5128_Ch10_pp530-561 1/13/06 3:50 PM Page 530 Chapter 10 Parametric, Vector, and Polar Functions I n 1935, air traffic control was conducted with a system of teletype machines, wall-sized blackboards, large table maps, and movable markers representing airplanes. Today’s radar data processing includes an automatic display of aircraft identification, speed, altitude, and velocity vectors. A DC-10 plane flying due west at 600 mph enters a region with a steady air current coming from the southwest at 100 mph. How should the pilot adjust the airplane’s course and speed to maintain its original velocity vector? This type of problem is covered in Section 10.2. 530 5128_Ch10_pp530-561 1/13/06 3:50 PM Page 531 Section 10.1 Parametric Functions 531 Chapter 10 Overview The material in this book is generally described as the calculus of a single variable, since it deals with functions of one independent variable (usually x or t). In this chapter you will apply your understanding of single-variable calculus in three kinds of two-variable contexts, enabling you to analyze some new kinds of curves (parametrically defined and polar) and to analyze motion in the plane that does not proceed along a straight line. Interestingly enough, this will not require the tools of multi-variable calculus, which you will probably learn in your next calculus course. We will simply use single-variable calculus in some new and interesting ways. 10.1 What you’ll learn about • Parametric Curves in the Plane • Slope and Concavity • Arc Length • Cycloids . . . and why Parametric equations enable us to define some interesting and important curves that would be difficult or impossible to define in the form y f (x). Parametric Functions Parametric Curves in the Plane We reviewed parametrically defined functions in Section 1.4. Instead of defining the points (x, y) on a planar curve by relating y directly to x, we can define both coordinates as functions of a parameter t. The resulting set of points may or may not define y as a function of x (that is, the parametric curve might fail the vertical line test). EXAMPLE 1 Reviewing Some Parametric Curves Sketch the parametric curves and identify those which define y as a function of x. In each case, eliminate the parameter to find an equation that relates x and y directly. (a) x cos t and y sin t for t in the interval [0, 2) (b) x 3 cos t and y 2 sin t for t in the interval [0, 4] (c) x t and y t 2 for t in the interval [0, 4] SOLUTION (a) This is probably the best-known parametrization of all. The curve is the unit circle (Figure 10.1a), and it does not define y as a function of x. To eliminate the parameter, we use the identity cos t2 sin t2 1 to write x2 y2 1. (b) This parametrization stretches the unit circle by a factor of 3 horizontally and by a factor of 2 vertically. The result is an ellipse (Figure 10.1b), which is traced twice as t covers the interval [0, 4]. (In fact, the point (3, 0) is visited three times.) It does not define y as a x 2 y 2 function of x. We use the same identity as in part (a) to write 1. 3 2 (c) This parametrization produces a segment of a parabola (Figure 10.1c). It does define y as a function of x. Since t x2, we write y x2 2. Now try Exercise 1. y y y 3 3 3 1 x –2 –4 –1 1 1 4 x –2 2 –1 –3 (a) –3 (b) –3 (c) Figure 10.1 A collection of parametric curves (Example 1). Each point (x, y) is determined by parametric functions of t, but only the parametrization in graph (c) determines y as a function of x. x 5128_Ch10_pp530-561 1/13/06 3:50 PM Page 532 532 Chapter 10 Parametric, Vector, and Polar Functions Slope and Concavity We can analyze the slope and concavity of parametric curves just as we can with explicitlydefined curves. The slope of the curve is still dydx, and the concavity still depends on d 2ydx 2, so all that is needed is a way of differentiating with respect to x when everything is given in terms of t. The required parametric differentiation formulas are straightforward applications of the Chain Rule. Parametric Differentiation Formulas If x and y are both differentiable functions of t and if dxdt 0, then dy dydt . dx dxdt If y dydx is also a differentiable function of t, then d 2y dydt d 2 (y) . dx dxdt dx EXAMPLE 2 Analyzing a Parametric Curve y 3 1 –6 0 1 6 x –3 Figure 10.2 The parametric curve defined in Example 2. Consider the curve defined parametrically by x t2 5 and y 2 sin t for 0 t . (a) Sketch a graph of the curve in the viewing window [7, 7] by [4, 4]. Indicate the direction in which it is traced. (b) Find the highest point on the curve. Justify your answer. (c) Find all points of inflection on the curve. Justify your answer. SOLUTION (a) The curve is shown in Figure 10.2. (b) We seek to maximize y as a function of t, so we compute dydt 2 cos t. Since dydt is positive for 0 t 2 and negative for 2 t , the maximum occurs when t 2. Substituting this t value into the parametrization, we find the highest point to be approximately (2.533, 2). (c) First we compute d 2ydx2. dy dy/dt 2 cos t cos t dx dx/dt 2t t (sin t)(t) (1)(cos t) t2 dydt t sin t cos t 2 dx dxdt 2t 2t3 d 2y A graph of [0, p] by [– 0.1, 0.1] Figure 10.3 The graph of d 2ydx 2 for the parametric curve in Example 2 shows a sign change at t 2.798386 ... , indicating a point of inflection on the curve. (Example 2) t sin t cos t y on the interval [0, ] (Figure 10.3) 2t3 shows a sign change at t 2.798386... . Substituting this t value into the parametrization, we find the point of inflection to be approximately (2.831, 0.673). Now try Exercise 19. Arc Length In Section 7.4 we derived two different formulas for arc length, each of them based on an 2 approximation of the curve by tiny straight line segments with length x yk2 k . (See Figure 10.4.) 5128_Ch10_pp530-561 1/13/06 3:50 PM Page 533 Section 10.1 Parametric Functions y 533 Here is a third formula based on the same approximation. (b, d) d (xk) 2 (yk) 2 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ √ Q Arc Length of a Parametrized Curve yk y f (x) P Let L be the length of a parametric curve that is traversed exactly once as t increases from t1 to t2. If dxdt and dydt are continuous functions of t, then x k (a, c) c 0 xk – 1 a xk ( ) ( ) t2 x b L 2 dx dy dt dt t1 Figure 10.4 The graph of f, approximated by line segments. 2 dt. EXAMPLE 3 Measuring a Parametric Curve x= cos3 t, y= sin3 t, 0 ≤ t ≤ 2p Find the length of the astroid (Figure 10.5) x cos 3 t, y sin 3 t, 0 t 2. SOLUTION Solve Analytically The curve is traced once as t goes from 0 to 2. Because of the curve’s symmetry with respect to the coordinate axes, its length is four times the length of the first quadrant portion. We have 2 ( ) ( ( ) ( ( ) ( ) Figure 10.5 The astroid in Example 3. dx dt 2 dy dt 2 dx dy dt dt 2 2 3 cos 2 tsin t ) 9 cos 4 t sin 2 t 2 3 sin 2 tcos t ) 9 sin 4 t cos 2 t 2 9co s 2tsin t co s 2t sin 2t 1 2 t 9co s 2tsin 3 cos t sin t . Thus, the length of the first quadrant portion of the curve is p 2 p 2 3 cos t sin t dt 3 0 cos t sin t dt 3 sin2 t 2 y cos t sin t 0, 0 t /2 0 ] p2 u sin t, du cos t dt 0 3 . 2 P(x, y) The length of the astroid is 432 6. t O at Support Numerically NINT 3 cos t sin t , t, 0, 2 6. a C(at, a) x Figure 10.6 The position of P(x, y) on the edge of the wheel when the wheel has turned t radians. (Example 4) Now try Exercise 29. Cycloids Suppose that a wheel of radius a rolls along a horizontal line without slipping (see Figure 10.6. The path traced by a point P on the wheel’s edge is a cycloid, where P is originally at the origin. 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 534 534 Chapter 10 Parametric, Vector, and Polar Functions EXAMPLE 4 Finding Parametric Equations for a Cycloid Huygens’s Clock The problem with a pendulum clock whose bob swings in a circular arc is that the frequency of the swing depends on the amplitude of the swing. The wider the swing, the longer it takes the bob to return to center. This does not happen if the bob can be made to swing in a cycloid. In 1673, Christiaan Huygens (1629–1695), the Dutch mathematician, physicist, and astronomer who discovered the rings of Saturn, designed a pendulum clock whose bob would swing in a cycloid. Driven by a need to make accurate determinations of longitude at sea, he hung the bob from a fine wire constrained by guards that caused it to draw up as it swung away from center. How were the guards shaped? They were cycloids, too. Find parametric equations for the path of the point P in Figure 10.6. SOLUTION We suppose that the wheel rolls to the right, P being at the origin when the turn angle t equals 0. Figure 10.6 shows the wheel after it has turned t radians. The base of the wheel is at distance at from the origin. The wheel’s center is at (at, a, and the coordinates of P are x at a cos , y a a sin . To express in terms of t, we observe that t 3 2 2k for some integer k, so 3 t 2k. 2 Thus, 3 cos cos t 2k sin t, 2 ( ( ) ) 3 sin sin t 2k cos t. 2 Therefore, x at a sin t at sin t, y a a cos t a1 cos t. Guard cycloid Guard cycloid Now try Exercise 41. EXPLORATION 1 Cycloid Investigating Cycloids Consider the cycloids with parametric equations x at sin t, 1. 2. 3. 4. 5. 6. y a1 cos t, a 0. Graph the equations for a 1, 2, and 3. Find the x-intercepts. Show that y 0 for all t. Explain why the arches of a cycloid are congruent. What is the maximum value of y? Where is it attained? Describe the graph of a cycloid. EXAMPLE 5 Finding Length Find the length of one arch of the cycloid x at sin t, y a1 cos t, a 0. SOLUTION [0, 3p] by [–2, 4] Figure 10.7 shows the first arch of the cycloid and part of the next for a 1. In Exploration 1 you found that the x-intercepts occur at t equal to multiples of 2 and that the arches are congruent. The length of the first arch is ( ) ( ) 2 Figure 10.7 The graph of the cycloid x t sin t, y 1 cos t, t 0. (Example 5) 0 dx dt 2 dy 2 dt. dt continued 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 535 Section 10.1 Parametric Functions 535 We have ( ) ( ) ( ) ( ) dx dt 2 dx dt 2 dy dt 2 dy dt 2 a1 cos t 2 a 2 1 2 cos t cos 2 t a sin t 2 a 2 sin 2 t a 2 2o cst. a 0, sin 2 t cos 2 t 1 Therefore, ( ) ( ) 2 0 dx 2 dy 2 dt a dt dt 2 2 2cost dt 8a. The length of one arch of the cycloid is 8a. Quick Review 10.1 Using NINT 0 Now try Exercise 43. (For help, go to Appendix A.1.) Use algebra or a trig identity to write an equation relating x and y. 6. x csc and y cot x2 1 y2 1. x t 1 and y 2t 3 y 2x 1 7. x cos and y cos(2) y 2x2 1 2. x 3t and y 54t3 3 y 2x3 3 8. x sin and y cos(2) y 1 2x2 3. x sin t and y cos t x2 y2 1 9. x cos and y sin 4. x sin t cos t and y sin(2t) y 2x 5. x tan and y sec y2 1 10. x cos and y sin (0 ) y 1 x2 ( 2) y 1 x2 x2 Section 10.1 Exercises In Exercises 1–6, sketch the parametric curves and identify those which define y as a function of x. In each case, eliminate the parameter to find an equation that relates x and y directly. 1. x 2t 3 and y 4t 3 for t in the interval [0, 3] t5 2. x t 2 and y for t in the interval [3, 11] 4 3. x tan t and y sec t for t in the interval [0, 4] In Exercises 17–22, (a) sketch the curve over the given t-interval, indicating the direction in which it is traced, (b) identify the requested point, and (c) justify that you have found the requested point by analyzing an appropriate derivative. 17. x t 1, 4. x sin t and y 2 cos t for t in the interval [0, ] 18. x 5. x sin t and y cos(2t) for t in the interval [0, 2] 19. x 2 sin t, 6. x sin 6t and y 2t for t in the interval [0, 2] 20. x tan t, In Exercises 7–16, find (a) dy dx and (b) d 2y dx2 in terms of t. 7. x 4 sin t, y 2 cos t , 1 9. x t 11. x t2 3t, 13. x tan t, y 3 t y t3 y sec t 8. x cos t, cos t y 3 10. x 1 t, y 2 ln t 12. x t2 t, 14. x 2cos t, y t2 t y cos(2t) 15. x ln(2t), y ln(3t)4 16. x ln(5t), y e5t 1 7. (a) 2 tan t 8. (a) 3 1 (b) 8 sec3 t (b) 0 3 3 9. (a) 3 (b) 3/2 t t t2 y t 2 t, 2t, 21. x 2 sin t, 22. x ln(5t), y y cos t, y 2 sec t, y cos(2t), y ln(4t2), Lowest point 2 t 3 Leftmost point 0t Rightmost point 1 t 1 Lowest point 1.5 t 4.5 0 t 10 Highest point Rightmost point In Exercises 23–26, find the points at which the tangent line to the 2 , or curve is (a) horizontal or (b) vertical. 25. (a) At t 3 23. x 2 cos t, 24. x sec t, y 1 sin t y tan t 26. x 2 3 cos t, 10. (a) t (b) t2 6t2 18t 3t2 11. (a) (b) 3 (2t 3) 2t 3 2t 1 4 12. (a) (b) 3 2t 1 (2t 1) 2 t 2 2t 3, t2 13. (a) sin t (b) cos3 t 14. (a) 2 cos t (b) 1 15. (a) 4 (b) 0 16. (a) 5t e5t (b) 25t2 e5t 5t e5t (0.845, 3.079) and (3.155, 3.079) (b) Nowhere 25. x 2 t, y t 3 4t y 1 3 sin t 23. (a) (2, 0) and (2, 2) (b) (1, 1) and (3, 1) 24. (a) Nowhere (b) (1, 0) and (1, 0) 26. (a) (2, 4) and (2, 2) (b) (1, 1) and (5, 1) 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 536 536 Chapter 10 Parametric, Vector, and Polar Functions In Exercises 27–34, find the length of the curve. (For an algebraic challenge, try evaluating the integrals without a calculator.) 27. x cos t, y sin t, 28. x 3 sin t, 0 t 2 2 y 3 cos t, 0 t 3 2a y 8 sin t 8t cos t, 0 t 2 2 29. x 8 cos t 8t sin t, 30. x 2 y2 0 t 2 12 2t 3 3 2 t2 31. x , y t , 0 t 3 21/2 3 2 (8t 8)32 32. x , y t2 t, 0 t 2 10 12 1 1 22 1 33. x t 3, y t 2, 0 t 1 0.609 3 3 2 34. x lnsec t tan t sin t, y cos t, 0 t 3 cos3 t, y sin3 t, 2ap 39. Find the area of the shaded region. (Hint: dx dx dt dt) 3a2 ln 2 35. Length is Independent of Parametrization To illustrate the fact that the numbers we get for length do not usually depend on the way we parametrize our curves, calculate the length of the semicircle y 1 x 2 with these two different parametrizations. (a) x cos 2t, y sin 2t, (b) x sin t, y cos t, 0 t 2 1 2 t 1 2 y 4 sin t, 0 t 2. 22.103 37. Cartesian Length Formula The graph of a function y f x over an interval a, b automatically has the parametrization x x, y f x, a x b. The parameter in this case is x itself. Show that for this parametrization, the length formula ( ) ( ) b L 2 dx dy dt dt a 2 dt reduces to the Cartesian formula ( ) b L a dy 1 dx 2 dx Just substitute x for t and note that dx/dx 1. derived in Section 7.4. 38. (Continuation of Exercise 37) Show that the Cartesian formula ( ) d L c dx 1 dy 2 dy for the length of the curve x gy, c y d, from Section 7.4 is a special case of the parametric length formula ( ) ( ) b L a 2 dx dy dt dt 2 dt. Exercises 39 and 40 refer to the region bounded by the x-axis and one arch of the cycloid x at sin t, y a1 cos t that is shaded in the figure shown at the top of the next column. 38. Use the parametrization x g(y), y y, c y d, substitute y for t and note dy/dy 1. 40. Find the volume swept out by revolving the region about the x-axis. (Hint: dV y 2 dx y 2 dx dt dt) 5 2 a3 41. Curtate Cycloid Modify Example 4 slightly to find the parametric equations for the motion of a point in the interior of a wheel of radius a as the wheel rolls along the horizontal line without slipping. Assume that the point is at distance b from the center of the wheel, where 0 b a. This curve, known as a curtate cycloid, has been used by artisans in designing the arches of violins (Source: mathworld.wolfram.com). x at b sin t and y a b cos t (0 a b) 36. Perimeter of an Ellipse Find the length of the ellipse x 3 cos t, x 42. Prolate Cycloid Modify Example 4 slightly to find the parametric equations for the motion of a point on the exterior of a wheel of radius a as the wheel rolls along the horizontal line without slipping. Assume that the point is at distance b from the center of the wheel, where a b 2a. This curve, known as a prolate cycloid, is traced out by a point on the outer edge of a train’s flanged wheel as the train moves along a track. (If you graph a prolate cycloid, you can see why they say that there is always part of a forward-moving train that is moving backwards!) x at b sin t and y a b cos t (a b 2a) 43. Arc Length Find the length of one arch (that is, the curve over one period) of the curtate cycloid defined parametrically by x 3t 2 sin t and y 3 2 cos t. 21.010 44. Arc Length Find the length of one arch (that is, the curve over one period) of the prolate cycloid defined parametrically by x 2t 3 sin t and y 2 3 cos t. 21.010 Standardized Test Questions You should solve the following problems without using a graphing calculator. 45. True or False In a parametrization, if x is a continuous function of t and y is a continuous function of t, then y is a continuous function of x. Justify your answer. 46. True or False If f is a function with domain all real numbers, then the graph of f can be defined parametrically by x t and y f (t) for t . Justify your answer. 47. Multiple Choice For which of the following parametrizations of the unit circle will the circle be traversed clockwise? B (A) x cos t, y sin t, 0 t 2 (B) x sin t, y cos t, 0 t 2 (C) x cos t, y sin t, 0 t 2 (D) x sin t, y cos t, 0 t 2 y cos t, 0 t 2 (E) x sin t, 45. False. Indeed, y may not even be a function of x. (See Example 1.) 46. True. The ordered pairs (x, f(x)) and (t, f(t)) are exactly the same. 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 537 Section 10.1 Parametric Functions 48. Multiple Choice A parametric curve is defined by x sin t and y csc t for 0 t 2. This curve is C 537 the string is tangent to the circle at Q, and t is the radian measure of the angle from the positive x-axis to the segment OQ. (a) Derive parametric equations for the involute by expressing the coordinates x and y of P in terms of t for t 0. x cos t (A) increasing and concave up. (B) increasing and concave down. t sin t, y sin t t cos t (b) Find the length of the involute for 0 t 2 . 2 2 (C) decreasing and concave up. (D) decreasing and concave down. (E) decreasing with a point of inflection. 49. Multiple Choice The parametric curve defined by x ln(t), y t for t 0 is identical to the graph of the function C (A) y ln x for all real x. (B) y lnx y for x 0. String Q (C) y e x for all real x. P(x, y) (D) y e x for x 0. t (E) y ln(e x) for x 0. O 1 x (1, 0) 50. Multiple Choice The curve parametrized by x 6 sin t 3 sin(7t) and y 6 cos t 3 cos(7t), as shown in the diagram below, is traversed exactly once as t increases from 0 to 2. The total length of the curve is given by D (A) 0 2 (6 (6 sin t 3 s in(7t)) cos t 3 cos (7t))2 dt (B) 0 (6 cos t 3 c os(7t) )2 (6 sin t 3 sin ( 7t))2 dt (C) 0 (6 cos t 21 cos(7 t))2 (6 sin t 21 sin(7t ))2 dt (D) 0 (6 cos t 21 cos(7 t))2 (6 si n t 2 1 sin(7 t))2 dt 2 2 2 2 (6 cos t 3 (6 sin t 3 (E) 0 7 2 cos(7t))2 sin (3t))2 dt 52. (Continuation of Exercise 51) Repeat Exercise 51 using the circle of radius a centered at the origin, x 2 y 2 a 2. (a) x a(cos t t sin t), y a(sin t t cos t) (b) 2a 2 In Exercises 53–56, a projectile is launched over horizontal ground at an angle with the horizontal and with initial velocity v0 ft sec. Its path is given by the parametric equations x v0 cos t, y v0 sin t 16t 2. (a) Find the length of the path traveled by the projectile. (b) Estimate the maximum height of the projectile. 53. 20°, v0 150 y 55. 60°, 10 54. 30°, v0 150 56. 90°, v0 150 (a) 641.236 ft (b) 5625/64 87.891 ft (a) 461.749 ft (b) 41.125 ft v0 150 (a) 840.421 ft (b) 16,875/64 263.672 ft Extending the Ideas 2 –10 –2 2 10 x If dx dt and dydt are continuous, the parametric curve defined by (x(t), y(t)) for a t b is called smooth. If the curve is traversed exactly once as t increases from a to b, and if y is a positive function of x, then the curve can be revolved about the x-axis to form a solid of revolution (see Section 7.3). The surface area of such a solid is given by ( ) ( ) b –10 S 2y a Explorations 51. Group Activity Involute of a Circle If a string wound around a fixed circle is unwound while being held taut in the plane of the circle, its end P traces an involute of the circle as suggested by the diagram below. In the diagram, the circle is the unit circle in the xy-plane, and the initial position of the tracing point is the point 1, 0 on the x-axis. The unwound portion of (a) 703.125 ft (b) 5625/16 351.5625 ft 2 dx dy dt dt 2 dt. Apply this formula in Exercises 57–60 to find the surface area when the parametric curve is revolved about the x-axis. 57. x cos t, y 2 sin t, 58. x 2t, y (2 3)t 3 2, 59. x t2 2, y t 1, 0 t 2 8 2 0 t 2 14.214 0 t 3 178.561 60. x ln(sec t tan t) sin t, y cos t, 0 t 3 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 538 538 Chapter 10 Parametric, Vector, and Polar Functions 10.2 What you’ll learn about • Two-Dimensional Vectors • Vector Operations • Modeling Planar Motion • Velocity, Acceleration, and Speed • Displacement and Distance Traveled . . . and why The jump from one to two dimensions (and eventually higher) is easier than one might think, thanks to the mathematics of vectors. Vectors in the Plane Two-Dimensional Vectors When an object moves along a straight line, its velocity can be determined by a single number that represents both magnitude and direction (forward if the number is positive, backward if it is negative). The speed of an object moving on a path in a plane can still be represented by a number, but how can we represent its direction when there are an infinite number of directions possible? Fortunately, we can represent both magnitude and direction with just two numbers, just as we can represent any point in the plane with just two coordinates (which is possible essentially for the same reason). This representation is what two-dimensional vectors were designed to do. While the pair (a, b) determines a point in the plane, it also determines a directed line segment (or arrow) with its tail at the origin and its head at (a, b) (Figure 10.8). The length of this arrow represents magnitude, while the direction in which it points represents direction. In this way, the ordered pair (a, b) represents a mathematical object with both magnitude and direction, called the position vector of (a, b). y y (a, b) (a, b) a, b O x O x Figure 10.8 The point represents the ordered pair (a, b). The arrow (directed line segment) represents the vector a, b. DEFINITION Two-Dimensional Vector A two-dimensional vector v is an ordered pair of real numbers, denoted in component form as a, b. The numbers a and b are the components of the vector v. The standard representation of the vector a, b is the arrow from the origin to the point (a, b). The magnitude (or absolute value) of v, denoted v, is the length of the arrow, and the direction of v is the direction in which the arrow is pointing. The vector 0 0, 0, called the zero vector, has zero length and no direction. The distance formula in the plane gives a simple computational formula for magnitude. Magnitude of a Vector The magnitude or absolute value of the vector a, b is the nonnegative real number a, b a2 b2. Direction can be quantified in several ways; for example, navigators use bearings from compass points. The simplest choice for us is to measure direction as we do with the trigonometric functions, using the usual position angle formed with the positive x-axis as the initial ray and the vector as the terminal ray. In this way, every nonzero vector determines a unique direction angle satisfying (in degrees) 0 360 or (in radians) 0 2. (See Figure 10.10 for an example.) 5128_Ch10_pp530-561 2/3/06 4:41 PM Page 539 Section 10.2 Vectors in the Plane 539 y R(–1, 6) Direction Angle of a Vector P(3, 4) The direction angle of a nonzero vector v is the smallest nonnegative angle formed with the positive x-axis as the initial ray and the standard representation of v as the terminal ray. Q(–4, 2) 1 x O 1 Figure 10.9 The arrows QR and OP both represent the vector 3, 4, as would any arrow with the same length pointing in the same direction. Such arrows are called equivalent. y (–1, √3−) 2 v u 2 –2 x –2 This textbook uses boldface variables to represent vectors (for example, u and v) to distinguish them from numbers. In handwritten form it is customary to distinguish vector variables by arrows (for example, u and v). We also use angled brackets to distinguish a vector x, y from a point (x, y) in the plane, although it is not uncommon to see (x, y) used for both, especially in handwritten form. It is often convenient in applications to represent vectors with arrows that begin at points other than the origin. The important thing to remember is that any two arrows with the same length and pointing in the same direction represent the same vector. In Figure 10.9, for ex an arrow with initial point Q and ample, the vector 3, 4 is shown represented by QR, terminal point R, as well as by its standard representation OP. Two arrows that represent the same vector are said to be equivalent. The quick way to associate arrows with the vectors they represent is to use the following rule. Head Minus Tail (HMT) Rule Figure 10.10 The vector v in Example 1 is represented by an arrow from the origin to the point 1, 3. If an arrow has initial point (x1, y1) and terminal point (x2, y2), it represents the vector x2 x1, y2 y1. y EXAMPLE 1 Finding Magnitude and Direction Find the magnitude and the direction angle of the vector v 1, 3 (Figure 10.10). SOLUTION 3 40˚ y = 3 sin 40˚ x x = 3 cos 40˚ The magnitude of v is v (1 0)2 (3 0)2 2. Using triangle ratios, we see that the direction angle satisfies cos 1 2 and sin 3 2, so 120º or 2 3 radians. Now try Exercise 5. 3 –1 EXAMPLE 2 Finding Component Form Figure 10.11 The vector in Example 2 is represented by an arrow from the origin to the point (3 cos 40º, 3 sin 40º). Why Not Use Slope for Direction? Notice that slope is inadequate for determining the direction of a vector, since two vectors with the same slope could be pointing in opposite directions. Moreover, vectors are still useful in dimensions higher than 2, while slope is not. Find the component form of a vector with magnitude 3 and direction angle 40º. SOLUTION The components of the vector, found trigonometrically, are x 3 cos 40º and y 3 sin 40º (Figure 10.11). The vector is 3 cos 40º, 3 sin 40º 2.298, 1.928. Now try Exercise 13. Vector Operations The algebra of vectors sometimes involves working with vectors and numbers at the same time. In this context, we refer to the numbers as scalars. The two most basic algebraic operations involving vectors are vector addition (adding a vector to a vector) and scalar multiplication (multiplying a vector by a number). Both operations are easily represented geometrically. 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 540 540 Chapter 10 Parametric, Vector, and Polar Functions DEFINITION Vector Addition and Scalar Multiplication Let u u1, u2 and v v1, v2 be vectors and let k be a real number (scalar). The sum (or resultant) of the vectors u and v is the vector u v u1 v1, u2 v2. The product of the scalar k and the vector u is ku ku1, u2 ku1, ku2. The opposite of a vector v is v (1)v. We define vector subtraction by u v u (v). v The vector is a vector of magnitude 1, called a unit vector. Its component form is v v cos , sin , where is the direction angle of v. For this reason, is sometimes called v the direction vector of v. The sum of two vectors u and v can be represented geometrically by arrows in two ways. In the tail-to-head representation, the arrow from the origin to (u1, u2) is the standard representation of u, the arrow from (u1, u2) to (u1 v1, u2 v2,) represents v (as you can verify by the HMT Rule), and the arrow from the origin to (u1 v1, u2 v2) then is the standard representation of u v (Figure 10.12a). In the parallelogram representation, the standard representations of u and v determine a parallelogram whose diagonal is the standard representation of u v (Figure 10.12b). y y v u u u+v u+v v x x (a) (b) –2 u Figure 10.12 Two ways to represent vector addition geometrically: (a) tail-to-head and (b) parallelogram. 2u u 1_ u 2 Figure 10.13 Representations of u and several scalar multiples of u. The product ku of the scalar k and the vector u can be represented by a stretch (or shrink) of u by a factor of k. If k 0, then ku points in the same direction as u; if k 0, then ku points in the opposite direction (Figure 10.13). EXAMPLE 3 Performing Operations on Vectors Let u 1, 3 and v 4, 7. Find the following. 1 (a) 2u 3v (b) u v (c) u 2 SOLUTION (a) 2u 3v 21, 3 34, 7 21 34, 23 37 10, 27 continued 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 541 Section 10.2 Vectors in the Plane 541 (b) u v 1, 3 4, 7 1 4, 3 7 5, 4 1 1 3 (c) u , 2 2 2 ( ) () 2 1 3 2 2 2 1 10 2 Now try Exercise 21. Vector operations have many of the properties of their real-number counterparts. Properties of Vector Operations Let u, v, w be vectors and a, b be scalars. 1. u v v u 2. u v w u v w 3. u 0 u 4. u u 0 5. 0u 0 6. 1u u 7. abu abu 8. au v au av 9. a bu au bu Modeling Planar Motion Although vectors are used in many other physical applications, our primary reason for introducing them into this course is to model the motion of objects moving in a coordinate plane. You may have seen vector problems of the following type in a physics or mechanics course. EXAMPLE 4 Finding Ground Speed and Direction A Boeing® 727 ® airplane, flying due east at 500 mph in still air, encounters a 70-mph tail wind acting in the direction 60º north of east. The airplane holds its compass heading due east but, because of the wind, acquires a new ground speed and direction. What are they? N SOLUTION If u the velocity of the airplane alone and v the velocity of the tail wind, then u 500 and v 70 (Figure 10.14). v 30˚ 70 u+v 500 u E We need to find the magnitude and direction of the resultant vector u v. If we let the positive x-axis represent east and the positive y-axis represent north, then the component forms of u and v are u 500, 0 NOT TO SCALE Figure 10.14 Vectors representing the velocities of the airplane and tail wind in Example 4. and v 70 cos 60º, 70 sin 60º 35, 35 3 . Therefore, u v 535, 35 3 , u and v 5352 35 3 2 538.4, 35 3 tan1 6.5º. 535 Interpret The new ground speed of the airplane is about 538.4 mph, and its new direction is about 6.5º north of east. Now try Exercise 25. 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 542 542 Chapter 10 Parametric, Vector, and Polar Functions Recall that if the position x of an object moving along a line is given as a function of time t, then the velocity of the object is dx dt and the acceleration of the object is d 2x dt2. It is almost as simple to relate position, velocity, and acceleration for an object moving in the plane, because we can model those functions with vectors and treat the components of the vectors as separate linear models. Example 5 shows how simple this modeling actually is. EXAMPLE 5 Doing Calculus Componentwise A particle moves in the plane so that its position at any time t 0 is given by (sin t, t2 2). (a) Find the position vector of the particle at time t. (b) Find the velocity vector of the particle at time t. (c) Find the acceleration of the particle at time t. (d) Describe the position and motion of the particle at time t 6. SOLUTION 6 cos 6 [–2, 2] by [0, 25] 0≤t≤6 Figure 10.15 The path of the particle in Example 5 from t 0 to t 6. The red arrow shows the velocity vector at t 6. (a) The position vector, which has the same components as the position point, is sin t, t22. In fact, it could also be represented as (sin t, t2 2), since the context would identify it as a vector. (b) Differentiate each component of the position vector to get cos t, t. (c) Differentiate each component of the velocity vector to get sin t, 1. (d) The particle is at the point (sin 6, 18), with velocity cos 6, 6 and acceleration sin 6, 1. You can graph the path of this particle parametrically, letting x sin(t) and y t2 2. In Figure 10.15 we show the path of the particle from t 0 to t 6. The red arrow at the point (sin 6, 18) represents the velocity vector (cos 6, 6). It shows both the magnitude and direction of the velocity at that moment in time. Now try Exercise 31. Velocity, Acceleration, and Speed We are now ready to give some definitions. DEFINITIONS Velocity, Speed, Acceleration, and Direction of Motion A Word About Differentiability Our definitions can be expanded to a calculus of vectors, in which (for example) d v dt a(t), but it is not our intention to get into that here. We have therefore finessed the fine point of vector differentiability by requiring the path of our particle to be “smooth.” The path can have vertical tangents, fail the vertical line test, and loop back on itself, but corners and cusps are still problematic. Suppose a particle moves along a smooth curve in the plane so that its position at any time t is (x(t)), y(t), where x and y are differentiable functions of t. 1. The particle’s position vector is r(t) x(t), y(t). dx dy 2. The particle’s velocity vector is v(t) , . dt dt 3. The particle’s speed is the magnitude of v, denoted v. Speed is a scalar, not a vector. d 2x d 2y 4. The particle’s acceleration vector is a(t) , . dt 2 dt 2 v 5. The particle’s direction of motion is the direction vector . v EXAMPLE 6 Studying Planar Motion A particle moves in the plane with position vector r (t) sin (3t), cos (5t). Find the velocity and acceleration vectors and determine the path of the particle. continued 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 543 Section 10.2 Vectors in the Plane SOLUTION 543 d d Velocity v(t) (sin(3t)), (cos(5t)) 3 cos(3t), 5 sin(5t). dt dt d d Acceleration a(t) (3 cos(3t)), (5 sin(5t)) 9 sin(3t), 25 cos(5t). dt dt The path of the particle is found by graphing (in parametric mode) the curve defined by x sin(3t) and y cos(5t) (Figure 10.16). Now try Exercise 33. EXAMPLE 7 Studying Planar Motion [–1.6, 1.6] by [–1.1, 1.1] 0 ≤ t ≤ 6.3 A particle moves in an elliptical path so that its position at any time t 0 is given by (4 sin t, 2 cos t). Figure 10.16 The path of the busy particle in Example 6. (a) Find the velocity and acceleration vectors. (b) Find the velocity, acceleration, speed, and direction of motion at t 4. (c) Sketch the path of the particle and show the velocity vector at the point (4, 0). (d) Does the particle travel clockwise or counterclockwise around the origin? SOLUTION y 4 (b) Velocity v( 4) 4 cos( 4), 2 sin( 4) 22, 2 Acceleration a( 4) 4 sin( 4), 2 cos( 4) 22, 2 1 –5 d d (a) Velocity v(t) (4 sin t), (2 cos t) 4 cos t, 2 sin t dt dt d d Acceleration a(t) (4 cos t), (2 sin t) 4 sin t, 2 cos t dt dt 1 x 0, –2 –4 Figure 10.17 The ellipse on which the particle travels in Example 7. The velocity vector at the point (4, 0) is 0, 2, represented by an arrow tangent to the ellipse at (4, 0) and pointing down. The direction of the velocity at that point indicates that the particle travels clockwise around the origin. Speed v( 4) 22, 2 (22)2 ( 2)2 10 (c) The ellipse defined parametrically by x 4 sin t and y 2 cos t is shown in Figure 10.17. At the point (4, 0), sin t 1 and cos t 0, so v(t) 4 cos t, 2 sin t 0, 2. The vector 0, 2 is drawn tangent to the curve at (4, 0). (d) As the vector in Figure 10.17 shows, the particle travels clockwise around the origin. Now try Exercise 35. Displacement and Distance Traveled Recall that when a particle moves along a line with velocity v(t), the displacement (or b net distance traveled) from time t a to time t b is given by a v(t) dt, while the (total) b distance traveled in that time interval is given by a v(t) dt. When a particle moves in the plane with velocity vector v(t), displacement and distance traveled can be found by applying the same integrals to the vector v, although in slightly different ways. DEFINITIONS Displacement and Distance Traveled Suppose a particle moves along a path in the plane so that its velocity at any time t is v(t) v1(t), v2(t), where v1 and v2 are integrable functions of t. The displacement from t a to t b is given by the vector b a v1(t) dt, b a v2(t) dt . continued 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 544 544 Chapter 10 Parametric, Vector, and Polar Functions The preceding vector is added to the position at time t a to get the position at time t b. The distance traveled from t a to t b is b b v(t)dt a v1(t)2 v2(t)2 dt. a There are two things worth noting about the formula for distance traveled. First of all, it is a nice example of the integral as an accumulator, since we are summing up bits of speed multiplied by bits of time, which equals bits of positive distance. Secondly, it is actually a new look at an old formula. Substitute dx dt for v1(t) and dy dt for v2(t) and you get the arc length formula for a curve defined parametrically (Section 10.1). This formula makes sense, since the distance the particle travels is precisely the length of the path along which it moves. EXAMPLE 8 Finding Displacement and Distance Traveled A particle moves in the plane with velocity vector v(t) (t 3 cos t, 2t sin t). At t 0, the particle is at the point (1, 5). (a) Find the position of the particle at t 4. (b) What is the total distance traveled by the particle from t 0 to t 4? SOLUTION (a) Displacement (t 3 cos t)dt, (2t sin t)dt 8, 16. 4 4 0 0 The particle is at the point (1 8, 5 16) (9, 21). (b) Distance traveled 0 (t 3 cos t )2 (2 t sin t)2 dt 33.533. 4 Now try Exercise 37. EXAMPLE 9 Finding the Path of the Particle Determine the path that the particle in Example 8 travels going from (1, 5) to (9, 21) . SOLUTION The velocity vector and the position at t 0 combine to give us the vector equivalent of an initial value problem. We simply find the components of the position vector separately. dx t 3 cos t dt t2 Antidifferentiate. x 3 sin t C 2 t2 x 3 sin t 1 x 1 when t 0. 2 dy 2t sin t dt Antidifferentiate. y t 2 cos t C [–5, 15] by [0, 23] 0≤t≤4 Figure 10.18 The path traveled by the particle in Example 8 as it goes from (1, 5) to (9, 21) in four seconds (Example 9). y t 2 cos t 4 We then graph the position 2 3 sin t 1, t 0 to t 4. The path is shown in Figure 10.18. t2 t2 y 5 when t 0. cos t 4 parametrically from Now try Exercise 41. 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 545 Section 10.2 Vectors in the Plane Quick Review 10.2 545 (For help, go to Sections 1.1, 4.3, and 10.1.) In Exercises 1–4, let P (1, 2) and Q (5, 3). 7. Find the velocity and acceleration of a particle moving along a line if its position at time t is given by x(t) t sin t. 17 1. Find the distance between the points P and Q. v(t) sin t t cos t; a(t) 2 cos t t sin t 2. Find the slope of the line segment PQ. 1/4 3. If R (3, b), determine b so that segments PQ and RQ are collinear. b 5/2 4. If R (3, b), determine b so that segments PQ and RQ are perpendicular. b 11 In Exercises 5 and 6, determine the missing coordinate so that the four points form a parallelogram ABCD. 5. A (0, 0), B (1, 3), C (5, 3), D (a, 0) a 4 8. A particle moves along the x-axis with velocity v(t) 3t2 12t for t 0. If its position is x 40 when t 0, where is the particle when t 4? x 8 9. A particle moves along the x-axis with velocity v(t) 3t2 12t for t 0. What is the total distance traveled by the particle from t 0 to t 4? 32 10. Find the length of the curve defined parametrically by x sin(2t) and y cos(3t) for 0 t 2. 15.289 6. A (1, 1), B (3, 5), C (8, b), D (6, 2) b 6 Section 10.2 Exercises In Exercises 1–4, find the component form of the vector. 1. the vector from the origin to the point A (2, 3) 2, 3 2. the vector from the point A (2, 3) to the origin 2, 3 where P (1, 3) and Q (2, 1) 1, 4 3. the vector PQ, where O is the origin and P is the midpoint of the 4. the vector OP, segment RS connecting R (2, 1) and S (4, 3). 1, 1 26. A river is flowing due east at 2 mph. A canoeist paddles across the river at 4 mph with his bow aimed directly northwest (a direction angle of 135º). What is the true direction angle of the canoeist’s path, and how fast is the canoe going? In Exercises 27–32, a particle travels in the plane with position vector r(t). Find (a) the velocity vector v(t) and (b) the acceleration vector a(t). 27. r(t) 3t2, 2t3 See page 547. 28. r(t) sin 2t, 2 cos t See page 547. In Exercises 5–10, find the magnitude of the vector and the direction angle it forms with the positive x-axis (0 360º). 5. 2, 2 6. 2, 2 8, 45° 7. 3, 1 2, 30° 9. 5, 0 5, 180° 8. 2, 23 10. 0, 4 2, 135° 4, 240° 4, 90° In Exercises 11–16, find the component form of the vector with the given magnitude that forms the given directional angle with the positive x-axis. 11. 4, 180º 4, 0 12. 6, 270º 13. 5, 100º 0.868, 4.924 14. 13, 200º 15. 32, 4 radians 3, 3 0, 6 12.216, 4.446 16. 23, 6 radians 3, 3 In Exercises 17–24, let u 3, 2 and v 2, 5. Find the (a) component form and (b) magnitude of the vector. 17. 3u (a) 9, 6 (b) 313 18. 2v (a) 4, 10 (b) 229 19. u v (a) 1, 3 (b) 10 20. u v (a) 5, 7 (b) 74 21. 2u 3v (a) 12, 19 22. 2u 5v (a) 16, 29 (b) 505 (b) 1097 5 3 4 12 23. u v (a) 1/5, 14/5 24. u v (a) 3, 70/13 13 5 5 13 (b) 197 5 13 (b) 6421 25. Navigation An airplane, flying in the direction 20º east of north at 325 mph in still air, encounters a 40-mph tail wind acting in the direction 40º west of north. The airplane maintains its compass heading but, because of the wind, acquires a new ground speed and direction. What are they? Speed 346.735 mph direction 14.266° east of north 29. r(t) tet, et See page 547. 30. r(t) 2 cos 3t, 2 sin 4t 31. r(t) t2 sin 2t, t2 cos 2t See page 547. See page 547. 32. r(t) t sin t, t cos t See page 547. 33. A particle moves in the plane with position vector cos 3t, sin 2t. Find the velocity and acceleration vectors and determine the path of the particle. 34. A particle moves in the plane with position vector sin 4t, cos 3t. Find the velocity and acceleration vectors and determine the path of the particle. 35. A particle moves in the plane so that its position at any time t 0 is given by x sin 4t cos t and y sin 2t. (a) Find the velocity and speed of the particle when t 5 4. (b) Draw the path of the particle and show the velocity vector at t 5 4. (c) Is the particle moving to the left or to the right when t 5 4? 36. A particle moves in the plane so that its position at any time t 0 is given by x et et and y et et. (a) Find the velocity vector. dydt (b) Find lim . t→ dxdt (c) Show algebraically that the particle moves on the hyperbola x2 y2 4. (d) Sketch the path of the particle, showing the velocity vector at t 0. 26. The velocity is 2 22, 22, so the true angle is about 106.3° and the true speed is about 2.95 miles per hour. 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 546 546 Chapter 10 Parametric, Vector, and Polar Functions In Exercises 37–40, the velocity v(t) of a particle moving in the plane is given, along with the position of the particle at time t 0. Find (a) the position of the particle at time t 3, and (b) the distance the particle travels from t 0 to t 3. 37. v(t) 3t2 2t, 1 cos t; (2, 6) (a) (20, 9) (b) 19.343 38. v(t) 2 cos 4 t, 4 sin 2 t; (7, 2) (a) (7, 2) (b) 28.523 39. v(t) (t 1)1, (t 2)2; (3, 2) (a) (3 ln 4, 1.7) (b) 1.419 40. v(t) et t, et t; (1, 1) (a) (15.586, 24.586) (b) 20.627 41. Sketch the path that the particle travels in Exercise 37. 42. Sketch the path that the particle travels in Exercise 38. 43. A point moves in the plane so that x 5 cos( t 6) and y 3 sin( t 6). (a) Find the speed of the point at t 2. 7/12 2.399 (b) Find the acceleration vector at t 2. 52/72, 2324 (c) Eliminate the parameter and find an equation in x and y that 2 2 defines the curve on which the point moves. x y 1 25 9 44. A particle moves with position vector sec t, tan t for 0 t 1 2. (a) Find the velocity and speed of the particle at t 1 4. (b) The particle moves along a hyperbola. Eliminate the parameter to find an equation of the hyperbola in terms of x and y. (c) Sketch the path of the particle over the time interval 0 t 1 2. 45. A particle moves on the circle x2 y2 1 so that its position vector (b) Is the particle ever at rest? Justify your answer. See page 547. (c) Give the coordinates of the point that the particle approaches as t increases without bound. See page 547. 46. A particle moves in the plane so that its position at any time t, 0 t 2, is given parametrically by x sin t and y cos(2t). (a) Find the velocity vector for the particle. (d) Sketch the path of the particle. 47. A particle moves in the plane so that its position at any time t, 0 t 2, is given parametrically by x et sin t and y et cos t. (a) Find the slope of the path of the particle at time t 2. 3.844 (c) Find the distance traveled by the particle along the path from t 0 to t 1. 2.430 48. The position of a particle at any time t 0 is given by 2 1088 32.985 (a) Find the magnitude of the velocity vector at t 4. (b) Find the total distance traveled by the particle from t 0 to t 4. 46.062 (c) Find dy dx as a function of x. (b) At time t 2, the value of dydt is 6. Write an equation for the line tangent to the curve at the point (x(2), y(2)). 2 (c) Find the speed of the object at time t 2. (d) For t 3, the line tangent to the curve at (x(t), y(t)) has a slope of 2t 1. Find the acceleration vector of the object at time t 4. 8 cos 16, 2(2 sin 16) 7(8)cos 16 7.661, 50.205 50. For 0 t 3, an object moving along a curve in the xy-plane has position (x(t), y(t)) with dxdt sin(t3) and dydt 3 cos(t2). At time t 2, the object is at position (4, 5). See page 547. (a) Write an equation for the line tangent to the curve at (4, 5). (b) Find the speed of the object at time t 2. See page 547. (c) Find the total distance traveled by the object over the time interval 0 t 1. 2.741 (d) Find the position of the object at time t 3. See page 547. Standardized Test Questions You may use a graphing calculator to solve the following problems. 51. True or False A scalar multiple of a vector v has the same direction as v. Justify your answer. False. For example, u and 53. Multiple Choice The position of a particle in the xy-plane is given by x t2 1 and y ln(2t 3) for all t 0. The acceleration vector of the particle is E 2 4 4 (A) 2t, . (B) 2t, 2 . (C) 2, 2 . 2t 3 (2t 3) (2t 3) 2 (D) 2, . (2t 3) 2 (b) For what values of t is the particle at rest? (c) Write an equation for the path of the particle in terms of x and y that does not involve trigonometric functions. x(t) t 2 3 and y(t) 3t 3. (a) Find the x-coordinate of the position of the object at time 4 t 4. 3 (2 sin(t2)) dt 3.942 52. True or False If a vector with direction angle 0º is added to a vector with direction angle 90º, the result is a vector with direction angle 45º. Justify your answer. See page 547. (b) Find the speed of the particle when t 1. 49. An object moving along a curve in the xy-plane has position (x(t), y(t)) at time t 0 with dxdt 2 sin(t2). The derivative dydt is not explicitly given. At time t 2, the object is at position (3, 5). 1(u) have opposite directions. 1 t2 2t at any time t 0 is , . 1 t2 1 t2 (a) Find the velocity vector. 6 (2 s in 4)2 (6)2 6.127 49. (b) y 5 (x 3) (c) 2 sin 4 dy/dx t 3x 4 (E) 2, . (2t 3) 2 54. Multiple Choice An object moving along a curve in the xy-plane has position (x(t), y(t)) with dxdt cos(t2) and dydt sin(t3). At time t 0, the object is at position (4, 7). Where is the particle when t 2? D (A) 0.654, 0.989 (B) 0.461, 0.452 (D) 4.461, 7.452 (E) 5.962, 8.962 (C) 3.346, 7.989 55. Multiple Choice A vector with magnitude 7 and direction angle 40º is added to a vector with magnitude 4 and direction angle 140º. The result is a vector with magnitude B (A) 4.684. (B) 7.435. (C) 8.062. (D) 9.369. (E) 11. 56. Multiple Choice The path of a particle moving in the plane is defined parametrically as a function of time t by x sin 2t and y cos 5t. What is the speed of the particle when t 2? B (A) 1.130 (B) 3.018 (D) 0.757, 0.839 (E) 1.307, 2.720 (C) 1.307, 2.720 , 1, which has a direction 52. False. For example, 3, 0 0, 1 3 angle of 30°. 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 547 Section 10.2 Vectors in the Plane Explorations Two nonzero vectors are said to be orthogonal if they are perpendicular to each other. The zero vector is considered to be orthogonal to every vector. 57. Orthogonal vectors A particle with coordinates (x, y) moves along a curve in the first quadrant in such a way that dxdt x and dydt 1 x2 for every t 0. Find the acceleration vector in terms of x and show that it is orthogonal to the corresponding velocity vector. 58. Orthogonal vectors A particle moves around the unit circle with position vector cos t, sin t. Use vectors to show that the particle’s velocity is always orthogonal to both its position and its acceleration. 59. Colliding particles The paths of two particles for t 0 are given by the position vectors r1(t) t 3, (t 3)2 3t 3t r2(t) 4, 2 . 2 2 (a) Determine the exact time(s) at which the particles collide. (b) Find the direction of motion of each particle at the time(s) of collision. 60. A Satellite in Circular Orbit A satellite of mass m is moving at a constant speed v around a planet of mass M in a circular orbit of radius r0 , as measured from the planet’s center of mass. Determine the satellite’s orbital period T (the time to complete one full orbit), as follows: (a) Coordinatize the orbital plane by placing the origin at the planet’s center of mass, with the satellite on the x-axis at t 0 and moving counterclockwise, as in the accompanying figure. y 547 Let rt be the satellite’s position vector at time t. Show that vt r0 and hence that vt vt rt r0 cos , r0 sin r0 r0 (b) Find the acceleration of the satellite. (c) According to Newton’s law of gravitation, the gravitational force exerted on the satellite by the planet is directed toward the origin and is given by ( ) Gm M r F , r02 r 0 where G is the universal constant of gravitation. Using Newton’s second law, F ma, show that v 2 GM r0 . (d) Show that the orbital period T satisfies vT 2 r0 . (e) From parts (c) and (d), deduce that 4 2 T 2 r03 ; GM that is, the square of the period of a satellite in circular orbit is proportional to the cube of the radius from the orbital center. Extending the Ideas Let u u1, u2 and v v1, v2 be vectors in the plane. The dot product or inner product u v is a scalar defined by u v u1, u2 v1, v2 u1v1 u2v2. 61. Using the Dot Product Show that the dot product of two perpendicular vectors is zero. 62. An Alternate Formula for Dot Product Let u u1, u2 and v v1, v2 be vectors in the plane, and let w u v. (a) Explain why w can be represented by the arrow in the accompanying diagram. m r(t) t 0 (c) Find the component form of w and use it to prove that M r0 w (b) Explain why |w|2 |u|2 |v|2 2|u||v| cos , u where is the angle between vectors u and v. x |u|2 |v|2 |w|2 u v 2(u1v1 u2v2). (d) Finally, prove that u v |u||v| cos , where is the angle between vectors u and v. 27. v(t) 6t, 6t2, a(t) 6, 12t 28. v(t) 2 cos 2t, 2sin t, a(t) 4 sin 2t, 2 cos t 29. v(t) et tet, et, a(t) 2et tet, et 30. v(t) 6 sin 3t, 8 cos 4t, a(t) 18 cos 3t, 32 sin 4t 31. v(t) 2t 2 cos 2t, 2t 2 sin 2t, a(t) 2 4 sin 2t, 2 4 cos 2t 32. v(t) sin t t cos t, cos t t sin t, 32. a(t) 2 cos t t sin t, 2 sin t t cos t 4t 2 2t2 45. (a) , 2 2 (1 t ) (1 t2)2 (b) No. The x-component of velocity is zero only if t 0, while the y-component of velocity is zero only if t 1. At no time will the velocity be 0, 0. 1 t2 2t 1, 0 (c) lim , t→ 1 t2 1 t2 3 cos 4 (3 cos 4)2 (sin 8)2 2.196 50. (a) y 5 (x 4) (b) sin 8 3 3 50. (d) 4 sin (t3) dt, 5 3 cos(t2) dt (4.004, 5.724) 2 2 1 x2 57. The velocity vector is x, 1 x2, which has slope . x d d The acceleration vector is (x), ( 1 x2) dt dt dx 2x dx , 2 dt 1 x dt x2 x x, 2 , which has slope . Since the slopes are negative 1 x 1 x2 reciprocals of each other, the vectors are orthogonal. 5128_Ch10_pp530-561 2/3/06 4:41 PM Page 548 548 Chapter 10 Parametric, Vector, and Polar Functions 10.3 What you’ll learn about • Polar Coordinates • Polar Curves • Slopes of Polar Curves • Areas Enclosed by Polar Curves • A Small Polar Gallery . . . and why Polar equations enable us to define some interesting and important curves that would be difficult or impossible to define in the form y f (x). Polar Functions Polar Coordinates If you graph the two functions y sin 3x and y cos 5x on the same pair of axes, you will get two sinusoids. But if you graph the curve defined parametrically by x sin 3t and y cos 5t, you will get the figure shown. Parametric graphing opens up a whole new world of curves that can be defined using our familiar basic functions. Another way to enter that world is to use a different coordinate system. In polar coordinates we identify the origin O as the pole and the positive x-axis as the initial ray of angles measured in the usual trigonometric way. We can then identify each point P in the plane by polar coordinates (r, ), where r gives the directed distance from O to In Figure 10.19 we see that P and gives the directed angle from the initial ray to the ray OP. the point P with rectangular (Cartesian) coordinates (2, 2) has polar coordinates (22, 4). y y P(2, 2) 2 P(2 2, p/4) 2 2 2 p/4 x x O O 2 Rectangular coordinates Polar coordinates Figure 10.19 Point P has rectangular coordinates (2, 2) and polar coordinates (22, 4). As you would expect, we can also coordinatize point P with the polar coordinates Less obvi(22, 94) or (22, 74), since those angles determine the same ray OP. ously, we can also coordinatize P with polar coordinates (22, 34), since the directed distance 22 in the 34 direction is the same as the directed distance 22 in the 4 direction (Figure 10.20). So, although each pair (r, ) determines a unique point in the plane, each point in the plane can be coordinatized by an infinite number of polar ordered pairs. y y –2 2 2 2 p/4 x x −3p/4 Figure 10.20 The directed negative distance 22 in the 34 direction is the same as the directed positive distance 22 in the 4 direction. Thus the polar coordinates (22 , 34) and (22, 4) determine the same point. 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 549 Section 10.3 Polar Functions 549 EXAMPLE 1 Rectangular and Polar Coordinates (a) Find rectangular coordinates for the points with given polar coordinates. (i) (4, 2) (ii) (3, ) (iii) (16, 56) (iv) (2, 4) (b) Find two different sets of polar coordinates for the points with given rectangular coordinates. (i) (1, 0) (ii) (3, 3) (iii) (0, 4) (iv) (1, 3) SOLUTION (a) (i) (0, 4) (ii) (3, 0) (iii) (83, 8) (iv) (1, 1) (b) A point has infinitely many sets of polar coordinates, so here we list just two typical examples for each given point. (i) (1, 0), (1, 2) (ii) (32, 34), (32, 4) (iii) (4, 2), (4, 32) (iv) (2, 3), (2, 43) Now try Exercises 1 and 3. EXAMPLE 2 Graphing with Polar Coordinates Graph all points in the plane that satisfy the given polar equation (a) r 2 (b) r 2 (c) 6 SOLUTION First, note that we do not label our axes r and . We are graphing polar equations in the usual xy-plane, not renaming our rectangular variables! (a) The set of all points with directed distance 2 units from the pole is a circle of radius 2 centered at the origin (Figure 10.21a). (b) The set of all points with directed distance 2 units from the pole is also a circle of radius 2 centered at the origin (Figure 10.21b). (c) The set of all points of positive or negative directed distance from the pole in the 6 direction is a line through the origin with slope tan(6) (Figure 10.21c). Now try Exercise 7. y y 3 y 3 2 3 –2 p/6 x –3 3 –3 (a) x x –3 3 –3 (b ) –3 3 –3 (c) Figure 10.21 Polar graphs of (a) r 2, (b) r 2, and (c) 6. (Example 2) Polar Curves The curves in Example 2 are a start, but we would not introduce a new coordinate system just to graph circles and lines; there are far more interesting polar curves to study. In the past it was hard work to produce reasonable polar graphs by hand, but today, thanks to graphing technology, it is 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 550 550 Chapter 10 Parametric, Vector, and Polar Functions just a matter of finding the right window and pushing the right buttons. Our intent in this section is to use the technology to produce the graphs and then concentrate on how calculus can be used to give us further information. EXAMPLE 3 Polar Graphing with Technology Find an appropriate graphing window and produce a graph of the polar curve. (a) r sin 6 (b) r 1 2 cos (c) r 4 sin SOLUTION For all these graphs, set your calculator to POLAR mode. (a) First we find the window. Notice that r sin 6 1 for all , so points on the graph are all within 1 unit from the pole. We want a window at least as large as [1, 1] by [1, 1], but we choose the window [1.5, 1.5] by [1, 1] in order to keep the aspect ratio close to the screen dimensions, which have a ratio of 3:2. We choose a -range of 0 2 to get a full rotation around the graph, after which we know that sin 6 will repeat the same graph periodically. Choose step 0.05. The result is shown in Figure 10.22a. (b) In this graph we notice that r 12 cos 3, so we choose [3, 3] for our y-range and, to get the right aspect ratio, [4.5, 4.5] for our x-range. Due to the cosine’s period, 0 2 again suffices for our -range. The graph is shown in Figure 10.22b. (c) Since r 4 sin 4, we choose [4, 4] for our y-range and [6, 6] for our x-range. Due to the sine’s period, 0 2 again suffices for our -range. The graph is shown in Figure 10.22c. Now try Exercise 13. [–1.5, 1.5] by [–1, 1] 0 ≤ u ≤ 2p [– 4.5, 4.5] by [–3, 3] 0 ≤ u ≤ 2p [–6, 6] by [– 4, 4] 0 ≤ u ≤ 2p (a) (b) (c) Figure 10.22 The graphs of the three polar curves in Example 3. The curves are (a) a 12-petaled rose, (b) a limaçon, and (c) a circle. A Rose is a Rose The graph in Figure 10.22a is called a 12-petaled rose, because it looks like a flower and some flowers are roses. The graph in Figure 10.22b is called a limaçon (LEE-ma-sohn) from an old French word for snail. We will have more names for you at the end of the section. With a little experimentation, it is possible to improve on the “safe” windows we chose in Example 3 (at least in parts (b) and (c)), but it is always a good idea to keep a 3:2 ratio of the x-range to the y-range so that shapes do not become distorted. Also, an astute observer may have noticed that the graph in part (c) was traversed twice as went from 0 to 2, so a range of 0 would have sufficed to produce the entire graph. From 0 to , the circle is swept out by positive r values; then from to 2, the same circle is swept out by negative r values. Although the graph in Figure 10.22c certainly looks like a circle, how can we tell for sure that it really is? One way is to convert the polar equation to a Cartesian equation and verify that it is the equation of a circle. Trigonometry gives us a simple way to convert polar equations to rectangular equations and vice versa. 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 551 Section 10.3 Polar Functions 551 Polar–Rectangular Conversion Formulas x r cos r2 x2 y2 y r sin y tan x EXAMPLE 4 Converting Polar to Rectangular Use the polar–rectangular conversion formulas to show that the polar graph of r 4 sin is a circle. SOLUTION To facilitate the substitutions, multiply both sides of the original equation by r. (This could introduce extraneous solutions with r 0, but the pole is the only such point, and we notice that it is already on the graph.) r 4 sin r2 4r sin x2 y2 4y Multiply by r. Polar–rectangular conversion x2 y2 4y 0 x2 y2 4y 4 4 x2 (y 2)2 22 Completing the square Circle in standard form Sure enough, the graph is a circle centered at (0, 2) with radius 2. Now try Exercise 25. The polar–rectangular conversion formulas also reveal the calculator’s secret to polar graphing: It is really just parametric graphing with as the parameter. Parametric Equations of Polar Curves The polar graph of r f () is the curve defined parametrically by: x r cos f () cos y r sin f () sin EXPLORATION 1 Graphing Polar Curves Parametrically Switch your grapher to parametric mode and enter the equations x sin (6t) cos t y sin (6t) sin t. 1. Set an appropriate window and see if you can reproduce the polar graph in Figure 10.22a. 2. Then produce the graphs in Figures 10.22b and 10.22c in the same way. 5128_Ch10_pp530-561 01/16/06 12:15 PM Page 552 552 Chapter 10 Parametric, Vector, and Polar Functions Slopes of Polar Curves Since polar curves are drawn in the xy-plane, the slope of a polar curve is still the slope of the tangent line, which is dydx. The polar–rectangular conversion formulas enable us to write x and y as functions of , so we can find dydx as we did with parametrically defined functions: dy dyd . dx dxd EXAMPLE 5 Finding Slope of a Polar Curve Find the slope of the rose curve r 2 sin 3 at the point where 6 and use it to find the equation of the tangent line (Figure 10.23). SOLUTION The slope is dy dyd dx 6 dxd 6 [–3, 3] by [–2, 2] 0≤u≤p Figure 10.23 The 3-petaled rose curve r 2 sin 3. Example 5 shows how to find the tangent line to the curve at 6. d (2 sin 3 sin) d d (2 sin 3 cos) d . 6 This expression can be computed by hand, but it is an excellent candidate for your calculator’s numerical derivative functionality (Section 3.2). NDERIV quickly gives an answer of 1.732050808, which you might recognize as 3. When 6, x 2 sin(2) cos(6) 3 and y 2 sin(2) sin(6) 1. So the tangent line has equation y 1 3 (x 3). Now try Exercise 39. Areas Enclosed by Polar Curves We would like to be able to use numerical integration to find areas enclosed by polar curves just as we did with curves defined by their rectangular coordinates. Converting the equations to rectangular coordinates is not a reasonable option for most polar curves, so we would like to have a formula involving small changes in rather than small changes in x. While a small change x produces a thin rectangular strip of area, a small change produces a thin circular sector of area (Figure 10.24). y y ∆u ∆x x x Figure 10.24 A small change in x produces a rectangular strip of area, while a small change in produces a thin sector of area. 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 553 Section 10.3 Polar Functions 553 1 Recall from geometry that the area of a sector of a circle is 2r 2, where r is the radius and is the central angle measured in radians. If we replace by the differential d, we get the area 1 differential dA 2r 2d (Figure 10.25), which is exactly the quantity that we need to integrate to get an area in polar coordinates. y 1 dA – r 2 d 2 P(r, ) r d x O Figure 10.25 The area differential dA. Area in Polar Coordinates The area of the region between the origin and the curve r f for is 1 1 2 A r 2 d f d. 2 2 EXAMPLE 6 Finding Area Find the area of the region in the plane enclosed by the cardioid r 21 cos . SOLUTION y We graph the cardioid (Figure 10.26) and determine that the radius OP sweeps out the region exactly once as runs from 0 to 2 . Solve Analytically The area is therefore r 2 (1 cos ) P(r, ) 2 r 4 O 0, 2 x 2 0 1 r 2 d 2 ( 2p 1 • 41 cos 2 d 2 0 2p 21 2 cos cos 2 d 0 2p Figure 10.26 The cardioid in Example 6. 0 ) 1 cos 2 2 4 cos 2 d 2 2p 3 4 cos cos 2 d 0 [ sin 2 3 4 sin 2 ] 2 6 0 6 . 0 Support Numerically NINT 21 cos 2, , 0, 2 18.84955592, which agrees with 6 to eight decimal places. Now try Exercise 43. 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 554 554 Chapter 10 Parametric, Vector, and Polar Functions y EXAMPLE 7 Finding Area r 2 cos 1 2– — 3 Find the area inside the smaller loop of the limaçon r 2 cos 1. SOLUTION 0 x After watching the grapher generate the curve over the interval 0 2 (Figure 10.27), we see that the smaller loop is traced by the point r, as increases from 2 3 to 4 3 (the values for which r 2 cos 1 0). The area we seek is 4 3 A 4– — 3 1 1 r 2 d 2 2 2 3 4 3 2 cos 1 2 d. 2 3 Solve Numerically Figure 10.27 The limaçon in Example 7. 1 NINT 2 cos 1 2, , 2 3, 4 3 0.544. 2 Now try Exercise 47. y r2 To find the area of a region like the one in Figure 10.28, which lies between two polar curves r1 r1() and r2 r2() from to , we subtract the integral of (1 2)r12 from the integral of (1 2)r 22. This leads to the following formula. r1 Area Between Polar Curves The area of the region between r1 and r2 for is A x O Figure 10.28 The area of the shaded region is calculated by subtracting the area of the region between r1 and the origin from the area of the region between r2 and the origin. 1 1 1 r22 d r12 d r22 r 12 d. 2 2 2 EXAMPLE 8 Finding Area Between Curves Find the area of the region that lies inside the circle r 1 and outside the cardioid r 1 cos . SOLUTION y r1 1 cos The region is shown in Figure 10.29. The outer curve is r2 1, the inner curve is r1 1 cos , and runs from 2 to 2. Using the formula for the area between polar curves, the area is Upper limit /2 r2 1 A 2 x 0 2 1 r 22 r12 d 2 2 2 0 1 r 22 r12 d 2 Symmetry 2 Lower limit – /2 Figure 10.29 The region and limits of integration in Example 8. 1 1 2 cos cos 2 d 0 2 2 cos cos 2 d 1.215. Using NINT 0 In case you are interested, the exact value is 2 4. Now try Exercise 53. 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 555 Section 10.3 Polar Functions 555 A SMALL POLAR GALLERY Here are a few of the more common polar graphs and the -intervals that can be used to produce them. CIRCLES y y y x x x r a cos 0 r a sin 0 r constant 0 2 ROSE CURVES y y x x r a sin n, n even 0 2 2n petals y-axis symmetry and x-axis symmetry r a sin n, n odd 0 n petals y-axis symmetry y y x r a cos n, n odd 0 n petals x-axis symmetry x r a cos n, n even 0 2 2n petals y-axis symmetry and x-axis symmetry 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 556 556 Chapter 10 Parametric, Vector, and Polar Functions LIMAÇON CURVES r a b sin or r a b cos with a 0 and b 0 (r a b sin has y-axis symmetry; r a b cos has x-axis symmetry.) y y x x a 1 b 0 2 Cardioid a 1 b 0 2 Limaçon with loop y y x x a 2 b 0 2 Convex limaçon a 1 2 b 0 Dimpled limaçon LEMNISCATE CURVES y y x x r2 a2 sin 2 0 r2 a2 cos 2 0 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 557 Section 10.3 Polar Functions 557 SPIRAL OF ARCHIMEDES y x r0 Quick Review 10.3 (For help, go to Sections 10.1 and 10.2.) 1. Find the component form of a vector with magnitude 4 and direction angle 30º. 23, 2 2. Find the area of a 30º sector of a circle of radius 6. Exercises 6–10 refer to the parametrized curve x 3 cos t, y 5 sin t, 0 t 2. 5 3 6. Find dydx. cot t 3 8. Find the points on the curve where the slope is zero. (0, 5) and (0, 5) 4. Find the rectangular equation of a circle of radius 5 centered at the origin. x2 y2 25 9. Find the points on the curve where the slope is undefined. (3, 0) and 5. Explain how to use your calculator in function mode to graph 2 4x 4 x2 1/2 the curve x2 3y2 4. Graph y and y 3 3 5 3 7. Find the slope of the curve at t 2. cot 2 0.763 3. Find the area of a sector of a circle of radius 8 that has a central angle of 8 radians. 4 10. Find the length of the curve from t 0 to t . (3, 0) 12.763 1/2 Section 10.3 Exercises In Exercises 1 and 2, plot each point with the given polar coordinates and find the corresponding rectangular coordinates. 1. (a) 2, 4 (c) 0, 2 (b) 1, 0 (d) 2, 4 2. (a) 3, 5 6 (b) 5, tan1 4 3 (c) 1, 7 (d) 2 3, 2 3 In Exercises 3 and 4, plot each point with the given rectangular coordinates and find two sets of corresponding polar coordinates. 3. (a) 1, 1 (c) 0, 3 4. (a) 3, 1 (c) 0, 2 (b) 1, 3 (d) 1, 0 (b) 3, 4 (d) 2, 0 In Exercises 5–10, graph the set of points whose polar coordinates satisfy the given equation. 5. r 3 7. r2 4 9. 6 6. r 3 8. 4 10. r2 8 6r 23. x y 1, a line (slope 1, y-intercept 1) 24. x2 y2 1, a circle (center (0, 0), radius 1) 25. y 2x 5, a line (slope 2, y-intercept 5) In Exercises 11–20, find an appropriate window and use a graphing calculator to produce the polar curve. Then sketch the complete curve and identify the type of curve by name. 11. r 1 cos 12. r 2 2 cos 13. r 2 cos 3 14. r 3 sin 2 15. r 1 2 sin 16. r 32 cos 17. r2 4 cos 2 19. r 4 sin 18. r 2 sin 2 20. r 3 cos In Exercises 21–30, replace the polar equation by an equivalent Cartesian (rectangular) equation. Then identify or describe the graph. 21. r 4 csc 22. r 3 sec 23. r cos r sin 1 5 25. r sin 2 cos 24. r 2 1 y 4, a horizontal line 27. cos 2 sin 2 x 3, a vertical line 26. r 2 sin 2 2 xy 1, a hyperbola 28. r 2 4r cos 29. r 8 sin 30. r 2 cos 2 sin 27. x2 y2, the union of two lines: y x 28. (x 2)2 y2 4, a circle (center (2, 0), radius 2) 29. x2 (y 4)2 16, a circle (center (0, 4), radius 4) 30. (x 1)2 (y 1)2 2, a circle (center (1, 1), radius 2) 5128_Ch10_pp530-561 2/3/06 4:41 PM Page 558 558 Chapter 10 Parametric, Vector, and Polar Functions 62. False. Integrating from 0 to 2 traverses the curve twice, giving twice the area. The correct upper limit of integration is . In Exercises 31–38, find an appropriate window and use a graphing calculator to produce the polar curve. Then sketch the complete curve and identify the type of curve by name. (Note: You won’t find these in the Polar Gallery.) 57. Sketch the polar curves r 3 cos and r 1 cos and find the area that lies inside the circle and outside the cardioid. 31. r sec tan 1 33. r 1 cos 14 35. r 5 9 cos 1 37. r 1 0.8 cos 32. r csc cot 2 34. r 1 sin 12 36. r 8 6 cos 1 38. r 1 1.3 cos 58. Sketch the polar curves r 2 and r 2(1 sin ) and find the area that lies inside the circle and outside the cardioid. 59. Sketch the polar curve r 2 sin 3. Find the area enclosed by the curve and find the slope of the curve at the point where 4. 60. The accompanying figure shows the parts of the graphs of the line 5 x 3 y and the curve x 1 y2 that lie in the first quadrant. Region R is enclosed by the line, the curve, and the x-axis. y In Exercises 39–42, find the slope of the curve at each indicated point. 39. r 1 sin , 0, At 0: 1; At : 1 40. r cos 2 , 0, 2, At 0: undefined; At 2: 0 At 2: 0; At : undefined 41. r 2 3 sin At (2, 0): 2/3 At (1: 2): 0 At (2, ): 2/3 At (5, 32): 0 y (2, p) (2, 0) x p –1, 2 ( R ( O 1 x 5y 1 y dy 0.347 3 3/4 2 0 (a) Set up and evaluate an integral expression with respect to y that gives the area of R. (5, 3p2 ( 42. r 31 cos At (1.5, 3): undefined At (4.5, 23): 0 At (6, ): undefined At (3, 32): 1 1 y2 can be described in polar (b) Show that the curve x 1 Let x r cos and 2 coordinates by r . cos2 sin2 y r sin and solve for r 2. y (4.5 2p 3 ( p 1.5 3 ( (c) Use the polar equation in part (b) to set up an integral expression with respect to that gives the area of R. 1 1 ( Let tan1 (3/5). Then the area is Standardized Test Questions x (6, p) 2 d. cos sin 0 2 2 You may use a graphing calculator to solve the following problems. (3 32p ( 61. True or False There is exactly one point in the plane with polar coordinates (2, 2). Justify your answer. True. Polar coordinates determine a unique point. In Exercises 43–56, find the area of the region described. 43. inside the convex limaçon r 4 2 cos 44. inside the cardioid r 2 2 sin 2 1 2 18 r sin 3 is 0 6 45. inside one petal of the four-petaled rose r cos 2 46. inside the eight-petaled rose r 2 sin 4 48. inside the six-petaled rose 2 sin 3 8 (C) 2 11 51. shared by the circles r 2 cos and r 2 sin 2 (3 cos) d (E) 3 cos d 2 0 2 (B) 0 3 cos d (D) 0 (3 cos) d 0 50. inside the inner loop of the limaçon r 2 sin 1 0.544 52. shared by the circles r 1 and r 2 sin 63. Multiple Choice The area of the region enclosed by the polar graph of r 3 co s is given by which integral? D 2 4 49. inside the dimpled limaçon r 3 2 cos sin2 3 d. Justify your answer. (A) 0 3 cos d 2 47. inside one loop of the lemniscate r2 4 cos 2 r2 62. True or False The total area enclosed by the 3-petaled rose (2) 1 (23) (3 2) 53. shared by the circle r 2 and the cardioid r 2(1 cos ) 5 8 54. shared by the cardioids r 2(1 cos ) and r 2(1 cos ) 6 16 64. Multiple Choice The area enclosed by one petal of the 3-petaled rose r 4 cos(3) is given by which integral? E 3 (A) 16 3 (C) 8 55. inside the circle r 2 and outside the cardioid r 2(1 sin ) 8 56. inside the four-petaled rose r 4 cos 2 and outside the circle r 2 43 (83) 3 (E) 8 3 6 6 cos(3) d cos2(3) d cos2(3) d 6 (B) 8 6 cos(3) d 6 (D) 16 6 cos2(3) d 5128_Ch10_pp530-561 01/16/06 12:43 PM Page 559 Section 10.3 Polar Functions 65. Multiple Choice If a 0 and 0, all of the following must necessarily represent the same point in polar coordinates except which ordered pair? B (A) (a, ) (B) (a, ) (C) (a, ) (D) (a, ) (E) (a, 2) 66. Multiple Choice Which of the following gives the slope of the polar curve r f() graphed in the xy-plane? D dr (A) d dy (B) d dyd (D) dxd dx (C) d dy dr (E) dx d Explorations 67. Rotating Curves Let r 1 31 cos and r 2 r 1 . (a) Graph r 2 for 6, 4, 3, and 2 and compare with the graph of r 1. (b) Graph r 2 for 6, 4, 3, and 2 and compare with the graph of r1. (c) Based on your observations in parts (a) and (b), describe the relationship between the graphs of r 1 f and r 2 f . 2 68. Let r . 1 k cos (a) Graph r in a square viewing window for k 0.1, 0.3, 0.5, 0.7, and 0.9. Describe the graphs. (b) Based on your observations in part (a), conjecture what happens to the graphs for 0 k 1 and k→0. 2 69. Let r . 1 k cos (a) Graph r in a square viewing window for k 1.1, 1.3, 1.5, 1.7, and 1.9. Describe the graphs. (b) Based on your observations in part (a), conjecture what happens to the graphs for k 1 and k→1. k 70. Let r . 1 cos 559 (a) Graph r in a square viewing window for k 1, 3, 5, 7, and 9. Describe the graphs. (b) Based on your observations in part (a), conjecture what happens to the graphs for k 0 and k→0. Extending the Ideas 71. Distance Formula Show that the distance between two points r1, 1 and r 2, 2 in polar coordinates is 2 2 r 1 r 2 r1 r 2cos d 2 1 2 . 72. Average Value If f is continuous, the average value of the polar coordinate r over the curve r f , , with respect to is 1 rav f d. Use this formula to find the average value of r with respect to over the following curves a 0. (a) the cardioid r a1 cos a (b) the circle r a a (c) the circle r a cos , 2 2 2a/ 73. Length of a Polar Curve The parametric form of the arc length formula (Section 10.1) gives the length of a polar curve as dx 2 dy 2 d. d d Assuming that the necessary derivatives are continuous, show that the substitutions x r cos and y r sin transform this expression into L L dr 2 r 2 d. d 74. Length of a Cardioid Use the formula in Exercise 73 to find the length of the cardioid r 1 cos . 8 Quick Quiz for AP* Preparation: Sections 10.1–10.3 You may use a graphing calculator to solve the following problems. 1. Multiple Choice Which of the following is equal to the area of the region inside the polar curve r 2 cos and outside the polar curve r cos ? A 2 cos2 d 3 (C) cos2 d 2 (A) 3 0 2 0 (B) 30 cos2 d (D) 3 2 0 cos d (E) 30 cos d (A) 0 4t 1 dt (B) 20 t2 1 dt (D) 0 4t2 1 dt (E) 20 4t2 1 dt 4 4 4 (C) 0 2t2 1 dt 4 4 4. Free Response A polar curve is defined by the equation r sin 2 for 0 . (a) Find the area bounded by the curve and the x-axis. 2. Multiple Choice For what values of t does the curve given by the parametric equations x t3 t2 1 and y t 4 2t 2 8t have a vertical tangent? C (A) 0 only (B) 1 only (C) 0 and 2 3 only (D) 0, 2 3, and 1 (E) No value 3. Multiple Choice The length of the path described by the parametric equations x t2 and y t from t 0 to t 4 is given by which integral? D (b) Find the angle that corresponds to the point on the curve where x 2. 2 dr (c) For , is negative. How can this be seen 3 d 3 from the graph? (d) At what angle in the interval 0 2 is the curve farthest away from the origin? Justify your answer. 5128_Ch10_pp530-561 01/16/06 12:43 PM Page 560 560 Chapter 10 Parametric, Vector, and Polar Functions Chapter 10 Key Terms Absolute value of a vector (p. 538) Acceleration vector (p. 542) Archimedes spiral (p. 557) Arc length of a parametrized curve (p. 533) Arc length of a polar curve (p. 553) Area between polar curves (p. 554) Area differential (p. 553) Area in polar coordinates (p. 553) Arrow (p. 538) Cardioid (p. 556) Cartesian equation of a curve (p. 550) Component form of a vector (p. 539) Components of a vector (p. 538) Convex limaçon (p. 556) Cycloid (p. 533) Dimpled limaçon (p. 556) Directed distance (p. 548) Directed line segment (p. 538) Direction angle of a vector (p. 539) Direction of motion (p. 542) Direction vector (p. 538) Displacement (p. 543) Distance traveled (p. 544) Pole (p. 548) Position of a particle (p. 544) Position vector (p. 538) Properties of vectors (p. 541) Rectangular coordinates (p. 548) Resultant vector (p. 540) Rose curve (p. 555) Scalar (p. 539) Scalar multiple of a vector (p. 540) Speed (p. 542) Standard representation of a vector (p. 538) Sum of vectors (p. 540) Tail-to-head representation of vector addition (p. 540) Terminal point of an arrow (p. 539) Unit vector (p. 540) Vector (p. 538) Vector addition (p. 539) Velocity vector (p. 542) Zero vector (p. 538) Dot product of vectors (p. 547) Equivalent arrows (p. 539) Head Minus Tail Rule (p. 539) Huygens’s clock (p. 534) Initial point of an arrow (p. 539) Initial ray of angle of direction (p. 548) Lemniscate (p. 556) Limaçon (p. 556) Limaçon with inner loop (p. 556) Magnitude of a vector (p. 538) Opposite of a vector (p. 540) Orthogonal vectors (p. 547) Parallelogram representation of vector addition (p. 540) Parametric equations of a polar curve (p. 551) Parametric formula for dydx (p. 532) Parametric formula for d 2 ydx2 (p. 532) Path of a particle (p. 542) Polar coordinates (p. 548) Polar equation of a curve (p. 549) Polar graphing (p. 549) Polar–rectangular conversion formulas (p. 551) Chapter 10 Review Exercises In Exercises 1–4, let u 3, 4 and v 2, 5. Find (a) the component form of the vector and (b) its magnitude. 1. 3u 4v 2. u v 3. 2u 4. 5v In Exercises 5–8, find the component form of the vector. 5. the vector obtained by rotating 0, 1 through an angle of 2p 3 radians 32, 1/2 [assuming counterclockwise] 6. the unit vector that makes an angle of 6 radian with the positive x-axis 32, 1/2 7. the vector 2 units long in the direction 4 i j 8/17 , 2/17 8. the vector 5 units long in the direction opposite to the direction of 3 5, 4 5 3, 4 In Exercises 9 and 10, (a) find an equation for the tangent to the curve at the point corresponding to the given value of t, and (b) find the value of d 2 ydx 2 at this point. 1. (a) 17, 32 (b) 1313 3. (a) 6, 8 (b) 10 2. (a) 1, 1 (b) 2 4. (a) 10, 25 (b) 725 529 9. x 1 2 tan t, 10. x 1 1 t 2, 3 1 y 1 2 sec t; t 3 (a) y 2x 4 (b) 1/4 y 1 3 t; t 2 (a) 1, 1 (b) 2 In Exercises 11–14, find the points at which the tangent to the curve is (a) horizontal; (b) vertical. 11. x 1 2 tan t, 12. x 2 cos t, 13. x cos t, (a) (0, 1/2) and (0, 1/2) (b) Nowhere y 2 sin t (a) (0, 2) and (0, 2) (b) (2, 0) and (2, 0) y 1 2 sec t y cos 2 t 14. x 4 cos t, y 9 sin t In Exercises 15–20, find an appropriate window and graph the polar curve on a graphing calculator. Then sketch the curve on paper and identify the type of curve. 15. r 1 sin 16. r 2 cos 17. r cos 2 18. r cos 1 sin 2 20. r sin 19. r2 13. (a) (0, 0) (b) Nowhere 14. (a) (0, 9) and (0, 9) (b) (4, 0) and (4, 0) 5128_Ch10_pp530-561 1/13/06 3:51 PM Page 561 23. Horizontal: y 0, y 0.443, y 1.739 Vertical: x 2, x 0.067, x 1.104 Chapter 10 Review Exercises 24. Horizontal: y 1/2, y 4 Vertical: x 0, x 2.598 561 34. (r cos 2)2 (r sin 5)2 16 In Exercises 21 and 22, find the slope of the tangent lines at the point where 3. (b) Find the x- and y-components of the acceleration of the particle at t 3. 21. r cos 2 (c) Find a single equation in x and y for the path of the particle. 22. r 2 cos 2 4.041 0.346 In Exercises 23 and 24, find equations for the horizontal and vertical tangent lines to the curves. 23. r 1 cos 2, 0 4 24. r 21 sin , 48. Particle Motion At time t, 0 t 4, the position of a particle moving along a path in the plane is given by the parametric equations x e t cos t, 0 2 y e t sin t. 25. Find equations for the lines that are tangent to the tips of the petals of the four-petaled rose r sin 2 . y x 2 and y x 2 (a) Find the slope of the path of the particle at time t . 1 26. Find equations for the lines that are tangent to the cardioid r 1 sin at the points where it crosses the x-axis. (c) Find the distance traveled by the particle along the path from t 0 to t 3. (e3 1)2 y x 1 and y x 1 In Exercises 27–30, replace the polar equation by an equivalent Cartesian equation. Then identify or describe the graph. 27. r cos r sin x y, a line 28. r 3 cos 29. r2 4 tan sec 30. r cos 3 2 3 x 4y, a parabola (b) Find the speed of the particle when t 3. e32 49. Particle Motion The position of a particle at any time t 0 is given by 2 xt t 2 2, yt t 3. 5 (a) Find the magnitude of the velocity vector at t 4. 104/5 In Exercises 31–34, replace the Cartesian equation by an equivalent polar equation. (b) Find the total distance traveled by the particle from t 0 to t 4. 4144/135 31. x 2 y 2 5y 0 32. x 2 y 2 2y 0 r 2 sin (c) Find dydx as a function of x. 33. x 2 4y 2 16 34. x 2 2 y 5 2 16 r 5 sin dy 3 5x 2 dx 36. enclosed by one petal of the three-petaled rose r sin 3 12 50. Navigation An airplane, flying in the direction 80º east of north at 540 mph in still air, encounters a 55-mph tail wind acting in the direction 100º east of north. The airplane holds its compass heading but, because of the wind, acquires a different ground speed and direction. What are they? 37. inside the “figure eight” r 1 cos 2 and outside the circle r 1 (4) 2 AP* Examination Preparation In Exercises 35–38, find the area of the region described. 35. enclosed by the limaçon r 2 cos 92 38. inside the cardioid r 21 sin and outside the circle r 2 sin 5 In Exercises 39 and 40, r(t) is the position vector of a particle moving in the plane at time t. Find (a) the velocity and acceleration vectors, and (b) the speed at the given value of t. 39. r(t) 4 cos t, 2 sin t, t 4 Speed 591.982 mph; You may use a graphing calculator to solve the following problems. 51. A particle moves along the graph of y cos x so that its x-component of acceleration is always 2. At time t 0, the particle is at the point (, 1) and the velocity of the particle is 0, 0. 40. r(t) 3 sec t, 3 tan t, t 0 (a) Find the position vector of the particle. 41. The position of a particle in the plane at time t is t 1 . Find the particle’s maximum speed. r , 1 t 2 1 t2 Direction 8.179° north of east (b) Find the speed of the particle when it is at the point (4, cos 4). 52. Two particles move in the xy-plane. For time t 0, the position of particle A is given by x t 2 and y (t 2)2, and the 3 3 42. Writing to Learning Suppose that r(t) et cos t, et sin t. Show that the angle between r and the acceleration vector a never changes. What is the angle? position of particle B is given by x 2t 4 and y 2t 2. In Exercises 43–46, find the position vector. (c) Determine the exact time when the particles collide. 43. v(t) sin t, cos t and r(0) 0, 1 r(t) cos t 1, sin t 1 1 t 44. v(t) , 2 and r(0) 1, 1 1 t 2 1 t t2 1 r(t) tan1 t 1, 45. a(t) 0, 2 and v(0) 0, 0 and r(0) 1, 0 r(t) 1, t2 (a) Find the velocity vector for each particle at time t 3. (b) Find the distance traveled by particle A from t 0 to t 3. 53. A region R in the xy-plane is bounded below by the x-axis 4 and above by the polar curve defined by r for 1 sin 0 . (a) Find the area of R by evaluating an integral in polar coordinates. 46. a(t) 2, 2 and v(1) 4, 0 and r(1) 3, 3 47. Particle Motion A particle moves in the plane in such a manner that its coordinates at time t are x 3 cos t, 4 (b) The curve resembles an arch of the parabola 8y 16 x2. Convert the polar equation to rectangular coordinates and prove that the curves are the same. y 5 sin t. 4 (c) Set up an integral in rectangular coordinates that gives the area of R. (a) Find the length of the velocity vector at t 3. 3 3 28. x2 y2 3x, a circle center , 0 , radius 2 2 x 4, a line 30. x 3 y 43 or y 3 46. r(t) t2 6t 2, t2 2t 2

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