. 90 divides into 630 seven times (notice that 7 is the only factor in 630 that is missing from 90), so 29 90 = 7·29 7·90 = 203 630. 126 divides into 630 ﬁve times (notice that 5 is the only factor in 630 that is missing from 126), so 13 126 = 5·13 5·126 = 65 630. Now we complete the problem: 29 90 −13 126 = 203 630 −65 630 = 138 630. This fraction simpliﬁes. To be sure of ﬁnding the simplest form for 138 630, we write out the prime factors of the numerator and denominator. We already know the prime factors of 630. The prime factors of 138 73 www.ck12.org are 2, 3 and 23. 138 630 = 2·3·23 2·3·3·5·7; one factor of 2 and one factor of 3 cancels out, leaving 23 3·5·7 or 23 105 as our answer. Identify and Apply Properties of Addition Three mathematical properties which involve addition are the commutative, associative, and the ad-ditive identity properties. Commutative property: When two numbers are added, the sum is the same even if the order of the items being added changes. Example: 3 + 2 = 2 + 3 Associative Property: When three or more numbers are added, the sum is the same regardless of how they are grouped. Example: (2 + 3) + 4 = 2 + (3 + 4) Additive Identity Property: The sum of any number and zero is the original number. Example: 5 + 0 = 5 Solve Real-World Problems Using Addition and Subtraction Example 6 Peter is hoping to travel on a school trip to Europe. The ticket costs $2400. Peter has several relatives who have pledged to help him with the ticket cost. His parents have told him that they will cover half the cost. His grandma Zenoviea will pay one sixth, and his grandparents in Florida will send him one fourth of the cost. What fraction of the cost can Peter count on his relatives to provide? The ﬁrst thing we need to do is extract the relevant information. Peter’s parents will provide 1 2 the cost; his grandma Zenoviea will provide 1 6; and his grandparents in Florida 1 4. We need to ﬁnd the sum of those numbers, or 1 2 + 1 6 + 1 4. To determine the sum, we ﬁrst need to ﬁnd the LCD. The LCM of 2, 6 and 4 is 12, so that’s our LCD. Now we can ﬁnd equivalent fractions: 1 2 = 6 · 1 6 · 2 = 6 12 1 6 = 2 · 1 2 · 6 = 2 12 1 4 = 3 · 1 3 · 4 = 3 12 Putting them all together: 6 12 + 2 12 + 3 12 = 11 12. Peter will get 11 12 the cost of the trip, or $2200 out of $2400, from his family. Example 7 A property management ﬁrm is buying parcels of land in order to build a small community of condominiums. It has just bought three adjacent plots of land. The ﬁrst is four-ﬁfths of an acre, the second is ﬁve-twelfths of an acre, and the third is nineteen-twentieths of an acre. The ﬁrm knows that it must allow one-sixth of an acre for utilities and a small access road. How much of the remaining land is available for development? The ﬁrst thing we need to do is extract the relevant information. The plots of land measure 4 5, 5 12, and 19 20 acres, and the ﬁrm can use all of that land except for 1 6 of an acre. The total amount of land the ﬁrm can www.ck12.org 74 use is therefore 4 5 + 5 12 + 19 20 −1 6 acres. We can add and subtract multiple fractions at once just by ﬁnding a common denominator for all of them. The factors of 5, 9, 20, and 6 are as follows: 5 5 12 2 · 2 · 3 20 2 · 2 · 5 6 2 · 3 We need a 5, two 2’s, and a 3 in our LCD. 2 · 2 · 3 · 5 = 60, so that’s our common denominator. Now to convert the fractions: 4 5 = 12 · 4 12 · 5 = 48 60 5 12 = 5 · 5 5 · 12 = 25 60 19 20 = 3 · 19 3 · 20 = 57 60 1 6 = 10 · 1 10 · 6 = 10 60 We can rewrite our sum as 48 60 + 25 60 + 57 60 −10 60 = 48+25+57−10 60 = 120 60 . Next, we need to reduce this fraction. We can see immediately that the numerator is twice the denominator, so this fraction reduces to 2 1 or simply 2. One is sometimes called the invisible denominator, because every whole number can be thought of as a rational number whose denominator is one. Solution The property ﬁrm has two acres available for development. Evaluate Change Using a Variable Expression When we write algebraic expressions to represent a real quantity, the diﬀerence between two values is the change in that quantity. Example 8 The intensity of light hitting a detector when it is held a certain distance from a bulb is given by this equation: Intensity = 3 d2 where d is the distance measured in meters, and intensity is measured in lumens. Calculate the change in intensity when the detector is moved from two meters to three meters away. We ﬁrst ﬁnd the values of the intensity at distances of two and three meters. 75 www.ck12.org Intensity (2) = 3 (2)2 = 3 4 Intensity (3) = 3 (3)2 = 3 9 = 1 3 The diﬀerence in the two values will give the change in the intensity. We move from two meters to three meters away. Change = Intensity (3) −Intensity (2) = 1 3 −3 4 To ﬁnd the answer, we will need to write these fractions over a common denominator. The LCM of 3 and 4 is 12, so we need to rewrite each fraction with a denominator of 12: 1 3 = 4 · 1 4 · 3 = 4 12 3 4 = 3 · 3 3 · 4 = 9 12 So we can rewrite our equation as 4 12 −9 12 = −5 12. The negative value means that the intensity decreases as we move from 2 to 3 meters away. Solution When moving the detector from two meters to three meters, the intensity falls by 5 12 lumens. Lesson Summary • Subtracting a number is the same as adding the opposite (or additive inverse) of the number. • To add fractions, rewrite them over the lowest common denominator (LCD). The lowest common denominator is the lowest (or least) common multiple (LCM) of the two denominators. • When adding fractions: a c + b c = a+b c • When subtracting fractions: a c −b c = a−b c • Commutative property: the sum of two numbers is the same even if the order of the items to be added changes. • Associative Property: When three or more numbers are added, the sum is the same regardless of how they are grouped. • Additive Identity Property: The sum of any number and zero is the original number. • The number one is sometimes called the invisible denominator, as every whole number can be thought of as a rational number whose denominator is one. • The diﬀerence between two values is the change in that quantity. Further Practice For more practice adding and subtracting fractions, try playing the math games at com/fractions_add.html and or the one at Review Questions 1. Write the sum that the following moves on a number line represent. www.ck12.org 76 (a) (b) 2. Add the following rational numbers. Write each answer in its simplest form. (a) 3 7 + 2 7 (b) 3 10 + 1 5 (c) 5 16 + 5 12 (d) 3 8 + 9 16 (e) 8 25 + 7 10 (f) 1 6 + 1 4 (g) 7 15 + 2 9 (h) 5 19 + 2 27 3. Which property of addition does each situation involve? (a) Whichever order your groceries are scanned at the store, the total will be the same. (b) However many shovel-loads it takes to move 1 ton of gravel, the number of rocks moved is the same. (c) If Julia has no money, then Mark and Julia together have just as much money as Mark by himself has. 4. Solve the problem posed in the Introduction to this lesson. 5. Nadia, Peter and Ian are pooling their money to buy a gallon of ice cream. Nadia is the oldest and gets the greatest allowance. She contributes half of the cost. Ian is next oldest and contributes one third of the cost. Peter, the youngest, gets the smallest allowance and contributes one fourth of the cost. They ﬁgure that this will be enough money. When they get to the check-out, they realize that they forgot about sales tax and worry there will not be enough money. Amazingly, they have exactly the right amount of money. What fraction of the cost of the ice cream was added as tax? 6. Subtract the following rational numbers. Be sure that your answer is in the simplest form. (a) 5 12 −9 18 (b) 2 3 −1 4 (c) 3 4 −1 3 (d) 15 11 −9 7 (e) 2 13 −1 11 (f) 7 27 −9 39 (g) 6 11 −3 22 (h) 13 64 −7 40 (i) 11 70 −11 30 7. Consider the equation y = 3x + 2. Determine the change in y between x = 3 and x = 7. 8. Consider the equation y = 2 3 x + 1 2. Determine the change in y between x = 1 and x = 2. 9. The time taken to commute from San Diego to Los Angeles is given by the equation time = 120 speed where time is measured in hours and speed is measured in miles per hour (mph). Calculate the change 77 www.ck12.org in time that a rush hour commuter would see when switching from traveling by bus to traveling by train, if the bus averages 40 mph and the train averages 90 mph. 2.3 Multiplying and Dividing Rational Numbers Learning Objectives • Multiply by negative one. • Multiply rational numbers. • Identify and apply properties of multiplication. • Solve real-world problems using multiplication. • Find multiplicative inverses. • Divide rational numbers. • Solve real-world problems using division. Multiplying Numbers by Negative One Whenever we multiply a number by negative one, the sign of the number changes. In more mathematical terms, multiplying by negative one maps a number onto its opposite. The number line below shows two examples: 3 · −1 = 3 and −1 · −1 = 1. When we multiply a number by negative one, the absolute value of the new number is the same as the absolute value of the old number, since both numbers are the same distance from zero. The product of a number “x” and negative one is −x. This does not mean that −x is necessarily less than zero! If x itself is negative, then −x will be positive because a negative times a negative (negative one) is a positive. When you multiply an expression by negative one, remember to multiply the entire expression by negative one. Example 1 Multiply the following by negative one. a) 79.5 b) π c) (x + 1) d) \|x\| Solution a) -79.5 b) −π c) −(x + 1) or −x −1 d) −\|x\| www.ck12.org 78 Note that in the last case the negative sign outside the absolute value symbol applies after the absolute value. Multiplying the argument of an absolute value equation (the term inside the absolute value symbol) does not change the absolute value. \|x\| is always positive. \| −x\| is always positive. −\|x\| is always negative. Whenever you are working with expressions, you can check your answers by substituting in numbers for the variables. For example, you could check part d of Example 1 by letting x = −3. Then you’d see that \| −3\| , −\|3\|, because \| −3\| = 3 and −\|3\| = −3. Careful, though—plugging in numbers can tell you if your answer is wrong, but it won’t always tell you for sure if your answer is right! Multiply Rational Numbers Example 2 Simplify 1 3 · 2 5. One way to solve this is to think of money. For example, we know that one third of sixty dollars is written as 1 3 ·$60. We can read the above problem as one-third of two-ﬁfths. Here is a visual picture of the fractions one-third and two-ﬁfths. If we divide our rectangle into thirds one way and ﬁfths the other way, here’s what we get: Here is the intersection of the two shaded regions. The whole has been divided into ﬁve pieces width-wise and three pieces height-wise. We get two pieces out of a total of ﬁfteen pieces. Solution 1 3 · 2 5 = 2 15 Notice that 1 · 2 = 2 and 3 · 5 = 15. This turns out to be true in general: when you multiply rational numbers, the numerators multiply together and the denominators multiply together. Or, to put it more formally: When multiplying fractions: a b · c d = ac bd This rule doesn’t just hold for the product of two fractions, but for any number of fractions. Example 4 Multiply the following rational numbers: a) 2 5 · 5 9 b) 1 3 · 2 7 · 2 5 c) 1 2 · 2 3 · 3 4 · 4 5 79 www.ck12.org Solution a) With this problem, we can cancel the ﬁves: 2 5 · 5 9 = 2·5 5·9 = 2 9. b) With this problem, we multiply all the numerators and all the denominators: 1 3 · 2 7 · 2 5 = 1 · 2 · 2 3 · 7 · 5 = 4 105 c) With this problem, we multiply all the numerators and all the denominators, and then we can cancel most of them. The 2’s, 3’s, and 4’s all cancel out, leaving 1 5. With multiplication of fractions, we can simplify before or after we multiply. The next example uses factors to help simplify before we multiply. Example 5 Evaluate and simplify 12 25 · 35 42. Solution We can see that 12 and 42 are both multiples of six, 25 and 35 are both multiples of ﬁve, and 35 and 42 are both multiples of 7. That means we can write the whole product as 6·2 5·5 · 5·7 6·7 = 6·2·5·7 5·5·6·7. Then we can cancel out the 5, the 6, and the 7, leaving 2 5. Identify and Apply Properties of Multiplication The four mathematical properties which involve multiplication are the Commutative, Associative, Multiplicative Identity and Distributive Properties. Commutative property: When two numbers are multiplied together, the product is the same regardless of the order in which they are written. Example: 4 · 2 = 2 · 4 We can see a geometrical interpretation of The Commutative Property of Multiplication to the right. The Area of the shape (length × width) is the same no matter which way we draw it. Associative Property: When three or more numbers are multiplied, the product is the same regardless of their grouping. Example: 2 · (3 · 4) = (2 · 3) · 4 Multiplicative Identity Property: The product of one and any number is that number. Example: 5 · 1 = 5 Distributive property: The multiplication of a number and the sum of two numbers is equal to the ﬁrst number times the second number plus the ﬁrst number times the third number. Example: 4(6 + 3) = 4 · 6 + 4 · 3 Example 6 A gardener is planting vegetables for the coming growing season. He wishes to plant potatoes and has a www.ck12.org 80 choice of a single 8 × 7 meter plot, or two smaller plots of 3 × 7 and 5 × 7 meters. Which option gives him the largest area for his potatoes? Solution In the ﬁrst option, the gardener has a total area of (8 × 7) or 56 square meters. In the second option, the gardener has (3 × 7) or 21 square meters, plus (5 × 7) or 35 square meters. 21 + 35 = 56, so the area is the same as in the ﬁrst option. Solve Real-World Problems Using Multiplication Example 7 In the chemistry lab there is a bottle with two liters of a 15% solution of hydrogen peroxide (H2O2). John removes one-ﬁfth of what is in the bottle, and puts it in a beaker. He measures the amount of H2O2 and adds twice that amount of water to the beaker. Calculate the following measurements. a) The amount of H2O2 left in the bottle. b) The amount of diluted H2O2 in the beaker. c) The concentration of the H2O2 in the beaker. Solution a) To determine the amount of H2O2 left in the bottle, we ﬁrst determine the amount that was removed. That amount was 1 5 of the amount in the bottle (2 liters). 1 5 of 2 is 2 5. The amount remaining is 2 −2 5, or 10 5 −2 5 = 8 5 liter (or 1.6 liters). There are 1.6 liters left in the bottle. b) We determined that the amount of the 15% H2O2 solution removed was 2 5 liter. The amount of water added was twice this amount, or 4 5 liter. So the total amount of solution in the beaker is now 2 5 + 4 5 = 6 5 liter, or 1.2 liters. There are 1.2 liters of diluted H2O2 in the beaker. c) The new concentration of H2O2 can be calculated. John started with 2 5 liter of 15% H2O2 solution, so the amount of pure H2O2 is 15% of 2 5 liters, or 0.15 × 0.40 = 0.06 liters. After he adds the water, there is 1.2 liters of solution in the beaker, so the concentration of H2O2 is 0.06 1.2 = 1 20 or 0.05. To convert to a percent we multiply this number by 100, so the beaker’s contents are 5% H2O2. Example 8 Anne has a bar of chocolate and she oﬀers Bill a piece. Bill quickly breaks oﬀ1 4 of the bar and eats it. Another friend, Cindy, takes 1 3 of what was left. Anne splits the remaining candy bar into two equal pieces which she shares with a third friend, Dora. How much of the candy bar does each person get? First, let’s look at this problem visually. 81 www.ck12.org Anne starts with a full candy bar. Bill breaks oﬀ1 4 of the bar. Cindy takes 1 3 of what was left. Dora gets half of the remaining candy bar. We can see that the candy bar ends up being split four ways, with each person getting an equal amount. Solution Each person gets exactly 1 4 of the candy bar. We can also examine this problem using rational numbers. We keep a running total of what fraction of the bar remains. Remember, when we read a fraction followed by of in the problem, it means we multiply by that fraction. We start with 1 bar. Then Bill takes 1 4 of it, so there is 1 −1 4 = 3 4 of a bar left. Cindy takes 1 3 of what’s left, or 1 3 · 3 4 = 1 4 of a whole bar. That leaves 3 4 −1 4 = 2 4, or 1 2 of a bar. That half bar gets split between Anne and Dora, so they each get half of a half bar: 1 2 · 1 2 = 1 4. So each person gets exactly 1 4 of the candy bar. Extension: If each person’s share is 3 oz, how much did the original candy bar weigh? Identity Elements An identity element is a number which, when combined with a mathematical operation on a number, leaves that number unchanged. For example, the identity element for addition and subtraction is zero, because adding or subtracting zero to a number doesn’t change the number. And zero is also what you get when you add together a number and its opposite, like 3 and -3. The inverse operation of addition is subtraction—when you add a number and then subtract that same number, you end up back where you started. Also, adding a number’s opposite is the same as subtracting it—for example, 4 + (−3) is the same as 4 −3. Multiplication and division are also inverse operations to each other—when you multiply by a number and then divide by the same number, you end up back where you started. Multiplication and division also have an identity element: when you multiply or divide a number by one, the number doesn’t change. www.ck12.org 82 Just as the opposite of a number is the number you can add to it to get zero, the reciprocal of a number is the number you can multiply it by to get one. And ﬁnally, just as adding a number’s opposite is the same as subtracting the number, multiplying by a number’s reciprocal is the same as dividing by the number. Find Multiplicative Inverses The reciprocal of a number x is also called the multiplicative inverse. Any number times its own multiplicative inverse equals one, and the multiplicative inverse of x is written as 1 x. To ﬁnd the multiplicative inverse of a rational number, we simply invert the fraction—that is, ﬂip it over. In other words: The multiplicative inverse of a b is b a, as long as a , 0. You’ll see why in the following exercise. Example 9 Find the multiplicative inverse of each of the following. a) 3 7 b) 4 9 c) 3 1 2 d) −x y e) 1 11 Solution a) When we invert the fraction 3 7, we get 7 3. Notice that if we multiply 3 7 · 7 3, the 3’s and the 7’s both cancel out and we end up with 1 1, or just 1. b) Similarly, the inverse of 4 9 is 9 4; if we multiply those two fractions together, the 4’s and the 9’s cancel out and we’re left with 1. That’s why the rule “invert the fraction to ﬁnd the multiplicative inverse” works: the numerator and the denominator always end up canceling out, leaving 1. c) To ﬁnd the multiplicative inverse of 31 2 we ﬁrst need to convert it to an improper fraction. Three wholes is six halves, so 31 2 = 6 2 + 1 2 = 7 2. That means the inverse is 2 7. d) Don’t let the negative sign confuse you. The multiplicative inverse of a negative number is also negative! Just ignore the negative sign and ﬂip the fraction as usual. The multiplicative inverse of −x y is −y x. e) The multiplicative inverse of 1 11 is 11 1 , or simply 11. Look again at the last example. When we took the multiplicative inverse of 1 11 we got a whole number, 11. That’s because we can treat that whole number like a fraction with a denominator of 1. Any number, even a non-rational one, can be treated this way, so we can always ﬁnd a number’s multiplicative inverse using the same method. Divide Rational Numbers Earlier, we mentioned that multiplying by a number’s reciprocal is the same as dividing by the number. That’s how we can divide rational numbers; to divide by a rational number, just multiply by that number’s reciprocal. In more formal terms: 83 www.ck12.org a b ÷ c d = a b × d c . Example 10 Divide the following rational numbers, giving your answer in the simplest form. a) 1 2 ÷ 1 4 b) 7 3 ÷ 2 3 c) x 2 ÷ 1 4y d) 11 2x ÷ ( −x y ) Solution a) Replace 1 4 with 4 1 and multiply: 1 2 × 4 1 = 4 2 = 2. b) Replace 2 3 with 3 2 and multiply: 7 3 × 3 2 = 7·3 3·2 = 7 2. c) x 2 ÷ 1 4y = x 2 × 4y 1 = 4xy 2 = 2xy 1 = 2xy d) 11 2x ÷ ( −x y ) = 11 2x × ( −y x ) = −11y 2x2 Solve Real-World Problems Using Division Speed, Distance and Time An object moving at a certain speed will cover a ﬁxed distance in a set time. The quantities speed, distance and time are related through the equation Speed = Distance Time . Example 11 Andrew is driving down the freeway. He passes mile marker 27 at exactly mid-day. At 12:35 he passes mile marker 69. At what speed, in miles per hour, is Andrew traveling? Solution To ﬁnd the speed, we need the distance traveled and the time taken. If we want our speed to come out in miles per hour, we’ll need distance in miles and time in hours. The distance is 69 −27 or 42 miles. The time is 35 minutes, or 35 60 hours, which reduces to 7 12. Now we can plug in the values for distance and time into our equation for speed. Speed = 42 7 12 = 42 ÷ 7 12 = 42 1 × 12 7 = 6 · 7 · 12 1 · 7 = 6 · 12 1 = 72 Andrew is driving at 72 miles per hour. Example 12 Anne runs a mile and a half in a quarter hour. What is her speed in miles per hour? Solution We already have the distance and time in the correct units (miles and hours), so we just need to write them as fractions and plug them into the equation. Speed = 11 2 1 4 = 3 2 ÷ 1 4 = 3 2 × 4 1 = 3 · 4 2 · 1 = 12 2 = 6 www.ck12.org 84 Anne runs at 6 miles per hour. Example 13 – Newton’s Second Law Newton’s second law (F = ma) relates the force applied to a body in Newtons (F), the mass of the body in kilograms (m) and the acceleration in meters per second squared (a). Calculate the resulting acceleration if a Force of 71 3 Newtons is applied to a mass of 1 5kg. Solution First, we rearrange our equation to isolate the acceleration, a. If F = ma, dividing both sides by m gives us a = F m. Then we substitute in the known values for F and m: a = 71 3 1 5 = 22 3 ÷ 1 5 = 22 3 × 5 1 = 110 3 The resultant acceleration is 36 2 3 m/s2. Lesson Summary When multiplying an expression by negative one, remember to multiply the entire expression by negative one. To multiply fractions, multiply the numerators and multiply the denominators: a b · c d = ac bd The multiplicative properties are: • Commutative Property: The product of two numbers is the same whichever order the items to be multiplied are written. Example: 2 · 3 = 3 · 2 • Associative Property: When three or more numbers are multiplied, the sum is the same regardless of how they are grouped. Example: 2 · (3 · 4) = (2 · 3) · 4 • Multiplicative Identity Property: The product of any number and one is the original number. Example: 2 · 1 = 2 • Distributive property: The multiplication of a number and the sum of two numbers is equal to the ﬁrst number times the second number plus the ﬁrst number times the third number. Example: 4(2 + 3) = 4(2) + 4(3) The multiplicative inverse of a number is the number which produces 1 when multiplied by the original number. The multiplicative inverse of x is the reciprocal 1 x. To ﬁnd the multiplicative inverse of a fraction, simply invert the fraction: a b inverts to b a. To divide fractions, invert the divisor and multiply: a b ÷ c d = a b × d c. Further Practice For more practice multiplying fractions, try playing the fraction game at fra66mx2.htm, or the one at For more prac-tice dividing fractions, try the game at or the one at http: //www.mathplayground.com/fractions_div.html. Review Questions 1. Multiply the following expressions by negative one. 85 www.ck12.org (a) 25 (b) -105 (c) x2 (d) (3 + x) (e) (3 −x) 2. Multiply the following rational numbers. Write your answer in the simplest form. (a) 5 12 × 9 10 (b) 2 3 × 1 4 (c) 3 4 × 1 3 (d) 15 11 × 9 7 (e) 1 13 × 1 11 (f) 7 27 × 9 14 (g) ( 3 5 )2 (h) 1 11 × 22 21 × 7 10 (i) 12 15 × 35 13 × 10 2 × 26 36 3. Find the multiplicative inverse of each of the following. (a) 100 (b) 2 8 (c) −19 21 (d) 7 (e) −z3 2xy2 4. Divide the following rational numbers. Write your answer in the simplest form. (a) 5 2 ÷ 1 4 (b) 1 2 ÷ 7 9 (c) 5 11 ÷ 6 7 (d) 1 2 ÷ 1 2 (e) −x 2 ÷ 5 7 (f) 1 2 ÷ x 4y (g) ( −1 3 ) ÷ ( −3 5 ) (h) 7 2 ÷ 7 4 (i) 11 ÷ −x 4 5. The label on a can of paint says that it will cover 50 square feet per pint. If I buy a 1 8 pint sample, it will cover a square two feet long by three feet high. Is the coverage I get more, less or the same as that stated on the label? 6. The world’s largest trench digger, ”Bagger 288”, moves at 3 8 mph. How long will it take to dig a trench 2 3 mile long? 7. A 2 7 Newton force applied to a body of unknown mass produces an acceleration of 3 10 m/s2. Calculate the mass of the body. 2.4 The Distributive Property Learning Objectives • Apply the distributive property. • Identify parts of an expression. • Solve real-world problems using the distributive property. www.ck12.org 86 Introduction At the end of the school year, an elementary school teacher makes a little gift bag for each of his students. Each bag contains one class photograph, two party favors and ﬁve pieces of candy. The teacher will distribute the bags among his 28 students. How many of each item does the teacher need? Apply the Distributive Property When we have a problem like the one posed in the introduction, The Distributive Property can help us solve it. First, we can write an expression for the contents of each bag: Items = (photo + 2 favors + 5 candies), or simply I = (p + 2 f + 5c). For all 28 students, the teacher will need 28 times that number of items, so I = 28(p + 2 f + 5c). Next, the Distributive Property of Multiplication tells us that when we have a single term multiplied by a sum of several terms, we can rewrite it by multiplying the single term by each of the other terms separately. In other words, 28(p+2 f +5c) = 28(p)+28(2f)+28(5c), which simpliﬁes to 28p+56 f +140c. So the teacher needs 28 class photos, 56 party favors and 140 pieces of candy. You can see why the Distributive Property works by looking at a simple problem where we just have numbers inside the parentheses, and considering the Order of Operations. Example 1 Determine the value of 11(2 - 6) using both the Order of Operations and the Distributive Property. Solution Order of Operations tells us to evaluate the amount inside the parentheses ﬁrst: 11(2 −6) = 11(−4) = −44 Now let’s try it with the Distributive Property: 11(2 −6) = 11(2) −11(6) = 22 −66 = −44 Note: When applying the Distributive Property you MUST take note of any negative signs! Example 2 Use the Distributive Property to determine the following. a) 11(2x + 6) b) 7(3x −5) c) 2 7(3y2 −11) d) 2x 7 ( 3y2 −11 xy ) Solution a) 11(2x + 6) = 11(2x) + 11(6) = 22x + 66 b) Note the negative sign on the second term. 7(3x −5) = 21x −35 c) 2 7(3y2 −11) = 2 7(3y2) + 2 7(−11) = 6y2 7 −22 7 , or 6y2−22 7 87 www.ck12.org d) 2x 7 ( 3y2 −11 xy ) = 2x 7 (3y2) + 2x 7 ( −11 xy ) = 6xy2 7 −22x 7xy We can simplify this answer by canceling the x’s in the second fraction, so we end up with 6xy2 7 −22 7y. Identify Expressions That Involve the Distributive Property The Distributive Property can also appear in expressions that don’t include parentheses. In Lesson 1.2, we saw how the fraction bar also acts as a grouping symbol. Now we’ll see how to use the Distributive Property with fractions. Example 3 Simplify the following expressions. a) 2x+8 4 b) 9y−2 3 c) z+6 2 Solution Even though these expressions aren’t written in a form we usually associate with the Distributive Property, remember that we treat the numerator of a fraction as if it were in parentheses, and that means we can use the Distributive Property here too. a) 2x+8 4 can be re-written as 1 4(2x + 8). Then we can distribute the 1 4: 1 4(2x + 8) = 2x 4 + 8 4 = x 2 + 2 b) 9y−2 3 can be re-written as 1 3(9y −2), and then we can distribute the 1 3: 1 3(9y −2) = 9y 3 −2 3 = 3y −2 3 c) Rewrite z+6 2 as 1 2(z + 6), and distribute the 1 2: 1 2(z + 6) = z 2 + 6 2 = z 2 + 3 Solve Real-World Problems Using the Distributive Property The Distributive Property is one of the most common mathematical properties used in everyday life. Any time we have two or more groups of objects, the Distributive Property can help us solve for an unknown. Example 4 Each student on a ﬁeld trip into a forest is to be given an emergency survival kit. The kit is to contain a ﬂashlight, a ﬁrst aid kit, and emergency food rations. Flashlights cost $12 each, ﬁrst aid kits are $7 each and emergency food rations cost $2 per day. There is $500 available for the kits and 17 students to provide for. How many days worth of rations can be provided with each kit? The unknown quantity in this problem is the number of days’ rations. This will be x in our expression. Each kit will contain one $12 ﬂashlight, one $7 ﬁrst aid kit, and x times $2 worth of rations, for a total cost of (12 + 7 + 2x) dollars. With 17 kits, therefore, the total cost will be 17(12 + 7 + 2x) dollars. We can use the Distributive Property on this expression: www.ck12.org 88 17(12 + 7 + 2x) = 204 + 119 + 34x Since the total cost can be at most $500, we set the expression equal to 500 and solve for x. (You’ll learn in more detail how to solve equations like this in the next chapter.) 204 + 119 + 34x = 500 323 + 34x = 500 323 + 34x −323 = 500 −323 34x = 177 34x 34 = 177 34 x ≈5.206 Since this represents the number of days’ worth of rations that can be bought, we must round to the next lowest whole number. We wouldn’t have enough money to buy a sixth day of supplies. Solution Five days worth of emergency rations can be purchased for each survival kit. Lesson Summary • Distributive Property The product of a number and the sum of two numbers is equal to the ﬁrst number times the second number plus the ﬁrst number times the third number. • When applying the Distributive Property you MUST take note of any negative signs! Further Practice For more practice using the Distributive Property, try playing the Battleship game at com/ba/15357.html. Review Questions 1. Use the Distributive Property to simplify the following expressions. (a) (x + 4) −2(x + 5) (b) 1 2(4z + 6) (c) (4 + 5) −(5 + 2) (d) x(x + 7) (e) y(x + 7) (f) 13x(3y + z) (g) x ( 3 x + 5 ) (h) xy ( 1 x + 2 y ) 2. Use the Distributive Property to remove the parentheses from the following expressions. (a) 1 2(x −y) −4 (b) 0.6(0.2x + 0.7) (c) 6 + (x −5) + 7 89 www.ck12.org (d) 6 −(x −5) + 7 (e) 4(m + 7) −6(4 −m) (f) −5(y −11) + 2y (g) −(x −3y) + 1 2(z + 4) (h) a b ( 2 a + 3 b + b 5 ) 3. Use the Distributive Property to simplify the following fractions. (a) 8x+12 4 (b) 9x+12 3 (c) 11x+12 2 (d) 3y+2 6 (e) −6z−2 3 (f) 7−6p 3 (g) 3d−4 6d (h) 12g+8h 4gh 4. A bookcase has ﬁve shelves, and each shelf contains seven poetry books and eleven novels. How many of each type of book does the bookcase contain? 5. Amar is making giant holiday cookies for his friends at school. He makes each cookie with 6 oz of cookie dough and decorates them with macadamia nuts. If Amar has 5 lbs of cookie dough (1 lb = 16 oz) and 60 macadamia nuts, calculate the following. (a) How many (full) cookies he can make? (b) How many macadamia nuts he can put on each cookie, if each is to be identical? (c) If 4 cups of ﬂour and 1 cup of sugar went into each pound of cookie dough, how much of each did Amar use to make the 5 pounds of dough? 2.5 Square Roots and Real Numbers Learning Objectives • Find square roots. • Approximate square roots. • Identify irrational numbers. • Classify real numbers. • Graph and order real numbers. Find Square Roots The square root of a number is a number which, when multiplied by itself, gives the original number. In other words, if a = b2, we say that b is the square root of a. Note: Negative numbers and positive numbers both yield positive numbers when squared, so each positive number has both a positive and a negative square root. (For example, 3 and -3 can both be squared to yield 9.) The positive square root of a number is called the principal square root. The square root of a number x is written as √x or sometimes as 2 √x. The symbol √ is sometimes called a radical sign. Numbers with whole-number square roots are called perfect squares. The ﬁrst ﬁve perfect squares (1, 4, 9, 16, and 25) are shown below. www.ck12.org 90 You can determine whether a number is a perfect square by looking at its prime factors. If every number in the factor tree appears an even number of times, the number is a perfect square. To ﬁnd the square root of that number, simply take one of each pair of matching factors and multiply them together. Example 1 Find the principal square root of each of these perfect squares. a) 121 b) 225 c) 324 d) 576 Solution a) 121 = 11 × 11, so √ 121 = 11. b) 225 = (5 × 5) × (3 × 3), so √ 225 = 5 × 3 = 15. c) 324 = (2 × 2) × (3 × 3) × (3 × 3), so √ 324 = 2 × 3 × 3 = 18. d) 576 = (2 × 2) × (2 × 2) × (2 × 2) × (3 × 3), so √ 576 = 2 × 2 × 2 × 3 = 24. For more practice matching numbers with their square roots, try the Flash games at com/jg/65631.html. When the prime factors don’t pair up neatly, we “factor out” the ones that do pair up and leave the rest under a radical sign. We write the answer as a √ b, where a is the product of half the paired factors we pulled out and b is the product of the leftover factors. Example 2 Find the principal square root of the following numbers. a) 8 b) 48 c) 75 d) 216 Solution a) 8 = 2 × 2 × 2. This gives us one pair of 2’s and one leftover 2, so √ 8 = 2 √ 2. b) 48 = (2 × 2) × (2 × 2) × 3, so √ 48 = 2 × 2 × √ 3, or 4 √ 3. c) 75 = (5 × 5) × 3, so √ 75 = 5 √ 3. d) 216 = (2 × 2) × 2 × (3 × 3) × 3, so √ 216 = 2 × 3 × √ 2 × 3, or 6 √ 6. Note that in the last example we collected the paired factors ﬁrst, then we collected the unpaired ones under a single radical symbol. Here are the four rules that govern how we treat square roots. 91 www.ck12.org • √a × √ b = √ ab • A √a × B √ b = AB √ ab • √a √ b = √ a b • A √a B √ b = A B √ a b Example 3 Simplify the following square root problems a) √ 8 × √ 2 b) 3 √ 4 × 4 √ 3 c) √ 12 ÷ √ 3 d) 12 √ 10 ÷ 6 √ 5 Solution a) √ 8 × √ 2 = √ 16 = 4 b) 3 √ 4 × 4 √ 3 = 12 √ 12 = 12 √ (2 × 2) × 3 = 12 × 2 √ 3 = 24 √ 3 c) √ 12 ÷ √ 3 = √ 12 3 = √ 4 = 2 d) 12 √ 10 ÷ 6 √ 5 = 12 6 √ 10 5 = 2 √ 2 Approximate Square Roots Terms like √ 2, √ 3 and √ 7 (square roots of prime numbers) cannot be written as rational numbers. That is to say, they cannot be expressed as the ratio of two integers. We call them irrational numbers. In decimal form, they have an unending, seemingly random, string of numbers after the decimal point. To ﬁnd approximate values for square roots, we use the √ or √x button on a calculator. When the number we plug in is a perfect square, or the square of a rational number, we will get an exact answer. When the number is a non-perfect square, the answer will be irrational and will look like a random string of digits. Since the calculator can only show some of the inﬁnitely many digits that are actually in the answer, it is really showing us an approximate answer—not exactly the right answer, but as close as it can get. Example 4 Use a calculator to ﬁnd the following square roots. Round your answer to three decimal places. a) √ 99 b) √ 5 c) √ 0.5 d) √ 1.75 Solution a) ≈9.950 b) ≈2.236 c) ≈0.707 d) ≈1.323 www.ck12.org 92 You can also work out square roots by hand using a method similar to long division. (See the web page at for an explanation of this method.) Identify Irrational Numbers Not all square roots are irrational, but any square root that can’t be reduced to a form with no radical signs in it is irrational. For example, √ 49 is rational because it equals 7, but √ 50 can’t be reduced farther than 5 √ 2. That factor of √ 2 is irrational, making the whole expression irrational. Example 5 Identify which of the following are rational numbers and which are irrational numbers. a) 23.7 b) 2.8956 c) π d) √ 6 e) 3.27 Solution a) 23.7 can be written as 23 7 10, so it is rational. b) 2.8956 can be written as 2 8956 10000, so it is rational. c) π = 3.141592654 . . . We know from the deﬁnition of π that the decimals do not terminate or repeat, so π is irrational. d) √ 6 = √ 2 × √ 3. We can’t reduce it to a form without radicals in it, so it is irrational. e) 3.27 = 3.272727272727 . . . This decimal goes on forever, but it’s not random; it repeats in a predictable pattern. Repeating decimals are always rational; this one can actually be expressed as 36 11. You can see from this example that any number whose decimal representation has a ﬁnite number of digits is rational, since each decimal place can be expressed as a fraction. For example, 0.439 can be expressed as 4 10 + 3 100 + 9 1000, or just 439 1000. Also, any decimal that repeats is rational, and can be expressed as a fraction. For example, 0.2538 can be expressed as 25 100 + 38 9900, which is equivalent to 2513 9900. Classify Real Numbers We can now see how real numbers fall into one of several categories. If a real number can be expressed as a rational number, it falls into one of two categories. If the denominator 93 www.ck12.org of its simplest form is one, then it is an integer. If not, it is a fraction (this term also includes decimals, since they can be written as fractions.) If the number cannot be expressed as the ratio of two integers (i.e. as a fraction), it is irrational. Example 6 Classify the following real numbers. a) 0 b) -1 c) π 3 d) √ 2 3 e) √ 36 9 Solution a) Integer b) Integer c) Irrational (Although it’s written as a fraction, π is irrational, so any fraction with π in it is also irrational.) d) Irrational e) Rational (It simpliﬁes to 6 9, or 2 3.) Lesson Summary • The square root of a number is a number which gives the original number when multiplied by itself. In algebraic terms, for two numbers a and b, if a = b2, then b = √a. • A square root can have two possible values: a positive value called the principal square root, and a negative value (the opposite of the positive value). • A perfect square is a number whose square root is an integer. • Some mathematical properties of square roots are: – √a × √ b = √ ab – A √a × B √ b = AB √ ab – √a √ b = √ a b – A √a B √ b = A B √ a b • Square roots of numbers that are not perfect squares (or ratios of perfect squares) are irrational numbers. They cannot be written as rational numbers (the ratio of two integers). In decimal form, they have an unending, seemingly random, string of numbers after the decimal point. • Computing a square root on a calculator will produce an approximate solution since the calculator only shows a ﬁnite number of digits after the decimal point. Review Questions 1. Find the following square roots exactly without using a calculator, giving your answer in the simplest form. (a) √ 25 (b) √ 24 www.ck12.org 94 (c) √ 20 (d) √ 200 (e) √ 2000 (f) √ 1 4 (Hint: The division rules you learned can be applied backwards!) (g) √ 9 4 (h) √ 0.16 (i) √ 0.1 (j) √ 0.01 2. Use a calculator to ﬁnd the following square roots. Round to two decimal places. (a) √ 13 (b) √ 99 (c) √ 123 (d) √ 2 (e) √ 2000 (f) √ .25 (g) √ 1.35 (h) √ 0.37 (i) √ 0.7 (j) √ 0.01 3. Classify the following numbers as an integer, a rational number or an irrational number. (a) √ 0.25 (b) √ 1.35 (c) √ 20 (d) √ 25 (e) √ 100 4. Place the following numbers in numerical order, from lowest to highest. √ 6 2 61 50 √ 1.5 16 13 5. Use the marked points on the number line and identify each proper fraction. 2.6 Problem-Solving Strategies: Guess and Check, Work Backward Learning Objectives • Read and understand given problem situations. • Develop and use the strategy “Guess and Check.” • Develop and use the strategy “Work Backward.” • Plan and compare alternative approaches to solving problems. • Solve real-world problems using selected strategies as part of a plan. 95 www.ck12.org Introduction In this section, you will learn about the methods of Guess and Check and Working Backwards. These are very powerful strategies in problem solving and probably the most commonly used in everyday life. Let’s review our problem-solving plan. Step 1 Understand the problem. Read the problem carefully. Then list all the components and data involved, and assign your variables. Step 2 Devise a plan – Translate Come up with a way to solve the problem. Set up an equation, draw a diagram, make a chart or construct a table. Step 3 Carry out the plan – Solve This is where you solve the equation you came up with in Step 2. Step 4 Look – Check and Interpret Check that the answer makes sense. Let’s now look at some strategies we can use as part of this plan. Develop and Use the Strategy “Guess and Check” The strategy for the method “Guess and Check” is to guess a solution and then plug the guess back into the problem to see if you get the correct answer. If the answer is too big or too small, make another guess that will get you closer to the goal, and continue guessing until you arrive at the correct solution. The process might sound long, but often you will ﬁnd patterns that you can use to make better guesses along the way. Here is an example of how this strategy is used in practice. Example 1 Nadia takes a ribbon that is 48 inches long and cuts it in two pieces. One piece is three times as long as the other. How long is each piece? Solution Step 1: Understand We need to ﬁnd two numbers that add up to 48. One number is three times the other number. Step 2: Strategy We guess two random numbers, one three times bigger than the other, and ﬁnd the sum. If the sum is too small we guess larger numbers, and if the sum is too large we guess smaller numbers. Then, we see if any patterns develop from our guesses. Step 3: Apply Strategy/Solve www.ck12.org 96 Guess 5 and 15 5 + 15 = 20 sum is too small Guess 6 and 18 6 + 18 = 24 sum is too small Our second guess gives us a sum that is exactly half of 48. What if we double that guess? 12 + 36 = 48 There’s our answer. The pieces are 12 and 36 inches long. Step 4: Check 12 + 36 = 48 The pieces add up to 48 inches. 36 = 3(12) One piece is three times as long as the other. The answer checks out. Develop and Use the Strategy “Work Backward” The “Work Backward” method works well for problems where a series of operations is done on an unknown number and you’re only given the result. To use this method, start with the result and apply the operations in reverse order until you ﬁnd the starting number. Example 2 Anne has a certain amount of money in her bank account on Friday morning. During the day she writes a check for $24.50, makes an ATM withdrawal of $80 and deposits a check for $235. At the end of the day she sees that her balance is $451.25. How much money did she have in the bank at the beginning of the day? Step 1: Understand We need to ﬁnd the money in Anne’s bank account at the beginning of the day on Friday. She took out $24.50 and $80 and put in $235. She ended up with $451.25 at the end of the day. Step 2: Strategy We start with an unknown amount, do some operations, and end up with a known amount. We need to start with the result and apply the operations in reverse. Step 3: Apply Strategy/Solve Start with $451.25. Subtract $235, add $80, and then add $24.50. 451.25 −235 + 80 + 24.50 = 320.75 Anne had $320.75 in her account at the beginning of the day on Friday. Step 4: Check Anne starts with $320.75 She writes a check for $24.50. $320.75 −24.50 = $296.25 She withdraws $80. $296.25 −80 = $216.25 She deposits $235. $216.25 + 235 = $451.25 The answer checks out. 97 www.ck12.org Plan and Compare Alternative Approaches to Solving Problems Most word problems can be solved in more than one way. Often one method is more straightforward than others, but which method is best can depend on what kind of problem you are facing. Example 3 Nadia’s father is 36. He is 16 years older than four times Nadia’s age. How old is Nadia? Solution This problem can be solved with either of the strategies you learned in this section. Let’s solve it using both strategies. Guess and Check Method Step 1: Understand We need to ﬁnd Nadia’s age. We know that her father is 16 years older than four times her age, or 4 × (Nadia’s age) + 16. We know her father is 36 years old. Step 2: Strategy We guess a random number for Nadia’s age. We multiply the number by 4 and add 16 and check to see if the result equals 36. If the answer is too small, we guess a larger number, and if the answer is too big, we guess a smaller number. We keep guessing until we get the answer to be 36. Step 3: Apply strategy/Solve Guess Nadia’s age 10 4(10) + 16 = 56 too big for her father’s age Guess a smaller number 9 4(9) + 16 = 52 still too big Guessing 9 for Nadia’s age gave us a number that is 16 years too great to be her father’s age. But notice that when we decreased Nadia’s age by one, her father’s age decreased by four. That suggests that we can decrease our ﬁnal answer by 16 years if we decrease our guess by 4 years. 4 years less than 9 is 5. 4(5) + 16 = 36, which is the right age. Answer: Nadia is 5 years old. Step 4: Check Nadia is 5 years old. Her father’s age is 4(5) + 16 = 36. This is correct. The answer checks out. Work Backward Method Step 1: Understand We need to ﬁnd Nadia’s age. We know her father is 16 years older than four times her age, or 4 × (Nadia’s age) + 16. We know her father is 36 years old. Step 2: Strategy To get from Nadia’s age to her father’s age, we multiply Nadia’s age by four and add 16. Working backwards means we start with the father’s age, subtract 16 and divide by 4. www.ck12.org 98 Step 3: Apply Strategy/Solve Start with the father’s age 36 Subtract 16 36 −16 = 20 Divide by 4 20 ÷ 4 = 5 Answer Nadia is 5 years old. Step 4: Check Nadia is 5 years old. Her father’s age is 4(5) + 16 = 36. This is correct. The answer checks out. You see that in this problem, the “Work Backward” strategy is more straightforward than the Guess and Check method. The Work Backward method always works best when we know the result of a series of operations, but not the starting number. In the next chapter, you will learn algebra methods based on the Work Backward method. Lesson Summary The four steps of the problem solving plan are: • Understand the problem • Devise a plan – Translate • Carry out the plan – Solve • Look – Check and Interpret Two common problem solving strategies are: Guess and Check Guess a solution and use the guess in the problem to see if you get the correct answer. If the answer is too big or too small, then make another guess that will get you closer to the goal. Work Backward This method works well for problems in which a series of operations is applied to an unknown quantity and you are given the resulting number. Start with the result and apply the operations in reverse order until you ﬁnd the unknown. Review Questions 1. Finish the problem we started in Example 1. 2. Nadia is at home and Peter is at school which is 6 miles away from home. They start traveling towards each other at the same time. Nadia is walking at 3.5 miles per hour and Peter is skateboarding at 6 miles per hour. When will they meet and how far from home is their meeting place? 3. Peter bought several notebooks at Staples for $2.25 each; then he bought a few more notebooks at Rite-Aid for $2 each. He spent the same amount of money in both places and he bought 17 notebooks in all. How many notebooks did Peter buy in each store? 4. Andrew took a handful of change out of his pocket and noticed that he was only holding dimes and quarters in his hand. He counted and found that he had 22 coins that amounted to $4. How many quarters and how many dimes does Andrew have? 5. Anne wants to put a fence around her rose bed that is one and a half times as long as it is wide. She uses 50 feet of fencing. What are the dimensions of the garden? 99 www.ck12.org 6. Peter is outside looking at the pigs and chickens in the yard. Nadia is indoors and cannot see the animals. Peter gives her a puzzle. He tells her that he can see 13 heads and 36 feet and asks her how many pigs and how many chickens are in the yard. Help Nadia ﬁnd the answer. 7. Andrew invests $8000 in two types of accounts: a savings account that pays 5.25% interest per year and a more risky account that pays 9% interest per year. At the end of the year he has $450 in interest from the two accounts. Find the amount of money invested in each account. 8. 450 tickets are sold for a concert: balcony seats for $35 each and orchestra seats for $25 each. If the total box oﬀice take is $13,000, how many of each kind of ticket were sold? 9. There is a bowl of candy sitting on our kitchen table. One morning Nadia takes one-sixth of the candy. Later that morning Peter takes one-fourth of the candy that’s left. That afternoon, Andrew takes one-ﬁfth of what’s left in the bowl and ﬁnally Anne takes one-third of what is left in the bowl. If there are 16 candies left in the bowl at the end of the day, how much candy was there at the beginning of the day? 10. Nadia can completely mow the lawn by herself in 30 minutes. Peter can completely mow the lawn by himself in 45 minutes. How long does it take both of them to mow the lawn together? 11. Three monkeys spend a day gathering coconuts together. When they have ﬁnished, they are very tired and fall asleep. The following morning, the ﬁrst monkey wakes up. Not wishing to disturb his friends, he decides to divide the coconuts into three equal piles. There is one left over, so he throws this odd one away, helps himself to his share, and goes home. A few minutes later, the second monkey awakes. Not realizing that the ﬁrst has already gone, he too divides the coconuts into three equal heaps. He ﬁnds one left over, throws the odd one away, helps himself to his fair share, and goes home. In the morning, the third monkey wakes to ﬁnd that he is alone. He spots the two discarded coconuts, and puts them with the pile, giving him a total of twelve coconuts. (a) How many coconuts did the ﬁrst two monkeys take? (b) How many coconuts did the monkeys gather in all? 12. Two prime numbers have a product of 51. What are the numbers? 13. Two prime numbers have a product of 65. What are the numbers? 14. The square of a certain positive number is eight more than twice the number. What is the number? 15. Is 91 prime? (Hint: if it’s not prime, what are its prime factors?) 16. Is 73 prime? 17. Alison’s school day starts at 8:30, but today Alison wants to arrive ten minutes early to discuss an assignment with her English teacher. If she is also giving her friend Sherice a ride to school, and it takes her 12 minutes to get to Sherice’s house and another 15 minutes to get to school from there, at what time does Alison need to leave her house? 18. At her retail job, Kelly gets a raise of 10% every six months. After her third raise, she now makes $13.31 per hour. How much did she make when she ﬁrst started out? 19. Three years ago, Kevin’s little sister Becky had her ﬁfth birthday. If Kevin was eight when Becky was born, how old is he now? 20. A warehouse is full of shipping crates; half of them are headed for Boston and the other half for Philadelphia. A truck arrives to pick up 20 of the Boston-bound crates, and then another truck carries away one third of the Philadelphia-bound crates. An hour later, half of the remaining crates are moved onto the loading dock outside. If there are 40 crates left in the warehouse, how many were there originally? 21. Gerald is a bus driver who takes over from another bus driver one day in the middle of his route. He doesn’t pay attention to how many passengers are on the bus when he starts driving, but he does notice that three passengers get oﬀat the next stop, a total of eight more get on at the next three stops, two get on and four get oﬀat the next stop, and at the stop after that, a third of the passengers get oﬀ. (a) If there are now 14 passengers on the bus, how many were there when Gerald ﬁrst took over www.ck12.org 100 the route? (b) If half the passengers who got on while Gerald was driving paid the full adult fare of $1.50, and the other half were students or seniors who paid a discounted fare of $1.00, how much cash was in the bus’s fare box at the beginning of Gerald’s shift if there is now $73.50 in it? (c) When Gerald took over the route, all the passengers currently on the bus had paid full fare. However, some of the passengers who had previously gotten on and oﬀthe bus were students or seniors who had paid the discounted fare. Based on the amount of money that was in the cash box, if 28 passengers had gotten on the bus and gotten oﬀbefore Gerald arrived (in addition to the passengers who had gotten on and were still there when he arrived), how many of those passengers paid the discounted fare? (d) How much money would currently be in the cash box if all the passengers throughout the day had paid the full fare? Texas Instruments Resources In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http: //www.ck12.org/flexr/chapter/9612. 101 www.ck12.org Chapter 3 Equations of Lines 3.1 One-Step Equations Learning Objectives • Solve an equation using addition. • Solve an equation using subtraction. • Solve an equation using multiplication. • Solve an equation using division. Introduction Nadia is buying a new mp3 player. Peter watches her pay for the player with a $100 bill. She receives $22.00 in change, and from only this information, Peter works out how much the player cost. How much was the player? In algebra, we can solve problems like this using an equation. An equation is an algebraic expression that involves an equals sign. If we use the letter x to represent the cost of the mp3 player, we can write the equation x + 22 = 100. This tells us that the value of the player plus the value of the change received is equal to the $100 that Nadia paid. Another way we could write the equation would be x = 100 −22. This tells us that the value of the player is equal to the total amount of money Nadia paid (100 −22). This equation is mathematically equivalent to the ﬁrst one, but it is easier to solve. In this chapter, we will learn how to solve for the variable in a one-variable linear equation. Linear equations are equations in which each term is either a constant, or a constant times a single variable (raised to the ﬁrst power). The term linear comes from the word line, because the graph of a linear equation is always a line. We’ll start with simple problems like the one in the last example. Solving Equations Using Addition and Subtraction When we work with an algebraic equation, it’s important to remember that the two sides have to stay equal for the equation to stay true. We can change the equation around however we want, but whatever www.ck12.org 102 we do to one side of the equation, we have to do to the other side. In the introduction above, for example, we could get from the ﬁrst equation to the second equation by subtracting 22 from both sides: x + 22 = 100 x + 22 −22 = 100 −22 x = 100 −22 Similarly, we can add numbers to each side of an equation to help solve for our unknown. Example 1 Solve x −3 = 9. Solution To solve an equation for x, we need to isolate x−that is, we need to get it by itself on one side of the equals sign. Right now our x has a 3 subtracted from it. To reverse this, we’ll add 3—but we must add 3 to both sides. x −3 = 9 x −3 + 3 = 9 + 3 x + 0 = 9 + 3 x = 12 Example 2 Solve z −9.7 = −1.026 Solution It doesn’t matter what the variable is—the solving process is the same. z −9.7 = −1.026 z −9.7 + 9.7 = −1.026 + 9.7 z = 8.674 Make sure you understand the addition of decimals in this example! Example 3 Solve x + 4 7 = 9 5. Solution To isolate x, we need to subtract 4 7 from both sides. x + 4 7 = 9 5 x + 4 7 −4 7 = 9 5 −4 7 x = 9 5 −4 7 Now we have to subtract fractions, which means we need to ﬁnd the LCD. Since 5 and 7 are both prime, their lowest common multiple is just their product, 35. 103 www.ck12.org x = 9 5 −4 7 x = 7 · 9 7 · 5 −4 · 5 7 · 5 x = 63 35 −20 35 x = 63 −20 35 x = 43 35 Make sure you’re comfortable with decimals and fractions! To master algebra, you’ll need to work with them frequently. Solving Equations Using Multiplication and Division Suppose you are selling pizza for $1.50 a slice and you can get eight slices out of a single pizza. How much money do you get for a single pizza? It shouldn’t take you long to ﬁgure out that you get 8×$1.50 = $12.00. You solved this problem by multiplying. Here’s how to do the same thing algebraically, using x to stand for the cost in dollars of the whole pizza. Example 4 Solve 1 8 · x = 1.5. Our x is being multiplied by one-eighth. To cancel that out and get x by itself, we have to multiply by the reciprocal, 8. Don’t forget to multiply both sides of the equation. 8 (1 8 · x ) = 8(1.5) x = 12 Example 5 Solve 9x 5 = 5. 9x 5 is equivalent to 9 5 · x, so to cancel out that 9 5, we multiply by the reciprocal, 5 9. 5 9 (9x 5 ) = 5 9(5) x = 25 9 Example 6 Solve 0.25x = 5.25. www.ck12.org 104 0.25 is the decimal equivalent of one fourth, so to cancel out the 0.25 factor we would multiply by 4. 4(0.25x) = 4(5.25) x = 21 Solving by division is another way to isolate x. Suppose you buy ﬁve identical candy bars, and you are charged $3.25. How much did each candy bar cost? You might just divide $3.25 by 5, but let’s see how this problem looks in algebra. Example 7 Solve 5x = 3.25. To cancel the 5, we divide both sides by 5. 5x 5 = 3.25 5 x = 0.65 Example 8 Solve 7x = 5 11. Divide both sides by 7. x = 5 11.7 x = 5 77 Example 9 Solve 1.375x = 1.2. Divide by 1.375 x = 1.2 1.375 x = 0.872 Notice the bar above the ﬁnal two decimals; it means that those digits recur, or repeat. The full answer is 0.872727272727272.... To see more examples of one - and two-step equation solving, watch the Khan Academy video series starting at Solve Real-World Problems Using Equations Example 10 In the year 2017, Anne will be 45years old. In what year was Anne born? The unknown here is the year Anne was born, so that’s our variable x. Here’s our equation: x + 45 = 2017 x + 45 −45 = 2017 −45 x = 1972 105 www.ck12.org Anne was born in 1972. Example 11 A mail order electronics company stocks a new mini DVD player and is using a balance to determine the shipping weight. Using only one-pound weights, the shipping department found that the following arrangement balances: How much does each DVD player weigh? Solution Since the system balances, the total weight on each side must be equal. To write our equation, we’ll use x for the weight of one DVD player, which is unknown. There are two DVD players, weighing a total of 2x pounds, on the left side of the balance, and on the right side are 5 1-pound weights, weighing a total of 5 pounds. So our equation is 2x = 5. Dividing both sides by 2 gives us x = 2.5. Each DVD player weighs 2.5 pounds. Example 12 In 2004, Takeru Kobayashi of Nagano, Japan, ate 53.5 hot dogs in 12 minutes. This was 3 more hot dogs than his own previous world record, set in 2002. Calculate: a) How many minutes it took him to eat one hot dog. b) How many hot dogs he ate per minute. c) What his old record was. Solution a) We know that the total time for 53.5 hot dogs is 12 minutes. We want to know the time for one hot dog, so that’s x. Our equation is 53.5x = 12. Then we divide both sides by 53.5 to get x = 12 53.5, or x = 0.224 minutes. We can also multiply by 60 to get the time in seconds; 0.224 minutes is about 13.5 seconds. So that’s how long it took Takeru to eat one hot dog. b) Now we’re looking for hot dogs per minute instead of minutes per hot dog. We’ll use the variable y instead of x this time so we don’t get the two confused. 12 minutes, times the number of hot dogs per minute, equals the total number of hot dogs, so 12y = 53.5. Dividing both sides by 12 gives us y = 53.5 12 , or y = 4.458 hot dogs per minute. c) We know that his new record is 53.5, and we know that’s three more than his old record. If we call his old record z, we can write the following equation: z + 3 = 53.5. Subtracting 3 from both sides gives us z = 50.5. So Takeru’s old record was 50.5 hot dogs in 12 minutes. Lesson Summary • An equation in which each term is either a constant or the product of a constant and a single variable is a linear equation. • We can add, subtract, multiply, or divide both sides of an equation by the same value and still have an equivalent equation. www.ck12.org 106 • To solve an equation, isolate the unknown variable on one side of the equation by applying one or more arithmetic operations to both sides. Review Questions 1. Solve the following equations for x. (a) x = 11 = 7 (b) x −1.1 = 3.2 (c) 7x = 21 (d) 4x = 1 (e) 5x 12 = 2 3 (f) x + 5 2 = 2 3 (g) x −5 6 = 3 8 (h) 0.01x = 11 2. Solve the following equations for the unknown variable. (a) q −13 = −13 (b) z + 1.1 = 3.0001 (c) 21s = 3 (d) t + 1 2 = 1 3 (e) 7 f 11 = 7 11 (f) 3 4 = −1 2 −y (g) 6r = 3 8 (h) 9b 16 = 3 8 3. Peter is collecting tokens on breakfast cereal packets in order to get a model boat. In eight weeks he has collected 10 tokens. He needs 25 tokens for the boat. Write an equation and determine the following information. (a) How many more tokens he needs to collect, n. (b) How many tokens he collects per week, w. (c) How many more weeks remain until he can send oﬀfor his boat, r. 4. Juan has baked a cake and wants to sell it in his bakery. He is going to cut it into 12 slices and sell them individually. He wants to sell it for three times the cost of making it. The ingredients cost him $8.50, and he allowed $1.25 to cover the cost of electricity to bake it. Write equations that describe the following statements (a) The amount of money that he sells the cake for (u). (b) The amount of money he charges for each slice (c). (c) The total proﬁt he makes on the cake (w). 5. Jane is baking cookies for a large party. She has a recipe that will make one batch of two dozen cookies, and she decides to make ﬁve batches. To make ﬁve batches, she ﬁnds that she will need 12.5 cups of ﬂour and 15 eggs. (a) How many cookies will she make in all? (b) How many cups of ﬂour go into one batch? (c) How many eggs go into one batch? (d) If Jane only has a dozen eggs on hand, how many more does she need to make ﬁve batches? (e) If she doesn’t go out to get more eggs, how many batches can she make? How many cookies will that be? 107 www.ck12.org 3.2 Two-Step Equations Learning Objectives • Solve a two-step equation using addition, subtraction, multiplication, and division. • Solve a two-step equation by combining like terms. • Solve real-world problems using two-step equations. Solve a Two-Step Equation We’ve seen how to solve for an unknown by isolating it on one side of an equation and then evaluating the other side. Now we’ll see how to solve equations where the variable takes more than one step to isolate. Example 1 Rebecca has three bags containing the same number of marbles, plus two marbles left over. She places them on one side of a balance. Chris, who has more marbles than Rebecca, adds marbles to the other side of the balance. He ﬁnds that with 29 marbles, the scales balance. How many marbles are in each bag? Assume the bags weigh nothing. Solution We know that the system balances, so the weights on each side must be equal. If we use x to represent the number of marbles in each bag, then we can see that on the left side of the scale we have three bags (each containing x marbles) plus two extra marbles, and on the right side of the scale we have 29 marbles. The balancing of the scales is similar to the balancing of the following equation. 3x + 2 = 29 “Three bags plus two marbles equals 29 marbles” To solve for x, we need to ﬁrst get all the variables (terms containing an x) alone on one side of the equation. We’ve already got all the x’s on one side; now we just need to isolate them. 3x + 2 = 29 3x + 2 −2 = 29 −2 Get rid of the 2 on the left by subtracting it from both sides. 3x = 27 3x 3 = 27 3 Divide both sides by 3. x = 9 There are nine marbles in each bag. We can do the same with the real objects as we did with the equation. Just as we subtracted 2 from both sides of the equals sign, we could remove two marbles from each side of the scale. Because we removed the same number of marbles from each side, we know the scales will still balance. www.ck12.org 108 Then, because there are three bags of marbles on the left-hand side of the scale, we can divide the marbles on the right-hand side into three equal piles. You can see that there are nine marbles in each. Three bags of marbles balances three piles of nine marbles. So each bag of marbles balances nine marbles, meaning that each bag contains nine marbles. Check out for more interactive balance beam activi-ties! Example 2 Solve 6(x + 4) = 12. This equation has the x buried in parentheses. To dig it out, we can proceed in one of two ways: we can either distribute the six on the left, or divide both sides by six to remove it from the left. Since the right-hand side of the equation is a multiple of six, it makes sense to divide. That gives us x+4 = 2. Then we can subtract 4 from both sides to get x = −2. Example 3 Solve x−3 5 = 7. It’s always a good idea to get rid of fractions ﬁrst. Multiplying both sides by 5 gives us x −3 = 35, and then we can add 3 to both sides to get x = 38. Example 4 Solve 5 9(x + 1) = 2 7. First, we’ll cancel the fraction on the left by multiplying by the reciprocal (the multiplicative inverse). 9 5 · 5 9(x + 1) = 9 5 · 2 7 (x + 1) = 18 35 Then we subtract 1 from both sides. ( 35 35 is equivalent to 1.) x + 1 = 18 35 x + 1 −1 = 18 35 −35 35 x = 18 −35 35 x = −17 35 109 www.ck12.org These examples are called two-step equations, because we need to perform two separate operations on the equation to isolate the variable. Solve a Two-Step Equation by Combining Like Terms When we look at a linear equation we see two kinds of terms: those that contain the unknown variable, and those that don’t. When we look at an equation that has an x on both sides, we know that in order to solve it, we need to get all the x−terms on one side of the equation. This is called combining like terms. The terms with an x in them are like terms because they contain the same variable (or, as you will see in later chapters, the same combination of variables). Table 3.1: Like Terms Unlike Terms 4x, 10x, −3.5x, and x 12 3x and 3y 3y, 0.000001y, and y 4xy and 4x xy, 6xy, and 2.39xy 0.5x and 0.5 To add or subtract like terms, we can use the Distributive Property of Multiplication. 3x + 4x = (3 + 4)x = 7x 0.03xy −0.01xy = (0.03 −0.01)xy = 0.02xy −y + 16y + 5y = (−1 + 16 + 5)y = 10y 5z + 2z −7z = (5 + 2 −7)z = 0z = 0 To solve an equation with two or more like terms, we need to combine the terms ﬁrst. Example 5 Solve (x + 5) −(2x −3) = 6. There are two like terms: the x and the −2x (don’t forget that the negative sign applies to everything in the parentheses). So we need to get those terms together. The associative and distributive properties let us rewrite the equation as x + 5 −2x + 3 = 6, and then the commutative property lets us switch around the terms to get x −2x + 5 + 3 = 6, or (x −2x) + (5 + 3) = 6. (x −2x) is the same as (1 −2)x, or −x, so our equation becomes −x + 8 = 6 Subtracting 8 from both sides gives us −x = −2. And ﬁnally, multiplying both sides by -1 gives us x = 2. Example 6 Solve x 2 −x 3 = 6. This problem requires us to deal with fractions. We need to write all the terms on the left over a common denominator of six. 3x 6 −2x 6 = 6 Then we subtract the fractions to get x 6 = 6. Finally we multiply both sides by 6 to get x = 36. www.ck12.org 110 Solve Real-World Problems Using Two-Step Equations The hardest part of solving word problems is translating from words to an equation. First, you need to look to see what the equation is asking. What is the unknown for which you have to solve? That will be what your variable stands for. Then, follow what is going on with your variable all the way through the problem. Example 7 An emergency plumber charges $65 as a call-out fee plus an additional $75 per hour. He arrives at a house at 9:30 and works to repair a water tank. If the total repair bill is $196.25, at what time was the repair completed? In order to solve this problem, we collect the information from the text and convert it to an equation. Unknown: time taken in hours – this will be our x The bill is made up of two parts: a call out fee and a per-hour fee. The call out is a ﬂat fee, and independent of x—it’s the same no matter how many hours the plumber works. The per-hour part depends on the number of hours (x). So the total fee is $65 (no matter what) plus $75x (where x is the number of hours), or 65 + 75x. Looking at the problem again, we also can see that the total bill is $196.25. So our ﬁnal equation is 196.25 = 65 + 75x. Solving for x: 196.25 = 65 + 75x Subtract 65 from both sides. 131.25 = 75x Divide both sides by 75. 1.75 = x The job took 1.75 hours. Solution The repair job was completed 1.75 hours after 9:30, so it was completed at 11:15AM. Example 8 When Asia was young her Daddy marked her height on the door frame every month. Asia’s Daddy noticed that between the ages of one and three, he could predict her height (in inches) by taking her age in months, adding 75 inches and multiplying the result by one-third. Use this information to answer the following: a) Write an equation linking her predicted height, h, with her age in months, m. b) Determine her predicted height on her second birthday. c) Determine at what age she is predicted to reach three feet tall. Solution a) To convert the text to an equation, ﬁrst determine the type of equation we have. We are going to have an equation that links two variables. Our unknown will change, depending on the information we are 111 www.ck12.org given. For example, we could solve for height given age, or solve for age given height. However, the text gives us a way to determine height. Our equation will start with “h =”. The text tells us that we can predict her height by taking her age in months, adding 75, and multiplying by 1 3. So our equation is h = (m + 75) · 1 3, or h = 1 3(m + 75). b) To predict Asia’s height on her second birthday, we substitute m = 24 into our equation (because 2 years is 24 months) and solve for h. h = 1 3(24 + 75) h = 1 3(99) h = 33 Asia’s height on her second birthday was predicted to be 33 inches. c) To determine the predicted age when she reached three feet, substitute h = 36 into the equation and solve for m. 36 = 1 3(m + 75) 108 = m + 75 33 = m Asia was predicted to be 33 months old when her height was three feet. Example 9 To convert temperatures in Fahrenheit to temperatures in Celsius, follow the following steps: Take the temperature in degrees Fahrenheit and subtract 32. Then divide the result by 1.8 and this gives the temperature in degrees Celsius. a) Write an equation that shows the conversion process. b) Convert 50 degrees Fahrenheit to degrees Celsius. c) Convert 25 degrees Celsius to degrees Fahrenheit. d) Convert -40 degrees Celsius to degrees Fahrenheit. a) The text gives the process to convert Fahrenheit to Celsius. We can write an equation using two variables. We will use f for temperature in Fahrenheit, and c for temperature in Celsius. First we take the temperature in Fahrenheit and subtract 32. f −32 Then divide by 1.8. f −32 1.8 This equals the temperature in Celsius. c = f −32 1.8 In order to convert from one temperature scale to another, simply substitute in for whichever temperature you know, and solve for the one you don’t know. b) To convert 50 degrees Fahrenheit to degrees Celsius, substitute f = 50 into the equation. c = 50 −32 1.8 c = 18 1.8 c = 10 www.ck12.org 112 50 degrees Fahrenheit is equal to 10 degrees Celsius. c) To convert 25 degrees Celsius to degrees Fahrenheit, substitute c = 25 into the equation: 25 = f −32 1.8 45 = f −32 77 = f 25 degrees Celsius is equal to 77 degrees Fahrenheit. d) To convert -40 degrees Celsius to degrees Fahrenheit, substitute c = −40 into the equation. −40 = f −32 1.8 −72 = f −32 −40 = f -40 degrees Celsius is equal to -40 degrees Fahrenheit. (No, that’s not a mistake! This is the one temperature where they are equal.) Lesson Summary • Some equations require more than one operation to solve. Generally it, is good to go from the outside in. If there are parentheses around an expression with a variable in it, cancel what is outside the parentheses ﬁrst. • Terms with the same variable in them (or no variable in them) are like terms. Combine like terms (adding or subtracting them from each other) to simplify the expression and solve for the unknown. Review Questions 1. Solve the following equations for the unknown variable. (a) 1.3x −0.7x = 12 (b) 6x −1.3 = 3.2 (c) 5x −(3x + 2) = 1 (d) 4(x + 3) = 1 (e) 5q −7 = 2 3 (f) 3 5 x + 5 2 = 2 3 (g) s −3s 8 = 5 6 (h) 0.1y + 11 = 0 (i) 5q−7 12 = 2 3 (j) 5(q−7) 12 = 2 3 (k) 33t −99 = 0 (l) 5p −2 = 32 (m) 10y + 5 = 10 (n) 10(y + 5) = 10 (o) 10y + 5y = 10 (p) 10(y + 5y) = 10 113 www.ck12.org 2. Jade is stranded downtown with only $10 to get home. Taxis cost $0.75 per mile, but there is an additional $2.35 hire charge. Write a formula and use it to calculate how many miles she can travel with her money. 3. Jasmin’s Dad is planning a surprise birthday party for her. He will hire a bouncy castle, and will provide party food for all the guests. The bouncy castle costs $150 for the afternoon, and the food will cost $3 per person. Andrew, Jasmin’s Dad, has a budget of $300. Write an equation and use it to determine the maximum number of guests he can invite. 4. The local amusement park sells summer memberships for $50 each. Normal admission to the park costs $25; admission for members costs $15. (a) If Darren wants to spend no more than $100 on trips to the amusement park this summer, how many visits can he make if he buys a membership with part of that money? (b) How many visits can he make if he does not? (c) If he increases his budget to $160, how many visits can he make as a member? (d) And how many as a non-member? 5. For an upcoming school ﬁeld trip, there must be one adult supervisor for every ﬁve children. (a) If the bus seats 40 people, how many children can go on the trip? (b) How many children can go if a second 40-person bus is added? (c) Four of the adult chaperones decide to arrive separately by car. Now how many children can go in the two buses? 3.3 Multi-Step Equations Learning Objectives • Solve a multi-step equation by combining like terms. • Solve a multi-step equation using the distributive property. • Solve real-world problems using multi-step equations. Solving Multi-Step Equations by Combining Like Terms We’ve seen that when we solve for an unknown variable, it can take just one or two steps to get the terms in the right places. Now we’ll look at solving equations that take several steps to isolate the unknown variable. Such equations are referred to as multi-step equations. In this section, we’ll simply be combining the steps we already know how to do. Our goal is to end up with all the constants on one side of the equation and all the variables on the other side. We’ll do this by collecting like terms. Don’t forget, like terms have the same combination of variables in them. Example 1 Solve 3x+4 3 −5x = 6. Before we can combine the variable terms, we need to get rid of that fraction. First let’s put all the terms on the left over a common denominator of three: 3x+4 3 −15x 3 = 6. Combining the fractions then gives us 3x+4−15x 3 = 6. Combining like terms in the numerator gives us 4−12x 3 = 6. Multiplying both sides by 3 gives us 4 −12x = 18. Subtracting 4 from both sides gives us −12x = 14. www.ck12.org 114 And ﬁnally, dividing both sides by -12 gives us x = −14 12, which reduces to x = −7 6. Solving Multi-Step Equations Using the Distributive Property You may have noticed that when one side of the equation is multiplied by a constant term, we can either distribute it or just divide it out. If we can divide it out without getting awkward fractions as a result, then that’s usually the better choice, because it gives us smaller numbers to work with. But if dividing would result in messy fractions, then it’s usually better to distribute the constant and go from there. Example 2 Solve 7(2x −5) = 21. The ﬁrst thing we want to do here is get rid of the parentheses. We could use the Distributive Property, but it just so happens that 7 divides evenly into 21. That suggests that dividing both sides by 7 is the easiest way to solve this problem. If we do that, we get 2x −5 = 21 7 or just 2x −5 = 3. Then all we need to do is add 5 to both sides to get 2x = 8, and then divide by 2 to get x = 4. Example 3 Solve 17(3x + 4) = 7. Once again, we want to get rid of those parentheses. We could divide both sides by 17, but that would give us an inconvenient fraction on the right-hand side. In this case, distributing is the easier way to go. Distributing the 17 gives us 51x + 68 = 7. Then we subtract 68 from both sides to get 51x = −61, and then we divide by 51 to get x = −61 51 . (Yes, that’s a messy fraction too, but since it’s our ﬁnal answer and we don’t have to do anything else with it, we don’t really care how messy it is.) Example 4 Solve 4(3x −4) −7(2x + 3) = 3. Before we can collect like terms, we need to get rid of the parentheses using the Distributive Property. That gives us 12x −16 −14x −21 = 3, which we can rewrite as (12x −14x) + (−16 −21) = 3. This in turn simpliﬁes to −2x −37 = 3. Next we add 37 to both sides to get −2x = 40. And ﬁnally, we divide both sides by -2 to get x = −20. Example 5 Solve the following equation for x: 0.1(3.2 + 2x) + 1 2 ( 3 −x 5 ) = 0 This function contains both fractions and decimals. We should convert all terms to one or the other. It’s often easier to convert decimals to fractions, but in this equation the fractions are easy to convert to decimals—and with decimals we don’t need to ﬁnd a common denominator! In decimal form, our equation becomes 0.1(3.2 + 2x) + 0.5(3 −0.2x) = 0. Distributing to get rid of the parentheses, we get 0.32 + 0.2x + 1.5 −0.1x = 0. Collecting and combining like terms gives us 0.1x + 1.82 = 0. Then we can subtract 1.82 from both sides to get 0.1x = −1.82, and ﬁnally divide by 0.1 (or multiply by 10) to get x = −18.2. 115 www.ck12.org Solving Real-World Problems Using Multi-Step Equations Example 6 A growers’ cooperative has a farmer’s market in the town center every Saturday. They sell what they have grown and split the money into several categories. 8.5% of all the money taken in is set aside for sales tax. $150 goes to pay the rent on the space they occupy. What remains is split evenly between the seven growers. How much total money is taken in if each grower receives a $175 share? Let’s translate the text above into an equation. The unknown is going to be the total money taken in dollars. We’ll call this x. “8.5% of all the money taken in is set aside for sales tax.” This means that 91.5% of the money remains. This is 0.915x. “$150 goes to pay the rent on the space they occupy.” This means that what’s left is 0.915x −150. “What remains is split evenly between the 7 growers.” That means each grower gets 0.915x−150 7 . If each grower’s share is $175, then our equation to ﬁnd x is 0.915x−150 7 = 175. First we multiply both sides by 7 to get 0.915x −150 = 1225. Then add 150 to both sides to get 0.915x = 1375. Finally divide by 0.915 to get x ≈1502.7322. Since we want our answer in dollars and cents, we round to two decimal places, or $1502.73. The workers take in a total of $1502.73. Example 7 A factory manager is packing engine components into wooden crates to be shipped on a small truck. The truck is designed to hold sixteen crates, and will safely carry a 1200 lb cargo. Each crate weighs 12 lbs empty. How much weight should the manager instruct the workers to put in each crate in order to get the shipment weight as close as possible to 1200 lbs? The unknown quantity is the weight to put in each box, so we’ll call that x. Each crate when full will weigh x + 12 lbs, so all 16 crates together will weigh 16(x + 12) lbs. We also know that all 16 crates together should weigh 1200 lbs, so we can say that 16(x + 12) = 1200. To solve this equation, we can start by dividing both sides by 16: x + 12 = 1200 16 = 75. Then subtract 12 from both sides: x = 63. The manager should tell the workers to put 63 lbs of components in each crate. Ohm’s Law The electrical current, I (amps), passing through an electronic component varies directly with the applied voltage, V (volts), according to the relationship V = I · R where R is the resistance measured in Ohms (Ω). Example 8 A scientist is trying to deduce the resistance of an unknown component. He labels the resistance of the unknown component x Ω. The resistance of a circuit containing a number of these components is (5x+20)Ω. If a 120 volt potential diﬀerence across the circuit produces a current of 2.5 amps, calculate the resistance of the unknown component. Solution www.ck12.org 116 To solve this, we need to start with the equation V = I · R and substitute in V = 120, I = 2.5, and R = 5x + 20. That gives us 120 = 2.5(5x + 20). Distribute the 2.5 to get 120 = 12.5x + 50. Subtract 50 from both sides to get 70 = 12.5x. Finally, divide by 12.5 to get 5.6 = x. The unknown components have a resistance of 5.6 Ω. Distance, Speed and Time The speed of a body is the distance it travels per unit of time. That means that we can also ﬁnd out how far an object moves in a certain amount of time if we know its speed: we use the equation “distance = speed × time.” Example 8 Shanice’s car is traveling 10 miles per hour slower than twice the speed of Brandon’s car. She covers 93 miles in 1 hour 30 minutes. How fast is Brandon driving? Solution Here, we don’t know either Brandon’s speed or Shanice’s, but since the question asks for Brandon’s speed, that’s what we’ll use as our variable x. The distance Shanice covers in miles is 93, and the time in hours is 1.5. Her speed is 10 less than twice Brandon’s speed, or 2x −10 miles per hour. Putting those numbers into the equation gives us 93 = 1.5(2x −10). First we distribute, to get 93 = 3x −15. Then we add 15 to both sides to get 108 = 3x. Finally we divide by 3 to get 36 = x. Brandon is driving at 36 miles per hour. We can check this answer by considering the situation another way: we can solve for Shanice’s speed instead of Brandon’s and then check that against Brandon’s speed. We’ll use y for Shanice’s speed since we already used x for Brandon’s. The equation for Shanice’s speed is simply 93 = 1.5y. We can divide both sides by 1.5 to get 62 = y, so Shanice is traveling at 62 miles per hour. The problem tells us that Shanice is traveling 10 mph slower than twice Brandon’s speed; that would mean that 62 is equal to 2 times 36 minus 10. Is that true? Well, 2 times 36 is 72, minus 10 is 62. The answer checks out. In algebra, there’s almost always more than one method of solving a problem. If time allows, it’s always a good idea to try to solve the problem using two diﬀerent methods just to conﬁrm that you’ve got the answer right. Speed of Sound The speed of sound in dry air, v, is given by the equation v = 331 + 0.6T, where T is the temperature in Celsius and v is the speed of sound in meters per second. Example 9 117 www.ck12.org Tashi hits a drainpipe with a hammer and 250 meters away Minh hears the sound and hits his own drainpipe. Unfortunately, there is a one second delay between him hearing the sound and hitting his own pipe. Tashi accurately measures the time between her hitting the pipe and hearing Mihn’s pipe at 2.46 seconds. What is the temperature of the air? This is a complex problem and we need to be careful in writing our equations. First of all, the distance the sound travels is equal to the speed of sound multiplied by the time, and the speed is given by the equation above. So the distance equals (331 + 0.6T) × time, and the time is 2.46 −1 (because for 1 second out of the 2.46 seconds measured, there was no sound actually traveling). We also know that the distance is 250 × 2 (because the sound traveled from Tashi to Minh and back again), so our equation is 250 × 2 = (331 + 0.6T)(2.46 −1), which simpliﬁes to 500 = 1.46(331 + 0.6T). Distributing gives us 500 = 483.26+0.876T, and subtracting 483.26 from both sides gives us 16.74 = 0.876T. Then we divide by 0.876 to get T ≈19.1. The temperature is about 19.1 degrees Celsius. Lesson Summary • Multi-step equations are slightly more complex than one - and two-step equations, but use the same basic techniques. • If dividing a number outside of parentheses will produce fractions, it is often better to use the Distributive Property to expand the terms and then combine like terms to solve the equation. Review Questions 1. Solve the following equations for the unknown variable. (a) 3(x −1) −2(x + 3) = 0 (b) 3(x + 3) −2(x −1) = 0 (c) 7(w + 20) −w = 5 (d) 5(w + 20) −10w = 5 (e) 9(x −2) −3x = 3 (f) 12(t −5) + 5 = 0 (g) 2(2d + 1) = 2 3 (h) 2 ( 5a −1 3 ) = 2 7 (i) 2 9 ( i + 2 3 ) = 2 5 (j) 4 ( v + 1 4 ) = 35 2 (k) g 10 = 6 3 (l) s−4 11 = 2 5 (m) 2k 7 = 3 8 (n) 7x+4 3 = 9 2 (o) 9y−3 6 = 5 2 (p) r 3 + r 2 = 7 (q) p 16 −2p 3 = 1 9 (r) m+3 2 −m 4 = 1 3 (s) 5 ( k 3 + 2 ) = 32 3 (t) 3 z = 2 5 (u) 2 r + 2 = 10 3 (v) 12 5 = 3+z z www.ck12.org 118 2. An engineer is building a suspended platform to raise bags of cement. The platform has a mass of 200 kg, and each bag of cement is 40 kg. He is using two steel cables, each capable of holding 250 kg. Write an equation for the number of bags he can put on the platform at once, and solve it. 3. A scientist is testing a number of identical components of unknown resistance which he labels xΩ. He connects a circuit with resistance (3x + 4)Ωto a steady 12 volt supply and ﬁnds that this produces a current of 1.2 amps. What is the value of the unknown resistance? 4. Lydia inherited a sum of money. She split it into ﬁve equal parts. She invested three parts of the money in a high-interest bank account which added 10% to the value. She placed the rest of her inheritance plus $500 in the stock market but lost 20% on that money. If the two accounts end up with exactly the same amount of money in them, how much did she inherit? 5. Pang drove to his mother’s house to drop oﬀher new TV. He drove at 50 miles per hour there and back, and spent 10 minutes dropping oﬀthe TV. The entire journey took him 94 minutes. How far away does his mother live? 3.4 Equations with Variables on Both Sides Learning Objectives • Solve an equation with variables on both sides. • Solve an equation with grouping symbols. • Solve real-world problems using equations with variables on both sides. Solve an Equation with Variables on Both Sides When a variable appears on both sides of the equation, we need to manipulate the equation so that all variable terms appear on one side, and only constants are left on the other. Example 1 Dwayne was told by his chemistry teacher to measure the weight of an empty beaker using a balance. Dwayne found only one lb weights, and so devised the following way of balancing the scales. Knowing that each weight is one lb, calculate the weight of one beaker. Solution We know that the system balances, so the weights on each side must be equal. We can write an algebraic expression based on this fact. The unknown quantity, the weight of the beaker, will be our x. We can see that on the left hand scale we have one beaker and four weights. On the right scale, we have four beakers and three weights. The balancing of the scales is analogous to the balancing of the following equation: x + 4 = 4x + 3 “One beaker plus 4 lbs equals 4 beakers plus 3 lbs” To solve for the weight of the beaker, we want all the constants (numbers) on one side and all the variables 119 www.ck12.org (terms with x in them) on the other side. Since there are more beakers on the right and more weights on the left, we’ll try to move all the x terms (beakers) to the right, and the constants (weights) to the left. First we subtract 3 from both sides to get x + 1 = 4x. Then we subtract x from both sides to get 1 = 3x. Finally we divide by 3 to get 1 3 = x. The weight of the beaker is one-third of a pound. We can do the same with the real objects as we did with the equation. Just as we subtracted amounts from each side of the equation, we could remove a certain number of weights or beakers from each scale. Because we remove the same number of objects from each side, we know the scales will still balance. First, we could remove three weights from each scale. This would leave one beaker and one weight on the left and four beakers on the right (in other words x + 1 = 4x ): Then we could remove one beaker from each scale, leaving only one weight on the left and three beakers on the right, to get 1 = 3x: Looking at the balance, it is clear that the weight of one beaker is one-third of a pound. Example 2 Sven was told to ﬁnd the weight of an empty box with a balance. Sven found some one lb weights and ﬁve lb weights. He placed two one lb weights in three of the boxes and with a fourth empty box found the following way of balancing the scales: Knowing that small weights are one lb and big weights are ﬁve lbs, calculate the weight of one box. We know that the system balances, so the weights on each side must be equal. We can write an algebraic expression based on this equality. The unknown quantity—the weight of each empty box, in pounds—will be our x. A box with two 1 lb weights in it weighs (x + 2) pounds. Our equation, based on the picture, is 3(x + 2) = x + 3(5). Distributing the 3 and simplifying, we get 3x + 6 = x + 15. www.ck12.org 120 Subtracting x from both sides, we get 2x + 6 = 15. Subtracting 6 from both sides, we get 2x = 9. And ﬁnally we can divide by 2 to get x = 9 2, or x = 4.5. Each box weighs 4.5 lbs. To see more examples of solving equations with variables on both sides of the equation, see the Khan Academy video at Solve an Equation with Grouping Symbols As you’ve seen, we can solve equations with variables on both sides even when some of the variables are in parentheses; we just have to get rid of the parentheses, and then we can start combining like terms. We use the same technique when dealing with fractions: ﬁrst we multiply to get rid of the fractions, and then we can shuﬄe the terms around by adding and subtracting. Example 3 Solve 3x + 2 = 5x 3 . Solution The ﬁrst thing we’ll do is get rid of the fraction. We can do this by multiplying both sides by 3, leaving 3(3x + 2) = 5x. Then we distribute to get rid of the parentheses, leaving 9x + 6 = 5x. We’ve already got all the constants on the left side, so we’ll move the variables to the right side by subtracting 9x from both sides. That leaves us with 6 = −4x. And ﬁnally, we divide by -4 to get −3 2 = x, or x = −1.5. Example 4 Solve 7x + 2 = 5x−3 6 . Solution Again we start by eliminating the fraction. Multiplying both sides by 6 gives us 6(7x + 2) = 5x −3, and distributing gives us 42x + 12 = 5x −3. Subtracting 5x from both sides gives us 37x + 12 = −3. Subtracting 12 from both sides gives us 37x = −15. Finally, dividing by 37 gives us x = −15 37. Example 5 Solve the following equation for x: 14x (x+3) = 7 Solution The form of the left hand side of this equation is known as a rational function because it is the ratio of two other functions: 14x and (x + 3). But we can solve it just like any other equation involving fractions. First we multiply both sides by (x + 3) to get rid of the fraction. Now our equation is 14x = 7(x + 3). Then we distribute: 14x = 7x + 21. Then subtract 7x from both sides: 7x = 21. And divide by 7: x = 3. 121 www.ck12.org Solve Real-World Problems Using Equations with Variables on Both Sides Here’s another chance to practice translating problems from words to equations. What is the equation asking? What is the unknown variable? What quantity will we use for our variable? The text explains what’s happening. Break it down into small, manageable chunks, and follow what’s going on with our variable all the way through the problem. More on Ohm’s Law Recall that the electrical current, I (amps), passing through an electronic component varies directly with the applied voltage, V (volts), according to the relationship V = I · R where R is the resistance measured in Ohms (Ω). The resistance R of a number of components wired in a series (one after the other) is simply the sum of all the resistances of the individual components. Example 6 In an attempt to ﬁnd the resistance of a new component, a scientist tests it in series with standard resistors. A ﬁxed voltage causes a 4.8 amp current in a circuit made up from the new component plus a 15Ωresistor in series. When the component is placed in a series circuit with a 50Ωresistor, the same voltage causes a 2.0 amp current to ﬂow. Calculate the resistance of the new component. This is a complex problem to translate, but once we convert the information into equations it’s relatively straightforward to solve. First, we are trying to ﬁnd the resistance of the new component (in Ohms, Ω). This is our x. We don’t know the voltage that is being used, but we can leave that as a variable, V. Our ﬁrst situation has a total resistance that equals the unknown resistance plus 15Ω. The current is 4.8 amps. Substituting into the formula V = I · R, we get V = 4.8(x + 15). Our second situation has a total resistance that equals the unknown resistance plus 50Ω. The current is 2.0 amps. Substituting into the same equation, this time we get V = 2(x + 50). We know the voltage is ﬁxed, so the V in the ﬁrst equation must equal the V in the second. That means we can set the right-hand sides of the two equations equal to each other: 4.8(x + 15) = 2(x + 50). Then we can solve for x. Distribute the constants ﬁrst: 4.8x + 72 = 2x + 100. Subtract 2x from both sides: 2.8x + 72 = 100. Subtract 72 from both sides: 2.8x = 28. Divide by 2.8: x = 10. The resistance of the component is 10Ω. Lesson Summary If an unknown variable appears on both sides of an equation, distribute as necessary. Then simplify the equation to have the unknown on only one side. Review Questions 1. Solve the following equations for the unknown variable. (a) 3(x −1) = 2(x + 3) www.ck12.org 122 (b) 7(x + 20) = x + 5 (c) 9(x −2) = 3x + 3 (d) 2 ( a −1 3 ) = 2 5 ( a + 2 3 ) (e) 2 7 ( t + 2 3 ) = 1 5 ( t −2 3 ) (f) 1 7 ( v + 1 4 ) = 2 ( 3v 2 −5 2 ) (g) y−4 11 = 2 5 · 2y+1 3 (h) z 16 = 2(3z+1) 9 (i) q 16 + q 6 = (3q+1) 9 + 3 2 (j) 3 x = 2 x+1 (k) 5 2+p = 3 p−8 2. Manoj and Tamar are arguing about a number trick they heard. Tamar tells Andrew to think of a number, multiply it by ﬁve and subtract three from the result. Then Manoj tells Andrew to think of a number, add ﬁve and multiply the result by three. Andrew says that whichever way he does the trick he gets the same answer. (a) What was the number Andrew started with? (b) What was the result Andrew got both times? (c) Name another set of steps that would have resulted in the same answer if Andrew started with the same number. 3. Manoj and Tamar try to come up with a harder trick. Manoj tells Andrew to think of a number, double it, add six, and then divide the result by two. Tamar tells Andrew to think of a number, add ﬁve, triple the result, subtract six, and then divide the result by three. (a) Andrew tries the trick both ways and gets an answer of 10 each time. What number did he start out with? (b) He tries again and gets 2 both times. What number did he start out with? (c) Is there a number Andrew can start with that will not give him the same answer both ways? (d) Bonus: Name another set of steps that would give Andrew the same answer every time as he would get from Manoj’s and Tamar’s steps. 4. I have enough money to buy ﬁve regular priced CDs and have $6 left over. However, all CDs are on sale today, for $4 less than usual. If I borrow $2, I can aﬀord nine of them. (a) How much are CDs on sale for today? (b) How much would I have to borrow to aﬀord nine of them if they weren’t on sale? 5. Five identical electronics components were connected in series. A ﬁxed but unknown voltage placed across them caused a 2.3 amp current to ﬂow. When two of the components were replaced with standard 10Ωresistors, the current dropped to 1.9 amps. What is the resistance of each component? 6. Solve the following resistance problems. Assume the same voltage is applied to all circuits. (a) Three unknown resistors plus 20Ωgive the same current as one unknown resistor plus 70Ω. (b) One unknown resistor gives a current of 1.5 amps and a 15Ωresistor gives a current of 3.0 amps. (c) Seven unknown resistors plus 18Ωgives twice the current of two unknown resistors plus 150Ω. (d) Three unknown resistors plus 1.5Ωgives a current of 3.6 amps and seven unknown resistors plus seven 12Ωresistors gives a current of 0.2 amps. 3.5 Ratios and Proportions Learning Objectives • Write and understand a ratio. 123 www.ck12.org • Write and solve a proportion. • Solve proportions using cross products. • Solve problems using scale drawings. Introduction Nadia is counting out money with her little brother. She gives her brother all the nickels and pennies. She keeps the quarters and dimes for herself. Nadia has four quarters and six dimes. Her brother has ﬁfteen nickels and ﬁve pennies and is happy because he has more coins than his big sister. How would you explain to him that he is actually getting a bad deal? Write a ratio A ratio is a way to compare two numbers, measurements or quantities. When we write a ratio, we divide one number by another and express the answer as a fraction. There are two distinct ratios in the problem above. For example, the ratio of the number of Nadia’s coins to her brother’s is 4+6 15+5, or 10 20 = 1 2. (Ratios should always be simpliﬁed.) In other words, Nadia has half as many coins as her brother. Another ratio we could look at is the value of the coins. The value of Nadia’s coins is (4 × 25) + (6 × 10) = 160 cents. The value of her brother’s coins is (15×5)+(5×1) = 80 cents. The ratio of the value of Nadia’s coins to her brother’s is 160 80 = 2 1. So the value of Nadia’s money is twice the value of her brother’s. Notice that even though the denominator is one, we still write it out and leave the ratio as a fraction instead of a whole number. A ratio with a denominator of one is called a unit rate. Example 1 The price of a Harry Potter Book on Amazon.com is $10.00. The same book is also available used for $6.50. Find two ways to compare these prices. Solution We could compare the numbers by expressing the diﬀerence between them: $10.00 −$6.50 = $3.50. We can also use a ratio to compare them: 10.00 6.50 = 100 65 = 20 13 (after multiplying by 10 to remove the decimals, and then simplifying). So we can say that the new book is $3.50 more than the used book, or we can say that the new book costs 20 13 times as much as the used book. Example 2 A tournament size shuﬄeboard table measures 30 inches wide by 14 feet long. Compare the length of the table to its width and express the answer as a ratio. Solution We could just write the ratio as 14 feet 30 inches. But since we’re comparing two lengths, it makes more sense to convert all the measurements to the same units. 14 feet is 14 × 12 = 168 inches, so our new ratio is 168 30 = 28 5 . Example 3 A family car is being tested for fuel eﬀiciency. It drives non-stop for 100 miles and uses 3.2 gallons of gasoline. Write the ratio of distance traveled to fuel used as a unit rate. Solution The ratio of distance to fuel is 100 miles 3.2 gallons. But a unit rate has to have a denominator of one, so to make www.ck12.org 124 this ratio a unit rate we need to divide both numerator and denominator by 3.2. 100 3.2 miles 3.2 3.2 gallons = 31.25 miles 1 gallon or 31.25 miles per gallon. Write and Solve a Proportion When two ratios are equal to each other, we call it a proportion. For example, the equation 10 5 = 6 9 is a proportion. We know it’s true because we can reduce both fractions to 2 3. (Check this yourself to make sure!) We often use proportions in science and business—for example, when scaling up the size of something. We generally use them to solve for an unknown, so we use algebra and label the unknown variable x. Example 4 A small fast food chain operates 60 stores and makes $1.2 million proﬁt every year. How much proﬁt would the chain make if it operated 250 stores? Solution First, we need to write a ratio: the ratio of proﬁt to number of stores. That would be $1,200,000 60 . Now we want to know how much proﬁt 250 stores would make. If we label that proﬁt x, then the ratio of proﬁt to stores in that case is x 250. Since we’re assuming the proﬁt is proportional to the number of stores, the ratios are equal and our proportion is 1,200,000 60 = x 250. (Note that we can drop the units – not because they are the same in the numerator and denominator, but because they are the same on both sides of the equation.) To solve this equation, ﬁrst we simplify the left-hand fraction to get 20, 000 = x 250. Then we multiply both sides by 250 to get 5, 000, 000 = x. If the chain operated 250 stores, the annual proﬁt would be 5 million dollars. Example 5 A chemical company makes up batches of copper sulfate solution by adding 250 kg of copper sulfate powder to 1000 liters of water. A laboratory chemist wants to make a solution of identical concentration, but only needs 350 mL (0.35 liters) of solution. How much copper sulfate powder should the chemist add to the water? Solution The ratio of powder to water in the ﬁrst case, in kilograms per liter, is 250 1000, which reduces to 1 4. In the second case, the unknown amount is how much powder to add. If we label that amount x, the ratio is x 0.35. So our proportion is 1 4 = x 0.35. To solve for x, ﬁrst we multiply both sides by 0.35 to get 0.35 4 = x, or x = 0.0875. The mass of copper sulfate that the chemist should add is 0.0875 kg, or 87.5 grams. Solve Proportions Using Cross Products One neat way to simplify proportions is to cross multiply. Consider the following proportion: 16 4 = 20 5 125 www.ck12.org If we want to eliminate the fractions, we could multiply both sides by 4 and then multiply both sides by 5. But suppose we just do both at once? 4 × 5 × 16 4 = 4 × 5 × 20 5 5 × 16 = 4 × 20 Now comparing this to the proportion we started with, we see that the denominator from the left hand side ends up being multiplied by the numerator on the right hand side. You can also see that the denominator from the right hand side ends up multiplying the numerator on the left hand side. In eﬀect the two denominators have multiplied across the equal sign: becomes 5 × 16 = 4 × 20. This movement of denominators is known as cross multiplying. It is extremely useful in solving propor-tions, especially when the unknown variable is in the denominator. Example 6 Solve this proportion for x: 4 3 = 9 x Solution Cross multiply to get 4x = 9 × 3, or 4x = 27. Then divide both sides by 4 to get x = 27 4 , or x = 6.75. Example 7 Solve the following proportion for x: 0.5 3 = 56 x Solution Cross multiply to get 0.5x = 56 × 3, or 0.5x = 168. Then divide both sides by 0.5 to get x = 336. Solve Real-World Problems Using Proportions Example 8 A cross-country train travels at a steady speed. It covers 15 miles in 20 minutes. How far will it travel in 7 hours assuming it continues at the same speed? Solution We’ve done speed problems before; remember that speed is just the ratio distance time , so that ratio is the one we’ll use for our proportion. We can see that the speed is 15 miles 20 minutes, and that speed is also equal to x miles 7 hours. To set up a proportion, we ﬁrst have to get the units the same. 20 minutes is 1 3 of an hour, so our proportion will be 15 1 3 = x 7. This is a very awkward looking ratio, but since we’ll be cross multiplying, we can leave it as it is. Cross multiplying gives us 7×15 = 1 3 x. Multiplying both sides by 3 then gives us 3×7×15 = x, or x = 315. The train will travel 315 miles in 7 hours. Example 9 In the United Kingdom, Alzheimer’s disease is said to aﬀect one in ﬁfty people over 65 years of age. If approximately 250000 people over 65 are aﬀected in the UK, how many people over 65 are there in total? Solution www.ck12.org 126 The ﬁxed ratio in this case is the 1 person in 50. The unknown quantity x is the total number of people over 65. Note that in this case we don’t need to include the units, as they will cancel between the numerator and denominator. Our proportion is 1 50 = 250000 x . Each ratio represents people with Alzheimer’s total people . Cross multiplying, we get 1 · x = 250000 · 50, or x = 12, 500, 000. There are approximately 12.5 million people over the age of 65 in the UK. For some more advanced ratio problems and applications, watch the Khan Academy video at youtube.com/watch?v=PASSD2OcU0c. Scale and Indirect Measurement One place where ratios are often used is in making maps. The scale of a map describes the relationship between distances on a map and the corresponding distances on the earth’s surface. These measurements are expressed as a fraction or a ratio. So far we have only written ratios as fractions, but outside of mathematics books, ratios are often written as two numbers separated by a colon (:). For example, instead of 2 3, we would write 2:3. Ratios written this way are used to express the relationship between a map and the area it represents. For example, a map with a scale of 1:1000 would be a map where one unit of measurement (such as a centimeter) on the map would represent 1000 of the same unit (1000 centimeters, or 10 meters) in real life. Example 10 Anne is visiting a friend in London, and is using the map below to navigate from Fleet Street to Borough Road. She is using a 1:100,000 scale map, where 1 cm on the map represents 1 km in real life. Using a ruler, she measures the distance on the map as 8.8 cm. How far is the real distance from the start of her journey to the end? Solution The scale is the ratio of distance on the map to the corresponding distance in real life. Written as a fraction, it is 1 100000. We can also write an equivalent ratio for the distance Anne measures on the map and the distance in real life that she is trying to ﬁnd: 8.8 x . Setting these two ratios equal gives us our proportion: 1 100000 = 8.8 x . Then we can cross multiply to get x = 880000. That’s how many centimeters it is from Fleet Street to Borough Road; now we need to convert to kilometers. There are 100000 cm in a km, so we have to divide our answer by 100000. 880000 100000 = 8.8. The distance from Fleet Street to Borough Road is 8.8 km. 127 www.ck12.org In this problem, we could have just used our intuition: the 1 cm = 1 km scale tells us that any number of cm on the map is equal to the same number of km in real life. But not all maps have a scale this simple. You’ll usually need to refer to the map scale to convert between measurements on the map and distances in real life! Example 11 Antonio is drawing a map of his school for a project in math. He has drawn out the following map of the school buildings and the surrounding area He is trying to determine the scale of his ﬁgure. He knows that the distance from the point marked A on the baseball diamond to the point marked B on the athletics track is 183 meters. Use the dimensions marked on the drawing to determine the scale of his map. Solution We know that the real-life distance is 183 m, and the scale is the ratio distance on map distance in real life. To ﬁnd the distance on the map, we use Pythagoras’ Theorem: a2 + b2 = c2, where a and b are the horizontal and vertical lengths and c is the diagonal between points A and B. 82 + 142 = c2 64 + 196 = c2 260 = c2 √ 260 = c 16.12 ≈c So the distance on the map is about 16.12 cm. The distance in real life is 183 m, which is 18300 cm. Now we can divide: Scale = 16.12 18300 ≈ 1 1135.23 The scale of Antonio’s map is approximately 1:1100. Another visual use of ratio and proportion is in scale drawings. Scale drawings (often called plans) are used extensively by architects. The equations governing scale are the same as for maps; the scale of a drawing is the ratio distance on diagram distance in real life . Example 12 Oscar is trying to make a scale drawing of the Titanic, which he knows was 883 ft long. He would like his drawing to be at a 1:500 scale. How many inches long does his sheet of paper need to be? Solution www.ck12.org 128 We can reason intuitively that since the scale is 1:500, the paper must be 883 500 = 1.766 feet long. Converting to inches means the length is 12(1.766) = 21.192 inches. Oscar’s paper should be at least 22 inches long. Example 13 The Rose Bowl stadium in Pasadena, California measures 880 feet from north to south and 695 feet from east to west. A scale diagram of the stadium is to be made. If 1 inch represents 100 feet, what would be the dimensions of the stadium drawn on a sheet of paper? Will it ﬁt on a standard 8.5 × 11 inch sheet of paper? Solution Instead of using a proportion, we can simply use the following equation: (distance on diagram) = (distance in real life) × (scale). (We can derive this from the fact that scale = distance on diagram distance in real life .) Plugging in, we get height on paper = 880 feet × 1 inch 100 feet = 8.8 inches width on paper = 695 feet × 1 inch 100 feet = 6.95 inches The scale diagram will be 8.8 in × 6.95 in. It will ﬁt on a standard sheet of paper. Lesson Summary • A ratio is a way to compare two numbers, measurements or quantities by dividing one number by the other and expressing the answer as a fraction. • A proportion is formed when two ratios are set equal to each other. • Cross multiplication is useful for solving equations in the form of proportions. To cross multiply, multiply the bottom of each ratio by the top of the other ratio and set them equal. For instance, cross multiplying results in 11 × 3 = 5x. • Scale is a proportion that relates map distance to real life distance. Review Questions 1. Write the following comparisons as ratios. Simplify fractions where possible. (a) $150 to $3 (b) 150 boys to 175 girls (c) 200 minutes to 1 hour (d) 10 days to 2 weeks 2. Write the following ratios as a unit rate. (a) 54 hotdogs to 12 minutes (b) 5000 lbs to 250 square inches (c) 20 computers to 80 students (d) 180 students to 6 teachers (e) 12 meters to 4 ﬂoors (f) 18 minutes to 15 appointments 3. Solve the following proportions. (a) 13 6 = 5 x 129 www.ck12.org (b) 1.25 7 = 3.6 x (c) 6 19 = x 11 (d) 1 x = 0.01 5 (e) 300 4 = x 99 (f) 2.75 9 = x ( 2 9) (g) 1.3 4 = x 1.3 (h) 0.1 1.01 = 1.9 x 4. A restaurant serves 100 people per day and takes in $908. If the restaurant were to serve 250 people per day, how much money would it take in? 5. The highest mountain in Canada is Mount Yukon. It is 298 67 the size of Ben Nevis, the highest peak in Scotland. Mount Elbert in Colorado is the highest peak in the Rocky Mountains. Mount Elbert is 220 67 the height of Ben Nevis and 11 12 the size of Mont Blanc in France. Mont Blanc is 4800 meters high. How high is Mount Yukon? 6. At a large high school it is estimated that two out of every three students have a cell phone, and one in ﬁve of all students have a cell phone that is one year old or less. Out of the students who own a cell phone, what proportion owns a phone that is more than one year old? 7. Use the map in Example 10. Using the scale printed on the map, determine the distances (rounded to the nearest half km) between: (a) Points 1 and 4 (b) Points 22 and 25 (c) Points 18 and 13 (d) Tower Bridge and London Bridge 3.6 Percent Problems Learning Objectives • Find a percent of a number. • Use the percent equation. • Find the percent of change. Introduction A percent is simply a ratio with a base unit of 100. When we write a ratio as a fraction, the percentage we want to represent is the numerator, and the denominator is 100. For example, 43% is another way of writing 43 100. 43 1000, on the other hand, is equal to 4.3 100, so it would be equivalent to 4.3%. 2 5 is equal to 40 100, or 40%. To convert any fraction to a percent, just convert it to an equivalent fraction with a denominator of 100, and then take the numerator as your percent value. To convert a percent to a decimal, just move the decimal point two spaces to the right: 67% = 0.67 0.2% = 0.002 150% = 1.5 And to convert a decimal to a percent, just move the decimal point two spaces to the left: www.ck12.org 130 2.3 = 230% 0.97 = 97% 0.00002 = 0.002% Finding and Converting Percentages Before we work with percentages, we need to know how to convert between percentages, decimals and fractions. Converting percentages to fractions is the easiest. The word “percent” simply means “per 100”—so, for example, 55% means 55 per 100, or 55 100. This fraction can then be simpliﬁed to 11 20. Example 1 Convert 32.5% to a fraction. Solution 32.5% is equal to 32.5 per 100, or 32.5 100 . To reduce this fraction, we ﬁrst need to multiply it by 10 10 to get rid of the decimal point. 325 1000 then reduces to 13 40. Converting fractions to percentages can be a little harder. To convert a fraction directly to a percentage, you need to express it as an equivalent fraction with a denominator of 100. Example 2 Convert 7 8 to a percent. Solution To get the denominator of this fraction equal to 100, we have to multiply it by 12.5. Multiplying the numerator by 12.5 also, we get 87.5 100 , which is equivalent to 87.5%. But what about a fraction like 1 6, where there’s no convenient number to multiply the denominator by to get 100? In a case like this, it’s easier to do the division problem suggested by the fraction in order to convert the fraction to a decimal, and then convert the decimal to a percent. 1 divided by 6 works out to 0.166666.... Moving the decimal two spaces to the right tells us that this is equivalent to about 16.7%. Why can we convert from decimals to percents just by moving the decimal point? Because of what decimal places represent. 0.1 is another way of representing one tenth, and 0.01 is equal to one hundredth—and one hundredth is one percent. By the same logic, 0.02 is 2 percent, 0.35 is 35 percent, and so on. Example 3 Convert 2.64 to a percent. Solution To convert to a percent, simply move the decimal two places to the right. 2.64 = 264%. Does a percentage greater than 100 even make sense? Sure it does—percentages greater than 100 come up in real life all the time. For example, a business that made 10 million dollars last year and 13 million dollars this year would have made 130% as much money this year as it did last year. The only situation where a percentage greater than 100 doesn’t make sense is when you’re talking about dividing up something that you only have a ﬁxed amount of—for example, if you took a survey and found that 56% of the respondents gave one answer and 72% gave another answer (for a total of 128%), you’d know something went wrong with your math somewhere, because there’s no way you could have gotten answers from more than 100% of the people you surveyed. 131 www.ck12.org Converting percentages to decimals is just as easy as converting decimals to percentages—simply move the decimal to the left instead of to the right. Example 4 Convert 58% to a decimal. Solution The decimal point here is invisible—it’s right after the 8. So moving it to the left two places gives us 0.58. It can be hard to remember which way to move the decimal point when converting from decimals to percents or vice versa. One way to check if you’re moving it the right way is to check whether your answer is a bigger or smaller number than you started out with. If you’re converting from percents to decimals, you should end up with a smaller number—just think of how a number like 50 percent, where 50 is greater than 1, represents a fraction like 1 2 (or 0.50 in decimal form), where 1 2 is less than 1. Conversely, if you’re converting from decimals to percents, you should end up with a bigger number. One way you might remember this is by remembering that a percent sign is bigger than a decimal point—so percents should be bigger numbers than decimals. Example 5 Convert 3.4 to a percent. Solution If you move the decimal point to the left, you get 0.034%. That’s a smaller number than you started out with, but you’re moving from decimals to percents, so you want the number to get bigger, not smaller. Move it to the right instead to get 340%. Now let’s try another fraction. Example 6 Convert 2 7 to a percent. Solution 2 7 doesn’t convert easily unless you change it to a decimal ﬁrst. 2 divided by 7 is approximately 0.285714..., and moving the decimal and rounding gives us 28.6%. The following Khan Academy video shows several more examples of ﬁnding percents and might be useful for further practice: Use the Percent Equation The percent equation is often used to solve problems. It goes like this: Rate × Total = Part or R% of Total is Part Rate is the ratio that the percent represents (R% in the second version). Total is often called the base unit. Part is the amount we are comparing with the base unit. Example 7 Find 25% of $80. www.ck12.org 132 Solution We are looking for the part. The total is $80. ‘of’ means multiply. R% is 25%, so we can use the second form of the equation: 25% of $80 is Part, or 0.25 × 80 = Part. 0.25 × 80 = 20, so the Part we are looking for is $20. Example 8 Express $90 as a percentage of $160. Solution This time we are looking for the rate. We are given the part ($90) and the total ($160). Using the rate equation, we get Rate × 160 = 90. Dividing both sides by 160 tells us that the rate is 0.5625, or 56.25%. Example 9 $50 is 15% of what total sum? This time we are looking for the total. We are given the part ($50) and the rate (15%, or 0.15). Using the rate equation, we get 0.15 × Total = $50. Dividing both sides by 0.15, we get Total = 50 0.15 ≈333.33. So $50 is 15% of $333.33. Find Percent of Change A useful way to express changes in quantities is through percents. You’ve probably seen signs such as “20% extra free,” or “save 35% today.” When we use percents to represent a change, we generally use the formula Percent change = ﬁnal amount - original amount original amount × 100% or percent change 100 = actual change original amount This means that a positive percent change is an increase, while a negative change is a decrease. Example 10 A school of 500 students is expecting a 20% increase in students next year. How many students will the school have? Solution First let’s solve this using the ﬁrst formula. Since the 20% change is an increase, we represent it in the formula as 20 (if it were a decrease, it would be -20.) Plugging in all the numbers, we get 20% = ﬁnal amount −500 500 × 100% Dividing both sides by 100%, we get 0.2 = ﬁnal amount−500 500 . Multiplying both sides by 500 gives us 100 = ﬁnal amount −500. Then adding 500 to both sides gives us 600 as the ﬁnal number of students. How about if we use the second formula? Then we get 20 100 = actual change 500 . (Reducing the ﬁrst fraction to 1 5 will make the problem easier, so let’s rewrite the equation as 1 5 = actual change 500 133 www.ck12.org Cross multiplying is our next step; that gives us 500 = 5 × (actual change). Dividing by 5 tells us the change is equal to 100. We were told this was an increase, so if we start out with 500 students, after an increase of 100 we know there will be a total of 600. Markup A markup is an increase from the price a store pays for an item from its supplier to the retail price it charges to the public. For example, a 100% mark-up (commonly known in business as keystone) means that the price is doubled. Half of the retail price covers the cost of the item from the supplier, half is proﬁt. Example 11 A furniture store places a 30% markup on everything it sells. It oﬀers its employees a 20% discount from the sales price. The employees are demanding a 25% discount, saying that the store would still make a proﬁt. The manager says that at a 25% discount from the sales price would cause the store to lose money. Who is right? Solution We’ll consider this problem two ways. First, let’s consider an item that the store buys from its supplier for a certain price, say $1000. The markup would be 30% of 1000, or $300, so the item would sell for $1300 and the store would make a $300 proﬁt. And what if an employee buys the product? With a discount of 20%, the employee would pay 80% of the $1300 retail price, or 0.8 × $1300 = $1040. But with a 25% discount, the employee would pay 75% of the retail price, or 0.75 × $1300 = $975. So with a 20% employee discount, the store still makes a $40 proﬁt on the item they bought for $1000—but with a 25% employee discount, the store loses $25 on the item. Now let’s use algebra to see how this works for an item of any price. If x is the price of an item, then the store’s markup is 30% of x, or 0.3x, and the retail price of the item is x + 0.3x, or 1.3x. An employee buying the item at a 20% discount would pay 0.8 × 1.3x = 1.04x, while an employee buying it at a 25% discount would pay 0.75 × 1.3x = 0.975x. So the manager is right: a 20% employee discount still allows the store to make a proﬁt, while a 25% employee discount would cause the store to lose money. It may not seem to make sense that the store would lose money after applying a 30% markup and only a 25% discount. The reason it does work out that way is that the discount is bigger in absolute dollars after the markup is factored in. That is, an employee getting 25% oﬀan item is getting 25% oﬀthe original price plus 25% oﬀthe 30% markup, and those two numbers together add up to more than 30% of the original price. Solve Real-World Problems Using Percents Example 12 In 2004 the US Department of Agriculture had 112071 employees, of which 87846 were Caucasian. Of the remaining minorities, African-American and Hispanic employees had the two largest demographic groups, with 11754 and 6899 employees respectively.∗ a) Calculate the total percentage of minority (non-Caucasian) employees at the USDA. b) Calculate the percentage of African-American employees at the USDA. c) Calculate the percentage of minority employees who were neither African-American nor Hispanic. Solution www.ck12.org 134 a) Use the percent equation Rate × Total = Part. The total number of employees is 112071. We know that the number of Caucasian employees is 87846, which means that there must be 112071 −87646 = 24225 non-Caucasian employees. This is the part. Plugging in the total and the part, we get Rate × 112071 = 24225. Divide both sides by 112071 to get Rate = 24225 112071 ≈0.216. Multiply by 100 to get this as a percent: 21.6%. 21.6% of USDA employees in 2004 were from minority groups. b) Here, the total is still 112071 and the part is 11754, so we have Rate × 112071 = 11754. Dividing, we get Rate = 11754 112071 ≈0.105, or 10.5%. 10.5% of USDA employees in 2004 were African-American. c) Here, our total is just the number of non-Caucasian employees, which we found out is 24225. Subtracting the African-American and Hispanic employees leaves 24225 −11754 −6899 = 5572 employees in the group we’re looking at. So with 24225 for the whole and 5572 for the part, our equation is Rate × 24225 = 5572, or Rate = 5572 24225 ≈ 0.230, or 23%. 23% of USDA minority employees in 2004 were neither African-American nor Hispanic. Example 13 In 1995 New York had 18136000 residents. There were 827025 reported crimes, of which 152683 were violent. By 2005 the population was 19254630 and there were 85839 violent crimes out of a total of 491829 reported crimes. (Source: New York Law Enforcement Agency Uniform Crime Reports.) Calculate the percentage change from 1995 to 2005 in: a) Population of New York b) Total reported crimes c) violent crimes Solution This is a percentage change problem. Remember the formula for percentage change: Percent change = ﬁnal amount - original amount original amount × 100% In these problems, the ﬁnal amount is the 2005 statistic, and the initial amount is the 1995 statistic. a) Population: Percent change = 19254630 −18136000 18136000 × 100% = 1118630 18136000 × 100% ≈0.0617 × 100% = 6.17% The population grew by 6.17%. b) Total reported crimes: 135 www.ck12.org Percent change = 491829 −827025 827025 × 100% = −335196 827025 × 100% ≈−0.4053 × 100% = −40.53% The total number of reported crimes fell by 40.53%. c) Violent crimes: Percent change = 85839 −152683 152683 × 100% = −66844 152683 × 100% ≈−0.4377 × 100% = −43.77% The total number of violent crimes fell by 43.77%. Lesson Summary • A percent is simply a ratio with a base unit of 100—for example, 13% = 13 100. • The percent equation is Rate × Total = Part, or R% of Total is Part. • The percent change equation is Percent change = ﬁnal amount - original amount original amount × 100%. A positive percent change means the value increased, while a negative percent change means the value de-creased. Review Questions 1. Express the following decimals as a percent. (a) 0.011 (b) 0.001 (c) 0.91 (d) 1.75 (e) 20 2. Express the following percentages in decimal form. (a) 15% (b) 0.08% (c) 222% (d) 3.5% (e) 341.9% 3. Express the following fractions as a percent (round to two decimal places when necessary). (a) 1 6 (b) 5 24 (c) 6 7 www.ck12.org 136 (d) 11 7 (e) 13 97 4. Express the following percentages as a reduced fraction. (a) 11% (b) 65% (c) 16% (d) 12.5% (e) 87.5% 5. Find the following. (a) 30% of 90 (b) 16.7% of 199 (c) 11.5% of 10.01 (d) y% of 3x 6. A TV is advertised on sale. It is 35% oﬀand now costs $195. What was the pre-sale price? 7. An employee at a store is currently paid $9.50 per hour. If she works a full year she gets a 12% pay raise. What will her new hourly rate be after the raise? 8. Store A and Store B both sell bikes, and both buy bikes from the same supplier at the same prices. Store A has a 40% mark-up for their prices, while store B has a 250% mark-up. Store B has a permanent sale and will always sell at 60% oﬀthe marked-up prices. Which store oﬀers the better deal? Texas Instruments Resources In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http: //www.ck12.org/flexr/chapter/9613. 137 www.ck12.org Chapter 4 Graphs of Equations and Functions 4.1 The Coordinate Plane Learning Objectives • Identify coordinates of points. • Plot points in a coordinate plane. • Graph a function given a table. • Graph a function given a rule. Introduction Lydia lives 2 blocks north and one block east of school; Travis lives three blocks south and two blocks west of school. What’s the shortest line connecting their houses? The Coordinate Plane We’ve seen how to represent numbers using number lines; now we’ll see how to represent sets of numbers using a coordinate plane. The coordinate plane can be thought of as two number lines that meet at right angles. The horizontal line is called the x−axis and the vertical line is the y−axis. Together the lines are called the axes, and the point at which they cross is called the origin. The axes split the coordinate plane into four quadrants, which are numbered sequentially (I, II, III, IV) moving counter-clockwise from the upper right. www.ck12.org 138 Identify Coordinates of Points When given a point on a coordinate plane, it’s easy to determine its coordinates. The coordinates of a point are two numbers - written together they are called an ordered pair. The numbers describe how far along the x−axis and y−axis the point is. The ordered pair is written in parentheses, with the x−coordinate (also called the abscissa) ﬁrst and the y−coordinate (or the ordinate) second. (1, 7) An ordered pair with an x −value of one and a y −value of seven (0, 5) An ordered pair with an x −value of zero and a y −value of ﬁve (−2.5, 4) An ordered pair with an x −value of -2.5 and a y −value of four (−107.2, −.005) An ordered pair with an x −value of -107.2 and a y −value of −.005 Identifying coordinates is just like reading points on a number line, except that now the points do not actually lie on the number line! Look at the following example. Example 1 Find the coordinates of the point labeled P in the diagram above Solution Imagine you are standing at the origin (the point where the x−axis meets the y−axis). In order to move to a position where P was directly above you, you would move 3 units to the right (we say this is in the positive x−direction). The x−coordinate of P is +3. Now if you were standing at the 3 marker on the x−axis, point P would be 7 units above you (above the axis means it is in the positive y direction). 139 www.ck12.org The y−coordinate of P is +7. The coordinates of point P are (3, 7). Example 2 Find the coordinates of the points labeled Q and R in the diagram to the right. Solution In order to get to Q we move three units to the right, in the positive x−direction, then two units down. This time we are moving in the negative y−direction. The x−coordinate of Q is +3, the y−coordinate of Q is −2. The coordinates of R are found in a similar way. The x−coordinate is +5 (ﬁve units in the positive x−direction) and the y−coordinate is again −2. The coordinates of Q are (3, −2). The coordinates of R are (5, −2). Example 3 Triangle ABC is shown in the diagram to the right. Find the coordinates of the vertices A, B and C. Point A: x −coordinate = −2 y −coordinate = +5 Point B: x −coordinate = +3 y −coordinate = −3 www.ck12.org 140 Point C: x −coordinate = −4 y −coordinate = −1 Solution A(−2, 5) B(3, −3) C(−4, −1) Plot Points in a Coordinate Plane Plotting points is simple, once you understand how to read coordinates and read the scale on a graph. As a note on scale, in the next two examples pay close attention to the labels on the axes. Example 4 Plot the following points on the coordinate plane. A(2, 7) B(−4, 6) D(−3, −3) E(0, 2) F(7, −5) Point A(2, 7) is 2 units right, 7 units up. It is in Quadrant I. Point B(−4, 6) is 4 units left, 6 units up. It is in Quadrant II. Point D(−3, −3) is 3 units left, 3 units down. It is in Quadrant III. Point E(0, 2) is 2 units up from the origin. It is right on the y−axis, between Quadrants I and II. Point F(7, −5) is 7 units right, 5 units down. It is in Quadrant IV. Example 5 Plot the following points on the coordinate plane. A(2.5, 0.5) B(π, 1.2) C(2, 1.75) D(0.1, 1.2) E(0, 0) 141 www.ck12.org Here we see the importance of choosing the right scale and range for the graph. In Example 4, our points were scattered throughout the four quadrants. In this case, all the coordinates are positive, so we don’t need to show the negative values of x or y. Also, there are no x−values bigger than about 3.14, and 1.75 is the largest value of y. We can therefore show just the part of the coordinate plane where 0 ≤x ≤3.5 and 0 ≤y ≤2. Here are some other important things to notice about this graph: • The tick marks on the axes don’t correspond to unit increments (i.e. the numbers do not go up by one each time). This is so that we can plot the points more precisely. • The scale on the x−axis is diﬀerent than the scale on the y−axis, so distances that look the same on both axes are actually greater in the x−direction. Stretching or shrinking the scale in one direction can be useful when the points we want to plot are farther apart in one direction than the other. For more practice locating and naming points on the coordinate plane, try playing the Coordinate Plane Game at Graph a Function Given a Table Once we know how to plot points on a coordinate plane, we can think about how we’d go about plotting a relationship between x−and y−values. So far we’ve just been plotting sets of ordered pairs. A set like that is a relation, and there isn’t necessarily a relationship between the x−values and y−values. If there is a relationship between the x−and y−values, and each x−value corresponds to exactly one y−value, then the relation is called a function. Remember that a function is a particular way to relate one quantity to another. If you’re reading a book and can read twenty pages an hour, there is a relationship between how many hours you read and how many pages you read. You may even know that you could write the formula as either n = 20h or h = n 20, where h is the number of hours you spend reading and n is the number of pages you read. To ﬁnd out, for example, how many pages you could read in 31 2 hours, or how many hours it would take you to read 46 pages, you could use one of those formulas. Or, you could make a graph of the function: www.ck12.org 142 Once you know how to graph a function like this, you can simply read the relationship between the x−and y−values oﬀthe graph. You can see in this case that you could read 70 pages in 31 2 hours, and it would take you about 2 1 3 hours to read 46 pages. Generally, the graph of a function appears as a line or curve that goes through all points that have the relationship that the function describes. If the domain of the function (the set of x−values we can plug into the function) is all real numbers, then we call it a continuous function. If the domain of the function is a particular set of values (such as whole numbers only), then it is called a discrete function. The graph will be a series of dots, but they will still often fall along a line or curve. In graphing equations, we assume the domain is all real numbers, unless otherwise stated. Often, though, when we look at data in a table, the domain will be whole numbers (number of presents, number of days, etc.) and the function will be discrete. But sometimes we’ll still draw the graph as a continuous line to make it easier to interpret. Be aware of the diﬀerence between discrete and continuous functions as you work through the examples. Example 6 Sarah is thinking of the number of presents she receives as a function of the number of friends who come to her birthday party. She knows she will get a present from her parents, one from her grandparents and one each from her uncle and aunt. She wants to invite up to ten of her friends, who will each bring one present. She makes a table of how many presents she will get if one, two, three, four or ﬁve friends come to the party. Plot the points on a coordinate plane and graph the function that links the number of presents with the number of friends. Use your graph to determine how many presents she would get if eight friends show up. Table 4.1: Number of Friends Number of Presents 0 4 1 5 2 6 3 7 4 8 5 9 The ﬁrst thing we need to do is decide how our graph should appear. We need to decide what the independent variable is, and what the dependant variable is. Clearly in this case, the number of friends can vary independently, but the number of presents must depend on the number of friends who show up. 143 www.ck12.org So we’ll plot friends on the x−axis and presents on the y−axis. Let’s add another column to our table containing the coordinates that each (friends, presents) ordered pair gives us. Table 4.2: Friends (x) Presents (y) Coordinates (x, y) 0 4 (0, 4) 1 5 (1, 5) 2 6 (2, 6) 3 7 (3, 7) 4 8 (4, 8) 5 9 (5, 9) Next we need to set up our axes. It is clear that the number of friends and number of presents both must be positive, so we only need to show points in Quadrant I. Now we need to choose a suitable scale for the x−and y−axes. We only need to consider eight friends (look again at the question to conﬁrm this), but it always pays to allow a little extra room on your graph. We also need the y−scale to accommodate the presents for eight people. We can see that this is still going to be under 20! The scale of this graph has room for up to 12 friends and 15 presents. This will be ﬁne, but there are many other scales that would be equally good! Now we proceed to plot the points. The ﬁrst ﬁve points are the coordinates from our table. You can see they all lie on a straight line, so the function that describes the relationship between x and y will be linear. To graph the function, we simply draw a line that goes through all ﬁve points. This line represents the function. This is a discrete problem since Sarah can only invite a positive whole number of friends. For instance, it would be impossible for 2.4 or -3 friends to show up. So although the line helps us see where the other values of the function are, the only points on the line that actually are values of the function are the ones with positive whole-number coordinates. The graph easily lets us ﬁnd other values for the function. For example, the question asks how many presents Sarah would get if eight friends come to her party. Don’t forget that x represents the number of friends and y represents the number of presents. If we look at the graph where x = 8, we can see that the function has a y−value of 12. Solution If 8 friends show up, Sarah will receive a total of 12 presents. www.ck12.org 144 Graph a Function Given a Rule If we are given a rule instead of a table, we can proceed to graph the function in either of two ways. We will use the following example to show each way. Example 7 Ali is trying to work out a trick that his friend showed him. His friend started by asking him to think of a number, then double it, then add ﬁve to the result. Ali has written down a rule to describe the ﬁrst part of the trick. He is using the letter x to stand for the number he thought of and the letter y to represent the ﬁnal result of applying the rule. He wrote his rule in the form of an equation: y = 2x + 5. Help him visualize what is going on by graphing the function that this rule describes. Method One - Construct a Table of Values If we wish to plot a few points to see what is going on with this function, then the best way is to construct a table and populate it with a few (x, y) pairs. We’ll use 0, 1, 2 and 3 for x−values. Table 4.3: x y 0 5 1 7 2 9 3 11 Next, we plot the points and join them with a line. This method is nice and simple—especially with linear relationships, where we don’t need to plot more than two or three points to see the shape of the graph. In this case, the function is continuous because the domain is all real numbers—that is, Ali could think of any real number, even though he may only be thinking of positive whole numbers. Method Two - Intercept and Slope 145 www.ck12.org Another way to graph this function (one that we’ll learn in more detail in a later lesson) is the slope-intercept method. To use this method, follow these steps: 1. Find the y value when y = 0. y(0) = 2 · 0 + 5 = 5, so our y−intercept is (0, 5). 2. Look at the coeﬀicient multiplying the x. Every time we increase x by one, y increases by two, so our slope is +2. 3. Plot the line with the given slope that goes through the intercept. We start at the point (0, 5) and move over one in the x−direction, then up two in the y−direction. This gives the slope for our line, which we extend in both directions. We will properly examine this last method later in this chapter! Lesson Summary • The coordinate plane is a two-dimensional space deﬁned by a horizontal number line (the x−axis) and a vertical number line (the y−axis). The origin is the point where these two lines meet. Four areas, or quadrants, are formed as shown in the diagram above. • Each point on the coordinate plane has a set of coordinates, two numbers written as an ordered pair which describe how far along the x−axis and y−axis the point is. The x−coordinate is always written ﬁrst, then the y−coordinate, in the form (x, y). • Functions are a way that we can relate one quantity to another. Functions can be plotted on the coordinate plane. www.ck12.org 146 Review Questions 1. Identify the coordinates of each point, A −F, on the graph below. 2. Draw a line on the above graph connecting point B with the origin. Where does that line intersect the line connecting points C and D? 3. Plot the following points on a graph and identify which quadrant each point lies in: (a) (4, 2) (b) (-3, 5.5) (c) (4, -4) (d) (-2, -3) 4. Without graphing the following points, identify which quadrant each lies in: (a) (5, 3) (b) (-3, -5) (c) (-4, 2) (d) (2, -4) 5. Consider the graph of the equation y = 3. Which quadrants does it pass through? 6. Consider the graph of the equation y = x. Which quadrants does it pass through? 7. Consider the graph of the equation y = x + 3. Which quadrants does it pass through? 8. The point (4, 0) is on the boundary between which two quadrants? 9. The point (0, -5) is on the boundary between which two quadrants? 10. If you moved the point (3, 2) ﬁve units to the left, what quadrant would it be in? 11. The following three points are three vertices of square ABCD. Plot them on a graph, then determine what the coordinates of the fourth point, D, would be. Plot that point and label it. A(−4, −4) B(3, −4) C(3, 3) 12. In what quadrant is the center of the square from problem 10? (You can ﬁnd the center by drawing the square’s diagonals.) 13. What point is halfway between (1, 3) and (1, 5)? 14. What point is halfway between (2, 8) and (6, 8)? 15. What point is halfway between the origin and (10, 4)? 16. What point is halfway between (3, -2) and (-3, 2)? 17. Becky has a large bag of M&Ms that she knows she should share with Jaeyun. Jaeyun has a packet of Starburst. Becky tells Jaeyun that for every Starburst he gives her, she will give him three M&Ms in return. If x is the number of Starburst that Jaeyun gives Becky, and y is the number of M&Ms he gets in return, then complete each of the following. (a) Write an algebraic rule for y in terms of x. (b) Make a table of values for y with x-values of 0, 1, 2, 3, 4, 5. 147 www.ck12.org (c) Plot the function linking x and y on the following scale: 0 ≤x ≤10, 0 ≤y ≤10. 4.2 Graphs of Linear Equations Learning Objectives • Graph a linear function using an equation. • Write equations and graph horizontal and vertical lines. • Analyze graphs of linear functions and read conversion graphs. Introduction You’re stranded downtown late at night with only $8 in your pocket, and your home is 6 miles away. Two cab companies serve this area; one charges $1.20 per mile with an additional $1 fee, and the other charges $0.90 per mile with an additional $2.50 fee. Which cab will be able to get you home? Graph a Linear Equation At the end of Lesson 4.1 we looked at ways to graph a function from a rule. A rule is a way of writing the relationship between the two quantities we are graphing. In mathematics, we tend to use the words formula and equation to describe the rules we get when we express relationships algebraically. Interpreting and graphing these equations is an important skill that you’ll use frequently in math. Example 1 A taxi costs more the further you travel. Taxis usually charge a fee on top of the per-mile charge to cover hire of the vehicle. In this case, the taxi charges $3 as a set fee and $0.80 per mile traveled. Here is the equation linking the cost in dollars (y) to hire a taxi and the distance traveled in miles (x). y = 0.8x + 3 Graph the equation and use your graph to estimate the cost of a seven-mile taxi ride. Solution We’ll start by making a table of values. We will take a few values for x (0, 1, 2, 3, and 4), ﬁnd the corresponding y−values, and then plot them. Since the question asks us to ﬁnd the cost for a seven-mile journey, we need to choose a scale that can accommodate this. First, here’s our table of values: Table 4.4: x y 0 3 1 3.8 2 4.6 3 5.4 4 6.2 www.ck12.org 148 And here’s our graph: To ﬁnd the cost of a seven-mile journey, ﬁrst we ﬁnd x = 7 on the horizontal axis and draw a line up to our graph. Next, we draw a horizontal line across to the y−axis and read where it hits. It appears to hit around half way between y = 8 and y = 9. Let’s call it 8.5. A seven mile taxi ride would cost approximately $8.50 ($8.60 exactly). Here are some things you should notice about this graph and the formula that generated it: • The graph is a straight line (this means that the equation is linear), although the function is discrete and really just consists of a series of points. • The graph crosses the y−axis at y = 3 (notice that there’s a + 3 in the equation—that’s not a coincidence!). This is the base cost of the taxi. • Every time we move over by one square we move up by 0.8 squares (notice that that’s also the coeﬀicient of x in the equation). This is the rate of charge of the taxi (cost per mile). • If we move over by three squares, we move up by 3 × 0.8 squares. Example 2 A small business has a debt of $500,000 incurred from start-up costs. It predicts that it can pay oﬀthe debt at a rate of $85,000 per year according to the following equation governing years in business (x) and debt measured in thousands of dollars (y). y = −85x + 500 Graph the above equation and use your graph to predict when the debt will be fully paid. Solution First, we start with our table of values: Table 4.5: x y 0 500 1 415 2 330 3 245 4 160 149 www.ck12.org Then we plot our points and draw the line that goes through them: Notice the scale we’ve chosen here. There’s no need to include any points above y = 500, but it’s still wise to allow a little extra. Next we need to determine how many years it takes the debt to reach zero, or in other words, what x−value will make the y−value equal 0. We know it’s greater than four (since at x = 4 the y−value is still positive), so we need an x−scale that goes well past x = 4. Here we’ve chosen to show the x−values from 0 to 12, though there are many other places we could have chosen to stop. To read the time that the debt is paid oﬀ, we simply read the point where the line hits y = 0 (the x−axis). It looks as if the line hits pretty close to x = 6. So the debt will deﬁnitely be paid oﬀin six years. To see more simple examples of graphing linear equations by hand, see the Khan Academy video on graphing lines at The narrator shows how to graph several linear equations, using a table of values to plot points and then connecting the points with a line. Graphs and Equations of Horizontal and Vertical Lines Example 3 “Mad-cabs” have an unusual oﬀer going on. They are charging $7.50 for a taxi ride of any length within the city limits. Graph the function that relates the cost of hiring the taxi (y) to the length of the journey in miles (x). To proceed, the ﬁrst thing we need is an equation. You can see from the problem that the cost of a journey doesn’t depend on the length of the journey. It should come as no surprise that the equation then, does not have x in it. Since any value of x results in the same value of y(7.5), the value you choose for x doesn’t matter, so it isn’t included in the equation. Here is the equation: y = 7.5 The graph of this function is shown below. You can see that it’s simply a horizontal line. www.ck12.org 150 Any time you see an equation of the form “y = constant,” the graph is a horizontal line that intercepts the y−axis at the value of the constant. Similarly, when you see an equation of the form x = constant, then the graph is a vertical line that intercepts the x−axis at the value of the constant. (Notice that that kind of equation is a relation, and not a function, because each x−value (there’s only one in this case) corresponds to many (actually an inﬁnite number) y−values.) Example 4 Plot the following graphs. (a) y = 4 (b) y = −4 (c) x = 4 (d) x = −4 (a) y = 4 is a horizontal line that crosses the y−axis at 4. (b) y = −4 is a horizontal line that crosses the y−axis at −4. (c) x = 4 is a vertical line that crosses the x−axis at 4. (d) x = −4 is a vertical line that crosses the x−axis at −4. Example 5 Find an equation for the x−axis and the y−axis. Look at the axes on any of the graphs from previous examples. We have already said that they intersect at the origin (the point where x = 0 and y = 0). The following deﬁnition could easily work for each axis. 151 www.ck12.org x−axis: A horizontal line crossing the y−axis at zero. y−axis: A vertical line crossing the x−axis at zero. So using example 3 as our guide, we could deﬁne the x−axis as the line y = 0 and the y−axis as the line x = 0. Analyze Graphs of Linear Functions We often use graphs to represent relationships between two linked quantities. It’s useful to be able to interpret the information that graphs convey. For example, the chart below shows a ﬂuctuating stock price over ten weeks. You can read that the index closed the ﬁrst week at about $68, and at the end of the third week it was at about $62. You may also see that in the ﬁrst ﬁve weeks it lost about 20% of its value, and that it made about 20% gain between weeks seven and ten. Notice that this relationship is discrete, although the dots are connected to make the graph easier to interpret. Analyzing graphs is a part of life - whether you are trying to decide to buy stock, ﬁgure out if your blog readership is increasing, or predict the temperature from a weather report. Many graphs are very complicated, so for now we’ll start oﬀwith some simple linear conversion graphs. Algebra starts with basic relationships and builds to more complicated tasks, like reading the graph above. Example 6 Below is a graph for converting marked prices in a downtown store into prices that include sales tax. Use the graph to determine the cost including sales tax for a $6.00 pen in the store. To ﬁnd the relevant price with tax, ﬁrst ﬁnd the correct pre-tax price on the x−axis. This is the point x = 6. Draw the line x = 6 up until it meets the function, then draw a horizontal line to the y−axis. This line hits at y ≈6.75 (about three fourths of the way from y = 6 to y = 7). www.ck12.org 152 The approximate cost including tax is $6.75. Example 7 The chart for converting temperature from Fahrenheit to Celsius is shown to the right. Use the graph to convert the following: a) 70◦Fahrenheit to Celsius b) 0◦Fahrenheit to Celsius c) 30◦Celsius to Fahrenheit d) 0◦Celsius to Fahrenheit Solution a) To ﬁnd 70◦Fahrenheit, we look along the Fahrenheit-axis (in other words the x−axis) and draw the line x = 70 up to the function. Then we draw a horizontal line to the Celsius-axis (y−axis). The horizontal line hits the axis at a little over 20 (21 or 22). 70◦Fahrenheit is approximately equivalent to 21◦Celsius. b) To ﬁnd 0◦Fahrenheit, we just look at the y−axis. (Don’t forget that this axis is simply the line x = 0.) The line hits the y−axis just below the half way point between −15 and −20. 0◦Fahrenheit is approximately equivalent to −18◦Celsius. c) To ﬁnd 30◦Celsius, we look up the Celsius-axis and draw the line y = 30 along to the function. When this horizontal line hits the function, we draw a line straight down to the Fahrenheit-axis. The line hits the axis at approximately 85. 30◦Celsius is approximately equivalent to 85◦Fahrenheit. d) To ﬁnd 0◦Celsius, we look at the Fahrenheit-axis (the line y = 0). The function hits the x−axis just right of 30. 0◦Celsius is equivalent to 32◦Fahrenheit. Lesson Summary • Equations with the variables y and x can be graphed by making a chart of values that ﬁt the equation and then plotting the values on a coordinate plane. This graph is simply another representation of the equation and can be analyzed to solve problems. • Horizontal lines are deﬁned by the equation y = constant and vertical lines are deﬁned by the equation x = constant. • Be aware that although we graph the function as a line to make it easier to interpret, the function 153 www.ck12.org may actually be discrete. Review Questions 1. Make a table of values for the following equations and then graph them. (a) y = 2x + 7 (b) y = 0.7x −4 (c) y = 6 −1.25x 2. “Think of a number. Multiply it by 20, divide the answer by 9, and then subtract seven from the result.” (a) Make a table of values and plot the function that represents this sentence. (b) If you picked 0 as your starting number, what number would you end up with? (c) To end up with 12, what number would you have to start out with? 3. Write the equations for the ﬁve lines (A through E) plotted in the graph below. 4. In the graph above, at what points do the following lines intersect? (a) A and E (b) A and D (c) C and D (d) B and the y−axis (e) E and the x−axis (f) C and the line y = x (g) E and the line y = 1 2 x (h) A and the line y = x + 3 5. At the airport, you can change your money from dollars into euros. The service costs $5, and for every additional dollar you get 0.7 euros. (a) Make a table for this and plot the function on a graph. (b) Use your graph to determine how many euros you would get if you give the oﬀice $50. (c) To get 35 euros, how many dollars would you have to pay? (d) The exchange rate drops so that you can only get 0.5 euros per additional dollar. Now how many dollars do you have to pay for 35 euros? 6. The graph below shows a conversion chart for converting between weight in kilograms and weight in pounds. Use it to convert the following measurements. www.ck12.org 154 (a) 4 kilograms into weight in pounds (b) 9 kilograms into weight in pounds (c) 12 pounds into weight in kilograms (d) 17 pounds into weight in kilograms 7. Use the graph from problem 6 to answer the following questions. (a) An employee at a sporting goods store is packing 3-pound weights into a box that can hold 8 kilograms. How many weights can she place in the box? (b) After packing those weights, there is some extra space in the box that she wants to ﬁll with one-pound weights. How many of those can she add? (c) After packing those, she realizes she misread the label and the box can actually hold 9 kilograms. How many more one-pound weights can she add? 4.3 Graphing Using Intercepts Learning Objectives • Find intercepts of the graph of an equation. • Use intercepts to graph an equation. • Solve real-world problems using intercepts of a graph Introduction Sanjit’s oﬀice is 25 miles from home, and in traﬀic he expects the trip home to take him an hour if he starts at 5 PM. Today he hopes to stop at the post oﬀice along the way. If the post oﬀice is 6 miles from his oﬀice, when will Sanjit get there? 155 www.ck12.org If you know just one of the points on a line, you’ll ﬁnd that isn’t enough information to plot the line on a graph. As you can see in the graph above, there are many lines—in fact, inﬁnitely many lines—that pass through a single point. But what if you know two points that are both on the line? Then there’s only one way to graph that line; all you need to do is plot the two points and use a ruler to draw the line that passes through both of them. There are a lot of options for choosing which two points on the line you use to plot it. In this lesson, we’ll focus on two points that are rather convenient for graphing: the points where our line crosses the x−and y−axes, or intercepts. We’ll see how to ﬁnd intercepts algebraically and use them to quickly plot graphs. Look at the graph above. The y−intercept occurs at the point where the graph crosses the y−axis. The y−value at this point is 8, and the x−value is 0. Similarly, the x−intercept occurs at the point where the graph crosses the x−axis. The x−value at this point is 6, and the y−value is 0. So we know the coordinates of two points on the graph: (0, 8) and (6, 0). If we’d just been given those two coordinates out of the blue, we could quickly plot those points and join them with a line to recreate the above graph. Note: Not all lines will have both an x−and a y−intercept, but most do. However, horizontal lines never cross the x−axis and vertical lines never cross the y−axis. For examples of these special cases, see the graph below. www.ck12.org 156 Finding Intercepts by Substitution Example 1 Find the intercepts of the line y = 13 −x and use them to graph the function. Solution The ﬁrst intercept is easy to ﬁnd. The y−intercept occurs when x = 0. Substituting gives us y = 13−0 = 13, so the y−intercept is (0, 13). Similarly, the x−intercept occurs when y = 0. Plugging in 0 for y gives us 0 = 13 −x, and adding x to both sides gives us x = 13. So (13, 0) is the x−intercept. To draw the graph, simply plot these points and join them with a line. Example 2 Graph the following functions by ﬁnding intercepts. a) y = 2x + 3 b) y = 7 −2x c) 4x −2y = 8 d) 2x + 3y = −6 Solution a) Find the y−intercept by plugging in x = 0 : y = 2 · 0 + 3 = 3 −the y −intercept is (0, 3) 157 www.ck12.org Find the x−intercept by plugging in y = 0 : 0 = 2x + 3 −subtract 3 from both sides : −3 = 2x −divide by 2 : −3 2 = x −the x −intercept is (−1.5, 0) b) Find the y−intercept by plugging in x = 0 : y = 7 −2 · 0 = 7 −the y −intercept is (0, 7) Find the x−intercept by plugging in y = 0 : 0 = 7 −2x −subtract 7 from both sides : −7 = −2x −divide by −2 : 7 2 = x −the x −intercept is (3.5, 0) c) Find the y−intercept by plugging in x = 0 : 4 · 0 −2y = 8 −2y = 8 −divide by −2 y = −4 −the y −intercept is (0, −4) Find the x−intercept by plugging in y = 0 : www.ck12.org 158 4x −2 · 0 = 8 4x = 8 −divide by 4 : x = 2 −the x −intercept is (2, 0) d) Find the y−intercept by plugging in x = 0 : 2 · 0 + 3y = −6 3y = −6 −divide by 3 : y = −2 −the y −intercept is (0, −2) Find the x−intercept by plugging in y = 0 : 2x + 3 · 0 = −6 2x = −6 −divide by 2 : x = −3 −the x −intercept is (−3, 0) Finding Intercepts for Standard Form Equations Using the Cover-Up Method Look at the last two equations in example 2. These equations are written in standard form. Standard form equations are always written “coeﬀicient times x plus (or minus) coeﬀicient times y equals value”. In other words, they look like this: 159 www.ck12.org ax + by = c where a has to be positive, but b and c do not. There is a neat method for ﬁnding intercepts in standard form, often referred to as the cover-up method. Example 3 Find the intercepts of the following equations: a) 7x −3y = 21 b) 12x −10y = −15 c) x + 3y = 6 Solution To solve for each intercept, we realize that at the intercepts the value of either x or y is zero, and so any terms that contain that variable eﬀectively drop out of the equation. To make a term disappear, simply cover it (a ﬁnger is an excellent way to cover up terms) and solve the resulting equation. a) To solve for the y−intercept we set x = 0 and cover up the x−term: −3y = 21 y = −7 (0, −7) is the y −intercept. Now we solve for the x−intercept: 7x = 21 x = 3 (3, 0) is the x −intercept. b) To solve for the y−intercept (x = 0), cover up the x−term: −10y = −15 y = 1.5 (0, 1.5) is the y −intercept. Now solve for the x−intercept (y = 0): 12x = −15 x = −5 4 (−1.25, 0) is the x −intercept. c) To solve for the y−intercept (x = 0), cover up the x−term: www.ck12.org 160 3y = 6 y = 2 (0, 2) is the y −intercept. Solve for the y−intercept: x = 6 (6, 0) is the x −intercept. The graph of these functions and the intercepts is below: To learn more about equations in standard form, try the Java applet at line/line.htm (scroll down and click the “click here to start” button.) You can use the sliders to change the values of a, b, and c and see how that aﬀects the graph. Solving Real-World Problems Using Intercepts of a Graph Example 4 Jesus has $30 to spend on food for a class barbecue. Hot dogs cost $0.75 each (including the bun) and burgers cost $1.25 (including the bun). Plot a graph that shows all the combinations of hot dogs and burgers he could buy for the barbecue, without spending more than $30. This time we will ﬁnd an equation ﬁrst, and then we can think logically about ﬁnding the intercepts. If the number of burgers that Jesus buys is x, then the money he spends on burgers is 1.25x If the number of hot dogs he buys is y, then the money he spends on hot dogs is 0.75y So the total cost of the food is 1.25x + 0.75y. The total amount of money he has to spend is $30, so if he is to spend it ALL, we can use the following equation: 1.25x + 0.75y = 30 We can solve for the intercepts using the cover-up method. First the y−intercept (x = 0): 0.75y = 30 y = 40 y −intercept: (0, 40) 161 www.ck12.org Then the x−intercept (y = 0): 1.25x = 30 x = 24 x −intercept: (24, 0) Now we plot those two points and join them to create our graph, shown here: We could also have created this graph without needing to come up with an equation. We know that if John were to spend ALL the money on hot dogs, he could buy 30 .75 = 40 hot dogs. And if he were to buy only burgers he could buy 30 1.25 = 24 burgers. From those numbers, we can get 2 intercepts: (0 burgers, 40 hot dogs) and (24 burgers, 0 hot dogs). We could plot these just as we did above and obtain our graph that way. As a ﬁnal note, we should realize that Jesus’ problem is really an example of an inequality. He can, in fact, spend any amount up to $30. The only thing he cannot do is spend more than $30. The graph above reﬂects this: the line is the set of solutions that involve spending exactly $30, and the shaded region shows solutions that involve spending less than $30. We’ll work with inequalities some more in Chapter 6. Lesson Summary • A y−intercept occurs at the point where a graph crosses the y−axis (where x = 0) and an x−intercept occurs at the point where a graph crosses the x−axis (where y = 0). • The y−intercept can be found by substituting x = 0 into the equation and solving for y. Likewise, the x−intercept can be found by substituting y = 0 into the equation and solving for x. • A linear equation is in standard form if it is written as “positive coeﬀicient times x plus coeﬀicient times y equals value”. Equations in standard form can be solved for the intercepts by covering up the x (or y) term and solving the equation that remains. Review Questions 1. Find the intercepts for the following equations using substitution. (a) y = 3x −6 (b) y = −2x + 4 (c) y = 14x −21 (d) y = 7 −3x www.ck12.org 162 (e) y = 2.5x −4 (f) y = 1.1x + 2.2 (g) y = 3 8 x + 7 (h) y = 5 9 −2 7 x 2. Find the intercepts of the following equations using the cover-up method. (a) 5x −6y = 15 (b) 3x −4y = −5 (c) 2x + 7y = −11 (d) 5x + 10y = 25 (e) 5x −1.3y = 12 (f) 1.4x −3.5y = 7 (g) 3 5 x + 2y = 2 5 (h) 3 4 x −2 3y = 1 5 3. Use any method to ﬁnd the intercepts and then graph the following equations. (a) y = 2x + 3 (b) 6(x −1) = 2(y + 3) (c) x −y = 5 (d) x + y = 8 4. At the local grocery store strawberries cost $3.00 per pound and bananas cost $1.00 per pound. (a) If I have $10 to spend on strawberries and bananas, draw a graph to show what combinations of each I can buy and spend exactly $10. (b) Plot the point representing 3 pounds of strawberries and 2 pounds of bananas. Will that cost more or less than $10? (c) Do the same for the point representing 1 pound of strawberries and 5 pounds of bananas. 5. A movie theater charges $7.50 for adult tickets and $4.50 for children. If the theater takes in $900 in ticket sales for a particular screening, draw a graph which depicts the possibilities for the number of adult tickets and the number of child tickets sold. 6. Why can’t we use the intercept method to graph the following equation? 3(x + 2) = 2(y + 3) 7. Name two more equations that we can’t use the intercept method to graph. 4.4 Slope and Rate of Change Learning Objectives • Find positive and negative slopes. • Recognize and ﬁnd slopes for horizontal and vertical lines. • Understand rates of change. • Interpret graphs and compare rates of change. Introduction Wheelchair ramps at building entrances must have a slope between 1 16 and 1 20. If the entrance to a new oﬀice building is 28 inches oﬀthe ground, how long does the wheelchair ramp need to be? We come across many examples of slope in everyday life. For example, a slope is in the pitch of a roof, the grade or incline of a road, or the slant of a ladder leaning on a wall. In math, we use the word slope to deﬁne steepness in a particular way. 163 www.ck12.org Slope = distance moved vertically distance moved horizontally To make it easier to remember, we often word it like this: Slope = rise run In the picture above, the slope would be the ratio of the height of the hill to the horizontal length of the hill. In other words, it would be 3 4, or 0.75. If the car were driving to the right it would climb the hill - we say this is a positive slope. Any time you see the graph of a line that goes up as you move to the right, the slope is positive. If the car kept driving after it reached the top of the hill, it might go down the other side. If the car is driving to the right and descending, then we would say that the slope is negative. Here’s where it gets tricky: If the car turned around instead and drove back down the left side of the hill, the slope of that side would still be positive. This is because the rise would be -3, but the run would be -4 (think of the x−axis - if you move from right to left you are moving in the negative x−direction). That means our slope ratio would be −3 −4, and the negatives cancel out to leave 0.75, the same slope as before. In other words, the slope of a line is the same no matter which direction you travel along it. Find the Slope of a Line A simple way to ﬁnd a value for the slope of a line is to draw a right triangle whose hypotenuse runs along the line. Then we just need to measure the distances on the triangle that correspond to the rise (the vertical dimension) and the run (the horizontal dimension). Example 1 Find the slopes for the three graphs shown. www.ck12.org 164 Solution There are already right triangles drawn for each of the lines - in future problems you’ll do this part yourself. Note that it is easiest to make triangles whose vertices are lattice points (i.e. points whose coordinates are all integers). a) The rise shown in this triangle is 4 units; the run is 2 units. The slope is 4 2 = 2. b) The rise shown in this triangle is 4 units, and the run is also 4 units. The slope is 4 4 = 1. c) The rise shown in this triangle is 2 units, and the run is 4 units. The slope is 2 4 = 1 2. Example 2 Find the slope of the line that passes through the points (1, 2) and (4, 7). Solution We already know how to graph a line if we’re given two points: we simply plot the points and connect them with a line. Here’s the graph: Since we already have coordinates for the vertices of our right triangle, we can quickly work out that the rise is 7 −2 = 5 and the run is 4 −1 = 3 (see diagram). So the slope is 7−2 4−1 = 5 3. If you look again at the calculations for the slope, you’ll notice that the 7 and 2 are the y−coordinates of the two points and the 4 and 1 are the x−coordinates. This suggests a pattern we can follow to get a general formula for the slope between two points (x1, y1) and (x2, y2): Slope between (x1, y1) and (x2, y2) = y2−y1 x2−x1 or m = ∆y ∆x 165 www.ck12.org In the second equation the letter m denotes the slope (this is a mathematical convention you’ll see often) and the Greek letter delta (∆) means change. So another way to express slope is change in y divided by change in x. In the next section, you’ll see that it doesn’t matter which point you choose as point 1 and which you choose as point 2. Example 3 Find the slopes of the lines on the graph below. Solution Look at the lines - they both slant down (or decrease) as we move from left to right. Both these lines have negative slope. The lines don’t pass through very many convenient lattice points, but by looking carefully you can see a few points that look to have integer coordinates. These points have been circled on the graph, and we’ll use them to determine the slope. We’ll also do our calculations twice, to show that we get the same slope whichever way we choose point 1 and point 2. For Line A: (x1, y1) = (−6, 3) (x2, y2) = (5, −1) (x1, y1) = (5, −1) (x2, y2) = (−6, 3) m = y2 −y1 x2 −x1 = (−1) −(3) (5) −(−6) = −4 11 ≈−0.364 m = y2 −y1 x2 −x1 = (3) −(−1) (−6) −(5) = 4 −11 ≈−0.364 For Line B (x1, y1) = (−4, 6) (x2, y2) = (4, −5) (x1, y1) = (4, −5) (x2, y2) = (−4, 6) m = y2 −y1 x2 −x1 = (−5) −(6) (4) −(−4) = −11 8 = −1.375 m = y2 −y1 x2 −x1 = (6) −(−5) (−4) −(4) = 11 −8 = −1.375 You can see that whichever way round you pick the points, the answers are the same. Either way, Line A has slope -0.364, and Line B has slope -1.375. Khan Academy has a series of videos on ﬁnding the slope of a line, starting at watch?v=hXP1Gv9IMBo. Find the Slopes of Horizontal and Vertical lines Example 4 Determine the slopes of the two lines on the graph below. www.ck12.org 166 Solution There are 2 lines on the graph: A(y = 3) and B(x = 5). Let’s pick 2 points on line A—say, (x1, y1) = (−4, 3) and (x2, y2) = (5, 3)—and use our equation for slope: m = y2 −y1 x2 −x1 = (3) −(3) (5) −(−4) = 0 9 = 0. If you think about it, this makes sense - if y doesn’t change as x increases then there is no slope, or rather, the slope is zero. You can see that this must be true for all horizontal lines. Horizontal lines (y = constant) all have a slope of 0. Now let’s consider line B. If we pick the points (x1, y1) = (5, −3) and (x2, y2) = (5, 4), our slope equation is m = y2−y1 x2−x1 = (4)−(−3) (5)−(5) = 7 0. But dividing by zero isn’t allowed! In math we often say that a term which involves division by zero is undeﬁned. (Technically, the answer can also be said to be inﬁnitely large—or inﬁnitely small, depending on the problem.) Vertical lines (x = constant) all have an inﬁnite (or undeﬁned) slope. Find a Rate of Change The slope of a function that describes real, measurable quantities is often called a rate of change. In that case the slope refers to a change in one quantity (y) per unit change in another quantity (x). (This is where the equation m = ∆y ∆x comes in—remember that ∆y and ∆x represent the change in y and x respectively.) Example 5 A candle has a starting length of 10 inches. 30 minutes after lighting it, the length is 7 inches. Determine the rate of change in length of the candle as it burns. Determine how long the candle takes to completely burn to nothing. Solution First we’ll graph the function to visualize what is happening. We have 2 points to start with: we know that at the moment the candle is lit (time = 0) the length of the candle is 10 inches, and after 30 minutes (time = 30) the length is 7 inches. Since the candle length depends on the time, we’ll plot time on the horizontal axis, and candle length on the vertical axis. 167 www.ck12.org The rate of change of the candle’s length is simply the slope of the line. Since we have our 2 points (x1, y1) = (0, 10) and (x2, y2) = (30, 7), we can use the familiar version of the slope formula: Rate of change = y2 −y1 x2 −x1 = (7 inches) −(10 inches) (30 minutes) −(0 minutes) = −3 inches 30 minutes = −0.1 inches per minute Note that the slope is negative. A negative rate of change means that the quantity is decreasing with time—just as we would expect the length of a burning candle to do. To ﬁnd the point when the candle reaches zero length, we can simply read the x−intercept oﬀthe graph (100 minutes). We can use the rate equation to verify this algebraically: Length burned = rate × time 10 = 0.1 × 100 Since the candle length was originally 10 inches, our equation conﬁrms that 100 minutes is the time taken. Example 6 The population of ﬁsh in a certain lake increased from 370 to 420 over the months of March and April. At what rate is the population increasing? Solution Here we don’t have two points from which we can get x−and y−coordinates for the slope formula. Instead, we’ll need to use the alternate formula, m = ∆y ∆x. The change in y−values, or ∆y, is the change in the number of ﬁsh, which is 420 −370 = 50. The change in x−values, ∆x, is the amount of time over which this change took place: two months. So ∆y ∆x = 50 ﬁsh 2 months, or 25 ﬁsh per month. Interpret a Graph to Compare Rates of Change Example 7 The graph below represents a trip made by a large delivery truck on a particular day. During the day the truck made two deliveries, one taking an hour and the other taking two hours. Identify what is happening at each stage of the trip (stages A through E). www.ck12.org 168 Solution Here are the stages of the trip: a) The truck sets oﬀand travels 80 miles in 2 hours. b) The truck covers no distance for 2 hours. c) The truck covers (120 −80) = 40 miles in 1 hour. d) The truck covers no distance for 1 hour. e) The truck covers -120 miles in 2 hours. Let’s look at each section more closely. A. Rate of change = ∆y ∆x = 80 miles 2 hours = 40 miles per hour Notice that the rate of change is a speed—or rather, a velocity. (The diﬀerence between the two is that velocity has a direction, and speed does not. In other words, velocity can be either positive or negative, with negative velocity representing travel in the opposite direction. You’ll see the diﬀerence more clearly in part E.) Since velocity equals distance divided by time, the slope (or rate of change) of a distance-time graph is always a velocity. So during the ﬁrst part of the trip, the truck travels at a constant speed of 40 mph for 2 hours, covering a distance of 80 miles. B. The slope here is 0, so the rate of change is 0 mph. The truck is stationary for one hour. This is the ﬁrst delivery stop. C. Rate of change = ∆y ∆x = (120−80) miles (4−3) hours = 40 miles per hour. The truck is traveling at 40 mph. D. Once again the slope is 0, so the rate of change is 0 mph. The truck is stationary for two hours. This is the second delivery stop. At this point the truck is 120 miles from the start position. E. Rate of change = ∆y ∆x = (0−120) miles (8−6) hours = −120 miles 2 hours = −60 miles per hour. The truck is traveling at negative 60 mph. Wait – a negative speed? Does that mean that the truck is reversing? Well, probably not. It’s actually the velocity and not the speed that is negative, and a negative velocity simply means that the distance from the starting position is decreasing with time. The truck is driving in the opposite direction – back to where it started from. Since it no longer has 2 heavy loads, it travels faster (60 mph instead of 40 mph), covering the 120 mile return trip in 2 hours. Its speed is 60 mph, and its velocity is -60 mph, because it is traveling in the opposite direction from when it started out. 169 www.ck12.org Lesson Summary • Slope is a measure of change in the vertical direction for each step in the horizontal direction. Slope is often represented as “m”. • Slope can be expressed as rise run, or ∆y ∆x. • The slope between two points (x1, y1) and (x2, y2) is equal to y2−y1 x2−x1 . • Horizontal lines (where y = a constant) all have a slope of 0. • Vertical lines (where x = a constant) all have an inﬁnite (or undeﬁned) slope. • The slope (or rate of change) of a distance-time graph is a velocity. Review Questions 1. Use the slope formula to ﬁnd the slope of the line that passes through each pair of points. (a) (-5, 7) and (0, 0) (b) (-3, -5) and (3, 11) (c) (3, -5) and (-2, 9) (d) (-5, 7) and (-5, 11) (e) (9, 9) and (-9, -9) (f) (3, 5) and (-2, 7) (g) (2.5, 3) and (8, 3.5) 2. For each line in the graphs below, use the points indicated to determine the slope. www.ck12.org 170 3. For each line in the graphs above, imagine another line with the same slope that passes through the point (1, 1), and name one more point on that line. 4. The graph below is a distance-time graph for Mark’s three and a half mile cycle ride to school. During this ride, he rode on cycle paths but the terrain was hilly. He rode slower up hills and faster down them. He stopped once at a traﬀic light and at one point he stopped to mend a punctured tire. The graph shows his distance from home at any given time. Identify each section of the graph accordingly. 4.5 Graphs Using Slope-Intercept Form Learning Objectives • Identify the slope and y−intercept of equations and graphs. • Graph an equation in slope-intercept form. • Understand what happens when you change the slope or intercept of a line. • Identify parallel lines from their equations. Introduction The total proﬁt of a business is described by the equation y = 15000x −80000, where x is the number of months the business has been running. How much proﬁt is the business making per month, and what were its start-up costs? How much proﬁt will it have made in a year? 171 www.ck12.org Identify Slope and y−intercept So far, we’ve been writing a lot of our equations in slope-intercept form—that is, we’ve been writing them in the form y = mx + b, where m and b are both constants. It just so happens that m is the slope and the point (0, b) is the y−intercept of the graph of the equation, which gives us enough information to draw the graph quickly. Example 1 Identify the slope and y−intercept of the following equations. a) y = 3x + 2 b) y = 0.5x −3 c) y = −7x d) y = −4 Solution a) Comparing , we can see that m = 3 and b = 2. So y = 3x + 2 has a slope of 3 and a y−intercept of (0, 2). b) has a slope of 0.5 and a y−intercept of (0, -3). Notice that the intercept is negative. The b−term includes the sign of the operator (plus or minus) in front of the number—for example, y = 0.5x −3 is identical to y = 0.5x + (−3), and that means that b is -3, not just 3. c) At ﬁrst glance, this equation doesn’t look like it’s in slope-intercept form. But we can rewrite it as y = −7x + 0, and that means it has a slope of -7 and a y−intercept of (0, 0). Notice that the slope is negative and the line passes through the origin. d) We can rewrite this one as y = 0x −4, giving us a slope of 0 and a y−intercept of (0, -4). This is a horizontal line. Example 2 Identify the slope and y−intercept of the lines on the graph shown below. www.ck12.org 172 The intercepts have been marked, as well as some convenient lattice points that the lines pass through. Solution a) The y−intercept is (0, 5). The line also passes through (2, 3), so the slope is ∆y ∆x = −2 2 = −1. b) The y−intercept is (0, 2). The line also passes through (1, 5), so the slope is ∆y ∆x = 3 1 = 3. c) The y−intercept is (0, -1). The line also passes through (2, 3), so the slope is ∆y ∆x = 4 2 = 2. d) The y−intercept is (0, -3). The line also passes through (4, -4), so the slope is ∆y ∆x = −1 4 = −1 4 or -0.25. Graph an Equation in Slope-Intercept Form Once we know the slope and intercept of a line, it’s easy to graph it. Just remember what slope means. Let’s look back at this example from Lesson 4.1. Ali is trying to work out a trick that his friend showed him. His friend started by asking him to think of a number, then double it, then add ﬁve to the result. Ali has written down a rule to describe the ﬁrst part of the trick. He is using the letter x to stand for the number he thought of and the letter y to represent the ﬁnal result of applying the rule. He wrote his rule in the form of an equation: y = 2x + 5. Help him visualize what is going on by graphing the function that this rule describes. In that example, we constructed a table of values, and used that table to plot some points to create our graph. We also saw another way to graph this equation. Just by looking at the equation, we could see that the y−intercept was (0, 5), so we could start by plotting that point. Then we could also see that the slope was 2, so we could ﬁnd another point on the graph by going over 1 unit and up 2 units. The graph would then be the line between those two points. 173 www.ck12.org Here’s another problem where we can use the same method. Example 3 Graph the following function: y = −3x + 5 Solution To graph the function without making a table, follow these steps: 1. Identify the y−intercept: b = 5 2. Plot the intercept: (0, 5) 3. Identify the slope: m = −3. (This is equal to −3 1 , so the rise is -3 and the run is 1.) 4. Move over 1 unit and down 3 units to ﬁnd another point on the line: (1, 2) 5. Draw the line through the points (0, 5) and (1, 2). Notice that to graph this equation based on its slope, we had to ﬁnd the rise and run—and it was easiest to do that when the slope was expressed as a fraction. That’s true in general: to graph a line with a particular slope, it’s easiest to ﬁrst express the slope as a fraction in simplest form, and then read oﬀthe numerator and the denominator of the fraction to get the rise and run of the graph. Example 4 Find integer values for the rise and run of the following slopes, then graph lines with corresponding slopes. a) m = 3 b) m = −2 c) m = 0.75 d) m = −0.375 Solution a) www.ck12.org 174 b) c) d) Changing the Slope or Intercept of a Line The following graph shows a number of lines with diﬀerent slopes, but all with the same y−intercept: (0, 3). 175 www.ck12.org You can see that all the functions with positive slopes increase as we move from left to right, while all functions with negative slopes decrease as we move from left to right. Another thing to notice is that the greater the slope, the steeper the graph. This graph shows a number of lines with the same slope, but diﬀerent y−intercepts. Notice that changing the intercept simply translates (shifts) the graph up or down. Take a point on the graph of y = 2x, such as (1, 2). The corresponding point on y = 2x + 3 would be (1, 5). Adding 3 to the y−intercept means we also add 3 to every other y−value on the graph. Similarly, the corresponding point on the y = 2x−3 line would be (1, -1); we would subtract 3 from the y−value and from every other y−value. Notice also that these lines all appear to be parallel. Are they truly parallel? To answer that question, we’ll use a technique that you’ll learn more about in a later chapter. We’ll take 2 of the equations—say, y = 2x and y = 2x + 3—and solve for values of x and y that satisfy both equations. That will tell us at what point those two lines intersect, if any. (Remember that parallel lines, by deﬁnition, are lines that don’t intersect.) So what values would satisfy both y = 2x and y = 2x + 3? Well, if both of those equations were true, then y would be equal to both 2x and 2x + 3, which means those two expressions would also be equal to each other. So we can get our answer by solving the equation 2x = 2x + 3. But what happens when we try to solve that equation? If we subtract 2x from both sides, we end up with 0 = 3. That can’t be true no matter what x equals. And that means that there just isn’t any value for x that will make both of the equations we started out with true. In other words, there isn’t any point where those two lines intersect. They are parallel, just as we thought. And we’d ﬁnd out the same thing no matter which two lines we’d chosen. In general, since changing the intercept of a line just results in shifting the graph up or down, the new line will always be parallel to the old line as long as the slope stays the same. www.ck12.org 176 Any two lines with identical slopes are parallel. Further Practice To get a better understanding of what happens when you change the slope or the y−intercept of a linear equation, try playing with the Java applet at 7.5/index.htm. Lesson Summary • A common form of a line (linear equation) is slope-intercept form: y = mx + b, where m is the slope and the point (0, b) is the y−intercept • Graphing a line in slope-intercept form is a matter of ﬁrst plotting the y−intercept (0, b), then ﬁnding a second point based on the slope, and using those two points to graph the line. • Any two lines with identical slopes are parallel. Review Questions 1. Identify the slope and y−intercept for the following equations. (a) y = 2x + 5 (b) y = −0.2x + 7 (c) y = x (d) y = 3.75 2. Identify the slope of the following lines. 3. Identify the slope and y−intercept for the following functions. 4. Plot the following functions on a graph. (a) y = 2x + 5 (b) y = −0.2x + 7 (c) y = x 177 www.ck12.org (d) y = 3.75 5. Which two of the following lines are parallel? (a) y = 2x + 5 (b) y = −0.2x + 7 (c) y = x (d) y = 3.75 (e) y = −1 5 x −11 (f) y = −5x + 5 (g) y = −3x + 11 (h) y = 3x + 3.5 6. What is the y−intercept of the line passing through (1, -4) and (3, 2)? 7. What is the y−intercept of the line with slope -2 that passes through (3, 1)? 8. Line A passes through the points (2, 6) and (-4, 3). Line B passes through the point (3, 2.5), and is parallel to line A (a) Write an equation for line A in slope-intercept form. (b) Write an equation for line B in slope-intercept form. 9. Line C passes through the points (2, 5) and (1, 3.5). Line D is parallel to line C, and passes through the point (2, 6). Name another point on line D. (Hint: you can do this without graphing or ﬁnding an equation for either line.) 4.6 Direct Variation Models Learning Objectives • Identify direct variation. • Graph direct variation equations. • Solve real-world problems using direct variation models. Introduction Suppose you see someone buy ﬁve pounds of strawberries at the grocery store. The clerk weighs the strawberries and charges $12.50 for them. Now suppose you wanted two pounds of strawberries for yourself. How much would you expect to pay for them? Identify Direct Variation The preceding problem is an example of a direct variation. We would expect that the strawberries are priced on a “per pound” basis, and that if you buy two-ﬁfths the amount of strawberries, you would pay two-ﬁfths of $12.50 for your strawberries, or $5.00. Similarly, if you bought 10 pounds of strawberries (twice the amount) you would pay twice $12.50, and if you did not buy any strawberries you would pay nothing. If variable y varies directly with variable x, then we write the relationship as y = k · x. k is called the constant of proportionality. If we were to graph this function, you can see that it would pass through the origin, because y = 0 when x = 0,whatever the value of k. So we know that a direct variation, when graphed, has a single intercept at (0, 0). www.ck12.org 178 Example 1 If y varies directly with x according to the relationship y = k · x, and y = 7.5 when x = 2.5, determine the constant of proportionality, k. Solution We can solve for the constant of proportionality using substitution. Substitute x = 2.5 and y = 7.5 into the equation y = k · x to get 7.5 = k(2.5). Then divide both sides by 2.5 to get k = 7.5 2.5 = 3. The constant of proportionality, k, is 3. We can graph the relationship quickly, using the intercept (0, 0) and the point (2.5, 7.5). The graph is shown below. It is a straight line with slope 3. The graph of a direct variation always passes through the origin, and always has a slope that is equal to the constant of proportionality, k. Example 2 The volume of water in a ﬁsh-tank, V, varies directly with depth, d. If there are 15 gallons in the tank when the depth is 8 inches, calculate how much water is in the tank when the depth is 20 inches. Solution This is a good example of a direct variation, but for this problem we’ll have to determine the equation of the variation ourselves. Since the volume, V, depends on depth, d, we’ll use an equation of the form y = k · x, but in place of y we’ll use V and in place of x we’ll use d: V = k · d We know that when the depth is 8 inches the volume is 15 gallons, so to solve for k, we plug in 15 for V and 8 for d to get 15 = k(8). Dividing by 8 gives us k = 15 8 = 1.875. Now to ﬁnd the volume of water at the ﬁnal depth, we use V = k · d again, but this time we can plug in our new d and the value we found for k: V = 1.875 × 20 V = 37.5 At a depth of 20 inches, the volume of water in the tank is 37.5 gallons. Example 3 The graph shown below is a conversion chart used to convert U.S. dollars (US$) to British pounds (GB£) in a bank on a particular day. Use the chart to determine: 179 www.ck12.org a) the number of pounds you could buy for $600 b) the number of dollars it would cost to buy £200 c) the exchange rate in pounds per dollar Solution We can read the answers to a) and b) right oﬀthe graph. It looks as if at x = 600 the graph is about one ﬁfth of the way between £350 and £400. So $600 would buy £360. Similarly, the line y = 200 appears to intersect the graph about a third of the way between $300 and $400. We can round this to $330, so it would cost approximately $330 to buy £200. To solve for the exchange rate, we should note that as this is a direct variation - the graph is a straight line passing through the origin. The slope of the line gives the constant of proportionality (in this case the exchange rate) and it is equal to the ratio of the y−value to x−value at any point. Looking closely at the graph, we can see that the line passes through one convenient lattice point: (500, 300). This will give us the most accurate value for the slope and so the exchange rate. y = k · x ⇒= y x And so rate = 300 pounds 500 dollors = 0.60 pounds per dollar Graph Direct Variation Equations We know that all direct variation graphs pass through the origin, and also that the slope of the line is equal to the constant of proportionality, k. Graphing is a simple matter of using the point-slope or point-point methods discussed earlier in this chapter. Example 4 Plot the following direct variations on the same graph. a) y = 3x b) y = −2x c) y = −0.2x d) y = 2 9 x Solution www.ck12.org 180 a) The line passes through (0, 0), as will all these functions. This function has a slope of 3. When we move across by one unit, the function increases by three units. b) The line has a slope of -2. When we move across the graph by one unit, the function falls by two units. c) The line has a slope of -0.2. As a fraction this is equal to −1 5. When we move across by ﬁve units, the function falls by one unit. d) The line passes through (0, 0) and has a slope of 2 9. When we move across the graph by 9 units, the function increases by two units. For more examples of how to plot and identify direct variation functions, see the video at http:// neaportal.k12.ar.us/index.php/2010/06/slope-and-direct-variation/. Solve Real-World Problems Using Direct Variation Models Direct variations are seen everywhere in everyday life. Any time one quantity increases at the same rate another quantity increases (for example, doubling when it doubles and tripling when it triples), we say that they follow a direct variation. Newton’s Second Law In 1687 Sir Isaac Newton published the famous Principia Mathematica. It contained, among other things, his second law of motion. This law is often written as F = m · a, where a force of F Newtons applied to a mass of m kilograms results in acceleration of a meters per second2. Notice that if the mass stays constant, then this formula is basically the same as the direct variation equation, just with diﬀerent variables—and m is the constant of proportionality. Example 5 If a 175 Newton force causes a shopping cart to accelerate down the aisle with an acceleration of 2.5 m/s2, calculate: a) The mass of the shopping cart. b) The force needed to accelerate the same cart at 6 m/s2. Solution a) We can solve for m (the mass) by plugging in our given values for force and acceleration. F = m · a becomes 175 = m(2.5), and then we divide both sides by 2.5 to get 70 = m. So the mass of the shopping cart is 70 kg. b) Once we have solved for the mass, we simply substitute that value, plus our required acceleration, back into the formula F = m · a and solve for F. We get F = 70 × 6 = 420. 181 www.ck12.org So the force needed to accelerate the cart at 6 m/s2 is 420 Newtons. Ohm’s Law The electrical current, I (amps), passing through an electronic component varies directly with the applied voltage, V (volts), according to the relationship V = I · R, where R is the resistance (measured in Ohms). The resistance is considered to be a constant for all values of V and I, so once again, this formula is a version of the direct variation formula, with R as the constant of proportionality. Example 6 A certain electronics component was found to pass a current of 1.3 amps at a voltage of 2.6 volts. When the voltage was increased to 12.0 volts the current was found to be 6.0 amps. a) Does the component obey Ohm’s law? b) What would the current be at 6 volts? Solution Ohm’s law is a simple direct proportionality law, with the resistance R as our constant of proportionality. To know if this component obeys Ohm’s law, we need to know if it follows a direct proportionality rule. In other words, is V directly proportional to I? We can determine this in two diﬀerent ways. Graph It: If we plot our two points on a graph and join them with a line, does the line pass through (0, 0)? Voltage is the independent variable and current is the dependent variable, so normally we would graph V on the horizontal axis and I on the vertical axis. However, if we swap the variables around just this once, we’ll get a graph whose slope conveniently happens to be equal to the resistance, R. So we’ll treat I as the independent variable, and our two points will be (1.3, 2.6) and (6, 12). Plotting those points and joining them gives the following graph: The graph does appear to pass through the origin, so yes, the component obeys Ohm’s law. Solve for R: If this component does obey Ohm’s law, the constant of proportionality (R) should be the same when we plug in the second set of values as when we plug in the ﬁrst set. Let’s see if it is. (We can quickly ﬁnd the value of R in each case; since V = I · R, that means R = V I .) www.ck12.org 182 Case 1: R = V I = 2.6 1.3 = 2 Ohms Case 2: R = V I = 12 6 = 2 Ohms The values for R agree! This means that we are indeed looking at a direct variation. The component obeys Ohm’s law. b) Now to ﬁnd the current at 6 volts, simply substitute the values for V and R into V = I · R. We found that R = 2, so we plug in 2 for R and 6 for V to get 6 = I(2), and divide both sides by 2 to get 3 = I. So the current through the component at a voltage of 6 volts is 3 amps. Lesson Summary • If a variable y varies directly with variable x, then we write the relationship as y = k · x, where k is a constant called the constant of proportionality. • Direct variation is very common in many areas of science. Review Questions 1. Plot the following direct variations on the same graph. (a) y = 4 3 x (b) y = −2 3 x (c) y = −1 6 x (d) y = 1.75x 2. Dasan’s mom takes him to the video arcade for his birthday. (a) In the ﬁrst 10 minutes, he spends $3.50 playing games. If his allowance for the day is $20, how long can he keep playing games before his money is gone? (b) He spends the next 15 minutes playing Alien Invaders. In the ﬁrst two minutes, he shoots 130 aliens. If he keeps going at this rate, how many aliens will he shoot in ﬁfteen minutes? (c) The high score on this machine is 120000 points. If each alien is worth 100 points, will Dasan beat the high score? What if he keeps playing for ﬁve more minutes? 3. The current standard for low-ﬂow showerheads is 2.5 gallons per minute. (a) How long would it take to ﬁll a 30-gallon bathtub using such a showerhead to supply the water? (b) If the bathtub drain were not plugged all the way, so that every minute 0.5 gallons ran out as 2.5 gallons ran in, how long would it take to ﬁll the tub? (c) After the tub was full and the showerhead was turned oﬀ, how long would it take the tub to empty through the partly unplugged drain? (d) If the drain were immediately unplugged all the way when the showerhead was turned oﬀ, so that it drained at a rate of 1.5 gallons per minute, how long would it take to empty? 4. Amin is using a hose to ﬁll his new swimming pool for the ﬁrst time. He starts the hose at 10 PM and leaves it running all night. (a) At 6 AM he measures the depth and calculates that the pool is four sevenths full. At what time will his new pool be full? (b) At 10 AM he measures again and realizes his earlier calculations were wrong. The pool is still only three quarters full. When will it actually be full? 183 www.ck12.org (c) After ﬁlling the pool, he needs to chlorinate it to a level of 2.0 ppm (parts per million). He adds two gallons of chlorine solution and ﬁnds that the chlorine level is now 0.7 ppm. How many more gallons does he need to add? (d) If the chlorine level in the pool decreases by 0.05 ppm per day, how much solution will he need to add each week? 5. Land in Wisconsin is for sale to property investors. A 232-acre lot is listed for sale for $200,500. (a) Assuming the same price per acre, how much would a 60-acre lot sell for? (b) Again assuming the same price, what size lot could you purchase for $100,000? 6. The force (F) needed to stretch a spring by a distance x is given by the equation F = k · x, where k is the spring constant (measured in Newtons per centimeter, or N/cm). If a 12 Newton force stretches a certain spring by 10 cm, calculate: (a) The spring constant, k (b) The force needed to stretch the spring by 7 cm. (c) The distance the spring would stretch with a 23 Newton force. 7. Angela’s cell phone is completely out of power when she puts it on the charger at 3 PM. An hour later, it is 30% charged. When will it be completely charged? 8. It costs $100 to rent a recreation hall for three hours and $150 to rent it for ﬁve hours. (a) Is this a direct variation? (b) Based on the cost to rent the hall for three hours, what would it cost to rent it for six hours, assuming it is a direct variation? (c) Based on the cost to rent the hall for ﬁve hours, what would it cost to rent it for six hours, assuming it is a direct variation? (d) Plot the costs given for three and ﬁve hours and graph the line through those points. Based on that graph, what would you expect the cost to be for a six-hour rental? 4.7 Linear Function Graphs Learning Objectives • Recognize and use function notation. • Graph a linear function. • Analyze arithmetic progressions. Introduction The highly exclusive Fellowship of the Green Mantle allows in only a limited number of new members a year. In its third year of membership it has 28 members, in its fourth year it has 33, and in its ﬁfth year it has 38. How many members are admitted a year, and how many founding members were there? Functions So far we’ve used the term function to describe many of the equations we’ve been graphing, but in mathematics it’s important to remember that not all equations are functions. In order to be a function, a relationship between two variables, x and y, must map each x−value to exactly one y−value. Visually this means the graph of y versus x must pass the vertical line test, meaning that a vertical line drawn through the graph of the function must never intersect the graph in more than one place: www.ck12.org 184 Use Function Notation When we write functions we often use the notation “ f(x) =” in place of “y =”. f(x) is pronounced “ f of x”. Example 1 Rewrite the following equations so that y is a function of x and is written f(x): a) y = 2x + 5 b) y = −0.2x + 7 c) x = 4y −5 d) 9x + 3y = 6 Solution a) Simply replace y with f(x) : f(x) = 2x + 5 b) Again, replace y with f(x) : f(x) = −0.2x + 7 c) First we need to solve for y. Starting with x = 4y −5, we add 5 to both sides to get x + 5 = 4y, divide by 4 to get x+5 4 = y, and then replace y with f(x) : f(x) = x+5 4 . d) Solve for y : take 9x + 3y = 6, subtract 9x from both sides to get 3y = 6 −9x, divide by 3 to get y = 6−9x 3 = 2 −3x, and express as a function: f(x) = 2 −3x. Using the functional notation in an equation gives us more information. For instance, the expression f(x) = mx + b shows clearly that x is the independent variable because you plug in values of x into the function and perform a series of operations on the value of x in order to calculate the values of the dependent variable, y. We can also plug in expressions rather than just numbers. For example, if our function is f(x) = x + 2, we can plug in the expression (x + 5). We would express this as f(x + 5) = (x + 5) + 2 = x + 7. Example 2 A function is deﬁned as f(x) = 6x −36. Evaluate the following: a) f(2) b) f(0) c) f(z) d) f(x + 3) e) f(2r −1) Solution a) Substitute x = 2 into the function f(x) : f(2) = 6 · 2 −36 = 12 −36 = −24 b) Substitute x = 0 into the function f(x) : f(0) = 6 · 0 −36 = 0 −36 = −36 185 www.ck12.org c) Substitute x = z into the function f(x) : f(z) = 6z + 36 d) Substitute x = (x + 3) into the function f(x) : f(x + 3) = 6(x + 3) + 36 = 6x + 18 + 36 = 6x + 54 e) Substitute x = (2r + 1) into the function f(x) : f(2r + 1) = 6(2r + 1) + 36 = 12r + 6 + 36 = 12r + 42 Graph a Linear Function Since the notations “ f(x) =” and “y =” are interchangeable, we can use all the concepts we have learned so far to graph functions. Example 3 Graph the function f(x) = 3x+5 4 . Solution We can write this function in slope-intercept form: f(x) = 3 4 x + 5 4 = 0.75x + 1.25 So our graph will have a y−intercept of (0, 1.25) and a slope of 0.75. Example 4 Graph the function f(x) = 7(5−x) 5 . Solution This time we’ll solve for the x−and y−intercepts. To solve for the y−intercept, plug in x = 0 : f(0) = 7(5−0) 5 = 35 5 = 7, so the x−intercept is (0, 7). To solve for the x−intercept, set f(x) = 0 : 0 = 7(5−x) 5 , so 0 = 35 −7x, therefore 7x = 35 and x = 5. The y−intercept is (5, 0). We can graph the function from those two points: www.ck12.org 186 Arithmetic Progressions You may have noticed that with linear functions, when you increase the x−value by 1 unit, the y−value increases by a ﬁxed amount, equal to the slope. For example, if we were to make a table of values for the function f(x) = 2x + 3, we might start at x = 0 and then add 1 to x for each row: Table 4.6: x f(x) 0 3 1 5 2 7 3 9 4 11 Notice that the values for f(x) go up by 2 (the slope) each time. When we repeatedly add a ﬁxed value to a starting number, we get a sequence like {3, 5, 7, 9, 11....}. We call this an arithmetic progression, and it is characterized by the fact that each number is bigger (or smaller) than the preceding number by a ﬁxed amount. This amount is called the common diﬀerence. We can ﬁnd the common diﬀerence for a given sequence by taking 2 consecutive terms in the sequence and subtracting the ﬁrst from the second. Example 5 Find the common diﬀerence for the following arithmetic progressions: a) {7, 11, 15, 19, ...} b) {12, 1, -10, -21, ...} c) {7, , 12, , 17, ...} Solution a) 11 −7 = 4; 15 −11 = 4; 19 −15 = 4. The common diﬀerence is 4. b) 1 −12 = −11. The common diﬀerence is -11. c) There are not 2 consecutive terms here, but we know that to get the term after 7 we would add the common diﬀerence, and then to get to 12 we would add the common diﬀerence again. So twice the common diﬀerence is 12 −7 = 5, and so the common diﬀerence is 2.5. 187 www.ck12.org Arithmetic sequences and linear functions are very closely related. To get to the next term in a arithmetic sequence, you add the common diﬀerence to the last term; similarly, when the x−value of a linear function increases by one, the y−value increases by the amount of the slope. So arithmetic sequences are very much like linear functions, with the common diﬀerence playing the same role as the slope. The graph below shows the arithmetic progression {-2, 0, 2, 4, 6...} along with the function y = 2x −4. The only major diﬀerence between the two graphs is that an arithmetic sequence is discrete while a linear function is continuous. We can write a formula for an arithmetic progression: if we deﬁne the ﬁrst term as a1 and d as the common diﬀerence, then the other terms are as follows: a1 a2 a3 a4 a5 an a1 a1 + d a1 + 2d a1 + 3d a1 + 4d . . . a1 + (n −1) · d The online calculator at will tell you the nth term in an arithmetic pro-gression if you tell it the ﬁrst term, the common diﬀerence, and what value to use for n (in other words, which term in the sequence you want to know). It will also tell you the sum of all the terms up to that point. Finding sums of sequences is something you will learn to do in future math classes. Lesson Summary • In order for an equation to be a function, the relationship between the two variables, x and y, must map each x−value to exactly one y−value. • The graph of a function of y versus x must pass the vertical line test: any vertical line will only cross the graph of the function in one place. • Functions can be expressed in function notation using f(x) = in place of y =. • The sequence of f(x) values for a linear function form an arithmetic progression. Each number is greater than (or less than) the preceding number by a ﬁxed amount, or common diﬀerence. Review Questions 1. When an object falls under gravity, it gains speed at a constant rate of 9.8 m/s every second. An item dropped from the top of the Eiﬀel Tower, which is 300 meters tall, takes 7.8 seconds to hit the ground. How fast is it moving on impact? 2. A prepaid phone card comes with $20 worth of calls on it. Calls cost a ﬂat rate of $0.16 per minute. (a) Write the value left on the card as a function of minutes used so far. (b) Use the function to determine how many minutes of calls you can make with the card. www.ck12.org 188 3. For each of the following functions evaluate: (a) f(x) = −2x + 3 (b) f(x) = 0.7x + 3.2 (c) f(x) = 5(2−x) 11 i. f(−3) ii. f(0) iii. f(z) iv. f(x + 3) v. f(2n) vi. f(3y + 8) vii. f ( q 2 ) 4. Determine whether the following could be graphs of functions. (a) (b) (c) (d) 5. The roasting guide for a turkey suggests cooking for 100 minutes plus an additional 8 minutes per pound. (a) Write a function for the roasting time the given the turkey weight in pounds (x). (b) Determine the time needed to roast a 10 lb turkey. 189 www.ck12.org (c) Determine the time needed to roast a 27 lb turkey. (d) Determine the maximum size turkey you could roast in 4.5 hours. 6. Determine the missing terms in the following arithmetic progressions. (a) {-11, 17, , 73} (b) {2, , -4} (c) {13, , , __, 0} 4.8 Problem-Solving Strategies - Graphs Learning Objectives • Read and understand given problem situations. • Use the strategy “Read a Graph.” • Use the strategy “Make a Graph.” • Solve real-world problems using selected strategies as part of a plan. Introduction In this chapter, we’ve been solving problems where quantities are linearly related to each other. In this section, we’ll look at a few examples of linear relationships that occur in real-world problems, and see how we can solve them using graphs. Remember back to our Problem Solving Plan: 1. Understand the Problem 2. Devise a Plan—Translate 3. Carry Out the Plan—Solve 4. Look—Check and Interpret Example 1 A cell phone company is oﬀering its costumers the following deal: You can buy a new cell phone for $60 and pay a monthly ﬂat rate of $40 per month for unlimited calls. How much money will this deal cost you after 9 months? Solution Let’s follow the problem solving plan. Step 1: The phone costs $60; the calling plan costs $40 per month. Let x = number of months. Let y = total cost. Step 2: We can solve this problem by making a graph that shows the number of months on the horizontal axis and the cost on the vertical axis. Since you pay $60 when you get the phone, the y−intercept is (0, 60). You pay $40 for each month, so the cost rises by $40 for 1 month, so the slope is 40. We can graph this line using the slope-intercept method. www.ck12.org 190 Step 3: The question was “How much will this deal cost after 9 months?” We can now read the answer from the graph. We draw a vertical line from 9 months until it meets the graph, and then draw a horizontal line until it meets the vertical axis. We see that after 9 months you pay approximately $420. Step 4: To check if this is correct, let’s think of the deal again. Originally, you pay $60 and then $40 a month for 9 months. Phone = $60 Calling plan = $40 × 9 = $360 Total cost = $420. The answer checks out. Example 2 A stretched spring has a length of 12 inches when a weight of 2 lbs is attached to the spring. The same spring has a length of 18 inches when a weight of 5 lbs is attached to the spring. What is the length of the spring when no weights are attached? 191 www.ck12.org Solution Step 1: We know: the length of the spring = 12 inches when weight = 2 lbs the length of the spring = 18 inches when weight = 5 lbs We want: the length of the spring when weight = 0 lbs Let x = the weight attached to the spring. Let y = the length of the spring. Step 2: We can solve this problem by making a graph that shows the weight on the horizontal axis and the length of the spring on the vertical axis. We have two points we can graph: When the weight is 2 lbs, the length of the spring is 12 inches. This gives point (2, 12). When the weight is 5 lbs, the length of the spring is 18 inches. This gives point (5, 18). Graphing those two points and connecting them gives us our line. Step 3: The question was: “What is the length of the spring when no weights are attached? We can answer this question by reading the graph we just made. When there is no weight on the spring, the x−value equals zero, so we are just looking for the y−intercept of the graph. On the graph, the y−intercept appears to be approximately 8 inches. Step 4: To check if this correct, let’s think of the problem again. You can see that the length of the spring goes up by 6 inches when the weight is increased by 3 lbs, so the slope of the line is 6 inches 3 lbs = 2 inches/lb. www.ck12.org 192 To ﬁnd the length of the spring when there is no weight attached, we can look at the spring when there are 2 lbs attached. For each pound we take oﬀ, the spring will shorten by 2 inches. If we take oﬀ2 lbs, the spring will be shorter by 4 inches. So, the length of the spring with no weights is 12 inches - 4 inches = 8 inches. The answer checks out. Example 3 Christine took 1 hour to read 22 pages of Harry Potter. She has 100 pages left to read in order to ﬁnish the book. How much time should she expect to spend reading in order to ﬁnish the book? Solution Step 1: We know - Christine takes 1 hour to read 22 pages. We want - How much time it takes to read 100 pages. Let x = the time expressed in hours. Let y = the number of pages. Step 2: We can solve this problem by making a graph that shows the number of hours spent reading on the horizontal axis and the number of pages on the vertical axis. We have two points we can graph: Christine takes 1 hour to read 22 pages. This gives point (1, 22). A second point is not given, but we know that Christine would take 0 hours to read 0 pages. This gives point (0, 0). Graphing those two points and connecting them gives us our line. Step 3: The question was: “How much time should Christine expect to spend reading 100 pages?” We can ﬁnd the answer from reading the graph - we draw a horizontal line from 100 pages until it meets the graph and then we draw the vertical until it meets the horizontal axis. We see that it takes approximately 4.5 hours to read the remaining 100 pages. Step 4: To check if this correct, let’s think of the problem again. We know that Christine reads 22 pages per hour - this is the slope of the line or the rate at which she is reading. To ﬁnd how many hours it takes her to read 100 pages, we divide the number of pages by the rate. In this case, 100 pages 22 pages/hour = 4.54 hours. This is very close to the answer we got from reading the graph. The answer checks out. Example 4 193 www.ck12.org Aatif wants to buy a surfboard that costs $249. He was given a birthday present of $50 and he has a summer job that pays him $6.50 per hour. To be able to buy the surfboard, how many hours does he need to work? Solution Step 1: We know - The surfboard costs $249. Aatif has $50. His job pays $6.50 per hour. We want - How many hours Aatif needs to work to buy the surfboard. Let x = the time expressed in hours Let y = Aatif’s earnings Step 2: We can solve this problem by making a graph that shows the number of hours spent working on the horizontal axis and Aatif’s earnings on the vertical axis. Aatif has $50 at the beginning. This is the y−intercept: (0, 50). He earns $6.50 per hour. This is the slope of the line. We can graph this line using the slope-intercept method. We graph the y−intercept of (0, 50), and we know that for each unit in the horizontal direction, the line rises by 6.5 units in the vertical direction. Here is the line that describes this situation. Step 3: The question was: “How many hours does Aatif need to work to buy the surfboard?” We ﬁnd the answer from reading the graph - since the surfboard costs $249, we draw a horizontal line from $249 on the vertical axis until it meets the graph and then we draw a vertical line downwards until it meets the horizontal axis. We see that it takes approximately 31 hours to earn the money. Step 4: To check if this correct, let’s think of the problem again. We know that Aatif has $50 and needs $249 to buy the surfboard. So, he needs to earn $249 −$50 = $199 from his job. His job pays $6.50 per hour. To ﬁnd how many hours he need to work we divide: $199 $6.50/hour = 30.6 hours. This is very close to the answer we got from reading the graph. The answer checks out. www.ck12.org 194 Lesson Summary The four steps of the problem solving plan when using graphs are: 1. Understand the Problem 2. Devise a Plan—Translate: Make a graph. 3. Carry Out the Plan—Solve: Use the graph to answer the question asked. 4. Look—Check and Interpret Review Questions Solve the following problems by making a graph and reading it. 1. A gym is oﬀering a deal to new members. Customers can sign up by paying a registration fee of $200 and a monthly fee of $39. (a) How much will this membership cost a member by the end of the year? (b) The old membership rate was $49 a month with a registration fee of $100. How much more would a year’s membership cost at that rate? (c) Bonus: For what number of months would the two membership rates be the same? 2. A candle is burning at a linear rate. The candle measures ﬁve inches two minutes after it was lit. It measures three inches eight minutes after it was lit. (a) What was the original length of the candle? (b) How long will it take to burn down to a half-inch stub? (c) Six half-inch stubs of candle can be melted together to make a new candle measuring 25 6 inches (a little wax gets lost in the process). How many stubs would it take to make three candles the size of the original candle? 3. A dipped candle is made by taking a wick and dipping it repeatedly in melted wax. The candle gets a little bit thicker with each added layer of wax. After it has been dipped three times, the candle is 6.5 mm thick. After it has been dipped six times, it is 11 mm thick. (a) How thick is the wick before the wax is added? (b) How many times does the wick need to be dipped to create a candle 2 cm thick? 4. Tali is trying to ﬁnd the thickness of a page of his telephone book. In order to do this, he takes a measurement and ﬁnds out that 55 pages measures 1 8 inch. What is the thickness of one page of the phone book? 5. Bobby and Petra are running a lemonade stand and they charge 45 cents for each glass of lemonade. In order to break even they must make $25. (a) How many glasses of lemonade must they sell to break even? (b) When they’ve sold $18 worth of lemonade, they realize that they only have enough lemons left to make 10 more glasses. To break even now, they’ll need to sell those last 10 glasses at a higher price. What does the new price need to be? 6. Dale is making cookies using a recipe that calls for 2.5 cups of ﬂour for two dozen cookies. How many cups of ﬂour does he need to make ﬁve dozen cookies? 7. To buy a car, Jason makes a down payment of $1500 and pays $350 per month in installments. (a) How much money has Jason paid at the end of one year? (b) If the total cost of the car is $8500, how long will it take Jason to ﬁnish paying it oﬀ? (c) The resale value of the car decreases by $100 each month from the original purchase price. If Jason sells the car as soon as he ﬁnishes paying it oﬀ, how much will he get for it? 195 www.ck12.org 8. Anne transplants a rose seedling in her garden. She wants to track the growth of the rose so she measures its height every week. On the third week, she ﬁnds that the rose is 10 inches tall and on the eleventh week she ﬁnds that the rose is 14 inches tall. Assuming the rose grows linearly with time, what was the height of the rose when Anne planted it? 9. Ravi hangs from a giant spring whose length is 5 m. When his child Nimi hangs from the spring its length is 2 m. Ravi weighs 160 lbs and Nimi weighs 40 lbs. Write the equation for this problem in slope-intercept form. What should we expect the length of the spring to be when his wife Amardeep, who weighs 140 lbs, hangs from it? 10. Nadia is placing diﬀerent weights on a spring and measuring the length of the stretched spring. She ﬁnds that for a 100 gram weight the length of the stretched spring is 20 cm and for a 300 gram weight the length of the stretched spring is 25 cm. (a) What is the unstretched length of the spring? (b) If the spring can only stretch to twice its unstretched length before it breaks, how much weight can it hold? 11. Andrew is a submarine commander. He decides to surface his submarine to periscope depth. It takes him 20 minutes to get from a depth of 400 feet to a depth of 50 feet. (a) What was the submarine’s depth ﬁve minutes after it started surfacing? (b) How much longer would it take at that rate to get all the way to the surface? 12. Kiersta’s phone has completely run out of battery power when she puts it on the charger. Ten minutes later, when the phone is 40% recharged, Kiersta’s friend Danielle calls and Kiersta takes the phone oﬀthe charger to talk to her. When she hangs up 45 minutes later, her phone has 10% of its charge left. Then she gets another call from her friend Kwan. (a) How long can she spend talking to Kwan before the battery runs out again? (b) If she puts the phone back on the charger afterward, how long will it take to recharge completely? 13. Marji is painting a 75-foot fence. She starts applying the ﬁrst coat of paint at 2 PM, and by 2:10 she has painted 30 feet of the fence. At 2:15, her husband, who paints about 2 3 as fast as she does, comes to join her. (a) How much of the fence has Marji painted when her husband joins in? (b) When will they have painted the whole fence? (c) How long will it take them to apply the second coat of paint if they work together the whole time? Texas Instruments Resources In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http: //www.ck12.org/flexr/chapter/9614. www.ck12.org 196 Chapter 5 Writing Linear Equations 5.1 Forms of Linear Equations Learning Objectives • Write equations in slope-intercept form. • Write equations in point-slope form. • Write equations in standard form. • Solve real-world problems using linear models in all three forms. Introduction We saw in the last chapter that many real-world situations can be described with linear graphs and equations. In this chapter, we’ll see how to ﬁnd those equations in a variety of situations. Write an Equation Given Slope and y−Intercept You’ve already learned how to write an equation in slope–intercept form: simply start with the general equation for the slope-intercept form of a line, y = mx + b, and then plug the given values of m and b into the equation. For example, a line with a slope of 4 and a y−intercept of -3 would have the equation y = 4x −3. If you are given just the graph of a line, you can read oﬀthe slope and y−intercept from the graph and write the equation from there. For example, on the graph below you can see that the line rises by 1 unit as it moves 2 units to the right, so its slope is 1 2. Also, you can see that the y−intercept is -2, so the equation of the line is y = 1 2 x −2. 197 www.ck12.org Write an Equation Given the Slope and a Point Often, we don’t know the value of the y−intercept, but we know the value of y for a non-zero value of x. In this case, it’s often easier to write an equation of the line in point-slope form. An equation in point-slope form is written as y −y0 = m(x −x0), where m is the slope and (x0, y0) is a point on the line. Example 1 A line has a slope of 3 5, and the point (2, 6) is on the line. Write the equation of the line in point-slope form. Solution Start with the formula y −y0 = m(x −x0). Plug in 3 5 for m, 2 for x0 and 6 for y0. The equation in point-slope form is y −6 = 3 5(x −2). Notice that the equation in point-slope form is not solved for y. If we did solve it for y, we’d have it in y−intercept form. To do that, we would just need to distribute the 3 5 and add 6 to both sides. That means that the equation of this line in slope-intercept form is y = 3 5 x −6 5 + 6, or simply y = 3 5 x + 24 5 . Write an Equation Given Two Points Point-slope form also comes in useful when we need to ﬁnd an equation given just two points on a line. For example, suppose we are told that the line passes through the points (-2, 3) and (5, 2). To ﬁnd the equation of the line, we can start by ﬁnding the slope. Starting with the slope formula, m = y2−y1 x2−x1 , we plug in the x−and y−values of the two points to get m = 2−3 5−(−2) = −1 7 . We can plug that value of m into the point-slope formula to get y −y0 = −1 7(x −x0). Now we just need to pick one of the two points to plug into the formula. Let’s use (5, 2); that gives us y −2 = −1 7(x −5). What if we’d picked the other point instead? Then we’d have ended up with the equation y−3 = −1 7(x+2), which doesn’t look the same. That’s because there’s more than one way to write an equation for a given line in point-slope form. But let’s see what happens if we solve each of those equations for y. Starting with y−2 = −1 7(x−5), we distribute the −1 7 and add 2 to both sides. That gives us y = −1 7 x+ 5 7 +2, www.ck12.org 198 or y = −1 7 x + 19 7 . On the other hand, if we start with y −3 = −1 7(x + 2), we need to distribute the −1 7 and add 3 to both sides. That gives us y = −1 7 x −2 7 + 3, which also simpliﬁes to y = −1 7 x + 19 7 . So whichever point we choose to get an equation in point-slope form, the equation is still mathematically the same, and we can see this when we convert it to y−intercept form. Example 2 A line contains the points (3, 2) and (-2, 4). Write an equation for the line in point-slope form; then write an equation in y−intercept form. Solution Find the slope of the line: m = y2−y1 x2−x1 = 4−2 −2−3 = −2 5 Plug in the value of the slope: y −y0 = −2 5(x −x0). Plug point (3, 2) into the equation: y −2 = −2 5(x −3). The equation in point-slope form is y −2 = −2 5(x −3). To convert to y−intercept form, simply solve for y: y −2 = −2 5(x −3) →y −2 = −2 5 x −6 5 →y = −2 5 x −6 5 + 2 = −2 5 x + 4 5. The equation in y−intercept form is y = −2 5 x + 4 5. Graph an Equation in Point-Slope Form Another useful thing about point-slope form is that you can use it to graph an equation without having to convert it to slope-intercept form. From the equation y −y0 = m(x −x0), you can just read oﬀthe slope m and the point (x0, y0). To draw the graph, all you have to do is plot the point, and then use the slope to ﬁgure out how many units up and over you should move to ﬁnd another point on the line. Example 5 Make a graph of the line given by the equation y + 2 = 2 3(x −2). Solution To read oﬀthe right values, we need to rewrite the equation slightly: y −(−2) = 2 3(x −2). Now we see that point (2, -2) is on the line and that the slope is 2 3. First plot point (2, -2) on the graph: 199 www.ck12.org A slope of 2 3 tells you that from that point you should move 2 units up and 3 units to the right and draw another point: Now draw a line through the two points and extend it in both directions: www.ck12.org 200 Linear Equations in Standard Form You’ve already encountered another useful form for writing linear equations: standard form. An equation in standard form is written ax + by = c, where a, b, and c are all integers and a is positive. (Note that the b in the standard form is diﬀerent than the b in the slope-intercept form.) One useful thing about standard form is that it allows us to write equations for vertical lines, which we can’t do in slope-intercept form. For example, let’s look at the line that passes through points (2, 6) and (2, 9). How would we ﬁnd an equation for that line in slope-intercept form? First we’d need to ﬁnd the slope: m = 9−6 0−0 = 3 0. But that slope is undeﬁned because we can’t divide by zero. And if we can’t ﬁnd the slope, we can’t use point-slope form either. If we just graph the line, we can see that x equals 2 no matter what y is. There’s no way to express that in slope-intercept or point-slope form, but in standard form we can just say that x + 0y = 2, or simply x = 2. Converting to Standard Form To convert an equation from another form to standard form, all you need to do is rewrite the equation so that all the variables are on one side of the equation and the coeﬀicient of x is not negative. Example 1 Rewrite the following equations in standard form: a) y = 5x −7 b) y −2 = −3(x + 3) c) y = 2 3 x + 1 2 Solution We need to rewrite each equation so that all the variables are on one side and the coeﬀicient of x is not negative. a) y = 5x −7 Subtract y from both sides to get 0 = 5x −y −7. Add 7 to both sides to get 7 = 5x −y. Flip the equation around to put it in standard form: 5x −y = 7. b) y −2 = −3(x + 3) Distribute the –3 on the right-hand-side to get y −2 = −3x −9. Add 3x to both sides to get y + 3x −2 = −9. Add 2 to both sides to get y + 3x = −7. Flip that around to get 3x + y = −7. c) y = 2 3 x + 1 2 Find the common denominator for all terms in the equation – in this case that would be 6. Multiply all terms in the equation by 6: 6 ( y = 2 3 x + 1 2 ) ⇒6y = 4x + 3 Subtract 6y from both sides: 0 = 4x −6y + 3 Subtract 3 from both sides: −3 = 4x −6y The equation in standard form is 4x −6y = −3. 201 www.ck12.org Graphing Equations in Standard Form When an equation is in slope-intercept form or point-slope form, you can tell right away what the slope is. How do you ﬁnd the slope when an equation is in standard form? Well, you could rewrite the equation in slope-intercept form and read oﬀthe slope. But there’s an even easier way. Let’s look at what happens when we rewrite an equation in standard form. Starting with the equation ax + by = c, we would subtract ax from both sides to get by = −ax + c. Then we would divide all terms by b and end up with y = −a b x + c b. That means that the slope is −a b and the y−intercept is c b. So next time we look at an equation in standard form, we don’t have to rewrite it to ﬁnd the slope; we know the slope is just −a b, where a and b are the coeﬀicients of x and y in the equation. Example 2 Find the slope and the y−intercept of the following equations written in standard form. a) 3x + 5y = 6 b) 2x −3y = −8 c) x −5y = 10 Solution a) a = 3, b = 5, and c = 6, so the slope is −a b = −3 5, and the y−intercept is c b = 6 5. b) a = 2, b = −3, and c = −8, so the slope is −a b = 2 3, and the y−intercept is c b = 8 3. c) a = 1, b = −5, and c = 10, so the slope is −a b = 1 5, and the y−intercept is c b = 10 −5 = −2. Once we’ve found the slope and y−intercept of an equation in standard form, we can graph it easily. But if we start with a graph, how do we ﬁnd an equation of that line in standard form? First, remember that we can also use the cover-up method to graph an equation in standard form, by ﬁnding the intercepts of the line. For example, let’s graph the line given by the equation 3x −2y = 6. To ﬁnd the x−intercept, cover up the y term (remember, the x−intercept is where y = 0): 3x = 6 ⇒x = 2 The x−intercept is (2, 0). To ﬁnd the y−intercept, cover up the x term (remember, the y−intercept is where x = 0): −2y = 6 ⇒y = −3 The y−intercept is (0, -3). We plot the intercepts and draw a line through them that extends in both directions: www.ck12.org 202 Now we want to apply this process in reverse—to start with the graph of the line and write the equation of the line in standard form. Example 3 Find the equation of each line and write it in standard form. a) 203 www.ck12.org b) c) Solution a) We see that the x−intercept is (3, 0) ⇒x = 3 and the y−intercept is (0, −4) ⇒y = −4 We saw that in standard form ax + by = c: if we “cover up” the y term, we get ax = c, and if we “cover up” the x term, we get by = c. So we need to ﬁnd values for a and b so that we can plug in 3 for x and -4 for y and get the same value for c in both cases. This is like ﬁnding the least common multiple of the x−and y−intercepts. In this case, we see that multiplying x = 3 by 4 and multiplying y = −4 by –3 gives the same result: (x = 3) × 4 ⇒4x = 12 and (y = −4) × (−3) ⇒−3y = 12 Therefore, a = 4, b = −3 and c = 12 and the equation in standard form is 4x −3y = 12. b) We see that the x−intercept is (3, 0) ⇒x = 3 and the y−intercept is (0, 3) ⇒y = 3 The values of the intercept equations are already the same, so a = 1, b = 1 and c = 3. The equation in standard form is x + y = 3. c) We see that the x−intercept is (3 2, 0 ) ⇒x = 3 2 and the y−intercept is (0, 4) ⇒y = 4 Let’s multiply the x−intercept equation by 2 ⇒2x = 3 www.ck12.org 204 Then we see we can multiply the x−intercept again by 4 and the y−intercept by 3, so we end up with 8x = 12 and 3y = 12. The equation in standard form is 8x + 3y = 12. Solving Real-World Problems Using Linear Models in Point-Slope Form Let’s solve some word problems where we need to write the equation of a straight line in point-slope form. Example 4 Marciel rented a moving truck for the day. Marciel only remembers that the rental truck company charges $40 per day and some number of cents per mile. Marciel drives 46 miles and the ﬁnal amount of the bill (before tax) is $63. What is the amount per mile the truck rental company charges? Write an equation in point-slope form that describes this situation. How much would it cost to rent this truck if Marciel drove 220 miles? Solution Let’s deﬁne our variables: x = distance in miles y = cost of the rental truck Peter pays a ﬂat fee of $40 for the day; this is the y−intercept. He pays $63 for 46 miles; this is the coordinate point (46,63). Start with the point-slope form of the line: y −y0 = m(x −x0) Plug in the coordinate point: 63 −y0 = m(46 −x0) Plug in the point (0, 40): 63 −40 = m(46 −0) Solve for the slope: 23 = 46m →m = 23 46 = 0.5 The slope is 0.5 dollars per mile, so the truck company charges 50 cents per mile ($0.5 = 50 cents). Plugging in the slope and the y−intercept, the equation of the line is y = 0.5x + 40. To ﬁnd out the cost of driving the truck 220 miles, we plug in x = 220 to get y−40 = 0.5(220) ⇒y = $150. Driving 220 miles would cost $150. Example 5 Anne got a job selling window shades. She receives a monthly base salary and a $6 commission for each window shade she sells. At the end of the month she adds up sales and she ﬁgures out that she sold 200 window shades and made $2500. Write an equation in point-slope form that describes this situation. How much is Anne’s monthly base salary? Solution Let’s deﬁne our variables: x = number of window shades sold y = Anne’s earnings We see that we are given the slope and a point on the line: Nadia gets $6 for each shade, so the slope is 6. She made $2500 when she sold 200 shades, so the point is (200, 2500). 205 www.ck12.org Start with the point-slope form of the line: y −y0 = m(x −x0) Plug in the slope: y −y0 = 6(x −x0) Plug in the point (200, 2500): y −2500 = 6(x −200) To ﬁnd Anne’s base salary, we plug in x = 0 and get y −2500 = −1200 ⇒y = $1300. Anne’s monthly base salary is $1300. Solving Real-World Problems Using Linear Models in Standard Form Here are two examples of real-world problems where the standard form of an equation is useful. Example 6 Nadia buys fruit at her local farmer’s market. This Saturday, oranges cost $2 per pound and cherries cost $3 per pound. She has $12 to spend on fruit. Write an equation in standard form that describes this situation. If she buys 4 pounds of oranges, how many pounds of cherries can she buy? Solution Let’s deﬁne our variables: x = pounds of oranges y = pounds of cherries The equation that describes this situation is 2x + 3y = 12. If she buys 4 pounds of oranges, we can plug x = 4 into the equation and solve for y: 2(4) + 3y = 12 ⇒3y = 12 −8 ⇒3y = 4 ⇒y = 4 3 Nadia can buy 1 1 3 pounds of cherries. Example 7 Peter skateboards part of the way to school and walks the rest of the way. He can skateboard at 7 miles per hour and he can walk at 3 miles per hour. The distance to school is 6 miles. Write an equation in standard form that describes this situation. If he skateboards for 1 2 an hour, how long does he need to walk to get to school? Solution Let’s deﬁne our variables: x = time Peter skateboards y = time Peter walks The equation that describes this situation is: 7x + 3y = 6 If Peter skateboards 1 2 an hour, we can plug x = 0.5 into the equation and solve for y: 7(0.5) + 3y = 6 ⇒3y = 6 −3.5 ⇒3y = 2.5 ⇒y = 5 6 Peter must walk 5 6 of an hour. www.ck12.org 206 Further Practice Now that you’ve worked with equations in all three basic forms, check out the Java applet at http: //www.ronblond.com/M10/lineAP/index.html. You can use it to manipulate graphs of equations in all three forms, and see how the graphs change when you vary the terms of the equations. Another applet at lets you create multiple lines and see how they intersect. Each line is deﬁned by two points; you can change the slope of a line by moving either of the points, or just drag the whole line around without changing its slope. To create another line, just click Duplicate and then drag one of the lines that are already there. Review Questions Find the equation of each line in slope–intercept form. 1. The line has a slope of 7 and a y−intercept of -2. 2. The line has a slope of -5 and a y−intercept of 6. 3. The line has a slope of −1 4 and contains the point (4, -1). 4. The line has a slope of 2 3 and contains the point ( 1 2, 1 ) . 5. The line has a slope of -1 and contains the point (4 5, 0 ) . 6. The line contains points (2, 6) and (5, 0). 7. The line contains points (5, -2) and (8, 4). 8. The line contains points (3, 5) and (-3, 0). 9. The line contains points (10, 15) and (12, 20). Write the equation of each line in slope-intercept form. 10. 207 www.ck12.org 11. Find the equation of each linear function in slope–intercept form. 12. m = 5, f(0) = −3 13. m = −7, f(2) = −1 14. m = 1 3, f(−1) = 2 3 15. m = 4.2, f(−3) = 7.1 16. f ( 1 4 ) = 3 4, f(0) = 5 4 17. f(1.5) = −3, f(−1) = 2 Write the equation of each line in point-slope form. 18. The line has slope −1 10 and goes through the point (10, 2). 19. The line has slope -75 and goes through the point (0, 125). 20. The line has slope 10 and goes through the point (8, -2). 21. The line goes through the points (-2, 3) and (-1, -2). 22. The line contains the points (10, 12) and (5, 25). 23. The line goes through the points (2, 3) and (0, 3). 24. The line has a slope of 3 5 and a y−intercept of -3. 25. The line has a slope of -6 and a y−intercept of 0.5. Write the equation of each linear function in point-slope form. 26. m = −1 5 and f(0) = 7 27. m = −12 and f(−2) = 5 28. f(−7) = 5 and f(3) = −4 29. f(6) = 0 and f(0) = 6 30. m = 3 and f(2) = −9 31. m = −9 5 and f(0) = 32 Rewrite the following equations in standard form. 32. y = 3x −8 33. y −7 = −5(x −12) 34. 2y = 6x + 9 www.ck12.org 208 35. y = 9 4 x + 1 4 36. y + 3 5 = 2 3(x −2) 37. 3y + 5 = 4(x −9) Find the slope and y−intercept of the following lines. 38. 5x −2y = 15 39. 3x + 6y = 25 40. x −8y = 12 41. 3x −7y = 20 42. 9x −9y = 4 43. 6x + y = 3 Find the equation of each line and write it in standard form. 44. 45. 209 www.ck12.org 46. 47. 48. Andrew has two part time jobs. One pays $6 per hour and the other pays $10 per hour. He wants to make $366 per week. Write an equation in standard form that describes this situation. If he is only allowed to work 15 hours per week at the $10 per hour job, how many hours does he need to work per week in his $6 per hour job in order to achieve his goal? 49. Anne invests money in two accounts. One account returns 5% annual interest and the other returns 7% annual interest. In order not to incur a tax penalty, she can make no more than $400 in interest per year. Write an equation in standard form that describes this problem. If she invests $5000 in the 5% interest account, how much money does she need to invest in the other account? 5.2 Equations of Parallel and Perpendicular Lines Learning Objectives • Determine whether lines are parallel or perpendicular • Write equations of perpendicular lines • Write equations of parallel lines • Investigate families of lines www.ck12.org 210 Introduction In this section you will learn how parallel lines and perpendicular lines are related to each other on the coordinate plane. Let’s start by looking at a graph of two parallel lines. We can clearly see that the two lines have diﬀerent y−intercepts: 6 and –4. How about the slopes of the lines? The slope of line A is 6−2 0−(−2) = 4 2 = 2, and the slope of line B is 0−(−4) 2−0 = 4 2 = 2. The slopes are the same. Is that signiﬁcant? Yes. By deﬁnition, parallel lines never meet. That means that when one of them slopes up by a certain amount, the other one has to slope up by the same amount so the lines will stay the same distance apart. If you look at the graph above, you can see that for any x−value you pick, the y−values of lines A and B are the same vertical distance apart—which means that both lines go up by the same vertical distance every time they go across by the same horizontal distance. In order to stay parallel, their slopes must stay the same. All parallel lines have the same slopes and diﬀerent y−intercepts. Now let’s look at a graph of two perpendicular lines. We can’t really say anything about the y−intercepts. In this example, the y−intercepts are diﬀerent, but if 211 www.ck12.org we moved the lines four units to the right, they would both intercept the y−axis at (0, -2). So perpendicular lines can have the same or diﬀerent y−intercepts. What about the relationship between the slopes of the two lines? To ﬁnd the slope of line A, we pick two points on the line and draw the blue (upper) right triangle. The legs of the triangle represent the rise and the run. We can see that the slope is 8 4, or 2. To ﬁnd the slope of line B, we pick two points on the line and draw the red (lower) right triangle. Notice that the two triangles are identical, only rotated by 90◦. Where line A goes 8 units up and 4 units right, line B goes 8 units right and 4 units down. Its slope is −4 8, or −1 2. This is always true for perpendicular lines; where one line goes a units up and b units right, the other line will go a units right and b units down, so the slope of one line will be a b and the slope of the other line will be −b a. The slopes of perpendicular lines are always negative reciprocals of each other. The Java applet at lets you drag around a pair of perpendicular lines to see how their slopes change. Click “Show Grid” to see the x−and y−axes, and click “Show Constructors” to see the triangles that are being used to calculate the slopes of the lines (you can then drag the circle to make it bigger or smaller, and click on a triangle to see the slope calculations in detail.) Determine When Lines are Parallel or Perpendicular You can ﬁnd whether lines are parallel or perpendicular by comparing the slopes of the lines. If you are given points on the lines, you can ﬁnd their slopes using the formula. If you are given the equations of the lines, re-write each equation in a form that makes it easy to read the slope, such as the slope-intercept form. Example 1 Determine whether the lines are parallel or perpendicular or neither. a) One line passes through the points (2, 11) and (-1, 2); another line passes through the points (0, -4) and (-2, -10). b) One line passes through the points (-2, -7) and (1, 5); another line passes through the points (4, 1) and (-8, 4). www.ck12.org 212 c) One lines passes through the points (3, 1) and (-2, -2); another line passes through the points (5, 5) and (4, -6). Solution Find the slope of each line and compare them. a) m1 = 2−11 −1−2 = −9 −3 = 3 and m2 = −10−(−4) −2−0 = −6 −2 = 3 The slopes are equal, so the lines are parallel. b) m1 = 5−(−7) 1−(−2) = 12 3 = 4 and m2 = 4−1 −8−4 = 3 −12 = −1 4 The slopes are negative reciprocals of each other, so the lines are perpendicular. c) m1 = −2−1 −2−3 = −3 −5 = 3 5 and m2 = −6−5 4−5 = −13 −1 = 13 The slopes are not the same or negative reciprocals of each other, so the lines are neither parallel nor perpendicular. Example 2 Determine whether the lines are parallel or perpendicular or neither: a) 3x + 4y = 2 and 8x −6y = 5 b) 2x = y −10 and y = −2x + 5 c) 7y + 1 = 7x and x + 5 = y Solution Write each equation in slope-intercept form: a) line 1: 3x + 4y = 2 ⇒4y = −3x + 2 ⇒y = −3 4 x + 1 2 ⇒slope = −3 4 line 2: 8x −6y = 5 ⇒8x −5 = 6y ⇒y = 8 6 x −5 6 ⇒y = 4 3 x −5 6 ⇒slope = 4 3 The slopes are negative reciprocals of each other, so the lines are perpendicular. b) line 1: 2x = y −10 ⇒y = 2x + 10 ⇒slope = 2 line 2: y = −2x + 5 ⇒slope = −2 The slopes are not the same or negative reciprocals of each other, so the lines are neither parallel nor perpendicular. c) line 1: 7y + 1 = 7x ⇒7y = 7x −1 ⇒y = x −1 7 ⇒slope = 1 line 2: x + 5 = y ⇒y = x + 5 ⇒slope = 1 The slopes are the same, so the lines are parallel. Write Equations of Parallel and Perpendicular Lines We can use the properties of parallel and perpendicular lines to write an equation of a line parallel or perpendicular to a given line. You might be given a line and a point, and asked to ﬁnd the line that goes through the given point and is parallel or perpendicular to the given line. Here’s how to do this: 1. Find the slope of the given line from its equation. (You might need to re-write the equation in a form such as the slope-intercept form.) 2. Find the slope of the parallel or perpendicular line—which is either the same as the slope you found in step 1 (if it’s parallel), or the negative reciprocal of the slope you found in step 1 (if it’s perpendicular). 3. Use the slope you found in step 2, along with the point you were given, to write an equation of the 213 www.ck12.org new line in slope-intercept form or point-slope form. Example 3 Find an equation of the line perpendicular to the line y = −3x + 5 that passes through the point (2, 6). Solution The slope of the given line is -3, so the perpendicular line will have a slope of 1 3. Now to ﬁnd the equation of a line with slope 1 3 that passes through (2, 6): Start with the slope-intercept form: y = mx + b. Plug in the slope: y = 1 3 x + b. Plug in the point (2, 6) to ﬁnd b: 6 = 1 3(2) + b ⇒b = 6 −2 3 ⇒b = 20 3 . The equation of the line is y = 1 3 x + 20 3 . Example 4 Find the equation of the line perpendicular to x −5y = 15 that passes through the point (-2, 5). Solution Re-write the equation in slope-intercept form: x −5y = 15 ⇒−5y = −x + 15 ⇒y = 1 5 x −3. The slope of the given line is 1 5, so we’re looking for a line with slope -5. Start with the slope-intercept form: y = mx + b. Plug in the slope: y = −5x + b. Plug in the point (-2, 5): 5 = −5(−2) + b ⇒b = 5 −10 ⇒b = −5 The equation of the line is y = −5x −5. Example 5 Find the equation of the line parallel to 6x −5y = 12 that passes through the point (-5, -3). Solution Rewrite the equation in slope-intercept form: 6x −5y = 12 ⇒5y = 6x −12 ⇒y = 6 5 x −12 5 . The slope of the given line is 6 5, so we are looking for a line with slope 6 5 that passes through the point (-5, -3). Start with the slope-intercept form: y = mx + b. Plug in the slope: y = 6 5 x + b. Plug in the point (-5, -3): n −3 = 6 5(−5) + b ⇒−3 = −6 + b ⇒b = 3 The equation of the line is y = 6 5 x + 3. Investigate Families of Lines A family of lines is a set of lines that have something in common with each other. Straight lines can belong to two types of families: one where the slope is the same and one where the y–intercept is the same. Family 1: Keep the slope unchanged and vary the y−intercept. The ﬁgure below shows the family of lines with equations of the form y = −2x + b: www.ck12.org 214 All the lines have a slope of –2, but the value of b is diﬀerent for each line. Notice that in such a family all the lines are parallel. All the lines look the same, except that they are shifted up and down the y−axis. As b gets larger the line rises on the y−axis, and as b gets smaller the line goes lower on the y−axis. This behavior is often called a vertical shift. Family 2: Keep the y−intercept unchanged and vary the slope. The ﬁgure below shows the family of lines with equations of the form y = mx + 2: All the lines have a y−intercept of two, but the slope is diﬀerent for each line. The steeper lines have higher values of m. Example 6 Write the equation of the family of lines satisfying the given condition. a) parallel to the x−axis b) through the point (0, -1) c) perpendicular to 2x + 7y −9 = 0 d) parallel to x + 4y −12 = 0 Solution a) All lines parallel to the x−axis have a slope of zero; the y−intercept can be anything. So the family of lines is y = 0x + b or just y = b. 215 www.ck12.org b) All lines passing through the point (0, -1) have the same y−intercept, b = −1. The family of lines is: y = mx −1. c) First we need to ﬁnd the slope of the given line. Rewriting 2x + 7y −9 = 0 in slope-intercept form, we get y = −2 7 x + 9 7. The slope of the line is −2 7, so we’re looking for the family of lines with slope 7 2. The family of lines is y = 7 2 x + b. www.ck12.org 216 d) Rewrite x + 4y −12 = 0 in slope-intercept form: y = −1 4 x + 3. The slope is −1 4, so that’s also the slope of the family of lines we are looking for. The family of lines is y = −1 4 x + b. Review Questions For questions 1-10, determine whether the lines are parallel, perpendicular or neither. 1. One line passes through the points (-1, 4) and (2, 6); another line passes through the points (2, -3) and (8, 1). 2. One line passes through the points (4, -3) and (-8, 0); another line passes through the points (-1, -1) and (-2, 6). 3. One line passes through the points (-3, 14) and (1, -2); another line passes through the points (0, -3) and (-2, 5). 4. One line passes through the points (3, 3) and (-6, -3); another line passes through the points (2, -8) and (-6, 4). 5. One line passes through the points (2, 8) and (6, 0); another line has the equation x −2y = 5. 217 www.ck12.org 6. One line passes through the points (-5, 3) and (2, -1); another line has the equation 2x + 3y = 6. 7. Both lines pass through the point (2, 8); one line also passes through (3, 5), and the other line has slope 3. 8. Line 1: 4y + x = 8 Line 2: 12y + 3x = 1 9. Line 1: 5y + 3x = 1 Line 2: 6y + 10x = −3 10. Line 1: 2y −3x + 5 = 0 Line 2: y + 6x = −3 11. Lines A, B,C, D, and E all pass through the point (3, 6). Line A also passes through (7, 12); line B passes through (8, 4); line C passes through (-1, -3); line D passes through (1, 1); and line E passes through (6, 12). (a) Are any of these lines perpendicular? If so, which ones? If not, why not? (b) Are any of these lines parallel? If so, which ones? If not, why not? 12. Find the equation of the line parallel to 5x −2y = 2 that passes through point (3, -2). 13. Find the equation of the line perpendicular to y = −2 5 x −3 that passes through point (2, 8). 14. Find the equation of the line parallel to 7y + 2x −10 = 0 that passes through the point (2, 2). 15. Find the equation of the line perpendicular to y + 5 = 3(x −2) that passes through the point (6, 2). 16. Line S passes through the points (2, 3) and (4, 7). Line T passes through the point (2, 5). If Lines S and T are parallel, name one more point on line T. (Hint: you don’t need to ﬁnd the slope of either line.) 17. Lines P and Q both pass through (-1, 5). Line P also passes through (-3, -1). If P and Q are perpendicular, name one more point on line Q. (This time you will have to ﬁnd the slopes of both lines.) 18. Write the equation of the family of lines satisfying the given condition. (a) All lines that pass through point (0, 4). (b) All lines that are perpendicular to 4x + 3y −1 = 0. (c) All lines that are parallel to y −3 = 4x + 2. (d) All lines that pass through the point (0, -1). 19. Name two lines that pass through the point (3, -1) and are perpendicular to each other. 20. Name two lines that are each perpendicular to y = −4x −2. What is the relationship of those two lines to each other? 21. Name two perpendicular lines that both pass through the point (3, -2). Then name a line parallel to one of them that passes through the point (-2, 5). 5.3 Fitting a Line to Data Learning Objectives • Make a scatter plot. • Fit a line to data and write an equation for that line. • Perform linear regression with a graphing calculator. • Solve real-world problems using linear models of scattered data. Introduction Katja has noticed that sales are falling oﬀat her store lately. She plots her sales ﬁgures for each week on a graph and sees that the points are trending downward, but they don’t quite make a straight line. How can she predict what her sales ﬁgures will be over the next few weeks? www.ck12.org 218 In real-world problems, the relationship between our dependent and independent variables is linear, but not perfectly so. We may have a number of data points that don’t quite ﬁt on a straight line, but we may still want to ﬁnd an equation representing those points. In this lesson, we’ll learn how to ﬁnd linear equations to ﬁt real-world data. Make a Scatter Plot A scatter plot is a plot of all the ordered pairs in a table. Even when we expect the relationship we’re analyzing to be linear, we usually can’t expect that all the points will ﬁt perfectly on a straight line. Instead, the points will be “scattered” about a straight line. There are many reasons why the data might not fall perfectly on a line. Small errors in measurement are one reason; another reason is that the real world isn’t always as simple as a mathematical abstraction, and sometimes math can only describe it approximately. Example 1 Make a scatter plot of the following ordered pairs: (0, 2); (1, 4.5); (2, 9); (3, 11); (4, 13); (5, 18); (6, 19.5) Solution We make a scatter plot by graphing all the ordered pairs on the coordinate axis: Fit a Line to Data Notice that the points look like they might be part of a straight line, although they wouldn’t ﬁt perfectly on a straight line. If the points were perfectly lined up, we could just draw a line through any two of them, and that line would go right through all the other points as well. When the points aren’t lined up perfectly, we just have to ﬁnd a line that is as close to all the points as possible. 219 www.ck12.org Here you can see that we could draw many lines through the points in our data set. However, the red line A is the line that best ﬁts the points. To prove this mathematically, we would measure all the distances from each data point to line A: and then we would show that the sum of all those distances—or rather, the square root of the sum of the squares of the distances—is less than it would be for any other line. Actually proving this is a lesson for a much more advanced course, so we won’t do it here. And ﬁnding the best ﬁt line in the ﬁrst place is even more complex; instead of doing it by hand, we’ll use a graphing calculator or just “eyeball” the line, as we did above—using our visual sense to guess what line ﬁts best. For more practice eyeballing lines of best ﬁt, try the Java applet at regression/. Click on the green ﬁeld to place up to 50 points on it, then use the slider to adjust the slope of the red line to try and make it ﬁt the points. (The thermometer shows how far away the line is from the points, so you want to try to make the thermometer reading as low as possible.) Then click “Show Best Fit” to show the actual best ﬁt line in blue. Refresh the page or click “Reset” if you want to try again. For more of a challenge, try scattering the points in a less obvious pattern. Write an Equation For a Line of Best Fit Once you draw the line of best ﬁt, you can ﬁnd its equation by using two points on the line. Finding the equation of the line of best ﬁt is also called linear regression. Caution: Make sure you don’t get caught making a common mistake. Sometimes the line of best ﬁt won’t pass straight through any of the points in the original data set. This means that you can’t just use two points from the data set – you need to use two points that are on the line, which might not be in the data set at all. In Example 1, it happens that two of the data points are very close to the line of best ﬁt, so we can just www.ck12.org 220 use these points to ﬁnd the equation of the line: (1, 4.5) and (3, 11). Start with the slope-intercept form of a line: y = mx + b Find the slope: m = 11−4.5 3−1 = 6.5 2 = 3.25. So y = 3.25x + b. Plug (3, 11) into the equation: 11 = 3.25(3) + b ⇒b = 1.25 So the equation for the line that ﬁts the data best is y = 3.25x + 1.25. Perform Linear Regression With a Graphing Calculator The problem with eyeballing a line of best ﬁt, of course, is that you can’t be sure how accurate your guess is. To get the most accurate equation for the line, we can use a graphing calculator instead. The calculator uses a mathematical algorithm to ﬁnd the line that minimizes the sum of the squares. Example 2 Use a graphing calculator to ﬁnd the equation of the line of best ﬁt for the following data: (3, 12), (8, 20), (1, 7), (10, 23), (5, 18), (8, 24), (11, 30), (2, 10) Solution Step 1: Input the data in your calculator. Press [STAT] and choose the [EDIT] option. Input the data into the table by entering the x−values in the ﬁrst column and the y−values in the second column. Step 2: Find the equation of the line of best ﬁt. Press [STAT] again use right arrow to select [CALC] at the top of the screen. Chose option number 4, LinReg(ax + b), and press [ENTER] 221 www.ck12.org The calculator will display LinReg(ax + b). Press [ENTER] and you will be given the a−and b−values. Here a represents the slope and b represents the y−intercept of the equation. The linear regression line is y = 2.01x + 5.94. Step 3. Draw the scatter plot. To draw the scatter plot press [STATPLOT] [2nd] [Y=]. Choose Plot 1 and press [ENTER]. Press the On option and set the Type as scatter plot (the one highlighted in black). Make sure that the X list and Y list names match the names of the columns of the table in Step 1. Choose the box or plus as the mark, since the simple dot may make it diﬀicult to see the points. Press [GRAPH] and adjust the window size so you can see all the points in the scatter plot. Step 4. Draw the line of best ﬁt through the scatter plot. Press [Y=] Enter the equation of the line of best ﬁt that you just found: y = 2.01x + 5.94. Press [GRAPH]. www.ck12.org 222 Solve Real-World Problems Using Linear Models of Scattered Data Once we’ve found the line of best ﬁt for a data set, we can use the equation of that line to predict other data points. Example 3 Nadia is training for a 5K race. The following table shows her times for each month of her training program. Find an equation of a line of ﬁt. Predict her running time if her race is in August. Table 5.1: Month Month number Average time (minutes) January 0 40 February 1 38 March 2 39 April 3 38 May 4 33 June 5 30 Solution Let’s make a scatter plot of Nadia’s running times. The independent variable, x, is the month number and the dependent variable, y, is the running time. We plot all the points in the table on the coordinate plane, and then sketch a line of ﬁt. 223 www.ck12.org Two points on the line are (0, 42) and (4, 34). We’ll use them to ﬁnd the equation of the line: m = 34 −42 4 −0 = −8 4 = −2 y = −2x + b 42 = −2(0) + b ⇒b = 42 y = −2x + 42 In a real-world problem, the slope and y−intercept have a physical signiﬁcance. In this case, the slope tells us how Nadia’s running time changes each month she trains. Speciﬁcally, it decreases by 2 minutes per month. Meanwhile, the y−intercept tells us that when Nadia started training, she ran a distance of 5K in 42 minutes. The problem asks us to predict Nadia’s running time in August. Since June is deﬁned as month number 5, August will be month number 7. We plug x = 7 into the equation of the line of best ﬁt: y = −2(7) + 42 = −14 + 42 = 28 The equation predicts that Nadia will run the 5K race in 28 minutes. In this solution, we eyeballed a line of ﬁt. Using a graphing calculator, we can ﬁnd this equation for a line of ﬁt instead: y = −2.2x + 43.7 If we plug x = 7 into this equation, we get y = −2.2(7) + 43.7 = 28.3. This means that Nadia will run her race in 28.3 minutes. You see that the graphing calculator gives a diﬀerent equation and a diﬀerent answer to the question. The graphing calculator result is more accurate, but the line we drew by hand still gives a good approximation to the result. And of course, there’s no guarantee that Nadia will actually ﬁnish the race in that exact time; both answers are estimates, it’s just that the calculator’s estimate is slightly more likely to be right. Example 4 Peter is testing the burning time of “BriteGlo” candles. The following table shows how long it takes to burn candles of diﬀerent weights. Assume it’s a linear relation, so we can use a line to ﬁt the data. If a candle burns for 95 hours, what must be its weight in ounces? Table 5.2: Candle weight (oz) Time (hours) 2 15 3 20 4 35 5 36 10 80 16 100 22 120 26 180 Solution Let’s make a scatter plot of the data. The independent variable, x, is the candle weight and the dependent variable, y, is the time it takes the candle to burn. We plot all the points in the table on the coordinate plane, and draw a line of ﬁt. www.ck12.org 224 Two convenient points on the line are (0,0) and (30, 200). Find the equation of the line: m = 200 30 = 20 3 y = 20 3 x + b 0 = 20 3 (0) + b ⇒b = 0 y = 20 3 x A slope of 20 3 = 6 2 3 tells us that for each extra ounce of candle weight, the burning time increases by 62 3 hours. A y−intercept of zero tells us that a candle of weight 0 oz will burn for 0 hours. The problem asks for the weight of a candle that burns 95 hours; in other words, what’s the x−value that gives a y−value of 95? Plugging in y = 95: y = 20 3 x ⇒95 = 20 3 x ⇒x = 285 20 = 57 4 = 141 4 A candle that burns 95 hours weighs 14.25 oz. A graphing calculator gives the linear regression equation as y = 6.1x + 5.9 and a result of 14.6 oz. Review Questions For problems 1-4, draw the scatter plot and ﬁnd an equation that ﬁts the data set by hand. 1. (57, 45); (65, 61); (34, 30); (87, 78); (42, 41); (35, 36); (59, 35); (61, 57); (25, 23); (35, 34) 2. (32, 43); (54, 61); (89, 94); (25, 34); (43, 56); (58, 67); (38, 46); (47, 56); (39, 48) 3. (12, 18); (5, 24); (15, 16); (11, 19); (9, 12); (7, 13); (6, 17); (12, 14) 4. (3, 12); (8, 20); (1, 7); (10, 23); (5, 18); (8, 24); (2, 10) 5. Use the graph from problem 1 to predict the y−values for two x−values of your choice that are not in the data set. 6. Use the graph from problem 2 to predict the x−values for two y−values of your choice that are not in the data set. 225 www.ck12.org 7. Use the equation from problem 3 to predict the y−values for two x−values of your choice that are not in the data set. 8. Use the equation from problem 4 to predict the x−values for two y−values of your choice that are not in the data set. For problems 9-11, use a graphing calculator to ﬁnd the equation of the line of best ﬁt for the data set. 9. (57, 45); (65, 61); (34, 30); (87, 78); (42, 41); (35, 36); (59, 35); (61, 57); (25, 23); (35, 34) 10. (32, 43); (54, 61); (89, 94); (25, 34); (43, 56); (58, 67); (38, 46); (47, 56); (95, 105); (39, 48) 11. (12, 18); (3, 26); (5, 24); (15, 16); (11, 19); (0, 27); (9, 12); (7, 13); (6, 17); (12, 14) 12. Graph the best ﬁt line on top of the scatter plot for problem 10. Then pick a data point that’s close to the line, and change its y−value to move it much farther from the line. (a) Calculate the new best ﬁt line with that one point changed; write the equation of that line along with the coordinates of the new point. (b) How much did the slope of the best ﬁt line change when you changed that point? 13. Graph the scatter plot from problem 11 and change one point as you did in the previous problem. (a) Calculate the new best ﬁt line with that one point changed; write the equation of that line along with the coordinates of the new point. (b) Did changing that one point seem to aﬀect the slope of the best ﬁt line more or less than it did in the previous problem? What might account for this diﬀerence? 14. Shiva is trying to beat the samosa-eating record. The current record is 53.5 samosas in 12 minutes. Each day he practices and the following table shows how many samosas he eats each day for the ﬁrst week of his training. Table 5.3: Day No. of samosas 1 30 2 34 3 36 4 36 5 40 6 43 7 45 (a) Draw a scatter plot and ﬁnd an equation to ﬁt the data. (b) Will he be ready for the contest if it occurs two weeks from the day he started training? (c) What are the meanings of the slope and the y−intercept in this problem? 15. Anne is trying to ﬁnd the elasticity coeﬀicient of a Superball. She drops the ball from diﬀerent heights and measures the maximum height of the ball after the bounce. The table below shows the data she collected. www.ck12.org 226 Table 5.4: Initial height (cm) Bounce height (cm) 30 22 35 26 40 29 45 34 50 38 55 40 60 45 65 50 70 52 (a) Draw a scatter plot and ﬁnd the equation. (b) What height would she have to drop the ball from for it to bounce 65 cm? (c) What are the meanings of the slope and the y−intercept in this problem? (d) Does the y−intercept make sense? Why isn’t it (0, 0)? 16. The following table shows the median California family income from 1995 to 2002 as reported by the US Census Bureau. Table 5.5: Year Income 1995 53,807 1996 55,217 1997 55,209 1998 55,415 1999 63,100 2000 63,206 2001 63,761 2002 65,766 (a) Draw a scatter plot and ﬁnd the equation. (b) What would you expect the median annual income of a Californian family to be in year 2010? (c) What are the meanings of the slope and the y−intercept in this problem? (d) Inﬂation in the U.S. is measured by the Consumer Price Index, which increased by 20% between 1995 and 2002. Did the median income of California families keep up with inﬂation over that time period? (In other words, did it increase by at least 20%?) 227 www.ck12.org 5.4 Predicting with Linear Models Learning Objectives • Interpolate using an equation. • Extrapolate using an equation. • Predict using an equation. Introduction Katja’s sales ﬁgures were trending downward quickly at ﬁrst, and she used a line of best ﬁt to describe the numbers. But now they seem to be decreasing more slowly, and ﬁtting the line less and less accurately. How can she make a more accurate prediction of what next week’s sales will be? In the last lesson we saw how to ﬁnd the equation of a line of best ﬁt and how to use this equation to make predictions. The line of “best ﬁt” is a good method if the relationship between the dependent and the independent variables is linear. In this section you will learn other methods that are useful even when the relationship isn’t linear. Linear Interpolation We use linear interpolation to ﬁll in gaps in our data—that is, to estimate values that fall in between the values we already know. To do this, we use a straight line to connect the known data points on either side of the unknown point, and use the equation of that line to estimate the value we are looking for. Example 1 The following table shows the median ages of ﬁrst marriage for men and women, as gathered by the U.S. Census Bureau. Table 5.6: Year Median age of males Median age of females 1890 26.1 22.0 1900 25.9 21.9 1910 25.1 21.6 1920 24.6 21.2 1930 24.3 21.3 1940 24.3 21.5 1950 22.8 20.3 1960 22.8 20.3 1970 23.2 20.8 1980 24.7 22.0 1990 26.1 23.9 2000 26.8 25.1 Estimate the median age for the ﬁrst marriage of a male in the year 1946. Solution We connect the two points on either side of 1946 with a straight line and ﬁnd its equation. Here’s how www.ck12.org 228 that looks on a scatter plot: We ﬁnd the equation by plugging in the two data points: m = 22.8 −24.3 1950 −1940 = −1.5 10 = −0.15 y = −0.15x + b 24.3 = −0.15(1940) + b b = 315.3 Our equation is y = −0.15x + 315.3. To estimate the median age of marriage of males in the year 1946, we plug x = 1946 into the equation we just found: y = −0.15(1946) + 315.3 = 23.4 years old Example 2 The Center for Disease Control collects information about the health of the American people and behaviors that might lead to bad health. The following table shows the percent of women who smoke during pregnancy. Table 5.7: Year Percent of pregnant women smokers 1990 18.4 1991 17.7 1992 16.9 1993 15.8 1994 14.6 1995 13.9 1996 13.6 2000 12.2 2002 11.4 2003 10.4 2004 10.2 Estimate the percentage of pregnant women that were smoking in the year 1998. 229 www.ck12.org Solution We connect the two points on either side of 1998 with a straight line and ﬁnd its equation. Here’s how that looks on a scatter plot: We ﬁnd the equation by plugging in the two data points: m = 12.2 −13.6 2000 −1996 = −1.4 4 = −0.35 y = −0.35x + b 12.2 = −0.35(2000) + b b = 712.2 Our equation is y = −0.35x + 712.2. To estimate the percentage of pregnant women who smoked in the year 1998, we plug x = 1998 into the equation we just found: y = −0.35(1998) + 712.2 = 12.9% For non-linear data, linear interpolation is often not accurate enough for our purposes. If the points in the data set change by a large amount in the interval you’re interested in, then linear interpolation may not give a good estimate. In that case, it can be replaced by polynomial interpolation, which uses a curve instead of a straight line to estimate values between points. But that’s beyond the scope of this lesson. Linear Extrapolation Linear extrapolation can help us estimate values that are outside the range of our data set. The strategy is similar to linear interpolation: we pick the two data points that are closest to the one we’re looking for, ﬁnd the equation of the line between them, and use that equation to estimate the coordinates of the missing point. Example 3 The winning times for the women’s 100 meter race are given in the following table. Estimate the winning time in the year 2010. Is this a good estimate? www.ck12.org 230 Table 5.8: Winner Country Year Time (seconds) Mary Lines UK 1922 12.8 Leni Schmidt Germany 1925 12.4 Gerturd Glasitsch Germany 1927 12.1 Tollien Schuurman Netherlands 1930 12.0 Helen Stephens USA 1935 11.8 Lulu Mae Hymes USA 1939 11.5 Fanny Blankers-Koen Netherlands 1943 11.5 Marjorie Jackson Australia 1952 11.4 Vera Krepkina Soviet Union 1958 11.3 Wyomia Tyus USA 1964 11.2 Barbara Ferrell USA 1968 11.1 Ellen Strophal East Germany 1972 11.0 Inge Helten West Germany 1976 11.0 Marlies Gohr East Germany 1982 10.9 Florence Griﬀith Joyner USA 1988 10.5 Solution We start by making a scatter plot of the data; then we connect the last two points on the graph and ﬁnd the equation of the line. m = 10.5 −10.9 1988 −1982 = −0.4 6 = −0.067 y = −0.067x + b 10.5 = −0.067(1988) + b b = 143.7 Our equation is y = −0.067x + 143.7. The winning time in year 2010 is estimated to be: y = −0.067(2010) + 143.7 = 9.03 seconds. Unfortunately, this estimate actually isn’t very accurate. This example demonstrates the weakness of linear extrapolation; it uses only a couple of points, instead of using all the points like the best ﬁt line method, so 231 www.ck12.org it doesn’t give as accurate results when the data points follow a linear pattern. In this particular example, the last data point clearly doesn’t ﬁt in with the general trend of the data, so the slope of the extrapolation line is much steeper than it would be if we’d used a line of best ﬁt. (As a historical note, the last data point corresponds to the winning time for Florence Griﬀith Joyner in 1988. After her race she was accused of using performance-enhancing drugs, but this fact was never proven. In addition, there was a question about the accuracy of the timing: some oﬀicials said that tail-wind was not accounted for in this race, even though all the other races of the day were aﬀected by a strong wind.) Here’s an example of a problem where linear extrapolation does work better than the line of best ﬁt method. Example 4 A cylinder is ﬁlled with water to a height of 73 centimeters. The water is drained through a hole in the bottom of the cylinder and measurements are taken at 2 second intervals. The following table shows the height of the water level in the cylinder at diﬀerent times. Table 5.9: Time (seconds) Water level (cm) 0.0 73 2.0 63.9 4.0 55.5 6.0 47.2 8.0 40.0 10.0 33.4 12.0 27.4 14.0 21.9 16.0 17.1 18.0 12.9 20.0 9.4 22.0 6.3 24.0 3.9 26.0 2.0 28.0 0.7 30.0 0.1 a) Find the water level at time 15 seconds. b) Find the water level at time 27 seconds c) What would be the original height of the water in the cylinder if the water takes 5 extra seconds to drain? (Find the height at time of –5 seconds.) Solution Here’s what the line of best ﬁt would look like for this data set: www.ck12.org 232 Notice that the data points don’t really make a line, and so the line of best ﬁt still isn’t a terribly good ﬁt. Just a glance tells us that we’d estimate the water level at 15 seconds to be about 27 cm, which is more than the water level at 14 seconds. That’s clearly not possible! Similarly, at 27 seconds we’d estimate the water to have all drained out, which it clearly hasn’t yet. So let’s see what happens if we use linear extrapolation and interpolation instead. First, here are the lines we’d use to interpolate between 14 and 16 seconds, and between 26 and 28 seconds. a) The slope of the line between points (14, 21.9) and (16, 17.1) is m = 17.1−21.9 16−14 = −4.8 2 = −2.4. So y = −2.4x + b ⇒21.9 = −2.4(14) + b ⇒b = 55.5, and the equation is y = −2.4x + 55.5. Plugging in x = 15 gives us y = −2.4(15) + 55.5 = 19.5 cm. b) The slope of the line between points (26, 2) and (28, 0.7) is m = 0.7−2 28−26 = −1.3 2 = −.65, so y = −.65x+b ⇒ 2 = −.65(26) + b ⇒b = 18.9, and the equation is y = −.65x + 18.9. Plugging in x = 27, we get y = −.65(27) + 18.9 = 1.35 cm. c) Finally, we can use extrapolation to estimate the height of the water at -5 seconds. The slope of the line between points (0, 73) and (2, 63.9) is m = 63.9−73 2−0 = −9.1 2 = −4.55, so the equation of the line is 233 www.ck12.org y = −4.55x + 73. Plugging in x = −5 gives us y = −4.55(−5) + 73 = 95.75 cm. To make linear interpolation easier in the future, you might want to use the calculator at ajdesigner.com/phpinterpolation/linear_interpolation_equation.php. Plug in the coordinates of the ﬁrst known data point in the blanks labeled x1 and y1, and the coordinates of the second point in the blanks labeled x3 and y3; then enter the x−coordinate of the point in between in the blank labeled x2, and the y−coordinate will be displayed below when you click “Calculate.” Review Questions 1. Use the data from Example 1 (Median age at ﬁrst marriage) to estimate the age at marriage for females in 1946. Fit a line, by hand, to the data before 1970. 2. Use the data from Example 1 (Median age at ﬁrst marriage) to estimate the age at marriage for females in 1984. Fit a line, by hand, to the data from 1970 on in order to estimate this accurately. 3. Use the data from Example 1 (Median age at ﬁrst marriage) to estimate the age at marriage for males in 1995. Use linear interpolation between the 1990 and 2000 data points. 4. Use the data from Example 2 (Pregnant women and smoking) to estimate the percentage of pregnant smokers in 1997. Use linear interpolation between the 1996 and 2000 data points. 5. Use the data from Example 2 (Pregnant women and smoking) to estimate the percentage of pregnant smokers in 2006. Use linear extrapolation with the ﬁnal two data points. 6. Use the data from Example 3 (Winning times) to estimate the winning time for the female 100-meter race in 1920. Use linear extrapolation because the ﬁrst two or three data points have a diﬀerent slope than the rest of the data. 7. The table below shows the highest temperature vs. the hours of daylight for the 15th day of each month in the year 2006 in San Diego, California. Table 5.10: Hours of daylight High temperature (F) 10.25 60 11.0 62 12 62 13 66 13.8 68 14.3 73 14 86 13.4 75 12.4 71 11.4 66 10.5 73 10 61 (a) What would be a better way to organize this table if you want to make the relationship between daylight hours and temperature easier to see? (b) Estimate the high temperature for a day with 13.2 hours of daylight using linear interpolation. (c) Estimate the high temperature for a day with 9 hours of daylight using linear extrapolation. Is the www.ck12.org 234 prediction accurate? (d) Estimate the high temperature for a day with 9 hours of daylight using a line of best ﬁt. The table below lists expected life expectancies based on year of birth (US Census Bureau). Use it to answer questions 8-15. Table 5.11: Birth year Life expectancy in years 1930 59.7 1940 62.9 1950 68.2 1960 69.7 1970 70.8 1980 73.7 1990 75.4 2000 77 8. Make a scatter plot of the data. 9. Use a line of best ﬁt to estimate the life expectancy of a person born in 1955. 10. Use linear interpolation to estimate the life expectancy of a person born in 1955. 11. Use a line of best ﬁt to estimate the life expectancy of a person born in 1976. 12. Use linear interpolation to estimate the life expectancy of a person born in 1976. 13. Use a line of best ﬁt to estimate the life expectancy of a person born in 2012. 14. Use linear extrapolation to estimate the life expectancy of a person born in 2012. 15. Which method gives better estimates for this data set? Why? The table below lists the high temperature for the ﬁst day of the month for the year 2006 in San Diego, California (Weather Underground). Use it to answer questions 16-21. Table 5.12: Month number Temperature (F) 1 63 2 66 3 61 4 64 5 71 6 78 7 88 8 78 9 81 10 75 11 68 12 69 16. Draw a scatter plot of the data. 17. Use a line of best ﬁt to estimate the temperature in the middle of the 4th month (month 4.5). 235 www.ck12.org 18. Use linear interpolation to estimate the temperature in the middle of the 4th month (month 4.5). 19. Use a line of best ﬁt to estimate the temperature for month 13 (January 2007). 20. Use linear extrapolation to estimate the temperature for month 13 (January 2007). 21. Which method gives better estimates for this data set? Why? 22. Name a real-world situation where you might want to make predictions based on available data. Would linear extrapolation/interpolation or the best ﬁt method be better to use in that situation? Why? Texas Instruments Resources In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http: //www.ck12.org/flexr/chapter/9615. www.ck12.org 236 Chapter 6 Linear Inequalities 6.1 Solving Inequalities Learning Objectives • Write and graph inequalities in one variable on a number line. • Solve inequalities using addition and subtraction. • Solve inequalities using multiplication and division. • Solve multi-step inequalities. Introduction Dita has a budget of $350 to spend on a rental car for an upcoming trip, but she wants to spend as little of that money as possible. If the trip will last ﬁve days, what range of daily rental rates should she be willing to consider? Like equations, inequalities show a relationship between two expressions. We solve and graph inequalities in a similar way to equations—but when we solve an inequality, the answer is usually a set of values instead of just one value. When writing inequalities we use the following symbols: > is greater than ≥is greater than or equal to < is less than ≤is less than or equal to Write and Graph Inequalities in One Variable on a Number Line Let’s start with the simple inequality x > 3. We read this inequality as “x is greater than 3.” The solution is the set of all real numbers that are greater than three. We often represent the solution set of an inequality with a number line graph. 237 www.ck12.org Consider another simple inequality: x ≤4. We read this inequality as “x is less than or equal to 4.” The solution is the set of all real numbers that are equal to four or less than four. We can graph this solution set on the number line. Notice that we use an empty circle for the endpoint of a strict inequality (like x > 3), and a ﬁlled circle for one where the equals sign is included (like x ≤4). Example 1 Graph the following inequalities on the number line. a) x < −3 b) x ≥6 c) x > 0 d) x ≤8 Solution a) The inequality x < −3 represents all numbers that are less than -3. The number -3 is not included in the solution, so it is represented by an open circle on the graph. b) The inequality x ≥6 represents all numbers that are greater than or equal to 6. The number 6 is included in the solution, so it is represented by a closed circle on the graph. c) The inequality x > 0 represents all numbers that are greater than 0. The number 0 is not included in the solution, so it is represented by an open circle on the graph. d) The inequality x ≤8 represents all numbers that are less than or equal to 8. The number 8 is included in the solution, so it is represented by a closed circle on the graph. Example 2 Write the inequality that is represented by each graph. a) b) c) www.ck12.org 238 d) Solution a) x ≤−12 b) x > 540 c) x < 6.5 d) x ≥85 Inequalities appear everywhere in real life. Here are some simple examples of real-world applications. Example 3 Write each statement as an inequality and graph it on the number line. a) You must maintain a balance of at least $2500 in your checking account to get free checking. b) You must be at least 48 inches tall to ride the “Thunderbolt” Rollercoaster. c) You must be younger than 3 years old to get free admission at the San Diego Zoo. d) The speed limit on the interstate is 65 miles per hour or less. Solution a) The words “at least” imply that the value of $2500 is included in the solution set, so the inequality is written as x ≥2500. b) The words “at least” imply that the value of 48 inches is included in the solution set, so the inequality is written as x ≥48. c) The inequality is written as x < 3. d) Speed limit means the highest allowable speed, so the inequality is written as x ≤65. Solving Inequalities Using Addition and Subtraction To solve an inequality we must isolate the variable on one side of the inequality sign. To isolate the variable, we use the same basic techniques used in solving equations. We can solve some inequalities by adding or subtracting a constant from one side of the inequality. Example 4 Solve each inequality and graph the solution set. a) x −3 < 10 b) x −20 ≥14 239 www.ck12.org c) x + 8 ≤−7 d) x + 4 > 13 Solution a) Starting inequality: x −3 < 10 Add 3 to both sides of the inequality: x −3 + 3 < 10 + 3 Simplify: x < 13 b) Starting inequality: x −20 ≤14 Add 20 to both sides of the inequality: x −20 + 20 ≤14 + 20 Simplify: x ≤34 c) Starting inequality: x + 8 ≤−7 Subtract 8 from both sides of the inequality: x + 8 −8 ≤−7 −8 Simplify: x ≤−15 d) Starting inequality: x + 4 > 13 Subtract 4 from both sides of the inequality: x + 4 −4 > 13 −4 Simplify: x > 9 Solving Inequalities Using Multiplication and Division We can also solve inequalities by multiplying or dividing both sides by a constant. For example, to solve the inequality 5x < 3, we would divide both sides by 5 to get x < 3 5. However, something diﬀerent happens when we multiply or divide by a negative number. We know, for example, that 5 is greater than 3. But if we multiply both sides of the inequality 5 > 3 by -2, we get −10 > −6. And we know that’s not true; -10 is less than -6. This happens whenever we multiply or divide an inequality by a negative number, and so we have to ﬂip the sign around to make the inequality true. For example, to multiply 2 < 4 by -3, ﬁrst we multiply the 2 and the 4 each by -3, and then we change the < sign to a > sign, so we end up with −6 > −12. The same principle applies when the inequality contains variables. Example 5 Solve each inequality. a) 4x < 24 b) −5x ≤21 c) x 25 < 3 2 www.ck12.org 240 d) x −7 ≥9 Solution a) Original problem: 4x < 24 Divide both sides by 4: 4x 4 < 24 4 Simplify: x < 6 b) Original problem: −5x ≤21 Divide both sides by -5 : −5x −5 ≥21 −5 Flip the inequality sign. Simplify: x ≥−21 5 c) Original problem: x 25 < 3 2 Multiply both sides by 25: 25 · x 25 < 3 2 · 25 Simplify: x < 75 2 or x < 37.5 d) Original problem: x −7 ≥9 Multiply both sides by -7: −7 · x −7 ≤9 · (−7) Flip the inequality sign. Simplify: x ≤−63 Solving Multi-Step Inequalities In the last two sections, we considered very simple inequalities which required one step to obtain the solution. However, most inequalities require several steps to arrive at the solution. As with solving equa-tions, we must use the order of operations to ﬁnd the correct solution. In addition, remember that when we multiply or divide the inequality by a negative number, the direction of the inequality changes. The general procedure for solving multi-step inequalities is almost exactly like the procedure for solving multi-step equations: 1. Clear parentheses on both sides of the inequality and collect like terms. 2. Add or subtract terms so the variable is on one side and the constant is on the other side of the inequality sign. 3. Multiply and divide by whatever constants are attached to the variable. Remember to change the direction of the inequality if you multiply or divide by a negative number. Example 6 Solve each of the following inequalities and graph the solution set. a) 9x 5 −7 ≥−3x + 12 b) −25x + 12 ≤−10x −12 Solution a) Original problem: 9x 5 −7 ≥−3x + 12 Add 3x to both sides: 9x 5 + 3x −7 ≥−3x + 3x + 12 Simplify: 24x 5 −7 ≥12 Add 7 to both sides: 24x 5 −7 + 7 ≥12 + 7 Simplify: 24x 5 ≥19 241 www.ck12.org Multiply 5 to both sides: 5 · 24x 5 ≥5 · 19 Simplify: 24x ≥95 Divide both sides by 24: 24x 24 ≥95 24 Simplify: x ≥95 24 Answer Graph: b) Original problem: −25x + 12 ≤−10x −12 Add 10x to both sides: −25x + 10x + 12 ≤−10x + 10x −12 Simplify: −15x + 12 ≤−12 Subtract 12: −15x + 12 −12 ≤−12 −12 Simplify: −15x ≤−24 Divide both sides by -15: −15x −15 ≥−24 −15 ﬂip the inequality sign Simplify: x ≥8 5 Answer Graph: Example 7 Solve the following inequalities. a) 4x −2(3x −9) ≤−4(2x −9) b) 5x−1 4 > −2(x + 5) Solution a) Original problem: 4x −2(3x −9) ≤−4(2x −9) Simplify parentheses: 4x −6x + 18 ≤−8x + 36 Collect like terms: −2x + 18 ≤−8x + 36 Add 8x to both sides: −2x + 8x + 18 ≤−8x + 8x + 36 Simplify: 6x + 18 ≤36 Subtract 18: 6x + 18 −18 ≤36 −18 Simplify: 6x ≤18 Divide both sides by 6: 6x 6 ≤18 6 Simplify: x ≤3 Answer b) Original problem: 5x−1 4 > −2(x + 5) Simplify parenthesis: 5x−1 4 > −2x −10 Multiply both sides by 4: 4 · 5x−1 4 > 4(−2x −10) Simplify: 5x −1 > −8x −40 Add 8x to both sides: 5x + 8x −1 > −8x + 8x −40 Simplify: 13x −1 > −40 Add 1 to both sides: 13x −1 + 1 > −40 + 1 Simplify: 13x > −39 www.ck12.org 242 Divide both sides by 13: 13x 13 > −39 13 Simplify: x > −3 Answer Further Practice For additional practice solving inequalities, try the online game at htm#section2. If you’re having a hard time with multi-step inequalities, the video at schooltube.com/video/aa66df49e0af4f85a5e9/MultiStep-Inequalities will walk you through a few. Lesson Summary • The answer to an inequality is usually an interval of values. • Solving inequalities works just like solving an equation. To solve, we isolate the variable on one side of the equation. • When multiplying or dividing both sides of an inequality by a negative number, you need to reverse the inequality. Review Questions 1. Write the inequality represented by the graph. 2. Write the inequality represented by the graph. 3. Write the inequality represented by the graph. 4. Write the inequality represented by the graph. Graph each inequality on the number line. 5. x < −35 6. x > −17 7. x ≥20 8. x ≤3 Solve each inequality and graph the solution on the number line. 9. x −5 < 35 10. x + 15 ≥−60 11. x −2 ≤1 12. x −8 > −20 13. x + 11 > 13 14. x + 65 < 100 243 www.ck12.org 15. x −32 ≤0 16. x + 68 ≥75 Solve each inequality. Write the solution as an inequality and graph it. 17. 3x ≤6 18. x 5 > −3 10 19. −10x > 250 20. x −7 ≥−5 Solve each multi-step inequality. 21. x −5 > 2x + 3 22. 2(x −3) ≤3x −2 23. x 3 < x + 7 24. 3(x−4) 12 ≤2x 3 25. 2 ( x 4 + 3 ) > 6(x −1) 26. 9x + 4 ≤−2 ( x + 1 2 ) 6.2 Using Inequalities Learning Objectives • Express answers to inequalities in a variety of ways. • Identify the number of solutions of an inequality. • Solve real-world problems using inequalities. Introduction Ms. Jerome wants to buy identical boxes of art supplies for her 25 students. If she can spend no more than $375 on art supplies, what inequality describes the price can she aﬀord for each individual box of supplies? Expressing Solutions of an Inequality The solution of an inequality can be expressed in four diﬀerent ways: 1. Inequality notation The answer is simply expressed as x < 15. 2. Set notation The answer is expressed as a set: {x\|x < 15}. The brackets indicate a set and the vertical line means “such that,” so we read this expression as “the set of all values of x such that x is a real number less than 15”. 3. Interval notation uses brackets to indicate the range of values in the solution. For example, the answer to our problem would be expressed as (−∞, 15), meaning “the interval containing all the numbers from −∞to 15 but not actually including −∞or 15”. (a) Square or closed brackets “[” and “]” indicate that the number next to the bracket is included in the solution set. www.ck12.org 244 (b) Round or open brackets “(” and “)” indicate that the number next to the bracket is not included in the solution set. When using inﬁnity and negative inﬁnity (∞and −∞), we always use open brackets, because inﬁnity isn’t an actual number and so it can’t ever really be included in an interval. 4. Solution graph shows the solution on the real number line. A closed circle on a number indicates that the number is included in the solution set, while an open circle indicates that the number is not included in the set. For our example, the solution graph is: Example 1 a) [-4, 6] means that the solutions is all numbers between -4 and 6 including -4 and 6. b) (8, 24) means that the solution is all numbers between 8 and 24 not including the numbers 8 and 24. c) [3, 12) means that the solution is all numbers between 3 and 12, including 3 but not including 12. d) (−10, ∞) means that the solution is all numbers greater than -10, not including -10. e) (−∞, ∞) means that the solution is all real numbers. Identify the Number of Solutions of an Inequality Inequalities can have: • A set that has an inﬁnite number of solutions. • A set that has a discrete number of solutions. • No solutions. The inequalities we have solved so far all have an inﬁnite number of solutions, at least in theory. For example, the inequality 5x−1 4 > −2(x + 5) has the solution x > −3. This solution says that all real numbers greater than -3 make this inequality true, and there are inﬁnitely many such numbers. However, in real life, sometimes we are trying to solve a problem that can only have positive integer answers, because the answers describe numbers of discrete objects. For example, suppose you are trying to ﬁgure out how many $8 CDs you can buy if you want to spend less than $50. An inequality to describe this situation would be 8x < 50, and if you solved that inequality you would get x < 50 8 , or x < 6.25. But could you really buy any number of CDs as long as it’s less than 6.25? No; you couldn’t really buy 6.1 CDs, or -5 CDs, or any other fractional or negative number of CDs. So if we wanted to express our solution in set notation, we couldn’t express it as the set of all numbers less than 6.25, or {x\|x < 6.25}. Instead, the solution is just the set containing all the nonnegative whole numbers less than 6.25, or {0, 1, 2, 3, 4, 5, 6}. When we’re solving a real-world problem dealing with discrete objects like CDs, our solution set will often be a ﬁnite set of numbers instead of an inﬁnite interval. An inequality can also have no solutions at all. For example, consider the inequality x −5 > x + 6. When we subtract x from both sides, we end up with −5 > 6, which is not true for any value of x. We say that this inequality has no solution. The opposite can also be true. If we ﬂip the inequality sign in the above inequality, we get x −5 < x + 6, which simpliﬁes to −5 < 6. That’s always true no matter what x is, so the solution to that inequality would be all real numbers, or (−∞, ∞). 245 www.ck12.org Solve Real-World Problems Using Inequalities Solving real-world problems that involve inequalities is very much like solving problems that involve equa-tions. Example 2 In order to get a bonus this month, Leon must sell at least 120 newspaper subscriptions. He sold 85 subscriptions in the ﬁrst three weeks of the month. How many subscriptions must Leon sell in the last week of the month? Solution Let x = the number of subscriptions Leon sells in the last week of the month. The total number of subscriptions for the month must be greater than 120, so we write 85 + x ≥120. We solve the inequality by subtracting 85 from both sides: x ≥35. Leon must sell 35 or more subscriptions in the last week to get his bonus. To check the answer, we see that 85 + 35 = 120. If he sells 35 or more subscriptions, the total number of subscriptions he sells that month will be 120 or more. The answer checks out. Example 3 Virena’s Scout troop is trying to raise at least $650 this spring. How many boxes of cookies must they sell at $4.50 per box in order to reach their goal? Solution Let x = number of boxes sold. Then the inequality describing this problem is 4.50x ≥650. We solve the inequality by dividing both sides by 4.50: x ≥144.44. We round up the answer to 145 since only whole boxes can be sold. Virena’s troop must sell at least 145 boxes. If we multiply 145 by $4.50 we obtain $652.50, so if Virena’s troop sells more than 145 boxes they will raise more than $650. But if they sell 144 boxes, they will only raise $648, which is not enough. So they must indeed sell at least 145 boxes. The answer checks out. Example 4 The width of a rectangle is 20 inches. What must the length be if the perimeter is at least 180 inches? Solution Let x = length of the rectangle. The formula for perimeter is Perimeter = 2 × length + 2 × width Since the perimeter must be at least 180 inches, we have 2x + 2(20) ≥180. Simplify: 2x + 40 ≥180 Subtract 40 from both sides: 2x ≥140 Divide both sides by 2: x ≥70 The length must be at least 70 inches. If the length is at least 70 inches and the width is 20 inches, then the perimeter is at least 2(70) + 2(20) = 180 inches. The answer checks out. www.ck12.org 246 Further Practice The videos at and youtube.com/watch?v=ArzPkaqym50 contain more examples of real-world problems using inequalities. Lesson Summary • Inequalities can have inﬁnite solutions, no solutions, or discrete solutions. • There are four ways to represent an inequality: Equation notation, set notation, interval notation, and solution graph. Review Questions Solve each inequality. Give the solution in inequality notation and interval notation. 1. x + 15 < 12 2. x −4 ≥13 3. 9x > −3 4 4. −x 15 ≤5 5. 620x > 2400 6. x 20 ≥−7 40 7. 3x 5 > 3 5 8. x + 3 > x −2 Solve each inequality. Give the solution in inequality notation and set notation. 9. x + 17 < 3 10. x −12 ≥80 11. −0.5x ≤7.5 12. 75x ≥125 13. x −3 > −10 9 14. x −15 < 8 15. x 4 > 5 4 16. 3x −7 ≥3(x −7) Solve the following inequalities, give the solution in set notation, and show the solution graph. 17. 4x + 3 < −1 18. 2x < 7x −36 19. 5x > 8x + 27 20. 5 −x < 9 + x 21. 4 −6x ≤2(2x + 3) 22. 5(4x + 3) ≥9(x −2) −x 23. 2(2x −1) + 3 < 5(x + 3) −2x 24. 8x −5(4x + 1) ≥−1 + 2(4x −3) 25. 9. 2(7x −2) −3(x + 2) < 4x −(3x + 4) 26. 2 3 x −1 2(4x −1) ≥x + 2(x −3) 247 www.ck12.org 27. At the San Diego Zoo you can either pay $22.75 for the entrance fee or $71 for the yearly pass which entitles you to unlimited admission. (a) At most how many times can you enter the zoo for the $22.75 entrance fee before spending more than the cost of a yearly membership? (b) Are there inﬁnitely many or ﬁnitely many solutions to this inequality? 28. Proteek’s scores for four tests were 82, 95, 86, and 88. What will he have to score on his ﬁfth and last test to average at least 90 for the term? 6.3 Compound Inequalities Learning Objectives • Write and graph compound inequalities on a number line. • Solve compound inequalities with “and.” • Solve compound inequalities with “or.” • Solve compound inequalities using a graphing calculator (TI family). • Solve real-world problems using compound inequalities. Introduction In this section, we’ll solve compound inequalities—inequalities with more than one constraint on the possible values the solution can have. There are two types of compound inequalities: 1. Inequalities joined by the word “and,” where the solution is a set of values greater than a number and less than another number. We can write these inequalities in the form “x > a and x < b,” but usually we just write “a < x < b.” Possible values for x are ones that will make both inequalities true. 2. Inequalities joined by the word “or,” where the solution is a set of values greater than a number or less than another number. We write these inequalities in the form “x > a or x < b.” Possible values for x are ones that will make at least one of the inequalities true. You might wonder why the variable x has to be greater than one number and/or less than the other number; why can’t it be greater than both numbers, or less than both numbers? To see why, let’s take an example. Consider the compound inequality “x > 5 and x > 3.” Are there any numbers greater than 5 that are not greater than 3? No! Since 5 is greater than 3, everything greater than 5 is also greater than 3. If we say x is greater than both 5 and 3, that doesn’t tell us any more than if we just said x is greater than 5. So this compound inequality isn’t really compound; it’s equivalent to the simple inequality x > 5. And that’s what would happen no matter which two numbers we used; saying that x is greater than both numbers is just the same as saying that x is greater than the bigger number, and saying that x is less than both numbers is just the same as saying that x is less than the smaller number. Compound inequalities with “or” work much the same way. Every number that’s greater than 3 or greater than 5 is also just plain greater than 3, and every number that’s greater than 3 is certainly greater than 3 or greater than 5—so if we say “x > 5 or x > 3,” that’s the same as saying just “x > 3.” Saying that x is greater than at least one of two numbers is just the same as saying that x is greater than the smaller number, and saying that x is less than at least one of two numbers is just the same as saying that x is less than the greater number. www.ck12.org 248 Write and Graph Compound Inequalities on a Number Line Example 1 Write the inequalities represented by the following number line graphs. a) b) c) d) Solution a) The solution graph shows that the solution is any value between -40 and 60, including -40 but not 60. Any value in the solution set satisﬁes both x ≥−40 and x < 60. This is usually written as −40 ≤x < 60. b) The solution graph shows that the solution is any value greater than 1 (not including 1) or any value less than -2 (not including -2). You can see that there can be no values that can satisfy both these conditions at the same time. We write: x > 1 or x < −2. c) The solution graph shows that the solution is any value greater than 4 (including 4) or any value less than -1 (including - 1). We write: x ≥4 or x ≤−1. d) The solution graph shows that the solution is any value that is both less than 25 (not including 25) and greater than -25 (not including -25). Any value in the solution set satisﬁes both x > −25 and x < 25. This is usually written as −25 < x < 25. Example 2 Graph the following compound inequalities on a number line. a) −4 ≤x ≤6 b) x < 0 or x > 2 c) x ≥−8 or x ≤−20 d) −15 < x ≤85 Solution a) The solution is all numbers between -4 and 6, including both -4 and 6. b) The solution is all numbers less than 0 or greater than 2, not including 0 or 2. c) The solution is all numbers greater than or equal to -8 or less than or equal to -20. 249 www.ck12.org d) The solution is all numbers between -15 and 85, not including -15 but including 85. Solve a Compound Inequality With “and” or “or” When we solve compound inequalities, we separate the inequalities and solve each of them separately. Then, we combine the solutions at the end. Example 3 Solve the following compound inequalities and graph the solution set. a) −2 < 4x −5 ≤11 b) 3x −5 < x + 9 ≤5x + 13 Solution a) First we re-write the compound inequality as two separate inequalities with and. Then solve each inequality separately. −2 < 4x −5 4x −5 ≤11 3 < 4x and 4x ≤16 3 4 < x x ≤4 Answer: 3 4 < x and x ≤4. This can be written as 3 4 < x ≤4. b) Re-write the compound inequality as two separate inequalities with and. Then solve each inequality separately. 3x −5 < x + 9 x + 9 ≤5x + 13 2x < 14 and −4 ≤4x x < 7 −1 ≤x Answer: x < 7 and x ≥−1. This can be written as: −1 ≤x < 7. Example 4 Solve the following compound inequalities and graph the solution set. a) 9 −2x ≤3 or 3x + 10 ≤6 −x b) x−2 6 ≤2x −4 or x−2 6 > x + 5 Solution a) Solve each inequality separately: 9 −2x ≤3 3x + 10 ≤6 −x −2x ≤−6 or 4x ≤−4 x ≥3 x ≤−1 www.ck12.org 250 Answer: x ≥3 or x ≤−1 b) Solve each inequality separately: x −2 6 ≤2x −4 x −2 6 > x + 5 x −2 ≤6(2x −4) x −2 > 6(x + 5) x −2 ≤12x −24 or x −2 > 6x + 30 22 ≤11x −32 > 5x 2 ≤x −6.4 > x Answer: x ≥2 or x < −6.4 The video at shows the pro-cess of solving and graphing compound inequalities in more detail. One thing you may notice in this video is that in the second problem, the two solutions joined with “or” overlap, and so the solution ends up being the set of all real numbers, or (−∞, ∞). This happens sometimes with compound inequalities that involve “or”; for example, if the solution to an inequality ended up being “x < 5 or x > 1,” the solution set would be all real numbers. This makes sense if you think about it: all real numbers are either a) less than 5, or b) greater than or equal to 5, and the ones that are greater than or equal to 5 are also greater than 1—so all real numbers are either a) less than 5 or b) greater than 1. Compound inequalities with “and,” meanwhile, can turn out to have no solutions. For example, the inequality “x < 3 and x > 4” has no solutions: no number is both greater than 4 and less than 3. If we write it as 4 < x < 3 it’s even more obvious that it has no solutions; 4 < x < 3 implies that 4 < 3, which is false. Solve Compound Inequalities Using a Graphing Calculator (TI-83/84 family) Graphing calculators can show you the solution to an inequality in the form of a graph. This can be especially useful when dealing with compound inequalities. Example 5 Solve the following inequalities using a graphing calculator. a) 5x + 2(x −3) ≥2 b) 7x −2 < 10x + 1 < 9x + 5 c) 3x + 2 ≤10 or 3x + 2 ≥15 Solution a) Press the [Y=] button and enter the inequality on the ﬁrst line of the screen. 251 www.ck12.org (To get the ≥symbol, press [TEST] [2nd] [MATH] and choose option 4.) Then press the [GRAPH] button. Because the calculator uses the number 1 to mean “true” and 0 to mean “false,” you will see a step function with the y−value jumping from 0 to 1. The solution set is the values of x for which the graph shows y = 1—in other words, the set of x−values that make the inequality true. Note: You may need to press the [WINDOW] key or the [ZOOM] key to adjust the window to see the full graph. The solution is x > 8 7, which is why you can see the y−value changing from 0 to 1 at about 1.14. b) This is a compound inequality: 7x −2 < 10x + 1 and 10x + 1 < 9x + 5. You enter it like this: www.ck12.org 252 (To ﬁnd the [AND] symbol, press [TEST], choose [LOGIC] on the top row and choose option 1.) The resulting graph should look like this: The solution are the values of x for which y = 1; in this case that would be −1 < x < 4. c) This is another compound inequality. (To enter the [OR] symbol, press [TEST], choose [LOGIC] on the top row and choose option 2.) The resulting graph should look like this: The solution are the values of x for which y = 1–in this case, x ≤2.7 or x ≥4.3. Solve Real-World Problems Using Compound Inequalities Many application problems require the use of compound inequalities to ﬁnd the solution. Example 6 The speed of a golf ball in the air is given by the formula v = −32t + 80. When is the ball traveling between 20 ft/sec and 30 ft/sec? 253 www.ck12.org Solution First we set up the inequality 20 ≤v ≤30, and then replace v with the formula v = −32t + 80 to get 20 ≤−32t + 80 ≤30. Then we separate the compound inequality and solve each separate inequality: 20 ≤−32t + 80 −32t + 80 ≤30 32t ≤60 and 50 ≤32t t ≤1.875 1.56 ≤t Answer: 1.56 ≤t ≤1.875 To check the answer, we plug in the minimum and maximum values of t into the formula for the speed. For t = 1.56, v = −32t + 80 = −32(1.56) + 80 = 30 ft/sec For t = 1.875, v = −32t + 80 = −32(1.875) + 80 = 20 ft/sec So the speed is between 20 and 30 ft/sec. The answer checks out. Example 7 William’s pick-up truck gets between 18 to 22 miles per gallon of gasoline. His gas tank can hold 15 gallons of gasoline. If he drives at an average speed of 40 miles per hour, how much driving time does he get on a full tank of gas? Solution Let t = driving time. We can use dimensional analysis to get from time per tank to miles per gallon: t hours 1 tank × 1 tank 15 gallons × 40 miles 1 hour × 40t 15 miles gallon Since the truck gets between 18 and 22 miles/gallon, we set up the compound inequality 18 ≤40t 15 ≤22. Then we separate the compound inequality and solve each inequality separately: 18 ≤40t 15 40t 15 ≤22 270 ≤40t and 40t ≤330 6.75 ≤t t ≤8.25 Answer: 6.75 ≤t ≤8.25. Andrew can drive between 6.75 and 8.25 hours on a full tank of gas. If we plug in t = 6.75 we get 40t 15 = 40(6.75) 15 = 18 miles per gallon. If we plug in t = 8.25 we get 40t 15 = 40(8.25) 15 = 22 miles per gallon. The answer checks out. Lesson Summary • Compound inequalities combine two or more inequalities with “and” or “or. ” • “And” combinations mean that only solutions for both inequalities will be solutions to the compound inequality. • “Or” combinations mean solutions to either inequality will also be solutions to the compound in-equality. www.ck12.org 254 Review Questions Write the compound inequalities represented by the following graphs. 1. 2. 3. 4. Solve the following compound inequalities and graph the solution on a number line. 5. −5 ≤x −4 ≤13 6. 1 ≤3x + 5 ≤4 7. −12 ≤2 −5x ≤7 8. 3 4 ≤2x + 9 ≤3 2 9. −2 ≤2x−1 3 < −1 10. 4x −1 ≥7 or 9x 2 < 3 11. 3 −x < −4 or 3 −x > 10 12. 2x+3 4 < 2 or −x 5 + 3 < 2 5 13. 2x −7 ≤−3 or 2x −3 > 11 14. 4x + 3 < 9 or −5x + 4 ≤−12 15. How would you express the answer to problem 5 as a set? 16. How would you express the answer to problem 5 as an interval? 17. How would you express the answer to problem 10 as a set? 18. (a) Could you express the answer to problem 10 as a single interval? Why or why not? (b) How would you express the ﬁrst part of the solution in interval form? (c) How would you express the second part of the solution in interval form? 19. Express the answers to problems 1 and 3 in interval notation. 20. Express the answers to problems 6 through 9 in interval notation. 21. Solve the inequality “x ≥−3 or x < 1” and express the answer in interval notation. 22. How many solutions does the inequality “x ≥2 and x ≤2” have? 23. To get a grade of B in her Algebra class, Stacey must have an average grade greater than or equal to 80 and less than 90. She received the grades of 92, 78, 85 on her ﬁrst three tests. (a) Between which scores must her grade on the ﬁnal test fall if she is to receive a grade of B for the class? (Assume all four tests are weighted the same.) (b) What range of scores on the ﬁnal test would give her an overall grade of C, if a C grade requires an average score greater than or equal to 70 and less than 80? (c) If an A grade requires a score of at least 90, and the maximum score on a single test is 100, is it possible for her to get an A in this class? (Hint: look again at your answer to part a.) 6.4 Absolute Value Equations and Inequalities Learning Objectives • Solve an absolute value equation. 255 www.ck12.org • Analyze solutions to absolute value equations. • Graph absolute value functions. • Solve absolute value inequalities. • Rewrite and solve absolute value inequalities as compound inequalities. • Solve real-world problems using absolute value equations and inequalities. Introduction Timmy is trying out his new roller skates. He’s not allowed to cross the street yet, so he skates back and forth in front of his house. If he skates 20 yards east and then 10 yards west, how far is he from where he started? What if he skates 20 yards west and then 10 yards east? The absolute value of a number is its distance from zero on a number line. There are always two numbers on the number line that are the same distance from zero. For instance, the numbers 4 and -4 are each a distance of 4 units away from zero. \|4\| represents the distance from 4 to zero, which equals 4. \| −4\| represents the distance from -4 to zero, which also equals 4. In fact, for any real number x: \|x\| = x if x is not negative, and \|x\| = −x if x is negative. Absolute value has no eﬀect on a positive number, but changes a negative number into its positive inverse. Example 1 Evaluate the following absolute values. a) \|25\| b) \| −120\| c) \| −3\| d) \|55\| e) −5 4 Solution a) \|25\| = 25 Since 25 is a positive number, the absolute value does not change it. b) \| −120\| = 120 Since -120 is a negative number, the absolute value makes it positive. c) \| −3\| = 3 Since -3 is a negative number, the absolute value makes it positive. d) \|55\| = 55 Since 55 is a positive number, the absolute value does not change it. e) −5 4 = 5 4 Since −5 4 is a negative number, the absolute value makes it positive. Absolute value is very useful in ﬁnding the distance between two points on the number line. The distance between any two points a and b on the number line is \|a −b\| or \|b −a\|. For example, the distance from 3 to -1 on the number line is \|3 −(−1)\| = \|4\| = 4. www.ck12.org 256 We could have also found the distance by subtracting in the opposite order: \| −1 −3\| = \| −4\| = 4. This makes sense because the distance is the same whether you are going from 3 to -1 or from -1 to 3. Example 2 Find the distance between the following points on the number line. a) 6 and 15 b) -5 and 8 c) -3 and -12 Solution Distance is the absolute value of the diﬀerence between the two points. a) distance = \|6 −15\| = \| −9\| = 9 b) distance = \| −5 −8\| = \| −13\| = 13 c) distance = \| −3 −(−12)\| = \|9\| = 9 Remember: When we computed the change in x and the change in y as part of the slope computa-tion, these values were positive or negative, depending on the direction of movement. In this discussion, “distance” means a positive distance only. Solve an Absolute Value Equation We now want to solve equations involving absolute values. Consider the following equation: \|x\| = 8 This means that the distance from the number x to zero is 8. There are two numbers that satisfy this condition: 8 and -8. When we solve absolute value equations we always consider two possibilities: 1. The expression inside the absolute value sign is not negative. 2. The expression inside the absolute value sign is negative. Then we solve each equation separately. Example 3 Solve the following absolute value equations. a) \|x\| = 3 b) \|x\| = 10 Solution a) There are two possibilities: x = 3 and x = −3. b) There are two possibilities: x = 10 and x = −10. 257 www.ck12.org Analyze Solutions to Absolute Value Equations Example 4 Solve the equation \|x −4\| = 5 and interpret the answers. Solution We consider two possibilities: the expression inside the absolute value sign is nonnegative or is negative. Then we solve each equation separately. x −4 = 5 and x −4 = −5 x = 9 x = −1 x = 9 and x = −1 are the solutions. The equation \|x −4\| = 5 can be interpreted as “what numbers on the number line are 5 units away from the number 4?” If we draw the number line we see that there are two possibilities: 9 and -1. Example 5 Solve the equation \|x + 3\| = 2 and interpret the answers. Solution Solve the two equations: x + 3 = 2 and x + 3 = −2 x = −1 x = −5 x = −5 and x = −1 are the answers. The equation \|x + 3\| = 2 can be re-written as: \|x −(−3)\| = 2. We can interpret this as “what numbers on the number line are 2 units away from -3?” There are two possibilities: -5 and -1. Example 6 Solve the equation \|2x −7\| = 6 and interpret the answers. Solution Solve the two equations: 2x −7 = 6 2x −7 = −6 2x = 13 and 2x = 1 x = 13 2 x = 1 2 Answer: x = 13 2 and x = 1 2. www.ck12.org 258 The interpretation of this problem is clearer if the equation \|2x −7\| = 6 is divided by 2 on both sides to get 1 2\|2x −7\| = 3. Because 1 2 is nonnegative, we can distribute it over the absolute value sign to get x −7 2 = 3. The question then becomes “What numbers on the number line are 3 units away from 7 2?” There are two answers: 13 2 and 1 2. Graph Absolute Value Functions Now let’s look at how to graph absolute value functions. Consider the function y = \|x −1\|. We can graph this function by making a table of values: Table 6.1: x y = \|x −1\| -2 y = \| −2 −1\| = \| −3\| = 3 -1 y = \| −1 −1\| = \| −2\| = 2 0 y = \|0 −1\| = \| −1\| = 1 1 y = \|1 −1\| = \|0\| = 0 2 y = \|2 −1\| = \|1\| = 1 3 y = \|3 −1\| = \|2\| = 2 4 y = \|4 −1\| = \|3\| = 3 You can see that the graph of an absolute value function makes a big “V”. It consists of two line rays (or line segments), one with positive slope and one with negative slope, joined at the vertex or cusp. We’ve already seen that to solve an absolute value equation we need to consider two options: 1. The expression inside the absolute value is not negative. 2. The expression inside the absolute value is negative. Combining these two options gives us the two parts of the graph. 259 www.ck12.org For instance, in the above example, the expression inside the absolute value sign is x −1. By deﬁnition, this expression is nonnegative when x −1 ≥0, which is to say when x ≥1. When the expression inside the absolute value sign is nonnegative, we can just drop the absolute value sign. So for all values of x greater than or equal to 1, the equation is just y = x −1. On the other hand, when x−1 < 0 — in other words, when x < 1 — the expression inside the absolute value sign is negative. That means we have to drop the absolute value sign but also multiply the expression by -1. So for all values of x less than 1, the equation is y = −(x −1), or y = −x + 1. These are both graphs of straight lines, as shown above. They meet at the point where x −1 = 0 — that is, at x = 1. We can graph absolute value functions by breaking them down algebraically as we just did, or we can graph them using a table of values. However, when the absolute value equation is linear, the easiest way to graph it is to combine those two techniques, as follows: 1. Find the vertex of the graph by setting the expression inside the absolute value equal to zero and solving for x. 2. Make a table of values that includes the vertex, a value smaller than the vertex, and a value larger than the vertex. Calculate the corresponding values of y using the equation of the function. 3. Plot the points and connect them with two straight lines that meet at the vertex. Example 7 Graph the absolute value function y = \|x + 5\|. Solution Step 1: Find the vertex by solving x + 5 = 0. The vertex is at x = −5. Step 2: Make a table of values: Table 6.2: x y = \|x + 5\| -8 y = \| −8 + 5\| = \| −3\| = 3 -5 y = \| −5 + 5\| = \|0\| = 0 -2 y = \| −2 + 5\| = \|3\| = 3 Step 3: Plot the points and draw two straight lines that meet at the vertex: www.ck12.org 260 Example 8 Graph the absolute value function: y = \|3x −12\| Solution Step 1: Find the vertex by solving 3x −12 = 0. The vertex is at x = 4. Step 2: Make a table of values: Table 6.3: x y = \|3x −12\| 0 y = \|3(0) −12\| = \| −12\| = 12 4 y = \|3(4) −12\| = \|0\| = 0 8 y = \|3(8) −12\| = \|12\| = 12 Step 3: Plot the points and draw two straight lines that meet at the vertex. Solve Real-World Problems Using Absolute Value Equations Example 9 A company packs coﬀee beans in airtight bags. Each bag should weigh 16 ounces, but it is hard to ﬁll each bag to the exact weight. After being ﬁlled, each bag is weighed; if it is more than 0.25 ounces overweight or underweight, it is emptied and repacked. What are the lightest and heaviest acceptable bags? Solution The weight of each bag is allowed to be 0.25 ounces away from 16 ounces; in other words, the diﬀerence between the bag’s weight and 16 ounces is allowed to be 0.25 ounces. So if x is the weight of a bag in ounces, then the equation that describes this problem is \|x −16\| = 0.25. Now we must consider the positive and negative options and solve each equation separately: x −16 = 0.25 and x −16 = −0.25 x = 16.25 x = 15.75 The lightest acceptable bag weighs 15.75 ounces and the heaviest weighs 16.25 ounces. We see that 16.25 −16 = 0.25 ounces and 16 −15.75 = 0.25 ounces. The answers are 0.25 ounces bigger and smaller than 16 ounces respectively. 261 www.ck12.org The answer checks out. The answer you just found describes the lightest and heaviest acceptable bags of coﬀee beans. But how do we describe the total possible range of acceptable weights? That’s where inequalities become useful once again. Absolute Value Inequalities Absolute value inequalities are solved in a similar way to absolute value equations. In both cases, you must consider the same two options: 1. The expression inside the absolute value is not negative. 2. The expression inside the absolute value is negative. Then you must solve each inequality separately. Solve Absolute Value Inequalities Consider the inequality \|x\| ≤3. Since the absolute value of x represents the distance from zero, the solutions to this inequality are those numbers whose distance from zero is less than or equal to 3. The following graph shows this solution: Notice that this is also the graph for the compound inequality −3 ≤x ≤3. Now consider the inequality \|x\| > 2. Since the absolute value of x represents the distance from zero, the solutions to this inequality are those numbers whose distance from zero are more than 2. The following graph shows this solution. Notice that this is also the graph for the compound inequality x < −2 or x > 2. Example 1 Solve the following inequalities and show the solution graph. a) \|x\| < 5 b) \|x\| ≥2.5 Solution a) \|x\| < 5 represents all numbers whose distance from zero is less than 5. This answer can be written as “−5 < x < 5” . b) \|x\| ≥2.5 represents all numbers whose distance from zero is more than or equal to 2.5 This answer can be written as “x ≤−2.5 or x ≥2.5” . www.ck12.org 262 Rewrite and Solve Absolute Value Inequalities as Compound In-equalities In the last section you saw that absolute value inequalities are compound inequalities. Inequalities of the type \|x\| < a can be rewritten as “−a < x < a”. Inequalities of the type \|x\| > b can be rewritten as “x < −b or x > b.” To solve an absolute value inequality, we separate the expression into two inequalities and solve each of them individually. Example 2 Solve the inequality \|x −3\| < 7 and show the solution graph. Solution Re-write as a compound inequality: −7 < x −3 < 7 Write as two separate inequalities: x −3 < 7 and x −3 > −7 Solve each inequality: x < 10 and x > −4 Re-write solution: −4 < x < 10 The solution graph is We can think of the question being asked here as “What numbers are within 7 units of 3?”; the answer can then be expressed as “All the numbers between -4 and 10.” Example 3 Solve the inequality \|4x + 5\| ≤13 and show the solution graph. Solution Re-write as a compound inequality: −13 ≤4x + 5 ≤13 Write as two separate inequalities: 4x + 5 ≤13 and 4x + 5 ≥−13 Solve each inequality: 4x ≤8 and 4x ≥−18 x ≤2 and x ≥−9 2 Re-write solution: −9 2 ≤x ≤2 The solution graph is Example 4 Solve the inequality \|x + 12\| > 2 and show the solution graph. Solution Re-write as a compound inequality: x + 12 < −2 or x + 12 > 2 Solve each inequality: x < −14 or x > −10 The solution graph is Example 5 Solve the inequality \|8x −15\| ≥9 and show the solution graph. 263 www.ck12.org Solution Re-write as a compound inequality: 8x −15 ≤−9 or 8x −15 ≥9 Solve each inequality: 8x ≤6 or 8x ≥24 x ≤3 4 or x ≥3 The solution graph is Solve Real-World Problems Using Absolute Value Inequalities Absolute value inequalities are useful in problems where we are dealing with a range of values. Example 6 The velocity of an object is given by the formula v = 25t −80, where the time is expressed in seconds and the velocity is expressed in feet per second. Find the times for which the magnitude of the velocity is greater than or equal to 60 feet per second. Solution The magnitude of the velocity is the absolute value of the velocity. If the velocity is 25t−80 feet per second, then its magnitude is \|25t −80\| feet per second. We want to ﬁnd out when that magnitude is greater than or equal to 60, so we need to solve \|25t −80\| ≥60 for t. First we have to split it up: 25t −80 ≥60 or 25t −80 ≤−60 Then solve: 25t ≥140 or 25t ≤20 t ≥5.6 or t ≤0.8 The magnitude of the velocity is greater than 60 ft/sec for times less than 0.8 seconds and for times greater than 5.6 seconds. When t = 0.8 seconds, v = 25(0.8) −80 = −60 ft/sec. The magnitude of the velocity is 60 ft/sec. (The negative sign in the answer means that the object is moving backwards.) When t = 5.6 seconds, v = 25(5.6) −80 = 60 ft/sec. To ﬁnd where the magnitude of the velocity is greater than 60 ft/sec, check some arbitrary values in each of the following time intervals: t ≤0.8, 0.8 ≤t ≤5.6 and t ≥5.6. Check t = 0.5 : v = 25(0.5) −80 = −67.5 ft/sec Check t = 2 : v = 25(2) −80 = −30 ft/sec Check t = 6 : v = 25(6) −80 = −70 ft/sec You can see that the magnitude of the velocity is greater than 60 ft/sec only when t ≥5.6 or when t ≤0.8. The answer checks out. Further Resources For a multimedia presentation on absolute value equations and inequalities, see com/viewVideo.php?video_id=124516. www.ck12.org 264 Lesson Summary • The absolute value of a number is its distance from zero on a number line. • \|x\| = x if x is not negative, and \|x\| = −x if x is negative. • An equation or inequality with an absolute value in it splits into two equations, one where the expression inside the absolute value sign is positive and one where it is negative. When the expression within the absolute value is positive, then the absolute value signs do nothing and can be omitted. When the expression within the absolute value is negative, then the expression within the absolute value signs must be negated before removing the signs. • Inequalities of the type \|x\| < a can be rewritten as “−a < x < a.” • Inequalities of the type \|x\| > b can be rewritten as “x < −b or x > b.” Review Questions Evaluate the absolute values. 1. \|250\| 2. \| −12\| 3. −2 5 4. 1 10 Find the distance between the points. 5. 12 and -11 6. 5 and 22 7. -9 and -18 8. -2 and 3 Solve the absolute value equations and interpret the results by graphing the solutions on the number line. 9. \|x −5\| = 10 10. \|x + 2\| = 6 11. \|5x −2\| = 3 12. \|x −4\| = −3 Graph the absolute value functions. 13. y = \|x + 3\| 14. y = \|x −6\| 15. y = \|4x + 2\| 16. y = x 3 −4 Solve the following inequalities and show the solution graph. 13. \|x\| ≤6 14. \|x\| > 3.5 15. \|x\| < 12 16. \|x\| > 10 265 www.ck12.org 17. \|7x\| ≥21 18. \|x −5\| > 8 19. \|x + 7\| < 3 20. x −3 4 ≤1 2 21. \|2x −5\| ≥13 22. \|5x + 3\| < 7 23. x 3 −4 ≤2 24. 2x 7 + 9 > 5 7 25. (a) How many solutions does the inequality \|x\| ≤0 have? (b) How about the inequality \|x\| ≥0? 26. A company manufactures rulers. Their 12-inch rulers pass quality control if they are within 1 32 inches of the ideal length. What is the longest and shortest ruler that can leave the factory? 27. A three month old baby boy weighs an average of 13 pounds. He is considered healthy if he is at most 2.5 lbs. more or less than the average weight. Find the weight range that is considered healthy for three month old boys. 6.5 Linear Inequalities in Two Variables Learning Objectives • Graph linear inequalities in one variable on the coordinate plane. • Graph linear inequalities in two variables. • Solve real-world problems using linear inequalities. Introduction Yasmeen is selling handmade bracelets for $5 each and necklaces for $7 each. How many of both does she need to sell to make at least $100? A linear inequality in two variables takes the form y > mx + b or y < mx + b. Linear inequalities are closely related to graphs of straight lines; recall that a straight line has the equation y = mx + b. When we graph a line in the coordinate plane, we can see that it divides the plane in half: The solution to a linear inequality includes all the points in one half of the plane. We can tell which half by looking at the inequality sign: > The solution set is the half plane above the line. www.ck12.org 266 ≥The solution set is the half plane above the line and also all the points on the line. < The solution set is the half plane below the line. ≤The solution set is the half plane below the line and also all the points on the line. For a strict inequality, we draw a dashed line to show that the points in the line are not part of the solution. For an inequality that includes the equals sign, we draw a solid line to show that the points on the line are part of the solution. Here are some examples of linear inequality graphs. This is a graph of y ≥mx + b; the solution set is the line and the half plane above the line. This is a graph of y < mx + b; the solution set is the half plane above the line, not including the line itself. Graph Linear Inequalities in One Variable in the Coordinate Plane In the last few sections we graphed inequalities in one variable on the number line. We can also graph inequalities in one variable on the coordinate plane. We just need to remember that when we graph an equation of the type x = a we get a vertical line, and when we graph an equation of the type y = b we get a horizontal line. Example 1 Graph the inequality x > 4 on the coordinate plane. Solution 267 www.ck12.org First let’s remember what the solution to x > 4 looks like on the number line. The solution to this inequality is the set of all real numbers x that are bigger than 4, not including 4. The solution is represented by a line. In two dimensions, the solution still consists of all the points to the right of x = 4, but for all possible y−values as well. This solution is represented by the half plane to the right of x = 4. (You can think of it as being like the solution graphed on the number line, only stretched out vertically.) The line x = 4 is dashed because the equals sign is not included in the inequality, meaning that points on the line are not included in the solution. Example 2 Graph the inequality \|x\| ≥2. Solution The absolute value inequality \|x\| ≥2 can be re-written as a compound inequality: x ≤−2 or x ≥2 In other words, the solution is all the coordinate points for which the value of x is smaller than or equal to -2 or greater than or equal to 2. The solution is represented by the plane to the left of the vertical line x = −2 and the plane to the right of line x = 2. Both vertical lines are solid because points on the lines are included in the solution. www.ck12.org 268 Example 3 Graph the inequality \|y\| < 5 Solution The absolute value inequality \|y\| < 5 can be re-written as −5 < y < 5. This is a compound inequality which can be expressed as y > −5 and y < 5 In other words, the solution is all the coordinate points for which the value of y is larger than -5 and smaller than 5. The solution is represented by the plane between the horizontal lines y = −5 and y = 5. Both horizontal lines are dashed because points on the lines are not included in the solution. Graph Linear Inequalities in Two Variables The general procedure for graphing inequalities in two variables is as follows: 1. Re-write the inequality in slope-intercept form: y = mx + b. Writing the inequality in this form lets you know the direction of the inequality. 2. Graph the line of the equation y = mx + b using your favorite method (plotting two points, using slope and y−intercept, using y−intercept and another point, or whatever is easiest). Draw the line as a dashed line if the equals sign is not included and a solid line if the equals sign is included. 3. Shade the half plane above the line if the inequality is “greater than.” Shade the half plane under the line if the inequality is “less than.” Example 4 Graph the inequality y ≥2x −3. Solution The inequality is already written in slope-intercept form, so it’s easy to graph. First we graph the line y = 2x −3; then we shade the half-plane above the line. The line is solid because the inequality includes the equals sign. 269 www.ck12.org Example 5 Graph the inequality 5x −2y > 4. Solution First we need to rewrite the inequality in slope-intercept form: −2y > −5x + 4 y < 5 2 x −2 Notice that the inequality sign changed direction because we divided by a negative number. To graph the equation, we can make a table of values: Table 6.4: x y -2 5 2(−2) −2 = −7 0 5 2(0) −2 = −2 2 5 2(2) −2 = 3 After graphing the line, we shade the plane below the line because the inequality in slope-intercept form is less than. The line is dashed because the inequality does not include an equals sign. www.ck12.org 270 Solve Real-World Problems Using Linear Inequalities In this section, we see how linear inequalities can be used to solve real-world applications. Example 8 A retailer sells two types of coﬀee beans. One type costs $9 per pound and the other type costs $7 per pound. Find all the possible amounts of the two diﬀerent coﬀee beans that can be mixed together to get a quantity of coﬀee beans costing $8.50 or less. Solution Let x = weight of $9 per pound coﬀee beans in pounds. Let y = weight of $7 per pound coﬀee beans in pounds. The cost of a pound of coﬀee blend is given by 9x + 7y. We are looking for the mixtures that cost $8.50 or less. We write the inequality 9x + 7y ≤8.50. Since this inequality is in standard form, it’s easiest to graph it by ﬁnding the x−and y−intercepts. When x = 0, we have 7y = 8.50 or y = 8.50 7 ≈1.21. When y = 0, we have 9x = 8.50 or x = 8.50 9 ≈0.94. We can then graph the line that includes those two points. Now we have to ﬁgure out which side of the line to shade. In y−intercept form, we shade the area below the line when the inequality is “less than.” But in standard form that’s not always true. We could convert the inequality to y−intercept form to ﬁnd out which side to shade, but there is another way that can be easier. The other method, which works for any linear inequality in any form, is to plug a random point into the inequality and see if it makes the inequality true. Any point that’s not on the line will do; the point (0, 0) is usually the most convenient. In this case, plugging in 0 for x and y would give us 9(0) + 7(0) ≤8.50, which is true. That means we should shade the half of the plane that includes (0, 0). If plugging in (0, 0) gave us a false inequality, that would mean that the solution set is the part of the plane that does not contain (0, 0). Notice also that in this graph we show only the ﬁrst quadrant of the coordinate plane. That’s because weight values in the real world are always nonnegative, so points outside the ﬁrst quadrant don’t represent real-world solutions to this problem. Example 9 Julius has a job as an appliance salesman. He earns a commission of $60 for each washing machine he sells and $130 for each refrigerator he sells. How many washing machines and refrigerators must Julius 271 www.ck12.org sell in order to make $1000 or more in commissions? Solution Let x = number of washing machines Julius sells. Let y = number of refrigerators Julius sells. The total commission is 60x + 130y. We’re looking for a total commission of $1000 or more, so we write the inequality 60x + 130y ≥1000. Once again, we can do this most easily by ﬁnding the x−and y−intercepts. When x = 0, we have 130y = 1000, or y = 1000 30 ≈7.69. When y = 0, we have 60x = 1000, or x = 1000 60 ≈16.67. We draw a solid line connecting those points, and shade above the line because the inequality is “greater than.” We can check this by plugging in the point (0, 0): selling 0 washing machines and 0 refrigerators would give Julius a commission of $0, which is not greater than or equal to $1000, so the point (0, 0) is not part of the solution; instead, we want to shade the side of the line that does not include it. Notice also that we show only the ﬁrst quadrant of the coordinate plane, because Julius’s commission should be nonnegative. The video at contains more ex-amples of real-world problems using inequalities in two variables. Review Questions Graph the following inequalities on the coordinate plane. 1. x < 20 2. y ≥−5 3. \|x\| > 10 4. \|y\| ≤7 5. y ≤4x + 3 6. y > −x 2 −6 7. 3x −4y ≥12 8. x + 7y < 5 9. 6x + 5y > 1 10. y + 5 ≤−4x + 10 11. x −1 2y ≥5 12. 6x + y < 20 www.ck12.org 272 13. 30x + 5y < 100 14. Remember what you learned in the last chapter about families of lines. (a) What do the graphs of y > x + 2 and y < x + 5 have in common? (b) What do you think the graph of x + 2 < y < x + 5 would look like? 15. How would the answer to problem 6 change if you subtracted 2 from the right-hand side of the inequality? 16. How would the answer to problem 7 change if you added 12 to the right-hand side? 17. How would the answer to problem 8 change if you ﬂipped the inequality sign? 18. A phone company charges 50 cents per minute during the daytime and 10 cents per minute at night. How many daytime minutes and nighttime minutes could you use in one week if you wanted to pay less than $20? 19. Suppose you are graphing the inequality y > 5x. (a) Why can’t you plug in the point (0, 0) to tell you which side of the line to shade? (b) What happens if you do plug it in? (c) Try plugging in the point (0, 1) instead. Now which side of the line should you shade? 20. A theater wants to take in at least $2000 for a certain matinee. Children’s tickets cost $5 each and adult tickets cost $10 each. (a) If x represents the number of adult tickets sold and y represents the number of children’s tickets, write an inequality describing the number of tickets that will allow the theater to meet their minimum take. (b) If 100 children’s tickets and 100 adult tickets have already been sold, what inequality describes how many more tickets of both types the theater needs to sell? (c) If the theater has only 300 seats (so only 100 are still available), what inequality describes the maximum number of additional tickets of both types the theater can sell? Texas Instruments Resources In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http: //www.ck12.org/flexr/chapter/9616. 273 www.ck12.org Chapter 7 Solving Systems of Equations and Inequalities 7.1 Linear Systems by Graphing Learning Objectives • Determine whether an ordered pair is a solution to a system of equations. • Solve a system of equations graphically. • Solve a system of equations graphically with a graphing calculator. • Solve word problems using systems of equations. Introduction In this lesson, we’ll discover methods to determine if an ordered pair is a solution to a system of two equations. Then we’ll learn to solve the two equations graphically and conﬁrm that the solution is the point where the two lines intersect. Finally, we’ll look at real-world problems that can be solved using the methods described in this chapter. Determine Whether an Ordered Pair is a Solution to a System of Equations A linear system of equations is a set of equations that must be solved together to ﬁnd the one solution that ﬁts them both. Consider this system of equations: y = x + 2 y = −2x + 1 Since the two lines are in a system, we deal with them together by graphing them on the same coordinate axes. We can use any method to graph them; let’s do it by making a table of values for each line. Line 1: y = x + 2 www.ck12.org 274 Table 7.1: x y 0 2 1 3 Line 2: y = −2x + 1 Table 7.2: x y 0 1 1 -1 We already know that any point that lies on a line is a solution to the equation for that line. That means that any point that lies on both lines in a system is a solution to both equations. So in this system: • Point A is not a solution to the system because it does not lie on either of the lines. • Point B is not a solution to the system because it lies only on the blue line but not on the red line. • Point C is a solution to the system because it lies on both lines at the same time. In fact, point C is the only solution to the system, because it is the only point that lies on both lines. For a system of equations, the geometrical solution is the intersection of the two lines in the system. The algebraic solution is the ordered pair that solves both equations—in other words, the coordinates of that intersection point. You can conﬁrm the solution by plugging it into the system of equations, and checking that the solution works in each equation. Example 1 Determine which of the points (1, 3), (0, 2), or (2, 7) is a solution to the following system of equations: y = 4x −1 y = 2x + 3 275 www.ck12.org Solution To check if a coordinate point is a solution to the system of equations, we plug each of the x and y values into the equations to see if they work. Point (1, 3): y = 4x −1 3 ? = ? 4(1) −1 3 = 3 solution checks y = 2x + 3 3 ? = ? 2(1) + 3 3 , 5 solution does not check Point (1, 3) is on the line y = 4x −1, but it is not on the line y = 2x + 3, so it is not a solution to the system. Point (0, 2): y = 4x −1 2 ? = ? 4(0) −1 2 , −1 solution does not check Point (0, 2) is not on the line y = 4x −1, so it is not a solution to the system. Note that it is not necessary to check the second equation because the point needs to be on both lines for it to be a solution to the system. Point (2, 7): y = 4x −1 7 ? = ? 4(2) −1 7 = 7 solution checks y = 2x + 3 7 ? = ? 2(2) + 3 7 = 7 solution checks Point (2, 7) is a solution to the system since it lies on both lines. The solution to the system is the point (2, 7). Determine the Solution to a Linear System by Graphing The solution to a linear system of equations is the point, (if there is one) that lies on both lines. In other words, the solution is the point where the two lines intersect. We can solve a system of equations by graphing the lines on the same coordinate plane and reading the intersection point from the graph. www.ck12.org 276 This method most often oﬀers only approximate solutions, so it’s not suﬀicient when you need an exact answer. However, graphing the system of equations can be a good way to get a sense of what’s really going on in the problem you’re trying to solve, especially when it’s a real-world problem. Example 2 Solve the following system of equations by graphing: y = 3x −5 y = −2x + 5 Solution Graph both lines on the same coordinate axis using any method you like. In this case, let’s make a table of values for each line. Line 1: y = 3x −5 Table 7.3: x y 1 -2 2 1 Line 2: y = −2x + 5 Table 7.4: x y 1 3 2 1 The solution to the system is given by the intersection point of the two lines. The graph shows that the lines intersect at point (2, 1). So the solution is x = 2, y = 1 or (2, 1). Example 3 Solve the following system of equations by graphing: 277 www.ck12.org 2x + 3y = 6 4x −y = −2 Solution Since the equations are in standard form, this time we’ll graph them by ﬁnding the x−and y−intercepts of each of the lines. Line 1: 2x + 3y = 6 x−intercept: set y = 0 ⇒2x = 6 ⇒x = 3 so the intercept is (3, 0) y−intercept: set x = 0 ⇒3y = 6 ⇒y = 2 so the intercept is (0, 2) Line 2: −4x + y = 2 x−intercept: set y = 0 ⇒−4x = 2 ⇒x = −1 2 so the intercept is ( −1 2, 0 ) y−intercept: set x = 0 ⇒y = 2 so the intercept is (0, 2) The graph shows that the lines intersect at (0, 2). Therefore, the solution to the system of equations is x = 0, y = 2. Solving a System of Equations Using a Graphing Calculator As an alternative to graphing by hand, you can use a graphing calculator to ﬁnd or check solutions to a system of equations. Example 4 Solve the following system of equations using a graphing calculator. x −3y = 4 2x + 5y = 8 To input the equations into the calculator, you need to rewrite them in slope-intercept form (that is, y = mx + b form). x −3y = 4 y = 1 3 x −4 3 ⇒ 2x + 5y = 8 y = −2 5 x + 8 5 www.ck12.org 278 Press the [y=] button on the graphing calculator and enter the two functions as: Y1 = x 3 −4 3 Y2 = −2x 5 + 8 5 Now press [GRAPH]. Here’s what the graph should look like on a TI-83 family graphing calculator with the window set to −5 ≤x ≤10 and −5 ≤y ≤5. There are a few diﬀerent ways to ﬁnd the intersection point. Option 1: Use [TRACE] and move the cursor with the arrows until it is on top of the intersection point. The values of the coordinate point will be shown on the bottom of the screen. The second screen above shows the values to be X = 4.0957447 and Y = 0.03191489. Use the [ZOOM] function to zoom into the intersection point and ﬁnd a more accurate result. The third screen above shows the system of equations after zooming in several times. A more accurate solution appears to be X = 4 and Y = 0. Option 2 Look at the table of values by pressing [2nd] [GRAPH]. The ﬁrst screen below shows a table of values for this system of equations. Scroll down until the Y−values for the two functions are the same. In this case this occurs at X = 4 and Y = 0. (Use the [TBLSET] function to change the starting value for your table of values so that it is close to the intersection point and you don’t have to scroll too long. You can also improve the accuracy of the solution by setting the value of ∆Table smaller.) Option 3 Using the [2nd] [TRACE] function gives the second screen shown above. Scroll down and select “intersect.” The calculator will display the graph with the question [FIRSTCURVE]? Move the cursor along the ﬁrst curve until it is close to the intersection and press [ENTER]. The calculator now shows [SECONDCURVE]? Move the cursor to the second line (if necessary) and press [ENTER]. The calculator displays [GUESS]? Press [ENTER] and the calculator displays the solution at the bottom of the screen (see the third screen above). The point of intersection is X = 4 and Y = 0. Note that with this method, the calculator works out the intersection point for you, which is generally more accurate than your own visual estimate. 279 www.ck12.org Solve Real-World Problems Using Graphs of Linear Systems Consider the following problem: Peter and Nadia like to race each other. Peter can run at a speed of 5 feet per second and Nadia can run at a speed of 6 feet per second. To be a good sport, Nadia likes to give Peter a head start of 20 feet. How long does Nadia take to catch up with Peter? At what distance from the start does Nadia catch up with Peter? Let’s start by drawing a sketch. Here’s what the race looks like when Nadia starts running; we’ll call this time t = 0. Now let’s deﬁne two variables in this problem: t = the time from when Nadia starts running d = the distance of the runners from the starting point. Since there are two runners, we need to write equations for each of them. That will be the system of equations for this problem. For each equation, we use the formula: distance = speed × time Nadia’s equation: d = 6t Peter’s equation: d = 5t + 20 (Remember that Peter was already 20 feet from the starting point when Nadia started running.) Let’s graph these two equations on the same coordinate axes. Time should be on the horizontal axis since it is the independent variable. Distance should be on the vertical axis since it is the dependent variable. We can use any method for graphing the lines, but in this case we’ll use the slope–intercept method since it makes more sense physically. To graph the line that describes Nadia’s run, start by graphing the y−intercept: (0, 0). (If you don’t see that this is the y−intercept, try plugging in the test-value of x = 0.) The slope tells us that Nadia runs 6 feet every one second, so another point on the line is (1, 6). Connecting these points gives us Nadia’s line: www.ck12.org 280 To graph the line that describes Peter’s run, again start with the y−intercept. In this case this is the point (0, 20). The slope tells us that Peter runs 5 feet every one second, so another point on the line is (1, 25). Connecting these points gives us Peter’s line: In order to ﬁnd when and where Nadia and Peter meet, we’ll graph both lines on the same graph and extend the lines until they cross. The crossing point is the solution to this problem. The graph shows that Nadia and Peter meet 20 seconds after Nadia starts running, and 120 feet from the starting point. These examples are great at demonstrating that the solution to a system of linear equations means the 281 www.ck12.org point at which the lines intersect. This is, in fact, the greatest strength of the graphing method because it oﬀers a very visual representation of system of equations and its solution. You can also see, though, that ﬁnding the solution from a graph requires very careful graphing of the lines, and is really only practical when you’re sure that the solution gives integer values for x and y. Usually, this method can only oﬀer approximate solutions to systems of equations, so we need to use other methods to get an exact solution. Review Questions Determine which ordered pair satisﬁes the system of linear equations. 1. y = 3x −2 y = −x (a) (1, 4) (b) (2, 9) (c) ( 1 2, −1 2 ) 2. y = 2x −3 y = x + 5 (a) (8, 13) (b) (-7, 6) (c) (0, 4) 3. 2x + y = 8 5x + 2y = 10 (a) (-9, 1) (b) (-6, 20) (c) (14, 2) 4. 3x + 2y = 6 y = 1 2 x −3 (a) ( 3, −3 2 ) (b) (-4, 3) (c) ( 1 2, 4 ) 5. 2x −y = 10 3x + y = −5 (a) (4, -2) (b) (1, -8) (c) (-2, 5) Solve the following systems using the graphing method. 6. y = x + 3 y = −x + 3 7. y = 3x −6 y = −x + 6 8. 2x = 4 y = −3 9. y = −x + 5 −x + y = 1 www.ck12.org 282 10. x + 2y = 8 5x + 2y = 0 11. 3x + 2y = 12 4x −y = 5 12. 5x + 2y = −4 x −y = 2 13. 2x + 4 = 3y x −2y + 4 = 0 14. y = 1 2 x −3 2x −5y = 5 15. y = 4 x = 8 −3y 16. Try to solve the following system using the graphing method: y = 3 5 x + 5 y = −2x −1 2. (a) What does it look like the x−coordinate of the solution should be? (b) Does that coordinate really give the same y−value when you plug it into both equations? (c) Why is it diﬀicult to ﬁnd the real solution to this system? 17. Try to solve the following system using the graphing method: y = 4x + 8 y = 5x + 1. Use a grid with x−values and y−values ranging from -10 to 10. (a) Do these lines appear to intersect? (b) Based on their equations, are they parallel? (c) What would we have to do to ﬁnd their intersection point? 18. Try to solve the following system using the graphing method: y = 1 2 x + 4 y = 4 9 x + 9 2. Use the same grid as before. (a) Can you tell exactly where the lines cross? (b) What would we have to do to make it clearer? Solve the following problems by using the graphing method. 19. Mary’s car has broken down and it will cost her $1200 to get it ﬁxed—or, for $4500, she can buy a new, more eﬀicient car instead. Her present car uses about $2000 worth of gas per year, while gas for the new car would cost about $1500 per year. After how many years would the total cost of ﬁxing the car equal the total cost of replacing it? 20. Juan is considering two cell phone plans. The ﬁrst company charges $120 for the phone and $30 per month for the calling plan that Juan wants. The second company charges $40 for the same phone but charges $45 per month for the calling plan that Juan wants. After how many months would the total cost of the two plans be the same? 21. A tortoise and hare decide to race 30 feet. The hare, being much faster, decides to give the tortoise a 20 foot head start. The tortoise runs at 0.5 feet/sec and the hare runs at 5.5 feet per second. How long until the hare catches the tortoise? 283 www.ck12.org 7.2 Solving Linear Systems by Substitution Learning Objectives • Solve systems of equations with two variables by substituting for either variable. • Manipulate standard form equations to isolate a single variable. • Solve real-world problems using systems of equations. • Solve mixture problems using systems of equations. Introduction In this lesson, we’ll learn to solve a system of two equations using the method of substitution. Solving Linear Systems Using Substitution of Variable Expres-sions Let’s look again at the problem about Peter and Nadia racing. Peter and Nadia like to race each other. Peter can run at a speed of 5 feet per second and Nadia can run at a speed of 6 feet per second. To be a good sport, Nadia likes to give Peter a head start of 20 feet. How long does Nadia take to catch up with Peter? At what distance from the start does Nadia catch up with Peter? In that example we came up with two equations: Nadia’s equation: d = 6t Peter’s equation: d = 5t + 20 Each equation produced its own line on a graph, and to solve the system we found the point at which the lines intersected—the point where the values for d and t satisﬁed both relationships. When the values for d and t are equal, that means that Peter and Nadia are at the same place at the same time. But there’s a faster way than graphing to solve this system of equations. Since we want the value of d to be the same in both equations, we could just set the two right-hand sides of the equations equal to each other to solve for t. That is, if d = 6t and d = 5t + 20, and the two d’s are equal to each other, then by the transitive property we have 6t = 5t + 20. We can solve this for t: 6t = 5t + 20 subtract 5t from both sides : t = 20 substitute this value for t into Nadia’s equation : d = 6 · 20 = 120 Even if the equations weren’t so obvious, we could use simple algebraic manipulation to ﬁnd an expression for one variable in terms of the other. If we rearrange Peter’s equation to isolate t: d = 5t + 20 subtract 20 from both sides : d −20 = 5t divide by 5 : d −20 5 = t We can now substitute this expression for t into Nadia’s equation (d = 6t) to solve: www.ck12.org 284 d = 6 (d −20 5 ) multiply both sides by 5 : 5d = 6(d −20) distribute the 6 : 5d = 6d −120 subtract 6d from both sides : −d = −120 divide by −1 : d = 120 substitute value for d into our expression for t : t = 120 −20 5 = 100 5 = 20 So we ﬁnd that Nadia and Peter meet 20 seconds after they start racing, at a distance of 120 feet away. The method we just used is called the Substitution Method. In this lesson you’ll learn several techniques for isolating variables in a system of equations, and for using those expressions to solve systems of equations that describe situations like this one. Example 1 Let’s look at an example where the equations are written in standard form. Solve the system 2x + 3y = 6 −4x + y = 2 Again, we start by looking to isolate one variable in either equation. If you look at the second equation, you should see that the coeﬀicient of y is 1. So the easiest way to start is to use this equation to solve for y. Solve the second equation for y: −4x + y = 2 add 4x to both sides : y = 2 + 4x Substitute this expression into the ﬁrst equation: 2x + 3(2 + 4x) = 6 distribute the 3 : 2x + 6 + 12x = 6 collect like terms : 14x + 6 = 6 subtract 6 from both sides : 14x = 0 and hence : x = 0 Substitute back into our expression for y: y = 2 + 4 · 0 = 2 As you can see, we end up with the same solution (x = 0, y = 2) that we found when we graphed these functions back in Lesson 7.1. So long as you are careful with the algebra, the substitution method can be a very eﬀicient way to solve systems. Next, let’s look at a more complicated example. Here, the values of x and y we end up with aren’t whole numbers, so they would be diﬀicult to read oﬀa graph! 285 www.ck12.org Example 2 Solve the system 2x + 3y = 3 2x −3y = −1 Again, we start by looking to isolate one variable in either equation. In this case it doesn’t matter which equation we use—all the variables look about equally easy to solve for. So let’s solve the ﬁrst equation for x: 2x + 3y = 3 subtract 3y from both sides : 2x = 3 −3y divide both sides by 2 : x = 1 2(3 −3y) Substitute this expression into the second equation: 2 · 1 2(3 −3y) −3y = −1 cancel the fraction and re −write terms : 3 −3y −3y = −1 collect like terms : 3 −6y = −1 subtract 3 from both sides : −6y = −4 divide by −6 : y = 2 3 Substitute into the expression we got for x: x = 1 2 ( 3 − 3 (2 3 )) x = 1 2 So our solution is x = 1 2, y = 2 3. You can see how the graphical solution ( 1 2, 2 3 ) might have been diﬀicult to read accurately oﬀa graph! Solving Real-World Problems Using Linear Systems Simultaneous equations can help us solve many real-world problems. We may be considering a pur-chase—for example, trying to decide whether it’s cheaper to buy an item online where you pay shipping or at the store where you do not. Or you may wish to join a CD music club, but aren’t sure if you would really save any money by buying a new CD every month in that way. Or you might be considering two diﬀerent phone contracts. Let’s look at an example of that now. Example 3 Anne is trying to choose between two phone plans. The ﬁrst plan, with Vendafone, costs $20 per month, with calls costing an additional 25 cents per minute. The second company, Sellnet, charges $40 per month, but calls cost only 8 cents per minute. Which should she choose? You should see that Anne’s choice will depend upon how many minutes of calls she expects to use each month. We start by writing two equations for the cost in dollars in terms of the minutes used. Since the number of minutes is the independent variable, it will be our x. Cost is dependent on minutes – the cost per month is the dependent variable and will be assigned y. www.ck12.org 286 For Vendafone: y = 0.25x + 20 For Sellnet: y = 0.08x + 40 By writing the equations in slope-intercept form (y = mx + b), you can sketch a graph to visualize the situation: The line for Vendafone has an intercept of 20 and a slope of 0.25. The Sellnet line has an intercept of 40 and a slope of 0.08 (which is roughly a third of the Vendafone line’s slope). In order to help Anne decide which to choose, we’ll ﬁnd where the two lines cross, by solving the two equations as a system. Since equation 1 gives us an expression for y(0.25x + 20), we can substitute this expression directly into equation 2: 0.25x + 20 = 0.08x + 40 subtract 20 from both sides : 0.25x = 0.08x + 20 subtract 0.08x from both sides : 0.17x = 20 divide both sides by 0.17 : x = 117.65 minutes rounded to 2 decimal places. So if Anne uses 117.65 minutes a month (although she can’t really do exactly that, because phone plans only count whole numbers of minutes), the phone plans will cost the same. Now we need to look at the graph to see which plan is better if she uses more minutes than that, and which plan is better if she uses fewer. You can see that the Vendafone plan costs more when she uses more minutes, and the Sellnet plan costs more with fewer minutes. So, if Anne will use 117 minutes or less every month she should choose Vendafone. If she plans on using 118 or more minutes she should choose Sellnet. Mixture Problems Systems of equations crop up frequently in problems that deal with mixtures of two things—chemicals in a solution, nuts and raisins, or even the change in your pocket! Let’s look at some examples of these. Example 4 Janine empties her purse and ﬁnds that it contains only nickels (worth 5 cents each) and dimes (worth 10 cents each). If she has a total of 7 coins and they have a combined value of 45 cents, how many of each coin does she have? Since we have 2 types of coins, let’s call the number of nickels x and the number of dimes y. We are given two key pieces of information to make our equations: the number of coins and their value. of coins equation: x + y = 7 (number o f nickels) + (number o f dimes) value equation: 5x + 10y = 55 (since nickels are worth 5c and dimes 10c) 287 www.ck12.org We can quickly rearrange the ﬁrst equation to isolate x: x = 7 −y now substitute into equation 2 : 5(7 −y) + 10y = 55 distribute the 5 : 35 −5y + 10y = 55 collect like terms : 35 + 5y = 55 subtract 35 from both sides : 5y = 20 divide by 5 : y = 4 substitute back into equation 1 : x + 4 = 7 subtract 4 from both sides : x = 3 Janine has 3 nickels and 4 dimes. Sometimes a question asks you to determine (from concentrations) how much of a particular substance to use. The substance in question could be something like coins as above, or it could be a chemical in solution, or even heat. In such a case, you need to know the amount of whatever substance is in each part. There are several common situations where to get one equation you simply add two given quantities, but to get the second equation you need to use a product. Three examples are below. Table 7.5: Type of mixture First equation Second equation Coins (items with $ value) total number of items (n1 + n2) total value (item value × no. of items) Chemical solutions total solution volume (V1 + V2) amount of solute (vol × concen-tration) Density of two substances total amount or volume of mix total mass (volume × density) For example, when considering mixing chemical solutions, we will most likely need to consider the total amount of solute in the individual parts and in the ﬁnal mixture. (A solute is the chemical that is dissolved in a solution. An example of a solute is salt when added to water to make a brine.) To ﬁnd the total amount, simply multiply the amount of the mixture by the fractional concentration. To illustrate, let’s look at an example where you are given amounts relative to the whole. Example 5 A chemist needs to prepare 500 ml of copper-sulfate solution with a 15% concentration. She wishes to use a high concentration solution (60%) and dilute it with a low concentration solution (5%) in order to do this. How much of each solution should she use? Solution To set this problem up, we ﬁrst need to deﬁne our variables. Our unknowns are the amount of concentrated solution (x) and the amount of dilute solution (y). We will also convert the percentages (60%, 15% and 5%) into decimals (0.6, 0.15 and 0.05). The two pieces of critical information are the ﬁnal volume (500 ml) and the ﬁnal amount of solute (15% of 500 ml = 75 ml). Our equations will look like this: Volume equation: x + y = 500 Solute equation: 0.6x + 0.05y = 75 To isolate a variable for substitution, we can see it’s easier to start with equation 1: www.ck12.org 288 x + y = 500 subtract y from both sides : x = 500 −y now substitute into equation 2 : 0.6(500 −y) + 0.05y = 75 distribute the 0.6 : 300 −0.6y + 0.05y = 75 collect like terms : 300 −0.55y = 75 subtract 300 from both sides : −0.55y = −225 divide both sides by −0.55 : y = 409 ml substitute back into equation for x : x = 500 −409 = 91 ml So the chemist should mix 91 ml of the 60% solution with 409 ml of the 5% solution. Further Practice For lots more practice solving linear systems, check out this web page: homework/coordinate/practice-linear-system.epl After clicking to see the solution to a problem, you can click the back button and then click Try Another Practice Linear System to see another problem. Review Questions 1. Solve the system: x + 2y = 9 3x + 5y = 20 2. Solve the system: x −3y = 10 2x + y = 13 3. Solve the system: 2x + 0.5y = −10 x −y = −10 4. Solve the system: 2x + 0.5y = 3 x + 2y = 8.5 5. Solve the system: 3x + 5y = −1 x + 2y = −1 6. Solve the system: 3x + 5y = −3 x + 2y = −4 3 7. Solve the system: x −y = −12 5 2x + 5y = −2 8. Of the two non-right angles in a right angled triangle, one measures twice as many degrees as the other. What are the angles? 289 www.ck12.org 9. The sum of two numbers is 70. They diﬀer by 11. What are the numbers? 10. A number plus half of another number equals 6; twice the ﬁrst number minus three times the second number equals 4. What are the numbers? 11. A rectangular ﬁeld is enclosed by a fence on three sides and a wall on the fourth side. The total length of the fence is 320 yards. If the ﬁeld has a total perimeter of 400 yards, what are the dimensions of the ﬁeld? 12. A ray cuts a line forming two angles. The diﬀerence between the two angles is 18◦. What does each angle measure? 13. I have $15 and wish to buy ﬁve pounds of mixed nuts for a party. Peanuts cost $2.20 per pound. Cashews cost $4.70 per pound. (a) How many pounds of each should I buy? (b) If I suddenly realize I need to set aside $5 to buy chips, can I still buy 5 pounds of nuts with the remaining $10? (c) What’s the greatest amount of nuts I can buy? 14. A chemistry experiment calls for one liter of sulfuric acid at a 15% concentration, but the supply room only stocks sulfuric acid in concentrations of 10% and 35%. (a) How many liters of each should be mixed to give the acid needed for the experiment? (b) How many liters should be mixed to give two liters at a 15% concentration? 15. Bachelle wants to know the density of her bracelet, which is a mix of gold and silver. Density is total mass divided by total volume. The density of gold is 19.3 g/cc and the density of silver is 10.5 g/cc. The jeweler told her that the volume of silver in the bracelet was 10 cc and the volume of gold was 20 cc. Find the combined density of her bracelet. 16. Jason is ﬁve years older than Becky, and the sum of their ages is 23. What are their ages? 17. Tickets to a show cost $10 in advance and $15 at the door. If 120 tickets are sold for a total of $1390, how many of the tickets were bought in advance? 18. The multiple-choice questions on a test are worth 2 points each, and the short-answer questions are worth 5 points each. (a) If the whole test is worth 100 points and has 35 questions, how many of the questions are multiple-choice and how many are short-answer? www.ck12.org 290 (b) If Kwan gets 31 questions right and ends up with a score of 86 on the test, how many questions of each type did she get right? (Assume there is no partial credit.) (c) If Ashok gets 5 questions wrong and ends up with a score of 87 on the test, how many questions of each type did he get wrong? (Careful!) (d) What are two ways you could have set up the equations for part c? (e) How could you have set up part b diﬀerently? 7.3 Solving Linear Systems by Elimination Learning Objectives • Solve a linear system of equations using elimination by addition. • Solve a linear system of equations using elimination by subtraction. • Solve a linear system of equations by multiplication and then addition or subtraction. • Compare methods for solving linear systems. • Solve real-world problems using linear systems by any method. Introduction In this lesson, we’ll see how to use simple addition and subtraction to simplify our system of equations to a single equation involving a single variable. Because we go from two unknowns (x and y) to a single unknown (either x or y), this method is often referred to by solving by elimination. We eliminate one variable in order to make our equations solvable! To illustrate this idea, let’s look at the simple example of buying apples and bananas. Example 1 If one apple plus one banana costs $1.25 and one apple plus 2 bananas costs $2.00, how much does one banana cost? One apple? It shouldn’t take too long to discover that each banana costs $0.75. After all, the second purchase just contains 1 more banana than the ﬁrst, and costs $0.75 more, so that one banana must cost $0.75. Here’s what we get when we describe this situation with algebra: a + b = 1.25 a + 2b = 2.00 Now we can subtract the number of apples and bananas in the ﬁrst equation from the number in the second equation, and also subtract the cost in the ﬁrst equation from the cost in the second equation, to get the diﬀerence in cost that corresponds to the diﬀerence in items purchased. (a + 2b) −(a + b) = 2.00 −1.25 →b = 0.75 That gives us the cost of one banana. To ﬁnd out how much one apple costs, we subtract $0.75 from the total cost of one apple and one banana. a + 0.75 = 1.25 →a = 1.25 −0.75 →a = 0.50 So an apple costs 50 cents. To solve systems using addition and subtraction, we’ll be using exactly this idea – by looking at the sum or diﬀerence of the two equations we can determine a value for one of the unknowns. 291 www.ck12.org Solving Linear Systems Using Addition of Equations Often considered the easiest and most powerful method of solving systems of equations, the addition (or elimination) method lets us combine two equations in such a way that the resulting equation has only one variable. We can then use simple algebra to solve for that variable. Then, if we need to, we can substitute the value we get for that variable back into either one of the original equations to solve for the other variable. Example 2 Solve this system by addition: 3x + 2y = 11 5x −2y = 13 Solution We will add everything on the left of the equals sign from both equations, and this will be equal to the sum of everything on the right: (3x + 2y) + (5x −2y) = 11 + 13 →8x = 24 →x = 3 A simpler way to visualize this is to keep the equations as they appear above, and to add them together vertically, going down the columns. However, just like when you add units, tens and hundreds, you MUST be sure to keep the x’s and y’s in their own columns. You may also wish to use terms like “0y” as a placeholder! 3x + 2y = 11 + (5x −2y) = 13 8x + 0y = 24 Again we get 8x = 24, or x = 3. To ﬁnd a value for y, we simply substitute our value for x back in. Substitute x = 3 into the second equation: 5 · 3 −2y = 13 since 5 × 3 = 15, we subtract 15 from both sides : −2y = −2 divide by −2 to get : y = 1 The reason this method worked is that the y−coeﬀicients of the two equations were opposites of each other: 2 and -2. Because they were opposites, they canceled each other out when we added the two equations together, so our ﬁnal equation had no y−term in it and we could just solve it for x. In a little while we’ll see how to use the addition method when the coeﬀicients are not opposites, but for now let’s look at another example where they are. Example 3 Andrew is paddling his canoe down a fast-moving river. Paddling downstream he travels at 7 miles per hour, relative to the river bank. Paddling upstream, he moves slower, traveling at 1.5 miles per hour. If he paddles equally hard in both directions, how fast is the current? How fast would Andrew travel in calm water? Solution www.ck12.org 292 First we convert our problem into equations. We have two unknowns to solve for, so we’ll call the speed that Andrew paddles at x, and the speed of the river y. When traveling downstream, Andrew speed is boosted by the river current, so his total speed is his paddling speed plus the speed of the river (x + y). Traveling upstream, the river is working against him, so his total speed is his paddling speed minus the speed of the river (x −y). Downstream Equation: x + y = 7 Upstream Equation: x −y = 1.5 Next we’ll eliminate one of the variables. If you look at the two equations, you can see that the coeﬀicient of y is +1 in the ﬁrst equation and -1 in the second. Clearly (+1) + (−1) = 0, so this is the variable we will eliminate. To do this we simply add equation 1 to equation 2. We must be careful to collect like terms, and make sure that everything on the left of the equals sign stays on the left, and everything on the right stays on the right: (x + y) + (x −y) = 7 + 1.5 ⇒2x = 8.5 ⇒x = 4.25 Or, using the column method we used in example 2: x + y = 7 + x −y = 1.5 2x + 0y = 8.5 Again we get 2x = 8.5, or x = 4.25. To ﬁnd a corresponding value for y, we plug our value for x into either equation and isolate our unknown. In this example, we’ll plug it into the ﬁrst equation: 4.25 + y = 7 subtract 4.25 from both sides : y = 2.75 Andrew paddles at 4.25 miles per hour. The river moves at 2.75 miles per hour. Solving Linear Systems Using Subtraction of Equations Another, very similar method for solving systems is subtraction. When the x−or y−coeﬀicients in both equations are the same (including the sign) instead of being opposites, you can subtract one equation from the other. If you look again at Example 3, you can see that the coeﬀicient for x in both equations is +1. Instead of adding the two equations together to get rid of the y’s, you could have subtracted to get rid of the x’s: (x + y) −(x −y) = 7 −1.5 ⇒2y = 5.5 ⇒y = 2.75 or... x + y = 7 −(x −y) = −1.5 0x + 2y = 5.5 So again we get y = 2.75, and we can plug that back in to determine x. The method of subtraction is just as straightforward as addition, so long as you remember the following: 293 www.ck12.org • Always put the equation you are subtracting in parentheses, and distribute the negative. • Don’t forget to subtract the numbers on the right-hand side. • Always remember that subtracting a negative is the same as adding a positive. Example 4 Peter examines the coins in the fountain at the mall. He counts 107 coins, all of which are either pennies or nickels. The total value of the coins is $3.47. How many of each coin did he see? Solution We have 2 types of coins, so let’s call the number of pennies x and the number of nickels y. The total value of all the pennies is just x, since they are worth 1 c each. The total value of the nickels is 5y. We are given two key pieces of information to make our equations: the number of coins and their value in cents. of coins equation : x + y = 107 (number o f pennies) + (number o f nickels) value equation : x + 5y = 347 pennies are worth 1 c, nickels are worth 5 c. We’ll jump straight to subtracting the two equations: x + y = 107 −(x + 5y) = −347 −4y = −240 y = 60 Substituting this value back into the ﬁrst equation: x + 60 = 107 subtract 60 from both sides : x = 47 So Peter saw 47 pennies (worth 47 cents) and 60 nickels (worth $3.00) making a total of $3.47. Solving Linear Systems Using Multiplication So far, we’ve seen that the elimination method works well when the coeﬀicient of one variable happens to be the same (or opposite) in the two equations. But what if the two equations don’t have any coeﬀicients the same? It turns out that we can still use the elimination method; we just have to make one of the coeﬀicients match. We can accomplish this by multiplying one or both of the equations by a constant. Here’s a quick review of how to do that. Consider the following questions: 1. If 10 apples cost $5, how much would 30 apples cost? 2. If 3 bananas plus 2 carrots cost $4, how mush would 6 bananas plus 4 carrots cost? If you look at the ﬁrst equation, it should be obvious that each apple costs $0.50. So 30 apples should cost $15.00. The second equation is trickier; it isn’t obvious what the individual price for either bananas or carrots is. Yet we know that the answer to question 2 is $8.00. How? www.ck12.org 294 If we look again at question 1, we see that we can write an equation: 10a = 5 (a being the cost of 1 apple). So to ﬁnd the cost of 30 apples, we could solve for a and then multiply by 30—but we could also just multiply both sides of the equation by 3. We would get 30a = 15, and that tells us that 30 apples cost $15. And we can do the same thing with the second question. The equation for this situation is 3b + 2c = 4, and we can see that we need to solve for (6b + 4c), which is simply 2 times (3b + 2c)! So algebraically, we are simply multiplying the entire equation by 2: 2(3b + 2c) = 2 · 4 distribute and multiply : 6b + 4c = 8 So when we multiply an equation, all we are doing is multiplying every term in the equation by a ﬁxed amount. Solving a Linear System by Multiplying One Equation If we can multiply every term in an equation by a ﬁxed number (a scalar), that means we can use the addition method on a whole new set of linear systems. We can manipulate the equations in a system to ensure that the coeﬀicients of one of the variables match. This is easiest to do when the coeﬀicient as a variable in one equation is a multiple of the coeﬀicient in the other equation. Example 5 Solve the system: 7x + 4y = 17 5x −2y = 11 Solution You can easily see that if we multiply the second equation by 2, the coeﬀicients of y will be +4 and -4, allowing us to solve the system by addition: 2 times equation 2: 10x −4y = 22 now add to equation one : + (7x + 4y) = 17 17x = 34 divide by 17 to get : x = 2 Now simply substitute this value for x back into equation 1: 7 · 2 + 4y = 17 since 7 × 2 = 14, subtract 14 from both sides : 4y = 3 divide by 4 : y = 0.75 Example 6 Anne is rowing her boat along a river. Rowing downstream, it takes her 2 minutes to cover 400 yards. Rowing upstream, it takes her 8 minutes to travel the same 400 yards. If she was rowing equally hard in 295 www.ck12.org both directions, calculate, in yards per minute, the speed of the river and the speed Anne would travel in calm water. Solution Step one: ﬁrst we convert our problem into equations. We know that distance traveled is equal to speed × time. We have two unknowns, so we’ll call the speed of the river x, and the speed that Anne rows at y. When traveling downstream, her total speed is her rowing speed plus the speed of the river, or (x + y). Going upstream, her speed is hindered by the speed of the river, so her speed upstream is (x −y). Downstream Equation: 2(x + y) = 400 Upstream Equation: 8(x −y) = 400 Distributing gives us the following system: 2x + 2y = 400 8x −8y = 400 Right now, we can’t use the method of elimination because none of the coeﬀicients match. But if we multiplied the top equation by 4, the coeﬀicients of y would be +8 and -8. Let’s do that: 8x + 8y = 1, 600 + (8x −8y) = 400 16x = 2, 000 Now we divide by 16 to obtain x = 125. Substitute this value back into the ﬁrst equation: 2(125 + y) = 400 divide both sides by 2 : 125 + y = 200 subtract 125 from both sides : y = 75 Anne rows at 125 yards per minute, and the river ﬂows at 75 yards per minute. Solving a Linear System by Multiplying Both Equations So what do we do if none of the coeﬀicients match and none of them are simple multiples of each other? We do the same thing we do when we’re adding fractions whose denominators aren’t simple multiples of each other. Remember that when we add fractions, we have to ﬁnd a lowest common denominator—that is, the lowest common multiple of the two denominators—and sometimes we have to rewrite not just one, but both fractions to get them to have a common denominator. Similarly, sometimes we have to multiply both equations by diﬀerent constants in order to get one of the coeﬀicients to match. Example 7 Andrew and Anne both use the I-Haul truck rental company to move their belongings from home to the dorm rooms on the University of Chicago campus. I-Haul has a charge per day and an additional charge per mile. Andrew travels from San Diego, California, a distance of 2060 miles in ﬁve days. Anne travels 880 miles from Norfolk, Virginia, and it takes her three days. If Anne pays $840 and Andrew pays $1845, what does I-Haul charge a) per day? www.ck12.org 296 b) per mile traveled? Solution First, we’ll set up our equations. Again we have 2 unknowns: the daily rate (we’ll call this x), and the per-mile rate (we’ll call this y). Anne’s equation: 3x + 880y = 840 Andrew’s Equation: 5x + 2060y = 1845 We can’t just multiply a single equation by an integer number in order to arrive at matching coeﬀicients. But if we look at the coeﬀicients of x (as they are easier to deal with than the coeﬀicients of y), we see that they both have a common multiple of 15 (in fact 15 is the lowest common multiple). So we can multiply both equations. Multiply the top equation by 5: 15x + 4400y = 4200 Multiply the lower equation by 3: 15x + 6180y = 5535 Subtract: 15x + 4400y = 4200 − (15x + 6180y) = 5535 −1780y = −1335 Divide by −1780 : y = 0.75 Substitute this back into the top equation: 3x + 880(0.75) = 840 since 880 × 0.75 = 660, subtract 660 from both sides : 3x = 180 divide both sides by 3 x = 60 I-Haul charges $60 per day plus $0.75 per mile. Comparing Methods for Solving Linear Systems Now that we’ve covered the major methods for solving linear equations, let’s review them. For simplicity, we’ll look at them in table form. This should help you decide which method would be best for a given situation. 297 www.ck12.org Table 7.6: Method: Best used when you... Advantages: Comment: Graphing ...don’t need an accu-rate answer. Often easier to see number and quality of intersections on a graph. With a graphing calculator, it can be the fastest method since you don’t have to do any computation. Can lead to impre-cise answers with non-integer solutions. Substitution ...have an explicit equa-tion for one variable (e.g. y = 14x + 2) Works on all systems. Reduces the system to one variable, making it easier to solve. You are not often given explicit functions in sys-tems problems, so you may have to do ex-tra work to get one of the equations into that form. Elimination by Addi-tion or Subtraction ...have matching coeﬀi-cients for one variable in both equations. Easy to combine equa-tions to eliminate one variable. Quick to solve. It is not very likely that a given system will have matching coeﬀicients. Elimination by Multi-plication and then Addi-tion and Subtraction ...do not have any vari-ables deﬁned explicitly or any matching coeﬀi-cients. Works on all systems. Makes it possible to combine equations to eliminate one variable. Often more algebraic manipulation is needed to prepare the equa-tions. The table above is only a guide. You might prefer to use the graphical method for every system in order to better understand what is happening, or you might prefer to use the multiplication method even when a substitution would work just as well. Example 8 Two angles are complementary when the sum of their angles is 90◦. Angles A and B are complementary angles, and twice the measure of angle A is 9◦more than three times the measure of angle B. Find the measure of each angle. Solution First we write out our 2 equations. We will use x to be the measure of angle A and y to be the measure of angle B. We get the following system: x + y = 90 2x = 3y + 9 First, we’ll solve this system with the graphical method. For this, we need to convert the two equations to y = mx + b form: x + y = 90 ⇒y = −x + 90 2x = 3y + 9 ⇒y = 2 3 x −3 www.ck12.org 298 The ﬁrst line has a slope of -1 and a y−intercept of 90, and the second line has a slope of 2 3 and a y−intercept of -3. The graph looks like this: In the graph, it appears that the lines cross at around x = 55, y = 35, but it is diﬀicult to tell exactly! Graphing by hand is not the best method in this case! Next, we’ll try solving by substitution. Let’s look again at the system: x + y = 90 2x = 3y + 9 We’ve already seen that we can start by solving either equation for y, so let’s start with the ﬁrst one: y = 90 −x Substitute into the second equation: 2x = 3(90 −x) + 9 distribute the 3 : 2x = 270 −3x + 9 add 3x to both sides : 5x = 270 + 9 = 279 divide by 5 : x = 55.8◦ Substitute back into our expression for y: y = 90 −55.8 = 34.2◦ Angle A measures 55.8◦; angle B measures 34.2◦. Finally, we’ll try solving by elimination (with multiplication): Rearrange equation one to standard form: x + y = 90 ⇒2x + 2y = 180 Multiply equation two by 2: 2x = 3y + 9 ⇒2x −3y = 9 Subtract: 299 www.ck12.org 2x + 2y = 180 − (2x −3y) = −9 5y = 171 Divide by 5 to obtain y = 34.2◦ Substitute this value into the very ﬁrst equation: x + 34.2 = 90 subtract 34.2 from both sides : x = 55.8◦ Angle A measures 55.8◦; angle B measures 34.2◦. Even though this system looked ideal for substitution, the method of multiplication worked well too. Once the equations were rearranged properly, the solution was quick to ﬁnd. You’ll need to decide yourself which method to use in each case you see from now on. Try to master all the techniques, and recognize which one will be most eﬀicient for each system you are asked to solve. The following Khan Academy video contains three examples of solving systems of equations using addition and subtraction as well as multiplication (which is the next topic): nok99JOhcjo (9:57). (Note that the narrator is not always careful about showing his work, and you should try to be neater in your mathematical writing.) For even more practice, we have this video. One common type of problem involving systems of equations (especially on standardized tests) is “age problems.” In the following video the narrator shows two examples of age problems, one involving a single person and one involving two people. Khan Academy Age Problems (7:13) Review Questions 1. Solve the system: 3x + 4y = 2.5 5x −4y = 25.5 2. Solve the system: 5x + 7y = −31 5x −9y = 17 3. Solve the system: 3y −4x = −33 5x −3y = 40.5 4. Nadia and Peter visit the candy store. Nadia buys three candy bars and four fruit roll-ups for $2.84. Peter also buys three candy bars, but can only aﬀord one additional fruit roll-up. His purchase costs $1.79. What is the cost of a candy bar and a fruit roll-up individually? 5. A small plane ﬂies from Los Angeles to Denver with a tail wind (the wind blows in the same direction as the plane) and an air-traﬀic controller reads its ground-speed (speed measured relative to the ground) at 275 miles per hour. Another, identical plane, moving in the opposite direction has a ground-speed of 227 miles per hour. Assuming both planes are ﬂying with identical air-speeds, calculate the speed of the wind. www.ck12.org 300 6. An airport taxi ﬁrm charges a pick-up fee, plus an additional per-mile fee for any rides taken. If a 12-mile journey costs $14.29 and a 17-mile journey costs $19.91, calculate: (a) the pick-up fee (b) the per-mile rate (c) the cost of a seven mile trip 7. Calls from a call-box are charged per minute at one rate for the ﬁrst ﬁve minutes, then a diﬀerent rate for each additional minute. If a 7-minute call costs $4.25 and a 12-minute call costs $5.50, ﬁnd each rate. 8. A plumber and a builder were employed to ﬁt a new bath, each working a diﬀerent number of hours. The plumber earns $35 per hour, and the builder earns $28 per hour. Together they were paid $330.75, but the plumber earned $106.75 more than the builder. How many hours did each work? 9. Paul has a part time job selling computers at a local electronics store. He earns a ﬁxed hourly wage, but can earn a bonus by selling warranties for the computers he sells. He works 20 hours per week. In his ﬁrst week, he sold eight warranties and earned $220. In his second week, he managed to sell 13 warranties and earned $280. What is Paul’s hourly rate, and how much extra does he get for selling each warranty? Solve the following systems using multiplication. 10. 5x −10y = 15 3x −2y = 3 11. 5x −y = 10 3x −2y = −1 12. 5x + 7y = 15 7x −3y = 5 13. 9x + 5y = 9 12x + 8y = 12.8 14. 4x −3y = 1 3x −4y = 4 15. 7x −3y = −3 6x + 4y = 3 Solve the following systems using any method. 16. x = 3y x −2y = −3 17. y = 3x + 2 y = −2x + 7 18. 5x −5y = 5 5x + 5y = 35 19. y = −3x −3 3x −2y + 12 = 0 20. 3x −4y = 3 4y + 5x = 10 21. 9x −2y = −4 2x −6y = 1 22. Supplementary angles are two angles whose sum is 180◦. Angles A and B are supplementary angles. The measure of Angle A is 18◦less than twice the measure of Angle B. Find the measure of each angle. 301 www.ck12.org 23. A farmer has fertilizer in 5% and 15% solutions. How much of each type should he mix to obtain 100 liters of fertilizer in a 12% solution? 24. A 150-yard pipe is cut to provide drainage for two ﬁelds. If the length of one piece is three yards less that twice the length of the second piece, what are the lengths of the two pieces? 25. Mr. Stein invested a total of $100,000 in two companies for a year. Company A’s stock showed a 13% annual gain, while Company B showed a 3% loss for the year. Mr. Stein made an 8% return on his investment over the year. How much money did he invest in each company? 26. A baker sells plain cakes for $7 and decorated cakes for $11. On a busy Saturday the baker started with 120 cakes, and sold all but three. His takings for the day were $991. How many plain cakes did he sell that day, and how many were decorated before they were sold? 27. Twice John’s age plus ﬁve times Claire’s age is 204. Nine times John’s age minus three times Claire’s age is also 204. How old are John and Claire? 7.4 Special Types of Linear Systems Learning Objectives • Identify and understand what is meant by an inconsistent linear system. • Identify and understand what is meant by a consistent linear system. • Identify and understand what is meant by a dependent linear system. Introduction As we saw in Section 7.1, a system of linear equations is a set of linear equations which must be solved together. The lines in the system can be graphed together on the same coordinate graph and the solution to the system is the point at which the two lines intersect. Or at least that’s what usually happens. But what if the lines turn out to be parallel when we graph them? If the lines are parallel, they won’t ever intersect. That means that the system of equations they represent has no solution. A system with no solutions is called an inconsistent system. And what if the lines turn out to be identical? www.ck12.org 302 If the two lines are the same, then every point on one line is also on the other line, so every point on the line is a solution to the system. The system has an inﬁnite number of solutions, and the two equations are really just diﬀerent forms of the same equation. Such a system is called a dependent system. But usually, two lines cross at exactly one point and the system has exactly one solution: A system with exactly one solution is called a consistent system. To identify a system as consistent, inconsistent, or dependent, we can graph the two lines on the same graph and see if they intersect, are parallel, or are the same line. But sometimes it is hard to tell whether two lines are parallel just by looking at a roughly sketched graph. Another option is to write each line in slope-intercept form and compare the slopes and y−intercepts of the two lines. To do this we must remember that: • Lines with diﬀerent slopes always intersect. • Lines with the same slope but diﬀerent y−intercepts are parallel. • Lines with the same slope and the same y−intercepts are identical. Example 1 Determine whether the following system has exactly one solution, no solutions, or an inﬁnite number of solutions. 2x −5y = 2 4x + y = 5 Solution We must rewrite the equations so they are in slope-intercept form 2x −5y = 2 −5y = −2x + 2 y = 2 5 x −2 5 ⇒ ⇒ 4x + y = 5 y = −4x + 5 y = −4x + 5 303 www.ck12.org The slopes of the two equations are diﬀerent; therefore the lines must cross at a single point and the system has exactly one solution. This is a consistent system. Example 2 Determine whether the following system has exactly one solution, no solutions, or an inﬁnite number of solutions. 3x = 5 −4y 6x + 8y = 7 Solution We must rewrite the equations so they are in slope-intercept form 3x = 5 −4y 4y = −3x + 5 y = −3 4 x + 5 4 ⇒ ⇒ 6x + 8y = 7 8y = −6x + 7 y = −3 4 x + 7 8 The slopes of the two equations are the same but the y−intercepts are diﬀerent; therefore the lines are parallel and the system has no solutions. This is an inconsistent system. Example 3 Determine whether the following system has exactly one solution, no solutions, or an inﬁnite number of solutions. x + y = 3 3x + 3y = 9 Solution We must rewrite the equations so they are in slope-intercept form x + y = 3 y = −x + 3 y = −x + 3 ⇒ ⇒ 3x + 3y = 9 3y = −3x + 9 y = −x + 3 The lines are identical; therefore the system has an inﬁnite number of solutions. It is a dependent system. Determining the Type of System Algebraically A third option for identifying systems as consistent, inconsistent or dependent is to just solve the system and use the result as a guide. Example 4 Solve the following system of equations. Identify the system as consistent, inconsistent or dependent. 10x −3y = 3 2x + y = 9 Solution Let’s solve this system using the substitution method. Solve the second equation for y: www.ck12.org 304 2x + y = 9 ⇒y = −2x + 9 Substitute that expression for y in the ﬁrst equation: 10x −3y = 3 10x −3(−2x + 9) = 3 10x + 6x −27 = 3 16x = 30 x = 15 8 Substitute the value of x back into the second equation and solve for y: 2x + y = 9 ⇒y = −2x + 9 ⇒y = −2 · 15 8 + 9 ⇒y = 21 4 The solution to the system is ( 15 8 , 21 4 ) . The system is consistent since it has only one solution. Example 5 Solve the following system of equations. Identify the system as consistent, inconsistent or dependent. 3x −2y = 4 9x −6y = 1 Solution Let’s solve this system by the method of multiplication. Multiply the ﬁrst equation by 3: 3(3x −2y = 4) 9x −6y = 12 ⇒ 9x −6y = 1 9x −6y = 1 Add the two equations: 9x −6y = 4 9x −6y = 1 0 = 13 This statement is not true. If our solution to a system turns out to be a statement that is not true, then the system doesn’t really have a solution; it is inconsistent. Example 6 Solve the following system of equations. Identify the system as consistent, inconsistent or dependent. 4x + y = 3 12x + 3y = 9 Solution 305 www.ck12.org Let’s solve this system by substitution. Solve the ﬁrst equation for y: 4x + y = 3 ⇒y = −4x + 3 Substitute this expression for y in the second equation: 12x + 3y = 9 12x + 3(−4x + 3) = 9 12x −12x + 9 = 9 9 = 9 This statement is always true. If our solution to a system turns out to be a statement that is always true, then the system is dependent. A second glance at the system in this example reveals that the second equation is three times the ﬁrst equation, so the two lines are identical. The system has an inﬁnite number of solutions because they are really the same equation and trace out the same line. Let’s clarify this statement. An inﬁnite number of solutions does not mean that any ordered pair (x, y) satisﬁes the system of equations. Only ordered pairs that solve the equation in the system (either one of the equations) are also solutions to the system. There are inﬁnitely many of these solutions to the system because there are inﬁnitely many points on any one line. For example, (1, -1) is a solution to the system in this example, and so is (-1, 7). Each of them ﬁts both the equations because both equations are really the same equation. But (3, 5) doesn’t ﬁt either equation and is not a solution to the system. In fact, for every x−value there is just one y−value that ﬁts both equations, and for every y−value there is exactly one x−value—just as there is for a single line. Let’s summarize how to determine the type of system we are dealing with algebraically. • A consistent system will always give exactly one solution. • An inconsistent system will yield a statement that is always false (like 0 = 13). • A dependent system will yield a statement that is always true (like 9 = 9). Applications In this section, we’ll see how consistent, inconsistent and dependent systems might arise in real life. Example 7 The movie rental store CineStar oﬀers customers two choices. Customers can pay a yearly membership of $45 and then rent each movie for $2 or they can choose not to pay the membership fee and rent each movie for $3.50. How many movies would you have to rent before the membership becomes the cheaper option? Solution Let’s translate this problem into algebra. Since there are two diﬀerent options to consider, we can write two diﬀerent equations and form a system. The choices are “membership” and “no membership.” We’ll call the number of movies you rent x and the total cost of renting movies for a year y. www.ck12.org 306 Table 7.7: ﬂat fee rental fee total membership $45 2x y = 45 + 2x no membership $0 3.50x y = 3.5x The ﬂat fee is the dollar amount you pay per year and the rental fee is the dollar amount you pay when you rent a movie. For the membership option the rental fee is 2x, since you would pay $2 for each movie you rented; for the no membership option the rental fee is 3.50x, since you would pay $3.50 for each movie you rented. Our system of equations is: y = 45 + 2x y = 3.50x Here’s a graph of the system: Now we need to ﬁnd the exact intersection point. Since each equation is already solved for y, we can easily solve the system with substitution. Substitute the second equation into the ﬁrst one: y = 45 + 2x ⇒3.50x = 45 + 2x ⇒1.50x = 45 ⇒x = 30 movies y = 3.50x You would have to rent 30 movies per year before the membership becomes the better option. This example shows a real situation where a consistent system of equations is useful in ﬁnding a solution. Remember that for a consistent system, the lines that make up the system intersect at single point. In other words, the lines are not parallel or the slopes are diﬀerent. In this case, the slopes of the lines represent the price of a rental per movie. The lines cross because the price of rental per movie is diﬀerent for the two options in the problem Now let’s look at a situation where the system is inconsistent. From the previous explanation, we can conclude that the lines will not intersect if the slopes are the same (and the y−intercept is diﬀerent). Let’s change the previous problem so that this is the case. Example 8 Two movie rental stores are in competition. Movie House charges an annual membership of $30 and charges $3 per movie rental. Flicks for Cheap charges an annual membership of $15 and charges $3 per 307 www.ck12.org movie rental. After how many movie rentals would Movie House become the better option? Solution It should already be clear to see that Movie House will never become the better option, since its membership is more expensive and it charges the same amount per movie as Flicks for Cheap. The lines on a graph that describe each option have diﬀerent y−intercepts—namely 30 for Movie House and 15 for Flicks for Cheap—but the same slope: 3 dollars per movie. This means that the lines are parallel and so the system is inconsistent. Now let’s see how this works algebraically. Once again, we’ll call the number of movies you rent x and the total cost of renting movies for a year y. Table 7.8: ﬂat fee rental fee total Movie House $30 3x y = 30 + 3x Flicks for Cheap $15 3x y = 15 + 3x The system of equations that describes this problem is: y = 30 + 3x y = 15 + 3x Let’s solve this system by substituting the second equation into the ﬁrst equation: y = 30 + 3x ⇒15 + 3x = 30 + 3x ⇒15 = 30 This statement is always false. y = 15 + 3x This means that the system is inconsistent. Example 9 Peter buys two apples and three bananas for $4. Nadia buys four apples and six bananas for $8 from the same store. How much does one banana and one apple costs? Solution We must write two equations: one for Peter’s purchase and one for Nadia’s purchase. Let’s say a is the cost of one apple and b is the cost of one banana. Table 7.9: cost of apples cost of bananas total cost Peter 2a 3b 2a + 3b = 4 Nadia 4a 6b 4a + 6b = 8 The system of equations that describes this problem is: 2a + 3b = 4 4a + 6b = 8 Let’s solve this system by multiplying the ﬁrst equation by -2 and adding the two equations: −2(2a + 3b = 4) −4a −6b = −8 www.ck12.org 308 ⇒ 4a + 6b = 8 4a + 6b = 8 0 + 0 = 0 This statement is always true. This means that the system is dependent. Looking at the problem again, we can see that we were given exactly the same information in both statements. If Peter buys two apples and three bananas for $4, it makes sense that if Nadia buys twice as many apples (four apples) and twice as many bananas (six bananas) she will pay twice the price ($8). Since the second equation doesn’t give us any new information, it doesn’t make it possible to ﬁnd out the price of each fruit. Review Questions Express each equation in slope-intercept form. Without graphing, state whether the system of equations is consistent, inconsistent or dependent. 1. 3x −4y = 13 y = −3x −7 2. 3 5 x + y = 3 1.2x + 2y = 6 3. 3x −4y = 13 y = −3x −7 4. 3x −3y = 3 x −y = 1 5. 0.5x −y = 30 0.5x −y = −30 6. 4x −2y = −2 3x + 2y = −12 7. 3x + y = 4 y = 5 −3x 8. x −2y = 7 4y −2x = 14 Find the solution of each system of equations using the method of your choice. State if the system is inconsistent or dependent. 9. 3x + 2y = 4 −2x + 2y = 24 10. 5x −2y = 3 2x −3y = 10 11. 3x −4y = 13 y = −3x −7 12. 5x −4y = 1 −10x + 8y = −30 13. 4x + 5y = 0 3x = 6y + 4.5 14. −2y + 4x = 8 y −2x = −4 15. x −1 2y = 3 2 3x + y = 6 309 www.ck12.org 16. 0.05x + 0.25y = 6 x + y = 24 17. x + 2 3y = 6 3x + 2y = 2 18. A movie theater charges $4.50 for children and $8.00 for adults. (a) On a certain day, 1200 people enter the theater and $8375 is collected. How many children and how many adults attended? (b) The next day, the manager announces that she wants to see them take in $10000 in tickets. If there are 240 seats in the house and only ﬁve movie showings planned that day, is it possible to meet that goal? (c) At the same theater, a 16-ounce soda costs $3 and a 32-ounce soda costs $5. If the theater sells 12,480 ounces of soda for $2100, how many people bought soda? (Note: Be careful in setting up this problem!) 19. Jamal placed two orders with an internet clothing store. The ﬁrst order was for 13 ties and 4 pairs of suspenders, and totaled $487. The second order was for 6 ties and 2 pairs of suspenders, and totaled $232. The bill does not list the per-item price, but all ties have the same price and all suspenders have the same price. What is the cost of one tie and of one pair of suspenders? 20. An airplane took four hours to ﬂy 2400 miles in the direction of the jet-stream. The return trip against the jet-stream took ﬁve hours. What were the airplane’s speed in still air and the jet-stream’s speed? 21. Nadia told Peter that she went to the farmer’s market and bought two apples and one banana, and that it cost her $2.50. She thought that Peter might like some fruit, so she went back to the seller and bought four more apples and two more bananas. Peter thanked Nadia, but told her that he did not like bananas, so he would only pay her for four apples. Nadia told him that the second time she paid $6.00 for the fruit. (a) What did Peter ﬁnd when he tried to ﬁgure out the price of four apples? (b) Nadia then told Peter she had made a mistake, and she actually paid $5.00 on her second trip. Now what answer did Peter get when he tried to ﬁgure out how much to pay her? (c) Alicia then showed up and told them she had just bought 3 apples and 2 bananas from the same seller for $4.25. Now how much should Peter pay Nadia for four apples? 7.5 Systems of Linear Inequalities Learning Objectives • Graph linear inequalities in two variables. • Solve systems of linear inequalities. • Solve optimization problems. Introduction In the last chapter you learned how to graph a linear inequality in two variables. To do that, you graphed the equation of the straight line on the coordinate plane. The line was solid for ≤or ≥signs (where the equals sign is included), and the line was dashed for < or > signs (where the equals sign is not included). Then you shaded above the line (if the inequality began with y > or y ≥) or below the line (if it began with y < or y ≤). In this section, we’ll see how to graph two or more linear inequalities on the same coordinate plane. The inequalities are graphed separately on the same graph, and the solution for the system is the common www.ck12.org 310 shaded region between all the inequalities in the system. One linear inequality in two variables divides the plane into two half-planes. A system of two or more linear inequalities can divide the plane into more complex shapes. Let’s start by solving a system of two inequalities. Graph a System of Two Linear Inequalities Example 1 Solve the following system: 2x + 3y ≤18 x −4y ≤12 Solution Solving systems of linear inequalities means graphing and ﬁnding the intersections. So we graph each inequality, and then ﬁnd the intersection regions of the solution. First, let’s rewrite each equation in slope-intercept form. (Remember that this form makes it easier to tell which region of the coordinate plane to shade.) Our system becomes 3y ≤−2x + 18 y ≤−2 3 x + 6 ⇒ −4y ≤−x + 12 y ≥x 4 −3 Notice that the inequality sign in the second equation changed because we divided by a negative number! For this ﬁrst example, we’ll graph each inequality separately and then combine the results. Here’s the graph of the ﬁrst inequality: The line is solid because the equals sign is included in the inequality. Since the inequality is less than or equal to, we shade below the line. 311 www.ck12.org And here’s the graph of the second inequality: The line is solid again because the equals sign is included in the inequality. We now shade above the line because y is greater than or equal to. www.ck12.org 312 When we combine the graphs, we see that the blue and red shaded regions overlap. The area where they overlap is the area where both inequalities are true. Thus that area (shown below in purple) is the solution of the system. The kind of solution displayed in this example is called unbounded, because it continues forever in at least one direction (in this case, forever upward and to the left). Example 2 There are also situations where a system of inequalities has no solution. For example, let’s solve this system. y ≤2x −4 y > 2x + 6 Solution We start by graphing the ﬁrst line. The line will be solid because the equals sign is included in the inequality. We must shade downwards because y is less than. 313 www.ck12.org Next we graph the second line on the same coordinate axis. This line will be dashed because the equals sign is not included in the inequality. We must shade upward because y is greater than. It doesn’t look like the two shaded regions overlap at all. The two lines have the same slope, so we know they are parallel; that means that the regions indeed won’t ever overlap since the lines won’t ever cross. So this system of inequalities has no solution. But a system of inequalities can sometimes have a solution even if the lines are parallel. For example, what happens if we swap the directions of the inequality signs in the system we just graphed? To graph the system y ≥2x −4 y < 2x + 6, we draw the same lines we drew for the previous system, but we shade upward for the ﬁrst inequality and downward for the second inequality. Here is the result: www.ck12.org 314 You can see that this time the shaded regions overlap. The area between the two lines is the solution to the system. Graph a System of More Than Two Linear Inequalities When we solve a system of just two linear inequalities, the solution is always an unbounded region—one that continues inﬁnitely in at least one direction. But if we put together a system of more than two inequalities, sometimes we can get a solution that is bounded—a ﬁnite region with three or more sides. Let’s look at a simple example. Example 3 Find the solution to the following system of inequalities. 3x −y < 4 4y + 9x < 8 x ≥0 y ≥0 Solution Let’s start by writing our inequalities in slope-intercept form. y > 3x −4 y < −9 4 x + 2 x ≥0 y ≥0 Now we can graph each line and shade appropriately. First we graph y > 3x −4 : 315 www.ck12.org Next we graph y < −9 4 x + 2 : Finally we graph x ≥0 and y ≥0, and we’re left with the region below; this is where all four inequalities overlap. www.ck12.org 316 The solution is bounded because there are lines on all sides of the solution region. In other words, the solution region is a bounded geometric ﬁgure, in this case a triangle. Notice, too, that only three of the lines we graphed actually form the boundaries of the region. Sometimes when we graph multiple inequalities, it turns out that some of them don’t aﬀect the overall solution; in this case, the solution would be the same even if we’d left out the inequality y > 3x −4. That’s because the solution region of the system formed by the other three inequalities is completely contained within the solution region of that fourth inequality; in other words, any solution to the other three inequalities is automatically a solution to that one too, so adding that inequality doesn’t narrow down the solution set at all. But that wasn’t obvious until we actually drew the graph! Solve Real-World Problems Using Systems of Linear Inequalities A lot of interesting real-world problems can be solved with systems of linear inequalities. For example, you go to your favorite restaurant and you want to be served by your best friend who happens to work there. However, your friend only waits tables in a certain region of the restaurant. The restaurant is also known for its great views, so you want to sit in a certain area of the restaurant that oﬀers a good view. Solving a system of linear inequalities will allow you to ﬁnd the area in the restaurant where you can sit to get the best view and be served by your friend. Often, systems of linear inequalities deal with problems where you are trying to ﬁnd the best possible situation given a set of constraints. Most of these application problems fall in a category called linear programming problems. Linear programming is the process of taking various linear inequalities relating to some situation, and ﬁnding the best possible value under those conditions. A typical example would be taking the limitations of materials and labor at a factory, then determining the best production levels for maximal proﬁts under those conditions. These kinds of problems are used every day in the organization and allocation of resources. These real-life systems can have dozens or hundreds of variables, or more. In this section, we’ll only work with the simple two-variable linear case. The general process is to: • Graph the inequalities (called constraints) to form a bounded area on the coordinate plane (called 317 www.ck12.org the feasibility region). • Figure out the coordinates of the corners (or vertices) of this feasibility region by solving the system of equations that applies to each of the intersection points. • Test these corner points in the formula (called the optimization equation) for which you’re trying to ﬁnd the maximum or minimum value. Example 4 If z = 2x + 5y, ﬁnd the maximum and minimum values of z given these constraints: 2x −y ≤12 4x + 3y ≥0 x −y ≤6 Solution First, we need to ﬁnd the solution to this system of linear inequalities by graphing and shading appropri-ately. To graph the inequalities, we rewrite them in slope-intercept form: y ≥2x −12 y ≥−4 3 x y ≥x −6 These three linear inequalities are called the constraints, and here is their graph: The shaded region in the graph is called the feasibility region. All possible solutions to the system occur in that region; now we must try to ﬁnd the maximum and minimum values of the variable z within that region. In other words, which values of x and y within the feasibility region will give us the greatest and smallest overall values for the expression 2x + 5y? Fortunately, we don’t have to test every point in the region to ﬁnd that out. It just so happens that the minimum or maximum value of the optimization equation in a linear system like this will always be found at one of the vertices (the corners) of the feasibility region; we just have to ﬁgure out which vertices. So www.ck12.org 318 for each vertex—each point where two of the lines on the graph cross—we need to solve the system of just those two equations, and then ﬁnd the value of z at that point. The ﬁrst system consists of the equations y = 2x −12 and y = −4 3 x. We can solve this system by substitution: −4 3 x = 2x −12 ⇒−4x = 6x −36 ⇒−10x = −36 ⇒x = 3.6 y = 2x −12 ⇒y = 2(3.6) −12 ⇒y = −4.8 The lines intersect at the point (3.6, -4.8). The second system consists of the equations y = 2x−12 and y = x−6. Solving this system by substitution: x −6 = 2x −12 ⇒6 = x ⇒x = 6 y = x −6 ⇒y = 6 −6 ⇒y = 6 The lines intersect at the point (6, 6). The third system consists of the equations y = −4 3 x and y = x −6. Solving this system by substitution: x −6 = −4 3 x ⇒3x −18 = −4x ⇒7x = 18 ⇒x = 2.57 y = x −6 ⇒y = 2.57 −6 ⇒y = −3.43 The lines intersect at the point (2.57, -3.43). So now we have three diﬀerent points that might give us the maximum and minimum values for z. To ﬁnd out which ones actually do give the maximum and minimum values, we can plug the points into the optimization equation z = 2x + 5y. When we plug in (3.6, -4.8), we get z = 2(3.6) + 5(−4.8) = −16.8. When we plug in (6, 0), we get z = 2(6) + 5(0) = 12. When we plug in (2.57, -3.43), we get z = 2(2.57) + 5(−3.43) = −12.01. So we can see that the point (6, 0) gives us the maximum possible value for z and the point (3.6, –4.8) gives us the minimum value. In the previous example, we learned how to apply the method of linear programming in the abstract. In the next example, we’ll look at a real-life application. Example 5 You have $10,000 to invest, and three diﬀerent funds to choose from. The municipal bond fund has a 5% return, the local bank’s CDs have a 7% return, and a high-risk account has an expected 10% return. To minimize risk, you decide not to invest any more than $1,000 in the high-risk account. For tax reasons, you need to invest at least three times as much in the municipal bonds as in the bank CDs. What’s the best way to distribute your money given these constraints? Solution Let’s deﬁne our variables: x is the amount of money invested in the municipal bond at 5% return y is the amount of money invested in the bank’s CD at 7% return 10000 −x −y is the amount of money invested in the high-risk account at 10% return z is the total interest returned from all the investments, so z = .05x + .07y + .1(10000 −x −y) or z = 1000 −0.05x −0.03y. This is the amount that we are trying to maximize. Our goal is to ﬁnd the values of x and y that maximizes the value of z. 319 www.ck12.org Now, let’s write inequalities for the constraints: You decide not to invest more than $1000 in the high-risk account—that means: 10000 −x −y ≤1000 You need to invest at least three times as much in the municipal bonds as in the bank CDs—that means: 3y ≤x Also, you can’t invest less than zero dollars in each account, so: x ≥0 y ≥0 10000 −x −y ≥0 To summarize, we must maximize the expression z = 1000 −.05x −.03y using the constraints: 10000 −x −y ≤1000 y ≥9000 −x 3y ≤x y ≤x 3 x ≥0 Or in slope-intercept form: x ≥0 y ≥0 y ≥0 10000 −x −y ≥0 y ≤10000 −x Step 1: Find the solution region to the set of inequalities by graphing each line and shading appropriately. The following ﬁgure shows the overlapping region: The purple region is the feasibility region where all the possible solutions can occur. Step 2: Next we need to ﬁnd the corner points of the feasibility region. Notice that there are four corners. To ﬁnd their coordinates, we must pair up the relevant equations and solve each resulting system. System 1: y = x 3 y = 10000 −x www.ck12.org 320 Substitute the ﬁrst equation into the second equation: x 3 = 10000 −x ⇒x = 30000 −3x ⇒4x = 30000 ⇒x = 7500 y = x 3 ⇒y = 7500 3 ⇒y = 2500 The intersection point is (7500, 2500). System 2: y = x 3 y = 9000 −x Substitute the ﬁrst equation into the second equation: x 3 = 9000 −x ⇒x = 27000 −3x ⇒4x = 27000 ⇒x = 6750 y = x 3 ⇒y = 6750 3 ⇒y = 2250 The intersection point is (6750, 2250). System 3: y = 0 y = 10000 −x. The intersection point is (10000, 0). System 4: y = 0 y = 9000 −x. The intersection point is (9000, 0). Step 3: In order to ﬁnd the maximum value for z, we need to plug all the intersection points into the equation for z and ﬁnd which one yields the largest number. (7500, 2500): z = 1000 −0.05(7500) −0.03(2500) = 550 (6750, 2250): z = 1000 −0.05(6750) −0.03(2250) = 595 (10000, 0): z = 1000 −0.05(10000) −0.03(0) = 500 (9000, 0): z = 1000 −0.05(9000) −0.03(0) = 550 The maximum return on the investment of $595 occurs at the point (6750, 2250). This means that: $6,750 is invested in the municipal bonds. $2,250 is invested in the bank CDs. $1,000 is invested in the high-risk account. Graphing calculators can be very useful for problems that involve this many inequalities. The video at shows a real-world linear programming problem worked through in detail on a graphing calculator, although the methods used there can also be used for pencil-and paper solving. Review Questions 1. Consider the system 321 www.ck12.org y < 3x −5 y > 3x −5. Is it consistent or inconsistent? Why? 2. Consider the system y ≤2x + 3 y ≥2x + 3. Is it consistent or inconsistent? Why? 3. Consider the system y ≤−x + 1 y > −x + 1. Is it consistent or inconsistent? Why? 4. In example 3 in this lesson, we solved a system of four inequalities and saw that one of the inequalities, y > 3x −4, didn’t aﬀect the solution set of the system. (a) What would happen if we changed that inequality to y < 3x −4? (b) What’s another inequality that we could add to the original system without changing it? Show how by sketching a graph of that inequality along with the rest of the system. (c) What’s another inequality that we could add to the original system to make it inconsistent? Show how by sketching a graph of that inequality along with the rest of the system. 5. Recall the compound inequalities in one variable that we worked with back in chapter 6. Compound inequalities with “and” are simply systems like the ones we are working with here, except with one variable instead of two. (a) Graph the inequality x > 3 in two dimensions. What’s another inequality that could be combined with it to make an inconsistent system? (b) Graph the inequality x ≤4 on a number line. What two-dimensional system would have a graph that looks just like this one? Find the solution region of the following systems of inequalities. 6. x −y < −6 2y ≥3x + 17 7. 4y −5x < 8 −5x ≥16 −8y 8. 5x −y ≥5 2y −x ≥−10 9. 5x + 2y ≥−25 3x −2y ≤17 x −6y ≥27 10. 2x −3y ≤21 x + 4y ≤6 3x + y ≥−4 11. 12x −7y < 120 7x −8y ≥36 5x + y ≥12 Solve the following linear programming problems. 12. Given the following constraints, ﬁnd the maximum and minimum values of z = −x + 5y: x + 3y ≤0 x −y ≥0 3x −7y ≤16 www.ck12.org 322 13. Santa Claus is assigning elves to work an eight-hour shift making toy trucks. Apprentice elves draw a wage of ﬁve candy canes per hour worked, but can only make four trucks an hour. Senior elves can make six trucks an hour and are paid eight candy canes per hour. There’s only room for nine elves in the truck shop, and due to a candy-makers’ strike, Santa Claus can only pay out 480 candy canes for the whole 8-hour shift. (a) How many senior elves and how many apprentice elves should work this shift to maximize the number of trucks that get made? (b) How many trucks will be made? (c) Just before the shift begins, the apprentice elves demand a wage increase; they insist on being paid seven candy canes an hour. Now how many apprentice elves and how many senior elves should Santa assign to this shift? (d) How many trucks will now get made, and how many candy canes will Santa have left over? 14. In Adrian’s Furniture Shop, Adrian assembles both bookcases and TV cabinets. Each type of furniture takes her about the same time to assemble. She ﬁgures she has time to make at most 18 pieces of furniture by this Saturday. The materials for each bookcase cost her $20 and the materials for each TV stand costs her $45. She has $600 to spend on materials. Adrian makes a proﬁt of $60 on each bookcase and a proﬁt of $100 on each TV stand. (a) Set up a system of inequalities. What x−and y−values do you get for the point where Adrian’s proﬁt is maximized? Does this solution make sense in the real world? (b) What two possible real-world x−values and what two possible real-world y−values would be closest to the values in that solution? (c) With two choices each for x and y, there are four possible combinations of x−and y−values. Of those four combinations, which ones actually fall within the feasibility region of the problem? (d) Which one of those feasible combinations seems like it would generate the most proﬁt? Test out each one to conﬁrm your guess. How much proﬁt will Adrian make with that combination? (e) Based on Adrian’s previous sales ﬁgures, she doesn’t think she can sell more than 8 TV stands. Now how many of each piece of furniture should she make, and what will her proﬁt be? (f) Suppose Adrian is conﬁdent she can sell all the furniture she can make, but she doesn’t have room to display more than 7 bookcases in her shop. Now how many of each piece of furniture should she make, and what will her proﬁt be? 15. Here’s a “linear programming” problem on a line instead of a plane: Given the constraints x ≤5 and x ≥−2, maximize the value of y where y = x + 3. Texas Instruments Resources In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http: //www.ck12.org/flexr/chapter/9617. 323 www.ck12.org Chapter 8 Exponential Functions 8.1 Exponent Properties Involving Products Learning Objectives • Use the product of a power property. • Use the power of a product property. • Simplify expressions involving product properties of exponents. Introduction Back in chapter 1, we brieﬂy covered expressions involving exponents, like 35 or x3. In these expressions, the number on the bottom is called the base and the number on top is the power or exponent. The whole expression is equal to the base multiplied by itself a number of times equal to the exponent; in other words, the exponent tells us how many copies of the base number to multiply together. Example 1 Write in exponential form. a) 2 · 2 b) (−3)(−3)(−3) c) y · y · y · y · y d) (3a)(3a)(3a)(3a) Solution a) 2 · 2 = 22 because we have 2 factors of 2 b) (−3)(−3)(−3) = (−3)3 because we have 3 factors of (-3) c) y · y · y · y · y = y5 because we have 5 factors of y d) (3a)(3a)(3a)(3a) = (3a)4 because we have 4 factors of 3a When the base is a variable, it’s convenient to leave the expression in exponential form; if we didn’t write x7, we’d have to write x · x · x · x · x · x · x instead. But when the base is a number, we can simplify the expression further than that; for example, 27 equals 2 · 2 · 2 · 2 · 2 · 2 · 2, but we can multiply all those 2’s to get 128. www.ck12.org 324 Let’s simplify the expressions from Example 1. Example 2 Simplify. a) 22 b) (−3)3 c) y5 d) (3a)4 Solution a) 22 = 2 · 2 = 4 b) (−3)3 = (−3)(−3)(−3) = −27 c) y5 is already simpliﬁed d) (3a)4 = (3a)(3a)(3a)(3a) = 3 · 3 · 3 · 3 · a · a · a · a = 81a4 Be careful when taking powers of negative numbers. Remember these rules: (negative number) · (positive number) = negative number (negative number) · (negative number) = positive number So even powers of negative numbers are always positive. Since there are an even number of factors, we pair up the negative numbers and all the negatives cancel out. (−2)6 = (−2)(−2)(−2)(−2)(−2)(−2) = (−2)(−2) \| {z } +4 · (−2)(−2) \| {z } +4 · (−2)(−2) \| {z } +4 = +64 And odd powers of negative numbers are always negative. Since there are an odd number of factors, we can still pair up negative numbers to get positive numbers, but there will always be one negative factor left over, so the answer is negative: (−2)5 = (−2)(−2)(−2)(−2)(−2) = (−2)(−2) \| {z } +4 · (−2)(−2) \| {z } +4 · (−2) \|{z} −2 = −32 Use the Product of Powers Property So what happens when we multiply one power of x by another? Let’s see what happens when we multiply x to the power of 5 by x cubed. To illustrate better, we’ll use the full factored form for each: (x · x · x · x · x) \| {z } x5 · (x · x · x) \| {z } x3 = (x · x · x · x · x · x · x · x) \| {z } x8 So x5 × x3 = x8. You may already see the pattern to multiplying powers, but let’s conﬁrm it with another example. We’ll multiply x squared by x to the power of 4: (x · x) \|{z} x2 · (x · x · x · x) \| {z } x4 = (x · x · x · x · x · x) \| {z } x6 So x2 × x4 = x6. Look carefully at the powers and how many factors there are in each calculation. 5 x’s times 3 x’s equals (5 + 3) = 8 x’s. 2 x’s times 4 x’s equals (2 + 4) = 6 x’s. 325 www.ck12.org You should see that when we take the product of two powers of x, the number of x’s in the answer is the total number of x’s in all the terms you are multiplying. In other words, the exponent in the answer is the sum of the exponents in the product. Product Rule for Exponents: xn · xm = x(n+m) There are some easy mistakes you can make with this rule, however. Let’s see how to avoid them. Example 3 Multiply 22 · 23. Solution 22 · 23 = 25 = 32 Note that when you use the product rule you don’t multiply the bases. In other words, you must avoid the common error of writing 22 · 23 = 45. You can see this is true if you multiply out each expression: 4 times 8 is deﬁnitely 32, not 1024. Example 4 Multiply 22 · 33. Solution 22 · 33 = 4 · 27 = 108 In this case, we can’t actually use the product rule at all, because it only applies to terms that have the same base. In a case like this, where the bases are diﬀerent, we just have to multiply out the numbers by hand—the answer is not 25 or 35 or 65 or anything simple like that. Use the Power of a Product Property What happens when we raise a whole expression to a power? Let’s take x to the power of 4 and cube it. Again we’ll use the full factored form for each expression: (x4)3 = x4 × x4 × x4 3 factors o f {x to the power 4} (x · x · x · x) · (x · x · x · x) · (x · x · x · x) = x · x · x · x · x · x · x · x · x · x · x · x = x12 So (x4)3 = x12. You can see that when we raise a power of x to a new power, the powers multiply. Power Rule for Exponents: (xn)m = x(n·m) If we have a product of more than one term inside the parentheses, then we have to distribute the exponent over all the factors, like distributing multiplication over addition. For example: (x2y)4 = (x2)4 · (y)4 = x8y4. Or, writing it out the long way: (x2y)4 = (x2y)(x2y)(x2y)(x2y) = (x · x · y)(x · x · y)(x · x · y)(x · x · y) = x · x · x · x · x · x · x · x · y · y · y · y = x8y4 Note that this does NOT work if you have a sum or diﬀerence inside the parentheses! For example, (x+y)2 , x2 +y2. This is an easy mistake to make, but you can avoid it if you remember what an exponent means: if you multiply out (x + y)2 it becomes (x + y)(x + y), and that’s not the same as x2 + y2. We’ll learn how we can simplify this expression in a later chapter. www.ck12.org 326 The following video from YourTeacher.com may make it clearer how the power rule works for a variety of exponential expressions: Example 5 Simplify the following expressions. a) 35 · 37 b) 26 · 2 c) (42)3 Solution When we’re just working with numbers instead of variables, we can use the product rule and the power rule, or we can just do the multiplication and then simplify. a) We can use the product rule ﬁrst and then evaluate the result: 35 · 37 = 312 = 531441. OR we can evaluate each part separately and then multiply them: 35 · 37 = 243 · 2187 = 531441. b) We can use the product rule ﬁrst and then evaluate the result: 26 · 2 = 27 = 128. OR we can evaluate each part separately and then multiply them: 26 · 2 = 64 · 2 = 128. c) We can use the power rule ﬁrst and then evaluate the result: (42)3 = 46 = 4096. OR we can evaluate the expression inside the parentheses ﬁrst, and then apply the exponent outside the parentheses: (42)3 = (16)3 = 4096. Example 6 Simplify the following expressions. a) x2 · x7 b) (y3)5 Solution When we’re just working with variables, all we can do is simplify as much as possible using the product and power rules. a) x2 · x7 = x2+7 = x9 b) (y3)5 = y3×5 = y15 Example 7 Simplify the following expressions. a) (3x2y3) · (4xy2) b) (4xyz) · (x2y3) · (2yz4) c) (2a3b3)2 Solution When we have a mix of numbers and variables, we apply the rules to each number and variable separately. a) First we group like terms together: (3x2y3) · (4xy2) = (3 · 4) · (x2 · x) · (y3 · y2) Then we multiply the numbers or apply the product rule on each grouping: = 12x3y5 b) Group like terms together: (4xyz) · (x2y3) · (2yz4) = (4 · 2) · (x · x2) · (y · y3 · y) · (z · z4) Multiply the numbers or apply the product rule on each grouping: = 8x3y5z5 327 www.ck12.org c) Apply the power rule for each separate term in the parentheses: (2a3b3)2 = 22 · (a3)2 · (b3)2 Multiply the numbers or apply the power rule for each term = 4a6b6 Example 8 Simplify the following expressions. a) (x2)2 · x3 b) (2x2y) · (3xy2)3 c) (4a2b3)2 · (2ab4)3 Solution In problems where we need to apply the product and power rules together, we must keep in mind the order of operations. Exponent operations take precedence over multiplication. a) We apply the power rule ﬁrst: (x2)2 · x3 = x4 · x3 Then apply the product rule to combine the two terms: x4 · x3 = x7 b) Apply the power rule ﬁrst: (2x2y) · (3xy2)3 = (2x2y) · (27x3y6) Then apply the product rule to combine the two terms: (2x2y) · (27x3y6) = 54x5y7 c) Apply the power rule on each of the terms separately: (4a2b3)2 · (2ab4)3 = (16a4b6) · (8a3b12) Then apply the product rule to combine the two terms: (16a4b6) · (8a3b12) = 128a7b18 Review Questions Write in exponential notation: 1. 4 · 4 · 4 · 4 · 4 2. 3x · 3x · 3x 3. (−2a)(−2a)(−2a)(−2a) 4. 6 · 6 · 6 · x · x · y · y · y · y 5. 2 · x · y · 2 · 2 · y · x Find each number. 6. 54 7. (−2)6 8. (0.1)5 9. (−0.6)3 10. (1.2)2 + 53 11. 32 · (0.2)3 Multiply and simplify: 12. 63 · 66 13. 22 · 24 · 26 14. 32 · 43 15. x2 · x4 16. (−2y4)(−3y) www.ck12.org 328 17. (4a2)(−3a)(−5a4) Simplify: 18. (a3)4 19. (xy)2 20. (3a2b3)4 21. (−2xy4z2)5 22. (−8x)3(5x)2 23. (4a2)(−2a3)4 24. (12xy)(12xy)2 25. (2xy2)(−x2y)2(3x2y2) 8.2 Exponent Properties Involving Quotients Learning Objectives • Use the quotient of powers property. • Use the power of a quotient property. • Simplify expressions involving quotient properties of exponents. Use the Quotient of Powers Property The rules for simplifying quotients of exponents are a lot like the rules for simplifying products. Let’s look at what happens when we divide x7 by x4: x7 x4 = x · x · x · x · x · x · x x · x · x · x = x · x · x 1 = x3 You can see that when we divide two powers of x, the number of x’s in the solution is the number of x’s in the top of the fraction minus the number of x’s in the bottom. In other words, when dividing expressions with the same base, we keep the same base and simply subtract the exponent in the denominator from the exponent in the numerator. Quotient Rule for Exponents: xn xm = x(n−m) When we have expressions with more than one base, we apply the quotient rule separately for each base: x5y3 x3y2 = x · x · x · x · x x · x · x · y · y · y y · y = x · x 1 · y 1 = x2y OR x5y3 x3y2 = x5−3 · y3−2 = x2y Example 1 Simplify each of the following expressions using the quotient rule. a) x10 x5 b) a6 a c) a5b4 a3b2 Solution a) x10 x5 = x10−5 = x5 329 www.ck12.org b) a6 a = a6−1 = a5 c) a5b4 a3b2 = a5−3 · b4−2 = a2b2 Now let’s see what happens if the exponent in the denominator is bigger than the exponent in the numerator. For example, what happens when we apply the quotient rule to x4 x7 ? The quotient rule tells us to subtract the exponents. 4 minus 7 is -3, so our answer is x−3. A negative exponent! What does that mean? Well, let’s look at what we get when we do the division longhand by writing each term in factored form: x4 x7 = x · x · x · x x · x · x · x · x · x · x = 1 x · x · x = 1 x3 Even when the exponent in the denominator is bigger than the exponent in the numerator, we can still subtract the powers. The x’s that are left over after the others have been canceled out just end up in the denominator instead of the numerator. Just as x7 x4 would be equal to x3 1 (or simply x3), x4 x7 is equal to 1 x3 . And you can also see that 1 x3 is equal to x−3. We’ll learn more about negative exponents shortly. Example 2 Simplify the following expressions, leaving all exponents positive. a) x2 x6 b) a2b6 a5b Solution a) Subtract the exponent in the numerator from the exponent in the denominator and leave the x’s in the denominator: x2 x6 = 1 x6−2 = 1 x4 b) Apply the rule to each variable separately: a2b6 a5b = 1 a5−2 · b6−1 1 = b5 a3 The Power of a Quotient Property When we raise a whole quotient to a power, another special rule applies. Here is an example: ( x3 y2 )4 = ( x3 y2 ) · ( x3 y2 ) · ( x3 y2 ) · ( x3 y2 ) = (x · x · x) · (x · x · x) · (x · x · x) · (x · x · x) (y · y) · (y · y) · (y · y) · (y · y) = x12 y8 Notice that the exponent outside the parentheses is multiplied by the exponent in the numerator and the exponent in the denominator, separately. This is called the power of a quotient rule: Power Rule for Quotients: ( xn ym )p = xn·p ym·p Let’s apply these new rules to a few examples. Example 3 Simplify the following expressions. a) 45 42 b) 53 57 c) ( 34 52 )2 Solution www.ck12.org 330 Since there are just numbers and no variables, we can evaluate the expressions and get rid of the exponents completely. a) We can use the quotient rule ﬁrst and then evaluate the result: 45 42 = 45−2 = 43 = 64 OR we can evaluate each part separately and then divide: 45 42 = 1024 16 = 64 b) Use the quotient rule ﬁrst and hen evaluate the result: 53 57 = 1 54 = 1 625 OR evaluate each part separately and then reduce: 53 57 = 125 78125 = 1 625 Notice that it makes more sense to apply the quotient rule ﬁrst for examples (a) and (b). Applying the exponent rules to simplify the expression before plugging in actual numbers means that we end up with smaller, easier numbers to work with. c) Use the power rule for quotients ﬁrst and then evaluate the result: ( 34 52 )2 = 38 54 = 6561 625 OR evaluate inside the parentheses ﬁrst and then apply the exponent: ( 34 52 )2 = ( 81 25 )2 = 6561 625 Example 4 Simplify the following expressions: a) x12 x5 b) ( x4 x )5 Solution a) Use the quotient rule: x12 x5 = x12−5 = x7 b) Use the power rule for quotients and then the quotient rule: ( x4 x )5 = x20 x5 = x15 OR use the quotient rule inside the parentheses ﬁrst, then apply the power rule: ( x4 x )5 = (x3)5 = x15 Example 5 Simplify the following expressions. a) 6x2y3 2xy2 b) ( 2a3b3 8a7b )2 Solution When we have a mix of numbers and variables, we apply the rules to each number or each variable separately. a) Group like terms together: 6x2y3 2xy2 = 6 2 · x2 x · y3 y2 Then reduce the numbers and apply the quotient rule on each fraction to get 3xy. b) Apply the quotient rule inside the parentheses ﬁrst: ( 2a3b3 8a7b )2 = ( b2 4a4 )2 Then apply the power rule for quotients: ( b2 4a4 )2 = b4 16a8 Example 6 Simplify the following expressions. a) (x2)2 · x6 x4 b) ( 16a2 4b5 )3 · b2 a16 Solution 331 www.ck12.org In problems where we need to apply several rules together, we must keep the order of operations in mind. a) We apply the power rule ﬁrst on the ﬁrst term: (x2)2 · x6 x4 = x4 · x6 x4 Then apply the quotient rule to simplify the fraction: x4 · x6 x4 = x4 · x2 And ﬁnally simplify with the product rule: x4 · x2 = x6 b) ( 16a2 4b5 )3 · b2 a16 Simplify inside the parentheses by reducing the numbers: (4a2 b5 )3 · b2 a16 Then apply the power rule to the ﬁrst fraction: (4a2 b5 )3 · b2 a16 = 64a6 b15 · b2 a16 Group like terms together: 64a6 b15 · b2 a16 = 64 · a6 a16 · b2 b15 And apply the quotient rule to each fraction: 64 · a6 a16 · b2 b15 = 64 a10b13 Review Questions Evaluate the following expressions. 1. 56 52 2. 67 63 3. 34 310 4. ( 3 8 )2 5. ( 22 33 )3 6. 22·32 52 7. 33·52 37 8. ( 23·42 24 )2 www.ck12.org 332 Simplify the following expressions. 9. a3 a2 10. x5 x9 11. ( a3b4 a2b )3 12. x6y2 x2y5 13. 6a3 2a2 14. 15x5 5x 15. ( 18a4 15a10 )4 16. 25yx6 20y5x2 17. ( x6y2 x4y4 )3 18. ( 6a2 4b4 )2 · 5b 3a 19. (3ab)2(4a3b4)3 (6a2b)4 20. (2a2bc2)(6abc3) 4ab2c 21. (2a2bc2)(6abc3) 4ab2c for a = 2, b = 1, and c = 3 22. ( 3x2y 2z )3 · z2 x for x = 1, y = 2, and z = −1 23. 2x3 xy2 · ( x 2y )2 for x = 2, y = −3 24. 2x3 xy2 · ( x 2y )2 for x = 0, y = 6 25. If a = 2 and b = 3, simplify (a2b)(bc)3 a3c2 as much as possible. 8.3 Zero, Negative, and Fractional Exponents Learning Objectives • Simplify expressions with negative exponents. • Simplify expressions with zero exponents. • Simplify expression with fractional exponents. • Evaluate exponential expressions. Introduction The product and quotient rules for exponents lead to many interesting concepts. For example, so far we’ve mostly just considered positive, whole numbers as exponents, but you might be wondering what happens when the exponent isn’t a positive whole number. What does it mean to raise something to the power of zero, or -1, or 1 2? In this lesson, we’ll ﬁnd out. Simplify Expressions With Negative Exponents When we learned the quotient rule for exponents ( xn xm = x(n−m)) , we saw that it applies even when the expo-nent in the denominator is bigger than the one in the numerator. Canceling out the factors in the numerator and denominator leaves the leftover factors in the denominator, and subtracting the exponents leaves a 333 www.ck12.org negative number. So negative exponents simply represent fractions with exponents in the denominator. This can be summarized in a rule: Negative Power Rule for Exponents: x−n = 1 xn , where x , 0 Negative exponents can be applied to products and quotients also. Here’s an example of a negative exponent being applied to a product: (x3y)−2 = x−6y−2 using the power rule x−6y−2 = 1 x6 · 1 y2 = 1 x6y2 using the negative power rule separately on each variable And here’s one applied to a quotient: (a b )−3 = a−3 b−3 using the power rule for quotients a−3 b−3 = a−3 1 · 1 b−3 = 1 a3 · b3 1 using the negative power rule on each variable separately 1 a3 · b3 1 = b3 a3 simplifying the division of fractions b3 a3 = (b a )3 using the power rule for quotients in reverse. That last step wasn’t really necessary, but putting the answer in that form shows us something useful: ( a b )−3 is equal to ( b a )3. This is an example of a rule we can apply more generally: Negative Power Rule for Fractions: ( x y )−n = ( y x )n, where x , 0, y , 0 This rule can be useful when you want to write out an expression without using fractions. Example 1 Write the following expressions without fractions. a) 1 x b) 2 x2 c) x2 y3 d) 3 xy Solution a) 1 x = x−1 b) 2 x2 = 2x−2 c) x2 y3 = x2y−3 d) 3 xy = 3x−1y−1 Example 2 Simplify the following expressions and write them without fractions. a) 4a2b3 2a5b b) ( x 3y2 )3 · x2y 4 Solution www.ck12.org 334 a) Reduce the numbers and apply the quotient rule to each variable separately: 4a2b3 2a5b = 2 · a2−5 · b3−1 = 2a−3b2 b) Apply the power rule for quotients ﬁrst: (2x y2 )3 · x2y 4 = 8x3 y6 · x2y 4 Then simplify the numbers, and use the product rule on the x’s and the quotient rule on the y’s: 8x3 y6 · x2y 4 = 2 · x3+2 · y1−6 = 2x5y−5 You can also use the negative power rule the other way around if you want to write an expression without negative exponents. Example 3 Write the following expressions without negative exponents. a) 3x−3 b) a2b−3c−1 c) 4x−1y3 d) 2x−2 y−3 Solution a) 3x−3 = 3 x3 b) a2b−3c−1 = a2 b3c c) 4x−1y3 = 4y3 x d) 2x−2 y−3 = 2y3 x2 Example 4 Simplify the following expressions and write the answers without negative powers. a) ( ab−2 b3 )2 b) x−3y2 x2y−2 Solution a) Apply the quotient rule inside the parentheses: ( ab−2 b3 )2 = (ab−5)2 Then apply the power rule: (ab−5)2 = a2b−10 = a2 b10 b) Apply the quotient rule to each variable separately: x−3y2 x2y−2 = x−3−2y2−(−2) = x−5y4 = y4 x5 Simplify Expressions with Exponents of Zero Let’s look again at the quotient rule for exponents ( xn xm = x(n−m)) and consider what happens when n = m. For example, what happens when we divide x4 by x4? Applying the quotient rule tells us that x4 x4 = x(4−4) = x0—so what does that zero mean? 335 www.ck12.org Well, we ﬁrst discovered the quotient rule by considering how the factors of x cancel in such a fraction. Let’s do that again with our example of x4 divided by x4: x4 x4 = x · x · x · x x · x · x · x = 1 So x0 = 1! You can see that this works for any value of the exponent, not just 4: xn xn = x(n−n) = x0 Since there is the same number of x’s in the numerator as in the denominator, they cancel each other out and we get x0 = 1. This rule applies for all expressions: Zero Rule for Exponents: x0 = 1, where x , 0 For more on zero and negative exponents, watch the following video at squidoo.com: com/url?sa=t&source=video&cd=4&ved=0CFMQtwIwAw&url=http%3A%2F%2Fwww.youtube. com%2Fwatch%3Fv%3D9svqGWwyN8Q&rct=j&q=negative%20exponents%20applet&ei=1fH6TP2IGoX4sAO 38;usg=AFQjCNHzLF4_2aeo0dMWsa2wJ_CwzckXNA&cad=rja. Simplify Expressions With Fractional Exponents So far we’ve only looked at expressions where the exponents are positive and negative integers. The rules we’ve learned work exactly the same if the powers are fractions or irrational numbers—but what does a fractional exponent even mean? Let’s see if we can ﬁgure that out by using the rules we already know. Suppose we have an expression like 9 1 2 —how can we relate this expression to one that we already know how to work with? For example, how could we turn it into an expression that doesn’t have any fractional exponents? Well, the power rule tells us that if we raise an exponential expression to a power, we can multiply the exponents. For example, if we raise 9 1 2 to the power of 2, we get ( 9 1 2 )2 = 92· 1 2 = 91 = 9. So if 9 1 2 squared equals 9, what does 9 1 2 itself equal? Well, 3 is the number whose square is 9 (that is, it’s the square root of 9), so 9 1 2 must equal 3. And that’s true for all numbers and variables: a number raised to the power of 1 2 is just the square root of the number. We can write that as √x = x 1 2 , and then we can see that’s true because ( √x )2 = x just as ( x 1 2 )2 = x. Similarly, a number to the power of 1 3 is just the cube root of the number, and so on. In general, x 1 n = n √x. And when we raise a number to a power and then take the root of it, we still get a fractional exponent; for example, 3 √ x4 = ( x4) 1 3 = x 4 3 . In general, the rule is as follows: Rule for Fractional Exponents: m √ an = a n m and ( m √a )n = a n m We’ll examine roots and radicals in detail in a later chapter. In this section, we’ll focus on how exponent rules apply to fractional exponents. Example 5 Simplify the following expressions. a) a 1 2 · a 1 3 b) ( a 1 3 )2 c) a 5 2 a 1 2 www.ck12.org 336 d) ( x2 y3 ) 1 3 Solution a) Apply the product rule: a 1 2 · a 1 3 = a 1 2 + 1 3 = a 5 6 b) Apply the power rule: ( a 1 3 )2 = a 2 3 c) Apply the quotient rule: a 5 2 a 1 2 = a 5 2−1 2 = a 4 2 = a2 d) Apply the power rule for quotients: ( x2 y3 ) 1 3 = x 2 3 y Evaluate Exponential Expressions When evaluating expressions we must keep in mind the order of operations. You must remember PEM-DAS: 1. Evaluate inside the Parentheses. 2. Evaluate Exponents. 3. Perform Multiplication and Division operations from left to right. 4. Perform Addition and Subtraction operations from left to right. Example 6 Evaluate the following expressions. a) 50 b) ( 2 3 )3 c) 16 1 2 d) 8−1 3 Solution a) 50 = 1 A number raised to the power 0 is always 1. b) ( 2 3 )3 = 23 33 = 8 27 c) 16 1 2 = √ 16 = 4 Remember that an exponent of 1 2 means taking the square root. d) 8−1 3 = 1 8 1 3 = 1 3 √ 8 = 1 2 Remember that an exponent of 1 3 means taking the cube root. Example 7 Evaluate the following expressions. a) 3 · 52 −10 · 5 + 1 b) 2·42−3·52 32−22 c) ( 33 22 )−2 · 3 4 Solution a) Evaluate the exponent: 3 · 52 −10 · 5 + 1 = 3 · 25 −10 · 5 + 1 Perform multiplications from left to right: 3 · 25 −10 · 5 + 1 = 75 −50 + 1 Perform additions and subtractions from left to right: 75 −50 + 1 = 26 337 www.ck12.org b) Treat the expressions in the numerator and denominator of the fraction like they are in parentheses: (2·42−3·52) (32−22) = (2·16−3·25) (9−4) = (32−75) 5 = −43 5 c) ( 33 22 )−2 · 3 4 = ( 22 33 )2 · 3 4 = 24 36 · 3 4 = 24 36 · 3 22 = 22 35 = 4 243 Example 8 Evaluate the following expressions for x = 2, y = −1, z = 3. a) 2x2 −3y3 + 4z b) (x2 −y2)2 c) ( 3x2y5 4z )−2 Solution a) 2x2 −3y3 + 4z = 2 · 22 −3 · (−1)3 + 4 · 3 = 2 · 4 −3 · (−1) + 4 · 3 = 8 + 3 + 12 = 23 b) (x2 −y2)2 = (22 −(−1)2)2 = (4 −1)2 = 32 = 9 c) ( 3x2y5 4z )−2 = ( 3·22·(−1)5 4·3 )−2 = ( 3·4·(−1) 12 )−2 = ( −12 12 )−2 = ( −1 1 )−2 = ( 1 −1 )2 = (−1)2 = 1 Review Questions Simplify the following expressions in such a way that there aren’t any negative exponents in the answer. 1. x−1y2 2. x−4 3. x−3 x−7 4. x−3y−5 z−7 5. (x 1 2 y −2 3 )(x2y 1 3 ) 6. ( a b )−2 7. (3a−2b2c3)3 8. x−3 · x3 Simplify the following expressions in such a way that there aren’t any fractions in the answer. 9. a−3(a5) a−6 10. 5x6y2 x8y 11. (4ab6)3 (ab)5 12. ( 3x y 1 3 )3 13. 3x2y 3 2 xy 1 2 14. (3x3)(4x4) (2y)2 15. a−2b−3 c−1 16. x 1 2 y 5 2 x 3 2 y 3 2 Evaluate the following expressions to a single number. www.ck12.org 338 17. 3−2 18. (6.2)0 19. 8−4 · 86 20. ( 16 1 2 )3 21. x2 · 4x3 · y4 · 4y2, if x = 2 and y = −1 22. a4(b2)3 + 2ab, if a = −2 and b = 1 23. 5x2 −2y3 + 3z, if x = 3, y = 2, and z = 4 24. ( a2 b3 )−2, if a = 5 and b = 3 25. ( x−2 y4 ) 1 2 , if x = −3 and y = 2 8.4 Scientiﬁc Notation Learning Objectives • Write numbers in scientiﬁc notation. • Evaluate expressions in scientiﬁc notation. • Evaluate expressions in scientiﬁc notation using a graphing calculator. Introduction Consider the number six hundred and forty three thousand, two hundred and ninety seven. We write it as 643,297 and each digit’s position has a “value” assigned to it. You may have seen a table like this before: hundred-thousands ten-thousands thousands hundreds tens units 6 4 3 2 9 7 We’ve seen that when we write an exponent above a number, it means that we have to multiply a certain number of copies of that number together. We’ve also seen that a zero exponent always gives us 1, and negative exponents give us fractional answers. Look carefully at the table above. Do you notice that all the column headings are powers of ten? Here they are listed: 100, 000 = 105 10, 000 = 104 1, 000 = 103 100 = 102 10 = 101 Even the “units” column is really just a power of ten. Unit means 1, and 1 is 100. If we divide 643,297 by 100,000 we get 6.43297; if we multiply 6.43297 by 100,000 we get 643, 297. But we have just seen that 100,000 is the same as 105, so if we multiply 6.43297 by 105 we should also get 643,297. In other words, 643, 297 = 6.43297 × 105 339 www.ck12.org Writing Numbers in Scientiﬁc Notation In scientiﬁc notation, numbers are always written in the form a × 10b , where b is an integer and a is between 1 and 10 (that is, it has exactly 1 nonzero digit before the decimal). This notation is especially useful for numbers that are either very small or very large. Here’s a set of examples: 1.07 × 104 = 10, 700 1.07 × 103 = 1, 070 1.07 × 102 = 107 1.07 × 101 = 10.7 1.07 × 100 = 1.07 1.07 × 10−1 = 0.107 1.07 × 10−2 = 0.0107 1.07 × 10−3 = 0.00107 1.07 × 10−4 = 0.000107 Look at the ﬁrst example and notice where the decimal point is in both expressions. So the exponent on the ten acts to move the decimal point over to the right. An exponent of 4 moves it 4 places and an exponent of 3 would move it 3 places. This makes sense because each time you multiply by 10, you move the decimal point one place to the right. 1.07 times 10 is 10.7, times 10 again is 107.0, and so on. Similarly, if you look at the later examples in the table, you can see that a negative exponent on the 10 means the decimal point moves that many places to the left. This is because multiplying by 10−1 is the same as multiplying by 1 10, which is like dividing by 10. So instead of moving the decimal point one place to the right for every multiple of 10, we move it one place to the left for every multiple of 1 10. That’s how to convert numbers from scientiﬁc notation to standard form. When we’re converting numbers to scientiﬁc notation, however, we have to apply the whole process backwards. First we move the decimal point until it’s immediately after the ﬁrst nonzero digit; then we count how many places we moved it. If we moved it to the left, the exponent on the 10 is positive; if we moved it to the right, it’s negative. So, for example, to write 0.000032 in scientiﬁc notation, we’d ﬁrst move the decimal ﬁve places to the right to get 3.2; then, since we moved it right, the exponent on the 10 should be negative ﬁve, so the number in scientiﬁc notation is 3.2 × 10−5. www.ck12.org 340 You can double-check whether you’ve got the right direction by comparing the number in scientiﬁc notation with the number in standard form, and thinking “Does this represent a big number or a small number?” A positive exponent on the 10 represents a number bigger than 10 and a negative exponent represents a number smaller than 10, and you can easily tell if the number in standard form is bigger or smaller than 10 just by looking at it. For more practice, try the online tool at activities/interactivities/sciNotation.swf. Click the arrow buttons to move the decimal point un-til the number in the middle is written in proper scientiﬁc notation, and see how the exponent changes as you move the decimal point. Example 1 Write the following numbers in scientiﬁc notation. a) 63 b) 9,654 c) 653,937,000 d) 0.003 e) 0.000056 f) 0.00005007 Solution a) 63 = 6.3 × 10 = 6.3 × 101 b) 9, 654 = 9.654 × 1, 000 = 9.654 × 103 c) 653, 937, 000 = 6.53937000 × 100, 000, 000 = 6.53937 × 108 d) 0.003 = 3 × 1 1000 = 3 × 10−3 e) 0.000056 = 5.6 × 1 100,000 = 5.6 × 10−5 f) 0.00005007 = 5.007 × 1 100,000 = 5.007 × 10−5 Evaluating Expressions in Scientiﬁc Notation When we are faced with products and quotients involving scientiﬁc notation, we need to remember the rules for exponents that we learned earlier. It’s relatively straightforward to work with scientiﬁc notation problems if you remember to combine all the powers of 10 together. The following examples illustrate this. Example 2 Evaluate the following expressions and write your answer in scientiﬁc notation. a) (3.2 × 106) · (8.7 × 1011) b) (5.2 × 10−4) · (3.8 × 10−19) c) (1.7 × 106) · (2.7 × 10−11) Solution The key to evaluating expressions involving scientiﬁc notation is to group the powers of 10 together and deal with them separately. a) (3.2 × 106)(8.7 × 1011) = 3.2 × 8.7 \| {z } 27.84 × 106 × 1011 \| {z } 1017 = 27.84 × 1017. But 27.84 × 1017 isn’t in proper scientiﬁc 341 www.ck12.org notation, because it has more than one digit before the decimal point. We need to move the decimal point one more place to the left and add 1 to the exponent, which gives us 2.784 × 1018. b) (5.2 × 10−4)(3.8 × 10−19) = 5.2 × 3.8 \| {z } 19.76 × 10−4 × 10−19 \| {z } 10−23 = 19.76 × 10−23 = 1.976 × 10−22 c) (1.7 × 106)(2.7 × 10−11) = 1.7 × 2.7 \| {z } 4.59 × 106 × 10−11 \| {z } 10−5 = 4.59 × 10−5 When we use scientiﬁc notation in the real world, we often round oﬀour calculations. Since we’re often dealing with very big or very small numbers, it can be easier to round oﬀso that we don’t have to keep track of as many digits—and scientiﬁc notation helps us with that by saving us from writing out all the extra zeros. For example, if we round oﬀ4,227, 457,903 to 4,200,000,000, we can then write it in scientiﬁc notation as simply 4.2 × 109. When rounding, we often talk of signiﬁcant ﬁgures or signiﬁcant digits. Signiﬁcant ﬁgures include • all nonzero digits • all zeros that come before a nonzero digit and after either a decimal point or another nonzero digit For example, the number 4000 has one signiﬁcant digit; the zeros don’t count because there’s no nonzero digit after them. But the number 4000.5 has ﬁve signiﬁcant digits: the 4, the 5, and all the zeros in between. And the number 0.003 has three signiﬁcant digits: the 3 and the two zeros that come between the 3 and the decimal point. Example 3 Evaluate the following expressions. Round to 3 signiﬁcant ﬁgures and write your answer in scientiﬁc notation. a) (3.2 × 106) ÷ (8.7 × 1011) b) (5.2 × 10−4) ÷ (3.8 × 10−19) c) (1.7 × 106) ÷ (2.7 × 10−11) Solution It’s easier if we convert to fractions and THEN separate out the powers of 10. a) (3.2 × 106) ÷ (8.7 × 1011) = 3.2 × 106 8.7 × 1011 −separate out the powers o f 10 : = 3.2 8.7 × 106 1011 −evaluate each fraction (round to 3 s. f.) : = 0.368 × 10(6−11) = 0.368 × 10−5 −remember how to write scienti fic notation! = 3.68 × 10−6 b) (5.2 × 10−4) ÷ (3.8 × 10−19) = 5.2 × 10−4 3.8 × 10−19 −separate the powers o f 10 : = 5.2 3.8 × 10−4 10−19 −evaluate each fraction (round to 3 s.f.) = 1.37 × 10((−4)−(−19)) = 1.37 × 1015 www.ck12.org 342 c) (1.7 × 106) ÷ (2.7 × 10−11) = 1.7 × 106 2.7 × 10−11 −next we separate the powers o f 10 : = 1.7 2.7 × 106 10−11 −evaluate each fraction (round to 3 s.f.) = 0.630 × 10(6−(−11)) = 0.630 × 1017 = 6.30 × 1016 Note that we have to leave in the ﬁnal zero to indicate that the result has been rounded. Evaluate Expressions in Scientiﬁc Notation Using a Graphing Calculator All scientiﬁc and graphing calculators can use scientiﬁc notation, and it’s very useful to know how. To insert a number in scientiﬁc notation, use the [EE] button. This is [2nd] [,] on some TI models. For example, to enter 2.6 × 105, enter 2.6 [EE] 5. When you hit [ENTER] the calculator displays 2.6E5 if it’s set in Scientiﬁc mode, or 260000 if it’s set in Normal mode. (To change the mode, press the ‘Mode’ key.) Example 4 Evaluate (2.3 × 106) × (4.9 × 10−10) using a graphing calculator. Solution Enter 2.3 [EE] 6 × 4.9 [EE] - 10 and press [ENTER]. The calculator displays 6.296296296E16 whether it’s in Normal mode or Scientiﬁc mode. That’s because the number is so big that even in Normal mode it won’t ﬁt on the screen. The answer displayed instead isn’t the precisely correct answer; it’s rounded oﬀto 10 signiﬁcant ﬁgures. Since it’s a repeating decimal, though, we can write it more eﬀiciently and more precisely as 6.296 × 1016. Example 5 Evaluate (4.5 × 1014)3 using a graphing calculator. 343 www.ck12.org Solution Enter (4.5 [EE] 14)∧3 and press [ENTER]. The calculator displays 9.1125E43. The answer is 9.1125 × 1043. Solve Real-World Problems Using Scientiﬁc Notation Example 6 The mass of a single lithium atom is approximately one percent of one millionth of one billionth of one billionth of one kilogram. Express this mass in scientiﬁc notation. Solution We know that a percent is 1 100, and so our calculation for the mass (in kg) is: 1 100 × 1 1, 000, 000 × 1 1, 000, 000, 000 × 1 1, 000, 000, 000 = 10−2 × 10−6 × 10−9 × 10−9 Next we use the product of powers rule we learned earlier: 10−2 × 10−6 × 10−9 × 10−9 = 10((−2)+(−6)+(−9)+(−9)) = 10−26 kg. The mass of one lithium atom is approximately 1 × 10−26 kg. Example 7 You could ﬁt about 3 million E. coli bacteria on the head of a pin. If the size of the pin head in question is 1.2 × 10−5 m2, calculate the area taken up by one E. coli bacterium. Express your answer in scientiﬁc notation Solution Since we need our answer in scientiﬁc notation, it makes sense to convert 3 million to that format ﬁrst: 3, 000, 000 = 3 × 106 Next we need an expression involving our unknown, the area taken up by one bacterium. Call this A. 3 × 106 · A = 1.2 × 10−5 −since 3 million of them make up the area o f the pin −head Isolate A: A = 1 3 × 106 · 1.2 × 10−5 −rearranging the terms gives : A = 1.2 3 · 1 106 × 10−5 −then using the de finition o f a negative exponent : A = 1.2 3 · 10−6 × 10−5 −evaluate & combine exponents using the product rule : A = 0.4 × 10−11 −but we can’t leave our answer like this, so . . . www.ck12.org 344 The area of one bacterium is 4.0 × 10−12 m2. (Notice that we had to move the decimal point over one place to the right, subtracting 1 from the exponent on the 10.) Review Questions Write the numerical value of the following. 1. 3.102 × 102 2. 7.4 × 104 3. 1.75 × 10−3 4. 2.9 × 10−5 5. 9.99 × 10−9 Write the following numbers in scientiﬁc notation. 6. 120,000 7. 1,765,244 8. 12 9. 0.00281 10. 0.000000027 How many signiﬁcant digits are in each of the following? 11. 38553000 12. 2754000.23 13. 0.0000222 14. 0.0002000079 Round each of the following to two signiﬁcant digits. 15. 3.0132 16. 82.9913 Perform the following operations and write your answer in scientiﬁc notation. 17. (3.5 × 104) · (2.2 × 107) 18. 2.1×109 3×102 19. (3.1 × 10−3) · (1.2 × 10−5) 20. 7.4×10−5 3.7×10−2 21. 12, 000, 000 × 400, 000 22. 3, 000, 000 × 0.00000000022 23. 17,000 680,000,000 24. 25,000,000 0.000000005 25. 0.0000000000042 0.00014 26. The moon is approximately a sphere with radius r = 1.08 × 103 miles. Use the formula Surface Area = 4πr2 to determine the surface area of the moon, in square miles. Express your answer in scientiﬁc notation, rounded to two signiﬁcant ﬁgures. 345 www.ck12.org 27. The charge on one electron is approximately 1.60×1019 coulombs. One Faraday is equal to the total charge on 6.02 × 1023 electrons. What, in coulombs, is the charge on one Faraday? 28. Proxima Centauri, the next closest star to our Sun, is approximately 2.5 × 1013 miles away. If light from Proxima Centauri takes 3.7 × 104 hours to reach us from there, calculate the speed of light in miles per hour. Express your answer in scientiﬁc notation, rounded to 2 signiﬁcant ﬁgures. 8.5 Geometric Sequences Learning Objectives • Identify a geometric sequence • Graph a geometric sequence. • Solve real-world problems involving geometric sequences. Introduction Consider the following question: Which would you prefer, being given one million dollars, or one penny the ﬁrst day, double that penny the next day, and then double the previous day’s pennies and so on for a month? At ﬁrst glance it’s easy to say ”Give me the million!” But why don’t we do a few calculations to see how the other choice stacks up? You start with a penny the ﬁrst day and keep doubling each day. Doubling means that we keep multiplying by 2 each day for one month (30 days). On the ﬁrst day, you get 1 penny, or 20 pennies. On the second day, you get 2 pennies, or 21 pennies. On the third day, you get 4 pennies, or 22 pennies. Do you see the pattern yet? On the fourth day, you get 8 pennies, or 23 pennies. Each day, the exponent is one less than the number of that day. So on the thirtieth day, you get 229 pennies, which is 536,870,912 pennies, or $5,368,709.12. That’s a lot more than a million dollars, even just counting the amount you get on that one day! This problem is an example of a geometric sequence. In this section, we’ll ﬁnd out what a geometric sequence is and how to solve problems involving geometric sequences. Identify a Geometric Sequence A geometric sequence is a sequence of numbers in which each number in the sequence is found by multiplying the previous number by a ﬁxed amount called the common ratio. In other words, the ratio between any term and the term before it is always the same. In the previous example the common ratio was 2, as the number of pennies doubled each day. The common ratio, r, in any geometric sequence can be found by dividing any term by the preceding term. Here are some examples of geometric sequences and their common ratios. www.ck12.org 346 4, 16, 64, 256, . . . r = 4 (divide 16 by 4 to get 4) 15, 30, 60, 120, . . . r = 2 (divide 30 by 15 to get 2) 11, 11 2 , 11 4 , 11 8 , 11 16, . . . r = 1 2 ( divide 11 2 by 11 to get 1 2 ) 25, −5, 1, −1 5, 1 25, . . . r = −1 5 ( divide 1 by -5 to get −1 5 ) If we know the common ratio r, we can ﬁnd the next term in the sequence just by multiplying the last term by r. Also, if there are any terms missing in the sequence, we can ﬁnd them by multiplying the term before each missing term by the common ratio. Example 1 Fill is the missing terms in each geometric sequence. a) 1, , 25, 125, ___ b) 20, , 5, ___, 1.25 Solution a) First we can ﬁnd the common ratio by dividing 125 by 25 to obtain r = 5. To ﬁnd the ﬁrst missing term, we multiply 1 by the common ratio: 1 · 5 = 5 To ﬁnd the second missing term, we multiply 125 by the common ratio: 125 · 5 = 625 Sequence (a) becomes: 1, 5, 25, 125, 625,... b) We need to ﬁnd the common ratio ﬁrst, but how do we do that when we have no terms next to each other that we can divide? Well, we know that to get from 20 to 5 in the sequence we must multiply 20 by the common ratio twice: once to get to the second term in the sequence, and again to get to ﬁve. So we can say 20 · r · r = 5, or 20 · r2 = 5. Dividing both sides by 20, we get r2 = 5 20 = 1 4, or r = 1 2 (because 1 2 · 1 2 = 1 4). To get the ﬁrst missing term, we multiply 20 by 1 2 and get 10. To get the second missing term, we multiply 5 by 1 2 and get 2.5. Sequence (b) becomes: 20, 10, 5, 2.5, 1.25,... You can see that if we keep multiplying by the common ratio, we can ﬁnd any term in the sequence that we want—the tenth term, the ﬁftieth term, the thousandth term.... However, it would be awfully tedious to keep multiplying over and over again in order to ﬁnd a term that is a long way from the start. What could we do instead of just multiplying repeatedly? Let’s look at a geometric sequence that starts with the number 7 and has common ratio of 2. The 1st term is: 7 or 7 · 20 We obtain the 2nd term by multiplying by 2 : 7 · 2 or 7 · 21 We obtain the 3rd term by multiplying by 2 again: 7 · 2 · 2 or 7 · 22 We obtain the 4th term by multiplying by 2 again: 7 · 2 · 2 · 2 or 7 · 23 We obtain the 5th term by multiplying by 2 again: 7 · 2 · 2 · 2 · 2 or 7 · 24 The nth term would be: 7 · 2n−1 347 www.ck12.org The nth term is 7·2n−1 because the 7 is multiplied by 2 once for the 2nd term, twice for the third term, and so on—for each term, one less time than the term’s place in the sequence. In general, we write a geometric sequence with n terms like this: a, ar, ar2, ar3, . . . , arn−1 The formula for ﬁnding a speciﬁc term in a geometric sequence is: nth term in a geometric sequence: an = a1rn−1 (a1 = ﬁrst term, r = common ratio) Example 2 For each of these geometric sequences, ﬁnd the eighth term in the sequence. a) 1, 2, 4,... b) 16, -8, 4, -2, 1,... Solution a) First we need to ﬁnd the common ratio: r = 2 1 = 2. The eighth term is given by the formula a8 = a1r7 = 1 · 27 = 128 In other words, to get the eighth term we start with the ﬁrst term, which is 1, and then multiply by 2 seven times. b) The common ratio is r = −8 16 = −1 2 The eighth term in the sequence is a8 = a1r7 = 16 · ( −1 2 )7 = 16 · (−1)7 27 = 16 · −1 27 = −16 128 = −1 8 Let’s take another look at the terms in that second sequence. Notice that they alternate positive, neg-ative, positive, negative all the way down the list. When you see this pattern, you know the common ratio is negative; multiplying by a negative number each time means that the sign of each term is opposite the sign of the previous term. Solve Real-World Problems Involving Geometric Sequences Let’s solve two application problems involving geometric sequences. Example 3 A courtier presented the Indian king with a beautiful, hand-made chessboard. The king asked what he would like in return for his gift and the courtier surprised the king by asking for one grain of rice on the ﬁrst square, two grains on the second, four grains on the third, etc. The king readily agreed and asked for the rice to be brought. (From Meadows et al. 1972, via Porritt 2005) How many grains of rice does the king have to put on the last square? Solution A chessboard is an 8 × 8 square grid, so it contains a total of 64 squares. The courtier asked for one grain of rice on the ﬁrst square, 2 grains of rice on the second square, 4 grains of rice on the third square and so on. We can write this as a geometric sequence: 1, 2, 4,... The numbers double each time, so the common ratio is r = 2. The problem asks how many grains of rice the king needs to put on the last square, so we need to ﬁnd the 64th term in the sequence. Let’s use our formula: www.ck12.org 348 an = a1rn−1, where an is the nth term, a1 is the ﬁrst term and r is the common ratio. a64 = 1 · 263 = 9, 223, 372, 036, 854, 775, 808 grains of rice. The problem we just solved has real applications in business and technology. In technology strategy, the Second Half of the Chessboard is a phrase, coined by a man named Ray Kurzweil, in reference to the point where an exponentially growing factor begins to have a signiﬁcant economic impact on an organization’s overall business strategy. The total number of grains of rice on the ﬁrst half of the chessboard is 1+2+4+8+16+32+64+128+ 256+512+1024 . . .+2, 147, 483, 648, for a total of exactly 4,294,967,295 grains of rice, or about 100,000 kg of rice (the mass of one grain of rice being roughly 25 mg). This total amount is about 1 1,000,000th of total rice production in India in the year 2005 and is an amount the king could surely have aﬀorded. The total number of grains of rice on the second half of the chessboard is 232 + 233 + 234 . . . + 263, for a total of 18, 446, 744, 069, 414, 584, 320 grains of rice. This is about 460 billion tons, or 6 times the entire weight of all living matter on Earth. The king didn’t realize what he was agreeing to—perhaps he should have studied algebra! [Wikipedia; GNU-FDL] Example 4 A super-ball has a 75% rebound ratio—that is, when it bounces repeatedly, each bounce is 75% as high as the previous bounce. When you drop it from a height of 20 feet: a) how high does the ball bounce after it strikes the ground for the third time? b) how high does the ball bounce after it strikes the ground for the seventeenth time? Solution We can write a geometric sequence that gives the height of each bounce with the common ratio of r = 3 4: 20, 20 · 3 4, 20 · (3 4 )2 , 20 · (3 4 )3 . . . a) The ball starts at a height of 20 feet; after the ﬁrst bounce it reaches a height of 20 · 3 4 = 15 feet. After the second bounce it reaches a height of 20 · ( 3 4 )2 = 11.25 feet. After the third bounce it reaches a height of 20 · ( 3 4 )3 = 8.44 feet. b) Notice that the height after the ﬁrst bounce corresponds to the second term in the sequence, the height after the second bounce corresponds to the third term in the sequence and so on. This means that the height after the seventeenth bounce corresponds to the 18th term in the sequence. You can ﬁnd the height by using the formula for the 18th term: a18 = 20 · (3 4 )17 = 0.15 feet. Here is a graph that represents this information. (The heights at points other than the top of each bounce are just approximations.) 349 www.ck12.org For more practice ﬁnding the terms in geometric sequences, try the browser game at tcd.ie/~jgilbert/maths_site/applets/sequences_and_series/geometric_sequences.html. Review Questions Determine the ﬁrst ﬁve terms of each geometric sequence. 1. a1 = 2, r = 3 2. a1 = 90, r = −1 3 3. a1 = 6, r = −2 4. a1 = 1, r = 5 5. a1 = 5, r = 5 6. a1 = 25, r = 5 7. What do you notice about the last three sequences? Find the missing terms in each geometric sequence. 8. 3, __ , 48, 192, __ 9. 81, __ , __ , __ , 1 10. 9 4 , , , 2 3 , 11. 2, __ , __ , -54, 162 Find the indicated term of each geometric sequence. 12. a1 = 4, r = 2; ﬁnd a6 13. a1 = −7, r = −3 4; ﬁnd a4 14. a1 = −10, r = −3; ﬁnd a10 15. In a geometric sequence, a3 = 28 and a5 = 112; ﬁnd r and a1. 16. In a geometric sequence, a2 = 28 and a5 = 112; ﬁnd r and a1. 17. As you can see from the previous two questions, the same terms can show up in sequences with diﬀerent ratios. (a) Write a geometric sequence that has 1 and 9 as two of the terms (not necessarily the ﬁrst two). (b) Write a diﬀerent geometric sequence that also has 1 and 9 as two of the terms. (c) Write a geometric sequence that has 6 and 24 as two of the terms. (d) Write a diﬀerent geometric sequence that also has 6 and 24 as two of the terms. (e) What is the common ratio of the sequence whose ﬁrst three terms are 2, 6, 18? (f) What is the common ratio of the sequence whose ﬁrst three terms are 18, 6, 2? www.ck12.org 350 (g) What is the relationship between those ratios? 18. Anne goes bungee jumping oﬀa bridge above water. On the initial jump, the bungee cord stretches by 120 feet. On the next bounce the stretch is 60% of the original jump and each additional bounce the rope stretches by 60% of the previous stretch. (a) What will the rope stretch be on the third bounce? (b) What will be the rope stretch be on the 12th bounce? 8.6 Exponential Functions Learning Objectives • Graph an exponential function. • Compare graphs of exponential functions. • Analyze the properties of exponential functions. Introduction A colony of bacteria has a population of three thousand at noon on Monday. During the next week, the colony’s population doubles every day. What is the population of the bacteria colony just before midnight on Saturday? At ﬁrst glance, this seems like a problem you could solve using a geometric sequence. And you could, if the bacteria population doubled all at once every day; since it doubled every day for ﬁve days, the ﬁnal population would be 3000 times 25. But bacteria don’t reproduce all at once; their population grows slowly over the course of an entire day. So how do we ﬁgure out the population after ﬁve and a half days? Exponential Functions Exponential functions are a lot like geometrical sequences. The main diﬀerence between them is that a geometric sequence is discrete while an exponential function is continuous. Discrete means that the sequence has values only at distinct points (the 1st term, 2nd term, etc.) Continuous means that the function has values for all possible values of x. The integers are included, but also all the numbers in between. The problem with the bacteria is an example of a continuous function. Here’s an example of a discrete function: An ant walks past several stacks of Lego blocks. There is one block in the ﬁrst stack, 3 blocks in the 2nd stack and 9 blocks in the 3rd stack. In fact, in each successive stack there are triple the number of blocks than in the previous stack. In this example, each stack has a distinct number of blocks and the next stack is made by adding a certain number of whole pieces all at once. More importantly, however, there are no values of the sequence between the stacks. You can’t ask how high the stack is between the 2nd and 3rd stack, as no stack exists at that position! As a result of this diﬀerence, we use a geometric series to describe quantities that have values at discrete points, and we use exponential functions to describe quantities that have values that change continuously. 351 www.ck12.org When we graph an exponential function, we draw the graph with a solid curve to show that the function has values at any time during the day. On the other hand, when we graph a geometric sequence, we draw discrete points to signify that the sequence only has value at those points but not in between. Here are graphs for the two examples above: The formula for an exponential function is similar to the formula for ﬁnding the terms in a geometric sequence. An exponential function takes the form y = A · bx where A is the starting amount and b is the amount by which the total is multiplied every time. For example, the bacteria problem above would have the equation y = 3000 · 2x. Compare Graphs of Exponential Functions Let’s graph a few exponential functions and see what happens as we change the constants in the formula. The basic shape of the exponential function should stay the same—but it may become steeper or shallower depending on the constants we are using. First, let’s see what happens when we change the value of A. Example 1 Compare the graphs of y = 2x and y = 3 · 2x. Solution Let’s make a table of values for both functions. Table 8.1: x y = 2x y = 3 · 2x -3 1 8 y = 3 · 2−3 = 3 · 1 23 = 3 8 -2 1 4 y = 3 · 2−2 = 3 · 1 22 = 3 4 -1 1 2 y = 3 · 2−1 = 3 · 1 21 = 3 2 0 1 y = 3 · 20 = 3 1 2 y = 3 · 21 = 6 2 4 y = 3 · 22 = 3 · 4 = 12 3 8 y = 3 · 23 = 3 · 8 = 24 www.ck12.org 352 Now let’s use this table to graph the functions. We can see that the function y = 3 · 2x is bigger than the function y = 2x. In both functions, the value of y doubles every time x increases by one. However, y = 3 · 2x “starts” with a value of 3, while y = 2x “starts” with a value of 1, so it makes sense that y = 3 · 2x would be bigger as its values of y keep getting doubled. Similarly, if the starting value of A is smaller, the values of the entire function will be smaller. Example 2 Compare the graphs of y = 2x and y = 1 3 · 2x. Solution Let’s make a table of values for both functions. Table 8.2: x y = 2x y = 1 3 · 2x -3 1 8 y = 1 3 · 2−3 = 1 3 · 1 23 = 1 24 -2 1 4 y = 1 3 · 2−2 = 1 3 · 1 22 = 1 12 -1 1 2 y = 1 3 · 2−1 = 1 3 · 1 21 = 1 6 0 1 y = 1 3 · 20 = 1 3 1 2 y = 1 3 · 21 = 2 3 2 4 y = 1 3 · 22 = 1 3 · 4 = 4 3 3 8 y = 1 3 · 23 = 1 3 · 8 = 8 3 Now let’s use this table to graph the functions. 353 www.ck12.org As we expected, the exponential function y = 1 3 · 2x is smaller than the exponential function y = 2x. So what happens if the starting value of A is negative? Let’s ﬁnd out. Example 3 Graph the exponential function y = −5 · 2x. Solution Let’s make a table of values: Table 8.3: x y = −5 · 2x -2 −5 4 -1 −5 2 0 -5 1 -10 2 -20 3 -40 Now let’s graph the function: This result shouldn’t be unexpected. Since the starting value is negative and keeps doubling over time, it makes sense that the value of y gets farther from zero, but in a negative direction. The graph is basically www.ck12.org 354 just like the graph of y = 5 · 2x, only mirror-reversed about the x−axis. Now, let’s compare exponential functions whose bases (b) are diﬀerent. Example 4 Graph the following exponential functions on the same graph: y = 2x, y = 3x, y = 5x, y = 10x. Solution First we’ll make a table of values for all four functions. Table 8.4: x y = 2x y = 3x y = 5x y = 10x -2 1 4 1 9 1 25 1 100 -1 1 2 1 3 1 5 1 10 0 1 1 1 1 1 2 3 5 10 2 4 9 25 100 3 8 27 125 1000 Now let’s graph the functions. Notice that for x = 0, all four functions equal 1. They all “start out” at the same point, but the ones with higher values for b grow faster when x is positive—and also shrink faster when x is negative. Finally, let’s explore what happens for values of b that are less than 1. Example 5 Graph the exponential function y = 5 · ( 1 2 )x. Solution Let’s start by making a table of values. (Remember that a fraction to a negative power is equivalent to its reciprocal to the same positive power.) Table 8.5: x y = 5 · ( 1 2 )x -3 y = 5 · ( 1 2 )−3 = 5 · 23 = 40 -2 y = 5 · ( 1 2 )−2 = 5 · 22 = 20 355 www.ck12.org Table 8.5: (continued) x y = 5 · ( 1 2 )x -1 y = 5 · ( 1 2 )−1 = 5 · 21 = 10 0 y = 5 · ( 1 2 )0 = 5 · 1 = 5 1 y = 5 · ( 1 2 )1 = 5 2 2 y = 5 · ( 1 2 )2 = 5 4 Now let’s graph the function. This graph looks very diﬀerent than the graphs from the previous example! What’s going on here? When we raise a number greater than 1 to the power of x, it gets bigger as x gets bigger. But when we raise a number smaller than 1 to the power of x, it gets smaller as x gets bigger—as you can see from the table of values above. This makes sense because multiplying any number by a quantity less than 1 always makes it smaller. So, when the base b of an exponential function is between 0 and 1, the graph is like an ordinary exponential graph, only decreasing instead of increasing. Graphs like this represent exponential decay instead of exponential growth. Exponential decay functions are used to describe quantities that decrease over a period of time. When b can be written as a fraction, we can use the Property of Negative Exponents to write the function in a diﬀerent form. For instance, y = 5 · ( 1 2 )x is equivalent to 5 · 2−x. These two forms are both commonly used, so it’s important to know that they are equivalent. Example 6 Graph the exponential function y = 8 · 3−x. Solution Here is our table of values and the graph of the function. www.ck12.org 356 Table 8.6: x y = 8 · 3−x -3 y = 8 · 3−(−3) = 8 · 33 = 216 -2 y = 8 · 3−(−2) = 8 · 32 = 72 -1 y = 8 · 3−(−1) = 8 · 31 = 24 0 y = 8 · 30 = 8 1 y = 8 · 3−1 = 8 3 2 y = 8 · 3−2 = 8 9 Example 7 Graph the functions y = 4x and y = 4−x on the same coordinate axes. Solution Here is the table of values for the two functions. Looking at the values in the table, we can see that the two functions are “backwards” of each other, in the sense that the values for the two functions are reciprocals. Table 8.7: x y = 4x y = 4−x -3 y = 4−3 = 1 64 y = 4−(−3) = 64 -2 y = 4−2 = 1 16 y = 4−(−2) = 16 -1 y = 4−1 = 1 4 y = 4−(−1) = 4 0 y = 40 = 1 y = 40 = 1 1 y = 41 = 4 y = 4−1 = 1 4 2 y = 42 = 16 y = 4−2 = 1 16 3 y = 43 = 1 64 y = 4−3 = 1 64 Here is the graph of the two functions. Notice that the two functions are mirror images of each other if the mirror is placed vertically on the y−axis. 357 www.ck12.org In the next lesson, you’ll see how exponential growth and decay functions can be used to represent situations in the real world. Review Questions Graph the following exponential functions by making a table of values. 1. y = 3x 2. y = 5 · 3x 3. y = 40 · 4x 4. y = 3 · 10x Graph the following exponential functions. 5. y = ( 1 5 )x 6. y = 4 · ( 2 3 )x 7. y = 3−x 8. y = 3 4 · 6−x 9. Which two of the eight graphs above are mirror images of each other? 10. What function would produce a graph that is the mirror image of the one in problem 4? 11. How else might you write the exponential function in problem 5? 12. How else might you write the function in problem 6? Solve the following problems. 13. A chain letter is sent out to 10 people telling everyone to make 10 copies of the letter and send each one to a new person. (a) Assume that everyone who receives the letter sends it to ten new people and that each cycle takes a week. How many people receive the letter on the sixth week? (b) What if everyone only sends the letter to 9 new people? How many people will then get letters on the sixth week? 14. Nadia received $200 for her 10th birthday. If she saves it in a bank account with 7.5% interest compounded yearly, how much money will she have in the bank by her 21st birthday? www.ck12.org 358 8.7 Applications of Exponential Functions Learning Objectives • Apply the problem-solving plan to problems involving exponential functions. • Solve real-world problems involving exponential growth. • Solve real-world problems involving exponential decay. Introduction For her eighth birthday, Shelley’s grandmother gave her a full bag of candy. Shelley counted her candy and found out that there were 160 pieces in the bag. As you might suspect, Shelley loves candy, so she ate half the candy on the ﬁrst day. Then her mother told her that if she eats it at that rate, the candy will only last one more day—so Shelley devised a clever plan. She will always eat half of the candy that is left in the bag each day. She thinks that this way she can eat candy every day and never run out. How much candy does Shelley have at the end of the week? Will the candy really last forever? Let’s make a table of values for this problem. Day 0 1 2 3 4 5 6 7 # of candies 160 80 40 20 10 5 2.5 1.25 You can see that if Shelley eats half the candies each day, then by the end of the week she only has 1.25 candies left in her bag. Let’s write an equation for this exponential function. Using the formula y = A · bx, we can see that A is 160 (the number of candies she starts out with and b is 1 2, so our equation is y = 160 · ( 1 2 )x.) Now let’s graph this function. The resulting graph is shown below. So, will Shelley’s candy last forever? We saw that by the end of the week she has 1.25 candies left, so there doesn’t seem to be much hope for that. But if you look at the graph, you’ll see that the graph never really gets to zero. Theoretically there will always be some candy left, but Shelley will be eating very tiny fractions of a candy every day after the ﬁrst week! This is a fundamental feature of an exponential decay function. Its values get smaller and smaller but never quite reach zero. In mathematics, we say that the function has an asymptote at y = 0; in other 359 www.ck12.org words, it gets closer and closer to the line y = 0 but never quite meets it. Problem-Solving Strategies Remember our problem-solving plan from earlier? 1. Understand the problem. 2. Devise a plan – Translate. 3. Carry out the plan – Solve. 4. Look – Check and Interpret. We can use this plan to solve application problems involving exponential functions. Compound interest, loudness of sound, population increase, population decrease or radioactive decay are all applications of exponential functions. In these problems, we’ll use the methods of constructing a table and identifying a pattern to help us devise a plan for solving the problems. Example 1 Suppose $4000 is invested at 6% interest compounded annually. How much money will there be in the bank at the end of 5 years? At the end of 20 years? Solution Step 1: Read the problem and summarize the information. $4000 is invested at 6% interest compounded annually; we want to know how much money we have in ﬁve years. Assign variables: Let x = time in years Let y = amount of money in investment account Step 2: Look for a pattern. We start with $4000 and each year we add 6% interest to the amount in the bank. Start: $4000 1st year: Interest = 4000 × (0.06) = $240 This is added to the previous amount: $4000 + $4000 × (0.06) = $4000(1 + 0.06) = $4000(1.06) = $4240 2nd year Previous amount + interest on the previous amount = $4240(1 + 0.06) = $4240(1.06) = $4494.40 The pattern is that each year we multiply the previous amount by the factor of 1.06. Let’s ﬁll in a table of values: Time (years) 0 1 2 3 4 5 Investments amount($) 4000 4240 4494.4 4764.06 5049.90 5352.9 www.ck12.org 360 We see that at the end of ﬁve years we have $5352.90 in the investment account. Step 3: Find a formula. We were able to ﬁnd the amount after 5 years just by following the pattern, but rather than follow that pattern for another 15 years, it’s easier to use it to ﬁnd a general formula. Since the original investment is multiplied by 1.06 each year, we can use exponential notation. Our formula is y = 4000 · (1.06)x, where x is the number of years since the investment. To ﬁnd the amount after 5 years we plug x = 5 into the equation: y = 4000 · (1.06)5 = $5352.90 To ﬁnd the amount after 20 years we plug x = 20 into the equation: y = 4000 · (1.06)20 = $12828.54 Step 4: Check. Looking back over the solution, we see that we obtained the answers to the questions we were asked and the answers make sense. To check our answers, we can plug some low values of x into the formula to see if they match the values in the table: x = 0 : y = 4000 · (1.06)0 = 4000 x = 1 : y = 4000 · (1.06)1 = 4240 x = 2 : y = 4000 · (1.06)2 = 4494.4 The answers match the values we found earlier. The amount of increase gets larger each year, and that makes sense because the interest is 6% of an amount that is larger every year. Example 2 In 2002 the population of schoolchildren in a city was 90,000. This population decreases at a rate of 5% each year. What will be the population of school children in year 2010? Solution Step 1: Read the problem and summarize the information. The population is 90,000; the rate of decrease is 5% each year; we want the population after 8 years. Assign variables: Let x = time since 2002 (in years) Let y = population of school children Step 2: Look for a pattern. Let’s start in 2002, when the population is 90,000. The rate of decrease is 5% each year, so the amount in 2003 is 90,000 minus 5% of 90,000, or 95% of 90,000. In 2003 : Population = 90, 000 × 0.95 In 2004 : Population = 90, 000 × 0.95 × 0.95 The pattern is that for each year we multiply by a factor of 0.95 Let’s ﬁll in a table of values: 361 www.ck12.org Year 2002 2003 2004 2005 2006 2007 Population 90, 000 85, 500 81, 225 77, 164 73, 306 69, 640 Step 3: Find a formula. Since we multiply by 0.95 every year, our exponential formula is y = 90000 · (0.95)x, where x is the number of years since 2002. To ﬁnd the population in 2010 (8 years after 2002), we plug in x = 8: y = 90000 · (0.95)8 = 59, 708 schoolchildren. Step 4: Check. Looking back over the solution, we see that we answered the question we were asked and that it makes sense. The answer makes sense because the numbers decrease each year as we expected. We can check that the formula is correct by plugging in the values of x from the table to see if the values match those given by the formula. Year 2002, x = 0 : Population = y = 90000 · (0.95)0 = 90, 000 Year 2003, x = 1 : Population = y = 90000 · (0.95)1 = 85, 500 Year 2004, x = 2 : Population = y = 90000 · (0.95)2 = 81, 225 Solve Real-World Problems Involving Exponential Growth Now we’ll look at some more real-world problems involving exponential functions. We’ll start with situa-tions involving exponential growth. Example 3 The population of a town is estimated to increase by 15% per year. The population today is 20 thousand. Make a graph of the population function and ﬁnd out what the population will be ten years from now. Solution First, we need to write a function that describes the population of the town. The general form of an exponential function is y = A · bx. Deﬁne y as the population of the town. Deﬁne x as the number of years from now. A is the initial population, so A = 20 (thousand). Finally we must ﬁnd what b is. We are told that the population increases by 15% each year. To calculate percents we have to change them into decimals: 15% is equivalent to 0.15. So each year, the population increases by 15% of A, or 0.15A. To ﬁnd the total population for the following year, we must add the current population to the increase in population. In other words, A + 0.15A = 1.15A. So the population must be multiplied by a factor of 1.15 each year. This means that the base of the exponential is b = 1.15. The formula that describes this problem is y = 20 · (1.15)x. Now let’s make a table of values. www.ck12.org 362 Table 8.8: x y = 20 · (1.15)x -10 4.9 -5 9.9 0 20 5 40.2 10 80.9 Now we can graph the function. Notice that we used negative values of x in our table of values. Does it make sense to think of negative time? Yes; negative time can represent time in the past. For example, x = −5 in this problem represents the population from ﬁve years ago. The question asked in the problem was: what will be the population of the town ten years from now? To ﬁnd that number, we plug x = 10 into the equation we found: y = 20 · (1.15)10 = 80, 911. The town will have 80,911 people ten years from now. Example 4 Peter earned $1500 last summer. If he deposited the money in a bank account that earns 5% interest compounded yearly, how much money will he have after ﬁve years? Solution This problem deals with interest which is compounded yearly. This means that each year the interest is calculated on the amount of money you have in the bank. That interest is added to the original amount and next year the interest is calculated on this new amount, so you get paid interest on the interest. Let’s write a function that describes the amount of money in the bank. The general form of an exponential function is y = A · bx. Deﬁne y as the amount of money in the bank. Deﬁne x as the number of years from now. A is the initial amount, so A = 1500. Now we have to ﬁnd what b is. 363 www.ck12.org We’re told that the interest is 5% each year, which is 0.05 in decimal form. When we add 0.05A to A, we get 1.05A, so that is the factor we multiply by each year. The base of the exponential is b = 1.05. The formula that describes this problem is y = 1500·1.05x. To ﬁnd the total amount of money in the bank at the end of ﬁve years, we simply plug in x = 5. y = 1500 · (1.05)5 = $1914.42 Solve Real-World Problems Involving Exponential Decay Exponential decay problems appear in several application problems. Some examples of these are half-life problems and depreciation problems. Let’s solve an example of each of these problems. Example 5 A radioactive substance has a half-life of one week. In other words, at the end of every week the level of radioactivity is half of its value at the beginning of the week. The initial level of radioactivity is 20 counts per second. Draw the graph of the amount of radioactivity against time in weeks. Find the formula that gives the radioactivity in terms of time. Find the radioactivity left after three weeks. Solution Let’s start by making a table of values and then draw the graph. Table 8.9: Time Radioactivity 0 20 1 10 2 5 3 2.5 4 1.25 5 0.625 www.ck12.org 364 Exponential decay ﬁts the general formula y = A · bx. In this case: y is the amount of radioactivity x is the time in weeks A = 20 is the starting amount b = 1 2 since the substance losses half its value each week The formula for this problem is y = 20 · ( 1 2 )x or y = 20· 2−x. To ﬁnd out how much radioactivity is left after three weeks, we plug x = 3 into this formula. y = 20 · (1 2 )3 = 20 · (1 8 ) = 2.5 Example 6 The cost of a new car is $32,000. It depreciates at a rate of 15% per year. This means that it loses 15% of each value each year. Draw the graph of the car’s value against time in year. Find the formula that gives the value of the car in terms of time. Find the value of the car when it is four years old. Solution Let’s start by making a table of values. To ﬁll in the values we start with 32,000 at time t = 0. Then we multiply the value of the car by 85% for each passing year. (Since the car loses 15% of its value, that means it keeps 85% of its value). Remember that 85% means that we multiply by the decimal .85. Table 8.10: Time Value (thousands) 0 32 1 27.2 2 23.1 3 19.7 4 16.7 5 14.2 Now draw the graph: 365 www.ck12.org Let’s start with the general formula y = A · bx In this case: y is the value of the car, x is the time in years, A = 32 is the starting amount in thousands, b = 0.85 since we multiply the amount by this factor to get the value of the car next year The formula for this problem is y = 32 · (0.85)x. Finally, to ﬁnd the value of the car when it is four years old, we plug x = 4 into that formula: y = 32 · (0.85)4 = 16.7 thousand dollars, or $16,704 if we don’t round. Review Questions Solve the following application problems. 1. Half-life: Suppose a radioactive substance decays at a rate of 3.5% per hour. (a) What percent of the substance is left after 6 hours? (b) What percent is left after 12 hours? (c) The substance is safe to handle when at least 50% of it has decayed. Make a guess as to how many hours this will take. (d) Test your guess. How close were you? 2. Population decrease: In 1990 a rural area has 1200 bird species. (a) If species of birds are becoming extinct at the rate of 1.5% per decade (ten years), how many bird species will be left in the year 2020? (b) At that same rate, how many were there in 1980? 3. Growth: Janine owns a chain of fast food restaurants that operated 200 stores in 1999. If the rate of increase is 8% annually, how many stores does the restaurant operate in 2007? 4. Investment: Paul invests $360 in an account that pays 7.25% compounded annually. (a) What is the total amount in the account after 12 years? (b) If Paul invests an equal amount in an account that pays 5% compounded quarterly (four times a year), what will be the amount in that account after 12 years? www.ck12.org 366 (c) Which is the better investment? 5. The cost of a new ATV (all-terrain vehicle) is $7200. It depreciates at 18% per year. (a) Draw the graph of the vehicle’s value against time in years. (b) Find the formula that gives the value of the ATV in terms of time. (c) Find the value of the ATV when it is ten years old. 6. A person is infected by a certain bacterial infection. When he goes to the doctor the population of bacteria is 2 million. The doctor prescribes an antibiotic that reduces the bacteria population to 1 4 of its size each day. (a) Draw the graph of the size of the bacteria population against time in days. (b) Find the formula that gives the size of the bacteria population in terms of time. (c) Find the size of the bacteria population ten days after the drug was ﬁrst taken. (d) Find the size of the bacteria population after 2 weeks (14 days). Texas Instruments Resources In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http: //www.ck12.org/flexr/chapter/9618. 367 www.ck12.org Chapter 9 Polynomials 9.1 Addition and Subtraction of Polynomials Learning Objectives • Write a polynomial expression in standard form. • Classify polynomial expression by degree. • Add and subtract polynomials. • Solve problems using addition and subtraction of polynomials. Introduction So far we’ve seen functions described by straight lines (linear functions) and functions where the variable appeared in the exponent (exponential functions). In this section we’ll introduce polynomial functions. A polynomial is made up of diﬀerent terms that contain positive integer powers of the variables. Here is an example of a polynomial: 4x3 + 2x2 −3x + 1 Each part of the polynomial that is added or subtracted is called a term of the polynomial. The example above is a polynomial with four terms. The numbers appearing in each term in front of the variable are called the coeﬀicients. The number appearing all by itself without a variable is called a constant. In this case the coeﬀicient of x3 is 4, the coeﬀicient of x2 is 2, the coeﬀicient of x is -3 and the constant is 1. www.ck12.org 368 Degrees of Polynomials and Standard Form Each term in the polynomial has a diﬀerent degree. The degree of the term is the power of the variable in that term. 4x3 has degree 3 and is called a cubic term or 3rd order term. 2x2 has degree 2 and is called a quadratic term or 2nd order term. −3x has degree 1 and is called a linear term or 1st order term. 1 has degree 0 and is called the constant. By deﬁnition, the degree of the polynomial is the same as the degree of the term with the highest degree. This example is a polynomial of degree 3, which is also called a “cubic” polynomial. (Why do you think it is called a cubic?). Polynomials can have more than one variable. Here is another example of a polynomial: t4 −6s3t2 −12st + 4s4 −5 This is a polynomial because all the exponents on the variables are positive integers. This polynomial has ﬁve terms. Let’s look at each term more closely. Note: The degree of a term is the sum of the powers on each variable in the term. In other words, the degree of each term is the number of variables that are multiplied together in that term, whether those variables are the same or diﬀerent. t4 has a degree of 4, so it’s a 4th order term −6s3t2 has a degree of 5, so it’s a 5th order term. −12st has a degree of 2, so it’s a 2nd order term. 4s4 has a degree of 4, so it’s a 4th order term. −5 is a constant, so its degree is 0. Since the highest degree of a term in this polynomial is 5, then this is polynomial of degree 5th or a 5th order polynomial. A polynomial that has only one term has a special name. It is called a monomial (mono means one). A monomial can be a constant, a variable, or a product of a constant and one or more variables. You can see that each term in a polynomial is a monomial, so a polynomial is just the sum of several monomials. Here are some examples of monomials: b2 −2ab2 8 1 4 x4 −29xy Example 1 For the following polynomials, identify the coeﬀicient of each term, the constant, the degree of each term and the degree of the polynomial. a) x5 −3x3 + 4x2 −5x + 7 b) x4 −3x3y2 + 8x −12 Solution a) x5 −3x3 + 4x2 −5x + 7 369 www.ck12.org The coeﬀicients of each term in order are 1, -3, 4, and -5 and the constant is 7. The degrees of each term are 5, 3, 2, 1, and 0. Therefore the degree of the polynomial is 5. b) x4 −3x3y2 + 8x −12 The coeﬀicients of each term in order are 1, -3, and 8 and the constant is -12. The degrees of each term are 4, 5, 1, and 0. Therefore the degree of the polynomial is 5. Example 2 Identify the following expressions as polynomials or non-polynomials. a) 5x5 −2x b) 3x2 −2x−2 c) x √x −1 d) 5 x3+1 e) 4x 1 3 f) 4xy2 −2x2y −3 + y3 −3x3 Solution a) This is a polynomial. b) This is not a polynomial because it has a negative exponent. c) This is not a polynomial because it has a radical. d) This is not a polynomial because the power of x appears in the denominator of a fraction (and there is no way to rewrite it so that it does not). e) This is not a polynomial because it has a fractional exponent. f) This is a polynomial. Often, we arrange the terms in a polynomial in order of decreasing power. This is called standard form. The following polynomials are in standard form: 4x4 −3x3 + 2x2 −x + 1 a4b3 −2a3b3 + 3a4b −5ab2 + 2 The ﬁrst term of a polynomial in standard form is called the leading term, and the coeﬀicient of the leading term is called the leading coeﬀicient. The ﬁrst polynomial above has the leading term 4x4, and the leading coeﬀicient is 4. The second polynomial above has the leading term a4b3, and the leading coeﬀicient is 1. Example 3 Rearrange the terms in the following polynomials so that they are in standard form. Indicate the leading term and leading coeﬀicient of each polynomial. a) 7 −3x3 + 4x b) ab −a3 + 2b c) −4b + 4 + b2 Solution www.ck12.org 370 a) 7 −3x3 + 4x becomes −3x3 + 4x + 7. Leading term is −3x3; leading coeﬀicient is -3. b) ab −a3 + 2b becomes −a3 + ab + 2b. Leading term is −a3; leading coeﬀicient is -1. c) −4b + 4 + b2 becomes b2 −4b + 4. Leading term is b2; leading coeﬀicient is 1. Simplifying Polynomials A polynomial is simpliﬁed if it has no terms that are alike. Like terms are terms in the polynomial that have the same variable(s) with the same exponents, whether they have the same or diﬀerent coeﬀicients. For example, 2x2y and 5x2y are like terms, but 6x2y and 6xy2 are not like terms. When a polynomial has like terms, we can simplify it by combining those terms. x2 + 6xy −4xy + y2 ↗ ↖ Like terms We can simplify this polynomial by combining the like terms 6xy and −4xy into (6 −4)xy, or 2xy. The new polynomial is x2 + 2xy + y2. Example 4 Simplify the following polynomials by collecting like terms and combining them. a) 2x −4x2 + 6 + x2 −4 + 4x b) a3b3 −5ab4 + 2a3b −a3b3 + 3ab4 −a2b Solution a) Rearrange the terms so that like terms are grouped together: (−4x2 + x2) + (2x + 4x) + (6 −4) Combine each set of like terms: −3x2 + 6x + 2 b) Rearrange the terms so that like terms are grouped together: (a3b3 −a3b3)+(−5ab4 +3ab4)+2a3b−a2b Combine each set of like terms: 0 −2ab4 + 2a3b −a2b = −2ab4 + 2a3b −a2b Adding and Subtracting Polynomials To add two or more polynomials, write their sum and then simplify by combining like terms. Example 5 Add and simplify the resulting polynomials. a) Add 3x2 −4x + 7 and 2x3 −4x2 −6x + 5 b) Add x2 −2xy + y2 and 2y2 −3x2 and 10xy + y3 Solution a) (3x2 −4x + 7) + (2x3 −4x2 −6x + 5) Group like terms: = 2x3 + (3x2 −4x2) + (−4x −6x) + (7 + 5) Simplify: = 2x3 −x2 −10x + 12 b) 371 www.ck12.org (x2 −2xy + y2) + (2y2 −3x2) + (10xy + y3) Group like terms: = (x2 −3x2) + (y2 + 2y2) + (−2xy + 10xy) + y3 Simplify: = −2x2 + 3y2 + 8xy + y3 To subtract one polynomial from another, add the opposite of each term of the polynomial you are sub-tracting. Example 6 a) Subtract x3 −3x2 + 8x + 12 from 4x2 + 5x −9 b) Subtract 5b2 −2a2 from 4a2 −8ab −9b2 Solution a) (4x2 + 5x −9) −(x3 −3x2 + 8x + 12) = (4x2 + 5x −9) + (−x3 + 3x2 −8x −12) Group like terms: = −x3 + (4x2 + 3x2) + (5x −8x) + (−9 −12) Simplify: = −x3 + 7x2 −3x −21 b) (4a2 −8ab −9b2) −(5b2 −2a2) = (4a2 −8ab −9b2) + (−5b2 + 2a2) Group like terms: = (4a2 + 2a2) + (−9b2 −5b2) −8ab Simplify: = 6a2 −14b2 −8ab Note: An easy way to check your work after adding or subtracting polynomials is to substitute a convenient value in for the variable, and check that your answer and the problem both give the same value. For example, in part (b) above, if we let a = 2 and b = 3, then we can check as follows: Given Solution (4a2 −8ab −9b2) −(5b2 −2a2) 6a2 −14b2 −8ab (4(2)2 −8(2)(3) −9(3)2) −(5(3)2 −2(2)2) 6(2)2 −14(3)2 −8(2)(3) (4(4) −8(2)(3) −9(9)) −(5(9) −2(4)) 6(4) −14(9) −8(2)(3) (−113) −37 24 −126 −48 −150 −150 Since both expressions evaluate to the same number when we substitute in arbitrary values for the variables, we can be reasonably sure that our answer is correct. Note: When you use this method, do not choose 0 or 1 for checking since these can lead to common problems. Problem Solving Using Addition or Subtraction of Polynomials One way we can use polynomials is to ﬁnd the area of a geometric ﬁgure. Example 7 Write a polynomial that represents the area of each ﬁgure shown. www.ck12.org 372 a) b) c) d) Solution a) This shape is formed by two squares and two rectangles. The blue square has area y × y = y2. The yellow square has area x × x = x2. The pink rectangles each have area x × y = xy. To ﬁnd the total area of the ﬁgure we add all the separate areas: Total area = y2 + x2 + xy + xy = y2 + x2 + 2xy b) This shape is formed by two squares and one rectangle. The yellow squares each have area a × a = a2. The orange rectangle has area 2a × b = 2ab. To ﬁnd the total area of the ﬁgure we add all the separate areas: 373 www.ck12.org Total area = a2 + a2 + 2ab = 2a2 + 2ab c) To ﬁnd the area of the green region we ﬁnd the area of the big square and subtract the area of the little square. The big square has area : y × y = y2. The little square has area : x × x = x2. Area of the green region = y2 −x2 d) To ﬁnd the area of the ﬁgure we can ﬁnd the area of the big rectangle and add the areas of the pink squares. The pink squares each have area a × a = a2. The blue rectangle has area 3a × a = 3a2. To ﬁnd the total area of the ﬁgure we add all the separate areas: Total area = a2 + a2 + a2 + 3a2 = 6a2 Another way to ﬁnd this area is to ﬁnd the area of the big square and subtract the areas of the three yellow squares: The big square has area 3a × 3a = 9a2. The yellow squares each have area a × a = a2. To ﬁnd the total area of the ﬁgure we subtract: Area = 9a2 −(a2 + a2 + a2) = 9a2 −3a2 = 6a2 Further Practice For more practice adding and subtracting polynomials, try playing the Battleship game at quia.com/ba/28820.html. (The problems get harder as you play; watch out for trick questions!) www.ck12.org 374 Review Questions Indicate whether each expression is a polynomial. 1. x2 + 3x 1 2 2. 1 3 x2y −9y2 3. 3x−3 4. 2 3t2 −1 t2 5. √x −2x 6. ( x 3 2 )2 Express each polynomial in standard form. Give the degree of each polynomial. 7. 3 −2x 8. 8 −4x + 3x3 9. −5 + 2x −5x2 + 8x3 10. x2 −9x4 + 12 11. 5x + 2x2 −3x Add and simplify. 12. (x + 8) + (−3x −5) 13. (−2x2 + 4x −12) + (7x + x2) 14. (2a2b −2a + 9) + (5a2b −4b + 5) 15. (6.9a2 −2.3b2 + 2ab) + (3.1a −2.5b2 + b) 16. ( 3 5 x2 −1 4 x + 4 ) + ( 1 10 x2 + 1 2 x −21 5 ) Subtract and simplify. 17. (−t + 5t2) −(5t2 + 2t −9) 18. (−y2 + 4y −5) −(5y2 + 2y + 7) 19. (−5m2 −m) −(3m2 + 4m −5) 20. (2a2b −3ab2 + 5a2b2) −(2a2b2 + 4a2b −5b2) 21. (3.5x2y −6xy + 4x) −(1.2x2y −xy + 2y −3) Find the area of the following ﬁgures. 22. 375 www.ck12.org 23. 24. 25. 9.2 Multiplication of Polynomials Learning Objectives • Multiply a polynomial by a monomial. • Multiply a polynomial by a binomial. • Solve problems using multiplication of polynomials. Introduction Just as we can add and subtract polynomials, we can also multiply them. The Distributive Property and the techniques you’ve learned for dealing with exponents will be useful here. Multiplying a Polynomial by a Monomial When multiplying polynomials, we must remember the exponent rules that we learned in the last chapter. Especially important is the product rule: xn · xm = xn+m. If the expressions we are multiplying have coeﬀicients and more than one variable, we multiply the coeﬀi-cients just as we would any number and we apply the product rule on each variable separately. Example 1 Multiply the following monomials. a) (2x2)(5x3) b) (−3y4)(2y2) c) (3xy5)(−6x4y2) www.ck12.org 376 d) (−12a2b3c4)(−3a2b2) Solution a) (2x2)(5x3) = (2 · 5) · (x2 · x3) = 10x2+3 = 10x5 b) (−3y4)(2y2) = (−3 · 2) · (y4 · y2) = −6y4+2 = −6y6 c) (3xy5)(−6x4y2) = −18x1+4y5+2 = −18x5y7 d) (−12a2b3c4)(−3a2b2) = 36a2+2b3+2c4 = 36a4b5c4 To multiply a polynomial by a monomial, we have to use the Distributive Property. Remember, that property says that a(b + c) = ab + ac. Example 2 Multiply: a) 3(x2 + 3x −5) b) 4x(3x2 −7) c) −7y(4y2 −2y + 1) Solution a) 3(x2 + 3x −5) = 3(x2) + 3(3x) −3(5) = 3x2 + 9x −15 b) 4x(3x2 −7) = (4x)(3x2) + (4x)(−7) = 12x3 −28x c) −7y(4y2 −2y + 1) = (−7y)(4y2) + (−7y)(−2y) + (−7y)(1) = −28y3 + 14y2 −7y Notice that when we use the Distributive Property, the problem becomes a matter of just multiplying monomials by monomials and adding all the separate parts together. Example 3 Multiply: a) 2x3(−3x4 + 2x3 −10x2 + 7x + 9) b) −7a2bc3(5a2 −3b2 −9c2) Solution a) 2x3(−3x4 + 2x3 −10x2 + 7x + 9) = (2x3)(−3x4) + (2x3)(2x3) + (2x3)(−10x2) + (2x3)(7x) + (2x3)(9) = −6x7 + 4x6 −20x5 + 14x4 + 18x3 b) −7a2bc3(5a2 −3b2 −9c2) = (−7a2bc3)(5a2) + (−7a2bc3)(−3b2) + (−7a2bc3)(−9c2) = −35a4bc3 + 21a2b3c3 + 63a2bc5 Multiplying Two Polynomials Let’s start by multiplying two binomials together. A binomial is a polynomial with two terms, so a product of two binomials will take the form (a + b)(c + d). We can still use the Distributive Property here if we do it cleverly. First, let’s think of the ﬁrst set of parentheses as one term. The Distributive Property says that we can multiply that term by c, multiply it by d, and then add those two products together: (a + b)(c + d) = (a + b) · c + (a + b) · d. We can rewrite this expression as c(a + b) + d(a + b). Now let’s look at each half separately. We can apply the distributive property again to each set of parentheses in turn, and that gives us c(a + b) + d(a + b) = ca + cb + da + db. 377 www.ck12.org What you should notice is that when multiplying any two polynomials, every term in one polynomial is multiplied by every term in the other polynomial. Example 4 Multiply and simplify: (2x + 1)(x + 3) Solution We must multiply each term in the ﬁrst polynomial by each term in the second polynomial. Let’s try to be systematic to make sure that we get all the products. First, multiply the ﬁrst term in the ﬁrst set of parentheses by all the terms in the second set of parentheses. Now we’re done with the ﬁrst term. Next we multiply the second term in the ﬁrst parenthesis by all terms in the second parenthesis and add them to the previous terms. Now we’re done with the multiplication and we can simplify: (2x)(x) + (2x)(3) + (1)(x) + (1)(3) = 2x2 + 6x + x + 3 = 2x2 + 7x + 3 This way of multiplying polynomials is called in-line multiplication or horizontal multiplication. Another method for multiplying polynomials is to use vertical multiplication, similar to the vertical multiplication you learned with regular numbers. Example 5 Multiply and simplify: a) (4x −5)(x −20) b) (3x −2)(3x + 2) c) (3x2 + 2x −5)(2x −3) d) (x2 −9)(4x4 + 5x2 −2) Solution a) With horizontal multiplication this would be (4x −5)(x −20) = (4x)(x) + (4x)(−20) + (−5)(x) + (−5)(−20) = 4x2 −80x −5x + 100 = 4x2 −85x + 100 To do vertical multiplication instead, we arrange the polynomials on top of each other with like terms in the same columns: 4x −5 x −20 −80x + 100 4x2 −5x 4x2 −85x + 100 Both techniques result in the same answer: 4x2 −85x + 100. We’ll use vertical multiplication for the other problems. www.ck12.org 378 b) 3x −2 3x + 2 6x −4 9x2 −6x 9x2 + 0x −4 The answer is 9x2 −4. c) It’s better to place the smaller polynomial on the bottom: 3x2 + 2x −5 2x −3 −9x2 −6x + 15 6x3 + 4x2 −10x 6x3 −5x2 −16x + 15 The answer is 6x3 −5x2 −16x + 15. d) Set up the multiplication vertically and leave gaps for missing powers of x: 4x4 + 5x2 −2 x2 −9 −36x4 −45x2 + 18 4x6 + 5x4 −2x2 4x6 −31x4 −47x2 + 18 The answer is 4x6 −31x4 −47x2 + 18. The Khan Academy video at shows how multiplying two binomials together is related to the distributive property. Solve Real-World Problems Using Multiplication of Polynomials In this section, we’ll see how multiplication of polynomials is applied to ﬁnding the areas and volumes of geometric shapes. Example 6 Find the areas of the following ﬁgures: a) 379 www.ck12.org b) Find the volumes of the following ﬁgures: c) d) Solution a) We use the formula for the area of a rectangle: Area = length × width. For the big rectangle: Length = b + 3, Width = b + 2 Area = (b + 3)(b + 2) = b2 + 2b + 3b + 6 = b2 + 5b + 6 b) We could add up the areas of the blue and orange rectangles, but it’s easier to just ﬁnd the area of the whole big rectangle and subtract the area of the yellow rectangle. Area of big rectangle = 20(12) = 240 Area of yellow rectangle = (12 −x)(20 −2x) = 240 −24x −20x + 2x2 = 240 −44x + 2x2 = 2x2 −44x + 240 The desired area is the diﬀerence between the two: www.ck12.org 380 Area = 240 −(2x2 −44x + 240) = 240 + (−2x2 + 44x −240) = 240 −2x2 + 44x −240 = −2x2 + 44x c) The volume of this shape = (area of the base)(height). Area of the base = x(x + 2) = x2 + 2x Height = 2x + 1 Volume = (x2 + 2x)(2x + 1) = 2x3 + x2 + 4x2 + 2x = 2x3 + 5x2 + 2x d) The volume of this shape = (area of the base)(height). Area of the base = (4a −3)(2a + 1) = 8a2 + 4a −6a −3 = 8a2 −2a −3 Height = a + 4 Volume = (8a2 −2a −3)(a + 4) Let’s multiply using the vertical method: 8a2 −2a −3 a + 4 32a2 −8a −12 8a3 −2a2 −3a 8a3 + 30a2 −11a −12 The volume is 8a3 + 30a2 −11a −12. Review Questions Multiply the following monomials. 1. (2x)(−7x) 2. (10x)(3xy) 3. (4mn)(0.5nm2) 4. (−5a2b)(−12a3b3) 5. (3xy2z2)(15x2yz3) Multiply and simplify. 381 www.ck12.org 6. 17(8x −10) 7. 2x(4x −5) 8. 9x3(3x2 −2x + 7) 9. 3x(2y2 + y −5) 10. 10q(3q2r + 5r) 11. −3a2b(9a2 −4b2) 12. (x −3)(x + 2) 13. (a + b)(a −5) 14. (x + 2)(x2 −3) 15. (a2 + 2)(3a2 −4) 16. (7x −2)(9x −5) 17. (2x −1)(2x2 −x + 3) 18. (3x + 2)(9x2 −6x + 4) 19. (a2 + 2a −3)(a2 −3a + 4) 20. 3(x −5)(2x + 7) 21. 5x(x + 4)(2x −3) Find the areas of the following ﬁgures. 22. 23. Find the volumes of the following ﬁgures. 24. 25. www.ck12.org 382 9.3 Special Products of Polynomials Learning Objectives • Find the square of a binomial • Find the product of binomials using sum and diﬀerence formula • Solve problems using special products of polynomials Introduction We saw that when we multiply two binomials we need to make sure to multiply each term in the ﬁrst binomial with each term in the second binomial. Let’s look at another example. Multiply two linear binomials (binomials whose degree is 1): (2x + 3)(x + 4) When we multiply, we obtain a quadratic polynomial (one with degree 2) with four terms: 2x2 + 8x + 3x + 12 The middle terms are like terms and we can combine them. We simplify and get 2x2 + 11x + 12. This is a quadratic, or second-degree, trinomial (polynomial with three terms). You can see that every time we multiply two linear binomials with one variable, we will obtain a quadratic polynomial. In this section we’ll talk about some special products of binomials. Find the Square of a Binomial One special binomial product is the square of a binomial. Consider the product (x + 4)(x + 4). Since we are multiplying the same expression by itself, that means we are squaring the expression. (x + 4)(x + 4) is the same as (x + 4)2. When we multiply it out, we get x2 + 4x + 4x + 16, which simpliﬁes to x2 + 8x + 16. Notice that the two middle terms—the ones we added together to get 8x—were the same. Is this a coincidence? In order to ﬁnd that out, let’s square a general linear binomial. (a + b)2 = (a + b)(a + b) = a2 + ab + ab + b2 = a2 + 2ab + b2 Sure enough, the middle terms are the same. How about if the expression we square is a diﬀerence instead of a sum? (a −b)2 = (a −b)(a −b) = a2 −ab −ab + b2 = a2 −2ab + b2 It looks like the middle two terms are the same in general whenever we square a binomial. The general pattern is: to square a binomial, take the square of the ﬁrst term, add or subtract twice the product of the terms, and add the square of the second term. You should remember these formulas: 383 www.ck12.org (a + b)2 = a2 + 2ab + b2 and (a −b)2 = a2 −2ab + b2 Remember! Raising a polynomial to a power means that we multiply the polynomial by itself however many times the exponent indicates. For instance, (a + b)2 = (a + b)(a + b). Don’t make the common mistake of thinking that (a + b)2 = a2 + b2! To see why that’s not true, try substituting numbers for a and b into the equation (for example, a = 4 and b = 3), and you will see that it is not a true statement. The middle term, 2ab, is needed to make the equation work. We can apply the formulas for squaring binomials to any number of problems. Example 1 Square each binomial and simplify. a) (x + 10)2 b) (2x −3)2 c) (x2 + 4)2 d) (5x −2y)2 Solution Let’s use the square of a binomial formula to multiply each expression. a) (x + 10)2 If we let a = x and b = 10, then our formula (a+b)2 = a2+2ab+b2 becomes (x+10)2 = x2+2(x)(10)+102, which simpliﬁes to x2 + 20x + 100. b) (2x −3)2 If we let a = 2x and b = 3, then our formula (a−b)2 = a2−2ab+b2 becomes (2x−3)2 = (2x2)−2(2x)(3)+(3)2, which simpliﬁes to 4x2 −12x + 9. c) (x2 + 4)2 If we let a = x2 and b = 4, then (x2 + 4)2 = (x2)2 + 2(x2)(4) + (4)2 = x4 + 8x2 + 16 d) (5x −2y)2 If we let a = 5x and b = 2y, then (5x −2y)2 = (5x)2 −2(5x)(2y) + (2y)2 = 25x2 −20xy + 4y2 Find the Product of Binomials Using Sum and Diﬀerence Pat-terns Another special binomial product is the product of a sum and a diﬀerence of terms. For example, let’s multiply the following binomials. www.ck12.org 384 (x + 4)(x −4) = x2 −4x + 4x −16 = x2 −16 Notice that the middle terms are opposites of each other, so they cancel out when we collect like terms. This is not a coincidence. This always happens when we multiply a sum and diﬀerence of the same terms. In general, (a + b)(a −b) = a2 −ab + ab −b2 = a2 −b2 When multiplying a sum and diﬀerence of the same two terms, the middle terms cancel out. We get the square of the ﬁrst term minus the square of the second term. You should remember this formula. Sum and Diﬀerence Formula: (a + b)(a −b) = a2 −b2 Let’s apply this formula to a few examples. Example 2 Multiply the following binomials and simplify. a) (x + 3)(x −3) b) (5x + 9)(5x −9) c) (2x3 + 7)(2x3 −7) d) (4x + 5y)(4x −5y) Solution a) Let a = x and b = 3, then: (a + b)(a −b) = a2 −b2 (x + 3)(x −3) = x2 −32 = x2 −9 b) Let a = 5x and b = 9, then: (a + b)(a −b) = a2 −b2 (5x + 9)(5x −9) = (5x)2 −92 = 25x2 −81 c) Let a = 2x3 and b = 7, then: (2x3 + 7)(2x3 −7) = (2x3)2 −(7)2 = 4x6 −49 d) Let a = 4x and b = 5y, then: (4x + 5y)(4x −5y) = (4x)2 −(5y)2 = 16x2 −25y2 385 www.ck12.org Solve Real-World Problems Using Special Products of Polynomi-als Now let’s see how special products of polynomials apply to geometry problems and to mental arithmetic. Example 3 Find the area of the following square: Solution The length of each side is (a + b), so the area is (a + b)(a + b). Notice that this gives a visual explanation of the square of a binomial. The blue square has area a2, the red square has area b2, and each rectangle has area ab, so added all together, the area (a + b)(a + b) is equal to a2 + 2ab + b2. The next example shows how you can use the special products to do fast mental calculations. Example 4 Use the diﬀerence of squares and the binomial square formulas to ﬁnd the products of the following numbers without using a calculator. a) 43 × 57 b) 112 × 88 c) 452 d) 481 × 319 Solution The key to these mental “tricks” is to rewrite each number as a sum or diﬀerence of numbers you know how to square easily. a) Rewrite 43 as (50 −7) and 57 as (50 + 7). Then 43 × 57 = (50 −7)(50 + 7) = (50)2 −(7)2 = 2500 −49 = 2451 b) Rewrite 112 as (100 + 12) and 88 as (100 −12). Then 112 × 88 = (100 + 12)(100 −12) = (100)2 −(12)2 = 10000 −144 = 9856 c) 452 = (40 + 5)2 = (40)2 + 2(40)(5) + (5)2 = 1600 + 400 + 25 = 2025 d) Rewrite 481 as (400 + 81) and 319 as (400 −81). Then 481 × 319 = (400 + 81)(400 −81) = (400)2 −(81)2 (400)2 is easy - it equals 160000. (81)2 is not easy to do mentally, so let’s rewrite 81 as 80 + 1. (81)2 = (80 + 1)2 = (80)2 + 2(80)(1) + (1)2 = 6400 + 160 + 1 = 6561 www.ck12.org 386 Then 481 × 319 = (400)2 −(81)2 = 160000 −6561 = 153439 Review Questions Use the special product rule for squaring binomials to multiply these expressions. 1. (x + 9)2 2. (3x −7)2 3. (5x −y)2 4. (2x3 −3)2 5. (4x2 + y2)2 6. (8x −3)2 7. (2x + 5)(5 + 2x) 8. (xy −y)2 Use the special product of a sum and diﬀerence to multiply these expressions. 9. (2x −1)(2x + 1) 10. (x −12)(x + 12) 11. (5a −2b)(5a + 2b) 12. (ab −1)(ab + 1) 13. (z2 + y)(z2 −y) 14. (2q3 + r2)(2q3 −r2) 15. (7s −t)(t + 7s) 16. (x2y + xy2)(x2y −xy2) Find the area of the lower right square in the following ﬁgure. 17. Multiply the following numbers using special products. 9. 45 × 55 10. 562 11. 1002 × 998 12. 36 × 44 13. 10.5 × 9.5 14. 100.2 × 9.8 15. −95 × −105 16. 2 × −2 387 www.ck12.org 9.4 Polynomial Equations in Factored Form Learning Objectives • Use the zero-product property. • Find greatest common monomial factors. • Solve simple polynomial equations by factoring. Introduction In the last few sections, we learned how to multiply polynomials by using the Distributive Property. All the terms in one polynomial had to be multiplied by all the terms in the other polynomial. In this section, you’ll start learning how to do this process in reverse. The reverse of distribution is called factoring. The total area of the ﬁgure above can be found in two ways. We could ﬁnd the areas of all the small rectangles and add them: ab + ac + ad + ae + 2a. Or, we could ﬁnd the area of the big rectangle all at once. Its width is a and its length is b + c + d + e + 2, so its area is a(b + c + d + e + 2). Since the area of the rectangle is the same no matter what method we use, those two expressions must be equal. ab + ac + ad + ae + 2a = a(b + c + d + e + 2) To turn the right-hand side of this equation into the left-hand side, we would use the distributive property. To turn the left-hand side into the right-hand side, we would need to factor it. Since polynomials can be multiplied just like numbers, they can also be factored just like numbers—and we’ll see later how this can help us solve problems. Find the Greatest Common Monomial Factor You will be learning several factoring methods in the next few sections. In most cases, factoring takes several steps to complete because we want to factor completely. That means that we factor until we can’t factor any more. Let’s start with the simplest step: ﬁnding the greatest monomial factor. When we want to factor, we always look for common monomials ﬁrst. Consider the following polynomial, written in expanded form: www.ck12.org 388 ax + bx + cx + dx A common factor is any factor that appears in all terms of the polynomial; it can be a number, a variable or a combination of numbers and variables. Notice that in our example, the factor x appears in all terms, so it is a common factor. To factor out the x, we write it outside a set of parentheses. Inside the parentheses, we write what’s left when we divide each term by x: x(a + b + c + d) Let’s look at more examples. Example 1 Factor: a) 2x + 8 b) 15x −25 c) 3a + 9b + 6 Solution a) We see that the factor 2 divides evenly into both terms: 2x + 8 = 2(x) + 2(4) We factor out the 2 by writing it in front of a parenthesis: 2( ) Inside the parenthesis we write what is left of each term when we divide by 2: 2(x + 4) b) We see that the factor of 5 divides evenly into all terms: 15x −25 = 5(3x) −5(5) Factor out the 5 to get: 5(3x −5) c) We see that the factor of 3 divides evenly into all terms: 3a + 9b + 6 = 3(a) + 3(3b) + 3(2) Factor 3 to get: 3(a + 3b + 2) Example 2 Find the greatest common factor: a) a3 −3a2 + 4a b) 12a4 −5a3 + 7a2 Solution a) Notice that the factor a appears in all terms of a3 −3a2 + 4a, but each term has a raised to a diﬀerent power. The greatest common factor of all the terms is simply a. So ﬁrst we rewrite a3 −3a2 + 4a as a(a2) + a(−3a) + a(4). Then we factor out the a to get a(a2 −3a + 4). b) The factor a appears in all the terms, and it’s always raised to at least the second power. So the greatest common factor of all the terms is a2. We rewrite the expression 12a4 −5a3 + 7a2 as (12a2 · a2) −(5a · a2) + (7 · a2) Factor out the a2 to get a2(12a2 −5a + 7). Example 3 Factor completely: 389 www.ck12.org a) 3ax + 9a b) x3y + xy c) 5x3y −15x2y2 + 25xy3 Solution a) Both terms have a common factor of 3, but they also have a common factor of a. It’s simplest to factor these both out at once, which gives us 3a(x + 3). b) Both x and y are common factors. When we factor them both out at once, we get xy(x2 + 1). c) The common factors are 5, x, and y. Factoring out 5xy gives us 5xy(x2 −3xy + 5xy2). Use the Zero-Product Property The most useful thing about factoring is that we can use it to help solve polynomial equations. For example, consider an equation like 2x2 +5x−42 = 0. There’s no good way to isolate x in this equation, so we can’t solve it using any of the techniques we’ve already learned. But the left-hand side of the equation can be factored, making the equation (x + 6)(2x −7) = 0. How is this helpful? The answer lies in a useful property of multiplication: if two numbers multiply to zero, then at least one of those numbers must be zero. This is called the Zero-Product Property. What does this mean for our polynomial equation? Since the product equals 0, then at least one of the factors on the left-hand side must equal zero. So we can ﬁnd the two solutions by setting each factor equal to zero and solving each equation separately. Setting the factors equal to zero gives us: (x + 6) = 0 OR (2x −7) = 0 Solving both of those equations gives us: x + 6 = 0 2x −7 = 0 x = −6 OR 2x = 7 x = 7 2 Notice that the solution is x = −6 OR x = 7 2. The OR means that either of these values of x would make the product of the two factors equal to zero. Let’s plug the solutions back into the equation and check that this is correct. Check : x = −6; Check : x = 7 2 (x + 6)(2x −7) = (x + 6)(2x −7) = (−6 + 6)(2(−6) −7) = (7 2 + 6 ) ( 2 · 7 2 −7 ) = (0)(−19) = 0 (19 2 ) (7 −7) = (19 2 ) (0) = 0 Both solutions check out. www.ck12.org 390 Factoring a polynomial is very useful because the Zero-Product Property allows us to break up the problem into simpler separate steps. When we can’t factor a polynomial, the problem becomes harder and we must use other methods that you will learn later. As a last note in this section, keep in mind that the Zero-Product Property only works when a product equals zero. For example, if you multiplied two numbers and the answer was nine, that wouldn’t mean that one or both of the numbers must be nine. In order to use the property, the factored polynomial must be equal to zero. Example 4 Solve each equation: a) (x −9)(3x + 4) = 0 b) x(5x −4) = 0 c) 4x(x + 6)(4x −9) = 0 Solution Since all the polynomials are in factored form, we can just set each factor equal to zero and solve the simpler equations separately a) (x −9)(3x + 4) = 0 can be split up into two linear equations: x −9 = 0 3x + 4 = 0 x = 9 or 3x = −4 x = −4 3 b) x(5x −4) = 0 can be split up into two linear equations: 5x −4 = 0 x = 0 or 5x = 4 x = 4 5 c) 4x(x + 6)(4x −9) = 0 can be split up into three linear equations: 4x = 0 x + 6 = 0 4x −9 = 0 x = 0 4 or x = −6 or 4x = 9 x = 0 x = 9 4 Solve Simple Polynomial Equations by Factoring Now that we know the basics of factoring, we can solve some simple polynomial equations. We already saw how we can use the Zero-Product Property to solve polynomials in factored form—now we can use that knowledge to solve polynomials by factoring them ﬁrst. Here are the steps: a) If necessary, rewrite the equation in standard form so that the right-hand side equals zero. b) Factor the polynomial completely. 391 www.ck12.org c) Use the zero-product rule to set each factor equal to zero. d) Solve each equation from step 3. e) Check your answers by substituting your solutions into the original equation Example 5 Solve the following polynomial equations. a) x2 −2x = 0 b) 2x2 = 5x c) 9x2y −6xy = 0 Solution a) x2 −2x = 0 Rewrite: this is not necessary since the equation is in the correct form. Factor: The common factor is x, so this factors as x(x −2) = 0. Set each factor equal to zero: x = 0 or x −2 = 0 Solve: x = 0 or x = 2 Check: Substitute each solution back into the original equation. x = 0 ⇒(0)2 −2(0) = 0 works out x = 2 ⇒(2)2 −2(2) = 4 −4 = 0 works out Answer: x = 0, x = 2 b) 2x2 = 5x Rewrite: 2x2 = 5x ⇒2x2 −5x = 0 Factor: The common factor is x, so this factors as x(2x −5) = 0. Set each factor equal to zero: x = 0 or 2x −5 = 0 Solve: x = 0 or 2x = 5 x = 5 2 Check: Substitute each solution back into the original equation. x = 0 ⇒2(0)2 = 5(0) ⇒0 = 0 works out x = 5 2 ⇒2 (5 2 )2 = 5 · 5 2 ⇒2 · 25 4 = 25 2 ⇒25 2 = 25 2 works out www.ck12.org 392 Answer: x = 0, x = 5 2 c) 9x2y −6xy = 0 Rewrite: not necessary Factor: The common factor is 3xy, so this factors as 3xy(3x −2) = 0. Set each factor equal to zero: 3 = 0 is never true, so this part does not give a solution. The factors we have left give us: x = 0 or y = 0 or 3x −2 = 0 Solve: x = 0 or y = 0 or 3x = 2 x = 2 3 Check: Substitute each solution back into the original equation. x = 0 ⇒9(0)y −6(0)y = 0 −0 = 0 works out y = 0 ⇒9x2(0) −6x(0) = 0 −0 = 0 works out x = 2 3 ⇒9 · (2 3 )2 y −6 · 2 3y = 9 · 4 9y −4y = 4y −4y = 0 works out Answer: x = 0, y = 0, x = 2 3 Review Questions Factor out the greatest common factor in the following polynomials. 1. 2x2 −5x 2. 3x3 −21x 3. 5x6 + 15x4 4. 4x3 + 10x2 −2x 5. −10x6 + 12x5 −4x4 6. 12xy + 24xy2 + 36xy3 7. 5a3 −7a 8. 3y + 6z 9. 10a3 −4ab 10. 45y12 + 30y10 11. 16xy2z + 4x3y 12. 2a −4a2 + 6 13. 5xy2 −10xy + 5y2 Solve the following polynomial equations. 14. x(x + 12) = 0 15. (2x + 1)(2x −1) = 0 393 www.ck12.org 16. (x −5)(2x + 7)(3x −4) = 0 17. 2x(x + 9)(7x −20) = 0 18. x(3 + y) = 0 19. x(x −2y) = 0 20. 18y −3y2 = 0 21. 9x2 = 27x 22. 4a2 + a = 0 23. b2 −5 3b = 0 24. 4x2 = 36 25. x3 −5x2 = 0 9.5 Factoring Quadratic Expressions Learning Objectives • Write quadratic equations in standard form. • Factor quadratic expressions for diﬀerent coeﬀicient values. Write Quadratic Expressions in Standard Form Quadratic polynomials are polynomials of the 2nd degree. The standard form of a quadratic polynomial is written as ax2 + bx + c where a, b, and c stand for constant numbers. Factoring these polynomials depends on the values of these constants. In this section we’ll learn how to factor quadratic polynomials for diﬀerent values of a, b, and c. (When none of the coeﬀicients are zero, these expressions are also called quadratic trinomials, since they are polynomials with three terms.) You’ve already learned how to factor quadratic polynomials where c = 0. For example, for the quadratic ax2 + bx, the common factor is x and this expression is factored as x(ax + b). Now we’ll see how to factor quadratics where c is nonzero. Factor when a = 1, b is Positive, and c is Positive First, let’s consider the case where a = 1, b is positive and c is positive. The quadratic trinomials will take the form x2 + bx + c You know from multiplying binomials that when you multiply two factors (x+m)(x+n), you get a quadratic polynomial. Let’s look at this process in more detail. First we use distribution: (x + m)(x + n) = x2 + nx + mx + mn Then we simplify by combining the like terms in the middle. We get: (x + m)(x + n) = x2 + (n + m)x + mn www.ck12.org 394 So to factor a quadratic, we just need to do this process in reverse. We see that x2 + (n + m)x + mn is the same form as x2 + bx + c This means that we need to ﬁnd two numbers m and n where n + m = b and mn = c The factors of x2 + bx + c are always two binomials (x + m)(x + n) such that n + m = b and mn = c. Example 1 Factor x2 + 5x + 6. Solution We are looking for an answer that is a product of two binomials in parentheses: (x )(x ) We want two numbers m and n that multiply to 6 and add up to 5. A good strategy is to list the possible ways we can multiply two numbers to get 6 and then see which of these numbers add up to 5: 6 = 1 · 6 and 1 + 6 = 7 6 = 2 · 3 and 2 + 3 = 5 This is the correct choice. So the answer is (x + 2)(x + 3). We can check to see if this is correct by multiplying (x + 2)(x + 3): x + 2 x + 3 3x + 6 x2 + 2x x2 + 5x + 6 The answer checks out. Example 2 Factor x2 + 7x + 12. Solution We are looking for an answer that is a product of two binomials in parentheses: (x )(x ) The number 12 can be written as the product of the following numbers: 12 = 1 · 12 and 1 + 12 = 13 12 = 2 · 6 and 2 + 6 = 8 12 = 3 · 4 and 3 + 4 = 7 This is the correct choice. 395 www.ck12.org The answer is (x + 3)(x + 4). Example 3 Factor x2 + 8x + 12. Solution We are looking for an answer that is a product of two binomials in parentheses: (x )(x ) The number 12 can be written as the product of the following numbers: 12 = 1 · 12 and 1 + 12 = 13 12 = 2 · 6 and 2 + 6 = 8 This is the correct choice. 12 = 3 · 4 and 3 + 4 = 7 The answer is (x + 2)(x + 6). Example 4 Factor x2 + 12x + 36. Solution We are looking for an answer that is a product of two binomials in parentheses: (x )(x ) The number 36 can be written as the product of the following numbers: 36 = 1 · 36 and 1 + 36 = 37 36 = 2 · 18 and 2 + 18 = 20 36 = 3 · 12 and 3 + 12 = 15 36 = 4 · 9 and 4 + 9 = 13 36 = 6 · 6 and 6 + 6 = 12 This is the correct choice. The answer is (x + 6)(x + 6). Factor when a = 1, b is Negative and c is Positive Now let’s see how this method works if the middle coeﬀicient is negative. Example 5 Factor x2 −6x + 8. Solution We are looking for an answer that is a product of two binomials in parentheses: (x )(x ) When negative coeﬀicients are involved, we have to remember that negative factors may be involved also. The number 8 can be written as the product of the following numbers: 8 = 1 · 8 and 1 + 8 = 9 but also 8 = (−1) · (−8) and −1 + (−8) = −9 and www.ck12.org 396 8 = 2 · 4 and 2 + 4 = 6 but also 8 = (−2) · (−4) and −2 + (−4) = −6 This is the correct choice. The answer is (x −2)(x −4). We can check to see if this is correct by multiplying (x −2)(x −4): x −2 x −4 −4x + 8 x2 −2x x2 −6x + 8 The answer checks out. Example 6 Factor x2 −17x + 16. Solution We are looking for an answer that is a product of two binomials in parentheses: (x )(x ) The number 16 can be written as the product of the following numbers: 16 = 1 · 16 and 1 + 16 = 17 16 = (−1) · (−16) and −1 + (−16) = −17 This is the correct choice. 16 = 2 · 8 and 2 + 8 = 10 16 = (−2) · (−8) and −2 + (−8) = −10 16 = 4 · 4 and 4 + 4 = 8 16 = (−4) · (−4) and −4 + (−4) = −8 The answer is (x −1)(x −16). In general, whenever b is negative and a and c are positive, the two binomial factors will have minus signs instead of plus signs. Factor when a = 1 and c is Negative Now let’s see how this method works if the constant term is negative. Example 7 Factor x2 + 2x −15. Solution We are looking for an answer that is a product of two binomials in parentheses: (x )(x ) Once again, we must take the negative sign into account. The number -15 can be written as the product of the following numbers: 397 www.ck12.org −15 = −1 · 15 and −1 + 15 = 14 −15 = 1 · (−15) and 1 + (−15) = −14 −15 = −3 · 5 and −3 + 5 = 2 This is the correct choice. −15 = 3 · (−5) and 3 + (−5) = −2 The answer is (x −3)(x + 5). We can check to see if this is correct by multiplying: x −3 x + 5 5x −15 x2 −3x x2 + 2x −15 The answer checks out. Example 8 Factor x2 −10x −24. Solution We are looking for an answer that is a product of two binomials in parentheses: (x )(x ) The number -24 can be written as the product of the following numbers: −24 = −1 · 24 and −1 + 24 = 23 −24 = 1 · (−24) and 1 + (−24) = −23 −24 = −2 · 12 and −2 + 12 = 10 −24 = 2 · (−12) and 2 + (−12) = −10 This is the correct choice. −24 = −3 · 8 and −3 + 8 = 5 −24 = 3 · (−8) and 3 + (−8) = −5 −24 = −4 · 6 and −4 + 6 = 2 −24 = 4 · (−6) and 4 + (−6) = −2 The answer is (x −12)(x + 2). Example 9 Factor x2 + 34x −35. Solution We are looking for an answer that is a product of two binomials in parentheses: (x )(x ) The number -35 can be written as the product of the following numbers: −35 = −1 · 35 and −1 + 35 = 34 This is the correct choice. −35 = 1 · (−35) and 1 + (−35) = −34 −35 = −5 · 7 and −5 + 7 = 2 −35 = 5 · (−7) and 5 + (−7) = −2 The answer is (x −1)(x + 35). www.ck12.org 398 Factor when a = - 1 When a = −1, the best strategy is to factor the common factor of -1 from all the terms in the quadratic polynomial and then apply the methods you learned so far in this section Example 10 Factor −x2 + x + 6. Solution First factor the common factor of -1 from each term in the trinomial. Factoring -1 just changes the signs of each term in the expression: −x2 + x + 6 = −(x2 −x −6) We’re looking for a product of two binomials in parentheses: −(x )(x ) Now our job is to factor x2 −x −6. The number -6 can be written as the product of the following numbers: −6 = −1 · 6 and −1 + 6 = 5 −6 = 1 · (−6) and 1 + (−6) = −5 −6 = −2 · 3 and −2 + 3 = 1 −6 = 2 · (−3) and 2 + (−3) = −1 This is the correct choice. The answer is −(x −3)(x + 2). Lesson Summary • A quadratic of the form x2+bx+c factors as a product of two binomials in parentheses: (x+m)(x+n) • If b and c are positive, then both m and n are positive. Example: x2 + 8x + 12 factors as (x + 6)(x + 2). • If b is negative and c is positive, then both m and n are negative. Example: x2 −6x + 8 factors as (x −2)(x −4). • If c is negative, then either m is positive and n is negative or vice-versa. Example: x2 + 2x −15 factors as (x + 5)(x −3). Example: x2 + 34x −35 factors as (x + 35)(x −1). • If a = −1, factor out -1 from each term in the trinomial and then factor as usual. The answer will have the form: −(x + m)(x + n) Example: −x2 + x + 6 factors as −(x −3)(x + 2). 399 www.ck12.org Review Questions Factor the following quadratic polynomials. 1. x2 + 10x + 9 2. x2 + 15x + 50 3. x2 + 10x + 21 4. x2 + 16x + 48 5. x2 −11x + 24 6. x2 −13x + 42 7. x2 −14x + 33 8. x2 −9x + 20 9. x2 + 5x −14 10. x2 + 6x −27 11. x2 + 7x −78 12. x2 + 4x −32 13. x2 −12x −45 14. x2 −5x −50 15. x2 −3x −40 16. x2 −x −56 17. −x2 −2x −1 18. −x2 −5x + 24 19. −x2 + 18x −72 20. −x2 + 25x −150 21. x2 + 21x + 108 22. −x2 + 11x −30 23. x2 + 12x −64 24. x2 −17x −60 25. x2 + 5x −36 9.6 Factoring Special Products Learning Objectives • Factor the diﬀerence of two squares. • Factor perfect square trinomials. • Solve quadratic polynomial equation by factoring. Introduction When you learned how to multiply binomials we talked about two special products. The sum and diﬀerence formula: (a + b)(a −b) = a2 −b2 The square of a binomial formulas: (a + b)2 = a2 + 2ab + b2 (a −b)2 = a2 −2ab + b2 In this section we’ll learn how to recognize and factor these special products. www.ck12.org 400 Factor the Diﬀerence of Two Squares We use the sum and diﬀerence formula to factor a diﬀerence of two squares. A diﬀerence of two squares is any quadratic polynomial in the form a2 −b2, where a and b can be variables, constants, or just about anything else. The factors of a2 −b2 are always (a + b)(a −b); the key is ﬁguring out what the a and b terms are. Example 1 Factor the diﬀerence of squares: a) x2 −9 b) x2 −100 c) x2 −1 Solution a) Rewrite x2 −9 as x2 −32. Now it is obvious that it is a diﬀerence of squares. The diﬀerence of squares formula is: a2 −b2 = (a + b)(a −b) Let’s see how our problem matches with the formula: x2 −32 = (x + 3)(x −3) The answer is: x2 −9 = (x + 3)(x −3) We can check to see if this is correct by multiplying (x + 3)(x −3): x + 3 x −3 −3x −9 x2 + 3x x2 + 0x −9 The answer checks out. Note: We could factor this polynomial without recognizing it as a diﬀerence of squares. With the methods we learned in the last section we know that a quadratic polynomial factors into the product of two binomials: (x )(x ) We need to ﬁnd two numbers that multiply to -9 and add to 0 (since there is no x−term, that’s the same as if the x−term had a coeﬀicient of 0). We can write -9 as the following products: −9 = −1 · 9 and −1 + 9 = 8 −9 = 1 · (−9) and 1 + (−9) = −8 −9 = 3 · (−3) and 3 + (−3) = 0 These are the correct numbers. We can factor x2 −9 as (x + 3)(x −3), which is the same answer as before. You can always factor using the methods you learned in the previous section, but recognizing special products helps you factor them faster. b) Rewrite x2 −100 as x2 −102. This factors as (x + 10)(x −10). c) Rewrite x2 −1 as x2 −12. This factors as (x + 1)(x −1). 401 www.ck12.org Example 2 Factor the diﬀerence of squares: a) 16x2 −25 b) 4x2 −81 c) 49x2 −64 Solution a) Rewrite 16x2 −25 as (4x)2 −52. This factors as (4x + 5)(4x −5). b) Rewrite 4x2 −81 as (2x)2 −92. This factors as (2x + 9)(2x −9). c) Rewrite 49x2 −64 as (7x)2 −82. This factors as (7x + 8)(7x −8). Example 3 Factor the diﬀerence of squares: a) x2 −y2 b) 9x2 −4y2 c) x2y2 −1 Solution a) x2 −y2 factors as (x + y)(x −y). b) Rewrite 9x2 −4y2 as (3x)2 −(2y)2. This factors as (3x + 2y)(3x −2y). c) Rewrite x2y2 −1 as (xy)2 −12. This factors as (xy + 1)(xy −1). Example 4 Factor the diﬀerence of squares: a) x4 −25 b) 16x4 −y2 c) x2y8 −64z2 Solution a) Rewrite x4 −25 as (x2)2 −52. This factors as (x2 + 5)(x2 −5). b) Rewrite 16x4 −y2 as (4x2)2 −y2. This factors as (4x2 + y)(4x2 −y). c) Rewrite x2y4 −64z2 as (xy2)2 −(8z)2. This factors as (xy2 + 8z)(xy2 −8z). Factor Perfect Square Trinomials We use the square of a binomial formula to factor perfect square trinomials. A perfect square trinomial has the form a2 + 2ab + b2 or a2 −2ab + b2. In these special kinds of trinomials, the ﬁrst and last terms are perfect squares and the middle term is twice the product of the square roots of the ﬁrst and last terms. In a case like this, the polynomial factors into perfect squares: a2 + 2ab + b2 = (a + b)2 a2 −2ab + b2 = (a −b)2 Once again, the key is ﬁguring out what the a and b terms are. www.ck12.org 402 Example 5 Factor the following perfect square trinomials: a) x2 + 8x + 16 b) x2 −4x + 4 c) x2 + 14x + 49 Solution a) The ﬁrst step is to recognize that this expression is a perfect square trinomial. First, we can see that the ﬁrst term and the last term are perfect squares. We can rewrite x2 + 8x + 16 as x2 + 8x + 42. Next, we check that the middle term is twice the product of the square roots of the ﬁrst and the last terms. This is true also since we can rewrite x2 + 8x + 16 as x2 + 2 · 4 · x + 42. This means we can factor x2 + 8x + 16 as (x + 4)2. We can check to see if this is correct by multiplying (x + 4)2 = (x + 4)(x + 4) : x + 4 x + 4 4x + 16 x2 + 4x x2 + 8x + 16 The answer checks out. Note: We could factor this trinomial without recognizing it as a perfect square. We know that a trinomial factors as a product of two binomials: (x )(x ) We need to ﬁnd two numbers that multiply to 16 and add to 8. We can write 16 as the following products: 16 = 1 · 16 and 1 + 16 = 17 16 = 2 · 8 and 2 + 8 = 10 16 = 4 · 4 and 4 + 4 = 8 These are the correct numbers So we can factor x2 + 8x + 16 as (x + 4)(x + 4), which is the same as (x + 4)2. Once again, you can factor perfect square trinomials the normal way, but recognizing them as perfect squares gives you a useful shortcut. b) Rewrite x2 + 4x + 4 as x2 + 2 · (−2) · x + (−2)2. We notice that this is a perfect square trinomial, so we can factor it as (x −2)2. c) Rewrite x2 + 14x + 49 as x2 + 2 · 7 · x + 72. We notice that this is a perfect square trinomial, so we can factor it as (x + 7)2. Example 6 Factor the following perfect square trinomials: a) 4x2 + 20x + 25 403 www.ck12.org b) 9x2 −24x + 16 c) x2 + 2xy + y2 Solution a) Rewrite 4x2 + 20x + 25 as (2x)2 + 2 · 5 · (2x) + 52. We notice that this is a perfect square trinomial and we can factor it as (2x + 5)2. b) Rewrite 9x2 −24x + 16 as (3x)2 + 2 · (−4) · (3x) + (−4)2. We notice that this is a perfect square trinomial and we can factor it as (3x −4)2. We can check to see if this is correct by multiplying (3x −4)2 = (3x −4)(3x −4): 3x −4 3x −4 −12x + 16 9x2 −12x 9x2 −24x + 16 The answer checks out. c) x2 + 2xy + y2 We notice that this is a perfect square trinomial and we can factor it as (x + y)2. For more examples of factoring perfect square trinomials, watch the videos at com/perfect-square-trinomial.html. Solve Quadratic Polynomial Equations by Factoring With the methods we’ve learned in the last two sections, we can factor many kinds of quadratic polynomials. This is very helpful when we want to solve them. Remember the process we learned earlier: 1. If necessary, rewrite the equation in standard form so that the right-hand side equals zero. 2. Factor the polynomial completely. 3. Use the zero-product rule to set each factor equal to zero. 4. Solve each equation from step 3. 5. Check your answers by substituting your solutions into the original equation We can use this process to solve quadratic polynomials using the factoring methods we just learned. Example 7 Solve the following polynomial equations. a) x2 + 7x + 6 = 0 b) x2 −8x = −12 c) x2 = 2x + 15 Solution a) Rewrite: We can skip this since the equation is in the correct form already. Factor: We can write 6 as a product of the following numbers: www.ck12.org 404 6 = 1 · 6 and 1 + 6 = 7 This is the correct choice. 6 = 2 · 3 and 2 + 3 = 5 x2 + 7x + 6 = 0 factors as (x + 1)(x + 6) = 0. Set each factor equal to zero: x + 1 = 0 or x + 6 = 0 Solve: x = −1 or x = −6 Check: Substitute each solution back into the original equation. x = −1 (−1)2 + 7(−1) + 6 = 1 −7 + 6 = 0 checks out x = −6 (−6)2 + 7(−6) + 6 = 36 −42 + 6 = 0 checks out b) Rewrite: x2 −8x = −12 is rewritten as x2 −8x + 12 = 0 Factor: We can write 12 as a product of the following numbers: 12 = 1 · 12 and 1 + 12 = 13 12 = −1 · (−12) and −1 + (−12) = −13 12 = 2 · 6 and 2 + 6 = 8 12 = −2 · (−6) and −2 + (−6) = −8 This is the correct choice. 12 = 3 · 4 and 3 + 4 = 7 12 = −3 · (−4) and −3 + (−4) = −7 x2 + 8x + 12 = 0 factors as (x −2)(x −6) = 0. Set each factor equal to zero: x −2 = 0 or x −6 = 0 Solve: x = 2 or x = 6 Check: Substitute each solution back into the original equation. x = 2 (2)2 −8(2) = 4 −16 = −12 checks out x = 6 (6)2 −8(6) = 36 −48 = −12 checks out c) Rewrite: x2 = 2x + 15 is rewritten as x2 −2x −15 = 0 Factor: We can write -15 as a product of the following numbers: 405 www.ck12.org −15 = 1 · (−15) and 1 + (−15) = −14 −15 = −1 · (15) and −1 + (15) = 14 −15 = −3 · 5 and −3 + 5 = 2 −15 = 3 · (−5) and 3 + (−5) = −2 This is the correct choice. x2 −2x −15 = 0 factors as (x + 3)(x −5) = 0 Set each factor equal to zero: x + 3 = 0 or x −5 = 0 Solve: x = −3 or x = 5 Check: Substitute each solution back into the original equation. x = −3 (−3)2 = 2(−3) + 15 ⇒9 = 9 checks out x = 5 (5)2 = 2(5) + 15 ⇒25 = 25 checks out Example 8 Solve the following polynomial equations: a) x2 −12x + 36 = 0 b) x2 −81 = 0 c) x2 + 20x + 100 = 0 Solution a) x2 −12x + 36 = 0 Rewrite: The equation is in the correct form already. Factor: Rewrite x2 −12x + 36 = 0 as x2 −2 · (−6)x + (−6)2. We recognize this as a perfect square. This factors as (x −6)2 = 0 or (x −6)(x −6) = 0 Set each factor equal to zero: x −6 = 0 or x −6 = 0 Solve: x = 6 or x = 6 Notice that for a perfect square the two solutions are the same. This is called a double root. Check: Substitute each solution back into the original equation. x = 6 62 −12(6) + 36 = 36 −72 + 36 = 0 checks out b) x2 −81 = 0 www.ck12.org 406 Rewrite: this is not necessary since the equation is in the correct form already Factor: Rewrite x2 −81 as x2 −92. We recognize this as a diﬀerence of squares. This factors as (x −9)(x + 9) = 0. Set each factor equal to zero: x −9 = 0 or x + 9 = 0 Solve: x = 9 or x = −9 Check: Substitute each solution back into the original equation. x = 9 92 −81 = 81 −81 = 0 checks out x = −9 (−9)2 −81 = 81 −81 = 0 checks out c) x2 + 20x + 100 = 0 Rewrite: this is not necessary since the equation is in the correct form already Factor: Rewrite x2 + 20x + 100 as x2 + 2 · 10 · x + 102. We recognize this as a perfect square. This factors as (x + 10)2 = 0 or (x + 10)(x + 10) = 0 Set each factor equal to zero: x + 10 = 0 or x + 10 = 0 Solve: x = −10 or x = −10 This is a double root. Check: Substitute each solution back into the original equation. x = 10 (−10)2 + 20(−10) + 100 = 100 −200 + 100 = 0 checks out Review Questions Factor the following perfect square trinomials. 1. x2 + 8x + 16 2. x2 −18x + 81 3. −x2 + 24x −144 4. x2 + 14x + 49 5. 4x2 −4x + 1 6. 25x2 + 60x + 36 7. 4x2 −12xy + 9y2 8. x4 + 22x2 + 121 Factor the following diﬀerences of squares. 407 www.ck12.org 9. x2 −4 10. x2 −36 11. −x2 + 100 12. x2 −400 13. 9x2 −4 14. 25x2 −49 15. −36x2 + 25 16. 4x2 −y2 17. 16x2 −81y2 Solve the following quadratic equations using factoring. 18. x2 −11x + 30 = 0 19. x2 + 4x = 21 20. x2 + 49 = 14x 21. x2 −64 = 0 22. x2 −24x + 144 = 0 23. 4x2 −25 = 0 24. x2 + 26x = −169 25. −x2 −16x −60 = 0 9.7 Factoring Polynomials Completely Learning Objectives • Factor out a common binomial. • Factor by grouping. • Factor a quadratic trinomial where a , 1. • Solve real world problems using polynomial equations. Introduction We say that a polynomial is factored completely when we can’t factor it any more. Here are some suggestions that you should follow to make sure that you factor completely: • Factor all common monomials ﬁrst. • Identify special products such as diﬀerence of squares or the square of a binomial. Factor according to their formulas. • If there are no special products, factor using the methods we learned in the previous sections. • Look at each factor and see if any of these can be factored further. Example 1 Factor the following polynomials completely. a) 6x2 −30x + 24 b) 2x2 −8 c) x3 + 6x2 + 9x www.ck12.org 408 Solution a) Factor out the common monomial. In this case 6 can be divided from each term: 6(x2 −5x −6) There are no special products. We factor x2 −5x + 6 as a product of two binomials: (x )(x ) The two numbers that multiply to 6 and add to -5 are -2 and -3, so: 6(x2 −5x + 6) = 6(x −2)(x −3) If we look at each factor we see that we can factor no more. The answer is 6(x −2)(x −3). b) Factor out common monomials: 2x2 −8 = 2(x2 −4) We recognize x2 −4 as a diﬀerence of squares. We factor it as (x + 2)(x −2). If we look at each factor we see that we can factor no more. The answer is 2(x + 2)(x −2). c) Factor out common monomials: x3 + 6x2 + 9x = x(x2 + 6x + 9) We recognize x2 + 6x + 9 as a perfect square and factor it as (x + 3)2. If we look at each factor we see that we can factor no more. The answer is x(x + 3)2. Example 2 Factor the following polynomials completely: a) −2x4 + 162 b) x5 −8x3 + 16x Solution a) Factor out the common monomial. In this case, factor out -2 rather than 2. (It’s always easier to factor out the negative number so that the highest degree term is positive.) −2x4 + 162 = −2(x4 −81) We recognize expression in parenthesis as a diﬀerence of squares. We factor and get: −2(x2 −9)(x2 + 9) If we look at each factor we see that the ﬁrst parenthesis is a diﬀerence of squares. We factor and get: −2(x + 3)(x −3)(x2 + 9) If we look at each factor now we see that we can factor no more. The answer is −2(x + 3)(x −3)(x2 + 9). b) Factor out the common monomial: x5 −8x3 + 14x = x(x4 −8x2 + 16) We recognize x4 −8x2 + 16 as a perfect square and we factor it as x(x2 −4)2. 409 www.ck12.org We look at each term and recognize that the term in parentheses is a diﬀerence of squares. We factor it and get ((x + 2)(x −2))2, which we can rewrite as (x + 2)2(x −2)2. If we look at each factor now we see that we can factor no more. The ﬁnal answer is x(x + 2)2(x −2)2. Factor out a Common Binomial The ﬁrst step in the factoring process is often factoring out the common monomials from a polynomial. But sometimes polynomials have common terms that are binomials. For example, consider the following expression: x(3x + 2) −5(3x + 2) Since the term (3x + 2) appears in both terms of the polynomial, we can factor it out. We write that term in front of a set of parentheses containing the terms that are left over: (3x + 2)(x −5) This expression is now completely factored. Let’s look at some more examples. Example 3 Factor out the common binomials. a) 3x(x −1) + 4(x −1) b) x(4x + 5) + (4x + 5) Solution a) 3x(x −1) + 4(x −1) has a common binomial of (x −1). When we factor out the common binomial we get (x −1)(3x + 4). b) x(4x + 5) + (4x + 5) has a common binomial of (4x + 5). When we factor out the common binomial we get (4x + 5)(x + 1). Factor by Grouping Sometimes, we can factor a polynomial containing four or more terms by factoring common monomials from groups of terms. This method is called factor by grouping. The next example illustrates how this process works. Example 4 Factor 2x + 2y + ax + ay. Solution There is no factor common to all the terms. However, the ﬁrst two terms have a common factor of 2 and the last two terms have a common factor of a. Factor 2 from the ﬁrst two terms and factor a from the last two terms: 2x + 2y + ax + ay = 2(x + y) + a(x + y) www.ck12.org 410 Now we notice that the binomial (x + y) is common to both terms. We factor the common binomial and get: (x + y)(2 + a) Example 5 Factor 3x2 + 6x + 4x + 8. Solution We factor 3 from the ﬁrst two terms and factor 4 from the last two terms: 3x(x + 2) + 4(x + 2) Now factor (x + 2) from both terms: (x + 2)(3x + 4). Now the polynomial is factored completely. Factor Quadratic Trinomials Where a 1 Factoring by grouping is a very useful method for factoring quadratic trinomials of the form ax2 + bx + c, where a , 1. A quadratic like this doesn’t factor as (x ± m)(x ± n), so it’s not as simple as looking for two numbers that multiply to c and add up to b. Instead, we also have to take into account the coeﬀicient in the ﬁrst term. To factor a quadratic polynomial where a , 1, we follow these steps: 1. We ﬁnd the product ac. 2. We look for two numbers that multiply to ac and add up to b. 3. We rewrite the middle term using the two numbers we just found. 4. We factor the expression by grouping. Let’s apply this method to the following examples. Example 6 Factor the following quadratic trinomials by grouping. a) 3x2 + 8x + 4 b) 6x2 −11x + 4 c) 5x2 −6x + 1 Solution Let’s follow the steps outlined above: a) 3x2 + 8x + 4 Step 1: ac = 3 · 4 = 12 Step 2: The number 12 can be written as a product of two numbers in any of these ways: 12 = 1 · 12 and 1 + 12 = 13 12 = 2 · 6 and 2 + 6 = 8 This is the correct choice. 12 = 3 · 4 and 3 + 4 = 7 411 www.ck12.org Step 3: Re-write the middle term: 8x = 2x + 6x, so the problem becomes: 3x2 + 8x + 4 = 3x2 + 2x + 6x + 4 Step 4: Factor an x from the ﬁrst two terms and a 2 from the last two terms: x(3x + 2) + 2(3x + 2) Now factor the common binomial (3x + 2): (3x + 2)(x + 2) This is the answer. To check if this is correct we multiply (3x + 2)(x + 2): 3x + 2 x + 2 6x + 4 3x2 + 2x 3x2 + 8x + 4 The solution checks out. b) 6x2 −11x + 4 Step 1: ac = 6 · 4 = 24 Step 2: The number 24 can be written as a product of two numbers in any of these ways: 24 = 1 · 24 and 1 + 24 = 25 24 = −1 · (−24) and −1 + (−24) = −25 24 = 2 · 12 and 2 + 12 = 14 24 = −2 · (−12) and −2 + (−12) = −14 24 = 3 · 8 and 3 + 8 = 11 24 = −3 · (−8) and −3 + (−8) = −11 This is the correct choice. 24 = 4 · 6 and 4 + 6 = 10 24 = −4 · (−6) and −4 + (−6) = −10 Step 3: Re-write the middle term: −11x = −3x −8x, so the problem becomes: 6x2 −11x + 4 = 6x2 −3x −8x + 4 Step 4: Factor by grouping: factor a 3x from the ﬁrst two terms and a -4 from the last two terms: 3x(2x −1) −4(2x −1) Now factor the common binomial (2x −1): (2x −1)(3x −4) This is the answer. c) 5x2 −6x + 1 www.ck12.org 412 Step 1: ac = 5 · 1 = 5 Step 2: The number 5 can be written as a product of two numbers in any of these ways: 5 = 1 · 5 and 1 + 5 = 6 5 = −1 · (−5) and −1 + (−5) = −6 This is the correct choice. Step 3: Re-write the middle term: −6x = −x −5x, so the problem becomes: 5x2 −6x + 1 = 5x2 −x −5x + 1 Step 4: Factor by grouping: factor an x from the ﬁrst two terms and a −1 from the last two terms: x(5x −1) −1(5x −1) Now factor the common binomial (5x −1): (5x −1)(x −1) This is the answer. Solve Real-World Problems Using Polynomial Equations Now that we know most of the factoring strategies for quadratic polynomials, we can apply these methods to solving real world problems. Example 7 One leg of a right triangle is 3 feet longer than the other leg. The hypotenuse is 15 feet. Find the dimensions of the triangle. Solution Let x = the length of the short leg of the triangle; then the other leg will measure x + 3. Use the Pythagorean Theorem: a2 + b2 = c2, where a and b are the lengths of the legs and c is the length of the hypotenuse. When we substitute the values from the diagram, we get x2 + (x + 3)2 = 152. In order to solve this equation, we need to get the polynomial in standard form. We must ﬁrst distribute, collect like terms and rewrite in the form “polynomial = 0.” x2 + x2 + 6x + 9 = 225 2x2 + 6x + 9 = 225 2x2 + 6x −216 = 0 Factor out the common monomial: 2(x2 + 3x −108) = 0 To factor the trinomial inside the parentheses, we need two numbers that multiply to -108 and add to 3. It would take a long time to go through all the options, so let’s start by trying some of the bigger factors: 413 www.ck12.org −108 = −12 · 9 and −12 + 9 = −3 −108 = 12 · (−9) and 12 + (−9) = 3 This is the correct choice. We factor the expression as 2(x −9)(x + 12) = 0. Set each term equal to zero and solve: x −9 = 0 x + 12 = 0 or x = 9 x = −12 It makes no sense to have a negative answer for the length of a side of the triangle, so the answer must be x = 9. That means the short leg is 9 feet and the long leg is 12 feet. Check: 92 + 122 = 81 + 144 = 225 = 152, so the answer checks. Example 8 The product of two positive numbers is 60. Find the two numbers if one numbers is 4 more than the other. Solution Let x = one of the numbers; then x + 4 is the other number. The product of these two numbers is 60, so we can write the equation x(x + 4) = 60. In order to solve we must write the polynomial in standard form. Distribute, collect like terms and rewrite: x2 + 4x = 60 x2 + 4x −60 = 0 Factor by ﬁnding two numbers that multiply to -60 and add to 4. List some numbers that multiply to -60: −60 = −4 · 15 and −4 + 15 = 11 −60 = 4 · (−15) and 4 + (−15) = −11 −60 = −5 · 12 and −5 + 12 = 7 −60 = 5 · (−12) and 5 + (−12) = −7 −60 = −6 · 10 and −6 + 10 = 4 This is the correct choice. −60 = 6 · (−10) and 6 + (−10) = −4 The expression factors as (x + 10)(x −6) = 0. Set each term equal to zero and solve: x + 10 = 0 x −6 = 0 or x = −10 x = 6 Since we are looking for positive numbers, the answer must be x = 6. One number is 6, and the other number is 10. www.ck12.org 414 Check: 6 · 10 = 60, so the answer checks. Example 9 A rectangle has sides of length x + 5 and x −3. What is x if the area of the rectangle is 48? Solution Make a sketch of this situation: Using the formula Area = length × width, we have (x + 5)(x −3) = 48. In order to solve, we must write the polynomial in standard form. Distribute, collect like terms and rewrite: x2 + 2x −15 = 48 x2 + 2x −63 = 0 Factor by ﬁnding two numbers that multiply to -63 and add to 2. List some numbers that multiply to -63: −63 = −7 · 9 and −7 + 9 = 2 This is the correct choice. −63 = 7 · (−9) and 7 + (−9) = −2 The expression factors as (x + 9)(x −7) = 0. Set each term equal to zero and solve: x + 9 = 0 x −7 = 0 or x = −9 x = 7 Since we are looking for positive numbers the answer must be x = 7. So the width is x −3 = 4 and the length is x + 5 = 12. Check: 4 · 12 = 48, so the answer checks. Resources The WTAMU Virtual Math Lab has a detailed page on factoring polynomials here: edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut7_factor.htm. This page contains many videos showing example problems being solved. Review Questions Factor completely. 415 www.ck12.org 1. 2x2 + 16x + 30 2. 5x2 −70x + 245 3. −x3 + 17x2 −70x 4. 2x4 −512 5. 25x4 −20x3 + 4x2 6. 12x3 + 12x2 + 3x Factor by grouping. 7. 6x2 −9x + 10x −15 8. 5x2 −35x + x −7 9. 9x2 −9x −x + 1 10. 4x2 + 32x −5x −40 11. 2a2 −6ab + 3ab −9b2 12. 5x2 + 15x −2xy −6y Factor the following quadratic trinomials by grouping. 13. 4x2 + 25x −21 14. 6x2 + 7x + 1 15. 4x2 + 8x −5 16. 3x2 + 16x + 21 17. 6x2 −2x −4 18. 8x2 −14x −15 Solve the following application problems: 19. One leg of a right triangle is 7 feet longer than the other leg. The hypotenuse is 13. Find the dimensions of the right triangle. 20. A rectangle has sides of x + 2 and x −1. What value of x gives an area of 108? 21. The product of two positive numbers is 120. Find the two numbers if one numbers is 7 more than the other. 22. A rectangle has a 50-foot diagonal. What are the dimensions of the rectangle if it is 34 feet longer than it is wide? 23. Two positive numbers have a sum of 8, and their product is equal to the larger number plus 10. What are the numbers? 24. Two positive numbers have a sum of 8, and their product is equal to the smaller number plus 10. What are the numbers? 25. Framing Warehouse oﬀers a picture framing service. The cost for framing a picture is made up of two parts: glass costs $1 per square foot and the frame costs $2 per foot. If the frame has to be a square, what size picture can you get framed for $20? www.ck12.org 416 Chapter 10 Quadratic Equations and Quadratic Functions 10.1 Graphs of Quadratic Functions Learning Objectives • Graph quadratic functions. • Compare graphs of quadratic functions. • Graph quadratic functions in intercept form. • Analyze graphs of real-world quadratic functions. Introduction The graphs of quadratic functions are curved lines called parabolas. You don’t have to look hard to ﬁnd parabolic shapes around you. Here are a few examples: • The path that a ball or a rocket takes through the air. • Water ﬂowing out of a drinking fountain. • The shape of a satellite dish. • The shape of the mirror in car headlights or a ﬂashlight. • The cables in a suspension bridge. Graph Quadratic Functions Let’s see what a parabola looks like by graphing the simplest quadratic function, y = x2. We’ll graph this function by making a table of values. Since the graph will be curved, we need to plot a fair number of points to make it accurate. Table 10.1: x y = x2 −3 (−3)2 = 9 417 www.ck12.org Table 10.1: (continued) x y = x2 –2 (−2)2 = 4 –1 (−1)2 = 1 0 (0)2 = 0 1 (1)2 = 1 2 (2)2 = 4 3 (3)2 = 9 Here are the points plotted on a coordinate graph: To draw the parabola, draw a smooth curve through all the points. (Do not connect the points with straight lines). Let’s graph a few more examples. Example 1 Graph the following parabolas. a) y = 2x2 + 4x + 1 b) y = −x2 + 3 c) y = x2 −8x + 3 Solution a) y = 2x2 + 4x + 1 Make a table of values: www.ck12.org 418 Table 10.2: x y = 2x2 + 4x + 1 −3 2(−3)2 + 4(−3) + 1 = 7 –2 2(−2)2 + 4(−2) + 1 = 1 –1 2(−1)2 + 4(−1) + 1 = −1 0 2(0)2 + 4(0) + 1 = 1 1 2(1)2 + 4(1) + 1 = 7 2 2(2)2 + 4(2) + 1 = 17 3 2(3)2 + 4(3) + 1 = 31 Notice that the last two points have very large y−values. Since we don’t want to make our y−scale too big, we’ll just skip graphing those two points. But we’ll plot the remaining points and join them with a smooth curve. b) y = −x2 + 3 Make a table of values: Table 10.3: x y = −x2 + 3 −3 −(−3)2 + 3 = −6 –2 −(−2)2 + 3 = −1 –1 −(−1)2 + 3 = 2 0 −(0)2 + 3 = 3 1 −(1)2 + 3 = 2 2 −(2)2 + 3 = −1 3 −(3)2 + 3 = −6 Plot the points and join them with a smooth curve. 419 www.ck12.org Notice that this time we get an “upside down” parabola. That’s because our equation has a negative sign in front of the x2 term. The sign of the coeﬀicient of the x2 term determines whether the parabola turns up or down: the parabola turns up if it’s positive and down if it’s negative. c) y = x2 −8x + 3 Make a table of values: Table 10.4: x y = x2 −8x + 3 −3 (−3)2 −8(−3) + 3 = 36 –2 (−2)2 −8(−2) + 3 = 23 –1 (−1)2 −8(−1) + 3 = 12 0 (0)2 −8(0) + 3 = 3 1 (1)2 −8(1) + 3 = −4 2 (2)2 −8(2) + 3 = −9 3 (3)2 −8(3) + 3 = −12 Let’s not graph the ﬁrst two points in the table since the values are so big. Plot the remaining points and join them with a smooth curve. Wait—this doesn’t look like a parabola. What’s going on here? Maybe if we graph more points, the curve will look more familiar. For negative values of x it looks like the values of y are just getting bigger and bigger, so let’s pick more positive values of x beyond x = 3. www.ck12.org 420 Table 10.5: x y = x2 −8x + 3 −1 (−1)2 −8(−1) + 3 = 12 0 (0)2 −8(0) + 3 = 3 1 (1)2 −8(1) + 3 = −4 2 (2)2 −8(2) + 3 = −9 3 (3)2 −8(3) + 3 = −12 4 (4)2 −8(4) + 3 = −13 5 (5)2 −8(5) + 3 = −12 6 (6)2 −8(6) + 3 = −9 7 (7)2 −8(7) + 3 = −4 8 (8)2 −8(8) + 3 = 3 Plot the points again and join them with a smooth curve. Now we can see the familiar parabolic shape. And now we can see the drawback to graphing quadratics by making a table of values—if we don’t pick the right values, we won’t get to see the important parts of the graph. In the next couple of lessons, we’ll ﬁnd out how to graph quadratic equations more eﬀiciently—but ﬁrst we need to learn more about the properties of parabolas. Compare Graphs of Quadratic Functions The general form (or standard form) of a quadratic function is: y = ax2 + bx + c Here a, b and c are the coeﬀicients. Remember, a coeﬀicient is just a number (a constant term) that can go before a variable or appear alone. Although the graph of a quadratic equation in standard form is always a parabola, the shape of the parabola depends on the values of the coeﬀicients a, b and c. Let’s explore some of the ways the coeﬀicients can aﬀect the graph. Dilation Changing the value of a makes the graph “fatter” or “skinnier”. Let’s look at how graphs compare for 421 www.ck12.org diﬀerent positive values of a. The plot on the left shows the graphs of y = x2 and y = 3x2. The plot on the right shows the graphs of y = x2 and y = 1 3 x2. Notice that the larger the value of a is, the skinnier the graph is – for example, in the ﬁrst plot, the graph of y = 3x2 is skinnier than the graph of y = x2. Also, the smaller a is, the fatter the graph is – for example, in the second plot, the graph of y = 1 3 x2 is fatter than the graph of y = x2. This might seem counterintuitive, but if you think about it, it should make sense. Let’s look at a table of values of these graphs and see if we can explain why this happens. Table 10.6: x y = x2 y = 3x2 y = 1 3 x2 −3 (−3)2 = 9 3(−3)2 = 27 (−3)2 3 = 3 –2 (−2)2 = 4 3(−2)2 = 12 (−2)2 3 = 4 3 –1 (−1)2 = 1 3(−1)2 = 3 (−1)2 3 = 1 3 0 (0)2 = 0 3(0)2 = 0 (0)2 3 = 0 1 (1)2 = 1 3(1)2 = 3 (1)2 3 = 1 3 2 (2)2 = 4 3(2)2 = 12 (2)2 3 = 4 3 3 (3)2 = 9 3(3)2 = 27 (3)2 3 = 3 From the table, you can see that the values of y = 3x2 are bigger than the values of y = x2. This is because each value of y gets multiplied by 3. As a result the parabola will be skinnier because it grows three times faster than y = x2. On the other hand, you can see that the values of y = 1 3 x2 are smaller than the values of y = x2, because each value of y gets divided by 3. As a result the parabola will be fatter because it grows at one third the rate of y = x2. Orientation As the value of a gets smaller and smaller, then, the parabola gets wider and ﬂatter. What happens when a gets all the way down to zero? What happens when it’s negative? Well, when a = 0, the x2 term drops out of the equation entirely, so the equation becomes linear and the graph is just a straight line. For example, we just saw what happens to y = ax2 when we change the value of a; if we tried to graph y = 0x2, we would just be graphing y = 0, which would be a horizontal line. So as a gets smaller and smaller, the graph of y = ax2 gets ﬂattened all the way out into a horizontal line. Then, when a becomes negative, the graph of y = ax2 starts to curve again, only it curves downward instead of upward. This ﬁts with what you’ve already learned: the graph opens upward if a is positive and downward if a is negative. www.ck12.org 422 For example, here are the graphs of y = x2 and y = −x2. You can see that the parabola has the same shape in both graphs, but the graph of y = x2 is right-side-up and the graph of y = −x2 is upside-down. Vertical Shift Changing the constant c just shifts the parabola up or down. The following plot shows the graphs of y = x2, y = x2 + 1, y = x2 −1, y = x2 + 2, and y = x2 −2. You can see that when c is positive, the graph shifts up, and when c is negative the graph shifts down; in either case, it shifts by \|c\| units. In one of the later sections we’ll learn about horizontal shift (i.e. moving to the right or to the left). Before we can do that, though, we need to learn how to rewrite quadratic equations in diﬀerent forms. Meanwhile, if you want to explore further what happens when you change the coeﬀicients of a quadratic equation, the page at has an applet you can use. Click on the “Click here to start” button in section A, and then use the sliders to change the values of a, b, and c. Graph Quadratic Functions in Intercept Form Now it’s time to learn how to graph a parabola without having to use a table with a large number of points. Let’s look at the graph of y = x2 −6x + 8. 423 www.ck12.org There are several things we can notice: • The parabola crosses the x−axis at two points: x = 2 and x = 4. These points are called the x− intercepts of the parabola. • The lowest point of the parabola occurs at (3, -1). – This point is called the vertex of the parabola. – The vertex is the lowest point in any parabola that turns upward, or the highest point in any parabola that turns downward. – The vertex is exactly halfway between the two x−intercepts. This will always be the case, and you can ﬁnd the vertex using that property. • The parabola is symmetric. If you draw a vertical line through the vertex, you see that the two halves of the parabola are mirror images of each other. This vertical line is called the line of symmetry. We said that the general form of a quadratic function is y = ax2 + bx + c. When we can factor a quadratic expression, we can rewrite the function in intercept form: y = a(x −m)(x −n) This form is very useful because it makes it easy for us to ﬁnd the x−intercepts and the vertex of the parabola. The x−intercepts are the values of x where the graph crosses the x−axis; in other words, they are the values of x when y = 0. To ﬁnd the x−intercepts from the quadratic function, we set y = 0 and solve: 0 = a(x −m)(x −n) Since the equation is already factored, we use the zero-product property to set each factor equal to zero and solve the individual linear equations: x −m = 0 x −n = 0 or x = m x = n www.ck12.org 424 So the x−intercepts are at points (m, 0) and (n, 0). Once we ﬁnd the x−intercepts, it’s simple to ﬁnd the vertex. The x−value of the vertex is halfway between the two x−intercepts, so we can ﬁnd it by taking the average of the two values: m+n 2 . Then we can ﬁnd the y−value by plugging the value of x back into the equation of the function. Example 2 Find the x−intercepts and the vertex of the following quadratic functions: a) y = x2 −8x + 15 b) y = 3x2 + 6x −24 Solution a) y = x2 −8x + 15 Write the quadratic function in intercept form by factoring the right hand side of the equation. Remember, to factor we need two numbers whose product is 15 and whose sum is –8. These numbers are –5 and –3. The function in intercept form is y = (x −5)(x −3) We ﬁnd the x−intercepts by setting y = 0. We have: 0 = (x −5)(x −3) x −5 = 0 x −3 = 0 or x = 5 x = 3 So the x−intercepts are (5, 0) and (3, 0). The vertex is halfway between the two x−intercepts. We ﬁnd the x−value by taking the average of the two x−intercepts: x = 5+3 2 = 4 We ﬁnd the y−value by plugging the x−value we just found into the original equation: y = x2 −8x + 15 ⇒y = 42 −8(4) + 15 = 16 −32 + 15 = −1 So the vertex is (4, -1). b) y = 3x2 + 6x −24 Re-write the function in intercept form. Factor the common term of 3 ﬁrst: y = 3(x2 + 2x −8) Then factor completely: y = 3(x + 4)(x −2) Set y = 0 and solve: x + 4 = 0 x −2 = 0 0 = 3(x + 4)(x −2) ⇒ or x = −4 x = 2 The x−intercepts are (-4, 0) and (2, 0). For the vertex, x = −4+2 2 = −1 and y = 3(−1)2 + 6(−1) −24 = 3 −6 −24 = −27 425 www.ck12.org The vertex is: (-1, -27) Knowing the vertex and x−intercepts is a useful ﬁrst step toward being able to graph quadratic functions more easily. Knowing the vertex tells us where the middle of the parabola is. When making a table of values, we can make sure to pick the vertex as a point in the table. Then we choose just a few smaller and larger values of x. In this way, we get an accurate graph of the quadratic function without having to have too many points in our table. Example 3 Find the x−intercepts and vertex. Use these points to create a table of values and graph each function. a) y = x2 −4 b) y = −x2 + 14x −48 Solution a) y = x2 −4 Let’s ﬁnd the x−intercepts and the vertex: Factor the right-hand side of the function to put the equation in intercept form: y = (x −2)(x + 2) Set y = 0 and solve: 0 = (x −2)(x + 2) x −2 = 0 x + 2 = 0 or x = 2 x = −2 The x−intercepts are (2, 0) and (-2, 0). Find the vertex: x = 2 −2 2 = 0 y = (0)2 −4 = −4 The vertex is (0, -4). Make a table of values using the vertex as the middle point. Pick a few values of x smaller and larger than x = 0. Include the x−intercepts in the table. Table 10.7: x y = x2 −4 −3 y = (−3)2 −4 = 5 –2 y = (−2)2 −4 = 0 x−intercept –1 y = (−1)2 −4 = −3 0 y = (0)2 −4 = −4 vertex 1 y = (1)2 −4 = −3 2 y = (2)2 −4 = 0 x−intercept 3 y = (3)2 −4 = 5 www.ck12.org 426 Then plot the graph: b) y = −x2 + 14x −48 Let’s ﬁnd the x−intercepts and the vertex: Factor the right-hand-side of the function to put the equation in intercept form: y = −(x2 −14x + 48) = −(x −6)(x −8) Set y = 0 and solve: 0 = −(x −6)(x −8) x −6 = 0 x −8 = 0 or x = 6 x = 8 The x−intercepts are (6, 0) and (8, 0). Find the vertex: x = 6 + 8 2 = 7 y = −(7)2 + 14(7) −48 = 1 The vertex is (7, 1). Make a table of values using the vertex as the middle point. Pick a few values of x smaller and larger than x = 7. Include the x−intercepts in the table. Table 10.8: x y = −x2 + 14x −48 4 y = −(4)2 + 14(4) −48 = −8 5 y = −(5)2 + 14(5) −48 = −3 6 y = −(6)2 + 14(6) −48 = 0 7 y = −(7)2 + 14(7) −48 = 1 8 y = −(8)2 + 14(8) −48 = 0 9 y = −(9)2 + 14(9) −48 = −3 10 y = −(10)2 + 14(10) −48 = −8 427 www.ck12.org Then plot the graph: Analyze Graphs of Real-World Quadratic Functions. As we mentioned at the beginning of this section, parabolic curves are common in real-world applications. Here we will look at a few graphs that represent some examples of real-life application of quadratic functions. Example 4 Andrew has 100 feet of fence to enclose a rectangular tomato patch. What should the dimensions of the rectangle be in order for the rectangle to have the greatest possible area? Solution Drawing a picture will help us ﬁnd an equation to describe this situation: If the length of the rectangle is x, then the width is 50 −x. (The length and the width add up to 50, not 100, because two lengths and two widths together add up to 100.) If we let y be the area of the triangle, then we know that the area is length × width, so y = x(50−x) = 50x−x2. Here’s the graph of that function, so we can see how the area of the rectangle depends on the length of the rectangle: www.ck12.org 428 We can see from the graph that the highest value of the area occurs when the length of the rectangle is 25. The area of the rectangle for this side length equals 625. (Notice that the width is also 25, which makes the shape a square with side length 25.) This is an example of an optimization problem. These problems show up often in the real world, and if you ever study calculus, you’ll learn how to solve them without graphs. Example 5 Anne is playing golf. On the 4th tee, she hits a slow shot down the level fairway. The ball follows a parabolic path described by the equation y = x −0.04x2, where y is the ball’s height in the air and x is the horizontal distance it has traveled from the tee. The distances are measured in feet. How far from the tee does the ball hit the ground? At what distance from the tee does the ball attain its maximum height? What is the maximum height? Solution Let’s graph the equation of the path of the ball: x(1 −0.04x) = 0 has solutions x = 0 and x = 25. From the graph, we see that the ball hits the ground 25 feet from the tee. (The other x−intercept, x = 0, tells us that the ball was also on the ground when it was on the tee!) We can also see that the ball reaches its maximum height of about 6.25 feet when it is 12.5 feet from the tee. 429 www.ck12.org Review Questions Rewrite the following functions in intercept form. Find the x−intercepts and the vertex. 1. y = x2 −2x −8 2. y = −x2 + 10x −21 3. y = 2x2 + 6x + 4 4. y = 3(x + 5)(x −2) Does the graph of the parabola turn up or down? 5. y = −2x2 −2x −3 6. y = 3x2 7. y = 16 −4x2 8. y = 3x2 −2x −4x2 + 3 The vertex of which parabola is higher? 9. y = x2 + 4 or y = x2 + 1 10. y = −2x2 or y = −2x2 −2 11. y = 3x2 −3 or y = 3x2 −6 12. y = 5 −2x2 or y = 8 −2x2 Which parabola is wider? 13. y = x2 or y = 4x2 14. y = 2x2 + 4 or y = 1 2 x2 + 4 15. y = −2x2 −2 or y = −x2 −2 16. y = x2 + 3x2 or y = x2 + 3 Graph the following functions by making a table of values. Use the vertex and x−intercepts to help you pick values for the table. 17. y = 4x2 −4 18. y = −x2 + x + 12 19. y = 2x2 + 10x + 8 20. y = 1 2 x2 −2x 21. y = x −2x2 22. y = 4x2 −8x + 4 23. Nadia is throwing a ball to Peter. Peter does not catch the ball and it hits the ground. The graph shows the path of the ball as it ﬂies through the air. The equation that describes the path of the ball is y = 4 + 2x −0.16x2. Here y is the height of the ball and x is the horizontal distance from Nadia. Both distances are measured in feet. (a) How far from Nadia does the ball hit the ground? (b) At what distance x from Nadia, does the ball attain its maximum height? (c) What is the maximum height? 24. Jasreel wants to enclose a vegetable patch with 120 feet of fencing. He wants to put the vegetable against an existing wall, so he only needs fence for three of the sides. The equation for the area is given by A = 120x −x2. From the graph, ﬁnd what dimensions of the rectangle would give him the greatest area. www.ck12.org 430 10.2 Quadratic Equations by Graphing Learning Objectives • Identify the number of solutions of a quadratic equation. • Solve quadratic equations by graphing. • Analyze quadratic functions using a graphing calculator. • Solve real-world problems by graphing quadratic functions. Introduction Solving a quadratic equation means ﬁnding the x−values that will make the quadratic function equal zero; in other words, it means ﬁnding the points where the graph of the function crosses the x−axis. The solutions to a quadratic equation are also called the roots or zeros of the function, and in this section we’ll learn how to ﬁnd them by graphing the function. Identify the Number of Solutions of a Quadratic Equation Three diﬀerent situations can occur when graphing a quadratic function: Case 1: The parabola crosses the x−axis at two points. An example of this is y = x2 + x −6: Looking at the graph, we see that the parabola crosses the x−axis at x = −3 and x = 2. We can also ﬁnd the solutions to the equation x2 + x −6 = 0 by setting y = 0. We solve the equation by factoring: (x + 3)(x −2) = 0, so x = −3 or x = 2. When the graph of a quadratic function crosses the x−axis at two points, we get two distinct solutions to the quadratic equation. Case 2: The parabola touches the x−axis at one point. An example of this is y = x2 −2x + 1: 431 www.ck12.org We can see that the graph touches the x−axis at x = 1. We can also solve this equation by factoring. If we set y = 0 and factor, we obtain (x −1)2 = 0, so x = 1. Since the quadratic function is a perfect square, we get only one solution for the equation—it’s just the same solution repeated twice over. When the graph of a quadratic function touches the x−axis at one point, the quadratic equation has one solution and the solution is called a double root. Case 3: The parabola does not cross or touch the x−axis. An example of this is y = x2 + 4: If we set y = 0 we get x2 + 4 = 0. This quadratic polynomial does not factor. When the graph of a quadratic function does not cross or touch the x−axis, the quadratic equation has no real solutions. Solve Quadratic Equations by Graphing So far we’ve found the solutions to quadratic equations using factoring. However, in real life very few functions factor easily. As you just saw, graphing a function gives a lot of information about the solutions. We can ﬁnd exact or approximate solutions to a quadratic equation by graphing the function associated with it. Example 1 Find the solutions to the following quadratic equations by graphing. a) −x2 + 3 = 0 www.ck12.org 432 b) 2x2 + 5x −7 = 0 c) −x2 + x −3 = 0 d) y = −x2 + 4x −4 Solution Since we can’t factor any of these equations, we won’t be able to graph them using intercept form (if we could, we wouldn’t need to use the graphs to ﬁnd the intercepts!) We’ll just have to make a table of arbitrary values to graph each one. a) Table 10.9: x y = −x2 + 3 −3 y = −(−3)2 + 3 = −6 –2 y = −(−2)2 + 3 = −1 –1 y = −(−1)2 + 3 = 2 0 y = −(0)2 + 3 = 3 1 y = −(1)2 + 3 = 2 2 y = −(2)2 + 3 = −1 3 y = −(3)2 + 3 = −6 We plot the points and get the following graph: From the graph we can read that the x−intercepts are approximately x = 1.7 and x = −1.7. These are the solutions to the equation. b) Table 10.10: x y = 2x2 + 5x −7 −5 y = 2(−5)2 + 5(−5) −7 = 18 –4 y = 2(−4)2 + 5(−4) −7 = 5 –3 y = 2(−3)2 + 5(−3) −7 = −4 –2 y = 2(−2)2 + 5(−2) −7 = −9 –1 y = 2(−1)2 + 5(−1) −7 = −10 0 y = 2(0)2 + 5(0) −7 = −7 433 www.ck12.org Table 10.10: (continued) x y = 2x2 + 5x −7 1 y = 2(1)2 + 5(1) −7 = 0 2 y = 2(2)2 + 5(2) −7 = 11 3 y = 2(3)2 + 5(3) −7 = 26 We plot the points and get the following graph: From the graph we can read that the x−intercepts are x = 1 and x = −3.5. These are the solutions to the equation. c) Table 10.11: x y = −x2 + x −3 −3 y = −(−3)2 + (−3) −3 = −15 –2 y = −(−2)2 + (−2) −3 = −9 –1 y = −(−1)2 + (−1) −3 = −5 0 y = −(0)2 + (0) −3 = −3 1 y = −(1)2 + (1) −3 = −3 2 y = −(2)2 + (2) −3 = −5 3 y = −(3)2 + (3) −3 = −9 We plot the points and get the following graph: www.ck12.org 434 The graph curves up toward the x−axis and then back down without ever reaching it. This means that the graph never intercepts the x−axis, and so the corresponding equation has no real solutions. d) Table 10.12: x y = −x2 + 4x −4 −3 y = −(−3)2 + 4(−3) −4 = −25 –2 y = −(−2)2 + 4(−2) −4 = −16 –1 y = −(−1)2 + 4(−1) −4 = −9 0 y = −(0)2 + 4(0) −4 = −4 1 y = −(1)2 + 4(1) −4 = −1 2 y = −(2)2 + 4(2) −4 = 0 3 y = −(3)2 + 4(3) −4 = −1 4 y = −(4)2 + 4(4) −4 = −4 5 y = −(5)2 + 4(5) −4 = −9 Here is the graph of this function: The graph just touches the x−axis at x = 2, so the function has a double root there. x = 2 is the only solution to the equation. 435 www.ck12.org Analyze Quadratic Functions Using a Graphing Calculator A graphing calculator is very useful for graphing quadratic functions. Once the function is graphed, we can use the calculator to ﬁnd important information such as the roots or the vertex of the function. Example 2 Use a graphing calculator to analyze the graph of y = x2 −20x + 35. Solution 1. Graph the function. Press the [Y=] button and enter “x2 −20x + 35” next to [Y1 =]. Press the [GRAPH] button. This is the plot you should see: If this is not what you see, press the [WINDOW] button to change the window size. For the graph shown here, the x−values should range from -10 to 30 and the y−values from -80 to 50. 2. Find the roots. There are at least three ways to ﬁnd the roots: Use [TRACE] to scroll over the x−intercepts. The approximate value of the roots will be shown on the screen. You can improve your estimate by zooming in. OR Use [TABLE] and scroll through the values until you ﬁnd values of y equal to zero. You can change the accuracy of the solution by setting the step size with the [TBLSET] function. OR Use [2nd] [TRACE] (i.e. ‘calc’ button) and use option ‘zero’. Move the cursor to the left of one of the roots and press [ENTER]. Move the cursor to the right of the same root and press [ENTER]. Move the cursor close to the root and press [ENTER]. The screen will show the value of the root. Repeat the procedure for the other root. Whichever technique you use, you should get about x = 1.9 and x = 18 for the two roots. 3. Find the vertex. There are three ways to ﬁnd the vertex: Use [TRACE] to scroll over the highest or lowest point on the graph. The approximate value of the roots www.ck12.org 436 will be shown on the screen. OR Use [TABLE] and scroll through the values until you ﬁnd values the lowest or highest value of y. You can change the accuracy of the solution by setting the step size with the [TBLSET] function. OR Use [2nd] [TRACE] and use the option ‘maximum’ if the vertex is a maximum or ‘minimum’ if the vertex is a minimum. Move the cursor to the left of the vertex and press [ENTER]. Move the cursor to the right of the vertex and press [ENTER]. Move the cursor close to the vertex and press [ENTER]. The screen will show the x−and y−values of the vertex. Whichever method you use, you should ﬁnd that the vertex is at (10, -65). Solve Real-World Problems by Graphing Quadratic Functions Here’s a real-world problem we can solve using the graphing methods we’ve learned. Example 3 Andrew is an avid archer. He launches an arrow that takes a parabolic path. The equation of the height of the ball with respect to time is y = −4.9t2 + 48t, where y is the height of the arrow in meters and t is the time in seconds since Andrew shot the arrow. Find how long it takes the arrow to come back to the ground. Solution Let’s graph the equation by making a table of values. Table 10.13: t y = −4.9t2 + 48t 0 y = −4.9(0)2 + 48(0) = 0 1 y = −4.9(1)2 + 48(1) = 43.1 2 y = −4.9(2)2 + 48(2) = 76.4 3 y = −4.9(3)2 + 48(3) = 99.9 4 y = −4.9(4)2 + 48(4) = 113.6 5 y = −4.9(5)2 + 48(5) = 117.6 6 y = −4.9(6)2 + 48(6) = 111.6 7 y = −4.9(7)2 + 48(7) = 95.9 8 y = −4.9(8)2 + 48(8) = 70.4 9 y = −4.9(9)2 + 48(9) = 35.1 10 y = −4.9(10)2 + 48(10) = −10 Here’s the graph of the function: 437 www.ck12.org The roots of the function are approximately x = 0 sec and x = 9.8 sec. The ﬁrst root tells us that the height of the arrow was 0 meters when Andrew ﬁrst shot it. The second root says that it takes approximately 9.8 seconds for the arrow to return to the ground. Further Practice Now that you’ve learned how to solve quadratic equations by graphing them, you can sharpen your skills even more by learning how to ﬁnd an equation from the graph alone. Go to the page linked in the previous section, and scroll down to section E. Read the example there to learn how to ﬁnd the equation of a quadratic function by reading oﬀa few key values from the graph; then click the “Click here to start” button to try a problem yourself. The “New graph” button will give you a new problem when you ﬁnish the ﬁrst one. Review Questions Find the solutions of the following equations by graphing. 1. x2 + 3x + 6 = 0 2. −2x2 + x + 4 = 0 3. x2 −9 = 0 4. x2 + 6x + 9 = 0 5. 10x −3x2 = 0 6. 1 2 x2 −2x + 3 = 0 Find the roots of the following quadratic functions by graphing. 7. y = −3x2 + 4x −1 8. y = 9 −4x2 9. y = x2 + 7x + 2 10. y = −x2 −10x −25 11. y = 2x2 −3x 12. y = x2 −2x + 5 Using your graphing calculator, ﬁnd the roots and the vertex of each polynomial. 13. y = x2 + 12x + 5 www.ck12.org 438 14. y = x2 + 3x + 6 15. y = −x2 −3x + 9 16. y = −x2 + 4x −12 17. y = 2x2 −4x + 8 18. y = −5x2 −3x + 2 19. Graph the equations y = 2x2 −4x + 8 and y = x2 −2x + 4 on the same screen. Find their roots and vertices. (a) What is the same about the graphs? What is diﬀerent? (b) How are the two equations related to each other? (Hint: factor them.) (c) What might be another equation with the same roots? Graph it and see. 20. Graph the equations y = x2 −2x + 2 and y = x2 −2x + 4 on the same screen. Find their roots and vertices. (a) What is the same about the graphs? What is diﬀerent? (b) How are the two equations related to each other? 21. Phillip throws a ball and it takes a parabolic path. The equation of the height of the ball with respect to time is y = −16t2 + 60t, where y is the height in feet and t is the time in seconds. Find how long it takes the ball to come back to the ground. 22. Use your graphing calculator to solve Ex. 3. You should get the same answers as we did graphing by hand, but a lot quicker! 10.3 Quadratic Equations by Square Roots Learning Objectives • Solve quadratic equations involving perfect squares. • Approximate solutions of quadratic equations. • Solve real-world problems using quadratic functions and square roots. Introduction So far you know how to solve quadratic equations by factoring. However, this method works only if a quadratic polynomial can be factored. In the real world, most quadratics can’t be factored, so now we’ll start to learn other methods we can use to solve them. In this lesson, we’ll examine equations in which we can take the square root of both sides of the equation in order to arrive at the result. Solve Quadratic Equations Involving Perfect Squares Let’s ﬁrst examine quadratic equations of the type x2 −c = 0 We can solve this equation by isolating the x2 term: x2 = c Once the x2 term is isolated we can take the square root of both sides of the equation. Remember that when we take the square root we get two answers: the positive square root and the negative square root: x = √c and x = −√c 439 www.ck12.org Often this is written as x = ± √c. Example 1 Solve the following quadratic equations: a) x2 −4 = 0 b) x2 −25 = 0 Solution a) x2 −4 = 0 Isolate the x2: x2 = 4 Take the square root of both sides: x = √ 4 and x = − √ 4 The solutions are x = 2 and x = −2. b) x2 −25 = 0 Isolate the x2: x2 = 25 Take the square root of both sides: x = √ 25 and x = − √ 25 The solutions are x = 5 and x = −5. We can also ﬁnd the solution using the square root when the x2 term is multiplied by a constant—in other words, when the equation takes the form ax2 −c = 0 We just have to isolate the x2: ax2 = b x2 = b a Then we can take the square root of both sides of the equation: x = √ b a and x = − √ b a Often this is written as: x = ± √ b a. Example 2 Solve the following quadratic equations. a) 9x2 −16 = 0 b) 81x2 −1 = 0 Solution a) 9x2 −16 = 0 Isolate the x2: 9x2 = 16 x2 = 16 9 www.ck12.org 440 Take the square root of both sides: x = √ 16 9 and x = − √ 16 9 Answer: x = 4 3 and x = −4 3 b) 81x2 −1 = 0 Isolate the x2: 81x2 = 1 x2 = 1 81 Take the square root of both sides: x = √ 1 81 and x = − √ 1 81 Answer: x = 1 9 and x = −1 9 As you’ve seen previously, some quadratic equations have no real solutions. Example 3 Solve the following quadratic equations. a) x2 + 1 = 0 b) 4x2 + 9 = 0 Solution a) x2 + 1 = 0 Isolate the x2: x2 = −1 Take the square root of both sides: x = √ −1 and x = − √ −1 Square roots of negative numbers do not give real number results, so there are no real solutions to this equation. b) 4x2 + 9 = 0 Isolate the x2: 4x2 = −9 x2 = −9 4 Take the square root of both sides: x = √ −9 4 and x = − √ −9 4 There are no real solutions. We can also use the square root function in some quadratic equations where one side of the equation is a perfect square. This is true if an equation is of this form: (x −2)2 = 9 Both sides of the equation are perfect squares. We take the square root of both sides and end up with two equations: x −2 = 3 and x −2 = −3. Solving both equations gives us x = 5 and x = −1. Example 4 Solve the following quadratic equations. 441 www.ck12.org a) (x −1)2 = 4 b) (x + 3)2 = 1 Solution a) (x −1)2 = 4 Take the square root of both sides : x −1 = 2 and x −1 = −2 Solve each equation : x = 3 and x = −1 Answer: x = 3 and x = −1 b) (x + 3)2 = 1 Take the square root of both sides : x + 3 = 1 and x + 3 = −1 Solve each equation : x = −2 and x = −4 Answer: x = −2 and x = −4 It might be necessary to factor the right-hand side of the equation as a perfect square before applying the method outlined above. Example 5 Solve the following quadratic equations. a) x2 + 8x + 16 = 25 b) 4x2 −40x + 25 = 9 Solution a) x2 + 8x + 16 = 25 Factor the right-hand-side : x2 + 8x + 16 = (x + 4)2 so (x + 4)2 = 25 Take the square root of both sides : x + 4 = 5 and x + 4 = −5 Solve each equation : x = 1 and x = −9 Answer: x = 1 and x = −9 b) 4x2 −20x + 25 = 9 Factor the right-hand-side : 4x2 −20x + 25 = (2x −5)2 so (2x −5)2 = 9 Take the square root of both sides : 2x −5 = 3 and 2x −5 = −3 Solve each equation : 2x = 8 and 2x = 2 Answer: x = 4 and x = 1 Approximate Solutions of Quadratic Equations We can use the methods we’ve learned so far in this section to ﬁnd approximate solutions to quadratic equations, when taking the square root doesn’t give an exact answer. Example 6 Solve the following quadratic equations. www.ck12.org 442 a) x2 −3 = 0 b) 2x2 −9 = 0 Solution a) Isolate the x2 : x2 = 3 Take the square root of both sides : x = √ 3 and x = − √ 3 Answer: x ≈1.73 and x ≈−1.73 b) Isolate the x2 : 2x2 = 9 so x2 = 9 2 Take the square root of both sides : x = √ 9 2 and x = − √ 9 2 Answer: x ≈2.12 and x ≈−2.12 Example 7 Solve the following quadratic equations. a) (2x + 5)2 = 10 b) x2 −2x + 1 = 5 Solution a) Take the square root of both sides : 2x + 5 = √ 10 and 2x + 5 = − √ 10 Solve both equations : x = −5 + √ 10 2 and x = −5 − √ 10 2 Answer: x ≈−0.92 and x ≈−4.08 b) Factor the right-hand-side : (x −1)2 = 5 Take the square root of both sides : x −1 = √ 5 and x −1 = − √ 5 Solve each equation : x = 1 + √ 5 and x = 1 − √ 5 Answer: x ≈3.24 and x ≈−1.24 Solve Real-World Problems Using Quadratic Functions and Square Roots Quadratic equations are needed to solve many real-world problems. In this section, we’ll examine problems about objects falling under the inﬂuence of gravity. When objects are dropped from a height, they have no initial velocity; the force that makes them move towards the ground is due to gravity. The acceleration of gravity on earth is given by the equation g = −9.8 m/s2 or g = −32 ft/s2 The negative sign indicates a downward direction. We can assume that gravity is constant for the problems we’ll be examining, because we will be staying close to the surface of the earth. The acceleration of gravity decreases as an object moves very far from the earth. It is also diﬀerent on other celestial bodies such as the moon. The equation that shows the height of an object in free fall is y = 1 2gt2 + y0 443 www.ck12.org The term y0 represents the initial height of the object, t is time, and g is the constant representing the force of gravity. You then plug in one of the two values for g above, depending on whether you want the answer in feet or meters. Thus the equation works out to y = −4.9t2 + y0 if you want the height in meters, and y = −16t2 + y0 if you want it in feet. Example 8 How long does it take a ball to fall from a roof to the ground 25 feet below? Solution Since we are given the height in feet, use equation : y = −16t2 + y0 The initial height is y0 = 25 feet, so : y = −16t2 + 25 The height when the ball hits the ground is y = 0, so : 0 = −16t2 + 25 Solve for t : 16t2 = 25 t2 = 25 16 t = 5 4 or t = −5 4 Since only positive time makes sense in this case, it takes the ball 1.25 seconds to fall to the ground. Example 9 A rock is dropped from the top of a cliﬀand strikes the ground 7.2 seconds later. How high is the cliﬀin meters? Solution Since we want the height in meters, use equation : y = −4.9t2 + y0 The time of ﬂight is t = 7.2 seconds : y = −4.9(7.2)2 + y0 The height when the ball hits the ground is y = 0, so : 0 = −4.9(7.2)2 + y0 Simplify : 0 = −254 + y0 so y0 = 254 The cliﬀis 254 meters high. Example 10 Victor throws an apple out of a window on the 10th ﬂoor which is 120 feet above ground. One second later Juan throws an orange out of a 6th ﬂoor window which is 72 feet above the ground. Which fruit reaches the ground ﬁrst, and how much faster does it get there? Solution Let’s ﬁnd the time of ﬂight for each piece of fruit. Apple: Since we have the height in feet, use this equation : y = −16t2 + y0 The initial height is y0 = 120 feet : y = −16t2 + 120 The height when the ball hits the ground is y = 0, so : 0 = −16t2 + 120 Solve for t : 16t2 = 120 t2 = 120 16 = 7.5 t = 2.74 or t = −2.74 seconds www.ck12.org 444 Orange: The initial height is y0 = 72 feet : 0 = −16t2 + 72 Solve for t : 16t2 = 72 t2 = 72 16 = 4.5 t = 2.12 or t = −2.12 seconds The orange was thrown one second later, so add 1 second to the time of the orange: t = 3.12 seconds The apple hits the ground ﬁrst. It gets there 0.38 seconds faster than the orange. Review Questions Solve the following quadratic equations. 1. x2 −1 = 0 2. x2 −100 = 0 3. x2 + 16 = 0 4. 9x2 −1 = 0 5. 4x2 −49 = 0 6. 64x2 −9 = 0 7. x2 −81 = 0 8. 25x2 −36 = 0 9. x2 + 9 = 0 10. x2 −16 = 0 11. x2 −36 = 0 12. 16x2 −49 = 0 13. (x −2)2 = 1 14. (x + 5)2 = 16 15. (2x −1)2 −4 = 0 16. (3x + 4)2 = 9 17. (x −3)2 + 25 = 0 18. x2 −6 = 0 19. x2 −20 = 0 20. 3x2 + 14 = 0 21. (x −6)2 = 5 22. (4x + 1)2 −8 = 0 23. x2 −10x + 25 = 9 24. x2 + 18x + 81 = 1 25. 4x2 −12x + 9 = 16 26. (x + 10)2 = 2 27. x2 + 14x + 49 = 3 28. 2(x + 3)2 = 8 29. Susan drops her camera in the river from a bridge that is 400 feet high. How long is it before she hears the splash? 30. It takes a rock 5.3 seconds to splash in the water when it is dropped from the top of a cliﬀ. How high is the cliﬀin meters? 31. Nisha drops a rock from the roof of a building 50 feet high. Ashaan drops a quarter from the top 445 www.ck12.org story window, 40 feet high, exactly half a second after Nisha drops the rock. Which hits the ground ﬁrst? 10.4 Solving Quadratic Equations by Completing the Square Learning Objectives • Complete the square of a quadratic expression. • Solve quadratic equations by completing the square. • Solve quadratic equations in standard form. • Graph quadratic equations in vertex form. • Solve real-world problems using functions by completing the square. Introduction You saw in the last section that if you have a quadratic equation of the form (x −2)2 = 5, you can easily solve it by taking the square root of each side: x −2 = √ 5 and x −2 = − √ 5 Simplify to get: x = 2 + √ 5 ≈4.24 and x = 2 − √ 5 ≈−0.24 So what do you do with an equation that isn’t written in this nice form? In this section, you’ll learn how to rewrite any quadratic equation in this form by completing the square. Complete the Square of a Quadratic Expression Completing the square lets you rewrite a quadratic expression so that it contains a perfect square trinomial that you can factor as the square of a binomial. Remember that the square of a binomial takes one of the following forms: (x + a)2 = x2 + 2ax + a2 (x −a)2 = x2 −2ax + a2 So in order to have a perfect square trinomial, we need two terms that are perfect squares and one term that is twice the product of the square roots of the other terms. Example 1 Complete the square for the quadratic expression x2 + 4x. Solution To complete the square we need a constant term that turns the expression into a perfect square trinomial. Since the middle term in a perfect square trinomial is always 2 times the product of the square roots of the other two terms, we re-write our expression as: www.ck12.org 446 x2 + 2(2)(x) We see that the constant we are seeking must be 22: x2 + 2(2)(x) + 22 Answer: By adding 4 to both sides, this can be factored as: (x + 2)2 Notice, though, that we just changed the value of the whole expression by adding 4 to it. If it had been an equation, we would have needed to add 4 to the other side as well to make up for this. Also, this was a relatively easy example because a, the coeﬀicient of the x2 term, was 1. When that coeﬀicient doesn’t equal 1, we have to factor it out from the whole expression before completing the square. Example 2 Complete the square for the quadratic expression 4x2 + 32x. Solution Factor the coeﬀicient of the x2 term : 4(x2 + 8x) Now complete the square of the expression in parentheses. Re-write the expression : 4(x2 + 2(4)(x)) We complete the square by adding the constant 42 : 4(x2 + 2(4)(x) + 42) Factor the perfect square trinomial inside the parenthesis : 4(x + 4)2 Answer The expression “completing the square” comes from a geometric interpretation of this situation. Let’s revisit the quadratic expression in Example 1: x2 + 4x. We can think of this expression as the sum of three areas. The ﬁrst term represents the area of a square of side x. The second expression represents the areas of two rectangles with a length of 2 and a width of x: We can combine these shapes as follows: We obtain a square that is not quite complete. To complete the square, we need to add a smaller square of side length 2. 447 www.ck12.org We end up with a square of side length (x + 2); its area is therefore (x + 2)2. Solve Quadratic Equations by Completing the Square Let’s demonstrate the method of completing the square with an example. Example 3 Solve the following quadratic equation: 3x2 −10x = −1 Solution Divide all terms by the coeﬀicient of the x2 term: x2 −10 3 x = −1 3 Rewrite: x2 −2 ( 5 3 ) (x) = −1 3In order to have a perfect square trinomial on the right-hand-side we need to add the constant ( 5 3 )2. Add this constant to both sides of the equation: x2 −2 (5 3 ) (x) + (5 3 )2 = −1 3 + (5 3 )2 Factor the perfect square trinomial and simplify: ( x −5 3 )2 = −1 3 + 25 9 ( x −5 3 )2 = 22 9 Take the square root of both sides: x −5 3 = √ 22 9 and x −5 3 = − √ 22 9 x = 5 3 + √ 22 9 ≈3.23 and x = 5 3 − √ 22 9 ≈0.1 Answer: x = 3.23 and x = 0.1 If an equation is in standard form (ax2 + bx + c = 0), we can still solve it by the method of completing the square. All we have to do is start by moving the constant term to the right-hand-side of the equation. Example 4 Solve the following quadratic equation: x2 + 15x + 12 = 0 Solution www.ck12.org 448 Move the constant to the other side of the equation: x2 + 15x = −12 Rewrite: x2 + 2 ( 15 2 ) (x) = −12 Add the constant ( 15 2 )2 to both sides of the equation: x2 + 2 (15 2 ) (x) + (15 2 )2 = −12 + (15 2 )2 Factor the perfect square trinomial and simplify: ( x + 15 2 )2 = −12 + 225 4 ( x + 15 2 )2 = 177 4 Take the square root of both sides: x + 15 2 = √ 177 4 and x + 15 2 = − √ 177 4 x = −15 2 + √ 177 4 ≈−0.85 and x = −15 2 − √ 177 4 ≈−14.15 Answer: x = −0.85 and x = −14.15 Graph Quadratic Functions in Vertex Form Probably one of the best applications of the method of completing the square is using it to rewrite a quadratic function in vertex form. The vertex form of a quadratic function is y −k = a(x −h)2 This form is very useful for graphing because it gives the vertex of the parabola explicitly. The vertex is at the point (h, k). It is also simple to ﬁnd the x−intercepts from the vertex form: just set y = 0 and take the square root of both sides of the resulting equation. To ﬁnd the y−intercept, set x = 0 and simplify. Example 5 Find the vertex, the x−intercepts and the y−intercept of the following parabolas: a) y −2 = (x −1)2 b) y + 8 = 2(x −3)2 Solution a) y −2 = (x −1)2 Vertex: (1, 2) To ﬁnd the x−intercepts, 449 www.ck12.org Set y = 0 : −2 = (x −1)2 Take the square root of both sides : √ −2 = x −1 and − √ −2 = x −1 The solutions are not real so there are no x−intercepts. To ﬁnd the y−intercept, Set x = 0 : y −2 = (−1)2 Simplify : y −2 = 1 ⇒y = 3 b) y + 8 = 2(x −3)2 Rewrite : y −(−8) = 2(x −3)2 Vertex : (3, −8) To ﬁnd the x−intercepts, Set y = 0 : 8 = 2(x −3)2 Divide both sides by 2 : 4 = (x −3)2 Take the square root of both sides : 4 = x −3 and −4 = x −3 Simplify : x = 7 and x = −1 To ﬁnd the y−intercept, Set x = 0 : y + 8 = 2(−3)2 Simplify : y + 8 = 18 ⇒y = 10 To graph a parabola, we only need to know the following information: • the vertex • the x−intercepts • the y−intercept • whether the parabola turns up or down (remember that it turns up if a > 0 and down if a < 0) Example 6 Graph the parabola given by the function y + 1 = (x + 3)2. Solution Rewrite : y −(−1) = (x −(−3))2 Vertex : (−3, −1) vertex : (−3, −1) To ﬁnd the x−intercepts, Set y = 0 : 1 = (x + 3)2 Take the square root of both sides : 1 = x + 3 and −1 = x + 3 Simplify : x = −2 and x = −4 x −intercepts : (−2, 0) and (−4, 0) www.ck12.org 450 To ﬁnd the y−intercept, Set x = 0 : y + 1 = (3)2 Simplify: y = 8 y −intercept : (0, 8) And since a > 0, the parabola turns up. Graph all the points and connect them with a smooth curve: Example 7 Graph the parabola given by the function y = −1 2(x −2)2. Solution: Rewrite y −(0) = −1 2(x −2)2 Vertex: (2, 0) vertex:(2, 0) To ﬁnd the x−intercepts, Set y = 0 : 0 = −1 2(x −2)2 Multiply both sides by −2 : 0 = (x −2)2 Take the square root of both sides : 0 = x −2 Simplify : x = 2 x −intercept:(2, 0) Note: there is only one x−intercept, indicating that the vertex is located at this point, (2, 0). To ﬁnd the y−intercept, Set x = 0 : y = −1 2(−2)2 Simplify: y = −1 2(4) ⇒y = −2 y −intercept:(0, −2) Since a < 0, the parabola turns down. 451 www.ck12.org Graph all the points and connect them with a smooth curve: Solve Real-World Problems Using Quadratic Functions by Com-pleting the Square In the last section you learned that an object that is dropped falls under the inﬂuence of gravity. The equation for its height with respect to time is given by y = 1 2gt2 + y0, where y0 represents the initial height of the object and g is the coeﬀicient of gravity on earth, which equals −9.8 m/s2 or −32 ft/s2. On the other hand, if an object is thrown straight up or straight down in the air, it has an initial vertical velocity. This term is usually represented by the notation v0y. Its value is positive if the object is thrown up in the air and is negative if the object is thrown down. The equation for the height of the object in this case is y = 1 2gt2 + v0yt + y0 Plugging in the appropriate value for g turns this equation into y = −4.9t2 + v0yt + y0 if you wish to have the height in meters y = −16t2 + v0yt + y0 if you wish to have the height in feet Example 8 An arrow is shot straight up from a height of 2 meters with a velocity of 50 m/s. a) How high will the arrow be 4 seconds after being shot? After 8 seconds? b) At what time will the arrow hit the ground again? c) What is the maximum height that the arrow will reach and at what time will that happen? Solution Since we are given the velocity in m/s, use: y = −4.9t2 + v0yt + y0 We know v0y = 50 m/s and y0 = 2 meters so: y = −4.9t2 + 50t + 2 a) To ﬁnd how high the arrow will be 4 seconds after being shot we plug in t = 4: www.ck12.org 452 y = −4.9(4)2 + 50(4) + 2 = −4.9(16) + 200 + 2 = 123.6 feet we plug in t = 8: y = −4.9(8)2 + 50(8) + 2 = −4.9(64) + 400 + 2 = 88.4 feet b) The height of the ball arrow on the ground is y = 0, so: 0 = −4.9t2 + 50t + 2 Solve for t by completing the square: −4.9t2 + 50t = −2 −4.9(t2 −10.2t) = −2 t2 −10.2t = 0.41 t2 −2(5.1)t + (5.1)2 = 0.41 + (5.1)2 (t −5.1)2 = 26.43 t −5.1 = 5.14 and t −5.1 = −5.14 t = 10.2 sec and t = −0.04 sec The arrow will hit the ground about 10.2 seconds after it is shot. c) If we graph the height of the arrow with respect to time we would get an upside down parabola (a < 0). The maximum height and the time when this occurs is really the vertex of this parabola: (t, h). We re-write the equation in vertex form: y = −4.9t2 + 50t + 2 y −2 = −4.9t2 + 50t y −2 = −4.9(t2 −10.2t) Complete the square: y −2 −4.9(5.1)2 = −4.9 ( t2 −10.2t + (5.1)2) y −129.45 = −4.9(t −5.1)2 The vertex is at (5.1, 129.45). In other words, when t = 5.1 seconds, the height is y = 129 meters. Another type of application problem that can be solved using quadratic equations is one where two objects are moving away from each other in perpendicular directions. Here is an example of this type of problem. Example 9 Two cars leave an intersection. One car travels north; the other travels east. When the car traveling north had gone 30 miles, the distance between the cars was 10 miles more than twice the distance traveled by the car heading east. Find the distance between the cars at that time. Solution Let x = the distance traveled by the car heading east Then 2x + 10 = the distance between the two cars Let’s make a sketch: 453 www.ck12.org We can use the Pythagorean Theorem to ﬁnd an equation for x: x2 + 302 = (2x + 10)2 Expand parentheses and simplify: x2 + 900 = 4x2 + 40x + 100 800 = 3x2 + 40x Solve by completing the square: 800 3 = x2 + 40 3 x 800 3 + (20 3 )2 = x2 + 2 (20 3 ) x + (20 3 )2 2800 9 = ( x + 20 3 )2 x + 20 3 = 17.6 and x + 20 3 = −17.6 x = 11 and x = −24.3 Since only positive distances make sense here, the distance between the two cars is: 2(11) + 10 = 32 miles Review Questions Complete the square for each expression. 1. x2 + 5x 2. x2 −2x 3. x2 + 3x 4. x2 −4x 5. 3x2 + 18x 6. 2x2 −22x 7. 8x2 −10x 8. 5x2 + 12x Solve each quadratic equation by completing the square. 9. x2 −4x = 5 www.ck12.org 454 10. x2 −5x = 10 11. x2 + 10x + 15 = 0 12. x2 + 15x + 20 = 0 13. 2x2 −18x = 3 14. 4x2 + 5x = −1 15. 10x2 −30x −8 = 0 16. 5x2 + 15x −40 = 0 Rewrite each quadratic function in vertex form. 17. y = x2 −6x 18. y + 1 = −2x2 −x 19. y = 9x2 + 3x −10 20. y = −32x2 + 60x + 10 For each parabola, ﬁnd the vertex; the x−and y−intercepts; and if it turns up or down. Then graph the parabola. 21. y −4 = x2 + 8x 22. y = −4x2 + 20x −24 23. y = 3x2 + 15x 24. y + 6 = −x2 + x 25. Sam throws an egg straight down from a height of 25 feet. The initial velocity of the egg is 16 ft/sec. How long does it take the egg to reach the ground? 26. Amanda and Dolvin leave their house at the same time. Amanda walks south and Dolvin bikes east. Half an hour later they are 5.5 miles away from each other and Dolvin has covered three miles more than the distance that Amanda covered. How far did Amanda walk and how far did Dolvin bike? 10.5 Solving Quadratic Equations by the Quadratic Formula Learning Objectives • Solve quadratic equations using the quadratic formula. • Identify and choose methods for solving quadratic equations. • Solve real-world problems using functions by completing the square. Introduction The Quadratic Formula is probably the most used method for solving quadratic equations. For a quadratic equation in standard form, ax2 + bx + c = 0, the quadratic formula looks like this: x = −b ± √ b2 −4ac 2a This formula is derived by solving a general quadratic equation using the method of completing the square that you learned in the previous section. 455 www.ck12.org We start with a general quadratic equation: ax2 + bx + c = 0 Subtract the constant term from both sides: ax2 + bx = −c Divide by the coeﬀicient of the x2 term: x2 + b a x = −c a Rewrite: x2 + 2 ( b 2a ) x = −c a Add the constant ( b 2a )2 to both sides: x2 + 2 ( b 2a ) x + ( b 2a )2 = −c a + b2 4a2 Factor the perfect square trinomial: ( x + b 2a )2 = −4ac 4a2 + b2 4a2 Simplify: ( x + b 2a )2 = b2 −4ac 4a2 Take the square root of both sides: x + b 2a = √ b2 −4ac 4a2 and x + b 2a = − √ b2 −4ac 4a2 Simplify: x + b 2a = √ b2 −4ac 2a and x + b 2a = − √ b2 −4ac 2a x = −b 2a + √ b2 −4ac 2a and x = −b 2a − √ b2 −4ac 2a x = −b + √ b2 −4ac 2a and x = −b − √ b2 −4ac 2a This can be written more compactly as x = −b± √ b2−4ac 2a . You can see that the familiar formula comes directly from applying the method of completing the square. Applying the method of completing the square to solve quadratic equations can be tedious, so the quadratic formula is a more straightforward way of ﬁnding the solutions. Solve Quadratic Equations Using the Quadratic Formula To use the quadratic formula, just plug in the values of a, b, and c. Example 1 Solve the following quadratic equations using the quadratic formula. a) 2x2 + 3x + 1 = 0 b) x2 −6x + 5 = 0 c) −4x2 + x + 1 = 0 Solution Start with the quadratic formula and plug in the values of a, b and c. a) www.ck12.org 456 Quadratic formula: x = −b ± √ b2 −4ac 2a Plug in the values a = 2, b = 3, c = 1 x = −3 ± √ (3)2 −4(2)(1) 2(2) Simplify: x = −3 ± √ 9 −8 4 = −3 ± √ 1 4 Separate the two options: x = −3 + 1 4 and x = −3 −1 4 Solve: x = −2 4 = −1 2 and x = −4 4 = −1 Answer: x = −1 2 and x = −1 b) Quadratic formula: x = −b ± √ b2 −4ac 2a Plug in the values a = 1, b = −6, c = 5 x = −(−6) ± √ (−6)2 −4(1)(5) 2(1) Simplify: x = 6 ± √ 36 −20 2 = 6 ± √ 16 2 Separate the two options: x = 6 + 4 2 and x = 6 −4 2 Solve: x = 10 2 = 5 and x = 2 2 = 1 Answer: x = 5 and x = 1 c) Quadratic formula: x = −b ± √ b2 −4ac 2a Plug in the values a = −4, b = 1, c = 1 x = −1 ± √ (1)2 −4(−4)(1) 2(−4) Simplify: x = −1 ± √1 + 16 −8 = −1 ± √ 17 −8 Separate the two options: x = −1 + √ 17 −8 and x = −1 − √ 17 −8 Solve: x = −.39 and x = .64 Answer: x = −.39 and x = .64 Often when we plug the values of the coeﬀicients into the quadratic formula, we end up with a negative number inside the square root. Since the square root of a negative number does not give real answers, we say that the equation has no real solutions. In more advanced math classes, you’ll learn how to work with “complex” (or “imaginary”) solutions to quadratic equations. Example 2 Use the quadratic formula to solve the equation x2 + 2x + 7 = 0. Solution 457 www.ck12.org Quadratic formula: x = −b ± √ b2 −4ac 2a Plug in the values a = 1, b = 2, c = 7 x = −2 ± √ (2)2 −4(1)(7) 2(1) Simplify: x = −2 ± √ 4 −28 2 = −2 ± √ −24 2 Answer: There are no real solutions. To apply the quadratic formula, we must make sure that the equation is written in standard form. For some problems, that means we have to start by rewriting the equation. Example 3 Solve the following equations using the quadratic formula. a) x2 −6x = 10 b) −8x2 = 5x + 6 Solution a) Re-write the equation in standard form: x2 −6x −10 = 0 Quadratic formula: x = −b ± √ b2 −4ac 2a Plug in the values a = 1, b = −6, c = −10 x = −(−6) ± √ (−6)2 −4(1)(−10) 2(1) Simplify: x = 6 ± √36 + 40 2 = 6 ± √ 76 2 Separate the two options: x = 6 + √ 76 2 and x = 6 − √ 76 2 Solve: x = 7.36 and x = −1.36 Answer: x = 7.36 and x = −1.36 b) Re-write the equation in standard form: 8x2 + 5x + 6 = 0 Quadratic formula: x = −b ± √ b2 −4ac 2a Plug in the values a = 8, b = 5, c = 6 x = −5 ± √ (5)2 −4(8)(6) 2(8) Simplify: x = −5 ± √ 25 −192 16 = −5 ± √ −167 16 Answer: no real solutions For more examples of solving quadratic equations using the quadratic formula, see the Khan Academy video at . This video isn’t necessarily diﬀerent from the examples above, but it does help reinforce the procedure of using the quadratic formula to solve equations. www.ck12.org 458 Figure 10.1: Quadratic Equation part 2 (Watch Youtube Video) Finding the Vertex of a Parabola with the Quadratic Formula Sometimes a formula gives you even more information than you were looking for. For example, the quadratic formula also gives us an easy way to locate the vertex of a parabola. Remember that the quadratic formula tells us the roots or solutions of the equation ax2 + bx + c = 0. Those roots are x = −b± √ b2−4ac 2a , and we can rewrite that as x = −b 2a ± √ b2−4ac 2a Recall that the roots are symmetric about the vertex. In the form above, we can see that the roots of a quadratic equation are symmetric around the x−coordinate −b 2a, because they are √ b2−4ac 2a units to the left and right (recall the ± sign) from the vertical line x = −b 2a. For example, in the equation x2 −2x −3 = 0, the roots -1 and 3 are both 2 units from the vertical line x = 1, as you can see in the graph below: Identify and Choose Methods for Solving Quadratic Equations. In mathematics, you’ll need to solve quadratic equations that describe application problems or that are part of more complicated problems. You’ve learned four ways of solving a quadratic equation: • Factoring • Taking the square root • Completing the square • Quadratic formula Usually you’ll have to decide for yourself which method to use. However, here are some guidelines as to which methods are better in diﬀerent situations. 459 www.ck12.org Factoring is always best if the quadratic expression is easily factorable. It is always worthwhile to check if you can factor because this is the fastest method. Many expressions are not factorable so this method is not used very often in practice. Taking the square root is best used when there is no x−term in the equation. Completing the square can be used to solve any quadratic equation. This is usually not any better than using the quadratic formula (in terms of diﬀicult computations), but it is very useful if you need to rewrite a quadratic function in vertex form. It’s also used to rewrite the equations of circles, ellipses and hyperbolas in standard form (something you’ll do in algebra II, trigonometry, physics, calculus, and beyond). Quadratic formula is the method that is used most often for solving a quadratic equation. When solving directly by taking square root and factoring does not work, this is the method that most people prefer to use. If you are using factoring or the quadratic formula, make sure that the equation is in standard form. Example 4 Solve each quadratic equation. a) x2 −4x −5 = 0 b) x2 = 8 c) −4x2 + x = 2 d) 25x2 −9 = 0 e) 3x2 = 8x Solution a) This expression if easily factorable so we can factor and apply the zero-product property: Factor: (x −5)(x + 1) = 0 Apply zero-product property: x −5 = 0 and x + 1 = 0 Solve: x = 5 and x = −1 Answer: x = 5 and x = −1 b) Since the expression is missing the x term we can take the square root: Take the square root of both sides: x = √ 8 and x = − √ 8 Answer: x = 2.83 and x = −2.83 c) Re-write the equation in standard form: −4x2 + x −2 = 0 It is not apparent right away if the expression is factorable so we will use the quadratic formula: Quadratic formula: x = −b ± √ b2 −4ac 2a Plug in the values a = −4, b = 1, c = −2 : x = −1 ± √ 12 −4(−4)(−2) 2(−4) Simplify: x = −1 ± √ 1 −32 −8 = −1 ± √ −31 −8 Answer: no real solution www.ck12.org 460 d) This problem can be solved easily either with factoring or taking the square root. Let’s take the square root in this case: Add 9 to both sides of the equation: 25x2 = 9 Divide both sides by 25 : x2 = 9 25 Take the square root of both sides: x = √ 9 25 and x = − √ 9 25 Simplify: x = 3 5 and x = −3 5 Answer: x = 3 5 and x = −3 5 e) Re-write the equation in standard form: 3x2 −8x = 0 Factor out common x term: x(3x −8) = 0 Set both terms to zero: x = 0 and 3x = 8 Solve: x = 0 and x = 8 3 = 2.67 Answer: x = 0 and x = 2.67 Solve Real-World Problems Using Quadratic Functions by any Method Here are some application problems that arise from number relationships and geometry applications. Example 5 The product of two positive consecutive integers is 156. Find the integers. Solution Deﬁne: Let x = the smaller integer Then x + 1 = the next integer Translate: The product of the two numbers is 156. We can write the equation: x(x + 1) = 156 Solve: x2 + x = 156 x2 + x −156 = 0 Apply the quadratic formula with: a = 1, b = 1, c = −156 x = −1 ± √ 12 −4(1)(−156) 2(1) x = −1 ± √ 625 2 = −1 ± 25 2 x = −1 + 25 2 and x = −1 −25 2 x = 24 2 = 12 and x = −26 2 = −13 461 www.ck12.org Since we are looking for positive integers, we want x = 12. So the numbers are 12 and 13. Check: 12 × 13 = 156. The answer checks out. Example 6 The length of a rectangular pool is 10 meters more than its width. The area of the pool is 875 square meters. Find the dimensions of the pool. Solution Draw a sketch: Deﬁne: Let x = the width of the pool Then x + 10 = the length of the pool Translate: The area of a rectangle is A = length × width, so we have x(x + 10) = 875. Solve: x2 + 10x = 875 x2 + 10x −875 = 0 Apply the quadratic formula with a = 1, b = 10 and c = −875 x = −10 ± √ (10)2 −4(1)(−875) 2(1) x = −10 ± √100 + 3500 2 x = −10 ± √ 3600 2 = −10 ± 60 2 x = −10 + 60 2 and x = −10 −60 2 x = 50 2 = 25 and x = −70 2 = −35 Since the dimensions of the pool should be positive, we want x = 25 meters. So the pool is 25 meters × 35 meters. Check: 25 × 35 = 875 m2. The answer checks out. Example 7 Suzie wants to build a garden that has three separate rectangular sections. She wants to fence around the whole garden and between each section as shown. The plot is twice as long as it is wide and the total area is 200 ft2. How much fencing does Suzie need? Solution Deﬁne: Let x = the width of the plot Then 2x = the length of the plot www.ck12.org 462 Translate: area of a rectangle is A = length × width, so x(2x) = 200 Solve: 2x2 = 200 Solve by taking the square root: x2 = 100 x = √ 100 and x = − √ 100 x = 10 and x = −10 We take x = 10 since only positive dimensions make sense. The plot of land is 10 feet × 20 feet. To fence the garden the way Suzie wants, we need 2 lengths and 4 widths = 2(20) + 4(10) = 80 feet of fence. Check: 10 × 20 = 200 ft2 and 2(20) + 4(10) = 80 feet. The answer checks out. Example 8 An isosceles triangle is enclosed in a square so that its base coincides with one of the sides of the square and the tip of the triangle touches the opposite side of the square. If the area of the triangle is 20 in2 what is the length of one side of the square? Solution Draw a sketch: Deﬁne: Let x = base of the triangle Then x = height of the triangle Translate: Area of a triangle is 1 2 × base × height, so 1 2 · x · x = 20 Solve: 1 2 x2 = 20 Solve by taking the square root: x2 = 40 x = √ 40 and x = − √ 40 x = 6.32 and x = −6.32 The side of the square is 6.32 inches. That means the area of the square is (6.32)2 = 40 in2, twice as big as the area of the triangle. Check: It makes sense that the area of the square will be twice that of the triangle. If you look at the ﬁgure you can see that you could ﬁt two triangles inside the square. 463 www.ck12.org Review Questions Solve the following quadratic equations using the quadratic formula. 1. x2 + 4x −21 = 0 2. x2 −6x = 12 3. 3x2 −1 2 x = 3 8 4. 2x2 + x −3 = 0 5. −x2 −7x + 12 = 0 6. −3x2 + 5x = 2 7. 4x2 = x 8. x2 + 2x + 6 = 0 Solve the following quadratic equations using the method of your choice. 9. x2 −x = 6 10. x2 −12 = 0 11. −2x2 + 5x −3 = 0 12. x2 + 7x −18 = 0 13. 3x2 + 6x = −10 14. −4x2 + 4000x = 0 15. −3x2 + 12x + 1 = 0 16. x2 + 6x + 9 = 0 17. 81x2 + 1 = 0 18. −4x2 + 4x = 9 19. 36x2 −21 = 0 20. x2 −2x −3 = 0 21. The product of two consecutive integers is 72. Find the two numbers. 22. The product of two consecutive odd integers is 1 less than 3 times their sum. Find the integers. 23. The length of a rectangle exceeds its width by 3 inches. The area of the rectangle is 70 square inches, ﬁnd its dimensions. 24. Angel wants to cut oﬀa square piece from the corner of a rectangular piece of plywood. The larger piece of wood is 4 feet × 8 feet and the cut oﬀpart is 1 3 of the total area of the plywood sheet. What is the length of the side of the square? 25. Mike wants to fence three sides of a rectangular patio that is adjacent the back of his house. The area of the patio is 192 ft2 and the length is 4 feet longer than the width. Find how much fencing Mike will need. 10.6 The Discriminant Learning Objectives • Find the discriminant of a quadratic equation. www.ck12.org 464 • Interpret the discriminant of a quadratic equation. • Solve real-world problems using quadratic functions and interpreting the discriminant. Introduction In the quadratic formula, x = −b± √ b2−4ac 2a , the expression inside the square root is called the discriminant. The discriminant can be used to analyze the types of solutions to a quadratic equation without actually solving the equation. Here’s how: • If b2 −4ac > 0, the equation has two separate real solutions. • If b2 −4ac < 0, the equation has only non-real solutions. • If b2 −4ac = 0, the equation has one real solution, a double root. Find the Discriminant of a Quadratic Equation To ﬁnd the discriminant of a quadratic equation we calculate D = b2 −4ac. Example 1 Find the discriminant of each quadratic equation. Then tell how many solutions there will be to the quadratic equation without solving. a) x2 −5x + 3 = 0 b) 4x2 −4x + 1 = 0 c) −2x2 + x = 4 Solution a) Plug a = 1, b = −5 and c = 3 into the discriminant formula: D = (−5)2 −4(1)(3) = 13 D > 0, so there are two real solutions. b) Plug a = 4, b = −4 and c = 1 into the discriminant formula: D = (−4)2 −4(4)(1) = 0 D = 0, so there is one real solution. c) Rewrite the equation in standard form: −2x2 + x −4 = 0 Plug a = −2, b = 1 and c = −4 into the discriminant formula: D = (1)2 −4(−2)(−4) = −31 D < 0, so there are no real solutions. Interpret the Discriminant of a Quadratic Equation The sign of the discriminant tells us the nature of the solutions (or roots) of a quadratic equation. We can obtain two distinct real solutions if D > 0, two non-real solutions if D < 0 or one solution (called a double root) if D = 0. Recall that the number of solutions of a quadratic equation tells us how many times its graph crosses the x−axis. If D > 0, the graph crosses the x−axis in two places; if D = 0 it crosses in one place; if D < 0 it doesn’t cross at all: 465 www.ck12.org Example 2 Determine the nature of the solutions of each quadratic equation. a) 4x2 −1 = 0 b) 10x2 −3x = −4 c) x2 −10x + 25 = 0 Solution Use the value of the discriminant to determine the nature of the solutions to the quadratic equation. a) Plug a = 4, b = 0 and c = −1 into the discriminant formula: D = (0)2 −4(4)(−1) = 16 The discriminant is positive, so the equation has two distinct real solutions. The solutions to the equation are: 0± √ 16 8 = ± 4 8 = ± 1 2 b) Re-write the equation in standard form: 10x2 −3x + 4 = 0 Plug a = 10, b = −3 and c = 4 into the discriminant formula: D = (−3)2 −4(10)(4) = −151 The discriminant is negative, so the equation has two non-real solutions. c) Plug a = 1, b = −10 and c = 25 into the discriminant formula: D = (−10)2 −4(1)(25) = 0 The discriminant is 0, so the equation has a double root. The solution to the equation is: 10± √ 0 2 = 10 2 = 5 If the discriminant is a perfect square, then the solutions to the equation are not only real, but also rational. If the discriminant is positive but not a perfect square, then the solutions to the equation are real but irrational. Example 3 Determine the nature of the solutions to each quadratic equation. a) 2x2 + x −3 = 0 b) 5x2 −x −1 = 0 Solution Use the discriminant to determine the nature of the solutions. a) Plug a = 2, b = 1 and c = −3 into the discriminant formula: D = (1)2 −4(2)(−3) = 25 The discriminant is a positive perfect square, so the solutions are two real rational numbers. The solutions to the equation are: −1± √ 25 4 = −1±5 4 , so x = 1 and x = −3 2. b) Plug a = 5, b = −1 and c = −1 into the discriminant formula: D = (−1)2 −4(5)(−1) = 21 www.ck12.org 466 The discriminant is positive but not a perfect square, so the solutions are two real irrational numbers. The solutions to the equation are: 1± √ 21 10 , so x ≈0.56 and x ≈−0.36. Solve Real-World Problems Using Quadratic Functions and In-terpreting the Discriminant You’ve seen that calculating the discriminant shows what types of solutions a quadratic equation possesses. Knowing the types of solutions is very useful in applied problems. Consider the following situation. Example 4 Marcus kicks a football in order to score a ﬁeld goal. The height of the ball is given by the equation y = −32 6400 x2 + x. If the goalpost is 10 feet high, can Marcus kick the ball high enough to go over the goalpost? What is the farthest distance that Marcus can kick the ball from and still make it over the goalpost? Solution Deﬁne: Let y = height of the ball in feet. Let x = distance from the ball to the goalpost. Translate: We want to know if it is possible for the height of the ball to equal 10 feet at some real distance from the goalpost. Solve: Write the equation in standard form: − 32 6400 x2 + x −10 = 0 Simplify: −0.005x2 + x −10 = 0 Find the discriminant: D = (1)2 −4(−0.005)(−10) = 0.8 Since the discriminant is positive, we know that it is possible for the ball to go over the goal post, if Marcus kicks it from an acceptable distance x from the goalpost. To ﬁnd the value of x that will work, we need to use the quadratic formula: x = −1 ± √ 0.8 −0.01 = 189.4 feet or 10.56 feet What does this answer mean? It means that if Marcus is exactly 189.4 feet or exactly 10.56 feet from the goalposts, the ball will just barely go over them. Are these the only distances that will work? No; those are just the distances at which the ball will be exactly 10 feet high, but between those two distances, the ball will go even higher than that. (It travels in a downward-opening parabola from the place where it is kicked to the spot where it hits the ground.) This means that Marcus will make the goal if he is anywhere between 10.56 and 189.4 feet from the goalposts. Example 5 Emma and Bradon own a factory that produces bike helmets. Their accountant says that their proﬁt per year is given by the function P = −0.003x2 +12x+27760, where x is the number of helmets produced. Their goal is to make a proﬁt of $40,000 this year. Is this possible? Solution We want to know if it is possible for the proﬁt to equal $40,000. 467 www.ck12.org 40000 = −0.003x2 + 12x + 27760 Write the equation in standard form: −0.003x2 + 12x −12240 = 0 Find the discriminant: D = (12)2 −4(−0.003)(−12240) = −2.88 Since the discriminant is negative, we know that it is not possible for Emma and Bradon to make a proﬁt of $40,000 this year no matter how many helmets they make. Review Questions Find the discriminant of each quadratic equation. 1. 2x2 −4x + 5 = 0 2. x2 −5x = 8 3. 4x2 −12x + 9 = 0 4. x2 + 3x + 2 = 0 5. x2 −16x = 32 6. −5x2 + 5x −6 = 0 7. x2 + 4x = 2 8. −3x2 + 2x + 5 = 0 Determine the nature of the solutions of each quadratic equation. 9. −x2 + 3x −6 = 0 10. 5x2 = 6x 11. 41x2 −31x −52 = 0 12. x2 −8x + 16 = 0 13. −x2 + 3x −10 = 0 14. x2 −64 = 0 15. 3x2 = 7 16. x2 + 30 + 225 = 0 Without solving the equation, determine whether the solutions will be rational or irrational. 17. x2 = −4x + 20 18. x2 + 2x −3 = 0 19. 3x2 −11x = 10 20. 1 2 x2 + 2x + 2 3 = 0 21. x2 −10x + 25 = 0 22. x2 = 5x 23. 2x2 −5x = 12 24. Marty is outside his apartment building. He needs to give his roommate Yolanda her cell phone but he does not have time to run upstairs to the third ﬂoor to give it to her. He throws it straight up with a vertical velocity of 55 feet/second. Will the phone reach her if she is 36 feet up? (Hint: the equation for the height is y = −32t2 + 55t + 4.) 25. Bryson owns a business that manufactures and sells tires. The revenue from selling the tires in the month of July is given by the function R = x(200 −0.4x) where x is the number of tires sold. Can Bryson’s business generate revenue of $20,000 in the month of July? www.ck12.org 468 10.7 Linear, Exponential and Quadratic Models Learning Objectives • Identify functions using diﬀerences and ratios. • Write equations for functions. • Perform exponential and quadratic regressions with a graphing calculator. • Solve real-world problems by comparing function models. Introduction In this course we’ve learned about three types of functions, linear, quadratic and exponential. • Linear functions take the form y = mx + b. • Quadratic functions take the form y = ax2 + bx + c. • Exponential functions take the form y = a · bx. In real-world applications, the function that describes some physical situation is not given; it has to be found before the problem can be solved. For example, scientiﬁc data such as observations of planetary motion are often collected as a set of measurements given in a table. Part of the scientist’s job is to ﬁgure out which function best ﬁts the data. In this section, you’ll learn some methods that are used to identify which function describes the relationship between the variables in a problem. Identify Functions Using Diﬀerences or Ratios One method for identifying functions is to look at the diﬀerence or the ratio of diﬀerent values of the dependent variable. For example, if the diﬀerence between values of the dependent variable is the same each time we change the independent variable by the same amount, then the function is linear. Example 1 Determine if the function represented by the following table of values is linear. Table 10.14: x y −2 –4 –1 –1 0 2 1 5 2 8 If we take the diﬀerence between consecutive y−values, we see that each time the x−value increases by one, the y−value always increases by 3. Since the diﬀerence is always the same, the function is linear. When we look at the diﬀerence of the y−values, we have to make sure that we examine entries for which 469 www.ck12.org the x−values increase by the same amount. For example, examine the values in this table: Table 10.15: x y 0 5 1 10 3 20 4 25 6 35 At ﬁrst glance this function might not look linear, because the diﬀerence in the y−values is not always the same. But if we look closer, we can see that when the y−value increases by 10 instead of 5, it’s because the x−value increased by 2 instead of 1. Whenever the x−value increases by the same amount, the y−value does too, so the function is linear. Another way to think of this is in mathematical notation. We can say that a function is linear if y2−y1 x2−x1 is always the same for any two pairs of x−and y−values. Notice that the expression we used here is simply the deﬁnition of the slope of a line. Diﬀerences can also be used to identify quadratic functions. For a quadratic function, when we increase the x−values by the same amount, the diﬀerence between y−values will not be the same. However, the diﬀerence of the diﬀerences of the y−values will be the same. Here are some examples of quadratic relationships represented by tables of values: In this quadratic function, y = x2, when we increase the x−value by one, the value of y increases by diﬀerent values. However, it increases at a constant rate, so the diﬀerence of the diﬀerence is always 2. www.ck12.org 470 In this quadratic function, y = 2x2 −3x + 1, when we increase the x−value by one, the value of y increases by diﬀerent values. However, the increase is constant: the diﬀerence of the diﬀerence is always 4. To identify exponential functions, we use ratios instead of diﬀerences. If the ratio between values of the dependent variable is the same each time we change the independent variable by the same amount, then the function is exponential. Example 2 Determine if the function represented by each table of values is exponential. a) b) 471 www.ck12.org a) If we take the ratio of consecutive y−values, we see that each time the x−value increases by one, the y−value is multiplied by 3. Since the ratio is always the same, the function is exponential. b) If we take the ratio of consecutive y−values, we see that each time the x−value increases by one, the y−value is multiplied by 1 2. Since the ratio is always the same, the function is exponential. Write Equations for Functions Once we identify which type of function ﬁts the given values, we can write an equation for the function by starting with the general form for that type of function. Example 3 Determine what type of function represents the values in the following table. Table 10.16: x y 0 5 1 1 2 -3 3 -7 4 -11 Solution Let’s ﬁrst check the diﬀerence of consecutive values of y. If we take the diﬀerence between consecutive y−values, we see that each time the x−value increases by one, the y−value always decreases by 4. Since the diﬀerence is always the same, the function is linear. To ﬁnd the equation for the function, we start with the general form of a linear function: y = mx+b. Since m is the slope of the line, it’s also the quantity by which y increases every time the value of x increases by one. The constant b is the value of the function when x = 0. Therefore, the function is y = −4x + 5. Example 4 Determine what type of function represents the values in the following table. www.ck12.org 472 Table 10.17: x y 0 0 1 5 2 20 3 45 4 80 5 125 6 180 Solution Here, the diﬀerence between consecutive y−values isn’t constant, so the function is not linear. Let’s look at those diﬀerences more closely. Table 10.18: x y 0 0 1 5 5 −0 = 5 2 20 20 −5 = 15 3 45 45 −20 = 25 4 80 80 −45 = 35 5 125 125 −80 = 45 6 180 180 −125 = 55 When the x−value increases by one, the diﬀerence between the y−values increases by 10 every time. Since the diﬀerence of the diﬀerences is constant, the function describing this set of values is quadratic. To ﬁnd the equation for the function that represents these values, we start with the general form of a quadratic function: y = ax2 + bx + c. We need to use the values in the table to ﬁnd the values of the constants: a, b and c. The value of c represents the value of the function when x = 0, so c = 0. Plug in the point (1, 5) : 5 = a + b Plug in the point (2, 20) : 20 = 4a + 2b ⇒10 = 2a + b To ﬁnd a and b, we solve the system of equations: 5 = a + b 10 = 2a + b Solve the ﬁrst equation for b : 5 = a + b ⇒b = 5 −a Plug the ﬁrst equation into the second: 10 = 2a + 5 −a Solve for a and b a = 5 and b = 0 Therefore the equation of the quadratic function is y = 5x2. Example 5 Determine what type of function represents the values in the following table. 473 www.ck12.org Table 10.19: x y 0 400 1 500 2 25 3 6.25 4 1.5625 Solution The diﬀerences between consecutive y−values aren’t the same, and the diﬀerences between those diﬀerences aren’t the same either. So let’s check the ratios instead. Each time the x−value increases by one, the y−value is multiplied by 1 4. Since the ratio is always the same, the function is exponential. To ﬁnd the equation for the function that represents these values, we start with the general form of an exponential function, y = a · bx. Here b is the ratio between the values of y each time x is increased by one. The constant a is the value of the function when x = 0. Therefore, the function is y = 400 ( 1 4 )x. Perform Exponential and Quadratic Regressions with a Graphing Calculator Earlier, you learned how to perform linear regression with a graphing calculator to ﬁnd the equation of a straight line that ﬁts a linear data set. In this section, you’ll learn how to perform exponential and quadratic regression to ﬁnd equations for curves that ﬁt non-linear data sets. Example 6 The following table shows how many miles per gallon a car gets at diﬀerent speeds. Table 10.20: Speed (mph) Miles per gallon 30 18 www.ck12.org 474 Table 10.20: (continued) Speed (mph) Miles per gallon 35 20 40 23 45 25 50 28 55 30 60 29 65 25 70 25 Using a graphing calculator: a) Draw the scatterplot of the data. b) Find the quadratic function of best ﬁt. c) Draw the quadratic function of best ﬁt on the scatterplot. d) Find the speed that maximizes the miles per gallon. e) Predict the miles per gallon of the car if you drive at a speed of 48 mph. Solution Step 1: Input the data. Press [STAT] and choose the [EDIT] option. Input the values of x in the ﬁrst column (L1) and the values of y in the second column (L2). (Note: in order to clear a list, move the cursor to the top so that L1 or L2 is highlighted. Then press [CLEAR] and then [ENTER].) Step 2: Draw the scatterplot. First press [Y=] and clear any function on the screen by pressing [CLEAR] when the old function is highlighted. Press [STATPLOT] [STAT] and [Y=] and choose option 1. Choose the ON option; after TYPE, choose the ﬁrst graph type (scatterplot) and make sure that the Xlist and Ylist names match the names on top of the columns in the input table. Press [GRAPH] and make sure that the window is set so you see all the points in the scatterplot. In this case, the settings should be 30 ≤x ≤80 and 0 ≤y ≤40. You can set the window size by pressing the [WINDOW] key at the top. Step 3: Perform quadratic regression. Press [STAT] and use the right arrow to choose [CALC]. Choose Option 5 (QuadReg) and press [ENTER]. You will see “QuadReg” on the screen. Type in L1, L2 after ‘QuadReg’ and press [ENTER]. The calculator shows the quadratic function: y = −0.017x2 + 1.9x −25 Step 4: Graph the function. Press [Y=] and input the function you just found. Press [GRAPH] and you will see the curve ﬁt drawn over the data points. 475 www.ck12.org To ﬁnd the speed that maximizes the miles per gallon, use [TRACE] and move the cursor to the top of the parabola. You can also use [CALC] [2nd] [TRACE] and option 4:Maximum, for a more accurate answer. The speed that maximizes miles per gallon is 56 mph. Finally, plug x = 48 into the equation you found: y = −0.017(48)2 + 1.9(48) −25 = 27.032 miles per gallon. Note: The image above shows our function plotted on the same graph as the data points from the table. One thing that is clear from this graph is that predictions made with this function won’t make sense for all values of x. For example, if x < 15, this graph predicts that we will get negative mileage, which is impossible. Part of the skill of using regression on your calculator is being aware of the strengths and limitations of this method of ﬁtting functions to data. Example 7 The following table shows the amount of money an investor has in an account each year for 10 years. Table 10.21: Year Value of account 1996 $5000 1997 $5400 1998 $5800 1999 $6300 2000 $6800 2001 $7300 2002 $7900 2003 $8600 2004 $9300 2005 $10000 2006 $11000 Using a graphing calculator: a) Draw a scatterplot of the value of the account as the dependent variable, and the number of years since www.ck12.org 476 1996 as the independent variable. b) Find the exponential function that ﬁts the data. c) Draw the exponential function on the scatterplot. d) What will be the value of the account in 2020? Solution Step 1: Input the data. Press [STAT] and choose the [EDIT] option. Input the values of x in the ﬁrst column (L1) and the values of y in the second column (L2). Step 2: Draw the scatterplot. First press [Y=] and clear any function on the screen. Press [GRAPH] and choose Option 1. Choose the ON option and make sure that the Xlist and Ylist names match the names on top of the columns in the input table. Press [GRAPH] make sure that the window is set so you see all the points in the scatterplot. In this case the settings should be 0 ≤x ≤10 and 0 ≤y ≤11000. Step 3: Perform exponential regression. Press [STAT] and use the right arrow to choose [CALC]. Choose Option 0 and press [ENTER]. You will see “ExpReg” on the screen. Press [ENTER]. The calculator shows the exponential function: y = 4975.7(1.08)x Step 4: Graph the function. Press [Y=] and input the function you just found. Press [GRAPH]. Finally, plug x = 2020 −1996 = 24 into the function: y = 4975.7(1.08)24 = $31551.81 In 2020, the account will have a value of $31551.81. Note: The function above is the curve that comes closest to all the data points. It won’t return y−values that are exactly the same as in the data table, but they will be close. It is actually more accurate to use the curve ﬁt values than the data points. If you don’t have a graphing calculator, there are resources available on the Internet for ﬁnding lines and curves of best ﬁt. For example, the applet at LRegression.html does linear regression on a set of data points; the one at edu/~plaval/applets/QRegression.html does quadratic regression; and the one at kennesaw.edu/~plaval/applets/ERegression.html does exponential regression. Also, programs like Microsoft Oﬀice or OpenOﬀice have the ability to create graphs and charts that include lines and curves of best ﬁt. Solve Real-World Problems by Comparing Function Models Example 8 The following table shows the number of students enrolled in public elementary schools in the US (source: US Census Bureau). Make a scatterplot with the number of students as the dependent variable, and the number of years since 1990 as the independent variable. Find which curve ﬁts this data the best and predict the school enrollment in the year 2007. 477 www.ck12.org Table 10.22: Year Number of students (millions) 1990 26.6 1991 26.6 1992 27.1 1993 27.7 1994 28.1 1995 28.4 1996 28.1 1997 29.1 1998 29.3 2003 32.5 Solution We need to perform linear, quadratic and exponential regression on this data set to see which function represents the values in the table the best. Step 1: Input the data. Input the values of x in the ﬁrst column (L1) and the values of y in the second column (L2). Step 2: Draw the scatterplot. Set the window size: 0 ≤x ≤10 and 20 ≤y ≤40. Here is the scatterplot: Step 3: Perform Regression. Linear Regression The function of the line of best ﬁt is y = 0.51x + 26.1. Here is the graph of the function on the scatterplot: www.ck12.org 478 Quadratic Regression The quadratic function of best ﬁt is y = 0.064x2 −.067x + 26.84. Here is the graph of the function on the scatterplot: Exponential Regression The exponential function of best ﬁt is y = 26.2(1.018)x. Here is the graph of the function on the scatterplot: 479 www.ck12.org From the graphs, it looks like the quadratic function is the best ﬁt for this data set. We’ll use this function to predict school enrollment in 2007. Plug in x = 2007 −1990 = 17 : y = 0.064(17)2 −.067(17) + 26.84 = 44.2 million students. Review Questions Determine whether the data in the following tables can be represented by a linear function. 1. Table 10.23: x y −4 10 -3 7 -2 4 -1 1 0 -2 1 -5 2. Table 10.24: x y −2 4 -1 3 0 2 www.ck12.org 480 Table 10.24: (continued) x y 1 3 2 6 3 11 3. Table 10.25: x y 0 50 1 75 2 100 3 125 4 150 5 175 Determine whether the data in the following tables can be represented by a quadratic function. 4. Table 10.26: x y −10 10 -5 2.5 0 0 5 2.5 10 10 15 22.5 5. Table 10.27: x y 1 4 2 6 3 6 4 4 5 0 6 -6 481 www.ck12.org 6. Table 10.28: x y −3 -27 -2 -8 -1 -1 0 0 1 1 2 8 3 27 Determine whether the data in the following tables can be represented by an exponential function. 7. Table 10.29: x y 0 200 1 300 2 1800 3 8300 4 25800 5 62700 8. Table 10.30: x y 0 120 1 180 2 270 3 405 4 607.5 5 911.25 9. www.ck12.org 482 Table 10.31: x y 0 4000 1 2400 2 1440 3 864 4 518.4 5 311.04 Determine what type of function represents the values in the following tables and ﬁnd the equation of each function. 10. Table 10.32: x y 0 400 1 500 2 625 3 781.25 4 976.5625 11. Table 10.33: x y −9 -3 -7 -2 -5 -1 -3 0 -1 1 1 2 12. Table 10.34: x y −3 14 -2 4 -1 -2 0 -4 483 www.ck12.org Table 10.34: (continued) x y 1 -2 2 4 3 14 13. As a ball bounces up and down, the maximum height that the ball reaches continually decreases from one bounce to the next. For a given bounce, this table shows the height of the ball with respect to time: Table 10.35: Time (seconds) Height (inches) 2 2 2.2 16 2.4 24 2.6 33 2.8 38 3.0 42 3.2 36 3.4 30 3.6 28 3.8 14 4.0 6 Using a graphing calculator: a) Draw the scatterplot of the data b) Find the quadratic function of best ﬁt c) Draw the quadratic function of best ﬁt on the scatterplot d) Find the maximum height the ball reaches on the bounce e) Predict how high the ball is at time t = 2.5 seconds 14. A chemist has a 250 gram sample of a radioactive material. She records the amount of radioactive material remaining in the sample every day for a week and obtains the data in the following table: Table 10.36: Day Weight (grams) 0 250 1 208 2 158 3 130 4 102 5 80 www.ck12.org 484 Table 10.36: (continued) Day Weight (grams) 6 65 7 50 Using a graphing calculator: a) Draw a scatterplot of the data b) Find the exponential function of best ﬁt c) Draw the exponential function of best ﬁt on the scatterplot d) Predict the amount of material after 10 days. 15. The following table shows the rate of pregnancies (per 1000) for US women aged 15 to 19. (source: US Census Bureau). (a) Make a scatterplot with the rate of pregnancies as the dependent variable and the number of years since 1990 as the independent variable. (b) Find which type of curve ﬁts this data best. (c) Predict the rate of teen pregnancies in the year 2010. Table 10.37: Year Rate of pregnancy (per 1000) 1990 116.9 1991 115.3 1992 111.0 1993 108.0 1994 104.6 1995 99.6 1996 95.6 1997 91.4 1998 88.7 1999 85.7 2000 83.6 2001 79.5 2002 75.4 485 www.ck12.org Chapter 11 Algebra and Geometry Connections 11.1 Graphs of Square Root Functions Learning Objectives • Graph and compare square root functions. • Shift graphs of square root functions. • Graph square root functions using a graphing calculator. • Solve real-world problems using square root functions. Introduction In this chapter you’ll learn about a diﬀerent kind of function called the square root function. You’ve seen that taking the square root is very useful in solving quadratic equations. For example, to solve the equation x2 = 25 we take the square root of both sides: √ x2 = ± √ 25, so x = ±5. A square root function is any function with the form: y = a √ f(x)+c —in other words, any function where an expression in terms of x is found inside a square root sign (also called a “radical” sign), although other terms may be included as well. Graph and Compare Square Root Functions When working with square root functions, you’ll have to consider the domain of the function before graphing. The domain is very important because the function is undeﬁned when the expression inside the square root sign is negative, and as a result there will be no graph in whatever region of x−values makes that true. To discover how the graphs of square root functions behave, let’s make a table of values and plot the points. Example 1 Graph the function y = √x. Solution www.ck12.org 486 Before we make a table of values, we need to ﬁnd the domain of this square root function. The domain is found by realizing that the function is only deﬁned when the expression inside the square root is greater than or equal to zero. Since the expression inside the square root is just x, that means the domain is all values of x such that x ≥0. This means that when we make our table of values, we should pick values of x that are greater than or equal to zero. It is very useful to include zero itself as the ﬁrst value in the table and also include many values greater than zero. This will help us in determining what the shape of the curve will be. Table 11.1: x y = √x 0 y = √ 0 = 0 1 y = √ 1 = 1 2 y = √ 2 = 1.4 3 y = √ 3 = 1.7 4 y = √ 4 = 2 5 y = √ 5 = 2.2 6 y = √ 6 = 2.4 7 y = √ 7 = 2.6 8 y = √ 8 = 2.8 9 y = √ 9 = 3 Here is what the graph of this table looks like: The graphs of square root functions are always curved. The curve above looks like half of a parabola lying on its side, and in fact it is. It’s half of the parabola that you would get if you graphed the expression y2 = x. And the graph of y = −√x is the other half of that parabola: 487 www.ck12.org Notice that if we graph the two separate functions on the same coordinate axes, the combined graph is a parabola lying on its side. Now let’s compare square root functions that are multiples of each other. Example 2 Graph the functions y = √x, y = 2 √x, y = 3 √x, and y = 4 √x on the same graph. Solution Here is just the graph without the table of values: If we multiply the function by a constant bigger than one, the function increases faster the greater the constant is. www.ck12.org 488 Example 3 Graph the functions y = √x, y = √ 2x, y = √ 3x, and y = √ 4x on the same graph. Solution Notice that multiplying the expression inside the square root by a constant has the same eﬀect as multi-plying by a constant outside the square root; the function just increases at a slower rate because the entire function is eﬀectively multiplied by the square root of the constant. Also note that the graph of √ 4x is the same as the graph of 2 √x. This makes sense algebraically since √ 4 = 2. Example 4 Graph the functions y = √x, y = 1 2 √x, y = 1 3 √x, and y = 1 4 √x on the same graph. Solution If we multiply the function by a constant between 0 and 1, the function increases more slowly the smaller the constant is. Example 5 Graph the functions y = 2 √x and y = −2 √x on the same graph. Solution If we multiply the whole function by -1, the graph is reﬂected about the x−axis. 489 www.ck12.org Example 6 Graph the functions y = √x and y = √−x on the same graph. Solution On the other hand, when just the x is multiplied by -1, the graph is reﬂected about the y−axis. Notice that the function y = √−x has only negative x−values in its domain, because when x is negative, the expression under the radical sign is positive. Example 7 Graph the functions y = √x, y = √x + 2 and y = √x −2. Solution When we add a constant to the right-hand side of the equation, the graph keeps the same shape, but shifts www.ck12.org 490 up for a positive constant or down for a negative one. Example 8 Graph the functions y = √x, y = √ x −2, and y = √x + 2. Solution When we add a constant to the argument of the function (the part under the radical sign), the function shifts to the left for a positive constant and to the right for a negative constant. Now let’s see how to combine all of the above types of transformations. Example 9 Graph the function y = 2 √ 3x −1 + 2. Solution We can think of this function as a combination of shifts and stretches of the basic square root function y = √x. We know that the graph of that function looks like this: If we multiply the argument by 3 to obtain y = √ 3x, this stretches the curve vertically because the value of y increases faster by a factor of √ 3. Next, when we subtract 1 from the argument to obtain y = √ 3x −1 this shifts the entire graph to the left by one unit. Multiplying the function by a factor of 2 to obtain y = 2 √ 3x −1 stretches the curve vertically again, because y increases faster by a factor of 2. Finally we add 2 to the function to obtain y = 2 √ 3x −1 + 2. This shifts the entire function vertically by 2 units. 491 www.ck12.org Each step of this process is shown in the graph below. The purple line shows the ﬁnal result. Now we know how to graph square root functions without making a table of values. If we know what the basic function looks like, we can use shifts and stretches to transform the function and get to the desired result. Solve Real-World Problems Using Square Root Functions Mathematicians and physicists have studied the motion of pendulums in great detail because this motion explains many other behaviors that occur in nature. This type of motion is called simple harmonic motion and it is important because it describes anything that repeats periodically. Galileo was the ﬁrst person to study the motion of a pendulum, around the year 1600. He found that the time it takes a pendulum to complete a swing doesn’t depend on its mass or on its angle of swing (as long as the angle of the swing is small). Rather, it depends only on the length of the pendulum. The time it takes a pendulum to complete one whole back-and-forth swing is called the period of the pendulum. Galileo found that the period of a pendulum is proportional to the square root of its length: T = a √ L. The proportionality constant, a, depends on the acceleration of gravity: a = 2π √g. At sea level on Earth, acceleration of gravity is g = 9.81 m/s2 (meters per second squared). Using this value of gravity, we ﬁnd a = 2.0 with units of s √m (seconds divided by the square root of meters). www.ck12.org 492 Up until the mid 20th century, all clocks used pendulums as their central time keeping component. Example 10 Graph the period of a pendulum of a clock swinging in a house on Earth at sea level as we change the length of the pendulum. What does the length of the pendulum need to be for its period to be one second? Solution The function for the period of a pendulum at sea level is T = 2 √ L. We start by making a table of values for this function: Table 11.2: L T = 2 √ L 0 T = 2 √ 0 = 0 1 T = 2 √ 1 = 2 2 y = 2 √ 2 = 2.8 3 y = 2 √ 3 = 3.5 4 y = 2 √ 4 = 4 5 y = 2 √ 5 = 4.5 Now let’s graph the function. It makes sense to let the horizontal axis represent the length of the pendulum and the vertical axis represent the period of the pendulum. We can see from the graph that a length of approximately 1 4 meters gives a period of 1 second. We can conﬁrm this answer by using our function for the period and plugging in T = 1 second: T = 2 √ L ⇒1 = 2 √ L Square both sides of the equation: 1 = 4L Solve for L : L = 1 4 meters For more equations that describe pendulum motion, check out edu/hbase/pend.html, where you can also ﬁnd a tool for calculating the period of a pendulum in diﬀerent gravities than Earth’s. Example 11 493 www.ck12.org “Square” TV screens have an aspect ratio of 4:3; in other words, the width of the screen is 4 3 the height. TV “sizes” are traditionally represented as the length of the diagonal of the television screen. Graph the length of the diagonal of a screen as a function of the area of the screen. What is the diagonal of a screen with an area of 180 in2? Solution Let d = length of the diagonal, x = width Then 4 × height = 3 × width Or, height = 3 4 x. The area of the screen is: A = length × width or A = 3 4 x2 Find how the diagonal length relates to the width by using the Pythagorean theorem: x2 + (3 4 x )2 = d2 x2 + 9 16 x2 = d2 25 16 x2 = d2 ⇒x2 = 16 25d2 ⇒x = 4 5d Therefore, the diagonal length relates to the area as follows: A = 3 4 ( 4 5d )2 = 3 4 · 16 25d2 = 12 25d2. We can also ﬂip that around to ﬁnd the diagonal length as a function of the area: d2 = 25 12A or d = 5 2 √ 3 √ A. Now we can make a graph where the horizontal axis represents the area of the television screen and the vertical axis is the length of the diagonal. First let’s make a table of values: Table 11.3: A d = 5 2 √ 3 √ A 0 0 25 7.2 50 10.2 75 12.5 100 14.4 125 16.1 150 17.6 175 19 200 20.4 www.ck12.org 494 From the graph we can estimate that when the area of a TV screen is 180 in2 the length of the diagonal is approximately 19.5 inches. We can conﬁrm this by plugging in A = 180 into the formula that relates the diagonal to the area: d = 5 2 √ 3 √ A = 5 2 √ 3 √ 180 = 19.4 inches. Review Questions Graph the following functions on the same coordinate axes. 1. y = √x, y = 2.5 √x and y = −2.5 √x 2. y = √x, y = 0.3 √x and y = 0.6 √x 3. y = √x, y = √ x −5 and y = √x + 5 4. y = √x, y = √x + 8 and y = √x −8 Graph the following functions. 5. y = √ 2x −1 6. y = √4x + 4 7. y = √ 5 −x 8. y = 2 √x + 5 9. y = 3 −√x 10. y = 4 + 2 √x 11. y = 2 √2x + 3 + 1 12. y = 4 + √ 2 −x 13. y = √x + 1 − √ 4x −5 14. The acceleration of gravity can also given in feet per second squared. It is g = 32 ft/s2 at sea level. (a) Graph the period of a pendulum with respect to its length in feet. (b) For what length in feet will the period of a pendulum be 2 seconds? 15. The acceleration of gravity on the Moon is 1.6 m/s2. (a) Graph the period of a pendulum on the Moon with respect to its length in meters. (b) For what length, in meters, will the period of a pendulum be 10 seconds? 16. The acceleration of gravity on Mars is 3.69 m/s2. (a) Graph the period of a pendulum on the Mars with respect to its length in meters. (b) For what length, in meters, will the period of a pendulum be 3 seconds? 495 www.ck12.org 17. The acceleration of gravity on the Earth depends on the latitude and altitude of a place. The value of g is slightly smaller for places closer to the Equator than places closer to the poles and the value of g is slightly smaller for places at higher altitudes that it is for places at lower altitudes. In Helsinki the value of g = 9.819 m/s2, in Los Angeles the value of g = 9.796 m/s2 and in Mexico City the value of g = 9.779 m/s2. (a) Graph the period of a pendulum with respect to its length for all three cities on the same graph. (b) Use the formula to ﬁnd for what length, in meters, will the period of a pendulum be 8 seconds in each of these cities? 18. The aspect ratio of a wide-screen TV is 2.39:1. (a) Graph the length of the diagonal of a screen as a function of the area of the screen. (b) What is the diagonal of a screen with area 150 in2? Graph the following functions using a graphing calculator. 19. y = √ 3x −2 20. y = 4 + √ 2 −x 21. y = √ x2 −9 22. y = √x −√x + 2 11.2 Radical Expressions Learning Objectives • Use the product and quotient properties of radicals. • Rationalize the denominator. • Add and subtract radical expressions. • Multiply radical expressions. • Solve real-world problems using square root functions. Introduction A radical reverses the operation of raising a number to a power. For example, the square of 4 is 42 = 4·4 = 16, and so the square root of 16 is 4. The symbol for a square root is √. This symbol is also called the radical sign. In addition to square roots, we can also take cube roots, fourth roots, and so on. For example, since 64 is the cube of 4, 4 is the cube root of 64. 3 √ 64 = 4 since 43 = 4 · 4 · 4 = 64 We put an index number in the top left corner of the radical sign to show which root of the number we are seeking. Square roots have an index of 2, but we usually don’t bother to write that out. 2 √ 36 = √ 36 = 6 The cube root of a number gives a number which when raised to the power three gives the number under the radical sign. The fourth root of number gives a number which when raised to the power four gives the number under the radical sign: www.ck12.org 496 4 √ 81 = 3 since 34 = 3 · 3 · 3 · 3 = 81 And so on for any power we can name. Even and Odd Roots Radical expressions that have even indices are called even roots and radical expressions that have odd indices are called odd roots. There is a very important diﬀerence between even and odd roots, because they give drastically diﬀerent results when the number inside the radical sign is negative. Any real number raised to an even power results in a positive answer. Therefore, when the index of a radical is even, the number inside the radical sign must be non-negative in order to get a real answer. On the other hand, a positive number raised to an odd power is positive and a negative number raised to an odd power is negative. Thus, a negative number inside the radical sign is not a problem. It just results in a negative answer. Example 1 Evaluate each radical expression. a) √ 121 b) 3 √ 125 c) 4 √ −625 d) 5 √ −32 Solution a) √ 121 = 11 b) 3 √ 125 = 5 c) 3 √ −625 is not a real number d) 5 √ −32 = −2 Use the Product and Quotient Properties of Radicals Radicals can be re-written as rational powers. The radical: m √ an is deﬁned as a n m . Example 2 Write each expression as an exponent with a rational value for the exponent. a) √ 5 b) 4 √a c) 3 √4xy d) 6 √ x5 Solution a) √ 5 = 5 1 2 b) 4 √a = a 1 4 497 www.ck12.org c) 3 √4xy = (4xy) 1 3 d) 6 √ x5 = x 5 6 As a result of this property, for any non-negative number a we know that n √ an = a n n = a. Since roots of numbers can be treated as powers, we can use exponent rules to simplify and evaluate radical expressions. Let’s review the product and quotient rule of exponents. Raising a product to a power: (x · y)n = xn · yn Raising a quotient to a power: ( x y )n = xn yn In radical notation, these properties are written as Raising a product to a power: m √x · y = m √x · m √y Raising a quotient to a power: m √x y = m √x m √y A very important application of these rules is reducing a radical expression to its simplest form. This means that we apply the root on all the factors of the number that are perfect roots and leave all factors that are not perfect roots inside the radical sign. For example, in the expression √ 16, the number 16 is a perfect square because 16 = 42. This means that we can simplify it as follows: √ 16 = √ 42 = 4 Thus, the square root disappears completely. On the other hand, in the expression √ 32, the number 32 is not a perfect square, so we can’t just remove the square root. However, we notice that 32 = 16 · 2, so we can write 32 as the product of a perfect square and another number. Thus, √ 32 = √ 16 · 2 If we apply the “raising a product to a power” rule we get: √ 32 = √ 16 · 2 = √ 16 · √ 2 Since √ 16 = 4, we get: √ 32 = 4 · √ 2 = 4 √ 2 Example 3 Write the following expressions in the simplest radical form. a) √ 8 b) √ 50 c) √ 125 72 Solution The strategy is to write the number under the square root as the product of a perfect square and another number. The goal is to ﬁnd the highest perfect square possible; if we don’t ﬁnd it right away, we just repeat the procedure until we can’t simplify any longer. www.ck12.org 498 a) We can write 8 = 4 · 2, so √ 8 = √ 4 · 2. With the “Raising a product to a power” rule, that becomes √ 4 · √ 2. Evaluate √ 4 and we’re left with 2 √ 2. b) We can write 50 = 25 · 2, so: √ 50 = √ 25 · 2 Use “Raising a product to a power” rule: = √ 25 · √ 2 = 5 √ 2 c) Use “Raising a quotient to a power” rule to separate the fraction: √ 125 72 = √ 125 √ 72 Re-write each radical as a product of a perfect square and another number: = √ 25 · 5 √ 36 · 2 = 5 √ 5 6 √ 2 The same method can be applied to reduce radicals of diﬀerent indices to their simplest form. Example 4 Write the following expression in the simplest radical form. a) 3 √ 40 b) 4 √ 162 80 c) 3 √ 135 Solution In these cases we look for the highest possible perfect cube, fourth power, etc. as indicated by the index of the radical. a) Here we are looking for the product of the highest perfect cube and another number. We write: 3 √ 40 = 3 √ 8 · 5 = 3 √ 8 · 3 √ 5 = 2 3 √ 5 b) Here we are looking for the product of the highest perfect fourth power and another number. Re-write as the quotient of two radicals: 4 √ 162 80 = 4 √ 162 4 √ 80 Simplify each radical separately: = 4 √ 81 · 2 4 √ 16 · 5 = 4 √ 81 · 4 √ 2 4 √ 16 · 4 √ 5 = 3 4 √ 2 2 4 √ 5 Recombine the fraction under one radical sign: = 3 2 4 √ 2 5 c) Here we are looking for the product of the highest perfect cube root and another number. Often it’s not very easy to identify the perfect root in the expression under the radical sign. In this case, we can factor the number under the radical sign completely by using a factor tree: 499 www.ck12.org We see that 135 = 3 · 3 · 3 · 5 = 33 · 5. Therefore 3 √ 135 = 3 √ 33 · 5 = 3 √ 33 · 3 √ 5 = 3 3 √ 5. (You can ﬁnd a useful tool for creating factor trees at factor_tree/. Click on “User Number” to type in your own number to factor, or just click “New Number” for a random number if you want more practice factoring.) Now let’s see some examples involving variables. Example 5 Write the following expression in the simplest radical form. a) √ 12x3y5 b) 4 √ 1250x7 405y9 Solution Treat constants and each variable separately and write each expression as the products of a perfect power as indicated by the index of the radical and another number. a) Re-write as a product of radicals: √ 12x3y5 = √ 12 · √ x3 · √ y5 Simplify each radical separately: ( √ 4 · 3 ) · ( √ x2 · x ) · ( √ y4 · y ) = ( 2 √ 3 ) · ( x √x ) · ( y2 √y ) Combine all terms outside and inside the radical sign: = 2xy2 √ 3xy b) Re-write as a quotient of radicals: 4 √ 1250x7 405y9 = 4 √ 1250x7 4 √ 405y9 Simplify each radical separately: = 4 √ 625 · 2 · 4 √ x4 · x3 4 √ 81 · 5 · 4 √ y4 · y4 · y = 5 4 √ 2 · x · 4 √ x3 3 4 √ 5 · y · y · 4 √y = 5x 4 √ 2x3 3y2 4 √5y Recombine fraction under one radical sign: = 5x 3y2 4 √ 2x3 5y Add and Subtract Radical Expressions When we add and subtract radical expressions, we can combine radical terms only when they have the same expression under the radical sign. This is a lot like combining like terms in variable expressions. For example, 4 √ 2 + 5 √ 2 = 9 √ 2 or 2 √ 3 − √ 2 + 5 √ 3 + 10 √ 2 = 7 √ 3 + 9 √ 2 It’s important to reduce all radicals to their simplest form in order to make sure that we’re combining all possible like terms in the expression. For example, the expression √ 8 −2 √ 50 looks like it can’t be simpliﬁed any more because it has no like terms. However, when we write each radical in its simplest form we get 2 √ 2 −10 √ 2, and we can combine those terms to get −8 √ 2. Example 6 Simplify the following expressions as much as possible. www.ck12.org 500 a) 4 √ 3 + 2 √ 12 b) 10 √ 24 − √ 28 Solution a) Simplify √ 12 to its simplest form: = 4 √ 3 + 2 √ 4 · 3 = 4 √ 3 + 6 √ 3 Combine like terms: = 10 √ 3 b) Simplify √ 24 and √ 28 to their simplest form: = 10 √ 6 · 4 − √ 7 · 4 = 20 √ 6 −2 √ 7 There are no like terms. Example 7 Simplify the following expressions as much as possible. a) 4 3 √ 128 − 3 √ 250 b) 3 √ x3 −4x √ 9x Solution a) Re-write radicals in simplest terms: = 4 3 √ 2 · 64 − 3 √ 2 · 125 = 16 3 √ 2 −5 3 √ 2 Combine like terms: = 11 3 √ 2 b) Re-write radicals in simplest terms: 3 √ x2 · x −12x √x = 3x √x −12x √x Combine like terms: = −9x √x Multiply Radical Expressions When we multiply radical expressions, we use the “raising a product to a power” rule: m √x · y = m √x · m √y. In this case we apply this rule in reverse. For example: √ 6 · √ 8 = √ 6 · 8 = √ 48 Or, in simplest radical form: √ 48 = √ 16 · 3 = 4 √ 3. We’ll also make use of the fact that: √a · √a = √ a2 = a. When we multiply expressions that have numbers on both the outside and inside the radical sign, we treat the numbers outside the radical sign and the numbers inside the radical sign separately. For example, a √ b · c √ d = ac √ bd. Example 8 Multiply the following expressions. a) √ 2 ( √ 3 + √ 5 ) b) 2 √x ( 3 √y −√x ) c) ( 2 + √ 5 ) ( 2 − √ 6 ) d) ( 2 √x + 1 ) ( 5 −√x ) 501 www.ck12.org Solution In each case we use distribution to eliminate the parentheses. a) Distribute √ 2 inside the parentheses: √ 2 ( √ 3 + √ 5 ) = √ 2 · √ 3 + √ 2 · √ 5 Use the “raising a product to a power” rule: = √ 2 · 3 + √ 2 · 5 Simplify: = √ 6 + √ 10 b) Distribute 2 √x inside the parentheses: = (2 · 3) ( √x · √y ) −2 · ( √x · √x ) Multiply: = 6 √xy −2 √ x2 Simplify: = 6 √xy −2x c) Distribute: (2 + √ 5)(2 − √ 6) = (2 · 2) − ( 2 · √ 6 ) + ( 2 · √ 5 ) − ( √ 5 · √ 6 ) Simplify: = 4 −2 √ 6 + 2 √ 5 − √ 30 d) Distribute: ( 2 √x −1 ) ( 5 −√x ) = 10 √x −2x −5 + √x Simplify: = 11 √x −2x −5 Rationalize the Denominator Often when we work with radicals, we end up with a radical expression in the denominator of a fraction. It’s traditional to write our fractions in a form that doesn’t have radicals in the denominator, so we use a process called rationalizing the denominator to eliminate them. Rationalizing is easiest when there’s just a radical and nothing else in the denominator, as in the fraction 2 √ 3. All we have to do then is multiply the numerator and denominator by a radical expression that makes the expression inside the radical into a perfect square, cube, or whatever power is appropriate. In the example above, we multiply by √ 3: 2 √ 3 · √ 3 √ 3 = 2 √ 3 3 Cube roots and higher are a little trickier than square roots. For example, how would we rationalize 7 3 √ 5? We can’t just multiply by 3 √ 5, because then the denominator would be 3 √ 52. To make the denominator a whole number, we need to multiply the numerator and the denominator by 3 √ 52: 7 3 √ 5 · 3 √ 52 3 √ 52 = 7 3 √ 25 3 √ 53 = 7 3 √ 25 5 Trickier still is when the expression in the denominator contains more than one term. For example, consider the expression 2 2+ √ 3. We can’t just multiply by √ 3, because we’d have to distribute that term and then the denominator would be 2 √ 3 + 3. Instead, we multiply by 2 − √ 3. This is a good choice because the product ( 2 + √ 3 ) ( 2 − √ 3 ) is a product of a sum and a diﬀerence, which means it’s a diﬀerence of squares. The radicals cancel each other out when we multiply out, and the denominator works out to ( 2 + √ 3 ) ( 2 − √ 3 ) = 22 − ( √ 3 )2 = 4 −3 = 1. When we multiply both the numerator and denominator by 2 − √ 3, we get: www.ck12.org 502 2 2 + √ 3 · 2 − √ 3 2 − √ 3 = 2 ( 2 − √ 3 ) 4 −3 = 4 −2 √ 3 1 = 4 −2 √ 3 Now consider the expression √x−1 √x−2 √y. In order to eliminate the radical expressions in the denominator we must multiply by √x + 2 √y. We get: √x−1 √x−2 √y · √x+2 √y √x+2 √y = ( √x−1)( √x+2 √y) ( √x−2 √y)( √x+2 √y) = x−2 √y−√x+2 √xy x−4y Solve Real-World Problems Using Radical Expressions Radicals often arise in problems involving areas and volumes of geometrical ﬁgures. Example 9 A pool is twice as long as it is wide and is surrounded by a walkway of uniform width of 1 foot. The combined area of the pool and the walkway is 400 square feet. Find the dimensions of the pool and the area of the pool. Solution Make a sketch: Let x = the width of the pool. Then: Area = length × width Combined length of pool and walkway = 2x + 2 Combined width of pool and walkway = x + 2 Area = (2x + 2)(x + 2) Since the combined area of pool and walkway is 400 ft2 we can write the equation (2x + 2)(x + 2) = 400 503 www.ck12.org Multiply in order to eliminate the parentheses: 2x2 + 4x + 2x + 4 = 400 Collect like terms: 2x2 + 6x + 4 = 400 Move all terms to one side of the equation: 2x2 + 6x −396 = 0 Divide all terms by 2: x2 + 3x −198 = 0 Use the quadratic formula: x = −b ± √ b2 −4ac 2a x = −3 ± √ 32 −4(1)(−198) 2(1) x = −3 ± √ 801 2 = −3 ± 28.3 2 x = 12.65 feet (The other answer is negative, so we can throw it out because only a positive number makes sense for the width of a swimming pool.) Check by plugging the result in the area formula: Area = (2(12.65) + 2)(12.65 + 2) = 27.3 · 14.65 = 400 ft2. The answer checks out. Example 10 The volume of a soda can is 355 cm3. The height of the can is four times the radius of the base. Find the radius of the base of the cylinder. Solution Make a sketch: Let x = the radius of the cylinder base. Then the height of the cylinder is 4x. The volume of a cylinder is given by V = πR2 · h; in this case, R is x and h is 4x, and we know the volume is 355. Solve the equation: 355 = πx2 · (4x) 355 = 4πx3 x3 = 355 4π x = 3 √ 355 4π = 3.046 cm www.ck12.org 504 Check by substituting the result back into the formula: V = πR2 · h = π(3.046)2 · (4 · 3.046) = 355 cm3 So the volume is 355 cm3. The answer checks out. Review Questions Evaluate each radical expression. 1. √ 169 2. 4 √ −81 3. 3 √ −125 4. 5 √ 1024 Write each expression as a rational exponent. 5. 3 √ 14 6. 4 √zw 7. √a 8. 9 √ y3 Write the following expressions in simplest radical form. 9. √ 24 10. √ 300 11. 5 √ 96 12. √ 240 567 13. 3 √ 500 14. 6 √ 64x8 15. 3 √ 48a3b7 16. 3 √ 16x5 135y4 Simplify the following expressions as much as possible. 17. 3 √ 8 −6 √ 32 18. √ 180 + √ 405 19. √ 6 − √ 27 + 2 √ 54 + 3 √ 48 20. √ 8x3 −4x √ 98x 21. √ 48a + √ 27a 22. 3 √ 4x3 + x · 3 √ 256 Multiply the following expressions. 23. √ 6 ( √ 10 + √ 8 ) 24. ( √a − √ b ) ( √a + √ b ) 505 www.ck12.org 25. ( 2 √x + 5 ) ( 2 √x + 5 ) Rationalize the denominator. 26. 7 √ 5 27. 9 √ 10 28. 2x √ 5x 29. √ 5 √3y 30. 12 2− √ 5 31. 6+ √ 3 4− √ 3 32. √x √ 2+ √x 33. 5y 2 √y−5 34. The volume of a spherical balloon is 950 cm3. Find the radius of the balloon. (Volume of a sphere = 4 3πR3). 35. A rectangular picture is 9 inches wide and 12 inches long. The picture has a frame of uniform width. If the combined area of picture and frame is 180 in2, what is the width of the frame? 11.3 Radical Equations Learning Objectives • Solve a radical equation. • Solve radical equations with radicals on both sides. • Identify extraneous solutions. • Solve real-world problems using square root functions. Introduction When the variable in an equation appears inside a radical sign, the equation is called a radical equation. To solve a radical equation, we need to eliminate the radical and change the equation into a polynomial equation. A common method for solving radical equations is to isolate the most complicated radical on one side of the equation and raise both sides of the equation to the power that will eliminate the radical sign. If there are any radicals left in the equation after simplifying, we can repeat this procedure until all radical signs are gone. Once the equation is changed into a polynomial equation, we can solve it with the methods we already know. We must be careful when we use this method, because whenever we raise an equation to a power, we could introduce false solutions that are not in fact solutions to the original problem. These are called extraneous solutions. In order to make sure we get the correct solutions, we must always check all solutions in the original radical equation. Solve a Radical Equation Let’s consider a few simple examples of radical equations where only one radical appears in the equation. www.ck12.org 506 Example 1 Find the real solutions of the equation √ 2x −1 = 5. Solution Since the radical expression is already isolated, we can just square both sides of the equation in order to eliminate the radical sign: ( √ 2x −1 )2 = 52 Remember that √ a2 = a so the equation simpliﬁes to: 2x −1 = 25 Add one to both sides: 2x = 26 Divide both sides by 2: x = 13 Finally we need to plug the solution in the original equation to see if it is a valid solution. √ 2x −1 = √ 2(13) −1 = √ 26 −1 = √ 25 = 5 The solution checks out. Example 2 Find the real solutions of 3 √ 3 −7x −3 = 0. Solution We isolate the radical on one side of the equation: 3 √ 3 −7x = 3 Raise each side of the equation to the third power: ( 3 √ 3 −7x )3 = 33 Simplify: 3 −7x = 27 Subtract 3 from each side: −7x = 24 Divide both sides by –7: x = −24 7 Check: 3 √ 3 −7x −3 = 3 √ 3 −7 ( −24 7 ) −3 = 3 √3 + 24 −3 = 3 √ 27 −3 = 3 −3 = 0. The solution checks out. Example 3 Find the real solutions of √ 10 −x2 −x = 2. Solution We isolate the radical on one side of the equation: √ 10 −x2 = 2 + x Square each side of the equation: ( √ 10 −x2)2 = (2 + x)2 Simplify: 10 −x2 = 4 + 4x + x2 Move all terms to one side of the equation: 0 = 2x2 + 4x −6 Solve using the quadratic formula: x = −4 ± √ 42 −4(2)(−6) 4 Simplify: x = −4 ± √ 64 4 Re-write √ 24 in simplest form: x = −4 ± 8 4 Reduce all terms by a factor of 2: x = 1 or x = −3 507 www.ck12.org Check: √ 10 −12 −1 = √ 9 −1 = 3 −1 = 2 This solution checks out. √ 10 −(−3)2 −(−3) = √ 1 + 3 = 1 + 3 = 4 This solution does not check out. The equation has only one solution, x = 1; the solution x = −3 is extraneous. Solve Radical Equations with Radicals on Both Sides Often equations have more than one radical expression. The strategy in this case is to start by isolating the most complicated radical expression and raise the equation to the appropriate power. We then repeat the process until all radical signs are eliminated. Example 4 Find the real roots of the equation √2x + 1 − √ x −3 = 2. Solution Isolate one of the radical expressions: √ 2x + 1 = 2 + √ x −3 Square both sides: ( √ 2x + 1 )2 = ( 2 + √ x −3 )2 Eliminate parentheses: 2x + 1 = 4 + 4 √ x −3 + x −3 Simplify: x = 4 √ x −3 Square both sides of the equation: x2 = ( 4 √ x −3 )2 Eliminate parentheses: x2 = 16(x −3) Simplify: x2 = 16x −48 Move all terms to one side of the equation: x2 −16x + 48 = 0 Factor: (x −12)(x −4) = 0 Solve: x = 12 or x = 4 Check: √ 2(12) + 1 − √ 12 −3 = √ 25 − √ 9 = 5 −3 = 2. The solution checks out. √ 2(4) + 1 − √ 4 −3 = √ 9 − √ 1 = 3 −1 = 2 The solution checks out. The equation has two solutions: x = 12 and x = 4. Identify Extraneous Solutions to Radical Equations We saw in Example 3 that some of the solutions that we ﬁnd by solving radical equations do not check out when we substitute (or “plug in”) those solutions back into the original radical equation. These are called extraneous solutions. It is very important to check the answers we obtain by plugging them back into the original equation, so we can tell which of them are real solutions. Example 5 Find the real solutions of the equation √ x −3 −√x = 1. Solution www.ck12.org 508 Isolate one of the radical expressions: √ x −3 = √x + 1 Square both sides: ( √ x −3 )2 = ( √x + 1 )2 Remove parenthesis: x −3 = ( √x )2 + 2 √x + 1 Simplify: x −3 = x + 2 √x + 1 Now isolate the remaining radical: −4 = 2 √x Divide all terms by 2: −2 = √x Square both sides: x = 4 Check: √ 4 −3 − √ 4 = √ 1 −2 = 1 −2 = −1 The solution does not check out. The equation has no real solutions. x = 4 is an extraneous solution. Solve Real-World Problems using Radical Equations Radical equations often appear in problems involving areas and volumes of objects. Example 6 Anita’s square vegetable garden is 21 square feet larger than Fred’s square vegetable garden. Anita and Fred decide to pool their money together and buy the same kind of fencing for their gardens. If they need 84 feet of fencing, what is the size of each garden? Solution Make a sketch: Deﬁne variables: Let Fred’s area be x; then Anita’s area is x + 21. Find an equation: Side length of Fred’s garden is √x Side length of Anita’s garden is √x + 21 The amount of fencing is equal to the combined perimeters of the two squares: 4 √x + 4 √ x + 21 = 84 Solve the equation: 509 www.ck12.org Divide all terms by 4: √x + √ x + 21 = 21 Isolate one of the radical expressions: √ x + 21 = 21 −√x Square both sides: ( √ x + 21 )2 = ( 21 −√x )2 Eliminate parentheses: x + 21 = 441 −42 √x + x Isolate the radical expression: 42 √x = 420 Divide both sides by 42: √x = 10 Square both sides: x = 100 ft2 Check: 4 √ 100 + 4 √100 + 21 = 40 + 44 = 84. The solution checks out. Fred’s garden is 10 ft × 10 ft = 100 ft2 and Anita’s garden is 11 ft × 11 ft = 121 ft2. Example 7 A sphere has a volume of 456 cm3. If the radius of the sphere is increased by 2 cm, what is the new volume of the sphere? Solution Make a sketch: Deﬁne variables: Let R = the radius of the sphere. Find an equation: The volume of a sphere is given by the formula V = 4 3πR3. Solve the equation: Plug in the value of the volume: 456 = 4 3πR3 Multiply by 3: 1368 = 4πR3 Divide by 4π : 108.92 = R3 Take the cube root of each side: R = 3 √ 108.92 ⇒R = 4.776 cm The new radius is 2 centimeters more: R = 6.776 cm The new volume is: V = 4 3π(6.776)3 = 1302.5 cm3 Check: Let’s plug in the values of the radius into the volume formula: V = 4 3πR3 = 4 3π(4.776)3 = 456 cm3. The solution checks out. Example 8 The kinetic energy of an object of mass m and velocity v is given by the formula: KE = 1 2mv2. A baseball has a mass of 145 kg and its kinetic energy is measured to be 654 Joules (kg · m2/s2) when it hits the catcher’s glove. What is the velocity of the ball when it hits the catcher’s glove? www.ck12.org 510 Solution Start with the formula: KE = 1 2mv2 Plug in the values for the mass and the kinetic energy: 654kg · m2 s2 = 1 2(145 kg)v2 Multiply both sides by 2: 1308kg · m2 s2 = 145 kg · v2 Divide both sides by 145 kg : 9.02m2 s2 = v2 Take the square root of both sides: v = √ 9.02 √ m2 s2 = 3.003 m/s Check: Plug the values for the mass and the velocity into the energy formula: KE = 1 2mv2 = 1 2(145 kg)(3.003 m/s)2 = 654 kg · m2/s2 (To learn more about kinetic energy, watch the video at .) Figure 11.1: Eureka! Episode 9 - Kinetic Energy (Watch Youtube Video) Review Questions Find the solution to each of the following radical equations. Identify extraneous solutions. 1. √x + 2 −2 = 0 2. √ 3x −1 = 5 3. 2 √ 4 −3x + 3 = 0 4. 3 √ x −3 = 1 5. 4 √ x2 −9 = 2 6. 3 √ −2 −5x + 3 = 0 7. √ x2 −3 = x −1 8. √x = x −6 9. √ x2 −5x −6 = 0 10. √ (x + 1)(x −3) = x 11. √x + 6 = x + 4 12. √x = √ x −9 + 1 13. √x + 2 = √ 3x −2 14. √3x + 4 = −6 511 www.ck12.org 15. 5 √x = √x + 12 + 6 16. √ 10 −5x + √ 1 −x = 7 17. √ 2x −2 −2 √x + 2 = 0 18. √2x + 5 −3 √ 2x −3 = √ 2 −x 19. 3 √x −9 = √ 2x −14 20. √x + 7 = √x + 4 + 1 21. The area of a triangle is 24 in2 and the height of the triangle is twice as long as the base. What are the base and the height of the triangle? 22. The length of a rectangle is 7 meters less than twice its width, and its area is 660 m2. What are the length and width of the rectangle? 23. The area of a circular disk is 124 in2. What is the circumference of the disk? (Area = πR2, Circumference = 2πR). 24. The volume of a cylinder is 245 cm3 and the height of the cylinder is one third of the diameter of the base of the cylinder. The diameter of the cylinder is kept the same but the height of the cylinder is increased by 2 centimeters. What is the volume of the new cylinder? (Volume = πR2 · h) 25. The height of a golf ball as it travels through the air is given by the equation h = −16t2 + 256. Find the time when the ball is at a height of 120 feet. 11.4 The Pythagorean Theorem and Its Converse Learning Objectives • Use the Pythagorean Theorem. • Use the converse of the Pythagorean Theorem. • Solve real-world problems using the Pythagorean Theorem and its converse. Introduction Teresa wants to string a clothesline across her backyard, from one corner to the opposite corner. If the yard measures 22 feet by 34 feet, how many feet of clothesline does she need? The Pythagorean Theorem is a statement of how the lengths of the sides of a right triangle are related to each other. A right triangle is one that contains a 90 degree angle. The side of the triangle opposite the 90 degree angle is called the hypotenuse and the sides of the triangle adjacent to the 90 degree angle are called the legs. If we let a and b represent the legs of the right triangle and c represent the hypotenuse then the Pythagorean Theorem can be stated as: www.ck12.org 512 In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. That is: a2 + b2 = c2. This theorem is very useful because if we know the lengths of the legs of a right triangle, we can ﬁnd the length of the hypotenuse. Also, if we know the length of the hypotenuse and the length of a leg, we can calculate the length of the missing leg of the triangle. When you use the Pythagorean Theorem, it does not matter which leg you call a and which leg you call b, but the hypotenuse is always called c. Although nowadays we use the Pythagorean Theorem as a statement about the relationship between distances and lengths, originally the theorem made a statement about areas. If we build squares on each side of a right triangle, the Pythagorean Theorem says that the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares formed by the legs of the triangle. Use the Pythagorean Theorem and Its Converse The Pythagorean Theorem can be used to verify that a triangle is a right triangle. If you can show that the three sides of a triangle make the equation a2 +b2 = c2 true, then you know that the triangle is a right triangle. This is called the Converse of the Pythagorean Theorem. Note: When you use the Converse of the Pythagorean Theorem, you must make sure that you substitute the correct values for the legs and the hypotenuse. The hypotenuse must be the longest side. The other two sides are the legs, and the order in which you use them is not important. Example 1 Determine if a triangle with sides 5, 12 and 13 is a right triangle. Solution The triangle is right if its sides satisfy the Pythagorean Theorem. If it is a right triangle, the longest side has to be the hypotenuse, so we let c = 13. We then designate the shorter sides as a = 5 and b = 12. We plug these values into the Pythagorean Theorem: a2 + b2 = c2 ⇒52 + 122 = c2 25 + 144 = 169 = c2 ⇒c = 13 The sides of the triangle satisfy the Pythagorean Theorem, thus the triangle is a right triangle. Example 2 Determine if a triangle with sides, √ 10, √ 15 and 5 is a right triangle. Solution The longest side has to be the hypotenuse, so c = 5. 513 www.ck12.org We designate the shorter sides as a = √ 10 and b = √ 15. We plug these values into the Pythagorean Theorem: a2 + b2 = c2 ⇒ ( √ 10 )2 + ( √ 15 )2 = c2 10 + 15 = 25 = c2 ⇒c = 5 The sides of the triangle satisfy the Pythagorean Theorem, thus the triangle is a right triangle. The Pythagorean Theorem can also be used to ﬁnd the missing hypotenuse of a right triangle if we know the legs of the triangle. (For a demonstration of this, see .) Figure 11.2: Pythagorean Theorem in 60 Seconds (Watch Youtube Video) Example 3 In a right triangle one leg has length 4 and the other has length 3. Find the length of the hypotenuse. Solution Start with the Pythagorean Theorem: a2 + b2 = c2 Plug in the known values of the legs: 32 + 42 = c2 Simplify: 9 + 16 = c2 25 = c2 Take the square root of both sides: c = 5 Use the Pythagorean Theorem with Variables Example 4 Determine the values of the missing sides. You may assume that each triangle is a right triangle. www.ck12.org 514 a) b) c) Solution Apply the Pythagorean Theorem. a) a2 + b2 = c2 x2 + 152 = 212 x2 + 225 = 441 x2 = 216 ⇒ x = √ 216 = 6 √ 6 b) a2 + b2 = c2 y2 + 32 = 72 y2 + 9 = 49 y2 = 40 ⇒ y = √ 40 = 2 √ 10 c) a2 + b2 = c2 182 + 152 = z2 324 + 225 = z z2 = 549 ⇒ z = √ 549 = 3 √ 61 Example 5 515 www.ck12.org One leg of a right triangle is 5 units longer than the other leg. The hypotenuse is one unit longer than twice the size of the short leg. Find the dimensions of the triangle. Solution Let x = length of the short leg. Then x + 5 = length of the long leg And 2x + 1 = length of the hypotenuse. The sides of the triangle must satisfy the Pythagorean Theorem. Therefore: x2 + (x + 5)2 = (2x + 1)2 Eliminate the parentheses: x2 + x2 + 10x + 25 = 4x2 + 4x + 1 Move all terms to the right hand side of the equation: 0 = 2x2 −6x −24 Divide all terms by 2 : 0 = x2 −3x −12 Solve using the quadratic formula: x = 3 ± √9 + 48 2 = 3 ± √ 57 2 x = 5.27 or x = −2.27 The negative solution doesn’t make sense when we are looking for a physical distance, so we can discard it. Using the positive solution, we get: short leg = 5.27, long leg = 10.27 and hypotenuse = 11.54. Solve Real-World Problems Using the Pythagorean Theorem and Its Converse The Pythagorean Theorem and its converse have many applications for ﬁnding lengths and distances. Example 6 Maria has a rectangular cookie sheet that measures 10 inches × 14 inches. Find the length of the diagonal of the cookie sheet. Solution Draw a sketch: Deﬁne variables: Let c = length of the diagonal. Write a formula: Use the Pythagorean Theorem: a2 + b2 = c2 Solve the equation: 102 + 142 = c2 100 + 196 = c2 c2 = 296 ⇒c = √ 296 ⇒c = 2 √ 74 or c = 17.2 inches www.ck12.org 516 Check: 102 + 142 = 100 + 196 = 296 and c2 = 17.22 = 296. The solution checks out. Example 7 Find the area of the shaded region in the following diagram: Solution Draw the diagonal of the square in the ﬁgure: Notice that the diagonal of the square is also the diameter of the circle. Deﬁne variables: Let c = diameter of the circle. Write the formula: Use the Pythagorean Theorem: a2 + b2 = c2. Solve the equation: 22 + 22 = c2 4 + 4 = c2 c2 = 8 ⇒c = √ 8 ⇒c = 2 √ 2 The diameter of the circle is 2 √ 2, therefore the radius R = √ 2. Area of a circle formula: A = π · R2 = π ( √ 2 )2 = 2π. The area of the shaded region is therefore 2π −4 = 2.28. Example 8 In a right triangle, one leg is twice as long as the other and the perimeter is 28. What are the measures of the sides of the triangle? Solution Make a sketch and deﬁne variables: 517 www.ck12.org Let: a = length of the short leg 2a = length of the long leg c = length of the hypotenuse Write formulas: The sides of the triangle are related in two diﬀerent ways. The perimeter is 28, so a + 2a + c = 28 ⇒3a + c = 28 The triangle is a right triangle, so the measures of the sides must satisfy the Pythagorean Theorem: a2 + (2a)2 = c2 ⇒a2 + 4a2 = c2 ⇒5a2 = c2 or c = a √ 5 = 2.236a Solve the equation: Plug the value of c we just obtained into the perimeter equation: 3a + c = 28 3a + 2.236a = 28 ⇒5.236a = 28 ⇒a = 5.35 The short leg is: a = 5.35 The long leg is: 2a = 10.70 The hypotenuse is: c = 11.95 Check: The legs of the triangle should satisfy the Pythagorean Theorem: a2 + b2 = 5.352 + 10.702 = 143.1, c2 = 11.952 = 142.80. The results are approximately the same. The perimeter of the triangle should be 28: a + b + c = 5.35 + 10.70 + 11.95 = 28 The answer checks out. Example 9 Mike is loading a moving van by walking up a ramp. The ramp is 10 feet long and the bed of the van is 2.5 feet above the ground. How far does the ramp extend past the back of the van? Solution Make a sketch: Deﬁne variables: Let x = how far the ramp extends past the back of the van. Write a formula: Use the Pythagorean Theorem: x2 + 2.52 = 102 www.ck12.org 518 Solve the equation: x2 + 6.25 = 100 x2 = 93.5 x = √ 93.5 = 9.7 ft Check by plugging the result in the Pythagorean Theorem: 9.72 + 2.52 = 94.09 + 6.25 = 100.34 ≈100. So the ramp is 10 feet long. The answer checks out. Review Questions Verify that each triangle is a right triangle. 1. a = 12, b = 9, c = 15 2. a = 6, b = 6, c = 6 √ 2 3. a = 8, b = 8 √ 3, c = 16 4. a = 2 √ 14, b = 5, c = 9 Find the missing length of each right triangle. 5. a = 12, b = 16, c =? 6. a =?, b = 20, c = 30 7. a = 4, b =?, c = 11 8. 9. 10. 11. One leg of a right triangle is 4 feet less than the hypotenuse. The other leg is 12 feet. Find the lengths of the three sides of the triangle. 12. One leg of a right triangle is 3 more than twice the length of the other. The hypotenuse is 3 times the length of the short leg. Find the lengths of the three legs of the triangle. 519 www.ck12.org 13. Two sides of a right triangle are 5 units and 8 units respectively. Those sides could be the legs, or they could be one leg and the hypotenuse. What are the possible lengths of the third side? 14. A regulation baseball diamond is a square with 90 feet between bases. How far is second base from home plate? 15. Emanuel has a cardboard box that measures 20 cm long × 10 cm wide × 8 cm deep. (a) What is the length of the diagonal across the bottom of the box? (b) What is the length of the diagonal from a bottom corner to the opposite top corner? 16. Samuel places a ladder against his house. The base of the ladder is 6 feet from the house and the ladder is 10 feet long. (a) How high above the ground does the ladder touch the wall of the house? (b) If the edge of the roof is 10 feet oﬀthe ground and sticks out 1.5 feet beyond the wall, how far is it from the edge of the roof to the top of the ladder? 17. Find the area of the triangle below if the area of a triangle is deﬁned as A = 1 2 base × height: 18. Instead of walking along the two sides of a rectangular ﬁeld, Mario decided to cut across the diagonal. He thus saves a distance that is half of the long side of the ﬁeld. (a) Find the length of the long side of the ﬁeld given that the short side is 123 feet. (b) Find the length of the diagonal. 19. Marcus sails due north and Sandra sails due east from the same starting point. In two hours Marcus’ boat is 35 miles from the starting point and Sandra’s boat is 28 miles from the starting point. (a) How far are the boats from each other? (b) Sandra then sails 21 miles due north while Marcus stays put. How far is Sandra from the original starting point? (c) How far is Sandra from Marcus now? 20. Determine the area of the circle below. (Hint: the hypotenuse of the triangle is the diameter of the circle.) 11.5 Distance and Midpoint Formulas Learning Objectives • Find the distance between two points in the coordinate plane. • Find the missing coordinate of a point given the distance from another known point. www.ck12.org 520 • Find the midpoint of a line segment. • Solve real-world problems using distance and midpoint formulas. Introduction In the last section, we saw how to use the Pythagorean Theorem to ﬁnd lengths. In this section, you’ll learn how to use the Pythagorean Theorem to ﬁnd the distance between two coordinate points. Example 1 Find the distance between points A = (1, 4) and B = (5, 2). Solution Plot the two points on the coordinate plane. In order to get from point A = (1, 4) to point B = (5, 2), we need to move 4 units to the right and 2 units down. These lines make the legs of a right triangle. To ﬁnd the distance between A and B we ﬁnd the value of the hypotenuse, d, using the Pythagorean Theorem. d2 = 22 + 42 = 20 d = √ 20 = 2 √ 5 = 4.47 Example 2 Find the distance between points C = (2, −1) and D = (−3, −4). Solution We plot the two points on the graph above. In order to get from point C to point D, we need to move 3 units down and 5 units to the left. We ﬁnd the distance from C to D by ﬁnding the length of d with the Pythagorean Theorem. d2 = 32 + 52 = 34 d = √ 34 = 5.83 The Distance Formula The procedure we just used can be generalized by using the Pythagorean Theorem to derive a formula for the distance between any two points on the coordinate plane. 521 www.ck12.org Let’s ﬁnd the distance between two general points A = (x1, y1) and B = (x2, y2). Start by plotting the points on the coordinate plane: In order to move from point A to point B in the coordinate plane, we move x2 −x1 units to the right and y2 −y1 units up. We can ﬁnd the length d by using the Pythagorean Theorem: d2 = (x1 −x2)2 + (y1 −y2)2 Therefore, d = √ (x1 −x2)2 + (y1 −y2)2. This is called the Distance Formula. More formally: Given any two points (x1, y1) and (x2, y2), the distance between them is d = √ (x1 −x2)2 + (y1 −y2)2. We can use this formula to ﬁnd the distance between any two points on the coordinate plane. Notice that the distance is the same whether you are going from point A to point B or from point B to point A, so it does not matter which order you plug the points into the distance formula. Let’s now apply the distance formula to the following examples. Example 2 Find the distance between the following points. a) (-3, 5) and (4, -2) b) (12, 16) and (19, 21) c) (11.5, 2.3) and (-4.2, -3.9) Solution Plug the values of the two points into the distance formula. Be sure to simplify if possible. a) d = √ (−3 −4)2 + (5 −(−2))2 = √ (−7)2 + (7)2 = √49 + 49 = √ 98 = 7 √ 2 b) d = √ (12 −19)2 + (16 −21)2 = √ (−7)2 + (−5)2 = √49 + 25 = √ 74 c) d = √ (11.5 + 4.2)2 + (2.3 + 3.9)2 = √ (15.7)2 + (6.2)2 = √ 284.93 = 16.88 We can also use the Pythagorean Theorem Example 3 Find all points on the line y = 2 that are exactly 8 units away from the point (-3, 7). Solution Let’s make a sketch of the given situation. Draw line segments from the point (-3, 7) to the line y = 2. www.ck12.org 522 Let k be the missing value of x we are seeking. Let’s use the distance formula: 8 = √ (−3 −k)2 + (7 −2)2 Square both sides of the equation: 64 = (−3 −k)2 + 25 Therefore: 0 = 9 + 6k + k2 −39 or 0 = k2 + 6k −30 Use the quadratic formula: k = −6 ± √36 + 120 2 = −6 ± √ 156 2 Therefore: k = 3.24 or k = −9.24 The points are (-9.24, 2) and (3.24, 2). Find the Midpoint of a Line Segment Example 4 Find the coordinates of the point that is in the middle of the line segment connecting the points A = (−7, −2) and B = (3, −8). Solution Let’s start by graphing the two points: We see that to get from point A to point B we move 6 units down and 10 units to the right. In order to get to the point that is halfway between the two points, it makes sense that we should move half the vertical distance and half the horizontal distance—that is, 3 units down and 5 units to the right from point A. 523 www.ck12.org The midpoint is M = (−7 + 5, −2 −3) = (−2, −5). The Midpoint Formula We now want to generalize this method in order to ﬁnd a formula for the midpoint of a line segment. Let’s take two general points A = (x1, y1) and B = (x2, y2) and mark them on the coordinate plane: We see that to get from A to B, we move x2 −x1 units to the right and y2 −y1 units up. In order to get to the half-way point, we need to move x2−x1 2 units to the right and y2−y1 2 up from point A. Thus the midpoint M is at ( x1 + x2−x1 2 , y1 + y2−y1 2 ) . This simpliﬁes to M = ( x1+x2 2 , y1+y2 2 ) . This is the Midpoint Formula: The midpoint of the line segment connecting the points (x1, y1) and (x2, y2) is ( x1+x2 2 , y1+y2 2 ) . It should hopefully make sense that the midpoint of a line is found by taking the average values of the x and y−values of the endpoints. For a graphic demonstration of the midpoint formula, watch this video: v=bcp9pJxaAOk . Figure 11.3: The Midpoint Formula (Watch Youtube Video) Example 5 Find the midpoint between the following points. a) (-10, 2) and (3, 5) b) (3, 6) and (7, 6) c) (4, -5) and (-4, 5) Solution Let’s apply the Midpoint Formula: ( x1+x2 2 , y1+y2 2 ) www.ck12.org 524 a) the midpoint of (-10, 2) and (3, 5) is ( −10+3 2 , 2+5 2 ) = ( −7 2 , 7 2 ) = (−3.5, 3.5) b) the midpoint of (3, 6) and (7, 6) is ( 3+7 2 , 6+6 2 ) = ( 10 2 , 12 2 ) = (5, 6) c) the midpoint of (4, -5) and (-4, 5) is ( 4−4 2 , −5+5 2 ) = (0 2, 0 2 ) = (0, 0) Example 6 A line segment whose midpoint is (2, -6) has an endpoint of (9, -2). What is the other endpoint? Solution In this problem we know the midpoint and we are looking for the missing endpoint. The midpoint is (2, -6). One endpoint is (x1, x2) = (9, −2). Let’s call the missing point (x, y). We know that the x−coordinate of the midpoint is 2, so: 2 = 9+x2 2 ⇒4 = 9 + x2 ⇒x2 = −5 We know that the y−coordinate of the midpoint is -6, so: −6 = −2 + y2 2 ⇒−12 = −2 + y2 ⇒y2 = −10 The missing endpoint is (-5, -10). Here’s another way to look at this problem: To get from the endpoint (9, -2) to the midpoint (2, 6), we had to go 7 units left and 4 units down. To get from the midpoint to the other endpoint, then, we would need to go 7 more units left and 4 more units down, which takes us to (-5, -10). Solve Real-World Problems Using Distance and Midpoint For-mulas The distance and midpoint formula are useful in geometry situations where we want to ﬁnd the distance between two points or the point halfway between two points. Example 7 Plot the points A = (4, −2), B = (5, 5), and C = (−1, 3) and connect them to make a triangle. Show that the triangle is isosceles. Solution Let’s start by plotting the three points on the coordinate plane and making a triangle: 525 www.ck12.org We use the distance formula three times to ﬁnd the lengths of the three sides of the triangle. AB = √ (4 −5)2 + (−2 −5)2 = √ (−1)2 + (−7)2 = √ 1 + 49 = √ 50 = 5 √ 2 BC = √ (5 + 1)2 + (5 −3)2 = √ (6)2 + (2)2 = √ 36 + 4 = √ 40 = 2 √ 10 AC = √ (4 + 1)2 + (−2 −3)2 = √ (5)2 + (−5)2 = √ 25 + 25 = √ 50 = 5 √ 2 Notice that AB = AC, therefore triangle ABC is isosceles. Example 8 At 8 AM one day, Amir decides to walk in a straight line on the beach. After two hours of making no turns and traveling at a steady rate, Amir is two miles east and four miles north of his starting point. How far did Amir walk and what was his walking speed? Solution Let’s start by plotting Amir’s route on a coordinate graph. We can place his starting point at the origin: A = (0, 0). Then his ending point will be at B = (2, 4). The distance can be found with the distance formula: d = √ (2 −0)2 + (4 −0)2 = √ (2)2 + (4)2 = √ 4 + 16 = √ 20 d = 4.47 miles Since Amir walked 4.47 miles in 2 hours, his speed is s = 4.47 miles 2 hours = 2.24 mi/h. Review Questions Find the distance between the two points. 1. (3, -4) and (6, 0) 2. (-1, 0) and (4, 2) 3. (-3, 2) and (6, 2) 4. (0.5, -2.5) and (4, -4) 5. (12, -10) and (0, -6) 6. (-5, -3) and (-2, 11) 7. (2.3, 4.5) and (-3.4, -5.2) www.ck12.org 526 8. Find all points having an x−coordinate of -4 whose distance from the point (4, 2) is 10. 9. Find all points having a y−coordinate of 3 whose distance from the point (-2, 5) is 8. 10. Find three points that are each 13 units away from the point (3, 2) but do not have an x−coordinate of 3 or a y−coordinate of 2. Find the midpoint of the line segment joining the two points. 11. (3, -4) and (6, 1) 12. (2, -3) and (2, 4) 13. (4, -5) and (8, 2) 14. (1.8, -3.4) and (-0.4, 1.4) 15. (5, -1) and (-4, 0) 16. (10, 2) and (2, -4) 17. (3, -3) and (2, 5) 18. An endpoint of a line segment is (4, 5) and the midpoint of the line segment is (3, -2). Find the other endpoint. 19. An endpoint of a line segment is (-10, -2) and the midpoint of the line segment is (0, 4). Find the other endpoint. 20. Find a point that is the same distance from (4, 5) as it is from (-2, -1), but is not the midpoint of the line segment connecting them. 21. Plot the points A = (1, 0), B = (6, 4),C = (9, −2) and D = (−6, −4), E = (−1, 0), F = (2, −6). Prove that triangles ABC and DEF are congruent. 22. Plot the points A = (4, −3), B = (3, 4),C = (−2, −1), D = (−1, −8). Show that ABCD is a rhombus (all sides are equal) 23. Plot points A = (−5, 3), B = (6, 0),C = (5, 5). Find the length of each side. Show that ABC is a right triangle. Find its area. 24. Find the area of the circle with center (-5, 4) and the point on the circle (3, 2). 25. Michelle decides to ride her bike one day. First she rides her bike due south for 12 miles and then the direction of the bike trail changes and she rides in the new direction for a while longer. When she stops Michelle is 2 miles south and 10 miles west from her starting point. Find the total distance that Michelle covered from her starting point. 527 www.ck12.org Chapter 12 Rational Equations and Functions 12.1 Inverse Variation Models Learning Objectives • Distinguish direct and inverse variation. • Graph inverse variation equations. • Write inverse variation equations. • Solve real-world problems using inverse variation equations. Introduction Many variables in real-world problems are related to each other by variations. A variation is an equation that relates a variable to one or more other variables by the operations of multiplication and division. There are three diﬀerent kinds of variation: direct variation, inverse variation and joint variation. Distinguish Direct and Inverse Variation In direct variation relationships, the related variables will either increase together or decrease together at a steady rate. For instance, consider a person walking at three miles per hour. As time increases, the distance covered by the person walking also increases, at the rate of three miles each hour. The distance and time are related to each other by a direct variation: distance = speed × time Since the speed is a constant 3 miles per hour, we can write: d = 3t. The general equation for a direct variation is y = kx, where k is called the constant of proportionality. You can see from the equation that a direct variation is a linear equation with a y−intercept of zero. The graph of a direct variation relationship is a straight line passing through the origin whose slope is k, the constant of proportionality. www.ck12.org 528 A second type of variation is inverse variation. When two quantities are related to each other inversely, one quantity increases as the other one decreases, and vice versa. For instance, if we look at the formula distance = speed × time again and solve for time, we obtain: time = distance speed If we keep the distance constant, we see that as the speed of an object increases, then the time it takes to cover that distance decreases. Consider a car traveling a distance of 90 miles, then the formula relating time and speed is: t = 90 s . The general equation for inverse variation is y = k x, where k is the constant of proportionality. In this chapter, we’ll investigate how the graphs of these relationships behave. Another type of variation is a joint variation. In this type of relationship, one variable may vary as a product of two or more variables. For example, the volume of a cylinder is given by: V = πR2 · h In this example the volume varies directly as the product of the square of the radius of the base and the height of the cylinder. The constant of proportionality here is the number π. In many application problems, the relationship between the variables is a combination of variations. For instance Newton’s Law of Gravitation states that the force of attraction between two spherical bodies varies jointly as the masses of the objects and inversely as the square of the distance between them: F = Gm1m2 d2 In this example the constant of proportionality is called the gravitational constant, and its value is given by G = 6.673 × 10−11 N · m2/kg2. Graph Inverse Variation Equations We saw that the general equation for inverse variation is given by the formula y = k x, where k is a constant of proportionality. We will now show how the graphs of such relationships behave. We start by making a table of values. In most applications, x and y are positive, so in our table we’ll choose only positive values of x. 529 www.ck12.org Example 1 Graph an inverse variation relationship with the proportionality constant k = 1. Solution Table 12.1: x y = 1 x 0 y = 1 0 = undeﬁned 1 4 y = 1 1 4 = 4 1 2 y = 1 1 2 = 2 3 4 y = 1 3 4 = 1.33 1 y = 1 1 = 1 3 2 y = 1 3 2 = 0.67 2 y = 1 2 = 0.5 3 y = 1 3 = 0.33 4 y = 1 4 = 0.25 5 y = 1 5 = 0.2 10 y = 1 10 = 0.1 Here is a graph showing these points connected with a smooth curve. Both the table and the graph demonstrate the relationship between variables in an inverse variation. As one variable increases, the other variable decreases and vice versa. Notice that when x = 0, the value of y is undeﬁned. The graph shows that when the value of x is very small, the value of y is very big—so it approaches inﬁnity as x gets closer and closer to zero. Similarly, as the value of x gets very large, the value of y gets smaller and smaller but never reaches zero. We will investigate this behavior in detail throughout this chapter. Write Inverse Variation Equations As we saw, an inverse variation fulﬁlls the equation y = k x. In general, we need to know the value of y at a particular value of x in order to ﬁnd the proportionality constant. Once we know the proportionality constant, we can then ﬁnd the value of y for any given value of x. www.ck12.org 530 Example 2 If y is inversely proportional to x, and if y = 10 when x = 5, ﬁnd y when x = 2. Solution Since y is inversely proportional to x, then: y = k x Plug in the values y = 10 and x = 5 : 10 = k 5 Solve for k by multiplying both sides of the equation by 5 : k = 50 The inverse relationship is given by: y = 50 x When x = 2 : y = 50 2 or y = 25 Example 3 If p is inversely proportional to the square of q, and p = 64 when q = 3, ﬁnd p when q = 5. Solution Since p is inversely proportional to q2, then: p = k q2 Plug in the values p = 64 and q = 3 : 64 = k 32 or 64 = k 9 Solve for k by multiplying both sides of the equation by 9 : k = 576 The inverse relationship is given by: p = 576 q2 When q = 5 : p = 576 25 or y = 23.04 To see some more variation problems worked out, including problems involving joint variation, watch the video at . Figure 12.1: Algebra: Direct, Inverse, Joint Variation Problem (Watch Youtube Video) Solve Real-World Problems Using Inverse Variation Equations Many formulas in physics are described by variations. In this section we’ll investigate some problems that are described by inverse variations. Example 4 The frequency, f, of sound varies inversely with wavelength, λ. A sound signal that has a wavelength of 34 meters has a frequency of 10 hertz. What frequency does a sound signal of 120 meters have? 531 www.ck12.org Solution The inverse variation relationship is: f = k λ Plug in the values: λ = 34 and f = 10 : 10 = k 34 Multiply both sides by 34 : k = 340 Thus, the relationship is given by: f = 340 λ Plug in λ = 120 meters: f = 340 120 ⇒f = 2.83 Hertz Example 5 Electrostatic force is the force of attraction or repulsion between two charges. The electrostatic force is given by the formula F = Kq1q2 d2 , where q1 and q2 are the charges of the charged particles, d is the distance between the charges and k is a proportionality constant. The charges do not change, so they too are constants; that means we can combine them with the other constant k to form a new constant K, so we can rewrite the equation as F = K d2 . If the electrostatic force is F = 740 Newtons when the distance between charges is 5.3 × 10−11 meters, what is F when d = 2.0 × 10−10 meters? Solution The inverse variation relationship is: F = K d2 Plug in the values F = 740 and d = 5.3 × 10−11 : 740 = K (5.3 × 10−11)2 Multiply both sides by (5.3 × 10−11)2 : K = 740 ( 5.3 × 10−11)2 K = 2.08 × 10−18 The electrostatic force is given by: F = 2.08 × 10−18 d2 When d = 2.0 × 10−10 : F = 2.08 × 10−18 (2.0 × 10−10)2 Use scientiﬁc notation to simplify: F = 52 Newtons Review Questions Graph the following inverse variation relationships. 1. y = 3 x 2. y = 10 x 3. y = 1 4x 4. y = 5 6x 5. If z is inversely proportional to w and z = 81 when w = 9, ﬁnd w when z = 24. 6. If y is inversely proportional to x and y = 2 when x = 8, ﬁnd y when x = 12. 7. If a is inversely proportional to the square root of b, and a = 32 when b = 9, ﬁnd b when a = 6. 8. If w is inversely proportional to the square of u and w = 4 when u = 2, ﬁnd w when u = 8. 9. If a is proportional to both b and c and a = 7 when b = 2 and c = 6, ﬁnd a when b = 4 and c = 3. 10. If x is proportional to y and inversely proportional to z, and x = 2 when y = 10 and z = 25, ﬁnd x www.ck12.org 532 when y = 8 and z = 35. 11. If a varies directly with b and inversely with the square of c, and a = 10 when b = 5 and c = 2, ﬁnd the value of a when b = 3 and c = 6. 12. If x varies directly with y and z varies inversely with x, and z = 3 when y = 5, ﬁnd z when y = 10. 13. The intensity of light is inversely proportional to the square of the distance between the light source and the object being illuminated. (a) A light meter that is 10 meters from a light source registers 35 lux. What intensity would it register 25 meters from the light source? (b) A light meter that is registering 40 lux is moved twice as far away from the light source illumi-nating it. What intensity does it now register? (Hint: let x be the original distance from the light source.) (c) The same light meter is moved twice as far away again (so it is now four times as far from the light source as it started out). What intensity does it register now? 14. Ohm’s Law states that current ﬂowing in a wire is inversely proportional to the resistance of the wire. If the current is 2.5 Amperes when the resistance is 20 ohms, ﬁnd the resistance when the current is 5 Amperes. 15. The volume of a gas varies directly with its temperature and inversely with its pressure. At 273 degrees Kelvin and pressure of 2 atmospheres, the volume of a certain gas is 24 liters. (a) Find the volume of the gas when the temperature is 220 Kelvin and the pressure is 1.2 atmo-spheres. (b) Find the temperature when the volume is 24 liters and the pressure is 3 atmospheres. 16. The volume of a square pyramid varies jointly with the height and the square of the side length of the base. A pyramid whose height is 4 inches and whose base has a side length of 3 inches has a volume of 12 in3. (a) Find the volume of a square pyramid that has a height of 9 inches and whose base has a side length of 5 inches. (b) Find the height of a square pyramid that has a volume of 49 in3 and whose base has a side length of 7 inches. (c) A square pyramid has a volume of 72 in3 and its base has a side length equal to its height. Find the height of the pyramid. 12.2 Graphs of Rational Functions Learning Objectives • Compare graphs of inverse variation equations. • Graph rational functions. • Solve real-world problems using rational functions. Introduction In this section, you’ll learn how to graph rational functions. Graphs of rational functions are very dis-tinctive, because they get closer and closer to certain values but never reach those values. This behavior is called asymptotic behavior, and we will see that rational functions can have horizontal asymptotes, vertical asymptotes or oblique (or slant) asymptotes. 533 www.ck12.org Compare Graphs of Inverse Variation Equations Inverse variation problems are the simplest example of rational functions. We saw that an inverse variation has the general equation: y = k x. In most real-world problems, x and y take only positive values. Below, we will show graphs of three inverse variation functions. Example 1 On the same coordinate grid, graph inverse variation relationships with the proportionality constants k = 1, k = 2, and k = 1 2. Solution We’ll skip the table of values for this problem, and just show the graphs of the three functions on the same coordinate axes. Notice that for larger constants of proportionality, the curve decreases at a slower rate than for smaller constants of proportionality. This makes sense because the value of y is related directly to the proportionality constants, so we should expect larger values of y for larger values of k. Graph Rational Functions Now we’ll extend the domain and range of rational equations to include negative values of x and y. First we’ll plot a few rational functions by using a table of values, and then we’ll talk about the distinguishing characteristics of rational functions that can help us make better graphs. As we graph rational functions, we need to always pay attention to values of x that will cause us to divide by 0. Remember that dividing by 0 doesn’t give us an actual number as a result. Example 2 Graph the function y = 1 x. Solution Before we make a table of values, we should notice that the function is not deﬁned for x = 0. This means that the graph of the function won’t have a value at that point. Since the value of x = 0 is special, we should make sure to pick enough values close to x = 0 in order to get a good idea how the graph behaves. Let’s make two tables: one for x−values smaller than zero and one for x−values larger than zero. Table 12.2: x y = 1 x x y = 1 x −5 y = 1 −5 = −0.2 0.1 y = 1 0.1 = 10 www.ck12.org 534 Table 12.2: (continued) x y = 1 x x y = 1 x -4 y = 1 −4 = −0.25 0.2 y = 1 0.2 = 5 -3 y = 1 −3 = −0.33 0.3 y = 1 0.3 = 3.3 -2 y = 1 −2 = −0.5 0.4 y = 1 0.4 = 2.5 -1 y = 1 −1 = −1 0.5 y = 1 0.5 = 2 -0.5 y = 1 −0.5 = −2 1 y = 1 1 = 1 -0.4 y = 1 −0.4 = −2.5 2 y = 1 2 = 0.5 -0.3 y = 1 −0.3 = −3.3 3 y = 1 3 = 0.33 -0.2 y = 1 −0.2 = −5 4 y = 1 4 = 0.25 -0.1 y = 1 −0.1 = −10 5 y = 1 5 = 0.2 We can see that as we pick positive values of x closer and closer to zero, y gets larger, and as we pick negative values of x closer and closer to zero, y gets smaller (or more and more negative). Notice on the graph that for values of x near 0, the points on the graph get closer and closer to the vertical line x = 0. The line x = 0 is called a vertical asymptote of the function y = 1 x. We also notice that as the absolute values of x get larger in the positive direction or in the negative direction, the value of y gets closer and closer to y = 0 but will never gain that value. Since y = 1 x, we can see that there are no values of x that will give us the value y = 0. The horizontal line y = 0 is called a horizontal asymptote of the function y = 1 x Asymptotes are usually denoted as dashed lines on a graph. They are not part of the function; instead, they show values that the function approaches, but never gets to. A horizontal asymptote shows the value of y that the function approaches (but never reaches) as the absolute value of x gets larger and larger. A 535 www.ck12.org vertical asymptote shows that the absolute value of y gets larger and larger as x gets closer to a certain value which it can never actually reach. Now we’ll show the graph of a rational function that has a vertical asymptote at a non-zero value of x. Example 3 Graph the function y = 1 (x−2)2 . Solution We can see that the function is not deﬁned for x = 2, because that would make the denominator of the fraction equal zero. This tells us that there should be a vertical asymptote at x = 2, so we can start graphing the function by drawing the vertical asymptote. Now let’s make a table of values. Table 12.3: x y = 1 (x−2)2 0 y = 1 (0−2)2 = 1 4 1 y = 1 (1−2)2 = 1 1.5 y = 1 (1.5−2)2 = 4 2 undeﬁned 2.5 y = 1 (2.5−2)2 = 4 3 y = 1 (3−2)2 = 1 4 y = 1 (4−2)2 = 1 4 Here’s the resulting graph: www.ck12.org 536 Notice that we didn’t pick as many values for our table this time, because by now we have a pretty good idea what happens near the vertical asymptote. We also know that for large values of \|x\|, the value of y could approach a constant value. In this case that value is y = 0: this is the horizontal asymptote. A rational function doesn’t have to have a vertical or horizontal asymptote. The next example shows a rational function with no vertical asymptotes. Example 4 Graph the function y = x2 x2+1. Solution We can see that this function will have no vertical asymptotes because the denominator of the fraction will never be zero. Let’s make a table of values to see if the value of y approaches a particular value for large values of x, both positive and negative. Table 12.4: x y = x2 x2+1 −3 y = (−3)2 (−3)2+1 = 9 10 = 0.9 -2 y = (−2)2 (−2)2+1 = 4 5 = 0.8 -1 y = (−1)2 (−1)2+1 = 1 2 = 0.5 0 y = (0)2 (0)2+1 = 0 1 = 0 1 y = (1)2 (1)2+1 = 1 2 = 0.5 2 y = (2)2 (2)2+1 = 4 5 = 0.8 3 y = (3)2 (3)2+1 = 9 10 = 0.9 Below is the graph of this function. 537 www.ck12.org The function has no vertical asymptote. However, we can see that as the values of \|x\| get larger, the value of y gets closer and closer to 1, so the function has a horizontal asymptote at y = 1. Finding Horizontal Asymptotes We said that a horizontal asymptote is the value of y that the function approaches for large values of \|x\|. When we plug in large values of x in our function, higher powers of x get larger much quickly than lower powers of x. For example, consider: y = 2x2 + x −1 3x2 −4x + 3 If we plug in a large value of x, say x = 100, we get: y = 2(100)2 + (100) −1 3(100)2 −4(100) + 3 = 20000 + 100 −1 30000 −400 + 2 We can see that the beginning terms in the numerator and denominator are much bigger than the other terms in each expression. One way to ﬁnd the horizontal asymptote of a rational function is to ignore all terms in the numerator and denominator except for the highest powers. In this example the horizontal asymptote is y = 2x2 3x2 , which simpliﬁes to y = 2 3. In the function above, the highest power of x was the same in the numerator as in the denominator. Now consider a function where the power in the numerator is less than the power in the denominator: y = x x2 + 3 As before, we ignore all the terms except the highest power of x in the numerator and the denominator. That gives us y = x x2 , which simpliﬁes to y = 1 x. For large values of x, the value of y gets closer and closer to zero. Therefore the horizontal asymptote is y = 0. To summarize: • Find vertical asymptotes by setting the denominator equal to zero and solving for x. • For horizontal asymptotes, we must consider several cases: – If the highest power of x in the numerator is less than the highest power of x in the denominator, then the horizontal asymptote is at y = 0. www.ck12.org 538 – If the highest power of x in the numerator is the same as the highest power of x in the denomi-nator, then the horizontal asymptote is at y = coef ficient of highest power of x coef ficient of highest power of x. – If the highest power of x in the numerator is greater than the highest power of x in the de-nominator, then we don’t have a horizontal asymptote; we could have what is called an oblique (slant) asymptote, or no asymptote at all. Example 5 Find the vertical and horizontal asymptotes for the following functions. a) y = 1 x−1 b) y = 3x 4x+2 c) y = x2−2 2x2+3 d) y = x3 x2−3x+2 Solution a) Vertical asymptotes: Set the denominator equal to zero. x −1 = 0 ⇒x = 1 is the vertical asymptote. Horizontal asymptote: Keep only the highest powers of x. y = 1 x ⇒y = 0 is the horizontal asymptote. b) Vertical asymptotes: Set the denominator equal to zero. 4x + 2 = 0 ⇒x = −1 2 is the vertical asymptote. Horizontal asymptote: Keep only the highest powers of x. y = 3x 4x ⇒y = 3 4 is the horizontal asymptote. c) Vertical asymptotes: Set the denominator equal to zero: 2x2 + 3 = 0 ⇒2x2 = −3 ⇒x2 = −3 2. Since there are no solutions to this equation, there is no vertical asymptote. Horizontal asymptote: Keep only the highest powers of x. y = x2 2x2 ⇒y = 1 2 is the horizontal asymptote. d) Vertical asymptotes: Set the denominator equal to zero: x2 −3x + 2 = 0 Factor: (x −2)(x −1) = 0 Solve: x = 2 and x = 1 are the vertical asymptotes. Horizontal asymptote. There is no horizontal asymptote because the power of the numerator is larger than the power of the denominator. Notice the function in part d had more than one vertical asymptote. Here’s another function with two vertical asymptotes. Example 6 Graph the function y = −x2 x2−4. Solution Let’s set the denominator equal to zero: x2 −4 = 0 539 www.ck12.org Factor: (x −2)(x + 2) = 0 Solve: x = 2, x = −2 We ﬁnd that the function is undeﬁned for x = 2 and x = −2, so we know that there are vertical asymptotes at these values of x. We can also ﬁnd the horizontal asymptote by the method we outlined above. It’s at y = −x2 x2 , or y = −1. So, we start plotting the function by drawing the vertical and horizontal asymptotes on the graph. Now, let’s make a table of values. Because our function has a lot of detail we must make sure that we pick enough values for our table to determine the behavior of the function accurately. We must make sure especially that we pick values close to the vertical asymptotes. Table 12.5: x y = −x2 x2−4 −5 y = −(−5)2 (−5)2−4 = −25 21 = −1.19 -4 y = −(−4)2 (−4)2−4 = −16 12 = −1.33 -3 y = −(−3)2 (−3)2−4 = −9 5 = −1.8 -2.5 y = −(−2.5)2 (−2.5)2−4 = −6.25 2.25 = −2.8 -1.5 y = −(−1.5)2 (−1.5)2−4 = −2.25 −1.75 = 1.3 -1 y = −(−1)2 (−1)2−4 = −1 −3 = 0.33 0 y = −02 (0)2−4 = 0 −4 = 0 1 y = −12 (1)2−4 = −1 −3 = 0.33 1.5 y = −1.52 (1.5)2−4 = −2.25 −1.75 = 1.3 2.5 y = −2.52 (2.5)2−4 = −6.25 2.25 = −2.8 3 y = −32 (3)2−4 = −9 5 = −1.8 4 y = −42 (4)2−4 = −16 12 = −1.33 5 y = −52 (5)2−4 = −25 21 = −1.19 Here is the resulting graph. www.ck12.org 540 To explore more graphs of rational functions, try the applets available at rational/rational1.html. Solve Real-World Problems Using Rational Functions Electrical circuits are commonplace is everyday life—for example, they’re in all the electrical appliances in your home. The ﬁgure below shows an example of a simple electrical circuit. It consists of a battery which provides a voltage (V, measured in Volts, V), a resistor (R, measured in ohms, Ω) which resists the ﬂow of electricity and an ammeter that measures the current (I, measured in amperes, A) in the circuit. Ohm’s Law gives a relationship between current, voltage and resistance. It states that I = V R Your light bulbs, toaster and hairdryer are all basically simple resistors. In addition, resistors are used in an electrical circuit to control the amount of current ﬂowing through a circuit and to regulate voltage levels. One important reason to do this is to prevent sensitive electrical components from burning out due to too much current or too high a voltage level. Resistors can be arranged in series or in parallel. For resistors placed in a series: the total resistance is just the sum of the resistances of the individual resistors: Rtot = R1 + R2 For resistors placed in parallel: 541 www.ck12.org the reciprocal of the total resistance is the sum of the reciprocals of the resistances of the individual resistors: 1 Rc = 1 R1 + 1 R2 Example 7 Find the quantity labeled x in the following circuit. Solution We use the formula I = V R. Plug in the known values:I = 2 A, V = 12 V : 2 = 12 R Multiply both sides by R : 2R = 12 Divide both sides by 2 : R = 6Ω Answer Example 8 Find the quantity labeled x in the following circuit. Solution Ohm’s Law also tells us that Itotal = Vtotal Rtotal Plug in the values we know, I = 2.5 A and E = 9 V : 2.5 = 9 Rtot Multiply both sides by R : 2.5Rtot = 9 www.ck12.org 542 Divide both sides by 2.5 : Rtot = 3.6Ω Since the resistors are placed in parallel, the total resistance is given by: 1 Rtot = 1 X + 1 20 ⇒1 36 = 1 X + 1 20 Multiply all terms by 72X : 1 3.6(72X) = 1 X(72X) + 1 20(72X) Cancel common factors: 20X = 72 + 3.6X Solve: 16.4X = 72 Divide both sides by 16.4 : X = 4.39Ω Answer Review Questions Find all the vertical and horizontal asymptotes of the following rational functions. 1. y = 4 x+2 2. y = 5x−1 2x−6 3. y = 10 x 4. y = x+1 x2 5. y = 4x2 4x2+1 6. y = 2x x2−9 7. y = 3x2 x2−4 8. y = 1 x2+4x+3 9. y = 2x+5 x2−2x−8 Graph the following rational functions. Draw dashed vertical and horizontal lines on the graph to denote asymptotes. 10. y = 2 x−3 11. y = 3 x2 12. y = x x−1 13. y = 2x x+1 14. y = −1 x2+2 15. y = x x2+9 16. y = x2 x2+1 17. y = 1 x2−1 18. y = 2x x2−9 19. y = x2 x2−16 20. y = 3 x2−4x+4 21. y = x x2−x−6 Find the quantity labeled x in each of the following circuits. 543 www.ck12.org 22. 23. 24. 25. 12.3 Division of Polynomials Learning Objectives • Divide a polynomial by a monomial. • Divide a polynomial by a binomial. • Rewrite and graph rational functions. Introduction A rational expression is formed by taking the quotient of two polynomials. Some examples of rational expressions are 2x x2 −1 4x2 −3x + 4 2x 9x2 + 4x −5 x2 + 5x −1 2x3 2x + 3 Just as with rational numbers, the expression on the top is called the numerator and the expression on the bottom is called the denominator. In special cases we can simplify a rational expression by dividing the numerator by the denominator. www.ck12.org 544 Divide a Polynomial by a Monomial We’ll start by dividing a polynomial by a monomial. To do this, we divide each term of the polynomial by the monomial. When the numerator has more than one term, the monomial on the bottom of the fraction serves as the common denominator to all the terms in the numerator. Example 1 Divide. a) 8x2−4x+16 2 b) 3x2+6x−1 x c) −3x2−18x+6 9x Solution a) 8x2−4x+16 2 = 8x2 2 −4x 2 + 16 2 = 4x2 −2x + 8 b) 3x3+6x−1 x = 3x3 x + 6x x −1 x = 3x2 + 6 −1 x c) −3x2−18x+6 9x = −3x2 9x −18x 9x + 6 9x = −x 3 −2 + 2 3x A common error is to cancel the denominator with just one term in the numerator. Consider the quotient 3x+4 4 . Remember that the denominator of 4 is common to both the terms in the numerator. In other words we are dividing both of the terms in the numerator by the number 4. The correct way to simplify is: 3x + 4 4 = 3x 4 + 4 4 = 3x 4 + 1 A common mistake is to cross out the number 4 from the numerator and the denominator, leaving just 3x. This is incorrect, because the entire numerator needs to be divided by 4. Example 2 Divide 5x3−10x2+x−25 −5x2 . Solution 5x3 −10x2 + x −25 −5x2 = 5x3 −5x2 −10x2 −5x2 + x −5x2 − 25 −5x2 The negative sign in the denominator changes all the signs of the fractions: −5x3 5x2 + 10x2 5x2 − x 5x2 + 25 5x2 = −x + 2 −1 5x + 5 x2 Divide a Polynomial by a Binomial We divide polynomials using a method that’s a lot like long division with numbers. We’ll explain the method by doing an example. Example 3 Divide x2+4x+5 x+3 . Solution 545 www.ck12.org When we perform division, the expression in the numerator is called the dividend and the expression in the denominator is called the divisor. To start the division we rewrite the problem in the following form: x + 3)x2 + 4x + 5 We start by dividing the ﬁrst term in the dividend by the ﬁrst term in the divisor: x2 x = x. We place the answer on the line above the x term: x x + 3)x2 + 4x + 5 Next, we multiply the x term in the answer by the divisor, x + 3, and place the result under the dividend, matching like terms. x times (x + 3) is x2 + 3x, so we put that under the divisor: x x + 3)x2 + 4x + 5 x2 + 3x Now we subtract x2 +3x from x2 +4x+5. It is useful to change the signs of the terms of x2 +3x to −x2 −3x and add like terms vertically: x x + 3)x2 + 4x + 5 −x2 −3x x Now, we bring down the 5, the next term in the dividend. x x + 3)x2 + 4x + 5 −x2 −3x x + 5 And now we go through that procedure once more. First we divide the ﬁrst term of x + 5 by the ﬁrst term of the divisor. x divided by x is 1, so we place this answer on the line above the constant term of the dividend: x + 1 x + 3)x2 + 4x + 5 −x2 −3x x + 5 Multiply 1 by the divisor, x + 3, and write the answer below x + 5, matching like terms. x + 1 x + 3)x2 + 4x + 5 −x2 −3x x + 5 x + 3 www.ck12.org 546 Subtract x + 3 from x + 5 by changing the signs of x + 3 to −x −3 and adding like terms: x + 1 x + 3)x2 + 4x + 5 −x2 −3x x + 5 −x −3 2 Since there are no more terms from the dividend to bring down, we are done. The quotient is x + 1 and the remainder is 2. Remember that for a division with a remainder the answer is quotient + remainder divisor . So the answer to this division problem is x2+4x+5 x+3 = x + 1 + 2 x+3. Check To check the answer to a long division problem we use the fact that (divisor × quotient) + remainder = dividend For the problem above, here’s how we apply that fact to check our solution: (x + 3)(x + 1) + 2 = x2 + 4x + 3 + 2 = x2 + 4x + 5 The answer checks out. To check your answers to long division problems involving polynomials, try the solver at com/webMathematica/long-divide.jsp. It shows the long division steps so you can tell where you may have made a mistake. Rewrite and Graph Rational Functions In the last section we saw how to ﬁnd vertical and horizontal asymptotes. Remember, the horizontal asymptote shows the value of y that the function approaches for large values of x. Let’s review the method for ﬁnding horizontal asymptotes and see how it’s related to polynomial division. When it comes to ﬁnding asymptotes, there are basically four diﬀerent types of rational functions. Case 1: The polynomial in the numerator has a lower degree than the polynomial in the denominator. Take, for example, y = 2 x−1. We can’t reduce this fraction, and as x gets larger the denominator of the fraction gets much bigger than the numerator, so the whole fraction approaches zero. The horizontal asymptote is y = 0. Case 2: The polynomial in the numerator has the same degree as the polynomial in the denominator. Take, for example, y = 3x+2 x−1 . In this case we can divide the two polynomials: 3 x −1)3x + 2 −3x + 3 5 547 www.ck12.org So the expression can be written as y = 3 + 5 x−1. Because the denominator of the remainder is bigger than the numerator of the remainder, the remainder will approach zero for large values of x. Adding the 3 to that 0 means the whole expression will approach 3. The horizontal asymptote is y = 3. Case 3: The polynomial in the numerator has a degree that is one more than the polynomial in the denominator. Take, for example, y = 4x2+3x+2 x−1 . We can do long division once again and rewrite the expression as y = 4x+7+ 9 x−1. The fraction here approaches zero for large values of x, so the whole expression approaches 4x + 7. When the rational function approaches a straight line for large values of x, we say that the rational function has an oblique asymptote. In this case, then, the oblique asymptote is y = 4x + 7. Case 4: The polynomial in the numerator has a degree that in two or more than the degree in the denominator. For example: y = x3 x−1. This is actually the simplest case of all: the polynomial has no horizontal or oblique asymptotes. Example 5 Find the horizontal or oblique asymptotes of the following rational functions. a) y = 3x2 x2+4 b) y = x−1 3x2−6 c) y = x4+1 x−5 d) y = x3−3x2+4x−1 x2−2 Solution a) When we simplify the function, we get y = 3 − 12 x2+4. There is a horizontal asymptote at y = 3. b) We cannot divide the two polynomials. There is a horizontal asymptote at y = 0. c) The power of the numerator is 3 more than the power of the denominator. There are no horizontal or oblique asymptotes. d) When we simplify the function, we get y = x−3+ 6x−7 x2−2. There is an oblique asymptote at y = x−3. Notice that a rational function will either have a horizontal asymptote, an oblique asymptote or neither kind. In other words, a function can’t have both; in fact, it can’t have more than one of either kind. On the other hand, a rational function can have any number of vertical asymptotes at the same time that it has horizontal or oblique asymptotes. Review Questions Divide the following polynomials: 1. 2x+4 2 2. x−4 x 3. 5x−35 5x 4. x2+2x−5 x 5. 4x2+12x−36 −4x www.ck12.org 548 6. 2x2+10x+7 2x2 7. x3−x −2x2 8. 5x4−9 3x 9. x3−12x2+3x−4 12x2 10. 3−6x+x3 −9x3 11. x2+3x+6 x+1 12. x2−9x+6 x−1 13. x2+5x+4 x+4 14. x2−10x+25 x−5 15. x2−20x+12 x−3 16. 3x2−x+5 x−2 17. 9x2+2x−8 x+4 18. 3x2−4 3x+1 19. 5x2+2x−9 2x−1 20. x2−6x−12 5x4 Find all asymptotes of the following rational functions: 21. x2 x−2 22. 1 x+4 23. x2−1 x2+1 24. x−4 x2−9 25. x2+2x+1 4x−1 26. x3+1 4x−1 27. x−x3 x2−6x−7 28. x4−2x 8x+24 Graph the following rational functions. Indicate all asymptotes on the graph: 29. x2 x+2 30. x3−1 x2−4 31. x2+1 2x−4 32. x−x2 3x+2 12.4 Rational Expressions Learning Objectives • Simplify rational expressions. • Find excluded values of rational expressions. 549 www.ck12.org Introduction A simpliﬁed rational expression is one where the numerator and denominator have no common factors. In order to simplify an expression to lowest terms, we factor the numerator and denominator as much as we can and cancel common factors from the numerator and the denominator. Simplify Rational Expressions Example 1 Reduce each rational expression to simplest terms. a) 4x−2 2x2+x−1 b) x2−2x+1 8x−8 c) x2−4 x2−5x+6 Solution a) Factor the numerator and denominator completely: 2(2x−1) (2x−1)(x+1) Cancel the common factor (2x −1) : 2 x+1 b) Factor the numerator and denominator completely: (x−1)(x−1) 8(x−1) Cancel the common factor (x −1) : x−1 8 c) Factor the numerator and denominator completely: (x−2)(x+2) (x−2)(x−3) Cancel the common factor(x −2) : x+2 x−3 When reducing fractions, you are only allowed to cancel common factors from the denominator but NOT common terms. For example, in the expression (x+1)·(x−3) (x+2)·(x−3), we can cross out the (x −3) factor because (x−3) (x−3) = 1. But in the expression x2+1 x2−5 we can’t just cross out the x2 terms. Why can’t we do that? When we cross out terms that are part of a sum or a diﬀerence, we’re violating the order of operations (PEMDAS). Remember, the fraction bar means division. When we perform the operation x2+1 x2−5 , we’re really performing the division (x2 + 1) ÷ (x2 −5) — and the order of operations says that we must perform the operations inside the parentheses before we can perform the division. Using numbers instead of variables makes it more obvious that canceling individual terms doesn’t work. You can see that 9+1 9−5 = 10 4 = 2.5 — but if we canceled out the 9’s ﬁrst, we’d get 1 −5, or -0.2, instead. For more examples of how to simplify rational expressions, watch the video at watch?v=B4bVlDgHF5I . Find Excluded Values of Rational Expressions Whenever there’s a variable expression in the denominator of a fraction, we must remember that the denominator could be zero when the independent variable takes on certain values. Those values, corre-sponding to the vertical asymptotes of the function, are called excluded values. To ﬁnd the excluded values, we simply set the denominator equal to zero and solve the resulting equation. Example 2 www.ck12.org 550 Figure 12.2: Simplifying Rational Expressions - YourTeacher.com - Math Help (Watch Youtube Video) Find the excluded values of the following expressions. a) x x+4 b) 2x+1 x2−x−6 c) 4 x2−5x Solution a) When we set the denominator equal to zero we obtain: x + 4 = 0 ⇒x = −4 So −4 is the excluded value. b) When we set the denominator equal to zero we obtain: x2 −x −6 = 0 Solve by factoring: (x −3)(x + 2) = 0 ⇒x = 3 and x = −2 So 3 and −2 are the excluded values. c) When we set the denominator equal to zero we obtain: x2 −5x = 0 Solve by factoring: x(x −5) = 0 ⇒x = 0 and x = 5 So 0 and 5 are the excluded values. Removable Zeros Notice that in the expressions in Example 1, we removed a division by zero when we simpliﬁed the problem. For instance, we rewrote 4x−2 2x2+x−1 as 2(2x−1) (2x−1)(x+1). The denominator of this expression is zero when x = 1 2 or when x = −1. However, when we cancel common factors, we simplify the expression to 2 x+1. This reduced form allows the value x = 1 2, so x = −1 is its only excluded value. Technically the original expression and the simpliﬁed expression are not the same. When we recduce a radical expression to its simplest form, we should specify the removed excluded value. In other words, we should write our ﬁnal answer as 4x−2 2x2+x−1 = 2 x+1, x , 1 2. Similarly, we should write the answer from Example 1, part b as x2−2x+1 8x−8 = x−1 8 , x , 1 and the answer from 551 www.ck12.org Example 1, part c as x2−4 x2−5x+6 = x+2 x−3 , x , 2. Review Questions Reduce each fraction to lowest terms. 1. 4 2x−8 2. x2+2x x 3. 9x+3 12x+4 4. 6x2+2x 4x 5. x−2 x2−4x+4 6. x2−9 5x+15 7. x2+6x+8 x2+4x 8. 2x2+10x x2+10x+25 9. x2+6x+5 x2−x−2 10. x2−16 x2+2x−8 11. 3x2+3x−18 2x2+5x−3 12. x3+x2−20x 6x2+6x−120 Find the excluded values for each rational expression. 13. 2 x 14. 4 x+2 15. 2x−1 (x−1)2 16. 3x+1 x2−4 17. x2 x2+9 18. 2x2+3x−1 x2−3x−28 19. 5x3−4 x2+3x 20. 9 x3+11x2+30x 21. 4x−1 x2+3x−5 22. 5x+11 3x2−2x−4 23. x2−1 2x2+x+3 24. 12 x2+6x+1 25. In an electrical circuit with resistors placed in parallel, the reciprocal of the total resistance is equal to the sum of the reciprocals of each resistance. 1 Rc = 1 R1 + 1 R2 . If R1 = 25 Ωand the total resistance is Rc = 10 Ω, what is the resistance R2? 26. Suppose that two objects attract each other with a gravitational force of 20 Newtons. If the distance between the two objects is doubled, what is the new force of attraction between the two objects? 27. Suppose that two objects attract each other with a gravitational force of 36 Newtons. If the mass of both objects was doubled, and if the distance between the objects was doubled, then what would be the new force of attraction between the two objects? 28. A sphere with radius R has a volume of 4 3πR3 and a surface area of 4πR2. Find the ratio the surface area to the volume of a sphere. 29. The side of a cube is increased by a factor of 2. Find the ratio of the old volume to the new volume. 30. The radius of a sphere is decreased by 4 units. Find the ratio of the old volume to the new volume. www.ck12.org 552 12.5 Multiplying and Dividing Rational Expres-sions Learning Objectives • Multiply rational expressions involving monomials. • Multiply rational expressions involving polynomials. • Multiply a rational expression by a polynomial. • Divide rational expressions involving polynomials. • Divide a rational expression by a polynomial. • Solve real-world problems involving multiplication and division of rational expressions. Introduction The rules for multiplying and dividing rational expressions are the same as the rules for multiplying and dividing rational numbers, so let’s start by reviewing multiplication and division of fractions. When we multiply two fractions we multiply the numerators and denominators separately: a b · c d = a · c b · d When we divide two fractions, we replace the second fraction with its reciprocal and multiply, since that’s mathematically the same operation: a b ÷ c d = a b · d c = a · d b · c Multiply Rational Expressions Involving Monomials Example 1 Multiply the following: a 16b8 · 4b3 5a2 . Solution Cancel common factors from the numerator and denominator. The common factors are 4, a, and b3. Canceling them out leaves 1 4b5 · 1 5a = 1 20ab5 . Example 2 Multiply 9x2 · 4y2 21x4 . Solution Rewrite the problem as a product of two fractions: 9x2 1 · 4y2 21x4 Then cancel common factors from the numerator and denominator. The common factors are 3 and x2. Canceling them out leaves 3 1 · 4y2 7x2 = 12y2 7x2 . Multiply Rational Expressions Involving Polynomials When multiplying rational expressions involving polynomials, ﬁrst we need to factor all polynomial ex-pressions as much as we can. Then we follow the same procedure as before. 553 www.ck12.org Example 3 Multiply 4x+12 3x2 · x x2−9. Solution Factor all polynomial expressions as much as possible: 4(x+3) 3x2 · x (x+3)(x−3) The common factors are x and (x + 3). Canceling them leaves 4 3x · 1 (x−3) = 4 3x(x−3) = 4 3x2−9x. Example 4 Multiply 12x2−x−6 x2−1 · x2+7x+6 4x2−27x+18. Solution Factor polynomials: (3x+2)(4x−3) (x+1)(x−1) · (x+1)(x+6) (4x−3)(x−6). The common factors are (x+1) and (4x−3). Canceling them leaves (3x+2) (x−1) · (x+6) (x−6) = (3x+2)(x+6) (x−1)(x−6) = 3x2+20x+12 x2−7x+6 Multiply a Rational Expression by a Polynomial When we multiply a rational expression by a whole number or a polynomial, we can write the whole number (or polynomial) as a fraction with denominator equal to one. We then proceed the same way as in the previous examples. Example 5 Multiply 3x+18 4x2+19x−5 · (x2 + 3x −10). Solution Rewrite the expression as a product of fractions: 3x+18 4x2+19x−5 · x2+3x−10 1 Factor polynomials: 3(x+6) (x+5)(4x−1) · (x−2)(x+5) 1 The common factor is (x + 5). Canceling it leaves 3(x+6) (4x−1) · (x−2) 1 = 3(x+6)(x−2) (4x−1) = 3x2+12x−36 4x−1 Divide Rational Expressions Involving Polynomials Just as with ordinary fractions, we ﬁrst rewrite the division problem as a multiplication problem and then proceed with the multiplication as outlined in the previous example. Note: Remember that a b ÷ c d = a b · d c. The ﬁrst fraction remains the same and you take the reciprocal of the second fraction. Do not fall into the common trap of ﬂipping the ﬁrst fraction. Example 6 Divide 4x2 15 ÷ 6x 5 . Solution First convert into a multiplication problem by ﬂipping the second fraction and then simplify as usual: 4x2 15 ÷ 6x 5 = 4x2 15 · 5 6x = 2x 3 · 1 3 = 2x 9 Example 7 Divide 3x2−15x 2x2+3x−14 ÷ x2−25 2x2+13x+21. Solution www.ck12.org 554 3x2 −15x 2x2 + 3x −14 · 2x2 + 13x + 21 x2 −25 = 3x(x −5) (2x + 7)(x −2) · (2x + 7)(x + 3) (x −5)(x + 5) = 3x (x −2) · (x + 3) (x + 5) = 3x2 + 9x x2 + 3x −10 Divide a Rational Expression by a Polynomial When we divide a rational expression by a whole number or a polynomial, we can write the whole number (or polynomial) as a fraction with denominator equal to one, and then proceed the same way as in the previous examples. Example 8 Divide 9x2−4 2x−2 ÷ (21x2 −2x −8). Solution Rewrite the expression as a division of fractions, and then convert into a multiplication problem by taking the reciprocal of the divisor: 9x2 −4 2x −2 ÷ 21x2 −2x −8 1 = 9x2 −4 2x −2 · 1 21x2 −2x −8 Then factor and solve: 9x2 −4 2x −2 · 1 21x2 −2x −8 = (3x −2)(3x + 2) 2(x −1) · 1 (3x −2)(7x + 4) = (3x + 2) 2(x −1) · 1 (7x + 4) = 3x + 2 14x2 −6x −8 For more examples of how to multiply and divide rational expressions, watch the video at youtube.com/watch?v=hVIol-6vocY . Figure 12.3: multiply or divide rational expressions (Watch Youtube Video) Solve Real-World Problems Involving Multiplication and Divi-sion of Rational Expressions Example 9 Suppose Marciel is training for a running race. Marciel’s speed (in miles per hour) of his training run each morning is given by the function x3 −9x, where x is the number of bowls of cereal he had for breakfast. Marciel’s training distance (in miles), if he eats x bowls of cereal, is 3x2 −9x. What is the function for Marciel’s time, and how long does it take Marciel to do his training run if he eats ﬁve bowls of cereal on Tuesday morning? 555 www.ck12.org Solution time = distance speed time = 3x2−9x x3−9x = 3x(x−3) x(x2−9) = 3x(x−3) x(x+3)(x−3) time = 3 x+3 If x = 5, then time = 3 5+3 = 3 8 Marciel will run for 3 8 of an hour. Review Questions Perform the indicated operation and reduce the answer to lowest terms. 1. x3 2y3 · 2y2 x 2. 2xy ÷ 2x2 y 3. 2x y2 · 4y 5x 4. 2x3 y ÷ 3x2 5. 2xy · 2y2 x3 6. 3x+6 y−4 ÷ 3y+9 x−1 7. 4y2−1 y2−9 · y−3 2y−1 8. 6ab a2 · a3b 3b2 9. x2 x−1 ÷ x x2+x−2 10. 33a2 −5 · 20 11a3 11. a2+2ab+b2 ab2−a2b ÷ (a + b) 12. 2x2+2x−24 x2+3x · x2+x−6 x+4 13. 3−x 3x−5 ÷ x2−9 2x2−8x−10 14. x2−25 x+3 ÷ (x −5) 15. 2x+1 2x−1 ÷ 4x2−1 1−2x 16. x x−5 · x2−8x+15 x2−3x 17. 3x2+5x−12 x2−9 ÷ 3x−4 3x+4 18. 5x2+16x+3 36x2−25 · (6x2 + 5x) 19. x2+7x+10 x2−9 · x2−3x 3x2+4x−4 20. x2+x−12 x2+4x+4 ÷ x−3 x+2 21. x4−16 x2−9 ÷ x2+4 x2+6x+9 22. x2+8x+16 7x2+9x+2 · 7x+2 x2+4x 23. Maria’s recipe asks for 2 1 2 times more ﬂour than sugar. How many cups of ﬂour should she mix in if she uses 31 3 cups of sugar? 24. George drives from San Diego to Los Angeles. On the return trip he increases his driving speed by 15 miles per hour. In terms of his initial speed, by what factor is the driving time decreased on the return trip? www.ck12.org 556 25. Ohm’s Law states that in an electrical circuit I = V Rc . The total resistance for resistors placed in parallel is given by: 1 Rtot = 1 R1 + 1 R2 . Write the formula for the electric current in term of the component resistances: R1 and R2. 12.6 Adding and Subtracting Rational Expres-sions Learning Objectives • Add and subtract rational expressions with the same denominator. • Find the least common denominator of rational expressions. • Add and subtract rational expressions with diﬀerent denominators. • Solve real-world problems involving addition and subtraction of rational expressions. Introduction Like fractions, rational expressions represent a portion of a quantity. Remember that when we add or subtract fractions we must ﬁrst make sure that they have the same denominator. Once the fractions have the same denominator, we combine the diﬀerent portions by adding or subtracting the numerators and writing that answer over the common denominator. Add and Subtract Rational Expressions with the Same Denomi-nator Fractions with common denominators combine in the following manner: a c + b c = a + b c and a c −b c = a −b c Example 1 Simplify. a) 8 7 −2 7 + 4 7 b) 4x2−3 x+5 + 2x2−1 x+5 c) x2−2x+1 2x+3 −3x2−3x+5 2x+3 Solution a) Since the denominators are the same we combine the numerators: 8 7 −2 7 + 4 7 = 8 −2 + 4 7 = 10 7 b) Since the denominators are the same we combine the numerators: 4x2−3+2x2−1 x+5 Simplify by collecting like terms: 6x2−4 x+5 c) Since the denominators are the same we combine the numerators. Make sure the subtraction sign is distributed to all terms in the second expression: x2 −2x + 1 −(3x2 −3x + 5) 2x + 3 = x2 −2x + 1 −3x2 + 3x −5 2x + 3 = −2x2 + x −4 2x + 3 557 www.ck12.org Find the Least Common Denominator of Rational Expressions To add and subtract fractions with diﬀerent denominators, we must ﬁrst rewrite all fractions so that they have the same denominator. In general, we want to ﬁnd the least common denominator. To ﬁnd the least common denominator, we ﬁnd the least common multiple (LCM) of the expressions in the denominators of the diﬀerent fractions. Remember that the least common multiple of two or more integers is the least positive integer that has all of those integers as factors. The procedure for ﬁnding the lowest common multiple of polynomials is similar. We rewrite each polynomial in factored form and we form the LCM by taking each factor to the highest power it appears in any of the separate expressions. Example 2 Find the LCM of 48x2y and 60xy3z. Solution First rewrite the integers in their prime factorization. 48 = 24 · 3 60 = 22 · 3 · 5 The two expressions can be written as: 48x2y = 24 · 3 · x2 · y 60xy3z = 22 · 3 · 5 · x · y3 · z To ﬁnd the LCM, take the highest power of each factor that appears in either expression. LCM = 24 · 3 · 5 · x2 · y3 · z = 240x2y3z Example 3 Find the LCM of 2x2 + 8x + 8 and x3 −4x2 −12x Solution Factor the polynomials completely: 2x2 + 8x + 8 = 2(x2 + 4x + 4) = 2(x + 2)2 x3 −4x2 −12x = x(x2 −4x −12) = x(x + 2)(x −6) To ﬁnd the LCM, take the highest power of each factor that appears in either expression. LCM = 2x(x + 2)2(x −6) It’s customary to leave the LCM in factored form, because this form is useful in simplifying rational expressions and ﬁnding any excluded values. Example 4 www.ck12.org 558 Find the LCM of x2 −25 and x2 + 3x + 2 Solution Factor the polynomials completely: x2 −25 = (x −5)(x + 5) x2 + 3x + 2 = (x + 1)(x + 2) Since the two expressions have no common factors, the LCM is just the product of the two expressions. LCM = (x −5)(x + 5)(x + 1)(x + 2) Add and Subtract Rational Expressions with Diﬀerent Denomi-nators Now we’re ready to add and subtract rational expressions. We use the following procedure. 1. Find the least common denominator (LCD) of the fractions. 2. Express each fraction as an equivalent fraction with the LCD as the denominator. 3. Add or subtract and simplify the result. Example 5 Perform the following operation and simplify: 2 x+2 − 3 2x−5 Solution The denominators can’t be factored any further, so the LCD is just the product of the separate denomi-nators: (x + 2)(2x −5). That means the ﬁrst fraction needs to be multiplied by the factor (2x −5) and the second fraction needs to be multiplied by the factor (x + 2): 2 x + 2 · (2x −5) (2x −5) − 3 2x −5 · (x + 2) (x + 2) Combine the numerators and simplify: 2(2x−5)−3(x+2) (x+2)(2x−5) = 4x−10−3x−6 (x+2)(2x−5) Combine like terms in the numerator: x−16 (x+2)(2x−5) Answer Example 6 Perform the following operation and simplify: 4x x−5 −3x 5−x. Solution Notice that the denominators are almost the same; they just diﬀer by a factor of -1. Factor out -1 from the second denominator: 4x x−5 − 3x −(x−5) The two negative signs in the second fraction cancel: 4x x−5 + 3x (x−5) Since the denominators are the same we combine the numerators: 7x x−5 Answer Example 7 Perform the following operation and simplify: 2x−1 x2−6x+9 −3x+4 x2−9 . 559 www.ck12.org Solution We factor the denominators: 2x−1 (x−3)2 − 3x+4 (x+3)(x−3) The LCD is the product of all the diﬀerent factors, each taken to the highest power they have in either denominator: (x −3)2(x + 3). The ﬁrst fraction needs to be multiplied by a factor of (x+3) and the second fraction needs to be multiplied by a factor of (x −3): 2x −1 (x −3)2 · (x + 3) (x + 3) − 3x + 4 (x + 3)(x −3) · (x −3) (x −3) Combine the numerators by subtracting: (2x−1)(x+3)−(3x+4)(x−3) (x−3)2(x+3) Eliminate parentheses in the numerator: 2x2+5x−3−(3x2−5x−12) (x−3)2(x+3) Distribute the negative sign: 2x2+5x−3−3x2+5x+12 (x−3)2(x+3) Combine like terms in the numerator: −x2+10x+9 (x−3)2(x+3) Answer For more examples of how to add and subtract rational expressions, watch the video at youtube.com/watch?v=FZdt73khrxA . Figure 12.4: Adding and Subtracting Rational Expressions (Watch Youtube Video) Solve Real-World Problems by Adding and Subtracting Rational Expressions Example 8 In an electrical circuit with two resistors placed in parallel, the reciprocal of the total resistance is equal to the sum of the reciprocals of each resistance: 1 Rtot = 1 R1 + 1 R2 . Find an expression for the total resistance, Rtot. Solution Let’s simplify the expression 1 R1 + 1 R2 . The lowest common denominator is R1R2, so we multiply the ﬁrst fraction by R2 R2 and the second fraction by R1 R1 : R2 R2 · 1 R1 + R1 R1 · 1 R2 Simplify: R2+R1 R1R2 www.ck12.org 560 The total resistance is the reciprocal of this expression: Rtot = R1R2 R1+R2 Answer Example 9 The sum of a number and its reciprocal is 53 14. Find the numbers. Solution Deﬁne variables: Let x be the number; then its reciprocal is 1 x. Set up an equation: The equation that describes the relationship between the numbers is x + 1 x = 53 14 Solve the equation: Find the lowest common denominator: LCM = 14x Multiply all terms by 14x : 14x · x + 14x · 1 x = 14x · 53 14 (Notice that we’re multiplying the terms by 14x instead of by 14x 14x. We can do this because we’re multiplying both sides of the equation by the same thing, so we don’t have to keep the actual values of the terms the same. We could also multiply by 14x 14x, but then the denominators would just cancel out a couple of steps later.) Cancel common factors in each term: 14x · x + 14x · 1 x = 14x · 53 14 Simplify: 14x2 + 14 = 53x Write all terms on one side of the equation: 14x2 −53x + 14 = 0 Factor: (7x −2)(2x −7) = 0 x = 2 7 and x = 7 2 Notice there are two answers for x, but they are really parts of the same solution. One answer represents the number and the other answer represents its reciprocal. Check: 2 7 + 7 2 = 4+49 14 = 53 14. The answer checks out. Work problems are problems where two people or two machines work together to complete a job. Work problems often contain rational expressions. Typically we set up such problems by looking at the part of the task completed by each person or machine. The completed task is the sum of the parts of the tasks completed by each individual or each machine. To determine the part of the task completed by each person or machine we use the following fact: Part of the task completed = rate of work × time spent on the task It’s usually useful to set up a table where we can list all the known and unknown variables for each person or machine and then combine the parts of the task completed by each person or machine at the end. Example 10 Mary can paint a house by herself in 12 hours. John can paint a house by himself in 16 hours. How long would it take them to paint the house if they worked together? Solution 561 www.ck12.org Deﬁne variables: Let t = the time it takes Mary and John to paint the house together. Construct a table: Since Mary takes 12 hours to paint the house by herself, in one hour she paints 1 12 of the house. Since John takes 16 hours to pain the house by himself, in one hour he paints 1 16 of the house. Mary and John work together for t hours to paint the house together. Using Part o f the task completed = rate of work · time spent on the task we can write that Mary completed t 12 of the house and John completed t 16 of the house in this time. This information is nicely summarized in the table below: Table 12.6: Painter Rate of work (per hour) Time worked Part of task Mary 1 12 t t 12 John 1 16 t t 16 Set up an equation: In t hours, Mary painted t 12 of the house and John painted t 16 of the house, and together they painted 1 whole house. So our equation is t 12 + t 16 = 1. Solve the equation: Find the lowest common denominator: LCM = 48 Multiply all terms in the equation by the LCM: 48 · t 12 + 48 · t 16 = 48 · 1 Cancel common factors in each term: 4 · t 1 + 3 · t 1 = 48 · 1 Simplify: 4t + 3t = 48 7t = 48 ⇒t = 48 7 = 6.86 hours Check: The answer is reasonable. We’d expect the job to take more than half the time Mary would take by herself but less than half the time John would take, since Mary works faster than John. Example 11 Suzie and Mike take two hours to mow a lawn when they work together. It takes Suzie 3.5 hours to mow the same lawn if she works by herself. How long would it take Mike to mow the same lawn if he worked alone? Solution Deﬁne variables: Let t = the time it takes Mike to mow the lawn by himself. Construct a table: www.ck12.org 562 Table 12.7: Painter Rate of work (per hour) Time worked Part of Task Suzie 1 3.5 = 2 7 2 4 7 Mike 1 t 2 2 t Set up an equation: Since Suzie completed 4 7 of the lawn and Mike completed 2 t of the lawn and together they mowed the lawn in 2 hours, we can write the equation: 4 7 + 2 t = 1 Solve the equation: Find the lowest common denominator: LCM = 7t Multiply all terms in the equation by the LCM: 7t · 4 7 + 7t · 2 t = 7t · 1 Cancel common factors in each term: t · 4 1 + 7 · 2 1 = 7t · 1 Simplify: 4t + 14 = 7t 3t = 14 ⇒t = 14 3 = 4 2 3 hours Check: The answer is reasonable. We’d expect Mike to work slower than Suzie, because working by herself it takes her less than twice the time it takes them to work together. Review Questions Perform the indicated operation and simplify. Leave the denominator in factored form. 1. 5 24 −7 24 2. 10 21 + 9 35 3. 5 2x+3 + 3 2x+3 4. 3x−1 x+9 −4x+3 x+9 5. 4x+7 2x2 −3x−4 2x2 6. x2 x+5 − 25 x+5 7. 2x x−4 + x 4−x 8. 10 3x−1 − 7 1−3x 9. 5 2x+3 −3 10. 5x+1 x+4 + 2 11. 1 x + 2 3x 12. 4 5x2 − 2 7x3 13. 4x x+1 − 2 2(x+1) 14. 10 x+5 + 2 x+2 15. 2x x−3 − 3x x+4 16. 4x−3 2x+1 + x+2 x−9 17. x2 x+4 −3x2 4x−1 18. 2 5x+2 −x+1 x2 563 www.ck12.org 19. x+4 2x + 2 9x 20. 5x+3 x2+x + 2x+1 x 21. 4 (x+1)(x−1) − 5 (x+1)(x+2) 22. 2x (x+2)(3x−4) + 7x (3x−4)2 23. 3x+5 x(x−1) − 9x−1 (x−1)2 24. 1 (x−2)(x−3) + 4 (2x+5)(x−6) 25. 3x−2 x−2 + 1 x2−4x+4 26. −x3 x2−7x+6 + x −4 27. 2x x2+10x+25 − 3x 2x2+7x−15 28. 1 x2−9 + 2 x2+5x+6 29. −x+4 2x2−x−15 + x 4x2+8x−5 30. 4 9x2−49 − 1 3x2+5x−28 31. One number is 5 less than another. The sum of their reciprocals is 13 36. Find the two numbers. 32. One number is 8 times more than another. The diﬀerence in their reciprocals is 21 20. Find the two numbers. 33. A pipe can ﬁll a tank full of Kool-Aid in 4 hours and another pipe can empty the tank in 8 hours. If the valves to both pipes are open, how long will it take to ﬁll the tank? 34. Stefan and Misha have a lot full of cars to wash. Stefan could wash the cars by himself in 6 hours and Misha could wash the cars by himself in 5 hours. Stefan starts washing the cars by himself, but he has to leave after 2.5 hours. Misha continues the task by himself. How long does it take Misha to ﬁnish washing the cars? 35. Amanda and her sister Chyna are shoveling snow to clear their driveway. Amanda can clear the snow by herself in 3 hours and Chyna can clear the snow by herself in 4 hours. After Amanda has been working by herself for one hour, Chyna joins her and they ﬁnish the job together. How long does it take to clear the snow from the driveway? 36. At a soda bottling plant one bottling machine can fulﬁll the daily quota in 10 hours and a second machine can ﬁll the daily quota in 14 hours. The two machines start working together, but after four hours the slower machine breaks and the faster machine has to complete the job by itself. How many more hours does the fast machine take to ﬁnish the job? 12.7 Solutions of Rational Equations Learning Objectives • Solve rational equations using cross products. • Solve rational equations using lowest common denominators. • Solve real-world problems with rational equations. Introduction A rational equation is one that contains rational expressions. It can be an equation that contains rational coeﬀicients or an equation that contains rational terms where the variable appears in the denominator. An example of the ﬁrst kind of equation is: 3 5 x + 1 2 = 4. An example of the second kind of equation is: x x−1 + 1 = 4 2x+3. The ﬁrst aim in solving a rational equation is to eliminate all denominators. That way, we can change a rational equation to a polynomial equation which we can solve with the methods we have learned this far. www.ck12.org 564 Solve Rational Equations Using Cross Products A rational equation that contains just one term on each side is easy to solve by cross multiplication. Consider the following equation: x 5 = x + 1 2 Our ﬁrst goal is to eliminate the denominators of both rational expressions. In order to remove the 5 from the denominator of the ﬁrst fraction, we multiply both sides of the equation by 5: 5 · x 5 = 5 · x + 1 2 x = 5(x + 1) 2 Now, we remove the 2 from the denominator of the second fraction by multiplying both sides of the equation by 2: 2 · x = 2 · 5(x + 1) 2 2x = 5(x + 1) Then we can solve this equation for x. Notice that this equation is what we would get if we simply multiplied each numerator in the original equation by the denominator from the opposite side of the equation. It turns out that we can always simplify a rational equation with just two terms by multiplying each numerator by the opposite denominator; this is called cross multiplication. Example 1 Solve the equation 2x x+4 = 5 x. Solution Cross-multiply. The equation simpliﬁes to: 2x2 = 5(x + 4) Simplify: 2x2 = 5x + 20 Move all terms to one side of the equation: 2x2 −5x −20 = 0 Solve using the quadratic formula: x = 5± √ 185 4 ⇒x = −2.15 or x = 4.65 It’s important to plug the answer back into the original equation when the variable appears in any denom-inator of the equation, because the answer might be an excluded value of one of the rational expressions. If the answer obtained makes any denominator equal to zero, that value is not really a solution to the equation. Check: 2x x+4 = 5 x ⇒2(−2.15) −2.15+4 ? = 5 −2.15 ⇒−4.30 1.85 ? = −2.3 ⇒−2.3 = −2.3. The answer checks out. 2x x+4 = 5 x ⇒2(4.65) 4.65+4 ? = 5 4.65 ⇒9.3 8.65 ? =1.08 ⇒1.08 = 1.08. The answer checks out. Solve Rational Equations Using Lowest Common Denominators Another way of eliminating the denominators in a rational equation is to multiply all the terms in the equation by the lowest common denominator. You can use this method even when there are more than 565 www.ck12.org two terms in the equation. Example 2 Solve 3 x+2 − 4 x−5 = 2 x2−3x−10. Solution Factor all denominators: 3 x+2 − 4 x−5 = 2 (x+2)(x−5) Find the lowest common denominator: LCD = (x + 2)(x −5) Multiply all terms in the equation by the LCD: (x + 2)(x −5) · 3 x + 2 −(x + 2)(x −5) · 4 x −5 = (x + 2)(x −5) · 2 (x + 2)(x −5) The equation simpliﬁes to: 3(x −5) −4(x + 2) = 2 Simplify: 3x −15 −4x −8 = 2 x = −25 Check: 3 x+2 − 4 x−5 = 2 x2−3x−10 ⇒ 3 −25+2 − 4 −25−5 ? = 2 (−25)2−3(−25)−10 ⇒.003 = .003. The answer checks out. Example 3 Solve 2x 2x+1 + x x+4 = 1. Solution Find the lowest common denominator: LCD = (2x + 1)(x + 4) Multiply all terms in the equation by the LCD: (2x + 1)(x + 4) · 2x 2x + 1 + (2x + 1)(x + 4) · x x + 4 = (2x + 1)(x + 4) Cancel all common terms. 2x(x + 4) + x(2x + 1) = (2x + 1)(x + 4) The simpliﬁed equation is: Eliminate parentheses: 2x2 + 8x + 2x2 + x = 2x2 + 9x + 4 Collect like terms: 2x2 = 4 x2 = 2 ⇒x = ± √ 2 Check: 2x 2x+1 + x x+4 = 2 √ 2 2 √ 2+1 + √ 2 √ 2+4 = 0.739 + 0.261 = 1. The answer checks out. 2x 2x+1 + x x+4 = 2(− √ 2) 2(− √ 2)+1 + − √ 2 − √ 2+4 = 1.547 −0.547 = 1. The answer checks out. Solve Real-World Problems Using Rational Equations A motion problem with no acceleration is described by the formula distance = speed×time. These problems can involve the addition and subtraction of rational expressions. www.ck12.org 566 Example 4 Last weekend Nadia went canoeing on the Snake River. The current of the river is three miles per hour. It took Nadia the same amount of time to travel 12 miles downstream as it did to travel 3 miles upstream. Determine how fast Nadia’s canoe would travel in still water. Solution Deﬁne variables: Let s = speed of the canoe in still water Then, s + 3 = the speed of the canoe traveling downstream s −3 = the speed of the canoe traveling upstream Construct a table: Table 12.8: Direction Distance (miles) Rate Time Downstream 12 s + 3 t Upstream 3 s −3 t Write an equation: Since distance = rate × time, we can say that time = distance rate . The time to go downstream is: t = 12 s+3 The time to go upstream is: t = 3 s−3 Since the time it takes to go upstream and downstream are the same, we have: 3 s−3 = 12 s+3 Solve the equation: Cross-multiply: 3(s + 3) = 12(s −3) Simplify: 3s + 9 = 12s −36 Solve: s = 5 mi/h Check: Upstream: t = 12 8 = 1 1 2 hour; downstream: t = 3 2 = 1 1 2 hour. The answer checks out. Example 5 Peter rides his bicycle. When he pedals uphill he averages a speed of eight miles per hour, when he pedals downhill he averages 14 miles per hour. If the total distance he travels is 40 miles and the total time he rides is four hours, how long did he ride at each speed? Solution Deﬁne variables: Let t = time Peter bikes at 8 miles per hour. Construct a table: 567 www.ck12.org Table 12.9: Direction Distance (miles) Rate (mph) Time (hours) Uphill d 8 t1 Downhill 40 −d 14 t2 Write an equation: We know that time = distance rate . The time to go uphill is: t1 = d 8 The time to go downhill is: t2 = 40−d 14 We also know that the total time is 4 hours: d 8 + 40−d 14 = 4 Solve the equation: Find the lowest common denominator: LCD = 56 Multiply all terms by the common denominator: 7d + 160 −4d = 224 Solve: d = 21.3 mi Check: Uphill: t = 21.3 8 = 2.67 hours; downhill: t = 40−21.3 14 = 1.33 hours. The answer checks out. Example 6 A group of friends decided to pool together and buy a birthday gift that cost $200. Later 12 of the friends decided not to participate any more. This meant that each person paid $15 more than their original share. How many people were in the group to begin with? Solution Deﬁne variables: Let x = the number of friends in the original group. Make a table: Table 12.10: Number of people Gift price Share amount Original group x 200 200 x Later group x −12 200 200 x−12 Write an equation: Since each person’s share went up by $15 after 2 people refused to pay, we write the equation 200 x−12 = 200 x +15 Solve the equation: Find the lowest common denominator: LCD = x(x −12) Multiply all terms by the LCD: x(x −12) · 200 x−12 = x(x −12) · 200 x + x(x −12) · 15 Cancel common factors and simplify: 200x = 200(x −12) + 15x(x −12) www.ck12.org 568 Eliminate parentheses: 200x = 200x −2400 + 15x2 −180x Get all terms on one side of the equation: 0 = 15x2 = 180x −2400 Divide all terms by 15 : 0 = x2 −12x −160 Factor: 0 = (x −20)(x + 8) Solve: x = 20, x = −8 The answer that makes sense is x = 20 people. Check: Originally $200 shared among 20 people is $10 each. After 12 people leave, $200 shared among 8 people is $25 each. So each person pays $15 more. The answer checks out. Review Questions Solve the following equations. 1. 2x+1 4 = x−3 10 2. 4x x+2 = 5 9 3. 5 3x−4 = 2 x+1 4. 7 x+3 = x+1 2x−3 5. 7x x−5 = x+3 x 6. 2 x+3 − 1 x+4 = 0 7. 3x2+2x−1 x2−1 = −2 8. x + 1 x = 2 9. −3 + 1 x+1 = 2 x 10. 1 x − x x−2 = 2 11. 3 2x−1 + 2 x+4 = 2 12. 2x x−1 − x 3x+4 = 3 13. x+1 x−1 + x−4 x+4 = 3 14. x x−2 + x x+3 = 1 x2+x−6 15. 2 x2+4x+3 = 2 + x−2 x+3 16. 1 x+5 − 1 x−5 = 1−x x+5 17. x x2−36 + 1 x−6 = 1 x+6 18. 2x 3x+3 − 1 4x+4 = 2 x+1 19. −x x−2 + 3x−1 x+4 = 1 x2+2x−8 20. Juan jogs a certain distance and then walks a certain distance. When he jogs he averages 7 miles/hour and when he walks he averages 3.5 miles per hour. If he walks and jogs a total of 6 miles in a total of 1.2 hours, how far does he jog and how far does he walk? 21. A boat travels 60 miles downstream in the same time as it takes it to travel 40 miles upstream. The boat’s speed in still water is 20 miles per hour. Find the speed of the current. 22. Paul leaves San Diego driving at 50 miles per hour. Two hours later, his mother realizes that he forgot something and drives in the same direction at 70 miles per hour. How long does it take her to catch up to Paul? 23. On a trip, an airplane ﬂies at a steady speed against the wind and on the return trip the airplane ﬂies with the wind. The airplane takes the same amount of time to ﬂy 300 miles against the wind as 569 www.ck12.org it takes to ﬂy 420 miles with the wind. The wind is blowing at 30 miles per hour. What is the speed of the airplane when there is no wind? 24. A debt of $420 is shared equally by a group of friends. When ﬁve of the friends decide not to pay, the share of the other friends goes up by $25. How many friends were in the group originally? 25. A non-proﬁt organization collected $2250 in equal donations from their members to share the cost of improving a park. If there were thirty more members, then each member could contribute $20 less. How many members does this organization have? www.ck12.org 570
16655	https://cdnx.uobabylon.edu.iq/lectures/1ySSz7pexECIHuIIhn1uA.pdf	45 Chapter 4Boolean algebra 4.1 Definition Boolean algebra is the mathematics of digital logic in which the values of the variables are the truth values true and false, usually denoted 1 and 0, respectively. Basic knowledge of Boolean algebra is indispensable to the study and analysis of logic circuits. Variable, complement, and literal are terms used in Boolean algebra. A variable is a symbol (usually an italic uppercase letter or word) used to represent an action, a condition, or data. Any single variable can have only a 1 or a 0 value. The complement is the inverse of a variable and is indicated by a bar over the variable (overbar). There are four connecting symbols used in Boolean algebra. 1. Equal sign (=): This refers to the sign of equality as in mathematics. 2. Multiplication sign (·): It refers to the AND operation. 3. Plus sign (+): This refers to the OR operation. 4. Inversion sign (‘) or (−): This operation performs a complement of the input given to the logic gate. 4.2 Laws of Boolean Algebra The basic laws of Boolean algebra—the commutative laws for addition and multiplication, the associative laws for addition and multiplication, and the distributive law—are the same as in ordinary algebra. Each of the laws is illustrated with two or three variables, but the number of variables is not limited to this. 46 18/2/2025 4.2.1 Commutative Laws The commutative law of addition for two variables is written as A + B = B + A Application of commutative law of addition. The commutative law of multiplication for two variables is AB = BA Application of commutative law of multiplication. 4.2.2 Associative Laws The associative law of addition is written as follows for three variables: A + (B + C) = (A + B) + C Application of associative law of addition. The associative law of multiplication is written as follows for three variables: A (BC) = (AB) C 47 18/2/2025 This law states that it makes no difference in what order the variables are grouped when ANDing more than two variables. Figure illustrates this law as applied to 2-input AND gates. 4.2.3 Distributive Law The distributive law is written for three variables as follows: A (B + C) = AB + AC 4.3 Rules of Boolean Algebra The table below lists 12 basic rules that are useful in manipulating and simplifying Boolean expressions. A, B, or C can represent a single variable or a combination of variables. 48 18/2/2025 Rule 1: A + 0 = A A variable ORed with 0 is always equal to the variable. If the input variable A is 1, the output variable X is 1, which is equal to A. If A is 0, the output is 0, which is also equal to A. Rule 2: A + 1 = 1 A variable ORed with 1 is always equal to 1. A 1 on an input to an OR gate produces a 1 on the output, regardless of the value of the variable on the other input. Rule 3: 𝑨 . 𝟎 = 𝟎 A variable ANDed with 0 is always equal to 0. Any time one input to an AND gate is 0, the output is 0, regardless of the value of the variable on the other input. 49 18/2/2025 Rule 4: 𝑨 . 𝟏 = 𝑨 A variable ANDed with 1 is always equal to the variable. If A is 0, the output of the AND gate is 0. If A is 1, the output of the AND gate is 1 because both inputs are now 1s. Rule 5: A + A = A A variable ORed with itself is always equal to the variable. If A is 0, then 0 + 0 = 0; and if A is 1, then 1 + 1 = 1. Rule 6: 𝑨 + 𝑨= 𝟏 A variable ORed with its complement is always equal to 1. If A is 0, then 0 + 0 = 0 + 1 = 1. If A is 1, then 1 + 1 = 1 + 0 = 1. 50 18/2/2025 Rule 7: 𝑨 ∙𝑨= 𝑨 A variable ANDed with itself is always equal to the variable. If A = 0, then 0 #0 = 0; and if A = 1, then 1 #1 = 1. Rule 8: 𝑨∙𝑨= 𝟎 A variable ANDed with its complement is always equal to 0. Either A or A will always be 0; and when a 0 is applied to the input of an AND gate, the output will be 0 also. Rule 9: 𝑨 ̅ = 𝑨 The double complement of a variable is always equal to the variable. If you start with the variable A and complement (invert) it once, you get A. If you then take A and complement (invert) it, you get A, which is the original variable. 51 18/2/2025 Rule 10: A + AB = A This rule can be proved by applying the distributive law, rule 2, and rule 4 as follows: 𝐴+ 𝐴𝐵= 𝐴 . 1 + 𝐴𝐵= 𝐴 (1 + 𝐵) Factoring (distributive law) = 𝐴 . 1 Rule 2: (1 + 𝐵) = 1 = 𝐴 Rule 4: 𝐴 . 1 = 𝐴 The proof is shown in below, which shows the truth table and the resulting logic circuit simplification. Rule 11: 𝑨+ 𝑨𝑩= 𝑨+ 𝑩 This rule can be proved as follows: 𝐴 + 𝐴𝐵 = (𝐴 + 𝐴𝐵) + 𝐴𝐵 Rule 10: A = A + AB =(𝐴𝐴 + 𝐴𝐵) + 𝐴𝐵 Rule 7: A = AA =AA + AB + AA + AB Rule 8: adding AA = 0 = (A + A)(A + B) Factoring = 1 . (A + B) Rule 6: A + A = 1 = A + B Rule 4: drop the 1 52 18/2/2025 The proof is shown in the table below, which shows the truth table and the resulting logic circuit simplification. Rule 12: (A + B) (A + C) = A + BC This rule can be proved as follows: (A + B)(A + C) = AA + AC + AB + BC Distributive law = A + AC + AB + BC Rule 7: AA = A = A(1 + C) + AB + BC Factoring (distributive law) = A .1 + AB + BC Rule 2: 1 + C = 1 = A(1 + B) + BC Factoring (distributive law) = A . 1 + BC Rule 2: 1 + B = 1 = A + BC Rule 4: A. 1 = A The proof is shown in the table below, which shows the truth table and the resulting logic circuit simplification. 53 18/2/2025 H.W: 1. Apply the associative law of addition to the expression A + (B + C + D). 2. Apply the distributive law to the expression A(B + C + D). 4.4 DeMorgan’s Theorems DeMorgan, a mathematician, proposed two theorems that are an important part of Boolean algebra. In practical terms, DeMorgan’s theorems provide mathematical verification of the equivalency of the NAND and negative-OR gates and the equivalency of the NOR and negative-AND gates. 4.4.1 DeMorgan’s first theorem The complement of a product of variables is equal to the sum of the complements of the variables. Stated another way, The complement of two or more ANDed variables is equivalent to the OR of the complements of the individual variables. The formula for expressing this theorem for two variables is 54 18/2/2025 𝑋𝑌 ̅ ̅ ̅ ̅ = 𝑋 ̅ + 𝑌 ̅ ……… (1) 4.4.2 DeMorgan’s second theorem The complement of a sum of variables is equal to the product of the complements of the variables. Stated another way, The complement of two or more ORed variables is equivalent to the AND of the complements of the individual variables. The formula for expressing this theorem for two variables is 𝑋+ 𝑌 ̅̅̅̅̅̅̅̅ = 𝑋 ̅𝑌 ̅ ……… (2) Figure below shows the gate equivalencies and truth tables for Equations 1 and 2. 55 18/2/2025 Example: Apply DeMorgan’s theorems to the expressions Sol. Example: Apply DeMorgan’s theorems to the expressions Sol. Example: Prove that A+AB = A Sol. A+AB = A (1+B) = A . 1 = A Example: Prove that 𝐴+ 𝐴̅𝐵= 𝐴+ 𝐵 Sol. 𝐴+ 𝐴̅𝐵= (𝐴+ 𝐴𝐵) + 𝐴̅𝐵 = 𝐴+ 𝐵(𝐴+ 𝐴̅) = 𝐴+ 𝐵. 1 = 𝐴+ 𝐵 Example: Apply DeMorgan’s theorems to each expression: 56 18/2/2025 Sol. Example: Prove that (𝐴+ 𝐵)(𝐴+ 𝐶) = 𝐴+ 𝐵𝐶 Sol. (A+B)(A+C) = AA + AC + AB + BC = A + AC + AB + BC = A + (1+C) + AB + BC = A.1 + AB + BC = A + AB + BC = 𝐴(1 + 𝐵) + 𝐵𝐶 = A.1 + BC = A + BC H.W.: Apply DeMorgan’s theorems to the following expressions: 1. 2. 3. 2. Apply DeMorgan’s theorems to the expression 57 18/2/2025 4.5 Logic Simplification Using Boolean Algebra A logic expression can be reduced to its simplest form or changed to a more convenient form to implement the expression most efficiently using Boolean algebra. Example: Using Boolean algebra techniques, simplify this expression: 000000000AB + A(B + C) + B(B + C) Sol. AB + AB + AC + BB + BC distributive law to the second and third terms AB + AB + AC + B + BC (BB = B) rule 7 AB + AC + B + BC (AB + AB = AB) rule 5 AB + AC + B (B + BC = B) rule 10 B + AC (AB + B = B) rule 10 The figure below shows that the simplification process significantly reduced the number of logic gates required to implement the expression. Part (a) shows that 5 gates are required to implement the expression in its original form; however, only 2 gates are needed for the simplified expression, shown in part (b). It is important to realize that these two gate circuits are equivalent. 58 18/2/2025 Example: Simplify the following Boolean expression: 𝐴𝐵+ 𝐴𝐶 ̅̅̅̅̅̅̅̅̅̅̅ + 𝐴̅𝐵 ̅𝐶 Sol. (𝐴𝐵 ̅̅̅̅)(𝐴𝐶 ̅ ̅ ̅ ̅) + 𝐴̅𝐵 ̅𝐶 DeMorgan’s theorem (𝐴̅ + 𝐵 ̅)(𝐴̅ + 𝐶̅) + 𝐴̅𝐵 ̅𝐶 DeMorgan’s theorem 𝐴̅𝐴̅ + 𝐴̅𝐶̅ + 𝐴̅𝐵 ̅ + 𝐵 ̅𝐶̅ + 𝐴̅𝐵 ̅𝐶 distributive law to the two terms 𝐴̅𝐵 ̅ + 𝐴̅𝐵 ̅𝐶= 𝐴̅𝐵 ̅(1 + 𝐶) = 𝐴̅𝐵 ̅ rule 7 (𝐴̅𝐴̅ = 𝐴̅) = A) to the first term 𝐴̅ + 𝐴̅𝐶̅ + 𝐴̅𝐵 ̅ + 𝐵 ̅𝐶̅ rule 10 to the third and last terms. 𝐴̅ + 𝐴̅𝐶̅ = 𝐴̅(1 + 𝐶̅) = 𝐴̅ rule 10 to the first and second terms 𝐴̅ + 𝐴̅𝐵 ̅ + 𝐵 ̅𝐶̅ 𝐴̅ + 𝐴̅𝐵 ̅ = 𝐴̅(1 + 𝐵 ̅) = 𝐴̅ rule 10 to the first and second terms. 𝐴̅ + 𝐵 ̅𝐶̅ H.W.: 4.6 Boolean Expressions For Truth Table All Boolean expressions, regardless of their form, can be converted into either of two standard forms: the sum-of-products form or the product-of-sums form. Standardization makes the evaluation, simplification, and implementation of Boolean expressions much more systematic and easier. 59 18/2/2025 4.6.1 The Sum-of-Products (SOP) Form (Minterm) This form is sometimes called "minterm". A product term that contains each of the n-variables factors in either complemented or uncomplemented form for output digits "1" only, is called SOP. For example for the truth table below: Input Output A B C F 0 0 0 1 𝐴̅𝐵 ̅𝐶̅ 0 0 1 0 𝐴̅𝐵 ̅𝐶 0 1 0 1 𝐴̅𝐵𝐶̅ 0 1 1 1 𝐴̅𝐵𝐶 1 0 0 0 𝐴𝐵 ̅𝐶̅ 1 0 1 0 𝐴𝐵 ̅𝐶 1 1 0 1 𝐴𝐵𝐶̅ 1 1 1 1 𝐴𝐵𝐶 The Logical SOP expression for the output digit "1" is written as" 𝐹= 𝐴̅𝐵 ̅𝐶̅ + 𝐴̅𝐵𝐶̅ + 𝐴̅𝐵𝐶+ 𝐴𝐵𝐶̅ + 𝐴𝐵𝐶 This function com be put in another form such as: 𝐹= ∑0, 2,3,6,7 Since F= 1 in rows 0, 2,3,6,7 only. The second form is called the Canonical Sum of Products (Canonical SOP). 4.6.2 The Product-of-Sum (POS) Form (Maxterm) A Logical equation can also be expressed as a product of sum (POS) form (sometimes this method is called "Maxterm". This is done by considering the combination for F=0 (output = 0). So for the above example from the truth table F=0 is in rows 1, 4, 5 hence: 60 18/2/2025 𝐹 ̅(𝐴, 𝐵, 𝐶) = 𝐴̅𝐵 ̅𝐶+ 𝐴𝐵 ̅𝐶̅ + 𝐴𝐵 ̅𝐶 𝐹(𝐴, 𝐵, 𝐶) = 𝐹 ̅ ̅(𝐴, 𝐵, 𝐶) = 𝐴̅𝐵 ̅𝐶+ 𝐴𝐵 ̅𝐶̅ + 𝐴𝐵 ̅𝐶 ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ = 𝐴̅𝐵 ̅𝐶 ̅̅̅̅̅̅ ∙𝐴𝐵 ̅𝐶̅ ̅̅̅̅̅̅ ∙𝐴𝐵 ̅𝐶 ̅̅̅̅̅̅ = (𝐴̿ + 𝐵 ̿ + 𝐶̅) ∙(𝐴̅ + 𝐵 ̿ + 𝐶̿) ∙(𝐴̅ + 𝐵 ̿ + 𝐶̅) 𝐹(𝐴, 𝐵, 𝐶) = (𝐴+ 𝐵+ 𝐶̅) ∙(𝐴̅ + 𝐵 ̿ + 𝐶̿) ∙(𝐴̅ + 𝐵 ̿ + 𝐶̅) This is POS form. POS form can be expressed as: 𝐹= ∏1, 4, 5 This form is called the Canonical Product of Sum (Canonical POS). Example: Put F in SOP and POS form and simplifying it: Sol. 𝑆𝑂𝑃: 𝐹(𝐴, 𝐵) = ∑0,1,3 = 𝐴̅𝐵 ̅ + 𝐴̅𝐵+ 𝐴𝐵 = 𝐴̅ (𝐵 ̅ + 𝐵 ) + 𝐴𝐵 = 𝐴̅ + 𝐴𝐵 𝐹(𝐴, 𝐵) = 𝐴̅ + 𝐵 𝑃𝑂𝑆: 𝐹(𝐴, 𝐵) = ∏2 𝐹(𝐴, 𝐵) = 𝐴̅ + 𝐵 A B F 0 0 1 0 1 1 1 0 0 1 1 1 61 18/2/2025 Example: Put in canonical SOP form 𝐹(𝐴, 𝐵, 𝐶) = 𝐴𝐵 ̅𝐶+ 𝐴̅𝐵𝐶+ 𝐴𝐵𝐶 Sol. 𝐹(𝐴, 𝐵, 𝐶) = 𝐴𝐵 ̅𝐶+ 𝐴̅𝐵𝐶+ 𝐴𝐵𝐶 101 011 111 𝐹(𝐴, 𝐵, 𝐶) = ∑3, 5, 7 Example: Put in canonical POS form and draw the truth table, then determine canonical SOP and SOP form 𝐹(𝐴, 𝐵, 𝐶) = (𝐴+ 𝐵+ 𝐶̅)(𝐴+ 𝐵 ̅ + 𝐶)(𝐴̅ + 𝐵 ̅ + 𝐶̅)(𝐴̅ + 𝐵 ̅ + 𝐶) Sol. 𝐹(𝐴, 𝐵, 𝐶) = 001 010 111 110 M1 M2 M3 M4 𝐹(𝐴, 𝐵, 𝐶) = ∏1,2,6,7 A B C F 0 0 0 1 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 0 1 1 1 0 𝐹(𝐴, 𝐵, 𝐶) = ∑0,3,4,5 𝐹(𝐴, 𝐵, 𝐶) = 𝐴̅𝐵 ̅𝐶̅ + 𝐴̅𝐵𝐶+ 𝐴𝐵 ̅𝐶̅ + 𝐴𝐵 ̅𝐶 62 18/2/2025 Example: Represent F1, F2 in SOP & POS forms then simplified F1 and F2 using Boolean algebra. A B C F1 F2 0 0 0 0 1 0 0 1 1 0 0 1 0 1 1 0 1 1 1 0 1 0 0 0 1 1 0 1 1 0 1 1 0 1 0 1 1 1 1 0 Sol. In SOP: 𝐹 1(𝐴, 𝐵, 𝐶) = ∑1,2,3,5,6,7 = 𝐴̅𝐵 ̅𝐶+ 𝐴̅𝐵𝐶̅ + 𝐴̅𝐵𝐶+ 𝐴𝐵 ̅𝐶+ 𝐴𝐵𝐶̅ + 𝐴𝐵𝐶 = 𝐴̅(𝐵 ̅𝐶+ 𝐵𝐶̅ + 𝐵𝐶) + 𝐴(𝐵 ̅𝐶+ 𝐵𝐶̅ + 𝐵𝐶) = 𝐴̅[𝐵 ̅𝐶+ 𝐵(𝐶̅ + 𝐶)] + 𝐴[𝐵 ̅𝐶+ 𝐵(𝐶̅ + 𝐶)] = 𝐴̅(𝐵 ̅𝐶+ 𝐵) + 𝐴(𝐵 ̅𝐶+ 𝐵) = (𝐴̅ + 𝐴) ∙(𝐵 ̅𝐶+ 𝐵) = 𝐵 ̅𝐶+ 𝐵 𝐹 1(𝐴, 𝐵, 𝐶) = 𝐵+ 𝐶 In POS: 𝐹(𝐴, 𝐵, 𝐶) = ∏0,4 = (𝐴+ 𝐵+ 𝐶) ∙(𝐴̅ + 𝐵+ 𝐶) = 𝐴𝐴̅ + 𝐴𝐵+ 𝐴𝐶+ 𝐴̅𝐵+ 𝐵𝐵+ 𝐵𝐶+ 𝐶𝐴̅ + 𝐶𝐵+ 𝐶𝐶 = 𝐴𝐵+ 𝐴𝐶+ 𝐴̅𝐵+ 𝐵+ 𝐵𝐶+ 𝐴̅𝐶+ 𝐵𝐶+ 𝐶 = 𝐴𝐵+ 𝐴𝐶+ 𝐴̅𝐵+ 𝐵(1 + 𝐶) + 𝐴̅𝐶+ 𝐶(1 + 𝐵) 63 18/2/2025 = 𝐴𝐵+ 𝐴𝐶+ 𝐴̅𝐵+ 𝐵+ 𝐴̅𝐶+ 𝐶 = 𝐵(𝐴+ 𝐴̅) + 𝐶(𝐴+ 𝐴̅) + 𝐵+ 𝐶 = 𝐵+ 𝐶+ 𝐵+ 𝐶 𝐹 1(𝐴, 𝐵, 𝐶) = 𝐵+ 𝐶 H.W.: Solution for F2 4.6.3 Converting SOP to POS and Vice Versa The binary values of the product terms in a given SOP expression aren't present in the equivalent POS expression. Therefore to convert from standard SOP to standard POS the following steps may be used: Step 1: Evaluate each product term in the SOP expression that determines the binary numbers representing the product term. Step 2: Determine all the binary numbers not included in the evaluation in step 1. Step 3: Write the equivalent sum term for each binary number from step 2 and express it in POS form. Note: A Standard SOP expression is one in which all the variables in the domain appear in each term of the expression. If any variable is missing from any term, we must add these missing variables to that term, by multiplying the term by the variables missing. For example, if variable B is missing from the term AC, we must multiply this term AC, by 𝐵+ 𝐵 ̅ to make the expression standard SOP. 𝐴𝐶(𝐵+ 𝐵 ̅) Note: using a similar procedure explained above (steps 1, 2, and 3) we can convert from standard POS to standard SOP. If there is missing any variable from any term, we must add the missing variable multiplied by its complement to that term. For example if variable A is missing from the term (𝐵+ 𝐶̅) we must add 𝐴𝐴̅ 64 18/2/2025 [(𝐵+ 𝐶̅) + 𝐴𝐴̅] = (𝐵+ 𝐶̅ + 𝐴)(𝐵+ 𝐶̅ + 𝐴̅) Example: Put in canonical POS form and draw the truth table, then determine canonical SOP and SOP form 𝐹(𝐴, 𝐵, 𝐶) = 𝐵+ 𝐴𝐶 Sol. 1st method 𝐹(𝐴, 𝐵, 𝐶) = 𝐵+ 𝐴𝐶 = 𝐵(𝐴+ 𝐴̅)(𝐶̅ + 𝐶) + 𝐴𝐶(𝐵+ 𝐵 ̅) = 𝐵(𝐴𝐶+ 𝐴𝐶̅ + 𝐴̅𝐶+ 𝐴̅𝐶̅) + 𝐴𝐵𝐶+ 𝐴𝐵 ̅𝐶 = 𝐴𝐵𝐶+ 𝐴𝐵𝐶̅ + 𝐴̅𝐵𝐶+ 𝐴̅𝐵𝐶̅ + 𝐴𝐵𝐶+ 𝐴𝐵 ̅𝐶 = 𝐴𝐵𝐶+ 𝐴𝐵𝐶̅ + 𝐴̅𝐵𝐶+ 𝐴̅𝐵𝐶̅ + 𝐴𝐵 ̅𝐶 111 110 011 010 101 ∴𝐹(𝐴, 𝐵, 𝐶) = ∑2,3,5,6,7 ∴𝐹(𝐴, 𝐵, 𝐶) = ∏0,1,4 𝐹(𝐴, 𝐵, 𝐶) = (𝐴+ 𝐵+ 𝐶)(𝐴+ 𝐵+ 𝐶̅)(𝐴̅ + 𝐵+ 𝐶) 2nd method: 𝐹(𝐴, 𝐵, 𝐶) = 𝐵+ 𝐴𝐶 A B C AC F= B+AC 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 1 1 0 1 1 0 0 0 0 1 0 1 1 1 1 1 0 0 1 1 1 1 1 1 65 18/2/2025 ∴𝐹(𝐴, 𝐵, 𝐶) = ∑2,3,5,6,7 ∴𝐹(𝐴, 𝐵, 𝐶) = ∏0,1,4 𝐹(𝐴, 𝐵, 𝐶) = (𝐴+ 𝐵+ 𝐶)(𝐴+ 𝐵+ 𝐶̅)(𝐴̅ + 𝐵+ 𝐶) H.W.: Convert the POS form to SOP form and find these canonical: 𝐹(𝐴, 𝐵, 𝐶) = (𝐴+ 𝐵)(𝐴̅ + 𝐶)(𝐴+ 𝐵+ 𝐶) 4.7 The Karnaugh Map (K-map) A K- map provides a systematic method for simplifying Boolean expressions and, if properly used, will produce the simplest SOP or POS expression. As you have seen, the effectiveness of algebraic simplification depends on your familiarity with all the laws, rules, and theorems of Boolean algebra and on your ability to apply them. The K-map is an array of cells in which each cell represents a binary value of the input variables. The cells are arranged in a way so that simplification of a given expression is simply a matter of properly grouping the cells. The K-maps can be used for expressions with two, three, four, and five variables, but we will discuss only 2, 3, and 4 variables. The number of cells in a K-map, as well as the number of rows in a truth table. For 2 input variables, the number of cells is 22 = 4 cells 66 18/2/2025 For 3 input variables, the number of cells is 23 = 8 cells And for 4 input variables, the number of cells is 24 = 16 cells 4.7.1 The 2-variebles K - map 1. 𝐹(𝐴, 𝐵) = 𝐵 ̅ 𝑆𝑂𝑃 𝐹(𝐴, 𝐵) = 𝐵 𝑃𝑂𝑆 67 18/2/2025 2. 𝐹(𝐴, 𝐵) = 𝐵 ̅ + 𝐴 𝑆𝑂𝑃 𝐹(𝐴, 𝐵) = 𝐴̅ 𝐵 𝑃𝑂𝑆 3. 𝐹(𝐴, 𝐵) = 1 4.7.2 The 3-variebles K - map 1. 𝐹(𝐴, 𝐵, 𝐶) = 𝐵 ̅ 𝑆𝑂𝑃 𝐹(𝐴, 𝐵, 𝐶) = 𝐵 𝑃𝑂𝑆 2. 𝐹(𝐴, 𝐵, 𝐶) = 𝐶̅ + 𝐵 ̅ 𝑆𝑂𝑃 𝐹(𝐴, 𝐵, 𝐶) = 𝐶𝐵 𝑃𝑂𝑆 68 18/2/2025 4.7.3 The 4-variebles K – map 1. 𝐹(𝐴, 𝐵) = 𝐵 ̅ 𝑆𝑂𝑃 𝐹(𝐴, 𝐵) = 𝐵 𝑃𝑂𝑆 2. 𝐹(𝐴, 𝐵) = 𝐵 ̅𝐷 ̅ 𝑆𝑂𝑃 𝐹(𝐴, 𝐵) = 𝐵+ 𝐷 𝑃𝑂𝑆 3. 𝐹(𝐴, 𝐵, 𝐶) = 𝐴+ 𝐶̅ + 𝐵 ̅ + 𝐷 ̅ 𝑆𝑂𝑃 𝐹(𝐴, 𝐵, 𝐶) = 𝐴̅𝐵𝐶𝐷 𝑃𝑂𝑆 Note: 1. Number of 1's or 0's in one group must be 1, 2, 4, 8, and 16. 2. We must take maximum number of 1's or 0's in one group. 69 18/2/2025 Example: Simplify the following SOP expression on a Karnaugh map: 𝐹= 𝐴̅𝐵 ̅𝐶̅𝐷 ̅ + 𝐴𝐵 ̅𝐶𝐷 ̅ + 𝐴𝐵 ̅𝐶̅𝐷 ̅ + 𝐴̅𝐶𝐷 + 𝐴𝐵 ̅𝐶𝐷 ̅ Sol. 𝐹= 𝐵 ̅𝐷 ̅ + 𝐴̅𝐶𝐷 Example: Determine the simply expression by the truth table below using Karnaugh map method. A B C F 0 0 0 1 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 1 70 18/2/2025 Sol. 𝐹= 𝐴𝐵+ 𝐵 ̅𝐶̅ HW: Implement the Logic function specified in the above example. Example: Simplify the following Boolean function in: (a) SOP form (b) POS form 𝐹(𝐴, 𝐵, 𝐶, 𝐷) = ∑(0, 1, 2, 5, 8, 9, 10) Sol. (a) The 1's marked in the map represent all minterm of the function. The cells marked with 0's represent the Maxterm not included in the function and therefore the function will be: 𝐹= 𝐵 ̅ 𝐶̅ + 𝐵 ̅𝐷 ̅ + 𝐴̅𝐶̅𝐷 71 18/2/2025 (b) If the squares marked with 0's are combined we obtain the simplified POS form or the complement of F: 𝐹 ̅ = 𝐴𝐵+ 𝐶𝐷+ 𝐵𝐷 ̅ Applying DeMorgan's theorem by taking the complement of each side, we obtain the simplified function in POS form: 𝐹 ̅ = 𝐴𝐵+ 𝐶𝐷+ 𝐵𝐷 ̅ 𝐹 ̅ = (𝐴𝐵) ∙(𝐶𝐷) ∙(𝐵𝐷 ̅) 𝐹 ̅ = (𝐴+ 𝐵) ∙(𝐶+ 𝐷) ∙(𝐵+ 𝐷 ̅) 𝐹 ̅ = (𝐴+ 𝐵) ∙(𝐶+ 𝐷) ∙(𝐵+ 𝐷) Note: To use K-map for simplification a function expressed in POS form, follow these rules: 1. Take the complement of the function. 2. From the results write "0" in the Squares of POS form. Or convert the POS to SOP form, then follow the standard rules used to enter the 1's in the cells of K-map. 4.7.4 Don't Care Conditions Sometimes a situation arises in which some input variable combinations are not allowed. For example, recall that in the BCD code, there are six invalid combinations: 1010, 1011, 1100, 1101, 1110, and 1111. Since these unallowed states will never occur in an application involving the BCD code, they can be treated as “don’t care” terms with respect to their effect on the output. That is, for these “don’t care” terms either a 1 or a 0 may be assigned to the output; it really does not matter since they will never occur. The “don’t care” terms can be used to advantage on the Karnaugh map. The figure below shows that for each “doesn’t care” term, an X is placed in the cell. When grouping the 1s, the Xs can be treated as 1s to make a larger grouping or as 0s if they cannot be used to advantage. The larger a group, the simpler the resulting term will be. 72 18/2/2025 The truth table describes a logic function that has a 1 output only when the BCD code for 7, 8, or 9 is present on the inputs. If the “don’t care” are used as 1s, the resulting expression for the function is A + BCD, as indicated in K-map. If the “don’t care” is not used as 1s, the resulting expression is ABC + ABCD; so you can see the advantage of using “don’t care” terms to get the simplest expression. Example: In a 7-segment display, each of the seven segments is activated for various digits. For example, segment-a is activated for the digits 0, 2, 3, 5, 6, 7, 8, and 9, as illustrated in the figure below. Since each digit can be represented by a BCD code, derive an SOP expression for segment-a using the variables ABCD and then minimize the expression using a K - map. 73 18/2/2025 Sol. The expression for segment-a is: Each term in the expression represents one of the digits in which segment-a is used. The Karnaugh map minimization is shown in the figure below. X’s (don’t care) are entered for those states that do not occur in the BCD code. From the K - map, the minimized expression for segment-a is:
16656	https://projecteuclid.org/journals/kodai-mathematical-journal/volume-23/issue-1/Uniqueness-of-entire-functions-that-share-some-small-functions/10.2996/kmj/1138044152.pdf	G. QIU KODAI MATH. J. 23 (2000), 1-11 UNIQUENESS OF ENTIRE FUNCTIONS THAT SHARE SOME SMALL FUNCTIONS GANGDI QIU Abstract In this paper we obtain a unicity theorem of an entire function and its derivative that share two small functions IM. So we generalize and improve some results given by Rubel-Yang, Mues-Steinmetz and J. H. Zheng etc. 1. Introduction and main results In this paper, we use the same signs as given in Nevanlinna theory of meromorphic functions (see ). By S(r,f) we denote any quantity satisfying S( rif) = °{T{r,f)} as r ^ o o , possibly outside a set of r with finite linear measure. Let / and g be two meromorphic functions. Then the meromorphic function α is said a small function of/if and only if Γ(r, α) = S{r,f). We say that / and g share a value a IM(CM) if / - a and g — a have the same zeros ignoring multiplicities (with the same multiplicity). When a is a small function of/and g, a is said a common small function of/and g IM(CM). In addition, we introduce the following denotations: S(m,n)(b) = {z\z is a common zero of / — b and f'-b with multiplicities m and n respectively}. N{m,n){r,\/{f — b)) denotes the counting function of/ with respect to the set S(m,ή){b). On the problems of uniqueness of an entire function and its derivative that share some values, Rubel-Yang (see ) proved that if the entire function / and / ' share two distinct finite values CM then / = /'. Mues-Steinmetz (see ) improved this result to the case when / and / ' share two values IM. In 1992, J. H. Zheng and S. P. Wang (see ) generalized this result to the/and / ' which share two small functions CM. In this paper, we generalize and improve above results to obtain the following result: THEOREM 1. Let f be a nonconstant entire function, a and b two distinct small functions of f with a ψ oo and b ψ oo. If f and f' share a and b IM, then f = f'. Supported by Fujian Provmcal Science Foundation. Keywords: Entire function, Meromorphic function, Small function, Uniqueness. Received August 24, 1998; revised July 5, 1999 1 GANGDI QIU 2. Some lemmas LEMMA 1. Let f be a nonconstant entire function, a\ and a-χ two distinct small functions of f with a\ ψ oo and a2 Ψ °0 Set Then / - a\ a\~a2 f'-a[ a[-a'2 (i) Δ(/)#0, f — a2 a\ — a2 a[-a'2 (iii) m r, '(/-«l)(/-«2) where β is an arbitrary small function off (v) (i) (2) (3) (4) (5) Proof Suppose that Δ(/) = 0, then from (1) we have a'2 f'-a[=a[-a'2 f-a\ a-a2 By integrating for above two side we get f = a\ + c(a\ — a2), (c φ 0 is a constant) which contradicts the fact that a\ and a2 are small functions of/ Hence Δ(/) φ 0. Again by (1) we have ( Δ ( Π \ if' — a f \ r, -7 j < m(r, Λ{ - a'2) + m(r, αi - a2) + / w ί r, — ί - j + log 2 = S(r,f), ( i = l , 2 ) i.e., (3) holds. Note that 1 1 Γ 1 1 (7) UNIQUENESS OF ENTIRE FUNCTIONS and Hf)(f-β) _ Δ(/) + (a2-β)A(f) (f-aι)(f-a2) f-a2 x (/ - ax)(f - a2)' w So (3) and (7) imply (4). (5) follows from (3), (4) and (8). Next, it is easy to see from (1) if any zero of / - at (i = 1,2) with mul-tiplicity p is not the pole of a\ and a2, as well as is not the zero of (a\ - a2), then it must be a zero of Δ(/) with multiplicity at least (p — 1). Thus (6) holds. This completes the proof of Lemma 1. LEMMA 2. Let f be a noneonstant entire function, a and b two distinct small functions off with aφ GO and b φ oo. Again let Ck — a + k(a — b), ( A : is a positive integer). (9) Then k=\ ' f -bj (10) Proof It is easy to see from (9) that cuφa and Ck Φ b {k = 1,2,...,«), and they are distinct small function of/. Let F = ΊΓ~^> ( n ) then T(r,F) = T(rJ') + S(r,f). (12) By the second fundamental theorem (/!+ )T{r,F) < N(r,F)+N(r,jJ +^(^^A This and (12) imply (10). This completes the proof of Lemma 2. LEMMA 3. Let f be a noneonstant entire function, a and b two distinct samll functions off with a ψ oo and b φ oo. If f and f f share a and b IM, and 4 GANGDI QIU T(r,f) = T(r,f') + S(r,f), (13) then f = /'. Proof. Assume that / # / ' . From the fact that /and / ' share a and b IM we know that -<m(rJ)+m(r,l-γj+S(r,f) <T(r,f) + S(r,f). (14) Now by the second fudamental theorem ^ ^ S ( r , f ) . (15) Combining (14) and (15) we have T(r,f)=N(r,j^z)+N[r,Ύ^-u)+S(r,f). (16) Set ψ {f-a){f-bY and {f'-a){f-by where Δ(/) and Δ(/') are defined by (1), a\ = a and ^2 = ί?. From (2) we know that Δ(/) # 0 and Δ(/ ;) # 0. Therefore, it follows that φ ψ 0 and / # 0. It is easy to see from (6) that N(r,φ) = S(r,f) and N(r,χ) = S(r,f). Again by (5) we get J - 7 Thus T(r,φ)=S(rJ). (19) Next, for any positive integer k, by (3), (4) and (5) we have UNIQUENESS OF ENTIRE FUNCTIONS m(r,χ) JΓ-Γ ) - N(r,f- ck) + S(r,f) Thus <T(r,f)-N(r,jr—)+S(rJ). T(r,χ) + S(r,f). (20) On the other hand, combining (10), (13) and (14) we get 2T(r,f'), m) f r, —L " ^ ("' j ^ ί ) + ( Γ' / } -for all positive integers m and n. Again by (16) we have ',/) UNIQUENESS OF ENTIRE FUNCTIONS It is impossible for this to hold, thus / = /'. This completes the proof of Lemma 3. 3. The proof of Theorem 1 Assume that / ψ /'. Let φ and χ be defined by (17) and (18) respectively. From (17) we have φ(f-a)(f-b)=A(f)(f-f), we rewrite this in the following form [φ - (a 1 - b')]f 2 = bxf + b2f + biff + b4f' 2 + b5, (25) where b\ = ab' - ba' + (a + b)φ, bι = ba' — ab', 63 = b + b' - a - a', 64 = a — b, bs = -abφ are all small functions of /. We discuss the following two cases: (I) Suppose that φ - (a'- b') φ 0. By (25) we have < m{r, f)+m \r, f (b, + b4 • =Q 1 + S(r, f) <m(r,f)+m(r, f') + S(r,f), which results in m(r,f) 2, we get by (17) φ(zι)=a'(zι)-a(zι)=a f(zι)-b'(zι), which results in a(z\) — b'(z\) = 0. If a — b' = 0, from (17) we get / ' - a 1 _ a' - b f — a a — b This implies that T{rJ) = T(rJ') + S{rJ). By Lemma 3 this is a contradiction again. Thus a-b 1 # 0. Hence P>2 Similary, we have P>2 Therefore (27) Setting H = φ — χ, we suppose that H = 0. It is easy to see from (17) and (18) that This is also a contradiction. Thus H ψ 0. By (17) and (18) we know that H(z2) = 0 for any z2 e S(l,l)(a)\JS(l,l)(b). Combining this, (20) and (27) we get T(r,f) < N\r,-+S(r,f) < T(r,χ)+S(r,f) <T(r,f)-N(r,-^—]+S(r,f), UNIQUENESS OF ENTIRE FUNCTIONS which results in This is also impossible. (II.2) Suppose that either a 1 Ξ Ξ a and V ψ b or a 1 # a and V = b. Without loss of generality, we can assume that a 1 = a and b' ψb. According to the discussion in (II. 1) we know that +S(r,f). (28) Since the zeros of / — a are all the zeros of / ' - a = f' — a f, it follows that It is easy to see from (17) that the counting function corresponding to the zeros of / — a and f' — a with multiplicities all larger than one equals to S(r, / ) . This derives that + S(rJ). (29) Set G = 2^—τ^ 2—^ h - . (30) It is easy to see from (28) that T(r, G) = S(r,f). When G = 0, by (30) we have This is also a contradiction. When G^O, combining (17), we get N(2,1) (r, j^-λ < N U ^) + 5(r, /) = S(r, /). Hence T(r,χ)+S(r,f) 10 GANGDI QIU i.e., N(r,7T±—)=S{rJ), (keN +). V f -CkJ This is also impossible. (II.3) Suppose that a' = a and b 1 = b. By the discussion in (II.2) we know that ( ^ ^ ^ S(r,/). (31) Now let z be a simple zero of / ' — a and a zero of / - a with multiplicity two but not a pole of a and & , also not a zero of a — b. Set If Gi Ξ 0, by (32) (/' - b) 2 = D2(a - b) 2(f - b). (D2 Φ 0 is a constant). This implies that z must be a zero of a — b — (I/D2). Since φ = a 1 — b' = a -b φθ, so a — b — (I/.D2) # 0, which results in If G\ # 0, from (17) and (32) it follows that G(z) = 0, also we have that Thus, from (31) we get T(r, f) = N(2,1) (r, - ^ ) + S(r, /) < ^ Γ(r, /) + S(r, f). This is also a contradiction. According to above all discussion we obtain that / = /'. This completes the proof of Theorem 1. REFERENCES [ 1 ] W K. HAYMAN, Meromorphic Functions, Clarendon Press, Oxford, 1964. [ 2 ] L. A. RUBEL AND C. C. YANG, Values shared by an entire function and its derivative, Lecture Notes in Math., 599, Springer, Berlin, 1977, 101-103. [ 3 ] E. MUES AND N. STEINMETZ, Meromorphe Funktionen, die mit ihrer Ableitung werte teilen, Manuscπpta Math., 29 (1979), 195-206. UNIQUENESS OF ENTIRE FUNCTIONS 11 [ 4 ] J. H. ZHENG AND S. P WANG, On the unicity of meromorphic functions and their derivatives, Adv. in Math. (China), 21 (1992), 334-341. [ 5 ] J. CLUNIE, On integral and meromorphic functions, J. London Math. Soc, 37 (1962), 17-27 DEPARTMENT OF MATHEMATICS NINGDE TEACHERS COLLEGE NINGDE, FUJIAN 352100 P.R. CHINA
16657	https://www.lung.org/lung-health-diseases/lung-disease-lookup/rsv/symptoms-diagnosis	Skip to main content Home Lung Health & Diseases Lung Disease Lookup Respiratory Syncytial Virus (RSV) RSV Symptoms and Diagnosis RSV Symptoms and Diagnosis Facebook Twitter LinkedIn Email Print RSV is the leading cause of hospitalizations in all infants. Know the signs and symptoms that may indicate worsening illness and when to contact your child’s healthcare provider. iframe video Watch Video When you're expecting a baby, it's natural to consult recent parents to learn their stories and learn helpful tips. You may be told about a disease called respiratory syncytial virus, or RSV. This virus is so common that nearly 100% of children catch it before the age of two. In addition to poor appetite, the initial symptoms of RSV are similar to those of a common cold: runny nose, fever, and cough. Very young babies may be fussy, tired, and have trouble breathing. Symptoms usually go away on their own within a week, but it's important to stay in close contact with your baby's healthcare provider and know what to do. The diagnosis of RSV is based on your child's medical history, the time of year, and a physical exam. You may order a white blood cell count and virus screening test to confirm the diagnosis. All babies can experience severe RSV with short, shallow, rapid breathing. Their chest may sink between the ribs and below the ribs; this is known as chest wall retractions. Nasal flaring occurs. The mouth, lips, and nails may take on a bluish color from lack of oxygen. Each year, about 58,000 children age five and younger are hospitalized due to RSV. RSV is highly contagious and spreads from person to person through close contact with secretions from coughs and sneezes or by touching objects where the virus is present. To help protect your newborn baby, take these steps: Avoid close contact with people who have respiratory infections. Do not share cups, bottles, or toys that may be contaminated by the virus. Wash your hands with soap and water after coming into contact with a person who has a respiratory infection. Learn more about RSV at Lung.org/rsv. Watch in Spanish What Are the Symptoms of RSV? People who become infected with RSV show symptoms within four to six days after the virus enter the body. Initial signs of RSV are similar to mild cold symptoms, including sneezing, runny nose, fever, cough and decrease in appetite. Very young infants may be irritable, fatigued and have breathing difficulties. These symptoms do not usually all show up at the same time, instead they come in stages. Normally these symptoms will clear up on their own in a few days. A barking or wheezing cough can be one of the first signs of a more serious illness. In these instances, the virus has spread to the lower respiratory tract, causing inflammation of the small airways entering the lungs. This can lead to pneumonia or bronchiolitis. RSV Symptoms that Require Urgent Care Difficult breathing Not drinking enough Decreased activity Nasal flaring Bluish lips & fingernails Infants with severe RSV will have short, shallow and rapid breathing. This can be identified by "caving-in" of the chest in between the ribs and under the ribs (chest wall retractions), "spreading-out" of the nostrils with every breath (nasal flaring), and abnormally fast breathing. In addition, their mouth, lips and fingernails may turn a bluish color due to lack of oxygen. How RSV Is Diagnosed Because RSV symptoms are similar to other respiratory viruses that spread during the fall in winter, your healthcare provider might order testing to determine which virus you have based on your medical history, time of year and a physical exam. This helps a healthcare provider determine the correct treatment if you fall into a high-risk category for specific respiratory viruses. If your healthcare provider does test to confirm the diagnosis, they will use a salt water solution that is inserted into your nose and then gently suctioned out or a nose swab to collect a sample and look for viruses. In severe RSV cases that require hospitalization, additional testing may be needed. Imaging tests, such as a chest X-ray or CT scan can check for lung complications. Blood and urine cultures may be necessary when infants are very sick, as RSV-related bronchiolitis can occur with a urinary tract infection in newborns. Parents Share Their RSV Stories My Infant Son Was Hospitalized Twice with RSV My Newborn was Hospitalized with RSV: How a much-anticipated holiday gathering quickly turned into a harrowing experience RSV and Infants: a respiratory disease that can be deadly When to See Your Doctor You should call your doctor if you or your child is having trouble breathing, has poor appetite or decreased activity levels, cold symptoms that become severe, a shallow cough that continues throughout the day and night. Previous: Learn About Respiratory Syncytial Virus (RSV) Next: RSV Prevention and Treatment Reviewed and approved by the American Lung Association Scientific and Medical Editorial Review Panel. Page last updated: February 21, 2025 Show A Breath of Fresh Air in Your Inbox Join over 700,000 people who receive the latest news about lung health, including research, lung disease, air quality, quitting tobacco, inspiring stories and more! Cloudflare currently validating... Thank You! You will now receive email updates from the American Lung Association. Show Make a Donation Your tax-deductible donation funds lung disease and lung cancer research, new treatments, lung health education, and more. Make a Donation Become a Lung Health Insider Join over 700,000 people who receive the latest news about lung health, including research, lung disease, air quality, quitting tobacco, inspiring stories and more! Cloudflare currently validating... Thank you! You will now receive email updates from the American Lung Association. Freedom From Smoking Clinic - Chardon, OH Chardon, OH \| Sep 10, 2025 LUNG FORCE Walk - Cleveland, OH Cleveland, OH \| Sep 28, 2025 See All Events Skip to main content Select Your Location Select your location to view local American Lung Association events and news near you. Or Change Language Powered by Translate Lung HelpLine Talk to our lung health experts at the American Lung Association. Our service is free and we are here to help you. 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16658	https://www.sciencedirect.com/topics/engineering/air-fuel-ratio	Skip to Main content My account Sign in Air-Fuel Ratio In subject area:Engineering Air-fuel (A/F) ratio is defined as the ratio of the mass of air to the mass of fuel in a combustion process, which is crucial for achieving complete combustion and optimal engine performance. AI generated definition based on: Biofuels and Bioenergy, 2022 How useful is this definition? Add to Mendeley Discover other topics Chapters and Articles You might find these chapters and articles relevant to this topic. Combustion of Hydrocarbons 2011, Handbook of Industrial Hydrocarbon ProcessesJames G. Speight PhD, DSc 7.1Air–fuel ratio The air–fuel ratio (AFR, λ) is the mass ratio of air to the mass of the fuel present during combustion: When all the fuel is combined with all the free oxygen, typically within a vehicle’s combustion chamber, the reaction is chemically balanced and this air–fuel ratio is a stoichiometric relationship. The air–fuel ratio is an important measure for anti-pollution and performance tuning reasons. Most practical air–fuel ratio devices actually measure the amount of residual oxygen (for lean mixes) or unburned hydrocarbons (for rich mixtures) in the exhaust gas. Lambda (λ) is the ratio of actual air–fuel ratio to stoichiometry for a given mixture. Lambda of 1.0 is at stoichiometry, rich mixtures are less than 1.0, and lean mixtures are greater than 1.0. There is a direct relationship between lambda and the air–fuel ratio and to calculate the air–fuel ratio from a given value of lambda, multiply the measured lambda by the stoichiometric air–fuel ratio for the fuel. Alternatively, to recover lambda from an air–fuel ratio, divide the air–fuel ratio by the stoichiometric air–fuel ratio for that fuel: For gasoline, the stoichiometric air–fuel mixture is approximately 14.7 times the mass of air to fuel. Any mixture less than 14.7 to 1 is considered to be a rich mixture, whereas more than 14.7 to 1 is a lean mixture – assuming a perfect (ideal) gasoline consisting of solely n-heptane and iso-octane. In reality, gasoline is more complex than a simple two-component mixture and the stoichiometric ratio is altered. Lean mixtures produce hotter combustion gases than does a stoichiometric mixture, so much so that pistons can melt as a result. Rich mixtures produce cooler combustion gases than does a stoichiometric mixture, primarily due to the excessive amount of carbon which oxidizes to form carbon monoxide, rather than carbon dioxide. The chemical reactionoxidizing carbon to form carbon monoxide releases significantly less heat than the similar reaction to form carbon dioxide. Carbon monoxide retains significant potential chemical energy because it is a fuel whereas carbon dioxide is not, being the result of complete combustion of a hydrocarbon or carbonaceous fuel. Lean mixtures, when consumed in an internal combustion engine, produce less power than does the stoichiometric mixture. Similarly, rich mixtures return poorer fuel efficiency than the stoichiometric mixture. The mixture for the best fuel efficiency is slightly different from the stoichiometric mixture. View chapterExplore book Read full chapter URL: Book2011, Handbook of Industrial Hydrocarbon ProcessesJames G. Speight PhD, DSc Chapter Experimental investigation of performance of biodiesel with different blends in a diesel engine 2022, Biofuels and BioenergyP. Saranya, ... M. Vichitra 22.3.2.4Variation of air-fuel ratio Air-fuel (A/F) Ratio is the ratio of mass of air to mass of fuel (Noor El-Din et al., 2019). For combustion process, certain quantities of air and fuel have to be supplied in the combustion chamber. For complete combustion all the fuel should be burnt, thus in exhaust gas there is no unburned fuel. Variation of A/F ratio is represented in Fig. 22.8. A/F ratio is inversely proportional to BP. A/F ratio decreases with increase in brake power. This is due to the fact that fuel content at initial stage is the maximum, so heat generated will also be maximum. As the engine runs for long time, fuel content begins to decrease hence output power will also reduce. Decreased A/F ratio may cause the problems of deteriorated combustion efficiency, high smoke, and high exhaust gas temperature. Diesel, CD10, CD20, and CD30 tend to have same variation trends. After 50% load condition, A/F ratio is nearly same for all the fuel blends. View chapterExplore book Read full chapter URL: Book2022, Biofuels and BioenergyP. Saranya, ... M. Vichitra Review article Barriers of commercial power generation using biomass gasification gas: A review 2014, Renewable and Sustainable Energy ReviewsMohammad Asadullah 5.4Air fuel ratio and equivalence ratio (ER) The mass ratio of air to fuel in any combustion unit is defined as the air–fuel ratio (AFR). The minimum ratio of air to fuel that is exactly enough to burn the fuel completely is termed as stoichiometric ratio. Combustion of fuel requires a minimum stoichiometric ratio of air to fuel, while gasification requires an air–fuel ratio lower than stoichiometric ratio. The equivalence ratio can be defined as the ratio between the air–fuel ratio of the gasification process and the air–fuel ratio for complete combustion. The mathematical representations of air–fuel ratio and equivalence ratio are as follows: (1) (2) From Eq. (2), it seems that the higher ER creates more oxidation environment in the gasifier, and thus attributed to lower calorific product gas. On the other hand, lower ER results in higher calorific product gas; however, the tar yield is considerably higher. The higher concentration of burnable gas composition and lower tar concentration in the product gas is of prime importance for downstream application. Therefore, the process optimization is the focus of biomass gasification research. The thermodynamic analysis to evaluate the effect of ER on energy efficiency in different biomass gasification was carried out and it was found that the efficiency decreased with increasing the ER . In one study, it was found that the energy efficiency of the gasification system increased until the optimum ER (0.25), while it was decreased at higher ER . View article Read full article URL: Journal2014, Renewable and Sustainable Energy ReviewsMohammad Asadullah Review article Impact of cold flow properties of biodiesel on engine performance 2014, Renewable and Sustainable Energy ReviewsGaurav Dwivedi, M.P. Sharma 6.2.1Starting During starting temperatures are very low and when fuel comes in contact with colder walls it undergoes condensation. Even though air fuel ratio is within normal combustion range, the ratio of vaporised fuel to air is less and hence during starting, very rich mixture must be supplied. The air to fuel ratio is around 3 to 5. View article Read full article URL: Journal2014, Renewable and Sustainable Energy ReviewsGaurav Dwivedi, M.P. Sharma Review article Biomass gasification for electricity generation: Review of current technology barriers 2013, Renewable and Sustainable Energy ReviewsJ.A. Ruiz, ... M.A. Mendívil 5.4Air–fuel ratio and equivalent ratio (ER) The air–fuel ratio is the ratio between the air and fuel used, which is considerably lower than in combustion processes, which operate with excess stoichiometric air, whereas gasification involves default air values: The equivalent ratio (ER) is the ratio between the air–fuel ratio for the current process and the air–fuel ratio for complete combustion. The equivalent ratio would have a value of 1 for combustion, and would be expressed as follows: The air–fuel ratio is considered to have the greatest influence on the final calorific power of the syngas generated . Suitable values of the ER for gasification fall within the 0.2–0.4 range, thereby enabling the generation of tars and char to be controlled . By increasing the ER and keeping the biomass flow constant, the gasifier's temperature increases, as there is more oxygen per volume of biomass for conducting the partial combustion reactions, which are the ones that generate the necessary energy . Hosseini et al. used thermodynamic analysis to demonstrate the effect on energy efficiency of increasing the ER with different biomass moisture levels. They found that efficiency decreased with the same trend regardless of whether air or steam was used as the gasifying agent. View article Read full article URL: Journal2013, Renewable and Sustainable Energy ReviewsJ.A. Ruiz, ... M.A. Mendívil Chapter Fundamentals of dynamic and static diesel engine system designs 2013, Diesel Engine System DesignQianfan Xin 4.1.3 The characteristics of engine air–fuel ratio and exhaust gas recirculation rate Engine air–fuel ratio is defined as the mass flow rate ratio of fresh air to fuel. The shape of the air–fuel ratio contours is largely affected by the type of turbocharger and EGR rate. An excessively high air–fuel ratio may produce high pumping loss, high peak cylinder pressure, and high compressor outlet temperature. An excessively low air–fuel ratio may produce the problems of deteriorated combustion efficiency, high smoke, and high exhaust gas temperature. Air–fuel ratio is affected by the engine air flow rate at a given engine speed and load mode, and the air flow rate is determined by the intake manifold boost pressure and engine volumetric efficiency. Unlike the fixed-geometry turbine or pneumatically controlled wastegated turbine, the electronically controlled wastegated turbine and VGT in modern diesel engines are able to flexibly regulate or reduce air–fuel ratio to a large extent or maintain it even at a nearly constant level, especially at medium or high loads. EGR rate is defined as the mass flow rate ratio of EGR to the sum of fresh air and EGR. The shape of the EGR rate contours is determined primarily by emissions calibration. EGR rate directly affects coolant heat rejection, air–fuel ratio, and exhaust temperature. View chapterExplore book Read full chapter URL: Book2013, Diesel Engine System DesignQianfan Xin Review article A review on low temperature combustion engines: Performance, combustion and emission characteristics 2019, Renewable and Sustainable Energy ReviewsM. Krishnamoorthi, ... Sabariswaran Kandasamy 2.2.8Stoichiometric air/fuel ratio Stoichiometric air/fuel ratio (SAFR) or theoretical air/fuel ratio is the amount of air required for complete combustion of fuel which is calculated from the equation of stoichiometry of air/fuel reaction. The equivalence ratio (φ) is widely used to define the air/fuel quality in engines. Generally, fuel/air equivalence ratio is denoted by Greek symbol φ and air/fuel equivalence ratio is denoted by λ is a ratio between the actual to stoichiometry AFR in the given mixture. SAFR for fossil fuels lies around 15:1 which is less than 10:1 for alcohol-based fuels and it becomes higher for gaseous hydrogen fuel (30:1) . Hence, fossil and hydrogen fuels need more oxygen to achieve complete combustion compared to that of alcohol fuels. The engine can be operated with higher charge dilutions for alcohol/biofuels based fuels/blends because it contains oxygen element in its chemical structure which enhances the combustion process [18,80]. View article Read full article URL: Journal2019, Renewable and Sustainable Energy ReviewsM. Krishnamoorthi, ... Sabariswaran Kandasamy Chapter Effect of engine operating parameters in NO reduction 2022, NOx Emission Control Technologies in Stationary and Automotive Internal Combustion EnginesA. Tamilvanan, ... R. Sakthivel 5.5Effect of air-fuel/equivalence ratio on NOx emissions Generally, NOx emissions greatly depend on how much air and fuel take part during combustion inside the cylinder. The relation between the air and fuel proportion is termed the equivalence ratio (ϕ), which is the ratio of the actual fuel-air ratio to the stoichiometric fuel-air ratio. Another term is the air-fuel ratio (λ), which is the ratio of the mass of air to the mass of fuel. During combustion with ϕ < 1 or λ > 14 is termed a lean mixture; otherwise, with ϕ > 1 or λ < 14, the mixture is called a rich mixture when more amount of fuel is taking part in combustion when compared to stoichiometric combustion (ϕ = 1 or λ = 14.7). The variation of NOx emissions with respect to different air-fuel/equivalence ratios is shown in Fig. 5.5. Most of the IC engines work under leaner mixture or stochiometric conditions because of high fuel economy as well as lower HC and CO emissions. However, it resulted in higher NOx emissions, which is mainly due to the presence of relatively larger oxygen content when compared to fuel content at lean mixture conditions. With the engine operating at stoichiometric conditions, the flame propagation time is very low. Therefore, the heat generated inside the combustion chamber will not get time to disperse as an adequate quantity of oxygen exists for complete combustion. This results in higher combustion temperature inside the cylinder, which is the key reason for NOx formation. Hence, it remains desirable for the SI engine to run in a little rich zone (ϕ > 1) with EGR. Here, EGR reduces the fraction of oxygen to take part in the combustion that overwhelms the combustion temperature, which brings about lesser NOx emissions. On other hand, engines with a rich air-fuel mixture or ϕ > 1 resulted in lower NOx emissions along with greater emissions of HC and CO. At an equivalence ratio greater than one, more fuel is injected relative to the quantity of oxygen for stoichiometric combustion. At that condition, the fuel should cool the combustion chamber environment owing to the latent heat of fuel, which reduces the flame speed as well as combustion temperature that caused lesser NOx emission. At high speed and high loads, the temperature of the engine is so high, which is the foremost reason for the formation of NOx emissions because more fuel takes part in the combustion. In this situation, the formation of NOx is reduced by operating the engine at a slightly rich fuel mixture. Here, the fuel automatically cools the combustion environment produced by the previous cycle, which results in a reduction of maximum combustion temperature that results in lower NOx emissions. The counterpart effect resulted in greater emissions of HC and CO due to incomplete combustion as a result of a relatively higher equivalence ratio. View chapterExplore book Read full chapter URL: Book2022, NOx Emission Control Technologies in Stationary and Automotive Internal Combustion EnginesA. Tamilvanan, ... R. Sakthivel Review article A review on control system architecture of a SI engine management system 2016, Annual Reviews in ControlB. Ashok, ... C. Ramesh Kumar 3Air fuel ratio (AFR) control module Air fuel ratio (AFR) is one of the important control modules in the EMS since the ratio has been varied according to the torque demand requirement originated from the torque structure by considering the engine demands (catalyst heating, AC, etc.) and vehicle demand (cruise control, transmission control, etc.) as well. The three way catalytic converter (TWC) achieves its best efficiency only if the engine is operated within a narrow band around the stoichiometric air/fuel ratio. Due to the TWC's ability to store oxygen and carbon monoxide on its surface, short excursions of the air/fuel ratio can be tolerated as long as they do not exceed the remaining storage capacity and the mean deviation. So, the AFR module needs to maintain the air–fuel ratio in a stoichiometric condition in order to ensure the maximum conversion efficiency of the TWC (Hongming, 1999; Rolf & Norbert, 2003; Sardarmehnia, Keighobada, Menhaj, & Rahmani, 2013). The main problems faced by the researchers in this area are, concern with the variety of engine operating regimes, nonlinear dynamics, complexity of physical & chemical processes in the engine, uncertainties, noise, disturbances, number of un-measurable variables which directly affect AFR and demanded torque (Gerasimov, Javaherian, & Nikiforov, 2011). The major elements of the AFR control module shown in Fig. 5, are mass air flow estimator, fuel film compensator, A/F ratio observer and the controller which makes use of all the information provided by the elements to produce the appropriate injector pulse width (Tseng & Cheng, 1999). 3.1Mass air flow estimator To meter the correct amount of fuel, it is necessary to know the air mass inducted into the cylinder. The base fuel mass required to maintain stoichiometric combustion based on the air flow and manifold air pressure is calculated by the engine control unit (Franceschi et al., 2007). In general two techniques are followed to estimate the air flow into the cylinder of an SI engine. A conventional technique which uses a manifold absolute pressure (MAP) sensor and other widely used technique is Mass Air Flow (MAF) sensor based, which measures the air mass directly (Alexander & Kolmanovsky, 2002; Haiping & Qian, 2010). Both of these techniques have their own advantages and disadvantages. The MAP sensor technique uses speed density equation relating the manifold pressure and the intake air temperature with the known volumetric efficiency (lookup table) characteristics of the engine, in order to calculate the airflow into the cylinder and thus makes it possible to calculate fueling requirements. In this method, density of the air is measured by the temperature of the inlet air and manifold pressure (MAP). With the density of intake air as a known value, the AFR control module then calculates, how much air is expects to be moving at a specific engine speed and manifold pressure (Alexander & Kolmanovsky, 2002). (1) Where Wcyl is the mean value of the flow into the engine cylinders, Vd is the engine displacement and ηe is the engine speed; T is the intake manifold temperature, p is the mean value of the intake manifold pressure and ηv is the volumetric efficiency of the engine. This is done in the volumetric efficiency table or VE table and needs to be mapped in the control unit during calibration. Traditionally the VE table is in 2D format, it has two axes, where one is engine speed (RPM) and other is the manifold pressure. However variations in the volumetric efficiency due to some factors such as engine aging and wear, combustion chamber deposit buildup etc., can introduce errors in the air flow estimation. In the MAF sensor type, air mass flow is directly measured in the intake. Air flow measurement by means of a MAF sensor (which is generally a hot wire anemometer) accurately estimates the flow in the cylinder only in steady state, while in transient state the intake manifold filling/empting dynamics play a significant role (Ferdinando & Lavorgna, 2006). Hence in MAF type air flow measurement an input estimator can be used to correct the air flow into the cylinder during both in transient and in steady state operation. In general there are lots of approaches followed for the estimator algorithm and in the following section we will discuss one such approach developed by Stotsky and Kolmanovsky. Input estimators are an important class of observer algorithms aimed at estimating unmeasured inputs to dynamic systems from state and output measurements. Practical issues such as the need to deal with MAF sensor time constant and filter out periodic noise at the engine firing frequency dictate that this input observer be combined with additional filters into a larger estimation scheme (Alain et al., 2000;Grizzle et al., 1994). The resulting, overall estimation scheme shown in Fig. 6, consists of three interconnected observers and the approach uses the signals from the intake manifold pressure sensor (in some system called as Boost Pressure Sensor-BPS) and throttle mass flow. The first observer estimates the flow through the throttle based on the signal from the MAF sensor thereby compensating for the MAF sensor time constant. The second observer estimates the intake manifold pressure using the ideal gas law and the signal from the intake manifold pressure sensor. This second observer is introduced to filter out the noise and periodic oscillations at the engine firing frequency contained in the intake manifold pressure and throttle mass flow signals. This second observer is of a state estimation type as opposed to input estimation type. The third observer is at the core of the estimation scheme, and it is the one that provides an on-line correction to the cylinder flow estimation (Alexander & Kolmanovsky, 2002). 3.2Fuel film compensator In the port fuel injection system, some of the fuel which is injected at the intake port does not enter the cylinder immediately, in fact, the fuel will impinge on the port walls, on the valve stem, and on the backside of the intake valve forming a fuel-film, causing a difference between the injected mass of fuel and that which is inducted within the cylinder. A fraction of the injected fuel mass remains vaporized and is mixed with the air before it is sucked into the combustion chamber. When this ‘fuel lag’ is not compensated, there are significant spikes in the A/F ratio response. A compensation action is therefore necessary to balance this fuel film mass and it is accomplished by a model based approach. A model describing the fuel mass flow into the cylinders is necessary, since not all of the injected fuel mass is in gaseous form when the intake valves opens. One of the most popular models describing the behavior of the fuel-film is the Aquino model; which is a simple first-order model and macroscopically tracks the liquid puddle dynamics inside the engine intake manifold. It has been widely used to develop fuel-metering strategies, which compensate for the fuel transport lag. The fueling model estimates the fuel puddle mass balance as a function of the injected fuel mass rate as the input to the model and the liquid fuel flow into the cylinder as output of the model (Franchek Matthew et al., 2006; Roberto et al., 2005). This fuel film compensator which is the part of AFR control module modifies the quantity of injected fuel quantity in order to balance the amount of fuel stored in and released by the film (Alain et al., 2000). The fuel film compensator model may not be required for the direct injection engine because of the fuel is injected directly inside the combustion chamber. 3.3A/F ratio observer The information required from the observer of a control loop concerns the air–fuel ratio in the individual cylinders. Most of the AFR observer approaches are based on the development of a simplified model for exhaust transport delay, mixing phenomena, and sensor dynamics. The transport delay mainly consists of two parts: the cycle delay due to the four strokes of the engine and the exhaust gas transport delay caused by the exhaust gas flowing from the exhaust valve to the tailpipe exhaust gas oxygen (EGO) sensor. A predetermined model or a time-delay look-up table (2D map) is used by the controller while computing the required time delay compensation. In addition, the time delay is largely dependent on the engine operating condition defined by the engine speed and the air mass flow. A 2D map yields the specific time-delay for any given combination of engine speed and load. Throughout the engine operating envelope, the time delay can change significantly (Feng, Grigoriadis, Franchek, & Makki, 2006). In practice, the estimated time delay does not exactly match the actual total engine delay (Rajagopalan et al., 2014; Yildiray, 2009; Yildiray et al., 2008). For engine transients, signal compensation is needed because of the sensors finite time response, time delay and mixing behavior between the exhaust gases from the different cylinders (Hajime et al., 2002; Tseng & Cheng, 1999). An observer is then applied to the model, in order to perform real-time state estimation of air–fuel ratio, and many approaches have been followed, such as Linear Quadratic Gaussian, sliding mode control, Kalman filter, static steady-state observer, nonlinear observer, etc. (Bin et al., 2008). In order to construct an observer for the lambda control loop, it is first necessary to model the dynamics of the injection to exhaust dynamics (Per et al., 1998). In the AFR observer the information sought is the fuel air equivalence ratio of the individual cylinders. For this purpose the basic measurement is provided by the EGO sensor signal which is linear and the necessary compensation action will be carried out for the time delay, mixing phenomena and sensor characteristics by a proper model. The exhaust oxygen sensor provides the information for close loop A/F ratio feedback control. This information is then converted into an injection time correction that is identical for all the engine cylinders, on a cycle-by-cycle basis (Al-Himyari et al., 2014; Nicolo et al., 2010). 3.4A/F ratio controller Most of the current production AFR controllers are based on the gain-scheduling approach to design feedforward and feedback control system by constructing lookup tables. The AFR module consists of estimation of the air and fuel path dynamics combined with appropriate compensations. The controller calculates the injector pulse width (IPW) based on air flow estimation, either by a manifold absolute pressure (MAP) or mass air flow (MAF) sensors approach with respect to the driver demand and engine speed. This predicted air mass is used to estimate the mass of fuel that must enter into the cylinder to achieve the air/fuel ratio specified in advance as per the driver torque demand. To deliver this mass of fuel, a fueling path model based on injector characteristics, fuel puddling dynamics, fuel vaporization, and fuel entrainment dynamics is used to estimate the needed injector pulse width command (Franchek Matthew et al., 2006; Winge et al., 1999). If the air-path model and fuel-path model are accurate, the application of the feedforward fueling strategy will result in a stoichiometric air–fuel ratio during constant throttle operation (Stroh David et al., 2001). A conventional air-to-fuel ratio control module as shown in Fig. 7 includes two nested control loop, a feedforward and feedback control. The feedforward controller is tuned by means of experimental study during the calibration phase of the engine development and obtained fuel injection map (lookup table) values are stored in the memory of the controller. Generally, the procedure of constructing the fuel injection map in the feedforward loop means the tuning of the feedforward controller for various operating conditions in order to obtain a final injection map. The look- up tables is generated empirically for different engine operating conditions and mappings are done for a large number of speeds and loads during calibration phase (Seungbum et al., 2003; Dickinson, 2009; Rajagopalan, Stephen Yurkovich, Dudek, Guezennec, & Meyer, 2014; Tianyu, Haiqiao, & Zhao, 2011). In vehicle applications, the driver prescribes a torque demand to the control system through the accelerator pedal. This information is directly transformed into AFR control module as desired air-mass set point, using a manifold model or a 2D look-up table. Thus, feedforward loop generates a reference AFR value based on the torque demand from the torque control module and by means of the controller the feedback loop maintains the air fuel ratio as close as possible to the desired AFR using the feedback output measured by the EGO sensor signal. The correction factor for the feedforward controller is accomplished by the feedback controller using the signal from EGO sensor. But in actual operation of engine, the transient condition tends to give an error in the feedforward controller due to modeling deficiencies, arising from say environmental factors, variation in fuel composition, manufacturing tolerances or mechanical wear. Thus the air/fuel ratio feedback control system compensates the unavoidable errors in the feedforward loop. Because the base fuel calculation has a tendency to drift off from the intended stoichiometric operating point and an active closed loop lambda control is necessary to adjust the base fuel calculation (Ohyama, 2001; Per et al., 1998). Feedforward control is fast which may not be accurate and also handles the transients, but the feedback control- loop is slow, due to the feed gas oxygen delay, but it ensures the required higher steady-state accuracy. For the feedforward control loop the time-varying delay is the key parameter in the AFR control that imposes a limitation on the bandwidth of the AFR feedback loop by decreasing the phase margin (Ivan et al., 2006). In order to stabilize the unstable internal dynamics of the system and reduce the effect of unmatched disturbances on the steady-state tracking error, various controllers are used. Lot of approaches dealing with AFR prediction and control methods has been proposed which are listed in Table 2. Table 2. Different controllers used in the air fuel ratio (AFR) module. \| S. no. \| Controller used \| Authors \| Year \| Outcomes \| --- --- \| Empty Cell \| for AFR module \| Empty Cell \| Empty Cell \| Empty Cell \| \| 1 \| PID \| (Franceschi et al., 2007) \| 2007 \| Discrete, time-based, delay-compensated, adaptive PID control algorithm for AFR control is employed. \| \| Empty Cell \| (Kwiatkowski et al., 2009) \| 2009 \| Using a hybrid evolutionary-algebraic synthesis approach that combines LMI techniques based on K-S iteration with evolutionary search, a scheduled PID controller is designed. \| \| Empty Cell \| (Behrouz et al., 2012) \| 2012 \| A new synthesis method for AFR control by the time-varying delay in the system dynamics is first approximated by Pade approximation with time-varying parameters. The associated error is then utilized to construct a filtered PID controller combined with a parameter-varying dynamic compensator. \| \| Empty Cell \| (Guo et al., 2013) \| 2013 \| PID controller is aimed to track the given value of fuel injection quantity. Tuning controller parameters are chosen by randomized algorithm according to the criteria of performance. \| \| Empty Cell \| (Mayr Christian et al., 2014) \| 2014 \| New approach for automated calibration of nonlinear PID controllers and feedforward maps is introduced. A dynamic local model network is used for actual physical process. \| \| 2 \| Adaptive control \| (Tseng & Cheng, 1999) \| 1999 \| Adaptive AFR control scheme based on a one- parameter port fuel dynamics model with the parameter being identified online. \| \| Empty Cell \| (Stroh David et al., 2001) \| 2001 \| The steady-state adaptive fueling controller presented in this paper incorporates a modular model structure which eliminates static maps and enables a plug and play feature for changes to the sensor set. \| \| Empty Cell \| (Yildiray et al., 2008) \| 2008 \| Two controllers for AFR, an Adaptive Feed-Forward Controller (AFFC) and an Adaptive Posicast Controller (APC), have been developed and implemented in a vehicle. \| \| Empty Cell \| (Rui et al., 2009) \| 2009 \| Nonlinear control approaches for multi-input multi-output (MIMO) engine models is developed, by developing adaptive control and learning control methods. \| \| Empty Cell \| (Yildiray et al., 2010) \| 2010 \| Two adaptive controller designs are considered. The designed AFR controller must reject disturbances due to canister vapor purge and inaccuracies in air charge estimation and wall-wetting compensation. \| \| Empty Cell \| (Kahveci Nazli & Jankovic, 2010) \| 2010 \| An adaptive control structure which consists of an adaptive PI controller and an adaptive Smith predictor for time-delay systems with unknown plant parameters \| \| 3 \| Neural network \| (Seungbum et al., 2003) \| 2003 \| The control is based on the feedback error learning. The controller consists of Neural Network (NN) with linear feedback. The NN are radial basis function network that are trained by using the feedback error. \| \| Empty Cell \| (Wang et al., 2005) \| 2005 \| Model predictive control (MPC) based on neural network model for air–fuel ratio, in which the model is adapted on-line to cope with nonlinear dynamics and parameter uncertainties. \| \| Empty Cell \| (Ivan et al., 2006) \| 2006 \| Recurrent Neural Networks for modeling and controlling AFR. The developed forward model used to generate a reference AFR signal to train another RNN model aimed at simulating the inverse AFR dynamics by evaluating the fuel injection time as function of AFR, manifold pressure and engine speed. \| \| Empty Cell \| (Zhai & Yu, 2009) \| 2009 \| MPC strategy is applied to air/fuel ratio control using neural network. The neural network uses information from multivariable and considers dynamics to do multi-step ahead prediction. \| \| Empty Cell \| (Zhai et al., 2011) \| 2011 \| MPC based on adaptive NN model attempted for AFR control. A Radial basis function (RBF) network employed and recursive least squares (RLS) algorithm used for weight updating. \| \| Empty Cell \| (Sardarmehni et al., 2013) \| 2013 \| MPC system is designed for robust control of lambda. Based on the simulation data, two neural networks models of the engine are generated. The identified Multi- Layer Perceptron (MLP) NN model yields small verification error compared with that of the adaptive Radial Base Function (RBF). \| \| 4 \| Fuzzy logic \| (Anurak & Sooraksa, 2012) \| 2012 \| Improvement of mean value model (MVEM) and effective nonlinear control for the AFR regulator. The regulator is designed by a discrete fuzzy PI algorithm, which provides easy tuning, robustness. \| \| Empty Cell \| (Farzin, Mansoorzadeh, Zare, Shahryarzadeh, & Akbari, 2013) \| 2013 \| Approach for fuel control combines the design technique from variable structure controller is based on Lyapunov& fuzzy estimator to estimate the nonlinearity of undefined dynamic in backstepping controller \| \| 5 \| Model based methods for non-linear control \| (Roberto et al., 2005) \| 2005 \| Model-based AFR control technique is proposed: this is based on a dynamical model of the air dynamics inside inlet manifolds and on the online identification of the fuel-film parameters. \| \| Empty Cell \| (Bin et al., 2008) \| 2008 \| From multirate sampling method, control-oriented model, combining fuel delivery and exhaust gas dynamics, is established. From estimated AFR, a decoupled PI compensator is designed. \| \| Empty Cell \| (Shuntaro et al., 2009) \| 2009 \| The control consists of a feedforward control using a fuel behavior model, a feedback control using an UEGO sensor and a feedback control HEGO sensor. \| \| 6 \| Sliding mode control (SMC) \| (Sei-Bum et al., 1994) \| 1994 \| An observer-based control algorithm based on sliding mode control technique is suggested for fast response and small amplitude chattering of the air to- fuel ratio. \| \| Empty Cell \| (Hajime et al., 2002) \| 2002 \| SMC-based AFR feedback section with the oxygen storage mass predicting and controlling section with a self tuning strategy. \| \| 7 \| Other controls in the AFR module \| (Holzmann et al., 1997) \| 1997 \| Neuro-fuzzy approach is discussed. For replacing 3D maps a modeling of engine characteristics for vehicle control and simulation by multi-layer Perceptron and radial-basis function networks is developed. \| \| Empty Cell \| (Winge et al., 1999) \| 1999 \| A new lambda (normalized AFR) control methodology (H∞-control) which has a somewhat larger bandwidth and guarantee robustness with respect to selected engine variable and parameter variations. \| \| Empty Cell \| (Alain et al., 2000) \| 2000 \| AFR control method with feedforward, feedback methods with the injector puddling models. \| \| Empty Cell \| (Ferdinando & Lavorgna, 2006) \| 2006 \| Soft computing mass air flow estimator which is able to estimate, by using the combustion pressure signal, the incoming mass air flow both in steady states and in transient conditions \| \| Empty Cell \| (Franchek Matthew et al., 2006) \| 2006 \| The feedforward fueling control is separated into steady state and transient phenomena. Nonlinear behavior associated with fueling is captured with nonlinear steady state models. \| \| Empty Cell \| (Feng et al., 2006) \| 2006 \| An approach to combine an input shaping method together with a linear parameter varying (LPV) feedback controller is proposed to solve the transient air–fuel ratio tracking problem. \| \| Empty Cell \| (Dickinson, 2009) \| 2009 \| Systematic calibration of fuelling and speed controller. Non-linear black-box parameter directly produces a dynamic inverse multivariable NARMA feedforward controller and linearizing feedback compensator. \| \| Empty Cell \| (Nicolo et al., 2010) \| 2010 \| A real time application of an original closed-loop individual cylinder AFR control system, based on a spectral analysis of the lambda sensor signal is proposed. \| \| Empty Cell \| (Gerasimov et al., 2011) \| 2011 \| Torque tracking and air-to-fuel ratio (AFR) stabilization at the stoichiometric level are addressed. A data driven approach based on the design of direct and inverse models is proposed \| \| Empty Cell \| (Rajagopalan et al., 2014) \| 2014 \| Control architecture for AFR control of designed to work with switching and/or wide range EGO sensors, for minimizing calibration effort while meeting performance requirements. \| \| Empty Cell \| (Efimov, Nikiforov, & Javaherian, 2014) \| 2014 \| The first control law is based on an a priori off-line identified engine model, while the second control law is adaptive; it provides on-line adaptive adjustment to closed loop system. The supervisor realizes a switching rule between these control laws providing better performance of regulation. \| However, most of the controllers in production vehicle fuel-injection systems consist of an open-loop feed-forward control which employs a look-up table and closed loop feedback control with PID, which works on the basis of a gain-scheduling approach. Due to their reliability and ease in implementation, PID controllers are far most dominant control design approach for the AFR control applications in the vehicles. This is due to its simple structure and robust performance over a wide range of operating conditions. In addition the major advantage of the PID controller over the other approaches is the familiarity of the algorithm and the relatively few tuning parameters and also there is no need of system model. These results in a significant reduction on the tuning and calibration effort required to implement the controller. View article Read full article URL: Journal2016, Annual Reviews in ControlB. Ashok, ... C. Ramesh Kumar Chapter Combustion-Related Emissions in SI Engines 1998, Handbook of Air Pollution From Internal Combustion EnginesSimone Hochgreb Relative Air-Fuel Ratio (λ) Hydrocarbon emission levels decrease with increasing air-fuel ratio (Figure 6.23), and are lower for lower-volatility, nonaromatic fuels. Although temperatures peak around stoichiometric or slightly rich values of air-fuel ratio, the excess oxygen contributes to a lower initial HC level in the charge, as well as additional postflame oxidation. For air-fuel ratios λ higher than about 1.3, however, misfires can become more frequent, leading to increased HC emissions. Lean operation is obviously desirable from the point of view of HC emissions; as mentioned in the previous sections, however, difficulties arise in achieving high NO removal efficiencies in the catalytic converter. View chapterExplore book Read full chapter URL: Book1998, Handbook of Air Pollution From Internal Combustion EnginesSimone Hochgreb Related terms: Energy Engineering Natural Gas Compressed Natural Gas Biodiesel Exhaust Gas Recirculation Compression Ratio Diesel Engine Equivalence Ratio Combustion Chamber Exhaust Gas View all Topics
16659	https://www.wyzant.com/resources/answers/926145/the-standard-cell-potential-e-cell-is-0-77-v-for-the-redox-reaction-2-fe-aq	The standard cell potential E°(cell) is +0.77 V for the redox reaction 2 Fe³⁺(aq) + H₂(g) → 2 H⁺(aq) + 2 Fe²⁺(aq). Determine E° for the half-reaction Fe³⁺(aq) + e⁻ → Fe²⁺(aq). \| Wyzant Ask An Expert Log inSign up Find A Tutor Search For Tutors Request A Tutor Online Tutoring How It Works For Students FAQ What Customers Say Resources Ask An Expert Search Questions Ask a Question Wyzant Blog Start Tutoring Apply Now About Tutors Jobs Find Tutoring Jobs How It Works For Tutors FAQ About Us About Us Careers Contact Us All Questions Search for a Question Find an Online Tutor Now Ask a Question for Free Login WYZANT TUTORING Log in Sign up Find A Tutor Search For Tutors Request A Tutor Online Tutoring How It Works For Students FAQ What Customers Say Resources Ask An Expert Search Questions Ask a Question Wyzant Blog Start Tutoring Apply Now About Tutors Jobs Find Tutoring Jobs How It Works For Tutors FAQ About Us About Us Careers Contact Us Subject ZIP Search SearchFind an Online Tutor NowAsk Ask a Question For Free Login Chemistry Halle C. asked • 04/14/23 The standard cell potential E°(cell) is +0.77 V for the redox reaction 2 Fe³⁺(aq) + H₂(g) → 2 H⁺(aq) + 2 Fe²⁺(aq). Determine E° for the half-reaction Fe³⁺(aq) + e⁻ → Fe²⁺(aq). Follow •1 Add comment More Report 1 Expert Answer Best Newest Oldest By: Grant L.answered • 04/14/23 Tutor 5(13) Chemistry Education Major \| Purdue University About this tutor› About this tutor› Since Fe3+ is being reduced, we know that it is the cathode. Hydrogen has a standard potential of zero, so we don't need to take Eoxidized into account with the formula Ecell = Ereduced - Eoxidized Therefore, Ecell = Ereduced - 0. Thus, .77 V = Ereduced. E for the half reaction of Fe3+ + e- -> Fe2+ is .77 V. Upvote • 0Downvote Add comment More Report Still looking for help? Get the right answer, fast. Ask a question for free Get a free answer to a quick problem. Most questions answered within 4 hours. OR Find an Online Tutor Now Choose an expert and meet online. 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16660	https://stackoverflow.com/questions/30441433/predict-y-value-for-a-given-x-in-r	linear regression - Predict y value for a given x in R - Stack Overflow Join Stack Overflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google Sign up with GitHub OR Email Password Sign up Already have an account? Log in Skip to main content Stack Overflow 1. About 2. Products 3. 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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Predict y value for a given x in R Ask Question Asked 10 years, 4 months ago Modified10 years, 4 months ago Viewed 25k times Part of R Language Collective This question shows research effort; it is useful and clear 6 Save this question. Show activity on this post. I have a linear model: r mod=lm(weight~age, data=f2) I would like to input an age value and have returned the corresponding weight from this model. This is probably simple, but I have not found a simple way to do this. R Language Collective r linear-regression predict Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications asked May 25, 2015 at 15:18 EricaErica 125 1 1 gold badge 2 2 silver badges 9 9 bronze badges 1 4 There is a predict method for lm, and you have to supply the new data as newdata argument. Look here: stat.ethz.ch/R-manual/R-patched/library/stats/html/…DatamineR –DatamineR 2015-05-25 15:20:54 +00:00 Commented May 25, 2015 at 15:20 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 10 Save this answer. Show activity on this post. Its usually more robust to use the predict method of lm: r f2<-data.frame(age=c(10,20,30),weight=c(100,200,300)) f3<-data.frame(age=c(15,25)) mod<-lm(weight~age,data=f2) pred3<-predict(mod,f3) This spares you from wrangling with all of the coefs when the models can be potentially large. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered May 25, 2015 at 18:04 chepylechepyle 996 1 1 gold badge 9 9 silver badges 16 16 bronze badges Comments Add a comment This answer is useful 6 Save this answer. Show activity on this post. If your purposes are related to just one prediction you can just grab your coefficient with r coef(mod) Or you can just build a simple equation like this. r coef(mod) + "Your_Value"coef(mod) Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered May 25, 2015 at 15:26 SabDeMSabDeM 7,200 3 3 gold badges 28 28 silver badges 38 38 bronze badges 1 Comment Add a comment gosuto gosutoOver a year ago Is this truly the most concise way to do this? Can predict() not do this, without creating a whole new data.frame? 2019-09-30T14:08:29.503Z+00:00 0 Reply Copy link Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions r linear-regression predict See similar questions with these tags. 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16661	https://www.quora.com/A-rectangular-prism-has-a-total-surface-area-of-48-Also-the-sum-of-all-the-edges-of-the-prism-is-40-What-is-the-length-of-the-diagonal-joining-one-corner-of-the-prism-to-the-opposite-corner	A rectangular prism has a total surface area of 48. Also, the sum of all the edges of the prism is 40. What is the length of the diagonal joining one corner of the prism to the opposite corner? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Geometry Mathematics Word Problems Rectangular Prism Total Surface Area Solid Space Diagonals Solid 3D Geometry Geometric Mathematics Prisms 5 A rectangular prism has a total surface area of 48. Also, the sum of all the edges of the prism is 40. What is the length of the diagonal joining one corner of the prism to the opposite corner? All related (33) Sort Recommended Assistant Bot · Sep 6 Let the edge lengths be a, b, c > 0. Equations: 1) Total surface area: 2(ab + bc + ca) = 48 ⇒ ab + bc + ca = 24. 2) Sum of all edges: 4(a + b + c) = 40 ⇒ a + b + c = 10. We need the space diagonal d = √(a^2 + b^2 + c^2). Compute a^2 + b^2 + c^2 from (a + b + c)^2: (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca). So a^2 + b^2 + c^2 = (a + b + c)^2 − 2(ab + bc + ca) = 10^2 − 2·24 = 100 − 48 = 52. Hence diagonal d = √52 = 2√13. Upvote · Related questions More answers below How do you solve for the diagonal length of a rectangular prism when only the length and width are given? How many corners does a rectangular prism have? How do I find the length of a rectangular prism? How many edges does a rectangular prism have? How does a rectangular prism look like? Jan Koning Studied Mathematics&Computer Science at Delft University of Technology (Graduated 1987) · Author has 798 answers and 646.5K answer views ·5y a rectangular prisma has sides l l, b b, and h h. So we know that 2(l h+l b+b h)=48 2(l h+l b+b h)=48 (1) And we know that 4∗(l+b+h)=40 4∗(l+b+h)=40, so l+b+h=10 l+b+h=10. (2) Goal: find √l 2+b 2+h 2 l 2+b 2+h 2, which is the length of the diagonal. So we have two equations and three variables, which has an infinite number of solutions. But we can find a solution without knowing each of the variables separately. Take (2) and square both sides: (l+b+h)2=100.(l+b+h)2=100. So l 2+h 2+b 2+2 l b+2 l h+2 b h=100 l 2+h 2+b 2+2 l b+2 l h+2 b h=100 (3) Combining (1) and (3) yields l 2+h 2+b 2+48=100,l 2+h 2+b 2+48=100, thus l 2+h 2+b 2=52.l 2+h 2+b 2=52. And so √l 2+b 2+h 2=√52.l 2+b 2+h 2=52. Upvote · 9 2 Ramaswamy Arunachalam Former PROFESSOR OF MECHANICAL ENGINEERING at Government College of Technology, Tamil Nadu, India (1983–1997) · Author has 13.9K answers and 2.7M answer views ·5y 2(LB+LH+BH) = 48 (L+B+H) = 40/4 = 10 (L+B+H)^2 = L^2 + B^2 +H^2 + 2LB + 2BH + 2 LH = 10^2 L^2+B^2+H^2 = 100–48 = 52. THE REQUIRED DIAGONAL =(L^2 + B^2 + H^2)^0.5 = 52^0.5 = 7.21 Upvote · Pradeep Hebbar Many years of Structural Engineering & Math enthusiasm · Author has 9.3K answers and 6.2M answer views ·4y Related If the sum of the edge of a prism is 120 cm, what is the surface area of the prism? There are many types of prisms. Which one are you talking about? Calculation for a square prism is shown below. Total no. of edges = 12 ; Given sum of edge lengths = 120 Let a be the side of the square. a= 120/12 = 10 Surface area = 6a² = 6(10)² = 600 cm² Continue Reading There are many types of prisms. Which one are you talking about? Calculation for a square prism is shown below. Total no. of edges = 12 ; Given sum of edge lengths = 120 Let a be the side of the square. a= 120/12 = 10 Surface area = 6a² = 6(10)² = 600 cm² Upvote · Sunil Kumar Rai Lives in Kanchanpur Gate Varanasi Blw near Anguthi Restaurant (2010–present) · Author has 325 answers and 58.1K answer views ·2y Related How do I solve a rectangular prism? A rectangular prism is a three-dimensional shape with two identical rectangular faces and one identical square face: Other examples of prisms include triangular prisms, square prisms, and polygon prisms: Let's learn how to find the surface area and volume of a rectangular prism. How to Find the Surface Area of a Rectangular Prism Let's review the surface area of a rectangle. The length, width, and height of a rectangular prism are made by six separate shapes that, when put together, creates a three-dimensional figure. The surface area of a rectangle in prism form is the sum of the areas of all six s Continue Reading A rectangular prism is a three-dimensional shape with two identical rectangular faces and one identical square face: Other examples of prisms include triangular prisms, square prisms, and polygon prisms: Let's learn how to find the surface area and volume of a rectangular prism. How to Find the Surface Area of a Rectangular Prism Let's review the surface area of a rectangle. The length, width, and height of a rectangular prism are made by six separate shapes that, when put together, creates a three-dimensional figure. The surface area of a rectangle in prism form is the sum of the areas of all six shapes. Here is the formula for the surface area of a prism: Let's use this formula to find the surface area of the rectangular prism below: Length: 12 feet Width: 4 feet Height: 4 feet If you need a refresher about the order of these calculations, remember the order of operations through the acronym PEMDAS (here’s how this works). For surface area, always keep your answer in square units. How to Find Rectangular Prism Volume Now that we know how to find the surface area, let's move on to finding the volume of the rectangular prism. Volume is the amount of space that a three-dimensional figure takes up. For finding volume, use these formulas: The area of the base is the length times the width. So to get the final volume, you must multiply the height of the rectangular prism by the area of the base. Let's use this formula to determine the volume of this prism, which must be expressed in cubic centimeters (cm³): Length: 17 centimeters Width: 7 centimeters Height: 7 centimeters With volume, your answer should be in terms of units squared. Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Upvote · Related questions More answers below The main diagonal of a rectangular prism is 31 units in length and each dimension of the rectangular prism is an integer. What's the maximum and minimum possible volume of the rectangular prism? Does a rectangular prism have 6 rectangular faces? What is a cube and rectangular prism? A rectangular prism has a square base with edges measuring 8 inches each. Its volume is 768 cubic inches. What is the height of the prism and the surface area of the Prism? How many vertices does a rectangular prism have? Sergei Michailov Retired System Analyst (2009–present) · Author has 5.5K answers and 3.5M answer views ·5y Related What is the diagonal of a rectangular prism if the total surface area is 13cm^2 and the sum of all the edges is 28cm? Nesta, please ignore my solution. Ramaswary Ar has a much better one. And ignore my comment about three different sides of the cuboid. It is wrong. Ramaswary’s solution takes care of this too. Sorry. If the side of the base is x and the height is y we have 2x^2+4xy=13 8x+4y=28 We need to find sqrt(x^2+x^2+y^2)=sqrt(2x^2+y^2) From the second equation 2x+y=7 y=7–2x Then the first equation becomes 2x^2+4x(7–2x)=13 2x^2+28x-8x^2=13 6x^2-28x+13=0 x1=(28+sqrt(472))/12 y1= find it, then find sqrt(2x^2+y^2) x2=(28+sqrt(472))/12 find y^2 and sqrt(2x^2+y^2) Upvote · 9 1 9 2 Ramaswamy Arunachalam Former PROFESSOR OF MECHANICAL ENGINEERING at Government College of Technology, Tamil Nadu, India (1983–1997) · Author has 13.9K answers and 2.7M answer views ·5y Related What is the diagonal of a rectangular prism if the total surface area is 13cm^2 and the sum of all the edges is 28cm? TOTAL SURFACE AREA = 2[ AB + BC + CA ] = 13 i.e AB+BC+CA = 6.5 SUM OF ALL THE EDGES = 4[A + B + C] = 28; i.e A+B+C = 7 DIAGONAL^2 = [ A^2 + B^2 + C^2 ]^0.5 [A+B+C]^2 = A^2+B^2+C^2 +2AB + 2BC+2CA = 7^2 = 49 A^2+B^2+C^2 = 49- 2[AB+BC+CA] = 49–26.5 = 36 DIAGONAL. = 36^0.5 = 6 cm Upvote · 9 2 9 2 Katherine Sanders Data Scientist · Author has 102 answers and 96K answer views ·1y Related What is the formula for the surface area of a rectangular prism? Part 1 of 2:Finding the Surface Area Label the length, width, and height of your rectangular prism. Each rectangular prism has a length, a width, and a height. Draw a picture of the prism, and write the symbols , , and next to three different edges of the shape. If you're not sure which sides to label, pick any corner. Label the three lines that meet at that corner. For example: A box has a base that measures 3 inches by 4 inches, and it stands 5 inches tall. The long side of the base is 4 inches, so = 4, = 3, and = 5. Look at the six faces of the prism. To cover the whole surface area, you'd Continue Reading Part 1 of 2:Finding the Surface Area Label the length, width, and height of your rectangular prism. Each rectangular prism has a length, a width, and a height. Draw a picture of the prism, and write the symbols , , and next to three different edges of the shape. If you're not sure which sides to label, pick any corner. Label the three lines that meet at that corner. For example: A box has a base that measures 3 inches by 4 inches, and it stands 5 inches tall. The long side of the base is 4 inches, so = 4, = 3, and = 5. Look at the six faces of the prism. To cover the whole surface area, you'd need to paint six different "faces." Think about each one — or find a box of cereal and look at them directly: There are a top and bottom face. Both are the same size. There are a front and a back face. Both are the same size. There are a left and right face. Both are the same size. If you have trouble picturing this, cut a box apart along the edges and lay it out. Find the area of the bottom face. To start out, let's find the surface area of just one face: the bottom. This is a rectangle, just like every face. One edge of the rectangle is labeled length and the other is labeled width. To find the area of the rectangle, just multiply the two edges together. Area (bottom edge) = length times width = . Going back to our example, the area of the bottom face is 4 inches x 3 inches = 12 square inches. Find the area of the top face. Wait a second — we already noticed that the top and bottom faces are the same size. This must also have an area of . In our example, the top area is also 12 square inches. Find the area of the front and back faces. Go back to your diagram and look at the front face: the one with one edge labeled width and one labeled height. The area of the front face = width times height = . The area of the back is also . In our example, w = 3 inches and h = 5 inches, so the area of the front is 3 inches x 5 inches =15 square inches. The area of the back face is also 15 square inches. Find the area of the left and right faces. We've just got two faces left, each the same size. One edge is the length of the prism, and one edge is the height of the prism. The area of the left face is and the area of the right face is also . In our example, l = 4 inches and h = 5 inches, so the area of the left face = 4 inches x 5 inches = 20 square inches. The area of the right face is also 20 square inches. Add the six areas together. Now you've found the area of each of the six faces. Add them all together to get the area of the whole shape: . You can use this formula for any rectangular prism, and you will always get the surface area. To finish our example, just add up all the blue numbers above: 12 + 12 + 15 + 15 + 20 + 20 = 94 square inches. Part 2 of 2:Making the Formula Shorter Simplify the formula. You now know enough to find the surface area of any rectangular prism. You can do it faster if you've learned some basic algebra. Start with our equation above: Area of a rectangular prism = lw + lw + wh + wh + lh + lh. If we combine all the terms that are the same, we get: Area of a rectangular prism = Factor out the two. If you know how to factor in algebra, you can make it even shorter: Area of a Rectangular Prism = 2lw + 2wh + 2lh = . Test it on an example. Let's go back to our example box, with length 4, width 3, and height 5. Plug these numbers into the formula: Area = 2(lw + wh + lh) = 2 x (lw + wh + lh) = 2 x (4x3 + 3x5 + 4x5) = 2 x (12 + 15 + 20) = 2 x (47) = 94 square inches. That's the same answer we got before. Once you've practiced doing these equations, this is a much faster way to find the surface area. Upvote · Stuart Herring Author has 11.7K answers and 8.2M answer views ·3y Related A rectangular prism has a base whose dimension is 10 cm by 12 cm. Its height is 20 cm. What is its diagonal length, lateral surface area, and volume? Q. A rectangular prism has a base whose dimension is 10 cm by 12 cm. Its height is 20 cm. What is its diagonal length, lateral surface area, and volume? The most difficult part of this homework problem is calculating the space diagonal (or: body diagonal) of the prism. Start by finding the diagonal of the base, using the Pythagorean relation: D=√L 2+W 2 D=L 2+W 2, where L and W are the length and width of the base. This diagonal is one leg of a right triangle whose other leg is the height of the prism, and whose hypotenuse is the desired space diagonal. Apply Pythagoras again. Continue Reading Q. A rectangular prism has a base whose dimension is 10 cm by 12 cm. Its height is 20 cm. What is its diagonal length, lateral surface area, and volume? The most difficult part of this homework problem is calculating the space diagonal (or: body diagonal) of the prism. Start by finding the diagonal of the base, using the Pythagorean relation: D=√L 2+W 2 D=L 2+W 2, where L and W are the length and width of the base. This diagonal is one leg of a right triangle whose other leg is the height of the prism, and whose hypotenuse is the desired space diagonal. Apply Pythagoras again. Computing the lateral surface area and the volume is simple. The formulas have already been discussed in your math book; check it out. . . . Please do not post homework questions here without doing some work, and showing what you have already done. We can give guidance if you have difficulty, but we are not here to simply do all of your work for you. Upvote · Mikayla Turner Former Android Developer ·2y Related What is the formula for the surface area of a rectangular prism? Part 1 of 2:Finding the Surface Area Label the length, width, and height of your rectangular prism. Each rectangular prism has a length, a width, and a height. Draw a picture of the prism, and write the symbols , , and next to three different edges of the shape. If you're not sure which sides to label, pick any corner. Label the three lines that meet at that corner. For example: A box has a base that measures 3 inches by 4 inches, and it stands 5 inches tall. The long side of the base is 4 inches, so = 4, = 3, and = 5. Look at the six faces of the prism. To cover the whole surface area, you'd Continue Reading Part 1 of 2:Finding the Surface Area Label the length, width, and height of your rectangular prism. Each rectangular prism has a length, a width, and a height. Draw a picture of the prism, and write the symbols , , and next to three different edges of the shape. If you're not sure which sides to label, pick any corner. Label the three lines that meet at that corner. For example: A box has a base that measures 3 inches by 4 inches, and it stands 5 inches tall. The long side of the base is 4 inches, so = 4, = 3, and = 5. Look at the six faces of the prism. To cover the whole surface area, you'd need to paint six different "faces." Think about each one — or find a box of cereal and look at them directly: There are a top and bottom face. Both are the same size. There are a front and a back face. Both are the same size. There are a left and right face. Both are the same size. If you have trouble picturing this, cut a box apart along the edges and lay it out. Find the area of the bottom face. To start out, let's find the surface area of just one face: the bottom. This is a rectangle, just like every face. One edge of the rectangle is labeled length and the other is labeled width. To find the area of the rectangle, just multiply the two edges together. Area (bottom edge) = length times width = . Going back to our example, the area of the bottom face is 4 inches x 3 inches = 12 square inches. Find the area of the top face. Wait a second — we already noticed that the top and bottom faces are the same size. This must also have an area of . In our example, the top area is also 12 square inches. Find the area of the front and back faces. Go back to your diagram and look at the front face: the one with one edge labeled width and one labeled height. The area of the front face = width times height = . The area of the back is also . In our example, w = 3 inches and h = 5 inches, so the area of the front is 3 inches x 5 inches =15 square inches. The area of the back face is also 15 square inches. Find the area of the left and right faces. We've just got two faces left, each the same size. One edge is the length of the prism, and one edge is the height of the prism. The area of the left face is and the area of the right face is also . In our example, l = 4 inches and h = 5 inches, so the area of the left face = 4 inches x 5 inches = 20 square inches. The area of the right face is also 20 square inches. Add the six areas together. Now you've found the area of each of the six faces. Add them all together to get the area of the whole shape: . You can use this formula for any rectangular prism, and you will always get the surface area. To finish our example, just add up all the blue numbers above: 12 + 12 + 15 + 15 + 20 + 20 = 94 square inches. Part 2 of 2:Making the Formula Shorter Simplify the formula. You now know enough to find the surface area of any rectangular prism. You can do it faster if you've learned some basic algebra. Start with our equation above: Area of a rectangular prism = lw + lw + wh + wh + lh + lh. If we combine all the terms that are the same, we get: Area of a rectangular prism = Factor out the two. If you know how to factor in algebra, you can make it even shorter: Area of a Rectangular Prism = 2lw + 2wh + 2lh = . Test it on an example. Let's go back to our example box, with length 4, width 3, and height 5. Plug these numbers into the formula: Area = 2(lw + wh + lh) = 2 x (lw + wh + lh) = 2 x (4x3 + 3x5 + 4x5) = 2 x (12 + 15 + 20) = 2 x (47) = 94 square inches. That's the same answer we got before. Once you've practiced doing these equations, this is a much faster way to find the surface area. Upvote · Vasudevan A.N.S. Former Retired Professor of Chemistry. (1977–2014) · Author has 8K answers and 13.3M answer views ·5y Related The length of the diagonal of a cube is 8√2cm. What is the surface area? Let The length of the diagonal of acube=√3a So √3a= 8√2 So a= 8 x √(2/3) Surface area of the cube= 6a² = 6x(8√(2/3))²=( 6 x 64 x 2/3) =256-cm² Continue Reading Let The length of the diagonal of acube=√3a So √3a= 8√2 So a= 8 x √(2/3) Surface area of the cube= 6a² = 6x(8√(2/3))²=( 6 x 64 x 2/3) =256-cm² Upvote · 9 5 9 1 Ravi Sharma Former Group A Officer From Indian Railways (1973–2009) · Author has 15.1K answers and 3.8M answer views ·3y Related In a parallelogram, the sum of two adjacent sides is 16 cm and one diagonal is 8 cm. The area of parallelograms is 48 sq. cm. What is the length of two sides? THE FIGURE OF PARALLELOGRAM IS SHOWN ABOVE. LET AD= BC = X AB=DC= 16-X AREA OF TRIANGLE ABD IS HALF OF THE AREA OF PARALLELOGRAM= 48/2= 24 AREA OF TRIANGLE ABD S= (X+16-X+8)/2= 24/2= 12 AREA=✓[{12(12–8)(12-X){12-(16-X)}]=24 =✓{12×4(12-X)(X-4)}=24 48(12-X)(X-4)=24×24 12X-X²-48+4X= 12 X²-16X+60=0 X²-6X-10X+60=0 X(X-6)—10(X-6)=0 (X-6)(X-10)=0 X=6 OR X= 10 ADJACENT SIDES ARE 10 AND 6 Continue Reading THE FIGURE OF PARALLELOGRAM IS SHOWN ABOVE. LET AD= BC = X AB=DC= 16-X AREA OF TRIANGLE ABD IS HALF OF THE AREA OF PARALLELOGRAM= 48/2= 24 AREA OF TRIANGLE ABD S= (X+16-X+8)/2= 24/2= 12 AREA=✓[{12(12–8)(12-X){12-(16-X)}]=24 =✓{12×4(12-X)(X-4)}=24 48(12-X)(X-4)=24×24 12X-X²-48+4X= 12 X²-16X+60=0 X²-6X-10X+60=0 X(X-6)—10(X-6)=0 (X-6)(X-10)=0 X=6 OR X= 10 ADJACENT SIDES ARE 10 AND 6 Upvote · Bheema Mudda Former Rtd Director at Government of India (2002–2007) · Author has 4.8K answers and 2.8M answer views ·4y Related The length, width, and height of a rectangular prism is in the ratio 2:3:5 respectively. If the sum of the edges of the prism is 120 cm, what is the surface area of the Prism? prism has 12 edges ,4 each of L,W&H ; 4 L+4 W+4 H =120 so L+W+H =30 [A] 2+3+5 =10 length =2/10 30 =6 c m s ; 3/1030 =9 c m; 5/1030 =15 c ms surface area of prism = 2 [ 69 + 915 +15 6] = 2[ 54+135+90 ] =558 cm^2 Answer : 558 cm^2 Upvote · 9 1 Related questions How do you solve for the diagonal length of a rectangular prism when only the length and width are given? How many corners does a rectangular prism have? How do I find the length of a rectangular prism? How many edges does a rectangular prism have? How does a rectangular prism look like? The main diagonal of a rectangular prism is 31 units in length and each dimension of the rectangular prism is an integer. What's the maximum and minimum possible volume of the rectangular prism? Does a rectangular prism have 6 rectangular faces? What is a cube and rectangular prism? A rectangular prism has a square base with edges measuring 8 inches each. Its volume is 768 cubic inches. What is the height of the prism and the surface area of the Prism? How many vertices does a rectangular prism have? How many corners does a normal prism have? How many vertices and edges does a rectangular prism have? What are the errors in triangular prism and rectangular glass prism? How do I make a rectangular prism? How can we change the length of a rectangular prism? Related questions How do you solve for the diagonal length of a rectangular prism when only the length and width are given? How many corners does a rectangular prism have? How do I find the length of a rectangular prism? How many edges does a rectangular prism have? How does a rectangular prism look like? The main diagonal of a rectangular prism is 31 units in length and each dimension of the rectangular prism is an integer. What's the maximum and minimum possible volume of the rectangular prism? 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16662	https://www.ck12.org/section/parallel-lines-and-transversals-%3A%3Aof%3A%3A-parallel-and-perpendicular-lines-%3A%3Aof%3A%3A-ck-12-geometry/	Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up HomeMathematicsParallel Lines and Transversals Parallel Lines and Transversals Difficulty Level: Basic \| Created by: CK-12 Last Modified: Aug 22, 2014 Learning Objectives Identify angles formed by two parallel lines and a non-perpendicular transversal. Identify and use the Corresponding Angles Postulate. Identify and use the Alternate Interior Angles Theorem. Identify and use the Alternate Exterior Angles Theorem. Identify and use the Consecutive Interior Angles Theorem. Introduction In the last lesson, you learned to identify different categories of angles formed by intersecting lines. This lesson builds on that knowledge by identifying the mathematical relationships inherent within these categories. Parallel Lines with a Transversal—Review of Terms As a quick review, it is helpful to practice identifying different categories of angles. Example 1 In the diagram below, two vertical parallel lines are cut by a transversal. Identify the pairs of corresponding angles, alternate interior angles, alternate exterior angles, and consecutive interior angles. Corresponding angles: Corresponding angles are formed on different lines, but in the same relative position to the transversal—in other words, they face the same direction. There are four pairs of corresponding angles in this diagram—\begin{align}\angle{6}\end{align}∠6 and \begin{align}\angle{8}, \angle{7}\end{align}∠8,∠7 and \begin{align}\angle{1}, \angle{5}\end{align}∠1,∠5 and \begin{align}\angle{3}\end{align}∠3, and \begin{align}\angle{4}\end{align}∠4 and \begin{align}\angle{2}\end{align}∠2. Alternate interior angles: These angles are on the interior of the lines crossed by the transversal and are on opposite sides of the transversal. There are two pairs of alternate interior angles in this diagram—\begin{align}\angle{7}\end{align}∠7 and \begin{align}\angle{3}\end{align}∠3, and \begin{align}\angle{8}\end{align}∠8 and \begin{align}\angle{4}\end{align}∠4. Alternate exterior angles: These are on the exterior of the lines crossed by the transversal and are on opposite sides of the transversal. There are two pairs of alternate exterior angles in this diagram—\begin{align}\angle{1}\end{align}∠1 and \begin{align}\angle{5}\end{align}∠5, and \begin{align}\angle{2}\end{align}∠2 and \begin{align}\angle{6}\end{align}∠6. Consecutive interior angles: Consecutive interior angles are in the interior region of the lines crossed by the transversal, and are on the same side of the transversal. There are two pairs of consecutive interior angles in this diagram—\begin{align}\angle{7}\end{align}∠7 and \begin{align}\angle{8}\end{align}∠8 and \begin{align}\angle{3}\end{align}∠3 and \begin{align}\angle{4}\end{align}∠4. Corresponding Angles Postulate By now you have had lots of practice and should be able to easily identify relationships between angles. Corresponding Angles Postulate: If the lines crossed by a transversal are parallel, then corresponding angles will be congruent. Examine the following diagram. You already know that \begin{align}\angle{2}\end{align}∠2 and \begin{align}\angle{3}\end{align}∠3 are corresponding angles because they are formed by two lines crossed by a transversal and have the same relative placement next to the transversal. The Corresponding Angles postulate says that because the lines are parallel to each other, the corresponding angles will be congruent. Example 2 In the diagram below, lines \begin{align}p\end{align} and \begin{align}q\end{align} are parallel. What is the measure of \begin{align}\angle{1}\end{align}? Because lines \begin{align}p\end{align} and \begin{align}q\end{align} are parallel, the \begin{align}120^\circ\end{align} angle and \begin{align}\angle{1}\end{align} are corresponding angles, we know by the Corresponding Angles Postulate that they are congruent. Therefore, \begin{align}m\angle{1} = 120^\circ\end{align}. Alternate Interior Angles Theorem Now that you know the Corresponding Angles Postulate, you can use it to derive the relationships between all other angles formed when two lines are crossed by a transversal. Examine the angles formed below. If you know that the measure of \begin{align}\angle{1}\end{align} is \begin{align}120^\circ,\end{align} you can find the measurement of all the other angles. For example, \begin{align}\angle{1}\end{align} and \begin{align}\angle{2}\end{align} must be supplementary (sum to \begin{align}180^\circ\end{align}) because together they are a linear pair (we are using the Linear Pair Postulate here). So, to find \begin{align}m\angle{2}\end{align}, subtract \begin{align}120^\circ\end{align} from \begin{align}180^\circ.\end{align} \begin{align}m \angle {2} & = 180^\circ - 120^\circ \ m \angle {2} & = 60^\circ\end{align} So, \begin{align}m\angle{2}=60^\circ\end{align}. Knowing that \begin{align}\angle{2}\end{align} and \begin{align}\angle{3}\end{align} are also supplementary means that \begin{align}m \angle {3}=120^\circ,\end{align} since \begin{align}120+60=180\end{align}. If \begin{align}m\angle{3}=120^\circ\end{align}, then \begin{align}m \angle {4}\end{align} must be \begin{align}60^\circ,\end{align} because \begin{align}\angle{3}\end{align} and \begin{align}\angle{4}\end{align} are also supplementary. Notice that \begin{align}\angle{1} \cong \angle{3}\end{align} (they both measure \begin{align}120^\circ\end{align}) and \begin{align}\angle{2} \cong \angle{4}\end{align} (both measure \begin{align}60^\circ\end{align}). These angles are called vertical angles. Vertical angles are on opposite sides of intersecting lines, and will always be congruent by the Vertical Angles Theorem, which we proved in an earlier chapter. Using this information, you can now deduce the relationship between alternate interior angles. Example 3 Lines \begin{align}l\end{align} and \begin{align}m\end{align} in the diagram below are parallel. What are the measures of angles \begin{align}\alpha\end{align} and \begin{align}\beta\end{align}? In this problem, you need to find the angle measures of two alternate interior angles given an exterior angle. Use what you know. There is one angle that measures \begin{align}80^\circ.\end{align} Angle \begin{align}\beta\end{align} corresponds to the \begin{align}80^\circ\end{align} angle. So by the Corresponding Angles Postulate, \begin{align}m \angle{\beta} = 80^\circ\end{align}. Now, because \begin{align}\angle{\alpha}\end{align} is made by the same intersecting lines and is opposite the \begin{align}80^\circ\end{align} angle, these two angles are vertical angles. Since you already learned that vertical angles are congruent, we conclude \begin{align}m \angle{\alpha} = 80^\circ\end{align}. Finally, compare angles \begin{align}\alpha\end{align} and \begin{align}\beta\end{align}. They both measure \begin{align}80^\circ,\end{align} so they are congruent. This will be true any time two parallel lines are cut by a transversal. We have shown that alternate interior angles are congruent in this example. Now we need to show that it is always true for any angles. Alternate Interior Angles Theorem Alternate interior angles formed by two parallel lines and a transversal will always be congruent. Given: \begin{align}\overleftrightarrow{AB}\end{align} and \begin{align}\overleftrightarrow{CD}\end{align} are parallel lines crossed by transversal \begin{align}\overleftrightarrow{XY}\end{align} Prove that Alternate Interior Angles are congruent Note: It is sufficient to prove that one pair of alternate interior angles are congruent. Let's focus on proving \begin{align}\angle{DWZ} \cong \angle{WZA}\end{align}. \| Statement \| Reason \| --- \| \| 1. \begin{align}\overleftrightarrow{AB} \\| \overleftrightarrow{CD}\end{align} \| 1.Given \| \| 2. \begin{align}\angle{DWZ} \cong \angle{BZX}\end{align} \| 2. Corresponding Angles Postulate \| \| 3. \begin{align}\angle{BZX} \cong \angle{WZA}\end{align} \| 3. Vertical Angles Theorem \| \| 4. \begin{align}\angle{DWZ} \cong \angle{WZA}\end{align} \| 4. Transitive property of congruence \| Alternate Exterior Angles Theorem Now you know that pairs of corresponding, vertical, and alternate interior angles are congruent. We will use logic to show that Alternate Exterior Angles are congruent—when two parallel lines are crossed by a transversal, of course. Example 4 Lines \begin{align}g\end{align} and \begin{align}h\end{align} in the diagram below are parallel. If \begin{align}m \angle{4} = 43^\circ\end{align}, what is the measure of \begin{align}\angle{5}\end{align}? You know from the problem that \begin{align}m \angle{4} = 43^\circ\end{align}. That means that \begin{align}\angle{4}'s\end{align} corresponding angle, which is \begin{align}\angle{3}\end{align}, will measure \begin{align}43^\circ\end{align} as well. The corresponding angle you just filled in is also vertical to \begin{align}\angle{5}\end{align}. Since vertical angles are congruent, you can conclude \begin{align}m \angle{5} = 43^\circ\end{align}. This example is very similar to the proof of the alternate exterior angles Theorem. Here we write out the theorem in whole: Alternate Exterior Angles Theorem If two parallel lines are crossed by a transversal, then alternate exterior angles are congruent. We omit the proof here, but note that you can prove alternate exterior angles are congruent by following the method of example 4, but not using any particular measures for the angles. Consecutive Interior Angles Theorem The last category of angles to explore in this lesson is consecutive interior angles. They fall on the interior of the parallel lines and are on the same side of the transversal. Use your knowledge of corresponding angles to identify their mathematical relationship. Example 5 Lines \begin{align}r\end{align} and \begin{align}s\end{align} in the diagram below are parallel. If the angle corresponding to \begin{align}\angle{1}\end{align} measures \begin{align}76^\circ,\end{align} what is \begin{align}m\angle{2}\end{align}? This process should now seem familiar. The given \begin{align}76^\circ\end{align} angle is adjacent to \begin{align}\angle{2}\end{align} and they form a linear pair. Therefore, the angles are supplementary. So, to find \begin{align}m\angle{2}\end{align}, subtract \begin{align}76^\circ\end{align} from \begin{align}180^\circ.\end{align} \begin{align}m\angle{2} & = 180 -76\ m\angle{2} & = 104^\circ\end{align} This example shows that if two parallel lines are cut by a transversal, the consecutive interior angles are supplementary; they sum to \begin{align}180^\circ.\end{align} This is called the Consecutive Interior Angles Theorem. We restate it here for clarity. Consecutive Interior Angles Theorem If two parallel lines are crossed by a transversal, then consecutive interior angles are supplementary. Proof: You will prove this as part of your exercises. Multimedia Link Now that you know all these theorems about parallel lines and transverals, it is time to practice. In the following game you use apply what you have learned to name and describe angles formed by a transversal. Interactive Angles Game. Lesson Summary In this lesson, we explored how to work with different angles created by two parallel lines and a transversal. Specifically, we have learned: How to identify angles formed by two parallel lines and a non-perpendicular transversal. How to identify and use the Corresponding Angles Postulate. How to identify and use the Alternate Interior Angles Theorem. How to identify and use the Alternate Exterior Angles Theorem. How to identify and use the Consecutive Interior Angles Theorem. These will help you solve many different types of problems. Always be on the lookout for new and interesting ways to analyze lines and angles in mathematical situations. Points To Consider You used logic to work through a number of different scenarios in this lesson. Always apply logic to mathematical situations to make sure that they are reasonable. Even if it doesn’t help you solve the problem, it will help you notice careless errors or other mistakes. Review Questions Solve each problem. Use the diagram below for Questions 1-4. In the diagram, lines \begin{align} \overleftrightarrow{AB}\end{align} and \begin{align} \overleftrightarrow{CD}\end{align} are parallel. What term best describes the relationship between \begin{align} \angle{AFG}\end{align} and \begin{align} \angle{CGH}?\end{align} alternate exterior angles consecutive interior angles corresponding angles alternate interior angles What term best describes the mathematical relationship between \begin{align} \angle{BFG}\end{align} and \begin{align}\angle{DGF}\end{align}? congruent supplementary complementary no relationship What term best describes the relationship between \begin{align} \angle{FGD}\end{align} and \begin{align} \angle{AFG}?\end{align} alternate exterior angles consecutive interior angles complementary alternate interior angles What term best describes the mathematical relationship between \begin{align} \angle{AFE}\end{align} and \begin{align}\angle{CGH}?\end{align} congruent supplementary complementary no relationship Use the diagram below for questions 5-7. In the diagram, lines \begin{align} l\end{align} and \begin{align} m\end{align} are parallel \begin{align} \gamma , \beta , \theta\end{align} represent the measures of the angles. What is \begin{align} \gamma?\end{align} What is \begin{align} \beta?\end{align} What is \begin{align} \theta?\end{align} The map below shows some of the streets in Ahmed’s town. Jimenez Ave and Ella Street are parallel. Use this map to answer questions 8-10. What is the measure of angle 1? What is the measure of angle 2? What is the measure of angle 3? Prove the Consecutive Interior Angle Theorem. Given \begin{align}r \|\| s\end{align}, prove \begin{align}\angle{1}\end{align} and \begin{align}\angle{2}\end{align} are supplementary. Review Answers c a d b \begin{align}73^\circ\end{align} \begin{align}107^\circ\end{align} \begin{align}107^\circ\end{align} \begin{align}65^\circ\end{align} \begin{align}65^\circ\end{align} \begin{align}115^\circ\end{align} Proof of Consecutive Interior Angle Theorem. Given \begin{align}r \|\| s\end{align}, prove \begin{align}\angle{1}\end{align} and \begin{align}\angle{2}\end{align} are supplementary. \| Statement \| Reason \| --- \| \| 1. \begin{align}r \|\| s\end{align} \| 1. Given \| \| 2. \begin{align}\angle{1} \cong \angle{3}\end{align} \| 2. Corresponding Angles Postulate \| \| 3. \begin{align}\angle{2}\end{align} and \begin{align}\angle{3}\end{align} are supplementary \| 3. Linear Pair Postulate \| \| 4. \begin{align}m\angle{2} + m\angle{3}=180^\circ\end{align} \| 4. Definition of supplementary angles \| \| 5. \begin{align}m\angle{2} + m\angle{1}=180^\circ\end{align} \| 5. Substitution \begin{align}(\angle{1} \cong \angle{3})\end{align} \| \| 6. \begin{align}\angle{2}\end{align} and \begin{align}\angle{1}\end{align} are supplementary \| 6. Definition of supplementary angles \| Notes/Highlights \| Color \| Highlighted Text \| Notes \| \| --- --- \| \| \| Please Sign In to create your own Highlights / Notes \| \| \| Currently there are no resources to be displayed. Description No description available here... Difficulty Level Basic Tags slope, Parallel Lines, slope-intercept form, standard form, perpendicular lines, converse, vertical angles, complementary angles, Euclidean Geometry, Non-Euclidean Geometry, linear angles, opposite reciprocal, Parallel Lines Property, Alternate Interior Angles Theorem, Alternate Exterior Angles Theorem, Corresponding Angles Postulate, alternate exterior angles, consecutive interior angles, alternate interior angles, corresponding angles, adjacent angles, skew lines, transversal, Perpendicular Line Postulate, Parallel Line Postulate, CK.MAT.ENG.SE.1.Geometry.3, (23 more) Subjects mathematics Grades 9, 10, 11 Standards Correlations - Concept Nodes License CC BY NC Language English \| Cover Image \| Attributions \| --- \| \| \| License: CC BY-NC \| Date Created Feb 23, 2012 Last Modified Aug 22, 2014 . \| Image \| Reference \| Attributions \| --- \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| Show Attributions Show Details ▼ Related Content Corresponding Angles Read + Corresponding Angles Principles - Basic Video Reviews Was this helpful? 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16663	https://www.youtube.com/watch?v=4rHox50vbRc	3-7 Evaluating Polynomials By Synthetic Substitution Mr. Dan Muscarella, NBCT 4630 subscribers 5 likes Description 505 views Posted: 6 Dec 2021 MathwithMusky #Polynomials #Synthetic How to use synthetic substitution to evaluate polynomial functions for a given x-value. Transcript: all right what's up i'm athletes that's right miss muscard coming at you in this video we're going to continue learning a little bit more about polynomial and evaluating them and then this one we're going to take a look at the technique known as synthetic substitution that's right last video we took a look at direct substitution where you just plug the number in and go in this video we are going to take a look at synthetic substitution now a couple things about synthetic substitution when you use whatever when you look at a polynomial like on this one my highest degree here is four so that means when i look at those terms i want to have something to the fourth power to the third power to the second power to the first power and then a constant a constant so if i look here like my constant right here is negative eight i've got something to the first degree now i'm missing a term that's squared so i don't have an x squared but i do have a something to the third power and of course we've got our term here the fourth power so when we do that that's going to play into our setup of this equation now the first thing you're going to do when you set this up after you evaluate that you're going to draw kind of like a little upside down division bar and whatever number they're being asked to evaluate which in this case is 2 we're going to write that on the outside and then each one of the terms that are given i'm going to take the coefficient the number in front and write down just the coefficient so the coefficient is set up in front of negative x to the fourth is just negative one so i'm gonna write that there my coefficient in front of negative two x cubed is negative two now i'm missing an x squared term so i have to put a zero in for a placeholder and that's very important if there's a term that is missing from the highest degree all the way down to the constant you've got to put a zero in as a placeholder so i put a zero in for my x squared term and then i put a 3 in for my linear term because that's 3x then negative 8 is my constant so that is this setup so the setup is super important because you want to make sure that we do that now we did this problem last one this was our last example so we'll kind of tie this back into this here momentarily so the first thing you want going to want to do is bring down that negative 1. now procedurally you're going to take the 2 and you're going to multiply it by the number that is we just brought down which in this case is negative one so two times negative one that's right it is negative two now the next pair of numbers negative two and two negative two you find yourself negative two plus negative two gives us negative four then again we're going to take that 2 on the outside and we're going to multiply that by negative 4 so 2 times negative 4 is negative 8. when i add up 0 and negative 8 my sum is negative 8. again i'm going to multiply 2 times negative 8. their product gives me negative 16. now when i add up 3 with negative 16 the sum of those two values 3 plus negative 16 gives me a total of negative 13. now when i multiply 2 times negative 13 i get negative 26 when i find the sum of negative 8 and negative 26 that sum is going to be negative 34. now this value right here at the end what that negative 34 that is my answer so it's like boom that's so cool so when you're done you don't just stop there and say okay put a little circle around you write this down next you write f of 2 equals negative 34. and that's it so that is the value of that function if i were to plug a 2 in the output value the answer that i would get would be negative 34. so if you think about that also from a coordinate point to negative 34 that'd be my input and output ruled in there together now on the previous example when we did that using direct substitution we came up with the same value so sometimes your teachers will ask you to direct sometimes they'll ask you to use synthetic but you can always use one to check the other to check the other so those are going to be some techniques and some options that you'll have as you go through this one all right so let's go ahead and take a look at example number two here now if you feel like you got it you can go ahead hit pause and come back and check to make sure that your mathematical writing is solid and that you are good to go my friend all right for everybody else let's go ahead and start rolling now again we're given p of two we've got to figure out what that is so if i were to plug 2 in for all the x's then we can figure out what our answer is so we're going to have 2 on the outside we draw a little upside down division symbol there and when our first coefficient is a 3 because we have 3x to the fourth power followed by a negative 7 followed by a negative 5 followed by a 9 and lastly we have a 10 so we have all the constants coefficients there from our fourth degree from our highest power of four all the way down to our constant so we have a fourth degree uh term right here and then we go all the way down into our very last little term right here our constant we've got everything so we don't have to put a zero in anywhere like we did on that last one now again bring down that 3 so that's going to be the first number we'll write down in that space then 2 times 3 of course is 6. when you add them up you get negative 1. multiplying 2 times negative 1 we get negative two the sum of negative five and negative two is negative seven when you multiply two times negative seven you get negative fourteen the sum of nine and negative fourteen is negative five but when you multiply two times negative 5 this gives us negative 10. when we add those guys up we get a sum of zero so the value so again remember our notations got to be good so we would say p of 2 so be careful there because not f now it's p so p of 2 would have a value of zero which tells me something else which is really really cool anytime so it's going to come into play later when we're doing that synthetic substitution and we plug in a value for x and we end up with right in this spot right here that means that 2 is a 0 an x intercept a solution to that polynomial function that we have given here which also means if 2 is a 0 then that would mean x minus 2 would be a factor for that polynomial so more on that coming later when we divide polynomials yes so more on that later but that's it for that example right there now we've got two other examples here examples three and four so i want you to go ahead and hit pause and then go ahead come back and let's see how you did all right so let's go ahead and check out how we did here with number three now number three man treat a little spot right there don't forget yes we're gonna have to add in a placeholder because we are missing a third degree so we don't have something to the third powers we have put a zero in that space and that is common for people to miss that so be careful take your time when you do that and then of course when you're done we get a value of negative 12. and then sliding over to example four same thing again try to see if you were paying attention there we have a zero in there for the term because if you look at all the terms from negative x to the fourth all the way down to our constant of negative two we are missing a squared term so in that space we gotta put a zero in which case we will end up with a value of four so f of three is four so give yourself pat on the back for getting all those for correct mundo now let's talk points if you were doing this on a quiz this is how we'd roll for it right so you so that's we can go back to this one right here it doesn't matter which one so you get one point right here for the setup all right you get one point for the setup and then you would get one point for all the numbers that are at the bottom making sure they're all good and then of course you have to write your answer f of two equals negative 34. so each one of these will be three points similar to that and then on this one of course we have one point for the setup one point here and then one point four the p of two equals zero and then our last two though tricky ones here we get a point for that first line that third line then our answer there and then a point a point and a point so that's it for this that way you know you guys are pretty solid with that so by now you should know how to and remember how to evaluate polynomials using this really cool technique called synthetic substitution much more simple for a lot of people than direct substitution so i hope you found this video helpful and i wish you the best day ever alright make it a great one and i'll see my mathlete soon peace out cub scout
16664	https://arxiv.org/html/2501.10203v1	The Erdős–Moser sum-free set problem via improved bounds for -configurations Adrian Beker111University of Zagreb, Faculty of Science, Department of Mathematics, Zagreb, Croatia. Email: adrian.beker@math.hr (January 17, 2025) Abstract A -configuration is a collection of distinct integers together with their pairwise arithmetic means for . Building on recent work of Filmus, Hatami, Hosseini and Kelman on binary systems of linear forms and of Kelley and Meka on Roth’s theorem on arithmetic progressions, we show that, for , any subset of density at least contains a -configuration. This improves on the previously best known bound , due to Shao. As a consequence, it follows that any finite non-empty set contains a subset of size at least such that for any distinct . This provides a new proof of a lower bound for the Erdős–Moser sum-free set problem of the same shape as the best known bound, established by Sanders. 1 Introduction The motivation for this paper comes from the following old question of Erdős and Moser [10, p. 187], which was recently reiterated by Green in his list of open problems [14, Problem 2] (see also [3, Problem #787]). Given a finite non-empty set of integers , what is the largest size of a subset which is sum-free with respect to ? Here, we say that is sum-free with respect to to mean that for any distinct . Letting denote the answer to this question, we are interested in the behaviour of , the minimum of as ranges over all sets of integers of size . Note that the above requirement that be distinct is necessary in order for the problem to be non-trivial. Indeed, the example shows that dropping the distinctness assumption would result in being constantly equal to . We briefly review the history of progress on this problem. A simple example showing that was given by Erdős . This was subsequently improved by Selfridge [10, p. 187] to . A sublinear upper bound on was first established by Choi [10, p. 190], who later improved this to a power-saving bound222In fact, Choi also observed that the present formulation of the problem is equivalent to the one initially proposed by Erdős and Moser, in which the reals were taken as the ambient group. of the form . The error term in the exponent was refined by Baltz, Schoen and Srivastav , who proved that . Ruzsa was the first to prove that grows subpolynomially in , namely . This is the best known upper bound to date. In the other direction, progress has been somewhat slower. Erdős and Moser showed that as and Klarner obtained the quantification (see ). However, the first published proof of a non-trivial lower bound seems to be that of Choi , who showed using a greedy argument that (see also [25, Theorem 6.2] for a proof of the slightly weaker bound ). Using a more nuanced greedy strategy, Ruzsa was able to obtain the constant factor improvement . Soon afterwards, a significant advance was made by Sudakov, Szemerédi and Vu , who proved the first superlogarithmic lower bound, namely with (here, denotes the -fold logarithm). This was later improved by Dousse to and Shao to . Finally, in a recent tour de force of additive combinatorics, Sanders showed that is permissible. We take a moment to further discuss the series of works starting from the Sudakov–Szemerédi–Vu breakthrough since these are particularly relevant to us. By making extensive use of the Balog–Szemerédi–Gowers–Freiman machinery, Sudakov, Szemerédi and Vu ultimately reduce the problem to that of finding -configurations in dense sets of integers. Herein, if is an abelian group such that the map , is an automorphism333In our case, either the additive group of the rationals or a finite abelian group of odd order. and is an integer, a -configuration is defined to be a collection of elements together with their pairwise arithmetic means for . We say that is generated by ; is said to be non-degenerate if are pairwise distinct. Given and , let denote the smallest positive integer such that any subset of of density at least contains a non-degenerate -configuration. Sudakov, Szemerédi and Vu observed that an arithmetic progression of length contains the -configuration generated by the elements at odd indices, so appeal to Gowers’ bound for Szemerédi’s theorem yields \| \| \| \| However, as was first observed by Dousse , this is overkill in that one doesn’t need the full strength Szemerédi’s theorem in order to obtain bounds for . Indeed, the system of linear forms defining a -configuration has complexity in the sense of Green and Tao [15, Definition 1.5]. Thus, to get a handle on -configurations, one can use the -norm (i.e. Fourier analysis) rather than the higher-order -norm. Implementing this idea analogously to Roth’s proof of his theorem on three-term progressions , Dousse obtains the bound \| \| \| \| Shao’s improvement arises, roughly speaking, from the use of Bohr sets in analogy to Bourgain’s proof of Roth’s theorem . He proves in that \| \| \| \| In contrast to previous works, Sanders’ bounds on aren’t obtained by improving the bounds on . Instead, Sanders bypasses the study of -configurations by exploiting certain weaknesses in the initial reduction to this problem. Hence, even though we now have a better understanding of the Erdős–Moser sum-free set problem, there remains the problem of improving bounds for sets lacking -configurations. In the meanwhile, there have been major advances in our understanding of sets lacking three-term progressions. In a remarkable breakthrough, Kelley and Meka established that any subset of with this property must have density at most444This has subsequently been improved to by Bloom and Sisask . . This matches the shape of Behrend’s lower bound and improves on a long line of incremental progress culminating in the work of Bloom and Sisask . Moreover, the techniques introduced by Kelley and Meka are fundamentally different to the ones previously used to study three-term progressions and have already seen success in applications to other problems in additive combinatorics. One such work is that of Filmus, Hatami, Hosseini and Kelman , which establishes bounds for sets lacking so-called binary systems of linear forms. By combining the ideas of Filmus et al. with those of Kelley and Meka, we are able to improve the bounds for sets lacking -configurations. Our main result is the following. Theorem 1.1. There is an absolute constant such that the following holds. Let and . If , then any subset of density at least contains a non-degenerate -configuration. Theorem 1.1 can be viewed as an extension of the main result of Kelley and Meka [16, Theorem 1.1], which corresponds to the case of our result. At the same time, it offers an improvement over [23, Theorem 1.3], the best bound previously available for general . We remark that, if we consider separately the dependence of on and , then the bound in Theorem 1.1 is optimal up to respective exponents of and . Indeed, the classical construction of Behrend implies and Green’s bound on the clique number of random Cayley graphs gives for some absolute constant (see [21, p. 5] for a more detailed discussion). Going back to the Erdős–Moser sum-free set problem, we establish the following as an application of Theorem 1.1. Theorem 1.2. Let be arbitrary. Then for any sufficiently large finite set , there exists a subset of size at least such that for any distinct . Theorem 1.2 implies that with , so we recover a bound of the same shape as in [21, Theorem 1.2]. We do not claim any improvement in the value of , though our constant potentially has the minor advantage of being simpler to calculate. We should note, however, that the approach to the Erdős–Moser problem via -configurations is a priori limited to the range . Our arguments could probably be optimised and as a result one could bring the value of closer to . It is not clear whether one can get arbitrarily close to ; one seemingly runs into similar obstacles as when attempting to improve the exponent in the Kelley–Meka bound for Roth’s theorem. For a summary of how the bounds on influence those on , we refer the reader to [21, §2]. The key technical input used in the proof of Theorem 1.1 is the graph counting lemma of Filmus et al. [11, Theorem 2.1]. Our improvement over the the work is twofold. First, deals with binary systems of linear forms, which are defined to be collections of linear forms in a fixed set of variables such that each form depends on exactly two variables and no two forms depend on the same pair of variables. On the other hand, each of the linear forms defining a -configuration is supported on two variables, but some of them actually depend on a single variable. At first, one might think that this requires merely a formal change to the arguments of , for instance allowing self-loops in the graphs being counted by [11, Theorem 2.1]. However, this is not the case. In a qualitative sense, we use [11, Theorem 2.1] as a black box, but to overcome this difficulty, we have to incorporate some further arguments in the style of and . Nevertheless, in order to obtain Theorem 1.1 as stated, we do have to make some improvements to the graph counting lemma of Filmus et al., but these happen purely at a quantitative level. To be specific, the dependence of the density increment parameter on the number of variables is not made explicit in [11, Theorem 2.1]. In fact, their argument yields an exponential dependence on , the primary aim of that paper being a good dependence on the density . However, for our application to the Erdős–Moser problem, a polynomial dependence on is paramount. Luckily, the arguments of can be modified so as to obtain a density increment of the required strength. The rest of the paper is organised as follows. In Section 2, we establish a version of the graph counting lemma which is suitable for our application. Section 3 is devoted to the proof of Theorem 1.1 and occupies the main bulk of the paper. In fact, we deduce Theorem 1.1 as a direct consequence of Theorem 3.1, which deals with general finite abelian groups. We first give a high-level overview of the argument in the finite field setting and then proceed to transfer these arguments to the general case using the machinery of Bohr sets. In Section 4, we establish Theorem 1.2 as a consequence of Theorem 1.1. In Appendix A, we collect some variants of the Kelley–Meka arguments that are needed in our proofs, but do not appear elsewhere. Finally, Appendix B contains the relevant background material on Bohr sets. Notational conventions. Notationwise, we mostly follow . In particular, we use Vinogradov asymptotic notation. Hence, given quantities and , we write to mean , that is to say there exists an absolute constant such that . This is equivalent to the notation , i.e. . We use to denote that and hold simultaneously. Given a positive integer , we abbreviate the set to . If is a finite non-empty set, we will use the averaging notation to denote . If has the form of a -fold Cartesian product , we may sometimes use instead of . Logarithmic factors are ubiquitous in the paper, so it will be convenient to use the abbreviation for the function on the interval . Given sets in an ambient abelian group , we define their sumset and difference set by \| \| \| \| respectively. If is an integer, we write . This is the image of under the multiplication-by- map \| \| \| \| and should not be confused with , which is defined to be the -fold sumset \| \| \| \| Unless otherwise stated, will denote a finite abelian group, written additively. By default, will be equipped with the uniform probability measure and this normalisation will be used to define -norms and inner products of complex-valued functions on . Given two such functions and , we also define their convolution and difference convolution to be \| \| \| \| respectively. Given a further function , we will frequently (sometimes without mention) make use of the adjoint property \| \| \| \| The Fourier transform plays a minor role in the paper – it makes an appearance only in Appendix A. For the sake of completeness, we define the Fourier transform of to be \| \| \| \| where denotes the group of characters of , which will always carry the counting measure. It will also be useful to consider various other probability measures on . Given a function such that , we will identify with a probability measure in the obvious way. Thus, it will be convenient to make the slight abuse of terminology of calling a probability measure555A more correct term would be probability density function. and writing for the corresponding measure of a set . Hence, we have \| \| \| \| and the corresponding -norms and inner products are given by \| \| \| \| Henceforth, unless specified otherwise, will denote the uniform probability measure on . A particularly important class of probability measures arises from uniform distributions on subsets of , so given a non-empty set , we write for the normalised indicator function of . Thus, for example, is the density of a subset in . 2 A quantitatively improved graph counting lemma The purpose of this section is to establish the following version of [11, Theorem 2.1]. This is achieved by uncovering and improving various quantitative dependencies on and . In what follows, by an oriented graph we mean a directed graph obtained by orienting the edges of a simple undirected graph. In particular, neither parallel edges nor cycles of length at most are allowed. The degree of a vertex in such a graph is defined to be its degree in the underlying undirected graph. Theorem 2.1. Let and let be a positive integer. Then the following holds with . Let and let be an oriented graph with edges such that for each we have . Let be finite non-empty sets and for each let be a subset of density . If \| \| \| \| then there exists an edge such that (i) either there exist of densities at least such that \| \| \| \| 2. (ii) or there exists of density at least such that for all we have \| \| \| \| To obtain Theorem 2.1, we have to make an improvement to the key technical ingredient in the proof of [11, Theorem 2.1], namely [11, Lemma 2.8]. We accomplish this by closely following the proof of [11, Lemma 2.8] and making careful choices of parameters where necessary. Before doing so, we recall some notation from [11, §2]. For a function on a finite non-empty set and a parameter , we define the -norm of to be \| \| \| \| In a similar vein, if is a function on the Cartesian product of finite non-empty sets and are positive integers, we define the corresponding grid-semi-norm of to be \| \| \| \| All relevant properties of grid-semi-norms can be found in [11, §2]. For an oriented graph , define , where is the number of vertices of degree in . Observe that is equal to the sum of the degrees of all vertices of degree greater than in . In particular, if all degrees in are at most , then , and otherwise . Lemma 2.2. Let and let be an acyclic oriented graph with edges. Let be finite non-empty sets and for each let be a function with mean . If \| \| \| \| then writing , there exist an edge and a positive integer such that (i) either there exists a positive integer such that and ; 2. (ii) or and . Proof. We will be fairly brief since the argument overlaps heavily with that of ; we indicate the key changes that have to be made. We proceed by induction on . As in , we may reduce to the case when has no isolated vertices/edges, so in particular . For each , consider the normalised variant of given by . Since is acyclic, it contains vertex of outdegree , call it . Let be the in-neighbours of , where . Let be the graph obtained by removing from . Since has no isolated edges, it follows that and hence \| \| \| \| Thus, we can make the assumption that \| \| \| --- \| \| \| (1) \| where we define666Note the change in the choice of here – in , the choice was made. \| \| \| \| Indeed, if (1) doesn’t hold, then we may finish by applying the induction hypothesis to . In particular, we may certainly assume that \| \| \| \| If we let and be such that , then we may use the above as in to bound \| \| \| --- \| \| \| (2) \| We now turn to the main part of the argument. Consider first the case when . Then (1) implies \| \| \| \| which can be rewritten as \| \| \| \| Using Hölder’s inequality and the estimate (2) as in , we obtain that \| \| \| \| which means that (ii) holds. Suppose now that . Let be the graph obtained from by adding a vertex , removing the edge and adding the edge ; we also let and . Then note that , so we have \| \| \| \| Thus, we may assume that \| \| \| \| as otherwise we are done by the induction hypothesis applied to . By the triangle inequality, it follows that \| \| \| \| This can be rewritten as \| \| \| \| where we define \| \| \| \| We may now use Hölder’s inequality, the Cauchy–Schwarz inequality, the Gowers–Cauchy–Schwarz inequality [11, Lemma 2.5] and the estimate (2) in the same way as in to bound the left-hand side. Hence, we obtain \| \| \| \| If there exists such that , then (i) holds with . Thus, we may assume that for all , so , where \| \| \| \| The last inequality holds since777Note that in , the weaker estimate is applied. \| \| \| \| Applying [11, Lemma 2.9], we obtain that with . If we now let , then the triangle inequality implies that \| \| \| \| so either or . In the former case, it follows that (i) holds with and playing the role of , so we are done. On the other hand, since doesn’t depend on , the latter case degenerates to \| \| \| \| so (ii) holds. This concludes the proof. ∎ With the ancillary results of [11, §2] at hand, it is now a short step from Lemma 2.2 to Theorem 2.1 – we mimic the deduction of [11, Theorem 2.1] from [11, Lemma 2.8]. Proof of Theorem 2.1.. We apply Lemma 2.2 with for . If (i) holds, then we are done by [11, Corollary 2.3] applied with and in place of and respectively. Otherwise, if (ii) holds, then we apply [11, Lemma 2.7] to the function \| \| \| \| to conclude that (a) either has density at least \| \| \| \| 2. (b) or has density at least \| \| \| \| In case (a), we see that the alternative (i) of Theorem 2.1 holds with , whereas in case (b), it follows that the alternative (ii) of Theorem 2.1 holds. ∎ 3 Kelley–Meka bounds for sets free of -configurations In this section, we establish Theorem 1.1. As mentioned in Section 1, our arguments naturally yield the following more general variant. It should be compared to [11, Theorem 1.5], which establishes an analogous bound for binary systems of linear forms. Our argument should likewise allow for more general coefficients in Theorem 3.1. However, since this is not our main goal and the paper is already quite technical in its current form, we do not pursue this here. Theorem 3.1. Let be a finite abelian group of odd order and let be an integer. Let be a subset of density . Then \| \| \| --- \| \| \| (3) \| In particular, if contains no non-degenerate -configuration, then \| \| \| --- \| \| \| (4) \| We begin by giving a quick deduction of Theorem 1.1 from Theorem 3.1. This is entirely standard: we embed the interval into a suitably large cyclic group of odd order and apply the bound (4). Proof of Theorem 1.1 assuming Theorem 3.1.. Assume to the contrary that doesn’t contain any non-degenerate -configuration. Set and let be the natural projection. Then contains no non-degenerate -configuration. Indeed, if this were not the case, there would exist distinct such that for all we have . But this means that for any such there is some such that . Since , we must have , i.e. . But then contains a non-degenerate -configuration, which contradicts our starting assumption. Therefore, by Theorem 3.1, we have \| \| \| \| Since and , we obtain that \| \| \| \| Taking to be sufficiently large now gives the desired contradiction. ∎ Thus, our main task for this section is to prove Theorem 3.1. Since the details are rather technical, we first give a sketch of a simplified version of the argument in the hope of illuminating the main ideas. Along the way, we also include a comparison with the work of Filmus et al. , highlighting the extra difficulties that arise in our case. For the purposes of this discussion, we work in the setting of vector spaces over finite fields. This allows one to express the main ideas much more cleanly, without having to go through all the technicalities of the general case. Outline of the argument. Suppose that , where is an odd prime. As is common with problems of this kind, we employ a density increment approach. The basic idea is that if the number of -configurations in deviates from what one would expect for a random set of the same density, then some translate of has increased density on a large subspace of . One then iterates this argument; since the density of is bounded above by , this process terminates in a bounded number of steps, thereby giving the conclusion. The starting point, therefore, is to apply Theorem 2.1 (though for the level of the discussion here the symmetric variant [11, Theorem 1.4] would also suffice). We take to be the transitive orientation of the complete graph , the sets to be copies of and each set to consist of the pairs such that . There is a serious obstacle, however, to the application of the graph counting lemma, which is that we a priori have no control over the density of the sets . Indeed, in the extreme case when is free of three-term arithmetic progressions, each is empty! This should be compared with the situation in , where such difficulties are absent. More concretely, in the case of the binary system given by for , one instead takes each of to be the whole of and to consist of the pairs such that . Observe that, in this case, the bipartite graph represented by has density . Moreover, it is regular – every vertex has degree exactly equal to . Even though our situation is more complicated, there is still a way out. Indeed, the case when there is a significant discrepancy in the density of from is precisely the starting assumption in the Kelley–Meka proof of Roth’s theorem. Therefore, this case can be shown to lead to a density increment of the desired kind. In the complementary case, the graph counting lemma provides us with suitably dense sets such that either has significantly increased density on or the degree in of each vertex from is significantly lower than average. The process of deducing a suitable density increment from the former is precisely the main concern of [11, §3]. Briefly, this can be achieved as follows. One rewrites the conclusion as \| \| \| \| and applies the adjoint property of convolutions to transform this into \| \| \| \| From there, an application of Hölder’s inequality implies an unusually large -norm of the difference convolution for a not too large value of the parameter . An asymmetric version of the dependent random choice argument of Kelley and Meka followed by almost-periodicity then yields a density increment of on a subspace of suitably small codimension. Rescaling, we obtain an adequate density increment of . We remark that, out of the main components of the Kelley–Meka proof of Roth’s theorem, we did not have to use spectral non-negativity since we already had an upwards deviation from an expected count. This is because spectral non-negativity (more precisely an analogue thereof, namely [11, Lemma 2.9]) has already been used in the argument leading to this alternative of the graph counting lemma. In the remaining case, is bounded above by everywhere on , whence averaging yields \| \| \| \| We note in passing that, in the context of , this cannot happen since the graph is regular. In our case, however, this means that \| \| \| \| whence an application of Hölder’s inequality followed by the decoupling inequality of Kelley and Meka [16, Proposition 2.12] implies that at least one of \| \| \| \| must be large for a not too large value of the parameter . From this, one can use spectral non-negativity together with the other Kelley–Meka techniques (sifting, almost-periodicity) to obtain a density increment of either or . By dilating if necessary, these two cases can be unified into a density increment of . We now begin the process of translating the above ideas into the framework of Bohr sets. In general finite abelian groups, Bohr sets serve as a substitute for subspaces. This complicates the arguments on a technical level, but the core ideas remain essentially the same. The reader unfamiliar with this topic may wish to consult Appendix B before reading the rest of this section. The proof of Theorem 3.1 is based on the following density increment result, which plays a role analogous to that of [11, Theorem 3.3]. Proposition 3.2. Let be a finite abelian group of odd order and let be an integer. Let be a regular Bohr set of rank in and let be a subset of density . Then (i) either the proportion of -tuples such that for all is at least ; 2. (ii) or there exist of size at most , and such that is a regular Bohr set and \| \| \| \| where for some . Theorem 3.1 follows from Proposition 3.2 by iteration. For the most part, this follows standard lines. Nevertheless, we have to make an additional observation about the behaviour of the rank of the ambient Bohr set. Indeed, per (ii) of Proposition 3.2, it may seem at first glance that at each step of the iteration, the frequency set doubles in size. This would cause the rank to grow exponentially, which would result in poor bounds in Theorem 3.1. Luckily, by inspecting the specific structure of the frequency set, one readily sees that the rank in fact grows quadratically. This kind of observation seems to have first appeared in the work of Pilatte (see [17, Remark 3.4] for a more detailed discussion). Proof of Theorem 3.1 assuming Proposition 3.2.. Consider pairs of sequences , such that the following hold for : (i) , ; 2. (ii) is a regular Bohr set of rank , where , ; 3. (iii) is a subset of density ; 4. (iv) if , then for some ; 5. (v) if , then , where is an absolute constant; 6. (vi) if , then for some and of size at most ; 7. (vii) if , then . In particular, the sequences and are increasing. Moreover, property (v) implies via induction that for all . Since , it follows that . In particular, we may choose such a pair of sequences whose length is maximal. We now estimate the rank and width of . First, using (vi), it is straightforward to show by induction that for all we have the inclusion \| \| \| \| where we specially define . Hence, by employing the union bound, we get \| \| \| \| Furthermore, (vii) implies that \| \| \| --- \| \| \| \| \| \| \| \| By maximality of , upon applying Proposition 3.2 to , we must end up in case (i). But this means that the proportion of -tuples generating a -configuration in is at least \| \| \| --- \| \| \| \| \| \| \| \| To conclude, note that by inductively applying (iv), it follows that is a subset of a translate of . Since -configurations are translation-invariant, we thus obtain the claimed bound (3). If contains no non-degenerate -configuration, then the left-hand side in (3) can be crudely upper bounded by \| \| \| \| from which (4) follows by rearranging. ∎ The goal for the rest of this section is to establish Proposition 3.2. To accomplish this, we will require some preparation. Specifically, we will have to adapt several intermediate results of and to our setting. The first result says that, in a suitable local setting, self-regularity implies mixing for three-variable linear equations. Here, we informally say that a set is self-regular if the -norm of the difference convolution of with itself is roughly equal to the ‘expected’ value (for a precise definition, see [16, Definition 2.3]). Hence, our result can be viewed as a local variant of [16, Theorem 2.11]. The proof consists of carrying out the Hölder lifting and spectral non-negativity/unbalancing steps of the Kelley–Meka proof as exposited by Bloom and Sisask . It also requires an intermediate application of the local decoupling inequality of Kelley and Meka [16, Lemma 5.8]. Proposition 3.3. There is an absolute constant such that the following holds. Let be regular Bohr sets of rank . Let be subsets of densities at least and let be a subset of density . Suppose that for some and let be non-empty subsets. If \| \| \| \| then there exist , a positive integer and such that \| \| \| \| Proof. Since is symmetric and the conclusion is invariant under replacing by , by considering in place of , we may instead assume that \| \| \| \| Hence, by Proposition A.1, there exists an even positive integer such that \| \| \| \| Then consider the probability measures \| \| \| \| Since is supported on , it follows by Lemma B.11 that . Furthermore, since is symmetric, we have . Hence, Lemma A.2 and [16, Lemma 5.8] imply that \| \| \| \| where we write , . Thus, by [6, Proposition 18] and the remarks following [6, Lemma 7], there exist and a positive integer \| \| \| \| such that \| \| \| \| The desired conclusion now follows by applying Lemma A.2. ∎ We next establish a bespoke variant of [6, Proposition 15]. The main difference is that we allow the sets to be different and we make explicit the frequency sets of our Bohr sets as well as the dependencies of all bounds on the parameter . We should stress that these differences are mainly cosmetic in nature and the proof encompasses the same techniques: dependent random choice and almost-periodicity. The treatment of these techniques is deferred to Appendix A and entails several optimisations of the arguments of . Proposition 3.4. There are absolute constants such that the following holds. Let and be regular Bohr sets of rank in such that . Let be subsets of translates of a non-empty set with respective densities . Suppose that \| \| \| \| for some , and integer . Then there exist of size and such that \| \| \| \| and is a regular Bohr set with the property that \| \| \| \| Proof. To begin, we may assume for convenience that . Indeed, the general case then follows on observing that \| \| \| \| and applying the special case with instead of . Hence, by Lemma A.3, we obtain subsets , of respective densities , such that \| \| \| \| where we write \| \| \| \| Letting , we have and \| \| \| \| Thus, on applying Theorem A.5 with in place of , we obtain a set of size and a parameter such that \| \| \| \| and is a regular Bohr satisfying \| \| \| \| as required. ∎ By chaining together Propositions 3.3 and 3.4, we immediately obtain the following corollary. Corollary 3.5. There is an absolute constant such that the following holds. Let be subsets of densities at least , where is a regular Bohr set of rank . Let and be such that is a regular Bohr set. Let be a subset of density . If \| \| \| \| then there exist , of size and such that \| \| \| \| and is a regular Bohr set with the property that \| \| \| \| Proof. By Lemma B.10, there exist such that the Bohr sets and are regular. In particular, has width . We may now conclude by applying Proposition 3.3 followed by Proposition 3.4. ∎ We are now in a position to prove Proposition 3.2. In doing so, the main technical difficulties arise from working with dilates of Bohr sets on different scales, of which there are linearly many in . Likewise, the need to keep track of how the density increment parameters depend on adds an extra layer of bookkeeping. This is in contrast to the situation in Roth’s theorem , where the corresponding constants can be taken to be absolute. Proof of Proposition 3.2.. Throughout the proof, we let denote a sufficiently small and sufficiently large absolute constant, respectively. Let be as in Theorem 2.1 and define the parameters \| \| \| \| In particular, we have . We introduce the Bohr set . Note that has width and frequency set \| \| \| \| so its rank is at most . By Lemma B.10, we can inductively choose parameters such that and for all we have that and is a regular Bohr set. In particular, for all we have . By choice of , we may apply [11, Lemma 3.4] to the family of Bohr sets \| \| \| \| to obtain an such that, writing , we either have for all or there exists such that . In the latter case, we obtain the desired density increment (ii) straight away, so we may assume that the former holds. We now set the stage for an application of the graph counting lemma. Define for each the sets and and for each the set \| \| \| \| In particular, our assumption implies that and are subsets of of densities belonging to the interval . Let be the density of inside and set . Then note that we have the expression \| \| \| --- \| \| \| (5) \| Assume to begin with that there exist such that . As , it follows from (5) that \| \| \| --- \| \| \| (6) \| Since we have \| \| \| --- \| \| \| (7) \| dividing both sides of (6) by gives \| \| \| --- \| \| \| (8) \| By Corollary 3.5 applied to in place of respectively and , , we obtain a set , a set of size and a parameter \| \| \| \| such that is a regular Bohr set and \| \| \| \| Recalling that are subsets of a translate of , we have \| \| \| \| \| \| \| \| This means that the alternative (ii) holds, so we are done in this case. Hence, we may assume from now on that for all we have . In other words, the density of is roughly what one would naively expect it to be. In particular, we have . Applying Theorem 2.1 with , , for and in place of , we obtain that either \| \| \| --- \| \| \| (9) \| or there exist such that either there exist subsets , of densities at least satisfying \| \| \| --- \| \| \| (10) \| or there exists a subset of density at least such that for each we have \| \| \| --- \| \| \| (11) \| We deal with these three cases in turn. First, if (9) holds, then by translation-invariance, we have \| \| \| \| By passing to subsets, we may bound this from below by \| \| \| \| Since for all and for all , we have the further lower bound \| \| \| \| Furthermore, by Lemma B.7, we have \| \| \| \| Since , we obtain a lower bound for the density of -configurations in of the same form as above, so (i) holds. Suppose now that we are in the case when (11) holds. Then for all we have \| \| \| \| so by averaging over and using , we obtain that \| \| \| \| Since , it follows that \| \| \| \| On dividing through by and using (7), we arrive at the conclusion that \| \| \| \| By applying Corollary 3.5 in the same way as before, except now with and playing the role of , we obtain a set , a set of size and a parameter \| \| \| \| such that is a regular Bohr set and \| \| \| \| By the same token as before, we conclude from this that the alternative (ii) holds. It remains to deal with the case when (10) holds. Since , we have the following lower bound on the relative densities of and : \| \| \| \| Moreover, since , we obtain that \| \| \| \| Starting from this assumption, a suitable density increment can be deduced in a similar fashion as in [11, pp. 29-31]. On dividing through by and recalling that , one finds that \| \| \| \| Hence, by Hölder’s inequality, for any we have \| \| \| --- \| \| \| \| \| \| \| \| whence we obtain that \| \| \| \| For we have and hence \| \| \| \| In order to apply Proposition 3.4, we must first pass to -norms with respect to a convolution of two Bohr sets. To this end, we first use Lemma B.10 to choose parameters such that and are regular Bohr sets. By Lemma B.11, we then have \| \| \| \| where we define . Hence, Lemma A.2 supplies us with a shift such that \| \| \| \| If is chosen so that , we will have and hence \| \| \| \| Invoking Proposition 3.4 with and in place of respectively, we obtain a set of size and a parameter \| \| \| \| such that is a regular Bohr set and \| \| \| \| Similarly as in the previous case, we infer that the alternative (ii) holds, thereby completing the proof. ∎ 4 The Erdős–Moser sum-free set problem In this short section we apply our results concerning -configurations to the Erdős–Moser problem. We deduce Theorem 1.2 from Theorem 1.1 much in the same way as [23, Corollary 1.4] is deduced from [23, Theorem 1.3]. It follows from the following result, which is [21, Proposition 2.7] with explicit values for the implied constants. Proposition 4.1. Let be a sufficiently large positive integer and let be such that and . Then there exists of size which is sum-free with respect to . Assuming Proposition 4.1, the proof of Theorem 1.2 is the same as that of [21, Theorem 1.2] given [21, Proposition 2.7] (cf. the proof of [23, Corollary 1.4] given [23, Proposition 6.1] and also of [24, Theorem 1.2] given [24, Theorem 1.1]). Hence, we focus on establishing Proposition 4.1. This will be accomplished by combining Theorem 1.1 with two extra ingredients, the first of which is the following. Proposition 4.2. Let , and be as in the statement of Proposition 4.1 and suppose that the conclusion of Proposition 4.1 fails. Then there exists such that , and . Proposition 4.2 is a variant of [23, Proposition 6.2] and its proof is contained in [24, §6]; we omit the details. The second ingredient needed for the proof of Proposition 4.1 is Ruzsa’s embedding lemma [19, Lemma 5.1]. Its use in the present context originates in the work of Shao ; a proof can be found in [25, Lemma 5.26]. Lemma 4.3. Let be a finite non-empty set and let be a positive integer. Then there exists a subset such that and is Freiman -isomorphic to a subset of , that is to say there exists a map such that \| \| \| --- \| \| \| (12) \| whenever . We are ready to deduce Proposition 4.1 from Theorem 1.1. In fact, it will be slightly quicker to apply Theorem 3.1, though the former would of course also suffice. Proof of Proposition 4.1.. Assuming otherwise, Proposition 4.2 provides us with a set such that , and . By the Plünnecke–Ruzsa inequality [25, Corollary 6.29], we have . Hence, by Lemma 4.3, there exist a subset , an odd positive integer and a map such that and (12) holds for all . In particular, is a subset of of density and we have \| \| \| \| Provided , where is a sufficiently large constant, Theorem 3.1 implies the existence of a non-degenerate -configuration in . By pulling back, we obtain a non-degenerate -configuration in , generated by say. Then for any we have and hence , so we may conclude by taking . ∎ Acknowledgements. This work was supported by the Croatian Science Foundation under the project number HRZZ-IP-2022-10-5116 (FANAP). The author would like to thank Rudi Mrazović for advisement and support, Zach Hunter and Rushil Raghavan for useful discussions and Tom Sanders for helpful remarks. Appendix A Auxiliary Kelley–Meka-type results We collect here several results of Kelley–Meka-type that are not present in the literature in a form suitable for our applications. That being said, each of these can be obtained by making minor modifications to known results. We begin by recording an asymmetric variant of the Hölder lifting step of the Kelley–Meka argument. Proposition A.1. There is an absolute constant such that the following holds. Let be a parameter and let be a regular Bohr set of rank . Let be subsets of densities at least . Let be a non-empty set and let be a subset of density . Then (i) either ; 2. (ii) or there exists such that . Proof. The proof is the same as for [6, Proposition 20]. The result we are referring to deals only with the case , but the proof carries over to the general setting without any difficulty. ∎ The following lemma encodes a simple averaging argument which is used repeatedly in the paper. Lemma A.2. Let and . Suppose that are probability measures on such that for some . Then there exists such that \| \| \| \| Proof. Observe that \| \| \| \| so by averaging, we obtain an such that \| \| \| \| as desired. ∎ We next give a version of the dependent random choice argument underpinning the sifting step of the Kelley–Meka proof of Roth’s theorem. In essence, it can be obtained by combining the proofs of [16, Lemma 4.9] and [6, Lemma 8] taken together with the remarks following its statement. We need the former for two reasons: it allows us to take different sets , as input to the sifting process and produces better bounds for the densities of the outputted sets (see Remark A.4). The latter is required in order to obtain a local variant of the result. For the convenience of the reader, we include a proof. Lemma A.3. Let and let be an integer. Let be non-empty subsets and let be subsets of respective densities . Let and \| \| \| \| Then for there exists a subset of density at least such that \| \| \| \| Proof. For each consider \| \| \| \| where is a -tuple chosen uniformly at random from . Write for the relative density of in . Then it is straightforward to see using linearity of expectation that \| \| \| \| A similar calculation reveals that for any we have \| \| \| \| Thus, for any weight function , linearity of expectation implies that \| \| \| --- \| \| \| (13) \| In particular, taking in the above equation, it follows that \| \| \| \| Now consider the event \| \| \| --- \| \| \| (14) \| and observe that \| \| \| --- \| \| \| \| \| \| \| \| Therefore, we have \| \| \| \| The condition guarantees that , so taking in (13), we get \| \| \| \| Hence, by averaging, we obtain a realisation of on such that \| \| \| \| For this choice of , we have the required lower bound on the densities, as well as \| \| \| \| This concludes the proof. ∎ Remark A.4. The choice (14) of the event is based on the argument from the proof of [16, Lemma 4.7]. We could have alternatively followed the proof of [6, Lemma 10] by considering \| \| \| \| with the choice , and using the Cauchy–Schwarz inequality to bound \| \| \| \| This would ultimately lead to the same conclusion, but with the lower bound on the relative density of in . In the case of three-term progressions, the sets and are the same, so this would make no difference. In our case, however, this bound is weaker. Indeed, in our application, and are roughly proportional to and respectively. Hence, we would get an upper bound of the form for the product occurring in Theorem A.5. Our argument instead delivers , thereby saving a factor of . The following almost-periodicity result is essentially [11, Theorem 3.6]. In particular, it can be regarded as an asymmetric variant of [5, Lemma 8]. However, in contrast to these results, we use separate parameters to keep track of the densities of the input sets , which is worthwhile in view of Remark A.4. We also spell out the dependencies of the rank and width of on as well as the (unsurprising) fact that its frequency set extends that of . As in the case of Lemma A.3, we present a full proof. Theorem A.5. There is an absolute constant such that the following holds. Let be a non-empty set and let be subsets of densities respectively. Let and be regular Bohr sets of rank in such that . Let , be subsets of densities respectively. Let and let be a set such that (i) ; 2. (ii) for all . Then there exist of size and such that is a regular Bohr set with the property that \| \| \| \| Proof. We start by using Lemma B.10 to select such that is a regular Bohr set. In particular, by Remark B.9, we have \| \| \| \| for . Similarly, since \| \| \| \| it follows that . By [22, Theorem 5.1] applied with and in place of and respectively, we obtain a subset of a translate of of density at least such that \| \| \| \| In particular, by (i), we have \| \| \| \| and hence (ii) implies that \| \| \| --- \| \| \| (15) \| Since the large spectrum is invariant under translations, the Chang–Sanders lemma [22, Proposition 5.3] and Lemma B.10 imply the existence of a set of size \| \| \| \| and a parameter such that is a regular Bohr set with the property that for all and . Here, is a parameter to be determined later. Writing for brevity, a standard Fourier-analytic calculation now shows that, for , \| \| \| --- \| \| \| (16) \| By the Cauchy–Schwarz inequality and Parseval’s identity, we may bound \| \| \| --- \| \| \| \| Thus, choosing and , it follows by combining (15), (16) and averaging that \| \| \| \| Finally, using the adjoint property of convolutions and Hölder’s inequality, we may rewrite and bound the left-hand side as \| \| \| \| whence the desired conclusion follows. ∎ Appendix B Bohr sets We record here the definitions and properties concerning Bohr sets that are important to us; a more complete account is available in one of the standard sources in the literature, e.g. [25, §4]. A large portion of what we need also appears in [6, Appendix]; we make appropriate references whenever possible. We begin by formally introducing the concept of a Bohr set. Definition B.1. Given a non-empty set and a parameter , we define the Bohr set \| \| \| \| is called the frequency set and its cardinality is called the rank of the corresponding Bohr set. We call the width of the Bohr set. Remark B.2. All Bohr sets are symmetric and contain . It is important to note that, when speaking of a Bohr set , we always implicitly fix a frequency set and a width , and not just the “physical” Bohr set determined by and as in Definition B.1. Thus, on a formal level, one can think of a Bohr set purely as an ordered pair . When omitted, the frequency set and width will always be clear from the context. The following two definitions capture several ways of obtaining new Bohr sets from old. Definition B.3. Let be a Bohr set as in Definition B.1. Given a parameter , we define the -dilate of to be . Given an automorphism , we define the image of under as888We suppress the function composition symbol in order to avoid confusion with difference convolution. \| \| \| \| In particular, if is an integer coprime to the order of , then . Remark B.4. It is a very simple matter to check that the notion of the image of a Bohr set introduced in Definition B.3 is compatible with the corresponding set-theoretic concept. Definition B.5. If are Bohr sets, their intersection is defined to be \| \| \| \| Remark B.6. The set-theoretic intersection of Bohr sets need not coincide with the physical Bohr set associated to the formal intersection. However, the former is easily seen to contain the latter. Furthermore, the operation of intersection of Bohr sets as per Definition B.5 is commutative and associative. The next lemma provides a guarantee on the size of a Bohr set given only its rank and width; it is a variant of [6, Lemma A.4]. Lemma B.7. Let be a Bohr set of rank and width . Then . In general, Bohr sets are not even approximately closed under addition. Indeed, if is a Bohr set of rank , then in principle grows exponentially in . However, on replacing a copy of by a narrow dilate , one can hope to recover an approximate form of closure under addition. The following definition identifies precisely the class of Bohr sets for which this is feasible. Definition B.8. A Bohr set of rank is called regular if \| \| \| \| for all . Remark B.9. If is a regular Bohr set of rank and , then . In addition to Remark B.9, regularity of Bohr sets is most often exploited via a result such as [6, Lemma A.5]. Even though it does not appear explicitly in our paper, we do use it indirectly since it features in the proofs of [6, Proposition 18] and [6, Proposition 20]. See also Lemma B.11 below for a result in a similar spirit. The content of the following lemma is that regular Bohr sets exist on all scales; it appears for example as [6, Lemma A.3]. Lemma B.10. For any Bohr set , there exists such that is regular. We end with a simple consequence of regularity, which is [6, Lemma A.6]. Lemma B.11. There is an absolute constant such that the following holds. Let be a regular Bohr set of rank and let be a positive integer. Let and let be a probability measure supported on . Then \| \| \| \| References Andreas Baltz, Tomasz Schoen, and Anand Srivastav. Probabilistic construction of small strongly sum-free sets via large Sidon sets. Colloq. Math., 86(2):171–176, 2000. F. A. Behrend. On sets of integers which contain no three terms in arithmetical progression. Proc. Nat. Acad. Sci. U.S.A., 32:331–332, 1946. Thomas Bloom. Erdős problems. Accessed January 17, 2025. Thomas F. Bloom and Olof Sisask. Breaking the logarithmic barrier in Roth’s theorem on arithmetic progressions. 2020. arXiv:2007.03528. Thomas F. Bloom and Olof Sisask. An improvement to the Kelley–Meka bounds on three-term arithmetic progressions. 2023. arXiv:2309.02353. Thomas F. Bloom and Olof Sisask. The Kelley–Meka bounds for sets free of three-term arithmetic progressions. Essent. Number Theory, 2(1):15–44, 2023. J. Bourgain. On triples in arithmetic progression. Geom. Funct. Anal., 9(5):968–984, 1999. S. L. G. Choi. On a combinatorial problem in number theory. Proc. London Math. Soc. (3), 23:629–642, 1971. Jehanne Dousse. On a generalisation of Roth’s theorem for arithmetic progressions and applications to sum-free subsets. Math. Proc. Cambridge Philos. Soc., 155(2):331–341, 2013. P. Erdős. Extremal problems in number theory. In Proc. Sympos. Pure Math., Vol. VIII, pages 181–189. Amer. Math. Soc., Providence, RI, 1965. Yuval Filmus, Hamed Hatami, Kaave Hosseini, and Esty Kelman. Sparse graph counting and Kelley–Meka bounds for binary systems. In 2024 IEEE 65th Annual Symposium on Foundations of Computer Science—FOCS 2024, pages 1559–1578. IEEE Computer Soc., Los Alamitos, CA, ©2024. W. T. Gowers. A new proof of Szemerédi’s theorem. Geom. Funct. Anal., 11(3):465–588, 2001. Ben Green. Counting sets with small sumset, and the clique number of random Cayley graphs. Combinatorica, 25(3):307–326, 2005. Ben Green. 100 open problems. 2024. manuscript. Benjamin Green and Terence Tao. Linear equations in primes. Ann. of Math. (2), 171(3):1753–1850, 2010. Zander Kelley and Raghu Meka. Strong bounds for 3-progressions. In 2023 IEEE 64th Annual Symposium on Foundations of Computer Science—FOCS 2023, pages 933–973. IEEE Computer Soc., Los Alamitos, CA, ©2023. Cédric Pilatte. New bound for Roth’s theorem with generalized coefficients. Discrete Anal., pages Paper No. 16, 21, 2022. K. F. Roth. On certain sets of integers. J. London Math. Soc., 28:104–109, 1953. I. Z. Ruzsa. Generalized arithmetical progressions and sumsets. Acta Math. Hungar., 65(4):379–388, 1994. Imre Z. Ruzsa. Sum-avoiding subsets. Ramanujan J., 9(1-2):77–82, 2005. Tom Sanders. The Erdős–Moser sum-free set problem. Canad. J. Math., 73(1):63–107, 2021. Tomasz Schoen and Olof Sisask. Roth’s theorem for four variables and additive structures in sums of sparse sets. Forum Math. Sigma, 4:Paper No. e5, 28, 2016. Xuancheng Shao. Finding linear patterns of complexity one. Int. Math. Res. Not. IMRN, 2015(9):2311–2327, 2015. B. Sudakov, E. Szemerédi, and V. H. Vu. On a question of Erdős and Moser. Duke Math. J., 129(1):129–155, 2005. Terence Tao and Van H. Vu. Additive combinatorics, volume 105 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, paperback edition, 2010.
16665	https://pmc.ncbi.nlm.nih.gov/articles/PMC5566842/	Genetic factors contributing to human primary ciliary dyskinesia and male infertility - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. PMC Search Update PMC Beta search will replace the current PMC search the week of September 7, 2025. Try out PMC Beta search now and give us your feedback. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Asian J Androl . 2016 Jun 7;19(5):515–520. doi: 10.4103/1008-682X.181227 Search in PMC Search in PubMed View in NLM Catalog Add to search Genetic factors contributing to human primary ciliary dyskinesia and male infertility Zhi-Yong Ji Zhi-Yong Ji 1 The Center for Reproductive Medicine, Xiamen Maternity and Child Care Hospital, No. 10 Zhenhai Road, Xiamen, China Find articles by Zhi-Yong Ji 1, Yan-Wei Sha Yan-Wei Sha 1 The Center for Reproductive Medicine, Xiamen Maternity and Child Care Hospital, No. 10 Zhenhai Road, Xiamen, China Find articles by Yan-Wei Sha 1, Lu Ding Lu Ding 1 The Center for Reproductive Medicine, Xiamen Maternity and Child Care Hospital, No. 10 Zhenhai Road, Xiamen, China Find articles by Lu Ding 1, Ping Li Ping Li 1 The Center for Reproductive Medicine, Xiamen Maternity and Child Care Hospital, No. 10 Zhenhai Road, Xiamen, China Find articles by Ping Li 1,✉ Author information Article notes Copyright and License information 1 The Center for Reproductive Medicine, Xiamen Maternity and Child Care Hospital, No. 10 Zhenhai Road, Xiamen, China ✉ Correspondence: Dr. P Li (saarc2001@sina.com) Received 2015 Oct 26; Revised 2016 Jan 3; Accepted 2016 Apr 18; Issue date 2017 Sep-Oct. Copyright: © The Author(s)(2017) This is an open access article distributed under the terms of the Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License, which allows others to remix, tweak, and build upon the work non-commercially, as long as the author is credited and the new creations are licensed under the identical terms. PMC Copyright notice PMCID: PMC5566842 PMID: 27270341 Abstract Primary ciliary dyskinesia (PCD) is an autosomal-recessive disorder resulting from the loss of normal ciliary function. Symptoms include neonatal respiratory distress, chronic sinusitis, bronchiectasis, situs inversus, and infertility. However, only 15 PCD-associated genes have been identified to cause male infertility to date. Owing to the genetic heterogeneity of PCD, comprehensive molecular genetic testing is not considered the standard of care. Here, we provide an update of the progress on the identification of genetic factors related to PCD associated with male infertility, summarizing the underlying molecular mechanisms, and discuss the clinical implications of these findings. Further research in this field will impact the diagnostic strategy for male infertility, enabling clinicians to provide patients with informed genetic counseling, and help to adopt the best course of treatment for developing directly targeted personalized medicine. Keywords: genetic factors, Kartagener syndrome, male infertility, primary ciliary dyskinesia INTRODUCTION Primary ciliary dyskinesia (PCD) (Mendelian Inheritance in Man, MIM 242650) is a rare genetic disorder affecting approximately one in 20 000 individuals worldwide, which is multisystemic and caused by motility defects in the cilia and flagella.1,2 Specifically, ineffective cilia movement results in limited mucociliary clearance in the upper and lower respiratory tract, leading to rhinitis, sinusitis, rhinorrhea, chronic cough, recurrent respiratory infections, infertility, and ultimately, scarring of the lungs in the form of bronchiectasis.3 PCD is associated with situs inversus in 50% of the patients, which is termed Kartagener syndrome (KS; MIM 244400).4 In humans, motile cilia have a common microtubule-based ultrastructure (axoneme), comprising nine peripheral microtubule doublets surrounding a central microtubule pair,5 which are linked to a variety of microtubule-associated proteins. These proteins include the inner dynein arm (IDA) and outer dynein arm (ODA) motor complexes, which project from the peripheral microtubule doublets; the radial spokes, which provide a radial scaffold between the central pair and peripheral microtubules to facilitate signal transduction from the center out to the dynein arms to govern ciliary beats and waveforms;6 and nexin–dynein regulatory complexes, which attach between adjacent peripheral doublets to facilitate IDA attachment and regulate dynein activity.7 Male infertility has often been described as part of the clinical symptoms of PCD, but this particular aspect of the PCD physiopathology has not been systematically explored and is generally poorly described in scientific reports. The sperm of infertile male PCD patients is usually immotile and presents various ultrastructural defects of the flagella, such as missing dynein arms, microtubular translocations, and lack of radial spokes. There has been recent research progress on PCD, contributing to the identification and characterization of the numerous proteins required for adequate axonemal molecular structure and assembly. In this review, we focus on the progress made in identifying the genes related to PCD that are directly or potentially associated with male infertility, provide an update of the genetic etiology, summarizing the proposed underlying molecular mechanisms, and discuss the clinical implications arising from these findings. COILED-COIL DOMAIN-CONTAINING 39 (CCCD39) FAP59, the Chlamydomonas ortholog of human CCDC39 (MIM 613798), which is located on chromosomal region 3q26.33 and consists of twenty exons, was predicted to be essential for motile ciliary function, as no orthologs have been found in nonciliated organisms (“CiliaCut”) or in Caenorhabditis elegans (“MotileCut”).8 Merveille et al.9 found that a substantial proportion of human PCD cases with axonemal disorganization and abnormal ciliary beating is associated with loss-of-function mutations in CCDC39. Furthermore, functional analyses indicated that CCDC39 localizes to the ciliary axonemes and is essential for the assembly of IDA and the dynein regulatory complex. Antony et al.10 sequenced the CCDC39 and CCDC40 genes in 54 “radial spoke defect” families, as these are the two genes identified so far to cause this defect. These findings highlighted a key role of both genes in the development of PCD with axonemal disorganization and IDA loss. IDA defects account for about 16% of all PCD cases, and were found in 14% up to 29% of all PCD cases in other cohorts. DYNEIN, AXONEMAL, ASSEMBLY FACTOR 1–3 (DNAAF1–3) The most frequent defects in PCD involve ODAs and IDAs, the large multiprotein complexes responsible for cilia-beat generation and regulation, respectively. Duquesnoy et al.11 investigated whether PCD might result from mutations in the human gene LRRC50 (also named DNAAF1; MIM 611088), which is located at chromosome 16q24.1 and encodes a member of the superfamily of leucine-rich repeat (LRR)-containing proteins. Functional analyses performed in Chlamydomonas reinhardtii and in another flagellated protist, Trypanosoma brucei, support a key role for DNAAF1 in the cytoplasmic preassembly of the dynein arms. Furthermore, Loges et al.12 demonstrated that DNAAF1 deficiency disrupts the assembly of distal and proximal dynein, axonemal, heavy chain 5 (DNAH5)- and dynein, axonemal, intermediate chain 1 (DNAI2)-containing ODA complexes, as well as dynein, axonemal, light chain 1 (DNALI1)-containing IDA complexes, resulting in immotile cilia. Kintoun (KTU, previously termed C14 or f104, also named DNAFF2; MIM 612517) is located on chromosome 14q21.3 and consists of three exons, encoding cDNAs of 2511 bp or 2367 bp (an in-frame splice variant lacking exon 2). This gene was first identified in a medaka mutant, and was found to be mutated in PCD patients from two affected families.13 In the absence of DNAFF2, both ODAs and IDAs are missing or defective in the axoneme, leading to general loss of motility. Biochemical and immunohistochemical studies have shown that KTU/PF13 is one of the long-sought after proteins involved in the preassembly of dynein arm complexes in the cytoplasm before being loaded for intraflagellar transport to the ciliary compartment.14 The gene DNAAF3 (MIM 614566) on human chromosome 19q13 (previously designated C19 or f51) encodes a 588-amino acid protein (GenBank NP_849159).15 Lunt et al.16 showed that PF22/DNAAF3 is essential for the preassembly of dyneins into complexes prior to their transport into cilia. Loss-of-function mutations were identified in the human DNAAF3 gene in patients from families with situs inversus, causing defects in the assembly of the inner and outer dynein arms. Moreover, DNAAF3 loss prevents correct assembly of the inner and outer dynein arms, abolishing the motility of respiratory cilia, and giving rise to classical PCD associated with defective left–right organ asymmetry and male infertility.17 DNAH5 DNAH5 (MIM 603335) was identified based on homozygosity mapping and a candidate gene approach.18 This 79-exon gene (with one alternative first exon) encodes a heavy-chain protein that localizes to the outer dynein arm and is the homolog of the dynein c-heavy chain of Chlamydomonas reinhardtii. A PCD-related locus was identified at chromosome 5p15.2, which contains DNAH5, the human ortholog of the Chlamydomonas ODA γ-heavy chain gene.19 Recessive mutations of DNAH5 result in nonfunctional DNAH5 proteins,20 and affected patients have dysmotile respiratory cilia with ODA defects. Mutations in DNAI1, encoding an ODA intermediate chain orthologous to Chlamydomonas IC78, also cause ODA defects in patients with PCD.21 In the normal ciliated airway epithelium, DNAH5 and DNAH9 show a specific regional distribution along the ciliary axoneme, indicating the existence of at least two distinct ODA types. Cilia with complete axonemal DNAH5 deficiency were immotile, whereas cilia with distal DNAH5 deficiency showed residual motility. In addition, the observation of the absence of DNAH5 within the respiratory ciliary compartment but a normal DNAH5 distribution within the sperm flagellum in a patient with DNAH5 mutations raised the possibility that the two organelle types are assembled via distinct mechanisms.22 DNAI1 The first gene in which mutations were found to be associated with PCD was DNAI1 (MIM 604366).23DNAI1 is an axonemal dynein intermediate chain gene that was isolated from Chlamydomonas reinhardtii, a unicellular alga with two flagella showing an axonemal structure similar to that of human respiratory cilia and sperm tails.24DNAI1 is localized on chromosome 9p13-p21 and is composed of twenty exons encoding a 699-amino acid protein. Guichard et al.25 demonstrated a link between ciliary function and situs determination, given that heterozygosity for a compound mutation in DNAI1 results in PCD with situs solitus or situs inversus (KS). Zariwala et al.26 found that mutations in DNAI1 cause PCD with ODA defects, and are likely the genetic origin of clinical disease in some PCD patients with ultrastructural defects in the ODA. DNAI2 DNAI2 (MIM 605483), the human ortholog of Chlamydomonas intermediate ODA chain IC69/IC2, which is located on chromosome 17q25, comprises 14 exons extending over a 39-kb genomic distance.27,28,29 Applying a positional and functional candidate gene approach, Loges et al.30 identified homozygous loss-of-function DNAI2 mutations (IVS11þ1G>A) in four individuals from a family with PCD and ODA defects. DNAI2 and DNAH5 mutations affect the assembly of proximal and distal ODA complexes, whereas DNAI1 mutations mainly disrupt the assembly of proximal ODA complexes. Mutations in an ODA intermediate dynein chain are reported in 2%–13% of all PCD patients with defined ODA defects. DYSLEXIA SUSCEPTIBILITY 1 CANDIDATE 1 (DYX1C1) DYX1C1 (MIM 608706) was initially identified as a candidate dyslexia gene due to a single-balanced translocation t(2;15)(q11;q21) that coincidentally segregates with dyslexia in a family, and was confirmed in subsequent single nucleotide polymorphism association studies.31DYX1C1 is located on chromosome 15q21.3 and comprises ten exons (translation starts at exon 2) encompassing 77.93 kb of genomic DNA. Follow-up gene association studies have provided both positive32,33,34 and negative35,36,37 support for the association with dyslexia. Molecular and cellular analyses of DYX1C1 have indicated potential functional roles with chaperonins,38,39 estrogen receptor trafficking,40 and neuronal migration.41,42 Tarkar et al.43 found that deletion of Dyx1c1 exons 2–4 in mice caused a phenotype resembling PCD. Ultrastructural and immunofluorescence analyses of DYX1C1-mutant motile cilia in mice and humans revealed disruptions of the ODA and IDA. DYX1C1 localizes to the cytoplasm of respiratory epithelial cells, its interactome is enriched for molecular chaperones, and it interacts with the cytoplasmic ODA/IDA assembly factor DNAAF2/KTU. HEAT REPEAT-CONTAINING 2 (HEATR2) HEATR2 (MIM 614864), located on chromosome 7p22.3, encodes a member of the family of ten other uncharacterized HEAT repeat-containing proteins in humans. Preliminary analyses have shown that HEATR2 gene and protein sequences are highly conserved, and that HEATR2 is enriched in organisms with motile cilia and flagella.44 Horani et al.45 demonstrated that airway epithelial cells isolated from PCD-affected individuals showed markedly reduced HEATR2 levels, absence of dynein arms, and loss of ciliary beating. Moreover, immunohistochemistry studies in human airway epithelial cells showed that HEATR2 was localized to the cytoplasm and not to the cilia, which suggests a role for this protein in either dynein arm transport or assembly. HYDIN, AXONEMAL CENTRAL PAIR APPARATUS PROTEIN (HYDIN) Homozygous-recessive Hydin mutations are lethal in the 1 st week of life in hy3 mice, due to hydrocephalus from abnormal ependymal ciliary motility.46,47,48,49 With homozygosity mapping, Olbrich et al.50 identified a PCD-associated locus at the chromosomal region 16q21-q23, which contains HYDIN (MIM 610812). However, a nearly identical 360-kb paralogous segment (HYDIN2) in the chromosomal region 1q21.1 complicated the mutational analysis. Electron microscopy tomography of HYDIN mutant respiratory cilia showed results consistent with the effects of loss-of-function mutations, in that the C2b projection of the central pair apparatus was lacking; similar findings were reported in Hydin-deficient Chlamydomonas and mice. High-speed video microscopy demonstrated markedly reduced beating amplitudes of the respiratory cilia and stiff sperm flagella. LRR-CONTAINING 6 (LRRC6) LRRC6 (MIM 614930) is expressed in the flagella of Chlamydomonas reinhardtii and in human cilia.51 In mice, Lrrc6 transcripts were found to be expressed in tissues that have flagella or motile cilia.52 In addition, a putative PCD-related locus containing LRRC6 was identified in a genome-wide linkage study performed in familial cases of PCD.53 Most importantly, the phenotypic features of several animal models with mutations in LRRC6 orthologs are consistent with ciliary defects. In zebrafish, seahorse mutants display a curved body with pronephric cysts,54 which was associated with left–right abnormalities in half of the cases.55 In Drosophila, a null allele of the LRRC6 ortholog called tilB was found to be responsible for the ciliary dysfunction of sensory neurons of the auditory organ and male sterility.56 In humans, LRRC6, which is located on chromosome 8q24.22, consists of 12 exons, and the only predicted transcript (RefSeq accession number NM_012472.3) encodes a 466-amino acid residue protein with five N-terminal LRR motifs. The LRRs of LRRC6 contain the consensus sequence LxxLxLxxNxIxxIxxLxzx Lxx (“z” indicates frequent deletions), which defines the SDS22-like subfamily of LRR-containing proteins.57 Each LRR is a beta-strand-turn/alpha-helix structure, and together, these motifs are known to form a solenoid (repeated structural units that form a continuous superhelix).58 LRRC6 also contains an LRR-cap that shields the solenoid, a coiled-coil (CC) domain, a polylysine motif, and a C-terminal alpha-crystallin p23-like domain. Kott et al.59 demonstrated that in spite of the structural and functional similarities between LRRC6 and DNAAF1, another LRR-containing protein was associated with the same PCD phenotype, the two proteins are not redundant. Therefore, the evolutionarily conserved LRRC6 emerges as an additional player in IDA assembly, a process that is essential for proper axoneme building and that appears to be much more complex than previously thought. Horani et al.60 revealed a single novel mutation in LRRC6 in PCD patients, which fits the model of autosomal-recessive genetic transmission, leading to a change of a highly conserved aminoacid from aspartic acid to histidine (Asp146His). These findings suggest that LRRC6 plays a role in dynein arm assembly or trafficking, and its mutation leads to PCD with laterality defects. RADIAL SPOKE HEAD HOMOLOGS In humans, RSPH1 (MIM609314; RSPH: radial spoke head homologs) is located on chromosomal region 21q22.3 and consists of nine exons; the only predicted transcript (RefSeq accession number NM_080860.2) encodes a 309-aminoacid residue protein with five N-terminal membrane occupation and recognition of nexus (MORN) repeats,61 followed by a linker and a sixth MORN repeat. RSPH1 is the homolog of Chlamydomonas RSP1. Kott et al.62 combined homozygosity mapping and whole-exome sequencing in a consanguineous individual with central complex defects, and identified a nonsense mutation in RSPH1. RSPH1 mutations appear to play an important role in the etiology of this PCD phenotype, which includes radial spoke defects, thereby unveiling the importance of RSPH1 in the proper building of the central complex and radial spokes in humans. Daniels et al.63 described a novel splice-site mutation (c.921+3_6delAAGT) in RSPH4A (MIM 612647), which leads to a premature translation termination signal, in nine subjects with PCD (seven families). Loss-of-function of this mutation was confirmed with quantitative ciliary ultrastructural analysis, measurement of ciliary beat frequency and waveform, and transcript analysis. All nine individuals carrying the c.921+3_6delAAGT splice-site mutation in RSPH4A were Hispanic with ancestry tracing to Puerto Rico. This mutation is considered to be a founder mutation and a common cause of PCD without situs abnormalities in patients of Puerto Rican descent. Castleman et al.64 identified mutations in two positional candidate genes, RSPH9 on chromosome 6p21.1 and RSPH4A on chromosome 6q22.1. Haplotype analysis identified a common ancestral founder effect of an RSPH4A mutation present in the United Kingdom-Pakistani pedigrees. Both RSPH9 (MIM 612648) and RSPH4A encode protein components of the axonemal radial spoke head. In situ hybridization of murine Rsph9 showed that gene expression was restricted to regions containing motile cilia. Investigation of the effect of knockdown or mutations of RSPH9 orthologs in zebrafish and Chlamydomonas indicates that radial spoke head proteins are important in maintaining normal movement in motile, “9 + 2”-structure cilia and flagella. This effect could be rescued by the reintroduction of gene expression for restoration of a normal beat pattern in zebrafish. Disturbance in the function of these genes was not associated with defects in left–right axis determination in humans or zebrafish. Onoufriadis et al.65 found that mutations in RSPH1, RSPH4A, and RSPH9, which all encode homologs of components of the “head” structure of ciliary radial spoke complexes identified in Chlamydomonas, cause clinical phenotypes that appear to be indistinguishable except at the gene level. Using high-resolution immunofluorescence, they identified loss of RSPH4A and RSPH9 associated with RSPH1-mutated cilia, suggesting that RSPH1 mutations may result in loss of the entire spoke head structure. ZINC FINGER, MYND-TYPE-CONTAINING 10 (ZMYND10) ZMYND10 (also known as BLU; MIM 607070), which is located on chromosome 3p21.3, encodes a protein containing a C-terminal myeloid, nervy, and DEAF-1 (MYND) domain. ZMYND10 is highly enriched in ciliated cells compared to nonciliated cells,66 but little is known about its function. ZMYND10 is known to act as a tumor suppressor, inhibiting the clonogenic growth of nasopharyngeal carcinoma cells, arresting the cell cycle at the G1 phase, downregulating c-Jun N-terminal kinase and cyclin D1 promoter activities, and inhibiting the phosphorylation of c-Jun.67 In mice, Zmynd10 mRNA is restricted to regions containing motile cilia. In a Drosophila model of PCD, ZMYND10 is exclusively expressed in cells with motile cilia, chordotonal sensory neurons, and sperm. In these cells, P-element-mediated gene silencing caused IDA and ODA defects, proprioception deficits, and sterility due to immotile sperm. Human ZMYND10 interacts with LRRC6, another cytoplasmically localized protein that is altered in PCD.68 Moore et al.68 concluded that ZMYND10 is a cytoplasmic protein required for IDA and ODA assembly, and that its variants cause ciliary dysmotility and PCD with laterality defects. Using whole-exome and candidate-gene Sanger resequencing in PCD-affected families afflicted with combined IDA and ODA defects, they found that 6/38 (16%) of the subjects carried biallelic mutations in ZMYND10. Zariwala et al.69 identified mutations in ZMYND10 that result in the absence of the axonemal protein components DNAH5 and DNALI1 from respiratory cilia. Animal models also support the association between ZMYND10 and human PCD, given that Zmynd10 knockdown in zebrafish caused ciliary paralysis, leading to cystic kidneys and otolith defects, and that knockdown in Xenopus interfered with ciliogenesis. Thus, a cytoplasmic protein complex containing ZMYND10 and LRRC6 is necessary for motile ciliary function. Summary and perspectives A male infertility phenotype has often been described as one of the clinical symptoms of PCD, but this particular aspect of the physiopathology has not yet been systematically explored and is often only scarcely described in scientific reports (Table 1). PCD/KS severely affects the quality of life of patients, and male infertility resulting from the disease should receive more attention. The gaining popularity and progress in assisted reproductive technology will certainly benefit infertile couples affected by this disease. However, full understanding of the underlying molecular mechanisms of PCD/KS will require further studies. This is a very exciting time in the field of the genetics of infertility. We have so far witnessed only the tip of the iceberg, but we are confident that the rest will come to light in the foreseeable future. Table 1. Primary ciliary dyskinesia gene mutations and their consequences for axonemal ultrastructure and sperm phenotype Open in a new tab AUTHOR CONTRIBUTIONS LD, YWS, and ZYJ drafted the manuscript, and PL revised it critically for important intellectual content. All authors have read and approved the final manuscript. COMPETING INTERESTS All authors declare no competing interests. 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ACTIONS View on publisher site PDF (231.9 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract INTRODUCTION COILED-COIL DOMAIN-CONTAINING 39 (CCCD39) DYNEIN, AXONEMAL, ASSEMBLY FACTOR 1–3 (DNAAF1–3) DNAH5 DNAI1 DNAI2 DYSLEXIA SUSCEPTIBILITY 1 CANDIDATE 1 (DYX1C1) HEAT REPEAT-CONTAINING 2 (HEATR2) HYDIN, AXONEMAL CENTRAL PAIR APPARATUS PROTEIN (HYDIN) LRR-CONTAINING 6 (LRRC6) RADIAL SPOKE HEAD HOMOLOGS ZINC FINGER, MYND-TYPE-CONTAINING 10 (ZMYND10) AUTHOR CONTRIBUTIONS COMPETING INTERESTS REFERENCES Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
16666	https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/14%3A_Conjugated_Compounds_and_Ultraviolet_Spectroscopy/14.03%3A_Kinetic_vs._Thermodynamic_Control_of_Reactions	Skip to main content 14.3: Kinetic vs. Thermodynamic Control of Reactions Last updated : Mar 18, 2024 Save as PDF 14.2: Electrophilic Additions to Conjugated Dienes- Allylic Carbocations 14.4: The Diels-Alder Cycloaddition Reaction Page ID : 31551 Steven Farmer, Dietmar Kennepohl, Natasha Singh, & Orthocresol (StackExchange) LibreTexts ( \newcommand{\kernel}{\mathrm{null}\,}) Objectives After completing this section, you should be able to explain the difference between thermodynamic and kinetic control of a chemical reaction; for example, the reaction of a conjugated diene with one equivalent of hydrogen halide. 2. draw a reaction energy diagram for a reaction which can result in both a thermodynamically controlled product and a kinetically controlled product. 3. explain how reaction conditions can determine the product ratio in a reaction in which there is competition between thermodynamic and kinetic control . Key Terms Make certain that you can define, and use in context, the key terms below. kinetic control thermodynamic control Like non-conjugated dienes, conjugated dienes are subject to attack by electrophiles. In fact, conjugated dienes experience relatively greater kinetic reactivity when reacted with electrophiles than non-conjugated dienes do. The reaction mechanism is similar to other electrophilic addition reactions to alkenes (Section 7.9). During the electrophilic addition of HBr to a 1,3-butadiene, However, there are two possible outcomes once the carbocation intermediate is formed. The allyl carbocation is stabilized by resonance structures that vary in the position of the carbocation . This allows the bromide ion to add to either of these carbons resulting in the formation 1,2 and 1,4 addition products. When this reaction is run at a temperature of 0 oC or lower, the 1,2 addition product dominates. However, if the same reaction is run at 40 oC the 1,4 addition product dominates. The reaction of one equivalent of hydrogen bromide with 1,3-butadiene yields different product ratios under different reaction conditions and is a classic example of the concept of thermodynamic versus kinetic control of a reaction. At lower temperatures the formation of the 1,2 and 1,4 addition products are irreversible and thus do not reach equilibrium. When a reaction is irreversible the major product is determined by the relative reaction rates and not by thermodynamic stability. Of the two products, the formation of the 1,2 addition product has a higher rate of reaction and forms faster making it the major product. The reaction product which forms with a higher rate of reaction is called the Kinetic Product and when the kinetic product dominates, the reaction is said to be under Kinetic Control. At higher temperatures the reaction to form both products becomes reversible and a reaction equilibrium is reached. When a reaction is reversible the major product is determined by thermodynamic stability. The 1,4 addition product is more stable and becomes the major product under these reaction conditions. The more stable product is called the Thermodynamic Product and when the thermodynamic product dominates, the reaction is said to be under Thermodynamic Control. The 1,4 addition product is more stable because it has an internal, disubstituted double bond , and we know that as a general rule that the thermodynamic stability of an alkene increases with increasing substitution. So, when compared to the terminal, monosubstituted alkene of the 1,2 addition product, the 1,4 addition product is expected to be more stable. A simple definition is that the kinetic product is the product that is formed faster, and the thermodynamic product is the product that is more stable. This is precisely what is happening here. The kinetic product is 3-bromobut-1-ene, and the thermodynamic product is 1-bromobut-2-ene (specifically, the trans isomer). An explanation for the temperature influence is shown in the following energy diagram for the addition of HBr to 1,3-butadiene. The initial step in which a proton bonds to carbon #1 is the rate determining step, as indicated by the large activation energy (light gray arrow). The second faster step is the product determining step, and there are two reaction paths (colored blue for 1,2-addition and magenta for 1,4-addition). At elevated temperatures, the products are more likely to have enough energy to overcome the reverse activation barrier for the second step allowing regeneration of the carbocation intermediate . Under these conditions, this step of the mechanism will be reversible and an equilibrium will be established. Since the system is no longer limited by temperature, the system will minimize its Gibbs free energy, which is the thermodynamic criterion for chemical equilibrium. This places the reaction under thermodynamic control and the most thermodynamically stable molecule , the 1,4 addition product, will be predominantly formed. If the reaction temperature is kept sufficiently low, the products will not have enough energy to overcome the reverse activation barrier to regenerate the carbocation intermediate making this step of the mechanism effectively irreversible. This reaction is under kinetic control meaning the product which forms faster, the kinetic product, will predominate. Of the two reaction pathways, the 1,2-addition has a smaller activation energy and would be expected to have a higher reaction rate than the 1,4-addition. The high reaction rate for 1,2-addition can be attributed to the formation of an ion pair during the reaction mechanism . This means that, after a double bond is protonated, the halide counterion remains in close proximity to the carbocation generated. Immediately following dissociation of HX, the chloride ion is going to be much closer to C−2 than it is to C−4, and therefore attack at C−2 is much faster. This ion pair mechanism is a pre-exponential constant effects that is attributed to the proximity and frequency of collision rather than a activation barrier effect. Experimental Evidence for the Ion-Pair Mechanism In 1979, Nordlander et al. carried out a similar investigation on the addition of DCl (deuterated hydrochloric acid) to a different substrate, 1,3-pentadiene.This experiment was ingenious, because it was designed to proceed via an almost symmetrical intermediate : Resonance forms 7a and 7b are both allylic and secondary. There is a very minor difference in their stabilities arising from the different hyperconjugative ability of C-D vs C-H bonds, but in any case, it is not very large. Therefore, if we adopt the explanation in the previous section, one would expect there not to be any major kinetic pathway, and both 1,2- and 1,4-addition products (8 and 9) would theoretically be formed roughly equally. Instead, it was found that the 1,2-addition product was favored over the 1,4-addition product. For example, at -78oC in the absence of solvent, there was a roughly (75:25) ratio of 1,2- to 1,4-addition products. Clearly, there is a factor that favors 1,2-addition that does not depend on the electrophilicity of the carbon being attacked! The authors attributed this effect to an ion pair mechanism . This means that, after the double bond is protonated (deuterated in this case), the chloride counterion remains in close proximity to the carbocation generated. Immediately following dissociation of DCl, the chloride ion is going to be much closer to C-2 than it is to C-4, and therefore attack at C-2 is much faster. In fact, normal electrophilic addition of HX to conjugated alkenes in polar solvents can also proceed via similar ion pair mechanisms. This is reflected by the greater proportion of syn addition products to such substrates. Exercise 14.3.1 1) Why is the 1,4-addition product the thermodynamically more stable product? 2) Addition of 1 equivalent of bromine to 2,4-hexadiene at 0 degrees C gives 4,5-dibromo-2-hexene plus an isomer. What is the structure of that isomer? 3) The kinetically controlled product in the below reaction is: 4) For the reaction above, which product is the result of 1,4-addition? 5) What would be the major product of the addition of HBr to 2,3-dimethyl-1,3-cyclohexadiene under thermodynamic conditions? 6) Consider the reaction with 1,3-buta-diene reacting with HCl. Propose a mechanism for the reaction. Also, Predict why the 1,4 adduct is the major product in this reaction compared to the 1,2. Answer : 1) The 1,4- product is more thermodynamically stable because there are two alkyl groups on each side of the double bond and more substituted alkenes are more stable. 2) 2,5-Dibromo-3-hexene 3) 3-Chloro-1-butene 4) 1-Chloro-2-butene Even though the cation would prefer to be in a secondary position in the transition state , the final product is less stable with a terminal alkene . Therefore the major product will be the 1,4 adduct. 14.2: Electrophilic Additions to Conjugated Dienes- Allylic Carbocations 14.4: The Diels-Alder Cycloaddition Reaction
16667	https://www.youmath.it/lezioni/fisica/gravitazione/3749-formule-gravitazione-universale.html	Formulario di Gravitazione Universale - [x] MENU Lezioni Algebra Geometria Analisi Matematica 1 Algebra Lineare Analisi 2 Probabilità Fisica Esercizi Esercizi di Algebra Problemi di Geometria Esercizi di Analisi 1 Esercizi di Algebra Lineare Esercizi di Analisi 2 Esercizi di Probabilità Risolutore Domande e risposte Forum Scuola Primaria Shop Giochi matematici Blog Formulario di Gravitazione Universale Home Lezioni Fisica Gravitazione universale In questa pagina riepiloghiamo tutte le principali formule di Gravitazione universaleche abbiamo proposto, spiegato e dimostrato nella sezione dedicata alla teoria della Gravitazione Universalepresente su YouMath.it. Formule della teoria della Gravitazione Universale Nota bene: non tutti gli argomenti, le definizioni e le applicazioni della teoria della Gravitazione Universale possono essere riassunti in una formula. ;) Vi consigliamo di utilizzare il formulario con cautela, e di consultarlo non prima di aver acquisito tutte le basi teoriche necessarie. Cliccando sui vari link potete accedere alle lezioni relative a ciascun argomento. Legge di gravitazione universale(in modulo) Legge di gravitazione universale (in forma vettoriale) Costante di gravitazione universale Accelerazione di gravità(in modulo) Accelerazione di gravità effettivasulla Terra Primo teorema dei gusci:un guscio sferico dimassauniforme attrae gravitazionalmente un punto materiale esterno come se tutta la massa del guscio fosse concentrata nel suo centro. Secondo teorema dei gusci:una particella collocata in punto qualsiasi all'interno di un guscio sferico di densità uniforme non risente di alcunaforza di attrazione gravitazionale. Prima legge di Keplero(legge delle orbite ellittiche): le orbite dei pianeti sono ellittiche e che il sole occupa uno dei due fuochi. Seconda legge di Keplero(legge delle aree): i pianeti in rivoluzione intorno al Sole spazzano aree uguali in tempi uguali. Terza legge di Keplero(legge dei periodi): il quadrato del periodo di rivoluzione di un pianeta intorno al Sole è proporzionale al cubo della distanza media del pianeta dal Sole. Campo gravitazionale Variazione di energia potenziale gravitazionale Energia potenziale gravitazionale Lavoro ed energia potenziale gravitazionale per un sistema di due corpi... ... Per un sistema di tre corpi Energia potenziale gravitazionale ed energia di legame Velocità di fuga Buchi neri e raggio di Schwarzschild Equazione di una conica in forma polare Equazione della traiettoria di un corpo rispetto a un altro corpo e relativa soluzione Energia in funzione dei parametri dell'orbita Flusso del campo gravitazionale attraverso una superficie eteorema di Gauss per il campo gravitazionale Alcuni grandezze utili (cfr: Sistema Solare. Per ulteriori spunti, cfr: Astronomia) Nome Tipo Massa (kg)Diametro medio (km)Distanza media dal Sole (in km \| in UA) Sole Stella 1,9891 x 10 30 1,39095 x 10 6 0 \| 0 MercurioPianeta terrestre 3,302 x 10 23 4,8794 x 10 3 57,9 x 10 6 \| 0,387 VenerePianeta terrestre 4,8685 x 10 24 12,1038 x 10 3 108,2 x 10 6 \| 0,723 Terra Pianeta terrestre 5,9726 x 10 24 12,7456 x 10 3 149,6 x 10 6 \| 1 LunaSatellite 7,342 x 10 22 3,476 x 10 3- \| - MartePianeta terrestre 6,4185 x 10 23 6,786 x 10 3 227,9 x 10 6 \| 1,523 GiovePianeta gassoso 1,8986 x 10 27 138,346 x 10 3 778,3 x 10 6 \| 5,203 SaturnoPianeta gassoso 5,6846 x 10 26 114,632 x 10 3 1427 x 10 6 \| 9,539 UranoPianeta gassoso 8,6832 x 10 25 50,532 x 10 3 2896,6 x 10 6 \| 19,36 NettunoPianeta gassoso 1,0243 x 10 26 49,106 x 10 3 4496,6 x 10 6 \| 30,058 PlutonePianeta nano 1,305 x 10 22 2,370 x 10 3 6089 x 10 6 \| 40,702 Il formulario di Gravitazione universale in versione pdf è disponibile qui - click per il download. Buona Fisica a tutti! Alessandro Catania (Alex) Tags: formule di teoria della gravitazione universale in pdf. Ultima modifica: 02/05/2023 Contattaci Il Team di YouMath Chi siamo Dicono di noi Termini e Privacy YouMath® - Didattica e divulgazione dal 2011. Corsi, lezioni, esercitazioni e approfondimenti per tutti. Contenuti originali ideati e scritti da matematici e fisici, non da algoritmi AI. © 2011-2025 - Math Industries Srl. P.IVA 07608320961. Informativa YouMath usa cookie tecnici (necessari per il funzionamento del sito) e cookie facoltativi, tra cui cookie analitici, cookie relativi a strumenti forniti da terze parti (necessari per il loro funzionamento) e cookie per la personalizzazione degli annunci in base alle tue preferenze di navigazione. Gli utilizzi e le finalità dei cookie sono regolati dalla privacy policy. Riguardo ai cookie facoltativi, puoi accettarli cliccando sul bottone qui di seguito e proseguire la navigazione. 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16668	https://pmc.ncbi.nlm.nih.gov/articles/PMC3354840/	Calcium and bone disorders in pregnancy - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Indian J Endocrinol Metab . 2012 May-Jun;16(3):358–363. doi: 10.4103/2230-8210.95665 Search in PMC Search in PubMed View in NLM Catalog Add to search Calcium and bone disorders in pregnancy Shriraam Mahadevan Shriraam Mahadevan 1 Department of Clinical Endocrinology, Dr. E. V. Kalyani Medical Centre, Chennai, India Find articles by Shriraam Mahadevan 1,✉, V Kumaravel V Kumaravel 1 Department of Clinical Endocrinology, Alpha Hospital and Research Centre, Madurai, India Find articles by V Kumaravel 1, R Bharath R Bharath 2 Department of Endocrinology, Chettinad Medical College and Research Institute, Chennai, India Find articles by R Bharath 2 Author information Copyright and License information 1 Department of Clinical Endocrinology, Dr. E. V. Kalyani Medical Centre, Chennai, India 1 Department of Clinical Endocrinology, Alpha Hospital and Research Centre, Madurai, India 2 Department of Endocrinology, Chettinad Medical College and Research Institute, Chennai, India ✉ Corresponding Author: Dr. Shriraam Mahadevan, Department of Clinical Endocrinology, Dr. E. V. Kalyani Medical Centre, 4, Radhakrishnan Salai, Mylapore, Chennai 600004, India. E-mail: mshriraam@gmail.com Copyright: © Indian Journal of Endocrinology and Metabolism This is an open-access article distributed under the terms of the Creative Commons Attribution-Noncommercial-Share Alike 3.0 Unported, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. PMC Copyright notice PMCID: PMC3354840 PMID: 22629499 Abstract Significant transplacental calcium transfer occurs during pregnancy, especially during the last trimester, to meet the demands of the rapidly mineralizing fetal skeleton. Similarly, there is an obligate loss of calcium in the breast milk during lactation. Both these result in considerable stress on the bone mineral homeostasis in the mother. The maternal adaptive mechanisms to conserve calcium are different in pregnancy and lactation. During pregnancy, increased intestinal absorption of calcium from the gut mainly due to higher generation of calcitriol (1,25 dihydroxy vitamin D) helps in maintaining maternal calcium levels. On the other hand, during lactation, the main compensatory mechanism is skeletal resorption due to increased generation of parathormone related peptide (PTHrP) from the breast. Previous studies suggest that in spite of considerable changes in bone mineral metabolism during pregnancy, parity and lactation are not significantly associated with future risk for osteoporosis. However, in India, the situation may not be the same as a significant proportion of pregnancies occur in the early twenties when peak bone mass is not yet achieved. Further, malnutrition, anemia and vitamin D deficiency are commonly encountered in this age group. This may have an impact on future bone health of the mother. It may also probably provide an opportunity for health care providers for prevention. Other metabolic bone diseases like hypoparathyroidism, hyperparathyroidism and pseudohypoparathyroidism are rarely encountered in pregnancy. Their clinical implications and management are also discussed. Keywords: Bone mineral, calcium, lactation, osteoporosis, pregnancy, vitamin D I NTRODUCTION In the non-pregnant state, calcium homeostasis is maintained mainly by the intricate inter-relationship between parathormone (PTH) and vitamin D. PTH secreted from the parathyroid glands in response to a fall in serum calcium level mediates skeletal resorption of calcium through its effect on osteoclasts. It also enhances renal tubular reabsorption of calcium by stimulating 1-alpha hydroxylase in the kidneys that promotes conversion of 25 hydroxy vitamin D to 1,25 dihydroxy vitamin D[1,25(OH)2 D]. This in turn helps in intestinal absorption of calcium. The calcium physiology and maternal adaptations to meet the rising demand of calcium from the fetus are discussed below. C ALCIUM H OMEOSTASIS D URING P REGNANCY: P HYSIOLOGY Calcium homeostasis in pregnancy is slightly different from that of non-pregnant state to meet the calcium demands of the mother and fetus. Calcium At birth, about 30 g of calcium is present in a term neonate.[1,2] This amount of calcium is actively transferred across the placenta and most of it occurs during the third trimester when the collagen matrix is rapidly ossified. Total calcium level is decreased during pregnancy due to hemodilution associated low albumin. Albumin-corrected calcium and ionized calcium values remain normal throughout pregnancy. Theoretically, the calcium demand may be met by increased resorption of maternal skeleton, increased absorption or decreased urinary excretion. Much of the calcium conservation observed during pregnancy is due to increased intestinal absorption of calcium. This occurs mainly due to the increased generation of 1,25 (OH)2 D.[2,3] Compared to non-pregnant state, the 24-hour urinary excretion of calcium during pregnancy is higher but fasting urinary calcium levels are similar. Hence, it is likely to be a reflection of increased absorption of calcium (absorptive hypercalciuria) and to a lesser extent due to higher calcitonin levels seen in pregnancy. Other minerals Serum levels of magnesium and phosphorus are usually within normal limits during pregnancy.[1,2] Parathormone Previously, the assays for PTH were less specific and the levels were found to be high during pregnancy, leading to an erroneous assumption of a state of “hyperparathyroidism” during pregnancy. However, the modern immunometric assays that are specific for two sites and measure the intact PTH clearly show that the PTH levels are relatively low during the first trimester and remain normal through the rest of pregnancy. This relatively low PTH may be due to the suppressive effect of raised 1,25 (OH)2 D levels.[2,4,5] Parathormone related peptide Parathormone related peptide (PTHrP) is a prohormone that produces multiple N-terminal, mid-molecule and C-terminal peptides which differ in their biological activities and specificities. However, none of these peptides have been systematically measured during pregnancy. The most studied is the large molecule which comprises 1–86 amino acids. The levels start rising usually around the mid-second to third trimester of pregnancy. It has several sources both from the fetus and the mother, including breast, myometrium, decidua, amnion and fetal parathyroids. Which one contributes more to the elevated level seen in pregnancy is unclear. Several roles of PTHrP are postulated from animal studies, including fetal calcium transfer and stimulating 1-alpha hydroxylase activity. Further, the carboxy terminal of PTHrP called “osteostatin” may suppress osteoclastic activity and may have a possible bone protection role in the mother.[1,4,5] Calcitonin Circulating levels of calcitonin are high during pregnancy. The most likely sources are hypertrophied C cells of thyroid and possibly from breast and placenta. Though postulated to affect the maternal bone, human studies have not convincingly shown any significant effect of calcitonin on calcium metabolism during pregnancy.[1,2] Vitamin D Serum 1,25 (OH)2 D level is increased during pregnancy up to twice the upper limit observed in non-pregnant state.[1,6] The total 1,25 (OH)2 D level (bound to vitamin D binding protein + free unbound fraction) may be elevated due to the increased binding globulins seen in pregnancy. However, studies have shown that free 1,25 (OH)2 D level is also raised and is likely to be a result of increased production rather than decreased clearance. Transplacental transfer of vitamin D takes place and the fetal levels are around 20% less than the maternal level as studied from cord blood. This rise in 1,25 (OH)2 D level is largely PTH independent and is mainly due to elevated 1-alpha hydroxylase activity in the maternal kidneys. High PTHrP, estrogen, prolactin and human placental lactogen may also augment the enzyme activity. Placenta and fetal kidneys may also be additional sources.[1,6,7] Other hormones During pregnancy, various other hormones like estrogen, progesterone, growth hormone (human placental lactogen) and insulin-like growth factor 1 (IGF-1) which affect bone mineral homeostasis are increased. However, systematic studies addressing the role of each in the feto-maternal calcium physiology have not been done.[1,2] B ONE C HANGES D URING P REGNANCY Bone histomorphometric studies in animals suggest a slight increase or no significant change in bone mineral content during pregnancy. Systematic studies of bone biopsy and histomorphometry in human pregnancy are not available.[1–3] Serum markers of bone turnover have been studied in human pregnancy, but as in non-pregnant state, they also have several caveats in interpretation during pregnancy, viz. lack of normal values during pregnancy, correction for hemodilution, increased glomerular filtration rate,diurnal variation and difficulty in obtaining fasted samples. Further, contribution of these markers from the fetus and placenta cannot be excluded. Placental secretion of alkaline phosphatase results in elevation of total alkaline phosphatase and hence cannot be used as a bone marker in pregnancy. Overall, the bone marker studies do not confirm significant resorption during pregnancy. This again points to the main adaptive mechanism of calcium conservation during pregnancy, being increased intestinal absorption. Bone mineral density (BMD) studies by Dual Energy X-ray Absorptiometry (DXA) scan or its other versions are contraindicated during pregnancy. Few studies in pregnant women where it was done immediately after delivery or after an abortion at various periods of gestation showed variable results precluding any concrete conclusions. Further, change in bone volume and body composition during pregnancy may also interfere with bone density estimation by DXA.[1,3] Ultrasound estimation of bone density at calcaneus has been done with variable results and whether it truly reflects the changes in the axial skeleton is unclear. Overall, the existing studies have insufficient power to clarify whether any significant bone loss occurs during pregnancy. It may be reasonable to decipher that pregnancy does not impair skeletal strength or density. Few epidemiological studies in post-menopausal osteoporosis have not demonstrated a relationship between parity and bone density.[1,8–11] C ALCIUM H OMEOSTASIS D URING L ACTATION Calcium During lactation, the mother undergoes a continued stress on calcium demand with production of breast milk. On an average, 300–400 mg of calcium is lost in breast milk daily. Except in mothers with vitamin D deficiency and low serum calcium levels, the calcium concentration in milk does not change.[1,12] In contrast to the pregnant state, this demand is met mainly by increased resorption of calcium from the bone and partly by increased reabsorption from the kidneys.[1–3] Both the effects are mediated by very high levels of PTHrP secreted by the breast and not by PTH. Intestinal absorption of calcium returns to pre-pregnant levels after delivery. Other minerals/ions Phosphorus level is marginally elevated above normal during lactation and is likely to be related to the dietary flux and decreased clearance seen during lactation. Magnesium level is not altered in lactation. Parathormone related peptide The level of this hormone increases by more than 1000-fold seen in pregnant state and the main source is the breast. Though it is biologically weak when compared to PTH, the high levels result in bone resorption and increased tubular reabsorption of calcium and also suppress PTH. Studies from hypoparathyroid mothers show that the calcium and active vitamin D requirement may significantly come down during lactational period.[1,2,4,5] Parathormone Due to high levels of PTHrP during pregnancy, PTH remains low during lactation and gradually rises to normal level once weaning is started.[1,4,5] Calcitonin The high serum calcitonin levels seen during pregnancy gradually fall to normal few weeks postpartum. Its role in bone physiology in human pregnancy is unclear.[1,2] B ONE C HANGES D URING L ACTATION Loss of bone mass during lactation occurs mainly due to elevated PTHrP as well as hypoestrogenic state associated with high prolactin levels. BMD data suggest a 2–3% loss of bone per month during lactation. This is very significant as compared to 1–3% bone loss per year that occurs in menopause.[1,3] Overall, observations suggest that during lactation, the obligate loss of calcium in the milk and PTHrP effect contribute more to bone resorption than hypoestrogenemia. Total alkaline phosphatase level drops to pre-pregnant levels immediately after delivery of the placenta but may be slightly elevated due to increased bone turnover.[1–3] Cross-sectional and longitudinal studies on bone turnover markers in lactation show an overall predominance of resorption markers. However, bone formation occurs rapidly and the density and turnover markers normalize rapidly within 2–6 months of weaning. Hence, in an otherwise healthy woman, the pregnancy and lactation induced bone changes may not have long-term effect on the skeletal health.[1,3,8,9,13] V ITAMIN D D EFICIENCY IN P REGNANCY Maternal vitamin D deficiency is associated with detrimental effects on the fetus/infant as well as complications for the mother during pregnancy. Fetal and neonatal risks include intrauterine growth retardation, neonatal hypocalcemic seizures, impaired postnatal growth, rickets in infancy and cardiomyopathy.[12,14,15] Future risk of immune-mediated conditions like atopy, asthma and type 1 diabetes may also have a relation to vitamin D deficiency. Maternal adverse outcomes include possible increased risk of preeclampsia, gestational diabetes and increased rate of cesarean section.[12,16] In spite of being a tropical country with abundant sunshine, vitamin D deficiency is quite prevalent in India.[17–19] A study on 207 mothers from rural and urban North India showed a prevalence of 83.6% and 84.3% of vitamin D deficiency (vitamin D deficiency defined in that study as <22.5 ng/ml), respectively. As discussed above, physiological studies mainly from the developed countries show that the impact of the huge calcium demand on the mother during pregnancy and lactation is reversible and usually does not affect the maternal skeleton significantly. Also, in a vitamin D and calcium replete healthy mother, it is unlikely to have an effect on the future bone heath, especially osteoporosis, post-menopause. However, in our Indian setting of poor maternal nutrition and early pregnancy before peak bone mass being achieved coupled with severe vitamin D deficiency, the scenario may be different and further studies are needed to clarify the long-term impact on osteoporosis.[17,18,20] After extensive review of literature, the recent Endo Society guidelines recommend that the 600 IU of vitamin D that was the previous recommended daily allowance for a pregnant mother is not sufficient. To meet the demands of the growing fetus and to maintain the vitamin D level above the currently accepted optimum of >30 ng/ml, the mother needs to take at least 1500–2000 units of vitamin D (cholecalciferol) per day. If she is already on a prenatal multivitamin pill containing 400 IU of vitamin D, an additional 1000 units may be supplemented. Similarly during lactation, if the child is not on any supplementation, the mother needs to be on at least 1500–2000 units of vitamin D to satisfy the needs of the suckling infant and protect her bones with an optimal vitamin D level of >30 ng/ml. O STEOPOROSIS IN P REGNANCY Osteoporosis is difficult to diagnose during pregnancy as DXA scan, the gold standard for diagnosing the condition, cannot be done during pregnancy. Occasionally, a pregnant woman may present with severe low back ache and may be diagnosed to have a vertebral fracture or osteoporosis. Pre-existing low bone mass or conditions which may affect bone health like low calcium intake and vitamin D deficiency cannot be ruled out in many such instances unless prior information is available.[8,9] In a series of 35 women with osteoporosis in pregnancy, majority had a maternal history of fracture, possibly implicating a genetic basis. However, pregnancy itself is a state of relatively high bone turnover and may thus cause further deterioration in bone density in an already predisposed individual. Another point in favor of the role of pregnancy in this condition is that most of the cases recover within few months of delivery.[1,9–11] Clinical manifestation usually includes severe back pain especially in the lumbar area and may be associated with collapse of the vertebrae. Usually, it presents in the last trimester or puerperal period of the first pregnancy. Management is conservative and most of the patients recover clinically and radiologically in 3–6 months postpartum. It usually does not recur in the subsequent pregnancies. Though case reports of the usage of potent anti-osteoporotic agents like bisphosphonates, PTH analogues, and calcitonin have been reported, the reversible nature of the condition does not justify their routine use. They may be reserved for the most severe cases if at all.[9–11,21] Transient osteoporosis of hip (TOH): This is a distinct condition seen in pregnancy, in contrast to that discussed above. It is also called algodystrophy of the hip.[1,10,22] The exact incidence of the condition is not known. The pathophysiology is more related to local factors and the various hypotheses include venous stasis due to gravid uterus, bone marrow edema/hypertrophy, reflex sympathetic dystrophy, compression of obturator nerve by the fetus, ischemia, trauma and viral infections. Clinical diagnosis is suspected when the patient presents with low back ache, unilateral or bilateral hip pain, limp or rarely with a hip fracture usually in the third trimester.[1,10,23] Vitamin D deficiency with or without calcipenia may also need to be considered in the differential diagnosis. Magnetic resonance imaging (MRI) shows reduced bone density, increased water content especially in the femoral head and even joint effusion occasionally. Radionuclide imaging in postpartum situation may be helpful. In TOH, the femoral head takes up the agent; but in osteonecrosis of the femoral head, there is no uptake. This condition also usually recovers in 3–6 months both clinically and radiologically. However, recurrence of up to 40% has been reported in subsequent pregnancies. Bisphosphonates are contraindicated in pregnancy. However, use of intravenous pamidronate/zoledronic acid or oral alendronate in the postpartum period in women who did not show good recovery have been reported.[21,25] Rarely, in severe cases, especially with fracture, surgical intervention including total hip arthroplasty may be required. From the obstetric point of view, in patients with bilateral involvement, cesarean section may be indicated. Infrequently, women may present in the early part of lactation with fragility fracture.[1,8,9] Like in pregnancy, a pre-existing low bone mass state cannot be ruled out. However, pathophysiologically, lactation is associated with more bone depletion mainly due to high levels of PTHrP as discussed above. Cessation of lactation helps in recovery along with adequate calcium and vitamin D supplementation as per the recent guidelines. Specific bone active agents like bisphosphonates may be used in the most severe cases. H YPERPARATHYROIDISM IN P REGNANCY Hypercalcemia is rarely encountered in pregnancy. The commonest cause of hypercalcemia in pregnancy is hyperparathyroidism.[26,27] It is associated with significant morbidity to the fetus and mother in more than two-thirds of the cases. Adverse fetal outcomes include increased rate of abortions, severe intrauterine growth retardation and still birth. Since the fetal parathyroids are suppressed during pregnancy in hyperparathyroidism, once the cord is clamped after delivery, the calcium level drops precipitously in the neonate and the suppressed parathyroids may not be able to respond well. This leads to severe hypocalcemic tetany and seizures requiring prolonged neonatal care. Conversely, mild to moderately severe hyperparathyroidism that is not diagnosed in the mother during pregnancy is detected only when neonatal hypocalcemia is evaluated.[2,3,26–28] Total calcium estimation may pose some diagnostic difficulties during pregnancy as it is low in normal pregnancy. Any value of ionized calcium or albumin-corrected calcium higher than normal in pregnancy may need to be evaluated further. In the setting of high prevalence of vitamin D deficiency, even high normal calcium may carry significance. As discussed above, PTH levels are low to mid-normal in pregnancy and higher than normal values in the background of high calcium may point to the diagnosis of primary hyperparathyroidism. Classical presentation of primary hyperparathyroidism like bone pains, fracture or renal stones occurs less commonly in pregnancy and many cases are picked up only during routine evaluation with calcium profile or rarely when they present with hypercalcemic crisis. Symptoms of hypercalcemia like nausea, vomiting, malaise and drowsiness may be attributed to those of hyperemesis gravidarum seen in pregnancy, unless very severe and investigated further. Since significant calcium transfer to fetus occurs during pregnancy, severe hypercalcemia may not occur. However, in the postpartum period, the patient may present with severe hypercalcemia which is further aggravated by the high PTHrP during lactation.[1,26–28] Surgical management of the parathyroid adenoma during second trimester is recommended by many authors due to the associated significant adverse fetal outcome. This may not be always feasible in many settings and medical management is necessary. In mild cases, hydration alone may ameliorate high calcium levels. In cases with more severe hypercalcemia, calcitonin may be tried as it does not cross the placenta.[1,2] Bisphosphonates are very effective in reducing calcium levels and are the agents of choice in non-pregnant setting. However, they are contraindicated in pregnancy. Oral phosphates or phosphate enema are modestly effective in reducing calcium levels. But side effects like diarrhea and hypokalemia need to be monitored. Use of cinacalcet, which is a calcium sensing receptor agonist, to suppress the parathyroids has been reported, but concerns regarding adverse effects on the fetus are not clear. Surgical removal of the parathyroid tumor in second trimester seems to be the most definitive treatment in severe cases. Women who have been conservatively managed during pregnancy also need to undergo surgical treatment in the postpartum period. H YPOPARATHYROIDISM IN P REGNANCY It is not uncommon to encounter hypoparathyroidism as a pre-existing condition during pregnancy.[1,2,30,31] Diagnosing hypoparathyroidism for the first time in pregnancy may pose some problems as calcium levels are usually low during normal pregnancy. However, ionized calcium whose levels remain normal during pregnancy may help in confirming the diagnosis. The principle of management of hypoparathyroidism during pregnancy is to maintain near normal calcium in the mother to prevent fetal hyperparathyroidism which has serious consequences including fetal death. In a patient on treatment for hypoparathyroidism, the requirement of calcitriol and calcium may come down during the latter half of pregnancy and more so during lactation due to the effects of PTHrP. Hence, close monitoring of calcium levels to titrate the dosage is mandatory to prevent adverse fetal consequences. Inadvertent excessive use of calcitriol may result in hypercalcemia. P SEUDOHYPOPARATHYROIDISM IN P REGNANCY Pseudohypoparathyroidism (PHP) is characterized by resistance to the action of PTH and is very rarely encountered in pregnancy. Hence, the management is less well documented. Reports on requirements of calcitriol and calcium are variable. In some cases, calcitriol requirement reduced presumably due to increased generation from non-PTH/PTHrP dependent sources like placenta. The management goal here again is to prevent maternal hypocalcemia which may cause fetal hyperparathyroidism. During lactation, since the placental source of 1,25 (OH)2 D is lost and PHP is associated with resistance to renal action of PTHrP, the dosage of calcium and calcitriol usually reverts to pre-pregnant levels.[1,34] C ONCLUSIONS Maternal adaptations differ between pregnancy and lactation to meet the mineral demands of the growing fetus. Increased intestinal absorption of calcium during pregnancy and skeletal resorption of calcium during lactation form the main maternal adaptive mechanisms to meet the raised requirement. In an otherwise healthy pregnant woman, the mild bone resorption which occurs during pregnancy and lactation is rapidly reversed after weaning, resulting in nearly no significant effect on the bones. However, it is likely that maternal malnutrition and vitamin D deficiency, as in our Indian scenario, may lead to more severe skeletal depletion during this reproductive period and may probably have long-term effects on bone health, including an increased risk of skeletal fragility. Further elucidation of the mechanisms of bone loss and restoration during pregnancy and lactation may help evolve newer avenues of management of metabolic bone diseases and osteoporosis. Currently, in our population, we need to focus our emphasis on maternal nutrition, especially adequate vitamin D and calcium intake, which may pave way in the long run for prevention of future bone health related conditions like osteoporosis. Footnotes Source of Support: Nil Conflict of Interest: None declared. R EFERENCES 1.Kovacs CS, Kronenberg HM. Maternal-fetal calcium and bone metabolism during pregnancy, puerperium, and lactation. Endocr Rev. 1997;18:832–72. doi: 10.1210/edrv.18.6.0319. 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[DOI] [PubMed] [Google Scholar] Articles from Indian Journal of Endocrinology and Metabolism are provided here courtesy of Wolters Kluwer -- Medknow Publications ACTIONS View on publisher site PDF (480.7 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract INTRODUCTION CALCIUM HOMEOSTASIS DURING PREGNANCY: PHYSIOLOGY BONE CHANGES DURING PREGNANCY CALCIUM HOMEOSTASIS DURING LACTATION BONE CHANGES DURING LACTATION VITAMIN D DEFICIENCY IN PREGNANCY OSTEOPOROSIS IN PREGNANCY HYPERPARATHYROIDISM IN PREGNANCY HYPOPARATHYROIDISM IN PREGNANCY PSEUDOHYPOPARATHYROIDISM IN PREGNANCY CONCLUSIONS Footnotes REFERENCES Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
16669	https://www.tiger-algebra.com/en/solution/linear-equation-one-unknown/m-1/2/1+1/2m=1/	Copyright Ⓒ 2013-2025 tiger-algebra.com This site is best viewed with Javascript. If you are unable to turn on Javascript, please click here. Solution - Linear equations with one unknown Other Ways to Solve Video Explanation Step-by-step explanation 1. Simplify the expression m+-12+12m=1 Group like terms: (m+12m)+-12=1 Group the coefficients: (1+12)m+-12=1 Convert the integer into a fraction: (22+12)m+-12=1 Combine the fractions: (2+1)2m+-12=1 Combine the numerators: 32m+-12=1 2. Group all constants on the right side of the equation 32m+-12=1 Add 12 to both sides: (32m+-12)+12=1+12 Combine the fractions: 32m+(-1+1)2=1+12 Combine the numerators: 32m+02=1+12 Reduce the zero numerator: 32m+0=1+12 Simplify the arithmetic: 32m=1+12 Convert the integer into a fraction: 32m=22+12 Combine the fractions: 32m=(2+1)2 Combine the numerators: 32m=32 3. Isolate the m 32m=32 Multiply both sides by inverse fraction 23: (32m)·23=(32)·23 Group like terms: (32·23)m=(32)·23 Multiply the coefficients: (3·2)(2·3)m=(32)·23 Simplify the fraction: m=(32)·23 Multiply the fraction(s): m=(3·2)(2·3) Simplify the fraction: m=1 How did we do? Why learn this Linear equations cannot tell you the future, but they can give you a good idea of what to expect so you can plan ahead. How long will it take you to fill your swimming pool? How much money will you earn during summer break? What are the quantities you need for your favorite recipe to make enough for all your friends? Linear equations explain some of the relationships between what we know and what we want to know and can help us solve a wide range of problems we might encounter in our everyday lives. Terms and topics Related links Latest Related Drills Solved Copyright Ⓒ 2013-2025 tiger-algebra.com
16670	https://www.youtube.com/watch?v=Pfoy2IW7oHw	Surface area of an octagonal pyramid \| Geometry (TX TEKS) \| Khan Academy Khan Academy 9090000 subscribers 30 likes Description 1569 views Posted: 6 Sep 2025 Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now! Given a right pyramid with a regular octagon as the base, let's find the lateral and total surface area. Since the base is a regular octagon, we can use trigonometric ratios to help find the areas of the base and the lateral faces. Khan Academy is a nonprofit organization with the mission of providing a free, world-class education for anyone, anywhere. We offer quizzes, questions, instructional videos, and articles on a range of academic subjects, including math, biology, chemistry, physics, history, economics, finance, grammar, preschool learning, and more. We provide teachers with tools and data so they can help their students develop the skills, habits, and mindsets for success in school and beyond. Khan Academy has been translated into dozens of languages, and 15 million people around the globe learn on Khan Academy every month. As a 501(c)(3) nonprofit organization, we would love your help! Donate or volunteer today! Donate here: Volunteer here: Transcript: We're told the base of the following right pyramid is a regular octagon and we see that in the base right over here. They want us to figure out what is the lateral the lateral surface area of the pyramid and what is the total surface area of the pyramid. So why don't you go ahead pause this video have a go at that and I'll I'll warn you it is a little bit hairy. All right now let's do this together. So first let's think about the lateral surface area. So if we could figure out the surface area of each of these faces right over here. There's eight of them. So that we can just multiply by eight. But how do we do that? Well, we know the base of each of those faces. It is 6 mters. So we need to figure out the height of each of those bases. So this is the height over here. I'm trying to handdraw it. Not so well. All right. So we need to figure out what that height is of each of the faces. And I'll call that h. Not to be confused with the height of the pyramid, which they actually gave us. This is the height of each of those faces. So, how do we figure that out? Well, you might imagine there's a little bit of trigonometry involved. If I knew if I knew what this length was, then I could use the Pythagorean theorem to figure out our h there. Now, how could I figure out this length that I just drew? Maybe I'll do it in a different color. I'll do it in deep blue right here. How could I figure out this blue length? Let me call that X. Well, I could imagine each of these triangles, each of these eight triangles that make up that regular octagon. So, let me just draw it right over here. So, this is a triangle that's making up that regular octagon. I want to figure out the height of this triangle. That is going to be x. I know the base of this triangle is 6 m. I also know what the some of the angles of this triangle. How do I figure that out? Well, if I were to go all the way around this octagon, it would be 360°. But this total angle right over here is 360 divided by 8. So if I take 8 into 360, 8 goes into 36 four times. 4 8 is 32. Subtract 36 - 32 is 40. 8 goes into 40 five times. 5 8 is 40. And I'm done. So this whole thing would be 45°. And this part right over, let me do this in a different color. This part right over here in red. That is going to be 45 over 2°. This is a regular octagon. Each of these triangles are going to be isoclesles triangles. If I were to drop this perpendicular like that, it directly splits the angle. that also splits this side into two. So now I can use a little bit of trigonometry. If I think about this angle, I know the opposite side here, it's going to be three. This is a right triangle. So what involves opposite and adjacent? Well, tangent involves opposite and adjacent. So, if I said the tangent of 45 /2, that's the same thing as 22.5°, is going to be equal to the opposite side, 3 overx. Well, if I wanted to solve for x, I could multiply both sides by x and divide both sides by tangent of 22.5. And so I would get x is equal to 3 over the tangent of 22.5°. And I'll leave it like that for now. We have figured out x, which I said I would do in blue, but I somehow switched into red right over there. Well, we know what x is. We know what the height of this pyramid is. So we can use a pagorean theorem to figure out our h. We could write that h2 h ^2 is going to be equal to 9^2. So that is going to be equal to 81 plus this business over here squared plus 9 over tangent of 22.5 squared. All I did is I squared this. I said this squared plus that squared is going to get equal that squared. Or I could say that our height is going to be equal to the square root, this is getting a little hairy, 81 + 9 over the tangent of 22.5° squared. Now, why was that useful? Well, if I know the base and the height and I multiply that times 1/2, I get the area of one of these triangles on the that's part of the lateral surface area. And if I multiply that by eight, then I'll get all of them. So let's do that. So the lateral surface area right over there, lateral surface area is going to be equal to I have eight of these triangles. And then each of their areas is 1/2 times the base. So times 6 meters times this height that's the hairy part times the square t of 81 + 9 over the tangent of 22.5 squared. That is the lateral surface area. Now if I wanted to find the total surface area, what would I do? Well, I'd have to add the area of this octagon right over there. And how do I figure that out? Well, I know the area or I can figure out the area of each of these triangles. I know its base and I know its height. Each of these triangles is going to have an area. So, let me just write this is going to be I'll just rewrite it over here. It's going to be 8 1/2 6. So, that's 4 6. So I'll just write 24 the square of 81 + 9 over tangent of 22.5° squared plus plus we're going to get 8 of these 8 the area of each of them is 12 times the base which is 6 times the height which we already figured out was 3 over tangent of 22.5°. So 8 12 6 is 24 which and then you multiply that times let me make sure I'm doing that 8 1/2 is 4 6 is 24 3 I get 72. So this part over here and I'm squinging it up is going to be 72 over the tangent of 22.5°. If you were to enter this expression for lateral surface area into your calculator and round to the nearest square meter, this is going to be approximately 277 square meters. And then if you were to take this total that has both the lateral and the area of the base over here and put that into your calculator, and I'm not doing this because it's just going to take so much time, but I think you know how to enter into a calculator. This is going to be approximately 451 square meters.
16671	https://www.instagram.com/reel/DNG7ZBDSl15/	Instagram Log In Sign Up rolisonpg • Follow Original audio rolisonpg7w Merriam-Webster definition for Camaraderie: a spirit of friendly good-fellowship #OTK#OTKUSA#Kosmic#KosmicKart#TonyKart#RolisonPerformanceGroup#RPG#AVPEngines No comments yet. Start the conversation. 285 likes August 8 Log in to like or comment. More posts from rolisonpg See more posts Meta About Blog Jobs Help API Privacy Consumer Health Privacy Terms Locations Instagram Lite Meta AI Meta AI Articles Threads Contact Uploading & Non-Users Meta Verified English © 2025 Instagram from Meta See more from rolisonpg See photos, videos and more from Rolison Performance Group. Sign up for Instagram Log in By continuing, you agree to Instagram's Terms of Use and Privacy Policy.
16672	https://ntrs.nasa.gov/api/citations/19880004710/downloads/19880004710.pdf	NASA Technical Memorandum 86705 Aerodynamic and Propeller Performance Characteristics of a Propfan-Powered, Semispan Model Alan D. Levin, Ronald C. Smith, and Richard D. Wood December 1985 Date for general release December 1987 National Aeronautics and Space Ad mi n 1 s t ration NOMENCLATURE t . CD CL C DSS ' m CN cP C C PProP T~~~ thrust removed drag coefficient, drag/qS thrust removed lift coefficient, lift/qS slipstream interference drag coefficient thrust removed pitching moment coefficient, pitching-moment/qSE normal force coefficient, normal-force/qS pressure coefficient, (p - & ) / q propeller power coefficient , power/pN3D5 propeller net thrust coefficient, net-thrust/qS propeller thrust coefficient, thrust/pN 2 4 D ' ' p r o p axial force coefficient, axial-force/qS CX hub-forebody axial force coefficient, forebody-axial-force/qS hub-base axial force coefficient, hub-base-axial-force/qS xH X~~ C C C mean aerodynamic chord, 0.702 m - D propeller diameter, 0.622 m EPR exhaust pressure ratio, pte/pse J advance ratio, 60V/ND HP horsepower, W M,MACH Mach number N propeller rotational speed, rpm P local static pressure, kPa average nozzle exit static pressure, kPa average nozzle exit total pressure, kPa pse ' t e P U J free-stream static pressure, kPa iii PREZEDINO PAGE BLANK NOT FILMED Q 9 RN I S V x/c I CD xN Ac I NET torque, N-m dynamic pressure, kPa - Reynolds number based on c reference area, 1.434 m 2 free-stream velocity, m/s local chord station, fraction of chord measured from the baseline wing leading edge nacelle station, m angle of attack, deg propeller blade pitch angle, deg total installation drag coefficient nacelle buoyancy axial force coefficient droop of propeller axis, deg toe-in of propeller axis, deg span station, fraction of semispan propeller net efficiency P density , kg/m3 6 nacelle azimuth station, positive clockwise (see fig. 2), deg Subscripts BAL JET jet TR thrust removed Configuration codes B body ba 1 an c e I I I F3 fillet iv c wing leading-edge extension (LEX) nacelle propeller strake wing V SUMMARY T A semispan wing/body model with a powered propeller has been tested to provide data on the total power plant installation drag penalty of advanced propfan-powered aircraft. The test conducted in the Ames Research Center's 11-Foot Transonic Wind Tunnel, i s a part of a National Aeronautics and Space Administration (NASA) program to develop the technology for fuel efficient, high-speed, propeller-driven air- craft. The test objectives were: ( 1 ) to determine the total powerplant installa- tion drag penalty on a representative propfan aircraft; (2) to study the effect of configuration modifications on the installed powerplant drag; and ( 3 ) to determine performance characteristics of an advanced-design propeller which was mounted on a representative nacelle in the presence of a wing. The semispan wing/body model consisted of a swept, supercritical wing mounted low on the body. tion. The model had instruments for measuring propeller forces, total configuration forces and moments, and pressure distributions over the wing and nacelle. Other measurements included unsteady propeller blade stresses and acoustic pressures on the body. Only the force data and pressure distributions are presented in this report. number range 0.6 to 0 . 8 . resulting in Reynolds numbers between 7.8 million and 9.5 million. was designed to assess jet-off and jet-on nacelle installation drag, propeller slipstream interference drag, and total powerplant installation drag. The nacelle was mated to the wing in an under-the-wing configura- The test was conducted at angles of attack from -3" to 4" over the Mach Tunnel total pressure was held constant at one atmosphere, The test program Test results indicated that the total powerplant installation drag penalty can be as high as 77 drag counts (0.0077). However, the penalty was reduced to 18 drag counts by the addition of a wing leading edge extension, between the nacelle and body, in combination with a fillet and strake at the wing-nacelle intersections. INTRODUCTION Jet fuel cost has become a dominant component of the direct operating cost of transport aircraft. As a result, there i s growing interest in alternate propulsion system concepts having improved fuel efficiency. One of the primary condidates i s the highly loaded, high-speed propeller, typically referred to as the propfan. Several system studies (refs. 1-6) have indicated that a propfan-powered aircraft operating at M = 0.8 could achieve a 10% to 30% saving in fuel relative to a comparable turbofan-powered aircraft. (Near M = 0 . 8 flight speeds are being considered to ensure compatibility with existing airline operations.) At these speeds, recently developed propeller designs can provide efficient performance. 1 However, the fuel saving suggested by the system studies can be realized only if the propulsion system is properly integrated with the airframe. In addition to generating the technology base from which airframe manufacturers can realiably design propfan aircraft, several technical issues must be resolved. Among these i s the aerodynamic integration of the powerplant with the wing. the concerns about the propfan installation i s the interference drag that may result from integration of the nacelle and propeller on a supercritical wing. tant interference effects are dependent upon the nacelle/wing interactions along with the Mach number and swirl increment generated in the slipstream. A slipstream simulator test was conducted in 1 9 7 8 to experimentally evaluate these interference effects (ref. 7 ) . A flow-through, ejector-powered nacelle located ahead of the wing was used to simulate the propeller slipstream flow. This approach permitted simula- tion of various propeller parameters to obtain a basic understanding of power effects on the wing. Although useful results were obtained from this investigation, a more accurate representation of the flow was required to adequately define the installed performance of these advanced propulsion systems. One of The resul- Accordingly, to assess the installation drag penalties of advanced propfan- powered aircraft, tests on a semispan wing/body model of a supercritical swept-wing transport have been conducted as a part of the NASA Advanced Turboprop Program (ref, 8 ) . Wind tunnel tests were conducted to investigate the installation losses for a wing-mounted powerplant involving interactions between the nacelle, propeller slipstream, and a swept supercritical wing. The tests reported herein were focused on the assessment of (1) the interference drag penalties of the nacelle and the propeller slipstream; (2) an understanding of the flows that cause the interference; and ( 3 ) the determination of what configuration modifications would be required to reduce the interference drag. The test was conducted in the Ames Research Center's 11-Foot Transonic Wind Tunnel at M = 0.6 to M = 0.8. Propeller performance characteristics, pressure distributions over the wing and nacelle, powerplant installation losses, and fluo- rescent oil flow visualizlation studies were made. Only selected portions of these data are presented for discussion herein. More extensive test results are provided in graphical form on microfiche affixed to the inside back cover of this report. MODEL DESCRIPTION AND INSTRUMENTATION The model configuration was derived from a supercritical wing which incorpor- ated 1975 advanced design technology for medium-range transports and which demon- strated good drag-rise characteristics near mated to the wing with little attempt to blend or shape the design for optimum aerodynamic integration, such as described in reference 9. M = 0.8. The turboprop nacelle was Photographs of the semispan wing/body wing-tunnel models mounted in the Ames 11-Foot Transonic Wind Tunnel are shown in figure 1. Figures l ( a ) and l(b) show front and rear views, respectively, of the under-the-wing nacelle mated to the L supercritical wing. or without the propeller). The propeller axis was aligned downward 3.75' (droop) and inboard 2 O (toe-in) to minimize the cyclic blade loads caused by wing sweep and angle of attack. The procedure used to achieve this i s reported in reference 10. This configuration is hereinafter denoted as the baseline (with Results of a previous powered test of the baseline model in the Ames 14-Foot Transonic Wind Tunnel indicated that the installation of the nacelle significantly affected the wing flow. The flow on the wing upper surface was separated near the leading edge during powered conditions. A s a result of this finding, a modification to the existing wing was developed in an attempt to reduce the upper surface pres- sure coefficients at powered conditions. The wing modification was performed by the Douglas Aircraft Company under an Ames contract (ref. 11) and was accomplished by adding a leading edge extension (LEX) to the baseline wing. Photographs of the LEX configuration are shown in figures l ( c ) through l ( e ) . Other modifications included the addition of a fillet and strake at the wing-nacelle junctures. These were tested as a consequence of findings from an earlier unpowered test (refs. 1 2 and 13). The fillet was added to the inboard wing-nacelle intersections, and the strake to the outboard, with the LEX on. A photograph of this configuration i s shown in figure 1 (f) . LEX Design Philosophy A brief discussion of the design philosophy of the LEX is presented below. A more comprehensive description, and the airfoil coordinate definition of the LEX, can be found in reference 11. The baseline wing was designed based on supercritical airfoil technology. Supercritical airfoils have a characteristic known as controlled supersonic flow. This implies that as the region of local supersonic flow develops and grows in extent over the airfoil, the shock wave terminating such a region remains weak. Integrating the nacelle with the wing caused the wing upper surface flow to separate inboard of the nacelle (ref. 12). To rectify the problem, the inboard section of the wing was redesigned with the objective of recovering the clean wing pressure distribution. The actual criteria used were that the final design, including the effects of power, should have upper surface pressure coefficients, chordwise pres- sure gradients, and spanwise upper surface isobar patterns similar to those exhib- ited by the clean wing (because the clean wing did not exhibit any flow separations and performed well). The airfoil sections between the nacelle and the body were modified to conform to these criteria to the fullest extent permitted by the existing hardware. Because no undersirable flow phenomena were identified outboard of the nacelle, no modifica- tions were made there. Inboard, the design constraint used to preserve wing struc- tural integrity was that the airfoil remain unchanged except for the forward or aft 20% to 25% of the chord. Modified airfoils that fit within the existing planform which conformed to these constraints could not be found. Therefore, the use of a leading or trailing edge extension was analytically investigated. A trailing edge 3 extension was ruled out because it caused an undesirable unsweeping of the wing isobar pattern. Therefore, a 15% chord leading edge extension was selected. The wing airfoil sections between the body and the nacelle for the modified planform were developed to have upper surface pressure coefficients and gradients similar to the clean wing at the transonic cruise condition (M = 0.8, CL = 0.5), and to fair smoothly into the existing airfoil shape at the 25% chord. The airfoil shapes were developed using a two-dimensional Garabedian transonic analysis method. Then a three-dimensional analysis was used to ensure that the pressure distribution over the entire airfoil conformed t o accepted design practice. The Jarneson computer code, 3s modified by Douglas Aircraft Co. (ref. 111, restricted to wing alone, was used for the three-dimensional analysis. ~ The most promising design developed i s denoted as MOD 3 in reference 11. MOD 3 had calculated pressure peaks near the leading edge which were about half those of the baseline wing. The transonic flow over the entire upper surface was free of shock waves even at the flow condition which corresponded to cruise power. With power, the pressure levels were less than for the clean baseline wing. the gradients were less, and constant levels of pressure coefficient occurred at corresponding chordwise stations (constant x/c). These results indicated properly swept isobar patterns. Moreover, Model Details I The model scale was 12% based on a 180 passenger transport. The powerplant was designed for testing at 1.5 atm. Baseline model planform details are shown in figure 2 ( a ) . turbulent boundary layer. On the wing upper surface, the grit was applied 11.7 cm (4.6 in.) from the leading edge, or at the 15% chord line, whichever was less. On the wing lower surface, it was applied 15.9 cm (6.25 in.) from the leading edge. The nacelle was scaled based on a M = 0 . 8 transport powered by two 30,000-hp engines (ref. 4). The LEX was added between the inboard side of the nacelle and the body. A fillet was added inboard of the nacelle, and a strake outboard, in an attempt to alleviate the sharp corners at the nacelle intersection with the wing. Nacelle details are shown in figure 2(b) and airfoil sections of the fillet at q = 0.396 and q = 0.424 are shown in figures 2(c) and 2(d), respectively. The coordinates of the fillet are presented in Table 1. Wing airfoil and nacelle coor- dinates are provided in reference 13 and the LEX coordinates are given in refer- ence 11. Details of the strake geometry are not available at the present time. Fine grit was applied near the wing leading edge to produce an all A sectioned profile of the model powerplant i s shown in figure 3. The hub contained a six-component strain gauge balance for measuring propeller forces and moments. fan geometry. posite and were designated SR-2C. The propeller diameter was 0.622 m (2.04 ft). Each of the eight blades was fitted with a gear sector at the end of the shank which meshed with a synchronizing ring gear in the hub. Thus, the pitch angles of all The propeller used the Hamilton Standard SR-2 (unswept) eight-blade prop- Blades for this model were fabricated from molded carbon-epoxy com- 4 blades were the same and could be manually changed by relocating a pin which locked the ring gear to the hub. The six-stage axial flow turbine motor was driven by compressed air. Thrust loads from the propeller and turbine were carried by an oil-cooled, deep-race front ball bearing. A 24-channel slip ring was located at the rear of the turbine. Instrumentation signals from the hub balance were passed through a hole in the motor shaft and out through the slip rings to the nonrotating system. In addition to the propeller balance signals, the slip ring carried signals from up to four propeller- blade strain gauges for measuring blade structural response. was ducted beneath the wing and exited the exhaust pipe at the wing midchord. Pitot probes (thrust rake) and static wall pressure taps were used to measure the exhaust thrust. A section of stainless steel honeycomb located upstream of this instrumen- tation removed any residual swirl from the motor exhaust. calibration of the thrust rake and exhaust nozzle. and instrumentation was done at Fluidyne, Minneapolis, MN, using a standard ASME nozzle and thrust stand. Turbine exhaust air This permitted accurate Calibration of the powerplant There was a gap of about 3.8 cm (1.5 in.) between the bottom of the body and This permitted some of the tunnel floor boundary-layer flow to the tunnel floor. pass beneath the body. A shroud was placed around the wing floor-mount to prevent this flow from impinging on the balance. A seal was placed at the wing/body junc- ture to prevent airflow beneath the body from passing over the wing. test the wind-tunnel floor slots were sealed. the body on the unpowered test results were described in reference 14. Throughout the The effect of flow passing beneath Balance and Pressure Instrumentation Total lift, drag, and pitching moment were obtained from a five-component force-moment floor balance. The balance capacity in the normal-force and axial- force directions was 53,400 N (12,000 lb) and 4,500 N (1,000 lb), respectively. Propeller forces and moments were obtained from a six-component rotating propeller balance. The balance capacity was 2,225 N (500 lb) in the normal- and side-force directions, 2,000 N (450 lb) in the axial-force direction, and 745 N-m (550 ft-lb) in rolling moment. The coordinates of all model pressure instrumentation are given in refer- ence 13, and consisted of the following: ( 1 ) 239 wing static orifices distributed over eight spanwise stations (fig. 2(a)); (2) 103 nacelle static orifices distrib- uted over 12 longitudinal stations; ( 3 ) 25 hub-cavity static orifices; ( 4 ) 6 inter- nal duct static orifices; (5) 6 nacelle-base static orifices; and (6) 2 1 internal duct total probes. assemblies located within a cavity in the wing. Other model instrumentation con- sisted of a temperature rake in the motor duct containing 10 total temperature probes, strain gauges for measuring unsteady blade stresses, 17 pressure transducers (Kulites) for measuring acoustic pressures on the wing and body, and probes to measure the motor plenum temperature and pressure. Additional instrumentation, such These pressures were recorded using two 6-module Scanivalve 5 as accelerometers and motor bearing temperatures, were provided to monitor the health of the turbine system for wind tunnel safety. TEST PROCEDURE The test variables were Mach number, angle of attack, propeller blade pitch angle, and nozzle exhaust pressure ratio (EPR). constant at one atmosphere throughout the test. aerodynamic chord then varied from 7 . 8 ~ 1 0 ~ at M = 0.6 t o 9 . 5 ~ 1 0 ~ at M = 0.8. For the jet effects configurations (blades off) at each test Mach number, EPR was varied at a fixed angle of attack. EPR was varied from 1.0 to approximately 1 . 9 in 10 increments. Note that EPR = 1.0 represents a jet-off point; that is, no air was flowing through the exhaust nozzle. For the blades-on configurations, propeller speed was varied from windmill (jet-off) to 9,000 rpm in approximately 250 rpm increments, or to the maximum speed possible within power limitations, whichever occured first. Tunnel total pressure was held The Reynolds number based on mean Note that propeller speed and EPR could not be varied independently. Model angle of attack was varied from -3" to 4" throughout most of the test. Towards the end of the test, time constraints necessitated reducing the test angle of attack range to 0" to 3". numbers of 0.75, 0.78, and 0 . 8 ; some results were obtained at Mach numbers of 0.6 and 0.7. station; the design value was 57". blade pitch angles of 55.2O and 59.2". Most of the blades-on data were obtained at Mach The measured propeller blade pitch angle was 57.2" at the 75% radius Limited data were also obtained at measured DATA REDUCTION The procedure used to obtain the total powerplant installation drag has been described in detail in reference 1 4 . A brief description of that procedure i s repeated here. The interference drag of the various components i s conceptually illustrated by the thrust removed drag polars in figure 4 ( a ) . drag i s the increase in drag between the nacelle-off configuration (WING-BODY), and nacelle-on configuration (W-NJET-oFF). A jet-off drag increment i s not representa- tive of the power-on case because there is a large region of separated flow at the base of the nacelle. This separation, with resulting lift-loss, gives the appear- ance of an unusually large nacelle-interference drag. lift-loss is restored, and the drag polar moves left (W-NCRuIsE EPR). reduction so generated is called C The jet-off nacelle interference With the jet on, much of this The drag DJET Some of the jet-on drag reduction i s lost when the propeller i s installed because of the interference drag caused by the propeller slipstream interaction, C . This results in the drag polar W-N-PCRuISE EPR. The total powerplant DSS 6 installation drag penalty is, therefore, the increase in drag between the WING-BODY and the blades-on configurations (W-N-PCRuIsE EPR) evaluated at constant lift. The procedures used to obtain propeller net thrust and to adjust the floor balance measurements for both propeller and jet thrust are described in the following para- graphs. The derivation of the net thrust obtained from balance readings and pressure measurements is illustrated in figure 4(b). The hub-base-axial-force, CXHBT is determined by integrating the pressure measurements in the hub-base cavity. The hub-forebody-axial-force, ments obtained with the blades removed. When a representative nacelle i s placed behind the propeller a buoyancy force, AC is induced on the propeller due to the disturbance of the flow field by the nacelle. The apparent thrust force on the , is determined from hub-balance and pressure measure- cXH xN ' blades in the presence of the nacelle is then equal to the net thrust, CTNET, Plus the buoyancy force, AC . According to inviscid theory, an equal and opposite force resulting from the buoyancy effect i s felt on the nacelle. The buoyancy force is determined by integrating the measured surface pressures over the nacelle for the blades-on and blades-off configurations and calculating the difference. The equa- tion for net thrust (fig. 4 ( b ) ) has four terms representing six independent measure- ments, each subject to experimental uncertainty. XN The derivation of the slipstream interference drag, C i s illustrated in DSS ' figure 4 ( c ) . blades-off, and blades-on configurations. Also shown i s the resultant drag indi- cated by the floor balance, C . The difference between the blades-on and blades- off expression i s the slipstream interference drag, C . This difference was obtained for blades-on/blades-off results at the same EPR. total enthalpy of the flow was matched as well as the EPR. for the slipstream interference lift i s The drag of the various components i s shown for the clean wing, DBAL DSS Where possible, the Similarly, the equation sin a sin e sin e , ) T c = ( c - sin a sin 0 - Lss L~~~ T - C ' - sin a sin 8 L~~~ c ' J E , where the prime indicates the blades-off value. The equation for the slipstream interference drag requires 10 independent measurements, each subject to experimental uncertainty. The uncertainty in C has been estimated to be +0.0013. The maximum uncertainties in the floor balance readings, based on repeatability, have been estimated to be ?0.0020 in lift coeffi- cient and t0.0002 in drag coefficient. DSS 1 7 The equations used to obtain the net thrust, thrust removed lift, drag and pitching moment, propeller performance parameters, and propulsion interference drags are detailed below. - cx on Nof f xN xN Propeller-nacelle buoyancy: AC = C Net thrust: - A C + C + C xN ‘ H ‘HB Thrust removed lift, drag, and pitching moment: = c + c sin €tD cos eT ‘NTR N~~~ ‘ ‘ J E T T~~~ COS e COS eT D C = c + c + c ‘ T R ‘BAL ’ JET T~~~ C = C cos a - C sin a N~~ ‘TR C - C cos a + C sin a - ‘TR N~~ (3) (4) + 0.20053 C - 0.10625 CN (5) ‘JET m~~~ JET cm = cm BAL Net efficiency: nNET = 9.5493 C qSV/QN T~~~ Propeller thrust coefficient: CT = 3600 CT q s / p ( N D ~ ) (7) Prop NET Propeller power coefficient: Cp = 0.216~10~ HP/pN3D5 (8) Prop The equations for the interference drags are obtained by differencing the drags of two different configurations at constant values of lift coefficient and EPR. Slipstream drag: C = c - c ( 10) DSS Dblades-on Dblades-off Total installation drag: CD = C - c Dblades-on DWB 8 RESULTS AND DISCUSSION Measured results, flow visualization studies, and pressure contours are pre- sented in figures 5 through 28. total propulsion installation drag are shown in figures 2 9 through 31. A limited sample of the data are presented for discussion herein. More extensive test results are provided in graphical form on microfiche which are affixed to the inside back cover of this report. A listing of the data shown on the microfiche can be found in the appendix. The variations of slipstream interference drag and Pressure Distributions Wing pressure distributions for three configurations are compared in figures 5 through 1 0 . Nacelle pressures are presented as a function of azimuth in figures 1 1 through 15. The discussion of the pressure results is limited to the nominal cruise condition of M = 0 . 8 at an angle of attack of 2 ' . At this condition, the varia- tions shown in EPR (figs. 5, 6, 10, 1 1 and 12) are not significant. Results at other Mach numbers and angles of attack are included on the microfiche. Note that along the nacelle centerline ( r l = 0 . 4 8 1 ) the nacelle pressures at 6 = Oo (upper surface) and 6 = 1 8 0 ° (lower surface) have been referenced to the baseline wing leading edge. At this semispan station, the pressures between -0.194 < x/c < 0.390 are on the nacelle centerline. A negative value for x/c represents a point ahead of the baseline wing leading edge. Blades off- Blades-off pressure distributions at cruise EPR are compared with the wing-body in figure 5. shown with the same symbol. changes in the pressure distributions in the vicinity of the nacelle. Just inboard of the nacelle, the wing shock wave was strengthened by the nacelle as indicated by a change in the peak upper surface shock wave was formed on the lower surface due to the addition of the nacelle. Integration of the pressures showed that the net effect was a substantial loss in section lift. The pressure distributions further inboard (n < 0.418) indicate small separations may occur due to the nacelle installation. peaks at 0 < 0.418 were increased, but not to the level that would cause large flow separation. Just outboard of the nacelle ( r l = 0.5441, both the upper and lower surface pressures were reduced slightly from the leading edge to approximately 70% of the wing chord. towards the tip. Note that both upper and lower surface pressures are The addition of the nacelle produced significant Cp from about -1 . 0 to -1.6. In addition, a The upper surface pressure The effect of the nacelle on the wing pressures diminished The addition of the LEX (fig. 5) reduced the upper surface pressure peak from -1.6 to about -1.2 at rl = 0 . 4 1 8 . The lower surface pressure peak has also been reduced and hence, the strength of the shock wave has diminished. The pressures outboard of the nacelle were unaffected by the LEX, as would be expected. 9 Blades on- A comparison of wing pressure distributions at cruise power are shown in figure 6 for the three test configurations. rotation was up-inboard, which increased the local angle of attack of the inboard airfoil sections and decreased the local angle outboard. At q = 0.418, the addi- tion of the LEX did not affect the magnitude of upper surface suction peak. location of the pressure peak on the LEX was approximately the same distance behind the LEX leading edge as the wing peak was behind the wing leading edge. The addi- tion of the LEX has not affected the upper surface pressure peak because the flow i s separated. However, the lower surface suction peak has been eliminated. Addition of the fillet and strake reduced the upper surface suction peak from about -1.6 to about -1.1 at q = 0.418. This indicates that adding the fillet eliminated the flow separation as indicated by the steep pressure gradient. Further inboard, the fillet increased the suction peak and moved the peak pressure location slightly behind that for the LEX configuration. surface pressures. ferences among the configurations. installation drag penalty, which will be discussed later. Because of this drag reduction, it i s suggested that the LEX and fillet may work as a more efficient stator for recovering some of the energy lost in the form of slipstream angular momentum. The removal of a portion of the slipstream swirl indicates it may be desirable to allow the wing t o d e p a r t from an elliptic span load distribution. The direction of propeller The The fillet and strake had a small effect on the lower Outboard of the nacelle (TI > 0.481) there were only small dif- The fillet and strake significantly reduced the Power effects- The effect of power on the wing pressures of the baseline con- figuration (WBNP) i s shown in figure 7. An EPR = 1.0 represents the propeller at windmill speed (jet-off); an EPR = 1.257 represents an intermediate power setting; and an EPR = 1.774 i s near cruise power. Inboard of the nacelle, within the propeller slipstream, increased power tends to increase regions of separated flow because of propeller swirl. gradients, the flow separates. On the lower surface, there was generally an increase in Cp with increasing power. The net effect of the increase in power was a small increase in section lift inboard from the nacelle. Just outboard of the nacelle, the effect of power was to decrease the upper and lower surface pressures forward of the 60% chord station. in local section angle of attack due to propeller rotation. ler slipstream ( q > 0.597) the effects of power were not significant. Because the wing cannot support the large pressure The effects shown are consistent with the changes Outboard of the propel- The effect of power on the WBNLP and WBNLF3SP configurations are shown in figures 8 and 9 , respectively. The trends resulting from power are similar to those described for the baseline configuration. Inboard, additional power decreases the upper surface pressure coefficients and increases them on the lower surface. Out- board, additional power increases the upper surface surface and decreases the lower cP . These results are consistent with an up-inboard propeller rotation. cP Figure 10 compares blades on/off pressure distributions at cruise power for the LEX configuration. with blades on, caused by the increase in local section angle of attack due to the propeller slipstream swirl. q = 0.418 as follows: ( 1 ) on the upper surface the blades-off shock wave has been Inboard of the nacelle, the upper surface pressures decrease The slipstream changes the nature of the flow at 10 replaced by a region of separated flow ( 0 < x/c < 0.3); and ( 2 ) on the lower surface the weak shock wave has been eliminated. Outboard of the nacelle ( q > 0.481), the effect of the slipstream is generally consistent with a decreased angle of attack. Nacelle pressures- The variation of nacelle surface pressures with an azimuth at each of the 12 nacelle stations i s shown in figures 11 through 15. The values of azimuth position i s at the top of the nacelle, increasing clockwise when viewed from the front. The first nacelle station (XNAC = 0.254 m) is 1.27 cm (0.5 in.) behind mately XNAC = 0.521 m for the LEX configuration. With blades-off (fig. 111, there were only small differences in near the wing/nacelle intersection (XNAC = 0.444 to 0.597 m ) . there was an increase in the negative pressure peak with the LEX caused by the wing upwash field. changes caused by the LEX. The effects of the LEX with blades on are similar to those with blades off. Near the winghacelle juncture, the variation in greatest with LEX on. Further downstream, the differences between blades on and blades off were about the same for the LEX configuration as for the baseline. Although slight, the effect of the slipstream extends over the entire length of the nacelle. I XNAC indicated on the figures i s the nacelle station in meters. The zero-degree- a the hub. The inboard wing, leading-edge/nacelle intersection i s located at approxi- XNAC = 0.597 m for the baseline configuration and at approximately between the baseline and LEX configurations, except cP At these stations, Behind the leading-edge intersection there were again only minor Cp was The effect of the fillet and strake on the nacelle pressures i s presented in figure 12. The nacelle pressures were affected more by the fillet ( $ = 270") than by the strake ( $ = go"), probably as a consequence of the greater spanwise extent of the fillet. from the strake was about equal to the increment provided by the inboard fillet. The reason for this i s not apparent from either the nacelle or wing pressure distributions. It was shown in reference 1 3 that with blades off the drag improvement The effect of power on the nacelle pressures for the WBNP, WBNLP, and WBNLF3SP configurations i s shown in figures 13 through 15, respectively. Applying cruise power slightly increased the surface pressures over the nacelle when compared with the windmilling propeller. wing/nacelle juncture. The effect of power on the WBNLF3SP configuration was less than for the other two configurations. The pressures generally decreased downstream from the Flow Visualization Studies Several configurations were studied using fluorescent oil to visualize the flow in the boundary layer. figure 16. Photographs taken under ultraviolet light are presented in Figure 1 6 ( a ) shows the wing (alone) upper surface. The effect of adding the This photograph The shock location agrees with that nacelle with propeller off and jet off i s shown in figure 1 6 ( b ) . shows the strong inward sidewash behind the nacelle and the strong normal shock adjacent to the inboard side of the nacelle. 11 indicated on the pressure distributions (fig. 5 ) . A small separation bubble can be seen just downstream of this shock. upper surface boundary layer flow. nacelle has been almost completely eliminated. A reduction in shock strength would be expected because the increased chord has effectively thinned the wing section, increasing the drag rise Mach number. inward sidewash or the small region of trailing edge separation behind the nacelle. Figure 1 6 ( d ) is a photograph of the underside of the WBNLP with jet off (propeller at windmill). It shows the separation behind the nacelle exhaust pipe which is primarily responsible for the jet-off lift loss described in refer- ence 1 2 . The effect of the jet (propeller off) i s shown in figure 1 6 ( e ) . It is evident that the large separated flow region behind the exhaust pipe has been greatly reduced. With the jet on, much of the wing lower surface lift i s restored, the major contribution to the "jet effect" benefit. Figure 16(c) shows the effect of the LEX on the The normal shock wave on the inboard side of the The LEX had little effect on the strong The effect of a windmilling propeller (power off) on the wing upper surface boundary layer flow i s shown in figure 1 6 ( f ) . the propeller- and jet-off case (fig. 16(b)) shows little or not effect of the wind- milling propeller. shown in figure 1 6 ( g ) . The inboard upper surface flow curves sharply forward in a highly irregular separated flow region. Across the top of the nacelle, the sidewash in the boundary layer has been greatly increased from the windmill condition (fig. 16(f)), indicating the presence of strong spanwise pressure gradients. This i s supported by the pressure distributions (fig. 8 ) and the pressure contours described in the next section. The lower surface boundary layer flow, with the propeller at cruise thrust, is shown in figure 1 6 ( h ) . Comparison of this photograph with the propeller-off, jet-on case (fig. 16(e)), shows little or no effect of the propeller on the lower surface flow. For this under-the-wing nacelle configuration, the lower surface flow appears to be dominated by nacelle geometry rather than by the propeller. Based on these findings, it would be expected that for an over-the- wing nacelle configuration, the wing lower surface boundary layer flow would be affected more by the propeller rather than by the nacelle. Comparison of this photograph with The upper surface flow with the propeller at cruise thrust is Pressure Contours Pressure contours (isobars) for the upper and lower wing surfaces are presented in figures 1 7 through 1 9 . The outline enclosing the contours on the figures extends from rl = 0.250 to rl = 0.849, the range for the rows of pressure taps. For those configurations that include the nacelle, the pressures on the nacelle at IJJ = 0" are used for the upper surface and 6 = 180" for the lower surface. Only those nacelle pressures which lie at or behind the wing leading edge were used to obtain the isobars. Isobar patterns serve as a useful guide to study configuration modifications and slipstream effects on the wing. Figures 17(a) and 17(b) show the isobars for the wing-body upper and lower surfaces, respectively. Note that the benefits of sweep are maintained over most of the wing planform. No undesirable features are 1 2 evident, such as unnecessarily high suction peaks or steep adverse pressure gra- dients. Adding the nacelle (figs. 17(c) and 1 7 ( d ) ) causes a concentration of the isobars inboard of the nacelle and toward the wing leading edge. Upper surface suction peaks are increased and the effectiveness of wing sweep i s lost in this region. wave as shown by the oil flow visualization studies. The isobar patterns for the WBNL are shown in figures 17(e) and 1 7 ( f ) . They show an improvement due to adding the LEX. The suction peaks have been reduced, the pressure gradients are less severe, and effectiveness of wing sweep has been restored to some extent. This indicates the presence of strong pressure gradients caused by a shock , For the WBNP configuration, in the presence of the slipstream (figs. 18(a) and 18(b)), there i s an unsweeping of the isobars. Also, the beneficial effects of wing sweep are lost when compared to the blades-off configuration (fig. 1 7 ( c ) ) . Inboard, the wing local angle of attack i s increased and the pressure peak moved forward. Toward the center of the propeller, the low-pressure isobars increase slightly (less negative (fig. 1 7 ( c ) ) . This causes the isobar pattern to be less severely unswept. Further outboard, within the propeller slipstream, the wing local angle of attack is decreased. Outside the slipstream, the isobars tend to assume a pattern similar to the wing alone. When the LEX was installed (figs. 18(c) and 1 8 ( d ) ) , there was a significant improvement in the isobars inboard from the nacelle compared to the baseline configuration. Cp) compared to the blades-off configuration The addition of the fillet and strake (figs. 19(a) and 19(b)) increases the suction pressure peak and gradients on the LEX. The strake causes the outboard isobars to be less densely spaced (reduced gradients) and thereby restores sweep effectiveness. Unfortunately, a motor bearing failure preevented obtaining boundary layer flow visualization photographs for the fillet-strake configuration. However, oil flow studies of an unpowered wing-nacelle-fillet configuration were made during a previous test (ref. 1 3 ) . The boundary layer flow visualization studies from that test showed the inboard shock wave had moved inward and was not normal to the flow as was the shock on the baseline configuration. Behind the shock, the flow had the appearance of being similar to the flow around a flat plate with stationary vortices rolled up at the edges. Because of the vortices, the flow behind the shock appeared to be quickly returned to the uniform streamwise flow. with the propeller on would be similar to that with the propeller off in those regions outboard of the slipstream. It is felt that the flow Longitudinal Aerodynamic Characteristics The effects of Reynolds number and power on the thrust-removed longitudinal aerodynamic parameters are presented in figures 20 through 22. The longitudinal force and moment coefficients for all configurations are referenced to the trapezoi- dal wing area of the baseline configuration. Wing-body characteristics- Reynolds number effects on the longitudinal aerody- I namic parameters of the wing/body configuration are shown in figure 20. At all test 13 I Mach numbers, both lift and pitching moment are affected only slightly by variations in Reynolds number. number on skin friction. ences in calculated skin friction drag, assuming an all turbulent boundary layer. Note that for semispan configurations the absolute drag levels are not representa- tive of a full span configuration. Additional contributions to the drag are caused by the tunnel-floor boundary layer acting on the underside of the body and possibly altering the reflection plane by the crossflow boundary layer with changes in angle of attack. However, it should be emphasized that reliable increments can be obtained from semispan testing. The variation in drag i s caused by the effect of Reynolds The differences shown are generally less than the differ- Baseline characteristics- The effects of jet exhaust on the thrust-removed aerodynamic parameters of the baseline configuration are compared with the wing-body parameters in figure 21. An EPR = 1 . 0 represents jet-off; an EPR = 1 . 7 5 is near cruise. without significant effect on the lift curve slope, the stability was reduced, and the drag coefficient increased significantly at all lift coefficients. The lift loss with jet-off was caused by flow separation at the base of the nacelle, evi- denced by the flow visualization photograph (fig. 1 6 ( d ) ) . restored much of the lift loss caused by this flow separation; the stability was I slightly increased and the drag was reduced. For instance, at CL = 0.5, the jet- off nacelle drag penalty, indicated by the difference between the circles and squares, is about 7 1 counts ( 0 . 0 0 7 1 ) . Turning on the jet reduced the installed nacelle drag penalty to about 3 8 counts. is the jet effect, With jet-off, the lift coefficient at all angles of attack was reduced ' l Turning on the jet , The difference between the two (33 counts) , discussed previously. ' % E T , When the nacelle was installed, the angle of attack had to be increased approx- imately 0.5" to maintain a CL = 0.5. With jet on, the angle of attack had to be increased only 0.15" to maintain this CL. This confirms the results described in reference 1 3 that the high jet-off nacelle drag and lift loss was caused primarily by separated flow at the base of the nacelle. Therefore, the jet-off nacelle drag increments do not represent the nacelle installed drag. LEX characteristics- The effect of the LEX on the thrust-removed longitudinal Adding the LEX had a negligible However, adding the LEX parameters at M = 0.8 i s shown in figure 22. effect on the lift when compared with the baseline (WBN). had a significant destabilizing effect due to an increase in area forward of the moment reference. The pressure distributions with LEX on (fig. 5) confirm the forward movement of the wing chordwise loading. The LEX reduced the drag at all lift coefficients when compared with the baseline wing-body-nacelle configuration. At CL = 0.5, the drag with LEX on was about 10 counts less than the baseline. net jet-on nacelle installation drag penalty was about 23 counts at for the LEX configuration (circles vs. diamonds). was to restore the inboard wing sections drag divergence characteristics to more nearly resemble the wing alone. The 23-count installed drag includes a friction drag increment for the LEX. Adding the LEX has reduced the nacelle installation drag penalty close to that of the isolated nacelle (20 counts), as reported in reference 12. The Ct = 0 . 5 The primary benefit of adding the LEX 14 Propeller performance- The variation in propeller power coefficient, and net efficiency as a function of advance ratio i s shown in figure 23. Results are shown for two propeller blade pitch angles at angles of attack from 0 ' to 4 ' . Propeller power and thrust coefficient increase with increased blade pitch angle because of the expected decrease in windmill speed with increased blade pitch. These charac- teristics were affected only slightly by changes in angle of attack. The maximum value of net efficiency increased slightly with increased blade pitch angle; the maximum occurred at a higher advance ratio (lower propeller speed). highest value of advance ratio represents the propeller at windmill. The lowest value represents a point beyond the estimated cruise condition. Note that the The effects of configuration modifications on propeller performance at the cruise condition are shown in figure 24. advance ratio are lower with LEX on than for the baseline. At a constant value of propeller thrust coefficient, configuration modifications cause a decrease in the advance ratio. The largest incremental decrease in advance ratio ( J ) occurred when the LEX was added. The difference in J due to configuration modifications i s equivalent to an increase of 0.25O in blade pitch angle (for a fixed configura- tion). with a further reduction when the fillet and strake are added. The net efficiency The power and thrust for a constant The net efficiency i s less for the LEX configuration than for the baseline, varies according to the relation nNET = JC . For a given thrust coeffi- cient, the decrease in J resulting from configura ion modifications results in a T p r o p ' ' P y -o p reduced net efficiency. a significant effect on propeller efficiency. It is not clear why the addition of the LEX, fillet, and strake have such an effect on propeller performance. From these data it appears that configuration geometry has The effect of Reynolds number on propeller performance for the baseline config- uration is shown in figure 25. At the lower Reynolds number, less power was required for the same propeller speed and less thrust was generated. At low advance ratios ( J < 3.2) the net efficiency shows no significant Reynolds number effect. There appears to be a greater effect of Reynolds number at the lower rpm. However, at high advance ratio and low Reynolds number the data i s more uncertain because the balance was so lightly loaded. A small Reynolds number effect would be expected as a consequence of a slight change in the flow characteristics over the propeller. The 87% efficiency point ( J = 3.43) at the low Reynolds number was due to the power being somewhat low, and the point is suspect. It was not known whether this was due to an error in the torque measurement or in the propeller speed. Thurst-removed characteristics- The thrust-removed longitudinal aerodynamic parameters are shown as a function of net thrust in figures 26 through 28. Figure 2 6 shows the variation for several angles of attack and blade pitch angles for the baseline configuration; figure 27 presents similar data for the LEX config- uration; and figure 28 compares results for different configurations. negative net thrust coefficient occurs when the propeller is windmilling. Note that a At this condition the propeller actually produces drag. i In general, the data show a significant increase in lift with net thrust at low I thrust levels. Only a small thrust (EPR about 1.05) was required to restore the 1 5 lift-loss due to separation at the nacelle base. ler speed, close to windmill. Approximately 80% of the increase in CT, from the This was obtained at a low propel- - jet-off value was achieved at an EPR = 1.05 or CT = 0.0080 (EPR = 1.75 i s NET required for cruise), and further increases in thrust produced little improvement in lift. higher angles of attack. increasing thrust, the slipstream interference drag continued to increase. Ini- tially, nose-down pitching moment increases at low thrust levels because of the restoration of the lift near the wing trailing edge. Further increases in thrust generally had only small effects on the moment. The drag generally increased with increasing thrust and at a faster rate at Even as the lift remained relatively constant with A comparison of the three configurations at an angle of attack of 2 ' i s shown At moderate and high thrust levels the LEX-fillet-strake configura- The low in figure 28. tion had the highest lift, and at all thrust levels had the lowest drag. drag levels with the fillet and strake are believed to be caused by a combination of configuration features. First, the strake may have reduced the strong inward cross flow around the nacelle, although this was not apparent from the pressure distribu- tions. confirmed. wing-nacelle intersection and effectively reduced the local thickness-to-chord ratio as well. The effectively thinner wing reduces the compressibility drag, supported by the pressure distributions previously described. This reduction in compressibil- ity drag was apparently much greater than the slight drag increase caused by an increase, in surface area. It is expected that the LEX-fillet-strake configuration would have the least slipstream interference drag because the reduced thickness perturbation permits higher lifting pressures without separation. However, jet effects data on this configuration were not obtained, and therefore the slipstream interference drag could not be determined. Because oil flow photographs were not available, this effect could not be Second, the addition of the fillet eliminated the sharp corner at the Installation Drag Increments The variation of slipstream interference drag with propeller power coefficient i s presented in figure 29 for a constant angle of attack of 2'. Each point repre- sents the difference between blades-on and blades-off drag polars evaluated at the EPR and CL for the blades-on configuration. When presented in this manner, there- fore, a portion of the slipstream drag includes a small drag increment caused by an increase in lift with power. The lift coefficients caused by power at constant angle of attack may differ by up to 0 . 0 3 in the range of net thrust coefficients of interest. The slipstream interference drag coefficient increases linearly with increasing power. The addition of the LEX (fig. 2 9 ( b ) ) substantially reduced the slipstream drag (note change in scale). configuration was approximately half the value for the baseline. In general, the slipstream drag for the LEX The variation of slipstream interference drag with Mach number at CL = 0 . 5 and cruise thrust i s shown in figure 30(a) for LEX off, and figure 30(b) for LEX on. At M = 0.80, adding the LEX reduced the slipstream drag to approximately 25% of the value for the baseline configuration. . The addition of the LEX has 16 effectively thinned the inboard wing sections thereby increasing the drag divergence Mach number. The variation of total powerplant installation drag as a function of Mach number is presented in figure 31. For the baseline configuration (fig. 31(a)), the installation drag generally increased with increasing Mach number; increased with increasing CL; and had a greater variation with Mach number at the high At M = 0 . 8 and CL = 0.5, the total installation drag of 77 counts i s approxi- mately 24% of a typical airplane cruise drag coefficient (estimated from and a lift-drag ratio of about 1 6 ) . CL's. CL = 0.5 The effect of adding the LEX on the total installation drag i s shown in figure 31(b). The total installation drag again increased with increasing Mach number, but in contrast to the LEX-off results, generally decreased with increas- ing CL. slipstream, at the cruise condition. These results clearly indicate that the design approach used (ref. 11) was quite successful. At the cruise condition, the instal- lation drag was reduced from 77 counts for the baseline configuration to 36 counts with the LEX. Thus, adding the LEX (WBNLP) reduced the installation drag from 24% (WBNP) to about 11% of a typical airplane cruise drag. The LEX was designed to provide best performance, in the presence of the The effect of adding the fillet and strake on the total powerplant installation drag is shown in figure 3 1 ( c ) . Eliminating the sharp corners at the wing-nacelle intersections reduced the installation drag to 1 8 counts at the cruise condition. This represents less than 6% of a typical airplane cruise drag and compares favora- bly with current turbofan installations. CL, the data indicate that there i s favorable interference. However, the result is within the experimental uncertainty of the measurements, and further testing i s required to substantiate these values. At the lowest test Mach number and CONCLUSIONS Tests were conducted on a semispan wing/body model with a powered propeller in the Ames 11-Foot Transonic Wind Tunnel. The swept supercritical wing was tested with an under-the-wing nacelle at Mach numbers from 0 . 6 to 0 . 8 and angles of attack from -3" to 4". sphere. The test results indicated the following conclusions: The test was conducted at a constant total pressure of one atmo- 1. The addition of the nacelle caused a strong acceleration of the flow, resulting in stronger shock waves on both the upper and lower wing surface. Addi- tion of a LEX between the nacelle and body reduced the strength of the shock waves. 2. The propeller increased the suction pressure peaks inboard from the nacelle because the up-inboard propeller rotation increased the local angle of attack of the wing. Adding the LEX had a negligible effect on the upper surface suction peaks, but eliminated the lower surface suction peak. significantly reduced the upper surface suction peak at the wing-nacelle juncture. The addition of a fillet and strake 1 7 3 . Low values of exhaust pressure ratio (about 1 . 0 5 ) restored much of the lift-loss that was caused by flow separation at the base of the nacelle when the jet was off. I 4. levels; it had the lowest drag at all thrust levels. The LEX-fillet-strake configuration had the highest lift at moderate thrust 5. The total powerplant installation drag penalty (M = 0.8, CL = 0.5) for the Adding the LEX reduced the drag penalty to about 36 drag counts (11% of the The LEX-fillet-strake configuration had the lowest installation drag baseline configuration was about 77 drag counts (24% of a typical airplane cruise drag). cruise drag). penalty of 1 8 drag counts (less than 6% of the cruise drag). 1 1 8 APPENDIX MICROFICHE DATA ORGANIZATION O n t h e microfiche a f f i x e d to the i n s i d e back cover of t h i s r e p o r t , t h e data are presented i n graphical format. 0.8, and angles of attack from -3" to 4". pages 1 through 1 1 , and force data are on pages 12 and 13. The data cover the test Mach number range 0.6 to The pressure data are on microfiche A page of microfiche contains a frame matrix o f 7 rows, and up to 9 columns. On each page, the first frame (upper left corner) is i n t e n t i o n a l l y l e f t blank. The f i g u r e s are arranged s e q u e n t i a l l y by column. L e t t i n g "1" denote t h e i t h row and 'lJ" t h e j t h column, then t h e f i g u r e order an any page is given by: The following list of f i g u r e s shows t h e microfiche l o c a t i o n of each f i g u r e by page, row, and column. For example, 7,3,6 i n d i c a t e s t h a t the f i g u r e is located on microfiche page 7, row 3, column 6. 19 I F I G . ' N O . t --- I i 6 1 5 j 7 I 8 9 1 0 - 1 1 1 2 1 3 14 1 5 B R I E F T I T L E M A C H N U M B E R W i n g P r e s s u r e Data B l a d e s - o f f , C r u i s e Power B l a d e s - o n , C r u i s e Power Pcwer E f f e c t s - WBNP Power E f f e c t s - WBNLP Power E f f e c t s - WBNLF3SP B l a d e s - o f f v s Blades-on N a c e l l e P r e s s u r e Data WBN,WBNL,WBNP,WBNLP E f f e c t o f F i l l e t and S t r a k e Power E f f e c t s - WBNP Power E f f e c t s - WBNLP Power E f f e c t s - WBNLF3SP T h r u s t Removed C h a r a c t e r i s t i c ! R e y n o l d s Number E f f e c t s N a c e l l e a n d J e t E f f e c t s Lex-on v s Lex-off P r o p e l l e r P e r f o r m a n c e d =oo d = 2 O d = 3 O d d O 0 d = 1 0 . 7 0 1 0.75 I 0.78 0.80 20 c __ T T Y B R Z E F T I T L E M A C H N U M B E R r - I , 0.60 , 0 . 7 0 j 0.75 0.78 0 . 8 0 i ---- i i I 1 3 , 6 , 2 1 3 . 1 . 3 1 3 , 3 . 3 1 3 . 5 . 3 i I Con f i gur a t i o n Ef f e c t s E f f e c t o f Reynolds Number ~ V a r i a t i o n -_______-____ w i t h Net T h r u s t i- WBNP, PITCHs55' 1 3 , 7 , 3 WBNP, PITCH=5g0 13v4.4 W B N L P , PITCH=5g0 1 3 . 5 . 5 WBNP, PITCH=57' 13.1.4 1 3 . 2 . 4 1 3 , 3 , 4 W B N L P , PITCH=55' 1 3 . 5 . 4 1 3 . 6 . 4 WBNLP. PITCH=57' 13.7.4 1 3 . 1 . 5 1 3 . 2 . 5 1 3 , 3 , 5 1 3 . 4 , 5 C o n f i g u r a t i o n E f f e c t s 13.6.5 1 3 . 7 , 5 1 3 . 1 . 5 ~- F I G . N O . M a c h V a r i a t i o n , WBNP Mach V a r i a t i o n , WBNLP Mach Variation, WBNLF3SP 24 25 13, 4.8 - 1 3 , 5 , 8 t - 13,6,8 - 26 27 28 29(A) 2 9 ( B ) Power E f f e c t , d = O 0 , WBNP Pouer E f f e c t , & = l o , WBNP Power E f f e c t , d = 2 ' , WBNP Power E f f e c t , d = 3 ' , WBNP Pouer E f f e c t , d=OO, WBNLP Pouer E f f e c t , & = l o , WBNLP Power E f f e c t , k = 2 ' , WBNLP Power E f f e c t , a i = 3 0 , WBNLP Mach V a r i a t i o n , CL=0.5, WBNP Mach V a r i a t i o n , CL=0.5, WBNLF T o t a l I n s t a l l a t i o n D r a g 1 3 , 3 9 8 - - 1 3 2 9 8 t 21 E , REFERENCES I 1. Hopkins, J. P.; and Wharton, H. E.: Study of the Cost/Benefit Tradeoffs for Reducing the Energy Consumption of the Commercial Air Transportation System, Lockheed-California Co. NASA CR-137927, 1976. 2. Kraus, E. F.; and Van Abkoude, J. C.: Cost/Benefit Tradeoffs for Reducing the Energy Consumption of the Commercial Air Transportation System, Vol. I: Technical Analysis, Douglas Aircraft Co. NASA CR-137923, 1976. 3. Hopkins, J. P.: Study of the Cost/Benefit Tradeoffs for Reducing the Energy Consumption of the Commercial Air Transportation System, Lockheed-California CO. NASA CR-137926, 1976. 4. Staff Paper, Energy Consumption Characteristics of Transports Using the Prop- Fan Concept, Boeing Commercial Airplane Co. NASA CR-137938, 1976. 5. Goldsmith, I. M.: A Study to Define the Research and Technology Requirements for Advanced Turbo/Propfan Transport Aircraft, Douglas Aircraft Co. NASA CR-166138, 1981. 6. Dugan, J. F . ; Bencze, D. P.; and Williams, L. J.: Advanced Turboprop Technol- ogy Development. AIAA Paper 77-1223, 1977. 7. Welge, H. R.; and Crowder, J. P . : Simulated Propeller Slipstream Effects on a Supercritical Wing, Douglas Aircraft Co. NASA CR-152138, 1978. 8 . Dugan, James F.: The NASA High Speed Turboprop Program. SAE Paper 80-1120, 1 9 8 0 . 9. Welge, H. Robert: Prop-Fan Integration at Cruise Speeds, AGARD Paper 33, presented at AGARD Symposium on Aerodynamics of Power Plant Installation, Toulouse, France, May 11-14, 1 9 8 1 . 1 0 . Mendoza, J. P.: Interference Effects of Aircraft Components on the Local Blade Angle of Attack of a Wing Mounted Propeller. NASA TM 78587, 1979. 11. Welge, H. Robert; Newhart, Dan H.; and Dahlin, John A.: Analysis of Mach Number 0 . 8 Turboprop Slipstream Wing/Nacelle Interactions, Douglas Aircraft CO. NASA CR-166216, 1 9 8 1 . 12. Smith, Ronald C.; and Levin, Alan D.: Propfan Installation Aerodynamics of a Supercritical Swept Wing Transport Configuration. AIAA Paper 81-1563, 1 9 8 1 . 22 13. Levin, Alan D.; and Smith, Ronald C.: Installed Nacelle Drag-Improvement Tests of a M = 0.8 Turboprop Transport Configuration. NASA TM 84302, 1983. 14. Levin, Alan D.; and Smith, Ronald C.: Propfan Propulsion Integration Test Techniques. AIAA Paper 82-0577, 1982. 23 TABLE 1.- FILLET COORDINATES y = 88.80 c m ( q = 0.396) x, c m 284.25 284.48 285.12 285.75 286.39 287.02 287.66 288.29 288.93 289.56 290.20 290.83 291.47 292.10 292.74 293.37 294.01 294.64 295.28 295.91 296.55 z , c m upper -12.09 -1 1.63 -1 1.30 -1 1.07 -10.95 -10.80 -10.67 -10.52 - 10 ..3 1 -10.13 -9.98 -9 - 83 -9.73 -9.58 -9.45 -9 - 32 -9.19 -9.09 -9.02 -8.89 -8.81 z, c m lower -12.09 -12.40 -12.57 -12.65 -12.73 -12.80 -12.90 -12.98 -13.06 -13.11 -13.16 -13.20 -13.26 -13.34 -13.39 -13.46 -13.51 -13.56 -13.59 -13.64 -13.72 F a i r s tangent to LEX Y = 95.07 c m (TI = 0.424) x, c m 279.83 279.81 280.04 280.67 281.31 281.94 283.21 284.48 285.75 287.02 288.29 289.56 290.83 292.10 293.37 294.64 295.91 297.18 298.45 299.72 300.35 300.99 301.63 z, c m upper -13.64 -13.46 -13.31 -12.95 -12.83 -12.57 -12.24 -1 1.86 -1 1.58 -1 1.33 -1 1.05 -10.75 -10.46 -10.21 -9.96 -9.80 -9.58 -9.37 -9.19 -9.04 -8.92 -8-86 -8.69 z, c m lower -13.64 -13.72 -13.89 -13.84 -13.71 -13.69 -13.64 -13.56 -13.49 -13.41 -13.36 -13.34 -13.34 -13.36 -13.39 -13.41 -13.46 -13.49 -13.51 -13.56 -13.59 -13.61 -13.64 -- F a i r s tangent to LEX x is i n the wing reference plane L 24 C E W E G I T i ! L L PACE 4s OF POOR QUALITY (a) Baseline, f r o n t view. Figure 1.- Model i n s t a l l e d i n the 11-ft wind tunnel. 25 ORIGINAL PAGE I S Q E I O I R QUALITY (b) Baseline, rear view. Figure 1.- Continued. 26 (c) LEX configuration, upper surface. Figure 1.- Continued. 27 (d) LEX configuration, lower surface. Figure 1.- Continued. 28 (e) Close-up of LEX, upper surface. Figure 1.- Continued. 29 (f) LEX, fillet, strake configuration. Figure 1.- Concluded. c 31 ch2 / / / / ' I , - I u , w a 0 z - 5\ 0 II N 0 m I d 32 -15 E N ' -10 - - UPPER SURFACE -5 1 I I I I 290 295 300 280 285 X, cm (c) Fillet geometry at q = 0.396. -15 rc Figure 2.- Continued. -5 1 I I I I I 290 295 300 305 280 285 X, cm (d) Fillet geometry at q = 0.424. Figure 2.- Concluded. 33 L , c (d 4 a L a , 5 0 [L I cn 34 I CL \ - WING-BODY ‘ D - (a) Interference drag increments. Figure 4.- Illustration of drag and thrust determination. (b) Derivation of net thrust. Figure 4.- Continued. 35 L ” L t; 2 7 .I- o + J -00 U I n n 2 0 ci, w n m 4 m II J Z 4 I - m o :n I o I- m v) 8 SI 8 L z I- u + I- m I - % s 0 G 0 u SI 00 g SI 8 t; 7 .I- o I LL LL v) u n 4 m 9 J I =r aJ c 4 rn w 7 ? 3 n o L I 0 C 0 .r( II c, (d > .d J 2 n -o L a , cl cn cn n o 0 v I CJ 2 - 3 3 n o T 3 n o II J 2 n o z a I I uj J o 2 z 0 o 4 36 m 0 a 3 o o m m m m o o m m ;~r m a ootctnrr . . . . . . a _ _ _ _ U z 0 - i - a V Ln x z 0 - c a a 3 ‘3 - I . . Z 0 0 zzlz’ mmmmmm X K X K K X V \ X I + w 0 I1 z . e L a , 3 0 a a , VI 11 L 0 v) c 0 .rl a , L 3 VI v) a , L a I vl a , a 37 L z z z z i z ’ m m m m m m K K K I J I I K 6 p 000003 r m I I I I c c c I d I I I I I I I 38 c a a 3 u - I e c Y I e c W 39 - d I I I I 4 d I I I o o c u n ~ m m I o o m w c u ~ u a o o ( ~ u ) r r a ---- . _ _ _ . . w + a- a a v m w I - 4 3 a 2 J ; 000000 > w C 0 u 40 + . a a u v , W 0 d p 000000 v , 41 c n 0 co al 3 0 a c ? m I al L 3 M I I I I 42 W Z 0 - t - a a V Ln w D z 0 4 - +- 0 d OD CD 0 ru I d I I I I d d d I I I I d3 0 u3 3- cu 0 cu I + - d d I I I I I I I I d3 0 \ X I V \ X I 4 w 43 r 4 rr 4 - I I I I - - I I I I a Q, -0 3 c 0 u 44 + 3 0 0 \ X I c W I I I d3 r( 0, v) (d . n I I 9 0 \ X c W L a , 3 0 a G-i 0 c , 0 cu w E I tc a, M .I4 fxl 5 45 a w a a , 3 C C 0 u I r- a , 46 -- m m - - MMhnU-In3f m m m m r r 5 m m r r m m ( L oor-rss o o m i n r r (L o o r ~ r ~ r r a _ I _ _ _ _ . . . . . . w z 0 - c d f c W c W L ) - C 0 u I t - 47 I “I???? 5 rrrrr? - F F L P m F l U , a -- m m - - m c x m m r r n n r l r m f r f o c o r r m m ( I o o r r - ~ r o o m m r r p : oocuarr . . . . . . c a ( I V Ln w D z E! d g rJ00000 r c d I I I I - I I I I E : 0 u I r- a , 48 a 661166 f f f f f f f a a r r m m a m m a a o o oocncnrr o o - - r r . . . . . . Z 0 - F a a 3 0 J H 000000 > Ln 4 d I I I I -4 I I I I 0 \ X m a c W d I I I I + + d I I I I 0 \ X C 0 0 x rd J + a , C c W C 0 L a , 3 0 a G-l 0 c, 0 a, r, L l w I a 3 a , L 3 b n Lrr .r( 49 I - n - I I I I - - - I I I I d3 -0 al 3 c .rl 50 m m l D l D 0 0 o o m m r r a 9977p.r Z 0 a z 0 - & - a n 3 0 LL z a a a n a a 0 JJJJJJ u z z z z z z mmmmmm K K X K K K J @ 000000 t 4 c w - d l I I I l I I I I 4 I - w 51 dd3Pd6 l t f f f f f f c o c o r r m m a mm 0088PP a 00--rr . . . . . P W d p 000000 >. cn I co W 52 I - o a 0 E W c a I OI Q, 1 s M I 53 I ggIIoo a . . . . . . @ 000000 0 \ X 3 ? I c w 54 J ------ a z 0 I I - n L u \ X 55 J 0 t cn OOOOGO 0 \ X 4 c W U - - - - I I I I I I I 0 \ X 4 c W 56 I ?? , " rr a - f f mo) f && a m m o o m m r r a rrrr a --- . . . . W z 0 - Z 0 I I I I -4 & d I I I I 0 - 1D f 0 R l I I I I I H H H # I I I I d3 57 V \ M o a 3 0 I1 L 0 I v) (u -a rl 3 c P 6 ( 0 V I \ E (u L 3 bD D c W f f m m E & & n m f f i o o m r u r t - a rrrr . . . . a ---- w c a . . . J 2 C 0 V 58 6 6 5 & & a d d d d I I I 1 m m o o I I I I r u r u r r r r r r U W n _ _ _ _ c w a a , 3 C 4 L , C 0 u I 0 c 59 I --- a 3.7 mm 3.7 f (Dm a J p 0000 + in I < I - w I ‘c1 W o ) a 7 , -I < I - 0; C 0 V 0 \ K 60 a m w w w a . . . _ f g&88 -3 r m I I I I c . - I I L u u) a , L 5 0 u) a , L a cn 0 co 0 II I : .- L a, 3 0 a a , u) 4 5 L 0 u (d u) r : 0 d 9 oooa > cn 61 -& a E i & a I I I I d I I B o a x - I - 0 ul c 3 m I c c al L 3 M .?I Lrr 62 d I I I I - I I Z 0 - c n a - V In w 0 I c c d - I I I I I I z 0 - c a ( L V LFI u l CI z 0 - c a a 3 0 - m Y t t ? 64 a f P o ) - E d d a 0” p oooa m d3 65 F r n E & m a ~ Y x r u o m r m r r u a r o - r a --- I . . . . I w Z 4: I I I I - - I I + -I 0 C 0 u I c c 66 a M K -l 4 lDY a a 1 9 -I -- a I I I I 4 I I - n z Ei Y I I I I I - I 0 II . . . VI Q, L 3 In VI ( I , L a 0 (d z: 67 f l n mm f && a - I I I I I c a a V I n w D I I I I - I c, C 0 V 68 OKI r ( u a 77 a _ _ W I I I I - I c c 4 L, C 0 u I N 0, L c 3 tro 69 00 rcu a 7 r a 4 - W z 0 Z 0 L J 0 I I I I d I I I I d I I 70 C 0 u 71 61 Q 616 E I N c e l I I I I 72 C 0 z c - .- ..- I I I I 3 M) C 0 0 a, 10 (d I d3 73 74 -- m i - corn M f L+ I I I I - - - I I d3 75 O f o r a o r a - w d d d e - I I I I I l d3 76 O f o r a 0': a -_ w I I I I - - I I I I I I - - I I 77 I O f or o r E 11 w 0 C 0 u 78 ... I ?? r r - lnln a f f 3.7 mm mm a M D O F a o r a _ _ w z 0 - L a ( r V m W 0 z 0 - J on X w r n J & t 6, C r( I I I I - I I z a, 79 M O o r o r E -1 W c , C 0 u 80 dd g B& c P C 0 u 81 I I I I d I 3 a a, 3 C c, s 2 0 V I = r a, L 7 bo . , -I crr . , -I T I I I 4 I I 82 a M O o r (r o r a -_ W z El L 0 JJ v z z 11 z a n mm N 0 9 I H a - I I I I I d3 84 I I I I d I 85 I m 0 d I I I I I d3 86 I I I I I d I - I I I I - 0 ru M 87 aa a m f f r m f a m ( L z 0 I I I I d I l I I I l P a, 3 89 -1 -- a lolo fa f f lorn a W I I I c I 90 (a) Wing; upper surface. (b) Wing/nacelle; upper surface. Figure 16.- Oil flow visualization; M = 0.8, a = 2 O . 91 ORIGINAL PAGE B OE POOR QUAL$I-yi ( c ) WBNL; upper s u r f a c e , jet off. (d) WBNLP; lower surface, jet off. Figure 16.- Continued. 92 (e) WBNL; lower surface, jet on. ( f ) WBNLP; upper surface, windmill. Figure 16.- Continued. 93 91tMNAL PAGE-IS 980R QUALITY 4 .. ' (g) WBNLP; upper s u r f a c e , c r u i s e power. (h) WBNLP; lower surface, c r u i s e power. Figure 16.- Concluded. 94 (a) Wing-body, upper surface; M = 0.80, a = 1.99". Figure 17.- Propeller-off pressure isobars. 95 ( b ) Wing-body, lower s u r f a c e ; M = 0.80, a = 1.99". Figure 17.- Continued. 96 (c) Wing-body-nacelle, upper surface; M = 0.80, a = 1.95", EPR = 1.68. Figure 17.- Continued. 97 (d) Wing-body-nacelle, lower surface; M = 0.80, a = 1.95", EPR = 1.68. Figure 17.- Continued. 98 (e) Wing-body-nacelle LEX, upper surface; M = 0.80, a 1.96", EPR = 1.73. Figure 17.- Continued. 99 .-----_____--- (f) Wing-body-nacelle-LEX, lower surface; M = 0.80, a 1.96", EPR = 1.73. Figure 17.- Concluded. 100 (a) Baseline configuration, upper surface; M = 0.87, a = 1.94", EPR = 1.77, rpm = 8471. Figure 1 8 . - Propeller-on pressure isobars. 101 (b) Baseline configuration, lower surface; M = 0.81, a = 1-94', EPR = 1.77, rpm = 8471. Figure 18.- Continued. 102 (c) Baseline + LEX, upper surface; M = 0.80, a 1.96O, EPR = 1.77, rpm = 8494. Figure 18.- Continued. (d) Baseline + LEX, lower surface; M = 0.80, a = 1.96O, EPR = 1.77, rpm = 8494. Figure 18.- Concluded. 104 (a) Upper surface; M = 0.80, a = 1.94O, EPR = 1.73, rpm = 8486. Figure 19.- Wing-body-nacelle-LEX-fillet-strake pressure isobars, propeller on. ( b ) Lower s u r f a c e ; M = 0.80, a = 1.94', EPR = 1.73, rpm = 8486. Figure 19.- Concluded. 106 t - a [L 3 . . . 0 c u v) P 4 0 c h Q) a; L L z v 111 o mmm I 0 c u E 0 1 0 N ' 3 109 U W 3 c z 0 L LL z u x x x o mmm J 0 > 000 i n 8 0 -1 3 111 lL z u x x x J o mmm p 000 a3 112 2 0 - I - a a 0 o mmm 2 L z z z v X X X d 000 > m d c, E r n 0 0 w a a, a I (v c 1 1 4 E 0 I '3 115 a3 116 Lo 0 L n o Lo r Ln cu I I 0 a l 0 II x 0 0 II t 3 (d - W a, 0 c L 0 G I L Q) a L a, 4 4 Q, a 0 L LL I M N Q, L 3 M 4 Lr, 2 13Nu 118 I I z 0 l - 3 a a 1 0 ' 13 d o J d -I LD M 0 4 M I I m (u I I 119 13No 120 > i n h 0 t : 0 N II t l 0 v 0 m -- I I 0 II I 121 Z 0 L a ! 3 '3 a a z z i o mm L - l u 1111 13Nu 122 o u c o u d o J d 13 --I-- L D C I I I . I - 0 a0 0 II f: - 0 a II 8 n Q) W I m (u Q) L 7 M . a a, a 3 7 4 0 C 0 u a, - v a, a 3 4 0 c 0 u I M N a, L 3 M Lrr . , -I 13Nu 126 . + d a L 13Nu 128 rum 8! f cnm d p 00 >- i n I I 13 d o J d o m 0 L n 0 Ln 0 Ln 0 m 0 mcu L n c u 0 r L n ru 0 r Ln ru cu W r u W ,-. d d I d d3 d o J d 129 13Nu a a 4 C 0 V L 130 cu 0 + W z I - o 0 0 0 0 0 0 0 0 d - I I I I I I I I I I cu 0 . e w z I - o 0 o r l n ru o r l n ru o r m r u o r m c u o a o r r r r U I co co C D L O L I - J l n l n 2 2 j j 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 . . . . a3 . L cu cu 0 X w 4 I I m 0 0 .I4 a , I1 c J J X c 0 cu 0 8 - I - d p oooa >. I n '3 I d p oaoad f 0 cu 0 . I - w Z I - o 0 0 I n r CD 0 0 I I w Z I - o 0 f f 0 0 L . I I I I I I I w3 I I I I I I U aJ U 3 4 0 C 0 u I W cu a J M r . L l 5 .rl L n 0 LD U J f f 0 0 0 0 3- 0 cu 0 . I - W z I - o 0 0 l r l 0 M cu cu z 0 a a 0 LL a a a a a z z z z z z o mmmmm 0 X X X X I Z J 0 I - 2 p o o o a i n '3 133 f 0 cu 0 . I - w z I - o 0 m r u o r m r u P P P ~ U I U I 0 0 0 0 0 0 o h o ~ o L t i o m o o r t n cu o r m r u o a m Ln m Ln 3- f f f 0 0 0 0 0 0 0 0 0 I d p oooad o n 134 Z 0 c d p oooa In 135 B - L I I I co Ln Lo Ln 0 0 I I I I I I I I I I '3 136 $000 t ( I , I o c o c SSa3 o m o m c u - 4 0 0 0 0 0 0 0 0 0 137 C 0 X w Cl n n U I cn N 138 z 0 - I - d T I 4 0 a , ru J2-L U C c 3 - 3 - f f f f j - 0 0 0 0 0 0 0 0 C 0 0 0 0 0 0 0 0 C 139 0 r i c u r c l 0 r 3 - 3 - a ' 0 0 0 0 k k 0 X W + l n (d v f I T 0 0 0 0 0 0 c L d a d p oooa >- VI 0 0 0 G; cw 0 X W J n (d v 141 P Q) 3 G c, C 0 u I m a , L l 3 M Crr .rl - .rl c . m 03V 142 J f 000 2 . m ORIGINAL PAGE I S Q B B X ’ B Q U w 1. Report No. 2. Government Accession No. NASA TM-86705 4. Title and Subtitle AERODYNAMIC AND PROPELLER PERFORMANCE CHARACTER- ISTICS OF A PROPFAN-POWERED, SEMISPAN MODEL 3. Recipient's Catalog No. 5. Report Date 7. Author(s) Alan D. Levin, Ronald C. Smith, and Richard D. Wood 9. Performing Organization Name and Address Ames Research Center 8. Performing Organization Report No. A-8.5175 10. Work Unit No. 11. Contract or Grant No. Moffett Field, CA 94035 12. Sponsoring Agency Name and Address National Aeronautics and Space Administration Washington, DC 20546 13. Type of Report and Period Covered Technical Memorandum 14. Sponsoring Agency code 535-03-1 1 17. Key Words (Suggested by Author(sl1 Turboprop Powerplant integration Drag reduction 19. Security Classif. (of this report1 Unclassified Subject category - 05 20. Security Classif. (of this page) 21. NO. of pages 22. Rice' Unclassified 145 A07
16673	https://pmc.ncbi.nlm.nih.gov/articles/PMC11324266/	Lung Adenocarcinoma Presenting as Early Cardiac Tamponade: A Case Report - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Case Rep Oncol . 2024 Jul 25;17(1):779–787. doi: 10.1159/000540183 Search in PMC Search in PubMed View in NLM Catalog Add to search Lung Adenocarcinoma Presenting as Early Cardiac Tamponade: A Case Report Volha Chapiolkina Volha Chapiolkina a SBH Health System, Department of Medicine, Internal Medicine, Bronx, NY, USA Find articles by Volha Chapiolkina a, Homa Saadati Homa Saadati a SBH Health System, Department of Medicine, Internal Medicine, Bronx, NY, USA Find articles by Homa Saadati a,✉, Nehemias Antonio Guevara-Rodriguez Nehemias Antonio Guevara-Rodriguez a SBH Health System, Department of Medicine, Internal Medicine, Bronx, NY, USA Find articles by Nehemias Antonio Guevara-Rodriguez a, Garry Francis-Morel Garry Francis-Morel b The Ohio State University College of Pharmacy, MS Translational Pharmacology, and Clinical Trial Design, Columbus, OH, USA Find articles by Garry Francis-Morel b Author information Article notes Copyright and License information a SBH Health System, Department of Medicine, Internal Medicine, Bronx, NY, USA b The Ohio State University College of Pharmacy, MS Translational Pharmacology, and Clinical Trial Design, Columbus, OH, USA ✉ Correspondence to: Homa Saadati, homasaadati@gmail.com ✉ Corresponding author. Received 2024 Jan 31; Accepted 2024 Jun 28; Collection date 2024 Jan-Dec. © 2024 The Author(s). Published by S. Karger AG, Basel This article is licensed under the Creative Commons Attribution-NonCommercial 4.0 International License (CC BY-NC) ( Usage and distribution for commercial purposes requires written permission. PMC Copyright notice PMCID: PMC11324266 PMID: 39144247 Abstract Introduction Lung cancer remains the most common cause of cancer death in the USA and worldwide despite continued advances in lung cancer screening and treatment. Pericardial effusion (PerF) has been found in up to 50% of postmortem patients with cancer; lung and breast cancers are the most frequent malignancies. Furthermore, it is a sign of poor outcomes with fewer than 5 months of survival. Nevertheless, PErF with or without tamponade as a presentation of lung cancer is uncommon. Case Presentation We present a 72-year-old male without medical history who presented with 1 month of cough with white sputum and shortness of breath, progressively worsening, associated with weight loss (20 pounds). Further studies demonstrated early cardiac tamponade secondary to malignancy. Conclusion Cardiac tamponade can arise secondarily from various etiologies and have different presentations depending on the cause. In general, it is a slowly developing and clinically silent disease process. Therefore, malignant PerFs can rarely present with hemodynamic instability and be the initial manifestation of an underlying malignancy. Our case review presents a rare case of metastatic lung adenocarcinoma manifesting as early symptomatic cardiac tamponade and as an emergency. The results might be life-threatening if this presentation is not recognized and managed appropriately. Clinicians must be aware of such atypical presentations of thoracic malignancies to take action adequately. Keywords: Lung cancer, Adenocarcinoma, Tamponade, Diagnosis, Treatment, Oncology Introduction Lung cancer remains the second most diagnosed cancer in females and males. However, it is still the most common cause of cancer death in the USA, accounting for 21% of deaths related to cancer. Furthermore, despite continued advances in lung cancer screening and treatment, it is the principal cause of death worldwide . Close to 237,000 people were diagnosed with lung cancer in 2022 . The majority of patients with lung cancer present for diagnostic evaluation because of suspicious symptoms such as cough, hemoptysis, dyspnea, and chest pain or an incidental finding on chest imaging . Lung cancer arises from the epithelium cells in the respiratory system and can be divided into two broad categories – small cell lung cancer and non-small lung cancer . Small cell lung cancer is a poorly prognostic malignant tumor derived from cells exhibiting neuroendocrine characteristics, accounting for approximately 15% of all lung cancer cases. NSCLC accounts for the remaining 85% of cases. Adenocarcinoma is one of the NSCLC subtypes of lung cancer, accounting for 38.5% of all lung cancer cases. Cardiac metastasis from cancer is not uncommon, but cardiac tamponade complicating malignant pericardial effusion (PerF) is a rare presentation of any kind of malignancy . It is a fairly rare presentation of lung adenocarcinoma and always points toward a grim prognosis. Malignant cardiac tamponade is a serious complication of cancer overall and has been recognized in 2–15% of cases of postmortem autopsy in patients with a lung cancer history . PerF or cardiac tamponade secondary to lung cancer is far less common and could be identified during life . The most common primary tumor that involves pericardium is lung cancer; other types include breast, lymphoma, leukemia, and esophageal cancer. The clinical manifestation of malignant pericardium can present as pericarditis, cardiac tamponade, malignant PerF, or pericardial constriction. In patients with known lung cancer diagnosis, the manifestation of symptomatic malignant PerF has been associated with poor outcomes and an average survival of less than 4 months . Malignant cardiac tamponade is an understudied oncological emergency. This case provides an unusual presentation of lung adenocarcinoma and shows the importance of intervention in a timely manner in malignancy associated with pleural effusions and malignant cardiac tamponade. Quick recognition can be crucial to saving a life, providing symptomatic relief, and helping to diagnose the etiology . Case Narrative We present a 72-year-old male without medical history who presented to the emergency department with 1 month of cough associated with white sputum and shortness of breast, progressively worsening, and weight loss of 20 pounds. Social history was significant for 25-year smokers, with 25 packs per year. Vital signs at arrival were blood pressure of 126/72 mm Hg, heart rate of 110 per minute, respiratory rate of 16 per minute, and temperature of 97.7 F. Initial blood work showed mild hyperkalemia (K: 5.7 mEq/L), hyponatremia (132 mEq/L), creatinine of 1.6 mg/dL, and mild transaminitis (Tables 1, 2). The electrocardiogram on presentation showed sinus rhythm and tachycardia of 100 beats per minute with low voltage complexes (Fig. 1). The chest X-ray demonstrated a mass-like structure in the right lower lung lobe with infiltrates. Therefore, computer tomography of the chest (CT chest) demonstrated moderate to large PerF, right pleural effusion, and a posteriorly layering component with multiple lymphadenopathies (Fig. 2). Because of these findings, pneumonia was considered, and the patient was started on antibiotics, which were discontinued later. Table 1. Pertinent laboratories upon admission \| Variable \| Results \| Reference range \| --- \| Potassium (K) \| 5.7 mEq/L \| 3.5–5.3 mEq/L \| \| Sodium (Na) \| 132 mEq/L \| 135–145 mEq/L \| \| Blood urea nitrogen (BUN) \| 40 mg/dL \| 8–23 mg/dL \| \| Creatinine \| 1.5 mg/dL \| 0.6–1.2 mg/dL \| \| AST \| 86 IU/L \| 8–33 IU/L \| \| ALT \| 110 IU/L \| 4–36 IU/L \| \| White blood cell (WBC) count \| 9.4 × 10 3/μL \| 4.2–9.1 × 10 3/μL \| \| Absolute lymphocyte count \| 1.01 × 10 3/μL \| 1.32–3.57 × 10 3/μL \| \| Red blood cell (RBC) count \| 4.25 × 10 6/μL \| 4.63–6.08 × 10 6/μL \| \| Hemoglobin \| 13.3 g/dL \| 13.7–17.5 g/dL \| \| Mean corpuscular volume (MCV) \| 95.8 fL \| 79.0–92.9 fL \| \| COVID-19 test \| Negative \| Negative \| \| Flu and RSV test \| Negative \| Negative \| Open in a new tab Table 2. Pertinent laboratories on the first day of admission \| Variable \| Results \| Reference range \| --- \| Potassium (K) \| 5.4 mEq/L \| 3.5–5.3 mEq/L \| \| Sodium (Na) \| 137 mEq/L \| 135–145 mEq/L \| \| Blood urea nitrogen (BUN) \| 44 mg/dL \| 8–23 mg/dL \| \| Creatinine \| 1.4 mg/dL \| 0.6–1.2 mg/dL \| \| AST \| 43 IU/L \| 8–33 IU/L \| \| ALT \| 87 IU/L \| 4–36 IU/L \| \| White blood cell (WBC) count \| 12.0 × 10 3/μL \| 4.2–9.1 × 10 3/μL \| \| Absolute lymphocyte count \| 1.01 × 10 3/μL \| 1.32–3.57 × 10 3/μL \| \| Red blood cell (RBC) count \| 3.91 × 10 6/μL \| 4.63–6.08 × 10 6/μL \| \| Hemoglobin \| 12.0 g/dL \| 13.7–17.5 g/dL \| \| Mean corpuscular volume (MCV) \| 96.4 fL \| 79.0–92.9 fL \| \| COVID-19 test \| Negative \| Negative \| \| Flu and RSV test \| Negative \| Negative \| Open in a new tab Fig. 1. Open in a new tab First ECG demonstrating sinus rhythm and tachycardia of 100 beats per minute with low voltage complexes. Fig. 2. Open in a new tab Sagittal and coronal view of the CT chest without contrast demonstrated moderate to large PerF, right pleural effusion, and a posteriorly layering component with multiple lymphadenopathies. During his stay at ED, the patient developed persistent tachycardia; therefore, an electrocardiogram (Fig. 3) was performed, showing atrial fibrillation with rapid ventricular response, therefore prompted an echocardiogram showing a large PerF collapsing the right atrium, an ejection fraction of 56%, and moderate left ventricular hypertrophy; furthermore, the patient was tachycardic during the test (Fig. 4). Hence, an emergent pericardiocentesis was performed by draining 60 mL of hemorrhagic pericardial fluid, and further fluid analysis revealed malignant epithelial cells consistent with metastatic lung adenocarcinoma. A pericardial window was placed per the recommendation of cardiothoracic surgery and was discontinued 4 days after placement; drainage was 1,040 mL in the first 24 h and 1,920 in total. Furthermore, the patient had a pigtail due to a significant left pleural effusion. Histopathology (Fig. 5) showed PDL1 70%, tumor mutational burden was high 14 Muts/Mb, MS stable with IHC + CK7 and Ber EP4, negative p40, TTF1, CK20, CALR, Napsin A. Further brain magnetic resonance imaging showed 5-mm enhancing lesions in the right superior cerebellum and 6 mm at the left inferior cerebellum, consistent with metastasis. Cardiology was consulted due to the new onset of atrial fibrillation, which was considered secondary to the malignant PerF. A CT abdomen pelvis with contrast was performed and was unremarkable. Fig. 3. Open in a new tab Second ECG showing the new onset of atrial fibrillation (Afib) with rapid ventricular response. Fig. 4. Open in a new tab Echocardiogram showing a large PerF collapsing the right atrium, an ejection fraction of 56%, and moderate left ventricular hypertrophy. Fig. 5. Open in a new tab Histopathology showed PDL1 70%, tumor mutational burden was high 14 Muts/Mb, MS stable with IHC + CK7 and Ber EP4, negative p40, TTF1, CK20, CALR, Napsin A. Nevertheless, he was never started on anticoagulation due to the high risk of bleeding. The patient was planning to start pembrolizumab for further treatment of metastatic adenocarcinoma of the lung. After the resolution of pericardial and pleural effusion, he was discharged to be followed up as a patient with oncology. Unfortunately, the patient died 1 month after the diagnosis due to complications of lung cancer. Discussion Lung cancer, a multifaceted disease with varying clinical presentations, remains a significant public health challenge and is a predominant malignancy in terms of incidence and mortality. Globally, it accounted for approximately 2.1 million new cases and 1.8 million deaths in 2018, making it the most commonly diagnosed cancer and the leading cause of cancer death . Lung cancers can be categorized into four primary histological subtypes: adenocarcinoma, squamous cell carcinoma, large-cell carcinoma, and small cell carcinoma. Adenocarcinoma represents approximately 50% of all lung cancers, marking it as the predominant type. Furthermore, it is the leading lung cancer type among nonsmokers, light smokers, and women . The incidence of lung cancer varies by region, with high rates observed in countries with high tobacco consumption or exposure to carcinogens like radon. Notably, in some developed countries, incidence rates have plateaued or even declined, likely due to effective public health campaigns against smoking and better occupational health practices . Disparities in lung cancer presentation are evident across sex and racial lines. Historically, males exhibited a higher incidence of lung cancer than females. However, recent trends show a narrowing of this gap, possibly due to changing smoking behaviors among women in the latter half of the 20th century . Regarding racial disparities, black men in the USA have consistently had higher lung cancer incidence rates compared to white men, even though the gap has narrowed over time. Differences in socioeconomic status, access to healthcare, and exposure to risk factors contribute to these disparities . The diagnosis of lung cancer is mostly made following evaluation for respiratory signs and symptoms or incidental finding of a nodule or mass during routine imaging. The most common presentations of lung cancer are cough (50–75%), hemoptysis (25–50%), dyspnea (25%), and chest pain (20%). Other symptoms include hemoptysis, hoarseness, superior vena cava, and Pancoast syndrome. It can metastasize to other organs in the body, including the liver, adrenal glands, bones, brain, and pericardium . In advanced malignant disease, cardiac and pericardial involvement is frequently observed. PerF is a common postmortem finding in such patients. A review of the literature demonstrates that lung cancer is the most prevalent cause of this complication, accounting for 37% of cases , followed by breast cancer . PerF could develop during the course of malignancies, or it could be the initial early presentation, leading to the diagnosis of underlying neoplasm. With a history of cough and weight loss and early onset of cardiac tamponade as an initial presentation of lung adenocarcinoma, our patient’s case was quite unique. The development of pericardial tamponade depends both on fluid and rate of accumulation. The pericardial space contains a thin layer of less than 50 mL of an ultrafiltrate of plasma called pericardial fluid . Furthermore, it can be present acutely, subacutely, or occultly. Malignancy-related PerF resulting in tamponade is subacute and is most commonly from cardiac metastases (e.g., lung cancer, breast cancer, lymphoma) rather than primary cardiac tumors. Our patient presented with dyspnea, which is the most common symptom of cardiac tamponade and sometimes can be a presenting sign of the neoplasm. The classic Beck triad of cardiac tamponade consists of hypotension, dilated neck veins, and muffled heart sounds. Nevertheless, one positive finding of the Beck triad is that it has 50% sensitivity to diagnose pericardial tamponade (28.0–72.0%) . Fluid accumulation in the pericardial cavity leads to intracardiac pressure change and impairs ventricular filling, resulting in ventricular diastolic collapse and decreased cardiac output. Cardiac tamponade is a cardiac emergency requiring prompt diagnosis and intervention in order to stabilize the patient. Properly understanding the spectrum of clinical and hemodynamic changes in patients with pericardial tamponade is essential for hospitalists to perform pericardiocentesis promptly . Malignant PerF is usually diagnosed by cytologic examination of pericardial fluid. PerF caused by neoplasms has the worst prognosis and highest mortality rate compared to other causes. The average survival when lung cancer manifests as cardiac tamponade is 1.5 months. The lymphatic drainage of the heart provides some anatomical explanations for the prevalence of carcinoma of the lung as the cause of cardiac tamponade and the discrepancy in finding tumor cells in the pericardial fluid but not in the pericardium . As our patient was found to be symptomatic on initial presentation due to the onset of cardiac tamponade, he had undergone pericardiocentesis, which was a diagnostic procedure as well as the treatment procedure. The evaluation of fluid cytology revealed malignant epithelial cells consistent with metastatic lung adenocarcinoma. The presence of malignant cells in the pericardial fluid analysis is usually consistent with an advanced stage of the neoplasm and usually has a negative impact on staging. In our case, the patient had advanced disease with brain mets, which is consistent with the current literature. Treatment of PerF or pericarditis secondary to malignancy, in general, is not straightforward. In cases where a patient presents with early cardiac tamponade or cardiac tamponade, emergency surgical drainage is the most proper approach [19, 20]. In most cases, the frequency of reaccumulation of the PerF is greater than 90%, even with aggressive treatment [19, 20]. Therefore, a pericardial window following drainage may be needed for durable palliation in suitably fit patients [21, 22]. Other studies have shown that the recurrence rate decreases with intrapericardial platinum bleomycin as sclerosants and nonoperative therapies. Nevertheless, the treatment is based on the burden and advance of the disease. With the new target treatments, there has been a noticeable improvement and an increase in monthly survival, but the prognosis continues to be poor [23, 24]. Colchicine and nonsteroidal anti-inflammatory treatments have no significant benefits [9, 24, 25]. This case is a good reminder that although PerF is not a commonly reported lung cancer presentation, the reverse is not valid. It has been reported that malignancy, most importantly primary lung cancer, is the cause of symptomatic PerF in up to a fifth of cases when basic workup is non-revealing. In our case, the initial basic evaluation did not reveal a potential cause for the large PerF; further, the workup with the chest CT scan revealed an underlying lung mass. The CARE Checklist has been completed by the authors for this case report, attached as supplementary material (for all online suppl. material, see Conclusion Cardiac tamponade can arise secondarily from various etiologies and have different presentations depending on the cause. In general, it is a slowly developing and clinically silent disease process. Therefore, malignant PerFs can rarely present with hemodynamic instability and be the initial manifestation of an underlying malignancy. Our case review presents a rare case of metastatic lung adenocarcinoma manifesting as early symptomatic cardiac tamponade and as an emergency. The results might be life-threatening if this presentation is not recognized and managed appropriately. Clinicians must be aware of such atypical presentations of thoracic malignancies to take action adequately. Statement of Ethics The study was approved by the Ethical Committee of St. Barnabas Hospital, City University of New York, New York, NY, USA. However as it was a case report, no IRB approval was obtained. Nevertheless, written consent was taken and is available for your review upon request. This retrospective review of patient data did not require ethical approval in accordance with local/national guidelines. Informed written consent was obtained from the patient for the publication of this case report and the accompanying images. Conflict of Interest Statement None of the authors have a financial and nonfinancial competing interest. Funding Sources This case report has not been financially supported by any entity. Author Contributions Volha Chapiolkina and Homa Saadati wrote the initial draft and revised the manuscript. Nehemias Antonio Guevara-Rodriguez contributed to the interpretation of the data and was involved in the management of the patient. Garry Francis-Morel supervised all phases of writing. All authors have read and approved of the manuscript for submission. Funding Statement This case report has not been financially supported by any entity. Data Availability Statement The data that support the findings of this study are not publicly available due to privacy reasons but are available from the corresponding author upon reasonable request. Supplementary Material. Supplementary_material-Suppl.1-s1.pdf (637.5KB, pdf) Supplementary Material. Download video file (3.8MB, mp4) Supplementary Material. Download video file (4.4MB, mp4) Supplementary Material. Download video file (6.8MB, mp4) Supplementary Material. Download video file (6MB, mp4) References Cancer facts and figures 2022. 1930. [Google Scholar] Hutchinson BD, Shroff GS, Truong MT, Ko JP. Spectrum of lung adenocarcinoma. Semin Ultrasound CT MR. 2019;40(3):255–64. [DOI] [PubMed] [Google Scholar] Dela Cruz CS, Tanoue LT, Matthay RA. Lung cancer: epidemiology, etiology, and prevention. Clin Chest Med. 2011;32(4):605–44. 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16674	https://artofproblemsolving.com/community/c2202h1036222?srsltid=AfmBOoq9snUouBTqDkN-YyIPAsmABrxGfKGJPu9Xa3GWWQVG6Xl2xB-U	Blog of the (Former) Rising Olympian : Brilliant.org Level 5 Sample #3 Community » Blogs » Blog of the (Former) Rising Olympian » Brilliant.org Level 5 Sample #3 Sign In • Join AoPS • Blog Info Blog of the (Former) Rising Olympian ==================================== Brilliant.org Level 5 Sample #3 by djmathman, Jan 22, 2013, 9:18 PM This is a brilliant problem (pun intended). Problem wrote: Given positive real values and such that has the form , where ,, and are integers and is not divisible by a square. What is ? Solution Note that the system of equations can be transformed into We can thus transform the problem into one of a geometrical nature by constructing segments of length ,, and inside a triangle as shown: Next, using the fact that the area of any triangle is , we get that so . Adding all three original equations together gives and plugging our value of in and simplifying gives . Finally, note that The requested answer is . This post has been edited 4 times. Last edited by djmathman, Feb 1, 2023, 6:49 PM Reason: align equation fixes, made diagram smaller 4 nerdy comments Contribute Something Comment 4 Comments The post below has been deleted. Click to close. This post has been deleted. Click here to see post. hmm this is a lot more motivated than my solution. But my solution uses almost no trig (just a LoC at the end to compute and only transformational geo so yay ^_^ This post has been edited 1 time. Last edited by dinoboy, Jan 22, 2013, 9:30 PM by dinoboy, Jan 22, 2013, 9:20 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. Does anyone have a pure algebraic solution? (Just wondering. Wow, I never thought of using geo.) by mathway, Jan 22, 2013, 9:42 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. Seeing the perfect squares on the you should be able to deduce that the equations probably came from an application of the Law of Cosines. A problem very similar to this appeared in PSS. I think it was also a Russian Olympiad problem, but not sure on that. by moose97, Jan 22, 2013, 10:44 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. 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Shouts Submit aural by centslordm, Jul 9, 2025, 9:01 PM whoaaaaaa you've used this blog for a while! by plum28, May 14, 2025, 6:36 PM dj so orz by Yiyj1, Mar 29, 2025, 1:42 AM legendary problem writer by Clew28, Jul 29, 2024, 7:20 PM orz by balllightning37, Jul 26, 2024, 1:05 AM hi dj by OronSH, Jul 23, 2024, 2:14 AM i wanna submit my own problems lol by ethanzhang1001, Jul 20, 2024, 9:54 PM hi dj, may i have the role of contributer? by lpieleanu, Feb 23, 2024, 1:31 AM This was helpful! by YIYI-JP, Nov 23, 2023, 12:42 PM waiting for a recap of your amc proposals for this year by ihatemath123, Feb 17, 2023, 3:18 PM also happy late bday man! i missed it by 2 days but hope you are enjoyed it by ab456, Dec 30, 2022, 10:58 AM Contrib? by MC413551, Nov 20, 2022, 10:48 PM tfw kakuro appears on amc by bissue, Aug 18, 2022, 4:32 PM Hi dj by 799786, Aug 10, 2022, 1:44 AM Roses are red, Wolfram is banned, The best problem writer is Djmathman by ihatemath123, Aug 6, 2022, 12:19 AM how did you do that cursor by mathgenius64, Sep 1, 2011, 10:14 PM Finally posted that problem! And hooray for pencil cursor! by djmathman, Aug 24, 2011, 6:49 PM Geo 2 is correct though by sjaelee, Aug 24, 2011, 12:21 AM Hello...Lol yesterday sjaelee posted Geomretry #2. Isn't it supposed to be Geometry #2? Lol. by mcqueen, Aug 23, 2011, 11:57 PM Delete geo #4! Plese! by sjaelee, Aug 22, 2011, 11:41 PM Hrm, there's this problem that I've been too lazy to post. I think I should soon. by djmathman, Aug 22, 2011, 7:02 PM Why hello...funny I'm saying this when I know you'll be unresponsive because you're V/LA for a week. by mcqueen, Aug 14, 2011, 10:47 AM But that's because those problems are the easiest to find. by djmathman, Aug 12, 2011, 3:49 AM Yea, I kind of keep forgetting to change that. by djmathman, Aug 11, 2011, 11:27 AM I noticed that only algebra and number theory come up (trig was basically algebra) by sjaelee, Aug 11, 2011, 11:22 AM There you go. by djmathman, Aug 7, 2011, 2:07 PM cough contrib by Mrdavid445, Aug 7, 2011, 12:40 PM Ah, i'm sure you will use this blog well. by tc1729, Aug 7, 2011, 2:10 AM First shout! by djmathman, Aug 5, 2011, 6:58 PM 365 shouts Contributors 54math • agbdmrbirdyface • ahaanomegas • AlcumusGuy • AwesomeToad • Binomial-theorem • bluephoenix • budu • CaptainFlint • chezbgone • cire_il • csmath • djmathman • droid347 • Einstein314 • El_Ectric • forthegreatergood • gamjawon • giratina150 • hwl0304 • infiniteturtle • IsabeltheCat • Lord.of.AMC • mathguy623 • mathman523 • MathSlayer4444 • mathwizard888 • Mrdavid445 • niraekjs • nsun48 • phi_ftw1618 • pinetree1 • Seedleaf • shiningsunnyday • sjaelee • ssilwa • sunny2000 • tc1729 • va2010 • W.Sun • wu2481632 • yugrey Tags Real LifealgebrageometryAMCnumber theorypuzzlesCMUAwesomeMathcombinatoricsNIMOtrigonometryAIMEcalculusCMIMCPutnamvector analysisVieta ssymmetryAoPSARMLHMMTlinear algebraOmoProblemsPuMACreal analysisSequencesUSAMO2016Acegikmoqsuwy2000analytic geometrycollegecontestFactoringIMOinequalitiesinternal tangentslength chasingmathNYSMLOlympiadPerfect SquaresSawayamatest-solvetractorTST About Owner Posts: 7961 Joined: Feb 23, 2011 Blog Stats Blog created: Aug 5, 2011 Total entries: 567 Total visits: 565166 Total comments: 1524 Search Blog Something appears to not have loaded correctly. 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16675	https://www.engineeringtoolbox.com/tractive-effort-d_1783.html	Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications! Car - Traction Force Adhesion and tractive force between car wheel and surface. The tractive force between a car wheel and the surface can be expressed as F = μt W = μt m ag (1) F = traction effort or force acting on the wheel from the surface (N, lbf) μt = traction - or friction - coefficient between the wheel and the surface W = weight or vertical force between wheel and surface (N, lbf)) m = mass on the wheel (kg, slugs) ag = acceleration of gravity (9.81 m/s2, 32.17405 ft/s2) Traction Coefficients for normal Car Tires Car Tires - Traction Coefficients \| Surface \| Traction Coefficient - μt - \| \| Wet Ice \| 0.1 \| \| Dry Ice/Snow \| 0.2 \| \| Loose Sand \| 0.3 - 0.4 \| \| Dry Clay \| 0.5 - 0.6 \| \| Wet rolled Gravel \| 0.3 - 0.5 \| \| Dry rolled Gravel \| 0.6 - 0.7 \| \| Wet Asphalt \| 0.6 \| \| Wet Concrete \| 0.6 \| \| Dry Asphalt \| 0.9 \| \| Dry Concrete \| 0.9 \| Example - Traction Force on an Accelerating Car The maximum traction force available from one of the two rear wheels on a rear wheel driven car - with mass 2000 kg equally distributed on all four wheels - on wet asphalt with adhesion coefficient 0.5 - can be calculated as Fone_wheel = 0.5 ((2000 kg) (9.81 m/s2) / 4) = 2453 N The traction force from both rear wheels Fboth_wheels = 2 (2452 N) = 4905 N Note! - that during acceleration the force from the engine creates a moment that tries to rotate the vehicle around the driven wheels. For a rear drive car this is beneficial by increased vertical force and increased traction on the driven wheels. For a front wheel driven car the traction force will be reduced during acceleration. The maximum acceleration of the car under these conditions can be calculated with Newton's Second Law as acar = F / m = (4904 N) / (2000 kg) = 2.45 m/s2 = (2.45 m/s2) / (9.81 m/s2) = 0.25 g where acar = acceleration of car (m/s2) The minimum time to accelerate from 0 km/h to 100 km/h can be calculated as dt = dv / acar = ((100 km/h) - (0 km/h)) (1000 m/km) (1/3600 h/s) / (2.4 m/s2) = 11.3 s where dt = time used (s) dv = change in velocity (m/s) Accelerating Car Calculator This calculator can be used to calculate the maximum acceleration and minimum accelation time for a car on different surfaces. Unit Converter . Cookie Settings \| \| \| --- \| \| \| \| \| \| \| --- \| \| \| \|
16676	https://www.alcula.com/conversion/length/centimeter-to-mil/	Centimeters to mils Calculators Conversions Centimeters to mils Converter \| cm: \| \| \| mil: \| \| \| \| This page allows you to convert length values expressed in centimeters to their equivalent in mils. Enter the value in centimeters in the top field (the one marked "cm"), then press the "Convert" button or the "Enter" key. The converter also works the other way round: if you enter the value in mils in the "mil" field, the equivalent value in centimeters is calculated and displayed in the top field. Centimeters to Mils Conversion Formula Where L centimeters is the length in centimeters and L mils is its equivalent in mils. Reverse formula (mils to centimeters) Centimeter Centimeter is a unit of length used by the metric system. One centimeter is 1/100 of a meter. That means there are 100 centimeters in 1 meter. One inch is equal to 2.54 cm. The following line should be approximately 1 centimeter long: Mil The mil is a unit of measure typically used in manufacturing and engineering for describing distance tolerances with high precision or for specifying the thickness of materials. One mil is equal to one thousandth of an inch, or 10-3 inches. The mil is also referred to as thou. The closest unit to the mil in the metric system is the micrometer: One mil is equal to 25.4 µm. © 2009-2014 Giorgio Arcidiacono\|Disclaimer Your Privacy Choices
16677	https://onlinelibrary.wiley.com/doi/10.1111/resp.70029	Letter From China - Lu - 2025 - Respirology - Wiley Online Library Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Skip to Article Content Skip to Article Information Search within Search term Advanced SearchCitation Search Search term Advanced SearchCitation Search Login / Register Individual login Institutional login REGISTER Respirology Volume 30, Issue 5 pp. 448-451 Letter from Asia-Pacific and Beyond Free Access Letter From China Chenyang Lu, Chenyang Lu State Key Laboratory of Respiratory Disease, National Clinical Research Center for Respiratory Disease, National Center for Respiratory Medicine, Department of Pulmonary and Critical Care Medicine, Guangzhou Institute of Respiratory Health, the First Affiliated Hospital of Guangzhou Medical University, Guangzhou, Guangdong, People's Republic of China Search for more papers by this author Qingling Zhang, Corresponding Author Qingling Zhang zqling@gzhmu.edu.cn orcid.org/0000-0002-2433-2849 State Key Laboratory of Respiratory Disease, National Clinical Research Center for Respiratory Disease, National Center for Respiratory Medicine, Department of Pulmonary and Critical Care Medicine, Guangzhou Institute of Respiratory Health, the First Affiliated Hospital of Guangzhou Medical University, Guangzhou, Guangdong, People's Republic of China Guangzhou National Laboratory, Bioland, Guangzhou, Guangdong, People's Republic of China Correspondence: Qingling Zhang (zqling@gzhmu.edu.cn) Search for more papers by this author Chenyang Lu, Chenyang Lu State Key Laboratory of Respiratory Disease, National Clinical Research Center for Respiratory Disease, National Center for Respiratory Medicine, Department of Pulmonary and Critical Care Medicine, Guangzhou Institute of Respiratory Health, the First Affiliated Hospital of Guangzhou Medical University, Guangzhou, Guangdong, People's Republic of China Search for more papers by this author Qingling Zhang, Corresponding Author Qingling Zhang zqling@gzhmu.edu.cn orcid.org/0000-0002-2433-2849 State Key Laboratory of Respiratory Disease, National Clinical Research Center for Respiratory Disease, National Center for Respiratory Medicine, Department of Pulmonary and Critical Care Medicine, Guangzhou Institute of Respiratory Health, the First Affiliated Hospital of Guangzhou Medical University, Guangzhou, Guangdong, People's Republic of China Guangzhou National Laboratory, Bioland, Guangzhou, Guangdong, People's Republic of China Correspondence: Qingling Zhang (zqling@gzhmu.edu.cn) Search for more papers by this author First published: 24 March 2025 Funding: The authors acknowledge funding from the Guangzhou Municipal Science and Technology Bureau Basic Research Program (2025A03J4397), the National Natural Science Foundation of China (82400031), ZHONGNANSHAN MEDICAL FOUNDATION OF GUANGDONG PROVINCE (ZNSA-2020013, ZNSXS-20220083, ZNSXS-20240005), the Clinical and Epidemiological Research Project of the State Key Laboratory of Respiratory Disease (SKLRD-L-202404), and the Major clinical research project of the Guangzhou Medical University Research Ability Enhancement Program (GMUCR2024-01010). About Figures ------- References ---------- Related ------- Information ----------- PDF Sections Acknowledgements Conflicts of Interest References PDF Tools Request permission Export citation Add to favorites Track citation ShareShare Give access Share full text access Close modal Share full-text access Please review our Terms and Conditions of Use and check box below to share full-text version of article. [x] I have read and accept the Wiley Online Library Terms and Conditions of Use Shareable Link Use the link below to share a full-text version of this article with your friends and colleagues. Learn more. Copy URL Share a link Share on Email Facebook x LinkedIn Reddit Wechat Bluesky Blood eosinophilia is not uncommon among patients in China. In this Letter, we report some observations in this regard that may be of interest to readers in other countries. A study from a general hospital in southern China revealed that among a total of 132,000 patients admitted between June 2018 and February 2021 across most departments, the proportion of blood eosinophilia (defined as a blood eosinophil count of > 0.5 × 10 9/L) was 6%, particularly among males and in children aged 0–6 years. The majority of patients with moderate to severe eosinophilia were found to have an associated infectious disease (59.9%, 372/621), followed by cancer (15.5%), skin diseases (10.0%), respiratory diseases (6.0%), leukaemia (4.2%), rheumatic diseases (2.6%), vasculitis (0.5%), lymphoma (0.5%), and parasitic diseases (0.3%) . These patients were not specifically investigated for their eosinophilia, and an accurate diagnosis can be difficult. The lack of thorough investigations and multidisciplinary consultation for patients with moderate to severe eosinophilia underscores the need for heightened attention to the clinical and pathophysiological implications of increased eosinophil levels. Diseases associated with eosinophilia, such as severe eosinophilic asthma, allergic bronchopulmonary aspergillosis (ABPA) and eosinophilic granulomatosis with polyangiitis (EGPA), exhibit distinct differences in epidemiological and clinical features between Chinese and European populations. In a prospective International Severe Asthma Registry study excluding Chinese cohorts, 69.7% of patients had peripheral blood eosinophil counts ≥ 300 cells/μL, compared with 38.4% in the Chinese registry [2, 3]. The annual incidence of ABPA in China is estimated at 86.33 per 100,000 population, compared with 9.3 per 100,000 population in Ireland [4, 5]. The global prevalence of EGPA is 15.27 cases per million, with a European prevalence of 12.13 cases per million . In Asia, prevalence has been rising, with studies showing an increase in Japan from 4.2 per million in 2005 to 58.6 per million in 2020 . In the Chinese EGPA cohort, ANCA positivity was less frequent compared to European studies; fever was more common, the percentage of asthma at baseline was lower, and EGPA-associated central nervous system involvement occurred more frequently . There is therefore a pressing need to elucidate the mechanisms underlying the role of eosinophils in various eosinophilia-related diseases and to investigate the factors contributing to the phenotypic differences observed between Chinese and European (mainly Caucasian) patients with eosinophilia . To improve the awareness of eosinophilic pulmonary disease (EPD) among Chinese physicians, a multidisciplinary panel of experts, including those from respiratory medicine, rheumatology, haematology, oncology, allergy medicine, infectious diseases, otorhinolaryngology, imaging, and pathology, collaborated nationwide to develop the “Expert Consensus on the Diagnosis and Treatment of Eosinophilic Pulmonary Diseases.” The objective of this document was to guide clinicians in formulating comprehensive and appropriate diagnostic and therapeutic strategies tailored to the available medical resources and the specific causes of EPD. Based on the consensus, EPD was defined as a group of diseases that met at least one of the following four criteria: 1. pulmonary infiltration with increased eosinophil (EOS) counts in peripheral blood (> 0.5 × 10 9/L); 2. increased EOS counts in bronchoalveolar lavage fluid (BALF) (> 10%); 3. tissue EOS infiltration confirmed by surgical biopsy or bronchoscopic biopsy; 4. increased EOS counts in pleural effusion (≥ 10%). However, the current definition of EPD requires further consideration of the dual roles of eosinophils. To address this gap, we propose the concept of Eosinophil-Driven Pulmonary Disease (EDPD), encompassing a spectrum of diseases where eosinophils play a pivotal role in disease onset and progression. The introduction of EDPD, along with an investigation of the underlying mechanisms, is expected to facilitate the identification of shared molecular pathways across different eosinophil-associated diseases. In severe eosinophilic asthma and EGPA, distinguishing between these conditions can be challenging, but IL-5-driven eosinophil activation remains a central shared pathogenic mechanism. The role of eosinophils in diseases characterised by eosinophil proliferation is multifaceted, embodying both protective and detrimental aspects. Clinically, the decision to intervene in elevated eosinophil counts and the selection of appropriate therapeutic strategies must account for this dual nature. On one hand, eosinophils play a protective role as effector cells in immune responses to various antigens, including allergens, helminths, and other pathogens. Conversely, eosinophils can exert harmful effects in certain diseases by releasing cytotoxic substances upon activation. In the context of asthmatic lung inflammation, eosinophils are recruited by Type-2 cytokines and chemokines. Once activated, eosinophils release a range of cytotoxic proteins, including major basic protein (MBP), eosinophil cationic protein (ECP), eosinophil peroxidase (EPO), and matrix metalloproteinases (MMPs), which contribute to epithelial damage and airway remodelling. Targeted therapies with anti-Type2 cytokine or cytokine receptors such as Mepolizumab and Reslizumab (anti-IL5 antibody), Benralizumab (anti-IL5Rα antibody), and Dupilumab (anti-IL4Rα antibody) have been developed to address these deleterious effects of eosinophils and are now available in China. (Figure1). FIGURE 1 Open in figure viewerPowerPoint An overview of eosinophil biology as pertaining to respiratory disease. (Created in BioRender. Qingling, Z. (2024) BioRender.com/l24e288). In addition to examining the diverse roles of eosinophils across various diseases, it is crucial to differentiate between eosinophils based on their origin and localisation. Recent work supports the concept that eosinophils located permanently in specific tissues are resident eosinophils with homeostatic properties, while those recruited to inflammatory or allergic sites are pro-inflammatory and termed inflammatory eosinophils . The distinct cellular functions and molecular markers of these subtypes highlight the need for further exploration of their differences at a molecular level. A nuanced understanding of these variations is essential for accurately establishing the role of eosinophils in various pathological conditions. Different types of eosinophils may differentially influence disease progression, suggesting that targeted therapies addressing specific pathogenic pathways could be effective even without identifying disease-specific phenotypes. This opens up the use of targeted treatments in EDPDs. Interestingly, the anti-IL5 antibody mepolizumab has been shown to deplete inflammatory eosinophils while preserving the homeostatic eosinophils in blood in patients with eosinophilic severe asthma . Moreover, differences in genetic background and environmental exposures likely underlie the phenotypic variations observed between Chinese and European patients with EDPD. For instance, linkage disequilibrium between the IL-13-1111T allele and the 2043A allele (Gln130) associated with the asthma phenotype is present in Chinese patients but not in those of European ethnicity . In contrast to Europe, the subtropical monsoon climate typical of southern China is marked by high temperatures and heavy rainfall during the summer, with mild, wet winters that may exacerbate the impact of air pollution and may have effects on eosinophil activation. Varying climatic conditions may also influence the distribution and concentration of allergens. In our unpublished study of asthma patients, plant-derived allergens were found to predominate in northern China, while dust mites were the primary allergens in southern regions. To accurately elucidate the pathogenesis of EDPDs and develop effective interventions, it is crucial to explore not only the common molecular mechanisms across various EDPDs but also to define the population-specific and geographic factors through population genetics and epidemiology. This approach will enable proactive intervention based on geographic, demographic, and molecular characteristics, potentially circumventing prolonged and challenging diagnostic processes. In summary, the validity and benefits of the Chinese EPD consensus require further validation, and we welcome constructive comments from other researchers active in this area. Acknowledgements We thank Kian Fan Chung (National Heart and Lung Institute, Imperial College London, London, UK), Xiaohui Wang (Guangzhou Institute of Paediatrics, Guangzhou Women and Children's Medical Center, Clinical Medical Research Center for Children's Health and Diseases, Guangzhou, China), Jie Yan (Guangdong Provincial Key Laboratory of Allergy and Immunology, The Second Affiliated Hospital, Guangzhou Medical University, Guangzhou Medical University, Guangzhou, China) and Changxing Ou (State Key Laboratory of Respiratory Disease, National Clinical Research Center for Respiratory Disease, National Center for Respiratory Medicine, Department of Pulmonary and Critical Care Medicine, Guangzhou Institute of Respiratory Health, the First Affiliated Hospital of Guangzhou Medical University, Guangzhou, China). Conflicts of Interest The authors declare no conflicts of interest. References 1 B. Chen, Q. Rong, Y. Fu, et al., “Characteristics of Patients With Incidental Eosinophilia Admitted to a Tertiary Hospital in Southern China,” Heliyon 9, no. 5 (2023): e15569. 10.1016/j.heliyon.2023.e15569 PubMedWeb of Science®Google Scholar 2 L. G. Heaney, L. Perez de Llano, M. Al-Ahmad, et al., “Eosinophilic and Noneosinophilic Asthma: An Expert Consensus Framework to Characterize Phenotypes in a Global Real-Life Severe Asthma Cohort,” Chest 160, no. 3 (2021): 814–830. 10.1016/j.chest.2021.04.013 CASPubMedWeb of Science®Google Scholar 3 Q. Zhang, X. Fu, C. Wang, et al., “Severe Eosinophilic Asthma in Chinese C-BIOPRED Asthma Cohort,” Clinical and Translational Medicine 12, no. 2 (2022): e710. 10.1002/ctm2.710 PubMedWeb of Science®Google Scholar 4 L. H. Zhou, Y. K. Jiang, R. Y. Li, et al., “Risk-Based Estimate of Human Fungal Disease Burden, China,” Emerging Infectious Diseases 26, no. 9 (2020): 2137–2147, 10.3201/eid2609.200016 PubMedWeb of Science®Google Scholar 5 D. W. Denning, A. Pleuvry, and D. C. Cole, “Global Burden of Allergic Bronchopulmonary Aspergillosis With Asthma and Its Complication Chronic Pulmonary Aspergillosis in Adults,” Medical Mycology 51, no. 4 (2013): 361–370. 10.3109/13693786.2012.738312 PubMedWeb of Science®Google Scholar 6 R. W. Jakes, N. Kwon, B. Nordstrom, et al., “Burden of Illness Associated With Eosinophilic Granulomatosis With Polyangiitis: A Systematic Literature Review and Meta-Analysis,” Clinical Rheumatology 40, no. 12 (2021): 4829–4836. 10.1007/s10067-021-05783-8 PubMedWeb of Science®Google Scholar 7 K. E. Sada, T. Suzuki, S. Joksaite, et al., “Trends in Prevalence, Treatment Use, and Disease Burden in Patients With Eosinophilic Granulomatosis With Polyangiitis in Japan: Real-World Database Analysis,” Modern Rheumatology 34, no. 5 (2024): 988–998. 10.1093/mr/road104 PubMedGoogle Scholar 8 S. Liu, L. Han, Y. Liu, et al., “Clinical Significance of MPO-ANCA in Eosinophilic Granulomatosis With Polyangiitis: Experience From a Longitudinal Chinese Cohort,” Frontiers in Immunology 13 (2022): 885198. 10.3389/fimmu.2022.885198 CASPubMedWeb of Science®Google Scholar 9 National Centre of Respiratory Medicine, the First Affiliated Hospital of Guangzhou Medical University, National Clinical Research Center for Respiratory Disease, Asthma Group of the Respiratory Disease Branch of the Chinese Medical Association (CMA), Chinese Expert Consensus on The Diagnosis and Treatment of Eosinophilic Pulmonary Diseases, “Chinese Expert Consensus on The Diagnosis and Treatment of Eosinophilic Pulmonary Diseases,” National Medical Journal of China 102, no. 1 (2022): 21–35. Google Scholar 10 C. Mesnil, S. Raulier, G. Paulissen, et al., “Lung-Resident Eosinophils Represent a Distinct Regulatory Eosinophil Subset,” Journal of Clinical Investigation 126, no. 9 (2016): 3279–3295, 10.1172/JCI85664 PubMedWeb of Science®Google Scholar 11 M. Fricker, J. Harrington, S. A. Hiles, and P. G. Gibson, “Mepolizumab Depletes Inflammatory but Preserves Homeostatic Eosinophils in Severe Asthma,” Allergy 79, no. 11 (2024): 3118–3128. 10.1111/all.16267 CASPubMedWeb of Science®Google Scholar 12 E. Tarazona-Santos and S. A. Tishkoff, “Divergent Patterns of Linkage Disequilibrium and Haplotype Structure Across Global Populations at the Interleukin-13 (IL13) Locus,” Genes and Immunity 6, no. 1 (2005): 53–65. 10.1038/sj.gene.6364149 CASPubMedWeb of Science®Google Scholar Volume 30, Issue 5 May 2025 Pages 448-451 Figures ------- References ---------- Related ------- Information ----------- Recommended Absence of airway inflammation in a large proportion of adolescents with asthma Collin R. Brooks,Christine J. van Dalen,Angela Zacharasiewicz,Jodie L. Simpson,Jacquie L. Harper,Graham Le Gros,Peter G. Gibson,Neil Pearce,Jeroen Douwes, Respirology Baseline airway hyperresponsiveness and its reversible component: role of airway inflammation and airway calibre M Ichinose,T Takahashi,H Sugiura,N Endoh,M Miura,Y Mashito,K Shirato, European Respiratory Journal Individual time‐courses of ECP and EPX during allergen provocation tests in asthmatic children I. Kleinau,B. Niggemann,U. Wahn, Pediatric Allergy and Immunology Role for Complement C5 in Eosinophilic Inflammation of Severe Asthma Cong Dong,Shaohua Lu,Zhenan Deng,Xuliang Cai,Huahao Shen,Guochao Shi,Changxing Ou,Zuofu Peng,Wei Jiang,Xiuhua Fu,Changzheng Wang,Meiling Jin,Zhongmin Qiu,Xiaoyang Wei,Wei Gu,Kewu Huang,Qiang Li,Xiangyan Zhang,Nanshan Zhong,Kian Fan Chung,Qingling Zhang,C-BIOPRED Consortium, Allergy A Comprehensive Review of Interstitial Lung Abnormalities Yuben Moodley,John A. Mackintosh, Respirology Metrics Full text views: 614 Full text views and downloads on Wiley Online Library. More metric information Details © 2025 Asian Pacific Society of Respirology. Check for updates Research funding Guangzhou Municipal Science and Technology Bureau Basic Research Program. Grant Number: 2025A03J4397 National Natural Science Foundation of China. Grant Number: 82400031 Clinical and Epidemiological Research Project of State Key Laboratory of Respiratory Disease. Grant Number: SKLRD-L-202404 Major clinical research project of the Guangzhou Medical University Research Ability Enhancement Program. Grant Number: GMUCR2024-01010 ZHONGNANSHAN MEDICAL FOUDATION OF GUANGDONG PROVINCE. Grant Numbers: ZNSA-2020013, ZNSXS-20220083, ZNSXS-20240005 Keywords eosinophil eosinophil-driven pulmonary disease eosinophilic pulmonary diseases Publication History Issue Online: 08 May 2025 Version of Record online: 24 March 2025 Manuscript accepted: 27 February 2025 Manuscript revised: 30 January 2025 Manuscript received: 11 November 2024 Close Figure Viewer Previous FigureNext Figure Caption Download PDF back Additional links About Wiley Online Library Privacy Policy Terms of Use About Cookies Manage Cookies Accessibility Wiley Research DE&I Statement and Publishing Policies Developing World Access Help & Support Contact Us Training and Support DMCA & Reporting Piracy Sitemap Opportunities Subscription Agents Advertisers & Corporate Partners Connect with Wiley The Wiley Network Wiley Press Room Copyright © 1999-2025 John Wiley & Sons, Inc or related companies. 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16678	https://emedicine.medscape.com/article/2087447-overview	For You News & Perspective Tools & Reference Edition English Medscape Editions About You Professional Information Newsletters & Alerts Your Watch List Formulary Plan Manager Log Out Log In Medscape Editions English Deutsch EspaÃ±ol FranÃ§ais PortuguÃªs UK X Univadis from Medscape About You Professional Information Newsletters & Alerts Your Watch List Formulary Plan Manager Log Out Register Log In close Please confirm that you would like to log out of Medscape. If you log out, you will be required to enter your username and password the next time you visit. Log out Cancel processing.... Tools & Reference>Laboratory Medicine Serum Calcium Updated: Aug 26, 2025 Author: Alina G Sofronescu, PhD, NRCC-CC, FAAC; Chief Editor: Sridevi Devaraj, PhD, DABCC, FAACC, FRSC, CCRP more...;) Share Print Feedback Close Facebook Twitter LinkedIn WhatsApp Email Sections Serum Calcium Sections Serum Calcium Reference Range Interpretation Collection and Panels Background Show All References;) Reference Range Calcium concentration, both total and free, is characterized by a high physiological variation, depending on age, sex, physiological state (e.g., pregnancy), and even season (owing to the seasonal variation of vitamin D, which is directly involved in the regulation of calcium concentration). Therefore, separate reference intervals have been established according to the age and sex of the individual being tested. Approximately 50% of the total calcium in the blood circulates in its free, biologically active (ionized) form, while the remaining calcium is bound primarily to proteins such as albumin. The serum calcium measurement reflects both the ionized and protein-bound fractions. Total calcium reference ranges are as follows : Umbilical: 9-11.5 mg/dL; 2.25-2.88 mmol/L < 10 days: 7.6-10.4 mg/dL; 1.9-2.6 mmol/L 10 days to 2 years: 9-10.6 mg/dL; 2.3-2.65 mmol/L Children: 8.8-10.8 mg/dL; 2.2-2.7 mmol/L Adults: 9-10.5 mg/dL; 2.25-2.62 mmol/L (values tend to be reduced in older adults) Possible critical values for total calcium are < 6 mg/dL or > 13 mg/dL. Possible critical values for ionized calcium are < 2.2 mg/dL or > 7 mg/dL. Next: Interpretation When serum calcium levels are elevated on at least three separate measurements, the patient is considered to have hypercalcemia. The most frequent cause of hypercalcemia is primary hyperparathyroidism. The second most common cause is malignancy, which raises calcium levels through two primary mechanisms : Bone metastases (commonly from multiple myeloma or lung, breast, or renal cell carcinomas) can induce osteolysis, releasing calcium into the circulation. Certain tumors (such as lung, breast, and renal cell carcinomas) may secrete a parathyroid hormone (PTH)related peptide (PTHrP) that mimics PTH activity and drives calcium levels higher (ectopic PTH production). Other contributors to hypercalcemia include excessive vitamin D intake, which enhances gastrointestinal and renal calcium absorption, and granulomatous diseases such as sarcoidosis and tuberculosis, which can also increase serum calcium. In contrast, hypocalcemia commonly occurs in the setting of hypoalbuminemia, where the total calcium level is reduced due to decreased protein binding, although ionized calcium may remain normal. Large-volume blood transfusions can lead to hypocalcemia because citrate preservatives in stored blood chelate free calcium in the recipient's plasma. Additional causes include intestinal malabsorption, chronic renal failure, rhabdomyolysis, alkalosis, and acute pancreatitis, where calcium is sequestered in areas of fat necrosis (saponification). Measurement of the ionized calcium component is generally obtained in specialized laboratories and through a special procedure. In most laboratories, autoanalyzers are used to measure the total serum calcium level accurately and reproducibly, although atomic absorption spectrophotometers probably provide even greater accuracy. Unless serum proteins contain abnormalities, total serum calcium concentration is normally about 9-10.5 mg/dL, or 2.25-2.62 mmol/L, of serum (though slight variations exist between laboratories as well as patient populations). Because ionized calcium is the only component of the total serum calcium level that is regulated by calciotropic hormones, decisions on the total serum calcium concentration should not be made unless changes in concentrations of plasma proteins, particularly Albumin, are considered. Total serum calcium is less difficult to measure than the ionized calcium component is, and ionized calcium measurements are rarely needed if serum protein concentrations can be measured. Traditional teaching advocates adjusting total calcium for hypoalbuminemia using various formulas (e.g., Payne and James). However, large-scale clinical studies demonstrate that adjusted calcium calculations do not reliably improve diagnostic accuracy over unadjusted total calcium and may introduce misclassification particularly underestimating true hypocalcemia in hypoalbuminemic patients. These findings are consistent across multiple contemporary populations and suggest that if there is doubt about calcium status, direct measurement of ionized calcium is preferred over formulaic adjustment. In patients with multiple myeloma, the globulin concentration is often increased, leading to excessive binding of calcium to the monoclonal paraprotein and occasional elevation of the total serum calcium concentration, yet the ionized calcium level may be normal in these individuals. Assessment of ionized calcium would be useful in such patients. It should be emphasized that these patients frequently have bone lesions, which lead to the release of calcium from the damaged bone tissue. Although serum calcium levels above 11.5 mg/dL commonly cause symptoms, patients may be asymptomatic at this level. Critical levels are reached above 12 mg/dL, with levels above 15 mg/dL (severe hypercalcemia) being a medical emergency. Previous Next: Collection and Panels Preferred specimens are as follows: Plasma, serum, whole blood Urine specimens (random collection or 24-h timed collection) Acceptable containers/tubes are as follows: Green top tube (sodium heparin, ammonium heparin, lithium heparin) Red top tube (clot activator) Serum separator tube Plasma separator tube Electrolytebalanced heparinized syringes (for whole blood sample testing) Urine containers (may contain 6 mol/L HCl, 20-23 mL in 24-h urine collection, to prevent precipitation of calcium salts) Specimen volumes are as follows: 0.5 mL plasma or serum (0.25 mL minimum volume) 0.5 mL whole blood Entire urine collection Specimen stability Whole blood specimens should be analyzed within 15-30 min of collection. If this is not possible, the specimen should be kept in ice. In ice, the specimen is stable for at least 2 h; however, if concurrent potassium testing is requested on the same specimen, the low temperature leads to a spurious increase in potassium within 1 h of collection. If the specimen cannot be analyzed within 1 h, the preferred specimen is serum. The specimen should be centrifuged, and the serum or plasma should be removed from the cells within 2 h of collection. Samples can be stored at room temperature for 8 h or refrigerated at 2-8° C for up to 48 h. If assays are not completed within 48 h or the separated sample is to be stored beyond 48 h, samples should be frozen at 15° C to 20° C. Frozen samples should be thawed only once. Analyte deterioration may occur in samples that are repeatedly frozen and thawed. Data confirm that calcium remains stable for 24 h at room temperature, 4 weeks refrigerated, and at least 9 weeks frozen (20°C), with up to three freeze-thaw cycles showing minimal effect on total calcium. However, repeated freeze-thaw cycles may affect ionized calcium more significantly. If drawing for more than total calcium, send the first tube drawn. Collection considerations and sources of preanalytical errors Fist-clenching or forearm exercise can lead to falsely elevated ionized (free) calcium levels. The sample should be drawn with the patient in a sitting position. Standing increases the total calcium concentration. Hemolysis and delayed plasma/serum separation lead to a decreased calcium concentration. Samples collected in tubes containing citrate, oxalate, or ethylenediaminetetraacetic acid are not suitable for calcium testing. Measurement of calcium Ionized calcium can be measured in whole blood using the ion-specific electrode (ISE) potentiometric method. This was designated by the International Federation of Clinical Chemistry and Laboratory Medicine (IFCC) as the reference method for ionized/free calcium evaluation. Total calcium also can be measured using the ISE potentiometric method, but the sample must be pre-acidified to release all bound and complexed calcium to a free form. However, total calcium is commonly measured with spectrophotometric methods, such as the o-Cresolphthalein Complexone method, Arsenazo III method, atomic absorption spectrometry, or, rarely, isotope dilution mass spectrometry (ID-MS). The atomic absorption spectrometry method was designated by the IFCC as the reference method for total serum calcium evaluation, while the ID-MS is considered "the definitive method" for total calcium evaluation developed by the National Institute of Standards and Technology. Previous Next: Background Description Calcium is the fifth most abundant element in the body. Adults have a calcium content of over 1 kg or approximately 2% of body weight. All but 1% is contained in the bones, where it exists as calcium hydroxyapatite. The extraosseous intracellular space and extracellular space contain a portion of the remaining 1%. A dynamic equilibrium is maintained between the extracellular space and the rapidly exchangeable fraction of bone calcium. In blood, virtually all calcium is present in the plasma. About 50% of the calcium present in circulation is free (also known as ionized calcium); 40% of serum calcium is bound to proteins, especially albumin (80%) and, secondary, to globulins (20%); and about 10% exists as various small diffusible inorganic and organic anions (e.g., bicarbonate, lactate, citrate). Heart and skeletal muscle contractility are affected by calcium ions; in addition, calcium ions are vital to nervous system function and are associated with blood clotting and bone mineralization. The concentration of serum calcium is tightly regulated by PTH and 1,25-hydroxy vitamin D. In patients with low serum albumin levels (such as those who are malnourished), total serum calcium may appear falsely low. The total calcium level decreases by about 0.8 mg/dL for every 1 g/dL reduction in serum albumin. For this reason, serum albumin should always be measured alongside serum calcium to ensure accurate interpretation. The clinical utility of "corrected" (albumin-adjusted) calcium is increasingly questioned, especially in hospitalized or critically ill patients. Large studies show that unadjusted total calcium often correlates more closely with ionized calcium than does albumin-corrected calcium, and that adjustment formulas may misclassify calcium status, particularly in hypoalbuminemia. Direct measurement of ionized calcium is recommended when accurate assessment is critical (e.g., in critical illness, renal failure, or when calcium-binding proteins are abnormal). For further reading, please see the Medscape Drugs and Diseases topics Hypocalcemia and Hypercalcemia. Indications/Applications The serum calcium test helps assess parathyroid function and overall calcium metabolism by measuring the total calcium concentration in the blood. It is commonly used to monitor patients with conditions such as renal failure, following renal transplantation, hyperparathyroidism, and certain types of cancer. Additionally, it is valuable for tracking calcium levels during and after treatment in these settings. Hypocalcemia Serum total calcium concentrations below the reference interval for the appropriate age and sex reflect a hypocalcemic status. Nevertheless, results of total serum calcium should be interpreted with caution, as they might not follow the same pattern (decreased concentration) of free (ionized) calcium. This is because total calcium is influenced by multiple factors, such as the concentration of albumin, the serum pH, and the concentration of immunoglobulins. As 80% of the bound calcium present in the circulation is carried by albumin (albumin has multiple binding sites for calcium), changes in serum albumin concentration trigger significant changes in the concentration of total calcium. Hypoalbuminemia is the most common cause of pseudohypocalcemia. Therefore, serum albumin changes are generally corrected with the addition of 0.8 mg/dL for every gram that the serum albumin level falls below 4 g/dL. In patients with suspected hypercalcemia or hypocalcemia, this principle can be expressed as the following equation: Corrected calcium (mg/dL) = measured total calcium (mg/dL) + 0.8 (4.0serum albumin [g/dL]) The corrected total serum calcium concentration is normally 8.5-10.2 mg/dL, but there is no sure means of predicting the serum calcium level, for either hypocalcemia or hypercalcemia, at which symptoms will occur. Direct measurement is preferred in critical illness, major protein abnormalities, or acid-base disturbances. The rapidity of change, as well as the absolute serum calcium concentration, impacts symptom development. It has been found, however, that hypocalcemic symptoms rarely occur if the corrected serum calcium concentration is above 8 mg/dL, while hypercalcemic symptoms rarely develop in patients with a corrected serum calcium level under 11 mg/dL. [6, 7, 8] Calcium binds to the negatively charged sites of proteins, especially albumin. Therefore, this binding is pH-dependent. Alkalosis (increased serum pH) leads to increased protein ionization, increased negative charge and calcium binding, and, consequently, increased bound calcium concentration and decreased free calcium. Acidosis (decreased pH) has the opposite effects. Serum calcium is decreased (hypocalcemia) in the following conditions: Hypoparathyroidism Vitamin D deficiency (either from intake deficiency or decreased conversion/activation) or resistance (osteomalacia and rickets) Chronic renal diseases (e.g., renal acidosis, Fanconi's syndrome) Pseudohypoparathyroidism Magnesium deficiency (PTH glandular release is magnesium-dependent) Hyperphosphatemia Massive transfusion Hypoalbuminemia Chronic liver disease and biliary obstructive diseases (from impaired absorption and conversion of vitamin D) Overexpression of fibroblast growth factor 23 (oncogenic osteomalacia) Severe calcium dietary deficiency Severe pancreatitis (calcium saponification) Hungry bone syndrome Fat embolism Hypocalcemia results when the parathyroid glands are either absent or impaired. Impaired vitamin D synthesis can also cause the condition. Owing to an associated decrease in vitamin D synthesis and the presence of hyperphosphatemia and skeletal resistance to PTH, chronic renal failure is a frequent source of hypocalcemia. Latent or manifest tetany and osteomalacia are characteristic signs of hypocalcemia. Hypercalcemia Serum total calcium concentrations above the reference interval for the appropriate age and sex reflect a hypercalcemic status. Serum calcium is increased in the following: Hyperparathyroidism (primary, such as multiple endocrine neoplasia type 1, hyperplasia, adenoma, or carcinoma; or secondary, from chronic kidney injury and hyperphosphatemia or after PTH administration during treatment for osteoporosis) Malignancies (humoral hypercalcemia of malignancy) that secrete PTHrP, especially squamous cell carcinoma of the lung and renal cell carcinoma Vitamin D excess Milk-Alkali Syndrome Multiple Myeloma, owing to bone lesions Paget Disease of bone with prolonged immobilization Sarcoidosis Other granulomatous disorders Familial hypocalciuria hypercalcemia Vitamin A intoxication Thyrotoxicosis Hypothyroidism, owing to prolongation of vitamin D action as its metabolism is slowed down Addison Disease Drug exposure - Some drugs that can increase serum calcium are as follows: antacids (some), calcium salts, long-term thiazide therapy, and lithium. Hypercalcemia results from increased mobilization of calcium from the skeletal system or increased intestinal absorption. The condition is usually caused by primary hyperparathyroidism or bone metastasis of carcinoma of the breast, prostate, thyroid gland, or lungs. In 10% of patients with malignancies, coexistent hyperparathyroidism is the source of hypercalcemia; this indicates that evaluation of serum PTH levels should be performed at initial presentation in all hypercalcemic patients. Surgical removal of one or more parathyroid glands is a consideration in patients with primary hyperparathyroidism and bone disease, renal stones or nephrocalcinosis, or other signs or symptoms. Severe hypercalcemia may cause cardiac arrhythmia. [5, 9, 10] Urinary Calcium Urinary calcium laboratory results should be always normalized to the glomerular filtrate (GF) and take into account the serum and urine concentration of creatinine, using the following formula: UCa [mg/100 mL GF] = (UCa [mg/dL] Ã SCr [mg/dL]) / UCr [mg/dL] UCa = Urinary calcium concentration SCr = Serum creatinine concentration UCr = Urine creatinine concentration Reference intervals for urinary calcium were established taking into account an unrestricted diet, as well as in fasting conditions. In diet-unrestricted conditions, males and females can excrete up to 300 mg calcium/24 h; while in fasting, less than 200 mg/24 h calcium will be excreted in the urine. Under fasting conditions, intestinal and renal calcium absorption are fixed, and the urinary excretion of calcium reflects the skeletal matrix resorption/release of calcium. A urinary calcium concentration of greater than 0.16 mg/100 mL GF is usually associated with osteoclastic bone resorption. Evaluation of urinary calcium is a useful tool for assessing renal stone diseases and high-turnover osteoporosis. Considerations During pregnancy, the serum albumin concentration can often be decreased. However, an increase in serum free calcium is not observed. However, pregnancy-specific reference intervals should be used for correct patient evaluation. Drugs (e.g., phenytoin) and bilirubin are commonly competing with calcium for albumin binding. Therefore, laboratory results from icteric patients and patients undergoing treatments with highly albumin-bonded drugs should be interpreted with caution. Since gadolinium can interfere with most metal tests, 48 h must elapse between administration of gadolinium-containing contrast media and specimen collection. Vitamin D toxicity can result in elevated serum calcium levels due to increased intestinal absorption, while excessive milk consumption, particularly in combination with antacids (as in milk-alkali syndrome), may also contribute to hypercalcemia. Fluctuations in serum pH significantly impact calcium measurements; acidosis (low pH) increases ionized calcium levels, whereas alkalosis lowers them. Prolonged tourniquet application during phlebotomy can reduce sample pH and falsely elevate calcium values. In addition, calcium exhibits mild diurnal variation, with peak levels typically occurring in the late evening around 9 PM. Hypoalbuminemia may artifactually lower total serum calcium concentrations because a substantial portion of calcium is protein-bound; therefore, ionized calcium or albumin-corrected calcium should be considered for accurate assessment in such cases. Studies continue to emphasize the need for trimester- and population-specific reference intervals for calcium, as general adult ranges may lead to misinterpretation. Previous References Pagana KD, Pagana TJ, Pagana TN. Mosby's Diagnostic and Laboratory Test Reference. 17th ed. Mosby's, Elsevier; 2024. DesgagnÃ©s N, King JA, Kline GA, Seiden-Long I, Leung AA. Use of Albumin-Adjusted Calcium Measurements in Clinical Practice. JAMA Netw Open. 2025 Jan 2. 8 (1):e2455251. [QxMD MEDLINE Link].[Full Text]. Hunsaker JJ, Laulu SL, Doyle K. A-073 Specimen Storage and Onboard Stability of Serum Ionized Calcium Using the Nova Prime ES Comp Plus Analyzer. Clin Chem. 2023. 69 (1):hvad097.070. [Full Text]. [Guideline] C31-A2 Ionized Calcium Determinations: Precollection Variables, Specimen Choice, Collection, and Handling; Approved Guideline - Second Edition. Vol.12, No.10. Clinical and Laboratory Standards Institute. Burtis CA, Ashwood ER, Bruns DE, eds. Tietz Textbook of Clinical Chemistry and Molecular Diagnostics. 4th ed. Philadelphia, PA: Saunders; 2005. Deng XR, Zhang YF, Wang TG, et al. Serum calcium level is associated with brachial-ankle pulse wave velocity in middle-aged and elderly Chinese. Biomed Environ Sci. 2014 Aug. 27 (8):594-600. [QxMD MEDLINE Link]. Sanchis-Gomar F, Salvagno GL, Lippi G. Inhibition of xanthine oxidase and exercise on serum uric acid, 25(OH)D3, and calcium concentrations. Clin Lab. 2014. 60 (8):1409-11. [QxMD MEDLINE Link]. Kanagal DV, Rajesh A, Rao K, et al. Levels of Serum Calcium and Magnesium in Pre-eclamptic and Normal Pregnancy: A Study from Coastal India. J Clin Diagn Res. 2014 Jul. 8 (7):OC01-4. [QxMD MEDLINE Link].[Full Text]. Jacobs TP, Bilezikian JP. Clinical review: Rare causes of hypercalcemia. J Clin Endocrinol Metab. 2005 Nov. 90 (11):6316-22. [QxMD MEDLINE Link]. Carroll MF, Schade DS. A practical approach to hypercalcemia. Am Fam Physician. 2003 May 1. 67 (9):1959-66. [QxMD MEDLINE Link]. Hanna B. The Role of Calcium Correction during Normal Pregnancy at Third Trimester in Mosul. Oman Med J. 2009 Jul. 24 (3):188-94. [QxMD MEDLINE Link].[Full Text]. Media Gallery Red-top Vacutainer tube. of 1 Tables Back to List Contributor Information and Disclosures Author Alina G Sofronescu, PhD, NRCC-CC, FAAC Associate Professor, Department of Pathology, Wake Forest University School of Medicine; Medical Director, Clinical Chemistry, Toxicology and Mass Spectrometry Lab, Atrium Health Wake Forest Baptist Medical CenterAlina G Sofronescu, PhD, NRCC-CC, FAAC is a member of the following medical societies: American Association for Clinical Chemistry, Canadian Society of Clinical ChemistsDisclosure: Nothing to disclose. Specialty Editor Board Disclosure: Nothing to disclose. Chief Editor Sridevi Devaraj, PhD, DABCC, FAACC, FRSC, CCRP Medical Director of Clinical Chemistry and Point of Care Technology (POCT), Texas Childrens Hospital and Clearlake Health Center; Director of Laboratories, TCH Centers for Women and Children; Professor of Pathology and Immunology, Director, Clinical Chemistry Fellowship and Clinical Chemistry Resident Rotation, Baylor College of Medicine; Associate Director, Texas Childrens Microbiome CenterDisclosure: Nothing to disclose. Close;) What would you like to print? What would you like to print? Print this section Print the entire contents of Print the entire contents of article Sections Serum Calcium Reference Range Interpretation Collection and Panels Background Show All References;) encoded search term (Serum Calcium) and Serum Calcium What to Read Next on Medscape Related Conditions and Diseases Serum Calcium Hypocalcemia Oliguria Hypercalciuria Pediatric Acute Tubular Necrosis Diabetic Ketoacidosis (DKA) Pediatric Hypercalcemia News & Perspective Elevated Serum Phosphate Could be a Risk Factor for Aortic Stenosis, Study Suggests Calcium and CV Risk: Are Supplements and Vitamin D to Blame? Higher Calcium Intake May Lower CRC Risk Relationship Between Exercise-Induced Cardiac Troponin Elevations and Occult Coronary Atherosclerosis in Middle-Aged Athletes How to Identify and Treat Hypomagnesemia Gout: Diagnosis and Management in Primary Care Drug Interaction Checker Pill Identifier Calculators Formulary 2002 2087447-overview Laboratory Medicine Laboratory Medicine Serum Calcium 2003 /viewarticle/1001783 Bone and Beyond in Hypoparathyroidism: Latest Best Practice Recommendations 1.25 CME / ABIM MOC Credits You are being redirected to Medscape Education Yes, take me there 1.25 CME / ABIM MOC Bone and Beyond in Hypoparathyroidism: Latest Best Practice Recommendations 2002 766906-overview Diseases & Conditions Diseases & Conditions Hyperparathyroidism in Emergency Medicine 2002 874690-overview Procedures Procedures Parathyroid Physiology
16679	https://flexbooks.ck12.org/cbook/ck-12-trigonometry-concepts/section/5.16/primary/lesson/resultant-of-two-displacements-trig/	CK12-Foundation AI Teacher Tools – Save Hours on Planning & Prep. Try it out! Skip to content Subject Math Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? Science Grade K to 5 Earth Science Life Science Physical Science Biology Chemistry Physics Advanced Biology FlexLets Math FlexLets Science FlexLets English Writing Spelling Social Studies Economics Geography Government History World History Philosophy Sociology More Astronomy Engineering Health Photography Technology College College Algebra College Precalculus Linear Algebra College Human Biology The Universe Adult Education Basic Education High School Diploma High School Equivalency Career Technical Ed English as 2nd Language Country Bhutan Brasil Chile Georgia India Translations Spanish Korean Deutsch Chinese Greek Polski Explore EXPLORE Flexi A FREE Digital Tutor for Every Student FlexBooks 2.0 Customizable, digital textbooks in a new, interactive platform FlexBooks Customizable, digital textbooks Schools FlexBooks from schools and districts near you Study Guides Quick review with key information for each concept Adaptive Practice Building knowledge at each student’s skill level Simulations Interactive Physics & Chemistry Simulations PLIX Play. Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign InSign Up CK-12 Foundation is a non-profit organization that provides free educational materials and resources. FLEXIAPPS ABOUT Our missionMeet the teamPartnersPressCareersSecurityBlogCK-12 usage mapTestimonials SUPPORT Certified Educator ProgramCK-12 trainersWebinarsCK-12 resourcesHelpContact us BYCK-12 Common Core MathK-12 FlexBooksCollege FlexBooksTools and apps CONNECT TikTokInstagramYouTubeTwitterMediumFacebookLinkedIn v2.11.10.20250923073248-4b84c670be © CK-12 Foundation 2025 \| FlexBook Platform®, FlexBook®, FlexLet® and FlexCard™ are registered trademarks of CK-12 Foundation. Terms of usePrivacyAttribution guide Curriculum Materials License Student Sign Up Are you a teacher? Sign up here Sign in with Google Having issues? Click here Sign in with Microsoft Sign in with Apple or Sign up using email By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? Sign In
16680	https://www.plannedparenthood.org/planned-parenthood-orange-san-bernardino/patients/services/sti-testing-treatment	STI Testing & Treatment \| Planned Parenthood of Orange & San Bernardino Counties, Inc. Go to Content Back to PlannedParenthood.org Español Planned Parenthood of Orange & San Bernardino Counties, Inc. Get Care Make an Appointment Locations & Hours Services Patient Portal Payment Options Patient Information Abortion Aid Primary Care Learn Sex Education Programs For Teens & Young Adults For Parents & Caregivers For Educators & Professionals Community Resource Guides Blog Get Involved Ways to Give Become an Advocate Attend an Event Share Your Story About Who We Are Facts About Planned Parenthood Careers News & Perspectives Contact Us Search Donate Donate Back to Planned Parenthood Get Care Overview Make an Appointment Locations & Hours Services Patient Portal Payment Options Patient Information Abortion Aid Primary Care Learn Overview Sex Education Programs For Teens & Young Adults For Parents & Caregivers For Educators & Professionals Community Resource Guides Blog Get Involved Overview Ways to Give Become an Advocate Attend an Event Share Your Story About Overview Who We Are Facts About Planned Parenthood Careers News & Perspectives Contact Us Search Español STI Testing & Treatment Most Sexually Transmitted Infections (STIs) don’t have symptoms, so the only way to know if you have an STI is through testing. If you’ve had sexual contact that can spread STIs — like vaginal, anal, or oral sex — consider talking to a health care provider about getting tested. make an appt STI Testing and Treatment Services What Are STDs and How Are They Transmitted \| Planned Parenthood Video STDs are super common and they usually don’t show any symptoms — so lots of people don’t even know they have one. STI Testing and Treatment Services Chlamydia Genital warts Gonorrhea Hepatitis B Herpes Pubic lice Syphilis Trichomoniasis HPV If you test positive for an STI, don’t panic. Most STIs are easily treatable, and some are even curable. Planned Parenthood will provide a treatment plan for you and your partners. learn more How Does STI Testing Work? STI testing is quick, easy, and generally painless. Each STI requires its own test, and your provider can help decide which tests you need. Testing options include: Urine test – you’ll pee into a cup. Oral swab – you’ll rub a soft swab inside your mouth or throat. 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16681	https://www.youtube.com/watch?v=UcYbFN49uGc&vl=en	Solving Systems of 3 Equations Elimination Mario's Math Tutoring 453000 subscribers 4638 likes Description 438573 views Posted: 19 Feb 2016 Learn how to Solve Systems of 3 Equations using the Elimination Method in this free math video tutorial by Mario's Math Tutoring. Timestamps: 00:00 Intro 0:09 Explanation of How the Equations Represent Planes 0:16 Choosing a Variable to Eliminate 0:31 Using the Elimination Method 1:20 Using the Elimination Method a Second Time Related Videos: General Solution With 3 Equations Organized List of My Video Lessons to Help You Raise Your Scores & Pass Your Class. Videos Arranged by Math Subject as well as by Chapter/Topic. (Bookmark the Link Below) ➡️JOIN the channel as a CHANNEL MEMBER at the "ADDITIONAL VIDEOS" level to get access to my math video courses(Algebra 1, Algebra 2/College Algebra, Geometry, and PreCalculus), midterm & final exam reviews, ACT and SAT prep videos and more! (Over 390+ videos) 312 comments Transcript: Intro use all three equations. I'm going to start by trying to eliminate one of the variables. We're going to pick Z because it looks like that's going to be the easiest to eliminate, but you could, you Explanation of How the Equations Represent Planes know, start with eliminating Y's or X's. But let's go ahead and eliminate uh the Z's first. And we'll do that by combining the first and the second Choosing a Variable to Eliminate equation. If we add those together, we get 3X minus Y, the Z's are going to cancel, equals 1. Okay, now we want to use all three equations. We use use the first Using the Elimination Method two. We want to use the the third one. So, let's go ahead and combine it with the second one. You can combine it with the first or the second, but you want to use that third equation. So, we use all three. What I'm going to do here is I'm going to multiply this middle equation times 2. So, that's going to give us 4x - 6 y + 2 z = -2. The third equation I'm just going to bring down and we're going to add these two together. So if we do that we get 9x - 7 y =5. Okay, now we're down to two equations with two variables, just x's and y's. So let's multiply this equation by -7. So we get a positive 7 y. We can add it to this equation and eliminate the y's. So if we do that, we get 21x. Okay. -21x Using the Elimination Method a Second Time + 7 y = -7. And we're going to add those together. We get -12x. The y's cancel. We get -12. Divide by -12. x = 1. So we went from three equations with three variables down to two equations with two variables down to one equation with one variable. Now we're going to work our way backwards. Okay? So we're going to put this into one of the equations that has just two variables. Let's put it into this one over here. So 3 1 - y = 1. We put in one for x. So we subtract three from both sides. That gives you -2. Divide by -1. Y = 2. So now we know what x and y are. Let's go ahead and put it back into one of the original equations to solve for z. So we'll just do this top one. So we have 1 + y is 2. So 2 2 is 4 - z = 2. That's 5 - z = 2. Subtract 5. That equals -3. Divide by -1. Z = 3. So now we have x, y, and z. And we're going to write this as a triple. XYZ. Alphabetical order. 1 2 3. XYZ. Okay. Okay. So you just write it as a triple
16682	https://artofproblemsolving.com/wiki/index.php/Semiperimeter?srsltid=AfmBOoqjSvjki7qMjPmWC8tcmGo7XzU_ve06aUWUtVd5A4aYI8Zi4Pkz	Art of Problem Solving Semiperimeter - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Semiperimeter Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Semiperimeter The semiperimeter of a geometric figure is one half of the perimeter, or , where is the total perimeter of a figure. It is typically denoted . In a triangle, it has uses in formulas for the lengths relating the excenter and incenter. This article is a stub. Help us out by expanding it. Applications The semiperimeter has many uses in geometric formulas. Perhaps the simplest is , where is the area of a triangle and is the triangle's inradius (that is, the radius of the circleinscribed in the triangle). Two other well-known examples of formulas involving the semiperimeter are Heron's formula and Brahmagupta's formula. Retrieved from " Category: Stubs Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
16683	https://pmc.ncbi.nlm.nih.gov/articles/PMC6838041/	The Effectiveness of Rehabilitation Interventions on the Employment and Functioning of People with Intellectual Disabilities: A Systematic Review - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. 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Learn more: PMC Disclaimer \| PMC Copyright Notice J Occup Rehabil . 2019 May 16;29(4):773–802. doi: 10.1007/s10926-019-09837-2 Search in PMC Search in PubMed View in NLM Catalog Add to search The Effectiveness of Rehabilitation Interventions on the Employment and Functioning of People with Intellectual Disabilities: A Systematic Review Nina Nevala Nina Nevala 1 Finnish Institute of Occupational Health, PO Box 40, 00032 Työterveyslaitos, Finland Find articles by Nina Nevala 1,✉, Irmeli Pehkonen Irmeli Pehkonen 1 Finnish Institute of Occupational Health, PO Box 40, 00032 Työterveyslaitos, Finland Find articles by Irmeli Pehkonen 1, Antti Teittinen Antti Teittinen 2 The Finnish Association On Intellectual and Developmental Disabilities, Viljatie 4 A, 007004 Helsinki, Finland Find articles by Antti Teittinen 2, Hannu T Vesala Hannu T Vesala 2 The Finnish Association On Intellectual and Developmental Disabilities, Viljatie 4 A, 007004 Helsinki, Finland Find articles by Hannu T Vesala 2, Pia Pörtfors Pia Pörtfors 3 National Institute for Health and Welfare, PO Box 30, 00271 Helsinki, Finland Find articles by Pia Pörtfors 3, Heidi Anttila Heidi Anttila 3 National Institute for Health and Welfare, PO Box 30, 00271 Helsinki, Finland Find articles by Heidi Anttila 3 Author information Article notes Copyright and License information 1 Finnish Institute of Occupational Health, PO Box 40, 00032 Työterveyslaitos, Finland 2 The Finnish Association On Intellectual and Developmental Disabilities, Viljatie 4 A, 007004 Helsinki, Finland 3 National Institute for Health and Welfare, PO Box 30, 00271 Helsinki, Finland ✉ Corresponding author. Issue date 2019. © The Author(s) 2019 Open AccessThis article is distributed under the terms of the Creative Commons Attribution 4.0 International License ( which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. PMC Copyright notice PMCID: PMC6838041 PMID: 31098847 Abstract Purpose This systematic review analyzed the effectiveness of rehabilitation interventions on the employment and functioning of people with intellectual disabilities (ID), as well as barriers and facilitators of employment. Methods This was a systematic review of quantitative, qualitative, and mixed methods studies. The outcomes were employment, transition to the open labor market and functioning. The review included qualitative studies of employment barriers and facilitators. The population comprised people with ID aged 16–68 years. Peer-reviewed articles published in English between January 1990 and February 2019 were obtained from the databases Cinahl, the Cochrane Library, Embase, Eric, Medic, Medline, OTseeker, Pedro, PsycInfo, PubMed, Socindex, and the Web of Science. We also searched Google Scholar and Base. The modified selection instrument (PIOS: participants, intervention, outcome, and study design) used in the selection of the articles depended on the selection criteria. Results Ten quantitative (one randomized controlled, one concurrently controlled, and eight cohort studies), six qualitative studies, one multimethod study, and 21 case studies met the inclusion criteria. The quantitative studies showed that secondary education increases employment among people with ID when it includes work experience and personal support services. Supported employment also increased employment in the open labor market, which sheltered work did not. The barriers to employment were the use of sheltered work, discrimination in vocational experience, the use of class teaching, and deficient work experience while still at school. The facilitators of employment were one’s own activity, the support of one’s family, job coaching, a well-designed work environment, appreciation of one’s work, support form one’s employer and work organization, knowledge and experience of employment during secondary education, and for entrepreneurs, the use of a support person. Conclusions The employment of people with ID can be improved through secondary education including proper teaching methods and personal support services, the use of supported work, workplace accommodations and support from one’s family and employer. These results can be utilized in the development of rehabilitation, education, and the employment of people with ID, to allow them the opportunity to work in the open labor market and participate in society. Electronic supplementary material The online version of this article (10.1007/s10926-019-09837-2) contains supplementary material, which is available to authorized users. Keywords: Employment, Intellectual Disability, Developmental Disability, Learning Disability, Rehabilitation, Systematic review, Barriers, Facilitators Background Employment is one of the primary goals of people with intellectual disabilities (ID) [1, 2]. Employment can lead to positive psychosocial and economic benefits for people with ID, including a sense of purpose, opportunities for new friendships [1, 2], health and better quality of life [4, 5]. People with ID seldom work in the open labor market and the proportion of people with ID in employment varies between countries. For example, in Finland, it is 3% of working-age people with ID , in England 5–11% [7–10], and in the USA, 10% . The prevalence of ID is about 1% of the population, but it differs between countries . In Finland, it is 1% [13, 14], as in most European countries . In Finland, 0.8% of working-aged people have ID, which means 25,000 people . Rehabilitation is a goal-oriented process, which here aims to enable people with ID to reach an optimum mental, physical and social functional level, thereby providing them with the tools required to change their lives . People with ID should receive services that support their functioning, self-determination and participation [2, 17]. In the last years, instead of using a system-centered approach, services for people with ID and other disabilities have begun to use a person-centered approach, tailoring services around the individual rather than enforcing a one-size-fits-all principle [2, 18, 19]. According to Austin and Lee , job-related services (job search assistance, job placement, job readiness training, on-the-job support services) significantly predicted the employment outcomes of people with ID. However, they found no significant relationship between personal services (diagnosis and treatment, counseling and guidance, transportation, maintenance, miscellaneous training) and employment outcomes. Sheltered work or workshops are one form of rehabilitation for people with ID. One of the main tenets often cited by supporters of sheltered workshops is that center-based programs provide employment opportunities for people with ID in a segregated environment by building up their vocational and social skills [21, 22]. However, young adults with mild to moderate ID also have difficulties transitioning from sheltered workshops to the open labor market. Moreover, parents’ overprotection, important friendships, and the attitudes of employers and co-workers can be barriers during this transition process . In this review, the term intellectual disability (ID) refers to people who have a significantly reduced ability to understand new or complex information and to learn new skills (impaired intelligence), and who have reduced ability to cope independently (restricted social functioning). In the United Kingdom (UK) the term ‘people with learning disabilities’ (LD) is used to describe those referred to elsewhere as people with ‘intellectual disabilities’ or ‘developmental disabilities’ . No reviews were found on the effects of rehabilitation on employment among people with ID. Some reviews, however, evaluated the effectiveness of rehabilitation on the functioning (activities of daily living, self-care skills, communication skills, and cognitive achievements) of people with ID , the effects of the person-centered planning of services for people with ID on different outcomes (i.e., employment) , and effects of technology use on employment . The research questions were: 1) How effective are rehabilitation interventions for the employment of people with ID, 2) what are the barriers to and facilitators of employment of people with ID, and 3) what kind of individual support measures have been used to increase the work ability and functioning of people with ID? The theoretical orientation guiding this study was the International Classification of Functioning, Disability and Health (ICF) framework (Fig.1). The ICF identifies and classifies the domain of environmental factors, including rehabilitation, as one of its health-related domains [26–28]. ICF provides a detailed framework for describing disease experiences as a dynamic interaction between its components, and ‘core sets’, comprising lists of ICF categories . The ICF identifies and classifies the component of environmental factors, including rehabilitation, as one of its health-related components [26, 27]. According to the ICF, these environmental factors can also be either barriers to or facilitators of employment for these individual . Fig.1. Open in a new tab The theoretical orientation that guided this study was the ICF (International Classification of Functioning, Disability and Health) framework Methods This systematic review analyzed the effectiveness of rehabilitation interventions on the employment and functioning of people with ID, as well as barriers and facilitators of employment. We included quantitative, qualitative, and multimethod studies, because they provided different knowledge regarding this phenomenon [29, 30]. Quantitative studies showed the effectiveness of rehabilitation; qualitative studies showed the process, facilitators and barriers; and multimethod studies combined these. We used the modified selection instrument (PIOS: participants, intervention, outcome, and study design) [29, 31] to select the titles, abstracts, and full papers according to the selection criteria. The inclusion criteria for studies were randomized controlled trials (RCTs) and non-randomized studies such as concurrently controlled trials (CCTs), case–control studies, cohort studies, follow-up studies, and case studies that investigated the effectiveness of rehabilitation among people with ID. We expected the outcomes of the studies to be employment (supported or non-supported employment, transition from school to work, transition from sheltered workshop to employment), or functioning (job performance), and the population to comprise people with ID (developmental disabilities, learning disabilities, cognitive disabilities, mental retardation, mental handicaps), in the age group of 16–68 years. We searched for articles published in English from January 1990 to February 2019. The databases we searched in February and October in 2016 and in February in 2019 included: Cinahl, the Cochrane Library, Embase, Eric, Medic, Medline, OTseeker, Pedro, PsycInfo, PubMed, Socindex and the Web of Science. In addition, we searched Google Scholar and Base. Our main search terms were: (1) intellectual disability, mental disability, mental retardation, mentally disabled persons; (2) employment, employability, sheltered work, work capacity, work ability, vocational status, career; (3) rehabilitation, treatment, therapy, training, career counseling, social support, education, program, intervention, assistive technology, counseling, mainstreaming; (4) outcomes, outcome assessment, randomized controlled trials, program evaluation, effectiveness, validation studies, evaluation studies, comparative studies, cost–benefit analysis, before-after and follow up. We also manually scanned reference lists of identified reviews for additional relevant articles. Full details of the search strategy are available (Supplementary material). The review team comprised five researchers, whose areas of expertise were disability, ergonomics, rehabilitation, social sciences, interventions, systematic reviews, and quantitative and qualitative methodology. First, pairs of researchers independently reviewed titles and abstracts and made a unanimous decision. If consensus was not reached, a third researcher was consulted. The full texts of all the eligible articles and those whose eligibility could not be discerned from reading the abstract were obtained. Two reviewers independently assessed the methodological quality of the included RCT and CCT studies using the checklist by van Tulder et al. , which has also been used in other reviews concerning rehabilitation [28, 33, 34] and workplace accommodation . The checklist consisted of 11 criteria. Each item scored two points for ‘Yes’, one point for ‘Don’t know’, and 0 points for ‘No’. The item scores were summed into a single total score (range 0–22). Studies were considered to be of high methodological quality if the score was at least 11 out of 22, and of low methodological quality if they achieved fewer than 11 points . The cohort studies and the mixed method study were assessed using the Newcastle–Ottawa Scale (NOS) . This assessment scale consists of eight items within the following three categories: selection of study groups (four items), comparability of groups (one item/two questions) and ascertainment of exposure/outcome (three items). The highest value for quality assessment is nine ‘stars’. One star is allocated for each item in the selection and outcome categories and two stars in the comparability category. Study quality was graded as follows: 1–3 stars (low quality), 4–6 stars (intermediate quality), and 7–9 stars (high quality) . Three pairs of reviewers independently assessed the methodological quality of the included qualitative studies using a modified version of the CASP method . Any difference in the item scoring was resolved through discussion with the third author until consensus was achieved. This assessment tool consisted of evaluation criteria that are commonly acknowledged as crucial in qualitative research in the social sciences. The assessment tool involved 10 questions based on the following four main principles: The research should (1) provide a defensible research strategy that can answer the questions posed, (2) demonstrate rigor through systematic and transparent data collection, analysis and interpretation, (3) contribute to advancing wider knowledge and understanding, and (4) demonstrate credibility with plausible arguments about the significance of the findings. Each item scored ‘Yes’ or ‘No, depending on whether the topic was described sufficiently. An additional score of ‘Partially’ was added, as the original checklist was not able to adequately differentiate between the quality of the studies . This addition resulted in three options: ‘Yes’ (2 points), ‘Partially’ (1 point), and ‘No’ (0 points). The higher the total score, the better the methodological quality, with a maximum score of 20. The studies were rated as being of high methodological quality if they achieved more than 10 points. The case studies were assessed using eight questions from the Joanna Briggs Institute (JBI) method : (1) Were the demographic characteristics of the person attending clearly described? (2) Was the history of the person attending clearly described and presented in the timeline? (3) Was the current clinical condition of the person clearly described, (4) Were diagnostic tests or assessment methods and the results clearly described? (5) Was the intervention(s) or treatment procedure(s) clearly presented? (6) Was the post-intervention clinical condition clearly described? (7) Were adverse events (harms) or unanticipated events clearly described? (8) Does the case report provide takeaway lessons? Each question scored one point for ‘Yes’, two points for ‘No’, three points for ‘Unclear’, and four points for ‘Not applicable’. All the reviewers participated in the data extraction, which was carried out separately to the quantitative, qualitative, and multimethod studies, and included details on the participants and descriptions of the rehabilitation and the outcomes. The meaningful concepts of each outcome were linked to the ICF categories [40, 41]. We descriptively synthesized the data using tables to describe the characteristics, results and quality of the included studies. Different tables were used to describe the quantitative, qualitative and case studies. We categorized the overall quality of the quantitative studies and their outcomes using the principles of GRADE (Grading of Recommendations, Assessment, Development, and Evaluation) . The GRADE approach classifies the quality of evidence as high, moderate, low or very low. Randomized controlled studies’ evidence begins with strong and cohort studies with scant evidence on the basis of the assumption that randomization is the best method for controlling unknown factors. Five factors can weaken evidence: (1) the risk of bias, which results from the weakness of the research method and its implementation, (2) differences and inconsistencies between different studies, (3) indirectness of the results in comparison to the PICO criteria, (4) the inaccuracy of the results (the number of events and participants and their effect on confidence intervals) and (5) publication bias (Chapter 12, Cochrane Handbook). Thus, strong research evidence required at least two-high quality studies, the results of which were parallel. Moderate research results required one or several high-quality studies, the results of which were only slightly contradictory, or several adequate studies, the results of which were parallel. Scant research evidence meant that the research results had significant contradictions. No research evidence meant that no studies existed or that they were methodically weak. Results The search strategy identified 2618 references (Fig.2). After removing duplicates, 1848 references remained. Two of the reviewer (altogether five) authors scrutinized all the titles and abstracts according to the inclusion criteria, and when information necessary for inclusion was lacking, they read the articles’ full texts. We obtained the full texts of 244 articles, and 38 studies met our inclusion criteria. Fig.2. Open in a new tab Flow chart of review identification and selection process Characteristics of the included studies Design and methods Ten of the 38 included studies were quantitative (Table1), six studies were qualitative (Table2), one was a mixed methods study (Table2), and 21 were case studies (Table3, 4). The quantitative studies included one RCT , one CCT and eight cohort studies [45–52]. The qualitative studies included three interview studies [53–55], one interview and document study and two ethnographic studies [57, 58]. The mixed method study used structured interviews, focus group interviews and register analysis. The 21 case studies included five qualitative case studies [60–64] and 16 quantitative experiments [65–80]. Table 1. Characteristics of included quantitative studies \| Authors, year, country \| Study design, methods \| Population, definitions/terms \| Intervention/rehabilitation/personally tailored/length of intervention/personal or environmental measures \| Outcome measures \| Findings \| :--- :--- :--- \| \| Arvidsson et al. , Sweden \| A cohort study. Regression analysis of register data from Halmstad University Register on Pupils with Intellectual Disability (based on complete national survey of students who graduated from Upper Secondary School for Pupils With Intellectual Disability in Sweden) merged with data from two national registers, the Integrated Database for Labor Market Research (LISA; which includes all people in Sweden over the age of 15) and the LSS register (which includes all those who receive services under the Act Concerning Support and Service for Persons with Certain Functional Impairments) \| N = 2745 (71% M/29% F) gainfully employed pupils with intellectual disability (22.4% of 12,269 students who graduated from Upper Secondary School for Pupils with Intellectual Disability between 2001 and 2011) \| Upper Secondary School for Pupils with Intellectual Disability programs: national program, specially designed program, vocational training, training activities, or graduated with incomplete grades \| Gainful employment i.e. work full- or part-time and have either unsubsidized or subsidized employment (they all had a job during 2011 that generated a salary reported to the Swedish tax agency and they neither had daily activities according to LSS nor were registered as students. The variable gainful employment is dichotomous and states whether a person was gainfully employed or not in 2011, according to the LISA register \| 339 people (72% M/28% F) were gainfully employed without any type of subsidies. 195 (57.5%) had attended a Upper Secondary School for Pupils With Intellectual Disability programs, 90 (26.5%) attended a specially designed program, 46 (13.6%) graduated from USSID with incomplete grades, and 8 people (2.4%) attended an individual program/vocational training Of the 96 women, 70.8% worked in the private sector and 24% in the public sector. The most frequent occupation (36.5%) was “personal care and related work.” The 243 men worked primarily in the private sector (93.4%), with the largest proportion (18.9%) in professions described as “other service work” (e.g. janitors, garbage collectors, newspaper and package deliverers) Probability of being gainfully employed: OR: 3.84 national program, OR: 2.87 specially designed program, and OR: 0.31 vocational training program, compared to those who graduated with incomplete grades. Men (OR: 4.06) were more likely to be employed than the women (OR: 3.51) \| \| Beyer and Kaehne , UK \| A cohort study. Carried out in six local authority areas in England, Scotland and Wales. A total of 14 special schools took part in this study. Logistic regression \| N = 87 (61%M/39%F), Dg: LD, additional diagnoses/problems: Autistic spectrum disorder (9%), Asperger’s syndrome (1%), emotional or behavioral difficulties (5%) IQ: nr Age:17.8 years Occupation: nr \| Work awareness training (e.g. watching videos showing work, talking about presenting one’s self at work, health and safety instruction), vocational qualifications courses, work experience (In special schools provided within the school e.g. assisting in class for younger children, helping the janitor, working in the kitchen and in external transition organizations primarily gained in the community with or without a job coach), vocational profiling, practical project placement (e.g. café, sandwich-making, and furniture-making enterprises) \| Full- or part-time employment six month post graduation \| Employment 18 (21%) of the young people were employed (56% from mainstream schools, 33% from colleges and 11% from special schools). 2 variables were significantly related to employment. Persons that had work experience (provided by external transition support organizations) were 1.01 times more likely to be employed, and those that had work awareness training (provided by the schools/colleges) were also 1.01 times more likely to be employed. These 2 variables explained 19.2% of the variance in predicting employment post graduation \| \| Bouck and Chamberlain , USA \| Secondary data analysis of the National Longitudinal Transition Study-2 (NLTS2). The data collection occurred from 2000 to 2009. This study included postschool data, including postschool services received and postschool outcomes. Survey via telephone, mailed survey. Frequency counts, F tests, cross-tabulations, logistic regression analyses \| N = 121,335 Age: 20.2 years (17–25). Gender: nr. Dg: mild ID. IQ: nr. Ethnicity: White 66.7%, African American 23.6%, Hispanic 8.8%, Asian/Pacific Islander 0.1% \| Postschool services: postsecondary education institution accommodations and services, job training services, and life skills services \| Paid employment, full-time employment \| Significantly more (p < .0.1) individuals who did not receive job training services were employed full-time (62.7%) compared with those who received job training services (16.5%). No significant differences were between those who did (56.2%) or did not (65.3%) receive job training services with regard to paid employment \| \| Cimera , USA \| Cohort study, two cohorts, matched pairs 2002–2006. Data comes from the Rehabilitation Services Administration’s 911-database. Two groups of supported employees were compared \| N = 9808. Sheltered workshops cohort: N = 4904 (41.7% F/58.3% M). Age: 38.93 years. Not in sheltered workshops cohort: N = 4904 (41.7% F/58.3% M). Age: 31.56 years \| Sheltered workshops \| The rate of competitive employment \| Employment There was no difference in the rates of employment for these two groups. 59.6% of sheltered workshop supported employees were competitively employed compared to 60.4% for individuals who did not participate in shetered workshops prior to entering supported employment. This difference was not statistically significant \| \| Cimera et al. , USA \| Cohort study. Matched pairs 2006–2009. Two groups were compared \| N = 15,040 (7520 in both groups) (54.6–56.7% M/43.3–45.4% F). Age (average): 20.28–20.33 years. Dg: Significant ID. Multiple disability 6.7–13.6%. IQ: nr. Ethnicity: White 46.8–50.6%, African American 48.5–52.3%, Asian 0.3–0.75, Native American 0.4–0.5%, Pacific Islanders < 0.1%, Hispanic 2.4–3.7%. Occupation: nr \| Transition services. Early transition group: transition services were addressed in individualized educational programs by age 14. Later transition group: transition services were addressed by age 16 \| Successful employment (e.g. employment in the community paying at least minimum wage) \| Employment Individuals from the early transition states were significantly more likely to become employed than their matched peers from the later transition states. For example, in 2006, 74.3% of young adults from the early transition states were employed compared with 57.8% for the later transition states \| \| Goldberg et al. , USA \| A randomized controlled trial (RCT). A Functional Assessment Scale. Chi square. Spearman correlation coefficient \| N = 49 (M 24/F 25), experimental group N = 24, control group N = 25 Age: mean 31.2 years (experimental group), 30 years (control group) Dg: DD since birth \| Experimental group (supported employment program): placement agents who made contacts with employers in the community, clients were immediately prepared for competitive employment with a variety of supportive services (i.e., job monitoring, social work with families). Control group (conventional workshop services): placement agents who made contacts with employers in the community, staff waited until clients were considered ready before placing them outside the workshop \| Employment in the competitive labor market (employed for 60 days) and in three categories: no work, some work (work at least 2 weeks), and currently working. Performance (Functional Assessment Scale) \| Employment Persons in the experimental group (supported work program) had a better change of obtaining employment than the control group X2 (2, N = 49) = 8.5, p < 0.02. of 24 persons in the experimental group, 20.8% were currently working, and 8.3% worked at some time. None of the control group had worked in competitive employment Performance There were no significant differences in performance between the groups. Pretest/posttest results were in the experimental group 54.2 (6.4)/51.4 (8.0), and in the control group 53.3 (7.6)/53.6 (7.4) \| \| Gray et al. , USA \| Cohort study. The data were provided by a State agency (South Carolina Department of Disabilities and Special Needs) and by the local boards. Registers, written survey. Logistic regression \| Employed in 1997: N = 431, 59.6%M/40.4%F, Age: 19–30 years 44.3%, 31–50 years 49.0%, 51–64 years 6.7%. Ethnicity: White 42.5%, nonwhite 57.5% IQ: 20–29 3.7%, 30–39 15.1%, 40–49 20.2%, 50–59 24.8%, 60–69 31.6%, 70–74 4.6% Unemployed in 1997: N = 6659, 48.7% M/45.3%F. Age: 19–30 years 35.1%, 31–50 years 51.3%, 51–64 years 13.7% Ethnicity: White 51.9%, nonwhite 48.1% IQ: 20–29 16.5%, 30–39 24.9%, 40–49 14.3%, 50–59 17.8%, 60–69 22.2%, 70–74 4.3% \| Number of job coaches \| Employment (earning at least $50 per week in a competitive job or enclave for at least six months in 1997) \| Employment The number of coaches varied from 0.00 to 2.48 per 100 individuals. Most individuals (82.6%) were served between 0.5 to 1.5 job coaches per individuals. Significant, positive effects of job coaches on employment were found as follows: The risk of employment as a function of coaches was approximately two times greater for individuals with low than for high IQ, approximately three times greater for individuals in counties with low or intermediate employment, and approximately ten times greater for individuals located in highly urban counties \| \| Joshi et al. , USA \| Cohort study. A study utilized data collected in the National Longitudinal Transition Study-2. Telephone interview, mail survey. Descriptive analysis, frequency counts, multiple regression and logistic regression analysis \| N = 62,513 (61.7%M, 38.3%F). Age: Average age for students in school was 18 years, average age for students out of school was 20.1 years. Dg: Mild ID. Ethnicity: Caucasian 70.2%, African American 24%, Hispanic 5.2% \| Students received special education services between 13 and 16 years of age. Employment-related transition activities identified in NLTS2 data were vocational assessment, career counseling, prevocational education, career technical education or vocational education, prevocational or job readiness training, instruction looking for jobs, job shadowing, job coach, special job skills training, placement support, internship or apprenticeship programs, tech prep programs, work experiences in school, and other paid work experiences \| Postschool employment (e.g. whether they ever engaged in employment, their current employment status, and whether they worked part- or full-time) \| 75.9% of the students reported they engaged in employment sometime after leaving school. 62.1% were currently employed and 42.6% were engaged in full-time employment. Participation in one additional transition activity represented within the employment activities summation variable resulted in students being 1.2 times likely to be currently employed post school. Students were 5.704 times as likely to ever engage in employment after school if they engaged in paid-employment experiences while in school, and 3.489 times as likely to ever engage in employment after school if they experienced a school-sponsored job. However, vocational education was not found to be significantly related to postschool employment \| \| Kilsby and Beyer , UK \| Intervention study. An independent groups design (a baseline and two intervention phases, three months each). Direct observations, questionnaire, taster sessions, job review forms. A one-way ANOVA \| N = 35. Three groups matched. Baseline: N = 12 (M 58%/F 42%). Age: 32 years (mean). Dg: 58% ID borderline, 33% mild, 8% moderate Intervention 1: N = 12 (M 50%/F 50%) Age: 26 (mean) Dg: 22% borderline, 45% mild, 33% moderate Intervention 2: N = 11 (M 50%/F 50%) Age: 32 (mean) Dg: 54% borderline, 38% mild, 6% moderate \| A job taster program (a two phase self-determination package) for adults with LD entering the job market for the first time. Job tasters were short, unpaid, time-limited work experiences which took place in the workplace in order to allow people to sample a variety of different jobs and work cultures. In each group, each participant was required to attend 3 job tasters for 6 sessions. Baseline: No specific instructions were provided to agency staff about how to conduct the job tasters. Intervention 1: Introduction to self-determination and systematic job taster reviews. A one-day package was implemented to introduce job coaches to the concepts of self-determination using Wehmeyer’s definition as a conceptual base; and job reviews were introduced. Job reviews were short meetings where job seekers were encouraged to evaluate the job taster sessions and overall program. Intervention 2: Introduction of pictorial job review profile. A second one-day package was implemented: job coaches were encouraged to support the job seekers to complete the job review forms for themselves. In order to help those who could not read, a pictorial questionnaire replaced the written version \| The rate of overall job coach assistance (instructions and questions). An average rate of assistance per minute was calculated by dividing the frequency of job coach assistance by the duration in minutes \| Performance Intervention 1 led to a reduction in rates of job coach assistance for both sessions and reviews. The rate of assistance fell significantly between Baseline (0.83 per min) and Intervention 1 (0.51 per min). The introduction of intervention 2 led to a further drop in rates of assistance during the reviews. The reviews in intervention 1 and 2 were generally accurate, and job seekers were consistent with their statements. This indicate that the job seekers had an awareness of their surroundings and a realistic grasp of their personal limitations and abilities \| \| Sannicandro et al. , USA \| A secondary data analysis of the RSA-911 datasets from 2008 through 2013. A quasi-experimental design Multilevel modeling, propensity score matching techniques. Multilevel logistic regressions and multilevel linear regressions, depending on whether the outcome was binary or continuous \| N = 11,280. Experimental group: N = 5640, (M 46.9%/F 53.1%). Age: 20.9 years. Dg: ID. Ethnicity: White 57.6%, Black 31.1%, American Indian 1.5%, Asian 2%, Native Hawiian or Pacific Islander 0.5%, Hispanic 7.3%. Control group: N = 5640, (M 57.2%/F 42.8%). Age: 20.4 years. Dg: ID. Ethnicity: White 52.6%, Black 36%, American Indian 1.3%, Asian 1.4%, Native Hawiian or Pacific Islander 0.4%, Hispanic 8.3% \| Postsecondary education was comprised of positive response to the following variables from the RSA-911 dataset (1) postsecondary education with no degree, (2) associate degree or vocational/technical certificate, (3) bachelor’s degree or, (3) master’s degree or higher \| Employed at closure (yes/no) \| Individuals who participated in postsecondary education were more likely to be employed (OR: 2.00, 95% CI:1.82–2.20) than their counterparts who had not participated in postsecondary education \| Open in a new tab N number, M males, F females, Dg diagnosis, IQ Intellectual Quantity, nr not reported, ID Intellectual Disability, IDD Intellectual and Developmental Disability, DD Developmental Disability, LD Learning Disability, MR Mental Retardation, SE Supported Employment Table 2. Characteristics of included qualitative studies and a multimethod study \| Authors, year, country \| Methods, study design \| Population, definitions, terms \| Intervention \| Findings, barriers and facilitators \| :--- :--- \| Alborno and Gaad , Dubai \| Observation, semi-structured interviews, document analysis. Data triangulation \| N = 27 (100% M) Age: 19–37 years Dg: ID and physical disabilities (nine with Down syndrome) IQ: nr. Average mental age 4–15 years. Most had attended rehabilitation centers where they had learned some basic life skills \| Employment either in administrative jobs in the head office or as gardeners in the nursery. Functional and vocational training programs, productivity monitoring, recreational and sports activities, support in psychological and behavioral issues by documenting the development of the behavioral, psychological, and professional skills of the employees with disabilities, and regular provision of individualized and group sessions for the employees to address any behavioral or psychological issues, and transportation services in dedicated buses supplied for the group \| All were successfully and fully integrated into their jobs: data entry, public relations, photography, and office mail distribution. Among their peers and the company’s management, they were recognized as loyal, efficient, and reliable staff The employer recognized that certain barriers between their employees with disabilities and the work environment had to be overcome The employees felt they were valued and respected, which resulted in higher self-esteem and self-confidence. They were also trusted through an open-door policy with management which allowed them to voice their opinions and needs. High job satisfaction Facilitators Recruitment with no formal testing of abilities Training: efficient, staged training programs Workplace accommodations (shading the area from the sun, shorter working hours in summer, fully paid two-month summer holiday, special chairs, modified equipment, absence allowed for rehabilitation/treatment, first aid training workshops, transportation services with company busses) Management, supervision, and performance monitoring (support of staff and management, behavioral tally sheets, positive behaviors such as greeting peers, cooperation, following orders, and speech) as well as negative behaviors (such as rude words, violence, disobedience, or stubbornness) were recorded, which enabled early detection of problems and effective problem-solving involving all stakeholders Incentives such as adequate remuneration, professional development, and recreational and social programs resulted in empowerment and positive attitudes, creating a friendly work environment Professional progress: opportunities for developing new skills \| \| Christensen and Richardson , USA \| Semi-structured interviews. Inductive content analysis approach \| N = 10 (M = 6) Age: 25–51 years Dg: ID IQ: nr \| Project SEARCH is a licensed transition-to-work model, which features total workplace immersion, and facilitates a seamless combination of classroom instruction, career exploration, and on-the-job training and support. The goal was the transition of individuals out of sheltered workshops to competitive employment \| Eight persons completed the course and five of them were employed. The program emphasized learning about other job opportunities in the community (e.g., touring local businesses). Individuals with significant disabilities can imagine themselves working in the community when they are exposed to a variety of career opportunities and can directly see and under-stand what the job entails The cohort model of Project SEARCH allowed ten individuals to exit the workshop and transition into a community setting with familiar peers. This alleviated some of the initial fears of losing friends as a result of leaving the workshop. During the course the participants became more independent and lived in the community, and their views about the meanings of employment changed. Participants demonstrated an increased awareness of employment as a means to take greater responsibility as a productive member of their house-hold and the community. Competitive employment represented the means to earn wages in order to pay their own bills and not have to rely on Medicaid benefits The sample size is small and localized. While the results of this study provide some level of insight regarding the experiences of individuals with IDD who are transitioning from a sheltered workshop into competitive, integrated employment, the views expressed by program participants are limited to their own unique experiences \| \| Devlieger and Trach , USA \| Ethnographic study. Extensive life history interviews with focal people, parents, and agency or school personnel \| N = 6 (M = 4) Age: 18–24 years Dg: Mild mental retardation IQ: nr \| Intervention not described, instead the impact of different actors (school, agency, parents, other people) on the transition process and employment was analyzed \| The involvement of parents and focal people was disproportionate to that of school and agency personnel. School and agency efforts most often resulted in sheltered employment, while personal or parent mediation resulted more often in self-employment and continuing education outcome \| \| Donelly et al. , Australia \| Ethnographic study. Interviews of participants and their social networks. On average, five people in each network were interviewed. Interviewees included parents, other family members, members of various support circles, support workers, friends and allies. Participant observations were conducted of the three participants’ work or work preparation environments. The fourth participant provided information about her work in interviews \| N = 4 (M = 1) Age: 21–48 years Dg: Intellectual disability IQ: nr \| No intervention was described \| The meaning of work to participants. A range of meanings including experience of meaningful occupation, the development and recognition of skills, experiences such as travelling to and from work, relationships with co-workers, financial remuneration and access to opportunities that extended beyond the workplace The meaning of work to families included skill development, time use (participant’s days filled with meaningful activity), the choices (or interests) of participants as indicators of potential employment opportunities; quality of performance was linked to level of interest; interests and skills were not always valued or recognized. Vocational preparations that failed: formal service providers fitting people into existing programs that did not meet their needs. Classroom-based vocational preparation vs. actual work experiences (failing in classroom may prevent work experience opportunities); Effective vocational preparation: helped the person get a job based on personal interests, skills and choices and provided on-the-job training and support Barriers to employment Inadequate identification of support needs Lack of responsiveness to individual needs Emphasis on fitting people into existing support models Loss of vocational skills or failure to develop vocational skills in sheltered employment Difficulty moving from sheltered to competitive employment Classroom-based vocational preparation Use of classroom-based learning experiences and written examinations as pre-requisites for more meaningful, effective employment opportunities Failure of employment agencies to find opportunities for work experience and on-the-job training \| \| Fashing , Austria \| Grounded theory, biographical interviews \| N = 3 (F) Age: nr Dg: ID Women who have similarly difficult post-school vocational orientation phases despite having been successful in their transition to vocational training or employment on the regular labor market \| ‘What experiences do women with intellectual disabilities have in their transition from school to vocational training and employment?’ The interviewed women initially went from compulsory school to a vocational preparation measure or directly into a sheltered workshop or occupational therap. \| Main category: ‘coming to terms with disability through vocational participation’ was formed on the basis of two key categories: discrimination at school due to learning difficulties and post-school orientation phase as a process of self-determination \| \| Winsor et al. , USA \| Multimethod study. Structured and in-person interviews, focus groups, register data \| N = 1452 Participants in PP countries, N = 390 (2008 N = 160, 2009 N = 230) Comparison groups: Non-participants in PP counties N = 656 (2008 N = 315, 2009 N = 341), no PP county students, N = 406, 2008 N = 212, 2009 N = 194) Sex: nr Age: 21 years IQ: nr High school students who were eligible for developmental disability services and had turned 21 during their year of graduation, and participated the project Dg nr. Division of Developmental Disability eligible students. Informants were staff members, stakeholders (county developmental disability staff, school administrators, teachers, employment vendors, family member, and young adults who had obtained jobs) \| The starting point of the ‘Jobs by 21 Partnership Project’ (PP) was, that postgraduate employment outcomes seem to be related to employment experiences prior to graduation. The aim of the project was to provide high school students with ID and their families with information about and experiences of employment and the adult service system prior to their graduation, thus making the transition from school to adult life more seamless. The project was implemented in 11 counties (9 in 2008 and 11 in 2009) by local Developmental Disability offices. The local offices gathered different stakeholders to collaborate: Division of Vocational Rehabilitation, the Work Source Center, adult employment vendors, local community colleges, and local businesses Partnership project (PP) goals: 1) a post-school outcome plan for students, 2) expansion and improvement of collaboration in country between Division of Developmental Disability Counties and schools to enable students to make use of available support, 3) establishment of a statewide partnership, 4) to make use of job training and job preparation opportunities, labor market guides, workforce development trends, and post-graduation outcome reports to achieve post-school employment for transition-aged students with developmental disabilities \| Research questions 1) What was the impact of connecting young adults with employment vendors prior to graduation? 2) How were resources maximized across the school and adult service systems? 3) What strategies were most effective in encouraging collaboration between school and adult service systems? Results Employment outcomes: PP participants were more likely to be employed following school exit (after three months) and had stronger employment outcomes (mean hours and mean wages) than students who were not participants. In 2008, 45% of participants were employed, compared to 6% of non-participants; in 2009 11% participants were employed, compared to 05% and 5.7% of non-participants System outcomes: The availability of project funds encouraged stakeholders from the school and adult service systems to contribute additional dollars and resources in-kind to the project Collaborative relationships: The process of leveraging resources helped bring together stakeholders to collectively commit to achieving employment outcomes and problem-solving \| Open in a new tab N number, M males, F females, Dg diagnosis, IQ Intellectual Quantity, nr not reported, ID Intellectual Disability, IDD Intellectual and Developmental Disability, DD Developmental Disability, LD Learning Disability, MR Mental Retardation, SE Supported Employment Table 3. Characteristics of included quantitative case studies \| Authors, year, country \| Study design, methods \| Population, definitions/terms \| Intervention/rehabilitation/personally tailored/length of intervention/personal or environmental measures \| Outcome measures \| Findings \| :--- :--- :--- \| \| Allen et al. , USA \| Quantitative case study. Interrupted time-series design. Observation \| N = 3 Ned: M, 18 years, dg: mild MR. Tracey: M, 17 years, dg: moderate MR Emma: F, 16 years, dg: moderate MR \| The two interventions (video modeling and audio cuing) were evaluated in an interrupted time series withdrawal design during training and then again in an actual job setting (at the factory and warehouse). During video modeling, the participants watched standard training videos on a laptop computer For the Audio Cuing condition, the participants wore a (Radio Shack TRC-508 s FM) transceiver with microphone and earphones that allowed hands-free operation. During Audio Cuing, the participants were told to ‘Listen to the attendant, who will give you ideas of things you can do to entertain and interact with customers’ \| Percentage occurrence of multiple skill during 2-min work samples across baseline, video modeling, audio cuing, and 10-min work samples while actually working in the job at 1 and 3 months \| Job skills Video modeling was not effective whereas audio cuing produced job performances well above the designated criteria At baseline, Tracey showed a decreasing trend in multiple skills use, Ned showed no evidence of multiple skill use, and Emma showed some evidence of multiple skill use. During video modeling, Tracey showed no evidence of multiple skill use, Ned showed an initial but unsustained increase, and Emma showed a modest increase in multiple skill use. With the introduction of audio cuing, the rates of multiple skills use changed immediately and substantially for Tracey (100%), Emma (90%), and Ned (80%) \| \| Bennet et al. , USA \| Experimental case study. The design consisted of three conditions including baseline, intervention Covert Audio Coaching (CAC) and follow-up (five consecutive data collection sessions followed by weekly probes. The study was conducted at the participants’ job sites \| N = 2 Andy: M, 30 years, dg: ID, IQ: 38 Language summary: speaks in partial sentences in Creole and English Worked as a custodial assistant at a public school Daniel: M, dg: ID, IQ: 55 Language summary: speaks in complete sentences Worked at a food bank \| Supported employees received feedback delivered via CAC, which consisted of delivering praise, guidance and correction statements The tasks selected for Andy were washing the windows of automatic sliding glass doors and collecting trash in the schoolyard, and for Daniel stacking bread crates. All the tasks were part of their regular job duties \| Work performance (intervention fidelity and impact of CAC on accuracy, durability and fluency): Rate of praise, guidance and corrections delivered to the worker; percentage of task steps completed, pieces collected (in trash collection), rate of correct responses per minute \| Work performance Accuracy, durability and fluency: Andy, window washing: baseline < 38%, final sessions 97–100%, slight reduction after intervention. Rate increased by 10.9%. Andy, trash collecting: baseline stable but low, final sessions 66–79%, slight reduction after intervention. Rate increased by 17.4%. Daniel, crate stacking, high but inconsistent, final sessions > 95%, maintained same level after intervention. Rate increased by 78.3% The CAC intervention was effective in increasing the work performance of supported employees. The changes lasted for several weeks after the end of the intervention. The intervention was effective across different participants and work tasks \| \| Carson et al. , USA \| Case study. Observation \| N = 3 Tim: M, 18 years, dg: moderate ID and ADHD Brian: M, 20 years 10 months, dg: moderate ID Hope: F, 20 years 3 months, dg: mild ID \| Participants worked using functional skills (e.g., meal preparation) and functional academics (e.g., telling time, using a calculator). The photo activity schedule book (PASB) was used to increase the number of independent task changes in vocational tasks and the rate of vocational task completion. Five color photographs were placed in the album: four showed the individual tasks to be completed and the last depicted the finish location \| Number of independent task changes and rate of vocational task completion \| Performance The use of the PASB resulted in high levels of independent task changes among all participants. It also increased the rate of completion for two of the three students: In Wal-Mart, for Brian from 1.75 to 1.83 and for Tim from 1.10 to 1.52, and in the cafeteria, for Brian from 5.25 to 7.54 and for Tim from 7.92 to 9.78. In both places, the use of the book had little effect on Hope’s rate of performance \| \| Chang et al. , Taiwan \| Quantitative experimental case. Observation, tape recording \| N = 1, Yvonne: F, 27 years, dg: IDD \| The Location-based task prompting system (Locompt) was used in a short-order snack shop. The smart phone was fastened onto the participant’s lower arm. The system was programmed to be able to generate task cues in text, sound, picture, or a combination of these. Three sets of task sets were performed by the participant. Each task set had nine task steps to carry out an order with desserts, beverages and cookies. Baseline: Task set was performed with no assistive technology Intervention phase: Task set was performed with assistive technology (Locompt) \| Percentage of correct task steps \| Performance During the intervention phase, the percentage of correct task steps was significantly greater (99%) than at baseline (55%). The results indicate that the Locompt system in conjunction with operant conditioning strategies may facilitate autonomous functioning of vocational jobs across multiple workstations \| \| Devlin , USA \| Case study. Data were collected at baseline, during the intervention phase, and during the maintenance phase \| N = 4, Fred: M, 32 years, dg: moderate ID Mat: M, 20 years, dg: moderate ID, attention deficit disorder Kevin: M, 21 years, dg: mild ID, gross motor deficits Steve: M, 30 years, dg: mild cerebral palsy All worked 20 h per week and had been employed for between 2 and 4 months. They had general cleaning duties on different floors of the designated buildings \| The Self-Determined Career Model was developed to enable adult service providers to help individuals become self-regulated problem-solvers, to self-direct in the career decision-making process, and to gain enhanced self-determination. The model included three phases: 1) set a career/job goal, 2) take action, and 3) assess/adjust goal or plan. The model used a generic set of questions, which represent steps in the problem-solving sequence. Questions that followed the basic framework allowed individuals to modify their own behavior and thus become self-directed in reaching their goal \| Job performance (percentage of correct responses) \| Job performance Fred: baseline 50–53%, during intervention session 70–80%, and during maintenance condition 77–96% Matt: baseline 34–48%, during intervention session 72–93%, and during maintenance condition 93–100% Kevin: baseline 14–36%, during intervention session 32–100%, and during maintenance condition 77–91% Steve: baseline 44%, during intervention session 64–88%, and during maintenance condition 79% The work-related performances of all four employees improved after the three phases of the Self-Determined Career Model. Positive changes were evident between baseline and intervention conditions and continued in the maintenance phase. \| \| Dotson et al. , USA \| Case study, observation \| N = 3, Candy: F, 23 years, dg: MR Leah: F, 21 years, dg: Down syndrome Ethan: M, 27 years, dg: Down syndrome \| Transition academy provided instructions for independent living and vocational skills as well as community-based opportunities to practice these skills through activities such as volunteering at a local food bank and going out to eat and shopping. Self-employment job skills were evaluated within the context of the recycling program. The Transition academy team created an analog business that the Transition Academy students were responsible for running. Participants were paid with tokens. The aim of the intervention was not to teach all the skills required to run a business without help or supervision, but to teach worker job skills (how to do the job), supervisor skills (how to supervise someone else doing the job), and office skills (how to keep records of work completed) \| Job skills: Percentage of job steps performed correctly and independently \| Candy performed job skills at levels well below 20% on average at baseline, improved substantially during teaching, and maintained a high level of performance of job skills during maintenance and while working shifts in the natural environment. Leah and Ethan performed job skills at lower levels at baseline, improved during teaching, and maintained a generally high level of performance of job skills during maintenance and while working shifts in the natural environment \| \| Furniss et al. , UK \| Quantitative case study, observation, interview of the co-workers and carers \| N = 6 Mr. P: M, 35 years, dg: severe DD, job: assembling and packing nut/bolt/washer kit for classic cars Mr. W: M, 31 years, dg: severe DD, job: assembling of box and packing with forage caps Ms. R: F, 34 years, dg: severe DD, job: assembling and packing nut/bolt/washer kit for classic cars Mr. V: M, 43 years, dg: severe DD, job: preparing clock cards (attaching name sticker, sorting by dept., date stamping) Mr. H: M, 36 years, dg: severe DD, assembling aqualung pillar valve Mr. S: M, 47 years, dg: severe DD, assembling aqualung pillar valve \| VICAID, palmtop-based job aid for workers with severe developmental disabilities. The system was controlled by a single large key that the worker used and a second, inconspicuous key that the supporting job coach or co-worker used. Three user prompt devices were used: a small free-standing loudspeaker unit, a smaller, portable device to which the user listened via an ear-piece, and a vibrating prompting device carried by the user on their belt or in their pocket. For each job task, a sequence of pictorial instructions created on a desktop PC was downloaded to the palmtop. The system enabled the worker to access pictorial instructions designed to help them accurately complete their tasks. It also provided reminders to access instructions, and/or alerted a job coach or supervisor if the workers had difficulty with a task \| The percentage of task steps correctly completed, total time taken by the participant to complete the task, amount of time during each session that the participant spent actively engaged in the task \| Job performance The computer-aided VICAID system was effective in enabling participants to perform work tasks with a high degree of accuracy following relatively brief periods of intensive training. The transition of intensive training by a job coach in the VICAID-supported maintenance condition, together with limited job coach input, led to four of the six participants showing further improvements in performance accuracy at or beyond the levels achieved with intensive training. Practice in job tasks with the computer-aided system led to both maintenance of previously learned skills and learning new job competencies \| \| Gilson and Carter , USA \| Quantitative case study. Single-case experimental design \| N = 1 Braxton, M, 20 years, dg: ID, IQ: nr. Ethnicity: African American. Previous volunteer and employment experiences at the thrift store, restaurant, and his local church \| The study was conducted the participant’s individual jobsite. Braxton worked 4 h weekly at a market within a residence hall on campus. Tasks included stocking grocery shelves, refilling supplies at the coffee station and breakfast buffet line, and marking items with price tags and expiration dates. Braxton typically had one direct supervisor, about 3 co-workers and 4–6 customers. The intervention package was comprised of a) reduced proximity, b) use of CVC (convert audio coaching, c) social-focused coaching, and d) task-related proximal coaching as needed. The participant and the coach were linked through a two-way radio with accompanying earpiece. CAC job coach was trained to focus majority of the prompts on encouraging the participant to seek assistance from others when completing a task, rather than relying on the job coach. In addition, she/he was taught to give explicit social prompts when someone was in proximity for an interaction \| Percentage of intervals with task engagement and interactions. \| Performance. At baseline, Braxton’s levels of social interaction were consistently low with a flat trend. He rarely initiated or responded to interactions with anyone except his job coach. After he was coached on initiating conversations with his co-workers an accelerating trend and change in level was observed (from 0.7 to 27%), and it maintained at 27% after completely removing the coach Task engagement. Braxton’s task engagement level was 98% at baseline and it slightly increased to 98 and 99% during CAC and withdrawal phases respectively \| \| Kemp and Carr , USA \| Quantitative, experimental case \| N = 3 (2 M/1F) dg: Autism and severe mental retardation; Severe problem behavior (aggressive, self-injury) Age: 26–30 years \| The intervention included three factors: a) Interventions were chosen on the basis of the hypotheses regarding the maintaining variables for problem behavior; b) the multicomponent intervention package included some combination of functional communication training, building rapport, making choices, embedding demands, and building tolerance of delay of reinforcement; c) Measures of latency of problem behavior and percentage of work steps completed were used as outcome measures. At the end of the intervention phase, the job coach trained the regular employees in the implementation of the intervention and gradually reduced their supervisory time while the regular employees increased theirs \| Outcomes \| Results/Outcomes: All three employees (participants) were able to complete tasks (a planting sequence in a community greenhouse) with no significant problem behavior following multicomponent intervention, and were able to work four hours at a time. In addition, the greenhouse managers reported considerable confidence in the job coach’s ability to deal with any behavior difficulties, keeping both coworkers and property safe from harm; reported that problem behavior was almost never severe following the intervention, and acknowledged the employees as productive members of the greenhouse team \| \| McGlashing-Johnson et al. , USA \| Quantitative case study. A multiple baseline across students was used to evaluate the effects of the intervention. Task analyses. Observations by trained observers 2–4 times per week. The students recorded their behaviors using self-monitoring cards. The Goal Attainment Scaling (GAS). Likert scale \| N = 4. Jessica: F, 17 years, dg: moderate MR Sam: M, 17 years, dg: moderate MR Lindsay: F, 20 years, dg: moderate/severe MR Milo: M, 16 years, dg: moderate/severe mental retardation \| All students were involved in a work-based learning program in the community, which was operated by the school district. The program involved a self-regulated problem-solving process in which students set their goals, developed and implemented action plans to enable them to achieve their goals, and evaluated their progress in achieving these goals. A job coach accompanied all students to their respective sites. Jessica: packing bread/buns for patients in the hospital, work in the dish room. Lindsay and Milo: work in the garage of a small metropolitan bus station and cleaning the interiors of the buses. Sam: work at a gardening center. He priced items, swept the floors, shoveled snow, and counted inventory \| The percentage of correct responses in the task analysis for each task \| Job performance The Self-Determined Learning Model of Instruction (SDLMI) represents an effective method for teaching problem-solving to people with cognitive disabilities. During baseline, after training sessions and during maintenance, the respective percentages of correct responses were for Jessica 50%, 80%, 93%, Sam 31%, 70%, 80%, Lindsay 15%, 79%, 80%, Milo 6%, 46% \| \| McMahon et al. , USA \| Quantitative case study. Navigation checks, interview \| N = 3 Catelyn: F, 23 years, IQ: 45 Jon: M, 24 years, IQ: 56 Arya: F, 20 years, IQ: 64 \| The students completed a postsecondary education program. Three treatment conditions were implemented: navigation skills including a) a paper map, b) a Google Map, and c) augmented reality (AR). During each navigation session, each student was randomly assigned to one of the three treatment conditions using a spinner. This study occurred in a downtown area of a city. Participants navigated city streets to locate businesses that offered potential employment opportunities. The starting and ending localities were within a 12- to 20-minute walking distance of one another \| The percentage of correct independent navigation decisions within 30 s during ‘navigation checks’ while walking to a targeted unknown business location. Responses Yes/No \| Functioning/performance The AR treatment condition was the most effective. At baseline, with a paper map, a Google map, and AR application, the respective percentages of correct responses were for Catelyn 11.5%, 20.14%, 45.75, 75%, Jon 16.13%, 20.47%, 40.95%, 75%, and Arya 13.6%, 19%, 31.4%, 85.7% \| \| Renzaglia et al. , USA \| Quantitative case study \| N = 1 Phil: M, 23 years, dg: Down syndrome, low-moderate level of functioning. Occupation: animal carer in a university vivarium (competitive employment) \| A five-stage process included job task analysis, pre-baseline assessment, baseline assessment, treatment, and post-treatment follow-up. 1) Task analysis involved direct observation of co-workers performing the job tasks, performance of the tasks by the principle investigator (i.e., job coach), and reviewing the recorded task sequences with Phil’s supervisor. 2) The second state was to evaluate Phil’s current performance and to identify the specific skill areas in which training and retraining was needed. 3) The baseline data on the nine skills were collected for a five-day period. 4) Instruction in the nine skill areas identified as deficient in baseline assessment. Systematic instruction procedure using a least-to-most intrusive prompt system. 5) Post-training assessment of the subject’s performance under the same non-training conditions as those in the baseline assessment \| The percentage of job tasks performed correctly in a nine-item task analysis \| Job performance The percentage of correct performance in the nine designated tasks was 44% across the five days baseline, 68.7% during treatment, and 92.5% in the post-treatment assessment \| \| Simmons and Flexer , USA \| Quantitative case study.Observation. \| N = 2 Sandy: F, 27 years, dg: moderate MR Donna: F, 30 years, dg: severe MR \| The intervention was conducted in a janitorial and maintenance supported employment program at a restaurant/hotel complex. The training was carried out by a community employment specialist employed by the supported employment program to perform individualized placement and training (on Monday through Friday, 9–11 am, weekly). The training tasks were those of cleaning a restroom \| The percentage of steps performed independently during baseline, training and follow-up (two months after training had ceased) \| Job performance Both participants increased their performance to the criterion performance of 80%. Sandy: Performance increased from baseline (13% in Phase 1.15% in Phase 2, 54% in Phase 3) to training (81, 83, and 81%, respectively) Donna: Performance increased from baseline (46% in Phase 1, 40% in Phase 2, 54% for Phase 3) to training (89, 92, and 88%, respectively) \| \| Taber et al. , USA \| Case study, quantitative. Video tape, observation. Wilcoxon Matched Pairs Signed-Ranks Test \| N = 5 Student 1: M, 18 years, dg: moderate MR, IQ: 41 Student 2: F, 17 years, dg: moderate MR, IQ: 43 Student 3: M, 18 years, dg: moderate MR, IQ: 42 Student 4: M, 16 years, dg: moderate MR, IQ = 41 Student 5: M, 18 years, dg: moderate MR, IQ = 40 \| A self-operated single-word auditory prompting system and a self-operated multiple-word (3 words) prompting system. Both systems were delivered via a tape recorder and headphones using separate, prerecorded cassette tapes with cues specific to each worker’s chain of vocational tasks in two vocational settings (church, pet store) \| Number of independent task transitions across sessions in two vocational settings Performance duration \| Performance A statistically significant difference was found between baseline and both auditory prompting systems in both vocational settings. The number of independent task changes ranged at baseline from 0 to 3, during intervention from 2 to 6 with the single-word and from 3 to 6 with the multiple-word prompting system. Performance duration differences between prompting systems were significant for Student 3 (T = 1.00, p < .05, N = 7) and Student 5 (T = 0.00, p < .05, n = 7). They transitioned through tasks in significantly less time using the multiple-word auditory prompts in the pet store \| \| Van Laarhoven et al. , USA \| Quantitative case study. Observation \| N = 2 Gerald: M, 17 years, dg: Autism and moderate ID, IQ: 44 Nick: M, 18 years, dg: Autism and moderate ID, IQ: nr \| The study was conducted within the faculty conference room. The participants were responsible for cleaning and preparing the room for different meetings: configure tables and chairs, clean the white board, and throw away garbage or place important items in the lost and found. The work comprised 3 decision points. Universally-designed prompting systems presented on iPads and HP Slates were compared to improve the independent vocational performance of participants \| Percentage of correct responses, percentage of media options selected by participants, the percentage of decision points correctly selected by participants across baseline and intervention phases \| Performance Both participants increased their vocational skills using mobile devices. There was not a large difference in correct responding when devices were compared, but participants performed slightly better when using their preferred device. The participants selected video prompts more often than other media prompts (i.e. picture prompts) during initial sessions, except Nick during the first session. The participants effectively self-selected and self-faded their reliance on media-prompts as they became more independent with the task \| \| West and Patton , USA \| Quantitative case study \| N = 4 Adam: M, 41 years, dg: severe ID Gena: F, 38 years, dg: ID and Rhett syndrome Alex: M, 35 years, dg: moderate ID Kylie: F, 34 years, dg: severe ID \| Participants attended a community-based habilitation agency for six hours a day, Monday to Friday. The setting contained several activities including vocational tasks (i.e., sorting and assembly) and other activities such as listening to music, reading books, watching television, and kitchen and laundry tasks \| The number of independent correct responses for each client during each job training session \| No independent correct responses were observed for any participant during baseline sessions across task and procedures. Gena performed all five steps independently and correctly in Session 15, Kylie in session 14, Adam in Session 15, and Alex in Session 15 \| Open in a new tab N number, M males, F females, Dg diagnosis, IQ Intellectual Quantity, nr not reported, ID Intellectual Disability, IDD Intellectual and Developmental Disability, DD Developmental Disability, LD Learning Disability, MR Mental Retardation, SE Supported Employment Table 4. Characteristics of included qualitative case studies \| Authors, year, country \| Methods, study design \| Population, definitions, terms \| Intervention \| Findings, barriers and facilitators \| :--- :--- \| Aspinall , UK \| Qualitative case study. Interview, questionnaire \| N = 10 (5 M/5F) Dg: Learning disabilities Age: nr IQ: nr \| The TATE project of assistive technology and telecare to improve quality of life, especially employment \| Assistive technology supported the independence of people with learning disabilities \| \| Grossi et al. , USA \| Case study. Observation \| N = 2 Dana: F, 28 years, dg: mild ID, IQ: 64 Rick, M, 28 years, dg: borderline, IQ: 70 \| The workplace was a restaurant. The aim was to enhance social interaction. The social interaction situations were tape-recorded prior to and during the work shifts. The recordings were analyzed together with the participants, and during these analysis sessions the participants received guidance in social interaction skills \| Prior to the intervention Dana did not pay attention or reacted negatively to her supervisor and workmates; she also had difficulties accepting critique Rick was very sensitive to critique. He tried to avoid other people and also behaved obsessively and was paranoid After the intervention Both Dana and Rick were more responsive to the initiatives of the others and in a socially acceptable manner \| \| Hagner and Davies , USA \| Case-study, interviews of entrepreneurs and support people \| N = 7 Rhonda: F, age: nr, dg: ID Pat: F, age: nr, dg: ID, physical disability. Jenny: F, age: nr, dg: ID, physical disability Paul: M, age: nr, dg: ID Maxwell: M, age: nr, dg: ID Peggy: F, age: nr, dg: ID Richard: M, age: nr, dg: ID, physical disability \| People with ID as entrepreneurs. Entrepreneurship was built according to each person’s own interests and values. Rhonda, Pat, Jenny and Maxwell received business training. The established firms had no personnel other than the entrepreneur. Each entrepreneur had their own support person, who took responsibility for the firm’s practical issues \| Factors enhancing employment: Only a small monetary investment was needed to start the business. Most of these people only worked part-time. Some of them had the support person with them all the time, others only occasionally. Work as an entrepreneur was flexible, independent, and they could do work that was interesting Barriers to employment: The income from the business was low, so other sources of income were needed. The support people did not have enough expertise or interest in business. It was difficult to obtain support for running a business, and to make social contacts \| \| Ham et al. , USA \| Case study \| N = 1 Kristen: F, age: nr (young adult), dg: Down syndrome, autism, and hearing disability \| Kristen worked in the hospital delivery ward. Her jobs were cleaning the babies’ nursing room, cleaning computers, phones and kitchens, and taking care of blankets and clothes in the nursery. The aim of the positive behavior program was to decrease her disruptive behavior and to help her better complete her duties. The program included self-assessment, a timetable in pictorial form and auditive reminders. The program was planned by a positive behavior support person and her workmates took part in its implementation \| Prior to the intervention: Kristen behaved disruptively at the workplace, and was unable to carry out her duties. She was in danger of being dismissed After the intervention: Kristen was more self-regulative and independent. In three months her refusals to comply with the timetables decreased by 79%, refusals to follow instructions decreased by 59% and refusals to handle the alarms decreased by 67%. Her workmates felt it was more comfortable to work with her. Kristen continued to work in the hospital for two more years. After that she moved to another city and obtained another job \| \| Jarhag et al. \| Case study \| N = 2 Anna: F, 30 years, dg: ID (50% disability allowance), IQ: nr, had attended to a school for children with ID Adam: M, 21 years, dg: ID, IQ: nr, had attended to a school for children with ID \| Model for special support and follow-up. The support person helped the employee plan work tasks and gave individual support to the person with ID according to the plan. The support person was present at the workplace and supervised the tasks, and also helped the employee modify working conditions. The amount of support given by the support person gradually diminished, and when the employee succeeded in their tasks independently, the support ended. The maximum duration of support was six months \| Prior to intervention: Anna worked as a wage-subsidized cleaner for two years. When she moved to another city the wage-subsidy ended. A year later, after her maternity leave, Anna enrolled as a job-seeker, and temporarily attended a sheltered workshop until a job was found for her. However, she did not find employment. The employees felt that Anna could not be employed because of her shyness, lack of initiative and low stress tolerance Adam was registered as a job-seeker in the employment office After the intervention Anna was employed with a wage subsidy in the hotel and restaurant industry. Her job tasks were tailored according to her skills. Later Anna returned to being a customer of the employment office. The support person helped her and again she obtained a wage-subsided job Adam: The support person helped him while he was in his last year of school for people with ID. After school, Adam worked as a trainee, but this did not lead to employment. The support person arranged another job for him, but this did not last very long. Following discussions with the support person, Adam again obtained a place as a trainee, which later turned into a wage-subsided post The factors enabling employment were sufficient duration of support, individually tailored tasks and wage subsidies \| \| Wehman et al. , USA \| Qualitative case study, observations \| N = 2. Karen, F, 18 years, dg: severe ID, autism, communication deficits, IQ: nr, at school Lisa, F, 22 years, dg: moderate/severe ID, IQ: nr \| Employment with support of family, job coach and workmates. The employees’ performance was evaluated at home and in job tasks, and on the grounds of these evaluations, suitable job tasks and workplaces were defined. The intervention included choosing the workplace and the employee, getting to know these; and guidance, training and support at the workplace \| Both Karen and Lisa obtained employment. Karen worked at a restaurant, where her tasks where first spreading flour and later also handling dirty dishes. Lisa’s first workplace was at a café at school, and after that a grocer’s, where she had simple tasks such as towel folding Factors promoting employment and work performance: support from family and job coach, support and opportunities to develop at work \| Open in a new tab N number, M males, F females, Dg diagnosis, IQ Intellectual Quantity, nr not reported, ID Intellectual Disability, IDD Intellectual and Developmental Disability, DD Developmental Disability, LD Learning Disability, MR Mental Retardation, SE Supported Employment Participants The total number of participants in the 38 studies was 2,41,080. The quantitative studies included 35–1,21,335 participants (n = 2,39,506), the qualitative studies 3–27 participants (n = 58), the mixed method study included 1452 participants and the case studies 1–10 participants (n = 64). The age of the participants varied between 16 and 51. However, two qualitative studies [53, 54] and two case studies [62, 64] did not report the age of the participants. The proportion of men and women differed in the studies. In the quantitative studies, the proportion of women varied from 29% to 58% . In one qualitative study all the participants were men and in another qualitative study all the participants were women. Two studies [51, 59] did not report the gender of the participants. Seven studies reported the ethnicity of the participants [45, 47–49, 51, 52, 79]. Sixteen of the 38 studies reported the degree (mild, moderate, severe, profound) of ID as a background factor of the participants. People with mild ID participated in two cohort studies [48, 51], one qualitative study , and one case study . People with either mild or moderate ID participated in three case studies [71, 75, 76], and people with moderate ID participated in two case studies [68, 80]. Renzaglia et al. , Simmons and Flexer , McGlashing-Johnson et al. , and West and Patton focused on people with moderate or severe disabilities. People with severe ID were participants in three case studies [61, 67, 69]. Two studies [45, 77] reported the participants ‘level of functional capacity. Seven of 38 studies concerned students [46, 48–52, 59], one study concerned people who worked in sheltered workshops, and one study concerned job-seekers. Four studies reported the occupation or work tasks of the participants as the background factors [69, 72, 75, 79]. Interventions All studies contained various mixtures of intervention components. The interventions were carried out during secondary education, the transition from education to work, job-seeking and sheltered work. In general, the content of the intervention was briefly described. However, the stages of the intervention process were poorly reported, and the term rehabilitation was only seldom used. During secondary education (Upper Secondary School for Pupils with ID) (ICF code d825), the students participated in different educational programs such as national programs, specially designed programs, vocational training and training activities or graduated with inadequate grades . National programs focused on different parts of the labor market, for example, vehicles and transportation, hotels and restaurants, or social and healthcare. Some specially designed programs or individually tailored education were also on offer. The educational programs aimed to improve vocational qualifications and work awareness. The vocational qualification courses included assisting smaller pupils during classes, helping the janitor and working in the school kitchen. Work awareness training courses included watching videos about work, talking about presenting oneself at work, and health and safety instructions at school . Some interventions included postschool and transition services (ICF code e5900) during education to improve the transition from education to work [43, 48, 49, 51, 55, 75]. Postschool services included postsecondary education institution accommodations and services, job training services, and life skills services . Employment-related transition services were, for example, vocational assessment, career counseling, pre-vocational education, career-related technical or vocational education, pre-vocational or job readiness training, instructions for job-seeking, job shadowing, job coaching, special job skills training, placement support, internship or apprenticeship programs, work experience at school and other paid work experience [44, 48, 55, 65]. Job tasting was a short, unpaid, time-limited work experience period at the workplace which allowed people to sample a variety of different jobs and work cultures . One qualitative study analyzed the different actors’ experiences (schools, agencies, parents, other people) of the transition process and of employment (e310), and Fasching’s study analyzed the experiences of people with ID of the transition from school to vocational training and employment. The interventions also included supported work (supported employment, SE) (d855) , the use of a job coach or other support person (e340, e360) [45, 63], and designing personal solutions (e5900) . One cohort study analyzed whether attending a sheltered workshop (e5900) improved the employment outcomes of supported employees with ID. Some interventions developed the independence of people with ID, applying the ICF model to the areas of body structures/functions, activities and participation. The interventions included support in psychological and behavioral issues (behavior) (b122) [56, 64] and communication abilities (d3) ; and a self-regulated problem-solving process in which students set their goals, developed and implemented their action plans (b164, d175) , made career-related decisions (d177) , were able to perform daily activities (d620, d630) , learned navigation skills such as finding the workplace and using public transportation (d4602) , and gradually reduced help in carrying out work tasks (d850) [72, 75, 77]. The use of digital solutions (e135) was intended to improve the daily work performance of people with ID in the open labor market. They could use these tools for receiving digital instructions for work tasks, work processes, work techniques or schedules. These solutions consisted of video modeling and audio coaching [72, 76], photo activity schedule books , smart phones , palmtop-based job aids , and self-operated auditory prompting systems [68, 80]. In the case studies, the interventions were carried out at workplaces such as restaurants [60, 66], hospitals , factories and warehouses , at markets , at schools [70, 72], and in conference rooms . Outcomes The outcomes were employment in the open labor market [43, 45, 47, 51, 52], transition from school to the open labor market [44, 46–48, 54, 55], and work performance [43, 44]. The outcomes in all 16 quantitative experiments were job skills and work performance [65–80]. Study quality The RCT study was considered to be of high methodological quality, with scores of 11 out of 22, and the CCT study of Kilsby and Beyer was of low methodological quality, with scores of 9 out of 22 according to van Tulder et al. . The methodological quality of four cohort studies [47, 49, 50, 52] out of six were high, with the maximum of nine ‘stars’ according to Wells et al. . Four cohort studies [45, 46, 48, 51] and one multimethod study were considered to be of intermediate quality. All six qualitative studies [53–58] were considered to be of high quality, with scores ranging from 12 to 20 out of 20, according to the modified CASP method. (Table 6–9, Supplementary Files). Effectiveness The quantitative studies showed that supported work increases the employment of people with ID in the open labor market (Table5). This result was based on one high-quality RCT study , one high-quality cohort study , and one moderate-quality cohort study of altogether 16,947 people with ID. The quantitative studies also showed that both secondary and postsecondary education, including support services and work training, increased the transition of people with ID from school to the open labor market. The result was based on two high-quality cohort studies [50, 52] and three moderate-quality cohort studies [46, 48, 51] with 2,07,484 participants altogether. Table 5. Summary of results \| Outcome \| Percentage and/or comparative risk (95% Confidence Interval) \| Number of subjects \| Evidence (GRADE) \| Study \| :--- :--- \| \| Intervention \| Control group \| \| \| \| \| Employment in open labor market \| Supported work 21% \| Sheltered employment 0% \| 49 \| Moderate \| High quality RCT study \| \| Employment to supported work after sheltered work 59.6% \| Employment to supported work without sheltered work 60.4% \| 9,808 \| Low \| High quality cohort study \| \| Employment after postsecondary education (OR 2.0) \| Employment without postsecondary education \| 11,280 \| Low \| High quality cohort study \| \| Support of work coach 6.1% \| No control group \| 7,090 \| Very low \| Moderate quality cohort study \| \| Transfer from school to open labor market \| Support services and work training during secondary education 22.4% \| No control group \| 12,269 \| Low \| High quality cohort study \| \| Support activities during education 75.9% One transition activity during education (OR 1.2) Work period during education (OR 3.5) Work experience during education (OR 5.7) \| No control group \| 62,513 \| Very low \| Moderate quality cohort study \| \| Work experience and training of work awareness 21%, (OR 1.01) \| No control group \| 87 \| Very low \| Moderate quality cohort study \| \| Work briefing and work training during secondary education 45%/2008, 11%/2009 \| No work training or work experience during education 6%/2008, 5%/2009 \| 1452 \| Very low \| Moderate quality multimethod study \| \| Personal educational program and transition services at age of 14 years 74.3% \| Personal educational program and transition services at age of 16 years 57.8% \| 15,040 \| Low \| High quality cohort study \| \| Work performance \| Supported work Before intervention: 54 (SD 6) After intervention: 51 (SD 8) \| Sheltered work Before intervention: 53 (SD 8) After intervention: 54 (SD 7) \| 49 \| Moderate \| High quality RCT study \| \| Short work tastings, self-evaluation, education of job coaches. Need for job coaches decreased: at beginning 4.79 > 4.04 or 2.80. \| No control group \| 35 \| Very low \| Moderate intervention study \| Open in a new tab However, on the basis of one high-quality RCT study and one high-quality cohort study covering a total of 15,089 participants, sheltered work did not increase the employment of people with ID in the open labor market (Table5). Barriers to and facilitators of employment The qualitative studies concerned both the barriers to and the facilitators of employment in the open labor market (Table2). The main barriers to employment were that the school and service system tried to guide people with ID towards traditional services such as sheltered work, sometimes against their own needs and interests . However, these people’s work skills were not developed or their needs for support were not noticed in sheltered work . Further barriers were discrimination in vocational experience after leaving school , poor experiences of class teaching and lack of work experience [58, 59]. The main facilitators of employment were people’s own activity and support from their families (e310) , effective job coaching (e360) , a well-designed work environment (e2) , appreciation of their work and support from their employer and work organization . Other facilitators were the knowledge and experience of work during education , and for entrepreneurs, the use of a support person . Work performance The case studies showed that the use of digital solutions (e135) improved the work performance of people with ID in the open labor market. Work performance was measured as the percentage and rate of the tasks completed correctly [69, 72], the number of movements measured by a costume , the number of independent task changes , task steps , and percentage of intervals with task engagement and interactions . Discussion People with ID still face specific difficulties in gaining employment, which leads to inequality and social exclusion. Personally-tailored support services, occupational education, work experience and digital solutions can be used to help these individuals become employed in the open labor market. This systematic review covered 38 studies (one RCT, one CCT, eight cohort studies, six qualitative studies, one multimethod study, and 21 case studies) that investigated the effectiveness of, barriers to or facilitators of employment among people with ID. The main outcome was employment in the open labor market through either supported or non-supported work, transition from education or from sheltered work to the open labor market, or functioning at work. The quantitative studies showed that secondary and postsecondary education increases the employment of people with ID in the open labor market when it includes work experience and personal support services. Supported employment also increases employment in the open labor market, but sheltered work does not. The results were in line with those of earlier studies [2, 22, 81] which have shown the support of a job coach to be important for people with ID to find work, start work and continue at work among. The qualitative studies showed the barriers to and facilitators of employment in the open labor market among people with ID. The barriers to employment were that the school and service system tried to guide people with ID towards traditional services such as sheltered work; that these people were discriminated against at school and had negative experiences of class teaching; and that they did not obtain work experience during education. These results support the findings of earlier studies [6, 22] which have shown low rates of employment when individuals come from sheltered work to the open labor market. The facilitators of employment were the people’s own activity and support from their families, job coaches, work environments with necessary accommodations, appreciation of their work, support from an employer and work organization, knowledge and experience of work during education; and for entrepreneurs, the use of a support person. These findings support earlier results that have shown the importance of job coaches, employers’ responsibility and personally designed support services [17, 18, 20]. The case studies showed that the work performance of people with ID could be improved through the use of digital solutions in daily work. However, digital solutions are only rarely used among people with ID. One reason may be the employees’ low competence in using digital systems, especially in sheltered work. Services in general are becoming increasingly digital, and this also requires more competence from people with ID . Methodological discussion of included studies Better reporting of basic methodological quality issues is generally required. A better description of participants, such as their gender, age, education, occupation and work experience is also needed when they are people with ID. We can conclude that people with ID are not seen as professionals because, although the studies described their diagnosis or disability well, they did not emphasize their competence and strengths, educational background or work experience. This is also true of studies of disability groups . The studies included several different concepts of disability, rehabilitation, education, performance, and service systems. The concept of ID was defined as ‘developmental disability’, ‘mental retardation’, and ‘mental disability’. In the UK, ‘learning disability’ is used in the same way as ID, but in other countries, ‘learning disability’ means difficulties in learning without ID. The articles failed to clearly report some elements of the interventions. They should have reported more information about the process and implementation schedule of rehabilitation, the initiator and the place. These shortfalls were found especially in cohort studies that focused on results and background factors such as the participants’ diagnoses. Only few randomized controlled interventions have been carried out to enhance the employment among people with ID. This is possibly because of the low employment rate of people with ID, the low number of implemented rehabilitation programs, the ethical aspects of the study designs, negative attitudes, and a lack of financing instruments for such studies. The outcomes in this review were employment in the open labor market in either supported or non-supported work, the transition from education or from sheltered work to the open labor market, or functioning at work. With the ICF model as a framework, these outcomes belong to ‘participation’. According to the ICF model, rehabilitation belongs to the ‘environmental factors’ that affect ‘activity’ (e.g., functioning) as do most of the synthesized themes from the analysis of qualitative studies. Obviously, the primary aim of rehabilitation on a personal level is to enhance the functioning and work ability of people with ID and to enable them to work in the open labor market. On the societal level, it is important to develop and implement solutions that enhance employment and participation and are simultaneously cost effective . Strengths and limitations of this review The strengths of this review included its multi-scientific research group, the comprehensiveness of the searches, the use of the ICF model as the theoretical framework, and the inclusion of a wide range of studies. The reviewers were experts in different scientific areas, including both quantitative and qualitative methodology and systematic reviews. Every effort was made to insure a comprehensive search. It is possible, however, that we did not find all the relevant studies. Another limitation was that the included studies used different concepts of ID. We included quantitative, qualitative, multimethod and case studies, which showed different kinds of knowledge regarding the process and the effectiveness of rehabilitation. However, this was also a shortcoming of this study because a meta-analyses of several types of results was complicated. Quality assessment This systematic review was of 38 studies (one RCT, one CCT, eight cohort studies, six qualitative studies, one multimethod study, and 21 case studies). The quality of the RCT study was assessed using the validated method of van Tulder et al. , which has also been used in several other reviews. The quality of the CCT study was low, mainly due to the fact that the study design did not include randomization, treatment allocation, blinding, or intention to treat the analysis. The quality of the cohort studies and the mixed method study were assessed using the Newcastle–Ottawa Scale . The validity and reliability of this method was only partly evaluated in that the content validity and inter-rater reliability were established, but the criterion validity and intra-rater reliability were still in progress . The original qualitative assessment tool CASP was not perceived as very powerful in differentiating between high- and low-quality studies. It only measured whether certain basic items that are essential identifiers of high-quality research were mentioned in the report. This type of measurement is crude and makes the scale difficult to use when some of the criteria are implicit in the study. Furthermore, a ‘Yes or No’ scale does not capture the fact that certain items in the CASP criteria may be more crucial to the quality of the study than others. Adding a third level of assessment, ‘Partially’, to the method, may solve the first problem. However, the second problem remains: Of the three problematic points of the qualitative studies evaluated, researcher effect is a self-evident fact connected to any study of social life, and is thus less informative than reporting the contribution of a particular study to existing knowledge. Although CASP offers a good basis for evaluating qualitative research reports, it can be further developed by giving different weights to different criteria. The comparison of the results of the assessment of the quantitative and qualitative studies revealed a bias in that the quality assessment of qualitative studies resulted in several high-quality studies, whereas the quantitative assessment yielded only a few. This was, of course, partly due to different evaluation methods, but it may also be an indication of the different nature of these two types of research. Qualitative studies report interesting new observations about the ways in which participants observe, understand, or experience the phenomenon studied, while quantitative studies aim to make generalizations about possible causes and effects, and reveal other connections between the variables describing the phenomenon being studied. As the knowledge gained through qualitative research is descriptive and not numerical by nature, ranking studies is also difficult. Conclusions More people with ID could be employed through personally tailored services and measures. Tailoring can mean secondary or postsecondary education, including proper teaching methods and personal support services, the use of supported work, workplace accommodations, and the support of one’s family and employer. Our results can be utilized in the development of rehabilitation, education and employment of people with ID, to provide them with opportunities to work in the open labor market and to participate in society. Electronic supplementary material Below is the link to the electronic supplementary material. Supplementary material 1 (DOCX 88 kb) (88.1KB, docx) Supplementary material 2 (DOCX 26 kb) (26.9KB, docx) Acknowledgements The authors wish to thank Researcher Simo Klem for his help in reviewing the titles, and Alice Lehtinen for editing the language of the manuscript. Financial support was provided by The Social Insurance Institution of Finland. Authors’ contributions NN: study plan and design, conceptual development of the variables, data collection, review of the titles and abstracts, quality assessment of the quantitative and case studies, analysis and interpretation of the data, draft and critical revision of the manuscript. IP: study plan and design, conceptual development of the variables, data collection, review of the titles and abstracts, quality assessment of the quantitative and case studies, analysis and interpretation of the data, draft and critical revision of the manuscript. AT: conceptual development of the variables, data collection, review of the titles and abstracts, quality assessment of the qualitative studies, analysis and interpretation of the data, draft and critical revision of the manuscript. HTV: conceptual development of the variables, data collection, review of the titles and abstracts, quality assessment of the qualitative studies, analysis and interpretation of the data, draft and critical revision of the manuscript. PP: performance of the searches in 12 databases and two search engines, definition of the search terms with the research group, management of the articles in Refworks, management of full details of the search strategy, and critical revision of the manuscript. 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16684	https://eli.thegreenplace.net/2018/affine-transformations/	Affine transformations This is a brief article on affine mappings and their relation to linear mappings, with some applications. Linear vs. Affine To start discussing affine mappings, we have to first address a common confusion around what it means for a function to be linear. According to Wikipedia the term linear function can refer to two distinct concepts, based on the context: In Calculus, a linear function is a polynomial function of degree zero or one; in other words, a function of the form for some constants a and b. In Linear Algebra, a linear function is a linear mapping, or linear transformation. In this article we're going to be using (2) as the definition of linear, and it will soon become obvious why (1) is confusing when talking about transformations. To avoid some of the jumble going forward, I'm goine to be using the term mapping instead of function, but in linear algebra the two are interchangeable (transformation is another synonym, which I'm going to be making less effort to avoid since it's not as overloaded ). Linear transformations Since we're talking about linear algebra, let's use the domain of vector spaces for the definitions. A transformation (or mapping) f is linear when for any two vectors and (assuming the vectors are in the same vector space, say ): for some scalar k For example, the mapping - where and are the components of - is linear. The mapping is not linear. In fact, it can be shown that for the kind of vector spaces we're mostly interested in , any linear mapping can be represented by a matrix that is multiplied by the input vector. This is because we can represent any vector in terms of the standard basis vectors: . Then, since f is linear: If we think of as column vectors, this is precisely the multiplication of a matrix by : This multiplication by a matrix can also be seen as a change of basis for from the standard base to a base defined by f. If you want a refresher on how changes of basis work, take a look at my older post on this topic. Let's get back to our earlier example of the mapping . We can represent this mapping with the following matrix: Meaning that: Representing linear mappings this way gives us a number of interesting tools for working with them. For example, the associativity of matrix multiplication means that we can represent compositions of mappings by simply multiplying the mapping matrices together. Consider the following mapping: In equational form: . This mapping stretches the input vector 2x in both dimensions. To visualize a mapping, it's useful to examine its effects on some standard vectors. Let's use the vectors (0,0), (0,1), (1,0), (1,1) (the "unit square"). In they represent four points that can be connected together as follows : It's easy to see that when transformed with , we'll get: It's also well known that rotation (relative to the origin) can be modeled with the following mapping with in radians: Transforming our unit square with this matrix we get: Finally, let's say we want to combine these transformations. To stretch and then rotate a vector, we would do: . Since matrix multiplication is associative, this can also be rewritten as: . In other words, we can find a matrix which represents the combined transformation, and we "find" it by simply multiplying R and S together : And when we multiply our unit by this matrix we get: Affine transformations Now that we have some good context on linear transformations, it's time to get to the main topic of this post - affine transformations. For an affine space (we'll talk about what this is exactly in a later section), every affine transformation is of the form where is a matrix representing a linear transformation and is a vector. In other words, an affine transformation combines a linear transformation with a translation. Quite obviously, every linear transformation is affine (just set to the zero vector). However, not every affine transformation is linear. For a non-zero , the linearity rules don't check out. Let's say that: Then if we try to add these together, we get: Whereas: The violation of the scalar multiplication rule can be checked similarly. Let's examine the affine transformation that stretches a vector by a factor of two (similarly to the S transformation we've discussed before) and translates it by 0.5 for both dimensions: Here is this transformation visualized: With some clever augmentation, we can represent affine transformations as a multiplication by a single matrix, if we add another dimension to the vectors : The translation vector is tacked on the right-hand side of the transform matrix, with a 1 for the extra dimension (the matrix gets 0s in that dimension). The result will always have a 1 in the final dimension, which we can ignore. Affine transforms can be composed similarly to linear transforms, using matrix multiplication. This also makes them associative. As an example, let's compose the scaling+translation transform discussed most recently with the rotation transform mentioned earlier. This is the augmented matrix for the rotation: The composed transform will be . Its matrix is: The visualization is: Affine subspaces The previous section defined affine transformation w.r.t. the concept of affine space, and now it's time to pay the rigor debt. According to Wikipedia, an affine space: ... is a geometric structure that generalizes the properties of Euclidean spaces in such a way that these are independent of the concepts of distance and measure of angles, keeping only the properties related to parallelism and ratio of lengths for parallel line segments. Since we've been using vectors and vector spaces so far in the article, let's see the relation between vector spaces and affine spaces. The best explanation I found online is the following. Consider the vector space , with two lines: The blue line can be seen as a vector subspace (also known as linear subspace) of . On the other hand, the green line is not a vector subspace because it doesn't contain the zero vector. The green line is an affine subspace. This leads us to a definition: A subset of a vector space is an affine space if there exists a such that is a vector subspace of . If you recall the definition of affine transformations from earlier on, this should seem familiar - linear and affine subspaces are related by using a translation vector. It can also be said that an affine space is a generalization of a linear space, in that it doesn't require a specific origin point. From Wikipedia, again: Any vector space may be considered as an affine space, and this amounts to forgetting the special role played by the zero vector. In this case, the elements of the vector space may be viewed either as points of the affine space or as displacement vectors or translations. When considered as a point, the zero vector is called the origin. Adding a fixed vector to the elements of a linear subspace of a vector space produces an affine subspace. One commonly says that this affine subspace has been obtained by translating (away from the origin) the linear subspace by the translation vector. When mathematicians define new algebraic structures, they don't do it just for fun (well, sometimes they do) but because such structures have some properties which can lead to useful generalizations. Affine spaces and transformations also have interesting properties, which make them useful. For example, an affine transformation always maps a line to a line (and not to, say, a parabola). Any two triangles can be converted one to the other using an affine transform, and so on. This leads to interesting applications in computational geometry and 3D graphics. Affine functions in linear regression and neural networks Here I want to touch upon the linear vs. affine confusion again, in the context of machine learning. Recall that Linear Regression attempts to fit a line onto data in an optimal way, the line being defined as the function: But as this article explained, is not actually a linear function; it's an affine function (because of the constant factor ). Should linear regression be renamed to affine regression? It's probably too late for that :-), but it's good to get the terminology right. Similarly, a single fully connected layer in a neural network is often expressed mathematically as: Where is the input vector, is the weight matrix and is the bias vector. This function is also usually referred to as linear although it's actually affine. Affine expressions and array accesses Pivoting from algebra to programming, affine functions have a use when discussing one of the most fundamental building blocks of computer science: accessing arrays. Let's start by defining an affine expression: An expression is affine w.r.t. variables if it can be expressed as where are constants. Affine expressions are interesting because they are often used to index arrays in loops. Consider the following loop in C that copies all elements in an MxN matrix "one to the left": ``` for (int i = 0; i < M; ++i) { for (int j = 1; j < N; ++j) { arr[i][j-1] = arr[i][j]; } } ``` Since C's memory layout for multi-dimensional arrays is row-major, the statement in the loop assigns a value to arr[iN + j - 1] at every iteration. iN + j - 1 is an affine expression w.r.t. variables i and j . When all expressions in a loop are affine, the loop is amenable to some advanced analyses and optimizations, but this is a topic for another post. \| \| Though it's also not entirely precise. Generally speaking, transformations are more limited than functions. A transformation is defined on a set as a binjection of the set to itself, whereas functions are more general (they can map between different sets, for example). \| \| \| Finite-dimensional vector spaces with a defined basis. \| \| \| Tossing a bit of rigor aside, we can imagine points and vectors to be isomophic since both are represented by pairs of numbers on the plane. Some resources will mention the Euclidean plane - when talking about points and lines, but the Euclidean plane can be modeled by a same-dimensional real plane so I'll just be using . \| \| \| I'll admit this result looks fairly obvious. But longer chains of transforms work in exactly the same way, and the fact that we can represent such chains with a single matrix is very useful. \| \| \| This trick has a geometrical explanation: translation in 2D can be modeled as adding a dimension and performing a 3D shear operation, then projecting the resulting object onto a 2D plane again. The object will appear shifted. \| \| \| It's actually only affine if N is a compile-time constant or can be proven to be constant throughout the loop. \| For comments, please send me an email.
16685	https://www.wallstreetmojo.com/histogram-formula/	📊 Master Excel & Finance Skills Join 100K+ Learners ⚡upto80% OFF! Instructor: Dheeraj Vaidya, CFA, FRM Table Of Contents What Is Histogram? A Histogram is the graphical or visual representation of the distribution of categorical numeric data in a dataset. We can create them using the Analysis ToolPak or the Pivot Table, and by using the formulas. We can create the Histogram Formula using the FREQUENCY and COUNTIFS formulas. For example, we can represent the student data, such as scores, height, weight, etc., graphically, as shown below. Key Takeaways Histogram Formula Publication Date : 13 Jul, 2023 Blog Author : WallStreetMojo Team Edited by : Sheeba M Reviewed by : Dheeraj Vaidya, CFA, FRM Share Share Video Explanations of Histogram Explanation of the formula of the Histogram We can derive the formula for the calculation of the Area of the Histogram by using the following seven steps: Firstly, it is to be decided on how we measure the process and what we should collect as data. Once determined, the data is gathered and presented in a tabular form, such as a spreadsheet. Now, count the number of data points gathered. Next, determine the range of the sample, which is the difference between the maximum and minimum values in the data sample. Range = Maximum value – Minimum value Next, determine the number of class intervals that can be based on either of the following two methods, - As a thumb rule, use 10 as the number of intervals, or - The number of intervals can be calculated by the square root of the number of data points, which is then rounded to the nearest whole number. Number of intervals = Now, determine the width of the interval class by dividing the range of the data sample by the number of intervals. Class width = Range / Number of intervals Next, develop a table or spreadsheet with frequencies for each interval. Then, derive the frequency density for each interval by dividing the frequency by the corresponding class width. Finally, the Area for the Histogram equation is calculated by adding the product of all the frequency density and their corresponding class width. How To Read the Histogram example and its interpretation? Now that we understand the basics and different types of this concept, it is vital to understand how to read and interpret the data after it is converted into a Histogram. Let us do so through the step-by-step guide mentioned below. Examples Let us understand the concept of Histogram Formula Excel with the help of the following examples. Let us consider the table below, which shows children’s weights in a class. From the above table, we can calculate the following: Again, For the Histogram Formula calculation, we will first need to calculate class width and frequency density, as shown above. Hence, Area of the Histogram = 0.4 5 + 0.7 10 + 4.2 5 + 3.0 5 + 0.2 10 So, the Area of the Histogram will be – A graphical representation of the weight of children is shown below: Relevance And Uses Of Histogram Let us understand the relevance and uses of a frequency density Histogram Formula and other related factors through the discussion below. Important Things To Note Frequently Asked Questions (FAQs) What are the different types of Histograms? Give some information on the Formula of the Histogram. How to enable Analysis ToolPak to use Histogram? Create a Full Dynamic Financial Model in 2 Days (6 hours) \| Any Graduate Or Professional is eligible \| Build & Forecast IS, BS, CF from Scratch. The Exact Training Used by Top Investment Banks \| FM, DCF, LBO, M&A, Accounting, Derivatives & More \| $2400+ in Exclusive Benefits \| 100+ Wall Street-Level Skills. Boost Productivity 10X with AI-Powered Excel \| Save Hours, Eliminate Errors \| $300+ in Exclusive Bonuses \| Advanced Data Analysis & Reporting with AI. Master Excel, VBA, PowerBI Like a Pro \| 70+ Hours of Expert Training \| Real-world Excel applications \| Earn Your Certification & Land High-Paying Roles! Company Our Policies Resources Partner with us Support Connect With Us Top Categories Blog Categories Free Courses Disclaimer: Copyright © 2024. CFA Institute Does Not Endorse, Promote Or Warrant The Accuracy Or Quality Of WallStreetMojo. CFA® And Chartered Financial Analyst® Are Registered Trademarks Owned By CFA Institute.
16686	https://www.quora.com/What-are-the-formulas-for-arccos-a-b-arcsin-a-b-and-arctan-a-b	Something went wrong. Wait a moment and try again. Applications of Inverse T... Arctangent (function) Formulas (functions) Mathematical Equations Inverse Functions (mathem... Range Trigonometric Inver... 5 What are the formulas for arccos (a + b), arcsin (a + b), and arctan (a + b)? Jan van Delden MSc Math and still interested · Author has 4.8K answers and 6.5M answer views · 6y I'd like to rephrase your question. There exists trigoniometric formulae to rewrite cos(α+β),sin(α+β),... Are there similar identities for expressions of the form arccos(a+b),arcsin(a+b),…? To answer your question it might be a good idea to investigate why we can rewrite cos(α+β). In order to do so I could stick to our ‘real’ world, but the argument is shorter, simpler, if I use the complex numbers as an intermediate step. I'll use the Euler identity: eiθ=cos(θ)+isin(θ)(1) With a little effort one could deduce: [math]\displaystyle \cos(\theta) = [/math] I'd like to rephrase your question. There exists trigoniometric formulae to rewrite cos(α+β),sin(α+β),... Are there similar identities for expressions of the form arccos(a+b),arcsin(a+b),…? To answer your question it might be a good idea to investigate why we can rewrite cos(α+β). In order to do so I could stick to our ‘real’ world, but the argument is shorter, simpler, if I use the complex numbers as an intermediate step. I'll use the Euler identity: eiθ=cos(θ)+isin(θ)(1) With a little effort one could deduce: cos(θ)=eiθ+e−iθ2(2) sin(θ)=eiθ−e−iθ2i Substitute θ=α+β into (2) and reorganise: cos(α+β)=eiα+e−iα2eiβ+e−iβ2+eiα−e−iα2eiβ−e−iβ2 which can be rewritten to: cos(α+β)=cos(α)cos(β)−sin(α)sin(β)(3) The derivation itself is not that important, that's why I skipped a few intermediate steps, important is that you realise that once you know α,β the rest of the steps follow. Let's try the other way around: arccos(a+b)=θ Thus cos(θ)=a+b We must split the left hand side into two parts, but we don't know how big each part must be! We could choose a, within reasonable bounds and define b by: b=cos(θ)−a Which tells us that multiple solutions exist. We need to have more information regarding a,b. Now you might argue that we could choose: θ=α−β and use (3) and pick: a=cos(α)cos(β)b=sin(α)sin(β) How to compute α,β? We do have two equations with two unknowns. Surely there is a solution. For particular a,b we found a system of equations to solve. But that doesn't mean we can express α,β into nice expressions using a,b in general. Expressions of type arccos(a)+arccos(b) do exist . This can be easily derived from (3). If cos(α)=a,cos(β)=b we have: sin(α)=√1−a2,sin(β)=√1−b2 where I assumed a positive sign for the root, and cos(α+β)=ab−√1−a2√1−b2 Thus arccos(a)+arccos(b)=arccos(ab−√(1−a2)(1−b2)) However you must be carefull regarding the sign in front of the root and account for the periodicity of the solutions. Footnotes List of trigonometric identities - Wikipedia Rushikesh Deshmukh In this space we are providing you answers of general question · 2y I'd like to rephrase your question. There exists trigoniometric formulae to rewrite cos(α+β),sin(α+β),... Are there similar identities for expressions of the form arccos(a+b),arcsin(a+b),…? To answer your question it might be a good idea to investigate why we can rewrite cos(α+β). In order to do so I could stick to our ‘real’ world, but the argument is shorter, simpler, if I use the complex numbers as an intermediate step. I'll use the Euler identity: (1)eiθ=cos(θ)+isin(θ) With a little effort one could deduce: (2)cos(θ)=eiθ+e−iθ2 and sin(θ)=eiθ−e−iθ2i Substitute θ=α+β into (2) and reorganise: I'd like to rephrase your question. There exists trigoniometric formulae to rewrite cos(α+β),sin(α+β),... Are there similar identities for expressions of the form arccos(a+b),arcsin(a+b),…? To answer your question it might be a good idea to investigate why we can rewrite cos(α+β). In order to do so I could stick to our ‘real’ world, but the argument is shorter, simpler, if I use the complex numbers as an intermediate step. I'll use the Euler identity: (1)eiθ=cos(θ)+isin(θ) With a little effort one could deduce: (2)cos(θ)=eiθ+e−iθ2 and sin(θ)=eiθ−e−iθ2i Substitute θ=α+β into (2) and reorganise: cos(α+β)=eiα+e−iα2eiβ+e−iβ2+eiα−e−iα2eiβ−e−iβ2 which can be rewritten to: (3)cos(α+β)=cos(α)cos(β)−sin(α)sin(β) The derivation itself is not that important, that's why I skipped a few intermediate steps, important is that you realise that once you know α,β the rest of the steps follow. Let's try the other way around: arccos(a+b)=θ Thus cos(θ)=a+b We must split the left hand side into two parts, but we don't know how big each part must be! We could choose a, within reasonable bounds and define b by: b=cos(θ)−a Which tells us that multiple solutions exist. We need to have more information regarding a,b. Now you might argue that we could choose: θ=α−β and use (3) and pick: a=cos(α)cos(β)b=sin(α)sin(β) How to compute α,β? We do have two equations with two unknowns. Surely there is a solution. For particular a,b we found a system of equations to solve. But that doesn't mean we can express α,β into nice expressions using a,b in general. Expressions of type arccos(a)+arccos(b) do exist What are the formulas for arccos (a + b), arcsin (a + b), and arctan (a + b)? . This can be easily derived from (3). If cos(α)=a,cos(β)=b we have: sin(α)=1−a2,sin(β)=1−b2 where I assumed a positive sign for the root, and cos(α+β)=ab−1−a21−b2 Thus arccos(a)+arccos(b)=arccos(ab−(1−a2)(1−b2)) However you must be carefull regarding the sign in front of the root and account for the periodicity of the solutions. Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.5M answer views · 1y Related What is the solution for arcsin(3/5) /arcran (24/7)? This is a very obscure question! Firstly, I just typed it into my calculator and I must say I was quite surprised! “Wow” I thought! Then I decided to draw some triangles! And I real... This is a very obscure question! Firstly, I just typed it into my calculator and I must say I was quite surprised! “Wow” I thought! Then I decided to draw some triangles! And I real... Enrico Gregorio Associate professor in Algebra · Author has 18.1K answers and 15.8M answer views · 1y Related Is arctan x = arcsin x arccos x ? No, that’s really wrong. Look at the graphs Red is the arctangent, blue is the fraction. Already the domains are very different, aren’t they? The domain of the arctangent is R, while the domain of the fraction is , where we need to exclude because It is true that but this doesn’t extend to the inverse functions (well, the arcsine, arccosine and arctangent aren’t really inverse of sine, cosine and tangent respectively, but “almost”). Let’s make a simple example: let and We can compute the inve No, that’s really wrong. Look at the graphs Red is the arctangent, blue is the fraction. Already the domains are very different, aren’t they? The domain of the arctangent is while the domain of the fraction is , where we need to exclude because It is true that but this doesn’t extend to the inverse functions (well, the arcsine, arccosine and arctangent aren’t really inverse of sine, cosine and tangent respectively, but “almost”). Let’s make a simple example: let and We can compute the inverse function of by solving for in that gives so and therefore whereas The quotient of the inverse functions is the negative of the actual inverse function. But this is a very special case and in general there’s no “easy” relation between the inverse function of a quotient and the quotient of the inverse functions. Assistant Bot · 1y The inverse trigonometric functions, such as arccos, arcsin, and arctan, do not have simple addition formulas like the regular trigonometric functions. However, we can express them in terms of the original trigonometric functions. Here are some useful identities and relationships involving these functions: Arccosine For , there isn't a direct formula. However, if you need to find for some , you can use the relationship: Arcsine Similarly, for , there isn't a direct addition formula. However, we can express it in terms The inverse trigonometric functions, such as arccos, arcsin, and arctan, do not have simple addition formulas like the regular trigonometric functions. However, we can express them in terms of the original trigonometric functions. Here are some useful identities and relationships involving these functions: Arccosine For , there isn't a direct formula. However, if you need to find for some , you can use the relationship: Arcsine Similarly, for , there isn't a direct addition formula. However, we can express it in terms of the cosine: Arctangent For , there is no straightforward formula, but there's a useful sum formula for tangent: From this, you can derive if you know and and their product is less than 1. Summary There are no direct formulas for , , or . Use the relationships between inverse functions and their corresponding trigonometric identities to work with these expressions. If you need a specific application or example, feel free to ask! Michael Jørgensen I love it! · Author has 1.1K answers and 3.2M answer views · 8y Related How do I calculate arcsine, arccosine and arctangent without a calculator? Before calculators became abundant (ca. 1975), people used to look up the result in huge books with tables of numbers. So on page 1 you’d have arcsine of numbers from 0.00 to 0.01 and on page 2 you’d have arcsine of numbers from 0.01 to 0.02, etc. The best of these books would give you an accuracy of 6 decimal places, which was quite sufficient for most needs. Using these books was quite easy and efficient, but begs the question of how they very made, which was a monumental task. Today, everything is done by computer hardware. The algorithms used vary a lot, because there are a lot of different m Before calculators became abundant (ca. 1975), people used to look up the result in huge books with tables of numbers. So on page 1 you’d have arcsine of numbers from 0.00 to 0.01 and on page 2 you’d have arcsine of numbers from 0.01 to 0.02, etc. The best of these books would give you an accuracy of 6 decimal places, which was quite sufficient for most needs. Using these books was quite easy and efficient, but begs the question of how they very made, which was a monumental task. Today, everything is done by computer hardware. The algorithms used vary a lot, because there are a lot of different methods available. The main parameter is the accuracy you require. If you only require two decimal places of accuracy, then a simple linear interpolation will do. If you require 6 or more decimal places, then other methods are needed. Another parameter is the amount of memory available. It’s quite possible, that the computer hardware internally contains a huge table of numbers, much like the old books I mentioned before. Then calculating the arcsine of a number amounts to a single quick table lookup. Usually, we want a compromise, i.e. a high degree of accuracy without too much memory. Since computers are very fast, most modern methods involve a lot of intermediate calculations to arrive at the answer. For instance, the CORDIC method is essentially a form a bisection, in that one repeatedly halves the interval containing the correct answer. Thus we’re adding one binary digit in each iteration. Twenty such iterations will give us 6 decimal digits of accuracy. There are other methods based on power series approximations. Related questions What is the point of arctan, arcsin and arccos? When do I have to use arcsin, arccos, and arctan? What's the domain and range of arcsin, arctan and arccos? How do I convert arcsin x to arctan and arctan x to arcsin? Is ? David Joyce Ph.D. in Mathematics, University of Pennsylvania (Graduated 1979) · Upvoted by Terry Moore , M.Sc. Mathematics, University of Southampton (1968) · Author has 9.9K answers and 68.1M answer views · 5y Related Is ? Arctangent is the function inverse to the tangent function. Likewise, arcsine is inverse to sine, and arccosine is inverse to cosine. Inverse functions aren’t like reciprocals. They’re something else entirely. Arctangent is inverse to tangent because it undoes with tangent does. If you start with an angle and take its tangent to get then you can apply arctangent to get back: at least if the angle is between and (that is, and ). Inverses of functions don’t have a lot of “nice” properties. Although tangent Arctangent is the function inverse to the tangent function. Likewise, arcsine is inverse to sine, and arccosine is inverse to cosine. Inverse functions aren’t like reciprocals. They’re something else entirely. Arctangent is inverse to tangent because it undoes with tangent does. If you start with an angle and take its tangent to get then you can apply arctangent to get back: at least if the angle is between and (that is, and ). Inverses of functions don’t have a lot of “nice” properties. Although tangent is the quotient of sine by cosine, arctangent is not the quotient of arcsine by arccosine. It would be if the inverse of the quotient of two functions were equal to the quotient of the inverses of those two functions, but that’s not a property that inverse functions have. Let’s take a particular example to see what happens. Take to be Then (30°) and (60°). But (26.6°) while In fact, the equation only holds when and when The equation is graphed in green below, while the equation is graphed in red. Brian Sittinger PhD in Mathematics, University of California, Santa Barbara (Graduated 2006) · Upvoted by Utkarsh Priyam , studied Mathematics at Harker School (2021) and Aditya Garg , M.Sc. Mathematics, Indian Institute of Technology, Delhi (2013) · Author has 8.4K answers and 20.8M answer views · 5y Related How do you solve ? Since , the equation reduces to Hence, we obtain Next, we take the cosine of both sides: Then, we rewrite the left side with the sum of angles identity: By using the Pythagorean theorem, this reduces to Since , the equation reduces to Hence, we obtain Next, we take the cosine of both sides: Then, we rewrite the left side with the sum of angles identity: By using the Pythagorean theorem, this reduces to Next, we clear the denominators and move the terms to one side: Squaring both sides and rearranging: By the quadratic formula, We ignore the minus sign above, because we need due to the inverse sines and cosines. Taking square roots one more time: Having squared both sides of the equation earlier, it is a good idea to check the solutions. We find immediately by substitution that is extraneous. Hence, the only solution is Brett Schmidt Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 2K answers and 3.2M answer views · 5y Related Is ? QUESTION: Is arctan(x) = arcsin(x)/arccos(x) ? ANSWER: No, not in general. Whilst it is true that tan(x) = sin(x)/cos(x) for all values of x, it is not true that arctan(x) = arcsin(x)/arccos(x) for all values of x. The simplest way to check this is simply to pick a value of x and try it. Let us try x = ½. Arcsin(½) = π/6 and arccos(½) = π/3. So Arcsin (½)/arccos(½) = (π/6)/(π/3) = π/6×3/π = 3/6 = ½. But arctan(½) ~ 0.4636 . So arctan(½) is not equal to sin(½)/cos(½). This counterexample shows that it is not the case that arctan(x) = arcsin(x)/arccos(x) for all values of x. Emad Noujeim Knowledge & science,reader,former teacher,multiple interests · Author has 2.2K answers and 10.7M answer views · Updated 6y Related What's the domain and range of arcsin, arctan and arccos? Each range of an inverse function is a proper subset of the domain of the original function. The domain of arcsin (x) is the range of sin (x) , which is [−1, 1] . The range of arcsin (x) is [− π /2 , π /2 ]. The domain of arcos(x) is −1 ≤ x ≤ 1 , the range of arcos(x) is [0 , π] , arcos(x) is the angle in [0, π] whose cosine is x. The domain of arctan(x) is all real numbers, the range of arctan is from −π/2 to π/2 radians exclusive . The arctangent function can be extended to the complex numbers. In this case the domain is all complex numbers. Related questions How are arcsin (a/b) and arctan(a/b) related? What is the formula for arctan(x+y)? What is the formula of tan (A - B)? What is the formula of tan (A + B + C)? What is the formula of (1+cos(x)) ^2? Mohammad Afzaal Butt B.Sc in Mathematics & Physics, Islamia College Gujranwala (Graduated 1977) · Author has 24.6K answers and 22.8M answer views · 6y Related How do we derive the derivative of arcsin(x), arccos(x), arctan(x), arcsec(x), arccsc(x), and arccot(x)? Steve Sparling BSc in Mathematics, The University of British Columbia · Author has 365 answers and 280.4K answer views · 5y Related When do I have to use arcsin, arccos, and arctan? arcsin is the inverse of the sine function. That means that it takes you back to the angle. So if you know the sine of an angle arcsin will give you the angle. An example. Suppose you measured the side opposite an angle and the hypotenuse and found the ratio of these measurements is equal to 0.5. If you operated on this with arcsin(0.5) you would get 30 degrees. You would use arccos and arctan in similar situations. On a more advanced situation. If you wanted to integrate 1/(1+x^2) you could set up a triangle (involving an angle theta) where the side opposite theta is x and the side adjacent is 1 arcsin is the inverse of the sine function. That means that it takes you back to the angle. So if you know the sine of an angle arcsin will give you the angle. An example. Suppose you measured the side opposite an angle and the hypotenuse and found the ratio of these measurements is equal to 0.5. If you operated on this with arcsin(0.5) you would get 30 degrees. You would use arccos and arctan in similar situations. On a more advanced situation. If you wanted to integrate 1/(1+x^2) you could set up a triangle (involving an angle theta) where the side opposite theta is x and the side adjacent is 1. If you understand substitution in calculus then you will know what to do for the rest of the problem and your answer will be an inverse trig function but I am not sure if you want to go this deep! Jasper Bhardwaj Former student · 6y Related What is the formula of tan (A + B + C)? Ihave explained it hope you understand … Plz upvote…. Hi…. Ihave explained it hope you understand … Plz upvote…. Related questions What is the point of arctan, arcsin and arccos? When do I have to use arcsin, arccos, and arctan? What's the domain and range of arcsin, arctan and arccos? How do I convert arcsin x to arctan and arctan x to arcsin? Is ? How are arcsin (a/b) and arctan(a/b) related? What is the formula for arctan(x+y)? What is the formula of tan (A - B)? What is the formula of tan (A + B + C)? What is the formula of (1+cos(x)) ^2? How can I solve arcsin (2a/ (1+a^2)) + arcsin (2b/ (1+b^2)) = arctan(x)? What is the formula of tan(a+b) tan(a-b)? What is the formula for half of arctan(x)? What is the formula of (cos^2) x-(cos^2) y? What is the formula of cos (ax+b)? Related questions What is the point of arctan, arcsin and arccos? When do I have to use arcsin, arccos, and arctan? What's the domain and range of arcsin, arctan and arccos? How do I convert arcsin x to arctan and arctan x to arcsin? Is ? How are arcsin (a/b) and arctan(a/b) related? What is the formula for arctan(x+y)? What is the formula of tan (A - B)? What is the formula of tan (A + B + C)? What is the formula of (1+cos(x)) ^2? 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16687	https://courses.physics.ucsd.edu/2011/Winter/physics4d/files/HW2.pdf	PHYS 4D Solution to HW 2 January 15, 2011 Problem Giancoli 31-22 (I) The E ﬁeld in an EM wave has a peak of 26.5mV/m. What is the average rate at which this wave carries energy across unit area per unit time? Solution: The average rate at which this wave carries energy across unit area per unit time is the average magnitude of the Poynting vector. ¯ S = 1 2ε0cE2 0 = 1 2 ( 8.85 × 10−12C2/N · m2) ( 3 × 108m/s ) (0.0265V/m) = 9.32 × 10−7W/m2. Problem Giancoli 31-24 (II) How much energy is transported across a 1.00cm2 area per hour by an EM wave whose E ﬁeld has an rms strength of 32.8mV/m? Solution: The energy transported across unit area per unit time is the magnitude of the Poynting vector. S = cε0E2 rms = ∆U A∆t ∆U ∆t = cε0E2 rmsA = ( 3 × 108m/s ) ( 8.85 × 10−12C2/N · m2) (0.0328V/m)2 ( 1 × 10−4m2) (3600s/h) = 1.03 × 10−6J/h. Problem Giancoli 31-26 (II) If the amplitude of the B ﬁeld of an EM wave is 2.5 × 10−7T, (a) what is the amplitude of the E ﬁeld? Solution: E0 = cB0 = ( 3 × 108m/s ) × 2.5 × 10−7T = 75V/m (b) What is the average power per unit area of the EM wave? Solution: The average power per unit area is given by the Poynting vector I = E0B0 2µ0 = 75V/m × 2.5 × 10−7T 2 × (4π × 10−7Ns2/C2) = 7.5W/m2. Problem Giancoli 31-28 (II) A 15.8 −mW laser puts out a narrow beam 2.00mm in diameter. What are the rms values of E and B in the beam? Solution: The power output per unit area is the intensity and also is the magnitude of the Poynting vector. Use Eq. 31-19a with rms values S = P A = cε0E2 rms ⇒ Erms = √ P Acε0 = √ 0.0158W π (10−3m)2 (3 × 108m/s) (8.85 × 10−12C2/N · m2) = 1376.3V/m. Brms = Erms c = 1376.3V/m (3 × 108m/s) = 4.59 × 10−6T. 1 Problem Giancoli 31-29 Estimate the average power output of the Sun, given that about 1350W/m2 reaches the upper atmosphere of the Earth. Solution: The radiation from the Sun has the same intensity in all directions, so the rate at which it reaches the Earth is the rate at which it passes through a sphere centered at the Sun with a radius equal to the Earth’s orbit radius. The 1350W/m2 is the intensity, or the magnitude of the Poynting vector. S = P A ⇒P = SA = 4π ( 1.496 × 1011m )2 ( 1350W/m2) = 3.80 × 1026W. Problem Giancoli 31-31 (II) How practical is solar power for various devices? Assume that on a sunny day, sunlight has an intensity of 1000W/m2 at the surface of Earth and that, when illuminated by that sunlight, a solar-cell panel can convert 10% of the sunlight’s energy into electric power. For each device given below, calculate the area A of solar panel needed to power it. (a) A calculator consumes 50mW. Find A in cm2. Is A small enough so that the solar panel can be mounted directly on the calculator that it is powering? Solution: A = P l = 50 × 10−3W 100W/m2 = 5 × 10−4m2 So yes. (b) A hair dryer consumes 1500W. Find A in m2. Assuming no other electronic devices are operating within a house at the same time, is A small enough so that the hair dryer can be powered by a solar panel mounted on the house’s roof? Solution: A = P l = 1500W 100W/m2 = 15m2 So yes. (c) A car requires 20hp for highway driving at constant velocity (this car would perform poorly in situations requiring acceleration). Find A in m2. Is A small enough so that this solar panel can be mounted directly on the car and power it in ”real time”? Solution: A = P l = 20hp (746W/hp) 100W/m2 = 149m2 This would require a square panel of side length about 12m. So no. Problem Giancoli 31-32 (III) (a) Show that the Poynting vector S points radially inward toward the center of a circular parallel-plate capacitor when it is being charged as in Example 31-1. Solution: In a cylindrical coordinate, (ˆ r, ˆ θ,ˆ z) are orthogonal coordinates. If we denote ˆ e1 = ˆ r,ˆ e2 = ˆ θ,ˆ e3 = ˆ z, they satisfy ˆ ek = εijkˆ ei × ˆ ej. Now, ˆ E = ˆ z, ˆ B = ˆ θ. So ˆ S = −ˆ r, that is S points radially inward toward the center of a circular parallel-plate capacitor. (b) Integrate S over the cylindrical boundary of the capacitor gap to show that the rate at which energy enters the capacitor is equal to the rate at which electrostatic energy is being stored in the electric ﬁeld of the capacitor (Section 24-4). Ignore fringing of E. Solution: We evaluate the Poynting vector, and then integrate it over the curved cylindrical surface between the capacitor plates. The magnetic ﬁeld (from Example 31-1) is B = 1 2µ0ε0r0 dE dt \|r=r0, E ⊥B, so S = 1 µ0 E × B = (−ˆ r) 1 2ε0r0E dE dt \|r=r0 2 In calculating ∫∫ S · dA for energy ﬂow into the capacitor volume, note that both S and dA pointing inward, and that S is constant over the surface, ∫∫ S · dA = ∫∫ SdA = S ∫∫ dA = S2πr0d = 1 2ε0r0E dE dt \|r=r02πr0d = ε0πdr2 0E dE dt \|r=r0 The energy stored in the capacitor U = the energy density u × the volume of the capacitor V . By Eq. 24-6 U = u · V = (1 2ε0E2 ) ( dπr2 0 ) . dU dt = ε0Edπr2 0 dE dt . Thus, we have ∫∫ S · dA = dU dt . Problem Giancoli 31-43 A 1.60-m-long FM antenna is oriented parallel to the electric ﬁeld of an EM wave. How large must the electric ﬁeld be to produce a 1.00-mV (rms) voltage between the ends of the antenna? What is the rate of energy transport per square meter? Solution: The electric ﬁeld is found from the desired voltage and the length of the antenna. Then use that electric ﬁeld to calculate the magnitude of the Poynting vector. Erms = Vrms d = 10−3V 1.60m = 6.25 × 10−4V/m. S = cε0E2 rms = cε0 V 2 rms d2 = ( 3 × 108m/s ) ( 8.85 × 10−12C2/N · m2) ( 6.25 × 10−4V/m )2 = 1.04 × 10−9W/m2. Problem Giancoli 31-48 Cosmic microwave background radiation ﬁlls all space with an average energy density of 4 × 10−14J/m3. (a) Find the rms value of the electric ﬁeld associated with this radiation. Solution: The rms value of the electric ﬁeld associated with this radiation is found from Eq. 24-6 u = 1 2ε0E2 = ε0E2 rms Erms = √u ε0 = √ 4 × 10−14J/m3 8.85 × 10−12C2/N · m2 = 0.07V/m. (b) How far from a 7.5-kW radio transmitter emitting uniformly in all directions would you ﬁnd a comparable value? Solution: A comparable value can be found using the magnitude of the Poynting vector ¯ S = cε0E2 rms = P 4πr2 ⇒ r = 1 Erms √ P 4πcε0 = 1 0.07V/m √ 7500W 4π (3 × 108m/s) (8.85 × 10−12C2/N · m2) = 7km. Problem Giancoli 31-52 How large an emf (rms) will be generated in an antenna that consists of a circular coil 2.2cm in diameter having 320 turns of wire, when an EM wave of frequency 810kHz transporting energy at an 3 average rate of 1.0 × 10−4W/m2 passes through it? [Hint: you can use Eq. 29-4 for a generator, since it could be applied to an observer moving with the coil so that the magnetic ﬁeld is oscillating with the frequency f = ω/2π.] Solution: Using Eq. 29-4, which says E = E0 sin ωt = NBAω sin ωt ⇒ Erms = NAωBrms = NAω √ µ0 ¯ S c = 320 × π × ( 1.1 × 10−2m )2 × 2π × 8.1 × 105Hz × √ (4π × 10−7Ns2/C2) × 1.0 × 10−4W/m2 (3 × 108m/s) = 4.0 × 10−4V. 4
16688	https://www.doubtnut.com/qna/277384869	If the radius of a sphere is halved, its volume becomestimes the volum Download Allen App Courses IIT-JEE Class 11 Class 12 Class 12+ NEET Class 11 Class 12 Class 12+ UP Board Class 9 Class 10 Class 11 Class 12 Bihar Board Class 9 Class 10 Class 11 Class 12 CBSE Board Class 9 Class 10 Class 11 Class 12 Other Boards MP Board Class 9 Class 10 Class 11 Class 12 Jharkhand Board Class 9 Class 10 Class 11 Class 12 Haryana Board Class 9 Class 10 Class 11 Class 12 Himachal Board Class 9 Class 10 Class 11 Class 12 Chhattisgarh Board Class 9 Class 10 Class 11 Class 12 Uttarakhand Board Class 9 Class 10 Class 11 Class 12 Rajasthan Board Class 9 Class 10 Class 11 Class 12 Ncert Solutions Class 6 Maths Physics Chemistry Biology English Class 7 Maths Physics Chemistry Biology English Class 8 Maths Physics Chemistry Biology English Class 9 Maths Physics Chemistry Biology English Class 10 Maths Physics Chemistry Biology English Reasoning Class 11 Maths Physics Chemistry Biology English Class 12 Maths Physics Chemistry Biology English Exam IIT-JEE NEET CBSE UP Board Bihar Board Results Study with ALLEN JEE NEET Class 6-10 My Profile Logout Ncert Solutions English Medium Class 6 Maths Physics Chemistry Biology English Class 7 Maths Physics Chemistry Biology English Class 8 Maths Physics Chemistry Biology English Class 9 Maths Physics Chemistry Biology English Class 10 Maths Physics Chemistry Biology English Reasoning Class 11 Maths Physics Chemistry Biology English Class 12 Maths Physics Chemistry Biology English Courses IIT-JEE Class 11 Class 12 Class 12+ NEET Class 11 Class 12 Class 12+ UP Board Class 9 Class 10 Class 11 Class 12 Bihar Board Class 9 Class 10 Class 11 Class 12 CBSE Board Class 9 Class 10 Class 11 Class 12 Other Boards MP Board Class 9 Class 10 Class 11 Class 12 Jharkhand Board Class 9 Class 10 Class 11 Class 12 Haryana Board Class 9 Class 10 Class 11 Class 12 Himachal Board Class 9 Class 10 Class 11 Class 12 Chhattisgarh Board Class 9 Class 10 Class 11 Class 12 Uttarakhand Board Class 9 Class 10 Class 11 Class 12 Rajasthan Board Class 9 Class 10 Class 11 Class 12 Exam IIT-JEE NEET CBSE UP Board Bihar Board Study with ALLEN JEE NEET Class 6-10 Books Class 12 Class 11 Class 10 Class 9 Class 8 Class 7 Class 6 Online Class Blog Select Theme Class 10 MATHS If the radius of a sphere is halved, its... If the radius of a sphere is halved, its volume becomes______times the volume of original sphere. Video Solution To view this video, please enable JavaScript and consider upgrading to a web browser thatsupports HTML5 video Video Player is loading. Play Video Play Skip Backward Mute Current Time 0:00 / Duration-:- Loaded: 0% Stream Type LIVE Seek to live, currently behind live LIVE Remaining Time-0:00 1x Playback Rate 2x 1.5x 1x, selected 0.5x 0.25x Chapters Chapters Descriptions descriptions off, selected Captions captions settings, opens captions settings dialog captions off, selected Audio Track Picture-in-Picture Fullscreen This is a modal window. Beginning of dialog window. Escape will cancel and close the window. Text Color Opacity Text Background Color Opacity Caption Area Background Color Opacity Font Size Text Edge Style Font Family Reset restore all settings to the default values Done Close Modal Dialog End of dialog window. More from this Exercise 9 videos Text Solution Generated By DoubtnutGPT The correct Answer is:1 8 To solve the problem of how the volume of a sphere changes when its radius is halved, we can follow these steps: Understand the Volume Formula: The volume V of a sphere is given by the formula: V=4 3 π r 3 where r is the radius of the sphere. Calculate the Original Volume: Let the original radius of the sphere be r. Therefore, the volume of the original sphere is: V original=4 3 π r 3 Determine the New Radius: If the radius is halved, the new radius R becomes: R=r 2 Calculate the Volume of the New Sphere: Using the new radius R, we can calculate the volume of the new sphere: V new=4 3 π R 3=4 3 π(r 2)3 Simplifying this: V new=4 3 π(r 3 8)=4 3⋅π r 3 8=1 6 π r 3 Relate the New Volume to the Original Volume: Now, we can express the new volume in terms of the original volume: V new=1 8(4 3 π r 3)=1 8 V original Conclusion: Therefore, the volume of the new sphere is 1 8 times the volume of the original sphere. Final Answer: If the radius of a sphere is halved, its volume becomes 1 8 times the volume of the original sphere. Show More 4 \| Share Save Class 10MATHSSURFACE AREAS AND VOLUMES Topper's Solved these Questions SURFACE AREA AND VOLUME Book:CBSE COMPLEMENTARY MATERIAL Chapter:SURFACE AREA AND VOLUME Exercise:VERY SHORT ANSWER TYPE QUESTIONS (TRUE OR FALSE) Explore 5 Videos SURFACE AREA AND VOLUME Book:CBSE COMPLEMENTARY MATERIAL Chapter:SURFACE AREA AND VOLUME Exercise:VERY SHORT ANSWER TYPE QUESTIONS (CHOOSE THE CORRECT ANSWER) Explore 10 Videos SURFACE AREA AND VOLUME Book:CBSE COMPLEMENTARY MATERIAL Chapter:SURFACE AREA AND VOLUME Exercise:PRACTICE TEST ( SECTION-D) Explore 1 Video STATISTICS Book:CBSE COMPLEMENTARY MATERIAL Chapter:STATISTICS Exercise:Practice -Test Explore 5 Videos TRIANGLE Book:CBSE COMPLEMENTARY MATERIAL Chapter:TRIANGLE Exercise:Short Answer Type Questions-II Explore 25 Videos Similar Questions If the radius of a sphere is doubled, its volume becomes _ times the volume of original sphere. View Solution If the radius of a sphere is doubled, its volume becomes View Solution Knowledge Check If the radius of a sphere is doubled, its volume becomes A a.double B b. four times C c.six times D d.eight times Submit If the radius of a sphere is 2r , them its volume will be A 4 3 π r 3 B 4 π r 3 C 8 π r 3 3 D 32 3 π r 3 Submit If the radius of a sphere becomes 3 times then its volume will become A 3 times B 6 times C 9 times D 27 times Submit If the edse of a cube is halved, then its volume reduces- to__ times the actual volume. View Solution If the sides of a cube are halved, its volume will become............times the original volume. View Solution If the radius of a sphere be douled, then the percentage increase in volume is View Solution The diameter of a sphere is 14 cm. Its volume is View Solution If the radius of a sphere is increased to three times its previous value, the volume of the sphere increases to : View Solution CBSE COMPLEMENTARY MATERIAL-SURFACE AREA AND VOLUME -VERY SHORT ANSWER TYPE QUESTIONS (FILL IN THE BLANKS) The total surface area of cuboid of dimension a xx a xx b is.. 03:09 The volume of right circular cylinder of base radius r and height 2r i... 01:01 The total surface area of a cylinder of base radius r and height h is. 04:39 The curved surface area of a cone of base radius r and height h is. 01:59 If the height of a cone is equal to diameter of its base, the volume o... 01:23 The total surface area of a hemisphere of radius r is pir^2 (b) 2p... 01:03 The lateral surface area of a hollow cylinder of outer radius R, inner... 02:35 If the radius of a sphere is doubled, its volume becomes times the vo... 02:05 If the radius of a sphere is halved, its volume becomestimes the volum... 02:00 Exams IIT JEE NEET UP Board Bihar Board CBSE Free Textbook Solutions KC Sinha Solutions for Maths Cengage Solutions for Maths DC Pandey Solutions for Physics HC Verma Solutions for Physics Sunil Batra Solutions for Physics Pradeep Solutions for Physics Errorless Solutions for Physics Narendra Awasthi Solutions for Chemistry MS Chouhan Solutions for Chemistry Errorless Solutions for Biology Free Ncert Solutions English Medium NCERT Solutions NCERT Solutions for Class 12 English Medium NCERT Solutions for Class 11 English Medium NCERT Solutions for Class 10 English Medium NCERT Solutions for Class 9 English Medium NCERT Solutions for Class 8 English Medium NCERT Solutions for Class 7 English Medium NCERT Solutions for Class 6 English Medium Free Ncert Solutions Hindi Medium NCERT Solutions NCERT Solutions for Class 12 Hindi Medium NCERT Solutions for Class 11 Hindi Medium NCERT Solutions for Class 10 Hindi Medium NCERT Solutions for Class 9 Hindi Medium NCERT Solutions for Class 8 Hindi Medium NCERT Solutions for Class 7 Hindi Medium NCERT Solutions for Class 6 Hindi Medium Boards CBSE UP Board Bihar Board UP Board Resources Unit Converters Calculators Books Store Seminar Results Blogs Class 12 All Books Class 11 All Books Class 10 All Books Class 9 All Books Class 8 All Books Class 7 All Books Class 6 All Books Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc NCERT solutions for CBSE and other state boards is a key requirement for students. 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16689	https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.02%3A_Galvanic_cells_and_Electrodes	Skip to main content 16.2: Galvanic cells and Electrodes Last updated : Mar 1, 2022 Save as PDF 16.1: Chemistry and Electricity 16.3: Cell Potentials and Thermodynamics Page ID : 260 Stephen Lower Simon Fraser University ( \newcommand{\kernel}{\mathrm{null}\,}) It is physically impossible to measure the potential difference between a piece of metal and the solution in which it is immersed. We can, however, measure the difference between the potentials of two electrodes that dip into the same solution, or more usefully, are in two different solutions. In the latter case, each electrode-solution pair constitutes an oxidation-reduction half cell, and we are measuring the sum of the two half-cell potentials. This arrangement is called a galvanic cell. A typical cell might consist of two pieces of metal, one zinc and the other copper, each immersed each in a solution containing a dissolved salt of the corresponding metal. The two solutions are separated by a porous barrier that prevents them from rapidly mixing but allows ions to diffuse through. If we connect the zinc and copper by means of a metallic conductor, the excess electrons that remain when Zn2+ ions emerge from the zinc in the left cell would be able to flow through the external circuit and into the right electrode, where they could be delivered to the Cu2+ ions which become "discharged", that is, converted into Cu atoms at the surface of the copper electrode. The net reaction is the oxidation of zinc by copper(II) ions: Zn(s)+Cu2+→Zn2++Cu(s)(16.2.1) but this time, the oxidation and reduction steps (half reactions) take place in separate locations: \| \| \| \| --- \| left electrode: \| Zn(s) → Zn2+ + 2e– \| oxidation \| \| right electrode: \| Cu2+ + 2e–→ Cu(s) \| reduction \| Electrochemical cells allow measurement and control of a redox reaction The reaction can be started and stopped by connecting or disconnecting the two electrodes. If we place a variable resistance in the circuit, we can even control the rate of the net cell reaction by simply turning a knob. By connecting a battery or other source of current to the two electrodes, we can force the reaction to proceed in its non-spontaneous, or reverse direction. By placing an ammeter in the external circuit, we can measure the amount of electric charge that passes through the electrodes, and thus the number of moles of reactants that get transformed into products in the cell reaction. Electric charge q is measured in coulombs. The amount of charge carried by one mole of electrons is known as the Faraday, which we denote by F. Careful experiments have determined that 1 F = 96467 C. For most purposes, you can simply use 96,500 Coulombs as the value of the faraday. When we measure electric current, we are measuring the rate at which electric charge is transported through the circuit. A current of one ampere corresponds to the flow of one coulomb per second. Charge Transport within the Cell For the cell to operate, not only must there be an external electrical circuit between the two electrodes, but the two electrolytes (the solutions) must be in contact. The need for this can be understood by considering what would happen if the two solutions were physically separated. Positive charge (in the form of Zn2+) is added to the electrolyte in the left compartment, and removed (as Cu2+) from the right side, causing the solution in contact with the zinc to acquire a net positive charge, while a net negative charge would build up in the solution on the copper side of the cell. These violations of electroneutrality would make it more difficult (require more work) to introduce additional Zn2+ ions into the positively-charged electrolyte or for electrons to flow into right compartment where they are needed to reduce the Cu2+ ions, thus effectively stopping the reaction after only a chemically insignificant amount has taken place. In order to sustain the cell reaction, the charge carried by the electrons through the external circuit must be accompanied by a compensating transport of ions between the two cells. This means that we must provide a path for ions to move directly from one cell to the other. This ionic transport involves not only the electroactive species Cu2+ and Zn2+, but also the counterions, which in this example are nitrate, NO3-. Thus an excess of Cu2+ in the left compartment could be alleviated by the drift of these ions into the right side, or equally well by diffusion of nitrate ions to the left. More detailed studies reveal that both processes occur, and that the relative amounts of charge carried through the solution by positive and negative ions depends on their relative mobilities, which express the velocity with which the ions are able to make their way through the solution. Since negative ions tend to be larger than positive ions, the latter tend to have higher mobilities and carry the larger fraction of charge. In the simplest cells, the barrier between the two solutions can be a porous membrane, but for precise measurements, a more complicated arrangement, known as a salt bridge, is used. The salt bridge consists of an intermediate compartment filled with a concentrated solution of KCl and fitted with porous barriers at each end. The purpose of the salt bridge is to minimize the natural potential difference, known as the junction potential, that develops (as mentioned in the previous section) when any two phases (such as the two solutions) are in contact. This potential difference would combine with the two half-cell potentials so as introduce a degree of uncertainty into any measurement of the cell potential. With the salt bridge, we have two liquid junction potentials instead of one, but they tend to cancel each other out. Cell description conventions In order to make it easier to describe a given electrochemical cell, a special symbolic notation has been adopted. In this notation the cell we described above would be Zn(s) \| Zn2+(aq) \|\| Cu2+(aq) \| Cu(s) There are several other conventions relating to cell notation and nomenclature that you are expected to know: The anode is where oxidation occurs, and the cathode is the site of reduction. In an actual cell, the identity of the electrodes depends on the direction in which the net cell reaction is occurring. If electrons flow from the left electrode to the right electrode (as depicted in the above cell notation) when the cell operates in its spontaneous direction, the potential of the right electrode will be higher than that of the left, and the cell potential will be positive. "Conventional current flow" is from positive to negative, which is opposite to the direction of the electron flow. This means that if the electrons are flowing from the left electrode to the right, a galvanometer placed in the external circuit would indicate a current flow from right to left. Electrodes and Electrode Reactions An electrode reaction refers to the net oxidation or reduction process that takes place at an electrode. This reaction may take place in a single electron-transfer step, or as a succession of two or more steps. The substances that receive and lose electrons are called the electroactive species. This process takes place within the very thin interfacial region at the electrode surface, and involves quantum-mechanical tunneling of electrons between the electrode and the electroactive species. The work required to displace the H2O molecules in the hydration spheres of the ions constitutes part of the activation energy of the process. In the example of the Zn/Cu cell we have been using, the electrode reaction involves a metal and its hydrated cation; we call such electrodes metal-metal ion electrodes. There are a number of other kinds of electrodes which are widely encountered in electrochemistry and analytical chemistry. Ion-ion Electrodes Many electrode reactions involve only ionic species, such as Fe2+ and Fe3+. If neither of the electroactive species is a metal, some other metal must serve as a conduit for the supply or removal of electrons from the system. In order to avoid complications that would arise from electrode reactions involving this metal, a relatively inert substance such as platinum is commonly used. Such a half cell would be represented as Pt(s) \| Fe3+(aq), Fe2+(aq) \|\| ... and the half-cell reaction would be Fe2+(aq)→Fe3+(aq)+e−(16.2.2) The reaction occurs at the surface of the electrode (Fig 4 above). The electroactive ion diffuses to the electrode surface and adsorbs (attaches) to it by van der Waals and Coulombic forces. In doing so, the waters of hydration that are normally attached to any ionic species must be displaced. This process is always endothermic, sometimes to such an extent that only a small fraction of the ions be able to contact the surface closely enough to undergo electron transfer, and the reaction will be slow. The actual electron-transfer occurs by quantum-mechanical tunnelling. Gas Electrodes Some electrode reactions involve a gaseous species such as H2, O2, or Cl2. Such reactions must also be carried out on the surface of an electrochemically inert conductor such as platinum. A typical reaction of considerable commercial importance is Cl−(aq)→½Cl2(g)+e−(16.2.3) Similar reactions involving the oxidation of Br2 or I2 also take place at platinum surfaces. Insoluble–salt Electrodes A typical electrode of this kind consists of a silver wire covered with a thin coating of silver chloride, which is insoluble in water. The electrode reaction consists in the oxidation and reduction of the silver: AgCl(s)+e–→Ag(s)+Cl–(aq)(16.2.4) The half cell would be represented as ...\|\|Cl–(aq)\|AgCl(s)\|Ag(s)(16.2.5) Although the usefulness of such an electrode may not be immediately apparent, this kind of electrode finds very wide application in electrochemical measurements, as we shall see later. Reference Electrodes In most electrochemical experiments our interest is concentrated on only one of the electrode reactions. Since all measurements must be on a complete cell involving two electrode systems, it is common practice to employ a reference electrode as the other half of the cell. The major requirements of a reference electrode are that it be easy to prepare and maintain, and that its potential be stable. The last requirement essentially means that the concentration of any ionic species involved in the electrode reaction must be held at a fixed value. The most common way of accomplishing this is to use an electrode reaction involving a saturated solution of an insoluble salt of the ion. One such system, the silver-silver chloride electrode has already been mentioned: Ag\|AgCl(s)\|Cl–(aq)\|\|...(16.2.6) Ag(s)+Cl–(aq)→AgCl(s)+e–(16.2.7) This electrode usually takes the form of a piece of silver wire coated with AgCl. The coating is done by making the silver the anode in an electrolytic cell containing HCl; the Ag+ ions combine with Cl– ions as fast as they are formed at the silver surface. The other common reference electrode is the calomel electrode; calomel is the common name for mercury(I) chloride. Such a half cell would be represented as Hg\|Hg2+(aq)\|KCl\|\|...(16.2.8) and the half-cell reaction would be Hg(l)+Cl–→½HgCl2(s)+e–(16.2.9) The potentials of both of these electrodes have been very accurately determined against the hydrogen electrode. The latter is seldom used in routine electrochemical measurements because it is more difficult to prepare; the platinum surface has to be specially treated by preliminary electrolysis. Also, there is need for a supply of hydrogen gas which makes it somewhat cumbersome and hazardous. Summary and additional notes Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic. A galvanic cell (sometimes more appropriately called a voltaic cell) consists of two half-cells joined by a salt bridge or some other path that allows ions to pass between the two sides in order to maintain electroneutrality. The conventional way of representing an electrochemical cell of any kind is to write the oxidation half reaction on the left and the reduction on the right. Thus for the reaction Zn(s) + Cu2+ → Zn2+ + Cu(s) we write Zn(s) \| Zn2+(aq) \|\| Cu2+(aq) \| Cu(s) in which the single vertical bars represent phase boundaries. The double bar denotes a liquid-liquid boundary which in laboratory cells consists of a salt bridge or in ion-permeable barrier. If the net cell reaction were written in reverse, the cell notation would become Cu(s) \| Cu2+(aq) \|\| Zn 2+(aq) \| Zn (s) Remember: the Reduction process is always shown on the Right. at the electrode surface. The energy required to displace water molecules from the hydration shell of an ion as it approaches the electrode surface constitutes an activation energy which can slow down the process. Even larger activation energies (and slower reactions) occur when a molecule such as O2 is formed or consumed. 16.1: Chemistry and Electricity 16.3: Cell Potentials and Thermodynamics
16690	https://www.education.ky.gov/curriculum/standards/kyacadstand/Documents/270304_Algebra_1_2019.pdf	Kentucky Department of Education KDE:OSAA:MD:CD:TC 5/2019 Page 1 Course Standards for 2019-2020 and Beyond Course Code: 270304 Course Name: Algebra 1 Grade Level: 9-11 Course standards documents are designed to show how specific standards align to courses. For instructional planning and assessment, please access the complete Kentucky Academic Standards for Mathematics for the full scope of what students should know and be able to do. Upon course completion students should be able to: Standards Standards for Mathematical Practice ➢ Make sense of problems and persevere in solving them. ➢ Reason abstractly and quantitatively. ➢ Construct viable arguments and critique the reasoning of others. ➢ Model with mathematics. ➢ Use appropriate tools strategically. ➢ Attend to precision. ➢ Look for and make use of structure. ➢ Look for and express regularity in repeated reasoning. Modeling Standards: Modeling links classroom mathematics and statistics to everyday life, work, and decision-making. Modeling is the process of choosing and using appropriate mathematics and statistics to analyze empirical situations, to understand them better, and to improve decisions. Quantities and their relationships in physical, economic, public policy, social, and everyday situations can be modeled using mathematical and statistical methods. When making mathematical models, technology is valuable for varying assumptions, exploring consequences, and comparing predictions with data. Modeling is best interpreted not as a collection of isolated topics but rather in relation to other standards. Making mathematical models is a Standard for Mathematical Practice, and specific modeling standards appear throughout the high school standards indicated by a star symbol (★). The high school standards are listed in conceptual categories: • Number and Quantity (N) • Algebra (A) • Functions (F) • Geometry (G) Kentucky Department of Education KDE:OSAA:MD:CD:TC 5/2019 Page 2 Standards • Statistics and Probability (SP) Conceptual Category Number and Quantity (N) – Standards KY.HS.N.1 Extend the properties of integer exponents to rational exponents, allowing for the expression of radicals in terms of rational exponents. KY.HS.N.2 Rewrite expressions involving radicals and rational exponents using the properties of exponents. KY.HS.N.4 Use units in context as a way to understand problems and to guide the solution of multi-step problems; ★ a. Choose and interpret units consistently in formulas; b. Choose and interpret the scale and the origin in graphs and data displays. KY.HS.N.5 Define appropriate units in context for the purpose of descriptive modeling. ★ KY.HS.N.6 Choose a level of accuracy appropriate to limitations on measurement when reporting quantities. ★ Conceptual Category Algebra (A) – Standards KY.HS.A.1 Interpret expressions that represent a quantity in terms of its context. ★ a. Interpret parts of an expression, such as terms, factors and coefficients. b. Interpret complicated expressions, given a context, by viewing one or more of their parts as a single entity. KY.HS.A.2 Use the structure of an expression to identify ways to rewrite it and consistently look for opportunities to rewrite expressions in equivalent forms. KY.HS.A.3 Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression. ★ a. Write the standard form of a given polynomial and identify the terms, coefficients, degree, leading coefficient and constant term. b. Factor a quadratic expression to reveal the zeros of the function it defines. c. Use the properties of exponents to rewrite exponential expressions. KY.HS.A.5 Add, subtract and multiply polynomials. Kentucky Department of Education KDE:OSAA:MD:CD:TC 5/2019 Page 3 Standards KY.HS.A.7 Identify roots of polynomials when suitable factorizations are available. Know these roots become the zeros (x-intercepts) for the corresponding polynomial function. KY.HS.A.12 Create equations and inequalities in one variable and use them to solve problems. KY.HS.A.13 Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. KY.HS.A.14 Create a system of equations or inequalities to represent constraints within a modeling context. Interpret the solution(s) to the corresponding system as viable or nonviable options within the context. KY.HS.A.15 Rearrange formulas to solve a literal equation, highlighting a quantity of interest, using the same reasoning as in solving equations. KY.HS.A.16 Understand each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method. KY.HS.A.18 Solve linear equations and inequalities in one variable, including literal equations with coefficients represented by letters. KY.HS.A.19 Solve quadratic equations in one variable. a. Solve quadratic equations by taking square roots, the quadratic formula and factoring, as appropriate to the initial form of the equation. KY.HS.A.20 Solve systems of linear equations in two variables. a. Understand a system of two equations in two variables has the same solution as a new system formed by replacing one of the original equations with an equivalent equation. b. Solve systems of linear equations with graphs, substitution and elimination, focusing on pairs of linear equations in two variables. Kentucky Department of Education KDE:OSAA:MD:CD:TC 5/2019 Page 4 Standards KY.HS.A.23 Understand that the graph of an equation in two variables is the set of all its solutions plotted in the coordinate plane. KY.HS.A.24 Justify that the solutions of the equations f(x) = g(x) are the x-coordinates of the points where the graphs of y = f(x) and y = g(x) intersect. Find the approximate solutions graphically, using technology or tables. ★ KY.HS.A.25 Graph linear inequalities in two variables. a. Graph the solutions to a linear inequality as a half-plane (excluding the boundary in the case of a strict inequality). b. Graph the solution set to a system of linear inequalities as the intersection of the corresponding half-planes. Conceptual Category Functions (F) – Standards KY.HS.F.1 Understand properties and key features of functions and the different ways functions can be represented. a. Understand that a function from one set (called the domain) to another set (called the range) assigns to each element of the domain exactly one element of the range. If f is a function and x is an element of its domain, then f(x) denotes the output of f corresponding to the input x. b. Using appropriate function notation, evaluate functions for inputs in their domains and interpret statements that use function notation in terms of a context. c. For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities and sketch graphs showing key features given a verbal description of the relationship. d. Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes. e. Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). KY.HS.F.2 Recognize that arithmetic and geometric sequences are functions, sometimes defined recursively, whose domain is a subset of the integers. KY.HS.F.3 Understand average rate of change of a function over an interval. Kentucky Department of Education KDE:OSAA:MD:CD:TC 5/2019 Page 5 Standards a. Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. b. Estimate the rate of change from a graph. ★ KY.HS.F.4 Graph functions expressed symbolically and show key features of the graph, with and without using technology (computer, graphing calculator). ★ a. Graph linear and quadratic functions and show intercepts, maxima and minima. KY.HS.F.5 Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function. a. Identify zeros, extreme values and symmetry of the graph within the context of a quadratic function. Use the properties of exponents to interpret expressions for exponential functions and classify the exponential function as representing growth or decay. KY.HS.F.6 Write a function that describes a relationship between two quantities. ★ a. Determine an explicit expression, a recursive process, or steps for calculation from a context. b. Combine standard function types using arithmetic operations. KY.HS.F.7 Use arithmetic and geometric sequences to model situations and scenarios. a. Use formulas (explicit and recursive) to generate terms for arithmetic and geometric sequences. b. Write formulas to model arithmetic and geometric sequences and apply those formulas in realistic situations. ★ KY.HS.F.11 Distinguish between situations that can be modeled with linear functions and with exponential functions. a. Recognize and justify that linear functions grow by equal differences over equal intervals and that exponential functions grow by equal factors over equal intervals. b. Recognize situations in which one quantity changes at a constant rate per unit interval relative to another. c. Recognize situations in which a quantity grows or decays by a constant percent rate per unit interval relative to another. Kentucky Department of Education KDE:OSAA:MD:CD:TC 5/2019 Page 6 Standards KY.HS.F.12 Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or two input-output pairs (include reading these from a table). KY.HS.F.13 Observe using graphs and tables that a quantity increasing exponentially eventually exceeds a quantity increasing linearly, quadratically, or (more generally) as a polynomial function. KY.HS.F.14 Interpret the parameters in a linear or exponential function in terms of a context. Conceptual Category Statistics and Probability (S) – Standards KY.HS.SP.6 Represent data on two quantitative variables on a scatter plot and describe how the explanatory and response variables are related. a. Calculate an appropriate mathematical model, or use a given mathematical model, for data to solve problems in context. b. Informally assess the fit of a model (through calculating correlation for linear data, plotting, calculating and/or analyzing residuals). KY.HS.SP.7 Interpret the slope (rate of change) and the intercept (constant term) of a linear model in the context of the data. KY.HS.SP.8 Understand the role and purpose of correlation in linear regression. a. Use technology to compute correlation coefficient of a linear fit. b. Interpret the meaning of the correlation within the context of the data. c. Describe the limitations of correlation when establishing causation.
16691	https://baike.baidu.com/item/%E7%94%B5%E8%AF%9D%E7%BA%BF/3133982	电话线_百度百科网页新闻贴吧知道网盘图片视频地图文库资讯采购百科百度首页登录注册进入词条全站搜索帮助进入词条全站搜索帮助播报编辑讨论 1收藏赞登录首页历史上的今天百科冷知识图解百科秒懂百科懂啦秒懂本尊答秒懂大师说秒懂看瓦特秒懂五千年秒懂全视界特色百科数字博物馆非遗百科恐龙百科多肉百科艺术百科科学百科知识专题观千年·见今朝中国航天古鱼崛起食品百科数字文物守护计划史记2024·科学100词加入百科新人成长进阶成长任务广场百科团队校园团分类达人团热词团繁星团蝌蚪团权威合作合作模式常见问题联系方式个人中心电话线播报编辑讨论 1上传视频电话的进户线快速了解电话线 02:14 517世界电信日硬核科普!从传统铜缆到光纤通信,电话线如何变身高速信息通道?带你解码网络底层密码! 00:55 拆解电话线路对号器,古老的电路怎么实现双向对讲和巡线的? 03:45 收藏查看我的收藏 287 有用+1 9 电话线是连接用户终端与通信网络的电话进户线路，由导体、绝缘层和护套构成，主要用于语音及数据传输。其导体材料分为铜包钢、铜包铝和全铜三类，其中全铜导体传输性能最优且符合国家标准。按芯数可分为2芯、4芯和6芯，家庭市话多采用2芯线，企业通信及数字电话需使用4芯或6芯线。常见型号如HYV、HSYV通过芯数、材质和线径命名，并采用扭绞结构降低干扰。产品执行YD/T630-93等行业标准，支持音频、模拟及数字信号传输 [4-5]。电话线技术标准随通信需求不断演进，早期以铜包钢、铜包铝材料为主，后全铜导体逐步普及以提升远距传输能力。20世纪90年代起执行的YD/T630-93标准规范了绝缘材料和扭绞工艺，2009年发布的YD/T1955-2009标准进一步扩展了对宽带信号传输的支持。随着数字电话发展，6芯线逐步应用于高频信号场景，同时抗腐蚀、低阻抗等性能持续优化。中文名电话线外文名 telephoneline 基本释义电话的进户线连接到电话机上直径 0.4mm、0.5mm 内导体单支退火裸铜丝目录 1说明 2应用 3规格 4材料 5型号解释 6辨别好坏说明播报编辑 1、产品执行标准：参考邮电局YD/T630-93标准和奋进达企业标准； 2、绝缘材料：高密度聚乙烯或聚丙烯，按照国标色谱标明绝缘线的颜色； 3、绝缘成对：把单根绝缘按照不同的节距扭绞成对，并采用规定的色谱组合以识别线对降低了线索之间的互相干扰串音，功率耗损小。应用播报编辑通信工程布线，室内电话通信电缆系统布线之间的连接，语音通信系统之间主干线，程控交换机，电话。规格播报编辑电话线常见规格有二芯和四芯，线径分别有0.4和0.5，部分地区有0.8和1.0。除了二芯和四芯之外，还有四、六、八、十芯。一般家庭如果是市话使用模式的话，2芯足够使用。如果是公司或部分集团电话使用的话，考虑到电话宽带使用需要，建议使用4芯电话线较好，如果使用的是数字电话，则建议用6芯的电话线。材料播报编辑常见材料（导体）有： 1、铜包钢这种线比较硬，不适合用于外部扯线，容易断芯。但是埋在墙里还是将就可以使用的。并且只能用于近距离使用，（例如楼道接线箱到用户） 2、铜包铝这种线比较软，比较容易断芯。可以埋在墙里，也可以墙外扯线。并且只能用于近距离使用，（例如楼道接线箱到用户） 3、全铜。这种线软，可以埋在墙里，也可以墙外扯线。可以用于远距离传输使用。其中，达到国家标准的只有采用全铜的电话线。型号解释播报编辑 HYV2x1/0.4 CCS HYV4x1/0.4 CCS HYV2x1/0.4 BC HYV2x1/0.5 BC HYV4x1/0.4 BC HYV4x1/0.5 BC HSYV 2x2x0.5 BC 备注：CCS为铜包钢、BC为全铜。 HYV2x1/0.4 CCS、其中HYV为电话线英文型号、2代表2芯、1/0.4 CCS 代表单支0.4mm直径的铜包钢导体。以此类推，HYV4x1/0.4 BC、其中HYV型号、4代表4芯、1/0.5 BC 代表单支0.5 mm直径的纯铜导体。特别说一下、HSYV 2x2x0.5、"S"代表双绞，2x2代表2对（4芯）双绞、0.5代表单支0.5 mm直径的导体。双绞电话线对比普通的平行电话线，主要作用在于传输速度的提高和功率方面更低的损耗。电话线辨别好坏播报编辑 1、上榜品牌值得信赖,产品品质有保证。 2、导体采用进口高纯度无氧铜的电话线,传输衰减小,信号损耗小,音质清晰无噪,通话无距离感。 3、护套采用高档透明外被料的电话线,耐酸碱腐蚀,防老化,使用寿命长。 4、护套采用透明材料的电话线,具有良好的机械物理性能、电气性能和热稳定性,铅、镉等重金属和重金属化合物的含量极低。 5、采用进口聚丙烯材料的电话线,纯度高,传输衰减更低,传输速度更高,适合用于宽带上网，能达到可视电话的高传输速度的要求。词条图册更多图册概述图册(1张) 词条图片(1张) 1/1 参考资料 1 杨强, 王博亮, 陈砚蒲, et al. 多参数电话线传输监护系统的研制[C]// 全国计算机应用联合学术会议. 1999. 2 蔡广平, 危韧勇. 基于电话线的农村用户电能表远程抄表系统[J]. 自动化仪表, 2005, 26(7):46-47. 3 电话线．百度汉语．2025-04-19 4 电话线．www.potel-group.com．2025-05-27 5 电话软线HRVB-20.4室内电话线．建材网．2021-03-27 电话线的概述图（1张）词条统计浏览次数：363913次编辑次数：36次历史版本最近更新：树懒挂在藤蔓上（2025-06-17）突出贡献榜 a0yang 1 说明2 应用3 规格4 材料5 型号解释6 辨别好坏相关搜索电话线品牌电话线规格型号黑色电话线电话线价格多芯电话线电话线接线盒电话险电话号码归属地查询电话400 法律咨询免费热线电话电话线选择朗读音色成熟女声成熟男声磁性男声年轻女声情感男声 0 0 2x 1.5x 1.25x 1x 0.75x 0.5x 分享到微信朋友圈打开微信“扫一扫”即可将网页分享至朋友圈新手上路成长任务编辑入门编辑规则本人编辑我有疑问内容质疑在线客服官方贴吧意见反馈投诉建议举报不良信息未通过词条申诉投诉侵权信息封禁查询与解封 ©2025 Baidu使用百度前必读\|百科协议\|隐私政策\|百度百科合作平台\|京ICP证030173号京公网安备11000002000001号
16692	https://slcc.pressbooks.pub/humanbiology/chapter/4-cell-structure/	Skip to content Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. 4 Chapter 4: Cell Structure and Transport Chapter Outline 4.1 Studying Cells 4.2 Prokaryotic Cells 4.3 Eurkaryotic Cells 4.4 The Endomembrane System and Proteins 4.5 The Cytoskeleton 4.6 Connections between Cells and Cellular Activities 4.7 Plasma Membrane Components and Structure 4.8 Passive Transport 4.9 Active Transport 4.10 Bulk Transport Introduction Close your eyes and picture a brick wall. What is the basic building block of that wall? A single brick, of course. Like a brick wall, your body is composed of basic building blocks, and the building blocks of your body are cells. Your body has many kinds of cells, each specialized for a specific purpose. Just as a home is made from a variety of building materials, the human body is constructed from many cell types. For example, epithelial cells protect the surface of the body and cover the organs and body cavities within. Bone cells help to support and protect the body. Cells of the immune system fight invading bacteria. Additionally, blood and blood cells carry nutrients and oxygen throughout the body while removing carbon dioxide. Each of these cell types plays a vital role during the growth, development, and day-to-day maintenance of the body. In spite of their enormous variety, however, cells from all organisms—even ones as diverse as bacteria, onion, and human—share certain fundamental characteristics. Figure 4.1 (a) Nasal sinus cells (viewed with a light microscope), (b) onion cells (viewed with a light microscope), and (c) Vibrio tasmaniensis bacterial cells (seen through a scanning electron microscope) are from very different organisms, yet all share certain characteristics of basic cell structure. (credit a: modification of work by Ed Uthman, MD; credit b: modification of work by Umberto Salvagnin; credit c: modification of work by Anthony D’Onofrio, William H. Fowle, Eric J. Stewart, and Kim Lewis of the Lewis Lab at Northeastern University; scale-bar data from Matt Russell) Learning Objectives You will be able to describe the basics of cells: Identify major cell components Differentiate the functions of cell part Be able to describe these characteristics of cellular transport: compare and contrast the different forms of membrane transport diffusion passive transport active transport vesicular transport 4.1 \| Studying Cells A cell is the smallest unit of a living thing. A living thing, whether made of one cell (like bacteria) or many cells (like a human), is called an organism. Thus, cells are the basic building blocks of all organisms. Several cells of one kind that interconnect with each other and perform a shared function form tissues, several tissues combine to form an organ (your stomach, heart, or brain), and several organs make up an organ system (such as the digestive system, circulatory system, or nervous system). Several systems that function together form an organism (like a human being). Here, we will examine the structure and function of cells. There are many types of cells, all grouped into one of two broad categories: prokaryotic and eukaryotic. For example, both animal and plant cells are classified as eukaryotic cells, whereas bacterial cells are classified as prokaryotic. Before discussing the criteria for determining whether a cell is prokaryotic or eukaryotic, let’s first examine how biologists study cells. Microscopy Cells vary in size. With few exceptions, individual cells cannot be seen with the naked eye, so scientists use microscopes (micro- = “small”; -scope = “to look at”) to study them. A microscope is an instrument that magnifies an object. Most photographs of cells are taken with a microscope, and these images can also be called micrographs.The optics of a microscope’s lenses change the orientation of the image that the user sees. A specimen that is right-side up and facing right on the microscope slide will appear upside-down and facing left when viewed through a microscope, and vice versa. Similarly, if the slide is moved left while looking through the microscope, it will appear to move right, and if moved down, it will seem to move up. This occurs because microscopes use two sets of lenses to magnify the image. Because of the manner by which light travels through the lenses, this system of two lenses produces an inverted image (binocular, or dissecting microscopes, work in a similar manner, but include an additional magnification system that makes the final image appear to be upright). Light Microscopes To give you a sense of cell size, a typical human red blood cell is about eight millionths of a meter or eight micrometers (abbreviated as eight μm) in diameter; the head of a pin of is about two thousandths of a meter (two mm) in diameter. That means about 250 red blood cells could fit on the head of a pin.Most student microscopes are classified as light microscopes (Figure 4.2a). Visible light passes and is bent through the lens system to enable the user to see the specimen. Light microscopes are advantageous for viewing living organisms, but since individual cells are generally transparent, their components are not distinguishable unless they are colored with special stains. Staining, however, usually kills the cells.Light microscopes commonly used in the undergraduate college laboratory magnify up to approximately 400 times. Two parameters that are important in microscopy are magnification and resolving power. Magnification is the process of enlarging an object in appearance. Resolving power is the ability of a microscope to distinguish two adjacent structures as separate: the higher the resolution, the better the clarity and detail of the image. When oil immersion lenses are used for the study of small objects, magnification is usually increased to 1,000 times. In order to gain a better understanding of cellular structure and function, scientists typically use electron microscopes. Figure 4.2 (a) Most light microscopes used in a college biology lab can magnify cells up to approximately 400 times and have a resolution of about 200 nanometers. (b) Electron microscopes provide a much higher magnification, 100,000x, and a have a resolution of 50 picometers. (credit a: modification of work by “GcG”/Wikimedia Commons; credit b: modification of work by Evan Bench) Electron Microscopes In contrast to light microscopes, electron microscopes (Figure 4.2b) use a beam of electrons instead of a beam of light. Not only does this allow for higher magnification and, thus, more detail (Figure 4.3), it also provides higher resolving power. The method used to prepare the specimen for viewing with an electron microscope kills the specimen. Electrons have short wavelengths (shorter than photons) that move best in a vacuum, so living cells cannot be viewed with an electron microscope. In a scanning electron microscope, a beam of electrons moves back and forth across a cell’s surface, creating details of cell surface characteristics. In a transmission electron microscope, the electron beam penetrates the cell and provides details of a cell’s internal structures. As you might imagine, electron microscopes are significantly more bulky and expensive than light microscopes. Figure 4.3 (a) These Salmonella bacteria appear as tiny purple dots when viewed with a light microscope. (b) This scanning electron microscope micrograph shows Salmonella bacteria (in red) invading human cells (yellow). Even though subfigure (b) shows a different Salmonella specimen than subfigure (a), you can still observe the comparative increase in magnification and detail. (credit a: modification of work by CDC/Armed Forces Institute of Pathology, Charles N. Farmer, Rocky Mountain Laboratories; credit b: modification of work by NIAID, NIH; scale-bar data from Matt Russell) For another perspective on cell size, try the How Big interactive at this site( . Cell Theory The microscopes we use today are far more complex than those used in the 1600s by Antony van Leeuwenhoek, a Dutch shopkeeper who had great skill in crafting lenses. Despite the limitations of his now-ancient lenses, van Leeuwenhoek observed the movements of protista (a type of single-celled organism) and sperm, which he collectively termed “animalcules.” In a 1665 publication called Micrographia, experimental scientist Robert Hooke coined the term “cell” for the box-like structures he observed when viewing cork tissue through a lens. In the 1670s, van Leeuwenhoek discovered bacteria and protozoa. Later advances in lenses, microscope construction, and staining techniques enabled other scientists to see some components inside cells. By the late 1830s, botanist Matthias Schleiden and zoologist Theodor Schwann were studying tissues and proposed the unified cell theory, which states that all living things are composed of one or more cells, the cell is the basic unit of life, and new cells arise from existing cells. Rudolf Virchow later made important contributions to this theory. 4.2 \| Prokaryotic Cells Cells fall into one of two broad categories: prokaryotic and eukaryotic. Only the predominantly single celled organisms of the domains Bacteria and Archaea are classified as prokaryotes (pro- = “before”; -kary- = “nucleus”). Cells of animals, plants, fungi, and protists are all eukaryotes (ceu- = “true”) and are made up of eukaryotic cells. Components of Prokaryotic Cells All cells share four common components: 1) a plasma membrane, an outer covering that separates the cell’s interior from its surrounding environment; 2) cytoplasm, consisting of a jelly-like cytosol within the cell in which other cellular components are found; 3) DNA, the genetic material of the cell; and 4) ribosomes, which synthesize proteins. However, prokaryotes differ from eukaryotic cells in several ways. A prokaryote is a simple, mostly single-celled (unicellular) organism that lacks a nucleus, or any other membrane-bound organelle. We will shortly come to see that this is significantly different in eukaryotes. Prokaryotic DNA is found in a central part of the cell: the nucleoid (Figure 4.5). Figure 4.5 This figure shows the generalized structure of a prokaryotic cell. All prokaryotes have chromosomal DNA localized in a nucleoid, ribosomes, a cell membrane, and a cell wall. The other structures shown are present in some, but not all, bacteria. Most prokaryotes have a peptidoglycan cell wall and many have a polysaccharide capsule (Figure 4.5). The cell wall acts as an extra layer of protection, helps the cell maintain its shape, and prevents dehydration. The capsule enables the cell to attach to surfaces in its environment. Some prokaryotes have flagella, pili, or fimbriae. Flagella are used for locomotion. Pili are used to exchange genetic material during a type of reproduction called conjugation. Fimbriae are used by bacteria to attach to a host cell. Cell Size At 0.1 to 5.0 μm in diameter, prokaryotic cells are significantly smaller than eukaryotic cells, which have diameters ranging from 10 to 100 μm (Figure 4.6). The small size of prokaryotes allows ions and organic molecules that enter them to quickly diffuse to other parts of the cell. Similarly, any wastes produced within a prokaryotic cell can quickly diffuse out. This is not the case in eukaryotic cells, which have developed different structural adaptations to enhance intracellular transport. Figure 4.6 This figure shows relative sizes of microbes on a logarithmic scale (recall that each unit of increase in a logarithmic scale represents a 10-fold increase in the quantity being measured). Small size, in general, is necessary for all cells, whether prokaryotic or eukaryotic. Let’s examine why that is so. First, we’ll consider the area and volume of a typical cell. Not all cells are spherical in shape, but most tend to approximate a sphere. You may remember from your high school geometry course that the formula for the surface area of a sphere is 4πr2, while the formula for its volume is 4πr3/3. Thus, as the radius of a cell increases, its surface area increases as the square of its radius, but its volume increases as the cube of its radius (much more rapidly). Therefore, as a cell increases in size, its surface area-to-volume ratio decreases. This same principle would apply if the cell had the shape of a cube (Figure 4.7). If the cell grows too large, the plasma membrane will not have sufficient surface area to support the rate of diffusion required for the increased volume. In other words, as a cell grows, it becomes less efficient. One way to become more efficient is to divide; another way is to develop organelles that perform specific tasks. These adaptations lead to the development of more sophisticated cells called eukaryotic cells. 4.3 \| Eukaryotic Cells Have you ever heard the phrase “form follows function?” It’s a philosophy practiced in many industries. In architecture, this means that buildings should be constructed to support the activities that will be carried out inside them. For example, a skyscraper should be built with several elevator banks; a hospital should be built so that its emergency room is easily accessible. Our natural world also utilizes the principle of form following function, especially in cell biology, and this will become clear as we explore eukaryotic cells ( Figure 4.8). Unlike prokaryotic cells, eukaryotic cells have: 1) a membrane-bound nucleus; 2) numerous membrane-bound organelles such as the endoplasmic reticulum, Golgi apparatus, chloroplasts, mitochondria, and others; and 3) several, rod-shaped chromosomes. Because a eukaryotic cell’s nucleus is surrounded by a membrane, it is often said to have a “true nucleus.” The word “organelle” means “little organ,” and, as already mentioned, organelles have specialized cellular functions, just as the organs of your body have specialized functions. At this point, it should be clear to you that eukaryotic cells have a more complex structure than prokaryotic cells. Organelles allow different functions to be compartmentalized in different areas of the cell. Before turning to organelles, let’s first examine two important components of the cell: the plasma membrane and the cytoplasm. The Plasma Membrane Like prokaryotes, eukaryotic cells have a plasma membrane ( Figure 4.9), a phospholipid bilayer with embedded proteins that separates the internal contents of the cell from its surrounding environment. A phospholipid is a lipid molecule with two fatty acid chains and a phosphate-containing group. The plasma membrane controls the passage of organic molecules, ions, water, and oxygen into and out of the cell. Wastes (such as carbon dioxide and ammonia) also leave the cell by passing through the plasma membrane. Figure 4.9 The eukaryotic plasma membrane is a phospholipid bilayer with proteins and cholesterol embedded in it. The plasma membranes of cells that specialize in absorption are folded into fingerlike projections called microvilli (singular = microvillus); ( Figure 4.10). Such cells are typically found lining the small intestine, the organ that absorbs nutrients from digested food. This is an excellent example of form following function. People with celiac disease have an immune response to gluten, which is a protein found in wheat, barley, and rye. The immune response damages microvilli, and thus, afflicted individuals cannot absorb nutrients. This leads to malnutrition, cramping, and diarrhea. Patients suffering from celiac disease must follow a gluten-free diet. Figure 4.10 Microvilli, shown here as they appear on cells lining the small intestine, increase the surface area available for absorption. These microvilli are only found on the area of the plasma membrane that faces the cavity from which substances will be absorbed. (credit “micrograph”: modification of work by Louisa Howard) The Cytoplasm The cytoplasm is the entire region of a cell between the plasma membrane and the nuclear envelope (a structure to be discussed shortly). It is made up of organelles suspended in the gel-like cytosol, the cytoskeleton, and various chemicals ( Figure 4.8). Even though the cytoplasm consists of 70 to 80 percent water, it has a semi-solid consistency, which comes from the proteins within it. However, proteins are not the only organic molecules found in the cytoplasm. Glucose and other simple sugars, polysaccharides, amino acids, nucleic acids, fatty acids, and derivatives of glycerol are found there, too. Ions of sodium, potassium, calcium, and many other elements are also dissolved in the cytoplasm. Many metabolic reactions, including protein synthesis, take place in the cytoplasm. The Nucleus Typically, the nucleus is the most prominent organelle in a cell ( Figure 4.8). The nucleus (plural = nuclei) houses the cell’s DNA and directs the synthesis of ribosomes and proteins. Let’s look at it in more detail ( Figure 4.11). Figure 4.11 The nucleus stores chromatin (DNA plus proteins) in a gel-like substance called the nucleoplasm. The nucleolus is a condensed region of chromatin where ribosome synthesis occurs. The boundary of the nucleus is called the nuclear envelope. It consists of two phospholipid bilayers: an outer membrane and an inner membrane. The nuclear membrane is continuous with the endoplasmic reticulum. Nuclear pores allow substances to enter and exit the nucleus. The Nuclear Envelope The nuclear envelope is a double-membrane structure that constitutes the outermost portion of the nucleus ( Figure 4.11). Both the inner and outer membranes of the nuclear envelope are phospholipid bilayers.The nuclear envelope is punctuated with pores that control the passage of ions, molecules, and RNA between the nucleoplasm and cytoplasm. The nucleoplasm is the semi-solid fluid inside the nucleus, where we find the chromatin and the nucleolus. Chromatin and Chromosomes To understand chromatin, it is helpful to first consider chromosomes. Chromosomes are structures within the nucleus that are made up of DNA, the hereditary material. You may remember that in prokaryotes, DNA is organized into a single circular chromosome. In eukaryotes, chromosomes are linear structures. Every eukaryotic species has a specific number of chromosomes in the nuclei of its body’s cells. For example, in humans, the chromosome number is 46, while in fruit flies, it is eight. Chromosomes are only visible and distinguishable from one another when the cell is getting ready to divide. When the cell is in the growth and maintenance phases of its life cycle, proteins are attached to chromosomes, and they resemble an unwound, jumbled bunch of threads. These unwound protein chromosome complexes are called chromatin ( Figure 4.12); chromatin describes the material that makes up the chromosomes both when condensed and decondensed. (a)(b)Figure 4.12 (a) This image shows various levels of the organization of chromatin (DNA and protein). (b) This image shows paired chromosomes. (credit b: modification of work by NIH; scale-bar data from Matt Russell) The Nucleolus We already know that the nucleus directs the synthesis of ribosomes, but how does it do this? Some chromosomes have sections of DNA that encode ribosomal RNA. A darkly staining area within the nucleus called the nucleolus (plural = nucleoli) aggregates the ribosomal RNA with associated proteins to assemble the ribosomal subunits that are then transported out through the pores in the nuclear envelope to the cytoplasm. Ribosomes Ribosomes are the cellular structures responsible for protein synthesis. When viewed through an electron microscope, ribosomes appear either as clusters (polyribosomes) or single, tiny dots that float freely in the cytoplasm. They may be attached to the cytoplasmic side of the plasma membrane or the cytoplasmic side of the endoplasmic reticulum and the outer membrane of the nuclear envelope ( Figure 4.8). Electron microscopy has shown us that ribosomes, which are large complexes of protein and RNA, consist of two subunits, aptly called large and small ( Figure 4.13). Ribosomes receive their “orders” for protein synthesis from the nucleus where the DNA is transcribed into messenger RNA (mRNA). The mRNA travels to the ribosomes, which translate the code provided by the sequence of the nitrogenous bases in the mRNA into a specific order of amino acids in a protein. Amino acids are the building blocks of proteins. Figure 4.13 Ribosomes are made up of a large subunit (top) and a small subunit (bottom). During protein synthesis, ribosomes assemble amino acids into proteins. Because protein synthesis is an essential function of all cells (including enzymes, hormones, antibodies, pigments, structural components, and surface receptors), ribosomes are found in practically every cell. Ribosomes are particularly abundant in cells that synthesize large amounts of protein. For example, the pancreas is responsible for creating several digestive enzymes and the cells that produce these enzymes contain many ribosomes. Thus, we see another example of form following function. Mitochondria Mitochondria (singular = mitochondrion) are often called the “powerhouses” or “energy factories” of a cell because they are responsible for making adenosine triphosphate (ATP), the cell’s main energy carrying molecule. ATP represents the short-term stored energy of the cell. Cellular respiration is the process of making ATP using the chemical energy found in glucose and other nutrients. In mitochondria, this process uses oxygen and produces carbon dioxide as a waste product. In fact, the carbon dioxide that you exhale with every breath comes from the cellular reactions that produce carbon dioxide as a byproduct. In keeping with our theme of form following function, it is important to point out that muscle cells have a very high concentration of mitochondria that produce ATP. Your muscle cells need a lot of energy to keep your body moving. When your cells don’t get enough oxygen, they do not make a lot of ATP. Instead, the small amount of ATP they make in the absence of oxygen is accompanied by the production of lactic acid. Mitochondria are oval-shaped, double membrane organelles ( Figure 4.14) that have their own ribosomes and DNA. Each membrane is a phospholipid bilayer embedded with proteins. The inner layer has folds called cristae. The area surrounded by the folds is called the mitochondrial matrix. The cristae and the matrix have different roles in cellular respiration. Figure 4.14 This electron micrograph shows a mitochondrion as viewed with a transmission electron microscope. This organelle has an outer membrane and an inner membrane. The inner membrane contains folds, called cristae, which increase its surface area. The space between the two membranes is called the intermembrane space, and the space inside the inner membrane is called the mitochondrial matrix. ATP synthesis takes place on the inner membrane. (credit: modification of work by Matthew Britton; scale-bar data from Matt Russell) Peroxisomes Peroxisomes are small, round organelles enclosed by single membranes. They carry out oxidation reactions that break down fatty acids and amino acids. They also detoxify many poisons that may enter the body. (Many of these oxidation reactions release hydrogen peroxide, H2O2, which would be damaging to cells; however, when these reactions are confined to peroxisomes, enzymes safely break down the H2O2 into oxygen and water.) For example, alcohol is detoxified by peroxisomes in liver cells. Glyoxysomes, which are specialized peroxisomes in plants, are responsible for converting stored fats into sugars. Vesicles and Vacuoles Vesicles and vacuoles are membrane-bound sacs that function in storage and transport. Other than the fact that vacuoles are somewhat larger than vesicles, there is a very subtle distinction between them: The membranes of vesicles can fuse with either the plasma membrane or other membrane systems within the cell. Additionally, some agents such as enzymes within plant vacuoles break down macromolecules. The membrane of a vacuole does not fuse with the membranes of other cellular components. Animal Cells versus Plant Cells At this point, you know that each eukaryotic cell has a plasma membrane, cytoplasm, a nucleus, ribosomes, mitochondria, peroxisomes, and in some, vacuoles, but there are some striking differences between animal and plant cells. While both animal and plant cells have microtubule organizing centers (MTOCs), animal cells also have centrioles associated with the MTOC: a complex called the centrosome. Animal cells each have a centrosome and lysosomes, whereas plant cells do not. Plant cells have a cell wall, chloroplasts and other specialized plastids, and a large central vacuole, whereas animal cells do not. The Centrosome The centrosome is a microtubule-organizing center found near the nuclei of animal cells. It contains a pair of centrioles, two structures that lie perpendicular to each other ( Figure 4.15). Each centriole is a cylinder of nine triplets of microtubules. Figure 4.15 The centrosome consists of two centrioles that lie at right angles to each other. Each centriole is a cylinder made up of nine triplets of microtubules. Non-tubulin proteins (indicated by the green lines) hold the microtubule triplets together. The centrosome (the organelle where all microtubules originate) replicates itself before a cell divides, and the centrioles appear to have some role in pulling the duplicated chromosomes to opposite ends of the dividing cell. However, the exact function of the centrioles in cell division isn’t clear, because cells that have had the centrosome removed can still divide, and plant cells, which lack centrosomes, are capable of cell division. Lysosomes Animal cells have another set of organelles not found in plant cells: lysosomes. The lysosomes are the cell’s “garbage disposal.” In plant cells, the digestive processes take place in vacuoles. Enzymes within the lysosomes aid the breakdown of proteins, polysaccharides, lipids, nucleic acids, and even worn-out organelles. These enzymes are active at a much lower pH than that of the cytoplasm. Therefore, the pH within lysosomes is more acidic than the pH of the cytoplasm. Many reactions that take place in the cytoplasm could not occur at a low pH, so again, the advantage of compartmentalizing the eukaryotic cell into organelles is apparent. The Cell Wall If you examine Figure 4.8b, the diagram of a plant cell, you will see a structure external to the plasma membrane called the cell wall. The cell wall is a rigid covering that protects the cell, provides structural support, and gives shape to the cell. Fungal and protistan cells also have cell walls. While the chief component of prokaryotic cell walls is peptidoglycan, the major organic molecule in the plant cell wall is cellulose ( Figure 4.16), a polysaccharide made up of glucose units. Have you ever noticed that when you bite into a raw vegetable, like celery, it crunches? That’s because you are tearing the rigid cell walls of the celery cells with your teeth. Figure 4.16 Cellulose is a long chain of β-glucose molecules connected by a 1-4 linkage. The dashed lines at each end of the figure indicate a series of many more glucose units. The size of the page makes it impossible to portray an entire cellulose molecule. Chloroplasts Like the mitochondria, chloroplasts have their own DNA and ribosomes, but chloroplasts have an entirely different function. Chloroplasts are plant cell organelles that carry out photosynthesis. Photosynthesis is the series of reactions that use carbon dioxide, water, and light energy to make glucose and oxygen. This is a major difference between plants and animals; plants (autotrophs) are able to make their own food, like sugars, while animals (heterotrophs) must ingest their food. Like mitochondria, chloroplasts have outer and inner membranes, but within the space enclosed by a chloroplast’s inner membrane is a set of interconnected and stacked fluid-filled membrane sacs called thylakoids ( Figure 4.17). Each stack of thylakoids is called a granum (plural = grana). The fluid enclosed by the inner membrane that surrounds the grana is called the stroma. Figure 4.17 The chloroplast has an outer membrane, an inner membrane, and membrane structures called thylakoids that are stacked into grana. The space inside the thylakoid membranes is called the thylakoid space. The light harvesting reactions take place in the thylakoid membranes, and the synthesis of sugar takes place in the fluid inside the inner membrane, which is called the stroma. Chloroplasts also have their own genome, which is contained on a single circular chromosome. The chloroplasts contain a green pigment called chlorophyll, which captures the light energy that drives the reactions of photosynthesis. Like plant cells, photosynthetic protists also have chloroplasts. Some bacteria perform photosynthesis, but their chlorophyll is not relegated to an organelle. The Central Vacuole Previously, we mentioned vacuoles as essential components of plant cells. If you look at Figure 4.8b, you will see that plant cells each have a large central vacuole that occupies most of the area of the cell. The central vacuole plays a key role in regulating the cell’s concentration of water in changing environmental conditions. Have you ever noticed that if you forget to water a plant for a few days, it wilts? That’s because as the water concentration in the soil becomes lower than the water concentration in the plant, water moves out of the central vacuoles and cytoplasm. As the central vacuole shrinks, it leaves the cell wall unsupported. This loss of support to the cell walls of plant cells results in the wilted appearance of the plant. The central vacuole also supports the expansion of the cell. When the central vacuole holds more water, the cell gets larger without having to invest a lot of energy in synthesizing new cytoplasm. 4.4 \| The Endomembrane System and Proteins The endomembrane system (endo = “within”) is a group of membranes and organelles (Figure 4.18) in eukaryotic cells that works together to modify, package, and transport lipids and proteins. It includes the nuclear envelope, lysosomes, and vesicles, which we’ve already mentioned, and the endoplasmic reticulum and Golgi apparatus, which we will cover shortly. Although not technically within the cell, the plasma membrane is included in the endomembrane system because, as you will see, it interacts with the other endomembranous organelles. The endomembrane system does not include the membranes of either mitochondria or chloroplasts. The Endoplasmic Reticulum The endoplasmic reticulum (ER) (Figure 4.18) is a series of interconnected membranous sacs and tubules that collectively modifies proteins and synthesizes lipids. However, these two functions are performed in separate areas of the ER: the rough ER and the smooth ER, respectively. The hollow portion of the ER tubules is called the lumen or cisternal space. The membrane of the ER, which is a phospholipid bilayer embedded with proteins, is continuous with the nuclear envelope. Rough ER The rough endoplasmic reticulum (RER) is so named because the ribosomes attached to its cytoplasmic surface give it a studded appearance when viewed through an electron microscope (Figure 4.19). Figure 4.19 This transmission electron micrograph shows the rough endoplasmic reticulum and other organelles in a pancreatic cell. (credit: modification of work by Louisa Howard) Ribosomes transfer their newly synthesized proteins into the lumen of the RER where they undergo structural modifications, such as folding or the acquisition of side chains. These modified proteins will be incorporated into cellular membranes—the membrane of the ER or those of other organelles—or secreted from the cell (such as protein hormones, enzymes). The RER also makes phospholipids for cellular membranes. If the phospholipids or modified proteins are not destined to stay in the RER, they will reach their destinations via transport vesicles that bud from the RER’s membrane (Figure 4.18). Since the RER is engaged in modifying proteins (such as enzymes, for example) that will be secreted from the cell, you would be correct in assuming that the RER is abundant in cells that secrete proteins. This is the case with cells of the liver, for example. Smooth ER The smooth endoplasmic reticulum (SER) is continuous with the RER but has few or no ribosomes on its cytoplasmic surface (Figure 4.18). Functions of the SER include synthesis of carbohydrates, lipids, and steroid hormones; detoxification of medications and poisons; and storage of calcium ions. In muscle cells, a specialized SER called the sarcoplasmic reticulum is responsible for storage of the calcium ions that are needed to trigger the coordinated contractions of the muscle cells. You can watch an excellent animation of the endomembrane system here ( l/endomembrane) . At the end of the animation, there is a short self-assessment. The Golgi Apparatus We have already mentioned that vesicles can bud from the ER and transport their contents elsewhere, but where do the vesicles go? Before reaching their final destination, the lipids or proteins within the transport vesicles still need to be sorted, packaged, and tagged so that they wind up in the right place. Sorting, tagging, packaging, and distribution of lipids and proteins takes place in the Golgi apparatus (also called the Golgi body), a series of flattened membranes (Figure 4.20). Figure 4.20 The Golgi apparatus in this white blood cell is visible as a stack of semicircular, flattened rings in the lower portion of the image. Several vesicles can be seen near the Golgi apparatus. (credit: modification of work by Louisa Howard) The receiving side of the Golgi apparatus is called the cis face. The opposite side is called the trans face. The transport vesicles that formed from the ER travel to the cis face, fuse with it, and empty their contents into the lumen of the Golgi apparatus. As the proteins and lipids travel through the Golgi, they undergo further modifications that allow them to be sorted. The most frequent modification is the addition of short chains of sugar molecules. These newly modified proteins and lipids are then tagged with phosphate groups or other small molecules so that they can be routed to their proper destinations. Finally, the modified and tagged proteins are packaged into secretory vesicles that bud from the trans face of the Golgi. While some of these vesicles deposit their contents into other parts of the cell where they will be used, other secretory vesicles fuse with the plasma membrane and release their contents outside the cell. In another example of form following function, cells that engage in a great deal of secretory activity (such as cells of the salivary glands that secrete digestive enzymes or cells of the immune system that secrete antibodies) have an abundance of Golgi. In plant cells, the Golgi apparatus has the additional role of synthesizing polysaccharides, some of which are incorporated into the cell wall and some of which are used in other parts of the cell. Lysosomes In addition to their role as the digestive component and organelle-recycling facility of animal cells, lysosomes are considered to be parts of the endomembrane system. Lysosomes also use their hydrolytic enzymes to destroy pathogens (disease-causing organisms) that might enter the cell. A good example of this occurs in a group of white blood cells called macrophages, which are part of your body’s immune system. In a process known as phagocytosis or endocytosis, a section of the plasma membrane of the macrophage invaginates (folds in) and engulfs a pathogen. The invaginated section, with the pathogen inside, then pinches itself off from the plasma membrane and becomes a vesicle. The vesicle fuses with a lysosome. The lysosome’s hydrolytic enzymes then destroy the pathogen (Figure 4.21). Figure 4.21 A macrophage has engulfed (phagocytized) a potentially pathogenic bacterium and then fuses with a lysosomes within the cell to destroy the pathogen. Other organelles are present in the cell but for simplicity are not shown. 4.5 \| The Cytoskeleton If you were to remove all the organelles from a cell, would the plasma membrane and the cytoplasm be the only components left? No. Within the cytoplasm, there would still be ions and organic molecules, plus a network of protein fibers that help maintain the shape of the cell, secure some organelles in specific positions, allow cytoplasm and vesicles to move within the cell, and enable cells within multicellular organisms to move. Collectively, this network of protein fibers is known as the cytoskeleton. There are three types of fibers within the cytoskeleton: microfilaments, intermediate filaments, and microtubules (Figure 4.22). Here, we will examine each. Figure 4.22 Microfilaments thicken the cortex around the inner edge of a cell; like rubber bands, they resist tension. Microtubules are found in the interior of the cell where they maintain cell shape by resisting compressive forces. Intermediate filaments are found throughout the cell and hold organelles in place. Microfilaments Of the three types of protein fibers in the cytoskeleton, microfilaments are the narrowest. They function in cellular movement, have a diameter of about 7 nm, and are made of two intertwined strands of a globular protein called actin (Figure 4.23). For this reason, microfilaments are also known as actin filaments. Figure 4.23 Microfilaments are made of two intertwined strands of actin. Actin is powered by ATP to assemble its filamentous form, which serves as a track for the movement of a motor protein called myosin. This enables actin to engage in cellular events requiring motion, such as cell division in animal cells and cytoplasmic streaming, which is the circular movement of the cell cytoplasm in plant cells. Actin and myosin are plentiful in muscle cells. When your actin and myosin filaments slide past each other, your muscles contract. Microfilaments also provide some rigidity and shape to the cell. They can depolymerize (disassemble) and reform quickly, thus enabling a cell to change its shape and move. White blood cells (your body’s infection-fighting cells) make good use of this ability. They can move to the site of an infection and phagocytize the pathogen. To see an example of a white blood cell in action, click here ( chasing_bcteria) and watch a short time-lapse video of the cell capturing two bacteria. It engulfs one and then moves on to the other. Intermediate Filaments Intermediate filaments are made of several strands of fibrous proteins that are wound together (Figure 4.24). These elements of the cytoskeleton get their name from the fact that their diameter, 8 to 10 nm, is between those of microfilaments and microtubules. Figure 4.24 Intermediate filaments consist of several intertwined strands of fibrous proteins. Intermediate filaments have no role in cell movement. Their function is purely structural. They bear tension, thus maintaining the shape of the cell, and anchor the nucleus and other organelles in place. Figure 4.22 shows how intermediate filaments create a supportive scaffolding inside the cell. The intermediate filaments are the most diverse group of cytoskeletal elements. Several types of fibrous proteins are found in the intermediate filaments. You are probably most familiar with keratin, the fibrous protein that strengthens your hair, nails, and the epidermis of the skin. Microtubules As their name implies, microtubules are small hollow tubes. The walls of the microtubule are made of polymerized dimers of α-tubulin and β-tubulin, two globular proteins (Figure 4.25). With a diameter of about 25 nm, microtubules are the widest components of the cytoskeleton. They help the cell resist compression, provide a track along which vesicles move through the cell, and pull replicated chromosomes to opposite ends of a dividing cell. Like microfilaments, microtubules can dissolve and reform quickly. Figure 4.25 Microtubules are hollow. Their walls consist of 13 polymerized dimers of α-tubulin and β-tubulin (right image). The left image shows the molecular structure of the tube. Microtubules are also the structural elements of flagella, cilia, and centrioles (the latter are the two perpendicular bodies of the centrosome). In fact, in animal cells, the centrosome is the microtubule organizing center. In eukaryotic cells, flagella and cilia are quite different structurally from their counterparts in prokaryotes, as discussed below. Flagella and Cilia To refresh your memory, flagella (singular = flagellum) are long, hair-like structures that extend from the plasma membrane and are used to move an entire cell (for example, sperm, Euglena). When present, the cell has just one flagellum or a few flagella. When cilia (singular = cilium) are present, however, many of them extend along the entire surface of the plasma membrane. They are short, hair-like structures that are used to move entire cells (such as paramecia) or substances along the outer surface of the cell (for example, the cilia of cells lining the Fallopian tubes that move the ovum toward the uterus, or cilia lining the cells of the respiratory tract that trap particulate matter and move it toward your nostrils.) Despite their differences in length and number, flagella and cilia share a common structural arrangement of microtubules called a “9 + 2 array.” This is an appropriate name because a single flagellum or cilium is made of a ring of nine microtubule doublets, surrounding a single microtubule doublet in the center (Figure 4.26). Figure 4.26 This transmission electron micrograph of two flagella shows the 9 + 2 array of microtubules: nine microtubule doublets surround a single microtubule doublet. (credit: modification of work by Dartmouth Electron Microscope Facility, Dartmouth College; scale-bar data from Matt Russell) You have now completed a broad survey of the components of prokaryotic and eukaryotic cells. For a summary of cellular components in prokaryotic and eukaryotic cells, see Table 4.1. Components of Prokaryotic and Eukaryotic Cells \| \| \| \| \| \| --- --- \| Cell Component \| Present in Prokaryotes Function \| \| Present in Animal Cells? \| Present in Plant Cells? \| \| Plasma membrane \| Separates cell from external environment; controls passage of organic molecules, ions, water, oxygen, and wastes into and out of cell \| Yes \| Yes \| Yes \| \| Cytoplasm \| Provides turgor pressure to plant cells as fluid inside the central vacuole; site of many metabolic reactions; medium in which organelles are found \| Yes \| Yes \| Yes \| \| Nucleolus \| Darkened area within the nucleus where ribosomal subunits are synthesized. \| No \| Yes \| Yes \| \| Nucleus \| Cell organelle that houses DNA and directs synthesis of ribosomes and proteins \| No \| Yes \| Yes \| \| Ribosomes \| Protein synthesis \| Yes \| Yes \| Yes \| \| Mitochondria \| ATP production/cellular respiration \| No \| Yes \| Yes \| \| Peroxisomes \| Oxidizes and thus breaks down fatty acids and amino acids, and detoxifies poisons \| No \| Yes \| Yes \| \| Vesicles and vacuoles \| Storage and transport; digestive function in plant cells \| No \| Yes \| Yes \| \| Centrosome \| Unspecified role in cell division in animal cells; source of microtubules in animal cells \| No \| Yes \| No \| \| Lysosomes \| Digestion of macromolecules; recycling of worn-out organelles \| No \| Yes \| No \| \| Cell wall \| Protection, structural support and maintenance of cell shape \| Yes, primarily peptidoglycan \| No \| Yes, primarily cellulose \| \| Chloroplasts \| Photosynthesis \| No \| No \| Yes \| \| Endoplasmic reticulum \| Modifies proteins and synthesizes lipids \| No \| Yes \| Yes \| \| Golgi apparatus \| Modifies, sorts, tags, packages, and distributes lipids and proteins \| No \| Yes \| Yes \| \| \| \| \| \| \| Table 4.1 Components of Prokaryotic and Eukaryotic Cells \| \| \| \| \| \| --- --- \| Cell Component \| Function \| Present in Prokaryotes? \| Present in Animal Cells? \| Present in Plant Cells? \| \| Cytoskeleton \| Maintains cell’s shape, secures organelles in specific positions, allows cytoplasm and vesicles to move within cell, and enables unicellular organisms to move independently \| Yes \| Yes \| Yes \| \| Flagella \| Cellular locomotion \| Some \| Some \| No, except for some plant sperm cells. \| \| Cilia \| Cellular locomotion, movement of particles along extracellular surface of plasma membrane, and filtration \| Some \| Some \| No \| \| \| \| \| \| \| Table 4.1 4.6 \| Connections between Cells and Cellular Activities You already know that a group of similar cells working together is called a tissue. As you might expect, if cells are to work together, they must communicate with each other, just as you need to communicate with others if you work on a group project. Let’s take a look at how cells communicate with each other. Extracellular Matrix of Animal Cells Most animal cells release materials into the extracellular space. The primary components of these materials are proteins, and the most abundant protein is collagen. Collagen fibers are interwoven with carbohydrate-containing protein molecules called proteoglycans. Collectively, these materials are called the extracellular matrix (Figure 4.27). Not only does the extracellular matrix hold the cells together to form a tissue, but it also allows the cells within the tissue to communicate with each other. How can this happen? Figure 4.27 The extracellular matrix consists of a network of proteins and carbohydrates. Cells have protein receptors on the extracellular surfaces of their plasma membranes. When a molecule within the matrix binds to the receptor, it changes the molecular structure of the receptor. The receptor, in turn, changes the conformation of the microfilaments positioned just inside the plasma membrane. These conformational changes induce chemical signals inside the cell that reach the nucleus and turn “on” or “off” the transcription of specific sections of DNA, which affects the production of associated proteins, thus changing the activities within the cell. Blood clotting provides an example of the role of the extracellular matrix in cell communication. When the cells lining a blood vessel are damaged, they display a protein receptor called tissue factor. When tissue factor binds with another factor in the extracellular matrix, it causes platelets to adhere to the wall of the damaged blood vessel, stimulates the adjacent smooth muscle cells in the blood vessel to contract (thus constricting the blood vessel), and initiates a series of steps that stimulate the platelets to produce clotting factors. Intercellular Junctions Cells can also communicate with each other via direct contact, referred to as intercellular junctions. There are some differences in the ways that plant and animal cells do this. Plasmodesmata are junctions between plant cells, whereas animal cell contacts include tight junctions, gap junctions, and desmosomes. Plasmodesmata In general, long stretches of the plasma membranes of neighboring plant cells cannot touch one another because they are separated by the cell wall that surrounds each cell (Figure 4.8b). How then, can a plant transfer water and other soil nutrients from its roots, through its stems, and to its leaves? Such transport uses the vascular tissues (xylem and phloem) primarily. There also exist structural modifications called plasmodesmata (singular = plasmodesma), numerous channels that pass between cell walls of adjacent plant cells, connect their cytoplasm, and enable materials to be transported from cell to cell, and thus throughout the plant (Figure 4.28). Figure 4.28 A plasmodesma is a channel between the cell walls of two adjacent plant cells. Plasmodesmata allow materials to pass from the cytoplasm of one plant cell to the cytoplasm of an adjacent cell. Tight Junctions A tight junction is a watertight seal between two adjacent animal cells (Figure 4.29). The cells are held tightly against each other by proteins (predominantly two proteins called claudins and occludins). Figure 4.29 Tight junctions form watertight connections between adjacent animal cells. Proteins create tight junction adherence. (credit: modification of work by Mariana Ruiz Villareal) This tight adherence prevents materials from leaking between the cells; tight junctions are typically found in epithelial tissues that line internal organs and cavities, and comprise most of the skin. For example, the tight junctions of the epithelial cells lining your urinary bladder prevent urine from leaking out into the extracellular space. Desmosomes Also found only in animal cells are desmosomes, which act like spot welds between adjacent epithelial cells (Figure 4.30). Short proteins called cadherins in the plasma membrane connect to intermediate filaments to create desmosomes. The cadherins join two adjacent cells together and maintain the cells in a sheet-like formation in organs and tissues that stretch, like the skin, heart, and muscles. Figure 4.30 A desmosome forms a very strong spot weld between cells. It is created by the linkage of cadherins and intermediate filaments. (credit: modification of work by Mariana Ruiz Villareal) Gap Junctions Gap junctions in animal cells are like plasmodesmata in plant cells in that they are channels between adjacent cells that allow for the transport of ions, nutrients, and other substances that enable cells to communicate (Figure 4.31). Structurally, however, gap junctions and plasmodesmata differ. Figure 4.31 A gap junction is a protein-lined pore that allows water and small molecules to pass between adjacent animal cells. (credit: modification of work by Mariana Ruiz Villareal) Gap junctions develop when a set of six proteins (called connexins) in the plasma membrane arrange themselves in an elongated donut-like configuration called a connexon. When the pores (“doughnut holes”) of connexons in adjacent animal cells align, a channel between the two cells forms. Gap junctions are particularly important in cardiac muscle: The electrical signal for the muscle to contract is passed efficiently through gap junctions, allowing the heart muscle cells to contract in tandem. To conduct a virtual microscopy lab and review the parts of a cell, work through the steps of this interactive assignment ( . Figure 4.32 Despite its seeming hustle and bustle, Grand Central Station functions with a high level of organization: People and objects move from one location to another, they cross or are contained within certain boundaries, and they provide a constant flow as part of larger activity. Analogously, a plasma membrane’s functions involve movement within the cell and across boundaries in the process of intracellular and intercellular activities. (credit: modification of work by Randy Le’Moine) 4.7 \| Plasma Membrane Components and Structure A cell’s plasma membrane defines the cell, outlines its borders, and determines the nature of its interaction with its environment (see Table 4.2 for a summary). Cells exclude some substances, take in others, and excrete still others, all in controlled quantities. The plasma membrane must be very flexible to allow certain cells, such as red blood cells and white blood cells, to change shape as they pass through narrow capillaries. These are the more obvious functions of a plasma membrane. In addition, the surface of the plasma membrane carries markers that allow cells to recognize one another, which is vital for tissue and organ formation during early development, and which later plays a role in the “self” versus “non-self” distinction of the immune response. Among the most sophisticated functions of the plasma membrane is the ability to transmit signals by means of complex, integral proteins known as receptors. These proteins act both as receivers of extracellular inputs and as activators of intracellular processes. These membrane receptors provide extracellular attachment sites for effectors like hormones and growth factors, and they activate intracellular response cascades when their effectors are bound. Occasionally, receptors are hijacked by viruses (HIV, human immunodeficiency virus, is one example) that use them to gain entry into cells, and at times, the genes encoding receptors become mutated, causing the process of signal transduction to malfunction with disastrous consequences. Fluid Mosaic Model The existence of the plasma membrane was identified in the 1890s, and its chemical components were identified in 1915. The principal components identified at that time were lipids and proteins. The first widely accepted model of the plasma membrane’s structure was proposed in 1935 by Hugh Davson and James Danielli; it was based on the “railroad track” appearance of the plasma membrane in early electron micrographs. They theorized that the structure of the plasma membrane resembles a sandwich, with protein being analogous to the bread, and lipids being analogous to the filling. In the 1950s, advances in microscopy, notably transmission electron microscopy (TEM), allowed researchers to see that the core of the plasma membrane consisted of a double, rather than a single, layer. A new model that better explains both the microscopic observations and the function of that plasma membrane was proposed by S.J. Singer and Garth L. Nicolson in 1972. The explanation proposed by Singer and Nicolson is called the fluid mosaic model. The model has evolved somewhat over time, but it still best accounts for the structure and functions of the plasma membrane as we now understand them. The fluid mosaic model describes the structure of the plasma membrane as a mosaic of components—including phospholipids, cholesterol, proteins, and carbohydrates—that gives the membrane a fluid character. Plasma membranes range from 5 to 10 nm in thickness. For comparison, human red blood cells, visible via light microscopy, are approximately 8 µm wide, or approximately 1,000 times wider than a plasma membrane. The membrane does look a bit like a sandwich (Figure 4.33). Figure 4.33 The fluid mosaic model of the plasma membrane describes the plasma membrane as a fluid combination of phospholipids, cholesterol, and proteins. Carbohydrates attached to lipids (glycolipids) and to proteins (glycoproteins) extend from the outward-facing surface of the membrane. The principal components of a plasma membrane are lipids (phospholipids and cholesterol), proteins, and carbohydrates attached to some of the lipids and some of the proteins. A phospholipid is a molecule consisting of glycerol, two fatty acids, and a phosphate-linked head group. Cholesterol, another lipid composed of four fused carbon rings, is found alongside the phospholipids in the core of the membrane. The proportions of proteins, lipids, and carbohydrates in the plasma membrane vary with cell type, but for a typical human cell, protein accounts for about 50 percent of the composition by mass, lipids (of all types) account for about 40 percent of the composition by mass, with the remaining 10 percent of the composition by mass being carbohydrates. However, the concentration of proteins and lipids varies with different cell membranes. For example, myelin, an outgrowth of the membrane of specialized cells that insulates the axons of the peripheral nerves, contains only 18 percent protein and 76 percent lipid. The mitochondrial inner membrane contains 76 percent protein and only 24 percent lipid. The plasma membrane of human red blood cells is 30 percent lipid. Carbohydrates are present only on the exterior surface of the plasma membrane and are attached to proteins, forming glycoproteins, or attached to lipids, forming glycolipids. Phospholipids The main fabric of the membrane is composed of amphiphilic, phospholipid molecules. The hydrophilic or “water-loving” areas of these molecules (which look like a collection of balls in an artist’s rendition of the model) (Figure 4.33) are in contact with the aqueous fluid both inside and outside the cell. Hydrophobic, or water-hating molecules, tend to be non-polar. They interact with other non-polar molecules in chemical reactions, but generally do not interact with polar molecules. When placed in water, hydrophobic molecules tend to form a ball or cluster. The hydrophilic regions of the phospholipids tend to form hydrogen bonds with water and other polar molecules on both the exterior and interior of the cell. Thus, the membrane surfaces that face the interior and exterior of the cell are hydrophilic. In contrast, the interior of the cell membrane is hydrophobic and will not interact with water. Therefore, phospholipids form an excellent two-layer cell membrane that separates fluid within the cell from the fluid outside of the cell. A phospholipid molecule (Figure 4.34) consists of a three-carbon glycerol backbone with two fatty acid molecules attached to carbons 1 and 2, and a phosphate-containing group attached to the third carbon. This arrangement gives the overall molecule an area described as its head (the phosphate-containing group), which has a polar character or negative charge, and an area called the tail (the fatty acids), which has no charge. The head can form hydrogen bonds, but the tail cannot. A molecule with this arrangement of a positively or negatively charged area and an uncharged, or non-polar, area is referred to as amphiphilic or “dual-loving.” Figure 4.34 This phospholipid molecule is composed of a hydrophilic head and two hydrophobic tails. The hydrophilic head group consists of a phosphate-containing group attached to a glycerol molecule. The hydrophobic tails, each containing either a saturated or an unsaturated fatty acid, are long hydrocarbon chains. This characteristic is vital to the structure of a plasma membrane because, in water, phospholipids tend to become arranged with their hydrophobic tails facing each other and their hydrophilic heads facing out. In this way, they form a lipid bilayer—a barrier composed of a double layer of phospholipids that separates the water and other materials on one side of the barrier from the water and other materials on the other side. In fact, phospholipids heated in an aqueous solution tend to spontaneously form small spheres or droplets (called micelles or liposomes), with their hydrophilic heads forming the exterior and their hydrophobic tails on the inside (Figure 4.35). Figure 4.34 In an aqueous solution, phospholipids tend to arrange themselves with their polar heads facing outward and their hydrophobic tails facing inward. (credit: modification of work by MarianaRuiz Villareal) Proteins Proteins make up the second major component of plasma membranes. Integral proteins (some specialized types are called integrins) are, as their name suggests, integrated completely into the membrane structure, and their hydrophobic membrane-spanning regions interact with the hydrophobic region of the the phospholipid bilayer (Figure 4.33). Single-pass integral membrane proteins usually have a hydrophobic transmembrane segment that consists of 20–25 amino acids. Some span only part of the membrane—associating with a single layer—while others stretch from one side of the membrane to the other, and are exposed on either side. Some complex proteins are composed of up to 12 segments of a single protein, which are extensively folded and embedded in the membrane (Figure 4.36). This type of protein has a hydrophilic region or regions, and one or several mildly hydrophobic regions. This arrangement of regions of the protein tends to orient the protein alongside the phospholipids, with the hydrophobic region of the protein adjacent to the tails of the phospholipids and the hydrophilic region or regions of the protein protruding from the membrane and in contact with the cytosol or extracellular fluid. Figure 4.36 Integral membranes proteins may have one or more alpha-helices that span the membrane (examples 1 and 2), or they may have beta-sheets that span the membrane (example 3). (credit: “Foobar”/Wikimedia Commons) Peripheral proteins are found on the exterior and interior surfaces of membranes, attached either to integral proteins or to phospholipids. Peripheral proteins, along with integral proteins, may serve as enzymes, as structural attachments for the fibers of the cytoskeleton, or as part of the cell’s recognition sites. These are sometimes referred to as “cell-specific” proteins. The body recognizes its own proteins and attacks foreign proteins associated with invasive pathogens. Carbohydrates Carbohydrates are the third major component of plasma membranes. They are always found on the exterior surface of cells and are bound either to proteins (forming glycoproteins) or to lipids (forming glycolipids) (Figure 4.33). These carbohydrate chains may consist of 2–60 monosaccharide units and can be either straight or branched. Along with peripheral proteins, carbohydrates form specialized sites on the cell surface that allow cells to recognize each other. These sites have unique patterns that allow the cell to be recognized, much the way that the facial features unique to each person allow him or her to be recognized. This recognition function is very important to cells, as it allows the immune system to differentiate between body cells (called “self”) and foreign cells or tissues (called “non-self”). Similar types of glycoproteins and glycolipids are found on the surfaces of viruses and may change frequently, preventing immune cells from recognizing and attacking them. These carbohydrates on the exterior surface of the cell—the carbohydrate components of both glycoproteins and glycolipids—are collectively referred to as the glycocalyx (meaning “sugar coating”). The glycocalyx is highly hydrophilic and attracts large amounts of water to the surface of the cell. This aids in the interaction of the cell with its watery environment and in the cell’s ability to obtain substances dissolved in the water. As discussed above, the glycocalyx is also important for cell identification, self/ non-self determination, and embryonic development, and is used in cell-cell attachments to form tissues Membrane Fluidity The mosaic characteristic of the membrane, described in the fluid mosaic model, helps to illustrate its nature. The integral proteins and lipids exist in the membrane as separate but loosely attached molecules. These resemble the separate, multicolored tiles of a mosaic picture, and they float, moving somewhat with respect to one another. The membrane is not like a balloon, however, that can expand and contract; rather, it is fairly rigid and can burst if penetrated or if a cell takes in too much water. However, because of its mosaic nature, a very fine needle can easily penetrate a plasma membrane without causing it to burst, and the membrane will flow and self-seal when the needle is extracted. The mosaic characteristics of the membrane explain some but not all of its fluidity. There are two other factors that help maintain this fluid characteristic. One factor is the nature of the phospholipids themselves. In their saturated form, the fatty acids in phospholipid tails are saturated with bound hydrogen atoms. There are no double bonds between adjacent carbon atoms. This results in tails that are relatively straight. In contrast, unsaturated fatty acids do not contain a maximal number of hydrogen atoms, but they do contain some double bonds between adjacent carbon atoms; a double bond results in a bend in the string of carbons of approximately 30 degrees (Figure 4.34). Thus, if saturated fatty acids, with their straight tails, are compressed by decreasing temperatures, they press in on each other, making a dense and fairly rigid membrane. If unsaturated fatty acids are compressed, the “kinks” in their tails elbow adjacent phospholipid molecules away, maintaining some space between the phospholipid molecules. This “elbow room” helps to maintain fluidity in the membrane at temperatures at which membranes with saturated fatty acid tails in their phospholipids would “freeze” or solidify. The relative fluidity of the membrane is particularly important in a cold environment. A cold environment tends to compress membranes composed largely of saturated fatty acids, making them less fluid and more susceptible to rupturing. Many organisms (fish are one example) are capable of adapting to cold environments by changing the proportion of unsaturated fatty acids in their membranes in response to the lowering of the temperature. Visit this site ( to see animations of the fluidity and mosaic quality of membranes. Animals have an additional membrane constituent that assists in maintaining fluidity. Cholesterol, which lies alongside the phospholipids in the membrane, tends to dampen the effects of temperature on the membrane. Thus, this lipid functions as a buffer, preventing lower temperatures from inhibiting fluidity and preventing increased temperatures from increasing fluidity too much. Thus, cholesterol extends, in both directions, the range of temperature in which the membrane is appropriately fluid and consequently functional. Cholesterol also serves other functions, such as organizing clusters of transmembrane proteins into lipid rafts. The Components and Functions of the Plasma Membrane \| \| \| --- \| \| Component \| Location \| \| Phospholipid \| Main fabric of the membrane \| \| Cholesterol \| Attached between phospholipids and between the two phospholipid layers \| \| Integral proteins (for example, integrins) \| Embedded within the phospholipid layer(s). May or may not penetrate through both layers \| \| Peripheral proteins \| On the inner or outer surface of the phospholipid bilayer; not embedded within the phospholipids \| \| Carbohydrates (components of glycoproteins and glycolipids) \| Generally attached to proteins on the outside membrane layer \| \| \| \| Table 4.2 4.8 \| Passive Transport Plasma membranes must allow certain substances to enter and leave a cell, and prevent some harmful materials from entering and some essential materials from leaving. In other words, plasma membranes are selectively permeable—they allow some substances to pass through, but not others. If they were to lose this selectivity, the cell would no longer be able to sustain itself, and it would be destroyed. Some cells require larger amounts of specific substances than do other cells; they must have a way of obtaining these materials from extracellular fluids. This may happen passively, as certain materials move back and forth, or the cell may have special mechanisms that facilitate transport. Some materials are so important to a cell that it spends some of its energy, hydrolyzing adenosine triphosphate (ATP), to obtain these materials. Red blood cells use some of their energy doing just that. All cells spend the majority of their energy to maintain an imbalance of sodium and potassium ions between the interior and exterior of the cell. The most direct forms of membrane transport are passive. Passive transport is a naturally occurring phenomenon and does not require the cell to exert any of its energy to accomplish the movement. In passive transport, substances move from an area of higher concentration to an area of lower concentration. A physical space in which there is a range of concentrations of a single substance is said to have a concentration gradient. Selective Permeability Plasma membranes are asymmetric: the interior of the membrane is not identical to the exterior of the membrane. In fact, there is a considerable difference between the array of phospholipids and proteins between the two leaflets that form a membrane. On the interior of the membrane, some proteins serve to anchor the membrane to fibers of the cytoskeleton. There are peripheral proteins on the exterior of the membrane that bind elements of the extracellular matrix. Carbohydrates, attached to lipids or proteins, are also found on the exterior surface of the plasma membrane. These carbohydrate complexes help the cell bind substances that the cell needs in the extracellular fluid. This adds considerably to the selective nature of plasma membranes (Figure 4.38). Figure 4.38 The exterior surface of the plasma membrane is not identical to the interior surface of the same membrane. Recall that plasma membranes are amphiphilic: They have hydrophilic and hydrophobic regions. This characteristic helps the movement of some materials through the membrane and hinders the movement of others. Lipid-soluble material with a low molecular weight can easily slip through the hydrophobic lipid core of the membrane. Substances such as the fat-soluble vitamins A, D, E, and K readily pass through the plasma membranes in the digestive tract and other tissues. Fat-soluble drugs and hormones also gain easy entry into cells and are readily transported into the body’s tissues and organs. Molecules of oxygen and carbon dioxide have no charge and so pass through membranes by simple diffusion. Polar substances present problems for the membrane. While some polar molecules connect easily with the outside of a cell, they cannot readily pass through the lipid core of the plasma membrane. Additionally, while small ions could easily slip through the spaces in the mosaic of the membrane, their charge prevents them from doing so. Ions such as sodium, potassium, calcium, and chloride must have special means of penetrating plasma membranes. Simple sugars and amino acids also need help with transport across plasma membranes, achieved by various transmembrane proteins (channels). Diffusion Diffusion is a passive process of transport. A single substance tends to move from an area of high concentration to an area of low concentration until the concentration is equal across a space. You are familiar with diffusion of substances through the air. For example, think about someone opening a bottle of ammonia in a room filled with people. The ammonia gas is at its highest concentration in the bottle; its lowest concentration is at the edges of the room. The ammonia vapor will diffuse, or spread away, from the bottle, and gradually, more and more people will smell the ammonia as it spreads. Materials move within the cell’s cytosol by diffusion, and certain materials move through the plasma membrane by diffusion (Figure 4.39). Diffusion expends no energy. On the contrary, concentration gradients are a form of potential energy, dissipated as the gradient is eliminated. Figure 4.39 Diffusion through a permeable membrane moves a substance from an area of high concentration (extracellular fluid, in this case) down its concentration gradient (into the cytoplasm). (credit: modification of work by Mariana Ruiz Villareal) Each separate substance in a medium, such as the extracellular fluid, has its own concentration gradient, independent of the concentration gradients of other materials. In addition, each substance will diffuse according to that gradient. Within a system, there will be different rates of diffusion of the different substances in the medium. Factors That Affect Diffusion Molecules move constantly in a random manner, at a rate that depends on their mass, their environment, and the amount of thermal energy they possess, which in turn is a function of temperature. This movement accounts for the diffusion of molecules through whatever medium in which they are localized. A substance will tend to move into any space available to it until it is evenly distributed throughout it. After a substance has diffused completely through a space, removing its concentration gradient, molecules will still move around in the space, but there will be no net movement of the number of molecules from one area to another. This lack of a concentration gradient in which there is no net movement of a substance is known as dynamic equilibrium. While diffusion will go forward in the presence of a concentration gradient of a substance, several factors affect the rate of diffusion. Extent of the concentration gradient: The greater the difference in concentration, the more rapid the diffusion. The closer the distribution of the material gets to equilibrium, the slower the rate of diffusion becomes. Mass of the molecules diffusing: Heavier molecules move more slowly; therefore, they diffuse more slowly. The reverse is true for lighter molecules. Temperature: Higher temperatures increase the energy and therefore the movement of the molecules, increasing the rate of diffusion. Lower temperatures decrease the energy of the molecules, thus decreasing the rate of diffusion. Solvent density: As the density of a solvent increases, the rate of diffusion decreases. The molecules slow down because they have a more difficult time getting through the denser medium. If the medium is less dense, diffusion increases. Because cells primarily use diffusion to move materials within the cytoplasm, any increase in the cytoplasm’s density will inhibit the movement of the materials. An example of this is a person experiencing dehydration. As the body’s cells lose water, the rate of diffusion decreases in the cytoplasm, and the cells’ functions deteriorate. Neurons tend to be very sensitive to this effect. Dehydration frequently leads to unconsciousness and possibly coma because of the decrease in diffusion rate within the cells. Solubility: As discussed earlier, nonpolar or lipid-soluble materials pass through plasma membranes more easily than polar materials, allowing a faster rate of diffusion. Surface area and thickness of the plasma membrane: Increased surface area increases the rate of diffusion, whereas a thicker membrane reduces it. Distance travelled: The greater the distance that a substance must travel, the slower the rate of diffusion. This places an upper limitation on cell size. A large, spherical cell will die because nutrients or waste cannot reach or leave the center of the cell, respectively. Therefore, cells must either be small in size, as in the case of many prokaryotes, or be flattened, as with many single celled eukaryotes. A variation of diffusion is the process of filtration. In filtration, material moves according to its concentration gradient through a membrane; sometimes the rate of diffusion is enhanced by pressure, causing the substances to filter more rapidly. This occurs in the kidney, where blood pressure forces large amounts of water and accompanying dissolved substances, or solutes, out of the blood and into the renal tubules. The rate of diffusion in this instance is almost totally dependent on pressure. One of the effects of high blood pressure is the appearance of protein in the urine, which is “squeezed through” by the abnormally high pressure. Facilitated transport In facilitated transport, also called facilitated diffusion, materials diffuse across the plasma membrane with the help of membrane proteins. A concentration gradient exists that would allow these materials to diffuse into the cell without expending cellular energy. However, these materials are ions are polar molecules that are repelled by the hydrophobic parts of the cell membrane. Facilitated transport proteins shield these materials from the repulsive force of the membrane, allowing them to diffuse into the cell. The material being transported is first attached to protein or glycoprotein receptors on the exterior surface of the plasma membrane. This allows the material that is needed by the cell to be removed from the extracellular fluid. The substances are then passed to specific integral proteins that facilitate their passage. Some of these integral proteins are collections of beta pleated sheets that form a pore or channel through the phospholipid bilayer. Others are carrier proteins which bind with the substance and aid its diffusion through the membrane. Channels The integral proteins involved in facilitated transport are collectively referred to as transport proteins, and they function as either channels for the material or carriers. In both cases, they are transmembrane proteins. Channels are specific for the substance that is being transported. Channel proteins have hydrophilic domains exposed to the intracellular and extracellular fluids; they additionally have a hydrophilic channel through their core that provides a hydrated opening through the membrane layers (Figure 4.40). Passage through the channel allows polar compounds to avoid the nonpolar central layer of the plasma membrane that would otherwise slow or prevent their entry into the cell. Aquaporins are channel proteins that allow water to pass through the membrane at a very high rate. Figure 4.40 Facilitated transport moves substances down their concentration gradients. They may cross the plasma membrane with the aid of channel proteins. (credit: modification of work by MarianaRuiz Villareal) Channel proteins are either open at all times or they are “gated,” which controls the opening of the channel. The attachment of a particular ion to the channel protein may control the opening, or other mechanisms or substances may be involved. In some tissues, sodium and chloride ions pass freely through open channels, whereas in other tissues a gate must be opened to allow passage. An example of this occurs in the kidney, where both forms of channels are found in different parts of the renal tubules. Cells involved in the transmission of electrical impulses, such as nerve and muscle cells, have gated channels for sodium, potassium, and calcium in their membranes. Opening and closing of these channels changes the relative concentrations on opposing sides of the membrane of these ions, resulting in the facilitation of electrical transmission along membranes (in the case of nerve cells) or in muscle contraction (in the case of muscle cells). Carrier Proteins Another type of protein embedded in the plasma membrane is a carrier protein. This aptly named protein binds a substance and, in doing so, triggers a change of its own shape, moving the bound molecule from the outside of the cell to its interior (Figure 4.41); depending on the gradient, the material may move in the opposite direction. Carrier proteins are typically specific for a single substance. This selectivity adds to the overall selectivity of the plasma membrane. The exact mechanism for the change of shape is poorly understood. Proteins can change shape when their hydrogen bonds are affected, but this may not fully explain this mechanism. Each carrier protein is specific to one substance, and there are a finite number of these proteins in any membrane. This can cause problems in transporting enough of the material for the cell to function properly. When all of the proteins are bound to their ligands, they are saturated and the rate of transport is at its maximum. Increasing the concentration gradient at this point will not result in an increased rate of transport. Figure 4.41 Some substances are able to move down their concentration gradient across the plasma membrane with the aid of carrier proteins. Carrier proteins change shape as they move molecules across the membrane. (credit: modification of work by Mariana Ruiz Villareal) An example of this process occurs in the kidney. Glucose, water, salts, ions, and amino acids needed by the body are filtered in one part of the kidney. This filtrate, which includes glucose, is then reabsorbed in another part of the kidney. Because there are only a finite number of carrier proteins for glucose, if more glucose is present than the proteins can handle, the excess is not transported and it is excreted from the body in the urine. In a diabetic individual, this is described as “spilling glucose into the urine.” A different group of carrier proteins called glucose transport proteins, or GLUTs, are involved in transporting glucose and other hexose sugars through plasma membranes within the body. Channel and carrier proteins transport material at different rates. Channel proteins transport much more quickly than do carrier proteins. Channel proteins facilitate diffusion at a rate of tens of millions of molecules per second, whereas carrier proteins work at a rate of a thousand to a million molecules per second. Osmosis Osmosis is the movement of water through a semipermeable membrane according to the concentration gradient of water across the membrane, which is inversely proportional to the concentration of solutes. While diffusion transports material across membranes and within cells, osmosis transports only water across a membrane and the membrane limits the diffusion of solutes in the water. Not surprisingly, the aquaporins that facilitate water movement play a large role in osmosis, most prominently in red blood cells and the membranes of kidney tubules. Mechanism Osmosis is a special case of diffusion. Water, like other substances, moves from an area of high concentration to one of low concentration. An obvious question is what makes water move at all? Imagine a beaker with a semipermeable membrane separating the two sides or halves (Figure 4.42). On both sides of the membrane the water level is the same, but there are different concentrations of a dissolved substance, or solute, that cannot cross the membrane (otherwise the concentrations on each side would be balanced by the solute crossing the membrane). If the volume of the solution on both sides of the membrane is the same, but the concentrations of solute are different, then there are different amounts of water, the solvent, on either side of the membrane. Figure 4.42 In osmosis, water always moves from an area of higher water concentration to one of lower concentration. In the diagram shown, the solute cannot pass through the selectively permeable membrane, but the water can. To illustrate this, imagine two full glasses of water. One has a single teaspoon of sugar in it, whereas the second one contains one-quarter cup of sugar. If the total volume of the solutions in both cups is the same, which cup contains more water? Because the large amount of sugar in the second cup takes up much more space than the teaspoon of sugar in the first cup, the first cup has more water in it. Returning to the beaker example, recall that it has a mixture of solutes on either side of the membrane. A principle of diffusion is that the molecules move around and will spread evenly throughout the medium if they can. However, only the material capable of getting through the membrane will diffuse through it. In this example, the solute cannot diffuse through the membrane, but the water can. Water has a concentration gradient in this system. Thus, water will diffuse down its concentration gradient, crossing the membrane to the side where it is less concentrated. This diffusion of water through the membrane—osmosis—will continue until the concentration gradient of water goes to zero or until the hydrostatic pressure of the water balances the osmotic pressure. Osmosis proceeds constantly in living systems. Tonicity Tonicity describes how an extracellular solution can change the volume of a cell by affecting osmosis. A solution’s tonicity often directly correlates with the osmolarity of the solution. Osmolarity describes the total solute concentration of the solution. A solution with low osmolarity has a greater number of water molecules relative to the number of solute particles; a solution with high osmolarity has fewer water molecules with respect to solute particles. In a situation in which solutions of two different osmolarities are separated by a membrane permeable to water, though not to the solute, water will move from the side of the membrane with lower osmolarity (and more water) to the side with higher osmolarity (and less water). This effect makes sense if you remember that the solute cannot move across the membrane, and thus the only component in the system that can move—the water—moves along its own concentration gradient. An important distinction that concerns living systems is that osmolarity measures the number of particles (which may be molecules) in a solution. Therefore, a solution that is cloudy with cells may have a lower osmolarity than a solution that is clear, if the second solution contains more dissolved molecules than there are cells. Hypotonic Solutions Three terms—hypotonic, isotonic, and hypertonic—are used to relate the osmolarity of a cell to the osmolarity of the extracellular fluid that contains the cells. In a hypotonic situation, the extracellular fluid has lower osmolarity than the fluid inside the cell, and water enters the cell. (In living systems, the point of reference is always the cytoplasm, so the prefix hypo– means that the extracellular fluid has a lower concentration of solutes, or a lower osmolarity, than the cell cytoplasm.) It also means that the extracellular fluid has a higher concentration of water in the solution than does the cell. In this situation, water will follow its concentration gradient and enter the cell. Hypertonic Solutions As for a hypertonic solution, the prefix hyper– refers to the extracellular fluid having a higher osmolarity than the cell’s cytoplasm; therefore, the fluid contains less water than the cell does. Because the cell has a relatively higher concentration of water, water will leave the cell. Isotonic Solutions In an isotonic solution, the extracellular fluid has the same osmolarity as the cell. If the osmolarity of the cell matches that of the extracellular fluid, there will be no net movement of water into or out of the cell, although water will still move in and out. Blood cells and plant cells in hypertonic, isotonic, and hypotonic solutions take on characteristic appearances (Figure 4.43). For a video illustrating the process of diffusion in solutions, visit this site ( l/dispersion) . Tonicity in Living Systems In a hypotonic environment, water enters a cell, and the cell swells. In an isotonic condition, the relative concentrations of solute and solvent are equal on both sides of the membrane. There is no net water movement; therefore, there is no change in the size of the cell. In a hypertonic solution, water leaves a cell and the cell shrinks. If either the hypo- or hyper- condition goes to excess, the cell’s functions become compromised, and the cell may be destroyed. A red blood cell will burst, or lyse, when it swells beyond the plasma membrane’s capability to expand. Remember, the membrane resembles a mosaic, with discrete spaces between the molecules composing it. If the cell swells, and the spaces between the lipids and proteins become too large, the cell will break apart. In contrast, when excessive amounts of water leave a red blood cell, the cell shrinks, or crenates. This has the effect of concentrating the solutes left in the cell, making the cytosol denser and interfering with diffusion within the cell. The cell’s ability to function will be compromised and may also result in the death of the cell. Various living things have ways of controlling the effects of osmosis—a mechanism called osmoregulation. Some organisms, such as plants, fungi, bacteria, and some protists, have cell walls that surround the plasma membrane and prevent cell lysis in a hypotonic solution. The plasma membrane can only expand to the limit of the cell wall, so the cell will not lyse. In fact, the cytoplasm in plants is always slightly hypertonic to the cellular environment, and water will always enter a cell if water is available. This inflow of water produces turgor pressure, which stiffens the cell walls of the plant (Figure 4.44). In nonwoody plants, turgor pressure supports the plant. Conversely, if the plant is not watered, the extracellular fluid will become hypertonic, causing water to leave the cell. In this condition, the cell does not shrink because the cell wall is not flexible. However, the cell membrane detaches from the wall and constricts the cytoplasm. This is called plasmolysis. Plants lose turgor pressure in this condition and wilt (Figure 4.46). Figure 4.44 The turgor pressure within a plant cell depends on the tonicity of the solution that it is bathed in. (credit: modification of work by Mariana Ruiz Villareal) Figure 4.45 Without adequate water, the plant on the left has lost turgor pressure, visible in its wilting; the turgor pressure is restored by watering it (right). (credit: Victor M. Vicente Selvas) Tonicity is a concern for all living things. For example, paramecia and amoebas, which are protists that lack cell walls, have contractile vacuoles. This vesicle collects excess water from the cell and pumps it out, keeping the cell from lysing as it takes on water from its environment (Figure 4.46). Figure 4.46 A paramecium’s contractile vacuole, here visualized using bright field light microscopy at 480x magnification, continuously pumps water out of the organism’s body to keep it from bursting in a hypotonic medium. (credit: modification of work by NIH; scale-bar data from Matt Russell) Many marine invertebrates have internal salt levels matched to their environments, making them isotonic with the water in which they live. Fish, however, must spend approximately five percent of their metabolic energy maintaining osmotic homeostasis. Freshwater fish live in an environment that is hypotonic to their cells. These fish actively take in salt through their gills and excrete diluted urine to rid themselves of excess water. Saltwater fish live in the reverse environment, which is hypertonic to their cells, and they secrete salt through their gills and excrete highly concentrated urine.In vertebrates, the kidneys regulate the amount of water in the body. Osmoreceptors are specialized cells in the brain that monitor the concentration of solutes in the blood. If the levels of solutes increase beyond a certain range, a hormone is released that retards water loss through the kidney and dilutes the blood to safer levels. Animals also have high concentrations of albumin, which is produced by the liver, in their blood. This protein is too large to pass easily through plasma membranes and is a major factor in controlling the osmotic pressures applied to tissues. 4.9 \| Active Transport Active transport mechanisms require the use of the cell’s energy, usually in the form of adenosine triphosphate (ATP). If a substance must move into the cell against its concentration gradient—that is, if the concentration of the substance inside the cell is greater than its concentration in the extracellular fluid (and vice versa)—the cell must use energy to move the substance. Some active transport mechanisms move small-molecular weight materials, such as ions, through the membrane. Other mechanisms transport much larger molecules. Electrochemical Gradient We have discussed simple concentration gradients—differential concentrations of a substance across a space or a membrane—but in living systems, gradients are more complex. Because ions move into and out of cells and because cells contain proteins that do not move across the membrane and are mostly negatively charged, there is also an electrical gradient, a difference of charge, across the plasma membrane. The interior of living cells is electrically negative with respect to the extracellular fluid in which they are bathed, and at the same time, cells have higher concentrations of potassium (K+) and lower concentrations of sodium (Na+) than does the extracellular fluid. So in a living cell, the concentration gradient of Na+ tends to drive it into the cell, and the electrical gradient of Na+ (a positive ion) also tends to drive it inward to the negatively charged interior. The situation is more complex, however, for other elements such as potassium. The electrical gradient of K+, a positive ion, also tends to drive it into the cell, but the concentration gradient of K+ tends to drive K+ out of the cell (Figure 4.47). The combined gradient of concentration and electrical charge that affects an ion is called its electrochemical gradient. Moving Against a Gradient To move substances against a concentration or electrochemical gradient, the cell must use energy. This energy is harvested from ATP generated through the cell’s metabolism. Active transport mechanisms, collectively called pumps, work against electrochemical gradients. Small substances constantly pass through plasma membranes. Active transport maintains concentrations of ions and other substances needed by living cells in the face of these passive movements. Much of a cell’s supply of metabolic energy may be spent maintaining these processes. (Most of a red blood cell’s metabolic energy is used to maintain the imbalance between exterior and interior sodium and potassium levels required by the cell.) Because active transport mechanisms depend on a cell’s metabolism for energy, they are sensitive to many metabolic poisons that interfere with the supply of ATP. Two mechanisms exist for the transport of small-molecular weight material and small molecules. Primary active transport moves ions across a membrane and creates a difference in charge across that membrane, which is directly dependent on ATP. Secondary active transport describes the movement of material that is due to the electrochemical gradient established by primary active transport that does not directly require ATP. Carrier Proteins for Active Transport An important membrane adaption for active transport is the presence of specific carrier proteins or pumps to facilitate movement: there are three types of these proteins or transporters (Figure 5.17). A uniporter carries one specific ion or molecule. A symporter carries two different ions or molecules, both in the same direction. An antiporter also carries two different ions or molecules, but in different directions. All of these transporters can also transport small, uncharged organic molecules like glucose. These three types of carrier proteins are also found in facilitated diffusion, but they do not require ATP to work in that process. Some examples of pumps for active transport are Na+-K+ ATPase, which carries sodium and potassium ions, and H+-K+ ATPase, which carries hydrogen and potassium ions. Both of these are antiporter carrier proteins. Two other carrier proteins are Ca2+ ATPase and H+ ATPase, which carry only calcium and only hydrogen ions, respectively. Both are pumps. Figure 5.17 A uniporter carries one molecule or ion. A symporter carries two different molecules or ions, both in the same direction. An antiporter also carries two different molecules or ions, but in different directions. (credit: modification of work by “Lupask”/Wikimedia Commons) Primary Active Transport The primary active transport that functions with the active transport of sodium and potassium allows secondary active transport to occur. The second transport method is still considered active because it depends on the use of energy as does primary transport (Figure 5.18). Figure 5.18 Primary active transport moves ions across a membrane, creating an electrochemical gradient (electrogenic transport). (credit: modification of work by Mariana Ruiz Villareal) One of the most important pumps in animal cells is the sodium-potassium pump (Na+-K+ ATPase), which maintains the electrochemical gradient (and the correct concentrations of Na+ and K+) in living cells. The sodium-potassium pump moves K+ into the cell while moving Na+ out at the same time, at a ratio of three Na+ for every two K+ ions moved in. The Na+-K+ ATPase exists in two forms, depending on its orientation to the interior or exterior of the cell and its affinity for either sodium or potassium ions. The process consists of the following six steps. With the enzyme oriented towards the interior of the cell, the carrier has a high affinity for sodium ions. Three ions bind to the protein. ATP is hydrolyzed by the protein carrier and a low-energy phosphate group attaches to it. As a result, the carrier changes shape and re-orients itself towards the exterior of the membrane. The protein’s affinity for sodium decreases and the three sodium ions leave the carrier. The shape change increases the carrier’s affinity for potassium ions, and two such ions attach to the protein. Subsequently, the low-energy phosphate group detaches from the carrier. With the phosphate group removed and potassium ions attached, the carrier protein repositions itself towards the interior of the cell. The carrier protein, in its new configuration, has a decreased affinity for potassium, and the two ions are released into the cytoplasm. The protein now has a higher affinity for sodium ions, and the process starts again. Several things have happened as a result of this process. At this point, there are more sodium ions outside of the cell than inside and more potassium ions inside than out. For every three ions of sodium that move out, two ions of potassium move in. This results in the interior being slightly more negative relative to the exterior. This difference in charge is important in creating the conditions necessary for the secondary process. The sodium-potassium pump is, therefore, an electrogenic pump (a pump that creates a charge imbalance), creating an electrical imbalance across the membrane and contributing to the membrane potential. Visit the site ( to see a simulation of active transport in a sodium-potassium ATPase. Secondary Active Transport (Co-transport) Secondary active transport brings sodium ions, and possibly other compounds, into the cell. As sodium ion concentrations build outside of the plasma membrane because of the action of the primary active transport process, an electrochemical gradient is created. If a channel protein exists and is open, the sodium ions will be pulled through the membrane. This movement is used to transport other substances that can attach themselves to the transport protein through the membrane (Figure 4.50). Many amino acids, as well as glucose, enter a cell this way. This secondary process is also used to store high-energy hydrogen ions in the mitochondria of plant and animal cells for the production of ATP. The potential energy that accumulates in the stored hydrogen ions is translated into kinetic energy as the ions surge through the channel protein ATP synthase, and that energy is used to convert ADP into ATP. 4.10 \| Bulk Transport In addition to moving small ions and molecules through the membrane, cells also need to remove and take in larger molecules and particles (see Table 4.3 for examples). Some cells are even capable of engulfing entire unicellular microorganisms. You might have correctly hypothesized that the uptake and release of large particles by the cell requires energy. A large particle, however, cannot pass through the membrane, even with energy supplied by the cell. Endocytosis Endocytosis is a type of active transport that moves particles, such as large molecules, parts of cells, and even whole cells, into a cell. There are different variations of endocytosis, but all share a common characteristic: The plasma membrane of the cell invaginates, forming a pocket around the target particle. The pocket pinches off, resulting in the particle being contained in a newly created intracellular vesicle formed from the plasma membrane. Phagocytosis Phagocytosis (the condition of “cell eating”) is the process by which large particles, such as cells or relatively large particles, are taken in by a cell. For example, when microorganisms invade the human body, a type of white blood cell called a neutrophil will remove the invaders through this process, surrounding and engulfing the microorganism, which is then destroyed by the neutrophil (Figure 4.51). Figure 4.51 In phagocytosis, the cell membrane surrounds the particle and engulfs it. (credit:Mariana Ruiz Villareal) In preparation for phagocytosis, a portion of the inward-facing surface of the plasma membrane becomes coated with a protein called clathrin, which stabilizes this section of the membrane. The coated portion of the membrane then extends from the body of the cell and surrounds the particle, eventually enclosing it. Once the vesicle containing the particle is enclosed within the cell, the clathrin disengages from the membrane and the vesicle merges with a lysosome for the breakdown of the material in the newly formed compartment (endosome). When accessible nutrients from the degradation of the vesicular contents have been extracted, the newly formed endosome merges with the plasma membrane and releases its contents into the extracellular fluid. The endosomal membrane again becomes part of the plasma membrane. Pinocytosis A variation of endocytosis is called pinocytosis. This literally means “cell drinking” and was named at a time when the assumption was that the cell was purposefully taking in extracellular fluid. In reality, this is a process that takes in molecules, including water, which the cell needs from the extracellular fluid. Pinocytosis results in a much smaller vesicle than does phagocytosis, and the vesicle does not need to merge with a lysosome (Figure 4.53). Figure 4.53 In pinocytosis, the cell membrane invaginates, surrounds a small volume of fluid, andpinches off. (credit: Mariana Ruiz Villareal) A variation of pinocytosis is called potocytosis. This process uses a coating protein, called caveolin, on the cytoplasmic side of the plasma membrane, which performs a similar function to clathrin. The cavities in the plasma membrane that form the vacuoles have membrane receptors and lipid rafts in addition to caveolin. The vacuoles or vesicles formed in caveolae (singular caveola) are smaller than those in pinocytosis. Potocytosis is used to bring small molecules into the cell and to transport these molecules through the cell for their release on the other side of the cell, a process called transcytosis. Receptor-mediated Endocytosis A targeted variation of endocytosis employs receptor proteins in the plasma membrane that have a specific binding affinity for certain substances (Figure 4.53). Figure 4.53 In receptor-mediated endocytosis, uptake of substances by the cell is targeted to a single type of substance that binds to the receptor on the external surface of the cell membrane.(credit: modification of work by Mariana Ruiz Villareal) In receptor-mediated endocytosis, as in phagocytosis, clathrin is attached to the cytoplasmic side of the plasma membrane. If uptake of a compound is dependent on receptor-mediated endocytosis and the process is ineffective, the material will not be removed from the tissue fluids or blood. Instead, it will stay in those fluids and increase in concentration. Some human diseases are caused by the failure of receptor-mediated endocytosis. For example, the form of cholesterol termed low-density lipoprotein or LDL (also referred to as “bad” cholesterol) is removed from the blood by receptor-mediated endocytosis. In the human genetic disease familial hypercholesterolemia, the LDL receptors are defective or missing entirely. People with this condition have life-threatening levels of cholesterol in their blood, because their cells cannot clear LDL particles from their blood. Although receptor-mediated endocytosis is designed to bring specific substances that are normally found in the extracellular fluid into the cell, other substances may gain entry into the cell at the same site. Flu viruses, diphtheria, and cholera toxin all have sites that cross-react with normal receptor-binding sites and gain entry into cells. See receptor-mediated endocytosis in action, and click on different parts ( l/endocytosis) for a focused animation. Exocytosis The reverse process of moving material into a cell is the process of exocytosis. Exocytosis is the opposite of the processes discussed above in that its purpose is to expel material from the cell into the extracellular fluid. Waste material is enveloped in a membrane and fuses with the interior of the plasma membrane. This fusion opens the membranous envelope on the exterior of the cell, and the waste material is expelled into the extracellular space (Figure 5.23). Other examples of cells releasing molecules via exocytosis include the secretion of proteins of the extracellular matrix and secretion of neurotransmitters into the synaptic cleft by synaptic vesicles. Figure 4.54 In exocytosis, vesicles containing substances fuse with the plasma membrane. The contents are then released to the exterior of the cell. (credit: modification of work by Mariana RuizVillareal) Methods of Transport, Energy Requirements, and Types of Material Transported \| \| \| \| --- \| Transport Method \| Active/ Passive \| Material Transported \| \| Diffusion \| Passive \| Small-molecular weight material \| \| Osmosis \| Passive \| Water \| \| Facilitated transport/ diffusion \| Passive \| Sodium, potassium, calcium, glucose \| \| Primary active transport \| Active \| Sodium, potassium, calcium \| \| Secondary active transport \| Active \| Amino acids, lactose \| \| Phagocytosis \| Active \| Large macromolecules, whole cells, or cellular structures \| \| Pinocytosis and potocytosis \| Active \| Small molecules (liquids/water) \| \| Receptor-mediated endocytosis \| Active \| Large quantities of macromolecules \| \| \| \| \| Table 4.3 KEY TERMS active transport method of transporting material that requires energy amphiphilic molecule possessing a polar or charged area and a nonpolar or uncharged area capable of interacting with both hydrophilic and hydrophobic environments antiporter transporter that carries two ions or small molecules in different directions aquaporin channel protein that allows water through the membrane at a very high rate carrier protein membrane protein that moves a substance across the plasma membrane by changing its own shape caveolin protein that coats the cytoplasmic side of the plasma membrane and participates in the process of liquid update by potocytosis channel protein membrane protein that allows a substance to pass through its hollow core across the plasma membrane clathrin protein that coats the inward-facing surface of the plasma membrane and assists in the formation of specialized structures, like coated pits, for phagocytosis concentration gradient area of high concentration adjacent to an area of low concentration diffusion passive process of transport of low-molecular weight material according to its concentration gradient electrochemical gradient gradient produced by the combined forces of an electrical gradient and a chemical gradient electrogenic pump pump that creates a charge imbalance endocytosis type of active transport that moves substances, including fluids and particles, into a cell exocytosis process of passing bulk material out of a cell facilitated transport process by which material moves down a concentration gradient (from high to low concentration) using integral membrane proteins fluid mosaic model describes the structure of the plasma membrane as a mosaic of components including phospholipids, cholesterol, proteins, glycoproteins, and glycolipids (sugar chains attached to proteins or lipids, respectively), resulting in a fluid character (fluidity) glycolipid combination of carbohydrates and lipids glycoprotein combination of carbohydrates and proteins hydrophilic molecule with the ability to bond with water; “water-loving” hydrophobic molecule that does not have the ability to bond with water; “water-hating” hypertonic situation in which extracellular fluid has a higher osmolarity than the fluid inside the cell, resulting in water moving out of the cell hypotonic situation in which extracellular fluid has a lower osmolarity than the fluid inside the cell, resulting in water moving into the cell integral protein protein integrated into the membrane structure that interacts extensively with the hydrocarbon chains of membrane lipids and often spans the membrane; these proteins can be removed only by the disruption of the membrane by detergents isotonic situation in which the extracellular fluid has the same osmolarity as the fluid inside the cell, resulting in no net movement of water into or out of the cell osmolarity total amount of substances dissolved in a specific amount of solution osmosis transport of water through a semipermeable membrane according to the concentration gradient of water across the membrane that results from the presence of solute that cannot pass through the membrane passive transport method of transporting material through a membrane that does not require energy peripheral protein protein found at the surface of a plasma membrane either on its exterior or interior side; these proteins can be removed (washed off of the membrane) by a high-salt wash pinocytosis a variation of endocytosis that imports macromolecules that the cell needs from theextracellular fluid plasmolysis detaching of the cell membrane from the cell wall and constriction of the cell membrane when a plant cell is in a hypertonic solution potocytosis variation of pinocytosis that uses a different coating protein (caveolin) on the cytoplasmic side of the plasma membrane primary active transport active transport that moves ions or small molecules across a membrane and may create a difference in charge across that membrane pump active transport mechanism that works against electrochemical gradients receptor-mediated endocytosis variation of endocytosis that involves the use of specific binding proteins in the plasma membrane for specific molecules or particles, and clathrin-coated pits that become clathrin-coated vesicles secondary active transport movement of material that is due to the electrochemical gradient established by primary active transport selectively permeable characteristic of a membrane that allows some substances through but not others solute substance dissolved in a liquid to form a solution symporter transporter that carries two different ions or small molecules, both in the same direction tonicity amount of solute in a solution transport protein membrane protein that facilitates passage of a substance across a membrane by binding it transporter specific carrier proteins or pumps that facilitate movement uniporter transporter that carries one specific ion or molecule cell theory see unified cell theory cell wall rigid cell covering made of cellulose that protects the cell, provides structural support, and gives shape to the cell central vacuole large plant cell organelle that regulates the cell’s storage compartment, holds water, and plays a significant role in cell growth as the site of macromolecule degradation centrosome region in animal cells made of two centrioles chlorophyll green pigment that captures the light energy that drives the light reactions of photosynthesis chloroplast plant cell organelle that carries out photosynthesis chromatin protein-DNA complex that serves as the building material of chromosomes chromosome structure within the nucleus that is made up of chromatin that contains DNA, the hereditary material cilium (plural = cilia) short, hair-like structure that extends from the plasma membrane in large numbers and is used to move an entire cell or move substances along the outer surface of the cell cytoplasm entire region between the plasma membrane and the nuclear envelope, consisting of organelles suspended in the gel-like cytosol, the cytoskeleton, and various chemicals cytoskeleton network of protein fibers that collectively maintain the shape of the cell, secure some organelles in specific positions, allow cytoplasm and vesicles to move within the cell, and enable unicellular organisms to move independently cytosol gel-like material of the cytoplasm in which cell structures are suspended desmosome linkages between adjacent epithelial cells that form when cadherins in the plasma membrane attach to intermediate filaments electron microscope an instrument that magnifies an object using a beam of electrons passed and bent through a lens system to visualize a specimen endomembrane system group of organelles and membranes in eukaryotic cells that work together modifying, packaging, and transporting lipids and proteins endoplasmic reticulum (ER) series of interconnected membranous structures within eukaryotic cells that collectively modify proteins and synthesize lipids eukaryotic cell cell that has a membrane-bound nucleus and several other membrane-bound compartments or sacs extracellular matrix material (primarily collagen, glycoproteins, and proteoglycans) secreted from animal cells that provides mechanical protection and anchoring for the cells in the tissue flagellum (plural = flagella) long, hair-like structure that extends from the plasma membrane and is used to move the cell gap junction channel between two adjacent animal cells that allows ions, nutrients, and low molecular weight substances to pass between cells, enabling the cells to communicate Golgi apparatus eukaryotic organelle made up of a series of stacked membranes that sorts, tags, and packages lipids and proteins for distribution intermediate filament cytoskeletal component, composed of several intertwined strands of fibrous protein, that bears tension, supports cell-cell junctions, and anchors cells to extracellular structures light microscope an instrument that magnifies an object using a beam visible light passed and bent through a lens system to visualize a specimen lysosome organelle in an animal cell that functions as the cell’s digestive component; it breaks down proteins, polysaccharides, lipids, nucleic acids, and even worn-out organelles microfilament narrowest element of the cytoskeleton system; it provides rigidity and shape to the cell and enables cellular movements microscope an instrument that magnifies an object microtubule widest element of the cytoskeleton system; it helps the cell resist compression, provides a track along which vesicles move through the cell, pulls replicated chromosomes to opposite ends of a dividing cell, and is the structural element of centrioles, flagella, and cilia mitochondria (singular = mitochondrion) cellular organelles responsible for carrying out cellular respiration, resulting in the production of ATP, the cell’s main energy-carrying molecule nuclear envelope double-membrane structure that constitutes the outermost portion of the nucleus nucleoid central part of a prokaryotic cell in which the chromosome is found nucleolus darkly staining body within the nucleus that is responsible for assembling the subunits of the ribosomes nucleoplasm semi-solid fluid inside the nucleus that contains the chromatin and nucleolus nucleus cell organelle that houses the cell’s DNA and directs the synthesis of ribosomes and proteins organelle compartment or sac within a cell peroxisome small, round organelle that contains hydrogen peroxide, oxidizes fatty acids and amino acids, and detoxifies many poisons plasma membrane phospholipid bilayer with embedded (integral) or attached (peripheral) proteins, and separates the internal content of the cell from its surrounding environment plasmodesma (plural = plasmodesmata) channel that passes between the cell walls of adjacent plant cells, connects their cytoplasm, and allows materials to be transported from cell to cell prokaryote unicellular organism that lacks a nucleus or any other membrane-bound organelle ribosome cellular structure that carries out protein synthesis rough endoplasmic reticulum (RER) region of the endoplasmic reticulum that is studded with ribosomes and engages in protein modification and phospholipid synthesis smooth endoplasmic reticulum (SER) region of the endoplasmic reticulum that has few or no ribosomes on its cytoplasmic surface and synthesizes carbohydrates, lipids, and steroid hormones; detoxifies certain chemicals (like pesticides, preservatives, medications, and environmental pollutants), and stores calcium ions tight junction firm seal between two adjacent animal cells created by protein adherence unified cell theory a biological concept that states that all organisms are composed of one or more cells; the cell is the basic unit of life; and new cells arise from existing cells vacuole membrane-bound sac, somewhat larger than a vesicle, which functions in cellular storage and transport vesicle small, membrane-bound sac that functions in cellular storage and transport; its membrane is capable of fusing with the plasma membrane and the membranes of the endoplasmic reticulum and Golgi apparatus CHAPTER SUMMARY 4.1 Studying Cells A cell is the smallest unit of life. Most cells are so tiny that they cannot be seen with the naked eye. Therefore, scientists use microscopes to study cells. Electron microscopes provide higher magnification, higher resolution, and more detail than light microscopes. The unified cell theory states that all organisms are composed of one or more cells, the cell is the basic unit of life, and new cells arise from existing cells. 4.2 Prokaryotic Cells Prokaryotes are predominantly single-celled organisms of the domains Bacteria and Archaea. All prokaryotes have plasma membranes, cytoplasm, ribosomes, and DNA that is not membrane-bound. Most have peptidoglycan cell walls and many have polysaccharide capsules. Prokaryotic cells range in diameter from 0.1 to 5.0 μm. As a cell increases in size, its surface area-to-volume ratio decreases. If the cell grows too large, the plasma membrane will not have sufficient surface area to support the rate of diffusion required for the increased volume. 4.3 Eukaryotic Cells Like a prokaryotic cell, a eukaryotic cell has a plasma membrane, cytoplasm, and ribosomes, but a eukaryotic cell is typically larger than a prokaryotic cell, has a true nucleus (meaning its DNA is surrounded by a membrane), and has other membrane-bound organelles that allow for compartmentalization of functions. The plasma membrane is a phospholipid bilayer embedded with proteins. The nucleus’s nucleolus is the site of ribosome assembly. Ribosomes are either found in the cytoplasm or attached to the cytoplasmic side of the plasma membrane or endoplasmic reticulum. They perform protein synthesis. Mitochondria participate in cellular respiration; they are responsible for the majority of ATP produced in the cell. Peroxisomes hydrolyze fatty acids, amino acids, and some toxins. Vesicles and vacuoles are storage and transport compartments. In plant cells, vacuoles also help break down macromolecules. Animal cells also have a centrosome and lysosomes. The centrosome has two bodies perpendicular to each other, the centrioles, and has an unknown purpose in cell division. Lysosomes are the digestive organelles of animal cells. Plant cells and plant-like cells each have a cell wall, chloroplasts, and a central vacuole. The plant cell wall, whose primary component is cellulose, protects the cell, provides structural support, and gives shape to the cell. Photosynthesis takes place in chloroplasts. The central vacuole can expand without having to produce more cytoplasm. 4.4 The Endomembrane System and Proteins The endomembrane system includes the nuclear envelope, lysosomes, vesicles, the ER, and Golgi apparatus, as well as the plasma membrane. These cellular components work together to modify, package, tag, and transport proteins and lipids that form the membranes. The RER modifies proteins and synthesizes phospholipids used in cell membranes. The SER synthesizes carbohydrates, lipids, and steroid hormones; engages in the detoxification of medications and poisons; and stores calcium ions. Sorting, tagging, packaging, and distribution of lipids and proteins take place in the Golgi apparatus. Lysosomes are created by the budding of the membranes of the RER and Golgi. Lysosomes digest macromolecules, recycle worn-out organelles, and destroy pathogens. 4.5 The Cytoskeleton The cytoskeleton has three different types of protein elements. From narrowest to widest, they are the microfilaments (actin filaments), intermediate filaments, and microtubules. Microfilaments are often associated with myosin. They provide rigidity and shape to the cell and facilitate cellular movements. Intermediate filaments bear tension and anchor the nucleus and other organelles in place. Microtubules help the cell resist compression, serve as tracks for motor proteins that move vesicles through the cell, and pull replicated chromosomes to opposite ends of a dividing cell. They are also the structural element of centrioles, flagella, and cilia. 4.6 Connections between Cells and Cellular Activities Animal cells communicate via their extracellular matrices and are connected to each other via tight junctions, desmosomes, and gap junctions. Plant cells are connected and communicate with each other via plasmodesmata. When protein receptors on the surface of the plasma membrane of an animal cell bind to a substance in the extracellular matrix, a chain of reactions begins that changes activities taking place within the cell. Plasmodesmata are channels between adjacent plant cells, while gap junctions are channels between adjacent animal cells. However, their structures are quite different. A tight junction is a watertight seal between two adjacent cells, while a desmosome acts like a spot weld. 4.7 Components and Structure The modern understanding of the plasma membrane is referred to as the fluid mosaic model. The plasma membrane is composed of a bilayer of phospholipids, with their hydrophobic, fatty acid tails in contact with each other. The landscape of the membrane is studded with proteins, some of which span the membrane. Some of these proteins serve to transport materials into or out of the cell. Carbohydrates are attached to some of the proteins and lipids on the outward-facing surface of the membrane, forming complexes that function to identify the cell to other cells. The fluid nature of the membrane is due to temperature, the configuration of the fatty acid tails (some kinked by double bonds), the presence of cholesterol embedded in the membrane, and the mosaic nature of the proteins and protein-carbohydrate combinations, which are not firmly fixed in place. Plasma membranes enclose and define the borders of cells, but rather than being a static bag, they are dynamic and constantly in flux. 4.8 Passive Transport The passive forms of transport, diffusion and osmosis, move materials of small molecular weight across membranes. Substances diffuse from areas of high concentration to areas of lower concentration, and this process continues until the substance is evenly distributed in a system. In solutions containing more than one substance, each type of molecule diffuses according to its own concentration gradient, independent of the diffusion of other substances. Many factors can affect the rate of diffusion, including concentration gradient, size of the particles that are diffusing, temperature of the system, and so on. In living systems, diffusion of substances into and out of cells is mediated by the plasma membrane. Some materials diffuse readily through the membrane, but others are hindered, and their passage is made possible by specialized proteins, such as channels and transporters. The chemistry of living things occurs in aqueous solutions, and balancing the concentrations of those solutions is an ongoing problem. In living systems, diffusion of some substances would be slow or difficult without membrane proteins that facilitate transport. 4.9 Active Transport The combined gradient that affects an ion includes its concentration gradient and its electrical gradient. A positive ion, for example, might tend to diffuse into a new area, down its concentration gradient, but if it is diffusing into an area of net positive charge, its diffusion will be hampered by its electrical gradient. When dealing with ions in aqueous solutions, a combination of the electrochemical and concentration gradients, rather than just the concentration gradient alone, must be considered. Living cells need certain substances that exist inside the cell in concentrations greater than they exist in the extracellular space. Moving substances up their electrochemical gradients requires energy from the cell. Active transport uses energy stored in ATP to fuel this transport. Active transport of small molecular sized materials uses integral proteins in the cell membrane to move the materials: These proteins are analogous to pumps. Some pumps, which carry out primary active transport, couple directly with ATP to drive their action. In co-transport (or secondary active transport), energy from primary transport can be used to move another substance into the cell and up its concentration gradient. 4.10 Bulk Transport Active transport methods require the direct use of ATP to fuel the transport. Large particles, such as macromolecules, parts of cells, or whole cells, can be engulfed by other cells in a process called phagocytosis. In phagocytosis, a portion of the membrane invaginates and flows around the particle, eventually pinching off and leaving the particle entirely enclosed by an envelope of plasma membrane. Vesicle contents are broken down by the cell, with the particles either used as food or dispatched. Pinocytosis is a similar process on a smaller scale. The plasma membrane invaginates and pinches off, producing a small envelope of fluid from outside the cell. Pinocytosis imports substances that the cell needs from the extracellular fluid. The cell expels waste in a similar but reverse manner: it pushes a membranous vacuole to the plasma membrane, allowing the vacuole to fuse with the membrane and incorporate itself into the membrane structure, releasing its contents to the exterior. ART CONNECTION QUESTIONS Figure 4.7 Prokaryotic cells are much smaller than eukaryotic cells. What advantages might small cell size confer on a cell? What advantages might large cell size have? Figure 4.8 If the nucleolus were not able to carry out its function, what other cellular organelles would be affected? Figure 4.18 If a peripheral membrane protein were synthesized in the lumen (inside) of the ER, would it end up on the inside or outside of the plasma membrane? REVIEW QUESTIONS When viewing a specimen through a light microscope, scientists use ________ to distinguish the individual components of cells. a. a beam of electrons b. radioactive isotopes c. special stains d. high temperatures The ________ is the basic unit of life. a. organism b. cell c. tissue d. organ Prokaryotes depend on ________ to obtain some materials and to get rid of wastes. a. ribosomes b. flagella c. cell division d. diffusion Which of the following is surrounded by two phospholipid bilayers? a. the ribosomes b. the vesicles c. the cytoplasm d. the nucleoplasm Peroxisomes got their name because hydrogen peroxide is: a. used in their detoxification reactions b. produced during their oxidation reactions c. incorporated into their membranes d. a cofactor for the organelles’ enzymes In plant cells, the function of the lysosomes is carried out by __________. a. vacuoles b. peroxisomes c. ribosomes d. nuclei Which of the following is found both in eukaryotic and prokaryotic cells? a. nucleus b. mitochondrion c. vacuole d. ribosomes Which of the following is not a component of the endomembrane system? a. mitochondrion b. Golgi apparatus c. endoplasmic reticulum d. lysosome The process by which a cell engulfs a foreign particle is known as: a. endosymbiosis b. phagocytosis c. hydrolysis d. membrane synthesis Which of the following is most likely to have the greatest concentration of smooth endoplasmic reticulum? a. a cell that secretes enzymes b. a cell that destroys pathogens c. a cell that makes steroid hormones d. a cell that engages in photosynthesis Which of the following sequences correctly lists in order the steps involved in the incorporation of a proteinaceous molecule within a cell? a. synthesis of the protein on the ribosome; modification in the Golgi apparatus; packaging in the endoplasmic reticulum; tagging in the vesicle b. synthesis of the protein on the lysosome; tagging in the Golgi; packaging in the vesicle; distribution in the endoplasmic reticulum c. synthesis of the protein on the ribosome; modification in the endoplasmic reticulum; tagging in the Golgi; distribution via the vesicle d. synthesis of the protein on the lysosome; packaging in the vesicle; distribution via the Golgi; tagging in the endoplasmic reticulum Which of the following have the ability to disassemble and reform quickly? a. microfilaments and intermediate filaments b. microfilaments and microtubules c. intermediate filaments and microtubules d. only intermediate filaments Which of the following do not play a role in intracellular movement? a. microfilaments and intermediate filaments b. microfilaments and microtubules c. intermediate filaments and microtubules d. only intermediate filaments Which of the following are found only in plant cells? a. gap junctions b. desmosomes c. plasmodesmata d. tight junctions The key components of desmosomes are cadherins and __________. a. actin b. microfilaments c. intermediate filaments d. microtubules Which characteristic of a phospholipid contributes to the fluidity of the membrane? its head a saturated fatty acid tail double bonds in the fatty acid tail What is the primary function of carbohydrates attached to the exterior of cell membranes? identification of the cell flexibility of the membrane strengthening the membrane channels through membrane Water moves via osmosis _________. throughout the cytoplasm from an area with a high concentration of other solutes to a lower one from an area with a high concentration of water to one of lower concentration from an area with a low concentration of water to one of higher concentration The principal force driving movement in diffusion is the __________. particle size concentration gradient membrane surface area What problem is faced by organisms that live in fresh water? Their bodies tend to take in too much water. They have no way of controlling their tonicity. Only salt water poses problems for animals that live in it. Their bodies tend to lose too much water to their environment. Active transport must function continuously because __________. plasma membranes wear out not all membranes are amphiphilic facilitated transport opposes active diffusion is constantly moving solutes in opposite directions How does the sodium-potassium pump make the interior of the cell negatively charged? by expelling anions by pulling in anions by expelling more cations than are taken in by taking in and expelling an equal number of cations What is the combination of an electrical gradient and a concentration gradient called? potential gradient electrical potential concentration potential electrochemical gradient What happens to the membrane of a vesicle after exocytosis? It leaves the cell. It is disassembled by the cell. It fuses with and becomes part of the plasma membrane. It is used again in another exocytosis event. Which transport mechanism can bring whole cells into a cell? facilitated transport primary active transport In what important way does receptor mediated endocytosis differ from phagocytosis? It transports only small amounts of fluid. It does not involve the pinching off of membrane. It brings in only a specifically targeted substance. It brings substances into the cell, while phagocytosis removes substances. CRITICAL THINKING QUESTIONS In your everyday life, you have probably noticed that certain instruments are ideal for certain situations. For example, you would use a spoon rather than a fork to eat soup because a spoon is shaped for scooping, while soup would slip between the tines of a fork. The use of ideal instruments also applies in science. In what situation(s) would the use of a light microscope be ideal, and why? Antibiotics are medicines that are used to fight bacterial infections. These medicines kill prokaryotic cells without harming human cells. What part or parts of the bacterial cell do you think antibiotics target? Why? Explain why not all microbes are harmful. You already know that ribosomes are abundant in red blood cells. In what other cells of the body would you find them in great abundance? Why? What are the structural and functional similarities and differences between mitochondria and chloroplasts? In the context of cell biology, what do we mean by form follows function? What are at least two examples of this concept? In your opinion, is the nuclear membrane part of the endomembrane system? Why or why not? Defend your answer. What are the similarities and differences between the structures of centrioles and flagella? How do cilia and flagella differ? How does the structure of a plasmodesma differ from that of a gap junction? Explain how the extracellular matrix function Why is it advantageous for the cell membrane to be fluid in nature? Why do phospholipids tend to spontaneously orient themselves into something resembling a membrane? Discuss why the following affect the rate of diffusion: molecular size, temperature, solution density, and the distance that must be traveled. Why does water move through a membrane? Both of the regular intravenous solutions administered in medicine, normal saline and lactated Ringer’s solution, are isotonic. Why is this important? Where does the cell get energy for active transport processes? How does the sodium-potassium pump contribute to the net negative charge of the interior of the cell? Why is it important that there are different types of proteins in plasma membranes for the transport of materials into and out of a cell? Chapter is adapted from: Biology 2e by Mary Ann Clark, Matthew Douglas, and Jung Choi, published by OpenStax, licensed under CC BY 4.0, click here to access for free.
16693	https://kito.wordpress.ncsu.edu/files/2018/04/funa3.pdf	Functional Analysis and Optimization Kazufumi Ito∗ November 29, 2016 Abstract In this monograph we develop the function space method for optimization problems and operator equations in Banach spaces. Optimization is the one of key components for mathematical modeling of real world problems and the solution method provides an accurate and essential description and validation of the mathematical model. Op-timization problems are encountered frequently in engineering and sciences and have widespread practical applications.In general we optimize an appropriately chosen cost functional subject to constraints. For example, the constraints are in the form of equal-ity constraints and inequality constraints. The problem is casted in a function space and it is important to formulate the problem in a proper function space framework in order to develop the accurate theory and the eﬀective algorithm. Many of the constraints for the optimization are governed by partial diﬀerential and functional equations. In order to discuss the existence of optimizers it is essential to develop a comprehensive treatment of the constraints equations. In general the necessary optimality condition is in the form of operator equations. Non-smooth optimization becomes a very basic modeling toll and enlarges and en-hances the applications of the optimization method in general. For example, in the classical variational formulation we analyze the non Newtonian energy functional and nonsmooth friction penalty. For the imaging/signal analysis the sparsity optimization is used by means of L1 and TV regularization. In order to develop an eﬃcient solu-tion method for large scale optimizations it is essential to develop a set of necessary conditions in the equation form rather than the variational inequality. The Lagrange multiplier theory for the constrained minimizations and nonsmooth optimization prob-lems is developed and analyzed The theory derives the complementarity condition for the multiplier and the solution. Consequently, one can derive the necessary optimality system for the solution and the multiplier for a general class of nonsmooth and noncovex optimizations. In the light of the theory we derive and analyze numerical algorithms based the primal-dual active set method and the semi-smooth Newton method. The monograph also covers the basic materials for real analysis, functional anal-ysis, Banach space theory, convex analysis, operator theory and PDE theory, which makes the book self-contained, comprehensive and complete. The analysis component is naturally connected to the optimization theory. The necessary optimality condition is in general written as nonlinear operator equations for the primal variable and La-grange multiplier. The Lagrange multiplier theory of a general class of nonsmooth and ∗Department of Mathematics, North Carolina State University, Raleigh, North Carolina, USA 1 non-convex optimization are based on the functional analysis tool. For example, the necessary optimality for the constrained optimization is in general nonsmooth and a psuedo-monontone type operator equation. To develop the mathematical treatment of the PDE constrained minimization we also present the PDE theory and linear and nonlinear C0-semigroup theory on Banach spaces and we have a full discussion of well-posedness of the control dynamics and optimization problems. That is, the existence and uniqueness of solutions to a general class of controlled linear and nonlinear PDEs and Cauchy problems for nonlinear evo-lution equations in Banach space are discussed. Especially the linear operator theory and the (psuedo-) monotone operator and mapping theory and the ﬁxed point the-ory.Throughout the monograph demonstrates the theory and algorithm using concrete examples and describes how to apply it for motivated applications 1 Introduction In this monograph we develop the function space approach for the optimization prob-lems, e.g., nonlinear programming in Banach spaces, convex and non-convex nonsmooth variational problems, control and inverse problems, image/signal analysis, material design, classiﬁcation, and resource allocation. We also develop the basic functional analysis tool for for the nonlinear equations and Cauchy problems in Banach spaces in order to establish the well-posedness of the constraint optimization. The function space optimization method allows us to study the well-posednees for a general class of optimization problems systematically using the functional analysis tool and it also enables us to develop and analyze the numerical optimization algorithm based on the calculus of variations and the Lagrange multiplier theory for the constrained opti-mization. The function space analysis provides the regularizing procedure to remeady the ill-posedness of the constraints and the necessary optimality system and thus one can develop a stable and eﬀective numerical algorithm. The convergence analysis for the function space optimization guides us to analyze and improve the stability and eﬀectiveness of algorithms for the associated large scale optimizations. A general class of constrained optimization problems in function spaces is consid-ered and we develop the Lagrange multiplier theory and eﬀective solution algorithms. Applications are presented and analyzed for various examples including control and design optimization, inverse problems, image analysis and variational problems. Non-convex and non-smooth optimization becomes increasing important for the advanced and eﬀective use of optimization methods and thus they are essential topics for the monograph. We introduce the basic function space and functional analysis tools needed for our analysis. The monograph provides an introduction to the functional analysis, real analysis and convex analysis and their basic concepts with illustrated examples. Thus, one can use the book as a basic course material for the functional analysis and nonlinear operator theory. The nonlinear operator theory and their applications to PDE’s problems are presented in details, including classical variational optimization problems in Newtonian and non-Newtonian mechanics and ﬂuid. An emerging appli-cation of optimizations include the imaging and signal analysis and the classiﬁcation and machine learning. It is often formulated as the PDE’s constrained optimization. There are numerous applications of the optimization theory and variational method in all sciences and to real world applications. For example, the followings are the 2 motivated examples. Example 1 (Quadratic programming) min 1 2xtAx −atx over x ∈Rn subject to Ex = b (equality constraint) Gx ≤c (inequality constraint). where A ∈Rn×n is a symmetric positive matrix, E ∈Rm×n, G ∈Rp×n, and a ∈Rn, b ∈Rm, c ∈Rp are given vectors. The quadratic programming is formulated on a Hilbert space X for the variational problem, in which (Ax, x)X deﬁnes the quadratic form on X. Example 2 (Linear programming) The linear programing minimizes (c, x)X subject to the linear constraint Ax = b and x ≥0. Its dual problem is to maximize (b, x) subject to A∗x ≤c. It has many important classes of applications. For example, the optimal transport problem is to ﬁnd the joint distribution function µ(x, y) ≥0 on X × Y that transports the probability density p0(x) to p1(y), i.e. Z Y µ(x, y) dx = p0(x), Z Y µ(x, y) dy = p1(y), with minimum energy Z X×Y c(x, y)µ(x, y) dxdy = (c, µ). Example 3 (Integer programming) Let F is a continuous function Rn. The integer programming is to minimize F(x) over the integer variables x, i.e., min F(x) subject to x ∈{0, 1, · · · N}. The binary optimizations to minimize F(x) over the binary variables x. The network optimization problem is the one of important applications. Example4 (Obstacle problem) Consider the variational problem; min J(u) = Z 1 0 (1 2\| d dxu(x)\|2 −f(x)u(x)) dx (1.1) subject to u(x) ≤ψ(x) over all displacement function u(x) ∈H1 0(0, 1), where the function ψ represents the upper bound (obstacle) of the deformation. The functional J represents the Newtonian deformation energy and f(x) is the applied force. Example 5 (Function Interpolation) Consider the interpolation problem; min Z 1 0 (1 2\| d2 dx2 u(x)\|2 dx over all functions u(x), (1.2) subject to the interpolation conditions u(xi) = bi, d dxu(xi) = ci, 1 ≤i ≤m. 3 Example 6 (Image/Signal Analysis) Consider the deconvolution problem of ﬁnding the image u that satisﬁes (Ku)(x) = f(x) for the convolution K (Ku)(x) = Z Ω k(x, y)u(y) dy. over a domain Ω. It is in general ill-posed and we use the (Tikhonov) regularization formulation to have a stable deconvolution solution u, i.e., for Ω= [0, 1] min Z 1 0 (\|Ku −f\|2 + α 2 \| d dxu(x)\|2 + β u(x)) dx over all possible image u ≥0, (1.3) where α, β ≥0 are the regularization parameters. The ﬁrst regularization term is the quadratic varation of image u. Since if u ≥0 then \|u\| = u, the regularization term for the L1 sparsity. The two regularization parameters are chosen so that the variation of image u is regularized. Example 7 (L0 optimization) Problem of optimal resource allocation and material dis-tribution and optimal time scheduling can be formulated as F(u) + Z Ω h(u(ω)) dω (1.4) with h(u) = α 2 \|u\|2 + β \|u\|0 where \|0\|0 = 0, \|u\|0 = 1, otherwise. Since Z Ω \|u(ω)\|0 dω = vol({u(ω) ̸= 0}), the solution to (1.4) provides the optimal distribution of u for minimizing the perfor-mance F(u) with appropriate choice of (regularization) parameters α, β > 0. Example 8 (Bayesian statistics) Consider the statical optimization; min −log (p(y\|x) + βp(x)) over all admissible parameters x, (1.5) where p(y\|x) is the conditional probability density of observation y given parameter x and the p(x) is the prior probability density for parameter x. It is the maximum posterior estimation of the parameter x given observation y. In the case of inverse problems it has the form φ(x\|y) + α p(x) (1.6) for x is in a function space X and φ(x\|y) is the ﬁdelity and p is a regularization semi-norm on X. A constant α > 0 is the Tikhonov regularization parameter. Example 9 (Optimal Control Problem) Consider the constrained optimization of the form min F(x) + H(u) subject to E(x, u) = 0 and u ∈C where x ∈X and u ∈U is the state and control variables in Hilbert spaces X and U, respectively. Equality constraint E represents the model equations for x = x(u) ∈X as a function of u ∈C, a closed convex set in U. The separable cost functionals are appropriately selected to achieve a speciﬁc performance. Such optimization problems 4 arise in the optimal control problem and inverse medium problem and optimal design problem. Example 8 (Machine Learning) Consider the constrained optimization of the form L(p, q) = Multi-dimensional versions of these examples and more illustrated examples will be discussed and analyzed throughout the monograph. Discretization An important aspect of the function space optimization is the approx-imation method and theory. For example, let Bn i (x) is the linear B-spline deﬁned by Bi(x) =      x−xi−1 xi−xi−1 on [xi−1, xi] x−xi+1 xi−xi+1 on [xi, xi+1] 0 otherwise (1.7) and let un(x) = n−1 X i=1 uiBi(x) ∼u(x). (1.8) Then, d dxun = ui −ui−1 xi −xi−1 , x ∈(xi−1, xi). Let xi = i n and ∆x = 1 n and one can deﬁne the discretized problem of Example 1: min n X i=1 (1 2\|ui −ui−1 ∆x \|2 −fiui) ∆x subject to u0 = un = 0, ui ≤ψi = ψ(xi). where fi = f(xi). The discretized problem is the quadratic programming for ⃗ u ∈Rn−1 with A = 1 ∆x        2 −1 −1 2 −1 ... ... ... −1 2 −1 −1 2        , b = ∆x        f1 f2 . . . fn−2 fn−1        Thus, we will discuss a convergence analysis of the approximate solutions {un} to the solution u as n →∞in an appropriate function space. In general, one has the optimization of the form min F0(x) + F1(x) over x ∈X subject to E(x) ∈K where X is a Banach space, F0 is C1 and F1 is nonsmooth functional and E(x) ∈K represents the constraint conditions with K is a closed convex cone. For the existence of a minimizer we require a coercivity condition and lower-semicontinuity and the 5 continuity of the constraint in a weak or weak star topology of X. For the necessary optimality condition, we develop a generalized Lagrange multiplier theory:            F ′(x∗) + λ + E′(x∗)∗µ = 0, E(x) ∈K C1(x∗, λ) = 0 C2(x∗, µ) = 0, (1.9) where λ is the Lagrange multiplier for a relaxed derivative of F1 and µ is the Lagrange multiplier for the constraint E(x) ∈K. We derive the complementarity conditions C1 and C2 so that (1.9) is a complete set of equation for determining (x∗, λ, µ). We discuss the solution method for linear and nonlinear equations in Banach spaces of the form f ∈A(x) (1.10) where f ∈X∗in the dual space of a Banach space and A is pseudo-monotone mapping X →X∗and the Cauchy problem: d dtx(t) ∈A(t)(x(t)) + f(t), x(0) = x0, (1.11) where A(t) is an m-dissipative mapping in X × X. Also, we note that the necessary optimality (1.9) is a nonlinear operator equation and we use the nonlinear operator theory to analyze solutions to (1.9). We also discuss the PDE constrained optimiza-tion problems, in which f and f(t) are a function of control and design and medium parameters. We minimize a proper performance index deﬁned for the state and control functions subject to the constraint (1.10) or (1.11). Thus, we combine the analysis of the PDE theory and the optimization theory to develop a comprehensive treatment and analysis of the PDE constrained optimization problems. The outline of the monograph is as follows. In Section 2 we introduce the basic function space and functional analysis tools needed for our analysis. It provides the introduction to the functional analysis and real analysis and their basic concepts with illustrated examples. For example, the dual space of normed space, Hahn-Banach theorem, open mapping theory, closed range theory, distribution theory and weak solution to PDEs, compact operator and spectral theory. In Section 3 we develop the Lagrange multiplier method for a general class of (smooth) constrained optimization problems in Banach spaces.The Hilbert space the-ory is introduced ﬁrst and the Lagrange multiplier is deﬁned as the limit of the penal-ized problem. The constraint qualiﬁcation condition is introduced and we derive the (normal) necessary optimality system for the primal and dual variables. We discuss the minimum norm problem and the duality principle in Banach spaces. In Section 4 we present the linear operator theory and C0-semigroup theory on Banach spaces. In Section 5 we develop the monotone operator theory and pseudo-monotone operator theory for nonlinear equations in Banach spaces and the nonlinear semigroup theory in Banach spaces. The ﬁxed point theory for nonlinear equations in Banach spaces. In Section 7 we develop the basic convex analysis tools and the Lagrange multiplier theory for a general class of convex non-smooth optimization. We introduce and analyze the augmented Lagrangian method for the constrained optimization and non-smooth optimization problems. 6 In Section 8 we develop the Lagrange multiplier theory for a general class of non-smooth and non-convex optimizations. In this course we discuss the well-posededness of the evolution equations in Ba-nach spaces. Such problems arise in PDEs dynamics and functional equations. We develop the linear and nonlinear theory for the corresponding solution semigroups. The lectures include for example the Hille-Yosiida theory, Lumer-Philiips theory for linear semigroup and Crandall-Liggett theory for nonlinear contractive semigroup and Crandall-Pazy theory for nonlinear evolution equations. Especially, (numerical) ap-proximation theory for PDE solutions are discussed based on Trotter-Kato theory and Takahashi-Oharu theory, Chernoﬀtheory and the operator splitting method. The the-ory and its applications are examined and demonstrated using many motivated PDE examples including linear dynamics (e.g. heat, wave and hyperbolic equations) and nonlinear dynamics (e.g. nonlinear diﬀusion, conservation law, Hamilton-Jacobi and Navier-Stokes equations). A new class of PDE examples are formulated and the de-tailed applications of the theory is carried out. The lecture also covers the basic elliptic theory via Lax-Milgram, Minty-Browder theory and convex optimization. Functional analytic methods are also introduced for the basic PDEs theory. The students are expected to have the basic knowledge in real and functional anal-ysis and PDEs. Lecture notes will be provided. Reference book: ”Evolution equations and Approx-imation” K. Ito and F. Kappel, World Scientiﬁc. 2 Basic Function Space Theory In this section we discuss the basic theory and concept of the functional analysis and introduce the function spaces, which is essential to our function space analysis of a gen-eral class of optimizations. We refer to the more comprehensive treatise and discussion on the functional analysis for e.g., Yosida []. 2.1 Complete Normed Space In this section we introduce the vector space, normed space and Banach spaces. Deﬁnition (Vector Space) A vector space over the scalar ﬁeld K is a set X, whose elements are called vectors, and in which two operations, addition and scalar multipli-cation, are deﬁned, with the following familiar algebraic properties. (a) To every pair of vectors x and y corresponds a vector x + y, in such a way that x + y = y + x and x + (y + z) = (x + y) + z; X contains a unique vector 0 (the zero vector) such that x+0 = 0+x for every x ∈X, and to each x ∈X corresponds a unique inverse vector −x such that x + (−x) = 0. (b) To every pair (α, x) with α ∈K and x ∈X corresponds a vector α x, in such a way that 1 x = x, α(βx) = (αβ) x, 7 and such that the two distributive laws α (x + y) = α x + α y, (α + β) x = α x + β x hold. The symbol 0 will of course also be used for the zero element of the scalar ﬁeld. It is easy to see that the inverse element −x of the additivity satisﬁes −x = (−1) x since 0 x = (1 −1) x = x + (−1) x for each x ∈X. In what follows we consider vector spaces only over the real number ﬁeld R or the complex number ﬁeld C. Deﬁnition (Subspace, Basis) (1) A subset S of X is a (linear) subspace of a vector space X if α x1 + β x2 ∈S for all x, x2 ∈S and and α, β ∈K. (2) A family of linear vectors {xi}n i=1 are linearly independent if Pn i=1 αi xi = 0 if and only if αi = 0 for all 1 ≤i ≤n. The dimension dim(X) is the largest number of independent vectors in X. (3) A basis {eα} of a vector space X is a set of linearly independent vectors that, in a linear combination, can represent every vector in X, i.e., x = P α aαeα. Example (Vector Space) (1) Let X = C[0, 1] be the vector space of continuous functions f on [0, 1]. X is a vector space by (α f + β g)(t) = α f(t) + β g(t), t ∈[0, 1] S= a space of polynomials of order less than n is a linear subspace of X and dim(X) = n. A family {tk}∞ k=0 of polynomials is a basis of X and dom(X) = ∞. (2) The vector space of real number sequences x = (x1, x2, x3, · · · ) is a vector space by (α x + β y)k = α xk + β yk, k ∈N. A family of unit vectors ek = (0, · · · , 1, 0, · · · ), k ∈N is a basis. (3) Let X be the vector space of square integrable functions on (0, 1). Let ek(x) = p (2) sin(kπx), 0 ≤x ≤1. Then {ek} are orthonomal basis, i.e. (ek, ej) = Z 1 0 ek(x)ej(x) dx = δk,j and f(x) = ∞ X k=1 fk ek(x), the Fourier sine series where fk = R 1 0 f(x)ek(x) dx. en+1 is not linearly dependent with (e1, · · · , en). If so, en+1 = Pn i=1 βiei but βi = (en+1, ei) = 0, which is the contradiction. Deﬁnition (Normed Space) A norm \| · \| on a vector space X is a real-valued func-tional on X which satisﬁes \|x\| ≥0 for all x ∈X with equality if and only if x = 0, \|α x\| = \|α\| \|x\| for every x ∈x and α ∈K \|x + y\| ≤\|x\| + \|y\| (triangle inequality) for every x, y ∈X. A normed space is a vector space X which is equipped with a norm \| · \| and will be denoted by (X, \| · \|). The norm of a normed space X will be denoted by \| · \|X or simply by \| · \| whenever the underlined space X is understood from the context. 8 Example (Normed space) Let X = C[0, 1] be the vector space of continuous functions f on [0, 1]. The followings are examples the two normed space for X; X1 is equipped with sup norm \|f\|X1 = max x∈[0,1] \|f(x)\|. and X2 is equipped with L2 norm \|f\|X2 = sZ 1 0 \|f(x)\|2 dx. Deﬁnition (Banach Space) Let X be a normed space. (1) A sequence {xn} in a normed space X converges to the limit x∗∈X if and only if limn→∞\|xn −x∗\| = 0 in R. (2) A subspace S of a normed space X is said to be dense in X if each x ∈X is the limit of a sequence of elements in S. (3) A sequence {xn} in a normed space X is called a Cauchy sequence if and only if limm, n→∞\|xm −xn\| = 0, i.e., for all ϵ > 0 there exists N such that for m, n ≥N such that \|xm −xn\| < ϵ. If every Cauchy sequence in X converges to a limit in X, then X is complete and X is called a Banach space. A Cauchy sequence {xn} in a normed space X has a unique limit if it has accumu-lation points. In fact, for any limit x∗ lim n→∞\|x∗−xn\| = lim n→∞ lim m→∞\|xm −xn\| = 0. Example (continued) Two norms \|x\|1 and \|x\|2 on a vector space X are equivalent if there exist 0 < c ≤¯ c < ∞such that c\|x\|1 ≤\|x\|2 ≤¯ c \|x\|1 for all x ∈X. (1) All norms on Rn is are equivalent. But for X = C[0, 1] consider xn(t) = tn. Then, \|xn\|X1 = 1 but \|xn\|X2 = q 1 2n+1 for all n. Thus, \|xn\|X2 →0 as n →∞and the two norms X1 and X2 are not equivalent. (2) Every bounded sequence in Rn has a convergent subsequence in Rn (Bolzano-Weierstrass theorem). Consider the sequence xn(t) = tn in X = C[0, 1]. Then xn is not convergent despite \|xn\|X1 = 1 is bounded. Deﬁnition (Hilbert Space) If X is a vector space, a functional (·, ·) deﬁned on X×X is called an inner product on X provided that for every x, y, z ∈X and α, β ∈C (x, y) = (y, x) (α x + β y, z) = α (x, z) + β (y, z), (x, x) ≥0 and (x, x) = 0 if and only if x = 0. where ¯ c denotes the complex conjugate of c ∈C. Given such an inner product, a norm on X can be deﬁned by \|x\| = (x, x) 1 2 . 9 A vector space (X, (·, ·)) equipped with an inner product is called a pre–Hilbert space. If X is complete with respect to the induced norm, then it is called a Hilbert space. Every inner product satisﬁes the Cauchy-Schwarz inequality \|(x, y)\| ≤\|x\| \|y\| for x, y ∈X In fact, for all t ∈R \|x + t(x, y) y\|2 = \|x\|2 + 2t (x, y) + t2 \|y\|2 ≥0. Thus, we must have \|(x, y)\|2 −\|x\|2\|y\|2 ≤0 which shows the Cauchy-Schwarz inequality. Example (Hilbert space) (1) Rn with inner product (x, y) = n X k=1 xkyk for x, y ∈Rn (2) Consider a space ℓ2 of square summable real number sequences x = (x1, x2, · · · ) deﬁne the inner product (x, y) = ∞ X k=1 xkyk for x, y ∈ℓ2 Then, ℓ2 is an Hilbert space. In fact, let {xn} is a Cauchy sequence in ℓ2. Since \|xn k −xm k \| ≤\|xn −xm\|ℓ2, {xn k} is a Cauchy sequence in R for every k. Thus, one can deﬁne a sequence x by xk = limn→∞xn k. Since lim m→∞\|xm −xn\|ℓ2 = \|x −xn\|ℓ2 for a ﬁxed n, x ∈ℓ2 and \|xn −x\|ℓ2 →as n →∞. (3) Consider a space X2 of continuously diﬀerentiable functions on [0, 1] vanishing at x = 0, x = 1 with inner product (f, g) = Z 1 0 d dxf d dxg dx for f, g ∈X2. Then X2 is a Pre-Hilbert space. Every normed space X is either a Banach space or a dense subspace of a Banach space Y whose norm \|x\|Y satisﬁes \|x\|Y = \|x\| for every x ∈X. In this latter case Y is called the completion of X. The completion of a normed space proceeds as in Cantor’s Construction of real numbers from rational numbers. Completion The set of Cauchy sequences {xn} of the normed space X can be classiﬁed according to the equivalence {xn} ∼{yn} if lim n→∞\|xn−yn\| = 0. Since \|\|xn\|−\|xm\|\| ≤ \|xn −xm\| and R is complete, it follows that limn→∞\|xn\| exists. We denote by {xn}′, 10 the class containing {xn}. Then the set Y of all such equivalent classes ˜ x = {xn}′ is a vector space by {xn}′ + {yn}′ = {xn + yn}′, α {xn}′ = {α xn}′ and we deﬁne \|{xn}′\|Y = lim n→∞\|xn\|. It is easy to see that these deﬁnitions of the vector sum and scalar multiplication and norm do not depend on the particular representations for the classes {xn}′ and {yn}′. For example, if {xm} ∼{yn}, then lim n→∞\|\|xn\| −\|yn\|\| ≤lim n→∞\|xn −yn\| = 0, and thus lim n→∞\|xn\| = lim n→∞\|yn\|. Since \|{xn}′\|Y = 0 implies that limn→∞\|xn −0\|, we have {0} ∈{xn}′ and thus \| · \|Y is a norm. Since an element x ∈X corresponds to the Cauchy sequence: ¯ x = {x, x, . . . , x, . . . }′ ∈Y, X is naturally contained in Y . To prove the completeness of Y , let {˜ xk} = {{x(k) n }} be a Cauchy sequence in Y . Then for each k, we can choose an integer nk such that \|x(k) m −x(k) nk \|X ≤k−1 for m > nk. We show that the sequence {˜ xk} converges to the class containing the Cauchy sequence {x(k) nk } of X. To this end, note that \|˜ xk −x(k) nk \|Y = lim m→∞\|x(k) m −x(k) nk \|X ≤k−1. Since \|x(k) nk −x(m) nm \|X = \|x(k) nk −x(m) nm \|Y ≤\|x(k) nk −˜ xk\|Y + \|˜ xk −˜ xm\|Y + \|˜ xm −x(m) nm \|Y ≤k−1 + m−1 + \|˜ xk −˜ xm\|Y , {x(k) nk } is a Cauchy sequence of X. Let ˜ x = {x(k) nk }′. Then, \|˜ x −˜ xk\|Y ≤\|˜ x −x(k) nk \|Y + \|x(k) nk −˜ xk\|Y ≤\|˜ x −x(k) nk \|Y + k−1. Since, as shown above \|˜ x −x(k) nk \|Y = lim m→∞\|xm nm −x(k) nk \|X ≤lim m→∞\|˜ xm −˜ xk\|Y + k−1 thus we prove that limk→∞\|˜ x −x(k) nk \|Y = 0 and so limk→∞\|˜ x −˜ xk\|Y = 0. Moreover, the above proof shows that the correspondence x ∈X to ¯ x ∈Y is isomorphic and isometric and the image of X by this correspondence is dense in Y , that is X is dense in Y . □ Example (Completion) Let X = C[0, 1] be the vector space of continuous functions f on [0, 1]. We consider the two normed space for X; X1 is equipped with sup norm \|f\|X1 = max x∈[0,1] \|f(x)\|. 11 and X2 is equipped with L2 norm \|f\|X2 = sZ 1 0 \|f(x)\|2 dx We claim that X1 is complete but X2 is not. The completion of X2 is L2(0, 1) =the space of square integrable functions. First, discuss X1. Let {fn} is a Cauchy sequence in X1. Since \|fn(x) −fm(x)\| ≤\|fn −fm\|X1, {fn(x)} is Cauchy sequence in R for every x ∈[0, 1]. Thus, one can deﬁne a function f by f(x) = limn→∞fn(x). Since \|f(x) −f(ˆ x)\| ≤\|fn(x) −fn(ˆ x)\| + \|fn(x) −f(x)\| + \|fn(ˆ x) −f(ˆ x)\| and \|fn(x) −f(x)\| →0 uniformly x ∈[0, 1], the candidate f is continuous. Thus, X1 is complete. Next, discuss X2. Deﬁne a Cauchy sequence {fn} by fn(x) =    1 2 + n (x −1 2) on [ 1 2 −1 n, 1 2 + 1 n] 0 on [0, 1 2 −1 n] 1 on [ 1 2 + 1 n, 1] The candidate limit f in this case is deﬁned by f(x) = 0, x < 1 2, f( 1 2) = 1 2 and f(x) = 1, x > 1 2 and is not in X. Consider a Cauchy sequences gn and hn in the equivalence class {fn}; gn(x) =    1 + n (x −1 2) on [ 1 2 −1 n, 1 2] 0 on [0, 1 2 −1 n] 1 on [ 1 2, 1] hn(x) =    n (x −1 2) on [ 1 2, 1 2 + 1 n] 0 on [0, 1 2] 1 on [1 2 + 1 n, 1]. Thus, the candidate limits g, (g(x) = 0, x < 1 2, g(x) = 1, x ≥1 2 and h, (h(x) = 0, x ≤ 1 2, h(x) = 1, x > 1 2 belong to the same equivalent class f = {fn}. Note that f, g and h diﬀer only at x = 1 2. In general functions in an equivalent class diﬀer at countably many points. The completion Y of X2 is denoted by L2(0, 1) = {the space of square integrable measurable functions on (0, 1)} with \|f\|2 L2 = Z 1 0 \|f(x)\|2 dx. Here, the integrable is deﬁned in the sense of Lebesgue as discussed in the next section. The concept of completion states, for every f ∈Y = ¯ X =the completion of X there exists a sequence fn in X such that \|fn −f\|Y →0 as n →∞and \|f\|Y = limn→∞\|fn\|X. 12 For the example X2, for all square integrable function f ∈L2(0, 1) = Y = ¯ X2 there exists a continuous function sequence fn ∈X2 such that \|f −fn\|L2(0,1) →0. Example H1 0(0, 1) The completion of C1 ( 0, 1), the pre-Hilberrt space of continuously diﬀerentiable functions on [0, 1] vanishing at x = 0 x = 1 with respect to H1 0 inner product (f, g)H1 0 = Z 1 0 d dxf(x) d dxg(x) dx is denoted by H1 0(0, 1). It can be proved that H1 0(0, 1) = {a space of absolutely continuous functions on [0, 1] with square integrable derivative and vanishing at x = 0, x = 1} with (f, g)H1 0 = Z 1 0 d dxf(x) d dxg(x) dx where the integrable is deﬁned in the sense of Lebesgue. Here, f is absolutely contin-uous function if there exits an integrable function g such that f(x) = f(0) + Z x 0 g(x) dx x ∈[0, 1]. and f is almost everywhere (a.e.) diﬀerentiable and d dxf = g a.e. in (0, 1). For the linear spline function Bn k (x), 1 ≤k ≤n −1 is in H1 0(0, 1). 2.2 Measure Theory In this section we discuss the measure theory. Deﬁnition (1) A topological space is a set E together with a collection τ of subsets of E, called open sets and satisfying the following axioms; The empty set and E itself are open. Any union of open sets is open. The intersection of any ﬁnite number of open sets is open. The collection τ of open sets is then also called a topology on E. We assume that a topological space (E, τ) satisﬁes the Hausdoroﬀ’s axiom of separation: for every disjoints x1, x2 ∈E there exist disjoint open sets G, G2 auch that x1 ∈G1, x2 ∈G2. (2) A metric space E is a topological space with metric d if for all x, y, z ∈E d(x, y) ≥for x, y ∈E and d(x, y) = 0 if and only if x = y d(x, y) = d(y, x) d(x, z) ≤d(x, y) + d(y, z). For x0 ∈E and r > 0 B(x0, r) = {x ∈E : d(x, x0) < r is an open ball of E with center x0 an radius r. A set O is called open if for any points x ∈E there exists an open ball B(x, r) contained in the set O. 13 (3) A collection of subsets F of E is σ-algebra if for all A ∈F, Ac = E \ A ∈F for arbitrary family {Ai} in F, countable union S∞ i=1 Ai ∈F (4) A measure µ on (E, F) assigns A ∈F the nonnegative real value µ(A) and satisﬁes σ-additivity; µ( ∞ [ i=1 Ai) = ∞ X i=1 µ(Ai) for all disjoint sets {Ai} in F. Theorem (Monotone Convergence) The measure µ is σ-additive if and only if for all sequence {Ak} of nondecreasing events and A = S k≥1 Ak, limn→∞µ(An) = µ(A). Proof: {Ac k} is a sequence of nonincreasing events and Ac = T k≥1 Ac k. Since \ k≥1 Ac k = Ac 1 + (Ac 2 \ Ac 1) + (Ac 3 \ Ac 2) + · · · we have µ(Ac) = µ(Ac 1) + µ(Ac 2 \ Ac 1) + µ(Ac 3 \ Ac 2) + · · · = µ(Ac 1) + µ(Ac 2) −µ(Ac 1) + µ(Ac 3) −µ(Ac 1) + · · · = limn→∞µ(Ac n) Thus, µ(A) = 1 −µ(Ac) = 1 −(1 −lim n→∞µ(An)) = lim n→∞µ(An). Conversely, let A1, A2, · · · ∈F be pairwise disjoint and let P∞ k=1 Ak ∈F. Then µ( ∞ X k=1 Ak) = µ( n X k=1 Ak) + µ( ∞ X k=n+1 Ak) Since P∞ k=n+1 Ak ↓∅, we have lim n→∞ ∞ X k=n+1 Ak = ∅ and thus µ( ∞ X k=1 Ak) = lim n→∞ n X k=1 µ(Ak) = ∞ X k=1 µ(Ak).□ Example (Metric space) A normed space X is a metric space with d(x, y) = \|x −y\|. A point x ∈E is the accumulation point of a sequence {xn} if d(xn, x) →0 as n →∞. The closure ¯ A of a set A is the collection of all accumulation points of A. The boundary of asset A is the collection points of x that any open ball at x contains both points in A and Ac. A set B in E is closed if B = ¯ B, i.e., every sequence in {xn} in B has a accumulation x. The complement Ac of an open set A is closed. Examples (σ-algebra) There are trivial σ-algebras; F0 = {Ω, ∅}, F∗= all subsets of Ω. 14 Let A be a subset of Ωand σ-algebra generated by A is FA = {Ω, ∅, A, Ac}. A ﬁnite set of subsets A1, A2, · · · , An of E which are pairwise disjoint and whose union is E. It is called a partition of Ω. It generates the σ-algebra: A = {A = S j∈J Aj} where J runs over all subsets of 1, · · · , n. This σ-algebra has 2n elements. Every ﬁnite σ-algebra is of this form. The smallest nonempty elements {A1, · · · , An} of this algebra are called atoms. Example (Countable measure) Let Ωhas a countable decomposition {Dk}, i.e., Ω= X Dk, Dj ∩Dj = ∅, i ̸= j. Let F = F∗and P(Dk) = αk > 0 and P k αk = 1. For the Poisson measure on nonnegative integers; Dk = {k}, m(Dk) = e−λ λk k! for λ > 0. Deﬁnition For any family C of subsets of E, we can deﬁne the σ-algebra σ(C) by the smallest σ-algebra A which contains C. The σ-algebra σ(C) is the intersection of all σ-algebras which contains C. It is again a σ-algebra, i.e., A = \ α Aα where Aα are all σ-algebras that contain C. The following construction of the measure space and the measure is essential for the triple (E, F, m) on uncountable space E. If (E, O) is a topological space, where O is the set of open sets in E, then the σ-algebra B(E) generated by O is called the Borel σ-algebra of the topological space E and (E, B(E)) deﬁnes the measure space and a set B in B(E) is called a Borel set. For example (Rn, B(Rn)) is the the measure space and B(Rn) is the σ-algebra generated by open balls in Rn. Caratheodory Theorem Let B = σ(A), the smallest σ-algebra containing an algebra A of subsets of E. Let µ0 is a σ additive measure of on (E, A). Then there exist a unique measure µ on (E, B) which is an extension of µ0, i.e., µ(A) = µ0(A), A ∈A. Deﬁnition A map f from a measure space (X, A) to an other measure space (Y, B) is called measurable, if f−1(B) = {x ∈X : f(x) ∈B} ∈A for all B ∈B. If f : (Rn, B(Rn)) →(Rm, B(Rm)) is measurable, we say f is a Borel function. In general every continuous function Rn →Rm is a Borel function since the inverse image of open sets in Rn are open in Rm. Example (Measure space) Let Ω= R and B(R) be the Borel σ-algebra. Note that (a, b] = \ n (a, b + 1 n), [a, b] = \ n (a −1 n, b + 1 n) ∈B(R). 15 Thus, B(R) coincides with the σ-algebra generated by the semi-closed intervals. Let A be the algebra of ﬁnite disjoint sum of semi-closed intervals (ai, bi] and deﬁne P0 by µ0( n X k=1 (ak, bk]) = n X k=1 (F(bk) −F(ak)) where x ∈R →F(x) ∈R+ is nondecreasing, right continuous and the left limit exists everywhere. We have the measure µ on (R, B(R)) by the Caratheodory Theorem if µ0 is σ-additive on A We now prove that µ0 is σ-additive on A. By the monotone convergence theorem it suﬃces to prove that P0(An) ↓0, An ↓∅, An ∈A. Without loss of the generality one can assume that An ⊂[−N, N]. Since F is the right continuous, for each An there exists a set Bn ∈A such that Bn ⊂An and µ0(An) −µ0(Bn) ≤ϵ2−n for all ϵ > 0. The collection of sets {[−N, N]\Bn} is an open covering of the compact set [−N, N] since ∩Bn = ∅. By the Heine-Borel theorem there exists a ﬁnite subcovering: n0 [ n=1 [−N, N] \ Bn = [−N, N]. and thus Tn0 n=1 Bn = 0. Thus, µ0(An0) = µ0(An0 \ Tn0 k=1 Bk) + µ0(Tn0 k=1 Bk) = µ0(An0 \ Tn0 k=1 Bk) µ0( n0 \ k=1 (Ak \ Bk)) ≤ n0 X k=1 µ0(Ak \ Bk) ≤ϵ. Since ϵ > 0 is arbitrary µ0(An) →0 as n →∞. Next, we deﬁne the integral of a measurable function f on (Ω, F, µ). Deﬁntion (Elementary function) An elementary function f is deﬁned by f(x) = n X k=1 xk IAk(x) where {Ak} is a partition of E, i.e, Ak ∈F are disjoint and ∪Ak = E. Then the integral of f is given by Z E f(x) dµ(x) = n X k=1 xkµ(Ak). Theorem (Approximation) For every measurable function f ≥0 on (E, F) there exists a sequence of elementary functions {fn} such that 0 ≤fn(x) ≤f(x) and fn(x) ↑ f(x) for all x ∈E. 16 Proof: For n ≥1, deﬁne a sequence of elementary functions by fn(ω) = n2n X k=1 k −1 2n Ik,n(ω) + nIf(x)>n where Ik,n is the indicator function of the set {x : k−1 2n < f(x) ≤ k 2n }. It is easy to verify that fn(x) is monotonically nondecreasing and fn(x) ≤f(x) and thus fn(x) →f(x) for all x ∈E. □ Deﬁnition (Integral) For a nonnegative measurable function f on (E, F) we deﬁne the integral by Z E f(x) dµ(x) = lim n→∞ Z E fn(x) dµ(x) where the limit exists since Z E fn(x) dµ(x) is an increasing number sequence. Note that f = f+ −f−with f+(x) = max(0, f(x)), f−(x) = max(0, f(x)). So, we can apply for the above Theorem and Deﬁnition for f+ and f−. Z E f(x) dµ(x) = Z E f+(x) dµ(x) − Z E f−(x) dµ(x) If Z E f+(x) dµ(x), Z E f−(x) dµ(x) < ∞, f is integrable and Z E \|f(x)\| dµ(x) = Z E f+(x) dµ(x) + Z E f−(x) dµ(x). Corollary Let f is a measurable function on (R, B(R), µ) with µ((a, b]) = F(b)−F(a), then we have as n →∞ Z ∞ 0 f(x) dµ(x) = n2n X k=0 f(k −1 2n )(F( k 2n )−F(k −1 2n ))+f(n)(1−F(n)) → Z ∞ 0 f(x) dF(x), which is the Lebesgue Stieljes integral with respect to measure dF. The space L of measurable functions on (E, F) is a complete metric space with metric d(f, g) = Z E \|f −g\| 1 + \|f −g\| dµ. In fact, d(f, g) = 0 if and if f = g almost everywhere and since \|f + g\| 1 + \|f + g\| ≤ \|f\| 1 + \|f\| + \|g\| 1 + \|g\|, d satisﬁes the triangle inequality. Theorem (completeness) (L, d) is a complete metric space. Proof: Let {fn} be a Cauchy sequence of (L, d). Select a subsequence fnk such that ∞ X k=1 d(fnk, fnk+1) < ∞ 17 Then, X k E[ \|fnk −fnk+1\| 1 + \|fnk −fnk+1\| < ∞ Since \|f\| ≤≤ 2\|f\| 1+\|f\| for \|f\| ≤1, ∞ X \|fnk −fnk+1\| < ∞a.e. and {fnk} almost everywhere converges to f for some f ∈L. Moreover, d(fnk, f) →0 and thus (L, d) is complete. Deﬁnition (Convergence in Measure) A sequence {fn} of mesurable functions on S con-verges to f in measure if for ϵ > 0, limn→∞m({s ∈S : \|fn(s) −f(s)\| ≥ϵ})=0. Theorem (Convergence in measure) {fn} converges in measure to f if and only if {fn} converges to f in d-metric. Proof: For f, g ∈L d(f, g) = Z \|f−g\|≥ϵ \|f −g\| 1 + \|f −g\| dµ + Z \|f−g\|<ϵ \|f −g\| 1 + \|f −g\| dµ ≤µ(\|f −g\| ≥ϵ) + ϵ 1 + ϵ holds for all ϵ > 0. Thus, fn converges in mesure to f, then fn converges to f in d-metric. Conversely, since d(f, g) ≥ ϵ 1 + ϵµ(\|f −g\| ≥ϵ) if fn converges to f in d-metric, then fn converges in measure to f. □ Corollary The completion of C(E) with respect to Lp(E)-norm: ( Z E \|f\|p dµ) 1 p is Lp(E) = {f ∈L : R E \|f\|p dµ < ∞}. 2.3 Baire’s Category Theory In this section we discuss the uniform boundedness principle and the open mapping theorem. The underlying idea of the proofs of these theorems is the Baire theorem for complete metric spaces. It is concerned with the decomposition of a space as a union of subsets. For instance, we have a countable union; R2 = [ k,j Sk,j, Sk = (k, k + 1] × (j, j + 1]. On the other hand, we have uncountable union; [ α∈R ℓα, ℓα = {α} × R 18 but ℓα has no interior points, the one dimensional subspace of R2. The question is can we represent R2 asa countable union of these sets? It turns out that the answer is no. Deﬁnition (Category) Let (X, d) be a metric space. A subset E of X is called nowhere dense if its closure does not contain any open set. X is said to be the ﬁrst category if X is the union of a countable number of nowhere dense sets. Otherwise, X is said to be the second category, i.e., suppose X = S k Ek, then the closure of at least one of Ek’s has non-empty interior. Baire-HausdorﬀTheorem A complete metric space is of the second category. Proof: Let {Ek} be a sequence of closed sets and assume ∪n En = X in a complete metric space X. We will show that there is at least one of Ek’s has non-empty interior. Assume otherwise, no En contains an interior point. Let On = Ec n and then On is open and dense in X for every n ≥1. Thus, O1 contains a closed ball B1 = {x : d(x, x1) ≤r1} with r1 ∈(0, 1 2). Since O2 is open and dense, there exists a closed ball B2 = {x : d(x, x2) ≤r2} with r2 ∈(0, 1 22 ) in B1. By repeating the same process, there exists a sequence {Bk} of closed ball Bk = {x : x : d(x, xk) ≤rk such that Bk+1 ⊂Bk, 0 < rk 1 2k , Bk ∩Ek = ∅. The sequence {xk} is a Cauchy sequence in X and since X is complete x∗= lim xn ∈X exists. Since d(xn, x∗) ≤d(xn, xm) + d(xm, x∗) ≤rn + d(xm, x∗) →rn as m →∞, x∈Bn for every n, i.e. x∗/ ∈En and thus x∗/ ∈∩nEn = X, which is a contradiction. □ Baire Category theorem Let M be a set of the ﬁrst category in a compact topolog-ical space. Then, the complement Mc is dense in X. Theorem (Totally bounded) A subset M in a complete metric space X is relatively compact if and only if it is totally bounded, i.e, for every ϵ > 0 there exists a fam-ily of ﬁnite many points {m1, m2, · · · , mn} in M such that for every point m ∈M, d(m, mk) < ϵ for at least one mk. Theorem (Banach-Steinhaus, uniform boundedness principle) Let X, Y be Banach spaces and let Tk, k ∈K be a family (not necessarily count-able) of continuous linear operators from X to Y . Assume that sup k∈K \|Tkx\|Y < ∞for all x ∈X. Then there exists a constant c such that \|Tkx\|Y ≤c \|x\|X for all k ∈K and x ∈X. Proof: Deﬁne Xn = {x ∈X : \|Tkx\| ≤n for all k ∈K}. Then, Xn is closed, and by the assumption we have X = ∪n Xn. It follows from the Baire category theorem that int(X0) is not empty for some n0 and select x0 ∈X and r > 0 such that B(x0, r) ⊂int(X0). We have Tk(x0 + r z) ≤n0 for all k and z ∈B(0, 1), 19 which implies r \|Tk\|L(X,Y ) ≤n0 + \|Tkx0\|.□ The uniform bounded principle is quite remarkable and surprising, since from point-wise estimates one derives a global (uniform) estimate. We have the following examples. Examples (1) Let X be a Banach space and B be a subset of X. If for every f ∈X∗ the set f(B) = {⟨f, x⟩, x ∈B} is bounded in R, then B is bounded. (2) Let X be a Banach space and B∗be a subset of X∗. If for every x ∈X the set ⟨B∗, x⟩= {⟨f, x⟩, f ∈B∗} is bounded in R, then B∗is bounded. Theorem (Banach closed graph theorem) A closed linear operator on a Banach space X to a Banach space Y is continuous. 2.4 Dual space and Hahn-Banach theorem In this section we discuss the dual space of a normed space and the Hahn-Banach theorem. Deﬁnition (Dual Space) If X be a normed space, let X∗denote the collection of all bounded linear functionals on X. X∗is called the dual space of X. We deﬁne the dual product ⟨x, f⟩X×X∗by ⟨f, x⟩X∗×X = f(x) for x ∈X f ∈X∗. For f ∈X∗we deﬁne the operator norm of f by \|f\|X∗= sup x̸=0 \|f(x)\| \|x\|X . Theorem (Dual space) The dual space X∗equipped with the operator norm is a Banach space. Proof: {fn} be a Cauchy sequence in X∗, i.e. \|fn(φ) −fm(φ)\| ≤\|fn −fm\|∗\|φ\| →0 as m →∞, for all φ ∈X. Thus, {fn(φ)} is Cauchy in R, deﬁne the functional f on X by f(φ) = lim n→∞fn(φ), φ ∈X. Then f is linear and for all ϵ > 0 there exists Nϵ such that for m, n ≥Nϵ \|fm(φ) −fn(φ)\| ≤ϵ \|φ\| Letting m →∞, we have \|f(φ) −fn(φ)\| ≤ϵ \|φ\| for all n ≥Nϵ. and thus f ∈X∗and \|fn −f\|∗→0 as n →∞. □ Examples (Dual space) (1) Let ℓp be the space of p-summable inﬁnite sequences x = (s1, s2, · · · ) with norm \|x\|p = (P \|sk\|p) 1 p . Then, (ℓp)∗= ℓq, 1 p + 1 q = 1 for 1 ≤p < 20 ∞. Let c = ℓ∞be the space of uniformly bounded sequences x = (ξ1, ξ2, · · · ) with norm \|x\|∞= maxk \|ξk\| and c0 be the space of uniformly bounded sequences with lim n →∞ξn = 0. Then, (ℓ1)′ = c0 (2) Let Lp(Ω) is p-integrable functions f on Ωwith \|f\|Lp = ( Z Ω \|f(x)\|p dx) 1 p Then, Lp(Ω)∗= Lq(Ω), 1 p + 1 q = 1 for p ≥1. Especially, L2(Ω)∗= L2(Ω) and L1(Ω)∗= L∞(Ω), where L∞(Ω) = the space of essentially bounded functions on Ω with \|f\|L∞= inf M where \|f(x)\| ≤M almost everywhere on Ω. (3) Consider a linear functional f(x) = x(1). Then, f is in X∗ 1 but not in X∗ 2 since for the sequence deﬁned by xn(t) = tn, t ∈[0, 1] we have f(xn) = 1 but \|xn\|X2 = q 1 2n+1 →0 as n →∞. Since for x ∈X2 \|x(t)\| = \|x(0) + Z t 0 x′(s) ds\| ≤ sZ 1 0 \|x′(t)\|2 dt, thus f ∈X∗ 2. (4) The total variation V 1 0 (g) of a function g on an interval [0, 1] is deﬁned by V 1 0 (g) = sup P∈P n−1 X i=0 \|g(xi+1) −g(xi)\|, where the supremum is taken over all partitions P = {P : 0 ≤t0 ≤· · · ≤tn ≤1} of the interval [0, 1]. Then, BV (0, 1) = {uniformly bounded function g with V 1 0 (f) < ∞} and \|g\|BV = V 1 0 (g). It is a normed space if we assume g(0) = 0. For example, every monotone functions is of bounded variation. Every real-valued BV-function can be expressed as the diﬀerence of two increasing functions (Jordan decomposition theorem). Hahn-Banach Theorem Suppose S is a subspace of a normed space X and f : S →R is a linear functional satisfying f(x) ≤\|x\| for all x ∈S. Then there exists an F ∈X∗ such that F(x) ≤\|x\| and F(s) = f(s) for all s ∈S. In general, the theorem holds for a vector space X and sub-additive and positive homogeneous function p, i.e., for all x, y ∈X and α > 0 p(x + y) ≤p(x) + p(y), p(αx) = αp(x). Every linear functional f on a subspace S, satisfying f(x) ≤p(x) has a extension F on X satisfying F(x) ≤p(x) for all x ∈X. 21 Proof: First, we show that there exists an one step extension of f, i.e., for x0 ∈X \ S, there exists an extension f1 of f on the subspace Y1 spanned by (S, x0) such that f1(x) ≤p(x) on Y1. Note that for arbitrary s1, s2 ∈S, f(s1) + f(s2) ≤p(s1 −x0) + p(s2 + x0). Hence, there exist a constant c such that sup s∈S {f(s) −p(s −x0)} ≤c ≤inf s∈S{p(s + x0) −f(s)}. Extend f by deﬁning, for s ∈S and α ∈R f1(s + αx0) = f(s) + α c, where f1(x0) = c, is determined as above. First note that the representation s+α x0 ∈ Y1 is unique since if s1 + α x0 = s2 + β x0 s2 −s1 = (α −β)x0 and thus s1 = s2 and α = β. Next, we claim that f1(s + αx0) ≤p(s + αx0) for all α ∈R and s ∈S, i.e., by the deﬁnition of c, for α > 0 f1(s + αx0) = α(c + f( s α) ≤α (p( s α + x0) −f( s α) + f( s α) = p(s + αx0), and for all α < 0 f1(s + αx0) = −α(−c + f( s −α) ≤−α(p( s −α −x0) −f( s −α) + f( s −α) = p(s + αx0). If X is a separable normed space. Let {x1, x2, . . . } be a countable dense base of X \ S. Select victors from this dense vectors, which is independent of S, recursively. Extend f to subspaces Yn = span{A, x1, ·, xn}, recursively, as described above. Thus, we have an extension F on Y = span{S, x1, x2, . . . } Since Y is dense in X, the ﬁnal extension F is deﬁned by the continuous extension F from Y to X = ¯ Y . In general, the remaining proof is done by using Zorn’s lemma. □ Remark If X is a Hilbert space, then we have the orthogonal decomposition theorem X = S ⊕S⊥. Thus, one can deﬁne the extension of f by f(x) = f(x1) for x = x1 + x2, ; x1 ∈S, x2 ∈S⊥. Corollary Let X be a normed space. For each x0 ∈X there exists an f ∈X∗such that f(x0) = \|f\|X∗\|x0\|X. Proof of Corollary: Let S = {α x0 : α ∈R} and deﬁne f(α x0) = α \|x0\|X. By Hahn-Banach theorem there exits an extension F ∈X∗of f such that F(x) ≤\|x\| for all x ∈X. Since −F(x) = F(−x) ≤\| −x\| = \|x\|, 22 we have \|F(x)\| ≤\|x\|, in particular \|F\|X∗≤1. On the other hand, F(x0) = f(x0) = \|x0\|, thus \|F\|X∗= 1 and F(x0) = f(x0) = \|F\|\|x0\|. □ Example (Hahn-Banach Theorem) (1) \|x\| = sup\|x∗\|=1 ⟨x∗, x⟩ (2) Next, we prove the geometric form of the Hahn-Banach theorem. It states that given a convex set K containing an interior point and x0 / ∈K, then x0 is separated from K by a hyperplane. Let K be a convex set and 0 is an interior point of K. A hyperplane is a maximal aﬃne space x0+M where M is a subspace of X. We introduce the Minkowski functional that deﬁnes the distance from the origin measured by K. Deﬁnition (Minkowski functional) Let K be a convex set in a normed space X and 0 is an interior point of K. Then the Minkowski functional p of K is deﬁned by p(x) = inf{r : x r ∈K, r > 0}. Note that if K is the unit sphere in X, p(x) = \|x\|. Lemma (Minkowski functional) (1) 0 ≤p(x) < ∞. (2) p(αx) = α p(x) for α ≥0. (3) p(x1 + x2) ≤p(x1) + p(x2). (4) p is continuous. (5) ¯ K = {x : p(x) ≤1}, int(K) = {x : p(x) < 1}. Proof: (1) Since K contains a sphere at 0, for every x ∈X there exists an r > 0 such that x r ∈K and thus p(x) is ﬁnite. (2) For α > 0 p(αx) = inf{r : αx r ∈K} = inf{αr′ : x r′ ∈K} = α inf{r′ : x r′ ∈K} = α p(x). (3) Let p(xi) < ri < p(xi)+ϵ, i = 1, 2 for arbitrary ϵ > 0. Let r = r1+r2. By convexity of K since r1 r x1 r1 + r2 r x2 r2 = x1+x2 r ∈K and thus p(x1 + x2) ≤r ≤p(x1) + p(x2) + 2ϵ. Since ϵ > 0 is arbitrary, p is subadditive. (4) Let ϵ be a radius of a closed sphere centered at 0 and contained in K. Since ϵ x \|x\| ∈K, p(ϵ x \|x\|) ≤1 and thus p(x) ≤1 ϵ\|x\|. This shows that p is continuous at 0. Since p is sub linear p(x) = p(x −y + y) ≤p(x −y) + p(y) and p(y) = p(y −x + x) ≤p(y −x) + p(x) and thus −p(y −x) ≤p(x) −p(y) ≤p(x −y). Hence, p is continuous since it is continuous at 0. (5) It follows from the continuity of p. □ Lemma (Hyperplane) Let H be a hyperplane in a vector space X if and only if there exist a linear functional f on X and a constant c such that H = {f(x) = c}. Proof: If H ia hyperplane, then H is the translation of a subspace M in X, i.e., H = x0 + M. If x0 ̸ inM, then [M + x0] = X. if for x = α x0 + m, m ∈M we let f(x) = α, we have H = {x : f(x) = 1}. Conversely, if f is a nonzero linear functional 23 on X, then M = {x : f(x) = 0} is a subspace. Let x0 ∈X such that f(x0) = 1. Since f(x −f(x)x0) = 0, X = span[{x0}, M] and {x : f(x) = c} = {f(x −c x0) = 0} is a hyperplane. Theorem (Mazur’s Theorem) Let K be convex set with a nonempty interior point in a normed space X. Suppose V is an aﬃne space in X that contains no interior points of K. Then, there exist x∗∈X∗and constant c such that the hyperplane langlex∗, v⟩= c has the separation: ⟨x∗, v⟩= c for all v ∈V and ⟨x∗, k⟩< c for all k ∈int(K). Proof: By an appropriate translation one may assume that 0 is an interior point of K. Let M is the subspace spanned by V . Since V does not contain 0, exists a linear functional f on M such that V = {x : f(x) = 1}. Let p be the Minkowski functional of K. Since V contains no interior points of K, f(x) = 1 ≤p(x) for all x ∈V . By the homogeneity of p, f(x) ≤p(x) for all x ∈V By the Hahn-Banach space, there exits an extension F of f to V from M to X with F(x) ≤p(x). From Lemma p is continuous and F is continuous and F(x) < 1 for x ∈int(K). Thus, H = {x : F(x) = 1} is the desired closed hyperplane. □ Example (Hilbert space) If X is a Hilbert space and S is a subspace of X. Then the extension F is deﬁned as F(s) = lim f(sn) for s ∈S 0 for s ∈S⊥ where X = S ⊕S⊥. Example (Integral) Let X = L2(0, 1) S = C(0, 1) and f(x) = R − R 1 0 x(t) dt be the Riemman integral of x ∈S . Then the extension F is F(x) = L − R 1 0 x(t) dt is the Lebesgue integral of x ∈X. Example (Alignment) (1) For Hilbert space X f = x0 is self aligned. (2) Let x0 ∈X = Lp(Ω), 1 ≤p < ∞. For f ∈X∗= Lq(Ω) deﬁned by by f(t) = \|x0(t)\|p−2x0(t), t ∈Ω, f(x) = Z Ω \|x0(t)\|p dt. Thus, x0 \|x0\|p−2 ∈X∗is aligned with x0. A function g is called right continuous if limh→0+ g(t + h) = g(t). Let BV0[0, 1] = {g ∈BV0[0, 1] : g is right continuous on [0, 1) and g(0) = 0}. Let C[0, 1] is the normed space of continuous functions on [0, 1] with the sup norm. For f ∈C[0, 1] and g ∈BV0[0, 1] deﬁne the Riemann-Stieltjes integral by Z t 0 x(t)dg(t) = lim ∆t R(f, g, P) = lim ∆t X k=1 f(zk)(g(tk) −g(tk−1) for the partition P = {0 = t0 < · · · < tn = 1} and zk ∈[tk−1, tk]. There is a version of the Riesz representation theorem. 24 Theorem (Dual of C[0, 1]) There is a norm-preserving linear isomorphism from C[0, 1] to V [0, 1], i.e., for F ∈C[0, 1]∗there exists a unique g ∈BV0[0, 1] such that F(x) = Z 1 0 x(t)dg(t) and \|F\|C[0,1]∗= \|g\|BV . That is, C[0, 1]∗∼ = BV (0, 1), the space of functions with bounded variation. Proof: Deﬁne the normed space of uniformly bound function on [0, 1] by B = {x : [0, 1] →R is bounded on [0, 1]} with norm \|x\|B = sup t∈[0,1] \|x(t)\|. By the Hahn-Banach theorem, an extension F of f from C[0, 1] to B exists and \|F\| = \|f\|. For any s ∈(0, 1], deﬁne χ[0,s] ∈B and deﬁne v(s) = F(χ[0,s]). Then, P k \|v(tk) −v(tk−1)\| = P k sign(v(tk) −v(tk−1))(v(tk) −v(tk−1)) = sign(v(tk) −v(tk−1)(F(χ[0,tk]) −F(χ[0,tk−1])) = F(P k sign(v(tk) −v(tk−1)(χ[0,tk] −χ[0,tk−1]) ≤\|F\|\| P k sign(v(tk) −v(tk−1)χ(tk−1,tk]\| ≤\|f\|. Thus, v ∈BV [0, 1] and \|v\|BV ≤\|F\|. Next, for any x ∈C[0, 1] deﬁne the function z(t) = X k x(tk−1(χ[0,tk] −χ[0,tk−1]). Note that the diﬀerence \|z −x\|B = max k max t∈[tk−1,tk] \|x(tk−1) −x(t)\| →0 as ∆t →0+. Since F is bounded F(z) →F(x) = f(x) and F(z) = X k x(tk−1)(v(tk) −v(tk−1)) → Z 1 0 x(t)dv(t) we have f(x) = Z 1 0 x(t)dv(t) and \|F\| ≤\|x\|C[0,1]\|v\|BV . Hence F\| = \|f\| = \|v\|BV . □ Riesz Representation Theorem For every bounded linear functional f on a Hilbert space X there exits a unique element yf ∈X such that f(x) = (x, yf) for all x ∈X, and \|f\|X∗= \|yf\|X. We deﬁne the Riesz map R : X∗→X by Rf = yf ∈X for f ∈X∗ 25 for a Hilbert space X. Proof: There exists a z ∈X such that (z, x) = 0 for all f(x) = 0 and \|z\| = 1. Otherwise, f(x) = 0 for all x and f = 0. Note that for every x ∈X we have f(f(x)z −f(z)x) = 0. Thus, 0 = (f(x) −f(z)x, z) = (f(x)z, z) −(f(z)x, z) which implies f(x) = (x, f(z)z) The existence then follows by taking yf = f(z)z. Let for u1, u2 f(x) = (x, u1) = (x, u2) for all x ∈X. Then, then (x, u1 −u2) = 0 and by taking x = u1 −u2 we obtain \|u1 −u2\| = 0, which implies the uniqueness. Deﬁnition (Reﬂexive Banach space) For a Banach space X we deﬁne an injection i of X into X∗∗= (X∗)∗(the double dual) as follows. If x ∈X then i x ∈X∗∗is deﬁned by i x(f) = f(x) for f ∈X∗. Note that \|i x\|X∗∗= sup f∈X∗ \|f(x)\| \|f\|X∗≤sup f∈X∗ \|f\|X∗\|x\|X \|f\|X∗ = \|x\|X an thus \|i x\|X∗∗≤\|x\|X. But by Hahn-Banach theorem there exists an f ∈X∗such that \|f\|X∗= 1 and f(x) = \|x\|X. Hence \|i x\|X∗∗= \|x\|X and i : X →X∗∗is an isometry. If i is also subjective, then we say that X is reﬂexive and we may identify X with X∗∗. Deﬁnition (Weak and Weak star Convergence) (1) A sequence {xn} in a normed space X is said to converge weakly to x ∈X if limn→∞f(xn) →f(x) for all f ∈X∗ (2) A sequence {fn} in the dual space X∗of a normed space X is said to converge weakly star to f ∈X∗if limn→∞fn(x) →f(x) for all x ∈X. Notes (1) If xn converges strongly to x ∈X then xn converges weakly to x, but not conversely. (2) If xn converges weakly to x, then {\|xn\|X} is bounded and \|x\| ≤lim inf \|xn\|X, i.e., the norm is weakly lower semi-continuous. (3) A sequence {xn} in a normed space X converges weakly to x ∈X if limn→∞f(xn) → f(x) for all f inS, where S is a dense set in X∗. (4) If X be a Hilbert space. If a sequence {xn} of X converges weakly to x ∈X, then xn converges strongly to x if and only if limn→∞\|xn\| = \|x\|. In fact, \|xn −x\|2 = (xn −x, xn −x) = \|xn\|2 −(xn, x) −(x, xn) + \|x\|2 →0 as n →∞if xn converges weakly to x. Example (Weak Convergence) (1) Let X = L2(0, 1) and consider a sequence xn = √ 2 sin(n πt), t ∈(0, 1). Then \|xn\| = 1 and xn converges weakly to 0. In fact, since a family {sin(kπt)} is dense in L2(0, 1) and (xn, sin(kπt))L2 →0 as n →∞ for all k and thus xn converges weakly to 0. Since \|0\| = 0, the sequence xn does not converge strongly to 0. 26 In general, if xn →x weakly and yn →y strongly in a Hilbert space, then (xn, yn) → (x, y) since (xn, yn) = (xn −x, y) + (xn, yn −y). But, note that (xn, xn) = 2 R 1 0 sin2(nπt) dt = 1 ̸= 0 and (xn, xn) does not converges to 0. In general (xn, yn) does not necessarily converge to (x, y) if xn →x and yn →y weakly in X. Alaoglu Theorem If X is a normed space and {x∗ n} is a bounded sequence in X∗, then there exists a subsequence that converges weakly star to an element of X∗. Proof: We prove the theorem for the case when X is separable. Let {x∗ n} be sequence in X∗such that \|x∗ n\| ≤1. Let {xk} is a dense sequence in X. The sequence {⟨x∗ n, x1⟩} is a bounded sequence in R and thus a convergent subsequence denoted by {⟨x∗ n1, x1⟩}. Similarly, the sequence {⟨x∗ n1, x2⟩} contains a a convergent subsequence {⟨x∗ n2, x1⟩}. Continuing in this manner we extract subsequences {⟨x∗ nk, xk⟩} that converges in R. Thus, the diagonal sequence {x∗ nn} on the dense subset {xk}, i.e., {⟨x∗ nn, xk⟩} converges for each xk. We will show that x∗ nn converges weakly start to an element x∗∈X∗. Let a x ∈X be ﬁxed. For arbitrary ϵ > 0 there exists N ≥Nϵ such that \|x − N X k=1 αk xk\| ≤ϵ 3. Thus, \|⟨x∗ nn, x⟩−⟨x∗ mm, x⟩\| ≤\|⟨x∗ nn, x −PN k=1 αkxk⟩\| +\|⟨x∗ nn −x∗ mm, PN k=1 αkxk⟩\| + \|⟨x∗ mm, x −PN k=1 αkxk⟩\| ≤ϵ for suﬃciently large nn, mm. Thus, {⟨x∗ nn, x⟩} is a Cauchy sequence in R and converges to a real number ⟨x∗, x⟩. The functional ⟨x∗, x⟩so deﬁned is obviously linear and \|x∗\| ≤1 since \|⟨x∗ nn, x⟩\| ≤\|x\|. □ Example (Weak Star Convergence) (1) Let X = L1(Ω) and L∞(Ω) = X∗. Thus, a bounded sequence in L∞(Ω) has a weakly star convergent subsequence. Let X = L∞(0, 1) and {xn} be a sequence in X deﬁned by xn(t) =            1 on [ 1 2 + 1 n, 1] n x on [ 1 2 −1 n, 1 2 −1 n] −1 on [0, 1 2 −1 n]; Then \|xn\|X = 1 and {xn} converges weakly star to the sign function x(t) = sign(t−1 2). But, since \|xn −x\| = 1 2, {xn} does not converge strongly to x. (2) Let X = c0, the space of inﬁnite sequences x = (x1, x2, · · · ) that converge to zero with norm \|x\|∞= maxk \|xk\|. Then X∗= ℓ1 and X∗∗= ℓ∞. Let {x∗ n} in ℓ1 = X∗ deﬁned by x∗ n = (0, · · · , 1, 0, · · · ) where 1 appears at n−th term only. Then, x∗ n converge weakly star to 0 in X∗but x∗ n does not converge weakly to 0 since ⟨x∗ n, x∗∗⟩does not converge for x∗∗= (1, 1, · · · ) ∈X∗∗. 27 Example (L1(Ω) bounded sequence) Let X = L1(0, 1) and let {xn} ∈X be a sequence deﬁned by xn(t) = n t ∈( 1 2 −1 2n, 1 2 + 1 2n) 0 otherwise. Then, \|xn\|L1 = 1 and Z 1 0 xn(t)φ(t) dt →φ(1 2) for all φ ∈C[0, 1] but a linear functional f(φ) = φ( 1 2) is not bounded on L∞(0, 1). Thus, xn does not converge weakly. Actually, sequence xn converges weakly star in C[0, 1]∗. Eberlein-Suhmulyan Theorem A Banach space X is reﬂexive if and only if and only if every closed bounded set is weakly sequentially compact, i.e., every bounded sequence of X contains a subsequence that converges weakly to an element of X. Next, we consider linear operator T : X →Y . Deﬁnition (Linear Operator) Let X, Y be normed spaces. A linear operator T on D(T) ⊂X into Y satisﬁes T(α x1 + β x2) = α Tx1 + β Tx2 where the domain D(T) is a subspace of X. Then a linear operator T is continuous if and only if there exists a constant M ≥0 such that \|Tx\|Y ≤M \|x\|X for all x ∈X. For a continuous linear operator we deﬁne the operator norm \|T\| by \|T\| = sup \|x\|X=1 \|Tx\|Y = inf M, where we assume D(T) = X. A continuous linear operator on a normed space X into Y is called a bounded linear operator on X into Y and will be denoted by T ∈L(X, Y ). The norm of T ∈L(X, Y ) is denoted either by \|T\|L(X,Y ) or simply by \|T\|. Note that the space L(X, Y ) is a Banach space with the operator norm and X∗= L(X, R). Open Mapping theorem Let T be a continuous linear operator from a Banach space X onto a Banach space Y . Then, T is an open map, i.e., maps every open set to of X onto an open set in Y . Proof: Suppose A : X →Y is a surjective continuous linear operator. In order to prove that A is an open map, it is suﬃcient to show that A maps the open unit ball in X to a neighborhood of the origin of Y . Let U = B1(0) and V = B1(0) be a open unit ball of 0 in X and Y , respectively. Then X = S k∈N kU. Since A is surjective, Y = A(X) = A [ k∈N kU ! = [ k∈N A(kU). Since Y is a Banach, by the Baire’s category theorem there exists a k ∈N such that A(kU) ◦ ̸= ∅. Let c ∈ A(kU) ◦ . Then, there exists r > 0 such that Br(c) ⊆ 28 A(kU) ◦ . If v ∈V , then c, c + rv ∈Br(c) ⊆ A(kU) ◦ ⊆A(kU). and the diﬀerence rv = (c + rv) −c of elements in A(kU) ◦ satisﬁes rv ∈⊆A(2kU). Since A is linear, V ⊆A 2k r U . Thus, it follows that for all y ∈Y and ϵ > 0, there exists x ∈X such that \|x\|X < 2k r \|y\|Y and \|y −Ax\|Y < ϵ. (2.1) Fix y ∈δV , by (2.1), there is some x1 ∈X with \|x1\| < 1 and \|y −Ax1\| < δ 2. By (2.1) one can ﬁnd a sequence {xn} inductively by \|xn\| < 2−(n−1) and \|y −A(x1 + x2 + · · · + xn)\| < δ2−n. (2.2) Let sn = x1+x2+· · ·+xn. From the ﬁrst inequality in (2.2), {sn} is a Cauchy sequence, and since X is complete, {sn} converges to some x ∈X. By (2.2), the sequence Asn tends to y in Y , and thus Ax = y by continuity of A. Also, we have \|x\| = lim n→∞\|sn\| ≤ ∞ X n=1 \|xn\| < 2. This shows that every y ∈δ V belongs to A(2U), or equivalently, that the image A(U) of U = B1(0) in X contains the open ball δ 2V ⊂Y . □ Corollary (Open Mapping theorem) A bounded linear operator T on Banach spaces is bounded invertible if and only if it is one-to-one and onto. Deﬁnition (Closed Linear Operator) (1) The graph G(T) of a linear operator T on the domain D(T) ⊂X into Y is the set (x, Tx) : x ∈D(T)} in the product space X × Y . Then T is closed if its graph G(T) is a closed linear subspace of X × Y , i.e., if xn ∈D(T) converges strongly to x ∈X and Txn converges strongly to y ∈Y , then x ∈D(T) and y = Tx. Thus the notion of a closed linear operator is an extension of the notion of a bounded linear operator. (2) A linear operator T is said be closable if xn ∈D(T) converges strongly to 0 and Txn converges strongly to y ∈Y , then y = 0. For a closed linear operator T, the domain D(T) is a Banach space if it is equipped by the graph norm \|x\|D(T) = (\|x\|2 X + \|Tx\|2 Y ) 1 2 . Example (Closed linear Operator) Let T = d dt with X = Y = L2(0, 1) is closed and dom (A) = H1(0, 1) = {f ∈L2(0, 1) : absolutely continuous functions on [0, 1] with square integrable derivative}. If yn = Txn, then xn(t) = xn(0) + Z t 0 yn(s) ds. If xn ∈dom(T) →x and yn →y in L2(0, 1), then letting n →∞we have x(t) = x(0) + Z t 0 y(s) ds, 29 i.e., x ∈dom (T) and Tx = y. In general if for λ I + T for some λ ∈R has a bounded inverse (λ I + T)−1, then T : dom (A) ⊂X →X is closed. In fact, Txn = yn is equivalent to xn = (λ I + T)−1(yn + λ xn Suppose xn →x and yn →y in X, letting n →∞in this, we have x ∈dom (T) and Tx = T(λ I + T)−1(λ x + y) = y. Deﬁnition (Dual Operator) Let T be a linear operator X into Y with dense domain D(T). The dual operator of T ∗of T is a linear operator on Y ∗into X∗deﬁned by ⟨y∗, Tx⟩Y ∗×Y = ⟨T ∗y∗, x⟩X∗×X for all x ∈D(T) and y∗∈D(T ∗). In fact, for y∗∈Y ∗x∗∈X∗satisfying ⟨y∗, Tx⟩= ⟨x∗, x⟩for all x ∈D(T) is uniquely deﬁned if and only if D(T) is dense. The only if part follows since if D(T) ̸= X then the Hahn-Banach theory there exits a nonzero x∗ 0 ∈X∗such that ⟨x∗ 0, x⟩= 0 for all D(T), which contradicts to the uniqueness assumption. If T is bounded with D(T) = X then T ∗is bounded with ∥T∥= ∥T ∗∥. Examples Consider the gradient operator T : L2(Ω) →L2(Ω)n as Tu = ∇u = (Dx1u, · · · Dxnu) with D(T) = H1(Ω). The, we have for v ∈L2(Ω)n T ∗v = −div v = − X Dxkvk with domain D(T ∗) = {v ∈L2(Ω)n : div v ∈L2(Ω) and n·v = 0 at the boundary ∂Ω}. In fact by the divergence theorem (Tu, v) = Z Ω ∇u · v Z ∂Ω (n · v)u ds − Z Ω u(div v) dx = (u, T ∗v) for all v ∈C1(Ω). First, let u ∈H1 0(Ω) we have T ∗v = −div v ∈L2(Ω) since H1 0(Ω) is dense in L2(Ω). Thus, n · v ∈L(∂Ω) and n · v = 0. Deﬁnition (Hilbert space Adjoint operator) Let X, Y be Hilbert spaces and T be a linear operator X into Y with dense domain D(T). The Hilbert self adjoint operator of T ∗of T is a linear operator on Y into X deﬁned by (y, Tx)Y = (T ∗y, x)X for all x ∈D(T) and y ∈D(T ∗). Note that if we let T ′ : Y ∗→X∗is the dual operator of T, then T ∗RY ∗→Y = RX∗→XT ′ where RX∗→X and RY ∗→Y are the Riesz maps. 30 Examples (self-adjoint operator) Let X = L2(Ω) and T be the Laplace operator Tu = ∆u = n X k=1 Dxkxku with domain D(T) = H2(Ω) ∩H1 0(Ω). Then T is sel-adjoint, i.e., T ∗= T. In fact (Tu, v)X = Z Ω ∆u v dx = Z ∂Ω ((n · ∇u)v −(n · ∇v)u) ds + Z Ω ∆v u dx = (x, T ∗v) for all v ∈C1(Ω). Closed Range Theorem Let X and Y be Banach spaces and T a closed linear op-erator on D(T) ⊂X into Y . Assume that D(T) is dense in X. The the following properties are all equivalent. (a) R(T) is closed in Y . (b) R(T ∗) is closed in X∗. (c) R(T) = N(T ∗)⊥= {y ∈Y : ⟨y∗, y⟩= 0 for all y∗∈N(T ∗)}. (d) R(T ∗) = N(T)⊥= {x∗∈X∗: ⟨x∗, x⟩= 0 for all x∗∈N(T)}. Proof: If y∗∈N(T ∗)⊥, then ⟨Tx, y∗⟩= ⟨x, T ∗y∗⟩ for all x ∈dom(T) and N(T ∗)⊥⊂R(T). If y ∈R(T), then ⟨y, y∗⟩= ⟨x, T ∗y∗⟩ for all y∗∈N(T ∗) and R(T) ⊂N(T ∗)⊥. Thus, R(T) = N(T ∗)⊥. Since T is closed, the graph G = G(T) = {(x, Tx)inX × Y : x ∈dom(T)} is closed. Thus, G is a Banach space equipped with norm \|{x, y}\| = \|x\| + \|y\| of X × Y . Deﬁne a continuous linear operator S from G into Y by S{x, Tx} = Tx Then, the dual operator S∗of S is continuous linear operator from Y ∗intoG∗and we have ⟨{x, Tx}, S∗y∗⟩= ⟨S{x, Tx}, y∗⟩= ⟨Tx, y∗⟩= ⟨{x, Tx}, {0, y∗}⟩ for x ∈dom(T) and y∗∈Y ∗. Thus, the functional ˜ S ˜ Sy∗= S ∗y∗−{0, y∗} : Y ∗→ X∗×Y ∗vanishes for all elements of G. But, since ⟨{x, Tx}, {x∗, y∗ 1} for all x ∈dom(T), is equivalent to ⟨x, x∗⟩= ⟨−Tx, y∗ 1⟩for all x ∈dom(T), i.e., −T ∗y∗ 1 = x∗, S∗y∗= {0, y∗} + {−T ∗y∗ 1, y∗ 1} = {−T ∗y∗ 1, y∗+ y∗ 1} for all y∗∈Y ∗. Since y∗ 1 is arbitrary, R(S∗) = R(T ∗) × Y ∗. Therefore, R(S∗) is closed in X∗× Y ∗if and only if R(T) is closed and since R(S) = R(T), R(S) is closed if and only if R(T) is closed in Y . Hence we have only prove the equivalency of (a) and (b) in the special case of a bounded operator S. If T is bounded, then (a) implies(b). Since R(T) a Banach space, by the Hahn-Banach theorem, one can assume R(T) = Y without loss of generality ⟨Tx, y∗ 1⟩= ⟨x, T ∗y∗ 1⟩ 31 By the open mapping theorem, there exists c > 0 such that for each y ∈Y there exists an x ∈X with Tx = y and \|x\| ≤c \|y\|. Thus, for y∗∈Y ∗, we have \|⟨y, y∗⟩\| = \|⟨Tx, y∗⟩\| = \|⟨x, T ∗y∗⟩\| ≤c\|y\|\|T ∗y∗\|. and \|y∗\| ≤sup \|y\|≤1 \|⟨y, y∗⟩\| ≤c\|T ∗y∗\| and so (T ∗)−1 exists and bounded and D(T ∗) = R(T ∗) is closed. Let T1 : X →Y1 = R(T) be deﬁned by T1x = Tx For y∗ 1 ∈Y ∗ 1 ⟨T1x, y∗ 1⟩= ⟨x, T ∗y∗⟩= 0, for all x ∈X and since R(T) is dense in R(T), y∗ 1 = 0. Since R(T ∗) = R(T ∗ 1 ) is closed, (T ∗ 1 )−1 exists on Y1 = (Y1)∗ 2.5 Distribution and Generalized Derivatives In this section we introduce the distribution (generalized function). The concept of distribution is very essential for deﬁning a generalized solution to PDEs and provides the foundation of PDE theory. Let D(Ω) be a vector space of all inﬁnitely many con-tinuously diﬀerentiable functions C∞ 0 (Ω) with compact support in Ω. For any compact set K of Ω, let DK(Ω) be the set of all functions f ∈C∞ 0 (Ω) whose support are in K. Deﬁne a family of seminorms on D(Ω) by pK,m(f) = sup x∈K sup \|s\|≤m \|Dsf(x)\| where Ds = ∂ ∂x1 s1 · · · ∂ ∂xn sn where s = (s1, · · · , sn) is nonnegative integer valued vector and \|s\| = P sk ≤m. Then, DK(Ω) is a locally convex topological space. Deﬁnition (Distribution) A linear functional T deﬁned on C∞ 0 (Ω) is a distribution if for every compact subset K of Ω, there exists a positive constant C and a positive integer k such that \|T(φ)\| ≤C sup\|s\|≤k, x∈K \|Dsφ(x)\| for all φ ∈DK(Ω). Deﬁnition (Generalized Derivative) A distribution S deﬁned by S(φ) = −T(Dxkφ) for all φ ∈C∞ 0 (Ω) is called the distributional derivative of T with respect to xk and we denote S = DxkT. In general we have S(φ) = DsT(φ) = (−1)\|s\| T(Dsφ) for all φ ∈C∞ 0 (Ω). 32 This deﬁnition is naturally followed from that for f is continuously diﬀerentiable Z Ω Dxkfφ dx = − Z Ω f ∂ ∂xk φ dx and thus Dxkf = DxkTf = T ∂ ∂xk f. Thus, we let Dsf denote the distributional deriva-tive of Tf. Example (Distribution) (1) For f is a locally integrable function on Ω, one deﬁnes the corresponding distribution by Tf(φ) = Z Ω fφ dx for all φ ∈C∞ 0 (Ω). since \|Tf(φ)\| ≤ Z K \|f\| dx sup x∈K \|φ(x)\|. (2) T(φ) = φ(0) deﬁnes the Dirac delta δ0 at x = 0, i.e., \|δ0(φ)\| ≤sup x∈K \|φ(x)\|. (3) Let H be the Heaviside function deﬁned by H(x) = 0 for x < 0 1 for x ≥0 Then, DTH(φ) = − Z ∞ −∞ H(x)φ′(x) dx = φ(0) and thus DTH = δ0 is the Dirac delta function at x = 0. (4) The distributional solution for −D2u = δx0 satisﬁes − Z ∞ −∞ uφ′′ dx = φ(x0) for all φ ∈C∞ 0 (R). That is, u = 1 2\|x −x0\| is the fundamental solution, i.e., − Z ∞ −∞ \|x −x0\|φ′′ dx = Z x0 ∞ φ′(x) dx − Z ∞ x0 φ′(x) dx = 2φ(x0). In general for d ≥2 let G(x, x0) =    1 4π log\|x −x0\| d = 2 cd \|x −x0\|2−d d ≥3. Then ∆G(x, x0) = 0, x ̸= x0. and u = G(x, x0) is the fundamental solution to to −∆in Rd, −∆u = δx0. 33 In fact, let Bϵ = {\|x−x0\| ≤ϵ} and Γ = {\|x−x0\| = ϵ} be the surface. By the divergence theorem Z Rd\Bϵ(x0) G(x, x0)∆φ(x) dx = Z Γ ∂ ∂ν φ(G(x, x0) −∂ ∂ν G(x, x0)φ(s)) ds = Z Γ (ϵ2−d ∂φ ∂ν −(2 −d)ϵ1−dφ(s)) ds →1 cd φ(x0) That is, G(x, x0) satisﬁes − Z Rd G(x, x0)∆φ dx = φ(x0). In general let L be a linear diﬀrenrtial operator and L∗denote the formal adjoint operator of L An locally integrable function u is said to be a distributional solution to Lu = T where L with a distribution T if Z Ω u(L∗φ) dx = T(φ) for all φ ∈C∞ 0 (Ω). Deﬁnition (Sovolev space) For 1 ≤p < ∞and m ≥0 the Sobolev space is W m,p(Ω) = {f ∈Lp(Ω) : Dsf ∈Lp(Ω), \|s\| ≤m} with norm \|f\|W m,p(Ω) =   Z Ω X \|s\|≤m \|Dsf\|p dx   1 p . That is, \|Dsf(φ)\| ≤c \|φ\|Lq with 1 p + 1 q = 1. Remark (1) X = W m,p(Ω) is complete. In fact If {fn} is Cauchy in X, then {Dsfn} is Cauchy in Lp(Ω) for all \|s\| ≤m. Since Lp(Ω) is complete, Dsfn →gs in Lp(Ω). But since lim n→∞ Z Ω fnDsφ dx = Z Ω fDsφ dx = Z gsφ dx, we have Dsf = gs for all \|s\| ≤m and \|fn −f\|X →0 as n →∞. (2) Hm,p ⊂W 1,p(Ω). Let Hm,p(Ω) be the completion of Cm(Ω) with respect to W I,p(Ω) norm. That is, f ∈Hm,p(Ω) there exists a sequence fn ∈Cm(Ω) such that fn →f and Dsfn →gs strongly in Lp(Ω) and thus Dsfn(φ) = (−1)\|s\| Z Ω Dsfnφ dx →(−1)\|s\| Z Ω gsφ dx which implies gs = Dsf and f ∈W 1,p(Ω). (3) If Ωhas a Lipschitz continuous boundary, then W m,p(Ω) = Hm,p(Ω). 34 2.6 Lax-Milgram Theorem ann Banach-Necas-Babuska The-orem Let X be a Hilbert space. Let σ be a (complex-valued) sesquilinear form on X × X satisfying σ(α x1 + β x2, y) = α σ(x1, y) + β σ(x2, y) σ(x, α y1 + β y2) = ¯ α σ(x, y1) + ¯ β σ(x, y2), \|σ(x, y)\| ≤M \|x\|\|y\| for all x, y ∈X (Bounded) and Re σ(x, x) ≥δ \|x\|2 for all x ∈X and δ > 0 (Coercive). Then for each f ∈X∗there exist a unique solution x ∈X to σ(x, y) = ⟨f, y⟩X∗×X for all y ∈X and \|x\|X ≤δ−1 \|f\|X∗. Proof: Let us deﬁne the linear operator S from X∗into X by Sf = x, f ∈X∗ where x ∈X satisﬁes σ(x, y) = ⟨f, y⟩ for all y ∈X. The operator S is well deﬁned since if x1, x2 ∈X satisfy the above, then σ(x1−x2, y) = 0 for all y ∈X and thus δ \|x1 −x2\|2 X ≤Re σ(x1 −x2, x1 −x2) = 0. Next we show that dom(S) is closed in X∗. Suppose fn ∈dom(S), i.e., there exists xn ∈X satisfying σ(xn, y) = ⟨fn, y⟩for all y ∈X and fn →f in X∗as n →∞. Then σ(xn −xm, y) = ⟨fn −fm, y⟩ for all y ∈X Setting y = xn −xm in this we obtain δ \|xn −xm\|2 X ≤Re σ(xn −xm, xn −xm) ≤\|fn −fm\|X∗\|xn −xm\|X. Thus {xn} is a Cauchy sequence in X and so xn →x for some x ∈X as n →∞. Since σ and the dual product are continuous, thus x = Sf. Now we prove that dom(S) = X∗. Suppose dom(S) ̸= X∗. Since dom(S) is closed there exists a nontrivial x0 ∈X such that ⟨f, x0⟩= 0 for all f ∈dom(S). Consider the linear functional F(y) = σ(x0, y), y ∈X. Then since σ is bounded F ∈X∗and x0 = SF. Thus F(x0) = 0. But since σ(x0, x0) = ⟨F, x0⟩= 0, by the coercivity of σ x0 = 0, which is a contradiction. Hence dom(S) = X∗. □ Assume that σ is coercive. By the Lax-Milgram theorem A has a bounded inverse S = A−1. Thus, dom ( ˜ A) = A−1H. Moreover ˜ A is closed. In fact, if xn ∈dom ( ˜ A) →x and fn = Axn →f in H, 35 then since xn = Sfn and S is bounded, x = Sf and thus x ∈dom ( ˜ A) and ˜ Ax = f. If σ is symmetric, σ(x, y) = (x, y)X deﬁnes an inner product on X. and SF coincides with the Riesz representation of F ∈X∗. Moreover, ⟨Ax, y⟩= ⟨Ay, x⟩for all x, y ∈X. and thus ˜ A is a self-adjoint operator in H. Example (Laplace operator) Consider X = H1 0(Ω), H = L2(Ω) and σ(u, φ) = (u, φ)X = Z Ω ∇u · ∇φ dx. Then, Au = −∆u = −( ∂2 ∂x2 1 u + ∂2 ∂x2 2 u) and dom ( ˜ A) = H2(Ω) ∩H1 0(Ω). for Ωwith C1 boundary or convex domain Ω. For Ω= (0, 1) and f ∈L2(0, 1) Z 1 0 d dxy d dxu dt = Z 1 0 f(x)y(x) dx is equivalent to Z 1 0 d dxy ( d dxu + Z 1 x f(s) ds) dx = 0 for all y ∈H1 0(0, 1). Thus, d dxu + Z 1 x f(s) ds = c (a constant) and therefore d dxu ∈H1(0, 1) and Au = −d2 dx2 u = f in L2(0, 1). Example (Elliptic operator) Consider a second order elliptic equation Au = −∇· (a(x)∇u) + b(x) · ∇u + c(x)u(x) = f(x), ∂u ∂ν = g at Γ1 u = 0 at Γ0 where Γ0 and Γ1 are disjoint and Γ0 ∪Γ1 = Γ. Integrating this against a test function φ, we have Z Ω Auφ dx = Z Ω (a(x)∇u · ∇φ + b(x) · ∇uφ + c(x)uφ) dx − Z Γ1 gφ dsx = Z Ω f(x)φ(x) dx, for all φ ∈C1(Ω) vanishing at Γ0. Let X = H1 Γ0(Ω) is the completion of C1(Ω) vanishing at Γ0 with inner product (u, φ) = Z Ω ∇u · ∇φ dx 36 i.e., H1 Γ0(Ω) = {u ∈H1(Ω) : u\|Γ0 = 0} Deﬁne the bilinear form σ on X × X by σ(u, φ) = Z Ω (a(x)∇u · ∇φ + b(x) · ∇uφ + c(x)uφ. Then, by the Green’s formula σ(u, u) = Z Ω (a(x)\|∇u\|2 + b(x) · ∇(1 2\|u\|2) + c(x)\|u\|2) dx = Z Ω (a(x)\|∇u\|2 + (c(x) −1 2∇· b) \|u\|2) dx + + Z Γ1 1 2n · b\|u\|2 dsx. If we assume 0 < a ≤a(x) ≤¯ a, c(x) −1 2∇· b ≥0, n · b ≥0 at Γ1, then σ is bounded and coercive with δ = a. The Banach space version of Lax-Milgram theorem is as follows. Banach-Necas-Babuska Theorem Let V and W be Banach spaces. Consider the linear equation for u ∈W a(u, v) = f(v) for all v ∈V (2.3) for given f ∈V ∗, where a is a bounded bilinear form on W × V . The problem is well-posed in if and only if the following conditions hold: inf u∈W sup v∈V a(u, v) \|u\|W \|v\|V ≥δ > 0 a(u, v) = 0 for all u ∈W implies v = 0 (2.4) Under conditions we have the unique solution u ∈W to (2.3) satisﬁes \|u\|W ≤1 δ \|f\|V ∗. Proof: Let A be a bounded linear operator from W to V ∗deﬁned by ⟨Au, v⟩= a(u, v) for all u ∈W, v ∈V. The inf-sup condition is equivalent to for any w?W \|Aw\|V ∗≥δ\|u\|W , and thus the range of A, R(A), is closed in V ∗and N(A) = 0. But since V is reﬂexive and ⟨Au, v⟩V ∗×V = ⟨u, A∗v⟩W×W ∗ 37 from the second condition N(A∗) = {0}. It thus follows from the closed range and open mapping theorems that A−1 is bounded. □ Next, we consider the generalized Stokes system. Let V and Q be Hilbert spaces. We consider the mixed variational problem for (u, p) ∈V × Q of the form a(u, v) + b(p, v) = f(v), b(u, q) = g(q) (2.5) for all v ∈V and q ∈Q, where a and b is bounded bilinear form on V × V and V × Q. If we deﬁne the linear operators A ∈L(V, V ∗) and B ∈L(V, Q∗) by ⟨Au, v⟩= a(u, v) and ⟨Bu, q⟩= b(u, q) then it is equivalent to the operator form:   A B∗ B 0     u p  =   f g  . Assume the coercivity on a a(u, u) ≥δ \|u\|2 V (2.6) and the inf-sup condition on b inf q∈P sup u∈V b(u, q) \|u\|V \|q\|Q ≥β > 0 (2.7) Note that inf-sup condition that for all q there exists u ∈V such that Bu = q and \|u\|V ≤1 β\|q\|Q. Also, it is equivalent to \|B∗p\|V ∗≥β \|p\|Q for all p ∈Q. Theorem (Mixed problem) Under conditions (2.6)-(2.7) there exits a unique solution (u, p) ∈V × Q to (2.5) and \|u\|V + \|p\|Q ≤c (\|f\|V ∗+ \|g\|Q∗) Proof: For ϵ > 0 consider the penalized problem a(uϵ, v) + b(v, pϵ) = f(v), for all v ∈V −b(uϵ, q) + ϵ(pϵ, q)Q = −g(q), for all q ∈Q. (2.8) By the Lax-Milgram theorem for every ϵ > 0 there exists a unique solution (uϵ, pϵ) ∈ V × Q. From the ﬁrst equation and (2.7), β \|pϵ\|Q ≤\|f −Auϵ\|V ∗≤\|f\|V ∗+ M \|uϵ\|V . Letting v = uϵ and q = pϵ in the ﬁrst and second equation and (2.7), we have δ \|uϵ\|2 V + ϵ \|pϵ\|2 Q ≤\|f\|V ∗\|\|uϵ\|V + \|pϵ\|Q\|g\|Q∗≤C (\|f\|V ∗+ \|g\|Q∗)\|uϵ\|V ), and thus \|uϵ\|V and thus \|pϵ\|Q are bounded uniformly in ϵ > 0. Thus, (uϵ, pϵ) has a weakly convergent subspace to (u, p) in V × Q and (u, p) satisﬁes (2.5). □ 38 2.7 Compact Operator A compact operator is a linear operator A from a Banach space X to another Banach space Y , such that the image under A of any bounded subset of X is a relatively compact subset of Y , i.e., its closure is compact. For example, the integral operator is a concrete example of compact operators. A Fredholm integral equation gives rise to a compact operator A on function spaces. A compact set in a Banach space is bounded. The converse doses not hold in general. The the closed unit sphere is weakly star compact but not strongly compact unless X is of ﬁnite dimension. Let C(S) be the space of continuous functions on a compact metric space S with norm \|x\| = maxs∈S \|x(s)\|. the Arzela-Ascoli theorem gives the necessary and suﬃcient condition for a subset {xα} being (strongly) relatively compact. Arzela-Ascoli theorem A family {xα} in C(S) is strongly compact if and only if the following conditions hold: sup α \|xα\| < ∞(equi-bouned) lim δ→0+ sup α, \|s′−s′′\|≤δ \|xα(s′) −xα(s′)\| = 0 (equi-continuous). Proof: By the Bolzano-Weirstrass theorem for a ﬁxed s ∈S {xα(s)} contains a con-vergent subsequence. Since S is compact, there is a countable dense subset {sn} in S such that for every ϵ > 0 there exists nϵ satisfying sup s∈S inf 1≤j≤nϵ dist(s, sj) ≤ϵ. Let us denote a convergent subsequence of {xn(s1)} by {xn1(s1)}. Similarly, the se-quence {⟨xn1(s2)} contains a a convergent subsequence {xn2(s2)}. Continuing in this manner we extract subsequences {xnk(sk)}. The diagonal sequence {x∗ nn(s)} converges for s = s1, s2, · · · , simultaneously for the dense subset {sk}. We will show that {xnn} is a Cauchy sequence in C(S). By the equi-continuity of {xn(s)} for all ϵ > 0 there exists δϵ such that \|xn(s′) −xn(s′′)\| ≤ϵ for all s′, s′ satisisfying dist(s′, s′′) ≤≤δ. Thus, for s ∈S there exists a j ≤kϵ such that \|xnn(s) −xmm(s)\| ≤\|xnn(s) −xmm(sj)\| + \|xnn(sj) −xmm(sj)\| + \|xmm(s) −xmm(sj)\| ≤2ϵ + \|xnn(sj) −xmm(sj)\| which implies that limm,m→∞maxs∈S \|xnn(s) −xmm(s)\| ≤2ϵ for arbitrary ϵ > 0. □ For the space Lp(Ω), we have Frechet-Kolmogorov theorem Let Ωbe a subset of Rd. A family {xα} of Lp(Ω) functions is strongly compact if and only if the following conditions hold: sup α Z Ω \|xα\|p ds < ∞(equi-bouned) lim \|t\|→0 sup α Z Ω \|xα(t + s) −xα(s)\|p ds = 0 (equi-continuous). 39 Example (Integral Operator) Let k(x, y) be the kernel function satisfying Z Ω Z Ω \|k(x, y)\|2 dxdy < ∞ Deﬁne the integral operator A by Au = Z Ω k(x, y)u(y) dy for u ∈X = L2(Ω). Then A ∈L(X) i compact. This class of operators A is called the Hilbert-Schmitz operator. Rellich-Kondrachov theorem Every uniformly bounded sequence in W 1,p(Ω) has a subsequence that converges in Lq(Ω) provided that q < p∗= dp d−p. Fredholm Alternative theorem Let A ∈L(X) be compact. For any nonzero λ ∈C, either 1) The equation Ax −λ x = 0 has a nonzero solution x, or 2) The equation Ax −λ, x = f has a unique solution x for any function f ∈X. In the second case, (λ I −A)−1 is bounded. Exercise Problem 1 T : D(T) ⊂X →Y is closable if and only if the closure of its graph G(T) in X × Y the graph of a linear operator S. Problem 2 For a closed linear operator T, the domain D(T) is a Banach space if it is equipped by the graph norm \|x\|D(T) = (\|x\|2 X + \|Tx\|2 Y ) 1 2 . 3 Constrained Optimization In this section we develop the Lagrange multiplier theory for the constrained minimiza-tion in Banach spaces. 3.1 Hilbert space theory In this section we develop the Hilbert space theory for the constrained minimization. Theorem 1 (Riesz Representation) Let X be a Hilbert space. Given F ∈X∗, consider the minimization J(x) = 1 2(x, x)X −F(x) over x ∈X There exists a unique solution x∗∈X and x∗satisﬁes (v, x∗) = F(v) for all v ∈X (3.1) The solution x∗∈X is the Riesz representation of F ∈X∗and it satisﬁes \|x∗\|X = \|F\|X∗. 40 Proof: Suppose x∗is a minimizer. Then 1 2(x∗+ tv, x∗+ tv) −F(x∗+ tv) ≥1 2(x∗, x∗) −F(x∗) for all t ∈R and v ∈X. Thus, for t > 0 (v, x∗) −F(v) + t 2\|v\|2 ≥0 Letting t →0+, we have (v, x∗) −F(v) ≥0 for all v ∈X. If v ∈X, then −v ∈X and this implies (3.1). (Uniqueness) Suppose x∗ 1, x∗ 2 are minimizers. Then, (v, .x∗ 1) −(v, x∗ 2) = F(v) −F(v) = 0 for all v ∈X Letting v = x∗ 1 −x∗ 2, we have \|x∗ 1 −x∗ 2\|2 = (x∗ 1 −x∗ 2, x∗ 1 −x∗ 2) = 0, i.e., x∗ 1 = x∗ 2. □ (Existence) Suppose {xn} is a minimizing sequence, i.e., J(xn) is decreasing and limn→∞J(xn) = δ = infx∈X J(x). Note that \|xn −xm\|2 + \|xn + xm\|2 = 2(\|xn\|2 + \|xm\|2) Thus, 1 4\|xn−xm\|2 = J(xm)+J(xn)−2J(xn + xm 2 )−F(xn)−F(xm)+2F(xn + xm 2 ) ≤J(xn)+J(xm)−2δ. This implies {xn} is a Cauchy sequence in X. Since X is complete there exists x∗= limn→∞xn in X and J(x∗) = δ, i.e, x∗is the minimizer. □ Remark The existence and uniqueness proof of the theorem holds if we assume J is uniformly convex functional with modulus φ, i.e., J(λx1 + (1 −λ)x2) ≤J(x1) + (1 −λ)J(x2) −λ(1 −λ)φ(\|x1 −x2\|) where φ is a function that is increasing and vanishes only at 0. In fact 1 4φ(\|xn −xm\|) ≤J(xm) + J(xn) −2J(xn + xm 2 ) ≤J(xn) + J(xm) −2δ. The necessary and suﬃcient optimality is given by 0 ∈∂J(x∗) where ∂J(x∗) is the subdiﬀerential of J at x∗. Example (Mechanical System) min Z 1 0 1 2\| d dxu(x)\|2 − Z 1 0 f(x)u(x) dx (3.2) over u ∈H1 0(0, 1). Here X = H1 0(0, 1) = {u ∈H1(0, 1) : u(0) = u(1) = 0} is a Hilbert space with the inner product (u, v)H1 0 = Z 1 0 d dxu d dxv dx. 41 H1(0, 1) is a space of absolute continuous functions on [0, 1] with square integrable derivative, i.e., u(x) = u(0) + Z x 0 g(x) dx, g(x) = d dxu(x), a.e. and g ∈L2(0, 1). The solution yf = x∗to (3.8) satisﬁes Z 1 0 d dxv d dxyf dx = Z 1 0 f(x)v(x) dx. By the integration by part Z 1 0 f(x)v(x) dx = Z 1 0 ( Z 1 x f(s) ds) d dxv(x) dx, and thus Z 1 0 ( d dxyf − Z 1 x f(s) ds) d dxv(x) dx = 0 for all v ∈H1 0(0, 1). This implies that d dxyf − Z 1 x f(s) ds = c = a constant Integrating this, we have yf = Z x 0 Z 1 x f(s) dsdx + cx Since yf(1) = 0 we have c = − Z 1 0 Z 1 x f(s) dsdx. Moreover, d dxyf ∈H1(0, 1) and −d2 dx2 yf(x) = f(x) a.e. with yf(0) = yf(1) = 0. Now, we consider the case when F(v) = v(1/2). Then, \|F(v)\| = \| Z 1/2 0 d dxv dx\| ≤( Z 1/2 0 \| d dxv\| dx)1/2( Z 1/2 0 1 dx)1/2 ≤ 1 √ 2\|v\|X. That is, F ∈X∗. Thus, we have Z 1 0 ( d dxyf −χ(0,1/2)(x)) d dxv(x) dx = 0 for all v ∈H1 0(0, 1), where χS is the characteristic function of a set S. It is equivalent to d dxyf −χ(0,1/2)(x) = c = a constant. 42 Integrating this, we have yf(x) =    (1 + c)x x ∈[0, 1 2] cx + 1 2 x ∈[ 1 2, 1]. Since yf(1) = 0 we have c = −1 2. Moreover yf satisﬁes −d2 dx2 yf = 0, for x ̸= 1 2 yf(0) = yf(1) = 0 yf(( 1 2)−) = yf( 1 2)+), d dxyf(( 1 2)−) −d dxyf(( 1 2)+) = 1. Or, equivalently −d2 dx2 yf = δ 1 2 (x) where δx0 is the Dirac delta distribution at x = x0. Theorem 2 (Orthogonal Decomposition X = M ⊕M⊥) Let X be a Hilbert space and M is a linear subspace of X. Let M be the closure of M and M⊥= {z ∈X : (z, y) = 0 for all y ∈M} be the orthogonal complement of M. Then every element u ∈X has the unique representation u = u1 + u2, u1 ∈M and u2 ∈M⊥. Proof: Consider the minimum distance problem min \|x −u\|2 over x ∈M. (3.3) As for the proof of Theorem 1 one can prove that there exists a unique solution x∗for problem (3.3) and x∗∈M satisﬁes (x∗−u, y) = 0, for all y ∈M i.e., u −x∗∈M⊥. If we let u1 = x∗∈M and u2 = u −x∗∈M⊥, then u = u1 + u2 is the desired decomposition. □ Example (Closed subspace) (1) M = span{(yk, k ∈= {x ∈X; x = P k∈K αk yk} is a subspace of X. If K is ﬁnite, M is closed. (2) Let X = L2(−1, 1) and M = {odd functions} = {f ∈X : f(−x) = f(x), a.e.} and then M is closed. Let M1 = span{sin(kπt)} and M2 = span{cos(kπt)}. L2(0, 1) = M1 ⊕M2. (3) Let E ∈L(X, Y ). Then M = N(E) = {x ∈X : Ex = 0} is closed. In fact, if xn ∈M and xn →x in X, then 0 = limn→∞Exn = E limn→∞xn = Ex and thus x ∈N(E) = M. Example (Hodge-Decomposition) Let X = L2(Ω)3 with where Ωis a bounded open set in R3 with Lipschitz boundary Γ. Let M = {∇u, u ∈H1(Ω)} is the gradient ﬁeld. Then, M⊥= {ψ ∈X : ∇· ψ = 0, n · ψ = 0 at Γ} 43 is the divergence free ﬁeld. Thus, we have the Hodge decomposition X = M ⊕M⊥= {grad u, u ∈H1(Ω)} ⊕{the divergence free ﬁeld}. In fact ψ ∈M⊥ Z ∇φ · ψ = Z Γ n · ψφ ds − Z Ω ∇ψφ dx = 0 and ∇u ∈M, u ∈H1(Ω)/R is uniquely determined by Z Ω ∇u · ∇φ dx = Z Ω f · ∇φ dx for all φ ∈H1(Ω). Theorem 3 (Decomposition) Let X, Y be Hilbert spaces and E ∈L(X, Y ). Let E∗∈L(Y, X) is the adjoint of E, i.e., (Ex, y)Y = (x, E∗y)X for all x ∈X and y ∈Y . Then N(E) = R(E∗)⊥and thus from the orthogonal decomposition theorem X = N(E) ⊕R(E∗). Proof: Suppose x ∈N(E), then (x, E∗y)X = (Ex, y)Y = 0 for all y ∈Y and thus N(E) ⊂R(E∗)⊥. Conversely, x∗∈R(E∗)⊥, i.e., (x∗, E∗y) = (Ex, y) = 0 for all y ∈Y Thus, Ex∗= 0 and R(E∗)⊥⊂N(E). □ Remark Suppose R(E) is closed in Y , then R(E∗) is closed in X and Y = R(E) ⊕N(E∗), X = R(E∗) ⊕N(E) Thus, we have b = b1 + b2, b1 ∈R(E), b2 ∈N(E∗) for all b ∈Y and \|Ex −b\|2 = \|Ex −b1\|2 + \|b2\|2. Example (R(E) is not closed) Consider X = L2(0, 1) and deﬁne a linear operator E ∈ L(X) by (Ef)(t) = Z t 0 f(s) ds, t ∈(0, 1) Since R(E) ⊂{absolutely continuous functions on [0, 1]}, R(E) = L2(0, 1) and R(E) is not a closed subspace of L2(0, 1). 3.2 Minimum norm problem We consider min \|x\|2 subject to (x, yi)X = ci, 1 ≤i ≤m. (3.4) Theorem 4 (Minimum Norm) Suppose {yi} are linearly independent. Then there exits a unique solution x∗to (3.4) and x∗= Pm i=1 βiyi where β = {βi}m i=1 satisﬁes Gβ = c, Gi,j = (yi, yj). 44 Proof: Let M = span{yi}m i=1. Since X = M ⊕M⊥, it is necessary that x∗∈M, i.e., x∗= m X i=1 βiyi. From the constraint (x∗, yi) = ci, 1 ≤i ≤m, the condition for β ∈Rn follows. In fact, x ∈X satisﬁes (x, yi) = ci, 1 ≤i ≤m if and only if x = x∗+ z, z ∈M⊥. But \|x\|2 = \|x∗\|2 + \|z\|2. □ Remark {yi} is linearly independent if and only if the symmetric matrix G ∈Rn,n is positive deﬁnite since (x, Gx)Rn = \| n X k=0 xk yk\|2 X for all x ∈Rn. Theorem 5 (Minimum Norm) Suppose E ∈L(X, Y ). Consider the minimum norm problem min \|x\|2 subject to Ex = c If R(E) = Y , then the minimum norm solution x∗exists and x∗= E∗y, (EE∗)y = c. Proof: Since R(E∗) is closed by the closed range theorem, it follows from the decom-position theorem that X = N(E) ⊕R(E∗). Thus, we x∗= E∗y if there exists y ∈Y such that (EE∗)y = c. Since Y = N(E∗) ⊕R(E), N(E∗) = {0}. Since if EE∗x = 0, then (x, EE∗x) = \|E∗x\| = 0 and E∗x = 0. Thus, N(EE∗) = N(E∗) = {0}. Since R(EE∗) = R(E) it follows from the open mapping theorem EE∗is bounded invertible and there exits a unique solution to (EE∗)y = c. □ Remark If R(E) is closed and c ∈R(E), we let Y = R(E) and Theorem 5 is valid. For example we deﬁne E ∈L(X, Rm) by Ex = ((y1, x)X, · · · , (ym, x)X) ∈Rm. If {yi}m i=1 are dependent, then span{yi}m i=1 is a closed subspace of Rm Example (DC-motor Control) Let (θ(t), ω(t))) satisfy d dtω(t) + ω(t) = u(t), ω(0) = 0 d dtθ(t) = ω(t), θ(0) = 0 (3.5) This models a control problem for the dc-motor and the control function u(t) is the current applied to the motor, θ is the angle and ω is the angular velocity. The objective to ﬁnd a control u on [0, 1] such the desired state (¯ θ, 0) is reached at time t = 1 and with minimum norm Z 1 0 \|u(t)\|2 dt. Let X = L2(0, 1). Given u ∈X we have the unique solution to (3.5) and ω(1) = R 1 0 et−1u(t) dt θ(t) = R 1 0 (1 −et−1)u(t) dt 45 Thus, we can formulate this control problem as problem (3.4) by letting X = L2(0, 1) with the standard inner product and y1(t) = 1 −et−1, y2(t) = et−1 c1 = ¯ θ, c2 = 0. Example (Linear Control System) Consider the linear control system d dtx(t) = Ax(t) + Bu(t), x(0) = x0 ∈Rn (3.6) where x(t) ∈Rn and u(t) ∈Rm are the state and the control function, respectively and system matrices A ∈Rn×n and B ∈Rn×m are given. Consider a control problem of ﬁnding a control that transfers x(0) = 0 to a desired state ¯ x ∈Rn at time T > 0, i.e., x(T) = ¯ x with minimum norm. The problem can be formulated as (3.4). First, we have the variation of constants formula: x(t) = eAtx0 + Z t 0 eA(t−s)Bu(s) ds where eAt is the matrix exponential and satisﬁes d dteAt = AeAt = eAtA. Thus, condition x(T) = ¯ x is written as Z T 0 eA(T−t)Bu(t) ds = ¯ x Let X = L2(0, T; Rm) and it then follows from Theorem 4 that the minimum norm solution is given by u∗(t) = BtEAt(T−t)β, Gβ = ¯ x, provided that the control Gramean G = Z T 0 eA(T−s)BBteAt(T−t) dt ∈Rn×n is positive on Rn. If G is positive, then control system (3.6) is controllable and it is equivalent to BteAtx = 0, t ∈[0, T] implies x = 0. since (x, Gx) = Z T 0 \|BteAttx\|2 dt Moreover, G is positive if and only if the Kalman rank condition holds; i.e., rank [B, AB, · · · An−1B] = n. In fact, (x, Gx) = 0 implies BteAttx = 0, t ≥0 and thus Bt(At)k = 0 for all k ≥0 and thus the claim follows from the Cayley-Hamilton theorem. Example (Function Interpolation) (1) Consider the minimum norm on X = H1 0(0, 1): min Z 1 0 \| d dxu\|2 dx over H1 0(0, 1) (3.7) 46 subject to u(xi) = ci, 1 ≤i ≤m, where 0 < x1 < · · · < xm < 1 are given nodes. Let X = H1 0(0, 1). From the mechanical system example we have the piecewise linear function yi(x) ∈X such that (u, yi)X = u(xi) for all u ∈X for each i. From Theorem 3 the solution to (3.7) is given by u∗(x) = m X i=1 βi yi(x). That is, u∗is a piecewise linear function with nodes {xi}, 1 ≤i ≤m and nodal values ci at xi, i.e., βi = ci yi(xi), 1 ≤i ≤n. (2) Consider the Spline interpolation problem (1.2): min Z 1 0 1 2\| d2 dx2 u(x)\|2 dx over all functions u(x) subject to the interpolation conditions u(xi) = bi, u′(xi) = ci, 1 ≤i ≤m. (3.8) Let X = H2 0(0, 1) be the completion of the pre-Hilbert space C2 0[0, 1] = the space of twice continuously diﬀerentiable function with u(0) = u′(0) = 0 and u(1) = u′(1) = 0 with respect to H2 0(0, 1) inner product; (f, g)H2 0 = Z 1 0 d2 dx2 f d2 dx2 g dx, i.e., H2 0(0, 1) = {u ∈L2(0, 1) : d dxu, d2 dx2 u ∈L2(0, 1) with u(0) = u′(0) = u(1) = u′(1) = 0}. Problem (3.8) is a minimum norm problem on x = H2 0(0, 1). For Fi ∈X∗deﬁned by Fi(v) = v(xi) we have (yi, v)X −Fi(v) = Z 1 0 ( d2 dx2 yi + (x −xi)χ[0,xi]) d2 dx2 v dx, (3.9) and for ˜ Fi ∈X∗deﬁned by ˜ Fi(v) = v′(xi) we have (˜ yi, v)X −˜ Fi(v) = Z 1 0 ( d2 dx2 ˜ yi −χ[0,xi]) d2 dx2 v dx. (3.10) Thus, d2 dx2 yi + (x −xi)χ[0,xi] = a + bx, d2 dx2 ˜ yi −χ[0,xi] = ˜ a + ˜ bx where a, ˜ a , b, ˜ b are constants and yi = RFi and ˜ yi = R ˜ Fi are piecewise cubic functions on [0, 1]. One can determine yi, ˜ yi using the conditions y(0) = y′(0) = y(1) = y′(0) = 0. From Theorem 3 the solution to (3.8) is given by u∗(x) = m X i=1 (βi yi(x) + ˜ βi ˜ yi(x)) 47 That is, u∗is a piecewise cubic function with nodes {xi}, 1 ≤i ≤m and nodal value ci and derivative bi at xi. Example (Diﬀerential form, Natural Boundary Conditions) (1) Let X be the Hilbert space deﬁned by X = H1(0, 1) = {u ∈L2(0, 1) : du dx ∈L2(0, 1)}. Given a, c ∈C[0, 1] with a > 0 c ≥0 and α ≥0 consider the minimization min Z 1 0 (1 2(a(x)\|du dx\|2 + c(x)\|u\|2) −uf) dx + α 2 \|u(1)\|2 −gu(1) over u ∈X = H1(0, 1). If u∗is a minimizer, it follows from Theorem 1 that u∗∈X satisﬁes Z 1 0 (a(x)du∗ dx dv dx + c(x) u∗v −fv) dx + α u(1)v(1) −gv(1) = 0 (3.11) for all v ∈X. By the integration by parts Z 1 0 (−d dx(a(x)du∗ dx )+c(x)u∗(x)−f(x))v(x) dx−a(0)du∗ dx (0)v(0)+(a(1)du∗ dx (1)+α u∗(1)−g)v(1) = 0, assuming a(x) du∗ dx ∈H1(0, 1). Since v ∈H1(0, 1) is arbitrary, −d dx(a(x)du∗ dx ) + c(x) u∗= f(x), x ∈(0, 1) a(0)du∗ dx (0) = 0, a(1)du∗ dx (1) + α u∗(1) = g. (3.12) (3.11) is the weak form and (3.12) is the strong form of the optimality condition. (2) Let X be the Hilbert space deﬁned by X = H2(0, 1) = {u ∈L2(0, 1) : du dx, d2u dx ∈L2(0, 1)}. Given a, b, c ∈C[0, 1] with a > 0 b, c ≥0 and α ≥0 consider the minimization on X = H2(0, 1) min Z 1 0 (1 2(a(x)\|d2u dx2 \|2+b(x)\|du dx\|2+c(x)\|u\|2)−uf) dx+α 2 \|u(1)\|2+β 2 \|u′(1)\|2−g1u(1)−g2u′(1) over u ∈X = H2(0, 1) satisfying u(0) = 1. If u∗is a minimizer, it follows from Theorem 1 that u∗∈X satisﬁes Z 1 0 (a(x)d2u∗ dx2 d2v dx2 +b(x)du∗ dx dv dx+c(x) u∗v−fv) dx+α u∗(1)v(1)+β (u∗)′(1)v′(1)−g1v(1)−g2v′(1) = 0 (3.13) for all v ∈X satisfying v(0) = 0. By the integration by parts Z 1 0 ( d2 dx2 (a(x) d2 dx2 u∗) −d dx(b(x) d dxu∗) + c(x)u∗(x) −f(x))v(x) dx +(a(1)d2u∗ dx2 (1) + β du∗ dx (1) −g2)v′(1) −a(0)d2u∗ dx2 (0)v′(0) +(−d dx(a(x)d2u∗ dx2 + b(x)du∗ dx + α u∗−g1)(1)v(1) −(−d dx(a(x)d2u∗ dx2 ) + b(x)du∗ dx )(0)v(0) = 0 48 assuming u∗∈H4(0, 1). Since v ∈H2(0, 1) satisfying v(0) = 0 is arbitrary, d2 dx2 (a(x) d2 dx2 u∗) −d dx(b(x)du∗ dx ) + c(x) u∗= f(x), x ∈(0, 1) u∗(0) = 1, a(0)d2u∗ dx2 (0) = 0 a(1)d2u∗ dx2 (1) + β du∗ dx (1) −g2 = 0, (−d dx(a(x)d2u∗ dx2 ) + b(x)du∗ dx )(1) + α u∗(1) −g1 = 0. (3.14) (3.13) is the weak form and (3.14) is the strong (diﬀerential) form of the optimality condition. Next consider the multi dimensional case. Let Ωbe a subdomain in Rd with Lips-chitz boundary and Γ0 is a closed subset of the boundary ∂Ωwith Γ1 = ∂Ω\ Γ0. Let Γ is a smooth hypersurface in Ω. H1 Γ0(Ω) = {u ∈H1(Ω) : u(s) = 0, s ∈Γ0} with norm \|∇u\|L2(Ω). Let Γ is a smooth hyper-surface in Ωand consider the weak form of equation for u ∈H1 Γ0(Ω): Z (σ(x)∇u · ∇φ + u(x) (⃗ b(x) · ∇φ) + c(x) u(x)φ(x)) dx + Z Γ1 α(s)u(s)φ(s) ds − Z Γ g0(s)φ(s) ds − Z Ω f(x)φ(x) dx − Z Γ1 g(s)u(s) ds = 0 for all H1 Γ0(Ω). Then the diﬀerential form for u ∈H1 Γ0(Ω) is given by              −∇· (σ(x)∇u +⃗ bu) + c(x)u(x) = f in Ω n · (σ(x)∇u +⃗ bu) + α u = g at Γ1 [n · (σ(x)∇u +⃗ bu)] = g0 at Γ. By the divergence theorem if u satisﬁes the strong form, then u is a weak solution. Conversely, letting φ ∈C0(Ω\ Γ) we obtain the ﬁrst equation, i.e., ∇· (σ∇u +⃗ bu) ∈ L2(Ω) Thus, n·(σ∇u+⃗ bu) ∈L2(Γ) and the third equation holds. Also, n·(σ∇u+⃗ bu) ∈ L2(Γ1) and the third equation holds. 3.3 Lax-Milgram Theory and Applications Let H be a Hilbert space with scalar product (φ, ψ) and X be a Hilbert space and X ⊂H with continuous dense injection. Let X∗denote the strong dual space of X. H is identiﬁed with its dual so that X ⊂H = H∗⊂X∗(i.e., H is the pivoting space). The dual product ⟨φ, ψ⟩on X∗× X is the continuous extension of the scalar product of H restricted to H × X. This framework is called the Gelfand triple. Let σ is a bounded coercive bilinear form on X ×X. Note that given x ∈X, F(y) = σ(x, y) deﬁnes a bounded linear functional on X. Since given x ∈X, y →σ(x, y) is a 49 bounded linear functional on X, say x∗∈X∗. We deﬁne a linear operator A from X into X∗by x∗= Ax. Equation σ(x, y) = F(y) for all y ∈X is equivalently written as an equation Ax = F ∈X∗. (3.15) It from the Lax-Milgram theorem that A has the bouned inverse A−1 ∈L(X∗, X) Also, ⟨Ax, y⟩X∗×X = σ(x, y), x, y ∈X, and thus A is a bounded linear operator. In fact, \|Ax\|X∗≤sup \|y\|≤1 \|σ(x, y)\| ≤M \|x\|. Let R be the Riesz operator X∗→X, i.e., \|Rx∗\|X = \|x∗\| and (Rx∗, x)X = ⟨x∗, x⟩for all x ∈X, then ˆ A = RA represents the linear operator ˆ A ∈L(X, X). Moreover, we deﬁne a linear operator ˜ A on H by ˜ Ax = Ax ∈H with dom ( ˜ A) = {x ∈X : \|σ(x, y)\| ≤cx \|y\|H for all y ∈X}. That is, ˜ A is a restriction of A on dom ( ˜ A). It thus follows from the Lax-Milgram theorem dom ( ˜ A) = A−1H. We will use the symbol A for all three linear operators as above in the lecture note and its use should be understood by the underlining context. 3.4 Quadratic Constrained Optimization In this section we discuss the constrained minimization. First, we consider the case of equality constraint. Let A be coercive and self-adjoint operator on X and E ∈L(X, Y ). Let σ : X × X →R is a symmetric, bounded, coercive bilinear form, i.e., σ(α x1 + β x2, y) = α σ(x1, y) + β σ(x2, y) for all x1, x2, y ∈X and α, β ∈R σ(x, y) = σ(y, x), \|σ(x, y)\| ≤M \|x\|X\|y\|X, σ(x, x) ≥δ \|x\|2 for some 0 < δ ≤M < ∞. Thus, σ(x, y) deﬁnes an inner product on X. In fact, since σ is bounded and coercive, p σ(x, x) = p (Ax, x) is an equivalent norm of X. Since y →σ(x, y) is a bounded linear functional on X, say x∗and deﬁne a linear operator from X to X∗by x∗= Ax. Then, A ∈L(X, X∗) with \|A\|X→X∗≤M and ⟨Ax, y⟩X∗×X = σ(x, y) for all x, y ∈X. Also, if RX∗→X is the Riesz map of X∗and deﬁne ˆ A = RX∗→XA ∈LX, X), then ( ˆ Ax, y)X = σ(x, y) for all x, y ∈X, 50 and thus ˆ A is a self-adjoint operator in X. Throughout this section, we assume X = X∗and Y = Y ∗, i.e., we identify A as the self-adjoint operator ˜ A. Consider the unconstrained minimization problem: for f ∈X∗ min F(x) = 1 2σ(x, x) −f(x). (3.16) Since for x, v ∈X and t ∈R F(x + tv) −F(x) t = σ(x, v) −f(v) + t 2σ(v, v), x ∈X →F(x) ∈R is diﬀerentiable and we have the necessary optimality for (3.16): σ(x, v) −f(v) = 0 for all v ∈X. (3.17) Or, equivalently Ax = f in X∗ Also, if RX∗→X is the Riesz map of X∗and deﬁne ˜ A = RX∗→XA ∈L(X, X), then ( ˜ Ax, y)X = σ(x, y) for all x, y ∈X, and thus ˜ A is a self-adjoint operator in X. Throughout this section, we assume X = X∗ and Y = Y ∗, i.e., we identify A as the self-adjoint operator ˜ A. Consider the equality constraint minimization; min 1 2(Ax, x)X −(a, x)X subject to Ex = b ∈Y, (3.18) or equivalently for f ∈X∗ min 1 2σ(x, x) −f(x) subject to Ex = b ∈Y. The necessary optimality for (3.18) is: Theorem 6 (Equality Constraint) Suppose there exists a ¯ x ∈X such that E¯ x = b then (3.18) has a unique solution x∗∈X and Ax∗−a ⊥N(E). If R(E∗) is closed, then there exits a Lagrange multiplier λ ∈Y such that Ax∗+ E∗λ = a, Ex∗= b. (3.19) If range(E) = Y (E is surjective), there exits a Lagrange multiplier λ ∈Y is unique. Proof: Suppose there exists a ¯ x ∈X such that E¯ x = b then (3.18) is equivalent to min 1 2 (A(z + ¯ x), z + ¯ x) −(a, z + ¯ x) over z ∈N(E). Thus, it follows from Theorem 2 that it has a unique minimizer z∗∈N(E) and (A(z∗+ ¯ x) −a, v) = 0 for all v ∈N(E) and thus A(z∗+ ¯ x) −a ⊥N(E). From the orthogonal decomposition theorem Ax∗−a ∈R(E∗). Moreover if R(E∗) is surjective, from the closed range theorem R(E∗) = R(E∗) and there exist exists a λ ∈Y satisfying 51 Ax∗+E∗λ = a. If E is surjective (R(E) = Y ), then N(E∗) = 0 and thus the multiplier λ is unique. □ Remark (Lagrange Formulation) Deﬁne the Lagrange functional L(x, λ) = J(x) + (λ, Ex −b). The optimality condition (3.19) is equivalent to ∂ ∂xL(x∗, λ) = 0, ∂ ∂λL(x∗, λ) = 0. Here, x ∈X is the primal variable and the Lagrange multiplier λ ∈Y is the dual variable. Remark (1) If E is not surjective, one can seek a solution using the penalty formula-tion: min 1 2(Ax, x)X −(a, x)X + β 2 \|Ex −b\|2 Y (3.20) for β > 0. The necessary optimality condition is Axβ + β E∗(Exβ −b) −a = 0 If we let λβ = β (Ex −b), then we have A E∗ E −I β xβ λβ = a b . the optimality system (3.19) is equivalent to the case when β = ∞. One can prove the monotonicity J(xβ) ↑, \|Exβ −b\|Y ↓as β →∞ In fact for β > ˆ β we have J(xβ) + β 2 \|Exβ −b\|2 Y ≤J(xˆ β) + β 2 \|Exˆ β −b\|2 Y J(xˆ β) + ˆ β 2 \|Exˆ β −b\|2 Y ≤J(xβ) + ˆ β 2 \|Exβ −b\|2 Y . Adding these inequalities, we obtain (β −ˆ β)(\|Exβ −b\|2 Y −\|Exˆ β −b\|2 Y ) ≤0, which implies the claim. Suppose there exists a ¯ x ∈X such that E¯ x = b then from Theorem 6 J(xβ) + β 2 \|Exβ −b\|2 ≤J(x∗). and thus \|xβ\|X is bounded uniformly in β > 0 and lim β→∞\|Exβ −b\| = 0 Since X is a Hilbert space, there exists a weak limit x ∈X with J(x) ≤J(x∗) and Ex = b since norms are weakly sequentially lower semi-continuous. Since the optimal solution x∗to (3.18) is unique, lim xβ = x∗as β →∞. Suppose λβ is bounded in Y , 52 then there exists a subsequence of λβ that converges weakly to λ in Y . Then, (x∗, λ) satisﬁes Ax∗+ E∗λ = a. Conversely, if there is a Lagrange multiplier λ∗∈Y , then λβ is bounded in Y . In fact, we have (A(xβ −x∗), xβ −x∗) + (xβ −x∗, E∗(λβ −λ∗)) = 0 where β (xβ −x∗, E∗(λβ −λ∗)) = β (Exβ −b, λβ −λ∗) = \|λβ\|2 −(λβ, λ∗). Thus we obtain β (xβ −x∗, A(xβ −x∗)) + \|λβ\|2 = (λβ, λ∗) and hence \|λβ\| ≤\|λ∗\|. Moreover, if R(E) = Y , then N(E∗) = {0}. Since λβ satisﬁes E∗λβ = a −Axβ ∈X, we have λβ = Esβ, sβ = (E∗E)−1(a −Axβ) and thus λβ ∈Y is bounded uniformly in β > 0. (2) If E is not surjective but R(E) is closed. We have λ ∈R(E) ⊂Y such that Ax∗+ E∗λ = a by letting Y = R(E) is a Hilbert space. Also, we have (xβ, λβ) →(x, λ) satisfying Ax + E∗λ = a Ex = PR(E)b since \|Ex −b\|2 = \|Ex −PR(E)b\|2 + \|PN(E∗)b\|2. Example (Parameterized constrained problem) min 1 2 ((Ay, y) + α (p, p)) −(a, y) subject to E1y + E2p = b where x = (y, p) ∈X1 × X2 = X and E1 ∈L(X1, Y ), E2 ∈L(X2, Y ). Assume Y = X1 and E1 is bounded invertible. Then we have the optimality            Ay∗+ E∗ 1λ = a α p + E∗ 2λ = 0 E1y∗+ E2p∗= b Example (Stokes system) Let X = H1 0(Ω) × H1 0(Ω) where Ωis a bounded open set in R2. A Hilbert space H1 0(Ω) is the completion of C0(Ω) with respect to the H1 0(Ω) inner product (f, g)H1 0(Ω) = Z Ω ∇f · ∇g dx 53 Consider the constrained minimization Z Ω (1 2(\|∇u1\|2 + \|∇u2\|2) −f · u) dx over the velocity ﬁeld u = (u1, u2) ∈X satisfying div u = ∂ ∂x1 u1 + ∂ ∂x2 u2 = 0. From Theorem 5 with Y = L2(Ω) and Eu = div u on X we have the necessary opti-mality −∆u + gradp = f, div u = 0 which is called the Stokes system. Here f ∈L2(Ω) × L2(Ω) is a applied force p = −λ∗∈L2(Ω) is the pressure. 3.5 Variational Inequalities Let Z be a Hilbert lattice with order ≤and Z∗= Z and G ∈L(X, Z). For exam-ple, Z = L2(Ω) with a.e. pointwise inequality constarint. Consider the quadratic programming: min 1 2(Ax, x)X −(a, x)X subject to Ex = b, Gx ≤c. (3.21) Note that the constrain set C = {x ∈X, Ex = b, Gx ≤c} is a closed convex set in X. In general we consider the constrained problem on a Hilbert space X: F0(x) + F1(x) over x ∈C (3.22) where C is a closed convex set, F : X →R is C1 and F1 : X →R is convex, i.e. F1((1 −λ)x1 + λ x2) ≥(1 −λ) F1(x1) + λ F(x2) for all x, x2 ∈X and 0 ≤λ ≤1. Then, we have the necessary optimality in the form of the variational inequality: Theorem 7 (Variational Inequality) Suppose x∗∈C minimizes (3.22), then we have the optimality condition F ′ 0(x∗)(x −x∗) + F1(x) −F1(x∗) ≥0 for all x ∈C. (3.23) Proof: Let xt = x∗+ t (x −x∗) for all x ∈C and 0 ≤t ≤1. Since C is convex, xt ∈C. Since x∗∈C is optimal, F0(xt) −F0(x∗) + F1(xt) −F1(x∗) ≥0, (3.24) where F1(xt) ≤(1 −t)F1(x∗) + tF(x) and thus F1(xt) −F1(x∗) ≤t (F1(x) −F1(x∗)). 54 Since F0(xt) −F0(x∗) t →F ′ 0(x∗)(x −x∗) as t →0+, the variational inequality (3.23) follows from (3.24) by letting t →0+. □ Example (Inequality constraint) For X = Rn F ′(x∗)(x −x∗) ≥0, x ∈C implies that F ′(x∗) · ν ≤0 and F ′(x∗) · τ = 0, where ν is the outward normal and τ is the tangents of C at x∗. Let X = R2, C = [0, 1] × [0, 1] and assume x∗= [1, s], 0 < s < 1. Then, ∂F ∂x1 (x∗) ≤0, ∂F ∂x2 (x∗) = 0. If x∗= [1, 1] then ∂F ∂x1 (x∗) ≤0, ∂F ∂x2 (x∗) ≤0. Example (L1-optimization) Let U be a closed convex subset in R. Consider the mini-mization problem on X = L2(Ω) and C = {u ∈U a.e. in Ω}; min 1 2\|Eu −b\|2 Y + α Z Ω \|u(x)\| dx. over u ∈C. (3.25) Then, the optimality condition is (Eu∗−b, E(u −u∗))Y + α Z Ω (\|u\| −\|u∗\|) dx ≥0 (3.26) for all u ∈C. Let p = −E∗(Eu∗−b) ∈X and u = v ∈U on \|x −¯ x\| ≤δ, otherwise u = u∗. From (4.31) we have 1 δ Z \|x−¯ x\|≤δ (−p(v −u∗(x)) + α (\|v\| −\|u∗(x)\|) dx ≥0. Suppose u∗∈L1(Ω) and letting δ →0+ we obtain −p(v −u∗(¯ x)) + α (\|v\| −\|u∗(¯ x)\|) ≥0 a.e. ¯ x ∈Ω(Lebesgue points of \|u∗\| and for all v ∈U. That is, u∗∈U minimizes −pu + α \|u\| over u ∈U. (3.27) Suppose U = R it implies that if \|p\| < α, then u∗= 0 and otherwise p = α u∗ \|u∗\| = α sign(u∗). Thus, we obtain the necessary optimality (E∗(Eu∗−b))(x) + α ∂\|u\|(u∗(x)) = 0, a.e. in Ω, where the sub-diﬀerential of \|u\| (see, Section ) is deﬁned by ∂\|u\|(s) =    1 s > 0 [−1, 1] s = 0 −1 s < 0. 55 3.5.1 Inequality constraint problem Consider the quadratic programing problem. Theorem 8 (Quadratic Programming) Let Z be a Hilbert space lattice with order ≤. For E ∈L(X, Y ) and G ∈L(X, Z) consider the quadratic programming (3.21): min 1 2(Ax, x)X −(a, x) subject to Ex = b, Gx ≤c. If there exists ¯ x ∈X such that E¯ x = b and G¯ x ≤c, then there exits unique solution to (3.21) and (Ax∗−a, x −x∗) ≥0 for all x ∈C, (3.28) where C = {Ex = b, Gx ≤c}. Let Z = Z+ + Z−with positive corn Z+ and negative cone Z−. Moreover, if we assume the regular point condition 0 ∈int {(E(x −x∗), G(x −x∗) −Z−+ G(x∗) −c) : x ∈X}, (3.29) then there exists λ ∈Y and µ ∈Z+ such that            Ax∗+ E∗λ + G∗µ = a Ex∗= b Gx∗−c ≤0, µ ≥0 and (Gx∗−c, µ)Z = 0. (3.30) Proof: First note that (3.21) is equivalent to min 1 2(A(z + ¯ x, z + ¯ x)X −(a, z + ¯ x)X subject to Ez = 0, Gz ≤c −G¯ x = ˆ c Since b C = {Ez = 0, Gz ≤ˆ c} is a closed convex set in the Hilbert space N(E). As in the proof of Theorem 1 the minimizing sequence zn ∈b C satisﬁes 1 4\|zm −zn\| ≤J(zn) + J(zm) −2 δ →0 as m ≥n →∞. Since b C is closed, there exists a unique minimizer z∗∈b C and z∗ satisﬁes (A(z∗+ ¯ x) −a, z −z∗) ≥0 for all z ∈b C, which is equivalent to (3.28). Moreover, it follows from the Lagrange Multiplier theory (below) if the regular point condition holds, then we obtain (3.30). □ Example Suppose Z = Rm. If (Gx∗−c)i = 0, then i is called an active index and let Ga the (row) vectors of G corresponding to the active indices. Suppose E Ga is surjective (Kuhn-Tuker condition), then the regular point condition holds. Remark (Lagrange Formulation) Deﬁne the Lagrange functional; L(x, λ, µ) = J(x) + (λ, Ex −b) + (µ, (Gx −c)+). 56 for (x, λ) ∈X × Y, µ ∈Z+; The optimality condition (3.30) is equivalent to ∂ ∂xL(x∗, λ, µ) = 0, ∂ ∂λL(x∗, λ, µ) = 0, ∂ ∂µL(x∗, λ, µ) = 0 Here x ∈X is the primal variable and the Lagrange multiplier λ ∈Y and µ ∈Z+ are the dual variables. Remark Let A ∈L(X, X∗) and E : X →Y is surjective. If we deﬁne G is natural injection X to Z. Deﬁne µ ∈X∗by −µ = Ax∗−a + E∗λ. Since ⟨t(x∗−c) −(x∗−c), µ⟩= (t −1) ⟨x∗−c, µ⟩≤0 for t > 0 we have ⟨x∗−c, µ⟩= 0. Thus, ⟨x −c, µ⟩≤0 for all x −c ∈Z+ ∩X If we deﬁne the dual cone X∗ + by X∗ + = {x∗∈X∗: ⟨x∗, x⟩≤0 for all x ∈X and x ≤0}, then µ ∈X∗ +. Thus, the regular point condition gives the stronger regularity µ ∈Z+ for the Lagrange multiplier µ. Example (Source Identiﬁcation) Let X = Z = L2(Ω) and S : L(X, Y ) and consider the constrained minimization: min 1 2\|Sf −y\|2 Y + Z Ω f dx + α 2 \|f\|2 subject to −f ≤0. The second term represents the L1(Ω) regularization for the sparsity of f. In this case, A = α I + S∗S. α > 0, a = S∗y −β and G = −I. The regular point condition (??) is given by 0 ∈int {(−Z + f∗) −Z−−f∗}, Since −Z + f∗contains Z−and the regular point condition holds. Hence there exists µ ∈Z+ such that α f∗+ S∗Sf∗+ β + µ = S∗y, µ ≥0 and (−f∗, µ) = 0. For example, the operator S is deﬁned for Poisson equation: −∆u = f in Ω with a boundary condition ∂ ∂u + u = 0 at Γ = ∂Ωand Sf = Cu with a measurement operator C, e.g., Cu = u\|Γ or Cu = u\|ˆ Ωwith ˆ Ω, a subdomain of Ω. Remark (Penalty Method) Consider the penalized problem J(x) + β 2 \|Ex −b\|2 Y + β 2 \| max(0, Gx −c)\|2 Z. (3.31) The necessary optimality condition is given by Ax + E∗λβ + G∗µβ = a, λβ = β (Exβ −b), µβ = β (Gxβ −c)+. 57 It follows from Theorem 7 that there exists a unique solution xβto (3.21) and satisﬁes J(xβ) + β 2 \|Exβ −b\|2 Y + β 2 \| max(0, Gxβ −c)\|2 Z ≤J(x∗) (3.32) Thus, {xβ} is bounded in X and has a weak convergence subsequence that converges to x in X. Moreover, we have Ex = b and max(0, Gx −c) = 0 and J(x) ≤J(x∗) since the norm is weakly sequentially lower semi-continuous. Since the minimizer to (3.21) is unique, x = x∗and limβ→∞J(xβ) = J(x∗) which implies that xβ →x∗strongly in X. Suppose λβ ∈Y and µβ ∈Z are bounded, there exists a subsequence that converges weakly to (λ, µ) in Y × Z and µ ≥0. Since limβ→∞J(xβ) = J(x∗) it follows from (3.32) that (µβ, Gxβ −c) →0 as β →∞ and thus (µ, Gx∗−c) = 0. Hence (x∗, λ, µ) satisﬁes (3.30). Conversely, if there is a Lagrange multiplier (λ∗, µ∗) ∈Y , then (λβ, µβ) is bounded in Y × Z. In fact, we have (xβ −x∗, A(xβ −x∗)) + (xβ −x∗, E∗(λβ −λ∗)) + (xβ −x∗, G∗(µβ −µ∗)) ≤0 where β (xβ −x∗, E∗(λc −λ∗)) = β (Exβ −b, λβ −λ∗) = \|λβ\|2 −(λβ, λ∗) and β (xβ −x∗, G∗(µβ −µ∗)) = β (Gxβ −c −(Gx∗−c), µβ −µ∗) ≥\|µβ\|2 −(µβ, µ∗). Thus we obtain β (xβ −x∗, A(xβ −x∗)) + \|λβ\|2 + \|µβ\|2 ≤(λβ, λ) + (µβ, µ∗) and hence \|(λβ, µβ)\| ≤\|(λ∗, µ∗)\|. Moreover, we have Lemma Consider the penalized problem J(x) + β 2 \|E(x)\|2 Y + β 2 \| max(0, G(x))\|2 Z. over x ∈C, where C is a closed convex set. Let xβ is a minimizer. Then λβ = β E(xβ), µβ = max(0, Gxβ) are uniformly bounded. Proof: The necessary optimality condition is given by (J′(xβ) + (E′(xβ))∗λβ + (G′(xβ))∗µβ, x −xβ) ≥0 for all x ∈C λβ = β E(xβ), µβ = β max(0, G(xβ)). 58 For all feasible solution ¯ x ∈C J(xβ) + β 2 \|E(xβ)\|2 Y + β 2 \| max(0, G(xβ))\|2 Z ≤J(¯ x). (3.33) Assume that {xβ} is bounded in X (for example, J is J(x) →∞as \|x\| →∞, x ∈ \|calC or C is bounded) and thus has a subsequence that converges weakly to x∗∈C. Moreover, if E(xβ) →E(x∗) and G(xβ) →G(x∗) weakly in Y and Z, respectively (i.e., E and G are weakly sequentially continuous), then E(x∗) = 0 and max(0, G(x∗)) = 0. and J(x∗) ≤J(¯ x). Since J is weakly sequentially lower semi-continuous. Hence x∗is a minimizer of (3.34) and limβ→∞J(xβ) = J(x∗). If J is convex, xβ →x∗in X. Suppose λβ ∈Y and µβ ∈Z are bounded, there exists a subsequence that converges weakly to (λ, µ) in Y × Z and µ ≥0. Since limβ→∞J(xβ) = J(x∗) it follows from (3.33) that (µβ, G(xβ)) →0 as β →∞ and thus (µ, G(x∗)) = 0. Hence (x∗, λ, µ) satisﬁes (3.39). Next we will show that λβ ∈Y and µβ ∈Z are bounded under the regular pint condition. Note that (λβ, E′(xβ)(x −xβ)) = (λβ, E′(x∗)(x −x∗) + (E′(xβ) −E′(x∗))(x −x∗) + E′(xβ)(x∗−xβ)) (µβ, G′(xβ)(x −xβ)) = (µβ, G′(x∗)(x −x∗) + (G′(xβ) −G′(x∗))(x −x∗) + G′(xβ)(x∗−xβ)) and it follows from the regular point condition that there exists ρ > 0 and x ∈C and z ≤0 such that −ρ λβ = E′(x∗)(x −x∗) −ρ µβ = G′(x∗)(x −x∗) −z + G(x∗). Since for all x ∈C (J′(xβ), x −xβ) + (λβ, E′(xβ)(x −xβ) + (µβ, G′(xβ)(x −xβ)) ≥0 we obtain ρ (\|λβ\|2 + \|µβ\|2) ≤(J′(xβ), x∗−xβ) −(λβ, (E′(xβ) −E′(x∗))(x −x∗)) −(µβ, (G′(xβ) −G′(x∗))(x −x∗)) −(µβ, G(xβ)) + (µβ, G(xβ) −G(x∗)). where we used (µβ, z) ≤0. Since \|x −x∗\| ≤ρ M and \|xβ −x∗\| →0, there exist a constant ˜ M > 0 such that for suﬃciently large β > 0 \|λβ\| + \|µβ\| ≤˜ M.□ Example (Quadratic Programming on Rn) Let X = Rn. Given a ∈Rn, b ∈Rm and c ∈Rp consider min 1 2(Ax, x) −(a, x) subject to Bx = b, Cx ≤c 59 where A ∈Rn×n is symmetric and positive, B ∈Rm×n and C ∈Rp×n. Let Ca the (row) vectors of C corresponding to the active indices and assuming B Ca is surjective, the necessary and suﬃcient optimality condition is given by            Ax∗+ Btλ + Ctµ = a Bx∗= b (Cx∗−c)j ≤0 µj ≥0 and (Cx∗−c)jµj = 0. Example (Minimum norm problem with inequality constraint) min \|x\|2 subject to (x, yi) ≤ci, 1 ≤i ≤m. Assume {yi} are linearly independent. Then it has a unique solution x∗and the necessary and suﬃcient optimality is given by x∗+ Pm i=1 µiyi = 0 (x∗, yi) −ci ≤0, µi ≥0 and ((x∗, yi) −ci)µi = 0, ; 1 ≤i ≤m. That is, if G is Grammian (Gij = (yi, yj)) then (−Gx∗−c)i ≤0 µi ≥0 and (−Gx∗−c)iµi = 0, 1 ≤i ≤m. Example (Pointwise Obstacle Problem) Consider min Z 1 0 1 2\| d dxu(x)\|2 − Z 1 0 f(x)u(x) dx subject to u(1 2) ≤c over u ∈X = H1 0(0, 1). Let f = 1. The Riesz representation of F1(u) = Z 1 0 f(t)u(t) dt, F2(u) = u(1 2) are y1(x) = 1 2x(1 −x). y2(x) = 1 2(1 2 −\|x −1 2\|), respectively. Assume that f = 1. Then, we have u∗−y1 + µ y2 = 0, u∗(1 2) = 1 8 −µ 4 ≤c, µ ≥0. Thus, if c ≥1 8, then u∗= y1 and µ = 0. If c ≤1 8, then u∗= y1 −µ y2 and µ = 4(c −1 8). 60 3.6 Constrained minimization in Banach spaces and La-grange multiplier theory In this section we discuss the general nonlinear programming. Let X, Y be Hilbert spaces and Z be a Hilbert lattice. Consider the constrained minimization; min J(x) subject to E(x) = 0 and G(x) ≤0. (3.34) over a closed convex set C in X. Here, where J : X →R, E : X →Y and G : X →Z are continuously diﬀerentiable. Deﬁnition (Derivatives) (1) The functional F on a Banach space X is Gateaux diﬀerentiable at u ∈X if for all d ∈X lim t→0 F(u + td) −F(u) t = F ′(u)(d) exists. If the G-derivative d ∈X →F ′(u)(d) ∈R is liner and bounded, then F is Frechet diﬀerentiable at u. (2) E : X →Y , Y ia Banach space is Frechet diﬀerentiable at u if there exists a linear bounded operator E′(u) such that lim \|d\|X→0 \|E(u + d) −E(u) −E′(u)d\|Y \|d\|X = 0. If Frechet derivative u ∈X →E′(u) ∈L(X, Y ) is continuous, then E is continuously diﬀerentiable. Deﬁnition (Lower semi-continuous) (1) A functional F is lower-semi continuous if lim inf n→∞F(xn) ≥F( lim n→∞xn) (2) A functional F is weakly lower-semi continuous if lim inf n→∞F(xn) ≥F(w −limn→∞xn) Theorem (Lower-semicontinuous) (1) Norm is weakly lower-semi continuous. (2) A convex lower-semicontinuous functional is weakly lower-semi continuous. Proof: Assume xn →x weakly in X. Let x∗∈F(x), i.e., ⟨x∗, x⟩= \|x∗\|\|x\|. Then, we have \|x\|2 = lim n→∞⟨x∗, xn⟩ and \|⟨x∗, xn⟩\| ≤\|xn\|\|x∗\|. Thus, lim inf n→∞\|xn\| ≥\|x\|. (2) Since F is convex, F( X k tkxk) ≤ X k tk F(xk) 61 for all convex combination of xk, i.e., P P k tk = 1, tk ≥0. By the Mazur lemma there exists a sequence of convex combination of weak convergent sequence ({xk}, {F(xk)}) to (x, F(x)) in X × R that converges strongly to (x, F(x)) and thus F(x) ≤lim inf n →∞F(xn).□ Theorem (Existence) (1) A lower-semicontinuous functional F on a compact set S has a minimizer. (2) A weak lower-semicontinuous, coercive functional F has a minimizer. Proof: (1) One can select a minimizing sequence {xn} such that F(xn) ↓η = infx∈S F(x). Since S is a compact, there exits a subsequaence that converges strongly to x∗∈S. Since F is lower-semicontinuous, F(x∗) ≤η. (2) Since F(x) →∞as \|x\| →∞, there exits a bounded minimizing sequence {xn} of F. A bounded sequence in a reﬂexible Banach space X has a weak convergent subsequence to x∗∈X. Since F is weakly lower-semicontinuous, F(x∗) ≤η. □ Next, we consider the constrained optimization in Banach spaces X, Y : min F(y) subject to G(y) ∈K (3.35) over y ∈C, where K is a closed convex cone of a Banach space Y , F : X →R and G : X →Y are C1 and C is a closed convex set of X. For example, if K = {0}, then G(y) = 0 is the equality constraint. If Y = Z is a Hilbert lattice and K = {z ∈Z : z ≤0}, then G(y) ≤0 is the inequality constraint. Then, we have the Lagrange multiplier theory: Theorem (Lagrange Multiplier Theory) Let y∗∈C be a solution to (3.35) and assume the regular point condition 0 ∈int {G′(y∗)(C −y∗) −K + G(y∗)} (3.36) holds. Then, there exists a Lagrange multiplier λ ∈K+ = {y ∈Y : (y, z) ≤0 for all z ∈K} satisfying (λ, G(y∗))Y = 0 such that (F ′(y∗) + G′(y∗)∗λ, y −y∗)X ≥0 for all y ∈C. (3.37) Proof: Let F = {y ∈C : G(y) ∈K} be the feasible set. The sequential tangent cone of F at y∗is deﬁned by T(F, y∗) = {x ∈X : x = lim n→∞ yn −y∗ tn , tn →0+, yn ∈F}. It is easy show that (F ′(y∗), y −y∗) ≥0 for all y ∈T(F, y∗). Let C(y∗) = ∪s≥0, y∈C {s(y −y∗)}, K(G(y∗)) = {K −t G(y∗), t ≥0}. 62 and deﬁne the linearizing cone L(F, y∗) of F at y∗by L(F, y∗) = {z ∈C(y∗) : G′(y∗)z ∈K(G(y∗))}. It follows the Listernik’s theorem [] that L(F, y∗) ⊆T(F, y∗), and thus (F ′(y∗), z) ≥0 for all z = y −y∗∈L(F, y∗), (3.38) Deﬁne the convex cone B by B = {(F ′(y∗)C(y∗) + R+, G′(y∗)C(y∗) −K(G(y∗))} ⊂R × Y. By the regular point condition B has an interior point. From (3.38) the origin (0, 0) is a boundary point of B and thus there exists a hyperplane in R × Y that supports B at (0, 0), i.e., there exists a nontrivial (α, λ) ∈R × Y such that α (F ′(y∗)z + r) + (λ, G′(y∗)z −y) ≥0 for all z ∈C(y∗), y ∈K(G(y∗)) and r ≥0. Setting (z, r) = (0, 0), we have (λ, y) ≤0 for all y ∈K(G(y∗)) and thus λ ∈K+ and (λ, G(y∗)) = 0. Letting (r, y) = (0, 0) and z ∈L(F, y∗), we have α ≥0. If α = 0, the regular point condition implies λ = 0, which is a contradiction. Without loss of generality we set α = 1 and obtain (3.37) by setting (r, y) = 0. □ Corollary (Nonlinear Programming ) If there exists ¯ x such that E(¯ x) = 0 and G(¯ x) ≤0, then there exits a solution x∗to (3.34). Assume the regular point condition 0 ∈int {E′(x∗)(C −x∗) × (G′(x∗)(C −x∗) −Z−+ G(x∗))}, then there exists λ ∈Y and µ ∈Z such that (J′(x∗) + (E′(x∗))∗λ + (G′(x∗)∗µ, x −x∗) ≥0 for all x ∈C, E(x∗) = 0 G(x∗) ≤0, µ ≥0 and (G(x∗), µ)Z = 0. (3.39) 3.7 Control problem Next, we consider the parameterized nonlinear programming. Let X, Y, U be Hilbert spaces and Z be a Hilbert lattice. We consider the constrained minimization of the form min J(x, u) = F(x) + H(u) subject to E(x, u) = 0 and G(x, u) ≤0 over u ∈ˆ C, (3.40) where J : X × U →R, E : X × U →Y and G : X × U →Z are continuously diﬀerentiable and ˆ C is a closed convex set in U. Here we assume that for given u ∈ˆ C 63 there exists a unique x = x(u) ∈X that satisﬁes E(x, u) = 0. Thus, one can eliminate x ∈X as a function of u ∈U and thus (3.40) can be reduced to the constrained minimization over u ∈ˆ C. But it is more advantageous to analyze (3.40) directly using Langrange Theorem as Theorem (Lagrange Caluclus) Assume F and H is C1. Suppose that for given v ∈ˆ C there exists a unique x(t) = x(u + t (v −u)) ∈X for u(t) = u + t(v −u) in a neighborhood B(x, δ) in X and t ∈[0, δ] that satisﬁes E(x(t), u + t (v −u)) = 0. Assume there exist a p satisﬁes the adjoint equation Ex(x, u)∗p + F ′(x) = 0 (3.41) and for all 0 < t < 1 and v ∈ˆ C (E(x(t), u(t)) −E(x, u) −Ex(x(t) −x(0)) −Eu(t(v −u)), p) +F(x(t)) −F(x) −F ′(x)(x(t) −x) = o(t) (3.42) Then, we have lim t→0 J(x(t), ut(v −u) −J(x, u) t = (Eu(x, u)∗p + H′(u), v −u). Proof For v ∈ˆ C let u(t) = u + t(v −u) and x(t) = x(u + t(v −u)) J(x(t), u(t)) −J(x, u) = F ′(x)(x(t) −x(0)) + H′(u)(th) + o(t)ϵ1 = o(t), ϵ1 = F(x(t)) −F(x) −F ′(x(t))(x(t) −F(x) (3.43) From (3.41) F ′(x)(x(t) −x) = −(Ex(x, u)(x(t) −x(0)), p) (3.44) Since E(x(t), u(t)) = E(x(0), u(0)) = 0 deﬁne Ex(x, u)(x(t) −x(0)) + Eu(x, u)(u(t) −u) = Ex(x, u)(x(t) −x(0)) + Eu(x, u)(u(t) −u) −E(x(t), u(t) −E(x(0, u(0)) = ϵ2 (3.45) and thus it follows from it follows from (3.43)–(3.45) that J(x(t), u(t) −J(x, u) = (Eu(x, u)(u(t) −u), p) + H(u)(t (v −u)) + o(t) + ϵ1 + ϵ2 and from the assumption (3.42) the claim holds. □ Corollary (Lagrange Multiplier theory (revised)) Let (x∗, u∗) be a minimizer of (3.40), The conditions in Theorem hols at (x∗, u∗). Then the necessary optimality condition is given by (Eu(x∗, u∗)∗p + H′(u), v −u∗) for all v ∈ˆ C where p satisﬁes the adjoint equation Ex(x∗, u∗)p + F ′(x∗) = 0. 64 Remark (1) If E(x, u) is C1 and Ex(¯ x, ¯ u) is bijective it follows from the implicit function theorem (see, Section??) there exists a neighborhood B((¯ x, ¯ u), δ) such that E(x, u) = 0 has a unique solution x given u. Moreover, t →x(t) ∈X is continuously diﬀerentiable and Ex(x, u) d dtx(0) + Eu(x, u)(v −u) = 0. Thus, assumption (3.42) holds. (2) Suppose E : Y →Y ∗and F are C2, p ∈Y and \|x(t)−x(0)\|2 t →0 as t →0, then assumption (3.42) holds. (3) The penalty formulation is given by min Jβ(x, u) = F(x) + H(u) + β 2 \|E(x, u)\|2 subject to u ∈ˆ C. Assume (xβ, uβ) is a solution. Then the necessary optimality condition is given by Ex(xβ, uβ)∗λβ + F ′(xβ) = 0 (Eu(xβ, uβ)∗λβ + H′(uβ), v −uβ) = 0 for all v ∈ˆ C E(xβ, uβ) −λβ β = 0. (3.46) Consider the iterative method for (xn, un) for as follows. If we apply E(x, u) ∼ E(xn, un) + Ex(xn, un)(x+ −xn) + Eu(u+ −u). the linearization of E at (xn, un) for solutions to (3.46), then we have the update formula for (x+, u+), given (xn, un) with 0 < β ≤∞by solving the system: E(xn, un) + Ex(xn, un)(x+ −xn) + Eu(u+ −u) −λ+ β = 0 Ex(xn, un)∗λ+ + F ′(x+) = 0 (Eu(xn, un)∗λ+ + H′(u+), v −u+) = 0 for all v ∈ˆ C, which is a sequential programming as in Section 10. In general the optimality condition is written as u=Ψ(λ+) and thus the system is for (x+, λ+). Corollary (General case) Let (x∗, u∗) be a minimizer of (3.40) and assume the reg-ular point holds. It follows from Theorem 8 that the necessary optimality condition is given by                    Jx(x∗, u∗) + Ex(x∗, u∗)∗λ + Gx(x∗, u∗)∗µ = 0 (Ju(x∗, u∗) + Eu(x∗, u∗)∗λ + Gu(x∗, u∗)∗µ, u −u∗) ≥0 for all u ∈ˆ C E(x∗, u∗) = 0 G(x∗, µ∗) ≤0, µ ≥0, and (G(x∗, u∗), µ) = 0. (3.47) 65 Example (Control problem) Consider the optimal control problem min Z T 0 (ℓ(x(t)) + h(u(t))) dt + G(x(T)) subject to d dtx(t) −f(x, u) = 0, x(0) = x0 u(t) ∈ˆ U = a closed convex set in Rm. (3.48) We let X = H1(0, T; Rn), U = L2(0, T : Rm), Y = L2(0, T; Rn) and J(x, u) = Z T 0 (ℓ(x) + h(u)) dt + G(x(T)), E(x, u) = f(x, u) −d dtx and ˆ C = {u ∈U, u(t) ∈ˆ U a.e in (0, T)}. From (3.47) we have Z T 0 (ℓ′(x∗(t))φ(t) + (p(t), fx(x∗, u∗)φ(t) −d dtφ) dt + G′(x(T))φ(T) = 0 for all φ ∈H1(0, T; Rn) with φ(0) = 0. By the integration by part Z T 0 ( Z T t (ℓ′(x∗) + fx(x∗, u∗)tp) ds + G′(x∗(T)) −p(t), dφ dt ) dt = 0. Since φ ∈H1(0, T; Rn) satisfying φ(0) = 0 is arbitrary, we have p(t) = G′(x∗(T)) + Z T t (ℓ′(x∗) + fx(x∗, u∗)tp) ds and thus the adjoint state p is absolutely continuous and satisﬁes −d dtp(t) = fx(x∗, u∗)tp(t) + ℓ′(x∗), p(T) = G′(x∗(T)). From the the optimality, we have Z T 0 (h′(u∗) + fu(x∗, u∗)tp(t))(u(t) −u∗(t)) dt for all u ∈ˆ C, which implies (h′(u∗(t)) + fu(x∗(t), u∗(t))tp(t))(v −u∗(t)) ≥0 for all v ∈ˆ U at Lebesgue points t of u∗(t). Hence we obtain the necessary optimality is of the form of Two-Point-Boundary value problem;            d dtx∗(t) = f(x∗, u∗), x(0) = x0 (h′(u∗(t)) + fu(x∗(t), u∗(t))tp(t), v −u∗(t)) ≥0 for all v ∈ˆ U −d dtp(t) = fx(x∗, u∗)tp(t) + lx(x∗), p(T) = Gx(x∗(T)). 66 Let ˆ U = [−1, 1]m coordinate-wise and h(u) = α 2 \|u\|2, f(x, u) = ˜ f(x) + Bu. Thus, the optimality condition implies (α u∗(t) + Btp(t), v −u∗(t))Rm ≥0 fur all \|v\|∞≤1. Or, equivalently u∗(t) minimizes α 2 \|u\|2 + (Btp(t), u) over \|u\|∞≤1 Thus, we have u∗(t) =              1 if −Btp(t) α > 1 −Btp(t) α if \| Btp(t) α \| ≤1 −1 if −Btp(t) α < −1. (3.49) Example (Coeﬃcient estimation) Let ω be a bounded open set in Rd and ˜ Ωbe a subset in Ω. Consider the parameter estimation problem min J(u, c) = Z ˜ Ω \|y −u\|2 dx + Z Ω \|∇c\|2 dx subject to −∆u + c(x)u = f, c ≥0 over (u, c) ∈H1 0(Ω) × H1(Ω). We let Y = (H1 0(Ω))∗and ⟨E(u, c), φ⟩= (∇u, ∇φ) + (c u, φ)L2(Ω) −⟨f, φ⟩ for all φ ∈H1 0(Ω). Since ⟨Jx + E∗ xλ, h⟩= (∇h, ∇λ) + (c h, λ)L2(ω) −((y −u) χ˜ Ω, h) = 0, for all h ∈H1 0(Ω), i.e., the adjoint λ ∈H1 0(Ω) satisﬁes −∆λ + c λ = (y −u) χ˜ Ω. Also, there exits µ ≥0 in L2(Ω) such that ⟨Jc + E∗ c λ, d⟩= α (∇c∗, ∇d) + (d u, λ)L2(Ω) −(µ, d) = 0 for all d ∈H1(Ω) and ⟨µ, c −c∗⟩= 0 and ⟨µ, d⟩≥0 for all d ∈H1(Ω) satisfying d ≥0. 3.8 Minimum norm solution in Banach space In this section we discuss the minimum distance and the minimum norm problem in Banach spaces. Given a subspace M of a normed space X, the orthogonal complement M⊥of M is deﬁned by M⊥= {x∗∈X∗: ⟨x∗, x⟩= 0 for all x ∈M}. 67 Theorem If M is a closed subspace of a normed space, then (M⊥)⊥= {x ∈X : ⟨x, x∗⟩= 0 for all x∗∈M⊥} = M Proof: If there exits a x ∈(M⊥)⊥but x / ∈M, deﬁne a linear functional f by f(α x + m) = α on the space spanned by x + M. Since x / ∈m and M is closed \|f\| = sup m∈M, α∈R \|f(α x + m)\| \|α x + m\|X = sup m∈M \|f(x + m)\| \|x + m\|X = 1 infm∈M \|x + m\|X < ∞, and thus f is bounded. By the Hahn-Banach theorem, f can be extended to x∗∈X∗. Since ⟨x∗, m⟩= 0 for all m ∈M, thus x∗∈M⊥. It is contradicts to the fact that ⟨x∗, x⟩= 1. □ Theorem (Duality I) Given x ∈X and a subspace M, we have d = inf m∈M \|x −m\|X = max \|x∗\|≤1, x∗∈M⊥⟨x∗, x⟩= ⟨x∗ 0, x⟩, where the maximum is attained by some x∗ 0 ∈M⊥with \|x∗ 0\| = 1. If the inﬁmum on the ﬁrst equality holds for some m0 ∈M, then x∗ 0 is aligned with x −m0. Proof: By the deﬁnition of inf, for any ϵ > 0, there exists mϵ ∈M such that \|x−mϵ\| ≤ d + ϵ. For any x∗∈M⊥, \|x∗\| ≤1, it follows that ⟨x∗, x⟩= ⟨x∗, x −mϵ⟩≤\|x∗\|\|x −mϵ\| ≤d + ϵ. Since ϵ > 0 is arbitrary, ⟨x∗, x⟩≤d. It only remains to show that ⟨x∗ 0, x⟩= d for some x∗ 0 ∈X∗. Deﬁne a linear functional f on the space spanned by x + M by f(α x + m) = α d Since \|f\| = sup m∈M, α∈R \|f(α x + m)\| \|α x + m\|X = d infm∈M \|x + m\|X = 1 f is bounded. By the Hahn-Banach theorem, there exists an extension x∗ 0 of f with \|x∗ 0\| = 1. Since ⟨x∗ 0, m⟩= 0 for all m ∈M, x∗ 0 ∈M⊥. By construction ⟨x∗ 0, x⟩= d. ???Moreover, if \|x −m0\| = d for some m0 ∈M, then ⟨x∗ 0, x −m0⟩= d = \|x∗ 0\|\|x −m0\|.□ 3.8.1 Duality Principle Corollary (Duality) If m0 ∈M attains the minimum if and only if there exists a nonzero x∗∈M⊥that is aligned with x −m0. Proof: Suppose that x∗∈M⊥is aligned with x −m0, \|x −m0\| = ⟨x∗, x −m0⟩= ⟨x∗, x⟩= ⟨x∗, x⟩≤\|x −m\| for all m ∈M. □ 68 Theorem (Duality principle) (1) Given x∗∈X∗, we have min m∗∈M⊥\|x∗−m∗\|X∗= sup x∈M, \|x\|≤1 ⟨x∗, x⟩. (2) The minimum of the left is achieved by some m∗ 0 ∈M⊥. (3) If the supremum on the right is achieved by some x0 ∈M, then x∗−m∗ 0 is aligned with x0. Proof: (1) Note that for m∗∈∈M⊥ \|x∗−m∗\| = sup \|x\|≤1 ⟨x∗−m∗, x⟩≥ sup \|x\|≤1, x∈M ⟨x∗−m∗, x⟩= sup \|x\|≤1, x∈M ⟨x∗, x⟩. Consider the restriction of x∗to M (with norm supx∈M, \|x\|≤1⟨x∗, x⟩. Let y∗be the Hahn-Banach extension of this restricted x∗. Let m∗ 0 = x∗−y∗. Then m∗ 0 ∈M⊥and \|x∗−m∗ 0\| = \|y∗\| = supx∈M, \|x\|≤1⟨x∗, x⟩. Thus, the minimum of the left is achieved by some m∗ 0 ∈M⊥. If the supremum on the right is achieved by some x0 ∈M, then x∗−m∗ 0 is aligned with x0. Obviously, \|x0\| = 1 and \|x∗−m∗ 0\| = ⟨x∗, x0⟩= ⟨x∗−m∗ 0, x0⟩. □ In all applications of the duality theory, we use the alignment properties of the space and its dual to characterize optimum solutions, guarantee the existence of a solution by formulating minimum norm problems in the dual space, and examine to see if the dual problem is easier than the primal problem. Example 1(Chebyshev Approximation) Problem is to to ﬁnd a polynomial p of degree less than or equal to n that ”best” approximates f ∈C[0, 1] in the the sense of the sup-norm over [a, b]. Let M be the space of all polynomials of degree less than or equal to n. Then, M is a closed subspace of C[0, 1] of dimension n + 1. The inﬁmum of \|f −p\|∞is achievable by some p0 ∈M since M is closed. The Chebyshev theorem states that the set Γ = {t ∈[0, 1] : \|f(t) −p0(t)\| = \|f −p0\|∞} contains at least n + 2 points. In fact, f −p0 must be aligned with some element in M⊥⊂C[0, 1]∗= BV [0, 1]. Assume Γ contains m < n + 2 points 0 ≤t1 < · · · < tm ≤1. If v ∈BV [0, 1] is aligned with f −p0, then v is a piecewise continuous function with jump discontinuities only at these tk’s. Let tk be a point of jump discontinuity of v. Deﬁne q(t) = Πm j̸=k(t −tj) Then, q ∈M but ⟨q, v⟩̸= 0 and hence v / ∈M⊥, which is a contradiction. Next, we consider the minimum norm problem; min \|x∗\|X∗ subject to ⟨x∗, yk⟩= ck, 1 ≤k ≤n. where y1, · · · , yn ∈X are given linearly independent vectors. Theorem Let ¯ x∗∈X∗satisfying the constraints ⟨¯ x∗, yk⟩= ck, 1 ≤k ≤n. Let M = span{yk}. Then, d = min ⟨¯ x∗,yk⟩=ck \|x∗\|X∗= min m∗∈M⊥\|¯ x∗−m∗\|. By the duality principle, d = min m∗ıM⊥\|¯ x∗−m∗\| = sup x∈M, \|x\|≤1 ⟨¯ x∗, x⟩. 69 Write x = Y a where Y = [y1, · · · , yn] and a ∈Rn. Then d = min \|x∗\| = max \|Y a\|≤1⟨¯ x∗, Y a⟩= max \|Y a\|≤1 (c, a). Thus, the optimal solution x∗ 0 must be aligned with Y a. Example (DC motor control) Let X = L1(0, 1) and X∗= L∞(0, 1). min \|u\|∞ subject to Z 1 0 et−1u(t) dt = 0, Z 1 0 (1 + et−1)u(t) = 1. Let y1 = et−1 and y2 = 1 + et−1. From the duality principle min ⟨yk,u⟩=ck \|u\|∞= max \|a1y1+a2y2\|1≤1 a2. The control problem now is reduced to ﬁnding constants a1 and a2 that maximizes a2 subject to R 1 0 \|(a1 −a2)et−1 + a2\| dt ≤1. The optimal solution u ∈L∞(0, 1) should be aligned with (a1 −a2)et−1 + a2 ∈L1(0, 1). Since (a1 −a2)et−1 + a2, can change sign at most once. The alignment condition implies that u must have values \|u\|∞and can change sign at most once, which is the so called bang-bang control. 4 Linear Cauchy problem and C0-semigroup the-ory In this section we discuss the Cauchy problem of the form d dtu(t) = Au(t) + f(t), u(0) = u0 ∈X in a Banach space X, where u0 ∈X is the initial condition and f ∈L1(0, T; X). Such problems arise in PDE dynamics and functional equations. We construct the mild solution u(t) ∈C(0, T; X): u(t) = S(t)u0 + Z t 0 S(t −s)f(s) ds (4.1) where a family of bounded linear operator {S(t), t ≥0} is C0-semigroup on X. Deﬁnition (C0 semigroup) (1) Let X be a Banach space. A family of bounded linear operators {S(t), t ≥0} on X is called a strongly continuous (C0) semigroup if S(t + s) = S(t)S(s) for t, s ≥0 with S(0) = I \|S(t)φ −φ\| →0 as t →0+ for all φ ∈X. (2) A linear operator A in X deﬁned by Aφ = lim t→0+ S(t)φ −φ t (4.2) 70 with dom(A) = {φ ∈X : the strong limit of lim t→0+ S(t)φ −φ t in X exists}. is called the inﬁnitesimal generator of the C0 semigroup S(t). In this section we present the basic theory of the linear C0-semigroup on a Banach space X. The theory allows to analyze a wide class of the physical and engineering dynamics using the uniﬁed framework. We also present the concrete examples to demonstrate the theory. There is a necessary and suﬃcient condition (Hile-Yosida Theorem) for a closed, densely deﬁned linear A in X to be the inﬁnitesimal generator of the C0 semigroup S(t). Moreover, we will show that the mild solution u(t) satisﬁes ⟨u(t), ψ⟩= ⟨u0, ψ⟩+ Z (⟨x(s), A∗ψ⟩+ ⟨f(s), ψ⟩ds (4.3) for all ψ ∈dom (A∗). Examples (1) For A ∈L(X), deﬁne a sequence of linear operators in X SN(t) = X k 1 k!(At)k. Then \|SN(t)\| ≤ X 1 k!(\|A\|t)k ≤e\|A\| t and d dtSN(t) = ASN−1(t) Since S(t) = eAt = limN→∞SN(t) in the operator norm, we have d dtS(t) = AS(t) = S(t)A. (2) Consider the the hyperbolic equation ut + ux = 0, u(0, x) = u0(x) in (0, 1). (4.4) Deﬁne the semigroup S(t) of translations on X = L2(0, 1) by S(t)u0 = ˜ u0(x −t), where ˜ u0(x) = 0, x ≤0, ˜ u0 = u0 on [0, 1]. Then, {S(t), t ≥0} is a C0 semigroup on X. If we deﬁne u(t, x) = S(t)u0 with u0 ∈H1(0, 1) with u0(0) = 0 satisﬁes (4.6) a.e.. The generator A is given by Aφ = −φ′ with dom(A) = {φ ∈H1(0, 1) with φ(0) = 0}. In fact S(t)u0 −u0 t = ˜ u0(x −t) −˜ u0(x) t = u′ 0(x), a.e. x ∈(0, 1). if u0 ∈dom(A). Thus, u(t) = S(t)u0 satisﬁes the Cauchy problem d dtu(t) = Au(t) if u0 ∈dom(A). 71 (3) Let Xt is a Markov process, i.e. Ex[g(Xt+h\|calF t] = E0,Xt[g(Xh)]. for all g ∈X = L2(Rn). Deﬁne the linear operator S(t) by (S(t)u0)(x) = E0,x[u0(Xt)], t ≥0, u0 ∈X. The semigroup property of S(t) follows from the Markov property, i.e. S(t+s)u0 = E0,x[u0(Xt+s)] = E[E[g(X0,x t+s\|Ft] = E[EX0,x t u0(Xs)]] = Ex[(S(t)u0)(Xs)] = S(s)(S(t)u0). The strong continuity follows from that X0,x t −x is a.s for all x ∈Rn. If Xt = Bt is a Brownian motion, then the semigroup S(t) is deﬁned by S(t)u0 = 1 ( √ 2πtσ)n Z Rn e−\|x−y\|2 2σ2 t u0(y) dy, (4.5) and u(t) = S(t)u0 satisﬁes the heat equation. ut = σ2 2 ∆u, u(0, x) = u0(x) in L2(Rn). (4.6) 4.1 Finite diﬀerence in time Let A be closed, densely deﬁned linear operator dom(A) →X. We use the ﬁnite diﬀerence method in time to construct the mild solution (4.1). For a stepsize λ > 0 consider a sequence {un} in X generated by un −un−1 λ = Aun + fn−1, (4.7) with fn−1 = 1 λ Z nλ (n−1)λ f(t) dt. Assume that for λ > 0 the resolvent operator Jλ = (I −λ A)−1 is bounded. Then, we have the product formula: un = Jn λ u0 + n−1 X k=0 Jn−k λ fk λ. (4.8) In order to un ∈X is uniformly bounded in n for all u0 ∈X (with f = 0), it is necessary that \|Jn λ \| ≤ M (1 −λω)n for λω < 1, (4.9) for some M ≥1 and ω ∈R. Hille’s Theorem Deﬁne a piecewise constant function in X by uλ(t) = uk−1 on [tk−1, tk) 72 Then, max t∈[0,T] \|uλ −u(t)\|X →0 as λ →0+ to the mild solution (4.1). That is, S(t)x = lim n→∞(I −t nA)[ t n ]x exists for all x ∈X and {S(t), t ≥0} is the C0 semigoup on X and its generator is A, where [s] is the largest integer less than s ∈R. Proof: First, note that \|Jλ\| ≤ M 1 −λω and for x ∈dom(A) Jλx −x = λJλAx, and thus \|Jλx −x\| = \|λJλAx\| ≤ λ 1 −λω\|Ax\| →0 as λ →0+. Since dom(A) is dense in X it follows that \|Jλx −x\| →0 as λ →0+ for all x ∈X. Deﬁne the linear operators Tλ(t) and Sλ(t) by Sλ(t) = Jk λ and Tλ(t) = Jk−1 λ + t −tk λ (Jk λ −Jk−1 λ ), on (tk−1, tk]. Then, d dtTλ(t) = ASλ(t), a.e. in t ∈[0, T]. Thus, Tλ(t)u0−Tµ(t)u0 = Z t 0 d ds(Tλ(s)Tµ(t−s)u0) ds = Z t 0 (Sλ(s)Tµ(t−s)−Tλ(s)Sµ(t−s))Au0 ds Since Tλ(s)u −Sλ(s)u = s −tk λ Tλ(tk−1)(Jλ −I)u on s ∈(tk−1, tk]. By the bounded convergence theorem \|Tλ(t)u0 −Tµ(t)u\|X →0 as λ, µ →0+ for all u ∈dom(A2). Thus, the unique limit deﬁnes the linear operator S(t) by S(t)u0 = lim λ→0+ Sλ(t)u0. (4.10) for all u0 ∈dom(A2). Since \|Sλ(t)\| ≤ M (1 −λω)[t/n] ≤Meωt 73 and dom(A2) is dense, (4.10) holds for all u0 ∈X. Moreover, we have S(t + s)u = lim λ→0+ Jn+m λ = Jn λ Jm λ u = S(t)S(s)u and limt→0+ S(t)u = limt→0+ Jtu = u for all u ∈X. Thus, S(t) is the C0 semigroup on X. Moreover, {S(t), t ≥0} is in the class G(M, ω), i.e., \|S(t)\| ≤Meωt. Note that Tλ(t)u0 −u0 = A Z t 0 Sλu0 ds. Since limλ→0+ Tλ(t)u0 = limλ→0+ Sλ(t)u0 = S(t)u0 and A is closed, we have S(t)u0 −u0 = A Z t 0 S(s)u0 ds, Z t 0 S(s)u0 ds ∈dom(A). If B is a generator of {S(t), t ≥0}, then Bx = lim t→0+ S(t)x −x t = Ax if x ∈dom(A). Conversely, if u0 ∈dom(B), then u0 ∈dom(A) since A is closed and t →S(t)u is continuous at 0 for all u ∈X and thus 1 t A Z t 0 S(s)u0 ds = Au0 as t →0+. Hence Au0 = S(t)u0 −u0 t = Bu0 That is, A is the generator of {S(t), t ≥0}. Similarly, we have n−1 X k=0 Jn−k λ fk = Z t 0 Sλ(t −s)f(s) ds → Z t 0 S(t −s)f(s) ds as λ →0+ by the Lebesgue dominated convergence theorem. □ The following theorem state the basic properties of C0 semigroups: Theorem (Semigroup) (1) There exists M ≥1, ω ∈R such that S ∈G(M, ω) class, i.e., \|S(t)\| ≤M eωt, t ≥0. (4.11) (2) If x(t) = S(t)x0, x0 ∈X, then x ∈C(0, T; X) (3) If x0 ∈dom (A), then x ∈C1(0, T; X) ∩C(0, T; dom(A)) and d dtx(t) = Ax(t) = AS(t)x0. (4) The inﬁnitesimal generator A is closed and densely deﬁned. For x ∈X S(t)x −x = A Z t 0 S(s)x ds. (4.12) 74 (5) λ > ω the resolvent is given by (λ I −A)−1 = Z ∞ 0 e−λsS(s) ds (4.13) with estimate \|(λ I −A)−n\| ≤ M (λ −ω)n . (4.14) Proof: (1) By the uniform boundedness principle there exists M ≥1 such that \|S(t)\| ≤ M on [0, t0] For arbitrary t = k t0+τ, k ∈N and τ ∈[0, t0) it follows from the semigroup property we get \|S(t)\| ≤\|S(τ)\|\|S(t0\|k ≤Mek log \|S(t0)\| ≤Meω t with ω = 1 t0 log \|S(t0)\|. (2) It follows from the semigroup property that for h > 0 x(t + h) −x(t) = (S(h) −I)S(t)x0 and for t −h ≥0 x(t −h) −x(t) = S(t −h)(I −S(h))x0 Thus, x ∈C(0, T; X) follows from the strong continuity of S(t) at t = 0. (3)–(4) Moreover, x(t + h) −x(t) h = S(h) −I h S(t)x0 = S(t)S(h)x0 −x0 h and thus S(t)x0 ∈dom(A) and lim h→0+ x(t + h) −x(t) h = AS(t)x0 = Ax(t). Similarly, lim h→0+ x(t −h) −x(t) −h = lim h→0+ S(t −h)S(h)φ −φ h = S(t)Ax0. Hence, for x0 ∈dom(A) S(t)x0 −x0 = Z t 0 S(s)Ax0 ds = Z t 0 AS(s)x0 ds = A Z t 0 S(s)x0 ds (4.15) If xn ∈don(A) →x and Axn →y in X, we have S(t)x −x = Z t 0 S(s)y ds Since lim t→0+ 1 t Z t 0 S(s)y ds = y x ∈dom(A) and y = Ax and hence A is closed. Since A is closed it follows from (4.15) that for x ∈X Z t 0 S(s)x ds ∈dom(A) 75 and (4.12) holds. For x ∈X let xh = 1 h Z h 0 S(s)x ds ∈dom(A) Since xh →x as h →0+, dom(A) is dense in X. (5) For λ > ω deﬁne Rt ∈L(X) by Rt = Z t 0 e−λsS(s) ds. Since A −λ I is the inﬁnitesimal generator of the semigroup eλtS(t), from (4.12) (λ I −A)Rtx = x −e−λtS(t)x. Since A is closed and \|e−λtS(t)\| →0 as t →∞, we have R = limt→∞Rt satisﬁes (λ I −A)Rφ = φ. Conversely, for φ ∈dom(A) R(A −λ I)φ = Z ∞ 0 e−λsS(s)(A −λ I)φ = lim t→∞e−λtS(tφ −φ = −φ Hence R = Z ∞ 0 e−λsS(s) ds = (λ I −A)−1 Since for φ ∈X \|Rx\| ≤ Z ∞ 0 \|e−λsS(s)x\| ≤M Z ∞ 0 e(ω−λ)s\|x\| ds = M λ −ω\|x\|, we have \|(λ I −A)−1\| ≤ M λ −ω, λ > ω. Note that (λ I −A)−2 = Z ∞ 0 e−λtS(t) ds Z ∞ 0 eλ sS(s) ds = Z ∞ 0 Z ∞ 0 e−λ(t+s)S(t + s) ds dt = Z ∞ 0 Z ∞ t e−λσS(σ) dσ dt = Z ∞ 0 σe−λσS(σ) dσ. By induction, we obtain (λ I −A)−n = 1 (n −1)! Z ∞ 0 tn−1e−λtS(t) dt. (4.16) Thus, \|(λ I −A)−n\| ≤ 1 (n −1)! Z ∞ 0 tn−1e−(λ−ω)t dt = M (λ −ω)n .□ We then we have the necessary and suﬃcient condition: 76 Hile-Yosida Theorem A closed, densely deﬁned linear operator A on a Banach space X is the inﬁnitesimal generator of a C0 semigroup of class G(M, ω) if and only if \|(λ I −A)−n\| ≤ M (λ −ω)n for all λ > ω (4.17) Proof: The suﬃcient part follows from the previous Theorem. In addition, we describe the Yosida construction. Deﬁne the Yosida approximation Aλ ∈L(X) of A by Aλ = Jλ −I λ = AJλ. (4.18) Deﬁne the Yosida approximation: Sλ(t) = eAλt = e−t λ eJλ t λ . Since \|Jk λ\| ≤ M (1 −λω)k we have \|Sλ(t)\| ≤e−t λ ∞ X k=0 M k! \|Jk λ\|( t λ)k ≤Me ω 1−λω t. Since d dsSλ(s)Sˆ λ(t −s) = Sλ(s)(Aλ −Aˆ λ)Sˆ λ(t −s), we have Sλ(t)x −Sˆ λ(t)x = Z t 0 Sλ(s)Sˆ λ(t −s)(Aλ −Aˆ λ)x ds Thus, for x ∈dom(A) \|Sλ(t)x −Sˆ λ(t)x\| ≤M2teωt \|(Aλ −Aˆ λ)x\| →0 as λ, ˆ λ →0+. Since dom(A) is dense in X this implies that S(t)x = lim λ→0+ Sλ(t)x exist for all x ∈X, deﬁnes a C0 semigroup of G(M, ω) class. The necessary part follows from (4.16) □ Theorem (Mild solution) (1) If for f ∈L1(0, T; X) deﬁne x(t) = x(0) + Z t 0 S(t −s)f(s) ds, then x(t) ∈C(0, T; X) and it satisﬁes x(t) = A Z t 0 x(s) ds + Z t 0 f(s) ds. (4.19) (2) If Af ∈L1(0, T; X) then x ∈C(0, T; dom(A)) and x(t) = x(0) + Z t 0 (Ax(s) + f(s)) ds. 77 (3) If f ∈W 1,1(0, T; X), i.e. f(t) = f(0) + R t 0 f′(s) ds, d dtf = f′ ∈L1(0, T; X), then Ax ∈C(0, T; X) and A Z t 0 S(t −s)f(s) ds = S(t)f(0) −f(t) + Z t 0 S(t −s)f′(s) ds. (4.20) Proof: Since Z t 0 Z τ 0 S(t −s)f(s) dsdτ = Z t 0 ( Z t s S(τ −s)d τ)f(s) ds, and A Z t 0 S(s) ds = S(t) −I we have x(t) ∈dom(A) and A Z t 0 x(s) ds = S(t)x −x + Z t 0 S(t −s)f(s) ds − Z t 0 f(s) ds. and we have (4.19). (2) Since for h > 0 x(t + h) −x(t) h = Z t 0 S(t −s)S(h) −I h f(s) ds + 1 h Z t+h t S(t + h −s)f(s) ds if Af ∈L1(0, T; X) lim h→0+ x(t + h) −x(t) h = Z t 0 S(t −s)Af(s) ds + f(t) a.e. t ∈(0, T). Similarly, x(t −h) −x(t) −h = Z t−h 0 S(t −h −s)S(h) −I h f(s) ds + 1 h Z t t−h S(t −s)f(s) ds → Z t 0 S(t −s)Af(s) ds + f(t) a.e. t ∈(0, T). (3) Since S(h) −I h x(t) = 1 h( Z h 0 S(th −s)f(s) ds − Z t+h t S(t + f −s)f(s) ds + Z t 0 S(t −s)f(s + h) −f(s) h ds, letting h →0+, we obtain(4.20). □ It follows from Theorems the mild solution x(t) = S(t)x(0) + Z t 0 S(t −s)f(s) ds 78 satisﬁes x(t) = x(0) + A Z t 0 x(s) + Z t 0 f(s) ds. Note that the mild solution x ∈C(0, T; X) depends continuously on x(0) ∈X and f ∈L1(0, T; X) with estimate \|x(t)\| ≤M(eωt\|x(0)\| + Z t 0 eω(t−s)\|f(s)\| ds). Thus, the mild solution is the limit of a sequence {xn} of strong solutions with xn(0) ∈ dom(A) and fn ∈W 1,1(0, T; X), i.e., since dom(A) is dense in X and W 1,1(0, T; X) is dense in L1(0, T; X), \|xn(t) −x(t)\|X →0 uniformly on [0, T] for \|xn(0) −x(0)\|x →0 and \|fn −f\|L1(0,T;X) →0 as n →∞. Moreover, the mild solution x ∈C(0, T : X) is a weak solution to the Cauchy problem d dtx(t) = Ax(t) + f(t) (4.21) in the sense of (4.3), i.e., for all ψ ∈dom(A∗) ⟨x(t), ψ⟩X×X∗is absolutely continues and d dt⟨x(t), ψ⟩= ⟨x(t), ψ⟩+ ⟨f(t), ψ⟩a.e. in (0, T). If x(0) ∈dom(A) and Af ∈L1(0, T; X), then Ax ∈C(0, T; X), x ∈W 1,1(0, T; X) and d dtx(t) = Ax(t) + f(t), a.e. in (0, T) If x(0) ∈dom(A) and f ∈W 1,1(0, T; X), then x ∈C(0, T; dom(A)) ∩C1(0, T; X) and d dtx(t) = Ax(t) + f(t), everywhere in [0, T]. 4.2 Weak-solution and Ball’s result Let A be a densely deﬁned, closed linear operator on a Banach space X. Consider the Cauchy equation in X: d dtu = Au + f(t), (4.22) where u(0) = x ∈X and f ∈L1(0, τ; X) is a weak solution to of (4.22) if for every ψ ∈dom(A∗) the function t →⟨u(t), ψ⟩is absolutely continuous on [0, τ] and d dt⟨u(t), ψ⟩= ⟨u(t), A∗ψ⟩+ ⟨f(t), ψ⟩, a.e. in [0, τ]. (4.23) It has been shown that the mild solution to (4.22) is a weak solution. Lemma B.1 Let A be a densely deﬁned, closed linear operator on a Banach space X. If x, y ∈X satisfy ⟨y, ψ⟩= ⟨x, A∗ψ⟩for all ψ ∈dom(A∗), then x ∈dom(A) and y = Ax. 79 Proof: Let G(A) ⊂X × X denotes the graph of A. Since A is closed G(A) is closed. Suppose y ̸= Ax. By Hahn-Banach theorem there exist z, z∗∈X∗such that ⟨Ax, z⟩+ ⟨x, z∗⟩= 0 and ⟨y, z⟩+ ⟨x, z∗⟩̸= 0. Thus z ∈dom(A∗) and z∗= A∗z. By the condition ⟨y, z⟩+ ⟨x, z∗⟩= 0, which is a contradiction. □ Then we have the following theorem. Theorem (Ball) There exists for each x ∈X and f ∈L1(0, τ; X) a unique weak solution of (4.22)satisfying u(0) = x if and only if A is the generator of a strongly continuous semigroup {T(t)} of bounded linear operator on X, and in this case u(t) is given by u(t) = T(t)x + Z t 0 T(t −s)f(s) ds. (4.24) Proof: Let A generate the strongly continuous semigroup {T(t)} on X. Then, for some M, \|T(t)\| ≤M on t ∈[0, τ]. Suppose x ∈dom(A) and f ∈W 1,1(0, τ; X). Then we have d dt⟨u(t), ψ⟩= ⟨Au(t) + f(t), ψ⟩= ⟨u(t), A∗ψ⟩+ ⟨f(t), ψ⟩. For (x, f) ∈X ×L1(0, τ; X) there exista a sequence (xn, fn) in dom(A)×W 1,1(0, τ; X) such that \|xn −x\|X + \|fn −f\|L1(0,τ;X) →0 as n →∞If we deﬁne un(t) = T(t)xn + Z t 0 T(t −s)fn(s) ds, then we have ⟨un(t), ψ⟩= ⟨x, ψ⟩+ Z t 0 (⟨un(s), A∗ψ⟩+ ⟨fn(s), ψ⟩) ds and \|un(t) −u(t)\|X ≤M (\|xn −x\|X + Z t 0 \|fn(s) −f(s)\|X ds). Passing limit n →]∞, we see that u(t) is a weak solution of (4.22). Next we prove that u(t) is the only weak solution to (4.22) satisfying u(0) = x. Let ˜ u(t) be another such weak solution and set v = u −˜ u. Then we have ⟨v(t), ψ⟩= ⟨ Z t 0 v(s) dt, A∗ψ⟩ for all ψ ∈dom(A∗) and t ∈[0, τ]. By Lemma B.1 this implies z(t) = R t 0 v(s) ds ∈dom(A) and d dtz(t) = Az(t) with z(0) = 0. Thus z = 0 and hence u(t) = ˜ u(t) on [0, τ]. Suppose that A such that (4.22) has a unique weak solution u(t) satisfying u(0) = x. For t ∈[0, τ] we deﬁne the linear operator T(t) on X by T(t)x = u(t) −u0(t), where u0 is the weak solution of (4.22) satisfying u(0) = 0. If for t = nT + s, where n is a nonnegative integer and s ∈[0, τ) we deﬁne T(t)x = T(s)T(τ)nx, then T(t) is a semigroup. The map θ : x →C(0, τ; X) deﬁned by θ(x) = T(·)x has a closed graph by the uniform bounded principle and thus T(t) is a strongly continuous semigroup. Let B be the generator of {T(t)} and x ∈dom(B). For ψ ∈dom(A∗) d dt⟨T(t)x, ψ⟩\|t=0 = ⟨Bx, ψ⟩= ⟨x, A∗ψ⟩. 80 It follows from Lemma that x ∈dom(A) and Ax = Bx. Thus dom(B) ⊂dom(A). The proof of Theorem is completed by showing dom(A) ⊂dom(B). Let x ∈dom(A). Since for z(t) = T(t)x ⟨z(t), ψ⟩= ⟨ Z t 0 z(s) dt, A∗ψ⟩ it follows from Lemma that R t 0 T(s)x ds and R t 0 T(s)Ax ds belong to dom(A) and T(t)x = x + A Z t 0 T(s)x ds T(t)Ax = Ax + A Z t 0 T(s)Ax ds (4.25) Consider the function w(t) = Z t 0 T(s)Ax ds −A Z t 0 T(s)x ds. It then follows from (4.25) that z ∈C(0, τ; X). Clearly w(0) = 0 and it also follows from (4.25) that d dt⟨w(t), ψ⟩= ⟨w(t), A∗ψ⟩. (4.26) for ψ ∈dom(A∗). But it follows from our assumptions that (4.26) has the unique solution w = 0. Hence from (4.25) T(t)x −x = A Z t 0 T(s)x ds and thus lim t→0+ T(t)x −x t = Ax which implies x ∈dom(B). □ 4.3 Lumer-Phillips Theorem The condition (4.17) is very diﬃcult to check for a given A in general. For the case M = 1 we have a very complete characterization. Lumer-Phillips Theorem The followings are equivalent: (a) A is the inﬁnitesimal generator of a C0 semigroup of G(1, ω) class. (b) A −ω I is a densely deﬁned linear m-dissipative operator,i.e. \|(λ I −A)x\| ≥(λ −ω)\|x\| for all x ∈don(A), λ > ω (4.27) and for some λ0 > ω R(λ0 I −A) = X. (4.28) Proof: It follows from the m-dissipativity \|(λ0 I −A)−1\| ≤ 1 λ0 −ω 81 Suppose xn ∈dom(A) →x and Axn →y in X, the x = lim n→∞xn = (λ0 I −A)−1 lim n→∞(λ0 xn −Axn) = (λ0 I −A)−1(λ0 x −y). Thus, x ∈dom(A) and y = Ay and hence A is closed. Since for λ > ω λ I −A = (I + (λ −λ0)(λ0 I −A)−1)(λ0 I −A), if \|λ−λ0\| λ0−ω < 1, then (λ I −A)−1 ∈L(X). Thus by the continuation method we have (λ I −A)−1 exists and \|(λ I −A)−1\| ≤ 1 λ −ω, λ > ω. It follows from the Hile-Yosida theorem that (b) →(a). (b) →(a) Since for x∗∈F(x), the dual element of x, i.e. x∗∈X∗satisfying ⟨x, x∗⟩X×X∗= \|x\|2 and \|x∗\| = \|x\| ⟨e−ωtS(t)x, x∗⟩≤\|x\|\|x∗\| = ⟨x, x∗⟩ we have for all x ∈dom(A) 0 ≥lim t→0+⟨e−ωtS(t)x −x t , x∗⟩= ⟨(A −ω I)x, x∗⟩for all x∗∈F(x). which implies A −ω I is dissipative. □ Theorem (Dissipative I) (1) A is a ω-dissipative \|λ x −Ax\| ≥(λ −ω)\|x\| for all x ∈dom (A). if and only if (2) for all x ∈dom(A) there exists x∗∈F(x) such that ⟨Ax, x∗⟩≤ω \|x\|2. (4.29) (2) →(1). Let x ∈dom(A) and choose x∗∈F(0) such that ⟨A, x∗⟩≤0. Then, for any λ > 0, \|x\|2 = ⟨x, x∗⟩= ⟨λ x −Ax + Ax, x∗⟩≤⟨λ x −Ax, x∗⟩−ω \|x\|2 ≤\|λ x −Ax\|\|x\| −ω \|x\|2, which implies (2). (1) →(2). From (1) we obtain the estimate −1 λ(\|x −λ Ax\| −\|x\|) ≤0 and ⟨Ax, x⟩−= lim λ→0+ = 1 λ(\|x\| −\|x −λ Ax\|) ≤0 which implies there exists x∗∈F(x) such that (4.29) holds since ⟨Ax, x⟩−= ⟨AX, x∗⟩ for some x∗∈F(x). □ Thus, Lumer-Phillips theorem says that if m-diisipative, then (4.29) hold for all x∗∈F(x). 82 Theorem (Dissipative II) Let A be a closed densely deﬁned operator on X. If A and A∗are dissipative, then A is m-dissipative and thus the inﬁnitesimal generator of a C −0-semigroup of contractions. Proof: Let y ∈R(I −A) be given. Then there exists a sequence xn ∈dom(A) such that y=xn −Axn →y as n →∞. By the dissipativity of A we obtain \|xn −xm\| ≤\|xn −xm −A(xn −xm)\| ≤\|y−ym\| Hence xn is a Cauchy sequence in X. We set x = limn→∞xn. Since A is closed, we see that x ∈dom (A) and x −Ax = y, i.e., y ∈R(I −A). Thus R(I −A) is closed. Assume that R(I −A) ̸= X. Then there exists an x∗∈X∗such that ⟨(I −A)x, x∗) = 0 for all x ∈dom (A). By deﬁnition of the dual operator this implies x∗∈dom (A∗) and (I −A)∗x∗= 0. Dissipativity of A implies \|x∗\| < \|x∗−A∗x∗\| = 0, which is a contradiction. □ Example (revisited example 1) Aφ = −φ′ in X = L2(0, 1) and for φ ∈H1(0, 1) (Aφ, φ)X = − Z 1 0 φ′(x)φ dx == 1 2(\|φ(0)\|2 −\|φ(1)\|2) Thus, A is dissipative if and only if φ(0) = 0, the in ﬂow condition. Deﬁne the domain of A by dom(A) = {φ ∈H1(0, 1) : φ(0) = 0} The resolvent equation is equivalent to λ u + d dxu = f and u(x) = Z x 0 e−λ (x−s)f(s) ds, and R(λ I −A) = X. By the Lumer-Philips theorem A generates the C0 semigroup on X = L2(0, 1). Example (Conduction equation) Consider the heat conduction equation: d dtu = Au = X i,j aij(x) ∂2u ∂xi∂xj + b(x) · ∇u + c(x)u, in Ω. Let X = C(Ω) and dom(A) ⊂C2(Ω). Assume that a ∈Rn×n ∈C(Omega) b ∈Rn,1 and c ∈R are continuous on ¯ Ωand a is symmetric and m I ≤a(x) ≤M I for 0 < m ≤M < ∞. 83 Then, if x0 is an interior point of Ωat which the maximum of φ ∈C2(Ω) is attained. Then, ∇φ(x0) = 0, X ij aij(x0) ∂2u ∂xi∂xj (x0) ≤0. and thus (λ φ −Aφ)(x0) ≤ω φ(x0) where ω ≤max x∈Ωc(x). Similarly, if x0 is an interior point of Ωat which the minimum of φ ∈C2(Ω) is attained, then (λ φ −Aφ)(x0) ≥0 If x0 ∈∂Ωattains the maximum, then ∂ ∂ν φ(x0) ≤0. Consider the domain with the Robin boundary condition: dom(A) = {u ∈α(x) u(x) + β(x) ∂ ∂ν u = 0 at ∂Ω} with α, β ≥0 and infx∈∂Ω(α(x) + β(x)) > 0. Then, \|λφ −Aφ\|X ≥(λ −ω)\|φ\|X. (4.30) for all φ ∈C2(Ω). It follows from the he Lax Milgram theory that (λ0 I −A)−1 ∈L(L2(Ω), H2(Ω), assuming that coeﬃcients (a, b, c) are suﬃciently smooth. Let dom(A) = {(λ0 I −A)−1C(Ω)}. Since C2(Ω) is dense in dom(A), (??) holds for all φ ∈dom(A), which shows A is dissipative. Example (Advection equation) Consider the advection equation ut + ∇· (⃗ b(x)u) = ν ∆u. Let X = L1(Ω). Assume ⃗ b ∈L∞(Ω) Let ρ ∈C1(R) be a monotonically increasing function satisfying ρ(0) = 0 and ρ(x) = sign(x), \|x\| ≥1 and ρϵ(s) = ρ( s ϵ) for ϵ > 0. For u ∈C1(Ω) (Au, u) = Z Γ (ν ∂ ∂nu −n ·⃗ b u, ρϵ(u)) ds + (⃗ b u −ν ∇u, 1 ϵ ρ′ ϵ(u) ∇u) + (c u, ρϵ(u)). where (⃗ b u, 1 ϵ ρ′ ϵ(u) ∇u) ≤ν (∇u, 1 ϵ ρ′ ϵ(u) ∇u) + ϵ 4ν meas({\|u\| ≤ϵ}). 84 Assume the inﬂow condition u on {s ∈∂Ω: n · b < 0}. Note that for u ∈L1(Rd) (u, ρϵ(u)) →\|u\|1 and (ψ, ρϵ(u)) →(ψ, sign0(u)) for ψ ∈L1(Ω) as ϵ →0. It thus follows (λ −ω) \|u\| ≤\|λ u −λ Au\|. (4.31) Since C2(Ω) is dense in W 2,1(Rd) (4.31) holds for u ∈W 2,1(Ω). Example (X = Lp(Ω)) Let Au = ν ∆u + b · ∇u with homogeneous boundary condition u = 0 on X = Lp(Ω). Since ⟨∆u, u∗⟩= Z Ω (∆u, \|u\|p−2u) = −(p −1) Z Ω (∇u, \|u\|p−2∇u) and (b · ∇u, \|u\|p−2u)L2 ≤(p −1)ν 2 \|(∇u, \|u\|p−2∇u)L2 + \|b\|2 ∞ 2ν(p −1) (\|u\|p, 1)L2 we have ⟨Au, u∗⟩≤ω\|u\|2 for some ω > 0. Example (Fractional PDEs I) In this section we consider the nonlocal diﬀusion equation of the form ut = Au = Z Rd J(z)(u(x + z) −u(x)) dz. Or, equivalently Au = Z (Rd)+ J(z)(u(x + z) −2u(x) + u(x −z)) dz for the symmetric kernel J in Rd. It will be shown that (Au, φ)L2 = Z Rd Z (Rd)+ J(z)(u(x + z) −u(x))(φ(x + z) −φ(x)) dz dx and thus A has a maximum extension. Also, the nonlocal Fourier law is given by Au = ∇· ( Z Rd J(z)∇u(x + z) dz). Thus, (Au, φ)L2 = Z Rd×Rd J(z)∇u(x + z) · ∇φ(x) dz dx Under the kernel J is completely monotone, one can prove that A has a maximal monotone extension. 85 4.4 Jump diﬀusion Model for American option In this section we discuss the American option for the jump diﬀusion model ut + (x −σ2 2 )ux + σ2x2 2 uxx + Bu + λ = 0, u(T, x) = ψ, (λ, u −ψ) = 0, λ ≤0, u ≥ψ where the generator B for the jump process is given by Bu = Z ∞ −∞ k(s)(u(x + s) −u(x) + (es −1)ux) ds. The CMGY model for the jump kernel k is given by k(s) =    Ce−M\|s\|\|s\|1+Y = k+(s) s ≥0 Ce−G\|s\|\|s\|1+Y = k−(s) s ≤0 Since Z ∞ −∞ k(s)(u(x + s) −u(x)) ds = Z ∞ 0 k+(s)(u(x + s) −u(x)) ds + Z ∞ 0 k−(s)(u(x −s) −u(x)) ds = Z ∞ 0 k+(s) + k−(s) 2 (u(x + s) −2u(x) + u(x −s)) ds + Z ∞ 0 k+(s) −k−(s) 2 (u(x + s) −u(x −s)) ds. Thus, Z ∞ −∞ ( Z ∞ −∞ k(s)(s)(u(x + s) −u(x) ds)φ dx = Z ∞ −∞ Z ∞ 0 ks(s)(u(x + s) −u(x))(φ(x + s) −φ(s)) ds dx + Z ∞ −∞ ( Z ∞ 0 ku(s)(u(x + s) −u(s)))φ(x) dx where ks(s) = k+(s) + k−(s) 2 , ku(s) = k+(s) −k−(s) 2 and hence (Bu, φ) = Z ∞ −∞ Z ∞ −∞ ks(s)(u(x + s) −u(x))(φ(x + s) −φ(s)) ds dx + Z ∞ −∞ ( Z ∞ −∞ ku(s)(u(x + s) −u(s)))φ(x) dx + ω Z ∞ −∞ uxφ dx. where ω = Z ∞ −∞ (es −1)k(s) ds. If we equip V = H1(R) by \|u\|2 V = Z ∞ −∞ Z ∞ −∞ ks(s)\|u(x + s) −u(x)\|2 ds dx + σ2 2 Z ∞ −∞ \|ux\|2 dx, then A + B ∈L(V, V ∗) and A + B generates the analytic semigroup on X = L2(R). 86 4.5 Numerical approximation of nonlocal operator In this section we describe our higher order integration method for the convolution; Z ∞ 0 k+(s) + k−(s) 2 (u(x+s)−2u(x)+u(x−s)) ds+ Z ∞ 0 k+(s) −k−(s) 2 (u(x+s)−u(x−s)) ds. For the symmetric part, Z ∞ −∞ s2ks(s) u(x + s) −2u(x) + u(x −s) s2 ds, where we have u(x + s) −2u(x) + u(x −s) s2 ∼uxx(x) + s2 12uxxxx(x) + O(s4) We apply the fourth order approximation of uxx by uxx(x) ∼u(x + h) −2u(x) + u(x −h) h2 −1 12 u(x + 2h) −4u(x) + 6u(x) −4u(x −h) + u(x −2h) h2 and we apply the second order approximation of uxxxx(x) by uxxxx(x) ∼u(x + 2h) −4u(x) + 6u(x) −4u(x −h) + u(x −2h) h4 . Thus, one can approximate Z h 2 −h 2 s2ks(s) u(x + s) −2u(x) + u(x −s) s2 ds by ρ0 (uk+1 −2uk + uk−1 h2 −1 12 uk+2 −4uk+1 + 6uk −4uk−1 + uk−2 h2 ) +ρ1 12 uk+2 −4uk+1 + 6uk −4uk−1 + uk−2 h2 , where ρ0 = Z h 2 −h 2 s2ks(s) ds and ρ1 = 1 h2 Z h 2 −h 2 s4ks(s) ds. The remaining part of the convolution Z (k+ 1 2 )h (k−1 2 )h u(xk+j + s)ks(s) ds can be approximated by three point quadrature rule based on u(xk+j + s) ∼u(xk+j) + u′(xk+j)s + s2 2 u′′(xk+j) with u′(xk+j) ∼uk+j+1 −uk+j−1 2h u′′(xk+j) ∼uk+j+1 −2uk+j + uk+j−1 h2 . 87 That is, Z (k+ 1 2 )h (k−1 2 )h u(xk+j + s)ks(s) ds ∼ρk 0uk+j + ρk 1 uk+j−1 −uk+j+1 2 + ρk 2 uj+k+1 −2uk+j + uj+k−1 2 where ρk 0 = R (k+ 1 2 )h (k−1 2 )h ks(s) ds ρk 1 = 1 h R (k+ 1 2 )h (k−1 2 )h (s −xk)ks(s) ds ρk 2 = 1 h2 R (k+ 1 2 )h (k−1 2 )h (s −xk)2ks(s) ds. For the skew-symmetric integral Z h 2 −h 2 ku(s)(u(x + s) −u(x −s)) ds ∼ρ2 ux(x) + ρ3 6 h2 uxxx(x) where ρ2 = Z h 2 −h 2 2sku(s) ds, ρ3 = 1 h2 Z h 2 −h 2 2s3ku(s) ds. We may use the forth order diﬀerence approximation ux(x) ∼u(x + h) −u(x −h) 2h −u(x + 2h) −2u(x + h) + 2u(x −h) −u(x −2h) 6h and the second order diﬀerence approximation uxxx(x) ∼u(x + 2h) −2u(x + h) + 2u(x −h) −u(x −2h) h3 and obtain Z h 2 −h 2 ku(s)(u(x + s) −u(x −s)) ds ∼ρ2 (uk+1 −uk−1 2h −uk+2 −2uk+1 + 2uk−1 −uk−1 6h ) + ρ3 6 uk+2 −2uk+1 + 2uk−1 −uk−1 h . Consider the nonlocal PDE equations of the form Example (Fractional PDEs II) Consider the fractional equation of the form Z 0 −t g(θ)u′(t + θ) dθ = Au, u(0) = u0, 88 where the kernel g satisﬁes g > 0, g ∈L1(−∞, 0) and non-decreasing. For example, the case of the Caputo (fractional) derivative has g(θ) = 1 \|Gamma(1 −α)\|θ\|−α. Deﬁne z(t, θ) = u(t + θ), θ ∈(−∞, 0]. Then, d dtz = ∂ ∂θz. Thus, we deﬁne the linear operator A on Z = C((−∞, 0]; X) by Az = z′(θ) with dom(A) = {z′ ∈X : Z 0 −∞ g(θ)z′(θ) dθ = Az(0)} Theorem 4.1 Assume A is m-dissipative in a Banach space X. Then, A is dissipative and R(λ I −A) = Z for λ > 0. Thus, A generates the C0-semigroup T(t) on Z = C((−∞, 0]; X). Proof: First we show that A is dissipative. For φ ∈dom (A) suppose \|φ(0)\| > \|φ(θ)\| for all θ < 0. Deﬁne gϵ(θ) = 1 ϵ Z θ θ−ϵ g(θ) dθ. For all x∗∈F(φ(0)) ⟨x∗, Z 0 −∞ gϵ(θ)(φ′)dθ⟩ = ⟨x∗, Z 0 −∞ g(θ) −g(θ −ϵ) ϵ ⟨φ(θ) −φ(0)), ⟩dθ ≤0 since ⟨x∗, φ(θ) −φ(0)⟩≤(\|φ(θ)\| −\|φ(0)\|)\|φ(0)\| < 0, θ < 0. Thus, lim ϵ→0+⟨x∗, Z 0 −∞ gϵ(θ)(φ′)dθ⟩= ⟨ Z 0 −∞ g(θ)φ′dθ, x∗⟩< 0. (4.32) But, since there exists a x∗∈F(φ(0)) such that ⟨x∗, Ax⟩≥0 which contradicts to (4.32). Thus, there exists θ0 such that \|φ(θ0)\| = \|φ\|Z. Since ⟨φ(θ), x∗⟩≤\|φ(θ)\| for x∗∈F(φ(θ0)), θ →⟨φ(θ), x∗⟩attains the maximum at θ0 and thus ⟨x∗φ′(θ0)⟩= 0 Hence, \|λ φ −φ′\|Z ≥⟨x∗, λ φ(θ0) −φ′(θ0), ⟩= λ \|φ(θ0)\| = λ \|φ\|Z. (4.33) For the range condition λφ −Aφ = f we note that φ(θ) = eλθφ(0) + ψ(θ) 89 where ψ(θ) = Z 0 θ eλ(θ−ξ)f(ξ) dξ. Thus, (∆(λ) I −A)φ(0) = Z 0 −∞ g(θ)ψ′(θ) dθ) where ∆(λ) = λ Z 0 −∞ g(θ)eλθ dθ > 0 Thus, φ(0) = (∆(λ) I −A)−1 Z 0 −∞ g′(θ)ψ(θ) dθ. Since A is dissipative and λψ −ψ′ = f ∈Z, ψ(0) = 0, thus \|ψ\|Z ≤1 λ\|f\|Z. Thus φ = (λ I −A)−1f ∈Z. □ Example (Renewable system) We discuss the renewable system of the form dp0 dt = −P i λi p0(t) + P i R L 0 µi(x)pi(x, t) dx (pi)t + (pi)x = −µi(x)p, p(0, t) = λi p0(t) for (p0, pi, 1 ≤i ≤d) ∈R × L1(0, T)d. Here, p0(t) ≥0 is the total utility and λi ≥0 is the rate for the energy conversion to the i-th process pi. The ﬁrst equation is the energy balance law and s is the source = generation −consumption. The second equation is for the transport (via pipeline and storage) for the process pi and µi ≥0 is the renewal rate and ¯ µ ≥0 is the natural loos rate. {λi ≥0} represent the distribution of the utility to the i-th process. Assume at the time t = 0 we have the available utility p0(0) = 1 and pi = 0. Then we have the following conservation p0(t) + Z t 0 pi(s) ds = 1 if t ≤L. Let X = R × L1(0, L)d. Let A(µ) deﬁned by Ax = (− X i λi p0 + X i Z L 0 µi(x) dx, −(pi)x −µi(x)pi) with domain dom(A) = {(p0, pi) ∈R × W 11(0, L)d : pi(0) = λi p0} Let signϵ(s) =    s \|s\| \|s\| > ϵ s ϵ \|s\| ≤ϵ 90 Then, (A(p0, p), (sign0(p0), signϵ(p)) ≤−(P i λi)\|p0\| + \| R L 0 µipi dx\| P i(Ψϵ(pi(0)) −Ψϵ(pi(L)) − R L 0 µipisignϵ dx) where Ψϵ(s) =    \|s\| \|s\| > ϵ s2 2ϵ + ϵ 2 \|s\| ≤ϵ Since signϵ →signϵ, Ψϵ →\|s\| by the Lebesgue dominated convergence theorem, we have (A(p0, p), (sign0(p0), sign0(p)) ≤0. The resolvent equation A(p0, p) = (s, f), (4.34) has a solution pi(x) = λip0e− R x 0 µi + R x 0 e− R x s µif(s) ds (P i λi)(1 −e R L 0 µi) p0 = s + R L 0 µipi(x) dx Thus, A generates the contractive C0 semigroup S(t) on X. Moreover, it is cone preserving S(t)C+ ⊂C+ since the resolvent is positive cone preserving. Example (Bi-domain equation) Example (Second order equation) Let V ⊂H = H∗⊂V ∗be the Gelfand triple. Let ρ be a bounded bilinear form on H × H, µ and σ be bounded bilinear forms on V × V . Assume ρ and σ are symmetric and coercive and µ(φ, φ) ≥0 for all φ ∈V . Consider the second order equation ρ(utt, φ) + µ(ut, φ) + σ(u, φ) = ⟨f(t), φ⟩for all φ ∈V. (4.35) Deﬁne linear operators M (mass), D (dampping), K and (stiﬀness) by (Mφ, ψ)H = ρ(φ, ψ), φ, ψ ∈H⟨Dφ, ψ⟩= µ(φ, ψ) φ, ψ ∈V ⟨Kφ, ψ⟩V ∗×V , φ, ψ ∈V Let v = ut and deﬁne A on X = V × H by A(u, v) = (v, −M−1(Ku + Dv)) with domain dom (A) = {(u, v) ∈X : v ∈V and Ku + Dv ∈H} The state space X is a Hilbert space with inner product ((u1, v1), (u,v2)) = σ(u1, u2) + ρ(v1, v2) and E(t) = \|(u(t), v(t))\|2 X = σ(u(t), u(t)) + ρ(v(t), v(t)) 91 deﬁnes the energy of the state x(t) = (u(t), v(t)). First, we show that A is dissipative: (A(u, v), (u, v))X = σ(u, v)+ρ(−M−1(Ku+Dv), v) = σ(u, v)−σ(u, v)−µ(v, v) = −µ(v, v) ≤0 Next, we show that R(λ I −A) = X. That is, for (f, g) ∈X the exists a solution (u, v) ∈dom (A) satisfying λ u −v = f, λMv + Dv + Ku = Mg, or equivalently v = λ u −f and λ2Mu + λDu + Ku = Mg + λ Mf + Df (4.36) Deﬁne the bilinear form a on V × V a(φ, ψ) = λ2 ρ(φ, ψ) + λ µ(φ, ψ) + σ(φ, ψ) Then, a is bounded and coercive and if we let F(φ) = (M(g + λ f)φ)H + µ(f, φ) then F ∈V ∗. It thus follows from the Lax-Milgram theory there exists a unique solution u ∈V to (4.36) and Dv + Ku ∈H. For example, consider the wave equation 1 c2(x)utt + κ(x)ut = ∆u [ ∂u ∂n] + αu = γ ut at Γ In this example we let V = H1(Ω)/R and H = L2(Ω) and deﬁne σ(φ, ψ) = Z Ω (∇φ, ∇ψ) dx + Z Γ αφψ ds µ(φ, ψ) = Z ∂Ω κ(x)φ, ψ dx + Z Γ γ)φ, ψ ds ρ(φ, ψ) = Z Ω 1 c2(x) φψ dx Example (Maxwell system for elecro-magnetic equations) ϵ Et = ∇× H, ∇E = ρ µ Ht = −∇× E, ∇· B = 0 with boundary condition E × n = 0 where E is Electric ﬁeld, Bµ H is Magnetic ﬁeld and D = ϵ E is dielectric with ϵ, µ is electric and magnetic permittivity, respectively Let X = L2(Ω)d × L2(Ω)d with the norm deﬁned by Z Ω (ϵ \|E\|2 + µ \|H\|2) dx. 92 The dissipativity follows from Z Ω (E · (∇× H) −H · (∇× E)) dx = Z Ω ∇· (E × H) dx = Z ∂Ω n · (E × H) ds = 0 Let ρ = 0 and thus ∇· E = 0. The range condition is equivalent to ϵ E + ∇× 1 µ(∇× E −g) = f The weak form is given by (ϵ E, ψ) + ( 1 µ∇× E, ∇× ψ) = (t, ψ) + (g, 1 µ∇× ψ). (4.37) for ψ ∈V = {H1(Ω) : ∇cot ψ = 0, n × ψ = 0 at ∂Ω} Since \|∇× ψ\|2 = \|∇ψ\|2 for ∇· ψ = 0. the right hand side of (4.37) deﬁnes the bounded coercive quadratic form on V × V , it follows from the Lax-Milgram equation that (??) has a unique solution in V . 4.6 Dual semigroup Theorem (Dual semigroup) Let X be a reﬂexive Banach space. The adjoint S∗(t) of the C0 semigroup S(t) on X forms the C0 semigroup and the inﬁnitesimal generator of S∗(t) is A∗. Let X be a Hilbert space and dom(A∗) be the Hilbert space with graph norm and X−1 be the strong dual space of dom(A∗), then the extension S(t) to X−1 deﬁnes the C0 semigroup on X−1. Proof: (1) Since for t, s ≥0 S∗(t + s) = (S(s)S(t))∗= S∗(t)S∗(s) and ⟨x, S∗(t)y −y⟩X×X∗= ⟨S(t)x −x, y⟩X×X∗→0. for x ∈X and y ∈X∗Thus, S∗(t) is weakly star continuous at t = 0 and let B is the generator of S∗(t) as Bx = w∗−lim S∗(t)x −x t . Since (S(t)x −x t , y) = (x, S∗(t)y −y t ), for all x ∈dom(A) and y ∈dom(B) we have ⟨Ax, y⟩X×X∗= ⟨x, By⟩X×X∗ and thus B = A∗. Thus, A∗is the generator of a w∗−continuous semigroup on X∗. (2) Since S∗(t)y −y = A∗ Z t 0 S∗(s)y ds for all y ∈Y = dom (A∗). Thus, S∗(t) is strongly continuous at t = 0 on Y . 93 (3) If X is reﬂexive, dom (A∗) = X∗. If not, there exists a nonzero y0 ∈X such that ⟨y0, x∗⟩X×X∗= 0 for all x∗∈dom(A∗). Thus, for x0 = (λ I −A)−1y0 ⟨λ x0−Ax0, x∗⟩= ⟨x0, λ x∗−A∗x∗⟩= 0. Letting x∗= (λ I −A∗)−1x∗ 0 for x∗ 0 ∈F(x0), we have x0 = 0 and thus y0 = 0, which yields a contradiction. (4) X1 = dom (A∗) is a closed subspace of X∗and is a invariant set of S∗(t). Since A∗ is closed, S∗(t) is the C0 semigroup on X1 equipped with its graph norm. Thus, (S∗(t))∗is the C0 semigroup on X−1 = X∗ 1 and deﬁnes the extension of S(t) to X−1. Since for x ∈X ⊂X−1 and x∗∈X∗ ⟨S(t)x, x∗⟩= ⟨x, S∗(t)x∗⟩, S(t) is the restriction of (S∗(t))∗onto X. □ 4.7 Stability Theorem (Datko 1970, Pazy 1972). A strongly continuous semigroup S(t), t ≥0 on a Banach space X is uniformly exponentially stable if and only if for p ∈[1, ∞) one has Z ∞ 0 \|S(t)x\|p dt < ∞for all x ∈X. Theorem. (Gearhart 1978, Pruss 1984, Greiner 1985) A strongly continuous semigroup on S(t), t ≥0 on a Hilbert space X is uniformly exponentially stable if and only if the half-plane {λ ∈C : Reλ > 0} is contained in the resolvent set ρ(A) of the generator A with the resolvent satisfying \|(λ I −A)−1\|∞< ∞ 4.8 Sectorial operator and Analytic semigroup In this section we have the representation of the semigroup S(t) in terms of the inverse Laplace transform. Taking the Laplace transform of d dtx(t) = Ax(t) + f(t) we have ˆ x = (λ I −A)−1(x(0) + ˆ f) where for λ > ω ˆ x = Z ∞ 0 e−λtx(t) dt is the Laplace transform of x(t). We have the following the representation theory (inverse formula). Theorem (Resolvent Calculus) For x ∈dom(A2) and γ > ω S(t)x = 1 2πi Z γ+i∞ γ−i∞ eλt(λ I −A)−1x dλ. (4.38) 94 Proof: Let Aµ be the Yosida approximation of A. Since Re σ(Aµ) ≤ ω0 1 −µω0 < γ, we have uµ(t) = Sµ(t)x = 1 2πi Z γ+i∞ γ−i∞ eλt(λ I −Aµ)−1x dλ. Note that λ(λ I −A)−1 = I + (λ I −A)−1A. (4.39) Since 1 2πi Z γ+i∞ γ−i∞ eλt λ dλ = 1 and Z γ+i∞ γ−i∞ \|(\|λ −ω\|−2\| dλ < ∞, we have \|Sµ(t)x\| ≤M \|A2x\|, uniformly in µ > 0. Since (λ I −Aµ)−1x −(λ I −A)−1x = µ 1 + λµ(ν I −A)−1(λ I −A)−1A2x, where ν = λ 1 + λµ, {uµ(t)} is Cauchy in C(0, T; X) if x ∈dom(A2). Letting µ →0+, we obtain (4.38). □ Next we consider the sectorial operator. For δ > 0 let Σδ ω = {λ ∈C : arg(λ −ω) < π 2 + δ} be the sector in the complex plane C. A closed, densely deﬁned, linear operator A on a Banach space X is a sectorial operator if \|(λ I −A)−1\| ≤ M \|λ −ω\| for all λ ∈Σδ ω. For 0 < θ < δ let Γ = Γω,θ be the integration path deﬁned by Γ± = {z ∈C : \|z\| ≥δ, arg(z −ω) = ±( π 2 + θ)}, Γ0 = {z ∈C : \|z\| = δ, \|arg(z −ω)\| ≤π 2 + θ}. For 0 < θ < δ deﬁne a family {S(t), t ≥0} of bounded linear operators on X by S(t)x = 1 2πi Z Γ eλt(λ I −A)−1x dλ. (4.40) Theorem (Analytic semigroup) If A is a sectorial operator on a Banach space X, then A generates an analytic (C0) semigroup on X, i.e., for x ∈X t →S(t)x is an analytic function on (0, ∞). We have the representation (4.40) for x ∈X and \|AS(t)x\|X ≤Mθ t \|x\|X 95 Proof: The elliptic operator A deﬁned by the Lax-Milgram theorem deﬁnes a sectorial operator on Hilbert space. Theorem (Sectorial operator) Let V, H are Hilbert spaces and assume H ⊂V ∗. Let ρ(u, v) is bounded bilinear form on H × H and ρ(u, u) ≥\|u\|2 H for all u ∈H Let a(u, v) to be a bounded bilinear form on V × V with σ(u, u) ≥δ \|u\|2 V for all u ∈V. Deﬁne the linear operator A by ρ(Au, φ) = a(u, φ) for all φ ∈V. Then, for Re λ > 0we have \|(λ I −A)−1\|L(V,V ∗) ≤1 δ \|(λ I −A)−1\|L(H) ≤M \|λ\| \|(λ I −A)−1\|L(V ∗,H) ≤ M √ \|λ\| \|(λ I −A)−1\|L(H,V ) ≤ M √ \|λ\| Proof: Let a(u, v) to be a bounded bilinear form on V × V . For f ∈V ∗and ℜλ > 0, Deﬁne MH, H by (Mu, v) = ρ(u, v) for all v ∈H and A0 ∈L(V, V ∗) by ⟨A0u, v⟩= σ(u, v) for v ∈V Then, A = M−1A0 and (λ I −A)u = M−1f is equivalent to λρ(u, φ) + a(u, φ) = ⟨f, φ⟩, for all φ ∈V. (4.41) It follows from the Lax-Milgram theorem that (4.41) has a unique solution, given f ∈V ∗and Re λρ(u, u) + a(u, u) ≤\|f\|V ∗\|u\|V . Thus, \|(λ I −A)−1\|L(V ∗,V ) ≤1 δ . Also, \|λ\| \|u\|2 H ≤\|f\|V ∗\|u\|V + M \|u\|2 V = M1 \|f\|2 V ∗ for M1 = 1 + M δ2 and thus \|(λ I −A)−1\|L(V ∗,H) ≤ √M1 \|λ\|1/2 . 96 For f ∈H ⊂V ∗ δ \|u\|2 V ≤Re λ ρ(u, u) + a(u, u) ≤\|f\|H\|u\|H, (4.42) and \|λ\|ρ(u, u) ≤\|f\|H\|u\|H + M\|u\|2 V ≤M1\|f\|H\|u\|H Thus, \|(λ I −A)−1\|L(H) ≤M1 \|λ\| . Also, from (4.42) δ \|u\|2 V ≤\|f\|H\|u\|H ≤M1 \|f\|2. which implies \|(λ I −A)−1\|L(H,V ) ≤ M2 \|λ\|1/2 . 4.9 Approximation Theory In this section we discuss the approximation theory for the linear C0-semigroup. Equiv-alence Theorem (Lax-Richtmyer) states that for consistent numerical approximations, stability and convergence are equivalent. In terms of the linear semigroup theory we have Theorem (Trotter-Kato theorem) Let X and Xn be Banach spaces and A and An be the inﬁnitesimal generator of C0 semigroups S(t) on X and Sn(t) on Xn of G(M, ω) class. Assume a family of uniformly bounded linear operators Pn ∈L(X, Xn) and En ∈L(Xn, X) satisfy PnEn = I \|EnPnx −x\|X →0 for all x ∈X (4.43) Then, the followings are equivalent. (1) there exist a λ0 > ω such that for all x ∈X \|En(λ0 I −An)−1Pnx −(λ0 I −A)−1x\|X →0 as n →∞, (4.44) (2) For every x ∈X and T ≥0 \|EnSn(t)Pnx −S(t)x\|X → as n →∞. uniformly on t ∈[0, T]. Proof: Since for λ > ω En(λ I −A)−1Pnx −(λ I −A)−1x = Z ∞ 0 EnSn(t)Pnx −S(t)x dt (1) follows from (2). Conversely, from the representation theory EnSn(t)Pnx −S(t)x = 1 2πi Z γ+i∞ γ−i∞ (En(λ I −A)−1Pnx −(λ I −A)−1x) dλ where (λ I −A)−1 −(λ0 I −A)−1 = (λ −λ0)(λ I −A)−1(λ0 I −A)−1. 97 Thus, from the proof of Theorem (Resolvent Calculus) (1) holds for x ∈dom(A2). But since dom(A2) is dense in X, (2) implies (1). □ Remark (Stability) If An is uniformly dissipative: \|λun −Anun\| ≥(λ −ω) \|un\| for all un ∈dom(An) and some ω ≥0, then An generates ω contractive semigroup Sn(t) on Xn. Remark (Consistency) (λI −An)Pnun = Pnf Pn(λI −A)u = Pnf we have (λ I −An)Pn(u −un) + PnAu −AnPnu = 0 Thus \|Pn(u −un)\| ≤M \|PnAu −AnPnu\| The consistency(4.44) foll0ws from \|PnAu −AnPnu\| →0 for all u in a dense subset of dom(A). Corollary Let the assumptions of Theorem hold. The statement (1) of Theorem is equivalent to (4.43) and the followings: (C.1) there exists a subset D of doma(A) such that D = X and (λ0 I −A)D = X. (C.2) for all u ∈D there exists a sequence ¯ un ∈dom(An) such that lim En¯ un = u and lim EnAn¯ un = Au. Proof: Without loss of generality we can assume λ0 = 0. First we assume that condition (1) hold. We set D = dom(A) and thus AD = X. For u ∈dom(A) we set ¯ un = A−1 n PnAu and En¯ un −u = EnA−1 n Pnx −A−1x →0 and EnAn¯ un −Au = EnAnA−1 n Pnx −AA−1x = EnPnx −x →0 as n →∞. Hence condition (C.2) hold. Conversely, we assume conditions (C.1)–(C.2) hold. EnA−1 n Pn −A−1 = En(A−1 n PnA −Pn)A−1 + (EnPn −I)A−1. For x ∈AD we choose u ∈D such that x = Au and set un = A−1 n Pnx = A−1 n PnAu. We then for u we choose ¯ un according to (C.2). Thus, we obtain \|¯ un −Pnu\| = \|Pn(En¯ un −u)\| ≤M \|En¯ un −u\| →0 as n →∞and \|¯ un −un\| ≤\|A−1 n \|\|An¯ un + PnAu\| ≤\|A−1 n \|\|Pn\|EnAn¯ un + Au\| →0 98 as n∞. It thus follows that \|EnA−1 n Pnx−A−1x\| ≤\|En(un−Pnu)\|+\|EnPnu−u\| ≤M (\|un−Pnu\|+\|EnPnu−u\| →0 as n →∞for all x ∈AD. □ Example 1 (Trotter-Kato theoarem) Consider the heat equation on Ω= (0, 1) × (0, 1): d dtu(t) = ∆u, u(0, x) = u0(x) with boundary condition u = 0 at the boundary ∂Ω. We use the central diﬀerence approximation on uniform grid points: (i h, j h) ∈Ωwith mesh size h = 1 n: d dtui,j(t) = ∆hu = 1 h(ui+1,j −ui,j h −ui,j −ui−1,j h ) + 1 h(ui,j+1 −ui,j h −ui,j −ui,j−1 h ) for 1 ≤i, j ≤n1, where ui,0 = ui,n = u1, j = un,j = 0 at the boundary node. First, let X = C(Ω) and Xn = R(n−1)2 with sup norm. Let Enui,j = the piecewise linear interpolation and (Pnu)i,j = u(i h, j h) is the point-wise evaluation. We will prove that ∆h is dissipative on Xn. Suppose uij = \|un\|∞. Then, since λ ui,j −(∆hu)i,j = fij and −(∆hu)i,j = 1 h2 (4ui,j −ui+1,j −ui,j+1 −ui−1,j −ui,j−1) ≥0 we have 0 ≤ui,j ≤fi,j λ . Thus, ∆h is dissipative on Xn with sup norm. Next X = L2(Ω) and Xn with ℓ2 norm. Then, (−∆hun, un) = X i,j \|ui,j −ui−1,j h \|2 + \|ui,j −ui,j−1 h \|2 ≥0 and thus ∆h is dissipative on Xn with ℓ2 norm. Example 2 (Galerkin method) Let V ⊂H = H∗⊂V ∗is the Gelfand triple. Consider the parabolic equation ρ( d dtun, φ) = a(un, φ) (4.45) for all φ ∈V , where the ρ is a symmetric mass form ρ(φ, φ) ≥c \|φ\|2 H and a is a bounded coercive bilnear form on V × V such that a(φ, φ) ≥δ \|φ\|2 V . Deﬁne A by ρ(Au, φ) = a(u, φ) for all φ ∈V. By the Lax-Milgram theorem (λI −A)u = f ∈H 99 has a unique solution satisfying λρ(u, φ) −a(u, φ) = (f, φ)H for all φ ∈V . Let dom(A) = (I −A)−1H. Assume Vn = {u = X akφn k, φn k ∈V } is dense in V Consider the Galerkin method, i.e. un(t) ∈Vn satisﬁes ρ( d dtun(t), φ) = a(un, φ). Since for u = (λ I −A)u and ¯ un ∈Vn λρ(un, φ) + a(un, φ) = (f, φ) for φ ∈Vn λρ(¯ un, φ) + a(¯ un, φ) = λρ(¯ un −u, φ) + a(¯ un −u, φ) + (f, φ) for φ ∈Vn λρ(un −¯ un, φ) + a(un −¯ un, φ) = λρ(¯ un −u, φ) + a(¯ un −u, φ). Thus, \|un −¯ un\| ≤M δ \| ¯ un −u\|V . Example 2 (Discontinuous Galerkin method) Consider the parabolic system for u = ⃗ u ∈L2(Ω)d ∂ ∂tu = ∇· (a(x)∇u) + c(x)u where a ∈Rd × d is symmetric and a\|ξ\|2 ≤(ξ, a(x), ξ)Rd ≤a\|ξ\|2, ξ ∈Rd for 0 < a ≤a < ∞. The region Ωis dived into n non-overlapping sub-domains Ωi with boundaries ∂Ωi such that Ω= ∪Ωi. At the interface Γij = ∂Ωi ∩∂Ωj deﬁne = u\|∂Ωi −u\|∂Ωj << u >>= 1 2(u\|∂Ωi + u\|∂Ωj ). The approximate solution uh(t) in Vh = {uh ∈L2(Ω) : uh is linear on Ωi}. Deﬁne the bilinear for on Vh × Vh ah(u, v) = X i Z Ωi (a(x)∇u, ∇v) dx− X i>j Z Γij (<< n·(a∇u) >> ± << n·(a∇v) >> +β h ds), whee h is the meshsize and β > 0 is suﬃciently large. If + on the third term ah is symmetric and for the case −then ah enjoys the coercivity ah(u, u) ≥ X i Z Ωi (a(x)∇u, ∇u) dx, ∈u ∈Vh, 100 regardless of β > 0. Example 3 (Population dynaims) The transport equation ∂p ∂t + ∂p ∂x + m(x)p(x, t) = 0 p(0, t) = R β(x)p(x, t) dx Deﬁne the diﬀerence approximation Anp = (−pi −pi−1 h −m(xi)pi, 1 ≤i ≤n), p0 = X i βi pi Then, (Anp, sign0(p)) ≤( X mi −βi)\|pi\| ≤0 Thus, An on L1(0, 1) is dissipative. (A, Enφ) −(PnAn, φ). Example 4 (Yee’s scheme) Consider the two dimensional Maxwell’s equation. Consider the staggered grid; i.e. E = (E1 i−1 2 ,j, E2 i,j−1 2 ) is deﬁned at the the sides and H = Hi−1 2 ,j−1 2 is deﬁned at the center of the cell Ωi,j = ((i −1)h, ih) × ((j −1)h, jh). ϵi−1 2 ,j d dtE1 i−1 2 ,j = − Hi−1 2 ,j+ 1 2 −Hi−1 2 ,j−1 2 h ϵi,j+ 1 2 d dtE2 i,j+ 1 2 = Hi+ 1 2 ,j+ 1 2 −Hi−1 2 ,j−+ 1 2 h µi−1 2 ,j−1 2 d dtHi−1 2 ,j−1 2 = E2 i,j−1 2 −E2 i−1,j−1 2 h − E1 i−1 2 ,j−E1 i−1 2 ,j−1 h , (4.46) where E1 i−1 2 ,j = 0, j = 0, j = N and E2 i,j−1 2 ,j = 0, i = 0, j = N. Since N X i=1 N X j=1 − Hi−1 2 ,j+ 1 2 −Hi−1 2 ,j−1 2 h E1 i−1 2 ,j + Hi+ 1 2 ,j+ 1 2 −Hi−1 2 ,j−+ 1 2 h E 2 i,j+ 1 2 +( E2 i,j−1 2 −E2 i−1,j−1 2 h − E1 i−1 2 ,j −E1 i−1 2 ,j−1 h )Hi−1 2 ,j−1 2 = 0 (4.46) is uniformly dissipative. The range condition λ I−Ah = (f, g) ∈Xh is equivalent to the minimization for E min 1 2(ϵi−1 2 ,jE1 i−1 2 ,j + ϵi,j+ 1 2 E2 i,j+ 1 2 ) + 1 2 1 µi,j (\| E2 i,j−1 2 −E2 i−1,j−1 2 h \|2 + \| E1 i−1 2 ,j −E1 i−1 2 ,j−1 h \|2) −(f1 i−1 2 ,j − 1 µi,j gi−1 2 ,j+ 1 2 −gi−1 2 ,j−1 2 h , E1 i−1 2 ,j −(f2 i,j+ 1 2 + 1 µi,j Hi+ 1 2 ,j+ 1 2 −Hi−1 2 ,j−+ 1 2 h )E2 i,j+ 1 2 . Example 5 Legende-Tau method 101 5 Dissipative Operators and Semigroup of Non-linear Contractions In this section we consider du dt ∈Au(t), u(0) = u0 ∈X for the dissipative mapping A on a Banach space X. Deﬁnition (Dissipative) A mapping A on a Banach space X is dissipative if \|x1 −x2 −λ (y1 −y2)\| ≥\|x1 −x2\| for all λ > 0 and [x1, y1], [x2, y2] ∈A, or equivalently ⟨y1 −y2, x1 −x2⟩−≤0 for all [x1, y1], [x2, y2] ∈A. and if in addition R(I −λA) = X, then A is m-dissipative. In particular, it follows that if A is dissipative, then for λ > 0 the operator (I − λ A)−1 is a single-valued and nonexpansive on R(I −λ A), i.e., \|(I −λ A)−1x −(I −λ A)−1y\| ≤\|x −y\| for all x, y ∈R(I −λ A). Deﬁne the resolvent and Yosida approximation A by Jλx = (I −λ A)−1x ∈dom (A), x ∈dom (Jλ) = R(I −λ A) Aλ = λ−1(Jλx −x), x ∈dom (Jλ). (5.1) We summarize some fundamental properties of Jλ and Aλ in the following theorem. Theorem 1.4 Let A be an ω−dissipative subset of X × X, i.e., \|x1 −x2 −λ (y1 −y2)\| ≥(1 −λω) \|x1 −x2\| for all 0 < λ < ω−1 and [x1, y1], [x2, y2] ∈A and deﬁne ∥Ax∥by ∥Ax∥= inf{\|y\| : y ∈Ax}. Then for 0 < λ < ω−1, (i) \|Jλx −Jλy\| ≤(1 −λω)−1 \|x −y\| for x, y ∈dom (Jλ). (ii) Aλx ∈AJλx for x ∈R(I −λ A). (iii) For x ∈dom(Jλ) ∩dom (A) \|Aλx\| ≤(1 −λω)−1 ∥Ax∥and thus \|Jλx −x\| ≤ λ(1 −λω)−1 ∥Ax∥. (iv) If x ∈dom (Jλ), λ, µ > 0, then µ λ x + λ −µ λ Jλx ∈dom (Jµ) and Jλx = Jµ µ λ x + λ −µ λ Jλx . 102 (v) If x ∈dom (Jλ)∩dom (A) and 0 < µ ≤λ < ω−1, then (1−λω)\|Aλx\| ≤(1−µω) \|Aµy\| (vi) Aλ is ω−1(1 −λω)−1-dissipative and for x, y ∈dom(Jλ) \|Aλx −Aλy\| ≤λ−1(1 + (1 −λω)−1) \|x −y\|. Proof: (i) −−(ii) If x, y ∈dom (Jλ) and we set u = Jλx and v = Jλy, then there exist ˆ u and ˆ v such that x = u −λ ˆ u and y = v −λ ˆ v. Thus, from (1.8) \|Jλx −Jλy\| = \|u −v\| ≤(1 −λω)−1\|u −v −λ (ˆ u −ˆ v)\| = (1 −λω)−1 \|x −y\|. Next, by the deﬁnition Aλx = λ−1(u −x) = ˆ u ∈Au = AJλx. (iii) Let x ∈dom (Jλ)∩dom (A) and ˆ x ∈Ax be arbitrary. Then we have Jλ(x−λ ˆ x) = x since x −λ ˆ x ∈(I −λ A)x. Thus, \|Aλx\| = λ−1\|Jλx−x\| = λ−1\|Jλx−Jλ(x−λ ˆ x)\| ≤(1−λω)−1λ−1 \|x−(x−λ ˆ x)\| = (1−λω)−1 \|ˆ x\|. which implies (iii). (iv) If x ∈dom (Jλ) = R(I −λ A) then we have x = u −λ ˆ u for [u, ˆ u] ∈A and thus u = Jλx. For µ > 0 µ λ x + λ −µ λ Jλx = µ λ (u −λ ˆ u) + λ −µ λ u = u −µ ˆ u ∈R(I −µ A) = dom (Jµ). and Jµ µ λ x + λ −µ λ Jλx = Jµ(u −µ ˆ u) = u = Jλx. (v) From (i) and (iv) we have λ \|Aλx\| = \|Jλx −x\| ≤\|Jλx −Jλy\| + \|Jµx −x\| ≤ Jµ µ λ x + λ −µ λ Jλx −Jµx + \|Jµx −x\| ≤(1 −µω)−1 µ λ x + λ −µ λ Jλx −x + \|Jµx −x\| = (1 −µω)−1(λ −µ) \|Aλx\| + µ \|Aµx\|, which implies (v) by rearranging. (vi) It follows from (i) that for ρ > 0 \|x −y −ρ (Aλx −Aλy)\| = \|(1 + ρ λ) (x −y) −ρ λ (Jλx −Jλy)\| ≥(1 + ρ λ) \|x −y\| −ρ λ \|Jλx −Jλy\| ≥((1 + ρ λ) −ρ λ(1 −λω)−1) \|x −y\| = (1 −ρ ω(1 −λω)−1) \|x −y\|. The last assertion follows from the deﬁnition of Aλ and (i). □ Theorem 1.5 (1) A dissipative set A ⊂X × X is m-dissipative, if and only if R(I −λ0 A) = X for some λ0 > 0. 103 (2) An m-dissipative mapping is maximal dissipative, i.e., all dissipative set containing A in X × X coincide with A. (3) If X = X∗= H is a Hilbert space, then the notions of the maximal dissipative set and m-dissipative set are equivalent. Proof: (1) Suppose R(I −λ0 A) = X. Then it follows from Theorem 1.4 (i) that Jλ0 is contraction on X. We note that I −λ A = λ0 λ I −(1 −λ0 λ ) Jλ0 (I −λ0 A) (5.2) for 0 < λ < ω−1. For given x ∈X deﬁne the operator T : X →X by Ty = x + (1 −λ0 λ ) Jλ0y, y ∈X Then \|Ty −Tz\| ≤\|1 −λ0 λ \| \|y −z\| where \|1 −λ λ0 \| < 1 if 2λ > λ0. By Banach ﬁxed-point theorem the operator T has a unique ﬁxed point z ∈X, i.e., x = (I −(1 −λ0 λ Jλ0)z. Thus, x ∈(I −(1 −λ0 λ ) Jλ0)(I −λ0 A)dom (A). and it thus follows from (5.2) that R(I −λ A) = X if λ > λ0 2 . Hence, (1) follows from applying the above argument repeatedly. (2) Assume A is m-dissipative. Suppose ˜ A is a dissipative set containing A. We need to show that ˜ A ⊂A. Let [x, ˆ x] ∈˜ A Since x −λ ˆ x ∈X = R(I −λ A), for λ > 0, there exists a [y, ˆ y] ∈A such that x −λ ˆ x = y −λ ˆ y. Since A ⊂˜ A it follows that [y, ˆ y] ∈A and thus \|x −y\| ≤\|x −y −λ (ˆ x −ˆ y)\| = 0. Hence, [x, ˆ x] = [y, ˆ y] ∈A. (3) It suﬃces to show that if A is maximal dissipative, then A is m-dissipative. We use the following extension lemma, Lemma 1.6. Let y be any element of H. By Lemma 1.6, taking C = H, we have that there exists x ∈H such that (ξ −x, η −x + y) ≤0 for all [ξ, η] ∈A. and thus (ξ −x, η −(x −y)) ≤0 for all [ξ, η] ∈A. Since A is maximal dissipative, this implies that [x, x −y] ∈A, that is x −y ∈Ax, and therefore y ∈R(I −H). □ Lemma 1.6 Let A be dissipative and C be a closed, convex, non-empty subset of H such that dom (A) ∈C. Then for every y ∈H there exists x ∈C such that (ξ −x, η −x + y) ≤0 for all [ξ, η] ∈A. Proof: Without loss of generality we can assume that y = 0, for otherwise we deﬁne Ay = {[ξ, η + y] : [ξ, η] ∈A} with dom (Ay) = dom (A). Since A is dissipative if and only if Ay is dissipative , we can prove the lemma for Ay. For [ξ, η] ∈A, deﬁne the set C([ξ, η]) = {x ∈C : (ξ −x, η −x) ≤0}. Thus, the lemma is proved if we can show that T [ξ,η]∈A C([ξ, η]) is non-empty. 104 5.0.1 Properties of m-dissipative operators In this section we discuss some properties of m-dissipative sets. Lemma 1.7 Let X∗be a strictly convex Banach space. If A is maximal dissipative, then Ax is a closed convex set of X for each x ∈dom (A). Proof: It follows from Lemma that the duality mapping F is single-valued. First, we show that Ax is convex. Let ˆ x1, ˆ x2 ∈Ax and set ˆ x = αˆ x1 + (1 −α) ˆ x2 for 0 ≤α ≤1. Then, Since A is dissipative, for all [y, ˆ y] ∈A Re ⟨ˆ x −ˆ y, F(x −y)⟩= α Re ⟨ˆ x1 −ˆ y, F(x −y)⟩+ (1 −α) Re ⟨ˆ x2 −ˆ y, F(x −y)⟩≤0. Thus, if we deﬁne a subset ˜ A by ˜ Az =    Az if z ∈dom (A) \ {x} Ax ∪{ˆ x} if z = x, then ˜ A is a dissipative extension of A and dom ( ˜ A) = dom (A). Since A is maximal dissipative, it follows that ˜ Ax = Ax and thus ˆ x ∈Ax as desired. Next, we show that Ax is closed. Let ˆ xn ∈Ax and limn→∞ˆ xn = ˆ x. Since A is dissipative, Re ⟨ˆ xn −ˆ y, x −y⟩≤0 for all [y, ˆ y] ∈A. Letting n →∞, we obtain Re ⟨ˆ x −ˆ y, x −y⟩≤0. Hence, as shown above ˆ x ∈Ax as desired. □ Deﬁnition 1.4 A subset A of X × X is said to be demiclosed if xn →x and yn ⇀y and [xn, yn] ∈A imply that [x, y] ∈A. A subset A is closed if [xn, yn], xn →x and yn →y imply that [x, y] ∈A. Theorem 1.8 Let A be m-dissipative. Then the followings hold. (i) A is closed. (ii) If {xλ} ⊂X such that xλ →x and Aλxλ →y as λ →0+, then [x, y] ∈A. Proof: (i) Let [xn, ˆ xn] ∈A and (xn, ˆ xn) →(x, ˆ x) in X × X. Since A is dissipative Re ⟨ˆ xn −ˆ y, xn −y⟩i ≤0 for all [y, ˆ y] ∈A. Since ⟨·, ·⟩i is lower semicontinuous, letting n →∞, we obtain Re ⟨ˆ x −ˆ y, x −y⟩i ≤for all [y, ˆ y] ∈A. Then A1 = [x, ˆ x] ∪A is a dissipative extension of A. Since A is maximal dissipative, A1 = A and thus [x, ˆ x] ∈A. Hence, A is closed. (ii) Since {Aλx} is a bounded set in X, by the deﬁnition of Aλ, lim \|Jλxλ −xλ\| →0 and thus Jλxλ →x as λ →0+. But, since Aλxλ ∈AJλxλ, it follows from (i) that [x, y] ∈A. □ Theorem 1.9 Let A be m-dissipative and let X∗be uniformly convex. Then the followings hold. (i) A is demiclosed. (ii) If {xλ} ⊂X such that xλ →x and {\|Aλx\|} is bounded as λ →0+, then x ∈ dom (A). Moreover, if for some subsequence Aλnxn ⇀y, then y ∈Ax. (iii) limλ→0+ \|Aλx\| = ∥Ax∥. Proof: (i) Let [xn, ˆ xn] ∈A be such that lim xn = x and w −lim ˆ xn = ˆ x as n →∞. Since X∗is uniformly convex, from Lemma the duality mapping is single-valued and uniformly continuous on the bounded subsets of X. Since A is dissipative Re ⟨ˆ xn − ˆ y, F(xn −y)⟩≤0 for all [y, ˆ y] ∈A. Thus, letting n →∞, we obtain Re ⟨ˆ x −ˆ y, F(x − y)⟩≤0 for all [y, ˆ y] ∈A. Thus, [x, ˆ x] ∈A, by the maximality of A. 105 Deﬁnition 1.5 The minimal section A0 of A is deﬁned by A0x = {y ∈Ax : \|y\| = ∥Ax∥} with dom (A0) = {x ∈dom (A) : A0x is non-empty}. Lemma 1.10 Let X∗be a strictly convex Banach space and let A be maximal dissi-pative. Then, the followings hold. (i) If X is strictly convex, then A0 is single-valued. (ii) If X reﬂexible, then dom (A0) = dom (A). (iii) If X strictly convex and reﬂexible, then A0 is single-valued and dom (A0) = dom (A). Theorem 1.11 Let X∗is a uniformly convex Banach space and let A be m-dissipative. Then the followings hold. (i) limλ→0+ F(Aλx) = F(A0x) for each x ∈dom (A). Moreover, if X is also uniformly convex, then (ii) limλ→0+ Aλx = A0x for each x ∈dom (A). Proof: (1) Let x ∈dom (A). By (ii) of Theorem 1.2 \|Aλx\| ≤∥Ax∥ Since {Aλx} is a bounded sequence in a reﬂexive Banach space (i.e., since X∗is uniformly convex, X∗is reﬂexive and so is X), there exists a weak convergent sub-sequence {Aλnx}. Now we set y = w −limn→∞Aλnx. Since from Theorem 1.2 Aλnx ∈AJλnx and limn→∞Jλnx = x and from Theorem 1.10 A is demiclosed, it follows that [x, y] ∈A. Since by the lower-semicontinuity of norm this implies ∥Ax∥≤\|y\| lim inf n→∞\|Aλnx\| ≤lim sup n→∞\|Aλnx\| ≤∥Ax∥, we have \|y\| = ∥Ax∥= limn→∞\|Aλnx\| and thus y ∈A0x. Next, since \|F(Aλnx)\| = \|Aλnx\| ≤∥Ax∥, F(Aλnx) is a bounded sequence in the reﬂexive Banach space X∗ and has a weakly convergent subsequence F(Aλkx) of F(Aλnx). If we set y∗= w − limk→∞F(Aλkx), then it follows from the dissipativity of A that Re ⟨Aλkx −y, F(Aλkx)⟩= λ−1 n Re ⟨Aλkx −y, F(Jλkxx)⟩≤0, or equivalently \|Aλkx\|2 ≤Re ⟨y, F(Aλnx)⟩. Letting k →∞, we obtain \|y\|2 ≤Re ⟨y, y∗⟩. Combining this with \|y∗\| ≤lim k→∞\|F(Aλkx)\| = lim k→∞\|Aλkx\| = \|y\|, we have \|y∗\| ≤Re ⟨y, y∗⟩≤\|⟨y, y∗⟩\| ≤\|y\|\|y∗\| ≤\|y\|2. Hence, ⟨y, y∗⟩= \|y\|2 = \|y∗\|2 and we have y∗= F(y). Also, limk→∞\|F(Aλkx)\| = \|y\| = \|F(y)\|. It thus follows from the uniform convexity of X∗that lim k→∞F(Aλkx) = F(y). 106 Since Ax is a closed convex set of X from Theorem , we can show that x →F(A0x) is single-valued. In fact, if C is a closed convex subset of X, the y is an element of minimal norm in C, if and only if \|y\| ≤\|(1 −α)y + α z\| for all z ∈C and 0 ≤α ≤1. Hence, ⟨z −y, y⟩+ ≥0. and from Theorem 1.10 0 ≤Re ⟨z −y, f⟩= Re ⟨z, f⟩−\|y\|2 (5.3) for all z ∈C and f ∈F(y). Now, let y1, y2 be arbitrary in A0x. Then, from (5.3) \|y1\|2 ≤Re ⟨y2, F(y1)⟩≤\|y1\|\|y2\| which implies that ⟨y2, F(y1)⟩= \|y2\|2 and \|F(y1)\| = \|y2\|. Therefore, F(y1) = F(y2) as desired. Thus, we have shown that for every sequence {λ} of positive numbers that converge to zero, the sequence {F(Aλx)} has a subsequence that converges to the same limit F(A0x). Therefore, limλ→0 F(Aλx) →F(A0x). Furthermore, we assume that X is uniformly convex. We have shown above that for x ∈dom (A) the sequence {Aλ} contains a weak convergent subsequence {Aλnx} and if y = w −limn→∞Aλnx then [x, y] ∈A0 and \|y\| = limn→∞\|Aλnx\|. But since X is uniformly convex, it follows from Theorem 1.10 that A0 is single-valued and thus y = A0x. Hence, w −limn→∞Aλnx = A0x and limn→∞\|Aλnx\| = \|A0x\|. Since X is uniformly convex, this implies that limn→∞Aλnx = A0x. □ Theorem 1.12 Let X is a uniformly convex Banach space and let A be m-dissipative. Then dom (A) is a convex subset of X. Proof: It follows from Theorem 1.4 that \|Jλx −x\| ≤λ ∥Ax∥ for x ∈dom (A) Hence \|Jλx −x\| →0 as λ →0+. Since Jλx ∈dom (A) for X ∈X, it follows that dom (A) = {x ∈X : \|Jλx −x\| →0 as λ →0+}. Let x1, x2 ∈dom (A) and 0 ≤α ≤1 and set x = α x1 + (1 −α) x2. Then, we have \|Jλx −x1\| ≤\|x −x1\| + \|Jλx1 −x1\| \|Jλx −x2\| ≤\|x −x2\| + \|Jλx2 −x2\| (5.4) where x −x1 = (1 −α) (x2 −x1) and x −x2 = α (x1 −x2). Since {Jλx} is a bounded set and a uniformly convex Banach space is reﬂexive, it follows that there exists a subsequence {Jλnx} that converges weakly to z. Since the norm is weakly lower semi-continuous, letting n →∞in (5.4) with λ = λn, we obtain \|z −x1\| ≤(1 −α) \|x1 −x2\| \|z −x2\| ≤α \|x1 −x2\| 107 Thus, \|x1 −x2\| = \|(x1 −z) + (z −x2)\| ≤\|x1 −z\| + \|z −x2\| ≤\|x1 −x2\| and therefore \|x1−z\| = (1−α) \|x1−x2\|, \|z−x2\| = α \|x1−x2\| and \|(x1−z)+(z−x2)\| = \|x1 −x2\|. But, since X is uniformly convex we have z = x and w −lim Jλnx = x as n →∞. Since we also have \|x −x1\| ≤lim inf n→∞\|Jλnx −Jλnx1\| ≤\|x −x1\| \|Jλnx −Jλnx1\| →\|x −x1\| and w −lim Jλnx −Jλnx1 = x −x1 as n →∞. Since X is uniformly convex, this implies that limλ→0+ Jλx = x and x ∈dom (A). □ 5.0.2 Generation of Nonlinear Semigroups In this section, we consider the generation of nonlinear semigroup by Crandall-Liggett on a Banach space X. Deﬁnition 2.1 Let X0 be a subset of X. A semigroup S(t), t ≥ of nonlinear contractions on X0 is a function with domain [0, ∞) × X0 and range in X0 satisfying the following conditions: S(t + s)x = S(t)S(s)x and S(0)x = x for x ∈X0, t, s ≥0 t →S(t)x ∈X is continuous \|S(t)x −S(t)y\| ≤\|x −y\| for t ≥0, x, y ∈X0 Let A be a ω-dissipative operator and Jλ = (I −λ A)−1 is the resolvent. The following estimate plays an essential role in the Crandall-Liggett generation theory. Lemma 2.1 Assume a sequence {an,m} of positive numbers satisﬁes an,m ≤α an−1,m−1 + (1 −α) an−1,m (5.5) and a0,m ≤m λ and an,0 ≤n µ for λ ≥µ > 0 and α = µ λ. Then we have the estimate an,m ≤[(mλ −nµ)2 + mλ2] 1 2 + [(mλ −nµ)2 + nλµ] 1 2 . (5.6) Proof: From the assumption, (7.2) holds for either m = 0 or n = 0. We will use the induction in n, m, that is if (7.2) holds for (n + 1, m) when (7.2) is true for (n, m) and (n, m −1), then (7.2) holds for all (n, m). Let β = 1 −α. We assume that (7.2) holds for (n, m) and (n, m −1). Then, by (7.1) and Cauchy-Schwarz inequality αx + βy ≤(α + β) 1 2 (αx2 + βy2) 1 2 an+1,m ≤α an,m−1 + β an,m ≤α ([((m −1)λ −nµ)2 + (m −1)λ2] 1 2 + [((m −1)λ −nµ)2 + nλµ] 1 2 ) +β ([(mλ −nµ)2 + mλ2] 1 2 + [(mλ −nµ)2 + nλµ] 1 2 ) = (α + β) 1 2 (α [((m −1)λ −nµ)2 + (m −1)λ2] + β [(mλ −nµ)2 + mλ2]) 1 2 +(α + β) 1 2 (α [((m −1)λ −nµ)2 + nλµ] + β [(mλ −nµ)2 + nλµ]) 1 2 ≤[(mλ −(n + 1)µ)2 + mλ2] 1 2 + [(λ −(n + 1)µ)2 + (n + 1)λµ] 1 2 . 108 Here, we used α + β = 1, αλ = µ and α [((m −1)λ −nµ)2 + (m −1)λ2] + β [(mλ −nµ)2 + mλ2]) ≤(mλ −nµ)2 + mλ2 −αλ(mλ −nµ) = (mλ −(n + 1)µ)2 + mλ2 −µ2 α [((m −1)λ −nµ)2 + nλµ] + β [(mλ −nµ)2 + nλµ] ≤(mλ −nµ)2 + (n + 1)λµ −2αλ(mλ −nµ) ≤(mλ −(n + 1)µ)2 + (n + 1)λµ −µ2.□ Theorem 2.2 Assume A be a dissipative subset of X × X and satisﬁes the range condition dom (A) ⊂R(I −λ A) for all suﬃciently small λ > 0. (5.7) Then, there exists a semigroup of type ω on S(t) on dom (A) that satisﬁes for x ∈ dom (A) S(t)x = lim λ→0+ (I −λ A)−[ t λ]x, t ≥0 (5.8) and \|S(t)x −S(t)x\| ≤\|t −s\| ∥Ax∥ for x ∈dom (A), and t, s ≥0. Proof: First, note that from (7.3) dom (A) ⊂dom (Jλ). Let x ∈dom (A) and set an,m = \|Jn µx −Jm λ x\| for n, m ≥0. Then, from Theorem 1.4 a0,m = \|x −Jm λ x\| ≤\|x −Jλx\| + \|Jλx −J2 λx\| + · · · + \|Jm−1 λ x −Jm λ x\| ≤m \|x −Jλx\| ≤mλ ∥Ax∥. Similarly, an,0 = \|Jn µx −x\| ≤nµ ∥Ax∥. Moreover, an,m = \|Jn µx −Jm λ x\| ≤\|Jn µx −Jµ µ λ Jm−1 λ x + λ −µ λ Jm λ x \| ≤µ λ \|Jn−1 µ x −Jm−1 λ x\| + λ −µ λ \|Jn−1 µ x −Jm λ x\| = α an−1,m−1 + (1 −α) an−1,m. It thus follows from Lemma 2.1 that \|J [ t µ ] µ x −J [ t λ] λ x\| ≤2(λ2 + λt) 1 2 ∥Ax∥. (5.9) Thus, J [ t λ] λ x converges to S(t)x uniformly on any bounded intervals, as λ →0+. Since J [ t λ ] λ is non-expansive, so is S(t). Hence (5.8) holds. Next, we show that S(t) satisﬁes the semigroup property S(t + s)x = S(t)S(s)x for x ∈dom (A) and t, s ≥0. Letting µ →0+ in (??), we obtain \|T(t)x −J [ t λ ] λ x\| ≤2(λ2 + λt) 1 2 . (5.10) for x ∈dom (A). If we let x = J [ s λ] λ z, then x ∈dom (A) and \|S(t)J [ s λ] λ z −J [ t λ ] λ J [ s λ] λ z\| ≤2(λ2 + λt) 1 2 ∥Az∥ (5.11) 109 where we used that ∥AJ [ s λ ] λ z∥≤∥Az∥for z ∈dom (A). Since [ t+s λ ] −([ t λ] + [ s λ]) equals 0 or 1, we have \|J [ t+s λ ] λ z −J [ t λ] λ J [ s λ ] λ z\| ≤\|Jλz −z\| ≤λ ∥Az∥. (5.12) It thus follows from (5.10)–(5.12) that \|S(t + s)z −S(t)S(s)z\| ≤\|S(t + s)x −J [ t+s λ ] λ z\| + \|J [ t+s λ ] λ z −J [ t λ ] λ J [ s λ] λ z\| +\|J [ t λ] λ J [ s λ ] λ z −S(t)J [ s λ ] λ z\| + \|S(t)J [ s λ ] λ z −S(t)S(s)z\| →0 as λ →0+. Hence, S(t + s)z = S(t)S(s)z. Finally, since \|J [ t λ ] λ x −x\| ≤[ t λ] \|Jλx −x\| ≤t ∥Ax∥ we have \|S(t)x −x\| ≤t ∥Ax∥for x ∈dom (A). From this we obtain \|S(t)x −x\| →0 as t →0+ for x ∈dom (A) and also \|S(t)x −S(s)x\| ≤\|S(t −s)x −x\| ≤(t −s) ∥Ax∥ for x ∈dom (A) and t ≥s ≥0. □ 6 Cauchy Problem Deﬁnition 3.1 Let x0 ∈X and ω ∈R. Consider the Cauchy problem d dtu(t) ∈Au(t), u(0) = x0. (6.1) (1) A continuous function u(t) : [0, T] →X is called a strong solution of (6.1) if u(t) is Lipschitz continuous with u(0) = x0, strongly diﬀerentiable a.e. t ∈[0, T], and (6.1) holds a.e. t ∈[0, T]. (2) A continuous function u(t) : [0, T] →X is called an integral solution of type ω of (6.1) if u(t) satisﬁes \|u(t) −x\| −\|u(ˆ t) −x\| ≤ Z t ˆ t (ω \|u(s) −x\| + ⟨y, u(s) −x⟩+) ds (6.2) for all [x, y] ∈A and t ≥ˆ t ∈[0, T]. Theorem 3.1 Let A be a dissipative subset of X × X. Then, the strong solution to (6.1) is unique. Moreover if the range condition (7.3) holds, then then the strong solution u(t) : [0, ∞) →X to (6.1) is given by u(t) = lim λ→0+ (I −λ A)−[ t λ]x for x ∈dom (A) and t ≥0. Proof: Let ui(t), i = 1, 2 be the strong solutions to (6.1). Then, t →\|u1(t) −u2(t)\| is Lipschitz continuous and thus a.e. diﬀerentiable t ≥0. Thus d dt\|u1(t) −u2(t)\|2 = 2 ⟨u′ 1(t) −u′ 2(t), u1(t) −u2(t)⟩i 110 a.e. t ≥0. Since u′ i(t) ∈Aui(t), i = 1, 2 from the dissipativeness of A, we have d dt\|u1(t) −u2(t)\|2 ≤0 and therefore \|u1(t) −u2(t)\|2 ≤ Z t 0 d dt\|u1(t) −u2(t)\|2 dt ≤0, which implies u1 = u2. For 0 < 2λ < s let uλ(t) = (I −λ A)−[ t λ ]x and deﬁne gλ(t) = λ−1(u(t) −u(t −λ)) − u′(t) a.e. t ≥λ. Since limλ→0+ \|gλ\| = 0 a.e. t > 0 and \|gλ(t)\| ≤2M for a.e. t ∈[λ, s], where M is a Lipschitz constant of u(t) on [0, s], it follows that limλ→0+ R s λ \|gλ(t)\| dt = 0 by Lebesgue dominated convergence theorem. Next, since u(t −λ) + λ gλ(t) = u(t) −λ u′(t) ∈(I −λ A)u(t), we have u(t) = (I −λ A)−1(u(t −λ) + λ gλ(t)). Hence, \|uλ(t) −u(t)\| ≤\|(I −λ A)−[ t−λ λ ]x −u(t −λ) −λ gλ(t)\| ≤\|uλ(t −λ) −u(t −λ)\| + λ \|gλ(t)\| a.e. t ∈[λ, s]. Integrating this on [λ, s], we obtain λ−1 Z s s−λ \|uλ(t) −u(t)\| dt ≤λ−1 Z λ 0 \|uλ(t) −u(t)\| dt + Z s λ \|gλ(t)\| dt. Letting λ →0+, it follows from Theorem 2.2 that \|S(s)x−u(s)\| = 0 since u is Lipschitz continuous, which shows the desired result. □ In general, the semigroup S(t) generated on dom (A) in Theorem 2.2 in not neces-sary strongly diﬀerentiable. In fact, an example of an m-dissipative A satisfying (7.3) is given, for which the semigroup constructed in Theorem 2.2 is not even weakly dif-ferentiable for all t ≥0. Hence, from Theorem 3.1 the corresponding Cauchy problem (6.1) does not have a strong solution. However, we have the following. Theorem 3.2 Let A be an ω-dissipative subset satisfying (7.3) and S(t), t ≥0 be the semigroup on dom (A), as constructed in Theorem 2.2. Then, the followings hold. (1) u(t) = S(t)x on dom (A) deﬁned in Theorem 2.2 is an integral solution of type ω to the Cauchy problem (6.1). (2) If v(t) ∈C(0, T; X) be an integral of type ω to (6.1), then \|v(t)−u(t)\| ≤eωt \|v(0)− u(0)\|. (3) The Cauchy problem (6.1) has a unique solution in dom (A) in the sense of Deﬁnition 2.1. Proof: A simple modiﬁcation of the proof of Theorem 2.2 shows that for x0 ∈dom (A) S(t)x0 = lim λ→0+ (I −λ A)−[ t λ ]x0 exists and deﬁnes the semigroup S(t) of nonlinear ω-contractions on dom (A), i.e., \|S(t)x −S(t)y\| ≤eωt \|x −y\| for t ≥0 and x, y ∈dom (A). 111 For x0 ∈dom (A) we deﬁne for λ > 0 and k ≥1 yk λ = λ−1(Jk λx0 −Jk−1 λ x0) = AλJk−1 λ x0 ∈AJk λ. (6.3) Since A is ω-dissipative, ⟨yλ,k −y, Jk λx0 −x)⟩−≤ω \|Jk λx0 −x)\| for [x, y] ∈A. Since from Lemma 1.1 (4) ⟨y, x⟩−−⟨z, x⟩+ ≤⟨y −z, x⟩−, it follows that ⟨yk λ, Jk λx0 −x⟩−≤ω \|Jk λx0 −x)\| + ⟨y, Jk λx0 −x⟩+ (6.4) Since from Lemma 1.1 (3) ⟨x + y, x⟩−= \|x\| + ⟨y, x⟩−, we have ⟨λyk λ, Jk λx0 −x)⟩−= \|Jk λx0 −x\| + ⟨−(Jk−1 λ −x), Jk λx0 −x⟩≥\|Jk λx0 −x)\| −\|Jk−1 λ x0 −x\|. It thus follows from (6.4) that \|Jk λx0 −x\| −\|Jk−1 λ x0 −x\| ≤λ (ω \|Jk λx0 −x\| + ⟨y, Jk λx0 −x⟩+). Since J[ t λ ] = Jk λ on t ∈[kλ, (k + 1)λ), this inequality can be written as \|Jk λx0 −x\| −\|Jk−1 λ x0 −x\| ≤ Z (k+1)λ kλ (ω \|J [ t λ λ x0 −x\| + ⟨y, J [ t λ ] λ x0 −x⟩+ dt. Hence, summing up this in k from k = [ ˆ t λ] + 1 to [ t λ] we obtain \|J [ t λ] λ x0 −x\| −\|J [ ˆ t λ ] λ x0 −x\| ≤ Z [ t λ ]λ [ ˆ t λ]λ (ω \|J [ s λ ] λ x0 −x\| + ⟨y, J [ s λ ] λ x0 −x⟩+ ds. Since \|J [ s λ ] λ x0\| ≤(1 −λω)−k \|x0\| ≤ekλω \|x0\|, by Lebesgue dominated convergence the-orem and the upper semicontinuity of ⟨·, ·⟩+, letting λ →0+ we obtain \|S(t)x0 −x\| −\|S(ˆ t)x0 −x\| ≤ Z t ˆ t (ω \|S(s)x0 −x\| + ⟨y, S(t)x0 −x⟩+) ds (6.5) for x0 ∈dom (A). Similarly, since S(t) is Lipschitz continuous on dom (A), again by Lebesgue dominated convergence theorem and the upper semicontinuity of ⟨·, ·⟩+, (6.5) holds for all x0 ∈dom (A). (2) Let v(t) ∈C(0, T; X) be an integral of type ω to (6.1). Since [Jλ k x0, yk λ] ∈A, it follows from (6.2) that \|v(t) −Jk λx0\| −\|v(ˆ t) −Jk λx0\| ≤ Z t ˆ t (ω \|v(s) −Jk λx0\| + ⟨yk λ, v(s) −Jk λx0⟩+) ds. (6.6) Since λ yk λ = −(v(s)−Jk λx0)+(v(s)−Jk−1 λ x0) and from Lemma 1.1 (3) ⟨−x+y, x⟩+ = −\|x\| + ⟨y, x⟩+, we have ⟨λyk λ, v(s)−Jk λx0⟩= −\|v(s)−Jk λx0\|+⟨v(s)−Jk−1 λ x0, v(s)−Jk λx0⟩+ ≤−\|v(s)−Jk λx0\|+\|v(s)−Jk−1 λ x0\|. Thus, from (6.6) (\|v(t)−Jk λx0\|−\|v(ˆ t)−Jk λx0\|)λ ≤ Z t ˆ t (ωλ \|v(s)−Jk λx0\|−\|v(s)−Jk λx0\|+\|v(s)−Jk−1 λ x0\|) ds 112 Summing up the both sides of this in k from [ ˆ τ λ] + 1 to [ τ λ], we obtain Z [ τ λ]λ [ ˆ τ λ ]λ (\|v(t) −J [ σ λ ] λ x0\| −\|v(ˆ t −J [ σ λ ] λ x0\| dσ ≤ Z t ˆ t (−\|v(s) −J [ τ λ ] λ x0\| + \|v(s) −J [ ˆ τ λ ] λ x0\| + Z [ τ λ]λ [ ˆ τ λ ]λ ω \|v(s) −J [ σ λ ] λ x0\| dσ) ds. Now, by Lebesgue dominated convergence theorem, letting λ →0+ Z τ ˆ τ (\|v(t) −u(σ)\| −\|v(ˆ t −u(σ)\|) dσ + Z t ˆ t (\|v(s) −u(τ)\| −\|v(s) −u(ˆ τ)\|) ds ≤ Z t ˆ t Z τ ˆ τ ω \|v(s) −u(σ)\| dσ ds. (6.7) For h > 0 we deﬁne Fh by Fh(t) = h−2 Z t+h t Z t+h t \|v(s) −u(σ)\| dσ ds. Then from (6.7) we have d dtFh(t) ≤ω Fh(t) and thus Fh(t) ≤eωtFh(0). Since u, v are continuous we obtain the desired estimate by letting h →0+. □ Lemma 3.3 Let A be an ω-dissipative subset satisfying (7.3) and S(t), t ≥0 be the semigroup on dom (A), as constructed in Theorem 2.2. Then, for x0 ∈dom (A) and [x, y] ∈A \|S(t)x0 −x\|2 −\|S(ˆ t)x0 −x\|2 ≤2 Z t ˆ t (ω \|S(s)x0 −x\|2 + ⟨y, S(s)x0 −x⟩s ds (6.8) and for every f ∈F(x0 −x) lim sup t→0+ Re ⟨S(t)x0 −x0 t , f⟩≤ω \|x0 −x\|2 + ⟨y, x0 −x⟩s. (6.9) Proof: Let yλ k be deﬁned by (6.3). Since A is ω-dissipative, there exists f ∈F(Jk λx0−x) such that Re (yk λ −y, f⟩≤ω \|Jk λx0 −x\|2. Since Re ⟨yk λ, f⟩= λ−1 Re ⟨Jk λx0 −x −(Jk−1 λ x0 −x), f⟩ ≥λ−1(\|Jk λx0 −x\|2 −\|Jk−1 λ x0 −x\|\|Jk λx0 −x\|) ≥(2λ)−1(\|Jk λx0 −x\|2 −\|Jk−1 λ x0 −x\|2), we have from Theorem 1.4 \|Jk λx0 −x\|2 −\|Jk−1 λ x0 −x\|2 ≤2λ Re ⟨yk λ, f⟩≤2λ (ω \|Jk λx0 −x\|2 + ⟨y, Jk λx0 −x⟩s. Since J [ t λ ] λ x0 = Jk λx0 on [kλ, (k + 1)λ), this can be written as \|Jk λx0 −x\|2 −\|Jk−1 λ x0 −x\|2 ≤ Z (k+1)λ kλ (ω \|J [ t λ λ x0 −x\|2 + ⟨y, J [ t λ ] λ x0 −x⟩s) dt. 113 Hence, \|J [ t λ ] λ x0 −x\|2 −\|J [ ˆ t λ ] λ x0 −x\|2 ≤2λ Z [ t λ]λ [ ˆ t λ ]λ (ω \|J [ s λ] λ x0 −x\|2 + ⟨y, J [ s λ ] λ x0 −x⟩s) ds. Since \|J [ s λ ] λ x0\| ≤(1 −λω)−k \|x0\| ≤ekλω \|x0\|, by Lebesgue dominated convergence the-orem and the upper semicontinuity of ⟨·, ·⟩s, letting λ →0+ we obtain (6.8). Next, we show (6.9). For any given f ∈F(x0 −x) as shown above 2Re ⟨S(t)x0 −x0, f⟩≤\|S(t)x0 −x\|2 −\|x0 −x\|2. Thus, from (6.8) Re ⟨S(t)x0 −x0, f⟩≤ Z t 0 (ω \|S(s)x0 −x\|2 + ⟨y, S(s)x0 −x⟩s) ds Since s →S(s)x0 is continuous, by the upper semicontinuity of ⟨·, ·⟩s, we have (6.9). □ Theorem 3.4 Assume that A is a close dissipative subset of X × X and satisﬁes the range condition (7.3) and let S(t), t ≥0 be the semigroup on dom (A), deﬁned in Theorem 2.2. Then, if S(t)x is strongly diﬀerentiable at t0 > 0 then S(t0)x ∈dom (A) and d dtS(t)x \|t=t0 ∈AS(t)x, and moreover S(t0)x ∈dom (A0) and d dtS(t)x t=t0 = A0S(t0)x. Proof: Let d dtS(t)x \|t=t0 = y. Then S(t0 −λ)x −(S(t0)x −λ y) = o(λ), where \|o(λ)\| λ → 0 as λ →0+. Since S(t0 −λ)x ∈dom (A), there exists a [xλ, yλ] ∈A such that S(t0 −λ)x = xλ −λ yλ and λ (y −yλ) = S(t0)x −xλ + o(λ). (6.10) If we let x = xλ, y = yλ and x0 = S(t0)x in (6.9), then we obtain Re ⟨y, f⟩≤ω \|S(t0)x −xλ\|2 + ⟨yλ, S(t0)x −xλ⟩s. for all f ∈(S(t0)x−xλ). It follows from Lemma 1.3 that there exists a g ∈F(S(t0)x− xλ) such that ⟨yλ, S(t)x −xλ⟩s = Re ⟨yλ, g⟩and thus Re ⟨y −yλ, g⟩≤ω \|S(t0)x −xλ\|2. From (6.10) λ−1\|S(t0)x −xλ\|2 ≤\|o(λ)\| λ \|S(t0)x −xλ\| + ω \|S(t0)x −xλ\|2 and thus (1 −λω) S(t0)x −xλ λ →0 as λ →0+. 114 Combining this with (6.10), we obtain xλ →S(t0)x and yλ →y as λ →0+. Since A is closed, it follows that [S(t0)x, y] ∈A, which shows the ﬁrst assertion. Next, from Theorem 2.2 \|S(t0 + λ)x −S(t0)x\| ≤λ ∥AS(t0)x∥ for λ > 0 This implies that \|y\| ≤∥AS(t0)x∥. Since y ∈AS(t0)x, it follows that S(t0)x ∈ dom (A0) and y ∈A0S(t0)x. □ Theorem 3.5 Let A be a dissipative subset of X × X satisfying the range condition (7.3) and S(t), t ≥0 be the semigroup on dom (A), deﬁned in Theorem 2.2. Then the followings hold. (1) For ever x ∈dom (A) lim λ→0+ \|Aλx\| = lim inf t→0+ \|S(t)x −x\|. (2) Let x ∈dom (A). Then, limλ→0+ \|Aλx\| < ∞if and only if there exists a sequence {xn} in dom (A) such that limn→∞xn = x and supn ∥Axn∥< ∞. Proof: Let x ∈dom (A). From Theorem 1.4 \|Aλx\| is monotone decreasing and lim \|Aλx\| exists (including ∞). Since \|J [ t λ ] λ −x\| ≤t \|Aλx\|, it follows from Theorem 2.2 that \|S(t)x −x\| ≤t lim λ→0+ \|Aλx\| thus lim inf t→0+ 1 t \|S(t)x −x\| ≤lim λ→0+ \|Aλx\|. Conversely, from Lemma 3.3 −lim inf t→0+ 1 t \|S(t)x −x\|\|x −u\| ≤⟨v, x −u⟩s for [u, v] ∈A. We set u = Jλx and v = Aλx. Since x −u = −λ Aλx, −lim inf t→0+ 1 t \|S(t)x −x\|λ\|Aλx\| ≤−λ \|Aλx\|2 which implies lim inf t→0+ 1 t \|S(t)x −x\| ≥\|Aλx\|. Theorem 3.7 Let A be a dissipative set of X × X satisfying the range condition (7.3) and let S(t), t ≥0 be the semigroup deﬁned on dom (A) in Theorem 2.2. (1) If x ∈dom (A) and S(t)x is diﬀerentiable a.e., t > 0, then u(t) = S(t)x, t ≥0 is a unique strong solution of the Cauchy problem (6.1). (2) If X is reﬂexive. Then, if x ∈dom (A) then u(t) = S(t)x, t ≥0 is a unique strong solution of the Cauchy problem (6.1). Proof: The assertion (1) follows from Theorems 3.1 and 3.5. If X is reﬂexive, then since an X-valued absolute continuous function is a.e. strongly diﬀerentiable, (2) follows from (1). □ 115 6.0.1 Inﬁnitesimal generator Deﬁnition 4.1 Let X0 be a subset of a Banach space X and S(t), t ≥0 be a semigroup of nonlinear contractions on X0. Set Ah = h−1(T(h) −I) for h > 0 and deﬁne the strong and weak inﬁnitesimal generators A0 and Aw by A0x = limh→0+ Ahx with dom (A0) = {x ∈X0 : limh→0+ Ahx exists} A0x = w −limh→0+ Ahx with dom (A0) = {x ∈X0 : w −limh→0+ Ahx exists}, (6.11) respectively. We deﬁne the set ˆ D by ˆ D = {x ∈X0 : lim inf h→0 \|Ahx\| < ∞} (6.12) Theorem 4.1 Let S(t), t ≥0 be a semigroup of nonlinear contractions deﬁned on a closed subset X0 of X. Then the followings hold. (1) ⟨Awx1 −Awx2, x∗⟩for all x1, x2 ∈dom (Aw) and x∗∈F(x1 −x2). In particular, A0 and Aw are dissipative. (2) If X is reﬂexive, then dom (A0) = dom (Aw) = ˆ D. (3) If X is reﬂexive and strictly convex, then dom(Aw) = ˆ D. In addition, if X is uniformly convex, then dom (Aw) = dom (A0) = ˆ D and Aw = A0. Proof: (1) For x1, x2 ∈X0 and x∗∈F(x1 −x2) we have ⟨Ahx1 −Ahx2, x∗⟩= h−1(⟨S(h)x1 −S(h)x2, x∗⟩−\|x1 −x2\|2) ≤h−1(\|S(h)x1 −S(h)x2\|\|x1 −x2\| −\|x1 −x2\|2) ≤0. Letting h →0+, we obtain the desired inequality. (2) Obviously, dom (A0) ⊂dom (Aw) ⊂ˆ D. Let x ∈ˆ D. It suﬃces to show that x ∈dom (A0). We can show that t →S(t)x is Lipschitz continuous. In fact, there exists a monotonically decreasing sequence {tk} of positive numbers and L > 0 such that tk →0 as k →∞and \|S(tk)x −x\| ≤L tk. Let h > 0 and nk be a nonnegative integer such that 0 ≤h −nktk < tk. Then we have \|S(t + h)x −S(t)x\| ≤\|S(t)x −x\| = \|S(h −nktk + nktk)x −x\| ≤\|S(h −nktk)x −x\| + L nktk ≤\|S(h −nktk)x −x\| + L h By the strong continuity of S(t)x at t = 0, letting k →∞, we obtain \|S(t + h)x − S(t)x\| ≤L h. Now, since X is reﬂexive this implies that S(t)x is a.e. diﬀerentiable on (0, ∞). But since S(t)x ∈dom (A0) whenever d dtS(t)x exists, S(t)x ∈dom (A0) a.e. t > 0. Thus, since \|S(t)x −x\| →0 as t →0+, it follows that x ∈dom (A0). (3) Assume that X is reﬂexive and strictly convex. Let x0 ∈ˆ D and Y be the set of all weak cluster points of t−1(S(t)x0 −x0) as t →0+. Let ˜ A be a subset of X × X deﬁned by ˜ A = A0 ∪[x0, co Y ] and dom ( ˜ A) = dom (A0) ∪{x0} where co Y denotes the closure of the convex hull of Y . Note that from (1) ⟨A0x1 −A0x2, x∗⟩≤0 for all x1, x2 ∈dom (A0) and x∗∈F(x1 −x2) 116 and for every y ∈Y ⟨A0x1 −y, x∗⟩≤0 for all x1 ∈dom (A0) and x∗∈F(x1 −x0). This implies that ˜ A is a dissipative subset of X×X. But, since X is reﬂexive, t →S(t)x0 is a.e. diﬀerentiable and d dtS(t)x0 = A0S(t)x0 ∈˜ AS(t)x0, a.e., t > 0. It follows from the dissipativity of ˜ A that ⟨d dt(S(t)x0 −x0), x∗⟩≤⟨y, x∗⟩, a.e, t > 0 and y ∈˜ Ax0 (6.13) for x∗∈F(S(t)x0 −x0). Note that for h > 0 ⟨h−1(S(t+h)x0−S(t)x0), x∗⟩≤h−1(\|S(t+h)x0−x0\|−\|S(t)x0−x0\|)\|x∗\| for x∗∈F(S(t)x0−x0). Letting h →0+, we have ⟨d dt(S(t)x0 −x0), x∗⟩≤\|S(t)x0 −x0\| d dt\|S(t)x0 −x0\|, a.e. t > 0 The converse inequality follows much similarly. Thus, we have \|S(t)x0 −x0\| d dt\|S(t)x0 −x0\| = ⟨d dt(S(t)x0 −x0), x∗⟩ for x∗∈F(S(t)x0 −x0). (6.14) It follows from (6.13)–(6.14) that d dt\|S(t)x0 −x0\| ≤\|y\| for y ∈Y and a.e. t > 0 and thus \|S(t)x0 −x0\| ≤t ∥˜ Ax0∥ for all t > 0. (6.15) Note that ˜ Ax0 = co Y is a closed convex subset of X. Since X is reﬂexive and strictly convex, there exists a unique element y0 ∈˜ Ax0 such that \|y0\| = ∥˜ Ax0∥. Hence, (6.15) implies that co Y = y0 = Awx0 and therefore x0 ∈dom (Aw). Next, we assume that X is uniformly convex and let x0 ∈dom(Aw) = ˆ D. Then w −lim S(t)x0 −x0 t = y0 as t →0+. From (6.15) \|t−1(S(t)x0 −x0)\| ≤\|y0\|, a.e. t > 0. Since X is uniformly convex, these imply that lim S(t)x0 −x0 t = y0 as t →0+. which completes the proof. □ Theorem 4.2 Let X and X∗be uniformly convex Banach spaces. Let S(t), t ≥ 0 be the semigroup of nonlinear contractions on a closed subset X0 and A0 be the inﬁnitesimal generator of S(t). If x ∈dom (A0), then 117 (i) S(t)x ∈dom (A0) for all t ≥0 and the function t →A0S(t)x is right continuous on [0, ∞). (ii) S(t)x has a right derivative d+ dt S(t)x for t ≥0 and d+ dt S(t)x = A0S(t)x, t ≥0. (iii) d dtS(t)x exists and is continuous except a countable number of values t ≥0. Proof: (i) −(ii) Let x ∈dom (A0). By Theorem 4.1, dom (A0) = ˆ D and thus S(t)x ∈ dom (A0) and d+ dt S(t)x = A0S(t)x for t ≥0. (6.16) Moreover, t →S(t)x a.e. diﬀerentiable and d dtS(t)x = A0S(t)x a.e. t > 0. We next prove that A0S(t)x is right continuous. For h > 0 d dt(S(t + h)x −S(t)x) = A0S(t + h)x −A0S(t)x, a.e. t > 0. From (6.14) \|S(t + h)x −S(t)x\| d dt\|S(t + h)x −S(t)x\| = ⟨A0S(t + h)x −A0S(t)x, x∗⟩≤0 for all x∗∈F(S(t + h)x −S(t)x), since A0 is dissipative. Integrating this over [s, t], we obtain \|S(t + h)x −S(t)x\| ≤\|S(s + h)x −S(s)x\| for 0 ≤s ≤t and therefore \|d+ dt S(t)x\| ≤\|d+ ds S(s)x\|. Hence t →\|A0S(t)x\| is monotonically non-increasing function and thus it is right continuous. Let t0 ≥0 and let {tk} be a decreasing sequence of positive numbers such that tk →t0. Without loss of generality, we may assume that w−limk→∞A0S(tk) = y0. The right continuity of \|A0S(t)x\| at t = t0, thus implies that \|y0\| ≤\|A0S(t0)x\| (6.17) since norm is weakly lower semicontinuous. Let ˜ A0 be the maximal dissipative exten-sion of A0. It then follows from Theorem 1.9 that ˜ A is demiclosed and thus y0 ∈˜ AS(t)x. On the other hand, for x ∈dom (A0) and y ∈˜ Ax, we have ⟨d dt(S(t)x −x), x∗⟩≤⟨y, x∗⟩ for all x∗∈F(S(t)x −x) a.e. t > 0, since ˜ A is dissipative and d dtS(t)x = A0S(t)x ∈˜ AS(t)x a.e. t > 0. From (6.14) we have t−1\|S(t)x −x\| ≤\| ˜ Ax\| = ∥˜ A0x∥ for t ≥0 where ˜ A0 is the minimal section of ˜ A. Hence A0x = ˜ A0x. It thus follows from (6.17) that y0 = A0S(t0)x and limk→∞A0S(tk)x = y0 since X is uniformly convex. Thus, we have proved the right continuity of A0S(t)x for t ≥0. (iii) Integrating (6.16) over [t, t + h], we have S(t + h)x −S(t)x = Z t+h t A0S(s)x ds 118 for t, h ≥0. Hence it suﬃces to prove that the function t →A0S(t)x is continuous except a countable number of t > 0. Using the same arguments as above, we can show that if \|A0S(t)x\| is continuous at t = t0, then A0S(t)x is continuous at t = t0. But since \|A0S(t)x\| is monotone non-increasing, it follows that it has at most countably many discontinuities, which completes the proof. □ Theorem 4.3 Let X and X∗be uniformly convex Banach spaces. If A be m-dissipative, then A is demiclosed, dom (A) is a closed convex set and A0 is single-valued operator with dom (A0) = dom (A). Moreover, A0 is the inﬁnitesimal generator of a semigroup of contractions on dom (A). Proof: It follows from Theorem 1.9 that A is demiclosed. The second assertion follows from Theorem 1.12. Also, from Theorem 3.4 d dtS(t)x = A0S(t)x, a.e. t > 0 and \|S(t)x −x\| ≤t \|A0x\|, t > 0 (6.18) for x ∈dom (A). Let A0 be the inﬁnitesimal generator of the semigroup S(t), t ≥0 generated by A deﬁned in Theorem 2.2. Then, (6.18) implies that by Theorem 4.1 x ∈dom (A0) and by Theorem 4.2 d+ dt S(t)x = A0S(t)x and A0S(t)x is right continuous in t. Since A is closed, A0x = lim t→0+ A0S(t)x ∈Ax. Hence, (6.17) implies that A0x = A0x. When X is a Hilbert space we have the nonlinear version of Hille-Yosida theorem as follows. Theorem 4.4 Let H be a Hilbert space. Then, (1) The inﬁnitesimal generator A0 of a semigroup of contractions S(t), t ≥0 on a closed convex set X0 has a dense domain in X0 and there exists a unique maximal dissipative operator A such that A0 = A0. Conversely, (2) If A0 is a maximal dissipative operator, then dom (A) is a closed convex set and A0 is the inﬁnitesimal generator of contractions on dom (A). Proof: (2) Since from Theorem 1.7 the maximal dissipative operator in a Hilbert space is m-dissipative, (2) follows from Theorem 4.3. □ Example (Nonlinear Diﬀusion) Consider the nonlinear diﬀusion equation of the form ut = Au = ∆γ(u) −β(u) on X = L1(Ω). Assume γ : R →R is maximal monotone. and γ : R →R is monotone. Let dom(A) = {there exists a v ∈W 1,1(Ω) such that v ∈γ(u) and ∆v ∈X} Thus, sign(x −y) = sign(γ(x) −γ(y)). Let ρ ∈C2(R) be a monotonically increasing function satisfying ρ(0) = 0 and ρ(x) = sign(x), \|x\| ≥1 and ρϵ(x) = ρ( x ϵ ) for ϵ > 0. Note that for u ∈X (u, ρϵ(u)) →\|u\| and (ψ, ρϵ(u)) →(ψ, sign0(u)) for ψ ∈X 119 as ϵ →0+. (Au1 −Au2, ρϵ(γ(u1) −γ(u2))) = −(∇(γ(u1) −γ(u2))ρ′ ϵ)∇(γ(u1) −γ(u2))) −(β(u1) −β(u2), ρϵ(γ(u1) −γ(u2)) ≤0 Letting ϵ →0∗we obtain (Au1 −Au2, sign0(u1 −u2)) ≤0 for all u1, u2 ∈dom(A). For the range condition: we consider v ∈γ(u) such that λ γ−1(v) −∆v + β(γ−1(v)) = f, (6.19) Let ∂j = λ γ−1(·) −β(γ−1(·)). For f ∈L2(Ω) consider the minimization 1 2 Z Ω (\|∇v\|2 + j(v) −f(x)v) dx over v ∈H1 0(Ω). It has a unique solution v ∈H2(Ω) ∩H1 0(Ω) such that ∆v + f ∈∂j(v). For f ∈X we choose fn ∈L2(Ω) such that \|fn −f\|X →0 as n →∞. As show above un = ∆vn + fn, \|un −um\|X ≤M \|fn −fm\|X Thus, there exists u ∈X such that ∆vn →u −f. Moreover, there exists a v ∈ W 1,q, 1 < q < d d−1 such that vn →v in X and ∆v = u −f. In fact, let p > d. For all h0 ∈Lp(Ω) and ⃗ h ∈Lp(Ω)d −∆φ = h0 + ∇· ⃗ h has a unique solution u ∈W 1,p(Ω) and \|φ\|W 1,p ≤M (\|h\|p + \|⃗ h\|p). By the Green’s formula \|(h0, vn) −(h, ∇vn)\| = \| −(∆vn, φ)\| ≤M (\|h\|p + \|⃗ h\|p\|)\|∆vn\|1. Since h0 ∈Lp(Ω) and h ∈Lp(Ω)d are arbitraly \|vn\|W 1,q ≤M \|∆vn\|, Since W 1,q(Omega) is compactly embedded into L1(Ω), vn →v, v ∈W 1,q 0 (Ω). Since ∂j is maximal monotone u ∈∂j(v) equivalently u ∈γ−1(v) and ∆v + f ∈∂j(v). 120 Example (Conservation law) We consider the scalar conservation law ut + (f(u))x + f0(x, u) = 0, t > 0 u(x, 0) = u0(x), x ∈Rd (6.20) where f : R →Rd is C1. Let X = L1(Rd) and deﬁne Au = −(f(u))x, where we assume f0 = 0 for the sake of simplicity of our presentation. Deﬁne Let C = {φ ∈Z : φ ≥0}. Since Ac = 0 for all constant c, it follows that φ −c ∈C = ⇒(I −λ A)−1φ −c ∈C. Similarly, c −φ ∈C = ⇒c −(I −λ A)−1φ ∈C. Thus, without loss of generality, one can assume f is bounded. We use the following lemma. Lemma 2.1 For ψ ∈H1(Rd) and φ ∈L2 div = {φ ∈(L2(Rd))d : ∇· φ ∈L2(Rd)} we have (φ, ∇ψ) + (∇· φ, ψ) = 0 Proof: Note that for ζ ∈C∞ 0 (Rd) (φ, ∇(ζ ψ)) + (∇· φ, ζ ψ) = 0 Let g ∈C∞(Rd) satisfying g = 1 for \|x\| ≤1 and g = 0 if \|x\| ≥2 and set ζ = g( x r ). Then we have (φ, ζ∇ψ + 1 r ψ∇g) + (∇· φ, ζ ψ) = 0. Since g( x r ) →1 a.e. in Rd as r →∞thus the lemma follows from Fatou’s lemma. □. Note that −(f(u1)x −f(u2)x, ρ(u1 −u2)) = (f(u1) −f(u2), ρ′(u1 −u2) (u1 −u2)x), If we deﬁne Ψ(x) = R x 0 σρ′(σ) dσ and ρϵ(x) = ρ( x ϵ ) for ϵ > 0. then \|(η (u1 −u2), ρ′ ϵ(u1 −u2) (u1 −u2)x)\| = ϵ (Ψ(u1 −u2 ϵ ), ηx) ≤M ϵ \|ηx\|1 →0 as ϵ →0. Note that for u ∈L1(Rd) (u, ρϵ(u)) →\|u\| and (ψ, ρϵ(u)) →(ψ, sign0(u)) for ψ ∈L1(Rd) as ϵ →0+. Thus, ⟨Au1 −Au2, sign0(u1 −u2)⟩≤0 and A is dissipative. It will be show that range(λ I −A) = X, 121 i.e., for any g ∈X there exists an entropy solution satisfying (sign(u −k)(λu −g), ψ) ≤(sing(u −k)(f(u) −f(k)), ψx) for all ψ ∈C1 0(Rd) and k ∈R. Hence A has a maximal monotone extension in L1(Rd). In fact, for ϵ > 0 consider the viscous equation λ u −ϵ ∆u + f(u)x = g (6.21) First, assume f is Lipschitz continuos, one can show that λ u −ϵ∆u −f(u)x : H1(Rd) →(H1(Rd))∗ is monotone, hemi-continuous and coercive. It thus follows from [] that (6.21) has a solution u ∈H1(Rd) for all g ∈L2(Rd) ∩L1(Rd). Since f(u)x ∈L2(Rd), u ∈H2(Rd). L∞(Rd) estimate From (6.21) (λ u −ϵ ∆u + f(u)x, \|u\|p−2u) = (g, \|u\|p−2u). Since λ\|u\|p p + (f′(u)ux, \|u\|p−2u) −ϵ(p −1)(\|u\| p 2 −1uux, \|u\| p 2 −1uux) ≤((1 −δ 2 −\|f′\|∞ 2ϵδ p −1)\|u\|p p, we have \|up ≤(1 2λ −\|f′\|∞ 2ϵλ p −1)−1\|g\|p. By letting p →∞ \|u\|∞≤1 λ\|g\|∞. Thus, without loss of generality, f is C1(R)d. W 1,1(Rd) estimate Assuming g ∈W 11(Rd), v = ux satisﬁes λ v −ϵ∆v + (f′(u)v)x = gx. Using the same arguments as above \|v\| ≤\|gx\|1. Entropy condition We show that for every k ∈R and nonnegative function ψ ∈C∞ 0 (Rd) (sign(u−k)(λu−g, , ψ)−ϵ(∆u, ψ) ≤(sign(u−k)(f(x)−f(k)), ψx)+ϵ (\|u−k\|, ∆ψ) ≥0 (6.22) It suﬃces to prove for u ∈C2 0(Rd). Note that (ρϵ(u −k)f(u)x, ψ) = (( Z u k ρϵ(s −k)f′(u) ds)x, ψ) = −( Z u k ρϵ(s −k)fu(t, x, u) ds, ψx) 122 Since R k+ϵ k ρϵ(s −k) ds = ϵ R 1 0 ρ(s) ds →0 as ϵ →0, letting ϵ →0 we obtain (sgn(u −k)f(u)x, ψ) = −(sgn(u −k)(f(u) −f(k)), ψx) Next, −(ρϵ(u −k)∆u, ψ) = (ρ′ ϵ(u −k)ux, ψ ux) + (ρϵ(u −k)ux, ψx) where (ρϵ(u −k)ux, ψx) = (−Ψϵ(u −k), De\|taψ) →−(\|u −k\|, ∆ψ) as ϵ →0. Thus, we obtain −(ρϵ(u −k)∆u, ψ) ≥−(\|u −k\|, ∆ψ) and (6.22). Example (Hamilton Jacobi equation) Let u is a solution to a scalar conservation in R1, then v = R x u dx satisﬁes the the Hamilton-Jacobi equation vt + f(vx) = 0. (6.23) Let X = C0(Rd) and Av = −f(vx) dom(A) = {f(vx) ∈X} Then, for v1, v2 ∈C1 ( Rd) ⟨A(v1 −v2), δx0⟩= −(f((v1)x(x0)) −f((v2)x(x0))) = 0 where x0 ∈Rn such that \|v\|X = \|v(x0)\|. We prove the range condition range(λ I −A) = X for λ > 0. That is, there exists a unique viscosity solution to λv −f(vx) = g; for all φ ∈C1(Ω) if v −φ attains a local maximum at x0 ∈Rd, then λ v(x0) −g(x0) + f(φx(x0)) ≤0 and if v −φ attains a local minimum at x0 ∈Rd, then λ v(x0) −g(x0) + f(φx(x0)) ≥0. Thus, A is maximal monotone and (6.23) has an intgral solution. Consider the equation of the form λ V + ˆ H(x, Vx) −ν ∆V = ω f. Assume that ˆ H is C1 and there exist ˜ c1 > 0, ˜ c2 > 0 and ˜ c3 ∈L2(Rd) such that \| ˆ H(x, p) −ˆ H(x, q)\| ≤˜ c1 \|p −q\| and ˆ H(t, x, 0) ∈L2(Rd) and \| ˆ Hx(x, p)\| ≤˜ c3(x) + ˜ c2 \|p\| 123 Deﬁne the Hilbert space H = H1(Rd) by H = {φ ∈L2(Rd) : φx ∈L2(Rd)} with inner product (φ, ψ)H = Z Rdφ(x)ψ(x) + φx(x) · ψx(x) dx. Deﬁne the single valued operator A on H by A V = −ˆ H(x, Vx) + ϵ ∆V with dom (A) = H3(Rd). We show that A −λ I is m-dissipative for some λ. First, A −ˆ ω I with λ = ˜ c2 1 2ϵ is monotone since (Aφ −Aψ, φ −ψ)H = −ϵ (\|φx −ψx\|2 2 + \|∆(φ −ψ)\|2 2) −( ˆ H(·, φx) −ˆ H(·, ψx), φ −ψ −∆(φ −ψ)) ≤λ \|φ −ψ\|2 H −ϵ 2 \|φx −ψx\|2 H, where we used the following lemma. Lemma 2.1 For ψ ∈H1(Rd) and φ ∈L2 div = {φ ∈(L2(Rd))d : ∇· φ ∈L2(Rd)} we have (φ, ∇ψ) + (∇· φ, ψ) = 0 Proof: Note that for ζ ∈C∞ 0 (Rd) (φ, ∇(ζ ψ)) + (∇· φ, ζ ψ) = 0 Let g ∈C∞(Rd) satisfying g = 1 for \|x\| ≤1 and g = 0 if \|x\| ≥2 and set ζ = g( x r ). Then we have (φ, ζ∇ψ + 1 r ψ∇g) + (∇· φ, ζ ψ) = 0. Since g( x r ) →1 a.e. in Rd as r →∞thus the lemma follows from Fatou’s lemma. □. Let us deﬁne the linear operator T on X by Tφ = ν 2 ∆. with dom (T) = H3(Rd). Then T is a self-adjoint operator in H. Moreover, if let ˜ X = H2(Rd) then ˜ X∗= L2(Rd) where H = H1(Rd) is the pivoting space and H = H∗and T ∈L( ˜ X, ˜ X∗) is hermite and coercive. Thus, T is maximal monotone. Hence equation ω V −AV = ω f in L2(Rd) has a unique solution V ∈H. Note that from (2.7) for φ ∈H2(Rd) ˆ H(x, φx)x = ˆ Hx(x, φx) + ˆ Hp(x, φx)φxx ∈L2(Rd). Thus, if f ∈H then the solution V ∈dom (A) and thus A is maximum monotone in H. Step 3: We establish W 1,∞(Rd) estimate of solutions to ω V + H(x, Vx) −ν ∆V = ω f, 124 when fx, Hx(x, 0) ∈L∞(Rd) and H satisﬁes (2.9) \|Hx(x, p) −Hx(x, 0)\| ≤M1 \|p\| and \|Hp(x, p)\| ≤M2 p 1 + \|x\|2. Consider the equation of the form (2.10) ω V + ψ H(x, Vx) −ν ∆V = ω f where ψ(x) = c2 c2 + \|x\|2 for c ≥1. Then ψ(x) H(x, p) satisﬁes (2.6)–(2.7) and thus there exists a unique solution V ∈ H3(Rd) to (2.10) for suﬃciently large ω > 0. Deﬁne U = Vx. Then, from (2.10) we have (2.11) ω (U, φ) + (ψx H(x, U) + ψ Hp(x, U) · Ux, φ) + ν (Ux, φx) = ω (fx, φ) for φ ∈H1(Rd)d. Let \|U\| = q U 2 1 + · · · + U 2 n and \|U\|p = Z Rd\|U(x)\|p dx 1 p Deﬁne the functions Ψ, Φ : R+ →R+ by Ψ(r) =    r p 2 for r ≤R2 Rp−2r for r ≥R2 and Φ(r) =    r p 2 −1 for r ≤R2 Rp−2 for r ≥R2. Setting φ = Φ(\|U\|2)U ∈H1(Rd)d in (2.11), we obtain (2.12) ω \|Ψ(\|U\|2)\|1 −( 2x · U c2 + \|x\|2 Φ(\|U\|2), ψ H(x, U)) +(ψ (Hx(x, U) −ω fx), Φ(\|U\|2)U) + (ψ Hp(x, U), Φ(\|U\|2)(1 2\|U\|2)x) +ν {2 (Φ′(\|U\|2)(1 2\|U\|2)x, (1 2\|U\|2)x) + (Φ(\|U\|2)Ux, Ux))} = 0. Since from (2.9) \|H(x, p)\| ≤const p 1 + \|x\|2 (1 + \|p\|), there exists constants k1, k2 independent of c ≥1 and R > 0 such that ( 2x · U c2 + \|x\|2 Φ(\|U\|2), ψ H(x, U)))) ≤k1 (ψ, Ψ(\|U\|2)) + k2 \|Φ(\|U\|2)U\|q where q = p p−1. It thus follows from (2.9) and (2.12) that ω \|Ψ(\|U\|2)\|1+ 1 ν (Φ(\|U\|2) Ux, Ux) ≤˜ ω\|Ψ(\|U\|2)\|1+(k2+\|Hx(x, 0)\|p+ω \|fx\|p) \|Φ(\|U\|2)U\|q. 125 for some constant ˜ ω > 0 independent of c ≥1 and R > 0. Since \|Ψ(\|U\|2)\|1 ≥ \|Φ(\|U\|2)U\|q q, it follows that for ω ≥˜ ω \|Ψ(\|U\|2)\|1 + 1 2ν (Φ(\|U\|2) Ux, Ux) is uniformly bounded in R > 0. Letting R →∞, it follows from Fatou’s lemma that U ∈Lp(Rd) and from (2.12) ω (\|U\|p −\|fx\|p) + (ψ 2x · U c2 + \|x\|2 \|U\|p−2, H(x, U) + (ψ Hp(x, U), \|U\|p−2(1 2\|U\|2)x) +(ψ Hx(x, U), \|U\|p−2U) + ν {(p −2) (\|U\|p−4, (1 2\|U\|2)x, (1 2\|U\|2)x) + (\|U\|p−2Ux, Ux)}. Thus we have \|U\|p ≤\|fx\|p + 1 ω(σp \|U\|p + k2 + \|Hx(x, 0\|p) where σp = k1 + M1 + \|ψ Hp(x, U)\|2 ∞ (p −2)ν . Letting p →∞, we obtain (2.13) \|U\|∞≤\|fx\|∞+ 1 ω((k1 + M1) \|U\|∞+ k2 + \|Hx(x, 0\|∞). For c ≥1 let us denote by V c, the unique solution of (2.10). Let ζ(x) = χ(x/r) ∈ C2(Rd), r ≥1 where χ = χ(\|x\|) ∈C2(Rd) is a nonincreasing function such that χ(s) = 1 for 0 ≤s ≤1 and χ(s) = exp(−\|s\|) for s ≥2. and we assume −∆χ ≤k3 χ. Then (2.14) ω (V c, ζ ξ) + (ψ H(x, Uc), ζ ξ) −ν (∆V c, ζ ξ) = ω (f, ζ ξ) for all ξ ∈L2(Rd), and U c = V c x satisﬁes (2.15) ω (U c, ζ φ) + (ψ H(x, Uc), ∇· (ζ φ)) −ν (tr U c x, ∇· ζ φ) = ω (fx, ζ φ) for all φ ∈H1(Rd)d. Setting φ = U c in (2.15), we obtain ω (ζ U c, Uc) + (ψ H(x, Uc), Uc · ζx + ζ∇· U c) +ν {(ζ U c x, Uc x) −1 2 (∆ζ, \|U c\|2)} = ω (fx, ζ U). Since \|U c\|∞is uniformly bounded in c ≥1, it follows that for any compact set Ωin Rd there exists a constant MΩindependent of c ≥1 such that ω (ζ U c, Uc) + ν (ζ U c x, Uc x) ≤MΩ. Hence for every compact set Ωof Rd Since if Ωr = {\|x\| < r}, then H2(Ωr) is compactly embedded into H1(Ωr), it follows that there exists a subsequence of {V c} which con-verges strongly in H1(Ωr). By a standard diagonalization process, we can construct 126 a subsequence {V ˆ c} which converges to a function V strongly in H1(ˆ Ω)) and weakly in H2(ˆ Ω) for every compact set ˆ Ωin Rd. Let U = Vx. Then, U ˆ c converges weakly in H1(ˆ Ω)d and strongly in L2(ˆ Ω). Since L2(ˆ Ω) convergent sequence has an a.e. pointwise convergent subsequence, without loss of generality we can assume that U ˆ c converges to U a.e. in Rd. Hence, by Lebesgue dominated convergence theorem Hp(·, U ˆ c) →Hp(·, U) and Hx(·, U ˆ c) →Hx(·, U) strongly in L2(ˆ Ω)d. It follows from (2.14)–(2.15) that the limit V satisﬁes (2.16) ω (V, ζ ξ) + (H(x, Vx), ζ ξ) −ν(∆V, ζ ξ) = ω (V, ζ ξ) for all ξ ∈L2 loc(Rd) and (2.17) ω (U, ζ φ) + (H(x, U), ∇· (ζ φ)) −ν (tr Ux, (ζ φ)x) = ω (fx, ζ φ) for all φ ∈H1 loc(Rd). Setting φ = \|U\|p−2U in (2.17), we obtain (2.18) ω (ζ, \|U\|p) + (ζ Hp(x, U), \|U\|p−2(1 2\|U\|2)x) + (ζ Hx(x, U), \|U\|p−2U) −ω (fx, ζ \|U\|p−2U) +ν {(p −2) (ζ \|U\|p−4(1 2\|U\|2)x, (1 2\|U\|2)x) + (ζ \|U\|p−2Ux, Ux)) −1 p(∆ζ, \|U\|p)} = 0 for all ζ = χ(x/r). Thus, (2.19) (ζ, \|U\|p) 1 p ≤(ζ, \|fx\|p) 1 p + 1 ω (cp (ζ, \|U\|p) 1 p + (ζ, \|Hx(x, 0)\|p) 1 p ) where cp = M1 + k3 p + \|ζ Hp(x,U)\|2 ∞ 2ν (p−2) and (2.20) (Hx(x, 0) −Hx(x, p), p) ≤M1 \|p\|2 all x ∈Rd. Now, letting p →∞we obtain from (2.19) (ω −M1) sup \|x\|≤r \|U\| ≤ω \|fx\|∞+ \|Hx(x, 0)\|∞. Since r ≥1 is arbitrary, we have (2.21) (ω −M1) \|U\|∞≤ω \|fx\|∞+ \|Hx(x, 0)\|∞ Setting ξ = V in (2.16), we obtain ω (ζ, \|V \|2) + (ζ, Hp(x, U)) −ω (f, ζ V ) + ν { (ζ \|V \|2)x, −1 2(∆ζ, \|U\|2)} = 0. Also, from (2.18) ω (ζ, \|U\|2) + (ζ Hp(x, U), U · Ux) + (ζ Hx(x, U), U) −ω (fx, ζ U) +ν {(ζ Ux, Ux)) −1 2(∆ζ, \|U\|2)} = 0 127 Since \|U\|∞is bounded, thus V ∈H2 ℓoc. In fact ω (ζ V, V ) + ν (ζ Ux, Ux) ≤a1 \|U\|2 ∞+ a2ω ((ζf, f) + (ζfx, fx) for some constants a1, a2. Step 4: Next we prove that for ω > max(M1, ωα) equation ω V −Aν(t)V = ω f has a unique solution satisfying (2.21). We deﬁne the sequence {Vk} in H2 ℓoc(Rd) by the successive iteration (2.22) ω Vk+1 −Aν(t)Vk+1 −(ω −ω0)Vk = ω0 f. From Step 3 (2.22) has a solution Vk+1 satisfying (2.23) (ω −M1) \|Uk+1\|∞≤(ω −ω0) \|Uk\|∞+ ω0 \|fx\|∞+ \|Hx(x, 0)\|∞ Thus, \|Uk\|∞≤(1 −ω −ω0 ω −M1 )−1 (ω0 \|fx\|∞+ \|Hx(x, 0)\|∞) ω −M1 = ω0 \|fx\|∞+ \|Hx(x, 0)\|∞ ω0 −M1 = α for all k ≥1. {Vk} is bounded sequence in W 1,∞(Rd) and thus in H2 ℓoc(Rd). Moreover, we have from (2.4) \|Vk+1 −Vk\|X ≤ω −ω0 ω −ωα \|Vk −Vk−1\|X. Thus {Vk} is a Cauchy sequence in X and {Vk} converges to V in X. Let us deﬁne the single-valued operator B on X by Bφ = −H(x, φx). Since {BVk} is bounded in L∞(Rd) we may assume that BVk converges weakly star to w in L∞(Rd). If we show that w = Bu, then V ∈H2(Rd) solves the desired equation. Since {Vk} is bounded in H2(Ωr), {Vk} is strongly precompact in H1(Ωr) for each r > 0. Hence BVk →BV a.e. in Ωr and thus w = BV . Step 5: Next, we consider the case when H ∈C1 is without the Lipschitz bound in p but satisﬁes (2.2). Consider the cut-oﬀfunction of H(x, p) by HM(x, p) =    Hp(x, p \|p\|)(p −M p \|p\|) + H(M p \|p\|) if \|p\| ≥M H(x, p) if \|p\| ≤M. From Step 3 and (2.20) equation ω V + HM(x, Vx) −ν ∆V = ω f has the unique solution V and U = Vx satisﬁes (2.21) with M1 = β M + a. Let M > 0 be so that (ω −(β M + a)) M ≤ω \|fx\|∞+ \|Hx(x, 0)\|∞. Then, \|U\|∞≤M. and thus V ∈H2 loc(Rd) ∩W 1,∞(Rd) satisﬁes (2.24) ω V + H(x, Vx) −ν ∆V = ω f and (2.25) (ω −(β M + a)) M ≤ω \|fx\|∞+ \|Hx(x, 0)\|∞ 128 for ω > M1. If β = 0, then ϕ(V ) −ϕ(f) ω ≤a ϕ(V ) + b where b = \|Hx(x, 0)\|∞. Step 6: We prove that the solution V ν to (2.24) converges to a viscosity solution V to (2.26) ω V + H(x, Vx) = ω f as ν →0+. First we show that for all φ ∈C2(Ω) if V ν −φ attains a local maximum (minimum, respectively) at x0 ∈Rn, then (2.27) ω (V (x0) −f(x0)) + H(x0, φx(x0)) −ν (∆V )(x0) ≤0 (≥0, respectively). For φ ∈C2(Ω) we assume that V ν −φ attains a local maximum at x0 ∈Rd. Let Ω= {x ∈Rd : \|x −x0\| < 1} and Γ denote its boundary. Then without loss of generality we can assume that V ν −φ attains the unique global maximum 1 at x0 and V ν −φ ≤0 on Γ. In fact we can choose ζ ∈C∞(Ω) such that V ν −(φ −ζ) attains the unique global maximum 1 at x0, V ν −(φ −ζ) ≤0 on Γ and ζx(x0) = 0. Let ψ = sup(0, V ν −φ) ∈W 1,∞ 0 (Ω). Multiplying ψp−1 to (2.26) and integrating over Ω, we obtain ω \|ψ\|p p + (η · (V ν −φ)x, ψp−1) + ν (p −1) (ψx, ψp−2ψx) = −(δ ψ, ψp−1), where η = Z 1 0 Hp(x, φx + θ (V ν −φ)x) dθ and δ = ω (φ −f) + H(x, φx) −ν ∆φ. Since \|(η · ψx, ψp−1)\| ≤ 1 4ν(p −1)\|η\|2 L∞(Ω) \|ψ\|p p + ν(p −1) \|ψ p 2 −1ψx\|2 2 it follows that (ω − \|η\|2 ∞ 4ν(p −1)) \|ψ\|p p ≤−(δψ, ψp−1). Letting p →∞, we can conclude that δ(x0) ≤0 since δ ∈C(Ω). Setting ψ = inf(0, V ν −φ)) and assuming V ν −φ has the unique global minimum −1 at x0 and V ν −φ ≥0 on Γ, the same argument as above shows the second assertion. Next, we show that there exists a subsequence of {V ν} that converges to a viscosity solution V to (2.28) ω V + H(x, Vx) = ω f, i.e., for all φ ∈C1(Ω) if V −φ attains a local maximum at x0 ∈Rd, then (2.29a) ω (V (x0) −f(x0)) + H(x0, φx(x0)) ≤0 and if V −φ attains a local minimum at x0 ∈Rh, then (2.29b) ω (V (x0) −f(x0)) + H(x0, φx(x0)) ≥0. 129 It follows from Step 5 that for some γ > 0 independent of ν \|V ν\|W 1,∞(Ω) ≤γ Thus there exists a subsequence of {V ν} (denoted by the same) that converges weakly star to V in W 1,∞(Rd), and thus the convergence is uniform in Ω. We prove (2.29a) ﬁrst for φ ∈C2(Ω). Assume that for φ ∈C2(Ω) V ν −φ has a local maximum at x0 ∈Ω. We can choose ζ ∈C∞(Ω) such that ζx(x0) = 0 and V ν −(φ −ζ) has a strict local maximum at x0. For ν > 0 suﬃciently small, V ν −(φ −ζ) has a local maximum at some xν ∈Ωand xν →x0 as ν →0+. From (2.27) ω (V ν(xν) −f(xν)) + H(xν, φx(xν)) −ν (∆φ)(xν) ≤0 We conclude (2.29a), since V ν(xν) →V (x0), φx(xν) −ζx(xν) →φx(x0) −ζx(x0) = φx(x0) and ν ∆φ(xν) →0 as ν →0+. For φ ∈C1(Ω) exactly the same argument is applied to the convergent sequence φn ∈C2(Ω) to φ in C1(Ω) to prove (2.29a). Step 7: We show that if V , W ∈Dα are viscosity solutions to ω (V −f)+H(x, Vx) = 0 and ω (W −g) + H(x, Wx) = 0, respectively, then (3.30) (ω −ωα) \|u −v\|X ≤ω \|f −g\|X. For δ > 0 let ψ(x) = 1 p 1 + \|x\|2+δ If u, v ∈Dα then (2.31) lim \|x\|→∞ψ(x)V (x) = lim \|x\|→∞ψ(x)W(x) = 0. We choose a function β ∈C∞(Rd) satisfying 0 ≤β ≤1, β(0) = 1, β(x) = 0 if \|x\| > 1. Let M = max (\|u\|Xδ, \|W\|Xδ). Deﬁne the function Φ : Rn × Rn →R by (2.32) Φ(x, y) = ψ(x)V (x) −ψ(y)W(y) + 3Mβϵ(x −y) where βϵ(x) = β(x ϵ ) for x ∈Rd. Oﬀthe support of βϵ(x −y), Φ ≤2M, while if \|x\| + \|y\| →∞on this support, then \|x\|, \|y\| →∞and thus from (2.30) lim\|x\|+\|y\|→∞Φ ≤3M. We may assume that V (¯ x) −W(¯ x) > 0 for some ¯ x. Then, Φ(¯ x, ¯ x) = ψ(¯ x)(V (¯ x) −W(¯ x)) + 3M βϵ(0) > 3M. Hence Φ attains its maximum value at some point (x0, y0) ∈Rd × Rd. Moreover, \|x0 −y0\| ≤ϵ since βϵ(x0 −y0) > 0. Now x0 is a maximum point of ψ(x) V (x) −ψ(y0)W(y0) −3Mβϵ(x −y0) + Φ(x0, y0) ψ(x) 130 and since ψ > 0 the function x →V (x) −ψ(y0)W(y0) −3Mβϵ(x −y0) + Φ(x0, y0) ψ(x) attains a maximum 0 at x0. Since ψ(y0)W(y0) −3Mβϵ(x0 −y0) + Φ(x0, y0) = ψ(x0)V (x0) and V is a viscosity solution (2.33) ψ(x0)(ω (V (x0) −f(x0)) + H(x0, p)) ≤0, where p = 2 + δ 2 ψ(x0)V (x0)ψ(x0)\|x0\|δ x0 −3Mβ′ ϵ(x0 −y0) ψ(x0) . and we used the fact that (\|x\|2+δ)′ = (2 + δ)\|x\|δx. Moreover since V ∈Dα (2.34) \|p\| ≤α Similarly, the function y →W(y) −ψ(x0)V (x0) + 3Mβϵ(x0 −y) −Φ(x0, y0) ψ(y) attains a minimum 0 at y0 and since W is a viscosity solution (2.35) ψ(y0)(ω (W(y0) −g(y0)) + H(y0, q)) ≥0, where q = 2 + δ 2 ψ(y0)W(y0)ψ(y0)\|y0\|δ y0 −3Mβ′ ϵ(x0 −y0) ψ(y0) and \|q\| ≤α. Thus by (2.33) and (2.35) we have (2.36) ω (ψ(x0)V (x0) −ψ(y0)W(y0)) ≤ψ(y0) H(y0, q) −ψ(x0) H(x0, p) + ω(ψ(x0)f(x0) −ψ(y0)g(y0)) Since Φ(x0, y0) ≥Φ(¯ x, ¯ x) we have ψ(x0)V (x0) −ψ(y0)W(y0) ≥ψ(¯ x)(V (¯ x) −W(¯ x)) + 3M (1 −βϵ(x0 −y0)) and thus (ψ(x0)−ψ(y0)) V (x0)+ψ(y0)(V (x0)−W(y0)) ≥ψ(¯ x)(V (¯ x)−W(¯ x))+3M (1−βϵ(x0−y0)). Since (2.37) \|(ψ(x0)−ψ(y0)) V (x0)\| = ψ(y0)ψ(x0)\|V (x0)\|( q 1 + \|y0\|2+δ− q 1 + \|x0\|2+δ) ≤const \|x0−y0\|, it follows that V (x0) ≥W(y0) for suﬃciently small ϵ > 0. Note that ψ(y0)H(y0, q) −ψ(x0)H(x0, p) = (ψ(y0) −ψ(x0))H(x0, p) +ψ(y0)(H(y0, p) −H(x0, p)) + ψ(y0)(H(y0, q) −H(y0, p)). 131 From (2.36)–(2.37) we have that (2.38) ω (ψ(x0)V (x0) −ψ(y0)W(y0) −(ψ(x0)f(x0) −ψ(y0)g(y0))) ≤O(ϵ) + ψ(y0) (c1(y0), p −q) + c2 \|p −q\|). where O(ϵ) →0 as ϵ →0. Now we evaluate p −q, i.e., p −q = 2 + δ 2 (ψ(x0)V (x0) −ψ(y0)W(y0))ψ(y0)\|y0\|δ y0 +2 + δ 2 ψ(x0)V (x0)(ψ(x0) \|x0\|δx0 −ψ(y0)\|y0\|δy0) +3Mβ′ ϵ(x0 −y0) ( q 1 + \|x0\|2+δ − q 1 + \|y0\|2+δ). Since \|ψ(x0)V (x0)ψ(x0)\|x0\|δ x0\| ≤\|u\|X \|x0\|δp 1 + \|x0\|2 1 + \|x0\|2+δ \|x0\| ≤M3 for some M3 > 0, it follows from (2.34) that 3\|β′ ϵ(x0 −y0) q 1 + \|x0\|2+δ\| ≤M4 for some M4 > 0. Thus, \|3Mβ′ ϵ(x0 −y0) \| q 1 + \|x0\|2+δ − q 1 + \|y0\|2+δ\| ≤const (2 + δ)MM4\|x0 −y0\|. and therefore 2 + δ 2 ψ(x0)V (x0)(ψ(x0)\|x0\|δx0−ψ(y0)\|y0\|δy0)+3Mβϵ(x0−y0) (\|x0\|2+δ−\|y0\|2+δ) = O(ϵ). In the right-hand side of (2.38) we have ψ(y0)((c1(y0), p −q) + c2 \|p −q\|) ≤O(ϵ) + 2 + δ 2 β \|y0\|2+δ + c2α \|y0\|1+δ 1 + \|y0\|2+δ (ψ(x0)u(x0) −ψ(y0)v(y0)). Hence from (2.38) we conclude (2.39) ωy0 (ψ(x0)u(x0) −ψ(y0)v(y0)) ≤ψ(x0)f(x0) −ψ(y0)g(y0) + O(ϵ) where ωδ = sup y0 (ω −2 + δ 2 β \|y0\|2+δ + c2α \|y0\|1+δ 1 + \|y0\|2+δ ). Assume that ω > λα. For x ∈Rd we have ψ(x)(u(x) −v(x)) + 3M = Φ(x, x) ≤Φ(x0, y0) ≤ψ(x0)u(x0) −ψ(y0)v(y0) + 3M 132 and so by (2.39) ωδ sup Rd ψ(x)(u(x) −v(x))+ ≤ ω (ψ(x0)u(x0) −ψ(y0)v(y0)) ≤ψ(x0)f(x0) −ψ(y0)g(y0) + O(ϵ) ≤ sup Rd ψ(f −g)+ + \|ψ(x0)g(x0) −ψ(y0)g(y0)\| + O(ϵ) ≤ sup Rd ψ(f −g)+ + ωψg(ϵ) + O(ϵ) where ωψg(·) is the modulus of continuity of ψg. Letting ϵ →0, we obtain (3.40) ωδ sup Rd ψ(x)(u(x) −v(x))+ ≤sup Rn ψ(f −g)+ Since \|ψ\|Xδ →\|ψ\|X as δ →0+ for ψ ∈X we obtain (2.30) by taking limit δ →0+ in (2.40). Example (Plastic equations) Consider the visco-plastic equation of the form vt + div(σ) = 0 where the stress σ(ϵ) = sigmat minimizes h(σ) −ϵ : σ and the strain ϵ is given by ϵi,j = 1 2( ∂ ∂xi vj + ∂ ∂xi vj That is, σ∂h∗(ϵ) where h∗is the conjugate of h h∗(ϵ) = sup σ∈C {ϵ : σ −h(σ)} For the case of the linear elastic system σ11 6.1 Evolution equations In this section we consider the evolution equation of the form d dtx(t) ∈A(t)u(t) (6.24) in a Banach space X ∩D, where D is a closed set. We assume the dissipativity: there exist a constant ω = ωD and continuous functions f : [0, T] →X and L : R+ →R+ independent of t, s ∈[0, T] such that (1 −λω) \|x1 −x2\| ≤\|x1 −x2 −λ (y1 −y2)\| + λ \|f(t) −f(s)\|L(x2)\|K(\|y2\|) 133 for all x1 ∈D ∩dom (A(t)) x2 ∈D ∩dom (A(s))and y1 ∈A(t)x1, y2 ∈A(s)x2, and A(t), t in[0, T] is m-dissipative and Jλ(t) = (λ I −A)−1 : D →D ∩dom(A(tk)). Thus, one constructs the mild solution as u(t) = lim λ→0+ Π[t/λ] k=1 Jλ(tk)u0, tk = kλ. Theorem 1.1 Let (A(t), X) satisfy (H.1)–(H.4). Then, U(t, s) = lim λ→0+ Π [ t−s λ ] i=1 Jλ(s + iλ)x exists for x ∈dom (A(0)) and 0 ≤s ≤t ≤T. The U(t, s) for 0 ≤s ≤t ≤T deﬁnes an evolution operator on dom A(0) and moreover satisﬁes \|U(t, s)x −U(t, s)y\| ≤eω (t−s) \|x −y\| for 0 ≤s ≤t ≤T and x, y ∈dom (A(0)). Proof: Let hi = λ and ti = s + iλ in (). Then xi = Jλ(ti)xi−1, where we dropped the superscript λ and xm = (Πm i=1Jλ(ti))x. Thus \|xi\| ≤(1 −ωλ)−m ≤e2ω(T−s) \|x\| = M1 for 0 < mλ ≤T −s. Let ˆ xj is the approximation solution corresponding to the stepsize ˆ h −j = µ and ˆ tj = s + jµ. Deﬁne am,n = \|xm −ˆ xn\|. We ﬁrst evaluate a0,n, am,0. am,0 = \|xm −x\| ≤ m X k=1 \|(Πm i=kJλ(ti))x −(Πm i=k+1Jλ(ti))x\| ≤ m X k=1 (1 −ωλ)−(m−k+1)λ \|\|\|A(ti)x\|\|\| ≤e2ω(T−s)mλ M(x). where M(x) = supt∈[0,T] \|\|\|A(t)x\|\|\|. Similarly, we have a0,n ≤e2ω(T−s)nµ M(x). Next, we establish the recursive formula for ai,j. For λ ≥µ > 0 ai,j = \|xi −ˆ xj\| ≤\|Jλ(ti)xi−1 −Jµ(ˆ tj)ˆ xj−1\| ≤\|Jλ(ti)xi−1 −Jµ(ti)ˆ xj−1\| + \|Jµ(ti)ˆ xj−1 −Jµ(ˆ tj)ˆ xj−1\| From Theorem 1.10 \|Jλ(ti)xi−1 −Jµ(ti)ˆ xj−1\| = \|Jµ(µ λxi−1 + λ −µ λ Jλ(ti)xi−1) −Jµ(ti)ˆ xj−1\| ≤(1 −ω µ)−1(µ λ\|xi−1 −ˆ xj−1\| + λ −µ λ \|xi −ˆ xj−1\|) 134 Hence for i, j ≥1 we have ai,j ≤(1 −ωµ)−1(α ai−1,j−1 + β ai,j−1) + bi,j where α = µ λ, and β = λ −µ λ and bi,j = \|Jµ(ti)ˆ xj−1 −Jµ(ˆ tj)ˆ xj−1\| ≤µ \|f(ti) −f(ˆ tj)\|L(\|ˆ xj\|)K(\|Aµ(ˆ tj)ˆ xj−1\|). We show that if yi = Aλ(ti)xi−1, then there exists a constant M2 = M2(x0, T, \|\|A(s)x0\|\|) such that \|yi\| ≤M2 Since xi = Jλ(ti)xi−1 and Aλ(ti)xi−1 ∈A(ti)xi, from (H.4) and (1.2)–(1.3) we have \|\|\|A(ti)xi\|\|\| = \|Aλ(ti)xi−1\| ≤\|Aλ(ti−1)xi−1\| + \|f(ti) −f(ti−1)\|L(M1)(1 + \|Ah(ti−1)xi−1\|)) ≤(1 −λ)−1(\|\|\|A(ti−1)xi−1\|\|\| + \|f(ti) −f(ti−1)\|L(M1)(1 + \|\|\|A(ti−1)xi−1\|\|\|) If we deﬁne ai = \|\|\|A(ti)xi\|\|\|, then (1 −ω λ) ai ≤ai−1 + bi(1 + ai−1). where bi = L(M1)\|f(ti) −f(ti−1)\|. Thus, it follows from the proof of Lemma 2.4 that \|\|\|A(ti)xi\|\|\| ≤M2, for some constant M2 = M(x0, T). Since (1.∗) \|yi\| ≤(1 −ωλ)−1(\|\|\|A(ti−1)xi−1\|\|\| + \|f(ti) −f(ti−1)\|L(M1)(1 + \|\|\|A(ti−1)xi−1\|\|\|), thus \|yi\| is uniformly bounded. It follows from Theorem and [] that am,n ≤e2ω(T−s)M(x) [((nµ −mλ)2 + nµ(λ −µ))1/2 + ((nµ −mλ)2 + mλ(λ −µ))1/2] +(1 −ωµ)−n n−1 X j=0 min(m−1,j) X i=0 βj−1αi j i bm−i,n−j, where mλ, nµ ≤T −s. et ρ be the modulus of continuity of f on [0, T], i.e., ρ(r) = sup {\|f(t) −f(τ) : 0 ≤t, τ ≤T and \|t −τ\| ≤r} Then ρ is is nondecreasing and subadditive; i.e., ρ(r1+r2) ≤ρ(r1)+ρ(r2) for r1, r2 ≥0. Thus J = n−1 X j=0 min(m−1,j) X i=0 βj−1αi j i bm−i,n−j ≤Cµ  n ρ(\|nµ −mλ)\|) + n−1 X j=0 X i=0 (m −1)j βj−1αi j i ρ(\|jµ −iλ)\|)  . 135 where we used () with C = L(M1)K(M2), the subadditivity of of ρ and the estimate min(m−1,j) X i=0 βj−1αi j i ≤1. Next, let δ > 0, be given and write n−1 X j=0 min(m−1,j) X i=0 βj−1αi j i ρ(\|jµ −iλ)\|) = I1 + I2 where I1 is the sum over indecies such that \|jµ −iλ\| < δ, while I2 is the sum over indecies satisfying \|jµ −iλ\| ≥δ. Clearly I1 ≤nρ(δ), but I2 ≤ρ(T) n−1 X j=0 min(m−1,j) X i=0 βj−1αi j i \|jµ −iλ\|2 δ2 = ρ(T) δ2 n(n−1)(λµ−µ2) ≤ρ(T)n2 δ2 µ(λ−µ) Therefore, J ≤Cnµ ρ(\|nµ −mλ\| + ρ(δ) + ρ(T) δ2 nµ(λ −µ) . Combining ()–(), we obtain am,n ≤e2ω(T−s)M(x) [((nµ −mλ)2 + nµ(λ −µ))1/2 + ((nµ −mλ)2 + mλ(λ −µ))1/2] +e2ω(T−s)C nµ ρ(\|nµ −mλ\| + ρ(δ) + ρ(T) δ2 nµ(λ −µ) . Now, we can choose, e.g, δ2 = √λ −µ and it follows from () that an,m as function of m, n and λ, µ, tends to zero as \|nµ −mλ\| →0 and n, m →∞, subject to 0 < nµ, mλ ≤T −s, and the convergence is uniform in s. Theorem 1.1 Let (A(t), X) satisfy (H.1)–(H.4). Then, U(t, s) = lim λ→0+ Π [ t−s λ ] i=1 Jλ(s + iλ)x exists for x ∈dom (A(0)) and 0 ≤s ≤t ≤T. The U(t, s) for 0 ≤s ≤t ≤T deﬁnes an evolution operator on A(0) and moreover satisﬁes \|U(t, s)x −U(t, s)y\| ≤eω (t−s) \|x −y\| for 0 ≤s ≤t ≤T and x, y ∈dom (A(0)). Moreover, we have the following lemma. Lemma 1.2 If x ∈ˆ D, then there exists a constant L such that \|U(s + r, s)x −U(ˆ s + r, ˆ s)x\| ≤L ρ(\|s −ˆ s\|) for r ≥0 and s, ˆ s, s + r, ˆ s + r ≤T. Proof: Let ak = \|xk −ˆ xk\| where xk = Πk i=1(I −λ A(s + i λ))−1x and ˆ xk = Πk i=1(I −λ A(ˆ s + i λ))−1x. 136 Then, for λ = r n ak = \|Jλ(s + k λ)xk−1 −Jλ(ˆ s + k λ)ˆ xk−1\| ≤\|Jλ(s + k λ)xk−1 −Jλ(ˆ s + k λ)xk−1\| + \|Jλ(ˆ s + k λ)xk−1 −Jλ(ˆ s + k λ)ˆ xk−1\| ≤λCρ (\|s −ˆ s\|) + (1 −λω)−1 ak−1 for some C. Thus, we obtain \|ak\| ≤e2ωr −1 2ω ρ(\|s −ˆ s\|), and letting λ →0 we obtain the desired result. □ 6.2 Applications We consider the Cauchy problem of the form (1.1) d dtu(t) = A(t, u(t))u(t) u(s) = u0 where A(t, u), t ∈[0, T] is a maximal dissipative linear operator in a Banach space X for each u belonging to D. Deﬁne the nonlinear operator A(t) in X by A(t)u = A(t, u)u. We assume that dom (A(t, u)) is independent of u ∈D and for each α ∈R there exists an ωα ∈R such that (1.2) ⟨A(t, u)x1−A(s, u)x2, x1−x2⟩−≤ωα \|x1−x2\|+\|f(t)−f(s)\|L(\|x2\|)K(\|A(s, u)x2\|) for u ∈Dα and x1 ∈dom (A(t, u)), x2 ∈dom (A(s, u)). Moreover, A(t) satisﬁes (C.2), i.e., (1.3) (1−λ ωα) \|u1−u2\| ≤\|(u1−u2)−λ (A(t)u1−A(s)u2)\|+λ \|f(t)−f(s)\|L(\|u2\|)K(\|A(s)u2\|) for u1 ∈dom (A(t)) ∩Dα, u2 ∈dom (A(s)) ∩Dα. We consider the ﬁnite diﬀerence approximation of (1.1); for suﬃciently small λ > 0 there exists a family {uλ i } in D such that (1.4) uλ i −uλ i−1 hλ i = A(tλ i , uλ i−1)uλ i with uλ 0 = u0 ϕ(uλ i ) −ϕ(uλ i−1) hλ i ≤a ϕ(uλ i ) + b. Then, it follows from Theorems 2.5–2.7 that if the sequence (1.5) ϵλ i = A(tλ i , uλ i )uλ i −A(tλ i , uλ i−1)uλ i satisfy PNλ i=1 hλ i \|ϵλ i \| →0 as λ →0+, then (1.1) has the unique integrable solution. We have the following theorem. Theorem 5.0 Assume (1.2) holds and for each α ∈R there exist a cα ≥0 such that (1.6) \|(A(t, u)u −A(t, v)u\| ≤cα \|u −v\|, for u, v ∈dom (A(t)) ∩Dα. 137 Let f be of bounded variation and u0 ∈dom (A(s)) ∩D. Then (1.3) and (1.5) are sat-isﬁed and thus (1.1) has a unique integrable solution u and limλ→0+ uλ = u uniformly on [s, T]. Proof: If yλ i = (hλ i )−1(uλ i −uλ i−1), then we have yλ i+1 −yλ i = A(tλ i+1, uλ i )uλ i+1 −A(tλ i , uλ i )uλ i +A(tλ i , uλ i )uλ i −A(tλ i , uλ i−1)uλ i and thus from (1.2) and (1.6) (1 −λωα) \|yλ i+1\| ≤(1 + λcα) \|yλ i \| + \|f(tλ i+1) −f(tλ i )\|L(\|uλ i \|)K(\|yλ i \|) with yλ 0 = A(s, u0)u0 = A(s)u0. Thus by the same arguments as in the proof of Lemma 2.4, we obtain \|yλ i \| ≤M for some constant M and therefore from (1.6) PNλ i=1 \|ϵλ i \|λ ≤MT λ →0 as λ →0+. We also note that (1.3) follows from (1.2) and (1.6). 6.3 Navier Stokes Equation, Revisited We consider the incompressible Navier-Stokes equations (5.1). We use exactly the same notation as in Section.. Deﬁne the evolution operator A(t, u) by w = A(t, u)v ∈H, where (5.1) (w, φ)H + ν σ(u, φ) + b(u, v, φ) −(f(t), φ)H = 0 for φ ∈V , with dom (A(t)) = dom (A0). We let D = V and deﬁne the functional as below. Theorem 5.1 The evolution operator (A(t, u), ϕ, D, H) deﬁned above satisﬁes the conditions (1.2)–(1.5) with g(r) = b, a suitably chosen positive constant. Proof: For u1, u2 ∈dom (A0). (A(t, v)u1 −A(s, v)u2, u1 −u2)H + ν \|u1 −u2\|2 V = (f(t) −f(s), u1 −u2) since b(u, v1 −v2, v1 −v2) = 0, which implies (1.2). The existence of uδ ∈V for the equation: δ−1(uδ −u0) = A(t, u0)uδ for u0 ∈V , δ > 0 and t ∈[0, T] follows from Step 2. of Section NS and we have (5.2) \|uδ\|2 H −\|u0\|2 H δ + ν \|uδ\|2 V ≤1 ν \|f(t)\|2 V ∗. We also have the estimate of \|uδ\|V . (5.3) 1 2δ (\|uδ\|2 V −\|u0\|2 V ) + ν 2 \|A0uδ\|2 ≤27M4 1 4ν3 \|u0\|2 H\|u0\|2 V \|uδ\|2 V + 1 ν \|Pf(t)\|2 H. for the two dimensional case. Multiplying the both side of (5.2) by \|uδ\|2 H + \|u0\|2 H, we have \|uδ\|4 H −\|u0\|4 H δ + ν (\|uδ\|2 H + \|u0\|2 H)\|uδ\|2 V ≤1 ν (\|uδ\|2 H + \|u0\|2 H) \|f(t)\|2 V ∗ 138 Since s →log(1 + s) is concave c0ν4 log(1 + \|uδ\|2 V ) −log(1 + \|u0\|2 V ) δ ≤c0ν4δ−1 \|uδ\|2 V −\|u0\|2 V 1 + \|u0\|2 V ≤2ν \|u0\|2 H\|u0\|2 V \|uδ\|2 V + 2ν−1\|Pf(t)\|2 H 1 + \|u0\|2 V where we set c0 = 4 27M4 1 . Thus, if we deﬁne ϕ(u) = 2 \|u\|2 H + c0ν4 log(1 + \|u\|2 V ) then for every δ > 0 ϕ(uδ) −ϕ(u0) δ ≤b for some constant b ≥0, since \|uλ i \|H is uniformly bounded. For (1.5) if yλ i = (hλ i )−1(uλ i − uλ i−1), then (yλ i+1 −yλ i , yλ i+1) + νhλ i+1 \|yλ i+1\|2 + hλ i b(yλ i , uλ i , yλ i+1) + (f(tλ i+1) = f(tλ i ), yλ i+1) Note that (5.4) \|b(yλ i , uλ i , yλ i+1)\| ≤M1\|uλ i \|V \|yλ i \| 1 2 H\|yλ i \| 1 2 V \|yλ i+1\| 1 2 H\|yλ i+1\| 1 2 V ≤ν 2 \|yλ i+1\|2 V + ν 2 \|yλ i+1\|2 V + M2 1 \|uλ i \|2 V 8ν \|yλ i+1\|2 V + M2 1 \|uλ i \|2 V 8ν \|yλ i+1\|2 V . For simplicity of our discussions we assume hλ i = h Then we have (5.5) (1−h ωα) \|yλ i+1\|2 H+hν \|yλ i+1\|2 V + ≤(1+h ωα) \|yλ i+1\|2 H+ν \|yλ i+1\|2 V +\|f(tλ i+1−f(tλ i )\|H\|yλ i+1\|H which implies that \|yλ i \|H ≤M for some constant independent of h > 0. Note that \|A(t, u)u −A(t, v)u\|H = sup \|phi\|H≤1 \|b(u, u, φ) −b(v, u, φ)\| ≤\|u −v\| 1 2 V \|u −v\| 1 2 H\|u\| 1 2 V \|A0u\| 1 2 H and from (5.3) ν 2 N X i=1 hλ i \|A0uλ i \|2 H ≤C for some constant C independent of N and hλ i . Since \|yλ i \|H ≤M, \|uλ i −uλ i−1\| ≤M hλ i . Thus (1.5) holds and (1.1) has a unique integrable solution u and limλ→0+ uλ = u uniformly on [s, T]. Next, we consider the three dimensional problem (d = 3). We show that there exists a locally deﬁned solution and a global solution exists when the data (u0, f(·)) are small. We have the corresponding estimate of (5.3): (5.6) 1 2δ (\|uδ\|2 V −\|u0\|2 V ) + ν 2 \|A0uδ\|2 ≤27M4 2 4ν3 \|u0\|4 V \|uδ\|2 V + 1 ν \|Pf(t)\|2 H. 139 We deﬁne the functional ϕ by ϕ(u) = 1 −(1 + \|u\|2 V )−1 Since s →1 −(1 + s)−1 is concave and \|uλ i \|H is uniformly bounded, it follows from (5.2) and (5.6) that for every δ > 0 (5.6) ϕ(uδ) −ϕ(u0) δ ≤(27M4 2 4ν3 \|u0\|4 V \|uδ\|2 V + 1 ν \|Pf(t)\|2 H)(1 + \|u0\|2)−2. Thus ϕ(uλ i ) −ϕ(uλ i−1) hλ i ≤(27M4 2 4ν3 \|u0\|4 V \|uδ\|2 V + 1 ν \|Pf(t)\|2 H)(1 + \|u0\|2)−2 ≤bi where bi = ( 27M4 2 4ν3 \|uλ i \|2 V + c1 ν and from (5.3) N X i=1 hλ i \|uλ i \|2 V ≤1 ν (\|u0\|2 H + c1T ν ). The estimate (5.4) is replaced by \|b(yλ i , uλ i , yλ i+1)\| ≤M1\|uλ i \|V \|yλ i \| 1 4 H\|yλ i \| 3 4 V \|yλ i+1\| 1 4 H\|yλ i+1\| 3 4 V ≤ν 2 \|yλ i+1\|2 V + ν 2 \|yλ i+1\|2 V + 3M2 1 \|uλ i \|2 V 32ν \|yλ i+1\|2 V + 3M2 1 \|uλ i \|2 V 32ν \|yλ i+1\|2 V . Thus, if hλ i = h then we obtain (5.5) and thus \|yλ i \|H ≤M for some constant independent of h > 0 and (1.5) holds. □ 6.4 Approximation theory In this section we discuss the approximation theory of the mild solution to (6.24). Consider an evolution equation in Xn: (3.1) d dtun(t) ∈An(t)un(t) t > s; un(s) = x where Xn is a linear closed subspace of X. We consider the family of approximating sequences (An(t), dom (An(t)), ϕ, D) on Xn satisfying the uniform dissipativity: there exist constant ω = ωα continous functions f : [0, T] →X and L : R+ →R+ indepen-dent of t, s ∈[0, T] and n such that (3.2) (1 −λωα) \|x1 −x2\| ≤\|x1 −x2 −λ (y1 −y2)\| + λ \|f(t) −f(s)\|L(x2)\|K(\|y2\|) for all x1 ∈Dα ∩dom (An(t)) x2 ∈Dα ∩dom (An(s))and y1 ∈An(t)x1, y2 ∈An(s)x2, and the consistency: (3.3) for β > 0, t ∈[0, T], and [x, y] ∈A(t) with x ∈Dβ, there exists [xn, yn] ∈An(t) with xn ∈Dα(β) such that lim \|xn −x\| + \|yn −y\| = 0 as n →∞. 140 where α(β) ≥β and α : R+ →R+ is an increasing function. Lemma 3.1 Let (A(t), X) satisfy (A.1)−(A.2) and either (R)−(C.1) or (R.1)−(C.2) on [0, T]. For each λ > 0, we assume that (tλ i , xλ i , yλ i , ϵλ i ) satisﬁes yλ j = xλ i −xλ i−1 tλ i −tλ i−1 −ϵλ i ∈A(tλ i )xλ i with tλ 0 = s and xλ 0 = x and xλ ∈Dα for 0 ≤i ≤Nλ. For the case of (R.1) −(C.2) we assume that x ∈D × dom (A(s)), f is of bounded variation, and \|ylambdai\| is uniformly bounded in 1 ≤i ≤Nλ and λ > 0. Then the step function uλ(t; s, x) deﬁned by uλ(t, s, x) = xλ i on (tλ i−1, tλ i ], satisﬁes \|uλ(t; s, x) −u(t; s, x)\| ≤e2ω(2t+dλ) (2\|x −u\| + dλ (\|v\| + ˜ Mρ(T)) + ˜ M(T −s)(c−1ρ(T)dλ + ρ(δ) + δλ). Proof: It follows from Lemms 2.6–2.7 that there exists a DS-approximation sequence (tµ j , xµ j , yµ j , ϵµ j ) as deﬁned in (2.12). For the case of (C.2) −(R.1), from Lemma 2.4 we have \|yµ j \| ≤˜ K for 1 ≤j ≤Nµ uniformly in µ. It thus follows from Theorem 2.5 that there exists a constnat ˜ M such that \|uλ(t; s, x) −uµ(t; s, x)\| ≤e2ω(2t+dλ) (2\|x −u\| + dλ (\|v\| + ˜ Mρ(T)) + ˜ M(T −s)(c−1ρ(T)dλ + ρ(δ) + δλ −δµ). Theorem 3.2 Let (An(t), dom (An(t)), ϕ, D) be approximating sequences satisfying (R) or (R.1) (resp.) and (3.2)–(3.3) and we assume (A(t), dom (A(t)), D, ϕ) satisﬁes (A.1)–(A.2) and (R) or (R.1) (resp.). Then for every x ∈D ∩dom (A(s)) and xn ∈ D ∩Xn such that lim xn = x as n →∞we have lim \|un(t; s, xn) −u(t; s, x)\| = 0, as n →∞ uniformly on [s, T], where u(t; s, x) and un(t; s, x) is the unique mild solution to (2.1) and (3.1), respectively. Proof: Let [xi, yi] ∈A(ti) and xi ∈Dα for i = 1, 2. From (3.3) we can choose [xn i , yn i ] ∈An(ti), i = 1, 2 with xi ∈Dα′ such that \|xn i −xi\| + \|yn i −yi\| →0 as n →∞ for i = 1, 2. Thus, letting n →∞in (3.2), we obtain (C.1) or (C.2). Let (tλ i , xλ i , yλ i ) be a DS–approximation sequence of (2.1). We assume that there exits a β > 0 such that xλ i ∈Dβ for all λ and 1 ≤i ≤Nλ. By the consistency (3.3) for any ϵ > 0 there exists an integer n = n(ϵ) such that for n ≥n(ϵ) (3.4) \|xλ,n i −xλ i \| ≤ϵ, \|yλ,n i −yλ i \| ≤ϵ and xλ,n i ∈Dα for 1 ≤i ≤Nλ. (3.5) PNλ i=1 \|xλ,n i −xλ,n i−1 −hλ i yλ,n i \| ≤PNλ i=1 \|xλ i −xλ i−1 −hλ i yλ i \| + (Nλ + T) ϵ + \|xn −x\| = δn,λ,ϵ 141 By Theorems 2.5 and 2.8 that (3.6) \|uλ,n(t, xn) −un(t; xn)\| ≤e2ω(2t+dλ) (2\|xn −un\| + dλ (\|vn\| + ˜ Mρ(T)) + ˜ M(T −s)(c−1ρ(T)dλ + ρ(δ) + δλ,n,ϵ) for all 0 < c < δ < T and [un, vn] ∈An(s) with un ∈Dα. From the deﬁnition of function uλ, uλ,n and (3.4) \|uλ(t; s, x) −uλ,n(t; s, xn)\| ≤ϵ, t ∈(s, T], n ≥n(ϵ) Thus, we have \|un(t; s, xn) −u(t; , s, x)\| ≤e2ω(2t+dλ) (2\|xn −un\| + 2\|x −u\| + dλ (\|vn\| + \|v\| + 2 ˜ Mρ(T)) +2 ˜ M(T −s)(c−1ρ(T)dλ + ρ(δ)) + δλ,n,ϵ) + ϵ for t ∈(s, T], n ≥n(ϵ), where [u, v] ∈A(s) with u ∈Dβ. From the consistency (3.3) we can take [un, vn] ∈An(s) such that un →u and vn →v as n →∞. It thus follows that limn→∞\|un(t; s, xn) −u(t; s, x)\| = e2ω(2t+dλ) (4\|x −u\| + 2dλ (\|v\| + ˜ Mρ(T)) +2 ˜ M(T −s)(c−1ρ(T)dλ + ρ(δ)) + δλ,ϵ) + ϵ for t ∈[s, T], where δλ,ϵ = (Nλ +T)ϵ. Now, letting ϵ →0+ and then λ →0+, we obtain lim n→∞\|un(t; s, xn) −u(t; s, x)\| = e4ωt (4\|x −u\| + ρ(δ)). Since u ∈Dβ∩dom (A(s)) and δ > 0 are arbitrary it follows that limn→∞\|un(t; s, xn)− u(t; s, x)\| = 0 uniformly on [s, T]. □ The following theorems give the equivalent characterization of the consistency con-dition (3.3). Theorem 3.2 Let (An, Xn), n ≥1 and (A, X) be dissipative operators satisfying the condition dom (An) ⊂R(I −λ An) and dom (A) ⊂R(I −λ A) and set Jn λ = (I −λ An)−1 and Jλ = (I −λ A)−1 for λ > 0. Also, let B be the operator that has {[Jλx, λ−1(Jλx −x)] : x ∈dom (A), 0 < λ < ω−1} as its graph. Then the following statements (i) and (ii) are equivalent. (i) B ⊂limn→∞An (i.e., for all [x, y] ∈B there exists a sequence {(xn, yn)}\| such that [xn, yn] ∈An and lim \|xn −x\| + \|yn −y\| →0 as n →∞.) (ii) dom (A) ⊂limn→∞dom (An) (equivalently, for all x ∈dom (A) there exists a sequence {xn} such that xn ∈dom (An) and lim \|xn −x\| →0 as n →∞.) and for all xn ∈dom (An) and x ∈dom (A) with x = limn→∞xn, we have limn→∞Jn λ xn →Jλx for each 0 < λ < ω−1. In particular, if dom (A) ⊂dom (An) for all n, then the above are equivalent to (iii) limn→∞Jn λ x →Jλx for all 0 < λ < ω−1 and x ∈dom (A). Proof: (i) →(ii). Assume (i) holds. Then it is easy to prove that dom (B) ⊂ limn→∞dom (An). Since Jλx →x as lambda →0+ for all x ∈dom (A), it follows that 142 dom (B) = dom (A). Thus, the ﬁrst assertion of (ii) holds. Next, we let xn ∈dom (An), x ∈dom (A) and limn→∞xn = x. From (i), we have I−λ B ⊂limn→∞I−λ An. Thus, R(I −λ B) ⊂limn→∞R(I −λ An). Since x = Jλx −λ{λ−1(Jλx −x)} ∈(I −λ B)Jλx for x ∈dom (A), we have dom (A) ⊂R(I −λ B). From (i) we can choose [un, vn] ∈An such that lim \|un−Jλx)\|+\|vn−λ−1(Jλx−x)\| →0 as n →∞. If zn = Jn λ xn, then there exists a [zn, yn] ∈An such that xn = zn −λyn. It thus follows from the dissipativity of An that \|un −zn\| ≤\|un −zn −λ(vn −yn)\| = \|un −λvn −xn\| →\|Jλx −(Jλx −x) −x\| = 0 as n →∞. Hence, we obtain lim Jλxn = lim zn = lim un = Jλx as n →∞. (ii) →(i). If [u, v] ∈B, then from the deﬁnition of B, there exist x ∈dom (A) and 0 < λ < ω−1 such that u = Jλx and v = λ−1(Jλx −x). Since x ∈dom (A) ⊂ limn→∞dom (An) we can choose xn ∈dom (An) such that lim xn = x. Thus from the assumption, we have lim Jn λ xn = jλx = u and lim λ−1(Jn λ xn −xn) = λ−1(Jλx−x) = v as n →∞. Hence we obtain [u, v] ∈limn→∞An and thus B ⊂limn→∞An. Finally, if dom (A) ⊂dom (An), then (ii) →(iii) is obvious. Conversely, if (iii) holds, then \|Jn λ xn −Jλx\| ≤\|xn −x\| + \|Jn λ x −Jλx\| →0 as n →∞ when lim xn = x, and thus (ii) holds. □ Theorem 3.3 Let (An, Xn), n ≥1 and (A, X) be m-dissipative operators, i.e., Xn = R(I −λ An) and X = R(I −λ A) and set Jn λ = (I −λ An)−1 and Jλ = (I −λ A)−1 for 0 < λ < ω−1. Then the following statements are equivalent. (i) A = limn→∞An (ii) A ⊂limn→∞An (iii) For all xn, x ∈X such that limn→∞xn = x, we have limn→∞Jn λ xn →Jλx for each 0 < λ < ω−1. (iv) For all x ∈X and 0 < λ < ω−1 limn→∞Jn λ x →Jλx. (v) For some 0 < λ0 < ω−1, limn→∞Jn λ x →Jλx for all x ∈X. Proof: (i) →(ii) and (iv) →(v) are obvious. (ii) →(iii) follows from the proof of (i) →(ii) in Theorem 3.2. (v) →(ii). If [x, y] ∈A, then from (v), lim Jn λ0(x −λ0 y) = Jλ0(x −λ0 y) = x and lim λ−1 0 (Jn λ0(x−λ0 y)−(x−λ0 y)) = y as n →∞. Since λ−1 0 (Jn λ0(x−λ0 y)−(x−λ0 y)) ∈ AnJn λ0(x −λ0 y), [x, y] ∈limn→∞An and thus (ii) holds. (ii) →(i). It suﬃces to show that limn→∞An ⊂A. If [x, y] ∈limn→∞An then there exists a [xn, yn] ∈An such that lim \|xn −x\| + \|yn −y\| = 0 as n →∞. Since xn −λ yn →x −λ y and n →∞, it follows from (iii) that xn = Jn λ (xn −λ yn) → Jλ(x −λ y) as n →∞and hence x = Jλ(x −λ y) ∈dom (A). But, since λ−1(x −(x −λ y)) = λ−1(Jλ(x −λ y) −(x −λ y)) ∈AJλ(x −λ y) = Ax we have [x, y] ∈A and thus (i) holds. □ 143 The following corollary is an immediate consequence of Theorems 3.1 and 3.3. Corollary 3.4 Let (An(t), Xn) and (A(t), X) be m-dissipative operators for t ∈[0, T], (i.e., Xn = R(I −λ An(t)) and X = R(I −λ A(t)) and (C.2) is satisﬁed), and let Un(t, s), U(t, s) be the nonlinear semigroups generated by An(t), A(t), respectively. We set Jn λ (t) = (I −λ An(t))−1 and Jλ(t) = (I −λ A(t))−1 for 0 < λ < ω−1. Then, if Jn λ0xn →Jλ0x as n →∞ for all sequence {xn} satisfying xn →x as n →∞, then \|Un(t, s)xn −U(t, s)x\| →0 as n →∞ where the convergence is uniform on arbitrary bounded subintervals. The next corollary is an extension of the Trotter-Kato theorem on the convergence of linear semigroups to the nonlinear evolution operators. Corollary 3.5 Let (An(t), X) be m-dissipative operators and {Un(t, s), t ≥s ≥0} be the semigroups generated by An(t). We set Jn λ (t) = (I −λ An(t))−1. For some )λ0 < ω−1, we assume that there exists lim Jn λ0(t)x exits as n →∞for all x ∈X and t ≥0 and denote the limit by Jλ0x. Then we have (i) The operator A(t) deﬁned by the graph {[Jλ0(t)x, λ−1 0 (Jλ0(t)x −x)] : x ∈X} is an m-dissipative operator. Hence (A(t), X) generates a nonlinear semigroup U(t, s) on X. (ii) For every x ∈R(Jλ0(s)) (= dom (A(s))), there exist a xn ∈dom (An(s)) such that lim xn = x and lim U(t, s)xn = U(t, s)x for t ≥s ≥0 as n →∞. Moreover, the above convergence holds for every xn ∈dom (An(s)) = X satisfying lim xn = x, and the convergence is uniform on arbitrary bounded intervals. Proof: (i) Let [ui, vi] ∈A(ti), i = 1, 2. Then from the deﬁnition of A(t) we have ui = Jλ0(ti)xi and vi = λ−1 0 (Jλ0(ti)xi −xi) for xi, i = 1, 2. By the assumption lim Jn λ0(ti)xi = Jλ0(ti)xi lim λ−1 0 (Jn λ0(ti)xi −xi) = λ−1 0 (Jλ0(ti)xi −xi) as n →∞for i = 1, 2. Since λ−1 0 (Jn λ0(ti)xi −xi) ∈An(ti)Jn(Jn λ0(ti)xi), it follows from (C.2) that (1 −λω) \|Jn λ0(t1)x1 −Jn λ0(t2)x2\| ≤\|Jn λ0(t1)x1 −Jn λ0(t2)x2 −λ (λ−1 0 (Jn λ0(t1)x1 −x1) −λ−1 0 (Jn λ0(t2)x2 −x2)\| +λ \|f(t) −f(s)\|L(\|Jn λ0(t2)x2\|)(1 + \|λ−1 0 (Jn λ0(t2)xs −x2)\|). Letting n →∞, we obtain (1 −λω) \|u1 −u2\| ≤\|u1 −u2 −λ (v1 −v2)\| + λ \|f(t) −f(s)\|L(\|u2\|)(1 + \|v2\|). 144 Hence, (A(t), X) is dissipative. Next, from the deﬁnition of A(t), for every x\|inX we have x = Jλ0(t)x −λ0(λ−1 0 (Jλ0(t)x −x)) ∈(I −λ0 A(t))Jλ0x, which implies Jλ0(t)x = (I −λ A(t))−1x. Hence, we obtain R(I −λ0 A(t)) = X and thus A(t) is m-dissipative. (ii) Since A(t) is an m-dissipative operator and Jλ0x = (I−λ0 A(t))−1 and dom (A(t)) = R((Jλ0(t)), it follows from Corollary 3.4 that Un(t, s)xn →U(t, s)x as n →∞if x ∈dom (A(s)) and xn →x as n →∞. □ 6.5 ChernoﬀTheorem In this section we discuss the Chernoﬀtheorem for the evolution equation (2.1). Lemma 2.12 Let {Tρ(t)}, t ∈[0, T] for ρ > 0 be a family of mapping from D into itself satisfying Tρ(t) : Dβ →Dψ(ρ,β), for t ∈[0, T] and β ≥0, \|Tρ(t)x −Tρ(t)y\| ≤(1 + ωαρ) \|x −y\| for x, y ∈Dα, and \|Aρ(t)x −Aρ(s)x\| ≤L(\|x\|)(1 + \|Aρ(s)x\|) \|f(t) −f(s)\| for x ∈D and t, s ∈[0, T], where Aρ(t)x = 1 ρ(Tρ(t)x −x), x ∈D Then, the evolution operator Aρ(t), t ∈[0, T] satisﬁes (C.2). Proof: For x1, x2 ∈Dα \|x1 −x2 −λ (Aρ(t)x1 −Aρ(s)x2)\| ≥(1 + λ ρ) \|x1 −x2\| −λ ρ (\|Tρ(t)x1 −Tρ(t)x2\|) + \|Tρ(t)x2 −Tρ(s)x2\| ≥(1 −λωα) \|x1 −x2\| −λ L(\|x2\|)(1 + \|Aρ(s)x2\|) \|f(t) −f(s)\|.□ Lemma 2.14 Let X0 be a closed convex subset of a Banach space X and α ≥1. Suppose C(t) : X0 →X0, t ∈[0, T] be a family of evolution operators satisfying (2.30) \|C(t)x −C(t)y\| ≤α \|x −y\| for x, y ∈X0, and (2.31) \|C(t)x −C(s)x\| ≤L(\|x\|)(δ + \|(C(s) −I)x\|) \|g(t) −g(s)\|. where δ > 0 and g : [0, T] →X is continuous. Then, there exists a unique function u = u(t; x) ∈C1(0, T; X0) satisfying (2.32) d dtu(t) = (C(t) −I)u(t), u(0) = x, 145 and we have (2.33) \|u(t; x) −u(t; y)\| ≤e(α−1)t \|x −y\|. Proof: It suﬃces to prove that there exists a unique function u ∈C(0, T; X0) satisfying (2.34) u(t) = e−tx + Z t 0 e−(t−s)C(s)u(s) ds, t ∈[0, T]. First, note that if v(t) : [0, T] →X0 is continuous, then for t ∈[0, T] e−tx + R t 0 e−(t−s)C(s)v(s) ds = (1 −λ)x + λ R t 0 e−(t−s)C(s)v(s) ds R t 0 e−(t−s) ds ! ∈X0 where λ = Z t 0 e−(t−s) ds, since X0 is closed and convex. Deﬁne a sequence of functions un ∈C(0, T; X0) by un(t) = e−tx + Z t 0 e−(t−s)C(s)un−1(s) ds with u0(t) = x ∈X0. By induction we can show that \|un+1(t) −un(t)\| ≤(αt)n n! Z t 0 \|C(s)x −x\| ds and thus {un} is a Cauchy sequence in C(0, T; X0). It follows that un converges to a unique limit u ∈C(0, T; X0) and u satisﬁes (2.34). Also, it is easy to show that there exists a unique continuous function that satisﬁes (2.34). Moreover since for x, y ∈X0 et \|u(t; x) −u(t; y)\| ≤ Z t 0 α es\|u(s; x) −u(s; y)\| ds, by Gronwall’s inequality we have (2.33). Since s →C(s)u(s) ∈X0 is continuous, u ∈C1(0, T; X0) satisﬁes (2.32). □ Theorem 2.15 Let α ≥1 and let C(t) : Dβ →Dβ, t ∈[0, T] be a family of evolution operators satisfying (2.30)–(2.31). Assume that Xβ is a closed convex subset of X and g is Lipschitz continuous with Lipschitz constant Lg. Then, if we let u(t) = u(t; x) be the solution to d dtu(t; x) = (C(t) −I)u(t; x), u(0; x) = x, then there exist some constants Mτ such that for t ∈[0, τ] (2.35) \|u(t; x) −Πn i=1Cix\| ≤αne(α−1)t [(n −αt)2 + αt] 1 2 (Cn,τ + \|C(0)x −x\|) +LgL(Mτ) Z t 0 e(α−1)(t−s) [(n −s) −α(t −s)2 + α(t −s)] 1 2 K(\|C(s)u(s) −u(s)\|) ds 146 where Ci = C(i), i ≥0 and Cn,τ = L(\|x\|)Lg(δ + \|C(0)x −x\|) max(n, τ). Proof: Note that (2.36) u(t; x) −x = Z t 0 es−t(C(s)u(s; x) −x) ds Thus, from (2.30)–(2.31) \|u(t; x) −x\| ≤ Z t 0 es−t(\|C(s)u(s; x) −C(0)u(s, x)\| + \|C(0)u(s; x) −C(0)x\| + \|C(0)x −x\|) ds ≤ Z t 0 es−t(L(\|u(s, x)\|)\|g(s) −g(0)\|K(\|C(0)x −x\|) + \|C(0)x −x\| + α \|u(s; x) −x\|) ds where K(s) = δ + s. Or, equivalently et \|u(t; x)−x\| ≤ Z t 0 es(L(M)\|g(s)−g(0)\|K(\|C(0)x−x\|)+\|C(0)x−x\|+α es\|u(s; x)−x\|) ds since \|u(t, x)\| ≤M on [0, τ] for some M = Mτ. Hence by Gronwall’s inequality (2.37) \|u(t; x)−x\| ≤ Z t 0 e(α−1)(t−s)(L(M)\|g(s)−g(0)\|K(\|C(0)x−x\|)+\|C(0)x−x\|) ds. It follows from (2.36) that u(t; x) −Πn i=1Cix = e−t(x −Πn i=1Cix) + Z t 0 es−t(C(s)u(s; x) −Πn i=1Cix) ds. Thus, by assumption \|u(t) −Πn i=1Cix\| ≤e−t\|x −Πn i=1Cix\| + Z t 0 es−t(\|C(s)u(s) −Cnu(s)\| + \|Cnu(s) −CnΠn−1 i=1 Cix\|) ds ≤e−tαnδn + Z t 0 es−t(L(M)\|g(s) −g(n)\| \|C(s)x −x\| + α \|u(s) −Πn−1 i=1 Cix\| ds where δn = Pn i=1 \|Cix −x\|. If we deﬁne ϕn(t) = α−net\|u(t, x) −Πn i=1Cix\| then ϕn(t) ≤δn + Z t 0 (Mesα−n\|g(s) −g(n)\| + ϕn−1(s)) ds By induction in n we obtain from (2.36)–(2.37) (2.38) ϕn(t) ≤ n−1 X k=0 δn−ktk k! + 1 (n −1)! Z t 0 (t −s)n−1αseαs ds (L(\|x\|)LgτK(\|C(0)x −x\|) + \|C(0)x −x\|) +L(M)α−n n X k=0 Z t 0 esαk (t −s)k k! \|C(s)u(s) −u(s)\| \|g(s) −g(n −k)\| ds 147 on t ∈[0, τ], where we used Z t 0 e(α−1)s ds ≤αte−teαt. Since Z t 0 (t −s)n−1sk+1 ds = tk+n+1 (k + 1)!(n −1)! (k + n + 1)! . we have Z t 0 (t −s)n−1seαs ds = ∞ X k=0 αk k! Z t 0 (t −s)n−1sk+1 ds = (n −1)! ∞ X k=0 (k + 1)αktk+n+1 (k + n + 1)! = (n −1)! ∞ X k=n+1 (k −n)αk−1tk k! . Note that from (2.31) \|C(k)x −x\| ≤L(\|x\|)LgkK(\|C(0)x −x\|) for 0 ≤k ≤n. Let Cn,τ = L(\|x\|)LgK(\|C(0)x −x\|)max(n, τ). Then we have (2.39) n−1 X k=0 δn−ktk k! + 1 (n −1)! Z t 0 (t −s)n−1αseαs ds (L(\|x\|)LgτK(\|C(0)x −x\|) + \|C(0)x −x\|) ≤ ∞ X k=0 \|k −n\|αktk k! ! (Cn,τ + \|C(0)x −x\|) ≤Ceαt [(n −αt)2 + αt] 1 2 (Cn,τ + \|C(0)x −x\|), where we used ∞ X k=0 \|k −n\|αktk k! ≤αt 2 ∞ X k=0 \|k −n\|2αktk k! ! 1 2 ≤eαt[(n −αt)2 + αt] 1 2 Moreover, we have (2.40) n X k=0 esαk (t −s)k k! \|g(s) −g(n −k)\| ≤Lges n X k=0 αk(t −s)k k! \|s −(n −k)\| ≤Lgese α(t−s) 2 ∞ X k=0 \|s −(n −k)\|2αk(t −s)k k! ! 1 2 ≤Lgeseα(t−s) [(n −s) −α(t −s)2 + α(t −s)] 1 2 Hence (2.35) follows from (2.36)–(2.40). □ 148 Theorem 2.16 Let {Tρ(t)}, t ∈[0, T] for ρ > 0 be a family of mapping from D into itself satisfying (2.41) Tρ(t) : Dβ →Dψ(ρ,β), for t ∈[0, T] and β ≥0, (2.42) \|Tρ(t)x −Tρ(t)y\| ≤(1 + ωαρ) \|x −y\| for x, y ∈Dα, and (2.43) \|Aρ(t)x −Aρ(s)x\| ≤L(\|x\|)(1 + \|Aρ(s)x\|) \|f(t) −f(s)\| for x ∈D and t, s ∈[0, T], where Aρ(t)x = 1 ρ(Tρ(t)x −x), x ∈D Assume that Dβ is a closed convex subset in X for each β ≥0 and f is Lipschitz continuous on [0, T] with Lipschitz constant Lf. Then, if we let uρ(t) = u(t; x) be the solution to (2.44) d dtu(t; x) = Aρ(t)u(t; x), u(0; x) = x ∈Dα, then there exist constant M and ω = ωα such that \|uρ(t) −Π [ t ρ ] k=1Tρ(kρ)x\| ≤≤eωt [(1 + ωt)2ρ + ωtρ + t] 1 2 (eωt (L(\|x\|Lft(1 + \|Aρ(0)x\|) + \|Aρ(0)x\|) +LfL(M) Z t 0 eω(t−σ)(1 + \|Aρ(σ)uρ(σ)\|) dσ) √ρ. for x ∈dom (A(0)) and t ∈[0, T]. Proof: Let α = 1 + ωρ and we let C(t) = Tρ(ρ t), g(t) = f(ρt) and δ = ρ. Then, C(t) satisﬁes (2.30)–(2.31). Next, note that \|uρ(t)\| ≤M. It thus follows from Lemma 2.14 and Theorem 2.15 that \|uρ(t) −Π [ t ρ ] k=1Tρ(kρ)x\| ≤e2ωt [(1 + ωt)2ρ + ωtρ + t] 1 2 √ρ ((\|Aρ(0)x\| + L(\|x\|)Lft(1 + \|Aρ(0)x\|)) +LfL(M)eωt Z t 0 eω(t−σ)[(1 + ω(t −σ))2ρ + ω(t −σ)ρ + (t −σ)] 1 2 √ρ(1 + \|Aρ(σ)uρ(σ)\|) dσ. where we set s = σ ρ, since Lg = Lfρ. □ Theorem 2.17 Assume that (A(t), dom (A(t)), D, ϕ) satisﬁes (A.1)−(A.2) and either (R) −(C.1) or (2.3) −(C.2) and that Dβ is a closed convex subset in X for each β ≥0 and f is Lipschitz continuous on [0, T]. Let {Tρ(t)}, t ∈[0, T] for ρ > 0 be a family of mapping from D into itself satisfying (2.41)–(2.43) and assume the consistency for β ≥0, t ∈[0, T], and [x, y] ∈A(t) with x ∈Dβ there exists xρ ∈Dα(β) such that lim \|xρ −x\| + \|Aρ(t)xρ −y\| = 0 as ρ →0. 149 Then, for 0 ≤s ≤t ≥T and x ∈D ∩dom (A(s)) (2.45) \|Π [ t−s ρ ] k=1 Tρ(kρ + s)x −u(t; s, x)\| →0 as ρ →0+ uniformly on [0, T], where u(t; s, x) is the mild solution to (2.1), deﬁned in Theorem 2.10. Proof: Without loss of generality we can assume that s = 0. It follows from Lemma 2.4 and Lemma 2.12 that (2.46) \|Aρ(t)u(t)\| ≤e(α−1) t+L(M) \|f\|BV (0,T ) (\|Aρ(0)x\| + L(M) \|f\|BV (0,T)) on [0, T]. Note that uρ(t) satisﬁes uρ(iλ) −uρ((i −1)λ) λ = Aρ(iλ)uρ(iλ) + ϵλ i where ϵλ i = 1 λ Z iλ (i−1)λ (Aρ(s)uρ(s) −Aρ(iλ)uρ(iλ)) ds. Since Aρ(t) : [0, T] × X →X and t →uρ(t) ∈X are Lipschitz continuous, it follows that PNλ i=1 λ \|ϵλ i \| →0 as λ →0+. Hence uρ(t) is the integral solution to (2.44). It thus follows from Theorem 3.2 that lim \|uρ(t) −u(t; s, x)\|X →0 as ρ →0 for x ∈D ∩dom (A(s)). Now, from Theorem 2.16 and (2.46) \|Π [ t−s ρ ] k=1 Tρ(kρ + s)x −uρ(t)\| ≤M√ρ for x ∈Dβ ∩dom (A(s)). Thus, (2.45) holds for all for x ∈Dβ ∩dom (A(s)). By the continuity of the right-hand side of (2.45) with respect to the initial condition x ∈X (2.45) holds for all x ∈Dβ ∩dom (A(s)). □ The next corollary follows from Theorem 2.17 and is an extension of the Chernoﬀ theorem of the autonomous nonlinear semigroups to the evolution case. Corollary 2.18 (ChernoﬀTheorem Let C be a closed convex subset of X. We assume that the evolution operator (A(t), X) satisﬁes (A.1) with Lipschtz continuous f and C ⊂R(I −λ A(t)) and (I −λA(t))−1 ∈C for 0 < λ ≤δ and t ≥0. Let {Tρ(t)}, t ≥0 for ρ > 0 be a family of mapping from C into itself satisfying \|Tρ(t)x −Tρ(t)y\| ≤(1 + ωρ) \|x −y\| and \|Aρ(t)x −Aρ(s)x\| ≤L(\|x\|)(1 + \|Aρ(s)x\|) \|f(t) −f(s)\| for x, y ∈C. If A(t) ⊂limρ→0+ Aρ(t), or equivalently (I −λ Tρ(t) −I ρ )−1x →(I −λ A(t))−1x 150 for all x ∈C and 0 < λ ≤δ, then for x ∈C and 0 ≤s ≤t \|Π [ t−s ρ ] k=1 Tρ(s + kρ)x −U(t, s)x\| →0 as ρ →0+, where U(t, s) is the nonlinear semigroup generated by A(t) and the convergence is uniform on arbitrary bounded intervals. Corollary 2.19 (ChernoﬀTheorem) Assume that (A(t), X) is m-dissipative and satisfy (A.1) with Lipschitz continuous f, and that dom (A(t)) are independent of t ∈[0, T] and convex. Let {Tρ(t)}, t ≥0 for ρ > 0 be a family of mapping from X into itself satisfying \|Tρ(t)x −Tρ(t)y\| ≤(1 + ωρ) \|x −y\| for x, y ∈X, and \|Aρ(t)x −Aρ(s)x\| ≤L(\|x\|)(1 + \|Aρ(s)x\|) \|f(t) −f(s)\| If A(t) ⊂limρ→0+ Aρ(t), or equivalently (I −λ0 Tρ(t) −I ρ )−1x →(I −λ0 A(t))−1x for all x ∈X, t ≥0 and some 0 < λ0 < ω−1, then for x ∈X and t ≥s ≥0 \|Π [ t−s ρ ] k=1 Tρ(s + kρ)x −U(t, s)x\| →0 as ρ →0+, where U(t, s) is the nonlinear semigroup generated by A(t) and the convergence is uniform on arbitrary bounded intervals. Theorem 2.20 Let (A(t), X) be m-dissipative subsets of X ×X and satisfy (A.1) with Lipschitz continuous f and assume that dom (A(t)) are independent of t ∈[0, T] and convex. Let Aλ(t) = λ−1(Jλ(t)−I) for λ > 0 and t ∈[0, T] and u(t; s, x) = U(t, s; Aλ)x be the solution to d dtu(t) = Aλ(t)u(t), u(s) = x ∈dom (A(s)) Then, we have U(t, s)x = lim λ→0+ Π [ t λ] k=1Jλ(s + k λ) = lim λ→0+ U(t, s; Aλ)x. 6.6 Operator Splitting Theorem 3.1 Let X and X∗be uniformly convex and let An, n ≥1 and A be m-dissipative subsets of X × X. If for all [x, y] ∈A0 there exists [xn, yn] ∈An such that \|xn −x\| + \|yn −y\| →0 as n →∞, then Sn(t)xn →S(t)x as n →∞ for every sequence xn ∈dom (An) satisfying xn →x ∈dom (A), where the convergence is uniform on arbitrary bounded intevals. 151 Proof: It follows from Theorem 1.4.2–1.4.3 that for x ∈dom (A), S(t)x ∈dom (A0), d+ dt S(t)x = A0S(t)x, and t →A0S(t)x ∈X is continuous except a countable number of values t ≥0. Hence if we deﬁne xλ i = S(tλ i )x, tλ i = i λ, then xλ i −xλ i−1 λ −A0xλ i = ϵλ i = λ−1 Z tλ i tλ i−1 (A0(t)x(t) −A(tλ i )x(tλ i )) dt where PNλ i=1 λ \|ϵλ i \| ≤ Z T 0 \|A0S(t)x −A0S(([ t λ] + 1)λ\| dt. Since \|A0S(t)x −A0S(([ t λ] + 1)λ)x\| →0 a.e. t ∈[0, T] as λ →0+, by Lebesgue dominated convergence theorem PNλ i=1 λ \|ϵλ i \| →0 as λ →0. Hence it follows from the proof of Theorem 2.3.2 that \|Sn(t)xn −S(t)x\| →0 as λ →0 for all xn ∈dom (A) satisfying xn →x ∈dom (A). The theorem follows from the fact that S(t) and Sn(t) are of cotractions. □ Theorem 3.2 Let X and X∗be uniformly convex and let A and B be two m-dissipative subests of X × X. Assume that A + B is m-dissipative and let S(t) be the semigroup generated by A + B. Then we have S(t)x = lim ρ→0+ (I −ρ A)−1(I −ρ B)−1[ t ρ ] x for x ∈dom (A) ∩dom (B), uniformly in any t-bounded intervals. □ Proof: Deﬁne Tρ = JA ρ JB ρ and let xρ = x−ρ b where b ∈Bx. Then, since JB ρ (x−ρ b) = x Tρxρ −xρ ρ = JA ρ x −x ρ + b. Hence ρ−1(Tρxρ −xρ) →A0x + b as ρ →0+. If we choose b ∈Bx such that A0x + b = (A + B)0x, then ρ−1(Tρxρ −xρ) →(A + B)0x as ρ →0+. Thus, the theorem follows from Theorem 3.1 and Corollay 2.3.6. □ Theorem 3.3 Let X and X∗be uniformly convex and let A and B be two m-dissipative subests of X × X. Assume that A + B is m-dissipative and let S(t) be the semigroup generated by A + B. Then we have S(t)x = lim ρ→0+ (2(I −ρ 2 A)−1 −I)(2(I −ρ 2 B)−1 −I) [ t ρ] x for x ∈dom (A) ∩dom (B), uniformly in any t-bounded intervals. Proof: Deﬁne T2ρ = (2JA ρ −I)(2JB ρ −I) and let xρ = x −ρ b where b ∈Bx. Then, since JB ρ (x −ρ b) = x, it follows that 2JB ρ (xρ) −xρ = x + ρ b and T2ρxρ −xρ 2ρ = 2JA ρ (x + ρ b) −(x + ρ b) −(x −ρ b) 2ρ = JA ρ (x + ρ b) −x ρ . If we deﬁne the subset E of X × X by Ex = Ax + b, then E is m-dissipative and JA ρ (x+ρ b) = JE ρ x. Thus, it follows from Theorem 1.6 that ρ−1(Tρxρ−xρ) →(A+B)0x as ρ →0+ and hence the theorem follows from Theorem 3.1 and Corollary 2.3.6. □ 152 Theorem 3.4 Let X and X∗be uniformly convex and let A and B be two m-dissipative subests of X ×X. Assume that A, B are single-valued and A+B is m-dissipative. Let SA(t), SB(t) and S(t) be the semigroups generated by A, B and A + B, respectively. Then we have S(t)x = lim ρ→0+ (SA(ρ)SB(ρ))[ t ρ ] x for x ∈dom (A) ∩dom (B), uniformly in any t-bounded intervals. Proof: Clearly Tρ = SA(ρ)SB(ρ) is nonexpansive on C = dom (A) ∩dom (B). We show that limρ→0+ h−1(Tρx −x) = Ax + B0x for every dom (A) ∩dom (B). Note that Tρx −x ρ = SA(ρ)x −x ρ + SA(ρ)SB(ρ)x −SA(ρ)x ρ . Since A is single-valued, it follows from Theorem 1.4.3 that limρ→0+ h−1(SA(ρ)x−x) = Ax. Thus, it suﬃces to show that yρ = SA(ρ)SB(ρ)x −SA(ρ)x ρ →B0x as ρ →0+. Since SA(ρ) is nonexpansive and B is dissipative, it follows from Theorem 1.4.3 that (3.1) \|yρ\| ≤\|SB(ρ)x −x ρ \| ≤\|B0x\| for all ρ > 0. On the other hand, (3.2) ⟨SA(ρ)u −u ρ + SB(ρ)x −x ρ −SA(ρ)x −x ρ −yρ, F(u −SB(ρ)x)⟩≥0 since SA(ρ) −I is dissipative and F is singled-valued. We choose a subsequence {yρn} that converges weakly to y in X. It thus follows from (3.2) that since A is single-valued ⟨Au + B0x −Ax −y, F(u −x)⟩≥0. Since A is maximal dissipstive, this implies that y = B0x and thus yρn converges weakly to B0x. Since X is uniformly convex, (3.2) implies that yρn converges strongly to B0x as ρn →0+. Now, since C = A + B is m-dissipative and C0x = Ax + B0x, the theorem follows from Corollary 2.3.6. □ 7 Monotone Operator Theory and Nonlinear semi-group In this section we discuss the nonlinear operator theory that extends the Lax-Milgram theory for nonlinear monotone operators and mappings from Banach space from X to X∗. Let us denote by F : X →X∗, the duality mapping of X, i.e., F(x) = {x∗∈X∗: ⟨x, x∗⟩= \|x\|2 = \|x∗\|2}. 153 By Hahn-Banach theorem, F(x) is non-empty. In general F is multi-valued. Therefore, when X is a Hilbert space, ⟨·, ·⟩coincides with its inner product if X∗is identiﬁed with X and F(x) = x. Let H be a Hilbert space with scalar product (φ, ψ) and X be a real, reﬂexive Banach space and X ⊂H with continuous dense injection. Let X∗denote the strong dual space of X. H is identiﬁed with its dual so that X ⊂H = H∗⊂X∗. The dual product ⟨φ, ψ⟩on X × X∗is the continuous extension of the scalar product of H restricted to X × H. The following proposition contains some further important properties of the duality mapping F. Theorem (Duality Mapping) (a) F(x) is a closed convex subset. (b) If X∗is strictly convex (i.e., balls in X∗are strictly convex), then for any x ∈X, F(x) is single-valued. Moreover, the mapping x →F(x) is demicontinuous, i.e., if xn →x in X, then F(xn) converges weakly star to F(x) in X∗. (c) Assume X be uniformly convex (i.e., for each 0 < ϵ < 2 there exists δ = δ(ϵ) > 0 such that if \|x\| = \|y\| = 1and \|x −y\| > ϵ, then \|x + y\| ≤2(1 −δ)). If xn converges weakly to x and lim supn→∞\|xn\| ≤\|x\|, then xn converges strongly to x in X. (d) If X∗is uniformly convex, then the mapping x →F(x) is uniformly continuous on bounded subsets of X. Proof: (a) Closeness of F(x) is an easy consequence of the follows from the continuity of the duality product. Choose x∗ 1, x∗ 2 ∈F(x) and α ∈(0, 1). For arbitrary z ∈X we have (using \|x∗ 1\| = \|x∗ 2\| = \|x\|) ⟨z, αx∗ 1 + (1 −α)x∗ 2⟩≤α\|z\| \|x∗ 1\| + (1 −α)\|z\| \|x∗ 2\| = \|z\| \|x\|, which shows \|αx∗ 1 + (1 −α)x∗ 2\| ≤\|x\|. Using ⟨x, x∗⟩= ⟨x, x∗ 1⟩= \|x\|2 we get ⟨x, αx∗ 1 + (1 −α)x∗ 2⟩= α⟨x, x∗ 1⟩+ (1 −α)⟨x, x∗ 2⟩= \|x\|2, so that \|αx∗ 1 + (1 −α)x∗ 2\| = \|x\|. This proves αx∗ 1 + (1 −α)x∗ 2 ∈F(x). (b) Choose x∗ 1, x∗ 2 ∈F(x), α ∈(0, 1) and assume that \|αx∗ 1+(1−α)x∗ 2\| = \|x\|. Since X∗ is strictly convex, this implies x∗ 1 = x∗ 2. Let {xn} be a sequence such that xn →x ∈X. From \|F(xn)\| = \|xn\| and the fact that closed balls in X∗are weakly star compact we see that there exists a weakly star accumulation point x∗of {F(xn)}. Since the closed ball in X∗is weakly star closed, thus ⟨x, x∗⟩= \|x\|2 ≥\|x∗\|2. Hence ⟨x, x∗⟩= \|x\|2 = \|x∗\|2 and thus x∗= F(x). Since F(x) is single-valued, this implies F(xn) converges weakly to F(x). (c) Since lim inf \|xn\| ≤\|x\|, thus limn→∞\|xn\| = \|x\|. We set yn = xn/\|xn\| and y = x/\|x\|. Then yn converges weakly to y in X. Suppose yn does not converge strongly to y in X. Then there exists an ϵ > 0 such that for a subsequence y˜ n \|y˜ n −y\| > ϵ. Since X∗is uniformly convex there exists a δ > 0 such that \|y˜ n + y\| ≤2(1 −δ). Since the norm is weakly lower semicontinuos, letting ˜ n →∞we obtain \|y\| ≤1 −δ, which is a contradiction. (d) Assume F is not uniformly continuous on bounded subsets of X. Then there exist constants M > 0, ϵ > 0 and sequences {un}, {vn} in X satisfying \|un\|, \|vn\| ≤M, \|un −vn\| →0, and \|F(un) −F(vn)\| ≥ϵ. Without loss of the generality we can assume that, for a constant β > 0, we have in 154 addition \|un\| ≥β, \|vn\| ≥β. We set xn = un/\|un\| and yn = vn/\|vn\|. Then we have \|xn −yn\| = 1 \|un\| \|vn\| \|\|vn\|un −\|un\|vn\| ≤1 β2 (\|vn\|\|un −vn\| + \|\|vn\| −\|un\|\| \|vn\|) ≤2M β2 \|un −vn\| →0 as n →∞. Obviously we have 2 ≥\|F(xn) + F(yn)\| ≥⟨xn, F(xn) + F(yn)⟩and this together with ⟨xn, F(xn) + F(yn)⟩= \|xn\|2 + \|yn\|2 + ⟨xn −yn, F(yn)⟩ = 2 + ⟨xn −yn, F(yn)⟩≥2 −\|xn −yn\| implies lim n→∞\|F(xn) + F(yn)\| = 2. Suppose there exists an ϵ0 > 0 and a subsequence {nk} such that \|F(xnk)−F(ynk)\| ≥ϵ0. Observing \|F(xnk)\| = \|F(ynk)\| = 1 and using uniform convexity of X∗we conclude that there exists a δ0 > 0 such that \|F(xnk) + F(ynk)\| ≤2(1 −δ0), which is a contradiction to the above. Therefore we have limn→∞\|F(xn) −F(yn)\| = 0. Thus \|F(un) −F(vn)\| ≤\|un\| \|F(xn) −F(yn)\| + \|\|un\| −\|vn\|\| \|F(yn)\| which implies F(un) converges strongly to F(vn). This contradiction proves the result. □ 7.1 Monotone equations and Minty–Browder Theorem Deﬁnition (Monotone Mapping) (a) A mapping A ⊂X × X∗be given. is called monotone if ⟨x1 −x2, y1 −y2⟩≥0 for all [x1, y1], [x2, y2] ∈A. (b) A monotone mapping A is called maximal monotone if any monotone extension of A coincides with A, i.e., if for [x, y] ∈X × X∗, ⟨x −u, y −v⟩≥0 for all [u, v] ∈A then [x, y] ∈A. (c) The operator A is called coercive if for all sequences [xn, yn] ∈A with limn→∞\|xn\| = ∞we have lim n→∞ ⟨xn, yn⟩ \|xn\| = ∞. (d) Assume that A is single-valued with dom(A) = X. The operator A is called hemicontinuouson X if for all x1, x2, x ∈X, the function deﬁned by t ∈R →⟨x, A(x1 + tx2)⟩ is continuous on R. 155 For example, let F be the duality mapping of X. Then F is monotone, coercive and hemicontinuous. Indeed, for [x1, y1], [x2, y2] ∈F we have ⟨x1 −x2, y1 −y2⟩= \|x1\|2 −⟨x1, y2⟩−⟨x2, y1⟩+ \|x2\|2 ≥(\|x1\| −\|x2\|)2 ≥0, (7.1) which shows monotonicity of F. Coercivity is obvious and hemicontinuity follows from the continuity of the duality product. Lemma 1 Let X be a ﬁnite dimensional Banach space and A be a hemicontinuous monotone operator from X to X∗. Then A is continuous. Proof: We ﬁrst show that A is bounded on bounded subsets. In fact, otherwise there exists a sequence {xn} in X such that \|Axn\| →∞and xn →x0 as n →∞. By monotonicity we have ⟨xn −x, Axn \|Axn\| − Ax \|Axn\|⟩≥0 for all x ∈X. Without loss of generality we can assume that Axn \|Axn\| →y0 in X∗as n →∞. Thus ⟨x0 −x, y0⟩≥0 for all x ∈X and therefore y0 = 0. This is a contradiction and thus A is bounded. Now, assume {xn} converges to x0 and let y0 be a cluster point of {Axn}. Again by monotonicity of A ⟨x0 −x, y0 −Ax⟩≥0 for all x ∈X. Setting x = x0 + t (u −x0), t > 0 for arbitrary u ∈X, we have ⟨x0 −u, y0 −A(x0 + t (u −x0)) ≥0⟩ for all u ∈X. Then, letting limit t →0+, by hemicontinuity of A we have ⟨x0 −u, y0 −Ax0⟩≥0 for all u ∈X, which implies y0 = Ax0. □ Lemma 2 Let X be a reﬂexive Banach space and A : X →X∗be a hemicontinuous monotone operator. Then A is maximumal monotone. Proof: For [x0, y0] ∈X × X∗ ⟨x0 −u, y0 −Au⟩≥0 for all u ∈X. Setting u = x0 + t (x −x0), t > 0 and letting t →0+, by hemicontinuity of A we have ⟨x0 −x, y0 −Ax0⟩≥0 for all x ∈X. Hence y0 = Ax0 and thus A is maximum monotone. □ The next theorem characterizes maximal monotone operators by a range condition. Minty–Browder Theorem Assume that X, X∗are reﬂexive and strictly convex. Let F denote the duality mapping of X and assume that A ⊂X × X∗is monotone. Then A is maximal monotone if and only if Range (λF + A) = X∗ 156 for all λ > 0 or, equivalently, for some λ > 0. Proof: Assume that the range condition is satisﬁed for some λ > 0 and let [x0, y0] ∈ X × X∗be such that ⟨x0 −u, y0 −v⟩≥0 for all [u, v] ∈A. Then there exists an element [x1, y1] ∈A with λFx1 + y1 = λFx0 + y0. (7.2) From these we obtain, setting [u, v] = [x1, y1], ⟨x1 −x0, Fx1 −Fx0⟩≤0. By monotonicity of F we also have the converse inequality, so that ⟨x1 −x0, Fx1 −Fx0⟩= 0. From (7.1) this implies that \|x1\| = \|x0\| and ⟨x1, Fx0⟩= \|x1\|2, ⟨x0, Fx1⟩= \|x0\|2. Hence Fx0 = Fx1 and ⟨x1, Fx0⟩= ⟨x0, Fx0⟩= \|x0\|2 = \|Fx0\|2. If we denote by F ∗the duality mapping of X∗(which is also single-valued), then the last equation implies x1 = x0 = F ∗(Fx0). This and (7.2) imply that [x0, y0] = [x1, y1] ∈A, which proves that A is maximal monotone. □ In stead of the detailed proof of ”only if’ part of Theorem, we state the following results. □ Corollary Let X be reﬂexive and A be a monotone, everywhere deﬁned, hemicontinous operator. If A is coercive, then R(A) = X∗. Proof: Suppose A is coercive. Let y0 ∈X∗be arbitrary. By the Appland’s renorming theorem, we may assume that X and X∗are strictly convex Banach spaces. It then follows from Theorem that every λ > 0, equation λ Fxλ + Axλ = y0 has a solution xλ ∈X. Multiplying this by xλ, λ \|xλ\|2 + ⟨xλ, Axλ⟩= ⟨y0, xλ⟩. and thus ⟨xλ, Axλ⟩ \|xλ\|X ≤\|y0\|X∗ Since A is coercive, this implies that {xλ} is bounded in X as λ →0+. Thus, we may assume that xλ converges weakly to x0 in X and Axλ converges strongly to y0 in X∗ as λ →0+. Since A is monotone ⟨xλ −x, y0 −λ Fxλ −Ax⟩≥0, and letting λ →0+, we have ⟨x0 −x, y0 −Ax⟩≥0, 157 for all x ∈X. Since A is maximal monotone, this implies y0 = Ax0. Hence, we conclude R(A) = X∗. □ Theorem (Galerkin Approximation) Assume X is a reﬂexive, separable Banach space and A is a bounded, hemicontinuous, coercive monotone operator from X into X∗. Let Xn = span{φ}n i=1 satisﬁes the density condition: for each ψ ∈X and any ϵ > 0 there exists a sequence ψn ∈Xn such that \|ψ −ψn\| →0 as n →∞. The xn be the solution to ⟨ψ, Axn⟩= ⟨ψ, f⟩ for all ψ ∈Xn, (7.3) then there exists a subsequence of {xn} that converges weakly to a solution to Ax = f. Proof: Since ⟨x, Ax⟩/\|x\|X →∞as \|x\|X →∞there exists a solution xn to (7.3) and \|xn\|X is bounded. Since A is bounded, thus Axn bounded. Thus there exists a subsequence of {n} (denoted by the same) such that xn converges weakly to x in X and Axn converges weakly in X∗. Since lim n→∞⟨ψ, Axn⟩= lim n→∞(⟨ψn, f⟩+ ⟨ψ −ψn, Axn⟩) = ⟨ψ, f⟩ Axn converges weakly to f. Since A is monotone ⟨xn −u, Axn −Au⟩≥0 for all u ∈X Note that lim n→∞⟨xn, Axn⟩= lim n→∞⟨xn, f⟩= ⟨x, f⟩. Thus taking limit n →∞, we obtain ⟨x −u, f −Au⟩≥0 for all u ∈X. Since A is maximum monotone this implies Ax = f. □ The main theorem for monotone operators applies directly to the model problem involving the p-Laplace operator −div(\|∇u\|p−2∇u) = f on Ω (with appropriate boundary conditions) and −∆u + c u = f, −∂ ∂nu ∈β(u). at ∂Ω with β maximal monotone on R. Also, nonlinear problems of non-variational form are applicable, e.g., Lu + F(u, ∇u) = f on Ω where L(u) = −div(σ(∇u) −⃗ b u) and we are looking for a solution u ∈W 1,p 0 (Ω), 1 < p < ∞. We assume the following conditions: (i) Monotonicity for the principle part L(u): (σ(ξ) −σ(η), ξ −η)Rn ≥0 for all ξ, η ∈Rn. 158 (ii) Monotonicity for F = F(u): (F(u) −F(v), u −v) ≥0 for all u, v ∈R. (iii) Coerciveness and Growth condition: for some c, d > 0 (σ(ξ), σ) ≥c \|ξ\|p, \|σ(ξ)\| ≤d (1 + \|ξ\|p−1) hold for all ξ ∈Rn. 7.2 Pseudo-Monotone operator Next, we examine a somewhat more general class of nonlinear operators, called pseudo-monotone operators. In applications, it often occurs that the hypotheses imposed on monotonicity are unnecessarily strong. In particular, the monotonicity assumption F(u) involves both the ﬁrst-order derivatives and the function itself. In general the compactness will take care of the lower-order terms. Deﬁnition (Pseudo-Monotone Operator) Let X be a reﬂexive Banach space. An operator T : X →X∗is said to be pseudo-monotone operator if T is a bounded operator (not necessarily continuous) and if whenever un converges weakly to u in X and lim sup n→∞⟨un −u, T(un)⟩≤0, it follows that, for all v ∈X, lim inf n→∞⟨un −v, T(un)⟩≥⟨u −v, T(u)⟩. (7.4) For example, a monotone and hemi-continuous operator is pseudo-monotone. Lemma If T is maximal monotone, then T is pseudo-monotone. Proof: Since T is monotone, ⟨un −u, T(un)⟩≥⟨un −u, T(u)⟩→0 it follows from (7.4) that limn ⟨un −u, T(un)⟩= 0. Assume un →u and T(un) →y weakly in X and X∗, respectively. Thus, 0 ≤⟨un −v, T(un) −T(v)⟩= ⟨un −u + u −v, T(un) −T(v)⟩→⟨u −v, y −T(v)⟩. Since T is maximal monotone, we have y = T(u). Now, lim n ⟨un −v, T(un)⟩= lim n ⟨un −u + u −v, T(un)⟩= ⟨u −v, T(u)⟩.□ A class of pseudo-monotone operators is intermediate between monotone and hemi-continuous. The sum of two pseudo-monotone operators is pseudo-monotone. More-over, the sum of a pseudo-monotone and a strongly continuous operator (xn converges weakly to x, then Txn →Tx strongly) is pseudo-monotone. Property of Pseudo-monotone operator (1) Letting v = u in (7.4) 0 ≤lim inf⟨un −u, T(un)⟩≤lim sup⟨un −u, T(un)⟩≤0 159 and thus limn⟨un −u, T(un)⟩= 0. (2) From (1) and (7.4) ⟨u −v, Tu⟩≤lim inf⟨un −v, T(un)⟩+ lim inf⟨u −un −u, T(un)⟩= lim inf⟨u −v, T(un)⟩ (3) T(un) →T(u) weakly. In fact, from(2) ⟨v, T(u)⟩≤lim inf⟨v, T(un)⟩≤lim sup⟨v, T(un)⟩ = lim sup⟨un −u, T(un)⟩+ lim sup⟨u + v −un, T(un)⟩ = −lim inf⟨un −(u + v), T(un)⟩≤⟨v, T(u)⟩. (4) ⟨un −u, T(un) −T(u)⟩→0 By the hypothesis 0 ≤lim inf⟨un −u, T(un)⟩= lim inf⟨un −u, T(un)⟩−lim inf⟨un −u, T(u)⟩ = lim inf⟨un −u, T(un) −T(u)⟩≤lim sup⟨un −u, T(un) −T(u)⟩ = lim sup⟨un −u, T(un) −T(u)⟩≤0 (5) ⟨un, T(un)⟩→⟨u, T(u)⟩. Since ⟨un, T(un)⟩= ⟨un −u, T(un) −u⟩−⟨u, T(un)⟩−⟨un, T(u)⟩+ ⟨un, T(un)⟩ the claim follows from (3)–(4). (6) Conversely, if T(un) →T(u) weakly and ⟨un, T(un)⟩→⟨u, T(u)⟩→0, then the hypothesis holds. Using a very similar proof to that of the Browder-Minty theorem, one can show the following. Theorem (Bresiz) Assume, the operator A : X →X∗is pseudo-monotone, bounded and coercive on the real separable and reﬂexive Banach space X. Then, for each f ∈X∗ the equation A(u) = f has a solution. 7.3 Generalized Pseudo-Monotone Mapping In this section we discuss pseudo-monotone mappings. We extend the notion of the maximal monotone mapping, i.e., the maximal monotone mapping is a generalized pseudo-monotone mappings. Deﬁnition (Pseudo-monotone Mapping) (a) The set T(u) is nonempty, bounded, closed and convex for all u ∈X. (b) T is upper semicontinuous from each ﬁnite-dimensional subspace F of X to the weak topology on X∗, i.e., to a given element u+ ∈F and a weak neighborhood V of T(u), in X∗there exists a neighborhood U of u0 in F such that T(u) ⊂V for all u ∈U. (c) If {un} is a sequence in X that converges weakly tou ∈X and if wn ∈T(un) is such that lim sup⟨un −u, wn⟩≤0, then to each element v ∈X there exists w ∈T(u) with the property that lim inf⟨un −v, wn⟩≥⟨u −v, w⟩. 160 Deﬁnition (Generalized Pseudo-monotone Mapping) A mapping T from X into X∗is said to be generalized pseudo-monotone if the following is satisﬁed. For any se-quence {un} in X and a corresponding sequence {wn} in X∗with wn ∈T(xn), if un converges weakly to u and wn weakly to w such that lim sup n→∞⟨un −u, wn)⟩≤0, then w ∈T(u) and ⟨un, wn⟩→⟨u, w⟩. Let X be a reﬂexive Banach space, T a pseudo-monotone mapping from X into X∗. Then T is generalized pseudo-monotone. Conversely, suppose T is a bounded generalized pseudo-monotone mapping from X into X∗. and assume that for each u ∈X, Tu is a nonempty closed convex subset of X∗. Then T is pseudo-monotone. PROPOSITION 2. A maximal monotone mapping T from the Banach space X into X∗is generalized pseudo-monotone. Proof: Let {un} be a sequence in X that converges weakly to u ∈X, {wn} a se-quence in X∗with wn ∈T(un) that converges weakly to w ∈X∗. Suppose that lim sup⟨un −u, wn⟩≤0 Let [x, y] ∈T be an arbitrary element of the graph G(T). By the monotonicity of T, ⟨un −x, wn −y⟩≥0 Since ⟨un, wn⟩= ⟨un −x, wn −y⟩+ ⟨x, wn⟩+ ⟨un, y⟩−⟨x, y⟩, where ⟨x, wn⟩+ ⟨un, y⟩→⟨x, w⟩+ ⟨u, y⟩−⟨x, y⟩, we have ⟨u, w⟩≤lim sup⟨un, wn⟩≥⟨x, w⟩+ ⟨u, y⟩−⟨x, y⟩, i.e., ⟨u −x, w −y⟩≥0 Since T is maximal, w ∈T(u). Consequently, ⟨un −u, wn −w⟩≥0. Thus, lim inf⟨un, wn⟩≥lim inf(⟨un, w⟩⟨u, wn⟩+ ⟨u, w⟩= ⟨u, w⟩, It follows that lim⟨un, wn⟩→⟨u, w⟩. □ DEFINITION 5 (1) A mapping T from X into X∗is coercive if there exists a function c : R+ →R+ with limr∞c(r) = ∞such that ⟨u, w⟩≥c(\|u\|)\|u\| for all [u, w] ∈G(T). (1) An operator T from X into X∗is called smooth if it is bounded, coercive, maximal monotone, and has eﬀective domain D(T) = X. (2) Let T be a generalized pseudo-monotone mapping from X into X∗. Then T is said to be regular if R(T + T2) = X∗for each smooth operator T2. 161 (3) T is said to be quasi-bounded if for each M > 0 there exists K(M) > 0 such that whenever [u, w] ∈G(T) of T and ⟨u, w⟩≤M \|u\|, \|u\| ≤M, then \|w\| ≤K(M). T is said to be strongly quasi-bounded if for each M > 0 there exists K(M) > 0 such that whenever [u, w] ∈G(T) of T and ⟨u, w⟩≤M \|u\| ≤M, then \|w\| ≤K(M). Theorem 1 Let T = T1 + T0, where T1 is maximal monotone with 0 ∈D(T), while T0 is a regular generalized pseudo-monotone mapping such that for a given constant k, ⟨u, w⟩k \|u\| for all [u, w] ∈G(T). If either T0 is quasi-bounded or T1 is strongly quasi-bounded, then T is regular. Theorem 2 (Browder). Let X be a reﬂexive Banach space and T be a generalized pseudo-monotone mapping from X into X∗. Suppose that R(ϵ F + T) = X∗for ϵ > 0 and T is coercive. Then R(T) = X∗. Lemma 1 Let T be a generalized pseudo-monotone mapping from the reﬂexive Banach space X into X∗, and let C be a bounded weakly closed subset of X. Then T(C) = {w ∈X∗: w ∈T(u)for some u ∈C} is closed in the strong topology of X∗. Proof: Let Let {wn} be a sequence in T(C) converging strongly to some w ∈X∗. For each n, there exists un ∈C such that wn ∈T(un). Since C is bounded and weakly closed, without loss of the generality we assume un →u ∈C, weakly in X. It follows that lim⟨un −u, wn⟩= 0. The generalized pseudo-monotonicity of T implies that w ∈T(u), i.e., w lies in T(C). □ Proof of Theorem 2: Let J denote the normalized Let F denote the duality mapping of X into X∗. It is known that F is a smooth operator. Let w0 be a given element of X∗. We wish to show that w0 ∈R(T). For each ϵ > 0, it follows from the regularity of T that there exists an element uϵ ∈X such that w0 −ϵ pϵ ∈T(uϵ) for some pϵ ∈F(uϵ). Let wϵ = w0 −ϵ pϵ. Since T is coercive and ⟨uϵ, wϵ⟩= \|uϵ\|2, we have for some k ϵ \|uϵ\|2 ≤k \|uϵ\| + \|uϵ\|\|w0\|. Hence ϵ \|uϵ\| = ϵ \|pϵ\| ≤\|w0\|. and \|wϵ\| ≤\|w0\| + ϵ \|pϵ\| ≤k + 2\|w0\| Since T is coercive, there exits M > 0 such that \|uϵ\| ≤M and thus \|wϵ −w0\| ≤ϵ \|pϵ\| = ϵ \|uϵ\| →0 as ϵ →0+ It thus follows from Lemma 1 that T(B(0, M)) is closed and thus w0 ∈T(B(0, M)), i.e., w0 ∈R(T). □ 162 THEOREM 3 Let X be a reﬂexive Banach space and T a pseudo-monotone mapping from X into X∗. Suppose that T is coercive, R(T) = X∗. PROPOSITION 10. Let F be a ﬁnite-dimensional Banach space, a mapping from F into F ∗such that for each u ∈F, T(u) is a nonempty bounded closed convex subset of F ∗. Suppose that T is coercive and upper semicontinuous from F to F ∗. Then R(T) = F ∗. Proof: Since for each w0 ∈F ∗deﬁne the mapping Tw0 : F →F ∗by Tw0(u) = T(u) −w0 satisﬁes the same hypotheses as the original mapping T. It thus suﬃces to prove that 0 ∈R(T). Suppose that 0 does not lie in R(T). Then for each u ∈F, there exists v(u) ∈F such that inf w∈T(u) ⟨v(u), w⟩> 0 Since T is coercive, there exists a positive constant R such that c(R) > 0, and hence for each u ∈F with \|u\| = R and each w ∈T(u), ⟨u, w⟩≥λ > 0 where λ = c(R)R, For such points u ∈F, we may take v(u) = u. For a given nonzero v0 in F let Wv0 = {u ∈F : inf w∈T(u)⟨v0, w⟩> 0}. The family {Wv0 : v0 ∈F, v0 ̸= 0} forms a covering of the space F. By the upper semicontinuity of T, from F to F ∗each Wv0 is open in F. Hence the family {Wv0 : v0 ∈F, v0 ̸= 0} forms an open covering of F. We now choose a ﬁnite open covering {V1, · · · , Vm} of the closed ball B(0, R) in F of radius R about the origin, with the property that for each k there exists an element vk in F such that Vk ⊂Wvk and the additional condition that if Vk intersects the boundary sphere S(0, R), then vk is a point of Vk ∩S(0, R) and the diameter of such Vk is less than R 2 .. We next choose a partition of unity {α1, · · · , αm} subordinated to the covering {V1, · · · , Vm} where each αk is a continuous mapping of F into [0, I] which vanishes outside the corresponding set Vk, and with the property that P k αk(x) = 1 for each x ∈B(0, R). Using this partition of unity, we may deﬁne a continuous mapping f of B(0, R) into F by setting f(x) = X k α(x)vk. We note that for each k for which alphak(x) > 0 and for any winT0x, we have ⟨vk, w⟩> 0 since x must lie in Vk, which is a subset of Wk. Hence for each x ∈B(0, R) and any w ∈T0x ⟨f(x), w⟩= X k α(x)⟨vk, w⟩> 0 Consequently, f(x) ̸= 0 for any x ∈B(0, R). This implies that the degree of the mapping f on B(0, R) with respect to 0 equals 0. For x ∈S(0, R), on the other hand, f(x) is a convex linear combination of points vk ∈S(0, R), each of which is at distance at most R 2 from the point x. We infer that \|f(x) −x\| ≤R 2 , x ∈S(0, R). 163 By this inequality, f considered as a mapping from B(0, R) into F is homotopic to the identity mapping. Hence the degree of f on B(0, R) with respect to 0 is 1. This contradiction proves that 0 must lie in T0u for some u ∈B(0, R). □ Proof of Theorem 3 Let Λ be the family of all ﬁnite-dimensional subspaces F of X, ordered by inclusion. For F ∈Λ, let jF : F →X denote the inclusion mapping of F into X, and j∗ F : X∗→F ∗the dual projection mapping of X∗onto F ∗. The operator TF = j∗ F TjF then maps F into F ∗. For each u ∈F, TF u is a weakly compact convex subset of X∗ and j∗ F is continuous from the weak topology on X∗to the (unique) topology on F ∗. Hence TF u is a nonempty closed convex subset of F ∗. Since T is upper semicontinuous from F to F ∗with X∗given its weak topology, TF is upper semicontinuous from F to F ∗. By Proposition 10, to each F ∈Λ there exists an element uF ∈F such that j∗ F w0 = wF for some wF ∈TF uF . The coerciveness of T implies that wF ∈TF uF , i.e. ⟨uF , j∗ F w0⟩= ⟨uF , wF ⟩≥c(\|uF \|)\|uF \|. Consequently, the elements {\|uF \|} are uniformly bounded by M for all F ∈Λ. For F ∈Λ, let VF = [ F ′∈Λ, F ′⊂F {uF ′}. Then the set VF is contained in the closed ball B(0, M) in X Since B(0, M) is weakly compact, and since the family {weak closure(VF )} has the ﬁnite intersection property, the intersection ∩F∈Λ{weak closure(VF )} is not empty. Let u0 be an element contained in this intersection. The proof will be complete if we show that 0 ∈Tw0u0. Let v ∈X be arbitrarily given. We choose F ∈Λ such that it contains u0, v ∈F Let {uFk}denote a sequence in VF converging weakly to u0. Since j∗ FkwFk = 0, we have ⟨uFk −u0, wFk⟩= 0 for all k. Consequently, by the pseudo-monotonicity of T, to given v ∈X there exists w(v) ∈ T(u0) with lim⟨uFk −v, wFk⟩≥⟨u0 −v, w⟩. (7.5) Suppose now that 0 / ∈T(u0). Then 0 can be separated from the nonempty closed convex set T(u0), i.e., there exists an element x = u −v ∈X such that inf z∈T(u0)⟨u0 −v, z⟩> 0. But this is a contradiction to (7.5). □ Navier-Stokes equations Example Consider the constrained minimization min F(y) subject to E(y) = 0. where E(y) = E0y + f(y). 164 The necessary optimality condition for x = (y, p) A(y, p) ∈ ∂F −E′(y)∗p E(y) . Let A0(y, p) ∈ ∂F −E∗ 0p E0y , and A1(y, p) = −f′(y)p f(y) . 7.4 Fixed Point Theorems In this section we discuss the ﬁxed point theorem. Banach Fixed Point Theorem. Let (X, d) be a non-empty complete metric space with a contraction mapping T : X →X. Then T admits a unique ﬁxed-point x∗in X (i.e. T(x∗) = x∗). Moreover, x∗is a ﬁxed point of the ﬁxed point iterate xn = T(xn−1) with arbitrary initial condition x0. Proof: d(xn+1, xn) = d(T(xn), T(xn−1) ≤ρ d(xn, xn−1. By the induction d(xn+1, xn) ≤ρn d(x1, x0) and d(xm, xn) ≤ρn−1 1 −ρ d(x1, x0) →0 as m ≥n →∞. Thus, {xn} is a Cauchy sequence and {xn} converges a ﬁxed point x∗ of T. Suppose x∗ 1, x∗ 2 are two ﬁxed point. Then, d(x∗ 1, x∗ 2) = d(T(x∗ 1), T(x∗ 2) ≤ρ d(x∗ 1, x∗ 2) Since ρ < 1, d(x∗ 1, x∗ 2) = 0 and thus x∗ 1 = x∗ 2. □ Brouwer ﬁxed point theorem: is a fundamental result in topology which proves the existence of ﬁxed points for continuous functions deﬁned on compact, convex subsets of Euclidean spaces. Kakutani’s theorem extends this to set-valued functions. Kakutani’s theorem states: Let S be a non-empty, compact and convex subset of some Euclidean space Rn. Let T : S →S be a set-valued function on S with a closed graph and the property that T(x) is non-empty and convex for all x ∈S. Then T has a ﬁxed point. The Schauder ﬁxed point theorem is an extension of the Brouwer ﬁxed point theorem to topological vector spaces. Schauder ﬁxed point theorem Let X be a locally convex topological vector space, and let K ⊂Xbe a compact, and convex set. Then given any continuous mapping T : K →K has a ﬁxed point. 165 Proof: Given ϵ > 0 there exists the family of open balls {Bϵ(x) : x ∈K} is open covering of K. Since K is compact, there exists a ﬁnite open sub-cover, i.e. there exist ﬁnite many points {xk} in K such that K = n [ k=1 Bϵ(xk). Deﬁne the functions {gk} by gk(x) = max(0, ϵ −\|x −xk\|). It is clear that each gk is continuous, gk(x) ≥0 and P k gk(x) > 1 for all x ∈K. Thus, if we deﬁne a function on K by g(x) = P k gk(x) xk P k gk(x) and then g is a continuous function from K to the convex hull K0 of points {xk}n k=1. It is easily shown that \|g(x) −x\| ≤ϵ for x ∈K. Then g maps K0 to K0. Since K0 is compact convex subset of a ﬁnite dimensional vector space, we can apply the Brouwer ﬁxed point theorem to assure the existence of zϵ ∈K0 such that zϵ = g(zϵ). We have the estimate \|zϵ −T(zϵ)\| = \|g(T(zϵ)) −T(zϵ)\| ≤ϵ. Since K is compact, the exits a subsequence ϵn →0 such that xϵn →x0 ∈K and thus \|xϵn −T(xϵn)\| ≤\|g(T(xϵn)) −T(xϵn)\| ≤ϵn. Since T is continuous, T(x0) = x0. □ Schaefer’s ﬁxed point theorem: Let T be a continuous and compact mapping of a Banach space X into itself, such that the set {x ∈X : x = λTx for some 0 ≤λ ≤1} is bounded. Then T has a ﬁxed point. Proof: Let \|x\| ≤M for {x ∈X : x = λTx for some 0 ≤λ ≤1}. Deﬁne ˜ T(x) = MT(x) max(M, \|T(x)\|) Note that ˜ T : B(0, M) →B(0, M).. Let K be the closed convex hull go ˜ T(B(0, M)). Since T is compact, K is a compact convex subset of X. Since ˜ TK →K, it follows from Shauder’s ﬁxed point theorem that there exits x ∈K such that x = ˜ T(x). We now claim that x is a ﬁxed point of T. If not, we should have \|T(x)\| > M and x = λ T(u) for λ = M \|T(x)\| < 1 But, since \|x\| = \| ˜ T(x)\| ≤M, which is a contradiction. □ The advantage of Schaefer’s theorem over Schauder’s for applications is that it is not necessary to determine an explicit convex, compact set K. Krasnoselskii theorem The sum of two operators T = A + B has a ﬁxed point in a nonempty closed convex subset C of a real Banach space X under conditions such 166 that T(C) ⊂C, A is continuous on C, A(C) is a relatively compact subset of X, and B is a strict contraction on C. Note that the proof of Kransnoselskiis ﬁxed point theorem combines the Banach contraction theorem and Schauders ﬁxed point theorem, i.e., y ∈C, Ay + Bx = x has the ﬁxed point map x = ψ(y) and use Schauders ﬁxed point theorem for ψ. Kakutani-Glicksberg-Fan theorem Let S be a non-empty, compact and convex subset of a locally convex topological vector space X. Let T : S →S be a Kakutani map, i.e., it is upper hemicontinuous, i.e., if for every open set W ⊂X, the set {x ∈X : T(x) ∈W} is open in X and T(x) is non-empty, compact and convex for all x ∈X. Then T has a ﬁxed point. The corresponding result for single-valued functions is the Tychonoﬀﬁxed-point theorem. 8 Convex Analysis and Duality This chapter is organized as follows. We present the basic convex analysis and du-ality theory for the optimization in Banach spaces. The subdiﬀerential operator and monotone operator equation in Banach spaces are presented. 8.1 Convex Functional and Subdiﬀerential Deﬁnition (Convex Functional) (1) A proper convex functional on a Banach space X is a function ϕ from X to (−∞, ∞], not identically +∞such that ϕ((1 −λ) x1 + λ x2) ≤(1 −λ) ϕ(x1) + λ ϕ(x2) for all x1, x2 ∈X and 0 ≤λ ≤1. (2) A functional ϕ : X →R is said to be lower-semicontinuous if ϕ(x) ≤lim inf y→x ϕ(y) for all x ∈X. (3) A functional ϕ : X →R is said to be weakly lower-semicontinuous if ϕ(x) ≤lim inf n→∞ϕ(xn) for all weakly convergent sequence {xn} to x. (4) The subset D(ϕ) = {x ∈X; ϕ(x) < ∞} of X is called the domain of ϕ. (5) The epigraph of ϕ is deﬁned by epi(ϕ) = {(x, c) ∈X × R : ϕ(x) ≤c}. Lemma 3 A convex functional ϕ is lower-semicontinuous if and only if it is weakly lower-semicontinuous on X. Proof: Since the level set {x ∈X : ϕ(x) ≤c} is a closed convex subset if ϕ is lower-semicontinuous. Thus, the claim follows the fact that a convex subset of X is closed if and only if it is weakly closed. Lemma 4 If ϕ be a proper lower-semicontinuous, convex functional on X, then ϕ is bounded below by an aﬃne functional, i.e., there exist x∗∈X∗and c ∈R such that ϕ(x) ≥⟨x∗, x⟩+ β, x ∈X. 167 Proof: Let x0 ∈X and β ∈R be such that ϕ(x0) > c. Since ϕ is lower-semicontinuous on X, there exists an open neighborhood V (x0) of X0 such that ϕ(x) > c for all x ∈V (x0). Since the ephigraph epi(ϕ) is a closed convex subset of the product space X ×R. It follows from the separation theorem for convex sets that there exists a closed hyperplane H ⊂X × R; H = {(x, r) ∈X × R : ⟨x∗ 0, x⟩+ r = α} with x∗ 0 ∈X∗, α ∈R, that separates epi(ϕ) and V (x0) × (−∞, c). Since {x0} × (−∞, c) ⊂{(x, r) ∈X × R : ⟨x∗ 0, x⟩+ r < α} it follows that ⟨x∗ 0, x⟩+ r > α for all (x, c) ∈epi(ϕ) which yields the desired estimate. Theorem C.6 If F : X →(−∞, ∞] is convex and bounded on an open set U, then F is continuous on U. Proof: We choose M ∈R such that F(x) ≤M −1 for all x ∈U. Let ˆ x be any element in U. Since U is open there exists a δ > 0 such that the open ball {x ∈X : \|x−ˆ x\| < δ is contained in U. For any epsilon ∈(0, 1), let θ = ϵ M −F(ˆ x). Then for x ∈X satisfying \|x −ˆ x\| < θ δ \|x −ˆ x θ + ˆ x −ˆ x\| = \|x −ˆ x\| θ < δ Hence x −ˆ x θ + ˆ x ∈U. By the convexity of F F(x) ≤(1 −θ)F(ˆ x) + θ F(x −ˆ x θ + ˆ x) ≤(1 −θ)F(ˆ x) + θ M and thus F(x) −F(ˆ x) < θ(M −F(ˆ x) = ϵ Similarly, ˆ x −x θ + ˆ x ∈U and F(ˆ x) ≤ θ 1 + θ F( ˆ x −x θ + ˆ x) + 1 1 + θ F(x) < θM 1 + θ + 1 1 + θ F(x) which implies F(x) −F(ˆ x) > −θ(M −F(ˆ x) = −ϵ Therefore \|F(x) −F(¯ x)\| < ϵ if \|x −ˆ x\| < θδ and F is continuous in U. □ Deﬁnition (Subdiﬀerential) Given a proper convex functional ϕ on a Banach space X the subdiﬀerential of ∂ϕ(x) is a subset in X∗, deﬁned by ∂ϕ(x) = {x∗∈X∗: ϕ(y) −ϕ(x) ≥⟨x∗, y −x⟩for all y ∈X}. Since for x∗ 1 ∈∂ϕ(x1) and x∗ 2 ∈∂ϕ(x2), ϕ(x1) −ϕ(x2) ≤⟨x∗ 2, x1 −x2⟩ ϕ(x2) −ϕ(x1) ≤⟨x∗ 1, x2 −x1⟩ 168 it follows that ⟨x∗ 1 −x∗ 2, x1 −x2⟩≥0. Hence ∂ϕ is a monotone operator from X into X∗. Example 1 Let ϕ be Gateaux diﬀerentiable at x. i.e., there exists w∗∈X∗such that lim t→0+ ϕ(x + t v) −ϕ(x) t = ⟨w∗, h⟩ for all h ∈X and w∗is the Gateaux diﬀerential of ϕ at x and is denoted by ϕ′(x). If ϕ is convex, then ϕ is subdiﬀerentiable at x and ∂ϕ(x) = {ϕ′(x)}. Indeed, for v = y −x ϕ(x + t (y −x)) −ϕ(x) t ≤ϕ(y) −ϕ(x), 0 < t < 1 Letting t →0+ we have ϕ(y) −ϕ(x) ≥⟨ϕ′(x), y −x⟩ for all y ∈X, and thus ϕ′(x) ∈∂ϕ(x). On the other hand if w∗∈∂ϕ(x) we have for y ∈X and t > 0 ϕ(x + t y) −ϕ(x) t ≥⟨w∗, y⟩. Taking limit t →0+, we obtain ⟨ϕ′(x) −w∗, y⟩≥0 for all y ∈X. This implies w∗= ϕ′(x). Example 2 If ϕ(x) = 1 2 \|x\|2 then we will show that ∂ϕ(x) = F(x), the duality map-ping. In fact, if x∗∈F(x), then ⟨x∗, x −y⟩= \|x\|2 −⟨y, x∗⟩≥1 2 (\|x\|2 −\|y\|2) for all y ∈X. Thus x∗∈∂ϕ(x). Conversely, if x∗∈∂ϕ(x), then 1 2 (\|y\|2 −\|x\|2) ≥⟨x∗, y −x⟩ for all y ∈X (8.1) We let y = t x, 0 < t < 1 and obtain 1 + t 2 \|x\|2 ≤⟨x, x∗⟩ and thus \|x\|2 ≤⟨x, x∗⟩. Similarly, if t > 1, then we conclude \|x\|2 ≥⟨x, x∗⟩and therefore \|x\|2 = ⟨x, x∗⟩and \|x∗\| ≥\|x\|. On the other hand, letting y = x + λ u, λ > 0 in (8.1), we have λ ⟨x∗, u⟩≤1 2 (\|x + λ u\|2 −\|x\|2) ≤λ \|u\|\|x\| + λ \|u\|2, which implies ⟨x∗, u⟩≤\|u\|\|x\|. Hence \|x∗\| ≤\|x\| and we obtain \|x\|2 = \|x∗\|2 = ⟨x∗, x⟩. Example 3 Let K be a closed convex subset of X and IK be the indicator function of K, i.e., IK(x) = 0 if x ∈K ∞ otherwise. 169 Obviously, IK is convex and lower-semicontinuous on X. By deﬁnition we have for x ∈K ∂IK(x) = {x∗∈X∗: ⟨x∗, x −y⟩≥0 for all y ∈K} Thus D(IK) = D(∂IK) = K and ∂K(x) = {0} for each interior point of K. Moreover, if x lies on the boundary of K, then ∂IK(x) coincides with the cone of normals to K at x. Note that ∂F(x) is closed and convex and may be empty. Theorem C.10 If a convex function F is continuous at ¯ x then ∂F(¯ x) is non empty. Proof: Since F is continuous at x for any ϵ > 0 there exists a neighborhood Uϵ of ¯ x such that F(x) ≤F(¯ x) + ϵ, x ∈Uϵ. Then Uϵ × (F(¯ x) + ϵ, ∞) is an open set in X × R and is contained in epi F. Hence (epi F)o is non empty. Since F is convex epi F is convex and (epi F)o is convex. For any neighborhood of O of (¯ x, F(¯ x)) there exists a t < 1 such that (¯ x, tF(¯ x)) ∈O. But, tF(¯ x) < F(¯ x) and so (¯ x, tF(¯ x)) / ∈epi F. Thus (¯ x, F(¯ x)) / ∈(epi F)o. By the Hahn Banach separation theorem, there exists a closed hyperplane S = {(x, a) ∈X × R : ⟨x∗, x⟩+ α a = β} for nontrivial (x∗, α) ∈X∗× R and β ∈R such that ⟨x∗, x⟩+ α a > β for all (x, a) ∈(epi F)o ⟨x∗, ¯ x⟩+ α F(¯ x) = β. (8.2) Since (epi F)o = epi F every neighborhood of (x, a) ∈epi F contains an element of (epi ϕ)o. Suppose ⟨x∗, x⟩+ α a < β. Then {(x′, a′) ∈X × R : ⟨x∗, x′⟩+ α a′ < β} is an neighborhood of (x, a) and contains an element of (epi F)o, which contradicts to (8.2). Hence ⟨x∗, x⟩+ α a ≥β for all (x, a) ∈epi F. (8.3) Suppose α = 0. For any u ∈Uϵ there is an a ∈R such that F(u) ≤a. Then from (8.3) ⟨x∗, u⟩= ⟨x∗, u⟩+ α a ≥β and thus ⟨x∗, u −¯ x⟩≥0 for all u ∈Uϵ. Choose a δ such that \|u −¯ x\| ≤δ implies u ∈U. For any nonzero element x ∈X let t = δ \|x\|. Then \|(tx + ¯ x) −¯ x\| = \|tx\| = δ so that tx + ¯ x ∈Uϵ. Hence ⟨x∗, x⟩= ⟨x∗, (tx + ¯ x) −¯ x⟩/t ≥0. Similarly, −t x + ¯ x ∈Uϵ and ⟨x∗, x⟩= ⟨x∗, (−tx + ¯ x) −¯ x⟩/(−t) ≤0. Thus, ⟨x∗, x⟩and x∗= 0, which is a contradiction. Therefore α is nonzero. It now follows from (8.2)–(8.3) that ⟨−x∗ α , x −¯ x⟩+ F(¯ x) ≤F(x) 170 for all x ∈X and thus −x∗ α ∈∂F(¯ x). □ Deﬁnition (Conjugate function) Given a proper functional ϕ on a Banach space X, the conjugate functional on X∗is deﬁned by ϕ∗(x∗) = sup x {⟨x∗, x⟩−ϕ(x)}. Examples (Conjugate functions) (1) If ϕ(x) = ⟨a, x⟩−b is a bounded aﬁne functional, then ϕ∗(x∗) = Ix∗=a. (2) If ϕ(x) = 1 p\|x\|p, then ϕ∗(x) = 1 q\|x∗\|p. (3) If ϕ(x) = \|x\|, then ϕ∗(x∗) = I\|x∗\|≤1. Theorem 2.1 The conjugate function ϕ∗is convex, lower semi-continuous and proper. Proof: For 0 < t < 1 and x∗ 1, x∗ 2 ∈X∗ ϕ∗(t x∗ 1 + (1 −t) x∗ 2) = supx{⟨t x∗ 1 + (1 −t) x∗ 2, x⟩−ϕ(x)} ≥t supx{⟨x∗ 1, x⟩−ϕ(x)} + (1 −t) supx{⟨x∗ 2, x⟩−ϕ(x)} = t ϕ∗(x∗ 1) + (1 −t) ϕ∗(x∗ 2). Thus, ϕ∗is convex. For ϵ > 0 arbitrary let y ∈X be ϕ∗(x∗) −ϵ ≤⟨x∗, y⟩−ϕ(y) For xn →x∗ ϕ∗(x∗ n) ≥⟨x∗ n, y⟩−ϕ(y) and lim inf n→∞ϕ∗(x∗ n) ≥⟨x∗, y⟩−ϕ(y) Since ϵ > arbitrary, ϕ∗is lower semi-continuous. □ Deﬁnition (Bi-Conjugate Function) For any proper functional ϕ on X the bi-conjugate function of ϕ is ϕ∗∗: X →(−∞, ∞] is deﬁned by ϕ∗∗(x) = sup x∗∈X∗{⟨x∗, x⟩−ϕ∗(x∗)}. Theorem C.3 For any F : X →(−∞, ∞] F ∗∗is convex and l.s.c. and F ∗∗≤F. Proof: Note that F ∗∗(¯ x) = sup x∗∈D(F ∗) {⟨x∗, ¯ x⟩−F ∗(x∗)}. Since for x∗∈D(F ∗) ⟨x∗, x⟩−F ∗(x∗) ≤F(x) for all x, F ∗∗(¯ x) ≤F(¯ x). □ Theorem If F is a proper l.s.c. convex function, then F = F ∗∗. If X is reﬂexive, then (F ∗)∗= F ∗∗. 171 Proof: By Theorem if F is a proper l.s.c. convex function, then F is bounded below by an aﬃne functional ⟨x∗, x⟩−a, i.e. a ≥⟨x∗, x⟩−F(x) for all x ∈X. Deﬁne a(x∗) = inf{a ∈R : a ≥⟨x∗, x⟩−F(x) for all x ∈X} Then for a ≥a(x∗) ⟨x∗, x⟩−a ≤⟨x∗, x⟩−a(x∗) for all x ∈X. By Theorem again F(¯ x) = sup x∗∈X∗{⟨x∗, ¯ x⟩−a(x∗)}. Since a(x∗) = sup x∈X {⟨x∗, x⟩−F(x)} = F ∗(x) we have F(x) = sup x∗∈X∗{⟨x∗, x⟩−F ∗(x∗)}.□ We have the duality property. Theorem 2.2 Let F be a proper convex function on X. (1) For F : X →(−∞, ∞] x∗∈∂F(¯ x) is if and only if F(¯ x) + F ∗(x∗) = ⟨x∗, ¯ x⟩. (2) Assume X is reﬂexive. If x∗∈∂F(¯ x), then ¯ x ∈∂F ∗(x∗). If F is convex and lower semi-continuous, then x∗∈∂F(¯ x) if and only if ¯ x ∈∂F ∗(x∗). Proof: (1) Note that x∗∈∂F(¯ x) is if and only if ⟨x∗, x⟩−F(x) ≤⟨x∗, ¯ x⟩−F(¯ x) (8.4) for all x ∈X. By the deﬁnition of F ∗this implies F ∗(x∗) = ⟨x∗, ¯ x⟩−F(¯ x). Conversely, if F(x∗) + F(¯ x) = ⟨x∗, ¯ x⟩then (8.4) holds for all x ∈X. (2) Since F ∗∗≤F by Theorem C.3, it follows from (1) F ∗∗(¯ x) ≤⟨x∗, ¯ x⟩−F ∗(x∗) But by the deﬁnition of F ∗∗ F ∗∗(¯ x) ≥⟨x∗, ¯ x⟩−F ∗(x∗). Hence F ∗(x∗) + F ∗∗(¯ x) = ⟨x∗, ¯ x⟩. Since X is reﬂexive, F ∗∗= (F ∗)∗. If we apply (1) to F ∗we have ¯ x ∈∂F ∗(x∗). In addition if F is convex and l.s.c, it follows from Theorem C.1 that F ∗∗= F. Thus if x∗∈∂F ∗(¯ x) then F(¯ x) + F ∗(x∗) = F ∗(x∗) + F ∗∗(x) = ⟨x∗, ¯ x⟩ by applying (1) for F ∗. Therefore x∗∈∂F(¯ x) again by (1). □ 172 Theorem(Rockafellar) Let X be real Banach space. If ϕ is lower-semicontinuous proper convex functional on X, then ∂ϕ is a maximal monotone operator from X into X∗. Proof: We prove the theorem when X is reﬂexive. By Apuland theorem we can assume that X and X∗are strictly convex. By Minty-Browder theorem ∂ϕ it suﬃces to prove that R(F + ∂ϕ) = X∗. For x∗ 0 ∈X∗we must show that equation x∗ 0 ∈Fx + ∂ϕ(x) has at least a solution x0 Deﬁne the proper convex functional on X by f(x) = 1 2 \|x\|2 X + ϕ(x) −⟨x∗ 0, x⟩. Since f is lower-semicontinuous and f(x) →∞as \|x\| →∞there exists x0 ∈D(f) such that f(x0) ≤f(x) for all x ∈X. Since F is monotone ϕ(x) −ϕ(x0) ≥⟨x∗ 0, x −x0, ⟩−⟨x −x0, F(x)⟩. Setting xt = x0 + t (u −x0) and since ϕ is convex, we have ϕ(u) −ϕ(x0) ≥1 t (ϕ(xt) −ϕ(x0)) ≥⟨x∗ 0, u −x0, ⟩−⟨F(xt), u −x0⟩. Taking limit t →0+, we obtain ϕ(u) −ϕ(x0) ≥⟨x∗ 0, u −x0⟩−⟨F(x0), u −x0⟩, which implies x∗ 0 −F(x0) ∈∂ϕ(x0). □ We have the perturbation result. Theorem Assume that X is a real Hilbert space and that A is a maximal monotone operator on X. Let ϕ be a proper, convex and lower semi-continuous functional on X satisfying dom(A) ∩dom(∂ϕ) is not empty and ϕ((I + λ A)−1x) ≤ϕ(x) + λ M, for all λ > 0, x ∈D(ϕ), where M is some non-negative constant. Then the operator A + ∂ϕ is maximal mono-tone. We use the following lemma. Lemma Let A and B be m-dissipative operators on X. Then for every y ∈X the equation y ∈−Ax −Bλx (8.5) has a unique solution x ∈dom (A). Proof: Equation (10.15) is equivalent to y = xλ −wλ −Bλxλ for some wλ ∈A(xλ). Thus, xλ − λ λ + 1wλ = λ λ + 1y + 1 λ + 1(xλ + λ Bλxλ) = λ λ + 1y + 1 λ + 1(I −λ B)−1. 173 Since A is m-dissipative, we conclude that (10.15) is equivalent to that xλ is the ﬁxed point of the operator Fλx = (I − λ λ + 1 A)−1( λ λ + 1 y + 1 λ + 1 (I −λ B)−1x). By m-dissipativity of the operators A and B their resolvents are contractions on X and thus \|Fλx1 −Fλx2\| ≤ λ λ + 1 \|x1 −x2\| for all λ > 0, x1, x2 ∈X. Hence, Fλ has the unique ﬁxed point xλ and xλ ∈dom (A) solves (10.15). □ Proof of Theorem: From Lemma there exists xλ for y ∈X such that y ∈xλ −(−A)λxλ + ∂ϕ(xλ) Moreover, one can show that \|xλ\| is bounded uniformly. Since y −xλ + (−A)λxλ ∈∂ϕ(xλ) for z ∈X ϕ(z) −ϕ(xλ) ≥(z −xλ, y −xλ + (−A)λxλ) Letting λ(I + λA)−1x, so that z −xλ = λ(−A)λxλ and we obtain (λ(−A)λxλ, y −xλ + (−A)λxλ) ≤ϕ((I + λ A)−1) −ϕ(xλ) ≤λ M, and thus \|(−A)λxλ\|2 ≤\|(−A)λxλ\|\|y −xλ\| + M. Since xλ\| is bounded and so that \|(−A)λxλ\|. □ 8.2 Duality Theory In this section we discuss the duality theory. We call (P) inf x∈X F(x) is the primal problem, where X is a Banach space and F : X →(−∞, ∞] is a proper l.s.c. convex function. We have the following result for the existence of minimizer. Theorem 2 Let X be reﬂexive and F be a lower semicontinuous proper convex func-tional deﬁned on X. Suppose F lim F(x) = ∞ as \|x\| →∞. (8.6) Then there exists an ¯ x ∈X such that F(¯ x) = inf {F(x); x ∈X}. Proof: Let η = inf{F(x); x ∈X} and let {xn} be a minimizing sequence such that lim F(xn) = η. The condition (8.6) implies that {xn} is bounded in X. Since X is reﬂexive there exists a subsequence that converges weakly to ¯ x in X and it follows from Lemma 1 that F(¯ x) = η. □ 174 We imbed (P) into a family of perturbed problem (Py) inf x∈X Φ(x, y) where y ∈Y , a Banach space is an imbedding variable and Φ : X × Y →(−∞, ∞] is a proper l.s.c. convex function with Φ(x, 0) = F(x). Thus (P0) = (P). For example in terms of (13.1) we let F(x) = f(x) + ϕ(Λx) (8.7) and with Y = H Φ(x, y) = f(x) + ϕ(Λx + y) (8.8) Deﬁnition The dual problem of (P) with respect to Φ is (P ∗) sup y∗∈Y (−Φ∗(0, y∗)). The value function of (Py) is deﬁned by h(y) = inf x∈X Φ(x, y), y ∈Y. We assume that h(y) > −∞for all y ∈Y . Then we have Theorem D.0 inf (P) ≥sup (P ∗). Proof. For any (x, y) ∈X × Y and (x∗, y∗) ∈X∗× Y ∗ Φ∗(x∗, y∗) ≥⟨x∗, x⟩+ ⟨y∗, y⟩−Φ(x, y). Thus, letting y = 0 and x∗= 0 F(x) + Φ∗(0, y∗) ≥⟨0, x⟩+ ⟨y∗, 0⟩= 0 for all x ∈X and y∗∈Y ∗. Therefore inf (P) = inf F(x) ≥sup y∗∈Y ∗(−Φ∗(0, y∗)) = sup (P ∗).□ In what follows we establish the equivalence of the primal problem (P) and the dual problem (P ∗). We start with the properties of h. Lemma D.1 h is convex. Proof. The proof is by contradiction. Suppose there exist y1, y2 ∈Y and θ ∈(0, 1) such that h(θ y1 + (1 −θ) y2) > θ h(y1) + (1 −θ) h(y2). Then there exist c and ϵ > 0 such that h(θ y1 + (1 −θ) y2) > c > c −ϵ > θ h(y1) + (1 −θ) h(y2). Let a1 = h(y1) + ϵ θ > h(y1) and a2 = c −θ a1 1 −θ = c −ϵ −θ h(y1) 1 −θ > h(y2). 175 By deﬁnition of h there exist x1, x2 ∈X such that h(y1) ≤Φ(x1, y1) ≤a1 and h(y2) ≤Φ(x2, y2) ≤a2. Thus h(θ y1 + (1 −θ) y2) ≤Φ(θ x1 + (1 −θ) x2, θ y1 + (1 −θ) y2) ≤θ Φ(x1, y1) + (1 −θ) Φ(x2, y2) ≤θ a1 + (1 −θ) a2 = c, which is a contradiction. Hence h is convex. □ Lemma D.2 For all y∗∈Y ∗, h∗(y∗) = Φ∗(0, y∗). Proof. h∗(y∗) = supy∈Y (⟨y∗, y⟩−h(y)) = supy∈Y (⟨y∗, y⟩−infx∈X Φ(x, y)) = sup y∈Y sup x∈X ((⟨y∗, y⟩−Φ(x, y)) = sup y∈Y sup x∈X (⟨0, x⟩+ ⟨y∗, y⟩−Φ(x, y)) = sup(x,y)∈X×Y (⟨(0, y∗), (x, y)⟩−Φ(x, y)) = Φ∗(0, y∗).□ Note that h is not necessarily l.s.c.. Theorem D.3 If h is l.s.c. at 0, then inf (P) = sup (P ∗). Proof. Since F is proper, h(0) = infx∈X F(x) < ∞. Since h is convex by Lemma D.1, it follows from Theorem C.4 that h(0) = h∗∗(0). Thus sup (P ∗) = supy∗∈Y ∗(−Φ(0, y∗)) = supy∗∈Y ∗((⟨y∗, 0⟩−h∗(y∗)) = h∗∗(0) = h(0) = inf (P)□ Theorem D.4 If h is subdiﬀerentiable at 0, then inf (P) = sup (P ∗) and ∂h(0) is the set of solutions of (P ∗). Proof. By Lemma D.2 ¯ y∗solves (P ∗) if and only if −h∗(¯ y∗) = −Φ(0, ¯ y∗) = supy∗∈Y ∗(−Φ(0, y∗)) = supy∗(⟨y∗, 0) −h∗(y∗)) = h∗∗(0). By Theorem 2.2 h∗∗(0) + h∗∗∗(¯ y∗) = ⟨¯ y∗, 0) = 0 if and only if ¯ y∗∈∂h∗∗(0). Since h∗∗∗= h∗by Theorem C.5, y∗solves (P ∗) if and only if y∗∈∂h∗∗(y∗). Since ∂h(0) is not empty, ∂h(0) = ∂h∗∗(0). Therefore ∂h(0) is the set of all solutions of (P ∗) and (P ∗) has at least one solution. Let y∗∈∂h(0). Then ⟨y∗, x⟩+ h(0) ≤h(x) for all x ∈X. Let {xn} is a sequence in X such that xn →0. lim inf n→∞h(xn) ≥lim⟨y∗, xn⟩+ h(0) = h(0) and h is l.s.c. at 0. By Theorem D.3 inf (P) = sup (P ∗). □ 176 Corollary D.6 If there exists an ¯ x ∈X such that Φ(¯ x, ·) is ﬁnite and continuous at 0, h is continuous on an open neighborhood U of 0 and h = h∗∗. Moreover, inf (P) = sup (P ∗) and ∂h(0) is the set of solutions of (P ∗). Proof. First show that h is continuous. Clearly, Φ( ¯ X, ·) is bounded above on an open neighborhood U of 0. Since for all y ∈Y h(y) ≤Φ(¯ x, y), his bounded above on U. Since h is convex by Lemma D.1 and h is continuous by Theorem C.6. Hence h = h∗∗by Theorem C. Now, h is subdiﬀerential at 0 by Theorem C.10. The conclusion follows from Theorem D.4. □ Example D.4 Consider the example (8.7)–(8.8), i.e., Φ(x, y) = f(x) + ϕ(Λx + y) (8.9) for (13.1). Let us calculate the conjugate of Φ. Φ∗(x∗, y∗) = supx∈X supy∈Y {⟨x∗, x⟩+ ⟨y∗, y⟩−Φ(x, y)} = supx∈X supy∈Y {⟨x∗, x⟩+ ⟨y∗, y⟩−f(x) −ϕ(Λx + y)} = supx∈X{⟨x∗, x⟩−f(x) supy∈Y [⟨y∗, y⟩−ϕ(Λx + y)]} where supy∈Y [⟨y∗, y⟩−ϕ(Λx + y)] = supy∈Y [⟨y∗, Λx + y⟩−ϕ(Λx + y) −⟨y∗, Λx⟩] = supz∈Y [⟨y∗, z⟩−G(z)] −⟨y∗, Λx⟩= ϕ(y∗) −⟨y∗, Λx⟩. Thus Φ∗(x∗, y∗) = supx∈X{⟨x∗, x⟩−⟨y∗, Λx⟩−f(x) + ϕ(y∗)} = sup x∈X {⟨x∗−Λ∗y∗, x⟩−f(x) + ϕ(y∗)} = f∗(x∗−Λy∗) + ϕ∗(y∗). Theorem D.7 For any ¯ x ∈X and ¯ y∗∈Y ∗, the following statements are equivalent. (1) ¯ x solves (P), ¯ y∗solves (P ∗), and min (P) = max (P ∗). (2) Φ(¯ x) + Φ∗(0, ¯ y∗) = 0. (3) (0, ¯ y∗) ∈∂Φ(¯ x, 0). Proof: Suppose (1) holds, then obviously (2) holds. Suppose (2) holds, then since Φ(¯ x, 0) = F(¯ x) ≥inf (P) ≥sup (P ∗) ≥−Φ(0, ¯ y∗) by Theorem D.2, thus Φ(¯ x, 0) = min (P) = sup (P ∗) = −Φ∗(0, ¯ y∗) 177 Therefore (2) implies (1). Since ⟨(0, ¯ y∗), (¯ x, 0)⟩= 0, (2) and (3) by Theorem C.7. □ Any solution y∗of (P ∗) is called a Lagrange multiplier associated with Φ. In fact for the example (8.9) the optimality condition implies 0 = Φ(¯ x, 0) + Φ∗(0, ¯ y∗) = f(¯ x) + f∗(−Λ∗¯ y∗) + ϕ(Λ¯ x) + ϕ(¯ y∗) = [f(¯ x) + f∗(−Λ∗¯ y∗) −⟨−Λ∗¯ y∗, ¯ x⟩] + [ϕ(Λ¯ x) + ϕ∗(¯ y∗) −⟨¯ y∗, Λ¯ x)] (8.10) Since each expression of (8.10) in square brackets is nonnegative, it follows that f(¯ x) + f∗(−Λ∗¯ y∗) −⟨−Λ∗¯ y∗, ¯ x⟩= 0 ϕ(Λ¯ x) + ϕ∗(¯ y∗) −⟨¯ y∗, Λ¯ x) = 0 By Theorem 2.2 0 ∈∂f(¯ x) + Λ∗¯ y∗ ¯ y∗∈∂ϕ(Λ¯ x). (8.11) The function L : X × Y ∗→(−∞, ∞] deﬁned by L(x, y∗) = inf y∈Y {Φ(x, y) −⟨y∗, y⟩} (8.12) is called the Lagrangian. Note that Φ∗(x∗, y∗) = sup x∈X, y∈Y {⟨x∗, x⟩+ ⟨y∗, y⟩−Φ(x, y)} = sup x∈X {⟨x∗, x⟩+ sup y∈Y {⟨y∗, y⟩−Φ(x, y)}} = sup x∈X {⟨x∗, x⟩−L(x, y∗)} Thus, −Φ∗(0, y∗) = inf x∈X L(x, y∗). (8.13) Hence the dual problem (P ∗) is equivalently written as sup y∗∈Y ∗inf x∈X L(x, y∗). Similarly, if Φ is a proper convex l.s.c. function, then the bi-conjugate of Φ in y, Φ∗∗ x = Φ and Φ(x, y) = Φ∗∗ x (y) = supy∗∈Y ∗{⟨y∗, y⟩−Φ∗ x(y∗)} = sup y∗∈Y ∗{⟨y∗, y⟩+ L(x, y∗)}. Hence Φ(x, 0) = sup y∗∈Y ∗L(x, y∗) (8.14) and the primal problem (P) is equivalently written as inf x∈X sup y∗∈Y L(x, y∗). 178 By introducing the Lagrangian L, the Problems (P) and (P ∗) are formulated as min-max problems, which arise in the game theory. By Theorem D.0 sup y∗inf x L(x, y∗) ≤inf x sup y∗ L(x, y∗). Theorem D.8 (Saddle Point) Assume Φ is a proper convex l.s.c. function. Then the followings are equivalent. (1) (¯ x, ¯ y∗) ∈X × Y ∗is a saddle point of L, i.e., L(¯ x, y∗) ≤L(¯ x, ¯ y∗) ≤L(x, ¯ y∗) for all x ∈X, y∗∈Y ∗. (8.15) (2) ¯ x solves (P), ¯ y∗solves (P ∗), and min (P) = max (P ∗). Proof: Suppose (1) holds. From (8.13) and (8.15) L(¯ x, ¯ y∗) = inf x∈X L(x, ¯ y∗) = −Φ∗(0, ¯ y∗) and form (8.14) and (8.15) L(¯ x, ¯ y∗) = sup y∗∈Y ∗L(¯ x, y∗) = Φ(¯ x, 0) Thus, Φ∗(¯ x, 0) + Φ(0, ¯ y∗) = 0 and (2) follows from Theorem D.7. Conversely, if (2) holds, then from (8.13) and (8.14) −Φ∗(0, ¯ y∗) = infx∈X L(x, ¯ y∗) ≤L(¯ x, ¯ y∗) Φ∗(¯ x, 0) = supy∗∈Y ∗L(¯ x, y) ≥L(¯ x, ¯ y∗). By Theorem D.7 −Φ∗(0, ¯ y∗) = Φ∗(¯ x, 0) and (8.15) holds. □ Theorem D.8 implies that no duality gap between (P) and (P ∗) is equivalent to the saddle point property of the pair (¯ x, ¯ y∗). For the example L(x, y∗) = inf y∈Y {f(x) + ϕ(Λx + y) −⟨y∗, y⟩} = f(x) + ⟨y∗, Λx⟩−ϕ∗(y∗) (8.16) If (¯ x, ¯ y∗) is a saddle point, then from (8.15) 0 ∈∂f(¯ x) + Λ∗¯ y∗ 0 ∈Λ¯ x −∂ϕ∗(¯ x). (8.17) It follows from Theorem 2.2 that the second equation is equivalent to ¯ y∗∈∂ϕ(Λ¯ x) and (8.17) is equivalent to (8.11). 8.3 p-Laplacian Φ(u, p) = Z ϕ(p) + (λ, ∆u −p) 179 9 Optimization Algorithms In this section we discuss iterative methods for ﬁnding a solution to the optimization problem. First, consider the unconstrained minimization: min J(x) over a Hilbert space X. Assume J : X →R+ is continuously diﬀerentiable. If u∗∈X minimizes J, then the necessarily optimality is given by J′(u∗) = 0. The gradient method is given by xn+1 = xn −α J′(xn) (9.1) where α > 0 is a stepsize chosen so that the decent J(xn+1) < J(xn) holds. Let gn = J′(xn) and since J(xn −α gn) −J(xn) = −α \|J′(xn)\|2 + o(α), (9.2) there exits a α > 0 satisﬁes the decent property. A stepsize α > 0 can be determined by min α>0 J(xn −α gn) (9.3) or by a line search method (e.g, Amijo-rule) Next, consider the constrained minimization min J(x) over x ∈C. (9.4) where C is a closed convex set in X. It follows from Theorem 7 that if J : X →R is C1 and x∗∈C is a minimizer, then (J′(x∗), x −x∗) ≥0 for all x ∈C. (9.5) Deﬁne the orthogonal projection operator PC of X onto C, i.e., z∗= PCx minimizes \|z −x\|2 over z ∈C. From Theorem 7, z∗∈C satisﬁes (z∗−x, z −z∗) ≥0 for all z ∈C. (9.6) Note that (9.5) is equivalent to (x∗−(x∗−α J′(x∗)), x −x∗) ≥0 for all x ∈C and α > 0. Thus, it is follows from (9.6) that (9.5) is equivalent to x∗= PC(x∗−α J′(x∗)). The projected gradient method for (9.4) is deﬁned by xn+1 = PC(xn −αn J′(xn)) (9.7) with a stepsize αn > 0. 180 Deﬁne the the Hessian H = J′′(x0) ∈L(X, X) of J at x0 by J(x) −J(x0) −(J′(x0), x −x0) = 1 2 (x −x0, H(x −x0)) + o(\|x −x0\|2). Thus, we minimize over x J(x) ∼J(x0) + (J′(x0), x −x0) + 1 2 (x −x0, H(x −x0)) we obtain the damped-Newton method for equation given by xn+1 = xn −α H(xn)−1J′(xn), (9.8) where α > 0 is a stepsize and α = 1 corresponds to the Newton method. In the case of min J(x) over x ∈C we obtain the projected Newton method xn+1 = ProjC(xn −α H(xn)−1J′(xn)). (9.9) Computing the Hessian can be expensive and numerical evaluation can also be less accurate. The ﬁnite rank scant update (BFGS and GFP) is used to substitute H (Quasi-Newton method). It is a variable merit method and such updates are pre-conditioner for the gradient method. Consider the (regularized) nonlinear least square problem min 1 2\|F(x)\|2 Y + α 2 (Px, x)X (9.10) where F : X →Y is C1 and P is a nonnegative self-adjoint operator on a Hilbert space X. The Gauss-Newton method is an iterative method of the form xn+1 = argmin{1 2\|F ′(xn)(x −xn) + F(xn)\|2 Y + α 2 (Px, x)X}, i.e., xn+1 = xn −(F ′(xn)∗F(xn) + α P)−1(F ′(xn)∗F(xn)). (9.11) 9.1 Constrained minimization and Lagrange multiplier method Consider the constrained minimization min J(x, u) = F(x) + H(u) subject to E(x, u) = 0 and u ∈C. (9.12) We use the implicit function theory for developing algorithms for (9.12). Implicit Function Theory Let E : X ×U →X is C1. Suppose a pair (¯ x, ¯ u) satisﬁes E(x, u) = 0 and Ex(¯ x, ¯ u) is bounded invertible. Then the exists a δ > 0 such that for \|u −¯ u\| < δ equation E(x, u) = 0 has a locally deﬁned unique solution x = Φ(u). Moreover, Φ : U →X is continuously diﬀerentiable at ¯ u and ˙ x = Φ′(¯ x)(d) satisﬁes Ex(¯ x, ¯ u) ˙ x + Eu(¯ x, ¯ u)d = 0. Theorem (Lagrange Calculus) Assume that E(x, u) = 0 and Ex(x, u) is bounded invertible. Then, by the implicit function theory x = Φ(u) and for J(u) = J(Φ(u), u) (J′(u), d) = (H′(u), d) + (Eu(x, u)∗λ, d) 181 where λ ∈Y satisﬁes Ex(x, u)∗λ + F ′(x) = 0. Proof: From the implicit function theory (J′, d) = (F ′(x), ˙ x) + (H′(u), d), where Ex(x, u) ˙ x + Eu(x, u) = 0. Thus, the claim follows from (F ′(x), ˙ x) = −(E∗ xλ, ˙ x) = −(λ, Ex ˙ x) = (λ, Eud).□ Hence we obtain the gradient method for equality constraint problem (9.12); Gradient method 1. Solve for xn E(xn, un) = 0 2. Solve for λn Ex(xn, un)∗λn + F ′(xn) = 0 3. Compute the gradient by gn = H′(un) + Eu(xn, un)∗λn 4. Gradient Step: Update un+1 = PC(un −α gn) with a line search for α > 0. 9.2 Conjugate gradient method In this section we develop the conjugate gradient and conjugate residual method for the saddle point problem. Consider the quadratic programing J(x) = 1 2(Ax, x)X −(b, x)X where A is positive self-adjioint operator on a Hilbert space X. Then, the negative gradient of J is given by r = −J′(x) = b −Ax. The gradient method is given by xk+1 = xk + αk pk, pk = rk, where the step size αk is selected by min J(xk + α rk) over α > 0. 182 That is, the Cauchy step is given by 0 = (J′(xk+1), pk) = α (Apk, pk) −(pk, rk) ⇒αk = (pk, rk) (Apk, pk). (9.13) It is known that the gradient method with Cauchy step searches the solution on a subspace spanned by the major and minor axes of A dominantly, and exhibits a zig-zag pattern of its iterates. If the condition number ρ = max σ(A) min σ(A) is large, then it is very slow convergent. In order to improve the convergence we introduce the conjugacy of the search directions pk and it can be achieved by the conjugate gradient method. For example, a new search direction pk is extracted for the residual rk = b −Axk by pk = rk −β pk−1 and the orthogonality 0 = (Apk, pk−1) = (rk, Apk−1) −β(Apk−1, pk−1) implies βk = (rk, Apk−1) (Apk−1, pk−1). (9.14) Let P is the pre-conditioner, self-adjiont operator so that the condition number of P −1 2 AP −1 2 is much smaller that the one for A. If we apply (9.13)(9.14) with the pre-conditioner P, we have the pre-conditioned conjugate gradient method: • Initialize x0 and set r0 = b −Ax0, z0 = P −1r0, p0 = z0 • Iterate on k until (rk, zk) is suﬃciently small αk = (rk, zk) (pk, Apk), xk+1 = xk + αk pk, rk+1 = rk −αk Apk. zk+1 = P −1rk+1, βk = (zk+1, rk+1) (zk, rk) , pk+1 := zk+1 + βk pk Here, zk is the pre-conditioned residual and (rk, zk) = (rk, P −1rk). The above formulation is equivalent to applying the conjugate gradient method without precon-ditioned system P −1 2 AP −1 2 ˆ x = P −1 2 b, where ˆ x = P 1 2 x. We have the conjugate gradient theory. Theorem (Conjugate gradient method) (1) We have conjugate property (rk, zj) = 0, (pk, Apj) = 0 for 0 ≤j < k. 183 (2) xm+1 minimizes J(x) over x ∈x0 + span{p0, · · · , pm}. Proof: For k = 1 (r1, z0) = (r0, z0) −α0 (Az0, z0) = 0 and since (r1, z1) = (r0, z1) −α0 (r1, Az0) (p1, Ap0) = (z1, Az0) + β0(z0, Az0) = −1 α0 (r1, z1) + β0(z0, Az0) = 0 Note that span{p0, · · · , pk} = span{z0, · · · , zk}. In induction in k, for j < k (rk+1, zj) = (rk, zj) −αk(Apk, pj) = 0 and ((rk+1, zk) = (rk, zk) −αk(Apk, pk) = 0 Also, for j < k (pk+1, Apj) = (zk+1, Apj) + βk(pk, Apj) = −1 αj (zk+1, rj+1 −rj) = 0 and (pk+1, Apk) = (zk+1, Apk) + βk(pk, Apk) = −1 αk (rk+1, zk) + βk(pk, Apk) = 0. Hence (1) holds. For (2) we note that (J′(xm+1), pj) = (rm+1, pj) = 0 for j ≤m and xm+1 = x0 + span{p0, · · · , pm}. Thus, (2) holds. □ Remark Note that rk = r0 − k−1 X j=0 αjP −1Apj Thus, \|rk\|2 = \|r0\|2 −(2 k−1 X αj=0P −1Apj, r0) + \|αj\|. 9.3 Conjugate residual method Consider the constrained quadratic programing min 1 2(Qy, y)X −(a, y)X subject to Ey = b in Y. Then the necessary optimality condition is written for x = (y, λ) ∈X × Y : A(y, λ) = (a, b) where A =   Q E∗ E 0  , where A is self-adjoint but is indeﬁnite. 184 In general we assume A is self-adjoint and invertible. Consider the least square problem 1 2\|Ax −b\|2 X = 1 2(AAx, x)X −(Ab, x)X + 1 2\|b\|2. We consider the conjugate directions {pk} satisfying (Apk, Apj) = 0, j < k. The pre-conditioned residual method may be derived in the same way as done for the conjugate gradient method: • Initialize x0 and set r0 = P −1(b −Ax0), p0 = r0, (Ap)0 = Ap0 • Iterate with k until (rk, rk) is suﬃciently small αk = (rk, Ark) ((Ap)k, P −1(Ap)k), xk+1 = xk + αk pk, rk+1 = rk −αk P −1(Ap)k. βk = (rk+1, Ark+1) (rk, Ark) , pk+1 = rk+1 + βk pk, (Ap)k+1 = Ark+1 + βk (Ap)k Theorem (Conjugate residual method) Assuming αk ̸= 0, (1) We have conjugate property (rj, Ark) = 0, (Apk, P −1Apj) = 0 for 0 ≤j < k. (2) xm+1 minimizes \|Ax −b\|2 over x ∈x0 + span{p0, · · · , pm}. Proof: For k = 1 (Ar0, r1) = (Ar0, r0) −α0 (Ap0, P −1Ap0) = 0 and since (Ap0, P −1Ap1) = (P −1Ap0, Ar1) + β0 (Ap0, P −1Ap0) and (P −1Ap0, Ar1) = 1 α0 (r0 −r1, Ar1) = −1 α0 (r1, Ar1), we have (Ap0, P −1Ap1) = −1 α0 (r1, Ar1) + β0(Ap0, P −1Ap0) = 0. Note that span{p0, · · · , pk} = span{r0, · · · , rk} and rk = r0+P −1A span{p0, · · · , pk−1}. In induction in k, for j < k (rk+1, Arj) = (rk, Arj) −αk(P −1Apk, Arj) = 0 and (Ark+1, rk) = (Ark, rk) −αk(Apk, P −1Apk) = 0 Also, for j < k (Apk+1, P −1Apj) = (Ark+1, P −1Apj) + βk(Apk, P −1Apj) = −1 αj (Ark+1, rj+1 −rj) = 0 and (Apk+1, P −1Apk) = (Ark+1, P −1Apk)+βk(Apk, P −1Apk) = −1 αk (rk+1, Ark+1)+βk(pk, Apk) = 0. Hence (1) holds. For (2) we note that (Axm+1 −b, P −1Apj) = −(rm+1, P −1Apj) = 0 for j ≤m and xm+1 = x0 + span{p0, · · · , pm}. Thus, (2) holds. □ 185 9.4 MINRES The conjugate gradient method can be viewed as a special variant of the Lanczos method for positive deﬁnite symmetric systems. The minimal residual method (MIN-RES) and symmetric LQ method (SYMMLQ) methods are variants that can be applied to symmetric indeﬁnite systems. The vector sequences in the conjugate gradient method correspond to a factorization of a tridiagonal matrix similar to the coeﬃcient matrix. Therefore, a breakdown of the algorithm can occur corresponding to a zero pivot if the matrix is indeﬁnite. The CG method is for the minimization min(x, Ax) −(b, x) Thus, for indeﬁnite matrix A the minimization property of the conjugate gradient method is no longer well-deﬁned. The MINRES methods is a variant of the conjugate gradient method that avoids the LU decomposition and does not suﬀer from breakdown. MINRES minimizes the residual in the 2-norm min \|Ax −b\|2 The convergence behavior of the conjugate gradient and MINRES methods for indeﬁ-nite systems was analyzed by Paige et al. (1995). When A is not positive deﬁnite, but symmetric, we can still construct an orthogonal basis for the Krylov subspace by three-term recurrence relations. Eliminating the search directions in the equations of the conjugate gradient method gives a recurrence Ari = ri+1ti+1,i + riti,i + ri−1ti−1,i, which can be written in matrix form as ARi = Ri+1Ti, where Ti is an (i+1)×i is the tridiagonal matrix. That is, given a symmetric matrix A and a vector b, the Lanczos process computes vectors rk and tridiagonal matrices Tk according to r0 = 0 and β1v1 = b, and then pi = Ari, αi = (ri, pi), βi+1ri+1 = pi −αi ri −βi ri−1 where ti+1,i = βi+1, ti,i = αi, ti−1,i = βi. In exact arithmetic, the columns of Ri are orthonormal and the process stops with i = ℓand βℓ+1 = 0. In this case we have the problem that (·, ·)A no longer deﬁnes an inner product. However we can still try to minimize the residual in the 2-norm by obtaining x(i) ∈Krylov subspace = {r0, Ar0, ..., Ai−1r0}, i.e. x(i) = Riy that minimizes \|Ax(i) −b\|2 = \|ARiy −b\|2 = \|Ri+1Tiy −b\|2. Now we exploit the fact that if Di+1 = diag(\|r(0)\|2, \|r(1)\|2, ..., \|r(i)\|2), 186 then Ri+1D−1 i+1 is an orthonormal transformation with respect to the current Krylov subspace \|Ax(i) −b\|2 = \|Di+1T iy −r(0)\|2\|e((1)\|2, and this ﬁnal expression can simply be seen as a minimum norm least squares problem. The element in the (i + 1, i) position of Ti can be annihilated by a simple Givens rotation and the resulting upper bidiagonal system (the other sub-diagonal elements having been removed in previous iteration steps) can simply be solved, which leads to the MINRES method (Paige and Saunders 1975). 9.5 DIIS method DIIS (direct inversion in the iterative subspace or direct inversion of the iterative sub-space), also known as Pulay mixing, is an extrapolation technique. DIIS was developed by Peter Pulay in the ﬁeld of computational quantum chemistry with the intent to ac-celerate and stabilize the convergence of the Hartree-Fock self-consistent ﬁeld method. Approximate error vectors ei = b −Api corresponding to the sequences of the trial vectors {pi}m i=1 is evaluated. With suﬃcient large m, we consider a linear combination {pi}m i=1 p = m X cipi. Then, we have r = b −Ap = X i ci(b −Api) = m X i=1 ciri for P i ci = 1. The DIIS method seeks to minimize the norm of r = b−Ap over {ci}m i=1, i.e. min 1 2 \|r\|2 subject to P i ci = 1. The necessary optimality condition is given by Bc = λ, X i ci = 1, where the symmetric matrix B is given by Bij = (ri, rj) and a constant λ is a constant Lagrange multiplier for the constraint P i ci = 1. 9.6 Nonlinear Conjugate Residual method One can extend the conjugate residual method for the nonlinear equation. Consider the equality constrained optimization min J(y) subject to E(y) = 0. The necessary optimality system for x = (y, λ) is F(y, λ) =   J′(y) + E′(y)∗λ E(y)  = 0. 187 Multiplication of vector by the Jacobian F ′ can be approximated by the diﬀerence of two residual vectors by F(xk + t y) −F(xk) t ∼Ay with t \|y\| ∼1. Thus, we have the nonlinear version of the conjugate residual method: Nonlinear Conjugate Residual method • Calculate sk for approximating Ark by sk = F(xk + t rk) −F(xk) t , t = 1 \|rk\|. • Update the direction by pk = P −1rk −βk pk−1, βk = (sk, rk) (sk−1, rk−1). • Calculate qk = sk −βk qk−1. • Update the solution by xk+1 = xk + αk pk, αk = (sk, rk) (qk, P −1qk). • Calculate the residual rk+1 = F(xk+1). 9.7 Newton method Next, we develop the Newton method for J′(u) = 0. Deﬁne the Lagrange functional L(x, u, λ) = F(x) + H(u) + (λ, E(x, u)). Note that J′(u) = Lu(x(u), u, λ(u)) where (x(u), λ(u)) satisﬁes E(x(u), u) = 0, E′(x(u))∗λ(u) + F ′(x(u)) = 0 where we assumed Ex(x, u) is bounded invertible. By the implicit function theory J′′(u) = Lux(x(u), u, λ(u)) ˙ x + Eu(x(u), u, λ(u))∗˙ λ + Luu(x(u), u, λ(u)), where ( ˙ x, ˙ λ) satisﬁes   Lxx(x, u, λ) Ex(x, u)∗ Ex(x, u) 0     ˙ x ˙ λ  +   Lxu(x, u, λ) Eu(x, u)  = 0. 188 Thus, ˙ x = −(Ex(x, u))−1Eu(x, u), ˙ λ = −(Ex(x, u)∗)−1(Lxx(x, u, λ) ˙ x + Lxu(x, u, λ)) and J′′(u) = Lux ˙ x −Eu(x, u)∗(Ex(x, u)∗)−1(Lux ˙ x + Lxu) + Luu(x, u, λ). Theorem (Lagrange Calculus II) The Newton step J′′(u)(u+ −u) + J′(u) = 0 is equivalently written as       Lxx(x, u, λ) Lxu(x, u, λ) Ex(x, u)∗ Lux(x, u, λ) Luu(x, u, λ) Eu(x, u)∗ Ex(x, u) Eu(x, u, λ) 0             ∆x ∆u ∆λ       +       0 Eu(x, u)∗λ + H′(u) 0       = 0 where ∆x = x+ −x, ∆u = u+ −u and ∆λ = λ+ −λ. Proof: Deﬁne the linear operator T by T(x, u) =   −Ex(x, u)−1Eu(x, u) I  . Then, the Newton method is equivalently written as T(x, u)∗L′′(x, u, λ) ∆x ∆u + T(x, u)∗ 0 E∗ uλ + H′(u) = 0. Since E′(x, u) is surjective, it follows from the closed range theorem N(T(x, u)∗) = R(T(x, u))⊥= N(E′(x, u)∗)⊥= R(E′(x, u)). Hence, we obtain the claimed update. □ Thus, we have the Newton method for the equality constraint problem (9.12). Newton method 1. Given u0, initialize (x0, λ0) by E(x0, u0) = 0, Ex(x0, u0)∗λ0 + F ′(x0) = 0 2. Newton Step: Solve for (∆x, ∆u, ∆λ)   L′′(xn, un, λn) E′(xn, un)∗ E′(xn, un) 0     ∆x ∆u ∆λ  +   0 Eu(xn, un)∗λn + H′(un) 0  = 0 and update un+1 = ProjC(un + α ∆u) with a stepsize α > 0. 189 3. Feasible Step: Solve for (xn+1, λn+1) E(xn+1, un+1) = 0, Ex(xn+1, un+1)∗λn+1 + F ′(xn+1) = 0. 4. Stop or set n = n + 1 and return to step 2. where L′′(xn, un, λn) =   F ′′(xn) + (λn, Exx(xn, un)) (λn, Exu(xn, un)) (λn, Eux(xn, un)) H′′(un) + (λn, Euu(xn, un))  . and E′(xn, un) = (Ex(xn, un), Eu(xn, un)). Consider the general equality constraint minimization; min J(y) subject to E(y) = 0 The necessary optimality condition is given by J′(y) + E′(y)∗λ = 0, E(y) = 0, (9.15) where we assume E′(y) is surjective. Problem (9.12) is a speciﬁc case of this, i.e., y = (x, u), J(y) = F(x) + H(u), E = E(x, u). The Newton method applied to the necessary optimality (9.15) for (y, λ) is as follows. One can apply the Newton method for the necessary optimality system (Lx(x, u, λ), Lu(x, u, λ, E(x, u)) = 0 and results in the SQP (sequential quadratic programing): Newton method applied to Necessary optimality system (SQP) 1. Newton direction: Solve for (∆y, ∆λ)   L′′(yn, λn) E′(yn)∗ E′(yn) 0     ∆y ∆λ  +   J′(yn) + E′(yn)∗λn E(yn)  = 0. (9.16) 2. Update yn+1 = yn + α ∆y and λn+1 = λm + α ∆λ. 3. Stop or set n = n + 1 and return to step 1. where L′′(y, λ) = J′′(y) + E′′(y)∗λ. The both Newton methods solve the same linear system (9.16) with the diﬀerent right hand side. But, the ﬁrst Newton method we iterate on the variable u only but each step uses the feasible step E(yn) = E(xn, un) = 0 and Lx(xn, un, λn) = 0 for (xn, λn), which are the steps for the gradient method. The, it computes the Newton direction for J′(u) and updates u. In this sense the second Newton’s method is a relaxed version of the ﬁrst one. 190 10 Sequential programing In this section we discuss the sequential programming. It is an intermediate method between the gradient method and the Newton method. Consider the constrained op-timization F(y) subject to E(y) = 0, y ∈C. We linearize the equality constraint and consider a sequence of the constrained opti-mization; min F(y) subject to E′(yn)(y −yn) + E(yn) = 0, y ∈C. (10.1) The necessary optimality (if E(yn) is surjective) is given by    (F ′(y) + E′(yn)∗λ, ˜ y −y) ≥0 for all ˜ y ∈C E′(yn)(y −yn) + E(yn) = 0. (10.2) Note that the necessary condition for (y∗, λ∗) is written as    (F ′(y∗) + E′(yn)∗λ∗−(E′(yn)∗−E′(y∗)∗)λ∗, ˜ y −y∗) ≥0 for all ˜ y ∈C E′(yn)(y∗−yn) + E(yn) = E′(yn)(y∗−yn) + E(yn) −E(y∗). (10.3) Suppose we have the Lipschitz continuity of solutions to (10.2) and (10.3) with respect to the perturbation; ∆1 = (E′(yn)∗−E′(y∗)∗)λ∗and ∆2 = E′(yn)(y∗−yn) + E(yn) − E(y∗), we have \|y −y∗\| ≤c1 \|(E′(yn)∗−E′(y∗)∗)λ∗\| + c2 \|E′(yn)(y∗−yn) + E(yn) −E(y∗)\|. Thus, for the (damped) update: yn+1 = (1 −α)yn + α y, α ∈(0, 1), (10.4) we have the estimate \|yn+1 −y∗\| ≤(1 −α + αγ)\|yn −y∗\| + cα \|yn −y∗\|2, (10.5) for some constants γ and c > 0. In fact, for the case C = X let zn = (yn, λn), z∗= (y∗, λ∗) and deﬁne Gn by Gn =   F ′′(yn) E′(yn)∗ E′(yn) 0  . For the update: zn+1 = (1 −α)zn + α z, α ∈(0, 1] we have zn+1 −z∗= (1 −α) (zn −z∗) + α G−1 n ( (E′(yn)∗−E′(y∗)∗)λ∗ 0 + δn) 191 with δn = F ′(y∗) −F ′(yn+1) −F ′′(yn)(y∗−yn+1) E(y∗) −E′(yn)(y∗−yn) −E(yn). . Note that if F ′′(yn) = Q, then G−1 n ∆1 0 = Q−1(∆1 −E′(yn)∗e) e , e = (E′(yn)Q−1E′(yn)∗)−1E′(yn)Q−1∆1 Thus, we have \|G−1 n (E′(yn)∗−E′(y∗)∗)λ∗ 0 \| ≤γ \|yn −y∗\| Since \|G−1 n δn\| ≤c \|yn −y∗\|2, we have \|zn+1 −z∗\| ≤(1 −α + αγ) \|zn −z∗\| + αc \|zn −z∗\|2 which imply (10.5). When either the quadratic variation of E is dominated by the linearization E′ of E, i.e., \|E′(y∗)†(E′(yn) −E′(y∗))\| ≤β \|yn −y\| with small β > 0 or the multiplier \|λ∗\| is small, γ > 0 is suﬃciently small. The estimate (10.5) implies that if \|y −y∗\|2 is dominating we use a small α > 0 to the damp out and if \|y −y∗\| is dominating we use α →1. 10.1 Second order version The second order sequential programming is given by min F(y) + ⟨λn, E(y) −(E′(yn)(y −yn) + E(yn))⟩ subject to E′(yn)(y −yn) + E(yn) = 0 and y ∈C. (10.6) The necessary optimality condition for (10.6) is given by    (F ′(y) + (E′(y)∗−E′(yn)∗)λn + E′(yn)∗λ, ˜ y −y) ≥0 for all ˜ y ∈C E′(yn)(y −yn) + E(yn) = 0. (10.7) It follows from (10.3) and (10.7) that \|y −y∗\| + \|λ −λ∗\| ≤c (\|yn −y∗\|2 + \|λn −λ∗\|2). In summary we have Sequential Programming I 1. Given yn ∈C, we solve for min F(y) subject to E′(yn)(y −yn) + E(yn) = 0, over y ∈C. 192 2. Update yn+1 = (1 −α) yn + α y, α ∈(0, 1). Iterate until convergence. Sequential Programming II 1. Given yn ∈C, λn and we solve for y ∈C: min F(y)+(λn, E(y)−(E′(yn)(y−yn)+E(yn)) subject to E′(yn)(y−yn)+E(yn) = 0, y ∈C. 2. Update (yn+1, λn+1) = (y, λ). Iterate until convergence. 10.2 Examples Consider the constrained optimization of the form min F(x) + H(u) subject to E(x, u) = 0 and u ∈C. (10.8) Suppose H(u) is nonsmooth. We linearize the equality constraint and consider a se-quence of the constrained optimization; min F(x) + H(u) + ⟨λn, E(x, u) −(E′(xn, un)(x −xn, u −un) + E(xn, un)⟩ subject to E′(xn, un)(x −xn, u −un) + E(xn, un) = 0 and u ∈C. (10.9) Note that if (x∗, u∗) is an optimizer of (10.8), then F ′(x∗) + (E′(x∗, u∗) −E′(xn, un))λn + E′(xn, un)λ∗= (E′(xn, un) −E′(x∗, u∗))(λ∗−λn) = ∆1 E′(xn, un)(x∗−xn, u∗−un) + E(xn, un) = ∆2, \|∆2\| ∼M (\|xn −x∗\|2 + \|un −u∗\|2). (10.10) Thus, (x∗, u∗) is the solution to the perturbed problem of (10.9): min F(x) + H(u) + ⟨λn, E(x, u) −(E′(xn, un)(x −xn, u −un) + E(xn, un)⟩ subject to E′(xn, un)(x −xn, u −un) + E(xn, un) = ∆and u ∈C. (10.11) Let (xn+1, un+1) be the solution of (10.9). One can assume the Lipschitz continuity of solutions to (10.11): \|xn+1 −x∗\| + \|un+1 −u∗\| + \|λn −λ∗\| ≤C \|∆\| (10.12) From (10.10) and assumption (10.12) we have the quadratic convergence of the sequen-tial programing method (10.15): \|xn+1 −x∗\| + \|un+1 −u∗\| + \|λ −λ∗\| ≤M (\|xn −x∗\|2 + \|un −u∗\|2 + \|λn −λ∗\|). (10.13) Assuming H is convex, the necessary optimality given by            F ′(x) + (Ex(x, u)∗−Ex(xn, un)∗)λn + Ex(xn, un)∗λ = 0 H(v) −H(u) + ((Eu(x, u)∗−Eu(xn, un)∗)λn + Eu(x, u)∗λ, v −u) ≥0 for all v ∈C Ex(xn, un)(x −xn) + Eu(xn, un)(u −un) + E(xn, un) = 0. (10.14) 193 As in the Newton method we use (Ex(x, u)∗−Ex(xn, un)∗)λn ∼E′′(xn, un)(x −xn, u −un)λn F ′(xn+1) ∼F ′′(xn)(xn+1 −xn) + F ′(xn) and obtain F ′′(xn)(x −xn) + F ′(xn) + λn(Exx(xn, un)(x −xn) + Exu(xn, un)(u −un)) + Ex(xn, un)∗λ = 0 Ex(xn, un)x + Eu(xn, un)u = Ex(xn, un)xn + Eu(xn, un)un −E(xn, un), which is a linear system of equations for (x, λ). Thus, one has x = x(u), λ = λ(u), a continuous aﬃne function in u;   x(u) λ(u)  =   F ′′(xn) + λnExx(xn, un) Ex(xn, un)∗ Ex(xn, un) 0   −1 RHS (10.15) with RHS = F ′′(xn)xn −F ′(xn) + λnExx(xn, un)xn −Exu(xn, un)(u −un))λn Ex(xn, un)xn −Eu(xn, un)(u −un) −E(xn, un) and then the second inequality of (10.14) becomes the variational inequality for u ∈C. Consequently, we obtain the following algorithm: Sequential Programming II 1. Given u ∈C, solve (10.15) for (x(u), λ(u)). 2. Solve the variational inequality for u ∈C: H(v) −H(u) + ((Eu(x(u), u)∗−Eu(xn, un)∗)λn + Eu(xn, un)∗λ(u), v −u) ≥0, (10.16) for all v ∈C. 3. Set un+1 = u and xn+1 = x(u). Iterate until convergence. Example (Optimal Control Problem) Let (x, u) ∈H1(0, T; Rn) × L2(0, T; Rm). Con-sider the optimal control problem: min Z T 0 (ℓ(x(t)) + h(u(t))) dt subject to the dynamical constraint d dtx(t) = f(x(t)) + Bu(t), x(0) = x0 and the control constraint u ∈C = {u(t) ∈U, a.e.}, 194 where U is a closed convex set in Rm. Then, the necessary optimality condition for (10.9) is E′(xn, un)(x −xn, u −un) + E(xn, un) = −d dtx + f′(xn)(x −xn) + Bu = 0 F ′(x) + Ex(xn, un)λ = d dtλ + f′(xn)tλ + ℓ′(x) = 0 u(t) = argminv∈U{h(v) + (Btλ(t), v)}. If h(u) = α 2 \|u\|2 and U = Rm, then u(t) = −Btλ(t) α . Thus, (10.9) is equivalent to solving the two point boundary value for (x, λ):          d dtx = f′(xn)(x −xn) −1 αBBtλ, x(0) = x0 −d dtλ = (f′(x) −f′(xn))∗λn + f′(xn)∗λ + ℓ′(x), λ(T) = 0. Example (Inverse Medium problem) Let (y, v) ∈H1(Ω) × H(Ω). Consider the inverse medium problem: min Z Γ 1 2\|y −z\|2 dsx + Z Ω (β \|v\| + α 2 \|∇v\|2) dx subject to −∆y + vy = 0 in Ω, ∂y ∂n = g at the boundary Γ and v ∈C = {0 ≤v ≤γ < ∞}, where Ωis a bounded open domain in R2, z is a boundary measurement of the voltage y and g ∈L2(Γ) is a given current. Problem is to determine the potential function v ∈C from z ∈L2(Γ) in a class of sparse and clustered media. Then, the necessary optimality condition for (10.9) is given by                      −∆y + vn(y −yn) + vyn = 0, ∂y ∂n = g −∆λ + vnλ = 0, ∂λ ∂n = y −z Z Ω (α (∇v, ∇φ −∇v) + β (\|φ\| −\|v\|) + (λyn, v)) dx ≥0 for all φ ∈C 10.3 Exact penalty method Consider the penalty method for the inequality constraint minimization min F(x), subject to Gx ≤c (10.17) 195 where G ∈L(X, L2(Ω)). If x∗is a minimizer of (10.17), the necessary optimality condition is given by F ′(x∗) + G∗µ = 0, µ = max(0, µ + Gx∗−c) a.e. in Ω. (10.18) For β > 0 the penalty method is deﬁned by min F(x) + β ψ(Gx −c) (10.19) where ψ(y) = Z Ω max(0, y) dω. The necessary optimality condition is given by −F ′(x) ∈βG∗∂ψ(Gx −c). (10.20) where ∂ψ(s) =    0 s < 0 [0, 1] s = 0 1 s > 0 Suppose the Lagrange multiplier µ satisﬁes sup ω∈Ω \|µ(ω)\| ≤β, (10.21) then µ ∈β ∂ψ(Gx∗−c) and thus x∗satisﬁes (10.20). Moreover, if (10.19) has a unique minimizer in a neighborhood of x∗, then x = x∗where x is a minimizer of (10.19). Due to the singularity and the non-uniqueness of the subdiﬀerential of ∂ψ, the direct treatment of the condition (10.19) many not be feasible for numerical computation. We deﬁne a regularized functional of max(0, s); for ϵ > 0 maxϵ(0, s) =            ϵ 2 s ≤0 1 2ϵ\|s\|2 + ϵ 2 0 ≤s ≤ϵ s s ≥0 . and consider the regularized problem of (10.19); min Jϵ(x) = F(x) + β ψϵ(Gx −c), (10.22) whee ψϵ(y) = Z Ω maxϵ(0, y) dω Theorem (Consistency) Assume F is weakly lower semi-continuous. For an arbi-trary β > 0, any weak cluster point of the solution xϵ, ϵ > 0 of the regularized problem (10.22) converges to a solution x of the non-smooth penalized problem (10.19) as ϵ →0. Proof: First we note that 0 ≤maxϵ(s) −max(0, s) ≤ϵ 2 for all s ∈R. Let x be a solution to (10.19) and xϵ be a solution of the regularized problem (10.22). Then we have F(xϵ) + β ψϵ(xϵ) ≤F(x) + β ψϵ(x) F(x) + β ψ(x) ≤F(xϵ) + β ψ(xϵ). 196 Adding these inequalities, ψϵ(xϵ) −ψ(xϵ) ≤ψϵ(x) −ψ(x). Thus, we have for all cluster points ¯ x of xϵ F(¯ x) + β ψ(¯ x) ≤F(x) + β ψ(x), since F is weakly lower semi-continuous and ψϵ(xϵ) −ψ(¯ x) = ψϵ(xϵ) −ψ(xϵ) + ψ(xϵ) −ψ(¯ x) ≤ψϵ(x) −ψ(x) + ψ(xϵ) −ψ(¯ x).□ and lim ϵ→0+(ψϵ(x) −ψ(x) + ψ(xϵ) −ψ(¯ x)) ≤0.□ As a consequence of Theorem, we have Corollary If (10.19) has the unique minimizer in a neighborhood of x∗and (10.21) holds, then xϵ converges to the solution x∗of (10.17) as ϵ →0. The necessary optimality for (10.22) is given by F ′(x) + G∗ψ′ ϵ(Gx −c) = 0. (10.23) Although the non-uniqueness for concerning subdiﬀerential in the optimality condition in (10.20) is now bypassed through regularization, the optimality condition (10.23) is still nonlinear. For this objective, we ﬁrst observe that max′ ϵ has an alternative representation, i.e., max′ ϵ(s) = χϵ(s)s, χϵ(s) = 0 s ≤0 1 max(ϵ,s) s > 0 (10.24) This suggests the following semi-implicit ﬁxed point iteration; α P(xk+1 −xk) + F ′(xk) + β G∗χk(Gxk −c) = 0, χk = χϵ(Gxk −c), (10.25) where P is positive, self-adjoint and serves a pre-conditioner for F ′ and α > 0 serves a stabilizing and acceleration stepsize (see, Theorem 2). Let dk = xk+1 −xk. Equation (10.25) for xk+1 is equivalent to the equation for dk α Pdk + F ′(xk) + βG∗χk(Gxk −c + Gdk) = 0, which gives us (α P + β G∗χkG)dk = −F ′ ϵ(xk). (10.26) The direction dk is a decent direction for Jϵ(x) at xk, indeed, (dk, J′ ϵ(xk)) = −((α P + β G∗χkG)dk, dk) = −α(Ddk, dk) −β(χkGdk, Gdk) < 0. Here we used the fact that P is strictly positive deﬁnite. So, the iteration (10.25) can be seen as a decent method: 197 Algorithm (Fixed point iteration) Step 0. Set parameters: β, α, ϵ, P and 0 < ρ ≤1 Step 1. Compute the direction by (α P + βG∗χkG)dk = −F ′(xk) Step 2. Update xk+1 =)xk + ρdk. Step 3. If \|J′ ϵ(xk)\| < TOL, then stop. Otherwise repeat Step 1 - Step 2. Let us make some remarks on the Algorithm. In many applications, the structure of F ′ and G are sparse block diagonals, and the resulting system (10.26) for the direction dk then becomes a linear system with a sparse symmetric positive-deﬁnite matrix; and can be eﬃciently solved by the Cholesky decomposition method, for example. If F(x) = 1 2(x, Ax) −(b, x), then we have F ′(x) = Ax −b. For this case we may use the alternative update α P(xk+1 −xk) + Axk+1 −b + β G∗χk(Gxk+1 −c) = 0, assuming that it doesn’t cost much to perform this fully implicit step. For the bilateral inequality constraint g ≤Gx ≤c the update (10.25) becomes α P(xk+1 −xk) + F ′(xk) + β G∗(χk(Gxk+1 −c)A+ k + (g −Gxk+1)A− k (10.27) where the active index set Ak and the diagonal matrix χk on Ak are deﬁned by if j ∈A+ k = {j : (Gxk −c)j ≥0}, (χk)jj = 1 max(ϵ,(Gxk−c)j). if j ∈A− k = {j : (Gxk −g)j ≤0}, (χk)jj = 1 max(ϵ,(g−Gxk)j) The damping parameter 0 < ρ ≤1 is used to globalizing the convergence of the proposed algorithm, i.e., we select ρ to have the decent property for Jϵ(x). The algo-rithm is globally convergent with ρ = 1 practically and the following results justify the fact. Theorem 3 Let R(x, ˆ x) = −(F(x) −F(ˆ x) −F ′(ˆ x)(x −ˆ x)) + α (P(x −ˆ x), x −ˆ x) ≥ω \|x −ˆ x\|2 for some ω > 0 and all x, ˆ x. Then, we have R(xk+1, xk) + F(xk+1) −F(xk) + β 2 ((χkGdk, Gdk) + X k=j′ (χk, \|Gxk+1 −c\|2 −\|Gxk −c\|2) + X k=j′′ (χk, \|Gxk+1 −g\|2 −\|Gxk −g\|2) = 0, where {j′ : (Gxk −c)j′ ≥0} and {j′′ : (Gxk −g)j′′ ≤0}. Proof: Multiplying (10.27) by dk = xk+1 −xk α (Pdk, dk) −(F(xk+1) −F(xk) −F ′(xk)dk) +F(xk+1) −F(xk) + Ek = 0, 198 where Ek = β(χk(G(j, :)xk+1 −c), G(j, :)dk) = β 2 ((χkG(j, :)tdk, G(j, :)dk) +β 2 X k=j′ (χk, \|Gxk+1 −c\|2 −\|Gxk −c\|2) + X k=j′′ (χk, \|Gxk+1 −g\|2 −\|Gxk −g\|2)), Corollary 2 Suppose all inactive indexes {χk = 0} remain inactive, then ω \|xk+1 −xk\|2 + Jϵ(xk+1) −Jϵ(xk) ≤0 and {xk} is globally convergent. Proof: Since s →ψ( p \|s −c\|) on s ≥c is and s →ψ( p \|s −g\|) on s ≤g are concave, we have (χk, \|(Gxk+1 −c)j′\|2 −\|(Gxk −c)j′\|2) ≥ψϵ((Gxk+1 −c)j′) −ψϵ((Gxk −c)j′) and (χk, \|(g −Gxk+1)j′′\|2 −\|(g −Gxk)j′′\|2) ≥ψϵ((g −Gxk+1)j′′) −ψϵ((g −Gxk)j′′) Thus, we obtain F(xk+1)+β ψϵ(xk+1)+R(xk+1, xk)+β 2 (χkG(j, :)(xk+1−xk), G(j, :)(xk+1−xk)) ≤F(xk)+β ψϵ(xk). If we assume R(x, ˆ x) ≥ω \|x −ˆ x\|2 for some ω > 0, then F(xk) + ψϵ(Gxk −c) is monotonically decreasing and X \|xk+1 −xk\|2 < ∞.□ Corollary 3 Suppose i ∈Ac k = Ik is inactive but ψϵ(Gxk+1 i −c) > 0. Assume β 2 (χkG(xk+1 −xk), G(xk+1 −xk)) −β X i∈Ik ψϵ(Gxk+1 i −c) ≥−ω′ \|xk+1 −xk\|2 (10.28) with 0 ≤ω′ < ω, then the algorithm is globally convergent. The condition (10.28) holds if n=5000; m=50; A=rand(n,n); A=A+A’; A=A+70eye(n); C=rand(m,n); f=[ones(n,1);.5ones(m,1)]; u=A\f(1:n); k=find(Cu(1:n)>=.5ones(m,1)); AA=[A C(k,:)’;C(k,:) -1e-8eye(size(k,1))]; uu=u; u=AA[f(1:n);.5ones(size(k,1),1)]; norm(u(1:n)-uu(1:n)) 199 11 Sensitivity analysis In this section the sensitivity analysis is discussed for the parameter-dependent opti-mization. Consider the parametric optimization problem; F(x, p) x ∈C. (11.1) For given p ∈P, a complete matrix space, let x = x(p) ∈C is a minimizer of (11.1). For ¯ p ∈P and p = ¯ p + t ˙ p ∈P with t ∈R and increment ˙ p we have F(x(p), p) ≤F(x(¯ p), p), F(x(¯ p), ¯ p) ≤F(x(p), ¯ p). Thus, we have F(x(p), p) −F(x(¯ p), ¯ p) = F(x(p), p) −F(x(¯ p), p) + F(x(¯ p), p) −F(x(¯ p), ¯ p) ≤F(x(¯ p), p) −F(x(¯ p), ¯ p) F(x(p), p) −F(x(¯ p), ¯ p) = F(x(p), ¯ p) −F(x(¯ p), ¯ p) + F(x(p), p) −F(x(p), ¯ p) ≥F(x(p), p) −F(x(p), ¯ p) and F(x(p), p) −F(x(p), ¯ p) ≤F(x(p), p) −F(x(¯ p), ¯ p) ≤F(x(¯ p), p) −F(x(¯ p), ¯ p). Hence for t > 0 we have F(x(¯ p + t ˙ p), ¯ p + t ˙ p) −F(x(¯ p + t ˙ p), ¯ p) t ≤F(x(¯ p + t ˙ p), ¯ p + t ˙ p) −F(x(¯ p), ¯ p) t ≤F(x(¯ p), ¯ p + t ˙ p) −F(x(¯ p), ¯ p) t F(x(¯ p), ¯ p) −F(x(¯ p), ¯ p −t ˙ p) t ≤F(x(¯ p), ¯ p) −F(x(¯ p −t ˙ p), ¯ p −t ˙ p) t ≤F(x(¯ p −t ˙ p), ¯ p) −F(x(¯ p −t ˙ p), ¯ p −t ˙ p) t . Based on this inequality we have Theorem (Sensitivity I) Assume (H1) there exists the continuous graph p →x(p) ∈C in a neighborhood of (x(¯ p), ¯ p) ∈ C × P and deﬁne the value function V (p) = F(x(p), p). (H2) in a neighborhood of (x(¯ p), ¯ p) (x, p) ∈C × Q →Fp(x, p) is continuous. Then, the G-derivative of V at ¯ p exists and is given by V ′(¯ p) ˙ p = Fp(x(¯ p, ¯ p) ˙ p. (11.2) Conversely, if V (p) is diﬀerentiable at ¯ p then (11.2) holds. Lemma (Sensitivity I) Suppose V (p) = infx∈C F(x, p) where C is a closed convex set. If p ∈F(x, p) is convex, then V (p) is concave and Fp(¯ x, ¯ p) ∈−∂V (¯ p). Proof: Since for 0 ≤t ≤1 F(x, t p1 + (1 −t) p2) ≥t F(x, p1) + (1 −t) F(t, p2) 200 we have V (t p1 + (1 −t) p2) ≥t V (p1) + (1 −t) V (p2). and V is concave. Since V (p) −V (¯ p) = F(x, p) −F(¯ x, p) + F(¯ x, p) −F(¯ x, ¯ p) ≤F(¯ x, p) −F(¯ x, ¯ p) we have Fp(¯ x, ¯ p) ∈−∂V (¯ p).□ Next, consider the implicit function case. Consider a functional deﬁned by V (p) = F(x(p), p), E(x(p), p) = 0 where we assume that the constraint E is C1 and E(x, p) = 0 deﬁnes a continuous solution graph p →x(p) in a neighborhood of (x(¯ p), ¯ p). Theorem (Sensitivity II) Assume there exists λ that satisﬁes the adjoint equation Ex(x(¯ p), ¯ p)∗λ + Fx(x(¯ p), ¯ p) = 0 and assume the Lagrange functional L(x, p, λ) = F(x) + (E(x), λ) satisﬁes L(x(p), ¯ p, λ) −L(x(¯ p), ¯ p, λ) −Lx(x(¯ p), ¯ p, λ)(x(p) −x(¯ p)) = o(\|p −¯ p\|). (11.3) Then, p →V (p) is diﬀerentiable at ¯ p and V ′(¯ p) ˙ p = (Lp(x(¯ p), ¯ p, λ), ˙ p). Proof: If we let F(x(p), p) −F(x(¯ p, (¯ p) −F ′(x(¯ p))(x(p) −x(ˆ p)) = ϵ1, we have V (p) −V (¯ p) −Fx(x(¯ p))(x(p) −x(ˆ p)) = ϵ1 + Fp(x(¯ p), ¯ p)(p −¯ p) + o(\|p −¯ p\|). Note that (Ex(x(¯ p), ¯ p)(x(p) −x(¯ p)), λ) + Fx(x(¯ p))(x(p) −x(¯ p)) = 0. Let ϵ2 = (E(x(p), ¯ p) −E(x(¯ p), ¯ p) −Ex(x(¯ p), ¯ p)(x(p) −x(¯ p)), λ), where 0 = (E(x(p), p) −E(x(¯ p), ¯ p), λ) = ϵ2 −Ep(x(¯ p), ¯ p)(p −¯ p)), λ) + o(\|p −¯ p\|) Thus, summing these we obtain V (p) −V (¯ p) = Lp(x(¯ p), ¯ p)(p −¯ p) + ϵ1 + ϵ2 + o(\|p −¯ p\|) 201 Since L(x(p), ¯ p, λ) −L(x(¯ p), ¯ p, λ) −Lx(x(¯ p), ¯ p, λ)(x(p) −x(¯ p)) = ϵ1 + ϵ2, p →V (x(p)) is diﬀerentiable at ¯ p and the desired result. □ Lemma (Sensitivity II) If F(x, p) = F(x), p →E(x, p) is linear and x →L(x, p, λ) is convex, then p →V (p) is concave and Lp(¯ x, ¯ p, λ) ∈−∂V (¯ p). If we assume that E and F is C2 in x, then it is suﬃcient to have the Holder continuity \|x(p) −x(¯ p)\|X ∼o(\|p −¯ p\| 1 2 Q) for condition (11.3) holding. Next, consider the parametric constrained optimization: F(x, p) subject to E(x, p) ∈K. We assume (H3) λ ∈Y satisﬁes the adjoint equation Ex(x(¯ p), ¯ p)∗λ + Fx(¯ p), ¯ p) = 0. (H4) Conditions (11.3) holds. (H5) In a neighborhood of (x(¯ p), ¯ p). (x, p) ∈C × P →Ep(x, p) ∈Y is continuous. Eigenvalue problem Let A(p) be a linear operator in a Hilbert space X. For x = (µ, y) ∈R × X F(x) = µ, subject to A(p)y = µ y. That is, (µ, y) is an eigen pair of A(p). A(p)∗λ −µ λ = 0, (y, λ) = 1 Thus, µ′( ˙ p) = ( d dpA(p) ˙ py, λ). Next, consider the parametric constrained optimization; given p ∈Q min x,u F(x, u, p) subject to E(x, u, p) = 0. (11.4) Let (¯ x, ¯ u) is a minimizer given ¯ p ∈Q. We assume (H3) λ ∈Y satisﬁes the adjoint equation Ex(¯ x, ¯ u, ¯ p)∗λ + Fx(¯ x, ¯ u, ¯ p) = 0. (H4) Conditions (11.3) holds. (H5) In a neighborhood of (x(¯ p), ¯ p). (x, p) ∈C × Q →Ep(x, p) ∈Y is continuous. 202 Theorem (Sensitivity III) Under assumptions (H1)–(H5) the necessary optimality for (13.18) is given by E∗ xλ + Fx(¯ x, ¯ u, ¯ p) = 0 E∗ uλ + Fu(¯ x, ¯ u, ¯ p) = 0 E(¯ x, ¯ u, ¯ p) = 0 and V ′(¯ p) ˙ p = Fp(¯ x, ¯ u, ¯ p) ˙ p + (λ, Ep(¯ x, ¯ u, ¯ p) ˙ p) = 0. Proof: F(x(p), p) −F(x(¯ p), ¯ p) = F(x(p), ¯ p) −F(x(¯ p), ¯ p) + F(x(p), p) −F(x(p), ¯ p) ≥Fx(x(¯ p), ¯ p)(x(p) −x(¯ p) + F(x(p), p) −F(x(p), ¯ p) F(x(p), p) −F(x(¯ p), ¯ p) = F(x(p), p) −F(x(¯ p), p) + F(x(¯ p), p) −F(x(¯ p), ¯ p) ≤Fx(x(p), p)(x(p) −x(¯ p)) + F(x(¯ p), p) −F(x(¯ p), ¯ p). Note that E(x(p), p)−E(x(¯ p), ¯ p) = Ex(x(¯ p))(x(p)−x(¯ p)+E(x(p), p)−E(x(p), ¯ p) = o(\|x(p)−x(¯ p\|) and (λ, Ex(¯ x, ¯ p)(x(p) −x(¯ p)) + Fx(¯ x, ¯ p)(x(p) −x(¯ p)⟩= 0 Combining the above inequalities and letting t →0, we have 12 Augmented Lagrangian Method In this section we discuss the augmented Lagrangian method for the constrained min-imization; min F(x) subject to E(x) = 0, G(x) ≤0 (12.1) over x ∈C. For the equality constraint case we consider the augmented Lagrangian functional Lc(x, λ) = F(x) + (λ, E(x)) + c 2\|E(x)\|2. for c > 0. The augmented Lagrangian method is an iterative method with the update xn+1 = argminx∈C Lc(x, λn) λn+1 = λn + c E(xn). It is a combination of the multiplier method and the penalty method. That is, if c = 0 and λn+1 = λn + α E(xn) with step size α > 0 is the multiplier method and if λn = 0 is the penalty method. The multiplier method converges (locally) if F is (locally) uniformly convex. The penalty method requires c →∞and may suﬀer if c > 0 is too 203 large. It can be shown that the augmented Lagrangian method is locally convergent provided that L′′ c(x, λ) is uniformly positive near the solution pair (¯ x, ¯ λ). Since L′′(¯ x, ¯ λ) = F ′′(¯ x) + (¯ λ, E′′(¯ x)) + c E′(¯ x)∗E′(¯ x), the augmented Lagrangian method has an enlarged convergent class compared to the multiplier method. Unlike the penalty method it is not necessary to c →∞and the Lagrange multiplier update speeds up the convergence. For the inequality constraint case G(x) ≤0 we consider the equivalent formulation: min F(x) + c 2\|G(x) −z\|2 + (µ, G(x) −z) over (x, z) ∈C × Z subject to G(x) = z, and z ≤0. For minimizing this over z ≤0 we have that z∗= min(0, µ + c G(x) c ) attains the minimum and thus we obtain min F(x) + 1 2c(\| max(0, µ + c G(x))\|2 −\|µ\|2) over x ∈C. For (12.1), given (λ, µ) and c > 0, deﬁne the augmented Lagrangian functional Lc(x, λ, µ) = F(x) + (λ, E(x)) + c 2 \|E(x)\|2 + 1 2c(\| max(0, µ + c G(x))\|2 −\|µ\|2). (12.2) The ﬁrst order augmented Lagrangian method is a sequential minimization of Lc(x, λ, µ) over x ∈C; Algorithm (First order Augmented Lagrangian method) 1. Initialize (λ0, µ0) 2. Let xn be a solution to min Lc(x, λn, µn) over x ∈C. 3. Update the Lagrange multipliers λn+1 = λn + c E(xn), µn+1 = max(0, µn + c G(xn)). (12.3) Remark (1) Deﬁne the value function Φ(λ, µ) = min x∈C Lc(x, λ, µ). Then, one can prove that Φλ = E(x), Φµ = max(0, µ + c G(x)) (2) The Lagrange multiplier update (12.3) is a gradient method for maximizing Φ(λ, µ). 204 (3) For the equality constraint case Lc(x, λ) = F(x) + (λ, E(x)) + c 2\|E(x)\|2 and L′ c(x, λ) = L′ 0(x, λ + c E(x)). The necessary optimality for max λ min x Lc(x, λ) is given by L′ c(x, λ) = L′ 0(x, λ + c E(x)) = 0, E(x) = 0 (12.4) If we apply the Newton method to system (12.4), we obtain   L′′ 0(x, λ + c E(x)) + c E′(x)∗E′(x) E′(x)∗ E′(x) 0     x+ −x λ+ −λ  = −   L′ 0(x, λ + c E(x)) E(x)  . The c > 0 adds the coercivity on the Hessian L′′ 0 term. 12.1 Primal-Dual Active set method For the inequality constraint G(x) ≤0 (e.g., Gx −˜ c ≤0); Lc(x, µ) = F(x) + 1 2c(\| max(0, µ + c G(x)\|2 −\|µ\|2) and the necessary optimality condition is F ′(x) + G′(x)∗µ, µ = max(0, µ + c G(x)) where c > 0 is arbitrary. Based on the complementarity condition we have primal dual active set method. Primal Dual Active Set method For the aﬃne case G(x) = Gx −˜ c we have the Primal Dual Active Set method as: 1. Deﬁne the active index and inactive index by A = {k ∈(µ + c (Gx −˜ c))k > 0}, I = {j ∈(µ + c (Gx −˜ c))j ≤0} 2. Let µ+ = 0 on I and Gx+ = ˜ c on A. 3. Solve for (x+, µ+ A) F ′′(x)(x+ −x) + F ′(x) + G∗ Aµ+ = 0, GAx+ −c = 0 4. Stop or set n = n + 1 and return to step 1, 205 where GA = {Gk}, k ∈A. For the nonlinear G G′(x)(x+ −x) + G(x) = 0 on A = {k ∈(µ + c G(x))k > 0}. The primal dual active set method is a semismooth Newton method. That is, if we deﬁne a generalized derivative of s →max(0, s) by 0 on (−∞, 0] and 1 on (0, ∞). Note that s →max(0, s) is not diﬀerentiable at s = 0 and we deﬁne the derivative at s = 0 as the limit of derivative for s < 0. So, we select the generalized derivative of max(0, s) as 0, s ≤0 and 1, s > 0 and thus the generalized Newton update µ+ −µ + 0 = µ if µ + c (Gx −c) ≤0 c G(x+ −x) + c (Gx −˜ c) = Gx+ −˜ c = 0 if µ + c (Gx −˜ c) > 0, which results in the active set strategy µ+ j = 0, j ∈I and (Gx+ −˜ c)k = 0, k ∈A. We will introduce the class of semismooth functions and the semismooth Newton method in Chapter 7 in function spaces. Speciﬁcally, the pointwise (coordinate) oper-ation s →max(0, s) deﬁnes a semismooth function from Lp(Ω) →Lq(Ω), p > q. 13 Lagrange multiplier Theory for Nonsmooth convex Optimization In this section we present the Lagrange multiplier theory for the nonsmmoth optimiza-tion of the form min f(x) + ϕ(Λx) over x ∈C (13.1) where X is a Banach space, H is a Hilbert space lattice, f : X →R is C1, Λ ∈L(X, H) and C is a closed convex set in X. The nonsmoothness is represented by ϕ and ϕ is a proper, lower semi continuous convex functional in H. This problem class encompasses a wide variety of optimization problems including variational inequalities of the ﬁrst and second kind [?, ?]. We will specify those to our speciﬁc examples. Moreover, the exact penalty formulation of the constraint problem is min f(x) + c \|E(x)\|L1 + c \|(G(x))+\|L1 (13.2) where (G(x))+ = max(0, G(x)). We describe the Lagrange multiplier theory to deal with the non-smoothness of ϕ. Note that (13.1) is equivalent to min f(x) + ϕ(Λx −u) subject to x ∈C and u = 0 in H. (13.3) Treating the equality constraint u = 0 in (13.3) by the augmented Lagrangian method, results in the minimization problem min x∈C,u∈H f(x) + ϕ(Λx −u) + (λ, u)H + c 2 \|u\|2 H, (13.4) 206 where λ ∈H is a multiplier and c is a positive scalar penalty parameter. By (13.6) we have min y∈C Lc(y, λ) = f(x) + ϕc(Λx, λ). (13.5) where ϕc(u, λ) = infv∈H {ϕ(u −v) + (λ, v)H + c 2 \|v\|2 H} = inf z∈H {ϕ(z) + (λ, u −z)H + c 2 \|u −z\|2 H} (u −v = z). (13.6) 13.1 Yosida-Moreau approximations and Monotone Op-erators For u, λ ∈H and c > 0, ϕc(u, λ) is called the generalized Yoshida-Moreau approxima-tions ϕ. It can be shown that u →ϕc(u, λ) is continuously Fr´ echet diﬀerentiable with Lipschitz continuous derivative. Then, the augmented Lagrangian functional [?, ?, ?] of (13.1) is given by Lc(x, λ) = f(x) + ϕc(Λx, λ). (13.7) Let ϕ∗is the convex conjugate of ϕ ϕc(u, λ) = infz∈H{supy∈H{(y, z) −ϕ∗(y)} + (λ, u −z)H + c 2 \|u −z\|2 H} = sup y∈H {−1 2c\|y −λ\|2 + (y, u) −ϕ∗(y)}. (13.8) where we used inf z∈H{(y, z)H + (λ, u −z)H + c 2 \|u −z\|2 H} = (y, u) −1 2c\|y −λ\|2 H Thus, from Theorem 2.2 ϕ′ c(u, λ) = yc(u, λ) = argmaxy∈H{−1 2c\|y −λ\|2 + (y, u) −ϕ∗(y)} = λ + c vc(u, λ) (13.9) where vc(u, λ) = argminv∈H{ϕ∗(u −v) + (λ, v) + c 2\|v\|2} = ∂ ∂λϕc(u, λ) Moreover, if pc(λ) = argminp∈H{ 1 2c\|p −λ\|2 + ϕ∗(p)}, (13.10) then (13.9) is equivalent to ϕ′ c(u, λ) = pc(λ + c u). (13.11) Theorem (Lipschitz complementarity) (1) If λ ∈∂ϕ(x) for x, λ ∈H, then λ = ϕ′ c(x, λ) for all c > 0. (2) Conversely, if λ = ϕ′ c(x, λ) for some c > 0, then λ ∈∂ϕ(x). Proof: If λ ∈∂ϕ(x), then from (13.6) and (13.8) ϕ(x) ≥ϕc(x, λ) ≥⟨λ, x⟩−ϕ∗(λ) = ϕ(x). 207 Thus, λ ∈H attains the supremum of (13.9) and we have λ = ϕ′ c(x, λ). Conversely, if λ ∈H satisﬁes λ = ϕ′ c(x, λ) for some c > 0, then vc(x, λ) = 0 by (13.9). Hence it follows from that ϕ(x) = ϕc(x, λ) = (λ, x) −ϕ∗(λ). which implies λ ∈∂ϕ(x). □ If xc ∈C denotes the solution to (13.5), then it satisﬁes ⟨f′(xc) + Λ∗λc, x −xc⟩X∗,X ≥0, for all x ∈C λc = ϕ′ c(Λxc, λ). (13.12) Under appropriate conditions [?, ?, ?] the pair (xc, λc) ∈C × H has a (strong-weak) cluster point (¯ x, ¯ λ) as c →∞such that ¯ x ∈C is the minimizer of (13.1) and that ¯ λ ∈H is a Lagrange multiplier in the sense that ⟨f′(¯ x) + Λ∗¯ λ, x −¯ x⟩X∗,X ≥0, for all x ∈C, (13.13) with the complementarity condition ¯ λ = ϕ′ c(Λ¯ x, ¯ λ), for each c > 0. (13.14) System (13.13)–(13.14) is for the primal-dual variable (¯ x, ¯ λ). The advantage here is that the frequently employed diﬀerential inclusion ¯ λ ∈∂ϕ(Λ¯ x) is replaced by the equivalent nonlinear equation (13.14). The ﬁrst order augmented Lagrangian method for (13.1) is given by First order augmented Lagrangian method • Select λ0 and set n = 0. • Let xn+1 = argminx∈CLc(x, λn). • Update the Lagrange multiplier by λn+1 = ϕ′ c(Λxn+1, λn). • Stop or set n = n + 1 and return to step 1. The step minx∈C Lc(x, λn) can be spitted into the coordinate-wise as Coordinate splitting method • Select λ0 and set n = 0 • zn+1 = argminz{ϕ(v) + (λn, Λxn −z)H + c 2 \|Λxn −z\|2 H}. • xn+1 = argminx∈C{f(x) + (λn, Λx −zn+1)H + c 2 \|Λx −zn+1\|2 H}. • Update the Lagrange multiplier by λn+1 = pc(λn + c Λxn+1). • Stop or set n = n + 1 and return to step 1. In many applications, the convex conjugate functional ϕ∗of ϕ is given by ϕ∗(v) = IK∗(v), where K∗is a closed convex set in H and IS is the indicator function of a set S. From (13.8) ϕc(u, λ) = sup y∈K∗{−1 2c\|y −λ\|2 + (y, u)} 208 and it follows from (13.10)–(13.11) that (13.14) is equivalent to ¯ λ = ProjK∗(¯ λ + c Λ¯ x). (13.15) which is the basis of our approach. 13.2 Examples In this section we discuss the applications of the Lagrange multiplier theorem and the complementarity. Example (Inequality) Let H = L2(Ω) and ϕ = IC with C = {s ∈L2(Ω) : z ≤0, a.e., i.e., min f(x) over x ∈Λx −c ≤0. (13.16) In this case ϕ∗(v) = IK∗(v) with K∗= −C and the complementarity (15.26) implies that f′(¯ x) + Λ∗¯ λ = 0 ¯ λ = max(0, ¯ λ + c (Λ¯ x −c)). Example (L1(Ω) optimization) Let H = L2(Ω) and ϕ(v) = Z Ω \|v\| dx, i.e. min f(y) + Z Ω \|Λy\|2 dx. (13.17) In this case ϕ∗(v) = IK∗(v) with K∗= {z ∈L2(Ω) : \|z\|2 ≤1, a.e.} and the complementarity (15.26) implies that f′(¯ x) + Λ∗¯ λ = 0 ¯ λ = ¯ λ + c Λ¯ x max(1, \|¯ λ + c Λ¯ x\|2) a.e.. Exaples (Constrained optimization) Let X be a Hilbert space and consider min f(x) subject to Λx ∈C (13.18) where f : X →R is C1 and C is a closed convex set in X. In this case we let f = J and Λ is the natural injection and ϕ = IC. ϕc(u, λ) = inf z∈C{(λ, u −z)H + c 2 \|u −z\|2 H}. (13.19) and z∗= ProjC(u + λ c ) (13.20) 209 attains the minimum in (13.19) and ϕ′ c(u, λ) = λ + c (u −z∗) (13.21) If H = L2(Ω) and C is the bilateral constraint {y ∈X : φ ≤Λy ≤ψ}, the complemen-tarity (13.20)–(13.21) implies that for c > 0 f′(x∗) + Λ∗µ = 0 µ = max(0, µ + c (Λx∗−ψ) + min(0, µ + c (Λx∗−φ)) a.e.. (13.22) If either φ = −∞or ψ = ∞, it is unilateral constrained and deﬁnes the generalized obstacle problem. The cost functional J(y) can represent the performance index for the optimal control and design problem, the ﬁdelity of data-to-ﬁt for the inverse problem and deformation and restoring energy for the variational problem. 13.3 Monotone Operators and Yosida-Morrey approxi-mations 13.4 Lagrange multiplier Theory In this action we introduce the generalized Lagrange multiplier theory. We establish the conditions for the existence of the Lagrange multiplier ¯ λ and derive the optimal-ity condition (1.7)–(1.8). In Section 1.5 it is shown that the both Uzawa and aug-mented Lagrangian method are ﬁxed point methods for the complementarity condition (??))and the convergence analysis is presented . In Section 1.6 we preset the concrete examples and demonstrate the applicability of the results in Sections 1.4–1.5. 14 Semismooth Newton method In this section we present the semismooth Newton method for the nonsmooth equation in Banach spaces, especially the necessary optimality condition, e.g., the complemen-tarity condition (13.12): λ = ϕ′ c(Λx, λ). Consider the nonlinear equation F(y) = 0 in a Banach space X. The generalized Newton update is given by yk+1 = yk −V −1 k F(yk), (14.1) where Vk is a generalized derivative of F at yk. In the ﬁnite dimensional space for a locally Lipschitz continuous function F let DF denote the set of points at which F is diﬀerentiable. For x ∈X = Rn we deﬁne the Bouligand derivative ∂BF(x) as ∂BF(x) = J : J = lim xi→x, xi∈DF ∇F(xi) . (14.2) Here, DF is dense by Rademacher’s theorem which states that every locally Lipschitz continuous function in the ﬁnite dimensional space is diﬀerentiable almost everywhere. Thus, we take Vk ∈∂BF(yk). 210 In inﬁnite dimensional spaces notions of generalized derivatives for functions which are not C1 cannot rely on Rademacher’s theorem. Here, instead, we shall mainly utilize a concept of generalized derivative that is suﬃcient to guarantee superlinear convergence of Newton’s method [?]. This notion of diﬀerentiability is called Newton derivative and is deﬁned below. We refer to [?, ?, ?, ?] for further discussion of the notions and topics. Let X, Z be real Banach spaces and let D ⊂X be an open set. Deﬁnition (Newton diﬀerentiable) (1) F : D ⊂X →Z is called Newton dif-ferentiable at x, if there exists an open neighborhood N(x) ⊂D and mappings G: N(x) →L(X, Z) such that lim \|h\|→0 \|F(x + h) −F(x) −G(x + h)h\|Z \|h\|X = 0. The family {G(y) : y ∈N(x)} is called a N-derivative of F at x. (2) F is called semismooth at y, if it is Newton diﬀerentiable at y and lim t→0+ G(y + t h)h exists uniformly in \|h\| = 1. Semi-smoothness was originally introduced in [?] for scalar-valued functions. Con-vex functions and real-valued C1 functions are examples for such semismooth functions [?, ?] in the ﬁnite dimensional space. For example, if F(y)(s) = ψ(y(s)), point-wise, then G(y)(s) ∈ψB(y(s)) is an N-derivative in Lp(Ω) →Lq(Ω) under appropriate conditions [?]. We often use ψ(s) = \|s\| and max(0, s) for the necessary optimality. Suppose F(y∗) = 0, Then, y∗= yk + (y∗−yk) and \|yk+1 −y∗\| = \|V −1 k (F(y∗) −F(yk) −Vk(y∗−yk)\| ≤\|V −1 k \|o(\|yk −y∗\|). Thus, the semismooth Newton method is q-superlinear convergent provided that the Jacobian sequence Vk is uniformly invertible as yk →y∗. That is, if one can select a sequence of quasi-Jacobian Vk that is consistent, i.e., \|Vk −G(y∗)\| as yk →y∗and invertible, (14.1) is still q-superlinear convergent. 14.1 Bilateral constraint Consider the bilateral constraint problem min f(x) subject to φ ≤Λx ≤ψ. (14.3) We have the optimality condition (13.22): f′(x) + Λ∗µ = 0 µ = max(0, µ + c (Λx −ψ) + min(0, µ + c (Λx −φ)) a.e.. Since ∂B max(0, s) = {0, 1}, ∂B min(0, s) = {−1, 0}, at s = 0, the semismooth Newton method for the bilateral constraint (14.3) is of the form of the primal-dual active set method: Primal dual active set method 211 1. Initialize y0 ∈X and λ0 ∈H. Set k = 0. 2. Set the active set A = A+ ∪A−and the inactive set I by A+ = {x ∈Ω: µk+c (Λyk−ψ) > 0}, A−= {x ∈Ω: µk+c (Λyk−φ) < 0}, I = Ω\A 3. Solve for (yk+1, λk+1) J′′(yk)(yk+1 −yk) + J′(yk) + Λ∗µk+1 = 0 Λyk+1(x) = ψ(x), x ∈A+, Λyk−1(x) = φ(x), x ∈A−, µk+1(x) = 0, x ∈I 4. Stop, or set k = k + 1 and return to the second seep. Thus the algorithm involves solving the linear system of the form   A Λ∗ A+ Λ∗ A− ΛA+ 0 0 ΛA− 0 0     y µA+ µA−  =   f ψ φ  . In order to gain the stability the following one-parameter family of the regularization can be used [?, ?] µ = α (max(0, µ + c (Λy −ψ)) + min(0, µ + c (Λy −φ))), α →1+. 14.2 L1 optimization As an important example, we discuss the case of ϕ(v) = R Ω\|v\|2 dx (L1-type optimiza-tion). The complementarity is reduced to the form λ max(1 + ϵ, \|λ + c v\|) = λ + c v (14.4) for ϵ ≥0. It is often convenient (but not essential) to use very small ϵ > 0 to avoid the singularity for the implementation of algorithm. Let ϕϵ(s) =      s2 2ϵ + ϵ 2, if \|s\| ≤ϵ, \|s\|, if \|s\| ≥ϵ. (14.5) Then, (14.4) corresponds to the regularized one of (4.31): min Jϵ(y) = F(y) + ϕϵ(Λy). The semismooth Newton method is given by F ′(y+) + Λ∗λ+ = 0              λ+ = v+ ϵ if \|λ + v\| ≤1 + ϵ, \|λ + c v\| λ+ + λ λ + c v \|λ + c v\| t! (λ+ + c v+) = λ+ + c v+ + \|λ + c v\| λ if \|λ + c v\| > 1 + ϵ. (14.6) 212 There is no guarantee that (14.6) is solvable for λ+ and is stable. In order to obtain the compact and unconditionally stable formula we use the damped and regularized algorithm with β ≤1;                      λ+ = v ϵ if \|λ + c v\| ≤1 + ϵ, \|λ + c v\| λ+ −β λ max(1, \|λ\|) λ + c v \|λ + c v\| t! (λ+ + c v+) = λ+ + c v+ + β \|λ + c v\| λ max(1, \|λ\|) if \|λ + c v\| > 1 + ϵ. (14.7) Here, the purpose of the regularization λ \|λ\|∧1 is to automatically constrain the dual variable λ into the unit ball. The damping factor β is automatically selected to achieve the stability. Let d = \|λ + c v\|, η = d −1, a = λ \|λ\| ∧1, b = λ + c v \|λ + c v\|, F = abt. Then, (14.7) is equivalent to λ+ = (η I + β F)−1((I −β F)(c v+) + βd a), where by Sherman–Morrison formula (η I + β F)−1 = 1 η (I − β η + β a · bF). Then, (η I + β F)−1βd a = β d η + β a · b a. Since F 2 = (a · b) F, (η I + β F)−1(I −β F) = 1 η(I − βd η + β a · b F). In order to achieve the stability, we let β d η + β a · b = 1, i.e., β = d −1 d −a · b ≤1. Consequently, we obtain a compact Newton step λ+ = 1 d −1(I −F)(c v+) + λ \|λ\| ∧1, (14.8) which results in; Primal-dual active set method (L1-optimization) 1. Initialize: λ0 = 0 and solve F ′(y0) = 0 for y0. Set k = 0. 2. Set inactive set Ik and active set Ak by Ik = {\|λk + c Λyk\| > 1 + ϵ}, and Ak = {\|λk + c Λyk\| ≤1 + ϵ}. 213 3. Solve for (yk+1, λk+1) ∈X × H:                      F ′(yk+1) + Λ∗λk+1 = 0 λk+1 = 1 dk −1(I −F k)(c Λyk+1) + λk \|λk\| ∧1 in Ak and Λyk+1 = ϵ λk+1 in Ik. 4. Convergent or set k = k + 1 and Return to Step 2. This algorithm is unconditionally stable and is rapidly convergent forour test examples. Remark (1) Note that (λ+, v+) = \|v+\|2 −(a · v+)(b · v+) d −1 + a · v+, (14.9) which implies stability of the algorithm. In fact, since          F ′(yk+1) −F ′(yk) + Λ∗(λk+1 −λk) λk+1 −λk = (I −F k)(c Λyk+1) + ( 1 \|λk\| −1) λk, we have (F ′(yk+1) −F ′(yk), yk+1 −yk) + c ((I −F k)Λyk+1, Λyk+1) + ( 1 \|λk\| −1) (λk, Λyk+1). Suppose F(y) −F(x) −(F ′(y), y −x) ≥ω \|y −x\|2 X, we have F(yk) + k X j=1 ω \|yj −yj−1\|2 + ( 1 \|λj−1\| ∧1 −1) (λj−1, vj) ≤F(y0). where vj = Λyj. Note that (λk, vk+1) > 0 implies that (λk+1, vk+1) > 0 and thus inactive at k+1-step, pointwise. This fact is a key step for proving a global convergence of the algorithm. (2) If a · b →1+, then β →1. Suppose (λ, v) is a ﬁxed point of (14.8), then 1 − 1 \|λ\| ∧1 1 − c d −1 λ + c v \|λ + c v\| · v λ = 1 d −1v. Thus, the angle between λ and λ + c v is zero and 1 − 1 \|λ\| ∧1 = c \|λ\| + c \|v\| −1 \|v\| \|λ\| − \|v\| \|λ\| ∧1 , which implies \|λ\| = 1. It follows that λ + c v = λ + c v max(\|λ + c v\|, 1 + ϵ). 214 That is, if the algorithm converges, a · b →1 and \|λ\| →1 and it is consistent. (3) Consider the substitution iterate:          F ′(y+) + Λ∗λ+ = 0, λ+ = 1 max(ϵ, \|v\|)v+, v = Λy. (14.10) Note that v \|v\|, v+ −v = \|v+\|2 −\|v\|2 + \|v+ −v\|2 2\|v\| dx = \|v+\| −\|v\| −(\|v+\| −\|v\|)2 + \|v+ −v\|2 2\|v\| . Thus, Jϵ(v+) ≤Jϵ(v), where the equality holds only if v+. This fact can be used to prove the iterative method (14.10) is globally convergent [?, ?]. It also suggests to use the hybrid method; for 0 < µ < 1 λ+ = µ max(ϵ, \|v\|)v+ + (1 −µ) ( 1 d −1(I −F)v+ + λ max(\|λ\|, 1)), in order to gain the global convergence property without loosing the fast convergence of the Newton method. 15 Non-convex nonsmooth optimization In this section we consider a general class of non-convex, nonsmooth optimization problems. Let X be a Banach space and continuously embed to H = L2(Ω)m. Let J : X →R+ is C1 and N is nonsmooth functional on H is given by N(y) = Z Ω h(y(ω)) dω. where h : Rm →R is lower semi-continuous. Consider the minimization on X; min J(x) + N(x) (15.1) subject to x ∈C = {x(ω) ∈U a.e. } where U is a closed convex set in Rm. For example, we consider h(u) = α 2 \|u\|2 + \|u\|0 where \|s\|0 = 1, s ̸= 0 and \|0\|0 = 0. For an integrable function u Z Ω \|u(ω)\|0 dω = meas({u ̸= 0}). Thus, one can formulate the volume control for material design and and time scheduling problem for the optimal control problem as (15.1). 215 15.1 Existence of Minimizer In this section we develop a existence method for a general class of minimizations min G(x) (15.2) on a Banach space X, including (15.1). Here, the constrained problem x ∈C is formulated as G(x) = G(x) + IC(x) using the indicator function of the constrained set C. Let X be a reﬂexible space or X = ˜ X∗for some normed space ˜ X. Thus xn ⇀x means the weak or weakly star convergence in X. To prove the existence of solutions to (15.2) we modify a general existence result given in [?]. Theorem (Existence) Let G be an extended real-valued functional on a Banach space X satisfying the following properties: (i) G is proper, weakly lower semi-continuous and G(x) + ϵ \|x\|p for each ϵ > 0 is coercive for some p. (ii) for any sequence {xn} in X with \|xn\| →∞, xn \|xn\| ⇀¯ x weakly in X, and {G(xn)} bounded from above, we have xn \|xn\| →¯ x strongly in X and there exists ρn ∈(0, \|xn\|] and n0 = n0({xn}) such that G(xn −ρn¯ x) ≤G(xn), (15.3) for all n ≥n0. Then minx∈X G(x) admits a minimizer. Proof: Let {ϵn} ⊂(0, 1) be a sequence converging to 0 from above and consider the family of auxiliary problems min x∈x G(x) + ϵn \|x\|p. (15.4) By (i), every minimizing sequence for (15.1) is bounded. Extracting a weakly conver-gent subsequence, the existence of a solution xn ∈X for (15.1) can be argued in a standard manner. Suppose {xn} is bounded. Then there exists a weakly convergent subsequence, denoted by the same symbols and x∗such that xn ⇀x∗. Passing to the limit ϵn →0+ in G(xn) + ϵn 2 \|xn\|p ≤G(x) + ϵn 2 \|x\|p for all x ∈X and using again (i) we have G(x∗) ≤G(x) for all x ∈X and thus x∗is a minimizer for G. We now argue that {xn} is bounded, and assume to the contrary that limn→∞\|xn\| = ∞. Then a weakly convergent subsequence can be extracted from xn \|xn\| such that, again dropping indices, xn \|xn\| ⇀¯ x, for some ¯ x ∈X. Since G(xn) is bounded from (ii) xn \|xn\| →¯ x strongly in X. (15.5) Next, choose ρ > 0 and n0 according to (15.3) and thus for all n ≥n0 G(xn) + ϵn \|xn\|p ≤G(xn −ρ¯ x) + ϵn \|xn −ρ¯ x\|p ≤G(xn) + ϵn \|xn −ρ¯ x\|p. 216 It follows that \|xn\| ≤\|xn −ρ¯ x\| = \|xn −ρn xn \|xn\| + ρn( xn \|xn\| −¯ x)\| ≤\|xn\|(1 − ρ \|xn\|) + ρn\| xn \|xn\| −¯ x\|. This implies that 1 ≤\| xn \|xn\| −¯ x\|, which give a contradiction to (15.5), and concludes the proof. □ The ﬁrst half of condition (ii) is a compactness assumption and (15.3) is a compat-ibility condition [?]. Remark (1) If G is not bounded below. Deﬁne the (sequential) recession functional G∞associated with G. It is the extended real-valued functional deﬁned by G∞(x) = inf xn⇀x,tn→∞lim inf n→∞ 1 tn G(tnxn). (15.6) and assume G∞(x) ≥0. Then, G(x) + ϵ \|x\|2 is coercive. If not, there exists a sequence {xn} satisfying G(xn) + ϵ \|xn\|2 is bounded but \|xn\| →∞. Let w −lim xn \|xn\| = ¯ x. 1 \|xn\|G(\|xn\| xn \|xn\|) + ϵ\|xn\| →0. Since G∞(¯ x) ≤ 1 \|xn\|G(\|xn\| xn \|xn\|), which yields a contradiction. Moreover, in the proof note that G∞(¯ x) ≤lim inf n→∞ 1 \|xn\|G(xn) = lim inf n→∞ 1 \|xn\|G( xn \|xn\|\|xn\|) ≤0. If we assume G∞≥0, then G∞(¯ x) = 0. Thus, (15.3) must hold, assuming G∞(¯ x) = 0. In [?] (15.3) is assumed for all x ∈X and G∞(¯ x) = 0, i.e., G(x −ρ¯ x) ≤G(x). Thus, our condition is much weaker. In [?] we ﬁnd a more general version of (15.3): there exists a ρn and zn such that ρn ∈(0, \|xn\|] and zn ∈X G(xn −ρn zn) ≤G(xn), \|zn −z\| →0 and \|z −¯ x\| < 1. (2) We have also proved that a sequence {xn} that minimizes the regularized problem for each ϵn has a weakly convergent subsequence to a minimizer of G(x). Example 1 If ker(G∞) = 0, then ¯ x = 0 and the compactness assumption implies \|¯ x\| = 1, which is a contradiction. Thus, the theorem applies. Example 2 (ℓ0 optimization) Let X = ℓ2 and consider ℓ0 minimization: G(x) = 1 2\|Ax −b\|2 2 + β \|x\|0, (15.7) where \|x\|0 = the number of nonzero elements of x ∈ℓ2, 217 the counting measure of x. Assume N(A) is ﬁnite and R(A) is closed. The, one can apply Theorem to prove the existence of solutions to (15.7). That is, suppose \|xn\| →∞, G(xn) is bounded from above and xn \|xn\| ⇀z. First, we show that z ∈N(A) and xn \|xn\| →z. Since {G(xn)} is bounded from above, here exists M such that J(xn) = 1 2\|Axn −b\|2 2 + β \|xn\|0 ≤M, for all n. (15.8) Since R(A) closed, by the closed range theorem X = R(A∗)+N(A) and let xn = x1 n+x2 n Consequently, 0 ≤\|Ax1 n\|2 −2(b, Ax1 n)2 + \|b\|2 2 ≤K and \|Ax1 n\| is bounded. 0 ≤ A( x1 n \|xn\|2 ) 2 2 −2 1 \|xn\|2 ( A∗b, xn \|xn\|2 )2 + \|b\|2 2 →0. (15.9) By the closed range theorem this implies that \|xn i is bounded and x1 n \|xn\|2 →¯ x1 = 0 in ℓ2. Since x2 n \|xn\|2 ⇀¯ x2 and by assumption dim N(A) < ∞it follows that xn \|xn\|2 →z = ¯ x2 strongly in ℓ2. Next, (15.3) holds. Since \|xn\|0 = \| xn \|xn\|\|0 and thus \|z\|0 < ∞(15.3) is equivalent to showing that \|(x1 n + x2 n −ρ z)i\|0 ≤\|(x1 n + x2 n)i\|0 for all i, and for all n suﬃciently large. Only the coordinates for which (x1 n + x2 n)i = 0 with zi ̸= 0 require our attention. Since \|z\|0 < ∞there exists ˜ i such that zi = 0 for all i > ˜ i. For i ∈{1, . . . ,˜ i} we deﬁne Ii = {n : (x1 n + x2 n)i = 0, zi ̸= 0}. These sets are ﬁnite. In fact, if Ii is inﬁnite for some i ∈{1, . . . ,˜ i}, then limn→∞,n∈Ii 1 \|xn\|2 (x1 n + x2 n)i = 0. Since limn→∞ 1 \|xn\|2 (x1 n)i = 0 this implies that zi = 0, which is a contradiction. Example 3 (Obstacle problem) Consider min G(u) = Z 1 0 (1 2\|ux\|2 −fu) dx (15.10) subject to u( 1 2) ≤1. Let X = H1(0, 1). If \|un\| →∞and vn = un \|un\| ⇀v. By the assumption in (ii) 1 2 Z 1 0 \|(vn)x\|2 dx ≤ 1 \|un\| Z 1 0 fvn dx + G(un) \|un\|2 →0 and thus (vn)x →0 and vn →v = c for some constant c. But, since vn( 1 2) ≤ 1 \|un\| →0, c = v( 1 2) ≤0 Thus, G(un −ρ v) = G(un) + Z 1 0 ρvf(x) dx ≤G(un) if R 1 0 f(x) dx ≥0. That is, if R 1 0 f(x) dx ≥0 (15.10) has a minimizer. 218 Example 4 (Friction Problem) min Z 1 0 (1 2\|ux\|2 −fu) dx + \|u(1)\| (15.11) Using exactly the same argument above. we have (vn) →0 and vn →v = c. But, we have G(un −ρv) = G(un) + ρc Z 1 0 f(x) dx + \|un(1) −ρc\| −\|un(1)\| where \|un(1) −ρc\| −\|un(1)\| \|un\| = \|vn(1) − ρ \|un\|c\| −\|vn(1)\| ≤0 for suﬃciently large n since \|un(1) −ρc\| −\|un(1)\| \|un\| →\|c −ˆ ρc\| −\|c\| = −ˆ ρ\|c\| where ˆ ρ = limn→∞ ρn \|un\|. Thus, if \| R 1 0 f(x) dx\| ≤1, then (15.11) has a minimizer. Example 5 (L∞Laplacian). min Z 1 0 \|u −f\| dx subject to \|ux\| ≤1 Similarly, we have (vn)x →0 and vn →v = c. G(un −ρc) −G(un) \|un\| = G(vn −ˆ ρc − f \|un\|) −G(vn − f \|un\|) ≤0 for suﬃciently large n. Example 6 (Elastic contact problem) Let X = H1(Ω)d and ⃗ u is the deformation ﬁeld. Given, the boundary body force ⃗ g, consider the elastic contact problem: min Z Ω 1 2ϵ : σ dx − Z Γ1 ⃗ g · ⃗ u dsx + Z Γ2 \|τ · ⃗ u\| dsx, subject to n · ⃗ u ≤ψ, (15.12) where we assume the linear strain: ϵ(u) = 1 2( ∂ui ∂xj + ∂uj ∂xi ) and the Hooke’s law: σ = µϵ + λ trϵI with positive Lame constants µ, λ. If \|⃗ un\| →∞and G(un) is bounded we have ϵ(vn) →0 ∇ϵ(vn) →0 for ⃗ vn = ⃗ un \| ⃗ un\| ⇀v. Thus, one can show that ⃗ vn →v = (v1, v2) ∈{(−Ax2 + C1, Ax1 + C2)} for the two dimensional problem. Consider the case when Ω= (0, 1)2 and Γ1 = {x2 = 1} and Γ2 = {x2 = 0}. As shown in the previous examples G(⃗ un −ρ v) −G(⃗ un) \|⃗ un\| = Z Γ1 ˆ ρn · ⃗ gv2 dsx + Z Γ1 ˆ ρτ · ⃗ gv1 dsx + Z Γ2 (\|(vn)1 −ˆ ρv1\| −\|(vn)1\|) dsx 219 Since v2 ∈C = {Ax1 + C2 ≤0} and for v1 = C1 \|(vn)1 −ˆ ρv1\| −\|(vn)1\| →\|(v)1 −ˆ ρv2\| −\|(v)1\| = −ˆ ρ\|v1\|, if we assume \| R Γ1 τ · ⃗ g dsx\| ≤1 and R Γ1 n · ⃗ gv2 dsx ≥0 for all v2 ∈C, then (15.12) has a minimizer. Example 7 min Z 1 0 (1 2(a(x)\|ux\|2 + \|u\|2) −fu) dx with a(x) > 0 except a( 1 2) = 0. 15.2 Necessary Optimality Let X be a Banach space and continuously embed to H = L2(Ω)m. Let J : X →R+ is C1 and N is nonsmooth functional on H = L2(Ω)m is given by N(y) = Z Ω h(y(ω)) dω. where h : Rm →R is a lower semi-continuous functional. Consider the minimization on H; min J(y) + N(y) (15.13) subject to y ∈C = {y(ω) ∈U a.e. } where U is a closed convex set in Rm. Given y∗∈C, deﬁne needle perturbations of (the optimal solution) y∗by y = y∗ on \|ω −s\| > δ u on \|ω −s\| ≤δ for δ > 0, s ∈ω, and u ∈U. Assume that \|J(y) −J(y∗) −(J′(y∗), y −y∗)\| ∼o(meas({\|s\| < δ}). (15.14) Let H is the Hamiltonian deﬁned by H(y, λ) = (λ, y) −h(y), y, λ ∈Rm. Then, we have the pointwise maximum principle. Theorem (Pointwise Optimality) If an integrable function y∗attains the minimum and (15.14) holds, then for a.e. ω ∈Ω J′(y∗) + λ = 0 H(u, λ(ω)) ≤H(y∗(ω), λ(ω)) for all u ∈U. (15.15) Proof: Since 0 ≥J(y) + N(y) −(J(y∗) + N(y∗)) = J(y) −J(y∗) −(J′(y∗), y −y∗) + Z \|ω−s\|<δ (J′(y∗), u −y∗) + h(u) −h(y∗)) dω 220 Since h is lower semicontinuos, it follows from (15.14) that at a Lubegue point s = ω of y∗(15.15) holds. □ For λ ∈Rm deﬁne a set-valued function by Φ(λ) = argmaxy∈U{(λ, y) −h(y)}. Then, (15.15) is written as y(ω) ∈Φ(−J′(y(ω)) The graph of Φ is monotone: (λ1 −λ2, y1 −y2) ≥0 for all y1 ∈Φ(λ1), y2 ∈Φ(λ2). but it not necessarily maximum monotone. Let h∗be the conjugate function of h deﬁned by h∗(λ) = max u∈U H(λ, y) = max u∈U {(λ, u) −h(u)}. Then, h∗is necessarily convex. Since if u ∈Φ(λ), then h∗(ˆ λ) ≥(ˆ λ, y) −h(u) = (ˆ λ −λ, y) + h∗(u), thus u ∈∂h∗(λ). Since h∗is convex, ∂h∗is maximum monotone and thus ∂h∗is the maximum monotone extension of Φ. Let h∗∗be the bi-conjugate function of h; h∗∗(x) = sup λ {(x, λ) −sup y H(y, λ)} whose epigraph is the convex envelope of the epigraph of h, i.e., epi(h∗∗) = {λ1 y1 + (1 −λ) y2 for all y1, y2 ∈epi(h) and λ ∈[0, 1]}. Then, h∗∗is necessarily convex and is the convexﬁcation of h and λ ∈∂h∗∗(u) if and only if u ∈∂h∗(λ) and h∗(λ) + h∗∗(u) = (λ, u). Consider the relaxed problem: min J(y) + Z Ω h∗∗(y(ω)) dω. (15.16) Since h∗∗is convex, N∗∗(y) = Z Ω h∗∗(y) dω is weakly lower semi-continuous and thus there exits a minimizer y of (15.16). It thus follows from Theorem (Pointwise Optimality) that the necessary optimality of (15.16) is given by −J′(y)(ω) ∈∂h∗∗(y(ω)), or equivalently y ∈∂h∗(−J′(y)). (15.17) 221 Lemma 2.2 Assume J(ˆ y) −J(y) −J′(y)(ˆ y −y) ≥0 for all ˆ y ∈C, where y(ω) ∈Φ(−J′(y(ω))). Then, y minimizes the original cost functional (15.13). Proof: Since Z Ω (J′(y), ˆ y −y) + h(ˆ y) −h(y) ≥0, we have J(ˆ y) + Z Ω h(ˆ y) dω ≥J(y) + Z Ω h(y) dω for all ˆ y ∈C.□ One can construct a C1 realization of h∗by h∗ ϵ(λ) = min p { ϵ 2\|p −λ\|2 + h∗(p)}. That is, h∗ ϵ is C1 and (h∗ ϵ)′ is Lipschitz and (h∗ ϵ)′ is the Yosida approximation of ∂h∗. If we let hϵ = (h∗ ϵ)∗, then hϵ is convex and hϵ(u) = min{ 1 2ϵ\|y −u\|2 + h∗∗(y)}. Consider the relaxed problem min J(y) + Z Ω hϵ(y(ω)) dω. (15.18) The necessary optimality condition becomes y(ω) = (h∗ ϵ)′(−J′(y(ω))). (15.19) Deﬁne the proximal approximation by pϵ(λ) = argmin{ ϵ 2\|p −λ\|2 + h∗(p)} Then, (h∗ ϵ)′ = ϵ (λ −pϵ(λ)). Thus, one can develop the semi smooth Newton method to solve (15.19). Since hϵ ≤h∗∗, one can argue that a sequence of minimizers yϵ of (15.18) to a minimizer of (15.16). In fact, J(yϵ) + Z Ω hϵ(yϵ) dω ≤J(y) + Z Ω hϵ(y) dω. for all y ∈C. Suppose \|yϵ\| is bounded, then yϵ has a weakly convergent subsequence to ¯ y ∈C and J(¯ y) + Z Ω h∗∗(¯ y) dω ≤J(y) + Z Ω h∗∗(y) dω. for all y ∈C. 222 Example (L0(Ω) optimization) Based on our analysis one can analyze when h involves the volume control by L0(Ω), i.e., Let h on R be given by h(u) = α 2 \|u\|2 + β\|u\|0, (15.20) where \|u\|0 = ( 0 if u = 0 1 if u ̸= 0. That is, Z Ω \|u\|0 dω = meas({u(ω) ̸= 0}). In this case Theorem applies with Φ(q) := argminu∈R(h(u) −qu) =    q α for \|q\| ≥√2αβ 0 for \|q\| < √2αβ. (15.21) and the conjugate function h∗of h; −h∗(q) = h(Φ(q)) −q Φ(q) =    −1 2α\|q\|2 + β for \|q\| ≥√2αβ 0 for \|q\| < √2αβ. The bi-conjugate function h∗∗(u) is given by h∗∗(u) =        α 2 \|u\|2 + β \|u\| > q 2β α √2αβ \|u\| \|u\| ≤ q 2β α . Clearly, Φ : R →R is monotone, but it is not maximal monotone. The maximal monotone extension ˜ Φ = (∂h∗∗)−1 = ∂h∗of Φ and is given by ˜ Φ(q) ∈                    q α for \|q\| > √2αβ 0 for \|q\| < √2αβ [ q α, 0] for q = −√2αβ [0, q α] for q = √2αβ. Thus, (h∗ ϵ)′ = λ −pϵ(λ) ϵ where pϵ(λ) = argmin{ 1 2ϵ\|p −λ\|2 + h∗(p)}. 223 We have pϵ(λ) =                      λ \|λ\| < √2αβ √2αβ √2αβ ≤λ ≤(1 + ϵ α)√2αβ −√2αβ √2αβ ≤−λ ≤(1 + ϵ α)√2αβ λ 1+ ϵ α \|λ\| ≥(1 + ϵ α)√2αβ. (h∗ ϵ)′(λ) =                      0 \|λ\| < √2αβ λ−√2αβ ϵ √2αβ ≤λ ≤(1 + ϵ α)√2αβ λ+√2αβ ϵ √2αβ ≤−λ ≤(1 + ϵ α)√2αβ λ α+ϵ \|λ\| ≥(1 + ϵ α)√2αβ. (15.22) Example (Authority optimization) We analyze the authority optimization of two un-knowns u1 and u2, i.e., the at least one of u1 and u2 is zero. We formulate such a problem by using h(u) = α 2 (\|u\|2 1 + \|u2\|2) + β\|u1u2\|0, (15.23) The, we have h∗(p1, p2) =            1 2α(\|p1\|2 + \|p2\|2) −β \|p1\|, \|p2\| ≥√2αβ 1 2α\|p1\|2 \|p1\| ≥\|p2\|, \|p2\| ≤√2αβ 1 2α\|p2\|2 \|p2\| ≥\|p1\|, \|p1\| ≤√2αβ 224 Also,we have pϵ(λ1, λ2) =                                                                  1 1+ ϵ α (λ1, λ2) \|λ1\|, \|λ2\| ≥(1 + ϵ α)√2αβ ( 1 1+ ϵ α λ1, ±√2αβ) \|λ1\| ≥(1 + ϵ α)√2αβ, √2αβ ≤\|λ2\| ≤(1 + ϵ α)√2αβ (±√2αβ, 1 1+ ϵ α λ2) \|λ2\| ≥(1 + ϵ α)√2αβ, √2αβ ≤\|λ1\| ≤(1 + ϵ α)√2αβ (λ2, λ2) \|λ2\| ≤\|λ1\| ≤(1 + ϵ α)√2αβ, √2αβ ≤\|λ2\| ≤(1 + ϵ α)√2αβ (λ1, λ1) \|λ1\| ≤\|λ2\| ≤(1 + ϵ α)√2αβ, √2αβ ≤\|λ1\| ≤(1 + ϵ α)√2αβ ( λ1 1+ ϵ α , λ2) \|λ1\| ≥(1 + ϵ α)\| ≥\|λ2\|, \|λ2\| ≤√2αβ (λ1, λ2 1+ ϵ α ) \|λ2\| ≥(1 + ϵ α)\| ≥\|λ1\|, \|λ1\| ≤√2αβ (λ2, λ2) \|λ2\| ≤\|λ1\| ≤(1 + ϵ α)\|λ2\|, \|λ2\| ≤√2αβ (λ1, λ1) \|λ1\| ≤\|λ2\| ≤(1 + ϵ α)\|λ1\|, \|λ1\| ≤√2αβ (15.24) 15.3 Augmented Lagrangian method Given λ ∈H and c > 0, consider the augmented Lagrangian functional; Lc(x, y, λ) = J(x) + (λ, x −y)H + c 2\|x −y\|2 H + N(y). Since Lc(x, y, λ) = J(x) + Z Ω (c 2\|x −y\|2 + (λ, x −y) + h(y)) dω it follows from Theorem that if (¯ x, ¯ y) minimizes Lc(x, y, λ), then the necessary opti-mality condition is given by −J′(¯ x) = λ + c (¯ x −¯ y) ¯ y ∈Φc(λ + c ¯ x). where Φc(p) = argmax{(p, y) −c 2\|y\|2 −h(y)} Thus, the augmented Lagrangian update is given by λn+1 = λn + c (xn −yn) where (xn, yn) solves −J′(xn) = λn + c (xn −yn) yn ∈Φc(λn + c xn). 225 Let L∗(λ) = inf(x,y) Lc(x, y, λ). Since L∗(λ) = −sup (x,y) {(x, −λ) + (y, λ) −J(x) −h(y) −c 2\|x −y\|2\|} and thus L∗(λ) is concave and y −x ∈∂L∗(λ) for (x, y) = argmin(x,y)Lc(x, y, λ). Let ¯ λ is a maximizer of L∗(λ). Since L∗(λ) ≤L∗(¯ λ) + (λ −¯ λ, ¯ x −¯ y) and L∗(λ) −L∗(¯ λ) ≤0 we have y −x = 0 and thus we obtain the necessary optimality −J′(x) = λ x ∈Φc(λ + c x). Note that supp{−1 2c\|p −λ\|2 + (x, p) −h∗(p)} = sup p {−sup y {(y, p) −h(y)} −1 2c\|p −λ\|2 + (x, p)} = inf y {h(y) + (λ, x −y) + c 2 \|y −x\|2} = ϕc(x, λ). Let Lc(x, y, λ) = J(x) + (λ, x −y)H + c 2\|y −x\|2 H + h(y). Then L∗(λ) = inf x inf p {J(x) + 1 2c\|p −λ\|2 + (x, p) + h∗(p)} If we deﬁne the proximal approximation by pc(λ) = argmin{ 1 2c\|p −λ\|2 + h∗(p)}, then (ϕc)′(x, λ) = pc(λ + c x). (15.25) If we deﬁne the proximal approximation by pc(λ) = argmin{ 1 2c\|p −λ\|2 + h∗(p)}, then (ϕc)′(x, λ) = pc(λ + c x). (15.26) Since −1 2c\|p −λ\|2 + (x, p) = −1 2c\|p −(λ + c x)\|2 + 1 2c(\|λ + c x)\|2 −\|λ\|2) 226 Also, we have (ϕc)′(λ) = λ −pc(λ) c . (15.27) Thus, the maximal monotone extension ˜ Φc of the graph Φc is given by (ϕc)′(λ), the Yosida approximation of ∂h∗(λ) and results in the update: −J′(xn) = λn + c (xn −yn) yn = (ϕ∗ c)′(λn + c xn). In the case c = 0, we have the multiplier method λn+1 = λn + α (xn −yn), α > 0 where (xn, yn) solves −J′(xn) = λn yn ∈Φ(λn) In the case λ = 0 we have the proximal iterate of the form xn+1 = (ϕc)′(xn −α J′(xn)). 15.4 Semismooth Newton method The necessary optimality condition is written as −J′(x) = λ λ = h′ c(x, λ) = pc(λ + c x). is the semi-smooth equation. For the case of h = α 2 \|x\|2 + β\|x\|0 we have from (15.22)-(15.26) h′ c(x, λ) =                    λ if \|λ + cx\| ≤√2αβ √2αβ if √2αβ ≤λ + cx ≤(1 + c α)√2αβ −√2αβ if √2αβ ≤−(λ + cx) ≤(1 + c α)√2αβ λ α+c if \|λ + cx\| ≥(1 + c α)√2αβ. Thus, the semi-smooth Newton update is −J′(xn+1) = λn+1,                    xn+1 = 0 if \|λn + cxn\| ≤√2αβ λn+1 = √2αβ if √2αβ < λn + cxn < (1 + c α)√2αβ λn+1 = −√2αβ if √2αβ < −(λn + cxn) < (1 + c α)√2αβ xn+1 = λn+1 α if \|λn + cxn\| ≥(1 + c α)√2αβ. 227 For the case h(u1, u2) = α 2 (\|u1\|2 + \|u2\|2) + β\|u1u2\|0. Let µn = λn + cun. From (15.24) the semi-smooth Newton update is                                                                      un+1 = λn+1 α \|µn 1\|, \|µn 2\| ≥(1 + ϵ α)√2αβ un+1 1 = λn+1 1 α , λn+1 2 = ±√2αβ \|µn 1\| ≥(1 + ϵ α)√2αβ, √2αβ ≤\|µn 2\| ≤(1 + ϵ α)√2αβ λn+1 = ±√2αβ, un+1 2 = λn+1 2 α \|µn 2\| ≥(1 + ϵ α)√2αβ, √2αβ ≤\|µn 1\| ≤(1 + ϵ α)√2αβ λn+1 1 , λn+1 2 , un+1 1 = λn+1 1 α \|µn 2\| ≤\|µn 1\| ≤(1 + ϵ α)√2αβ, √2αβ ≤\|µn 2\| ≤(1 + ϵ α)√2αβ λn+1 1 = λn+1 2 , un+1 2 = λn+1 2 α \|λ1\| ≤\|λ2\| ≤(1 + ϵ α)√2αβ, √2αβ ≤\|λ1\| ≤(1 + ϵ α)√2αβ un+1 1 = λn+1 1 α , un+1 2 = 0 \|µn 1\| ≥(1 + ϵ α)\| ≥\|µn 2\|, \|µn 2\| ≤√2αβ un+1 2 = λn+1 2 α , un+1 1 = 0 \|µn 2\| ≥(1 + ϵ α)\| ≥\|µn 1\|, \|µn 1\| ≤√2αβ λn+1 2 = λn+1 2 un+1 2 = 0 \|µn 2\| ≤\|µn 1\| ≤(1 + ϵ α)\|µn 2\|, \|µn 2\| ≤√2αβ λn+1 1 = λn+1 2 , un+1 1 = 0 \|µn 1\| ≤\|µn 2\| ≤(1 + ϵ α)\|µn 1\|, \|µn 1\| ≤√2αβ. 15.5 Constrained optimization Consider the constrained minimization problem of the form; min J(x, u) = Z Ω (ℓ(x) + h(u)) dω over u ∈U, (15.28) subject to the equality constraint ˜ E(x, u) = Ex + f(x) + Bu = 0. (15.29) Here U = {u ∈L2(Ω)m : u(ω) ∈U a.e. in Ω}, where U is a closed convex subset of Rm. Let X is a closed subspace of L2(Ω)n and E : X × U →X∗with E ∈L(X, X∗) bounded invertible, B ∈L(L2(Ω)m, L2(Ω)n) and f : L2(Ω)n →L2(Ω)n Lipschitz. We assume ˜ E(x, u) = 0 has a unique solution x = x(u) ∈X, given u ∈U. We assume there exist a solution (¯ x, ¯ x) to Problem (15.33). Also, we assume that the adjoint equation: (E + fx(¯ x))∗p + ℓ′(¯ x) = 0. (15.30) has a solution in X. It is a special case of (15.1) when J(u) is the implicit functional of u: J(u) = Z Ω ℓ(x(ω)) dω, where x = x(u) ∈X for u ∈U is the unique solution to Ex + f(x) + Bu = 0. To derive a necessary condition for this class of (nonconvex) problems we use a maximum principle approach. For arbitrary s ∈Ω, we shall utilize needle perturbations 228 of the optimal solution ¯ u deﬁned by v(ω) =    u on {ω : \|ω −s\| < δ} ¯ u(ω) otherwise, (15.31) where u ∈U is constant and δ > 0 is suﬃciently small so that {\|ω −s\| < δ} ⊂Ω. The following additional properties for the optimal state ¯ x and each perturbed state x(v) will be used:              \|x(v) −¯ x\|2 L2(Ω) = o(meas({\|ω −s\| < δ})) 1 2 R Ω ℓ(·, x(v)) −ℓ(·, ¯ x) −ℓx(·, ¯ x)(x(v) −¯ x) dω = O(\|x(v) −¯ x\|2 L2(Ω)) ⟨f(·, x(v)) −f(·, ¯ x) −fx(·, ¯ x)(x(v) −¯ x), p⟩X∗,X = O(\|x(v) −¯ x\|2 X) (15.32) Theorem Suppose (¯ x, ¯ u) ∈X ×U is optimal for problem (15.33) , that p ∈X satisﬁes the adjoint equation (15.30) and that (??), (??), and (15.32) hold. Then we have the necessary optimality condition that ¯ u(ω) minimizes h(u) + (p, Bu) pointwise a.e. in Ω. Proof: By the second property in (15.32) we have 0 ≤J(v) −J(¯ u) = R Ω ℓ(·, x(v)) −ℓ(·, x(¯ u)) + h(v) −h(¯ u) dω = R Ω(ℓx(·, ¯ x)(x −¯ x) + h(v) −h(¯ u)) dω + O(\|x −¯ x\|2), where (ℓx(·, ¯ x)(x −¯ x), p) = −⟨(E + fx(·, ¯ x)(x −¯ x)), p⟩ and 0 = ⟨E(x −¯ x) + f(·, x) −f(·, ¯ x) + B(v −¯ u), p⟩ = ⟨E(x −¯ x) + fx(·, ¯ x)(x −¯ x) + B(v −¯ u), p⟩+ O(\|x −¯ x\|2) Thus, we obtain 0 ≤J(v) −J(¯ u) ≤ Z Ω ((−B∗p, v −¯ u) + h(v) −h(¯ u)) dω. Dividing this by meas({\|ω −s\| < δ}) > 0 , letting δ →0, and using the ﬁrst property in (15.32) we obtain the desired result at a Lebesgue point s ω →(−B∗p, v −¯ u) + h(v) −h(¯ u)) since h is lower semi continuous. □ Consider the relaxed problem of (15.33): min J(x, u) = Z Ω (ℓ(x) + h∗∗(u)) dω over u ∈U, (15.33) 229 We obtain the pointwise necessary optimality            E¯ x + f′(¯ x) + B¯ u = 0 (E + f′(¯ x))∗p + ℓ′(¯ x) = 0 ¯ u ∈∂h∗(−B∗p), or equivalently λ = (h∗ c)′(u, λ), λ = B∗p. where h∗ c is the Yoshida-M approximation of h∗∗. 15.6 Algoritm Based the complementarity condition (??) we have the Primal dual active method for Lp optimization (h(u) = α 2 \|u\|2 + β \|u\|p.): Primal-Dual Active method (Lp(Ω)-optimization) 1. Initialize u0 ∈X and λ0 ∈H. Set n = 0. 2. Solve for (yn+1, un+1, pn+1) Eyn+1 + Bun+1 = g, E∗pn+1 + ℓ′(yn+1) = 0, λn+1 = B∗pn+1 and un+1 = − λn+1 α + βp\|un\|p−2 if ω ∈{\|λn + c un\| > (µp + cup} λn+1 = µp if ω ∈{µp ≤λn + cun ≤µp + cun}. λn+1 = −µp if ω ∈{µp ≤−(λn + cun) ≤µp + cun}. un+1 = 0, if ω ∈{\|λn + cun\| ≤µp}. 3. Stop, or set n = n + 1 and return to the second seep. 16 ℓ0 optimization Consider ℓ0 minimization in ℓ2 J(x) = F(x) + β N0(x) where N0(x) = X i \|xi\|0 counts the number of nonzero elements of ℓ2. Necessary optimality: Let ¯ x be a minimizer. Then for each coordinate Gi( ¯ xi) + β \| ¯ xi\|0 ≤Gi(xi) + β \|xi\|0 for all xi ∈R, 230 where Gi(xi) = F(¯ x + (xi −¯ xi) ei) Suppose xi →Gi(xi) is strict convex. If zi is a minimizer of Gi, then Fxi(z) = 0 and Gi(z) + β < Gi(0), then zi = ¯ xi is a minimizer of Gi(xi) + β \|xi\|0. Otherwise ¯ xi = 0 is a minimizer Thus, we obtain the necessary optimality F ′(¯ x) + λ = 0, λi¯ xi = 0 ¯ xi = 0 if Gi(0) −Gi(¯ xi) ≤β, λi = 0 if Gi(0) −Gi(¯ xi) > β Note that Gi(0) −Gi(zi) = Z zi 0 (z −s)G′′ i (s) ds ∼1 2FG′′ i (zi 2 )\|zi\|2. and λi = −G′ i(0) = −(G′ i(0) −G′ i(zi)) = Z z 0 G′′ i (s) ds ∼G′′ i (zi 2 )zi. Thus, we have approximate complementarity \|λi\|2 < 2βG′′ i ( zi 2 ) ⇒¯ xi = 0 λi = 0 ⇒\|¯ xi\|2 ≥ 2β G′′ i ( zi 2 ) Equivalently, λk = 0 if k ∈{k : \|λk + Λkxk\|2 > 2βΛk} λj = 0 if j ∈{j : \|λj + Λjxj\|2 ≤2βΛj} (16.1) where Λi = G′′ i ( zi 2 ). For the quadratic case F(x) = 1 2(\|Ax −b\|2 + α (x, Px) with nonnegative P and α ≥0. We have G′′ i ( zi 2 ) = \|Ai\|2 + Pii and thus A∗(Ax −b) + Px + λ = 0, λixi = 0 with complementarity condition (16.1) is the necessary optimality condition. 17 Exercise Problem 1 If A is a symmetric matrix on Rn and b ∈Rn is a vector. Let F(x) = 1 2xtAx −btx. Show that F ′(x) = Ax −b. Problem 2 Sequences {fn}, {gn}, {hn} are Cauchy in L2(0, 1) but not on C[0, 1]. Problem 3 Show that \|\|x\| −\|y\|\| ≤\|x −y\| and thus the norm is continuous. Problem 4 Show that all norms are equivalent on a ﬁnite dimensional vector space. 231 Problem 5 Show that for a closed linear operator T, the domain D(T) is a Banach space if it is equipped by the graph norm \|x\|D(T) = (\|x\|2 X + \|Tx\|2 Y )1/2. Problem 6 Complete the DC motor problem. Problem 7 Consider the spline interpolation problem (3.8). Find the Riesz representa-tions Fi and ˜ Fj in H2 0(0, 1). Complete the spline interpolation problem (3.8). Problem 8 Derive the necessary optimality for L1 optimization (3.27) with U = [−1, 1]. Problem 9 Solve the constrained minimization in a Hilbert space X; min 1 2\|x\|2 −(a, x) subject to (yi, x) = bi, 1 ≤i ≤m1, (˜ yj, x) ≤cj, 1 ≤j ≤m2. over x ∈X. We assume {yi}m1 i=1, {˜ yj}m2 j=1 are linearly independent. Problem 10 Solve the pointwise obstacle problem: min Z 1 0 (1 2\|du dx\|2 −f(x)u(x)) dx subject to u(1 3) ≤c1, u(2 3) ≤c2, over a Hilbert space X = H1 0(0, 1) for f = 1. Problem 11 Find the diﬀerential form of the necessary optimality for min Z 1 0 (1 2(a(x)\|du dx\|2+c(x)\|u\|2)−f(x)u(x)) dx+1 2(α1\|u(0)\|2+α2\|u(1)\|2)−(g1, u(0))−(g2, u(1)) over a Hilbert space X = H1(0, 1). Problem 12 Let B ∈Rm,n. Show that R(B) = Rm if and only if BBt is positive deﬁnite on Rm. Show that x∗= Bt(BBt)−1c deﬁnes a minimum norm solution to Bx = c. Problem 13 Consider the optimal control problem (3.48) (Example Control problem) with ˆ U = {u ∈Rm : \|u\|∞≤1}. Derive the optimality condition (3.49). Problem 14 Consider the optimal control problem (3.48) (Example Control problem) with ˆ U = {u ∈Rm : \|u\|2 ≤1}. Find the orthogonal projection PC and derive the necessary optimality and develop the gradient method. Problem 15 Let J(x) = R 1 0 \|x(t)\| dt is a functional on C(0, 1). Show that J′(d) = R 1 0 sign(x(t))d(t) dt. Problem 16 Develop the Newton methods for the optimal control problem (3.48) with-out the constraint on u (i.e., ˆ U = Rm). 232 Problem 17 (1) Show that the conjugate functional h∗(p) = sup u {(p, u) −h(u)} is convex. (2) The set-valued function Φ Φ(p) = argmaxu{(p, u) −h(u)} is monotone. Problem 18 Find the graph Φ(p) for max \|u\|≤γ{pu −\|u\|0} and max \|u\|≤γ{pu −\|u\|1}. 233
16694	https://artofproblemsolving.com/wiki/index.php/Multinomial_Theorem?srsltid=AfmBOooO0T7hj5aWZpyH7fvgaqYa3ks1p2rwomofKAq-ytjVEYXIq6Ig	Art of Problem Solving Multinomial Theorem - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Multinomial Theorem Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Multinomial Theorem The Multinomial Theorem states that where is the multinomial coefficient. Note that this is a direct generalization of the Binomial Theorem, when it simplifies to Contents [hide] 1 Proof 1.1 Proof by Induction 1.2 Combinatorial proof 2 Problems 2.1 Intermediate 2.2 Olympiad Proof Proof by Induction Proving the Multinomial Theorem by Induction For a positive integer and a non-negative integer , When the result is true, and when the result is the binomial theorem. Assume that and that the result is true for When Treating as a single term and using the induction hypothesis: By the Binomial Theorem, this becomes: Since , this can be rewritten as: Combinatorial proof This article is a stub. Help us out by expanding it. Problems Intermediate The expression is simplified by expanding it and combining like terms. How many terms are in the simplified expression? (Source: 2006 AMC 12A Problem 24) Olympiad Retrieved from " Categories: Stubs Theorems Combinatorics Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
16695	https://www.educationadvanced.com/blog/7-practical-teacher-productivity-tips-to-achieve-learning-outcomes	Request Demo Demo request Thank you! Your submission has been received! Oops! Something went wrong while submitting the form. 7 Practical Teacher Productivity Tips to Achieve Learning Outcomes November 15, 2022 No items found. No items found. Increasing teacher productivity is challenging because teaching is hard work and no matter how much work you invest, it will never be completely done. Especially for those of us who are always looking to get a little bit better all the time, the continuous improvement cycle creates endless work. When coupled with solid decisions about the most important work, efficiency leads to better productivity. Productivity in the world of law is more billable hours, in medicine is better patient health, and in business is greater profit. In education, higher productivity leads to better student outcomes. So efficiency, when tied to prioritization of work, is critical to achieving our mission of high achievement for all students. Here are some tips to consider that will help you reflect on your productivity: 1. Use More Organizational Tools Use physical, organizational systems like drawers, folders, bins, or binders. These are classroom essentials that help create a sense of order, as long as teachers can follow the filing system with fidelity. Practice doing the tasks immediately, so they don't compile or add to the number of priority jobs. This is especially true for easy-to-accomplish tasks and activities. Some examples of groupings to use might be: Homework / papers that need to be graded Homework that needs to be distributed later Materials needed for the next class Materials that have already been used, but can be used for next year Organizing miscellaneous materials such as permission slips Organizing materials based on subject areas or individual classes all together in one large bin or folder so you can pull relevant information all at one time Use folder dividers and labels in order to find information quickly Most importantly, take advantage of the digital world. Everything is one click away, so making the most of these tools and optimizing tasks is crucial. For example, school administrators can support their teachers by organizing information across multiple teachers in the same subject area to store information in a centralized location. Teacher productivity tools, such as Google Drive, help teachers quickly access shared files for collaboration. Tools such as Google Drive allow teachers to quickly search for relevant information within a large pool of data. In addition to Google Drive, also consider using technology such as: Electronic grading Electronic templates that organize information and make it easy to run lessons Email features such as autoreplies or contact information in signatures 2. Make Plans Ahead of Time Blocking a specific time slot for teachers to work on the upcoming week's planning can make sure teachers are well organized for their week. Teachers should use this time to do weekly lesson planning, outlining what materials need to be printed, copied, or organized, or organize what lessons should be coming up given the progress of the current classroom. On top of that, administrators can also hold weekly or bi-weekly quick follow-ups to make sure teachers are prioritizing the right tasks and getting the support they need to run an effective classroom. In addition to blocking off specific times to organize and prepare for lessons, also encourage teachers to block off times to take regular breaks. This helps teachers recover from the day and rest before getting back into work. 3. Grade Using Personalized Stamps There are many sites online or through craft sorts that sell personalized rubric rubber stamps that can be used. If teachers find that they are giving the same feedback over and over, one method is to create a mapping of the feedback to a specific stamp type. If the students know the mapping, every time they see the stamp, they know what the comment means. For example, teachers could use a stamp for “irrelevant content” or “needs to cite source.” Instead of writing personalized comments each time, using a stamp will help teachers grade their content quicker. 4. Maintain a To-Do List Creating a habit of keeping a to-do list helps increase efficiency. Teachers can display a physical to-do list on their table or classroom corner. A digital task list is also an option in many software suites such as GSuite or MS365. Frequently reviewing these lists and reordering them as needed or according to importance and urgency is vital. According to an article by Harvard Business Review, to-do lists can help free up brain space by offloading some tasks from our working memory, while still keeping track of important goals and tasks. 5. Establish Templates to Automate tasks To save time in creating teaching materials and disseminating information, teachers can take advantage of templates. Teachers can create grading bank stamps on a computer and utilize their learning management system's (LMS's) usage of criteria and scoring. This practice simplifies providing feedback, monitoring, and following up on students. On top of that, teachers can also utilize readily available messages and letter templates for information dissemination to their students and families, especially for important reminders. Suppose there are readily available templates for learning activities and resources, such as rubric guides and grading criteria. In that case, it will also be easier to track the students' progress while working on a task or project, lessening the time spent on clerical work. Templates can also be used in the classroom for things like providing students feedback, running brainstorming sessions, or setting up group projects. Once templates are established, teachers should be encouraged to share their templates with other teachers across different departments and years so together all teachers can revise and benefit from the templates. 6. Plan Out Quick Work & Extras In addition to lesson planning, plan out some quick work or “do-now” that students can do to begin learning the lesson before teachers start teaching. This can be done in the first few minutes of each class and gives teachers some time to prep or organize materials before class while activating prior knowledge for students and getting them ready for class. For example, before each math lesson, perhaps students do some problem sets from the last class or examples of problems that they will need to solve later in the class. 7. Borrow Lessons From Online The internet allows teachers access to lesson information taught all around the world. If teachers are having trouble coming up with the right strategy to explain a new concept, simply using Google to search the topic or looking up the topic on Youtube will give examples of how others have taught that lesson in the past. Websites such as TeacherVision or KhanAcademy show lesson templates across a variety of different topics that can be used to teach students in the classroom. Conclusion Increasing productivity can be difficult, but there are many ways for teachers to become more efficient so that they avoid burnout from the role. While it's still often the teacher's intrinsic motivation that can help them fully develop teacher productivity, administrators can also help by providing structure to automate tasks and organize information better. By implementing these practices collaboratively, this should foster a productive culture that will ultimately help lead to better student outcomes. More Great Content We know you'll love ### Unlocking Growth Through Job Embedded Professional Development by Kim Tunnell, Ed.D. August 29, 2025 ### Thoughtful, Not Fearful: How We’re Embracing AI at Education Advanced by Krista Endsley July 28, 2025 ### School Counseling and Its Impact on Student Success by Mia Finch, M.Ed. July 21, 2025 Stay In The Know Subscribe to our newsletter today! Sign Up No items found.
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Cite this lesson A Perfect Square is found when a number is the square of another integer. Learn the notation of radicals in the language of math, and how to calculate the square-roots of Perfect Squares. Create an account Table of Contents The Language of Math Squares and Square Roots Perfect Square Finding the Square Root of a Perfect Square Lesson Summary Learning Outcomes Show The Language of Math -------------------- Mathematics has its own special language. To understand and succeed in math, you need to be able to translate the language of math into English and back again. Just like you wouldn't move to France without knowing how to communicate at least the basics, you shouldn't venture into the study of math without knowing how to translate the terms you are using. To unlock this lesson you must be a Study.com Member. Create your account Click for sound 5:08 You must c C reate an account to continue watching Register to view this lesson Are you a student or a teacher? 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Your next lesson will play in 10 seconds 0:05 The Language of Math 0:23 Squares and Square Roots 1:31 Perfect Square 1:56 Finding the Square… 4:37 Lesson Summary View Video Only Save Timeline 83K views Video Quiz Course Video Only 83K views Squares and Square Roots ------------------------ The section of math dealing with radicals , or squares and square roots, is one place where knowing the translations of the terms is critical to understanding. To square a number means to multiply that number by itself. It is notated by a superscript number 2 after the main number. It can also be written with a caret between the number being squared and the 2. When you see the carat symbol (^), you say that the number is squared. For example: 3^2 is equal to 33 or 9 and 10^2 is equal to 1010 or 100 The square root operation is the inverse of the squared function and is notated by this symbol: The circled symbol indicates square root. This means that finding the square root of a number is the same as finding the opposite of a number squared. For example: √100 is 10 because 10^2 is 100 and √ 9 is 3 because 3^2 is 9 To unlock this lesson you must be a Study.com Member. Create your account Perfect Square -------------- A number is a perfect square if it is an integer that is the square of another integer. An integer is a number that does not contain a fraction or a decimal. For example: 5^2 is 25 So, 25 is a perfect square. and 17^2 is 289, which means 289 is also a perfect square. To unlock this lesson you must be a Study.com Member. Create your account Finding the Square Root of a Perfect Square ------------------------------------------- You can use a calculator to find the square root of a perfect square, but, if a calculator is not available, there is a way to calculate the square root. Here are the steps to calculating the square root of a perfect square: Step one: factor the number completely. An easy way to factor a number is by using a factor tree. A factor tree can be created by writing down the number you want to factor and drawing two lines coming down from that number. Then, write two factors of that number under the lines. Continue on until only prime numbers remain. A prime number is one that cannot be reduced any smaller. The purpose of the factor tree is so you can easily find the square root of large numbers if you don't have a calculator handy. By looking at the factor tree below, you can see that the factors of 225 are 3 3 5 5. Using this factor tree, you can see that the factors of 225 are 3 3 5 5. Step number 2 is to match up pairs of factors. In this case, we have a pair of 3s and a pair of 5s: (3 3) (5 5) Then, step number 3 to calculating the square root of a perfect square is to multiply one number from each pair of factors together to get the answer. In this case: 3 5 = 15 Fifteen is the square root of 225. For example, let's find the square root of 144. Your first step is to factor 144. The prime factors of 144 are (22) (22) (33) Your second step is to match up the pairs. There are 2 pairs of 2s, and 1 pair of 3s. Then, for step 3, take out 1 number from each pair, and multiply them together: 223, which equals 12. Therefore, the square root of 144 is equal to 12. Let's try another one. Solve √400. Again, your first step is to factor 400 by using a factor tree. The result you get is (22) (22) (55). Secondly, match up pairs of numbers. There are 2 pairs of 2s and 1 pair of 5s. Finally, multiply together the 225 to get 20. So, the √400 = 20 To unlock this lesson you must be a Study.com Member. Create your account Lesson Summary -------------- Squares and square roots are inverse operations. A perfect square is the square of a number that does not contain a fraction or decimal. To determine the square root of a perfect square, all you need to do is factor the number, combine pairs of factors and then multiply one number from each pair together. You can always check your work by performing the inverse operation, or squaring the number to see if your answer matches up with the original square root. To unlock this lesson you must be a Study.com Member. Create your account Learning Outcomes ----------------- After you have finished with this lesson, you should be able to: Describe the relationship between squares and square roots Define perfect square Explain how to determine the square root of a perfect square To unlock this lesson you must be a Study.com Member. Create your account Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Unlock Your Education See for yourself why 30 million people use Study.com Become a Study.com member and start learning now. Become a Member Already a member? 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16697	https://ui.adsabs.harvard.edu/abs/2015arXiv150100159S/abstract	A solution of the Erdos-Ulam problem on rational distance sets assuming the Bombieri-Lang conjecture - Astrophysics Data System Skip to main content Now on article abstract page Toggle navigation ads === Feedback Submit Updates General Feedback ORCID Sign in to ORCID to claim papers in the ADS. About About ADS What's New ADS Blog ADS Help Pages ADS Legacy Services Careers@ADS Sign Up Log In 2025-09-26: ADS servers will be undergoing scheduled maintenance starting 4:00PM EDT 2025-09-26. During this period, we will be unable to send any user emails, including new user registration and password reset emails. The outage is expected to last at least 24 hours, and this banner will be updated when email capabilities are restored. Close Search Bar to Enter New Query quick field:Author First Author Abstract Year Fulltext Select a field or operator All Search Terms Your search returned 0 results Your search returned 0 results Show Menu Full Text Sources view AbstractCitationsReferencesCo-ReadsSimilar PapersVolume ContentGraphicsMetricsExport Citation A solution of the Erdos-Ulam problem on rational distance sets assuming the Bombieri-Lang conjecture Show affiliations Loading affiliations affiliations loading Shaffaf, Jafar (0) Abstract A rational distance set in the plane is a point set which has the property that all pairwise distances between its points are rational. Erd\H os and Ulam conjectured in 1945 that there is no dense rational distance set in the plane. In this paper we associate an algebraic surface in $\mathbb{P}^3$, that we call a distance surface, to any finite rational distance set in the plane. Under a mild condition, we prove that a distance surface is always a surface of general type. From this, we deduce that the Bombieri-Lang conjecture in arithmetic algebraic geometry (restricted to the classes of surfaces) implies an answer to the Erd\H os-Ulam problem. Combined with the results of Solymosi and de Zeeuw, our proofs lead to the following stronger statement: for $S$ a rational distance set with infinitely many points, we have Either, all but at most four points of $S$ are on a line, Or, all but at most three points of $S$ are on a circle. Publication: eprint arXiv:1501.00159 Pub Date:December 2014 DOI: 10.48550/arXiv.1501.00159 arXiv:arXiv:1501.00159Bibcode: 2015arXiv150100159S Copied!Keywords: Mathematics - Number Theory; Mathematics - Algebraic Geometry; 11G99; 14G05; 14J29; 30D35; 52C10 E-Print Comments: 13 pages, analysis of singularity at infinity was added. Also a conjecture was added at the end of the paper; Discrete and Computational Geometry, 2018; doi:10.1007/s00454-018-0003-3 Feedback/Corrections? full text sources Preprint Preprint PDF \| Preprint article © The SAO Astrophysics Data System adshelp[at]cfa.harvard.edu The ADS is operated by the Smithsonian Astrophysical Observatory under NASA Cooperative Agreement 80NSSC25M7105 The material contained in this document is based upon work supported by a National Aeronautics and Space Administration (NASA) grant or cooperative agreement. Any opinions, findings, conclusions or recommendations expressed in this material are those of the author and do not necessarily reflect the views of NASA. Resources About ADS ADS Help System Status What's New Careers@ADS Web Accessibility Policy Social @adsabs ADS Blog Project Switch to basic HTML Privacy Policy Terms of Use Smithsonian Astrophysical Observatory Smithsonian Institution NASA 🌓 × Back How may we help you? #### Missing/Incorrect Record Submit a missing record or correct an existing record.#### Missing References Submit missing references to an existing ADS record.#### Associated Articles Submit associated articles to an existing record (e.g. arXiv / published paper).#### General Feedback Send your comments and suggestions for improvements. Name Email Feedback Submit You can also reach us at adshelp [at] cfa.harvard.edu This site is protected by reCAPTCHA and the Google Privacy Policy and Terms of Service apply. Close without submitting Alert
16698	https://en.wikipedia.org/wiki/Rabies_immunoglobulin	Jump to content Rabies immunoglobulin العربية Español Français Bahasa Indonesia 日本語 ଓଡ଼ିଆ Русский Српски / srpski 中文 Edit links From Wikipedia, the free encyclopedia Medication made up of antibodies against the rabies virus Pharmaceutical compound Rabies immunoglobulin \| Clinical data \| \| \| Trade names \| Imogam Rabies-HT, Kedrab, Hyperrab, others \| \| AHFS/Drugs.com \| Monograph \| \| Pregnancy category \| AU: B2 \| \| Routes of administration \| Intramuscular injection \| \| ATC code \| J06BB05 (WHO) \| \| Legal status \| \| \| Legal status \| AU: S4 (Prescription only) CA: ℞-only / Schedule D US: ℞-only \| \| Identifiers \| \| \| DrugBank \| DB11603 \| \| ChemSpider \| none \| \| UNII \| 95F619ATQ2 \| Rabies immunoglobulin (RIG) is a medication made up of antibodies against the rabies virus. It is used to prevent rabies following exposure. It is given after the wound is cleaned with soap and water or povidone-iodine and is followed by a course of rabies vaccine. It is given by injection into the site of the wound and into a muscle. It is not needed in people who have been previously vaccinated against rabies. Rabies, a viral zoonotic neglected tropical disease, poses a severe public health threat in over 150 countries and territories, primarily in Asia and Africa. Each year, this disease results in tens of thousands of fatalities, with children under 15 accounting for 40% of these deaths. Rabies infects mammals and is spread to humans and other animals by contact with saliva, most commonly through bites and scratches. Worldwide, nearly all human rabies cases are caused by dog bites and scratches. However, in the United States, bats are now the primary source of human rabies due to the diligent vaccination of dogs against rabies. Common side effects include pain at the site of injection, fever, and headache. Severe allergic reactions such as anaphylaxis may rarely occur. Use during pregnancy is not known to harm the fetus. It works by binding to the rabies virus before it can enter nerve tissue. After the virus has entered the central nervous system, rabies immunoglobulin is no longer useful. The use of rabies immunoglobulin in the form of blood serum dates from 1891. Use became common within medicine in the 1950s. It is on the World Health Organization's List of Essential Medicines. Rabies immunoglobulin is expensive and hard to come by in the developing world. In the United States it is estimated to be more than US$1,000.00 per dose, and around £600 in the United Kingdom. It is made by isolating rabies immunoglobulin from donated blood plasma of humans or horses who have high levels of the immunoglobulin. The equine preparation is less expensive but has a higher rate of side effects. Medical uses [edit] Rabies immunoglobulin (RIG) is indicated for the passive, transient post-exposure prophylaxis of rabies infection, when given immediately after contact with a rabid or possibly rabid animal and in combination with a rabies vaccine. Post-exposure prophylaxis (PEP) regimen [edit] The administration of Rabies Immunoglobulin (RIG) as part of post-exposure prophylaxis (PEP) depends on whether the individual has been previously vaccinated against rabies. Regardless of immunization status, PEP should begin immediately with thorough wound cleansing using soap and water, or preferably, irrigation with a povidone-iodine solution to reduce the viral load. Exposure risk categories and PEP indications [edit] According to the World Health Organization, PEP recommendations are based on the severity of exposure: Category I: Touching or feeding animals, or licks on intact skin — No PEP required, only washing of the exposed skin. Category II: Nibbling of uncovered skin, minor scratches, or abrasions without bleeding—Wound washing and immediate vaccination required. Category III: Transdermal bites, deep scratches, mucous membrane contamination with saliva, or any direct contact with bats—Wound washing, immediate vaccination, and administration of rabies immunoglobulin (RIG) or monoclonal antibodies required. PEP for non-immunized individuals [edit] For individuals who have never received rabies vaccination, PEP consists of both rabies vaccine and RIG. The rabies vaccine should be administered intramuscularly in the deltoid area on days 0, 3, 7, and 14. A fifth dose on day 28 is recommended for individuals with confirmed or suspected immune disorders. A full dose of rabies immunoglobulin (RIG) should be infiltrated around all identified wounds as much as anatomically possible, with any remaining volume administered intramuscularly at a site distant from the rabies vaccine injection. Mechanism of action [edit] Rabies Immune Globulin (RIG) binds to the rabies virus, thereby neutralizing it and preventing its spread to the central nervous system. This provides short-term passive immunity, allowing time for the rabies vaccine to stimulate an adaptive immune response capable of eradicating the virus. RIG is only effective if administered within the first eight days post-exposure, as the host’s immune system typically begins generating its own antibodies one week after exposure. Repeat doses should be avoided, as they may interfere with the body's natural immune response induced by the vaccine. PEP for previously immunized individuals [edit] For individuals who have been previously vaccinated against rabies, RIG should not be administered. Instead, the rabies vaccine is given intramuscularly in the deltoid area on days 0 and 3. Society and culture [edit] Names [edit] There are three versions of rabies immunoglobulin licensed and available in the US. Imogam Rabies-HT is produced by Sanofi Pasteur. Kedrab is produced by Kedrion Biopharma. Hyperrab is produced by Grifols. Imogam Rabies-HT and Kedrab have a nominal potency of 150 IU/mL while Hyperrab has a nominal potency of 300 IU/mL and requires smaller dosing. All three versions are used for post-exposure and indicate local infusion at the wound site with additional amount intramuscularly at a site distant from vaccine administration. Kamrab is approved for medical use in Australia. References [edit] ^ Jump up to: a b c d "Kamrab". Therapeutic Goods Administration (TGA). 23 August 2021. Retrieved 10 September 2021. ^ "Kamrab PI". Health Canada. 25 April 2012. Retrieved 10 September 2021. ^ "Imogam PI". Health Canada. 25 April 2012. Retrieved 10 September 2021. ^ "Hyperrab S/D PI". Health Canada. 25 April 2012. Retrieved 10 September 2021. ^ "Drug and medical device highlights 2018: Helping you maintain and improve your health". Health Canada. 14 October 2020. Retrieved 17 April 2024. ^ Jump up to: a b "Imogam Rabies-HT - human rabies virus immune globulin injection, solution". DailyMed. Retrieved 24 March 2020. ^ Jump up to: a b "Kedrab- human rabies virus immune globulin injection, solution". DailyMed. Retrieved 24 March 2020. ^ Jump up to: a b "Hyperrab (rabies immune globulin- human injection, solution". DailyMed. Retrieved 24 March 2020. ^ "Hyperrab S/D (rabies immune globulin- human injection". DailyMed. Retrieved 10 September 2021. ^ Jump up to: a b c d e f g h i "Rabies Immune Globulin". The American Society of Health-System Pharmacists. Archived from the original on 18 March 2011. Retrieved 8 January 2017. ^ World Health Organization (2009). Stuart MC, Kouimtzi M, Hill SR (eds.). WHO Model Formulary 2008. World Health Organization. p. 398. hdl:10665/44053. ISBN 9789241547659. ^ Jump up to: a b c d e "Rabies". World Health Organization (WHO). 5 June 2024. Retrieved 12 March 2025. ^ British national formulary : BNF 69 (69 ed.). British Medical Association. 2015. p. 869. ISBN 9780857111562. ^ Rupprecht CE, Plotkin SA (2013). "Rabies Vaccines". In Plotkin SA, Orenstein WA, Offit PA (eds.). Vaccines (6th ed.). [Edinburgh]: Elsevier/Saunders. p. 659. ISBN 978-1455700905. Archived from the original on 9 January 2017. ^ Jump up to: a b Jong EC, Zuckerman JN (2004). Travelers' Vaccines. PMPH-USA. p. 205. ISBN 9781550092257. Archived from the original on 9 January 2017. ^ World Health Organization (2019). World Health Organization model list of essential medicines: 21st list 2019. Geneva: World Health Organization. hdl:10665/325771. WHO/MVP/EMP/IAU/2019.06. License: CC BY-NC-SA 3.0 IGO. ^ Tintinalli JE (2010). Emergency Medicine: A Comprehensive Study Guide (Emergency Medicine (Tintinalli)) (7 ed.). New York: McGraw-Hill Companies. p. 1054. ISBN 978-0-07-148480-0. ^ Jump up to: a b c Research Advances in Rabies. Academic Press. 2011. p. 351. ISBN 9780123870414. Archived from the original on 9 January 2017. ^ Kliff, Sarah (7 February 2018). "Why a simple, lifesaving rabies shot can cost $10,000 in America". Vox. Retrieved 26 September 2024. ^ Jump up to: a b "Kedrab". U.S. Food and Drug Administration (FDA). 21 March 2018. Retrieved 7 June 2020. This article incorporates text from this source, which is in the public domain. ^ "Summary Basis for Regulatory Action - Kedrab". FDA. 23 August 2017. ^ Jump up to: a b c d e f "Rabies Post-exposure Prophylaxis". Centers for Disease Control and Prevention (CDC). 20 June 2024. Retrieved 12 March 2025. ^ Jump up to: a b c d "PubChem Compound Summary for Human Rabies Virus Immune Globulin". National Center for Biotechnology Information. 2025. Retrieved 12 March 2025. ^ "Vaccine and Immune Globulin Availability". Centers for Disease Control and Prevention (CDC). 26 February 2020. Retrieved 24 March 2020. ^ "WHO Guide for Rabies Pre and Post Exposure Prophylaxis in Humans" (PDF). World Health Organization (WHO). 2014. ^ "Rabies Biologics \| Specific Groups \| CDC". www.cdc.gov. 15 January 2021. Retrieved 18 November 2022. Further reading [edit] Meeting of the Antimicrobial Drugs Advisory Committee (Briefing Document). U.S. Food and Drug Administration (FDA). 25 April 2019. Archived from the original (PDF) on 14 December 2019. \| Immune sera and immunoglobulins (J06) \| \| --- \| \| Polyclonal antibodies \| \| \| \| --- \| \| IVIG \| Anthrax immune globulin Rho(D) immune globulin Hepatitis B immune globulin Zoster-immune globulin \| \| Antiserum \| \| \| \| Monoclonal antibodies \| Bezlotoxumab Clesrovimab Motavizumab Nebacumab Nirsevimab Obiltoxaximab Palivizumab Raxibacumab Sipavibart Tixagevimab/cilgavimab \| Retrieved from " Category: World Health Organization essential medicines Hidden categories: CS1:Vancouver names with accept markup Source attribution Articles with short description Short description is different from Wikidata Use dmy dates from September 2021 Chemicals that do not have a ChemSpider ID assigned Infobox drug articles without a structure image Chemical articles without CAS registry number Articles without EBI source Articles without KEGG source Articles without InChI source Articles containing unverified chemical infoboxes Wikipedia medicine articles ready to translate
16699	https://aviation.stackexchange.com/questions/89353/what-exactly-is-the-pressure-thrust-component-and-why-does-it-exist	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams What exactly is the pressure thrust component and why does it exist? Ask Question Asked Modified 4 years ago Viewed 3k times 10 $\begingroup$ I'm teaching PowerPlants on a ATPL theory course and honestly have some trouble wrapping my head around this one: Most literature describes the term "pressure thrust" as opposed to "momentum thrust" by explaining that during high thrust conditions in a choked exhaust, the exhaust gas reaches the speed of sound and cannot be accelerated further and thus the static pressure of the gas increases beyond atmospheric pressure, this pressure thrust is added to the momentum thrust to make up the total thrust. Firstly it is well known that exhaust gas velocities of aircraft can easily go supersonic, take a fighter jet for example... why is this a restriction in per say turbofan engines? Secondly it is often emphasized in training material that you should avoid thinking of thrust as acting on / pushing on the ambient air behind the engine, as opposed to thinking in terms of Newton's 3rd law... but isn't that exactly what pressure thrust is? It's defined by pressure $\times$ the area that it is acting on... jet-engine supersonic thrust Share Improve this question edited Sep 23, 2021 at 18:17 user14897 asked Sep 23, 2021 at 17:15 A.PilotA.Pilot 16911 silver badge66 bronze badges $\endgroup$ 0 Add a comment \| 3 Answers 3 Reset to default 9 $\begingroup$ Indeed, exhaust gases can be expanded beyond Mach 1, and doing so will result in higher thrust from higher propulsion efficiency. The problem is that pressurised gas is expanded by constricting the cross-section area when lower than M1, and expanding it when higher than M1. So in order to achieve full expansion to ambient pressure $p_0$ at supersonic exhaust speed, we need a cross-section that narrows, then expands, for all flight circumstances. The figure below illustrates how this is done in supersonic fighter jets: with an ejector jet exhaust. The primary exhaust is mounted inside a pipe, and the expanding exhaust gas sucks in a secondary flow, which dampens the primary flow expansion so that it takes place gradually. The secondary flow can be considered as the diverging part of the exhaust, and protects the actual metal exhaust from afterburner heat. In supersonic fighter jets both the primary and secondary exhausts are adjustable: pic left for subsonic speed, right for supersonic speed. So full expansion is possible yet complicated, with the necessity of constantly varying exhaust nozzles. Firstly it is well known that exhaust gas velocities of aircraft can easily go supersonic, take a fighter jet for example... why is this a restriction in per say turbofan engines? Because the turbofan's main flow is through the fan, which is only a compressor, no combustion takes place in this flow. In most circumstances the bypass flow is fully expanded at sub-sonic speeds - if it would not be the case, constructing in total four concentric regulated exhaust nozzles like in the pic above would be complicated and heavy, for very limited gain. Only above M1.5 will there be a significant thrust gain from full supersonic expansion. Secondly it is often emphasized in training material that you should avoid thinking of thrust as acting on / pushing on the ambient air behind the engine, as opposed to thinking in terms of Newton's 3rd law... but isn't that exactly what pressure thrust is? It's defined by pressure Ã the area that it is acting on... Exactly right! Share Improve this answer edited Sep 24, 2021 at 4:15 answered Sep 24, 2021 at 4:10 KoyovisKoyovis 63.5k1111 gold badges180180 silver badges301301 bronze badges $\endgroup$ 1 1 $\begingroup$ considering the fans.. not only would it be heavy and cumbersome, but to actually reach sonic speed in a nozzle, you need critical pressure ratio (roughly 1.9), and that's not provided by the fans. $\endgroup$ Apfelsaft – Apfelsaft 2021-09-24 19:43:34 +00:00 Commented Sep 24, 2021 at 19:43 Add a comment \| 6 $\begingroup$ Choked nozzle Whether flying supersonic or subsonic, a jet engine's intake, compressor, and diffuser, all slow down the air to slow subsonic velocity so combustion can take place. When the temperature rises sharply, part of this energy is converted to velocity. Because of the high combustion temperature, the speed of sound is much higher, so the mass flow on its way to the nozzle is subsonic; cool right? In subsonic flow, as air is squeezed, it speeds up, and its pressure drops. The designed pressure drop at the nozzle results in a certain mass flow to pass through due to the pressure difference. Drop it more, and at the narrowest part of the nozzle, the throat, the velocity reaches Mach 1. Now the flow is choked. Drop it further, and no more acceleration from subsonic by way of squeezing at that fixed throat is physically possible. The mass flow rate has reached its maximum (even for rockets). (Reference and further reading: Virginia Tech) Turbofan nozzles in jetliners (co-annular) do not even do that, purposefully for efficiency reasons the closer the exhaust velocity is to the free stream, the more efficient the propulsion is. Also, see: Wikipedia: Propulsive efficiency § Jet engines Fighters and high performance jets Fighters have two tricks: Hotter exhaust, and even hotter exhaust by using an afterburner. The hotter the air, the faster the speed of sound, allowing an exhaust's Mach 1 to be faster than the free stream Mach 1; cool again right? The exit temperature determines the exit speed of sound, which determines the exit velocity. NASA: Nozzle Design Variable geometry nozzles: an expanding throat allows more mass flow through the choked throat. At higher flight Mach numbers, the inlet provides more compression and therefore more mass flow rate becomes possible. That is why throats are widened when the plane is faster and/or using higher-thrust. Pressure thrust Because of the varying operating conditions in the fixed nozzles in rockets, "pressure thrust" is needed for the thrust equation, along with the change in momentum it's not one or the other, it's only that the pressure thrust is negligible for turbine engines: The nozzle of a turbine engine is usually designed to make the exit pressure equal to free stream. In that case, the pressure-area term in the general equation is equal to zero. NASA: General Thrust Equation Share Improve this answer edited Sep 24, 2021 at 20:40 answered Sep 24, 2021 at 4:36 user14897user14897 $\endgroup$ 4 $\begingroup$ The variable nozzle geometry doesn't really get more mass flow through a choked cross area. That m flow is restricted by the area and the inlet density, really. It's job is to provide a divergent nozzle for supersonic flow conditions, and increase the exhaust velocity even further than Mach 1. $\endgroup$ Apfelsaft – Apfelsaft 2021-09-24 19:56:13 +00:00 Commented Sep 24, 2021 at 19:56 $\begingroup$ @CarlBerger: But with higher thrust and faster flight Mach number, the inlet does provide more compression and therefore mass flow rate increases. If the throat is fixed, no additional mass flow can go through past the design point. We have a few topics here on inlet compression. $\endgroup$ user14897 – user14897 2021-09-24 20:06:31 +00:00 Commented Sep 24, 2021 at 20:06 $\begingroup$ ok. I wasn't very precise here, by inlet I meant "inlet to the nozzle" (because that was the part we're actually speaking about). The the flow state in the exhausts nozzle is defined by the pressure just before /around the nozzle, the exhaust nozzle gets choked when the pressure ratio over the nozzle gets supercritical. $\endgroup$ Apfelsaft – Apfelsaft 2021-09-24 21:10:23 +00:00 Commented Sep 24, 2021 at 21:10 $\begingroup$ Thanks, you got me actually digging again through one of the books I think that presents it best: Rogers / Cohen / Straznicky / Saravanamuttoo: Gas Turbine Theory $\endgroup$ Apfelsaft – Apfelsaft 2021-09-24 21:26:22 +00:00 Commented Sep 24, 2021 at 21:26 Add a comment \| 2 $\begingroup$ it's really two questions- 1) Compressible flow in ducts The first part of the question addresses how to accelerate a flow in ducts to supersonic speeds. In simple terms, for compressible flow in a pipe with varying diameter: when subsonic, it accelerates along the pipe, when the diameter of the pipe reduces when supersonic, it accelerates along the pipe, when the pipe widens. in order to produce supersonic flow, you need a duct that first contracts, to accelerate the flow to Mach 1 at the throat, then a duct that expands from there, because otherwise no further acceleration for the flow will take place - and it will remain at Mach 1 (super)critical pressure ratio, otherwise the flow will just accelerate to something below Mach1 in the throat, but since it's not supersonic, the divergent part will decelerate the flow again. a bit more more in-depth, if you want: or The cannot be accelerated further part refers to non-divergent nozzles. For supersonic designs, then there's a part about adapting the nozzle - meaning that the nozzle ideally should expand just enough so that the exit static pressure matches the atmospheric ambient, otherwise you get over- or under-expansion with a reduction in efficiency. 2) Momentum Conservation Thrust does in fact consist of two parts. The idea is, that when drawing a control volume somwhere, the reaction forces on this control volume are The reaction forces to balance to the change of momentum in the flow the sum of all pressures on the surfaces of that control volume (or mathematically more correct, the integral of the pressure normal to the control volume surface A: $\int_A p \vec{n} dA $ ) As a simplification, we could assume that the differences in pressures before and after the engine - far enough - are negligible, and then only the differences in momentum (stream velocity density) provide the thrust. But if the pressure differences are large enough, this cannot be neglected any more. The mathematical formulation gets a bit involved, then... but you can see the pressure terms. Share Improve this answer answered Sep 24, 2021 at 20:53 ApfelsaftApfelsaft 70244 silver badges55 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions jet-engine supersonic thrust See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked 12 Why does the J58 engine of the SR-71 have a diffuser after the inlet spike? what is best approach to make high thrust from nozzle side? higher velocity or higher pressure? 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