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1, rv2 = k (r v1, v2) = k(rv) Deﬁnition of Scalar Multiplication Deﬁnition of Scalar Multiplication 11.8 Vectors 1019 The remaining properties are proved similarly and are left as exercises. Our next example demonstrates how Theorem 11.19 allows us to do the same kind of algebraic manipulations with vectors as we do with variables – multiplication and division of vectors notwithstanding. If the pedantry seems familiar, it should. This is the same treatment we gave Example 8.3.1 in Section 8.3. As in that example, we spell out the solution in excruciating detail to encourage the reader to think carefully about why each step is justiﬁed. Example 11.8.3. Solve 5v − 2 (v + 1, −2) = 0 for v. Solution. 5v − 2 (v + 1, −2) = 0 5v + (−1) [2 (v + 1, −2)] = 0 5v + [(−1)(2)] (v + 1, −2) = 0 5v + (−2) (v + 1, −2) = 0 5v + [(−2)v + (−2) 1, −2] = 0 5v + [(−2)v + (−2)(1), (−2)(−2)] = 0 [5v + (−2)v] + −2, 4 = 0 (5 + (−2))v + −2, 4 = 0 3v + −2, 4 = 0 (3v + −2, 4) + (− −2, 4) = 0 + (− −2, 4) 3v + [−2, 4 + (− −2, 4)] = 0 + (−1) −2, 4 3v + 0 = 0 + (−1)(−2), (−1)(4) 3v = 2, −4 1 3 1 3 (3v) = 1 (3) v = 1 3 1v = 2 v = 2 3 (2, −4) (2), (−4) A vector whose initial point is (0, 0) is said to be in standard position. If v = v1, v2 is plotted in standard position, then its terminal point is necessarily (v1, v2). (Once more, think about this before reading on.) y (v1
, v2) v = v1, v2 in standard position. x 1020 Applications of Trigonometry 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 (v1, v2) in rectangular coordinates to a pair (r, θ) in polar coordinates where r ≥ 0. The magnitude of v, which we said earlier was length 2 and is denoted by v. From Section 11.4, we of the directed line segment, is r = know v1 = r cos(θ) = v cos(θ) and v2 = r sin(θ) = v sin(θ). From the deﬁnition of scalar multiplication and vector equality, we get 1 + v2 v2 v = v1, v2 = v cos(θ), v sin(θ) = v cos(θ), sin(θ) This motivates the following deﬁnition. Deﬁnition 11.8. Suppose v is a vector with component form v = v1, v2. Let (r, θ) be a polar representation of the point with rectangular coordinates (v1, v2) with r ≥ 0. The magnitude of v, denoted v, is given by v = r = 1 + v2 v2 2 If v = 0, the (vector) direction of v, denoted ˆv is given by ˆv = cos(θ), sin(θ) Taken together, we get v = v cos(θ), v sin(θ). A few remarks are in order. First, we note that if v = 0 then even though there are inﬁnitely many angles θ which satisfy Deﬁnition 11.8, the stipulation r > 0 means that all of the angles are coterminal. Hence, if θ and θ both satisfy the conditions of Deﬁnition 11.8, then cos(θ) = cos(θ) and sin(θ) = sin(θ), and as such, cos(θ), sin(θ) = cos(θ), sin(θ) making ˆv is well-deﬁned.10 If v = 0, then v = 0, 0, and we
know from Section 11.4 that (0, θ) is a polar representation for the origin for any angle θ. For this reason, ˆ0 is undeﬁned. The following theorem summarizes the important facts about the magnitude and direction of a vector. Theorem 11.20. Properties of Magnitude and Direction: Suppose v is a vector. v ≥ 0 and v = 0 if and only if v = 0 For all scalars k, k v = \|k\|v. If v = 0 then v = vˆv, so that ˆv = 1 v v. 1 + v2 v2 v2 1 + v2 1 + v2 2 = 0 if and only of v2 The proof of the ﬁrst property in Theorem 11.20 is a direct consequence of the deﬁnition of v. If v = v1, v2, then v = 2 which is by deﬁnition greater than or equal to 0. Moreover, 2 = 0 if and only if v1 = v2 = 0. Hence, v = 0 if and only if v = 0, 0 = 0, as required. The second property is a result of the deﬁnition of magnitude and scalar multiplication along with a propery of radicals. If v = v1, v2 and k is a scalar then 10If this all looks familiar, it should. The interested reader is invited to compare Deﬁnition 11.8 to Deﬁnition 11.2 in Section 11.7. 11.8 Vectors 1021 k v = k v1, v2 = kv1, kv2 Deﬁnition of scalar multiplication 2 = = = √ (kv1)2 + (kv2)2 Deﬁnition of magnitude 1 + k2v2 k2v2 1 + v2 k2(v2 2 ) k2 v2 1 + v2 = = \|k\| 1 + v2 v2 = \|k\|v Product Rule for Radicals k2 = \|k\| Since √ 2 2 The equation v = vˆv in Theorem 11.20 is a consequence of the deﬁnitions of v and ˆv and was worked out in the discussion just prior to Deﬁnition 11.8 on page 10
20. In words, the equation v = vˆv says that any given vector is the product of its magnitude and its direction – an important v is a result of concept to keep in mind when studying and using vectors. The equation ˆv = solving v = vˆv for ˆv by multiplying11 both sides of the equation by 1 v and using the properties of Theorem 11.19. We are overdue for an example. 1 v Example 11.8.4. 1. Find the component form of the vector v with v = 5 so that when v is plotted in standard position, it lies in Quadrant II and makes a 60◦ angle12 with the negative x-axis. 2. For v = 3, −3 √ 3, ﬁnd v and θ, 0 ≤ θ < 2π so that v = v cos(θ), sin(θ). 3. For the vectors v = 3, 4 and w = 1, −2, ﬁnd the following. (a) ˆv (b) v − 2 w (c) v − 2 w (d) ˆw Solution. 1. We are told that v = 5 and are given information about its direction, so we can use the formula v = vˆv to get the component form of v. To determine ˆv, we appeal to Deﬁnition 11.8. We are told that v lies in Quadrant II and makes a 60◦ angle with the negative x-axis, so the polar form of the terminal point of v, when plotted in standard position is (5, 120◦). (See the diagram below.) Thus ˆv = cos (120◦), sin (120◦) =, so v = vˆv = √ −. 11Of course, to go from v = vˆv to ˆv = 1 v v, we are essentially ‘dividing both sides’ of the equation by the scalar v. The authors encourage the reader, however, to work out the details carefully to gain an appreciation of the properties in play. 12Due to the utility of vectors in ‘real-world’ applications, we will usually use degree measure for the angle when giving the vector’s direction. However, since Carl doesn’t want you to forget about radians, he
’s made sure there are examples and exercises which use them. 1022 Applications of Trigonometry y 5 4 3 2 1 v 60◦ θ = 120◦ −3 −2 −1 1 2 3 x 2. For v = 3, −3 √ 3, we get v = (3)2 + (−3 √ ﬁnd the θ we’re after by converting the point with rectangular coordinates (3, −3 form (r, θ) where r = v > 0. From Section 11.4, we have tan(θ) = −3 (3, −3 may check our answer by verifying v = 3, −3 3) is a point in Quadrant IV, θ is a Quadrant IV angle. Hence, we pick θ = 5π √ 3)2 = 6. In light of Deﬁnition 11.8, we can 3) to polar √ 3. Since 3. We √ 3 = −. 3 = 6 cos 5π 3, sin 5π 3 √ √ 3 3. (a) Since we are given the component form of v, we’ll use the formula ˆv = 5 3, 4, we have v = 25 = 5. Hence, ˆv = 1 32 + 42 = √ √ 5 1 v. v. For (b) We know from our work above that v = 5, so to ﬁnd v−2 w, we need only ﬁnd w. 12 + (−2)2 = Since w = 1, −2, we get w = 5. Hence. (c) In the expression v −2 w, 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 v − 2 w. We get v − 2 w = 3, 4 − 2 1, −2 = 1, 8. Hence, v − 2 w = 1, 8 = 65. √ 12 + 82 = √ √ (d) To ﬁnd ˆw, we ﬁrst need ˆw. Using the formula ˆw = 1 w which we found the in the previous problem, we get ˆw = 1√ 5 √ √ 5 5.
Hence, ˆw = 25 + 20. 25 = w along with w = 5, 1, −2 = 1√ 5, − 2√ 5 = The process exempliﬁed by number 1 in Example 11.8.4 above by which we take information about the magnitude and direction of a vector and ﬁnd the component form of a vector is called resolving a vector into its components. As an application of this process, we revisit Example 11.8.1 below. Example 11.8.5. A plane leaves an airport with an airspeed of 175 miles per hour with bearing 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 11.8.1 and let v denote the plane’s velocity and w denote the wind’s velocity, and set about determining v + w. If we regard the airport as being 11.8 Vectors 1023 at the origin, the positive y-axis acting as due north and the positive x-axis acting as due east, we see that the vectors v and w are in standard position and their directions correspond to the angles 50◦ and −30◦, respectively. Hence, the component form of v = 175 cos(50◦), sin(50◦) = 175 cos(50◦), 175 sin(50◦) and the component form of w = 35 cos(−30◦), 35 sin(−30◦). 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.13 Adding corresponding components, we ﬁnd the resultant vector v + w = 175 cos(50◦) + 35 cos(−30◦), 175 sin(50◦) + 35 sin(−30◦). To ﬁnd the ‘true’ speed of the plane, we compute the magnitude of this resultant vector v + w = (175 cos(50◦) + 35 cos(−30◦))2 + (175 sin(50◦) + 35 sin(−30◦))2 ≈ 184 Hence, the ‘true�
� speed of the plane is approximately 184 miles per hour. To ﬁnd the true bearing, we need to ﬁnd the angle θ which corresponds to the polar form (r, θ), r > 0, of the point (x, y) = (175 cos(50◦) + 35 cos(−30◦), 175 sin(50◦) + 35 sin(−30◦)). Since both of these coordinates are positive,14 we know θ is a Quadrant I angle, as depicted below. Furthermore, tan(θ) = y x = 175 sin(50◦) + 35 sin(−30◦) 175 cos(50◦) + 35 cos(−30◦), so using the arctangent function, we get θ ≈ 39◦. Since, for the purposes of bearing, we need the angle between v + w and the positive y-axis, we take the complement of θ and ﬁnd the ‘true’ bearing of the plane to be approximately N51◦E. y (N) y (N) v v v + w 40◦ 50◦ −30◦ w 60◦ x (E) θ w x (E) In part 3d of Example 11.8.4, we saw that ˆw = 1. Vectors with length 1 have a special name and are important in our further study of vectors. Deﬁnition 11.9. Unit Vectors: Let v be a vector. If v = 1, we say that v is a unit vector. 13Keeping things ‘calculator’ friendly, for once! 14Yes, a calculator approximation is the quickest way to see this, but you can also use good old-fashioned inequalities and the fact that 45◦ ≤ 50◦ ≤ 60◦. 1024 Applications of Trigonometry 1 v If v is a unit vector, then necessarily, v = vˆv = 1 · ˆv = ˆv. Conversely, we leave it as an exercise15 In practice, if v is a unit v is a unit vector for any nonzero vector v. to show that ˆv = vector we write it as ˆv as opposed to v because we have reserved the ‘ˆ’ notation for unit vectors. 1 The process of multiplying a non
zero vector by the factor v to produce a unit vector is called ‘normalizing the vector,’ and the resulting vector ˆv is called the ‘unit vector in the direction of v’. The terminal points of unit vectors, when plotted in standard position, lie on the Unit Circle. (You should take the time to show this.) As a result, we visualize normalizing a nonzero vector v as shrinking16 its terminal point, when plotted in standard position, back to the Unit Circle. y v 1 ˆv −1 1 x −1 Visualizing vector normalization ˆv = 1 v v Of all of the unit vectors, two deserve special mention. Deﬁnition 11.10. The Principal Unit Vectors: The vector ˆı is deﬁned by ˆı = 1, 0 The vector ˆ is deﬁned by ˆı = 0, 1 We can think of the vector ˆı as representing the positive x-direction, while ˆ represents the positive y-direction. We have the following ‘decomposition’ theorem.17 Theorem 11.21. Principal Vector Decomposition Theorem: Let v be a vector with component form v = v1, v2. Then v = v1ˆı + v2ˆ. The proof of Theorem 11.21 is straightforward. Since ˆı = 1, 0 and ˆ = 0, 1, we have from the deﬁnition of scalar multiplication and vector addition that v1ˆı + v2ˆ = v1 1, 0 + v2 0, 1 = v1, 0 + 0, v2 = v1, v2 = v 15One proof uses the properties of scalar multiplication and magnitude. If v = 0, consider ˆv = 1 v. Use v the fact that v ≥ 0 is a scalar and consider factoring. 16... if v > 1... 17We will see a generalization of Theorem 11.21 in Section 11.9. Stay tuned! 11.8 Vectors 1025 Geometrically, the situation looks like this: y v2ˆ v = v1, v2 ˆ ˆ
ı v1ˆı x v = v1, v2 = v1ˆı + v2ˆ. 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 Netwons (N) in the SI system or pounds (lbs.) in the English system.18 The following example uses all of the concepts in this section, and should be studied in great detail. Example 11.8.6. 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 represent the problem schematically below and then provide the corresponding vector diagram. 30◦ 60◦ 30◦ 60◦ T1 T2 30◦ 60◦ w 50 lbs. We have three forces acting on the speaker: the weight of the speaker, which we’ll call w, pulling the speaker directly downward, and the forces on the support rods, which we’ll call T1 and T2 (for ‘tensions’) acting upward at angles 60◦ and 30◦, respectively. We are looking for the tensions on the support, which are the magnitudes T1 and T2. In order for the speaker to remain stationary,19 we require w + T1 + T2 = 0. Viewing the common initial point of these vectors as the 18See also Section 11.1.1. 19This is the criteria for ‘static equilbrium’. 1026 Applications of Trigonometry origin and the dashed line as the x-axis, we use Theorem 11.20 to get component representations for the three vectors involved. We can model the weight of the speaker as a vector pointing directly downwards with a magnitude of 50 pounds. That is, w = 50 and ˆw = −ˆ = 0, −1. Hence, w = 50 0, −1 = 0, −50. For the force in the ﬁrst support, we get T1 = T1 cos (60◦), sin (60◦)
T1 2, √ T1 2 3 = For the second support, we note that the angle 30◦ is measured from the negative x-axis, so the angle needed to write T2 in component form is 150◦. Hence T2 = T2 cos (150◦), sin (150◦) √ = − T2 2 3, T2 2 The requirement w + T1 + T2 = 0 gives us this vector equation. 0, −50 +, T1 2 T1 2 − T1 2 T2 2 √ 3 + √ 3, T1 2 − √, 3 T2 2 T2 2 + 3 w + T1 + T2 = 0 √ T2 2 = 0, 0 − 50 = 0, 0 Equating the corresponding components of the vectors on each side, we get a system of linear equations in the variables T1 and T2.    (E1) (E2) T1 2 3 √ + T1 2 √ T2 2 3 − = 0 T2 2 − 50 = 0 From (E1), we get T1 = T2 which yields 2 T2 − 50 = 0. Hence, T2 = 25 pounds and T1 = T2 3. Substituting that into (E2) gives ( T2 √ 2 3 = 25 + T2 3 √ 3 pounds. 2 − 50 = 0 √ √ √ 3) 11.8 Vectors 11.8.1 Exercises 1027 In Exercises 1 - 10, use the given pair of vectors v and w to ﬁnd the following quantities. State whether the result is a vector or a scalar. • v + w • w − 2v • − wv • wˆv Finally, verify that the vectors satisfy the Parallelogram Law v2 + w2 = 1 2 v + w2 + v − w2 1. v = 12, −5, w = 3, 4 2. v = −7, 24, w = −5, −12 3. v = 2, −1, w = −2, 4 √ √ 3, 1, w = 2 5. v = − √ 7. v = 3. v = 4. v = 10, 4, w = −2,
5 61, − 3 2 √ 3 9. v = 3ˆı + 4ˆ, w = −2ˆ 10. v = 1 2 (ˆı + ˆ), w = 1 2 (ˆı − ˆ) In Exercises 11 - 25, ﬁnd the component form of the vector v using the information given about its magnitude and direction. Give exact values. 11. v = 6; when drawn in standard position v lies in Quadrant I and makes a 60◦ angle with the positive x-axis 12. v = 3; when drawn in standard position v lies in Quadrant I and makes a 45◦ angle with the positive x-axis 13. v = 2 3 ; when drawn in standard position v lies in Quadrant I and makes a 60◦ angle with the positive y-axis 14. v = 12; when drawn in standard position v lies along the positive y-axis 15. v = 4; when drawn in standard position v lies in Quadrant II and makes a 30◦ angle with the negative x-axis √ 16. v = 2 3; when drawn in standard position v lies in Quadrant II and makes a 30◦ angle with the positive y-axis 17. v = 7 2 ; when drawn in standard position v lies along the negative x-axis √ 18. v = 5 6; when drawn in standard position v lies in Quadrant III and makes a 45◦ angle with the negative x-axis 19. v = 6.25; when drawn in standard position v lies along the negative y-axis 1028 20. v = 4 √ 3; when drawn in standard position v lies in Quadrant IV and makes a 30◦ angle Applications of Trigonometry with the positive x-axis 21. v = 5 √ 2; when drawn in standard position v lies in Quadrant IV and makes a 45◦ angle with the negative y-axis 22. v = 2 √ 5; when drawn in standard position v lies in Quadrant I and makes an angle measuring arctan(2) with the positive x-axis √ 23. v = 10; when drawn in standard position v lies in Quadrant II and makes an angle measuring arctan(3) with the negative x-axis 24. v = 5; when drawn in standard position v lies in Quadrant III
and makes an angle measuring arctan 4 3 with the negative x-axis 25. v = 26; when drawn in standard position v lies in Quadrant IV and makes an angle measuring arctan 5 12 with the positive x-axis In Exercises 26 - 31, approximate the component form of the vector v using the information given about its magnitude and direction. Round your approximations to two decimal places. 26. v = 392; when drawn in standard position v makes a 117◦ angle with the positive x-axis 27. v = 63.92; when drawn in standard position v makes a 78.3◦ angle with the positive x-axis 28. v = 5280; when drawn in standard position v makes a 12◦ angle with the positive x-axis 29. v = 450; when drawn in standard position v makes a 210.75◦ angle with the positive x-axis 30. v = 168.7; when drawn in standard position v makes a 252◦ angle with the positive x-axis 31. v = 26; when drawn in standard position v makes a 304.5◦ angle with the positive x-axis In Exercises 32 - 52, for the given vector v, ﬁnd the magnitude v and an angle θ with 0 ≤ θ < 360◦ so that v = v cos(θ), sin(θ) (See Deﬁnition 11.8.) Round approximations to two decimal places. 32. v = 1, √ 35. v = − √ 2, 3 √ 2 33. v = 5, 5 36 38. v = 6, 0 41. v = −10ˆ 39. v = −2.5, 0 40. v = 0, 42. v = 3, 4 43. v = 12, 5 34. v = −2 √ 37. v = − 1 3 11.8 Vectors 1029 44. v = −4, 3 45. v = −7, 24 46. v = −2, −1 47. v = −2, −6 48. v = ˆı + ˆ 49. v = ˆı − 4ˆ 50. v = 123.4, −77.05 51. v = 965.15, 831.6 52. v = −114.1, 42.
3 53. 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. 54. 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. 55. If the captain of the HMS Sasquatch in Exercise 54 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 account 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. HINT: If v denotes the velocity of the HMS Sasquatch and w denotes the velocity of the current, what does v + w need to be to reach Chupacabra Cove in three hours? 56. 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. 57. 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. 58. 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. 59. 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. 1030 Applications of Trigonometry 60. 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 a 18◦ angle with the ceiling. What is the tension on each of the cables? Round your answers to three decimal places. 61. 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. 62. Emboldened by the success of their late night keg pull in Exercise 61 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? 63. Let v = v1, v2 be any non-zero vector. Show that 1 v v has length 1. 64. We say that two non-zero vectors v and w are parallel if they have same or opposite directions. That is, v = 0 and w = 0 are parallel if either ˆv = ˆw or ˆv = − ˆw. Show that this means v = k w for some non-zero scalar k and that k > 0 if the vectors have the same direction and k < 0 if they point in opposite directions. 65. The goal of this exercise is to use vectors to describe non-vertical lines in the plane. To that end, consider the line y = 2x − 4. Let v0 = 0, −4 and let s = 1, 2. Let t be any real number. Show that the vector deﬁned by v = v0 + ts, when drawn in standard position, has its terminal point on the line y = 2x − 4. (Hint: Show that v0 + ts = t, 2t − 4 for any real number t.) Now consider the non-vertical line y = mx + b. Repeat the previous analysis with v0 = 0, b and let s = 1, m. Thus any non-vertical line can be thought of as a collection of terminal points of the vector sum of 0, b (the position vector of the y-intercept) and a scalar multiple of the slope vector s = 1, m. 66. Prove the associative and identity properties of vector addition in Theorem 11.18. 67. Prove the properties of scalar multiplication in Theorem 11.19. 11.8 Vectors 11.8.2 Answers 1. v + w = 15, −1, vector √ v + w = 226, scalar 1031 w − 2v = −21, 14, vector v + w = 18, scalar v w − wv = −21, 77, vector wˆv = 60 13, − 25 13, vector 2. v + w = −12, 12, vector w − 2v = 9, −60, vector v + w = 12 2, scalar √ v w − wv = −34, −612, vector v + w = 38, scalar w
ˆv = − 91 25, 312 25, vector 3. v + w = 0, 3, vector v + w = 3, scalar w − 2v = −6, 6, vector v + w = 3 √ 5, scalar v w − wv = −6 √ √ 5, 6 5, vector wˆv = 4, −2, vector 4. v + w = 8, 9, vector 5. 6. v + w = √ v w − wv = −14 v + w = √ 145, scalar √ 3, 3, vector √ 3, scalar v + w = 2 √ 29, vector 29, 6 v w − wv = 8 √ 3, 0, vector, scalar, vector v w − wv = − 7 5, − 1 5, vector w − 2v = −22, −3, vector v + w = 3 √ 29, scalar wˆv = 5, 2, vector w − 2v = 4 √ 3, 0, vector v + w = 6, scalar wˆv = −2 √ 3, 2, vector w − 2v = −2, −1, vector v + w = 2, scalar wˆv = 3 5 5, 4, vector √ √ − 3 2 2, 3 2 2 7. v + w = 0, 0, vector w − 2v =, vector v + w = 0, scalar v w − wv = − √ √ 2, 2, vector v + w = 2, scalar wˆv = √ 2 2, − √ 2 2, vector 1032 Applications of Trigonometry 8, vector w − 2v = −2, −2 √ 3, vector v + w = 1, scalar v w − wv = −2, −2 √ 3, vector v + w = 3, scalar wˆv = 1, √ 3, vector 9. v + w = 3, 2, vector v + w = √ 13, scalar w − 2v = −6, −10, vector v + w = 7, scalar v w − wv = −6, −18, vector 10. v + w = 1, 0, vector v + w = 1, scalar v w − wv = 0, −
√ 2 2, vector, vector wˆv = 6 5, 8 5 w − 2v = −, scalar, vector wˆv = 1 2, 1 2, vector 11. v = 3, 3 √ 3 14. v = 0, 12 17. v = − 7 2, 0 √ 20. v = 6, −2 3 12. v = √ 3 2 2, 3 √ 15. v = −2 3, 2 √ 2 2 13. v = √ 3 3 3, 1 √ 16. v = − 3, 3 18. v = −5 √ √ 3 3, −5 19. v = 0, −6.25 21. v = 5, −5 22. v = 2, 4 23. v = −1, 3 24. v = −3, −4 25. v = 24, −10 26. v ≈ −177.96, 349.27 27. v ≈ 12.96, 62.59 28. v ≈ 5164.62, 1097.77 29. v ≈ −386.73, −230.08 32. v = 2, θ = 60◦ 30. v ≈ −52.13, −160.44 √ 33. v = 5 2, θ = 45◦ 31. v ≈ 14.73, −21.43 34. v = 4, θ = 150◦ 35. v = 2, θ = 135◦ 36. v = 1, θ = 225◦ 38. v = 6, θ = 0◦ 39. v = 2.5, θ = 180◦ 41. v = 10, θ = 270◦ 42. v = 5, θ ≈ 53.13◦ 44. v = 5, θ ≈ 143.13◦ 45. v = 25, θ ≈ 106.26◦ 37. v = 1, θ = 240◦ √ 40. v = 7, θ = 90◦ 43. v = 13, θ ≈ 22.62◦ √ 46. v = 5, θ ≈ 206.57◦ 11.8 Vectors 1033 47. v = 2 √ 10, θ ≈ 251.57◦ 48. v = √ 2, θ �
� 45◦ √ 49. v = 17, θ ≈ 284.04◦ 50. v ≈ 145.48, θ ≈ 328.02◦ 51. v ≈ 1274.00, θ ≈ 40.75◦ 52. v ≈ 121.69, θ ≈ 159.66◦ 53. The boat’s true speed is about 10 miles per hour at a heading of S50.6◦W. 54. The HMS Sasquatch’s true speed is about 41 miles per hour at a heading of S26.8◦E. 55. She should maintain a speed of about 35 miles per hour at a heading of S11.8◦E. 56. She should ﬂy at 83.46 miles per hour with a heading of N22.1◦E 57. The current is moving at about 10 miles per hour bearing N54.6◦W. 58. The tension on each of the cables is about 346 pounds. 59. The maximum weight that can be held by the cables in that conﬁguration is about 133 pounds. 60. The tension on the left hand cable is 285.317 lbs. and on the right hand cable is 92.705 lbs. 61. The weaker student should pull about 60 pounds. The net force on the keg is about 153 pounds. 62. The resultant force is only about 296 pounds so the couch doesn’t budge. Even if it did move, the stronger force on the third rope would have made the couch drift slightly to the south as it traveled down the street. 1034 Applications of Trigonometry 11.9 The Dot Product and Projection In Section 11.8, we learned how add and subtract vectors and how to multiply vectors by scalars. In this section, we deﬁne a product of vectors. We begin with the following deﬁnition. Deﬁnition 11.11. Suppose v and w are vectors whose component forms are v = v1, v2 and w = w1, w2. The dot product of v and w is given by v · w = v1, v2 · w1, w2 = v1w1 + v2w2 For example, let v = 3, 4 and w = 1, −2. Then v · w = 3, 4 · 1, −2 =
(3)(1) + (4)(−2) = −5. Note that the dot product takes two vectors and produces a scalar. For that reason, the quantity v· w is often called the scalar product of v and w. The dot product enjoys the following properties. Theorem 11.22. Properties of the Dot Product Commutative Property: For all vectors v and w, v · w = w · v. Distributive Property: For all vectors u, v and w, u · (v + w) = u · v + u · w. Scalar Property: For all vectors v and w and scalars k, (kv) · w = k(v · w) = v · (k w). Relation to Magnitude: For all vectors v, v · v = v2. Like most of the theorems involving vectors, the proof of Theorem 11.22 amounts to using the deﬁnition of the dot product and properties of real number arithmetic. To show the commutative property for instance, let v = v1, v2 and w = w1, w2. Then v · w = v1, v2 · w1, w2 = v1w1 + v2w2 Deﬁnition of Dot Product = w1v1 + w2v2 = w1, w2 · v1, v2 Deﬁnition of Dot Product = w · v Commutativity of Real Number Multiplication The distributive property is proved similarly and is left as an exercise. For the scalar property, assume that v = v1, v2 and w = w1, w2 and k is a scalar. Then (kv) · w = (k v1, v2) · w1, w2 = kv1, kv2 · w1, w2 = (kv1)(w1) + (kv2)(w2) Deﬁnition of Dot Product Deﬁnition of Scalar Multiplication = k(v1w1) + k(v2w2) Associativity of Real Number Multiplication = k(v1w1 + v2w2) = k v1, v2 · w1, w2 = k(v · w) Distributive Law of Real Numbers Deﬁnition of Dot Product We leave the
proof of k(v · w) = v · (k w) as an exercise. 11.9 The Dot Product and Projection 1035 For the last property, we note that if v = v1, v2, then v · v = v1, v2 · v1, v2 = v2 where the last equality comes courtesy of Deﬁnition 11.8. The following example puts Theorem 11.22 to good use. As in Example 11.8.3, we work out the problem in great detail and encourage the reader to supply the justiﬁcation for each step. 2 = v2, 1 + v2 Example 11.9.1. Prove the identity: v − w2 = v2 − 2(v · w) + w2. Solution. We begin by rewriting v − w2 in terms of the dot product using Theorem 11.22. v − w2 = (v − w) · (v − w) = (v + [− w]) · (v + [− w]) = (v + [− w]) · v + (v + [− w]) · [− w] = v · (v + [− w]) + [− w] · (v + [− w]) = v · v + v · [− w] + [− w] · v + [− w] · [− w] = v · v + v · [(−1) w] + [(−1) w] · v + [(−1) w] · [(−1) w] = v · v + (−1)(v · w) + (−1)( w · v) + [(−1)(−1)]( w · w) = v · v + (−1)(v · w) + (−1)(v · w(v · w) + w · w = v2 − 2(v · w) + w2 Hence, v − w2 = v2 − 2(v · w) + w2 as required. If we take a step back from the pedantry in Example 11.9.1, we see that the bulk of the work is needed to show that (v − w)·(v − w) = v ·v −2(v · w)+ w· w. If this looks familiar, it should. Since the dot product enjoys many of the same properties enjoyed by real numbers
, the machinations required to expand (v − w) · (v − w) for vectors v and w match those required to expand (v − w)(v − w) for real numbers v and w, and hence we get similar looking results. The identity veriﬁed in Example 11.9.1 plays a large role in the development of the geometric properties of the dot product, which we now explore. Suppose v and w are two nonzero vectors. If we draw v and w with the same initial point, we deﬁne the angle between v and w to be the angle θ determined by the rays containing the vectors v and w, as illustrated below. We require 0 ≤ θ ≤ π. (Think about why this is needed in the deﬁnition.) The following theorem gives us some insight into the geometric role the dot product plays. Theorem 11.23. Geometric Interpretation of Dot Product: vectors then v · w = v w cos(θ), where θ is the angle between v and w. If v and w are nonzero 1036 Applications of Trigonometry We prove Theorem 11.23 in cases. If θ = 0, then v and w have the same direction. It follows1 that there is a real number k > 0 so that w = kv. Hence, v · w = v · (kv) = k(v · v) = kv2 = kvv. Since k > 0, k = \|k\|, so kv = \|k\|v = kv by Theorem 11.20. Hence, kvv = v(kv) = vkv = v w. Since cos(0) = 1, we get v · w = kvv = v w = v w cos(0), proving that the formula holds for θ = 0. If θ = π, we repeat the argument with the diﬀerence being w = kv where k < 0. In this case, \|k\| = −k, so kv = −\|k\|v = −kv = − w. Since cos(π) = −1, we get v · w = −v w = v w cos(π), as required. Next, if 0 < θ < π, the vectors v, w and v − w determine a triangle with side lengths v, w and v − w, respectively
, as seen below The Law of Cosines yields v − w2 = v2 + w2 − 2v w cos(θ). From Example 11.9.1, we know v − w2 = v2 − 2(v · w) + w2. Equating these two expressions for v − w2 gives v2 + w2 −2v w cos(θ) = v2 −2(v· w)+ w2 which reduces to −2v w cos(θ) = −2(v· w), or v · w = v w cos(θ), as required. An immediate consequence of Theorem 11.23 is the following. Theorem 11.24. Let v and w be nonzero vectors and let θ the angle between v and w. Then θ = arccos v · w v w = arccos(ˆv · ˆw) We obtain the formula in Theorem 11.24 by solving the equation given in Theorem 11.23 for θ. Since v and w are nonzero, so are v and w. Hence, we may divide both sides of v · w = v w cos(θ) by v w to get cos(θ) = v· w v w. Since 0 ≤ θ ≤ π by deﬁnition, the values of θ exactly match the. Using Theorem 11.22, we can rewrite v· w range of the arccosine function. Hence, θ = arccos v w = ˆv · ˆw, giving us the alternative formula θ = arccos(ˆv · ˆw). v· w v w = 1 vv 1 w w · We are overdue for an example. Example 11.9.2. Find the angle between the following pairs of vectors. 1. v = 3, −3 √ 3, and w = − 3, 1 √ 2. v = 2, 2, and w = 5, −5 3. v = 3, −4, and w = 2, 1 Solution. We use the formula θ = arccos v· w v w from Theorem 11.24 in each case below. 1Since v = vˆv and w = w ˆw, if ˆv = ˆw then w = wˆv = w v (
vˆv) = w v v. In this case, k = w v > 0. 1037 √ 3)2 = 11.9 The Dot Product and Projection 1. We have v· w = 3, −3 √ 3·− √ 3, 1 = −3 √ 36 = 6 and w = (− 3)2 + 12 = √ √ √ √ 3−3 3 = −6 √ 4 = 2, θ = arccos 3. Since v = −6 12 = arccos √ 3 32 + (−3 √ − 3 2 = 5π 6. 2. For v = 2, 2 and w = 5, −5, we ﬁnd v · w = 2, 2 · 5, −5 = 10 − 10 = 0. Hence, it doesn’t v· w matter what v and w are,2 θ = arccos = arccos(0) = π 2. v w 3. We ﬁnd v · w = 3, −4 · 2, 1 = 6 − 4 = 2. Also v = √ √ w = 22 + 12 = 5, so θ = arccos = arccos isn’t the cosine of one 32 + (−4)2 = √ 25 = 5 and 2 √ 5 5 of the common angles, we leave our answer as θ = arccos 2 √ 5 25 √ 5 25. Since 2 2 √ 5 25. The vectors v = 2, 2, and w = 5, −5 in Example 11.9.2 are called orthogonal and we write v ⊥ w, because the angle between them is π 2 radians = 90◦. Geometrically, when orthogonal vectors are sketched with the same initial point, the lines containing the vectors are perpendicular. w v v and w are orthogonal, v ⊥ w We state the relationship between orthogonal vectors and their dot product in the following theorem. Theorem 11.25. The Dot Product Detects Orthogonality: Let v and w be nonzero vectors. Then v ⊥ w if and only if v · w = 0. To prove Theorem 11.25, we ﬁrst assume v and w are nonzero vectors with v ⊥ w. By de�
�nition, = 0. Conversely, the angle between v and w is π v· w if v and w are nonzero vectors and v · w = 0, then Theorem 11.24 gives θ = arccos = 2. By Theorem 11.23, v · w = v w cos π 2 v w arccos to provide a diﬀerent proof about the relationship between the slopes of perpendicular lines.3 2, so v ⊥ w. We can use Theorem 11.25 in the following example v w = arccos(0) = π 0 Example 11.9.3. Let L1 be the line y = m1x + b1 and let L2 be the line y = m2x + b2. Prove that L1 is perpendicular to L2 if and only if m1 · m2 = −1. Solution. Our strategy is to ﬁnd two vectors: v1, which has the same direction as L1, and v2, which has the same direction as L2 and show v1 ⊥ v2 if and only if m1m2 = −1. To that end, we substitute x = 0 and x = 1 into y = m1x + b1 to ﬁnd two points which lie on L1, namely P (0, b1) 2Note that there is no ‘zero product property’ for the dot product since neither v nor w is 0, yet v · w = 0. 3See Exercise 2.1.1 in Section 2.1. 1038 Applications of Trigonometry −−→ P Q = 1 − 0, (m1 + b1) − b1 = 1, m1, and note that since v1 is and Q(1, m1 + b1). We let v1 = determined by two points on L1, it may be viewed as lying on L1. Hence it has the same direction as L1. Similarly, we get the vector v2 = 1, m2 which has the same direction as the line L2. Hence, L1 and L2 are perpendicular if and only if v1 ⊥ v2. According to Theorem 11.25, v1 ⊥ v2 if and only if v1 · v2 = 0. Notice that v1 · v2 = 1, m1 · 1, m2 = 1 +
m1m2. Hence, v1 · v2 = 0 if and only if 1 + m1m2 = 0, which is true if and only if m1m2 = −1, as required. While Theorem 11.25 certainly gives us some insight into what the dot product means geometrically, there is more to the story of the dot product. Consider the two nonzero vectors v and w drawn with a common initial point O below. For the moment, assume that the angle between v and w, which we’ll denote θ, is acute. We wish to develop a formula for the vector p, indicated below, which is called the orthogonal projection of v onto w. The vector p is obtained geometrically as follows: drop a perpendicular from the terminal point T of v to the vector w and call the point −−→ of intersection R. The vector p is then deﬁned as p = OR. Like any vector, p is determined by its magnitude p and its direction ˆp according to the formula p = pˆp. Since we want ˆp to have the same direction as w, we have ˆp = ˆw. To determine p, we make use of Theorem 10.4 as applied to the right triangle ORT. We ﬁnd cos(θ) = p v, or p = v cos(θ). To get things in terms of just v and w, we use Theorem 11.23 to get p = v cos(θ) = v w cos(θ) w. Using Theorem 11.22, we rewrite v· w = v · ˆw. Hence, p = v · ˆw, and since ˆp = ˆw, we now have a formula for p completely in terms of v and w, namely p = pˆp = (v · ˆw) ˆw. w = v · = v −−→ OR p = v w θ R p O Now suppose that the angle θ between v and w is obtuse, and consider the diagram below. In this case, we see that ˆp = − ˆw and using the triangle ORT, we ﬁnd p = v cos(θ). Since θ +θ = π, it follows that cos(
θ) = − cos(θ), which means p = v cos(θ) = −v cos(θ). Rewriting this last equation in terms of v and w as before, we get p = −(v · ˆw). Putting this together with ˆp = − ˆw, we get p = pˆp = −(v · ˆw)(− ˆw) = (v · ˆw) ˆw in this case as well. 11.9 The Dot Product and Projection 1039 T v w θ O θ −−→ OR p = R If the angle between v and w is π v · w = 0. It follows that v · ˆw = 0 and p = 0 = 0 ˆw = (v · ˆw) ˆw in this case, too. This gives us 2 then it is easy to show4 that p = 0. Since v ⊥ w in this case, Deﬁnition 11.12. Let v and w be nonzero vectors. The orthogonal projection of v onto w, denoted proj w(v) is given by proj w(v) = (v · ˆw) ˆw. Deﬁnition 11.12 gives us a good idea what the dot product does. The scalar v · ˆw is a measure of how much of the vector v is in the direction of the vector w and is thus called the scalar projection of v onto w. While the formula given in Deﬁnition 11.12 is theoretically appealing, because of the presence of the normalized unit vector ˆw, computing the projection using the formula proj w(v) = (v · ˆw) ˆw can be messy. We present two other formulas that are often used in practice. Theorem 11.26. Alternate Formulas for Vector Projections: If v and w are nonzero vectors then proj w(v) = (v · ˆw) ˆw = v · w w2 w = v · w w · w w The proof of Theorem 11.26, which we leave to the reader as an exercise, amounts to using the formula ˆw = w and properties of the dot product. It is
time for an example. 1 w Example 11.9.4. Let v = 1, 8 and w = −1, 2. Find p = proj w(v), and plot v, w and p in standard position. Solution. We ﬁnd v · w = 1, 8 · −1, 2 = (−1) + 16 = 15 and w · w = −1, 2 · −1, 2 = 1 + 4 = 5. Hence, p = v· w 5 −1, 2 = −3, 6. We plot v, w and p below. w· w w = 15 4In this case, the point R coincides with the point O, so p = −→ OR = −−→ OO = 0. 1040 Applications of Trigonometry 3 −2 −1 1 Suppose we wanted to verify that our answer p in Example 11.9.4 is indeed the orthogonal projection of v onto w. We ﬁrst note that since p is a scalar multiple of w, it has the correct direction, so what remains to check is the orthogonality condition. Consider the vector q whose initial point is the terminal point of p and whose terminal point is the terminal point of v3 −2 −1 1 From the deﬁnition of vector arithmetic, p + q = v, so that q = v − p. In the case of Example 11.9.4, v = 1, 8 and p = −3, 6, so q = 1, 8−−3, 6 = 4, 2. Then q· w = 4, 2·−1, 2 = (−4)+4 = 0, which shows q ⊥ w, as required. This result is generalized in the following theorem. Theorem 11.27. Generalized Decomposition Theorem: Let v and w be nonzero vectors. There are unique vectors p and q such that v = p + q where p = k w for some scalar k, and q · w = 0. Note that if the vectors p and q in Theorem 11.27 are nonzero, then we can say p is parallel 5 to w and q is orthogonal to w. In this case, the vector p is sometimes called the ‘vector component of v parallel to w’ and q is called the ‘vector component of v orthogonal to w.’ To prove Theorem 11.
27, we take p = proj w(v) and q = v − p. Then p is, by deﬁnition, a scalar multiple of w. Next, we compute q · w. 5See Exercise 64 in Section 11.8. 11.9 The Dot Product and Projection 1041 q · w = (v − p) · w Deﬁnition of q. = v · w − p · w Properties of Dot Product · w Since p = proj w(v). ( w · w) Properties of Dot Product Hence, q · w = 0, as required. At this point, we have shown that the vectors p and q guaranteed by Theorem 11.27 exist. Now we need to show that they are unique. Suppose v = p + q = p + q where the vectors p and q satisfy the same properties described in Theorem 11.27 as p and q. Then p − p = q − q, so w · (p − p ) = w · (q − q. Hence, w · (p − p ) = 0. Now there are scalars k and k so that p = k w and p = k w. This means w · (p − p ) = w · (k w − k w) = w · ([k − k ] w) = (k − k )( w · w) = (k − k ) w2. Since w = 0, w2 = 0, which means the only way w · (p − p ) = (k − k ) w2 = 0 is for k − k = 0, or k = k. This means p = k w = k w = p. With, it must be that q = q as well. Hence, we have shown there is only one way to write v as a sum of vectors as described in Theorem 11.27. We close this section with an application of the dot product. In Physics, if a constant force F is exerted over a distance d, the work W done by the force is given by W = F d. Here, we assume 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 F is applied to move an object from the point P to the point Q. F F θ �
� P Q To ﬁnd the work W done in this scenario, we need to ﬁnd how much of the force F is in the −−→ P Q. This is precisely what the dot product F · P Q represents. Since direction of the motion −−→ the distance the object travels is P QP Q, −−→ −−→ W = ( F · P Q) P Q = F · ( P Q cos(θ), where θ is the angle between the applied force F and the trajectory of the motion −−→ P Q, we get W = ( F · P Q) −−→ P Q = F −−→ P Q. We have proved the following. −−→ P QP Q) = F · −−→ P Q. Since −−→ P Q = 1042 Applications of Trigonometry Theorem 11.28. Work as a Dot Product: Suppose a constant force F is applied along the vector −−→ P Q. The work W done by F is given by where θ is the angle between F and −−→ P Q = F −−→ P Q cos(θ), W = F · −−→ P Q. Example 11.9.5. 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. 30◦ −−→ Solution. There are two ways to attack this problem. One way is to ﬁnd the vectors F and P Q −−→ mentioned in Theorem 11.28 and compute W = F · P Q. 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 F = 10. Since it is being applied at a constant angle of θ = 30◦ with respect to the positive x-axis, Deﬁnition √ 11.8 gives us F = 10 cos(30◦, sin(30◦) = 5 3, 5. Since the wagon is being pulled along 50 −−→ P Q =
50ˆı = 50 1, 0 = 50, 0. We get feet in the positive direction, the displacement vector is W = F · 3. Since force is measured in pounds and distance is measured in feet, we get W = 250 3 foot-pounds. Alternatively, we can use the formulation W = F −−→ P Q cos(θ) to get W = (10 pounds)(50 feet) cos (30◦) = 250 3, 5 · 50, 0 = 250 3 foot-pounds of work. −−→ P Q = 5 √ √ √ √ 11.9 The Dot Product and Projection 1043 11.9.1 Exercises In Exercises 1 - 20, use the pair of vectors v and w to ﬁnd the following quantities. v · w The angle θ (in degrees) between v and w proj w(v) q = v − proj w(v) (Show that q · w = 0.) 1. v = −2, −7 and w = 5, −9 2. v = −6, −5 and w = 10, −12 3. v = 1, √ 3 and w = 1, − √ 3 5. v = −2, 1 and w = 3, 6 4. v = 3, 4 and w = −6, −8 6. v = −3 √ 3, 3 and w = − 3, −1 √ 7. v = 1, 17 and w = −1, 0 8. v = 3, 4 and w = 5, 12 9. v = −4, −2 and w = 1, −5 10. v = −5, 6 and w = 4, −7 11. v = −8, 3 and w = 2, 6 12. v = 34, −91 and w = 0, 1 13. v = 3ˆı − ˆ and w = 4ˆ 14. v = −24ˆı + 7ˆ and w = 2ˆı 15. v = 3 17. v = 19. v = 2ˆı + 3 1 2, 2 ˆ and w = ˆı − ˆ √ √ 2 2, and and 16. v = 5ˆı + 12ˆ
and w = −3ˆı + 4ˆ 18. v = 20 and w = 1 2, − √ 3 2 and 21. A force of 1500 pounds is required to tow a trailer. Find the work done towing the trailer along a ﬂat stretch of road 300 feet. Assume the force is applied in the direction of the motion. 22. 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. 23. 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. 24. In Exercise 61 in Section 11.8, 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. 1044 Applications of Trigonometry 25. 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! 26. Prove the distributive property of the dot product in Theorem 11.22. 27. Finish the proof of the scalar property of the dot product in Theorem 11.22. 28. Use the identity in Example 11.9.1 to prove the Parallelogram Law v2 + w2 = 1 2 v + w2 + v − w2 29. We know that \|x + y\| ≤ \|x\| + \|y\| for all real numbers x and y by the Triangle Inequality established in Exercise 36 in Section 2.2. We can now
establish a Triangle Inequality for vectors. In this exercise, we prove that u + v ≤ u + v for all pairs of vectors u and v. (a) (Step 1) Show that u + v2 = u2 + 2u · v + v2. (b) (Step 2) Show that \|u · v\| ≤ uv. This is the celebrated Cauchy-Schwarz Inequality.6 (Hint: To show this inequality, start with the fact that \|u · v\| = \| uv cos(θ) \| and use the fact that \| cos(θ)\| ≤ 1 for all θ.) (c) (Step 3) Show that u + v2 = u2 + 2u · v + v2 ≤ u2 + 2\|u · v\| + v2 ≤ u2 + 2uv + v2 = (u + v)2. (d) (Step 4) Use Step 3 to show that u + v ≤ u + v for all pairs of vectors u and v. (e) As an added bonus, we can now show that the Triangle Inequality \|z + w\| ≤ \|z\| + \|w\| holds for all complex numbers z and w as well. Identify the complex number z = a + bi with the vector u = a, b and identify the complex number w = c + di with the vector v = c, d and just follow your nose! 6It is also known by other names. Check out this site for details. 11.9 The Dot Product and Projection 1045 11.9.2 Answers 1. v = −2, −7 and w = 5, −9 2. v = −6, −5 and w = 10, −12 v · w = 53 θ = 45◦ proj w(v = 90◦ proj w(v) = 0, 0 q = −6, −5 3. v = 1, √ 3 and w = 1, − √ 3 4. v = 3, 4 and w = −6, −8 v · w = −2 θ = 120◦ proj w(v50 θ = 180◦ proj w(v) = 3, 4 q = 0, 0 5. v = −2, 1 and w = 3, 6 v · w = 0 θ = 90◦ proj w(v) = 0, 0
q = −2, 1 3, 3 and w = − 3, −1 √ √ 6. v = −3 v · w = 6 θ = 60◦ proj w(v. v = 1, 17 and w = −1, 0 8. v = 3, 4 and w = 5, 12 v · w = −1 θ ≈ 93.37◦ proj w(v) = 1, 0 q = 0, 17 v · w = 63 θ ≈ 14.25◦ proj w(v) = 315 169, 756 q = 192 169, − 80 169 169 9. v = −4, −2 and w = 1, −5 10. v = −5, 6 and w = 4, −7 v · w = 6 θ ≈ 74.74◦ proj w(v) = 3 q = − 55 13, − 11 13 13, − 15 13 v · w = −62 θ ≈ 169.94◦ proj w(v) = − 248 q = − 77 65, − 44 65 65, 434 65 1046 Applications of Trigonometry 11. v = −8, 3 and w = 2, 6 12. v = 34, −91 and w = 0, 1 v · w = 2 θ ≈ 87.88◦ proj w(v) = 1 10, 3 q = − 81 10, 27 10 10 v · w = −91 θ ≈ 159.51◦ proj w(v) = 0, −91 q = 34, 0 13. v = 3ˆı − ˆ and w = 4ˆ 14. v = −24ˆı + 7ˆ and w = 2ˆı v · w = −4 θ ≈ 108.43◦ proj w(v) = 0, −1 q = 3, 0 15. v = 3 2 ˆ and w = ˆı − ˆ 2ˆı + 3 v · w = 0 θ = 90◦ proj w(v) = 048 θ ≈ 163.74◦ proj w(v) = −24, 0 q = 0, 7 16. v = 5ˆı + 12ˆ and w
= −3ˆı + 4ˆ v · w = 33 θ ≈ 59.49◦ proj w(v) = − 99 q = 224 25, 168 25 25, 132 25 17 = 75◦ and − 4 √ √ 3−1 4 3, proj w(v) = 1+ 4 q = √ 3 1− 4 √, 1+ 4 3 18 = 105◦ and − 4 6 proj w(v) = √ 3 √ q = 2+ 8 √ √ 2− 8 6 √ √ 2+ 8 6, √ 6 2 20. v = 1 and 19. v = √ 3 2, 1 2 √ and w = √ v · w = − θ = 165◦ 2 6+ 4 √ proj w(v) = √ q = 3−1 4, 1− 4 3+1 4 √ 3 √, 3++ 4 v · w = θ = 15◦ √ proj w(v) = 1− 4 q = √ 3, 1− 4 3+1 4 √ 3 √ 3+1 4, − 21. (1500 pounds)(300 feet) cos (0◦) = 450, 000 foot-pounds 22. (10 pounds)(3 feet) cos (0◦) = 30 foot-pounds 11.9 The Dot Product and Projection 1047 23. (13 pounds)(25 feet) cos (15◦) ≈ 313.92 foot-pounds 24. (100 pounds)(42 feet) cos (13◦) ≈ 4092.35 foot-pounds 25. (200 pounds)(10 feet) cos (77.5◦) ≈ 432.88 foot-pounds 1048 Applications of Trigonometry 11.10 Parametric Equations As we have seen in Exercises 53 - 56 in Section 1.2, Chapter 7 and most recently in Section 11.5, 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. y 5 4 3 2 1 P (x, y) = (f (t), g(t)) Q O x 1 2 3 4 5 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 P (x, y) 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. (Traditionally, f (t) is used for x and g(t) is used for y.) The independent variable t in this case is called a parameter and the system of equations x = f (t) y = g(t) is called a system of parametric equations or a parametrization of the curve C.1 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 {x = cos(t), y = sin(t) parametrizes the Unit Circle, giving it a counter-clockwise orientation. More generally, the equations of circular motion {x = r cos(ωt), y = r sin(ωt) developed on page 732 in Section 10.2.1 are parametric equations which trace out a circle of radius r centered at the origin. If ω > 0, the orientation is counterclockwise; if ω < 0, the orientation is clockwise. The angular frequency ω determines ‘how fast’ the 1Note the use of the indeﬁnite article
‘a’. As we shall see, there are inﬁnitely many diﬀerent parametric represen- tations for any given curve. 2Here, 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.) 11.10 Parametric Equations 1049 object moves around the circle. In particular, the equations x = 2960 cos π 12 t that model the motion of Lakeland Community College as the earth rotates (see Example 10.2.7 in Section 10.2) parameterize a circle of radius 2960 with a counter-clockwise rotation which completes one revolution as t runs through the interval [0, 24). It is time for another example. 12 t, y = 2960 sin π Example 11.10.1. Sketch the curve described by x = t2 − 3 y = 2t − 1 for t ≥ −2. Solution. We follow the same procedure here as we have time and time again when asked to graph anything new – choose friendly values of t, plot the corresponding points and connect the results in a pleasing fashion. Since we are told t ≥ −2, 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. t −2 −1 0 1 2 3 x(t) y(t) 1 −5 −2 −3 −3 −1 −2 1 3 1 5 6 (x(t), y(t)) (1, −5) (−2, −3) (−3, −1) (−2, 1) (1, 3) (6, 5) y 5 4 3 2 1 −2−1 −1 1 2 3 4 5 6 x −2 −3 −5 2. Substituting this into the equation x = t2 − 3 yields x = The curve sketched out in Example 11.10.1 certainly looks like a parabola, and the presence of the t2 term in the equation x = t2 − 3 reinforces this hunch. Since the parametric equations x = t2 − 3, y = 2t − 1 given to describe this curve are a system of equations, we can use the technique of substitution as
described in Section 8.7 to eliminate the parameter t and get an equation involving just x and y. To do so, we choose to solve the equation y = 2t − 1 for t to get t = y+1 − 3 or, after some rearrangement, (y + 1)2 = 4(x + 3). Thinking back to Section 7.3, we see that the graph of this equation is a parabola with vertex (−3, −1) which opens to the right, as required. Technically speaking, the equation (y + 1)2 = 4(x + 3) describes the entire parabola, while the parametric equations x = t2 − 3, y = 2t − 1 for t ≥ −2 describe only a portion of the parabola. In this case,3 we can remedy this situation by restricting the bounds on y. Since the portion of the parabola we want is exactly the part where y ≥ −5, the equation (y + 1)2 = 4(x + 3) coupled with the restriction y ≥ −5 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. y+1 2 2 Eliminating the parameter and obtaining an equation in terms of x and y, whenever possible, can be a great help in graphing curves determined by parametric equations. If the system of parametric equations contains algebraic functions, as was the case in Example 11.10.1, then the usual techniques of substitution and elimination as learned in Section 8.7 can be applied to the 3We will have an example shortly where no matter how we restrict x and y, we can never accurately describe the curve once we’ve eliminated the parameter. 1050 Applications of Trigonometry system {x = f (t), y = g(t) to eliminate the parameter. If, on the other hand, the parametrization In this case, it is often best involves the trigonometric functions, the strategy changes slightly. to solve for the trigonometric functions and relate them using an identity. We demonstrate these techniques in the following example. Example 11.10.2. Sketch the curves described by the following parametric equations. x = t3 y = 2t2 x = e−t y = e−2t 1. 2. for −1 ≤ t ≤ 1 for t ≥ 0 Solution. 3. 4
. for 0 < t < π x = sin(t) y = csc(t) x = 1 + 3 cos(t) y = 2 sin(t) for 0 ≤ t ≤ 3π 2 1. To get a feel for the curve described by the system x = t3, y = 2t2 we ﬁrst sketch the graphs of x = t3 and y = 2t2 over the interval [−1, 1]. We note that as t takes on values in the interval [−1, 1], x = t3 ranges between −1 and 1, and y = 2t2 ranges between 0 and 2. This means that all of the action is happening on a portion of the plane, namely {(x, y) \| − 1 ≤ x ≤ 1, 0 ≤ y ≤ 2}. Next, we plot a few points to get a sense of the position and orientation of the curve. Certainly, t = −1 and t = 1 are good values to pick since these are the extreme values of t. We also choose t = 0, since that corresponds to a relative minimum4 on the graph of y = 2t2. Plugging in t = −1 gives the point (−1, 2), t = 0 gives (0, 0) and t = 1 gives (1, 2). More generally, we see that x = t3 is increasing over the entire interval [−1, 1] whereas y = 2t2 is decreasing over the interval [−1, 0] and then increasing over [0, 1]. Geometrically, this means that in order to trace out the path described by the parametric equations, we start at (−1, 2) (where t = −1), then move to the right (since x is increasing) and down (since y is decreasing) to (0, 0) (where t = 0). We continue to move to the right (since x is still increasing) but now move upwards (since y is now increasing) until we reach (1, 2) (where t = 1). Finally, to get a good sense of the shape of the curve, we √ eliminate the parameter. Solving x = t3 for t, we get t = 3 x. Substituting this into y = 2t2 √ x)2 = 2x2/3. Our experience in Section 5.3 yields the graph of our ﬁnal answer gives y
= 2( 3 below. x 1 −1 1 t −1 y 2 1 y 2 1 −1 1 t −1 1 x x = t3, −1 ≤ t ≤ 1 y = 2t2, −1 ≤ t ≤ 1 x = t3, y = 2t2, −1 ≤ t ≤ 1 4You should review Section 1.6.1 if you’ve forgotten what ‘increasing’, ‘decreasing’ and ‘relative minimum’ mean. 11.10 Parametric Equations 1051 9 3, 1 and for t = ln(3) we get 2 2. For the system x = 2e−t, y = e−2t for t ≥ 0, we proceed as in the previous example and graph x = 2e−t and y = e−2t over the interval [0, ∞). We ﬁnd that the range of x in this case is (0, 2] and the range of y is (0, 1]. 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. Starting with t = ln(1) = 0 we get5 (2, 1), for t = ln(2) we get 1, 1. Since t is ranging over the unbounded interval [0, ∞), 4 we take the time to analyze the end behavior of both x and y. As t → ∞, x = 2e−t → 0+ and y = e−2t → 0+ as well. This means the graph of x = 2e−t, y = e−2t approaches the point (0, 0). Since both x = 2e−t and y = e−2t are always decreasing for t ≥ 0, we know that our ﬁnal graph will start at (2, 1) (where t = 0), and move consistently to the left (since x is decreasing) and down (since y is decreasing) to approach the origin. To eliminate the parameter, one way to proceed is to solve x = 2e−t for t to get t = − ln x. Substituting 2 = x2 this for t in y = e−2t gives y = e−2(− ln(x/2))
= e2 ln(x/2) = eln(x/2)2 4. Or, we could recognize that y = e−2t = e−t2 2 2, we get y = x = x2 4 this way as well. Either way, the graph of x = 2e−t, y = e−2t for t ≥ 0 is a portion of the parabola y = x2 4 which starts at the point (2, 1) and heads towards, but never reaches,6 (0, 0). = x 2, and since x = 2e−t means e− = 2e−t, t ≥ 0 y = e−2t, t ≥ 0 x = 2e−t, y = e−2t, t ≥ 0 6 gives the point 1 3. For the system {x = sin(t), y = csc(t) for 0 < t < π, we start by graphing x = sin(t) and y = csc(t) over the interval (0, π). We ﬁnd that the range of x is (0, 1] while the range of 2, 2, t = π y is [1, ∞). Plotting a few friendly points, we see that t = π 2, 2. Since t = 0 and t = π aren’t included in the 6 returns us to 1 gives (1, 1) and t = 5π domain for t, (because y = csc(t) is undeﬁned at these t-values), we analyze the behavior of the system as t approaches 0 and π. We ﬁnd that as t → 0+ as well as when t → π−, we get x = sin(t) → 0+ and y = csc(t) → ∞. Piecing all of this information together, we get that for t near 0, we have points with very small positive x-values, but very large positive y-values. As t ranges through the interval 0, π, x = sin(t) is increasing and y = csc(t) is decreasing. 2 This means that we are moving to the right and downwards, through 1 6 to (1, 1) when t = π 2, the orientation reverses, and we start to head to the left, since x = sin(t
) is now decreasing, and up, since y = csc(t) is now increasing. We pass back through 1 6 back to the points with small positive x-coordinates and large 2, 2 when t = 5π 2, 2 when t = π 2. Once t = π 2 5The reader is encouraged to review Sections 6.1 and 6.2 as needed. 6Note the open circle at the origin. See the solution to part 3 in Example 1.2.1 on page 22 and Theorem 4.1 in Section 4.1 for a review of this concept. 1052 Applications of Trigonometry positive y-coordinates. To better explain this behavior, we eliminate the parameter. Using a reciprocal identity, we write y = csc(t) = 1 sin(t). Since x = sin(t), the curve traced out by this parametrization is a portion of the graph of y = 1 x. We now can explain the unusual behavior as t → 0+ and t → π− – for these values of t, we are hugging the vertical asymptote x = 0 of the graph of y = 1 x. We see that the parametrization given above traces out the portion of y = 1 x for 0 < x ≤ 1 twice as t runs through the interval (0, π). = sin(t), 0 < t < π y = csc(t), 0 < t < π {x = sin(t), y = csc(t), 0 < t < π 4. Proceeding as above, we set about graphing {x = 1 + 3 cos(t), y = 2 sin(t) for 0 ≤ t ≤ 3π 2 by. We see that x ranges ﬁrst graphing x = 1 + 3 cos(t) and y = 2 sin(t) on the interval 0, 3π 2 from −2 to 4 and y ranges from −2 to 2. Plugging in t = 0, π 2, π and 3π 2 gives the points (4, 0), (1, 2), (−2, 0) and (1, −2), respectively. As t ranges from 0 to π 2, x = 1 + 3 cos(t) is decreasing, while y = 2 sin(t) is increasing. This means that we start tracing out our answer at (4, 0) and continue
moving to the left and upwards towards (1, 2). For π 2 ≤ t ≤ π, x is decreasing, as is y, so the motion is still right to left, but now is downwards from (1, 2) to (−2, 0). On the interval π, 3π, x begins to increase, while y continues to decrease. Hence, the motion becomes left 2 to right but continues downwards, connecting (−2, 0) to (1, −2). To eliminate the parameter here, we note that the trigonometric functions involved, namely cos(t) and sin(t), are related by the Pythagorean Identity cos2(t) + sin2(t) = 1. Hence, we solve x = 1 + 3 cos(t) for cos(t) 3, and we solve y = 2 sin(t) for sin(t) to get sin(t) = y to get cos(t) = x−1 2. Substituting these expressions into cos2(t)+sin2(t) = 1 gives x−1 9 + y2 4 = 1. From Section 3 7.4, we know that the graph of this equation is an ellipse centered at (1, 0) with vertices at (−2, 0) and (4, 0) with a minor axis of length 4. Our parametric equations here are tracing out three-quarters of this ellipse, in a counter-clockwise direction. = 1, or (x−1)1 π 2 π t 3π 2 −2 x = 1 + 3 cos(t), 0 ≤ t ≤ 3π 2 y 2 1 π 2 π t 3π 2 −1 1 2 3 x 4 −1 −2 y 2 1 −1 −2 y = 2 sin(t), 0 ≤ t ≤ 3π 2 {x = 1 + 3 cos(t), y = 2 sin(t), 0 ≤ t ≤ 3π 2 11.10 Parametric Equations 1053 Now that we have had some good practice sketching the graphs of parametric equations, we turn to the problem of ﬁnding parametric representations of curves. We start with the following. Parametrizations of Common Curves To parametrize y = f (x) as x runs through some interval I, let x = t and y = f (t) and let t run through I. To param
etrize x = g(y) as y runs through some interval I, let x = g(t) and y = t and let t run through I. To parametrize a directed line segment with initial point (x0, y0) and terminal point (x1, y1), let x = x0 + (x1 − x0)t and y = y0 + (y1 − y0)t for 0 ≤ t ≤ 1. To parametrize (x−h)2 a2 + (y−k)2 b2 = 1 where a, b > 0, let x = h + a cos(t) and y = k + b sin(t) for 0 ≤ t < 2π. (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 example. Example 11.10.3. Find a parametrization for each of the following curves and check your answers. 1. y = x2 from x = −3 to x = 2 2. y = f −1(x) where f (x) = x5 + 2x + 1 3. The line segment which starts at (2, −3) and ends at (1, 5) 4. The circle x2 + 2x + y2 − 4y = 4 4 + y2 5. The left half of the ellipse x2 9 = 1 Solution. 1. Since y = x2 is written in the form y = f (x), we let x = t and y = f (t) = t2. Since x = t, the bounds on t match precisely the bounds on x so we get x = t, y = t2 for −3 ≤ t ≤ 2. The check is almost trivial; with x = t we have y = t2 = x2 as t = x runs from −3 to 2. 2. We are told to parametrize y = f −1(x) for f (x) = x5 + 2x + 1 so it is safe to assume that f is one-to-one. (Otherwise, f −1 would not exist.) To ﬁnd a formula y = f −1(x), we follow the procedure outlined on page 384 – we start with the equation y = f
(x), interchange x and y and solve for y. Doing so gives us the equation x = y5 + 2y + 1. While we could attempt to solve this equation for y, we don’t need to. We can parametrize x = f (y) = y5 + 2y + 1 by setting y = t so that x = t5 + 2t + 1. We know from our work in Section 3.1 that since f (x) = x5 + 2x + 1 is an odd-degree polynomial, the range of y = f (x) = x5 + 2x + 1 is (−∞, ∞). Hence, in order to trace out the entire graph of x = f (y) = y5 + 2y + 1, we need to let y run through all real numbers. Our ﬁnal answer to this problem is x = t5 + 2t + 1, y = t for −∞ < t < ∞. As in the previous problem, our solution is trivial to check.7 7Provided you followed the inverse function theory, of course. 1054 Applications of Trigonometry 3. To parametrize line segment which starts at (2, −3) and ends at (1, 5), we make use of the formulas x = x0 +(x1 −x0)t and y = y0 +(y1 −y0)t for 0 ≤ t ≤ 1. While these equations at ﬁrst glance are quite a handful,8 they can be summarized as ‘starting point + (displacement)t’. To ﬁnd the equation for x, we have that the line segment starts at x = 2 and ends at x = 1. This means the displacement in the x-direction is (1 − 2) = −1. Hence, the equation for x is x = 2 + (−1)t = 2 − t. For y, we note that the line segment starts at y = −3 and ends at y = 5. Hence, the displacement in the y-direction is (5 − (−3)) = 8, so we get y = −3 + 8t. Our ﬁnal answer is {x = 2 − t, y = −3 + 8t for 0 ≤ t ≤ 1. To check, we can solve x = 2 − t for t to get
t = 2 − x. Substituting this into y = −3 + 8t gives y = −3 + 8t = −3 + 8(2 − x), or y = −8x + 13. We know this is the graph of a line, so all we need to check is that it starts and stops at the correct points. When t = 0, x = 2 − t = 2, and when t = 1, x = 2 − t = 1. Plugging in x = 2 gives y = −8(2) + 13 = −3, for an initial point of (2, −3). Plugging in x = 1 gives y = −8(1) + 13 = 5 for an ending point of (1, 5), as required. 9 + (y−2)2 9 + (y−2)2 4. In order to use the formulas above to parametrize the circle x2 +2x+y2 −4y = 4, we ﬁrst need to put it into the correct form. After completing the squares, we get (x + 1)2 + (y − 2)2 = 9, or (x+1)2 9 = 1. Once again, the formulas x = h + a cos(t) and y = k + b sin(t) can be a challenge to memorize, but they come from the Pythagorean Identity cos2(t) + sin2(t) = 1. In the equation (x+1)2 3. Rearranging these last two equations, we get x = −1 + 3 cos(t) and y = 2 + 3 sin(t). In order to complete one revolution around the circle, we let t range through the interval [0, 2π). We get as our ﬁnal answer {x = −1 + 3 cos(t), y = 2 + 3 sin(t) for 0 ≤ t < 2π. To check our answer, we could eliminate the parameter by solving x = −1 + 3 cos(t) for cos(t) and y = 2 + 3 sin(t) for sin(t), invoking a Pythagorean Identity, and then manipulating the resulting equation in x and y into the original equation x2 + 2x + y2 − 4y = 4. Instead, we opt for a more direct approach. We substitute x = −1 + 3 cos(t)
and y = 2 + 3 sin(t) into the equation x2 + 2x + y2 − 4y = 4 and show that the latter is satisﬁed for all t such that 0 ≤ t < 2π. 9 = 1, we identify cos(t) = x+1 3 and sin(t) = y−2 (−1 + 3 cos(t))2 + 2(−1 + 3 cos(t)) + (2 + 3 sin(t))2 − 4(2 + 3 sin(t)) 1 − 6 cos(t) + 9 cos2(t) − 2 + 6 cos(t) + 4 + 12 sin(t) + 9 sin2(t) − 8 − 12 sin(t) 9 cos2(t) + 9 sin2(t) − 5 9 cos2(t) + sin2(t) − 5 9 (1) − 5 x2 + 2x + y2 − 4y = Now that we know the parametric equations give us points on the circle, we can go through the usual analysis as demonstrated in Example 11.10.2 to show that the entire circle is covered as t ranges through the interval [0, 2π). 8Compare and contrast this with Exercise 65 in Section 11.8. 11.10 Parametric Equations 1055 5. In the equation x2 4 + y2 9 = 1, we can either use the formulas above or think back to the Pythagorean Identity to get x = 2 cos(t) and y = 3 sin(t). The normal range on the parameter in this case is 0 ≤ t < 2π, 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 2 ≤ t ≤ 3π π 2. Substituting x = 2 cos(t) and y = 3 sin(t) into x2 = 1, which reduces to the Pythagorean Identity cos2(t) + sin2(t) = 1. This proves that the points generated by the parametric equations {x = 2 cos(t), y = 3 sin(t) lie on the ellipse x2 9 = 1. Employing 2 ≤ t ≤ 3π the techniques demonstrated in Example 11.10.2, we ﬁnd that the restriction π 2 generates the left half of the
ellipse, as required. 2. Our ﬁnal answer is {x = 2 cos(t), y = 3 sin(t) for π 4 + y2 9 = 1 gives 4 cos2(t) 2 ≤ t ≤ 3π + 9 sin2(t) 9 4 + y2 4 We note that the formulas given on page 1053 oﬀer only one of literally inﬁnitely many ways to parametrize the common curves listed there. At times, the formulas oﬀered there need to be altered to suit the situation. Two easy ways to alter parametrizations are given below. Adjusting Parametric Equations Reversing Orientation: Replacing every occurrence of t with −t 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 (t − c) 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 11.10.4. Find a parametrization for the following curves. 1. The curve which starts at (2, 4) and follows the parabola y = x2 to end at (−1, 1). Shift the parameter so that the path starts at t = 0. 2. The two part path which starts at (0, 0), travels along a line to (3, 4), then travels along a line to (5, 0). 3. The Unit Circle, oriented clockwise, with t = 0 corresponding to (0, −1). Solution. 1. We can parametrize y = x2 from x = −1 to x = 2 using the formula given on Page 1053 as x = t, y = t2 for −1 ≤ t ≤ 2. This parametrization, however, starts at (−1, 1) and ends at (2, 4). Hence, we need to reverse the orientation. To do so, we replace every occurrence of t with −t to get x = −t, y = (−t)2 for −1 ≤ −t ≤ 2. After simplifying, we get x = −t, y = t2 for −2 ≤ t ≤ 1. We would like t to begin at t = 0 instead
of t = −2. 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 (t − 2) gives x = −(t − 2), y = (t − 2)2 for −2 ≤ t − 2 ≤ 1. Simplifying yields x = 2 − t, y = t2 − 4t + 4 for 0 ≤ t ≤ 3. 1056 Applications of Trigonometry 2. When parameterizing line segments, we think: ‘starting point + (displacement)t’. For the ﬁrst part of the path, we get {x = 3t, y = 4t for 0 ≤ t ≤ 1, and for the second part we get {x = 3 + 2t, y = 4 − 4t for 0 ≤ t ≤ 1. Since the ﬁrst parametrization leaves oﬀ at t = 1, we shift the parameter in the second part so it starts at t = 1. Our current description of the second part starts at t = 0, so we introduce a ‘time delay’ of 1 unit to the second set of parametric equations. Replacing t with (t − 1) in the second set of parametric equations gives {x = 3 + 2(t − 1), y = 4 − 4(t − 1) for 0 ≤ t − 1 ≤ 1. Simplifying yields {x = 1 + 2t, y = 8 − 4t for 1 ≤ t ≤ 2. Hence, we may parametrize the path as {x = f (t), y = g(t) for 0 ≤ t ≤ 2 where f (t) = 3t, 1 + 2t, for 0 ≤ t ≤ 1 for 1 ≤ t ≤ 2 and g(t) = 4t, 8 − 4t, for 0 ≤ t ≤ 1 for 1 ≤ t ≤ 2 3. We know that {x = cos(t), y = sin(t) for 0 ≤ t < 2π gives a counter-clockwise parametrization of the Unit Circle with t = 0 corresponding to (1, 0), so the ﬁrst order of business is to reverse the orientation. Replacing t with −t gives {x = cos(−t), y = sin(−t
) for 0 ≤ −t < 2π, which simpliﬁes9 to {x = cos(t), y = − sin(t) for −2π < t ≤ 0. This parametrization gives a clockwise orientation, but t = 0 still corresponds to the point (1, 0); the point (0, −1) is reached when t = − 3π 2. Our strategy is to ﬁrst get the parametrization to ‘start’ at the point (0, −1) and then shift the parameter accordingly so the ‘start’ coincides with t = 0. We know that any interval of length 2π will parametrize the entire circle, so we keep the equations {x = cos(t), y = − sin(t), but start the parameter t at − 3π 2, and ﬁnd the upper bound by adding 2π so − 3π 2. The reader can verify that {x = cos(t), y = − sin(t) for − 3π 2 traces out the Unit Circle clockwise starting at the point (0, −1). We now shift the parameter by introducing a ‘time delay’ of 3π 2 units by replacing every occurrence of t with t − 3π 2. This 2 simpliﬁes10 to {x = − sin(t), y = − cos(t) for 0 ≤ t < 2π, as required.. We get x = cos t − 3π 2, y = − sin t − 3π 2 2 ≤ t < π 2 ≤ t − 3π 2 ≤ t < π for − 3π 2 < π We put our answer to Example 11.10.4 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 θ be the angle in radians which measures the amount of clockwise rotation experienced by the radius highlighted in the ﬁgure. y r P (x, y) θ 9courtesy of the Even/Odd Identities 10courtesy of the Sum/Diﬀerence Formulas x 11.10 Parametric Equations 1057 Our goal is to ﬁnd parametric equations for the coordinates of the point P (x, y) in terms of θ. From our work
in Example 11.10.4 number 3, we know that clockwise motion along the Unit Circle starting at the point (0, −1) can be modeled by the equations {x = − sin(θ), y = − cos(θ) for 0 ≤ θ < 2π. (We have renamed the parameter ‘θ’ to match the context of this problem.) To model this motion on a circle of radius r, all we need to do11 is multiply both x and y by the factor r which yields {x = −r sin(θ), y = −r cos(θ). We now need to adjust for the fact that the circle isn’t stationary with center (0, 0), but rather, 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 vt 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 (vt, r). From Section 10.1.1, we know v = rθ t, or vt = rθ. Hence, the center of the circle, in terms of the parameter θ, is (rθ, r). As a result, we need to modify the equations {x = −r sin(θ), y = −r cos(θ) by shifting the x-coordinate to the right rθ units (by adding rθ to the expression for x) and the y-coordinate up r units12 (by adding r to the expression for y). We get {x = −r sin(θ) + rθ, y = −r cos(θ) + r, which can be written as {x = r(θ − sin(θ)), y = r(1 − cos(θ)). Since the motion starts at θ = 0 and proceeds indeﬁnitely, we set θ ≥ 0. We end the section with a demonstration of the graphing calculator. Example 11.10.5. Find the parametric equations of a cycloid which results from a circle of radius 3 rolling down
the positive x-axis as described above. Graph your answer using a calculator. Solution. We have r = 3 which gives the equations {x = 3(t − sin(t)), y = 3(1 − cos(t)) for t ≥ 0. (Here we have returned to the convention of using t as the parameter.) Sketching the cycloid by hand is a wonderful exercise in Calculus, but for the purposes of this book, we use a graphing utility. Using a calculator to graph parametric equations is very similar to graphing polar equations on a calculator.13 Ensuring that the calculator is in ‘Parametric Mode’ and ‘radian mode’ we enter the equations and advance to the ‘Window’ screen. As always, the challenge is to determine appropriate bounds on the parameter, t, as well as for x and y. We know that one full revolution of the circle occurs over the interval 0 ≤ t < 2π, so 11If we replace x with x 2 = 1 which reduces to x2 + y2 = r2. In the language of Section 1.7, we are stretching the graph by a factor of r in both the x- and y-directions. Hence, we multiply both the x- and y-coordinates of points on the graph by r. r in the equation for the Unit Circle x2 + y2 = 1, we obtain x r and y with y 2 + y r r 12Does this seem familiar? See Example 11.1.1 in Section 11.1. 13See page 959 in Section 11.5. 1058 Applications of Trigonometry it seems reasonable to keep these as our bounds on t. The ‘Tstep’ seems reasonably small – too large a value here can lead to incorrect graphs.14 We know from our derivation of the equations of the cycloid that the center of the generating circle has coordinates (rθ, r), or in this case, (3t, 3). Since t ranges between 0 and 2π, we set x to range between 0 and 6π. The values of y go from the bottom of the circle to the top, so y ranges between 0 and 6. Below we graph the cycloid with these settings, and then extend t to range from 0 to 6π which forces x to range from 0 to 18π yielding three arches of the cycloid. (It is instructive to note that keeping the y settings between
0 and 6 messes up the geometry of the cycloid. The reader is invited to use the Zoom Square feature on the graphing calculator to see what window gives a true geometric perspective of the three arches.) 14Again, see page 959 in Section 11.5. 11.10 Parametric Equations 1059 11.10.1 Exercises In Exercises 1 - 20, plot the set of parametric equations by hand. Be sure to indicate the orientation imparted on the curve by the parametrization. 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. x = 4t − 3 y = 6t − 2 for 0 ≤ t ≤ 1 x = 2t y = t2 for − 1 ≤ t ≤ 2 x = t2 + 2t + 1 y = t + 1 for t ≤ 1 x = t y = t3 for − ∞ < t < ∞ x = cos(t) y = sin(t) for − 1 + 3 cos(t) y = 4 sin(t) for 0 ≤ t ≤ 2π x = 2 cos(t) y = sec(t) for 0 ≤ t < π 2 x = sec(t) y = tan(t) for − π 2 < t < π 2 x = tan(t) y = 2 sec(t) for − π 2 < t < π 2 x = cos(t) y = t for 0 ≤ t ≤ π 2. 4. 6. 8. 10. 12. 14. 16. 18. 20. x = 4t − 1 y = 3 − 4t for + 2t − t2 18 − t2 x = 1 9 y = 1 3 t for 0 ≤ t ≤ 3 for t ≥ −3 x = t3 y = t for − ∞ < t < ∞ x = 3 cos(t) y = 3 sin(t) for 0 ≤ t ≤ π x = 3 cos(t) y = 2 sin(t) + 1 for π 2 ≤ t ≤ 2π x = 2 tan(t) y = cot(t) for 0 < t < π 2 x = sec(t) y = tan(t) for π 2 < t < 3π 2 x = tan(t) y = 2 sec(t) for π 2 < t < 3π 2 x = sin(
t) y = t for − π 2 ≤ t ≤ π 2 In Exercises 21 - 24, 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. 21. 23. x = t3 − 3t y = t2 − 4 x = et + e−t y = et − e−t for − 2 ≤ t ≤ 2 for − 2 ≤ t ≤ 2 22. 24. x = 4 cos3(t) y = 4 sin3(t) x = cos(3t) y = sin(4t) for 0 ≤ t ≤ 2π for 0 ≤ t ≤ 2π 1060 Applications of Trigonometry In Exercises 25 - 39, ﬁnd a parametric description for the given oriented curve. 25. the directed line segment from (3, −5) to (−2, 2) 26. the directed line segment from (−2, −1) to (3, −4) 27. the curve y = 4 − x2 from (−2, 0) to (2, 0). 28. the curve y = 4 − x2 from (−2, 0) to (2, 0) (Shift the parameter so t = 0 corresponds to (−2, 0).) 29. the curve x = y2 − 9 from (−5, −2) to (0, 3). 30. the curve x = y2 − 9 from (0, 3) to (−5, −2). (Shift the parameter so t = 0 corresponds to (0, 3).) 31. the circle x2 + y2 = 25, oriented counter-clockwise 32. the circle (x − 1)2 + y2 = 4, oriented counter-clockwise 33. the circle x2 + y2 − 6y = 0, oriented counter-clockwise 34. the circle x2 + y2 − 6y = 0, oriented clockwise (Shift the parameter so t begins at 0.) 35. the circle (x − 3)2 + (y + 1)2 = 117, oriented counter-clockwise 36. the ellipse (x − 1)2 + 9y2 = 9, oriented counter-clockwise 37. the ellipse 9x2 + 4y2 + 24y = 0, oriented counter-clockwise 38. the ellipse 9x2 +
4y2 + 24y = 0, oriented clockwise (Shift the parameter so t = 0 corresponds to (0, 0).) 39. the triangle with vertices (0, 0), (3, 0), (0, 4), oriented counter-clockwise (Shift the parameter so t = 0 corresponds to (0, 0).) 40. Use parametric equations and a graphing utility to graph the inverse of f (x) = x3 + 3x − 4. 41. Every polar curve r = f (θ) can be translated to a system of parametric equations with parameter θ by {x = r cos(θ) = f (θ) cos(θ), y = r sin(θ) = f (θ) sin(θ). Convert r = 6 cos(2θ) to a system of parametric equations. Check your answer by graphing r = 6 cos(2θ) by hand using the techniques presented in Section 11.5 and then graphing the parametric equations you found using a graphing utility. 42. Use your results from Exercises 3 and 4 in Section 11.1 to ﬁnd the parametric equations which model a passenger’s position as they ride the London Eye. 11.10 Parametric Equations 1061 Suppose an object, called a projectile, is launched into the air. Ignoring everything except the force gravity, the path of the projectile is given by15    x = v0 cos(θ) t 1 2 y = − gt2 + v0 sin(θ) t + s0 for 0 ≤ t ≤ T where v0 is the initial speed of the object, θ is the angle from the horizontal at which the projectile is launched,16 g is the acceleration due to gravity, s0 is the initial height of the projectile above the ground and T is the time when the object returns to the ground. (See the ﬁgure below.) y θ s0 (x(T ), 0) x 43. 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, ﬁ
nd the parametric equations for the ﬂight of the hammer. (Here, 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. 44. Eliminate the parameter in the equations for projectile motion to show that the path of the projectile follows the curve y = − g sec2(θ) 2v2 0 x2 + tan(θ)x + s0 Use the vertex formula (Equation 2.4) to show the maximum height of the projectile is y = 0 sin2(θ) v2 2g + s0 when x = v2 0 sin(2θ) 2g 15A nice mix of vectors and Calculus are needed to derive this. 16We’ve seen this before. It’s the angle of elevation which was deﬁned on page 753. 1062 Applications of Trigonometry 45. 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 part 44 to determine how fast the sheaf was launched into the air. (Once again, use g = 32 ft. s2.) 46. Suppose θ = π formula given for y(t) above using g = 9.8 m in Exercise 25 in Section 2.3. What is x(t) in this case? 2. (The projectile was launched vertically.) Simplify the general parametric s2 and compare that to the formula for s(t) given In Exercises 47 - 52, we explore the hyperbolic cosine function, denoted cosh(t), and the hyperbolic sine function, denoted sinh(t), deﬁned below: cosh(t) = et + e−t 2 and sinh(t) = et − e−t 2 47. Using a graphing utility as needed, verify that the domain of cosh(t) is (−∞, ∞) and the range of cosh(t) is [1, ∞). 48. Using a graphing utility as needed, verify that the domain and range of sinh(t) are both (−∞,
∞). 49. Show that {x(t) = cosh(t), y(t) = sinh(t) parametrize the right half of the ‘unit’ hyperbola x2 − y2 = 1. (Hence the use of the adjective ‘hyperbolic.’) 50. Compare the deﬁnitions of cosh(t) and sinh(t) to the formulas for cos(t) and sin(t) given in Exercise 83f in Section 11.7. 51. Four other hyperbolic functions are waiting to be deﬁned: the hyperbolic secant sech(t), the hyperbolic cosecant csch(t), the hyperbolic tangent tanh(t) and the hyperbolic cotangent coth(t). Deﬁne these functions in terms of cosh(t) and sinh(t), then convert them to formulas involving et and e−t. Consult a suitable reference (a Calculus book, or this entry on the hyperbolic functions) and spend some time reliving the thrills of trigonometry with these ‘hyperbolic’ functions. 52. If these functions look familiar, they should. Enjoy some nostalgia and revisit Exercise 35 in Section 6.5, Exercise 47 in Section 6.3 and the answer to Exercise 38 in Section 6.4. 11.10 Parametric Equations 1063 11.10.2 Answers x = 4t − 3 y = 6t − 2 1. for 3 −2 −1 −1 −2 2. x = 4t − 1 y = 3 − 4t y for 0 ≤ t ≤ 1 3 2 1 −1 −1 1 2 3 x x = 2t y = t2 3. for − + 2t − t2 4. for 3 −2 −1 1 2 3 4 x −1 1 2 x x = t2 + 2t + 1 y = t + 1 5. for t ≤ 1 6. 18 − t2 x = 1 9 y = 1 3 t for t ≥ −3 1 2 3 4 5 x −3 −2 −1 −1 −2 Applications of Trigonometry x = t3 y = t 8. for − ∞ < t < ∞ y 1 −4 −3 −2 −1 −1 1 2 3 4 x 1064 7. x
= t y = t3 y for − ∞ < t < ∞ 4 3 2 1 −1 −1 −2 −3 −4 1 x 9. x = cos(t) y = sin(t) y 1 for − π 2 ≤ t ≤ π 2 −1 x 1 −1 10. x = 3 cos(t) y = 3 sin(t) y for 0 ≤ t ≤ π 3 2 1 −3 −2 −1 1 2 3 x 11. x = −1 + 3 cos(t) y = 4 sin(t) for 0 ≤ t ≤ 2π 12. x = 3 cos(t) y = 2 sin(t) + 1 for π 2 ≤ t ≤ 2π y 4 3 2 1 1 2 x −4 −3 −2 −1 −1 −2 −3 −4 y 3 2 1 −3 −1 −1 1 3 x 11.10 Parametric Equations 1065 13. x = 2 cos(t) y = sec(t) y for 0 ≤ t < π 2 14. x = 2 tan(t) y = cot(t) y for 15. x = sec(t) y = tan(t) for − π 2 < t < π 2 y 16. x = sec(t) y = tan(t) < t < 3π 2 π 2 for y 4 3 2 1 −1 −2 −3 −4 −3 −2 −1 −1 −2 −3 −4 1066 Applications of Trigonometry 17. x = tan(t) y = 2 sec(t) π 2 for − y < t < π 2 18. x = tan(t) y = 2 sec(t) < t < 3π 2 π 2 for y 4 3 2 1 −2 −1 1 2 x −2 −1 1 2 x −1 −2 −3 −4 19. x = cos(t) y = t for 0 < t < π 20. x = sin(t) y = t for − 1 1 x y π 2 −1 1 x − π 2 21. x = t3 − 3t y = t2 − 4 for − 2 ≤ t ≤ 2 22. x = 4 cos3(t) y = 4 sin3(t) y for 0 ≤ t ≤ 2π y 1 2 x −2 −1 −1 −
2 −3 −4 4 3 2 1 −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 11.10 Parametric Equations 1067 23. x = et + e−t y = et − e−t for − 2 ≤ t ≤ 2 24. x = cos(3t) y = sin(4t) for 0 ≤ t ≤ 2π 1 1 x y 7 5 3 1 −1 −3 −5 −7 26. 28. 30. 32. 34. −1 x = 5t − 2 y = −1 − 3t x = t − 2 y = 4t − t2 x = t2 − 6t y = 3 − t for 0 ≤ t ≤ 1 for 0 ≤ t ≤ 4 for cos(t) y = 2 sin(t) x = 3 cos(t) y = 3 − 3 sin(t) x = 1 + 3 cos(t) y = sin(t) for 0 ≤ t < 2π for 0 ≤ t < 2π for 0 ≤ t < 2π x = 3 − 5t y = −5 + 7t for 0 ≤ t ≤ 1 for − 2 ≤ t ≤ 2 for − 2 ≤ t ≤ 3 for 0 ≤ t < 2π x = t y = 4 − t2 x = t2 − 9 y = t x = 5 cos(t) y = 5 sin(t) x = 3 cos(t) x = 3 + √ y = −1 + x = 2 cos(t) 25. 27. 29. 31. 33. 35. 37. 38. y = 3 + 3 sin(t) for 0 ≤ t < 2π 117 cos(t) √ 117 sin(t) for 0 ≤ t < 2π 36. for 0 ≤ t < 2π y = 3 sin(t) − 3    x = 2 cos t − y = −3 − 3 sin π 2 = 2 sin(t) π 2 t − = −3 + 3 cos(t) for 0 ≤ t < 2π 39. {x(t), y(t) where: x(t) =    3t, 0 ≤ t ≤ 1 6 − 3t, 1 ≤ t ≤ 2 0, 2 ≤ t ≤ 3 y(t) =    0
, 0 ≤ t ≤ 1 4t − 4, 1 ≤ t ≤ 2 12 − 4t, 2 ≤ t ≤ 3 1068 Applications of Trigonometry 40. The parametric equations for the inverse are x = t3 + 3t − 4 y = t for − ∞ < t < ∞ 41. r = 6 cos(2θ) translates to x = 6 cos(2θ) cos(θ) y = 6 cos(2θ) sin(θ) for 0 ≤ θ < 2π. 42. The parametric equations which describe the locations of passengers on the London Eye are x = 67.5 cos π y = 67.5 sin π 15 t − π 15 t − π 2 = 67.5 sin π 15 t 2 + 67.5 = 67.5 − 67.5 cos π 15 t for − ∞ < t < ∞ x = 33 cos(42◦)t 43. The parametric equations for the hammer throw are y = −16t2 + 33 sin(42◦)t + 6 for t ≥ 0. To ﬁnd when the hammer hits the ground, we solve y(t) = 0 and get t ≈ −0.23 or 1.61. Since t ≥ 0, the hammer hits the ground after approximately t = 1.61 seconds after it was launched into the air. To ﬁnd how far away the hammer hits the ground, we ﬁnd x(1.61) ≈ 39.48 feet from where it was thrown into the air. 45. We solve y = 0 sin2(θ) v2 2g + s0 = 0 sin2(85◦) v2 2(32) + 5 = 31.5 to get v0 = ±41.34. The initial speed of the sheaf was approximately 41.34 feet per second. Index nth root of a complex number, 1000, 1001 principal, 397 nth Roots of Unity, 1006 u-substitution, 273 x-axis, 6 x-coordinate, 6 x-intercept, 25 y-axis, 6 y-coordinate, 6 y-intercept, 25 abscissa, 6 absolute value deﬁnition of, 173 inequality, 211 properties of, 173 acidity of a solution pH, 432 acute angle, 694 adjoint of a matrix
, 622 alkalinity of a solution pH, 432 amplitude, 794, 881 angle acute, 694 between two vectors, 1035, 1036 central angle, 701 complementary, 696 coterminal, 698 decimal degrees, 695 deﬁnition, 693 degree, 694 DMS, 695 initial side, 698 measurement, 693 negative, 698 obtuse, 694 of declination, 761 of depression, 761 of elevation, 753 of inclination, 753 oriented, 697 positive, 698 quadrantal, 698 radian measure, 701 reference, 721 right, 694 standard position, 698 straight, 693 supplementary, 696 terminal side, 698 vertex, 693 angle side opposite pairs, 896 angular frequency, 708 annuity annuity-due, 667 ordinary deﬁnition of, 666 future value, 667 applied domain of a function, 60 arccosecant calculus friendly deﬁnition of, 831 graph of, 830 properties of, 831 1069 Index 1070 trigonometry friendly deﬁnition of, 828 graph of, 827 properties of, 828 arccosine deﬁnition of, 820 graph of, 819 properties of, 820 arccotangent deﬁnition of, 824 graph of, 824 properties of, 824 arcsecant calculus friendly deﬁnition of, 831 graph of, 830 properties of, 831 trigonometry friendly deﬁnition of, 828 graph of, 827 properties of, 828 arcsine deﬁnition of, 820 graph of, 820 properties of, 820 arctangent deﬁnition of, 824 graph of, 823 properties of, 824 argument of a complex number deﬁnition of, 991 properties of, 995 of a function, 55 of a logarithm, 425 of a trigonometric function, 793 arithmetic sequence, 654 associative property for function composition, 366 matrix addition, 579 matrix multiplication, 585 scalar multiplication, 581 vector addition, 1015 scalar multiplication, 1018 asymptote horizontal formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 308 of a hyperbola, 531 sl
ant determination of, 312 formal deﬁnition of, 311 slant (oblique), 311 vertical formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 306 augmented matrix, 568 average angular velocity, 707 average cost, 346 average cost function, 82 average rate of change, 160 average velocity, 706 axis of symmetry, 191 back substitution, 560 bearings, 905 binomial coeﬃcient, 683 Binomial Theorem, 684 Bisection Method, 277 BMI, body mass index, 355 Boyle’s Law, 350 buﬀer solution, 478 cardioid, 951 Cartesian coordinate plane, 6 Cartesian coordinates, 6 Cauchy’s Bound, 269 center of a circle, 498 of a hyperbola, 531 of an ellipse, 516 Index 1071 central angle, 701 change of base formulas, 442 characteristic polynomial, 626 Charles’s Law, 355 circle center of, 498 deﬁnition of, 498 from slicing a cone, 495 radius of, 498 standard equation, 498 standard equation, alternate, 519 circular function, 744 cis(θ), 995 coeﬃcient of determination, 226 cofactor, 616 Cofunction Identities, 773 common base, 420 common logarithm, 422 commutative property function composition does not have, 366 matrix addition, 579 vector addition, 1015 dot product, 1034 complementary angles, 696 Complex Factorization Theorem, 290 complex number nth root, 1000, 1001 nth Roots of Unity, 1006 argument deﬁnition of, 991 properties of, 995 conjugate deﬁnition of, 288 properties of, 289 deﬁnition of, 2, 287, 991 imaginary part, 991 imaginary unit, i, 287 modulus deﬁnition of, 991 properties of, 993 polar form cis-notation, 995 principal argument, 991 real part, 991 rectangular form, 991 set of, 2 complex plane, 991 component form of a vector, 1013 composite function deﬁnition of, 360 properties of, 367 compound interest, 470 conic sections deﬁnition, 495 conjugate axis of a hyperbola, 5
32 conjugate of a complex number deﬁnition of, 288 properties of, 289 Conjugate Pairs Theorem, 291 consistent system, 553 constant function as a horizontal line, 156 formal deﬁnition of, 101 intuitive deﬁnition of, 100 constant of proportionality, 350 constant term of a polynomial, 236 continuous, 241 continuously compounded interest, 472 contradiction, 549 coordinates Cartesian, 6 polar, 919 rectangular, 919 correlation coeﬃcient, 226 cosecant graph of, 801 of an angle, 744, 752 properties of, 802 cosine graph of, 791 of an angle, 717, 730, 744 properties of, 791 1072 cost average, 82, 346 ﬁxed, start-up, 82 variable, 159 cost function, 82 cotangent graph of, 805 of an angle, 744, 752 properties of, 806 coterminal angle, 698 Coulomb’s Law, 355 Cramer’s Rule, 619 curve orientated, 1048 cycloid, 1056 decibel, 431 decimal degrees, 695 decreasing function formal deﬁnition of, 101 intuitive deﬁnition of, 100 degree measure, 694 degree of a polynomial, 236 DeMoivre’s Theorem, 997 dependent system, 554 dependent variable, 55 depreciation, 420 Descartes’ Rule of Signs, 273 determinant of a matrix deﬁnition of, 614 properties of, 616 Diﬀerence Identity for cosine, 771, 775 for sine, 773, 775 for tangent, 775 diﬀerence quotient, 79 dimension of a matrix, 567 direct variation, 350 directrix of a conic section in polar form, 981 of a parabola, 505 discriminant Index of a conic, 979 of a quadratic equation, 195 trichotomy, 195 distance deﬁnition, 10 distance formula, 11 distributive property matrix matrix multiplication, 585 scalar multiplication, 581 vector dot product, 1034 scalar multiplication, 1018 DMS, 695 domain applied, 60 deﬁnition of, 45 implied, 58 dot product commutative property of, 1034 deﬁn
ition of, 1034 distributive property of, 1034 geometric interpretation, 1035 properties of, 1034 relation to orthogonality, 1037 relation to vector magnitude, 1034 work, 1042 Double Angle Identities, 776 earthquake Richter Scale, 431 eccentricity, 522, 981 eigenvalue, 626 eigenvector, 626 ellipse center, 516 deﬁnition of, 516 eccentricity, 522 foci, 516 from slicing a cone, 496 guide rectangle, 519 major axis, 516 minor axis, 516 Index 1073 reﬂective property, 523 standard equation, 519 vertices, 516 ellipsis (... ), 31, 651 empty set, 2 end behavior of f (x) = axn, n even, 240 of f (x) = axn, n odd, 240 of a function graph, 239 polynomial, 243 entry in a matrix, 567 equation contradiction, 549 graph of, 23 identity, 549 linear of n variables, 554 linear of two variables, 549 even function, 95 Even/Odd Identities, 770 exponential function algebraic properties of, 437 change of base formula, 442 common base, 420 deﬁnition of, 418 graphical properties of, 419 inverse properties of, 437 natural base, 420 one-to-one properties of, 437 solving equations with, 448 extended interval notation, 756 Factor Theorem, 258 factorial, 654, 681 ﬁxed cost, 82 focal diameter of a parabola, 507 focal length of a parabola, 506 focus of a conic section in polar form, 981 focus (foci) of a hyperbola, 531 of a parabola, 505 of an ellipse, 516 free variable, 552 frequency angular, 708, 881 of a sinusoid, 795 ordinary, 708, 881 function (absolute) maximum, 101 (absolute, global) minimum, 101 absolute value, 173 algebraic, 399 argument, 55 arithmetic, 76 as a process, 55, 378 average cost, 82 circular, 744 composite deﬁnition of, 360 properties of, 367 constant, 100, 156 continuous, 241 cost, 82 decreasing, 100 deﬁnition as a relation, 43 dependent variable of, 55 di�
��erence, 76 diﬀerence quotient, 79 domain, 45 even, 95 exponential, 418 Fundamental Graphing Principle, 93 identity, 168 increasing, 100 independent variable of, 55 inverse deﬁnition of, 379 properties of, 379 solving for, 384 uniqueness of, 380 linear, 156 local (relative) maximum, 101 local (relative) minimum, 101 logarithmic, 422 1074 Index notation, 55 odd, 95 one-to-one, 381 periodic, 790 piecewise-deﬁned, 62 polynomial, 235 price-demand, 82 product, 76 proﬁt, 82 quadratic, 188 quotient, 76 range, 45 rational, 301 revenue, 82 smooth, 241 sum, 76 transformation of graphs, 120, 135 zero, 95 fundamental cycle of y = cos(x), 791 Fundamental Graphing Principle for equations, 23 for functions, 93 for polar equations, 938 Fundamental Theorem of Algebra, 290 Gauss-Jordan Elimination, 571 Gaussian Elimination, 557 geometric sequence, 654 geometric series, 669 graph hole in, 305 horizontal scaling, 132 horizontal shift, 123 of a function, 93 of a relation, 20 of an equation, 23 rational function, 321 reﬂection about an axis, 126 transformations, 135 vertical scaling, 130 vertical shift, 121 greatest integer function, 67 growth model limited, 475 logistic, 475 uninhibited, 472 guide rectangle for a hyperbola, 532 for an ellipse, 519 Half-Angle Formulas, 779 harmonic motion, 885 Henderson-Hasselbalch Equation, 446 Heron’s Formula, 914 hole in a graph, 305 location of, 306 Hooke’s Law, 350 horizontal asymptote formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 308 horizontal line, 23 Horizontal Line Test (HLT), 381 hyperbola asymptotes, 531 branch, 531 center, 531 conjugate axis, 532 deﬁnition of, 531 foci, 531 from slicing a cone, 496 guide rectangle, 532 standard equation horizontal, 534 vertical, 534 transverse axis, 531 vertices, 531 hyperbolic cosine, 1062 hyperbolic sine, 1062 hyperbol
oid, 542 identity function, 367 matrix, additive, 579 Index 1075 matrix, multiplicative, 585 statement which is always true, 549 imaginary axis, 991 imaginary part of a complex number, 991 imaginary unit, i, 287 implied domain of a function, 58 inconsistent system, 553 increasing function interval deﬁnition of, 3 notation for, 3 notation, extended, 756 inverse matrix, additive, 579, 581 matrix, multiplicative, 602 of a function formal deﬁnition of, 101 intuitive deﬁnition of, 100 independent system, 554 independent variable, 55 index of a root, 397 induction base step, 673 induction hypothesis, 673 inductive step, 673 inequality absolute value, 211 graphical interpretation, 209 non-linear, 643 quadratic, 215 sign diagram, 214 inﬂection point, 477 information entropy, 477 initial side of an angle, 698 instantaneous rate of change, 161, 472, 707 integer deﬁnition of, 2 greatest integer function, 67 set of, 2 intercept deﬁnition of, 25 location of, 25 interest compound, 470 compounded continuously, 472 simple, 469 Intermediate Value Theorem polynomial zero version, 241 interrobang, 321 intersection of two sets, 4 deﬁnition of, 379 properties of, 379 solving for, 384 uniqueness of, 380 inverse variation, 350 invertibility function, 382 invertible function, 379 matrix, 602 irrational number deﬁnition of, 2 set of, 2 irreducible quadratic, 291 joint variation, 350 Kepler’s Third Law of Planetary Motion, 355 Kirchhoﬀ’s Voltage Law, 605 latus rectum of a parabola, 507 Law of Cosines, 910 Law of Sines, 897 leading coeﬃcient of a polynomial, 236 leading term of a polynomial, 236 Learning Curve Equation, 315 least squares regression line, 225 lemniscate, 950 lima¸con, 950 line horizontal, 23 least squares regression, 225 linear function, 156 of best ﬁt, 225 parallel, 166 1076 Index perpendicular, 167 point-slope form, 155 slope of, 151 slope-intercept form, 155 vertical, 23 linear equation n variables,
554 two variables, 549 linear function, 156 local maximum formal deﬁnition of, 102 intuitive deﬁnition of, 101 local minimum formal deﬁnition of, 102 intuitive deﬁnition of, 101 logarithm algebraic properties of, 438 change of base formula, 442 common, 422 general, “base b”, 422 graphical properties of, 423 inverse properties of, 437 natural, 422 one-to-one properties of, 437 solving equations with, 459 logarithmic scales, 431 logistic growth, 475 LORAN, 538 lower triangular matrix, 593 main diagonal, 585 major axis of an ellipse, 516 Markov Chain, 592 mathematical model, 60 matrix addition associative property, 579 commutative property, 579 deﬁnition of, 578 properties of, 579 additive identity, 579 additive inverse, 579 adjoint, 622 augmented, 568 characteristic polynomial, 626 cofactor, 616 deﬁnition, 567 determinant deﬁnition of, 614 properties of, 616 dimension, 567 entry, 567 equality, 578 invertible, 602 leading entry, 569 lower triangular, 593 main diagonal, 585 matrix multiplication associative property of, 585 deﬁnition of, 584 distributive property, 585 identity for, 585 properties of, 585 minor, 616 multiplicative inverse, 602 product of row and column, 584 reduced row echelon form, 570 rotation, 986 row echelon form, 569 row operations, 568 scalar multiplication associative property of, 581 deﬁnition of, 580 distributive properties, 581 identity for, 581 properties of, 581 zero product property, 581 size, 567 square matrix, 586 sum, 578 upper triangular, 593 maximum formal deﬁnition of, 102 intuitive deﬁnition of, 101 measure of an angle, 693 1077 Index midpoint deﬁnition of, 12 midpoint formula, 13 minimum formal deﬁnition of, 102 intuitive deﬁnition of, 101 minor, 616 minor axis of an ellipse, 516 model mathematical, 60 modulus
of a complex number deﬁnition of, 991 properties of, 993 multiplicity eﬀect on the graph of a polynomial, 245, 249 of a zero, 244 natural base, 420 natural logarithm, 422 natural number deﬁnition of, 2 set of, 2 negative angle, 698 Newton’s Law of Cooling, 421, 474 Newton’s Law of Universal Gravitation, 351 oblique asymptote, 311 obtuse angle, 694 odd function, 95 Ohm’s Law, 350, 605 one-to-one function, 381 ordered pair, 6 ordinary frequency, 708 ordinate, 6 orientation, 1048 oriented angle, 697 oriented arc, 704 origin, 7 orthogonal projection, 1038 orthogonal vectors, 1037 overdetermined system, 554 parabola axis of symmetry, 191 deﬁnition of, 505 directrix, 505 focal diameter, 507 focal length, 506 focus, 505 from slicing a cone, 496 graph of a quadratic function, 188 latus rectum, 507 reﬂective property, 510 standard equation horizontal, 508 vertical, 506 vertex, 188, 505 vertex formulas, 194 paraboloid, 510 parallel vectors, 1030 parameter, 1048 parametric equations, 1048 parametric solution, 552 parametrization, 1048 partial fractions, 628 Pascal’s Triangle, 688 password strength, 477 period circular motion, 708 of a function, 790 of a sinusoid, 881 periodic function, 790 pH, 432 phase, 795, 881 phase shift, 795, 881 pi, π, 700 piecewise-deﬁned function, 62 point of diminishing returns, 477 point-slope form of a line, 155 polar coordinates conversion into rectangular, 924 deﬁnition of, 919 equivalent representations of, 923 polar axis, 919 pole, 919 1078 Index polar form of a complex number, 995 polar rose, 950 polynomial division dividend, 258 divisor, 258 factor, 258 quotient, 258 remainder, 258 synthetic division, 260 polynomial function completely factored over the complex numbers, 291 over the real numbers, 291 constant term, 236 deﬁnition of, 235 degree, 236 end behavior,
239 leading coeﬃcient, 236 leading term, 236 variations in sign, 273 zero lower bound, 274 multiplicity, 244 upper bound, 274 positive angle, 698 Power Reduction Formulas, 778 power rule for absolute value, 173 for complex numbers, 997 for exponential functions, 437 for logarithms, 438 for radicals, 398 for the modulus of a complex number, 993 price-demand function, 82 principal, 469 principal nth root, 397 principal argument of a complex number, 991 principal unit vectors, ˆı, ˆ, 1024 Principle of Mathematical Induction, 673 product rule for absolute value, 173 for complex numbers, 997 for exponential functions, 437 for logarithms, 438 for radicals, 398 for the modulus of a complex number, 993 Product to Sum Formulas, 780 proﬁt function, 82 projection x−axis, 45 y−axis, 46 orthogonal, 1038 Pythagorean Conjugates, 751 Pythagorean Identities, 749 quadrantal angle, 698 quadrants, 8 quadratic formula, 194 quadratic function deﬁnition of, 188 general form, 190 inequality, 215 irreducible quadratic, 291 standard form, 190 quadratic regression, 228 Quotient Identities, 745 quotient rule for absolute value, 173 for complex numbers, 997 for exponential functions, 437 for logarithms, 438 for radicals, 398 for the modulus of a complex number, 993 radian measure, 701 radical properties of, 398 radicand, 397 radioactive decay, 473 radius of a circle, 498 range deﬁnition of, 45 rate of change average, 160 Index 1079 instantaneous, 161, 472 slope of a line, 154 rational exponent, 398 rational functions, 301 rational number deﬁnition of, 2 set of, 2 Rational Zeros Theorem, 269 ray deﬁnition of, 693 initial point, 693 real axis, 991 Real Factorization Theorem, 292 real number deﬁnition of, 2 set of, 2 real part of a complex number, 991 Reciprocal Identities, 745 rectangular coordinates also known as Cartesian coordinates, 919 conversion into polar, 924 rectangular form of a complex number, 991 recursion
equation, 654 reduced row echelon form, 570 reference angle, 721 Reference Angle Theorem for cosine and sine, 722 for the circular functions, 747 reﬂection of a function graph, 126 of a point, 10 regression coeﬃcient of determination, 226 correlation coeﬃcient, 226 least squares line, 225 quadratic, 228 total squared error, 225 relation algebraic description, 23 deﬁnition, 20 Fundamental Graphing Principle, 23 Remainder Theorem, 258 revenue function, 82 Richter Scale, 431 right angle, 694 root index, 397 radicand, 397 Roots of Unity, 1006 rotation matrix, 986 rotation of axes, 974 row echelon form, 569 row operations for a matrix, 568 scalar multiplication matrix associative property of, 581 deﬁnition of, 580 distributive properties of, 581 properties of, 581 vector associative property of, 1018 deﬁnition of, 1017 distributive properties of, 1018 properties of, 1018 scalar projection, 1039 secant graph of, 800 of an angle, 744, 752 properties of, 802 secant line, 160 sequence nth term, 652 alternating, 652 arithmetic common diﬀerence, 654 deﬁnition of, 654 formula for nth term, 656 sum of ﬁrst n terms, 666 deﬁnition of, 652 geometric common ratio, 654 deﬁnition of, 654 formula for nth term, 656 sum of ﬁrst n terms, 666 1080 recursive, 654 series, 668 set deﬁnition of, 1 empty, 2 intersection, 4 roster method, 1 set-builder notation, 1 sets of numbers, 2 union, 4 verbal description, 1 set-builder notation, 1 Side-Angle-Side triangle, 910 Side-Side-Side triangle, 910 sign diagram algebraic function, 399 for quadratic inequality, 214 polynomial function, 242 rational function, 321 simple interest, 469 sine graph of, 792 of an angle, 717, 730, 744 properties of, 791 sinusoid amplitude, 794, 881 baseline, 881 frequency angular, 881 ordinary, 881 graph of,
795, 882 period, 881 phase, 881 phase shift, 795, 881 properties of, 881 vertical shift, 881 slant asymptote, 311 slant asymptote determination of, 312 formal deﬁnition of, 311 slope deﬁnition, 151 Index of a line, 151 rate of change, 154 slope-intercept form of a line, 155 smooth, 241 sound intensity level decibel, 431 square matrix, 586 standard position of a vector, 1019 standard position of an angle, 698 start-up cost, 82 steady state, 592 stochastic process, 592 straight angle, 693 Sum Identity for cosine, 771, 775 for sine, 773, 775 for tangent, 775 Sum to Product Formulas, 781 summation notation deﬁnition of, 661 index of summation, 661 lower limit of summation, 661 properties of, 664 upper limit of summation, 661 supplementary angles, 696 symmetry about the x-axis, 9 about the y-axis, 9 about the origin, 9 testing a function graph for, 95 testing an equation for, 26 synthetic division tableau, 260 system of equations back-substitution, 560 coeﬃcient matrix, 590 consistent, 553 constant matrix, 590 deﬁnition, 549 dependent, 554 free variable, 552 Gauss-Jordan Elimination, 571 Gaussian Elimination, 557 Index 1081 inconsistent, 553 independent, 554 leading variable, 556 linear n variables, 554 two variables, 550 linear in form, 646 non-linear, 637 overdetermined, 554 parametric solution, 552 triangular form, 556 underdetermined, 554 unknowns matrix, 590 tangent graph of, 804 of an angle, 744, 752 properties of, 806 terminal side of an angle, 698 Thurstone, Louis Leon, 315 total squared error, 225 transformation non-rigid, 129 rigid, 129 transformations of function graphs, 120, 135 transverse axis of a hyperbola, 531 Triangle Inequality, 183 triangular form, 556 underdetermined system, 554 uninhibited growth, 472 union of two sets, 4 Unit Circle deﬁnition of, 501 important points, 724 unit vector, 1023 Upper
and Lower Bounds Theorem, 274 upper triangular matrix, 593 variable dependent, 55 independent, 55 variable cost, 159 variation constant of proportionality, 350 direct, 350 inverse, 350 joint, 350 variations in sign, 273 vector x-component, 1012 y-component, 1012 addition associative property, 1015 commutative property, 1015 deﬁnition of, 1014 properties of, 1015 additive identity, 1015 additive inverse, 1015, 1018 angle between two, 1035, 1036 component form, 1012 Decomposition Theorem Generalized, 1040 Principal, 1024 deﬁnition of, 1012 direction deﬁnition of, 1020 properties of, 1020 dot product commutative property of, 1034 deﬁnition of, 1034 distributive property of, 1034 geometric interpretation, 1035 properties of, 1034 relation to magnitude, 1034 relation to orthogonality, 1037 work, 1042 head, 1012 initial point, 1012 magnitude deﬁnition of, 1020 properties of, 1020 relation to dot product, 1034 normalization, 1024 orthogonal projection, 1038 1082 Index multiplicity of, 244 of a function, 95 upper and lower bounds, 274 orthogonal vectors, 1037 parallel, 1030 principal unit vectors, ˆı, ˆ, 1024 resultant, 1013 scalar multiplication associative property of, 1018 deﬁnition of, 1017 distributive properties, 1018 identity for, 1018 properties of, 1018 zero product property, 1018 scalar product deﬁnition of, 1034 properties of, 1034 scalar projection, 1039 standard position, 1019 tail, 1012 terminal point, 1012 triangle inequality, 1044 unit vector, 1023 velocity average angular, 707 instantaneous, 707 instantaneous angular, 707 vertex of a hyperbola, 531 of a parabola, 188, 505 of an angle, 693 of an ellipse, 516 vertical asymptote formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 306 vertical line, 23 Vertical Line Test (VLT), 43 whole number deﬁnition of, 2 set of, 2 work, 1041 wrapping function, 704 zeroferent
times and see how they differ (i.e., how they have changed). The two times of comparison are usually called the ﬁnal time and the initial time. We really need to be precise about this, so here is what we mean: ∆ quantity = (value of quantity at ﬁnal time) − (value of quantity at initial time) ∆ time = (ﬁnal time) − (initial time). For example, suppose that on June 4 we measure that the temperature at 8:00 am is 65◦ F and at 10:00 am it is 71◦ F. So, the ﬁnal time is 10:00 am, the initial time is 8:00 am and the temperature is changing according to the rate = ∆ quantity ∆ time = = ﬁnal value of quantity − initial value of quantity ﬁnal time − initial time 71 − 65 degrees 10:00 − 8:00 hours = 3 deg/hr. As a second example, suppose on June 5 the temperature at 8:00 am is 71◦ and at 10:00 am it is 65◦. So, the ﬁnal time is 10:00 am, the initial 1.2. TOTAL CHANGE = RATE TIME × 5 time is 8:00 am and the temperature is changing according to the rate = ∆ quantity ∆ time = = ﬁnal value of quantity − initial value of quantity ﬁnal time − initial time 65 − 71 degrees 10:00 − 8:00 hours = −3 deg/hr. These two examples illustrate that a rate can be either a positive or a negative number. More importantly, it highlights that we really need to be careful when making a rate computation. In both examples, the initial and ﬁnal times are the same and the two temperatures involved are the same, BUT whether they occur at the initial or ﬁnal time is interchanged. If we accidently mix this up, we will end up being off by a minus sign. There are many situations where the rate is the same for all time periods. In a case like this, we say we have a constant rate. For example, imagine you are driving down the freeway at a constant speed of 60 mi/hr. The fact that the speedometer needle indicates a steady speed of 60 mi/hr means the rate your distance is changing is constant. In
cases when we have a constant rate, we often want to ﬁnd the total amount of change in the quantity over a speciﬁc time period. The key principle in the background is this: Total Change in some Quantity = Rate Time × (1.2) It is important to mention that this formula only works when we have a constant rate, but that will be the only situation we encounter in this course. One of the main goals of calculus is to develop a version of (1.2) that works for non-constant rates. Here is another example; others will occur throughout the text. It is Example 1.2.1. A water pipe mounted to the ceiling has a leak. dripping onto the ﬂoor below and creates a circular puddle of water. The surface area of this puddle is increasing at a constant rate of 4 cm2/hour. Find the surface area and dimensions of the puddle after 84 minutes. Solution. The quantity changing is “surface area” and we are given a “rate” and “time.” Using (1.2) with time t = 84 minutes, Total Surface Area = Rate Time × cm2 hr = 4 = 5.6 cm2. 84 60 hr × The formula for the area of a circular region of radius r is given at the 5.6 π = 1.335 cm at back of this text. Using this, the puddle has radius r = time t = 84 minutes. q 6 CHAPTER 1. WARMING UP 1.3 The Modeling Process Modeling is a method used in disciplines ranging from architecture to zoology. This mathematical technique will crop up any time we are problem solving and consciously trying to both “describe” and “predict.” Inevitably, mathematics is introduced to add structure to the model, but the clean equations and formulas only arise after some (or typically a lot) of preliminary work. A model can be thought of as a caricature in that it will pick out certain features (like a nose or a face) and focus on those at the expense of others. It takes a lot of experience to know which models are “good” and “bad,” in the sense of isolating the right features. In the beginning, modeling will lead to frustration and confusion, but by the end of this course our comfort level will dramatically increase. Let’s look at an illustration of
the problem solving process. Example 1.3.1. How much time do you anticipate studying precalculus each week? Solution. One possible response is simply to say “a little” or “way too much!” You might not think these answers are the result of modeling, but they are. They are a consequence of modeling the total amount of study time in terms of categories such as “a little,” “some,” “lots,” “way too much,” etc. By drawing on your past experiences with math classes and using this crude model you arrived at a preliminary answer to the question. Let’s put a little more effort into the problem and try to come up with a numerical estimate. If T is the number of hours spent on precalculus a given week, it is certainly the case that: T = (hours in class) + (hours reading text) + (hours doing homework) Our time in class each week is known to be 5 hours. However, the other two terms require a little more thought. For example, if we can comfortably read and digest a page of text in (on average) 15 minutes and there are r pages of text to read during the week, then (hours reading text) = 15 60 r hours. As for homework, if a typical homework problem takes (on average) 25 minutes and there are h homework problems for the week, then (hours doing homework) = 25 60 h hours. We now have a mathematical model for the weekly time commitment to precalculus: T = 5 + 15 60 r + 25 60 h hours. 1.3. THE MODELING PROCESS 7 Is this a good model? Well, it is certainly more informative than our original crude model in terms of categories like “a little” or “lots.” But, the real plus of this model is that it clearly isolates the features being used to make our estimated time commitment and it can be easily modiﬁed as the amount of reading or homework changes. So, this is a pretty good model. However, it isn’t perfect; some homework problems will take a lot more than 25 minutes! 8 CHAPTER 1. WARMING UP 1.4 Exercises Problem 1.1. (a) Verify that 7685.33 seconds is 2 hours 8 minutes 5.33 seconds. (b) Allyson has a pace of 6
min/mile; what is her speed? (b) Which is faster: 100 mph or 150 ft/s? (c) Gina’s salary is 1 cent/second for a 40 hour work week. Tiare’s salary is $1400 for a 40 hour work week. Who has a higher salary? (d) Suppose it takes 180 credits to get a You accumubaccalaureate degree. late credit at the rate of one credit per quarter for each hour that the class meets per week. For instance, a class that meets three hours each week of the quarter will count for three credits. In addition, suppose that you spend 2.5 hours of study outside of class for each hour in class. A quarter is 10 weeks long. How many total hours, including time spent in class and time spent studying out of class, must you invest to get a degree? Problem 1.2. Sarah can bicycle a loop around the north part of Lake Washington in 2 hours and 40 minutes. If she could increase her average speed by 1 km/hr, it would reduce her time around the loop by 6 minutes. How many kilometers long is the loop? Problem 1.3. The is 11.34 g/cm3 and the density of aluminum is 2.69 g/cm3. Find the radius of lead and aluminum spheres each having a mass of 50 kg. density lead of Problem 1.4. The Eiffel Tower has a mass of 7.3 million kilograms and a height of 324 meIts base is square with a side length of ters. 125 meters. The steel used to make the Tower occupies a volume of 930 cubic meters. Air has a density of 1.225 kg per cubic meter. Suppose the Tower was contained in a cylinder. Find the mass of the air in the cylinder. Is this more or less than the mass of the Tower? Problem 1.5. Marathon runners keep track of their speed using units of pace = minutes/mile. (a) Lee has a speed of 16 ft/sec; what is his pace? (c) Adrienne and Dave are both running a race. Adrienne has a pace of 5.7 min/mile and Dave is running 10.3 mph. Who is running faster? Problem 1.6. Convert each of the following sentences into “pseudo-equations.” For example, suppose you start with the sentence: “The cost of the book was more than $10 and
the cost of the magazine was $4.” A ﬁrst step would be these “pseudo-equations”: (Book cost) > $10 and (Magazine cost) = $4. (a) John’s salary is $56,000 a year and he pays no taxes. (b) John’s salary is at most $56,000 a year and he pays 15% of his salary in taxes. (c) John’s salary is at least $56,000 a year and he pays more than 28% of his salary in taxes. (d) The number of students taking Math 120 at the UW is somewhere between 1500 and 1800 each year. (e) The cost of a new red Porsche is more than three times the cost of a new Ford F-150 pickup truck. (f) Each week, students spend at least two but no more than three hours studying for each credit hour. (g) Twice the number of happy math students exceeds ﬁve times the number of happy chemistry students. However, all of the happy math and chemistry students combined is less than half the total number of cheerful biology students. (h) The difference between Cady’s high and low midterm scores was 10%. Her ﬁnal exam score was 97%. (i) The vote tally for Gov. Tush was within one-hundredth of one percent of onehalf the total number of votes cast. Problem 1.7. Which is a better deal: A 10 inch diameter pizza for $8 or a 15 inch diameter pizza for $16? 1.4. EXERCISES 9 Problem 1.8. The famous theory of relativity predicts that a lot of weird things will happen when you approach the speed of light 108 m/sec. For example, here is a forc = 3 mula that relates the mass mo (in kg) of an object at rest and its mass when it is moving at a speed v: × m = mo 1 − v2 c2. q (a) Suppose the object moving is Dave, who has a mass of mo = 66 kg at rest. What is Dave’s mass at 90% of the speed of light? At 99% of the speed of light? At 99.9% of the speed of light? (b) How fast should Dave be moving to have a mass of 500 kg? Problem 1.9. During a typical evening in Seattle, Pag
liacci receives phone orders for pizza delivery at a constant rate: 18 orders in a typical 4 minute period. How many pies are sold in 4 hours? Assume Pagliacci starts taking orders at 5 : 00 pm and the proﬁt is a constant rate of $11 on 10 orders. When will phone order proﬁt exceed $1,000? Problem 1.10. Aleko’s Pizza has delivered a beautiful 16 inch diameter pie to Lee’s dorm room. The pie is sliced into 8 equal sized pieces, but Lee is such a non-conformist he cuts off an edge as pictured. John then takes one of the remaining triangular slices. Who has more pizza and by how much? John’s part Lee’s part Problem 1.11. A typical cell in the human body contains molecules of deoxyribonucleic acid, referred to as DNA for short. In the cell, this DNA is all twisted together in a tight little packet. But imagine unwinding (straightening out) all of the DNA from a single typical cell and laying it “end-to-end”; then the sum total length will be approximately 2 meters. isolate DNA from nucleus cell nucleus lay out end−to−end 2 m Assume the human body has 1014 cells containing DNA. How many times would the sum total length of DNA in your body wrap around the equator of the earth? Problem 1.12. A water pipe mounted to the ceiling has a leak and is dripping onto the ﬂoor below, creating a circular puddle of water. The area of the circular puddle is increasing at a constant rate of 11 cm2/hour. (a) Find the area and radius of the puddle after 1 minute, 92 minutes, 5 hours, 1 day. (b) Is the radius of the puddle increasing at a constant rate? Problem 1.13. During the 1950s, Seattle was dumping an average of 20 million gallons of sewage into Lake Washington each day. (a) How much sewage went into Lake Wash- ington in a week? In a year? (b) In order to illustrate the amounts inimagine a rectangular prism volved, whose base is the size of a football ﬁeld (100 yards 50 yards) with height h yards. What are the dimensions of such a rectangular prism containing the × 10 CHAPTER 1. WARMING UP sewage dumped into Lake Washington in a
single day? (Note: There are 7.5 gallons in one cubic foot. Dumping into Lake Washington has stopped; now it goes into the Puget Sound.) ter 1, 2, 5 and 20 years. Would tax rates increase or decrease over time? Congress claims that this law would ultimately cut peoples’ tax rates by 75 %. Do you believe this claim? Problem 1.14. Dave has inherited an apple orchard on which 60 trees are planted. Under these conditions, each tree yields 12 bushels of apples. According to the local WSU extension agent, each time Dave removes a tree the yield per tree will go up 0.45 bushels. Let x be the number of trees in the orchard and N the yield per tree. (a) Find a formula for N in terms of the unknown x. (Hint: Make a table of data with one column representing various values of x and the other column the corresponding values of N. After you complete the ﬁrst few rows of the table, you need to discover the pattern.) (b) What possible reason(s) might explain why the yield goes up when you remove trees? Problem 1.15. Congress is debating a proposed law to reduce tax rates. If the current tax rate is r %, then the proposed rate after x years is given by this formula Rewrite this formula as a simple fraction. Use your formula to calculate the new tax rate af- Problem 1.16. (a) The temperature at 7:00 am is 44◦F and the temperature at 10:00 am is 50◦F. What are the initial time, the ﬁnal time, the initial temperature and the ﬁnal temperature? What is the rate of change in the temperature between 7:00 am and 10:00 am? (b) Assume it is 50◦F at 10:00 am and the rate of change in the temperature between 10:00 am and 2:00 pm is the same as the rate in part (a). What is the temperature at 2:00 pm? (c) The temperature at 4:30 pm is 54◦F and the temperature at 6:15 pm is 26◦F. What are the initial time, the ﬁnal time, the initial temperature and the ﬁnal temperature? What is the rate of change in the temperature between 4:30 pm and 6:15 pm? Problem 1.
17. (a) Solve for t: 3t −7 = 11+ t. (b) Solve for a: 1 + 1 a = 3. q (c) Solve for x: √x2 + a2 = 2a + x. (d) Solve for t: 1 − t > 4 − 2t. (e) Write as a single fraction: 2 x − 1 x + 1 Chapter 2 Imposing Coordinates You ﬁnd yourself visiting Spangle, WA and dinner time is approaching. A friend has recommended Tiff’s Diner, an excellent restaurant; how will you ﬁnd it? Of course, the solution to this simple problem amounts to locating a “point” on a two-dimensional map. This idea will be important in many problem solving situations, so we will quickly review the key ideas. Q P Figure 2.1: Two points in a plane. 2.1 The Coordinate System If we are careful, we can develop the ﬂow of ideas underlying two-dimensional coordinate systems in such a way that it easily generalizes to three-dimensions. Suppose we start with a blank piece of paper and mark two points; let’s label these two points “P ” and “Q.” This presents the basic problem of ﬁnding a foolproof method to reconstruct the picture. The basic idea is to introduce a coordinate system for the plane (analogous to the city map grid of streets), allowing us to catalog points in the plane using pairs of real numbers (analogous to the addresses of locations in the city). Here are the details. Start by drawing two perpendicular lines, called the horizontal axis and the vertical axis, each of which looks like a copy of the real number line. We refer to the intersection point of these two lines as the origin. Given P in the plane, the plan is to use these two axes to obtain a pair of real numbers (x,y) that will give us the exact location of P. With this in mind, the horizontal axis is often called the x-axis and the vertical axis is often called the y-axis. Remember, a typical real number line (like the x-axis or the y-axis) is divided into three parts: the positive numbers, the negative numbers, and the number zero (see Figure 2.2(a)). This allows us to specify positive and negative portions of the x-
axis and y-axis. Unless we say otherwise, we will always adopt the convention that the positive x-axis consists of those numbers to the right of the origin on 11 12 CHAPTER 2. IMPOSING COORDINATES the x-axis and the positive y-axis consists of those numbers above the origin on the y-axis. We have just described the xy-coordinate system for the plane: Negative real numbers Positive real numbers Positive y-axis Origin Zero Negative x-axis Positive x-axis Negative y-axis (a) Number line. (b) xy-coordinate system. Figure 2.2: Coordinates. 2.1.1 Going from P to a Pair of Real Numbers. y-axis y P ℓ x-axis x ℓ∗ Figure 2.3: Coordinate pairs. Imagine a coordinate system had been drawn on our piece of paper in Figure 2.1. Let’s review the procedure of going from a point P to a pair of real numbers: 1. First, draw two new lines passing through P, one parallel to the x-axis and the other parallel to the y-axis; call these ℓ and ℓ∗, as pictured in Figure 2.3. 2. Notice that ℓ will cross the y-axis exactly once; the point on the y-axis where these two lines cross will be called “y.” Likewise, the line ℓ∗ will cross the x-axis exactly once; the point on the x-axis where these two lines cross will be called “x.” 3. If you begin with two different points, like P and Q in Figure 2.1, you will see that the two pairs of points you obtain will be different; i.e., if Q gives you the pair (x∗,y∗), then either x = y∗. This shows that two different points in the plane give two different pairs of real numbers and describes the process of assigning a pair of real numbers to the point P. = x∗ or y The great thing about the procedure we just described is that it is reversible! In other words, suppose you start with a pair of real numbers, say (x,y). Locate the number x on the x-axis and the number y on the y-axis. Now draw two lines: a line ℓ parallel to the x-axis passing through the
number y on the y-axis and a line ℓ∗ parallel to the y-axis passing through the number x on the x-axis. The two lines ℓ and ℓ∗ will intersect 6 6 2.2. THREE FEATURES OF A COORDINATE SYSTEM 13 in exactly one point in the plane, call it P. This procedure describes how to go from a given pair of real numbers to a point in the plane. In addition, if you start with two different pairs of real numbers, then the corresponding two points in the plane are going to be different. In the future, we will constantly be going back and forth between points in the plane and pairs of real numbers using these ideas. Deﬁnition 2.1.1. Coordinate System: Every point P in the xy-plane corresponds to a unique pair of real numbers (x, y), where x is a number on the horizontal x-axis and y is a number on the vertical y-axis; for this reason, we commonly use the notation “P = (x,y).” Having speciﬁed positive and negative directions on the horizontal and vertical axes, we can now divide our two dimensional plane into four quadrants. The ﬁrst quadrant corresponds to all the points where both coordinates are positive, the second quadrant consists of points with the ﬁrst coordinate negative and the second coordinate positive, etc. Every point in the plane will lie in one of these four quadrants or on one of the two axes. This quadrant terminology is useful to give a rough sense of location, just as we use the terminology “Northeast, Northwest, Southwest and Southeast” when discussing locations on a map. y-axis Second Quadrant First Quadrant Third Quadrant Fourth Quadrant x-axis Figure 2.4: Quadrants in the xy-plane. 2.2 Three Features of a Coordinate System A coordinate system involves scaling, labeling and units on each of the axes. 2.2.1 Scaling Sketch two xy coordinate systems. In the ﬁrst, make the scale on each axis the same. In the second, assume “one unit” on the x axis has the same length as “two units” on the y axis. Plot the points (1,1), (−1,1), 5, 16 − 4, (1,1)., (
0,0), Both pictures illustrate how the points lie on a parabola in the xy- coordinate system, but the aspect ratio has changed. The aspect ratio is deﬁned by this fraction: 5, 16 5, 4 5, 9 5, 1 5, 1 5, 4 5, 9 − 1 − 2 − 3 25 25 25 25 25 25 25 25,,,,,, 1 2 3 4 aspect ratio def= length of one unit on the vertical axis length of one unit on the horizontal axis. Figure 2.5(a) has aspect ratio 1, whereas Figure 2.5(b) has aspect ratio 1 2. In problem solving, you will often need to make a rough assumption about the relative axis scaling. This scaling will depend entirely on the 14 CHAPTER 2. IMPOSING COORDINATES y-axis 1.0 0.8 0.6 0.4 0.2 −1.0 −0.5 0.0 0.5 x-axis 1.0 y-axis 1.0 0.8 0.6 0.4 0.2 x-axis −1.0 −0.5 0.0 0.5 1.0 (a) Aspect ratio = 1. (b) Aspect ratio = 1 2. Figure 2.5: Coordinates. information given in the problem. Most graphing devices will allow you to specify the aspect ratio. 2.2.2 Axes Units Sometimes we are led to coordinate systems where each of the two axes involve different types of units (labels). Here is a sample, that illustrates the power of using pictures. Example 2.2.1. As the marketing director of Turbowebsoftware, you have been asked to deliver a brief message at the annual stockholders meeting on the performance of your product. Your staff has assembled this tabular collection of data; how can you convey the content of this table most clearly? TURBOWEB SALES (in $1000’s) week 1 2 3 4 5 6 7 8 9 10 sales 11.0517 12.214 13.4986 14.9182 16.4872 18.2212 20.1375 22.2554 24.596 27.1828 week 11 12 13 14 15 16 17 18 19 20 sales 30.0417 33.2012 36.693 40.552 44.8169 49.5303 54.7395 60.4965 66.8
589 73.8906 week 21 22 23 24 25 26 27 28 29 30 sales 81.6617 90.2501 99.7418 110.232 121.825 134.637 148.797 164.446 181.741 200.855 week 31 32 33 34 35 36 37 38 39 40 sales 221.98 245.325 271.126 299.641 331.155 365.982 404.473 447.012 494.024 545.982 week 41 42 43 44 45 46 47 48 49 50 sales 603.403 666.863 736.998 814.509 900.171 994.843 1099.47 1215.1 1342.9 1,484.131 One idea is to simply ﬂash an overhead slide of this data to the audience; this can be deadly! A better idea is to use a visual aid. Suppose we let the variable x represent the week and the variable y represent the gross sales (in thousands of dollars) in week x. We can then plot the points (x,y) in the xy-coordinate system; see Figure 2.6. Notice, the units on the two axes are very different: y-axis units are “thousands of dollars” and x-axis units are “weeks.” In addition, the aspect ratio of this coordinate system is not 1. The beauty of this picture is the visual impact it gives your audience. From the coordinate plot we can get a sense of how the sales ﬁgures are dramatically increasing. In fact, this plot is good evidence you deserve a big raise! 2.3. A KEY STEP IN ALL MODELING PROBLEMS 15 Mathematical modeling is all about relating concrete phenomena and symbolic equations, so we want to embrace the idea of visualization. Most typically, visualization will involve plotting a collection of points in the plane. This can be achieved by providing a “list” or a “prescription” for plotting the points. The material we review in the next couple of sections makes the transition from symbolic mathematics to visual pictures go more smoothly. y-axis (Thousands of Dollars) 1,400 1,200 1,000 800 600 400 200 x-axis (Weeks) 10 20 30 40 50 Figure 2.6: Turboweb sales. 2.3 A Key Step in all Modeling Problems The initial problem solving or modeling step of deciding on a choice of
xy-coordinate system is called imposing a coordinate system: There will often be many possible choices; it takes problem solving experience to develop intuition for a “natural” choice. This is a key step in all modeling problems. Example 2.3.1. Return to the tossed ball scenario on page 1. How do we decide where to draw a coordinate system in the picture? Figure 2.7 on page 16 shows four natural choices of xy-coordinate system. To choose a coordinate system we must specify the origin. The four logical choices for the origin are either the top of the cliff, the bottom of the cliff, the launch point of the ball or the landing point of the ball. So, which choice do we make? The answer is that any of these choices will work, but one choice may be more natural than another. For example, Figure 2.7(b) is probably the most natural choice: in this coordinate system, the motion of the ball takes place entirely in the ﬁrst quadrant, so the x and y coordinates of any point on the path of the ball will be non-negative. Example 2.3.2. Michael and Aaron are running toward each other, beginning at opposite ends of a 10,000 ft. airport runway, as pictured in Figure 2.8 on page 17. Where and when will these guys collide? Solution. This problem requires that we ﬁnd the “time” and “location” of the collision. Our ﬁrst step is to impose a coordinate system. We choose the coordinate system so that Michael is initially located at the point M = (0, 0) (the origin) and Aaron is initially located at the point A = (10,000, 0). To ﬁnd the coordinates of Michael after t seconds, we need to think about how distance and time are related. Since Michael is moving at the rate of 15 ft/second, then after one second he is located 15 feet right of the origin; i.e., at the point (15, 0). After 2 seconds, Michael has moved an additional 15 feet, for a total of 30 feet; so he is located at the point (30, 0), etc. Conclude Michael has traveled 15t ft. to the right after t seconds; i.e., his location is the point 16 CHAPTER 2. IMPOSING COORDINATES y-axis Path of tossed ball. x
-axis Cliff. y-axis Path of tossed ball. Cliff. x-axis (a) Origin at the top of the ledge. (b) Origin at the bottom of the ledge. Path of tossed ball. y-axis y-axis Path of tossed ball. Cliff. x-axis Cliff. x-axis (c) Origin at the landing point. (d) Origin at the launch point. Figure 2.7: Choices when imposing an xy-coordinate system. M(t) = (15t, 0). Similarly, Aaron is located 8 ft. left of his starting location after 1 second (at the point (9,992, 0)), etc. Conclude Aaron has traveled 8t ft. to the left after t seconds; i.e., his location is the point A(t) = (10,000 − 8t, 0). The key observation required to solve the problem is that the point of collision occurs when the coordinates of Michael and Aaron are equal. Because we are moving along the horizontal axis, this amounts to ﬁnding where and when the x-coordinates of M(t) and A(t) agree. This is a straight forward algebra problem: 15t = 10,000 − 8t 23t = 10,000 t = 434.78 (2.1) To the nearest tenth of a second, the runners collide after 434.8 seconds. Plugging t = 434.78 into either expression for the position: M(434.8) = (15(434.8), 0) = (6,522, 0). 2.4. DISTANCE Michael: 15 ft sec Aaron: 8 ft sec y-axis 17 x-axis 10,000 ft M = (0,0) A = (10,000, 0) (a) The physical picture. (b) The xy-coordinate picture. y-axis NOT TO SCALE! Michael starts here. Aaron starts here. (0,0) M(t) = (15t, 0) A(t) = (10,000 − 8t, 0) M after t seconds. A after t seconds. x-axis (c) Building a visual model Michael: 10 mph. Aaron: 8 mph. 6,522 feet to collision point. (d) Michael and Aaron’s collision point. Figure 2.8: Michael and Aaron running head-on. 2.4 Distance We end
this Chapter with a discussion of direction and distance in the plane. To set the stage, think about the following analogy: Example 2.4.1. You are in an airplane ﬂying from Denver to New York. How far will you ﬂy? To what extent will you travel north? To what extent will you travel east? Consider two points P = (x1,y1) and Q = (x2,y2) in the xy coordinate system, where we assume that the units on each axis are the same; for example, both in units of “feet.” Imagine starting at the location P (Denver) and ﬂying to the location Q (New York) along a straight line segment; see Figure 2.9(a). Now ask yourself this question: To what overall extent have the x and y coordinates changed? To answer this, we introduce visual and notational aides into this ﬁgure. We have inserted an “arrow” pointing from the starting position P to the ending position Q; see Figure 2.9(b). To simplify things, introduce the notation ∆x to keep track of the change in the x-coordinate and ∆y 18 CHAPTER 2. IMPOSING COORDINATES y-axis y-axis???? P Q End (stop) here. Begin (start) here. x-axis?? ∆y y2 y1 Q = (x2, y2) Ending point. d P = (x1, y1) Beginning point. x1 x2 ∆x x-axis (a) Starting and stopping points. (b) Coordinates for P and Q. Figure 2.9: The meaning of ∆x and ∆y. to keep track of the change in the y-coordinate, as we move from P to Q. Each of these quantities can now be computed: ∆x = change in x-coordinate going from P to Q (2.2) = (x-coord of ending point) − (x-coord of beginning point) = x2 − x1 ∆y = change in y-coordinate going from P to Q = (y-coord of ending point) − (y-coord of beginning point) = y2 − y1. We can interpret ∆x and ∆y using the right triangle in Figure 2.9(b). This means we can use the Pyth
agorean Theorem to write: d2 = (∆x)2 + (∆y)2; that is, d = (∆x)2 + (∆y)2, p which tells us the distance d from P to Q. In other words, d is the distance we would ﬂy if we had ﬂown along that line segment connecting the two points. As an example, if P = (1, 1) and Q = (5, 4), then ∆x = 5 − 1 = 4, ∆y = 4 − 1 = 3 and d = 5. There is a subtle idea behind the way we deﬁned ∆x and ∆y: You need to specify the “beginning” and “ending” points used to do the calculation in Equations 2.2. What happens if we had reversed the choices in Figure 2.9? Then the quantities ∆x and ∆y will both be negative and the lengths of the sides of the right triangle are computed by taking the absolute value 2.4. DISTANCE 19 of ∆x and ∆y. As far as a distance calculation is concerned, the previous formula still works because of this algebra equality: d = = (∆x)2 + (∆y)2 (\|∆x\|)2 + (\|∆y\|)2. p p We will sometimes refer to ∆x and ∆y as directed distances in the x and y directions. The notion of directed distance becomes important in our discussion of lines in Chapter 4 and later when you learn about vectors; it is also very important in calculus. For example, if P = (5, 4) and Q = (1, 1), then ∆x = 1 − 5 = −4, ∆y = 1 − 4 = −3 and d = 5. Important Fact 2.4.2 (Distance formula). If P = (x1, y1) and Q = (x2, y2) are two points in the plane, then the straight line distance between the points (in the same units as the two axes) is given by the formula (∆x)2 + (∆y)2 (x2 − x1)2 + (y2 − y1)2. d = = p p (2.3) y-axis \|∆y\| y
1 y2 P = (x1, y1) Beginning point. d Q = (x2, y2) Ending point. x2 x1 x-axis \|∆x\| Figure 2.10: A different direction. If your algebra is a little rusty, a very common mistake may crop up when you are using the distance formula. For example, 32 + 42? = √32 + √42? p √9 + 16 = 3 + 4 = 7. 5 Notice, you have an impossible situation: 5 is never equal to 7.!!! CAUTION!!! 6 20 CHAPTER 2. IMPOSING COORDINATES Example 2.4.3. Two cars depart from a four way intersection at the same time, one heading East and the other heading North. Both cars are traveling at the constant speed of 30 ft/sec. Find the distance (in miles) between the two cars after 1 hour 12 minutes. In addition, determine when the two cars would be exactly 1 mile apart. (0,b) y-axis (North) d x-axis (East) (a,0) Figure 2.11: Two departing cars. Solution. Begin with a picture of the situation. We have indicated the locations of the two vehicles after t seconds and the distance d between them at time t. By the distance formula, the distance between them is d = (a − 0)2 + (0 − b)2 = a2 + b2. p p This formula is a ﬁrst step; the difﬁculty is that we have traded the mystery distance d for two new unknown numbers a and b. To ﬁnd the coordinate a for the Eastbound car, we know the car is moving at the rate of 30 ft/sec, so it will travel 30t feet after t seconds; i.e., a = 30t. Similarly, we ﬁnd that b = 30t. Substituting into the formula for d we arrive at d = (30t)2 + (30t)2 2t2(30)2 = p = 30t√2. p First, we need to convert 1 hour and 12 minutes into seconds so that our formula can be used: 1 hr 12 min = 1 + 12/60 hr = 1.2 hr = (1.2hr) = 4,320 sec. 60 min hr 60 sec min Substituting t = 4
,320 sec and recalling that 1 mile = 5,280 feet, we arrive at d = 129,600√2 feet = 183,282 feet = 34.71 miles. For the second question, we specify the distance being 1 mile and want to ﬁnd when this occurs. The idea is to set d equal to 1 mile and solve for 2.4. DISTANCE 21 t. However, we need to be careful, since the units for d are feet: 30t√2 = d = 5,280 Solving for t: t = 5,280 30√2 = 124.45 seconds = 2 minutes 4 seconds. The two cars will be 1 mile apart in 2 minutes, 4 seconds. 22 CHAPTER 2. IMPOSING COORDINATES 2.5 Exercises Problem 2.1. In the following four cases, let P be the initial (starting) point and Q the ending point; recall Equation 2.2 and Figure 2.10 on Page 19. Compute d = the distance from P to Q, ∆x and ∆y. Give your answer in exact form; eg. √2 is an exact answer, whereas 1.41 is an approximation of √2. (a) P = (0,0), Q = (1,1). (b) P = (2,1), Q = (1, − 1). (c) P = (−1,2), Q = (4, − 1). (d) P = (1,2), Q = (1 + 3t,3 + t), where t is a constant. Problem 2.2. Start with two points M = (a,b) and N = (s,t) in the xy-coordinate system. Let d be the distance between these two points. Answer these questions and make sure you can justify your answers: (a) TRUE or FALSE: d = (a − s)2 + (b − t)2. (b) TRUE or FALSE: d = p (a − s)2 + (t − b)2. (c) TRUE or FALSE: d = p (s − a)2 + (t − b)2. (d) Suppose M is the beginning point and N is the ending point; recall Equation 2.2 and Figure 2.10 on Page 19. What is ∆x? What is ∆y? p (e)
Suppose N is the beginning point and M is the ending point; recall Equation 2.2 and Figure 2.10 on Page 19. What is ∆x? What is ∆y? (f) If ∆x=0, what can you say about the relationship between the positions of the two points M and N? If ∆y=0, what can you say about the relationship between the positions of the two points M and N? (Hint: Use some speciﬁc values for the coordinates and draw some pictures to see what is going on.) Problem 2.3. Steve and Elsie are camping in the desert, but have decided to part ways. Steve heads North, at 6 AM, and walks steadily at 3 miles per hour. Elsie sleeps in, and starts walking West at 3.5 miles per hour starting at 8 AM. When will the distance between them be 25 miles? Problem 2.4. Erik’s disabled sailboat is ﬂoating at a stationary location 3 miles East and 2 miles North of Kingston. A ferry leaves Kingston heading due East toward Edmonds at 12 mph. At the same time, Erik leaves the sailboat in a dinghy heading due South at 10 ft/sec (hoping to intercept the ferry). Edmonds is 6 miles due East of Kingston. sailboat Kingston Edmonds North Ballard UDub (a) Compute Erik’s speed in mph and the Ferry speed in ft/sec. (b) Impose a coordinate system and complete this table of data concerning locations (i.e., coordinates) of Erik and the ferry. Insert into the picture the locations of the ferry and Erik after 7 minutes. Time Ferry Erik Distance Between 0 sec 30 sec 7 min t hr (c) Explain why Erik misses the ferry. (d) After 10 minutes, a Coast Guard boat leaves Kingston heading due East at a speed of 25 ft/sec. Will the Coast Guard boat catch the ferry before it reaches Edmonds? Explain. Problem 2.5. Suppose two cars depart from a four way intersection at the same time, one heading north and the other heading west. The car heading north travels at the steady speed of 30 ft/sec and the car heading west travels at the steady speed of 58 ft/sec. 2.5. EXERCISES 23 (a) Find an expression for the distance between the two cars after t seconds. Brooke ocean (b) Find the distance in miles between the two
cars after 3 hours 47 minutes. 5 mi (c) When are the two cars 1 mile apart? kayak reaches shore here Problem 2.6. Allyson and Adrian have decided to connect their ankles with a bungee cord; one end is tied to each person’s ankle. The cord is 30 feet long, but can stretch up to 90 feet. They both start from the same location. Allyson moves 10 ft/sec and Adrian moves 8 ft/sec in the directions indicated. 20 ft Building 30 ft Allyson Adrian start (a) Where are the two girls located after 2 seconds? (b) After 2 seconds, will the slack in the bungee cord be used up? (c) Determine when the bungee cord ﬁrst becomes tight; there is no slack i.e. in the line. Where are the girls located when this occurs? (d) When will the bungee cord ﬁrst touch the corner of the building? (Hint: Use a fact about “similar triangles”.) Problem 2.7. Brooke is located 5 miles out from the nearest point A along a straight shoreline in her seakayak. Hunger strikes and she wants to make it to Kono’s for lunch; see picture. Brooke can paddle 2 mph and walk 4 mph. shore A Kono’s 6 mi (a) If she paddles along a straight line course to the shore, ﬁnd an expression that computes the total time to reach lunch in terms of the location where Brooke beaches the boat. (b) Determine the total time to reach Kono’s if she paddles directly to the point “A”. (c) Determine the total time to reach Kono’s if she paddles directly to Kono’s. (d) Do you think your answer to (b) or (c) is the minimum time required for Brooke to reach lunch? (e) Determine the total time to reach Kono’s if she paddles directly to a point on the shore half way between point “A” and Kono’s. How does this time compare to the times in parts (b) and (c)? Do you need to modify your answer to part (d)? Problem 2.8. A spider is located at the position (1,2) in a coordinate system, where the units on each axis are feet.
An ant is located at the position (15,0) in the same coordinate system. Assume the location of the spider after t minutes is s(t) = (1 + 2t,2 + t) and the location of the ant after t minutes is a(t) = (15 − 2t,2t). (a) Sketch a picture of the situation, indicating the locations of the spider and ant at times t = 0,1,2,3,4,5 minutes. Label the locations of the bugs in your picture, using the notation s(0), s(1),...,s(5), a(0), a(1),..., a(5). (b) When will the x-coordinate of the spider equal 5? When will the y-coordinate of the ant equal 5? (c) Where is the spider located when its y-coordinate is 3? (d) Where is each bug located when the y-coordinate of the spider is twice as large as the y-coordinate of the ant? 24 CHAPTER 2. IMPOSING COORDINATES (e) How far apart are the bugs when their x-coordinates coincide? Draw a picture, indicating the locations of each bug when their x-coordinates coincide. (f) A sugar cube is located at the position (9,6). Explain why each bug will pass through the position of the sugar cube. Which bug reaches the sugar cube ﬁrst? (g) Find the speed of each bug along its line of motion; which bug is moving faster? Problem 2.9. A Ferrari is heading south at a constant speed on Broadway (a north/south street) at the same time a Mercedes is heading west on Aloha Avenue (an east/west street). The Ferrari is 624 feet north of the intersection of Broadway and Aloha, at the same time that the Mercedes is 400 feet east of the intersection. Assume the Mercedes is traveling at the constant speed of 32 miles/hour. Find the speed of the Ferrari so that a collision occurs in the intersection of Broadway and Aloha. Problem 2.10. Two planes ﬂying opposite directions (North and South) pass each other 80 miles apart at the same altitude. The Northbound plane is ﬂying 200 mph (miles per hour) and the Southbound plane is ﬂying 150 mph. How far apart are the planes in 20 minutes? When
are the planes 300 miles apart? Problem 2.11. Here is a list of some algebra problems with ”solutions.” Some of the solutions are correct and some are wrong. For each problem, determine: (i) if the answer is correct, (ii) if the steps are correct, (iii) identify any incorrect steps in the solution (noting that the answer may be correct but some steps may not be correct). (a) If x = 1, x2 − 1 x + 1 = x2 + (−1)1 x + 1 −1 1 + = x2 x = x − 1 (b) (x + y)2 − (x − y)2 = (x2 + y2) − x2 − y2 = 0 (c) If x = 4, 9(x − 4)2 3x − 12 = 32(x − 4)2 3x − 12 (3x − 12)2 3x − 12 = 3x − 12. = Problem 2.12. Assume α, β are nonzero constants. Solve for x. (a) αx + β = 1 αx−β (b) 1 α + 1 (c Problem 2.13. Simplify as far as possible. (a) (1 − t)2 + (2 + 2t)2 (b) (t + 1)2 + (−t − 1)2 − 2 (c) (d) 1 t−1 − 1 t+1 (write as a single fraction) (2 + t)2 + 4t2 p 6 6 Chapter 3 Three Simple Curves Before we discuss graphing, we ﬁrst want to become acquainted with the sorts of pictures that will arise. This is surprisingly easy to accomplish: Impose an xy-coordinate system on a blank sheet of paper. Take a sharp pencil and begin moving it around on the paper. The resulting picture is what we will call a curve. For example, here is a sample of the sort of “artwork” we are trying to visualize. A number of examples in the text will involve basic curves in the plane. When confronted with a curve in the plane, the fundamental question we always try to answer is this: Can we give a condition (think of it as a “test”) that will tell us preciselywhen a pointin the plane lies on a curve? y-axis A typical curve. x-
axis Figure 3.1: A typical curve. Typically, the kind of condition we will give involves an equation in two variables (like x and y). We consider the three simplest situations in this chapter: horizontal lines, vertical lines and circles. 3.1 The Simplest Lines Undoubtedly, the simplest curves in the plane are the horizontal and vertical lines. For example, sketch a line parallel to the x-axis passing through 2 on the y-axis; the result is a horizontal line ℓ, as pictured. This means the line ℓ passes through the point (0, 2) in our coordinate system. A concise symbolic prescription for ALL of the points on ℓ can be given using “set notation”: ℓ = {(x, 2)\|x is any real number}. y-axis (−1, 2) (0, 2) (2, 2) (x, 2), a typical point. ℓ x-axis Figure 3.2: The points (x, y). We read the right-hand side of this expression as “the set of all points (x, 2) where x = any real number.” Notice, the points (x, y) on the line ℓ are EXACTLY the ones that lead to solutions of the equation y = 2; i.e., take any point on this line, plug the coordinates into the equation y = 2 25 26 CHAPTER 3. THREE SIMPLE CURVES and you get a true statement. Because the equation does not involve the variable x and only constrains y to equal 2, we see that x can take on any real value. In short, we see that plotting all of the solutions (x, y) to the equation y = 2 gives the line ℓ. We usually refer to the set of all solutions of the equation y = 2 as the graph of the equation y = 2. y-axis m (3, y), a typical point. (3, 3) As a second example, sketch the vertical line m passing through 3 on the x-axis; this means the line m passes through the point (3, 0) in our coordinate system. A concise symbolic prescription for ALL of the points on m can be given using “set notation”: m = {(3, y)\|y is any real number}. (3, 0) (3, −2) Figure 3.3: Stacked points
. x-axis Notice, the points (x, y) on the line m are EXACTLY the ones that lead to solutions of the equation x = 3; i.e., take any point on this line, plug the coordinates into the equation x = 3 and you get a true statement. Because the equation does not involve the variable y and only speciﬁes that x = 3, y can take on any real number value. Plotting all of the solutions (x, y) to the equation x = 3 gives the line m. We usually refer to the set of all solutions of the equation x = 3 as the graph of the equation x = 3. These two simple examples highlight our ﬁrst clear connection between a geometric ﬁgure and an equation; the link is achieved by plotting all of the solutions (x, y) of the equation in the xy-coordinate system. These observations work for any horizontal or vertical line. Deﬁnition 3.1.1. Horizontal and Vertical Lines: A horizontal line ℓ passing through k on the y-axis is precisely a plot of all solutions (x, y) of the equation y = k; i.e., ℓ is the graph of y = k. A vertical line m passing through h on the x-axis is precisely a plot of all solutions (x, y) of the equation x = h; i.e., m is the graph of x = h. 3.2 Circles Another common curve in the plane is a circle. Let’s see how to relate a circle and an equation involving the variables x and y. As a special case of the distance formula (2.3), suppose P = (0, 0) is the origin and Q = (x, y) is any point in the plane; then distance from P to Q = (x − 0)2 + (y − 0)2 x2 + y2. This calculation tells us that a point (x, y) is of distance r from the origin x2 + y2 or, squaring each side, that x2 + y2 = r2. This if and only if r = shows p p = {(x, y)\|distance (x, y) to origin is r} = {(x, y)\|x2 + y2 = r2 }. (3.1) p 3.2. CIRCLES
What is the left-hand side of Equation 3.1? To picture all points in the plane of distance r from the origin, fasten a pencil to one end of a non-elastic string (a string that will not stretch) of length r and tack the other end to the origin. Holding the string tight, the pencil point will locate a point of distance r from the origin. We could visualize all such points by simply moving the pencil around the origin, all the while keeping the string tight. 27 Pencil. r Start. r Draw with a tight string. Figure 3.4: Drawing a circle. What is the right-hand side of Equation 3.1? A point (x, y) in the right-hand set is a solution to the equation x2 + y2 = r2; i.e., if we plug in the coordinates we get a true statement. For example, in Figure 3.5 we plot eight solutions (r, 0), (−r, 0), (0, r), (0, −r), A =,, B =, r √2 r √2 r √2, −r √2 C = −r √2, −r √2, D = −r √2, r √2, of the equation. To see that the last point is a solution, here is the sample calculation: 2 + −r √2 2 r √2 = r2 2 + r2 2 = r2. Since the two sides of Equation 3.1 are equal, drawing the circle of radius r is the same as plotting all of the solutions of the equation x2 + y2 = r2. The same reasoning can be used to show that drawing a circle of radius r centered at a point (h, k) is the same as plotting all of the solutions of the equation: (x − h)2 + (y − k)2 = r2. We usually refer to the set of all solutions of the equation as the graph of the equation. (−r, 0) D C (0, r) A (r, 0) B (0, −r) Figure 3.5: Computing points. Deﬁnition 3.2.1 (Circles). Let (h, k) be a given point in the xy-plane and r > 0 a given positive real number. The circle of radius r centered at (h
, k) is precisely all of the solutions (x,y) of the equation (x − h)2 + (y − k)2 = r2; i.e., the circle is the graph of this equation. y-axis (x, y) r (h, k) x-axis We refer to the equation in the box as the standard form of the equation of a circle. From this equation you know both the center and radius of the circle described. Figure 3.6: Deﬁning a circle. 28 CHAPTER 3. THREE SIMPLE CURVES!!! CAUTION!!! Be very careful with the minus signs “−” in the standard form for a circle equation. For example, the equation (x + 3)2 + (y − 1)2 = 7 is NOT in standard form. We can rewrite it in standard form: (x − (−3))2 + (y − 1)2 = (√7)2; so, this equation describes a circle of radius √7 centered at (−3, 1). Examples 3.2.2. Here are some of the ways we can discuss circles: 1. The circle of radius 1 centered at the origin is the graph of the equation x2 + y2 = 1. This circle is called the unit circle and will be used extensively. 2. A circle of radius 3 centered at the point (h, k) = (1, −1) is the graph of the equation (x−1)2 +(y−(−1))2 = 32; or, equivalently (x−1)2+(y+1)2 = 32; or, equivalently x2 + y2 − 2x + 2y = 7. 3. The circle of radius √5 centered at (2, −3) does not pass through the origin; this is because (0, 0) is not a solution of the equation (x − 2)2 + (y + 3)2 = 5. 3.3 Intersecting Curves I In many problem solving situations, we will have two curves in the plane and need to determine where the curves intersect one another. Before we discuss a general procedure, let’s make sure we really understand the meaning of the word “intersect.” From Latin, the word “inter” means “within or in between” and the word “sectus” means
“to cut.” So, the intersection of two curves is the place where the curves “cut into” each other; in other words, where the two curves cross one another. If the pictures of two curves are given to us up front, we can often visually decide whether or not they intersect. This is one good reason for drawing a picture of any physical problem we are trying to solve. We will need a small bag of tricks used for ﬁnding intersections of curves. We begin with intersections involving the curves studied in this section. Two different horizontal lines (or two different vertical lines) will never intersect. However, a horizontal line always intersects a vertical line exactly once; Figure 3.7(a). Given a circle and a horizontal or vertical line, we may or may not have an intersection. Looking at Figure 3.7(b), you can convince yourself a given horizontal or vertical line will intersect a circle in either two points, one point or no points. This analysis is all pictorial; how do you ﬁnd the explicit coordinates of an intersection point? Let’s look at a sample problem to isolate the procedure used. 3.3. INTERSECTING CURVES I 29 y-axis Point: (h, k). Horizontal line: y = k. x-axis Vertical line: x = h. (a) Line equations. (b) Possible intersections. Figure 3.7: Circles and lines. Example 3.3.1. Glo-Tek Industries has designed a new halogen street light ﬁxture for the city of Seattle. According to the product literature, when placed on a 50 foot light pole, the resulting useful illuminated area is a circular disc 120 feet in diameter. Assume the light pole is located 20 feet east and 40 feet north of the intersection of Parkside Ave. (a north/south street) and Wilson St. (an east/west street). What portion of each street is illuminated? y-axis (Parkside) P Illuminated zone. x-axis (Wilson) R S Q Figure 3.8: Illuminated street. Solution. The illuminated area is a circular disc whose diameter and center are both known. Consequently, we really need to study the intersection of this circle with the two streets. Begin by imposing the pictured coordinate system; we will use units of feet for each axis. The illuminated region will be a circular disc centered at the point (20
, 40) in the coordinate system; the radius of the disc will be r = 60 feet. We need to ﬁnd the points of intersection P, Q, R, and S of the circle with the x-axis and the y-axis. The equation for the circle with r = 60 and center (h, k) = (20, 40) is (x − 20)2 + (y − 40)2 = 3600. To ﬁnd the circular disc intersection with the y-axis, we have a system of two equations to work with: (x − 20)2 + (y − 40)2 = 3,600; x = 0. To ﬁnd the intersection points we simultaneously solve both equations. To do this, we replace x = 0 in the ﬁrst equation (i.e., we impose the 30 CHAPTER 3. THREE SIMPLE CURVES conditions of the second equation on the ﬁrst equation) and arrive at (0 − 20)2 + (y − 40)2 = 3,600; 400 + (y − 40)2 = 3,600; (y − 40)2 = 3,200; (y − 40) = √3,200 ± y = 40 √3,200 ± = −16.57 or 96.57. Notice, we have two solutions. This means that the circle and y-axis intersect at the points P = (0,96.57) and Q = (0, − 16.57). Similarly, to ﬁnd the circular disc intersection with the x-axis, we have a system of two equations to work with: (x − 20)2 + (y − 40)2 = 3,600; y = 0. Replace y = 0 in the ﬁrst equation (i.e., we impose the conditions of the second equation on the ﬁrst equation) and arrive at (x − 20)2 + (0 − 40)2 = 3,600; (x − 20)2 = 2,000; (x − 20) = √2,000; ± x = 20 √2,000 ± = −24.72 or 64.72. Conclude the circle and x-axis intersect at the points S = (64.72,0) and R = (−24.72,0). The procedure we used in the solution of Example 3
.3.1 gives us a general approach to ﬁnding the intersection points of circles with horizontal and vertical lines; this will be important in the exercises. 3.4 Summary • • • Every horizontal line has equation of the form y = c. Every vertical line has equation of the form x = c. Every circle has equation of the form (x − h)2 + (y − k)2 = r2 where (h,k) is the center of the circle, and r is the circle’s radius. 3.5. EXERCISES 3.5 Exercises Problem 3.1. This exercise emphasizes the “mechanical aspects” of circles and their equations. (a) Find an equation whose graph is a circle of radius 3 centered at (−3,4). (b) Find an equation whose graph is a cir2 centered at the point cle of diameter 1 (3, − 11 3 ). (c) Find four different equations whose graphs are circles of radius 2 through (1,1). (d) Consider the equation (x − 1)2 + (y + 1)2 = 4. Which of the following points lie on the graph of this equation: (1,1), (1, − 1), (1, − 3), (1 + √3,0), (0, − 1 − √3), (0,0). Problem 3.2. Find the center and radius of each of the following circles. (a) x2 − 6x + y2 + 2y − 2 = 0 (b) x2 + 4x + y2 + 6y + 9 = 0 (c) x2 + 1 3 y = 127 3 x + y2 − 10 9 (d) x2 + y2 = 3 2 x − y + 35 16 Problem 3.3. Water is ﬂowing from a major broken water main at the intersection of two streets. The resulting puddle of water is circular and the radius r of the puddle is given by the equation r = 5t feet, where t represents time in seconds elapsed since the the main broke. (a) When the main broke, a runner was located 6 miles from the intersection. The runner continues toward the intersection at the constant speed of 17 feet per second. When will the runner’s feet get wet? (b) Suppose, instead, that when the main broke, the runner was 6 miles east, and

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