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SalasSV_09_07_ex - 9.7 TANGENTS TO CURVES GIVEN...

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Unformatted text preview: 9.7 TANGENTS TO CURVES GIVEN PARAMETRICALLY 565 Example 5 is vertical. Find the points of the cardioid r = 1 − cos θ at which the tangent line SOLUTION Since the cosine function has period 2π , we need only concern ourselves with θ in [0, 2π ). The parametric equations for the curve are: x(θ ) = (1 − cos θ ) cos θ , Differentiating and simplifying, we find that x (θ ) = (2 cos θ − 1) sin θ , y(θ ) = (1 − cos θ ) sin θ . [2, π ] [, ] 1 2 π 3 polar axis y (θ ) = (1 − cos θ )(1 + 2 cos θ ). [, ] 1 5π 23 The only numbers in the interval [0, 2π ) at which x is zero and y is not zero are 1 π , π , 3 and 5 π . The tangent line is vertical at 3 [ 1 , 5 π ]. 23 √ √ These points have rectangular coordinates ( 1 , 1 3), (−2, 0), ( 1 , − 1 3). See 44 4 4 Figure 9.7.6 EXERCISES 9.7 Find an equation in x and y for the line tangent to the curve. 1. x(t ) = t , 2. x(t ) = t , 2 [ 1 , 1 π ], 23 [2, π ], Figure 9.7.6 y(t ) = t 3 − 1; y(t ) = t + 5; y(t ) = cos πt ; y (t ) = t ; 4 t = 1. t = 2. t = 0. t = 1. t = 1. 2 t = 1. t = 1π. 4 t = 0. Find the points (x, y) at which the curve has: (a) a horizontal tangent; (b) a vertical tangent. Then sketch the curve. 19. 20. 21. 22. 23. 24. 25. 26. 27. x(t ) = 3t − t 3 , y(t ) = t + 1. x(t ) = t 2 − 2t , y(t ) = t 3 − 12t . x(t ) = 3 − 4 sin t , y(t ) = 4 + 3 cos t . x(t ) = sin 2t , y(t ) = sin t . x(t ) = t 2 − 2t , y(t ) = t 3 − 3t 2 + 2t . x(t ) = 2 − 5 cos t , y(t ) = 3 + sin t . x(t ) = cos t , y(t ) = sin 2t . x(t ) = 3 + 2 sin t , y(t ) = 2 + 5 sin t . Find the tangent(s) to the curve x(t ) = −t + 2 cos 1 πt , 4 at the point (2, 0). 28. Find the tangent(s) to the curve x (t ) = t 3 − t , y(t ) = t sin 1 πt 2 y(t ) = t 4 − 4t 2 3. x(t ) = 2t , 5. x(t ) = t 2 , 6. x(t ) = 1/t , 3 4. x(t ) = 2t − 1, y(t ) = (2 − t )2 ; y(t ) = t 2 + 1; y(t ) = sin t 3 7. x(t ) = cos t , 8. x(t ) = et , y(t ) = 3e−t ; Find an equation in x and y for the line tangent to the polar curve. 9. r = 4 − 2 sin θ , 10. r = 4 cos 2θ , 4 , 5 − cos θ 5 12. r = , 4 − cos θ 11. r = 13. r = 14. r = θ = 0. θ = 1π. 2 θ = 1π. 2 θ = 1π. 6 θ = 0. θ = 1π. 2 sin θ − cos θ , sin θ + cos θ sin θ + cos θ , sin θ − cos θ y = y (t ) at the point (0, 1). 29. Let P = [r1 , θ ] be a point on a polar curve r = f (θ ) as in the figure. Show that, if f (θ1 ) = 0 but f (θ1 ) = 0, Parametrize the curve by a pair of differentiable functions x = x(t ), with [x (t )]2 + [ y (t )]2 = 0. r1 P Sketch the curve and determine the tangent line at the origin by the method of this section. 15. y = x3 . 17. y5 = x3 . 16. x = y3 . 18. y3 = x5 . θ1 O polar axis 566 CHAPTER 9 THE CONIC SECTIONS; POLAR COORDINATES; PARAMETRIC EQUATIONS then the tangent line at P is perpendicular to the line segment OP . 30. If 0 < a < 1, the polar curve r = a − cos θ is a limaçon with an inner loop. Choose a so that the curve will intersect itself at the pole in a right angle. In Exercises 31–35, verify that x (0) = y (0) = 0 and that the given description holds at the point where t = 0. Sketch the graph. 31. x(t ) = t 3 , 32. x(t ) = t , 3 5 Calculate d 2 y/dx2 at the indicated point without eliminating the parameter. 37. x(t ) = cos t , y(t ) = sin t ; t = 1π. 6 38. x(t ) = t 3 , y(t ) = t − 2; t = 1. 39. x(t ) = et , y(t ) = e−t ; t = 0. 40. x(t ) = sin2 t , y(t ) = cos t ; t = 1 π . 4 c 41. Let x = 2 + cos t , y = 2 − sin t . Use a CAS to find d 2 y/dx2 . c 42. Use a CAS to find an equation in x and y for the line tangent to the curve: x = sin2 t , y = cos2 t ; t = 1π. 4 y (t ) = t 2 ; y (t ) = t ; 5 3 cusp. horizontal tangent. vertical tangent. tangent with slope 2. no tangent line. 33. x(t ) = t , 35. x(t ) = t 2 , y (t ) = t ; 34. x(t ) = t 3 − 1, y ( t ) = 2t 3 ; y(t ) = t 2 + 1; Then use a graphing utility to draw the curve and the tangent line together. 36. Suppose that x = x(t ), y = y(t ) are twice differentiable functions that parametrize a curve. Take a point on the curve at which x (t ) = 0 and d 2 y/dx2 exists. Show that d 2y x (t )y (t ) − y (t )x (t ) = 2 dx [x (t )]3 c 43. Repeat Exercise 42 with x = e−3t , y = et ; t = ln 2. (9.7.5) c 44. Use a CAS to find an equation in x and y for the line tangent to the polar curve: 4 ; θ = 1π. r= 3 2 + sin θ Then use a graphing utility to draw the curve and the tangent line together. 9.8 ARC LENGTH AND SPEED Arc Length In Figure 9.8.1 we have sketched the path C traced out by a pair of continuously differentiable functions† x = x(t ), C y = y(t ), t ∈ [a, b]. What is the length of this curve? Figure 9.8.1 We want to determine the length of C . Here our experience in Chapter 5 can be used as a model. To decide what should be meant by the area of a region , we approximated by the union of a finite number of rectangles. To decide what should be meant by the length of C , we approximate C by the union of a finite number of line segments. Each number t in [a, b] gives rise to a point P = P (x(t ), y(t )) that lies on C . By choosing a finite number of points in [a, b], a = t0 < t1 < · · · < ti−1 < ti < · · · < tn−1 < tn = b, we obtain a finite number of points on C , P0 , P1 , . . . , Pi−1 , Pi , . . . , Pn−1 , Pn . We join these points consecutively by line segments and call the resulting path, γ = P0 P1 ∪ · · · ∪ Pi−1 Pi ∪ · · · ∪ Pn−1 Pn , a polygonal path inscribed in C. (See Figure 9.8.2.) † By this we mean functions that have continuous first derivatives. ...
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