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vww23,571,2wEN17535 40 60 100 c 396 To visualize this sum, we draw with its initial point at, for convenience, so that its terminal point is. Next, we graph with its initial point at. Moving one to the right and two down, we find the terminal point of to be. We see that the vector has initial point and terminal point so its component form is as required. In order for vector addition to enjoy the same kinds of properties as real number addition, it is necessary to extend our definition of vectors to include a zero vector,. Geometrically, represents a point, which we can think of as a directed line segment with the same initial and terminal points. The reader may well object to the inclusion of since, after all, vectors are supposed to have both a magnitude (length) and a direction. While it seems clear that the magnitude of should be 0, it is not clear what its direction is. As we shall see, the direction of is in fact undefined, but this minor hiccup in the natural flow of things is worth the benefits we reap by including in our discussions. We have the following theorem. Theorem 9.1. Properties of Vector Addition. Commutative Property: For all vectors and,. Associative Property: For all vectors, and,. Identity Property: The vector. all vectors, acts as the additive identity for vector addition. That is, for Inverse Property: Every vector, there is a vector every vector has a unique additive inverse, denoted so that.. That is, for The properties in Theorem 9.1 are easily verified using the definition of vector addition. For the commutative property, we note that if and then 3,41,231,424,2vwv0,03,4w3,4w4,2vw0,04,24,2=0, |
0000000vwvwwvuvwuvwuvw0vv00vvv-vv-vv-v-vv012,vvv12,wwwxy vw w v v 397 Geometrically, we can ‘see’ the commutative property by realizing that the sums and are the same directed diagonal determined by the parallelogram. The proofs of the associative and identity properties proceed similarly, and the reader is encouraged to verify them and provide accompanying diagrams. The Additive Inverse The existence and uniqueness of the additive inverse is yet another property inherited from the real numbers. Given a vector, suppose we wish to find a vector so that. By the definition of vector addition, we have Hence, and, from which and, with the result that. Hence, has an additive inverse, and moreover it is unique and can be obtained by the formula. Geometrically, the vectors and have the same length but opposite directions. As a result, when adding the vectors geometrically, the sum results in starting at the initial point of and ending back at the initial point of. Or, in other words, the net result of moving then is not moving at all.7 7 An interesting property of a vector and its inverse is that the two vectors are ‘parallel’. In fact, we say two nonzero vectors are parallel when they have the same or opposite directions. That is, v is parallel to w if v=kw for some real, non-zero, number k. 1212112211221212,,,,,,vvwwvwvwwvwvwwvv definition of vector addition commutative property of real number addition definition of vector additionvwwvvwwv12,vv |
v12,wwwvw01122,0,0vwvwvw0110vw220vw11wv22wv12,vvwv12,vv-v12,vvv12,vv-vv-vvvv-v v v w w vw wv v -v 398 Using the additive inverse of a vector, we can define vector subtraction, or the difference of two vectors, as. If and then In other words, like vector addition, vector subtraction works component-wise. Below, we observe a geometrical interpretation of vector addition and vector subtraction. To interpret the vector geometrically, we note This means that the net result of moving along, then moving along, is just itself. From the diagram, we see that may be interpreted as the vector whose initial point is the terminal point of and whose terminal point is the terminal point of. It is also worth mentioning that in the parallelogram determined by the vectors and, the vector is one of the diagonals, the other being. Scalar Multiplication Next, we discuss scalar multiplication, which is taking a real number times a vector. We define scalar multiplication for vectors in the same way we defined it for matrices. vwv-w12,vvv12,www121211221122,,,,vvwwvwvwvwvwvwv-wvw� |
�� definition of vector subtraction commutativity of vector addition associativity of vector addition definition of additive inverse wvwwv-ww-wvw-wv0vv definition of additive identitywvwvvwwvvwvwvw v vw v-w -w w v vw v v w w vw 399 Definition. If k is a real number and, we define by Scalar multiplication of vectors by k can be understood geometrically as scaling the vector (if ) or scaling the vector and reversing its direction (if ) as demonstrated to the right. Note that, by definition, This and other properties of scalar multiplication are summarized below. Theorem 9.2. Properties of Scalar Multiplication. Associative Property: For every vector and scalars k and r,. Identity Property: For all vectors,. Additive Inverse Property: For all vectors,. Distributive Property of Scalar Multiplication over Scalar Addition: For every vector and scalars k and r,. Distributive Property of Scalar Multiplication over Vector Addition: For all vectors. and scalars k, and Zero Product Property: If is a vector and k is a scalar, then if and only if or. The proof of Theorem 9.2, like the proof of Theorem 9.1, ultimately boils down to the definition of scalar multiplication and properties of real numbers. For example, to prove the associative property, we let. If k and r are scalars, then 12,vvvkv1212,,kkvvkvkvv0k0k12121211,1,1,vvvvvv |
v-vvkrkrvvv1vvv1-vvvkrkrvvvvwkkkvwvwvkv00kv012,vvv 2v 2v v 12v 400 The remaining properties are proved similarly and are left as exercises. Our next example demonstrates how Theorem 9.2 allows us to do the same kind of algebraic manipulations with vectors as we do with variables, multiplication and division of vectors notwithstanding. Example 9.1.4. Solve for. Solution. Vectors in Standard Position A vector whose initial point is is said to be in standard position. If is plotted in standard position, then its terminal point is necessarily. (Once more, think about this before reading on.) 1212121212,,(),(),,krkrvvkrvkrvkrvkrvkrvrvkrvv definition of scalar multiplication associative property of real number multiplication definition of scalar multiplication devkrfinition of scalar multiplicationv521,2vv0v521,2521,25221,2321,232,4 distributive property over vector addition distributive property over scalar addition devv0vv0vv0v0v0++++32,42,42,430, |
02,432,41132,433241,33finition of scalar multiplication definition of vector addition property of additive identity associative property, scalar multv0v0vvv++24,33iplication property of multiplicative identityv0,012,vvv12,vv 401 in standard position Plotting a vector in standard position enables us to more easily quantify the concepts of magnitude and direction of the vector. We can convert the point in rectangular coordinates to a pair in polar coordinates where. The magnitude of, which we said earlier was the length of the directed line segment, is denoted. We can see from the right triangle corresponding to the vector that, for, and we use the Pythagorean Theorem to find. Using right triangle trigonometry, we also find that From the definition of scalar multiplication and vector equality, we get This motivates the following definition. Definition. Suppose is a vector with component form. Let be a polar representation of the point with rectangular coordinates with The magnitude of If, denoted, is given by, the direction angle of is given by. Taken together, we get... 12,vvv12,vv,r0rvv12,vvvrv0r2212rvv12coscossinsinvrvrvv12,cos,sincos,sinvvvvvvv12,vvv,r12,vv0rvv |
2212rvvvv0vcos,sinvvvxy 12,vv 12,vvv 0,0xy 1v 2v ,r v A few remarks are in order. 402 We note that if then even though there are infinitely many angles θ which satisfy the preceding definition, the stipulation means that all of the angles are coterminal. Hence, if and both satisfy the conditions of the definition, then and, and as such,, making well-defined. If, then, and we know from Section 8.1 that is a polar representation for the origin for any angle θ. For this reason, the direction of is undefined. The following theorem summarizes the important facts about the magnitude and direction of a vector. Theorem 9.3. Properties of Magnitude and Direction: Suppose is a vector. Then, and if and only if. For all scalars k,. If then The proof of the first property in Theorem 9.3 is a direct consequence of the definition of. If, then which is by definition greater than or equal to 0. Moreover, if and only if, if and only if. Hence, if and only if, as required. The second property is a result of the definition of magnitude and scalar multiplication along with a property of radicals. If and k is a scalar then v00r'coscos'sinsin'cos,sincos',sin'vv00,0v=0,0v0v0vv0kkvvv |
0cos,sinvvv12,vvv2212vvv22120vv22120vv120vv0v0,0v012,vvv21212221222221222212222122212,,kkkkvvkvkvkvkvkvkvkvvkvvkvv definition of scalar multiplication definition of magnitude product rule for radicals since =vk definition of magnitude v 403 The equation vectors, in Theorem 9.3 is a consequence of the component definition for, and was worked out prior to that definition. In words, the equation says that any given vector is the product of its magnitude and direction, an important concept to keep in mind when studying and using vectors. Example 9.1.5. Find the component form of the vector with so that when is plotted in standard position, it lies in Quadrant II and makes a 60° angle8 with the negative x-axis. Solution. We are told that and are given information about its direction, so we can use the formula to get the component form of. To determine, we are told that lies in Quadrant II and makes a 60° angle with the negative x-axis, so a polar form of the terminal point of, when plotted in standard position, is. (See the diagram below.) Thus, Example 9.1.6. For, find and θ, so that. 8 Due to the utility of vectors in real-world applications, we will usually use degree measure for the angle when giving the vector’s direction. cos,sinvvcos,sinvvvcos,sinvv |
v5vv5vcos,sinvvvvv5,120cos120,sin120135,22553,22vv3,33vv02cos,sinvvxy 60 120 v 404 Solution. For, we get. We can find the θ we’re after by converting the point with rectangular coordinates to polar form, where. From Section 8.1, we have Since is a point in Quadrant IV, θ is a Quadrant IV angle. Hence, we pick. We may check our answer by verifying that. Example 9.1.7. For the vectors and, find the following: 1. Solution. 2. 1. For, we have. The magnitude of is. Hence,. 2. In the expression, notice that the arithmetic on the vectors comes first, then the magnitude. Hence, our first step is to find the component form of the vector. Then 3,33v223336v3,33,r0rvtan3333yx3,3353553,336cos,sin33v3,4v1,2w2v |
w2vw3,4v22345v1,2w22125w2522vw2vw2vw23,421,23,42,41,8vw2221,81865vwxy v 3,33 405 9.1 Exercises In Exercises 1 – 3, sketch, and. 1. 2. 3. In Exercises 4 – 6, sketch, and. 5. 4. 6. In Exercises 7 – 8, use the given pair of vectors to compute, and. 7., 8. In Exercises 9 – 16, use the given pair of vectors to find the following quantities. State whether the result is a vector or a scalar. • • • • • • Finally, verify that the vectors satisfy the Parallelogram Law:. 9. 11.,, 10. 12.,, v3v12v2,1v1,4v3,2vuv+uv2uuvuv23uv2,3u1,5v3,4u2,1vvw2wvvwvwvwwvwv222212vw |
vwvw12,5v3,4w7,24v5,12w2,1v2,4w10,4v2,5w u v u v u v 406 13., 14., 15., 16., 17. Given a vector with initial point and terminal point, find an equivalent vector whose initial point is. Write the vector in component form 18. Given a vector with initial point and terminal point initial point is. Write the vector in component form 19. Given a vector with initial point and terminal point initial point is. Write the vector in component form..., find an equivalent vector whose, find an equivalent vector whose In Exercises 20 – 34, find the component form of the vector using the information given about its magnitude and direction. Give exact values. 20. ; when drawn in standard position lies in Quadrant I and makes a 60° angle with the positive x-axis. 21. ; when drawn in standard position lies in Quadrant I and makes a 45° angle with the positive x-axis. 22. ; when drawn in standard position lies in Quadrant I and makes a 60° angle with the positive x-axis. 23. ; when drawn in standard position lies along the positive y-axis. 24. ; when drawn in standard position lies in Quadrant II and makes a 30° angle with the negative x-axis. 25. ; when drawn in standard position lies in Quadrant II and makes a 30° angle with the positive y-axis. 26. ; when drawn in standard position lies along the negative x-axis. 3,1v23,2w34,55v43,55w22,22v22,22w13,22v1,3w5,21,30,0,ab4,23,3 |
0,0,ab7,11,70,0,abv6vv3vv23vv12vv4vv23vv72vv 407 27. ; when drawn in standard position lies in Quadrant III and makes a 45° angle with the 28. 29. negative x-axis. ; when drawn in standard position lies along the negative y-axis. ; when drawn in standard position lies in Quadrant IV and makes a 30° angle with the positive x-axis. 30. ; when drawn in standard position lies in Quadrant IV and makes a 45° angle with the negative y-axis. 31. ; when drawn in standard position lies in Quadrant I and makes an angle measuring with the positive x-axis. 32. ; when drawn in standard position lies in Quadrant II and makes an angle measuring with the negative x-axis. 33. ; when drawn in standard position lies in Quadrant III and makes an angle measuring with the negative x-axis. 34. ; when drawn in standard position lies in Quadrant IV and makes an angle measuring with the positive x-axis. In Exercises 35 – 40, approximate the component form of the vector using the information given about its magnitude and direction. Round your approximations to two decimal places. 35. 36. 37. 38. 39. ; when drawn in standard position makes a 117° angle with the positive x-axis. ; when drawn in standard position makes a 78.3° angle with the positive x-axis. ; when drawn in standard position makes a 12° angle with the positive x-axis. ; when drawn in standard position makes a 210.75° angle with the positive x-axis. ; when drawn in standard position makes a 252° angle with the positive x-axis. 56vv6.25vv43vv52vv25vvarctan210vvarctan35vv4arctan326vv5arctan12� |
��v392vv63.92vv5280vv450vv168.7vv 408 40. ; when drawn in standard position makes a 304.5° angle with the positive x-axis. In Exercises 41 – 58, for the given vector, find the magnitude and an angle θ with so that (See definition of magnitude and direction.) Round approximations to two decimal places. 41. 44. 47. 50. 53. 56. 42. 45. 48. 51. 54. 57. 43. 46. 49. 52. 55. 58. 59. A small boat leaves the dock at Camp DuNuthin and heads across the Nessie River at 17 miles per hour (that is, with respect to the water) at a bearing of S68°W. The river is flowing due east at 8 miles per hour. What is the boat’s true speed and heading? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. 60. The HMS Sasquatch leaves port with bearing S20°E maintaining a speed of 42 miles per hour (that is, with respect to the water). If the ocean current is 5 miles per hour with a bearing of N60°E, find the HMS Sasquatch’s true speed and bearing. Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. 61. The goal of this exercise is to use vectors to describe non-vertical lines in the plane. To that end, consider the line. Let and let. Let t be any real number. Show that the vector defined by, when drawn in standard position, has its terminal point on the line. (Hint: Show that for any real number t.) Now consider the non-vertical line. Repeat the previous analysis with and let. Thus, any non-vertical line can be thought of as a collection of terminal points of the 26vvvv0360cos,sinvv1,3v5,5v23,2v2,2 |
v22,22v13,22v6,0v2.5,0v0,7v3,4v12,5v4,3v7,24v2,1v2,6v123.4,77.05v965.15,831.6v114.1,42.3v24yx0,40v=1,2st0vvs24yx,24ttt0vs=ymxb0,b0v1,ms 409 vector sum of (the position vector of the y-intercept) and a scalar multiple of the slope vector. 62. Prove the associative and identity properties of vector addition in Theorem 9.1. 63. Prove the properties of scalar multiplication in Theorem 9.2. 0,b1,ms 410 9.2 The Unit Vector and Vector Applications Learning Objectives In this section you will: Use vectors in component form to solve applications. Find the unit vector in a given direction. Perform operations on vectors in terms of i and j. Use vectors to model forces. Using Vectors in Component Form to Solve Applications We continue our discussion of component forms of vectors from Section 9.1 and resume the process of resolving vectors into their components. This next example revisits Example 9.1.2, making use of component forms and vector algebra to solve this problem. Example 9.2.1. A plane leaves an airport with an airspeed of 175 miles per hour at a bearing of N40°E. A 35 mile per hour wind is blowing at a bearing of S60°E. Find the true speed of the plane, rounded to the nearest mile per hour, and the true bearing of the plane, rounded to the nearest degree. Solution |
. We proceed as we did in Example 9.1.2 and let denote the plane’s velocity and denote the wind’s velocity, and set about determining. If we regard the airport as being at the origin, the positive y-axis as acting as due north and the positive x-axis acting as due east, we see that the vectors and are in standard position and their directions correspond to the angles and, respectively. Hence, the component forms: Since we have no convenient way to express the exact values of cosine and sine of 50°, we leave both vectors in terms of cosines and sines. Adding corresponding components, we find the resultant vector: To find the true speed of the plane, we compute the magnitude of the resultant vector: vwvwvw5030175cos50,sin50175cos50,175sin50v35cos30,sin3035cos30,35sin30w175cos5035cos30,175sin5035sin30vwxy(E)(N) 40 60 v w 50 30 411 Hence, the true speed of the plane is approximately 184 miles per hour. To find the true bearing, we need to find the angle which corresponds to the polar form,, of the point. Since both of these coordinates are positive1, we know is a Quadrant I angle, as depicted below. Furthermore, Using the arctangent function, we get. Since, for the purposes of bearing, we need the angle between and the positive y-axis, we take the complement of and find the true bearing of the plane to be approximately N51°E. The Unit Vector In addition to finding a vector’s components, it is also useful in solving problems to find a vector in the same direction as the given vector, but of magnitude 1. We call a vector with a magnitude of 1 a unit vector. |
1 Yes, a calculator approximation is the quickest way to see this, but you can also use good old-fashioned inequalities and the fact that. 22175cos5035cos30175sin5035sin30184vw,r0r,175cos5035cos30,175sin5035sin30xytan175sin5035sin30175cos5035cos30yx39vw455060xy(E)(N) v w vw 412 Definition. Let be a vector. If, we say that is a unit vector. Any nonzero vector divided by its magnitude is a unit vector. If is a nonzero vector, then is a ‘unit vector in the direction of ’. Noting that magnitude is always a scalar, and that dividing by a scalar is the same as multiplying by its reciprocal, a unit vector for any nonzero vector can be found through multiplication by. The process of multiplying a nonzero vector by the reciprocal of its magnitude is called ‘normalizing the vector’. We leave it as an exercise to show that is a unit vector for any nonzero vector. The terminal points of unit vectors, when plotted in standard position, lie on the Unit Circle. (You should take time to show this.) As a result, we visualize normalizing a nonzero vector as shrinking2 its terminal point, when plotted in standard position, back to the Unit Circle. Example 9.2.2. Find a unit vector in the same direction as. |
Solution. We begin by finding the magnitude. 2 …if the magnitude of v is greater than 1… v1vvvvvvv1v1vvvv5,12v2251216913vxy 1 1 1 1 v vv 413 Next, we divide by. We can check that is indeed a unit vector by verifying that its magnitude is 1. Try it! Note that since a unit vector has length 1, multiplying a unit vector by the magnitude of a vector results in the vector itself, provided the unit vector has the same direction as. (Try this with the unit vector we found in Example 9.2.2.) As a rule of thumb, for any nonzero vector, The Principal Unit Vectors Of all of the unit vectors, two deserve special mention. Definition. The Principal Unit Vectors: The vector The vector is defined by is defined by.. 5,12v13v115,1213512,1313vvvv512,1313vvvvvvmagnitude ofunit vector in the direction ofvvv × =i1,0ij0,1jxy11 1,0i 0,1jxy 512,1313vv 5,12v 414 We can think of the vector as representing the positive x-direction while represents the positive y-direction. We have the following ‘decomposition’ theorem.3 Theorem 9.4. Principal Vector Decomposition Theorem: Let be a vector with component form. Then. The proof of Theorem 9.4 is straightforward. Since and, we have from the definition of scalar multiplication and vector addition that Geometrically, the situation looks like this: In Section 9.1 |
, we found the component form of a vector with initial point and terminal point to be. It follows from Theorem 9.4 that may also be written in terms of and as. Example 9.2.3. Given a vector with initial point and terminal point, write the vector in terms of and. 3 We will see a generalization of Theorem 9.4 in Section 9.3. Stay tuned! ijv12,vvv12vvvij1,0i0,1j121212121,00,1,00,,vvvvvvvv definition of i and j scalar multiplication vector additionijv1212,vvvvvijPQ00,Pxy11,Qxy1010,PQxxyyPQij1010PQxxyyijv2,6P6,6Qijxy i j 1vi 2vj 12,vvv Solution. 415 Performing Operations on Vectors in Terms of i and j When vectors are written in terms of and, we carry out addition, subtraction and scalar multiplication by performing operations on corresponding components. Operations on Vectors Written in terms of and : Given and, then for any scalar k These results can be verified using definitions of addition, subtraction and scalar multiplication from Section 9.1 along with Theorem 9.4. Their verification is left to the student. Example 9.2.4. Use vectors and to find. Solution. Using Vectors to Model Forces We conclude this section with a classic example which demonstrates how vectors are used to model forces. A force is defined as a ‘push’ or a ‘pull’. The intensity of the push or pull is the magnitude of the force, and is measured in Newtons (N) in the SI system or pounds (lbs.) in the English system. The following example should be studied in great detail |
. 6266812vijijijij12vvvij12wwwij1122vwvwvwij1122vwvwvwij12kkvkvvij42vij3wij3vw33423342312631236195 scalar multiplication vector additionvwijijijijijijijij 416 Example 9.2.5. A 50 pound speaker is suspended from the ceiling by two support braces. If one of them makes a 60° angle with the ceiling and the other makes a 30° angle with the ceiling, what are the tensions on each of the supports? Solution. We first represent the problem schematically. We have three forces acting on the speaker: the weight of the speaker, which we’ll call, pulling the speaker directly downward, and the forces on the support rods, which we’ll call and (for ‘tensions’) acting upward at angles 60° and 30°, respectively. We provide the corresponding vector diagram below. Note that we have used alternate |
interior angles to determine the added angle measures in the above diagram. We are looking for the tensions on the supports, which are the magnitudes and. In order for the speaker to remain stationary4, we require. Viewing the common initial point of these vectors as the origin and the dashed line as the x-axis, we find component representations for the three vectors involved. 4 This is the criteria for ‘static equilibrium’. w1T2T1T2T12wTT050 lbs. 60 30 60 30 w 1T 2T 30 60 417 We can model the weight of the speaker as a vector pointing directly downward with a magnitude of 50 pounds. That is,. Since the vector is directed strictly downward, is a unit vector in the direction of. Hence, For the force in the first support, applying Theorem 9.3, we get For the second support, we note that the angle 30° is measured from the negative x-axis, so the angle needed to write in component form is 150°. Hence, The requirement gives us 50ww0,1jw500,10,50w11111cos60,sin6013,223,22TTTTT2T22222cos150,sin15031,223,22TTTTT12wTT0xy w 1T 2T 30 60 418 Equating the corresponding components of the vectors on each side, we get a system of linear equations in the variables and. From (E1) we get. Substituting into (E2) gives Hence, pounds and pounds. 11221212330,50,,0,0222233,500,02222TTTTTTTT1T2T12123E1 0223E2 50022� |
�TTTT123TT222223350022350225TTTTT225T123253TT 419 9.2 Exercises 1. Given initial point and terminal point, write the vector in terms of and. 2. Given initial point and terminal point, write the vector in terms of and. In Exercises 3 – 4, use the vectors, and to find the following. 3. 4. 5. Let. Find a vector that is half the length and points in the same direction as. 6. Let. Find a vector that is twice the length and points in the opposite direction as. In Exercises 7 – 10, find a unit vector in the same direction as the given vector. 7. 8. 9. 10. In Exercises 11 – 12, use the given pair of vectors to find the following quantities. State whether the result is a vector or scalar. • • • • • • 11., 12., In Exercises 13 – 15, for the given vector, find the magnitude and an angle θ with so that 13.. Round approximations to two decimal places. 14. 15. 16. Let be any non-zero vector. Show that has length 1. 17. A woman leaves her home and walks 3 miles west, then 2 miles southwest. How far from home is she, and in what direction must she walk to head directly home? 18. If the captain of the HMS Sasquatch wishes to reach Chupacabra Cove, an island 100 miles away at a bearing of S20°E from port, in three hours, what speed and heading should she set to take into 13,1P25,2Pvij16,0P21,3Pvij+5uij23vij4wijuvw42vu43� |
��vijv52vijv=34aij25bij10cij1532dijvw2wvvwvwvwwvwv34vij2wj12vij12wijvv0360cossinvvij10vjvij4vij12,vvv1vv 420 account an ocean current of 5 miles per hour? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. HINT: If denotes the velocity of the HMS Sasquatch and denotes the velocity of the current, what does need to be to reach Chupacabra Cove in three hours? 19. In calm air, a plane flying from the Pedimaxus International Airport can reach Cliffs of Insanity Point in two hours by following a bearing of N8.2°E at 96 miles an hour. (The distance between the airport and the cliffs is 192 miles.) If the wind is blowing from the southeast at 25 miles per hour, what speed and bearing should the pilot take so that she makes the trip in two hours along the original heading? Round the speed to the nearest hundredth of a mile per hour and your angle to the nearest tenth of a degree. 20. The SS Bigfoot leaves Yeti Bay on a course of N37°W at a speed of 50 miles per hour. After traveling half an hour, the captain determines he is 30 miles from the bay and his bearing back to the bay is S40°E. What is the speed and bearing of the ocean current? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. 21. A 600 pound Sasquatch statue is suspended by two cables |
from a gymnasium ceiling. If each cable makes a 60° angle with the ceiling, find the tension on each cable. Round your answer to the nearest pound. 22. Two cables are to support an object hanging from a ceiling. If the cables are each to make a 42° angle with the ceiling, and each cable is rated to withstand a maximum tension of 100 pounds, what is the heaviest object that can be supported? Round your answer down to the nearest pound. 23. A 300 pound metal star is hanging on two cables which are attached to the ceiling. The left hand cable makes a 72° angle with the ceiling while the right hand cable makes an 18° angle with the ceiling. What is the tension on each of the cables? Round your answers to three decimal places. 24. Two drunken college students have filled an empty beer keg with rocks and tied ropes to it in order to drag it down the street in the middle of the night. The stronger of the two students pulls with a force of 100 pounds at a heading of N77°E and the other pulls at a heading of S68°E. What force should the weaker student apply to his rope so that the keg of rocks heads due east? What resultant force is applied to the keg? Round your answer to the nearest pound. vwvw 421 25. Emboldened by the success of their late night keg pull in Exercise 24 above, our intrepid young scholars have decided to pay homage to the chariot race scene from the movie ‘Ben-Hur’ by tying three ropes to a couch, loading the couch with all but one of their friends and pulling it due west down the street. The first rope points N80°W, the second points due west and the third points S80°W. The force applied to the first rope is 100 pounds, the force applied to the second rope is 40 pounds and the force applied (by the non-riding friend) to the third rope is 160 pounds. They need the resultant force to be at least 300 pounds; otherwise the couch won’t move. Does it move? If so, is it heading due west? 422 9.3 The Dot Product Learning Objectives In this section you will: Find the dot product of two vectors. Learn properties of the dot product. Determine the angle between two vectors. Determine whether or not two vectors are orthogonal. |
Solve applications using the dot product. Thus far in Chapter 9, we have learned how to add and subtract vectors and how to multiply vectors by scalars. In this section, we define a product of vectors. Definition and Algebraic Properties of the Dot Product We begin with the following definition. Definition. Suppose The dot product of and and are vectors whose component forms are is given by and. Example 9.3.1. Find the dot product of and. Solution. We have Note that the dot product takes two vectors and produces a scalar. For that reason, the quantity is often called the scalar product of and. The dot product enjoys the following properties. vw12,vvv12,wwwvw12121122,,vvwwvwvwvw3,4v1,2w3,41,231425vwvwvw 423 Theorem 9.5. Properties of the Dot Product: Commutative Property: For all vectors and,. Distributive Property: For all vectors, and,. Scalar Property: For all vectors and, and scalars k,. Relation to Magnitude: For all vectors,. Like most of the theorems involving vectors, the proof of Theorem 9.5 amounts to using the definition of the dot product and properties of real number arithmetic. To show the commutative property, for instance, we let and. Then The distributive property is proved similarly and is left as an exercise. For the scalar property, assume that, and k is a scalar. Then We leave the proof of as an exercise. For the last property, we note that if, then The following example puts Theorem 9.5 to good use. vwvwwvuvwuvwuvuwvw� |
�kkkvwvwvwv2vvv12,vvv12,www1212112211221212,,,,vvwwvwvwwvwvwwvv definition of dot product commutativity of real number multiplication definition of dot productvwwv12,vvv12,www12121212112211221122,,,,kkvvwwkvkvwwkvwkvwkvwkvwkvwvw definition of scalar multiplication definition of dot product associativity of real number multiplication vw1212,,kvvwwk distributive law for real numbers definition of dot productvwkkvwvw12,vvv121222122,,vvvvvv definition of magnitudevvv 424 Example 9.3.2. Prove the identity:. Solution. We begin by using Theorem 9.5 to rewrite in terms of the dot product. Hence, as required. If we take a step back from the pedantry in Example 9.3.2, we see that the bulk of the work is needed to show that. If this looks familiar, it should. Since the dot product enjoys many of the same properties as real numbers, the machinations required to expand for vectors and match those required to expand for real numbers v and w, and hence we get similar looking results. The identity verified in Example |
9.3.2 plays a large role in the development of the geometric properties of the dot product, which we now explore. Geometric Properties of the Dot Product Suppose and are two nonzero vectors. If we draw and with the same initial point, we define the angle between and to be the angle determined by the rays containing the vectors and, as illustrated below. We choose to define. The following theorem gives us some insight into the geometric role the dot product plays. 2222vwvvww2vw2 relation to magnitude property distributive propertyvwvwvwv-wvv-w-wvv-w-w11112 commutative property distributive property scalar & commutative propertiesv-wvvv-w-wv-w-wvvvwvwwwvvvw222 relation to |
magnitude propertywwvvww2222vwvvww2vwvwvvvwwwvwvwvwvwvwvwvwvwvw000 v w v w v w 425 Theorem 9.6. Geometric Interpretation of the Dot Product: If and are nonzero vectors then, where is the angle between and. We prove Theorem 9.6 in cases. Case 1: If, then and have the same direction. Thus, the unit vector in the direction of is also the unit vector in the direction of, and we have Thus, we have We note that, from which we get. It follows, from Theorem 9.3, that. Hence, We get This proves the formula holds for. vwcosvwvwvw0vwvwkk unit vectors in directions of and are the same scalar multiplication associative property of scalar multiplication letting wwvwvvwwvwwvvvwvwv2kkkk scalar property of dot productrelation to magnitude propertyvwvvvvv vv0kkkkkkvvvkkk |
scalar property of dot productvvvvvvvwcos01cos0k since vwvvvwvw0 426 Case 2: If, we repeat the argument with the difference being that, so that. Thus, where. It follows that,, resulting in We have Thus, the formula holds for. Case 3: Next, if, the vectors, and determine a triangle with side lengths, and, respectively, as seen below. The Law of Cosines yields. From Example 9.3.2, we know. Equating these two expressions for gives Thus,, as required. Determining the Angle Between Two Vectors An immediate consequence of Theorem 9.6 is the following. wvwvkwvkwv0kkkkkkvvvwcos1cos since vwvwvw0vwvwvwvw2222cosvwvwvw2222vwvvww2vw22222cos22cos2cosvwvwvvwwvwvwvwvwcosvwvw v w � |
�vw v w vw 427 Theorem 9.7. Let and be nonzero vectors and let be the angle between and. Then We obtain the formula in Theorem 9.7 by solving the equation given in Theorem 9.6 for. Since and are nonzero, so are and. Hence, we may divide both sides of by to get Since by definition, the values of exactly match the range of the arccosine function. Hence, We are long overdue for an example. Example 9.3.3. Find the angle between the following pairs of vectors. 1. 2. 3. and and and Solution. We use the formula from Theorem 9.7 in each case below. 1. For and, we have Then vwvwarccosvwvwvwvwcosvwvwvwcosvwvw0arccosvwvw3,33v3,1w2,2v5,5w3,4v2,1warccosvwvw3,33v3,1w3,333,1333363vw22333366v223142� |
��w 428 2. We have and, so that It follows that 3. We find, for and, Then arccos63arccos123arccos256vwvw2,2v5,5w2,25,510100vwarccos0arccosarccos0200 and and v0w0vvwwvwvw3,4v2,1w3,42,1642vw2234255v22215w2arccosarccos5525arccos25vwvw 429 Since isn’t the cosine of one of the common angles, we leave our answer as. Orthogonal Vectors The vectors and are called orthogonal, and |
we write, because the angle between them is radians, or. Geometrically, when orthogonal vectors are sketched with the same initial point, the lines containing the vectors are perpendicular. In the illustration to the right, and are orthogonal, and we write. We state the relationship between orthogonal vectors and their dot product in the following theorem. Theorem 9.8. The Dot Product Detects Orthogonality: Let and be nonzero vectors. Then if and only if. To prove Theorem 9.8, we first assume and are nonzero vectors with. By definition, the angle between and is. By Theorem 9.6,. Conversely, if and are nonzero vectors and, then Theorem 9.7 gives While Theorem 9.8 certainly gives us some insight into what the dot product means geometrically, there is more to the story of the dot product. 252525arccos252,2v5,5wvw290vwvwvwvw0vwvwvwvw2cos02vwvwvw0vwarccos0arccosarccos02 thus verifying vwvwwvvw v w 430 Orthogonal Projection Consider the two nonzero vectors and drawn with a common initial point below. For the moment, assume that the angle between and, which we’ll denote, is acute. We wish to develop a formula for the vector, indicated below, which is called the orthogonal projection of onto. The vector is obtained geometrically as |
follows: drop a perpendicular from the terminal point T of to the vector and call the point of intersection R. The vector is then defined as. Like any vector, is determined by its magnitude and its direction. To determine the magnitude, we observe that We determine the direction of by finding the unit vector in the direction of, which is the same as the unit vector in the direction of ; that is,. It follows that vwOvwpvwpvwpORpppcssoscoco from Theorem 9.6pvpvvwpvvwvwpvwvw,wppwww v w OTRO p vTRO v w ORp 431 Thus, we have a formula for the orthogonal projection of onto for an acute angle. Suppose next that the angle between and is obtuse, and consider the diagram below. In this case we see that, from which Thus, from, we have The unit vector in the direction of is the unit vector in the direction of, which is. We note that reversing the direction of does not affect the magnitude. Finally, 2 magnitude of times unit vector in direction of vwwpwwvwwwpvwpwwwvwvw'� |
�cos'coscoscossinsin1cos0sincos difference identity for cosinecos'pvcos'coscoscos'cos since from Theorem 9.6: vwvwpvvvwvvwvwwpwwwwwwTRO w v ' ORp 432 This formula for orthogonal projection when is obtuse matches the formula for an acute angle. If the angle between and is, then 1. Since in this case,. It follows that Finally, we have the following theorem. Theorem 9.9. If and are nonzero vectors, then the orthogonal projection of onto, denoted, is given by It is time for an example. Example 9.3.4. Let and. Find, and plot, and in standard position. Solution. We find 1 In this case, the point R coincides with the point O, so that. 2� |
�� magnitude times directionvwwpwwvwwwwvwwwvw2p0vw0vw2200p0wwwvwwwvwvwprojwv2projwvwvww1,8v1,2wprojwpvvwpOROOp0 433 Hence,. We plot, and below. Suppose we wanted to verify that our answer in Example 9.3.4 is indeed the orthogonal projection of onto. We first note that, since, is a scalar multiple of and so it has the correct direction. It remains to check the orthogonality condition. Consider the vector whose initial point is the terminal point of and whose terminal point is the terminal point of. 222222proj1,81,21,21,211821,2121161,2531,2� |
�wvwvwwproj3,6wpvvwppvw3pwpwqpvxy v w pxy v w p q 434 From the definition of vector arithmetic,, so that. In the case of Example 9.3.4, and, so. Then This shows as required. Work We close this section with an application of the dot product. In Physics, if a constant force moves an object a distance d, then the work, W, done by the force is given by the magnitude of the force times the amount of displacement, or, where the force is being applied in the direction of the motion. If the force applied is not in the direction of the motion, we can use the dot product to find the work done. Consider the scenario below where the constant force is applied to move an object from the point P to the point Q. To determine the work W done in this scenario, we find that the magnitude of the force in the direction of is, where is the angle between and. The distance the object travels is. Since work is the magnitude of the force in the direction of times the distance traveled from P to Q, we get We have proved the following. pqvqvp1,8v3,6p1,83,64,2q4,21,2440 |
qvqwFWdFFFPQcosFFPQPQFPQcoscosWPQPQPQ from Theorem 9.6FFFPQ F F 435 Theorem 9.10. Work as a Dot Product: Suppose a constant force is applied to move an object along the vector, from P to Q. The work W done by is given by where is the angle between and., Example 9.3.5. Taylor exerts a force of 10 pounds to pull her wagon a distance of 50 feet over level ground. If the handle of the wagon makes a 30° angle with the horizontal, how much work did Taylor do pulling the wagon? Assume Taylor exerts the force of 10 pounds at a 30° angle for the duration of the 50 feet. Solution. There are two ways to attack this problem. One way is to find the vectors and mentioned in Theorem 9.10 and compute. To do this, we assume the origin is at the point where the handle of the wagon meets the wagon and the positive x-axis lies along the dashed line in the figure above. Since the force applied is a constant 10 pounds, we have. Since it is being applied at a constant angle of with respect to the positive x-axis, Theorem 9.3 gives us Since the wagon is being pulled along 50 feet in the positive direction, the displacement vector is We get FPQFcosWPQPQFFFPQFPQWPQF10F3010cos30,sin303110,2253,5F50501,050,0PQi 436 Since force is measured in pounds and distance is measured in feet, we get foot- pounds. Alternately, we can use the formulation to get 53,550,02503WPQ |
F2503WcosWPQF10 pounds50 feetcos303500 foot-pounds22503 foot-pounds of workW 437 9.3 Exercises 1. Given and, calculate. 2. Given and, calculate. 3. Given and, calculate 4. Given and, calculate.. In Exercises 5 – 24, use the give pair of vectors, and, to find the following quantities. • • the angle θ (in degrees) between and 5. 7. 9. 11. 13. 15. 17. 19. and and and and and and and and 21. and 23. and • • 6. 8. 10. 12. 14. 16. 18. 20. 22. 24. (Show that.) and and and and and and and and and and 34uij23vijuvuij5vijuv2,4u3,1vuv1,6u6,1vuvvwvwprojwvvwprojwqvv0qw2,7v5,9w6,5v10,12w1,3v1,3w3,4v6,8w2,1v3,6w33,3v3,1w1,17 |
v1,0w3,4v5,12w4,2v1,5w5,6v4,7w8,3v2,6w34,91v0,1w3vij4wj247vij2wi3322vijwij512vij34wij13,22v22,22w22,22v13,22w31,22v22,22w13,22v22,22w 438 25. A force of 1500 pounds is required to tow a trailer. Find the work done towing the trailer 300 feet along a flat stretch of road Assume the force is applied in the direction of the motion. 26. Find the work done lifting a 10 pound book 3 feet straight up into the air. Assume the force of gravity is acting straight downwards. 27. Suppose Taylor fills her wagon with rocks and must exert a force of 13 pounds to pull her wagon across the yard. If she maintains a 15° angle between the handle of the wagon and the horizontal, compute how much work Taylor does pulling her wagon 25 feet. Round your answer to two decimal places. 28. In Exercise 24 in Section 9.2, two drunken college students have filled an empty beer keg with rocks which they drag down the street by pulling on two attached ropes. The stronger of the two students pulls with a force of 100 pounds on a rope which makes a 13° angle with the direction of motion. (In this case, the keg was being pulled due east and the student’s heading was N77°E.) Find the work done by this student if the keg is dragged 42 feet. 29. Find the work done pushing a 200 pound barrel 10 feet up a 12 |
.5° incline. Ignore all forces acting on the barrel except gravity, which acts downwards. Round your answer to two decimal places. HINT: Since you are working to overcome gravity only, the force being applied acts directly upwards. This means that the angle between the applied force in this case and the motion of the object is not the 12.5° of the incline! 30. Prove the distributive property of the dot product in Theorem 9.5. 31. Finish the proof of the scalar property of the dot product in Theorem 9.5. 32. Use the identity in Example 9.3.2,, to prove the Parallelogram Law: 33. We know that for all real numbers x and y by the Triangle Inequality from College Algebra. We can now establish a Triangle Inequality for vectors. In this exercise, we prove that for all pairs of vectors and. a) (Step 1) Show that. 2222vwvvww222212vwvwvwxyxyuvuvuv2222uvuuvv 439 b) (Step 2) Show that. This is the celebrated Cauchy-Schwarz Inequality. (Hint: To show this inequality, start with the fact that and use the fact that for all θ.) c) (Step 3) Show that d) (Step 4) Use Step 3 to show that for all pairs of vectors and. e) As an added bonus, we can now show that the Triangle Inequality holds for all complex numbers z and w as well. Identify the complex number with the vector and identify the complex number with the vector and just follow your nose! uvuv cos uvuvcos122222222222� |
�uvuuvvuuvvuuvvuvuvuvuvzwzwzabi,abuwcdi,cdv 440 9.4 Sketching Curves Described by Parametric Equations Learning Objectives In this section you will: Graph plane curves described by parametric equations. Analyze behavior in the graphs of parametric equations. As we have seen, most recently in Section 8.3, there are scores of interesting curves which, when plotted in the xy-plane, neither represent y as a function of x nor x as a function of y. In this section, we present a new concept which allows us to use functions to study these kinds of curves. To motivate the idea, we imagine a bug crawling across a table top starting at the point O and tracing out a curve C in the plane, as shown below. The curve C does not represent y as a function of x because it fails the Vertical Line Test and it does not represent x as a function of y because it fails the Horizontal Line Test. However, since the bug can be in only one place at any given time t, we can define the x-coordinate of P as a function of t and the y-coordinate of P as a (usually but not necessarily) different function of t. Curves Described by Parametric Equations The functions describing the curve C, traditionally, use to represent x and to represent y. The independent variable t in this case is called a parameter and the system of equations is called a system of parametric equations or a parametrization of the curve C.1 1 Note the use of the indefinite article ‘a’. As we shall see, there are infinitely many different parametric representations for any given curve. ,Pxyftgtxftygt 441 The parametrization of C endows it with an orientation and the arrows on C indicate motion in the direction of increasing values of t. In this case, our bug starts at the point O |
, travels upwards to the left, then loops back around to cross its path2 at the point Q and finally heads off into the first quadrant. It is important to note that the curve itself is a set of points and as such is devoid of any orientation. The parametrization determines the orientation and as we shall see, different parametrizations can determine different orientations. If all of this seems hauntingly familiar, it should. By definition, the system of equations parametrizes the Unit Circle, giving it a counter-clockwise orientation. It is time for an example. Example 9.4.1. Sketch the curve described by for. Solution. We follow the same procedure here as we have time and time again when asked to graph anything new. We choose friendly values of t, plot the corresponding points and connect the results in a pleasing fashion. Since we are told, we start there and as we plot successive points, we draw an arrow to indicate the direction of the path for increasing values of t. The curve sketched out in Example 9.4.1 certainly looks like a parabola and the presence of the term in the equation reinforces this hunch. In Section 9.5, we will use the technique of substitution to eliminate the parameter t and get an equation involving just x and y. As we will see, the resulting Cartesian equation describes a parabola with vertex. 2 Here, the bug reaches the point Q at two different times. While this does not contradict our claim that f (t) and g(t) are functions of t, it shows that neither f nor g can be one-to-one. (Think about this before reading on,) cossinxtyt2321xtyt2t2t2t23xt2143yx3,1t23xtt21ytt� |
�,xtyt2151,51232,30313,11212,12131,33656,5 442 Graphing Parametric Equations Example 9.4.2. Sketch the curve described by the parametric equations for. Solution. To get a feel for the curve described by the system, we first sketch the graphs of each equation, and, over the interval.,, We note that as t takes on values on the interval, ranges between and 1, and ranges between 0 and 2. This means that all of the action is happening on a portion of the plane, namely. More generally, we see that is increasing over the entire interval whereas is decreasing over the interval and then increasing over. Next, we plot a few points to get a sense of the position and orientation of the curve. 322xtyt11t3xt22yt1,13xt11t22yt11t1,13xt122yt,|11, 02xyxy3xt1,122yt1,00,1ty 1,2 0,0 1,2t3xtt |
22ytt,xtyt1121,20000,01121,2txty 443 To trace out the path described by the parametric equations: We start at, where, then move to the right (since x is increasing) and down (since y is decreasing) to. We continue to move to the right (since x is still increasing) but now move upwards (since y is now increasing) until we reach, where. for Example 9.4.3. Sketch the curve described by the parametric equations for. Solution. We proceed as in the previous example and graph and over the interval.,, We find that the range of x in this case is and the range of y is. Since t is ranging over the unbounded interval, we take the time to analyze the behavior of both x and y. As, and as well. This means the graph of the resulting function approaches the point. Since both and are always decreasing for, we know that our final graph will start at, where, and move consistently to the left (since x is decreasing) and down (since y is decreasing) to approach the origin. 1,21t0,01,21t322xtyt11t22ttxeye0t2txe2tye0,2txe0t2tye0t� |
��0,20,10,t20txe20tye0,02txe2tye0t2,10txytxty 444 Next, we plug in some friendly values of t to get a sense of the orientation of the curve. Since t lies in the exponent here, friendly values of t involve natural logarithms. for Example 9.4.4. Sketch the curve described by the parametric equations for. Solution. We start by graphing and over the interval.,, We find that the range of x is while the range of y is. Before moving on, we take a closer look at these two graphs. Since and aren’t included in the domain for t, we analyze the behavior of the system as t approaches each of these values. We find that as, and when, we get and. Piecing this information together, we get that for t near 0, and for t near, we have points with very small positive x-values, but very large y-values. 22ttxeye0tsincscxtyt0tsinxtcscyt0,sinxt0tcscyt0t0,11,� |
�0tt0ttsin0xtcscytxyt2txte2tyte,xtytln10212,1ln211411,4ln3231921,39tytx 445 As t ranges through the interval, is increasing and is decreasing. This means that we are moving to the right and downwards. Once, the orientation reverses, and we start to head to the left, since is now decreasing, and up, since is now increasing. We plot a few points before sketching the curve. We combine the above information to first graph the system of equations on the interval, followed by the interval, and finally on the combined interval. for We see that the parametrization given above traces out this portion of the curve twice as t runs through the interval. 0,2sinxtcscyt2tsinxtcscyttsinxttcscytt,xtyt61sin62 |
csc261,222sin12csc121,15651sin625csc261,220,2,20,02t2tsincscxtyt0t0,xyxyxy 446 Example 9.4.5. Sketch the curve described by the parametric equations for. Solution. Proceeding as above, we set about graphing the system of parametric equations by first graphing and on the interval.,, We see that x ranges from to 4 and y ranges from to 2. The direction of the curve is as follows. As t ranges from 0 to, x is decreasing while y is increasing, resulting in a movement left and upwards. For, x is decreasing as is |
y, so the motion is still right to left but is now downwards. On the interval, x begins to increase while y continues to decrease. Hence, the motion becomes left to right but continues downwards. Plugging in the values,, and gives the following coordinates. 13cos2sinxtyt302t13cosxt2sinyt30,213cosxt302t2sinyt302t2222t3,20t232,xytytx 447 We put all of our information together to get the following graph. for If this graph looks suspiciously like that of an ellipse, we will find in the next section that this is indeed the case. The next section is devoted to converting between parametric and Cartesian equations. t13cosxtt2sinytt,xtyt013cos04� |
��2sin004,0213cos122sin221,213cos22sin02,032313cos1232sin221,213cos2sinxtyt302txy432112 1 1 2 448 9.4 Exercises In Exercises 1 – 6, graph each set of parametric equations by making a table of values. Include the orientation on the graph. 1. 3. 5. 2. 4. 6. 21xttytt21xttytttxy3210123txy321012232xttytt223xttytttxy2� |
�10123txy3210132xttytt23xttytttxy21012txy21012 449 In Exercises 7 – 30, plot the set of parametric equations by hand. Be sure to indicate the orientation imparted on the curve by the parametrization. 7. for 8. for 9. for 10. for 11. for 12. for 13. for 14. for 15. for 16. for 17. for 18. for 19. for 21. for 23. for 25. for 27. for 20. 22. 24. 26. 28. for for for for for 29. for 30. for 4362xtyt01t4134xtyt01t22xtyt12t23xtyt05t2132xtytt03t2211xttyt1t2118913xtyt3t3xtytt3xtyt� |
�t225xtyt05txtyt0tcossinxtyt22t3cos3sinxtyt0t2cos6sinxtyt0t13cos4sinxtyt02t3cos2sin1xtyt22t2cossecxtyt02t2tancotxtyt02tsectanxtyt22tsectanxtyt322t� |
��tan2secxtyt22ttan2secxtyt322tcosxtyt0tsinxtyt22t 450 In Exercises 31 – 34, plot the set of parametric equations with the help of a graphing utility. Be sure to indicate the orientation imparted on the curve by the parametrization. 31. for 33. for 32. 34. for for In Exercises 35 – 38, use a graphing utility to view the graph of each of the four sets of parametric equations. Although they look unusual and beautiful, they are so common they have names, as indicated in each exercise. 35. An epicycloid: 36. A hypocycloid: 37. A hypotrochoid: for for for 38. A rose: for 3234xttyt22t334cos4sinxtyt02tttttxeeyee22tcos3sin4xtyt02t14cos |
()cos1414sinsin14xttytt02t6sin2sin66cos2cos6xttytt02t2sin5cos65cos()2sin(6)xttytt02t5sin2sin5sin2cosxttytt02t 451 9.5 Finding Parametric Descriptions for Oriented Curves Learning Objectives In this section you will: Eliminate the parameter in a pair of parametric equations. Parametrize curves given in Cartesian coordinates. Reverse orientation and shift starting point of a curve described by parametric equations. Now that we have had some good practice sketching the graphs of parametric equations, we turn to the problem of eliminating the parameter in a pair of parametric equations to get an equation involving just x and y. Eliminating the Parameter in Parametric Equations Recall that several curves in the examples from Section 9.4 resembled graphs of functions we’ve seen before. We revisit these examples, eliminating the parameter to determine a Cartesian equation. Example 9.5.1. Eliminate the parameter t in the system of equations from Example 9.4.1 to determine a Cartesian equation. Recall in this example. Solution. We use the technique of substitution to eliminate the parameter in the system of equations. The first step is to solve for t. Substituting this result into the equation yields We see that the graph of this equation is a parabola with vertex which opens to the right. 2321xtyt2 |
t21yt211212ytytyt23xt222132134143yxyxyx3,1 452 Technically speaking, the equation describes the entire parabola, while the parametric equations describe only a portion of the parabola. In this case, we can remedy the situation by restricting the bounds on y. Since the portion of the parabola we want is exactly the part where, the equation coupled with the restriction describes the same curve as the given parametric equations. The one piece of information we can never recover after eliminating the parameter is the orientation of the curve. Example 9.5.2. Eliminate the parameter t in the system of equations from Example 9.4.3, where the restriction on t was. Solution. To eliminate the parameter, one way to proceed is to solve for t to get. Substituting this for t in gives The parametrized curve is the portion of the parabola which starts at the point and heads toward, but never reaches,. Example 9.5.3. Eliminate the parameter t in the system of equations from Example 9.4.5, where the restriction on t was. Solution. To eliminate the parameter here, we note that the trigonometric functions involved, namely and, are related by the Pythagorean identity. Hence, we solve 2143yx5y2143yx5y22ttxeye0t2txeln2xt� |
��2tye22ln22ln2ln22224xxxyeeexx24xy2,10,013cos2sinxtyt302tcostsint22cossin1ttxy 22 for 0ttxetye 453 for to get and we solve for to get. Substituting these expressions into gives This is the equation of an ellipse centered at with vertices at and, and with a minor axis of length 4. The parametric equations trace out three-quarters of this ellipse in a counter-clockwise direction. We next turn to the problem of finding parametric representations of curves. Parametrizing Curves We start with the following. To parametrize through I. To parametrize through I. Parametrizations of Common Curves as x runs through some interval I, let,, and let t run as y runs through the interval I, let,, and let t run To parametrize a directed line segment with initial point and terminal point, let and for. To parametrize where and, let and for. (This will impart a counter-clockwise orientation.) The reader is |
encouraged to verify the above formulas by eliminating the parameter and, when indicated, checking the orientation. We put these formulas to good use in the following examples. Example 9.5.4. Find a parametrization for the curve from to. 13cosxtcost1cos3xt2sinytsintsin2yt22cossin1tt222211321194xyxy1,02,04,0yfxxtyftxgyxgtyt00,xy11,xy010xxxxt010yyyyt01t22221xhykab0a0bcosxhatsinykbt02t2yx3x2xxy432112 1 1 2 13cos3 for 022sinxttyt� |
�� 454 Solution. Since is written in the form, we let and. Since, the bounds on t match precisely the bounds on x so we get Example 9.5.5. Find a parametrization for the curve. Solution. While we could attempt to solve this equation for y, we don’t need to. We can parametrize by setting so that. Since and there are no bounds placed on y, it follows that there are no bounds placed on t. Our final answer is Example 9.5.6. Find a parametrization for the line segment which starts at and ends at. Solution. We make use of the formulas and for. These formulas can be summarized as To find the equation for x, we have that the line segment starts at and ends at. This means that the displacement in the x-direction is. Hence, the equation for x is, or. For y, we note that the line segment starts at and ends at. Thus, the displacement in the y-direction is, so we get. Our final answer is for 2yxyfxxt2yfttxt2 for 32xttyt521xyy521xfyyyyt521xttyt521 for xtttyt2,31,5010xxxxt010yyyyt01tstarting point displacementt2x1x121� |
��21xt2xt3y5y53838yt238xtyt01txy43xy 455 line segment line segment for Example 9.5.7. Find a parametrization for the circle. Solution. In order to use the formulas and to parametrize the circle, we first need to put it into the correct form,. The formulas and can be a challenge to memorize, but they come from the Pythagorean identity. By writing the equation as, we identify and. Rearranging these last two equations, we get and. In order to complete one revolution around the circle, we let t range through the interval. Our final answer is for 2xt01t38yt01t238xtyt01t22244xxyycosxhatsinykbt22244xxyy22221xhykab |
22222222244214441412912199xxyyxxyyxyxycosxhatsinykbt22cossin1tt2212199xy2212133xy1cos3xt2sin3yt13cosxt23sinyt0,213cos23sinxtyt02txy 2,3 1,5tx 0,2 1,1ty 0,3 1,5xy 0; 2tt 456 Example 9.5.8. Find a parametrization for the left half of the ellipse. Solution. In the equation, we can either use the |
formulas or think back to the Pythagorean identity, along with, to get The normal range on the parameter in this case is, but since we are interested in only the left half of the ellipse, we restrict t to the values which correspond to Quadrant II and Quadrant III angles, namely. Our final answer is for In the last two examples we avoided the formulas by instead associating the circle and ellipse equations with the Pythagorean identity. By getting a feel for the mechanics behind each of the previous five examples, reliance on formulas can be minimized. We note that the formulas provided prior to these examples offer only one of literally infinitely many ways to parametrize the common curves listed there. Adjusting Parametric Equations At times, the formulas that define a parametric curve need to be altered to suit the situation. Two easy ways to alter parametrizations are given below. 22149xy22149xy22cossin1tt22123xycos22cosxtxtsin33sinytyt02t322t2cos3sinxtyt322t22cossin1ttxy 2t 32t 457 Adjusting Parametric Equations Reversing Orientation: Replacing every occurrence of t with in a parametric description for a curve (including any inequalities which describe the bounds on t) reverses the orientation of the curve. � |
� Shift of Parameter: Replacing every occurrence of t with in a parametric description for a curve (including any inequalities which describe the bounds on t) shifts the start of the parameter t ahead by c units. We demonstrate these techniques in the following example. Example 9.5.9. Find a parametrization for the following curves. 1. The curve which starts at and follows the parabola to end at. Shift the parameter so the path starts at. 2. The two part path which starts at, travels along a line to, then travels along a line to. 3. The Unit Circle, oriented clockwise, with corresponding to. Solution. 1. We can parametrize from to as for. This parametrization, however, starts at and ends at. Hence, we need to reverse the orientation. To do so, we replace every occurrence of t with to get and for. After simplifying, we have for We would next like t to begin at instead of. The problem here is that the parametrization we have starts 2 units too soon, so we need to introduce a time delay of 2. Replacing every occurrence of t with gives and for. Simplifying yields for ttc2,42yx1,10t0,03,45,00t0,12yx1x2x2xtyt12t1,12,4txt2yt12t2xtyt21t0t2t2t2xt |
22yt221t2244xtytt03txy 458 2. When parameterizing line segments, we think:. For the first part of the path, which starts at and travels along a line to, we get For the second part we get for for Since the first parametrization leaves off at, we shift the parameter in the second part so it starts at. Our current description of the second part starts at, so we introduce a time delay of 1 unit to the second set of parametric equations. Replacing t with in the second set of parametric equations gives and for. Simplifying yields Hence, we may parametrize the path as for for, where and 3. To parametrize the Unit Circle with a clockwise orientation and ‘starting point’ of corresponding to, we first note that a counter-clockwise orientation is given by for starting point displacementt0,03,434xtyt01t3244xtyt01t1t1t0t1t321xt441yt011t1284xtyt12txftygt02t3 for 0112 for 12ttfttt� |
��4 for 0184 for 12ttgttt0,10tcossinxtyt02txy 459 The first order of business is to reverse the orientation. Replacing t with gives and for, which simplifies1 to for This parametrization gives a clockwise orientation, but still corresponds to the point ; the point is reached when. Our strategy is to first get the parametrization to start at the point and then shift the parameter accordingly so the start coincides with. We know that any interval of length will parametrize the entire circle, so we keep the equations and, but start the parameter t at, and find the upper bound by adding so that for The reader can verify that the Unit Circle is traced out clockwise starting at the point. To shift the parameter so that the start coincides with, we introduce a time delay of units by replacing each occurrence of t with. We get and for. This simplifies2 to for We put our answer to Example 9.5.9 number 3 to good use to derive the equation of a cycloid. Suppose a circle of radius r rolls along the positive x-axis at a constant velocity v as pictured below. Let be the 1 courtesy of the even/odd identities 2 courtesy of the sum/difference formulas tcosxtsinyt02tcossinxtyt20 |
t0t1,00,132t0,10t2cosxtsinyt322cossinxtyt322t0,10t3232t3cos2xt3sin2yt33222tsincosxtyt02txy 0,1 460 angle in radians which measures the amount of clockwise rotation experienced by the radius highlighted in the figure. Our goal is to find parametric equations for the coordinates of the point in terms of. From our work in Example 9.5.9 number 3, we know that clockwise motion along the Unit Circle starting at the point can be modeled by the equations for (We have renamed the parameter as to match the context of this problem.) To model this motion on a circle of radius r, all we need to do3 is multiply both x and y by the factor r which yields We now need to adjust for the fact that the circle isn’t stationary with center, but is rolling along the positive x-axis. Since the velocity v is constant, we know |
that at time t, the center of the circle has traveled a distance down the positive x-axis. Furthermore, since the radius of the circle is r and the circle isn’t moving vertically, we know that the center of the circle is always r units above the x-axis. Putting these two facts together, we have that at time t, the center of the circle is at the point. 3 If we replace x with and y with in the equation for the Unit Circle,, we obtain, which reduces to directions. Hence, we multiply both the x- and y-coordinates of points on the graph by r.. Note that we are ‘stretching’ the graph by a factor of r in both the x- and y- ,Pxy0,1sincosxy02sincosxryr0,0vt,vtrxryr221xy221xyrr222xyr 461 From Section 1.3, we know the angular velocity is and the linear velocity is. Putting these together, we have, or. Hence, the center of the circle, in terms of the parameter, is. As a result, we need to modify the equations and by shifting the x- coordinates to the right units (by adding to the expression for x) and the y-coordinate up r units (by adding r to the expression for y). We get and, which can be written as Since the motion starts at and proceeds indefinitely, we set. Example 9.5.10. Find the parametric equations of a cycl |
oid which results from a circle of radius 3 rolling down the positive x-axis. Solution. We completed the major part of our work above. With, we have the equations for (Here we have returned to the convention of using t as the parameter.) We know that one full revolution of the circle occurs over the interval. As t ranges between 0 and, we see that x ranges between 0 and. The values of y range between 0 and 6. Above, the first graph of the cycloid is over the interval. For the second graph, we extend t to range from 0 to which forces x to range from 0 to yielding three arches of the cycloid. tvrrvtvtr,rrsinxrcosyrrrsinxrrcosyrrsin1cosxryr003r3sin31cosxttyt0t02t2602t06t02t618 |
xyxy 462 9.5 Exercises In Exercises 1 – 18, eliminate the parameter t to rewrite the parametric equation as a Cartesian equation. 1. 4. 7. 10. 13. 16. 2. 5. 8. 11. 14. 17. 3. 6. 9. 12. 15. 18. In Exercises 19 – 22, parameterize (write parametric equations for) each Cartesian equation by setting or by setting. 19. 20. 21. 22. In Exercises 23 – 26, parameterize (write parametric equations for) each Cartesian equation by using and. Identify the curve. 23. 24. 25. 26. In Exercises 27 – 41, find a parametric description for the given oriented curve. 27. the directed line segment from to 28. the directed line segment from to 29. the curve from to 582xtyt6310xtyt213xtyt2312xtyt215txeyt22ttxeye4log32xtytlog21xtyt312xtyt42xttyt |
26ttxeye510xtyt4cos4sinxtyt3sin6cosxtyt22cossinxtyt2cos42sinxtyt21xtyt31xtytxtyt233yx2sin1yx3logxyy2xyycosxatsinybt22149xy2211636xy2216xy2210xy3,52,22,13,424yx2,02,0 463 30. the curve from to (Shift the parameter so corresponds to.) 31. the curve from to 32. the curve from to (Shift the parameter so corresponds to.) 33. the circle, oriented counter-clockwise 34. the circle, oriented counter-clockwise 35. the circle, oriented counter-clockwise |
36. the circle, oriented clockwise (Shift the parameter so t begins at 0.) 37. the circle, oriented counter-clockwise 38. the ellipse, oriented counter-clockwise 39. the ellipse, oriented counter-clockwise 40. the ellipse, oriented clockwise (Shift the parameter so correspond to.) 41. the triangle with vertices,,, oriented counter-clockwise (Shift the parameter so corresponds to.) 42. Use parametric equations and a graphing utility to graph the inverse of. 43. Every polar curve can be translated to a system of parametric equations with parameter θ by. Convert to a system of parametric equations. Check your answer by graphing by hand using the techniques presented in Section 8.3 and then graphing the parametric equations you found using a graphing utility. 24yx2,02,00t2,029xy5,20,329xy0,35,20t0,32225xy2214xy2260xyy2260xyy2231117xy22199xy2294240xyy2294240xyy0t0,00,03,00,40t0,0334fxxxrf� |
�coscos, sinsinxrfyrf6cos2r6cos2r 464 44. A dart is thrown upward with an initial velocity of 65 feet/second at an angle of elevation of 52°. Consider the position of the dart at any time t. Neglect air resistance. a) Find parametric equations and that model the position of the dart. Use g=32 ft./s2. b) Find all possible values of x that represent the situation. c) When will the dart hit the ground? d) Find the maximum height of the dart. e) At what time will the dart reach maximum height? 45. Carl’s friend Jason competes in Highland Games Competitions across the country. In one event, the ‘hammer throw’, he throws a 56 pound weight for distance. If the weight is released 6 feet above the ground at an angle of 42° with respect to the horizontal, with an initial speed of 33 feet per second, find the parametric equations for the flight of the hammer. (Use g=32 ft./s2.) When will the hammer hit the ground? How far away will it hit the ground? Check your answer using a graphing utility. 46. Eliminate the parameter in the equations for projectile motion to show that the path of the projectile follows the curve Recall that for a quadratic function, the vertex can be determined using the formula. Use this formula to show the maximum height of the projectile is 47. In another event, the ‘sheaf toss’, Jason throws a 20 pound weight for height. If the weight is released 5 feet above the ground at an angle of 85° with respect to the horizontal and the sheaf reaches a maximum height of 31.5 feet, use your results from Exercise 48 to determine how fast the sheaf was launched into the air. (Once again, use g=32 ft./s2.) xtyt22020 |
sectan2gyxxsv2fxaxbxc,22bbfaa222000sinsin2 when 22vvysxgg 465 In Exercises 48 – 51, we explore the hyperbolic cosine function, denoted, and the hyperbolic sine function, denoted, defined below: 48. Using a graphing utility as needed, verify that the domain of is and the range of is. 49. Using a graphing utility as needed, verify that the domain and range of are both. 50. Show that parameterize the right half of the ‘unit’ hyperbola. (Hence the use of the adjective ‘hyperbolic’.) 51. Four other hyperbolic functions are waiting to be defined: the hyperbolic secant the hyperbolic cosecant, the hyperbolic tangent and the hyperbolic cotangent. Define these functions in terms of and, then convert them to formulas involving and. Consult a suitable reference and spend some time reliving the thrills of trigonometry with these hyperbolic functions. coshtsinhtcosh and sinh22tttteeeettcosht,cosht1,sinht,cosh,sinhxttytt221xy� |
��sechtcschttanhtcothtcoshtsinhttetehat takes the input n and gives the output Q Table 4 Table 5 below displays the age of children in years and their corresponding heights. This table displays just some of the data available for the heights and ages of children. We can see right away that this table does not represent a function because the same input value, 5 years, has two different output values, 40 in. and 42 in. Age in years, a (input) Height in inches, h (output) 5 40 5 42 6 44 7 47 8 50 9 52 10 54 Table 5 How To… Given a table of input and output values, determine whether the table represents a function. 1. Identify the input and output values. 2. Check to see if each input value is paired with only one output value. If so, the table represents a function. Example 5 Identifying Tables that Represent Functions Which table, Table 6, Table 7, or Table 8, represents a function (if any)? Input Output Input Output Input Output 2 5 8 1 3 6 −3 0 4 5 1 5 Table 6 Table 7 1 5 5 0 2 4 Table 8 Solution Table 6 and Table 7 define functions. In both, each input value corresponds to exactly one output value. Table 8 does not define a function because the input value of 5 corresponds to two different output values. When a table represents a function, corresponding input and output values can also be specified using function notation. The function represented by Table 6 can be represented by writing Similarly, the statements f (2) = 1, f (5) = 3, and f (8) = 6 g (−3) = 5, g (0) = 1, and g (4) = 5 represent the function in table Table 7. Table 8 cannot be expressed in a similar way because it does not represent a function. Try It #3 Does Table 9 represent a function? Input Output 1 2 3 10 100 1000 Table 9 SECTION 1.1 Functions and Function notation 7 Finding Input and Output Values of a Function When we know an input value and want to determine the corresponding output value for a function, we evaluate the function. Evaluating will always produce one result because each input value of a function corresponds to exactly one output value. When we know an output value and want to |
determine the input values that would produce that output value, we set the output equal to the function’s formula and solve for the input. Solving can produce more than one solution because different input values can produce the same output value. Evaluation of Functions in Algebraic Forms When we have a function in formula form, it is usually a simple matter to evaluate the function. For example, the function f (x) = 5 − 3x 2 can be evaluated by squaring the input value, multiplying by 3, and then subtracting the product from 5. How To… Given the formula for a function, evaluate. 1. Replace the input variable in the formula with the value provided. 2. Calculate the result. Example 6 Evaluating Functions at Specific Values Evaluate f(x) = x 2 + 3x − 4 at: a. 2 b. a c. a + h d. f (a + h) − f (a) __ h Solution Replace the x in the function with each specified value. a. Because the input value is a number, 2, we can use simple algebra to simplify. f (2) = 22 + 3(2. In this case, the input value is a letter so we cannot simplify the answer any further. c. With an input value of a + h, we must use the distributive property. f (a) = a2 + 3a − 4 f (a + h) = (a + h)2 + 3(a + h) − 4 = a2 + 2ah + h2 + 3a + 3h −4 d. In this case, we apply the input values to the function more than once, and then perform algebraic operations on the result. We already found that and we know that f (a + h) = a2 + 2ah + h2 + 3a + 3h − 4 f(a) = a2 + 3a − 4 Now we combine the results and simplify. f (a + h) − f(a) __ = h (a2 + 2ah + h2 + 3a + 3h − 4) − (a2 + 3a − 4) ____ h 2ah + h2 + 3h ___________ h h(2a + h + 3) ___________ h Factor out h. = = = 2a + h + 3 Simplify. 8 CHAPTER 1 Functions Example 7 Evaluating Functions Given the function h(p) |
= p2 + 2p, evaluate h(4). Solution To evaluate h(4), we substitute the value 4 for the input variable p in the given function. h(p) = p2 + 2p h(4) = (4)2 +2 (4) = 16 + 8 = 24 Therefore, for an input of 4, we have an output of 24. Try It #4 Given the function g(m) = √ — m − 4. Evaluate g(5). Example 8 Solving Functions Given the function h(p) = p2 + 2p, solve for h(p) = 3. Solution h(p) = 3 p2 + 2p = 3 Substitute the original function h(p) = p2 + 2p. p2 + 2p − 3 = 0 Subtract 3 from each side. (p + 3)(p − 1) = 0 Factor. If (p + 3)(p − 1) = 0, either (p + 3) = 0 or (p − 1) = 0 (or both of them equal 0). We will set each factor equal to 0 and solve for p in each case. (p + 3) = 0, p = −3 (p − 1) = 0, p = 1 This gives us two solutions. The output h(p) = 3 when the input is either p = 1 or p = −3. We can also verify by graphing as in Figure 6. The graph verifies that h(1) = h(−3) = 3 and h(4) = 24. h(p) 35 30 25 20 15 10 5 1 2 3 4 5 p p –3 –2 h(p) 3 0 0 0 1 3 4 24 Figure 6 Try It #5 Given the function g(m) = √ — m − 4, solve g(m) = 2. Evaluating Functions Expressed in Formulas Some functions are defined by mathematical rules or procedures expressed in equation form. If it is possible to express the function output with a formula involving the input quantity, then we can define a function in algebraic form. For example, the equation 2n + 6p = 12 expresses a functional relationship between n and p. We can rewrite it to decide if p is a function of n. SECTION 1.1 Functions and Function notation 9 How To… Given a function in equation form, write its algebraic formula. 1. Solve the |
equation to isolate the output variable on one side of the equal sign, with the other side as an expression that involves only the input variable. 2. Use all the usual algebraic methods for solving equations, such as adding or subtracting the same quantity to or from both sides, or multiplying or dividing both sides of the equation by the same quantity. Example 9 Finding an Equation of a Function Express the relationship 2n + 6p = 12 as a function p = f (n), if possible. Solution To express the relationship in this form, we need to be able to write the relationship where p is a function of n, which means writing it as p = [expression involving n]. 2n + 6p = 12 Subtract 2n from both sides. Divide both sides by 6 and simplify. p = 6p = 12 − 2n 12 − 2n _______ 6 − 12 2n __ ___ 6 6 1 __ p = 2 − n 3 p = Therefore, p as a function of n is written as 1 __ p = f (n) = 2 − n 3 Analysis with a formula. It is important to note that not every relationship expressed by an equation can also be expressed as a function Example 10 Expressing the Equation of a Circle as a Function Does the equation x2 + y2 = 1 represent a function with x as input and y as output? If so, express the relationship as a function y = f (x). Solution First we subtract x2 from both sides. We now try to solve for y in this equation. y2 = 1 − x2 y = ± √ = + √ — 1 − x2 1 − x2 and − √ — — 1 − x2 We get two outputs corresponding to the same input, so this relationship cannot be represented as a single function y = f (x). Try It #6 If x − 8y 3 = 0, express y as a function of x. Q & A… Are there relationships expressed by an equation that do represent a function but which still cannot be represented by an algebraic formula? Yes, this can happen. For example, given the equation x = y + 2y, if we want to express y as a function of x, there is no simple algebraic formula involving only x that equals y. However, each x does determine a unique value for y, and there are mathematical procedures by which y can be found to any desired accuracy. In this case, we say that the equation gives an |
implicit (implied) rule for y as a function of x, even though the formula cannot be written explicitly. 10 CHAPTER 1 Functions Evaluating a Function Given in Tabular Form As we saw above, we can represent functions in tables. Conversely, we can use information in tables to write functions, and we can evaluate functions using the tables. For example, how well do our pets recall the fond memories we share with them? There is an urban legend that a goldfish has a memory of 3 seconds, but this is just a myth. Goldfish can remember up to 3 months, while the beta fish has a memory of up to 5 months. And while a puppy’s memory span is no longer than 30 seconds, the adult dog can remember for 5 minutes. This is meager compared to a cat, whose memory span lasts for 16 hours. The function that relates the type of pet to the duration of its memory span is more easily visualized with the use of a table. See Table 10.[2] Pet Puppy Adult dog Cat Goldfish Beta fish Memory span in hours 0.008 0.083 16 2160 3600 Table 10 At times, evaluating a function in table form may be more useful than using equations. Here let us call the function P. The domain of the function is the type of pet and the range is a real number representing the number of hours the pet’s memory span lasts. We can evaluate the function P at the input value of “goldfish.” We would write P(goldfish) = 2160. Notice that, to evaluate the function in table form, we identify the input value and the corresponding output value from the pertinent row of the table. The tabular form for function P seems ideally suited to this function, more so than writing it in paragraph or function form. How To… Given a function represented by a table, identify specific output and input values. 1. Find the given input in the row (or column) of input values. 2. Identify the corresponding output value paired with that input value. 3. Find the given output values in the row (or column) of output values, noting every time that output value appears. 4. Identify the input value(s) corresponding to the given output value. Example 11 Evaluating and Solving a Tabular Function Using Table 11, a. Evaluate g(3) b. Solve g(n) = 6. n g (n Table 11 Solution a. Evaluating g |
(3) means determining the output value of the function g for the input value of n = 3. The table output value corresponding to n = 3 is 7, so g (3) = 7. b. Solving g (n) = 6 means identifying the input values, n, that produce an output value of 6. Table 11 shows two solutions: 2 and 4. When we input 2 into the function g, our output is 6. When we input 4 into the function g, our output is also 6. 2 http://www.kgbanswers.com/how-long-is-a-dogs-memory-span/4221590. Accessed 3/24/2014. SECTION 1.1 Functions and Function notation 11 Try It #7 Using Table 11, evaluate g(1). Finding Function Values from a Graph Evaluating a function using a graph also requires finding the corresponding output value for a given input value, only in this case, we find the output value by looking at the graph. Solving a function equation using a graph requires finding all instances of the given output value on the graph and observing the corresponding input value( s ). Example 12 Reading Function Values from a Graph Given the graph in Figure 7, a. Evaluate f (2). b. Solve f (x) = 4. f (x) 7 6 5 4 3 2 1 –1 –1 2 – – 3 –5 –4 –3 –2 21 3 4 5 Figure 7 Solution a. To evaluate f (2), locate the point on the curve where x = 2, then read the y-coordinate of that point. The point has coordinates (2, 1), so f (2) = 1. See Figure 8. f (x) 7 6 5 4 3 2 1 –1 –1 2 – – 3 –5 –4 –3 –2 (2, 1) f (2) = 1 21 3 4 5 Figure 8 b. To solve f (x) = 4, we find the output value 4 on the vertical axis. Moving horizontally along the line y = 4, we locate two points of the curve with output value 4: (−1, 4) and (3, 4). These points represent the two solutions to f (x) = 4: −1 or 3. This means f (−1) = 4 and f (3) = 4, or when the input is −1 or 3, the output is 4. See Figure 9. f |
(x) 7 6 5 4 3 2 1 (−1, 4) (3, 4) 21 3 4 5 –5 –4 –3 –2 –1 –1 2 – – 3 Figure 9 12 CHAPTER 1 Functions Try It #8 Using Figure 7, solve f (x) = 1. Determining Whether a Function is One-to-One Some functions have a given output value that corresponds to two or more input values. For example, in the stock chart shown in Figure 1 at the beginning of this chapter, the stock price was $1,000 on five different dates, meaning that there were five different input values that all resulted in the same output value of $1,000. However, some functions have only one input value for each output value, as well as having only one output for each input. We call these functions one-to-one functions. As an example, consider a school that uses only letter grades and decimal equivalents, as listed in Table 12. Letter grade A B C D Grade point average 4.0 3.0 2.0 1.0 Table 12 This grading system represents a one-to-one function, because each letter input yields one particular grade point average output and each grade point average corresponds to one input letter. To visualize this concept, let’s look again at the two simple functions sketched in Figure 1(a) and Figure 1(b). The function in part ( a) shows a relationship that is not a one-to-one function because inputs q and r both give output n. The function in part (b ) shows a relationship that is a one-to-one function because each input is associated with a single output. one-to-one function A one-to-one function is a function in which each output value corresponds to exactly one input value. Example 13 Determining Whether a Relationship Is a One-to-One Function Is the area of a circle a function of its radius? If yes, is the function one-to-one? Solution A circle of radius r has a unique area measure given by A = πr 2, so for any input, r, there is only one output, A. The area is a function of radius r. If the function is one-to-one, the output value, the area, must correspond to a unique input value, the radius. Any area measure A is given by the formula A = πr2. Because areas and radii are positive |
numbers, there is exactly one solution: r = √ ___ A __ π So the area of a circle is a one-to-one function of the circle’s radius. Try It #9 a. Is a balance a function of the bank account number? b. Is a bank account number a function of the balance? c. Is a balance a one-to-one function of the bank account number? Try It #10 a. If each percent grade earned in a course translates to one letter grade, is the letter grade a function of the percent grade? b. If so, is the function one-to-one? SECTION 1.1 Functions and Function notation 13 Using the Vertical line Test As we have seen in some examples above, we can represent a function using a graph. Graphs display a great many input-output pairs in a small space. The visual information they provide often makes relationships easier to understand. By convention, graphs are typically constructed with the input values along the horizontal axis and the output values along the vertical axis. The most common graphs name the input value x and the output value y, and we say y is a function of x, or y = f (x) when the function is named f. The graph of the function is the set of all points (x, y) in the plane that satisfies the equation y = f (x). If the function is defined for only a few input values, then the graph of the function is only a few points, where the x-coordinate of each point is an input value and the y-coordinate of each point is the corresponding output value. For example, the black dots on the graph in Figure 10 tell us that f (0) = 2 and f (6) = 1. However, the set of all points (x, y) satisfying y = f (x) is a curve. The curve shown includes (0, 2) and (6, 1) because the curve passes through those points. y Figure 10 x The vertical line test can be used to determine whether a graph represents a function. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because a function has only one output value for each input value. See Figure 11. Function Not a Function Not a Function Figure 11 How To… Given a graph, use the vertical line test to determine if the graph represents a function. 1. Inspect the graph to see if any vertical |
line drawn would intersect the curve more than once. 2. If there is any such line, determine that the graph does not represent a function. 14 CHAPTER 1 Functions Example 14 Applying the Vertical Line Test Which of the graphs in Figure 12 represent( s ) a function y = f (x)? f (x) (a) y 5 4 3 2 1 f (x) x 4 5 6 7 8 9 10 11 12 x x (b) Figure 12 (c) Solution If any vertical line intersects a graph more than once, the relation represented by the graph is not a function. Notice that any vertical line would pass through only one point of the two graphs shown in parts ( a) and (b) of Figure 12. From this we can conclude that these two graphs represent functions. The third graph does not represent a function because, at most x-values, a vertical line would intersect the graph at more than one point, as shown in Figure 13. Try It #11 Does the graph in Figure 14 represent a function? y Figure 13 y Figure 14 x x SECTION 1.1 Functions and Function notation 15 Using the Horizontal line Test Once we have determined that a graph defines a function, an easy way to determine if it is a one-to-one function is to use the horizontal line test. Draw horizontal lines through the graph. If any horizontal line intersects the graph more than once, then the graph does not represent a one-to-one function. How To… Given a graph of a function, use the horizontal line test to determine if the graph represents a one-to-one function. 1. Inspect the graph to see if any horizontal line drawn would intersect the curve more than once. 2. If there is any such line, determine that the function is not one-to-one. Example 15 Horizontal Line Test Consider the functions shown in Figure 12(a) and Figure 12(b). Are either of the functions one-to-one? Solution The function in Figure 12(a) is not one-to-one. The horizontal line shown in Figure 15 intersects the graph of the function at two points (and we can even find horizontal lines that intersect it at three points.) f (x) Figure 15 x The function in Figure 12(b) is one-to-one. Any horizontal line will intersect a diagonal line at most once. Try It #12 Is the graph shown here one-to-one? y x Identifying Basic Toolkit Functions |
In this text, we will be exploring functions—the shapes of their graphs, their unique characteristics, their algebraic formulas, and how to solve problems with them. When learning to read, we start with the alphabet. When learning to do arithmetic, we start with numbers. When working with functions, it is similarly helpful to have a base set of building-block elements. We call these our “toolkit functions,” which form a set of basic named functions for which 16 CHAPTER 1 Functions we know the graph, formula, and special properties. Some of these functions are programmed to individual buttons on many calculators. For these definitions we will use x as the input variable and y = f (x) as the output variable. We will see these toolkit functions, combinations of toolkit functions, their graphs, and their transformations frequently throughout this book. It will be very helpful if we can recognize these toolkit functions and their features quickly by name, formula, graph, and basic table properties. The graphs and sample table values are included with each function shown in Table 13. Name Function Toolkit Functions Graph f (x) Constant f (x) = c, where c is a constant Identity f (x) = x Absolute value f (x) = ∣ x ∣ Quadratic f (x) = x2 Cubic f (x) = x3 f (x) f (x) f (x) f (x) x x x x x x –2 0 2 x –2 0 2 x –2 0 2 x –2 –1 0 1 2 f(x) 2 2 2 f(x) –2 0 2 f(x) 2 0 2 f(x) 4 1 0 1 4 x –1 –0.5 0 0.5 1 f(x) –1 –0.125 0 0.125 1 SECTION 1.1 Functions and Function notation 17 Reciprocal f (x) = 1 __ x Reciprocal squared f (x) = 1 _ x 2 Square root f (x) = √ — x Cube root f (x) = 3 √ — x f (x) f (x) f (x) f (x) x –2 –1 –0.5 0.5 1 2 x –2 –1 –0.5 0.5 1 2 x 0 1 4 f(x) –0.5 –1 –2 2 1 0.5 f(x) 0.25 |
1 4 4 1 0.25 f(x) 0 1 2 x –1 f(x) –1 –0.125 –0.5 0 0.125 1 0 0.5 1 x x x x Table 13 Access the following online resources for additional instruction and practice with functions. • Determine if a Relation is a Function (http://openstaxcollege.org/l/relationfunction) • Vertical line Test (http://openstaxcollege.org/l/vertlinetest) • Introduction to Functions (http://openstaxcollege.org/l/introtofunction) • Vertical line Test of Graph (http://openstaxcollege.org/l/vertlinegraph) • One-to-one Functions (http://openstaxcollege.org/l/onetoone) • Graphs as One-to-one Functions (http://openstaxcollege.org/l/graphonetoone) 18 CHAPTER 1 Functions 1.1 SeCTIOn exeRCISeS VeRBAl 1. What is the difference between a relation and 2. What is the difference between the input and the a function? output of a function? 3. Why does the vertical line test tell us whether the 4. How can you determine if a relation is a one-to-one graph of a relation represents a function? function? 5. Why does the horizontal line test tell us whether the graph of a function is one-to-one? AlGeBRAIC For the following exercises, determine whether the relation represents a function. 6. {(a, b), (c, d), (a, c)} 7. {(a, b),(b, c),(c, c)} For the following exercises, determine whether the relation represents y as a function of x. 9. y = x 2 8. 5x + 2y = 10 10. x = y 2 11. 3x 2 + y = 14 1 __ 14. y = x 3x + 5 ______ 7x − 1 17. y = 20. x = y 3 23. x = ± √ — 1 − y 26. y 3 = x 2 12. 2x + y 2 = 6 15. x = 3y + 5 ______ 7y − 1 18. x 2 + y 2 = 9 21. y = x 3 24. y = ± √ — 1 − x 13. y = −2x |
2 + 40x 16. x = √ — 1 − y 2 19. 2xy = 1 22. y = √ — 1 − x 2 25. y 2 = x 2 For the following exercises, evaluate the function f at the indicated values f (−3), f (2), f (−a), −f (a), f (a + h). 27. f (x) = 2x − 5 28. f (x) = −5x 2 + 2x − 1 29. f (x) = √ — 2 − x + 5 30. f (x) = 6x − 1 ______ 5x + 2 32. Given the function g(x) = 5 − x 2, evaluate 33. Given the function g(x) = x 2 + 2x, evaluate 31. f (x(x + h) − g(x) __ h g(x) − g(a 34. Given the function k(t) = 2t − 1: 35. Given the function f (x) = 8 − 3x: a. Evaluate k(2). b. Solve k(t) = 7. a. Evaluate f (−2). b. Solve f (x) = −1. 36. Given the function p(c) = c 2 + c: 37. Given the function f (x) = x2 − 3x a. Evaluate p(−3). b. Solve p(c) = 2. 38. Given the function f (x) = √ — x + 2 : a. Evaluate f (7). b. Solve f (x) = 4 a. Evaluate f (5). b. Solve f (x) = 4 39. Consider the relationship 3r + 2t = 18. a. Write the relationship as a function r = f (t). b. Evaluate f (−3). c. Solve f (t) = 2. SECTION 1.1 section exercises 19 GRAPHICAl For the following exercises, use the vertical line test to determine which graphs show relations that are functions. 40. y 41. y 42. y 43. y 46. y 49. y x x x x 44. y 47. y 50. y x x x x 45. y 48. y 51. y x x x x 20 CHAPTER 1 Functions 52. Given the following graph 53. Given the following graph 54. Given the following graph |
a. Evaluate f (−1). b. Solve for f (x) = 3. a. Evaluate f (0). b. Solve for f (x) = −3. a. Evaluate f (4). b. Solve for f (x) = 1. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x For the following exercises, determine if the given graph is a one-to-one function. 56. 21 3 4 5 x –5 –4 –3 –2 59. 21 1 –1 –2 –3 –4 – 55. 58. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 –5 –4 –3 –2 nUMeRIC 57. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x x π For the following exercises, determine whether the relation represents a function. 60. {(−1, −1),(−2, −2),(−3, −3)} 61. {(3, 4),(4, 5),(5, 6)} 62. {(2, 5),(7, 11),(15, 8),(7, 9)} For the following exercises, determine if the relation represented in table form represents y as a function of x. 63. x y 5 3 10 8 15 14 64. x y 5 3 10 8 15 8 65. x y 5 3 10 8 10 14 For the following exercises, use the function f represented in Table 14 below. x f (x) 0 74 1 28 2 1 3 53 4 56 Table 14 5 3 6 36 7 45 8 14 9 47 66. Evaluate f (3). 67. Solve f (x) = 1 SECTION 1.1 section exercises 21 For the following exercises, evaluate the function f at the values f |
(−2), f (−1), f (0), f (1), and f (2). 68. f (x) = 4 − 2x 69. f (x) = 8 − 3x 70. f (x) = 8x2 − 7x + 3 71. f (x) = 3 + √ — x + 3 72. f (x) = x − 2 _____ x + 3 73. f (x) = 3x For the following exercises, evaluate the expressions, given functions f, g, and h: f (x) = 3x − 2 g(x) = 5 − x2 h(x) = −2x2 + 3x − 1 74. 3f (1) − 4g(−2) TeCHnOlOGY 7 __ − h(−2) 75. f 3 For the following exercises, graph y = x2 on the given viewing window. Determine the corresponding range for each viewing window. Show each graph. 76. [−0.1, 0.1] 77. [−10, 10] 78. [−100, 100] For the following exercises, graph y = x3 on the given viewing window. Determine the corresponding range for each viewing window. Show each graph. 79. [−0.1, 0.1] 81. [−100, 100] 80. [−10, 10] For the following exercises, graph y = √ viewing window. Show each graph. — x on the given viewing window. Determine the corresponding range for each 82. [0, 0.01] 83. [0, 100] 84. [0, 10,000] For the following exercises, graph y = viewing window. Show each graph. 85. [−0.001, 0.001] — 3 √ x on the given viewing window. Determine the corresponding range for each 86. [−1,000, 1,000] 87. [−1,000,000, 1,000,000] ReAl-WORlD APPlICATIOnS 88. The amount of garbage, G, produced by a city with population p is given by G = f (p). G is measured in tons per week, and p is measured in thousands of people. a. The town of Tola has a population of 40,000 and produces 13 tons of garbage each week. Express this information |
in terms of the function f. b. Explain the meaning of the statement f (5) = 2. 89. The number of cubic yards of dirt, D, needed to cover a garden with area a square feet is given by D = g(a). a. A garden with area 5,000 ft2 requires 50 yd3 of dirt. Express this information in terms of the function g. b. Explain the meaning of the statement g(100) = 1. 90. Let f (t) be the number of ducks in a lake t years after 1990. Explain the meaning of each statement: a. f (5) = 30 b. f (10) = 40 91. Let h(t) be the height above ground, in feet, of a rocket t seconds after launching. Explain the meaning of each statement: a. h(1) = 200 b. h(2) = 350 92. Show that the function f (x) = 3(x − 5)2 + 7 is not one-to-one. 22 CHAPTER 1 Functions leARnInG OBjeCTIVeS In this section, you will: • • Find the domain of a function defined by an equation. Graph piecewise-defined functions. 1. 2 DOMAIn AnD RAnGe If you’re in the mood for a scary movie, you may want to check out one of the five most popular horror movies of all time—I am Legend, Hannibal, The Ring, The Grudge, and The Conjuring. Figure 1 shows the amount, in dollars, each of those movies grossed when they were released as well as the ticket sales for horror movies in general by year. Notice that we can use the data to create a function of the amount each movie earned or the total ticket sales for all horror movies by year. In creating various functions using the data, we can identify different independent and dependent variables, and we can analyze the data and the functions to determine the domain and range. In this section, we will investigate methods for determining the domain and range of functions such as these. Top-Five Grossing Horror Movies for years 2000–2013, fl 350 300 250 200 150 100 50 0 Market Share of Horror Moveis, by Year 8% 7% 6% 5% 4% 3% 2% 1% 0% I am Legend (2007) Hannibal (2001) The Ring (2002) The Grudge (2004) The Conjuring (2013) 2005 Figure |
1 Based on data compiled by www.the-numbers.com.[3] 2000 2001 2004 2002 2003 2006 2007 2008 2009 2010 2011 2012 2013 Finding the Domain of a Function Defined by an equation In Functions and Function Notation, we were introduced to the concepts of domain and range. In this section, we will practice determining domains and ranges for specific functions. Keep in mind that, in determining domains and ranges, we need to consider what is physically possible or meaningful in real-world examples, such as tickets sales and year in the horror movie example above. We also need to consider what is mathematically permitted. For example, we cannot include any input value that leads us to take an even root of a negative number if the domain and range consist of real numbers. Or in a function expressed as a formula, we cannot include any input value in the domain that would lead us to divide by 0. We can visualize the domain as a “holding area” that contains “raw materials” for a “function machine” and the range as another “holding area” for the machine’s products. See Figure 2. Domain a b c Function machine Figure 2 Range x y z We can write the domain and range in interval notation, which uses values within brackets to describe a set of numbers. In interval notation, we use a square bracket [when the set includes the endpoint and a parenthesis (to indicate that the endpoint is either not included or the interval is unbounded. For example, if a person has $100 to spend, he or she would need to express the interval that is more than 0 and less than or equal to 100 and write (0, 100]. We will discuss interval notation in greater detail later. 3 The Numbers: Where Data and the Movie Business Meet. “Box Office History for Horror Movies.” http://www.the-numbers.com/market/genre/Horror. Accessed 3/24/2014 SECTION 1.2 domain and range 23 Let’s turn our attention to finding the domain of a function whose equation is provided. Oftentimes, finding the domain of such functions involves remembering three different forms. First, if the function has no denominator or an even root, consider whether the domain could be all real numbers. Second, if there is a denominator in the function’s equation, exclude values in the domain that force the denominator to be zero. Third, if there is an even root, consider |
excluding values that would make the radicand negative. Before we begin, let us review the conventions of interval notation: • The smallest term from the interval is written first. • The largest term in the interval is written second, following a comma. • • Parentheses, (or), are used to signify that an endpoint is not included, called exclusive. Brackets, [or], are used to indicate that an endpoint is included, called inclusive. See Figure 3 for a summary of interval notation. Inequality Interval Notation Graph on Number Line a, ∞) (−∞, a) [a, ∞) (−∞, a] (a, b) [a, b) (a, b] [a, b Figure 3 Description x is greater than a x is less than a x is greater than or equal to a x is less than or equal to a x is strictly between a and b x is between a and b, to include a x is between a and b, to include b x is between a and b, to include a and b Example 1 Finding the Domain of a Function as a Set of Ordered Pairs Find the domain of the following function: {(2, 10), (3, 10), (4, 20), (5, 30), (6, 40)}. Solution First identify the input values. The input value is the first coordinate in an ordered pair. There are no restrictions, as the ordered pairs are simply listed. The domain is the set of the first coordinates of the ordered pairs. {2, 3, 4, 5, 6} Try It #1 Find the domain of the function: {(−5, 4), (0, 0), (5, −4), (10, −8), (15, −12)} How To… Given a function written in equation form, find the domain. 1. Identify the input values. 2. Identify any restrictions on the input and exclude those values from the domain. 3. Write the domain in interval form, if possible. 24 CHAPTER 1 Functions Example 2 Finding the Domain of a Function Find the domain of the function f (x) = x2 − 1. Solution The input value, shown by the variable x in the equation, is squared and then the result is lowered by one. Any real number may be squared and then be lowered by one, so there are no restrictions on the domain of this function. The domain is the set of real numbers. In |
interval form, the domain of f is (−∞, ∞). Try It #2 Find the domain of the function: f (x) = 5 − x + x 3. How To… Given a function written in an equation form that includes a fraction, find the domain. 1. Identify the input values. 2. Identify any restrictions on the input. If there is a denominator in the function’s formula, set the denominator equal to zero and solve for x. If the function’s formula contains an even root, set the radicand greater than or equal to 0, and then solve. 3. Write the domain in interval form, making sure to exclude any restricted values from the domain. Example 3 Finding the Domain of a Function Involving a Denominator Find the domain of the function f (x) = x + 1 _____. 2 − x Solution When there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to 0 and solve for x. 2 − x = 0 −x = −2 x = 2 Now, we will exclude 2 from the domain. The answers are all real numbers where x < 2 or x > 2. We can use a symbol known as the union, ∪, to combine the two sets. In interval notation, we write the solution: (−∞, 2) ∪ (2, ∞). –3 –2 2 3 1 –1 0 x < 2 or x > 2 ↓ ↓ (−∞, 2) ∪ (2, ∞) Figure 4 In interval form, the domain of f is (−∞, 2) ∪ (2, ∞). Try It #3 Find the domain of the function: f (x) = 1 + 4x ______ 2x − 1. How To… Given a function written in equation form including an even root, find the domain. 1. Identify the input values. 2. Since there is an even root, exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for x. 3. The solution(s) are the domain of the function. If possible, write the answer in interval form. SECTION 1.2 domain and range 25 Example 4 Finding the Domain of a Function with an Even Root Find the domain of the |
function f (x) = √ — 7 − x. Solution When there is an even root in the formula, we exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for x. 7 − x ≥ 0 −x ≥ −7 x ≤ 7 Now, we will exclude any number greater than 7 from the domain. The answers are all real numbers less than or equal to 7, or (−∞, 7]. Try It #4 Find the domain of the function f (x) = √ — 5 + 2x. Q & A… Can there be functions in which the domain and range do not intersect at all? Yes. For example, the function f (x) = − 1 _ — x √ negative real numbers as its range. As a more extreme example, a function’s inputs and outputs can be completely different categories (for example, names of weekdays as inputs and numbers as outputs, as on an attendance chart), in such cases the domain and range have no elements in common. has the set of all positive real numbers as its domain but the set of all Using notations to Specify Domain and Range In the previous examples, we used inequalities and lists to describe the domain of functions. We can also use inequalities, or other statements that might define sets of values or data, to describe the behavior of the variable in set-builder notation. For example, {x | 10 ≤ x < 30} describes the behavior of x in set-builder notation. The braces { } are read as “the set of,” and the vertical bar | is read as “such that,” so we would read {x | 10 ≤ x < 30} as “the set of x-values such that 10 is less than or equal to x, and x is less than 30.” Figure 5 compares inequality notation, set-builder notation, and interval notation. 5 5 5 5 5 Inequality Notation Set-builder Notation Interval Notation 5 < h ≤ 10 {h | 5 < h ≤ 10} (5, 10] 5 ≤ h < 10 {h | 5 ≤ h < 10} [5, 10) 5 < h < 10 {h | 5 < h < 10} (5, 10) h < 10 h ≥ 10 {h | h < 10} (−∞, 10) {h | h ≥ 10} [ |
10, ∞) All real numbers 핉 (−∞, ∞) Figure 5 10 10 10 10 10 10 26 CHAPTER 1 Functions To combine two intervals using inequality notation or set-builder notation, we use the word “or.” As we saw in earlier examples, we use the union symbol, ∪, to combine two unconnected intervals. For example, the union of the sets {2, 3, 5} and {4, 6} is the set {2, 3, 4, 5, 6}. It is the set of all elements that belong to one or the other (or both) of the original two sets. For sets with a finite number of elements like these, the elements do not have to be listed in ascending order of numerical value. If the original two sets have some elements in common, those elements should be listed only once in the union set. For sets of real numbers on intervals, another example of a union is {x | | x | ≥ 3} = (−∞, −3] ∪ [3, ∞) set-builder notation and interval notation Set-builder notation is a method of specifying a set of elements that satisfy a certain condition. It takes the form {x | statement about x} which is read as, “the set of all x such that the statement about x is true.” For example, {x | 4 < x ≤ 12} Interval notation is a way of describing sets that include all real numbers between a lower limit that may or may not be included and an upper limit that may or may not be included. The endpoint values are listed between brackets or parentheses. A square bracket indicates inclusion in the set, and a parenthesis indicates exclusion from the set. For example, (4, 12] How To… Given a line graph, describe the set of values using interval notation. 1. Identify the intervals to be included in the set by determining where the heavy line overlays the real line. 2. At the left end of each interval, use [with each end value to be included in the set (solid dot) or (for each excluded end value (open dot). 3. At the right end of each interval, use] with each end value to be included in the set (filled dot) or) for each excluded end value (open dot). 4. Use the union symbol ∪ to combine all intervals into one set. Example 5 Describing Sets on the Real-Number |
Line Describe the intervals of values shown in Figure 6 using inequality notation, set-builder notation, and interval notation. –2 –1 0 1 2 3 4 5 6 7 Figure 6 Solution To describe the values, x, included in the intervals shown, we would say, “x is a real number greater than or equal to 1 and less than or equal to 3, or a real number greater than 5.” Inequality 1 ≤ x ≤ 3 or x > 5 Set-builder notation { x |1 ≤ x ≤ 3 or x > 5 } Interval notation [1, 3] ∪ (5, ∞) Remember that, when writing or reading interval notation, using a square bracket means the boundary is included in the set. Using a parenthesis means the boundary is not included in the set. SECTION 1.2 domain and range 27 Try It #5 Given this figure, specify the graphed set in a. words b. set-builder notation c. interval notation Finding Domain and Range from Graphs –5 –4 –3 –2 –1 0 1 2 3 4 5 Figure 7 Another way to identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis. Keep in mind that if the graph continues beyond the portion of the graph we can see, the domain and range may be greater than the visible values. See Figure 8. Domain y 7 6 5 4 3 2 1 –6 –5 –4 –3 –2 –1 –1 –2 –3 –4 –5 –6 –7 – 8 – 9 Range 21 3 4 5 6 x Figure 8 We can observe that the graph extends horizontally from −5 to the right without bound, so the domain is [−5, ∞). The vertical extent of the graph is all range values 5 and below, so the range is (−∞, 5]. Note that the domain and range are always written from smaller to larger values, or from left to right for domain, and from the bottom of the graph to the top of the graph for range. Example 6 Finding Domain and Range from a Graph Find the domain and range of the function f whose graph is shown in Figure 9. –5 –4 –3 –2 –1 f y 1 –1 –2 – 3 – |
4 –5 1 2 3 4 5 x Figure 9 28 CHAPTER 1 Functions Solution We can observe that the horizontal extent of the graph is −3 to 1, so the domain of f is (−3, 1]. The vertical extent of the graph is 0 to −4, so the range is [−4, 0]. See Figure 10. y Domain 1 –5 –4 –3 –2 –1 f –1 –2 – 3 – 4 –5 1 2 3 4 5 x Range Figure 10 Example 7 Finding Domain and Range from a Graph of Oil Production Find the domain and range of the function f whose graph is shown in Figure 11. Alaska Crude Oil Production Th 2200 2000 1800 1600 1400 1200 1000 800 600 400 200 0 1975 1980 1985 1990 1995 2000 2005 Figure 11 (credit: modification of work by the U.S. energy Information Administration) [4] Solution The input quantity along the horizontal axis is “years,” which we represent with the variable t for time. The output quantity is “thousands of barrels of oil per day,” which we represent with the variable b for barrels. The graph may continue to the left and right beyond what is viewed, but based on the portion of the graph that is visible, we can determine the domain as 1973 ≤ t ≤ 2008 and the range as approximately 180 ≤ b ≤ 2010. In interval notation, the domain is [1973, 2008], and the range is about [180, 2010]. For the domain and the range, we approximate the smallest and largest values since they do not fall exactly on the grid lines. Try It #6 Given Figure 12, identify the domain and range using interval notation. World Population Increase 100 90 80 70 60 50 40 30 20 10 0 1950 1960 1970 1980 1990 2000 Year Figure 12 Q & A… Can a function’s domain and range be the same? Yes. For example, the domain and range of the cube root function are both the set of all real numbers. 4 http://www.eia.gov/dnav/pet/hist/LeafHandler.ashx?n=PET&s=MCRFPAK2&f=A. SECTION 1.2 domain and range 29 Finding Domains and Ranges of the Toolkit Functions We will now return to our set of toolkit functions to determine the domain and range of each. f (x) f (x) = c Domain: (−∞, ∞) Range: [c, |
c] x For the constant function f(x) = c, the domain consists of all real numbers; there are no restrictions on the input. The only output value is the constant c, so the range is the set {c} that contains this single element. In interval notation, this is written as [c, c], the interval that both begins and ends with c. Figure 13 f (x) Figure 14 f (x) Figure 15 f (x) Figure 16 f (x) Figure 17 For the identity function f(x) = x, there is no restriction on x. Both the domain and range are the set of all real numbers. Domain: (−∞, ∞) Range: (−∞, ∞) x Domain: (−∞, ∞) Range: [0, ∞) x For the absolute value function f(x) = ∣ x ∣, there is no restriction on x. However, because absolute value is defined as a distance from 0, the output can only be greater than or equal to 0. Domain: (−∞, ∞) Range: [0, ∞) x For the quadratic function f(x) = x 2, the domain is all real numbers since the horizontal extent of the graph is the whole real number line. Because the graph does not include any negative values for the range, the range is only nonnegative real numbers. Domain: (−∞, ∞) Range: (−∞, ∞) x For the cubic function f(x) = x 3, the domain is all real numbers because the horizontal extent of the graph is the whole real number line. The same applies to the vertical extent of the graph, so the domain and range include all real numbers. 30 CHAPTER 1 Functions f (x) Domain: (−∞, 0) ∪ (0, ∞) Range: (−∞, 0) ∪ (0, ∞) x For the reciprocal function f(x) = 1 _ x, we cannot divide by 0, so we must exclude 0 from the domain. Further, 1 divided by any value can never be 0, so the range also will not include 0. In set-builder notation, we could also write {x | x ≠ 0}, the set of all real numbers that are not zero. Figure 18 f (x) Figure 19 f (x) Figure 20 f (x) Figure 21 Domain: (−∞, |
0) ∪ (0, ∞) Range: (0, ∞) x For the reciprocal squared function f(x) = 1 __ x 2, we cannot divide by 0, so we must exclude 0 from the domain. There is also no x that can give an output of 0, so 0 is excluded from the range as well. Note that the output of this function is always positive due to the square in the denominator, so the range includes only positive numbers. Domain: [0, ∞) Range: [0, ∞) x Domain: (−∞, ∞) Range: (−∞, ∞) x — x, we cannot take the For the square root function f(x) = √ square root of a negative real number, so the domain must be 0 or greater. The range also excludes negative numbers because the square root of a positive number x is defined to be positive, even though the square of the negative number − √ — x also gives us x. — 3 √ x the domain and range For the cube root function f(x) = include all real numbers. Note that there is no problem taking a cube root, or any odd-integer root, of a negative number, and the resulting output is negative (it is an odd function). How To… Given the formula for a function, determine the domain and range. 1. Exclude from the domain any input values that result in division by zero. 2. Exclude from the domain any input values that have nonreal (or undefined) number outputs. 3. Use the valid input values to determine the range of the output values. 4. Look at the function graph and table values to confirm the actual function behavior. Example 8 Finding the Domain and Range Using Toolkit Functions Find the domain and range of f (x) = 2x3 − x. Solution There are no restrictions on the domain, as any real number may be cubed and then subtracted from the result. The domain is (−∞, ∞) and the range is also (−∞, ∞). SECTION 1.2 domain and range 31 Example 9 Finding the Domain and Range Find the domain and range of f (x) = 2 ____. x + 1 Solution We cannot evaluate the function at −1 because division by zero is undefined. The domain is (−∞, −1) ∪ (−1, ∞). Because the function is never zero, we exclude 0 from the |
range. The range is (−∞, 0) ∪ (0, ∞). Example 10 Finding the Domain and Range Find the domain and range of f (x) = 2 √ — x + 4. Solution We cannot take the square root of a negative number, so the value inside the radical must be nonnegative. The domain of f (x) is [−4, ∞). x + 4 ≥ 0 when x ≥ −4 We then find the range. We know that f (−4) = 0, and the function value increases as x increases without any upper limit. We conclude that the range of f is [0, ∞). Analysis Figure 22 represents the function f. f (x) 5 4 3 2 1 f –5 –4 –3 –2 –1 –1 21 3 4 5 x Figure 22 Try It #7 Find the domain and range of f (x) = − √ — 2 − x. Graphing Piecewise-Defined Functions Sometimes, we come across a function that requires more than one formula in order to obtain the given output. For example, in the toolkit functions, we introduced the absolute value function f (x) = |x|. With a domain of all real numbers and a range of values greater than or equal to 0, absolute value can be defined as the magnitude, or modulus, of a real number value regardless of sign. It is the distance from 0 on the number line. All of these definitions require the output to be greater than or equal to 0. If we input 0, or a positive value, the output is the same as the input. f (x) = x if x ≥ 0 If we input a negative value, the output is the opposite of the input. f (x) = −x if x < 0 Because this requires two different processes or pieces, the absolute value function is an example of a piecewise function. A piecewise function is a function in which more than one formula is used to define the output over different pieces of the domain. We use piecewise functions to describe situations in which a rule or relationship changes as the input value crosses certain “boundaries.” For example, we often encounter situations in business for which the cost per piece of a certain item is discounted once the number ordered exceeds a certain value. Tax brackets are another real-world example of piecewise functions. For example, consider a simple tax system in which incomes up to $10,000 are |
taxed at 10%, and any additional income is taxed at 20%. The tax on a total income S would be 0.1S if S ≤ $10,000 and $1000 + 0.2(S − $10,000) if S > $10,000. 32 CHAPTER 1 Functions piecewise function A piecewise function is a function in which more than one formula is used to define the output. Each formula has its own domain, and the domain of the function is the union of all these smaller domains. We notate this idea like this: f (x) = { formula 1 formula 2 formula 3 if x is in domain 1 if x is in domain 2 if x is in domain 3 In piecewise notation, the absolute value function is x −x { |x| = if x ≥ 0 if x < 0 How To… Given a piecewise function, write the formula and identify the domain for each interval. 1. Identify the intervals for which different rules apply. 2. Determine formulas that describe how to calculate an output from an input in each interval. 3. Use braces and if-statements to write the function. Example 11 Writing a Piecewise Function A museum charges $5 per person for a guided tour with a group of 1 to 9 people or a fixed $50 fee for a group of 10 or more people. Write a function relating the number of people, n, to the cost, C. Solution Two different formulas will be needed. For n-values under 10, C = 5n. For values of n that are 10 or greater, C = 50. { C(n) = 5n if 0 < n < 10 50 if n ≥ 10 Analysis The function is represented in Figure 23. The graph is a diagonal line from n = 0 to n = 10 and a constant after that. In this example, the two formulas agree at the meeting point where n = 10, but not all piecewise functions have this property. C (n) 50 40 30 20 10 C (n) – 25 – 20 – 15 – 10 – 5 –10 5 10 15 20 25 n Figure 23 Example 12 Working with a Piecewise Function A cell phone company uses the function below to determine the cost, C, in dollars for g gigabytes of data transfer. { C(g) = 25 if 0 < g < 2 25 + 10(g − 2) if g ≥ 2 Find the cost of using 1.5 gigabytes of data and the cost of using 4 |
gigabytes of data. Solution To find the cost of using 1.5 gigabytes of data, C(1.5), we first look to see which part of the domain our input falls in. Because 1.5 is less than 2, we use the first formula. C(1.5) = $25 To find the cost of using 4 gigabytes of data, C(4), we see that our input of 4 is greater than 2, so we use the second formula. C(4) = 25 + 10(4 − 2) = $45 SECTION 1.2 domain and range 33 Analysis The function is represented in Figure 24. We can see where the function changes from a constant to a shifted and stretched identity at g = 2. We plot the graphs for the different formulas on a common set of axes, making sure each formula is applied on its proper domain. C (g) 50 40 30 20 10 C (g10 1 2 3 4 5 g Figure 24 How To… Given a piecewise function, sketch a graph. 1. Indicate on the x-axis the boundaries defined by the intervals on each piece of the domain. 2. For each piece of the domain, graph on that interval using the corresponding equation pertaining to that piece. Do not graph two functions over one interval because it would violate the criteria of a function. Example 13 Graphing a Piecewise Function Sketch a graph of the function. f (x) = { x ≤ 1 x2 if 3 x if 1 < x ≤ 2 x > 2 if Solution Each of the component functions is from our library of toolkit functions, so we know their shapes. We can imagine graphing each function and then limiting the graph to the indicated domain. At the endpoints of the domain, we draw open circles to indicate where the endpoint is not included because of a less-than or greater-than inequality; we draw a closed circle where the endpoint is included because of a less-than-or-equal-to or greater-than-or-equal-to inequality. Figure 25 shows the three components of the piecewise function graphed on separate coordinate systems. f (x) 5 4 3 2 1 –1 (ax) 5 4 3 2 1 –1 (b) f (x) 5 4 3 2 1 –1 (c Figure 25 ( a) f (x) = x2 if x ≤ 1; ( b) f (x) = 3 if 1 < x ≤ 2; ( |
c) f (x) = x if x > 2 34 CHAPTER 1 Functions Now that we have sketched each piece individually, we combine them in the same coordinate plane. See Figure 26. f (x1 1 2 3 4 x Figure 26 Analysis Note that the graph does pass the vertical line test even at x = 1 and x = 2 because the points (1, 3) and (2, 2) are not part of the graph of the function, though (1, 1) and (2, 3) are. Try It #8 Graph the following piecewise function. f (x) = { x3 −2 — √ x x < −1 if if −1 < x < 4 x > 4 if Q & A… Can more than one formula from a piecewise function be applied to a value in the domain? No. Each value corresponds to one equation in a piecewise formula. Access these online resources for additional instruction and practice with domain and range. • Domain and Range of Square Root Functions (http://openstaxcollege.org/l/domainsqroot) • Determining Domain and Range (http://openstaxcollege.org/l/determinedomain) • Find Domain and Range Given the Graph (http://openstaxcollege.org/l/drgraph) • Find Domain and Range Given a Table (http://openstaxcollege.org/l/drtable) • Find Domain and Range Given Points on a Coordinate Plane (http://openstaxcollege.org/l/drcoordinate) SECTION 1.2 section exercises 35 1.2 SeCTIOn exeRCISeS VeRBAl 1. Why does the domain differ for different functions? 2. How do we determine the domain of a function defined by an equation? 3. Explain why the domain of f (x) = from the domain of f (x is different 4. When describing sets of numbers using interval notation, when do you use a parenthesis and when do you use a bracket? 5. How do you graph a piecewise function? AlGeBRAIC For the following exercises, find the domain of each function using interval notation. 6. f (x) = −2x(x − 1)(x − 2) 7. f (x) = 5 − 2x2 8. f (x) = 3 √ — x − 2 9. f (x) = 3 − √ — 6 − |
2x 10. f (x) = √ — 4 − 3x 11. f (x) = √ — x 2 + 4 12. f (x) = 3 √ — 1 − 2x 13. f (x) = 3 √ — x − 1 15. f (x) = 3x + 1 ______ 4x + 2 18. f (x) = 1 ________ x2 − x − 6 21. f (x) = 2x + 1 _ 5 − x √ — 16. f (x) = — x + 4 √ _______ x − 4 19. f (x) = 2x3 − 250 __________ x2 − 2x − 15 22. f (x √ — 14. f (x) = 9 _____ x − 6 17. f (x) = x − 3 __________ x2 + 9x − 22 20. f (x) = 5 _ — x − 3 √ 23. f (x √ — 24. f (x) = x __ x 25. f (x) = x2 − 9x _ x2 − 81 26. Find the domain of the function f (x) = √ — 2x 3 − 50x by: a. using algebra. b. graphing the function in the radicand and determining intervals on the x-axis for which the radicand is nonnegative. GRAPHICAl For the following exercises, write the domain and range of each function using interval notation. 27. y 28. y 29. –10 –8 –6 –4 10 8 6 4 2 –2 –2 –4 –6 –8 –10 42 6 8 10 x –10 –8 –6 –4 10 8 6 4 2 –2 –2 –4 –6 –8 –10 42 6 8 10 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x 36 30. 33. 36. CHAPTER 1 Functions y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 5 4 3 2 1 y –1 –1 –2 –3 –4 –5 10 8 6 4 2 –2 –2 –4 –6 –8 –10 –5 –4 –3 –2 –5 –4 –3 –2 –10 –8 –6 –4 31. 21 3 |
4 5 x –5 –4 –3 –2 34. 21 3 4 5 x –5 –4 –3 –2 37. 42 6 8 10 x –10 –8 –6 –4 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 5 4 3 2 1 y –1 –1 –2 –3 –4 –5 10 8 6 4 2 –2 –2 –4 –6 –8 –10 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 32. 21 3 4 5 x –5 –4 –3 –2 35. 21 3 4 5 x –6 –5 –4 –3 –2 1 6 –6, – ) ( ), –6 1 6 ( – 42 6 8 10 x 21, ) 1 6 21 3 4 5 6 x For the following exercises, sketch a graph of the piecewise function. Write the domain in interval notation. x + 1 if x < −2 { 38. f (x) = −2x − 3 if x ≥ −2 { 39. f (x) = 2x − 1 if x < 1 if x ≥ 1 1 + x { 40. f (x) = x + 1 if x < 0 x − 1 if x > 0 { 41. f (x) = 3 √ if x < 0 x if x ≥ 0 — { 42. f (x) = if x < 0 x2 1 − x if x > 0 { 43. f (x) = if x < 0 x2 x + 2 if x ≥ 0 { 44. f (x) = x + 1 if x < 1 if x ≥ 1 x3 { 45. f (x) = | x | if x < 2 if x ≥ 2 1 SECTION 1.2 section exercises 37 nUMeRIC For the following exercises, given each function f, evaluate f (−3), f (−2), f (−1), and f (0). { 46. f (x) = x + 1 if x < −2 −2x − 3 if x ≥ −2 { 47. f (x) = 1 if x ≤ −3 0 if x > −3 { 48. f (x) = −2x 2 + 3 if x ≤ −1 if x > −1 5x − 7 For the following exercises |
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